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Chapterwise Topicwise
Solved Papers
2021-1979
IITJEE
JEE Main & Advanced
Mathematics
Amit M Agarwal
Arihant Prakashan (Series), Meerut
Arihant Prakashan (Series), Meerut
All Rights Reserved
© Author
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Published by Arihant Publications (India) Ltd.
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CONTENTS
1. Complex Numbers
1-28
2. Theory of Equations
29-50
3. Sequences and Series
51-75
4. Permutations and Combinations
76-88
5. Binomial Theorem
89-102
6. Probability
103-134
7. Matrices and Determinants
135-170
8. Functions
171-187
9. Limit, Continuity and Differentiability
188-238
10. Application of Derivatives
239-277
11. Indefinite Integration
278-293
12. Definite Integration
294-328
13. Area
329-353
14. Differential Equations
354-378
15. Straight Line and Pair of Straight Lines
379-404
16. Circle
405-439
17. Parabola
440-457
18. Ellipse
458-473
19. Hyperbola
474-485
20. Trigonometrical Ratios and Identities
486-498
21. Trigonometrical Equations
499-514
22. Inverse Circular Functions
515-525
23. Properties of Triangles
526-547
24. Vectors
548-579
25. 3D Geometry
580-605
26. Miscellaneous
606-632
JEE Advanced Solved Paper 2021
1-16
SYLLABUS
JEE MAIN
UNIT I Sets, Relations and Functions
UNIT V Mathematical Induction
Sets and their representation, Union, intersection and
complement of sets and their algebraic properties, Power
set, Relation, Types of relations, equivalence relations,
functions, one-one, into and onto functions, composition
of functions.
Principle of Mathematical Induction and its simple
applications.
UNIT VI Binomial Theorem and its Simple
Applications
UNIT II Complex Numbers and
Quadratic Equations
Binomial theorem for a positive integral index, general
term and middle term, properties of Binomial coefficients
and simple applications.
Complex numbers as ordered pairs of reals,
Representation of complex numbers in the form a+ib and
their representation in a plane, Argand diagram, algebra
of complex numbers, modulus and argument (or
amplitude) of a complex number, square root of a
complex number, triangle inequality, Quadratic
equations in real and complex number system and their
solutions. Relation between roots and co-efficients,
nature of roots, formation of quadratic equations with
given roots.
UNIT III Matrices and Determinants
Matrices, algebra of matrices, types of matrices,
determinants and matrices of order two and three.
Properties of determinants, evaluation of deter-minants,
area of triangles using determinants. Adjoint and
evaluation of inverse of a square matrix using
determinants and elementary transformations, Test of
consistency and solution of simultaneous linear
equations in two or three variables using determinants
and matrices.
UNIT VII Sequences and Series
Arithmetic and Geometric progressions, insertion of
arithmetic, geometric means between two given
numbers. Relation between AM and GM Sum upto n
terms of special series: ∑ n, ∑ n2, ∑n3. Arithmetico Geometric progression.
UNIT VIII Limit, Continuity and Differentiability
Real valued functions, algebra of functions, polynomials,
rational, trigonometric, logarithmic and exponential
functions, inverse functions. Graphs of simple functions.
Limits, continuity and differenti-ability. Differentiation of
the sum, difference, product and quotient of two
functions. Differentiation of trigonometric, inverse
trigonometric, logarithmic, exponential, composite and
implicit functions; derivatives of order upto two. Rolle's
and Lagrange's Mean Value Theorems. Applications of
derivatives: Rate of change of quantities, monotonic increasing and decreasing functions, Maxima and minima
of functions of one variable, tangents and normals.
UNIT IV Permutations and Combinations
UNIT IX Integral Calculus
Fundamental principle of counting, permutation as an
arrangement and combination as selection, Meaning of
P(n,r) and C (n,r), simple applications.
Integral as an anti - derivative. Fundamental integrals
involving algebraic, trigonometric, exponential and
logarithmic functions. Integration by substitution, by
parts and by partial fractions. Integration using
trigonometric identities. Evaluation of simple integrals of
the type
dx
dx
,
x2 ± a2
x ±a
dx
,
ax 2 + bx + c
(px + q) dx
ax 2 + bx + c
,
2
2
dx
dx
,
a2 – x2
dx
,
,
2
a – x2
(px + q) dx
,
ax 2 + bx + c
ax 2 + bx + c
circle when the end points of a diameter are given, points
of intersection of a line and a circle with the centre at the
origin and condition for a line to be tangent to a circle,
equation of the tangent. Sections of cones, equations of
conic sections (parabola, ellipse and hyperbola) in
standard forms, condition for y=mx + c to be a tangent
and point (s) of tangency.
UNIT XII Three Dimensional Geometry
,
2
2
a ± x dx
and
2
2
x – a dx
Integral as limit of a sum. Fundamental Theorem of
Calculus. Properties of definite integrals. Evaluation of
definite integrals, determining areas of the regions
bounded by simple curves in standard form.
UNIT X Differential Equations
Ordinary differential equations, their order and degree.
Formation of differential equations. Solution of
differential equations by the method of separation of
variables, solution of homogeneous and linear differential
equations
of the type
UNIT XI Coordinate Geometry
Cartesian system of rectangular coordinates in a plane,
distance formula, section formula, locus and its equation,
translation of axes, slope of a line, parallel and
dy lines,
perpendicular
of a line on the coordinate
+ p(x)yintercepts
= q(x)
dx
axes.
Straight lines
Various forms of equations of a line, intersection of lines,
angles between two lines, conditions for concurrence of
three lines, distance of a point from a line, equations of
internal and external bisectors of angles between two
lines, coordinates of centroid, orthocentre and
circumcentre of a triangle, equation of family of lines
passing through the point of intersection of two lines.
Circles, Conic sections
Standard form of equation of a circle, general form of the
equation of a circle, its radius and centre, equation of a
Coordinates of a point in space, distance between two
points, section formula, direction ratios and direction
cosines, angle between two intersecting lines. Skew lines,
the shortest distance between them and its equation.
Equations of a line and a plane in different forms,
intersection of a line and a plane, coplanar lines.
UNIT XIII Vector Algebra
Vectors and scalars, addition of vectors, components of a
vector in two dimensions and three dimensional space,
scalar and vector products, scalar and vector triple
product.
UNIT XIV Statistics and Probability
Measures of Dispersion: Calculation of mean, median,
mode of grouped and ungrouped data. Calculation of
standard deviation, variance and mean deviation for
grouped and ungrouped data.
Probability: Probability of an event, addition and
multiplication theorems of probability, Baye's theorem,
probability distribution of a random variate, Bernoulli
trials and Binomial distribution.
UNIT XV Trigonometry
Trigonometrical identities and equations.
Trigonometrical functions. Inverse trigonometrical
functions and their properties. Heights and Distances.
UNIT XVI Mathematical Reasoning
Statements, logical operations And, or, implies, implied
by, if and only if. Understanding of tautology,
contradiction, converse and contra positive.
JEE ADVANCED
Algebra
Algebra of complex numbers, addition, multiplication, conjugation, polar representation, properties of modulus and
principal argument, triangle inequality, cube roots of unity, geometric interpretations.
Quadratic equations with real coefficients, relations between roots and coefficients, formation of quadratic equations
with given roots, symmetric functions of roots.
Arithmetic, geometric and harmonic progressions, arithmetic, geometric and harmonic means, sums of finite
arithmetic and geometric progressions, infinite geometric series, sums of squares and cubes of the first n natural
numbers.
Logarithms and their Properties, Permutations and combinations, Binomial theorem for a positive integral index,
properties of binomial coefficients.
Matrices
Matrices as a rectangular array of real numbers, equality of matrices, addition, multiplication by a scalar and product
of matrices, transpose of a matrix, determinant of a square matrix of order up to three, inverse of a square matrix of
order up to three, properties of these matrix operations, diagonal, symmetric and skew-symmetric matrices and their
properties, solutions of simultaneous linear equations in two or three variables.
Probability
Addition and multiplication rules of probability, conditional probability, independence of events, computation of
probability of events using permutations and combinations.
Trigonometry
Trigonometric functions, their periodicity and graphs, addition and subtraction formulae, formulae involving
multiple and sub-multiple angles, general solution of trigonometric equations.
Relations between sides and angles of a triangle, sine rule, cosine rule, half-angle formula and the area of a triangle,
inverse trigonometric functions (principal value only).
Analytical Geometry
Two Dimensions Cartesian oordinates, distance between two points, section formulae, shift of origin.
Equation of a straight line in various forms, angle between two lines, distance of a point from a line. Lines through
the point of intersection of two given lines, equation of the bisector of the angle between two lines, concurrency of
lines, centroid, orthocentre, incentre and circumcentre of a triangle.
Equation of a circle in various forms, equations of tangent, normal and chord.
Parametric equations of a circle, intersection of a circle with a straight line or a circle, equation of a circle through the
points of intersection of two circles and those of a circle and a straight line.
Equations of a parabola, ellipse and hyperbola in standard form, their foci, directrices and eccentricity, parametric
equations, equations of tangent and normal.
Locus Problems, Three Dimensions Direction cosines and direction ratios, equation of a straight line in space,
equation of a plane, distance of a point from a plane.
Differential Calculus
Real valued functions of a real variable, into, onto and one-to-one functions, sum, difference, product and
quotient of two functions, composite functions, absolute value, polynomial, rational, trigonometric, exponential
and logarithmic functions.
Limit and continuity of a function, limit and continuity of the sum, difference, product and quotient of two
functions, l'Hospital rule of evaluation of limits of functions.
Even and odd functions, inverse of a function, continuity of composite functions, intermediate value property of
continuous functions.
Derivative of a function, derivative of the sum, difference, product and quotient of two functions, chain rule,
derivatives of polynomial, rational, trigonometric, inverse trigonometric, exponential and logarithmic functions.
Derivatives of implicit functions, derivatives up to order two, geometrical interpretation of the derivative, tangents
and normals, increasing and decreasing functions, maximum and minimum values of a function, applications of
Rolle's Theorem and Lagrange's Mean Value Theorem.
Integral Calculus
Integration as the inverse process of differentiation, indefinite integrals of standard functions, definite integrals
and their properties, application of the Fundamental Theorem of Integral Calculus.
Integration by parts, integration by the methods of substitution and partial fractions, application of definite
integrals to the determination of areas involving simple curves.
Formation of ordinary differential equations, solution of homogeneous differential equations, variables separable
method, linear first order differential equations.
Vectors
Addition of vectors, scalar multiplication, scalar products, dot and cross products, scalar triple products and their
geometrical interpretations.
1
Complex Numbers
Topic 1 Complex Number in Iota Form
Objective Questions I (Only one correct option)
2z − n
1. Let z ∈ C with Im (z ) = 10 and it satisfies
= 2i − 1
2z + n
for some natural number n, then (2019 Main, 12 April II)
(a) n = 20 and Re(z ) = − 10 (b) n = 40 and Re(z ) = 10
(c) n = 40 and Re(z ) = − 10 (d) n = 20 and Re(z ) = 10
α + i

2. All the points in the set S = 
: α ∈ R (i = −1 ) lie
α − i

on a
(2019 Main, 9 April I)
(a) circle whose radius is 2.
(b) straight line whose slope is −1.
(c) circle whose radius is 1.
(d) straight line whose slope is 1.
3. Let z ∈ C be such that|z|< 1. If ω =
5 + 3z
, then
5(1 − z )
3
1
x + iy
(i = −1 ), where x and y are real

3 
27
(2019 Main, 11 Jan I)
numbers, then y − x equals

(c) – 85
(d) – 91

π  3 + 2i sin θ
, π :
is purely imaginary 
 2  1 − 2i sin θ

5. Let A = θ ∈  −

Then, the sum of the elements in A is (2019 Main, 9 Jan I)
(a)
3π
4
(b)
5π
6
(c) π
(b)
6i
7. If 4
–3 i
20
3
 3
−1  1 
(c) sin −1 
 (d) sin 

 3
4


π
6
1
–1 = x + iy, then
3i
(1998, 2M)
i
(a) x = 3, y = 1 (b) x = 1, y = 1 (c) x = 0, y = 3 (d) x = 0, y = 0
13
(a) i
(b) 5 Re (ω) > 1
(d) 5 Re(ω) > 4
(b) 85
π
3
2 + 3i sin θ
is purely imaginary, is
1 − 2i sin θ
(2016 Main)
∑ (i n + i n + 1 ), where i =
−1, equals
n =1
4. Let  −2 − i =
(a) 91
(a)
8. The value of sum
(2019 Main, 9 April II)
(a) 4 Im(ω) > 5
(c) 5 Im (ω) < 1
6. A value of θ for which
(d)
2π
3
(b) i − 1
(1998, 2M)
(c) − i
(d) 0
n
 1 + i
 = 1, is
1 − i
9. The smallest positive integer n for which 
(a) 8
(c) 12
(b) 16
(d) None of these
(1980, 2M)
Objective Question II
(One or more than one correct option)
10. Let a , b, x and y be real numbers such that a − b = 1 and
y ≠ 0. If the complex number z = x + iy satisfies
 az + b
Im 
 = y, then which of the following is(are)
 z+1
possible value(s) of x?
(2017 Adv.)
(a) 1 − 1 + y2
(b) − 1 − 1 − y2
(c) 1 + 1 + y2
(d) − 1 +
1 − y2
Topic 2 Conjugate and Modulus of a Complex Number
Objective Questions I (Only one correct option)
1. The equation|z − i| = |z − 1|, i = −1, represents
1
(2019 Main, 12 April I)
2
(b) the line passing through the origin with slope 1
(c) a circle of radius 1
(d) the line passing through the origin with slope − 1
(a) a circle of radius
2. If a > 0 and z =
equal to
1 3
(a)
− i
5 5
1 3
(c) − + i
5 5
(1 + i )2
2
, has magnitude
, then z is
a−i
5
(2019 Main, 10 April I)
1
(b) − −
5
3
(d) − −
5
3
i
5
1
i
5
2 Complex Numbers
3. Let z1 and z2 be two complex numbers satisfying| z1 | = 9
and | z2 − 3 − 4i | = 4. Then, the minimum value of
(2019 Main, 12 Jan II)
| z1 − z2|is
(a) 1
(b) 2
(c)
(d) 0
2
z −α
4. If
(α ∈ R) is a purely imaginary number and
z+α
(2019 Main, 12 Jan I)
|z| = 2, then a value of α is
(a) 2
1
(b)
2
(c) 1
(where i = − 1).
Then,| z |is equal to
34
3
(b)
(2019 Main, 11 Jan II)
5
3
41
4
(c)
(d)
5
4
6. A complex number z is said to be unimodular, if z ≠ 1.
z – 2 z2
is
If z1 and z2 are complex numbers such that 1
2 – z1z2
unimodular and z2 is not unimodular.
Then, the point z 1 lies on a
(2015 Main)
(a) straight line parallel to X-axis
(d) circle of radius 2
7. If z is a complex number such that |z| ≥ 2, then the
1
2
(c) is strictly greater than 5/2
(x − x0 )2 + ( y − y0 )2 = r 2 and (x − x0 )2 + ( y − y0 )2 = 4r 2,
respectively.
If z0 = x0 + iy0 satisfies the equation 2|z0|2 = r 2 + 2, then
(2013 Adv.)
|α |is equal to
(c)
1
7
(d)
1
3
9. Let z be a complex number such that the imaginary part
of z is non-zero and a = z + z + 1 is real. Then, a cannot
take the value
(2012)
2
(b)
1
3
(c)
1
2
(d)
3
4
10. Let z = x + iy be a complex number where, x and y are
integers. Then, the area of the rectangle whose vertices
are the root of the equation zz3 + zz3 = 350, is
(2009)
(a) 48
(c) 40
 1 
1
2
(c)
(d)
⋅
2
z
+
1
|z + 1|2

 |z + 1|
1
|z + 1|
2
14. For all complex numbers z1 , z2 satisfying |z1| = 12 and
|z2 − 3 − 4i| = 5, the minimum value of|z1 − z2|is
(a) 0
(c) 7
(b) 2
(d) 17
(2002, 1M)
15. If z1 , z2 and z3 are complex numbers such that
1
1
1
+ = 1, then |z1 + z2 + z3|is
|z1| = |z2| = |z3| =  +
z1 z2 z3
(a) equal to 1
(c) greater than 3
(b) less than 1
(d) equal to 3
(2000, 2M)
(a) n1 = n2 + 1
(c) n1 = n2
(b) n1 = n2 − 1
(d) n1 > 0, n2 > 0
complex
numbers
sin x + i cos 2x
cos x − i sin 2x are conjugate to each other, for
(b) x = 0
(d) no value of x
and
(1988, 2M)
vertices of a parallelogram taken in order, if and only if
8. Let complex numbers α and 1 /α lies on circles
(a) − 1
(b)
18. The points z1 , z2, z3 and z4 in the complex plane are the
(d) is strictly greater than 3/2 but less than 5/2
1
2
(a) 0
z −1
(where, z ≠ − 1), then Re (w) is
z+1
(2003, 1M)
(a) x = nπ
(c) x = (n + 1/2) π
(b) lies in the interval (1, 2)
(b)
13. If|z| = 1 and w =
17. The
(2014 Main)
(a) is equal to 5/2
1
2
(b)| z | = 1 and z ≠ 1
(d) None of these
(1 + i )n 1 + (1 + i3 )n1 + (1 + i5 )n 2 + (1 + i7 )n 2 , here
(1996, 2M)
i = −1 is a real number, if and only if
(c) circle of radius 2
(a)
(a)| z | = 1, z ≠ 2
(c) z = z
16. For positive integers n1 , n2 the value of expression
(b) straight line parallel toY -axis
minimum value of z +
 w − wz 
condition that 
 is purely real, then the set of
 1−z 
values of z is
(2006, 3M)
(d) 2
5. Let z be a complex number such that | z | + z = 3 + i
(a)
12. If w = α + iβ, where β ≠ 0 and z ≠ 1, satisfies the
(b) 32
(d) 80
11. If|z|= 1 and z ≠ ± 1, then all the values of
(a) a line not passing through the origin
(b)|z|= 2
(c) the X-axis
(d) the Y-axis
(a) z1 + z4 = z2 + z3
(c) z1 + z2 = z3 + z4
(b) z1 + z3 = z2 + z4 (1983, 1M)
(d) None of these
19. If z = x + iy and w = (1 − iz ) / (z − i ), then |w| = 1 implies
that, in the complex plane
(1983, 1M)
(a) z lies on the imaginary axis (b) z lies on the real axis
(c) z lies on the unit circle
(d) None of these
20. The inequality |z − 4| < |z − 2| represents the region
given by
(1982, 2M)
(a) Re (z ) ≥ 0
(c) Re (z ) > 0
(b) Re (z ) < 0
(d) None of these
5
5
 3 i
 3 i
21. If z = 
+  +
−  , then
2
2


 2 2
(a) Re (z ) = 0
(c) Re (z ) > 0, Im (z ) > 0
(1982, 2M)
(b) Im (z ) = 0
(d) Re (z ) > 0, Im (z ) < 0
22. The complex numbers z = x + iy which satisfy the
z
lie on
1 − z2
(2007, 3M)
z − 5i 
 = 1, lie on
equation
z + 5i 
(a) the X-axis
(b) the straight line y = 5
(c) a circle passing through the origin
(d) None of the above
(1981, 2M)
Complex Numbers 3
29. Let z be any point in A ∩ B ∩ C and let w be any point
Objective Questions II
(One or more than one correct option)
satisfying |w − 2 − i| < 3.
between
23. Let S be the set of all complex numbers z satisfying
(a) − 6 and 3
(c) − 6 and 6
| z 2 + z + 1| = 1. Then which of the following statements
is/are TRUE?
(2020 Adv.)
1 1
≤ for all z ∈S
2 2
1 1
(c) z +
≥ for all z ∈S
2 2
(a) z +
solutions z = x + iy (x, y ∈ R, i = − 1 ) of the equation
sz + tz + r = 0, where z = x − iy. Then, which of the
following statement(s) is (are) TRUE?
(2018 Adv.)
(a) If L has exactly one element, then| s|≠ |t |
(b) If|s|=|t |, then L has infinitely many elements
(c) The number of elements in L ∩ {z :| z − 1 + i| = 5} is at most 2
(d) If L has more than one element, then L has infinitely many
elements
25. Let z1 and z2 be complex numbers such that z1 ≠ z2 and
30. Let z be any point in A ∩ B ∩ C.
The|z + 1 − i|2 + |z − 5 − i|2 lies between
(a) 25 and 29
(c) 35 and 39
(a) 0
(c) 2
that |z1| = |z2| = 1 and Re (z1z2) = 0, then the pair of
complex numbers w1 = a + ic and w2 = b + id satisfies
(1985, 2M)
Passage Based Problems
32. Match the statements of Column I with those of
Column II.
Here, z takes values in the complex plane and Im (z )
and Re (z ) denote respectively, the imaginary part and
(2010)
the real part of z
A.
The set of points z satisfying
| z − i| z|| = | z + i | z|| is
contained in or equal to
p.
an ellipse with
eccentricity 4/5
B.
The set of points z satisfying
| z + 4| + | z − 4| = 0 is
contained in or equal to
q.
the set of points z
satisfying Im ( z) = 0
C.
If| w| = 2 , then the set of
1
points z = w − is contained
w
in or equal to
r.
the set of points z
satisfying|Im( z) |≤ 1
D.
If| w| = 1, then the set of points s.
1
z = w + is contained in or
t.
w
equal to
Passage I
27. min|1 − 3i − z|is equal to
z ∈s
2− 3
(a)
2
2+ 3
(b)
2
3− 3
(c)
2
3+ 3
(d)
2
28. Area of S is equal to
(a)
10 π
3
(b)
20 π
3
(c)
16 π
3
(d)
32 π
3
Column II
Column I
Read the following passages and answer the questions
that follow.
Let A, B, C be three sets of complex number as defined
below
A = { z : lm (z ) ≥ 1}
B = { z :|z − 2 − i| = 3}
(2008, 12M)
C = { z : Re((1 − i )z ) = 2 }
(b) 1
(d) ∞
Match the Columns
(b) real and positive
(d) purely imaginary
26. If z1 = a + ib and z2 = c + id are complex numbers such
(b) 30 and 34
(d) 40 and 44
31. The number of elements in the set A ∩ B ∩ C is
|z1| = |z2|. If z1 has positive real part and z2 has negative
z + z2
imaginary part, then 1
may be
(1986, 2M)
z1 − z2
(b)|w2| = 1
(d) None of these
(b) − 3 and 6
(d) − 3 and 9
Let S = S1 ∩ S 2 ∩ S3 , where


z − 1 + 3 i 
S1 = { z ∈ C :|z | < 4}, S 2 = z ∈ C : lm 
 > 0
 1− 3i 


and S3 : { z ∈ C : Re z > 0}
(2008)
(d) The set S has exactly four elements
(a)|w1| = 1
(c) Re (w1 w2 ) = 0
lies
Passage II
(b)|z|≤ 2 for all z ∈S
24. Let s, t, r be non-zero complex numbers and L be the set of
(a) zero
(c) real and negative
Then, |z | − |w| + 3
the set of points
satisfying|Re( z)|≤ 2
the set of points z
satisfying| z| ≤ 3
Fill in the Blanks
33. If α , β, γ are the cube roots of p, p < 0, then for any x, y
and z then
xα + yβ + zγ
= ... .
xβ + yγ + zα
(1990, 2M)
34. For any two complex numbers z1 , z2 and any real
numbers a and b,|az1 − bz2|2+ |bz1 + az2|2 = K .
(1988, 2M)
4 Complex Numbers


 x
 x
sin  2 + cos  2 − i tan (x)

 is real,
35. If the expression

 x 
1 + 2 i sin  2 


then the set of all possible values of x is… .
(1987, 2M)
41. If z1 and z2 are two complex numbers such that
|z1| < 1 < |z2|, then prove that
1 − z1z2
< 1.
z1 − z2
(2003, 2M)
42. For complex numbers z and w, prove that
| z |2 w − |w|2 z = z − w, if and only if z = w or z w = 1.
(1999, 10M)
43. Find all non-zero complex numbers z satisfying
True/False
z = iz 2.
36. If three complex numbers are in AP. Then, they lie on a
circle in the complex plane
(1985 M)
37. If the complex numbers, z1 , z2 and z3 represent the
vertices of an equilateral triangle
| z1 | = | z2| = | z3 |, then z1 + z2 + z3 = 0.
such
that
(1984, 1M)
38. For complex numbers z1 = x1 + iy1 and z2 = x2 + iy2, we
write z1 ∩ z2, if x1 ≤ x2 and y1 ≤ y2. Then, for all complex
1−z
numbers z with 1 ∩ z, we have
∩ 0.
(1981, 2M)
1+ z
Analytical & Descriptive Questions
39. Find the centre and radius of the circle formed by all the
points represented by z = x + iy satisfying the relation
z − α 


= k (k ≠ 1), where α and β are the constant
z −β
complex numbers given by α = α 1 + iα 2, β = β1 + iβ 2.
(2004, 2M)
40. Prove that there exists no complex number z such that
|z| < 1 /3 and
n
∑
44. If
(1996, 2M)
iz + z − z + i = 0,
3
2
then
show
that
|z| = 1.
(1995, 5M)
45. A relation R on the set of complex numbers is defined
z1 − z2
is real.
z1 + z2
Show that R is an equivalence relation.
by z1 R z2, if and only if
(1982, 2M)
46. Find the real values of x and y for which the following
equation is satisfied
(1 + i ) x − 2i (2 − 3i ) y + i
+
= i.
3+ i
3−i
47. Express
( 1980, 2M)
1
in the form A + iB.
(1 − cos θ ) + 2i sin θ
(1979, 3M)
48. If x + iy =
a + ib
a 2 + b2
, prove that (x2 + y2)2 = 2
c + id
c + d2
(1978, 2M)
Integer & Numerical Answer Type Question
49. If z is any complex number satisfying | z − 3 − 2i | ≤ 2,
then the maximum value of|2z − 6 + 5i |is …… (2011)
a rz r = 1, where|a r|< 2.
r =1
(2003, 2M)
Topic 3 Argument of a Complex Number
Objective Questions I (Only one correct option)
1. If z1 , z2 are complex numbers such that Re(z1 ) = |z1 − 1|,
Re(z2) = |z2 − 1| and arg(z1 − z2) =
π
, then Im(z1 + z2) is
6
equal to
(a)
(2020 Main, 3 Sep II)
3
2
(b)
1
3
(c)
2
3
(d) 2 3
2. Let z1 and z2 be any two non-zero complex numbers such
that 3|z1| = 4|z2|. If z =
(a) |z| =
1 17
2 2
3z1 2z2
+
, then
2z2 3z1
(2019 Main, 10 Jan I)
(d) |z| =
5
2
(a) − θ
(b)
π
−θ
2
(c) θ
(2000, 2M)
(b) − π
(d) π /2
5. Let z and w be two complex numbers such that|z| ≤ 1,
|w| ≤ 1 and|z + i w|= |z − iw| = 2 , then z equals
(1995, 2M)
(a) 1 or i
(c) 1 or −1
(b) i or −i
(d) i or −1
6. Let z and w be two non-zero complex numbers
such that |z| = |w| and arg (z ) + arg (w) = π, then z
(1995, 2M)
equals
(b) −w
(d) −w
7. If z1 and z2 are two non-zero complex numbers such
3. If z is a complex number of unit modulus and argument
 1 + z
θ, then arg 
 is equal to
1 + z
(a) π
(c) − π/2
(a) w
(c) w
(b) Im(z ) = 0
(c) Re(z) = 0
4. If arg (z ) < 0 , then arg (−z ) − arg (z ) equals
(2013 Main)
(d) π − θ
that|z1 + z2| = |z1| + |z2|, then arg (z1 ) − arg (z2) is equal
to
(a) − π
(c) 0
π
(b) −
2
π
(d)
2
(1987, 2M)
Complex Numbers 5
8. If a , b, c and u , v, w are the complex numbers
10. Let z1 and z2 be two distinct complex numbers and let
representing the vertices of two triangles such that
c = (1 − r ) a + rb and w = (1 − r ) u + rv, where r is a
(1985, 2M)
complex number, then the two triangles
z = (1 − t ) z1 + tz2 for some real number t with 0 < t < 1. If
arg (w) denotes the principal argument of a non-zero
(2010)
complex number w, then
(a) have the same area
(c) are congruent
(a) |z − z1| + |z − z2|= |z1 − z2|(b) arg (z − z1 ) = arg (z − z2 )
(b) are similar
(d) None of these
(c)
Objective Questions II
(One or more than one correct option)
principal argument with − π < arg(z ) ≤ π . Then, which of
the following statement(s) is (are) FALSE ? (2018 Adv.)
π
, where i =
4
z − z1
z2 − z1
z2 − z1
=0
(d) arg (z − z1 ) = arg (z2 − z1 )
Match the Columns
9. For a non-zero complex number z, let arg(z ) denote the
(a) arg (−1 − i ) =
z − z1
11. Match the conditions/expressions in Column I with
statement in Column II (z ≠ 0 is a complex number)
Column II
Column I
−1
Re ( z) = 0
π
arg ( z) =
4
A.
(b) The
function
defined
by
f : R → (− π, π],
f (t ) = arg (−1 + it ) for all t ∈ R, is continuous at all
points of R, where i = −1.
(c) For any two non-zero complex numbers z1 and z2,
z 
arg  1  − arg (z1 ) + arg (z2 ) is an integer multiple of
 z2 
B.
p.
Re ( z2 ) = 0
q.
Im ( z2 ) = 0
r.
Re ( z2 ) = Im ( z2 )
Analytical & Descriptive Questions
12. |z| ≤ 1,|w|≤ 1, then show that
|z − w|2 ≤ (|z| − |w|)2 + (arg z − arg w)2
2π.
(d) For any three given distinct complex numbers z1 , z2 and
z3 , the locus of the point z satisfying the condition
 (z − z1 ) (z2 − z3 ) 
arg 
 = π, lies on a straight line.
 (z − z3 ) (z2 − z1 ) 
(1995, 5M)
13. Let z1 = 10 + 6i and z2 = 4 + 6i. If z is any complex
number such that the argument of (z − z1 ) / (z − z2) is
(1991, 4M)
π /4, then prove that|z − 7 − 9i| = 3 2.
Topic 4 Rotation of a Complex Number
Objective Questions I (Only one correct option)
1. Let S be the set of all complex numbers z satisfying
| z − 2 + i | ≥ 5. If the complex number z0 is such that

 1
1
is the maximum of the set 
: z ∈ S , then
| z0 − 1 |
|
z
−
1
|


the principal argument of
(a)
π
4
(b)
3π
4
5
4 − z0 − z0
is
z0 − z0 + 2 i
(c) −
π
2
(2019 Adv.)
(d)
π
2
5
 3 i
 3 i
+  +
−  . If R(z ) and I (z )
2
2
 2
 2
2. Let z = 
respectively denote the real and imaginary parts of z,
then
(2019 Main, 10 Jan II)
(a) R (z ) > 0 and I (z ) > 0
(c) R (z ) < 0 and I (z ) > 0
(b) I (z ) = 0
(d) R (z ) = − 3
3. A particle P starts from the point z0 = 1 + 2 i, where
i = −1. It moves first horizontally away from origin by
5 units and then vertically away from origin by 3 units
to reach a point z1. From z1 the particle moves 2 units
in the direction of the vector $i + $j and then it moves
π
in anti-clockwise direction on a
through an angle
2
circle with centre at origin, to reach a point z2. The point
(2008, 3M)
z2 is given by
(a) 6 + 7i
(b) −7 + 6i
(c) 7 + 6i
(d) − 6 + 7i
4. A man walks a distance of 3 units from the origin
towards the North-East (N 45° E) direction. From there,
he walks a distance of 4 units towards the North-West
(N 45° W) direction to reach a point P. Then, the
position of P in the Argand plane is
(2007, 3M)
(a) 3ei π/ 4 + 4 i
(b) (3 − 4 i ) ei π / 4
(c) (4 + 3 i )ei π / 4
(d) (3 + 4 i ) ei π / 4
5. The shaded region, where P = (−1, 0), Q = (−1 + 2 , 2 )
R = (−1 + 2 , − 2 ), S = (1, 0) is represented by (2005, 1M)
π
4
π
(b)|z + 1|< 2,| arg (z + 1)|<
2
π
(c)|z + 1|> 2,| arg (z + 1)|>
4
π
(d)|z − 1|< 2,| arg (z + 1)|>
2
Y
(a)|z + 1|> 2,| arg (z + 1)|<
Q
X′
P
S
O
X
R
Y′
π
is a fixed angle. If P = (cos θ ,sin θ ) and
2
Q = {cos(α − θ ),sin(α − θ )}, then Q is obtained from P by
6. If 0 < α <
(2002, 2M)
(a) clockwise rotation around origin through an angle α
(b) anti-clockwise rotation around origin through an angleα
(c) reflection in the line through origin with slope tan α
α
(d) reflection in the line through origin with slope tan
2
6 Complex Numbers
7. The complex numbers z1 , z2 and z3
satisfying
z1 − z3 1 − i 3
are the vertices of a triangle which is
=
z2 − z3
2
(2001, 1M)
(a) of area zero
(c) equilateral
(b) right angled isosceles
(d) obtuse angled isosceles
(2016 Adv.)
1
1 
and centre 
(a) the circle with radius
, 0 for
 2a 
2a
a > 0, b ≠ 0
1 
1
and centre  −
(b) the circle with radius −
, 0 for
 2a 
2a
a < 0, b ≠ 0
(c) the X-axis for a ≠ 0, b = 0
(d) the Y-axis for a = 0, b ≠ 0
3+i
and P = {W n: n = 1, 2, 3,... }.
2
1

Further H 1 = z ∈ C : Re (z ) > 
2

1
−


and H 2 = z ∈ C : Re (z ) <
, where C is the set of all

2 
complex numbers. If z1 ∈ P ∩ H 1, z2 ∈ P ∩ H 2 and O
represents the origin, then ∠ z1Oz2 is equal to
9. Let W =
π
6
(c)
14. Let bz + bz = c, b ≠ 0, be a line in the complex plane,
(1997C, 5M)
15. Let z1 and z2 be the roots of the equation z + pz + q = 0,
2


1
Suppose S = z ∈ C : z =
, t ∈ R, t ≠ 0, where
a
+
i
bt


i = − 1. If z = x + iy and z ∈ S, then (x, y) lies on
(b)
circle |z − 1| = 2 is 2 + 3i. Find the other vertices of
(2005, 4M)
square.
that c = z1b + z2b.
8. Let a , b ∈ R and a 2 + b2 ≠ 0.
π
2
13. If one of the vertices of the square circumscribing the
where b is the complex conjugate of b. If a point z1 is the
reflection of the point z2 through the line, then show
Objective Questions II
(One or more than one correct option)
(a)
Analytical & Descriptive Questions
2π
3
(2013 JEE Adv.)
(d)
5π
6
Fill in the Blanks
10. Suppose z1 , z2, z3 are the vertices of an equilateral
triangle inscribed in the circle |z| = 2. If z1 = 1 + i 3,
then z2 = K, z3 = … .
(1994, 2M)
11. ABCD is a rhombus. Its diagonals AC and BD intersect
at the point M and satisfy BD = 2 AC. If the points D and
M represent the complex numbers 1 + i and 2 − i
respectively, then A represents the complex number
…or…
(1993, 2M)
where the coefficients p and q may be complex numbers.
Let A and B represent z1 and z2 in the complex plane. If
∠ AOB = α ≠ 0 and OA = OB, where O is the origin prove
α
that p2 = 4q cos 2  .
 2
(1997, 5M)
16. Complex numbers z1 , z2, z3 are the vertices A , B, C
respectively of an isosceles right angled triangle with
right angle at C. Show that
(z1 − z2)2 = 2(z1 − z3 ) (z3 − z2).
(1986, 2 1 M)
2
17. Show that the area of the triangle on the argand
diagram formed by the complex number z , iz and z + iz
1
is |z|2.
2
(1986, 2 1 M)
2
18. Prove that the complex numbers z1 , z2 and the origin
form an equilateral triangle only if z12 + z22 − z1z2 = 0.
(1983, 2M)
19. Let the complex numbers z1 , z2 and z3 be the vertices of
an equilateral triangle. Let z0 be the circumcentre of the
triangle. Then, prove that z12 + z22 + z32 = 3z02. (1981, 4M)
Integer & Numerical Answer Type Questions
20. For a complex number z, let Re(z ) denote the real part of
z. Let S be the set of all complex numbers z satisfying
z 4 − |z|4 = 4 i z 2, where i = −1. Then the minimum
possible value of|z1 − z2|2, where z1 , z2 ∈ S with Re(z1 ) > 0
and Re(z2) < 0 is …… .
(2020 Adv.)
kπ 
 kπ 
 + i sin   , where
 7
 7
i = −1. The value of the expression
21. For any integer k, let α k = cos 
12
∑|α k + 1 − α k|
k =1
12. If a and b are real numbers between 0 and 1 such that
the points z1 = a + i , z2 = 1 + bi and z3 = 0 form an
equilateral triangle, then a = K and b = K . (1990, 2M)
3
∑|α 4k − 1 − α 4k − 2|
k =1
is
(2016 Adv.)
Complex Numbers 7
Topic 5 De-Moivre’s Theorem, Cube Roots and nth Roots of Unity
Objective Questions I (Only one correct option)
9. Let z1 and z2 be nth roots of unity which subtend a right
angled at the origin, then n must be of the form
(2001, 1M)
(where, k is an integer)
3
2π
2π 

+ i cos

 1 + sin
9
9  is
1. The value of 
 1 + sin 2π − i cos 2π 

9
9 
(a) 4k + 1
[2020 Main, 2 Sep I]
1
(a) − ( 3 − i )
2
1
(c) ( 3 − i )
2
1
(b) − (1 − i 3 )
2
1
(d) (1 − i 3 )
2
π
, then
2
3. If z =
to
(2019 Main, 10 April II)
(a) 1
(c) −1
(d) 0
(d)
5. Let z = cos θ + i sin θ . Then, the value of
π
3
∑ Im (z
θ = 2° is
2m − 1
) at
6. The minimum value of |a + bω + cω |, where a, b and c
are all not equal integers and ω (≠ 1) is a cube root of
unity, is
(2005, 1M)
(c) 1
(d) 0
7. If ω (≠ 1) be a cube root of unity and (1 + ω 2)n = (1 + ω 4 )n,
then the least positive value of n is
(a) 2
(b) 3
(c) 5
(2004, 1M)
(d) 6
1
3
, then value of the determinant
+i
2
2
1
1
1
1 −1 − ω 2 ω 2 is
(2002, 1M)
ω2
ω
1
8. Let ω = −
(a) 3 ω
(a) – 1
(b) 3 ω (ω − 1) (c) 3 ω2
3
(d) −128 ω2
B
are
respectively
(b) 1, 1

∑ sin
(c) 1, 0
(d) 3 ω (1 − ω)
(d) −1, 1
2 πk
2 πk 
– i cos
 is
7
7 
(b) 0
(c) – i
(1998, 2M)
(d) i
Match the Columns
2 kπ 
 2 kπ 
 ; k = 1, 2, …9.
 + i sin 
 10 
 10 
14. Let zk = cos 
Column II
P.
For each zk, there exists a z j such that
zk ⋅ z j = 1
(i)
True
Q.
There exists a k ∈ { 1, 2, … , 9 } such that
z1 ⋅ z = zk has no solution z in the set of
complex numbers
(ii)
False
R.
|1 − z1||1 − z2| … |1 − z9|
equal
10
9
2 kπ 
1 − ∑ cos 
 equals
 10 
k =1
(iii)
1
(iv)
2
S.
2
1
2
and
A
k =1
(2009)
1
(b)
3 sin 2°
1
(d)
4 sin 2°
(b)
(d) −i
(1998, 2M)
(c) 128 ω2
Column I
m =1
1
(a)
sin 2°
1
(c)
2 sin 2°
3
12. If ω (≠ 1) is a cube root of unity and (1 + ω )7 = A + Bω,
(2019 Main, 9 Jan II)
(c) 0
(b) −128 ω
6
15
(a) 3
(a) 128 ω
13. The value of
z = 3 + 6iz081 − 3iz093 , then arg z is equal to
π
(b)
6
(c) i
(1995, 2M)
4. Let z0 be a root of the quadratic equation, x2 + x + 1 = 0, If
π
(a)
4
365
(1999, 2M)
(b) −1 + i 3
3
(a) 0, 1
(2019 Main, 8 April II)
(b) (−1 + 2i )
 1 i 3
+ 3 − +

2 
 2
(1 + ω − ω 2)7 is equal to
then
3 i
+ (i = −1 ), then (1 + iz + z5 + iz 8 )9 is equal
2
2
9
(d) 4k
11. If ω is an imaginary cube root of unity, then
(b) zw =
(c) zw = i
i 3

2 
334
is equal to
1− i
2
− 1+ i
(d) zw =
2
(a) zw = − i
 1
 2
(c) 4k + 3
10. If i = −1, then 4 + 5  − +
(a) 1 − i
2. If z and w are two complex numbers such that | zw| = 1
and arg(z ) − arg(w) =
(b) 4k + 2
(2011)
Codes
P
(a) (i)
(b) (ii)
(c) (i)
(d) (ii)
Q
(ii)
(i)
(ii)
(i)
R
(iv)
(iii)
(iii)
(iv)
S
(iii)
(iv)
(iv)
(iii)
Fill in the Blanks
2π
2π
+ i sin . Then
3
3
the number of distinct complex number z satisfying
15. Let ω be the complex number cos
ω
ω2
z+1
2
ω
1 = 0 is equal to ... .
z+ω
ω2
1
z+ω
(2010)
8 Complex Numbers
16. The value of the expression
19. If 1, a1 , a 2, ... , a n − 1 are the n roots of unity, then show
1 ( 2 − ω ) (2 − ω 2) + 2(3 − ω ) (3 − ω 2) + ...
+ (n − 1) ⋅ (n − ω ) (n − ω 2) ,
where, ω is an imaginary cube root of unity, is….
that (1 − a1 ) (1 − a 2) (1 − a3 ) K (1 − a n − 1 ) = n
(1984, 2M)
20. It is given that n is an odd integer greater than 3, but n
(1996, 2M)
True/False
is not a multiple of 3. Prove that x3 + x2 + x is a factor of
(1980, 3M)
(x + 1)n − xn − 1 .
21. If x = a + b, y = aα + bβ, z = aβ + bα , where α , β are
17. The cube roots of unity when represented on Argand
diagram form the vertices of an equilateral triangle.
complex cube
xyz = a3 + b3 .
roots
of
unity,
then
show
that
(1979, 3M)
(1988, 1M)
Integer & Numerical Answer Type Questions
Analytical & Descriptive Questions
18. Let a complex number α , α ≠ 1, be a root of the equation
z p + q − z p − zq + 1 = 0
where, p and q are distinct primes. Show that either
1 + α + α 2 + ... + α p − 1 = 0
or
1 + α + α 2 + ... + α q − 1 = 0
but not both together.
(2002, 5M)
22. Let ω ≠ 1 be a cube root of unity. Then the minimum of
the set {| a + bω + cω 2|2 : a , b, c distinct non-zero
integers} equals ............
(2019 Adv.)
23. Let ω = eiπ /3 and a , b, c, x, y, z be non-zero complex
numbers such that a + b + c = x, a + bω + cω 2 = y,
a + bω 2 + cω = z.
| x|2 + | y|2 + | z |2
Then, the value of
(2011)
is ……
| a |2 + | b|2 + | c|2
Answers
Topic 1
1. (c)
5. (d)
9. (d)
2. (c)
6. (d)
10. (b, d)
3. (b)
7. (d)
4. (a)
8. (b)
1
cot (θ / 2 )
−i
θ


1 + 3 cos2 (θ / 2 )
2 1 + 3 cos2 

2
49. 5
Topic 3
1. (d)
Topic 2
1. (b)
47. A + iB =
2. (b)
3. (d)
4. (d)
5. (c)
9. (a, b, d)
2. (*)
6. (d)
10. (a, c, d)
3. (c)
4. (a)
7. (c)
8. (b)
11. A → q ; B → p
5. (b)
6. (c)
7. (b)
8. (c)
9. (d)
10. (a)
11. (d)
12. (b)
13. (a)
14. (b)
15. (a)
16. (d)
1. (c)
2. (b)
3. (d)
4. (d)
17. (d)
18. (b)
19. (b)
20. (d)
5. (a)
6. (d)
7. (c)
8. (a,c,d)
21. (b)
22. (a)
23. (b,c)
24. (a, c, d)
25. (a,d)
26. (a, b, c)
27. (c)
28. (b)
29. (d)
30. (c)
31. (b)
32. A → q, r ; B → p; C → p, s, t ; D → q, r, s, t
33. ω
2
34. (a + b )(| z1| + | z 2| )
2
2
2
2
−1
35. x = 2nπ + 2α , α = tan k, where k ∈(1, 2 ) or x = 2nπ
36. False
37. True
38. True
k (α − β )
α − k 2β

39. Centre =
, Radius = 
2
2
1 −k
 1 −k 
Topic 4
9. (c, d)
10. z 2 = − 2, z 3 = 1 − i 3
i
3i
11. 3 − or 1 −
12. a = b = 2 ± 3
2
2
13. z 2 = − 3 i , z 3 = (1 − 3 ) + i and z 4 = (1 + 3 ) − i
20. (8)
21. (4)
Topic 5
1. (a)
2. (a)
3. (c)
3. (a)
5. (d)
6. (c)
7. (b)
8. (b)
9. (d)
10. (c)
11. (d)
12. (b)
14. (c)
15. (1)
13. (d)

3 i
43. z = i , ±
– 
2
2

 n (n + 1 )
16. 
 −n

2 
46. (x = 3 and y = −1)
23. (3)
2
17. True
22. (3)
Hints & Solutions
Topic 1 Complex Number in Iota Form
1. Let z = x + 10i, as Im (z ) = 10 (given).
Since z satisfies,
2z − n
= 2i − 1, n ∈ N ,
2z + n
∴ (2x + 20i − n ) = (2i − 1) (2x + 20i + n )
⇒ (2x − n ) + 20i = (− 2x − n − 40) + (4x + 2n − 20)i
On comparing real and imaginary parts, we get
2x − n = − 2x − n − 40 and 20 = 4x + 2n − 20
⇒
4x = − 40 and 4x + 2n = 40
⇒
x = − 10 and − 40 + 2n = 40 ⇒ n = 40
So,
n = 40 and x = Re (z ) = − 10
α+i
2. Let x + iy =
α −i
(α + i )2 (α 2 − 1) + (2α )i α 2 − 1  2α 
⇒ x + iy = 2
=
+
= 2
i
α +1
α2 + 1
α + 1  α 2 + 1
3
 3 + 2i sin θ   1 + 2 i sin θ 
 ×

 1 − 2i sin θ   1 + 2 i sin θ 
5. Let z = 
(rationalising the denominator)
=
2α
α2 −1
and y = 2
α +1
α2 + 1
2
2
 α 2 − 1
 2α 
Now, x + y =  2
 + 2

 α + 1
 α + 1
2
=
2
α 4 + 1 − 2α 2 + 4α 2 (α 2 + 1)2
= 2
=1
(α 2 + 1)2
(α + 1)2
⇒
x2 + y2 = 1
Which is an equation of circle with centre (0, 0) and
radius 1 unit.
3 − 4 sin θ + 8i sin θ
1 + 4 sin 2 θ
2
[Q a 2 − b2 = (a + b)(a − b) and i 2 = − 1]
 3 − 4 sin θ   8 sin θ 
=
 +
i
 1 + 4 sin 2 θ   1 + 4 sin 2 θ 
On comparing real and imaginary parts, we get
x=
2
As z is purely imaginary, so real part of z = 0
3 − 4 sin 2 θ
= 0 ⇒ 3 − 4 sin 2 θ = 0
∴
1 + 4 sin 2 θ
3
⇒
sin 2 θ =
4
3
⇒
sin θ = ±
2
Y
1
√3/2

α + i
So, S = 
; α ∈ R lies on a circle with radius 1.

α − i
X′
3. Given complex number
5 ω − 5 ω z = 5 + 3z
(3 + 5 ω )z = 5 ω − 5
…(i)
|3 + 5 ω||z| = |5 ω − 5|
[applying modulus both sides and |z1z2| = |z1||z2|]
Q
|z| < 1
[from Eq. (i)]
∴ |3 + 5 ω| > |5 ω − 5|
3

⇒
ω + > |ω − 1|
5

2
3

Let ω = x + iy, then  x +  + y2 > (x − 1)2 + y2

5
9
6
2
2
⇒
x +
+ x > x + 1 − 2x
25 5
1
16x 16
⇒
>
⇒ x > ⇒ 5x > 1
5
5
25
⇒
5 Re( ω ) > 1
y=sin θ
–π/2 –π/3
O π/3
5 + 3z
ω=
5(1 − z )
⇒
⇒
⇒
3
1   –1
x + iy 

=  − 2 − i =
(6 + i )


27
3   3
1
x + iy
⇒
=−
(216 + 108i + 18i 2 + i3 )
27
27
1
=−
(198 + 107i )
27
[Q (a + b)3 = a3 + b3 + 3a 2b + 3ab2, i 2 = − 1, i3 = − i]
On equating real and imaginary part, we get
x = − 198 and y = − 107
⇒ y − x = − 107 + 198 = 91
4. We have,
−1
2π/3
π
X
–√3/2
Y′
 π π 2π 
θ ∈ − , ,

 3 3 3 
2π
.
Sum of values of θ =
3
2 + 3i sin θ
6. Let z =
is purely imaginary. Then, we have
1 − 2i sin θ
Re(z ) = 0
2 + 3i sin θ
Now, consider z =
1 − 2i sin θ
(2 + 3i sin θ) (1 + 2i sin θ )
=
(1 − 2i sin θ ) (1 + 2i sin θ)
⇒
=
2 + 4i sin θ + 3i sin θ + 6i 2 sin 2 θ
12 − (2i sin θ) 2
10 Complex Numbers
=
2 + 7i sin θ − 6 sin 2 θ
1 + 4 sin 2 θ
=
2 − 6 sin 2 θ
7 sin θ
+i
1 + 4 sin 2 θ
1 + 4 sin 2 θ
Re(z ) = 0
2 − 6 sin 2 θ
= 0 ⇒ 2 = 6 sin 2 θ
1 + 4 sin 2 θ
1
sin 2 θ =
3
1
sin θ = ±
3
1
−1 
−1  1 
θ = sin  ±
 = ± sin  


 3
3
Q
∴
⇒
⇒
⇒
7. Given,
 6i
4
20
 6i
−3 i  4
 20
⇒
−3 i
3i
3
1
−1  = x + iy
i
1
−1
i
1
−1  = x + i y
i
⇒
x = 0, y = 0
13
8.
∑ (i
+i
n
n=1
n+1
13
) = ∑ i (1 + i ) = (1 + i )
n
n=1
1. Let the complex number z = x + iy
| z − i | =| z − 1|
Also given,
⇒
| x + iy − i | = | x + iy − 1|
⇒
x2 + ( y − 1)2 = (x − 1)2 + y2
[Q| z | = (Re(z ))2 + (Im(z ))2 ]
On squaring both sides, we get
x2 + y2 − 2 y + 1 = x2 + y2 − 2x + 1
⇒ y = x, which represents a line passing through the
origin with slope 1.
(1 + i )2
a−i
(1 − 1 + 2i ) (a + i )
=
a2 + 1
2i (a + i ) −2 + 2ai
=
=
a2 + 1
a2 + 1
2. The given complex number z =
x + iy = 0 [Q C 2 and C3 are identical]
⇒
Topic 2 Conjugate and Modulus of
Complex Number
13
∑
i
n
n =1
 i − (1 − i13 ) 
= (1 + i ) (i + i 2 + i3 + K + i13 ) = (1 + i ) 


 1−i
 i (1 − i ) 
= (1 + i ) 
 = (1 + i ) i = i − 1
 1−i 
Alternate Solution
4 + 4a 2
2
2
2
⇒
=
=
2
5
5
(a 2 + 1)2
1+ a
4
2
= ⇒ a 2 + 1 = 10 ⇒ a 2 = 9 ⇒ a = 3 [Qa > 0]
⇒
1 + a2 5
–2 + 6i
[From Eq. (i)]
∴
z=
10
1 3
 −2 + 6 i   1 3 
So, z = 
 =  − + i ⇒ z = − − i
 10   5 5 
5 5
+ (i 2 + i3 + K + i14 ) = i + i 2 = i − 1
n
 1 + i

 =1
1 − i
n
 1 + i 1 + i
×
⇒ 
 =1
 1 − i 1 + i
n
⇒
 2i 
  =1
 2
⇒
in = 1
The smallest positive integer n for which i n = 1 is 4.
∴
n =4
az + b ax + b + aiy (ax + b + aiy)((x + 1) − iy)
10.
=
=
z+1
(x + 1) + iy
(x + 1)2 + y2
∴
 az + b − (ax + b) y + ay(x + 1)
Im 
=
 z+1
(x + 1)2 + y2
⇒
(a − b) y
=y
(x + 1)2 + y2
Q
∴
a − b =1
(x + 1)2 + y2 = 1
∴
[Qif z = x + iy, then z = x − iy]
3. Clearly|z1|= 9, represents a circle having centre C1 (0, 0)
n=1
9. Since,
x = − 1 ± 1 − y2
[given]
⇒
13
∑ (i n + i n + 1 ) = (i + i 2 + K + i13 )
…(i)
z = 2 /5
Q
Since, sum of any four consecutive powers of iota is zero.
∴
[Q i 2 = − 1]
and radius r1 = 9.
and |z2 − 3 − 4i|= 4 represents a circle having centre
C 2(3, 4) and radius r2 = 4.
The minimum value of |z1 − z2| is equals to minimum
distance between circles|z1|= 9 and|z2 − 3 − 4i|= 4.
Q
C1C 2 = (3 − 0)2 + (4 − 0)2 = 9 + 16 = 25 = 5
and
|r1 − r2|=|9 − 4|= 5 ⇒ C1C 2 =|r1 − r2|
∴ Circles touches each other internally.
Hence,
|z1 − z2|min = 0
z −α
4. Since, the complex number
(α ∈ R) is purely
z+α
imaginary number, therefore
z −α
z −α
[Qα ∈ R]
+
=0
z+α z+α
⇒
zz − αz + αz − α 2 + zz − αz + αz − α 2 = 0
⇒
2 z
⇒
α2 = z
⇒
α=±2
2
− 2 α2 = 0
2
=4
2
[Qzz = z ]
[| z | = 2 given]
Complex Numbers 11
5. We have,|z | + z = 3 + i
∴
x2 + y2 + x + iy = 3 + i
⇒
(x + x + y ) + iy = 3 + i
⇒
x + x2 + y2 = 3 and y = 1
2
⇒
∴
⇒
= |z1|2 − z1z2 − z2z1 + |z2|2
Here, (x − x0 ) + ( y − y0 ) = r 2
2
|z − z0|2 = r 2 and |z − z0|2 = 4r 2
Since, α and
∴
⇒
(α − z0 ) (α − z0 ) = r 2
⇒
|α|2 − z0α − z0α + |z0|2 = r 2
⇒ (z1 − 2z2)(z1 − 2z2) = (2 − z1z2) (2 − z1z2)
[zz = |z|2 ]
⇒ |z1|2+4|z2|2−2z1z2 − 2z1z2
⇒

 1
1
2
 − z0   − z0  = 4 r

 α
α
⇒
z
z
1
− 0 − 0 + |z0|2 = 4r 2
α
|α|2 α
|α|2 = α ⋅ α
Since,
= 4+|z1|2|z2|2−2z1z2 − 2z1z2 ⇒ (|z2|2−1)(|z1|2−4) = 0
Q
|z2|≠ 1
∴
|z1|= 2
Let
z1 = x + iy ⇒ x2 + y2 = (2)2
∴ Point z1 lies on a circle of radius 2.
⇒
z ⋅α
z
1
− 0
− 0 ⋅ α + |z0|2 = 4r 2
|α|2 |α|2 |α|2
1 − z0α − z0α + |α|2|z0|2 = 4r 2|α|2
⇒
(|α|2 − 1) + |z0|2 (1 − |α|2 ) = r 2 (1 − 4|α|2 )
(0, 0) and radius is 2.
1
is distance of z, which lie on circle
Minimum z +
2
| z | = 2 from (−1 / 2, 0).
1
 1 
= Distance of  − , 0 from (−2, 0)
∴ Minimum z +
 2 
2
⇒
(|α|2 − 1) (1 − |z0|2 ) = r 2(1 − 4|α|2 )
⇒

r 2 + 2
(|α|2 − 1) 1 −
 = r 2(1 − 4|α|2 )
2 

|z0|2 =
Given,
2
1
3
3
 −1

+ 2 + 0 =
 +0= = 
 2

2
2
2
⇒
|α|2 − 1 = − 2 + 8|α|2 ⇒ 7|α|2 = 1
|α| = 1 / 7
∴
9.
A
1 , (0,0)
(– —
)
2 0
(2,0)
1
= AD
2
1
Hence, minimum value of z +
lies in the interval
2
(1, 2).
Geometrically Min z +
PLAN If ax 2 + bx + c = 0 has roots α, β, then
α, β =
X
Y′
r2 + 2
2
 − r 2
(|α|2 − 1) ⋅ 
 = r 2(1 − 4|α|2 )
 2 
⇒
Y
D
(–2,0)
…(ii)
On subtracting Eqs. (i) and (ii), we get
7. |z| ≥ 2 is the region on or outside circle whose centre is
X′
…(i)
1
− z0 = 4 r 2
α
and
z1 − 2z2
= 1 ⇒ |z1 − 2z2|2 =|2 − z1z2|2
2 − z1z2
2
1
lies on first and second respectively.
α
2
1
− z0 = 4 r 2
|α − z0|2 = r 2 and
α
2
Given, z2 is not unimodular i.e.|z2|≠ 1
z − 2 z2
is unimodular.
and 1
2 − z1z2

=  −2 +

2
(x − x0 )2 + ( y − y0 )2 = 4r 2 can be written as,
and
PLAN If z is unimodular, then| z| = 1. Also, use property of modulus
i.e. z z =| z|2
⇒
|z|2 = z ⋅ z
|z1 − z2|2 = (z1 − z2) (z1 − z2)
and
x2 + 1 = 9 − 6x + x2
4
6x = 8 ⇒ x =
3
4
z= +i
3
5
16
25
|z | =
+1=
⇒ |z | =
9
9
3
⇒
PLAN Intersection of circles, the basic concept is to solve the
equations simultaneously and using properties of modulus of
complex numbers.
Formula used
2
x2 + 1 = 3 − x
Now,
6.
8.
z = x + iy
Let
− b ± b 2 − 4ac
2a
For roots to be real b 2 − 4ac ≥ 0.
Description of Situation
z = x + iy is non-zero.
⇒
y ≠0
As imaginary part of
Method I Let z = x + iy
∴
a = (x + iy)2 + (x + iy) + 1
⇒
(x2 − y2 + x + 1 − a ) + i (2xy + y) = 0
⇒
(x2 − y2 + x + 1 − a ) + iy (2x + 1) = 0,
…(i)
12 Complex Numbers
It is purely real, if y (2x + 1) = 0
On squaring both sides, we get
1 + |w|2 − 2|w| Re (w) = 1 + |w|2 + 2|w| Re (w)
but imaginary part of z, i.e. y is non-zero.
[using|z1 ± z2|2 = |z1|2 + |z2|2 ± 2|z1||z2| Re (z1z2)]
⇒
2x + 1 = 0 or x = − 1 / 2
1
1
From Eq. (i),
− y2 − + 1 − a = 0
4
2
3
2 3
a =− y +
⇒ a<
⇒
4
4
Method II
4|w|Re|w| = 0
⇒
Re (w) = 0
14. We know, |z1 − z2| = |z1 − (z2 − 3 − 4i ) − (3 + 4i )|
≥ |z1| − |z2 − 3 − 4i | − |3 + 4i|
Here, z 2 + z + (1 − a ) = 0
∴
z=
− 1 ± 1 − 4 (1 − a )
2 ×1
⇒
z=
− 1 ± 4a − 3
2
For z do not have real roots, 4 a − 3 < 0 ⇒
≥ 12 − 5 − 5
∴
Alternate Solution
a<
3
4
Clearly from the figure|z1 − z2|is minimum when z1 , z2
lie along the diameter.
Y
2
2
2
,
(3
x2 − y2 = 7
x2 = 16, y2 = 9 ⇒ x = ± 4, y = ± 3
Y′
∴
|z1 − z2| ≥ C 2B − C 2A ≥ 12 − 10 = 2
|z1| = |z2| = |z3| = 1
|z1| = 1
Now,
11. Let z = cos θ + i sin θ
z
cos θ + i sin θ
=
1 − z 2 1 − (cos 2 θ + i sin 2 θ )
cos θ + i sin θ
=
2 sin 2 θ − 2i sin θ cos θ
cos θ + i sin θ
i
=
=
− 2i sin θ (cos θ + i sin θ ) 2 sin θ
z
lies on the imaginary axis i.e. Y-axis.
Hence,
1 − z2
⇒
Alternate Solution
z
z
1
Let E =
which is an imaginary.
=
=
1 − z 2 zz − z 2 z − z
w − wz
12. Let z1 =
be purely real ⇒ z1 = z1
1−z
w − wz w − wz
=
∴
1− z
1−z
w − wz − wz + wz ⋅ z = w − zw − wz + wz ⋅ z
⇒
⇒ (w − w ) + (w − w)| z |2 = 0
(w − w ) (1 − | z |2 ) = 0
⇒
| z |2 = 1 [as w − w ≠ 0, since β ≠ 0]
⇒
| z | = 1 and z ≠ 1
z −1
13. Since,|z| = 1 and w =
z+1
|1 + w|
1+ w
⇒ z − 1 = wz + w ⇒ z =
⇒ |z| =
|1 − w|
1−w
|1 − w| = |1 + w|
[ Q|z| = 1]
X
12
C2
15. Given,
∴ Area of rectangle = 8 × 6 = 48
Z1
Z2
4)
C1
X′
... (ii)
From Eqs. (i) and (ii),
⇒
B
A
2 (x + y ) (x − y ) = 350
(x2 + y2) (x2 − y2) = 175
2
Since, x, y ∈ I, the only possible case which gives
integral solution, is
... (i)
x2 + y2 = 25
⇒
[using|z1 − z2| ≥ |z1| − |z2|]
|z1 − z2| ≥ 2
zz (z 2 + z 2) = 350
10. Since,
⇒
⇒
⇒
⇒
|z1|2 = 1
⇒
z1z1 = 1
Similarly,
z2z2 = 1, z3 z3 = 1

1 + 1 + 1 
= 1
z1 z2 z3
Again now,
⇒ | z1 + z2 + z3 |= 1 ⇒ |z1 + z2 + z3|= 1
⇒
|z1 + z2 + z3| = 1
16. (1 + i )n1 + (1 − i )n1 + (1 + i )n2 + (1 − i )n2
= [n1 C 0 +
n1
C1i +
+ [n1 C 0 −
+ [n2 C 0 +
+ [n2 C 0 –
= 2 [n1 C 0 +
n1
= 2 [ C0 −
n1
n1
C 2i 2 +
C3 i3 + K ]
C1 i + n1 C 2i 2 − n1 C3 i3 + ... ]
C1i + n2C 2i 2 + n2C3 i3 + K ]
n2
C1 i + n2C 2i 2 – n2C3 i3 + .. ]
n1
n2
C2 i2 +
n1
C 4i 4 + K ]
+ 2 [n2 C 0 +
n1
n1
C2 +
n1
n2
C 2i 2 +
n2
C 4i 4 + K ]
C4 − K ] + 2 [ C0 −
n2
n2
+
C2
n2
C 4 −... ]
This is a real number irrespective of the values of n1 and
n2.
Alternate Solution
{(1 + i )n1 + (1 − i )n1 } + {(1 + i )n2 + (1 − i )n2 }
⇒ A real number for all n1 and n2 ∈ R.
[Q z + z = 2 Re (z ) ⇒ (1 + i )n1 + (1 − i )n1
is real number for all n ∈ R]
Complex Numbers 13
17. Since, (sin x + i cos 2x) = cos x − i sin 2x
⇒
sin x − i cos 2x = cos x − i sin 2x
⇒
sin x = cos x and cos 2x = sin 2x
⇒
tan x = 1 and tan 2x = 1
⇒ x = π / 4 and x = π / 8 which is not possible at same
time.
Alternate Solution
We know that,
z + z = 2 Re(z )
5
5
 3 i
 3 i
If
z=
+  +
−  , then
2
2
 2
 2
z is purely real. i.e. Im (z ) = 0
z − 5i 
 = 1 ⇒ |z − 5i| = |z + 5i|
z + 5i 
22. Given, 
Hence, no solution exists.
[Q if|z − z1| = |z − z2|, then it is a perpendicular
bisector of z1 and z2 ]
18. Since, z1 , z2, z3 , z4 are the vertices of parallelogram.
D(z4)
C(z3)
Y
(0, 5)
O
X′
X
(0, –5)
Y′
A(z1)
B(z2)
∴ Perpendicular bisector of (0, 5) and (0, – 5) is X-axis.
∴ Mid-point of AC = mid-point of BD
z1 + z3 z2 + z4
=
⇒
2
2
⇒
z1 + z3 = z2 + z4
23. It is given that the complex number Z, satisfying
|z 2 + z + 1| = 1
 1 − iz 
= 1
z−i 
19. Since,|w| = 1 ⇒ 
⇒
⇒
|z − i| = |1 − iz|
|z − i | = |z + i |

z +

⇒
Q
|z1 − z2| ≥ ||z1| − |z2||
∴
1
1
3

 3
− −
z +  − −  ≥ z +




2
4
2
4
2
[Q |1 − iz | = | − i || z + i | = | z + i |]
∴It is a perpendicular bisector of (0, 1) and (0, − 1)
i.e. X-axis. Thus, z lies on the real axis.
2
2
⇒
z+
20. Given,|z − 4| < |z − 2|
Since, |z − z1| > |z − z2| represents the region on right
side of perpendicular bisector of z1 and z2.
∴
|z − 2| > |z − 4|
⇒
Re (z ) > 3 and Im (z ) ∈ R
Y
X′
O
(2, 0) (3, 0) (4, 0)
5
 3 i
 3 i
+  +
− 
2
2
 2
 2
X
⇒
1
3
− ≤1
2
4
2
⇒
−
1
1
7
≤ z+
≤
4
2
4
 −1 + i 3 
3+i
= −i 
 = − iω
2
2


 −1 − i 3 
3−i
=i
 = iω 2
2
2


z = (− iω )5 + (iω 2)5 = − iω 2 + iω
= i(ω − ω 2) = i (i 3 ) = − 3
Re(z ) < 0 and lm (z ) = 0
0≤ z+
1
7
≤
2
4
⇒
0≤ z+
1
7
≤
2
2
Q
z+
1
7
≤
2
2
5

−1 + i 3
−1 − i 3 
and ω 2 =

Q ω =
2
2


∴
−1 ≤ z +
2
21. Given, z = 
and
1
3
− ≤1
2
4
2
⇒
⇒
Y′
Now,
2
1
3
 + =1

2
4
Q
∴
|z1 + z2| ≥ ||z1| − |z2||
1
1
z + ≥ |z| −
2
2
1
1
7
≤ z+ ≤
2
2
2
⇒
|z| −
⇒
7
1
7
≤ |z| − ≤
2
2
2
1− 7
7+1
≤ |z| ≤
2
2
1+ 7
|z| ≤
2
|z| ≤ 2
⇒
⇒
∴
−
{Q|z| ≥ 0}
… (i)
14 Complex Numbers

z +

Q
2
1≤ z+
2
⇒
|w2| = b2 + d 2 = a 2 + b2 = 1
1
3
1
3
+ ≥1
+
⇒ z+
2
4
2
4
1
1
z+
≥
2
4
|w1| = a 2 + c2 = a 2 + b2 = 1
Now,
2
2
⇒
Also, given w1 = a + ic and w2 = b + id
2
1
3
1
3
+
 + ≤ z+
2
4
2
4
⇒
1 1
z+ ≥
2 2
… (ii)
and Re(w1 w2) = ab + cd = (bλ )b + c(− λc) [from Eq. (i)]
= λ (b2 − c2) = 0
27. min|1 − 3 i − z|= perpendicular distance of point (1, − 3)
Z ∈S
From Eqs. (i) and (ii), we get
1
1
7
≤ z+ ≤
2
2
2
from the line
3x + y = 0 ⇒
28. Since, S = S1 ∩ S 2 ∩ S3
24. We have,
sz + tz + r = 0
On taking conjugate
sz + tz + r = 0
On solving Eqs. (i) and (ii), we get
rt − rs
z= 2
|s| − |t|2
Y
…(i)
…(ii)
150°
X′
For unique solutions of z
|s|2 − |t|2 ≠ 0 ⇒ |s| ≠ |t|
It is true
(b) If|s| = |t|, then rt − rs may or may not be zero.
So, z may have no solutions.
∴ L may be an empty set.
It is false.
(c) If elements of set L represents line, then this line
and given circle intersect at maximum two point.
Hence, it is true.
(d) In this case locus of z is a line, so L has infinite
elements. Hence, it is true.
O
25. Given,|z1| = |z2|
y = –3√x
∴
S=
1 2
1
5π 20π
r θ = × 42 ×
=
2
2
3
6
|w − (2 + i )| < 3 ⇒ |w| − |2 + i| < 3
29. Since,
⇒
−3 + 5 < |w| < 3 + 5
⇒
−3 − 5 < − |w| < 3 − 5
…(i)
|z − (2 + i )| = 3
Also,
−3 + 5 ≤ |z| ≤ 3 + 5
⇒
∴
…(ii)
−3 < |z| − |w| + 3 < 9
= (x + 1)2 + ( y − 1)2 + (x − 5)2 + ( y − 1)2
= 2(x2 + y2 − 4x − 2 y) + 28
= 2(4) + 28 = 36
[Q|z1|2 = |z2|2 ]
[Q x2 + y2 − 4x − 2 y = 4]
31. Let z = x + iy
Set A corresponds to the region y ≥ 1
…(i)
Set B consists of points lying on the circle, centre at
(2, 1) and radius 3.
i.e.
…(ii)
x2 + y 2 − 4 x − 2 y = 4
As, we know z − z = 2i Im (z )
∴
∴
30. |z + 1 − i| + |z − 5 − i|2
|z |2 + (z2 z1 − z1 z2) − |z2|2
= 1
|z1 − z2|2
z2z1 − z1z2
|z1 − z2|2
Y′
Clearly, the shaded region represents the area of sector
2
z1 + z2 z1 − z2 z1z1 − z1z2 + z2z1 − z2 z2
=
×
z1 − z2 z1 − z2
|z1 − z2|2
=
X
(4, 0)
(a)
Now,
| 3 − 3| 3 − 3
=
2
3+1
z2z1 − z1z2 = 2i Im (z2z1 )
z1 + z2 2i Im (z2 z1 )
=
z1 − z2
|z1 − z2|2
Set C consists of points lying on the x + y = 2
…(iii)
Y
which is purely imaginary or zero.
26. Since, z1 = a + ib and
P
z2 = c + id
⇒ |z1|2 = a 2 + b2 = 1 and |z2|2 = c2 + d 2 = 1
(0,√2)
…(i)
[Q|z1|=|z2| = 1]
Also, Re (z1z2) = 0 ⇒ ac + bd = 0
a
d
⇒
=− =λ
b
c
From Eqs. (i) and (ii), b2λ2 + b2 = c2 + λ2c2
⇒
b2 = c2 and a 2 = d 2
X′
(√2,0)
(2,1)
y=1
X
[say]…(ii)
Y′
Clearly, there is only one point of intersection of the line
x + y = 2 and circle x2 + y2 − 4x − 2 y = 4.
Complex Numbers 15
z = x + iy
32. A. Let
y x 2 + y2 = 0
⇒ we get
⇒
y = 0 ⇒ Im (z ) = 0
B. We have
2ae = 8, 2a = 10 ⇒ 10e = 8
4
16

e = ⇒ b2 = 25 1 −  = 9

5
25
⇒
C. Let
∴
z = 2 (cos θ + i sin θ ) −
= 2 (cos θ + i sin θ ) −
=
⇒
2
⇒
x2
y2
+
=1
9 / 4 25 / 4
e= 1−
⇒
Let
and
Then,
and
9 /4
4
=
25 / 4 5
1
cos θ + i sin θ
x = 2 cos θ , y = 0
ω 2 (x + yω + z ω 2)
ω 2 (xω + yω 2 + z )
ω 2 (x + yω + zω 2)
= ω2
x + yω + zω 2
34. |az1 − bz2| + |bz1 + az2|
2
= [a 2|z1|2 + b2|z2|2 − 2ab Re (z1z2)]
+ [b2|z1|2 + a 2|z2|2 + 2ab Re (z1z2)]
= (a 2 + b2) (|z1|2 + |z2|2 )
35.
... (i)
x

 

2 x
2 x
 tan + 1 1 − tan  + 1 + tan  = 0

 
2
2 
2
xα + yβ + z γ
x( p)1/3 + y( p)1/3 ω + z ( p)1/3 ω 2
33.
=
xβ + yγ + zα x( p)1/3 ω 2 + y( p)1/3 ω3 + z ( p)1/3 ω
2
x = 2 nπ
x
x 
x

2x
2 x
sin + cos   cos − sin  + cos = 0




2
2
2
2
2
x
On dividing by cos3 , we get
2
= 2 cos θ
=
x
=0
2
1
(cos θ − i sin θ )
2
z = x + iy = cos θ + i sin θ +
⇒
sin
or
w = cos θ + i sin θ
Then,
x 
x
x 
x
2x
2 x
sin + cos   cos − sin  + cos  = 0
2 
2
2 
2
2
2
1
2 (cos θ + i sin θ )
2
⇒
D. Let
⇒ sin
3
5
cos θ + i sin θ
2
2
 2x
 2 y
  +   =1
 3
5
∴
2 sin
⇒
z = x + iy
3
5
x = cos θ and y = sin θ
2
2
Let
x
x
x
x
x
sin + cos  cos x + 2 sin cos = 0
2
2
2
2
2
⇒
∴
x2 y2
+
=1
25 9
w = 2 (cos θ + i sin θ )
∴
Since, it is real, so imaginary part will be zero.
x
x
x
−2 sin sin + cos  − tan x = 0
∴
2
2
2
x
x

sin + cos  − i tan x

2
2
∈R
x
1 + 2 i sin
2
x
x
x


sin + cos − i tan x 1 − 2i sin 



2
2
2
=
x
1 + 4 sin 2
2
tan3
x
x
− tan − 2 = 0
2
2
x
tan = t
2
f (t ) = t3 − t − 2
f (1) = − 2 < 0
f (2) = 4 > 0
Thus, f (t ) changes sign from negative to positive in the
interval (1, 2).
∴ Let t = k be the root for which
f (k) = 0 and k ∈(1, 2)
x
∴
t = k or tan = k = tan α
2
⇒
x/2 = nπ + α
x = 2nπ + 2α , α = tan −1 k, where k ∈ (1, 2)
⇒

or x = 2nπ

36. Since, z1 , z2, z3 are in AP.
⇒
2z2 = z1 + z3
i.e. points are collinear, thus do not lie on circle. Hence,
it is a false statement.
37. Since, z1 , z2, z3 are vertices of equilateral triangle and
|z1| = |z2| = |z3|
⇒ z1 , z2, z3 lie on a circle with centre at origin.
⇒ Circumcentre = Centroid
z + z2 + z3
⇒
0= 1
3
∴
z1 + z2 + z3 = 0
38. Let z = x + iy ⇒ 1 ∩ z gives 1 ∩ x + iy
or
Given,
⇒
1 ≤ x and 0 ≤ y
1−z
1 − x − iy
∩0 ⇒
∩0
1+ z
1 + x + iy
(1 − x − iy) (1 + x − iy)
∩ 0 + 0i
(1 + x + iy) (1 + x − iy)
…(i)
16 Complex Numbers
⇒
1 − x2 − y2
2iy
−
∩ 0 + 0i
2
2
(1 + x) + y
(1 + x)2 + y2
⇒
⇒
x2 + y2 ≥ 1 and −2 y ≤ 0
⇒
⇒
or x2 + y2 ≥ 1 and y ≥ 0 which is true by Eq. (i).
39. As we know,|z| = z ⋅ z
|z − α|2
= k2
|z − β|2
⇒
1 − |z1|2 − |z2|2+ | z1|2|z2|2 < 0
2
2

1 − z1z2
< 1
 z1 − z2 
⇒
⇒ |z| − αz − αz + |α| = k (|z| − βz − β z + |β| )
2
2
2
Hence proved.
42. Given,|z|2 w − |w|2 z = z − w
⇒ |z|2 (1 − k2) − (α − k2β )z − (α − β k2) z
+ (|α|2 − k2|β|2 ) = 0
|α|2 − k2|β|2
(α − k2β )
(α − β k2)
z+
z−
= 0 …(i)
2
2
(1 − k2)
(1 − k )
(1 − k )
⇒
zz w − ww z = z − w
[Q |z|2 = zz ] …(i)
Taking modulus of both sides, we get
|zw||z − w| = |z − w|
On comparing with equation of circle,
⇒
|zw||z − w| = |z − w|
|z| + az + az + b = 0
whose centre is (− a ) and radius = |a|2 − b
⇒
| zw|| z − w | = |z − w |
2
∴ Centre for Eq. (i) =
α −kβ
1 − k2
2
 α − k β  α − k β  αα − k ββ
and radius = 
 −
 
1 − k2
 1 − k2   1 − k2 
2
2
2
∴
⇒
1
3
|a1z + a 2z 2 + a3 z3 + K + a nz n| = 1
|a1z| + |a 2z 2| + |a3 z3| + K + |a nz n| ≥ 1
⇒
2{(|z| + |z| + |z| + K + |z|n } > 1
|z| <
…(i)
[using|z1 + z2| ≤ |z1| + |z2|]
⇒
⇒
⇒
⇒
⇒
[using|a r| < 2]
3
2|z|(1 − |z|n )
>1
1 − |z|
[using sum of n terms of GP]
2|z| − 2|z|n + 1 > 1 − |z|
3|z| > 1 + 2|z|n + 1
1 2
|z| > + |z|n + 1
3 3
1
|z| > , which contradicts
3
|zw| − 1 = 0
⇒
|z − w| = 0
or
|zw| = 1
⇒
z − w=0
or
|z w|= 1
⇒
z=w
or
|zw| = 1



z1
 |z1|
using z = |z |


2
2 

…(ii)
(1 − z1z2)(1 − z1z2) < (z1 − z2)(z1 − z2) [using|z|2 = zz ]
1 iφ
e
r
On putting these values in Eq. (i), we get
1
1
 1
r 2 ei φ  − 2 (reiθ ) = reiθ − eiφ
r
 r
r
1 iθ
1 iφ
iφ
iθ
re − e = re − e
⇒
r
r
1  iφ 
1  iθ

⇒
=
+
r
+
e
r
 e





r
r
NOTE
…(i)
[say]
z = reiθ and w =
Let
⇒
r =1
|1 − z1z2| < |z1 − z2|
⇒
On squaring both sides, we get,
or
…(ii)
n

1 − z1z2
< 1
 z1 − z2 
|z − w| = 0
Therefore,
|z| < 1 / 3 and ∑ a rz r = 1
Then, to prove
|z − w|(|zw| − 1) = 0
⇒
∴ There exists no complex number z such that
41. Given,|z1| < 1 and |z2| > 1
⇒
⇒
Then,|zw| = 1 or|z||w| = 1
1
|z| =
=r
⇒
|w|
40. Given, a1z + a 2z 2 + K + a nz n = 1
2
[∴ |z| = |z| ]
Now, suppose z ≠ w
k(α − β ) 

=
2
1−k 
and
…(iii)
∴ Eq. (iii) is true whenever Eq. (ii) is true.
(z − α )(z − α ) = k2(z − β ) (z − β )
⇒ |z|2 −
1 + |z1|2|z2|2 <|z1|2 + |z2|2
⇒
(1 − |z1|2 )(1 − |z2|2 ) < 0
which is true by Eq. (i) as|z1| < 1 and|z2| > 1
∴
(1 − |z1|2 ) > 0 and (1 − |z2|2 ) < 0
2
Given,
1 − z1z2 − z1z2 + z1z1z2z2 < z1z1 − z1z2 − z2z1 + z2z2
eiφ = eiθ ⇒ φ = θ
1
z = reiθ and w = eiθ
r
1
zw = reiθ . e−iθ = 1
r
‘If and only if ’ means we have to prove the relation in
both directions.
Conversely
Assuming that z = w or z w = 1
If
z = w, then
LHS = zz w − w wz = |z|2⋅z − |w|2⋅z
= |z|2⋅z − |z|2⋅z = 0
and
RHS = z − w = 0
If
zw = 1, then zw = 1 and
LHS = zz w − ww z = z ⋅ 1 − w ⋅ 1
= z − w = z − w = 0 = RHS
Hence proved.
Complex Numbers 17
Alternate Solution
We have,
|z|2 w − |w|2 z = z − w
2
⇔
|z| w − |w|2 z − z + w = 0
⇔
45. Here, z1Rz2 ⇔
(i) Reflexive z1Rz1 ⇔
(|z|2 + 1)w − (|w|2 + 1)z = 0
⇔
(|z|2 + 1)w = (|w|2 + 1)z
z |z|2 + 1
=
w |w|2 + 1
⇔
z
is purely real.
∴
w
⇔
Again,
⇔
⇔
z
z
=
w w
⇒
zw = zw
(z − w)(zw − 1) = 0
⇔
z = w or zw = 1
(ii) Symmetric z1Rz2 ⇔
…(i)
⇒
⇒
⇒
NOTE
[from Eq. (i)]
Here, let z1 = x1 + iy1 , z2 = x2 + iy2 and z3 = x3 + iy3
z − z2
(x − x2) + i ( y1 − y2)
is real ⇒ 1
is real
∴ 1
z1 + z2
(x1 + x2) + i ( y1 + y2)
⇒
It is a compound equation, therefore we can generate
from it more than one primary equations.
⇒
x + 2xy = 0 and x2 − y2 + y = 0
x(1 + 2 y) = 0
x = 0 or y = − 1 / 2
Similarly,
0 − y2 + y = 0
⇒ y = 0 or y = 1
3 i
−
2
2
44. Given, iz3 + z 2 − z + i = 0
⇒
z = i, ±
⇒
iz3 − i 2z 2 − z + i = 0
⇒
iz 2(z − i ) − 1(z − i ) = 0
⇒
(iz 2 − 1)(z − i ) = 0
⇒
⇒
If
z − i = 0 or iz 2 − 1 = 0
1
z = i or z 2 = = − i
i
z = i, then|z| = |i| = 1
If z 2 = − i, then |z 2| = |− i| = 1
⇒
|z|2 = 1 ⇒ |z| = 1
2 x2y1 − 2 y2x1 = 0
x1 x2
=
y1 y2
⇒
⇒
When x = 0, x2 − y2 + y = 0 ⇒
Therefore, z = 0 + i 0 , 0 + i ; ±
{(x1 − x2) + i ( y1 − y2)}{(x1 + x2) − i ( y1 + y2)}
(x1 + x2)2 + ( y1 + y2)2
⇒ ( y1 − y2) (x1 + x2) − (x1 − x2) ( y1 + y2) = 0
and − y = x2 − y2
x2 =
z1Rz2
z1 − z2
is real
z1 + z2
z2Rz3
z2 − z3
is real
z2 + z3
⇒
(x + iy) = i (x + i y)2
x − iy = i (x2 − y2 + 2i xy)
x − iy = − 2xy + i (x2 − y2)
⇒
y(1 − y) = 0
When, y = − 1 / 2 , x2 − y2 + y = 0
1 1
⇒
x2 − − = 0 ⇒
4 2
3
⇒
x=±
2
z1Rz2 ⇒ z2Rz1
and
On equating real and imaginary parts, we get
⇒
⇒
∴
⇒
z = iz 2
x = − 2xy
− (z2 − z1 )
is real ⇒ z2Rz1
z1 + z2
Therefore, it is symmetric.
z = x + iy.
Given,
[purely real]
z1 − z2
is real
z1 + z2
⇒
(iii) Transitive
Therefore, |z|2 w − |w|2 z = z − w if and only if z = w or
zw = 1.
43. Let
z1 − z1
=0
z1 + z2
∴ z1Rz1 is reflexive.
|z|2 w − |w|2 z = z − w
z ⋅ zw − w ⋅ wz = z − w
z (zw − 1) − w (zw − 1) = 0
⇔
z1 − z2
is real
z1 + z2
⇒
z2Rz3
x2 x3
=
y2 y3
From Eqs. (i) and (ii), we have
3
4
... (i)
... (ii)
x1 x3
=
⇒ z1Rz3
y1 y3
Thus, z1Rz2 and z2Rz3 ⇒ z1Rz3 .
[transitive]
Hence, R is an equivalence relation.
3 i
−
2
2
46.
[Q z ≠ 0]
[Q i 2 = − 1]
(1 + i ) x − 2i (2 − 3i ) y + i
+
=i
3+ i
3−i
⇒ (1 + i ) (3 − i ) x − 2i (3 − i ) + (3 + i ) (2 − 3i ) y
+ i (3 + i ) = 10i
⇒
4x + 2ix − 6i − 2 + 9 y − 7iy + 3i − 1 = 10i
⇒
4x + 9 y − 3 = 0 and 2x − 7 y − 3 = 10
⇒
x = 3 and y = − 1
1
1
47. Now,
=
θ
(1 − cos θ ) + 2i sin θ 2 sin 2 + 4i sin θ cos θ
2
2
2
θ
θ
sin − 2 i cos
1
2
2
×
=
θ
θ
θ 
θ
θ
2 sin sin + 2 i cos  sin − 2 i cos 
2
2
2 
2
2
18 Complex Numbers
θ
θ
− 2 i cos
2
2
=
θ  2θ
θ
+ 4 cos 2 
2 sin sin

2
2
2
θ
θ
sin − 2 i cos
2
2
=
θ
θ
2 sin 1 + 3 cos 2 
2
2
y1 + y2 = 2 3 ⇒ Im(z1 + z2) = 2 3
⇒
Hence, option (d) is correct.
|z | 4
2. (*) Given, 3|z1| = 4|z2|⇒ 1 =
[Q z2 ≠ 0 ⇒|z2| ≠ 0]
|z2| 3
sin
⇒ A + iB =
1
θ

2 1 + 3 cos 2 

2
48. Since, (x + iy)2 =
⇒
|x + iy|2 =
⇒
(x2 + y2) =
⇒
(x2 + y2)2 =
−i
cot
∴
θ
2
1 + 3 cos 2
[Q z =|z|(cos θ + i sin θ) = |z| eiθ ]
z1 4 iθ
z
3
⇒
= e and 2 = e−iθ
z2 3
z1 4
3 z1
2 z2 1 −iθ
= e
= 2eiθ and
⇒
2 z2
3 z1 2
θ
2
a + ib
c + id
 z1 |z1|
Q z = |z |
2
2

|a + ib|
|c + id|
a 2 + b2
c2 + d 2
a + b2
c2 + d 2
2
On adding these two, we get
1
3 z1 2 z2
z=
= 2eiθ + e−iθ
+
2
2 z2 3 z1
1
1
= 2 cos θ + 2i sin θ + cos θ − i sin θ
2
2
[Q e± iθ = (cos θ ± i sin θ)]
5
3
= cos θ + i sin θ
2
2
2
Hence proved.
49. Given,|z − 3 − 2 i| ≤ 2
…(i)
To find minimum of|2z − 6 + 5 i|
5
or 2 z − 3 + i , using triangle inequality
2
= (z − 3 − 2 i ) +
9
i
2
≥ |z − 3 − 2 i| −
9
9
5
≥ 2−
≥
2
2
2
Note that z is neither purely imaginary and nor purely
real.
‘*’ None of the options is correct.
z = e–iθ ⇒ z =
∴


 1 + z
 1 + z
arg 
 = arg (z ) = θ
 = arg 
1 + z
1 + 1

z
4. Since, arg (z ) < 0 ⇒
arg (z ) = − θ
Y
(–z)
π−θ
r
X′
X
–θ
O
Topic 3 Argument of a Complex Number
r
1. Let the complex numbers
From Eqs. (i) and (ii), we get
y12 − y22 = 2(x1 − x2)
y1 − y2
1
⇒
( y1 + y2) = 2
( y1 + y2) = 2 ⇒
x1 − x2
3
1
z
∴
5
5
z − 3 + i ≥ or|2z − 6 + 5 i| ≥ 5
2
2
z1 = x1 + iy1 and z2 = x2 + iy2
Now, it is given that
Re (z1 ) = |z1 − 1| ⇒ x12 = (x1 − 1)2 + y12
⇒
y12 + 1 = 2x1
and
Re(z2) =|z2 − 1| ⇒ x22 = (x2 − 1)2 + y22
⇒
y22 + 1 = 2x2
π
y − y2
1
and arg (z1 − z2) =
⇒ 1
=
6
x1 − x2
3
2
34
17
 3
 5
=
|z| =   +   =
 2
 2
4
2
⇒
3. Given,|z| = 1 , arg z = θ∴z = eiθ
i.e. ||z1| − |z2|| ≤ |z1 + z2|
5
5
∴
z −3 + i = z −3 −2i + 2i + i
2
2
⇒
z1 z1 iθ
z
z
e and 2 = 2 e−iθ
=
z2 z2
z1 z1
Y′
⇒
(z)
z = r cos (−θ ) + i sin (−θ )
= r (cos θ − i sin θ )
…(i)
…(ii)
…(iii)
and
− z = − r [cos θ − i sin θ ]
= r [cos (π − θ ) + i sin (π − θ )]
∴
arg (− z ) = π − θ
Thus, arg (− z ) − arg (z )
= π − θ − (− θ ) = π
Alternate Solution
 − z
arg (− z ) − arg z = arg 
 = arg (− 1) = π
 z 
 z 
and also arg z − arg (− z ) = arg 
 = arg (− 1) = π
 − z
Reason
Complex Numbers 19
5. Given,
|z + iw| = |z − iw |= 2
⇒
|z − (− iw)| = |z − (iw )| = 2
⇒
|z − (− iw)| = |z − (− iw )|
Consider
∴ z lies on the perpendicular bisector of the line joining
− iw and − iw. Since, − iw is the mirror image of − iw in
the X-axis, the locus of z is the X-axis.
Let
z = x + iy and y = 0.
Now,
|z | ≤ 1 ⇒ x2 + 02 ≤ 1
⇒
−1 ≤ x ≤ 1.
|z + iw| ≤ 2
⇒
arg z − arg i w = 0
z
arg
=0
iw
|z − i w| = 2, then
This function is discontinuous at t = 0.
z
is purely
w
imaginary
Now, given relation
|z + iw| = |z − iw| = 2
Put w = i, we get
⇒
Put w = − i , we get
|z − i 2| = |z − i 2| = 2
⇒
|z + 1| = 2 ⇒ z = 1
z 
arg  1  − arg (z1 ) + arg (z2)
 z2 
∴
|z − 1| = 2
z = −1
(c) We have,
z 
Now,arg  1  = arg (z1 ) − arg (z2) + 2nπ
 z2 
|z + i 2| = |z + i 2| = 2
⇒
z = − 1 − i and arg(z) = θ
(b) We have, f (t ) = arg (−1 + it )
 π − tan −1 t , t ≥ 0
arg (−1 + it ) = 
−1
− (π + tan t ), t < 0
z
is purely real.
⇒
iw
z
is purely imaginary.
⇒
w
Similarly, when
[from Eq. (i)]
Now,
≤1 +1 =2
|z + i w| = 2 holds when
1
1 = 0
0 
 im (z ) 
  −1 
=1
tan θ = 
=
 Re(z )   −1 
π
θ=
⇒
4
Since, x < 0, y < 0
π
3π

arg (z ) = −  π −  = −
∴

4
4
|z + i w| ≤ |z| + |iw|= |z| + |w|
⇒
u
v
0
9. (a) Let
Alternate Solution
∴
Applying R3 → R3 – {(1 − r ) R1 + rR2}
a
u
1
=
b
v
1
c − (1 − r ) a − rb w − (1 − r ) u − rv 1 − (1 − r ) − r
a
= b
0
∴ z may take values given in option (c).
 a u 1
b v 1 
c w 1 
[Q|z| ≤ 1]
[Q|z| ≤ 1]
∴ z = 1 or − 1 is the correct option.
w = re iθ , then w = re–iθ
∴
z = rei ( π − θ ) = re iπ ⋅ e−iθ = − re−iθ = − w
= arg (z1 ) − arg (z2) + 2nπ − arg (z1 ) + arg (z2) = 2nπ
So, given expression is multiple of 2π.
 (z − z1 ) (z2 − z3 )
(d) We have, arg 
 =π
 (z − z3 ) (z2 − z1 )
 z − z1   z2 − z3 
⇒

 is purely real
 z − z3   z2 − z1 
6. Since,|z|=|w|and arg (z ) = π − arg (w)
Let
z 
arg  1  − arg (z1 ) + arg (z2)
 z2 
Thus, the points A (z1 ), B(z2), C (z3 ) and D (z ) taken in
order would be concyclic if purely real.
Hence, it is a circle.
7. Given,|z1 + z2| = |z1| + |z2|
C(z3)
On squaring both sides, we get
|z1|2 + |z2|2 + 2|z1||z2| cos (arg z1 − arg z2)
= |z1|2 + |z2|2 + 2|z1||z2|
⇒
⇒
⇒
D(z)
2|z1||z2|cos (arg z1 − arg z2) = 2|z1||z2|
cos (arg z1 − arg z2) = 1
arg (z1 ) − arg (z2) = 0
8. Since a , b, c and u , v, w are the vertices of two triangles.
Also,
c = (1 − r ) a + rb
and
w = (1 − r ) u + rv
…(i)
B(z2)
A(z1)
∴(a), (b), (d) are false statement.
20 Complex Numbers
(1 − t ) z1 + t z2
(1 − t ) + t
P
z
t : (1 - t)
B
z2
≤ |r1 − r2|2 + |θ1 − θ 2|2
⇒
Clearly, z divides z1 and z2 in the ratio of t : (1 − t ),
0 < t <1
⇒
AP + BP = AB i.e. |z − z1|+ |z − z2|=|z1 − z2|
⇒ Option (a) is true.
and
arg (z − z1 ) = arg (z2 − z ) = arg (z2 − z1 )
⇒ Option (b) is false and option (d) is true.
Also,
arg (z − z1 ) = arg (z2 − z1 )
 z − z1 
⇒
arg 
 =0
 z2 − z1 
∴
z − z1
is purely real.
z2 − z1
z − z1
z − z1
z − z1
or
=
z2 − z1
z2 − z1 z2 − z1
⇒
z − z1
=0
z2 − z1
Option (c) is correct.
11. Let z = a + ib.
Given, Re(z ) = 0 ⇒ a = 0
Then, z = ib ⇒
z 2 = − b2 or lm (z 2) = 0
Therefore, A → q
π
Also, given, arg (z ) = .
4
π
π

Let
z = r  cos + i sin 

4
4
π
π
π
π

Then,
z 2 = r 2 cos 2 − sin 2  + 2 ir 2 cos sin

4
4
4
4
2
2
= ir sin π /2 = ir
Therefore, Re (z 2) = 0 ⇒ B → p.
a = b =2− 3
[Q a , b ← (0, 1)]
⇒
12. Let z = r1 (cos θ1 + i sin θ1 ) and w = r2(cos θ 2 + i sin θ 2)
We have,|z| = r1 ,|w| = r2, arg (z ) = θ1 and arg (w) = θ 2
Given,|z| ≤ 1,|w| < 1
⇒
r1 ≤ 1 and r2 ≤ 1
Now,
z − w = (r1 cos θ1 − r2 cos θ 2) + i (r1 sin θ1 − r2 sin θ 2)
|z − w| ≤ (|z| − |w|)2 + (arg z − arg w)2
2
Alternate Solution
|z − w|2 = |z|2 + |w|2 − 2|z||w|cos (arg z − arg w)
= |z|2 + |w|2 − 2|z||w| + 2|z||w|
− 2|z||w|cos (arg z − arg w)
 arg z − arg w
= (|z| − |w|)2 + 2|z||w|⋅ 2 sin 2
 …(i)


2
∴
 arg z − arg w
|z − w|2 ≤ (|z| − |w|)2 + 4 ⋅ 1⋅ 1 



2
⇒
|z − w|2 ≤ (|z| − |w|)2 + (arg z − arg w)2
[Q sin θ ≤ θ ]
13. Since, z1 = 10 + 6i , z2 = 4 + 6i
 z − z1  π
and arg 
represents locus of z is a circle
 =
 z − z2  4
shown as from the figure whose centre is (7, y) and
∠ AOB = 90°, clearly OC = 9 ⇒ OD = 6 + 3 = 9
6
=3 2
∴ Centre = (7, 9) and radius =
2
z
C
Y
6
B (10, 6)
z2
z1
(4, 0)
D (7, 0) (10, 0)
=
+ sin θ1 ) +
r22(cos 2 θ 2
+ sin θ 2)
Topic 4 Rotation of a Complex Number
1. The complex number z satisfying|z − 2 + i| ≥ 5, which
represents the region outside the circle (including the
circumference) having centre (2, − 1) and radius 5 units.
Y
z0 (x,y)
(1,0)
2
−2r1r2(cos θ1 cos θ 2 + sin θ1 sin θ 2)
X
⇒ Equation of circle is| z − (7 + 9i )| = 3 2
+ r12 sin 2 θ1 + r22 sin 2 θ 2 − 2r1r2 sin θ1 sin θ 2
2
O
(7, y)
(4, 6) A
⇒|z − w|2 = (r1 cos θ1 − r2 cos θ 2)2 + (r1 sin θ1 − r2 sin θ 2)2
= r12 cos 2 θ1 + r22 cos 2 θ 2 − 2r1r2 cos θ1 cos θ 2
r12(cos 2 θ1
2
°
A
z1
2
θ − θ 2
Therefore, |z − w |2 ≤ |r1 − r2|2 + 4 1
 2 
45
10. Given, z =
X¢
¾
O Ö5
X
(2,-1)
= r12 + r22 − 2r1r2 cos (θ1 − θ 2)
= (r1 − r2)2 + 2r1r2[1 − cos (θ1 − θ 2)]
Y¢
 θ − θ 2
= (r1 − r2)2 + 4r1r2 sin 2  1

 2 
Now, for z0 ∈ S
2
 θ − θ 2 

≤ |r1 − r2|2 + 4sin  1

 2 

and |sin θ| ≤ |θ|, ∀ θ ∈ R
[Q r1 , r2 ≤ 1]
1
is maximum.
|z0 − 1|
When |z0 − 1| is minimum. And for this it is required
that z0 ∈ S, such that z0 is collinear with the points
(2, − 1) and (1, 0) and lies on the circumference of the
circle|z − 2 + i|= 5.
Complex Numbers 21
So let z0 = x + iy, and from the figure 0 < x < 1 and y > 0.
4 − z0 − z0
4 − x − iy − x + iy
So,
=
z0 − z0 + 2i x + iy − x + iy + 2i
5. Since, |PQ | = |PS | = |PR| = 2
∴ Shaded part represents the
external part of circle having
centre (−1, 0) and radius 2.
As we know equation of circle
having centre z0 and radius r,
is|z − z0| = r
∴
|z − (−1 + 0i )| > 2
⇒
|z + 1| > 2
2(2 − x)
 2 − x
=
= −i 

 y + 1
2i ( y + 1)
Q
2−x
is a positive real number, so
y+1
4 − z0 − z0
is purely negative imaginary number.
z0 − z0 + 2 i
5
5
∴
π
i( π / 6 )
 =e
6
i
5π
−i
3
X′
π/4
X
O
Y′
π
∴ I (z ) = 0 and R(z ) = −2 cos = − 3 < 0
6
5π
π
π


Q cos 6 = cos  π − 6  = − cos 6 


7.
z1 − z3 1 − i 3 (1 − i 3 )(1 + i 3 )
=
=
z2 − z3
2
2 (1 + i 3 )
1 − i 23
2 (1 + i 3 )
4
2
z3
=
=
2 (1 + i 3 ) (1 + i 3 )
z2 − z3 1 + i 3
π
π
=
= cos + i sin
z1 − z3
2
3
3


−
z
z
−
π
z
z


3
3
2
= 1 and arg  2
 =


−
z
z
−
3
z
z
1
3
1
3
z2
=
⇒
⇒
Imaginary axis
…(ii)
represented by e i( α − θ )
Now, rotation about a line with angle α is given by
e θ → e (α − θ ). Therefore, Q is obtained from P by reflection
in the line making an angle α /2.
[Q eiθ = cos θ + i sin θ]
5π
= 2 cos
6
− π / 4 ≤ arg (z + 1)
6. In the Argand plane, P is represented by e i0 and Q is
5π
z = (eiπ / 6 )5 + (e−iπ / 6 )5 = e 6 + e 6
5π
5π  
5π
5π 

=  cos
+ i sin  +  cos
− i sin 



6
6
6
6
3.
3e iπ/4
A
From Eqs. (i) and (ii),
|arg (z + 1)|≤ π / 4
3 i
 π
 −π 
and
− = cos   + i sin  −  = e−iπ / 6
 6


2 2
6
So,
4
and argument of z + 1 in anticlockwise direction is −π /4.
2. Given, z = 
Q Euler’s form of
3 i 
π
+ =  cos + i sin
2 2 
6
Y
Also, argument of z + 1 with respect to positive direction
of X-axis is π/4.
π
…(i)
arg (z + 1) ≤
∴
4
 4 − z0 − z0 
π
⇒ arg 
 =−
 z0 − z0 + 2 i 
2
 3 i
 3 i
+  +
− 
2
2


 2 2
P
π/3
Hence, the triangle is an equilateral.
z2
z'2 (7,6)
(1
,2
1
)
90° z 0
5
z2′ = (6 + 2 cos 45° , 5 +
By rotation about (0, 0),
 iπ 
z2
= ei π/ 2 ⇒ z2 = z2′  e 2 
 
z2′
 
π

= (7 + 6i )  cos + i sin

2
3 1
(6,2)
Real axis
2 sin 45° ) = (7, 6) = 7 + 6i
Alternate Solution
z1 − z3 1 − i 3
=
∴
z2 − z3
2
z2 − z3
2
1 +i 3
π
π
⇒
= cos + i sin
=
=
z1 − z3 1 − i 3
2
3
3
 z2 − z3  π
z2 − z3
and also
arg 
=1
⇒
 =
 z1 − z3  3
z1 − z3
Therefore, triangle is equilateral.
8. Here, x + iy =
π
 = (7 + 6i ) (i ) = − 6 + 7i
2
4. Let OA = 3, so that the complex number associated with
A is 3e iπ / 4 . If z is the complex number associated with P,
then
z − 3eiπ / 4 4 − iπ / 2
4i
= e
=−
iπ / 4
3
3
0 − 3e
⇒ 3z − 9eiπ / 4 = 12 ieiπ / 4 ⇒ z = (3 + 4 i ) eiπ / 4
∴
Let
∴
⇒
x + iy =
a − ibt
1
×
a + ibt a − ibt
a − ibt
a 2 + b2t 2
a ≠ 0, b ≠ 0
− bt
a
and y = 2
x= 2
a + b2t 2
a + b2t 2
ay
y − bt
=
⇒t =
bx
x
a
z1
22 Complex Numbers
On putting x =
a
, we get
a 2 + b2t 2
So,

a 2 y2 
x  a 2 + b2 ⋅ 2 2  = a ⇒
bx 

x2 + y2 −
or
Since, |z2| = |z3| = 2
2
2
x
=0
a
z1
… (i)
z2
1
1

2
x −
 + y =


2a
4a 2
P (2, 0)
O
P (–1, 0)
∴Option (a) is correct.
For
a ≠ 0 and b = 0,
1
1
x + iy = ⇒ x = , y = 0
a
a
⇒ z lies on X-axis.
∴ Option (c) is correct.
1
1
For a = 0 and b ≠ 0, x + iy =
⇒ x = 0, y = −
ibt
bt
⇒ z lies on Y-axis.
∴ Option (d) is correct.
9.
[given]
Y-axis
a (x + y ) = ax
2
2
or
z1 = 2 (cos π / 3 + sin π / 3)
X-axis
z3
PLAN It is the simple representation of points on Argand plane and
to find the angle between the points.
Now, the triangle z1 , z2 and z3 being an equilateral and
the sides z1z2 and z1z3 make an angle 2π / 3 at the centre.
π 2π
Therefore,
∠POz2 = +
=π
3
3
π 2π 2π 5π
and
∠POz3 = +
+
=
3
3
3
3
Therefore, z2 = 2 (cos π + i sin π ) = 2 (− 1 + 0) = − 2

1
5π
5π 
3
 =2  − i
and z3 = 2  cos
+ i sin
 =1 − i 3
3
3 
2
2

n
nπ
nπ
π
π

Here, P = W n =  cos + i sin  = cos
+ i sin

6
6
6
6
1

H 1 = z ∈ C : Re(z ) > 
2

∴ P ∩ H 1 represents those points for which cos
nπ
is + ve.
6
Hence, it belongs to I or IV quadrant.
π
π
11π
11π
+ i sin
⇒ z1 = P ∩ H 1 = cos + i sin or cos
6
6
6
6
∴
z1 =
3 i
3 i
+ or
−
2
2
2
2
…(i)
Similarly, z2 = P ∩ H 2 i.e. those points for which
nπ
cos
<0
6
–√3 , 1
—– — Z2
2 2
Alternate Solution
Whenever vertices of an equilateral triangle having
centroid is given its vertices are of the form z , zω , zω 2.
∴ If one of the vertex is z1 = 1 + i 3 , then other two
vertices are (z1ω ), (z1ω 2).
(−1 + i 3 )
(−1 − i 3 )
, (1 + i 3 )
⇒
(1 + i 3 )
2
2
− (1 + 3) (1 + i 2( 3 )2 + 2i 3 )
,−
⇒
2
2
(−2 + 2i 3 )
⇒
−2 , −
=1 − i 3
2
∴
z2 = − 2 and z3 = 1 − i 3
11. Given, D = (1 + i ), M = (2 − i )
and diagonals of a rhombus bisect each other.
Let B ≡ (a + ib), therefore
√3 , 1
Z1 —–
—
2 2
(–1, 0) Z2
⇒
a+1
b+1
= 2,
= − 1 ⇒ a + 1 = 4, b + 1 = − 2
2
2
a = 3, b = − 3 ⇒ B ≡ (3 − 3i )
O
∴ z2 = cos π + i sin π , cos
− 3
⇒ z2 = − 1 ,
+
2
2π
Thus, ∠z1Oz2 =
3
i − 3 i
,
−
2
2
2
5π
,
,π
6
r cos θ = 1, r sin θ = 3
r = 2 and θ = π /3
D (1+i )
M
7π
5π
5π cos 7π
+ i sin
+ i sin
,
6
6
6
6
10. z1 = 1 + i 3 = r (cos θ + i sin θ )
⇒
⇒
A
√3 –1
Z1 —– , —
2 2
–√3 , –1 Z
—– — 2
2 2
C
B
Again,
[let]
(2−i )
But
and
⇒
⇒
DM = (2 − 1) + (− 1 − 1)2 = 1 + 4 = 5
2
BD = 2DM ⇒ BD = 2 5
2 AC = BD
⇒ 2 AC = 2 5
AC = 5 and
AC = 2 AM
5
5 = 2 AM ⇒ AM =
2
Complex Numbers 23
Now, let coordinate of A be (x + iy).
But in a rhombus AD = AB, therefore we have
AD 2 = AB2
2
⇒
(x − 1) + ( y − 1)2 = (x − 3)2 + ( y + 3)2
2
⇒ x + 1 − 2x + y2 + 1 − 2 y = x2 + 9 − 6x+ y2 + 9 + 6 y
⇒
4x − 8 y = 16
⇒
x − 2y = 4
⇒
x = 2y + 4
…(i)
5
5
⇒ AM 2 =
2
4
5
2
2
⇒
(x − 2) + ( y + 1) =
4
5
[from Eq. (i)]
(2 y + 2)2 + ( y + 1)2 =
⇒
4
5
5 y2 + 10 y + 5 =
⇒
4
⇒
20 y2 + 40 y + 15 = 0
⇒
4 y2 + 8 y + 3 = 0
⇒
(2 y + 1) (2 y + 3) = 0
⇒
2 y + 1 = 0,2 y + 3 = 0
1
3
y=− , y=−
⇒
2
2
On putting these values in Eq. (i), we get
 1
 3
x = 2  −  + 4, x = 2  −  + 4
 2
 2
AM =
Again,
⇒
2 (a + b) = ab + 1
and
⇒
(a = b or a + b = 1)
2 (a + b) = ab + 1
and
If a = b ,
2 (2a ) = a 2 + 1
⇒
a − 4a + 1 = 0
4 ± 16 − 4
⇒
a=
=2 ± 3
2
If a + b = 1, 2 = a (1 − a ) + 1 ⇒ a 2 − a + 1 = 0
1 ± 1 −4
, but a and b ∈ R
⇒a=
2
∴ Only solution when
a=b
⇒
a = b =2± 3
a = b =2− 3
[Q a , b ∈ (0, 1)]
⇒
2
13. Here, centre of circle is (1, 0) is also the mid-point of
diagonals of square
Y
z3
z1(2, 3)
(1, 0)
O z
0
z2
X
z4
x = 3, x = 1
i
3i 


Therefore, A is either 3 −  or 1 −  .


2
2
⇒
Alternate Solution
Since, M is the centre of rhombus.
⇒
∴ By rotating D about M through an angle of ± π /2 , we
get possible position of A.
and
C
B
⇒
z1 + z2
= z0
2
z2 = − 3 i
z3 − 1
= e± iπ/ 2
z1 − 1
[where, z0 = 1 + 0 i ]
π
π

z3 = 1 + (1 + 3i ) ⋅  cos ± i sin  [Q z1 = 2 + 3i ]

2
2
= 1 ± i (1 + 3i ) = (1 + 3 ) ± i = (1 − 3 ) + i
and z4 = (1 + 3 ) − i
M
(2– i) z2
A (z3)
D z1(1+i )
z − (2 − i ) 1
z − (2 − i ) 1
⇒ 3
= (± i ) ⇒ 3
= (± i )
−1 + 2 i
2
−1 + 2 i
2
1
1
⇒ z3 = (2 − i ) ± i (2i − 1) = (2 − i ) ± (−2 − i )
2
2
(4 − 2i − 2 − i ) 4 − 2i + 2 + i
3
i
=
,
= 1 − i, 3 −
2
2
2
2
3 
i


∴ A is either 1 − i or 3 −  .



2
2
12. Since, z1 , z2 and z3 form an equilateral triangle.
⇒
⇒
⇒
z12 + z22 + z32 = z1z2 + z2z3 + z3 z1
(a + i ) + (1 + ib) + (0) = (a + i ) (1 + ib) + 0 + 0
2
2
2
a − 1 + 2ia + 1 − b2 + 2ib = a + i (ab + 1) − b
2
⇒
(a 2 − b2) + 2i (a + b) = (a − b) + i (ab + 1)
⇒
a 2 − b2 = a − b
14. Let Q be z2 and its reflection be the point P (z1 ) in the
given line. If O (z ) be any point on the given line then by
definition OR is right bisector of QP.
∴
OP = OQ or |z − z1| = |z − z2|
⇒
⇒
⇒
|z − z1|2 = |z − z2|2
(z − z1 ) (z − z1 ) = (z − z2) (z − z2)
z (z1 − z2) + z (z1 − z2) = z1z1 − z2z2
Comparing with given line zb + zb = c
z1 − z2 z1 − z2 z1z1 − z2z2
=
=
= λ,
b
c
b
z1 − z2
z − z2
z z − z2z2
= b, 1
= b, 1 1
=c
λ
λ
λ
 z − z2 
 z − z2 
∴ z1b + z2b = z1  1

 + z2  1
 λ 
 λ 
zz − z2z2
= 1
=c
λ
[say]
…(i)
[from Eq. (i)]
24 Complex Numbers
15. Since, z1 + z2 = − p and z1z2 = q
Now,
18. Since, z1 , z2 and origin form an equilateral triangle.
B (z2)
z1 |z1|
=
(cos α + i sin α )
z2 |z2|
z1 cos α + i sin α
=
O
z2
1
[Q|z1|=|z2|]
⇒
Q if z1 , z2, z3 from an equilateral triangle, then 



z12 + z22 + z32 = z1z2 + z2z3 + z3 z1
A (z1)
triangle.
 z + z2 + z3 
∴ Circumcentre (z0 ) = Centroid  1



3
z12 + z22 + z32 = z1z2 + z2z3 + z3 z1
9z02 = z12 + z22 + z32 + 2 (z1z2 + z2z3 + z3 z1 )
⇒
p
= − cot2(α /2)
(z1 + z2)2 − 4z1z2
⇒
p2
= − cot2(α /2)
p − 4q
2
p2 = − p2 cot2(α /2) + 4q cot2(α /2)
p (1 + cot α /2) = 4q cot2(α /2)
2
p cosec2(α /2) = 4q cot2(α /2) ⇒ p2 = 4q cos 2 α /2
2
16. Since, triangle is a right angled isosceles triangle.
∴ Rotating z2 about z3 in anti-clockwise direction
through an angle of π / 2 , we get
A (z1)
where,|z2 − z3|= |z1 − z3|
⇒
(z2 − z3 ) = i (z1 − z3 )
On squaring both sides, we get
B (z3)
(z2 − z3 ) = − (z1 − z3 )
2
C (z2)
2
⇒
z22 + z32 − 2z2z3 = − z12 − z32 + 2z1z3
⇒
z12 + z22 − 2z1z2 = 2z1z3 + 2z2z3 − 2z32 − 2z1z2
⇒
(z1 − z2)2 = 2{(z1z3 − z32 ) + (z2z3 − z1z2)}
⇒
(z1 − z2)2 = 2(z1 − z3 )(z3 − z2)
⇒ 9z02 = z12 + z22 + z32 + 2 (z12 + z22 + z32 )
⇒
z + iz
Y
X′
B
iz
z
O
A
X
+
z22
+
1
1
1
OA × OB = |z ||iz | = |z |2
2
2
2
[from Eq. (ii)]
z32
 2 kπ 
 2 kπ 
= cos 

 + i sin 
 14 
 14 
∴ α k are vertices of regular polygon having 14 sides.
Let the side length of regular polygon be a.
∴ α k + 1 − α k = length of a side of the regular polygon
…(i)
=a
and α 4k−1 − α 4k− 2 = length of a side of the regular
polygon
…(ii)
=a
12
Y′
Area of ∆ OAB =
=
z12
z 4 − | z |4 = 4iz 2
z 4 − z 2z 2 = 4 iz 2
⇒
2
⇒
z (z − z ) (z + z ) = 4iz 2
2
So, either z = 0 or (z − z ) (z + z ) = 4i
Now, Case-I, if z 2 = 0 and z = x + iy
So,
x2 − y2 + 2ixy = 0 ⇒ x2 − y2 = 0
and
xy = 0 ⇒ x = y = 0
⇒ z = 0, which is not possible according to given
conditions.
Case-II, if (z − z ) (z + z ) = 4i and z = x + iy
So,
(2iy) (2x) = 4i
⇒ xy = 1 is an equation of rectangular hyperbola and for
minimum value of | z1 − z2|2, the z1 and z2 must be
vertices of the rectangular hyperbola.
Therefore, z1 = 1 + i and z2 = − 1 − i
∴Minimum value of| z1 − z2|2 = (1 + 1)2 + (1 + 1)2
4 + 4 =8
kπ
 kπ 
21. Given, α k = cos   + i sin  
 7
 7
. This implies that iz is the vector
obtained by rotating vector z in anti-clockwise direction
through 90°. Therefore, OA ⊥ AB. So,
3z02
20. For a complex number z, it is given that
i π/2
17. We have, iz = ze
... (ii)
On squaring Eq. (i), we get
p2
= − cot2(α /2)
(z1 − z2)2
z2 − z 3 | z 2 − z 3 | iπ / 2
=
e
z1 − z3 | z1 − z 3 |
...(i)
Also, for equilateral triangle
2
2
z12 + z22 = z1z2 ⇒ z12 + z22 − z1z2 = 0
19. Since, z1 , z2, z3 are the vertices of an equilateral
2 cos 2(α /2) + 2i sin (α /2) cos (α /2)
=
−2 sin 2(α /2) + 2i sin (α /2) cos (α /2)
2 cos (α /2) [cos (α /2) + i sin (α /2)]
=
2i sin (α /2)[cos (α /2) + i sin (α /2)]
−p
cot (α /2)
= − i cot (α /2)
=
= − i cot α /2 ⇒
z1 − z2
i
⇒
⇒
⇒
z12 + z22 + 02 = z1z2 + z2 ⋅ 0 + 0 ⋅ z1
⇒
Applying componendo and dividendo, we get
z1 + z2 cos α + i sin α + 1
=
z1 − z2 cos α + i sin α − 1
On squaring both sides, we get
⇒
∴
∑
k =1
3
∑
k =1
αk+1 −αk
α 4k−1 − α 4k− 2
=
12 (a )
=4
3 (a )
Complex Numbers 25
Topic 5 De-Moivre’s Theorem, Cube Roots
and nth Roots of Unity
1. Given expression
2π
2π 

+ i cos

 1 + sin
9
9 

π
π
2
2

 1 + sin
− i cos

9
9 
3
π
Given, z =
i
3  1
π
π
+   i = cos + i sin = e 6
 2
2
6
6
π
5π
8π

i
i
i 
= 1 + ie 6 + e 6 + ie 6 




3
2π
5π
11 π 

i
i
i
= 1 + e 3 + e 6 + e 6 




3

= 1 +

3
O
9
2π
2π  
5π
5π 

+ i sin 
+ i sin  +  cos
 cos

3
3 
6
6
9
π
2
π

i 
Q i = e 2 



1 i 3
3 1
3 i
= 1 − +
−
+ i+
− 
2
2
2
2
2
2

B (z2)

 2π 
 2π  
= − i cos  −  + i sin  −  


 9 
9

2π
2π 

= − i cos
− i sin

3
3 
9
11π
11π  

+  cos
+ i sin


6
6  
3
2π
2π 

= i3 cos
− i sin

9
9 
9
π
π
π
4π
5π

i
i
i
i
i 
= 1 + e 2 ⋅ e 6 + e 6 + e 2 ⋅ e 3 




2π
2π  
2π
2π


 sin 9 + i cos 9  ⋅ sin 9 − i cos 9 + 1 
=

2π
2π


− i cos
1 + sin
9
9


3
3
2
2
2
2
π
π
π
π




= sin
+ i cos  =  − i 2 sin
+ i cos 


9
9
9
9
1
3i 
π

= +
 =  cos + i sin

2 
3
2
A (z1)
[according the De-Movier’s theorem]
  1

3
= − i  −  − i
2 
  2
1
3 i
=−
+ = − ( 3 − i)
2
2
2
2. It is given that, there are two complex numbers z and w,
such that| z w| = 1 and arg (z ) − arg (w) = π / 2
∴
|z || w| = 1
[Q| z1 z2| = | z1 || z2|]
π
and arg (z ) = + arg (w)
2
1
Let| z | = r, then| w| =
…(i)
r
π
and let arg(w) = θ, then arg(z ) = + θ
…(ii)
2
So, we can assume
…(iii)
z = rei ( π / 2 + θ )
[Q if z = x + iy is a complex number, then it can be
written as z = reiθ where, r =|z|and θ = arg (z )]
1
…(iv)
and
w = ei θ
r
1
Now,
z ⋅ w = re− i( π / 2 + θ ) ⋅ eiθ
r
and
Key Idea Use, e i θ = cos θ + i sin θ
so, (1 + iz + z5 + iz 8 )9
2π
2π  
  2 2π

2
2 2π 
 sin 9 − i cos 9  + sin 9 + i cos 9  
=

2π
2π


1 + sin
− i cos
9
9


= ei( − π / 2 − θ + θ ) = e− i( π / 2) = − i
3.
[Qe− i θ = cos θ − i sin θ]
1
z w = re i( π / 2 + θ ) ⋅ e− iθ
r
= ei( π / 2 + θ − θ ) = e i ( π / 2) = i
π

3
9
9
9
[Q for any natural number ‘n’
= cos 3π + i sin 3π
(cos θ + i sin θ )n = cos(nθ ) + i sin(nθ )]
= −1
4. Given, x2 + x + 1 = 0
− 1 ± 3i
2
[Q Roots of quadratic equation ax2 + bx + c = 0
⇒
x=
are given by x =
⇒ z0 = ω , ω 2 [where ω =
−b±
−1 + 3 i
and
2
b2 − 4ac
]
2a
−1 − 3 i
2
are the cube roots of unity and ω, ω 2 ≠ 1)
Now consider z = 3 + 6i z081 − 3i z093
(Qω3 n = (ω 2)3 n = 1)
= 3 + 6i − 3i
= 3 + 3i = 3(1 + i )
If ‘θ’ is the argument of z, then
Im(z )
[Qz is in the first quadrant]
tan θ =
Re(z )
3
π
= =1⇒ θ=
4
3
ω2 =
5. Given that, z = cos θ + i sin θ = e iθ
∴
15
15
15
µ =1
µ =1
µ =1
∑ Ιµ(ζ 2 µ −1 ) = ∑ Ιµ( ε ιθ )2 µ −1 = ∑ Ιµ
ε ι ( 2 µ −1 ) θ
= sin θ + sin 3 θ + sin 5 θ + ... + sin 29 θ
26 Complex Numbers
 15 × 2 θ 
 θ + 29 θ 
sin 

 sin 



2
2 
=
 2 θ
sin 

 2 
=
n
 z2 
n
  =i
 z1 
⇒
∴ z1 and z2 are nth roots of unity.
z1n = z2n = 1
n
 z2 
⇒
  =1
 z1 
sin (15 θ )sin (15 θ )
1
=
sin θ
4 sin 2°
6. Let z = |a + bω + cω 2|
⇒ z 2 = |a + bω + cω 2|2 = (a 2 + b2 + c2 − ab − bc − ca )
1
…(i)
or z 2 = {(a − b)2 + (b − c)2 + (c − a )2}
2
⇒
⇒
in = 1
n = 4k, where k is an integer.
10. We know that,
ω=−
Since, a , b, c are all integers but not all simultaneously
equal.
⇒ If a = b then a ≠ c and b ≠ c
Because difference of integers = integer
⇒ (b − c)2 ≥ 1 {as minimum difference of two consecutive
integers is (± 1)} also (c − a )2 ≥ 1
and we have taken a = b ⇒ (a − b)2 = 0
1
From Eq. (i),
z 2 ≥ (0 + 1 + 1) ⇒ z 2 ≥ 1
2
Hence, minimum value of|z | is 1 .
 1 i 3
∴ 4 + 5 − +

2 
 2
⇒
(− ω ) = (− ω )
2 n
⇒
⇒
[Qω = 1 and 1 + ω + ω = 0]
2
1
8. Let ∆ = 1 −1 − ω
1
ω2
[Q 1 + ω + ω 2 = 0]
= (−2ω 2)7 = (−2)7ω14= − 128 ω 2
12. (1 + ω )7 = (1 + ω ) (1 + ω )6
= (1 + ω ) (−ω 2)6 = 1 + ω
ω
ω
⇒
A + Bω = 1 + ω
⇒
A = 1, B = 1
6
13.

∑ sin
k =1
z1n
= (i )n ⇒ i n = 1
z2n
2 kπ
2 kπ 
− i cos

7
7 
=
6

∑ − i  cos
k =1
= (−2 − ω 2)(ω − 1) − (ω 2 − 1)2
= − 2ω + 2 − ω3 + ω 2 − (ω 4 − 2ω 2 + 1)
= 3 ω 2 − 3 ω = 3ω (ω − 1)
[Q ω 4 = ω ]
z
π
9. Since, arg 1 =
z2 2
z1
π
π
⇒
= cos + i sin = i
z2
2
2
⇒
[Q 1 + ω + ω 2 = 0]
11. (1 + ω − ω 2)7 = (− ω 2 − ω 2)7
2
Applying R2 → R2 − R1 ; R3 → R3 − R1
1
1 
1
2
2
0
=
−
−
−1 
2
ω
ω


2
−
0
1 
ω
−
1
ω

∴
[Q ω3 = 1]
= 1 + 3 + 2 ω + 3 ω + 3 ω2
= 1 + 2 ω + 3 (1 + ω + ω 2) = 1 + 2 ω + 3 × 0
1
2
365
= 4 + 5 ω + 3 ω2
= 1 + (−1 + 3i ) = 3i
3
ωn = 1
n = 3 is the least positive value of n.
1
 1 i 3
+ 3 − +

2 
 2
= 4 + 5 ⋅ (ω3 )111 ⋅ ω + 3 ⋅ (ω3 )121 ⋅ ω 2
4 n
n
334
= 4 + 5 ω334 + 3 ω365
7. Given, (1 + ω ) = (1 + ω )
2 n
1
3
+
i
2
2
2 kπ
2 kπ 
+ i sin

7
7 
 6 i2kπ 
= − i  ∑ e 7  = − i { ei2π / 7 + e i4 π / 7 + e i6π /7
k = 1

+ e i8π / 7 + e i10π / 7 + e i12π / 7 }

(1− ei12 π / 7 )
= − i  ei 2 π / 7

1− ei 2 π / 7 

 ei2π / 7 − ei14π / 7 
= −i

i2π / 7
 1−e

[Q|z2| = |z1| = 1]
[Q e i14π /7= 1]
 e i2π / 7 − 1 
= −i 
=i
i2π / 7 
1 − e

n = 4k
Alternate Solution
z
π
Since,
arg 2 =
z1 2
14. (P) PLAN
∴
z2 z2
e
=
z1 z1
⇒
z2
=i
z1
i
π
2
Given
e iθ⋅ e
iα
=e
i
2 kπ
10
zk = e
i( θ + α )
 2π 
i
 ( k + j)
 10 
⇒ zk ⋅ z j = e
zk is 10th root of unity.
[Q|z1| = |z2| = 1]
⇒ zk will also be 10th root of unity.
Taking, z j as zk, we have zk ⋅ z j = 1 (True)
Complex Numbers 27
e iθ
(Q) PLAN
e
iα
= e i( θ − α )
z = zk / z1 = e
i
For k = 2; z = e
π
5
2kπ
2π 
−
i 

 10
10 
=
π
i ( k − 1)
e 5
which is in the given set (False)
Now,
(R) PLAN
(i) 1 − cos 2 θ = 2 sin2 θ
(ii) sin 2 θ = 2 sin θ cos θ and
5 −1
(i) cos 36° =
4
5 + 1| 1 − z1 || 1 − z2 | K | 1 − z9 |
(ii) cos 108° =
4
10
2 πk
2 πk 

NOTE | 1 − zk | = 1 − cos
− i sin
10
10 

π
k
π
k
πk 
πk
 sin
− i cos
=  2 sin
= 2 sin
10 
10
10 
10

Now, required product is
π
2π
3π
8π
9π
⋅ sin
⋅ sin
K sin
⋅ sin
29 sin
10
10
10
10
10
10
2
π
2π
3π
4π 
5π

sin
sin
sin
29 sin
 sin

10
10
10
10 
10
=
10
2
π
π
2π
2π 

cos
cos
29 sin
⋅ sin
 ⋅1

10
10
10
10 
=
10
2
π 1
2π 
1
29  sin ⋅ sin

2
5 2
5
=
10
25 (sin 36°⋅ sin 72° )2
=
10
25
(2 sin 36° sin 72° )2
= 2
2 × 10
22
=
(cos 36° − cos 108° )2
5
2
22 5
2 2   5 − 1  5 + 1 
=
 +
 = ⋅ =1

5 4
5  4   4  
(S) Sum of nth roots of unity = 0
1 + α + α 2 + α3 + K + α 9 = 0
1+
9
∑
α =0
k
k =1
1+
9
∑
k =1
2 kπ
2 kπ 

+ i sin
 cos
 =0

10
10 
1+
So,
15. Let
1−
9
∑
cos
2 kπ
=0
10
∑
cos
2 kπ
=2
10
k =1
9
k =1
(P) → (i), (Q) → (ii), (R) → (iii), (S) → (iv)
A3 = 0
 1 ω ω 2


A =  ω ω2 1 
ω 2 1 ω 


0 0 0
A 2 = 0 0 0 and Tr ( A ) = 0, A = 0


0 0 0
⇒
ω2
z+1
ω
ω
ω2
z+ω
1
2
1 = [ A + zl] = 0
z+ω
⇒
z3 = 0
the
number
of
z
satisfying
the given equation
⇒ z = 0,
is 1.
16. Here, Tr = (r − 1) (r − ω ) (r − ω )2] = (r3 − 1)
∴
Sn =
n
∑ (r3 − 1)
r =1
2
=
 n (n + 1) 
−n
2


17. Since, cube root of unity are 1, ω , ω 2 given by
 1
3  1
3
A (1, 0), B − , −
 ,C  − , −

2  2
2
 2
⇒ AB = BC = CA = 3
Hence, cube roots of unity form an equilateral triangle.
18. Given,
z p + q − z p − zq + 1 = 0
…(i)
⇒
(z p − 1)(z q − 1) = 0
Since, α is root of Eq. (i), either α p − 1 = 0 or α q − 1 = 0
αp −1
αq − 1
[as α ≠ 1]
⇒ Either
= 0 or
=0
α −1
α −1
⇒ Either
1 + α + α 2 + ... + α p − 1 = 0
or
1 + α + K + αq −1 = 0
But α p − 1 = 0 and α q − 1 = 0 cannot
occur
simultaneously as p and q are distinct primes, so
neither p divides q nor q divides p, which is the
requirement for 1 = α p = α q.
19. Since, 1, a1 , a 2, ... , a n − 1 are nth roots of unity.
⇒ (xn − 1) = (x − 1) (x − a1 ) (x − a 2) .... (x − a n − 1 )
xn − 1
⇒
= (x − a1 ) (x − a 2) ..... (x − a n − 1 )
x−1
⇒ xn − 1 + xn − 2 + ..... + x2 + x + 1
= (x − a1 ) (x − a 2) ..... (x − a n − 1 )

 xn − 1
n −1
+ xn− 2 + ... + x + 1
Q x − 1 = x


On putting x = 1 , we get 1 + 1 + ... n times
= (1 − a1 ) (1 − a 2) ..... (1 − a n − 1 )
⇒ (1 − a1 ) (1 − a 2)... (1 − a n − 1 ) = n
28 Complex Numbers
20. Since, n is not a multiple of 3, but odd integers and
x3 + x2 + x = 0 ⇒ x = 0, ω , ω 2
Now, when x = 0
⇒ (x + 1)n − xn − 1 = 1 − 0 − 1 = 0
∴
x = 0 is root of (x + 1)n − xn − 1
Again, when x = ω
⇒
(x + 1)n − xn − 1 = (1 + ω )n − ω n − 1
= −ω 2n − ω n − 1 = 0
[as n is not a multiple of 3 and odd]
Similarly, x = ω 2 is root of {( x + 1)n − xn − 1 }
Hence, x = 0, ω , ω 2 are the roots of (x + 1)n − xn − 1
Thus, x3 + x2 + x divides (x + 1)n − xn − 1 .
21. Since, α, β are the complex cube roots of unity.
∴ We take α = ω and β = ω 2.
Now,
xyz = (a + b)(aα + bβ )(aβ + bα )
= (a + b)[a 2αβ + ab(α 2 + β 2) + b2αβ ]
= (a + b)[a 2(ω ⋅ ω 2) + ab(ω 2 + ω 4 ) + b2(ω ⋅ ω 2)]
= (a + b)(a 2 − ab + b2)
[Q1 + ω + ω 2 = 0 and ω3 = 1]
= a3 + b3
22. Given, ω ≠ 1 be a cube root of unity, then|a + bω + cω 2|2
= (a + bω + cω 2) (a + bω + cω 2), (Q zz = |z|2 )
= (a + bω + cω 2) (a + bω + 2 cω 2 )
[Q ω = ω 2 and ω 2 = ω]
= (a + bω + cω 2) (a + bω 2 + cω )
= a 2 + abω 2 + acω + abω + b2ω3 + bcω 2 + acω 2
+ bcω 4 + c2ω3
2
2
2
2
2
4
= a + b + c + ab(ω + ω ) + bc(ω + ω ) + ac(ω + ω 2)
[as ω3 = 1]
2
2
2
= a + b + c + ab(−1) + bc(−1) + ac(−1)
[as ω + ω 2 = −1, ω 4 = ω]
2
2
2
= a + b + c − ab − bc − ca
1
= {(a − b)2 + (b − c)2 + (c − a )2}
2
Q a , b and c are distinct non-zero integers.
For minimum value a = 1, b = 2 and c = 3
1
∴|a + bω + cω 2|2min = {12 + 12 + 2 2}
2
6
= = 3.00
2
i
23. Printing error = e
Then,
2π
3
|x|2|+ | y|2 + |z|2
=3
(a )2 + (b)2 + |c|2
NOTE Here, w = e
i
2π
3
, then only integer solution exists.
2
Theory of Equations
Topic 1 Quadratic Equations
Objective Questions I (Only one correct option)
1. Suppose a , b denote the distinct real roots of the
quadratic polynomial x2 + 20x − 2020 and suppose c, d
denote the distinct complex roots of the quadratic
polynomial x2 − 20x + 2020. Then the value of
ac(a − c) + ad (a − d ) + bc(b − c) + bd (b − d ) is (2020 Adv.)
(a) 0
(c) 8080
(b) 8000
(d) 16000
2. Let λ ≠ 0 be in R. If α and β are the roots of the equation,
x2 − x + 2λ = 0 and α and γ are the roots of the equation,
βγ
is equal to
3x2 − 10x + 27λ = 0, then
λ
(2020 Main, 4 Sep II)
(a) 36
(c) 27
(b) 9
(d) 18
equations, (λ + 1)x − 4λx + 2 = 0 always have exactly
(2020 Main, 3 Sep II)
one root in the interval (0, 1) is
2
(b) (− 3, − 1)
(d) (1, 3]
(a) (0, 2)
(c) (2, 4]
4. If α and β are the roots of the quadratic equation,
 π
x2 + x sin θ − 2 sin θ = 0, θ ∈ 0,  , then
 2
α 12 + β12
is equal to
(α −12 + β −12 )(α − β )24
(2019 Main, 10 April I)
(a)
212
(sin θ + 8)12
(b)
26
(c)
2
(sin θ − 4)
12
(d)
(sin θ − 8)
(a) q2 − 4 p − 16 = 0
(c) p 2 − 4q + 12 = 0
6
(2019 Main, 9 April I)
(b) p 2 − 4q − 12 = 0
(d) q2 + 4 p + 14 = 0
6. If
m is chosen in the quadratic equation
(m2 + 1)x2 − 3x + (m2 + 1)2 = 0 such that the sum of its
roots is greatest, then the absolute difference of the
cubes of its roots is
(2019 Main, 9 April II)
(a) 10 5
(b) 8 5
(c) 8 3
(b) 5
(d) 3
(2019 Main, 8 April I)
8. The number of integral values of m for which the
equation (1 + m2)x2 − 2(1 + 3m)x + (1 + 8m) = 0, has no
real root is
(2019 Main, 8 April II)
(a) 3
(c) 1
(b) infinitely many
(d) 2
9. The number of integral values of m for which the
quadratic
expression,
(1 + 2m) x2 − 2(1 + 3m)
x + 4(1 + m), x ∈ R, is always positive, is
(a) 6
(b) 8
(c) 7
(d) 3
10. If λ be the ratio of the roots of the quadratic equation in
x, 3m2x2 + m(m − 4)x + 2 = 0, then the least value of m for
1
which λ + = 1, is
(2019 Main, 12 Jan I)
λ
(a) − 2 + 2
(c) 4 − 3 2
(b) 4 − 2 3
(d) 2 − 3
11. If
one real root of the quadratic equation
81x2 + kx + 256 = 0 is cube of the other root, then a value
(2019 Main, 11 Jan I)
of k is
(a) 100
5
(a)
4
5. Let p, q ∈R. If 2 − 3 is a root of the quadratic equation,
x + px + q = 0, then
(a) 2
(c) 4
(b) 144
(c) −81
r cannot be equal to
212
2
n
α
then the least value of n for which   = 1 is
 β
(d) −300
12. If 5, 5r , 5r 2 are the lengths of the sides of a triangle, then
(sin θ + 8)12
12
If α and β are the roots of the equation x2 − 2x + 2 = 0,
(2019 Main, 12 Jan II)
3. The set of all real values of λ for which the quadratic
2
7.
(d) 4 3
7
(b)
4
(2019 Main, 10 Jan I)
3
(c)
2
(d)
3
4
13. The value of λ such that sum of the squares of the roots
of the quadratic equation, x2 + (3 − λ )x + 2 = λ has the
least value is
(2019 Main, 10 Jan II)
(a)
4
9
(b) 1
(c)
15
8
(d) 2
14. The number of all possible positive integral values of α
for which the roots of the quadratic equation,
6x2 − 11x + α = 0 are rational numbers is
(2019 Main, 9 Jan II)
(a) 5
(b) 2
(c) 4
(d) 3
30 Theory of Equations
15. Let α and β be two roots of the equation x2 + 2x + 2 = 0,
then α
15
+β
15
is equal to
(a) 256
(c) −256
(2019 Main, 9 Jan I)
(b) 512
(d) −512
16. Let S = { x ∈ R : x ≥ 0 and 2| x − 3| + x ( x − 6) + 6 = 0 .
Then, S
(2018 Main)
(a) is an empty set
(b) contains exactly one element
(c) contains exactly two elements
(d) contains exactly four elements
x2 − x + 1 = 0, then α 101 + β107 is equal to
(b) 0
(c) 1
(2018 Main)
(d) 2
x(x + 1) + (x + 1) (x + 2) + ... + (x + n − 1) (x + n ) = 10n
has two consecutive integral solutions, then n is
equal to
(2017 Main)
(b) 9
(c) 10
(d) 11
19. The sum of all real values of x satisfying the equation
(x2 − 5x + 5)x
2
(a) 3
(b) − 4
+ 4 x − 60
= 1 is
(2016 Main)
(c) 6
(b) 2secθ
(d) 0
a − 2a 8
is
a n = α n − β n, for n ≥ 1, then the value of 10
2a 9
(2015 Main)
(b) – 6
(c) 3
(d) – 3
22. In the quadratic equation p(x) = 0 with real coefficients
has purely imaginary roots. Then, the equation
(2014 Adv.)
p[ p(x)] = 0 has
(a) only purely imaginary roots
(b) all real roots
(c) two real and two purely imaginary roots
(d) neither real nor purely imaginary roots
(c)
61
9
34
9
(2014 Main)
2 17
9
2 13
(d)
9
(b)
a n = α n − β n for n ≥ 1 , then the value of
(b) 2
(c) 3
27. If a , b, c are the sides of a triangle ABC such that
x2 − 2 (a + b + c) x + 3λ (ab + bc + ca ) = 0 has real roots,
then
(2006, 3M)
(a) λ <
4
3
(b) λ >
5
3
4 5
1 5
(c) λ ∈  ,  (d) λ ∈  , 
 3 3
 3 3
28. If one root is square of the other root of the equation
x2 + px + q = 0, then the relation between p and q is
(a) p3
(b) p3
(c) p3
(d) p3
− q(3 p − 1) + q2 = 0
− q(3 p + 1) + q2 = 0
+ q(3 p − 1) + q2 = 0
+ q(3 p + 1) + q2 = 0
(2004, 1M)
is
(2002, 1M)
(a) (− ∞ , − 2) ∪ (2, ∞ )
(c) (− ∞ , − 1) ∪ (1, ∞ )
(b) (− ∞ , − 2 ) ∪ ( 2 , ∞ )
(d) ( 2 , ∞ )
30. The number of solutions of log 4 (x − 1) = log 2(x − 3) is
(a) 3
(c) 2
(b) 1
(d) 0
(2001, 2M)
31. For the equation 3x2 + px + 3 = 0, p > 0, if one of the root
is square of the other, then p is equal to
(b) 1
(c) 3
(2000, 1M)
(d) 2 /3
x2 + bx + c = 0, where c < 0 < b, then
(a) 0 < α < β
(c) α < β < 0
a10 − 2a 8
is
2a 9
(d) 4
(2011)
(2000, 1M)
(b) α < 0 < β < |α|
(d) α < 0 < |α|< β
33. The equation x + 1 − x − 1 = 4x − 1 has
(1997C, 2M)
(a) no solution
(b) one solution
(c) two solutions
(d) more than two solutions
3
24. Let α and β be the roots of x2 − 6x − 2 = 0, with α > β. If
(a) 1
2
(q − p ) (2 p − q)
9
2
(d) (2 p − q) (2q − p )
9
(b)
32. If α and β (α < β) are the roots of the equation
1 1
+ = 4, then the
α β
value of|α − β|is
(a)
2
( p − q) (2q − p )
9
2
(c) (q − 2 p ) (2q − p )
9
(a) 1 /3
23. Let α and β be the roots of equation px2 + qx + r = 0,
p ≠ 0. If p, q and r are in AP and
α
, 2 β be the roots of the equation x2 − qx + r = 0 . Then,
2
(2007, 3M)
the value of r is
29. The set of all real numbers x for which x2 − |x + 2| + x > 0
21. Let α and β be the roots of equation x2 − 6x − 2 = 0. If
(a) 6
( p3 + q) x2 − ( p3 + 2q) x + ( p3 + q) = 0
( p3 + q) x2 − ( p3 − 2q) x + ( p3 + q) = 0
( p3 − q) x2 − (5 p3 − 2q) x + ( p3 − q) = 0
( p3 − q) x2 − (5 p3 + 2q) x + ( p3 − q) = 0
(d) 5
π
π
20. Let − < θ < − . Suppose α 1 and β1 are the roots of the
6
12
equation x2 − 2x secθ + 1 = 0 , and α 2 and β 2 are the roots
of the equation x2 + 2x tan θ − 1 = 0. If α 1 > β1 and
(2016 Adv.)
α 2 > β 2, then α 1 + β 2 equals
(a) 2(secθ − tan θ)
(c) −2tanθ
(a)
(b)
(c)
(d)
(a)
18. For a positive integer n, if the quadratic equation,
(a) 12
p3 ≠ − q. If α and β are non-zero complex numbers
satisfying α + β = − p and α 3 + β3 = q, then a quadratic
α
β
equation having and as its roots is
(2010)
β
α
26. Let α, β be the roots of the equation x2 − px + r = 0 and
17. If α , β ∈ C are the distinct roots of the equation
(a) −1
25. Let p and q be real numbers such that p ≠ 0, p3 ≠ q and
34. The equation x 4
(log 2 x )2 + log 2 x −
5
4
= 2 has
(a) atleast one real solution
(b) exactly three real solutions
(c) exactly one irrational solution
(d) complex roots
(1989; 2M)
Theory of Equations 31
35. If α and β are the roots of x2 + px + q = 0 and α 4 , β 4are
the roots of x − rx + s = 0, then
x2 − 4qx + 2q2 − r = 0 has always
2
the
equation
(1989, 2M)
(a) two real roots
(b) two positive roots
(c) two negative roots
(d) one positive and one negative root
2
2
36. The equation x −
has
=1 −
x−1
x−1
(a) no root
(c) two equal roots
(1984, 2M)
(b) one root
(d) infinitely many roots
37. For real x, the function
(x − a )(x − b)
will assume all real
(x − c)
values provided
(a) a > b > c
(1984, 3M)
(b) a < b < c
(c) a > c < b
(d) a ≤ c ≤ b
38. The number of real solutions of the equation
|x|2−3|x| + 2 = 0 is
(a) 4
(1982, 1M)
(b) 1
(c) 3
(d) 2
39. Both the roots of the equation
(x − b) (x − c) + (x − a ) (x − c) + (x − a ) (x − b) = 0
are always
(1980, 1M)
(a) positive
(c) real
(b) negative
(d) None of these
equation ax + bx + c = 0
p and q are real, then ( p, q) = (…,…).
(1979, 1M)
(a) are real and negative
(b) have negative real parts
(c) have positive real parts (d) None of the above
45. The coefficient of x in the polynomial
(x − 1)(x − 2)... (x − 100) is....
(1982, 2M)
True/False
46. If P (x) = ax2 + bx + c and Q (x) = − ax2 + bx + c, where
ac ≠ 0, then P (x) Q (x) has atleast two real roots.
(1985, 1M)
47. The equation 2x + 3x + 1 = 0 has an irrational root.
2
(1983, 1M)
Analytical & Descriptive Questions
48. If x2 − 10ax − 11b = 0 have roots c and d. x2 − 10cx − 11d = 0
have roots a and b, then find a + b + c + d.
(2006, 6M)
49. If α , β are the roots of ax + bx + c = 0, (a ≠ 0) and
2
α + δ , β + δ are the roots of Ax2 + Bx + C = 0, ( A ≠ 0) for
some constant δ, then prove that
b2 − 4ac B2 − 4 AC
(2000, 4M)
=
a2
A2
numbers. prove that if f (x) is an integer whenever x is
an integer, then the numbers 2 A , A + B and C are all
integers. Conversely, prove that if the numbers
2 A , A + B and C are all integers, then f (x) is an integer
(1998, 3M)
whenever x is an integer.
51. Find the set of all solutions of the equation
Assertion and Reason
For the following question, choose the correct answer
from the codes (a), (b), (c) and (d) defined as follows :
(a) Statement I is true, Statement II is also true;
Statement II is the correct explanation of Statement I
(b) Statement I is true, Statement II is also true;
Statement II is not the correct explanation of
Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
41. Let a , b, c, p, q be the real numbers. Suppose α , β are the
roots of the equation x + 2 px + q = 0.
1
and α , are the roots of the equation ax2 + 2bx + c = 0,
β
2
2|y| − |2y − 1 − 1| = 2y − 1 + 1
2x
1
>
x
+1
2x + 5x + 2
(1987, 3M)
53. Solve| x2 + 4x + 3 | + 2x + 5 = 0
(1987, 5M)
2
54. For a ≤ 0, determine all real roots of the equation
x2 − 2a|x − a|− 3a 2 = 0
55. Solve for x : (5 + 2 6 )x
Statement I ( p − q) (b − ac) ≥ 0
Statement II b ∉ pa or c ∉ qa .
−3
+ (5 − 2 6 )x
2
(1986, 5M)
−3
= 10 (1985, 5M)
56. If one root of the quadratic equation ax + bx + c = 0 is
equal to the nth power of the other, then show that
1
n n +1
1
+ (a c)
n
n +1
+ b =0
(1983, 2M)
57. If α and β are the roots of x + px + q = 0 and γ , δ are the
2
(2008, 3M)
Fill in the Blanks
roots of x2 + rx + s = 0, then evaluate (α − γ ) (β − γ )
(1979, 2M)
(α − δ ) (β − δ ) in terms of p, q, r and s.
58. Solve 2 log x a + log ax a + 3 logb a = 0 ,
42. The sum of all the real roots of the equation
|x − 2|2 + |x − 2| − 2 = 0 is…… .
2
2
(ac )
2
(1997 C, 3M)
52. Find the set of all x for which
where β 2 ∉ { −1, 0, 1}.
2
(1982, 2M)
99
50. Let f (x) = Ax2 + Bx + C where, A , B, C are real
40. Let a > 0, b > 0 and c > 0. Then, both the roots of the
2
44. If 2 + i 3 is a root of the equation x2 + px + q = 0, where
(1997, 2M)
43. If the products of the roots of the equation
x2 − 3kx + 2e2log k − 1 = 0 is 7, then the roots are real for
(1984, 2M)
k =K .
where a > 0, b = a 2 x
59. If α
(1978 , 3M )
and β are the roots of the equation
x2 + px + 1 = 0 ; γ , δ are the roots of x2 + qx + 1 = 0, then
(1978, 2M)
q2 − p2 = (α − γ )(β − γ )(α + δ )(β + δ )
32 Theory of Equations
Passage Type Questions
60. a12 =
Let p, q be integers and let α , β be the roots of the equation,
x2 − x − 1 = 0 where α ≠ β. For n = 0, 1, 2, …… , let
a n = pα n + qβ n.
FACT : If a and b are rational numbers and a + b 5 = 0, then
(2017 Adv.)
a = 0 = b.
61. If a 4 = 28, then p + 2q =
(a) a11 + 2a10
(c) a11 − a10
(b) 2a11 + a10
(d) a11 + a10
(a) 14
(c) 21
(b) 7
(d) 12
Topic 2 Common Roots
Objective Questions I (Only one correct option)
1. If α , β and γ are three consecutive terms of a
non-constant
GP
such
that
the
equations
αx2 + 2βx + γ = 0 and x2 + x − 1 = 0 have a common root,
then, α(β + γ) is equal to
(2019 Main, 12 April II)
3. A value of b for which the equations x2 + bx − 1 = 0,
x2 + x + b = 0 have one root in common is
(a) −
2
(b) −i 3
(c) i 5
(2011)
(d) 2
Fill in the Blanks
the quadratic equations x2 + ax + b = 0 and
x + bx + a = 0 (a ≠ b) have a common root, then the
numerical value of a + b is… .
(1986, 2M)
(a) 0
(b) αβ
(c) αγ
(d) βγ
4. If
2
2. If the equations x2 + 2x + 3 = 0 and ax2 + bx + c = 0,
5. If x − r is a factor of the polynomial f (x) = a n xn,
a , b, c ∈ R have a common root, then a : b : c is
(a) 1 : 2 : 3
(c) 1 : 3 : 2
(b) 3 : 2 : 1
(d) 3 : 1 : 2
True/False
(2013 Main)
+ a n − 1 x n−1 + K + a 0 repeated m times (1 < m ≤ n ), then
(1983, 1M)
r is a root of f ′ (x) = 0 repeated m times.
Topic 3 Transformation of Roots
Objective Question I (Only one correct option)
1. Let α , β be the roots of the equation,(x − a )(x − b) = c, c ≠ 0.
Then the roots of the equation (x − α ) (x − β ) + c = 0 are
(a) a , c
(b) b , c
(c) a , b (d) a + c, b + c
(1992, 2M)
Analytical & Descriptive Question
2. Let a, b and c be real number s with a ≠ 0 and let α , β be
the roots of the equation ax2 + bx + c = 0. Express the
roots of a3 x2 + abcx + c3 = 0 in terms of α , β. (2001, 4M)
Topic 4 Graph of Quadratic Expression
Objective Questions I (Only one correct option)
1. Let P(4, − 4) and Q(9, 6) be two points on the parabola,
y2 = 4x and let X be any point on the arc POQ of this
parabola, where O is the vertex of this parabola, such
that the area of ∆PXQ is maximum. Then, this
(2019 Main, 12 Jan I)
maximum area (in sq units) is
125
2
625
(c)
4
(a)
75
2
125
(d)
4
(b)
2. Consider the quadratic equation, (c − 5) x2− 2cx + (c − 4)
= 0, c ≠ 5. Let S be the set of all integral values of c for
which one root of the equation lies in the interval (0, 2)
and its other root lies in the interval (2, 3). Then, the
number of elements in S is
(2019 Main, 10 Jan I)
(a) 11
(c) 12
(b) 10
(d) 18
3. If
both the roots of the quadratic equation
x2 − mx + 4 = 0 are real and distinct and they lie in the
interval [1, 5] then m lies in the interval
(2019 Main, 9 Jan II)
(b) (−5, − 4)
(d) (3, 4)
(a) (4, 5)
(c) (5, 6)
4. If a ∈ R and the equation −3 (x − [x])2 + 2 (x − [x])
+ a 2 = 0 (where, [x] denotes the greatest integer ≤ x) has
no integral solution, then all possible values of a lie in
the interval
(2014 Main)
(a) (−1, 0) ∪ (0, 1)
(c) (−2 , − 1)
(b) (1, 2)
(d) (−∞ , − 2) ∪ (2 , ∞ )
5. For all ‘x’, x2 + 2ax + (10 − 3a ) > 0, then the interval in
which ‘a’ lies is
(a) a < − 5
(b) −5 < a < 2 (c) a > 5
(2004, 1M)
(d) 2 < a < 5
Theory of Equations 33
6. If b > a, then the equation (x − a ) (x − b) − 1 = 0 has
(a) both roots in (a , b)
(b) both roots in ( − ∞ , a )
(c) both roots in (b, + ∞ )
(d) one root in (−∞ , a ) and the other in (b, ∞ )
(2000, 1M)
9. If x2 + (a − b)x + (1 − a − b) = 0 where a , b ∈ R, then find
the values of a for which equation has unequal real
(2003, 4M)
roots for all values of b.
7. If the roots of the equation x2 − 2ax + a 2 + a − 3 = 0 are
real and less than 3, then
Analytical & Descriptive Questions
(1999, 2M)
(a) a < 2
(b) 2 ≤ a ≤ 3
(c) 3 < a ≤ 4
(d) a> 4
10. Let a , b, c be real. If ax2 + bx + c = 0 has two real roots α
and β, where α < − 1 and β > 1, then show that
c
b
1 + +  < 0
(1995, 5M)
a a 
11. Find all real values of x which satisfy x2 − 3x + 2 > 0 and
x2 − 2x − 4 ≤ 0.
(1983, 2M)
8. Let f (x) be a quadratic expression which is positive for
all real values of x. If g (x) = f (x) + f ′ (x) + f ′ ′ (x), then for
(1990, 2M)
any real x
(a) g (x) < 0
(c) g (x) = 0
(b) g (x) > 0
(d) g (x) ≥ 0
Integer & Numerical Answer Type Question
12. The smallest value of k, for which both the roots of the
equation x2 − 8kx + 16 (k2 − k + 1) = 0 are real, distinct
(2009)
and have values atleast 4, is …… .
Topic 5 Some Special Forms
Objective Questions I (Only one correct option)
x12 − x9 + x4 − x + 1 > 0 is
1. The number of real roots of the equation
5 + |2 − 1| = 2 (2 − 2) is
x
x
x
(a) 1
(c) 4
(2019 Main, 10 April II)
(b) 3
(d) 2
2. All the pairs (x, y) that satisfy the inequality
2
sin 2 x − 2 sin x + 5
⋅
1
sin 2 y
≤ 1 also satisfy the equation
4
(a) 2|sin x| = 3 sin y
(c) sin x = 2 sin y
(2019 Main, 10 April I)
(b) sin x = |sin y|
(d) 2 sin x = sin y
(b) 12
(d) 10
real number k for which the equation,
2x3 + 3x + k = 0 has two distinct real roots in [0, 1]
(2013 Main)
(b) lies between 2 and 3
(d) does not exist
5. Let a , b, c be real numbers, a ≠ 0. If α is a root of
a 2x2 + bx + c = 0, β is the root of a 2x2 − bx − c = 0 and
0 < α < β, then the equation a 2x2 + 2bx + 2c = 0 has a root
(1989, 2M)
γ that always satisfies
β
2
(d) α < γ < β
(b) γ = α +
6. If a + b + c = 0, then the quadratic equation
3ax2 + 2bx + c = 0 has
(a) at least one root in (0, 1)
(b) one root in (2, 3) and the other in (−2, − 1)
(c) imaginary roots
(d) None of the above
8
x)(ax2 + bx + c)dx
(1981, 2M)
0
4. The
α+β
2
(c) γ = α
1
∫0 (1 + cos
Then, the quadratic equation ax2 + bx + c = 0 has
(2019 Main, 8 April I)
(a) γ =
8. Let a , b, c be non-zero real numbers such that
= ∫ (1 + cos 8 x)(ax2 + bx + c)dx
| x − 2| + x ( x − 4) + 2 = 0 (x > 0) is equal to
(a) lies between 1 and 2
(c) lies between − 1and 0
(1982, 2M)
(a) −4 < x ≤ 0
(b) 0 < x < 1
(c) −100 < x < 100
(d) −∞ < x < ∞
2
3. The sum of the solutions of the equation
(a) 9
(c) 4
7. The largest interval for which
(a) no root in (0, 2)
(b) atleast one root in (1, 2)
(c) a double root in (0, 2)
(d) two imaginary roots
Objective Questions II
(One or more than one correct option)
9. Let S be the set of all non-zero real numbers α such that
the quadratic equation αx2 − x + α = 0 has two distinct
real roots x1 and x2 satisfying the inequality x1 – x2 < 1.
Which of the following interval(s) is/are a subset of S?
(2015 Adv.)
1
1 
(a)  – , –

 2
5
1 
(c)  0,


5
1
(b)  –
, 0

5 
1 1
(d) 
, 
 5 2
10. Let a ∈ R and f : R → R be given by f (x) = x5 − 5x + a.
(1983, 1M)
Then,
(a)
(b)
(c)
(d)
f (x) has three real roots, if a > 4
f (x) has only one real root, if a > 4
f (x) has three real roots, if a < − 4
f (x) has three real roots, if −4 < a < 4
34 Theory of Equations
Passage Based Problems
Read the following passage and answer the questions.
Passage I
Consider the polynomial f ( x ) = 1 + 2x + 3x 2+ 4x3 . Let s be
the sum of all distinct real roots of f ( x ) and let t =| s|.
(2010)
11. The real numbers s lies in the interval
1
(a)  − , 0
 4 
(d)  0,

3
3
1
(b)  − 11, −  (c)  − , − 

 4
4
2
1

4
12. The area bounded by the curve y = f (x) and the lines
x = 0, y = 0 and x = t , lies in the interval
3
(a)  , 3
4 
21 11
(b)  , 
 64 16 
21
(d)  0, 
 64 
(c) (9, 10)
13. The function f ′ (x) is
1
1
(a) increasing in  − t, −  and decreasing in  − , t 

 4 
4
1
1
(b) decreasing in  − t, −  and increasing in  − , t 

 4 
4
(c) increasing in (−t , t )
(d) decreasing in (−t , t )
Passage II
If a continuous function f defined on the real line R,
assumes positive and negative values in R, then the
equation f ( x ) = 0 has a root in R. For example, if it is
known that a continuous function f on R is positive at some
point and its minimum values is negative, then the
equation f ( x ) = 0 has a root in R. Consider f ( x ) = kex − x for
all real x where k is real constant.
(2007, 4M)
14. The line y = x meets y = kex for k ≤ 0 at
(a) no point
(b) one point
(c) two points
(d) more than two points
15. The positive value of k for which kex − x = 0 has only one
root is
1
e
(c) e
(b) 1
(a)
(d) log e 2
16. For k > 0, the set of all values of k for which kex − x = 0
has two distinct roots, is
1
(a)  0, 
 e
1
(c)  , ∞ 
e 
1
(b)  , 1
e 
(d) (0, 1)
True/False
17. If a < b < c < d, then the roots of the equation (x − a )
(x − c) + 2 (x − b) (x − d ) = 0 are real and distinct.
(1984, 1M)
Analytical & Descriptive Question
18. Let −1 ≤ p < 1. Show that the equation 4x3 − 3x − p = 0
has a unique root in the interval [1/2, 1] and identify it.
(2001, 4M)
Answers
Topic 1
Topic 2
1. (d)
2. (d)
5. (b)
9.
13.
17.
21.
25.
29.
33.
37.
41.
6. (b)
10.
14.
18.
22.
26.
30.
34.
38.
42.
(c)
(d)
(c)
(c)
(b)
(b)
(a)
(d)
(b)
45. –5050
(c)
(d)
(d)
(d)
(d)
(b)
(b)
(a)
4
46. True
3. (d)
7. (c)
11.
15.
19.
23.
27.
31.
35.
39.
43.
4. (a)
1. (d)
8. (a)
5. False
12.
16.
20.
24.
28.
32.
36.
40.
44.
(d)
(c)
(a)
(d)
(a)
(c)
(a)
(c)
k =2
(b)
(c)
(c)
(c)
(a)
(b)
(a)
(b)
( −4, 7 )
51. y ∈ { −1 } ∪ [1, ∞ )
48. 1210
1
 2
52. x ∈ ( −2, − 1 ) ∪  − , − 
 3
2
53. −4 and ( − 1 − 3 )
54. x = {a (1 − 2 ), a ( 6 − 1 )}
55. ±2, ± 2
58. x = a
61. (d)
−1/ 2
47. False
57. (q − s ) 2 − rqp − rsp + sp 2 + qr 2
or a
− 4 /3
59. q − p
2
2
60. (d)
2. (a)
3. (b)
4. –1
2. (a)
3. (a)
4. (a)
6. (d)
7. (a)
8. (b)
Topic 3
1. (c)
2. x = α 2β, αβ 2
Topic 4
1. (d)
5. (b)
9. a > 1
11. x ∈ [1 − 5, 1 ) ∪ [1 + 5, 2 )
12. k = 2
Topic 5
1. (a)
2. (b)
3. (d)
4. (d)
5. (d)
6. (a)
7. (d)
8. (b)
10. (b, d)
11. (c)
12. (a)
15. (a)
1

18. x = cos cos−1 p
3

16. (a)
9. (a,d)
13. (b)
17. True
14. (b)
Hints & Solutions
Topic 1 Quadratic Equations
1. Given quadratic polynomials, x + 20x − 2020 and
2
x2 − 20x + 2020 having a , b distinct real and c, d distinct
complex roots respectively.
So,
a + b = − 20, ab = − 2020
and
c + d = 20, cd = 2020
Now, ac(a − c) + ad (a − d ) + bc(b − c) + bd (b − d )
= a 2(c + d ) − a (c2 + d 2) + b2(c + d ) − b(c2 + d 2)
= (c + d ) (a 2 + b2) − (c2 + d 2)(a + b)
= (c + d )[(a + b)2 − 2ab] − (a + b) [(c + d )2 − 2cd ]
= 20 [(20)2 + 4040] + 20 [(20)2 − 4040]
= 2 × 20 × (20)2 = 40 × 400 = 16000
2. It is given that α is a common roots of given quadratic
equations
x2 – x + 2λ = 0 and 3x2 – 10x + 27λ = 0
2
∴ 3 α – 10 α + 27λ = 0
3 α 2 − 3 α + 6λ = 0
–
+
–
0 – 7α + 21λ = 0 ⇒ α = 3λ
So, 9λ2 – 3λ + 2λ = 0
1
1
[Qλ ≠ 0]
λ= ⇒α=
⇒
9
3
1
2×
2λ
9 =2
As αβ = 2λ ⇒ β =
=
1
α
3
3
1
9×
9λ
9 =3
and αγ = 9λ ⇒ γ =
=
1
α
3
2
×3
βγ 3
=
= 18
∴
1
λ
9
3. Given quadratic equations
f (x) = (λ2 + 1)x2 − 4λx + 2 = 0 have exactly one root in
the interval (0, 1).
So,
D > 0 ⇒ 16λ2 − 4(λ2 + 1)2 > 0
2
⇒
8λ − 8 > 0 ⇒ λ2 > 1
…(i)
⇒
λ ∈ (−∞ , − 1) ∪ (1, ∞ )
and f (0) f (1) < 0
⇒ 2(λ2 + 1 − 4λ + 2) < 0 ⇒ λ2 − 4λ + 3 < 0
…(ii)
⇒
(λ − 3)(λ − 1) < 0 ⇒ λ ∈ (1, 3)
From Eqs, (i) and (ii), we get
λ ∈ (1, 3)
And if λ = 3, then the quadratic equation is
10x2 − 12x + 2 = 0
⇒
5 x2 − 6 x + 1 = 0
1
x = 1,
⇒
5
Q The root x =
1
∈ (0, 1) for λ = 3
5
∴ λ ∈ (1 , 3].
Hence, option (d) is correct.
4. Given quadratic equation is

x2 + x sin θ − 2 sin θ = 0, θ ∈ 0,

π

2
and its roots are α and β.
So, sum of roots = α + β = − sin θ
and product of roots = αβ = − 2 sin θ
⇒
αβ = 2(α + β )
α 12 + β12
Now, the given expression is −12
(α + β −12)(α − β)24
=
…(i)
α 12 + β12
α 12 + β12
= 12
 1
 β + α 12
1
24
 12 12  (α − β)24
 12 + 12 (α − β)
α
β 
 α β 
12
 αβ 


αβ
=
=

2
2
 (α + β ) − 4αβ
 (α − β ) 
12
12


2(α + β)
=

2
 (α + β ) − 8 (α + β) 


2
=

 (α + β ) − 8
=
12


2
=

 − sin θ − 8
[from Eq. (i)]
12
[Q α + β = − sin θ]
212
(sin θ + 8)12
5. Given quadratic equation is x2 + px + q = 0, where
p, q ∈R having one root 2 − 3 , then other root is 2 + 3
(conjugate of 2 − 3 ) [Q irrational roots of a quadratic
equation always occurs in pairs]
So, sum of roots = − p = 4 ⇒ p = −4
and product of roots = q = 4 − 3 ⇒ q = 1
Now, from options p2 − 4q − 12 = 16 − 4 − 12 = 0
6. Given quadratic equation is
(m2 + 1)x2 − 3x + (m2 + 1)2 = 0
…(i)
Let the roots of quadratic Eq. (i) are α and β, so
3
and αβ = m2 + 1
m2 + 1
α+β=
According to the question, the sum of roots is greatest
and it is possible only when ‘‘(m2 + 1) is minimum’’ and
‘‘minimum value of m2 + 1 = 1, when m = 0’’.
∴α + β = 3 and αβ = 1, as m = 0
Now, the absolute difference of the cubes of roots
= |α 3 − β3|
= |α − β||α 2 + β 2 + αβ|
= (α + β )2 − 4αβ |(α + β )2 − αβ|
= 9 − 4 |9 − 1|= 8 5
36 Theory of Equations
7. Given, α and β are the roots of the quadratic equation,
x − 2x + 2 = 0
⇒
(x − 1)2 + 1 = 0
⇒
(x − 1)2 = − 1
⇒
x−1 = ± i
⇒
x = (1 + i ) or (1 − i )
Clearly, if α = 1 + i, then β = 1 − i
2
10. Let the given quadratic equation in x,
[where i = −1]
n
n
 1 + i
α
According to the question   = 1 ⇒ 
 =1
 β
1 − i
n
 (1 + i )(1 + i )
[by rationalization]
⇒

 =1
 (1 − i )(1 + i ) 
n
⇒
 1 + i 2 + 2i 
 =1⇒

 1 − i2 
n
 2i 
n
  =1⇒ i =1
 2
So, minimum value of n is 4.
8.
From inequalities Eqs. (iii) and (iv), the integral values
of m are 0, 1, 2, 3, 4, 5, 6
Hence, the number of integral values of m is 7.
⇒
⇒
α 2 + β 2 = αβ
(α + β )2 = 3αβ
2
2
m (m − 4)2
=3
3m2
9m4
(m − 4)2 = 18
m − 4 = ±3 2
m =4±3 2
⇒
[Q i = 1]
4
Key Idea
(i) First convert the given equation in quadratic equation.
(ii) Use, Discriminant, D = b 2 − 4 ac < 0
Given quadratic equation is
(1 + m2)x2 − 2(1 + 3m)x + (1 + 8m) = 0
Now, discriminant
D = [−2(1 + 3m)]2 − 4(1 + m2)(1 + 8m)
= 4 [(1 + 3m)2 − (1 + m2)(1 + 8m)]
= 4 [1 + 9m2 + 6m − (1 + 8m + m2 + 8m3 )]
= 4 [−8m3 + 8m2 − 2m]
= − 8m(4m2 − 4m + 1) = − 8m(2m − 1)2
3m2x2 + m(m − 4)x + 2 = 0, m ≠ 0 have roots α and β, then
m(m − 4)
2
and αβ =
α +β = −
2
3m
3m2
α
Also, let
=λ
β
α β
1
Then,
(given)
λ + =1 ⇒ + =1
λ
β α
⇒
⇒
⇒
[Qm ≠ 0]
The least value of m = 4 − 3 2
11. Given quadratic equation is
…(i)
According to the question there is no solution of the
quadratic Eq. (i), then
D <0
∴ −8m(2m − 1)2 < 0 ⇒ m > 0
81x2 + kx + 256 = 0
Let one root be α, then other is α 3 .
k
256
and α ⋅ α 3 =
Now, α + α 3 = −
81
81
b
a
c
and product of roots = ]
a
[Q for ax2 + bx + c = 0, sum of roots = −
4
 4
α4 =  
 3
4
α=±
3
k = − 81 (α + α 3 )
⇒
⇒
∴
So, there are infinitely many values of ‘m’ for which,
there is no solution of the given quadratic equation.
= − 81 α (1 + α 2)
16
 4 
= − 81  ±  1 +
 = ± 300
 3 
9
9. The quadratic expression
ax2 + bx + c, x ∈ R is always positive,
if a > 0 and D < 0.
So, the quadratic expression
(1 + 2m) x2 − 2 (1 + 3m)x + 4(1 + m), x ∈ R will be
always positive, if 1 + 2m > 0
…(i)
and D = 4(1 + 3m)2 − 4(2m + 1) 4(1 + m) < 0
…(ii)
From inequality Eq. (i), we get
1
…(iii)
m>−
2
From inequality Eq. (ii), we get
1 + 9m2 + 6m − 4 (2m2 + 3m + 1) < 0
⇒
m2 − 6m − 3 < 0
⇒ [m − (3 + 12 )][m − (3 − 12 )] < 0
6 ± 36 + 12
= 3 ± 12]
[Q m2 − 6m − 3 = 0 ⇒ m =
2
…(iv)
3 − 12 < m < 3 + 12
⇒
12. Let a = 5, b = 5r and c = 5r 2
We know that, in a triangle sum of 2 sides is always
greater than the third side.
∴ a + b > c; b + c > a and c + a > b
Now,
a+ b>c
⇒ 5 + 5r > 5r 2
⇒ 5r 2 − 5r − 5 < 0
⇒ r − r −1 <0

1 + 5 
1 − 5  
⇒ r − 
 <0
  r − 
 2 
 2  

2
[Q roots of ax2 + bx + c = 0 are given by
x=
− b ± b2 − 4ac
and r 2 − r − 1 = 0
2a
⇒r =
1± 1+4 1± 5
]
=
2
2
Theory of Equations 37
1 − 5 1 + 5
r ∈
,

2 
 2
⇒
+
–
...(i)
+
1– √ 5
2
1+ √ 5
2
Similarly,
b+ c>a
⇒
5r + 5r 2 > 5
⇒
r2 + r − 1 > 0

 −1 + 5  
 − 1 − 5  
 >0
  r − 
r − 
2
2






 2
−1 ± 1 + 4 − 1 ± 5 
=

Q r + r − 1 = 0 ⇒ r =
2
2



 −1 + 5

−1 − 5 
r ∈  − ∞,
, ∞
 ∪
2
2




⇒
+
–1+ √ 5
2
⇒ x=
c+ a >b
5r 2 + 5 > 5r
r2 − r + 1 > 0
and
⇒
⇒
2
and
… (iii)
b2 − 4ac
]
2a
1+ √ 5
2
= − [(1 − i )15 + (1 + i )15 ]
15
15
  1
  1
i 
i  
+
= −  2 
−
 
  +  2
2 
2   
  2
  2


= − 

+

∞
15

π
π 

2  cos − i sin  


4
4 

15
π
π  
 
 2  cos + i sin   
4
4  

15π
15π  
15π
15π  

= − ( 2 )15  cos
− i sin
+ i sin
 +  cos

4
4  
4
4  

7
is the only value that does not satisfy.
4
[using De’ Moivre’s theorem
(cos θ ± i sin θ )n = cos nθ ± i sin nθ, n ∈ Z ]
13. Given quadratic equation is
x2 + (3 − λ )x + 2 = λ
2
x + (3 − λ )x + (2 − λ ) = 0
−b±
Then, α 15 + β15 = (−1 + i )15 + (− 1 − i )15
2
–1+ √ 5
2
[Q roots of ax2 + bx + c = 0 are
⇒
x = −1 ± i
Let α = − 1 + i and β = − 1 − i.
2
1
3

⇒
r −  + > 0

2
4
⇒
r ∈R
From Eqs. (i), (ii) and (iii), we get
 −1 + 5 1 + 5
r ∈
,

2
2 

1– √ 5
2
−2 ± 4 −8
2
given by x =
1
 1
 1
⇒ r2 − 2 ⋅ r +   + 1 −   > 0
 2
 2
2
–∞ –1– √ 5
2
rational D = (b2 − 4ac) should be perfect square.
In the equation 6x2 − 11x + α = 0
a = 6, b = − 11 and c = α
∴For roots to be rational
D = (− 11)2 − 4(6) (α) should be a perfect square.
⇒ D(α) = 121 − 24α should be a perfect square
Now,
D(1) = 121 − 24 = 97 is not a perfect square.
D(2) = 121 − 24 × 2 = 73 is not a perfect square.
D(3) = 121 − 24 × 3 = 49 is a perfect square.
D(4) = 121 − 24 × 4 = 25 is a perfect square.
D(5) = 121 − 24 × 5 = 1 is a perfect square.
and for α ≥ 6, D(α) < 0, hence imaginary roots.
∴For 3 values of α (α = 3, 4, 5), the roots are rational.
15. We have, x2 + 2x + 2 = 0
+
–
–1– √ 5
2
…(ii)
14. For the roots of quadratic equation ax2 + bx + c = 0 to be
… (i)
Let Eq. (i) has roots α and β, then α + β = λ − 3 and
αβ = 2 − λ
b
[Q For ax2 + bx + c = 0, sum of roots = −
a
c
and product of roots = ]
a
Now, α 2 + β 2 = (α + β )2 − 2αβ
= (λ − 3)2 − 2(2 − λ )
= λ2 − 6λ + 9 − 4 + 2λ
= λ2 − 4λ + 5 = (λ2 − 4λ + 4) +1
= (λ − 2)2 + 1
2
Clearly, a + β 2 will be least when λ = 2.
15π 
1 


= − ( 2 )15 2 cos
= − ( 2 )15 2 ×
4 
2 


15π
π
π
1 


Q cos 4 = cos 4π − 4  = cos 4 = 2 


= − ( 2 )16 = − 28
= − 256.
Alternate Method
α 15 + β15 = (−1 + i )15 + (−1 − i )15
= − [(1 − i )15 + (1 + i )15 ]
 (1 − i )16 (1 + i )16 
=−
+
1 + i 
 1−i
 [(1 − i )2]8 [(1 + i )2]8 
=−
+
1 + i 
 1−i
38 Theory of Equations
 [1 + i 2 − 2i ]8 [1 + i 2 + 2 i ]8 
=−
+

1−i
1+ i


 (−2 i )8 (2 i )8 
=−
+
1 + i 
 1−i
 1
1 
[Q i 4n = 1, n ∈ Z ]
= − 28 
+

−
+
1
i
1
i


 2 
2 
= − 256 
= − 256
= − 256
2
2 
1 − (i ) 
16. We have, 2| x − 3| +
x ( x − 6) + 6 = 0
Let x − 3 = y
x = y+3
⇒
∴ 2| y| + ( y + 3)( y − 3) + 6 = 0
⇒
2| y| + y2 − 3 = 0
⇒
| y|2 + 2| y| − 3 = 0
⇒
(| y| + 3)(| y| − 1) = 0
⇒
| y| ≠ − 3 ⇒ | y| = 1
⇒
y= ±1 ⇒
x −3 = ±1
x = 4, 2 ⇒ x = 16, 4
⇒
17. We have, α , β are the roots of x2 − x + 1 = 0
Q Roots of x2 − x + 1 = 0 are −ω ,−ω 2
∴ Let α = − ω and β = − ω 2
⇒ α 101 + β107 = (− ω )101 + (− ω 2)107 = − (ω101 + ω 214 )
= − (ω 2 + ω )
[Q ω3 = 1]
[Q1 + ω + ω 2 = 0]
= − (−1)
=1
or
III. x2 − 5x + 5 = − 1 and x2 + 4x − 60 = Even integer.
Case I
When x2 + 4x − 60 = 0
⇒
x2 + 10x − 6x − 60 = 0
⇒
x(x + 10) − 6(x + 10) = 0
⇒
(x + 10) (x − 6) = 0
⇒
x = − 10or x = 6
Note that, for these two values of x, x2 − 5x + 5 ≠ 0
Case II When
x2 − 5 x + 5 = 1
⇒
x2 − 5 x + 4 = 0
2
⇒
x − 4x − x + 4 = 0
⇒
x(x − 4) − 1 (x − 4) = 0
⇒
(x − 4) (x − 1) = 0
⇒
x = 4or x = 1
Case III When
x2 − 5 x + 5 = − 1
⇒
x2 − 5 x + 6 = 0
2
⇒
x − 2x − 3x + 6 = 0
⇒
x(x − 2) − 3(x − 2) = 0
⇒
(x − 2) (x − 3) = 0
⇒
x = 2 or x = 3
Now, when x = 2, x2 + 4x − 60 = 4 + 8 − 60 = − 48, which is
an even integer.
When x = 3, x2 + 4x − 60 = 9 + 12 − 60 = − 39, which is not
an even integer.
Thus, in this case, we get x = 2.
Hence, the sum of all real values of
x = − 10 + 6 + 4 + 1 + 2 = 3
18. Given quadratic equation is
x(x + 1) + (x + 1)(x + 2) + ... + (x + n − 1) (x + n ) =10n
⇒ (x2 + x2 + ... + x2) + [(1 + 3 + 5 + ... + (2n − 1)]x
+ [(1 ⋅ 2 + 2 ⋅ 3 + ... + (n − 1)n ] = 10n
n (n 2 − 1)
2
2
nx + n x +
− 10n = 0
⇒
3
2
n −1
⇒
x2 + nx +
− 10 = 0
3
⇒
3x2 + 3nx + n 2 − 31 = 0
Let α and β be the roots.
Since, α and β are consecutive.
∴
|α − β| = 1
⇒ (α − β )2 = 1
2
2
Again, (α − β ) = (α + β ) − 4αβ
2
 n 2 − 31
 − 3n 
⇒
1=

 − 4
 3 
 3 
4
⇒ 1 = n 2 − (n 2 − 31) ⇒ 3 = 3n 2 − 4n 2 + 124
3
⇒
n 2 = 121 ⇒ n = ± 11
[Qn > 0]
∴
n = 11
19. Given, (x − 5x + 5)
2
x 2 + 4 x − 60
=1
Clearly, this is possible when
I. x2 + 4x − 60 = 0 and x2 − 5x + 5 ≠ 0
or
II. x2 − 5x + 5 = 1
20. Here, x − 2x secθ + 1 = 0 has roots α 1 and β1.
2
2 sec θ ± 4 sec2 θ − 4
2 ×1
2 sec θ ± 2|tan θ|
=
2
π
 π
Since,
θ ∈  − ,−  ,
 6 12
2 sec θ m 2 tan θ
i.e. θ ∈ IV quadrant =
2
∴
α 1 = sec θ − tan θ and β1 = sec θ + tan θ [as α 1 > β1]
and x2 + 2x tan θ − 1 = 0 has roots α 2 and β 2 .
∴
i.e.
∴
and
Thus,
α 1 , β1 =
−2 tan θ ± 4 tan 2 θ + 4
2
α 2 = − tan θ + sec θ
[as α 2 > β 2]
β 2 = − tan θ − sec θ
α 1 + β 2 = −2 tan θ
α 2, β 2 =
21. Given, α and β are the roots of the equation
x2 − 6x − 2 = 0.
∴
a n = α n − β n for n ≥ 1
a10 = α 10 − β10
a 8 = α 8 − β8
a 9 = α 9 − β9
Theory of Equations 39
Now, consider
a10 − 2a 8 α
=
2a 9
10
8
8
and
α 8 (α 2 − 2) − β 8 (β 2 − 2)
=
2(α 9 − β 9 )
=
−q 4r 4r
4
=
=
=−
p
p −9 r
9
r
r
1
αβ = =
=
p −9 r −9
16 4 16 + 36
(α − β )2 = (α + β )2 − 4 αβ =
+ =
81 9
81
52
(α − β )2 =
81
2
|α − β| =
13
9
α+β=
Now,
− β − 2(α − β )
2(α 9 − α 9 )
10
∴
α 8 ⋅ 6 α − β 86 β 6 α 9 − 6 β 9 6
= =3
=
2(α 9 − β 9 )
2(α 9 − 6 β 9 ) 2
⇒

Q α and β are the roots of
x2 − 6x − 2 = 0 or x2 = 6x + 2


2
2
⇒
α = 6 α + 2 ⇒ α − 2 = 6 α
and β 2 = 6 β + 2 ⇒ β 2 − 2 = 6 β 
⇒
24.
a10 − 2a 8 (α 10 − β10 ) − 2 (α 8 − β 8 )
=
2a 9
2 (α 9 − β 9 )
Alternate Solution
=
Since, α and β are the roots of the equation
x2 − 6x − 2 = 0.
Qα is root of x2 − 6 x − 2 = 0 ⇒ α 2 − 2 = 6α
x2 = 6 x + 2
or
∴
α 2 = 6α + 2
⇒
α 10 = 6 α 9 + 2 α 8
[and β is root of x2 − 6 x − 2 = 0 ⇒ β 2 − 2 = 6 β]
…(ii)
25. Sum of roots =
On subtracting Eq. (ii) from Eq. (i), we get
α 10 − β10 = 6(α 9 − β 9 ) + 2(α 8 − β 8 )
⇒
=
...(i)
β10 = 6 β 9 + 2 β 8
Similarly,
(Q a n = α n − β n)
a10 = 6a 9 + 2a 8
α 2 + β2
and product = 1
αβ
(α + β ) (α 2 − αβ + β 2) = q
−q
α 2 + β 2 − αβ =
p
∴
22. If quadratic equation has purely imaginary roots, then
α 8 (6 α ) − β 8 (6 β ) 6 (α 9 − β 9 )
=3
=
2 (α 9 − β 9 )
2 (α 9 − β 9 )
Given, α + β = − p and α 3 + β3 = q
⇒
⇒ a10 − 2a 8 = 6a 9
a − 2a 8
⇒ 10
=3
2a 9
⇒
Let p(x) = ax + b with a, b of same sign and a , b ∈ R.
From Eqs. (i) and (ii), we get
Then,
p[ p(x)] = a (ax2 + b)2 + b
α 2 + β 2 + 2αβ = p2
α 2 + β2 =
p(x) has imaginary roots say ix.
Then, also ax2 + b ∈ R and (ax2 + b)2 > 0
∴
Thus,
PLAN
If ax 2 + bx + c = 0 has roots α and β, then α + β = − b / a
c
and α β = . Find the values of α + β and αβ and then put
a
in (α − β )2 = (α + β )2 − 4αβ to get required value.
Given, α and β are roots of px + qx + r = 0, p ≠ 0.
−q
r
, αβ =
α+β=
∴
p
p
2
2 (−4r ) = p + r ⇒
p = − 9r
p3 + q
3p
( p3 − 2q) x
+1 =0
( p3 + q)
…(i)
...(ii)
[from Eq. (i)]
26. The equation x2 − px + r = 0 has roots α, β and the
equation x2 − qx + r = 0 has roots
⇒
α
, 2β.
2
r = αβ and α + β = p,
α
2q − p
2 (2 p − q)
and
and α =
+ 2β = q ⇒ β =
2
3
3
2
⇒
αβ = r = (2q − p) (2 p − q)
9
27. Since, roots are real, therefore D ≥ 0
⇒
4 (a + b + c)2 − 12λ (ab + bc + ca ) ≥ 0
⇒
(a + b + c)2 ≥ 3λ (ab + bc + ca )
⇒
⇒
On putting the value of q in Eq. (ii), we get
⇒
and αβ =
...(ii)
⇒ ( p3 + q) x2 − ( p3 − 2q) x + ( p3 + q) = 0
p [ p(x)] ≠ 0, ∀ x
Since, p, q and r are in AP.
∴
2q = p + r
1 1
α+β
Also,
+ =4 ⇒
=4
α β
αβ
−q 4r
=
⇒
α + β = 4 αβ ⇒
p
p
⇒
q = − 4r
p3 − 2q
3p
∴ Required equation is, x2 −
a (ax2 + b)2 + b ≠ 0, ∀ x
...(i)
(α + β )2 = p2
and
coefficient of x must be equal to zero.
2
23.
α 8 (α 2 − 2) − β 8 (β 2 − 2)
2(α 9 − β 9 )
Also,
a 2 + b2 + c2 ≥ (ab + bc + ca ) (3λ − 2)
a 2 + b2+ c2
3λ − 2 ≤
ab + bc + ca
cos A =
b2 + c2 − a 2
<1
2bc
…(i)
40 Theory of Equations
⇒
b2 + c2 − a 2< 2bc
Similarly,
c + a − b < 2ca
⇒
⇒
and
a 2 + b2 − c2 < 2ab
32. Given,
⇒
a 2 + b2 + c2 < 2 (ab + bc + ca )
2
2
2
a 2 + b2 + c2
<2
ab + bc + ca
⇒
…(ii)
From Eqs. (i) and (ii), we get
3λ − 2 < 2
⇒ λ<
4
3
28. Let the roots of x2 + px + q = 0 be α and α 2.
⇒
α + α2 = − p
⇒
and α 3 = q
α (α + 1) = − p
⇒ α {α + 1 + 3α (α + 1)} = − p3
3
⇒
⇒
29. Given,
3
[cubing both sides]
q (q + 1 − 3 p) = − p3
p3 − (3 p − 1)q + q2 = 0
x2 − |x + 2| + x > 0
Case I When
…(i)
x+ 2 ≥0
∴
x2 − x − 2 + x > 0 ⇒ x2 − 2 > 0
⇒
x < − 2 or x > 2
− 1 = − p/3
p=3
c<0 < b
Since,
…(i)
α + β = −b
and
…(ii)
αβ = c
From Eq. (ii), c < 0 ⇒ α β < 0
⇒ Either α is –ve, β is −ve or α is + ve, β is – ve.
From Eq. (i), b > 0 ⇒ − b < 0 ⇒ α + β < 0 ⇒ the sum is
negative.
⇒ Modulus of negative quantity is > modulus of positive
quantity but α < β is given.
Therefore, it is clear that α is negative and β is positive
and modulus of α is greater than modulus of β
⇒
α < 0 < β < |α|
NOTE This question is not on the theory of interval in which root
lie, which appears looking at first sight. It is new type and first
time asked in the paper. It is important for future. The actual
type is interval in which parameter lie.
x + 1 − x − 1 = 4x − 1
33. Since,
⇒
(x + 1) + (x − 1) − 2 x2 − 1 = 4x − 1
Case II When x + 2 < 0
1 − 2 x = 2 x2 − 1 ⇒ 1 + 4 x2 − 4 x = 4 x 2 − 4
5
⇒
4x = 5 ⇒ x =
4
But it does not satisfy the given equation.
∴
x2 + x + 2 + x > 0
Hence, no solution exists.
⇒
x2 + 2 x + 2 > 0
⇒
(x + 1)2 + 1 > 0
⇒
x ∈ [−2, − 2 ) ∪ ( 2 , ∞ )
⇒
…(ii)
34. Given,
⇒
which is true for all x.
∴
x ≤ − 2 or x ∈ (−∞ , − 2)
…(iii)
From Eqs. (ii) and (iii), we get
x ∈ (−∞ , − 2 ) ∪ ( 2 , ∞ )
30. Given, log 4 (x − 1) = log 2(x − 3) = log 41/ 2 (x − 3)
⇒
log 4 (x − 1) = 2 log 4 (x − 3)
⇒
log 4 (x − 1) = log 4 (x − 3)2
⇒
⇒
(x − 3)2 = x − 1
x2 + 9 − 6x = x − 1
⇒
⇒
x2 − 7x + 10 = 0
(x − 2)(x − 5) = 0
x = 2, or x = 5
⇒
⇒
x = 5 [Q x = 2 makes log (x − 3) undefined].
Hence, one solution exists.
31. Let α , α 2 be the roots of 3x2 + px + 3 = 0
Now,
⇒
Now,
⇒
S = α + α 2 = − p /3,
P = α3 = 1
α = 1, ω , ω 2
α + α 2 = − p/3
ω + ω 2 = − p/3
⇒
x
3
5
(log 2 x )2 + log 2 x −
4
4
= 2
3
5
(log 2 x)2 + log 2 x − = log x 2
4
4
3
5
1
2
(log 2 x) + log 2 x − =
4
4 2 log 2 x
⇒
3(log 2 x)3 + 4(log 2 x)2 − 5(log 2 x) − 2 = 0
Put
∴
⇒
⇒
⇒
log 2 x = y
3 y3 + 4 y2 − 5 y − 2 = 0
( y − 1) ( y + 2) (3 y + 1) = 0
y = 1, −2 , −1 / 3
log 2 x = 1, − 2, − 1 / 3
1 1
x = 2, 1/3 ,
4
2
⇒
35. Since, α , β are the roots of x2 + px + q = 0 and α 4 , β 4 are
the roots of x2 − rx + s = 0.
⇒ α + β = − p ; α β = q; α 4 + β 4 = r and α 4β 4 = s
Let roots of x2 − 4qx + (2q2 − r ) = 0 be α ′ and β′
Now,
α ′ β ′ = (2q2 − r ) = 2 (αβ )2 − (α 4 + β 4 )
= − (α 4 + β 4 − 2α 2β 2) = − (α 2 − β 2)2 < 0
⇒ Roots are real and of opposite sign.
2
2
36. Given, x −
=1 −
⇒ x=1
x−1
x−1
But at x = 1, the given equation is not defined.
Hence, no solution exist.
Theory of Equations 41
37. Let y =
⇒
x2 − (a + b)x + ab
x−c
41. Given,
∴
yx − cy = x2 − (a + b)x + ab
⇒ x2 − (a + b + y)x + (ab + cy) = 0
And
For real roots, D ≥ 0
⇒
(a + b + y)2 − 4(ab + cy) ≥ 0
⇒ (a + b)2 + y2 + 2(a + b) y − 4ab − 4cy ≥ 0
⇒
x2 + 2 px + q = 0
y2 + 2(a + b − 2c) y + (a − b)2 ≥ 0
and
... (i)
αβ = q
... (ii)
ax2 + 2bx + c = 0
1
2b
α+ =−
β
a
α c
=
β a
D ≤0
 α + β 2

= 
 − αβ 


 −2

4 (a + b − 2c) − 4 (a − b) ≤ 0
2
2
⇒ 4 (a + b − 2c + a − b)(a + b − 2c − a + b) ≤ 0
⇒
(2a − 2c)(2b − 2c) ≤ 0
⇒
(a − c)(b − c) ≤ 0
⇒
(c − a )(c − b) ≤ 0
⇒ c must lie between a and b
⇒
| x|2 − 3|x| + 2 = 0
(|x| − 1) (|x| − 2) = 0
⇒
=
(α − β )2
16
x = 1, − 1, 2, − 2
39. (x − a ) (x − b) + (x − b) (x − c) + (x − c) (x − a ) = 0
3x2 − 2 (a + b + c) x + (ab + bc + ca ) = 0
Now, discriminant = 4 (a + b + c) − 12 (ab + bc + ca )
2
= 4 { a 2 + b2 + c2 − ab − bc − ca }
= 2 {(a − b)2 + (b − c)2 + (c − a )2}
which is always positive.
Hence, both roots are real.
40. Since, a , b, c > 0 and ax2 + bx + c = 0
−b
x=
±
2a
⇒
b2 − 4ac
2a
Case I When b2 − 4ac > 0
x=
⇒
and
−b
+
2a
−b
−
2a
2
1

2
α −  . a ≥ 0

β
a
1
α + 

β
2
Since,
pa ≠ b
⇒ α+
⇒
β 2 ≠ 1, β ≠ { −1, 0, 1}, which is correct.
Hence, four real solutions exist.
⇒
2


1

 α + 
α
β

−  a2
β
 2 






b=−
and
|x| = 1, 2
∴
... (iv)
∴ Statement I is true.
a
 α + β
Again, now
pa = − 
 a = − (α + β )
 2 
2
i.e. a ≤ c ≤ b or b ≤ c ≤ a
38. Since,
... (iii)
Now, ( p2 − q) (b2 − ac)
which is true for all real values of y.
∴
∴
α + β = −2 p
b2 − 4ac
2a
b2 − 4ac
both roots, are negative.
2a
1
≠α +β
β
Similarly, if c ≠ qa
α
1

⇒
a ≠ a α β ⇒ α β −  ≠ 0

β
β
⇒
α ≠ 0 and β −
⇒
β ≠ { −1 , 0 , 1 }
Statement II is true.
Both Statement I and Statement II are true. But
Statement II does not explain Statement I.
42. Given,|x − 2|2 + |x − 2| − 2 = 0
Case I When x ≥ 2
⇒
(x − 2)2 + (x − 2) − 2 = 0
⇒
x2 + 4 − 4 x + x − 2 − 2 = 0
⇒
x2 − 3 x = 0
⇒
x (x − 3) = 0
Case II When b − 4ac = 0
−b
, i.e. both roots are equal and negative
⇒ x=
2a
⇒
x = 0, 3
⇒
x=3
Case III When b2 − 4ac < 0
⇒
{ − (x − 2)}2 − (x − 2) − 2 = 0
⇒
(x − 2)2 − x + 2 − 2 = 0
⇒
x2 + 4 − 4 x − x = 0
have negative real part.
⇒
x2 − 4x − (x − 4) = 0
∴ From above discussion, both roots have negative real
parts.
⇒
x(x − 4) − 1 (x − 4) = 0
⇒
(x − 1) (x − 4) = 0
2
⇒
x=
4ac − b2
−b
±i
2a
2a
1
≠0
β
Case II When x < 2
[0 is rejected]
…(i)
42 Theory of Equations
⇒
x = 1, 4
⇒
x=1
[4 is rejected]
…(ii)
Hence, the sum of the roots is 3 + 1 = 4.
⇒ (| x − 2| + 2) (| x − 2| − 1) = 0
| x − 2| = − 2, 1
⇒
⇒
| x − 2| = 1
43. Since, x − 3kx + 2e
⇒
⇒
⇒
⇒
⇒
[neglecting –2]
⇒ x = 3, 1
Sum of the roots = 4
2
2 log k
− 1 = 0 has product of roots 7.
2e2 log k − 1 = 7
e2 log e k = 4
k2 = 4
k =2
[neglecting −2]
D1 = b2 − 4ac and D2 = b2 + 4ac
D1 + D2 = 2b2 ≥ 0
∴ Atleast one of D1 and D2 is (+ ve). Hence, atleast two
real roots.
Hence, statement is true.
…(i)
…(ii)
…(i)
2y + (2y − 1 − 1) = 2y − 1 + 1
2y = 2
y = 1 ∈ (0, 1]
2y − 2y − 1 + 1 = 2y − 1 + 1
2y − 2 ⋅ 2y − 1 = 0
⇒
2y − 2y = 0 true for all y > 1
From Eqs. (i), (ii) and (iii), we get
…(iii)
…(iv)
[from Eq. (i)]
…(ii)
Case III When y ∈ (1, ∞ )
⇒
On subtracting Eq. (iv) from Eq. (ii), we get
= 10 (c + a ) = 1210
y = − 1 ∈ (−∞ , 0]
∴
a 2 − 10ca − 11d = 0
c + a = 121
2−y = 2
⇒
x − 10cx − 11d = 0
a + b + c + d = 10c + 10a
2− y + (2y − 1 − 1) = 2y − 1 + 1
⇒
a + b = 10c and c + d = 10a
2
⇒
Case I When y ∈ (−∞ , 0]
∴
∴
Similarly, ‘a’ is the root of
∴
51. Given, 2| y | − |2y − 1 − 1| = 2y − 1 + 1
Case II When y ∈(0, 1]
∴ Roots are rational.
Hence, statement is false.
(c + a ) (c − a ) = 11 × 11 (c − a )
Let n be any integer. We have,
 n (n − 1) 
+ ( A + B) n + C
f (n ) = An 2 + Bn + C = 2 A
2


⇒
Here, D = (3)2 − 4 ⋅ 2 ⋅ 1 = 1 which is a perfect square.
∴
C , A + B + C , A − B + C are integers.
C , A + B, A − B are integers.
C , A + B, ( A + B) − ( A − B) = 2 A are integers.
⇒
47. Given, 2x2 + 3x + 1 = 0
(c2 − a 2) = 11 (b − d )
is an integer.
Since, n is an integer, n (n − 1) / 2 is an integer. Also,
2 A , A + B and C are integers.
We get f (n ) is an integer for all integer n.
46. P (x) ⋅ Q (x) = (ax2 + bx + c) (−ax2 + bx + c)
⇒
50. Suppose f (x) = Ax2 + Bx + C is an integer, whenever x
Conversely, suppose 2 A , A + B and C are integers.
= − (1 + 2 + 3 + ... + 100)
100
=−
(1 + 100) = −50(101) = −5050
2
⇒ (a − c) + (b − d ) = 10 (c − a )
⇒
(b − d ) = 11 (c − a )
Since, ‘c’ is the root of x2 − 10ax − 11b = 0
⇒
c2 − 10ac − 11b = 0
2
b2 4c B2 4C
−
= 2−
A
a2 a
A
2
2
b − 4ac B − 4 AC
=
a2
A2
⇒
⇒
⇒
⇒
45. The coefficient of x99 in (x − 1)(x − 2)... (x − 100)
48. Here,
2
4c  B 
4C
 b
= −  −
−  −
 a


a
A
A
⇒
∴ f (0), f (1), f (−1) are integers.
other root is 2 − i 3.
⇒
− p =2 + i 3 + 2 − i 3 = 4
and
q = (2 + i 3 ) (2 − i 3 ) = 7
⇒
( p, q) = (−4, 7)
Clearly,
(α + β )2 − 4αβ = (α + δ + β + δ )2 − 4(α + δ ) (β + δ )
⇒
44. If 2 + i 3 is one of the root of x2 + px + q = 0. Then,
Now,
B
C
, (α + δ ) (β + δ ) =
A
A
α − β = (α + δ ) − (β + δ )
(α − β )2 = [(α + δ ) − (β + δ )]2
Now,
⇒
Given,| x − 2|2 + | x − 2| − 2 = 0
b
c
, αβ =
a
a
α+δ+β+δ=−
and
Alternate Solution
∴
49. Since, α + β = −
52. Given,
y ∈{ −1} ∪ [1, ∞ ).
2x
1
>
2 x2 + 5 x + 2 x + 1
⇒
2x
1
−
>0
(2x + 1) (x + 2) (x + 1)
⇒
2x (x + 1) − (2x + 1) (x + 2)
>0
(2x + 1) (x + 2) (x + 1)
…(iii)
Theory of Equations 43
⇒
− (3x + 2)
> 0 ; using number line rule
(2x + 1) (x + 1) (x + 2)
–
+
–
–2
+
–
–1
1
 2
x ∈ (−2, − 1) ∪  − , − 
 3
2
∴
Given,
⇒
– 1
2
– 2
3
56. Let α , β are roots of ax2 + bx + c = 0
α = βn
αβ =
c
c
 c
⇒ β= 
⇒ βn + 1 =
 a
a
a
1/( n + 1 )
It must satisfy ax2 + bx + c = 0
i.e.
53. Given,| x + 4x + 3|+ 2x + 5 = 0
2
Case I
x2 + 4x + 3 > 0 ⇒ (x < − 3 or x > − 1)
2
∴
x + 4x + 3 + 2x + 5 = 0
⇒
x2 + 6x + 8 = 0 ⇒ (x + 4) (x + 2) = 0
[but x < − 3 or x > − 1]
⇒
x = − 4, − 2
…(i)
∴
x = − 4 is the only solution.
Case II
x2 + 4x + 3 < 0 ⇒ (−3 < x < − 1)
∴
− x2 − 4 x − 3 + 2 x + 5 = 0
⇒
x2 + 2x − 2 = 0 ⇒ (x + 1)2 = 3
⇒
| x + 1| = 3
⇒
x = − 1 − 3, −1 + 3
∴
x = − 1 − 3 is the only solution.
[but x ∈ (−3, − 1)]
…(ii)
 c
a 
 a
⇒
⇒
c1/( n + 1)
a1/( n + 1)
2 /( n + 1 )
 c
+ b 
 a
1 /( n + 1 )
+ c=0
a . c2/( n + 1) b. c1/( n + 1)
+ 1/( n + 1) + c = 0
a 2/( n + 1)
a
+
1
/(
n
1
)
 a. c
c. a1/( n + 1) 
 1/( n + 1) + b + 1/( n + 1)  = 0
c
 a

⇒
a n/( n + 1)c1/( n + 1) + b + cn/( n + 1)a1 / ( n + 1) = 0
⇒
(a nc)1/( n + 1) + (cna )1/( n + 1) + b = 0
57. Since, α , β are the roots of x2 + px + q = 0
and γ , δ are the roots of x2 + rx + s = 0
∴
α + β = − p, αβ = q and γ + δ = − r, γδ = s
Now, (α − γ )(α − δ )(β − γ )(β − δ )
From Eqs. (i) and (ii), we get
= [α 2 − (γ + δ ) α + γδ ][β 2 − (γ + δ ) β + γδ ]
x = − 4 and (−1 − 3) are the only solutions.
= (α 2 + rα + s)(β 2 + rβ + s)
a ≤0
Given,
x2 − 2 a | x − a | − 3 a 2 = 0
Case I When x ≥ a
= (αβ )2 + r (α + β )αβ + s(α 2 + β 2) + αβr 2 + rs(α + β ) + s2
54. Here,
⇒
x2 − 2a (x − a ) − 3a 2 = 0
⇒
x2 − 2ax − a 2 = 0
= q2 − rqp + s( p2 − 2q) + qr 2 − rsp + s2
= (q − s)2 − rqp − rsp + sp2 + qr 2
58. The given equation can be rewritten as
2
1
3
+
+
= 0 [Q b = a 2x, given]
log a x log a ax log a a 2x
x = a ± 2a
⇒
[as a (1 + 2 ) < a and a (1 − 2 ) > a]
⇒
∴ Neglecting x = a (1 + 2 ) as x ≥ a
⇒
x = a (1 − 2 )
Case II
⇒
…(i)
⇒
When x < a ⇒ x + 2a (x − a ) − 3a = 0
2
2
x2 + 2 ax − 5a 2 = 0 ⇒ x = − a ± 6a
[as a ( 6 − 1) < a and a (−1 − 6 ) > a]
∴ Neglecting x = a (−1 − 6 ) ⇒ x = a ( 6 − 1)
...(ii)
From Eqs. (i) and (ii), we get
55. Given, (5 + 2 6 )x
2
Put y = (5 + 2 6 )
From Eq. (i), y +
−3
+ (5 − 2 6 )x
x2 − 3
= 10
x2 − 3
…(i)
1
=
y
1
= 10
y
y2 − 10 y + 1 = 0 ⇒
⇒
(5 + 2 6 )x
or
(5 + 2 6 )x
x2 − 3 = 1
−3
⇒ (5 − 2 6 )
⇒
⇒
2
or
2
⇒
6 t 2 + 11 t + 4 = 0
⇒
(2 t + 1) (3 t + 4) = 0
1
4
t=−
or −
2
3
1
4
log a x = −
or log a x = −
2
3
∴
⇒
x = a −1/ 2 or
x =a
−4/3
59. Since, α + β = − p, αβ = 1 and γ + δ = − q, γδ = 1
y=5±2 6
2
2
1
3
+
+
= 0, where t = log a x
t 1+ t 2+ t
⇒ 2 (1 + t ) (2 + t ) + 3 t (1 + t ) + t (2 + t ) = 0
⇒
x = { a (1 − 2 ), a ( 6 − 1)}
2
1
3
+
+
=0
log a x 1 + log a x 2 + log a x
−3
=5 + 2 6
−3
=5 −2 6
x2 − 3 = − 1
⇒
x = ± 2 or x = ± 2
⇒
x = ± 2, ± 2
Now, (α − γ )(β − γ )(α + δ )(β + δ )
= {αβ − γ (α + β ) + γ 2}{αβ + δ (α + β ) + δ 2}
= {1 − γ (− p) + γ 2}{1 + δ ( − p) + δ 2}
= (1 + γ 2 + γp)(1 − δp + δ 2) = (− qγ + γp)(−δp − δq)
[Q γ 2 + qγ + 1 = 0 and δ 2 + qδ + 1 = 0]
= (q2 − p2)(γδ ) = q2 − p2
[Q γ δ = 1]
44 Theory of Equations
α2 = α + 1
60.
β =β + 1
a n = pα + qβ
n
and a 2x2 + b2x + c2 = 0
have a common real root, then
n
= p(α n − 1 + α n − 2) + q(β n − 1 + β n − 2)
⇒
= an − 1 + an − 2
α=
1+ 5
1− 5
,β =
2
2
a 4 = a3 + a 2
= 2a 2 + a1
= 3a1 + 2a 0
⇒
⇒
(1 + b)2 = (b2 + 1) (1 − b)
⇒
b2 + 2b + 1 = b2 − b3 + 1 − b
⇒
b3 + 3b = 0
∴
b (b2 + 3) = 0
x2 + bx + a = 0 have common root
p− q =0
7
( p + q) × = 28
2
On subtracting above equations, we get
(a − b) x + (b − a ) = 0
x=1
⇒
∴ x = 1 is the common root.
p+ q =8
⇒
p= q =4
⇒
∴
p + 2q = 12
⇒
1 Given α, β and γ are three consecutive terms of a
non-constant GP.
Let
α = α, β = αr , γ = αr 2, { r ≠ 0, 1}
and given quadratic equation is
αx2 + 2 βx + γ = 0
On putting the values of α,β, γ in Eq. (i), we get
αx2 + 2αrx + αr 2 = 0
⇒
x2 + 2rx + r 2 = 0
⇒
(x + r )2 = 0
⇒
x=−r
…(i)
f (x) = a nxn + a n − 1xn − 1 + ... + a 0
Then, x = r is root of f ′ (x) = 0 repeated (m − 1) times.
Topic 3 Transformation of Roots
1. Given, α , β are the roots of (x − a )(x − b) − c = 0
⇒
⇒
(x − a )(x − b) − c = (x − α ) (x − β )
(x − a )(x − b) = (x − α )(x − β ) + c
⇒ a , b are the roots of equation (x − α )(x − β ) + c = 0
2. Given equations are x2 + 2x + 3 = 0
…(i)
ax + bx + c = 0
…(ii)
Since, Eq. (i) has imaginary roots, so Eq. (ii) will also
have both roots same as Eq. (i).
a b c
Thus,
= =
1 2 3
Hence, a : b : c is 1 : 2 : 3.
a + b = −1
Hence, statement is false.
Q The quadratic equations αx2 + 2 βx + γ = 0 and
x2 + x − 1 = 0 have a common root, so x = − r must be root
of equation x2 + x −1 = 0, so
…(ii)
r2 − r − 1 = 0
Now,
α (β + γ ) = α (αr + αr 2)
= α 2 (r + r 2)
From the options,
βγ = αr ⋅ αr 2 = α 2r3 = α 2 (r + r 2)
[Q r 2 − r − 1 = 0 ⇒ r3 = r + r 2]
∴
α (β + γ ) = βγ
2
1 + a + b =0
5. Since, (x − r ) is a factor of the polynomial
Topic 2 Common Roots
and
b = 0, ± 3 i
4. Given equations are x2 + ax + b = 0 and
3 5
3

28 = ( p + q) + 2 + ( p − q)

2

 2 
and
x2 + bx − 1 = 0
 have a common root.
x2 + x + b = 0 
⇒
28 = p(3α + 2) + q(3β + 2)
∴
(a1c2 − a 2c1 )2 = (b1c2 − b2c1 ) (a1b2 − a 2b1 )
∴
∴ a12 = a11 + a10
61.
a1x2 + b1x + c1 = 0
3. If
2
2. Since, ax2 + bx + c = 0 has roots α and β.
⇒
α + β = − b /a
αβ = c / a
and
Now, a x + abcx + c3 = 0
3 2
On dividing the equation by c2, we get
a3 2 abcx c3
x + 2 + 2 =0
c
c
c2
2
⇒
⇒
⇒
 ax
 ax
a   + b   + c=0
 c
 c
ax
= α , β are the roots
c
c
c
x = α , β are the roots
a
a
⇒
x = α β α , α β β are the roots
⇒
x = α 2β , αβ 2 are the roots
…(i)
Theory of Equations 45
1
[4(2t − 6) + 4(t 2 − 9) + 1(6t 2 − 18t ]
2
1
= |[8t − 24 + 4t 2 − 36 + 6t 2 − 18t ]|
2
Divide the Eq. (i) by a3 , we get
=
3
x2 +
b c
 c
⋅ ⋅ x+   =0
 a
a a
⇒
x2 − (α + β ) ⋅ (αβ ) x + (αβ )3 = 0
⇒
x2 − α 2βx − αβ 2 x + (αβ )3 = 0
⇒
x (x − α 2β ) − αβ 2 (x − α 2β ) = 0
⇒
= |5t 2 − 5t − 30| = |5(t + 2) (t − 3)|
(x − α 2β )(x − αβ 2) = 0
⇒ x = α β , αβ which is the required answer.
2
2
Alternate Solution
a3 x2 + abcx + c3 = 0
Since,
⇒ x=
⇒ x=
⇒ x=
− abc ±
(abc)2 − 4 . a3 ⋅ c3
Now, as X is any point on the arc POQ of the parabola,
therefore ordinate of point X, 2t ∈ (− 4, 6) ⇒ t ∈ (− 2, 3).
∴ Area of ∆PXQ = − 5(t + 2) (t − 3) = − 5t 2 + 5t + 30
[Q| x − a | = − (x − a ), if x < a]
The maximum area (in square units)
25 − 4(− 5) (30)  125
=−
= 4
4(− 5)


[Q Maximum value of quadratic expression
D
]
ax2 + bx + c, when a < 0 is −
4a
2 a3
− (b/a )(c/a ) ±
(b/a ) (c/a ) − 4(c/a )
2
2
(α + β ) (α β ) ±
2
3
(α + β ) (α β ) − 4(α β )
2
2
2
3
2. Let f (x) = (c − 5)x2 − 2 cx + (c − 4) = 0.
Then, according to problem, the graph of y = f (x) will be
either of the two ways, shown below.
(α + β )(αβ ) ± α β (α + β )2 − 4 αβ
2
 (α + β ) ± (α − β )2 
⇒ x = αβ 

2


 (α + β ) ± (α − β ) 
⇒ x = αβ


2
⇒ x=
α + β + α − β α + β − α + β 
,
⇒ x = αβ


2
2
2α 2β 
,
⇒ x = αβ
 2 2 
⇒
x = α 2 β , α β 2 which is the required answer.
Topic 4 Graph of Quadratic Expression
O
O
X (t 2,2t)
Q(9,6)
y 2=4x
2
f (0) f (2) < 0
Now, consider
⇒
(c − 4) [4 (c − 5) − 4c + (c − 4)] < 0
⇒
(c − 4) (c − 24) < 0
⇒
c ∈ (4, 24)
+
–
4
⇒
P(4,–4)
Y′
1
∴ Area of ∆PXQ =
2
4
2
t
9
−4 1
2t
6
1
1
24
–
49/4
X
O
… (i)
+
Similarly, f (2) ⋅ f (3) < 0
⇒ [4 (c − 5) − 4c + (c − 4)]
[9(c − 5) − 6c + (c − 4)] < 0
⇒
(c − 24) (4c − 49) < 0
+
X′
3
In both cases f (0). f (2) < 0 and f (2) f (3) < 0
1. Given parabola is y = 4x,
Y
3
Or
2
Since, X lies on the parabola, so let the coordinates of X
be (t 2, 2t ). Thus, the coordinates of the vertices of the
triangle PXQ are P (4,–4), X (t 2,2t ) and Q (9, 6).
2
+
24
 49

c ∈
, 24
4

From Eqs. (i) and (ii), we get
 49

c ∈
, 24
4

∴Integral values of c are 13, 14, ……, 23.
Thus, 11 integral values of c are possible.
…(ii)
46 Theory of Equations
3. According to given information, we have the following
5. As we know, ax2 + bx + c > 0 for all x ∈ R, iff
graph
a > 0 and D < 0.
Given equation is x2 + 2ax + (10 − 3a ) > 0, ∀ x ∈ R
Y
D <0
Now,
O
X
5
1
⇒
4a 2 − 4(10 − 3a ) < 0
⇒
4(a 2 + 3a − 10) < 0
⇒
(a + 5)(a − 2) < 0 ⇒ a ∈ (−5, 2)
Now, the following conditions should satisfy
(i) D > 0 ⇒ b2 − 4ac > 0
⇒
m2 − 4 × 1 × 4 > 0
⇒
m2 − 16 > 0
⇒
(m − 4) (m + 4) > 0
⇒
m ∈ (− ∞ , − 4) ∪ (4, ∞ )
(ii) The vertex of the parabola should lie
between x = 1and x = 5
b
m
−
∈ (1, 5) ⇒1 < < 5 ⇒ m∈ (2, 10)
∴
2a
2
(iii) f (1) > 0 ⇒1 − m + 4 > 0
⇒ m < 5 ⇒ m ∈ (−∞ , 5)
y = (x – a)(x – b) –1
6.
α a
b β
1
From graph, it is clear that one of the roots of
(x − a )(x − b) − 1 = 0 lies in (− ∞ , a ) and other lies in
(b, ∞ ).
7. Let f (x) = x2 − 2ax + a 2 + a − 3
Since, both root are less than 3.
(iv) f (5) > 0 ⇒ 25 − 5m + 4 > 0
29

⇒ 5m < 29 ⇒ m ∈  − ∞, 

5
⇒
α < 3, β < 3
⇒
From the values of m obtained in (i), (ii), (iii) and (iv), we
get m ∈ (4, 5).
⇒
Sum, S = α + β < 6
α+β
<3
2
2a
<3
2
⇒
–∞
–4
2
4
5 29/5
⇒
∞
a <3
Again, product, P = αβ
⇒
4. Put t = x − [x] = { X }, which is a fractional part function
and lie between 0 ≤ { X } < 1 and then solve it.
−3 { x − [x]} + 2 { x − [x]} + a = 0
2
Let t = x − [x], then equation is
1 ± 1 + 3a 2
t=
3
Q
t = x − [x] = { X }
∴
0 ≤ t ≤1
[fractional part]
1 ± 1 + 3a 2
0≤
≤1
3
1 + 1 + 3a 2
0≤
<1
3
⇒
a 2 + a − 12 < 0
(a − 3) (a + 4) < 0
⇒
−4 < a < 3 …(ii)
Again, D = B − 4 AC ≥ 0
2
α
⇒
4a 2 − 4a 2 − 4a + 12 ≥ 0
⇒
−4a + 12 ≥ 0 ⇒ a ≤ 3
Again,
a f (3) > 0
2
⇒ 1 [(3) − 2a (3) + a 2 + a − 3] > 0
⇒
9 − 6a + a 2 + a − 3 > 0
⇒
a 2 − 5a + 6 > 0
∴
[Q{ x } > 0]
1 + 3a 2 < 2 ⇒ 1 + 3a 2 < 4
a 2 − 1 < 0 ⇒ (a + 1) (a − 1) < 0
∴ a ∈ (−1, 1), for no integer solution of a, we consider
(−1, 0) ∪ (0, 1)
a + a −3 <9
⇒
⇒
Taking positive sign, we get
⇒
⇒ αβ < 9
2
β
3
⇒ (−2a )2 − 4 ⋅ 1 (a 2 + a − 3) ≥ 0
−3 t 2 + 2 t + a 2 = 0
⇒
P <9
⇒
⇒
Given, a ∈ R and equation is
2
…(i)
…(iii)
(a − 2) (a − 3) > 0
a ∈ (−∞ , 2) ∪ (3, ∞ )
…(iv)
From Eqs. (i), (ii), (iii) and (iv), we get
a ∈ (−4, 2).
NOTE There is correction in answer a < 2 should be −4 < a < 2 .
8. Let f (x) = ax2 + bx + c > 0, ∀ x ∈ R
⇒
and
a >0
b2 − 4ac < 0
…(i)
Theory of Equations 47
∴
⇒
⇒
g (x) = f (x) + f ′ (x) + f ′ ′ (x)
g (x) = ax2 + bx + c + 2ax + b + 2a
g (x) = ax2 + x (b + 2a ) + (c + b + 2a )
11. Since, x2 − 3x + 2 > 0 and x2 − 2x − 4 ≤ 0
⇒
⇒
whose discriminant
= (b + 2a )2 − 4a (c + b + 2a )
= b2 + 4a 2 + 4ab − 4ac − 4ab − 8a 2
[from Eq. (i)]
= b2 − 4a 2 − 4ac = (b2 − 4ac) − 4a 2 < 0
∴ g (x) > 0 ∀ x, as a > 0 and discriminant < 0.
Thus, g (x) > 0, ∀ x ∈ R.
9. Given,
x2 + (a − b) x + (1 − a − b) = 0 has real and unequal roots.
⇒
D >0
⇒
(a − b)2 − 4(1) (1 − a − b) > 0
⇒ a 2 + b2 − 2ab − 4 + 4a + 4b > 0
Now, to find the values of ‘a’ for which equation has
unequal real roots for all values of b.
i.e. Above equation is true for all b.
or b2 + b(4 − 2a ) + (a 2 + 4a − 4) > 0, is true for all b.
∴ Discriminant, D < 0
⇒
(4 − 2a )2 − 4 (a 2 + 4a − 4) < 0
⇒
16 − 16a + 4a − 4a − 16a + 16 < 0
2
⇒
2
−32a + 32 < 0
10.
⇒
a >1
Y
a<0
y = ax2 + bx + c
–1
α
1
0
X
β
Y
a>0
y = ax2 + bx + c
α
–1
0 1
β
X
From figure, it is clear that, if a > 0, then f (−1) < 0 and
f (1) < 0 and if a < 0, f (−1) > 0 and f (1) > 0. In both
cases, af (−1) < 0 and af (1) < 0.
⇒
a (a − b + c) < 0
and
a (a + b + c) < 0
On dividing by a 2, we get
b
c
1 − + < 0 and
a a
On combining both, we get
b
c
1 ± + <0
a a
b
c
1 +  + < 0
⇒
a a
(x − 1) (x − 2) > 0
x ∈ [1 − 5 , 1) ∪ [1 + 5 , 2)
(i) Given, x2 − 8kx + 16 (k2 − k + 1) = 0
Now, D = 64 { k2 − (k2 − k + 1)} = 64 (k − 1) > 0
k >1
b
8k
(ii) −
>4 ⇒
>4
⇒ k >1
2a
2
(iii) f (4) ≥ 0
⇒
16 − 32k + 16 (k2 − k + 1) ≥ 0
⇒
k2 − 3k + 2 ≥ 0
⇒
(k − 2) (k − 1) ≥ 0
⇒
k ≤1
or k ≥ 2
k =2
Hence,
Topic 5 Some Special Forms
1. Given equation 5 + |2x − 1| = 2x (2x − 2)
Case I
If 2x − 1 ≥ 0 ⇒ x ≥ 0 ,
then 5 + 2x − 1 = 2x (2x − 2)
Put 2x = t, then
5 + t − 1 = t 2 − 2t ⇒ t 2 − 3t − 4 = 0
2
⇒ t − 4t + t − 4 = 0 ⇒ t (t − 4) + 1(t − 4) = 0
⇒
t = 4 or − 1 ⇒ t = 4
(Q t = 2x > 0)
x
⇒
2 =4⇒x=2 >0
⇒ x = 2 is the solution.
Case II
If 2x − 1 < 0 ⇒ x < 0 ,
then 5 + 1 − 2x = 2x (2x − 2)
Put 2x = y, then 6 − y = y2 − 2 y
⇒
y2 − y − 6 = 0 ⇒ y2 − 3 y + 2 y − 6 = 0
⇒ ( y + 2) ( y − 3) = 0 ⇒ y = 3 or − 2
⇒
y = 3(as y = 2x > 0) ⇒ 2x = 3
⇒ x = log 2 3 > 0
So, x = log 2 3 is not a solution.
Therefore, number of real roots is one.
2. Given, inequality is
2
sin 2 x − 2sin x + 5
⋅
1
sin 2 y
≤1
4
⇒2
b
c
1 + + <0
a a
x2 − 2 x + 1 ≤ 5
(x < 1 or x > 2) and (1 − 5 ≤ x ≤ 1 + 5 )
∴
12.
and
⇒2
⇒
(sin x − 1 )2 + 4
2
(sin x − 1 ) + 4
⋅ 2−2sin
2
≤2
2
y
≤1
2sin y
(sin x − 1)2 + 4 ≤ 2 sin 2 y
[if a > 1 and am ≤ a n ⇒ m ≤ n]
Q Range of (sin x − 1)2 + 4 is [2, 2 2 ]
and range of 2 sin 2 y is [0, 2].
48 Theory of Equations
∴ By Rolle’s theorem, f ′ (α ) = 0 for 0 < α < 1
∴The above inequality holds, iff
(sin x − 1)2 + 4 = 2 = 2 sin 2 y
⇒ sin x = 1 and sin 2 y = 1
⇒ sin x = |sin y|
3.
[from the options]
Key Idea Reduce the given equation into quadratic equation.
Now,
f ′ (x) = 3ax2 + 2bx + c
⇒
f ′ (α ) = 3aα 2 + 2bα + c = 0
∴ Eq. (i) has exist atleast one root in the interval (0, 1).
Thus, f ′ (x) must have root in the interval (0, 1) or
3ax2 + 2bx + c = 0 has root ∈ (0, 1).
7. Given, x12 − x9 + x4 − x + 1 > 0
Given equation is
| x − 2| + x ( x − 4) + 2 = 0
⇒ | x − 2| + x − 4 x + 4 = 2
Here, three cases arises:
Case I When x ≤ 0 ⇒ x12 > 0, − x9 > 0, x4 > 0, − x > 0
…(i)
∴
x12 − x9 + x4 − x + 1 > 0, ∀ x ≤ 0
⇒ | x − 2| + ( x − 2)2 = 2
⇒ (| x − 2|)2 + | x − 2| − 2 = 0
Let| x − 2| = y, then above equation reduced to
y2 + y − 2 = 0 ⇒ y2 + 2 y − y − 2 = 0
⇒ y( y + 2) − 1( y + 2) = 0 ⇒ ( y + 2)( y − 1) = 0
⇒
y = 1, − 2
[Q y = | x − 2| ≥ 0]
∴
y=1
⇒
| x − 2| = 1
x −2 = ±1
⇒
x = 3 or 1
⇒
⇒
x = 9 or 1
∴ Sum of roots = 9 + 1 = 10
4. Let f (x) = 2x3 + 3x + k
Case II When 0 < x ≤ 1
x9 < x4 and x < 1 ⇒ − x9 + x4 > 0
∴
9
4
f ′ (x) = 6x2 + 3 > 0, ∀ x ∈ R
⇒ f (x) = 0 has only one real root, so two roots are not
possible.
and β is a root of
a x − bx − c = 0
⇒
a 2β 2 − bβ − c = 0
f (x) = a 2x2 + 2bx + 2c
∴
f (α ) = a 2α 2 + 2bα + 2c
... (i)
8. Consider,
1
f (x) = ∫ (1 + cos 8 x)(ax2 + bx + c) dx
0
f (1) = f (2)
... (ii)
f ′ (x) = (1 + cos 8 x)(ax2 + bx + c)
Now,
f (β ) = α 2β 2 + 2bβ + 2c
f (α ) f (β ) < 0
f (x) must have a root lying in the open interval (α , β ).
∴
α < γ <β
6. Let
f (x) = ax3 + bx2 + cx + d
∴
f (0) = d and
∴
f (0) = f (1)
…(i)
f (1) = a + b + c + d = d
[Q a + b + c = 0]
f is continuous in the closed interval [0, 1] and f is
derivable in the open interval (0, 1).
Also,
f (0) = f (1).
⇒ (1 + cos k)(ak + bk + c) = 0
8
2
ak2 + bk + c = 0
[as (1 + cos 8 k) ≠ 0]
∴ x = k is root of ax2 + bx + c = 0,
where
= a 2β 2 + 2a 2β 2 = 3a 2β 2 [from Eq. (ii)]
⇒
[given]
∴ By Rolle’s theorem, there exist atleast one point
k ∈ (1, 2), such that f ′ (k) = 0.
⇒
= a 2α 2 − 2a 2α 2 = − a 2α 2
[from Eq. (i)]
and
…(iii)
f ′ (k) = 0
2 2
Let
4
From Eqs. (i), (ii) and (iii), the above equation holds for
all x ∈ R.
5. Since, α is a root of a 2x2 + bx + c = 0
a α + bα + c = 0
9
∴
Also,
⇒ f (x) is strictly increasing function.
⇒
…(ii)
Case III When x > 1 ⇒ x > x and x > x
x12 − x9 + x4 − x + 1 > 0, ∀ x > 1
12
Obviously, f (x) is continuous and differentiable in the
interval [1, 2].
On differentiating w.r.t. x, we get
2 2
and 1 − x > 0
x − x + x − x + 1 > 0, ∀ 0 < x ≤ 1
12
k ∈ (1, 2)
9. Given, x1 and x2 are roots of αx2 − x + α = 0.
∴
Also,
⇒
⇒
or
⇒
⇒
1
and x1x2 = 1
α
x1 − x2 < 1
|x1 − x2|2 < 1
(x1 − x2)2 < 1
(x1 + x2)2 − 4x1x2 < 1
1
1
− 4 < 1 or
<5
α2
α2
x1 + x2 =
5α 2 − 1 > 0 or ( 5 α − 1) ( 5 α + 1) > 0
+
–1/√5
∴
–
+
1/√5
1  1


α ∈  −∞ , −
, ∞
 ∪

5  5 
…(i)
Theory of Equations 49
D >0
1 − 4α 2 > 0
 1 1
α ∈ − , 
 2 2
Also,
⇒
or
1
 3
f (x) changes its sign in  − , − 
 4
2
From Eqs. (i) and (ii), we get
 1 −1   1 1 
, 
α ∈ − ,  ∪ 
 2 5   5 2
10.
 3 1
Hence, f (x) = 0 has a root in  − , −  .
 4 2
…(ii)
PLAN
(i) Concepts of curve tracing are used in this question.
12.
1/ 2
∫0
t
3/ 4
0
0
f (x) dx < ∫ f (x) dx < ∫
f (x) dx
Now, ∫ f (x) dx = ∫ (1 + 2x + 3x2 + 4x3 ) dx
= x + x2 + x3 + x4
(ii) Number of roots are taken out from the curve traced.
Let y = x5 − 5x
(i) As x → ∞, y → ∞ and as x → − ∞, y → − ∞
(ii) Also, at x = 0, y = 0, thus the curve passes through
the origin.
dy
(iii)
= 5x4 − 5 = 5 (x4 − 1) = 5 (x2 − 1) (x2 + 1)
dx
= 5 (x − 1) (x + 1) (x2 + 1)
+
–
+
–1
1
dy
> 0 in (− ∞ , − 1) ∪ (1, ∞ ), thus f (x) is
dx
increasing in these intervals.
dy
Also,
< 0 in (− 1, 1), thus decreasing in (− 1, 1).
dx
(iv) Also, at x = − 1, dy /dx changes its sign from + ve to
–ve.
∴ x = − 1 is point of local maxima.
Similarly, x = 1 is point of local minima.
Local maximum value, y = (− 1)5 − 5 (−1) = 4
Local minimum value, y = (1)5 − 5(1) = − 4
1
1 t 3
2
4
f (x) dx =
15 3
> ,
16 4
3/ 4
∫0
f (x) dx =
1
and
4
1
f ′ (x) < 0, when x < − .
4
f ′ (x) > 0, when x > −
∴ It could be shown as
1
S
–3
4
–1
2
1
2
3
4
14. Let y = x intersect the curve y = kex at exactly one point
when k ≤ 0 .
–1
Y
(1, – 4)
Now, let y = − a
As evident from the graph, if − a ∈ (− 4, 4)
i.e.
a ∈ (− 4, + 4)
Then, f (x) has three real roots and if − a > 4
or − a < − 4, then f (x) has one real root.
i.e. for a < − 4 or a > 4, f (x) has one real root.
11. Given, f (x) = 4x3 + 3x2 + 2x + 1
D = 9 − 24 < 0
Hence, f (x) = 0 has only one real root.
3 4
 1
f −  = 1 −1 + − > 0
 2
4 8
6 27 108
 3
−
f −  = 1 − +
 4
4 16 64
64 − 96 + 108 − 108
=
<0
64
X
X′
Y′
15. Let
f ′ (x) = 2 (6x2 + 3x + 1)
⇒
530
<3
256
13. As, f ′′ (x) = 2 (12x + 3)
Now,
(–1, 4)
1/ 2
∫0
⇒
f (x) = ke − x
f ′ (x) = kex − 1 = 0
x
⇒
x = − ln k
f ′ ′ (x) = kex
∴
[ f ′ ′ (x)] x = − ln k = 1 > 0
Hence,
f (− ln k) = 1 + ln k
For one root of given equation
⇒
1 + ln k = 0
1
k=
e
50 Theory of Equations
Which is ≤ 0 , ∀ p ∈ [−1, 1].
16. For two distinct roots, 1 + ln k < 0 (k > 0)
ln k < −1 ⇒ k <
Hence,
1
e
∴ f (x) has atleast one root in
 1
k ∈ 0, 
 e
Now, f ′ (x) = 12x2 − 3 = 3 (2x − 1) (2x + 1)
3
1 
1
1 
=  x −   x +  > 0 in
,1
2 
4
2 
2
17. Let f (x) = (x − a ) (x − c) + 2 (x − b) (x − d )
Here,
⇒ f (x) is an increasing function in [1 /2, 1]
f (a ) = + ve
Therefore, f (x) has exactly one root in [1 /2, 1] for any
p ∈ [−1, 1].
f (b) = − ve
f (c) = − ve
Now, let x = cos θ
f (d ) = + ve
∴ There exists two real and distinct roots one in the
interval (a , b) and other in (c, d ).
Hence, statement is true.
18. Let f (x) = 4x3 − 3x − p
…(i)
3
Now,
4 3
 1
 1
 1
f   =4   −3   − p= − − p
 2
 2
 2
8 2
= − (1 + p)
f (1) = 4(1)3 − 3(1) − p = 1 − p
⇒
1 
,1 .
2 
 1
f   . f (1) = − (1 + p)(1 − p)
 2
= ( p + 1)( p − 1) = p − 1
2
∴
x∈
1
,
2
1


 π
⇒ θ ∈ 0,
 3 
From Eq. (i),
4 cos3 θ − 3 cos θ = p
⇒
⇒
⇒
⇒
⇒
cos 3θ = p
3θ = cos −1 p
1
θ = cos −1 p
3
1

cos θ = cos  cos −1 p
3


1
x = cos  cos −1 p

3
3
Sequences and Series
Topic 1 Arithmetic Progression (AP)
Objective Questions I (Only one correct option)
1. If a1 , a 2, a3 , ... , a n are in AP and a1 + a 4 + a7 + ... + a16
= 114 , then a1 + a 6 + a11 + a16 is equal to
(2019 Main, 10 April I)
(a) 64
(c) 98
(b) 76
(d) 38
Analytical and Descriptive Question
5. If a1 , a 2 ..... , a n are in arithmetic progression, where
ai > 0, ∀ i, then show that
1
1
+
+ ...
a1 + a 2
a 2 + a3
+
2. If 19th term of a non-zero AP is zero, then its (49th
term) : (29th term) is
(a) 1 : 3
(c) 2 : 1
(2019 Main, 11 Jan II)
(b) 4 : 1
(d) 3 : 1
3. For any three positive real numbers a , b and c, if
9 (25a 2 + b2) + 25 (c2 − 3ac) = 15b (3a + c), then (2017 Main)
(a) b, c and a are in GP
(b) b, c and a are in AP
(c) a , b and c are in AP
(d) a , b and c are in GP
1
an − 1 +
an
=
n −1
a1 + a n
(1982, 2M)
True/False
6. n1 , n2, K , n p are p positive integers, whose sum is an
even number, then the number of odd integers among
them is odd.
(1985, 1M)
Integer & Numerical Answer Type Questions
4. If Tr is the r th term of an AP, for r = 1, 2, 3, .... . If for
7. Let AP (a ; d ) denote the set of all the terms of an
1
some positive integers m and n, we have Tm = and
n
1
Tn = , then Tmn equals
m
(1998, 2M)
infinite arithmetic progression with first term a and
common difference d > 0. If
1
mn
(c) 1
(a)
1
1
+
m n
(d) 0
(b)
AP (1 ; 3) ∩ AP (2 ; 5) ∩ AP (3 ; 7) = AP (a ; d )
then a + d equals ................
(2019 Adv.)
8. The sides of a right angled triangle are in arithmetic
progression. If the triangle has area 24, then what is the
length of its smallest side?
(2017 Adv.)
Topic 2 Sum of n Terms of an AP
Objective Questions I (Only one correct option)
1. If a1 , a 2, a3 , ... are in AP such that a1 + a7 + a16 = 40,
then the sum of the first 15 terms of this AP is
(2019 Main, 12 April II)
(a) 200
(c) 120
(b) 280
(d) 150
S 4 = 16 and S 6 = − 48, then S10 is equal to
(2019 Main, 12 April I)
(b) − 410
(d) − 380
sum of the series
1 
 1  1
+
+ − −
−
 3   3 100 
2 
 1
+…+
− −
 3 100 
 1 99 
is
− −
 3 100 
(2019 Main, 12 April I)
2. Let S n denote the sum of the first n terms of an AP. If
(a) − 260
(c) − 320
3. For x ∈ R, let [x] denote the greatest integer ≤ x, then the
(a) − 153
(c) − 131
(b) − 133
(d) − 135
4. If the sum and product of the first three terms in an AP
are 33 and 1155, respectively, then a value of its 11th
term is
(2019 Main, 9 April II)
(a) 25
(b) –36
(c) –25
(d) –35
52 Sequences and Series
5. Let the sum of the first n terms of a non-constant AP
n (n − 7)
A, where A is a constant.
a1 , a 2, a3 .....be 50n +
2
If d is the common difference of this AP, then the
ordered pair (d , a50 ) is equal to
(2019 Main, 9 April I)
(a) (A, 50 + 46A)
(c) (50, 50 + 46A)
(b) (50, 50 + 45A)
(d) (A, 50 + 45A)
6. The sum of all two digit positive numbers which when
divided by 7 yield 2 or 5 as remainder is
(2019 Main, 10 Jan I)
(a) 1256
(b) 1465
(c) 1356
(d) 1365
7. Let a1 , a 2, ..... a30 be an AP, S = ∑ ai and
∑ a( 2 i − 1). If a5 = 27 and S − 2T = 75,
i =1
(2019 Main, 9 Jan I)
(a) 42
(c) 52
(b) 57
(d) 47
... , log e b101 are in AP with the common difference log e 2
. Suppose a1 , a 2, ... , a101 are in AP, such that a1 = b1 and
If
and
a51 = b51.
t = b1 + b2 + ... + b51
(2016 Adv.)
s = a1 + a 2 + ... + a51, then
(a) s > t and a101 > b101
(b) s > t and a101 < b101
(c) s < t and a101 > b101
(d) s < t and a101 < b101
(a) Q1 , Q2 , Q3 ,... are in an AP with common difference 5
(b) Q1 , Q2 , Q3 ,... are in an AP with common difference 6
(c) Q1 , Q2 , Q3 ,... are in an AP with common difference 11
(d) Q1 = Q2 = Q3 = ...
of squares of these n terms is
15. Let p and q be the roots of the equation x2 − 2x + A = 0
and let r and s be the roots of the equation
x2 − 18x + B = 0. If p < q < r < s are in arithmetic
(1997, 2M)
progression, then A = … and B = … .
16. The sum of the first n terms of the series
n (n + 1)2
, when
2
n is even. When n is odd, the sum is .... .
(1988, 2M)
12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + K is
9. If the sum of first n terms of an AP is cn 2, then the sum
(2009)
n (4n 2 + 1) c2
(b)
3
n (4n 2 + 1) c2
(d)
6
17. The sum of integers from 1 to 100 that are divisible by 2
or 5 is ……
(1984, 2M)
Analytical & Descriptive Questions
18. The fourth power of the common difference of an
10. If the sum of the first 2n terms of the AP series 2,5,8,...,
is equal to the sum of the first n terms of the AP series
57, 59, 61,..., then n equals
(2001, 1M)
(a) 10
(c) 11
(b) an even number
(d) a composite number
Fill in the Blanks
8. Let bi > 1 for i = 1, 2, ... , 101 . Suppose log e b1 , log e b2,
n (4n 2 − 1) c2
(a)
6
n (4n 2 − 1) c2
(c)
3
13. Tr is always
14. Which one of the following is a correct statement ?
i =1
15
then a10 is equal to
1
n (n + 1) (3n 2 − n + 1)
12
1
(b)
n (n + 1) (3n 2 + n + 2)
12
1
(c) n (2n 2 − n + 1)
2
1
(d) (2n3 − 2n + 3)
3
(a)
(a) an odd number
(c) a prime number
30
T=
12. The sum V1 + V 2 + ... + V n is
arithmetic progression with integer entries is added to
the product of any four consecutive terms of it. Prove
that resulting sum is the square of an integer.(2000, 4M)
(b) 12
(d) 13
19. The real numbers x1 , x2, x3 satisfying the equation
Objective Question II
(One or more than one correct option)
20. The interior angles of a polygon are in arithmetic
4n
11. If S n = ∑
k( k + 1 )
(−1) 2 k2. Then, S
n
x3 − x2 + βx + γ = 0 are in AP. Find the intervals in
(1996, 3M)
which β and γ lie.
can take value(s)
(2013 Adv.)
2
(a) 1056
(c) 1120
(b) 1088
(d) 1332
Passage Based Problems
Read the following passage and answer the questions.
Passage
Let V r denotes the sum of the first r terms of an
arithmetic progression (AP) whose first term is r and
the common difference is (2r − 1). Let Tr = V r + 1 − V r and
(2007, 8M)
Qr = Tr + 1 − Tr for r = 1, 2, . . .
progression. The smallest angle is 120° and the common
difference is 5°. Find the number of sides of the polygon.
(1980, 3M)
Integer & Numerical Answer Type Questions
21. Suppose that all the terms of an arithmetic progression
are natural numbers. If the ratio of the sum of the first
seven terms to the sum of the first eleven terms is 6 : 11
and the seventh term lies in between 130 and 140, then
the common difference of this AP is
(2015 Adv.)
22. A pack contains n cards numbered from 1 to n. Two
consecutive numbered cards are removed from the pack
and the sum of the numbers on the remaining cards is
1224. If the smaller of the numbers on the removed
cards is k, then k − 20 is equal to
(2013 Adv.)
Sequences and Series 53
23. Let a1 , a 2, a3 , K , a100 be an arithmetic progression with
p
a1 = 3 and S p =
∑
24. Let a1 , a 2, a3 , ... , a11 be real numbers satisfying a1 = 15,
27 − 2a 2 > 0 and a k = 2a k − 1 − a k − 2 for k = 3, 4, ... , 11.
2
a 2 + a 22 + K + a11
If 1
= 90, then the value of
11
a1 + a 2 + K + a11
(2010)
is……
11
ai , 1 ≤ p ≤ 100. For any integer n with
i =1
1 ≤ n ≤ 20, let m = 5n. If
Sm
does not depend on n, then a 2
Sn
is equal to ……
(2011)
Topic 3 Geometric Progression (GP)
Objective Questions I (Only one correct option)
1. Let a , b and c be in GP with common ratio r, where a ≠ 0
1
and 0 < r ≤ . If 3a, 7b and 15care the first three terms of
2
an AP, then the 4th term of this AP is
(2019 Main, 10 April II)
(a) 5a
2
(b) a
3
7
(d) a
3
(c) a
2. If three distinct numbers a , b and c are in GP and the
equations ax + 2bx + c = 0 and dx + 2ex + f = 0 have a
common root, then which one of the following
(2019 Main, 8 April II)
statements is correct?
2
2
(a) d , e and f are in GP
(c) d , e and f are in AP
d e
f
, and are in AP
a b
c
d e
f
(d) , and are in GP
a b
c
(b)
3. The product of three consecutive terms of a GP is 512. If
4 is added to each of the first and the second of these
terms, the three terms now form an AP. Then, the sum
of the original three terms of the given GP is
(2019 Main, 12 Jan I)
(a) 36
(b) 28
(c) 32
4. Let a1 , a 2, .... , a10 be a GP. If
a9
equals
a5
(a) 53
(d) 24
(c) 4(52 )
(d) 54
5. Let a , b and c be the 7th, 11th and 13th
terms
respectively of a non-constant AP. If these are also the
a
three consecutive terms of a GP, then is equal to
c
(2019 Main, 9 Jan II)
(a) 2
(b)
7
13
(c) 4
(d)
1
2
6. If a , b and c be three distinct real numbers in GP and
a + b + c = xb, then x cannot be
(a) 4
(b) 2
(c) −2
(2019 Main, 9 Jan I)
(d) −3
7. If the 2nd, 5th and 9th terms of a non-constant AP are
in GP, then the common ratio of this GP is
8
(a)
5
4
(b)
3
(c) 1
α 2 + β 2 and α 3 + β3 are in GP, then
(a) ∆ ≠ 0
(b) b∆ = 0
(c) c∆ = 0
(2005, 1M)
(d) bc ≠ 0
9. Let a , b, c be in an AP and a 2, b2, c2 be in GP. If a < b < c
3
and a + b + c = , then the value of a is
2
(a)
1
2 2
(b)
1
2 3
(c)
1
1
−
2
3
(2002, 1M)
(d)
(2016 Main)
7
(d)
4
1
1
−
2
2
10. Let α , β be the roots of x2 − x + p = 0 and γ, δ be the roots
of x2 − 4x + q = 0. If α , β , γ, δ are in GP, then the integer
(2001, 1M)
values of p and q respectively are
(a) − 2, − 32
(b) − 2,3
(c) − 6, 3
(d) − 6, − 32
11. If a , b, c, d and p are distinct real numbers such that
(a 2 + b2 + c2) p2 − 2 (ab + bc + cd ) p
+ (b2 + c2 + d 2) ≤ 0, then a , b, c, d
(a) are in AP
(c) are in HP
(b) are in GP
(d) satisfy ab = cd
(1987, 2M)
12. If a , b, c are in GP, then the equations ax2 + 2bx + c = 0
and dx2 + 2ex + f = 0 have a common root, if
in
a3
= 25, then
a1
(2019 Main, 11 Jan I)
(b) 2(52 )
8. Let f (x) = ax2 + bx + c, a ≠ 0 and ∆ = b2 − 4ac. If α + β,
(a) AP
(c) HP
d e f
, , are
a b c
(1985, 2M)
(b) GP
(d) None of these
13. The third term of a geometric progression is 4. The
product of the first five terms is
(a) 43
(c) 44
(1982, 2M)
(b) 45
(d) None of these
Analytical & Descriptive Questions
14. Find three numbers a , b, c between 2 and 18 such that
(i) their sum is 25. (ii) the numbers 2, a , b are
consecutive terms of an AP. (iii) the numbers b, c, 18 are
(1983, 2M)
consecutive terms of a GP.
15. Does there exist a geometric progression containing
27,8 and 12 as three of its term? If it exists, then how
many such progressions are possible?
(1982, 2M)
16. If the mth, nth and pth terms of an AP and GP are equal
and are x, y, z, then prove that xy − z ⋅ yz − x ⋅ z x − y = 1.
(1979, 3M)
54 Sequences and Series
Topic 4 Sum of n Terms & Infinite Terms of a GP
Objective Questions I (Only one correct option)
1. If|x|< 1,| y|< 1 and x ≠ y, then the sum to infinity of the
following
series
(x + y) + (x2 + xy + y2)
(2020 Main, 2 Sep I)
+ (x3 + x2y + xy2 + y3 ) + … is
x + y + xy
(1 + x) (1 + y)
x + y + xy
(c)
(1 − x) (1 − y)
x + y − xy
(1 − x) (1 − y)
x + y − xy
(d)
(1 + x) (1 + y)
(a)
20
2. The sum
(2019 Main, 8 April II)
k =1
(a) 2 −
(c) 2 −
(b) 1 −
19
2
3
(d) 2 −
217
11
∞
(a)
n
2
n
(b) 202
(d) 299
terms is 3 and the sum of the cubes of its terms is
27
.
19
Then, the common ratio of this series is
(2019 Main, 11 Jan I)
4
9
(b)
2
3
(c)
2
9
(d)
1
3
2
3
(2014 Main)
6. If (10)9 + 2(11)1 (10)8 + 3(11)2(10)7 + ... + 10(11)9 = k(10)9,
then k is equal to
(a)
121
10
(b)
441
100
(b) bn = α n + β n for all n ≥ 1
(c) a1 + a2 + a3 + K + an = an +
∞
10
a
(d) ∑ nn =
89
n = 1 10
2
− 1for all n ≥ 1
(2014 Main)
(c) 100
length of a side of S n equals the length of a diagonal of
S n + 1. If the length of a side of S1 is 10 cm, then for which
of the following values of n is the area of S n less than
(1999, 3M)
1 sq cm?
(a) 7
(b) 8
(c) 9
2
middle term in this GP is doubled, then new numbers
are in AP. Then, the common ratio of the GP is
(b) 3 +
(d) 2 +
(2019 Adv.)
n =1
(d) 10
Analytical & Descriptive Questions
5. Three positive numbers form an increasing GP. If the
(a) 2 + 3
(c) 2 − 3
8
b
∑ 10nn = 89
12. Let S1 , S 2, ... be squares such that for each n ≥ 1 the
4. The sum of an infinite geometric series with positive
(a)
all positive integers n, define
α n − βn
, n ≥ 1,
an =
α −β
Then which of the following options is/are correct?
220
 q + 1
 q + 1  q + 1
Tn = 1 + 
 , where q is a
 +K+
 +
 2 
 2   2 
real number and q ≠ 1. If
101
C1 + 101C 2 ⋅ S1 + K + 101C101 ⋅ S100 = αT100, then α is
(2019 Main, 11 Jan II)
equal to
(a) 2100
(c) 200
(1988, 2M)
(c) n + 2 − n − 1 (d) 2n + 1
b1 = 1 and bn = a n − 1 + a n + 1, n ≥ 2
220
21
3. Let S n = 1 + q + q + K + q and
2
is equal to
(a) 2n − n − 1 (b) 1 − 2−n
11. Let α and β be the roots of x2 − x − 1 = 0, with α > β. For
1
11
3 7 15
+ +
+ ...
4 8 16
Objective Question II
(One or more than one correct option)
(b)
∑ k 2k is equal to
1
2
10. Sum of the first n terms of the series +
(d) 110
3
n
3
3
3
3
13. Let An =   −   +   + ... + (− 1)n − 1   ,
   
 
 
4
4
4
4
Bn = 1 − An. Find a least odd natural number n0 , so that
(2006, 6M)
Bn > An , ∀ n ≥ n0.
14. If S1 , S 2, S3 , ... , S n are the sums of infinite geometric
series, whose first terms are 1, 2, 3,..., n and whose
1 1 1
1
common ratios are , , ,... ,
respectively, then
2 3 4
n+1
find the values of S12 + S 22 + S32 + ... + S 22n − 1. (1991, 4M)
7. The sum of first 20 terms of the sequence
15. The sum of the squares of three distinct real numbers,
(2013 Main)
which are in GP, is S 2. If their sum is a S, then show that
1 
a 2 ∈  , 1 ∪ (1, 3)
3 
(1986, 5M)
0.7, 0.77, 0.777,… , is
7
(a)
(179 − 10− 20 )
81
7
(c)
(179 + 10− 20 )
81
7
(b)
(99 − 10− 20 )
9
7
(d)
(99 + 10− 20 )
9
8. An infinite GP has first term x and sum 5, then x
belongs to
(a) x < − 10
(2004, 1M)
(b) −10 < x < 0 (c) 0 < x < 10
(d) x > 10
Integer & Numerical Answer Type Question
16. Let S k , where k = 1, 2, , K , 100, denotes the sum of the
infinite geometric series whose first term is
9. Consider an infinite geometric series with first term a
and common ratio r. If its sum is 4 and the second term
is 3 /4, then
(2000, 2M)
(a) a = 4/7, r = 3/7
(c) a = 3/2, r = 1/2
(b) a = 2, r = 3 /8
(d) a = 3, r = 1/4
the
common
2
100
+
100 !
ratio
is
1
.
k
Then,
the
k −1
and
k!
value
of
100
∑|(k2 − 3k + 1) Sk |is ……
k =1
(2010)
Sequences and Series 55
Topic 5 Harmonic Progression (HP)
Objective Questions I (Only one correct option)
5. Suppose four distinct positive numbers a1 , a 2, a3 , a 4
are in GP. Let b1 = a1 , b2 = b1 + a 2, b3 = b2 + a3 and
b4 = b3 + a 4.
Statement I The numbers b1 , b2, b3 , b4 are neither in
AP nor in GP.
1. If a1 , a 2, a3 ,… are in a harmonic progression with a1 = 5
and a 20 = 25. Then, the least positive integer n for which
(2012)
a n < 0, is
(a) 22
(c) 24
(b) 23
(d) 25
Statement II The numbers b1 , b2, b3 , b4 are in HP.
(2008, 3M)
2. If the positive numbers a , b, c,d are in AP. Then,
abc, abd , acd , bcd are
(2001, 1M)
(a) not in AP / GP / HP
(c) in GP
(b) in AP
(d) in HP
Fill in the Blank
6. If cos (x − y), cos x and cos (x + y)’ are in HP. Then
 y
cos x ⋅ sec   = K .
 2
3. Let a1 , a 2, ..., a10 be in AP and h1 , h2, equal to ..... , h10 be
in HP. If a1 = h1 = 2 and a10 = h10 = 3, then a 4h7 is
(1999, 2M)
(a) 2
(b) 3
(c) 5
(d) 6
4. If x > 1, y > 1, z > 1 are in GP, then
1
are in
1 + ln z
1
1
,
,
1 + ln x 1 + ln y
(1997C, 2M)
Analytical & Descriptive Questions
7. If a , b, c are in AP, a 2, b2, c2 are in HP, then prove that
either a = b = c or a , b, −
(1998, 2M)
(a) AP
(c) GP
(b) HP
(d) None of these
(a) Statement I is true, Statement II is also true;
Statement II is the correct explanation of Statement I
(b) Statement I is true, Statement II is also true;
Statement II is not the correct explanation of
Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
(2003, 4M)
8. Let a and b be positive real numbers. If a , A1 , A2, b are
in arithmetic progression, a , G1 , G2, b are in geometric
progression and a , H 1 , H 2, B are in harmonic
progression, then show that
G1G2
A + A2 (2a + b)(a + 2b)
= 1
=
9ab
H 1H 2 H 1 + H 2
(2002, 5M)
Assertion and Reason
For the following question, choose the correct
answer from the codes (a), (b), (c) and (d) defined as
follows:
c
form a GP.
2
9.
(i) The value of x + y + z is 15. If a , x, y, z , b are in AP
1 1 1 5
while the value of + + is . If a , x, y, z , b are in
x y z 3
HP, then find a and b.
(ii) If x, y, z are in HP, then show that
log (x + z ) + log (x + z − 2 y) = 2 log (x − z ). (1978, 3M)
Topic 6 Relation between AM, GM, HM and Some Special Series
Objective Questions I (Only one correct option)
1 +2
1 +2 +3
+
+ ...
1+2
1+2+3
13 + 23 + 33 + K + 153 1
+
− (1 + 2 + 3 + K + 15) is
1 + 2 + 3 + K + 15
2
(2019 Main, 10 April II)
equal to
1. The sum of series 1 +
(a) 620
(b) 660
3
3
3
(c) 1240
3
3
(d) 1860
3 × 13 5 × (13 + 23 )
+
12
12 + 22
3
3
3
7 × (1 + 2 + 3 )
+ .......... + upto 10th term, is
+
12 + 22 + 32
(2019 Main, 10 April I)
2. The sum of series
(a) 680
(b) 600
(c) 660
(d) 620
3. The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 +... upto
11th term is
(a) 915
(2019 Main, 9 April II)
(b) 946
(c) 916
(d) 945
4. If the sum of the first 15 terms of the series
3
3
3
3
 3
 1
 1
 3
3
  + 1  +  2  + 3 +  3  + ...
 4
 4
 2
 4
is equal to 225 k, then k is equal to
(2019 Main, 12 Jan II)
(a) 108
(b) 27
(c) 54
(d) 9
5
1 + 2 + 3 + ... + k
2
5. Let S k =
. If S12 + S 22 + ... + S10
A,
=
k
12
(2019 Main, 12 Jan I)
then A is equal to
(a) 156
(b) 301
(c) 283
(d) 303
6. Let x, y be positive real numbers and m, n positive
integers. The maximum value of the expression
xm yn
is
2m
(1 + x ) (1 + y2n )
(2019 Main, 11 Jan II)
(a)
1
2
(b) 1
(c)
1
4
(d)
m+ n
6mn
56 Sequences and Series
18. If a , b and c are distinct positive numbers, then the
7. The sum of the following series
expression (b + c − a ) (c + a − b) (a + b − c) − abc is
9 (1 + 2 + 3 ) 12 (1 + 2 + 3 + 4 )
+
7
9
2
2
2
15 (1 + 2 + ... + 5 )
+
+ ... up to 15 terms is
11
2
1+6+
2
2
2
2
2
2
(a) positive
(c) non-positive
(2019 Main, 9 Jan II)
(a) 7510
(b) 7820
(c) 7830
(d) 7520
∑ a 4k + 1 = 416
k=0
2
and a 9 + a 43 = 66. If a12 + a 22 + … + a17
= 140 m, then m
is equal to
(2018 Main)
(a) 66
(b) 68
(c) 34
(d) 33
9. Let A be the sum of the first 20 terms and B be the sum
of the first 40 terms of the series
12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + …
If B − 2 A = 100λ, then λ is equal to
(a) 232
(b) 248
(c) 464
(2018 Main)
(d) 496
10. If the sum of the first ten terms of the series
2
2
2
2
16
 4
 3
 2
 1
2
m, then
1  + 2  + 3  + 4 + 4  + K, is
 5
 5
 5
 5
5
(2016 Main)
m is equal to
(a) 102
(b) 101
(c) 100
(d) 99
11. If m is the AM of two distinct real numbers l and
n (l, n > 1) and G1 , G2 and G3 are three geometric means
between l and n, then G14 + 2G24 + G34 equals
(2015)
(a) 4l2mn
(b) 4lm2n
(c) lmn 2
12. The
sum of first 9 terms
13 13 + 23 13 + 23 + 33
+
+
+ ... is
1
1+3
1+3+5
(a) 71
(b) 96
13. If α ∈ 0,

π
 , then
2
(c) 142
x2 + x +
(d) l2m2n 2
of
the
(2015)
(d) 192
tan α
2
x2 + x
is always greater
than or equal to
(a) 2 tan α
(b) 1
series
(2003, 2M)
(c) 2
(d) sec2 α
14. If a1 , a 2,... , a n are positive real numbers whose product
is a fixed number c, then the minimum value of
(2002, 1M)
a1 + a 2 + ... + a n − 1 + 2a n is
(b) (n + 1)c1/ n
(d) (n + 1) (2c)1/ n
(a) n (2c)1/ n
(c) 2nc1/ n
a + b + c + d = 2 , then M = (a + b) (c + d ) satisfies the
(2000, 2M)
relation
(a) 0 < M ≤ 1
(b) 1 ≤ M ≤ 2
(c) 2 ≤ M ≤ 3
(d) 3 ≤ M ≤ 4
16. The harmonic mean of the roots of the equation
(5 + 2 ) x2 − (4 + 5 ) x + 8 + 2 5 = 0 is
(a) 2
(c) 6
(1999, 2M)
(b) 4
(d) 8
17. The product of n positive numbers is unity, then their
sum is
(a) a positive integer
1
(c) equal to n +
n
integer, then


(a) n ∑ xi2 <  ∑ xi 


i =1 
i =1
n
n
(c) n ∑
i =1
xi2
(1991, 2M)
(b) divisible by n
(d) never less than n
(1982, 1M)
2
 n 
≥ n  ∑ xi 


i =1 


(b) n ∑ xi2 ≥  ∑ xi 


i =1 
i =1
n
n
2
2
(d) None of these
Passage Based Problems
Passage
Let A1 , G1 , H 1 denote the arithmetic, geometric and
harmonic means, respectively, of two distinct positive
numbers. For n ≥ 2, let An − 1 and H n − 1 has arithmetic,
geometric and harmonic means as An , Gn , H n ,
respectively.
(2007, 8M)
20. Which one of the following statements is correct?
(a) G1 > G2 > G3 > ...
(b) G1 < G2 < G3 < ...
(c) G1 = G2 = G3 = ...
(d) G1 < G3 < G5 < ... and G2 > G4 > G6 >...
21. Which of the following statements is correct?
(a) A1 > A2 > A3 >...
(b) A1 < A2 < A3 <...
(c) A1 > A3 > A5 >... and A2 < A4 < A6 <...
(d) A1 < A3 < A5 <... and A2 > A4 > A6 >...
22. Which of the following statements is correct ?
(a) H1 > H 2 > H3 >...
(b) H1 < H 2 < H3 <...
(c) H1 > H3 > H5 >... and H 2 < H 4 < H 6 <...
(d) H1 < H3 < H5 <... and H 2 > H 4 > H 6 >...
Objective Questions II
(One or more than one correct option)
23. For a positive integer n let
a (n ) = 1 +
15. If a , b, c are positive real numbers such that
(1991, 2M)
19. If x1 , x2,... , xn are any real numbers and n is any positive
n
12
8. Let a1 , a 2, a3 , …, a 49 be in AP such that
(b) negative
(d) non-negative
1 1 1
1
+ + + ... + n
, then
2 3 4
(2 ) − 1
(a) a (100) ≤ 100
(c) a (200)≤ 100
(1999, 3M)
(b) a (100) > 100
(d) a (200) > 100
24. If the first and the (2n − 1)th term of an AP, GP and HP
are equal and their nth terms are a , b and c
(1988, 2M)
respectively, then
(a) a = b = c
(c) a + c = b
(b) a ≥ b ≥ c
(d) ac − b2 = 0
Fill in the Blanks
25. If x be is the arithmetic mean and y, z be two geometric
means between any two positive numbers, then
y3 + z3
= ...
xyz
(1997C, 2M)
Sequences and Series 57
26. If the harmonic mean and geometric mean of two
32. If a > 0, b > 0 and c > 0, then prove that
positive numbers are in the ratio 4 : 5. Then, the two
numbers are in the ratio … .
(1992, 2M)
True/False
(a + b + c)
(1984, 2M)
Integer & Numerical Answer Type Questions
27. If x and y are positive real numbers and m, n are any
33. Let
m be the minimum possible value of
log3 (3y1 + 3y2 + 3y3 ), where y1 , y2, y3 are real numbers for
which y1 + y2 + y3 = 9. Let M be the maximum possible
value of (log3 x1 + log3 x2 + log3 x3 ), where x1 , x2, x3 are
positive real numbers for which x1 + x2 + x3 = 9. Then the
(2020 Adv.)
value of log 2(m3 ) + log3 (M 2) is ……… .
n m
positive
1 1 1
+ + ≥9
a b c
integers,
1
x y
> .
(1 + x2n )(1 + y2m ) 4
then
(1989, 1M)
28. For 0 < a < x, the minimum value of function
log a x + log x a is 2.
34. Let a1 , a 2, a3 , .... be a sequence of positive integers in
arithmetic progression with common difference 2. Also,
let b1 , b2, b3 , .... be a sequence of positive integers in
geometric progression with common ratio 2. If a1 = b1 = c,
then the number of all possible values of c, for which the
equality
Analytical & Descriptive Questions
29. If a , b, c are positive real numbers, then prove that
{(1 + a ) (1 + b) (1 + c)}7 > 77 a 4b4c4
(2004, 4M)
2(a1 + a 2 + K + a n ) = b1 + b2 + K + bn
holds for some positive integer n, is ……
30. Let a1 , a 2,.. be positive real numbers in geometric
(2020 Adv.)
progression. For each n, if An , Gn , H n are respectively,
the arithmetic mean, geometric mean and harmonic
mean of a1 , a 2, .... , a n. Then, find an expression for the
geometric mean of G1 , G2, ... , Gn in terms of
(2001, 5M)
A1 , A2, ... , An , H 1 , H 2, ... , H n.
35. If m arithmetic means (AMs) and three geometric means
31. If p is the first of the n arithmetic means between two
If a,b,c are in geometric progression and the arithmetic
a 2 + a − 14
mean of a,b,c is b + 2, then the value of
is
a+1
numbers and q be the first on n harmonic means
between the same numbers. Then, show that q does not
2
 n + 1
lie between p and 
(1991, 4M)
 p.
 n − 1
(GMs) are inserted between 3 and 243 such that 4th AM
is equal to 2nd GM, then m is equal to
[2020 Main, 3 Sep II]
36. Let a,b,c be positive integers such that b /a is an integer.
(2014 Adv.)
37. The minimum value of the sum of real numbers
a − 5 , a − 4 , 3a − 3 , 1, a 8 and a10 with a > 0 is ……
(2011)
Answers
Topic 1
1. (b)
6. False
Topic 4
2. (d)
7. (157.00)
3. (b)
8. (6)
4. (c)
(a)
2.
(a)
6.
(c)
10.
(d)
14.
2
n (n + 1 ) 
16. 
 17.
2


1

19. β ∈ – ∞, and γ

3 
22. (5)
23.
(c)
(c)
(c)
(b)
3.
7.
11.
15.
(b)
4. (c)
(c)
8. (b)
(a, d)
12. (b)
A = – 3 , B = 77
3050
 1 
∈ – , ∞ 20. (9)
 27 
(9)
24. (0)
(c)
(c)
(d)
(b)
3. (a)
4. (b)
7. (c)
8. (c)
2. (b)
6. (b)
10. (a)
14. (a = 5 ) (b = 8 ) (c = 12 )
3. (b)
7. (b)
11. (b)
15. Yes, infinite
13. (7)
10. (c)
11. (b, c, d)
1
14. (2n )(2n + 1 )( 4n + 1 ) – 1
6
12. (b,c,d)
16. (4)
Topic 5
1. (d)
5. (c)
2. (d)
6. ± 2
3. (d)
4. (b)
9. (i) a = 1, b = 9
Topic 6
21. (9)
Topic 3
1.
5.
9.
13.
2. (a)
6. (c)
9. (d)
Topic 2
1.
5.
9.
13.
1. (b)
5. (d)
4. (d)
8. (c)
12. (a)
1.
5.
9.
13.
17.
21.
25.
33.
(a)
(d)
(b)
(a)
(d)
(a)
2
(8)
36. (4)
2.
6.
10.
14.
18.
22.
26.
34.
(c)
(c)
(b)
(a)
(b)
(b)
4:1
C =12 for n
37. (8)
3.
7.
11.
15.
19.
23.
27.
=3
(b)
(b)
(b)
(a)
(b)
(a, d)
False
4.
8.
12.
16.
20.
24.
28.
35.
(b)
(c)
(b)
(b)
(c)
(a, b, d)
False
(39)
Hints & Solutions
Topic 1 Arithmetic Progression (AP)
1.
Key Idea Use nth term of an AP i.e. an = a + ( n − 1) d , simplify the
given equation and use result.
Given AP is a1 , a 2, a3 , … , a n
Let the above AP has common difference ‘d’, then
a1 + a 4 + a7 + … + a16
= a1 + (a1 + 3d ) + (a1 + 6d ) + … + (a1 + 15d )
= 6a1 + (3 + 6 + 9 + 12 + 15)d
(given)
∴
6a1 + 45d = 114
…(i)
⇒
2a1 + 15d = 38
Now, a1 + a 6 + a11 + a16
= a1 + (a1 + 5d ) + (a1 + 10d ) + (a1 + 15d )
= 4a1 + 30d = 2(2a1 + 15d )
[from Eq. (i)]
= 2 × 38 = 76
2. Let tn be the nth term of given AP. Then, we have t19 = 0
⇒ a + (19 − 1)d = 0
⇒
a + 18d = 0
t49 a + 48d
Now,
=
t29 a + 28d
− 18d + 48d
=
− 18d + 28d
30d
=
= 3 :1
10d
[Qtn = a + (n − 1)d ]
…(i)
[using Eq. (i)]
225a 2 + 9b2 + 25c2 − 75ac − 45ab − 15bc = 0
⇒ (15a )2 + (3b)2 + (5c)2 − (15a )(5c) − (15a )(3b)
− (3b)(5c) = 0
1
2
2
2
⇒
[(15a − 3b) + (3b − 5c) + (5c − 15a ) ] = 0
2
⇒ 15a = 3b, 3b = 5c and 5c = 15a
a b c
= = =λ
∴ 15a = 3b = 5c ⇒
1 5 3
(say)
⇒
a = λ , b = 5λ , c = 3λ
∴ b, c, a are in AP.
and
1
n
1
Tn = a + (n − 1) d =
m
Tm = a + (m − 1) d =
On subtracting Eq. (ii) from Eq. (i), we get
1 1 m−n
(m − n ) d = −
=
n m
mn
1
d=
⇒
mn
Again, Tmn = a + (mn − 1) d
= a + (mn − n + n − 1) d
5. Since, a1 , a 2, ... , a n are in an AP.
∴ (a 2 − a1 ) = (a3 − a 2) = ... = (a n − a n − 1 ) = d
1
1
1
Thus,
+
+ ... +
a1 + a 2
a 2 + a3
a n −1 + a n
 a 2 − a1   a3 − a 2 
+
 + ... +
=
d
d

 

=
1
1 (a n − a1 )
( a n − a1 ) =
=
d
d a n + a1
 a n − a n −1 




d


(n − 1)
a n + a1
6. Since, n1 , n2,... , n p are p positive integers, whose sum is
even and we know that, sum of any two odd integers is
even.
∴Number of odd integers must be even.
Hence, it is a false statement.
7. Given that, AP (a ; d ) denote the set of all the terms of an
3. We have,
4. Let
= a + (n − 1) d + (mn − n ) d
(m − 1)
1
1
= Tn + n (m − 1)
=
+
=1
mn m
m
…(i)
…(ii)
infinite arithmetic progression with first term ‘a’ and
common difference d > 0.
Now, let mth term of first progression
…(i)
AP(1 ; 3) = 1 + (m − 1)3 = 3m − 2
and nth term of progression
…(ii)
AP (2; 5) = 2 + (n − 1)5 = 5n − 3
and rth term of third progression AP (3; 7)
…(iii)
= 3 + (r − 1)7 = 7r − 4
are equal. Then
3m − 2 = 5n − 3 = 7r − 4
Now, for AP ( 1 ; 3 ) ∩ AP (2 ; 5 ) ∩ AP (3; 7), the common
5n − 1
terms of first and second progressions, m =
3
⇒ n = 2, 5, 11, … and the common terms of second and
5n + 1
the third progressions, r =
⇒ n = 4, 11,…
7
Now, the first common term of first, second and third
progressions (when n = 11), so a = 2 + (11 − 1)5 = 52
and
d = LCM(3, 5, 7) = 105
So, AP (1 ; 3) ∩ AP (2 ; 5) ∩ AP (3 ; 7) = AP (52 ; 105)
So,
a = 52 and d = 105 ⇒ a + d = 157.00
8. Let the sides are a − d , a and a + d. Then,
a (a − d ) = 48
and a − 2ad + d 2 + a 2 = a 2 + 2ad + d 2
⇒
a 2 = 4ad
⇒
a = 4d
Thus,
a = 8, d = 2
Hence,
a − d =6
2
Sequences and Series 59
= − 2 − 2 − 2 − 2 K 33 times
Topic 2 Sum of n Terms of an AP
= (−2) × 33 = − 66
1   1
2 
 1  1
+ − −
+ − −
+K+
∴ −
 3   3 100   3 100 
1. Let the common difference of given AP is ‘d’.
Since,
a1 + a7 + a16 = 40
∴ a1 + a1 + 6d + a1 + 15d = 40 [Q a n = a1 + (n − 1) d ]
…(i)
⇒
3a1 + 21d = 40
Now, sum of first 15 terms is given by
15
S15 =
[2a1 + (15 − 1) d ]
2
15
=
[2a1 + 14d ] = 15 [a1 + 7d ]
2
From Eq. (i), we have
40
a1 + 7d =
3
40
So,
S15 = 15 ×
3
= 5 × 40 = 200
= (− 67) + (− 66) = − 133.
Alternate Solution
Q [− x] = − [x] − 1, if x ∉Integer,
1 
2
n − 1


and [x] + x +
+ x+
+K+ x+
= [nx],
n  
n 
n 


n ∈N.
So given series
1 
 1  1
−
+ − −
+
 3   3 100 
∴
S 4 = 2[2a + 3d ] = 16
(given)
n


Q S n = [2a + (n − 1)d ]


2
… (i)
⇒
2a + 3d = 8
and
[given]
S 6 = 3[2a + 5d ] = − 48
… (ii)
⇒
2a + 5d = − 16
On subtracting Eq. (i) from Eq. (ii), we get
2d = − 24
⇒
d = − 12
So,
[put d = −12 in Eq. (i)]
2a = 44
Now,
S10 = 5[2a + 9d ]
= 5[44 + 9(− 12)] = 5[44 − 108]
= 5 × (− 64) = − 320
2  
99 
 1

 1
− 1
− 1 + K +  −
+
+ −
+
 3 100 

 3 100  
1   1
2 
 1  1
 1 99 
−
+ − −
+ − −
+ K ... + − −
 3   3 100   3 100 
 3 100 
[where, [x] denotes the greatest integer ≤ x]
Now,
1   1
2 
 1  1
 1 66 
, …+ − −
− , − −
, − −
 3   3 100   3 100 
 3 100 
all the term have value − 1
 1 67   1 68 
 1 99 
and − −
, − −
, …, − −
all the term
 3 100   3 100 
 3 100 
have value − 2.
1   1
2 
 1  1
 1 66 
So, −
+ − −
+ − −
+ ... + − −
 3   3 100   3 100 
 3 100 
= − 1 − 1 − 1 − 1 K 67 times.
= (− 1) × 67 = − 67
 1 67 
and − −
+
 3 100 
1

× 100 = − 100 − 33 = − 133.
3

4. Let first three terms of an AP as a − d, a, a + d.
= (− 1) × 100 −
3a = 33 ⇒ a = 11
[given sum of three terms = 33
and product of terms = 1155]
[given]
⇒ (11 − d )11(11 + d ) = 1155
⇒
112 − d 2 = 105
⇒
d 2 = 121 − 105 = 16
⇒
d = ±4
So the first three terms of the AP are either 7, 11, 15 or
15, 11, 7.
So, the 11th term is either 7 + (10 × 4) = 47
or
15 + (10 × (−4)) = − 25.
So,
5.
3. Given series is
 1 68 
− −
+K+
 3 100 
 1 99 
− −
 3 100 
2 
 1
 − 1 99 
− −
−
+…K+
 3 100 
 3
100 
1 

 1 
  1
− 1
− 1 +  −
+
= −

 3 
  3 100 
2. Given S n denote the sum of the first n terms of an AP.
Let first term and common difference of the AP be ‘a’
and ‘d’, respectively.
 1 99 
− −
 3 100 
Key Idea Use the formula of sum of first n terms of AP, i.e
Sn =
n
[2 a + ( n − 1) d ]
2
Given AP, is
a1 , a 2, a3 ,… having sum of first n-terms
n
= [2a1 + (n − 1)d ]
2
[where, d is the common difference of AP]
n (n − 7)
(given)
= 50n +
A
2
1
n−7
⇒
[2a1 + (n − 1)d ] = 50 +
A
2
2
1
7  n

[2a1 + nd − d ] = 50 − A + A
⇒

2
2  2
d  nd 
7  n

⇒
= 50 − A + A
 a1 −  +


2
2
2  2
On comparing corresponding term, we get
d
7
d = A and a1 − = 50 − A
2
2
60 Sequences and Series
A
7
= 50 − A
2
2
⇒
a1 = 50 − 3 A
So
a50 = a1 + 49d
= (50 − 3 A ) + 49 A
= 50 + 46 A
Therefore, (d , a50 ) = ( A , 50 + 46 A )
⇒
a1 −
[Qd = A]
Also, a1 , a 2, ... , a101 are in AP.
[Q d = A]
⇒
⇒
Now,
12
[16 + 93] = 654
2
n


QS n = ( a + l )


2
Similarly, the two digit number which leaves remainder
5 when divided by 7 is of the form N = 7k + 5
For k = 1, N = 12
k = 2, N = 19
M
k = 13, N = 96
∴13 such numbers are possible and these numbers also
forms an AP.
Now,
Total sum = S + S′ = 654 + 702 = 1356
s = a1 + a 2 + K + a51
51
=
(2a1 + 50 D )
2
…(iv)
t = a1 (251 − 1)
[Q a1 = b1 ]
…(v)
t = 251 a1 − a1 < 251 a1
51
[from Eq. (ii)]
and
s=
[a1 + (a1 + 50 D )]
2
51
=
[a1 + 250 a1 ]
2
51
51 50
=
a1 +
2 a1
2
2
…(vi)
∴
s > 251 a1
From Eqs. (v) and (vi), we get s > t
Also,
a101 = a1 + 100 D and b101 = 2100 b1
 250 a1 − a1 
∴
a101 = a1 + 100 
 and b101 = 2100 a1
50


∴ 12 such numbers are possible and these numbers
forms an AP.
13
[12 + 96] = 702
S′ =
2
n

QS n = ( a +

2
…(iii)
∴
or
k = 2, N = 16
k = 3, N = 23
M
M
k = 13, N = 93
S=
[Q a1 = b1 ] …(ii)
t = b1 + b2 + K + b51
(251 − 1)
t = b1
2 −1
and
when divided by 7 is of the form N = 7k + 2 [by Division
Algorithm]
Now,
a1 + 50 D = 250 b1
a1 + 50 D = 250 a1
⇒
6. Clearly, the two digit number which leaves remainder 2
For,
a1 = b1 and a51 = b51
Given,
⇒
⇒
⇒
a101 = a1 + 251 a1 − 2a1 = 251 a1 − a1
a101 < 251 a1 and b101 > 251 a1
b101 > a101
9. Let S n = cn 2
S n − 1 = c (n − 1)2 = cn 2 + c − 2cn

l )

∴
Tn2
…(i)
= 15[2a1 + 29d ]
(where d is the common difference)
n


Q S n = [2a + (n − 1)d ]


2
T = a1 + a3 + … + a 29
15
=
[2a1 + 14 × 2d )]
2
(Q common difference is 2d)
…(ii)
⇒
2T = 15[2a1 + 28d ]
From Eqs. (i) and (ii), we get
[QS − 2T = 75]
S − 2T = 15d = 75
⇒
d =5
Now,
a10 = a5 + 5d
= 27 + 25 = 52
2
2 2
4c2 ⋅ n (n + 1) (2n + 1)
+ nc2 − 2c2n (n + 1)
6
2 c2n (n + 1) (2n + 1) + 3nc2 − 6c2n (n + 1)
3
nc2(4n 2 + 6n + 2 + 3 − 6n − 6) nc2(4n 2 − 1)
=
=
3
3
=
10. According to given condition,
S 2n = S′n
2n
n
[2 × 2 + (2n − 1) × 3] = [2 × 57 + (n − 1) × 2]
⇒
2
2
1
⇒
(4 + 6n − 3) = (114 + 2n − 2)
2
⇒
6n + 1 = 57 + n − 1 ⇒ 5n = 55
∴
n = 11
11.
PLAN
Convert it into differences and use sum ofn terms of an AP,
n
S n = [2a + ( n − 1 )d ]
2
i.e.
8. If log b1 , log b2, ... , log b101 are in AP, with common
difference log e 2 , then b1 , b2, ... , b101 are in GP, with
common ratio 2.
…(i)
∴ b1 = 20 b1 , b2 = 21 b1 , b3 = 22b1,…, b101 = 2100 b1
[Q Tn = S n − S n − 1 ]
= (2 cn − c) = 4c n + c2 − 4c2n
∴ Sum = Σ Tn2 =
7. We have, S = a1 + a 2 + … + a30
and
Tn = 2cn − c
Now,
Sn =
4n
∑
k =1
(−1)
k( k + 1)
2
⋅ k2
Sequences and Series 61
= − (1)2 − 22 + 32 + 42 − 52 − 62 + 72 + 82 + K
∴
p = a − 3d = 5 − 6 = − 1
= (3 − 1 ) + (4 − 2 ) + (7 − 5 ) + (8 − 6 ) + K
q = a − d =5 −2 =3
= 2{(4 + 6 + 12 + K ) + (6 + 14 + 22 + K )}
144424443 144424443
r = a + d =5 + 2 = 7
2
2
2
2
2
2
n terms
2
2
s = a + 3d = 5 + 6 = 11
and
n terms
n
n

=2
{2 × 4 + (n − 1) 8} + {2 × 6 + (n − 1) 8}
 2

2
= 2 [n (4 + 4n − 4) + n (6 + 4n − 4)]
Therefore, A = pq = − 3 and B = rs = 77
16. Here, 12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + ... upto n terms
=
= 2 [4n + 4n + 2n ] = 4n (4n + 1)
2
2
Here, 1056 = 32 × 33, 1088 = 32 × 34,
1056 and 1332 are possible answers.
r
1
12. Here, V r = [ 2r + (r − 1) ( 2r − 1)] = ( 2r3 − r 2 + r )
2
2
1
∴ ΣV r = [ 2 Σr3 − Σr 2 + Σr ]
2
2
1   n (n + 1)  n (n + 1) ( 2n + 1) n (n + 1) 
+
= 2 
 −

2 
2
6
2


13. V r + 1
n (n + 1)
[ 3n (n + 1) − ( 2n + 1) + 3]
12
1
=
n (n + 1) ( 3n 2 + n + 2)
12
1
1
− V r = (r + 1)3 − r3 − [(r + 1)2 − r 2] + = 3r 2 + 2r − 1
2
2
=
∴
which is a composite number.
Tr = 3r 2 + 2r − 1
∴
Qr = Tr+1 − Tr = 3 [ 2r + 1] + 2 [1]
⇒
Qr = 6 r + 5
⇒
(n + 1)
n −1

= n2 
+ 1 = n 2
 2

2
∴ 12 + 2⋅ 22 + 32 + 2 ⋅ 42 + ... upto n terms, when n is odd
=
17. Integers divisible by 2 are {2,4,6,8,10, ...,100}.
Integers divisible by 5 are {5,10,15, ...,100}.
Thus, sum of integers divisible by 2
50
=
(2 + 100) = 50 × 51 = 2550
2
∴ Sum of integers from 1 to 100 divisible by 2 or 5
= 2550 + 1050 − 550
Qr+ 1 = 6(r + 1) + 5
= 2550 + 500 = 3050
Common difference = Qr+ 1 − Qr = 6
18. Let four consecutive terms of the AP are a − 3d , a − d ,
15. Given, p + q = 2, pq = A
and
a + d , a + 3d, which are integers.
Again, required product
r + s = 18, rs = B
and it is given that p, q, r , s are in an AP.
P = (a − 3d )(a − d )(a + d )(a + 3d ) + (2d )4
[by given condition]
Therefore, let p = a − 3d , q = a − d , r = a + d
s = a + 3d
and
We have,
d >0
Now,
2 = p + q = a − 3d + a − d = 2a − 4d
⇒
Again,
= (a 2 − 9d 2)(a 2 − d 2) + 16d 4
p<q<r<s
Since,
a − 2d = 1
= a 4 − 10a 2d 2 + 9d 4 + 16d 4 = (a 2 − 5d 2)2
Now, a − 5d = a 2 − 9d 2 + 4d 2
2
= I ⋅ I + I2
18 = r + s = a + d + a + 3d
9 = a + 2d
On subtracting Eq. (i) from Eq. (ii), we get
8 = 4d ⇒ d = 2
On putting in Eq. (ii), we get a = 5
2
= (a − 3d )(a + 3d ) + (2d )2
…(i)
=I +I =I
2
18 = 2a + 4d
⇒
n 2 (n + 1)
2
Sum of integers divisible by 10
10
=
(10 + 100) = 5 × 110 = 550
2
Tr + 1 = 3 (r + 1)2 + 2 (r + 1) − 1
and
= {12 + 2⋅22 + 32 + 2 ⋅ 42 + ... + 2 (n − 1)2} + n 2
 (n − 1) (n )2 
2
[from Eq. (i)]
=
+ n
2


Sum of integers divisible by 5
20
=
(5 + 100) = 10 × 105 = 1050
2
Tr = 3r 2 + 2r − 1 = (r + 1) ( 3r − 1)
14. Since,
[when n is even] …(i)
When n is odd, 12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 ... + n 2
1120 = 32 × 35, 1332 = 36 × 37
⇒
n (n + 1)2
2
…(ii)
2
[given]
2
=I
[where, I is any integer]
Therefore, P = (I ) = Integer
2
19. Since, x1 , x2, x3 are in an AP. Let x1 = a − d , x2 = a and
x3 = a + d and x1 , x2, x3 be the roots of x3 − x2 + βx + γ = 0
∴
Σα = a − d + a + a + d = 1
62 Sequences and Series
⇒
a = 1 /3
…(i)
Σαβ = (a − d ) a + a (a + d ) + (a − d ) (a + d ) = β …(ii)
αβγ = (a − d ) a (a + d ) = − γ
and
∴
…(iii)
∴
From Eq. (ii), 3a − d = β
2
⇒
3 (1 / 3)2 − d 2 = β
⇒
1 /3 − β = d
[from Eq. (i)]
2
NOTE In this equation, we have two variables β and γ but we have
only one equation. So, at first sight it looks that this equation
cannot solve but we know that d 2 ≥ 0, ∀ d ∈ R, then β can
be solved. This trick is frequently asked in IIT examples.
1
−β ≥0
[Q d 2 ≥ 0]
3
1
β≤
⇒ β ∈ [− ∞ , 1 / 3]
3
⇒
⇒
23. Given, a1 = 3,m = 5n and a1 , a 2, …, is an AP.
∴
∴
a 2 = a1 + d = 3 + 6 = 9
S
or If d = 0, then m is independent of n.
Sn
∴ a2 = 9
24. a k = 2a k − 1 − a k − 2
Hence, β ∈ (− ∞ , 1 / 3] and γ ∈ [−1 / 27, ∞ )
⇒ a1 , a 2, . . . , a11 are in an AP.
2
a12 + a 22 + . . . + a11
11a 2 + 35 × 11d 2 + 10ad
=
= 90
∴
11
11
20. Since, angles of polygon are in an AP.
∴ Sum of all angles
= (n − 2) × 180° =
n
{2 (120° ) + (n − 1) 5° }
2
⇒ 225 + 35 d 2 + 150 d = 90
5n 2 − 125n + 720 = 0 ⇒ n 2 − 25n + 144 = 0
⇒ 35 d 2 + 150 d + 135 = 0
(n − 9) (n − 16) = 0 ⇒ n = 9, 16
Again, if n = 16, the n largest angle
Given,
= a + 15d = 120° + 75 = 195°
Also,
130 < a + 6d < 140
⇒
130 < 9d + 6d < 140
⇒
130 < 15d < 140
27
2
9
7
a1 + a 2 + . . . + a11 11
=
[ 30 − 10 × 3] = 0
2
11
Topic 3 Geometric Progression (GP)
1.
Key Idea Use nth term of AP i.e., an = a + ( n − 1) d , If a, A , b are in
AP, then 2A = a + b and nth term of G.P. i.e., an = ar n − 1.
…(i)
It is given that, the terms a , b, c are in GP with common
1
ratio r, where a ≠ 0 and 0 < r ≤ .
2
So, let, b = ar and c = ar 2
[from Eq. (i)]
Now, the terms 3a, 7b and 15c are the first three terms
of an AP, then
130 < t7 < 140
⇒
⇒
[neglecting n = 16]
7
[2a + 6d ]
7 (2a + 6d )
6
2
=6
=
⇒
11
11
(2a + 10d )
[2a + 10d ]
2
a = 9d
a2 <
9
7
∴ d = − 3 and d ≠ −
which is not possible.
[since, any angle of polygon cannot be > 180°]
⇒
d = − 3, −
⇒
If n = 9, then largest angle = a + 8d = 160°
Hence, n = 9
S
6
21. Given, 7 = and 130 < t7 < 140
S11 11
Sm S5 n
is independent of n.
=
Sn Sn
5n
[2 × 3 + (5n − 1) d ]
5 {(6 − d ) + 5n }
,
= 2
=
n
(6 − d ) + n
[2 × 3 + (n − 1) d ]
2
independent of n
If
6 − d =0 ⇒ d =6
From Eq. (iii), a (a 2 − d 2) = − γ
1 1
1 1 2
2
− d = −γ
⇒
 − d  = −γ ⇒

3 9
27 3
1 1 2
⇒
γ+
= d
27 3
1
γ+
≥0
⇒
27
⇒
γ ≥ − 1 / 27
 1

γ∈ −
⇒
,∞
 27 
⇒
d =9
n (n + 1)
− k − (k + 1) = 1224
2
⇒
n 2 + n − 4k = 2450 ⇒ n 2 + n − 2450 = 4k
⇒
(n + 50) (n − 49) = 4k
∴
n > 49
Let
n = 50
∴
100 = 4k ⇒ k = 25
Now
k − 20 = 5
3a = 1 ⇒ a = 1 / 3
2
⇒
[since, d is a natural number]
22. Let number of removed cards be k and (k + 1).
From Eq. (i),
⇒
26
28
<d<
3
3
⇒
Sequences and Series 63
2(7b) = 3a + 15 c
⇒
14ar = 3a + 15ar 2
⇒
14r = 3 + 15r 2
2
⇒
15r − 14r + 3 = 0
⇒ 15r 2 − 5r − 9r + 3 = 0
⇒ 5r (3r − 1) − 3(3r − 1) = 0
⇒
(3r − 1) (5r − 3) = 0
1
3
r = or
⇒
3
5
1
 1
as, r ∈ 0, , so r =
 2 
3
[as b = ar, c = ar ]
[as a ≠ 0]
2
So, 4
⇒
Now, according to the question, we have
a
⋅ a ⋅ ar = 512
r
⇒
a3 = 512
⇒
a=8
…(i)
and the given quadratic equations
ax2 + 2bx + c = 0
dx2 + 2ex + f = 0
…(ii)
…(iii)
For quadratic Eq. (ii),
the discriminant D = ( 2b)2 − 4ac
[from Eq. (i)]
= 4(b2 − ac) = 0
⇒ Quadratic Eq. (ii) have equal roots, and it is equal to
b
x = − , and it is given that quadratic Eqs. (ii) and (iii)
a
have a common root, so

d−

⇒
⇒
⇒
⇒
⇒
⇒
2
b
 b
 + 2e  −  + f = 0
 a

a
db − 2eba + a f = 0
[Q b2 = ac]
d (ac) − 2eab + a 2f = 0
[Q a ≠ 0]
dc − 2eb + af = 0
2eb = dc + af
e dc af
2 = 2+ 2
b b
b
[dividing each term by b2]
 e d f
[Q b2 = ac]
2  = +
 b a
c
2
[Q c ≠ 0]
a
r
2. Given, three distinct numbers a , b and c are in GP.
b2 = ac
d (− r )2 + 2e(− r ) + f = 0
dr 2 − 2er + f = 0
 c
 c
d   − 2e  + f = 0
 b
 a
d 2e f
−
+ =0
a b
c
d f 2e
+ =
a
c b
3. Let the three consecutive terms of a GP are , a and ar.
 −2a 
term of AP = 3a + 3
 =a
 3 
∴
⇒
⇒
Now, the common difference of AP = 7b − 3a
2a

7
= 7ar − 3a = a  − 3 = −

3
3
th
∴
⇒
2
d e f
, , are in AP.
a b c
Alternate Solution
So,
Given, three distinct numbers a , b and c are in GP. Let
a = a, b = ar, c = ar 2 are in GP, which satisfies
ax2 + 2bx + c = 0
∴
ax2 + 2(ar )x + ar 2 = 0
[Q a ≠ 0]
⇒
x2 + 2rx + r 2 = 0
⇒
(x + r )2 = 0 ⇒ x = − r.
According to the question, ax2 + 2bx + c = 0 and
dx2 + 2ex + f = 0 have a common root.
So,
x = − r satisfies dx2 + 2ex + f = 0
… (i)
Also, after adding 4 to first two terms, we get
8
+ 4, 8 + 4, 8r are in AP
r
8
⇒
2 (12) = + 4 + 8r
r8

2
⇒
24 = + 8r + 4
⇒ 20 = 4  + 2r

r
r
2
2
5 = + 2r
⇒ 2r − 5r + 2 = 0
⇒
r
2
⇒
2r − 4r − r + 2 = 0
⇒ 2r (r − 2) − 1(r − 2) = 0
⇒
(r − 2) (2r − 1) = 0
1
⇒
r = 2,
2
Thus, the terms are either 16, 8, 4 or 4, 8, 16. Hence,
required sum = 28.
4. Let r be the common ratio of given GP, then we have the
following sequence a1 , a 2 = a1r , a3 = a1r 2, ... , a10 = a1r 9
Now,
a3 = 25 a1
⇒
a1r 2 = 25 a1
⇒
Consider,
r 2 = 25
a 9 a1r 8
=
= r 4 = (25)2 = 54
a5 a1r 4
5. Let A be the Ist term of AP and d be the common
difference.
∴
7th term = a = A + 6d
[Q nth term = A + (n − 1)d]
11th term = b = A + 10d
13th term = c = A + 12d
Q a, b, c are also in GP
∴
b2 = ac
⇒
( A + 10d )2 = ( A + 6d ) ( A + 12d )
2
⇒ A + 20 Ad + 100d 2 = A 2 + 18 Ad + 72d 2
⇒
2 Ad + 28d 2 = 0
⇒
2d ( A + 14d ) = 0
⇒
d = 0 or A + 14d = 0
But
[Q the series is non constant AP]
d ≠0
⇒
A = − 14d
64 Sequences and Series
∴ a = A + 6d = − 14d + 6d = − 8d
and
c = A + 12d = − 14d + 12d = − 2d
a − 8d
⇒
=
=4
c − 2d
6. Let b = ar and c = ar 2, where r is the common ratio.
Then,
⇒
⇒
⇒
a + b + c = xb
a + ar + ar 2 = xar
1 + r + r 2 = xr
1 + r + r2
1
x=
=1 + r +
r
r
… (i) [Q a ≠ 0]
1
We know that, r + ≥ 2 (for r > 0)
r
1
and
r + ≤ − 2 (for r < 0) [using AM ≥ GM]
r
1
∴
1+r + ≥ 3
r
1
or
1 + r + ≤ −1
r
⇒
x ≥ 3 or x ≤ −1
⇒
x ∈ ( − ∞ ,−1] ∪ [3, ∞ )
Hence, x cannot be 2.
Alternate Method
From Eq. (i), we have
1 + r + r 2 = xr ⇒ r 2 + (1 − x)r + 1 = 0
For real solution of r , D ≥ 0.
⇒
(1 − x )2 − 4 ≥ 0
⇒
x 2 − 2x − 3 ≥ 0
⇒
( x − 3)( x + 1) ≥ 0
+
–
–1
+
3
Then, we have a + d, a + 4d, a + 8 d in GP,
i.e.
(a + 4d ) 2 = (a + d ) (a + 8 d )
2
⇒
a + 16 d 2 + 8ad = a 2 + 8ad + ad + 8 d 2
⇒
8 d 2 = ad
[Q d ≠ 0]
⇒
8d = a
Now, common ratio,
a + 4d 8 d + 4d 12 d 4
r=
=
=
=
a+d
9d 3
8d + d
8. Since, (α + β), (α 2 + β 2), (α 3 + β3 ) are in GP.
(α 2 + β 2)2 = (α + β ) (α 3 + β3 )
α 4 + β 4 + 2α 2β 2 = α 4 + β 4 + αβ3 + βα 3
⇒
αβ (α 2 + β 2 − 2αβ ) = 0
⇒
αβ (α − β )2 = 0
⇒
α β =0
c
=0
a
c∆ = 0
⇒
⇒
2
2
1
 1 1

Also,  − D , ,  + D are in GP.
2
 4 2

2
2
2
2
1 1

 1
1

 1
⇒
=  − D 2
  =  − D  + D

 2
2

 4
16  4
1
1
1
⇒
− D2 = ±
⇒ D2 =
4
4
2
1
1
1
D=±
⇒ a= ±
⇒
2
2
2
1
1
is the right
So, out of the given values, a = −
2
2
choice.
α + β =1
λ + δ = 4
10.

 and
αβ = p
λ δ = q
∴
Let r be the common ratio.
Since, α , β , γ and δ are in GP.
Therefore,
β = αr, γ = αr 2
and
δ = αr3
α + αr = 1 ⇒ α (1 + r ) = 1
Then,
and
αr 2 + αr3 = 4 ⇒ αr 2(1 + r ) = 4
…(i)
…(ii)
α ⋅ αr = p and αr 2 ⋅ αr3 = q
Now,
7. Let a be the first term and d be the common difference.
⇒
a = A − D, b = A, c = A + D
3
Given,
a + b+ c=
2
3
⇒ ( A − D) + A + ( A + D) =
2
3
1
3A = ⇒ A =
⇒
2
2
1
1 1
∴ The number are − D , , + D.
2
2 2
Let
From Eqs. (i) and (ii), r 2 = 4 ⇒ r = ± 2
⇒ x ∈ ( −∞ , − 1] ∪ [3, ∞ )
∴ x cannot be 2.
⇒
9. Since, a, b and c are in an AP.
On putting
r = − 2, we get
α = − 1, p = − 2 and q = − 32
2
Again putting r = 2, we get α = 1 / 3 and p = −
9
Since, q and p are integers.
Therefore, we take p = − 2 and q = − 32.
11. Here, (a 2 + b2 + c2) p2 − 2 (ab + bc + cd ) p
+ (b2 + c2 + d 2) ≤ 0
⇒ (a 2p2 − 2abp + b2) + (b2p2 − 2bcp + c2)
+ (c2p2 − 2cdp + d 2) ≤ 0
⇒ (ap − b) + (bp − c) + (cp − d ) ≤ 0
2
2
2
[since, sum of squares is never less than zero]
Since, each of the squares is zero.
or
α =β
or
∆ =0
∴
⇒
∴
(ap − b)2 = (bp − c)2 = (cp − d )2 = 0
b c d
p= = =
a b c
a , b, c,d are in GP.
Sequences and Series 65
12. Since, a , b, c are in GP.
⇒
1
3
+
=0
n− p m−n
and
b = ac
2
Given,
⇒
⇒
ax + 2bx + c = 0
2
ax + 2 ac x + c = 0
⇒
( a x+
3 (n − p) = n − m
2n = 3 p – m
and
2
c )2 = 0 ⇒ x = −
Hence, there exists infinite GP for which 27, 8 and 12 as
three of its terms.
c
a
Since, ax2 + 2bx + c = 0 and dx2 + 2ex + f = 0 have
common root.
∴ x = − c/a must satisfy.
16. Let a , d be the first term and common difference of an
AP and b, r be the first term and common ratio of a GP.
Then,
x = a + (m − 1) d and x = brm−1
y = a + (n − 1)d and y = br n−1
dx2 + 2ex + f = 0
d
f
2e
c
c
d ⋅ − 2e
+ f =0 ⇒ −
+ =0
a
a
a
ac c
2e d f
= +
[Q b2 = ac]
b a
c
d e f
, , are in an AP.
a b c
⇒
⇒
Hence,
z = a + ( p − 1)d and z = br p−1
Now, x − y = (m − n )d, y − z = (n − p)d
and
Again now, xy − z ⋅ yz − x ⋅ z x − y
= [br m − 1 ]( n − p)d ⋅ [br n − 1 ]( p − m) d ⋅ [br p − 1 ](m − n)d
= b[ n − p + p − m + m − n] d ⋅ r [(m − 1)( n − p) + ( n − 1)( p − m)+ ( p − 1)(m − n)]d
= b0 ⋅ r 0 = 1
13. Here, t3 = 4 ⇒ ar 2 = 4
∴ Product of first five terms = a ⋅ ar ⋅ ar 2 ⋅ ar3 ⋅ ar 4
Topic 4 Sum of n Terms and Infinite
Terms of a GP
= a5 r10 = (a r 2)5 = 45
14. If a , b, c ∈ (2, 18), then
1. Given series of infinite terms, if|x| < 1 ,| y| < 1 and x ≠ y is
a + b + c = 25
Since, 2, a , b are in AP.
....(i)
⇒
2a = b + 2
and b, c, 18 are in GP.
⇒
c2 = 18b
From Eqs. (i), (ii) and (iii),
b+2
+ b + 18b = 25
2
⇒
3b + 2 + 6 2 b = 50
⇒
3b + 6 2 b − 48 = 0
⇒
b + 2 2 b − 16 = 0
⇒
b + 4 2 b − 2 2 b − 16 = 0
⇒
b ( b + 4 2) − 2 2 ( b + 4 2) = 0
⇒
( b − 2 2) ( b + 4 2) = 0
⇒
.... (ii)
S = (x + y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3 ) + …
On multiplying (x − y) both sides, we get
⇒ (x − y)S = (x2 − y2) + (x3 − y3 ) + (x4 − y4 ) + ....
= [x2 + x3 + x4 + ... ] − [ y2 + y3 + y4 + .... ]
... (iii)
=
x2(1 − y) − y2(1 − x) (x2 − y2) − (x2y − y2x)
=
(1 − x)(1 − y)
(1 − x)(1 − y)
⇒ (x − y) S =
(x − y) [(x + y) − xy]
x + y − xy
⇒S=
(1 − x) (1 − y)
(1 − x) (1 − y)
20
k =1
S=
b = 8, a = 5 and c = 12
S 1
2
3
19
20
= 2 + 3 + 4 + … + 20 + 21
2 2
2
2
2
2
and AR p − 1 = 12
 27
 
8
⇒
⇒
⇒
1/(m − n)
and
1/(m − n)
8
R= 
 12
 2
= 
 3
1/( n − p)
3
2
1
3
+
n− p m− n
=1 ⇒
…(i)
3
1
+
=0
m−n n− p
…(ii)
On subtracting Eq. (ii) from Eq. (i), we get
S−
1/( n − p)
33/(m − n) ⋅ 31/( n − p) = 21/( n − p) ⋅ 23/(m − n)
3
1
+
m− n n− p
1 2
3
4
20
+
+
+
+ … + 20
2 22 23 24
2
 1
On multiplying by   both sides, we get
 2
tm = 27, tn = 8 and t p = 12
 27
R= 
8
 1
∑ k 2k 
AR m − 1 = 27, ARn − 1 = 8
∴
x2
y2
as|x| < 1 and| y| < 1
−
1−x 1−y
⇒ (x − y) S =
2. Let S =
15. Let 27, 8, 12 be three terms of a GP.
⇒
z − x = ( p − m)d
⇒
1
1
20
S 1 1
= +
+
+ … + 20 − 21
2 2 22 23
2
2
1 
1
1 − 20 
S 2
2  − 20
=
1
2
221
1−
2


a (1 − r n )
Q
sum
of
GP
=
, r < 1

1−r


66 Sequences and Series
11
1
10
1
20
S
= 1 − 20 − 21 = 1 − 20 − 20 = 1 − 20
2
2
2
2
2
2
⇒
S =2 −
11
219
⇒
⇒
⇒
⇒
3. We have, S n = 1 + q + q + … + q and
2
n
2
 q + 1
 q + 1  q + 1
Tn = 1 + 

 +…+ 
 +
 2 
 2   2 
⇒
n
∴
Also, we have
101
⇒
C1 +
C2S1 +
101
C1 +
101
+…+
101
101
C2(1 + q ) +
C101S100 = αT100
101
C3 (1 + q + q 2 )
101
C101 (1 + q + q2 + … + q100)
= α ⋅ T100
⇒ 101C1 +
C3S 2 + … +
101
101
C2
(1 − q 2 )
+
1− q
+
 1 − q3 
C3 

 1− q 
101
 1 − q4 
C4 
 +…+
 1−q
101
 1 − q101 
C101 

 1−q 
101
 1 − rn
[Qfor a GP, S n = a 
,r ≠1]
1−r
= α ⋅ T100
1
⇒
[{ 101C1 + 101C2 + … + 101C101 }
1− q
− { 101C1q + 101C2q 2 + … +101 C101q101 } = α ⋅ T100
1
[( 2101 − 1) − ((1 + q )101 − 1)] = αT100
⇒
(1 − q )
⇒
2101 − ( q + 1)101
=α
1− q
[Q nC 0 + nC1 + … + nC n = 2n]
2
100

q + 1  q + 1
 q + 1 
+
 
 +…+ 
1 +
 2 
 2 
2


⇒
101
2
− ( q + 1)
1− q
101
101

 q + 1 
 1− 
 
 2  
= α 1 ⋅
q+1 

1−


2


[Q q ≠ 1 ⇒ q + 1 ≠ 2 ⇒
=
α [2
− (q + 1)
(1 − q) ⋅ 2100
101
101
]
q+1
≠ 1]
2
⇒ α = 2100
4. Let the GP be a , ar , ar 2, ar3 , .... ∞; where a > 0 and
0 < r < 1.
Then, according the problem, we have
a
3=
1−r
27
and
= a3 + (ar )3 + (ar 2)3 + (ar3 )3 + ...
19

27
a3
a 
=
⇒
Q S∞ =

3
19 1 − r
1 − r 

3

a
27 (3 (1 − r )) 
⇒
=
Q3 =
⇒ a = 3 (1 − r )
1−r
19
1 − r3 

27 27 (1 − r ) (1 + r 2 − 2r )
=
19
(1 − r ) (1 + r + r 2)
[Q (1 − r )3 = (1 − r ) (1 − r )2]
r 2 + r + 1 = 19 (r 2 − 2r + 1)
18r 2 − 39r + 18 = 0
6r 2 − 13r + 6 = 0
(3r − 2) (2r − 3) = 0
2
3
[Q0 < r < 1]
r = or r = (reject)
3
2
5. Let a , ar , ar 2 are in GP, where (r > 1).
On multiplying middle term by 2, we have
a , 2ar , ar 2 are in an AP.
⇒
4ar = a + ar 2
2
⇒
r − 4r + 1 = 0
4 ± 16 − 4
⇒
r=
=2 ± 3
2
[since, AP is increasing]
r =2 + 3
⇒
6. Given,
k ⋅ 109 = 109 + 2 (11)1 (10)8 + 3(11)2(10)7 + ... + 10(11)9
2
 11
 11
 11
k = 1 + 2   + 3  + ... + 10  
 10
 10
 10
⇒
 11
 k =1
 10
9
2
9
 11
 11
 11
 11
  + 2   + ... + 9   + 10  
 10
 10
 10
 10
...(i)
10
…(ii)
On subtracting Eq. (ii) from Eq. (i), we get
2
9
11
11  11
 11
 11

+   + ... +   − 10  
k 1 −  = 1 +
 10
 10

10
10  10
10
 11 10

1   − 1
10
 
 10 − 11
 10
 − 10  11
⇒ k
 =
 
 10 
 10
 11

− 1

 10



a (r n − 1)
Q In GP,sum of n terms = r − 1 , when r > 1


10 
  11 10
 11
⇒
− k = 10 10   − 10 − 10   
 10
 10


∴
k = 100
7. Let S = 0.7 + 0.77 + 0.777 + …
7
77
777
+
+
+ … upto 20 terms
10 102 103
11 111
1

=7
+
+
+ … upto 20 terms
10 102 103

79
99
999

=
+
+
+… upto 20 terms

9 10 100 1000
=
=
=
7
9
1 
1  
1 

1 − 10 + 1 − 102 + 1 − 103 

+ … + upto 20 terms]
7
[(1 + 1 +…+ upto 20 terms)
9
1
1
1

−  + 2 + 3 + … + upto 20 terms 
 10 10

10
Sequences and Series 67
20

1 
 1  

1 −    
10 
10  
7
= 20 −

1
9
1−


10





 20
Q ∑ = 20 and sum of n terms of 

 i =1
a (1 − r n )


GP, S n = 1 − r when (r < 1) 


=
20
7 
1
 1  
20 − 1 −   
 10
9 
9


=
7
9
179 1  1  20  7
+   =
[179 + (10)− 20 ]

 10
81
9
9


8. We know that, the sum of infinite terms of GP is
∴
or
⇒
i.e.
⇒
 a
, |r| < 1
S ∞ = 1 − r
 ∞ ,|r| ≥ 1
x
S∞ =
=5
1−r
[|r| < 1]
x
5
5−x
exists only when|r| < 1.
r=
5
5−x
−1 <
< 1 or −10 < − x < 0
5
1−r=
0 < x < 10
3
4
9. Since, sum = 4 and second term = .
It is given first term a and common ratio r.
3
3
a
⇒
= 4, ar = ⇒ r =
4
4a
1−r
⇒
⇒
a
1−
3
4a
=4 ⇒
4a 2
=4
4a − 3
(a − 1) (a − 3) = 0 ⇒ a = 1 or 3
When a = 1, r = 3 / 4
and when a = 3, r = 1 / 4
10. Sum of the n terms of the series
1 3 7 15
+ + +
+ ...
2 4 8 16
upto n terms can be written as
1 
1 
1 
1

1 −  + 1 −  + 1 −  + 1 −  + ... upto n terms

2 
4 
8 
16
1 1 1

= n −  + + + ... + n terms
2 4 8

1
1
1 − n 

2
2
= n + 2−n − 1
=n−
1
1−
2
11. Given quadratic equation x2 − x − 1 = 0 having roots α
and β, (α > β )
1+ 5
1− 5
So, α =
and β =
2
2
and α + β = 1, αβ = −1
α n − βn
, n ≥1
an =
Q
α −β
α n+ 1 − β n+ 1
So, a n+ 1 =
= α n + α n−1 β + α n− 2 β 2+ ....
α −β
+ αβ n−1 + β n
n
n− 2
n−3
n−2
n
= α −α
−α
β −....−β
+β
[as αβ = −1]
= α n + β n − (α n− 2 + α n−3β + ...+ β n− 2)
= α n + β n − a n−1


α n−1 − β n−1
= α n− 2 + α n−3 β + ...+ β n− 2
 as a n−1 =
−
α
β


⇒ a n+ 1 + a n−1 = α n + β n = bn , ∀n ≥ 1
So, option (b) is correct.
∞
∞
α n + βn
b
Now, ∑ nn = ∑
[as, bn = α n + β n]
n
10
10
n=1
n=1
n
∞
∞
α
β
=∑  +∑ 
 10
 10
n=1
n
n=1
β
 α

Q 10 < 1 and 10 < 1


α
β
α
β
10
10
=
+
=
+
α
β
10
−
α
10
−β
1−
1−
10
10
10 α − αβ + 10 β − αβ
10(α + β ) − 2 αβ
=
=
(10 − α ) (10 − β )
100 − 10(α + β ) + αβ
10(1) − 2(−1) 12
[as α + β = 1 and αβ = −1]
=
=
100 − 10(1) − 1 89
So, option (a) is not correct.
Q
α 2 = α + 1 and β 2 = β + 1
n+ 2
⇒
α
= α n+ 1 + α n and β n+ 2 = β n+ 1 + β n
n+ 2
n+ 2
⇒ (α
+ β ) = (α n+ 1 + β n+ 1 ) + (α n + β n )
⇒
a n+ 2 = a n+ 1 + a n
Similarly, a n+ 1 = a n + a n−1
a n = a n−1 + a n−2
………
………
a3 = a 2 + a1
On adding, we get
a n+ 2 = (a n + a n−1 + a n− 2+ ....+ a 2 + a1 ) + a 2


α 2 − β2
Q
a
=
= α + β = 1
2

α −β


So, a n+ 2 − 1 = a1 + a 2 + a3 + .....+ a n
So, option (c) is also correct.
∞
∞
an
α n − βn
And, now ∑
=∑
n
n
n=1 (α − β ) 10
n=1 10
=
n
n
∞
1 ∞ α
β 
∑   − ∑   
α − β n=1 10
10 
n=1
68 Sequences and Series
β 
 α
1  10
  α
10
−
=
, as

α
β
α −β 1 −
1 −   10


10
10 
1  α
β 
1
=
−

 =
α − β  10 − α 10 − β α − β
β

< 1 and
< 1
10

 10 α − αβ − 10 β + αβ 
100 − 10 (α + β ) + αβ 


10(α − β )
10
10
=
=
=
(α − β ) [100 − 10 (α + β ) + αβ ] 100 − 10 − 1 89
Hence, options (b), (c) and (d) are correct.
∴
S1 =
1
=2
1 − 1 /2
S2 =
2
=3
1 − 1 /3
M
S 2n − 1
∴
M
M
2n − 1
=
= 2n
1 − 1 / 2n
S12 + S 22 + S32 + ... + S 22n − 1
= 22 + 32 + 42 + ... + (2n )2
1
= (2n ) (2n + 1) (4n + 1) − 1
6
12. Let a n denotes the length of side of the square S n.
We are given, a n = length of diagonal of S n + 1.
a
⇒
an = 2 an + 1 ⇒ an + 1 = n
2
This shows that a1 , a 2, a3 , K form a GP with common
ratio 1 / 2.
n −1
 1
Therefore, a n = a1  
 2
 1
a n = 10  
 2
⇒
 1
⇒ a n2 = 100  
 2
2( n − 1)
⇒
100
≤1
2n − 1
[Q a n2 ≤ 1, given]
13. Bn = 1 − An > An

3
1 −  −  

4  1
1
3
An < ⇒
<
3
2
2
4
1+
4
n
n
1
 3
−  > −
 4
6
a 2 (1 + r 2 + r 4 )
S2
(1 + r 2)2 − r 2 1
= 2 2 ⇒
=
2
2 2
a (1 + r + r )
aS
(1 + r + r 2)2 a 2
1
r+ +1
(1 + r 2 − r ) 1
2
r
⇒ a =
=
1
(1 + r 2 + r ) a 2
r + −1
r
1
y+ 1
2
r+ =y ⇒
=a
r
y−1
⇒
⇒
y + 1 = a 2y − a 2
a2 + 1
a2 − 1
⇒
y=
⇒
a2 + 1
>2
a2 −1
⇒
1


Q | y | = r + r > 2


[where , (a 2 − 1) ≠ 0]
|a 2 + 1| > 2|a 2 − 1|
⇒ (a + 1) − {2 (a 2 − 1)}2 > 0
2
2
⇒ {(a 2 + 1) − 2 (a 2 − 1)}{(a 2 + 1) + 2 (a 2 − 1)} > 0
⇒
But for
∴
3
1
<−
4
6
3
27
1
 3
<−
n = 3,  −  = −
 4
64
6
5
243
1
 3
n = 5,  −  = −
<−
 4
1024
6
and for n = 7,
7
2187
1
 3
>−
−  = −
 4
12288
6
Hence, minimum odd natural number n0 = 7.
1 1 1
1
, , , ... ,
.
2 3 4
n+1
(− a 2 + 3) (3a 2 − 1) > 0
1
< a2 < 3
3
1 
a 2 ∈  , 1 ∪ (1, 3)
[Q a 2 ≠ 1]
3 
∴
k −1
1
16. We have, S k = k ! =
1 (k − 1)!
1−
k
1
Now, (k2 − 3k + 1) S k = {(k − 2) (k − 1) − 1} ×
(k − 1)!
1
1
=
−
(k − 3)! (k − 1)!
100
1 
1002
 1
+
⇒ ∑|(k2 − 3k + 1) S k|= 1 + 1 + 2 − 
 =4−
 99 ! 98 !
100 !
k =1
14. Consider an infinite GP with first term 1, 2, 3, ..., n and
common ratios
...(ii)
On dividing Eq. (i) by Eq. (ii) after squaring it, we get
Obviously, it is true for all even values of n.
n = 1, −
...(i)
a + ar + ar 2 = a S
and
Put
Hence, (b), (c), (d) are the correct answers.
⇒
a 2 + a 2r 2 + a 2r 4 = S 2
[ Q a1 = 10, given]
This is possible for n ≥ 8.
⇒
∴
n −1
100 ≤ 2n − 1
⇒
15. Let three numbers in GP be a , ar , ar 2.
100
⇒
1002
+ |(k2 − 3k + 1) S k|= 4
100 ! k∑
=1
Sequences and Series 69
Topic 5 Harmonic Progression (HP)
1.
PLAN
Here,
∴
nth term of HP, t n =
1
a + ( n − 1) n
a1 = 5, a 20 = 25 for HP
1
1
= 5 and
= 25
a
a + 19d
1 1
4
1
1
− =−
+ 19d =
⇒ 19d =
⇒
25 5
25
5
25
−4
d=
∴
19 × 25
1
Since,
an < 0 ⇒
+ (n − 1) d < 0
5
95
1
4
⇒
−
(n − 1) < 0 ⇒ (n − 1) >
4
5 19 × 25
⇒
n >1 +
95
or n > 24.75
4
∴ Least positive value of n = 25
2. Since, a , b, c, d are in AP.
a
b
c
d
are in AP.
,
,
,
abcd abcd abcd abcd
1
1
1
1
are in AP.
,
,
,
⇒
bcd cda abd abc
⇒ bcd , cda , abd , abc are in HP.
⇒ abc, abd , cda , bcd are in HP.
⇒
3. Since, a1 , a 2, a3 , K , a10 are in AP.
Now,
⇒
⇒
a10 = a1 + 9d ⇒ 3 = 2 + 9d
d = 1 / 9 and a 4 = a1 + 3d
a 4 = 2 + 3(1 / 9) = 2 + 1 / 3 = 7 / 3
Also, h1 , h2, h3 , K , h10 are in HP.
1 1 1
1
are in AP.
⇒
,
,
,K,
h1 h2 h3
h10
Given,
h1 = 2, h10 = 3
1
1
1
1 1
∴
=
+ 9d1 ⇒ = + 9d1 ⇒ − = 9d1
6
h10 h1
3 2
⇒
⇒
⇒
∴
1
1
1
and
=
+ 6d1
h7 h1
54
1 1 6 ×1
= +
h7 2 − 54
1 1 1
18
= − ⇒ h7 =
h7 2 9
7
7 18
a 4h7 = ×
=6
3
7
d1 = −
1
1
1
=
=
1 + ln z 1 + ln x + 2 ln r A + 2D
and
Therefore,
1
1
1
are in HP.
,
,
1 + ln x 1 + ln y 1 + ln z
a1 = 1, a 2 = 2 , ⇒ a3 = 4 , a 4 = 8
5. Let
∴
b1 = 1, b2 = 3, b3 = 7, b4 = 15
Clearly, b1 , b2, b3 , b4 are not in HP.
Hence, Statement II is false.
Statement I is already true.
6. Since, cos (x − y), cos x and cos (x + y) are in HP.
∴
cos x =
2 cos (x − y) cos (x + y)
cos (x − y) + cos (x + y)
⇒ cos x (2 cos x ⋅ cos y) = 2 {cos 2 x − sin 2 y}
⇒
⇒
cos 2 x ⋅ cos y = cos 2 x − sin 2 y
cos x (1 − cos y) = sin 2 y
y
y
y
cos 2 x ⋅ 2 sin 2 = 4 sin 2 ⋅ cos 2
2
2
2
y
2
2 y
cos x ⋅ sec = 2 ⇒ cos x ⋅ sec = ± 2
2
2
2
⇒
⇒
7. Since, a , b, c are in an AP.
∴
2b = a + c
and a 2, b2, c2 are in HP.
⇒
b2 =
⇒
2a 2c2
⇒
a 2 + c2
2
2a 2c2
 a + c

 = 2 2
 2 
a +c
(a 2 + c2)(a 2 + c2 + 2ac) = 8a 2c2
⇒
(a 2 + c2) + 2ac(a 2 + c2) = 8a 2c2
⇒ (a + c ) + 2ac(a 2 + c2) + a 2c2 = 9a 2c2
2
2
⇒
(a 2 + c2 + ac)2 = 9a 2c2
⇒
a 2 + c2 + ac = 3ac
⇒
⇒
a 2 + b2 – 2ac = 0
(a – c)2 = 0 ⇒ a = c
and if a = c ⇒ b = c or a 2 + c2 + ac = – 3ac
⇒
a 2 + c2 + 2ac = –2ac
⇒
(a + c)2 = –2ac
ac
4b2 = –2ac ⇒ b2 = –
⇒
2
c
Hence,
a , b, – are in GP.
2
c
∴ Either a = b = c or a , b, − are in GP.
2
8. Since, a , A1 , A2, b are in AP.
⇒
A1 + A2 = a + b
a , G1 , G2, b are in GP ⇒ G1G2 = ab
⇒ ln y = ln x + ln r and ln z = ln x + 2 ln r
and
A = 1 + ln x, D = ln r
1
1
1
1
1
Then,
= ,
=
=
1 + ln x A 1 + ln y 1 + ln x + ln r A + D
⇒
a , H 1 , H 2, b are in HP.
3ab
3ab
H1 =
, H2 =
b + 2a
2b + a
4. Let the common ratio of the GP be r. Then,
y = xr and z = xr
Let
2
∴
1
1
1 1
+
= +
H1 H 2 a b
70 Sequences and Series
H 1 + H 2 A1 + A2 1 1
=
= +
H 1H 2
G1G2
a b
⇒
G1G2
ab
=
H 1H 2  3ab   3ab 

 

 2b + a   b + 2a 
Now,
=
(2a + b) (a + 2b)
9ab
9. (i) Now,
a + b = 10
…(i)
1 1 1 1 1
Since, a , x, y, z , b are in HP, then , , , ,
a x y z b
are in AP.
Now,
a + b 10
5  1 1 5
=
 +  − ⇒
ab
9
2  a b 3
9 × 10
[from Eq. (i)]
ab =
10
=
⇒
ab = 9
…(ii)
On solving Eqs. (i) and (ii), we get
a = 1, b = 9
(ii) LHS = log (x + z ) + log (x + z − 2 y)

 2 xz  
= log (x + z ) + log x + z − 2 

 x + z 

= log (x + z ) + log
n =1
1 15 × 16 × 31 15 × 16 
+
2 
6
2 
1
[(5 × 8 × 31) + (15 × 8)]
2
= (5 × 4 × 31) + (15 × 4) = 620 + 60 = 680
1 15 × 16
1
and S 2 = (1 + 2 + 3 + K + 15) = ×
= 60
2
2
2
=
Therefore, S = S1 − S 2 = 680 − 60 = 620.
1 1  1 1 1 1 1   1 1 1
+ = + + + +  − + + 
a b  a x y z b  x y z 
⇒
15
∑ (n 2 + n )
 n
n (n + 1) (2n + 1) 
2

Q ∑ r =
6

 r = 1
[since, a , x, y, z are in AP]
⇒
=
n( n + 1) 1
=
2
2
n =1
15
∑
5
(a + b) − 15
2
5
Sum = (a + b)
2
∴
=
a + b = (a + x + y + z + b) − (x + y + z )
=
2
2
n
 n
n (n + 1) 
 n (n + 1)
Q ∑ r3 = 
 and ∑ r =



2
2
 r = 1

r =1
…(ii)
From Eqs. (i) and (ii), we get
G1G2
A + A2 ( 2a + b) (a + 2b)
=
= 1
9ab
H 1H 2 H 1 + H 2
2xz 

Qy=

x + z 
(x − z )2
(x + z )
= 2 log (x − z ) = RHS
Topic 6 Relation between AM, GM, HM
and Some Special Series
1. Given series,
S =1+
 n( n + 1)

15
15 


13 + 23 + K + n3
2
= ∑
= ∑
+
n
n
(
)
1
+
+
+
K
n
1
2
n =1
n =1
2
…(i)
13 + 23 13 + 23 + 33
+ ... +
+
1+ 2
1+ 2+ 3
13 + 23 + 33 + K + 153 1
− (1 + 2 + 3 + K + 15)
1 + 2 + 3 + K + 15
2
= S1 − S 2 (let)
where,
13 + 23 13 + 23 + 33
S1 = 1 +
+
+K+
1+ 2
1+ 2+ 3
13 + 23 + 33 + K + 153
1 + 2 + 3 + K + 15
2. Given series is
3 × 13 5 × (13 + 23 )
7 × (13 + 23 + 33 )
+ ...
+
+
2
2
2
1
1 +2
12 + 22 + 33
So, nth term
(3 + (n − 1)2)(13 + 23 + 33 ... + n3 )
Tn =
12 + 22 + 32 + K + n 2
 n (n + 1)
(2n + 1) × 



2
=
n (n + 1)(2n + 1)
6
2
n
 n 3  n (n + 1)  2
n (n + 1)(2n + 1) 
and Σ r 2 =
Q rΣ= 1 r = 


r =1
2
6



3n (n + 1) 3 2
So,
Tn =
= (n + n )
2
2
Now, sum of the given series upto n terms
3
S n = ΣTn = [Σn 2 + Σn ]
2
3  n (n + 1)(2n + 1) n (n + 1) 
=
+

2 
6
2
3 10 × 11 × 21 10 × 11 
+
S10 =
∴
2 
6
2 
3
[(5 × 11 × 7) + (5 × 11)]
2
3
3
= × 55(7 + 1) = × 55 × 8 = 3 × 55 × 4
2
2
= 12 × 55 = 660
=
3. (b) Given series is
1 + (2 × 3) + (3 × 5) + (4 × 7) + …upto
11 terms.
Now, the rth term of the series is a r = r (2r − 1)
∴Sum of first 11-terms is
Sequences and Series 71
S11 =
11
11
11
11
r =1
r =1
r =1
r =1
( x m + x − m ) ≥ 2 and ( y n + y − n ) ≥ 2
∑ r (2r − 1) = ∑ (2r 2 − r ) = 2 ∑ r 2 − ∑ r
11 × (11 + 1)(2 × 11 + 1) 11 × (11 + 1)
−
6
2
n
 n
n (n + 1)(2n + 1)
n (n + 1) 
2
and ∑ r =
Q ∑ r =

2
6
 r = 1

r =1
 11 × 12 × 23  11 × 12
=

 −
  2 

3
=2
= (11 × 4 × 23) − (11 × 6) = 11(92 − 6) = 11 × 86 = 946
4. Given series is
3
3
3
3
 3
 1
 1
 3
3
  + 1  +  2  + 3 +  3  + ...
 4
 4
 2
 4
3
3
3
3
 12
 9
 6
 3
Let S =   +   +   +  
 4
 4
 4
 4
3
 15
+   + … + upto 15 terms
 4
3
 3
=   [13 + 23 + 33 + 43 + 53 + ... + 153 ]
 4
3
2
 3  15 × 16
=  

 4  2 
2
 3

 n ( n + 1)
3
3
3
Q1 + 2 + 3 + ... + n = 
 , n ∈N


2


27 225 × 256
= 27 × 225
=
×
64
4
⇒ S = 27 × 225 = 225 k
⇒ k = 27.
1 + 2 + 3 + ... + k
5. Since, S k =
k
=
[given]
k ( k + 1) k + 1
=
2k
2
2
1
 k + 1
2
S 2k = 
 = (k + 1)
 2 
4
So,
10
5
2
A = S 12 + S 22 + S 23 + ... S 10
= ∑ S 2k
12
k =1
Now,
⇒
… (i)
5
1
A=
12
4
10
1
∑ (k + 1)2 = 4 [22 + 32 + 42 + ... 112]
k =1
1 11 × (11 + 1) (2 × 11 + 1) 2
=
−1
4 
6

[Q ∑ n 2 =
1
4
1
=
4
=
n ( n + 1) ( 2n + 1)
]
6
11 × 12 × 23
 1
− 1 = [(22 × 23) − 1]

 4
6
1
5
505
[506 − 1] = [505] ⇒
A=
⇒ A = 303
4
12
4
1
xm yn
6. Consider,
= m
2m
2n
−m
(1 + x )(1 + y ) (x + x )( yn + y−n )
By using AM ≥ GM (because x , y ∈ R + ), we get
[Q If x > 0, then x +
( x m + x − m )( y n + y − n ) ≥ 4
1
1
⇒
≤
( x m + x − m )( y n + y − n ) 4
1
∴ Maximum value = .
4
1
≥ 2]
x
⇒
7. General term of the given series is
3r (12 + 22 + K + r 2) 3r [r (r + 1) (2r + 1)]
=
2r + 1
6(2r + 1)
1 3
2
= (r + r )
2
15
1 15
Now, required sum = ∑ Tr = ∑ (r3 + r 2)
2 r =1
r =1
Tr =
 n (n + 1)  2 n (n + 1) (2n + 1) 
+



2
6
n = 15

1  n (n + 1)  n 2 + n 2n + 1 
= 
+

 2
2
2
3 

n = 15
1  n (n + 1) (3n 2 + 7n + 2) 
= 

2
2
6
n = 15
1 15 × 16 (3 × 225 + 105 + 2)
= ×
×
= 7820
2
2
6
8. We have, a1 , a 2, a3 , … a 49 are in AP.
=
1
2
12
∑ a 4k + 1 = 416 and
a 9 + a 43 = 66
k=0
Let a1 = a and d = common difference
Q
a1 + a5 + a 9 + L + a 49 = 416
∴ a + (a + 4d ) + (a + 8d ) + … (a + 48d ) = 416
13
…(i)
(2a + 48d ) = 41633a + 24d = 32
⇒
2
Also a 9 + a 43 = 66 ⇒ a + 8d + a + 42d = 66
…(ii)
⇒
2a + 50d = 66 ⇒ a + 25d = 33
Solving Eqs. (i) and (ii), we get
a = 8and d = 1
2
2
2
2
Now, a1 + a 2 + a3 + L + a17
= 140m
2
2
2
8 + 9 + 10 + … + 242 = 140m
⇒
(12 + 22 + 32 + … + 242) − (12 + 22
2
2
+ 3 + … + 7 ) = 140m
24 × 25 × 49 7 × 8 × 15
−
= 140m
⇒
6
6
3 × 7 ×8 ×5
⇒
(7 × 5 − 1) = 140m
6
⇒
7 × 4 × 5 × 34 = 140m
⇒
140 × 34 = 140m ⇒ m = 34
9. We have,
12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + …
A = sum of first 20 terms
B = sum of first 40 terms
72 Sequences and Series
∴A = 12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + … + 2 ⋅ 202
A = (12 + 22 + 32 + … + 202) + (22 + 42 + 62 + … + 202)
A = (12 + 22 + 32 + … + 202) + 4(12 + 22 + 32 + …+102)
20 × 21 × 41 4 × 10 × 11 × 21
A=
+
6
6
20 × 21
20 × 41 × 63
A=
(41 + 22) =
6
6
Similarly
B = (12 + 22 + 32 + … + 402) + 4(12 + 22 +… + 202)
40 × 41 × 81 4 × 20 × 21 × 41
B=
+
6
6
40 × 41
40 × 41 × 123
B=
(81 + 42) =
6
6
Now, B − 2 A = 100λ
40 × 41 × 123 2 × 20 × 21 × 63
∴
−
= 100λ
6
6
40
40
⇒
(5043 − 1323) = 100λ ⇒
× 3720 = 100λ
6
6
40 × 620
= 248
⇒
40 × 620 = 100λ ⇒ λ =
100
10. Let S10 be the sum of first ten terms of the series. Then,
we have
2
2
2
2
 3
 2
 1
 4
S10 = 1  + 2  + 3  + 42 + 4 
 5
 5
 5
 5
+ ... to 10 terms
2
2
2
2
 8
 12
 16
 24
=   +   +   + 42 +   + ... to 10 terms
 5
5
5
5
1 2
2
2
2
2
= 2 (8 + 12 + 16 + 20 + 24 + ... to 10 terms)
5
42
= 2 (22 + 32 + 42 + 52 + ... to 10 terms)
5
42
= 2 (22 + 32 + 42 + 52 + ... + 112)
5
16
=
((12 + 22 + ... + 112) − 12)
25
16  11 ⋅ (11 + 1) (2 ⋅ 11 + 1)

=
− 1


25 
6
16
16
16
16
m=
× 505 = 101
=
× 505 ⇒
(506 − 1) =
5
25
25
25
Write the nth term of the given series and simplify it to get its
lowest form. Then, apply, S n = ∑ Tn
PLAN
13 13 + 23 13 + 23 + 33
+
+
+ ...
1
1+3
1+3+5
Let Tn be the nth term of the given series.
13 + 23 + 33 + ... + n3
∴ Tn =
1 + 3 + 5 + ... + upto n terms
Given series is
2
 n (n + 1) 


2
2
 = (n + 1)
=
2
4
n
9
(n + 1)2 1 2
S9 = ∑
= (2 + 32 + ... + 102) + 12 – 12]
4
4
n =1
=
1 10(10 + 1)(20 + 1)
 384
= 96
–1 =

4 
6
4
π
2
13. Here, α ∈ (0, ) ⇒ tan α > 0
x2 + x +
∴
⇒
tan 2 α
x2 + x
2
x +x+
2
tan 2 α
x2 + x
l + n = 2m
[using AM ≥ GM]
≥ 2 tan α
15. Since, AM ≥ GM, then
Also,
(a + b) + (c + d )
≥ (a + b)(c + d ) ⇒ M ≤ 1
2
(a + b) + (c + d ) > 0
[Q a , b, c, d > 0]
∴
0 < M ≤1
α+β=
4+ 5
8+2 5
and α β =
5+ 2
5+ 2
Let H be the harmonic mean between α and β, then
∴ G1 = lr, G2 = lr 2, G3 = lr3 , n − lr 4
⇒
 n
r= 
 l
= (lr ) + 2(lr ) + (lr )
4
2 4
2
H =
1
4
3 4
= l × r 4 (1 + 2r 4 + r 6 ) = l4 × r 4 (r 4 + 1)2
= l4 ×
x2 + x
…(i)
a1 a 2 a3 ... (a n − 1 )(2a n ) = 2c
a1 + a 2 + a3 + ... + 2a n
1/ n
≥ (a1 ⋅ a 2 ⋅ a3 ... 2a n )
∴
n
[using AM ≥ GM]
[from Eq. (i)]
⇒ a1 + a 2 + a3 + ... + 2a n ≥ n (2c)1/ n
⇒ Minimum value of
a1 + a 2 + a3 + ... + 2a n = n (2c)1/ n
Let r be the common ratio of this GP.
+ G34
4
tan 2 α
⇒
... (i)
l, G1 , G2, G3 , n are in GP.
+ 2G24
x2 + x ⋅
14. Given, a1 a 2 a3 ... a n = c
and G1 , G2, G3 are geometric means between l and n.
Now, G14
≥
16. Let α , β be the roots of given quadratic equation. Then,
11. Given, m is the AM of l and n.
∴
12.
n  n + l
2
2

 = ln × 4 m = 4lm n
l  l 
2 αβ 16 + 4 5
=4
=
α+β
4+ 5
17. Since, product of n positive numbers is unity.
⇒
... (i)
x1 ⋅ x2 ⋅ x3 ... xn = 1
x1 + x2 + ... + xn
1/ n
Using AM ≥ GM,
≥ (x1 ⋅ x2 ... xn )
n
⇒
x1 + x2 + ... + xn ≥ n (1)1/ n
[from Eq. (i)]
Hence, sum of n positive numbers is never less than n.
Sequences and Series 73
18. Since, AM > GM
Again, a (n ) = 1 +
(b + c − a ) + (c + a − b)
> (b + c − a )(c + a − b)1/ 2
∴
2
⇒
c > [(b + c − a )(c + a − b)]1/ 2
…(i)
Similarly
b > [(a + b − c)(b + c − a )]1/ 2
…(ii)
and
a > [(a + b − c)(c + a − b)]
1/ 2
1  1 1  1
1
+  +  +  +K+ 
2  3 4  5
8
1
1 1

+ K +  n −1
+ K + n − n
2
2  2
+1
>1 +
…(iii)
1  1 1  1 1
1
+  +  +  + +K+ 



2
4 4
8 8
8
1 1
 1
+ K +  n + K + n − n
2
2  2
On multiplying Eqs. (i), (ii) and (iii), we get
abc > (a + b − c)(b + c − a )(c + a − b)
1 2 4
2n − 1
1
+ + +K+ n − n
2 4 8
2
2
1 1 1
1 1 
1 n
= 1 + + + + K + − n = 1 − n  +

 2
2
2
2
2
2
2
14442444
3
n times
Hence, (a + b − c)(b + c − a )(c + a − b) − abc < 0
=1 +
19. Since, x1 , x2 ,…, xn are positive real numbers.
∴ Using nth power mean inequality
x12 + x22 + ... + xn2  x1 + x2 + ... + xn 
≥



n
n
⇒
2
2
 n   n 
n  n 2  n 
 ∑ xi2 ≥  ∑ xi 
x
x
⇒
n
≥
i

∑ i 
 

n  i∑



 i =1   i =1 
=1
i =1
2
2
20. Let a and b are two numbers. Then,
a+b
2ab
; G1 = ab ; H 1 =
2
a+b
An − 1 + H n − 1
, G n = An − 1 H n − 1 ,
An =
2
2 An − 1 H n − 1
Hn =
An − 1 + H n − 1
A1 =
Clearly, G1 = G2 = G3 = ... = ab .
21. A2 is AM of A1 and H 1 and A1 > H 1
⇒
A1 > A2 > H 1
A3 is AM of A2 and H 2 and A2 > H 2
⇒
A2 > A3 > H 2
M
M
M
A1 > A2 > A3 > ...
∴
22. As above, A1 > H 2 > H 1 , A2 > H 3 > H 2
∴
H 1 < H 2 < H 3 < ...
1 1 1
1
+ + +K+ n
2 3 4
2 −1
1  1
1
 1 1  1
=1 +  +  +  + K +  +  + K+ 
 2 3  4


7
8
15
 1
1 
+ K +  n −1 + K + n

2
2 − 1
1  1 1
1
 1 1  1 1
<1 +  +  +  + + K+  +  + + K + 
 2 2  4 4
4  8 8
8
1 
 1
+ K +  n − 1 + ... + n − 1 
2
2 
23. Given, a (n ) = 1 +
2 4 8
2n −1
= 1 + + + + K + n −1
2 4 8
2
= 1 + 1 + 1 + 1 + ... + 1 = n
1444
424444
3
Thus, a(100) ≤ 100
(n ) times
Therefore, (a) is the answer.
1  200

Therefore, a (200) > 1 − 200  +
> 100

2
2 
Therefore, (d) is also the answer.
24. Since, first and (2n − 1)th terms are equal.
Let first term be x and (2n − 1) th term be y,
whose middle term is tn.
x+ y
Thus, in arithmetic progression, tn =
=a
2
In geometric progression, tn = xy = b
2xy
In harmonic progression, tn =
=c
x+ y
[using AM > GM > HM]
⇒ b2 = ac and a > b > c
Here, equality holds (i. e. a = b = c) only if all terms are
same. Hence, options (a), (b) and (d) are correct.
25. Let the two positive numbers be a and b.
∴
and
a+b
[since, x is AM between a and b] … (i)
2
a y z
[since, y, z are GM’s between a and b]
= =
y z b
x=
∴
a=
y2
z
and b =
z2
y
On substituting the values of a and b in Eq. (i), we get
2x =
y2 z 2
y3 + z3
y3 + z3
+
⇒
= 2x ⇒
=2
z
y
yz
xyz
26. Let the two positive numbers be ka and a , a > 0.
Then, G = ka ⋅ a = k ⋅ a and H =
Again,
⇒
[given]
⇒
2 k 4
=
k+1 5
5 k = 2k + 2 ⇒ 2k − 5 k + 2 = 0
⇒
⇒
H 4
=
G 5
2ka
k+1 4
=
ka 5
2(ka )a
2ka
=
ka + a k + 1
k=
5 ± 25 − 16 5 ± 3
1
=
= 2, ⇒ k = 4, 1 / 4.
4
4
2
Hence, the required ratio is 4 : 1.
74 Sequences and Series
27. Using AM ≥ GM,
= a12n ⋅ r1 + 2 + K + ( n − 1) = a12nr
1 + x2 n
1 + x2n
≥ 1 ⋅ x2 n ⇒
≥ xn
2
2
1
1
xn
xn ⋅ ym
≤
≤ ⇒
2n
2n
2
(1 + x ) (1 + y2m ) 4
1+x
⇒
⇒
Gm = ( A1 A2 K AnH 1H 2 K H n )1/ 2n
31. Let two numbers be a and b and A1 , A2, ... , An be n
arithmetic means between a and b. Then,
a , A1 , A2, ... , An , b are in AP with common difference
b−a
b−a
d=
⇒ p = A1 = a + d = a +
n+1
n+1
p=
⇒
29. Here, (1 + a )(1 + b)(1 + c)
…(i)
= 1 + a + b + c + ab + bc + ca + abc
a + b + c + ab + bc + ca + abc
4 4 4 1/7
Since,
≥ (a b c )
7
[using AM ≥ GM]
a + b + c + ab + bc + ca + abc ≥ 7(a 4b4c4 )1/7
1 + a + b + c + ca + abc > 7(a 4b4c4 )1/7
…(ii)
From Eqs. (i) and (ii), we get
{(1 + a )(1 + b)(1 + c)}7 > 77 (a 4b4c4 )
30. Let Gm be the geometric mean of G1 , G2, ... , Gn.
Gm = (G1 ⋅ G2 K Gn )1/ n
K (a1 ⋅ a1r ⋅ a1r K a1r
∴
1 1
1 1
(a − b)
= +D ⇒
= +
q a
q a (n + 1) ab
⇒
( n + 1) ab
1
nb + a
... (ii)
=
⇒ q=
nb + a
q (n + 1) ab
From Eq. (i),
b = (n + 1) p − na.
q { n (n + 1) p − n 2a + a } = (n + 1) a {(n + 1) p − na }
n − 1 1/ n 1/ n
) ]
⇒ n (n + 1) a 2 − {(n + 1)2 p + (n 2 − 1)q}a
where, r is the common ratio of GP a1 , a 2, K , a n.
= [(a1 ⋅ a1. K n times ) (r1/ 2 ⋅ r3/3 ⋅ r 6/ 4 K r
=
( n − 1 )n
2n
+ n (n + 1) pq = 0
)]1/ n
{(n + 1) p + (n − 1)q}2 − 4n 2pq > 0
⇒ (n + 1) p + (n − 1)2q2 + 2 (n 2 − 1) pq − 4n 2pq > 0
2 2
1/ n
 1  ( n − 1)n  
 n −1 
= a1 r 2  2   = a1 r 4 






An =
Hn =
and
=
∴
⇒
An ⋅ H n =
…(i)
a1 + a 2 + ... + a n a1 (1 − r n )
=
n
n (1 − r )
1
1
1
+
+K+


 a1 a 2
an
n
1 
1
1 
1 + + K + n − 1 

a1 
r
r
=
a1n (1 − r ) r n − 1
1 − rn
a1 (1 − r n ) a1n (1 − r )r n − 1
= a12r n − 1
×
n (1 − r )
(1 − r n )
n
k =1
k =1
⇒
∏ AkH k = ∏ (a12r n − 1 )
= (a12 ⋅ a12 ⋅ a12K n times ) × r 0 ⋅ r1 ⋅ r 2K r n − 1
(n + 1)2 p2 + (n − 1)2q2 − 2 (n 2 + 1) pq > 0
2
⇒
⇒
n
n
⇒ na − {(n + 1) p + (n − 1)q} a + npq = 0
2
Since, a is real, therefore
n −1
1
3
+ 1 + + L+
2
2 ]1/ n
[a1n ⋅ r 2
Now,
…(i)
Putting it in Eq. (ii), we get
= [(a1 ) ⋅ (a1 ⋅ a1r )1/ 2 ⋅ (a1 ⋅ a1r ⋅ a1r 2)1/3
2
na + b
n+1
Let H 1 , H 2, ... , H n be n harmonic means between a and
b.
1 1
1
1 1
is an AP with common
∴
,
,
, ... ,
,
a H1 H 2
Hn b
(a − b)
difference, D =
.
(n + 1) ab
(1 + a )(1 + b)(1 + c) > 7(a 4b4c4 )1/7
⇒
[from Eq. (i)]
 n

Gm =  ∏ AkH k 
k = 1

Hence, it is a false statement.
or
n −1
4 ]2 n
1/ 2 n
Here, equality holds only when x = a which is not
possible. So, log a x + log x a is greater than 2.
⇒
= [a1r
= [Gm ]2n
Hence, it is false statement.
1
log a x +
log a x
28. Since,
> 1, using AM > GM
2
⇒
n ( n − 1)
2
⇒
q2 −
 n + 1
2 (n 2 + 1)
2
pq + 
 p >0
 n − 1
(n − 1)2
2
2

 n + 1 
 n + 1) 2
q2 − 1 + 
  pq + 
 p >0
 n − 1 
 n −1


2

 n + 1 
(q − p) q − 
 p > 0
 n − 1 


2
⇒
Q
 n + 1
q < p or q > 
 p
 n − 1
 n + 1 2

 p > p

 n − 1 

2
 n + 1
Hence, q cannot lie between p and 
 p.
 n − 1
Sequences and Series 75
⇒
2n 2 + 1 ≥ 2n ⇒ n ≤ 6
and, also
C >0⇒n >2
∴The possible values of n are 3, 4, 5, 6
(2 × 9) − 6
So, at n = 3, C =
= 12
8 − 6 −1
32 − 8
24 8
at, n = 4, C =
=
= ∉N
16 − 8 − 1 9 3
50 − 10
40
at, n = 5, C =
=
∉N
32 − 10 − 1 21
72 − 12
60
and at, n = 6, C =
=
∉N
64 − 12 − 1 51
32. Since a , b, c > 0
⇒
Also,
(a + b + c)
> (abc)1/3
3
…(i)
[using AM ≥ GM]
1 1 1
+ +
1/3
a b c ≥  1 ⋅ 1 ⋅ 1
…(ii)


 a b c
3
[using AM ≥ GM]
On multiplying Eqs. (i) and (ii), we get
 1 1 1
(a + b + c)  + + 
 a b c
1
≥ (abc)1/3
9
(abc)1/3
∴
 1 1 1
(a + b + c)  + +  ≥ 9
 a b c
∴ The required value of C = 12 for n = 3.
35. The 4th AM out of m AMs inserted between 3 and 243 is
y1
33. For real numbers y1 , y2, y3 , the quantities 3 , 3
A4 = 3 + 4
y2
and
3y3 are positive real numbers, so according to the
AM-GM inequality, we have
2
1 

  243 3 + 1 
1/ 2
G2 = 3  

 = 3(81) = 27
 3 


1
1
3y1 + 3y2 + 3y3 ≥ 3(3y1 . 3y2 . 3y3 )3
 240 
⇒ 4 
 = 24 ⇒ m + 1 = 40 ⇒ m = 39
 m + 1
Hence, answer is 39.00
36.
Plan
(i) If a, b, c are in GP, then they can be taken as a, ar, ar 2
where r, (r ≠ 0) is the common ratio.
x + x2 + K + xn
(ii) Arithmetic mean of x1, x2, K , xn = 1
n
1
x1 + x2 + x3
≥ (x1x2x3 )3
3
On applying logarithm with base ‘3’, we get
 x + x2 + x3  1
log3  1
 ≥ (log3 x1 + log3 x2 + log3 x3 )
 3

3
1
⇒
1 ≥ (log3 x1 + log3 x2 + log3 x3 )
3
{Q x1 + x2 + x3 = 9}
∴
M =3
Now, log 2(m3 ) + log3 (M 2) = 3 log 2(4) + 2 log3 (3 )
=6 + 2 =8
34. Given arithmetic progression of positive integers terms
a1 , a 2, a3 , ..... having common difference ‘2’ and
geometric progression of positive integers terms
b1 , b2, b3 , .... having common ratio ‘2’ with a1 = b1 = c,
such that
2 (a1 + a 2 + a3 + ... + a n ) = b1 + b2 + b3 + .... + bn
 2 n − 1
n
2 × [2C + (n − 1)2] = C 
⇒

2
 2 −1
⇒
⇒
Q
2nC + 2n − 2n = 2 . C − C
C [2n − 2n − 1] = 2n 2 − 2n
C ∈ N ⇒ 2n 2 − 2n ≥ 2n − 2n − 1
2
n
 240 
A4 = G 2 ⇒ 3 + 4 
 = 27
 m + 1
Q
On applying logarithm with base ‘3’, we get
1


log3 (3y1 + 3y2 + 3y3 ) ≥ 1 + ( y1 + y2 + y3 )


3
=1 + 3
=4
{Q y1 + y2 + y3 = 9}
∴
m =4
Now, for positive real numbers x1 , x2 and x3 , according to
AM-GM inequality, we have
…(i)
and the 2nd GM out of three GMs inserted between 3
and 243 is
3y1 + 3y2 + 3y3
≥ (3y1 . 3y2 . 3y3 )3
3
⇒
243 − 3
m+1
Let a , b, c be a , ar , ar 2, where r ∈ N
a+ b+ c
Also,
=b+2
3
⇒
a + ar + ar 2 = 3 (ar ) + 6
⇒
ar 2 − 2ar + a = 6
6
(r − 1)2 =
⇒
a
Since, 6 /a must be perfect square and a ∈ N .
So, a can be 6 only.
⇒
r −1 = ± 1 ⇒ r =2
a + a − 14 36 + 6 − 14
=
=4
a+1
7
2
and
37. Using AM ≥ GM,
a −5 + a −4 + a −3 + a −3 + a −3 + 1 + a 8 + a10
8
1
≥ (a − 5 ⋅ a − 4 ⋅ a − 3 ⋅ a − 3 ⋅ a − 3 ⋅ 1 ⋅ a 8 ⋅ a10 ) 8
⇒ a −5 + a −4 + 3a −3 + 1 + a 8 + a10 ≥ 8 ⋅ 1
Hence, minimum value is 8.
4
Permutations
and Combinations
Topic 1 General Arrangement
Objective Questions I (Only one correct option)
1. The number of four-digit numbers strictly greater than
4321 that can be formed using the digits 0, 1, 2, 3, 4, 5
(repetition of digits is allowed) is
(2019 Main, 8 April II)
(a) 306
(c) 360
(b) 310
(d) 288
are there, for which the sum of the diagonal entries of
(2017 Adv.)
M T M is 5 ?
(b) 162
(c) 126
(d) 135
3. The number of integers greater than 6000 that can be
formed using the digits 3, 5, 6, 7 and 8 without
repetition is
(2015 Main)
(a) 216
(c) 120
(b) 192
(d) 72
4. The number of seven-digit integers, with sum of the
digits equal to 10 and formed by using the digits 1, 2 and
3 only, is
(2009)
(a) 55
(b) 66
(c) 77
(d) 88
5. How many different nine-digit numbers can be formed
from the number 22 33 55 888 by rearranging its digits
so that the odd digits occupy even positions? (2000, 2M)
(a) 16
(c) 60
(b) 36
(d) 180
6. An n-digit number is a positive number with exactly n
digits. Nine hundred distinct n-digit numbers are to be
formed using only the three digits 2,5 and 7. The smallest
value of n for which this is possible, is
(1998, 2M)
(a) 6
(b) 7
(c) 8
5 newspapers and every newspaper is read by 60
students. The number of newspapers is
(1998, 2M)
(a) atleast 30
(b) atmost 20
(c) exactly 25
(d) None of the above
8. A five digits number divisible by 3 is to be formed using
2. How many 3 × 3 matrices M with entries from {0, 1, 2}
(a) 198
7. In a collage of 300 students, every student reads
(d) 9
the numbers 0, 1 , 2, 3 , 4 and 5, without repetition. The
total number of ways this can be done, is
(1989, 2M)
(a) 216
(c) 600
(b) 240
(d) 3125
9. Eight chairs are numbered 1 to 8. Two women and
three men wish to occupy one chair each.
First the women choose the chairs from amongst the
chairs marked 1 to 4 and then the men select the chairs
from amongst the remaining. The number of possible
arrangements is
(a) 6C3 × 4 C2
(c) 4 C2 + 4 P3
(b) 4 P2 × 4 P3
(d) None of these
(1982, 2M)
10. The different letters of an alphabet are given. Words
with five letters are formed from these given letters.
Then, the number of words which have at least one
letter repeated, is
(1980, 2M)
(a) 69760
(c) 99748
(b) 30240
(d) None
Analytical & Descriptive Question
11. Eighteen guests have to be seated half on each side of a
long table. Four particular guests desire to sit on one
particular side and three other on the other side.
Determine the number of ways in which the sitting
arrangements can be made.
(1991, 4M)
Permutations and Combinations 77
Match the Columns
Match the conditions/expressions in Column I with statement in Column II.
12. Consider all possible permutations of the letters of the word ENDEANOEL.
Column I
(2008, 6M)
Column II
A.
The number of permutations containing the word
ENDEA, is
p.
5!
B.
The number of permutations in which the letter E
occurs in the first and the last positions, is
q.
2 × 5!
C.
The number of permutations in which none of the
letters D, L, N occurs in the last five positions, is
r.
7 × 5!
D.
The number of permutations in which the letters A, E,
O occur only in odd positions, is
s.
21 × 5!
Topic 2 Properties of Combinational and General Selections
Objective Questions I (Only one correct option)
6. If n C 4 , n C 5 and n C 6 are in AP, then n can be
(2019 Main, 12 Jan II)
1. There are 3 sections in a question paper and each
section contains 5 questions. A candidate has to answer
a total of 5 questions, choosing at least one question
from each section. Then the number of ways, in which
the candidate can choose the questions, is
[2020 Main, 5 Sep II]
(a) 3000
(c) 2255
(b) 1500
(d) 2250
objects of which 10 are identical and the remaining 21
are distinct, is
(2019 Main, 12 April I)
(a) 2
−1
21
(b) 2
20
(c) 2
20
(d) 2
+1
3. Suppose that 20 pillars of the same height have been
erected along the boundary of a circular stadium. If the
top of each pillar has been connected by beams with the
top of all its non-adjacent pillars, then the total number
of beams is
(2019 Main, 10 April II)
(a) 180
(b) 210
(c) 170
(d) 190
4. Some identical balls are arranged in rows to form an
equilateral triangle. The first row consists of one ball,
the second row consists of two balls and so on. If 99 more
identical balls are added to the total number of balls
used in forming the equilateral triangle, then all these
balls can be arranged in a square whose each side
contains exactly 2 balls less than the number of balls
each side of the triangle contains. Then, the number of
balls used to form the equilateral triangle is
(2019 Main, 9 April II)
(a) 262
(b) 190
(c) 225
(d) 157
5. There are m men and two women participating in a
chess tournament. Each participant plays two games
with every other participant. If the number of games
played by the men between themselves exceeds the
number of games played between the men and the
women by 84, then the value of m is (2019 Main, 12 Jan II)
(a) 12
(b) 11
(c) 9
(b) 11
25
7. If
∑{
50
Cr ⋅
(d) 7
(c) 14
(d) 12
50 − r
C 25 − r } = K (50C 25 ),
r = 0
then, K is equal to
(a) 224
8. If
(2019 Main, 10 Jan II)
(b) 225 − 1
(c) 225
(d) (25)2
3
20


Ci − 1
k

 = , then k equals
∑
20
20
Ci + Ci − 1 
21
i=1
(2019 Main, 10 Jan I)
20
2. The number of ways of choosing 10 objects out of 31
20
(a) 9
(a) 100
(b) 400
(c) 200
(d) 50
9. A man X has 7 friends, 4 of them are ladies and 3 are
men. His wife Y also has 7 friends, 3 of them are ladies
and 4 are men. Assume X and Y have no common
friends. Then, the total number of ways in which X and
Y together can throw a party inviting
3 ladies and 3 men, so that 3 friends of each of X and Y
are in this party, is
(2017 Main)
(a) 485
(b) 468
(c) 469
(d) 484
10. Let S = {1, 2, 3, …… , 9}. For k = 1, 2 , …… 5, let N k be the
number of subsets of S, each containing five elements
out
of
which
exactly
odd.
Then
k are
(2017 Adv.)
N1 + N 2 + N 3 + N 4 + N 5 =
(a) 210
(b) 252
(c) 126
(d) 125
11. A debate club consists of 6 girls and 4 boys. A team of
4 members is to be selected from this club including the
selection of a captain (from among these 4 members) for
the team. If the team has to include atmost one boy, the
number of ways of selecting the team is
(2016 Adv.)
(a) 380
(b) 320
(c) 260
(d) 95
12. Let Tn be the number of all possible triangles formed by
joining vertices of an n-sided regular polygon. If
(2013 Main)
Tn + 1 − Tn = 10, then the value of n is
(a) 7
(c) 10
(b) 5
(d) 8
78 Permutations and Combinations
13. If r , s, t are prime numbers and p, q are the positive
17. Let n ≥ 2 be an integer. Take n distinct points on a circle
integers such that LCM of p, q is r 2s4 t 2 ,then the number
of ordered pairs ( p, q) is
(2006, 3M)
and join each pair of points by a line segment. Colour the
line segment joining every pair of adjacent points by blue
and the rest by red. If the number of red and blue line
segments are equal, then the value of n is
(2014 Adv.)
(a) 252
(b) 254
(c) 225
(d) 224
14. The value of the expression 47 C 4 +
5
∑
52− j
C 3 is
j =1
(1980, 2M)
47
(a) C5
(b) 52C5
(c) 52C4
(d) None of these
Fill in the Blanks
18. Let A be a set of n distinct elements. Then, the total
number of distinct functions from A to A is…and out of
these… are onto functions.
(1985, 2M)
19. In a certain test, a i students gave wrong answers to at
least i questions, where i = 1, 2, K , k. No student gave
more that k wrong answers. The total number of wrong
answers given is … .
(1982, 2M)
Match the Columns
15. In a high school, a committee has to be formed from a
group of 6 boys M1 , M 2, M 3, M 4 , M 5, M 6 and 5 girls G1 ,
G2, G 3, G 4 , G 5.
(i) Let α1 be the total number of ways in which the
committee can be formed such that the committee
has 5 members, having exactly 3 boys and 2 girls.
(ii) Let α 2 be the total number of ways in which the
committee can be formed such that the committee
has at least 2 members, and having an equal
number of boys and girls.
(iii) Let α 3 be the total number of ways in which the
committee can be formed such that the committee
has 5 members, at least 2 of them being girls.
(iv) Let α 4 be the total number of ways in which the
committee can be formed such that the committee
has 4 members, having at least 2 girls such that both
M1 and G1 are NOT in the committee together.
(2018 Adv.)
List-I
List-II
P.
The value of α1 is
1.
136
Q.
The value of α 2 is
2.
189
R.
The value of α 3 is
3.
192
S.
The value of α 4 is
4.
200
5.
381
6.
461
The correct option is
(a) P → 4; Q → 6; R → 2; S → 1
(b) P → 1; Q → 4; R → 2; S → 3
(c) P → 4; Q → 6; R → 5; S → 2
(d) P → 4; Q → 2; R → 3; S → 1
Integer & Numerical Answer Type Questions
16. Five persons A , B, C , D and E are seated in a circular
arrangement. If each of them is given a hat of one of the
three colours red, blue and green, then the number of
ways of distributing the hats such that the persons
seated in adjacent seats get different coloured hats is
………
(2019 Adv.)
True/False
20. The product of any r consecutive natural numbers is
always divisible by r !.
(1985, 1M)
Analytical & Descriptive Questions
21. A committee of 12 is to be formed from 9 women and 8
men. In how many ways this can be done if at least five
women have to be included in a committee ? In how
many of these committees
(i) the women are in majority?
(ii) the men are in majority?
(1994, 4M)
22. A student is allowed to select atmost n books from n
collection of (2n + 1) books. If the total number of ways
in which he can select at least one books is 63, find the
value of n.
(1987, 3M)
23. A box contains two white balls, three black balls and
four red balls. In how many ways can three balls be
drawn from the box, if at least one black ball is to be
included in the draw ?
(1986, 2 12 M)
24. 7 relatives of a man comprises 4 ladies and 3
gentlemen, his wife has also 7 relatives ; 3 of them are
ladies and 4 gentlemen. In how many ways can they
invite a dinner party of 3 ladies and 3 gentlemen so
that there are 3 of man’s relative and 3 of the wife's
relatives?
(1985, 5M)
25. m men and n women are to be seated in a row so that no
two women sit together. If m > n, then show that the
number of ways in which they can be seated, is
m ! (m + 1) !
.
(m − n + 1) !
(1983, 2M)
26. mn squares of equal size are arranged to form a
rectangle of dimension m by n where m and n are
natural numbers. Two squares will be called
‘neighbours’ if they have exactly one common side. A
natural number is written in each square such that the
number in written any square is the arithmetic mean of
the numbers written in its neighbouring squares. Show
that this is possible only if all the numbers used are
(1982, 5M)
equal.
27. If n C r −1 = 36, n C r = 84 and n C r +1 = 126, then find the
values of n and r.
(1979, 3M)
Permutations and Combinations 79
Topic 3 Multinomial, Repeated Arrangement and Selection
Objective Question I (Only one correct option)
Integer & Numerical Answer Type Questions
1. The number of 6 digits numbers that can be formed
8. An engineer is required to visit a factory for exactly four
using the digits 0, 1, 2,5, 7 and 9 which are divisible by
11 and no digit is repeated, is
(2019 Main, 10 April I)
days during the first 15 days of every month and it is
mandatory that no two visits take place on consecutive
days. Then the number of all possible ways in which
such visits to the factory can be made by the engineer
during 1-15 June 2021 is ………
(2020 Adv.)
(a) 60
(c) 48
(b) 72
(d) 36
2. A committee of 11 members is to be formed from 8 males
and 5 females. If m is the number of ways the committee
is formed with at least 6 males and n is the number of
ways the committee is formed with atleast 3 females,
then
(2019 Main, 9 April I)
(a) m = n = 68
(c) m = n = 78
(b) m + n = 68
(d) n = m − 8
3. Consider three boxes, each containing 10 balls labelled
1, 2, …, 10. Suppose one ball is randomly drawn from
each of the boxes. Denote by ni , the label of the ball
drawn from the ith box, (i = 1, 2, 3). Then, the number of
ways in which the balls can be chosen such that
(2019 Main, 12 Jan I)
n1 < n2 < n3 is
(a) 82
(c) 240
(b) 120
(d) 164
4. The number of natural numbers less than 7,000 which
can be formed by using the digits 0, 1, 3, 7, 9 (repitition
of digits allowed) is equal to
(2019 Main, 9 Jan II)
(a) 374
(c) 372
(b) 375
(d) 250
5. Consider a class of 5 girls and 7 boys. The number of
different teams consisting of 2 girls and 3 boys that can
be formed from this class, if there are two specific boys
A and B, who refuse to be the members of the same
team, is
(2019 Main, 9 Jan I)
(a) 350
(c) 200
(b) 500
(d) 300
6. If all the words (with or without meaning) having five
letters, formed using the letters of the word SMALL and
arranged as in a dictionary, then the position of the
word SMALL is
(2016 Main)
(a) 46th
(c) 52nd
(b) 59th
(d) 58th
7. The letters of the word COCHIN are permuted and all
the permutations are arranged in an alphabetical order
as in an English dictionary. The number of words that
appear before the word COCHIN, is
(2007, 3M)
(a) 360
(c) 96
(b) 192
(d) 48
9. If the letters of the word ‘MOTHER’ be permuted and
all the words so formed (with or without meaning) be
listed as in dictionary, then the position of the word
‘MOTHER’ is…………
(2020 Main, 2 Sep I)
10. The number of 5 digit numbers which are divisible by 4,
with digits from the set {1, 2, 3, 4, 5} and the repetition of
(2018 Adv.)
digits is allowed, is ..................... .
11. Words of length 10 are formed using the letters A, B, C,
D, E, F, G, H, I, J. Let x be the number of such words
where no letter is repeated; and let y be the number of
such words where exactly one letter is repeated twice
y
and no other letter is repeated. Then,
=
(2017 Adv.)
9x
12. Let n be the number of ways in which 5 boys and 5 girls
can stand in a queue in such a way that all the girls
stand consecutively in the queue. Let m be the number
of ways in which 5 boys and 5 girls can stand in a queue
in such a way that exactly four girls stand consecutively
m
in the queue. Then, the value of is
n
(2015 Adv.)
13. Let n1 < n2 < n3 < n4 < n5 be positive integers such that
n1 + n2 + n3 + n4 + n5 = 20. The number of such distinct
(2014 Adv.)
arrangements (n1 , n2 , n3 , n4 , n5 ) is
Fill in the Blanks
14. Let n and k be positive integers such that n ≥
The number of solutions (x1 , x2 ,... , xk ),
x1 ≥ 1, x2 ≥ 2, ... , xk ≥ k for all integers
x1 + x2 + ... + xk = n is …
k(k + 1)
.
2
satisfying
(1996, 2M)
15. Total number of ways in which six ‘+’ and four ‘–’ signs
can be arranged in a line such that no two ‘–’signs occur
together is… .
(1988, 2M)
Analytical & Descriptive Question
16. Five balls of different colours are to be placed in three
boxes of different sizes. Each box can hold all five. In
how many different ways can we place the balls so that
no box remains empty?
(1981, 4M)
80 Permutations and Combinations
Topic 4 Distribution of Object into Group
Objective Questions I (Only one correct option)
4. The total number of ways in which 5 balls of different
1. A group of students comprises of 5 boys and n girls. If
the number of ways, in which a team of 3 students can
randomly be selected from this group such that there is
at least one boy and at least one girl in each team, is
(2019 Main, 12 April II)
1750, then n is equal to
(a) 28
(c) 25
(b) 27
(d) 24
(a) 75
having one vertex at the origin and the other two
vertices lie on coordinate axes with integral
coordinates. If each triangle in S has area 50 sq. units,
then the number of elements in the set S is
(2019 Main, 9 Jan II)
(b) 32
(d) 9
3.. From 6 different novels and 3 different dictionaries, 4
novels and 1 dictionary are to be selected and arranged
in a row on a shelf, so that the dictionary is always in
the middle. The number of such arrangements is
(2018 Main)
(a) atleast 1000
(b) less than 500
(c) atleast 500 but less than 750
(d) atleast 750 but less than 1000
(b) 150
(c) 210
(d) 243
5. The number of arrangements of the letters of the word
BANANA in which the two N’s do not appear adjacently,
is
(2002, 1M)
(a) 40
2. Let S be the set of all triangles in the xy-plane, each
(a) 36
(c) 18
colours can be distributed among 3 persons so that each
person gets at least one ball, is
(2012)
(b) 60
(c) 80
(d) 100
Analytical & Descriptive Questions
6. Using permutation or otherwise, prove that
integer, where n is a positive integer.
n2 !
is an
(n !)n
(2004, 2M)
7. In how many ways can a pack of 52 cards be
(i) divided equally among four players in order
(ii) divided into four groups of 13 cards each
(iii) divided in 4 sets, three of them having 17 cards each
and the fourth just one card?
(1979, 3M)
Integer & Numerical Answer Type Question
8. In a hotel, four rooms are available. Six persons are to
be accommodated in these four rooms in such a way that
each of these rooms contains at least one person and at
most two persons. Then the number of all possible ways
in which this can be done is ………
(2020 Adv.)
Topic 5 Dearrangement and Number of Divisors
Objective Question I (Only one correct option)
Fill in the Blank
1. Number of divisors of the form (4n + 2), n ≥ 0 of the
integer 240 is
(1998, 2M)
(a) 4
(c) 10
(b) 8
(d) 3
2. There are four balls of different colours and four boxes of
colours, same as those of the balls. The number of ways in
which the balls, one each in a box, could be placed such
that a ball does not go to a box of its own colour is.... .
(1992, 2M)
Answers
Topic 1
1. (b)
Topic 3
2. (a)
5. (c)
6. (b)
9. (d)
10. (a)
3. (b)
4. (c)
7. (c)
8. (a)
11.
9
P4 × 9 P3 (11 )!
13. (7)
Topic 2
(d)
(a)
(a)
(c)
2. (c)
3. (b)
6. (d)
7. (c)
10. (625)
11. (5)
1
2
14. (2n − k + k − 2 )
2
4. (a)
8. (495)
12. (5)
15. (35 ways)
16. (300)
2.
6.
10.
14.
3.
7.
11.
15.
(c)
(c)
(c)
(c)
n
17. (5)
5. (d)
9. (309)
12. ( A → p; B → s; C → q ; D → q )
1.
5.
9.
13.
1. (a)
18. n , ∑ ( −1 )
n
n −r n
4.
8.
12.
16.
(c)
(c)
(a)
(c)
n
Cr (r )
(b)
(a)
(b)
(30)
19. 2 − 1
21. 6062, (i) 2702 (ii) 1008
23. (64)
24. (485)
1. (c)
5. (a)
2. (a)
3. (a)
4. (b)
(52 )!
(52 )!
(52 )!
7. (i)
(ii)
(iii)
4
4
(13 !)
4 ! (13 !)
3 ! (17 ) 3
n
r =1
20. (True)
Topic 4
22. n = 3
27. (n = 9 and r = 3 )
8. (1080)
Topic 5
1. (a)
2. (9)
Hints & Solutions
Topic 1 General Arrangement
1. Following are the cases in which the 4-digit numbers
strictly greater than 4321 can be formed using digits 0,
1, 2, 3, 4, 5 (repetition of digits is allowed)
Case-I
4
3
∴ Total number of numbers = 42 + 35 = 77
5. X  X  X  X  X . The four digits 3, 3, 5, 5 can be
2
4!
= 6 ways.
2 !2 !
The five digits 2, 2, 8, 8, 8 can be arranged at (X ) places
5!
in
ways = 10 ways.
2 !3 !
Total number of arrangements = 6 × 10 = 60
[since, events A and B are independent, therefore
A ∩ B = A × B]
arranged at () places in
2/3/4/5
4 ways
4 numbers
Case-II
4
Case II Four 1’s, three 2’s
7!
Number of numbers =
= 35
4!3!
3
3/4/5 0/1/2/3/4/5
3 ways
3×6=18 numbers
6 ways
6. Distinct n-digit numbers which can be formed using
Case-III
4
4/5
0/1/2/3/4/5
2 ways
6 ways
2×6×6=72 numbers
Case-IV
digits 2, 5 and 7 are 3n . We have to find n, so that
3n ≥ 900
n− 2
⇒
3
≥ 100
⇒
n −2 ≥5
⇒ n ≥ 7, so the least value of n is 7.
7. Let n be the number of newspapers which are read by
5
6×6×6=216 numbers
the students.
Then,
0/1/2/3/4/5
⇒
6 ways
So, required total numbers = 4 + 18 + 72 + 216 = 310
2. Sum of diagonal entries of M M is ∑ a .
T
9
∑a
2
i
2
i
=5
i =1
Possibilities
9!
matrices
7!
9!
matrices
II. 1, 1, 1, 1, 1, 0, 0, 0, 0, which gives
4! × 5!
I. 2, 1, 0, 0, 0, 0, 0, 0, 0, which gives
Total matrices = 9 × 8 + 9 × 7 × 2 = 198
3. The integer greater than 6000 may be of 4 digits or
5 digits. So, here two cases arise.
Case I When number is of 4 digits.
Four-digit number can start from 6, 7 or 8.
Thus, total number of 4-digit numbers, which are
greater than 6000 = 3 × 4 × 3 × 2 = 72
Case II When number is of 5 digits.
Total number of five-digit numbers which are greater
than 6000 = 5 ! = 120
∴ Total number of integers = 72 + 120 = 192
4. There are two possible cases
Case I Five 1’s, one 2’s, one 3’s
7!
Number of numbers =
= 42
5!
60n = (300) × 5
n = 25
8. Since, a five-digit number is formed using the digits
{0, 1, 2, 3, 4 and 5} divisible by 3 i.e. only possible when
sum of the digits is multiple of three.
Case I Using digits 0, 1, 2, 4, 5
Number of ways = 4 × 4 × 3 × 2 × 1 = 96
Case II Using digits 1, 2, 3, 4, 5
Number of ways = 5 × 4 × 3 × 2 × 1 = 120
∴ Total numbers formed = 120 + 96 = 216
9. Since, the first 2 women select the chairs amongst 1 to 4
in 4 P2 ways. Now, from the remaining 6 chairs, three
men could be arranged in 6 P3.
∴ Total number of arrangements = 4 P2 × 6P3.
10. Total number of five letters words formed from ten
different letters = 10 × 10 × 10 × 10 × 10 = 105
Number of five letters words having no repetition
= 10 × 9 × 8 × 7 × 6 = 30240
∴ Number of words which have at least one letter
repeated = 105 − 30240 = 69760
11. Let the two sides be A and B. Assume that four
particular guests wish to sit on side A. Four guests who
wish to sit on side A can be accommodated on nine
chairs in 9 P4 ways and three guests who wish to sit on
side B can be accommodated in 9 P3 ways. Now, the
remaining guests are left who can sit on 11 chairs on
both the sides of the table in (11!) ways. Hence, the total
number of ways in which 18 persons can be seated
= 9P4 × 9P3 × (11)!.
82 Permutations and Combinations
12.
A. If ENDEA is fixed word, then assume this as a
single letter. Total number of letters = 5
Total number of arrangements = 5 !.
B. If E is at first and last places, then total number of
permutations = 7 !/ 2 ! = 21 × 5 !
C. If D, L, N are not in last five positions
← D, L, N, N → ← E, E, E, A, O →
4! 5!
Total number of permutations = × = 2 × 5 !
2! 3!
D. Total number of odd positions = 5
5!
Permutations of AEEEO are .
3!
Total number of even positions = 4
4!
∴ Number of permutations of N, N, D, L =
2!
5! 4!
⇒ Total number of permutations = × = 2 × 5 !
3! 2!
Topic 2 Properties of Combinational
and General Selections
1. As each section has 5 questions, so number of ways to
select 5 questions are
S1 S 2 S 3
1 1 3
1 3 1
1 2 2
2 1 2
2 2 1
0 identical + 10 distincts, number of ways = 1 ×
21
1 identical + 9 distincts, number of ways = 1 ×
C9
21
2 identicals + 8 distincts, number of ways = 1 ×
. . . . . .
. . . . . .
C9 +
⇒ C11 +
21
21
C8 + K +
C12 +
21
21
21
C 0 = x (let)
C13 + K +
21
C 21 = x
⇒
2x = 2
⇒ x=2
20
two
games
with
every
other
and the number of games played between the men and
the women = 2 × mC1 × 2C1
Now, according to the question,
⇒
C8
… (i)
… (ii)
[Q nC r = nC n − r ]
On adding both Eqs. (i) and (ii), we get
2x = 21C 0 + 21C1 + 21C 2 + K + 21C10
+ 21C11 + 21C12 + K +
21
triangle.
According to the question, we have
n ( n + 1)
+ 99 = ( n − 2 )2
2
⇒
n 2 + n + 198 = 2 [ n 2 − 4 n + 4 ]
⇒ n 2 − 9n − 190 = 0
⇒ n 2 − 19n + 10n − 190 = 0
⇒
( n − 19 )( n + 10 ) = 0
⇒
n = 19, − 10
⇒
n = 19 [Qnumber of balls n > 0]
C10
So, total number of ways in which we can choose 10
objects is
21
4. Let there are n balls used to form the sides of equilateral
∴ Number of games played by the men between
themselves = 2 × mC 2
remaining 21 are distinct, so in following ways, we can
choose 10 objects.
21
Q 20C 2 is selection of any two vertices of 20-sided polygon
which included the sides as well.
So, required number of total beams = 20C 2 − 20
[Q the number of diagonals in a n-sided closed
polygon = nC 2 − n]
20 × 19
=
− 20 = 190 − 20 = 170
2
participant plays
participant.
2. Given that, out of 31 objects 10 are identical and
C10 +
Now, the top of each pillar has been connected by beams
with the top of all its non-adjacent pillars, then total
number of beams = number of diagonals of 20-sided
polygon.
5. Since, there are m-men and 2-women and each
∴Total number of selection of 5 questions
= 3 × ( 5C1 × 5C1 × 5C 3 ) + 3 × ( 5C1 × 5C 2 × 5C 2 )
= 3(5 × 5 × 10) + 3(5 × 10 × 10)
= 750 + 1500 = 2250.
21
have been erected along the boundary of a circular
stadium.
Now, number of balls used to form an equilateral
n (n + 1)
triangle is
2
19 × 20
=
= 190.
2
3 1 1
and
3. It is given that, there are 20 pillars of the same height
21
C 21
⇒
⇒
⇒
⇒
⇒
⇒
2 mC 2 = 2 mC1 2C1 + 84
m!
= m × 2 + 42
2 !(m − 2)!
m(m − 1) = 4m + 84
m2 − m = 4m + 84
m2 − 5m − 84 = 0
m2 − 12m + 7m − 84 = 0
m(m − 12) + 7 (m − 12) = 0
m = 12
n
n
[Q m > 0]
n
6. If C 4 , C 5 and C 6 are in AP, then
2 ⋅n C 5 = n C 4 + n C 6
[If a , b, c are in AP , then 2b = a + c]
n!
n!
n!
=
+
⇒ 2
5 !(n − 5)! 4 !(n − 4)! 6 !(n − 6)!
n!
 n

Q C r = r !(n − r )! 
Permutations and Combinations 83
⇒
2
5 ⋅ 4 !(n − 5) (n − 6)!
9. Given, X has 7 friends, 4 of them are ladies and 3 are
men while Y has 7 friends, 3 of them are ladies and 4 are
men.
1
1
+
4 !(n − 4) (n − 5) (n − 6)! 6 ⋅ 5 ⋅ 4 ! (n − 6)!
2
1
1
=
+
5(n − 5) (n − 4) (n − 5) 30
2
30 + (n − 4) (n − 5)
=
5(n − 5)
30 (n − 4) (n − 5)
=
⇒
⇒
∴ Total number of required ways
= 3C 3 × 4C 0 × 4C 0 × 3C 3 + 3C 2 × 4C1 × 4C1 × 3C 2
+ 3C1 × 4C 2 × 4C 2 × 3C1 + 3C 0 × 4C 3 × 4C 3 × 3C 0
= 1 + 144 + 324 + 16 = 485
10. N i = 5C k × 4C 5 − k
⇒ 12 (n − 4) = 30 + n − 9n + 20
⇒
n 2 − 21n + 98 = 0
2
⇒
n − 14n − 7n + 98 = 0
⇒ n (n − 14) − 7(n − 14) = 0
⇒
(n − 7) (n − 14) = 0
⇒
n = 7 or 14
2
N1 = 5 × 1
N 2 = 10 × 4
N 3 = 10 × 6
N4 = 5 × 4
N5 = 1
N 1 + N 2 + N 3 + N 4 + N 5 = 126
25
7. Given, Σ { 50C r .50− r C 25− r } = K 50C 25
r=0
50 !
(50 − r )! 

50
⇒ Σ 
×
 = K C 25
r = 0  r !(50 − r )!
(25 − r )! 25 !
25 
50 !
25 ! 
50
Σ 
×
⇒
 = K C 25
r = 0  25 ! 25 !
r !(25 − r )!
[on multiplying 25 ! in
numerator and denominator.]
25
⇒
50
25
C 25 Σ
Cr = K
25
r = 0
 50
50 ! 
Q C 25 = 25 ! 25 ! 


50
C 25
25
⇒
K= Σ
Cr = 2
25
r = 0
K =2
∑
i=1
⇒
∴
C 3 − C 3 = 10
[given]
[Q C r + C r
n
n
C3
n
n
n
+1
=
n +1
Cr
+1
]
C 2 = 10 ⇒ n = 5
∴ Selection of p and q are as under
3


Ci − 1
k
 20
 =
20
21
 Ci + Ci − 1 
20
∑
20
3
 20C i − 1 
k
 21
 =
21
C

i 
(Q nC r + nC r − 1 =
∑


 20C i − 1 
k

 =
21
21
20

Ci − 1 
 i

3
20
k
 i
  =
∑


21
21
i=1
1
(21)3
⇒
⇒
− Tn =
n +1
13. Since, r , s, t are prime numbers.
i=1
⇒
+1
n
⇒
n +1
Cr )
3
⇒
Tn
n +1
⇒ C 2 + C 3 − C 3 = 10
25
20
i=1
⇒
12. Given, Tn = nC 3 ⇒ Tn + 1 =
n
8. Given,
20
(atmost one boy)
i.e. (1 boy and 3 girls) or (4 girls) = 6 C 3 ⋅4 C1 + 6C 4 …(i)
Now, selection of captain from 4 members = 4 C1 …(ii)
∴ Number of ways to select 4 members (including the
selection of a captain,
from these 4 members)
= ( 6C 3 ⋅4 C1 + 6C 4 ) 4C1
= (20 × 4 + 15) × 4 = 380
∴
25
[Q nC 0 + nC1 + n C 2 + ....+ nC n = 2n ]
⇒
11. We have, 6 girls and 4 boys. To select 4 members
1
(21)3
20
∑i
i=1
2
3
=
n
 n
Q C r =

r
n−1

Cr − 1 

k
21
k
 n (n + 1) 
=


2
 n = 20 21
2
 3
 n (n + 1)  
3
3
Q 1 + 2 + K + n = 
 
2



2
21  20 × 21
k=

 = 100
(21)3  2 
k = 100
p
r0
r1
r2
q
r2
r2
r 0 , r1 , r 2
Number of ways
1 way
1 way
3 ways
∴ Total number of ways to select, r = 5
Selection of s as under
s0
s1
s2
s3
s4
s4
s4
s4
s4
1 way
1 way
1 way
1 way
5 ways
∴ Total number of ways to select s = 9
Similarly, the number of ways to select t = 5
∴ Total number of ways = 5 × 9 × 5 = 225
5
14. Here, 47 C 4 + ∑
52 − j
C3
j =1
= 47C 4 +
= ( C4 +
47
51
C 3 + 50C 3 + 49C 3 + 48C 3 +
47
47
C3 ) + C3 + C3 + C3 +
48
49
50
C3
51
C3
[using C r + C r − 1 = n +1C r ]
n
n
84 Permutations and Combinations
= (48C 4 + 48C 3 ) + 49C 3 + 50C 3 +
50
= ( C4 + C3 ) + C3 +
51
= ( C4 + C3 ) +
51
51
C3 = C4 +
49
50
49
50
51
18. Let A = { x1 , x2 , ... , xn }
C3
∴ Number of functions from A to A is n n and out of these
C3
51
C3 = C4
52
15. Given 6 boys M1 , M 2 , M 3 , M 4 , M 5 , M 6 and
5 girls G1 , G2 , G3 , G4 , G5
(ii) α 2 → Total number of ways selecting at least 2
member and having equal number of boys and girls
5
5
5
5
5
i.e., 6C1 C1 + 6C 2 C 2 + 6C 3 C 3 + 6C 4 C 4 + 6C 5 C 5
= 30 + 150 + 200 + 75 + 6 = 461
⇒
α 2 = 461
(iii) α 3 → Total number of ways of selecting 5 members
in which at least 2 of them girls
6
6
6
6
i.e., 5C 2 C 3 + 5C 3 C 2+ 5C 4 C1 + 5C 5 C 0
= 200 + 150 + 30 + 1 = 381
α 3 = 381
(iv) α 4 → Total number of ways of selecting 4 members
in which at least two girls such that M1 and G1 are
not included together.
G1 is included → 4 C1 ⋅ 5C 2 + 4C 2 ⋅ 5C1 + 4C 3
= 40 + 30 + 4 = 74
M1 is included → 4 C 2 ⋅ 5C1 + 4C 3 = 30 + 4 = 34
G1 and M1 both are not included
4
C 4 + 4 C 3 ⋅ 5C1 + 4C 2 ⋅ 5C 2
1 + 20 + 60 = 81
∴ Total number = 74 + 34 + 81 = 189
α 4 = 189
Now, P → 4; Q → 6; R → 5; S → 2
Hence, option (c) is correct.
16. Given that, no two persons sitting adjacent in circular
arrangement, have hats of same colour. So, only
possible combination due to circular arrangement is
2 + 2 + 1.
So, there are following three cases of selecting hats are
2R + 2B + 1G or 2B + 2G + 1R or 2G + 2R + 1B.
To distribute these 5 hats first we will select a person
which we can done in 5C1 ways and distribute that hat
which is one of it’s colour. And, now the remaining four
hats can be distributed in two ways.
So, total ways will be 3 × 5C1 × 2 = 3 × 5 × 2 = 30
PLAN Number of line segment joining pair of adjacent point = n
Number of line segment obtained joining n points
on a circle = nC 2
Number of red line segments = nC 2 − n
Number of blue line segments = n
∴
⇒
⇒
n−r n
C r (r )n are onto functions.
r =1
19. The
(i) α 1 → Total number of ways of selecting 3 boys and 2
girls from 6 boys and 5 girls.
i..e, 6C 3 × 5C 2 = 20 × 10 = 200
∴
α 1 = 200
17.
n
∑ (−1)
C2 − n = n
n (n − 1)
= 2n
2
n
n =5
number of students answering exactly
k (1 ≤ k ≤ n − 1) questions wrongly is 2n − k − 2n − k − 1 . The
number of students answering all questions wrongly
is 20.
Thus, total number of wrong answers
= 1 (2n − 1 − 2n − 2 ) + 2 (2n − 2 − 2n − 3 ) + K
+ (n − 1) (21 − 20 ) + 20 ⋅ n
= 2n − 1 + 2n − 2 + 2n − 3 + K + 21 + 20 = 2n − 1
20. Let r consecutive integers be x + 1, x + 2, K , x + r.
∴ (x + 1) (x + 2) K (x + r ) =
(x + r )! r !
⋅ =
(x)! r !
=
(x + r )(x + r − 1) K (x + 1) x !
x!
x + r
C r ⋅ (r )!
Thus, (x + 1) (x + 2) K (x + r ) = x + rC r ⋅ (r )! , which is
clearly divisible by (r )! . Hence, it is a true statement.
21. Given that, there are 9 women and 8 men, a committee
of 12 is to be formed including at least 5 women.
This can be done in
= (5 women and 7 men) + (6 women and 6 men)
+ (7 women and 5 men) + (8 women and 4 men)
+ (9 women and 3 men) ways
Total number of ways of forming committee
= (9C 5 . 8C7 ) + (9C 6. 8C 6 ) + (9C7 . 8C 5 )
+ (9C 8 . 8C 4 ) + (9C 9. 8C 3 )
= 1008 + 2352 + 2016 + 630 + 56 = 6062
(i) The women are in majority = 2016 + 630 + 56
= 2702
(ii) The man are in majority = 1008 ways
22. Since, student is allowed to select at most n books out of
(2n + 1) books.
2 n +1
∴
C1 +
We know
⇒ 2(
⇒
2n +1
2n +1
2 n +1
C0 +
C0 +
2n + 1
2n +1
C1 +
C 2 + .... +
2n +1
C1 +
2n +1
2 n +1
C1 + ..... +
2n + 1
C 2 + ... +
C 2 + ... +
2n +1
C n = 63
2n +1
C2 n + 1 = 2
2n +1
... (i)
2n +1
C n ) = 22 n + 1
C n = (22 n − 1)
.... (ii)
From Eqs. (i) and (ii), we get
22 n − 1 = 63
⇒
22 n = 64
⇒
2n = 6
⇒
n =3
23. Case I
When one black and two others balls are
drawn.
Number of ways = 3C1 ⋅6 C 2 = 45
⇒
Case II When two black and one other balls are drawn
⇒
Number of ways = 3C 2 ⋅6 C1 = 18
Permutations and Combinations 85
n
Case III When all three black balls are drawn
⇒
∴
Also given,
Number of ways = C 3 = 1
Total number of ways = 45 + 18 + 1 = 64
3
⇒
24. The possible cases are
⇒
Case I A man invites 3 ladies and women invites 3
gentlemen.
Number of ways = 4C 3 ⋅4 C 3 = 16
Case II A man invites (2 ladies, 1 gentleman) and
women invites (2 gentlemen, 1 lady).
Number of ways = ( C 2 ⋅ C1 ) ⋅ ( C1 ⋅ C 2 ) = 324
4
3
3
4
Case III A man invites (1 lady, 2 gentlemen) and
women invites (2 ladies, 1 gentleman).
Number of ways = (4 C1 ⋅3 C 2 ) ⋅ (3C 2 ⋅ 4C1 ) = 144
Case IV A man invites (3 gentlemen) and women
invites (3 ladies).
Number of ways = 3C 3 ⋅3 C 3 = 1
∴ Total number of ways,
= 16 + 324 + 144 + 1 = 485
25. Since, m men and n women are to be seated in a row so
that no two women sit together. This could be shown as
× M1 × M 2 × M 3 × ... × M m ×
which shows there are (m + 1) places for n women.
∴ Number of ways in which they can be arranged
(m)! ⋅ (m + 1)!
= (m) ! m + 1 Pn =
(m + 1 − n )!
26. Let mn squares of equal size are arrange to form a
rectangle of dimension m by n. Shown as, from figure.
neighbours of x1 are {x2 , x3 , x4 , x5} x5 are {x1 , x6 , x7 } and
x7 are {x5 , x4 }.
x + x3 + x4 + x5
x + x6 + x7
x1 = 2
, x5 = 1
⇒
4
3
x + x5
and
x7 = 4
2
x + x6 + x7
∴
4x1 = x2 + x3 + x4 + 1
3
x + x5
⇒ 12x1 = 3x2 + 3x3 + 3x4 + x1 + x6 + 4
2
⇒
24x1 = 6x2 + 6x3 + 6x4 + 2x1 + 2x6 + x4 + x5
⇒ 22x1 = 6x2 + 6x3 + 7x4 + x5 + 2x6
where, x1 , x2 , x3 , x4 , x5 , x6 are all the natural numbers
and x1 is linearly expressed as the sum of x2 , x3 , x4 , x5 , x6
where sum of coefficients are equal only if, all
observations are same.
⇒
x2 = x3 = x4 = x5 = x6
⇒ All the numbers used are equal.
Cr
n−r+1
=
C r −1
r
n
27. We know that,
⇒
⇒
n
84 7 n − r + 1
= =
36 3
r
3n − 10r + 3 = 0
[given]
…(i)
Cr
84
=
C r +1 126
r+1 2
=
n−r 3
n
2n − 5r − 3 = 0
…(ii)
On solving Eqs. (i) and (ii), we get
r = 3 and n = 9
Topic 3 Multinomial, Repeated
Arrangement and Selection
1.
Key Idea Use divisibility test of 11 and consider different situation
according to given condition.
Since, the sum of given digits
0 + 1 + 2 + 5 + 7 + 9 = 24
Let the six-digit number be abcdef and to be divisible by
11, so the difference of sum of odd placed digits and sum
of even placed digits should be either 0 or a multiple of
11 means|(a + c + e) − (b + d + f )|should be either 0 or
a multiple of 11.
Hence, possible case is a + c + e = 12 = b + d + f (only)
Now, Case I
set { a , c, e} = {0, 5, 7} and set { b, d , f } = {1, 2, 9}
So, number of 6-digits numbers = (2 × 2 !) × (3 !) = 24
[Q a can be selected in ways only either 5 or 7].
Case II
Set { a , c, e} = {1, 2, 9} and set { b, d , f } = {0, 5, 7}
So, number of 6-digits numbers = 3 ! × 3 ! = 36
So, total number of 6-digits numbers = 24 + 36 = 60
2. Since there are 8 males and 5 females. Out of these 13
members committee of 11 members is to be formed.
According to the question, m = number of ways when
there is at least 6 males
= (8C 6 × 5C 5 ) + (8C7 × 5 C 4 ) + (8C 8 × 5C 3 )
= (28 × 1) + (8 × 5)+ (1 × 10) = 28 + 40 + 10 = 78
and n = number of ways when there is at least 3 females
= ( 5C 3 × 8 C 8 ) + ( 5C 4 × 8 C7 ) + ( 5C 5 × 8 C 6 )
= 10 × 1 + 5 × 8 + 1 × 28 = 78
So, m = n = 78
3. Given there are three boxes, each containing 10 balls
labelled 1, 2, 3, … , 10.
Now, one ball is randomly drawn from each boxes, and
ni denote the label of the ball drawn from the ith box,
(i = 1, 2, 3).
Then, the number of ways in which the balls can be
chosen such that n1 < n2 < n3 is same as selection of 3
different numbers from numbers {1, 2, 3, … , 10} = 10C 3
= 120.
4. Using the digits 0, 1, 3, 7, 9
number of one digit natural numbers that can be formed
= 4,
86 Permutations and Combinations
number of two digit natural numbers that can be
formed = 20,
4×5
Number of words starting with CH, CI, CN is 4! each.
Similarly, number of words before the first word
starting with CO = 4! + 4! + 4! + 4! = 96.
The word starting with CO found first in the dictionary
is COCHIN. There are 96 words before COCHIN.
(Q 0 can not come in Ist box)
8. Let the engineer visits the factory first time after x1 days
number of three digit natural numbers that can be
formed = 100
to 1 June, second time after x2 days to first visit and so
on.
∴
x1 + x2 + x3 + x4 + x5 = 11
where x1 , x5 ≥ 0 and x2 , x3 , x4 ≥ 1 according to the
requirement of the question.
Now, let x2 = a + 1, x3 = b + 1 and x4 = c + 1 where
a , b, c ≥ 0
∴New equation will be
x1 + a + b + c + x5 = 8
Now, the number of all possible ways in which the
engineer can made visits is equals to the non-negative
integral solution of equation
x1 + a + b + c + x5 = 8, and it is equal to
12 × 11 × 10 × 9
8 + 5−1
C 5 − 1 = 12C 4 =
= 495.
4 ×3 ×2
4×5× 5
and number of four digit natural numbers less than
7000, that can be formed = 250
2×5× 5×5
(Q only 1 or 3 can come in Ist box)
∴Total number of natural numbers formed
= 4 + 20 + 100 + 250 = 374
5. Number of girls in the class = 5 and number of boys in
the class = 7
9. Given word in MOTHER, now alphabetical order
Now, total ways of forming a team of 3 boys and 2 girls
= 7C 3 ⋅5 C 2 = 350
But, if two specific boys are in team, then number of
ways = 5C1 ⋅5 C 2 = 50
Required ways, i.e. the ways in which two specific boys
are not in the same team = 350 − 50 = 300.
Alternate Method
Number of ways when A is selected and B is not
= 5C 2 ⋅5 C 2 = 100
Number of ways when B is selected and A is not
= 5C 2 ⋅5 C 2 = 100
Number of ways when both A and B are not selected
= 5C 3 ⋅5 C 2 = 100
∴ Required ways = 100 + 100 + 100 = 300.
4!
6. Clearly, number of words start with A = = 12
2!
Number of words start with L = 4 ! = 24
4!
Number of words start with M =
= 12
2!
3!
Number of words start with SA =
=3
2!
Number of words start with SL = 3 ! = 6
Note that, next word will be “SMALL”.
Hence, the position of word “SMALL” is 58th.
7. Arrange the letters of the word COCHIN as in the order
of dictionary CCHINO.
Consider the words starting from C.
There are 5! such words. Number of words with the two
C’s occupying first and second place = 4 ! .
of letters is EHMORT, so number of words start with
letter.
E ------ is 5!
H ------ is 5!
M E ------ is 4!
M H ------ is 4!
M O E ------ is 3!
M O H ------ is 3!
M O R ------ is 3!
M O T E ------ is 2!
M O T H E R is 1
So, position of the word ‘MOTHER’ is
5! + 5! + 4! + 4! + 3! + 3! + 3! + 2! + 1
= 120 + 120 + 24 + 24 + 6 + 6 + 6 + 2 + 1 = 309
10. A number is divisible by 4 if last 2 digit number is
divisible by 4.
∴ Last two digit number divisible by 4 from (1, 2, 3, 4, 5)
are 12, 24, 32, 44, 52
∴ The number of 5 digit number which are divisible by
4, from the digit (1, 2, 3, 4, 5) and digit is repeated is
5 × 5 × 5 × (5 ×1) = 625
11. x = 10 !
y = 10C1 × 9C 8 ×
12. Here,
B1
10 !
10 !
y 10
= 10 × 9 ×
⇒
=
=5
2
2!
9x 2
B2
B3
B4
B5
Out of 5 girls, 4 girls are together and 1 girl is
separate. Now, to select 2 positions out of 6
…(i)
positions between boys = 6C 2
…(ii)
4 girls are to be selected out of 5 = 5C 4
Now, 2 groups of girls can be arranged in 2 !ways. …(iii)
Also, the group of 4 girls and 5 boys is arranged in 4 ! × 5 !
ways .
…(iv)
Permutations and Combinations 87
Now, total number of ways
∴
and
⇒
13.
= 6C 2 × 5C 4 × 2 ! × 4 ! × 5 !
[from Eqs. (i), (ii), (iii) and (iv)]
m = 6C 2 × 5C 4 × 2 ! × 4 ! × 5 !
n = 5! × 6!
m 6C 2 × 5C 4 × 2 ! × 4 ! × 5 ! 15 × 5 × 2 × 4 !
=
=
=5
n
6! × 5!
6 × 5 × 4!
PLAN Reducing the equation to a newer equation, where sum of
variables is less. Thus, finding the number of arrangements
becomes easier.
As,
n1 ≥ 1, n2 ≥ 2 , n3 ≥ 3, n4 ≥ 4, n5 ≥ 5
Let
n1 − 1 = x1 ≥ 0, n2 − 2 = x2 ≥ 0, ..., n5 − 5 = x5 ≥ 0
⇒ New equation will be
x1 + 1 + x2 + 2 + ... + x5 + 5 = 20
⇒
x1 + x2 + x3 + x4 + x5 = 20 − 15 = 5
x1
x2
x3
x4
x5
0
0
0
0
0
0
1
0
0
0
0
0
1
1
0
0
0
1
1
1
1
0
1
2
1
2
1
1
5
4
3
3
2
2
1
So, 7 possible cases will be there.
14. The number of solutions of x1 + x2 + ... + xk = n
= Coefficient of t n in (t + t 2 + t 3 + ... )(t 2 + t 3 + ... )...
(t k + t k +1 + ... )
n
1 + 2 + ... + k
2
= Coefficient of t in t
(1 + t + t + ... )k
k(k + 1)
[say]
Now,
1 + 2 + ... + k =
=p
2
1
and
1 + t + t 2 + ... =
1−t
Thus, the number of required solutions
= Coefficient of t n − p in (1 − t )− k
= Coefficient of t n − p in [1 + k C1 t + k +1 C 2t 2 + k + 2 C 3t 3 + ... ]
= k + n − p −1 C n − p = r C n − p
where, r = k + n − p − 1 = k + n − 1 −
1
k(k + 1)
2
1
= (2k + 2n − 2 + k2 − k)
2
1
= (2n − k2 + k − 2)
2
15. Since, six ‘+’ signs are + + + + + +
∴ 4 negative sign has seven places to be arranged in
7
C 4 ways = 35 ways
⇒
16. Since, each box can hold five balls.
∴ Number of ways in which balls could be distributed so
that none is empty, are (2, 2, 1) or (3, 1, 1).
i.e.
( 5C 2 3C 2 1C1 + 5C 3 C1 1C1 ) × 3 !
2
= (30 + 20) × 6 = 300
1. It is given that a group of students comprises of 5 boys
and n girls. The number of ways, in which a team of 3
students can be selected from this group such that each
team consists of at least one boy and at least one girls, is
= (number of ways selecting one boy and 2 girls) +
(number of ways selecting two boys and 1 girl)
= ( C1 × nC 2 ) ( C 2 × nC1 ) = 1750 [given]
n (n − 1)  5 × 4


⇒ 5 ×
× n = 1750
 +

  2

2
2
⇒ n (n − 1) + 4n = × 1750 ⇒ n 2 + 3n = 2 × 350
5
5
5
⇒ n 2 + 3n − 700 = 0 ⇒ n 2 + 28n − 25n − 700 = 0
⇒ n (n + 28) − 25(n + 28) = 0 ⇒ (n + 28) (n − 25) = 0
⇒ n = 25
x1 ≤ x2 ≤ x3 ≤ x4 ≤ x5
Now,
Topic 4 Distribution of Object into Group
[Q n ∈ N ]
2. According to given information, we have the following
figure.
Y
B
(0, b)
O
A (a, 0)
X
(Note that as a and b are integers so they can be
negative also). Here O(0, 0), A (a , 0) and B(0, b) are the
three vertices of the triangle.
Clearly, OA =| a |and OB =| b|.
1
∴Area of ∆OAB = | a ||b|.
2
But area of such triangles is given as 50 sq units.
1
|a || b| = 50
∴
2
⇒
|a || b| = 100 = 22 ⋅ 52
Number of ways of distributing two 2’s in|a |and| b| = 3
| a|
0
1
2
| b|
2
1
0
⇒ 3 ways
Similarly, number of ways of distributing two 5’s in| a |
and|b|= 3 ways.
∴ Total number of ways of distributing 2’s and 5’s
= 3 × 3 = 9 ways
Note that for one value of | a | , there are 2 possible
values of a and for one value of|b|, there are 2 possible
values of b.
∴Number of such triangles possible = 2 × 2 × 9 = 36.
So, number of elements in S is 36.
3. Given 6 different novels and 3 different dictionaries.
Number of ways of selecting 4 novels from 6 novels is
6!
6
C4 =
= 15
2 !4 !
88 Permutations and Combinations
Number of ways of selecting 1 dictionary is from 3
3!
dictionaries is 3C1 =
=3
1 !2 !
∴ Total number of arrangement of 4 novels and 1
dictionary where dictionary is always in the middle, is
15 × 3 × 4 ! = 45 × 24 = 1080
4. Objects
Distinct
Distinct
Groups
Distinct
Identical
Objects
Identical
Identical
Groups
Identical
Distinct
Description of Situation Here, 5 distinct balls are
distributed amongst 3 persons so that each gets at least
one ball. i.e. Distinct → Distinct
So, we should make cases
A B C 
A B C 
Case I
Case II


1 1 2 
1 2 2 
Number of ways to distribute 5 balls
3 ! 
3 !

=  5C1 ⋅4 C1 ⋅3 C 3 ×  +  5C1 ⋅4 C 2 ⋅2 C 2 × 

2 ! 
2 !
= 60 + 90 = 150
5. Total number of arrangements of word BANANA
6!
=
= 60
3!2!
The number of arrangements of words BANANA in
5!
which two N’s appear adjacently =
= 20
3!
Required number of arrangements = 60 − 20 = 40
6. Here, n 2 objects are distributed in n groups, each group
containing n identical objects.
∴ Number of arrangements
=
n
2
Cn .
n − n
2
Cn . n
2
− 2n
Cn .
n − 3n
2
Cn .
n − 2n
2
C n K nC n
(n 2 )!
(n 2 − n )!
n!
(n 2 )!
=
.
K
=
2
2
n ! (n − n )! n ! (n − 2n )!
n ! ⋅ 1 (n !)n
⇒ Integer (as number of arrangements has to be integer).
7.
(i) The number of ways in which 52 cards be divided
equally among four players in order
(52)!
= 52C13 × 39C13 × 26C13 × 13C13 =
(13 !)4
(ii) The number of ways in which a pack of 52 cards
can be divided equally into four groups of 13 cards
52
(52)!
C13 × 39C13 × 26C13 × 13C13
each =
=
4!
4 !(13 !)4
(iii) The number of ways in which a pack of 52 cards be
divided into 4 sets, three of them having 17 cards
each and the fourth just one card
=
52
(52)!
C17 × 35C18 × 18C17 × 1C1
=
3 !(17)3
3!
8. The groups of persons can be made only in 2, 2, 1, 1
∴ So the number of required ways is equal to number of
ways to distribute the 6 distinct objects in group sizes 1,
1, 2 and 2
6!
=
(4 !)
(2 !)2 (1 !)2 (2 !) (2 !)
= 360 × 3 = 1080.
Topic 5 Dearrangement and Number of
Divisors
1. Since, 240 = 24 .3.5
∴ Total number of divisors = (4 + 1)(2)(2) = 20
Out of these 2, 6, 10, and 30 are of the form 4n + 2.
2. The number of ways in which the ball does not go its
1
1
1
1

own colour box = 4 ! 1 − +
−
+ 

1 ! 2 ! 3 ! 4 !
1
1 1
= 4!  − +

 2 6 24
 12 − 4 + 1
= 24 
 =9


24
5
Binomial Theorem
Topic 1 Binomial Expansion and General Term
Objective Questions I (Only one correct option)
1. If the constant term in the binomial expansion of
k

 x − 2

x 
10
is 405, then|k|equals
(a) 9
(b) 1
(2020 Main, 6 Sep II)
(c) 3
(d) 2
2. If α and β be the coefficients of x and x respectively in
4
the expansion of (x +
2
x2 − 1 )6 + (x − x2 − 1 )6, then
(2020 Main, 8 Jan II)
(a) α + β = − 30
(c) α + β = 60
(b) − 126
(b) 104
(a) 100
(d) 103
(c) 10
10. The sum of the coefficients of all even degree terms is x
in the expansion of
(2019 Main, 8 April I)
x − 1 ) + (x − x − 1 ) , (x > 1) is equal to
3
6
(a) 29
3. The coefficient of x18 in the product
(a) 84
6
  1 
1
  1+ log10 x 

12
+ x  is equal to 200, and x > 1, then the
 x




(2019 Main, 8 April II)
value of x is
(x +
(b) α − β = − 132
(d) α − β = 60
(1 + x)(1 − x)10 (1 + x + x2)9 is
9. If the fourth term in the binomial expansion of
3
(b) 32
6
(c) 26
(d) 24
11. The total number of irrational terms in the binomial
(2019 Main, 12 April I)
(c) − 84
(d) 126
expansion of (71/5 − 31/10 )60 is
(a) 49
(b) 48
(2019 Main, 12 Jan II)
(c) 54
(d) 55
4. If the coefficients of x2 and x3 are both zero, in the
12. The ratio of the 5th term from the beginning to the 5th
(2019 Main, 10 April I)
term from the end in the binomial expansion of
10
 1

 3
1 
is
2 +
(2019 Main, 12 Jan I)
1


3


2(3)
expansion of the expression (1 + ax + bx2) (1 − 3x)15 in
powers of x, then the ordered pair (a , b) is equal to
(b) (− 21, 714)
(d) (− 54, 315)
(a) (28, 315)
(c) (28, 861)
5. The term independent of x in the expansion of
8
1
x
3

−  . 2x2 − 2 is equal to


x 
 60 81
(a) − 72
6
(c) − 36
(b) 36
(2019 Main, 12 April II)
(d) − 108
6. The smallest natural number n, such that the
n
1

coefficient of x in the expansion of  x2 + 3  is nC 23 , is

x 
(2019 Main, 10 April II)
(a) 35
(b) 23
(c) 58
(d) 38
7. If some three consecutive coefficients in the binomial
expansion of (x + 1) in powers of x are in the ratio 2 : 15 :
70, then the average of these three coefficients is
n
(2019 Main, 9 April II)
(a) 964
(c) 232
(b) 227
(d) 625
1
1
(a) 1 : 2(6)3
(b) 1 : 4(16)3
1
8
 x3 3
term in the binomial expansion of 
+  equals
x
3
5670 is
(2019 Main, 11 Jan I)
(a) 4
(b) 0
(c) 6
(b) 83
(d) 82
(d) 8
14. The positive value of λ for which the coefficient of x2 in
λ

the expression x2  x + 2

x 
(a) 3
(b) 5
10
is 720, is
(2019 Main, 10 Jan II)
(c) 2 2
(d) 4
15. If the third term in the binomial expansion of
(1 + xlog 2 x )5 equals 2560, then a possible value of x is
(2019 Main, 10 Jan I)
(a) 4 2
(b)
1
4
(c)
1
8
6
(a) 8−2
(c) 8
(d) 2(36)3 : 1
13. The sum of the real values of x for which the middle
8. If the fourth term in the binomial expansion of
log x 
2
7
 + x 8  (x > 0) is 20 × 8 , then the value of x is
x

(2019 Main, 9 April I)
1
(c) 4(36)3 : 1
(d) 2 2
3
 1 − t6 
 is
1−t
16. The coefficient of t 4 in the expansion of 
(2019 Main, 9 Jan II)
(a) 12
(b) 10
(c) 15
(d) 14
90 Binomial Theorem
17. The sum of the coefficients of all odd degree terms in the
expansion of  x +
(a) −1
5
5
x3 − 1 +  x − x3 − 1 , (x > 1) is
(2018 Main)
(b) 0
(c) 1
value
of
(21C1 − 10C1 ) + (21C 2 − 10C 2)
21
10
21
+ ( C3 − C3 ) + ( C 4 − 10C 4 ) + ... + (21C10 − 10C10 ) is
(2017 Main)
(c) 220 − 29
(d) 220 − 210
(d) 729
20. The sum of coefficients of integral powers of x in the
binomial expansion (1 − 2 x )
50
(a)
1 50
1
(3 + 1) (b) (350 )
2
2
(c)
is
(2015 Main)
1 50
(3 − 1)
2
(d)
1 50
(2 + 1)
2
(1 + x ) (1 + x ) (1 + x ) is
(a) 1051
22. The
term
4 12
(b) 1106

x+1
x−1 
−
 2/ 3

1 /3
x
− x + 1 x − x1 / 2
(a) 4
(2014 Adv.)
(c) 1113
independent
(b) 120
of
x
(d) 1120
in
expansion
is
(2013 Main)
(b)
n−4
5
r =1
2n
2n
r=0
r=0
2n + 1
∑ a r (x − 2)r = ∑ br (x − 3)r
show that bn =
(c) 210
(d) 310
5
n−4
(d)
Cn+ 1
(1992, 6M)
r

 1 3r
7r 15
r n
C
(
−
1
)
+
+
+
... upto m terms .

r
∑
r
2r
3r
4r
2
2
2
2
r=0

(1985, 5M)
37. Given, sn = 1 + q + q + K + q
2
(2001, 1M)
(c)
and a k = 1 , ∀ k ≥ n, then
n
(d) 12C6 + 2
(c) 12C6
(1993, 5M)
36. Find the sum of the series
the 5th and 6th terms is zero. Then, a / b equals
n−5
6
34. Prove that ∑ (−3)r − 1 3 nC 2r − 1 = 0, where k = (3n )/ 2 and n
35. If
10
24. In the binomial expansion of (a − b)n , n ≥ 5 the sum of
(a)
33. The larger of 9950 + 10050 and 10150 is ... .
of
(2003, 1M)
(b) 12C6 + 1
+ (− 1)n−113 = K
is an even positive integer.
23. Coefficient of t 24 in (1 + t 2)12 (1 + t12) (1 + t 24 ) is
(a) 12C6 + 3
32. For any odd integer n ≥ 1, n − (n − 1) + K
3
k
21. Coefficient of x in the expansion of
3 7
(1983, 2M)
3
Analytical & Descriptive Questions
11
2 4
Fill in the Blanks
31. If (1 + a x)n = 1 + 8x + 24x2 + … , then a = … and n = K .
(2016 Main)
(c) 243
272 
(d)  16,


3 
and 4th terms in the expansion of (1 + x)n are in AP,
then the value of n is… .
(1994, 2M)
n
2 4

1 − + 2 , x ≠ 0, is 28, then the sum of the

x x 
coefficients of all the terms in this expansion, is
(b) 2187
272 
(c)  14,


3 
30. Let n be a positive integer. If the coefficients of 2nd, 3rd,
19. If the number of terms in the expansion of
(a) 64
(1 + ax + bx2)(1 − 2x)18 in powers of x are both zero, then
(a , b) is equal to
251
251

(a)  16,
 (b)  14,



3 
3 
(d) 2
18. The
(a) 221 − 211 (b) 221 − 210
29. If the coefficients of x3 and x4 in the expansion of
6
n−5
25. If in the expansion of (1 + x)m (1 − x)n, the coefficients of
x and x are 3 and −6 respectively, then m is euqal to
2
n
2
Sn = 1 +
Prove that
n+ 1
n
q + 1  q + 1
 q + 1
+
 + ... + 
 ,q ≠1
 2 
 2 
2
C1 +
n+ 1
C 2 s1 +
n+ 1
C3 s2
+ ... +
n+ 1
C n+ 1 sn = 2nS n
(1984, 4M)
(1999, 2M)
(a) 6
(b) 9
(c) 12
(d) 24
26. The expression [x + (x3 − 1) 1/ 2]5 + [x − (x3 − 1)1/ 2]5 is a
polynomial of degree
(a) 5
(b) 6
(1992, 2M)
(c) 7
x 3
27. The coefficient of x in  − 2
2 x 
4
405
256
450
(c)
263
(a)
(b)
10
(1983, 1M)
504
259
(d) None of these
28. Given positive integers r > 1, n > 2 and the coefficient of
(3r )th and (r + 2)th terms in the binomial expansion of
(1980, 2M)
(1 + x)2n are equal. Then,
(a) n = 2r
(c) n = 3r
(b) n = 2r + 1
(d) None of these
38. The natural number m, for which the coefficient of x in
1

the binomial expansion of  xm + 2

x 
(d) 8
is
Integer & Numerical Answer Type Questions
22
is 1540, is …… .
[2020 Main, 5 Sep I]
39. Let m be the smallest positive integer such that
the coefficient of x2 in the expansion of
(1 + x)2 + (1 + x)3 + K + (1 + x)49 + (1 + mx)50is (3n + 1)
51
C3 for some positive integer n. Then, the value of n is
(2016 Adv.)
40. The coefficient of x9 in the expansion of
(1 + x) (1 + x2) (1 + x3 ) ... (1 + x100 ) is
(2015 Adv.)
41. The coefficients of three consecutive terms of (1 + x)n + 5
are in the ratio 5 : 10 : 14. Then, n is equal to (2013 Adv.)
Binomial Theorem 91
Topic 2 Properties of Binomial Coefficient
Objective Questions I (Only one correct option)
10. If C r stands for nC r, then the sum of the series
1. Let (x + 10)50 + (x − 10)50 = a 0 + a1x + a 2x2 + K + a50x50,
for all x ∈ R; then
a2
is equal to
a0
(a) 12.25
(c) 12.00
(2019 Main, 11 Jan II)
where n is an even positive integer, is
(b) 12.50
(d) 12.75
Cr
C0 +
20
20
Cr−1
C1 +
20
Cr− 2 20C2 + .... + 20C 020C r
20
is maximum, is
(2019 Main, 11 Jan I)
(a) 15
(c) 11
(b) 10
(d) 20
2
k
is , then k is
15
15
equal to
(2019 Main, 9 Jan I)
(a) 14
(c) 4
(b) 6
(d) 8
4. For r = 0, 1, ... , 10, if Ar, Br and C r denote respectively
the coefficient of xr in the expansions of (1 + x)10, (1 + x)20
10
and (1 + x)30. Then,
∑
Ar (B10Br − C10 Ar ) is equal to
r =1
(a) B10 − C10
(c) 0
(b)
− C10 A10 )
(d) C10 − B10
2
A10 (B10
(2010)
30 30
 30 30 30 30 30
5. 30
0  10 −  1  11 +  2  12 + K + 20 30
equal to
6. If
n −1
(b)
(d)
C10
65
C55
m
i=0
maximum when m is equal to
is
(2002, 1M)
 n + 1
(b) 2 

 r + 1
 n + 2
(c) 2 

 r 
 n + 2
(d) 

 r 
9. If a n = ∑
r=0
1
, then
Cr
n
(a) (n − 1) an
(c)
1
n an
2
n
r
∑ nC
r=0
 = 0 holds for some

∑ C k3 
k=0

n n
Ck
positive integer n. Then ∑
equals ………
k
+1
k=0
k=0
n
k
n
(2019 Adv.)
+ ... + 10(10C10 )2,
where 10C r, r ∈{1, 2,... , 10} denote binomial coefficients.
1
Then, the value of
X is .......... .
1430
(2018 Adv.)
2
10
2
10
2
14. The sum of the coefficients of the polynomial
(1 + x − 3x2)2163 is …. .
(1982, 2M)
Analytical & Descriptive Questions
15. Prove that
r
(d) None of these
∑ nC kk2 
Fill in the Blank
equals
(b) n an

n
13. Let X = ( C1 ) + 2( C 2) + 3( C3 )
 n   n 
+
 is equal to
 r − 1  r − 2
(2000, 2M)
 n + 1
(a) 

 r − 1
 n
 ∑k
k=0
det  n

n
 ∑ C kk
k = 0
10
8. For 2 ≤ r ≤ n,   + 2 
n
(a) g (m, n ) = g (n , m) for all positive integers m, n
(b) g (m, n + 1) = g (m + 1, n ) for all positive integers m, n
(c) g (2m, 2n ) = 2 g (m, n ) for all positive integers m, n
(d) g (2m, 2n ) = ( g (m, n ))2 for all positive integers m, n
12. Suppose
(b) 10
(d) 20
 n
 r
(2020 Adv.)
Then which of the following statements is/are TRUE?
(2004, 1M)
10 20  , where  p = 0 if p > q,
 
 i  m − i
 q
(a) 5
(c) 15
where for any non-negative integer p,
p  m  n + i  p + n
f (m, n , p) = Σ   
 
.
i =0  i   p   p − i 
Integer & Numerical Answer Type Questions
(b) [2, ∞ )
(d) ( 3 , 2]
∑
sum
s!

 s 
if r ≤ s,
  =  r ! (s − r )!
 r 
0
if r > s

For positive integers m and n, let
m+ n
f (m, n , p)
g (m, n ) = Σ
p = 0  n + p


 p 
60
C r = (k2 − 3) nC r + 1, then k belongs to
(a) (− ∞ , − 2]
(c) [ − 3 , 3 ]
7. The
is
(2005, 1M)
(a) 30 C11
(c) 30 C10
Objective Question II
(One or more than one correct option)
11. For non-negative integers s and r, let
403
3. If the fractional part of the number
(1986, 2M)
(b) (−1)n (n + 1)
(d) None of these
(a) (−1)n/ 2 (n + 2)
(c) (−1)n/ 2 (n + 1)
2. The value of r for which
20
 n  n
2  !  !
 2  2
[C 02 − 2 C12 + 3 C 22 − ... + (−1)n (n + 1) C n2 ],
n!
(1998, 2M)
 n  n
 n  n − 1
k − 2  n  n − 2
2k     − 2k −1   
 − ...
 +2   
 2  k − 2
 0  k
 1  k − 1
 n  n − k  n
+ (−1)k   
 =   (2003, 4 M)
 k  0   k
92 Binomial Theorem
16. For any positive integers m, n (with n ≥ m),
18. If n is a positive integer and
(1 + x + x2)n = a 0 + a1x + ... + a 2n x2n.
 n
If   = nCm. Prove that
m
 n   n − 1  n − 2
m  n + 1
 +
+ 
 + ... +   = 

m  m   m 
m m + 1
Then, show that, a 02 − a12 + ... + a 22n = a n.
(1994, 5M)
19. Prove that C 0 − 2 2 ⋅ C1 + 3 2 ⋅ C 2 − ... + (−1)n (n + 1) 2⋅ C n
= 0 , n > 2, where C r = nC r.
(1989, 5M)
20. If (1 + x) = C 0 + C1x + C 2 x + ... + C nx , then show that
or
2
n
Prove that
 n
 n − 1
 n − 2
  +2
 +3
 + ... + (n − m + 1)
m
 m 
 m 
m  n + 2
(IIT JEE 2000, 6M)
  =

m m + 2
n
the sum of the products of the Ci’s taken two at a time
represented by Σ Σ CiC j is equal to
(2n !)
(1983, 3M)
.
0 ≤ i < j ≤ n 22 n −1 −
2 (n !)2
21. Prove that C12 − 2 ⋅ C 22 + 3 ⋅ C32 − ...−2n ⋅ C 22n = (−1)n n ⋅ C n
(1979, 4M)
n
 nC 
3!
17. Prove that
= ∑ (−1)r  r + 3 r  .
2(n + 3) r = 0
Cr 

22. Prove that ( C 0 ) − ( C1 ) + ( C 2) − ... + (2 nC 2 n )2
2n
2
2n
2
2n
2
= (−1)n ⋅2n C n.
(1997C, 5M)
(1978, 4M)
Answers
Topic 1
32.
1. (c)
2. (b)
3. (a)
4. (a)
5. (c)
6. (d)
7. (c)
8. (d)
9. (c)
10. (d)
11. (c)
12. (c)
13. (b)
14. (d)
15. (b)
16. (c)
17. (d)
18. (d)
19. (d)
20. (a)
21. (c)
25. (c)
22. (c)
26. (c)
23. (d)
27. (a)
24. (b)
28. (a)
29. (d)
30. (n = 7)
1
(n + 1 ) 2 (2n − 1 )
4
33. (101 ) 50
 2mn − 1 
36.  mn n

2 (2 − 1 ) 
38. (13)
39. (5)
40. (8)
41. (n = 6 )
2. (d)
6. (d)
10. (a)
3. (d)
7. (c)
11. (a,b,d)
4. (d)
8. (d)
12. (6.20)
Topic 2
1. (a)
5. (c)
9. (c)
31. (a = 2, n = 4)
13. (646)
14. ( − 1 )
Hints & Solutions
Topic 1 Binomial Expansion and
General Term
1. Since, the general term in the expansion of binomial
k

 x − 2

x 
10
Tr + 1 =
is
10
Cr x
 10 − r 


 2 
10 − 5 r
2
(− k)r x− 2r = 10C r (− k)r x
Now, coefficient of x18 in the product
(1 + x) (1 − x)10 (1 + x + x2)9
= coefficient of x18 in the product (1 − x2) (1 − x3 )9
x2 − 1 )6 + (x − x2 − 1 )6
= 2 [ C 0x + C 2 x (x − 1) + C 4x (x − 1)
+ 6C 6 (x2 − 1)3 ]
4
The coefficients of x in above expansion
α = 2 [6C 2(−1) + 6C 42C1 (−1) + 6C 63C1 (−1)]
6
6
4
3. Given expression is (1 + x) (1 − x)10 (1 + x + x2)9
= (1 − x2) (1 − x3 )9
2. Given expression is
6
∴
α − β = − 96 − 36 = − 132
Hence, option (b) is correct.
= (1 + x) (1 − x) [(1 − x) (1 + x + x2)]9
Q Term is constant, so r = 2.
10 × 9 2
k = 405
∴ 10C 2(− k)2 = 405 ⇒
2
⇒
k2 = 9 ⇒ | k| = 3
(x +
= 2 [−15 − (15 × 2) − (1 × 3)] = − 96
and the coefficient of x2 in the expansion
β = 2 [6C 44C 0 + 6C 63C 2] = 2 [15 + 3] = 36
2
6
2
2
2
= coefficient of x18 in (1 − x3 )9
− coefficient of x16 in (1 − x3 )9
Since, (r + 1)th term in the expansion of
(1 − x3 )9 is 9C r (− x3 )r = 9C r (− 1)r x3 r
Binomial Theorem 93
= nC rx2n − 2r − 3 r = nC rx2n − 5 r
For the coefficient of x ,
…(i)
2n − 5r = 1 ⇒ 2n = 5r + 1
As coefficient of x is given as nC 23 , then either r = 23 or
n − r = 23 .
If r = 23, then from Eq. (i), we get
2n = 5(23) + 1
⇒ 2n = 115 + 1 ⇒ 2n = 116 ⇒ n = 58.
Now, for x18, 3r = 18 ⇒ r = 6
16
and for x16, 3r = 16 ⇒ r =
∉N.
3
9 ×8 × 7
9!
=
= 84
∴Required coefficient is 9C 6 =
3 ×2
6 !3 !
4. Given expression is (1 + ax + bx2)(1 − 3x)15 . In the
expansion of binomial (1 − 3x)15 , the (r + 1) th term is
Tr + 1 = 15C r (−3x)r = 15C r (−3)r xr
Now, coefficient of x2,
(1 + ax + bx2)(1 − 3x)15 is
15
⇒
in
the
expansion
C 2(−3) + a C1 (−3) + b C 0 (−3) = 0 (given)
(105 × 9) − 45 a + b = 0 ⇒ 45a − b = 945
2
15
1
15
⇒ 2n = 5n − 115 + 1 ⇒ 3n = 114 ⇒ n = 38
So, the required smallest natural number n = 38.
0
…(i)
Similarly, the coefficient of x3 , in the expansion of
(1 + ax + bx2)(1 − 3x)15 is
C3 (−3)3 + a 15C 2(−3)2 + b 15C1 (−3)1 = 0
⇒ − 12285 + 945a − 45b = 0
15
⇒
If n − r = 23, then from Eq. (i) on replacing the value of
‘r’, we get 2n = 5(n − 23) + 1
of
(given)
63a − 3b = 819 ⇒ 21a − b = 273
…(ii)
7.
Key Idea Use general term of Binomial expansion ( x + a) n i.e.
T r + 1 = nC r 1 x n − r a r
Given binomial is (x + 1)n, whose general term, is
Tr + 1 = nC r xr
According to the question, we have
C r − 1 : nC r : nC r + 1 = 2 : 15 : 70
n
From Eqs. (i) and (ii), we get
n
24a = 672 ⇒ a = 28
b = 315 ⇒ (a , b) = (28, 315)
So,
Key Idea Use the general term (or (r + 1)th term) in the
5.
⇒
expansion of binomial (a + b) n
Tr + 1 = C r a
n
i.e.
n−r
b
r
⇒
6
3

Let a binomial 2x2 − 2 , it’s (r + 1)th term

x 
= Tr + 1 = C r (2x )
 3
 − 2
 x 
6−r
12 − 2r − 2r
2 6−r
6
⇒
r
Now, the
1
x8 
− 

 60 81
r
…(i)
term independent of x in the expansion of
6
 2 3
 2x − 2

x 
= the term independent of x in the expansion of
6
1  2 3
2x − 2 + the term independent of x in the
60 
x 
6
x8  2 3 
expansion of −
 2x − 2
81 
x 
=
6
C3
(− 3)3 (2)6 − 3 x12 − 4 (3 )
60
 1
+  −  6C5 (− 3)5 (2)6 − 5 x12 − 4 (5 ) x8
 81
[put r = 3]
[put r = 5]
1
35 × 2(6)
= (− 3)3 23 +
= 36 − 72 = − 36
3
81
Tr + 1
n
1
th
 , its (r + 1) term, is

x3 
r
1
 1
= nC r (x2)n − r  3  = nC rx2n − 2r 3 r
x 
x
6. Given binomial is  x2 +
n
Cr
⇒
⇒
2
15
=
n!
(r − 1)!(n − r + 1)! 2
=
n!
15
r !(n − r )!
r
2
=
⇒ 15r = 2n − 2r + 2
n − r + 1 15
2n − 17r + 2 = 0
n
C
Similarly, n r
Cr + 1
= C r (− 3) (2)
x
6
r
6 − r 12 − 4 r
= C r (−3) (2)
x
6
Cr − 1
Now,
…(i)
n!
3
15
r !(n − r )!
=
⇒
=
n!
14
70
(r + 1)!(n − r − 1)!
r+1
3
=
⇒ 14r + 14 = 3n − 3r
n − r 14
3n − 17r − 14 = 0
…(ii)
On solving Eqs. (i) and (ii), we get
n − 16 = 0 ⇒ n = 16 and r = 2
C 2 + 16C3
3
16 + 120 + 560
696
=
=
= 232
3
3
Now, the average =
16
C1 +
log 8 x 
8. Given binomial is  + x
2
x
16
6


Since, general term in the expansion of (x + a )n is
Tr+ 1 = n C r xn− ra r
6 −3
 2
(xlog 8 x )3 = 20 × 87 (given)
T4 = T3 + 1 = 6 C3  
∴
 x
3
 2
[Q 6C3 = 20]
⇒ 20   x3 log 8 x = 20 × 87
 x

3
 log 2 x −3 

⇒ 23 x [3(log 8 x )−3 ]= (23 )7 ⇒ x 3
= (23 )6
94 Binomial Theorem
1


Q log an (x) = log a x for x > 0; a > 0, ≠ 1


n
(x +
x3 − 1 )6 + (x − x3 − 1 )6
= 2 [6C 0 − 6C 2 + 6 C 4 + 6C 4 − 1 − 3]
log x − 3 )
⇒ x( 2
= 218
= 2 [1 − 15 + 15 + 15 − 1 − 3] = 2(15 − 3) = 24
On taking log 2 both sides, we get
11. The general term in the binomial expansion of (a + b)n
(log 2 x − 3) log 2 x = 18
⇒
is Tr + 1 = nC r a n − rbr.
(log 2 x) − 3 log 2 x − 18 = 0
2
So, the general term in the binomial expansion of
(71/5 − 31/10 )60 is
⇒ (log 2 x)2 − 6 log 2 x + 3 log 2 x − 18 = 0
⇒ log 2 x(log 2 x − 6) + 3 (log 2 x − 6) = 0
⇒
log 2 x = −3, 6
1
⇒ x = 2 −3 , 2 6 ⇒ x = , 8 2
8
6
⇒
⇒
=
60
6
Cr 7
3
 2
1
1 3

C3  x1 + log10 x   x12 = 200

  



3
1
+ 

 2 (1 + log10 x ) 4 
20 × x
= 200 ⇒ x

3
1
2(1 + log x) + 4  log10 x = 1


10
3
1
+
2(1 + log10 x ) 4
= 10
⇒ [6 + (1 + log10 x)] log10 x = 4(1 + log10 x)
t 2 + 7t = 4 + 4t
t + 3t − 4 = 0
t = 1 , −4 = log10 x
10.
[let log10 x = t]
2
⇒
Since,
x = 10, 10−4
x = 10
x>1
Given expression is (x +
binomial (x + a )n is same as the (n − r + 2)th term from
the beginning in the expansion of same binomial.
T4 + 1
T5
T
= 5 =
∴Required ratio =
T10 − 5 + 2 T7 T6 + 1
⇒
T5
=
T10 − 5 + 2
x3 − 1 )6 + (x − x3 − 1 )6
= 2 [nC 0a n + nC 2a n − 2b2 + nC 4a n − 4b4 + …]}
= 2 [ C 0x + C 2x4 (x3 − 1) + 6C 4x2(x3 − 1)2 + 6C 6 (x3 − 1)3 ]
6
= 2 [ C 0x + C 2(− x ) + C 4x + C 4x + C 6 (−1 − 3x )]
4
6
8
6
2
6
6
= 2 [ C 0x − C 2x + C 4x + C 4x − 1 − 3x ]
6
6
4
6/3
[QTr + 1 = n Cr x n − r a r ]
1/3 6
2 (2(3) )
24/3 (2(3)1/3 )4
[Q 10C 4 = 10C 6 ]
= 26/3 − 4/3 (2(3)1/3 )6 − 4
/
/
= 223
⋅ 22 ⋅ 323
23
/
= 4(6) = 4(36)1/3
So, the required ratio is 4(36)1/3 : 1.
8
6
8
6
2
 x3   3 4 8 ⋅ 7 ⋅ 6 ⋅ 5 8
x
∴ 5670 = 8C4     =
1⋅ 2⋅ 3⋅ 4
 3   x
8− r

 x3 
QTr + 1 = 8Cr  
 3


8
4
⇒
x = 3 ⇒x= ± 3
r
 3 
 
 x 

So, sum of all values of x i.e + 3 and − 3 = 0
The sum of the terms with even power of x
6
 1 
C 6 (21/3 )10− 6 

 2(3)1/3 
6
10
4
{Q (a + b)n + (a − b)n
6
 1 
C 4 (21/3 )10− 4 

 2(3)1/3 
4
10
[Q Here, n = 8, which is even, therefore middle term
 n + 2
=
 th term]
 2 
+ 6C 4x2( x3 − 1 )4 + 6C 6 ( x3 − 1 )6 ]
6
r
310
3
+  , the middle term is T4 + 1.
x
3
= 2 [6C 0x6 + 6C 2x4 ( x3 − 1 )2
6
r
5
12. Since, rth term from the end in the expansion of a
 x3
Key Idea Use formula :
6
12 −
Cr 7
13. In the expansion of 
( a + b) n + ( a − b) n =
2 [ n C 0 a n + nC 2a n − 2b 2 + nC 4 a n − 4 b 4 + ...... ]
6
60
∴There are 7 rational terms in the binomial expansion
and remaining 61 − 7 = 54 terms are irrational terms.
=
(7 + log10 x) log10 x = 4 + 4 log10 x
⇒
⇒
⇒
r
310 = (−1)r
The possible non-negative integral values of ‘r’ for
r
r
and
are integer, where r ≤ 60, are
5
10
r = 0, 10, 20, 30, 40, 50, 60.
[applying log10 both sides]
⇒
60 − r
5 (−1 )r
which


 

1
1


9. Given binomial is   1 + log10 x 

+ x12
 x


Since, the fourth term in the given expansion is 200.
∴
60
(log 2 x − 6) (log 2 x + 3) = 0
⇒
C r (71/5 )60 − r (−31/10 )r
Tr + 1 =
6
Now, the required sum of the coefficients of even powers
of x in
14. The general term in the expansion of binomial
expression (a + b)n is Tr+ 1 = nC r a n− rbr, so the general
term in the expansion of binomial expression
10
λ

x2 x + 2 is

x 
Binomial Theorem 95
10 − r
r

 λ 
Tr+ 1 = x2  10C r ( x )10− r  2  =10C r x2 ⋅ x 2 λr x−2r
x  

= 10C r λr x
2+
10 − r
− 2r
2
Now, for the coefficient of x2, put 2 +
10 − r
− 2r = 2
2
10 − r
− 2r = 0
2
⇒
10 − r = 4r ⇒ r = 2
So, the coefficient of x2 is 10C 2 λ2 = 720 [given]
10 ! 2
10 ⋅ 9 ⋅ 8 ! 2
λ = 720 ⇒
λ = 720
⇒
2!8!
2⋅ 8!
⇒
45 λ2 = 720
⇒
λ2 = 16 ⇒ λ = ± 4
∴
λ =4
[λ > 0]
⇒
15. The (r + 1)th term in the expansion of (a + x)n is given
by Tr + 1 = nC ra n − rxr
∴ 3rd term in the expansion of (1 + xlog 2 x )5 is
5
⇒ 5 C 2(1)5 − 2(xlog 2 x )2 = 2560 (given)
log 2 x2log 2 x = log 2 256
(taking log 2 on both sides)
(Q log 2 256 = log 2 28 = 8)
⇒
2(log 2 x)(log 2 x) = 8
(log 2 x)2 = 4
⇒
log 2 x = ± 2
⇒ log 2 x = 2 or log 2 x = − 2
1
⇒
x = 4 or x = 2−2 =
4
3
6
1 − t 
16. Clearly, 
 = (1 − t 6 )3 (1 − t )− 3
1−t
∴ Coefficient of t 4 in (1 − t 6 )3 (1 − t )−3
= Coefficient of t 4 in (1 − t18 − 3t 6 + 3t12) (1 − t )− 3
= Coefficient of t 4 in (1 − t )− 3
= 3 + 4 − 1C 4 = 6C 4 = 15
(Q coefficient of xr in (1 − x)− n = n + r − 1C r)
We have, (x +
18. ( 21C1 − 10C1 ) + (21C 2 − 10C 2) + (21C3 − 10C3 )
+ ... + ( C10 −
21
C10 ) − (10C1 +
21
Alternate Solution
We have,
(1 − 2 x )50 = C o − C12 x + C 2( 2x)2 + ... + C50 (2 x )50 ...(i)
(1 + 2 x )50 = C o + C12 x + C 2(2 x )2 + ... + C50 (2 x )50...(ii)
On adding Eqs. (i) and (ii), we get
(1 − 2 x )50 + (1 + 2 x )50
= 2 [C 0 + C 2(2 x )2 + ... + C50 (2 x )50 ] ...(iii)
(1 − 2 x )50 + (1 + 2 x )50
2
(−1)50 + (3)50
= C 0 + C 2(2)2 + ... + C50 (2)50
2
1 + 350
= C 0 + C 2(2)2 + ... + C50 (2)50
2
21. Coefficient of xr in (1 + x)n is nC r.
2 (1 − 10 + 5 + 5) = 2
C 2 + ... +
r= 0
1
1
= [(1 + 2)50 + (1 − 2)50 ] = (350 + 1)
2
2
⇒
Sum of coefficients of all odd degree terms is
21
25
∴ Sum of coefficients = ∑ 50C 2r (2)2r
⇒
= 2(x5 + 10x6 − 10x3 + 5x7 − 10x4 + 5x)
C1 +
For the integral power of x, r should be even integer.
(1 − 2 1 )50 + (1 + 2 1 )50
= C 0 + C 2 + ... + C50 (2)50
2
5
= 2(x5 + 10x3 (x3 − 1) + 5x(x3 − 1)2)
21
Tr+ 1 = 50C r (1)50− r (−2x1/ 2)r = 50C r2r xr/ 2(−1)r
On putting x = 1, we get
x − 1 ) + (x − x − 1 ) , x > 1
3
= 2(5 C 0x5 + 5C 2x3 ( x3 − 1 )2 + 5C 4x( x3 − 1 )4 )
=(
∴
= C 0 + C 2(2 x )2 + ... + C50 (2 x )50
= 2 ( n C 0 a n + nC 2a n − 2b 2 + nC 4 a n − 4 b 4 + ...)
5
1
1
and 2 are functions of same variables, therefore
x
x
number of dissimilar terms will be 2 n + 1, i.e. odd, which is not
possible. Hence, it contains error.
⇒
Key Idea Use formula :
= ( a + b) n + ( a − b) n
3
Hence, sum of coefficients = (1 − 2 + 4)6 = 36 = 729
(1 − 2 x )50
x( 2log 2 x ) = 256
⇒
17.
n
(n + 2) (n + 1)
2 4

or n + 2C 2.
1 − + 2 is

2
x x 
1
1
[assuming and 2 distinct]
x
x
(n + 2) (n + 1)
= 28
∴
2
⇒ (n + 2) (n + 1) = 56 = (6 + 1) (6 + 2) ⇒ n = 6
20. Let Tt +1 be the general term in the expension of
10 (xlog 2 x )2 = 2560
⇒
19. Clearly, number of terms in the expansion of
Note As
C 2(1)5 − 2(xlog 2 x )2
⇒
1 21
( C1 + 21C 2 + ... + 21C 20 ) − (210 − 1)
2
1
= (21C1 + 21C 2 + ... + 21C 21 − 1) − (210 − 1)
2
1
= (221 − 2) − (210 − 1) = 220 − 1 − 210 + 1 = 220 − 210
2
=
C 2 + ... +
10
10
C10 )
10
C10 )
In this type of questions, we find different composition of
terms where product will give us x11.
Now, consider the following cases for x11 in
(1 + x2)4 (1 + x3 )7 (1 + x4 )12.
96 Binomial Theorem
Coefficient of x0 x3 x8; Coefficient of x2 x9 x0
Coefficient of x4 x3 x4; Coefficient of x8 x3 x0
= 4C 0 × 7C1 × 12C 2 + 4C1 × 7C3 × 12C 0 + 4C 2 × 7C1
× 12C1 + 4C 4 × 7C1 × 12C 0
= 462 + 140 + 504 + 7 = 1113

22. 
x+1
x
2/3
− x1/ 3 + 1
−
( x − 1) 

x − x1/ 2 
10
Therefore, the given expression is a polynomial of
degree 7.
27. The general term in  −
x
2
10
C r (x )
(− x
− 1/ 2 r
) =
C r (− 1)
10
r
tr + 1 = (−1)r
23. Here, Coefficient of t
10 − r r
−
2
x 3
12
2 12
= Coefficient of t 24 = (12C12 +
12
C6 +
12
C0 ) = 2 +
12
36
12
C6
24. Given, T5 + T6 = 0
⇒
⇒
C 4a n − 4b4 = nC5 a n − 5 b5 ⇒
n
3r = 6
r =2
10
32
28
45 × 9 405
=
=
256
256
= (−1)2. 10C 2.
tr + 2 =
C r + 1 (x)r + 1
2n
Since, binomial coefficients of t3 r and tr + 2 are equal.
∴
C3 r −1 = 2nC r + 1
2n
⇒
⇒
⇒
But
3r − 1 = r + 1 or 2n = (3r − 1) + (r + 1)
or 2n = 4r
2r = 2
or
r =1
n = 2r
r >1
In expansion of (1 + ax + bx2)(1 − 2x)18.
term containing power of x ≥ 3.
⇒
⇒
⇒
⇒
⋅ x10 − 3 r
coefficient of x r in (1 − x)n is (−1)r nCr and then simplify.
m (m − 1) 2

x +K

2

n (n − 1) 2


1 − nx +
x − ...


2

m(m − 1) n (n − 1)
= 1 + (m − n ) x +
+
− mn x2 + K


2
2
and
210 − r
29. To find the coefficient of x3 and x4, use the formula of
a nC5 n − 4
=
=
5
b nC 4
25. (1 + x)m (1 − x)n = 1 + mx +
m − n =3
C r.
∴ We take, n = 2r
C 4a n − 4b4 − nC5 a n − 5 b5 = 0
n
Now,
⇒
⇒
and
24
24
3r
10
28. In the expansion (1 + x)2n, t3 r = 2nC3 r − 1 (x)3 r − 1
= Coefficient of t in {(1 + t ) ⋅ (1 + t + t + t )}
= Coefficient of t 24 in
{(1 + t 2)12 + t12(1 + t 2)12 + t 24 (1 + t 2)12};
[neglecting t36 (1 + t 2)12]
24
10 − r
x 3
∴ Coefficient of x4 in  − 2
2 x 
in {(1 + t ) (1 + t )(1 + t )}
2 12
 x
Cr  
 2
10
10
3
 is
x2 
r
 3
r
 2 = (−1)
x 
For coefficient of x4, we put 10 − 3r = 4
For independent of x , put
10 − r r
− = 0 ⇒ 20 − 2r − 3r = 0
3
2
⇒
20 = 5r ⇒ r = 4
10 × 9 × 8 × 7
∴
T5 = 10C 4 =
= 210
4 ×3 ×2 ×1
24
= 2 [a5 + 10a3 b2 + 5ab4 ]
1/ 2 5
= 2 [x5 + 10x3 (x3 − 1) + 5x (x3 − 1)2]
 (x1/ 3 + 1)(x2/ 3 + 1 − x1/3 ) {( x )2 − 1} 
−
=

/
x ( x − 1) 
x23
− x1/3 + 1

10

( x + 1) 
1/3
− 1/ 2 10
= (x1/3 + 1) −
 = (x − x )
x 

∴ The general term is
Tr + 1 =
+ 5C3 a 2b3 + 5C 4ab4 + 5C5 b5 + 5C 0a5 − 5C1a 4b
+ 5C 2a3 b2 − 5C3 a 2b3 + 5C 4ab4 − 5C5 b5
∴ [x + (x − 1) ] + [x − (x3 − 1)1/ 2]5
 (x1/ 3 )3 + 13
{( x )2 − 1} 
−
=  23

1/3
/
x ( x − 1) 
x − x + 1
1/3 10 − r
(a + b)5 + (a − b)5 = 5C 0a5 + 5C1a 4b + 5C 2a3 b2
3
10
10
26. We know that,
…(i)
[Q coefficient of x = 3, given]
1
1
m(m − 1) + n (n − 1) − mn = − 6
2
2
m(m − 1) + n (n − 1) − 2mn = − 12
m2 − m + n 2 − n − 2mn = − 12
(m − n )2 − (m + n ) = − 12
m + n = 9 + 12 = 21
On solving Eqs. (i) and (ii), we get m = 12
Coefficient of x3 = Coefficient of x3 in (1 − 2x)18
+ Coefficient of x2 in a (1 − 2x)18
+ Coefficient of x in b(1 − 2x)18
=
C3 ⋅ 23 + a 18C 2 ⋅ 22 − b 18C1 ⋅ 2
Given, coefficient of x3 = 0
C3 ⋅ 23 + a 18C 2 ⋅ 22 − b 18C1 ⋅ 2 = 0
18 × 17 × 16
18 × 17 2
⋅8 + a ⋅
⋅ 2 − b ⋅ 18 ⋅ 2 = 0
⇒ −
3 ×2
2
34 × 16
⇒
17a − b =
3
Similarly, coefficient of x4 = 0
⇒
⇒
∴
…(ii)
18
18
18
..(i)
C 4 ⋅ 24 − a ⋅18 C3 23 + b ⋅18 C 2 ⋅ 22 = 0
32a − 3b = 240
On solving Eqs. (i) and (ii), we get
a = 16, b =
272
3
…(ii)
Binomial Theorem 97
30. Let the coefficients of 2nd, 3rd and 4th terms in the
expansion of (1 + x) is C1, C 2, C3 .
According to given condition,
n
n
n
n
2 (nC 2) = nC1 + nC3
n (n − 1)
n (n − 1) (n − 2)
2
=n+
⇒
1 ⋅2
1 ⋅2 ⋅3
(n − 1) (n − 2)
⇒
n −1 =1 +
6
n 2 − 3n + 2
n −1 =1 +
⇒
6
⇒
6n − 6 = 6 + n 2 − 3n + 2
⇒
n 2 − 9n + 14 = 0 ⇒ (n − 2) (n − 7) = 0
⇒
n =2 ⇒ n = 7
n
But C3 is true for n ≥ 3, therefore n = 7 is the answer.
3m
Also, k =
(1 + ax) = 1 + 8x + 24x + ...
n (n − 1) 2 2
a x + ... = 1 + 8x + 24x2 + ...
⇒ 1 + anx +
2!
n (n − 1)
= 24 ⇒ 8 (8 − a ) = 48
∴ an = 8 and a 2
2
⇒
8 − a =6 ⇒ a =2
a =2
and
n =4
32. Since, n is an odd integer, (− 1)n−1 = 1
C1 + (−3)
(1 + x)6 m =
C0 +
6m
(1 − x)
6m
=
C0 +
6m
[Q n − 1, n − 3, ... , are even integers]
3

 n − 1 
= Σ n3 − 16 Σ 
 
 2 


2
1  n − 1  n − 1

 n (n + 1) 
+ 1 
− 16  
=
 






2
2
2
2

2
1
1
(n + 1)2 [n 2 − (n − 1)2] = (n + 1)2(2n − 1)
4
4
33. Consider, (101)50 − (99)50 − (100)50
= (100)50 {2 [50C3 (0.01)3 +
∴
⇒
50
(101)
− {(99)
50
C3 (0.01)3 + K ] − 1}
50
C5 (0.01)5 + ... ]}
50
+ (100) } > 0
50
(101)50 > (99)50 + (100)50
34. Since, n is an even positive integer, we can write
n = 2 m ,m = 1, 2, 3, K
+
6m
C 6 m x6 m
…(ii)
C 2(− x) + K
6m
2
C 6 m (− x)6 m
6m
+
(1 + x)
6m
…(iii)
− (1 − x)
2x
6m
6m
6m
C3 x3
C5 x5 + K + 6 m C 6 m − 1x6 m − 1 ]
6m
= 6mC1 +
C3 x2 +
6m
x =y
(1 +
⇒
C5 x4 + ...
6m
+ 6 mC 6 m − 1 x6 m − 2
2
Let
y )6 m − (1 −
2 y
y )6 m
+
=
C1 +
6m
6m
C3 y
C5 y2 + K + 6 m C 6 m − 1 y 3 m − 1
6m
For the required sum we have to put y = − 3 in RHS.
S=
=
(1 +
−3 )6m − (1 − −3 )6m
2 −3
(1 + i 3 )6m − (1 − i 3 )6m
2i 3
…(iv)
z = 1 + i 3 = r (cos θ + i sin θ )
Let
⇒
r = |z| = 1 + 3 = 2
and
θ = π /3
z 6 m = [r (cos θ + i sin θ )]6 m
Now,
= r 6 m (cos 6m θ + i sin 6m θ )
z = r (cos θ − i sin θ )
Again,
(z )6 m = r 6 m (cos 6m θ − i sin 6m θ )
and
⇒
z 6 m − z 6 m = r 6 m (2i sin 6m θ )
…(v)
From Eq. (i),
S=
z 6 m − z 6 m r 6 m (2i sin 6 m θ ) 26 m sin 6 m θ
=
=
2i 3
2i 3
3
= 0 as m ∈ z , and θ = π /3
35. Let y = (x − a )m, where m is a positive integer, r ≤ m
= (100 + 1)50 − (100 − 1)50 − (100)50
= (100)50 {2 ⋅ [50C1 (0.01) +
C1 (− x) +
6m
x2 + K
(1 + x)6 m − (1 − x)6 m = 2 [6 mC1x +
Now,
= {(100)50 (1 + 0.01)50 − (1 − 0.01)50 − 1)}
…(i)
C3m − 1
On subtracting Eq. (iii) from Eq. (ii), we get
1
16 (n − 1)2(n + 1)2
= n 2 (n + 1)2 −
4
4 ×4 ×4
=
6m
C2
6m − 1
C 6 m − 1 (− x)6 m − 1 +
+
3
− 2 [(n − 1)3 + (n − 3)3 + K + 23 )]
  n − 1 3  n − 3 3

3
= Σ n3 − 2 × 23 
 +K+1 
 +
 2 
 2 


⋅
C 6 m − 1x
∴
= n + (n − 1) + (n − 2) + K + 1
3
C1x +
6m
6m
n3 − (n − 1)3 + (n − 2)3 − (n − 3)3 + K + (− 1)n−1 ⋅ 13
3
6m
+ (−3)
and n − 1, n − 3, n − 5, etc., are even integers, then
3
C3 + K
3 m − 1 6m
6m
From the binomial expansion, we write
2
n
Hence,
S = (−3)0
i.e.
⇒
31. Given,
3n 3(2m)
=
= 3m ∴ S = ∑ (−3)r − 1 ⋅6mC 2r − 1
2
2
r =1
⇒
dy
d 2y
= m(m − 1) (x − a )m − 2
= m(x − a )m − 1 ⇒
dx
dx2
d3 y
= m(m − 1)(m − 2)(m − 3)(x − a )m − 4
dx3
…………………………………
…………………………………
On differentiating r times, we get
dr y
= m(m − 1) ... (m − r + 1)(x − a )m − r
dxr
m!
=
(x − a )m − r = r !(mC r )(x − a )m − r
(m − r )!
98 Binomial Theorem
and for r > m,
2n
n +1
dr y
=0
dxr
2n
∑ a r (x − 2)r = ∑ br (x − 3)r
Now,
r=0
[given]
r=0
On differentiating both sides n times w.r.t. x, we get
2n
∑ a r (n !) C n (x − 2)
r
r−n
2n
∑ br (n !) C n (x − 3)
=
r=n
r
∑ a r (n !)rC n = (bn )n !
r=n
[since, all the terms except first on RHS become zero]
⇒
bn = nC n + n + 1C n + n + 2C n + K + 2nC n
n+3
[Q a r = 1, ∀ r ≥ n ]
C n ) + K + 2n C n
=
2n
n
∑ (−1)r nC r
r=0
n
=
∑ (−1)
Cn + 1 +
Cn =
2n
= ....
2n
Cn
2n + 1
Cn + 1

1
3
7
15
 r + 2r + 3 r + 4r + ... upto m terms 
2
2
2
2


r
r
r
 1
Cr   +
 2
r n
r=0
n+ 2
Cn + 1 + K +
r
r
n
∑ (−1)
 3
Cr   +
 4
r n
r=0
n
r
 7
∑ (−1)r nC r  8  + ... upto m terms
r=0
n
n
n
1
3
7



= 1 −  + 1 −  + 1 −  + ... upto m terms



2
4
8

∑ (−1)r nC rxr = (1 − x)n 

r= 0

using

n
n
n
n
 1
 1
 1
=   +   +   + ... upto m terms
 2
 4
 8
m

 1 
1 −  n  
mn
 1
2
 = 2 −1
=  
mn n
1
 2 
1 − n  2 (2 − 1)


2


n+ 1
C1 +
n+ 1
C 2 s1 +
n+ 1
C n+ 1sn
 1 − qr 
1
1 − qn + 1
1−q
 n +1
n+ 1

 r =1
r =1

∑ n + 1C r  1 − q  = 1 − q  ∑ n + 1C r − ∑ n + 1C r qr
r =1
C3 s2 + ... +
n+ 1
C n + 1sn = 2nS n
22
is
Tr + 1 = 22C r (xm )22 − r x−2r
Q The coefficient 22C r = 1540
⇒
r = 3 or 19
and
22m − mr − 2r = 1
at r = 3, 22m − 3m − 6 = 1 ⇒ 18m = 7
7
m=
∉N
⇒
16
Now, at r = 19, 22m − 19m − 38 = 1
⇒
3m = 39 ⇒ m = 13 ∈ N
So, the natural value of m = 13
[given]
39. Coefficient of x2 in the expansion of
{(1 + x)2 + (1 + x)3 + K + (1 + x)49 + (1 + mx)50}
⇒ 2C 2 + 3C 2 + 4C 2 + K + 49C 2 + 50C 2 ⋅ m2 = (3n + 1) ⋅51 C3
50
⇒
C3 + 50C 2m2 = (3n + 1) ⋅51 C3
[QrC r + r + 1C r + K+ nC r = n + 1C r + 1]
51 × 50 × 49
50 × 49 × 48 50 × 49
⇒
+
× m2 = (3n + 1)
3 ×2 ×1
3 ×2 ×1
2
⇒
m2 = 51n + 1
∴ Minimum value of m2 for which (51n + 1) is integer
(perfect square) for n = 5.
∴ m2 = 51 × 5 + 1 ⇒ m2 = 256 ⇒ m = 16 and n = 5
Hence, the value of n is 5.
n+ 5
where sn = 1 + q + q2 + ... + qn =
n+ 1
n+ 1
tr , tr + 1 , tr + 2 having coefficients
∑ n + 1C rsr −1,
r =1
∴
C 2 s1 +
41. Let the three consecutive terms in (1 + x)n + 5 be
n+ 1
=
n+ 1
(1 + x)(1 + x2)(1 + x3 ) K (1 + x100 ) =Terms having x9
= [199 ⋅ x9 , 198 ⋅ x ⋅ x8 , 198 ⋅ x2 ⋅ x7 , 198 ⋅ x3 ⋅ x6 ,
198 ⋅ x4 ⋅ x5 , 197 ⋅ x ⋅ x2 ⋅ x6 , 197 ⋅ x ⋅ x3 ⋅ x5 ,197 ⋅ x2 ⋅ x3 ⋅ x4]
∴ Coefficient of x9 = 8
n +1
C3 s2 + ... +
...(ii)
40. Coefficient of x9 in the expansion of
n
37.
C1 +
1

binomial  xm + 2

x 
2n
=
2n + 1 − (q + 1)n + 1
2n (1 − q)
38. The general term (i.e. (r + 1)th term) in the expansion of
r=n
= (n + 2C n + 1 +
=
From Eqs. (i) and (ii),
n+ 1
r−n
On putting x = 3, we get
36.
 q + 1
1−

 2 
=
 q + 1
1−

 2 
1
[(1 + 1)n + 1 − (1 + q)n + 1 ]
1 −q
1
…(i)
=
[2n + 1 − (1 + q)n + 1 ]
1 −q
=
2
 q + 1
 q + 1  q + 1
Also, S n = 1 + 
 + ... + 
 +
 2 
 2   2 
n
C r − 1 , n + 5C r , n + 5C r + 1.
Given, n + 5 C r − 1 : n + 5C r : n + 5C r + 1 = 5 : 10 : 14
n+ 5
n+ 5
C r + 1 14
10
Cr
and n + 5
=
∴
=
n+ 5
5
10
C r −1
Cr
⇒
n + 5 − (r − 1)
n−r+5 7
= 2 and
=
r
r+1
5
⇒
n − r + 6 =2r
and 5n − 5 r + 25 = 7r + 7
⇒
n + 6 = 3 r and 5n + 18 = 12r
n + 6 5n + 18
∴
=
3
12
⇒
4n + 24 = 5n + 18 ⇒ n = 6
Binomial Theorem 99
∴ When (16)100 is divided by 15, gives remainder 1100 = 1
Topic 2 Properties of Binomial Coefficient
and when 8(16)100 is divided by 15, gives remainder 8.
1. We have,
2403  8

= .
 15  15
(x + 10)50 + (x − 10)50 = a 0 + a1x + a 2x2 + … + a50x50
∴a 0 + a1x + a 2x2 + … + a50x50
∴
= [(50C 0x50 +
(where {⋅} is the fractional part function)
C1x49 10 +
50
C 2 x48 ⋅ 102 + … +
50
+ ( C 0x − C1x 10 + C 2 x 10 − … +
= 2 [50C 0 x50 + 50C 2x48 ⋅ 102 + 50C 4x46 ⋅ 104
50
50
50
49
50
48
2
50
C50 1050 )
50
50
⇒
C50 10 )]
+…+
C50 ⋅ 1050 ]
50
4. A r = Coefficient of xr in (1 + x)10 = 10C r
B r = Coefficient of xr in (1 + x)20 =
By comparing coefficients, we get
∴
a
50 ⋅ 49 (10)48
2(50C 2)(10)48
=2
∴ 2 =
50
a0
1 ⋅ 2 2 ⋅ (10)50
2 (10)
C 48 = C 2]
50
=
20
C0 +
C1x +
C r − 1 xr − 1 +
C rxr + ... +
20
20
20
C 20x20
∴ (1 + x)20 ⋅ (1 + x)20 = (20C 0 +
20
C 2x2 + ... +
× (20C 0 +
C r − 1 xr − 1 +
20
20
20
C rxr
+ ....+ C 20x )
⇒ (1 + x)
= ( C0 .
Cr +
20
20
20
C1
C120C r − 1 + ... +
20
40
The maximum value of
C r − 1 ...
∴
C r20C 0 ) xr + ...
20
Cr
20
C0 =
C10
Cr
C r is possible only when r = 20
C 2(15)2 + … +
100
C10 ( 30C10 − 1) − 30C10 (20C10 − 1)
20
100
C100 (15)100 )
C 2(15)2 + … +
∴
Alternate Method
2403 = 8 ⋅ 2400 = 8(16)100
Note that, when 16 is divided by 15, gives remainder 1.
C 20 ⋅30 C30
30
∑ (−1)r
30
C r (x2)r
[for coefficient of x20, put r = 10]
30
n −1
6. Given ,
⇒
C10
n −1
C r = (k2 − 3) nC r + 1
C r = (k2 − 3)
n
r+1
n −1
C r ⇒ k2 − 3 =
r+1
n
⇒
r+1
≤ 1 and n , r > 0]
n
0 < k2 − 3 ≤ 1 ⇒ 3 < k2 ≤ 4
⇒
k ∈ [−2, − 3 ) ∪ ( 3 , 2]
[since, n ≥ r ⇒
99
k =8
C 2 ⋅30 C12
30
= 30C10
C100 (15)100 )
(where {⋅} is the fractional part function)
30 30
20 30
r=0
100
C1 +......+ C100 (15) ∈ N
2403  8
2403 8 + 8 × 15λ
8
∴
=
= 8λ +
⇒ 
=
15
15
15
 15  15
30 30
 2  12
30
= 8 + 8 × 15λ
where λ =
30 30
 1  11
A = 30C 0 ⋅30 C10 − 30C1 ⋅30 C11 +
(1 + x)n = nC 0 + nC1x + nC 2x2 + … nC nxn , n ∈ N ]
100
C10 = C10 − B10
20
= Coefficient of x20 in
[By binomial theorem,
100
r =1
= (−1)10
2403 = 2400 + 3 = 8 ⋅ 2400 = 8 ⋅ (24 )100 = 8 (16)100= 8(1 + 15)100
100
r =1
= Coefficient of x20 in (1 − x2)30
3. Consider,
= 8 + 8 (100C1 (15) +
10
∑ 10C10 − r ⋅ 20C r − 30C10 ∑ 10C10 − r 10C r
= Coefficient of x20 in (1 + x)30 (1 − x)30
40
Thus, required value of r is 20.
C1 (15) +
20
−K +
[Q C n/2 is maximum when n is even]
100
r =1
10
30 30
 0  10
n
= 8 (1 +
10
5. Let A =     −     +     − ... +    
On comparing the coefficient of xr of both sides, we get
C 020C r +
10
∑ 10C10 − r 20C10 20C r − ∑ 10C10 − r 30C10 10C r
20
20
20
20
r =1
∑ 10C r 20C10 20C r − ∑ 10C r 30C10 10C r
= 30C10 −
20
20
40
=
c20x20 )
r =1
r =1
C1x +
C rxr + ... +
r =1
10
20
C r − 1 xr − 1 +
C1x + ...+
20
=
20
Cr
10
10
r =1
C 2x2 + ... +
20
Cr
30
∑ Ar (B10 Br − C10 Ar ) = ∑ Ar B10 Br − ∑ Ar C10 Ar
=
2. We know that,
=
20
10
10
r =1
50
[Q
50 × 49
5 × 49 245
=
=
= 12 .25
=
20
2 ⋅ (10 × 10)
20
20
(1 + x)
C r = Coefficient of x in
30
r
a 2 = 2 50C 48 (10)48 ; a 0 = 2 50C50 (10)50 = 2(10)50
(1 + x)20 =
k =8
m
7.
10  20 
∑  i  m − i is the coefficient of xm in the expansion of
i=0
(1 + x)10 (x + 1)20,
m
⇒
∑
i=0
10  20 
m
 
 is the coefficient of x in the
 i  m − i
expansion of (1 + x)30
100 Binomial Theorem
m
10  20  30
30
 
 = Cm =  
 i  m − i
 m
∑
i.e.
i=0
p
p=0
m+ n n+ p
n
[Q C r + C r − 1 =
9. Let b =
∑
n
r=0
n
=n
∑
=
n
∑
r=0
n
Σ
n
an
2
m+ n
m+ n
∑
C p = 2m + n
p=0
= [C 02 − C12 + C 22 − C32 + ... + (−1)nC n2 ]
− [ C12 − 2 C 22 + 3 C32 − ... + (−1)n nC n2 ]
n
−1 n
n!
n!
= (−1)n/ 2
− (−1) 2
2  n  n
 n  n
  !  !
 !  !
 2  2
 2  2
n

1 + 

2
n
Σ
n
Σ
n
C k k2
k=0
n
n
Ck . k
=0
C k3k
k=0
n
n
n (n + 1)


,
Σ nC k k = n.2 n − 1 ,
 Q k Σ= 0 k =
k=0
2

 n
n
 Σ nC k k2 = n (n + 1)2n − 2 and Σ nC k3k = 4n 
=
k
0
0
=
k


n (n + 1) n
2
2n − 3
=0
4 − n (n + 1) 2
2
⇒
C 02 − 2C12 + 3C 22 − 4C32 + ... + (−1)n (n + 1) C n2
n!
 n  n
 !  !
 2  2
=
n (n + 1)
n (n + 1)2 n − 2
=0
2
4n
n.2 n − 1
[Q nC r = nC n − r ]
10. We have,
= (−1)n/ 2
Cp
n
Σ k
⇒
= na n − b ⇒ 2b = na n ⇒ b =
C p × m + nC p
k=0
k=0
n−r
n
Cn − r
Cp
12. It is given that
Cr ]
n
= na n −
m+ n
, and
∴
g (m, n ) = g (n , m) = 2
g (m, n + 1) = 2m + n + 1 = g (m + 1, n ), and
g (2m; 2n ) = 22(m + n) = (2(m + n) )2 = ( g (m, n ))2.
n+1
n−r
1
− ∑
n
C r r = 0 nC r
r=0
Cp
m+ n
r
n − (n − r )
= ∑
n
n
Cr
Cr
r=0
∑
n+ p
f (m, n , p)
 n + p


 p 
n+ p
p=0
 n   n   n  n  
+
=  +

 r − 1  r − 2  r   r − 1 
 n   n    n + 1  n + 1  n + 2
+ 
+
+
= 
 =

 r − 1  r − 2   r   r − 1   r 
n
∑
Since, g (m, n ) =
8.   + 2 
n
(n + p)!
∑ mCi × nC p − i =
n ! p! i = 0
m+ n
 n
and we know that,   is maximum, when
 r
n

if n ∈ even.
 n
 r=2,
=
 
n±1
 r  max 
.
 r = 2 , if n ∈ odd
30
Hence,   is maximum when m = 15.
 m
 n
 r
=
…(i)
4n
4n−1
−n
=0⇒n =4
2
2
⇒
4
4
Ck
Ck
1 4 5
1
Σ C k + 1 = (25 − 1)
= Σ
=
k + 1 k=0 k + 1 5 k=0
5
n
n
∴Σ
k=0
=
1
31
= 6.20
(32 − 1) =
5
5
13. We have,
X = (10C1 )2 + 2(10C 2)2+ 3(10C3 )2 + ... + 10 (10C10 )2
 n  n
2 !  !
 2  2
[C 02 − 2 C12 + 3 C 22 − ... + (− 1)r (n + 1) C n2 ]
∴
n!
 n  n
2  !  !
 2  2
n!
(n + 2)
(−1)n/ 2
=
= (−1)n/ 2(n + 2)
n!
 n  n
2
 !  !
 2  2
⇒
X=
10
∑ r (10C r )2 ⇒ X =
r =1
⇒
X=
10
∑r×
r =1
⇒
10 9
Cr − 1
r
X = 10
10
∑ r 10C r 10C r
r =1
n
 n
Q Cr =
r

10
Cr
n −1
Cr − 1


10
∑ 9C r − 1 10C r
r =1
11. Since,
m  n + i  p + n
∑  i   p   p − i 
i=0
p
f (m, n , p) =
p
=
(n + i )!
m!
⇒
( p + n )!
i=0
=
10
∑ 9C r − 1 10C10 − r
[Q nC r = nC n − r ]
r =1
∑ i !(m − i )! × p !(n + i − p)! × ( p − i )! (n + i )!
p
X = 10
(n + p)!
1
∑ mCi ( p − i )! (n + i − p)!
p! i = 0
⇒ X = 10 ×
19
[Q
C9
C r − 1nC n − r =
C9
C9
1
10 × C 9
=
=
X=
143 11 × 13
1430
1430
19 × 17 × 16
=
= 19 × 34 = 646
8
19
Now,
n−1
19
19
2n − 1
Cn − 1 ]
Binomial Theorem 101
 n   n − 1  n − 2
m 
=  + 
+
 + ... +   
m  m   m 
m

n
2
−
n
1
−

 

m 
+
+
 + ... +   
 m   m 
m

m  n − m + 1 rows
 n − 2
...
+
+
+
 

m
 m 

......
m 
+  
m 
14. Sum of coefficients is obtained by putting x = 1
i.e.
(1 + 1 − 3)2163 = − 1
Thus, sum of the coefficients of the polynomial
(1 + x − 3x2)2163 is −1.
15. To show that
2k.n C 0.n C k − 2k − 1.n C1.n − 1 C k − 1
+ 2k − 2.n C 2.n − 2 C k − 2 − K + (−1)k nC kn − kC 0 = nC k
Taking LHS
2k.n C 0.n C k − 2k −1.n C1 ⋅n − 1 C k − 1 + K + (−1)k.n C k.n − k C 0
k
=
∑ (−1)
r. k − r . n
∑ (−1)
r
∑ (−1)
r. k − r.
2
r=0
k
=
r=0
k
=
2
k − r.
2
r=0
k
=
 n + 1
Now, sum of the first row is 
.
m + 1
C r.n − r C k − r
 n 
Sum of the second row is 
.
m + 1
n!
(n − r )!
.
r !(n − r )! (k − r )!(n − k)!
 n − 1
Sum of the third row is 
,
m + 1
…………………………
m m + 1
Sum of the last row is   = 
.
m m + 1
n!
k!
⋅
(n − k)!. k ! r !(k − r )!
∑ (−1)r. 2k − r nC k.kC r
r=0
 k
1
= 2k.n C k  ∑ (−1)r. r .k C r
2
r = 0
=2
k.n
1

C k 1 − 

2



m + 1  n + 1 + 1  n + 2
+K+
= 
 =

m + 1  m + 2  m + 2
k
[from Eq. (i) replacing n by n + 1 and m by m + 1]
= n C k = RHS
 n
m
 n − 1  n − 2
m  n + 1
+
 + ... +   = 
 …(i)
 m   m 
m m + 1
16. Let S =   + 
It is obvious that, n ≥ m.
NOTE
 n + 1  n   n − 1 
S=
+
+

m + 1 m + 1 m + 1
Thus,
[given]
This question is based upon additive loop.
m + 1 m + 1  m + 2
 n
= 
+
+ 
 + K 






m
+
m
1
m
m


 m
m + 1 
Q m = 1 = m + 1 


m + 2 m + 2
 n
=
+
 + ... +  
m + 1  m 
m
[Q nC r + nC r + 1 =
= ...............................
 n   n   n + 1
=
 +   =
 , which is true.
m + 1 m m + 1
17.
∑ (−1)r
r=0
n
= ∑ (−1)r
r=0
m m + 1 m + 2
 n
Now, S =   + 
+
 + ... +  
m  m   m 
m
m + 3
 n
=
 + ... +  
m + 1
m
n
n+1
Cr + 1 ]
n
Cr
Cr
r +3
n
n !⋅ 3 !
n!
= 3 ! ∑ (−1)r
(n − r ) ! (r + 3) !
(n − r ) ! (r + 3) !
r=0
n
(−1)r. (n + 3)!
∑ (n − r )!(r + 3)!
r=0
=
3!
(n + 1)(n + 2)(n + 3)
=
n
3!
⋅ ∑ (−1)r ⋅n + 3C r + 3
(n + 1)(n + 2)(n + 3) r = 0
=
3 ! (− 1)3
(n + 1)(n + 2)(n + 3)
n+3
∑ (−1)s ⋅n + 3 C3
s =3

n + 3
−3!
 ∑ (−1)s ⋅n + 3 C s
=

(n + 1)(n + 2)(n + 3)  s =` 0

C1 − n + 3C 2
(n + 3)(n + 2) 


0 − 1 + (n + 3) −
2!


−
…(ii)
Again, we have to prove that
 n
 n − 1
 n − 2
m  n + 2
  +2
 +3
 + ... + (n − m + 1)   = 

m
 m 
 m 
m m + 2
 n
 n − 1
 n − 2
m
Let S1 =   + 2 
 +3
 + ... + (n − m + 1)  
m
 m 
 m 
m
n+3
C0 +
n+3
=
−3!
(n + 1)(n + 2)(n + 3)
=
3!
−3!
(n + 2)(2 − n − 3)
=
⋅
2(n + 3)
(n + 1)(n + 2)(n + 3)
2
18. (1 + x + x2)n = a 0 + a1x + K + a 2nx2n
…(i)
Replacing x by −1 / x, we get
n
a 2n
a
a
a
1 1

1 − + 2 = a 0 − 1 + 22 − 33 + K + 2n

x x 
x
x
x
x
…(ii)
102 Binomial Theorem
Now, a 02 − a12 + a 22 − a32 + K + a 22n = coefficient of the
term independent of x in
[a 0 + a1x + a 2x + K + a 2nx ]
a 2n 

a
a
× a 0 − 1 + 22 − K + 2n 
x
x
x 

2
2n
= Coefficient of the term independent of x in
n
1 1

(1 + x + x2)n 1 − + 2

x x 
n
1
1


Now, RHS = (1 + x + x2)n 1 − + 2

x x 
(1 + x + x2)n (x2 − x + 1)n [(x2 + 1)2 − x2]n
=
=
x2 n
x2 n
2
4
2 n
2
4 n
(1 + 2x + x − x )
(1 + x + x )
=
=
2n
x
x2n
2
2
2
2
2
Thus, a 0 − a1 + a 2 − a3 + K+ a 2n
= Coefficient of the term independent of x in
1
(1 + x2 + x4 )n
x2n
= Coefficient of x2n in (1 + x2 + x4 )n
= Coefficient of t n in (1 + t + t 2)n = a n
=
=
n
∑ (−1) r(r + 1)2 nC r =
r=0
n
∑ (−1)r (r 2 + 2r + 1) nC r
n
∑ (−1)
n
r=0
r (r − 1) ⋅ C r + 3 ⋅ ∑ (−1) r ⋅ nC r
n
r.
r=0
+
n
∑ (−1)
r n
Cr
r=0
=
n
∑ (−1)r n (n − 1) n − 2C r − 2
r=2
n
+ 3 ∑ (−1)r n . n − 1C r − 1
r =1
+
n
∑ (−1)r nC r
r=0
= n (n − 1){ n − 2C 0 − n − 2C1 + n − 2C 2−... + (−1)n n − 2C n − 2}
+ 3n { − n−1C 0 + n − 1C1 − n −1C 2 + ...+ (−1)n n − 1C n − 1}
+ { nC 0 − nC1 + nC 2 + ... + (−1)n nC n }
= n (n − 1) . 0 + 3n . 0 + 0, ∀n > 2 = 0, ∀n > 2
20. We know that,
n
2 ∑ ∑ Ci C j = ∑
0 ≤i < j ≤ n
i=0
=
n
n
n
∑ Ci C j − ∑ ∑ CiC j
j=0
i = 0 j=0
n
n
n
i=0
j=0
i=0
∑ Ci ∑ C j − ∑ Ci2
0 ≤i< j≤n
(2n )!
22 n − 2 nC n
= 22 n−1 −
2
2 (n !)2
21. We know that, (1 + x)2 n = C 0 + C1x + C 2 x2 + ... + C 2 n x 2 n
On differentiating both sides w.r.t. x, we get
2n (1 + x)2 n−1 = C1 + 2 ⋅ C 2 x + 3 ⋅ C3 x2
+ ... + 2nC 2n x2 n−1 …(i)
2n
1
1
1
1

and 1 −  = C 0 − C1 ⋅ + C 2 ⋅ 2 − C3 ⋅ 3


x
x
x
x
1
…(ii)
+ ... + C 2n ⋅ 2 n
x
On multiplying Eqs. (i) and (ii), we get
1

2n (1 + x)2 n−1 1 − 

x
2n
= [C1 + 2 ⋅ C 2x + 3 ⋅ C3 x2 + ... + 2n ⋅ C 2n x2n−1 ]

 1
 1
 1 
× C 0 − C1   + C 2 2 − ..... + C 2 n  2 n  




x 
x
x

1
 1 
in 2n  2 n  (1 + x)2 n−1 (x − 1)2 n
x 
x
= 2n (−1)n−1 ⋅ (2n − 1) C n−1 (−1)
(2n )!
(2n − 1)!
= (−1)n (2n )
= (−1)n n
⋅n
(n − 1)! n !
(n !)2
r=0
r=0
r.
CiC j =
= Coefficient of x2 n−1 in 2n (1 − x2)2 n−1 (1 − x)
r=0
∑ (−1)r r 2 ⋅ nC r + 2 ∑ (−1)r r ⋅ nC r + ∑ (−1)r. nC r
r=0
n
∑∑
= Coefficient of
n
n
∴
 1
Coefficient of   on the LHS
 x
19. C 0 − 22 ⋅ C1 + 32 ⋅ C 2 − ... + (−1)n (n + 1)2 ⋅ C n
=
= 2n 2n − (2 nC n ) = 22 n − 2 nC n
= − (−1)n n ⋅ C n
 1
Again, the coefficient of   on the RHS
 x
= − (C12 − 2 ⋅ C 22 + 3 ⋅ C32 − ... − 2n C 22n )
…(iii)
…(iv)
From Eqs. (iii) and (iv),
C12 − 2 ⋅ C 22 + 3 ⋅ C32 − ... − 2n ⋅ C 22n = (−1)n n ⋅ C n
1
22. (1 + x)2 n 1 − 

2n
x
= [ C 0 + (2nC1 )x + (2nC 2)x2 + ...+ (2 nC 2 n )x2 n ]
1
1
1 

× 2 nC 0 − (2 nC1 ) + (2 nC 2) 2 + ... + (2 nC 2n ) 2n

x
x
x 
2n
Independent terms of x on RHS
= (2nC 0 )2 − (2nC1 )2 + (2nC 2)2 − ...+ (2nC 2n )2
 x − 1
LHS = (1 + x)2n 

 x 
2n
=
1
(1 − x2)2n
x2n
Independent term of x on the LHS = (−1)n ⋅2n C n.
6
Probability
Topic 1 Classical Probability
Objective Questions I (Only one correct option)
1. A person throws two fair dice. He wins
` 15 for throwing a doublet (same numbers on the two
dice), wins ` 12 when the throw results in the sum of
9, and loses ` 6 for any other outcome on the throw.
Then, the expected gain/loss (in `) of the person is
(2019 Main, 12 April II)
(a)
1
gain
2
(b)
1
loss
4
(c)
1
loss
2
(d) 2 gain
2. In a random experiment, a fair die is rolled until two
fours are obtained in succession. The probability that
the experiment will end in the fifth throw of the die is
(2019 Main, 12 Jan I)
equal to
(a)
(c)
175
(b)
65
200
(d)
65
225
65
150
(2019 Main, 12 April I)
3
(c)
10
3
(d)
20
4. Let S = {1, 2, K , 20}. A subset B of S is said to be
“nice”, if the sum of the elements of B is 203. Then,
the probability that a randomly chosen subset of S is
‘‘nice’’, is
(2019 Main, 11 Jan II)
(a)
6
220
(b)
4
220
(c)
7
220
(d)
5
220
5. If two different numbers are taken from the set {0, 1,
2, 3, …, 10}, then the probability that their sum as
well as absolute difference are both multiple of 4, is
(2017 Main)
(a)
6
55
(b)
12
55
(c)
14
45
(a)
55  2 
 
3  3
11
2
(b) 55  
 3
10
1
(c) 220  
 3
12
1
(d) 22  
 3
11
8. Three boys and two girls stand in a queue. The probability
that the number of boys ahead of every girl is atleast one
more that the number of girls ahead of her, is (2014 Adv.)
(a) 1 /2
(b) 1 /3
(c) 2 /3
(d) 3 /4
9. Four fair dice D1 , D2, D3 and D4 each having six faces
numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously. The
probability that D4 shows a number appearing on one of
(2012)
D1 , D2 and D3 , is
91
216
(b)
108
216
(c)
125
216
(d)
127
216
10. Let ω be a complex cube root of unity with ω ≠ 1. A fair die is
65
chosen at random, then the probability that the
triangle formed with these chosen vertices is
equilateral is
1
(b)
5
then the probability that one of the boxes contains excatly
(2015 Main)
3 balls, is
(a)
3. If there of the six vertices of a regular hexagon are
1
(a)
10
7. If 12 identical balls are to be placed in 3 different boxes,
(d)
7
55
thrown three times. If r1, r2 and r3 are the numbers obtained
on the die, then the probability that ω r 1 + ω r2 + ω r3 = 0, is
(2010)
(a) 1/18
(b) 1/9
(c) 2/9
(d) 1/36
11. If three distinct numbers are chosen randomly from the
first 100 natural numbers, then the probability that all
three of them are divisible by both 2 and 3, is (2004, 1M)
(a)
4
55
(b)
4
35
(c)
4
33
(d)
4
1155
12. Two numbers are selected randomly from the set
S = {1, 2, 3, 4, 5, 6} without replacement one by one. The
probability that minimum of the two numbers is less than
(2003, 1M)
4, is
(a) 1/15
(b) 14/15
(c) 1/5
(d) 4/5
13. If the integers m and n are chosen at random between 1
and 100, then the probability that a number of the form
(1999, 2M)
7m + 7n is divisible by 5, equals
(a)
1
4
(b)
1
7
(c)
1
8
(d)
1
49
6. Three randomly chosen non-negative integers x, y
14. Seven white balls and three black balls are randomly
and z are found to satisfy the equation x + y + z = 10.
(2017 Adv.)
Then the probability that z is even, is
placed in a row. The probability that no two black balls are
placed adjacently, equals
(1998, 2M)
(a)
1
2
(b)
36
55
(c)
6
11
(d)
5
11
(a)
1
2
(b)
7
15
(c)
2
15
(d)
1
3
104 Probability
15. Three of the six vertices of a regular hexagon are chosen
at rondom. The probability that the triangle with three
vertices is equilateral, equals
(1995, 2M)
(a) 1/2
(b) 1/5
(c) 1/10
(d) 1/20
16. Three identical dice are rolled. The probability that the
same number will appear on each of them, is (1984, 2M)
(a)
1
6
(b)
1
36
(c)
1
18
(d)
3
28
17. Fifteen coupons are numbered 1, 2, ..., 15, respectively.
Seven coupons are selected at random one at a time
with replacement. The probability that the largest
number appearing on a selected coupon is 9, is
9
(a)  
 16 
6
8
(b)  
 15 
7
3
(c)  
 5
7
(d) None of these
For the following questions, choose the correct answer
from the codes (a), (b), (c) and (d) defined as follows.
(a) Statement I is true, Statement II is also true;
Statement II is the correct explanation of Statement I
(b) Statement I is true, Statement II is also true;
Statement II is not the correct explanation of
Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
18. Consider the system of equations
ax + by = 0, cx + dy = 0,
where
a , b, c, d ∈ {0, 1}.
Statement I The probability that the system of
equations has a unique solution, is 3/8.
Statement II The probability that the system of
equations has a solution, is 1.
(2008, 3M)
Passage Based Problems
Passage
Box I contains three cards bearing numbers 1, 2, 3 ; box II
contains five cards bearing numbers 1, 2, 3, 4, 5; and box III
contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card
is drawn from each of the boxes. Let xi be the number on the
card drawn from the i th box i = 1, 2, 3.
(2014 Adv.)
19. The probability that x1 + x2 + x3 is odd, is
29
105
(b)
53
105
(c)
57
105
(d)
1
2
20. The probability that x1 , x2 and x3 are in an arithmetic
progression, is
9
(a)
105
10
(b)
105
21. Three faces of a fair die are yellow, two faces red and
one face blue. The die is tossed three times. The
probability that the colours, yellow, red and blue,
appear in the first, second and the third tosses
respectively, is…… .
(1992, 2M)
1 + 3p 1 − p
1 − 2p
22. If
are the probabilities of three
and
,
3
4
2
mutually exclusive events, then the set of all values of p
is… .
(1986, 2M)
23. A determinant is chosen at random from the set of all
determinants of order 2 with elements 0 or 1 only. The
probability that the value of the determinant chosen is
positive, is… .
(1982, 2M)
True/False
Assertion and Reason
(a)
Fill in the Blanks
11
(c)
105
7
(d)
105
24. If the letters of the word ‘ASSASSIN’ are written down
at random in a row, the probability that no two S’s occur
together is 1/35.
Analytical and Descriptive Questions
25. An unbiased die, with faces numbered 1, 2, 3, 4, 5 and 6
is thrown n times and the list of n numbers showing up
is noted. What is the probability that among the
numbers 1, 2, 3, 4, 5 and 6 only three numbers appear in
(2001, 5M)
this list?
26. If p and q are chosen randomly from the set {1, 2, 3, 4, 5,
6, 7, 8, 9 and 10} with replacement, determine the
probability that the roots of the equation x2 + px + q = 0
(1997, 5M)
are real.
27. In how many ways three girls and nine boys can be
seated in two vans, each having numbered seats, 3 in
the front and 4 at the back? How many seating
arrangements are possible if 3 girls should sit together
in a back row on adjacent seats? Now, if all the seating
arrangements are equally likely, what is the probability
of 3 girls sitting together in a back row on adjacent
(1996, 5M)
seats?
28. A box contains 2 fifty paise coins, 5 twenty five paise
coins and a certain fixed number n (≥ 2) of ten and five
paise coins. Five coins are taken out of the box at
random. Find the probability that the total value of
these 5 coins is less than one rupee and fifty paise.
(1988, 3M)
29. Six boys and six girls sit in a row at random. Find the
probability that
(i) the six girls sit together.
(ii) the boys and girls sit alternatively.
(1978, 3M)
Probability 105
Topic 2 Addition and Subtraction Law of Probability
Objective Questions I (Only one correct option)
1. The probabilities of three events A , B and C are given
by P ( A ) = 0.6, P (B) = 0.4 and P (C ) = 0.5. If
P ( A ∪ B) = 0.8, P ( A ∩ C ) = 0.3, P ( A ∩ B ∩ C ) = 0.2,
and
where
P (B ∩ C ) = β
P(A ∪ B ∪ C ) = α ,
0.85 ≤ α ≤ 0.95, then β lies in the interval
(2020 Main, 6 Sep II)
(a) [ 0.35, 0.36]
(c) [0.20, 0.25]
(b) [0.25, 0.35]
(d) [0.36, 0.40]
2. For three events A , B and C, if P (exactly one of A or B
occurs) = P(exactly one of B or C occurs) = P (exactly
1
one of C or A occurs) = and P (all the three events
4
1
occur simultaneously) = , then the probability that
16
atleast one of the events occurs, is
(2017 Main)
(a)
7
32
(b)
7
16
3
4
then P (B ∩ C ) is equal to
(c)
3. If P (B) = , P ( A ∩ B ∩ C ) =
(a)
1
12
(b)
1
6
7
64
3
16
(d)
1
1
and P ( A ∩ B ∩ C ) = ,
3
3
(2002, 3M)
(c)
1
15
(d)
1
9
4. If E and F are events with P (E ) ≤ P (F ) and
P (E ∩ F ) > 0, then which one is not correct? (1998, 2M)
(a) occurrence of E ⇒ occurrence of F
(b) occurrence of F ⇒ occurrence of E
(c) non-occurrence of E ⇒ non-occurrence of F
(d) None of the above
3p + 2p
2
p + 3p2
(c)
2
p + 3p
4
3p + 2p2
(d)
4
(b)
6. If 0 < P ( A ) < 1, 0 < P (B) < 1 and P ( A ∪ B) = P ( A )
(1995, 2M)
(a) P (B / A ) = P (B ) − P (A )
(b) P (A ′ − B ′ ) = P (A ′ ) − P (B ′ )
(c) P (A ∪ B ) ′ = P (A ) ′ P (B )′
(d) P (A / B ) = P (A ) − P (B )
occurs is 0.6. If A and B occur simultaneously with
probability 0.2, then P ( A ) + P (B ) is equal to
(1987, 2M)
(b) 0.8
(c) 1.2
Objective Questions II
(One or more than one correct option)
9. For two given events A and B, P ( A ∩ B) is
(1988, 2M)
(a) not less than P (A ) + P (B ) − 1
(b) not greater than P (A ) + P (B )
(c) equal to P (A ) + P (B ) − P (A ∪ B )
(d) equal to P (A ) + P (B ) + P (A ∪ B )
10. If M and N are any two events, then the probability that
exactly one of them occurs is
(1984, 3M)
(a) P (M ) + P (N ) − 2 P (M ∩ N )
(b) P (M ) + P (N ) − P (M ∪ N )
(c) P (M ) + P (N ) − 2 P (M ∩ N )
(d) P (M ∩ N ) − P (M ∩ N )
Fill in the Blanks
11. Three
numbers are chosen at random without
replacement from {1, 2,…, 10}. The probability that the
minimum of the chosen number is 3, or their maximum is
7, is … .
(1997C, 2M)
(1985, 2M)
13. If the probability for A to fail in an examination is 0.2 and
that of B is 0.3, then the probability that either A or B fails
is 0.5.
(1989, 1M)
Analytical and Descriptive Questions
14. In a certain city only two newspapers A and B are
published, it is known that 25% of the city population
reads A and 20% reads B, while 8% reads both A and B. It
is also known that 30% of those who read A but not B look
into advertisements and 40% of those who read B but not
A look into advertisements while 50% of those who read
both A and B look into advertisements. What is the
percentage of the population reads an advertisement?
(1984, 4M)
15. A, B, C are events such that
Pr ( A ) = 0.3, Pr (B) = 0.4, Pr (C ) = 0.8,
Pr ( AB) = 0.08, Pr ( AC ) = 0.28 and Pr ( ABC ) = 0.09
7. The probability that at least one of the events A and B
(a) 0.4
(b) 0.25
(d) None of these
True/False
2
+ P (B) − P ( A ) P (B), then
(a) 0.39
(c) 0.11
P ( A ) and P (B) is… .
events A or B occurs) = P(exactly one of the events B
or C occurs) = P(exactly one of the events C or A
occurs) = p and P(all the three events occurs
1
simultaneously) = p2, where 0 < p < . Then, the
2
probability of atleast one of the three events A, B and
C occurring is
(1996, 2M)
(a)
respectively. The probability that both A and B occur
simultaneously is 0.14. Then, the probability that neither
A nor B occurs, is
(1980, 1M)
12. P ( A ∪ B) = P ( A ∩ B) if and only if the relation between
5. For the three events A, B and C, P(exactly one of the
2
8. Two events A and B have probabilities 0.25 and 0.50,
(d) 1.4
If Pr ( A ∪ B ∪ C ) ≥ 0.75, then show that Pr (BC ) lies in the
(1983, 2M)
interval [ 0.23, 0.48 ].
16. A and B are two candidates seeking admission in IIT. The
probability that A is selected is 0.5 and the probability
that both A and B are selected is atmost 0.3. Is it possible
that the probability of B getting selected is 0.9? (1982, 2M)
106 Probability
Pragraph Based Questions
There are five students S1 , S 2, S3 , S 4 and S5 in a music class
and for them there are five seats R1 , R2, R3 , R4and R5
arranged in a row, where initially the seat Ri is allotted to the
student Si , i = 1, 2, 3, 4, 5. But, on the examination day, the
five students are randomly allotted the five seats.
(There are two questions based on Paragraph, the question
given below is one of them)
(2018 Adv.)
17. The probability that, on the examination day, the
NONE of the remaining students gets the seat
previously allotted to him/her is
(a)
3
40
(b)
1
8
(c)
7
40
(d)
1
5
18. For i = 1, 2, 3, 4, let Ti denote the event that the students
Si and Si+1 do NOT sit adjacent to each other on the day
of the examination. Then, the probability of the event
T1 ∩ T2 ∩ T3 ∩ T4 is
(a)
student S1 gets the previously allotted seat R1, and
1
15
(b)
1
10
(c)
7
60
(d)
1
5
Topic 3 Independent and Conditional Probability
Objective Questions I (Only one correct option)
1. Assume that each born child is equally likely to be a boy
or a girl. If two families have two children each, then the
conditional probability that all children are girls given
that at least two are girls; is
(2019 Main, 10 April I)
(a)
1
17
(b)
1
12
(c)
1
10
(d)
1
11
2. Four persons can hit a target correctly with
1 1 1
1
, , and respectively. If all hit at the
2 3 4
8
target independently, then the probability that the
target would be hit, is
(2019 Main, 9 April I)
probabilities
(a)
1
192
(b)
25
32
(c)
7
32
(d)
25
192
3. Let A and B be two non-null events such that A ⊂ B.
Then, which of the following statements is always
correct.
(2019 Main, 8 April I)
(a) P (A /B ) = P (B ) − P (A )
(c) P (A/B ) ≤ P (A )
(b) P (A/B ) ≥ P (A )
(d) P (A/B ) = 1
4. Two integers are selected at random from the set { 1, 2,
…… , 11}. Given that the sum of selected numbers is
even, the conditional probability that both the numbers
are even is
(2019 Main, 11 Jan I)
(a)
2
5
(b)
1
2
(c)
7
10
(d)
3
5
5. An unbiased coin is tossed. If the outcome is a head,
then a pair of unbiased dice is rolled and the sum of the
numbers obtained on them is noted. If the toss of the
coin results in tail, then a card from a well-shuffled
pack of nine cards numbered 1, 2, 3, …, 9 is randomly
picked and the number on the card is noted. The
probability that the noted number is either 7 or 8 is
(2019 Main, 10 Jan I)
(a)
15
72
(b)
13
36
(c)
19
72
(d)
19
36
6. Let two fair six-faced dice A and B be thrown
simultaneously. If E1 is the event that die A shows up
four, E 2 is the event that die B shows up two and E3 is
the event that the sum of numbers on both dice is odd,
then which of the following statements is not true?
(2016 Main)
(a) E1 and E2 are independent
(b) E2 and E3 are independent
(c) E1 and E3 are independent
(d) E1 , E2 and E3 are independent
1
6
7. Let A and B be two events such that P ( A ∪ B) = ,
1
1
and P ( A ) = , where A stands for the
4
4
complement of the event A. Then , the events A and B
(2014 Main)
are
P ( A ∩ B) =
(a) independent but not equally likely
(b) independent and equally likely
(c) mutually exclusive and independent
(d) equally likely but not independent
8. Four persons independently solve a certain problem
1 3 1 1
, , , . Then, the
2 4 4 8
probability that the problem is solved correctly by
(2013 Adv.)
atleast one of them, is
correctly with probabilities
235
256
3
(c)
256
21
256
253
(d)
256
(a)
(b)
9. An experiment has 10 equally likely outcomes. Let A
and B be two non-empty events of the experiment. If A
consists of 4 outcomes, then the number of outcomes
that B must have, so that A and B are independent, is
(a) 2, 4 or 8
(c) 4 or 8
(b) 3, 6 or 9
(d) 5 or 10
(2008, 3M)
10. Let E c denotes the complement of an event E. If E, F, G
are pairwise independent events with P (G ) > 0 and
P (E ∩ F ∩ G ) = 0 . Then, P (E c ∩ F c|G ) equals(2007, 3M)
(a) P (E c ) + P (F c )
(c) P (E c ) − P (F )
(b) P (E c ) − P (F c )
(d) P (E ) − P (F c )
11. One Indian and four American men and their wives are
to be seated randomly around a circular table. Then, the
conditional probability that Indian man is seated
adjacent to his wife given that each American man is
seated adjacent to his wife, is
(2007, 3M)
(a)
1
2
(b)
1
3
(c)
2
5
(d)
1
5
Probability 107
12. A fair die is rolled. The probability that the first time
1 occurs at the even throw, is
(a) 1/6
(b) 5/11
(2005, 1M)
(c) 6/11
(d) 5/36
13. There are four machines and it is known that exactly
two of them are faulty. They are tested, one by one, in
a random order till both the faulty machines are
identified. Then, the probability that only two tests
are needed, is
(1998, 2M)
(a)
1
3
(b)
1
6
(c)
1
2
(d)
1
4
14. A fair coin is tossed repeatedly. If tail appears on first
four tosses, then the probability of head appearing on
fifth toss equals
(1998, 2M)
(a)
1
2
(b)
1
32
(c)
31
32
(d)
1
5
15. If from each of the three boxes containing 3 white and
1 black, 2 white and 2 black, 1 white and 3 black balls,
one ball is drawn at random, then the probability that
2 white and 1 black balls will be drawn, is
(1998, 2M)
(a)
13
32
(b)
1
4
(c)
1
32
(d)
3
16
16. The probability of India winning a test match against
West Indies is 1/2. Assuming independence from
match to match the probability that in a 5 match
series India’s second win occurs at third test, is
(1995, 2M)
(a) 1/8
(b) 1/4
(c) 1/2
(d) 2/3
17. An unbiased die with faces marked 1, 2, 3, 4, 5 and 6 is
rolled four times. Out of four face values obtained, the
probability that the minimum face value is not less
than 2 and the maximum face value is not greater
than 5, is
(1993, 1M)
(a) 16/81
(b) 1/81
(c) 80/81
(d) 65/81
18. A student appears for tests I, II and III. The student is
successful if he passes either in tests I and II or tests I
and III. The probabilities of the student passing in
1
tests I, II and III are p, q and , respectively. If the
2
1
probability that the student is successful, is , then
2
1
2
(a) p = q = 1
(b) p = q =
(c) p = 1, q = 0
(d) p = 1, q =
(1986, 2M)
1
2
P ( A ) > 0, and P (B) ≠ 1, then P ( A / B ) is equal to
(c)
1 − P (A ∪ B )
P (B )
(b) 1 − P (A / B )
(d)
(1982, 2M)
P (A )
P (B )
20. The probability that an event A happens in one trial of
an experiment, is 0.4. Three independent trials of the
experiments are performed. The probability that the
event A happens atleast once, is
(1980, 1M)
(a) 0.936
(c) 0.904
(b) 0.784
(d) None of these
1
3
21. Let X andY be two events such that P (X ) = , P (X /Y ) =
2
and P (Y /X ) = . Then
5
(a) P (Y ) =
4
15
(c) P (X ∪Y ) =
22. If
(2017 Adv.)
(b) P (X ′/Y ) =
2
5
1
2
1
2
(d) P (X ∩ Y ) =
1
5
and Y
are
two
events
such
that
1
1
1
, P (Y /X ) = and P (X ∩ Y ) . Then, which of
2
3
6
(2012)
the following is/are correct?
X
P (X / Y ) =
(a) P (X ∪ Y ) = 2/3
(b) X and Y are independent
(c) X and Y are not independent
(d) P (X c ∩ Y ) = 1/3
23. Let E and F be two independent events. The probability
that exactly one of them occurs is
11
and the probability of
25
2
. If P (T ) denotes the
25
(2011)
probability of occurrence of the event T, then
none of them occurring is
4
3
, P (F ) =
5
5
2
1
(c) P (E ) = , P (F ) =
5
5
(a) P (E ) =
1
, P (F ) =
5
3
(d) P (E ) = , P (F ) =
5
(b) P (E ) =
2
5
4
5
24. The probabilities that a student passes in Mathematics,
Physics and Chemistry are m, p and c, respectively. Of
these subjects, the students has a 75% chance of passing
in atleast one, a 50% chance of passing in atleast two and
a 40% chance of passing in exactly two. Which of the
following relations are true?
(1999, 3M) (2011)
(a) p + m + c =
(c) pmc =
19
20
1
10
(b) p + m + c =
(d) pmc =
27
20
1
4
25. If E and F are the complementary events of E and F
respectively and if 0 < P (F ) < 1, then
(a) P (E / F ) + P (E / F ) = 1
(c) P (E / F ) + P (E / F ) = 1
(1998, 2M)
(b) P (E / F ) + P (E / F ) = 1
(d) P (E / F ) + P (E / F ) = 1
26. Let E and F be two independent events. If the probability
19. If A and B are two independent events such that
(a) 1 − P (A / B )
Objective Questions II
(One or more than one correct option)
that both E and F happen is 1/12 and the probability that
neither E nor F happen is 1/2. Then,
(a) P (E ) = 1 / 3, P (F ) = 1 / 4
(b) P (E ) = 1 / 2, P (F ) = 1 / 6
(c) P (E ) = 1 / 6, P (F ) = 1 / 2
(d) P (E ) = 1 / 4, P (F ) = 1 / 3
(1993, 2M)
27. For any two events A and B in a sample space
(1991, 2M)
P (A ) + P (B ) − 1
A
, P (B ) ≠ 0 is always true
(a) P   ≥
 B
P (B )
(b) P (A ∩ B ) = P (A ) − P (A ∩ B ) does not hold
(c) P (A ∪ B ) = 1 − P (A )P (B ), if A and B are independent
(d) P (A ∪ B ) = 1 − P (A )P (B ), if A and B are disjoint
108 Probability
28. If E and F are independent events such that 0 < P (E ) < 1
and 0 < P (F ) < 1, then
(1989, 2M)
(a) E and F are mutually exclusive
(b) E and F c (the complement of the event F) are
independent
(c) E c and F c are independent
(d) P (E / F ) + P (E c / F ) = 1
result is a tail, a card from a well-shuffled pack of eleven
cards numbered 2, 3, 4, …, 12 is picked and the number
on the card is noted. What is the probability that the
noted number is either 7 or 8?
(1994, 5M)
39. A lot contains 50 defective and 50 non-defective bulbs.
Fill in the Blanks
29. If two events A and B are such that P ( A c ) = 0.3,
P (B) = 0.4 and P ( A ∩ Bc ) = 0.5, then P [B / ( A ∪ Bc )] = K .
(1994, 2M)
30. Let A and B be two events such that P ( A ) = 0.3 and
P ( A ∪ B) = 0.8. If A and B are independent events,
(1990, 2M)
then P (B) = … .
31. A pair of fair dice is rolled together till a sum of either 5
or 7 is obtained. Then, the probability that 5 comes
before 7, is… .
(1989, 2M)
32. Urn A contains 6 red and 4 black balls and urn B
contains 4 red and 6 black balls. One ball is drawn at
random from urn A and placed in urn B. Then, one ball
is drawn at random from urn B and placed in urn A. If
one ball is drawn at random from urn A, the probability
that it is found to be red, is….
(1988, 2M)
33. A box contains 100 tickets numbered 1, 2, …,100. Two
tickets are chosen at random. It is given that the
maximum number on the two chosen tickets is not more
than 10. The maximum number on them is 5 with
(1985, 2M)
probability… .
Analytical and Descriptive Questions
34. If A and B are two independent events, prove that
P ( A ∪ B) ⋅ P ( A′ ∩ B ′ ) ≤ P (C ), where C is an event
defined that exactly one of A and B occurs. (2004, 2M)
35. A is targeting to B, B and C are targeting to A.
2 1
Probability of hitting the target by A, B and C are ,
3 2
1
and , respectively. If A is hit, then find the probability
3
that B hits the target and C does not.
(2003, 2M)
36. For a student to qualify, he must pass atleast two out of
three exams. The probability that he will pass the 1st
exam is p. If he fails in one of the exams, then the
p
probability of his passing in the next exam, is
2
otherwise it remains the same. Find the probability
(2003, 2M)
that he will qualify.
37. A coin has probability p of showing head when tossed. It
is tossed n times. Let pn denotes the probability that no
two (or more) consecutive heads occur. Prove that p1 = 1,
p2 = 1 − p2 and pn = (1 − p). pn − 1 + p(1 − p) pn − 2, ∀ n ≥ 3.
(2000, 5M)
38. An unbiased coin is tossed. If the result in a head, a pair
of unbiased dice is rolled and the number obtained by
adding the numbers on the two faces is noted. If the
Two bulbs are drawn at random, one at a time, with
replacement. The events A, B, C are defined as :
A = ( the first bulb is defective)
B = (the second bulb is non-defective)
C = (the two bulbs are both defective or both
non-defective).
Determine whether
(i) A, B, C are pairwise independent.
(1992, 6M)
(ii) A, B, C are independent.
40. In a multiple-choice question there are four alternative
answers, of which one or more are correct. A candidate
will get marks in the question only if he ticks the correct
answers. The candidates decide to tick the answers at
random, if he is allowed upto three chances to answer
the questions, find the probability that he will get
(1985, 5M)
marks in the question.
41. A and B are two independent events. The probability
that both A and B occur is
1
and the probability that
6
1
neither of them occurs is . Find the probability of the
3
occurrence of A.
(1984, 2M)
42. Cards are drawn one by one at random from a well
shuffled full pack of 52 playing cards until 2 aces are
obtained for the first time. If N is the number of cards
required to be drawn, then show that
(n − 1)(52 − n )(51 − n )
Pr { N = n } =
50 × 49 × 17 × 13
where, 2 < n ≤ 50.
(1983, 3M)
43. An anti-aircraft gun can take a maximum of four shots
at an enemy plane moving away from it. The
probabilities of hitting the plane at the first, second,
third and fourth shot are 0.4, 0.3, 0.2, and 0.1,
respectively. What is the probability that the gun hits
the plane?
(1981, 2M)
44. A box contanis 2 black, 4 white and 3 red balls. One ball
is drawn at random from the box and kept aside. From
the remaining balls in the box, another ball is drawn at
random and kept beside the first. This process is
repeated till all the balls are drawn from the box. Find
the probability that the balls drawn are in the sequence
of 2 black, 4 white and 3 red.
(1979, 2M)
Integer & Numerical Answer Type Questions
45. Let S be the sample space of all 3 × 3 matrices with
entries from the set {0, 1}. Let the events E1 and E 2 be
given by
E1 = { A ∈ S : det A = 0} and
E 2 = { A ∈ S : sum of entries of A is 7}.
If a matrix is chosen at random from S, then the
conditional probability P (E1 | E 2) equals ...........
Probability 109
46. Of the three independent events E1 , E 2 and E3 , the
probability that only E1 occurs is α , only E 2 occurs is β
and only E3 occurs is γ. Let the probability p that none of
events E1 , E 2 or E3 occurs satisfy the equations (α − 2 β ),
p = αβ and (β − 3γ ) p = 2 βγ. All the given probabilities
are assumed to lie in the interval (0, 1).
Then,
1 1
1
losing a game against T2 are , and , respectively. Each
2 6
3
team gets 3 points for a win, 1 point for a draw and 0 point for
a loss in a game. Let X andY denote the total points scored by
(2016 Adv.)
teams T1 and T2, respectively, after two games.
47. P (X > Y ) is
1
4
1
(c)
2
probability of occurrence of E1
is equal to
probability of occurrence of E3
(a)
Passage Type Questions
5
12
7
(d)
12
(b)
48. P (X = Y ) is
Passage
11
36
13
(c)
36
(a)
Football teams T1 and T2 have to play two games against each
other. It is assumed that the outcomes of the two games are
independent. The probabilities of T1 winning, drawing and
1
3
1
(d)
2
(b)
Topic 4 Law of Total Probability and Baye’s Theorem
Objective Question I (Only one correct option)
1. A pot contain 5 red and 2 green balls. At random a ball
is drawn from this pot. If a drawn ball is green then put
a red ball in the pot and if a drawn ball is red, then put a
green ball in the pot, while drawn ball is not replace in
the pot. Now we draw another ball randomnly, the
probability of second ball to be red is (2019 Main, 9 Jan II)
(a)
27
49
(b)
26
49
(c)
21
49
(d)
32
49
2. A bag contains 4 red and 6 black balls. A ball is drawn at
random from the bag, its colour is observed and this ball
along with two additional balls of the same colour are
returned to the bag. If now a ball is drawn at random
from the bag, then the probability that this drawn ball
is red, is
(2018 Main)
3
10
1
(c)
5
(a)
2
5
3
(d)
4
(b)
3. A computer producing factory has only two plants T1
and T2. Plant T1 produces 20% and plant T2 produces
80% of the total computers produced. 7% of computers
produced in the factory turn out to be defective. It is
known that P(computer turns out to be defective, given
that it is produced in plant T1) = 10P (computer turns
out to be defective, given that it is produced in plant T2),
where P (E ) denotes the probability of an event E. A
computer produced in the factory is randomly selected
and it does not turn out to be defective. Then, the
probability that it is produced in plant T2, is (2016 Adv.)
36
73
78
(c)
93
(a)
47
79
75
(d)
83
(b)
4. A signal which can be green or red with probability
and
transmitted to station B. The probability of each station
3
receiving the signal correctly is . If the signal received
4
at station B is green, then the probability that the
(2010)
original signal green is
3
5
20
(c)
23
(a)
6
7
9
(d)
20
(b)
Objective Question II
(One or more than one correct option)
5. There are three bags B1, B2 and B3 . The bag B1 contains
5 red and 5 green balls, B2 contains 3 red and 5 green
balls, and B3 contains 5 red and 3 green balls. Bags B1,
3 3
4
and
respectively
B2 and B3 have probabilities ,
10 10
10
of being chosen. A bag is selected at random and a ball is
chosen at random from the bag. Then which of the
following options is/are correct?
(a) Probability that the chosen ball is green, given that
3
the selected bag is B3 , equals .
8
(b) Probability that the selected bag is B3 , given that the
5
chosen ball is green, equals .
13
39
(c) Probability that the chosen ball is green equals .
80
(d) Probability that the selected bag is B3 and the chosen
3
ball is green equals .
10
6. A ship is fitted with three engines E1 , E 2 and E3 . The
4
5
1
respectively, is received by station A and then
5
engines function independently of each other with
respective probabilities 1/2, 1/4 and 1/4. For the ship to
be operational atleast two of its engines must function.
Let X denotes the event that the ship is operational and
let X1, X 2 and X3 denote, respectively the events that the
engines E1, E 2 and E3 are functioning.
110 Probability
Which of the following is/are true?
(a) P
[X1c| X ] =
(2012)
is
3 / 16
(b) P [exactly two engines of the ship are functioning] =
(c) P [X | X 2 ] =
10. The probability of the drawn ball fromU 2 being white,
5
16
(d) P [X | X1 ] =
7
8
7
16
(a)
(a) Statement I is true, Statement II is also true;
Statement II is the correct explanation of Statement I
(b) Statement I is true, Statement II is also true;
Statement II is not the correct explanation of
Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
P (H i ) > 0, i = 1, 2,... , n . Let E be any other event with
0 < P (E ) < 1.
Statement I P (H i/E ) > P (E/H i ) ⋅ P (H i ) for
19
30
11
30
(d)
17
23
(b)
11
23
(c)
15
23
(d)
12
23
Passage III
12. The probability that X = 3 equals
(a)
25
216
(b)
25
36
(c)
5
36
(d)
125
216
(d)
25
216
13. The probability that X ≥ 3 equals
(a)
125
216
(b)
25
36
(c)
5
36
14. The conditional probability that X ≥ 6 given X > 3
equals
(a)
i = 1, 2, . . . , n
125
216
(b)
25
216
(c)
5
36
(d)
25
36
Passage IV
(2007, 3M)
i =1
Passage Based Problems
Passage I
Let n1 and n2 be the number of red and black balls, respectively
in box I. Let n3 and n4 be the number of red and black balls,
respectively in box II.
(2015 Adv.)
8. One of the two boxes, box I and box II was selected at
random and a ball was drawn randomly out of this box.
The ball was found to be red. If the probability that this
1
red ball was drawn from box II, is , then the correct
3
option(s) with the possible values of n1 , n2, n3 and n4
is/are
(a) n1 = 3, n2 = 3, n3 = 5, n4 = 15
(b) n1 = 3, n2 = 6, n3 = 10, n4 = 50
(c) n1 = 8, n2 = 6, n3 = 5, n4 = 20
(d) n1 = 6, n2 = 12, n3 = 5, n4 = 20
9. A ball is drawn at random from box I and transferred to
box II. If the probability of drawing a red ball from box I,
1
after this transfer, is , then the correct option(s) with
3
the possible values of n1 and n2 is/are
(a) n1 = 4 and n2 = 6
(c) n1 = 10 and n2 = 20
(c)
A fair die is tossed repeatedly until a six is obtained. Let X
(2009)
denote the number of tosses required.
7. Let H 1 , H 2,... , H n be mutually exclusive events with
∑ P (Hi ) = 1
23
30
probability that head appeared on the coin is
For the following questions, choose the correct answer
from the codes (a), (b), (c) and (d) defined as follows.
Statement II
(b)
11. Given that the drawn ball from U 2 is white, the
(a)
Assertion and Reason
n
13
30
(b) n1 = 2 and n2 = 3
(d) n1 = 3 and n2 = 6
Passage II
LetU 1 andU 2 be two urns such thatU 1 contains 3 white and 2
red balls andU 2 contains only 1 white ball. A fair coin is tossed.
If head appears then 1 ball is drawn at random from U 1 and
put intoU 2. However, if tail appears then 2 balls are drawn at
random from U 1 and put into U 2. Now, 1 ball is drawn at
random from U 2.
(2011)
There are n urns each containing (n + 1) balls such that the
ith urn contains ‘i’white balls and (n + 1 − i) red balls. Let ui
be the event of selecting ith urn, i = 1, 2, 3, ... , n and W
(2006, 5M)
denotes the event of getting a white balls.
15. If P (ui ) ∝ i, where i = 1, 2, 3,... , n , then lim P (W ) is
n→ ∞
equal to
(a) 1
2
(b)
3
1
(c)
4
(d)
3
4
16. If P (ui ) = c , where c is a constant, then P (un / W ) is
equal to
2
n+1
n
(c)
n+1
(a)
1
n+1
1
(d)
2
(b)
17. If n is even and E denotes the event of choosing even
1

numbered urn P (ui )=
, then the value of P ( W /E ) is

n 
(a) n + 2
2n + 1
n
(c)
n+1
n+ 2
2 (n + 1)
1
(d)
n+1
(b)
Analytical and Descriptive Questions
18. A person goes to office either by car, scooter, bus or
1 3 2
1
and ,
, ,
7 7 7
7
respectively. Probability that he reaches offices late, if
2 1 4
1
he takes car, scooter, bus or train is , , and ,
9 9 9
9
respectively. Given that he reached office in time,
then what is the probability that he travelled by a car ?
train probability of which being
(2005, 2M)
Probability 111
19. A bag contains 12 red balls and 6 white balls. Six balls are
drawn one by one without replacement of which at least
4 balls are white. Find the probability that in the next two
drawn exactly one white ball is drawn. (Leave the answer
in nC r).
(2004, 4M)
20. A box contains N coins, m of which are fair and the rest are
biased. The probability of getting a head when a fair coin is
tossed, is 1/2, while it is 2/3 when a biased coin is tossed. A
coin is drawn from the box at random and is tossed twice.
The first time it shows head and the second time it shows
tail. What is the probability that the coin drawn is fair?
(2002, 5M)
21. An urn contains m white and n black balls. A ball is drawn
at random and is put back into the urn along with k
additional balls of the same colour as that of the ball
drawn. A ball is again drawn at random. What is the
probability that the ball drawn now is white? (2001, 5M)
22. Eight players P1 , P2, K , P8 play a knock-out tournament. It
is known that whenever the players Pi and Pj play, the
player Pi will win if i < j. Assuming that the players are
paired at random in each round, what is the probability
(1999, 10M)
that the player P4 reaches the final?
23. Three players, A, B and C, toss a coin cyclically in that
order (i.e. A, B, C, A, B, C, A, B, …) till a head shows. Let p
be the probability that the coin shows a head. Let α, β and
γ be, respectively, the probabilities that A, B and C gets
the first head. Prove that β = (1 − p) α . Determine α, β and
(1998, 8M)
γ (in terms of p).
24. Sixteen players S1 , S 2, K , S16 play in a tournament. They
are divided into eight pairs at random from each pair a
winner is decided on the basis of a game played between
the two players of the pair. Assume that all the players
are of equal strength.
(i) Find the probability that the player S1 is among the
eight winners.
(ii) Find the probability that exactly one of the two
players S1 and S2 is among the eight winners.
(1997C, 5M)
25. In a test an examinee either guesses or copies of
knows the answer to a multiple choice question with
1
four choices. The probability that he make a guess is
3
1
and the probability that he copies the answer is . The
6
probability that his answer is correct given that he
1
copied it, is . Find the probability that he knew the
8
answer to the question given that he correctly
answered it.
(1991, 4M)
26. An urn contains 2 white and 2 blacks balls. A ball is
drawn at random. If it is white it is not replaced into
the urn. Otherwise it is replaced along with another
ball of the same colour. The process is repeated. Find
the probability that the third ball drawn is black.
(1987, 4M)
27. A lot contains 20 articles. The probability that the lot
contains exactly 2 defective articles is 0.4 and the
probability that the lot contains exactly 3 defective
articles is 0.6. Articles are drawn from the lot at
random one by one without replacement and are
tested till all defective articles are found. What is the
probability that the testing procedure ends at the
twelfth testing?
(1986, 5M)
Topic 5 Probability Distribution and Binomial Distribution
Objective Questions I (Only one correct option)
1. Let C1 and C 2 be two biased coins such that the
2
1
and ,
3
3
respectively. Suppose α is the number of heads that
appear when C1 is tossed twice, independently, and
suppose β is the number of heads that appear when C 2 is
tossed twice, independently. Then the probability that
the roots of the quadratic polynomial x2 − ax + β are real
(2020 Adv.)
and equal, is
probabilities of getting head in a single toss are
40
81
1
(c)
2
(a)
20
81
1
(d)
4
(b)
2. Four fair dice are thrown independently 27 times. Then,
the expected number of times, at least two dice show up a
three or a five, is …… .
[2020 Main, 5 Sep I]
3. In a workshop, there are five machines and the
probability of any one of them to be out of service on a day
1
is . If the probability that at most two machines will be
4
3
 3
out of service on the same day is   k, then k is equal
 4
to
[2020 Main, 7 Jan II]
(a) 4
(b)
17
4
17
(c)
8
(d)
17
2
4. For an initial screening of an admission test, a
candidate is given fifty problems to solve. If the
probability that the candidate can solve any problem
4
is , then the probability that he is unable to solve less
5
than two problem is
(2019 Main, 12 April II)
(a)
(c)
201  1 
 
5  5
54  4 
 
5  5
49
48
(b)
316  4 
 
25  5 
(d)
164  1 
 
25  5 
49
48
5. Let a random variable X have a binomial distribution
with mean 8 and variance 4. If P (X ≤ 2) =
equal to
(a) 17
k
, then k is
216
(2019 Main, 12 April I)
(b) 121
(c) 1
(d) 137
112 Probability
6. Minimum number of times a fair coin must be tossed so
that the probability of getting atleast one head is more
than 99% is
(2019 Main 10 April II)
(a) 8
(b) 6
(c) 7
coin so that the probability of observing atleast one head
(2019 Main, 8 April II)
is atleast 90% is
(b) 3
15. If the mean and the variance of a binomial variate X are
2 and 1 respectively, then the probability that X takes a
value greater than one is equal to… .
(1991, 2M)
(d) 5
7. The minimum number of times one has to toss a fair
(a) 2
Fill in the Blanks
(c) 5
16. For a biased die the probabilities for the different faces
to turn up are given below
(d) 4
8. In a game, a man wins ` 100 if he gets 5 or 6 on a throw
of a fair die and loses ` 50 for getting any other number
on the die. If he decides to throw the die either till he
gets a five or a six or to a maximum of three throws,
then his expected gain/loss (in rupees) is
(2019 Main, 12 Jan II)
(a)
400
400
loss (b)
loss
3
9
(c) 0
(d)
400
gain
3
(a) 6
(b) 3
(c) 5
(d) 4
10. Two cards are drawn successively with replacement
from a well shuffled deck of 52 cards. Let X denote the
random variable of number of aces obtained in the two
drawn cards. Then, P (X = 1) + P (X = 2) equals
(2019 Main, 9 Jan I)
25
(a)
169
52
(b)
169
49
(c)
169
24
(d)
169
11. A box contains 15 green and 10 yellow balls. If 10 balls
are randomly drawn one-by-one with replacement, then
the variance of the number of green balls drawn is
(2017 Main)
12
(a)
5
(b) 6
6
(d)
25
(c) 4
1
2
3
4
5
6
Probability
0.1
0.32
0.21
0.15
0.05
0.17
This die is tossed and you are told that either face 1 or
face 2 has turned up. Then, the probability that it is face
1, is… .
(1981, 2M)
Analytical & Descriptive Questions
17. Numbers are selected at random, one at a time, from the
two-digit numbers 00, 01, 02, …, 99 with replacement.
An event E occurs if and only if the product of the two
digits of a selected number is 18. If four numbers are
selected, find probability that the event E occurs at least
(1993, 5M)
3 times.
9. If the probability of hitting a target by a shooter in any
1
shot, is , then the minimum number of independent
3
shots at the target required by him so that the
probability of hitting the target at least once is greater
5
than , is
(2019 Main, 10 Jan II)
6
Face
18.
A is a set containing n elements. A subset P of A is
chosen at random. The set A is reconstructed by
replacing the elements of P. A subset Q of A is again
chosen at random. Find the probability that P and Q
(1991, 4M)
have no common elements.
19. Suppose the probability for A to win a game against B is
0.4. If A has an option of playing either a ‘best of 3
games’ or a ‘best of 5 games’’ match against B, which
option should choose so that the probability of
his winning the match is higher? (no game ends in a
(1989, 5M)
draw).
20. A man takes a step forward with probability 0.4 and
backwards with probability 0.6. Find the probability
that at the end of eleven steps he is one step away from
the starting point.
(1987, 3M)
12. A multiple choice examination has 5 questions. Each
question has three alternative answers of which exactly
one is correct. The probability that a student will get 4
or more correct answers just by guessing is (2013 Main)
(a)
17
3
5
(b)
13
3
5
(c)
11
3
5
(d)
10
3
5
13. India plays two matches each with West Indies and
Australia. In any match the probabilities of India
getting points 0, 1 and 2 are 0.45, 0.05 and 0.50,
respectively. Assuming that the outcomes are
independent. The probability of India getting at least 7
points, is
(1992, 2M)
(a) 0.8750
(b) 0.0875
(c) 0.0625
(d) 0.0250
14. One hundred identical coins, each with probability p, of
showing up heads are tossed once. If 0 < p < 1 and the
probability of heads showing on 50 coins is equal to that
of heads showing on 51 coins, then the value of p is
(1988, 2M)
(a) 1/2
(b) 49/101
(c) 50/101
(d) 51/101
Integer & Numerical Answer Type Questions
21. Two fair dice, each with faces numbered 1, 2, 3, 4, 5 and
6, are rolled together and the sum of the numbers on the
faces is observed. This process is repeated till the sum is
either a prime number or a perfect square. Suppose the
sum turns out to be a perfect square before it turns out to
be a prime number. If p is the probability that this
perfect square is an odd number, then the value of 14 p is
……… .
22. The probability that a missile hits a target successfully
is 0.75. In order to destroy the target completely, at
least three successful hits are required. Then the
minimum number of missiles that have to be fired so
that the probability of completely destroying the target
is NOT less than 0.95, is ……… .
23. The minimum number of times a fair coin needs to be
tossed, so that the probability of getting atleast two
(2015 Adv.)
heads is atleast 0.96, is
Probability 113
Answers
Topic 1
39. (i) A, B and C are pairwise independent
1. (c)
2. (a)
5.
9.
13.
17.
(a)
6. (c)
(a)
10. (c)
(a)
14. (b)
(d)
18. (b)
1
1
1
21.
22. ≤ p ≤
36
3
2
(3n − 3. 2n + 3 ) × 6C 3
25.
6n
10 (n + 2 )
28. 1 − n + 7
C5
3. (a)
7.
11.
15.
19.
(a)
(d)
(c)
(b)
3
23.
16
8.
12.
16.
20.
26. 0.62
27.
29. (i)
(a)
(d)
(b)
(c)
24. False
1
91
1
1
(ii)
132
462
41.
2. (b)
5. (a)
6. (c)
9. (a, b, c)
13. False
3. (a)
4. (c)
7. (c)
11
40
11.
12. P ( A ∩ B )
14. 13.9%
16. No
17. (a)
18. (c)
Topic 3
1. (d)
5. (a,c)
9. (c,d)
2. (b)
6. (d)
3. (b)
7. (a)
4. (a)
8. (a)
9. (d)
10. (c)
11. (c)
12. (b)
13. (b)
14. (a)
15. (a)
16. (b)
17. (a)
18. (c)
19. (b)
20. (b)
21. (a, b)
22. (a,b)
23. (a, d)
24. (b, c)
25. (a, d)
1
29.
4
1
33.
9
26. (a, d)
5
30.
7
1
35.
2
27. (a, c)
2
31.
5
28. (b, c, d)
32
32.
55
193
38.
792
36. 2 p 2 – p 3
44.
47. (b)
48. (c)
2. (b)
3. (c)
45. 0.50
6. (b, d)
10. (b)
4. (c)
7. (d)
11. (d)
8. (a,b)
12. (a)
13. (b)`
17.
19.
21.
23.
24.
1. (d)
5. (c)
1
1260
43. 0.6976
1
5
Topic 4
8. (a)
10. (a, c)
1 1
or
3 2
46. (6)
Topic 2
1. (b)
40.
4. (d)
14. (d)
15. (b)
16.
1
(b)
18.
7
10
2
12
6
C 2 ⋅ C 4 C1 ⋅ C1 12C1 ⋅6 C 5 11C1 ⋅1 C1
20.
⋅ 12
+ 18
⋅ 12
18
C6
C2
C6
C2
m
4
22.
m+n
35
p
p (1 − p )
p − 2p 2 + p 3
,
,
=
α=
β
=
γ
1 − (1 − p ) 3
1 − (1 − p ) 3
1(1 − p ) 3
1
8
24
23
(i) (ii)
25.
26.
27.
2
15
29
30
(a)
9m
8N + m
99
1900
Topic 5
1. (b)
2. (11)
3. (c)
4. (c)
5. (d)
6. (c)
7. (d)
8. (c)
9. (c)
10. (a)
13. (b)
14. (d)
11. (a)
11
15.
16
12. (c)
5
16.
21
17.
97
25 4
20.
11
 3
18.  
 4
C 6( 0 . 24 ) 5
21. (8)
n
19. Best of 3 games
22. (6)
23. (8)
Hints & Solutions
Topic 1 Classical Probability
1. It is given that a person wins
`15 for throwing a doublet (1, 1) (2, 2), (3, 3),
(4, 4), (5, 5), (6, 6) and win `12 when the throw results in
sum of 9, i.e., when (3, 6), (4, 5),
(5, 4), (6, 3) occurs.
Also, losses ` 6 for throwing any other outcome, i.e.,
when any of the rest 36 − 6 − 4 = 26 outcomes occurs.
Now, the expected gain/loss
= 15 × P (getting a doublet) + 12 × P
(getting sum 9) − 6 × P (getting any
of rest 26 outcome)
6 
4 
26

= 15 ×  + 12 ×  − 6 × 

36 
36 
36
5 4 26 15 + 8 − 26
+ −
=
2 3 6
6
23 − 26
3
1
1
=
= − = − , means loss of `
6
6
2
2
=
2. Since, the experiment should be end in the fifth throw of
the die, so total number of outcomes are 65 .
Now, as the last two throws should be result in two
4 4
fours
(i) (ii) (iii) (iv) (v)
So, the third throw can be 1, 2, 3, 5 or 6 (not 4). Also,
throw number (i) and (ii) can not take two fours in
succession, therefore number of possibililites for throw
(i) and (ii) = 62 − 1 = 35
[Q when a pair of dice is thrown then
(4, 4) occur only once].
114 Probability
Hence, the required probability =
5 × 35 175
= 5
65
6
7. We have mentioned that boxes are different and one
particular box has 3 balls.
3. Since, there is a regular hexagon, then the number of
6
ways of choosing three vertices is C3 . And, there is only
two ways i.e. choosing vertices of a regular hexagon
alternate, here A1, A3 , A5 or A2, A4, A6 will result in an
equilateral triangle.
A2
A5
A3
A4
∴Required probability
2
2
2 ×3 ×2 ×3 ×2
1
= 6
=
=
=
6
!
6 × 5 × 4 × 3 × 2 × 1 10
C3
3 !3 !
4. Number of subset of S = 220
20(21)
Sum of elements in S is 1 + 2 + .....+20 =
= 210
2
n (n + 1) 

Q 1 + 2+ ...... + n =


2
Clearly, the sum of elements of a subset would be 203, if
we consider it as follows
S − { 7}, S − {1, 6} S − {2, 5}, S − {3, 4}
S − { 1, 2, 4 }
∴ Number of favourables cases = 5
5
Hence, required probability = 20
2
5. Total number of ways of selecting 2 different numbers
from {0, 1, 2, ..., 10} = 11C 2 = 55
Let two numbers selected be x and y.
Then,
x + y = 4m
and
x − y = 4n
⇒
2x = 4(m + n ) and 2 y = 4(m − n )
⇒
x = 2(m + n )and y = 2(m − n )
∴x and y both are even numbers.
x
y
0
4, 8
2
6, 10
4
0, 8
6
2, 10
8
0, 4
10
2, 6
∴Required probability =
6. Sample space → 12C 2
6
55
Number of possibilities for z is even.
z = 0 ⇒ 11C1; z = 2 ⇒ 9C1
z = 4 ⇒ 7C1; z = 6 ⇒ 5C1
z = 8 ⇒ 3C1; z = 10 ⇒ 1C1
Total = 36
36 6
∴ Probability =
=
66 11
…(i)
…(ii)
12
According to given condition, following cases may arise.
B
G
G
B
B
G
G
B
B
B
G
B
G
B
B
G
B
B
G
B
B
G
B
G
B
So, number of favourable ways = 5 × 3 ! × 2 ! = 60
60 1
Required probability =
=
∴
120 2
A1
A6
11
C3 × 29 55  2
=
 
3  3
312
8. Total number of ways to arrange 3 boys and 2 girls
are 5!.
Then, number of ways =
9.
PLAN As one of the dice shows a number appearing on one of P1, P2
and P3.
Thus, three cases arise.
(i) All show same number.
(ii) Number appearing on D 4 appears on any one of
D1, D 2 and D 3.
(iii) Number appearing on D 4 appears on any
two of D1, D 2 and D 3.
Sample space = 6 × 6 × 6 × 6 = 64 favourable events
= Case I or Case II or Case III
Case I First we should select one number for D4
which appears on all i.e. 6C1 × 1.
Case II For D4 there are 6C1 ways. Now, it appears
on any one of D1 , D2 and D3 i.e. 3 C1 × 1.
For other two there are 5 × 5 ways.
6
⇒
C1 × 3C1 × 1 × 5 × 5
Case III For D4 there are 6C1 ways now it appears on
any two of D1 , D2 and D3
3
⇒
C 2 × 12
For other one there are 5 ways.
6
⇒
C1 × 3C 2 × 12 × 5
6
C1 + 6C1 × 3C1 × 52 + 6C1 × 3C 2 × 5
Thus, probability =
64
6 (1 + 75 + 15)
=
64
91
=
216
10. Sample space A dice is thrown thrice, n (s) = 6 × 6 × 6.
r
r
Favorable events ω r 1 + ω 2 + ω 3 = 0
i.e. (r1 , r2, r3 ) are ordered 3 triples which can take
values,
(1, 2 , 3), (1, 5, 3), (4, 2 , 3), (4, 5, 3)
 i.e. 8 ordered pairs
(1, 2 , 6), (1, 5, 6), (4, 2 , 6), (4, 5, 6)
and each can be arranged in 3 ! ways = 6
8 ×6
2
∴
=
n (E ) = 8 × 6 ⇒ P (E ) =
6 ×6 ×6 9
Probability 115
Only two equilateral triangles can be formed
∆AEC and ∆BFD.
11. Since, three distinct numbers are to be selected from
first 100 natural numbers.
⇒
n (S ) = 100C3
D
E(favourable events) = All three of them are divisible by both
2 and 3 .
⇒ Divisible by 6 i.e. {6, 12, 18, …, 96}
Thus, out of 16 we have to select 3.
∴
n (E ) = 16C3
16
C
4
∴ Required probability = 100 3 =
C3 1155
13. 71 = 7, 72 = 49, 73 = 343, 74 = 2401, …
Therefore, for 7r, r ∈ N the number ends at unit place
7, 9, 3, 1, 7, …
∴ Total number of cases S = 6 × 6 × 6 = 216
and the same number appear on each of them = 6C1 = 6
6
1
∴ Required probability =
=
216 36
17. Since, there are 15 possible cases for selecting a coupon
and seven coupons are selected, the total number of
cases of selecting seven coupons = 157
It is given that the maximum number on the selected
coupon is 9, therefore the selection is to be made from
the coupons numbered 1 to 9. This can be made in 97
ways. Out of these 97 cases, 87 does not contain the
number 9.
Thus, the favourable number of cases = 97 − 87.
97 − 87
∴ Required probability =
157
For this m and n should be as follows
n
4r − 2
4r − 3
4r
4r − 1
18. The number of all possible determinants of the form
For any given value of m, there will be 25 values of n.
Hence, the probability of the required event is
100 × 25 1
=
100 × 100 4
a
b
c
d
0 1 0 0
,
0 0 1 0
14. The number of ways of placing 3 black balls without any
restriction is 10C3 . Since, we have total 10 places of
putting 10 balls in a row. Now, the number of ways in
which no two black balls put together is equal to the
number of ways of choosing 3 places marked ‘—’ out of
eight places.
Vanish and remaining six determinants have non-zero
6 3
values. Hence, the required probability =
=
16 8
Statement I is true.
Statement II is also true as the homogeneous equations
have always a solution and Statement II is not the
correct explanation of Statement I.
—W—W—W—W —W—W—W—
This can be done in 8C3 ways.
8
C3
8 × 7 ×6
7
=
=
C3 10 × 9 × 8 15
10
15. Three vertices out of 6 can be chosen in 6C3 ways.
So, total ways = 6C3 = 20
= 24 = 16
Out of which only 10 determinants given by
1 1 0 0 1 1 0 0 0 1 1 0 1 0
,
,
,
,
,
,
,
1 1 0 0 0 0 1 1 0 1 1 0 0 0
NOTE Power of prime numbers have cyclic numbers in their unit
place.
∴ Required probability =
2
1
=
20 10
16. Since, three dice are rolled.
Also for end at 0.
4r
4r − 1
4r − 2
4r − 3
B
So, required probability =
But it cannot end at 5.
m
F
∴ Favourable ways = 2
∴ 7m + 7n will be divisible by 5 if it end at 5 or 0.
1
2
3
4
C
A
12. Here, two numbers are selected from {1, 2, 3, 4, 5, 6}
⇒ n (S ) = 6 × 5 {as one by one without replacement}
Favourable events = the minimum of the two numbers
is less than 4. n (E ) = 6 × 4 {as for the minimum of the
two is less than 4 we can select one from (1, 2, 3, 4) and
other from (1, 2, 3, 4, 5, 6)
n (E ) 24 4
=
=
∴ Required probability =
n (S ) 30 5
E
19.
PLAN Probability =
Number of favourable outcomes
Number of total outcomes
As, x1 + x2 + x3 is odd.
So, all may be odd or one of them is odd and other two
are even.
116 Probability
∴ Required probability
2
C1 × 3C1 × 4C1 + 1C1 × 2C1 × 4C1 + 2C1 × 2C1 × 3C1
+ 1C1 × 3C1 × 3C1
=
5
7
3
C1 × C1 × C1
24 + 8 + 12 + 9
105
53
=
105
=
∴
x1 + x3 = 2x2
So, x1 + x3 should be even number.
B: [1, 2, 3], where r1 , r2, K , rn are the readings of n
throws and 1, 2, 3 are the numbers that appear in the n
throws.
Number of such functions, M = N − [n (1) − n (2) + n (3)]
C1 × 4C1 + 1C1 × 3C1
3
C1 × 5C1 × 7C1
2
where, N = total number of functions
and n (t ) = number of function having exactly t
elements in the range.
Now,
N = 3n, n (1) = 3 . 2n, n(2) = 3, n(3) = 0
11
=
105
21. According to given condition,
3 1
=
6 2
2 1
P (red at the second toss) = =
6 3
1
and P (blue at the third toss) =
6
Therefore, the probability of the required event
1 1 1 1
= × × =
2 3 6 36
1 + 3p 1 − p
1 − 2p
22. Since,
are the probability of
,
and
3
4
2
mutually exclusive events.
1 + 3p 1 − p 1 −2p
+
+
≤1
∴
3
4
2
P ( yellow at the first toss) =
4 + 12 p + 3 − 3 p + 6 − 12 p ≤ 12
⇒
13 − 3 p ≤ 12
1
⇒
p≥
3
1 + 3p
1− p
1 −2p
and 0 ≤
≤ 1, 0 ≤
≤ 1, 0 ≤
≤1
3
4
2
⇒
⇒
0 ≤ 1 + 3 p ≤ 3, 0 ≤ 1 − p ≤ 4, 0 ≤ 1 − 2 p ≤ 2
1
2
1
1
− ≤ p ≤ , 1 ≥ p ≥ −3 , ≥ p ≥ −
3
3
2
2
...(i)
...(ii)
From Eqs. (i) and (ii), 1 / 3 ≤ p ≤ 1 / 2
23. Since, determinant is of order 2 × 2 and each element is
0 or 1 only.
∴
n (S ) = 24 = 16
and the determinant is positive are
1 0
1 1 1 0
,
,
0 1
0 1 1 1
∴
n (E ) = 3
Thus, the required probability =
First we fix the position ⊗ A ⊗ A ⊗ I ⊗ N ⊗.
Number of ways in which no two S’s occur together
4! 5
=
× C4
2!
4! × 5 × 4! × 2! 1
∴ Required probability =
=
2! × 8!
14
25. Let us define a onto function F from A : [ r1 , r2, K , rn] to
Either both x1 and x3 are odd or both are even.
⇒
8!
.
4 !⋅ 2 !
Hence, it is a false statement.
20. Since, x1 , x2, x3 are in AP.
∴ Required probability =
24. Total number of ways to arrange ‘ASSASSIN’ is
3
16
⇒
M = 3n − 3 . 2n + 3
Hence, the total number of favourable cases
= (3n − 3 . 2n + 3) . 6C3
(3n − 3 . 2n + 3) × 6C3
∴ Required probability =
6n
26. The required probability = 1 − (probability of the event
that the roots of x2 + px + q = 0 are non-real).
The roots of x2 + px + q = 0 will be non-real if and only if
p2 − 4q < 0, i.e. if p2 < 4 q
The possible values of p and q can be possible according
to the following table.
Value of q
Value of p
Number of pairs of p, q
1
1
1
2
1, 2
2
3
1, 2, 3
3
4
1, 2, 3
3
5
1, 2, 3, 4
4
6
1, 2, 3, 4
4
7
1, 2, 3, 4, 5
5
8
1, 2, 3, 4, 5
5
9
1, 2, 3, 4, 5
5
1, 2, 3, 4, 5, 6
6
10
Therefore, the number of possible pairs = 38
Also, the total number of possible pairs is 10 × 10 = 100
38
∴ The required probability = 1 −
= 1 − 0.38 = 0.62
100
27. We have 14 seats in two vans and there are 9 boys and 3
girls. The number of ways of arranging 12 people on 14
seats without restriction is
14 !
14
P12 =
= 7(13 !)
2!
Probability 117
Now, the number of ways of choosing back seats is 2.
and the number of ways of arranging 3 girls on adjacent
seats is 2(3!) and the number of ways of arranging 9
boys on the remaining 11 seats is 11 P9 ways.
Therefore, the required number of ways
4 ⋅ 3 ! 11 !
= 12 !
= 2. (2 .3 !).11 P9 =
2!
Hence, the probability of the required event
12 !
1
=
=
7 ⋅ 13 ! 91
28. There are (n + 7) coins in the box out of which five coins
can be taken out in
n+7
C5 ways.
The total value of 5 coins can be equal to or more than
one rupee and fifty paise in the following ways.
(i) When one 50 paise coin and four 25 paise coins are
chosen.
(ii) When two 50 paise coins and three 25 paise coins
are chosen.
(iii) When two 50 paise coins, 2 twenty five paise coins
and one from n coins of ten and five paise.
∴ The total number of ways of selecting five coins so
that the total value of the coins is not less than one
rupee and fifty paise is
(2C1 ⋅5 C5 ⋅n C 0 ) + (2C 2 ⋅5 C3 ⋅n C 0 ) + (2C 2 ⋅5 C 2 ⋅n C1 )
= 10 + 10 + 10n = 10 (n + 2)
2. We have, P (exactly one of A or B occurs)
= P ( A ∪ B) − P ( A ∩ B)
= P ( A ) + P (B) − 2P ( A ∩ B)
According to the question,
1
…(i)
P ( A ) + P (B) − 2P ( A ∩ B) =
4
1
…(ii)
P (B) + P (C ) − 2P (B ∩ C ) =
4
1
…(iii)
and
P (C ) + P ( A ) − 2P (C ∩ A ) =
4
On adding Eqs. (i), (ii) and (iii), we get
2 [P ( A ) + P (B) + P (C ) − P ( A ∩ B) − P (B ∩ C )
3
− P (C ∩ A )] =
4
⇒ P ( A ) + P (B) + P (C ) − P ( A ∩ B) − P (B ∩ C )
3
− P (C ∩ A ) =
8
∴P (atleast one event occurs)
= P(A ∪ B ∪ C )
= P ( A ) + P (B) + P (C ) − P ( A ∩ B) − P (B ∩ C )
− P (C ∩ A ) + P ( A ∩ B ∩ C )
3
1
7
1

= +
=
Q P(A ∩ B ∩ C ) =

16 
8 16 16
3
1
3. Given, P (B) = , P ( A ∩ B ∩ C ) =
4
3
B
A
So, the number of ways of selecting five coins, so
that the total value of the coins is less than one
rupee and fifty paise is n + 7C5 − 10(n + 2)
(A ∩ B ∩ C)
n+7
(B ∩ C)
C5 − 10(n + 2)
n+7
C5
10 (n + 2)
=1 − n+7
C5
∴ Required probability =
29. (i) The total number of arrangements of six boys and
six girls = 12 !
6! × 7!
1
=
(12)!
132
[since, we consider six girls at one person]
2 ×6! ×6!
1
(ii) Required probability =
=
(12)!
462
∴ Required probability =
Topic 2 Addition and Subtraction Law of
Probability
1. As, we know that
P ( A ∩ B) = P ( A ) + P (B) − P ( A ∪ B)
⇒
P ( A ∩ B) = 0.6 + 0.4 − 0.8 ⇒ P ( A ∩ B) = 0.2
and, as
P ( A ∪ B ∪ C ) = P ( A ) + P (B) + P (C )
− P ( A ∩ B) − P (B ∩ C ) − P (C ∩ A ) + P ( A ∩ B ∩ C )
⇒
α = 0.6 + 0.4 + 0.5 − 0.2 − β − 0.3 + 0.2
⇒
α = 1.2 − β
Q 0.85 ≤ α ≤ 0.95 ⇒ 0.85 ≤ 1.2 − β ≤ 0.95
⇒ 0.25 ≤ β ≤ 0.35
(A ∩ B ∩ C)
C
and
P(A ∩ B ∩ C ) =
1
3
which can be shown in Venn diagram.
∴ P (B ∩ C ) = P (B) − { P ( A ∩ B ∩ C + P ( A ∩ B ∩ C ))}
3  1 1 3 2 1
= − +  = − =
4  3 3 4 3 12
4. It is given that, P (E ) ≤ P (F ) ⇒ E ⊆ F
…(i)
P (E ∩ F ) > 0 ⇒ E ⊂ F
…(ii)
(a) occurrence of E ⇒ occurrence of F
[from Eq. (i)]
(b) occurrence of F ⇒ occurrence of E
[from Eq. (ii)]
and
(c) non-occurrence of E ⇒ occurrence of F
Hence, option (c) is not correct.
[from Eq. (i)]
5. We know that,
P (exactly one of A or B occurs)
= P ( A ) + P (B) − 2P ( A ∩ B)
∴
P ( A ) + P (B) − 2P ( A ∩ B) = p
…(i)
Similarly,
P (B) + P (C ) − 2P (B ∩ C ) = p
…(ii)
and
P (C ) + P ( A ) − 2P (C ∩ A ) = p
…(iii)
118 Probability
On adding Eqs. (i), (ii) and (iii), we get
= P (M ∪ N ) − P (M ∩ N ) = P (M ) + P (N ) − 2P (M ∩ N )
2 [P ( A ) + P (B) + P (C ) − P ( A ∩ B)
− P (B ∩ C ) − P (C ∩ A )] = 3 p
⇒ P ( A ) + P (B) + P (C ) − P ( A ∩ B)
3p
…(v)
− P (B ∩ C ) − P (C ∩ A ) =
2
…(v)
It also given that, P ( A ∩ B ∩ C ) = p2
∴ P(at least one of the events A, B, and C occurs)
= P ( A ) + P (B) + P (C ) − P ( A ∩ B)
Hence, (a) and (c) are correct answers.
− P (B ∩ C ) − P (C ∩ A ) + P ( A ∩ B ∩ C )
3p
[from Eqs. (iv) and (v)]
=
+ p2
2
=
3 p + 2 p2
2
6. Since, P ( A ∩ B) = P ( A ) ⋅ P (B)
It means A and B are independent events, so A ′ and B ′
are also independent.
∴
P ( A ∪ B) ′ = P ( A ′∩ B ′ ) = P ( A )′ ⋅ P (B)′
11. Let E1 be the event getting minimum number 3 and E 2
be the event getting maximum number 7.
Then, P (E1 ) = P (getting one number 3 and other two
from numbers 4 to 10)
1
C1 × 7C 2 7
= 10
=
40
C3
P (E 2) = P(getting one number 7 and other two from
numbers 1 to 6)
1
C1 × 6C 2 1
= 10
=
8
C3
and P (E1 ∩ E 2) = P(getting one number 3, second
number 7 and third from 4 to 6)
1
C1 × 1C1 × 3C1 1
=
=
10
40
C3
∴
Alternate Solution
P ( A ∪ B)′ = 1 − P ( A ∪ B) = 1 − { P ( A ) + P (B) − P ( A ) ⋅ P (B)}
= {1 − P ( A )}{1 − P (B)} = P ( A )′ P (B)′
7. Given,
P ( A ∪ B) = 0.6 , P ( A ∩ B) = 0.2
∴ P ( A ) + P (B ) = [1 − P ( A )] + [1 − P (B)]
= 2 − [P ( A ) + P (B)]
= 2 − [P ( A ∪ B) + P ( A ∩ B)] = 2 − [0.6 + 0.2] = 1.2
8. Given, P ( A ) = 0.25, P (B) = 0.50, P ( A ∩ B) = 0.14
P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B)
= 0.25 + 0.50 – 0.14 = 0.61
Now, P ( A ∪ B) = 1 − P ( A ∪ B) = 1 − 0.61 = 0.39
P (E1 ∪ E 2) = P (E1 ) + P (E 2) − P (E1 ∩ E 2)
7
1 1 11
=
+ −
=
40 8 40 40
12. P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B)
If
P ( A ∪ B) = P ( A ∩ B),
then
P ( A ) and P (B) are equals.
Since, P ( A ∪ B) = P ( A ∩ B) ⇒ A and B are equals sets
Thus, P ( A ) and P (B) is equal to P ( A ∩ B).
13. Given, P (A fails in examination) = 0.2
and
P (B fails in examination) = 0.3
P ( A ∩ B) = P ( A )P (B) = (0.2) (0.3)
∴
∴
P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B)
= 0.2 + 0.3 − 0.06 = 0.44
Hence, it is a false statement.
9. We know that,
P ( A ∩ B) = P ( A ) + P (B) − P ( A ∪ B)
14. Let P ( A ) and P (B) denote respectively the percentage of
Also,
P ( A ∪ B) ≤ 1
city population that reads newspapers A and B.
∴
P ( A ∩ B)min. , when P ( A ∪ B)max = 1
Then,
⇒
P ( A ∩ B) ≥ P ( A ) + P (B) − 1
25 1
20 1
= , P (B) =
= ,
100 4
100 5
8
2
P ( A ∩ B) =
=
,
100 25
1 2
17
P ( A ∩ B) = P ( A ) − P ( A ∩ B) = −
=
,
4 25 100
1 2
3
P ( A ∩ B) = P (B) − P ( A ∩ B) = −
=
5 25 25
Let P (C ) be the probability that the population who
reads advertisements.
∴ P (C ) = 30% of P ( A ∩ B) + 40% of P ( A ∩ B)
∴ Option (a) is true.
Again,
P ( A ∪ B) ≥ 0
∴
P ( A ∩ B)max. , when P ( A ∪ B)min. = 0
⇒
P ( A ∩ B) ≤ P ( A ) + P (B)
∴ Option (b) is true.
Also, P ( A ∩ B) = P ( A ) + P (B) − P ( A ∪ B), Thus, (c) is
also correct.
Hence, (a), (b), (c) are correct options.
10. P(exactly one of M, N occurs)
= P{(M ∩ N ) ∪ (M ∩ N )} = P (M ∩ N ) + P (M ∩ N )
= P (M ) − P (M ∩ N ) + P (N ) − P (M ∩ N )
= P (M ) + P (N ) − 2P (M ∩ N )
Also, P(exactly one of them occurs)
= {1 − P (M ∩ N )}{1 − P (M ∪ N )}
P ( A) =
+ 50% of P ( A ∩ B)
[since, A ∩ B, A ∩ B and A ∩ B are all mutually
exclusive]
3
17
2 3
1 2
139
⇒ P (C ) =
×
+ ×
+ ×
=
= 13 . 9%
10 100 5 25 2 25 1000
Probability 119
 1
 
 2
15. We know that,
P ( A ) + P (B) + P (C ) − P ( A ∩ B) − P (B ∩ C )
− P (C ∩ A ) + P ( A ∩ B ∩ C ) = P ( A ∪ B ∪ C )
⇒ 0.3 + 0.4 + 0.8 – {0.08 + 0.28 + P (BC )} + 0.09
= P(A ∪ B ∪ C )
⇒ 1.23 − P (BC ) = P ( A ∪ B ∪ C )
where, 0.75 ≤ P ( A ∪ B ∪ C ) ≤ 1
⇒ 0.75 ≤ 1.23 − P (BC ) ≤ 1 ⇒ − 0.48 ≤ − P (BC ) ≤ − 0.23
⇒
0.23 ≤ P (BC ) ≤ 0.48
16. Given, P ( A ) = 0.5 and P ( A ∩ B) ≤ 0.3
⇒
P ( A ) + P (B) − P ( A ∪ B) ≤ 0.3
⇒
P (B) ≤ 0.3 + P ( A ∪ B) − P ( A ) ≤ P ( A ∪ B) − 0.2
[since, P ( A ∪ B) ≤ 1 ⇒ P ( A ∪ B) − 0.2 ≤ 0.8 ]
∴
P (B) ≤ 0.8
⇒
P (B) cannot be 0.9.
17. Here, five students S1 , S 2, S3 , S 4 and S5 and five seats
R1 , R2, R3 , R4 and R5
∴ Total number of arrangement of sitting five students
is 5 ! = 120
Here, S1 gets previously alloted seat R1
∴S 2, S3 , S 4 and S5 not get previously seats.
Total number of way S 2, S3 , S 4 and S5 not get previously
seats is
1
1
1
1
1 1
1


4 ! 1 −
+
−
+  = 24 1 − 1 + − +



1 ! 2 ! 3 ! 4 !
2 6 24
 12 − 4 + 1
= 24 
 =9


24
∴ Required probability =
9
3
=
120 40
18. Here, n (T1 ∩ T2 ∩ T3 ∩ T4 )
Total = − n (T1 ∪ T2 ∪ T3 ∪ T4 )
⇒ n (T1 ∩ T2 ∩ T3 ∩ T4 )
= 5 ! − [4C1 4 ! 2 ! − (3 C1 3 ! 2 ! + 3C1 3 ! 2 ! 2 !)
+ (2C1 2 ! 2 ! + 4C1 2 ⋅ 2 !) − 2]
⇒ n (T1 ∩ T2 ∩ T3 ∩ T4 )
= 120 − [192 − (36 + 72) + (8 + 16) − 2]
∴
= 120 − [192 − 108 + 24 − 2] = 14
14
7
Required probability =
=
120 60
Topic 3 Independent and Conditional
Probability
1. Let event B is being boy while event G being girl.
1
2
Now, required conditional probability that all children
are girls given that at least two are girls, is
All 4 girls
=
(All 4 girls ) + (exactly 3 girls + 1 boy)
+ (exactly 2 girls + 2 boys)
According to the question, P (B) = P (G ) =
=
2.
4
3
4
2
 1
 1  1 4  1  1
4
  + C3     + C 2   
 2
 2  2
 2  2
2
=
1
1
=
1 + 4 + 6 11
Key Idea Use P ( A) = 1 − P ( A) and condition of independent
events i.e P ( A ∩ B) = P ( A) ⋅ P ( B)
Given that probability of hitting a target independently
by four persons are respectively
1
1
1
1
P1 = , P2 = , P3 = and P4 =
2
3
4
8
Then, the probability of not hitting the target is
1 
1 
1 
1

= 1 −  1 −  1 −  1 − 

2 
3 
4 
8
[Q events are independent]
1 2 3 7
7
× × × =
2 3 4 8 32
Therefore, the required probability of hitting the target
= 1 − (Probability of not hitting the target)
7 25
=1−
=
32 32
P ( A ∩ B)
3. We know that, P( A / B) =
P (B)
[by the definition of conditional probability]
Q
A⊂B
⇒
A∩B= A
P ( A)
…(i)
P( A / B) =
∴
P (B)
As we know that, 0 ≤ P (B) ≤ 1
1
P ( A)
1≤
< ∞ ⇒ P ( A) ≤
<∞
∴
P (B)
P (B)
P ( A)
…(ii)
⇒
≥ P ( A)
P (B)
Now, from Eqs. (i) and (ii), we get
P( A/B) ≥ P(A)
=
4. In {1, 2, 3, ...., 11} there are 5 even numbers and 6 odd
numbers. The sum even is possible only when both are
odd or both are even.
Let A be the event that denotes both numbers are even
and B be the event that denotes sum of numbers is even.
Then, n ( A ) = 5C 2 and n (B) = 5C 2 + 6C 2
Required probability
5
P ( A ∩ B)
C / 11C
= 6 2 5 2
P ( A / B) =
P (B)
( C 2 + C 2)
11
C2
5
C2
10
2
= 6
=
=
C 2 + 5C 2 15 + 10 5
1
2
1
P (T ) = Probability of getting tail =
2
5. Clearly, P (H ) = Probability of getting head =
and
Now, let E1 be the event of getting a sum 7 or 8, when a
pair of dice is rolled.
120 Probability
Then, E1 = {(6, 1), (5, 2), (4, 3), (3, 4), (2, 5),
(1, 6), (6, 2), (5, 3), (4, 4), (3, 5), (2, 6)}
⇒ P (E1 ) = Probability of getting 7 or 8 when a pair of
11
36
dice is thrown =
Also, let P (E 2) = Probability of getting 7 or 8 when a
2
9
∴Probability that the noted number is 7 or 8
= P ((H ∩ E1 ) or (T ∩ E 2))
= P (H ∩ E1 ) + P (T ∩ E 2)
[Q (H ∩ E1 ) and (T ∩ E 2) are mutually exclusive]
= P (H ) ⋅ P (E1 ) + P (T ) ⋅ P (E 2)
[Q{ H , E1 } and {T , E 2} both are sets of
independent events]
1 11 1 2 19
= ×
+ × =
2 36 2 9 72
card is picked from cards numbered 1, 2, ...., 9 =
6. Clearly, E1 = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}
E 2 = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2)}
E3 = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5),
(3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5),
(5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)}
6 1
6 1
⇒ P (E1 ) =
= , P (E 2) =
=
36 6
36 6
18 1
and P (E3 ) =
=
36 2
Now, P (E1 ∩ E 2) = P (getting 4 on die A and 2 on die B)
1
=
= P (E1 ) ⋅ P (E 2)
36
P (E 2 ∩ E3 ) = P (getting 2 on die B and sum of numbers
on both dice is odd)
3
=
= P (E 2) ⋅ P (E3 )
36
P (E1 ∩ E3 ) = P (getting 4 on die A and sum of numbers
on both dice is odd)
3
=
= P (E1 ) ⋅ P (E3 )
36
and P (E1 ∩ E 2 ∩ E3 ) = P [getting 4 on die A, 2 on die B
and sum of numbers is odd]
= P (impossible event) = 0
Hence, E1, E 2 and E3 are not independent.
1
1
1
7. Given, P ( A ∪ B) = , P ( A ∩ B) = , P ( A ) =
6
4
4
1 5
∴
P ( A ∪ B) = 1 − P ( A ∪ B) = 1 − =
6 6
1 3
and
P ( A) = 1 − P ( A ) = 1 − =
4 4
and
∴
P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B)
5 3
1
= + P (B) −
6 4
4
1
P (B) = ⇒ A and B are not equally likely
⇒
3
1
P ( A ∩ B) = P ( A ) ⋅ P (B) =
4
So, events are independent.
8.
PLAN It is simple application of independent event, to solve a
certain problem or any type of compitition each event in
independent of other.
Formula used
P ( A ∩ B) = P ( A ) ⋅ P (B), when A and B are independent
events.
Probability that the problem is solved correctly by
atleast one of them = 1 − (problem is not solved by all)
∴ P (problem is solved) = 1 − P (problem is not solved)
= 1 − P ( A ) ⋅ P (B ) ⋅ P (C ) ⋅ P (D )
21 235
 1 1 3 7
=1 −  ⋅ ⋅ ⋅  =1 −
=
 2 4 4 8
256 256
2
5
For independent events,
P ( A ∩ B) = P ( A )P (B)
2
1 2 3 4
⇒
P ( A ∩ B) ≤ ⇒ P ( A ∩ B) =
,
,
,
5
10 10 10 10
[maximum 4 outcomes may be in A ∩ B]
1
(i) Now,
P ( A ∩ B) =
10
1
P ( A ) . P (B) =
⇒
10
1 5 1
P (B) =
× = , not possible
⇒
10 2 4
2
2
2
(ii) Now,
P ( A ∩ B) =
⇒
× P (B) =
10
5
10
5
P (B) =
, outcomes of B = 5
⇒
10
3
(iii) Now,
P ( A ∩ B) =
10
3
2
3
⇒
P ( A )P (B) =
⇒
× P (B) =
10
5
10
3
P (B) = , not possible
4
4
4
(iv) Now,
P ( A ∩ B) =
⇒ P ( A ) . p(B) =
10
10
9. Since, P ( A ) =
⇒
P (B) = 1 , outcomes of B = 10.
E ∩F
P (E c ∩ F c ∩ G )
 =
G
P (G )


P (G ) − P (E ∩ G ) − P (G ∩ F )
=
P (G )
P (G ) [1 − P (E ) − P (F )]
[Q P (G ) ≠ 0]
=
P (G )
10. P 
c
c
= 1 − P (E ) − P (F ) = P (E c ) − P (F )
11. Let E = event when each American man is seated
adjacent to his wife
and A = event when Indian man is seated adjacent
to his wife
Now,
n ( A ∩ E ) = (4 !) × (2 !)5
Even when each American man is seated adjacent to his
wife.
Again,
n (E ) = (5 !) × (2 !)4
Probability 121
∴
5
2
 A  n ( A ∩ E ) (4 !) × (2 !)
P  =
=
=
4
 E
n (E )
5
(5 !) × (2 !)
18. Let A, B and C denote the events of passing the tests I,
II and III, respectively.
Alternate Solution
Evidently A, B and C are independent events.
Fixing four American couples and one Indian man in
between any two couples; we have 5 different ways in
which his wife can be seated, of which 2 cases are
favourable.
2
∴ Required probability =
5
According to given condition,
1
= P [( A ∩ B) ∪ ( A ∩ C )]
2
= P ( A ∩ B) + P ( A ∩ C ) − P ( A ∩ B ∩ C )
= P ( A ) P (B) + P ( A ) ⋅ P (C ) − P ( A ) ⋅ P (B) ⋅ P (C )
1
1
= pq + p ⋅ − pq ⋅
2
2
12. Let E be the event of getting 1 on a die.
⇒ 1 = 2 pq + p − pq ⇒ 1 = p(q + 1)
1
5
and P (E ) =
6
6
∴ P (first time 1 occurs at the even throw)
= t2 or t4 or t6 or t8 ... and so on
⇒ P (E ) =
[Infact, Eq. (i) is satisfied for infinite number of values
of p and q. If we take any values of q such that 0 ≤ q ≤ 1,
1
. It is evident that,
then, p takes the value
q+1
1
0<
≤ 1 i.e. 0 < p ≤ 1. But we have to choose correct
q+1
answer from given ones.]
= { P (E )P (E )} + { P (E ) P (E ) P (E ) P (E )} + K ∞
5
3
5
5
5
1
5
1
5
1
          
36
=    +     +     +K∞ =
=
25 11
 6  6  6  6  6  6
1−
36
13. Probability that only two tests are needed = Probability
that the first machine tested is faulty × Probability that
2 1 1
the second machine tested is faulty = × =
4 3 6
19. Since, P ( A / B ) + P ( A / B ) = 1
∴
P(the event A happens at least once)
= 1 − P (none of the event happens)
= 1 − (0.6) (0.6) (0.6) = 1 − 0.216 = 0.784
1
3
 X  P (X ∩ Y ) 1
=
P  =
Y 
2
P (Y )
 Y  P (X ∩ Y ) 2
P  =
=
 X
P (X )
5
2
P (X ∩ Y ) =
15
4
P (Y ) =
15
21. P (X ) =
15. P (2 white and 1 black) = P (W1W 2B3 or W1B2W3 or
B1W 2W3 )
= P (W1W 2B3 ) + P (W1B2W3 ) + P (B1W 2W3 )
= P (W1 )P (W 2)P (B3 ) + P (W1 )P (B2)P (W3 )
+ P (B1 )P (W 2)P (W3 )
3.2.3 3.2.1 1.2.1
1
13
=
+
+
=
(9 + 3 + 1) =
4 4 4 4 4 4 4 4 4 32
32
16. Given, P (India wins) = 1/2
∴
P (India losses ) = 1 / 2
Out of 5 matches India’s second win occurs at third test.
⇒ India wins third test and simultaneously it has won
one match from first two and lost the other.
∴ Required probability = P (LWW ) + P (WLW )
3
3
1
 1
 1
=  +  =
 2
 2
4
17. Let A = getting not less than 2 and not greater than 5
⇒
4
A ={2, 3, 4, 5} ⇒ P ( A ) =
6
But die is rolled four times, therefore the probability in
getting four throws
 4  4  4  4 16
=     =
 6  6  6  6 81
P(A / B) = 1 − P(A / B)
20. Given that, P ( A ) = 0.4, P ( A ) = 0.6
14. The event that the fifth toss results in a head is
independent of the event that the first four tosses result
in tails.
∴ Probability of the required event = 1 / 2
…(i)
The values of option (c) satisfy Eq. (i).
4
2
−
 X ′  P (Y ) − P (X ∩ Y ) 15 15 1
P  =
=
=
4
Y 
P (Y )
2
15
1
4
2
7
7
P (X ∪ Y ) = +
−
=
=
3 15 15 15 15
22.
PLAN
(i) Conditional probability, i.e. P( A / B) =
P ( A ∩ B)
P( B)
(ii) P ( A ∪ B) = P( A ) + P( B) − P ( A ∩ B)
(iii) Independent event, then P ( A ∩ B) = P( A ) ⋅ P( B)
1
Y  1
,P  =
 X 3
2
and P (X ∩ Y ) = 6
 X  P (X ∩ Y )
P  =
∴
Y 
P (Y )
Here, P (X /Y ) =
122 Probability
1
1
1 /6
…(i)
=
⇒ P (Y ) =
3
2 P (Y )
P (X ∩ Y ) 1
Y  1
P  =
⇒
=
 X 3
P (X )
3
1 1
⇒
= P (X )
6 3
1
…(ii)
P (X ) =
∴
2
P (X ∪ Y ) = P (X ) + P (Y ) − P (X ∩ Y )
1 1 1 2
…(iii)
= + − =
2 3 6 3
1
1 1 1
and P (X ) ⋅ P (Y ) = ⋅ =
P (X ∩ Y ) =
6
2 3 6
⇒ P (X ∩ Y ) = P (X ) ⋅ P (Y )
⇒
i.e.
∴
independent events
P (X c ∩ Y ) = P (Y ) − P (X ∩ Y )
1 1 1
= − =
3 6 6
23.
student passes in Maths, Physics and Chemistry.
It is given,
P ( A ) = m, P (B) = p and P (C ) = c and
P (passing atleast in one subject)
= P ( A ∪ B ∪ C ) = 0.75
⇒
1 − P ( A′ ∩ B ′ ∩ C ′ ) = 0.75
[P ( A ) = 1 − P ( A )
Q
and
[P ( A ∪ B ∪ C ] = P ( A′ ∩ B′ ∩ C′ )]
⇒ 1 − P ( A′ ) . P (B ′ ) . P (C′ ) = 0.75
Q A, B and C are independent events, therefore A′, B′
and C ′ are independent events.
⇒
0.75 = 1 − (1 − m) (1 − p) (1 − c)
⇒
0 .25 = (1 − m) (1 − p) (1 − c)
…(i)
Also, P (passing exactly in two subjects)= 0.4
⇒ P ( A ∩ B ∩ C ∪ A ∩ B ∩ C ∪ A ∩ B ∩ C ) = 0.4
F
E
24. Let A, B and C respectively denote the events that the
⇒ P ( A ∩ B ∩ C ) + P ( A ∩ B ∩ C ) + P ( A ∩ B ∩ C ) = 0.4
⇒ P ( A ) P (B) P (C ) + P ( A )P (B ) P (C )
+ P ( A ) P (B) P (C ) = 0.4
P (E ∪ F ) − P (E ∩ F ) =
11
25
⇒
pm (1 − c) + p(1 − m) c + (1 − p) mc = 0.4
⇒
pm − pmc + pc − pmc + mc − pmc = 0.4
…(ii)
…(i)
[i.e. only E or only F]
Again, P (passing atleast in two subjects) = 0.5
⇒ P(A ∩ B ∩ C ) + P(A ∩ B ∩ C )
+ P ( A ∩ B ∩ C ) + P ( A ∩ B ∩ C ) = 0.5
⇒
E
F
pm(1 − c) + pc(1 − m) + cm(1 − p) + pcm = 0.5
⇒ pm − pcm + pc − pcm + cm − pcm + pcm = 0.5
⇒
( pm + pc + mc) − 2 pcm = 0.5 …(iii)
From Eq. (ii),
2
25
2
P (E ∩ F ) =
25
pm + pc + mc − 3 pcm = 0.4 …(iv)
Neither of them occurs =
⇒
11
From Eq. (i), P (E ) + P (F ) − 2 P (E ∩ F ) =
25
2
From Eq. (ii),
( 1 − P (E )) ( 1 − P (F )) =
25
2
1 − P (E ) − P (F ) + P (E ) ⋅ P (F ) =
⇒
25
From Eq. (i),
…(ii)
…(iii)
…(iv)
From Eqs. (iii) and (iv),
7
12
P (E ) + P (F ) = and P (E ) ⋅ P (F ) =
5
25
7
12
7
 12
− P (E ) =
⇒ (P (E ))2 − P (E ) +
=0
∴ P (E ) ⋅
 5
 25
5
25
⇒
3 
4

P (E ) −
P (E ) −
=0
5  
5 

∴
P (E ) =
4
3
3
4
or
⇒ P (F ) = or
5
5
5
5
0.25 = 1 − (m + p + c) + ( pm + pc + cm) − pcm
On solving Eqs. (iii), (iv) and (v), we get
p + m + c = 1.35 = 27 / 20
Therefore, option (b) is correct.
Also, from Eqs. (ii) and (iii), we get pmc = 1 / 10
Hence, option (c) is correct.
25. (a) P (E / F ) + P (E / F ) =
=
P (E ∩ F ) P (E ∩ F )
+
P (F )
P (F )
P (E ∩ F ) + P (E ∩ F ) P (F )
=
=1
P (F )
P (F )
Therefore, option (a) is correct.
P (E ∩ F ) P (E ∩ F )
(b) P (E / F ) + P (E / F ) =
+
P (F )
P (F )
=
P (E ∩ F ) P (E ∩ F )
+
≠1
P (F )
1 − P (F )
Therefore, option (b) is not correct.
…(v)
Probability 123
(c) P (E / F ) + P (E / F ) =
=
P (E ∩ F ) P (E ∩ F )
+
P (F )
P (F )
Now, P (E ∩ F ) = P (E ) − P (E ∩ F ) = P (E ) − P (E ) ⋅ P (F )
= P (E ) [1 − P (F )] = P (E ) ⋅ P (F )
P (E ∩ F ) P (E ∩ F )
+
≠1
P (F )
1 − P (F )
and P (E ∩ F ) = P (E ∪ F ) = 1 − P (E ∪ F )
Therefore, option (c) is not correct.
P (E ∩ F ) P (E ∩ F )
(d) P (E / F ) + P (E / F ) =
+
P (F )
P (F )
P (E ∩ F ) + P (E ∩ F ) P (F )
=
=
=1
P (F )
P (F )
Therefore, option (d) is correct.
26. Both E and F happen ⇒ P (E ∩ F ) =
1
12
1
and neither E nor F happens ⇒ P (E ∩ F ) =
2
But for independent events, we have
1
…(i)
P (E ∩ F ) = P (E ) P (F ) =
12
and
P (E ∩ F ) = P (E ) P (F )
= {1 − P (E )}{(1 − P (F )}
= 1 − P (E ) − P (F ) + P (E )P (F )
1
1
⇒
= 1 − { P (E ) + P (F )} +
2
12
1
1
7
…(ii)
P (E ) + P (F ) = 1 − +
=
⇒
2 12 12
On solving Eqs. (i) and (ii), we get
1
1
and P (F ) =
either
P (E ) =
3
4
1
1
or
and P (F ) =
P (E ) =
4
3
27. We know that,
 A  P ( A ∩ B) P ( A ) + P (B) − P ( A ∪ B)
=
P  =
 B
P (B)
P (B)
Since,
⇒
⇒
⇒
⇒
P ( A ∪ B) < 1
− P ( A ∪ B ) > −1
P ( A ) + P (B) − P ( A ∪ B) > P ( A ) + P (B) − 1
P ( A ) + P (B) − P ( A ∪ B) P ( A ) + P (B) − 1
>
P (B)
P (B)
 A  P ( A ) + P (B) − 1
P  >
 B
P (B)
Hence, option (a) is correct.
The choice (b) holds only for disjoint i.e. P ( A ∩ B) = 0
Finally, P ( A ∪ B) = P ( A ) + P (B) − P ( A ∩ B)
= P ( A ) + P (B) − P ( A ) ⋅ P (B),
if A , B are independent
= 1 − {1 − P ( A )} {1 − P (B)}
= 1 − P ( A ) ⋅ P (B )
Hence, option (c) is correct, but option (d) is not correct.
28. Since, E and F are independent events. Therefore,
P (E ∩ F ) = P (E ) ⋅ P (F ) ≠ 0, so E and F are not mutually
exclusive events.
= 1 − [1 − P (E ) ⋅ P (F )]
[Q E and F are independent]
= P (E ) ⋅ P (F )
So, E and F as well as E and F are independent events.
P (E ∩ F ) + P (E ∩ F )
Now, P (E / F ) + P (E / F ) =
P (F )
P (F )
=
=1
P (F )
29. P ( A c ) = 0.3
[given]
⇒
P ( A ) = 0.7
[given]
P (B) = 0.4
[given]
⇒
P (Bc ) = 0.6 and P ( A ∩ Bc ) = 0.5
Now, P ( A ∪ Bc ) = P ( A ) + P (Bc ) − P ( A ∩ Bc )
= 0.7 + 0.6 − 0.5 = 0.8
P{ B ∩ ( A ∪ Bc )}
c
∴ P [B / ( A ∪ B ] =
P ( A ∪ Bc )
P{(B ∩ A ) ∪ (B ∩ Bc )} P{(B ∩ A ) ∪ φ } P (B ∩ A )
=
=
0.8
0.8
0.8
1
c
=
[P ( A ) − P ( A ∩ B )]
0.8
0.7 − 0.5 0.2 1
=
=
=
0.8
0.8 4
=
30. P ( A ∪ B) = P ( A ) + P (B) − P ( A ) P (B), as A and B are
independent events.
⇒
0.8 = (0.3) + P (B) − (0.3) P (B)
5
⇒
0.5 = (0.7) P (B) ⇒ P (B) =
7
31. 5 can be thrown in 4 ways and 7 can be thrown in 6
ways, hence number of ways of throwing neither 5 nor 7
is
36 − (4 + 6) = 26
∴ Probability of throwing a five in a single throw with a
4 1
pair of dice =
= and probability of throwing neither
36 9
26 13
5 nor 7 =
=
36 18
Hence, required probability
1
2
2
 1  13  1  13  1
=   +     +     + ... = 9 =
13 5
 9  18  9  18  9
1−
18
32. Let R be drawing a red ball and B for drawing a black
ball, then required probability
= RRR + RBR + BRR + BBR
5
6 6
6
5
6
×
× 
=
×
× +
 10 11 10  10 11 10
4
7  4
7
6
4
+
×
× +
×
× 
 10 11 10  10 11 10
640 32
=
=
1100 55
124 Probability
33. Let A be the event that the maximum number on the
two chosen tickets is not more than 10, and B be the
event that the minimum number on them is 5
5
C
∴
P ( A ∩ B) = 100 1
C2
10
C
P ( A ) = 100 2
C2
and
[Q P ( A′ ) ≤ 1 and P (B ′ ) ≤ 1]
.
⇒ P ( A ∪ B) P ( A ′ ∩ B ′ ) ≤ P ( A ) . P (B ′ ) + P (B) . P ( A′ )
⇒ P ( A ∪ B) . P ( A′ ∩ B ′ ) ≤ P (C )
[Q P (C ) = P ( A ) . P (B ′ ) + P (B). P ( A′ )]
2
3
P (B) = probability that B will hit A =
1.2
P (B) . P (C ) 2 3 1
= P (B ∩ C / E ) =
=
=
2
2
P (E )
3
36. Let Ei denotes the event that the students will pass the
ith exam, where i = 1, 2, 3
and E denotes the student will qualify.
∴
P (E ) = [P (E1 ) × P (E 2 / E1 )]
+ [P (E1 ) × P (E 2′ /E1 ) × P (E3 / E 2′ )]
+ [P (E1′ ) × P (E 2 / E ′1 ) × P (E3 / E 2)]
p
p
2
= p + p(1 − p) . + (1 − p) . . p
2
2
⇒
P (E ) =
⇒
pn = (1 − p) pn − 1 + p(1 − p) pn − 2
Hence proved.
∴ By law of total probability,
⇒ { P ( A ) + P (B) − P ( A ∩ B)}{ P ( A′ ) . P (B ′ )}
[since A, B are independent, so A ′ , B ′ are independent]
∴ P ( A ∪ B) . P ( A ′ ∩ B ′ ) ≤ { P ( A ) + P (B)}. { P ( A′ ) . P (B′ )}
= P ( A ) . P ( A′ ) . P (B′ ) + P (B) . P ( A′ ) . P (B′ )
…(i)
≤ P ( A ) . P (B′ ) + P (B) . P ( A′ )
1
2
1
P (C ) = probability that C will hit A =
3
P (E ) = probability that A will be hit
1 2 2
P (E ) = 1 − P (B ) ⋅ P (C ) = 1 − ⋅ =
⇒
2 3 3
Probability if A is hit by B and not by C
For n ≥ 3, pn = pn − 1 (1 − p) + pn − 2(1 − p) p
E 2 = the event noted number is 8
H = getting head on coin
T = be getting tail on coin
34. Here, P ( A ∪ B) . P ( A′ ∩ B ′ )
35. Given, P ( A ) = probability that A will hit B =
= 1 − p(HH ) = 1 − pp = 1 − p2
38. Let, E1 = the event noted number is 7
 B  P ( A ∩ B)
P  =
 A
P ( A)
5
C1 1
= 10
=
C2 9
Then
For n = 2, p2 = 1 − p (two heads simultaneously occur).
2 p2 + p2 − p3 + p2 − p3
= 2 p2 − p3
2
37. Since, pn denotes the probability that no two (or more)
consecutive heads occur.
⇒ pn denotes the probability that 1 or no head occur.
For n = 1 , p1 = 1 because in both cases we get less than
two heads (H, T).
P (E1 ) = P (H ) ⋅ P (E1 / H ) + P (T ) ⋅ P (E1 / T )
and P (E 2) = P (H ) ⋅ P (E 2 / H ) + P (T ) ⋅ P (E 2 / T )
where, P (H ) = 1 / 2 = P (T )
P (E1/H ) = probability of getting a sum of 7 on two dice
Here, favourable cases are
∴
{(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}.
6 1
=
P (E1 / H ) =
36 6
Also,
P (E1 / T ) = probability of getting 7 numbered
card out of 11 cards
1
=
11
P (E 2 / H ) = probability of getting a sum of 8 on two dice
Here, favourable cases are
{(2, 6), (6, 2), (4, 4), (5, 3), (3, 5)}.
5
P (E 2 / H ) =
36
∴
P (E 2 / T ) = probability of getting ‘8’ numbered
card out of 11 cards
= 1 / 11
1
1
17
 1 1  1 1 
P (E1 ) =  ×  +  ×  =
+
=
 2 6  2 11 12 22 132
∴
and
1 5  1 1 
P (E 2) =  ×  +  × 
 2 36  2 11
=
1
2
91
 91 

 =
 396 729
Now, E1 and E 2 are mutually exclusive events.
Therefore,
P (E1 or E 2) = P (E1 ) + P (E 2) =
17
91 193
+
=
132 792 792
39. Let D1 denotes the occurrence of a defective bulb in Ist
draw.
Therefore, P (D1 ) =
50 1
=
100 2
and let D2 denotes the occurrence of a defective bulb in
IInd draw.
50 1
Therefore, P (D2) =
=
100 2
and let N 1 denotes the occurrence of non-defective bulb
in Ist draw.
Probability 125
50 1
=
100 2
Again, let N 2 denotes the occurrence of non-defective
bulb in IInd draw.
50 1
Therefore, P (N 2) =
=
100 2
Now, D1 is independent with N 1and D2 is independent
with N 2 .
According to the given condition,
A = {the first bulb is defective} = { D1D2, D1N 2}
B = {the second bulb is non-defective} = { D1N 2, N 1N 2}
and C = {the two bulbs are both defective}
= { D1D2, N 1N 2}
Again, we know that,
Therefore, P (N 1 ) =
⇒
6 x2 − 5 x + 1 = 0
⇒
(3x − 1)(2x − 1) = 0
1
1
x = and
3
2
1
1
P ( A ) = or
3
2
⇒
∴
42. P (N th draw gives 2nd ace)
= P{ 1 ace and (n − 2) other cards are drawn in (N − 1)
draws} × P { N th draw is 2nd ace}
4 ⋅ (48)! ⋅ (n − 1)! (52 − n )!
3
=
⋅
(52)! ⋅ (n − 2)! (50 − n )! (53 − n )
4(n − 1)(52 − n )(51 − n ) ⋅ 3
52 ⋅ 51 ⋅ 50 ⋅ 49
(n − 1) (52 − n ) (51 − n )
=
50 ⋅ 49 ⋅ 17 ⋅ 13
=
A ∩ B = { D1N 2}, B ∩ C = { N 1N 2}.
C ∩ A = { D1D2} and A ∩ B ∩ C = φ
P ( A ) = P{ D1D2} + P{ D1N 2}
Also,
= P (D1 )P (D2) + P (D1 )P (N 2)
 1  1  1  1 1
=   +   =
 2  2  2  2 2
1
1
Similarly, P (B) = and P (C ) =
2
2
 1  1 1
Also, P ( A ∩ B) = P (D1N 2) = P (D1 )P (N 2) =     =
 2  2 4
1
1
Similarly, P (B ∩ C ) = , P (C ∩ A ) =
4
4
and P ( A ∩ B ∩ C ) = 0.
Since,
P ( A ∩ B) = P ( A )P (B), P (B ∩ C ) = P (B)P (C )
and
P (C ∩ A ) = P (C )P ( A ).
Therefore, A, B and C are pairwise independent.
Also, P ( A ∩ B ∩ C ) ≠ P ( A )P (B)P (C ) therefore A, B and
C cannot be independent.
40. The total number of ways to answer the question
= C1 + C 2 + C3 + C 4 = 2 − 1 = 15
4
4
4
4
4
P(getting marks) = P( correct answer in I chance)
+ P(correct answer in II chance)
+ P( correct answer in III chance)
1
3 1
 14 1   14 13 1 
=
+ ⋅  + ⋅ ⋅ =
=




15
15 14
15 14 13
15 5
1
6
41. Given, P ( A ) ⋅ P (B) = , P ( A ) ⋅ P (B ) =
∴
Let
⇒
⇒
⇒
⇒
[1 − P ( A )] [1 − P (B)] =
1
3
1
3
P ( A ) = x and P (B) = y
1
1
and xy =
(1 − x)(1 − y) =
3
6
1
1
and xy =
1 − x − y + xy =
3
6
5
1
and xy =
x+ y=
6
6
5
 1
x  − x =
6
 6
43. Let P (H 1 ) = 0.4, P (H 2) = 0.3, P (H 3 ) = 0.2, P (H 4 ) = 0.1
P (gun hits the plane)
= 1 − P(gun does not hit the plane)
= 1 − P (H 1 ) ⋅ P (H 2) ⋅ P (H 3 ) ⋅ P (H 4 )
= 1 − (0.6) (0.7) (0.8) (0.9) = 1 − 0.3024 = 0.6976
44. Since, the drawn balls are in the sequence black, black,
white, white, white, white, red, red and red.
Let the corresponding probabilities be
Then,
p1 , p2,... , p9
2
1
4
3
2
p1 = , p2 = , p3 = , p4 = , p5 =
9
8
7
6
5
1
3
2
p6 = , p7 = , p8 = , p9 = 1
4
3
2
∴ Required probabilitie
p1 . p2 . p3 ⋅ K ⋅ p9
1
 2  1  4   3   2  1   3  2 
=                 (1) =
 9  8  7  6   5  4   3  2 
1260
45. Given sample space (S) of all 3 × 3 matrices with entries
from the set {0, 1} and events
E1 = { A ∈ S : det( A ) = 0} and
E 2 = { A ∈ S : sum of entries of A is 7}.
For event E 2, means sum of entries of matrix A is 7,
then we need seven 1s and two 0s.
9!
∴Number of different possible matrices =
7!2!
⇒
n (E 2) = 36
For event E1 ,| A|= 0, both the zeroes must be in same
row/column.
∴ Number of matrices such that their determinant is
zero
3!
= 6 × = 18 = n (E1 ∩ E 2)
2!
 E  n (E1 ∩ E 2)
∴Required probability, P  1  =
 E 2
n (E 2)
=
18 1
= = 0.50
36 2
126 Probability
46.
3. Let x = P (computer turns out to be defective, given that
PLAN
Forthe events to be independent,
it is produced in plant T2)
P( E1 ∩ E 2 ∩ E 3 ) = P( E1 ) ⋅ P( E 2 ) ⋅ P( E 3 )
P( E1 ∩ E 2 ∩ E 3 ) = P(only E1 occurs)
= P( E1 ) ⋅ (1 − P( E 2 )) (1 − P( E 3 ))
Let x, y and z be probabilities of E1 , E 2 and E3 ,
respectively.
…(i)
∴
α = x (1 − y) (1 − z )
…(ii)
β = (1 − x) ⋅ y(1 − z )
…(iii)
γ = (1 − x) (1 − y)z
…(iv)
⇒
p = (1 − x) (1 − y) (1 − z )
Given, (α − 2 β ) p = αβ and ( β − 3γ ) p = 2 βγ
…(v)
From above equations, x = 2 y and y = 3z
∴
x = 6z
x
=6
z
⇒
47. Here, P (X > Y ) = P (T1win ) P (T1 win )
+ P (T1 win ) P (draw ) + P (draw ) P (T1 win )
 1 1  1 1  1 1 5
= ×  + ×  + ×  =
 2 2  2 6  6 2 12
48. P [X = Y ] = P (draw ) ⋅ P (draw )
+ P (T1 win ) P (T2 win ) + P (T2 win ) ⋅ P (T1 win )
= (1 / 6 × 1 / 6) + (1 / 2 × 1 / 3) + (1 / 3 × 1 / 2) = 13 / 36
Topic 4 Law of Total Probability and
Baye’s Theorem
1. Let A be the event that ball drawn is given and B be the
event that ball drawn is red.
2
5
P ( A ) = and P (B) =
∴
7
7
Again, let C be the event that second ball drawn is red.
∴
2.
 D
x= P  
 T2
⇒
where, D = Defective computer
∴ P (computer turns out to be defective given that is
produced in plant T1) = 10x
 D
…(ii)
i.e.
P   = 10x
 T1 
20
80
and P (T2) =
100
100
7
Given, P (defective computer) =
100
7
i.e.
P (D ) =
100
P (T1 ) =
Also,
Using law of total probability,
 D
 D
P (D ) = 9(T1 ) ⋅ P   + P (T2) ⋅ P  
 T1 
 T2
∴
⇒
∴
7
 20 
 80 
=
 ⋅ 10x + 
⋅x
 100
100  100
1
7 = (280)x ⇒ x =
40
 D
 D  10
1
and P   =
P  =
 T2 40
 T1  40
Using Baye’s theorem,
P (T2 ∩ D )
T 
P  2 =
 D  P (T1 ∩ D ) + P (T2 ∩ D )
 D
P (T2) ⋅ P  
 T2
=
 D
 D
P (T1 ) ⋅ P   + P (T2) ⋅ P  
 T2
 T1 
80 39
⋅
78
100
40
=
=
20 30
80 39 93
⋅
+
⋅
100 40 100 40
Key idea Use the theorem of total probability
⇒
P (E 2) =
=
4. From the tree diagram, it follows that
S
4
6
6
4 24 + 24 48 2
×
+
×
=
=
=
10 12 10 12
120
120 5
1
5
4
5
 A
4
6
, P  =
10  E 2 12
By law of total probability
 A
 A
P ( A ) = P (E1 ) × P   + P (E 2) × P  
 E1 
 E 2
…(iii)
 D
 D
1 39
10 30
and P   = 1 −
…(iv)
=
=
⇒ P   =1−
40 40
40 40
 T2
 T1 
P (C ) = P ( A ) P (C / A ) + P (B) P (C / B)
2 6 5 4
= × + ×
7 7 7 7
12 + 40 32
=
=
49
49
Let E1 = Event that first ball drawn is red
E 2 = Event that first ball drawn is black
A = Event that second ball drawn is red
 A
4
6
P (E1 ) = , P   =
10  E1  12
…(i)
G
3
4
R
1
4
AR
AR
AG
1
4
3
4
3
4
1 3
4 4
11
44
BG
BR
BG
1 AG
3 4
4
BR
3
4
BG
Probability 127
46
80
10 5
P (BG|G ) =
=
16 8
5 4 1
P (BG ∩ G ) = × =
8 5 2
1
1 80 20
P (G|BG ) = 2 = ×
=
P (BG ) 2 46 23
P (BG ) =
∴
5.
= 1 /4
Now, (a) P
5 R
5 G
B1
3 R
5 G
5 R
3 G
B2
B3
Now, probability that the chosen ball is green, given
G
3
that selected bag is B3 = P   =
 B3 
8
Now, probability that the selected bag is B3 , given that
B 
the chosen ball is green = P  3 
G
=
G
P   P (B3 )
 B3 
G
G
G
P   P (B1 ) + P   P (B2) + P   P (B3 )
 B3 
 B1 
 B2
[by Baye’s
theorem]
=
3
5
× 

 10 10
3 4 
1
 × 
 8 10
4
2
=
=
 5 3   3 4  1 + 5 + 1 13
+ ×  + × 
 8 10  8 10 2 8 2
Now, probability that the chosen ball is green
G
G
G
= P (G ) = P (B1 )P   + P (B2)P   + P (B3 )P  
 B1 
 B2
 B3 
[By using theorem of total probability]
5   3 5  4 3
3
=
×  +
×  +
× 
 10 10  10 8  10 8
3
3
3 12 + 15 + 12 39
=
+
+
=
=
20 16 20
80
80
Now, probability that the selected bag is B3 and the
G
4 3 3
chosen ball is green = P (B3 ) × P   =
× =
 B3  10 8 20
Hence, options (a) and (c) are correct.
PLAN It is based on law of total probability and Bay’s Law.
Description of Situation It is given that ship would
work if atleast two of engines must work. If X be event
that the ship works. Then, X ⇒ either any two of
E1 , E 2, E3 works or all three engines E1 , E 2, E3 works.
(X1c
/X)
1 1 1
 X1c ∩ X  P (E1 ∩ E 2 ∩ E3 ) 2 ⋅ 4 ⋅ 4 1
=
=P
=
=
1
P (X )
8
 P (X ) 
4
Key Idea Use conditional probability, total probability and Baye’s
theorem.
It is given that there are three bags B1 , B2 and B3 and
probabilities of being chosen B1 , B2 and B3 are
respectively
3
3
4
and P (B3 ) = .
∴ P (B1 ) = , P (B2) =
10
10
10
6.
1
1
1
Given, P (E1 ) = , P (E 2) = , P (E3 ) =
2
4
4
P (E1 ∩ E 2 ∩ E3 ) + P (E1 ∩ E 2 ∩ E3 ) 
P(X) = 
∴

+ P (E1 ∩ E 2 ∩ E3 ) + P (E1 ∩ E 2 ∩ E3 ) 
 1 1 3 1 3 1 1 1 1  1 1 1
= ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ 
 2 4 4 2 4 4 2 4 4  2 4 4
(b) P (exactly two engines of the ship are functioning)
=
P (E1 ∩ E 2 ∩ E3 ) + P (E1 ∩ E 2 ∩ E3 ) + P (E1 ∩ E 2 ∩ E3 )
P (X )
1 1 3 1 3 1 1 1 1
⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅
7
=2 4 4 2 4 4 2 4 4=
1
8
4
 X  P (X ∩ X 2)
(c) P   =
 X 2
P (X 2)
=
P (ship is operating with E 2 function )
P (X 2)
P (E1 ∩ E 2 ∩ E3 ) + P (E1 ∩ E 2 ∩ E3 ) + P (E1 ∩ E 2 ∩ E3 )
P (E 2)
1 1 3 1 1 1 1 1 1
⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅
5
= 2 4 4 2 4 4 2 4 4 =
1
8
4
P (X ∩ X1 )
(d) P (X / X1 ) =
P (X1 )
1 1 1 1 3 1 1 1 3
⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅
=2 4 4 2 4 4 2 4 4
1 /2
7
=
16
=
7. Statement I If P (H i ∩ E ) = 0 for some i, then
 E
H 
P  i = P   = 0
 E
 Hi 
If P (H i ∩ E ) ≠ 0, ∀ i = 1, 2, K , n , then
 H  P (H i ∩ E ) P (H i )
×
P  i =
 E
P (H i )
P (E )
=
 E
P   × P (H i )
 Hi 
P (E )
 E
> P   ⋅ P (H i )
 Hi 
[Q0 < P (E ) < 1]
Hence, Statement I may not always be true.
Statement II Clearly, H 1 ∪ H 2 ∪ . . . ∪ H n = S
[sample space]
⇒
P (H 1 ) + P (H 2) + . . . + P (H n ) = 1
Hence, Statement II is ture.
128 Probability
Head appears
Passage I
2W
2W
8.
n1
n2
Red
n3
Black
n4
Box I
2R
Red
or
Black
1R
∴
1
3
P (B2) ⋅ P (B2 ∩ A ) 1
=
P ( A)
3
⇒
3W
1W
2W
2R
U2
1W
2W
2W
0R
2R
1R
1R
U1
U2
U1
U2
3W
3 Cases
10. Now, probability of the drawn ball from U 2 being white
⇒
1  n3 


2  n3 + n4 
1
=
3




1
n1
1
n3

 + 

2  n1 + n2 2  n3 + n4 
⇒
n3 (n1 + n2)
1
=
n1 (n3 + n4 ) + n3 (n1 + n2) 3
is
1
2
2
 3C
C
C
C 
P (white / U 2) = P (H ) ⋅  5 1 × 2 1 + 5 1 × 2 1 
C1
C1
C1 
 C1
Now, check options, then clearly options (a) and (b)
satisfy.
1
2
3
3
 3C
C
C
C
C1 ⋅ 2C1
+ P (T )  5 2 × 3 2 + 5 2 × 3 1 +
5
C2
C2
C2
C2
 C2
1 3
2 1
=  ×1 + × 
2 5
5 2
+
9.
1 Red
(n1 – 1)
n2
Red
(n3 + 1)
Black
Red
n1
= P (H ) .
Box II
Red
(n2 – 1) Black
(n4 + 1) Black
Box I
Box II
1
3
 n1 − 1   n1   n2  
 1
n1
⇒
 
+
 
=
 n1 + n2 − 1  n1 + n2  n1 + n2  n1 + n2 − 1 3
∴ P (drawing red ball from B1) =
n12 + n1n2 − n1
1
=
(n1 + n2) (n1 + n2 − 1) 3
⇒
Clearly, options (c) and (d) satisfy.
Passage II
3W
1W
2R
U1
U2
Initial
1
2
 3C1 2C1
C
C 
× 2
+ 5 1 × 2 1
5
C1
C1
C1 
 C1
23 / 30
2 1
3
 ×1 + × 
1 5
5 2
= 
20 / 30
2
12
=
23
Red
n3
C1 

C2 
2
3
11. P (Head appeared/white from U 2)
1 Black
or
×
13
1 1
6 2  23
× +
× =
 ×1 +
2 10
10 3 10 3  30
Black
n4
Box I
1R
U2
Tail appears
U1
P (B2 / A ) =
Given,
1W
1R
U1
P ( A ) = P (B1 ) ⋅ P ( A / B1 ) + P (B2) ⋅ P ( A / B2)
1  n1  1  n3 
= 
+ 

2  n1 + n2 2  n3 + n4 
2 Cases
U2
U1
3W
Box II
A = Drawing red ball
Let
1W
Passage III
5 5 1 25
6 6 6 216
5 5
25
13. P (X ≥ 3) = ⋅ ⋅ 1 =
6 6
36
P {(X > 3) / (X ≥ 6)} ⋅ P (X ≥ 6)
14. P {(X ≥ 6) / (X > 3)} =
P (X > 3)
5
  5  1  5 6  1

1 ⋅   ⋅   +   ⋅   + ... ∞ 
 6  6  6  6
 = 25
= 
36
  5 3 1  5 4 1

  ⋅ +   ⋅ + ... ∞ 
6
6
6
 6

12. P (X = 3) = ⋅ ⋅ =
Probability 129
Passage IV
2
15. Here, P (ui ) = ki, Σ P (ui ) = 1 ⇒ k =
n (n + 1)
lim P (W ) = lim
n→ ∞
n→ ∞
n
2i
∑ n (n + 1)2
i=1
n
2
u
+1
n
16. P  n  =
=
Σi
W 
n+1
n+1
W  2 + 4 + 6 + ...
n+2
=
 =
n (n + 1)
 E
2 (n + 1)
2
17. P 
18. As, the statement shows problem is to be related to
Baye’s law.
Let C , S , B, T be the events when person is going by car,
scooter, bus or train, respectively.
1
3
2
1
∴ P (C ) = , P (S ) = , P (B) = , P (T ) =
7
7
7
7
Again, L be the event of the person reaching office late.
∴ L be the event of the person reaching office in time.
 L 7
 L 8
 L 5
Then, P   = , P   = , P   =
C 9
S 9
 B 9
∴
12
=
2
2n (n + 1)(2n + 1)
= 2 /3
= lim
n→∞
6n (n + 1)2
and
 B
 B
P (B) = P   P ( A1 ) + P   P ( A2)
 A1 
 A2 
∴
 L 8
P  =
T 9
 L
P   ⋅ P (C )
C
C
P  =
 L
 L
 L
 L
P   ⋅ P (C ) + P   ⋅ P (S ) + P   ⋅ P (B)
S
C
 B
 L
+ P   ⋅ P (T )
T
7 1
×
1
9
7
=
=
7 1 8 3 5 2 8 1 7
× + × + × + ×
9 7 9 7 9 7 9 7
19. Let A1 be the event exactly 4 white balls have been
drawn. A2 be the event exactly 5 white balls have been
drawn.
A3 be the event exactly 6 white balls have been drawn.
B be the event exactly 1 white ball is drawn from two
draws. Then,
 B
 B
 B
P (B) = P   P ( A1 ) + P   P ( A2) + P   P ( A3 )
 A1 
 A2 
 A3 
 B
But P   = 0
 A3 
[since, there are only 6 white balls in the bag]
C 2.6 C 4
18
C6
10
.
C1.2 C1
+
12
C2
C1.6 C5 . 11C1.1 C1
12
18
C2
C6
12
20. Let E be the event that coin tossed twice, shows head at
first time and tail at second time and F be the event that
coin drawn is fair.
P (E / F ) ⋅ P (F )
P (F / E ) =
P (E / F ) ⋅ P (F ) + P (E / F ′ ) ⋅ P (F ′ )
1 1 m
⋅ ⋅
2 2 N
=
1 1 m 2 1 N −m
⋅ ⋅
+ ⋅ ⋅
2 2 N 3 3
N
m
9m
4
=
=
m 2 (N − m) 8N + m
+
4
9
21. Let W1 = ball drawn in the first draw is white.
B1 = ball drawn in the first draw in black.
W 2 = ball drawn in the second draw is white.
Then , P (W 2) = P (W1 ) P (W 2 / W1 ) + P (B1 )P (W 2 / B1 )
 m   m+ k   n  

m
=
 
 +
 

 m + n   m + n + k  m + n   m + n + k
=
m(m + k) + mn
m (m + k + n )
m
=
=
(m + n ) (m + n + k) (m + n ) (m + n + k) m + n
22. The number of ways in which P1 , P2, K , P8 can be paired
in four pairs
1 8
=
[( C 2)(6C 2)(4C 2)(2C 2)]
4!
1
8!
6!
4!
=
×
×
×
×1
4 ! 2 !6 ! 2 !4 ! 2 !2 !
1
8×7
6 ×5
4 ×3
=
×
×
×
4 ! 2 ! ×1 2 ! × 1 2 ! × 1
8 × 7 ×6 ×5
=
= 105
2 .2 .2 .2
Now, atleast two players certainly reach the second
round between P1, P2 and P3 and P4 can reach in final if
exactly two players play against each other between P1,
P2, P3 and remaining player will play against one of the
players from P5 , P6, P7, P8 and P4 plays against one of the
remaining three from P5 …P8.
This can be possible in
3
C 2 × 4C1 × 3C1 = 3 . 4 . 3 = 36 ways
∴ Probability that P4 and exactly one of P5 ... P8 reach
36 12
second round
=
=
105 35
If P1 , Pi , P4 and Pj , where i = 2 or 3 and j = 5 or 6 or 7
reach the second round, then they can be paired in 2
1 4
pairs in
( C 2) (2C 2) = 3 ways. But P4 will reach the
2!
final, if P1 plays against Pi and P4 plays against Pj .
130 Probability
Hence, the probability that P4 will reach the final round
1
from the second =
3
12 1 4
× = .
∴ Probability that P4 will reach the final is
35 3 35
∴
The probability of any one wining in the pairs of
S1 , S 2 = P (certain event) = 1
∴ The pairs of S1 , S 2 being in two pairs separately
and S1 wins, S 2 loses + The probability of S1 , S 2 being
in two pairs separately and S1 loses, S 2 wins.
α = probability of A getting the head on tossing firstly
= P (H 1 or T1T2T3 H 4 or T1T2T3T4T5T6H 7 or …)
= P (H ) + P (H )P (T )3 + P (H )P (T )6 + …
P (H )
p
=
=
3
1 − P (T )
1 − q3
(14)! 
(14)! 


7



1
1
2
(
!)7 ⋅ 7 !  1 1
(2 !) ⋅ 7 !
× × + 1 −
= 1 −
× ×

(16)!
(16)!  2 2
 2 2 



(2 !)8 ⋅ 8 ! 
(2 !)8 ⋅ 8 ! 
1 14 × (14)! 7
= ×
=
2 15 × (14)! 15
Also,
β = probability of B getting the head on tossing secondly
= P (T1H 2 or T1T2T3T4H 5 or T1T2T3T4T5T6T7H 8 or …)
= P (H ) [P (T ) + P (H )P (T )4 + P (H )P (T )7 + K ]
= P (T )[P (H ) + P (H )P (T )3 + P (H )P (T )6 + ... ]
p(1 − p)
= q α = (1 − p) α =
1 − q3
∴ Required probability =
p + p(1 − p)
=1 −
1 − (1 − p)3
=
1 − (1 − p)3 − p − p(1 − p)
1 − (1 − p)3
1 − (1 − p)3 − 2 p + p2 p − 2 p2 + p3
=
γ=
1 − (1 − p)3
1 − (1 − p)3
Also,
α=
p(1 − p)
p
, β=
1 − (1 − p)3
1 − (1 − p)3
24. (i) Probability of S1 to be among the eight winners
= (Probability of S1 being a pair )
× (Probability of S1 winning in the group)
1 1
[since, S1 is definitely in a group]
=1 × =
2 2
(ii) If S1 and S 2 are in the same pair, then exactly one
wins.
If S1 and S 2 are in two pairs separately, then exactly
one of S1 and S 2 will be among the eight winners. If
S1 wins and S 2 loses or S1 loses and S 2 wins.
Now, the probability of S1 , S 2 being in the same pair
and one wins
= (Probability of S1 , S 2 being the same pair)
× (Probability of anyone winning in the pair).
and the probability of S1 , S 2 being the same pair
n (E )
=
n (S )
where, n (E ) = the number of ways in which 16
persons can be divided in 8 pairs.
1
7
8
+
=
15 15 15
25. Let E1 , E 2, E3 and A be the events defined as
E1 = the examinee guesses the answer
p + p(1 − p)
γ = 1 − (α + β ) = 1 −
1 − q3
⇒
(14)!
(16)!
and n (S ) =
(2 !)7 ⋅ 7 !
(2 !)8 ⋅ 8 !
∴ Probability of S1 and S 2 being in the same pair
(14)! ⋅ (2 !)8 ⋅ 8 ! 1
=
=
(2 !)7 ⋅ 7 !⋅ (16)! 15
23. Let q = 1 − p = probability of getting the tail. We have,
Again, we have
α + β + γ =1
n (E ) =
E 2 = the examinee copies the answer
E3 = the examinee knows the answer
and A = the examinee answer correctly
1
1
We have,
P (E1 ) = , P (E 2) =
3
6
Since, E1 , E 2, E3 are mutually exclusive and exhaustive
events.
1 1 1
∴ P (E1 ) + P (E 2) + P (E3 ) = 1 ⇒ P (E3 ) = 1 − − =
3 6 2
If E1 has already occured, then the examinee guesses.
Since, there are four choices out of which only one is
correct, therefore the probability that he answer
correctly given that he has made a guess is 1/4.
1
i.e.
P ( A / E1 ) =
4
1
It is given that,
P ( A / E 2) =
8
and P ( A / E3 ) = probability that he answer correctly
given that he know the answer = 1
By Baye’s theorem, we have
P (E3 ) ⋅ P ( A / E3 )
P (E3 / A ) =


P (E1 ) ⋅ P ( A / E1 ) + P (E 2) ⋅ P ( A / E 2) 

+ P (E3 ) ⋅ P ( A / E3 ) 

1
×1
24
2
=
∴ P (E3 / A ) =
 1 1  1 1  1
 29
 ×  +  ×  +  × 1
 3 4  6 8  2

26. Let
and
Bi = ith ball drawn is black.
Wi = ith ball drawn is white, where i = 1, 2
A = third ball drawn is black.
Probability 131
We observe that the black ball can be drawn in the third
draw in one of the following mutually exclusive ways.
(i) Both first and second balls drawn are white and
third ball drawn is black.
i.e.
(W1 ∩ W 2) ∩ A
(ii) Both first and second balls are black and third ball
drawn is black.
i.e.
(B1 ∩ B2) ∩ A
(iii) The first ball drawn is white, the second ball drawn
is black and the third ball drawn is black.
i.e.
(W1 ∩ B2) ∩ A
(iv) The first ball drawn is black, the second ball drawn
is white and the third ball drawn is black.
i.e.
∴
(B1 ∩ W 2) ∩ A
P ( A ) = P [{(W1 ∩ W 2) ∩ A } ∪{(B1 ∩ B2) ∩ A }
∪ {(W1 ∩ B2) ∩ A } ∪ {(B1 ∩ W 2) ∩ A }]
= P{(W1 ∩ W 2) ∩ A } + P{(B1 ∩ B2) ∩ A }
+ P{(W1 ∩ B2) ∩ A } + P{(B1 ∩ W 2) ∩ A }
= P (W1 ∩ W 2) ⋅ P ( A / (W1 ∩ W 2)) + P (B1 ∩ B2)
∴
P ( A / (B1 ∩ B2)) + P (W1 ∩ B2) ⋅ P ( A / (W1 ∩ B2))
+ P (B1 ∩ W 2) ⋅ P ( A / (B1 ∩ W 2))
 2 3 4
 2 1
=  ×  ×1 +  ×  ×
 4 5 6
 4 3
 2 2 3  2 2 3
+ ×  × + ×  ×
 4 3 4  4 5 4
1 1 1 3 23
= + + +
=
6 5 4 20 30
27. The testing procedure may terminate at the twelfth
testing in two mutually exclusive ways.
I : When lot contains 2 defective articles.
II : When lot contains 3 defective articles.
Let A = testing procedure ends at twelth testing
A1 = lot contains 2 defective articles
A2 = lot contains 3 defective articles
∴ Required probability
= P ( A1 ) ⋅ P ( A / A1 ) + P ( A2) ⋅ P ( A / A2)
Here, P ( A / A1 ) = probability that first 11 draws contain
10 non-defective and one-defective and twelfth draw
contains a defective article.
18
C10 × 2C1 1
…(i)
=
×
20
9
C11
P ( A / A2) = probability that first 11 draws contains 9
non-defective and 2-defective articles and twelfth draw
17
C 9 × 3C 2 1
… (ii)
contains defective =
×
20
9
C11
∴ Required probability
= (0.4)P ( A / A1 ) + 0.6 P ( A / A2)
=
0.4 × 18C10 × 2C1 1 0.6 × 17C 9 × 3C 2 1
99
× +
× =
20
20
9 1900
9
C11
C11
Topic 5 Probability Distribution and
Binomial Distribution
1. It is given that α is the number of heads that appear
when C1 is tossed twice, the probability distribution of
random variable α is
α
0
P(α)
 1
 
 3
2
1
2
2 1
2    
 3  3
2
 
 3
2
Similarly, it is given that β is the number of heads that
appear when C 2 is tossed twice, so probability
distribution of random variable β is
β
P(β)
0
2
 
 3
2
1
2
4
9
1
9
Now, as the roots of quadratic polynomial x2 − αx + β are
real and equal, so D = α 2 − 4β = 0 and it is possible if
(α,β) = (0, 0)
or (2, 1)
2
2
2
 2  4
 1  2
∴ Required probability =     +    
 3  9
 3  3
4
16 20
=
+
=
81 81 81
2. As we know that the probability of show up a three or a
five on a throw of a dice is
2 1
P (E ) = =
6 3
2
∴
P (E ) =
3
Now, as four fair dice are thrown independently 27
times, so probability of getting at least two 3’s or 5’s in
one trial
2
2
3
4
 1  2
 1  2
 1
p = 4C 2    + 4C3     + 4C 4  
 3  3
 3  3
 3
2
1
 4
= 6  4 + 4 4 + 4
3 
3
3
24 + 8 + 1 33 11
=
= 4 =
27
34
3
Therefore, the expected number of times in 27 trials
= 27 p = 11
3. It is given that out of the five machines in a workshop,
the probability of any one of them to be out of service on
1
a day is .
4
So, the probability that at most two machines will be
out of service on the same day
= probability that no machine is out of service
+ probability that exactly one machine is out of service +
probability that exactly two of machines are out of
service
3
5
4
2
3
1
1
1
1
 3

 1 
5
5
1 −  + C1 1 −  + C 2  1 −  =   k
 4

 4 
4
4
4
4
(given)
132 Probability
5
3
4
5  3
33  3
 3
⇒   +   + 10 5 =   k
 4
 4
4  4
4
According to the question,
n
n
99
99
 1
 1
1−  >
⇒  <1 −
 2
 2
100
100
n
1
 1
⇒ 2n > 100
⇒
  <
 2
100
[for minimum]
⇒
n=7
2
15 10
 3
k=  + 2 + 2
 4
4
4
9 + 15 + 10 34
=
=
16
16
17
k=
8
⇒
⇒
7. The required probability of observing atleast one head
4. Given that, there are 50 problems to solve in an
admission test and probability that the candidate can
4
solve any problem is = q (say). So, probability that the
5
4 1
candidate cannot solve a problem is p = 1 − q = 1 − = .
5 5
Now, let X be a random variable which denotes the
number of problems that the candidate is unable to
solve. Then, X follows binomial distribution with
1
parameters n = 50 and p = .
5
Now, according to binomial probability distribution
concept
r
50 − r
 1  4
, r = 0, 1, ... , 50
P (X = r ) = 50C r    
 5  5
∴Required probability
= P (X < 2) = P (X = 0) + P (X = 1)
49
49
50
449
 4
 4  4 50 54  4
= 50C 0   + 50C1
=   +  =
 
50
 5
 5  5 5 
5  5
(5)
5. Let for the given random variable ‘X’ the binomial
probability distribution have n-number of independent
trials and probability of success and failure are p and q
respectively. According to the question,
Mean = np = 8 and variance = npq = 4
1
1
q = ⇒ p =1−q =
∴
2
2
1
Now, n × = 8 ⇒ n = 16
2
 1
P (X = r ) = 16C r  
 2
∴
16
 1
+ 16 C1  
 2
16
 1
+ 16 C 2 
 2
k
1 + 16 + 120 137
=
= 16 = 16
216
2
2
⇒
16
failure in a trial respectively. Then,
2 1
4 2
p = P (5 or 6) = = and q = 1 − p = = .
6 3
6 3
Now, as the man decides to throw the die either till he
gets a five or a six or to a maximum of three throws, so
he can get the success in first, second and third throw or
not get the success in any of the three throws.
So, the expected gain/loss (in `)
= ( p × 100) + qp(− 50 + 100)
+ q2p(− 50 − 50 + 100) + q3 (− 50 − 50 − 50)
3
2
1
  2 1
 2  1
 2
=  × 100 +  ×  (50) +     (0) +   (− 150)
3
  3 3
 3  3
 3
100 100
1200
=
+
+0−
3
9
27
900 + 300 − 1200 1200 − 1200
=
=
=0
27
27
9. The probability of hitting a target at least once
[by using binomial distribution]
Here,
(given)
k = 137
6. As we know probability of getting a head on a toss of a
fair coin is P (H ) =
8. Let p and q represents the probability of success and
= 1 − (probability of not hitting the target in any trial)
= 1 − nC 0 p0qn
where n is the number of independent trials and p and q
are the probability of success and failure respectively.
16
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
 1
=16 C 0  
 2
= 1 − P (no head)
1
[let number of toss are n]
=1 − n
2
1

Q P (Head) = P (Tail) =

2 
1
90
According to the question, 1 − n ≥
100
2
1
1
⇒ 2n ≥ 10 ⇒ n ≥ 4
⇒ n ≤
10
2
So, minimum number of times one has to toss a fair coin
so that the probability of observing atleast one head is
atleast 90% is 4.
1
= p (let)
2
Now, let n be the minimum numbers of toss required to
get at least one head, then required probability
= 1 − (probability that on all ‘n’ toss we are getting tail)
n
1

 1
=1 −  
Q P (tail) = P (Head ) =

 2
2 
1
p=
3
and
q =1− p =1−
0
1 2
=
3 3
n
5
 1  2
According to the question, 1 − nC 0     >
 3  3
6
n
n
1
5
 2
 2
⇒   <1− ⇒   <
 3
 3
6
6
Clearly, minimum value of n is 5.
10. Let p = probability of getting an ace in a draw
= probability of success
and q = probability of not getting an ace in a draw
= probability of failure
Probability 133
Then,
p=
⇒
∴
Now,
4
1
=
52 13
1 12
=
13 13
Here, number of trials, n = 2
and
q =1 − p=1 −
p = 1 /2
n = 4, p = q = 1 / 2
P (X > 1) = 1 − { P (X = 0) + P (X = 1)}
0
4
1
Clearly, X follows binomial distribution with parameter
1
n = 2 and p = .
13
x
2− x
 1   12
Now, P (X = x) = 2C x    
, x = 0, 1, 2
 13  13
∴
17. Let E be the event that product of the two digits is 18,
P (X = 1) + P (X = 2)
1
2
 1   12
 1   12
= C1     + 2C 2   
 13  13
 13  13
0
therefore required numbers are 29 , 36, 63 and 92.
4
Hence, p = P (E ) =
100
2
1
 12 
=2
 +
 169 169
24
1
25
=
+
=
169 169 169
and probability of non-occurrence of E is
4
96
q = 1 − P (E ) = 1 −
=
100 100
11. Given box contains 15 green and 10 yellow balls.
∴ Total number of balls = 15 + 10 = 25
15 3
P(green balls) =
= = p = Probability of success
25 5
10 2
P(yellow balls) =
= = q = Probability of unsuccess
25 5
and n = 10 = Number of trials.
3 2 12
∴Variance = npq = 10 × × =
5 5 5
1
12. Probability of guessing a correct answer, p = and
3
probability of guessing a wrong answer, q = 2 /3
∴ The probability of guessing a 4 or more correct
4
5
2
1 11
 1 2
 1
answers = 5C 4   ⋅ + 5C5   = 5 ⋅ 5 + 5 = 5
 3 3
 3
3
3
3
13. India play 4 matches and getting at least 7 points. It can
only be possible in WWWD or WWWW position, where W
represents two points and D represents one point.
Therefore, the probability of the required event
= [4(0.05) + 0.5] (0.5)
C50 p50 (1 − p)50 = 100C51 ( p)51 (1 − p)49
(100) ! (51 !) × (49 !)
p
p
51
⋅
=
⇒
=
(50 !) (50 !)
100 !
1− p
1 − p 50
100
51
101
15. For Binomial distribution, mean = np
and
∴
⇒
variance = npq
np = 2 and
npq = 1
q = 1 / 2 and p + q = 1
4
18. Since, set A contains n elements. So, it has 2n subsets.
∴ Set P can be chosen in 2n ways, similarly set Q can be
chosen in 2n ways.
∴ P and Q can be chosen in (2n )(2n ) = 4n ways.
Suppose, P contains r elements, where r varies from 0 to
n. Then, P can be chosen in nC r ways, for 0 to be disjoint
from A, it should be chosen from the set of all subsets of
set consisting of remaining (n − r ) elements. This can be
done in 2n − r ways.
∴ P and Q can be chosen in nC r ⋅ 2n − r ways.
But, r can vary from 0 to n.
∴ Total number of disjoint sets P and Q
n
∑ nC r2n − r = (1 + 2)n = 3n
Hence, required probability =
binomial variate with parameters n = 100 and p.
Since,
P (X = 50) = P (X = 51)
p=
3
97
 4   96   4 
=4 
 = 4
 +
 
 100  100  100
25
r=0
14. Let X be the number of coins showing heads. Let X be a
⇒
P = 4C3 p3 q + 4C 4 p4
3
= 0.0875
⇒
Out of the four numbers selected, the probability that
the event E occurs atleast 3 times, is given as
=
= 4C3 (0.05) (0.5)3 + 4C 4 (0.5)4
⇒
 1  1
   
 2  2
3
 1  1
= 1 − 4C 0     − 4C1
 2  2
1
4 11
=1 −
−
=
16 16 16
0 .1
0 .1
5
16. Probability (face 1) =
=
=
0 .1 + 0 .32 0 .42 21
[given]
3 n  3
= 
4 n  4
n
19. Case I When A plays 3 games against B.
In this case, we have n = 3, p = 0.4 and q = 0.6
Let X denote the number of wins. Then,
P (X = r ) = 3C r (0.4)r (0.6)3 − r; r = 0, 1, 2, 3
∴ P1 = probability of winning the best of 3 games
= P (X ≥ 2)
= P (X = 2) + P (X = 3)
= 3C 2(0.4)2(0.6)1 + 3C3 (0.4)3 (0.6)0
= 0.288 + 0.064 = 0.352
Case II When A plays 5 games against B.
In this case, we have
n = 5, p = 0.4
and
q = 0.6
134 Probability
Let X denotes the number of wins in 5 games.
Then,
P (X = r ) = 5C r (0.4)r (0.6)5 − r , where r = 0, 1, 2K , 5
∴ P2 = probability of winning the best of 5 games
= P (X ≥ 3)
= P (X = 3) + P (X = 4) + P (X = 5)
= 5C3 (0.4)3 (0.6)2 + 5C 4 (0.4)4 (0.6) + 5C5 (0.4)5 (0.6)0
= 0.2304 + 0.0768 + 0.1024 = 0.31744
Clearly, P1 > P2. Therefore, first option i.e. ‘best of 3
games’ has higher probability of winning the match.
20. The man will be one step away from the starting point,
if
(i) either he is one step ahead or (ii) one step behind the
starting point.
The man will be one step ahead at the end of eleven
steps, if he moves six steps forward and five steps
backward. The probability of this event is
11
C 6 (0.4)6 (0.6)5 .
The man will be one step behind at the end of eleven
steps, if he moves six steps backward and five steps
forward. The probability of this event is 11C 6 (0.6)6 (0.4)5 .
∴ Required probability
= 11C 6 (0.4)6 (0.6)5 +
C 6 (0.6)6 (0.4)5 = 11C 6 (0.24)5
11
21. Let an event E of sum of outputs are perfect square
(i.e., 4 or 9), so
E = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3)}
and an event F of sum of outputs are prime numbers
(i.e., 2, 3, 5, 7, 11) so
F = {(1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6),
(2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6), (6, 5)}
and event T of sum of outputs are odd numbers
(i.e., 3, 5, 7, 9, 11)
T = {(1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6),
(2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (3, 6), (4, 5), (5, 4),
(6, 3), (5, 6), (6, 5)}
Now, required probability p = P (T / E )
P (T ∩ E )
=
P (E )
where, P (T ∩ E ) = probability of occuring perfect square
odd number before prime
 4   14  4 
=  +   +
 36  36  36
4
4
2
36
=
=
=
14 22 11
1−
36
2
 14  4 
    +…∞
 36  36
and P (E ) = probability of occuring perfect square before
prime
2
 7   14  7   14  7 
=   +     +     +…∞
 36  36  36  36  36
7
7
36
=
=
14 22
1−
36
2
4
∴ P (T / E ) = 11 = = p
7
7
22
⇒
14 p = 8
22. It is given that the probability, a missile hits a target
3
successfully p = , so the probability to not hits the
4
1
target is q = .
4
And it is also given that to destroy the target
completely, at least three successful hits are required.
Now, according to the question, let the minimum
number of missiles required to fired is n, so
n
C3 p3 qn − 3 + nC 4 p4qn − 4 + … + nC n pn ≥ 0.95
⇒
n − 2
2
n
n −1

 1
 3  1
 3  1
1 −  nC 0   + nC1    
+ nC 2   
 ≥ 0.95
 4
 4  4
 4  4


95
1
3n n (n − 1) 9
⇒
≥
+
+
1−
100 4n 4n
2
4n
n
2
4
2 + 6n + 9n − 9n
≥
⇒
20
2
⇒
10(9n 2 − 3n + 2 ) ≤ 4n
Now, at n = 3, LHS = 720, RHS = 64
at n = 4, LHS = 1340, RHS = 256
at n = 5, LHS = 2120, RHS = 1024
at n = 6, LHS = 3080, RHS = 4096
Hence, n = 6 missiles should be fired.
23. Using Binomial distribution,
P (X ≥ 2) = 1 − P (X = 0) − P (X = 1)
n
n −1

 1  1
 1
= 1 −   −  nC1 ⋅   ⋅  

 2  2
 2


1
1
 1 + n
= 1 − n − nC1 ⋅ n = 1 −  n 
 2 
2
2
Given, P (X ≥ 2) ≥ 0. 96
n+1 1
(n + 1) 24
∴
≤
⇒ n =8
1−
≥
⇒
25
25
2n
2n
7
Matrices and Determinants
Topic 1 Types of Matrices, Addition, Subtraction,
Multiplication and Transpose of a Matrix
Objective Questions I (Only one correct option)
1. If A is a symmetric matrix and B is a skew-symmetric
2 3 
matrix such that A + B = 
, then AB is equal to
5 −1
(2019 Main, 12 April I)
 −4 −2
(a) 

 −1 4 
 4 −2
(c) 

 1 −4
4
(b) 
 −1
 −4
(d) 
1
−2
−4
(a) 52
(c) 201
2
4
 0 2y 1 
(x, y ∈ R, x ≠ y) for which AT A = 3I3 is


2x − y 1 
(2019 Main, 9 April II)
(b) 4
(d) 6
0 −1
cos α − sin α 
3. Let A = 
, (α ∈ R) such thatA32 = 
.

1 0 
sin α cos α 
(2019 Main, 8 April I)
Then, a value of α is
(a)
π
32
(b) 0
(c)
π
64
(d)
π
16
1 0 0
4. Let P = 3 1 0 and Q = [qij ] be two 3 × 3 matrices such


9 3 1
(b) 135
(d) 15
0 2q r 
5. Let A =  p q −r . If AAT = I3 , then| p|is


 p − q r 
(2019 Main, 11 Jan I)
(a)
1
5
(b)
1
2
(c)
1
3
(d)


 c 2 b 
AAT = 9 I, where, I is 3 × 3 identity matrix, then the
ordered pair (a, b) is equal to
(2015 Main)
(a) (2, − 1)
(c) (2, 1)
1
6
(b) (−2, 1)
(d) (−2, − 1)
 3 /2
8. If P = 
 −1 /2
PTQ 2005 P is
1 1
1 /2 
T
, A = 0 1 and Q = PAP , then
3 / 2


 1 2005
(a) 
1 
0
1
0

(c) 

 2005 1

A 2 = B,is
(2019 Main, 12 Jan I)
2
7. If A = 2 1 −2 is a matrix satisfying the equation
9. If A = α
1
q + q31
that Q − P 5 = I3 . Then, 21
is equal to
q32
(a) 10
(c) 9
(b) 103
(d) 205
1 2
2. The total number of matrices A = 2x y −1 ,
(a) 2
(c) 3
0 0
1 0 and I be the identity matrix of order


16 4 1
3. If Q = [qij ] is a matrix, such that P50 − Q = I, then
q31 + q32
equals
q21
(2016 Adv.)
1
6. Let P =  4
1
0
and B = 
1
5
(a) 1
(c) 4
(2005, 1M)
2005
 1
(b) 
2005
1 

1
0


(d) 

 0 1
0
, then value of α for which
1
(2003, 1M)
(b) –1
(d) no real values
10. If A and B are square matrices of equal degree, then
which one is correct among the following?
(a) A + B = B + A
(b) A + B = A − B
(c) A − B = B − A
(d) AB = BA
(1995, 2M)
136 Matrices and Determinants
15. If the point P (a , b, c), with reference to Eq. (i), lies on
Objective Questions II
One or more than one correct option)
1 0 0
11. Let P1 = I = 0 1 0,


0 0 1
0
P3 = 1

0
0
P5 = 1

0
and
the plane 2x + y + z = 1, then the value of 7a + b + c is
(a) 0
1 0 0
P2 = 0 0 1,


0 1 0
0 1
P4 = 0 0

1 0
0 0
P6 = 0 1

1 0
2 1 3
6
X = ∑ Pk 1 0 2 PkT


k=1
3 2 1
1 0
0 0,

0 1
0 1
0 0,

1 0
0
1,

0
1
0

0
(a) 6
(c)
6
7
(d) ∞
with b and c satisfying Eq. (i) then the value of
3
1
3
+ b + c is
a
ω
ω
ω
(a) − 2
(c) 3
(b) 2
(d) − 3
Passage II
Let p be an odd prime number and T p be the following set
of 2 × 2 matrices
a b


Tp =  A = 
; a , b, c ∈ { 0, 1, 2, K , p − 1}
 c a
 (2010)

18. The number of A in T p such that det (A) is not divisible
by p, is
(b) p3 − 5 p
(d) p3 − p 2
(a) 2 p 2
(c) p3 − 3 p
12. Let X and Y be two arbitrary, 3 × 3, non-zero,
skew-symmetric matrices and Z be an arbitrary, 3 × 3,
non-zero, symmetric matrix. Then, which of the
following matrices is/are skew-symmetric?
(2015 Adv.)
(b) X 44 + Y 44
(d) X 23 + Y 23
19. The number of A in T p such that the trace of A is not
divisible by p but det ( A ) is divisible by p is
(a) ( p − 1) ( p 2 − p + 1)
(c) ( p − 1)2
(b) p3 − ( p − 1)2
(d) ( p − 1) ( p 2 − 2)
20. The number of A in T p such that A is either symmetric or
skew-symmetric or both and det (A) is divisible by p is
13. For 3 × 3 matrices M and N , which of the following
(2013 Adv.)
(a) N T M N is symmetric or skew-symmetric, according as
M is symmetric or skew-symmetric
(b) MN − NM is symmetric for all symmetric matrices M
and N
(c) M N is symmetric for all symmetric matrices M and N
(d) (adj M ) (adj N ) = adj (MN ) for all invertible matrices M
and N
14. Let ω be a complex cube root of unity with ω ≠ 0 and
P = [ pij ] be an n × n matrix with pij = ωi+ j . Then, P 2 ≠ 0
when n is equal to
(2013 Adv.)
(b) 55
(d) 56
(b) 2 ( p − 1)
(d) 2 p − 1
(a) ( p − 1)2
(c) ( p − 1)2 + 1
NOTE
The trace of a matrix is the sum of its diagonal entries.
Analytical and Descriptive Questions

 , where a , b, c are real

 c
positive numbers, abc = 1 and AT A = I , then find the
(2003, 2M)
value of a3 + b3 + c3 .
a
21. If matrix A =  b
b
c
a
c
a
b
Integer & Numerical Answer Type Questions
22. The trace of a square matrix is defined to be the sum of
its diagonal entries. If A is a 2 × 2 matrix such that the
trace of A is 3 and the trace of A3 is − 18, then the value
(2020 Adv.)
of the determinant of A is ……… .
Passage Based Problems
−1 + 3 i
, where i = −1, and r , s ∈ {1, 2, 3}. Let
2
(− z )r z 2s 
P =  2s
 and I be the identity matrix of order 2.
zr 
 z
Then, the total number of ordered pairs (r , s) for which
(2016 Adv.)
P 2 = − I is
23. Let z =
Passage I
Let a , b and c be three real numbers satisfying
1 9 7
[a b c] 8 2 7 = [0 0 0]


7 3 7
(b) 7
17. Let ω be a solution of x3 − 1 = 0 with Im (ω ) > 0. If a = 2
(a) X is a symmetric matrix
(b) The sum of diagonal entries of X is 18
(c) X − 30 I is an invertible matrix
1
1
(d) If X 1 = α 1, then α = 30


1
1
(a) 57
(c) 58
(d) 6
the roots of the quadratic equation ax2 + bx + c = 0, then
n
∞
 1 1
∑  α + β is equal to
n= 0
where,
denotes the transpose of the matrix Pk. Then
which of the following option is/are correct? (2019 Adv.)
statement(s) is/are not correct ?
(c) 7
16. Let b = 6, with a and c satisfying Eq. (i). If α and β are
PkT
(a)Y 3 Z 4 − Z 4Y 3
(c) X 4 Z 3 − Z 3 X 4
(b) 12
...(i)
(2011)
Matrices and Determinants 137
Topic 2 Properties of Determinants
Objective Questions I (Only one correct option)
1. If the minimum and the maximum values of the
function f :
π π 
,
→ R,
 4 2 
If det( A ) ∈ [2, 16], then c lies in the interval
− sin 2 θ −1 − sin 2 θ
defined by f (θ ) = − cos 2 θ −1 − cos 2 θ
12
1
1 are m and M
−2
10
respectively, then the ordered pair (m, M ) is equal to
(2020 Main, 5 Sep I)
(a) (0, 2 2 )
(b) (−4, 4)
(d) (−4, 0)
(c) (0, 4)
2. A value of θ ∈ (0, π / 3), for which
π
9
(b)
π
18
(c)
7π
24
(2019 Main, 12 April II)
(d)
7π
36
3. The sum of the real roots of the equation
−6
−1
− 3x x − 3 = 0, is equal to
− 3 2x x + 2
x
2
(b) − 4
(a) 0
4. If
(2019 Main, 10 April II)
(c) 6
(d) 1
sin θ cos θ
x
∆1 = − sin θ − x
1
cos θ
1
x
sin 2θ cos 2θ
x
and ∆ 2 = − sin 2θ
, x ≠ 0,
−x
1
cos 2θ
1
x
 π
then for all θ ∈ 0, 
 2
(2019 Main, 10 April I)
1 n − 1 1 78
=
,then the
1  0 1 
5. If 
.
.
 ... 
0 1 0 1 0 1 0
1 n 
inverse of 
 is
0 1 
 1 0
(a) 

12 1
(2019 Main, 9 April I)
 1 −12
(d) 

0 1 
 1 0
 1 −13
(b) 
 (c) 13 1
0
1




6. Let α and β be the roots of the equation x2 + x + 1 = 0.
Then, for y ≠ 0 in R,
α
β
y+1
α
1 is equal to
y+β
β
1
y+α
(a) y( y2 − 1)
(c) y3 − 1
(2019 Main, 9 April I)
(b) y ( y2 − 3)
(d) y3
(b) (2 + 23 / 4 , 4)
(d) [2, 3)
sin θ
1 
 3π 5π 
1
, ,
sin θ ; then for all θ ∈ 
 4 4


− sin θ
1 
 − 1
(2019 Main, 12 Jan II)
det( A ) lies in the interval

1
8. If A = − sin θ
5
(b)  , 4
 2 
3
(c)  0, 
 2 
5
(d)  1, 
 2 
2a
2a
a−b−c
9. If
2b
2b
b−c−a
2c
2c
c−a −b
= (a + b + c) (x + a + b + c)2, x ≠ 0 and a + b + c ≠ 0, then
(2019 Main, 11 Jan II)
x is equal to
(a) − (a + b + c)
(b) − 2(a + b + c)
(d) abc
(c) 2(a + b + c)
10. Let a1 , a 2, a3 ..... , a10 be in GP with ai > 0 for
i = 1, 2, ..... ,10 and S be the set of pairs (r , k), r , k ∈ N
(the set of natural numbers) for which
log e a1r a 2k
log e a 2ra3k
log e a 4r a5k
log e a7r a 8k
log e a5r a 6k
log e a 8r a 9k
log e a3r a 4k
log e a 6r a7k = 0
k
log e a 9r a10
Then, the number of elements in S, is (2019 Main, 10 Jan II)
(a) 4
(b) 2
(c) 10
(d) infinitely many
(a) ∆1 + ∆ 2 = − 2(x3 + x − 1)
(b) ∆1 − ∆ 2 = − 2x3
(c) ∆1 + ∆ 2 = − 2x3
(d) ∆1 − ∆ 2 = x(cos 2θ − cos 4θ)
1 1 1 2 1 3
(2019 Main, 8 April II)
(a) [3, 2 + 23 / 4 ]
(c) [4, 6]
3
(a)  , 3
 2 
sin 2 θ
1 + cos 2 θ
4 cos 6θ
2
cos θ
= 0, is
1 + sin 2 θ
4 cos 6θ
2
2
cos θ
sin θ
1 + 4 cos 6θ
(a)
1 1 1 
7. Let the numbers 2, b, c be in an AP and A = 2 b c .


2
2
4 b c 
b
1
2
2

11. Let A = b b + 1 b, where b > 0. Then, the minimum


b
2 
1
det ( A )
is
value of
b
(2019 Main, 10 Jan II)
(a) − 3
(b) −2 3
(c) 2 3
(d)
3
12. Let d ∈ R, and
(sin θ ) − 2
4+ d

 −2
 , θ ∈ [θ , 2π ]. If

(sin θ ) + 2
A= 1
d


 5 (2 sin θ ) − d (− sin θ ) + 2 + 2d 
the minimum value of det(A) is 8, then a value of d is
(2019 Main, 10 Jan I)
(a) −5
(b) −7
(c) 2( 2 + 1)
(d) 2( 2 + 2)
2x 
x − 4 2x
2

13. If  2x x − 4 2x 
 = ( A + Bx)(x − A ) ,
2x x − 4
 2x
ordered pair ( A , B) is equal to
(a) (−4, − 5)
(b) (−4, 3)
(c) (−4, 5)
then
the
(2018 Main)
(d) (4, 5)
138 Matrices and Determinants
14. Let ω be a complex number such that 2ω + 1 = z, where
1
1
1
2
z = − 3. If 1 −ω − 1 ω 2 = 3 k, then k is equal to
ω2
ω7
1
(2017 Main)
(a) − z
(b) z
(c) − 1
(d) 1
15. If α, β ≠ 0 and f (n ) = α n + β n and
1 + f (1) 1 + f (2)
3
1 + f (1) 1 + f (2) 1 + f (3)
1 + f (2) 1 + f (3) 1 + f (4)
= K (1 − α )2(1 − β )2 (α − β )2, then K is equal to
(2014 Main)
1
(b)
αβ
(a) αβ
(d) −1
(c) 1
16. Let P = [aij ] be a 3 × 3 matrix and let Q = [bij ], where
bij = 2i + j aij for 1 ≤ i , j ≤ 3. If the determinant of P is 2,
(2012)
then the determinant of the matrix Q is
(a) 210
(b) 211
(c) 212
(d) 213
2
17. If A = α
and| A3| = 125, then the value of α is
2 α


(a) ± 1
(b) ± 2
(c) ± 3
(d) ± 5
(2001, 1M)
cos x cos x 
π
π
sin x cos x  = 0 in the interval − ≤ x ≤ is
4
4
cos x sin x 
(a) 0
(c) 1
(b) 2
(d) 3
matrix with real entries?
1 0 0 
(a)  0 1 0 


 0 0 −1
(2017 Adv.)
1 0 0 
(b)  0 − 1 0 


 0 0 −1
− 1 0 0 
(c)  0 − 1 0 


 0 0 −1
 1 0 0
(d)  0 1 0


 0 0 1
24. Which of the following values of α satisfy the equation
(1 + α )2 (1 + 2α )2 (1 + 3 α )2
(2 + α )2 (2 + 2α )2 (2 + 3 α )2 = − 648 α ?
(3 + α )2 (3 + 2α )2 (3 + 3 α )2
(a) −4
(c) −9
(b) 9
(2015 Adv.)
(d) 4
25. Let M and N be two 3 × 3 matrices such that MN = NM .
Further, if M ≠ N 2 and M 2 = N 4, then
(2014 Adv.)
(a) determinant of (M 2 + MN 2 ) is 0
(b) there is a 3 × 3 non-zero matrixU such that (M 2 + MN 2 ) U
is zero matrix
(c) determinant of (M 2 + MN 2 ) ≥ 1
(d) for a 3 × 3 matrix U, if (M 2 + MN 2 ) U equals the zero
matrix, then U is the zero matrix

x
x+1
1

x (x − 1)
(x + 1) x
2x
,
 3x (x − 1) x (x − 1) (x − 2) (x + 1) x (x − 1) 

19. If f (x) = 
then f (100) is equal to
(a) 0
(c) 100
23. Which of the following is(are) NOT the square of a 3 × 3
(2004, 1M)
18. The number of distinct real roots of
 sin x
 cos x
 cos x
Objective Questions II
(One or more than one correct option)
(1999, 2M)
(b) 1
(d) –100
a
26. The determinant  b
aα + b
aα + b
bα + c 
0 
b
c
bα + c
is equal to zero, then
(1986, 2M)
(a) a, b, c are in AP
(b) a, b, c are in GP
(c) a, b, c are in HP
(d) (x − α ) is a factor of ax2 + 2bx + c
20. The parameter on which the value of the determinant
Integer & Numerical Answer Type Questions


1
a
a2
cos
(
p
d
)
x
cos
cos
(
px
−
p
+ d) x 



sin ( p − d ) x sin px sin ( p + d ) x 
does not depend upon, is
(a) a
(b) p
(1997, 2M)
(c) d
xp + y
x
21. The determinant yp + z
y
0
xp + y
(a) x, y, z are in AP
(c) x, y, z are in HP
27. Let P be a matrix of order 3 × 3 such that all the entries in
(d) x
y
= 0, if
z
yp + z (1997C, 2M)
(b) x, y, z are in GP
(d) xy, yz , zx are in AP
22. Consider the set A of all determinants of order 3 with
entries 0 or 1 only. Let B be the subset of A consisting
of all determinants with value 1. Let C be the subset
of A consisting of all determinants with value –1.
Then,
(a) C is empty
(b) B has as many elements as C
(c) A = B ∪ C
(d) B has twice as many elements as C
(1981, 2M)
P are from the set { − 1, 0, 1}. Then, the maximum possible
value of the determinant of P is ......... .
Fill in the Blanks
28. For positive numbers x, y and z, the numerical value of the
log x y log x z
1
determinant log y x
log y z is…… .
1
log z x log z y
1
1
29. The value of the determinant 
1

1

a
b
c
(1993, 2M)
a − bc
b2 − ca is … .

c2 − ab 
2
(1988, 2M)
x
30. Given that x = − 9 is a root of 2
7
roots are... and... .
3
x
6
7
2  = 0, the other two
x
(1983, 2M)
Matrices and Determinants 139
1
4
1
2x
20 
5 
 = 0 is… .
5x2 (1981, 2M)
31. The solution set of the equation
1 − 2
 λ2 + 3λ
32. Let pλ + qλ + rλ + sλ + t = 
 λ+1
 λ −3
be an identity in λ , where p,q,r,s and
Then, the value of t is…. .
4
3
2
λ − 1 λ + 3
− 2λ λ − 4 

λ + 4 3λ 
t are constants.
(1981, 2M)
True/False
1 a
1 a
bc
a2
33. The determinants 1 b ca and 1 b b
1
c
1
ab
c
2
are not
2
(1983, 1M)
Analytical and Descriptive Questions
34. If M is a 3 × 3 matrix, where M T M = I and det (M ) = 1,
then prove that det (M − I ) = 0
(2004, 2M)
35. Let a , b, c be real numbers with a + b + c = 1 . Show
2
2
 cos ( A − P )

 cos (B − P )
 cos (C − P )
2
that the equation
bx + ay
cx + a
 ax − by − c

cy + b  = 0 represents
− ax + by − c
 bx + ay
cy + b
− ax − by + c 
 cx + a
(2001, 6M)
a straight line.
cos θ
2π 

sin θ +


3
2π 

sin θ −


3
2π 

cos θ +


3
2π 

cos θ −


3
(n + 2)! 
(n + 3)! ,
(n + 4)! 
 D

then show that 
− 4 is divisible by n.
3
 (n !)

(1992, 4M)
 p b c
a b r
p
q
r
Then, find the value of
. (1991, 4M)
+
+
p−a q−b r−c
43. Let the three digit numbers A28, 3B9 and 62C, where A,
B and C are integers between 0 and 9, be divisible by a
fixed integer k. Show that the determinant
 A 3 6
(1990, 4M)
 8 9 C is divisible by k.
2 B 2 
 a −1
n
2n 2
3n3
44. Let ∆ a = 
 (a − 1)2
3
 (a − 1)

6
4n − 2 

3n 2 − 3n 
n
sin 2θ
4π 

sin 2θ +
 =0

3
4π 

sin 2 θ −


3  (2000, 3M)
(a) f (0) = 2, f (1) = 1
(b) f has a minimum value at x = 5 / 2, and
2ax − 1
 2ax
(c) for all x, f ′ (x) = 
b
b+ 1

 2(ax + b) 2ax + 2b + 1
∑ ∆ a = c ∈ constant.
(1989, 5M)
a =1
conditions
45. Show that
 xCr
 yC
z r
C
 r
x
x
x+1
x+ 2
y
y
y+1
y+ 2
Cr + 2   xCr
C r + 2 =  y C r
 z
z
Cr + 2   Cr
Cr + 1
Cr + 1
z
Cr + 1
Cr + 1
Cr + 1
z +1
Cr + 1
C r + 2
C r + 2

z+2
C r + 2
(1985, 3M)
46. If α be a repeated root of a quadratic equation f (x) = 0
2ax + b +
−1
2ax + b
1



where a, b are some constants. Determine the constants
a, b and the function f (x).
(1998, 3M)
bc ca ab
38. Find the value of the determinant p q r , where
1 1 1
a , b and c are respectively the pth , qth and rth terms of a
harmonic progression.
(1997C, 2M)
39. Let a > 0, d > 0. Find the value of the determinant
1
a (a + d )
1
(a + d ) (a + 2d )
1
(a + 2d ) (a + 3d )
(n + 1)!
(n + 2)!
(n + 3)!
 n!
D =  (n + 1)!
 (n + 2)!
Show that
37. Suppose, f (x) is a function satisfying the following
1
a
1
(a + d )
1
(a + 2d )
(1994, 4M)
cos ( A − R) 
cos (B − R) 
= 0
cos (C − R) 
41. For a fixed positive integer n, if
36. Prove that for all values of θ
sin θ
cos ( A − Q )
cos (B − Q )
cos (C − Q )
42. If a ≠ p, b ≠ q, c ≠ r and a q c  = 0
c
identically equal.
40. For all values of A , B, C and P , Q , R, show that
1
(a + d ) (a + 2d )
1
(a + 2d ) (a + 3d )
1
(a + 3d ) (a + 4d )
and A (x), B (x) and C (x) be polynomials of degree 3, 4
and 5 respectively, then show that
B (x) C (x) 
 A (x)
 A (α ) B (α ) C (α ) 
 A′ (α ) B′ (α ) C′ (α ) 
is divisible by
derivatives.
f (x), where prime denotes the
(1984, 3M)
47. Without expanding a determinant at any stage, show
that
 x2 + x
 2x2 + 3x − 1
 2
 x + 2x + 3
x+1
3x
2x − 1
x−2 
3x − 3  = xA + B

2x − 1

where A and B are determinants of order 3 not
involving x.
(1982, 5M)
48. Let a, b, c be positive and not all equal. Show that the
 a b c
value of the determinant  b c a is negative.
 c a b
(1981, 4M)
140 Matrices and Determinants
Topic 3 Adjoint and Inverse of a Matrix
Objective Questions I (Only one correct option)
− 1 − sin 2 θ 
−1
 = α I + βM ,
2
cos 4 θ 
1 + cos θ
 sin 4 θ
1. Let M = 
where α = α (θ ) and β = β (θ ) are real numbers, and I is the
2 × 2 identity matrix. If α * is the minimum of the set
{α (θ ): θ ∈ [0, 2π )} and β * is the minimum of the set
{ β(θ ) : θ ∈ [0, 2π) }, then the value of α * + β * is
(2019 Adv.)
17
16
37
(c) −
16
31
16
29
(d) −
16
(a) −
(b) −
(2019 Main, 12 April I)

et
e− t sin t
e− t cos t
 t
−t
−t
−t
−t
3. If A = e −e cos t − e sin t − e sin t + e cos t  then A

et
−2e− t cos t
2e− t sin t


(2019 Main, 9 Jan II)
(a) invertible only when t = π
(b) invertible for every t ∈ R
(c) not invertible for any t ∈ R
π
(d) invertible only when t =
2
4. Let A and B be two invertible matrices of order 3 × 3. If
det( ABAT ) = 8 and det( AB− 1 ) = 8, then det(BA − 1BT ) is
equal to
(2019 Main, 11 Jan II)
(a) 1
(c)
(b)
1
16
1
4
cos θ − sin θ 
, then the matrix
cos θ 

π
when θ = , is equal to
12
(2019 Main, 9 Jan I)
 1

(a)  2
− 3
 2
 3

(c)  2
− 1
 2
3
2 
1 
2 
1 
2 
3
2 


(b) 




(d) 


− b
and A adj A = AAT , then 5a + b is equal
2 
(2016 Main)
(b) 5
(d) 13
8. If A is a 3 × 3 non-singular matrix such that AAT = AT A
and B = A −1 AT , then BBT is equal to
(a) I + B
(c) B −1
(2014 Main)
(b) I
(d) (B −1 )T
1 α 3
9. If P = 1 3 3 is the adjoint of a 3 × 3 matrix A and


2 4 4
| A | = 4 , then α is equal to
(a) 4
(c) 5
(2013 Main)
(b) 11
(d) 0
10. If P is a 3 × 3 matrix such that PT = 2P + I, where PT is
the transpose of P and I is the 3 × 3 identity matrix, then
 x  0
there exists a column matrix, X =  y ≠ 0 such that
   
 z  0
(2012)
 0
(a) PX =  0
 
 0
(c) PX = 2X
(b) PX = X
(d) PX = − X
11. Let ω ≠ 1 be a cube root of unity and S be the set of all
(d) 16
5. If A = 
sin θ
A −50
5a
3
7. If A = 
 51 63
(b) 

 84 72
 72 − 63
(d) 

 − 84 51 
to
(b) −1
(d) 2
is
 72 − 84
(a) 

 − 63 51 
 51 84
(c) 

 63 72
(a) − 1
(c) 4
5 2α 1 
2. If B = 0 2 1  is the inverse of a 3 × 3 matrix A, then


α 3 −1
the sum of all values of α for which det ( A ) + 1 = 0, is
(a) 0
(c) 1
 2 −3 
2
 , then adj (3 A + 12 A ) is equal to
 −4 1 
(2017 Main)
6. If A = 
3
2
1
2
1
2
3
2
1
− 
2
3
2 
3
−
2 
1 
2 
 1 a b
non-singular matrices of the form  ω 1 c , where
 2

ω ω 1 
each of a , b and c is either ω or ω 2. Then, the number of
(2011)
distinct matrices in the set S is
(a) 2
(c) 4
(b) 6
(d) 8
12. Let M and N be two 3 × 3 non-singular
skew-symmetric matrices such that MN = NM . If PT
denotes the transpose of P, then
(2011)
M 2N 2(M T N )−1 (MN −1 )T is equal to
(a) M 2
(c) −M 2
(b) −N 2
(d) MN
Matrices and Determinants 141
1 0 0
13. If A = 0 1 1, 6 A −1 = A 2 + cA + dI , then (c, d ) is


0 −2 4
(2005, 1M)
(a) (− 6, 11)
(c) (11, 6)
(b) (− 11, 6)
(d) (6, 11)
identity matrix of order 3. If q23 = −
then
(2016 Adv.)
(a) α = 0, k = 8
(c) det (P adj (Q )) = 29
(b) 4α − k + 8 = 0
(d) det (Q adj (P )) = 213
17. Let M be a 2 × 2 symmetric matrix with integer entries.
Objective Questions II
(One or more than one correct option)
Then, M is invertible, if
let I denote the 3 × 3 identity matrix. If
M −1 = adj(adj M ), then which of the following
(2020 Adv.)
statements is/are ALWAYS TRUE?
(b) det M = 1
(d) (adj M )2 = I
0 1 a 
− 1 1 − 1
15. Let M = 1 2 3  and adj M =  8 − 6 2 




3 b 1 
− 5 3 − 1
(2014 Adv.)
(a) the first column of M is the transpose of the second row
of M
(b) the second row of M is the transpose of the first column
of M
(c) M is a diagonal matrix with non-zero entries in the
main digonal
(d) the product of entries in the main diagonal of M is not
the square of an integer
14. Let M be a 3 × 3 invertible matrix with real entries and
(a) M = I
(c) M 2 = I
k
k2
and det (Q ) = ,
8
2
1 4 4 
18. If the adjoint of a 3 × 3 matrix P is 2 1 7, then the


1 1 3 
possible value(s) of the determinant of P is/are
(a) − 2
where a and b are real numbers. Which of the following
(2019 Adv.)
options is/are correct?
(b) − 1
(c) 1
(d) 2
Integer & Numerical Answer Type Question
(a) det (adj M 2 ) = 81
α   1
(b) If M β  =  2, then α − β + γ = 3
   
 γ   3
19. Let k be a positive real number and let
 2k − 1 2 k 2 k 


A= 2 k
− 2k  and
1
 − 2 k 2k
− 1 
(c) (adj M )− 1 + adj M − 1 = − M
(d) a + b = 3
 0
2k − 1
k

0
2 k
B =  1 − 2k
 − k − 2 k
0
3 −1 −2
16. Let P = 2 0 α , where α ∈ R. Suppose Q = [qij ] is a


3 −5 0 
matrix such that PQ = kI , where k ∈ R, k ≠ 0 and I is the




If det (adj A ) + det (adj B) = 106, then [k] is equal to……
(2010)
Topic 4 Solving System of Equations
Objective Questions I (Only one correct option)
1. If [x] denotes the greatest integer ≤ x , then the system of
liner equations [sin θ ]x + [− cos θ ] y = 0, [cot θ ]x + y = 0
(2019 Main, 12 April II)
π 2π 
(a) have infinitely many solutions if θ ∈  ,

2 3
7π 
and has a unique solution if θ ∈  π ,
.

6
(b) has a unique solution if
π 2π   7π 
θ ∈  ,

 ∪  π,
2 3 
6
π 2π 
(c) has a unique solution if θ ∈  ,

2 3
7π 
and have infinitely many solutions if θ ∈  π ,


6
(d) have infinitely many solutions if
7π 
π 2π
θ ∈  ,  ∪  π ,

2 3 
6
2. Let λ be a real number for which the system of linear
equations
x + y + z = 6, 4x + λy − λz = λ − 2 and
3x + 2 y − 4z = − 5
has infinitely many solutions. Then λ is a root of the
quadratic equation
(2019 Main, 10 April II)
(b) λ2 + 3λ − 4 = 0
(a) λ2 − 3λ − 4 = 0
(c) λ2 − λ − 6 = 0
(d) λ2 + λ − 6 = 0
3. If the system of linear equations
x+ y+ z =5
x + 2 y + 2z = 6
x + 3 y + λz = µ,(λ , µ ∈ R), has infinitely many solutions,
then the value of λ + µ is
(2019 Main, 10 April I)
(a) 7
(b) 12
(c) 10
(d) 9
142 Matrices and Determinants
4. If the system of equations 2x + 3 y − z = 0, x + ky − 2z = 0
and 2x − y + z = 0 has a non-trivial solution (x, y, z ), then
x y z
+ + + k is equal to
y z x
(2019 Main, 9 April II)
(a) −4
(b)
1
2
(c) −
1
4
(d)
3
4
5. If the system of linear equations
(b) 3x − 4 y − 1 = 0
(d) 4x − 3 y − 1 = 0
6. The greatest value of c ∈ R for which the system of
linear equations x −cy − cz = 0, cx − y + cz = 0,
cx + cy − z = 0 has a non-trivial solution, is
(2019 Main, 8 April I)
(a) −1
1
(b)
2
(c) 2
(d) 0
7. The set of all values of λ for which the system of linear
equations x − 2 y − 2z = λx,
− x − y = λz
x + 2 y + z = λyand
has a non-trivial solution
(2019 Main, 12 Jan II)
(a)
(b)
(c)
(d)
contains exactly two elements.
contains more than two elements.
is a singleton.
is an empty set.
(2019 Main, 12 Jan I)
(1 + α )x + βy + z = 2
αx + (1 + β ) y + z = 3
ax + βy + 2z = 2
has a unique solution, is
(b) a + b + c = 0
(d) b + c − a = 0
10. The number of values of θ ∈ (0, π ) for which the system
(c) four
(d) one
(2019 Main, 10 Jan I)
(c) 21
has a non-zero solution ( x , y , z ), then
(a) −10
(b) 10
(c) −30
xz
y2
is equal to
(2018 Main)
(d) 30
15. The system of linear equations
x + λy − z = 0; λx − y − z = 0; x + y − λz = 0
has a non-trivial solution for
(2016 Main)
(a) infinitely many values of λ (b) exactly one value of λ
(c) exactly two values of λ
(d) exactly three values of λ
equations 2x1 − 2x2 + x3 = λx1, 2x1 − 3x2 + 2x3 = λx2 and
(2015 Main)
− x1 + 2x2 = λx3 has a non-trivial solution
(a) is an empty set
(b) is a singleton set
(c) contains two elements
9d) contains more than two elements
(b) 1
(c) 2
(d) 3
 x  1
0 or 1 and for which the system A  y = 0 has exactly
   
 z  0
two distinct solutions, is
(2010)
(a) 0
(b) 29 − 1
(c) 168
(d) 2
19. Given, 2x − y + 2z = 2, x − 2 y + z = − 4, x + y + λz = 4,
(b) 1
(c) 0
(d) –3
20. The number of values of k for which the system of
(2019 Main, 10 Jan II)
x+ y+z=5
x + 2 y + 3z = 9
x + 3 y + αz = β
has infinitely many solutions, then β − α equals
(b) 18
x + ky + 3z = 0, 3x + ky − 2z = 0, 2x + 4 y − 3z = 0
(a) 3
11. If the system of equations
(a) 8
14. If the system of linear equations
then the value of λ such that the given system of
(2004, 1M)
equations has no solution, is
of linear equations
(b) three
(2019 Main, 9 Jan I)
has infinitely many solutions for a = 4
is inconsistent when a = 4
has a unique solution for|a| = 3
is inconsistent when|a| = 3
18. The number of 3 × 3 matrices A whose entries are either
where a , b, c are non-zero real numbers, has more than
one solution, then
(2019 Main, 11 Jan I)
(a) two
(a)
(b)
(c)
(d)
(a) infinite
2x + 2 y + 3z = a
3x − y + 5z = b
x − 3 y + 2z = c
x + 3 y + 7z = 0,
− x + 4 y + 7z = 0,
(sin 3θ )x + (cos 2θ ) y + 2z = 0
has a non-trivial solution, is
x + y + z = 2,
2x + 3 y + 2z = 5
2x + 3 y + ( a 2 − 1)z = a + 1
equation
(k + 1) x + 8 y = 4 y ⇒ kx + (k + 3) y = 3k − 1
has no solution, is
(2013 Main)
9. If the system of linear equations
(a) b − c − a = 0
(c) b − c + a = 0
(b) g + 2h + k = 0
(d) g + h + 2k = 0
17. The number of value of k, for which the system of
(b) (− 4, 2)
(d) (−3, 1)
(a) (2, 4)
(c) (1, − 3)
(a) 2 g + h + k = 0
(c) g + h + k = 0
16. The set of all values of λ for which the system of linear
8. An ordered pair (α , β) for which the system of linear
equations
x − 4 y + 7z = g,
3 y − 5z = h,
− 2x + 5 y − 9z = k
is consistent, then
(2019 Main, 9 Jan II)
13. The system of linear equations
x − 2 y + kz = 1 , 2x + y + z = 2 ,3x − y − kz = 3
has a solution (x, y, z ), z ≠ 0, then (x, y) lies on the
straight line whose equation is
(2019 Main, 8 April II)
(a) 3x − 4 y − 4 = 0
(c) 4x − 3 y − 4 = 0
12. If the system of linear equations
(d) 5
equations (k + 1) x + 8 y = 4k and kx + (k + 3) y = 3k − 1
has infinitely many solutions, is/are
(2002, 1M)
(a) 0
(c) 2
(b) 1
(d) ∞
21. If the system of equations x + ay = 0, az + y = 0 and
ax + z = 0 has infinite solutions, then the value of a is
(a) –1
(c) 0
(b) 1
(d) no real values
Matrices and Determinants 143
22. If the system of equations x − ky − z = 0, kx − y − z = 0,
x + y − z = 0 has a non-zero solution, then possible values
(2000, 2M)
of k are
(a) –1, 2
(b) 1, 2
(c) 0, 1
(d) –1, 1
Assertion and Reason
(a) x + 2 y + 3z = b1 , 4 y + 5z = b2 and x + 2 y + 6z = b3
(b) x + y + 3z = b1 , 5x + 2 y + 6z = b2 and − 2x − y − 3z = b3
(c) − x + 2 y − 5z = b1 , 2x − 4 y + 10z = b2 and x − 2 y + 5z = b3
(d) x + 2 y + 5z = b1 , 2x + 3z = b2 and x + 4 y − 5z = b3
Fill in the Blank
For the following questions, choose the correct answer from
the codes (a), (b), (c) and (d) defined as follows.
(a) Statement I is true, Statement II is also true; Statement
II is the correct explanation of Statement I
(b) Statement I is true, Statement II is also true; Statement
II is not the correct explanation of Statement I
(c) Statement I is true; Statement II is false.
(d) Statement I is false; Statement II is true.
x − 2 y + 3 z = −1 ,
x − 3 y + 4z = 1 and − x + y − 2z = k
Statement I The system of equations has no solution for
k ≠ 3 and
1
3 −1
Statement II The determinant −1 −2
k ≠ 0 , for
23. Consider the system of equations
1
k ≠ 0.
4
1
26. The system of equations λx + y + z = 0 , − x + λy + z = 0
and − x − y + λz = 0 will have a non-zero solution, if
(1982, 2M)
real values of λ are given by ...
Analytical and Descriptive Questions
a
27. A = 1
1
0
c
d
1
a
b, B = 0
 f
b
1
d
g
a 2
1
f 
c , U =  g , V = 0 
0 
h 
h 
 
If there is a vector matrix X, such that AX = U has
infinitely many solutions, then prove that BX = V
cannot have a unique solution. If a f d ≠ 0. Then, prove
that BX = V has no solution.
(2004, 4M)
28. Let λ and α be real. Find the set of all values of λ for
which the system of linear equations
(2008, 3M)
λx + (sin α ) y + (cos α )z = 0,
x + (cos α ) y + (sin α )z = 0
Objective Questions II (One or more than one correct)
and
24. Let x ∈ R and let
1 1 1
2 x x
P = 0 2 2, Q = 0 4 0 and R = PQP −1,




0 0 3
x x 6
the which of the following options is/are correct?
(2019 Adv.)
(a) There exists a real, number x such that PQ = QP
 1
 1
(b) For x = 0, if R  a  = 6  a , then a + b = 5
 
 
 b
 b
(c) For x = 1, there exists a unit vector α $i + β$j + γk$ for
α   0
which R β  =  0
   
 γ   0
 2 x x
(d) det R = det  0 4 0 + 8, for all x ∈ R


 x x 5
 b1 
25. Let S be the set of all column matrices  b2  such that b1,
 
b3 
b2, b3 ∈R and the system of equations (in real variables)
− x + 2 y + 5z = b1
2x − 4 y + 3z = b2
x − 2 y + 2z = b3
has at least one solution. Then, which of the following
system(s) (in real variables) has (have) at least one
 b1 
solution for each  b2  ∈ S ?
 
 b3 
− x + (sin α ) y − (cos α )z = 0
has a non-trivial solution.
For λ = 1, find all values of α.
(1993, 5M)
29. Let α 1 , α 2, β1 , β 2 be the roots of ax + bx + c = 0 and
2
px2 + qx + r = 0, respectively. If the system of equations
α 1 y + α 2z = 0 and β1 y + β 2z = 0 has a non-trivial solution,
b2 ac
then prove that 2 =
.
(1987, 3M)
pr
q
30. Consider the system of linear equations in x, y, z
(sin 3θ ) x − y + z = 0, (cos 2θ ) x + 4 y + 3z = 0 and
2x + 7 y + 7z = 0
Find the values of θ for which this system has
non-trivial solution.
(1986, 5M)
31. Show that the system of equations, 3x − y + 4z = 3,
x + 2 y − 3z = − 2 and 6x + 5 y + λz = − 3 has atleast one
solution for any real number λ ≠ − 5. Find the set of
solutions, if λ = − 5.
(1983, 5M)
32. For what values of m, does the system of equations
3x + my = m and 2x − 5 y = 20 has a solution satisfying
(1979, 3M )
the conditions x > 0, y > 0?
33. For what value of k, does the following system of
equations possess a non-trivial solution over the set of
rationals
x + y − 2z = 0, 2x − 3 y + z = 0, and x − 5 y + 4z = k
Find all the solutions.
(1979, 3M )
34. Given, x = cy + bz , y = az + cx, z = bx + ay, where x, y, z
are not all zero, prove that a 2 + b2 + c2 + 2ab = 1.
(1978, 2M)
144 Matrices and Determinants
Integer & Numerical Answer Type Questions
37. For a real number α , if the system
 1 α α 2  x   1 
   

 α 1 α   y = −1
2
α α 1   z   1 


35. If the system of equations
x − 2 y + 3z = 9
2x + y + z = b
x − 7 y + az = 24, has infinitely many solutions, then a − b
is equal to ............. .
(2020 Main, 4 Sep I)
36. If the system of linear equations,
of linear equations, has infinitely many solutions, then
(2017 Adv.)
1 + α + α2 =
0
 −1 
 
0
 
 3 
38. Let M be a 3 × 3 matrix satisfying M 1 =  2 ,
x+ y+ z =6
x + 2 y + 3z = 10
1 1
M −1 =  1 , and M
   
 0  −1
3 x + 2 y + λz = µ
has more than two solutions, then µ − λ2 is equal to
(2020 Main, 7 Jan II)
………… .
1  0 
1 =  0  ,
   
1 12
Then, the sum of the diagonal entries of M is … (2011)
Answers
Topic 3
Topic 1
1. (b)
5. (b)
2. (b)
6. (b)
3. (c)
7. (d)
4. (a)
8. (a)
1. (d)
2. (c)
3. (b)
4. (c)
5. (c)
6. (b)
7. (b)
8. (b)
9. (b)
9. (d)
10. (a)
11. (a,b,d)
12. (c, d)
10. (d)
11. (a)
12. (c)
13. (c, d)
14. (b, c, d)
15. (d)
16. (b)
13. (a)
14. (b,c,d)
15. (b,c,d)
16. (b,c)
17. (a)
18. (d)
19. (c)
20. (d)
17. (c, d)
18. (a,d)
19. (4)
21. (4)
22. (5)
23. (1)
2. (a)
3. (a)
2.
6.
10.
14.
18.
22.
26.
3.
7.
11.
15.
19.
23.
Topic 4
Topic 2
21. (b)
22. (b)
23. (a, c)
24. (b, c)
1.
5.
9.
13.
17.
21.
25.
25. (a, b)
26. (b,d)
27. (4)
28. (0)
28. − 2 ≤ λ ≤ 2 , α = nπ , nπ +
29. (0)
30. (2 and 7) 31. {–1,2}
32. (0)
1
5
1 2 5


37. a = , b = − and f ( x ) = x − x + 2


4
4
4
4
1. (d)
4. (c)
5. (b)
6. (d)
7. (c)
8. (a)
9. (b)
10. (d)
11. (c)
12. (a)
13. (c)
14. (a)
15. (c)
16. (d)
17. (c)
18. (c)
19. (a)
20. (b)
33. False
38. (0)
42. (2)


4d 4
39. 

2
3
2
 a (a + d ) (a + 2d ) (a + 3d ) (a + 4d )
(a)
(c)
(a)
(d)
(d)
(a)
(a,d)
(c)
(b)
(a)
(b)
(a)
(d)
λ=0
(c)
(c)
(a)
(d)
(b)
(a)
4.
8.
12.
16.
20.
24.
(b)
(a)
(a)
(c)
(b)
(b,d)
π
4
π
, n ∈Z
6
4 − 5k
13k − 9
15
31. x =
,y =
, z = k 32. m < − or m > 30
7
7
2
33. (k = 0, the given system has infinitely many solutions)
30. θ = nπ , nπ + ( −1 )n
35. (5)
36. (13)
37. (1)
38. (9)
Hints & Solutions
Topic 1 Types of Matrices, Addition,
Subtraction and Transpose of a
Matrix
1. Given matrix A is a symmetric and matrix B is a
skew-symmetric.
∴
A = A and B = − B
2 3 
Since, A + B = 

5 − 1
T
T
(given) …(i)
T
2 3 
( A + B)T = 

5 − 1
2 5 
AT + BT = 

3 − 1
T
T
Given, A = A and B = − B
2 5 
⇒ A−B=

3 − 1
… (ii)
On solving Eqs. (i) and (ii), we get
2 4 
0 − 1
A=
 and B = 1 0 
−
4
1




2 4  0 − 1  4 − 2
So, AB = 

=

4 − 1 1 0  − 1 − 4
2. Given matrix
 0 2y 1 
A = 2x y −1 , (x, y ∈ R, x ≠ y)


2x − y 1 
for which
AT A = 3I3
⇒
 0 2x 2x   0 2 y 1  3 0 0
2 y y − y 2x y −1 = 0 3 0

 


 1 −1 1  2x − y 1  0 0 3
⇒
8x2 0 0 3 0 0

 
2

 0 6 y 0 = 0 3 0
 0
0 3 0 0 3

cos α
∴ A2 = 
sin α

cos2 α − sin 2 α
=
sin
cosα + cosα sin α
α

cos 2 α
=
sin 2 α
4. Given matrix
1
P = 3

9
⇒
P=X +
Now, P5 = (I +
and
3
8
1
y=±
2
Q There are 2 different values of x and y each.
So, 4 matrices are possible such that AT A = 3I3 .
0 0 0 0 0
1 0 = 3 0 0 +
 

3 1 9 3 0
I (let)
X )5
1 0 0
0 1 0


0 0 1
= I + 5C1 (X ) + 5C 2(X 2) + 5C3 (X 3 ) + …
[Q I n = I , I ⋅ A = A and (a + x)n = nC 0a n +
n
C1a n − 1x + ...+ nC nxn]
0 0 0 0 0 0 0 0 0
Here, X 2 = 3 0 0 3 0 0 = 0 0 0


 

9 3 0 9 3 0 9 0 0
0 0 0 0 0 0 0 0 0
and
X 3 = X 2 ⋅ X = 0 0 0 3 0 0 = 0 0 0
 



9 0 0 9 3 0 0 0 0
0 0 0
⇒
X 4 = X 5 = 0 0 0


0 0 0
0 0 0
0 0 0
So,
P5 = I + 5 3 0 0 + 10 0 0 0




9 3 0
9 0 0
8x2 = 3 and 6 y2 = 3
x=±
− sin 2 α 
cos 2 α 
cos(32 α ) − sin(32 α ) 0 −1
⇒ A32 = 
=
 (given)
sin(32 α ) cos(32 α )  1 0 
So, cos(32 α ) = 0 and sin(32 α ) = 1
π
π
32 α = ⇒ α =
⇒
2
64
Here, two matrices are equal, therefore equating the
corresponding elements, we get
⇒
− cosα sin α − sin α cosα 

− sin 2 α + cos2 α

Similarly,
cos(nα ) − sin(nα )
An = 
, n ∈ N
sin(nα ) cos(nα ) 
On taking transpose both sides, we get
⇒
cos α − sin α 

sin α cos α 
− sin α  cos α − sin α 
cos α  sin α cos α 
3. Given, matrix A = 
and
⇒
0 0
 1

= 15 1 0


135 15 1
0 0
 2
Q = I + P5 =  15 2 0 = [q ij ]


135 15 2
q21 = 15, q31 = 135 and q 32 = 15
q + q31 15 + 135 150
Hence, 21
=
=
= 10
q32
15
15
146 Matrices and Determinants
5. Given, AAT = I
p p  1 0 0
0 2q r   0



⇒ p q − r 2q q − q = 0 1 0

 


 p − q r   r − r r  0 0 1
0 + 4q2 + r 2 0 + 2q2 − r 2 0 − 2q2 + r 2  1 0 0


⇒ 0 + 2q2 − r 2 p2 + q2 + r 2 p2 − q2 − r 2  = 0 1 0


2
2
2
2
2
2
2
2
0 − 2q + r
p −q −r
p + q + r  0 0 1

We know that, if two matrices are equal, then
corresponding elements are also equal, so
4q2 + r 2 = 1 = p2 + q2 + r 2,
2q2 − r 2 = 0 ⇒ r 2 = 2q2
and
p2 − q 2 − r 2 = 0
Using Eqs. (ii) and (iii), we get
…(i)
…(ii)
…(iii)
…(iv)
p2 = 3 q 2
Using Eqs. (ii) and (iv) in Eq. (i), we get
4q2 + 2q2 = 1
⇒
6q2 = 1
[using Eq. (iv)]
⇒
2 p2 = 1
1
1
2
p = ⇒ | p| =
2
2
 1 0 0
6. Here, P =  4 1 0


16 4 1
 1 0 0  1 0 0  1
2 
∴
P = 4 1 0  4 1 0 =  4 + 4
 


16 4 1 16 4 1 16 + 32
1
0
0


4 ×2
1
0
=


16 (1 + 2) 4 × 2 1
1
0
0  1 0 0

and P3 =  4 × 2
1
0  4 1 0



16 (1 + 2) 4 × 2 1 16 4 1
1
0
0

4 ×3
1
0
=


16 (1 + 2 + 3) 4 × 3 1
From symmetry,
1
0

50

P =
4 × 50
1

16 (1 + 2 + 3 + ... + 50) 4 × 50
P50 − Q = I
1 2
2
2 a
1 2



2 b 
2 −2 b 
2 2  1 2 a 
1 −2 2 1 2 


2 b  2 −2 b 
a + 4 + 2b 
9
0
0
9
2a + 2 − 2b

+ 4 + 2b 2a + 2 − 2b a 2 + 4 + b2 
1
7. Given, A = 2 1 −2, AT = 2

a
1
T
and AA = 2

a

=

a
It is given that, AAT = 9I
9
0
a + 4 + 2b 

1 0 0
⇒ 
0
9
2a + 2 − 2b = 9 0 1 0




2
2
a + 4 + 2b 2a + 2 − 2b a + 4 + b 
0 0 1
a + 4 + 2b  9 0 0
9
0

⇒ 
0
9
2a + 2 − 2b = 0 9 0

 

2
2
a + 4 + 2b 2a + 2 − 2b a + 4 + b  0 0 9
On comparing, we get
a + 4 + 2b = 0 ⇒ a + 2b = − 4
...(i)
2a + 2 − 2b = 0 ⇒ a − b = − 1
0
1
0

4 + 4 1
0
…(i)
a = − 2, b = − 1
This satisfies Eq. (iii)
(a , b) ≡ (−2,−1)
Hence,
 3 /2 −1 /2   3 /2
PT P = 

3 /2  −1 /2
 1 /2
1 0
PT P = 

0 1
⇒
1 /2 

3 /2
PT P = I
PT = P −1
⇒
Q = PAPT
Since,
∴ P Q
T
0
0

1
…(iii)
On solving Eqs. (i) and (ii), we get
⇒
...(ii)
…(ii)
and a 2 + 4 + b2 = 9
8. Now,
[given]


− q12
− q13  1 0 0
1 − q11

∴ 
− q23  = 0 1 0
200 − q21
1 − q22

 

50
(51) − q31 200 − q32 1 − q33  0 0 1
16 ×


2
16 × 50 × 51
⇒
200 − q21 = 0,
− q31 = 0,
2
200 − q32 = 0
∴
q21 = 200, q32 = 200, q31 = 20400
Q
q31 + q32 20400 + 200 20600
=
= 103
=
200
q21
200
Thus,
2005
P = PT [(PAPT )(PAPT ) K 2005 times ] P
= (PT P ) A (PT P ) A (PT P ) K (PT P ) A (PT P )
1444444442444444443
2005 times
∴
= IA 2005 = A 2005
1 1
A=

0 1
1 1 1 1 1 2
A2 = 

=

0 1 0 1 0 1
1 2 1 1 1 3
A3 = 

=

0 1 0 1 0 1
……………………
……………………
Matrices and Determinants 147
∴
1 2005
A 2005 = 
1 
0
1 2005
PTQ 2005 P = 
1 
0
9. Given, A = α
1

⇒
α
A2 = 
1
1
0
, B=
1
5
0
1
⇒
Which is not possible at the same time.
10. If A and B are square matrices of equal degree, then
A+ B=B+ A
0 0
0
0 1, P3 = 1


1 0
0
0 1
0
0 0, P6 = 0


1 0
1
1 0
0 0

0 1
0 1
1 0

0 0
2 1 3
and X = Σ Pk 1 0 2 PKT


K =1
3 2 1
Q P1T = P1 , P2T = P2, P3T = P3 , P4T = P5 , P5T = P4 and
2 1 3
P6T = P6 and Let Q = 1 0 2 and Q QT = Q


3 2 1
Now, X = (P1QP1T ) + (P2QP2T ) + (P3QP3T ) + (P4QP4T )
+ (P5QP5T ) + (P6QP6T )
T
T T
T T
So, X = (P1QP1 ) + (P2QP2 ) + (P3QP3T )T + (P4QP4T )T
+ (P5QP5T )T + (P6QP6T )T
6
= P1QP1T + P2QP2T + P3QP3T + P4QP4T + P5QP5T + P6QP6T
[Q ( ABC )T = CT BT AT and ( AT )T = A and QT = Q]
T
⇒X = X
⇒ X is a symmetric matrix.
The sum of diagonal entries of X = Tr (X )
6
= Σ Tr (Pi Q PiT )
i =1
[QTr ( ABC ) = Tr (BCA )]
6
= Σ Tr (QI )
i =1
6
[QPi’s are orthogonal matrices]
= Σ Tr ( Q ) = 6 Tr (Q ) = 6 × 3 = 18
i =1
[Q PKT R = R]
6 2 2 2 6
6
6
6
= Σ PK 3 = Σ PK 3 = 2 2 2 3
  
  K =1
 
K =1
6 2 2 2 6
6
30
1
1




XR = 30 ⇒ XR = 30R ⇒ X 1 = 30 1
 
 
 
30
1
1
⇒ (X − 30 I ) R = 0 ⇒|X − 30 I| = 0
So, (X − 30I ) is not invertible and value of α = 30.
Hence, options (a), (b) and (d) are correct.
12. Given,
11. Given matrices,
i =1
K =1
K =1
0 
1
∴ No real values of α exists.
6
K =1
= Σ (PK Q R)
⇒ α 2 = 1 and α + 1 = 5
= Σ Tr (Q PiT Pi )
6
6
0
1
1
1 0 0
P1 = I = 0 1 0, P2 = 0



0
0 0 1
0
0 1 0
P4 = 0 0 1, P5 = 1



0
1 0 0
6
XR = Σ (PK Q PKT )R = Σ (PK QPKT R)
0  α 2
=
1 α + 1
0  α
1 1
Also, given, A 2 = B
α 2
0 1
=
⇒
α + 1 1 5
1
Now, Let R = 1, then
 
1
X T = − X , Y T = −Y , Z T = Z
P = Y 3 Z 4 − Z 4Y 3
PT = (Y 3 Z 4 )T − (Z 4 Y 3 )T
= (Z T )4 (Y T )3 − (Y T )3 (Z T )4
= − Z 4Y 3 + Y 3 Z 4 = P
∴ P is symmetric matrix.
(b) Let
P = X 44 + Y 44
Then,
PT = (X T )44 + (Y T )44
= X 44 + Y 44 = P
∴ P is symmetric matrix.
(c) Let
P = X 4Z 3 − Z 3 X 4
Then,
PT = (X 4Z 3 )T − (Z 3 X 4 )T
= (Z T )3 (X T )4 − (X T )4 (Z T )3
= Z 3 X 4 − X 4Z 3 = − P
∴ P is skew-symmetric matrix.
(d) Let
P = X 23 + Y 23
Then,
PT = (X T )23 + (Y T )23 = − X 23 − Y 23 = − P
∴ P is skew-symmetric matrix.
(a) Let
Then,
13. (a) (N T MN )T = N T M T (N T )T = N T M T N , is symmetric
if M is symmetric and skew-symmetric, if M is
skew-symmetric.
(b) (MN − NM )T = (MN )T − (NM )T
= NM − MN = − (MN − NM )
∴ Skew-symmetric, when M and N are symmetric.
(c) (MN )T = N T M T = NM ≠ MN
∴ Not correct.
(d) (adj MN ) = (adj N ) ⋅ (adj M )
∴ Not correct.
14. Here,
P = [ pij ]n × n with pij = wi + j
∴ When n = 1
⇒
∴ When
P = [ pij ]1 × 1 = [ω 2]
P 2 = [ω 4 ] ≠ 0
n =2
148 Matrices and Determinants
 p11 p12  ω 2 ω3  ω 2 1 
=
P = [ pij ]2× 2 = 

4
= 3
 p21 p22 ω ω   1 ω 
ω 2 1  ω 2 1 
P2 = 


 1 ω  1 ω
ω 4 + 1 ω 2 + ω 
≠0
⇒ P2 =  2
2
ω + ω 1 + ω 
When n = 3
P = [ pij ]3 × 3
ω 2 ω3 ω 4  ω 2 1 ω 


 
= ω3 ω 4 ω5  =  1 ω ω 2
ω 4 ω5 ω 6   ω ω 2 1 


 
ω 2 1 ω  ω 2 1 ω  0 0 0



2
P =  1 ω ω 2  1 ω ω 2 = 0 0 0 = 0


2
2
ω ω
1  0 0 0
1   ω ω

∴
P 2 = 0, when n is a multiple of 3.
P 2 ≠ 0, when n is not a multiple of 3.
⇒
n = 57 is not possible.
∴
n = 55, 58, 56 is possible.
15. As (a , b, c) lies on 2x + y + z = 1 ⇒ 2a + b + c = 1
⇒
2a + 6a − 7a = 1
⇒
a = 1, b = 6, c = − 7
∴ 7a + b + c = 7 + 6 − 7 = 6
a
c
20. Given, A = 
If A is skew-symmetric matrix, then a = 0, b = − c
∴
| A|= − b2
Thus, P divides| A|, only when b = 0.
...(i)
Again, if A is symmetric matrix, then b = c and
| A|= a 2 − b2
Thus, p divides| A|, if either p divides (a − b) or p
divides (a + b).
p divides (a − b), only when a = b,
a = b ∈ {0, 1, 2 ,... , ( p − 1)}
i.e.
i.e. p choices
ax2 + bx + c = 0 ⇒ x2 + 6x − 7 = 0
⇒
(x + 7) (x − 1) = 0
∴
x = 1, − 7
∞
⇒ p choices, including a = b = 0 included in Eq. (i).
∴ Total number of choices are ( p + p − 1) = 2 p − 1
a
21. Given, A = b
c
Now, AT A = I
a b
⇒
b c
c a
1
6  6
 1 1
⇒∑  −  =1 + +   + L+ ∞ =
6
 1 7
 7
7
n= 0
1−
7
1
=
=7
1 /7
17. If a = 2, b = 12, c = − 14
c  a
a  b
b  c
 a 2 + b2 + c2
ab + bc + ca
ab + bc + ca

⇒
1
= 0
0
0
1
0
3
1
3
+
+
ω a ωb ω c
3
3
1
3
=
+ 1 + 3ω 2 = 3ω + 1 + 3ω 2
+
+
⇒
ω 2 ω12 ω −14 ω 2
= 1 + 3 (ω + ω 2) = 1 − 3 = − 2
18. The number of matrices for which p does not divide
Tr ( A ) = ( p − 1) p2 of these ( p − 1)2 are such that p
divides| A|. The number of matrices for which p divides
Tr ( A ) and p does not divides| A|are ( p − 1)2.
| A|= a 2 − bc, for (a 2 − bc) to be divisible by p. There are
exactly ( p − 1) ordered pairs (b, c) for any value of a.
∴ Required number is ( p − 1)2.
c  1 0 0
a  = 0 1 0
b  0 0 1
ab + bc + ca ab + bc + ca 
a 2 + b2 + c2 ab + bc + ca 
ab + bc + ca a 2 + b2 + c2 
b
c
a
0
0
1
and
3
ab + bc + ca = 0
…(ii)
3
= (a + b + c)(a 2 + b2 + c2 − ab − bc − ca )
⇒ a3 + b3 + c3 = (a + b + c) (1 − 0) + 3
[from Eqs. (i) and (ii)]
∴ a3 + b3 + c3 = (a + b + c) + 3
Now,
…(iii)
(a + b + c)2 = a 2 + b2 + c2 + 2(ab + bc + ca )
=1
…(iv)
From Eq. (iii), a3 + b3 + c3 = 1 + 3 ⇒ a3 + b3 + c3 = 4
22. Let a square matrix ‘A’ of order 2 × 2 , such that
tr ( A ) = 3, is
y 
x
A=

z 3 − x
So,
Required number = ( p − 1) p2 − ( p − 1)2 + ( p − 1)2
= p3 − p2
19. Trace of A = 2a, will be divisible by p, iff a = 0.
…(i)
We know, a + b + c − 3abc
3
∴
∴
c
a , abc = 1 and AT A = I
b 
b
c
a
⇒ a 2 + b2 + c2 = 1
2
n
...(ii)
p divides (a + b).
16. If b = 6 ⇒ a = 1 and c = − 7
∴
b
, a, b, c ∈ {0, 1, 2 ,... , p − 1}
a 
∴
Q
y  x
y 
x
A2 = 



z 3 − x z 3 − x
 x2 + yz
xy + 3 y − xy 
=

xz
z
xz
yz + (3 − x)2
+
−
3

 x
x2 + yz
y 
3y
A3 = 
 z 3 − x
2
yz + 9 + x − 6x 

 + 3z
tr ( A3 ) = x3 + xyz + 3 yz + 3 yz + 3 yz − xyz
+ 27 − 9x + 3x2 − x3 − 18x + 6x2
Matrices and Determinants 149
= 9 yz + 27 − 27x + 9x2 = − 18
yz + 3 − 3x + x2 = − 2
3x − x2 − yz = 5
x
y
= 3x − x2 − yz = 5.
| A| =
z 3−x
⇒
⇒
Q
23. Here,
Q
(given)
 π
f (θ ) minimum = f   = − 4 = m
 2
 π
and f (θ ) maximum = f   = 0 = M
 4
∴
∴The ordered pair (m, M ) is (−4, 0).
−1 + i 3
z=
=ω
2
(−ω )r ω 2s 
P =  2s

ωr 
 ω
(− ω )r ω 2s  (− ω )r ω 2s 
P 2 =  2s


ω r   ω 2s
ωr 
 ω

ω 2r + ω 4s
ω r + 2s [(− 1)r + 1]
=  r + 2s

ω 4s + ω 2r
[(− 1)r + 1]

ω
Given, P = − I
∴ ω 2r + ω 4s = − 1 and ω r + 2s [(− 1)r + 1] = 0
Since,
r ∈{1, 2, 3} and (− 1)r + 1 = 0
⇒
r = {1, 3}
Also,
ω 2r + ω 4s = − 1
If
r = 1, then ω 2 + ω 4s = − 1
which is only possible, when s = 1.
As,
ω2 + ω4 = − 1
∴
r = 1, s = 1
Again, if r = 3, then
ω 6 + ω 4s = − 1
[never possible]
⇒
ω 4 s = −2
∴
r ≠3
⇒ (r , s) = (1, 1) is the only solution.
Hence, the total number of ordered pairs is 1.
2
Topic 2 Properties of Determinants
1. Given function
 − sin 2 θ −1 − sin 2 θ 1 
f (θ ) =  − cos 2 θ −1 − cos 2 θ 1 


10
−2 
 12
− sin 2 θ −1 − sin 2 θ 1 
= 2− cos 2 θ −1 − cos 2 θ 1 


5
−1 
 6
On applying
R1 → R1 + R3 and R2 → R2 + R3 , we get
sin 2 θ
1 + cos 2 θ
4 cos 6 θ
2
2. Let ∆ = cos θ
1 + sin 2 θ
4 cos 6 θ
=0
2
2
1 + 4 cos 6 θ
cos θ
sin θ
Applying C1 → C1 + C 2, we get
4 cos 6 θ
sin 2 θ
2
∆ = 2 1 + sin θ
4 cos 6 θ
=0
1 + 4 cos 6 θ
1
sin 2 θ
2
Applying R1 → R1 − 2R3 and R2 → R2 − 2R3 , we get
− sin 2 θ
− 2 − 4 cos 6 θ
∆ = 0 1 − sin 2 θ − 2 − 4 cos 6 θ = 0
1 + 4 cos 6 θ
1
sin 2 θ
0
On expanding w.r.t. C1, we get
⇒ sin 2 θ (2 + 4 cos 6 θ ) + (2 + 4 cos 6 θ ) (1 − sin 2 θ ) = 0
1
2π
⇒ 2 + 4 cos 6 θ = 0 ⇒ cos 6 θ = − = cos
2
3
2π
π

 π 
⇒
6θ =
⇒θ =
Q θ ∈ 0, 3  
9
3


−6
−1
x
3. Given equation 2 − 3x x − 3 = 0
− 3 2x x + 2
On expansion of determinant along R1, we get
x [(− 3x) (x + 2) − 2x(x − 3)] + 6 [2(x + 2) + 3(x − 3)]
− 1 [2(2x) − (− 3x) (− 3)] = 0
⇒ x [− 3x2 − 6x − 2x2 + 6x] + 6[2x + 4 + 3x − 9]
− 1 [4x − 9x] = 0
⇒ x(− 5x2) + 6(5x − 5) − 1(− 5x) = 0
⇒
−5x3 + 30x − 30 + 5x = 0
⇒
5x3 − 35x + 30 = 0 ⇒ x3 − 7x + 6 = 0.
Since all roots are real
∴ Sum of roots = −
coefficient of x2
coefficient of x3
=0
4. Given determinants are
sin θ cos θ
x
∆1 = − sin θ − x
1
cos θ
1
x
6 − sin θ 4 − sin θ 0 
f (θ ) = 26 − cos 2 θ 4 − cos 2 θ 0 


6
5
−1 

= − x3 + sin θ cos θ − sin θ cos θ + x cos 2 θ − x + x sin 2 θ
= 2(−1)[(6 − sin 2 θ )(4 − cos 2 θ ) − (4 − sin 2 θ )(6 − cos 2 θ )]
= − x3
2
2
= − 2 [24 − 6 cos 2 θ − 4 sin 2 θ + sin 2 θ cos 2 θ −
x
24 + 4 cos θ + 6 sin θ − sin θ cos θ ]
2
2
= − 2 [−2 cos θ + 2 sin θ ] = 4 cos 2 θ
π π 
π

As θ ∈
,
⇒2θ ∈
,π
 4 2 
 2

2
2
2
2
and ∆ 2 = − sin 2θ
cos 2θ
sin 2θ cos 2θ
−x
1
= − x3 (similarly as ∆1)
1
x
,x≠0
So, according to options, we get ∆1 + ∆ 2 = − 2x3
150 Matrices and Determinants
= y [ y2 − (α + β )2 + 2αβ + 2αβ − 1 + (α + β ) − αβ ]
5. Given
1 1 1 2 1 3 ... 1 n − 1 1 78
=
0 1 0 1 0 1 0
1  0 1 



 
1 1 1 2 1 2 + 1
,
0 1 0 1 = 0
1 


 
Q
1 2 + 1
0
1 

1 3 1 3 + 2 + 1
0 1 = 0
,
1

 

:
:
:
:
:
:
1 1 1 2 1 3 1 n − 1
∴ 


 ... 
1 
0 1 0 1 0 1 0
1 (n − 1) + (n − 2)+ ...+3 + 2 + 1
=

1
0

n (n − 1)  1 78


=
2

0
1
 

1

1
=
0
Since, both matrices are equal, so equating
corresponding element, we get
n (n − 1)
= 78 ⇒ n (n − 1) = 156 = 13 × 12 = 13(13 − 1)
2
⇒
n = 13
1 13
1 −13
A=
= A −1 = 


0 1 
0 1 
So,
 d − b 
a b
−1
[Qif|A|= 1and A = 
 , then A =  − c a  


 c d
6. Given, quadratic equation is x2 + x + 1 = 0 having roots
α , β.
Then, α + β = −1 and αβ = 1
Now, given determinant
∆=
y+1
α
β
α
β
y+β
1
1
y+α
On applying R1 → R1 + R2 + R3 , we get
y+1+α +β y+1+α +β y+1+α +β
∆=
α
y+β
1
β
y
= α
β
1
y
y
y+β
1
1
y+α
y+α
[Qα + β = −1]
On applying C 2 → C 2 − C1 and C3 → C3 − C1, we get
0
0
y
1 −α
∆ = α y + β −α
β
1 −β
y + α −β
= y[( y + (β − α )) ( y − (β − α )) − (1 − α ) (1 − β )]
[expanding along R1]
= y [ y2 − (β − α )2 − (1 − α − β + αβ )]
= y [ y2 − β 2 − α 2 + 2αβ − 1 + (α + β ) − αβ ]
= y[ y2 − 1 + 3 − 1 − 1] = y3 [Qα + β = −1 and αβ = 1]
1 1 1 
7. Given, matrix A = 2 b c , so


2
2
4 b c 
1 1 1
det( A ) = 2 b c


2
2
4 b c 
On applying, C 2 → C 2 − C1 and C3 → C3 − C1,
0
0 
1
b − 2 c − 2

we get det( A ) = 2 b − 2 c − 2=  2
 b − 4 c2 − 4

2
2
4 b − 4 c − 4
b −2
c−2
=
(b − 2)(b + 2) (c − 2)(c + 2)
1 
 1

= (b − 2)(c − 2)
+
+
2
2
b
c

[taking common (b − 2) from C1 and
(c − 2) from C 2]
= (b − 2)(c − 2)(c − b)
Since, 2, b and c are in AP, if assume common difference
of AP is d, then
b = 2 + d and c = 2 + 2d
[given]
So,
| A| = d (2d )d = 2d3 ∈ [2, 16]
⇒
d3 ∈ [1, 8] ⇒ d ∈ [1, 2]
∴ 2 + 2d ∈ [2 + 2, 2 + 4]
= [4, 6] ⇒ c ∈ [4, 6]
1 
sin θ
 1

8. Given matrix A = − sin θ
1
sin θ 


− sin θ 1 
 −1
1
sin θ
1
1
sin θ
⇒ det( A ) =| A|= − sin θ
−1
− sin θ
1
= 1(1 + sin 2 θ ) − sin θ (− sin θ + sin θ ) + 1(sin 2 θ + 1)
…(i)
⇒| A| = 2 (1 + sin 2 θ )
 3π 5π 
As we know that, for θ ∈ 
, 
 4 4
1
 1
 1
2
sin θ ∈  −
,
 ⇒ sin θ ∈ 0, 


2
2
 2
1

 3

⇒ 1 + sin 2 θ ∈ 0 + 1, + 1 ⇒ 1 + sin 2 θ ∈ 1, 
 2


2
3
⇒ 2(1 + sin 2θ ) ∈ [2, 3) ⇒| A| ∈ [2, 3) ⊂  ,
2
2a
2a
a−b−c
9. Let ∆ =
2b
2b
b−c−a
2c
2c
c−a −b
Applying R1 → R1 + R2 + R3 , we get
a+ b+ c a+ b+ c a+ b+ c
2b
b−c−a
2b
∆=
2c
2c
c−a −b

3

Matrices and Determinants 151
1
1
1
2b
= (a + b + c) 2b b − c − a
2c
2c
c−a −b
(taking common (a + b + c) from R1)
Applying C 2 → C 2 − C1 and C3 → C3 − C1 , we get
1
0
0
∆ = (a + b + c) 2b − (a + b + c)
0
2c
− (a + b + c)
0
Now, expanding along R1, we get
∆ = (a + b + c) 1. {(a + b + c)2 − 0 }
= (a + b + c)3 = (a + b + c)(x + a + b + c)2 (given)
⇒ (x + a + b + c)2 = (a + b + c)2
⇒
x + a + b + c = ± (a + b + c)
⇒
x = − 2(a + b + c)
[Q x ≠ 0]
10. Given,
log e a1ra 2k
log e a 2ra3k
log e a3r a 4k
log e a 4r a5k
log e a7ra 8k
log e a5r a 6k
log e a 8r a 9k
log e a 6r a7k = 0
k
log e a 9r a10
On applying elementary operations
C 2 → C 2 − C1 and C3 → C3 − C1, we get
log e a1ra 2k
log e a 4r a5k
log e a7ra 8k
log e a 2ra3k
log e a5r a 6k
log e a 8r a 9k
−
−
−
log e a1ra 2k
log e a 4r a5k
log e a7ra 8k
log e a3r a 4k − log e a1ra 2k
log e a 6r a7k − log e a 4r a5k = 0
k
− log e a7ra 8k
log e a 9r a10
log e a1ra 2k
⇒ log e a 4r a5k
log e a7ra 8k
 a ra k 
 arak 
log e  2r 3k  log e  3r k4 
 a1 a 2 
 a1 a 2 
r
k
a a 
 arak 
log e  5r 6k  log e  6r 7k  = 0
 a 4a5 
 a 4a5 
k 
 a 8r a 9k 
 a 9r a10
log e  r k  log e  r k 
 a7 a 8 
 a7 a 8 

 m 
Q log e m − log e n = log e  n  


[Q a1 , a 2, a3 ....... , a10 are in GP, therefore put
a1 = a , a 2 = aR, a3 = aR2, ... , a10 = aR9 ]
log e a r + kRk
⇒ log e a r + kR3 r + 4k
log e a r + kR6r + 7k
 a r + kRr + 2k 
log e 

 a r + k Rk 
 a r + k R4 r + 5 k 
log e  r + k 3 r + 4k 
R
a

 a r + kR7r + 8k 
log e  r + k 6r + 7k 
R

a
 a r + kR2r + 3 k 
log e 

 a r + k Rk 
 a r + kR5 r + 6k 
log e  r + k 3 r + 4k  = 0
R

a
 a r + k R8 r + 9 k 
log e  r + k 6r + 7k 
R

a
log e (a r + kRk ) log e Rr + k log e R2r + 2k
⇒ log e a r + kR3 r + 4k log e Rr + k log e R2r + 2k = 0
log e a r + kR6r + 7k log e Rr + k log e R2r + 2k
log e (a r + kRk ) log e Rr + k 2 log e Rr + k
⇒ log e (a r + kR3 r + 4k ) log e Rr + k 2 log e Rr + k = 0
log e (a r + kR6r + 7k ) log e Rr + k 2 log e Rr + k
[Q log mn = n log m and here
log e R2r + 2k = log e R2( r + k) = 2 log e Rr + k ]
Q Column C 2 and C3 are proportional,
So, value of determinant will be zero for any value of
(r , k), r , k ∈ N .
∴ Set ‘S’ has infinitely many elements.
b
1
2
11. Given matrix, A = b b2 + 1 b, b > 0


b
2 
1
2
b
1
So, det ( A ) =| A| = b b2 + 1 b
b
1
2
= 2 [2(b2 + 1) − b2] − b(2b − b)
+1(b2 − b2 − 1)
2
2
2
= 2[2b + 2 − b ] − b − 1
= 2b2 + 4 − b2 − 1 = b2 + 3
det( A ) b2 + 3
3
=
= b+
b
b
b
Now, by AM ≥ GM, we get
3
1/ 2
b+
b ≥  b × 3 

b
2
3
⇒
b+ ≥2 3
b
det ( A )
So, minimum value of
=2 3
b
⇒
{Q b > 0 }
12. Given,
(sin θ ) − 2 
4+ d
 −2

A =  1 (sin θ ) + 2
d


 5 (2 sin θ ) − d (− sin θ ) + 2 + 2d 
4+ d
−2
(sin θ ) − 2
∴ | A| = 1
(sin θ ) + 2
d
5 (2 sin θ ) − d (− sin θ ) + 2 + 2d
(sin θ ) − 2
−2
4+ d
d
= 1 (sin θ ) + 2
1
0
(R3 → R3 − 2R2 + R1 )
= 1 [(4 + d )d − (sin θ + 2) (sin θ − 2)]
(expanding along R3 )
= (d 2 + 4d − sin 2 θ + 4)
= (d 2 + 4d + 4) − sin 2 θ
= (d + 2)2 − sin 2 θ
Note that| A|will be minimum if sin 2 θ is maximum i.e. if
sin 2 θ takes value 1.
Q
0
| A|min = 8,
152 Matrices and Determinants
therefore
(d + 2)2 − 1 = 8
⇒
⇒
⇒
15.
(d + 2)2 = 9
d+2=±3
d = 1, − 5
PLAN Use the property that, two determinants can be multiplied
column-to-row or row-to-column, to write the given
determinant as the product of two determinants and then
expand.
Given, f (n ) = α n + β n,
f (3) = α 3 + β3 ,
3
1+
Let ∆ = 1 + f (1) 1 +
1 + f (2) 1 +
13. Given,
2x 
x − 4 2x
 2x x − 4 2x  = ( A + Bx)(x − A )2


2x x − 4
 2x
⇒ Apply C1 → C1 + C 2 + C3
2x 
 5x − 4 2x
 5x − 4 x − 4 2x  = ( A + Bx)(x − A )2


 5x − 4 2x x − 4
3
⇒
∆= 1+α +β
1 + α 2 + β2
2ω + 1 = z
2ω + 1 = − 3
− 1 + 3i
⇒
ω=
2
Since, ω is cube root of unity.
− 1 − 3i
and ω3 n = 1
ω2 =
∴
2
1
1
1
2
Now, 1 − ω − 1 ω 2 = 3k
14. Given,
⇒
1
⇒
ω
3
⇒
⇒
⇒
∴
0
0
ω ω 2 = 3k
1 ω2 ω
1
3(ω 2 − ω 4 ) = 3k
(ω 2 − ω ) = k
 − 1 − 3i   − 1 + 3i 
k=
 −
 = − 3i = − z
2
2

 

1 ⋅ 1 + α 2 ⋅ α + β2 ⋅ β
1 ⋅ 1 + 1 ⋅ α 2 + 1 ⋅ β2
1 ⋅ 1 + α ⋅ α 2 + β ⋅ β2
1
α β
1 α 2 β2
1
1
α
1 α2
1
1 1 1
β = 1 α β
1 α 2 β2
β2
2
On expanding, we get ∆ = (1 − α )2(1 − β )2(α − β )2
∆ = K (1 − α )2(1 − β )2(α − β )2
Hence, K (1 − α )2(1 − β )2(α − β )2 = (1 − α )2(1 − β )2(α − β )2
∴
K =1
16.
PLAN It is a simple question on scalar multiplication, i.e.
k a1 k a2 k a3
a1 a2 a3
b1 b 2 b 3 = k b1 b 2 b 3
c1 c 2 c 3
c1 c 2 c 3
Description of Situation Construction of matrix,
i.e.
ω7
ω2
1
=1
[Q z = − 3]
[Q 1 + ω + ω 2 = 0 and ω7 = (ω3 )2 ⋅ ω = ω]
On applying R1 → R1 + R2 + R3 , we get
3 1 + ω + ω2 1 + ω + ω2
ω
ω2
= 3k
1
1
1
But given,
1 1 1
1 ω ω 2 = 3k
1 ω2 ω
1 ⋅1 + 1 ⋅α + 1 ⋅β
1 ⋅1 + α ⋅α + β ⋅β
1 ⋅ 1 + α 2 ⋅ α 2 + β2 ⋅ β2
(5x − 4)(x + 4)2 = ( A + Bx)(x − A )2
Equating, we get, A = − 4 and B = 5
ω2
1 + α + β 1 + α 2 + β2
1 + α 2 + β 2 1 + α 3 + β3
1 + α 3 + β3 1 + α 4 + β 4
x − 4
2x
Apply R2 → R2 − R1 and R3 → R3 − R1
2x
0 
1
2
∴ (5x − 4)
0
4
0 
−
x
−
 = ( A + Bx)(x − A )

−x − 4
0
0
Expanding along C1, we get
1
f (3) 1 + f (4)
1 ⋅1 + 1 ⋅1 + 1 ⋅1
= 1 ⋅1 + 1 ⋅α + 1 ⋅β
1 ⋅ 1 + 1 ⋅ α 2 + 1 ⋅ β2
Taking common (5x − 4) from C1, we get
2x 
1 2x
2
x
−
(5x − 4)
1
4
2x 
 = ( A + Bx)(x − A )

1
f (1) = α + β, f (2) = α 2 + β 2,
f (4) = α 4 + β 4
f (1) 1 + f (2)
f (2) 1 + f (3)
Here,
 a11 a12 a13 
if a = [aij ]3 × 3 =  a21 a22 a23 
 a31 a32 a33 
 a11 a12 a13 
P = [aij ]3 × 3 =  a 21 a 22 a 23 
 a31 a32 a33 
 b11 b12 b13 
Q = [bij ]3 × 3 =  b21 b22 b23 
 b31 b32 b33 
where, bij = 2i + j aij
4 a11 8 a12 16 a13
∴ Q = 8 a 21 16 a 22 32 a 23
16 a31 32 a32 64 a33
a11
a12
a13
= 4 × 8 × 16 2 a 21 2 a 22 2 a 23
4 a31 4 a32 4 a33
a11 a12
= 29 × 2 × 4 a 21 a 22
a31 a32
= 212 ⋅ P = 212 ⋅ 2 = 213
a13
a 23
a33
Matrices and Determinants 153
17. We know,
xp + y
| A n| = | A|n
| A | = 125 ⇒ | A| = 125
α 2 = 5
⇒ α2 − 4 = 5 ⇒ α = ± 3
2 α 
3
Since,
⇒
3
 sin x cos x cos x 
18. Given,  cos x sin x cos x  = 0
 cos x cos x sin x 
Applying C1 → C1 + C 2 + C3
 sin x + 2 cos x cos x
=  sin x + 2 cos x sin x
 sin x + 2 cos x cos x
1
= (2 cos x + sin x) 1
1
0
cos x
sin x − cos x
0
− (xp + yp + yp + z ) xp + y
⇒
cos x 
cos x 
sin x 
cos x
sin x
cos x
cos x 
cos x  = 0
sin x 
=0
yp + z
− (xp + 2 yp + z ) (xz − y2) = 0
2
∴ Either xp + 2 yp + z = 0 or y2 = xz
2
⇒ x, y, z are in GP.
22. Since, A is the determinant of order 3 with entries 0 or
1 only.
Also, B is the subset of A consisting of all
determinants with value 1.
cos x

0
= 0
sin x − cos x 
(2 cos x + sin x) (sin x − cos x)2 = 0
2 cos x + sin x = 0 or sin x − cos x = 0
2 cos x = − sin x or sin x = cos x
π
π
⇒ cot x = − 1 / 2 gives no solution in − ≤ x ≤
4
4
and sin x = cos x ⇒ tan x = 1 ⇒ x = π /4
[since, if we interchange any two rows or columns,
then among themself sign changes]
Given, C is the subset having determinant with
value −1.
∴ B has as many elements as C.
23. For a matrix to be square of other matrix its
determinant should be positive.
(a) and (c) → Correct, (b) and (d) → Incorrect
24. Given determinant could be expressed as product of two
19. Given,
Applying C3 → C3 − (C1 + C 2)
x
1

x (x − 1)
2x
=
 3x (x − 1) x (x − 1)(x − 2)
determinants.
(1 + α )2 (1 + 2α )2 (1 + 3α )2
i.e. (2 + α )2 (2 + 2α )2 (2 + 3α )2 = − 648 α
(3 + α )2 (3 + 2α )2 (3 + 3α )2
x+1

(x + 1) x

(x + 1) x (x − 1) 
1
x

f (x) =
2x
x (x − 1)
 3x (x − 1) x (x − 1) (x − 2)
20. Let
y
z
2
⇒
⇒
⇒
∴
y
=0
z
yp + z
Applying C1 → C1 − ( p C 2 + C3 )
x
0
y
⇒
0
Applying R2 → R2 − R1 , R3 → R3 − R1
1
⇒ (2 cos x + sin x) 0
0
x
y
xp + y
21. Given, yp + z
0
0 = 0
0
1 + 2α + α2
⇒ 4 + 4α + α2


a2
a
1
p
px
cos
(
p
d
)
x
cos
cos
(
∆ =
−
+ d) x 


sin ( p − d ) x sin px sin ( p + d) x 
⇒
1 + a2
a
a2
⇒ ∆ = cos ( p − d ) x + cos ( p + d ) x cos px cos ( p + d )x
sin( p − d ) x + sin( p + d ) x sin px sin ( p + d )x
1 1 1 1 1 1
⇒ α 3 4 2 1 ⋅ 2 4 6 = − 648 α
9 3 1 1 4 9
1 + a2
a
a2
⇒ ∆ = 2 cos px cos dx cos px cos ( p + d ) x
2 sin px cos dx sin px sin ( p + d ) x
a
cos px
sin px

a2
cos ( p + d ) x 

sin ( p + d ) x 
= − 648 α
α α2 1 1 1
4 2 α α 2 ⋅ 2 4 6 = − 648 α
9 3α α2 1 4 9
1
Applying C1 → C1 + C3
1 + a 2 − 2a cos dx
⇒ ∆ =
0

0

1 + 6α + 9 α2
4 + 12 α + 9 α 2
9 + 6 α + α 2 9 + 12 α + 4 α 2 9 + 18 α + 9 α 2
f (x) = 0 ⇒ f (100) = 0
Applying C1 → C1 − 2 cos dx C 2
1 + 4α + 4α2
4 + 8α + 4α2
⇒
⇒
∴
25.
−8 α 3 = − 648 α ⇒ α 3 − 81α = 0
α (α − 81) = 0
α = 0, ± 9
2
PLAN (i) If A and B are two non-zero matrices and AB = BA, then
( A − B)( A + B) = A 2 − B 2.
(ii) The determinant of the product of the matrices is equal to
product of their individual determinants, i.e. | A B | = | A || B |.
Given, M 2 = N 4
⇒ ∆ = (1 + a 2 − 2a cos dx) [sin ( p + d ) x cos px
− sin px cos ( p + d ) x]
⇒ ∆ = (1 + a 2 − 2a cos dx) sin dx
⇒
which is independent of p.
⇒
⇒
M2 − N 4 = 0
(M − N ) (M + N 2) = 0
[as MN = NM ]
2
M ≠N
Also,
M + N2 =0
2
154 Matrices and Determinants
⇒
det (M + N 2) = 0
Also, det (M 2 + MN 2) = (det M) (det M + N 2)
1 a a 2 − bc
1 a a2
1 a bc
2
2
1 b b − ca = 1 b b − 1 b ca
1 c c2 − ab
1 c c2
1 c ab
29.
= (det M) (0) = 0
As,
det (M 2 + MN 2) = 0
Thus, there exists a non-zero matrix Usuch that
(M 2 + MN 2) U = 0
b
aα + b
 a
26. Given,
c
bα + c  = 0
 b
0 
 aα + b bα + c
Now,
1 a
1 b
1
c
a a 2 abc
1
b b2 abc
ca =
abc
c c2 abc
ab
bc
Applying R1 → aR1 , R2 → bR2, R3 → cR3
1 a a2
a a2 1
1
2
=
⋅ abc b b 1 = 1 b b2
abc
c c2 1
1 c c2
Applying C3 → C3 − (α C1 + C 2)
 a

b
0
c
0


 b
= 0
2
 aα + b bα + c − (aα + 2bα + c)
⇒
⇒
− (aα 2 + 2bα + c) (ac − b2) = 0
aα 2 + 2bα + c = 0 or b2 = ac
∴
1 a a 2 − bc
1 b b2 − ca = 0
1 c c2 − ab
⇒ x − α is a factor of ax2 + 2bx + c or a , b, c are in GP.
a1
27. Let Det (P ) = a 2
b1
b2
c1
c2
a3
b3
c3
7 6 x
= a1 (b2c3 − b3 c2) − a 2 (b1c3 − b3 c1 ) + a3 (b1c2 − b2c1 )
Now, maximum value of Det (P ) = 6
If a1 = 1, a 2 = − 1, a3 = 1, b2c3 = b1c3 = b1c2 = 1
and b3 c2 = b3 c1 = b2c1 = − 1
But it is not possible as
(b2c3 ) (b3 c1 ) (b1c2) = − 1 and (b1c3 ) (b3 c2) (b2c1 ) = 1
i.e., b1b2b3 c1c2c3 = 1 and − 1
Similar contradiction occurs when
a1 = 1, a 2 = 1, a3 = 1, b2c1 = b3 c1 = b1c2 = 1
and b3 c2 = b1c3 = b1c2 = − 1
Now, for value to be 5 one of the terms must be zero but
that will make 2 terms zero which means answer
cannot be 5
1
1 1
Now, −1
1 1 =4
1
−1 1
Hence, maximum value is 4.
log x y log x z
1
28. Let
log y z
∆ = log y x
1
log z x log z y
1
=
log x
log y
log x
log z
log y
log x
1
log y
log z
x 3 7
2 x 2 =0
30. Given,
1
log z
log x
log z
log y
1
On dividing and multiplying R1 , R2, R3 by log x,
log y, log z, respectively.
log x log y log z
1
log x log y log z = 0
=
log x log y log z
log x log y log z
Applying R1 → R1 + R2 + R3
x+9 x+9 x+9
1 1 1
2
2 = 0 ⇒ (x + 9) 2 x 2 = 0
⇒
x
7
6
x
7 6 x
Applying C 2 → C 2 − C1
1
0
0
and C3 → C3 − C1
0 = 0 ⇒ (x + 9) (x − 2) (x − 7) = 0
⇒ (x + 9) 2 x − 2
7 −1 x − 7
⇒
x = − 9, 2, 7 are the roots.
∴ Other two roots are 2 and 7.
1
4
31. Given, 1 −2
20
=0
5
1 2x 5x
2
⇒ 1 (− 10 x2 − 10x) − 4 (5x2 − 5) + 20 (2x + 2) = 0
⇒
−30x2 + 30x + 60 = 0
⇒
(x − 2) (x + 1) = 0
⇒
x = 2, − 1
Hence, the solution set is { −1, 2}.
λ 2 + 3λ
32. Given,
λ+1
λ −3
λ −1
λ+3
−2 λ λ − 4
λ + 4 3λ
= pλ 4 + qλ 3 + rλ 2 + sλ + t
Thus, the value of t is obtained by putting λ = 0.
0 −1 3
1
0 −4 = t
⇒
−3 4
0
⇒
t =0
[Q determinants of odd order skew-symmetric matrix
is zero]
Matrices and Determinants 155
1
[(x2 + y2 + 1) (acx + a 2x2 + abxy)] = 0
ax
1
⇒
[ax(x2 + y2 + 1) (c + ax + by)] = 0
ax
⇒
(x2 + y2 + 1)(ax + by + c) = 0
⇒
ax + by + c = 0
which represents a straight line.
sin 2θ
sin θ
cos θ
4π 
2π 
2π 



sin θ +

 sin 2θ +
 cos θ +





3
3
3
36. Let ∆ =
a a 2 abc
 1 a bc 
1
2

33. Let ∆ = 
 1 b ca  = abc b b abc
2
c c abc
 1 c ab 
⇒
Applying R1 → aR1 , R2 → bR2, R3 → cR3
a a2 1
1 a a2
1
=
⋅ abc b b2 1 = 1 b b2
abc
c c2 1
1 c c2
1 a
1 b
∴
1
c
1 a
a2
ca = 1 b
ab
1 c
b2
c2
bc
2π 
2π 


sin θ −

 cos θ −


3
3
4π 

sin 2θ −


3
Applying R2 → R2 + R3
Hence, statement is false.
34. Since, M M = I and|M | = 1
T
∴
|M − I | = |IM − M T M |
[Q IM = M ]
⇒ |M − I | = |(I − M T )M |= |(I − M )T ||M |= | I − M |
= (− 1)3 |M − I |[Q I − M is a 3 × 3 matrix]
= − |M − I |
⇒
2|M − I| = 0
⇒
|M − I | = 0
bx + ay
cx + a
 ax − by − c
35. Given,  bx + ay −ax + by − c
cy + b
cy + b
− ax − by +
 cx + a

= 0
c
a x − aby − ac
cx + a
bx + ay
− ax + by − c
cy + b
abx + a 2y
=0
cy + b
− ax − by + c
acx + a 2
2
⇒
1
a
⇒
 (a 2 + b2 + c2) x
cx + a 
ay + bx
1 2
b + cy  = 0
(a + b2 + c2) y by − c − ax

a
2
2
2
c
ax − by
−
+
b
cy
 a +b +c

⇒
ay + bx
cx + a 
1x
b + cy  = 0
 y by − c − ax
a 1
b + cy
c − ax − by 
[Q a 2 + b2 + c2 = 1]
Applying C 2 → C 2 − bC1 and C3 → C3 − cC1
ay
a

1x
b
⇒
= 0
 y − c − ax
a 1
cy
− ax − by 
x2
axy
ax
1
y − c − ax
b
=0
ax
1
cy
− ax − by
Applying R1 → R1 + yR2 + R3
⇒


0
0
1 x + y +1
y
b
− c − ax



= 0
ax
1
cy
− ax − by 

⇒
1
[(x2 + y2 + 1) {(− c − ax)(− ax − by) − b(cy)}] = 0
ax
⇒
1
[(x2 + y2 + 1) (acx + bcy + a 2x2 + abxy − bcy)] = 0
ax
2
cos θ
2π 

cos θ +


3
2π 

+ cos θ −


3
2π 

cos θ −


3
sin 2θ
4π 

sin 2θ +


3
4π 

+ sin 2θ −


3
4π 

sin 2θ −


3
2π 
2π 


Now, sin θ +
 + sin θ − 


3
3
2π
2π 
2π
2π 


+θ−
−θ +


θ +
θ +
3
3  cos 
3
3 
= 2 sin 
2
2








π
2π

= 2 sin θ cos  π − 

3
3
π
= − 2 sin θ cos = − sin θ
3
2
π
2π 



and cos θ +
 + cos θ −




3
3
= 2 sin θ cos
Applying C1 → C1 + bC 2 + cC3
⇒
sin θ
2π 

sin θ +


3
=
2π 

+ sin θ −


3
2π 

sin θ −


3
2
2π
2π 
2π
2π 


−θ +
+θ−

θ +

θ +
3
3
3
3 
= 2 cos 
 cos 
2
2








 2π 
= 2 cos θ cos  
 3
 1
= 2 cos θ  −  = − cos θ
 2
4π 
4π 


and sin 2θ +

 + sin 2θ −


3
3
4π
4π 


+ 2θ −
 2θ +

 2θ +
3
3
= 2 sin 
 cos 
2






4π

= 2 sin 2θ cos
= 2 sin 2θ cos  π +

3
π
= − 2 sin 2θ cos = − sin 2θ
3
4π
4π 
− 2θ +

3
3 
2


π

3
156 Matrices and Determinants




sin 2 θ
cos θ
sin θ
∴ ∆ =  − sin θ
− cos θ
− sin 2θ = 0

sin θ − 2π  cos θ − 2π  sin 2θ − 4π  




3
3
3 

39. Given, a > 0, d > 0 and let
1
a
1
∆=
(a + d )
1
(a + 2d )
[since, R1 and R2 are proportional]
2ax − 1
b+1
2ax
37. Given, f ′ (x) =
b
2ax + b + 1
−1
2 (ax + b) 2ax + 2b + 1
2ax + b
 2ax 2ax − 1 2ax + b + 1 
f ′ (x) =  b
b+1
−1 
1
0
 0
2ax − 1   2ax −1 
=
b+1   b
1
⇒
[C 2 → C 2 − C1]
a

 (a + d )(a + 2d ) (a + 2d )
 (a + 2d )(a + 3d ) (a + 3d ) (a + d ) 
 (a + 3d )(a + 4d ) (a + 4d ) (a + 2d ) 
1
⇒ ∆=
∆′
a (a + d )2(a + 2d )3 (a + 3d )2(a + 4d )
f ′ (x) = 2ax + b
On integrating, we get f (x) = ax + bx + c,
2
where c is an arbitrary constant.
Since, f has maximum at x = 5 / 2.
⇒
f ′ (5 / 2) = 0 ⇒ 5a + b = 0
Also,
…(i)
f (0) = 2
⇒
c = 2 and f (1) = 1
⇒
a + b + c=1
…(ii)
On solving Eqs. (i) and (ii) for a , b, we get
1
5
a = ,b = −
4
4
1 2 5
Thus,
f (x) = x − x + 2
4
4
1 1 1
, , are in an AP.
a b c
1

= A + ( p − 1) D

a

1
= A + ( q − 1) D 
b

1
= A + (r − 1) D 
c

⇒
bc ca
Let ∆ = p
1
q
1
1
a
r = abc p
1
1
1
b
q
1
ab
A + ( p − 1) D
= abc
p
…(i)
∴
1
c
r
1
40. Let
[from Eq. (i)]
⇒
A + (r − 1) D
r
1
1
Applying R1 → R1 − ( A − D ) R3 − DR2
0 0 0
= abc p q r = 0 ⇒
1 1 1
ab
r =0
1
1
∆′ = (2d 2)(d )(a + 2d − a ) = 4d 4
4d 4
∆=
2
a (a + d ) (a + 2d )3 (a + 3d )2(a + 4d )
cos ( A − P )
∆ = cos (B − P )
cos (C − P )
cos A cos P
∆ = cos B cos P
cos C cos P
cos ( A − Q ) cos ( A − R)
cos (B − Q ) cos (B − R)
cos (C − Q ) cos (C − R)
+ sin A sin P cos ( A − Q )
+ sin B sin P cos (B − Q )
+ sin C sin P cos (C − Q )
cos ( A − R)
cos (B − R)
cos (C − R)
⇒
bc ca
p q
1
Applying R2 → R2 − R1 , R3 → R3 − R2
(a + d ) (a + 2d ) (a + 2d ) a
⇒
d
d
∆′ = (a + 2d ) (2d )
d
d
(a + 3d ) (2d )
Expanding along R3 , we get
a + 2d a 
∆′ = 2d 2
d
 d
A + (q − 1) D
q
1
a

 (a + d )(a + 2d ) (a + 2d )
where, ∆′ =  (a + 2d )(a + 3d ) (a + 3d ) (a + d ) 
 (a + 3d )(a + 4d ) (a + 4d ) (a + 2d ) 
Applying R3 → R3 − R2
(a + d )(a + 2d ) (a + 2d ) a
∆′ =
(a + 2d )2d
d
d
2d 2
0
0
38. Since, a , b, c are pth , qth and rth terms of HP.
⇒
1
common from R1 ,
a (a + d ) (a + 2d )
1
from R2,
(a + d )(a + 2d )(a + 3d )
1
from R3
(a + 2d ) (a + 3d )(a + 4d )
1
⇒ ∆=
a (a + d )2(a + 2d )3 (a + 3d )2(a + 4d )
Taking
Applying R3 → R3 − R1 − 2R2, we get
2ax
=
 b
1
1
(a + d ) (a + 2d )
a (a + d )
1
1
(a + d ) (a + 2d ) (a + 2d ) (a + 3d )
1
1
(a + 2d ) (a + 3d ) (a + 3d ) (a + 4d )
 cos A cos P cos ( A − Q ) cos ( A − R) 
∆ =  cos B cos P cos (B − Q ) cos (B − R) 
 cos C cos P cos (C − Q ) cos (C − R) 
 sin A sin P cos ( A − Q ) cos ( A − R) 
+  sin B sin P cos (B − Q ) cos (B − R) 
 sin C sin P cos (C − Q ) cos (C − R) 
Matrices and Determinants 157
⇒
 cos A cos ( A − Q ) cos ( A − R) 
∆ = cos P  cos B cos (B − Q ) cos (B − R) 
 cos C cos (C − Q ) cos (C − R) 
 sin A
+ sin P  sin B
 sin C
cos ( A − Q )
cos (B − Q )
cos (C − Q )
cos ( A − R) 
cos (B − R) 
cos (C − R) 
Applying C 2 → C 2 − C1 cos Q , C3 → C3 − C1 cos R in
first determinant and C 2 → C 2 − C1 sin Q and in
second determinant
⇒
cos A sin A sin Q sin A sin R
∆ = cos P cos B sin B sin Q sin B sin R
cos C sin C sin Q sin C sin R
 sin A cos A cos Q cos A cos R 
+ sin P sin B cos B cos Q cos B cos R 
 sin C cos C cos Q cos C cos R 
 cos A sin A sin A 
∆ = cos P sin Q sin R cos B sin B sin B 
 cos C sin C sin C 
 sin A cos A cos A 
+ sin P cos Q cos R sin B cos B cos B 
 sin C cos C cos C 

n!
 (n + 2)!
(n + 1)!
(n + 2)!
(n + 3)!
(n + 2)! 
(n + 3)! 
(n + 4)! 
Taking n !, (n + 1)! and (n + 2)! common from R1 , R2
and R3 , respectively.
∴
a
1
D = n ! (n + 1)! (n + 2) ! 1
1
(n + 1)
(n + 2)
(n + 3)
(n + 1) (n + 2) 
(n + 2) (n + 3) 
(n + 3) (n + 4) 
Applying R2 → R2 − R1 and R3 → R3 − R2, we get
1
D = n !(n + 1)!(n + 2)! 0
0
(n + 1)
1
1
(n + 1) (n + 2) 
2n + 4

2n + 6

Expanding along C1 , we get
D = (n !)(n + 1)!(n + 2)![(2n + 6) − (2n + 4)]
D = (n !)(n + 1)! (n + 2)! [2]
On dividing both side by (n !)3
D
(n !)(n !)(n + 1)(n !)(n + 1)(n + 2)2
=
⇒
3
(n !)
(n !)3
D
= 2(n + 1)(n + 1)(n + 2)
⇒
(n !)3
D
= 2(n3 + 4n 2 + 5n + 2)
⇒
(n !)3
Applying R1 → R2 − R1 and R3 → R3 − R1 , we get
p
b
c
0
∆= a− p q−b
a−p
0
r−c
=c
a− p q−b
a−p
D
− 4 = 2n (n 2 + 4n + 5)
(n !)3
 D

which shows that 
− 4 is divisible by n.
3
(
n
!)


0
+ (r − c)
p
b
a− p q−b
= − c (a − p) (q − b) + (r − c) [ p(q − b) − b(a − p)]
= − c (a − p) (q − b) + p(r − c) (q − b) − b(r − c)(a − p)
Since, ∆ = 0
⇒ − c (a − p) (q − b) + p(r − c) (q − b) − b(r − c) (a − p) = 0
c
p
b
+
+
=0
⇒
r−c p−a q−b
⇒
[on dividing both sides by (a − p)(q − b)(r − c)]
p
b
c
+
+1+
+ 1 =2
p−a q−b
r−c
p
q
r
+
+
=2
p−a q−b r−c
43. We know, A 28 = A × 100 + 2 × 10 + 8
3B9 = 3 × 100 + B × 10 + 9
and
62 C = 6 × 100 + 2 × 10 + C
Since, A 28, 3B 9 and 62 C are divisible by k, therefore
there exist positive integers m1 , m2 and m3 such that,
100 × A + 10 × 2 + 8 = m1k , 100 × 3 + 10 × B + 9 = m2k
and 100 × 6 + 10 × 2 + C = m3 k
… (i)
A 3 6
∴
∆= 8
2
9 C
B 2
Applying R2 → 100R1 + 10R3 + R2
3
A
⇒
∆ = 100 A + 2 × 10 + 8 100 × 3 + 10 × B + 9
2
B
6
100 × 6 + 10 × 2 + C
2
3
6
A
= A28 3B9 62C
2
2
B
A
= 2n (n 2 + 4n + 5) + 4
⇒
b r
⇒
∆ =0 + 0 =0
41. Given, D =  (n + 1)!
p b c
42. Let ∆ = a q c
3
6
[from Eq. (i)]
A
3
6
= m1k m2k m3 k = k m1 m2 m3
2
B
2
2
B
2
∴
∆ = mk
Hence, determinant is divisible by k.
158 Matrices and Determinants
 a −1
n
 (a − 1)
3n

6
4n − 2 

3n 2 − 3n 
2
2
44. Given, ∆ a = 
 (a − 1) 2n
3
3
46. Since, α is repeated root of f (x) = 0.
∴ f (x) = a (x − α )2, a ∈ constant (≠ 0)
B(x)
C (x) 
 A (x)
Let
φ (x) =  A (α )
B(α ) C (α ) 
 A ′ (α ) B ′ (α ) C ′ (α ) 
 n

n
 ∑ (a − 1)

6
 an=1

n
2
2


=
a
−
n
n
−
∆
1
2
4
2
(
)
∑ a ∑

a=1
a=1
 n

2
3
 ∑ (a − 1)3 3n 3n − 3n 
 a=1

∴
To show φ (x) is divisible by (x − α )2, it is sufficient to
show that φ (α ) and φ ′ (α ) = 0.
B(α ) C (α ) 
 A (α )
∴
φ (α ) =  A (α )
B(α ) C (α ) 
 A ′ (α ) B ′ (α ) C ′ (α ) 
n (n − 1)
6
n
2
n (n − 1)(2n − 1)
2n 2 4n − 2
=
6
n 2(n − 1)2
3n3 3n 2 − 3n
4
=0
[Q R1 and R2 are identical]
 A ′ (x) B ′ (x) C ′ (x) 
Again, φ ′ (x) =  A (α )
B(α ) C (α ) 
 A ′ (α ) B ′ (α ) C ′ (α ) 
 A ′ (α ) B ′ (α ) C ′ (α ) 
φ ′ (α ) =  A (α )
B(α ) C (α ) 
 A ′ (α ) B ′ (α ) C ′ (α ) 
1
1
6
n 2(n − 1) (2n − 1)
=
2n
4n − 2
2
3
n (n − 1)
3n 2 3n 2 − 3n
2
1
1
6
n3 (n − 1)
=
2n − 1 6n 12n − 6
12
n − 1 6n 6n − 6
=0
Thus, α is a repeated root of φ (x) = 0.
Hence, φ (x) is divisible by f (x).
 x2 + x
x+1
x−2
47. Let ∆ =  2x2 + 3x − 1
3x
3x − 3
 2
2x − 1 2x − 1
 x + 2x + 3
Applying R2 → R2 − (R1 + R3 ), we get
Applying C3 → C3 − 6 C1
1
1
 x2 + x
x+ 1 x−2 
0
4
0 
∆ =
−


2
 x + 2x + 3 2x − 1 2x − 1 
0
n3 (n − 1)
2n − 1 6n 0 = 0
=
12
n − 1 6n 0
⇒
n
∑ ∆a = c
[c = 0, i.e. constant]
Applying R1 → R1 +
a=1
 xCr
45. Let ∆ =  y C r
z
C
 r
x
Cr + 1
y
Cr + 1
z
Cr + 1
x2
R2, we get
4
x+ 1 x−2
 x

∆=
0
0
 −4
−
−1
2
2
x
1
x
 2x + 3
Applying C3 → C3 + C 2
x
x+1
y
y+1
Cr + 1
Cr + 1
z
Cr + 1
Cr + 2 
Cr + 2 

z +1
Cr + 2 
[Q nC r + nC r − 1 =
 x + 0 x + 1 x − 2
Applying R3 → R3 − 2R1 =  − 4
0
0 
−3
3 
 3
x
x  0
1 −2 
 x
=  −4
0
0
0
0  +  −4
3
3   3 −3
 3 −3
1
1
−
1
0
1
2

 

= x − 4
0 0  +  −4
0
0
3
 3 −3 3   3 −3
n+1
Cr ]
Applying C 2 → C 2 + C1
 xCr
∆ = y C r
z
C
 r
x+1
x+1
y+1
y+1
Cr + 1
Cr + 1
z+1
Cr + 1
Cr + 2 
Cr + 2 

z+1
Cr + 2 
Applying C3 → C3 + C 2
 xCr
⇒ ∆ = y C r
z
C
 r
x+1
x+ 2
y+1
y+ 2
Cr + 1
Cr + 1
z +1
Cr + 1
x2
R2
4
and R3 → R3 +
Cr + 2 
y
Cr + 2 

z
Cr + 2 
x
 xCr
∆ = y C r
z
C
 r
[Q R1 and R3 are identical]
Cr + 2 
Cr + 2 

z+ 2
Cr + 2 
Hence proved.
⇒
∆ = Ax + B
1 1
 1
where, A =  −4
0 0
 3 −3 3 
and
 0
B =  −4
 3
1
0
−3
−2 
0
3
Matrices and Determinants 159
a
b
c
c
a
b
2. Given matrix B is the inverse matrix of 3 × 3 matrix A,
48. Let ∆ =  b c a
Applying C1 → C1 + C 2 + C3
1 b c
 a + b + c b c
∆ =  a + b + c c a = (a + b + c) 1 c a
 a + b + c a b
1 a b
Applying R2 → R2 − R1 and R3 → R3 − R1 , we get
b
c 
1
= (a + b + c) 0 c − b a − c 
0 a − b b − c 
= (a + b + c) [− (c − b)2 − (a − b) (a − c)]
= − (a + b + c) (a 2 + b2 + c2 − ab − bc − ca )
1
= − (a + b + c) (2a 2 + 2b2 + 2c2 − 2ab − 2bc − 2ca )
2
1
= − (a + b + c)[(a − b)2 + (b − c)2 + (c − a )2]
2
which is always negative.
Topic 3 Adjoint and Inverse of a Matrix
1. It is given that matrix.0
 sin 4 θ
−1 − sin 2 θ 
−1
M =
 = α I + βM , where
2
4
cos
cos
1
θ
θ
+


α = α (θ ) and β = β (θ ) are real numbers and I is the 2 × 2
identity matrix.
Now, det(M ) =|M|= sin 4 θ cos 4 θ + 1 + sin 2 θ + cos 2 θ
+ sin 2 θ cos 2 θ
4
4
2
2
= sin θ cos θ + sin θ cos θ + 2
 sin 4 θ
−1 − sin 2 θ  α 0 
β
and 
 = 0 α  + |M| (adj M )
2
4
1
+
θ
θ
cos
cos




adj M 

Q M −1 =

|M| 
 sin 4 θ
−1 − sin 2 θ  α 0 
⇒ 
=
2
cos 4 θ  0 α 
1 + cos θ
a
1 + sin 2 θ  
β  cos 4 θ
+
Q adj 


2
4
|M| −1 − cos θ
sin θ  
c
b   d −b
=

d  − c a 
⇒ β = −|M|and α = sin 4 θ + cos 4 θ
1
⇒ α = α (θ) = 1 − sin 2(2 θ ), and
2
2

1
7
β = β (θ ) = − sin 2 θ cos 2 θ +  + 


2
4


2
 sin 2(2 θ ) 1
7 

= − 
+  + 
4
2
4



1
37
and β * = β min = −
2
16
2
Q α is minimum at sin (2 θ ) = 1 and β is minimum at
sin 2(2 θ ) = 1
1 37
29
So, α * + β * = −
=−
2 16
16
Now, α * = α min =
5 2α
where B = 0 2

α 3
1 
1 

− 1
We know that,
det( A ) =
1 

−1
Q det( A ) = det( A ) 
1
det(B)
det( A ) + 1 = 0
1
+ 1 =0
det(B)
Since,
(given)
⇒
det(B) = − 1
⇒
5(− 2 − 3) − 2α (0 − α ) + 1 (0 − 2α ) = − 1
⇒
− 25 + 2α 2 − 2α = − 1
⇒
2α 2 − 2α − 24 = 0
⇒
α 2 − α − 12 = 0
⇒
(α − 4) (α + 3) = 0
⇒
α = − 3, 4
So, required sum of all values of α is 4 − 3 = 1
et
3. | A | = et
et
−t
−e
e− t cos t
cos t − e− t sin t
−e
2e− t sin t
1
−t
e− t sin t
sin t + e− t cos t
− 2e− t cos t
cos t
= (et ) (e− t ) (e− t ) 1 − cos t − sin t
1
2 sin t
sin t
− sin t + cos t
− 2 cos t
(taking common from each column)
Aplying R2 → R2 − R1 and R3 → R3 − R1, we get
[Qet − t = e0 = 1]
cos t
sin t
1
= e− t 0 − 2 cos t − sin t − 2 sin t + cos t
0 2 sin t − cos t − 2 cos t − sin t
= e− t ((2 cos t + sin t )2 + (2 sin t − cos t )2)
(expanding along column 1)
= e− t (5 cos 2 t + 5 sin 2 t )
(Qcos 2 t + sin 2 t = 1)
= 5e− t
−t
for all t ∈ R
⇒| A | = 5e ≠ 0
∴ A is invertible for all t ∈ R
[QIf| A | ≠ 0, then A is invertible]
4. Given,| ABAT| = 8
⇒
| A||B|| AT| = 8
∴
| A|2|B| = 8
Also, we have | AB−1| = 8
⇒
| A||B−1| = 8
| A|
⇒
=8
|B|
[Q|XY | = |X ||Y |]
…(i) [Q| AT| = | A|]
1 

…(ii) Q| A −1|=| A|−1 =

| A|
On multiplying Eqs. (i) and (ii), we get
| A|3 = 8 ⋅ 8 = 43
⇒
| A| = 4
160 Matrices and Determinants
⇒
Now,
| A| 4 1
= =
8
8 2
1
−1 T
|BA B | = |B| |B|
| A|
1
 1 1  1
=    =
 2 4  2 16
|B| =
Now,
cos θ − sin θ 

sin θ cos θ 
5. We have, A = 
| A| = cos 2 θ + sin 2 θ = 1
 cos θ sin θ 
and
adj A = 

− sin θ cos θ 
a b 
 d − b
[Q If A = 
, then adj A = − c a  ]
c
d




cos
θ
sin
θ



adj A 
−1
⇒
A −1 = 
Q A =


−
sin
θ
cos
θ

| A| 


51 63 
∴ adj (3 A 2 + 12 A ) = 

84 72
∴
− b
and A adj A = AAT
2 
Clearly, A (adj A ) = A I 2
[Q if A is square matrix of order n,
then A (adj A ) = (adj A ) ⋅ A = A I n ]
5a − b
I2
=
3
2
5a
3
7. Given, A = 
Note that, A −50 = ( A −1 )50
Now,
A −2 = ( A −1 )( A −1 )
 cos θ sin θ   cos θ sin θ 
⇒ A −2 = 


− sin θ cos θ  − sin θ cos θ 

cos 2 θ − sin 2 θ
cos θ sin θ + sin θ cos θ 
=

θ
θ
θ
θ
−
cos
sin
−
cos
sin
− sin 2 θ + cos 2 θ 

 cos 2 θ sin 2 θ 
=

− sin 2 θ cos 2 θ 
= (10a + 3b) I 2
1 0
= (10a + 3b) 

0 1
0
10a + 3b

=

+
0
10
a
3
b


and
Also, A −3 = ( A −2)( A −1 )
 cos 2 θ sin 2 θ   cos θ sin θ 
A −3 = 


− sin 2 θ cos 2 θ  − sin θ cos θ 
 cos 3 θ sin 3 θ 
=

− sin 3 θ cos 3 θ 
Q
∴
 2 − 3

− 4 1 
 2 − 3  2 − 3
A2 = A ⋅ A = 


− 4 1  − 4 1 
 4 + 12 − 6 − 3  16 − 9
=

=
− 8 − 4 12 + 1  − 12 13 
∴
5a − b  5a 3
AAT = 
2  − b 2
3
25a 2 + b2 15a − 2b
...(ii)
=

13 
 15a − 2b
A (adj A ) = AAT
0
10a + 3b
 25a 2 + b2 15a − 2b
=


0
10a + 3b  15a − 2b
13 

⇒
1 
2 

3
2 
6. We have, A = 
...(i)
[using Eqs. (i) and (ii)]
15a − 2b = 0
2b
...(iii)
a=
⇒
15
and
...(iv)
10a + 3b = 13
On substituting the value of ‘a ’ from Eq. (iii) in
Eq. (iv), we get
 2b
10 ⋅   + 3b = 13
 15 
20b + 45b
= 13
⇒
15
65b
⇒
= 13 ⇒ b = 3
15
Now, substituting the value of b in Eq. (iii), we get
5a = 2
Hence, 5a + b = 2 + 3 = 5
 cos 50 θ sin 50 θ 
Similarly, A
=

− sin 50 θ cos 50 θ 
25
25 

π sin
π
cos
π


6
6 
=
 when θ = 
25
25 


12
π cos
π
− sin

6
6 
π
π   Q cos  25π  = cos 4π + π  = cos π 





cos
sin

 6 


6
6 
6
6
=


π
π
25π 
π
π

cos  and sin 
 = sin 4π +  = sin 
− sin
 6 


6
6  
6
6 
−50
 3

= 2
 −1
 2
 16 − 9
 2 − 3
3 A 2 + 12 A = 3 
 + 12 − 4 1 
−
12
13




 48 − 27  24 − 36
=
+ 

− 36 39  − 48 12 
 72 − 63
=

− 84 51 
8.
PLAN Use the following properties of transpose
( AB)T = BT AT , ( AT )T = A and A −1 A = I and simplify. If A is
non-singular matrix, then| A | ≠ 0.
Given,
AAT = AT A and B = A −1 AT
BBT = ( A −1 AT )( A −1 AT )T
= A −1 AT A ( A −1 )T
[Q ( AB)T = BT AT ]
Matrices and Determinants 161
= A −1 AAT ( A −1 )T
= IAT ( A −1 )T
= AT ( A −1 )T = ( A −1 A )T
[Q AAT = AT A]
[Q A −1 A = I ]
[Q ( AB)T = BT AT ]
= IT = I
1 α
3
∴
A3 + cA 2 + dA − 6I = O
c = − 6 and d = 11
14. It is given that matrix M be a 3 × 3 invertible matrix,
| P | = 1(12 − 12) − α (4 − 6) + 3 (4 − 6) = 2 α − 6
P = adj ( A )
[given]
[Q|adj A| = | A|n −1 ]
|P| = |adj A|= | A|2 = 16
such that
⇒
M −1 = adj(adj M )
M −1 = |M| M
(Q for a matrix A of order ‘n’ adj(adj A ) = | A|n − 2 A}
∴
2 α − 6 = 16
⇒
2 α = 22
⇒
M −1M =|M|M 2 ⇒ M 2|M|= I
Q
⇒
α = 11
det(M 2|M|) = det(I ) = 1
⇒
10. Given, PT = 2 P + I
∴
…(i)
(PT )T = (2 P + I )T = 2 PT + I
⇒
P = 2 PT + I
⇒
P = 2 (2 P + I ) + I
⇒
P = 4P + 3I
⇒
PX = − IX = –X
or
1
11. | A| ≠ 0, as non-singular ω
ω2
⇒
a
b
⇒
(1 − cω ) (1 − aω ) ≠ 0
1
1
a ≠ ,c≠
ω
ω
12. Given, M T = − M , N T = − N and MN = NM
...(i)
∴ M 2N 2 (M T N )− 1 (MN −1 )T
⇒ M 2 N 2N −1 (M T )−1 (N −1 )T ⋅ M T
⇒
⇒
|M| = −1(6 − 6) − 1(−8 + 10) − 1(24 − 30) = −2 + 6 = 4
|M|= ±2
M = |M|(adj M )−1
⇒
Q
⇒ − M2
13. Every square matrix satisfied its characteristic
−2
0
1
So
=0
4−λ
⇒
(1 − λ ) {(1 − λ ) (4 − λ ) + 2} = 0
⇒
λ3 − 6λ2 + 11λ − 6 = 0
⇒
A − 6 A 2 + 11 A − 6I = O
3
⇒
…(i)
−2 −6 
0
1
 −2 −4 −2 

|adj M| 
−4 −6 −2
 0 −2 −4 
1
=
−2 −4 −6 

4
−6 −2 −2
−4 
−2
a
0
|M| 
−6 
−2 −4
3 =


4 
−2
1 
−6 −2
(adj M )−1 =
Here, non-singular word should not be used, since there
is no non-singular 3 × 3 skew-symmetric matrix.
0
…(i)
T
⇒ − MN (NM −1 )N −1 M = − M (N N −1 )M
i.e. | A − λI | = 0 ⇒
−1
As we know A (adj A ) = | A| I
M 2 N (NN −1 )(− M )−1 (N T )−1 (− M )
M 2 N I (− M −1 ) ( − N )−1 (− M )
1−λ
0
0
1−λ
3
2
∴det (adj M 2) =|M 2|2 =|M|4 = 16
⇒ − M 2 NM −1N −1 M
⇒ − M ⋅ (MN )M −1N −1 M = − M (NM )M −1N −1 M
⇒
(adj M )2 = M 2
−5
⇒ a = ω , c = ω and b ∈{ω , ω 2} ⇒ 2 solutions
equation,
… (ii)
(adj M )2 = I
0 1 a 
15. Given square matrix M = 1 2 3 


3 b 1 
1
−1 
 −1
and
2
adj (M ) =  8
−6


3
−1
−5
−1 1
−1
2
Q adj (M ) =|M|2 = 8 −6
1 c ≠0
ω 1
1 − cω − aω + acω 2 ≠ 0
NOTE
⇒
|M| = 1
from Eqs. (i) and (ii), we get
M2 = I
−1
As, adj M =|M|M = M
⇒
⇒
⇒
⇒
… (i)
|M|3 |M|2 = 1
⇒
3P = − 3I
1 (1 − cω ) − a (ω − cω 2) + b (ω 2 − ω 2) ≠ 0
⇒
…(ii)
On comparing Eqs. (i) and (ii), we get


2 4 4
Q
6I = A3 + cA 2 + dA
⇒
9. Given, P = 1 3 3
∴
Given, 6 A −1 = A 2 + cA + dI , multiplying both sides by
A, we get
0 1
1 2

3 b
|M|= −2, a = 2and b = 1
0 1 2
M = 1 2 3


3 1 1
162 Matrices and Determinants
0 1 2 α  1
α  1
M β  = 2 ⇒ 1 2 3 β  = 2
   

   
3 1 1  γ  3
 γ  3
⇒ B + 2 γ = 1, α + 2 β + 3γ = 2 and 3 α + β + γ = 3
⇒
α = 1, β = −1and γ = 1
⇒
α −β + γ = 3
And (adj M )−1 + adj (M −1 )
[Qadj (M −1 ) = (adj M )−1]
= 2(adj M )−1
M
 M
[Q(adj M )−1 =
from Eq. (i)]
= 2 −  = −M
 2
|M|
17.
and Q a = 2 and b = 1, so a + b = 3
Hence, options (b), (c) and (d) are correct.
3 − 1 − 2
16. Here,
α 
P = 2 0


3 − 5 0 
Now,
|P| = 3(5α ) + 1(− 3α ) − 2(− 10)
…(i)
= 12α + 20
T
2α
− 10
 5α
6
12 
∴
adj (P ) = − 10


2 
 − α − (3α + 4)
−α 
 5α − 10
…(ii)
6
=  2α
− 3α − 4


10
12
2
−


As,
PQ = kI ⇒ |P||Q| = |kI|
⇒
|P||Q| = k3

 k2
k2 
⇒
|P|  = k3
given,|Q| = 

2
 2

⇒
Q
…(iii)
|P| = 2k
PQ = kI ⇒ Q = kp−1I
adjP k(adj P )
[from Eq. (iii)]
= k⋅
=
|P|
2k
−α 
 5 α − 10
adj P 1 
=
=
− 3α − 4
2α
6

2
2
2

− 10 12
k
− 3α − 4

q23 =
given, q23 = −
∴
8 
2

(3α + 4)
k
−
=−
⇒
2
8
⇒
(3α + 4) × 4 = k
…(iv)
⇒
12α + 16 = k
From Eq. (iii),
|P|= 2k
[from Eq. (i)] …(v)
⇒
12α + 20 = 2k
On solving Eqs. (iv) and (v), we get
…(vi)
α = − 1 and k = 4
∴
4α − k + 8 = − 4 − 4 + 8 = 0
∴ Option (b) is correct.
Now, |P adj (Q )| = |P||adj Q|
2
 k2
k5 210
=
= 29
= 2k  =
2
2
 2
∴ Option (c) is correct.
PLAN A square matrix M is invertible, iff dem (M) or| M| ≠ 0.
a b
M = 
 b c
a  b
(a) Given,   =   ⇒ a = b = c = α
[let]
 b  c
α α 
⇒ M =
 ⇒ |M |= 0 ⇒ M is non-invertible.
α α 
Now, If
Let
(b) Given, [b c] = [a b]
⇒
a = b = c=α
[let]
Again,|M | = 0
⇒ M is non-invertible.
a 0
(c) As given M = 
 ⇒|M |= ac ≠ 0
 0 c
[Q a and c are non-zero]
⇒ M is invertible.
a b
(d) M =   ⇒|M |= ac − b2 ≠ 0
 b c
Q ac is not equal to square of an integer.
M is invertible.
18.
PLAN If| A n × n| = ∆, then|adj A| = ∆A − 1
1 4 4 
Here, adj P3 × 3 = 2 1 7


1 1 3 
⇒
|adj P | = | P |2
1 4 4
∴
| adj P | = 2 1 7 = 1 (3 − 7) − 4 (6 − 7) + 4 (2 − 1)
1 1 3
= − 4 + 4 + 4 = 4 ⇒ |P |= ± 2
19. | A | = (2k + 1) ,| B| = 0
3
But det (adj A) + det (adj B) = 106
⇒
(2k + 1)6 = 106
9
k=
⇒
2
⇒
[k] = 4
Topic 4 Solving System of Equations
1. Given system of linear equations is
[sin θ ] x + [− cos θ ] y = 0
…(i)
[cot θ ] x + y = 0
…(ii)
and
where, [x] denotes the greatest integer ≤ x.
[sin θ ] [− cos θ ]
Here,
∆=
[cot θ ]
1
⇒ ∆ = [sin θ ] − [− cos θ ] [cot θ ]
 π 2π 
When
θ ∈ , 
2 3 
 3 
sin θ ∈ 
, 1
 2

Matrices and Determinants 163
⇒
[sin θ ] = 0
 1
− cos θ ∈ 0, 
 2
⇒
[− cos θ ] = 0
 1

cot θ ∈  −
, 0

3 
and
…(iii)
4. Given system of linear equations
2x + 3 y − z = 0,
x + ky − 2z = 0
…(iv)
and 2x − y + z = 0 has a non-trivial solution (x, y, z ).
2 3 −1

∴ ∆ = 0 ⇒
1 k −2 = 0
⇒
So,
…(v)
[cot θ ] = − 1
∆ = [sin θ ] − [− cos θ ] [cot θ ]
− (0 × (− 1)) = 0 [from Eqs. (iii), (iv) and (v)]
 π 2π 
Thus, for θ ∈  ,  , the given system have infinitely
2 3 
many solutions.
 7π 
 1 
When
θ ∈  π,
 , sin θ ∈  − , 0 ⇒ [sin θ ] = − 1

 2 
6
2 −1
2(k − 2) − 3(1 + 4) − 1(−1 − 2k) = 0
⇒
So, system of linear equations is
2x + 3 y − z = 0
2x + 9 y − 4z = 0
and
2x − y + z = 0
cot θ ∈ ( 3 , ∞ ) ⇒ [cot θ ] = n , n ∈ N .
∆ = − 1 − (0 × n ) = − 1
 7π 
Thus, for θ ∈  π,
 , the given system has a unique

6
solution.
From Eqs. (i) and (iii), we get
x
1
4x + 2 y = 0 ⇒ = −
y
2
2. Given, system of linear equations
1 1
⇒ 4 λ
3 2
… (i)
…(ii)
…(iii)
1
− λ =0
So,
∴
x x y
z
1
= × = − ⇒ = −4
z y z
x
4
x
1
 y 1
Q z = 2 and y = − 2 
x y z
1 1
9 1
+ + + k= − + −4 + = .
y z x
2 2
2 2
5. Given system of linear equations
…(i)
x − 2 y + kz = 1
…(ii)
2x + y + z = 2
and
…(iii)
3x − y − kz = 3
has a solution (x, y, z ), z ≠ 0.
On adding Eqs. (i) and (iii), we get
x − 2 y + kz + 3x − y − kz = 1 + 3
4x − 3 y = 4 ⇒ 4x − 3 y − 4 = 0
This is the required equation of the straight line in
which point (x, y) lies.
−4
⇒ 1(− 4λ + 2λ ) − 1(− 16 + 3λ) + 1(8 − 3λ) = 0
⇒ − 8λ + 24 = 0 ⇒ λ = 3
From, the option λ = 3, satisfy the quadratic equation
λ2 − λ − 6 = 0.
3. Given system of linear equations
x+ y+ z =5
x + 2 y + 2z = 6
x + 3 y + λz = µ
…(i)
…(ii)
…(iii)
(λ, µ ∈ R)
The above given system has infinitely many solutions,
then the plane represented by these equations
intersect each other at a line, means (x + 3 y + λz − µ )
= p(x + y + z − 5) + q (x + 2 y + 2z − 6)
= ( p + q)x + ( p + 2q) y + ( p + 2q)z − (5 p + 6q)
On comparing, we get
p + q = 1, p + 2q = 3, p + 2q = λ
and
5 p + 6q = µ
So,
( p, q) = (−1, 2)
⇒
λ = 3 and µ = 7
⇒
λ + µ = 3 + 7 = 10
…(i)
…(ii)
…(iii)
From Eqs. (i) and (ii), we get
y 1
6 y − 3z = 0, =
z 2
So,
x+ y+ z =6
4 x + λy − λz = λ − 2
and
3x + 2 y − 4z = − 5
has infinitely many solutions, then ∆ = 0
2k − 4 − 15 + 1 + 2k = 0
9
4 k = 18 ⇒ k =
2
⇒
 3 
− cos θ ∈ 
, 1 ⇒ [cos θ ] = 0
 2

and
1
6.
Key Idea A homogeneous system of linear equations have
non-trivial solutions iff ∆ = 0
Given system of linear equations is
x − cy − cz = 0,
cx − y + cz = 0
and
cx + cy − z = 0
We know that a homogeneous system of linear
equations have non-trivial solutions iff
∆ =0
1 − c − c
c −1
c
⇒
= 0

c
c
−
1


164 Matrices and Determinants
⇒ 1(1 − c2) + c(− c − c2) − c(c2 + c) = 0
⇒
1 − c2 − c2 − c3 − c3 − c2 = 0
⇒
−2c3 − 3c2 + 1 = 0
⇒
2c3 + 3c2 − 1 = 0
⇒
(c + 1)[2c2 + c − 1] = 0
⇒
(c + 1)[2c2 + 2c − c − 1] = 0
⇒
(c + 1)(2c − 1)(c + 1) = 0
⇒
has more than one solution, then
D1 = D2 = D3 = 0. In the given problem,
a 2 3
D1 = 0 ⇒ b − 1 5 = 0
c −3 2
c = − 1 or
1
2
10. We know that,
7. The given system of linear equations is
x − 2 y − 2 z = λx
x + 2 y + z = λy
− x − y − λz = 0,
which can be rewritten as
(1 − λ )x − 2 y − 2z = 0
⇒
x + (2 − λ ) y + z = 0
x + y + λz = 0
Now, for non-trivial solution, we should have
1 − λ −2 −2
1
2 − λ 1 =0
1
λ
[Q If a1x + b1 y + c1z = 0; a 2x + b2y + c2z = 0
a3 x + b3 y + c3 z = 0]

has a non-trivial solution, then a 2 b2 c2 = 0

a3 b3 c3

⇒ (1 − λ ) [(2 − λ )λ − 1] + 2 [λ − 1] − 2 [1 − 2 + λ ] = 0
⇒ (λ − 1)[λ2 − 2λ + 1 + 2 − 2] = 0
⇒
(λ −1)3 = 0
⇒
λ =1
a1
b1
c1
8. Given system of linear equations,
(1 + α )x + βy + z = 2
αx + (1 + β ) y + z = 3
αx + βy + 2z = 2
has a unique solution, if
1+α
β
1
α
α
(1 + β ) 1 ≠ 0
β
2
Apply R1 → R1 − R3 and R2 → R2 − R3
1 0 −1
0 1 −1 ≠0
α β
2
⇒ 1(2 + β ) − 0(0 + α ) − 1(0 − α ) ≠ 0 ⇒ α + β + 2 ≠ 0 … (i)
Note that, only (2, 4) satisfy the Eq. (i).
9. We know that, if the system of equations
a1x + b1 y + c1z = d1
a 2x + b2y + c2z = d2
a3 x + b3 y + c3 z = d3
and
⇒ a (− 2 + 15) − 2(2b − 5c) + 3(− 3b + c) = 0
⇒ 13a − 4b + 10c − 9b + 3c = 0
⇒
13a − 13b + 13c = 0
⇒ a − b + c=0⇒b − a − c=0
1
Clearly, the greatest value of c is .
2
1
D =0
the system of linear equations
a1x + b1 y + c1z = 0
a 2x + b2y + c2z = 0
a3 x + b3 y + c3 z = 0
has a non-trivial solution, if
a1 b1 c1
a 2 b2 c2 = 0
a3 b3 c3
Now, if the given system of linear equations
x + 3 y + 7z = 0
− x + 4 y + 7z = 0,
and (sin 3 θ )x + (cos 2 θ ) y + 2z = 0
has non-trivial solution, then
1
3
7
−1
4
7 =0
sin 3 θ cos 2 θ 2
⇒ 1(8 − 7 cos 2 θ ) − 3 (− 2 − 7 sin 3 θ )
+ 7 (− cos 2 θ − 4 sin 3 θ ) = 0
⇒ 8 − 7 cos 2 θ + 6 + 21 sin 3 θ
− 7 cos 2 θ − 28 sin 3 θ = 0
⇒ − 7 sin 3 θ − 14 cos 2 θ + 14 = 0
⇒ − 7 (3 sin θ − 4 sin3 θ ) − 14 (1 − 2 sin 2 θ ) +14 = 0
[ Q sin 3 A = 3 sin A − 4 sin3 A and
cos 2 A = 1 − 2 sin 2 A]
3
2
⇒ 28 sin θ + 28 sin θ − 21 sin θ − 14 + 14 = 0
⇒ 7 sin θ [4 sin 2 θ + 4 sin θ − 3] = 0
⇒ sin θ [4 sin 2 θ + 6 sin θ − 2 sin θ − 3] = 0
⇒ sin θ [2 sin θ (2 sin θ + 3) − 1 (2 sin θ + 3)] = 0
⇒ (sin θ ) (2 sin θ − 1) (2 sin θ + 3) = 0
1
Now, either sin θ = 0 or
2
3


Q sin θ ≠ − as − 1 ≤ sin θ ≤ 1


2
In given interval (0, π ),
1
sin θ =
2
π 5π
θ= ,
⇒
6 6
Hence, 2 solutions in (0, π )
[Q sin θ ≠ 0, θ ∈ (0, π )]
Matrices and Determinants 165
11. Since, the system of equations has infinitely many
solution, therefore D = D1 = D2 = D3 = 0
Here,
1 1 1
D = 1 2 3 = 1 (2α − 9) − 1 (α − 3) + 1(3 − 2) = α − 5
1 3 α
1 1 5
and D3 = 1 2 9 = 1 (2 β − 27) − 1(β − 9) + 5 (3 − 2)
1 3 β
= β − 13
Now,
⇒
and
∴
12. Here,
D =0
α −5 =0 ⇒ α =5
D3 = 0 ⇒ β − 13 = 0 ⇒ β = 13
β − α = 13 − 5 = 8
1 −4 7
3 −5
D= 0
−2
−9
5
= 1(− 27 + 25) + 4(0 − 10) + 7(0 + 6)
[expanding along R1]
= − 2 − 40 + 42 = 0
∴The system of linear equations have infinite many
solutions.
[Q system is consistent and does not have unique
solution as D = 0]
⇒
D1 = D2 = D3 = 0
g −4 7
Now, D1 = 0 ⇒ h 3 − 5 = 0
k
5
−9
⇒ g (− 27 + 25) + 4(− 9h + 5k) + 7(5h − 3k) = 0
⇒ − 2 g − 36h + 20k + 35h − 21k = 0
⇒ − 2g − h − k = 0 ⇒2g + h + k = 0
13 According to Cramer’s rule, here
1 1
1
1 0
0
2
0
D= 2 3
= 2 1
2 3 a2 − 1
2 1 a2 − 3
(Applying C 2 → C 2 − C1 and C3 → C3 − C1)
(Expanding along R1)
= a −3
2
1
1
2
1
0
and D1 = 5
3
2 = 5
3
−1
2
a + 1 3 a2 − 1
a + 1 3 a2 − 1 −3
(Applying C3 → C3 − C 2)
2
0
0
5
= 5
−1
3−
2
a + 1 3 − (a + 1) a 2 − 1 − 3
2
1
(Applying C 2 → C 2 − C1)
2
2
5
0
1
2
−1
a+1
5 a
−
2 2
a2 − 4
=
0
1
5 a 
[Expanding along R1]
= 2  (a 2 − 4) +  −  
2 2 
2

a2
5 a
=2
− 2 + −  = a2 − 4 + 5 − a = a2 − a + 1
2 2
2
Clearly, when a = 4, then D = 13 ≠ 0 ⇒ unique solution
and
when|a| = 3, then D = 0 and D1 ≠ 0.
∴When |a| = 3, then the system has no solution i.e.
system is inconsistent.
14. We have,
x + ky + 3z = 0; 3x + ky − 2z = 0; 2x + 4 y − 3z = 0
System of equation has non-zero solution, if
1 k 3 
 3 k −2 = 0


 2 4 −3
⇒ (−3k + 8) − k(−9 + 4) + 3(12 − 2k) = 0
⇒
− 3k + 8 + 9k − 4k + 36 − 6k = 0
⇒
− 4k + 44 = 0 ⇒ k = 11
Let
z = λ , then we get
x + 11 y + 3λ = 0
3x + 11 y − 2λ = 0
and
2x + 4 y − 3λ = 0
Solving Eqs. (i) and (ii), we get
5λ
−λ
5λ2
xz
, y=
,z=λ ⇒ 2=
x=
= 10
2
2
2
y
 λ
2 × − 
 2
…(i)
…(ii)
…(iii)
15. Given, system of linear equation is
x + λy − z = 0; λx − y − z = 0; x + y − λz = 0
Note that, given system will have a non-trivial solution
only if determinant of coefficient matrix is zero,
i.e.
1
λ
1
λ −1
−1 −1 = 0
1 −λ
⇒
1 (λ + 1) − λ(− λ2 + 1) − 1(λ + 1) = 0
⇒
λ + 1 + λ3 − λ − λ − 1 = 0
⇒
λ3 − λ = 0 ⇒ λ(λ2 − 1) = 0
⇒
λ = 0or λ = ± 1
Hence, given system of linear equation has a
non-trivial solution for exactly three values of λ.
16. Given system of linear equations
2x1 − 2x2 + x3 = λx1
⇒
(2 − λ )x1 − 2x2 + x3 = 0
2x1 − 3x2 + 2x3 = λx2
…(i)
…(ii)
…(iii)
166 Matrices and Determinants
⇒ 2x1 − ( 3 + λ )x2 + 2x3 = 0
− x1 + 2x2 = λx3 ⇒ − x1 + 2x2 − λx3 = 0
Since, the system has non-trivial solution.
−2
1
2 − λ
 2
∴
− (3 + λ ) 2  = 0


2
λ 
 −1
⇒ (2 − λ )(3λ + λ2 − 4) + 2(−2λ + 2) + 1(4 − 3) − λ ) = 0
⇒
(2 − λ )(λ2 + 3λ − 4) + 4(1 − λ ) + (1 − λ ) = 0
⇒
(2 − λ )(λ + 4)(λ − 1) + 5(1 − λ ) = 0
⇒
(λ − 1)[(2 − λ )(λ + 4) − 5] = 0
⇒
(λ − 1)(λ2 + 2λ − 3) = 0
⇒
(λ − 1)[(λ − 1)(λ + 8)] = 0
⇒
(λ − 1)2(λ + 3) = 0
⇒
λ = 1, 1, − 3
Hence, λ contains two elements.
17. Given equations can be written in matrix form
AX = B
where, A = 

k+1
k
4k
x
8 
, X =   and B =
k + 3 
3k − 1
 y
For no solution, A = 0 and (adj A) B ≠ 0
8 
k+1
Now, A = 
= 0 ⇒ (k2 + 1)(k + 3) − 8k = 0
k + 3 
 k
⇒
k = 1, 3
8
4.1
If k –1, then
, false
≠
1+3 2
And, if k = 3, then
(k − 1)(k − 3) = 0 ⇒ k = 1, k = 3,
k + 3 −8 
adj A = 
k + 1 
 − k
Now
k + 3 − 8   4k

(adj A)B = 
k + 1   3 k + 1 
 − k
 (k + 3)(4k) − 8 (3k − 1)
=
 − 4k2 + (k + 1)(3k − 1) 
Now,
 4k2 − 12k + 8 
=
2

 − k + 2k − 1 
Hence, only one value of k exist.
 x
Put k = 3
36 − 36 + 8  8 
(adj A) B = 
≠ 0 true
=
−9 + 6 − 1 −4
Hence, required value of k is 3.
   
 z  0
and that could have only unique, no solution or
infinitely many solution.
∴It is not possible to have two solutions.
Hence, number of matrices A is zero.
19. Since, given system has no solution.
∴ ∆ = 0 and any one amongst ∆ x , ∆ y , ∆ z is non-zero.
2
 2 −1
 2 −1 2 
Let 1 −2 1  = 0 and ∆ z =  1 −2 −4  = 6 ≠ 0
1 λ
1
4
1
1
⇒
8
4k
k+1
=
=
⇒ k =1
k
k + 3 3k − 1
21. Given equations x + ay = 0, az + y = 0, ax + z = 0 has
infinite solutions.
1 a 0 
∴
0 1 a = 0
a 0 1 
⇒
∴
−k
−1
1
−1 
−1  = 0
−1 
− C 2, C 2 → C 2 + C3
−1 
−1  = 0
−1 
⇒
2(k + 1) − (k + 1)2 = 0
⇒
(k + 1)(2 − k − 1) = 0
⇒
8
k+1
=
⇒ k2 + 4k + 3 = 8k
k
k+3
1
k
1
Applying C1 → C1
 1 + k −k − 1
−2
⇒ 1 + k
0
 0
NOTE
Take
1 + a3 = 0 or a = − 1
22. Since, the given system has non-zero solution.
Condition for the system of equations has no solution is
a1 b1 c1
= ≠
a 2 b2 c2
8
4k
k+1
=
≠
k
k + 3 3k − 1
λ =1
20. For infinitely many solutions, we must have
Alternate Solution
∴
1
18. Since, A  y = 0 is linear equation in three variables
Put k = 1
4 − 12 + 8 0
not true
(adj A) B = 
=
−1 + 2 − 1  0
8
4.3
, true
≠
6
9 −1
Therefore, k = 3
k2 + 4k + 3 − 8k = 0 ⇒ k2 − 4k × 3 = 0
⇒
k2 − 4k + 3 ⇒ (k –1) (k –3) = 0
k=±1
There is a golden rule in determinant that n one’s ⇒
( n − 1) zero’s or n (constant) ⇒ ( n − 1) zero’s for all
constant should be in a single row or a single column.
23. The given system of equations can be expressed as
3  x 
 1 −2
 1 −3
4  y =
 

−1 1 −2  z 
 −1 
 1
 
 k 
Matrices and Determinants 167
2 2 + 2x 2 + 5x 
= 0
8
8 


3x
3x + 18
x
Applying R2 → R2 − R1 , R3 → R3 + R1
1 −2 3
~ 0 −1 1


0 −1 1
 x   −1 
 y =  2 
  

 z  k − 1
Applying R3 → R3 − R2
1 −2 3  x   −1 
~ 0 −1 1  y =  2 

   

0 0 0  z  k − 3
When k ≠ 3 , the given system of equations has no
solution.
⇒ Statement I is true. Clearly, Statement II is also
true as it is rearrangement of rows and columns of
 1 −2 3 
 1 −3 4 .


−1 1 −2
24. It is given, that matrices
∴
2 x x
1 1 1
P = 0 2 2, Q = 0 4 0




x x 6
0 0 3
adj (P )
P −1 =
|P|
6
as|P| = 6 and adj P = −3

 0
6 −3
1
P −1 = 0 3
⇒
6
0 0
∴
⇒
T
0
3 0

−2 2
0
−2 

2 
0
−1
−1
[QR = PQP (given]
|R| = |P Q P |
−1
[Q|P||P −1| = |I| = 1]
|R| = |P||Q||P | = |Q|
2 x 0
2 x x
2 x x
= 0 4 0 = 0 4 0 + 0 4 0
x x 6
x x 5
x x 1
2 x x
= 0 4 0 + 2 (4 − 0) − x (0 − 0) + 0(0 − 4x)
x x 5
2 x x
= 0 4 0 + 8, for all x ∈ R
x x 5
Q
and
1 1
PQ = 0 2

0 0
2 + x
=  2x

 3x
1 2 x x
2 0 4 0


3 x x 6
4 + 2x x + 6
2x + 8 12 

3x
18 
2 x x 1 1 1
QP = 0 4 0 0 2 2



x x 6 0 0 3
There is no common value of ‘x’, for which each
corresponding element of matrices PQ and QP is equal.
2 0 0
For x = 0, Q = 0 4 0


0 0 6
1 
1 
then, if R a  = 6 a 
 
 
 b 
 b 
1 
1 
[Q R = PQP −1 ]
⇒ PQP −1 a  = 6 a 
 
 
 b 
 b 
1 
1 1 1 2 0 0 6 −3 0  1 
1
⇒ 0 2 2 0 4 0 0 3 −2 a  = 6 a 
 
 


6
2   b 
 b 
0 0 3 0 0 6 0 0
2
4
6
6
3
0
1
−
1


 
 
1
0 8 12 0 3 −2 a  = 6 a 
⇒

 
 
6
2   b 
0 0 18 0 0
 b 
12 6 4  1 
1 
 0 24 8  a  = 36 a 
⇒

 
 
 0 0 36  b 
 b 
+
+
a
b
36
12
6
4

 

0 + 24a + 8b = 36a 
⇒

 

 0 + 0 + 36b  36b 
⇒
⇒
6a + 4b = 24 and 12a = 8b
3a + 2b = 12 and 3a = 2b
⇒ a = 2 and b = 3
So a + b = 5.
α  0
Now, R β  = 0 and αi$ + β$j + γk$ is a unit vector, so det
   
 γ  0
(R) = 0
⇒
det(Q ) = 0 [Q R = PQP −1 So,|R|=|Q|]
2 x x
⇒
0 4 0 =0
x x 6
⇒
⇒
2 (24 − 0) − x (0 − 0) + x(0 − 4x)
48 − 4x2 = 0
⇒
⇒
x2 = 12
x=±2 3
So, for x = 1, there does not exist a unit vector
α  0
αi$ + β$j + γk$ , for which R β  = 0
   
 γ  0
Hence, options (b) and (d) are correct.
168 Matrices and Determinants
26. Given system λx + y + z = 0, − x + λy + z = 0
25. We have,
− x + 2 y + 5z = b1
2x − 4 y + 3z = b2
x − 2 y + 2z = b3
has at least one solution.
−1 2 5
D= 2
1
∴
and
⇒
will have non-zero solution, if
λ
1
⇒ λ (λ 2 + 1) − 1 (− λ + 1) + 1(1 + λ ) = 0
⇒
λ3 + λ + λ − 1 + 1 + λ = 0
⇒
λ 3 + 3λ = 0
⇒
λ (λ 2 + 3) = 0 ⇒
D1 = D2 = D3 = 0
b1 2 5
D1 = b2 − 4 3
b3 − 2 2
...(i)
= 1(− 6 + 6) − 1 (− 15 + 12) + 3 (− 5 + 4) = 0
For atleast one solution
D1 = D2 = D3 = 0
b1 1
3
Now, D1 = b2 2
6
5
⇒
a
| A2| = 1
1
and
f
| A1| = g
h
−2
3
BX = V
a
|B| = 0
f
1
d
g
= 1(0 − 12) − 2 (− 10 − 3) + 5 (8 − 0)
1 4 −5
= 54
D ≠0
It has unique solution for any b1, b2, b3 .
1
b =0 ⇒
b
g=h
1
b =0 ⇒
b
g=h
…(i)
1
c =0
h
[from Eq. (i)]
[since, C 2 and C3 are equal]
BX = V has no solution.
a2 1 1
[from Eq. (i)]
|B1|= 0
d c =0
0
g h
a
|B2| = 0
f
Since,
a2
0
0
[since, c = d and g = h]
1
[Q c = d ]
c = a 2cf = a 2df
h
adf ≠ 0 ⇒ |B2| ≠ 0
|B| = 0
5
= − 1(− 20 + 20) − 2(10 − 10) − 5(− 4 + 4) = 0
Here, b2 = − 2b1 and b3 = − b1 satisfies the Eq. (i)
Planes are parallel.
1 2 5
(d) D = 2 0
∴
f
g
h
0
c
d
g = h , c = d and ab = 1
Now,
−3
= − b1 (− 15 + 12) + b2(− 3 + 6) − b3 (6 − 15)
= 3b1 + 3b2 + 9b3 = 0 ⇒ b1 + b2 + 3b3 = 0
not satisfies the Eq. (i)
It has no solution.
−1 2 −5
(c) D = 2 − 4 10
1
∴
6
b2
1
b =0
b
a (bc − bd ) + 1(d − c) = 0 ⇒ (d − c)(ab − 1) = 0
ab = 1 or d = c
a 0
f
Again, | A3|= 1 c
g =0 ⇒ g = h
1 d h
−1 −3
= b1 (− 6 + 6) − b2(− 3 + 3) + b3 (6 − 6) = 0
1 b1 3
0
c
d
⇒
∴
−2 −1 −3
− 2 b3
a
1
1
⇒| A| = 0 ⇒
= 14 − 2 = 12 ≠ 0
Here, D ≠ 0 ⇒ unique solution for any b1, b2, b3 .
1
1
3
(b) D = 5
2
6
D2 =
λ =0
27. Since, AX = U has infinitely many solutions.
1 2 6
b3
1
−1 λ 1 = 0
−1 −1 λ
−4 3
−2 2
= − 2b1 − 14b2 + 26b3 = 0
⇒ b1 + 7b2 = 13b3
1 2 3
(a) D = 0 4 5 = 1(24 − 10) + 1(10 − 12)
− x − y + λz = 0
and
∴
and
|B2| ≠ 0
BX = V has no solution.
28. Given, λx + (sin α ) y + (cos α ) z = 0
x + (cos α ) y + (sin α ) z = 0
and − x + (sin α ) y − (cos α ) z = 0 has non-trivial
solution.
∴
⇒
λ
1
sin α
cos α
−1 sin α
∆ =0
cos α
sin α = 0
− cos α
Matrices and Determinants 169
⇒
λ (− cos 2 α − sin 2 α ) − sin α (− cos α + sin α )
+ cos α (sin α + cos α ) = 0
⇒
− λ + sin α cos α + sin α cos α − sin 2 α + cos 2 α = 0
⇒
λ = cos 2α + sin 2α
Q − a 2 + b2 ≤ a sin θ + b cos θ ≤ a 2 + b2 


∴
− 2≤λ≤ 2
…(i)
Again, when λ = 1, cos 2α + sin 2α = 1
1
1
1
cos 2α +
sin 2α =
⇒
2
2
2
⇒
⇒
2α − π / 4 = 2n π ± π / 4
2α = 2nπ − π / 4 + π / 4 or 2α = 2nπ + π / 4 + π / 4
∴
⇒
⇒
31. The given system of equations
has atleast one solution, if ∆ ≠ 0.
3 −1 4
∴
∆ = 1 2 −3 ≠ 0
α = nπ or nπ + π /4
6
29. Since, α 1 , α 2 are the roots of ax + bx + c = 0.
2
⇒
sin θ (4 sin 2 θ + 4 sin θ − 3) = 0
sin θ (2 sin θ − 1) (2 sin θ + 3) = 0
1
sin θ = 0, sin θ =
2
[neglecting sin θ = − 3 / 2]
π
θ = nπ , nπ + (−1)n , n ∈ Z
6
3x − y + 4z = 3
x + 2 y − 3z = − 2
6 x + 5 y + λz = − 3
cos (2α − π / 4) = cos π / 4
∴
⇒
⇒
b
α1 + α 2 = −
a
c
and α 1α 2 =
a
Also, β1 , β 2 are the roots of px2 + qx + r = 0.
q
r
and β1β 2 =
⇒
β1 + β 2 = −
p
p
...(i)
...(ii)
Given system of equations
α 1 y + α 2z = 0
⇒ 3 (2λ + 15) + 1 (λ + 18) + 4 (5 − 12) ≠ 0
⇒
7 (λ + 5) ≠ 0 ⇒ λ ≠ −5
Let z = − k, then equations become
3x − y = 3 − 4k and x + 2 y = 3k − 2
On solving, we get
4 − 5k
13k − 9
x=
,y=
,z=k
7
7
32. Given system of equations are
and β1 y + β 2 z = 0, has non-trivial solution.
α1 α 2
α 1 β1
=
∴
=0 ⇒
β1 β 2
α 2 β2
Applying componendo-dividendo,
α 1 + α 2 β1 + β 2
=
α 1 − α 2 β1 − β 2
3x + my = m and 2x − 5 y = 20
3 m
Here,
∆=
= −15 − 2m
2 −5
∆x =
and
⇒ (α 1 + α 2) (β1 − β 2) = (α 1 − α 2) (β1 + β 2)
∆y =
⇒ (α 1 + α 2)2 {(β1 + β 2)2 − 4 β 2 β 2}
= (β1 + β 2)2{(α 1 + α 2)2 − 4 α 1α 2}
From Eqs. (i) and (ii), we get
b2  q2 4r  q2  b2 4c
− =
− 


a 2  p2 p  p2  a 2 a 
⇒
⇒
b2q2 4b2r b2q2 4q2c
−
=
−
a 2p2 a 2p a 2p2 ap2
b2r q2c
b2 ac
=
⇒ 2=
a
p
pr
q
30. The system of equations has non-trivial solution, if
∆ = 0.
sin 3θ −1 1
cos 2θ 4 3 = 0
y=
and
For x > 0 ,
∆y
∆
=
sin 3θ ⋅ (28 − 21) − cos 2 θ (−7 − 7) + 2 (−3 − 4) = 0
⇒
⇒
7
7
7 sin 3θ + 14 cos 2θ − 14 = 0
⇒
sin 3θ + 2 cos 2 θ − 2 = 0
⇒ 3 sin θ − 4 sin3 θ + 2 (1 − 2 sin 2 θ ) − 2 = 0
m
3 m
2 20
= −25m
= 60 − 2m
or
60 − 2m
2m − 60
=
− 15 − 2m 15 + 2m
25m
>0 ⇒ m >0
15 + 2m
Expanding along C1 , we get
2
m
20 −5
If ∆ = 0, then system is inconsistent, i.e. it has no
solution.
15
If ∆ ≠ 0, i.e. m ≠
, the system has a unique solution
2
for any fixed value of m.
∆
− 25m
25m
We have,
x= x =
=
∆ − 15 − 2 m 15 + 2m
15
2
2m − 60
and y > 0,
>0
2m + 15
⇒
λ
5
m<−
m > 30 or m < −
…(i)
15
2
From Eqs. (i) and (ii), we get m < −
…(ii)
15
or m > 30
2
170 Matrices and Determinants
…(ii)
x + 2 y = 10
and
…(iii)
3x + 2 y = µ
On solving Eqs. (i) and (ii), we are getting
y = 4 and x = 2
From Eq. (iii), we get
µ = 3(2) + 2(4) = 14
Now, on putting y = 0 and µ = 14, in the given system of
linear Eq., we get
…(iv)
x+ z =6
…(v)
x + 3z = 10
and
…(vi)
3x + λz = 14
On solving Eqs. (iv) and (v), we get z = 2 and x = 4
From Eq. (vi), we get
3(4) + λ (2) = 14
⇒
λ =1
∴
µ − λ2 = 14 − 1 = 13
33. Since, the given system of equations posses non-trivial
0
1
solution, if 0 −3
k −5
−2
1
4
=0 ⇒ k =0
On solving the equations x = y = z = λ
[say]
∴ For k = 0, the system has infinite solutions of λ ∈R.
34. Given systems of equations can be rewritten as
− x + cy + by = 0, cx − y + az = 0 and bx + ay − z = 0
Above system of equations are homogeneous equation.
Since, x, y and z are not all zero, so it has non-trivial
solution.
Therefore, the coefficient of determinant must be zero.
−1 c
b
c −1 a = 0
∴
b
a
−1
⇒ − 1 (1 − a 2) − c (− c − ab) + b (ca + b) = 0
⇒
a 2 + b2 + c2 + 2abc − 1 = 0
⇒
a 2 + b2 + c2 + 2abc = 1
35. System of given linear equations is
x − 2 y + 3z = 9
2x + y + z = b
and x − 7 y + az = 24, has infinitely many solution, so
∆ = 0 = ∆1 = ∆ 2 = ∆3
So,
∆ =0
1 −2 3
2 1 1= 0
⇒


1 −7 a
⇒ 1(a + 7) + 2(2a − 1) + 3(−14 − 1) = 0
⇒
5a = 40 ⇒ a = 8
1 −2 9 
and
b
∆3 = 0 ⇒
= 0
2 1
1 −7 24
⇒ 1(24 + 7b) + 2(48 − b) + 9(−14 − 1) = 0
⇒ 5b + 120 − 135 = 0 ⇒ b = 3
∴ a − b =8 −3=5
Hence, answer is 5.00.
36. The given system of linear equations
x + y + z = 6,
x + 2 y + 3z = 10,
and 3x + 2 y + λz = µ has more than two solutions, so
system must have infinite number of solutions.
Now, on putting z = 0 in above equation, we get
…(i)
x+ y=6
1
37.
α
α2
α α2
1 α =0
α 1
⇒ α 4 − 2α 2 + 1 = 0
⇒ α2 = 1 ⇒α = ± 1
But α = 1 not possible
[Not satisfying equation]
∴
α = −1
Hence, 1 + α + α 2 = 1
a1 a 2 a3 
38. Let
M =  b1 b2 b3 


 c1 c2 c3 
0 − 1
 1  1
∴
M 1 =  2 ⇒ M − 1 =  1
   
   
0  3
 0 − 1
0
1
a 2 − 1 a1 − a 2  1
   




M ⋅ 1 = 0 ⇒  b2  =  2,  b1 − b2  =  1,
    
  
   
 c2   3  c1 − c2  − 1
1 12
a1 + a 2 + a3   0
 b + b + b  =  0
2
3
  
 1
 c1 + c2 + c3  12
⇒
a 2 = − 1, b2 = 2, c2 = 3, a1 − a 2 = 1,
b1 − b2 = 1, c1 − c2 = − 1
⇒ a1 + a 2 + a3 = 0, b1 + b2 + b3 = 0
c1 + c2 + c3 = 12
∴
a1 = 0, b2 = 2, c3 = 7
⇒ Sum of diagonal elements = 0 + 2 + 7 = 9
8
Functions
Topic 1 Classification of Functions, Domain and
Range and Even, Odd Functions
Objective Questions I (Only one correct option)
7. The domain of definition of f (x) =
1. If R = {(x, y): x, y ∈ Z , x2 + 3 y2 ≤ 8} is a relation on the set
of integers Z, then the domain of R−1 is
(2020 Main, 2 Sep I)
(a) {−1, 0,1}
(c) {−2, − 1, 0,1, 2}
(b) {− 2, − 1,1, 2}
(d) {0,1}
2. The domain of the definition of the function
f (x) =
1
+ log10 (x3 − x) is
4 − x2
(a) (−1, 0) ∪ (1, 2) ∪ (3, ∞ )
(c) (−1, 0) ∪ (1, 2) ∪ (2, ∞ )
(2019 Main, 9 April II)
(b) (−2, − 1) ∪ (−1, 0) ∪ (2, ∞ )
(d) (1, 2) ∪ (2, ∞ )
where f1 (x) is an even function and f2(x) is an odd
function. Then f1 (x + y) + f1 (x − y) equals
(2019 Main, 8 April II)
(b) 2f1 (x + y) ⋅ f1 (x − y)
(d) 2f1 (x) ⋅ f1 ( y)
4. Domain of definition of the function
f (x) = sin −1 (2x) +
1 1
(a)  − , 
 4 2 
1 1
(c)  − , 
 2 9
π
for real valued x, is
6
(2003, 2M)
1 1
(b)  − , 
 2 2 
1 1
(d)  − , 
 4 4 
minimum value of f (x). As b varies, the range of m (b) is
(2000, 1M)
(b) 0 ≤ x ≤ 1
(d) −∞ < x < 1
9. Let f (θ ) = sin θ (sin θ + sin 3 θ ). Then, f (θ )
(a) ≥ 0, only when θ ≥ 0
(c) ≥ 0, for all real θ
(2000, 1M)
(b) ≤ 0, for all real θ
(d) ≤ 0, only when θ ≤ 0
10. The domain of definition of the function
y=
1
+
log10 (1 − x)
x + 2 is
(1983, 1M)
(a) (− 3, − 2) excluding − 2. 5
(b) [0, 1] excluding 0.5
(c) (−2, 1) excluding 0
(d) None of these
Match the conditions / expressions in Column I with
statement in Column II.
11. Let f (x) =
x2 − 6 x + 5
.
x2 − 5 x + 6
(2007, 6M)
Column I
Column II
A.
If −1 < x < 1, then f ( x ) satisfies
p.
0 < f (x ) < 1
1
(b)  0, 
 2 
B.
If 1 < x < 2 , then f ( x ) satisfies
q.
f (x ) < 0
C.
If 3 < x < 5, then f ( x ) satisfies
r.
f (x ) > 0
(d) (0, 1]
D.
If x > 5, then f ( x ) satisfies
s.
f (x ) < 1
(2001, 1M)
1
(c)  , 1
 2 
(a) 0 < x ≤ 1
(c) − ∞ < x ≤ 0
(b) (1, 11/7)
(d) (1, 7/5)
6. Let f (x) = (1 + b2) x2 + 2bx + 1 and let m (b) be the
(a) [0, 1]
the equation 2x + 2y = 2 , is
Match the Columns
x2 + x + 2
5. Range of the function f (x) = 2
; x ∈ R is
x + x+1
(2003, 2M)
(a) (1, ∞)
(c) (1, 7/3]
(2001, 1M)
(a) R / {− 1, − 2}
(b) (− 2, ∞ )
(c) R / {− 1, − 2, − 3}
(d) (− 3, ∞ ) / {− 1, − 2}
8. The domain of definition of the function y (x) is given by
3. Let f (x) = a x (a > 0) be written as f (x) = f1 (x) + f2(x),
(a) 2f1 (x + y) ⋅ f2 (x − y)
(c) 2f1 (x) ⋅ f2 ( y)
log 2(x + 3)
is
x2 + 3x + 2
172 Functions
True/False
Objective Question II
(One or more than one correct option)
2x − 1
is
12. If S is the set of all real x such that 3
2 x + 3 x2 + x
positive, then S contains
(1986, 2M)
3
(a)  − ∞ , − 

2
3
1
(b)  − , − 
 2
4
1 1
(c)  − , 
 4 2
1
(d)  , 3
2 
16. If f1 (x) and f2(x) are defined on domains D1 and D2
respectively, then f1 (x) + f2(x) is defined on D1 ∩ D2.
(1988, 1M)
Analytical & Descriptive Questions
17. Find the range of values of t for which
2 sin t =
Fill in the Blanks
y=
18. Let
 4 − x2 
 , then the domain of
 1−x 


13. If f (x) = sin log 
f (x) is ……… .
1 − 2 x + 5 x2
 π π
.
, t∈ − ,
 2 2 
3 x2 − 2 x − 1
(2005, 2M)
(x + 1) (x − 3)
.
(x − 2)
Find all the real values of x for which y takes real
values.
(1980, 2M)
(1985, 2M)

x2 
14. The domain of the function f (x) = sin −1  log 2  is
2

given by ...
(1984, 2M)
 π2

15. The values of f (x) = 3 sin 
− x2 lie in the

 16

interval ………
(1983, 2M)
Integer & Numerical Answer Type Question
19. The value of the limit
4 2 (sin 3x + sin x)
3x
5x 
3x

+ cos  −  2 + 2 cos 2x + cos 
2 sin 2x sin

2
2 
2
(2020 Adv.)
is ………
lim
π
x→
2
Topic 2 Composite of Functions
Objective Questions I (Only one correct option)
10
4. Let
1. For a suitable chosen real constant a, let a functin
a −x
. Further
a+x
suppose that for any real number x ≠ − a and f (x) ≠ − a,
 1
( fof )(x) = x. Then, f  −  is equal to
 2
[2020 Main, 6 Sep II]
f : R − { a } → R be defined by f (x) =
1
(b) −
3
(d) 3
1
(a)
3
(c) −3
3
2. For x ∈ 0,  , let f (x) = x , g (x) = tan x and
 2
1 − x2
 π
. If φ(x) = ((hof )og )(x), then φ   is equal to
h (x) =
 3
1 + x2
(2019 Main, 12 April I)
π
(a) tan
12
7π
(c) tan
12
11π
(b) tan
12
5π
(d) tan
12
For
any
define
A ⊆ R,
f (x) = x , x ∈ R.
g ( A ) = { x ∈ R : f (x) ∈ A }. If S = [0, 4], then which one of
the following statements is not true?
3. Let
∑ f (a + k) = 16(210 − 1),
where the function f
k =1
satisfies f (x + y) = f (x) f ( y) for all natural numbers x, y
and f (1) = 2. Then, the natural number ‘a’ is
(2019 Main, 9 April I)
(a) 2
(b) 4
(c) 3
 1 − x
 ,|x| < 1, then
 1 + x
5. If f (x) = log e 
 2x 
f
 is equal to
 1 + x2 
(b) 2f (x2 )
(d) −2f (x)
(a) 2f (x)
(c) (f (x))2
x ∈ R − {0, 1},
6. For
(d) 16
let
f1 (x) =
(2019 Main, 8 April I)
1
, f2(x) = 1 − x
x
and
1
be three given functions. If a function, J (x)
1 −x
satisfies ( f2° J ° f1 )(x) = f3 (x), then J (x) is equal to
f3 (x) =
(2019 Main, 9 Jan I)
(a) f2 (x)
(b) f3 (x)
1
(d) f3 (x)
x
(c) f1 (x)
2
(2019 Main, 10 April I)
(a) f ( g (S)) = S
(c) g (f (S)) = g (S)
(b) g (f (S)) ≠ S
(d) f(g(S)) ≠ f (S)
7. Let a , b, c ∈ R. If f (x) = ax2+ bx + c be such that
a + b + c = 3 and f (x + y) = f (x) + f ( y) + xy, ∀ x, y ∈ R,
10
then
∑ f (n ) is
equal to
n =1
(a) 330
(b) 165
(2017 Main)
(c) 190
(d) 255
Functions 173
8. Let f (x) = x2 and g (x) = sin x for all x ∈ R. Then, the set of
all x satisfying
( fog )(x) = f ( g (x)), is
(a)
(b)
(c)
(d)
( fogogof )(x) = ( gogof )(x) ,
where
(2011)
± nπ, n ∈ {0, 1, 2, K }
± nπ, n ∈ {1, 2, K }
π /2 + 2nπ, n ∈ {...,− 2, − 1, 0, 1, 2, K }
2nπ, n ∈ {..., − 2, − 1, 0, 1, 2, K }
αx
, x ≠ − 1. Then, for what value of α is
x+1
(2001, 1M)
f [ f (x)] = x ?
(b) −
2
(d) −1
(c) 1
− 1 , x < 0

10. Let g (x) = 1 + x − [x] and f (x) =  0, x = 0 , then for all
 1, x > 0

(2001, 1M)
x, f [ g (x)] is equal to
(a) x
(b) 1
(c) f (x)
(d) g (x)
11. If g { f (x) } = |sin x| and f { g (x) } = (sin x )2, then
(1998, 2M)
(a) f
(b) f
(c) f
(d) f
(x) = sin 2 x, g (x) = x
(x) = sin x, g (x) = | x|
(x) = x2 , g (x) = sin x
and g cannot be determined

1   x


has the value
(1983, 1M)
1
2
(d) None of these
(b)
(c) − 2
13. Let f (x) = | x − 1|. Then,
(a) f (x ) = {f (x) }
(c) f (| x|) = | f (x)|
2
2
(1991, 2M)
( π / 2) = − 1
( π) = 1
(− π ) = 0
( π / 4) = 1
the equilateral triangle with two of its vertices at (0, 0)
and [x, g (x)] is 3 / 4, then the function g (x) is
(1989, 2M)
(a) g (x) = ±
1 − x2
(b) g (x) = 1 − x2
(c) g (x) = − 1 − x2
(d) g (x) = 1 + x2
17. If y = f (x) =
x+2
, then
x−1
(1984, 3M)
(a) x = f ( y)
(b) f (1) = 3
(c) y increases with x for x < 1
(d) f is a rational function of x
Fill in the Blank
π
π


f (x) = sin 2 x + sin 2 x +  + cos x cos  x +  and


3
3
 5
g   = 1, then (g o f ) (x) = ... .
 4
(1996, 2M)
18. If
12. If f (x) = cos(log x), then f (x) ⋅ f ( y) −  f   + f (xy)
2  y
(a) −1
greatest integer function, then
(a) f
(b) f
(c) f
(d) f
16. Let g (x) be a function defined on [− 1, 1]. If the area of
9. Let f (x) =
(a) 2
15. If f (x) = cos [π 2] x + cos [− π 2] x, where [x] stands for the
True/False
19. If f (x) = (a − xn )1/ n, where a > 0 and n is a positive
integer, then f [ f (x)] = x.
(1983, 1M)
(b) f (x + y) = f (x) + f ( y)
(d) None of the above
Objective Questions II
(One or more than one correct option)
π
π

f (x) = sin  sin  sin x  for all x ∈ R and


2
6
π
g (x) = sin x for all x ∈ R. Let ( fog )(x) denotes f { g (x)}
2
and (gof ) (x) denotes g{f (x)}. Then, which of the
following is/are true?
(2015 Adv.)
14. Let
1 1
1 1
(a) Range of f is  − ,  (b) Range of fog is  − , 
 2 2 
 2 2 
f (x ) π
(c) lim
=
x → 0 g (x )
6
(d) There is an x ∈ R such that ( gof ) (x) = 1
(1983, 1M)
Analytical & Descriptive Question
20. Find the natural number a for which
n
∑
f (a + k) = 16 (2n − 1),
k =1
where the function f satisfies the relation
f (x + y) = f (x) f ( y) for all natural numbers x , y and
further f (1) = 2.
(1992, 6M)
Integer & Numerical Answer Type Question
21. Let the function f : [0, 1] → R be defined by
f (x) =
4x
4x + 2
Then the value of
 39
3
2
1
 1
f   + f   + f   + ... + f   − f   is ………
 40
 40
 40
 40
 2
(2020 Adv.)
174 Functions
Topic 3 Types of Functions
Objective Questions I (Only one correct option)
1. If the function f : R → R is defined by f (x) =|x|(x − sin x),
then which of the following statements is TRUE?
(a) f is one-one, but NOT onto
(b) f is onto, but NOT one-one
(c) f is BOTH one-one and onto
(d) f is NEITHER one-one NOR onto
2. If the function
f (x) =
(2020 Adv.)
f : R − {1, − 1} → A
defined
by
 2 2
x
, is surjective, then A is equal to
1 − x2
(2019 Main, 9 April I)
(2017 Main)
9. The function f: [0, 3] → [1, 29], defined by
f (x) = 2x3 − 15 x2 + 36x + 1, is
0, if x is rational
x, if x is rational
, g (x) = 
0
if
is
irrational
,
x
x, if x is irrational

1
. Then, f is
x
(2019 Main, 11 Jan II)
injective only
both injective as well as surjective
not injective but it is surjective
neither injective nor surjective
2, 3, … , 20} such that f (k) is a multiple of 3, whenever k
(2019 Main, 11 Jan II)
is a multiple of 4, is
(b) 56 × 15
(d) 65 × (15)!
5. Let f : R → R be defined by f (x) =
x ∈ R. Then, the range of f is
1 1
(a)  − , 
 2 2 
1 1
(c) R −  − , 
 2 2 
x
,
1 + x2
(2019 Main, 11 Jan I)
(b) (−1, 1) − {0}
(d) R − [−1, 1]
6. Let N be the set of natural numbers and two functions f
and g be defined as f , g : N → N such that
n + 1
; if n is odd

f (n ) =  2
n

if n is even
;
 2
and g (n ) = n − (−1)n. Then, fog is
(2019 Main, 10 Jan II)
(a) one-one but not onto
(c) both one-one and onto
Then, f − g is
(2005, 1M)
(a) one-one and into
(c) many one and onto
(b) onto but not one-one
(d) neither one-one nor onto
(b) neither one-one nor onto
(d) one-one and onto
x
, then f is
1+ x
(2003, 2M)
11. If f : [0, ∞ ) → [0, ∞ ) and f (x) =
(a) one-one and onto
(c) onto but not one-one
4. The number of functions f from {1, 2, 3, … , 20} onto {1,
(a) (15)! × 6!
(c) 5! × 6!
(2012)
(a) one-one and onto
(b) onto but not one-one
(c) one-one but not onto (d) neither one-one nor onto
10. f (x) = 
3. Let a function f : (0, ∞ ) → (0, ∞ ) be defined by
f (x) = 1 −
x
is
1 + x2
(a) invertible
(b) injective but not surjective
(c) surjective but not injective
(d) neither injective nor surjective
2
(a) R − {−1}
(b) [0, ∞ )
(c) R − [−1, 0)
(d) R − (−1, 0)
(a)
(b)
(c)
(d)
1 1
8. The function f : R → − ,  defined as f (x) =
(b) one-one but not onto
(d) neither one-one nor onto
12. Let function f : R → R be defined by f (x) = 2 x + sin x
for x ∈ R . Then, f is
(2002, 1M)
(a) one-to-one and onto
(b) one-to-one but not onto
(c) onto but not one-to-one (d) neither one-to-one nor onto
13. Let E = {1, 2, 3, 4} and F = {1, 2}. Then, the number of
onto functions from E to F is
(a) 14
(b) 16
(2001, 1M)
(c) 12
(d) 8
Match the Columns
Match the conditions/expressions in Column I with
statement in Column II.
14. Let f1 : R → R, f2 : [0, ∞ ] → R, f3 : R → R and
f4 : R → [0, ∞ ) be defined by
(2014 Adv.)
sin x, if x < 0
|x|, if x < 0
; f2(x) = x2; f3 (x) = 
f1 (x) =  x
if
e
,
x
≥
0
 x, if x ≥ 0

 f [ f (x)], if x < 0
and
f4 (x) =  2 1
 f2[ f1 (x)] − 1, if x ≥ 0
Column I
Column II
A.
f4 is
p.
onto but not one-one
7. Let A = { x ∈ R : x is not a positive integer}. Define a
B.
f3 is
q.
neither continuous nor one-one
2x
, then f is
x−1
(2019 Main, 9 Jan II)
C.
f2of1 is
r.
differentiable but not one-one
D.
f2 is
s.
continuous and one-one
function f : A → R as f (x) =
(a) injective but not surjective
(b) not injective
(c) surjective but not injective
(d) neither injective nor surjective
Codes
A B C D
(a) r p s q
(c) r p q s
A B C D
(b) p r s q
(d) p r q s
Functions 175
15. Let the functions defined in Column I have domain
(−π /2, π /2) and range (− ∞ , ∞ )
Column I
A. 1 + 2 x
B. tan x
(1992, 2M)
18. The function f (x) =
Column II
p.
q.
r.
s.
True/False
onto but not one-one
one-one but not onto
one-one and onto
neither one-one nor onto
x2 + 4x + 30
is not one-to-one.
x2 − 8x + 18
(1983, 1M)
Analytical & Descriptive Question
19. A function f : IR → IR, where IR, is the set of real
Objective Question II
(One or more than one correct option)
numbers, is defined by f (x) =
π π
16. Let f :  − ,  → R be given by
αx2 + 6x − 8
.
α + 6 x − 8 x2
Find the interval of values of α for which is onto. Is the
functions one-to-one for α = 3 ? Justify your answer.
 2 2
f (x) = [log(sec x + tan x)]3 . Then,
(1996, 5M)
(a) f (x) is an odd function (b) f (x) is a one-one function
(c) f (x) is an onto function (d) f (x) is an even function
Fill in the Blanks
17. There are exactly two distinct linear functions, …,
and… which map {– 1, 1} onto {0, 2}.
20. Let A and B be two sets each with a finite number of
elements. Assume that there is an injective mapping
from A to B and that there is an injective mapping
from B to A. Prove that there is a bijective mapping
from A to B.
(1981, 2M)
(1989, 2M)
Topic 4 Inverse and Periodic Functions
Objective Questions I (Only one correct option)
whose graph is reflection of the graph of f (x) with
respect to the line y = x, then g (x) equals
1. The inverse function of
f (x) =
−2x
8 −8
, x ∈ (− 1, 1) is
82x + 8−2x
2x
(a)
1− x
1
(log 8 e) log e 

4
1 + x
(b)
1−
1
log e 
4
1+
(2002, 1M)
[2020 Main 8 Jan I]
(a) −
x − 1, x ≥ 0
(c) x + 1 , x ≥ − 1
(b)
1
(x + 1)2
,x> −1
(d) x − 1, x ≥ 0
1
x
x

x
1 + x
1
(c) (log 8 e) log e 

4
1− x
4. Suppose f (x) = (x + 1)2 for x ≥ − 1. If g (x) is the function
5. If f : [1, ∞ ) → [2, ∞ ) is given by f (x) = x + , then f −1(x)
equals
1 + x
1
(d) log e 

4
1− x
(a)
2. If X and Y are two non-empty sets where f : X → Y , is
function is defined such that
f (C ) = { f (x) : x ∈ C } for C ⊆ X
(2005, 1M)
3. If f (x) = sin x + cos x, g (x) = x2 − 1, then g { f (x) } is
(2004, 1M)
π
(a)  0, 
 2 
π π
(b)  − , 
 4 4 
π π
(c)  − , 
 2 2 
(d) [0 , π ]
x−
x2 − 4
2
(b)
x
1 + x2
(d) 1 +
x2 − 4
f : [1, ∞ ) → [1, ∞ ) is
f (x) = 2x ( x − 1), then f −1 (x) is
(a) f −1 {f (A )} = A
(b) f −1 {f (A )} = A, only if f (X ) = Y
(c) f {f −1 (B )} = B, only if B ⊆ f (x)
(d) f {f −1 (B )} = B
invertible in the domain
x −4
2
6. If the function
and f −1 (D ) = { x : f (x) ∈ D } for D ⊆ Y ,
for any A ⊆ Y and B ⊆ Y , then
(c)
x+
(2001, 1M)
2
x ( x − 1)
1
(a)  
 2
1
(c) (1 − 1 + 4 log2 x )
2
(b)
1
(1 +
2
defined
by
(1999, 2M)
1 + 4 log2 x )
(d) not defined
7. If f (x) = 3x − 5, then f −1 (x)
1
(a) is given by
3x − 5
x+ 5
(b) is given by
3
(c) does not exist because f is not one-one
(d) does not exist because f is not onto
(1998, 2M)
176 Functions
8. Which of the following functions is periodic?
(1983, 1M)
(a) f (x) = x − [ x ], where [x] denotes the greatest integer
less than or equal to the real number x
(b) f (x) = sin (1 /x) for x ≠ 0, f (0) = 0
(c) f (x) = x cos x
(d) None of the above
Objective Question II
(One or more than one correct option)
9. Let f : (0, 1) → R be defined by f (x) =
constant such that 0 < b < 1. Then,
b−x
, where b is a
1 − bx
(b) Statement I is true, Statement II is also true;
Statement II is not the correct explanation of
Statement I.
(c) Statement I is true; Statement II is false.
(d) Statement I is false; Statement II is true.
10. Let F (x) be an indefinite integral of sin 2 x.
Statement I The function F (x) satisfies
F (x + π ) = F (x) for all real x.
Because
Statement II sin 2(x + π ) = sin 2 x, for all real x.
(2011)
(2007, 3M)
(a) f is not invertible on (0, 1)
1
f ′ (0)
1
on (0, 1) and f ′ (b) =
f ′ (0)
Analytical & Descriptive Question
(b) f ≠ f −1 on (0, 1) and f ′ (b) =
(c) f = f −1
(d) f
−1
11. Let f be a one-one function with domain { x, y, z } and
range {1, 2, 3}. It is given that exactly one of the
following statements is true and the remaining two
are false f (x) = 1, f ( y) ≠ 1, f (z ) ≠ 2 determine f −1 (1).
is differentiable on (0, 1)
(1982, 2M)
Assertion and Reason
For the following questions, choose the correct answer
from the codes (a), (b), (c) and (d) defined as follows.
(a) Statement I is true, Statement II is also true;
Statement II is the correct explanation of Statement I.
12. If f is an even function defined on the interval (− 5, 5),
then four real values of x satisfying the equation
 x + 1
f (x) = f 
 are ………. .
 x + 2
(1996, 1M)
Answers
Topic 1
Topic 3
1. (a)
2. (c)
3. (d)
4. (a)
1. (c)
2. (c)
3. (d)
4. (a)
5. (c)
6. (d)
7. (d)
8. (d)
5. (a)
6. (b)
7. (a)
8. (c)
12. (a)
9. (c)
12. (a,d)
15.
 3 
0,
2 

10. (c)
11. A→ p; B→ q; C→ q; D→ p
9. (b)
10. (d)
11. (b)
13. (–2,1)
14. Domain ∈ [ −2,−1 ] ∪ [1, 2 ]
13. (a)
14. (d)
15. A → q; B → r
16. True
 π π  3π π 
17. t ∈ − ,
∪
,
 2 10   10 2 
16. (a, b, c)
19. (8)
18. True
18. x ∈ [ −1, 2 ) ∪ [3, ∞ )
19. 2 ≤ α ≤ 14, No
Topic 2
1. (d)
2. (b)
3. (c)
4. (c)
5. (a)
6. (b)
7. (a)
8. (b)
9. (d)
10. (b)
11. (a)
12. (d)
13. (d)
14. (a,b,c)
15. (a, c)
16. (b, c)
17. (a, d)
18. 1
19. True
20. (a = 3)
21. (19)
17. y = x + 1 and y = − x + 1
Topic 4
1. (c)
2. (c)
3. (b)
4. (d)
5. (a)
6. (b)
7. (b)
8. (a)
9. (b)
11. f
−1
(1 ) = y
10. (d)
 ± 3 ± 5
12. 

2


Hints & Solutions
–1
1
π
 – π
≤ 2x ≤
sin 
⇒
 ≤ 2x ≤ sin
 6 
2
2
2
–1
1
≤x≤
4
2
– 1 1
x∈
,
Q
 4 2 
Topic 1 Classification of Functions,
Domain and Range
1. Given relation
R = {(x, y) : x, y ∈ Z , x2 + 3 y2 ≤ 8}
For, y2 = 0, x2 = 0, 1, 4
For, y2 = 1, x2 = 0, 1, 4
For, y2 = 4, x2 ∈ φ
∴ Range of R is possible values of y = { − 1, 0, 1}
5. Let y = f (x) =
∴ Domain of R−1 = Range of R = { − 1, 0, 1}
1
2. Given function f (x) =
+ log10 (x3 − x)
4 − x2
For domain of f (x)
4 − x2 ≠ 0 ⇒ x ≠ ± 2
and
x3 − x > 0
⇒
x(x − 1)(x + 1) > 0
From Wavy curve method,
+
–∞
–
–1
+
0
–
+1
x ∈ (−1, 0) ∪ (1, ∞ )
x2 + x + 2
x2 + x + 1
1
y=1 + 2
x + x+1
∴
…(i)
+∞
…(ii)
From Eqs. (i) and (ii), we get the domain of f (x) as
(−1, 0) ∪ (1, 2) ∪ (2, ∞ ).
y=
So, f1 (x + y) + f1 (x − y)
1
1
= [a x + y + a − ( x + y ) ] + [a x − y + a − ( x − y ) ]
2
2
x

1
1
a
ay 
= a x a y + x y + y + x 
2
a a
a
a 
2
1
is positive
1 + b2
and m (b) varies from 1 to 0, so range = (0, 1]
log (x + 3)
log 2 (x + 3)
7. Given, f (x) = 2 2
=
(x + 3x + 2) (x + 1) (x + 2)
m (b) = minimum value of f (x) =
For numerator, x + 3 > 0
⇒
x> −3
⇒
From Eqs. (i) and (ii),
Domain is (− 3 , ∞ ) /{ − 1, − 2}
π
, to find domain we must
6
have,
π
sin (2x) + ≥ 0
6
π
π
− ≤ sin −1 (2x) ≤
6
2
−1
π
π

but − ≤ sin −1 θ ≤

2
2 
…(ii)

b 
b2
= (1 + b2)  x +
 +1−
2

1+ b 
1 + b2
1
1 
1
= ax + x  ay + y 
2
a 
a 
4. Here, f (x) = sin −1 (2x) +
…(i)
6. Given, f (x) = (1 + b2) x2 + 2bx + 1
1 
1
1  1

= a x  a y + y  + x  y + a y  

2 
a  a a
 a x + a −x   a y + a −y 
=2
 = 2 f1 (x) ⋅ f1 ( y)

2
2



[i.e. y > 1]
⇒
yx2 + yx + y = x2 + x + 2
2
⇒ x ( y − 1) + x ( y − 1) + ( y − 2) = 0, ∀ x ∈ R
Since, x is real, D ≥ 0
⇒
( y − 1 )2 − 4 ( y − 1 ) ( y − 2 ) ≥ 0
⇒
( y − 1) {( y − 1) − 4 ( y − 2)} ≥ 0
⇒
( y − 1) (− 3 y + 7) ≥ 0
7
1≤ y≤
⇒
3
 7
From Eqs. (i) and (ii), Range ∈ 1 ,
 3 
3. Given, function f (x) = a x , a > 0 is written as sum of an
even and odd functions f1 (x) and f2(x) respectively.
a x + a −x
a x − a −x
and f2(x) =
Clearly, f1 (x) =
2
2
x2 + x + 2
, x ∈R
x2 + x + 1
…(i)
and for denominator, (x + 1) (x + 2) ≠ 0
x ≠ − 1, − 2
8. Given, 2x + 2y = 2, ∀ x , y ∈ R
2x , 2y > 0, ∀ x , y ∈ R
But
Therefore, 2x = 2 − 2y < 2 ⇒ 0 < 2x < 2
Taking log on both sides with base 2, we get
log 2 0 < log 2 2x < log 2 2 ⇒ − ∞ < x < 1
9. It is given,
f (θ ) = sin θ (sin θ + sin 3 θ )
= (sin θ + 3 sin θ − 4 sin3 θ ) sin θ
…(ii)
178 Functions
= (4 sin θ − 4 sin3 θ ) sin θ = sin 2 θ (4 − 4 sin 2 θ )
= 4 sin 2 θ cos 2 θ = (2 sin θ cos θ )2
= (sin 2θ )2 ≥ 0
which is true for all θ.
⇒
1 x2
≤
≤2
2 2
⇒
1 ≤ x2 ≤ 4
⇒
1 ≤ |x| ≤ 2
⇒
10. For domain of y,
and
1 − x > 0, 1 − x ≠ 1
and
⇒
x < 1, x ≠ 0
⇒
−2 < x < 1 excluding 0
⇒
x ∈ (−2, 1) − {0}
(x − 1) (x − 5)
11. Given, f (x) =
(x − 2) (x − 3)
x+ 2 >0
x > −2
Domain ∈ [−2, − 1] ∪ [1, 2]
15. Given, f (x) = 3 sin
 π π
⇒ Domain ∈ − ,
 4 4 
 π2

− x2  ⋅
∴ For range, f ′ (x) = 3 cos 
 16



The graph of f (x) is shown as :
Y
Where,
y=1
X'
0
1
2
X
5
3
Y'
⇒
C. If 3 < x < 5 ⇒ f (x) < 0
(2x − 1)
>0
x(2x2 + 3x + 1)
⇒
(2x − 1)
>0
x (2x + 1) (x + 1)
−∞
−1
−1/2
0
+
1/2
∞
x ∈ (−∞ , − 1) ∪ (−1/2, 0) ∪ (1 / 2, ∞ )
Hence, (a) and (d) are the correct options.
 4 − x2 

13. Given, f (x) = sin log 
 1−x 


and
x<1
and
Thus, domain ∈ (−2, 1).

14. Given, f (x) = sin −1  log 2

For domain,
x2 

2
−1 ≤ log 2
17. Given, 2 sin t =
x2
≤ 1
2
∴
1 − 2 x + 5 x2
 π π
,t ∈ − ,
 2 2 
3 x2 − 2 x − 1
2 sin t = y ⇒ − 2 ≤ y ≤ 2
y=
1 − 2 x + 5 x2
3 x2 − 2 x − 1
⇒ (3 y − 5)x2 − 2x( y − 1) − ( y + 1) = 0
[as, 3x2 − 2x − 1 ≠ 0 ⇒ (x − 1)(x + 1 / 3) ≠ 0]
4−x
> 0 , 4 − x2 > 0 and 1 − x ≠ 0
1−x
(1 − x) > 0
 3 
range ∈ 0,

2 
Since, x ∈ R − {1, − 1 / 3}
2
⇒
π
3
=
4
2
Thus, domain of [ f1 (x) + f2(x)] is D1 ∩ D2 .
Hence, given statement is true.
Put
⇒
x=0
16. Since, domains of f1 (x) and f2(x) are D1 and D2 .
−
+
=0
 π
 π
f −  = f   =0
 4
 4
Hence,
Hence, the solution set is,
For domain,
f (0) = 3 sin
and
D. If x > 5 ⇒ 0 < f (x) < 1
2x − 1
12. Since,
>0
3
2 x + 3 x2 + x
⇒
or
π2
2
− x2
16
x =0
Thus,
B. If 1 < x < 2 ⇒ f (x) < 0
−
 π2

cos 
− x2  = 0
 16



1(−2x)

 π2

π2
π2
− x2  = 0 ⇒
− x2 =
neglecting cos 


16
4

 16



3π 2
2
, never possible
⇒ x = −



16
A. If − 1 < x < 1 ⇒ 0 < f (x) < 1
+
π2
− x2
16
4 − x >0
2
| x| < 2 ⇒ −2 < x < 1
∴
D ≥0
⇒
⇒
4( y − 1) + 4(3 y − 5) ( y + 1) ≥ 0
y2 − y − 1 ≥ 0
⇒
1
5

y−  − ≥0

2
4
⇒

1
5 
1
5
y− −
 y− +
 ≥0
2
2 
2
2

2
2
⇒
y≤
1− 5
2
Functions 179
1+ 5
2
1− 5
2 sin t ≤
2
1+ 5
2 sin t ≥
2
 π
sin t ≤ sin  − 
 10
Topic 2 Composite of Functions and
Even, Odd Functions
y≥
or
⇒
or
⇒
1. For a given function f : R − { −a } → R defined by
f (x) =
⇒
 3π 
sin t ≥ sin  
 10 
or
π
3π
t≤−
t≥
or
⇒
10
10
π  3π π 
 π
.
Hence, range of t is − , −
∪
,
 2
10   10 2 
18. Since, y =
(x + 1) (x − 3)
≥0
(x − 2)
when
−
−∞
⇒
−1 ≤ x < 2
∴
−
2
or
+
3
x≥3
19. The limit
π
x→
2
4 2 (sin 3x + sin x)
3x
3x
5x 

+ cos  −  2 + 2 cos 2x + cos 
2 sin 2x sin

2
2
2 
= lim
π
x→
2
4 2 (2 sin 2x cos x)
3x 
5x
3x
2 sin 2x sin
+  cos
− cos  − 2 (1 + cos 2x)

2
2
2
8 2 sin 2x cos x
x
3x
− 2 sin 2x sin − 2 (2 cos 2 x)
2 sin 2x sin
2
2
4 2 sin 2x cos x
= lim
π
3x
x

x → sin 2 x sin
− sin  − 2 cos 2 x

2

2
2
= lim
π
x→
2
4 2 sin 2x cos x
x
2 sin 2x cos x sin − 2 cos 2 x
2
4 2 sin 2x
= lim
π
x
x → 2 sin 2 x sin
− 2 cos x
2
2
8 2 sin x
= lim
π
x
x → 4 sin x sin
− 2
2
2
8 2
16
=
=
= 8.
4
−2
4
− 2
2
= lim
π
x→
2
a−x
a+x
=x
a−x
a+
a+x
a−
a 2 + ax − a + x
=x
a 2 + ax + a − x
⇒
a 2 + ax − a + x = a 2x + ax2 + ax − x2
⇒
So,
∞
x ∈ [−1, 2) ∪ [3, ∞ ).
lim
 a − x
f
 =x ⇒
 a + x
a (a − 1) = (a 2 − 1)x + x2(a − 1)
(a − 1) [x2 + (a + 1)x − a ] = 0 ⇒ a = 1
1−x
 1 1 + (1 /2)
f (x) =
=3
∴ f −  =
 2 1 − (1 /2)
1+ x
2. Given, for x ∈ (0, 3 / 2), functions
+
−1
( fof ) (x) = x
⇒
⇒
(x + 1) (x − 3)
takes all real values only
(x − 2)
a−x
. Q
a+x
and
f (x) = x
g (x) = tan x
1 − x2
h (x) =
1 + x2
… (i)
… (ii)
… (iii)
Also given, φ(x) = ((hof )og )(x) = (hof ) ( g (x))
= h ( f ( g (x))) = h ( f (tan x))
1 − ( tan x )2
= h ( tan x ) =
1 + ( tan x )2
1 − tan x
π

=
= tan  − x
4

1 + tan x
Now,
 π
 π π
φ   = tan  − 
 3
 4 3
 3π − 4π 
 π
= tan 
 = tan  − 
 12 
 12
π
π

 11π 
= − tan   = tan  π −  = tan 

 12

 12 
12
3. Given, functions f (x) = x2, x ∈ R
and g ( A ) = { x ∈ R : f (x) ∈ A }; A ⊆ R
Now, for S = [0, 4]
g (S ) = { x ∈ R : f (x) ∈ S = [0, 4]}
= { x ∈ R : x2 ∈ [0, 4]}
= { x ∈ R: x ∈ [−2, 2]}
⇒
g (S ) = [−2, 2]
So, f ( g (S )) = [0, 4] = S
Now, f (S ) = { x2 : x ∈ S = [0, 4]} = [0, 16]
and g ( f (S )) = { x ∈ R : f (x) ∈ f (S ) = [0, 16]}
= { x ∈ R : f (x) ∈ [0, 16]}
= { x ∈ R: x2 ∈ [0, 16]}
= { x ∈ R : x ∈ [−4, 4]} = [−4 ,4]
From above, it is clear that g ( f (S )) = g (S ).
180 Functions
4. Given,
−1
J (X ) = X
1
1−
X
−1
1
=
=
X −1 1 − X
f (x + y) = f (x) ⋅ f ( y)
f (x) = λ
f (1) = 2
[where λ > 0]
(given)
x
Let
Q
∴ λ =2
10
 10 
Σ f (a + k) = Σ λa+ k = λa  Σ λk
 k=1 
k =1
k =1
10
So,
= 2a [21 + 22 + 23 + ......+210 ]
10
2(2 − 1) 
= 2a 

 2 −1 
[by using formula of sum of n-terms of a GP having
first term ‘a’ and common ratio ‘r’, is

a (r n − 1)
Sn =
, where r > 1
r −1

⇒
⇒
7. We have, f (x) = ax2 + bx + c
Now, f (x + y) = f (x) + f ( y) + xy
Put
y = 0 ⇒ f (x) = f (x) + f (0) + 0
⇒
f (0) = 0
⇒
c=0
Again, put y = − x
∴
f (0) = f (x) + f (− x) − x2
⇒
0 = ax2 + bx + ax2 − bx − x2
⇒
2ax2 − x2 = 0
1
a=
⇒
2
Also, a + b + c = 3
5
1
⇒
+ b + 0 =3⇒ b =
2
2
x2 + 5 x
∴
f (x) =
2
 1 − x
 ,|x| < 1, then
 1 + x
5. Given, f (x) = log e 
  2x  
 < 1
Q
2
 1 + x  
 1 + x2 − 2 x 


2
 (1 − x)2 
 1 − x
1 + x2 
= log e 
=
=
log
log




e
e
 1 + x2 + 2x 
 1 + x
 (1 + x)2


2
 1+ x

∑
f (n ) =
n 2 + 5n 1 2 5
= n + n
2
2
2
f (n ) =
1
2
[Q f2(x) = 1 − x and f3 (x) =
1
 1
1 − J  =
 x 1 − x
1
 1
J  = 1 −
 x
1−x
=
1
= X , then
x
1 − x−1
−x
=
1−x
1−x
1
]
1−x
1
[Q f1 (x) = ]
x
∑ n2 +
n =1
5
2
10
∑n
n =1
1 10 × 11 × 21 5 10 × 11
⋅
+ ×
2
6
2
2
385 275 660
=
+
=
= 330
2
2
2
f (x) = x2, g (x) = sin x
8.
1
1
f1 (x) = , f2(x) = 1 − x and f3 (x) =
x
1−x
Also, we have ( f2 o J o f1 )(x) = f3 (x)
⇒
f2((J o f1 )(x)) = f3 (x)
⇒
f2(J ( f1 (x)) = f3 (x)
1
⇒
1 − J ( f1 (x)) =
1−x
10
=

 1 − x 
Q f (x) = log e  1 + x 


6. We have,
Now, put
10
∴
[Q log e| A|m = m log e| A|]
= 2 f (x)
⇒
Now,
n =1
 1 − x
= 2 log e 

 1 + x
⇒
J (X ) = f3 (X ) or J (x) = f3 (x)
⇒
2a+ 1 (210 − 1) = 16 (210 − 1) (given)
2a+ 1 = 16 = 24 ⇒ a + 1 = 4 ⇒ a = 3
2x 

1 −

 2x 
1
+ x2 
f
 = log e 
2
 1 + 2x 
1 + x 



1 + x2 
1

Qx=

X 
( gof )(x) = sin x2
go( gof ) (x) = sin (sin x2)
( fogogof ) (x) = (sin (sin x2))2
…(i)
Again, ( gof ) (x) = sin x2
( gogof ) (x) = sin (sin x2)
Given,
⇒
…(ii)
( fogogof ) (x) = ( gogof ) (x)
(sin (sin x2))2 = sin (sin x2)
⇒ sin (sin x2) {sin (sin x2) − 1} = 0
⇒ sin (sin x2) = 0 or sin (sin x2) = 1
π
⇒
sin x2 = 0 or sin x2 =
2
∴
x2 = n π
[sin x2 =
x=±
π
is not possible as − 1 ≤ sin θ ≤ 1]
2
nπ
Functions 181
9. Given, f (x) =
αx
x+1
 αx 
α

 x + 1
 αx 
f [ f (x) ] = f 
 =
αx
 x + 1
+1
x+1
α 2x
α 2x
x+1
=
=
= x [given] …(i)
α x + (x + 1) (α + 1) x + 1
x+1
2
⇒
α x = (α + 1) x2 + x
⇒
x [α 2 − (α + 1) x − 1] = 0
⇒
x(α + 1)(α − 1 − x) = 0
⇒
α − 1 = 0 and α + 1 = 0
⇒
α = −1
But α = 1 does not satisfy the Eq. (i).
10. g (x) = 1 + x − [x] is greater than 1
since x − [x] > 0
f [ g (x)] = 1, since f (x) = 1 for all x > 0
f (x) = sin 2 x and g (x) = x
11. Let
Now,
fog (x) = f [ g (x)] = f ( x ) = sin 2 x
and
gof (x) = g [ f (x)] = g (sin 2 x) = sin 2 x = |sin x|
Again, let f (x) = sin x , g (x) = | x|
fog (x) = f [ g (x)] = f (| x|)
= sin| x|≠ (sin x )2
f (x) = x2, g (x) = sin x
When
fog (x) = f [ g (x)] = f (sin x ) = (sin x )2
( gof ) (x) = g [ f (x)] = g (x2) = sin x2
and
= sin| x|≠|sin x|
12. Given, f (x) = cos (log x)
∴ f (x) ⋅ f ( y) −

1   x
f   + f (xy)
2   y

1
[cos (log x − log y)
2
+ cos(log x + log y)]
1
= cos (log x) ⋅ cos (log y) − [(2 cos (log x) ⋅ cos (log y)]
2
= cos (log x) ⋅ cos (log y) − cos (log x) ⋅ cos (log y) = 0
= cos (log x) ⋅ cos(log y) −
13.
Given,
∴
and
⇒
f (x) = |x − 1|
f (x ) = |x − 1|
2
2
{ f (x)}2 = (x − 1)2
f (x2) ≠ ( f (x))2, hence (a) is false.
Also, f (x + y) = |x + y − 1|
and
⇒
f (x) = |x − 1|,
f ( y) = | y − 1|
f (x + y) ≠ f (x) + f ( y), hence (b) is false.
f (|x|) = ||x| − 1|
and
| f (x)| = ||x − 1|| = |x − 1|
∴
f (|x|) ≠| f (x)|, hence (c) is false.
π
6
π
2


14. f (x) = sin  sin  sin x  , x ∈ R
π
π

 π π
= sin  sin θ , θ ∈ − , , where θ = sin x
 2 2 
6

2
π
 π π
,where α = sin θ
= sin α, α ∈ − ,
 6 6 
6
 1 1
f (x) ∈ − ,
∴
 2 2 
 1 1
Hence, range of f (x) ∈ − ,
 2 2 
So, option (a) is correct.
 π π
 1 1
(b) f { g (x)} = f (t ), t ∈ − ,
⇒ f (t ) ∈ − ,
 2 2 
 2 2 
∴
Option (b) is correct.
π

π
sin  sin  sin x 

2
f (x)
6

(c) lim
= lim
π
x → 0 g (x)
x→ 0
(sin x)
2
π
 π
π

π
sin  sin  sin x 
sin  sin x
 6


2
6
2

= lim
⋅
x→ 0
π

π
π

sin  sin x
 sin x

2
2

6
π
π
=1 × ×1 =
6
6
∴Option (c) is correct.
(d) g{ f (x)} = 1
π
sin { f (x)} = 1
⇒
2
2
...(i)
⇒
sin { f (x)} =
π
 1 1  π π 
But
f (x) ∈ − ,
⊂ − ,
 2 2   6 6 
 1 1
sin { f (x)} ∈ − ,
 2 2 
2
⇒ sin { f (x)} ≠ ,
π
i.e. No solution.
∴ Option (d) is not correct.
∴
...(ii)
[from Eqs. (i) and (ii)]
15. Since, f (x) = cos [π 2] x + cos [−π 2] x
⇒
∴
f (x) = cos (9) x + cos (−10) x
[using [π 2] = 9 and [− π 2] = − 10]
9π
 π
f   = cos
+ cos 5π = − 1
 2
2
f (π ) = cos 9π + cos 10π = − 1 + 1 = 0
f (− π ) = cos 9π + cos 10π = − 1 + 1 = 0
9π
10π
1
1
 π
f   = cos
+ cos
=
+0=
 4
4
4
2
2
Hence, (a) and (c) are correct options.
182 Functions
16. Since, area of equilateral triangle =
⇒
1
π


= 2 sin (2x + π / 3) ⋅
− sin 2x +  = 0


2 
3
3
(BC )2
4
3
3
=
⋅ [x2 + g 2(x)] ⇒ g 2(x) = 1 − x2
4
4
⇒ f (x) = c, where c is a constant.
But f (0) = sin 2 0 + sin 2(π / 3) + cos 0 cos π / 3
A
2
 3
1 3 1 5
=  + = + =
2 4 2 4
 2
Therefore, ( gof ) (x) = g [ f (x)] = g(5 / 4) = 1
C
(x,g(x))
B
(0,0)
⇒
g (x) = 1 − x2 or − 1 − x2
Hence, (b) and (c) are the correct options.
x+2
17. Given , y = f (x) =
x−1
⇒
⇒
yx − y = x + 2 ⇒ x( y − 1) = y + 2
y+2
x=
⇒ x = f ( y)
y−1
f (x) = (a − xn )1/ n
19. Given,
⇒
f [ f (x)] = [a − {(a − xn )1/ n }n ]1/ n = (xn )1/ n = x
∴
f [ f (x)] = x
Hence, given statement is true.
20. Let f (n ) = 2n for all positive integers n.
Now, for n = 1, f (1) = 2 = 2 !
⇒ It is true for n = 1.
Again, let f (k) is true.
⇒
f (k) = 2k, for some k ∈ N .
Here, f (1) does not exist, so domain ∈ R − {1}
dy (x − 1) ⋅ 1 − (x + 2) ⋅ 1
3
=
=−
dx
(x − 1)2
(x − 1)2
Again, f (k + 1) = f (k) ⋅ f (1)
⇒ f (x) is decreasing for all x ∈ R − {1}.
Also, f is rational function of x.
Hence, (a) and (d) are correct options.
Therefore, the result is true for n = k + 1. Hence, by
principle of mathematical induction,
18. f (x) = sin 2 x + sin 2(x + π / 3) + cos x cos (x + π / 3)
⇒
⇒
sin 2 x 3 cos 2 x 2 ⋅ 3
+
+
sin x cos x
4
4
4
3
cos 2 x
− cos x sin x ⋅
2
2
sin 2 x 3 cos 2 x cos 2 x
= sin 2 x +
+
+
4
4
2
5
5
5
2
2
= sin x + cos x =
4
4
4
+
and
gof (x) = g { f (x)} = g (5 / 4) = 1
Alternate Solution
f (x) = sin x + sin (x + π / 3) + cos x cos (x + π / 3)
2
2
⇒ f ′ (x) = 2 sin x cos x + 2 sin (x + π / 3) cos (x + π / 3)
− sin x cos (x + π / 3) − cos x sin (x + π / 3)
= sin 2x + sin (2x + 2π / 3) − [sin (x + x + π / 3)]
 2x − 2x − 2π / 3
 2x + 2x + 2π / 3
= 2 sin 

 ⋅ cos 




2
2
− sin (2x + π / 3)
= 2 [sin (2x + π / 3) ⋅ cos π / 3] − sin (2x + π / 3)
[from induction assumption]
f (n ) = 2n , ∀ n ∈ N
Now,
+ cos x cos (x + π / 3)
2
.

sin x ⋅ 1 cos x 3 
f (x) = sin 2 x + 
+

2
 2

f (x) = sin 2 x +
[by definition]
= 2k + 1
f (x) = sin 2 x + (sin x cos π / 3 + cos x sin π / 3)2
+ cos x (cos x cos π / 3 − sin x sin π / 3)
⇒
=2 ⋅2
k
n
n
k =1
k =1
n
∑ f (a + k) = ∑ f (a ) f (k) = f (a ) ∑ 2k
= f (a ) ⋅
k =1
2 (2 − 1)
2 −1
n
= 2a ⋅ 2 (2n − 1) = 2a + 1 (2n − 1)
n
But
∑ f (a + k) = 16 (2n − 1) = 24 (2n − 1)
k =1
Therefore,
a + 1 =4 ⇒ a =3
21. The given function f : [0, 1] → R be define by
f (x) =
∴
So,
∴
2
41 − x
4x
−
=
⇒
(
1
)
f
x
=
1−x
x
+ 2 2 + 4x
4
4 +2
f (x) + f (1 − x) =
4x
2
4x + 2
+
=
4x + 2 2 + 4x 4x + 2
f (x) + f (1 − x) = 1
1
2
 
 
3
 39
 1
f  + f  + f  +… + f  − f 
 40
 40
 40
 40
 2
 1
 39 
= f   + f    +


 40 
40

  18
+ … + f  
  40
… (i)
 2
 38 
 f  40 + f  40 


 22    19
 21 
+ f    + f   + f   
 40    40
 40 
20
 1
 
+ f  − f 
 2
 40
 1
 1
= {1 + 1 + K + 1 + 1} + f   − f   {from Eq. (i)}
 2
 2
19 − times = 19.
Functions 183
Topic 3 Types of Functions
1. The given function f : R → R is
f (x) = |x|(x − sin x)
… (i)
Q The function ‘f’ is a odd and continuous function
and as lim f (x) = ∞ and lim f (x) = − ∞, so range is R,
x→ ∞
x→ −∞
∴ f (x) is neither injective nor surjective
4. According to given information, we have if
k ∈{4, 8, 12, 16, 20}
Then, f (k) ∈ {3, 6, 9, 12, 15, 18}
[Q Codomain ( f ) = {1, 2, 3, …, 20}]
therefore
‘f’ is a onto function.
 x(x − sin x), x ≥ 0
Q
f (x) = 
− x (x − sin x), x < 0
∴
Clearly, f (x) is not injective because if f (x) < 1, then f is
many one, as shown in figure.
Also, f (x) is not surjective because range of f (x) is [0, ∞ [
and but in problem co-domain is (0, ∞ ), which is wrong.
 2x − sin x − x cos x, x > 0
f ′ (x) = 
− 2x + sin x + x cos x, x < 0
 (x − sin x) + x(1 − cos x), x > 0
(− x + sin x) − x(1 − cos x), x < 0

Q for x > 0, x − sin x > 0 and x (1 − cos x) > 0
∴ f ′ (x) > 0 ∀ x ∈ (0, ∞ )
⇒ f is strictly increasing function, ∀ x ∈ (0, ∞ ).
Similarly, for x < 0, − x + sin x > 0
and (− x) (1 − cos x) > 0, therefore, f ′ (x) > 0 ∀ x ∈ (− ∞ , 0)
⇒ f is strictly increasing function, ∀ x ∈ (0, ∞ )
Now, we need to assign the value of f (k) for
k ∈{4, 8, 12, 16, 20} this can be done in 6C5 ⋅ 5 ! ways
= 6 ⋅ 5 ! = 6 ! and remaining 15 element can be associated
by 15 ! ways.
∴Total number of onto functions = = 15 ! 6 !
x
5. We have, f (x) =
, x ∈R
1 + x2
Ist Method f (x) is an odd function and maximum
occur at x = 1
Y
x2
=y
1 − x2
x2 = y(1 − x2)
f (x) =
⇒
⇒
Q
(let)
[Q x2 ≠ 1]
2
Since, for surjective function, range of f = codomain
∴ Set A should be R − [−1, 0).
 (x − 1)
|x − 1| − x , if 0 < x ≤ 1
3. We have, f (x) =
=
x−1
x

if x > 1
,

x
1
− 1, if 0 < x ≤ 1

= x
1
1 − ,
if x > 1

x
Now, let us draw the graph of y = f (x)
Note that when x → 0, then f (x) → ∞, when x = 1, then
f (x) = 0, and when x → ∞, then f (x) → 1
Y
y=1
y=–
1
y=0
X
1
2
 1 1
From the graph it is clear that range of f (x) is − ,
 2 2 
IInd Method f (x) =
1
x+
1
x
If x > 0, then by AM ≥ GM, we get x +
1
⇒
1
x+
x
≤
1
⇒
1
x+
x
1
≥2
x
1
1
⇒ 0 < f (x) ≤
2
2
If x < 0, then by AM ≥ GM, we get x +
≥−
1
2
⇒–
1
≤ −2
x
1
≤ f (x) < 0
2
0
1
1
= 0. Thus, − ≤ f (x) ≤
1+0
2
2
 1 1
Hence, f (x) ∈ − ,
 2 2 
If x = 0, then f (x) =
IIIrd Method
x
Let y =
⇒ yx2 − x + y = 0
1 + x2
+
–
O
X
O1
Q x ∈ R, so D ≥ 0 ⇒ 1 − 4 y2 ≥ 0
 1 1
⇒ (1 − 2 y) (1 + 2 y) ≥ 0 ⇒ y ∈ − ,
 2 2 
x=0
1
2
(–1, 1/2)
y
[provided y ≠ −1]
x (1 + y) = y ⇒ x =
1+ y
y
≥ 0 ⇒ y ∈ (−∞ , − 1) ∪ [0, ∞ )
x2 ≥ 0 ⇒
1+ y
2
y=
–1
Therefore ‘f’ is a strictly increasing function for x ∈ R
and it implies that f is one-one function.
2. Given, function f : R – {1, − 1} → A defined as
(1, 1/2)
–1/2
–
1/2
184 Functions
 1 1
So, range is − , .
 2 2 
∴ (− 1) − 4 ( y)( y) ≥ 0 ⇒ 1 − 4 y2 ≥ 0 ⇒ y ∈
2
n + 1
, if n is odd

6. Given, f (n ) =  n 2
 , if n is even,
2
 n + 1 , if n is odd
and
g (n ) = n − (−1)n = 
n − 1, if n is even
1
if
f
(
n
+
),
n is odd

Now, f ( g (n )) = 
 f (n − 1), if n is even
n + 1
, if n is odd

=  n 2− 1 + 1 n
= , if n is even


2
2
= f (x)
[Q if n is odd, then (n + 1) is even and
if n is even, then (n − 1) is odd]
Clearly, function is not one-one as f (2) = f (1) = 1
But it is onto function.
[Q If m ∈ N (codomain) is odd, then 2m ∈ N (domain)
such that f (2m) = m and
if m ∈ N codomain is even, then
2m − 1 ∈ N (domain) such that f (2m − 1) = m]
∴Function is onto but not one-one
2x
7. We have a function f : A → R defined as, f (x) =
x −1
One-one Let x1, x2 ∈ A such that
2x1
2x2
=
f (x1 ) = f (x2) ⇒
x1 − 1 x2 − 1
⇒
2x1x2 − 2x1 = 2x1x2 − 2x2 ⇒ x1 = x2
Thus, f (x1 ) = f (x2) has only one solution, x1 = x2
∴ f (x) is one-one (injective)
2 ×2
Onto Let x = 2, then f (2) =
=4
2 −1
But x = 2 is not in the domain, and f (x) is one-one
function
∴f (x) can never be 4.
Similarly, f (x) can not take many values.
Hence, f (x) is into (not surjective).
∴f (x) is injective but not surjective.
x
8. We have, f (x) =
1 + x2
1
x
 1
f  = x =
∴
= f (x)
1 1 + x2
 x
1+ 2
x
 1
 1
f   = f (2)or f   = f (3)and so on.
∴
 2
 3
So, f (x) is many-one function.
x
Again, let
y = f (x) ⇒ y =
1 + x2
⇒
y + x2y = x ⇒ yx2 − x + y = 0
x ∈R
As,
∴ Range = Codomain =
− 1 1
,
 2 2 
− 1 1
,
 2 2 
So, f (x) is surjective.
Hence, f (x) is surjective but not injective.
9.
PLAN To check nature of function.
(i) One-one To check one-one, we must check whether
f ′ ( x )> 0 or f ′ ( x )< 0 in given domain.
(ii) Onto To check onto, we must check
Range = Codomain
Description of Situation To find range in given
domain [a , b], put f ′ (x) = 0 and find x = α 1, α 2, …,
α n ∈[a , b]
Now, find { f (a ), f (α 1 ), f (α 2), K , f (α n ), f (b)}
its greatest and least values gives you range.
Now, f : [0, 3] → [1, 29]
f (x) = 2x3 − 15x2 + 36x + 1
∴
f ′ (x) = 6x2 − 30x + 36 = 6 (x2 − 5x + 6)
= 6 (x − 2) (x − 3)
−
+
+
3
2
For given domain [0, 3], f (x) is increasing as well as
decreasing ⇒ many-one
Now, put
f ′ (x) = 0 ⇒ x = 2, 3
Thus, for range
f (0) = 1, f (2) = 29, f (3) = 28
⇒
Range ∈[1, 29]
∴ Onto but not one-one.
 x, x ∈ Q
− x, x ∉ Q
10. Let φ (x) = f (x) − g (x) = 
Now, to check one-one.
Take any straight line parallel to X-axis which will
intersect φ(x) only at one point.
⇒ φ(x) is one-one.
To check onto
As
 x, x ∈ Q
, which shows
f (x) = 
− x, x ∉ Q
y = x and
y = − x for rational and irrational values
⇒ y ∈ real numbers.
∴
Range = Codomain ⇒ onto
Thus, f − g is one-one and onto.
11. Given, f : [0, ∞ ) → [0, ∞ )
Here, domain is [0, ∞ ) and codomain is [0, ∞ ). Thus, to
check one-one
x
1
Since, f (x) =
> 0, ∀ x ∈ [0, ∞ )
⇒ f ′ (x) =
1+ x
(1 + x)2
∴ f (x) is increasing in its domain. Thus, f (x) is one-one
in its domain. To check onto (we find range)
Functions 185
x
⇒ y + yx = x
1+ x
Again,
y = f (x) =
⇒
y
y
x=
⇒
≥0
1− y
1− y
Y
y = 1 + 2x
X′
Since, x ≥ 0, therefore 0 ≤ y < 1
−π
2
O
π
2
X
i.e. Range ≠ Codomain
∴ f (x) is one-one but not onto.
Y′
f (x) = 2x + sin x
12. Given,
⇒
f ' (x) = 2 + cos x
⇒
f ' (x) > 0 , ∀x ∈ R
which shows f (x) is one-one, as f (x) is strictly increasing.
Since, f (x) is increasing for every x ∈ R,
∴ f (x) takes all intermediate values between (−∞ , ∞ ).
Range of f (x) ∈ R.
Hence, f (x) is one-to-one and onto.
13. The number of onto functions from
It is clear from the graph that y = tan x is one-one and
onto, therefore (B) → (r).
16. PLAN
(i) If f ′ ( x ) > 0, ∀x ∈ ( a, b ), then f( x ) is an increasing function in
( a, b ) and thus f( x ) is one-one function in ( a, b ) .
(ii) If range of f( x ) = codomain of f( x ) , then f( x ) is an onto
function.
(iii) A function f( x ) is said to be an odd function, if
f( − x ) = − f( x ), ∀x ∈ R, i.e.
f( − x ) + f( x ) = 0, ∀ x ∈ R
E = {1, 2, 3, 4} to F = {1, 2}
= Total number of functions which map E to F
f (x) = [ln (sec x + tan x)]3
3 [ln (sec x + tan x)]2 (sec x tan x + sec 2x)
f ′ (x) =
(sec x + tan x)
− Number of functions for which map f (x) = 1 and
f (x) = 2 for all x ∈ E = 24 − 2 = 14
 −π π 
, 
f ′ (x) = 3 sec x [ln (sec x + tan x)]2 > 0, ∀x ∈ 
 2 2
14. PLAN
(i) For such questions, we need to properly define the
functions and then we draw their graphs.
(ii) From the graphs, we can examine the function for continuity,
differentiability, one-one and onto.
f (x) is an increasing function.
∴ f (x) is an one-one function.
 π x
 π π
(sec x + tan x) = tan  +  , as x ∈  − ,  , then
 4 2
 2 2
 π x
0 < tan  +  < ∞
 4 2
− x, x < 0
f1 (x) =  x
e , x ≥ 0
f2(x) = x2, x ≥ 0
sin x, x < 0
f3 (x) = 
x≥0
x,
0 < sec x + tan x < ∞ ⇒ − ∞ < ln (sec x + tan x) < ∞
− ∞ < [ln (sec x + tan x)]3 < ∞ ⇒ −∞ < f (x) < ∞
Range of f (x) is R and thus f (x) is an ont function.
x<0
 f ( f (x)),
f4 (x) =  2 1
f
f
x
x≥0
(
(
))
−
1
,
 2 1
 x2 , x < 0
 x2 ,
x<0
Now, f2( f1 (x)) =  2x
⇒ f4 =  2x
e , x ≥ 0
e − 1 , x ≥ 0
x<0
2x,
As f4 (x) is continuous, f ′ 4 (x) =  2x
2
,
e
x>0

 

1
f (− x) = [ln (sec x − tan x)]3 = ln 



sec
x
+
tan
x

3
f (− x) = − [ln (sec x + tan x)]
f (x) + f (− x) = 0
⇒ f (x) is an odd function.
17. Let y = ax + b and y = cx + d be two linear functions.
When
f4(x)
∴
O
x
Graph for f4 (x)
f4′ (0) is not defined. Its range is [0, ∞ ).
Thus, range = codomain = [0, ∞ ), thus f4 is onto.
Also, horizontal line (drawn parallel to X-axis) meets
the curve more than once, thus function is not one-one.
x = − 1, y = 0 and x = 1, y = 2 , then
0 = − a + b and a + b = 2 ⇒ a = b = 1
...(i)
y=x+1
Again, when x = − 1, y = 2 and x = 1, y = 0, then
− c + d = 2 and c + d = 0
…(ii)
⇒
d = 1 and c = − 1 ⇒ y = − x + 1
Hence, two linear functions are y = x + 1 and y = − x + 1
18. Given, f (x) =
15. y = 1 + 2x is linear function, therefore it is one-one and
its range is (− π + 1, π + 1). Therefore, (1 + 2x) is one-one
but not onto so (A) → (q). Again, see the figure.
3
⇒
x2 + 4x + 30
x2 − 8x + 18
 (x2 − 8x + 18) (2x + 4) 


− (x2 + 4x + 30) (2x − 8)
f ′ (x) = 
(x2 − 8x + 18)2
186 Functions
=
2 (−6x2 − 12x + 156)
(x2 − 8x + 18)2
Topic 4 Inverse and Periodic Functions
1. Given function
−12 (x2 + 2x − 26)
=
(x2 − 8x + 18)2
f (x) =
which shows f ′ (x) is positive and negative both.
On applying componendo and dividendo law, we get
1+ y
84x =
1− y
∴ f (x) is many one.
Hence, given statement is true.
19. Let y =
αx2 + 6x − 8
α + 6 x − 8 x2
On applying logarithm having base ‘8’ both sides, we get
 1 + y
4x = log 8 

 1 − y
⇒
αy + 6xy − 8x2y = αx2 + 6x − 8
⇒
− αx2 − 8x2y + 6xy − 6x + αy + 8 = 0
⇒
αx2 + 8x2y − 6xy + 6x − αy − 8 = 0
⇒
x2 (α + 8 y) + 6x (1 − y) − (8 + αy) = 0
Since, x is real.
⇒
⇒
⇒
B2 − 4 AC ≥ 0
36 (1 − y) + 4 (α + 8 y) (8 + αy) ≥ 0
2
y2 (9 + 8α ) + y (46 + α 2) + 9 + 8α ≥ 0
⇒
and
…(i)
A > 0, D ≤ 0, ⇒ 9 + 8 α > 0
(46 + α 2)2 − 4 (9 + 8α )2 ≤ 0
⇒
α > − 9 /8
and [46 + α − 2 (9 + 8α )][46 + α + 2 (9 + 8α )] ≤ 0
2
2
⇒
and
α > − 9 /8
α > − 9 /8
and [(α − 2) (α − 14)] (α + 8)2 ≤ 0 ⇒ α > − 9 /8
and
⇒
(α − 2) (α − 14) ≤ 0
[Q (α + 8)2 ≥ 0]
α > − 9 / 8 and 2 ≤ α ≤ 14 ⇒ 2 ≤ α ≤ 14
Thus, f (x) =
αx2 + 6x − 8
will be onto, if 2 ≤ α ≤ 14
α + 6 x − 8 x2
Again, when α = 3
3 x2 + 6 x − 8
, in this case f (x) = 0
f (x) =
3 + 6 x − 8 x2
⇒
⇒
3 x2 + 6 x − 8 = 0
− 6 ± 36 + 96 − 6 ± 132 1
x=
=
= (− 3 ± 33 )
6
6
3
This shows that
1

f
(− 3 + 33) = f
3

 1 + y
 1 + y 1
1
log 8 

 = (log 8 e) log e 
 1 − y
 1 − y 4
4
By interchanging the variables x and y, we get the
inverse function of f (x) and it is
 1 + x
1
f −1 (x) = (log 8 e) log e 
.
 1 − x
4
Hence, option (c) is correct.
2. Since, only (c) satisfy given definition
f { f −1 (B)} = B
i.e.
B ⊆ f (x)
Only, if
3. By definition of composition of function,
(α 2 − 16α + 28) (α 2 + 16α + 64) ≤ 0
⇒
x=
{by base change property of logarithm
log a b = log a e ⋅ log e b}
⇒ 9 (1 − 2 y + y2) + [8α + (64 + α 2) y + 8 αy2] ≥ 0
⇒
82x − 8−2x
84x − 1
,
x
∈
(
−
1
,
1
)
= y (let)
=
82x + 8−2x
84x + 1
1

(− 3 − 33) = 0
3

Therefore, f is not one-to-one.
20. Since, there is an injective mapping from A to B, each
element of A has unique image in B.
Similarly, there is also an injective mapping from B to
A, each element of B has unique image in A or in other
words there is one to one onto mapping from A to B.
Thus, there is bijective mapping from A to B.
g ( f (x) ) = (sin x + cos x)2 − 1, is invertible (i.e. bijective)
⇒ g { f (x) } = sin 2x is bijective.
 π π
We know, sin x is bijective, only when x ∈ − , .
 2 2 
π
π
Thus, g { f (x) } is bijective, if − ≤ 2x ≤
2
2
π
π
− ≤x≤
⇒
4
4
4. It is only to find the inverse.
y = f (x) = (x + 1)2, for x ≥ − 1
Let
±
y = x + 1,
⇒
y = x+1
⇒
x=
⇒
f
−1
x≥ −1
⇒
y ≥ 0, x + 1 ≥ 0
y − 1 ⇒ f −1 ( y) =
(x) = x − 1 ⇒
y −1
x≥0
x2 + 1
⇒ y=
⇒ xy = x2 + 1
x
y ± y2 − 4
⇒ x2 − xy + 1 = 0 ⇒ x =
2
1
5. Let y = x +
x
⇒
f −1 ( y) =
y±
y2 − 4
x ± x2 − 4
⇒ f −1 (x) =
2
2
Since, the range of the inverse function is [1, ∞), then
we take
f −1 (x) =
x+
x2 − 4
2
Functions 187
x − x2 − 4
, then f −1 (x) > 1
2
This is possible only if (x − 2)2 > x2 − 4
⇒
x2 + 4 − 4x > x2 − 4 ⇒ 8 > 4x
⇒
x < 2, where x > 2
Therefore, (a) is the answer.
If we consider
f −1 (x) =
log 2 y = log 2 2x ( x − 1) ⇒ log 2 y = x (x − 1)
1 + 4 log 2 y ≥ 1 ⇒ − 1 + 4 log 2 y ≤ − 1
⇒
1 − 1 + 4 log 2 y ≤ 0
Let
⇒
Case III When f (z ) ≠ 2 is true.
If f (z ) ≠ 2 is true, then remaining statements are false.
∴ If
Thus, we have
[given]
⇒
⇒
For function to be invertible, it should be one-one onto.
∴ Check Range :
b−x
Let
f (x) = y
⇒ y=
1 − bx
⇒
y − bxy = b − x ⇒ x (1 − by) = b − y
b− y
, where 0 < x < 1
⇒ x=
1 − by
1
b
(b − 1) ( y + 1)
1
<0−1 < y<
1 − by
b
From Eqs. (i) and (ii), we get
1

y ∈  − 1,  ⊂ Codomain

b
…(ii)
 − x + 1
f (− x) = f 

 − x + 2
 − x + 1
f ( x) = f 

 − x + 2
[Q f (− x) = f (x)]
3 ± 9 −4 3 ± 5
=
2
2
 x + 1
f (x) = f 

 x + 2
⇒ x2 − 3 x + 1 = 0 ⇒ x =
⇒
…(i)
…(i)
Taking f −1 on both sides, we get
−x+1
x=
⇒ − x2 + 2 x = − x + 1
−x+2
Again,
b− y
b− y
> 0 and
<1
1 − by
1 − by
y>
f (x) = 2, f ( y) = 1 and f (z ) = 3
f −1 (1) = y
then f (− x) = f (x), ∀ x ∈ (− 5, 5)
 x + 1
Given ,
f (x) = f 

 x + 2
1
, x cos x are non-periodic functions.
x
b−x
9. Here, f (x) =
, where 0 < b < 1, 0 < x < 1
1 − bx
y < b or
Hence,
12. Since, f is an even function,
And sin
⇒
f (x) ≠ 1 and f ( y) = 1
But f is injective.
8. Clearly, f (x) = x − [x] = { x} which has period 1.
b− y
<1 ⇒
1 − by
f (x) ≠ 1 and f (z ) = 2
i.e. both x and y are not mapped to 1. So, either both
associate to 2 or 3. Thus, it is not injective.
y = f (x) = 3x − 5 ⇒ y + 5 = 3x
y+5
y+5
x+5
x=
⇒ f −1 ( y) =
⇒ f −1 (x) =
3
3
3
∴ 0<
Case II When f ( y) ≠ 1 is true.
If f ( y) ≠ 1 is true, then the remaining statements are
false.
∴
But
x≥1
So, x = 1 − 1 + 4 log 2 y is not possible.
1
Therefore, we take x = (1 + 1 + 4 log 2y )
2
1
−1
⇒
f ( y) = (1 + 1 + 4 log 2 y )
2
1
−1
⇒
f (x) = (1 + 1 + 4 log 2 x )
2
7. Given, f (x) = 3x − 5
f ( y) = 1 and f (z ) = 2
This means x and y have the same image, so f (x) is not
an injective, which is a contradiction.
1 ± 1 + 4 log 2 y
2
For y ≥ 1, log 2 y ≥ 0 ⇒ 4 log 2 y ≥ 0 ⇒ 1 + 4 log 2 y ≥ 1
⇒
But Statement II is true assin 2 x is periodic with period π.
∴
x2 − x − log 2 y = 0
x=
1
(2x − sin 2x) + C ⇒ F (x + π ) ≠ F (x)
4
Hence, Statement I is false.
F (x) =
Case I When f (x) = 1 is true.
In this case, remaining two are false.
Taking log 2 on both sides, we get
⇒
1 − cos 2x
dx
2
11. It gives three cases
6. Let y = 2x ( x − 1), where y ≥ 1 as x ≥ 1
⇒
10. Given, F (x) = ∫ sin 2 x dx = ∫
 x + 1
f (− x) = f 

 x + 2
[Q f (− x) = f (x)]
Taking f −1 on both sides, we get
x+1
−x=
⇒ x2 + 3 x + 1 = 0
x+2
⇒
x=
−3 ± 9 −4 −3 ± 5
=
2
2
Therefore, four values of x are
±3± 5
.
2
9
Limit, Continuity
and Differentiability
Topic 1
∞
0
and Form
∞
0
8.
Objective Questions I (Only one correct option)
x + 2 sin x
1. lim
x→ 0
x2 + 2 sin x + 1 − sin 2 x − x + 1
is
(2019 Main, 12 April II)
(a) 6
(b) 2
(c) 3
(d) 1
x2 − ax + b
2. If lim
= 5, then a + b is equal to
x→1
x−1
(2019 Main, 10 April II)
(a) − 4
3. If lim
x→1
(c) − 7
(b) 1
(d) 5
x4 − 1
x3 − k3
, then k is
= lim 2
x − 1 x → k x − k2
(2019 Main, 10 April I)
4
(a)
3
3
(b)
8
3
(c)
2
8
(d)
3
2
4. lim
x→ 0
sin x
equals
2 − 1 + cos x
(a) 4 2
(c) 2 2
5. lim
x→
π
4
x cot(4x)
x → 0 sin 2 x cot 2 (2 x)
(a) 0
7. lim
y→ 0
(c) 8
is equal to
(b) 1
(c) 4
1
4 2
(b) does not exist
(d) exists and equals
(d) 8 2
(2019 Main, 11 Jan II)
(d) 2
(2019 Main, 9 Jan I)
y4
(c) exists and equals
1
16
1
(d)
4
(a)
(b)
sin(π cos 2 x)
is equal to
x→ 0
x2
9. lim
(a)
π
2
10. lim
x→ 0
(b) 1
(2014 Main)
(c) − π
(d) π
(1 – cos 2x)(3 + cos x)
is equal to
x tan 4x
(a) 4
(b) 3
(c) 2
(2013 Main)
(d)
1
2

 x2 + x + 1
− ax − b = 4, then

 x+1
11. If lim 
x→ ∞
(a) a = 1, b = 4
(c) a = 2, b = − 3
(2012)
(b) a = 1, b = − 4
(d) a = 2, b = 3
(a) does not exist
(c) is equal to 3/2
(b) is equal to −3/2
(d) is equal to 3
{(a − n ) nx − tan x} sin nx
= 0, where n is
x2
non-zero real number, then a is equal to
(2003, 2M)
13. If
(a) 0
lim
x→ 0
(b)
n+1
n
(c) n
(d) n +
1
n
(cos x − 1) (cos x − ex )
is a
x→ 0
xn
finite non-zero number, is
(2002, 2M)
14. The integer n for which lim
1 + 1 + y4 − 2
(a) exists and equals
1
24
1
(c)
8
(2017 Main)
f (2h + 2 + h 2) − f (2)
, given that f ′ (2) = 6 and
h → 0 f (h − h 2 + 1 ) − f (1 )
(2003, 2M)
f ′ (1) = 4 ,
(2019 Main, 12 Jan I)
(b) 4
cot x − cos x
equals
(π − 2x)3
12. lim
(b) 2
(d) 4
cot3 x − tan x
is
π

cos  x + 

4
(a) 4 2
6. lim
(2019 Main, 8 April I)
lim
x → π/ 2
1
2 2
1
2 2 ( 2 + 1)
(a) 1
(c) 3
15. lim
x→ 0
(a) 2
1
(c)
2
(b) 2
(d) 4
x tan 2x − 2x tan x
is
(1 − cos 2x)2
(1999, 2M)
(b) −2
1
2
(d) −
Limit, Continuity and Differentiability 189
1 − cos 2 (x − 1)
x −1
16. lim
x→1
(a)
(b)
(c)
(d)
(1998, 2M)
exists and it equals 2
exists and it equals − 2
does not exist because x − 1 → 0
does not exist because left hand limit is not equal to
right hand limit
17. The value of lim
x→ 0
1
(1 − cos 2 x)
2
is
x
where, [x] denotes the greatest integer less than or
(1985, 2M)
equal to x, then lim f (x) equals
(a) 1
(c) −1
(b) 0
(d) None of these

n 
1
2
+
+ ... +
 is equal to (1984, 2M)
n → ∞  1 − n2
1 − n2
1 − n 2
19. lim 
(b) −
(a) 0
(c)
1
2
1
2
(d) None of these
20. If f (a ) = 2, f ′ (a ) = 1, g (a ) = − 1, g ′ (a ) = 2,
then the value of lim
x→ a
g (x) f (a ) − g (a ) f (x)
is (1983, 1M)
x−a
1
5
(d) None of these
(a) − 5
(b)
(c) 5
21. If G (x) = − 25 − x2, then lim
x→1
1
24
(c) − 24
G (x) − G (1)
has the value
x−1
(1983, 1M)
1
5
(d) None of these
(a)
(b)
n


1
fn (x) = Σ tan − 1 
 for all x ∈ (0, ∞ ).
j =1
 1 + (x + j) (x + j − 1)
(Here, the inverse trigonometric function tan − 1 x
 π π
assumes values in  − ,  ). Then, which of the
 2 2
(2018 Adv.)
following statement(s) is (are) TRUE?
j =1
(b) ∑
j =1
10
log (1 + 2h ) − 2 log (1 + h )
=K .
h2




25. If f (x) = sin x, x ≠ nπ , n = 0, ± 1, ± 2, ... 

other wise
 2,
24. lim
tan 2 (f j (0)) = 55
(1+ f ′ j (0)) sec2 (f j (0)) = 10
(1996, 2M)
26. ABC is an isosceles triangle inscribed in a circle of radius
r. If AB = AC and h is the altitude from A to BC, then the
∆ABC has perimeter P = 2( 2hr − h 2 + 2hr ) and area
A
A = K . Also, lim 3 = K
(1989, 2M)
h→ 0 P
 4
 1
2 
  x sin  x + x  
= …
27. lim 
x→ − ∞ 
(1 + |x|3 )





(1987, 2M)
28. Let f (x) = (x + x − 16x + 20) / (x − 2) , if x ≠ 2
3
2
2
.
if x = 2
k ,
(1981, 2M)
If f (x) is continuous for all x, then k = … .
πx
29. lim (1 − x) tan
=….
x→1
2
(1978, 2M)
True/False
lim [ f (x) g (x)] exists, then both
x→ a
lim f (x) and
x→ a
x→ a
1
x →∞
n
(d) For any fixed positive integer n, lim sec2 (fn (x)) = 1
x →∞
(1981, 2M)
Analytical & Descriptive Questions
ax −1
= log e a, to find
x→ 0
x
.
(1983, 3M)
31. Use the formula lim
2x − 1
x → 0 (1 + x)1/ 2 − 1
lim
(1982, 2M)
(a + h )2 sin (a + h ) − a 2 sin a
32. Evaluate lim
. (1980, 3M)
h→ 0
h
x − sin x
.
x + cos 2 x
33. Evaluate lim
(c) For any fixed positive integer n, lim tan(fn (x)) =
(1997C, 2M)
 x2 + 1 , x ≠ 0 , 2 


and g (x) =  4,
x = 0 , then lim g [f(x)] is ………
x→0
 5,
x = 2 

lim g (x) exist.
22. For any positive integer n, define fn : (0, ∞ ) → R as
5
(2009)
Fill in the Blanks
30. If
Objective Questions II
(One or more than one correct option)
(a) ∑
(a) a = 2
(b) a = 1
1
(c) L =
64
1
(d) L =
32
h→ 0
sin[x] [x] ≠ 0
,

18. If f (x) =  [x]
[x] ≠ 0
 0,
x→ 0
x4
x→ 0
x2
4 , a > 0 . If L is finite, then
(1991, 2M)
(b) −1
(d) None of these
(a) 1
(c) 0
23. Let L = lim
a − a 2 − x2 −
x→ 0
x −1

.
 2 x2 − 7 x + 5 
34. Evaluate lim 
x→1
(1979, 3M)
(1978, 3M)
190 Limit, Continuity and Differentiability
Integer & Numerical Answer Type Questions
x2 sin (βx)
35. Let α , β ∈ R be such that lim
= 1 . Then,
x → 0 αx − sin x
(2016 Adv.)
6 (α + β ) equals
36. Let m and n be two positive integers greater than 1. If
 ecos ( α n ) − e
 = −  e , then the value of m is
lim 
m

 2
α→ 0 
n
α


(2015 Adv.)
Topic 2 1∞ Form, RHL and LHL
Objective Questions I (Only one correct option)
1. Let f : R → R be a differentiable function satisfying
1
 1 + f (3 + x) − f (3) x
f ′ (3) + f ′ (2) = 0. Then lim
 is equal
x→ 0  1 + f (2 − x) − f (2 ) 
to
(2019 Main, 8 April II)
−1
(a) e
2.
(b) e
(c) e
π − 2 sin − 1 x
is equal to
1−x
lim
x→1 −
(a)
2
π
2
(b)
2
π
(c)
(d) 1
(2019 Main, 12 Jan II)
(d)
1
2π
3. Let [x] denote the greatest integer less than or equal to
(a) equals π
(c) equals 0
(2019 Main, 11 Jan I)
4. For each t ∈ R, let [t ] be the greatest integer less than or
equal to t. Then,
π

(1 − |x| + sin|1 − x|) sin  [1 − x]
2

lim
x → 1+
|1 − x|[1 − x]
(2019 Main, 10 Jan I)
(b) does not exist
(d) equals 1
(a) 0
(c) − sin 1
(2019 Main, 9 Jan II)
  1  2
15
lim x   +   + … +  
  x  x
 x 
x→ 0 +
(2018 Main)
(b) is equal to 15
(d) does not exist (in R)
 1 
1 − x (1 + |1 − x|)
cos 

 1 − x
|1 − x|
(b) limx→ 1 − f (x) does not exist
(c) limx→ 1 − f (x) = 0
(d) limx→ 1 + f (x) does not exist
1
and −1
2
9
(d) − and 3
2
(b) −
1
10. If lim [1 + x log (1 + b2)] x = 2b sin 2 θ, b > 0
and θ ∈ (− π , π ], then the value of θ is
π
4
π
(c) ±
6
π
3
π
(d) ±
2
(a) ±
(2011)
(b) ±

11. For x > 0, lim (sin x)1/ x +  
 x
x→0

1
(a) 0
(c) 1
sin x 
 is

(2006, 3M)
(b) – 1
(d) 2
1/ x
 f (1 + x) 
lim
x→ 0
 f (1) 
equals
(2002, 2M)
(b) e 2
(d) e3
(a) 1
(c) e2
x
equal to t. Then,
for x ≠ 1. Then
(a) limx→ 1 + f (x) = 0
a→ 0 +
5
(a) − and 1
2
7
(c) − and 2
2
x−3 
 is equal to
x→ ∞ x + 2 
13. For x ∈ R , lim 
6. For each t ∈R, let [t ] be the greatest integer less than or
7. Let f (x) =
( 3 1 + a − 1) x2 − ( 1 + a − 1) x + (6 1 + a − 1) = 0, where
(2012)
a > − 1. Then, lim α (a ) and lim β (a ) are
1
(b) sin 1
(d) 1
(a) is equal to 0
(c) is equal to 120
9. Let α (a ) and β (a ) be the roots of the equation
12. Let f : R → R be such that f (1) = 3 and f ′ (1) = 6. Then,
5. For each x ∈ R, let [x] be the greatest integer less than or
equal to x. Then,
x([x] + |x|) sin [x]
is equal to
lim
|x|
x → 0−
(2016 Main)
(b) 1
1
(d)
4
x→ 0
(b) equals π + 1
(d) does not exist
(a) equals 0
(c) equals − 1
x→ 0
(a) 2
1
(c)
2
a→ 0+
π
x. Then,
tan(π sin 2 x) + (|x| − sin(x[x]))2
lim
x→ 0
x2
8. Let p = lim+ (1 + tan 2 x )1/ 2x , then log p is equal to
(2000, 2M)
(b) e−1
(d) e5
(a) e
(c) e−5
Fill in the Blanks
 1 + 5 x2 
14. lim 

x → 0  1 + 3 x2 
 x + 6

+ 1
15. lim 
x → ∞ x
1/ x 2
=K .
(1996, 1M)
x+ 4
= …… .
(1991, 2M)
Analytical & Descriptive Question

16. Find lim tan 

x→ 0

1/ x
π
 
+ x  .
 
4
(1993, 2M)
Limit, Continuity and Differentiability 191
Integer & Numerical Answer Type Questions
17. Let e denote the base of the natural logarithm. The value of the real number a for which the right hand limit
1
(1 − x) x − e−1
is equal to a non zero real number, is ……… .
lim
+
x→ 0
xa
(2020 Adv.)
1−x
 − ax + sin(x − 1) + a 1−

x→1  x + sin(x − 1 ) − 1 
18. The largest value of the non-negative integer a for which lim 
x
=
1
is
4
(2014 Adv.)
Topic 3 Squeeze, Newton-Leibnitz’s Theorem and Limit Based on
Converting infinite Series into Definite Integrals
Objective Question II
(One more than one correct option)
Objective Questions I (Only one correct option)
1. If α and β are the roots of the equation
375x2 − 25x − 2 = 0, then lim
n→ ∞
n
n
∑ α r + nlim
∑ βr is equal
→∞
r =1
r =1
to
(2019 Main, 12 April I)
21
346
1
(c)
12
29
358
7
(d)
116
(a)
(b)
sec 2 x
2.
∫
lim 2
π
x→
4
1
(a) f   ≥ f (1)
 2
f (t ) dt
x2 −
(2007, 3M)
8
(a) f (2)
π
2
(b) f (2)
π
2  1
(c) f  
π  2
n→ ∞
1
n
(a) 1 +
2n
∑
r =1
5
(2016 Adv.)
Integer & Numerical Answer Type Question
5. For each positive integer n, let
1
1
((n + 1) (n + 2) ... (n + n )) n .
n
For x ∈ R, let [x] be the greatest integer less than or
equal to x. If lim yn = L, then the value of [L ] is
n→ ∞
................. .
(2018 Adv.)
yn =
(d) 4f (2)
3. lim
x
n
n
n


 ...  x + 

n
2
 ,
n 2  2 n 2 
 ...  x + 2 
4 
n  
1
2
(b) f   ≤ f  
 3
 3
f ′ (3) f ′ (2)
(d)
≥
f (3)
f (2)
(c) f ′ (2) ≤ 0
equals
π2
16


 n n (x + n )  x +


4. Letf (x) = lim
n→ ∞ 

2
2
2
 n ! (x + n )  x +


for all x = 0. Then
r
n2 + r2
Fill in the Blank
equals
(b) 5 − 1
(1999, 2M)
(c) −1 +
(d) 1 +
2
2
x2
6. lim
∫0
x→ 0
cos 2t dt
x sin x
=K
(1997C, 2M)
Topic 4 Continuity at a Point
Objective Questions I (Only one correct option)
π π
1. If the function f defined on  ,  by
2. The function f (x) = [x]2 − [x2] (where, [x] is the greatest
integer less than or equal to x), is discontinuous at
 6 3
 2 cos x − 1
, x≠

f (x) =  cot x − 1

k,
x=

then k is equal to
(a)
1
2
(c) 1
π
4 is continuous,
π
4
(2019 Main, 9 April I)
(b) 2
(d)
1
2
(1999, 2M)
(a) all integers
(b) all integers except 0 and 1
(c) all integers except 0
(d) all integers except 1
3. Let [.] denotes the greatest integer function and
f (x) = [tan 2 x], then
(a) lim f (x) does not exist
x→ 0
(b) f (x) is continuous at x = 0
(c) f (x) is not differentiable at x = 0
(d) f ′ (0) = 1
(1993, 1M)
192 Limit, Continuity and Differentiability
2x − 1
 π, [⋅] denotes the
 2 
greatest integer function, is discontinuous at
4. The function f (x) = [x] cos 
(a) all x
(b) all integer points
(c) no x
(d) x which is not an integer
5. If f (x) = x ( x + (x + 1), then
(1993, 1M)
(1985, 2M)
(a) f (x) is continuous but not differentiable at x = 0
(b) f (x) is differentiable at x = 0
(c) f (x) is not differentiable at x = 0
(d) None of the above
log (1 + ax) − log (1 − bx)
x
is not defined at x = 0. The value which should be
assigned to f at x = 0, so that it is continuous at x = 0, is
6. The function f (x) =
(a) a − b
(b) a + b
(c) log a + log b
(d) None of the above
(1994, 4M)

 1 − cos 4x ,
x<0

x2
12. Let f (x) = 
a,
x=0

x
, x>0

 16 + x − 4
Determine the value of a if possible, so that the function
(1990, 4M)
is continuous at x = 0.
(a) x = − 1
(b) x = 1
(c) x = 0
(d) x = 2
 x + a 2 sin x,
0 ≤ x ≤ π /4

π /4 ≤ x ≤ π /2
f (x) =  2x cot x + b,
a cos 2x − b sin x, π / 2 < x ≤ π

x≤0
 g (x),
1/ x

defined by f (x) =  (1 + x) 
 (2 + x)  , x > 0


Find the continuous
f ′ (1) = f (− 1).
8. For every pair of continuous function f , g : [0, 1] → R
such that max { f (x): x ∈ [0, 1]} = max { g (x): x ∈ [0,1]}.
The correct statement(s) is (are)
(1989)
is continuous for 0 ≤ x ≤ π.
Then, at which of the following point(s) the function
(2017 Adv.)
f (x) = x cos (π (x + [x])) is discontinuous ?
(2014 Adv.)
[f (c)] + 3f (c) = [ g (c)] + 3 g (c) for some c ∈[0,1]
[f (c)]2 + f (c) = [ g (c)]2 + 3 g (c) for some c ∈[0,1]
[f (c)]2 + 3f (c) = [ g (c)]2 + g (c) for some c ∈[0,1]
[f (c)]2 = [ g (c)]2 for some c ∈[0,1]
2
9. For every integer n, let a n and bn be real numbers. Let
function f : R → R be given by
a n + sin πx, for x ∈ [2n , 2n + 1]
f (x) = 
,
bn + cos πx, for x ∈ (2n − 1, 2n )
for all integers n.
If f is continuous, then which of the following hold(s) for
(2012)
all n ?
(a) an−1 − bn−1 = 0
(b) an − bn = 1
(c) an − bn + 1 = 1
(d) an − 1 − bn = −1
Fill in the Blank
10. A discontinuous function y = f (x) satisfying x2 + y2 = 4 is
given by f (x) = .... .
Determine a and b such that f (x) is continuous at x = 0.
14. Let g (x) be a polynomial of degree one and f (x) be
7. Let [x] be the greatest integer less than or equals to x.
(a)
(b)
(c)
(d)
π

{1 + |sin x|}a/|sin x |,
< x<0

6

11. Let f (x) = 
b,
x=0
π

tan 2 x / tan 3 x
e
,
0<x<

6
13. Find the values of a and b so that the function
(1983, 1M)
Objective Questions II
(One or more than one correct option)
2
Analytical & Descriptive Questions
(1982, 2M)
function
f (x)
satisfying
(1987, 6M)
15. Determine the values a, b, c, for which the function
 sin (a + 1) x + sin x
, for x < 0

x

for x = 0
f (x) =  c,
 (x + bx2) 1/ 2 − x1/ 2
, for x > 0


bx3/ 2
is continuous at x = 0.
(1982, 3M)
Match the Columns
π π
,  → R, f3 : (− 1, eπ / 2 − 2) → R and
 2 2
f4 : R → R be functions defined by
16. Let f1 : R → R, f2 :  −
2
(i) f1 (x) = sin( 1− e− x ),
 |sin x| if x ≠ 0

, where the inverse
(ii) f2 (x) =  tan − 1 x

if x = 0
1

trigonometric function tan − 1 x assumes values in
− π , π.


 2 2
(iii) f3 (x) = [sin(log e (x + 2))], where for t ∈R, [t ] denotes the
greatest integer less than or equal to t,
x2 sin  1  if x ≠ 0

 
(iv) f4 (x) = 
 x
if x = 0
0

Limit, Continuity and Differentiability 193
List-I
List-II
P.
The function f1 is
1.
NOT continuous at x = 0
Q.
The function f2 is
2.
continuous at x = 0 and NOT differentiable at x = 0
R.
The function f3 is
3.
differentiable at x = 0 and its derivative is NOT continuous at x = 0
S.
The function f4 is
4.
differentiable at x = 0 and its derivative is continuous at x = 0
The correct option is
(a) P → 2; Q → 3; R → 1; S → 4
(c) P → 4; Q → 2; R → 1; S → 3
(b) P → 4; Q → 1; R → 2; S → 3
(d) P → 2; Q → 1; R → 4; S → 3
Topic 5 Continuity in a Domain
Objective Questions I (Only one correct option)
1. Let f : R → R be a continuously differentiable
1
. If
48
f ( x) 3
∫ 4t dt = (x − 2) g(x), then lim g (x) is equal to
function such that f (2) = 6 and f′ (2) =
x→ 2
6
(a) 18
(b) 24
(2019 Main, 12 April I)
(c) 12
(d) 36
sin ( p + 1) x + sin x
,x<0

x

2. If f (x) = 
x=0
q,
2

x>0
x+ x − x

,

x3/ 2
(b)  −

(d)  −

1
,
2
3
,
2
3

2
1

2
 a|π − x|+1, x ≤ 5
is continuous at
b|x − π|+3, x > 5
(2019 Main, 9 April II)
x = 5, then the value of a − b is
3. If the function f (x) = 
−2
π+ 5
2
(c)
π −5
2
π+ 5
2
(d)
5− π
(b)
(a)
x
4. If f (x) = [x] −  , x ∈ R where [x] denotes the greatest
4 
(2019 Main, 9 April II)
integer function, then
(a) lim f (x) exists but lim f (x) does not exist
x→ 4 +
x→ 4 −
(b) f is continuous at x = 4
x→ 4 −
x→ 4 +
(d) lim f (x) exists but lim f (x) does not exist
x→ 4 −
x→ 4 +
5. Let f : [−1, 3] → R be defined as
|x| + [x], −1 ≤ x < 1

f (x) =  x + |x|, 1 ≤ x < 2
 x + [x], 2 ≤ x ≤ 3 ,

(2019 Main, 8 April II)
where, [t ] denotes the greatest integer less than or
equal to t. Then, f is discontinuous at
is continuous at x = 0 , then the ordered pair ( p, q) is
equal to
(2019 Main, 10 April I)
3
1
(a)  − , − 
 2
2
5 1

(c)  , 
 2 2
(c) Both lim f (x) and lim f (x) exist but are not equal
(a)
(b)
(c)
(d)
four or more points
only two points
only three points
only one point
6. Let f: R → R be a function defined as
 5,
a + bx,

f (x) = 
 b + 5x,
 30,
Then, f is
(a)
(b)
(c)
(d)
if
x≤1
if 1 < x < 3
if 3 ≤ x < 5
if
x≥5
(2019 Main, 9 Jan I)
continuous if a = − 5 and b = 10
continuous if a = 5 and b = 5
continuous if a = 0 and b = 5
not continuous for any values of a and b
7. If f (x) =
(a) tan [f
(b) tan [f
(c) tan [f
(d) tan [f
1
x − 1, then on the interval [0, π ]
2
(1989, 2M)
(x)] and 1/ f (x) are both continuous
(x)] and 1/ f (x) are both discontinuous
(x)] and f −1 (x) are both continuous
(x)] is continuous but 1/ f (x) is not continuous
194 Limit, Continuity and Differentiability
the points of discontinuity of f in the domain are…… .
Objective Questions II
(One or more than one correct option)
(1996, 2M)
Analytical & Descriptive Question
8. The following functions are continuous on (0, π)
(a) tan x
 1, 0 ≤ x≤ 3 π /4
(c) 
2
3π
< x< π
2 sin x,

9
4
1
(b) ∫ t sin dt
0
t
xπ sin x, 0 < x ≤
(d) 
sin ( π + x ),
 2
x
(1991, 2M)
π /2
π
< x< π
2

x2
0 ≤ x<1
,

2
11. Let f (x) = 
2x2 − 3x + 3 , 1 ≤ x ≤ 2

2
Discuss the continuity of f , f ′ and f ′ ′ on [0, 2].
9. Let [x] denotes the greatest integer less than or equal to
x. If f (x) = [x sin πx], then f (x) is
(a) continuous at x = 0
(c) differentiable at x = 1
(1983, 2M)
(1986, 2M)
(b) continuous in (−1, 0)
(d) differentiable in (−1, 1)
Fill in the Blank
Integer & Numerical Answer Type Question
1 1
12. If the function f defined on  − ,  by
 π 
 , where [⋅] denotes the
 [x + 1]
greatest integer function. The domain of f is …… and
10. Let f (x) = [x] sin 
 3 3
1
 1 + 3x

f (x) =  x log e  1 − 2x  , when x ≠ 0 is continuous, then k
k
, when x = 0
is equal to ………… .
(2020 Main, 7 Jan II)
Topic 6 Continuity for Composition and Function
Objective Question II
(One or more than one correct option)
1
x
1. For the function f (x) = x cos , x ≥ 1,
(2009)
(a) for atleast one x in the interval
[1, ∞ ), f (x + 2) − f (x) < 2
(b) lim f ′(x) = 1
where, a and b are non-negative real numbers.
Determine the compositie function gof. If ( gof ) (x) is
continuous for all real x determine the values of a and b.
Further, for these values of a and b, is gof differentiable
(2002, 5M)
at x = 0 ? Justify your answer.
3. Let f (x) be a continuous and g (x) be a discontinuous
x→ ∞
(c) for all x in the interval [1, ∞ ), f (x + 2) − f (x) > 2
(d) f ′(x) is strictly decreasing in the interval [1, ∞ )
Analytical & Descriptive Questions
 x + a , if x < 0
2. Let f (x) = 
|x − 1|, if x ≥ 0
if x < 0
 x + 1,
and g (x) = 
2
1
if x ≥ 0
(
x
−
)
+
b
,

function. Prove that f (x) + g (x) is a discontinuous
function.
(1987, 2M)
1 + x, 0 ≤ x ≤ 2
3 − x, 2 < x ≤ 3
4. Let f (x) = 
Determine the form of g (x) = f [ f (x)] and hence find the
points of discontinuity of g, if any
(1983, 2M)
5. Let f (x + y) = f (x) + f ( y) for all x and y. If the function
f (x) is continuous at x = 0, then show that f (x) is
(1981, 2M)
continuous at all x.
Topic 7 Differentiability at a Point
Objective Questions I (Only one correct option)
1. Let f : R → R be differentiable at c ∈ R and f (c) = 0. If
g (x) = | f (x)|, then at x = c, g is
(a)
(b)
(c)
(d)
not differentiable
differentiable if f ′ (c) ≠ 0
not differentiable if f ′ (c) = 0
differentiable if f ′ (c) = 0
2. If f : R → R is a differentiable function and
f ( x)
f (2) = 6, then lim
x→ 2
(2019 Main, 10 April I)
(a) 12f ′ (2)
(b) 0
∫
6
2t dt
is
(x − 2)
(2019 Main, 9 April II)
(c) 24f ′ (2)
(d) 2f ′ (2)
3. Let f (x) = 15 − x − 10 ; x ∈ R. Then, the set of all values
of x, at which the function, g (x) = f ( f (x)) is not
differentiable, is
(2019 Main, 9 April I)
(a) {5, 10, 15, 20}
(c) {10}
(b) {5, 10, 15}
(d) {10, 15}
Limit, Continuity and Differentiability 195
4. Let S be the set of all points in (− π , π ) at which the
function, f (x) = min {sin x, cos x} is not differentiable.
Then, S is a subset of which of the following?
(2019 Main, 12 Jan I)
π
π
(a) − , 0, 
4
 4
3π
π 3π π 

(c) −
,− ,
, 
4 4 4
 4
(b) −

(d) −

π
π π π
, − , , 
2
4 4 2
3π
π π 3π 
,− , ,

4
2 2 4
5. Let K be the set of all real values of x, where the function
f (x) = sin| x| − | x| + 2(x − π ) cos| x|is not differentiable.
(2019 Main, 11 Jan II)
Then, the set K is equal to
(b) φ (an empty set)
(d) {0, π }
(a) {0}
(c) { π }
−2 ≤ x < 0
 −1 ,
and
2
1
0 ≤ x≤2
x
−
,

6. Let f (x) = 
g (x) = | f (x)| + f (|x|). Then, in the interval (−2, 2), g is
(2019 Main, 11 Jan I)
(a)
(b)
(c)
(d)
not differentiable at one point
not differentiable at two points
differentiable at all points
not continuous
7. Let f : (−1, 1) → R be a function defined by
f (x) = max { − x , − 1 − x2 }. If K be the set of all points at
which f is not differentiable, then K has exactly
(2019 Main, 10 Jan II)
(a) three elements
(c) two elements
(b) five elements
(d) one element
max {|x|, x2}, |x| ≤ 2
2 < |x| ≤ 4
 8 − 2|x|,
Let S be the set of points in the interval (−4, 4) at which f
is not differentiable. Then, S
(2019 Main, 10 Jan I)
8. Let f (x) = 
(a) equals {−2, − 1, 0, 1, 2}
(b) equals {−2, 2}
(c) is an empty set
(d) equals {−2,−1, 1, 2}
π

 2
13. Let f (x) = x cos x , x ≠ 0, x ∈ R, then f is
 0,
(a)
(b)
(c)
(d)
3
differentiable both at x = 0 and at x = 2
differentiable at x = 0 but not differentiable at x = 2
not differentiable at x = 0 but differentiable at x = 2
differentiable neither at x = 0 nor at x = 2
(x − 1)n
; 0 < x < 2, m and n are
log cosm (x − 1)
integers, m ≠ 0, n > 0 and let p be the left hand
derivative of|x − 1| at x = 1 . If lim g (x) = p , then
14. Let
g (x) =
x→1 +
(a) n = 1, m = 1
(c) n = 2 , m = 2
(b) n = 1, m = −1
(d) n > 2, m = n
f 2(x) dx is equal to
(b)
1
2
(2019 Main, 9 Jan II)
(c) 1
(d) 0
10. Let S = (t ∈ R : f (x) = |x − π |(
⋅ e − 1)sin| x| is not
|x|
differentiable at t}.Then, the set S is equal to(2018 Main)
(a) φ (an empty set)
(c) { π }
(2008, 3M)
15. If f is a differentiable function satisfying
 1
f   = 0, ∀ n ≥ 1, n ∈ I ,then
 n
(2005, 2M)
(a) f (x) = 0, x ∈ (0, 1]
(b) f ′ (0) = 0 = f (0)
(c) f (0) = 0 but f ′ (0) not necessarily zero
(d)|f (x)|≤ 1, x ∈ (0, 1]
16. Let f (x) = ||x|− 1|, then points where, f (x) is not
differentiable is/are
(a) 0, ± 1
(c) 0
(2005, 2M)
(b) ± 1
(d) 1
17. The domain of the derivative of the functions
 tan −1 x ,
if | x| ≤ 1

is
f (x) = 1
(| x| − 1), if | x| > 1
2
(a) R − {0}
(c) R − {−1}
(2002, 2M)
(b) R − {1}
(d) R − {−1, 1}
(2001, 2M)
(a) cos (| x|) + | x|
(c) sin (| x|) + | x|
(b) cos (| x|) − | x|
(d) sin (| x|) − | x|
19. The left hand derivative of f (x) = [x] sin (π x) at x = k, k
is an integer, is
1
(a) 2
(2012)
x =0
at x = 0 ?
| f (x) − f ( y)| ≤ 2|x − y|2 , for all x, y ∈ R. If f (0) = 1, then
0
(b) 2 f ′(c) = 3 g ′(c)
(d) f ′(c) = 2 g ′(c)
18. Which of the following functions is differentiable
9. Let f be a differentiable function from R to R such that
∫
(a) 2f ′ (c) = g ′(c)
(c) f ′(c) = g ′(c)
(b) {0}
(d) {0, π }
11. For x ∈ R, f (x) = |log 2 − sin x|and g (x) = f ( f (x)), then
(a) g is not differentiable at x = 0
(2016 Main)
(b) g′ (0) = cos (log 2)
(c) g′ (0) = − cos (log 2)
(d) g is differentiable at x = 0 and g′ (0) = − sin (log 2)
12. If f and g are differentiable functions in (0, 1) satisfying
f (0) = 2 = g (1), g(0) = 0 and f (1) = 6, then for some
(2014 Main)
c ∈] 0, 1 [
(a) (−1)k (k − 1) π
(c) (−1)k kπ
(2001, 2M)
(b)(−1)k − 1 (k − 1) π
(d) (−1)k − 1 kπ
20. Let f : R → R be a function defined by f (x) = max { x, x3 }.
The set of all points, where f (x) is not differentiable, is
(a) {−1,1 }
(c) {0,1 }
(b) {−1, 0 }
(d) {−1, 0,1 }
(2001, 2M)
21. Let f : R → R be any function. Define g : R → R by
g (x) =| f (x)|, ∀ x. Then, g is
(2000, 2M)
(a) onto if f is onto
(b) one-one if f is one-one
(c) continuous if f is continuous
(d) differentiable if f is differentiable
22. The function f (x) = (x2 − 1)| x2 − 3x + 2| + cos (| x|) is
not differentiable at
(a) −1
(b) 0
(1999, 2M)
(c) 1
(d) 2
196 Limit, Continuity and Differentiability
23. The set of all points, where the function f (x) =
differentiable, is
x
is
1 + | x|
(1987, 2M)
(a) (− ∞ , ∞ )
(c) (− ∞ , 0) ∪ (0, ∞ )
(b) [0, ∞ )
(d) (0, ∞ )
24. There exists a function f (x) satisfying f (0) = 1,
f ′ (0) = − 1, f (x) > 0, ∀ x and
(1982, 2M)
(a) f ′′ (x) < 0 , ∀ x
(b) − 1 < f ′′ (x) < 0 , ∀ x
(c) − 2 ≤ f ′′ (x) ≤ − 1 , ∀ x
(d) f ′′ (x) < − 2 , ∀ x
25. For a real number y, let [ y] denotes the greatest
integer less than or equal to y. Then, the function
tan π [(x − π )]
(1981, 2M)
is
f (x) =
1 + [x]2
(a) discontinuous at some x
(b) continuous at all x, but the derivative f ′ (x) does not
exist for some x
(c) f ′(x) exists for all x, but the derivative f ′ ′ (x) does not
exist for some x
(d) f ′(x) exists for all x
Objective Questions II
(One or more than one correct option)
26. Let f : R → R and g : R → R be functions satisfying
f (x + y) = f (x) + f ( y) + f (x) f ( y)and f (x) = xg (x)
29. For every twice differentiable function f : R → [−2, 2]
with ( f (0))2 + ( f ′ (0))2 = 85, which of the following
(2018 Adv.)
statement(s) is (are) TRUE ?
(a) There exist r , s ∈ R , where r < s, such that f is one-one on
the open interval (r , s)
(b) There exists x0 ∈ (−4, 0) such that|f ′ (x0 )|≤ 1
(c) lim f (x) = 1
x→ ∞
(d) There exists α ∈ (−4, 4) such that f (α ) + f ′′(α ) = 0 and
f ′ (α ) ≠ 0
30. Let f : (0, π)→ R be a twice differentiable function such
that lim
t→ x
f (x) sin t − f (t )sin x
= sin 2 x for all x ∈ (0, π).
t−x
π
 π
If f   = − , then which of the following statement(s)
 6
12
is (are) TRUE?
(2018 Adv.)
π
π
(a) f   =
 4 4 2
(b) f (x)<
x4
− x2 for all x∈ (0, π)
6
(c) There exists α ∈(0, π) such that f ′ (α) = 0
π
π
(d) f ′′  + f   = 0
 2
 2
31. Let f : R → R, g : R → R and h : R → R be differentiable
for all x, y ∈ R. If lim g (x) = 1, then which of the
functions such that f (x) = x3 + 3x + 2, g ( f (x)) = x and
(2016 Adv.)
h ( g ( g (x))) = x for all x ∈ R. Then,
following statements is/are TRUE?
(a) g ′(2) =
x→ 0
(2020 Adv.)
(a) f is differentiable at every x ∈ R
(b) If g(0) = 1, then g is differentiable at every x∈ R
(c) The derivative f ′ (1) is equal to 1
(d) The derivative f ′ (0) is equal to 1
the function
f : R → R be defined by
3
2
f (x) = x − x + (x − 1)sin x and let g : R → R be an
arbitrary function. Let fg : R → R be the product
function defined by ( fg )(x) = f (x) g (x). Then which of the
(2020 Adv.)
following statements is/are TRUE?
27. Let
(a) If g is continuous at x = 1, then fg is differentiable at
x =1
(b) If f g is differentiable at x = 1, then g is continuous at
x=1
(c) If g is differentiable at x = 1, then f g is differentiable
at x = 1
(d) If f g is differentiable at x = 1, then g is differentiable
at x = 1
28. Let f : R be a function. We say that f has
f (h ) − f (0)
exists and is finite, and
|h|
f (h ) − f (0)
PROPERTY 2 if lim
exists and is finite.
h→ 0
h2
Then which of the following options is/are correct?
PROPERTY 1 if lim
h→ 0
(2019 Adv.)
(a) f (x) = sin x has PROPERTY 2
/
(b) f (x) = x23
has PROPERTY 1
(c) f (x) = |x|has PROPERTY 1
(d) f (x) = x|x|has PROPERTY 2
1
15
(c) h(0) = 16
(b) h ′(1) = 666
(d) h ( g (3)) = 36
32. Let a , b ∈ R and f : R → R be defined by
f (x) = a cos (|x3 − x|) + b|x|sin (|x3 + x|). Then, f is
(2016 Adv.)
(a) differentiable at x = 0, if a = 0 and b = 1
(b) differentiable at x = 1, if a = 1 and b = 0
(c) not differentiable at x = 0, if a = 1and b = 0
(d) not differentiable at x = 1, if a = 1and b = 1
33. Let f : − , 2 → R and g : − , 2 → R be functions
 2 
 2 
1
1
defined by f (x) = [x2 − 3] and g (x) =|x| f (x) + |4x − 7| f (x),
where [ y] denotes the greatest integer less than or equal
(2016 Adv.)
to y for y ∈ R. Then,
1
(a) f is discontinuous exactly at three points in  − , 2
 2 
1
(b) f is discontinuous exactly at four points in  − , 2
 2 
1
(c) g is not differentiable exactly at four points in  − , 2
 2 
1
(d) g is not differentiable exactly at five points in  − , 2
 2 
34. Let g : R → R be a differentiable function with
g (0) = 0, g′ (0) = 0 and g′ (1) =/ 0.
 x
g (x), x =/ 0

Let f (x) = |x|
x=0
 0 ,
(2015 Adv.)
Limit, Continuity and Differentiability 197
and h (x) = e|x| for all x ∈ R. Let ( foh ) (x) denotes f { h (x)}
and (hof )(x) denotes h { f (x)}. Then, which of the
following is/are true?
(a) f is differentiable at x = 0
(b) h is differentiable at x = 0
39. If f (x) = min { 1, x2, x3 }, then
(c) foh is differentiable at x = 0
(d) hof is differentiable at x = 0
35. Let f, g : [−1 , 2] → R be continuous functions which are
twice differentiable on the interval (−1, 2). Let the
values of f and g at the points −1, 0 and 2 be as given in
the following table:
x= −1
x=0
x=2
f (x)
3
6
0
g (x)
0
1
−1
In each of the intervals (−1, 0) and (0, 2), the function
( f − 3 g )″ never vanishes. Then, the correct statement(s)
is/are
(2015 Adv.)
(a) f ′ (x) − 3 g ′ (x) = 0 has exactly three solutions in
(−1, 0) ∪ (0, 2)
(b) f ′ (x) − 3 g ′ (x) = 0 has exactly one solution in (−1 , 0)
(c) f ′ (x) − 3 g ′ (x) = 0 has exactly one solution in (0, 2)
(d) f ′ (x) − 3 g ′ (x) = 0 has exactly two solutions in (−1, 0)
and exactly two solutions in (0, 2)
36. Let f : [a , b] → [1, ∞ ) be a continuous function and

, if
0
 x
g : R → R be defined as g (x) = ∫ f (t )dt , if
a
 b
∫ f (t )dt , if
 a
Then,
(a)
(b)
(c)
(d)
(a) f (x) is differentiable only in a finite interval containing
zero
(b) f (x) is continuous for all x ∈ R
(c) f ′ (x) is constant for all x ∈ R
(d) f (x) is differentiable except at finitely many points
x< a
a ≤ x ≤ b.
x> b
(2013)
g (x) is continuous but not differentiable at a
g (x) is differentiable on R
g (x) is continuous but not differentiable at b
g (x) is continuous and differentiable at either a or b
but not both
π
π

−x− , x≤−

2
2

π
37. If f (x) = − cos x, − < x ≤ 0, then
2

0 < x≤1
 x − 1,
 ln x,
x>1
π
2
(b) f (x) is not differentiable at x = 0
(c) f (x) is differentiable at x = 1
3
(d) f (x) is differentiable at x = −
2
(2006, 3M)
(a) f (x) is continuous everywhere
(b) f (x) is continuous and differentiable everywhere
(c) f (x) is not differentiable at two points
(d) f (x) is not differentiable at one point
40. Let h (x) = min { x, x2} for every real number of x, then
(a) h is continuous for all x
(b) h is differentiable for all x
(c) h ′ (x) = 1, ∀ x > 1
(d) h is not differentiable at two values of x
| x − 3|,
3x 13
 4 − 2 + 4 ,
41. The function f (x) =  x2
(1998, 2M)
x≥1
x<1
is (1988, 2M)
(a) continuous at x = 1
(b) differentiable at x = 1
(c) discontinuous at x = 1
(d) differentiable at x = 3
42. The function f (x) = 1 + |sin x|is
(1986, 2M)
(a) continuous no where
(b) continuous everywhere
(c) differentiable at x = 0
(d) not differentiable at infinite number of points
43. If x + | y| = 2 y, then y as a function of x is
(1984, 2M)
(a) defined for all real x
(b) continuous at x = 0
(c) differentiable for all x
dy 1
(d) such that
= for x < 0
dx 3
Assertion and Reason
For the following questions, choose the correct answer
from the codes (a), (b), (c) and (d) defined as follows.
(2011)
(a) f (x) is continuous at x = −
38. Let f : R → R be a function such that
f (x + y) = f (x) + f ( y), ∀x, y ∈ R. If f (x) is differentiable
(2011)
at x = 0, then
(a) Statement I is true, Statement II is also true;
Statement II is the correct explanation of Statement I
(b) Statement I is true, Statement II is also true;
Statement II is not the correct explanation of
Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
44. Let f and g be real valued functions defined on interval
(−1, 1) such that g′′ (x) is continuous, g (0) ≠ 0, g′ (0) = 0,
g′′ (0) ≠ 0, and f (x) = g (x)sin x .
Statement I lim [ g (x) cos x − g (0) cosec x] = f ′′ (0). and
x→ 0
Statement II f ′ (0) = g (0).
(2008, 3M)
198 Limit, Continuity and Differentiability
Match the Columns
45. In the following, [x] denotes the greatest integer less
than or equal to x.
Column I
A.
1
52. If f : [−1, 1] → R and f ′ (0) = lim nf   and f (0) = 0.
 
n→ ∞
n
2
−1  1 
Find the value of lim
(n + 1) cos   − n, given that
 n
n→ ∞π
Column II
x| x|
p.
continuous in (– 1, 1)
B.
| x|
q.
differentiable in (– 1, 1)
C.
x + [x ]
r.
strictly increasing (– 1, 1)
D.
| x − 1| +| x + 1|,
in ( −1, 1)
s.
not differentiable atleast at one
point in (– 1, 1)
 1  π

0 <  lim cos −1    < .
 n  2
n → ∞
53. Let α ∈ R. Prove that a function
(2004, 2M)
f : R → R is
differentiable at α if and only if there is a function
g : R → R which is continuous at α and satisfies
(2001, 5M)
f (x) − f (α ) = g (x) (x − α ) , ∀ x ∈ R.
(2007, 6M)
54. Determine the values of x for which the following
46. Match the conditions/expressions in Column I with
function fails to be continuous or differentiable
1 − x,
x<1


f (x) = (1 − x) (2 − x), 1 ≤ x ≤ 2 . Justify your answer.

3 − x,
x>2

(1997, 5M)
statement in Column II
(1992, 2M)
Column I
Column II
A.
sin (π [x])
p.
differentiable everywhere
B.
sin{ π (x − [x])}
q.
no where differentiable
r.
not differentiable at 1 and −1
 −  1 + 1 

55. Let f (x) = x e  |x | x  , x ≠ 0

, x=0
0

Test whether
Fill in the Blanks
47. Let F (x) = f (x) g (x) h (x) for all real x, where f (x), g (x)
and h (x) are differentiable functions. At same point
x0 , F ′ (x0 ) = 21 F (x0 ), f ′ (x0 ) = 4 f (x0 ), g ′ (x0 ) = − 7 g (x0 )
and h ′ (x0 ) = kh (x0 ), then k = K .
(1997C, 2M)
x
0
, x=0
the derivative from the right, f ′ (0+ ) = … and the
(1983, 2M)
derivative from the left, f ′ (0− ) = … .
1

2
49. Let f (x) = (x − 1) sin (x − 1) − |x|, if x ≠ 1 be a real
if x = 1
− 1,
valued function. Then, the set of points, where f (x) is
not differentiable, is …. .
(1981, 2M)
True/False
50. The derivative of an even function is always an odd
function.
(1997C, 5M)
56. Let f [(x + y) / 2] = { f (x) + f ( y)} / 2 for all real x and y, if
f ′ (0) exists and equals −1 and f (0) = 1 , find f (2).
(1995, 5M)
function
f : R → R satisfies the equation
f (x + y) = f (x) f ( y) , ∀ x , y in R and f (x) ≠ 0 for any x in
R. Let the function be differentiable at x = 0 and
f ′ (0) = 2 . Show that f ′ (x) = 2 f (x) , ∀ x in R. Hence,
determine f (x).
(1990, 4M)
57. A


48. For the function f (x) = 1 + e1/ x , x ≠ 0 ;

(i) f (x) is continuous at x = 0.
(ii) f (x) is differentiable at x = 0.
(1983, 1M)
Analytical & Descriptive Questions
1

−1  x + c
b sin  2  , − 2 < x < 0

1
51. f (x) = 
,
x=0
2

a x/ 2
1
−1
 e
0<x<
,

2
x
1
If f (x) is differentiable at x = 0 and|c| < , then find the
2
value of a and prove that 64b2 = (4 − c2).
(2004, 4M)
58. Draw a graph of the function
y = [x] + |1 − x|, − 1 ≤ x ≤ 3.
Determine the points if any, where this function is not
differentiable.
(1989, 4M)
59. Let R be the set of real numbers and f : R → R be such
that for all x and y in R, f (x) − f ( y)|2 ≤ (x − y)3 . Prove
(1988, 2M)
that f (x) is a constant.
60. Let f (x) be a function satisfying the condition
f (− x) = f (x), ∀ x. If f ′ (0) exists, find its value. (1987, 2M)
61. Let f (x) be defined in the interval [− 2, 2] such that
 −1 , −2 ≤ x ≤ 0
f (x) = 
x − 1 , 0 < x ≤ 2
g (x) = f (| x|) + | f (x)|.
and
Test the differentiability of g (x) in (− 2, 2).
(1986, 5M)
62. Let f (x) = x − x − x + 1
3
2
= max { f (t ) ; 0 ≤ t ≤ x}, 0 ≤ x ≤ 1
g (x) 
= 3 − x, 1 < x ≤ 2
Discuss the continuity and differentiability of the
function g (x) in the interval (0, 2).
(1985, 5M)
and
Limit, Continuity and Differentiability 199
x −1

, when x ≠ 1
2x2 − 7x + 5
63. Find f ′ (1) , if f (x) = 
.
1

when x = 1
− ,

3
(1979, 3M)
64. If f (x) = x tan −1 x , find f ′ (1) from first principle.
(1978, 3M)
Integer & Numerical Answer Type Questions
65. Let the functions f : (− 1, 1) → R and g : (− 1, 1) → (− 1, 1)
be defined by
f (x) = |2x − 1| + |2x + 1|and g (x) = x − [x],
where [x] denotes the greatest integer less than or equal
to x. Let f o g: (− 1, 1) → R be the composite function
defined by ( f o g )(x) = f ( g (x)). Suppose cis the number of
points in the interval (− 1, 1) at which f o g is NOT
continuous, and suppose d is the number of points in the
interval (− 1, 1) at which f o g is NOT differentiable.
(2020 Adv.)
Then the value of c + d is ……… .
66. Let f : R → R be a differentiable function such that
 π
f (0) = 0, f   = 3 and f′ (0) = 1.
 2
If g (x) = ∫
π
2
x
[ f ′ (t ) cosec t − cot t cosec t f (t )] dt
 π
for x ∈ 0, , then lim g (x) =
 2 
x→ 0
(2017 Adv.)
67. Let f : R → R and g : R → R be respectively given by
f (x) =|x|+ 1 and g (x) = x2 + 1. Define h : R → R by
max{ f (x), g (x)}, if x ≤ 0 .
h (x) = 
 min{ f (x), g (x)}, if x > 0 .
The number of points at which h (x) is not differentiable
(2014 Adv.)
is
68. Let p (x) be a polynomial of degree 4 having extremum
p (x) 

at x = 1 , 2 and lim 1 + 2 = 2 . Then, the value of p(2)
x→ 0 
x 

is ……… .
(2010)
Topic 8 Differentiation
Objective Questions I (Only one correct option)
1. If (a + 2b cos x) (a − 2b cos y) = a 2 − b2, where a > b >
0, the
dx  π π 
at  ,  is
dy  4 4 
a−b
(a)
a+ b
a+ b
(b)
a−b
(2020 Main, 4 Sep I)
2a + b
(c)
2a − b
(d)
a − 2b
a + 2b
C1 + (22) 20C 2 + (32) 20C3 + ..... + (202)20C 20 = A (2β )
, then the ordered pair ( A , β) is equal to
2. If
20
(2019 Main, 12 April II)
(a) (420, 19) (b) (420, 18) (c) (380, 18) (d) (380, 19)
 sin x − cos x 
x
3. The derivative of tan −1 
 , with respect to ,
 sin x + cos x
2

 π
where  x ∈ 0,   is
 2

(2019 Main, 12 April II)
(a) 1
(b)
2
3
(c)
1
2
(d) 2
 dy d 2y
4. If e + xy = e, the ordered pair  , 2  at x = 0 is equal
 dx dx 
y
to
(2019 Main, 12 April I)
1 1
1
1
1 1
(a)  , − 2  (b)  − , 2  (c)  , 2 
 e e 
e
e e 
e 
f (1) = 1, f′ (1) = 3, then
f ( f ( f (x))) + ( f (x))2 at x = 1 is
5. If
(a) 12
(b) 9
1
1
(d)  − , − 2 
 e
e 
the
derivative
of
(2019 Main, 8 April II)
(c) 15
(d) 33
2

 3 cos x + sin x 
dy
 π
6. If 2 y =  cot−1 
is
  , x ∈ 0,  then
 2
dx
 cos x − 3 sin x 

equal to
(a)
π
−x
6
(b) x −
π
6
(c)
π
−x
3
(2019 Main, 8 April I)
(d) 2x −
π
3
7. For x > 1, if (2x)2y = 4e2x − 2y , then (1 + log e 2x)2
equal to
(2019 Main, 12 Jan I)
x log e 2x + log e 2
(a)
x
(c) x log e 2x
x log e 2x − log e 2
(b)
x
(d) log e 2x
8. If x log e (log e x) − x2 + y2 = 4( y > 0), then
equal to
(a)
(c)
dy
is
dx
e
(b)
4 + e2
(1+ 2e )
(d)
4 + e2
(2e − 1)
dy
at x = e is
dx
(2019 Main, 11 Jan I)
2 4 + e2
(1+ 2e )
2 4 + e2
be
a
function
such
that
f :R→ R
f (x) = x3 + x2f ′ (1) + xf ′ ′ (2) + f ′′′ (3), x ∈ R.
(2019 Main, 10 Jan I)
Then, f (2) equals
9. Let
(b) − 4
(d) 8
(a) 30
(c) − 2
10. If x = 3 tan t and y = 3 sec t, then the value of
t=
π
, is
4
1
(a)
6
d 2y
at
dx2
(2019 Main, 9 Jan II)
(b)
1
6 2
(c)
1
3 2
(d)
3
2 2
1
11. For x ∈ 0,  , if the derivative of
 4
 6x x 
tan −1 
 is x ⋅ g (x), then g (x) equals
 1 − 9x3 
(2017 Main)
(a)
9
1 + 9x3
(b)
3x x
1 − 9x3
(c)
3x
1 − 9x3
(d)
3
1 + 9x3
200 Limit, Continuity and Differentiability
12. If g is the inverse of a function f and f ′ (x) =
1
, then
1 + x5
g′ (x) is equal to
(a) 1 + x5
(b) 5 x4
13. If y = sec (tan
(a)
(2015)
1
2
−1
(c)
1
(d) 1 + { g (x)}5
1 + { g (x)}5
dy
at x = 1 is equal to
x), then
dx
1
(b)
2
(c) 1
(2013)
(d)
2
14. Let g (x) = log f (x), where f (x) is a twice differentiable
positive function on (0, ∞ ) such that f (x + 1) = x f (x).
1
 1

Then, for N = 1, 2, 3,... , g′′  N +  − g′′   is equal to
 2

2


1 1
1
(a) − 4 1 + +
+ ... +

9 25
(2 N − 1)2 



1 1
1
(b) 4 1 + +
+ ... +
2
9 25
(2 N − 1) 

(2008, 3M)
 d 2y 
(b) −  2 
 dx 
 d 2 y  dy − 2
(c)  2   
 dx   dx 
−1
 dy 
 
 dx 
−3
differentiable function and g (x) = f ′ (x).
(c) 10
(2006, 3M)
f (1) = 1, f (2) = 4, f (3) = 9, then
18. If y is a function of x and log (x + y) = 2xy, then the value
of y′ (0) is
(2004, 1M)
(c) 2
19. If x2 + y2 = 1 , then
3
x
20. Let f (x) = 6
p
Then,
(d) 0
(2000, 1M)
(a) yy′ ′ − 2 ( y ′ )2 + 1 = 0
(c) yy′ ′ + ( y ′ )2 − 1 = 0
(b) yy′ ′ + ( y ′ )2 + 1 = 0
(d) yy′ ′ + 2 ( y ′ )2 + 1 = 0
sin x cos x
−1
p2
0
p3
d3
f (x) at x = 0 is
dx3
22. If x exy = y + sin 2 x, then at x = 0,
dy
= ...… .
dx
(1996, 2M)
23. Let f (x) = x|x|. The set of points, where f (x) is twice
differentiable, is … .
(1992, 2M)
24. If f (x) = |x − 2|and g (x) = f [ f (x)] , then g ′ (x) = K… for
−1 
(1990, 2M)
that fr (a ) = gr (a ) = hr (a ), r = 1, 2, 3
f1 (x) f2(x) f3 (x)
and
F (x) = g1 (x) g2(x) g3 (x) ,
(1985, 2M)
 2x − 1
dy
28. If y = f  2
= ....... .
 and f ′ (x) = sin 2 x, then
dx
 x + 1
(1982, 2M)
29. If y =
ax2
bx
c
+
+ 1,
+
(x − a ) (x − b) (x − c) (x − b) (x − c) (x − c)
Prove that
(2005, 2M)
(a) f ′ ′ (x) = 2, ∀ x ∈ (R )
(b) f ′ (x) = 5 = f ′ ′ (x), for some x ∈ (1, 3)
(c) there exists atleast one x ∈ (1, 3) such that f ′ ′ (x) = 2
(d) None of the above
(b) −1
Fill in the Blanks
(d) 25
17. Let f be twice differentiable function satisfying
(a) 1
(b) P ′ ′ (x) ⋅ P ′ ′ ′ (x)
(d) a constant
Analytical & Descriptive Questions
2
  x 
  x 
If F (x) =  f    +  g    and F (5) = 5,
  2 
  2 
then F (10) is
(b) 5
(a) P ′ ′ ′ (x) + P ′ (x)
(c) P (x) P ′ ′ ′ (x)
then F ′ (x) at x = a is …… .
16. If f ′′ (x) = − f (x), where f (x) is a continuous double
(a) 0
(1988, 2M)
h1 (x) h2(x) h3 (x)
 d 2 y  dy − 3
(d) −  2   
 dx   dx 
2
d  3 d 2y
 equals
y
dx 
dx2 
27. If fr (x), gr (x), hr (x), r = 1, 2, 3 are polynomials in x such
(2007, 3M)
−1
2
1 
25. The derivative of sec  − 2  with respect to
 2x − 1
1
(1986, 2M)
1 − x2at x = is …… .
2
26. If f (x) = log x (log x), then f ′ (x) at x = e is ...... . (1985, 2M)
d 2x
equals
dy2
 d 2y 
(a)  2 
 dx 
21. If y2 = P (x) is a polynomial of degree 3, then
x > 2.


1 1
1
(c) − 4 1 + +
+ ... +
2
9 25
(2 N + 1) 



1 1
1
(d) 4 1 + +
+ ... +
2
9
25
(2 N + 1) 

15.
(b) p + p 2
(d) independent of p
(a) p
(c) p + p3
, where p is constant.
30. Find
y′ 1  a
b
c 
= 
+
+
.
y x  a − x b − x c − x
(1998, 8M)
dy
at x = − 1, when
dx
sin
(sin y)
π
x
2
+
3
sec−1 (2x) + 2x tan ln (x + 2) = 0.
2
(1991, 4M)
31. If x = sec θ − cos θ and y = secn θ − cos nθ, then show that
 dy
(x2 + 4)  
 dx
2
= n 2 ( y2 + 4).
(1989, 2M)
32. If α be a repeated roots of a quadratic equation f (x) = 0
and A (x), B(x) and C (x) be polynomials of degree 3, 4 and
A (x)
B(x)
C (x)
5 respectively, then show that A (α )
B(α ) C (α ) is
A ′ (α ) B ′ (α ) C ′ (α )
divisible by f (x), where prime denotes the derivatives.
(1997, 2M)
(1984, 4M)
Limit, Continuity and Differentiability 201
33. Find the derivative with respect to x of the function

 2x  
y = (log cos x sin x) (log sin x cos x)−1 + sin −1 
 
 1 + x2  

π
at x = .
(1984, 4M)
4
34. If (a + bx) ey/ x = x, then prove that
3
36. Let y = ex sin x + (tan x)x , find
dy
.
dx
(1981, 2M)
5x
dy
37. Given, y =
+ cos 2 (2x + 1), find .
2
dx
3 (1 − x)
(1980)
Integer & Numerical Answer Type Question
38. Let f : R → R be a continuous odd function, which
2
d 2y  dy

x
= x
− y .

dx2  dx
1
vanishes exactly at one point and f (1) = .
2
3
(1983, 3M)
Suppose that F (x) = ∫
35. Let f be a twice differentiable function such that
(1983, 3M)
x
–1
x
G (x) = ∫ t| f { f (t )}| dt
f ′ ′ (x) = − f (x) , f ′ (x) = g (x) and
h (x) = [ f (x) ]2 + [ g (x)]2
Find h (10), if h (5) = 11.
for
–1
lim
x→1
f (t ) dt for all x ∈ [−1 , 2] and
all
x ∈ [−1 , 2].
F (x) 1
 1
= , then the value of f   is
 2
G (x) 14
(2015 Adv.)
Answers
Topic 1
1.
5.
9.
13.
17.
21.
25.
28.
32.
34.
3. (d)
7. (a)
11. (b)
15. (c)
19. (b)
23. (a, c)
1
2
1
26. h 2hr − h ,
128r
2
7
29.
30. False
π
33. 0
a 2 cosα + 2a sin α
−1
35. (7)
36. (2)
3
(b)
(c)
(d)
(d)
(d)
(a)
2.
6.
10.
14.
18.
22.
(c)
(b)
(c)
(c)
(d)
(d)
4.
8.
12.
16.
20.
24.
(a)
(b)
(d)
(c)
(c)
–1
27. − 1
(d)
(c)
(b)
(c)
(1)
2.
6.
10.
14.
18.
(b)
(c)
(d)
e2
(2)
3.
7.
11.
15.
(d)
(d)
(c)
e5
2. (a)
6. 1
3. (b)
4.
8.
12.
16.
(a)
(c)
(c)
e2
4. (b,c)
2. (b)
6. (b)
1. (a)
2. (d)
3. (d)
4. (b)
5. (c)
6. (d)
7. (b)
8. (b, c)
9. (a,b, d)
10. x ∈ (– ∞,– 1 ) ∪ [ 0, ∞ ),[ − 1, 0 )
11. f and f ′′ are continuous and f ′ is discontinuous at x = {1, 2 }.
Topic 6
x + a + 1,
( x + a − 1 ) 2,

2. g{ f ( x )} =  2
x + b ,
( x − 2 ) 2 + b,
if x < −a
if a ≤ x < c
if 0 ≤ x ≤ 1
if x > 1
a = 1, b = 0
gof is differentiable at x = 0
4 − x , 2 < x ≤ 3

4. g ( x ) = 2 + x, 0 ≤ x ≤ 1, discontinuous at x = {1, 2 }
2 − x, 1 < x ≤ 2

Discontinuity of g at x = {1, 2 }
Topic 7
Topic 4
1. (a)
5. (c)
16. (d)
Topic 5
1. (b, c, d)
Topic 3
1. (c)
5. (1)
–3
1
, c = and b ∈ R
2
2
12. (5)
31. loge 4
Topic 2
1.
5.
9.
13.
17.
15. a =
3. (b)
7. (a,b,d)
4. (c)
8. (a, d)
9. (b,d)
10. f ( x ) = 4 − x 2
2
π
–π
11. a = , b = e 2/3 12. a = 8
13. a = ,b =
3
6
12
2 
 2 1
3  log  3 – 6 x, x ≤ 0
14. f ( x ) = 
1/ x
  1 + x  ,
x>0
  2 + x 
1.
5.
9.
13.
17.
21.
25.
29.
32.
36.
40.
(b)
(b)
(c)
(b)
(d)
(c)
(d)
(a,b,d)
(a,b)
(b, c)
(a, c, d)
2.
6.
10.
14.
18.
22.
26.
30.
33.
37.
41.
(a)
(a)
(a)
(c)
(d)
(d)
(a,b,d)
(b,c,d)
(b,c)
(a, b, c, d)
(a, b)
3.
7.
11.
15.
19.
23.
27.
31.
34.
38.
42.
(b)
(a)
(b)
(b)
(a)
(a)
(b,c,d)
(b,c)
(a,d)
(b, c)
(b, d)
If
4.
8.
12.
16.
20.
24.
28.
(c)
(a)
(d)
(a)
(d)
(a)
(b,c)
35. (b,c)
39. (a, d)
43. (a, b, d)
202 Limit, Continuity and Differentiability
44.
45.
46.
48.
9.
13.
17.
21.
(b)
(A) → p, q, r, s; (B) → p, s; (C) → r, s; (D) → p, s
(A) → p; (B) → r
47. (24)
+
−
f ′( 0 ) = 0, f ′( 0 ) = 1
49. x = 0
51. (a = 1 )
50. True
54. (1, 2 )
55. (i) Yes (ii) No
56. ( − 1 )
57. e 2x
58. { 0, 1, 2 )
2

52. 1 − 

π
60. f ′( 0 ) = 0
61. g ( x ) is differentiable for all x ∈ ( − 2, 2 ) − { 0, 1 )
2. (b)
6. (b)
3. (d)
7. (b)
(a)
(d)
(b)
x ∈ R − { 0}
12.
16.
20.
24.
(d)
(b)
(d)
1
27. 0
34. (11)
3
36. e x sin x (3 x 3 cos x 3 + sin x 3 ) + (tan x ) x [2 x cosec 2 x + log (tan x )]
5

– 2 sin ( 4 x + 2 ), x < 1
3 (1 – x ) 2
37.  –5
– 2 sin ( 4 x + 2 ), x > 1

2
3 ( x – 1 )
Topic 8
1. (b)
5. (d)
11.
15.
19.
23.
3
π π2 –3
−8
32
33.
+
loge 2 16 + π 2
68. p (2 ) = 0
67. (3)
(b)
(a)
(a)
1
1
25. –4
26.
e
–2 ( x 2 – x – 1 )
2  2x – 1 
28.
⋅ sin  2

 x + 1
(x 2 + 1)2
30.
62. g ( x ) is continuous for all x ∈ ( 0, 2 ) − {1 ] and g ( x ) is
differentiable for all x ∈ ( 0, 2 ) − {1 }
 2
1 π
64.  + 
63.  − 
65. (4)
 9
2 4
66. (2)
10.
14.
18.
22.
(c)
(a)
(c)
(c)
4. (b)
8. (b)
38. (7)
Hints & Solutions
Topic 1
0
∞
and Form
0
∞
1. Let
P = lim
x→ 0
x + 2 sin x
x + 2 sin x + 1 − sin x − x + 1
2
2
0

form
0

On rationalization, we get
(x + 2 sin x)
P = lim 2
x → 0 x + 2 sin x + 1 − sin 2 x + x − 1
× ( x2 + 2 sin x + 1 + sin 2 x − x + 1 )
…(iii)
⇒
1 − b =5 ⇒ b = −4
On putting value of ‘b’ from Eq. (iii) to Eq. (ii), we get
a = −3 ⇒ a + b = − 7
= lim ( x2 + 2 sin x + 1 + sin 2 x − x + 1 )
x→ 0
× lim
x→ 0
= 2 × lim
x→ 0
x + 2 sin x
x − sin 2 x + 2 sin x + x
2
x + 2 sin x
x2 − sin 2 x + 2 sin x + x
0

form

0
Now applying the L′ Hopital’s rule, we get
1 + 2 cos x
P = 2 × lim
x → 0 2 x − sin 2 x +2 cos x + 1
(1 + 2)
[on applying limit]
=2
0 −0 + 2 + 1
3
=2 × =2
3
x + 2 sin x
=2
⇒ lim
x→ 0
x2 + 2 sin x + 1 − sin 2 x − x + 1
x2 − ax + b
=5
x→1
x−1
2. It is given that lim
Since, limit exist and equal to 5 and denominator is
zero at x = 1 , so numerator x2 − ax + b should be zero at
x = 1,
So
…(ii)
1 − a + b =0 ⇒ a =1 + b
On putting the value of ‘a’ from Eq. (ii) in
Eq. (i), we get
x2 − (1 + b) x + b
(x2 − x) − b(x − 1)
lim
= 5 ⇒ lim
=5
x→1
x→1
x−1
x−1
(x − 1) (x − b)
⇒
lim
= 5 ⇒ lim (x − b) = 5
x→1
x→1
x−1
…(i)
3. Given,
lim
x→1
= lim
x→ k
(x − 1)(x + 1)(x2 + 1)
x4 − 1
x3 − k3
⇒ lim
= lim 2
2
x→1
x −1
x − 1 x→ k x − k
8
3k2
(x − k)(x2 + k2 + xk)
⇒ k=
⇒ 2 ×2 =
2k
(x − k)(x + k)
3
4. Given limit is lim
x→ 0
= lim
sin 2 x
x
x→ 0
2 − 2 cos
2
sin 2 x
= lim
x→ 0
x

2 1 − cos 

2
sin 2 x
2 − 1 + cos x
0

form
0

x

Q 1 + cos x = 2 cos 2

2 
Limit, Continuity and Differentiability 203
sin 2 x
= lim
 x
2 × 2 sin 2 
 4
x→ 0
= lim
x→ 0
x
x

Q 1 − cos = 2 sin 2

2
4 
x2
 x
2 2 
 4
16
=
=4 2
2 2
2
[lim sin x = lim x]
x→ 0
x→ 0
cot3 x − tan x
x → π /4
π

cos  x + 

4
5. Given, limit = lim
1 − tan 4 x
= lim
y→ 0
y ( 1 + 1 + y + 2 ) ( 1 + y + 1)
4
4
1
= lim ⋅
h→ 0 8
=
2 (sec2 x)
cos 2 x − sin 2 x
×
x → π / 4 cos x − sin x
cos 2 x tan3 x
[Q 1 + tan 2 x = sec2 x]
1
= lim
8 h→ 0
2 sec4 x
(cos x − sin x) (cos x + sin x)
= lim
×
x → π /4
(cos x − sin x)
tan3 x
2
2
[Q (a − b ) = (a − b) (a + b)]
x → π /4
=
2 sec4 x
(cos x + sin x)
tan3 x
2 ( 2)  1
1
+


 2
2
(1)3
=
2
×
1 + 1 + y4 + 2
11.
PLAN
1 + 1 + y4 + 2
[rationalising the numerator]
= lim
y→ 0
= lim
y→ 0
(1 + 1 + y4 ) − 2
y4 ( 1 + 1 + y 4 + 2 )
1 + y4 − 1
y4 ( 1 + 1 + y4 + 2 )
[Q (a + b) (a − b) = a − b ]
2
×
sin θ


Q lim
=1

 θ→ 0 θ
2 sin 2 x(3 + cos x)
(1 – cos 2x)(3 + cos x)
= lim
tan 4x
x→ 0
x→ 0
x tan 4x
x×
× 4x
4x
(3 + cos x)
2 sin 2 x
1
= lim
× lim
×
tan 4x
x→ 0
x→ 0
4
x2
lim
x→ 0
4x
4
sin
θ
tan


= 2 × ×1
Q lim
= 1 and lim
=1


θ
θ
→
0
→
0
4
θ
θ

=2
y4
y
[Q sin (π − θ ) = sin θ ]
10. We have, lim
1 + 1 + y4 − 2
4
h3 cos h
 sin 2 x
sin π sin 2 x
× (π )  2  = π
2
x→ 0
π sin x
 x 
x
tan x 

Q lim
= 1 = lim
 x→ 0 sin x
x→ 0
x 
1 + 1 + y4 − 2
 h
sin h ⋅ sin 2 
 2
sin(π cos 2 x)
sin π (1 − sin 2 x)
= lim
2
x→0
x→ 0
x
x2
sin(π − π sin 2 x)
= lim
x→ 0
x2
= lim
 x   tan 2x 4

 .
 .
 sin x  2x  1
y→ 0
y→ 0
cos h ⋅ h3
sin(π sin 2 x)
x→ 0
x2
2
1
4
⋅ 1 ⋅1 ⋅1 ⋅ = 1
4
1
= lim
1
= lim
4 h→ 0
= lim
1 4x
4 (tan 4x)
7. Clearly, lim
h

sin h 2 sin 2 

2
h

sin 
1
 sin h  
2 ⋅ 1 ⋅ 1 = 1 × 1 = 1
= lim 
 
4 h → 0 h   h  cos h 4 4 4 16
 2 
[on applying limit]
x cot 4x
x
1 tan 2 2x
= lim
lim
.
2
2
x→0 sin x. cot 2 x
x→0 tan 4 x sin 2 x
1
2
2
1 4x
x
tan 2x
= lim
.
x→0 4 (tan 4 x) sin 2 x
x2
x→ 0
π

π

cos  − h 1 − sin  − h 
2

2

3

π
 π π
sin  − h  − + h

2
 2 2
9. lim
4
= lim
1
2 2 ×2
2
 2
= 4 2   = 8.
 2
6.
=
sin h (1 − cos h )
1
lim
8 h→ 0
cos h ⋅ h3
= lim
= lim
4
(by cancelling y4 and then by direct substitution).
1
.
=
4 2
1 cos x(1 − sin x)
cot x − cos x
8. lim
= lim ⋅
3
x → π/ 2 8
x→ π / 2 (π − 2 x)3
π

sin x − x
2

1 

Q cot x =

tan x 
1
×
= lim
3
x → π /4 1
(cos x − sin x) tan x
2
2 (1 + tan 2 x)
(1 − tan 2 x)
= lim
×
x → π / 4 cos x − sin x
tan3 x
y4
2
1 + y4 + 1
1 + y4 + 1
[again, rationalising the numerator]
 ∞  form
 
∞
a0x + a1x
n
lim
x→∞
n−1
+ K + an
b 0x m + b1x m − 1 + K + b m
 0,
 a0
 ,
=  b0
 + ∞,
 − ∞,

if n< m
if n = m
if n> mand a0b 0 > 0
if n>m and a0b 0 < 0
Description of Situation As to make degree of
numerator equal to degree of denominator.

 x2 + x + 1
lim 
− ax − b = 4
∴
x→ ∞ 
x+1

204 Limit, Continuity and Differentiability
⇒
lim
x→ ∞
x2 + x + 1 − ax2 − ax − bx − b
=4
x+1
= lim
x→ 0
x2(1 − a ) + x(1 − a − b) + (1 − b)
lim
=4
x→ ∞
x+1
⇒
1 − 1 + tan 2 x 
2x tan x 

2
 1 − tan x 
= lim
x→ 0
4 sin 4 x
Here, we make degree of numerator
= degree of denominator
∴
3
1−a =0
⇒ a =1
x (1 − a − b) + (1 − b)
lim
=4
x→ ∞
x+1
and
⇒
2x tan3 x
1
= lim
x → 0 2 sin 4 x (1 − tan 2 x)
x → 0 2 sin 4 x (1 − tan 2 x)
b = − 4 [Q (1 − a ) = 0]
12. Here, lim
h→ 0
x→ 0

x
x2
1 + x +
+ K
2
3!

2
 x
2   xn − 3
 2
In trigonometry try to make all trigonometric functions in
same angle. It is called 3rd Golden rule of trigonometry.
= lim
x→ 0
2 tan x
− 2x tan x
1 − tan 2 x
(2 sin 2 x)2
16. LHL = lim−
x→1
= lim
x → 1−
1 − cos 2 (x − 1)
x −1
2 sin 2 (x − 1)
|sin (x − 1)|
= 2 lim
x −1
x −1
x → 1−
Put x = 1 − h , h > 0, for x → 1− , h → 0
sin h
|sin (− h )|
= 2 lim
= 2 lim
=− 2
h
→
0
h→ 0
−h
−h
Again, RHL = lim
x→1
+
= lim
x →1+
1 − cos 2 (x − 1)
x−1
|sin (x − 1)|
2
x−1
x = 1 + h, h > 0
= lim 2
4
Above limit is finite, if n − 3 = 0, i.e. n = 3.
x tan 2x − 2x tan x
15. lim
x→ 0
(1 − cos 2x)2
x
1
1 ⋅ (1)3
1
lim
=
=
4
2 x → 0  sin x 4
2(1) (1 − 0) 2
2
 (1 − tan x)

 x 
For x → 1 , h → 0



x
x
− ...

 1 − +
2! 4 !



2 x 
 −2 sin  



x2 x3
2 
+
+ ... 
− 1 + x +

2! 3!
 

= lim
n
x→ 0
x
2


x
x
x3
2


2
−
− ...
 − 2 sin   − x −

2 
2! 3!

= lim
2
x→ 0
 x
4   xn − 2
 2
= lim
3
+
2
sin 2
 tan x


 x 
Put
(cos x − 1) (cos x − ex )
x→ 0
xn
14. lim
NOTE
=
f (2h + 2 + h 2) − f (2)
f (h − h 2 + 1) − f (1)
[Q f ′ (2) = 6 and f ′ (1) = 4, given]
Applying L’Hospital’s rule,
{ f ′ (2h + 2 + h 2)} ⋅ (2 + 2h ) − 0 f ' (2) ⋅ 2
=
= lim
h → 0 { f ′ (h − h 2 + 1 )} ⋅ (1 − 2 h ) − 0
f ' (1) ⋅ 1
6 .2
[using f ′ (2) = 6 and f ′ (1) = 4]
=
=3
4 .1
{(a − n ) nx − tan x} sin nx
13. Given, lim
=0
x→ 0
x2
tan x  sin n x

⇒ lim (a − n ) n −
×n =0

x→ 0
x  nx
1
⇒ {(a − n ) n − 1} n = 0 ⇒ (a − n ) n = 1 ⇒ a = n +
n
 tan x
3
x
 ⋅x
 x 
= lim
1−a −b =4
⇒
1


−1
1 − tan 2 x 
4 sin 4 x
2x tan x
h→ 0
∴
LHL ≠ RHL.
Hence,
17. lim
x→ 0
At
and
|sin h |
sin h
= lim 2
= 2
h→ 0
h
h
lim
x→1
f (x) does not exist.
1
(1 − cos 2 x)
1 .|sin x|
2
= lim
x→ 0 2
x
x
x=0
1 . sin h
1
RHL = lim
=
h→ 0 2
h
2
1
1 . sin h
LHL = lim
=−
h→ 0 2
−h
2
Here,
RHL ≠ LHL
∴ Limit does not exist.
sin [x]
, [x] ≠ 0

18. Since, f (x) =  [x]
[x] = 0
 0,
⇒
sin [x]
, x ∈ R − [0, 1)

f (x) =  [x]
 0,
0 ≤ x<1
At x = 0,
RHL = lim 0 = 0
x→ 0 +
Limit, Continuity and Differentiability 205
sin [x]
sin [0 − h ]
= lim
h→ 0
[x]
[0 − h ]
x→ 0
sin (−1)
= lim
= sin 1
h→ 0
−1
Since, RHL ≠ LHL
(d) lim sec2( fn (x)) = lim (1 + tan 2 fn (x))
LHL = lim
x→ ∞
−
x→ ∞
∴ (d) statement is true.
23. L = lim
∴ Limit does not exist.
 1
2
n 
19. lim 
+
+K+

n→∞  1 − n 2
1 − n2
1 − n 2
1+2+3+K+ n
n (n + 1)
= lim
= lim
n→∞
n→∞ 2 (1 − n ) (1 + n )
(1 − n 2)
n
1
=−
⇒ lim
n→∞ 2 (1 − n )
2
20. Given, f (a ) = 2, f ′ (a ) = 1, g (a ) = − 1, g ′ (a ) = 2
∴
lim
x→ a
g (x) f (a ) − g (a ) f (x)
x−a
g ′ (x) f (a ) − g (a ) f ′ (x)
,
x→ a
1 −0
[using L’ Hospital’s rule]
= g ′ (a ) f (a ) − g (a ) f ′ (a ) = 2 (2) − (−1) (1) = 5
= lim
G (x) = − 25 − x2
21. Given,
G (x) − G (1)
G ′ (x) − 0
= lim
x→1
x−1
1 −0
[using L’ Hospital’s rule]
1
= G′ (1) =
24


2x

Q G (x) = − 25 − x2 ⇒ G ′ (x) =

2 25 − x2 
22. We have,
n


1
fn (x) = ∑ tan − 1 
 for all x ∈ (0, ∞ )
 1 + (x + j) (x + j − 1)
j=1
∴ lim
x→1
⇒
⇒
n
 (x + j) − (x + j − 1) 
fn (x) = ∑ tan − 1 

 1 + (x + j) (x + j − 1)
j=1
fn (x) = ∑ [tan − 1 (x + j) − tan − 1 (x + j − 1)]
−1
x2 1 x4
x2
+ ⋅ 3 + ... −
4
= lim 2a 8 a 4
x→ 0
x
Since, L is finite
⇒
2a = 4 ⇒ a = 2
1
1
L = lim
=
x → 0 8 ⋅ a3
64
∴
log (1 + 2h ) − 2 log (1 + h )
h2
Applying L’Hospital’s rule, we get
2
2
−
1 + 2h 1 + h
= lim
h→ 0
2h
2 + 2h − 2 − 4h
= lim
h → 0 2 h (1 + 2 h ) (1 + h )
24. lim
h→ 0
= lim
h→ 0
n = 0, ± 1, ± 2, K
sin x, x ≠ nπ ,
otherwise
 2,
25. Given, f (x) = 
{ f (x)}2 + 1 , f (x) ≠ 0, 2

4
g [ f (x)] = 
, f (x) = 0

5
, f (x) = 2

2
g [ f (x)] = (sin x) + 1, x ≠ , nπ = 0, ± 1, K
5 , x = nπ

lim g [ f (x)] = lim (sin 2 x) + 1 = 1
x→ 0
x→ 0
26. Given, P = 2 ( 2hr − h 2 + 2hr )
−1
A
h
−1
(c) fn (x) = tan (x + n ) − tan x
lim tan( fn (x)) = lim tan(tan − 1 (x + n ) − tan − 1 x)
x→ ∞
x→ ∞


n
⇒ lim tan( fn (x)) = lim tan  tan − 1

x→ ∞
x→ ∞

1 + nx + x2
n
= lim
=0
x → ∞ 1 + nx + x2
∴ (c) statement is false.
0

form
0

−1
= −1
(1 + 2h ) (1 + h )
+ (tan (x + 3) − tan (x + 2))
+ ... + (tan − 1 (x + n ) − tan − 1 (x + n − 1))
⇒ fn (x) = tan − 1 (x + n ) − tan − 1 x
This statement is false as x ≠ 0. i.e., x ∈ (0, ∞ ).
(b) This statement is also false as 0 ∉ (0, ∞ )
−1
x2
4 , a >0
x4
1 1 


 x2
 1 x2 2  2 − 1 x4
⋅ 4 − ... −
a − a ⋅ 1 − ⋅ 2 +
2
a
 2 a
 4


= lim
x→ 0
x4
Now,
fn (x) = (tan − 1 (x + 1) − tan − 1 x)
+ (tan − 1 (x + 2) − tan − 1 (x + 1))
a − a 2 − x2 −
x→ 0
∴
n
j=1
⇒
x→ ∞
= 1 + lim tan 2( fn (x)) = 1 + 0 = 1
r
B
h–r
and
D
C
Here, BD = r 2 − (h − r )2 = 2hr − h 2
1
∴
A = . 2BD. h = ( 2hr − h 2) h
2
206 Limit, Continuity and Differentiability
∴
lim
h→ 0
h
h

a 2 ⋅ 2 cos  a +  ⋅ sin

2
2
= lim
+ (2a + h ) sin (a + h )
h
h→ 0
2⋅
2
= a 2 cos a + 2a sin a
h 2hr − h 2
A
= lim
3
h→ 0
P
8 ( 2hr − h 2 + 2hr )3
= lim
h→ 0
h3/ 2 ( 2r − h )
8 h3/ 2 ( 2r − h + 2r )3
lim(x − sin x)1/ 2
x − sin x
33. lim
= x→ 0
x→ 0 x + cos 2 x
lim(x + cos 2 x)1/ 2
1
1
2r
=
= ⋅
8 ( 2r + 2r )3 128 r
x→ 0
1/ 2
sin x 
 
lim x 1 −

x→ 0  
x  
=
lim(0 + 1)1/ 2
1
1
 4
2
x4 sin + x2
 x sin + x 
x
x
27. lim 
 = lim
1 − x3
1 + |x |3  x → −∞
x → −∞ 


x→ 0
 x3 + x2 − 16x + 20

(x − 2)2

k
1
1
x −1

=−
 = lim
x
→
1
(
)(
)
(
x
−
)
3
x
x
2
5
−
−
1
2
5


34. lim 
x→1
x2 sin (βx)
=1
x → 0 αx − sin x
35. Here, lim


(βx)3
(βx)5
+
− K
x2  β x −
3!
5!


lim
=1
x→ 0


x3
x5
αx −  x −
+
− K
3! 5!


⇒
, if x ≠ 2
, if x = 2
Since, continuous at x = 2.
x3 + x2 − 16x + 20
,[using L’Hospital’s rule]
x→ 2
(x − 2)2
⇒ f (2) = lim
6x + 2
3x2 + 2x − 16
= lim
=7
x→ 2
x→ 2
2
2 (x − 2)
= lim
∴
k=7
29. lim (1 − x) tan
x→1
πx
2
x−1 = y
π

π 
∴ − lim y tan ( y + 1) = − lim y − cot  y 
2 
y→ 0
y→ 0 
2
π 

 2 2
 y
2
= lim 
⋅ =
y→ 0
 tan π y π π

2 
Put
x→ a
x→ 0
x→ 0
x→ a
(2x − 1)( 1 + x + 1)
x
= log e (2) ⋅ (2) = 2 log e 2 = log e 4
(a + h )2 sin (a + h ) − a 2 sin a
h→ 0
h
a 2[sin (a + h ) − sin a ]
= lim
h→ 0
h
h [2a sin (a + h ) + h sin (a + h )]
+
h
32. Here, lim
lim
x→ 0
(α − 1)x +
 ecos ( α n ) − e 
e
=−
m
α→ 0
2
α


2x − 1
1+ x+1
×
1 + x −1
1+ x+1
= lim
=1
x3
x5
+
−K
3! 5!
Limit exists only, when α − 1 = 0
⇒
α =1
3 2


x
β
β5 x4
+
− K
x3 β −
3!
5!


lim
∴
=1
2
x→ 0


x
1
3
x  −
− K
3! 5!

⇒
36. Given, lim 
may or may not exist. Hence, it is a false statement.
31. lim


β3 x2 β 5 x4
+
− K
x3 β −
3!
5!


⇒
6β = 1
From Eqs. (i) and (ii), we get
6(α + β ) = 6α + 6 β = 6 + 1 = 7
30. If lim [ f (x) g (x)] exists, then both lim f (x) and lim g (x)
x→ a
0 .0
=0
1

On dividing by x3 , we get
sin (1 / x) 1
+
1
x
1+0
x
=
= −1
lim
1
x → −∞
0 −1
−1
3
x
28. f (x) = 
=
n
⇒
lim
α→ 0
e { ecos( α ) −1 − 1} cos(α n ) − 1 − e
⋅
=
2
cos(α n ) − 1
αm
αn
−2 sin 2
 ecos( α n ) −1 − 1 
2 = − e /2
⇒ lim e 
 ⋅ lim
m
α→ 0  cos(α n ) − 1  α→ 0
α


 α n
sin 2 
 2  α 2n
−e
⋅ m=
e × 1 × (−2) lim
⇒
2n
α→ 0
2
4α
α
4
α 2n − m − e
⇒ e × 1 × − 2 × 1 × lim
=
α→ 0
4
2
For this to be exists, 2n − m = 0
m
⇒
=2
n
…(i)
…(ii)
Limit, Continuity and Differentiability 207
Topic 2 1∞ Form, RHL and LHL
 1 + f (3 + x) − f (3)
1. Let l = lim

x→ 0  1 + f (2 − x) − f (2 ) 
3.
[1 form]
 1 + f ( 2 − x ) − f ( 2) − 1 − f (3 + x ) + f (3 ) 
lim 

x(1 + f ( 2 − x ) − f ( 2))

+
x→ 0
= lim
x→ 0 +
 f ( 2 − x ) − f (3 + x ) + f (3 ) − f ( 2) 
lim 

x(1 + f ( 2 − x ) − f ( 2))

x→ 0 
On applying L’Hopital rule, we get

lim 


− f ′( 2 − x )− f ′(3 + x )
x→ 0  1 − xf ′( 2 − x ) + f ( 2 − x ) − f ( 2) 
l=e
On applying limit, we get
 − f ′( 2) − f ′(3 ) 


 1 − 0 + f ( 2) − f ( 2) 
l=e
 1 + f (3 + x) − f (3)
So, lim

x→ 0  1 + f (2 − x) − f (2 ) 
2. Let L =
L=
π − 2 sin
1−x
lim
x → 1−
lim
x → 1−
1
x
= e0 = 1
=
lim
x → 1−
−1
x
, then
π + 2 sin −1 x
=
lim
2 cos −1 x
×
−
x→ 1
x → 1−
1−x
=
lim 2 cos −1 x
1
−
1−x
2 π x→ 1
lim
×
1
2 2 =
2 π
2
π

 tan(π sin 2 x) π sin 2 x
.
= lim 
+ 1
2
2
+
π sin x
x→ 0 
x

π + 2 sin −1 x
π

Q sin −1 x + cos −1 x =

2 
1
π + 2 sin −1 x
lim

π
−1
Q x → 1− sin x = 2 


lim


θ
Q x → 0+ sin θ = 1


+
sin 2 x
tan (π sin 2 x)
. lim
+1
2
+
x→ 0
x2
π sin x
tan x

Q
lim
=1

x→ 0
x
sin x

=1
and xlim
→0
x

=π+1





and LHL
= lim
x→ 0
−
x → 0−
tan (π sin 2 x) + (|x| − sin (x [x])2
x2
tan (π sin 2 x) + (− x − sin(x (− 1))2
x2
 Q|x| = − x for x < 0

and [x] = − 1 for − 1 < x < 0


1
Put x = cos θ, then as x → 1− , therefore θ → 0+
lim
2θ
1
Now, L =
+
1 − cos θ
2 π θ→0
lim
2θ
θ
1

=
Q 1 − cos θ = 2 sin 2
+

θ
 
2 
2 π θ→0

2 sin  
 2
 θ
2⋅  
 2
1
=
⋅ 2 lim
 θ
2 π
θ→ 0 +
sin  
 2
=
x → 0+
= lim
1
1−x
tan (π sin 2 x) + x2
x2
= lim
x→ 0
=1
[on rationalization]
π

π − 2 − cos −1 x
2

tan(π sin 2 x) + (x − sin(x ⋅ 0))2
x2
 Q|x| = x for x > 0 
and [x] = 0 for 0 < x < 1


= π lim
π − 2 sin −1 x
π + 2 sin −1 x
×
1−x
π + 2 sin −1 x
lim π − 2 sin −1 x
×
=
x → 1−
1−x
tan(π sin 2 x) + (|x| − sin(x [x]))2
x2
RHL = lim
x→ 0 
=e
x → a−
At x = 0,
l=e
=e
lim f ( x) = lim f ( x)
x → a+
∞
1 + f (3 + x ) − f (3 ) 
1

lim  1 −
1 + f ( 2 − x ) − f ( 2) 
x→ 0 x 
⇒
Key Idea lim f ( x) exist iff
x→a
1
x
= lim
x → 0−
tan(π sin 2 x) + (x + sin(− x))2
x2
tan(π sin 2 x) + (x − sin x)2
x → 0−
x2
[Qsin (− θ ) = − sin θ]
2
2
2
 tan(π sin x) + x + sin x − 2x sin x
= lim 

x → 0− 
x2

= lim
 tan(π sin 2 x)
sin 2 x 2x sin x
= lim 
+1+
−

2
x → 0− 
x
x2
x2 
 tan (π sin 2 x) π sin 2 x
sin 2 x
sin x
.
+ 1+
= lim 
−2

2
2
x 
π sin x
x2
x → 0− 
x
π sin 2 x
tan(π sin 2 x)
. lim
+
2
−
π sin x
x→ 0
x→ 0
x2
sin 2 x
sin x
− 2 lim
1 + lim
−
x
x→ 0
x2
x → 0−
= π + 1 + 1 −2 = π
= lim
−
Q RHL ≠ LHL
∴ Limit does not exist
208 Limit, Continuity and Differentiability
 n  n n 
= −
 x  x  x 
15 15 
 1 1  2 2 
∴Given limit = lim x −   + −   + … −  
+ x
x  x 
x→ 0
x x x
 1  2 
15 
= lim (1 + 2 + 3+ ...+15) − x   +   + ... +  
+

x→ 0
 x 
x x
4. Given,
Similarly,

π
(1 − |x| + sin|1 − x|) sin  [1 − x]

2
lim
|1 − x|[1 − x]
x →1+
Put x = 1 + h , then
x → 1+ ⇒ h → 0+

π
(1 − |x| + sin|1 − x|) sin  [1 − x]

2
∴ lim
= lim
h→ 0
= 120 − 0 = 120
|1 − x|[1 − x]
x →1+
π

(1 − |h + 1| + sin|− h|) sin  [− h ]
2

7.
|− h|[− h ]
+
π

(1 − (h + 1) + sin h ) sin  [− h ]
2

= lim
+
[
−
]
h
h
h→ 0
 1 
1 − x( 1 + |1 − x|)
cos 

 1 − x
|1 − x|
f (x) =
Now, lim f (x) = lim
x→1 −
x→1 −
π

( − h + sin h ) sin  (− 1)
2

= lim
h (− 1)
h→ 0 +
and
h→ 0 +
8. Given,
x→ 0
−
x(− 1 − x) sin (− 1)
(Q lim [x] = − 1)
−x
x→ 0
x → 0−
− x(x + 1) sin(− 1)
= lim
= lim (x + 1)sin(− 1)
−x
x → 0−
x → 0−
= lim
−
= (0 + 1) sin (− 1) (by direct substitution)
(Qsin(− θ) = − sin θ)
= − sin 1
Key Idea Use property of greatest integer function [ x ] = x − { x }.
 1 
lim x
+
+   x
x→ 0
 
15 
2 
+ …+
 x 
 x 
We know, [x] = x − { x}
1  1 1 
∴
= −
 x  x  x 
 1 
1 − x(1 − 1 + x)
cos 

 1 − x
x−1
x → 0+
=e
x([x] − x) sin [x]
x([x] + | x|) sin [x]
= lim
−
| x|
−x
x→ 0
+
p = lim (1 + tan 2 x )
lim
(Q| x| = − x, if x < 0)
6.
x→1
x → 0+
tan 2 x
2x
1


sin h
 h
 sin h 
= lim 
= 1
 − lim+   = 1 − 1 = 0 Q lim+
+ 



h
h
h
h→ 0
h→ 0

 h→ 0
lim
+
 1 
= lim − (x + 1) ⋅ cos 
 , which does not exist.
+
 x + 1
x→1
 h
 sin h 
= lim 
 − lim+  
h→ 0 +  h 
h → 0  h
5.
lim f (x) = lim
x→1
(Q [x] = − 1 for − 1 < x < 0 and h → 0+ ⇒ − h → 0− )
(− h + sinh)
 − π
sin 

 2 
−h
 1 
1 − x(1 + 1 − x)
cos 

 1 − x
1−x
 1 
= lim (1 − x) cos 
 =0
 1 − x
x→1 −
(Q|− h| = h and|h + 1| = h + 1 as h > 0)
= lim


n 

Q 0 ≤  x  < 1, therefore


n 
n 
0 ≤ x  < x ⇒ lim+ x  = 0
x→ 0
x
x


∴
9.
PLAN
log p = log e2 =
1
2x
(1∞ form)
1
 tan x 
lim 

2x → 0+ 
x 
2
=e
1
= e2
1
2
To make the quadratic into simple form we should
eliminate radical sign.
Description of Situation As for given equation,
when a → 0 the equation reduces to identity in x.
i.e. ax2 + bx + c = 0, ∀ x ∈ R or a = b = c → 0
Thus, first we should make above equation independent
from coefficients as 0.
Let a + 1 = t 6. Thus, when a → 0, t → 1.
∴
(t 2 − 1) x2 + (t3 − 1)x + (t − 1) = 0
⇒ (t − 1) {(t + 1) x2 + (t 2 + t + 1) x + 1} = 0, as t → 1
2 x2 + 3 x + 1 = 0
⇒
2 x2 + 2 x + x + 1 = 0
⇒
(2x + 1) (x + 1) = 0
x = − 1, − 1 / 2
Thus,
or
and
lim α (a ) = − 1 / 2
a→ 0 +
lim β (a ) = − 1
a→ 0 +
Limit, Continuity and Differentiability 209
[1∞ form]
10. Here, lim {1 + x log (1 + b2)}1/ x
x→ 0
1
lim {x log (1 + b )} ⋅
x
x→ 0
 x + 6

x → ∞  x + 1
x+4
15. lim 
2

5 
= lim 1 +

x→ ∞ 
x + 1
=e
log (1 + b 2 )
=e
= (1 + b )
2
⇒
1 + b2
2b
sin 2 θ =
…(ii)
1
1/ 2
b ≥  b ⋅ 1 
 b
2
b+
By AM ≥ GM,
b2 + 1
≥1
2b
⇒

π

lim tan  + x 
4

x→ 0
1/ x
π


1/ x
tan + tan x
1 + tan x 


4
= lim 
= lim 


π
x → 0  1 − tan x 
x→ 0
1 − tan tan x 


4
[(1 + tan x)1/tan x ]tan x/ x
e1
= lim
=
= e2
x → 0 [(1 − tan x)−1/tan x ]− tan x/ x
e−1
17. The right hand limit
lim
x →0 +
sin 2 θ = 1
(1 − x)1/ x − e−1
xa
π
θ = ± , as θ ∈ (− π , π ]
2
1
11. Here, lim (sin x)1/ x + lim  
 
x→ 0
x→ 0
1
log  
 x
sin x
= 0 + lim e
 1 
x −

 x2 
x → 0 − cosec x cot x
e
=e
= lim
= e0 = 1
18.
PLAN lim
x→0
[using L’ Hospital’s rule]
e
−1
xa
sin x
=1
x
(1 +
x ) (1 − x )
1− x
1+

 sin (x − 1)
−a

 (x − 1)
lim 
sin(x − 1) 
x→1

1 +
(x − 1) 

lim y = e2
x
 x − 3
(1 − 3 / x)x e−3
= 2 = e−5
 = lim
x→∞ (1 + 2 / x)x
 x + 2
e
13. For x ∈ R , lim 
 1 + 5x
lim 

x → 0  1 + 3 x2 
 x x2 x3

−  +
+
+ ... 
3
4
2

sin (x − 1) + a (1 − x) 
Given, lim 

x → 1  (x − 1 ) + sin (x − 1 ) 


⇒ log  lim y = 2
 x→ 0 
14.
x
− e−1
a
The above limit will be non-zero, if a = 1. And at a = 1,
the value of the limit is
1
 1
= e−1  −  = −
 2
2e
f (1) 6
=
=
f (1) 3
2  1/ x
e−1 . e
x →0 +
1
⇒ log y = [log f (1 + x) − log f (1)]
x


1
⋅ f ′ (1 + x)
⇒ lim log y = lim 
x→ 0
x → 0  f (1 + x)

x→ ∞

 x x2 x3

− −
 2 3 − 4 − ...... 


+
= e−1 lim
1/ x
x→ 0
xa
x →0 +
x →0
sin x
tan x
x→ 0 x
lim
− e−1
e
= lim
 lim (sin x)1/ x → 0 
x → 0

∞
as, (decimal) → 0
 f (1 + x) 
12. Let y = 
 f (1) 
⇒
x

1 
x2 x3
−x−
−
− K
2
3
x 

x
log (1/ x )
lim
x → 0 cosec x
− e−1
a
x →0 +
Applying L’Hospital’s rule, we get
lim
1

 log e (1 − x )

e x
= lim
sin x
=e
x→ 0
= e5
…(iii)
From Eqs. (ii) and (iii),
⇒
5 ( x + 4)
x +1
[1∞ form]
1/ x
16.
(1 + b2) = 2b sin 2 θ
∴
=e
…(i)
Given, lim {1 + x log (1 + b2)}1/ x = 2b sin 2 θ
x→ 0
lim
x→ ∞
x+ 4
2 1/5 x 2 5
2
=
lim [(1 + 5x )
x→ 0
]
2 1/3 x 2 3
lim [(1 + 3x )
x→ 0
]
=
e5
= e2
e3
=
1
4
=
1
4
x
2
⇒
1
1 − a

 =
 2 
4
⇒
a = 2 or 0
⇒ (a − 1)2 = 1
Hence, the maximum value of a is 2.
210 Limit, Continuity and Differentiability
Topic 3 Squeeze, Newton-Leibnitz’s Theorem and Limit Based
on Coverting infinite Serie s into Definite Integrals
1. Given α and β are roots of quadratic equation
375x2 − 25x − 2 = 0
25
1
=
375 15
2
αβ = −
375
∴
α+β=
and
n
n
r =1
r =1
… (i)
… (ii)
∑ α r + nlim
∑ βr
n→ ∞
→∞
= (α + α 2 + α 3 + K + upto infinite terms)+
(β + β 2 + β3 + K + upto infinite terms)


a
Q S ∞ = 1 − r for GP


α
β
+
1 −α 1 −β
=
α (1 − β) + β (1 − α )
(1 − α ) (1 − β )
=
α − αβ + β − αβ
1 − α − β + αβ
=
(α + β ) − 2αβ
1 − (α + β ) + αβ
On substituting the value α + β =
1
−2
and αβ =
from
15
375
Eqs. (i) and (ii) respectively,
we get
1
4
+
1
29
29
= 15 375 =
=
=
1
2
−
−
375
25
2
348
12
1−
−
15 375

0
form
2

0
π
x2 −
16
f (sec2 x) 2 sec x sec x tan x
= lim
x→ π /4
2x
[using L’ Hospital’s rule]
2 f (2) 8
=
= f (2)
π /4
π
2
π
x→
4
= lim
n→ ∞
=∫
2
0
1
n
2n
∑
r =1
2n
∑
r =1
x
1 + x2
1
= lim
2
2
n→ ∞ n
n +r
r
2n
∑
r =1
r
n 1 + (r / n )2
1 + (r / n )
∴
⇒
∑
r =1

 r

⋅x + 1 

log  n2

r
2
 2 ⋅ x + 1

n
x
 1 + z  dz
log e { f (x)} = x ∫ log 

0
 1 + z 2 x
x
 1+ z
log e{ f (x)} = ∫ log 
 dz
0
 1 + z 2
Using Newton-Leibnitz formula, we get
 1 + x
1
⋅ f ′ (x) = log 

f (x)
 1 + x2 
2
dx = [ 1 + x2 ]2 = 5 − 1
0
n
tx=z
xdt = dz
Put,
⇒
Here, at x = 1 ,
r /n
x
Converting summation into definite integration, we get
1
 xt + 1 
log e{ f (x)} = x ∫ log  2 2
 dt
0
 x t + 1
∫ f (t ) dt
1
3. Let I = lim
n→ ∞ n
x
n
n
n 

 … x +  

2
n 
,x>0
2

n
n 2 
 K  x2 + 2  
4
n  

n

n
n 


 n n (x + n )  x +  K  x +  


2
n 
log e { f (x)} = lim log 
2

n →∞

n 
n 2 
2
2  2
 K  x2 + 2  
 n !(x + n )  x +
4
n  



n


1 

Π x +




r =1 
x
r /n

= lim ⋅ log 
n→ ∞ n
1  n

 n  2
Π (r / n ) 
x +
 rΠ
2 r = 1
=1 

(
/
)
r
n




n


x+
1 n
r


log
= x lim
∑
2


n→ ∞ n

n
r
2
r =1
x + 2

n
r 

 
1
= x lim
n→ ∞ n
sec 2 x
2. lim


 n n (x + n )  x +

f (x) = lim 
n→ ∞ 
2
2  2
 n ! (x + n )  x +


Taking log on both sides, we get
Now, lim
=
4. Here,
∴
f ′ (1)
= log (1) = 0
f (1)
f′ (1) = 0
… (i)
Limit, Continuity and Differentiability 211
Now, sign scheme of f ′ (x) is shown below
+
x2
–
6. lim
∫0
x→ 0
Applying L’Hospital’s rule, we get
∴ At x = 1 , function attains maximum.
Since, f (x) increases on (0, 1).
∴
f (1) > f (1 / 2)
∴ Option (a) is incorrect.
f (1 / 3) < f (2 / 3)
∴Option (b) is correct.
2 ⋅ cos 2(x2)
2
cos 2(x2) ⋅ 2x − 0
=
=1
= lim
sin x 1 + 1
x→ 0
x → 0 x cos x + sin x
cos x +
x
= lim
Topic 4 Continuity at a Point
1. Given function is
 2 cos x − 1

f (x) =  cot x − 1

k

Also,
f ′ (x) < 0, when x > 1
⇒
f′ (2) < 0
∴ Option (c) is correct.
 1 + x
f ′ (x)
Also,
= log 

f (x)
 1 + x2
= log (2 / 3) < 0
f ′ (3) f ′ (2)
<
f (3)
f (2)
4
lim
π
k=
x→
4
⇒
∴ Option (d) is incorrect.
5. We have,
1
yn = [(n + 1) (n + 2) … (n + n )]1/ n
n
and lim yn = L
n→ ∞
1
⇒ L = lim [(n + 1) (n + 2) (n + 3) … (n + n )]1/ n
n→ ∞ n
Put x =
2 cos x − 1
cot x − 1
π
π
+ h, when x → , then h → 0
4
4

π
2 cos  + h − 1
lim

4
k=
h→0

π
cot + h − 1

4
1
⇒
1 
2 
3
n  n


L = lim 1 +  1 +  1 +  ... 1 +  







n→∞ 
n
n
n
n 
⇒
log L = lim
n→ ∞
1
n
=
n 
1
2




log 1 + n  + log 1 + n  … log 1 + n  


⇒ log L = lim
n→ ∞
1
n
n
∑ log 1 + n 
r =1
=
1
⇒ log L = ∫ 1 × log (1 + x) dx
0 II
I

(log(1 + x) ∫ dx dx

[by using integration by parts]
1 x
1
dx
⇒ log L = [x log(1 + x)]0 − ∫
01 + x
⇒ log L = (x ⋅ log (1 + x))10 − ∫
1
d
0
 dx
1 x + 1
1 
⇒ log L = log 2 − ∫ 
−
 dx
0x+ 1
x + 1
⇒ log L = log 2 − [x]10 + [log(x + 1)]10
⇒ log L = log 2 − 1 + log 2 − 0
4
4
⇒ L=
⇒ log L = log 4 − log e = log
e
e
4 
=1
[L ] =
 e 
⇒
lim
h→0
2
1
 1
cos h −
sin
 2
2
cot h − 1
−1
cot h + 1

h −1

[Q cos (x + y) = cos x cos y − sin x sin y and
cot x cot y − 1
]
cot (x + y) =
cot y + cot x
r

π
4
π
, x=
4
,x≠
Q Function f (x) is continuous, so it is continuous at
π
x= .
4
 π
∴
f   = lim f (x)
 4  x→ π
f ′ (3) f ′ (2)
4
 3
−
= log   − log  
 10
 5
f (3)
f (2)
⇒
0

form
0

x sin x
x=1
∴
cos 2tdt
=
lim cos h − sin h − 1
−2
h→0
1 + cot h

lim  (1 − cos h ) + sin h
(sin h + cos h )

h→0 
2 sin h

h
h


2h
lim 2 sin 2 + 2 sin 2 cos 2

(sin h + cos h )
=
h
h
h→0
4 sin cos




2
2
h
h


lim sin 2 + cos 2

=
× (sin h + cos h )
h
h→0
 2 cos



2
⇒k=
1
2
212 Limit, Continuity and Differentiability
2.
NOTE All integers are critical point for greatest integer function.
6. For f (x) to be continuous, we must have
Case I When x ∈ I
f (0) = lim f (x)
x→ 0
log (1 + ax) − log (1 − bx)
= lim
x→ 0
x
a log (1 + ax) b log (1 − bx)
= lim
+
x→ 0
− bx
ax
log (1 + x)
[using lim
= a ⋅1 + b ⋅1
= 1]
x→ 0
x
f (x) = [x]2 − [x2] = x2 − x2 = 0
Case II When x ∉ I
If
0 < x < 1, then [x] = 0
and
0 < x2 < 1, then [x2] = 0
Next, if
1 ≤ x2 < 2 ⇒ 1 ≤ x < 2
⇒
[x] = 1
and
[x2] = 1
=a+b
Therefore, f (x) = [x]2 − [x2] = 0, if 1 ≤ x < 2
Therefore,
f (x) = 0, if 0 ≤ x < 2
This shows that f (x) is continuous at x = 1.
Therefore, f (x) is discontinuous in (−∞ , 0) ∪ [ 2 , ∞ ) on
many other points. Therefore, (b) is the answer.
∴
7. f (x) = x cos(π (x + [x]))
At x = 0
lim f (x) = lim x cos(π (x + [x]) = 0
x→ 0
3. Given, f (x) = [tan x]
Now, − 45° < x < 45°
tan (− 45° ) < tan x < tan (45° )
− tan 45° < tan x < tan (45° )
−1 < tan x < 1
0 < tan 2 x < 1 ⇒ [tan 2 x] = 0
8.
Case I Let f and g attain their common maximum
value at p.
⇒
f ( p) = g ( p),
where p ∈ [0, 1]
Case II Let f and g attain their common maximum
value at different points.
, −1 ≤ x < 0
⇒
f (a ) = M and g (b) = M
⇒
f (a ) − g (a ) > 0 and f (b) − g (b) < 0
, 0 ≤ x<1
⇒ f (c) − g (c) = 0 for some c ∈ [0, 1] as f and g are
continuous functions.
, 1 ≤ x<2
⇒ f (c) − g (c) = 0 for some c ∈ [0, 1] for all cases.
, 2 ≤ x<3
Also,
f (1) = 0
∴ Continuous at x = 1.
Similarly, when x = 2,
lim f (x) = lim f (x) = 0
x → 2−
which is true from Eq. (i).
Option (d) ⇒ f 2(c) − g 2(c) = 0 which is true from Eq. (i)
Now, if we take f (x) = 1 and g (x) = 1, ∀x ∈ [0, 1]
Options (b) and (c) does not hold. Hence, options (a) and
(d) are correct.
f (2n ) = a n , f (2n + ) = a n
9.
f (2n − ) = bn + 1
⇒
a n − bn = 1
Thus, function is discontinuous at no x.
f (2n + 1) = a n
Hence, option (c) is the correct answer.
f {(2n + 1)− } = a n
f {(2n + 1)+ } = bn + 1 − 1
5. Given, f (x) = x ( x + x + 1 )
⇒ f (x) would exists when x ≥ 0 and x + 1 ≥ 0.
⇒ f (x) would exists when x ≥ 0.
∴ f (x) is not continuous at x = 0,
because LHL does not exist.
Hence, option (c) is correct.
...(i)
Option (a) ⇒ f 2(c) − g 2(c) + 3 [ f (c) − g (c)] = 0
which shows RHL = LHL at x = n ∈ Integer as if x = 1
 2x − 1
lim cos 
⇒
 π = 0 and lim− 0 = 0
 2 
x →1+
x→1
x → 2+
PLAN If a continuous function has values of opposite sign inside an
interval , then it has a root in that interval.
f,g : [0, 1] → R
We take two cases.
i.e. f (x) is zero for all values of x from x = − 45° to 45°.
Thus, f (x) exists when x → 0 and also it is continuous at
x = 0 . Also, f (x) is differentiable at x = 0 and has a value
of zero.
Therefore, (b) is the answer.
2x − 1
4. Here, f (x) = [x] cos 
π
 2 

 2x − 1
− cos  2  π
0

∴
f (x) =  cos  2x − 1 π


 2 

x
2
−
1



2 cos  2  π

x →0
f (x) = 0
∴It is continuous at x = 0 and clearly discontinuous at
other integer points.
and
2
⇒
⇒
⇒
⇒
f (0) = (a + b)
⇒
a n = bn + 1 − 1 or a n − bn + 1 = − 1
a n − 1 − bn = − 1
or
10. Given,
or
x + y =4
2
2
⇒
f (x) = 4 − x2
y = 4 − x2
Limit, Continuity and Differentiability 213
 {1 + |sin x|} a/|sin x | , π/6 < x < 0

11. f (x) = 
b
x=0
,
tan 2x /tan 3 x

e
,
0
x < π /6
<

1/ x
⇒
⇒
Since, f (x) is continuous at x = 0.
∴
RHL (at x = 0) = LHL (at x = 0) = f (0)
⇒
lim etan 2 h/tan 3 h = lim {1 + |sin h |}a /|sin h| = b
⇒
∴
and
h→ 0
On differentiating both sides, we get
⇒
h→ 0
/
e 23
= ea = b
a = 2 /3
/
b = e 23
∴
f (0) = LHL
1 − cos 4h
a = lim
h→ 0
h2
⇒
f (−1) = − a + b = − a
2 1
 2 
− a =  − log   
∴
 3 
3 6
2 
2
  1
3  log  3 − 6 x, x ≤ 0

Thus, f (x) = 
1/ x
 1 + x

x>0

 ,

 2 + x
and
⇒
2 sin 2 2h 4
×
a = lim
h→ 0
4
h2
⇒
a =8
13. Since, f (x) is continuous for 0 ≤ x ≤ π
∴
⇒
⇒
Also,
π
π


RHL at x =  = LHL at x = 


4
4
π
π
 π
 π
2 ⋅ cot + b =  + a 2 ⋅ sin 
 4
 4
4
4
π
π
π
+ b= + a ⇒ a−b=
4
2
4
π
π


RHL at x =  = LHL at x = 


2
2
∴
…(ii)
⇒
b =0
Also,
f ′ (1) = f (−1)

 2 1 
log  3 − 6  x = 0


sin (a + 1) x + sin x
, x<0

x

15. Given, f (x) = 
, x=0
c
 (x + bx2)1/ 2 − x1/ 2
, x>0


bx3/ 2
is continuous at x = 0.
(1 + bx)1/ 2 − 1
=c
x→ 0
bx
= lim
(LHL at x = 0) = (RHL at x = 0)
 x + 1
lim (ax + b) = lim 

x→ 0
x→ 0  x + 2
2
3
=0
(LHL at x = 0) = (RHL at x = 0) = f (0)
 sin (a + 1) x sin x 
+
⇒ lim
x→ 0 
x
x 

x≤0
 ax + b,
1/ x

f (x) =  1 + x
 2 + x , x > 0

⇒
x→ 0
1/ x
⇒
Since, f (x) is continuous and f ′ (1) = f (−1)
∴
LHL = lim
x

x
Hence, f (x) is continuous for all x.
14. Let g (x) = ax + b be a polynomial of degree one.
⇒
1 +
RHL = lim 
x → 0 2 +
…(i)
−a−b=b
⇒
a + 2b = 0
On solving Eqs. (i) and (ii), we get
π
π
and b = −
a=
6
12
[from Eq. (i)]
Now, to check continuity of f (x) (at x = 0).
2π
π  π
π


− b sin  = 2 ⋅ ⋅ cot + b
⇒  a cos


2
2  2
2
⇒


 1 + x 
− 1 log 


 2 + x 
x
f ′ (x)

=
f (x)
x2

 1 + x 
x
− log 

1/ x 
 2 + x 
 1 + x
(1 + x)(2 + x)

f ′ (x) = 
 
 2 + x 

x2




2 1
 2 
f ′ (1) =  − log   
 3 
3 6
 1
1
x
−
+
x
+
1
2

12. Since, f (x) is continuous at x = 0.
∴
 1 + x
f (x) = 
 ,x>0
 2 + x
1
log f (x) = [log (1 + x) − log (2 + x)]
x
1/ x
…(i)
⇒
(a + 1) + 1 = lim
⇒
a+2=
∴
and
x→ 0
bx
1
⋅
=c
bx 1 + bx + 1
1
=c
2
3
1
a =− ,c=
2
2
b ∈R
214 Limit, Continuity and Differentiability
2x sin  1  − cos  1  , x ≠ 0

 
 
f4 ′ (x) = 
 x
 x

0,
x=0
2
16. (i) Given, f1 : R → R and f1 (x) = sin ( 1 − e−x )
Thus,
∴ f1 (x) is continuous at x = 0
1
2
Now, f1 ′ (x) = cos 1 − e− x .
2
−x
2 1− e
2
(2xe− x )
1
1 

Again, lim f ′ (x) = lim  2x sin   − cos   
x→ 0 
x→ 0
 x
 x 
At x = 0
does not exists.
f1 ′ (x) does not exists.
1
Since, lim cos   does not exists.
x→ 0
 x
∴ f1 (x) is not differential at x = 0
Hence, f ′ (x) is not continuous at x = 0.
Hence, option (2) for P.
∴ Option (3) for S.
 |sin x| , if x ≠ 0

(ii) Given, f2 (x) =  tan −1 x

if x = 0
1,
⇒
 − sin x
 tan −1 x

 sin x
f2 (x) = 
−1
 tan x

1


Topic 5 Continuity in a Domain
x< 0
1. Given ∫
x> 0
f ( x)
6
4t3 dt = (x − 2) g (x)
f ( x)
∫
g (x) = 6
⇒
x=0
4t3 dt
f ( x)
Clearly, f2 (x) is not continuous at x = 0.
So, lim g (x) = lim
x→ 2
∫6
x→ 2
4t3 dt
x−2
 0

Q form as x → 2 ⇒ f (2) = 6
 0

∴ Option (1)for Q.
(iii) Given, f3 (x) = [sin (log e (x + 2))] , where [ ] is G.I.F.
and f3 : (−1, eπ / 2 − 2) → R
It is given
− 1 < x < eπ / 2 − 2
⇒
⇒
⇒
x2 sin  1  , if x ≠ 0

 
(iv) Given, f4 (x) = 
 x
if x = 0
0,

 1
Now, lim f4 (x) = lim x2 sin   = 0
 x
x→ 0
x→ 0
 1
 1
f4′ (x) = 2x sin   − cos  
 x
 x
f (0 + h ) − f (0)
For x = 0, f4′ (x) = lim
h→ 0
h
1

h 2 sin   − 0
 h
⇒
f4′ (x) = lim
h→ 0
h
 1
f4′ (x) = lim h sin   = 0
⇒
 h
h→ 0
4( f (x))3 f ′ (x)
x→ 2
1
lim g (x) = lim
x→ 2
π/ 2
−1 + 2 < x+ 2 < e −2 + 2
1 < x + 2 < eπ/ 2
log e 1 < log e (x + 2) < log e eπ / 2
π
⇒
0 < log e (x + 2) <
2
π
⇒
sin 0 < sin log e (x + 2) < sin
2
⇒
0 < sin log e (x + 2) < 1
[sin log e (x + 2)] = 0
∴
∴ f3 (x) = 0, f ′3 (x) = f3 ′ ′ (x) = 0
It is differentiable and continuous at x = 0.
∴Option (4) for R
[provided x ≠ 2]
(x − 2)
 d
Q
 dx
φ 2( x )
∫φ
1 ( x)

f (t ) dt = f (φ 2(x)), φ 2′ (x) − f (φ1 (x)) ⋅ φ1′ (x)

On applying limit, we get
lim g (x) = 4( f (2))3 f ′ (2) = 4 × (6)3
x→ 2
1 4 × 216
=
= 18
48
48
1

Q f (2) = 6 and f ′ (2) =
48 

2. Given function
sin( p + 1)x + sin x

, x<0
x

f (x) = 
q
, x =0

2
x+x − x

, x>0

x3 / 2
is continuous at x =`0, then
f (0) = lim f (x) = lim f (x) …(i)
x→ 0 −
x→ 0 +
sin( p + 1)x + sin x
x
sin(ax)


Q lim
=a
= p+1+1=p+2
 x→ 0

x
lim f (x) = lim
x→ 0 −
x→ 0 −
and lim f (x) = lim
x→ 0 +
x→ 0 +
= lim
x→ 0 +
x + x2 − x
x3/ 2
x [(1 + x)1/ 2 − 1]
x x
Limit, Continuity and Differentiability 215
1 1



 − 1


 2
2
1
2
1 + x +
x + .... − 1
2
2!






= lim
+
x
x→ 0
5. Given function f : [−1, 3] → R is defined as
|x| + [x], −1 ≤ x < 1

f (x) =  x + |x|, 1 ≤ x < 2
 x + [x ], 2 ≤ x ≤ 3

− x − 1 , − 1 ≤ x < 0
 x,
0 ≤ x<1

1 ≤ x<2
=  2x,
 x + 2, 2 ≤ x < 3

x=3
 6,
[Q (1 + x)n
n (n − 1) 2 n (n − 1(n − 2)) 3
x +
x + ... ,|x|< 1]
1 ⋅2
1 ⋅2 ⋅3
1 1



 − 1


2

1
1
2
= lim  +
x + ... =
+ 2
2!
 2
x→ 0




= 1 + nx +
From Eq. (i), we get
1
1
f (0) = q = and lim f (x) = p + 2 =
2
2
x→ 0 −
3
p=−
2
 3 1
( p, q ) =  − , 
 2 2
⇒
So,
Q
x = 5.
…(i)
…(ii)
h→ 0
…(iii)
∴f (5) = lim f (x) = lim f (x)
x→5 −
[Q lim f (x) = lim f (x) = f (1)]
x→1 −
lim (b + 5x) = b + 25 = 30
[Q lim f (x) = lim
x →5−
[Q put x = 4 + h, when x → 4+ , then h → 0]
= lim(4 − 1) = 3
−
[Q put x = 4 − h, when x → 4 then h → 0]
= lim(3 − 0) = 3
h→ 0
4 
f (4) = [4] −
=4 −1 =3
4 
x→ 4 +
So, function f (x) is continuous at x = 4.
x→5 +
…(ii)
f (x) = f (5)]
On solving Eqs. (i) and (ii), we get b = 5 and a = 0
Now, let us check the continuity of f (x) at x = 3.
Here,
lim (a + bx) = a + 3b = 15
x →3−
and
h→ 0
x→ 4 −
x→1 +
x →5−
4 

4 + h 
Now, lim f (x) = lim  [4 + h ] −

h→ 0 
x→ 4 +
 4 
Q lim f (x) = f (4) = lim f (x) = 3
if
x≤1
if 1 < x < 3
if 3 ≤ x < 5
if
x≥5
x →1+
x
4. Given function f (x) = [x] −  , x ∈ R
and
[Q f (3) = 6]
Clearly, for f (x) to be continuous, it has to be continuous
at x = 1, x = 3 and x = 5
[Q In rest portion it is continuous everywhere]
…(i)
∴
lim (a + bx) = a + b = 5
⇒ a (5 − π ) + 1 = b(5 − π ) + 3
⇒
(a − b)(5 − π ) = 2
2
⇒
a−b=
5−π

4 − h 
and lim f (x) = lim [4 − h ] −
 4 
h→ 0 
x→ 4 −
lim f (x) = 5 ≠ f (3)
x→ 3 −
 5
a + bx
f (x) = 
 b + 5x
 30
and lim f (x) = lim[b| (5 + h ) − π| + 3]
x→5 +
[Q f (2) = 4]
x→ 2 +
We have,
h→ 0
= b(5 − π ) + 3
Q Function f (x) is continuous at x = 5.
lim f (x) = 4 = f (2) = lim f (x) = 4
x→ 2−
Key Idea A function is said to be continuous if it is continuous at
each point of the domain.
lim f (x) = lim[a|π − (5 − h )| + 1]
x→5 +
[Q f (1) = 2]
∴ Function f (x) is discontinuous at points 0, 1 and 3.
6.
= a(5 − π ) + 1
lim f (x) = 1 ≠ f (1)
x→1 −
and
a|π − x| + 1, x ≤ 5
f (x) = 
 b|x − π| + 3, x > 5
x→5 −
x→ 0 −
Q
3. Given function
and it is also given that f (x) is continuous at
Clearly,
f (5) = a (5 − π ) + 1
[Q if n ≤ x < n + 1, ∀ n ∈ Integer, [x] = n]
[Q f (0) = 0]
lim f (x) = − 1 ≠ f (0)
Q
lim (b + 5x) = b + 15 = 20
x→3 +
Hence, for a = 0 and b = 5, f (x) is not continuous at x = 3
∴f (x) cannot be continuous for any values of a and b.
1
7. Given, f (x) = x − 1 for 0 ≤ x ≤ π
2
− 1 , 0 ≤ x < 2
∴
[ f (x)] = 
 0, 2 ≤ x ≤ π
⇒
tan (−1), 0 ≤ x < 2
tan [ f (x)] = 
2≤x≤π
 tan 0,
216 Limit, Continuity and Differentiability
∴
lim tan [ f (x)] = − tan 1
10.
x→ 2−
lim tan [ f (x)] = 0
and
x→ 2 +
So, tan f (x) is not continuous at x = 2.
1
x−2
1
2
Now, f (x) = x − 1 ⇒ f (x) =
⇒
=
2
2
f (x) x − 2
Clearly, 1 /f (x) is not continuous at x = 2 .
 1 
are both discontinuous at x = 2 .
So, tan [ f (x)] and 
 f (x) 
π
8. The function f (x) = tan x is not defined at x = , so
2
f (x) is not continuous on (0, π ).
1
(b) Since, g (x) = x sin is continuous on (0, π ) and the
x
integral function of a continuous function is
continuous,
x 
1
∴ f (x) = ∫ t sin  dt is continuous on (0, π ).
0 
t
3π

1,
0<x≤

4
(c) Also, f (x) = 
 2x 3π
2 sin   ,
<x<π
 9
4

We have,
lim
x→
3π−
4
f (x) = 1
 2x
lim f (x) = lim 2 sin   = 1
+
3
π
 9
3π
x→
x→
4
4
So, f (x) is continuous at x = 3π / 4.
⇒
f (x) is continuous at all other points.
π
π
 π
(d) Finally, f (x) = sin (x + π ) ⇒ f   = −


2
2
2
lim
π
x→  
 2
and
−
π
π

 3π
 π
f (x) = lim f  − h = lim sin 
− h =
2
 h→ 0 2
 2
 2
h→ 0

π
f (x) = lim f  + h

2
0
h
→
x→ ( π / 2)
lim
+
π
 3π
 π
= lim sin 
+ h =

 2
h→ 0 2
2
So, f (x) is not continuous at x = π / 2.
Thus, the function is defined in the interval.

x2
, 0 ≤ x<1

2
11. Given, f (x) = 
…(i)
2x2 − 3x + 3 , 1 ≤ x ≤ 2

2
Clearly, RHL (at x = 1) = 1 / 2 and LHL (at x = 1) = 1 / 2
Also,
f (x) = 1 / 2
∴ f (x) is continuous for all x ∈[0, 2].
On differentiating Eq. (i), we get
, 0 ≤ x<1
 x
…(ii)
f ′ (x) = 
4
3
, 1 ≤ x≤2
x
−

Clearly, RHL (at x = 1) for f ′ (x) = 1
and LHL (at x = 1) for f ′ (x) = 1
Also,
f (1) = 1
Thus, f ′ (x) is continuous for all x ∈[0, 2].
Again, differentiating Eq. (ii), we get
1 , 0 ≤ x < 1
f ′ ′ (x) = 
4 , 1 ≤ x ≤ 2
Clearly, RHL (at x = 1) ≠ LHL (at x = 1)
Thus, f ′ ′ (x) is not continuous at x = 1.
or f ′ ′ (x) is continuous for all x ∈ [0, 2] − {1}.
1 1
12. The given function ‘f’ defined on  − ,  by
 3 3
1
 1 + 3x
 log e 
 , where x ≠ 0
f (x) =  x
 1 − 2x 

k,
where x = 0

is continuous, then
 1 + 3x
1
k = lim log e 

x→ 0 x
 1 − 2x 
= lim
At x = 1, f (x) is discontinuous, since lim (1 − h ) = 0
lim (1 + h ) = − 1
h→ 0
h→ 0
∴ f (x) is not differentiable at x = 1.
Hence, (a), (b) and (d) are correct answers.
 log e (1 + 3x) log e (1 − 2x) 
−

x
x
x→ 0 

= 3 − (−2) = 5 ⇒ k = 5
[x sin πx] = 0
Also, x sin πx becomes negative and numerically less
than 1 when x is slightly greater than 1 and so by
definition of [x].
f (x) = [x sin π x] = − 1, when 1 < x < 1 + h
Thus, f (x) is constant and equal to 0 in the closed
interval [–1, 1] and so f (x) is continuous and
differentiable in the open interval (–1, 1).
and
We know that, [x] is continuous on R ~ I, where I
 π 
denotes the set of integers and sin 
 is
 [x + 1]
discontinuous for [x + 1] = 0.
⇒
0 ≤ x+ 1 <1 ⇒ −1 ≤ x<0
−1 < x < 1 ⇒ 0 ≤ x sin π x ≤ 1 / 2
9. We have, for
∴
 π 
f (x) = [x] sin 

 [x + 1]
Topic 6 Continuity for Composition and
Function
1
x
1
x
1. Given, f (x) = x cos , x ≥ 1 ⇒ f ′ (x) = sin
⇒
f ′ ′ (x) = −
1
 1
cos  
3
 x
x
1
1
+ cos
x
x
lim f ′ (x) = 0 + 1 = 1 ⇒ Option (b) is correct.
1
Now, x ∈ [1 , ∞ ) ⇒
∈ (0 , 1] ⇒ f ′ ′ (x) < 0
x
Option (d) is correct.
Now,
x→ ∞
Limit, Continuity and Differentiability 217
As f ′ (1) = sin 1 + cos 1 > 1
f ′ (x) is strictly decreasing and lim f ′ (x) = 1
1 + x, 0 ≤ x ≤ 2
3 − x, 2 < x ≤ 3
4. Given, f (x) = 
x→ ∞
1 + f (x), 0 ≤ f (x) ≤ 2
fof (x) = f [ f (x)] = 
3 − f (x), 2 < f (x) ≤ 3
1 + f (x), 0 ≤ f (x) ≤ 1 1 + (3 − x), 2 < x ≤ 3


⇒ fof = 1 + f (x), 1 < f (x) ≤ 2 = 1 + (1 + x), 0 ≤ x ≤ 1
3 − f (x), 2 < f (x) ≤ 3 3 − (1 + x), 1 < x ≤ 2


So, graph of f ′ (x) is shown as below.
∴
Y
(1, sin1+cos1)
1
X
0 1
⇒
Now, in [x , x + 2], x ∈ [1 , ∞ ), f (x) is continuous and
differentiable so by LMVT,
f (x + 2) − f (x)
f ′ (x) =
2
As,
f ′ (x) > 1
For all x ∈ [1 , ∞ )
f (x + 2) − f (x)
> 1 ⇒ f (x + 2) − f (x) > 2
⇒
2
For all
x ∈ [1 , ∞ )
if f (x) < 0
 f (x) + 1 ,
2. gof (x) = 
2
if f (x) ≥ 0
{ f (x) − 1} + b ,
x+ a +1,


=  (x + a − 1)2 + b,
(| x − 1| − 1)2 + b,

gof (0) = gof (0+ ) = gof (0− )
⇒
⇒
b = b = (a − 1)2 + b ⇒ a = 1
if x < −1
x + 2,


if − 1 ≤ x < 0
gof (x) = 
x2 ,
(| x − 1| − 1)2,
if x ≥ 0

Hence,
Now, RHL (at x = 2) = 2 and LHL (at x = 2) = 0
Also, RHL (at x = 1) = 1 and LHL (at x = 1) = 3
Therefore, f (x) is discontinuous at x = 1, 2
∴ f [ f (x)] is discontinuous at x = {1, 2}.
5. Since, f (x) is continuous at x = 0.
⇒
h→ 0
= f (k) + f (0)
h→ 0
∴
x→ k
Topic 7 Differentiability at a Point
1. Given function, g (x) = | f (x)|
where f : R → R be differentiable at c ∈ R and f (c) = 0,
then for function ‘g’ at x = c
g (c + h ) − g (c)
[where h > 0]
g′ (c) = lim
h→ 0
h
| f (c + h )| − | f (c)|
| f (c + h )|
= lim
= lim
h→ 0
h
→
0
h
h
= f ′ (c)
x→ k
x→ k
φ (x) = f (x) + g (x).
but
φ (k) = f (k) + g (k) ≠ lim { f (x) + g (x)}
x→ k
∴
φ (x) = f (x) + g (x) is discontinuous,
whenever g (x) is discontinuous.
[Qh > 0]
[Q f is differentiable at x = c]
Now, if f ′ (c) = 0, then g (x) is differentiable at x = c,
otherwise LHD (at x = c) and RHD
(at x = c) is different.
Case II g (x) is discontinuous as, lim g (x) ≠ g (k).
quantity
[as f (c) = 0 (given)]
f (c + h ) − f (c)
= lim
h→ 0
h
f (c + h ) − f (c)
= lim
h→ 0
h
∴ φ (x) is discontinuous.
x→ k
h→ 0
= f (k) + f (0− ) = f (k) + f (0)
lim f (x) = f (k)
⇒ f (x) is continuous for all x ∈ R.
⇒ lim φ (x) = lim { f (x) + g (x)} = does not exist.
x→ k
h→ 0
LHL = lim f (k − h ) = lim [ f (k) + f (− h )]
Case I g (x) is discontinuous as limit does not exist at
x = k.
∴
φ (x) = f (x) + g (x)
⇒ lim φ (x) = lim { f (x) + g (x)} = exists and is a finite
…(i)
RHL = lim f (k + h ) = lim [ f (k) + f (h )] = f (k) + f (0+ )
3. As, f (x) is continuous and g (x) is discontinuous.
∴
x→ 0
⇒
f (0 ) = f (0− ) = f (0) = 0
To show, continuous at x = k
In the neighbourhood of x = 0 , gof (x) = x2, which is
differentiable at x = 0.
x→ k
lim f (x) = f (0)
+
if x < − a
if − a ≤ x < 0
if x ≥ 0
As gof (x) is continuous at x = − a
gof (− a ) = gof (− a + ) = gof (− a − )
⇒
1 + b =1 + b =1 ⇒ b =0
Also, gof (x) is continuous at x = 0
4 − x, 2 < x ≤ 3

( fof ) (x) = 2 + x, 0 ≤ x ≤ 1
2 − x, 1 < x ≤ 2

2.
Key Idea (i) First use L’ Hopital rule
(ii) Now, use formula
d
dx
φ2 ( x )
∫ f (t ) dt = f [ φ2( x)] ⋅ φ′2 ( x) − f [ φ1( x)] ⋅ φ′1( x)
φ1 ( x )
218 Limit, Continuity and Differentiability
Here, Rf ′ (0) = lim (3 cos x − 1 − 2(x − π )sin x)
f ( x)
f ( x)
Let l = lim
x→ 2
∫
6
∫ 2tdt
x→ 0 +

0
form, as f (2) = 6

0
2tdt
= lim 6
(x − 2) x→ 2 (x − 2)
=3 −1 −0 =2
and
Lf ′ (0) = lim (cos x + 1 − 2(x − π )sin x)
x → 0−
On applying the L’ Hopital rule, we get
φ ( x)
2 f (x) f ′ (x)  d 2
Q
∫ f (t )dt = f (φ 2(x)) ⋅ φ2′ (x)
x→ 2
1
 dx φ1 ( x )
− f (φ1 (x)) ⋅ φ1′ (x)]
So, l = 2 f (2) ⋅ f ′ (2) = 12 f ′ (2)
[Q f (2) = 6]
=1 + 1 −0 =2
Q
l = lim
f ( x)
∴ lim
x→ 2
∫
6
So, f (x) is differentiable at all values of x.
⇒
K =φ
6.
2tdt
= 12 f ′ (2)
x−2
g (x) = | f (x)| + f (|x|)
−2 ≤ x<0
 1,
Clearly, | f (x)| =  2
1
0 ≤ x≤2
|
x
−
|,

1,
−2 ≤ x<0


= − (x2 − 1), 0 ≤ x < 1
 x2 − 1 ,
1 ≤ x≤2

and
g (x) = f ( f (x))
= f (15 −|x − 10|) = 15 − |15 − |x − 10| − 10|
= 15 − |5 − |x − 10||
15 − |5 − (x − 10)| , x ≥ 10
=
15 − |5 + (x − 10)| , x < 10
15 − |15 − x| , x ≥ 10
=
 15 − | x − 5| , x < 10
x<5
 15 + (x − 5) = 10 + x ,
 15 − (x − 5) = 20 − x , 5 ≤ x < 10

=
 15 + (x − 15) = x , 10 ≤ x < 15
x ≥ 15
15 − (x − 15) = 30 − x ,
f (|x|) = |x|2 − 1, 0 ≤ |x| ≤ 2
[Q f (|x|) = − 1 is not possible as|x| </ 0]
[Q |x|2 = x2]
= x2 − 1 , |x| ≤ 2
2
= x −1, −2 ≤ x≤2
∴ g (x) = | f (x)| + f (|x|)
and

1 + x2 − 1 ,
−2 ≤ x<0

= − (x2 − 1) + x2 − 1, 0 ≤ x < 1
 x2 − 1 + x2 − 1,
1 ≤ x≤2

From the above definition it is clear that g (x) is not
differentiable at x = 5, 10, 15.
4. Let us draw the graph of y = f (x), as shown below
–p
–3p
4
p
2
y= cos x
p/4
–1
 x2,
−2 ≤ x<0

=
0,
0 ≤ x<1
2 (x2 − 1), 1 ≤ x ≤ 2

y= sin x
p
O
Key Idea This type of problem can be solved graphically.
 − 1, − 2 ≤ x < 0
We have, f (x) =  2
x − 1 , 0 ≤ x ≤ 2
3. Given function is f (x) = 15 − |x − 10|, x ∈ R and
1
Rf ′ (0) = Lf ′ (0)
X
Now, let us draw the graph of y = g (x), as shown in the
figure.
y= min {sin x, cos x}
Clearly, the function f (x) = min {sin x, cos x} is not
− 3π
π
[these are point of
differentiable at x =
and
4
4
intersection of graphs of sin x and cos x in (− π , π ), on
 −3 π π 
which function has sharp edges]. So, S = 
, ,
 4 4
 −3 π − π 3 π π 
which is a subset of 
,
,
, 
4 4 4
 4
Y
(–2,4)
X′
–2
f (x) = sin|x| − |x| + 2 (x − π ) cos|x|
− sin x + x + 2(x − π ) cos x, if x < 0
f (x) = 
 sin x − x + 2(x − π ) cos x, if x ≥ 0
Clearly, f (x) is differentiable everywhere except
possibly at x = 0
[Q f ′ (x) exist for x < 0 and x > 0]
y=x2 y=0
y=2 (x2 –1)
–1 O
X
1
2
Y′
5. We have,
[Q sin(−θ ) = − sin θ and cos(−θ ) = cos θ ]
− cos x + 1 + 2 cos x − 2(x − π )sin x ; if x < 0
∴ f ′ (x) = 
 cos x − 1 + 2 cos x − 2(x − π ) sin x , if x > 0
(2,6)
1
( y + 2) represent a parabola
2
with vertex (0, − 2) and it open upward]
[ Here, y = 2 (x2 − 1) or x2 =
Note that there is a sharp edge at x = 1 only, so g (x) is
not differentiable at x = 1 only.
7.
Key Idea This type of questions can be solved graphically.
Given, f : (−1, 1) → R, such that
f (x) = max { −|x|, − 1 − x2 }
Limit, Continuity and Differentiability 219
On drawing the graph, we get the follwong figure.
4
Y
y=x2
O
–1 , –1 ,
√2 √2
1
y=f(x)
–1
1
1 , –1 ,
√2 √2
–2
y=–|x|
–1
2
1
y = |x|
[ Q graph of y = −| x|is
For| x| ∈ (2, 4]
Y
x ∈ (2, 4]
8 − 2x,
f (x) = 8 − 2|x|= 
8
2
x
,
x
[− 4, − 2)
+
∈

Q 2 < |x| ≤ 4


⇒|x| > 2 and |x| ≤ 4




X
and graph of y = − 1 − x2
Y
Y
5
4
X
3
2
y=8+2x
⇒
[Q x2 + y2 = 1 represent a complete circle]
1

− 1 − x2 ,
−1 < x ≤ −

2

1
1
f (x) =  −|x|,
−
<x≤
2
2

1
 − 1 − x2 ,
< x<1
2

X′
–5 –4 –3 –2 –1 O
1
2
3
4
5
X
Y′
Hence, the graph of y = f (x) is
4
8+
2x
1
y=
–4
2x
8–
y=
From the figure, it is clear that function have sharp
1
1
edges, at x = −
, 0,
2
2
∴ Function is not differentiable at 3 points.
8.
y=8–2x
1
–2
–1
1
2
4
Key Idea This type of problem can be solved graphically
max {|x|, x2},
|x| ≤ 2
We have, f (x) = 
2 < |x| ≤ 4
 8 − 2|x|,
From the graph it is clear that at x = − 2, − 1, 0, 1, 2 the
curve has sharp edges and hence at these points f is not
differentiable.
3
Let us draw the graph of y = f (x)
For| x| ≤ 2 f (x) = max{|x|, x2}
9. Given,| f (x) − f ( y)|≤ 2| x − y|2 , ∀ x, y ∈ R
1
Let us first draw the graph of y =|x|and y = x2 as shown
in the following figure.
y=x2
y=|x|
2
| f (x) − f ( y)|
≤ 2| x − y|2
| x − y|
(dividing both sides by|x − y|)
Put x = x + h and y = x, where h is very close to zero.
⇒
1
⇒
1
lim
h→ 0
f (x + h ) − f (x)
≤ lim 2|(x + h ) − x|2
h→ 0
(x + h ) − x
1
–2
0
–1
1
2
⇒
–1
–2
⇒
f (x + h ) − f (x)
≤ lim 2| h |2
h→ 0
h→ 0
h
f (x + h ) − f (x)
lim
≤0
h→ 0
h
lim
[substituting limit directly on right hand
Clearly, y =|x|and y = x2 intersect at x = − 1, 0, 1
Now, the graph of y = max {|x|, x2} for|x| ≤ 2 is
side and using lim | f (x)| = lim f (x) ]
x→ a
x→ a
220 Limit, Continuity and Differentiability
f (x + h ) − f (x)


= f ′ (x)
Q lim
 h→ 0

h
⇒| f ′ (x)| ≤ 0
(Q| f ′ (x)|can not be less than zero)
⇒| f ′ (x)| = 0
[Q| x| = 0 ⇔ x = 0]
⇒ f ′ (x) = 0
⇒ f (x) is a constant function.
Since, f (0) = 1, therefore f (x) is always equal to 1.
1
1
0
0
f (x) = |x − π|(
⋅ e − 1)sin|x|
 (x − π )(e− x − 1)sin x,
x<0

f (x) = − (x − π )(ex − 1)sin x, 0 ≤ x < π
 (x − π )(ex − 1)sin x,
x≥π

|x|
and
Description of Situation
As,
h→ 0
= lim
h→ 0
lim f ′ (x) = 0 = lim f ′ (x)
= lim
x→ π +
 π 
(2 + h )2 cos 
 −0
 2 + h
h→ 0
11. We have, f (x) = log 2 − sin x and g (x) = f ( f (x)), x ∈ R
Note that, for x → 0, log 2 > sin x
∴
f (x) = log 2 − sin x
⇒
g (x) = log 2 − sin ( f (x))
= log 2 − sin (log 2 − sin x)
Clearly, g (x) is differentiable at x = 0 as sin x is
differentiable.
Now,
g′ (x) = − cos (log 2 − sin x) (− cos x)
= cos x ⋅ cos (log 2 − sin x)
⇒
g′ (0) = 1 ⋅ cos (log 2)
h
f and g are differentiable in (0, 1).
…(i)
h (1) = f (1) − 2 g (1) = 6 − 2(2) = 2
h (0) = h (1) = 2
Hence, using Rolle’s theorem,
and
h′ (c) = 0, such that c ∈ (0, 1)
Differentiating Eq. (i) at c , we get
⇒
f ′ (c) − 2 g′ (c) = 0
⇒
f ′ (c) = 2 g′ (c)
π
h 2 cos  −  − 0
 h
−h
=0
 π 
(2 + h )2 ⋅ cos 

 2 + h
= lim
h→ 0
h
π
π 
(2 + h )2 ⋅ sin  −

 2 2 + h
= lim
h→ 0
h
 πh 
(2 + h )2 ⋅ sin 

 2 (2 + h )
π
⋅
=π
= lim
π
h→ 0
(
+ h)
2
2
h⋅
2 (2 + h )
and L { f ′ (2)} = lim
h→ 0
f (2 − h ) − f (2)
−h
(2 − h )2 ⋅ cos
= lim
h→ 0
12. Given, f (0) = 2 = g (1), g(0) = 0 and f (1) = 6
h (0) = f (0) − 2 g (0) = 2 − 0 = 2
π
=0
h
To check differentiability at x = 2,
f (2 + h ) − f (2)
R{ f ′ (2)} = lim
h→ 0
h
x→ 0 +
h (x) = f (x) − 2 g (x)
= lim h ⋅ cos
h→ 0
h
So, f (x) is differentiable at x = 0.
∴ f is differentiable at x = 0 and x = π
Hence, f is differentiable for all x.
Let
π
−0
h
f (0 − h ) − f (0)
= lim
L {f ′ (0)} = lim
h→ 0
h→ 0
−h
lim f ′ (x) = 0 = lim f ′ (x)
x→ π −
f (x + h ) − f (x)
h
f (x − h ) − f (x)
L{ f ′ (x)} = lim
h→ 0
−h
R{ f ′ (x)} = lim
h 2 cos
We check the differentiability at x = 0 and π.
We have,
(x − π ) (e−x − 1) cos x + (e−x − 1) sin x

+ (x − π ) sin xe−x (−1), x < 0
− [(x − π )(ex − 1) cos x + (ex − 1) sin x
f ′ (x) = 
+ (x − π ) sin xex ],0 < x < π

x
x
(x − π )(e − 1) cos x + (e − 1) sin x x
+
(x − π ) sin xe , x > π

x→ 0 −
To check differentiability at a point we use RHD and LHD at
a point and if RHD = LHD, then f( x ) is differentiable at the
point.
Here, students generally gets confused in defining
modulus. To check differentiable at x = 0,
f (0 + h ) − f (0)
R{ f ′ (0)} = lim
h→ 0
h
We have,
Clearly,
PLAN
and
Now, ∫ ( f (x))2dx = ∫ dx = [x]10 = (1 − 0) = 1
10.
13.
= lim
h→ 0
π
π
− 22 ⋅ cos
2−h
2
−h

π 
(2 − h )2 −  − cos
 −0

2 − h
−h
π
π 
− (2 − h )2 ⋅ sin  −

 2 2 − h
= lim
h→ 0
h

πh 
(2 − h )2 ⋅ sin  −

 2 (2 − h )
−π
×
= lim
=−π
−π
h→ 0
(
− h)
2
2
h×
2 (2 − h )
Thus, f (x) is differentiable at x = 0 but not at x = 2.
Limit, Continuity and Differentiability 221
(x − 1)n
; 0 < x < 2, m ≠ 0,
log cosm (x − 1)
x − 1 , x ≥ 1
integers and| x − 1| = 
1 − x, x < 1
14. Given,
g (x) =
n
are
The left hand derivative of|x − 1|at x = 1 is p = − 1.
Also, lim g (x) = p = − 1
x →1+
⇒
lim
h→ 0
(1 + h − 1)n
= −1
log cosm (1 + h − 1)
⇒
hn
= −1
lim
h → 0 log cosm h
⇒
hn
= −1
h → 0 m log cos h
lim
n ⋅ hn − 1
= −1
1
h→ 0
m
(− sin h )
cos h
[using L’Hospital’s rule]
⇒
lim
n−2
hn − 2
 n h
 n
⇒ lim  −  ⋅
=1
= − 1 ⇒   lim
 m h → 0  tan h 
h → 0  m  tan h 




 h 
 h 
n
⇒
n = 2 and
=1 ⇒ m = n =2
m
In function,
y = || x| − 1|we have 3 sharp edges at x = − 1, 0, 1.
Hence, f (x) is not differentiable at {0, ± 1}.
1
if x < − 1
(– x − 1),
2

−1
17. Given, f (x) =  tan x,
if − 1 ≤ x ≤ 1
1
if x > 1
 2 (x − 1),

f (x) is discontinuous at x = − 1 and x = 1.
∴ Domain of f ′ (x) ∈ R − { −1, 1}
sin h − h
18. RHD of sin (| x|) − | x| = lim
= 1 −1 = 0
h→ 0
h
[Q f (0) = 0]
LHD of sin (| x|) − | x|
sin| − h | − | − h | sin h − h
= lim
=
=0
h→ 0
−h
−h
Therefore, (d) is the answer.
19. Given, f (x) = [x] sin π x
If x is just less than k, [x] = k − 1
∴
f (x) = (k − 1) sin π x.
(k − 1) sin π x − k sin π k
LHD of f (x) = lim
x→ k
x−k
1
1
1
15. Given, f (1) = f   = f   = K = lim f   = 0
 n
 3
 2
n→ ∞
= lim
(k − 1) sin π x
,
x−k
 1
f   = 0; n ∈ integers and n ≥ 1.
 n
 1
lim f   = 0 ⇒ f (0) = 0
 n
n→ ∞
= lim
(k − 1) sin π (k − h )
−h
as
⇒
Since, there are infinitely
neighbourhood of x = 0.
∴
f (x) = 0
⇒
f ′ (x) = 0
⇒
f ′ (0) = 0
x→ k
h→ 0
many
points
in
f (x) = max { x, x3 } considering
separately, y = x3 and y = x
20. Given,
Now,
16. Using graphical transformation.
As, we know that, the function is not differentiable at
sharp edges.
Y
⇒
Y
y = |x| – 1
O
X
1
O
–1
–1
1
x
 3
f (x) = x
x
x3

1
 2
f ′ (x) = 3x
1
3x2

in (− ∞ , − 1]
in (−1, 0]
in (0, 1]
in (1, ∞ )
in (− ∞ , − 1]
in (−1, 0]
in (0, 1]
in (1, ∞ )
Y
X
–1
X′
(i) y = x – 1
(ii) y = |x| – 1
(–1, –1) B
Y
–1
O 1
the
NOTE y = x 3 is odd order parabola and y = x is always
intersect at (1, 1) and ( −1, − 1).
f (0) = f ′ (0) = 0
Hence,
[where x = k − h ]
(k − 1) (−1)k − 1 ⋅ sin h π
k
= (−1) (k − 1) π
= lim
h→ 0
−h
X
(iii) y = ||x| – 1|
O
y =x
A (1, 1)
y =x 3
X
Y′
The point of consideration are
f ′ (−1− ) = 1
and
f ' (−1+ ) = 3
f ′ (−0− ) = 0
and
f ′ (0+ ) = 1
−
f ′ (1 ) = 1
and
f ' (1+ ) = 3
Hence, f is not differentiable at −1, 0, 1.
graph
222 Limit, Continuity and Differentiability
21. Let h (x) = | x|, then g (x) = | f (x)| = h { f (x)}
Since, composition of two continuous functions is
continuous, g is continuous if f is continuous. So, answer
is (c).
(a) Let f (x) = x ⇒ g (x) = | x|
Now, f (x) is an onto function. Since, co-domain of x
is R and range of x is R. But g (x) is into function.
Since, range of g (x) is [0, ∞ ) but co-domain is given R.
Hence, (a) is wrong.
(b) Let f (x) = x ⇒ g (x) = | x|. Now, f (x) is one-one
function but g (x) is many-one function. Hence, (b) is
wrong.
(d) Let f (x) = x ⇒ g (x) = | x|. Now, f (x) is differentiable
for all x ∈ R but g (x) =|x|is not differentiable at x = 0
Hence, (d) is wrong.
22. Function f (x) = (x2 − 1)| x2 − 3x + 2| + cos (|x|)
…(i)
NOTE In differentiable of| f ( x )| we have to consider critical
points for which f ( x ) = 0.
| x|is not differentiable at x = 0
but
⇒
cos (− x), if x < 0
cos| x|= 
if x ≥ 0
 cos x,
if x < 0
 cos x,
cos| x| = 
if x ≥ 0
cos
x
,

Q Lf ′ (1) = Rf ′ (1). Therefore, function is differentiable
at x = 1.
f (x) − f (2)
x−2

cos x − cos 2 
= lim − (x2 − 1) (x − 1) +

−
x−2
x→ 2 

= − (4 − 1) (2 − 1) − sin 2 = − 3 − sin 2
f (x) − f (2)
R f ′ (2) = lim
x → 2+
x−2
Again, Lf ′ (2) = lim
x → 2−
and

cos x − cos 2 
= lim (x2 − 1) (x − 1) +

+
x−2
x→ 2 

= (22 − 1) (2 − 1) − sin 2 = 3 − sin 2
So,
L f ′ (2) ≠ R f ′ (2), f is not differentiable at x = 2
Therefore, (d) is the answer.
23. Given,
 x
, x≥0

x
f (x) =
=  1 x+ x
1 + | x| 
, x<0
1 − x
∴
 (1 + x) ⋅ 1 − x ⋅ 1
, x≥0
2

f ′ (x) =  (1 −(1x)+⋅ 1x)− x (−1)
, x<0

(1 − x)2

⇒
 1
,x≥0
2

f ′ (x) =  (1 +1 x)
,x<0

2
 (1 − x)
Therefore, it is differentiable at x = 0 .
Now, | x2 − 3x + 2| = |(x − 1) (x − 2)|
(x − 1) (x − 2),

=  (x − 1) (2 − x),
(x − 1) (x − 2),

Therefore,
 (x2 − 1) (x − 1) (x − 2) + cos x,

f (x) = − (x2 − 1) (x − 1) (x − 2) + cos x,
 (x2 − 1) (x − 1) (x − 2) + cos x,

if x < 1
if 1 ≤ x < 2
if 2 ≤ x
if − ∞ < x < 1
if 1 ≤ x < 2
if 2 ≤ x < ∞
Now, x = 1, 2 are critical point for differentiability.
Because f (x) is differentiable on other points in its
domain.
Differentiability at x = 1
f (x) − f (1)
L f ′ (1) = lim
x−1
x → 1−

cos x − cos 1 
= lim (x2 − 1) (x − 2) +

−
x −1
x→1 

and
= 0 − sin 1 = − sin 1
cos x − cos 1 d
[ Q lim
(cos x) at x = 1 − 0
=
x−1
dx
x → 1−
= − sin x at x = 1 − 0 = − sin x at x = 1 = − sin 1]
f (x) − f (1)
Rf ′ (1) = lim
x−1
x →1+

cos x − cos 1 
= lim − (x2 − 1) (x − 2) +

+
x−1
x→1 

= 0 − sin 1 = − sin 1
[same approach]
∴
RHD at x = 0
⇒ lim
1
=1
(1 + x)2
and
LHD at x = 0
⇒
1
=1
(1 − x)2
x→ 0
lim
x→ 0
Hence, f (x) is differentiable for all x.
24. Since, f (x) is continuous and differentiable where
f (0) = 1 and f′ (0) = − 1, f (x) > 0, ∀ x.
Thus, f (x) is decreasing for x > 0 and concave down.
⇒
f ′ ′ (x) < 0
Therefore, (a) is answer.
tan π [(x − π )]
25. Here, f (x) =
1 + [x]2
Since, we know π [(x − π )] = nπ and tan nπ = 0
Q
∴
1 + [x]2 ≠ 0
f (x) = 0, ∀ x
Thus, f (x) is a constant function.
∴ f ′ (x), f ′ ′ (x) ,... all exist for every x , their value
being 0.
⇒ f ′ (x) exists for all x.
26. The given function f : R → R is satisfying as
f (x + y) = f (x) + f ( y) + f (x) f ( y).
Limit, Continuity and Differentiability 223
So,
Q
⇒
∴
∴
⇒
f (x + h ) − f (x)
h
f (x) + f (h ) + f (x) f (h ) − f (x)
= lim
h→ 0
h
f (h )
= lim
(1 + f (x))
h→ 0 h
f (x) = xg (x)
f (x)
g (x) =
x
f (x)
(given)
lim
= lim g (x) = 1
x→ 0 x
x→ 0
f ′ (x) = 1 + f (x)
f ′ (x)
=1
1 + f (x)
f ′ (x) = lim
h→ 0
⇒ log e (1 + f (x)) = x + c
⇒
1 + f (x) = ex + c
⇒
f (x) = ex + c − 1
Q
f (0) = 0
∴
c=0
Therefore, f (x) = ex − 1 is differentiable at every x ∈ R.
And
f ′ (x) = ex ⇒ f′ (0) = 1
(x) ex − 1
Now,
and if g(0) = 1
g (x) = f
=
x
x
LHD (at x = 0) of
g (0 − h ) − g (0)
g (x) = lim
h→ 0
−h
e− h − 1
−1
−h
= lim
h→ 0
−h
e− h − 1 + h 1
=
h→ 0
2
h2
and, RHD (at x = 0) of
g (0 + h ) − g (0)
eh − 1 − h 1
g (x) = lim
= lim
=
h→ 0
h→ 0
h
2
h2
So, if g(0) = 1, then g is differentiable at every x ∈ R.
= lim
27. Given functions f : R → R be defined by
f (x) = x3 − x2 + (x − 1)sin x and g : R → R be an
arbitrary function.
Now, let g is continuous at x = 1, then
( fg )(x) − ( fg )(1)
( fg )(1 − h ) − ( fg )(1)
lim
= lim
h→ 0
x −1
1 − h −1
x → 1−
Q
(given)
( fg )(x) = f (x) . g (x)
f (1 − h ) . g (1 − h ) − f (1) . g (1)
= lim
h→ 0
−h
f (1 − h ) . g (1)
= lim
h→ 0
−h
{Q f (1) = 0 and g is continuous at x = 1, so
g (1 − h ) = g (1)}
2
(1 − h ) (− h ) + (− h )sin (1 − h )
= g (1) lim
h→ 0
−h
= (1 + sin 1) g (1)
Similarly,
( fg )(x) − ( fg )(1)
f (1 + h ) . g (1)
lim
= lim
h→ 0
x −1
h
x →1+
(1 + h )2(h ) + h sin(1 + h )
h→ 0
h
= (1 + sin 1) g (1)
Q RHD and LHD of function fg at x = 1 is finitely exists
and equal, so fg is differentiable at x = 1
Now, let ( fg )(x) is differentiable at x = 1, so
( fg )(x) − ( fg )(1)
( fg )(x) − ( fg )(1)
lim
= lim
x−1
x−1
x → 1−
x →1+
f (x) g (x) − f (1) g (1)
f (x) g (x) − f (1) g (1)
= lim
⇒ lim
x−1
x−1
x → 1−
x →1+
f (x) g (x)
f (x) g (x)
⇒
= lim
{Q f (1) = 0}
lim
x−1
x−1
x → 1−
x →1+
f (1 − h ) g (1 − h )
f (1 + h ) g (1 + h )
= lim
⇒ lim
h→ 0
h→ 0
h
−h
= g (1) lim
[(1 − h )2(− h ) + (− h )sin(1 − h )] g (1 − h )
h→ 0
−h
2
[(1 + h ) (h ) + (h )sin(1 + h )] g (1 + h )
= lim
h→ 0
h
⇒ lim [(1 − h )2 + sin(1 − h )] g (1 − h ) = lim
⇒ lim
h→ 0
h→ 0
[(1 + h )2 + sin(1 + h )] g (1 + h )
that g (x) is continuous or
It does not mean
differentiable at x = 1.
But if g is differentiable at x = 1, then it must be
continuous at x = 1 and so fg is differentiable at x = 1.
28. It is given, that f : R → R and
f (h ) − f (0)
exists and finite, and
|h|
f (h ) − f (0)
Property 2 : lim
exists and finite.
h→ 0
h2
Option a,
sin h − sin 0
1  sin h 
P2 : lim
= lim 
 = doesn’t exist.
2
h→ 0
h
→
0
h h 
h
Property 1 : lim
h→ 0
Option b,
/
−0
h 23
/ − 1/ 2
P1 : lim
= lim h 23
= lim h1/ 6 = 0
h→ 0
h→ 0
h→ 0
|h|
exists and finite.
Option c,
|h| − 0
P1 : lim
= lim |h| = 0, exists and finite.
h→ 0
|h| h → 0
Option d,
h|h| − 0
|h|
P2 : lim
= lim
2
h→ 0
h
→
0
h
h
 1, if h → 0+
=
−
−1, if h → 0
f (h ) − f (0)
does not exist.
h2
Hence, options (b) and (c) are correct.
So lim
h→ 0
224 Limit, Continuity and Differentiability
29. We have,
( f (0)) + ( f ′ (0)) = 85
2
2
f : R → [− 2, 2]
and
(a) Since, f is twice differentiable function, so f is
continuous function.
∴This is true for every continuous function.
Hence, we can always find x ∈ (r , s), where f (x) is
one-one.
∴This statement is true.
(b) By L.M.V.T
f ′ (c) =
f (b) − f (a )
f (b) − f (a )
⇒| f ′ (c)| =
b−a
b−a
| f ′ (x0 )| =
⇒
f (0) − f (− 4)
f (0) − f (− 4)
=
0+4
4
Range of f is [− 2, 2]
∴
− 4 ≤ f (0) − f (−4) ≤ 4 ⇒ 0 ≤
f (0) − f (− 4)
≤1
4
Hence,| f ′ (x0 )| ≤ 1.
Hence, statement is true.
(c) As no function is given, then we assume
 85 x
f (x) = 2 sin 

 2 
 85 x
f ′ (x) = 85 cos 

 2 
∴
Now, ( f (0))2 + ( f ′ (0))2 = (2 sin 0)2 + ( 85 cos 0)2
( f (0))2 + ( f ′ (0))2 = 85
and lim f (x) does not exists.
x→ ∞
Hence, statement is false.
30. Given, lim
t→ x
Using L′ Hospital rules
f (x) cos t − f ′ (t ) sin x
lim
= sin 2 x
t→ x
1
⇒
f (x) cos x − f ′ (x) sin x = sin 2 x
⇒
f ′ (x) sin x − f (x) cos x = − sin 2 x
f ′ (x) sin x − f (x) cos x
= −1
⇒
sin 2 x
 f (x) 
⇒
d
 = −1
 sin x
On integrating, we get
f (x)
= − x + C ⇒ f (x) = − x sin x + C sin x
sin x
π  π
π
It is given that x = , f   = −
12
6  6
π
π
π
π
π
1
 π
∴ f   = − sin + C sin = −
=−
+ C
 6
6
6
6
12
12 2
⇒
C=0
∴
f (x) = − x sin x
(a) f (x) = − x sin x
π
π
π
π
false
f   = − sin = −
 4
4
4
4 2
(b) f (x) = − x sin x
x3
x4
sin x > x −
, ∀ x ∈ (0, π ) ⇒ − x sin x < − x2 +
6
6
x4
2
f (x ) <
− x , ∀ x ∈ (0, π )
⇒
6
It is true
(c) f (x) = − x sin x
f ′ (x) = − sin x − x cos x
f ′ (x ) = 0
⇒ − sin x − x cos x = 0
tan x = − x
(d) From option b,| f ′ (x0 )| ≤ 1 and x0 ∈ (− 4, 0)
Y
∴
( f ′ (x0 ))2 ≤ 1
Hence, g (x0 ) = ( f (x0 ))2 + ( f ′ (x0 )2 ≤ 4 + 1
[Q f (x0 ) ∈ [−2, 2]]
⇒
X′
g (x0 ) ≤ 5
Now, let p ∈ (−4, 0) for which g ( p) = 5
Hence, by Rolle’s theorem is ( p, q)
g′ (c) = 0 for α ∈ (−4, 4) and since g (x) is greater than
5 as we move form x = p to x = q
⇒
So, f (α ) + f ′ ′ (α ) = 0 and f′ (α ) ≠ 0
Hence, statement is true.
π1
X
3π/2
y=–x
f ′ (x) = − sin x − x cos x
f ′ ′ (x) = − 2 cos x + x sin x
π
 π π
 π
f′ ′   = , f   = −
 2 2
 2
2
g′ (c) = 0
f′ f + f′ f′ ′ = 0
π2
⇒ Their exists α ∈ (0, π ) for which f′ (α ) = 0
It is true
(d) f (x) = − x sin x
and f (x))2 ≤ 4 ⇒ ( f ′ (x))2 ≥ 1in ( p, q)
Thus,
o
Y
Similarly, let q be smallest positive number q ∈ (0, 4)
such that g (q) = 5
f (x) sin t − f (t ) sin x
= sin 2 x
t −x
∴
 π
 π
f′ ′   + f   = 0
 2
 2
It is true.
Limit, Continuity and Differentiability 225
31. As, g ( f (x)) = x
and
g (x) = | x| f (x) + |4x − 7| f (x)
= (| x| + |4x − 7|) f (x)
Thus, g (x) is inverse of f (x).
⇒
g ( f (x)) = x ⇒ g′ ( f (x)) ⋅ f ′ (x) = 1
1
g′ ( f (x)) =
f ′ (x)
∴
= (| x| + |4x − 7|) [x2 − 3]
(− x − 4x − 7) (− 3), − 1 / 2 ≤ x < 0
 (x − 4x + 7) (− 3),
0 ≤ x<1

1≤x< 2
 (x − 4x + 7) (− 2),

=  (x − 4x + 7) (− 1),
2 ≤x< 3
 (x − 4x + 7) (0) ,
3 ≤ x < 7 /4

7 /4 ≤ x < 2
 (x + 4x − 7) (0),
 (x + 4x − 7) (1),
x=2

…(i)
[where, f ′ (x) = 3x2 + 3]
When
f (x) = 2, then
x3 + 3x + 2 = 2 ⇒ x = 0
i.e. when x = 0, then f (x) = 2
1
1
at (0, 2) ⇒ g′ (2) =
∴
g′ ( f (x)) = 2
3
3x + 3
∴ Option (a) is incorrect.
Now,
h ( g ( g (x))) = x ⇒ h ( g ( g ( f (x))) = f (x)
…(ii)
⇒
h ( g (x)) = f (x)
As
g ( f (x)) = x
∴
h ( g (3)) = f (3) = 33 + 3(3) + 2 = 38
∴ Option (d) is incorrect.
From Eq. (ii),
h ( g (x)) = f (x)
⇒
h ( g ( f (x))) = f ( f (x))
…(iii)
⇒
h (x) = f ( f (x))
[using g ( f (x)) = x]
…(iv)
⇒
h′ (x) = f ′ ( f (x)) ⋅ f ′ (x)
Putting x = 1, we get
h′ (1) = f ′ ( f (1)) ⋅ f ′ (1) = (3 × 36 + 3) × (6) = 111 × 6 = 666
∴ Option (b) is correct.
Putting x = 0 in Eq. (iii), we get
h (0) = f ( f (0)) = f (2) = 8 + 6 + 2 = 16
∴ Option (c) is correct.
∴
15x + 21,
9x − 21,

6x − 14,
g (x) = 
3x − 7,
0,

5x − 7,
∴
|x|sin (|x + x|) = x sin (x + x), ∀ x ∈ R
⇒
3
3
3 ≤ x<2
x=2
Now, the graphs of f (x)and g (x)are shown below.
Graph for f ( x)
Y
1
X′
–1/2
X
0
1
√2
√3
2
–1
–2
–3
32. Here, f (x) = a cos (|x3 − x|) + b|x|sin (|x3 + x|)
If x3 − x ≥ 0
⇒
cos|x3 − x| = cos (x3 − x)
x3 − x ≤ 0
⇒
cos|x3 − x| = cos (x3 − x)
∴
cos (|x3 − x|) = cos (x3 − x), ∀ x ∈ R
Again, if x3 + x ≥ 0
⇒
|x|sin (|x3 + x|) = x sin (x3 + x)
x3 + x ≤ 0
⇒
|x|sin (|x3 + x|) = − x sin { − (x3 + x )}
− 1 /2 ≤ x < 0
0 ≤ x<1
1≤x< 2
2 ≤x< 3
Y′
Clearly, f (x) is discontinuous at 4 points.
∴ Option (b) is correct.
Graph for g ( x)
…(i)
Y
21
27/2
3
…(ii)
f (x) = a cos (|x3 − x|) + b|x|sin (|x3 + x|)
…(iii)
f (x) = a cos (x3 − x) + bx sin (x3 + x)
∴
which is clearly sum and composition of differential
functions.
X′
–1/2
0
X
3 √ 3–71
3 √ 2–7
√2
√3
2
–8
Hence, f (x) is always continuous and differentiable.
33. Here,
− 3 , − 1 / 2 ≤ x < 1
− 2 , 1 ≤ x < 2

2
2
2 ≤x< 3
f (x) = [x − 3] = [x ] − 3 = − 1,
 0,
3 ≤ x<2

 1,
x=2
–12
–21
Y′
Clearly, g (x) is not differentiable at 4 points, when
x ∈ (− 1 / 2, 2).
∴ Option (c) is correct.
226 Limit, Continuity and Differentiability
x>0
x=0
 g (x),

34. Here, f (x) =  0,
Given that, f : [a , b] → [1, ∞ )
− g (x), x < 0

 g′ (x), x ≥ 0
f ′ (x) = 
− g′ (x), x < 0
an
∴ Option (a) is correct.
h (x) = e
(b)
x
 ex , x ≥ 0
=  −x
e , x < 0
 ex ,
x≥0
h′ (x) =  − x
,
e
x<0
−

+
⇒
h′ (0 ) = 1and h′ (0− ) = − 1
So, h (x) is not differentiable at x = 0.
∴Option (b) is not correct.
(c)
( foh )(x) = f { h (x)}, as h (x) > 0
 g (ex ), x ≥ 0
=  −x
 g (e ), x < 0
⇒
 ex g′ (ex ),
x≥0
( foh )′ (x) =  − x
−x
− e g′ (e ), x < 0
+
⇒
( foh )′ (0 ) = g′ (1), ( foh )′ (0− ) = − g′ (1)
So, ( foh )(x) is not differentiable at x = 0.
∴ Option (c) is not correct.
 e g( x ) , x ≠ 0
f ( x)
(d)
(hof )(x) = e
= 0
e = 1 , x = 0
Now, (hof )′ (0) = lim
= lim
e
h→ 0
g( x )
−1
x
h→ 0
g( x )
− 1 g (x)
⋅
g (x)
x
g (x) − 0
−1
|x|
⋅ lim
⋅ lim
h→ 0
h→ 0 x
g (x)
|x|
|x|
= 1 ⋅ g′ (0) ⋅ lim
= 0, as g′ (0) = 0
h→ 0 x
∴ Option (d) is correct.
= lim
e
g( x )
h→ 0
35. Let F (x) = f (x) − 3 g (x)
∴ F (−1) = 3, F (0) = 3 and F (2) = 3
So, F ′ (x) will vanish atleast twice in (−1, 0) ∪ (0, 2).
Q F ′′ (x) > 0 or < 0, ∀ x ∈ (−1, 0) ∪ (0, 2)
Hence, f ′ (x) − 3 g′ (x) = 0 has exactly one solution in
one solution in (0, 2).
(−1, 0) and
36. A function f (x) is continuous at x = a,
if lim f (x) = lim f (x) = f (a ).
x→ a−
x→ a +
Also, a function f (x) is differentiable at x = a, if
lim
x→ a−
i.e.
f (x) − f (a )
f (x) − f (a )
= lim
x−a
x−a
x→ a +
−
+
f ′ (a ) = f ′ (a )
g (a − ) = 0 = g (a + ) = g (a )
Now,
[as g (a + ) = lim
x→ a +
x
∫a f (t )dt = 0
a
and g (a ) = ∫ f (t )dt = 0]
a
b
g (b− ) = g (b+ ) = g (b) = ∫ f (t )dt
a
⇒ g is continuous for all x ∈ R.
,
x<a
 0

Now,
g′ (x) =  f (x) , a < x < b
 0
,
x>b

g′ (a − ) = 0
⇒
e

,
x<a
 x0

g (x) = ∫ f (t )dt , a ≤ x ≤ b
a
 b
∫ f (t )dt , x > b
 a
but
g′ (a + ) = f (a ) ≥ 1
[Q range of f (x) is [1, ∞ ), ∀ x ∈ [a , b]]
⇒ g is non-differentiable at x = a
and
g′ (b+ ) = 0
but
g′ (b− ) = f (b) ≥ 1
⇒ g is not differentiable at x = b.
π
π

−x− ,
x≤−

2
2

π
37. f (x) =  − cos x, − < x ≤ 0
2
 x − 1,
0 < x≤1

 log x,
x>1
π
,
2
 π π
 π
f −  = − −  − = 0
 2 2
 2
Continuity at x = −
 π

RHL = lim − cos  − + h = 0
 2

h→ 0
 π
∴ Continuous at x =  −  .
 2
Continuity at x = 0
f (0) = − 1
RHL = lim (0 + h ) − 1 = − 1
h→ 0
∴ Continuous at x = 0.
Continuity at x = 1,
f (1) = 0
RHL = lim log (1 + h ) = 0
h→ 0
∴ Continuous at x = 1
Limit, Continuity and Differentiability 227
π

− 1,
x≤−

2

π
sin x, − < x ≤ 0
f ′ (x) = 
2

1,
0 < x≤1
 1
x>1
,

 x
40. From the figure,
Y
X′
y = min {x, x 2}
X
O 1
Differentiable at x = 0, LHD = 0, RHD = 1
Y′
∴ Not differentiable at x = 0
Differentiable at x = 1, LHD = 1, RHD = 1
h (x) is continuous all x, but h (x) is not differentiable at
two points x = 0 and x = 1. (due to sharp edges). Also
h′ (x) = 1, ∀ x > 1.
∴ Differentiable at x = 1.
3
Also, for x = −
2
π
⇒
f (x) = − x −
2
∴ Differentiable at x = −
Hence, (a), (c) and (d) is correct answers.
, x≥1
 |x − 3|

41. Here, f (x) =  x2 3x 13
 4 − 2 + 4 , x < 1
∴ RHL at x = 1, lim |1 + h − 3| = 2
3
2
⇒
f ′ (0) = k
Now,
f ′ (x) = lim
…(i)
f (x + h ) − f (x)
h→ 0
h
f (x) + f (h ) − f (x)
= lim
h→ 0
h
f (h )
= lim
h→ 0 h
0

form
0

Given, f (x + y) = f (x) + f ( y), ∀ x, y
∴
f (0) = f (0) + f (0),
x= y=0
when
⇒ f (0) = 0)
Using L’Hospital’s rule,
f ′ (h )
= lim
= f ′ (0) = k
h→ 0
1
…(ii)
⇒ f ′ (x) = k, integrating both sides,
f (x) = kx + C , as f (0) = 0
⇒
C =0 ∴
f (x) = kx
∴ f (x) is continuous for all x ∈ R and f ′ (x) = k, i.e.
constant for all x ∈ R.
h→ 0
LHL at x = 1,
38. f (x + y) = f (x) + f ( y), as f (x) is differentiable at x = 0.
lim
h→ 0
(1 − h )2 3 (1 − h ) 13 1 3 13 14 3
−
+
= − +
=
− =2
4
2
4 4 2
4
4 2
∴ f (x) is continuous at x = 1

1 ≤ x<3
 − (x − 3),

Again,
(x − 3),
f (x) = 
x≥3
 x2 3x 13
+
,
x<1
 −
4
2
4

1 ≤ x<3
 −1 ,

∴
f ′ (x) =  1,
x≥3
x 3
x<1
2 − 2 ,
RHD at x = 1 ⇒
−1

1 3
 differentiable at x = 1.
∴
LHD at x = 1 ⇒ − = − 1

2 2
RHD at x = 3 ⇒ 1 
not differentiable at x = 3.
Again,
LHD at x = 3 ⇒ − 1
42. We know that, f (x) = 1 + |sin x|could be plotted as,
(1) y = sin x
…(i)
Y
Hence, (b) and (c) are correct.
39. Here, f (x) = min { 1, x2, x3 } which could be graphically
shown as
X′
y = x3
Y
1
y=
−π
x2
O
X
1
2π
3π
y = sin x
X
−1
y=1
X′
π
O
Y′
y =|sin x|
(2)
…(ii)
Y
1
Y′
⇒f (x) is continuous for x ∈ R and not differentiable at
x = 1 due to sharp edge.
Hence, (a) and (d) are correct answers.
X′
–2 π
–π
O
Y′
y=|sin x|
π
2π
3π
X
228 Limit, Continuity and Differentiability
y = 1+|sin x|
(3)
…(iii)
Y
–2
y=1 +|sin x|
–1
X′
–π
O
2π
π
X
3π
Y′
Therefore, sin (π [x]) = 0, ∀ x ∈ R. By theory, we know
that sin (π [x]) is differentiable everywhere, therefore
(A) ↔ (p).
Again,
f (x) = sin{ π (x − [x])}
Now,
x − [x] = { x} then π (x − [x]) = π { x}
which is not differentiable at x ∈ I.
Therefore, (B) ↔ (r ) is the answer.
47. Given, F (x) = f (x) ⋅ g (x) ⋅ h (x)
Clearly, y = 1 + |sin x| is continuous for all x, but not
differentiable at infinite number of points..
x + y = 2 y, when y > 0
43. Since, x + | y| = 2 y ⇒ 
 x − y = 2 y, when y < 0
⇒
46. We know, [x] ∈ I , ∀ x ∈ R.
 y = x, when y > 0 ⇒ x > 0

 y = x /3, when y < 0 ⇒ x < 0
On differentiating at x = x0, we get
F ′ (x0 ) = f ′ (x0 ) ⋅ g (x0 ) h (x0 ) + f (x0 ) ⋅ g′ (x0 ) h (x0 )
+ f (x0 ) g (x0 )h′ (x0 )
g′ (x0 ) = − 7 g (x0 ) and h ′ (x0 ) = k h (x0 )
On substituting in Eq. (i), we get
21 F (x0 ) = 4 f (x0 ) g (x0 )h (x0 ) − 7 f (x0 ) g (x0 ) h (x0 )
+ k f (x) g (x0 )h (x0 )
⇒
y = x, x > 0
45°
O
y =x , x < 0
3
F ′ (x0 ) = 21 F (x0 ), f ′ (x0 ) = 4 f (x0 )
where,
which could be plotted as,
Y
X′
…(i)
X
Y′
Clearly, y is continuous for all x but not differentiable at
x = 0.
dy  1, x > 0
Also,
=
dx 1 /3, x < 0
Thus, f (x) is defined for all x, continuous at x = 0,
dy 1
differentiable for all x ∈ R − {0},
= for x < 0.
dx 3
g (x) cos x − g (0)
0

44. We have, lim
form
0

x→ 0
sin x
g′ (x) cos x − g (x)sin x
= lim
=0
x→ 0
cos x
Since, f (x) = g (x)sin x
f ′ (x) = g′ (x) sin x + g (x) cos x
and f ′ ′ (x) = g′ ′ (x) sin x + 2 g′ (x) cos x − g (x) sin x
⇒
f ′ ′ (0) = 0
Thus, lim [ g (x) cos x − g (0) cosec x] = 0 = f ′ ′ (0)
∴
increasing in (−1, 1) .
B. |x| is continuous in (−1, 1) and not differentiable at
x = 0.
C. x + [x] is strictly increasing in (−1, 1) and
discontinuous at x = 0
⇒ not differentiable at x = 0.
D. |x − 1| + |x + 1| = 2 in (−1, 1)
⇒ The function is continuous and differentiable in
(−1, 1).


x

0,
, x ≠0
x=0
h
−0
1 + e1/ h
∴ Rf ′ (0) = f ′ (0+ ) = lim
h→ 0
h
1
= lim
=0
h → 0 1 + e1/ h
−h
−0
1 + e−1/ h
and Lf ′ (0) = f ′ (0− ) = lim
h→ 0
−h
1
1
= lim
=
=1
1
h→ 0
1+0
1 + 1/ h
e
∴
f ′ (0+ ) = 0 and f ′ (0− ) = 1
49. Given,
1

2
− | x|, if x ≠ 1
(x − 1) sin
f (x) = 
(x − 1)

−1 ,
if x = 1
As,
1

(x − 1)2 sin
− x, 0 ≤ x − {1}

(x − 1)

1
f (x) = (x − 1)2 sin
+ x,
x<0
(
x
− 1)

−1
,
x=1


Statement II is not a correct explanation of Statement I.
45. A. x|x| is continuous, differentiable and strictly
k = 24
48. Given, f (x) = 1 + e1/ x
x→ 0
⇒ Statement I is true.
Statement II f ′ (x) = g′ (x) sin x + g (x) cos x
⇒
f ′ (0) = g (0)
21 = 4 − 7 + k , [using F (x0 ) = f (x0) g (x0 ) h (x0 )]
Here, f (x) is not differentiable at x = 0 due to|x|.
Thus, f (x) is not differentiable at x = 0.
50. It is always true that differential of even function is and
odd function.
51. Since, f (x) is differentiable at x = 0.
⇒ It is continuous at x = 0.
i.e.
lim f (x) = lim f (x) = f (0)
x → 0+
x → 0−
Limit, Continuity and Differentiability 229
eah/ 2 − 1
eah/ 2 − 1 a a
= lim
⋅ =
h
h→ 0
h→ 0
2 2
h
x → 0+
a
2
 c − h
−1 c
Also, lim f (x) = lim b sin −1 
 = b sin
−


→
0
h
2
2
x→ 0
Here, lim f (x) = lim
∴
⇒
a =1
Also, it is differentiable at x = 0
R f ′ (0+ ) = L f ′ (0– )
eh/ 2 − 1 1
−
2
h
Rf ′ (0+ ) = lim
h→ 0
h
f ′ (α ) = lim g (x) ⇒
⇒
f (x) is differentiable at x = α.
[Q a = 1]
⇒
4 − c2
⇒
=
a =1
52. Here, lim
n→ ∞
c2
4
and
LHL = lim f (x) = lim (1 − x)
x → 1−
64b2 = (4 − c2)
h→ 0
and

 1
  − 1 = lim nf
 n
 n→ ∞
 1
cos −1   − 1 = f ′ (0)
 n
 1
 
 n
⇒
⇒
f ′ (0) =
x →1+
h→ 0
= lim − h ⋅ (1 − h ) = 0
∴ From Eqs. (i) and (ii), we get
2
2
 1
lim
(n + 1) cos −1   − n = 1 −
 n
n→ ∞ π
π
LHL = RHL = f (1) = 0
Therefore, f is continuous at x = 1
Differentiability at x = 1,
f (1 − h ) − f (1)
L f ′ (1) = lim
h→ 0
−h
1 − (1 − h ) − 0
 h 
= lim
= lim 
 = −1
h→ 0
h→ 0  − h 
−h
and
…(ii)
f (1 + h ) − f (1)
h→ 0
h
[1 − (1 + h )] [(2 − (1 + h )] − 0
= lim
h→ 0
h
− h (1 − h )
= lim
= lim (h − 1) = − 1
h→ 0
h→ 0
h
R f ′ (1) = lim
Since, L [ f ′ (1)] = Rf ′ (1), therefore f is differentiable at
x = 1.
Continuity at x = 2,
53. Since, g (x) is continuous at x = α ⇒ lim g (x) = g (α )
x→ α
and f (x) − f (α ) = g (x) (x − α ) , ∀x ∈ R
f (x) − f (α )
⇒
= g (x)
(x − α )
x →1+
h→ 0


−1
+ cos −1 x
(1 + x)
2
1−x


2 
π
2
− 1 +  = 1 −
2
π 
π
h→ 0
RHL = lim f (x) = lim (1 − x) (2 − x)
= lim [1 − (1 + h )] [2 − (1 + h )]
∴
2
(1 + x) cos −1 x − 1, f (0) = 0
π
2
f ′ (x) =
π
x → 1−
= lim [1 − (1 − h )] = lim h = 0

 1 
nf   
given, f ′ (0) = nlim
 n 
→
∞

2
1
…(i)
lim (n + 1) cos −1 − n = f ′ (0)
n→ ∞ π
n
where, f (x) =
is
Continuity at x = 1,
2
 1
(n + 1) cos −1   − n
 n
π
1
 1 2 
where, f   = 1 + 
 n π 
n
g (x) is continuous at x = α.
continuous and differentiable at all points except
possibly at x = 1 and 2.
2
1
2 
= lim n  1 +  cos −1
n→ ∞ π 
n
∴
1−
64b = (4 − c )
⇒
lim g (x) = f ′ (α )
x→ α
x<1
(1 − x),

f (x) = (1 − x)(2 − x), 1 ≤ x ≤ 2
(3 − x),
x>2

b /2
1
8
2
f ′ (α ) = g (α )
54. It is clear that the given function
− h 1
b sin 
 −
 2  2
–
Lf ′ (0 ) = lim
=
h→ 0
−h
b
x→ α
Hence, f (x) is differentiable at x = α, iff g (x)
continuous at x = α.
−1  c
∴
f (x) − f (α )
= lim g (x)
x→ α
x−α
⇒
Clearly,
2eh/ 2 − 2 − h 1
= lim
=
h→ 0
8
2h 2
and
lim
x→ α
Conversely, suppose f is differentiable at α, then
f (x) − f (α )
exists finitely.
lim
x→ α
x−α
 f (x) − f (α )
, x≠α

Let
g (x) = 
x−α
 f ′ (α ) ,
x=α
c a 1
= =
2 2 2
b sin −1
⇒
[given]
LHL = lim f (x) = lim (1 − x) (2 − x)
x → 2−
x → 2−
= lim [1 − (2 − h )] [(2 − (2 − h )]
h→ 0
= lim (− 1 + h ) h = 0
h→ 0
230 Limit, Continuity and Differentiability
RHL = lim f (x) = lim (3 − x)
and
x → 2+
Again, f ′ (x) = lim
x → 2+
h→ 0
= lim [3 − (2 + h )]
h→ 0
= lim
= lim (1 − h ) = 1
h→ 0
h
f (2x) + f (2h )
− f (x)
2
= lim
h→ 0
h
1   2x

 2h 
2 f   − 1 + 2 f   − 1 − f (x)





2
2
2

= lim 
h→ 0
h
h→ 0
Since, LHL ≠ RHL, therefore f is not continuous at x = 2
as such f cannot be differentiable at x = 2.
Hence, f is continuous and differentiable at all points
except at x = 2 .
 − 1 + 1 


 xe  x x 
, x>0

 1 1
 − − + 
55. Given, f (x) = xe  x x  , x < 0

0
, x=0



 −2
xe x , x > 0

, x<0
= x
 0
, x=0


[from Eq. (i)]
1
[2 f (x) − 1 + 2 f (h ) − 1] − f (x)
= lim 2
h→ 0
h
f (x) + f (h ) − 1 − f (x)
= lim
h→ 0
h
f (h ) − 1
[from Eq. (ii)]
= lim
= −1
h→ 0
h
∴
⇒
(i) To check continuity at x = 0,
LHL (at x = 0) = lim − h = 0
h→ 0
RHL = lim
h→ 0
Also,
h
=0
e2/ h
Since,
⇒
f ′ (0) = −1, we get
f (0 + h ) − f (0)
lim
= −1
h→ 0
h
f (h ) − 1
lim
= −1
h→ 0
h
f (x) = − x + k, where, k is a constant.
But
f (0) = 1,
1=k
f (x) = 1 − x, ∀x ∈ R ⇒
f (2) = −1
∴ f (0) = f (0) ⋅ f (0) ⇒ f (0) { f (0) − 1} = 0
⇒
f (0) = 1
[Q f (0) ≠ 0]
f (0 + h ) − f (0)
Since, f ′ (0) = 2 ⇒ lim
=2
h→ 0
h
f (h ) − 1
[Q f (0) = 1] …(i)
⇒
lim
=2
h→ 0
h
f (x + h ) − f (x)
Also,
f ′ (x) = lim
h→ 0
h
f (x) ⋅ f (h ) − f (x)
,
= lim
h→ 0
h
[using, f (x + y) = f (x) ⋅ f ( y)]
f (h ) − 1 

= f (x) lim

→
0
h
h


[Q f (0) = 1]
∴
⇒
 x
2 f   = f (x) + 1
 2
 x
f (x) = 2 f   − 1, ∀ x, y ∈ R
 2
− 1 dx
57. We have, f (x + y) = f (x) ⋅ f ( y), ∀ x, y ∈ R.
(ii) To check differentiability at x = 0,
f (0 − h ) − f (0)
L f ′ (0) = lim
h→ 0
−h
(0 − h ) − 0
= lim
=1
h→ 0
−h
f (0 + h ) − f (0)
R f ′ (0) = lim
h→ 0
h
he−2 / h − 0
= lim
=0
h→ 0
h
∴ f (x) is not differentiable at x = 0.
x + y f (x) + f ( y)
56. Given, f 
, ∀x, y ∈ R
 =
 2 
2
⇒
∫ f ′ (x)dx = ∫
⇒
⇒
⇒
∴ f (x) is continuous at x = 0
⇒
f ′ (x) = −1, ∀x ∈ R
therefore f (0) = −0 + k
f (0) = 0
On putting y = 0, we get
 x f (x) + f (0) 1
f   =
= [1 + f (x)]
 2
2
2
f (x + h ) − f (x)
h
 2x + 2h 
f
 − f (x)
 2 
…(i)
…(ii)
f ′ (x) = 2 f (x)
f ′ (x)
=2
f (x)
[from Eq. (i)]
On integrating both sides between 0 to x, we get
x f ′ (x)
∫ 0 f (x) dx = 2x
⇒
log e | f (x)| − log e | f (0)| = 2x
⇒
log e| f (x)| = 2x
⇒
log e | f (0)| = 0
⇒
f (x) = e2 x
[Q f (0) = 1]
Limit, Continuity and Differentiability 231
−2 ≤ x ≤ 0
 −1 ,
1
0 < x≤2
−
(
),
x

Since, x ∈ [−2, 2]. Therefore,|x|∈ [0, 2]
58. y = [x] + |1 − x|, − 1 ≤ x ≤ 3
61. Given that, f (x) = 
−1 + 1 − x, −1 ≤ x < 0
 0 + 1 − x, 0 ≤ x < 1
y=
 1 + x − 1, 1 ≤ x < 2
 2 + x − 1, 2 ≤ x ≤ 3
 − x, −1 ≤ x < 0
 1 − x, 0 ≤ x < 1

y=
1 ≤ x<2
 + x,
 x + 1, 2 ≤ x < 3
⇒
⇒
⇒
f (|x|) = |x| − 1, ∀ x ∈ [−2, 2]
 x − 1, 0 ≤ x ≤ 2
f (|x|) = 
− x − 1 , − 2 ≤ x ≤ 0
⇒
−2 ≤ x < 0
 1,

| f (x)| = 1 − x, 0 ≤ x < 1
x − 1 , 1 ≤ x ≤ 2

Also,
g (x) = f (|x|) + | f (x)|
 − x − 1 + 1 , −2 ≤ x ≤ 0

= x − 1 + 1 − x, 0 ≤ x < 1
x − 1 + x − 1 , 1 ≤ x ≤ 2

Also,
which could be shown as,
Y
4
1+
3
2
– 1
x
X′
–1
x
−2 ≤ x ≤ 0
 − x,

g (x) =  0,
0 ≤ x<1
2(x − 1), 1 ≤ x ≤ 2

x
1
–
x
O
1
2
3
∴
Y′
Clearly, from above figure, y is not continuous and not
differentiable at x = {0, 1, 2} .
59. Since,
| f ( y) − f (x)| ≤ (x − y)
⇒
| f ( y) − f (x)|2
≤ (x − y)
( y − x)2
2
3
2
⇒

f ( y) − f (x)
 ≤x− y
 y−x 
…(i)
2
⇒
f ( y) − f (x)
 ≤ lim (x − y)
lim 
y → x y − x 
y→ x
⇒
 − 1 , −2 ≤ x ≤ 0

g ′ (x) =  0,
0 ≤ x<1
 2,
1 ≤ x≤2

X
| f ′ (x)|2 ≤ 0
which is only possible, if| f ′ (x)| = 0
∴ f ′ (x) = 0
or f ′ (x) = Constant
60. Since, f (−x) = f (x)
∴f (x) is an even function.
f (0 + h ) − f (0)
∴ f ′ (0) = lim
h→ 0
h
f (0 − h ) − f (0)
= lim
h→ 0
−h
Since, f ′ (0) exists.
∴
R f ′ (0) = L f ′ (0)
f (h ) − f (0)
f (h ) − f (0)
⇒ lim
= lim
h→ 0
h→ 0
h
−h
f (h ) − f (0)
=0
⇒ 2 lim
h→ 0
h
f (h ) − f (0)
⇒ lim
=0
h→ 0
h
∴
f ′ (0) = 0
∴ RHD (at x = 1) = 2, LHD (at x = 1) = 0
⇒ g (x) is not differentiable at x = 1.
Also, RHD (at x = 0) = 0, LHD at (x = 0) = − 1
⇒ g (x) is not differentiable at x = 0.
Hence, g (x) is differentiable for all x ∈ (−2, 2) − {0, 1}
f (x) = x3 − x2 − x + 1
⇒
f ′ (x) = 3x2 − 2x − 1
= (3x + 1) (x − 1)
∴ f (x) is increasing for x ∈ (−∞ , − 1 /3) ∪ (1, ∞ )
and decreasing for x ∈ (−1 /3, 1)
max { f (t ); 0 ≤ t ≤ x}, 0 ≤ x ≤ 1
Also, given g (x) = 
3 − x,
1 < x≤2

62. Given,
⇒
 f (x), 0 ≤ x ≤ 1
g (x) = 
3 − x, 1 < x ≤ 2
⇒
x3 − x2 − x + 1, 0 ≤ x ≤ 1
g (x) = 
3 − x,
1 < x≤2

At x = 1,
[Q f (− h ) = f (h )]
RHL = lim (3 − x) = 2
x→1
and
LHL = lim (x3 − x2 − x + 1) = 0
x→1
∴ It is discontinuous at x = 1.
3x2 − 2x − 1, 0 ≤ x ≤ 1
Also, g ′ (x) = 
1 < x≤2
−1 ,

⇒
g ′ (1+ ) = − 1
and g ′ (1− ) = 3 − 2 − 1 = 0
∴ g (x) is continuous for all x ∈ (0, 2) − {1} and g (x) is
differentiable for all x ∈ (0, 2) − {1}.
232 Limit, Continuity and Differentiability
x −1

, when x ≠ 1
2x2 − 7x + 5
63. Given that, f (x) = 
1
−
, when x = 1
 3
f (1 + h ) − f (1)
RHD = lim
h→ 0
h

1 + h −1
 1 
2(1 + h )2 − 7(1 + h ) + 5 −  − 3 

= lim 
h→ 0
h
3h + 2 (1 + h )2 − 7(1 + h ) + 5 
= lim 

h→ 0 3 h {2 (1 + h )2 − 7 (1 + h ) + 5 }


2


2h
2
= lim 
 =−
h→ 0  3 h (−3 h + 2 h 2 )
9
f (1 − h ) − f (1)
h→ 0
−h

1 − h −1
 1 
2(1 − h )2 − 7(1 − h ) + 5 −  − 3 

= lim 
h→ 0
−h
−3h + 2(1 + h 2 − 2h ) − 7(1 − h ) + 5
= lim
h→ 0
−3h [2(1 − h )2 − 7(1 − h ) + 5]
And, since ( fog ) (x) is not differentiable at the point
where g (x) is not differentiable as well as at those points
also where g (x) attains the values so that f ( g (x)) is
non-differentiable.
Since g (x) is not continuous at x = 0 ∈ (− 1, 1) so fog (x) is
not differentiable and as f (x) = |2x − 1| + |2x + 1| is not
differentiable at x = − 1 / 2 and 1/2, so ( fog ) (x) is not
differentiable for those x, for which g (x) = − 1 / 2 or 1/2.
But g (x) ≥ 0, so g (x) can be 1/2 only and for x = − 1 / 2 and
1
1/2, g (x) = .
2
So, ( fog ) (x) is not differentiable at x = − 1 / 2, 0, 1 / 2 ,
therefore value of d = 3
∴
c + d = 1 + 3 = 4.
LHD = lim
= lim
h→ 0
2
.
9
64. Given, f (x) = x tan −1 x
Using first principle,
 f (1 + h ) − f (1) 
f ′ (1) = lim

h→ 0 
h

π
2
x
(f ′ (t ) cosec t − cot t cosec t f (t ))dt
π
 π
g (x) = f   cosec − f (x) cosec x
 2
2
f (x)
⇒ g (x) = 3 −
sin x
 3 sin x − f (x)
lim g (x) = lim


x→ 0
x→ 0 
sin x
3 cos x − f ′ (x)
= lim
x→ 0
cos x
3 −1
=
=2
1
∴
2h 2
2
= − ∴ LHD = RHD
9
−3h (2h 2 + 3h )
Hence, required value of f ′ (1) = −
g (x) = ∫
66.
67.
PLAN
(i) In these type of questions, we draw the graph of the function.
(ii) The points at which the curve taken a sharp turn, are the
points of non-differentiability.
Curve of f (x) and g (x) are
 (1 + h ) tan −1 (1 + h ) − tan −1 (1) 
= lim 

h→ 0
h


 tan −1 (1 + h ) − tan −1 (1) h tan −1 (1 + h ) 
= lim 
+

h→ 0
h
h


1

 h 
−1
= lim  tan −1 
 + tan (1 + h )
 2 + h
h→ 0  h


−1  h  
 tan  2 + h   π
= lim 
+
h→ 0
 (2 + h ) ⋅ h  4
2 + h 


−1  h  

 tan 
1 
2 + h  π 1 π
= lim
+ = +
h
h→ 0 2 + h 
 4 2 4


(2 + h ) 

given
functions
and
f : (− 1, 1) → R
g : (− 1, 1) → (− 1, 1) be defined by
f (x) = |2x − 1| + |2x + 1|and g (x) = x − [x].
As, we know the composite function ( fog ) (x) is
discontinuous at the points, where g (x) is discontinuous
for given domain. And, since g (x) is discontinuous at
x = 0 lies in interval (− 1, 1), so value of c = 1.
65. The
h (x) is not differentiable at x = ± 1 and 0.
As, h (x) take sharp turns at x = ± 1 and 0.
Hence, number of points of non-differentiability of h (x)
is 3.
68. Let
p (x) = ax4 + bx3 + cx2 + dx + e
⇒
p′ (x) = 4ax3 + 3bx2 + 2cx + d
∴
p′ (1) = 4a + 3b + 2c + d = 0
and p′ (2) = 32a + 12b + 4c + d = 0
p (x)

Since,
lim 1 + 2  = 2
x→ 0 
x 
ax4 + bx3 + (c + 1) x2 + dx + e
=2
x→ 0
x2
∴ lim
⇒
⇒
c + 1 = 2, d = 0, e = 0
c=1
... (i)
... (ii)
[given]
Limit, Continuity and Differentiability 233
 sin x − cos x 
−1  tan x − 1 

 = tan 
 tan x + 1
 sin x + cos x
From Eqs. (i) and (ii), we get
3. Let f (x) = tan −1 
4a + 3b = − 2
and
⇒
32a + 12b = − 4
1
a = and b = − 1.
4
[dividing numerator and denominator

by cos x > 0, x ∈ 0,

x4
− x3 + x2
4
16
p(2) =
−8 + 4
4
p(x) =
∴
⇒
⇒
= tan
Topic 8 Differentiation
1. It is given
(a + 2b cos x)(a − 2b cos y) = a 2 − b2
where
a > b > 0,
On differentiating w.r.t. ‘y’, we get
dx

(a − 2b cos y)0 + 2b (− sin x) 

dy
π
π


f (x) = tan −1  tan  x −   = x −


4
4

 π π 
−1
Q tan tan θ = θ, for θ ∈  − 2 , 2  


x
Now, derivative of f (x) w.r.t. is
2
d ( f (x))
df (x)
=2
d (x / 2)
d (x)
+ 2b(sin y)(a + 2 b cos x) = 0
 π π
At  ,  , we get
 4 4
dx

(a − b) − b  + b (a + b) = 0

dy
dx a + b
=
dy a − b
=2 ×
4.
Hence, option (b) is correct.
2. We know,
(1 + x)n = nC 0 + nC1x + nC 2 x2 + ... + nC n xn
On differentiating both sides w.r.t. x, we get
n (1 + x)n − 1 = nC1 + 2 nC 2 x + ... + n nC n xn − 1
On multiplying both sides by x, we get
n x(1 + x)n − 1 = nC1x + 2nC 2x2 + ... + n nC nxn
Again on differentiating both sides w.r.t. x,
we get
n [(1 + x)n − 1 + (n − 1) x (1 + x)n − 2]
= nC1 + 22 nC 2x + ... + n 2 nC nxn − 1
Now putting x = 1 in both sides, we get
C1 + (22) nC 2 + (32) nC3 + ... + (n 2) nC n
n
= n (2n − 1 + (n − 1) 2n − 2)
For n = 20, we get
C1 + (22) 20C 2 + (32) 20C3 + ... + (20)2
= 20(219 + (19) 218 )
= 20 (2 + 19) 218 = 420 (218 )
= A (2B ) (given)
On comparing, we get
( A , B) = (420, 18)
20
C 20
π 


tan  x − 4  


Then,
+ (a + 2b cos x)(0 − 2b (− sin y)) = 0
dx

⇒ (a − 2b cos y) − 2b (sin x) 

dy
20

π 
 tan x − tan

4  = tan −1

π



 1 +  tan  (tan x)


4
 tan A − tan B

Q 1 + tan A tan B = tan ( A − B)


 π
Since, it is given that x ∈ 0,  , so
 2
π  π π
x − ∈− , 
4  4 4
π  π π

Also, for  x −  ∈  − ,  ,

4  4 4
p(2) = 0
⇒
−1
π
]
2
d 
π
x −  = 2
4
dx 
Key Idea Differentiating the given equation twice w.r.t. ‘x’.
Given equation is
ey + xy = e
…(i)
On differentiating both sides w.r.t. x, we get
dy
dy
ey
+x
+ y=0
dx
dx
 y 
dy
=− y
⇒

dx
 e + x
Again differentiating Eq. (ii) w.r.t. ‘x’, we get
2
d 2y
d 2y dy dy
y  dy
e
ey
+
=0
+
+x 2+


2
 dx
dx dx
dx
dx
Now, on putting x = 0 in Eq. (i), we get
ey = e1
⇒
y=1
On putting x = 0, y = 1 in Eq. (iii), we get
dy
1
1
=−
=−
dx
e+0
e
Now, on putting x = 0, y = 1 and
Eq. (iv), we get
dy
1
= − in
dx
e
…(ii)
…(iii)
…(iv)
234 Limit, Continuity and Differentiability
e1
2
 d 2y  1  1
d 2y
1  1
+
−
+
0
e
 2 +  −  +  −  = 0


 e
dx2
 dx   e  e
7. Given equation is
2
1
d y
=
dx2 ( 0, 1) e2
⇒
 dy d 2y
 1 1
,
 at (0, 1) is  − , 2

 e e
 dx dx2 
So,
5. Let y = f ( f ( f (x))) + ( f (x))2
On differentiating both sides w.r.t. x, we get
dy
= f ′ ( f ( f (x))) ⋅ f ′ ( f (x)) ⋅ f ′ (x) + 2 f (x) f ′ (x)
dx
[by chain rule]
So, dy
dx at
∴
(2x)2y = 4 ⋅ e2x − 2y
On applying ‘ log e ’ both sides, we get
log e (2x)2y = log e 4 + log e e2x − 2y
2 y log e (2x) = log e (2)2 + (2x − 2 y)
[Q log e nm = m log e n and log e ef ( x ) = f (x)]
⇒ (2 log e (2x) + 2) y = 2x + 2 log e (2)
x + log e 2
y=
⇒
1 + log e (2x)
On differentiating ‘y’ w.r.t. ‘x’, we get
dy
=
dx
x=1
dy
= f ′ ( f (1)) ⋅ f ′ (1) ⋅ (3) + 2(1)(3)
dx x = 1
= f ′ (1) ⋅ (3) ⋅ (3) + 6
= (3 × 9) + 6 = 27 + 6 = 33
So, (1 + log e (2x))2
dy  x log e (2x) − log e 2
=


dx 
x
written as
2


 3 cot x + 1 
 3 cos x + sin x 
2 y =  cot−1 

  =  cot−1 
 cot x − 3  
 cos x − 3 sin x  


π




 cot cot x + 1 
−1
6
=  cot 


 cot x − cot π  


6 

 π

=  cot−1  cot − x  

 6

2
y2 = 4 + x2 − x log e (log e x)
2
[dividing each term of numerator and
denominator by sin x]
2
π


Q cot = 3


6

cot A cot B + 1 
Q cot( A − B) = cot B − cot A 


π
6
2
π
π

π
<x<
+  − x  ,

6
6
2
0<x<

π + θ , − π < θ < 0 
Q cot−1 (cot θ ) =  θ ,
0 <θ < π 



θ − π , π < θ < 2π 



2
 π
π

, 0<x<
x
−




 6
6
⇒ 2y = 
2
π
π
π
7



<x<
− x ,
 6

6
2
π
 π

2 − x (−1), 0 < x <

dy   6
6
⇒2
=
π
π

dx   7π
2
<x<
− x (−1),

  6
2
6
π
π

dy  x − 6 , 0 < x < 6
=
⇒
dx x − 7π , π < x < π

6
6
2
… (i)
Now, differentiating Eq. (i) w.r.t. x, we get
dy
1
1
2y
= 2x − x
. − 1 ⋅ log e (log e x)
dx
log e x x
[by using product rule of derivative]
1
2x −
− log e (log e x)
log e x
 dy
… (ii)
⇒
  =
 dx
2y
Now, at x = e, y2 = 4 + e2 − e log e (log e e)
[using Eq. (i)]
= 4 + e2 − e log e (1) = 4 + e2 − 0
= e2 + 4
2

π
 − x ,

6
2
2x
8. We have, x log e (log e x) − x2 + y2 = 4, which can be
6. Given expression is



=
 π

(1 + log e (2x))1 − (x + log e 2)
(1 + log e (2x))2
1
1 + log e (2x) − 1 − log e 2
x
=
(1 + log e (2x))2
= f ′ ( f ( f (1))) ⋅ f ′ ( f (1)) ⋅ f ′ (1) + 2 f (1) f ′ (1)
[Q f (1) = 1 and f′ (1) = 3]
... (i)
⇒
y = e2 + 4
[Q y > 0]
∴ At x = e and y = e2 + 4,
dy 2e − 1 − 0
2e − 1
=
=
dx 2 e2 + 4 2 e2 + 4
[using Eq. (ii)]
9. We have,
f (x) = x3 + x2f ′ (1) + xf ′ ′ (2) + f ′ ′ ′ (3)
⇒
f ′ (x) = 3x2 + 2xf ′ (1) + f ′ ′ (2)
⇒ f ′ ′ (x) = 6x + 2 f ′ (1)
⇒ f ′ ′ ′ (x) = 6
⇒ f′ ′ ′ (3) = 6
Putting x = 1 in Eq. (i), we get
f ′ (1) = 3 + 2 f ′ (1) + f ′ ′ (2)
and putting x = 2 in Eq. (ii), we get
f ′ ′ (2) = 12 + 2 f ′ (1)
From Eqs. (iv) and (v), we get
f ′ (1) = 3 + 2 f ′ (1) + (12 + 2 f ′ (1))
⇒
3 f′ (1) = − 15
… (i)
… (ii)
… (iii)
… (iv)
…(v)
Limit, Continuity and Differentiability 235
⇒
⇒
∴
⇒
⇒
f′ (1) = − 5
[using Eq. (v)]
f′ ′ (2) = 12 + 2 (− 5) = 2
3
2
f (x) = x + x f ′ (1) + xf ′ ′ (2) + f ′ ′ ′ (3)
f (x) = x3 − 5x2 + 2x + 6
f (2) = 23 − 5(2)2 + 2(2) + 6 = 8 − 20 + 4 + 6 = − 2
10. We have, x = 3 tan t and y = 3 sec t
dy
d
(3 sec t )
dy dt dt
Clearly, =
=
dx dx d (3 tan t )
dt dt
3 sec t tan t tan t
=
=
= sin t
sec t
3 sec2 t
... (i)
g (x + 1) − g (x) = log x
1
Replacing x by x − , we get
2
1
1
1



g  x +  − g  x −  = log  x −  = log (2x − 1) − log 2



2
2
2
dx
1
 dy
=
= 
dy dy / dx  dx
15. Since,
d  dx
d  dy
 
  =
dy  dy dx  dx
⇒
 2 ⋅ (3x3/ 2) 
 6x x 
 = tan − 1 
3/ 2 2 
3
 1 − 9x 
1 − (3x ) 
⇒
d
{ f ′ (x)} = − f (x)
dx
⇒
g′ (x) = − f (x)
  x 
  x 
F (x) =  f    +  g   


2
  2 


Also,
  x 
 x 1
+ 2  g    ⋅ g′   ⋅ = 0 [from Eq.(i)]
 2 2
  2 
fog (x) = x
∴
1
f ′{ g (x)} × g′ (x) = 1 ⇒ g′ (x) =
f ′ ( g (x))

1 
Q f ′ (x) =

1 + x5 

13. Given, y = sec (tan− 1 x )
Let
⇒
∴
tan
x=θ
x = tan θ
y = sec θ = 1 + x2
On differentiating w.r.t. x, we get
dy
1
=
⋅ 2x
dx 2 1 + x2
At x = 1,
dy
1
=
dx
2
F (x) is constant ⇒ F (10) = F (5) = 5
17. Let, g (x) = f (x) − x2
g′ (x) = 1 + { g (x)}5
−1
2
  x 
 x 1
F ′ (x) = 2  f    ⋅ f ′   ⋅
 2 2
  2 
⇒
On differentiating w.r.t. x, we get
⇒
[Q g (x) = f ′ (x), given]…(i)
2
12. Here, g is the inverse of f (x).
1
=
1
1 + { g (x)}5
dx
dy
f ′′ (x) = − f (x)
16. Since,

2x 
= 2 tan − 1 (3x3/ 2) Q 2 tan − 1 x = tan − 1

1
− x2 

9
dy
3
1
⋅ x
= 2⋅
⋅ 3 × (x)1/ 2 =
dx
2
1 + 9x3
1 + (3x3/ 2)2
9
g (x) =
1 + 9x3
−1
−1
 d 2y  dy −3
 d 2y  dy − 2 dx
d 2x
=
−
=
−
 2  






dy2
 dx   dx
 dx2   dx  dy
⇒
11. Let y = tan − 1 
⇒
... (ii)
On substituting, x = 1, 2, 3,... , N in Eq. (ii) and adding,
we get

 1 1
1
1
 1

.
+ ... +
g′′  N +  − g′′   = − 4 1 + +
2
 2

2
9
25
(
2
1
)
−
N


d 2y d  dy
d  dy dt
=
  =
 ⋅
2


dx dx
dt  dx dx
dx
d  dy
d
 
(sin t )
cos t
cos3 t
dt  dx
= dt
=
=
=
2
dx
d
3
(3 tan t ) 3 sec t
dt
dt
π
cos3
1
π
d 2y 
4 = 1
Now,
=
 at t =  =
3
4
3(2 2 ) 6 2
dx2 
∴
g (x + 1) = log x + g (x)
and
i.e.
1
1
−4


∴ g′′  x +  − g′′  x −  =


2
2 (2x − 1)2
and
∴
14. Since, f (x) = e g ( x ) ⇒ e g ( x + 1)= f (x + 1) = xf (x) = xe g ( x )
18. Given that, log (x + y) = 2xy
2
x
1+
θ
1
⇒ g (x) has atleast 3 real roots which are x = 1, 2, 3
[by mean value theorem]
⇒ g ′ (x) has atleast 2 real roots in x ∈ (1, 3)
⇒ g ′ ′ (x) has atleast 1 real roots in x ∈ (1, 3)
⇒ f ′ ′ (x) – 2 . 1 = 0. for atleast 1 real root in x ∈ (1, 3)
⇒ f ′ ′ (x) = 2, for atleast one root in x ∈ (1, 3)
x
∴ At x = 0, ⇒ log ( y) = 0 ⇒ y = 1
dy
at (0, 1)
∴ To find
dx
On differentiating Eq. (i) w.r.t. x, we get
1
x+ y
⇒
dy
dy

+ 2y ⋅ 1
1 +
 = 2x

dy
dx
dy 2 y (x + y) − 1
 dy
=1
=
⇒  
 dx ( 0, 1)
dx 1 − 2 (x + y) x
…(i)
236 Limit, Continuity and Differentiability
19. Given, x2 + y2 = 1
22. Given, xexy = y + sin 2 x
2x + 2 yy′ = 0 ⇒ x + yy′ = 0.
Again, differentiating w.r.t. x, we get
1 + y ′ y ′ + yy ′′ = 0 ⇒ 1 + ( y′ )2 + yy′′ = 0
x3
20. Given,
sin x cos x
f (x) = 6
p
−1
p2
0
p3
⇒
y=0
On differentiating Eq. (i) both sides w.r.t. x, we get
 dy
 dy
+ 2 sin x cos x
1 . exy + x ⋅ exy  x ⋅
+ y =
 dx
 dx
On putting x = 0, y = 0, we get
e0 + 0(0 + 0) =
On differentiating w.r.t. x, we get
3x2 cos x − sin x
f ′ (x) = 6
p
−1
p2
x3
sin x cos x
+ 0
p
0
p3
0
p2
x3
+ 6
0
⇒
⇒
p2
−1
0
0
0
p3
p2
6
0
p3
∴
+0+0
0
0 = 0 = independent of p
p
p2
p3
g (x) = f [ f (x)] =||x − 2|− 2|
x>2
∴
g′ (x) = 1 when x > 2
1
25. Let u = sec−1  − 2  and v = 1 − x2
 2x − 1
∴
x = cos θ
u = sec−1 (− sec 2 θ ) and v = sin θ
⇒
u = π − 2θ
Put
On differentiating both sides, we get
2 yy1 = P ′ (x),
du
dv
= − 2 and
= cos θ
dθ
dθ
du
2
 du 
,  
=−
= −4
dv
cos θ  dv  θ = π /3
⇒
2 yy2 + 2 y12 = P ′ ′ (x)
2 y3 y2 + 2 y2y12 = y2P ′ ′ (x)
⇒
2 y y2 = y P ′ ′ (x) − 2 ( yy1 )
⇒
2 y3 y2 = P (x) ⋅ P ′ ′ (x) −
2
{ P ′ (x)}2
2
Again, differentiating, we get
d 3
2
( y y2) = P ′ (x) ⋅ P ′ ′ (x) + P (x) ⋅ P ′ ′ ′ (x)
dx
2P ′ (x) ⋅ P ′ ′ (x)
−
2
d 3
⇒
2
( y y2) = P (x) ⋅ P ′ ′ ′ (x)
dx
d  3 d 2y
⇒ 2
 y ⋅ 2  = P (x) ⋅ P ′ ′ ′ (x)
dx 
dx 
26. Given,
∴
[Q sec–1 (− x) = π − sec−1 x]
v = sin θ
and
Again, differentiating, we get
⇒
x>0
x<0
g (x) = |(x − 2) − 2| = | x − 4| = x − 4
−1
−1
2
x>0
x<0
f (x) = | x − 2|
When,
6
f ′ ′ ′ (0) = 6
3
f ′ ′ (x) = 2,
− 2 ,
24. Given,
21. Since, y2 = P (x)
⇒
f ′ (x) = 2x,
 − 2x,
Therefore, f (x) is twice differentiable for all x ∈ R − {0}.
p3
−1
p2
 2
if x ≥ 0
f (x) = x , 2
if x < 0
− x ,
f (x) is not differentiable at x = 0 but all R − {0} .
⇒
⇒
− cos x sin x
and f ′ ′ ′ (x) = 6
p
23. Given, f (x) = x|x|
Therefore,
6x − sin x − cos x
f ′ ′ (x) = 6
−1
0
+0+0
p
∴
sin x cos x
 dy 
+ 2 sin 0 cos 0
 dx 
( 0, 0)
 dy 
=1
 dx 
0, 0
⇒
0
p3
3x2 cos x − sin x
f ′ (x) = 6
−1
0
p
…(i)
On putting x = 0, we get
0 . e0 = y + 0
On differentiating w.r.t. x, we get
f (x) = log x (log x)
log (log x)
f (x) =
log x
On differentiating both sides, we get
1
 1 1
⋅  − log (log x) ⋅
(log x) 
 log x x
x
f ′ (x) =
(log x)2
∴
f ′ (e) =
1
 1 1
1 ⋅  ⋅  − log (1) ⋅
 1 e
e
(1)2
⇒ f ′ (e) =
1
e
Limit, Continuity and Differentiability 237
f1 (x)
27. Given, F (x) = g1 (x)
f2 (x)
g2 (x)
f3 (x)
g3 (x)
h1 (x) h2 (x) h3 (x)
f1 ′ (x) f2 ′ (x) f3 ′ (x)
F ′ (x) = g1 (x) g2 (x) g3 (x)
∴
h1 (x)
F ′ (a ) = 0 + 0 + 0 = 0
[Q fr (a ) = gr (a ) = hr (a ) ; r = 1, 2, 3]
28. Given,
and
∴
 2x − 1
y= f  2

 x + 1
 2x − 1 −2x2 + 2x + 2
= sin 2  2
⋅
 x + 1
(x2 + 1)2
=
y=
 2x − 1
−2 (x2 − x − 1)
sin 2  2

2
2
 x + 1
(x + 1)
ax2
bx
c
+
+1
+
(x − a ) (x − b) (x − c) (x − b) (x − c) (x − c)
=
x
ax2
bx
+
+
(x − a ) (x − b) (x − c) (x − b) (x − c) (x − c)
=

ax
x  b
+
+ 1


(x − a ) (x − b) (x − c) (x − c)  x − b
2
ax2
x
x
=
+
⋅
(x − a ) (x − b) (x − c) (x − c) (x − b)
y=
x3
(x − a ) (x − b) (x − c)
⇒
log y = log x3 − log (x − a ) (x − b) (x − c)
⇒
log y = 3 log x − log (x − a ) − log (x − b) − log(x − c)
On differentiating, we get
1
1
1
y′ 3
= −
−
−
y x x − a x−b
x− c
⇒
⇒
⇒
⇒
y ′ 1
1  1
1  1
1 
= −
+ −
+ −

y  x x − a   x x − b  x x − c
y′
−a
b
c
=
−
−
y x (x − a ) x (x − b) x (x − c)
y′
a
b
c
=
+
+
y x (a − x) x (b − x) x (c − x)
y′ 1  a
b
c 
= 
+
+

y x  a − x b − x c − x
π 

+ 2x log 2 ⋅ tan {log (x + 2)} = 0

3
Putting  x = − 1, y = −
 , we get
π 

dy
=
dx

3

−
 π 
2
 3
1− 
 π 
2
=
3
π
π2 − 3
x = sec θ − cos θ and y = secn θ − cos n θ
On differentiating w.r.t. θ respectively, we get
dx
= sec θ tan θ + sin θ
dθ
dy
and
= n secn − 1 θ ⋅ sec θ tan θ − n cos n − 1 θ ⋅ (− sin θ )
dθ
dx
⇒
= tan θ (sec θ + cos θ )
dθ
dy
and
= n tan θ (secn θ + cos n θ )
dθ
31. Given,
⇒
dy n (secnθ + cos n θ )
=
dx
sec θ + cos θ
2
n 2 (secnθ + cos n θ )2
 dy
∴   =
 dx
(sec θ + cos θ )2
 a

x2
=
+ 1


(x − c) (x − b)  x − 1
⇒
+
 sin x  −1
dy
π 

+ sin x (sin y) 2  ⋅ cos y ⋅


2
dx
3
2
2x ⋅ sec2 {log (x + 2)}
+
⋅
+
(x + 2)
2 (2|x|) 4x2 − 1
f ′ (x) = sin 2 x
 2x − 1 d  2x − 1
dy
= f′  2


⋅
dx
 x + 1 dx  x2 + 1
 2x − 1  (x2 + 1) ⋅ 2 − (2x − 1) (2x) 
= sin 2 2
 ⋅

 x + 1 
(x2 + 1)2

29.
3
sec−1 (2x) + 2x tan {log (x + 2)} = 0
2
On differentiating both sides, we get
π
sin x
π π
(sin y) 2 ⋅ log (sin y) ⋅ cos x ⋅
2 2
h1 (x) h2 (x) h3 (x)
f1 (x)
f2 (x)
f3 (x)
f2 (x)
f3 (x)
g2 ′ (x) g3 ′ (x) + g1 (x) g2 (x) g3 (x)
h1 ′ (x) h2 ′ (x) h3 ′ (x)
h2 (x) h3 (x)
f1 (x)
+ g1 ′ (x)
⇒
π
sin x
2
30. Here, (sin y)
=
n 2 {(secn θ − cos n θ )2 + 4} n 2 ( y2 + 4)
=
(x2 + 4)
{(sec θ − cos θ )2 + 4}
2
 dy
⇒ (x2 + 4)   = n 2( y2 + 4)
 dx
A (x)
32. Let φ (x) = A (α )
B(x)
B (α )
C (x)
C (α )
…(i)
A ′ (α ) B ′ (α ) C ′ (α )
Given that, α is repeated root of quadratic equation
f (x) = 0.
∴ We must have f (x) = (x − α )2 ⋅ g (x)
A ′ (x) B ′ (x) C ′ (x)
∴
φ ′ (x) = A (α ) B (α ) C (α )
A ′ (α ) B ′ (α ) C ′ (α )
⇒
⇒
A ′ (α ) B ′ (α ) C ′ (α )
φ ′ (α ) = A (α ) B (α ) C (α ) = 0
A ′ (α ) B ′ (α ) C ′ (α )
x = α is root of φ′ (x).
238 Limit, Continuity and Differentiability
⇒
(x − α ) is a factor of φ′ (x) also.
or we can say (x − α )2 is a factor of f (x).
⇒
3
Here, u = ex sin x and log v = x log (tan x)
φ (x) is divisible by f (x).

33. Given,y = (log cos xsin x) ⋅ (logsin x cos x)−1 + sin −1 


2
 log e (sin x)
−1  2 x 
y=

 + sin 
 log e (cos x)
 1 + x2 
(log e (cos x) ⋅ cot x
loge ( in x) ⋅ tan x )
+

log (sin x)
dy
=2  e
⋅
dx
{log e (cos x)}2
 log e (cos x)
∴
⇒
 dy
⇒  
 dx  x = π 

4

 1 
 2 ⋅ log   
 2 
2

+
= 2 1 ⋅
2
2 
1
  1+ π
  log


 
16
2  

8
32
=−
+
log e 2 16 + π 2
2x  

1 + x2  


2
+
1
x2
+


 x 

 a + bx
y = x [log (x) − log (a + bx)]
…(i)
On differentiating both sides, we get
1
dy
b 
=x −
 + 1 [log (x) − log (a + bx)]
 x a + bx
dx
⇒ x


dy
a
ax
+ y …(ii)
= x2 
 + y ⇒ x y1 =
 x (a + bx)
dx
a + bx
From Eqs. (i), (ii) and (iii), wet get
3
dy
= ex sin x (3x3 ⋅ cos x3 + sin x3 ) + (tan x)x
dx
[2x cosec 2x + log (tan x)]
5x
37. Given, y =
+ cos 2(2x + 1)
3|1 − x|
⇒ x3 y2 =
 ax 
a 2x2
⇒ x3 y2 = 

2
 (a + bx)
(a + bx)
⇒
2
[from Eq. (ii)]
⇒
2
As
3
and
⇒
h ′ x = 2 f (x) ⋅ f ′ (x) + 2 g (x) ⋅ g ′ (x)
∴
= 2 [ f (x) ⋅ g (x) − g (x) ⋅ f (x)]
=0
F (x) 1
=
G (x) 14
F ′ (x) 1
lim
=
x→1 G′ (x)
14
x→1
35. Given, h (x) = [ f (x)]2 + [ g (x)]2
⇒
5

− 2 sin (4x + 2), x < 1
dy  3 (1 − x)2
=
5
dx −
− 2 sin (4x + 2), x > 1
 3 (x − 1)2
38. Here, lim
2
d 2y  dy

= x
− y
⇒ x y2 = (xy1 − y) ⇒ x

dx2  dx
3
 5x
+ cos 2(2x + 1), x < 1
3(1 − x)
y=
5x

+ cos 2(2x + 1), x > 1
3(x − 1)
The function is not defined at x = 1.
5  (1 − x) − x(−1) 
 
 − 2 sin (4x + 2), x < 1
(1 − x)2
dy 3 

=
⇒
dx  5  (x − 1) − x(1) 
− 2 sin (4x + 2), x > 1
 3  (x − 1)2 

 
Again, differentiating both sides, we get
 (a + bx) ⋅ 1 − x ⋅ b 
x y2 + y1 = a 
 + y1
(a + bx)2


[Q f ′ (x) = g (x) and g′ (x) = − f (x)]
∴ h (x) is constant.
⇒ h (10) = h (5) = 11
Given,
3
36. Since, y = ex sin x + (tan x)x , then
3
y = u + v, where u = ex sin x and v = (tan x)x
⇒
dy  du dv
=
+

dx  dx dx
…(ii)
1 dv x ⋅ sec2x
⋅
=
+ log (tan x)
v dx
tan x
dv
= (tan x)x [2x ⋅ cosec (2x) + log (tan x)] …(iii)
dx
and
⇒
34. Given, (a + bx) ey/ x = x ⇒ y = x log 
⇒
On differentiating both sides w.r.t. x, we get
3
du
= ex sin x ⋅ (3x3 cos x3 + sin x3 )
dx
∴
…(i)
F (x) = ∫
x
–1
[using L’Hospital’s rule]…(i)
f (t ) dt ⇒ F ′ (x) = f (x)
…(ii)
x
G (x) = ∫ t f { f (t )} dt
–1
G′ (x) = x f { f (x)}
F (x)
F ′ (x)
f (x)
lim
= lim
= lim
x→1 G (x)
x→1 G′ (x)
x→1 x f { f (x)}
f (1)
1 /2
=
=
1 f { f (1)}
f (1 / 2)
F (x) 1
lim
=
x→1 G (x)
14
1
2 = 1 ⇒ f  1 = 7
 2
 1 14
f 
 2
…(iii)
...(iv)
10
Application of Derivatives
Topic 1 Equations of Tangent and Normal
Objective Questions I (Only one correct option)
x
1. If the tangent to the curve y = 2 , x ∈ R, (x ≠ ± 3 ), at
x −3
a point (α , β ) ≠ (0, 0) on it is parallel to the line
(2019 Main, 10 April II)
2x + 6 y − 11 = 0, then
(a) | 6α + 2β | = 19
(c) | 2α + 6β | = 19
(b) | 6α + 2β | = 9
(d) | 2α + 6β | = 11
2. Let S be the set of all values of x for which the tangent to
the curve y = f (x) = x − x − 2x at (x, y) is parallel to the
line segment joining the points (1, f (1)) and (−1, f (−1)),
then S is equal to
(2019 Main, 9 April I)
3
1
(a)  , − 1

3
1
(b)  , 1
3 
2
1
(c) − , 1
 3 
1
(d) − , − 1

 3
3. If the tangent to the curve, y = x3 + ax − b at the point
(1, − 5) is perpendicular to the line, − x + y + 4 = 0, then
which one of the following points lies on the curve ?
(2019 Main, 9 April I)
(a) (−2, 2)
(c) (−2, 1)
(b) (2, − 2)
(d) (2, − 1)
4. The tangent to the curve y = x − 5x + 5, parallel to the
2
line 2 y = 4x + 1, also passes through the point
(2019 Main, 12 Jan II)
1
(a)  ,
4
7

2
7
(b)  ,
2
1

4
1
(c)  − , 7
 8 
1
(d)  , − 7
8

5. A helicopter is flying
along the curve given by
y − x3/ 2 = 7, (x ≥ 0). A soldier positioned at the point
1 
 , 7 wants to shoot down the helicopter when it is
2 
nearest to him. Then, this nearest distance is
(2019 Main, 10 Jan II)
(a)
1 7
3 3
(b)
5
6
(c)
1 7
6 3
(d)
1
2
6. If θ denotes the acute angle between the curves,
y = 10 − x2 and y = 2 + x2 at a point of their intersection,
then|tan θ|is equal to
(2019 Main, 9 Jan I)
(a)
7
17
(b)
8
15
(c)
4
9
(d)
8
17
7. If the curves y2 = 6x, 9x2 + by2 = 16 intersect each other
at right angles, then the value of b is
(a) 6
7
(b)
2
(c) 4
(2018 Main)
9
(d)
2
8. The normal to the curve y(x − 2) (x − 3) = x + 6 at the
point, where the curve intersects the Y -axis passes
(2017 Main)
through the point
1 1
(b)  , 
 2 2
1 1
(d)  , 
 2 3
1
1
(a)  − , − 
 2
2
1
1

(c)  , − 
2
3
 1 + sin x 
 π
 , x ∈ 0,  .
 2
 1 − sin x 
9. Consider f (x) = tan −1 
A normal to y = f (x) at x =
point
(a) (0, 0)
π
(c)  , 0
6 
π
also passes through the
6
(2016 Main)
2π 
(b)  0,


3
π
(d)  , 0
4 
10. The normal to the curve x2 + 2xy − 3 y2 = 0 at (1,1)
(a) does not meet the curve again
(2015 Main)
(b) meets in the curve again the second quadrant
(c) meets the curve again in the third quadrant
(d) meets the curve again in the fourth quadrant
11. The point(s) on the curve y3 + 3x2 = 12 y, where the
tangent is vertical, is (are)
(2002, 2M)
 11 
4
(a)  ±
, 0 (c) (0, 0)
, − 2 (b)  ±


3 
3

4
(d)  ±
, 2

3 
12. If the normal to the curve y = f (x) at the point
(3, 4) makes an angle
f ′ (3) is equal to
(a) –1
(c) 4/3
3π
with the positive X-axis, then
4
(2000, 1M)
(b) –3/4
(d) 1
normal to the curve x = a (cos θ + θ sin θ ),
y = a (sin θ − θ cos θ ) at any point ‘ θ ’ is such that
13. The
(a) it makes a constant angle with the X-axis
(b) it passes through the origin
(c) it is at a constant distance from the origin
(d) None of the above
(1983, 1M)
240 Application of Derivatives
Analytical & Descriptive Questions
Objective Questions II
(One or more than one correct option)
14. On the ellipse 4x + 9 y = 1, the point at which the
2
2
tangents are parallel to the line 8x = 9 y, are (1999, 2M)
2 1
(a)  , 
 5 5
2 1
(b)  − , 
 5 5
2
1
(c)  − , − 
 5
5
2
1

(d)  , − 
5
5
17. If | f (x1 ) − f (x2)| ≤ (x1 − x2)2, ∀x1 , x2 ∈ R.
Find the
equation of tangent to the curve y = f (x) at the point
(2005, 4M)
(1, 2).
18. The curve y = ax3 + bx2 + cx + 5, touches the X-axis at
P(−2, 0) and cuts the Y-axis at a point Q, where its
(1994, 5M)
gradient is 3. Find a, b, c.
19. Tangent at a point P1 {other than (0, 0)} on the curve
y = x3 meets the curve again at P2. The tangent at P2
meets the curve at P3 and so on.
15. If the line ax + by + c = 0 is a normal to the curve xy = 1,
then
(1986, 2M)
(a) a > 0, b > 0
(b) a > 0, b < 0
(c) a < 0, b > 0
(d) a < 0, b < 0
Show that the abscissa of P1, P2, P3 , …, Pn, form a GP.
Also, find the ratio of
[area (∆P1P2P3 )]/[area (∆P2P3 P4 )].
(1993, 5M)
20. Find the equation of the normal to the curve
y = (1 + x)y + sin −1 (sin 2 x) at x = 0.
(1993, 4M)
21. Find all the tangents to the curve y = cos (x + y),
−2π ≤ x ≤ 2π , that are parallel to the line x + 2 y = 0.
Fill in the Blank
(1985, 5M)
16. Let C be the curve y − 3xy + 2 = 0. If H is the set of
3
points on the curve C, where the tangent is horizontal
and V is the set of points on the curve C, where the
tangent is vertical, then H = K and V = K . (1994, 2M)
Integer & Numerical Answer Type Question
22. The slope of the tangent to the curve ( y − x5 )2 = x (1 + x2)2
at the point (1, 3) is
(2014 Adv.)
Topic 2 Rate Measure, Increasing and Decreasing Functions
Objective Questions I (Only one correct option)
1. A spherical iron ball of radius 10 cm is coated with a
layer of ice of uniform thickness that melts at a rate of
50 cm3 /min. When the thickness of the ice is 5 cm, then
the rate at which the thickness (in cm/min) of the ice
decreases, is
(2020 Main, 9 Jan I)
1
9π
1
(c)
36 π
5
6π
1
(d)
18 π
(b)
(a)
of all x ∈ R, where the function h (x) = ( fog ) (x) is
(2019 Main, 10 April II)
increasing, is
−1
1
(b)  −1,  ∪  , ∞ 

2   2 
−1
(d)  , 0 ∪ [1, ∞ )
 2 
(c) [0, ∞ )
3. A water tank has the shape of an inverted right circular
−1  1 
cone, whose semi-vertical angle is tan   . Water is
 2
poured into it at a constant rate of 5 cu m/min. Then, the
rate (in m/min) at which the level of water is rising at
the instant when the depth of water in the tank is 10 m
is
(2019 Main, 9 April II)
(a)
2
π
(b)
1
5π
(c)
1
15π
that f ′ ′ (x) > 0, for all x ∈ (0, 2). If φ (x) = f (x) + f (2 − x) ,
then φ is
(2019 Main, 8 April I)
(a) increasing on (0, 1) and decreasing on (1, 2)
(b) decreasing on (0, 2)
(c) decreasing on (0, 1) and increasing on (1, 2)
(d) increasing on (0, 2)
5. If the function f given by
2. Let f (x) = ex − x and g (x) = x2 − x, ∀ x ∈ R. Then, the set
1
(a)  0,  ∪ [1, ∞ )
 2 
4. Let f : [0, 2] → R be a twice differentiable function such
(d)
1
10π
f (x) = x3 − 3(a − 2) x2 + 3ax + 7,
for some a ∈ R is increasing in (0, 1] and decreasing in
f (x) − 14
[1, 5), then a root of the equation,
= 0 (x ≠ 1) is
(x − 1)2
(2019 Main, 12 Jan II)
(a) − 7
(b) 6
(c) 7
x
d−x
(d) 5
, x ∈ R, where a, b
−
a 2 + x2
b2 + (d − x)2
and d are non-zero real constants. Then,
6. Let f (x) =
(a)
(b)
(c)
(d)
(2019 Main, 11 Jan II)
f is an increasing function of x
f ′ is not a continuous function of x
f is a decreasing function of x
f is neither increasing nor decreasing function of x
π π
7. If the function g : (−∞ , ∞ ) →  − ,  is given by
 2 2
π
g (u ) = 2 tan (e ) − . Then, g is
2
−1
u
(2008, 3M)
Application of Derivatives 241
(a)
(b)
(c)
(d)
even and is strictly increasing in (0, ∞ )
odd and is strictly decreasing in (−∞ , ∞ )
odd and is strictly increasing in (−∞ , ∞ )
neither even nor odd but is strictly increasing in (−∞ , ∞ )
18. If f : (0, ∞ ) → R be given by
 1
x −  t + 
t
f (x) = ∫ e
1/ x
3 sin x − 4 sin3 x is increasing, is
(a)
π
3
(b)
π
2
(c)
(2002, 2M)
3π
2
Then,
(2001, 2M)
(2000, 1M)
(a) e < 1 + x
(c) sin x > x
(b) log e (1+ x) < x
(d) log ε x > x
Fill in the Blanks
20. The set of all x for which log (1 + x) ≤ x is equal to ..... .
12. Let f (x) = ∫ ex (x − 1) (x − 2) dx. Then, f decreases in the
interval
(1987, 2M)
function y = 2x − log|x| is monotonically
increasing for values of x ( ≠ 0), satisfying the
inequalities… and monotonically decreasing for values
of x satisfying the inequalities… .
(1983, 2M)
(d) (2, ∞ )
(c) (1, 2)
13. The function f (x) = sin x + cos x increases, if (1999, 2M)
4
(a) 0 < x <
π
8
3π
5π
(c)
< x<
8
8
4
π
3π
< x<
4
8
5π
3π
(d)
< x<
8
4
(b)
2
21. The
(2000, 2M)
(a) (−∞ , − 2) (b) (−2, − 1)
(1998, 2M)
(a) h is increasing, whenever f is increasing
(b) h is increasing, whenever f is decreasing
(c) h is decreasing, whenever f is decreasing
(d) Nothing can be said in general
(b) decreasing in R
(d) decreasing in [−1 / 2, 1]
11. For all x ∈ (0, 1)
x
19. If h (x) = f (x) − f (x)2 + f (x)3 for every real number x.
(d) π
10. If f (x) = xex (1 − x ) , then f (x) is
(a) increasing in [−1 / 2, 1]
(c) increasing in R
(a) f (x) is monotonically increasing on [1, ∞ )
(b) f (x) is monotonically decreasing on [0,1)
1
(c) f (x) + f   = 0, ∀x ∈ (0, ∞ )
 x
(d) f (2x ) is an odd function of x on R
(2004, 2M)
9. The length of a longest interval in which the function
dt
.
t
Then,
8. If f (x) = x3 + bx2 + cx + d and 0 < b2 < c, then in (−∞ , ∞ )
(a) f (x) is strictly increasing function
(b) f (x) has a local maxima
(c) f (x) is strictly decreasing function
(d) f (x) is bounded
(2014 Adv.)
Match the Columns
Directions (Q.Nos. 22-24) by appropriately matching
the information given in the three columns of the
following table.
x
x
and g (x) =
, where 0 < x ≤ 1, then in
sin x
tan x
this interval
(1997, 2M)
14. If f (x) =
(a) both f (x) and g (x) are increasing functions
(b) both f (x) and g (x) are decreasing functions
(c) f (x) is an increasing function
(d) g (x) is an increasing function
log (π + x)
15. The function f (x) =
is
log (e + x)
(1995, 1M)
(a) increasing on (0, ∞ )
(b) decreasing on (0, ∞ )
(c) increasing on (0, π / e), decreasing on ( π / e, ∞ )
(d) decreasing on (0, π / e), increasing on ( π / e, ∞ )
Let f (x) = x + log e x − x log e x, x ∈ (0, ∞ )
Column 1 contains information about zeros of f (x), f ′ (x)
and f ′′ (x).
Column 2 contains information about the limiting
behaviour of f (x), f ′ (x) and f ′′ (x) at infinity.
Column 3 contains information about increasing/decreasing
nature of f (x) and f ′ (x).
Column-1
(I)
f( x ) = 0 for some
x ∈ (1, e 2 )
Column-2
(i)
Column-3
lim f( x ) = 0
(P) f is increasing
in (0, 1)
lim f( x ) = − ∞
(Q) f is decreasing
in (e, e 2)
x→ ∞
(II) f ′ ( x ) = 0 for some
x ∈ (1, e )
(ii)
respectively from [0, ∞ ) to [0, ∞ ) and h (x) = f { g (x)}. If
(1987, 2M)
h(0) = 0, then h (x) − h (1) is
(III) f ′ ( x ) = 0 for some
x ∈ ( 0, 1)
(iii) lim f ′ ( x ) = − ∞
(R) f′ is increasing
in (0, 1)
(a) always negative
(c) strictly increasing
(IV) f ′′( x ) = 0 for some
x ∈ (1, e )
(iv) lim f ′′( x ) = 0
(S) f′ is decreasing
in (e, e 2)
16. Let f and g be increasing and decreasing functions
(b) always positive
(d) None of these
x→ ∞
x→ ∞
22. Which of the following options is the only INCORRECT
Objective Questions II
(One or more than one correct option)
combination?
17. If f : R → R is a differentiable function such that
f ′ (x) > 2 f (x) for all x ∈ R, and f (0) = 1 then
(2017 Adv.)
(a) f (x) > e in (0, ∞ )
(b) f ′ (x) < e in (0, ∞ )
(c) f (x) is increasing in (0, ∞ ) (d) f (x) is decreasing in (0 ∞ )
2x
x→ ∞
2x
(a) (I) (iii) (P)
(c) (II) (iii) (P)
(2017 Adv.)
(b) (II) (iv) (Q)
(d) (III) (i) (R)
23. Which of the following options is the only CORRECT
combination?
(a) (I) (ii) (R)
(c) (II) (iii) (S)
(b) (III) (iv) (P)
(d) (IV) (i) (S)
242 Application of Derivatives
24. Which of the following options is the only CORRECT
combination?
27. Using the relation 2 (1 − cos x) < x2, x ≠ 0 or prove that
sin (tan x) ≥ x , ∀ x ∈ [0, π / 4].
(a) (III) (iii) (R)
(c) (II) (ii) (Q)
(b) (IV) (iv) (S)
(d) (I) (i) (P)
statements in Column II.
Let the functions defined in Column I have domain
(−π / 2, π / 2).
Column II
A. x + sin x
p. increasing
B. sec x
q. decreasing
−1 ≤ p ≤ 1, then show that the equation
4x3 − 3x − p = 0 has a unique root in the interval [1 / 2, 1]
(2001, 5M)
and identify it.
28. If
25. Match the conditions/expressions in Column I with
Column I
(2003, 4M)

x≤0
2
3
x
ax
x
x >0
+
−
,

29. Let f (x) = 
xeax ,
where, a is a positive constant. Find the interval in
(1996, 3M)
which f ′ (x) is increasing.
30. Show that 2 sin x + 2 tan x ≥ 3x, where 0 ≤ x < π /2.
r. neither increasing nor decreasing
(1990, 4M)
31. Show that 1 + x log (x + x2 + 1 ) ≥ 1 + x2 ∀x ≥ 0.
Analytical & Descriptive Questions
26. Prove that sin x + 2x ≥
3x (x + 1)
 π
, ∀ x ∈ 0,
 2 
π
(Justify the inequality, if any used).
(2004, 4M)
(1983, 2M)
π
π
32. Given A = x : ≤ x ≤  and f (x) = cos x – x (1 + x). Find
3
 6
(1980, 2M)
f ( A ).
Topic 3 Rolle’s and Lagrange’s Theorem
Objective Question I (Only one correct option)
1. If f : R → R is a twice differentiable function such that
 1 1
f ′′ (x) > 0 for all x ∈ R, and f   = , f (1) = 1, then
 2 2
(2017 Adv.)
(a) f ′ (1) ≤ 0
(c) 0 < f ′ (1) ≤
1
2
(b) f ′ (1) > 1
1
< f ′ (1) ≤ 1
2
(d)
(a) Statement I is correct, Statement II is also correct;
Statement II is the correct explanation of Statement I
(b) Statement I is correct, Statement II is also correct;
Statement II is not the correct explanation of
Statement I
(c) Statement I is correct; Statement II is incorrect
(d) Statement I is incorrect; Statement II is correct
Analytical & Descriptive Question
Assertion and Reason
2. Let f (x) = 2 + cos x, for all real x.
Statement I For each real t, there exists a point c in
[t , t + π ], such that f ' (c) = 0. Because
Statement II f (t ) = f (t + 2π ) for each real t. (2007, 3M)
3. If f (x) and g (x) are differentiable functions for 0 ≤ x ≤ 1,
f (0) = 2, g (0) = 0
such that
f (1) = 6, g (1) = 2
Then, show that there exists c satisfying 0 < c < 1 and
(1982, 2M)
f ′ (c) = 2 g′ (c).
Topic 4 Maxima and Minima
Objective Questions I (Only one correct option)
1. Consider the rectangles lying in the region
π


(x, y) ∈ R × R : 0 ≤ x ≤ and 0 ≤ y ≤ 2 sin(2x)
2


and having one side on the X-axis. The area of the
rectangle which has the maximum perimeter among all
(2020 Adv.)
such rectangles, is
3π
(a)
2
π
(c)
2 3
(b) π
(d)
π 3
2
2. Let f (x) = 5 −|x − 2| and g (x) = |x + 1|, x ∈ R. If f (x)
attains maximum value at α and g (x) attains minimum
(x − 1) (x2 − 5x + 6)
is equal to
value of β, then lim
x → − αβ
x2 − 6x + 8
(2019 Main, 12 April II)
(b) − 3 / 2
(a) 1/2
(c) − 1 / 2
(d) 3/2
3. If the volume of parallelopiped formed by the vectors
$i + λ$j + k
$ , $j + λk
$ and λ$i + k
$ is minimum, then λ is equal
to
(2019 Main, 12 April I)
(a) −
1
3
(b)
1
3
(c)
3
(d) −
3
Application of Derivatives 243
4. If m is the minimum value of k for which the function
14. If 20 m of wire is available for fencing off a flower-bed in
f (x) = x kx − x2 is increasing in the interval [0, 3] and M
is the maximum value of f in the interval [0, 3] when
k = m, then the ordered pair (m, M ) is equal to
the form of a circular sector, then the maximum area (in
sq. m) of the flower-bed is
(2017 Main)
(2019 Main, 12 April I)
(a) (4, 3 2 )
(c) (3, 3 3 )
(b) (4, 3 3 )
(d) (5, 3 6 )
local extreme points at x = − 1, 0, 1, then the set
S = { x ∈ R : f (x) = f (0)} contains exactly
four rational numbers
(2019 Main, 9 April I)
two irrational and two rational numbers
four irrational numbers
two irrational and one rational number
6. The height of a right circular cylinder of maximum
volume inscribed in a sphere of radius 3 is
(2019 Main, 8 April II)
(a)
6
(b) 2 3
(c)
3
(d)
2
3
3
7. If S1 and S 2 are respectively the sets of local
minimum and local maximum points of the function,
f (x) = 9x4 + 12x3 − 36x2 + 25, x ∈ R, then
(a)
(b)
(c)
(d)
S1 = {−2} ; S2 = {0,1}
S1 = {−2, 0} ; S2 = {1}
S1 = {−2,1} ; S2 = {0}
S1 = {−1} ; S2 = {0, 2}
(2019 Main, 8 April I)
(2019 Main, 8 April I)
(c)
7
4 2
11
(d)
4 2
9. The maximum area (in sq. units) of a rectangle having
its base on the X-axis and its other two vertices on the
parabola, y = 12 − x2 such that the rectangle lies inside
the parabola, is
(2019 Main, 12 Jan I)
(a) 36
(b) 20 2
(c) 32
(d) 18 3
f (x) = 3x3 − 18x2 + 27x − 40
on the set S = { x ∈ R : x2 + 30 ≤ 11x} is (2019 Main, 11 Jan I)
(b) − 122
(c) − 222
(d) 222
11. Let A (4, − 4) and B(9, 6) be points on the parabola,
y2 = 4x. Let C be chosen on the arc AOB of the parabola,
where O is the origin, such that the area of ∆ACB is
maximum. Then, the area (in sq. units) of ∆ACB, is
(2019 Main, 9 Jan II)
(a) 31
1
4
(b) 32
(c) 31
3
4
(d) 30
1
2
12. The maximum volume (in cu.m) of the right circular
cone having slant height 3m is
(a)
4
π
3
(b) 2 3π
13. Let f (x) = x2 +
(c) 3 3π
(2019 Main, 9 Jan I)
(d) 6π
1
1
and g (x) = x − , x ∈ R − { −1, 0, 1}. If
x
x2
f (x)
, then the local minimum value of h (x) is
h (x) =
g (x)
(2018 Main)
(a) 3
(b) −3
(b) (4 − π )x = πr
(d) 2x = r
16. The least value of α ∈ R for which 4αx2 +
x > 0, is
(c) −2 2
(d) 2 2
1
≥ 1 , for all
x
(2016 Adv.)
1
(a)
64
1
(b)
32
1
(c)
27
1
(d)
25
17. Let f (x) be a polynomial of degree four having extreme
values at x = 1 and x = 2. If
lim 
x→ 0 1+

f (x) 
= 3, then f (2) is
x2 
equal to
(a) −8
(c) 0
(b) −4
(d) 4
1
2
1
(c) α = 2, β = −
2
(a) α = − 6, β =
extreme
points
of
(2014 Main)
(b) α = − 6, β = −
(d) α = 2, β =
1
2
1
2
19. The number of points in (− ∞ , ∞ ) for which
x2 − x sin x − cos x = 0, is
(a) 6
(2013 Adv.)
(b) 4
(c) 2
(d) 0
20. Let f, g and h be real-valued functions defined on the
2
2
2
2
interval [0, 1] by f (x) = ex + e− x , g (x) = x ex + e− x and
2
10. The maximum value of the function
(a) 122
(a) 2x = ( π + 4)r
(c) x = 2r
x = − 1 and x = 2 are
f (x) = α log|x| + βx2 + x, then
curve y2 = x − 2 is
(a) 2
are bent respectively to form a square of side = x units
and a circle of radius = r units. If the sum of the
areas of the square and the circle so formed is
(2016 Main)
minimum, then
18. If
8. The shortest distance between the line y = x and the
7
(b)
8
(b) 10
(d) 30
15. A wire of length 2 units is cut into two parts which
5. If f (x) is a non-zero polynomial of degree four, having
(a)
(b)
(c)
(d)
(a) 12.5
(c) 25
2
h (x) = x2ex + e− x . If a, b and c denote respectively, the
absolute maximum of f, g and h on [0, 1], then (2010)
(a) a = b and c ≠ b
(c) a ≠ b and c ≠ b
(b) a = c and a ≠ b
(d) a = b = c
21. The total number of local maxima and local minima of
(2 + x)3 , − 3 < x ≤ −1
the function f (x) =  2
is
3
−1 < x < 2
x ,
(a) 0
(c) 2
(2008, 3M)
(b) 1
(d) 3
22. If f (x) = x2 + 2bx + 2c2 and g (x) = − x2 − 2cx + b2, such
that min f (x) > max g (x), then the relation between b
and c, is
(2003, 2M)
(a) No real value of b and c (b) 0 < c < b 2
(d)| c|> | b| 2
(c)| c|< | b| 2
23. If f (x) = | x|, for 0 < | x| ≤ 2 . Then, at x = 0, f has
1
, for
(a) a local maximum
(c) a local minimum
x=0
(2000, 1M)
(b) no local maximum
(d) no extremum
244 Application of Derivatives
x2 − 1
, for every real number x, then the
x2 + 1
(1998, 2M)
minimum value of f
24. If f (x) =
(a) does not exist because f is unbounded
(b) is not attained even though f is bounded
(c) is 1
(d) is –1
25. The number of values of x, where the function
f (x) = cos x + cos ( 2x) attains its maximum, is(1998, 2M)
(a) 0
(b) 1
(c) 2
26. On the interval [0,1], the function x
(d) infinite
25
(1 − x)
75
maximum value at the point
(a) 0
(b) 1/4
(c) 1/2
takes its
(1995, 1M)
(d) 1/3
27. Find the coordinates of all the points P on the ellipse
2
2
x
y
+
= 1, for which the area of the ∆PON is
a 2 b2
maximum, where O denotes the origin and N is the foot
of the perpendicular from O to the tangent at P.
(1990, 10M)
 ± a2
± b2 

(a) 
,

 a 2 + b2 a 2 + b2 
 ± a2
± b2 

(c) 
,

2
2
 a +b
a 2 – b2 
 ± a2
± b2 

(b) 
,

 a 2 – b2 a 2 – b2 
 ± a2
± b2 

(d) 
,

2
2
 a –b
a 2 + b2 
28. If P (x) = a 0 + a1x2 + a 2x4 +... + a nx2n is a polynomial in
a real variable x with 0 < a 0 < a1 < a 2 < K < a n. Then,
the function P (x) has
(1986, 2M)
(a) neither a maximum nor a minimum
(b) only one maximum
(c) only one minimum
(d) only one maximum and only one minimum
29. If y = a log x + bx2 + x has its extremum values at x = − 1
and x = 2 , then
(a) a = 2, b = − 1
(c) a = − 2, b =
1
2
x5 + 5x4 + 10x3 + 10x2 + 3x + 1,
x < 0;

2
0 ≤ x < 1;
x − x + 1,

8
2 3
2
f (x) = 
1 ≤ x < 3;
x − 4x + 7x − ,
3
3

10

x ≥ 3;
 (x − 2) log e (x − 2) − x + 3 ,
Then which of the following options is/are correct?
(a) f is increasing on (− ∞ , 0)
1
(b) a = 2, b = −
2
cos(2x) cos(2x) sin(2x)
, then
cos x − sin x 

sin x
cos x 
 sin x
33. If f (x) = 
− cos x
(1982, 1M)
(a) max ( p , q) < max ( p , q, r )
1
(b) min ( p , q) = ( p + q − | p − q|)
2
(c) max ( p , q) < min ( p , q, r )
(d) None of the above
34. Let f : R → (0, ∞ ) and g : R → R be twice differentiable
functions such that f′ ′ and g′ ′ are continuous functions
on R. Suppose f ′ (2) = g (2) = 0, f ′ ′(2) ≠ 0 and g′ (2) ≠ 0.
f (x) g (x)
If lim
(2016 Adv.)
= 1, then
x → 2 f ′ (x) g′ (x)
(a) f has a local minimum at x = 2
(b) f has a local maximum at x = 2
(c) f ′ ′ (2) > f (2)
(d) f (x) − f ′ ′ (x) = 0, for atleast one x ∈ R
35. The function f (x) = 2|x| + |x + 2| − ||x + 2| − 2|x|| has a
local minimum or a local maximum at x is equal to
(b)
−2
3
(b) 32


x
Define F (x) = ∫ f (t )dt, x > 0
0
Then which of the following options is/are correct?
(2019 Adv.)
F (x) ≠ 0 for all x ∈ (0, 5)
F has a local maximum at x = 2
F has two local maxima and one local minimum in (0, ∞ )
F has a local minimum at x = 1
(d) 2/3
having their lengths in the ratio 8 : 15 is converted into
an open rectangular box by folding after removing
squares of equal area from all four corners. If the total
area of removed squares is 100, the resulting box has
maximum volume. The lengths of the sides of the
rectangular sheet are
(2013 Adv.)
x
e
,
37. If f (x) = 2 − ex − 1 ,
31. Let f : R → R be given by f (x) = (x − 1)(x − 2)(x − 5).
(2013 Adv.)
(c) 2
36. A rectangular sheet of fixed perimeter with sides
(a) 24
Objective Questions II
(One or more than one correct option)
(2017 Adv.)
(a) f (x) attains its minimum at x = 0
(b) f (x) attains its maximum at x = 0
(c) f ′ (x) = 0 at more than three points in (− π , π )
(d) f ′ (x) = 0 at exactly three points in (− π , π )
(a) −2
(d) None of the above
(2019 Adv.)
(b) f ′ is NOT differentiable at x = 1
(c) f is onto
(d) f ′ has a local maximum at x = 1
(1983, 1M)
30. If p, q and r are any real numbers, then
(a)
(b)
(c)
(d)
32. Let f : R → R be given by
 x−e ,

x ∈ [1, 3], then
(c) 45
0 ≤ x≤1
1 < x≤2
2 < x≤3
(d) 60
and g (x) = ∫
x
0
f (t ) dt ,
(2006, 3M)
(a) g (x) has local maxima at x = 1 + log e 2 and local
minima at x = e
(b) f (x) has local maxima at x = 1and local minima at x = 2
(c) g (x) has no local minima
(d) f (x) has no local maxima
38. If f (x) is a cubic polynomial which has local maximum at
x = − 1. If f (2) = 18, f (1) = − 1
minimum at x = 0, then
and f ′ (x) has local
(2006, 3M)
Application of Derivatives 245
(a) the distance between (– 1, 2) and (a , f (a )), where x = a
is the point of local minima, is 2 5
(b) f (x) is increasing for x ∈[1, 2 5 ]
(c) f (x) has local minima at x = 1
(d) the value of f (0) = 5
39. The function
f (x) = ∫
x
t (et − 1) (t − 1) (t − 2)3 (t − 3)5 dt has a local
−1
minimum at x equals
(a) 0
(1999, 3M)
(b) 1
(c) 2
(d) 3
2
40. If f (x) = 3x + 12x − 1, − 1 ≤ x ≤ 2, then
37
2 < x≤3
 − x,
(1993, 3M)
arc of the parabola y = 16x, 0 ≤ y ≤ 6 at the point
F (x0 , y0 ). The tangent to the parabola at F (x0 , y0 )
intersects the Y -axis at G (0, y1 ). The slope m of the
line L is chosen such that the area of the ∆EFG has a
local maximum
2
Match List I with List II and select the correct answer
using the codes given below the list.
Column I
m=
1.
1/2
Q.
Maximum area of ∆EFG is
2.
4
R.
y0 =
3.
2
S.
y1 =
4.
1
Analytical & Descriptive Questions
45. If f (x) is twice differentiable function such that f (a ) = 0,
f (b) = 2, f (c) = 1, f (d ) = 2, f (e) = 0, where a < b < c < d < e,
then
the
minimum
number
of
zeroes
of
g (x) = { f ′ (x)}2 + f ′′ (x) ⋅ f (x) in the interval [a, e] is
from the line x + y = 7, is minimum.
(2003, 2M)
48. Let f (x) is a function satisfying the following conditions
(i) f (0) = 2, f (1) = 1
(ii) f (x) has a minimum value at x = 5 / 2 and
2 ax
2ax − 1
2 ax + b + 1
(iii) For all x, f ′ (x) =
b
b+ 1
−1
2 (ax + b) 2 ax + 2b + 1 2 ax + b
where, a and b are some constants. Determine the
constants a , b and the function f (x).
(1998, 8M)
49. Let C1 and C 2 be respectively, the parabolas x2 = y – 1
Codes
S
3
4
(b)
(d)
P
3
1
Q
4
3
R
1
4
S
2
2
Passage Based Problems
Consider the function f : (−∞ , ∞ ) → (−∞ , ∞ ) defined by
x2 − ax + 1
; 0 < a < 2.
f (x) = 2
(2008, 12M)
x + ax + 1
42. Which of the following is true ?
(a) (2 + a ) f ′ ′ (1) + (2 − a ) f ′ ′ (−1) = 0
(b) (2 − a )2 f ′ ′ (1) − (2 + a )2 f ′ ′ (−1) = 0
(c) f ′ (1) f ′ (−1) = (2 − a )2
(d) f ′ (1) f ′ (−1) = − (2 + a )2
2
(a) g ′ (x) is positive on (− ∞ , 0) and negative on (0, ∞ )
(b) g ′ (x) is negative on (− ∞ , 0) and positive on (0, ∞ )
(c) g ′ (x) changes sign on both (− ∞ , 0) and (0, ∞ )
(d) g ′ (x) does not change sign (− ∞ , ∞ )
Column II
P.
R
2
2
f ′ (t )
dt. Which of the following is true?
1 + t2
47. Find a point on the curve x2 + 2 y2 = 6 whose distance
41. A line L : y = mx + 3 meets Y -axis at E (0, 3) and the
Q
1
3
0
the area enclosed by the tangents drawn from the point
P(6, 8) to the circle and the chord of contact is
maximum.
(2003, 2M)
Match the Columns
P
4
1
ex
46. For the circle x2 + y2 = r 2, find the value of r for which
(a) f (x) is increasing on [–1, 2]
(b) f (x) is continuous on [–1, 3]
(c) f ′(2) does not exist
(d) f (x) has the maximum value at x = 2
(a)
(c)
44. Let g (x) = ∫
2
43. Which of the following is true ?
(a) f (x) is decreasing on (−1, 1) and has a local minimum at
x=1
(b) f (x) is increasing on (−1, 1) and has a local maximum at
x=1
(c) f (x) is increasing on (−1, 1) but has neither a local
maximum nor a local minimum at x = 1
(d) f (x) is decreasing on (−1, 1) but has neither a local
maximum nor a local minimum at x = 1
and y2 = x – 1. Let P be any point on C1 and Q be any
point on C 2. If P1 and Q1 is the reflections of P and Q,
respectively, with respect to the line y = x. Prove that P1
lies on C 2 Q1 lies on C1 and PQ ≥ min (PP1 , QQ1 ). Hence,
determine points P0 and Q0 on the parabolas C1 and C 2
respectively such that P0Q0 ≤ PQ for all pairs of points
(P , Q ) with P on C1 and Q on C 2.
50. If S is a square of unit area. Consider any quadrilateral
which has one vertex on each side of S. If a , b, c and d
denote the length of the sides of the quadrilateral, then
prove that 2 ≤ a 2 + b2 + d 2 ≤ 4.
(1997, 5M)
51. Determine the points of maxima and minima of the
functionf (x) =
constant.
1
ln x − bx + x2, x > 0, where b ≥ 0 is a
8
(1996, 5M)
52. Let (h , k) be a fixed point, where h > 0 , k > 0. A straight
line passing through this point cuts the positive
directions of the coordinate axes at the points P and Q.
Find the minimum area of the ∆OPQ, O being the
origin.
(1995, 5M)
53. The circle x2 + y2 = 1 cuts the X-axis at P and Q. Another
circle with centre at Q and variable radius intersects
the first circle at R above the X-axis and the line
segment PQ at S. Find the maximum area of the ∆QSR.
(1994, 5M)
246 Application of Derivatives
 3 (b3 − b2 + b − 1)
− x +
, 0 ≤ x≤1
54. Let f (x) = 
(b2 + 3b + 2)

2x − 3,
1 ≤ x≤3

Find all possible real values of b such that f (x) has the
smallest value at x = 1.
(1993, 5M)
55. What normal to the curve y = x2 forms the shortest
chord?
(1992, 6M)
56. A window of perimeter (including the base of the arch) is
in the form of a rectangle surmounted by a semi-circle.
The semi-circular portion is fitted with coloured glass
while the rectangular part is fitted with clear glass. The
clear glass transmits three times as much light per
square meter as the coloured glass does. What is the
ratio for the sides of the rectangle so that the window
transmits the maximum light?
(1991, 4M)
57. A point P is given on the circumference of a circle of
radius r. Chord QR is parallel to the tangent at P.
Determine the maximum possible area of the ∆PQR.
(1990, 4M)
58. Find the point on the curve 4x + a y = 4a , 4 < a 2 < 8
2
2 2
that is farthest from the point (0, –2).
2
(1987, 4M)
59. Let A ( p2, − p) B(q2, q), C (r 2, − r ) be the vertices of the
triangle ABC. A parallelogram AFDE is drawn with
vertices D, E and F on the line segments BC, CA and AB,
respectively. Using calculus, show that maximum area
1
of such a parallelogram is ( p + q)(q + r )( p − r ).
4
(1986, 5M)
π
π
60. Let f (x) = sin3 x + λ sin 2 x, − < x < ⋅ Find the
2
2
intervals in which λ should lie in the order that f (x) has
exactly one minimum and exactly one maximum.
(1985, 5M)
x
,
1 + x2
where the tangent to the curve has the greatest slope.
61. Find the coordinates of the point on the curve y =
66. For a polynomial g (x) with real coefficients, let mg
denote the number of distinct real roots of g (x). Suppose
S is the set of polynomials with real coefficients defined
by
S = {(x2 − 1)2(a 0 + a1x + a 2x2 + a3 x3 ) : a 0 , a1 , a 2, a3 ∈R};
For a polynomial f, let f′ and f′ ′ denote its first and
second order derivatives, respectively. Then the
minimum possible value of (mf ′ + mf ′ ′ ), where f ∈ S, is
………
(2020 Adv.)
67. Let AD and BC be two vertical poles at A and B
respectively on a horizontal ground. If AD = 8 m,
BC = 11 m and AB = 10 m; then the distance (in meters)
of a point M on AB from the point A such that
(2020 Main, 6 Sep I)
MD 2 + MC 2 is minimum is ........
68. A cylindrical container is to be made from certain solid
material with the following constraints : It has a fixed
inner volume of V mm3 , has a 2 mm thick solid wall and
is open at the top. The bottom of the container is a solid
circular disc of thickness 2 mm and is of radius equal to
the outer radius of the container.
If the volume of the material used to make the container
is minimum, when the inner radius of the container is
V
10 mm, then the value of
is
(2015 Adv.)
250 π
69. A vertical line passing through the point (h , 0)
x2 y 2
+
= 1 at the points P and Q. If
4
3
the tangents to the ellipse at P and Q meet at the point
R.
intersects the ellipse
If ∆(h ) = area of the ∆ PQR, ∆1 = max ∆ (h ) and
1/ 2 ≤ h ≤ 1
8
∆1 − 8∆ 2 is equal to (2013 Adv.)
∆ 2 = min ∆ (h ), then
1/ 2 ≤ h ≤ 1
5
70. Let f : R → R be defined as f (x) = | x| + | x2 − 1|. The total
(1984, 4M)
number of points at which f attains either a local
(2012)
maximum or a local minimum is
62. A swimmer S is in the sea at a distance d km from the
71. Let p(x) be a real polynomial of least degree which has a
closest point A on a straight shore. The house of the
swimmer is on the shore at a distance L km from A. He
can swim at a speed of u km/h and walk at a speed of
v km/h (v > u ). At what point on the shore should be land
so that he reaches his house in the shortest possible
time?
(1983, 2M)
63. If ax2 + b / x ≥ c for all positive x where a > 0 and b > 0,
then show that 27ab ≥ 4c .
2
3
(1982, 2M)
64. If x and y be two real variables such that x > 0 and
xy = 1. Then, find the minimum value of x + y. (1981, 2M)
Integer & Numerical Answer Type Questions
65. Let the function f : (0, π) → R be defined by
f (θ ) = (sin θ + cos θ ) + (sin θ − cos θ )
Suppose the function f has a local minimum at θ
precisely when θ ∈{ λ 1π, ... , λ rπ }, where 0 < λ 1 < ... λ r < 1.
(2020 Adv.)
Then the value of λ 1 + ... + λ r is ....... .
2
4
local maximum at x = 1 and a local minimum at x = 3. If
(2012)
p(1) = 6 and p(3) = 2, then p′ (0) is equal to
72. The number of distinct real roots of
x4 − 4x3 + 12x2 + x − 1 = 0 is……
73. Let f be a function defined on R (the set of all real
numbers) such that f ′ (x) = 2010 (x − 2009) (x − 2010)2
(x − 2011)3 (x − 2012)4, ∀x ∈ R. If g is a function defined
on R with values in the interval (0, ∞ ) such that
f (x) = ln ( g (x)), ∀ x ∈ R, then the number of points in R
at which g has a local maximum is……
(2010)
74. The maximum value of the expression
1
is ……
sin 2 θ + 3 sin θ cos θ + 5 cos 2 θ
(2010)
75. The maximum value of the function
f (x) = 2x3 − 15x2 + 36x − 48 on the set
A = { x | x2 + 20 ≤ 9x} is ……… .
(2009)
Answers
Topic 1
9. (c)
10. (a)
11. (a)
12. (b)
1. (a)
5. (c)
2. (c)
6. (b)
3. (b)
7. (d)
4. (d)
8. (b)
13. (d)
14. (c)
15. (c)
16. (c)
17. (c)
18. (c)
19. (c)
20. (d)
9. (b)
13. (c)
10. (d)
14. (b, d)
11. (d)
15. (b, c)
12. (d)
21. (c)
22. (d)
23. (a)
24. (d)
25. (b)
26. (b)
27. (a)
28. (c)
29. (b)
30. (b)
31. (a,b,d)
32. (b,c,d)
33. (b, c)
34. (a, d)
35. (a, b)
36. (a, c)
37. (a, b)
38. (b, c)
39. (b, d)
40. (a, b, c, d)
41. (a) P → 4 Q → 1 R → 2 S → 3
42. (a)
43. (a)
16. H = φ, V = {1, 1 }
1
3
18. a = – , b = – , c = 3
2
4
17. y − 2 = 0
19. 1 : 16
20. y + x − 1 = 0
π
− 3π
21. x + 2y = and x + 2y =
2
2
22. (8)
Topic 2
1. (d)
2. (a)
3. (b)
4. (c)
5. (c)
6. (a)
7. (c)
8. (a)
9. (a)
10. (a)
11. (b)
12. (c)
13. (b)
17. (a, c)
 1 
21. x ∈  − , 0
 2 
22. (d)
14. (c)
15. (b)
18. (c, d)
19. (a, c)
1
1 

∪  , ∞ , x ∈  − ∞, −  ∪
2 

2
16. (d)
20. x > − 1
 1
 0, 
 2
54. b ∈ ( −2, − 1 ) ∪ [1, ∞ ]
55.
2. (b)
6. (b)
7. (c)
3 3 2
r sq units
4
58. (0, 2)
ud
v − u2
2
64. (2)
65. 0.50
66. (5)
67. (5)
4. (b)
68. (4)
69. (9)
70. (5)
71. (9)
8. (c)
72. (2)
73. (1)
74. (2)
75. (7)
Topic 4
5. (d)
57.
 3 3
60. λ ∈  − ,  61. x = 0, y = 0
 2 2
62.
3. (b)
2 x − 2y + 2 = 0, 2 x + 2y − 2 = 0
56. 6 : 6 + π
Topic 3
2. (a)
(b − b 2 − 1 )
1
and minima at x = (b + b 2 − 1 )
4
4
4 3
53.
9
52. 2hk
24. (c)
 2 a
25. A → p, B → r
29. – ,
 a 3 
1 π 
π 3 π 
π 
32.  − 1 +  ,
− 1 +  



2
3
3
2
6
6

1. (c)
45. 6
51. Maxima at x =
23. (c)
1. (b)
44. (a)
46. 5 units
47. (2, 1)
1
−5
1
5
48. a = , b = ; f ( x ) = x 2 − x + 2
4
4
4
4
 1 5
 5 1
49. P0 =  ,  and Q0 =  , 
 2 4
 4 2
Hints & Solutions
Topic 1 Equations of Tangent and Normal
∴ Slope of this line = −
1. Equation of given curve is
y=
x
, x ∈ R, (x ≠ ± 3 )
2
x −3
…(i)
On differentiating Eq. (i) w.r.t. x, we get
dy (x2 − 3) − x(2x) (− x2 − 3)
= 2
=
dx
(x − 3)2
(x2 − 3)2
It is given that tangent at a point (α , β ) ≠ (0, 0) on it is
parallel to the line
2x + 6 y − 11 = 0.
⇒
−
2 dy
=
6 dx ( α , β )
1
α2 + 3
=−
3
(α 2 − 3)2
⇒
3α 2 + 9 = α 4 − 6 α 2 + 9
⇒
α4 − 9 α2 = 0
⇒
α = 0, − 3, 3
⇒
α = 3 or − 3,
Now, from Eq. (i),
3
α
1
1
−3
or
⇒ β=
β= 2
= or −
9 −3
2
9 −3 2
α −3
[Q α ≠ 0]
248 Application of Derivatives
According to the options,|6 α + 2 β | = 19
1

at (α , β ) =  ± 3, ± 

2
2. Given curve is y = f (x) = x3 − x2 − 2x
...(i)
So,
f (1) = 1 − 1 − 2 = −2
and
f (−1) = −1 − 1 + 2 = 0
Since, slope of a line passing through (x1 , y1 ) and
y −y
(x2, y2) is given by m = tan θ = 2 1
x2 − x1
∴Slope of line joining points (1, f (1)) and
5. The helicopter is nearest to the soldier, if the tangent to
f (1) − f (−1) −2 − 0
(−1, f (−1)) is m =
=
= −1
1 − (−1)
1+1
dy
Now,
= 3x2 − 2x − 2 [differentiating Eq. (i), w.r.t. ‘x’]
dx
According to the question,
dy
=m
dx
⇒
3 x2 − 2 x − 2 = − 1 ⇒ 3 x2 − 2 x − 1 = 0
1
⇒
(x − 1) (3x + 1) = 0 ⇒ x = 1 ,−
3
 1 
Therefore, set S = − , 1.
 3 
3. Given curve is y = x + ax − b
3
…(i)
passes through point P(1, − 5).
…(ii)
∴
−5 =1+ a −b⇒b−a =6
and slope of tangent at point P(1, − 5) to the curve (i), is
dy
m1 =
= [3x2 + a ](1, −5 ) = a + 3
dx (1, −5 )
Q The tangent having slope m1 = a + 3 at point P(1, − 5)
is perpendicular to line − x + y + 4 = 0 , whose slope is
m2 = 1.
∴
a + 3 = −1 ⇒ a = −4
[Q m1m2 = −1]
Now, on substituting a = −4 in Eq. (ii), we get b = 2
On putting a = −2 and b = 2 in Eq. (i), we get
y = x3 − 4x − 2
Now, from option (2, − 2) is the required point which lie
on it.
4. The given curve is y = x2 − 5x + 5
Now, equation of tangent to the curve (i) at point
 7 1
 ,−  and having slope 2, is
 2 4
1
7
1

y + = 2  x −  ⇒ y + = 2x − 7

4
2
4
29
…(iii)
y = 2x −
⇒
4

1
On checking all the options, we get the point  , − 7

8
satisfy the line (iii).
…(i)
Now, slope of tangent at any point (x, y) on the curve is
dy
…(ii)
= 2x − 5
dx
[on differentiating Eq. (i) w.r.t. x]
Q It is given that tangent is parallel to line
2 y = 4x + 1
dy
So,
= 2 [Q slope of line 2 y = 4x + 1 is 2]
dx
7
⇒
2x − 5 = 2 ⇒ 2x = 7 ⇒ x =
2
7
On putting x = in Eq. (i), we get
2
49 35
69 35
1
y=
−
+5 =
−
=−
4
2
4
2
4
the path, y = x3/ 2 + 7, (x ≥ 0) of helicopter at point (x, y) is
perpendicular to the line joining (x, y) and the position
1 
of soldier  , 7 .
2 
(x, y)
y=x3/2+7
(1/2, 7)
Q Slope of tangent at point (x, y) is
dy 3 1/ 2
= x = m1 (let )
dx 2
1 
and slope of line joining (x, y) and  , 7 is
2 
y−7
m2 =
1
x−
2
Now,
m1 ⋅ m2 = −1
3 1/ 2  y − 7 
x 
 = −1
 x − (1 /2)
2
⇒
…(i)
…(ii)
[from Eqs. (i) and (ii)]
3 1/ 2 x3/ 2
[Q y = x3/ 2 + 7]
= −1
x
1
2
x−
2
3 2
1
x = −x +
⇒
2
2
⇒
3x2 + 2x − 1 = 0 ⇒ 3x2 + 3x − x − 1 = 0
⇒ 3x(x + 1) − 1(x + 1) = 0
1
1
⇒
x = , −1 ⇒ x ≥ 0 ⇒ x =
3
3
3/ 2
 1
and So,
[Q y = x3/ 2 + 7]
y=  +7
 3
⇒
 1  1 3 / 2

Thus, the nearest point is  ,   + 7


3
3


Now, the nearest distance
2
2
3
2 
3/ 2

 1
 1 1
 1
 1
=  −  +  7 −   − 7 =   +  
 3
 6
 2 3
 3


=
1
1
3+4
7
1
+
=
=
=
36 27
108
108 6
7
3
Application of Derivatives 249
6.
Given equation of curves are
y = 10 − x2
and
y = 2 + x2
…(i)
…(ii)
For point of intersection, consider
10 − x2 = 2 + x2
⇒
2 x2 = 8
⇒
x2 = 4
⇒
x=±2
Clearly, when x = 2 , then y = 6 (using Eq. (i)) and when
x = − 2, then y = 6
Thus, the point of intersection are (2, 6) and
(−2, 6).
Let m1 be the slope of tangent to the curve (i) and m2 be
the slope of tangent to the curve (ii)
dy
dy
For curve (i)
= −2x and for curve (ii)
= 2x
dx
dx
∴ At (2, 6), slopes m1 = − 4 and m2 = 4, and in that case
|tan θ| =
m2 − m1
4+4
8
=
=
1 + m1m2 1 − 16 15
At (−2, 6), slopes m1 = 4 and m2 = − 4 and in that case
|tan θ| =
m2 − m1
−4 − 4
8
=
=
1 + m1m2 1 − 16 15
7. We have, y2 = 6x
⇒
2y
dy
=6
dx
⇒
dy 3
=
dx y
Slope of tangent at ( x1 , y1 ) is m1 =
Also,
9x 2 + by 2 = 16
dy
=0
18x + 2by
dx
3
y1
dy −9x
=
dx
by
−9x1
Slope of tangent at ( x1 , y1 ) is m2 =
by1
Since, these are intersection at right angle.
27x1
∴ m1m2 = − 1 ⇒
=1
by12
27x1
[Q y12 = 6x1]
=1
⇒
6bx1
9
b=
⇒
2
⇒
⇒
8. Given curve is
∴ Equation of normal at (0, 1) is given by
−1
y−1 =
(x − 0) ⇒ x + y − 1 = 0
1
 1 1
which passes through the point  ,  .
 2 2
9. We have, f (x) = tan − 1
f (x) = tan − 1
⇒
Put x = 0 in Eq. (i), we get
y(− 2) (− 3) = 6 ⇒ y = 1
So, point of intersection is (0, 1).
x+6
Now, y =
(x − 2)(x − 3)
dy 1 (x − 2)(x − 3) − (x + 6)(x − 3 + x − 2)
=
dx
(x − 2)2(x − 3)2
…(i)
1 + sin x
 π
, x ∈ 0, 
 2
1 − sin x
x
x

 cos + sin 

2
2
2
x

 cos − sin

2
2
x

2
x
x

 cos + sin 
2
2
= tan 

 cos x − sin x 

2
2
x
x
x π

Q cos > sin for 0 < < 

2
2
2 4
x

 1 + tan 
−1
2
= tan 

 1 − tan x 

2

 π x  π x
= tan − 1 tan  +   = +
 4 2  4 2

1
 π 1
f ′ (x) = ⇒ f′   =
⇒
 6 2
2
π
Now, equation of normal at x = is given by
6
π

 π

 y − f    = − 2  x − 

6 
6
−1
π
π 
π 4π π 


 π π
⇒  y −  = − 2  x −  Q f   = +
=
=


 6  4 12 12 3 
3
6 
 2π 
which passes through 0,  .
 3
10. Given equation of curve is
x2 + 2xy − 3 y2 = 0
…(i)
On differentiating w.r.t x, we get
2x + 2xy′ + 2 y − 6 yy′ = 0 ⇒ y′ =
At
y(x − 2)(x − 3) = x + 6
⇒
6 + 30 36
 dy
=
=1
=
 
 dx ( 0, 1)
36
4 ×9
⇒
Key Idea Angle between two curves is the angle between the
tangents to the curves at the point of intersection.
i.e.
x+ y
3y − x
x = 1, y = 1, y′ = 1
 dy
=1
 
 dx (1, 1)
Equation of normal at (1, 1) is
1
y − 1 = − (x − 1) ⇒ y − 1 = − (x − 1)
1
⇒
x+ y=2
On solving Eqs. (i) and (ii) simultaneously, we get
⇒
x2 + 2x(2 − x) − 33(2 − x)2 = 0
…(ii)
250 Application of Derivatives
⇒ x2 + 4x − 2x2 − 3(4 + x2 − 4x) = 0
⇒
− x2 + 4x − 12 − 3x2 + 12x = 0
⇒
− 4x2 + 16x − 12 = 0
⇒
4x2 − 16x + 12 = 0
⇒
x2 − 4 x + 3 = 0
⇒
(1 − 1)(x − 3) = 0 ⇒ x = 1, 3
Now, when x = 1, then y = 1
and when x = 3, then y = − 1
∴
P = (1, 1) and Q = (3, − 1)
Hence, normal meets the curve again at (3, –1) in fourth
quadrant.
Alternate Solution
Given,
x2 + 2xy − 3 y2 = 0
⇒
(x − y)(x + 3 y) = 0
⇒
x − y = 0 or x + 3 y = 0
Equation of normal at (1, 1) is
y − 1 = − 1(x − 1) ⇒ x + y − 2 = 0
It intersects x + 3 y = 0 at (3, –1) and hence normal meets
the curve in fourth quadrant.
x + y = 2 Y'
x + 3y = 0
y=x
X
O
(3,–1)
Y'
11. Given,
y + 3x = 12 y
3
2
...(i)
On differentiating w.r.t. x, we get
dy
dy
⇒
3 y2
+ 6x = 12
dx
dx
dy
6x
dx 12 − 3 y2
=
⇒
⇒
=
dx 12 − 3 y2
dy
6x
dx
For vertical tangent,
=0
dy
⇒ 12 − 3 y2 = 0
⇒
y=±2
4
and again
3
2
putting y = − 2 in Eq. (i), we get 3x = − 16, no real
solution.
 4

So, the required point is  ±
, 2 .

3 
On putting, y = 2 in Eq. (i), we get x = ±
12. Slope of tangent y = f (x) is
dy
= f ′ (x)(3 , 4)
dx
Therefore, slope of normal
1
1
=−
=−
f ′ (x)(3 , 4)
f ′ (3)
But
⇒
∴
−
1
 3π 
= tan  
 4
f ′ (3)
−1
 π π
= tan  +  = − 1
 2 4
f ′ (3)
f ′ (3) = 1
y = a (sin θ − θ cos θ )
dx
∴
= a (− sin θ + sin θ + θ cos θ ) = a θ cos θ
dθ
dy
and
= a (cos θ − cos θ + θ sin θ )
dθ
dy
dy
= a θ sin θ ⇒
= tan θ
dθ
dx
Thus, equation of normal is
y − a (sin θ − θ cos θ ) − cos θ
=
x − a (cos θ + θ sin θ )
sin θ
and
⇒ − x cos θ + a θ sin θ cos θ + a cos 2 θ
= y sin θ + θ a sin θ cos θ − a sin 2 θ
⇒ x cos θ + y sin θ = a
whose distance from origin is
|0 + 0 − a |
=a
cos 2 θ + sin 2 θ
14. Given, 4x2 + 9 y2 = 1
…(i)
On differentiating w.r.t. x, we get
dy
dy
8x
4x
8x + 18 y
=0 ⇒
=−
=−
dx
dx
18 y
9y
(1,1)
X'
13. Given, x = a (cos θ + θ sin θ )
The tangent at point (h , k) will be parallel to 8x = 9 y,
then
4h 8
−
= ⇒ h = − 2k
9k 9
Point (h , k) also lies on the ellipse.
...(ii)
∴
4h 2 + 9k2 = 1
On putting value of h in Eq. (ii), we get
4(− 2k)2 + 9 k2 = 1 ⇒ 16k2 + 9k2 = 1
1
⇒
25k2 = 1 ⇒ k2 =
25
1
⇒
k=±
5
Thus, the point, where the tangents are parallel to
1
2
 2 1
8x = 9 y are  − ,  and  , −  .
5
 5 5
5
Therefore, options (b) and (d) are the answers.
1
15. Given,
xy = 1 ⇒ y =
x
dy
1
⇒
=− 2
dx
x
Thus, slope of normal = x2 (which is always positive) and
a
it is given ax + by + c = 0 is normal, whose slope = − .
b
a
a
⇒
− > 0 or
<0
b
b
Hence, a and b are of opposite sign.
[given]
16. Given,
y3 − 3xy + 2 = 0
On differentiating w.r.t. x, we get
dy
dy
3 y2
− 3x
− 3y = 0
dx
dx
Application of Derivatives 251
⇒
dy
(3 y2 − 3x) = 3 y ⇒
dx
dy
3y
=
dx 3 y2 − 3x
For the points where tangent is horizontal, the slope of
tangent is zero.
dy
3y
i.e.
=0
=0 ⇒
2
dx
3 y − 3x
⇒ y = 0 but y = 0 does not satisfy the given equation of
the curve, therefore y cannot lie on the curve.
So,
[null set]
H =φ
dy
For the points where tangent is vertical,
=∞
dx
y
⇒
=∞
y2 − x
⇒
y2 − x = 0 ⇒ y2 = x
On putting this value in the given equation of the curve,
we get
y3 − 3 ⋅ y2 ⋅ y + 2 = 0 ⇒ − 2 y3 + 2 = 0
⇒
y3 − 1 = 0 ⇒ y3 = 1
⇒
y=1, x=1
Then,
V = {1, 1}
17. As | f (x1) − f (x2)| ≤ (x1 − x2)2, ∀x1 , x2 ∈ R
⇒
3 = 3a (0)2 + 2b (0) + c
⇒
3=c
[as x2 =|x|2 ]
f (x1 ) − f (x2)
≤ |x1 − x2|
x1 − x2
0 = a (−2)3 + b(−2)2 + c(−2) + 5
⇒
0 = − 8a + 4b − 2c + 5
From Eqs. (i) and (ii),
12a − 4b = − 3
… (iv)
From Eqs. (ii) and (iii),
−8 a + 4 b = 1
… (v)
On adding Eqs. (iv) and (v), we get
4a = − 2 ⇒ a = − 1 / 2
On putting a = − 1 / 2 in Eq. (iv), we get
12(−1 / 2) − 4b = − 3 ⇒ − 6 − 4b = − 3
⇒
− 3 = 4b ⇒ b = − 3 / 4
∴
a = − 1 / 2 , b = − 3 / 4 and c = 3
19. Let any point P1 on y = x3 be (h , h3 ).
Then, tangent at P1 is
y − h3 = 3h 2(x − h )
...(i)
It meets y = x at P2.
On putting the value of y in Eq. (i), we get
⇒
 f (x1 ) − f (x2) 
 ≤ lim |x1 − x2|
lim 
x1 → x2 
x1 − x2  x1 → x2
(x − h ) (x2 + xh + h 2) = 3h 2(x − h )
⇒
x2 + xh + h 2 = 3h 2 or x = h
⇒
x2 + xh − 2h 2 = 0
⇒
(x − h ) (x + 2h ) = 0
⇒
∴ | f ′ (x)| ≤ 0, which shows| f ′ (x)| = 0
[as modulus is non negative or| f ′ (x)| ≥ 0]
x = h or x = − 2h
Therefore, x = − 2h is the point P2,
∴ f ′ (x) = 0 or f (x) is constant function.
which implies y = − 8h3
⇒ Equation of tangent at (1, 2) is
y−2
= f ′ (x) or y − 2 = 0
x−1
Hence, point P2 ≡ (−2h , − 8h3 )
[Q as f ′ (x) = 0]
⇒ y − 2 = 0 is required equation of tangent.
18. Given, y = ax3 + bx2 + cx + 5 touches X-axis at P(−2, 0)
which implies that X-axis is tangent at (–2, 0) and the
curve is also passes through (–2, 0).
The curve cuts Y-axis at (0, 5) and gradient at this point
is given 3, therefore at (0, 5) slope of the tangent is 3.
dy
Now,
= 3ax2 + 2bx + c
dx
Since, X-axis is tangent at (–2, 0).
 dy 
∴
=0
 dx x = − 2
0 = 3a (−2)2 + 2b (−2) + c
…(i)
0 = 12a − 4b + c
Again, slope of tangent at (0, 5) is 3.
dy
=3
∴
dx ( 0, 5 )
… (iii)
x3 − h3 = 3h 2(x − h )
⇒ | f ′ (x1 )| ≤ 0, ∀x1 ∈ R
⇒
⇒
…(ii)
Since, the curve passes through (–2, 0).
3
⇒ | f (x1 ) − f (x2)| ≤ |x1 − x2|2
∴
⇒
Again, tangent at P2 is y + 8h3 = 3 (−2h )2(x + 2h ).
y = x3 at P3
It meets
⇒
x3 + 8h3 = 12h 2(x + 2h )
⇒
⇒
x − 2hx − 8h 2 = 0
2
(x + 2h ) (x − 4h ) = 0 ⇒ x = 4h
⇒
y = 64h3
Therefore,
P3 ≡ (4h , 64h3 )
Similarly, we get
P4 ≡ (− 8h , − 83 h3 )
Hence, the abscissae are h, –2h, 4h, –8h,…, which form
a GP.
Let D ′ = ∆ P1 P2 P3 and D ′ ′ = ∆ P2 P3 P4
D ′ ∆ P1P2 P3
=
=
D ′ ′ ∆ P2 P3 P4
h
h3
1
−2h −8h3
2
4h 64h3
−2 h
−8 h
1
4h
64h3
2
−8h −512h3
3
1
1
1
1
1
1
252 Application of Derivatives
=
h
h3
1
−2h −8h3
2
4h 64h3
Since, tangent is parallel to x + 2 y = 0 ,
dy
1
then slope
=−
dx
2
1
1

From Eq. (i), − = − sin (x + y) ⋅ 1 − 

2
2
1
1
1
h
h3
1
× (−2) × (−8) −2h −8h3
2
4h 64h3
1
1
⇒
∴
1
D′
1
=
= 1 : 16 which is the required ratio.
D′ ′ 16
⇒
20. Given, y = (1 + x)y + sin −1 (sin 2 x)
Let
y = u + v, where u = (1 + x)y , v = sin −1 (sin 2 x).
On differentiating w.r.t.x, we get
dy du dv
…(i)
=
+
dx dx dx
u = (1 + x)y
Now,
On taking logarithm both sides, we get
⇒
⇒
Again,
⇒
⇒
⇒
∴ Equation of tangents are
y−0
y−0
1
1
= − and
=−
x − π /2
x + 3π / 2
2
2
⇒
log e u = y log e (1 + x)
1 du
y
dy
=
+
{log e (1 + x)}
u dx 1 + x dx
 y

du
dy
= (1 + x)y 
+
log e (1 + x)
dx
+
x
dx
1


sin (x + y) = 1, which shows cos (x + y) = 0.
y=0
π
3π
or −
x+ y=
⇒
2
2
π
3π
or −
∴
x=
2
2
 3π 
π 
Thus, required points are  , 0 and  −
, 0

 2
2 
⇒
π
3π
and 2 y = − x −
2
2
π
3π
x + 2 y = and x + 2 y = −
2
2
2y = − x +
are the required equations of tangents.
…(ii)
v = sin −1 (sin 2 x) ⇒ sin v = sin 2 x
dv
cos v
= 2 sin x cos x
dx
dv
1
=
(2 sin x cos x)
dx cos v
dv 2 sin x cos x 2 sin x cos x
…(iii)
=
=
dx
1 − sin 2 v
1 − sin 4 x
From Eq. (i),
 y
 2 sin x cos x
dy
dy
= (1 + x)y 
+
log e (1 + x) +
dx
+
x
dx
1


1 − sin 4 x
22. Slope of tangent at the point (x1 , y1 ) is 
dy

 dx ( x
.
1 , y1 )
Given curve, ( y − x5 )2 = x (1 + x2)2
 dy

− 5x4 = (1 + x2)2 + 2x (1 + x2) ⋅ 2x
⇒ 2 ( y − x5 ) 
 dx

Put x = 1 and y = 3, dy /dx = 8
Topic 2 Increasing and Decreasing
Functions
1. Let the thickness of layer of ice is x cm, the volume of
Again, the slope of the normal is
1
m=−
= −1
dy / dx
spherical ball (only ice layer) is
4
…(i)
V = π [(10 + x)3 − 103 ]
3
On differentiating Eq. (i) w.r.t. ‘t’, we get
dV 4
dx
[given]
= π (3(10 + x)2)
= − 50
dt 3
dt
[− ve sign indicate that volume is decreasing as
time passes].
dx
4π (10 + x)2
= − 50
⇒
dt
At x = 5 cm
dx
[4π (10 + 5)2] = − 50
dt
1
1
dx
50
⇒
=−
=−
cm /min
=−
9(2π )
18π
dt
225(4π )
Hence, the required equation of the normal is
y − 1 = (−1) (x − 0) i.e. y + x − 1 = 0
So, the thickness of the ice decreases at the rate of
1
cm /min.
18π
⇒
y −1
+ 2 sin x cos x / 1 − sin 4 x
dy y(1 + x)
=
dx
1 − (1 + x)y log e (1 + x)
At x = 0,
y = (1 + 0)y + sin −1sin (0) = 1
∴
1 −1
+ 2 sin 0 ⋅ cos 0 / (1 − sin 4 0)
dy 1 (1 + 0)
=
dx
1 − (1 + 0)1 log e (1 + 0)
⇒
dy
=1
dx
21. Given, y = cos (x + y)
dy
 dy

⇒   = − sin (x + y) ⋅ 1 +

 dx

dx
2. The given functions are
f (x) = ex − x,
…(i)
and
g (x) = x2 − x, ∀ x ∈ R
Application of Derivatives 253
Then, h (x) = ( fog )(x) = f ( g (x))
Now, h′ (x) = f ′ ( g (x)) ⋅ g′ (x)
2
= (eg( x ) − 1) ⋅ (2x − 1) = (e( x − x ) − 1) (2x − 1)
= (ex( x − 1) − 1) (2x − 1)
Q It is given that h (x) is an increasing function, so
h′ (x) ≥ 0
⇒
(ex( x − 1) − 1)(2x − 1) ≥ 0
Case I (2x − 1) ≥ 0 and (ex( x − 1) − 1) ≥ 0
1
⇒
x ≥ and x(x − 1) ≥ 0
2
⇒ x ∈ [1 / 2, ∞ ) and x ∈ (− ∞ , 0] ∪ [1, ∞ ), so x ∈ [1, ∞ )
Case II (2x − 1) ≤ 0 and [ex( x − 1) − 1] ≤ 0
1
1

x ≤ and x(x − 1) ≤ 0 ⇒ x ∈  −∞ , and x ∈ [0, 1]
⇒

2
2 
 1
x ∈ 0,
 2 
So,
 1
From, the above cases, x ∈ 0,
∪ [1, ∞ ).
 2 
3.
Key Idea Use formula :
1
Volume of cone = πr 2h, where r = radius and h = height of the
3
cone.
⇒
⇒
tan α =
1
2
r 1
=
h 2
1
r= h
2
f (x) = x3 − 3 (a − 2) x2 + 3ax + 7, for some
increasing in (0, 1] and decreasing in [1, 5).
r
]
h
…(i)
r
h
α
1 2
πr h
3
2
1
1 1 
[from Eq. (i)]
V = π  h (h ) =
πh3
∴
3 2 
12
On differentiating both sides w.r.t. ‘t’, we get
dV
dh
dh
1
4 dV
=
π (3h 2)
⇒
=
dt 12
dt πh 2 dt
dt
dh
dV
4
[Q given
⇒
= 5 m3 /min]
=
×5
dt πh 2
dt
Q Volume of cone is (V ) =
…(i)
⇒
φ′ (x) = f ′ (x) − f ′ (2 − x)
Also, we have f ′ ′ (x) > 0 ∀ x ∈ (0, 2)
⇒ f ′ (x) is a strictly increasing function
∀ x ∈ (0, 2).
Now, for φ(x) to be increasing,
φ′ (x) ≥ 0
[using Eq. (i)]
⇒
f ′ (x) − f ′ (2 − x) ≥ 0
⇒
f ′ (x) ≥ f ′ (2 − x) ⇒ x > 2 − x
[Q f′ is a strictly increasing function]
⇒
2x > 2 ⇒ x > 1
Thus, φ(x) is increasing on (1, 2).
Similarly, for φ(x) to be decreasing,
φ′ (x) ≤ 0
⇒
f ′ (x) − f ′ (2 − x) ≤ 0
[using Eq. (i)]
⇒
f ′ (x) ≤ f ′ (2 − x)
[Q f′ is a strictly increasing
⇒
x<2 − x
function]
⇒
2x < 2
⇒
x < 1 Thus, φ(x) is decreasing on (0, 1).
a ∈ R is
f′ (1) = 0
[from fig. tan α =
l
4. Given, φ (x) = f (x) + f (2 − x), ∀ x ∈ (0, 2)
5. Given that function,
Given, semi-vertical angle of right circular cone
 1
= tan −1  
 2
1


Let
α = tan −1  
 2
⇒
Now, at h = 10 m, the rate at which height of water
dh
4
1
level is rising =
m/min
=
×5=
dt h = 10 π (10)2
5π
[Q tangent at x = 1 will be
parallel to X-axis]
⇒ (3x2 − 6(a − 2) x + 3a )x = 1 = 0
⇒
3 − 6(a − 2) + 3a = 0
⇒
3 − 6a + 12 + 3a = 0
⇒
15 − 3a = 0 ⇒ a = 5
So,
f (x) = x3 − 9x2 + 15x + 7
⇒ f (x) − 14 = x3 − 9x2 + 15x − 7
⇒ f (x) − 14 = (x − 1) (x2 − 8x + 7) = (x − 1) (x − 1)(x − 7)
f (x) − 14
…(i)
⇒
= (x − 7)
(x − 1)2
f (x) − 14
Now,
= 0, (x ≠ 1)
(x − 1)2
⇒
⇒
x− 7 =0
x=7
[from Eq. (i)]
6. We have,
f (x) =
x
(d − x)
−
(a 2 + x2)1/ 2 (b2 + (d − x)2)1/ 2
Differentiating above w.r.t. x, we get
1
2x
(a 2 + x2)1/ 2 − x
2 (a 2 + x2)1/ 2
f ′ (x) =
(a 2 + x2)
(b2 + (d − x ) 2 )1/ 2 (−1) − (d − x )
−
2(d − x )(−1)
2(b2 + (d − x ) 2 )1/ 2
( b2 + ( d − x ) 2 )
[by using quotient rule of derivative]
254 Application of Derivatives
=
a 2 + x2 − x2 b2 + (d − x)2 − (d − x)2
+
(a 2 + x2)3/ 2
(b2 + (d − x)2)3/ 2
=
a2
b2
+ 2
> 0,
2 3/ 2
(a + x )
(b + (d − x)2)3/ 2
2
11.
Option (a) Let f (x) = ex − 1 − x.
∀ x ∈R
⇒
g (x) decreases
for 0 < x < 1
⇒
g (x) < g (0)
for 0 < x < 1
⇒ log e (1 + x) − x < 0
for 0 < x < 1
log e (1 + x) < x
for 0 < x < 1
Option (c) sin x > x
h (x) = sin x − x ⇒ h′ (x) = cos x − 1
Let
u
2e
1 + e 2u
For x ∈(0, 1), cos x − 1 < 0
[Q eu > 0]
So, g′ (u ) is increasing for all x ∈ R.
f (x) = x3 + bx2 + cx + d
f ′ (x) = 3x2 + 2bx + c
As we know that, if ax2 + bx + c > 0, ∀x, then a > 0 and
D < 0.
Now,
D = 4b2 − 12c = 4(b2 − c) − 8c
[where, b2 − c < 0 and c > 0]
D = (–ve) or D < 0
⇒ f ′ (x) = 3x2 + 2bx + c > 0 ∀x ∈ (− ∞ , ∞ )
[as D < 0 and a > 0]
Hence, f (x) is strictly increasing function.
9. Let f (x) = 3 sin x − 4 sin3 x = sin 3x
The longest interval in which sin x is increasing is of
length π .
So, the length of largest interval in which f (x) = sin 3x is
π
increasing is .
3
10. Given, f (x) = xex (1 − x )
⇒
f (x) > f (0) for 0 < x < 1
Therefore, option (b) is the answer.
g′ (u ) > 0, ∀ x ∈ R
∴
⇒
or
We have, g (u ) = 2 tan −1 (eu ) − π 2
⇒
f (x) increase in (0, 1)
Option (b) Let g (x) = log e (1 + x) − x, 0 < x < 1
x
1
g′ (x) =
−1 = −
< 0 for 0 < x < 1
1+ x
1+ x
⇒ g (u ) is an odd function.
8. Given,
⇒
⇒ ex − 1 − x > 0 or ex > 1 + x for 0 < x < 1
= π / 2 − 2 tan −1 (eu ) = − g (u )
g (− u ) = − g (u )
g′ (u ) =
f ′ (x) = ex − 1 > 0, ∀x ∈ (0, 1)
then
Hence, f (x) is an increasing function of x.
π
7. Given, g (u ) = 2 tan −1 (eu ) − for u ∈ (−∞ , ∞ )
2
π
g (− u ) = 2 tan −1 (e− u ) −
2
π
−1 u
= 2 (cot (e )) −
2
π
 π
= 2  − tan −1 (eu ) −
2
 2
∴
PLAN Inequation based upon uncompatible function. This type
of inequation can be solved by calculus only.
f ′ (x) = ex (1 − x ) + xex (1 − x ) (1 − 2x)
= ex (1 − x ) [1 + x (1 − 2x)]
= ex (1 − x ) (1 + x − 2x2)
= − ex (1 − x ) (2x2 − x − 1)
⇒ h (x) is decreasing function.
⇒
h (x) < h (0)
⇒
sin x − x < 0
⇒
sin x < x,which is not true.
Option (d)
Therefore, p′ (x) is an increasing function.
⇒
p(0) < p(x) < p(1)
⇒
− ∞ < log x − x < −1
⇒
log x − x < 0
⇒
log x < x
Therefore, option (d) is not the answer.
f (x) = ∫ ex (x − 1) (x − 2) dx
12. Let
⇒
f ′ (x) = ex (x − 1) (x − 2)
+
−
1
2
∴ f ′ (x) < 0 for 1 < x < 2
⇒ f (x) is decreasing for x ∈(1, 2).
13. Given, f (x) = sin 4 x + cos 4 x
On differentiating w.r.t. x, we get
f ′ (x) = 4 sin3 x cos x − 4 cos3 x sin x
= − ex (1 − x ) (x − 1) (2x + 1)
 1 
which is positive in  − , 1 .
 2 
 1 
Therefore, f (x) is increasing in − , 1 .
 2 
p(x) = log x − x
1
p′ (x) = − 1 > 0, ∀x ∈ (0, 1)
x
= 4 sin x cos x (sin 2 x − cos 2 x)
= 2 sin 2x (− cos 2x) = − sin 4x
Now,
⇒
f ′ (x) > 0 , if sin 4x < 0
π < 4x < 2π
+
Application of Derivatives 255
π
π
<x<
4
2
⇒
…(i)
⇒ Option (a) is not proper subset of Eq. (i), so it is not
correct.
π
3π
Now,
<x<
4
8
Since, option (b) is the proper subset of Eq. (i), so it is
correct.
x
14. Given, g (x) =
, where 0 < x ≤ 1
tan x
Now, g (x) is continuous in [0, 1] and differentiable in
]0, 1[.
0 < x < 1,
For
g′ (x) =
tan x − x sec2 x
tan 2 x
H ′ (x) = sec2 x − sec2 x − 2x sec2 x tan x
= − 2x sec2 x tan x < 0
Hence, H (x) is decreasing function in [0, 1].
Thus,
H (x) < H (0)
H (x) < 0
for
0 < x<1
⇒
g′ (x) < 0
for
0 < x<1
g (x) is decreasing function in (0, 1].
x
is a decreasing function in
Therefore, g (x) =
tan x
0 < x ≤ 1.
g (x) < g (0) for 0 < x ≤ 1
x
< 1 for 0 < x ≤ 1
tan x
⇒
⇒
Now, let
x < tan x for
x /sin x
f (x) = 
 1
∴
x > 0, π + x > e + x
F ′ (x) = f ′ { g (x)} ⋅ g ′ (x)
= (+ ) (− ) = − ve
f ′ (x) is –ve.
When
0 ≤ x < 1 ⇒ h (x) − h (1) = + ve
When
x≥1
Hence, for
x > 0,
⇒ h (x) − h (1) = − ve
h (x) − h (1) is neither always positive nor always
negative, so it is not strictly increasing throughout.
f ( x)
dy
> 2dx ⇒
y
x
dy
> 2∫ dx
y
0
∫
1
ln( f (x)) > 2x ⇒ f (x) > e2x
Also, as f ′ (x) > 2 f (x)
⇒
f ′ (x) > 2c2x > 0
18. Given, f (x) = ∫
x
1
x
As
 1
− t + 
 t
e
dt
t
1

− x + 

x
e
x
1

− x + 

x
e
x
+
1 
− + x 


 −1  e x
−  2
 x  1 /x
1

− x + 

x
e
1

− x + 

x
2e
=
x
x
f ′ (x) > 0, ∀x ∈ (0, ∞ )
∴ f (x) is monotonically increasing on (0, ∞ ).
⇒ Option (a) is correct and option (b) is wrong.
 1
− t + 
t
x e 
 1
Now, f (x) + f   = ∫
1/x
 x
t
dt + ∫
 1
− t + 
 t
e
1/ x
t
x
dt
= 0, ∀x ∈ (0, ∞ )
Therefore, option (c) is the answer.
log (π + x)
15. Given, f (x) =
log(e + x)
When
16. Let F (x) = h (x) − h (1) = f { g (x)} − f { g (1)}
=
⇒ f (x) increases in [0, 1].
x
Thus, f (x) =
increases in 0 < x ≤ 1.
sin x
f ′ (x) =
From Eqs. (i) and (iv), f ′ (x) < 0
f ′ (x) = 1 ⋅
For 0 < x < 1,
sin x − x cos x (tan x − x) cos x
f ′ (x) =
=
> 0 for 0 < x < 1
sin 2 x
sin 2 x
1
1
− log (π + x) ⋅
e+ x
π+x
[log (e + x)]2
…(iv)
∴ f (x) is decreasing for x ∈ (0, ∞ ).
0 < x≤1
for
0 < x≤1
for
x=0
Now, f is continuous in [0, 1] and differentiable in ]0, 1[.
log (e + x) ⋅
On multiplying Eqs. (ii) and (iii), we get
log (π + x) log (e + x)
>
π+x
e+ x
17. f ′ (x) > 2 f (x) ⇒
⇒
Also,
... (iii)
Therefore, option (d) is the answer.
for 0 < x < 1
⇒
... (ii)
∴ f (x) is decreasing function.
0 < x < 1, H (x) = tan x − x sec2x, 0 ≤ x ≤ 1
⇒
and
Since,
Now, H (x) is continuous in [0, 1] and differentiable in
]0, 1[.
For
log (π + x) > log (e + x)
1
1
>
e+ x π + x
[since, f (x) is an increasing function f ′ ( g (x)) is + ve and
g (x) is decreasing function g ′ ( f (x)) is −ve ]
H (x) = tan x − x sec2 x, 0 ≤ x ≤ 1
Again,
∴
Now, let
…(i)
g (x) = f (2x ) = ∫
−x
2x
2− x
g (− x) = f (2 ) = ∫
∴ f (2x ) is an odd function.
 1
− t + 
 t
e
2− x
2x
dt
t
 1
− t + 
 t
e
t
dt = − g (x)
256 Application of Derivatives
f ′ (x) > 0 for (0, 1)
19. Given, h (x) = f (x) − f (x)2 + f (x)3
f ′ (x) < 0 for (e, ∞ )
On differentiating w.r.t. x, we get
h ′ (x) = f ′ (x) − 2 f (x) ⋅ f ′ (x) + 3 f 2(x) ⋅ f ′ (x)
∴ P and Q are correct, II is correct, III is incorrect.
−1 1
f ′′ (x) = 2 − ⇒ f ′′ (x) < 0 for (0, ∞ )
x
x
= f ′ (x)[1 − 2 f (x) + 3 f 2(x)]
2
1

= 3 f ′ (x) ( f (x))2 − f (x) +

3
3 
∴ S, is correct, R is incorrect.

1
1 1
= 3 f ′ (x)  f (x) −  + − 


3
3 9

IV is incorrect.

1
3 − 1
= 3 f ′ (x)  f (x) −  +



3
9 

∴ ii, iii, iv are correct.
2
lim f (x) = − ∞ ⇒ lim f ′ (x) = − ∞ ⇒ lim f ′′ (x) = 0
x→∞
2

1
2
= 3 f ′ (x)  f (x) −  + 


3
9

2
h ′ ( x ) < 0, if f ′ ( x ) < 0 and h ′ ( x ) > 0, if f ′ ( x ) > 0
NOTE
Therefore, h (x) is an increasing function, if f (x) is
increasing function and h (x) is decreasing function, if
f (x) is decreasing function.
Therefore, options (a) and (c) are correct answers.
20. Let f (x) = log (1 + x) − x
⇒
f ′ (x) =
⇒
when
when
x
1
−1 = −
1+ x
1+ x
f ′ (x) > 0
− 1 < x < 0 and f ′ (x) < 0
x>0
∴ f (x) is increasing for −1 < x < 0.
⇒
f (x) < f (0) ⇒ log (1 + x) < x
Again, f (x) is decreasing for x > 0.
⇒
f (x) < f (0) ⇒ log (1 + x) < x
∴
log (1 + x) ≤ x, ∀x > − 1
21. Here,
⇒
 2x2 − log x,
x >0
y= 2
x
−
−
x
x
<0
2
log
(
),

1

dy 4x − x , x > 0
=
dx 4x − 1 , x < 0

x
4x2 − 1
(2x − 1) (2x + 1)
, x ∈ R − {0} =
x
x
1
1

 

Increasing when x ∈  − , 0 ∪  , ∞
 2  2 
=
∴
1  1

and decreasing when x ∈  −∞ , −  ∪ 0,  .

2  2
Solutions. (22-24)
f (x) = x + ln x − x ln x ⇒ f (1) = 1 > 0
f (e2) = e2 + 2 − 2e2 = 2 − e2 < 0
⇒ f (x) = 0 for some x ∈ (1, e2)
∴ I is correct
f ′ (x) = 1 +
1
1
− ln x − 1 = − ln x
x
x
22. (d)
x→∞
x→∞
23. (c)
24. (c)
d
x
π
π
(x + sin x) = 1 + cos x = 2 cos 2 > 0 for − < x < .
dx
2
2
2
Therefore, x + sin x is increasing in the given interval.
Therefore, (A)→ (p) is the answer.
d
Again,
(sec x) = sec x tan x which is > 0 for 0 < x < π / 2
dx
−π
and
< 0 for
< x<0
2
Therefore, sec x is neither increasing nor decreasing in
the given interval. Therefore, (B)→(r) is the answer.
3x (x + 1)
26. Let
f (x) = sin x + 2x −
π
25.
On differentiating w.r.t. x, we get
(6x + 3)
f ′ (x) = cos x + 2 −
⇒
π
6
 π
⇒
f ′ ′ (x) = − sin x − < 0, ∀x ∈ 0,
 2 
π
 π
.
∴
f ′ (x) is decreasing for all x ∈ 0,
 2 
⇒
f ′ (x) > 0
[ Q x < π /2]
⇒
f ′ (x) > f ′ (π / 2)
∴ f (x) is increasing.
Thus, when x ≥ 0, f (x) ≥ f (0)
3x(x + 1)
3x (x + 1)
⇒ sin x + 2x −
≥ 0 ⇒ sin x + 2x ≥
π
π
27. Let
f (x) = sin (tan x) − x
f ′ (x) = cos (tan x) ⋅ sec2x − 1
= cos (tan x) (1 + tan 2 x) − 1
= tan 2 x {cos (tan x)} + cos (tan x) − 1
tan 2 x
2

x2 
2
Q2
1
0
1
(
cos
)
,
cos
−
<
≠
⇒
>
−
x
x
x
x

2

tan 2 x 
cos (tan x) > 1 −

 ⇒


2
1


f ′ (x) > tan 2 x cos (tan x) −

2 
> tan 2 x cos (tan x) −
> tan 2 x [cos (tan x) − cos (π / 3)] > 0
Application of Derivatives 257
⇒

eax − 1 
= lim aeax +
x 
x → 0− 
f (x) is increasing function, for all x ∈ [0, π / 4]
f (0) = 0 ⇒ f (x) ≥ 0, for all x ∈ [0, π / 4]
As
⇒
sin (tan x) ≥ x
−1 ≤ p ≤ 1
28. Given,
Let
Now,
= lim aeax + a ⋅ lim
x → 0−
f (x) = 4x − 3x − p = 0
1 3
f (1 / 2) = − − p = − 1 − p ≤ 0
2 2
= ae0 + a (1) = 2a
f ′ (x) − f ′ (0)
Rf ′ ′ (0) = lim
+
x+0
x→ 0
3
f (1) = 4 − 3 − p = 1 − p ≥ 0
Also,
[Q p ≥ − 1]
and
[Q p ≤ 1]
∴ f (x) has atleast one real root between [1 /2, 1].
f ′ (x) = 12x − 3 > 0 on [1 / 2, 1]
⇒
f ′ (x) increasing on [1 /2, 1]
⇒
f has only one real root between [1 /2, 1].
To find a root, we observe f (x) contains 4x3 − 3x, which is
multiple angle formula for cos 3θ.
∴ Put
x = cos θ
−1
⇒
p = cos 3 θ ⇒ θ = (1 / 3) cos ( p)

1
∴ Root is cos  cos −1 ( p) .

3
NOTE This type is asked in 1983 and repeat after 13 years.
At x = 0, LHL = lim f (x) = lim xeax = 0
and
x → 0−
x → 0−
+
+
RHL = lim f (x) = lim (x + ax2 − x3 ) = 0
x→ 0
x→ 0
Therefore, LHL = RHL = 0 = f (0)
So, f (x) is continuous at x = 0.
ax
 ax
Also,
f ′ (x) = 1 ⋅ e + axe 2 , if x < 0
1 + 2ax − 3x , if x > 0
and
Lf ′ (0) = lim
x → 0−
= lim
x→0
and
−
Rf ′ (0) = lim
x→ 0
+
= lim
x →0 +
f (x) − f (0)
x−0
xeax − 0
= lim eax = e0 = 1
x
x → 0−
f (x) − f (0)
x+0
x + ax2 − x3 − 0
x
= lim 1 + ax − x = 1
2
Therefore, Lf ′ (0) = Rf ′ (0) = 1 ⇒ f ′ (0) = 1
(ax + 1) eax ,
if x < 0
Hence,
f ′ (x) = 
1,
if x = 0
2
1 + 2ax − 3x , if x > 0
Now, we can say without solving that, f ′ (x) is
continuous at x = 0 and hence on R. We have,
ax
ax
f ′ ′ (x) = ae + a (ax + 1) e , if x < 0
2
6
,
if x > 0
−
a
x

f ′ (x) − f ′ (0)
and
Lf ′ ′ (0) = lim
x−0
x → 0−
x → 0−
= lim
2ax − 3x2
= lim 2a − 3x = 2a
x
x → 0+
x→ 0
+
Therefore, Lf ′ ′ (0) = Rf ′ ′ (0) = 2a
a (ax + 2) eax ,
Hence,
f ′ ′ (x) = 2a ,
2a − 6x,
⇒
if x < 0
if x = 0
if x > 0
For x < 0, f ′ ′ (x) > 0, if x > − 2 / a
2
f ′ (x) > 0, if − < x < 0
a
⇒
and for x > 0, f ′ ′ (x) > 0,
⇒ for x > 0,
(ax + 1)eax − 1
x
if 2a − 6x > 0
f ′ ′ (x) > 0, if x < a /3
Thus, f (x) increases on [–2/a, 0] and on [0, a/3].
 2 a
.
Hence, f (x) increases on − ,
 a 3 
y = f (x) = 2 sin x + 2 tan x − 3x
30. Let
⇒
f ′ (x) = 2 cos x + 2 sec2 x − 3
0 ≤ x < π / 2, f ′ (x) > 0
For
Thus, f (x) is increasing.
When
x ≥ 0, f (x) ≥ f (0)
⇒
2 sin x + 2 tan x − 3x ≥ 0 + 0 − 0
⇒
2 sin x + 2 tan x ≥ 3x
31. Let f (x) = 1 + x log (x + x2 + 1 ) − 1 + x2
∴ f ′ (x) = x ⋅
x → 0+
= lim
(1 + 2ax − 3x2) − 1
x
+
Now, for x < 0, f ′ ′ (x) > 0, if ax + 2 > 0
⇒ 4 cos θ − 3 cos θ − p = 0
3
29.
= lim
x→ 0
2
Also,
x → 0−
eax − 1
ax
−
⇒
x
x +1
2

1 +




x2 + 1 
x
x+
=
x +1
2
x
x +1
2
f ′ (x) = log (x +
⇒
+ log (x +
+ log (x +
x2 + 1 )
x2 + 1 ) −
[Q log (x +
∴ f (x) is increasing for x ≥ 0.
f (x) ≥ f (0)
⇒ 1 + x log (x + 1 + x2 ) − 1 + x2 ≥ 1 + 0 − 1
⇒ 1 + x log (x + 1 + x2 ) ≥ 1 + x2 , ∀ x ≥ 0
32. Given, A = x :

x +1
x2 + 1 )
f ′ (x) ≥ 0
⇒
x
2
π
π
≤x≤ 
6
3
x2 + 1 ) ≥ 0]
258 Application of Derivatives
∴ Perimeter of the rectangle
π

y=4
− x + sin 2x
 4

dy
For maximum,
=0
dx
⇒
− 1 + 2 cos 2x = 0
1
cos 2x =
⇒
2
π
π
⇒
2x =
⇒ x=
3
6
and f (x) = cos x − x − x2
⇒
f ' (x) = − sin x − 1 − 2x = − (sin x + 1 + 2x)
π π 
which is negative for x ∈
,
⇒ f ' (x) < 0
 6 3 
or f (x) is decreasing.
  π  π 
Hence, f ( A ) =  f   , f   
  3  6 
1 π 
π 3 π 
π 
− 1 +  
=  − 1 +  ,
3 2 6 
6 
2 3 
and
Topic 3 Rolle’s and Lagrange’s Theorem
π
∴ At x = , the rectangle PQRS have maximum
6
perimeter.
So length of sides
 π π π
PS = QR = 2  −  =
 4 6 6
1 
For some x in  , 1
2 
f ′ (x) = 1
f′ (1) > 1
[LMVT]
2. Given, f (x) = 2 + cos x, ∀x ∈ R
There exists a point ∈ [t , t + r ], where
Statement I
f ′ (c) = 0
and
∴
Hence, Statement I is true.
Statement II f (t ) = f (t + 2π ) is true. But statement II
is not correct explanation for statement I.
0 ≤ x ≤ 1.
f ′ (c) =
Using Lagrange’s Mean Value theorem,
6 −2
=4
1 −0
g (1) − g (0) 2 − 0
=
=2
1 −0
1 −0
g ′ (c) =
⇒
f ′ (c) = 2 g′ (c)
Topic 4 Maxima and Minima
π
and 0 ≤ y ≤ 2 sin(2x)
2
On drawing the diagram,
Let the side PS on the X-axis, such that P (x, 0), and
π

Q (x, 2 sin(2x)), so length of the sides PS = QR = 2  − x
4

and PQ = RS = 2 sin 2x.
{(x, y) ∈ R × R : 0 ≤ x ≤
Y
2
Q
O
P
R
π/4
S π/2
X
π
π
× 3=
6
2 3
Key Idea Volume of parallelopiped formed by the vectors a,
b and c is V = [ a b c].
3.
1. Given region is
Required area =
and g (x) = | x + 1|, where x ∈ R.
Clearly, maximum of f (x) occurred at x = 2, so α = 2.
and minimum of g (x) occurred at x = − 1, so β = − 1.
⇒
αβ = − 2
(x − 1) (x2 − 5x + 6)
Now, lim
x → − αβ
x2 − 6 x + 8
(x − 1) (x − 3) (x − 2)
[Qαβ = − 2]
= lim
x→ 2
(x − 4) (x −2)
(x − 1) (x − 3) (2 − 1) (2 − 3) 1 × (− 1) 1
= lim
=
=
=
x→ 2
(2 − 4)
(x − 4)
(− 2)
2
f (1) − f (0)
1 −0
and
 π
PQ = RS = 2 sin   = 3
 3
2. Given functions are f (x) = 5 − |x − 2|
3. Since, f (x) and g (x) are differentiable functions for
∴
d 2y
= − 4 sin 2x| π < 0
x=
dx2 x = π
6
6
1. f ′ (x) is increasing
∴

 π 
Q x ∈ 0, 
 2 

$ , $j + λk
$ and λ$i + k
$ , which
Given vectors are $i + λ$j + k
forms a parallelopiped.
∴Volume of the parallelopiped is
⇒
1
λ
1
V = 0
λ
1
0
λ = 1 + λ3 − λ
1
V = λ3 − λ + 1
On differentiating w.r.t. λ, we get
dV
= 3 λ2 − 1
dλ
dV
For maxima or minima,
=0
dλ
1
⇒
λ=±
3
Application of Derivatives 259
1

2 3 > 0 , for λ =

d 2V
3
= 6λ = 
1
dλ2
2 3 < 0 , for λ = −
3

and
6.
1
d 2V
, so volume ‘V ’ is minimum
is positive for λ =
3
dλ2
1
for λ =
3
Key Idea
(i) Use formula of volume of cylinder, V = πr 2h
where, r = radius and h = height
(ii) For maximum or minimum, put first
derivative of V equal to zero
Q
4. Given function f (x) = x kx − x2
… (i)
From the figure,
the function f (x) is defined if kx − x ≥ 0
2
⇒
x2 − kx ≤ 0
⇒
Let a sphere of radius 3, which inscribed a right circular
cylinder having radius r and height is h, so
x ∈ [0, k]
… (ii)
⇒
and
h
= 3 cos θ
2
h = 6 cos θ
r = 3 sin θ
…(i)
r
because it is given that f (x) is increasing in interval
x ∈ [0, 3], so k should be positive.
Now, on differentiating the function f (x) w.r.t. x, we get
x
f ′ (x) = kx − x2 +
× (k − 2x)
2 kx − x2
=
2(kx − x ) + kx − 2x
2
2 kx − x
2
2
=
3kx − 4x
h
2
2 kx − x
θ
h/2
2
r
as f (x) is increasing in interval x ∈ [0, 3], so
f ′ (x) ≥ 0 ∀ x ∈ (0, 3)
⇒
3
3kx − 4x2 ≥ 0
⇒
4x2 − 3kx ≤ 0
3k
 3k 

(as k is positive)
⇒ 4x x −  ≤ 0 ⇒ x ∈ 0,
 4 

4
Q Volume of cylinder V = πr 2h
= π (3 sin θ )2(6 cos θ ) = 54π sin 2 θ cos θ .
dV
For maxima or minima,
=0
dθ
2
3
⇒
54π [2 sin θ cos θ − sin θ ] = 0
⇒
sin θ [2 cos 2 θ − sin 2 θ ] = 0
3k
⇒k ≥4
4
⇒ Minimum value of k = m = 4
and the maximum value of f in [0, 3] is f (3).
⇒
tan 2 θ = 2
⇒
tan θ = 2 ⇒ sin θ =
Q f is increasing function in interval x ∈ [0, 3]
and
So,
3≤
Q M = f (3) = 3 4 × 3 − 3 = 3 3
2
Therefore, ordered pair (m, M ) = (4, 3 3 )
5. The non-zero four degree polynomial
f (x) has
extremum points at x = −1, 0, 1, so we can assume
f ′ (x) = a (x + 1)(x − 0) (x − 1) = ax(x2 − 1)
where, a is non-zero constant.
f ′ (x) = ax3 − ax
a
a
f (x) = x4 − x2 + C
⇒
4
2
[integrating both sides]
where, C is constant of integration.
Now, since f (x) = f (0)
a 4 a 2
x 4 x2
=
⇒
x − x +C =C ⇒
4
2
4
2
⇒
x2(x2 − 2) = 0 ⇒ x = − 2 , 0, 2
Thus, f (x) = f (0) has one rational and two irrational
roots.
2
3
1
3
From Eqs. (i) and (ii), we get
1
h =6
=2 3
3
cos θ =
…(ii)
7. Given function is
f (x) = 9x4 + 12x3 − 36x2 + 25 = y (let)
dy
For maxima or minima put
=0
dx
dy
= 36x3 + 36x2 − 72x = 0
⇒
dx
⇒
x3 + x2 − 2x = 0
⇒
x[x2 + x − 2] = 0
2
⇒
x[x + 2x − x − 2] = 0
⇒
x[x(x + 2) − 1(x + 2)] = 0
⇒
x(x − 1)(x + 2) = 0
⇒
x = − 2, 0, 1
By sign method, we have following
–
+
–2
–
0

 π 
Q θ ∈ 0, 2  


+
1
260 Application of Derivatives
dy
changes it’s sign from negative to positive at
dx
x = ‘−2 ’ and ‘1’, so x = − 2, 1 are points of local minima.
dy
changes it’s sign from positive to negative at
Also,
dx
x = 0, so x = 0 is point of local maxima.
Y
Since,
(0, 12)
M
R
XN
∴ S1 = { −2, 1} and S 2 = {0}.
X
y=12–x2
YN
y2 = x − 2
and the equation of line is
…(i)
y=x
…(ii)
y=x
y2=x–2
M
P(t2+2, t)
O
P(a, 0)
O
S
8. Given equation of curve is
Y
Q(a, 12–a2)
X
(2, 0)
Then, area of rectangle PQRS
= 2 × (Area of rectangle PQMO)
[due to symmetry aboutY -axis]
= 2 × [a (12 − a 2)] = 24a − 2a3 = ∆ (let).
The area function ∆ a will be maximum, when
d∆
= 0 ⇒ 24 − 6a 2 = 0
da
⇒
a2 = 4 ⇒ a = 2
[Q a > 0]
So, maximum area of rectangle
PQRS = (24 × 2) − 2 (2)3
= 48 − 16 = 32 sq units
10. We have,
f (x) = 3x3 − 18x2 + 27x − 40
Consider a point P (t 2 + 2, t ) on parabola (i).
For the shortest distance between curve (i) and line
(ii), the line PM should be perpendicular to line (ii) and
parabola (i), i.e. tangent at P should be parallel to
y = x.
∴
dy
= Slope of tangent at point P to curve (i)
dx at point P
=1
⇒
[Q tangent is parallel
to line y = x]
1
=1
2yP
[differentiating the curve (i), we get2 y
⇒
1
1
=1⇒t =
2
2t
dy
=1]
dx
[Q P (x, y) = P (t 2 + 2, t )]
⇒ f ′ (x) = 9x2 − 36x + 27
= 9(x2 − 4x + 3) = 9 (x − 1) (x − 3)
Also, we have S = { x ∈ R : x2 + 30 ≤ 11 x}
Clearly,
x2 + 30 ≤ 11x
⇒
x2 − 11x + 30 ≤ 0
⇒
(x − 5) (x − 6) ≤ 0 ⇒ x ∈ [5, 6]
So,
S = [5, 6]
Note that f (x) is increasing in [5, 6]
…(i)
[Qf ′ (x) > 0 for x ∈[5, 6]
∴f (6) is maximum, where
f (6) = 3(6)3 − 18(6)2 + 27(6) − 40 = 122
11. According to given information, we have the following
figure.
C(t2, 2t)
B(9, 6)
 9 1
So, the point P is  ,  .
 4 2
9 − 1
4 2
Now, minimum distance = PM =
2
[Q distance of a point P (x1 , y1 ) from a line
7
|ax1 + by1 + c |
units
ax + by + c = 0 is
=
2
2
a + b
 4 2
9. Equation of parabola is given, y = 12 − x2
or
x2 = − ( y − 12).
Note that vertex of parabola is (0, 12) and its open
downward.
Let Q be one of the vertices of rectangle which lies on
parabola. Then, the coordinates of Q be (a, 12− a 2)
A(4, –4)
For y2 = 4ax, parametric coordinates of a point is (at 2,
2at).
∴For y2 = 4x, let coordinates of C be (t 2, 2t).
1
Then, area of ∆ABC =
2
t2
9
2t
6
4
−4 1
1
1
1
= |t 2(6 − (− 4)) − 2t (9 − 4) + 1(− 36 − 24)|
2
1
10 2
= |10t 2 − 10t − 60| =
|t − t − 6|= 5| t 2 − t − 6|
2
2
Application of Derivatives 261
Let,
A (t ) = 5| t 2 − t − 6|
Clearly, A (4, − 4) ≡ A (t12, 2t1 ) ⇒ 2t1 = − 4
⇒
t1 = − 2
...(i)
and B(9, 6) ≡ B(t22,2t2) ⇒ 2t2 = 6 ⇒ t2 = 3
Since, C is on the arc AOB, the parameter ‘t’ for point
C ∈ (− 2, 3).
Let f (t ) = t 2 − t − 6 ⇒ f ′ (t ) = 2t − 1
1
Now,
f ′ (t ) = 0 ⇒ t =
2
Thus, for A (t ), critical point is at t =
1
2
2
125
1
1
 1
 1
Now,A   = 5   − − 6 =
= 31 [Using Eq. (i)]
 2
 2
4
4
2
12. Let h = height of the cone, r = radius of circular base
= (3)2 − h 2
= 9−h
[Q l2 = h 2 + r 2]
2
…(i)
1
2
1

∈ ( −∞ , 2 2 ]
< 0,  x −  +

x
x x − 1
x
∴ Local minimum value is 2 2.
x−
14. Total length = 2r + r θ = 20
⇒
l=
3
Now, volume (V ) of cone =
V (h ) =
13. We have,
f(x) = x2 +
1
x2
and g( x ) = x −
2
1
f(x)
⇒ h( x ) =
g( x )
x
1
 +2
2
x
x =
∴ h( x ) =
1
1
x−
x−
x
x
1
2

⇒
h( x ) =  x −  +

x x − 1
x
1
2
1

∈ [2 2 , ∞ )
x − > 0,  x −  +

x x − 1
x
x
x2 +
1

x −

r
r
rθ
For maxima or minima, put
dA
= 0.
dr
10 − 2r = 0 ⇒ r = 5
1
20 − 2 (5) 
Amax = (5)2


2
5
1
= × 25 × 2 = 25 sq. m
2
15. According to given information, we have Perimeter of
1 2
πr h
3
1
[From Eq. (i)]
π (9 − h 2)h
3
1
…(ii)
= π [9h − h3 ]
3
For maximum volume V ′ (h ) = 0 and V ′′ (h ) < 0.
Here,
V ′ (h ) = 0 ⇒ (9 − 3h 2) = 0
[Q h </ 0]
⇒
h= 3
1
and V ′′ (h ) = π (−6h ) < 0 for h = 3
3
Thus, volume is maximum when h = 3
Now, maximum volume
1
V ( 3 ) = π (9 3 − 3 3 ) [from Eq. (ii)]
3
= 2 3π
⇒
θ
A = 10r − r 2
dA
= 10 − 2r
dr
∴
∴
r
20 − 2r
r
Now, area of flower-bed,
1
A = r 2θ
2
1 2 20 − 2r 
A= r 
⇒

2 
r 
⇒
h
θ=
⇒
square + Perimeter of circle = 2 units
⇒
4 x + 2 πr = 2
1 − 2x
...(i)
⇒
r=
π
Now, let A be the sum of the areas of the square and the
circle. Then,
(1 − 2x) 2
A = x 2 + π r 2 = x2 + π
π2
2
(1 − 2x)
A (x) = x2 +
⇒
π
dA
Now, for minimum value of A (x),
=0
dx
2 (1 − 2x)
2 − 4x
⇒ 2x +
⋅ (− 2) = 0 ⇒ x =
π
π
2
...(ii)
⇒
πx + 4x = 2 ⇒ x =
π+4
Now, from Eq. (i), we get
2
1 −2⋅
1
π + 4 π + 4 −4
r=
=
=
π(π + 4) π + 4
π
From Eqs. (ii) and (iii), we get
...(iii)
x = 2r
16. Here, to find the least value of α ∈ R, for which
1
≥ 1, for all x > 0.
x
i.e. to find the minimum value of α
1
y = 4αx2 + ; x > 0 attains minimum value of α.
x
dy
1
= 8αx − 2
∴
dx
x
4 αx2 +
when
…(i)
262 Application of Derivatives
2
d 2y
…(ii)
= 8α + 3
2
dx
x
dy
When
= 0, then 8x3α = 1
dx
1/3
d 2y
 1
At x =   ,
= 8α + 16α = 24α, Thus, y attains
 8α 
dx2
1/3
 1
minimum when x =   ; α > 0.
 8α 
Now,
 1
∴ y attains minimum when x =  
 8α 
i.e.
 1
4α  
 8α 
PLAN The given equation contains algebraic and trigonometric
functions called transcendental equation. To solve
transcendental equations we should always plot the graph for
LHS and RHS.
Here, x2 = x sin x + cos x
X′
Y′
Let f (x) = x2 and
23
/
+ (8α )1/3 ≥ 1
To plot,
PLAN Any function have extreme values (maximum or minimum) at
its critical points, where f ′ ( x ) = 2.
∴
g (x) = x sin x + cos x
g′ (x) = x cos x + sin x − sin x
g′ (x) = x cos x
lim f (x)
x→ 0 2
x
/2
–5 π/ 2
f (x)
=2
x2
f (x) will be of the form ax4 + bx3 + 2x2.
–π –π/2 O
lim
x→ 0
f (x) = ax4 + bx3 + 2x2
f ′ (x) = 4ax3 + 3bx2 + 4x
f ′ (1) = 4a + 3b + 4 = 0
f ′ (2) = 32a + 12b + 8 = 0
8a + 3b + 2 = 0
π/2 π
⇒
Y′
...(i)
So, graph of f (x) and g (x) are shown as
Y
...(ii)
f (x)
X′
x4
− 2x3 + 2x2
2
f (2) = 8 − 16 + 8 = 0
Y′
18. Here, x = − 1 and x = 2 are extreme points of
f (x) = α log|x| + βx2 + x, then
α
f ′ (x) = + 2 βx + 1
x
…(i)
f ′ (−1) = − α − 2 β + 1 = 0
[at extreme point, f ′ (x) = 0]
α
…(ii)
f ′ (2) = + 4 β + 1 = 0
2
g (x)
X
–π/2 O π/2
f (x) =
On solving Eqs. (i) and (ii), we get
1
α = 2, β = −
2
5π/2
3π 7π
At x = 0,
,
,... , f ′′ (x) > 0, so minimum
2 2
π 5π 9π
At x = ,
,
,... , f ′ (x) < 0, so maximum
2 2 2
On solving Eqs. (i) and (ii),
1
we get
a = , b = −2
2
∴
…(ii)
g′ (x) = 0 ⇒ x cos x = 0
π 3π 5π 7π
x = 0, ,
,
,
2 2 2 2
g (x)
=3
[Q f (x) is of four degree polynomial]
Let
⇒
⇒
and
⇒
…(i)
g′′ (x) = − x sin x + cos x
Put
Since, the function have extreme values at x = 1 and
x = 2.
∴
f ′ (x) = 0 at x = 1 and x = 2
⇒
f′ (1) = 0 and f′ (2) = 0
⇒
g (x) = x sin x + cos x
We know that, the graph for f (x) = x2
α 1/3 + 2α 1/3 ≥ 1
1
⇒
3α 1/3 ≥ 1 ⇒ α ≥
27
1
Hence, the least value of α is
.
27
⇒
X
O
.
Also, it is given that,
f (x) 
lim 
=3 ⇒ 1 +
x→ 0 1 +
x2 

x2
Y
1/3
⇒
17.
19.
So, number of solutions are 2.
2
2
2
2
20. Given function, f (x) = ex + e− x , g (x) = xex + e− x and
2
2
h (x) = x2ex + e− x are strictly increasing on [0, 1].
Hence, at x = 1, the given function attains absolute
maximum all equal to e + 1 / e .
⇒
a=b=c
21. Given,
(2 + x)3 , if − 3 < x ≤ −1
f (x) = 
x 2 / 3 , if − 1 < x < 2

Application of Derivatives 263
⇒
3 (x + 2)2 , if − 3 < x ≤ −1

f ′ (x) = 2 − 1
3
, if − 1 < x < 2
3 x
26. Let
⇒
f (x) = x25 (1 − x)75 , x ∈[0, 1]
f ′ (x) = 25 x24 (1 − x)75 − 75x25 (1 − x)74
= 25x24 (1 − x)74 [(1 − x) − 3x]
Y
= 25x24 (1 − x)74 (1 − 4x)
For maximum value of f (x), put f ′ (x) = 0
X′
(–3,0)
X
(–1,0)
⇒
25x24 (1 − x)74 (1 − 4x) = 0
1
x = 0, 1,
4
⇒
Y′
x = 0, y = 0
Also, at
Clearly, f ′ (x) changes its sign at x = −1 from +ve to −ve
and so f (x) has local maxima at x = −1 .
Also, f ′ (0) does not exist but f ′ (0 − ) < 0 and f ′ (0 + ) < 0.
It can only be inferred that f (x) has a possibility of a
minima at x = 0 . Hence, the given function has one local
maxima at x = −1 and one local minima at x = 0 .
22. Given f (x) = x2 + 2bx + 2c2 and g (x) = − x2 − 2cx + b2
At
x = 1, y = 0
and at
x = 1 / 4, y > 0
∴ f (x) attains maximum at x = 1 / 4.
27. Let the coordinates of P be (a cos θ , b sin θ )
Equations of tangents at P is
Y
Then, f (x) is minimum and g (x) is maximum at

−b
−D 
and f (x) =
x =
 , respectively.

4a
4a 
∴
and
min f (x) =
− (4b2 − 8c2)
= (2c2 − b2)
4
max g (x) = −
N
X'
O
(4c2 + 4b2)
= (b2 + c2)
4(−1)
Y'
x
y
cos θ + sin θ = 1
a
b
2c2 − b2 > b2 + c2
⇒
c2 > 2b2
⇒
| c| > 2| b|
Again, equation of normal at point P is
23. It is clear from figure that at x = 0, f (x) is not continuous.
Y
y=x
y = –x
X′
X
O
Y′
Here, f (0) > RHL at x = 0 and f (0) > LHL at x = 0
ax secθ − by cosec θ = a 2 − b2
Let M be foot of perpendicular from O to PK, the normal
at P.
1
Area of ∆OPN = (Area of rectangle OMPN )
2
1
= ON ⋅ OM
2
1
ab
Now,
ON =
=
cos 2 θ sin 2 θ
b2 cos 2 θ + a 2 sin 2 θ
+
a2
b2
[perpendicular from O, to line NP]
So, local maximum at x = 0.
2
x2 − 1
24. Given, f (x) = 2
=1 − 2
x +1
x +1
f (x) will be minimum, when
X
M
K
Now, min f (x) > max g (x)
⇒
P
2
is maximum,
x +1
2
i.e.
when x2 + 1 is minimum,
i.e.
at x = 0.
∴ Minimum value of f (x) is f (0) = − 1
and OM =
a 2 sec2 θ + b2cosec2θ
Thus, area of ∆OPN =
Let f (θ ) =
f ' (θ ) =
=
(a 2 – b2) ⋅ cos θ ⋅ sin θ
a 2 sin 2 θ + b2 cos 2 θ
ab(a 2 − b2) cos θ ⋅ sin θ
2(a 2 sin 2 θ + b2 cos 2 θ )
=
25. The maximum value of f (x) = cos x + cos ( 2x) is 2
which occurs at x = 0. Also, there is no other value of x
for which this value will be attained again.
a − b2
2
tan θ
α 2 tan 2 θ + b2
ab(a 2 − b2) tan θ
2(a 2 tan 2 θ + b2)
[0 < θ < π / 2]
sec2 θ( a 2 tan 2 θ + b2) − tanθ (2a 2 tanθ sec2θ)
(a 2 tan 2 θ + b2)2
264 Application of Derivatives
=
sec2 θ (a 2 tan 2 θ + b2 – 2a 2 tan 2 θ )
(a 2 tan 2 θ + b2)2
∴
sec2 θ( a tan θ + b)(b – a tan θ )
=
(a 2 tan 2 θ + b2)2
For maximum or minimum, we put
f′ (θ ) = 0 ⇒ b – a tan θ = 0
[sec2 θ ≠ 0, a tan θ + b ≠ 0, 0 < θ < π / 2 ]
⇒
tan θ = b / a
31. Given, f : R → R and f (x) = (x − 1) (x − 2 ) (x − 5 )
−1

Also, f ′ (θ )> 0, if 0 < θ−1< tan (b / a )
< 0, if tan (b / a ) < θ < π / 2
x
Since, F (x) = ∫ f (t ) dt , x > 0
0
So, F ′ (x) = f (x) = (x − 1)(x − 2 )(x − 5 )
According to wavy curve method
Therefore, f (θ ) has maximum, when
b
 b
θ = tan −1   ⇒ tan θ =
 a
a
Again,
sin θ =
b
a 2 + b2
, cos θ =
-
2
…(i)
where, (a1 + 2a 2 x2 + 3a3 x4 + K + na nx2n − 2) > 0, ∀ x ∈ R.
P ′ (x) > 0, when x > 0
Thus,

P ′ (x) < 0, when x < 0
i.e. P ′ (x) changes sign from (–ve) to (+ve) at x = 0.
P (x) attains minimum at x = 0.
Hence, it has only one minimum at x = 0 .
29. y = a log x + bx2 + x has extremum at x = − 1 and x = 2.
∴
⇒
dy
= 0, at x = − 1
dx
a
x=2 ⇒
+ 2bx + 1 = 0, at x = − 1
x
x=2
− a − 2b + 1 = 0 and
a = 2 and b = −
a
+ 4b + 1 = 0
2
1
2
 p, if p > q
 q, if q > p
30. Since, max ( p, q) = 
 p, if p is greatest.

and max ( p, q, r ) =  q, if q is greatest.
 r , if r is greatest.

∴ max ( p, q) < max ( p, q, r ) is false.
 p − q , if p ≥ q
We know that, | p − q| = 
q − p , if p < q
0
2

t
t
t2
=  −8
+ 17 − 10t 
4
3
2
0

64
124
10
=4 −
+ 34 − 20 = 38 −
=−
3
3
3
Q At the point of maxima x = 2 , the functional value
10
F (2 ) = − , is negative for the interval x ∈ (0, 5 ), so
3
F (x) ≠ 0 for any value of x ∈ (0, 5 ),
Hence, options (a), (b) and (d) are correct.
4
⇒ P ′ (x) = 2a1x + 4a 2 x3 + ... + 2na nx2n − 1
and
5
2
0
where, a n > a n − 1 > a n − 2 > K > a 2 > a1 > a 0 > 0
and
2
+
Q F (2 ) = ∫ f (t ) dt = ∫ (t3 − 8t 2 + 17t − 10)dt
28. Given, P (x) = a 0 + a1x2 + a 2 x4 + K+ a nx2n
∴
-
F ′ (x) changes, it’s sign from negative to positive at x = 1
and 5, so, F (x) has minima at x = 1 and 5 and as F ′ (x)
changes, it’s sign from positive to negative at x = 2 , so
F (x) has maxima at x = 2 .
a 2 + b2
= 2x (a1 + 2a 2 x2 + K + na nx2n − 2)
+
1
a
By using symmetry, we get the required points
 ±a2
± b2 
,


 a 2 + b2
a 2 + b2 
∴
⇒
1
( p + q − p + q), if p ≥ q
1
( p + q −| p − q|) = 12
2
 ( p + q + p − q), if p < q
2
 q, if p ≥ q
=
 p, if p < q
1
{ p + q − | p − q|} = min ( p, q)
2
3
32. Given function f : R → R is
x5 + 5x4 + 10x3 + 10x2 + 3x + 1 ,
x<0

2
1
x
−
x
+
,
0
≤
x<1

2 3
8
f (x) = 
x − 4 x2 + 7 x −
, 1 ≤ x<3
3
3

 (x − 2 ) log (x − 2 ) − x + 10
,
x≥3
e

3
5x4 + 20x3 + 30x2 + 20x + 3 ,
x<0

, 0 < x<1
2
1
x
−
So, f ′ (x) = 
, 1 < x<3

2 x2 − 8 x + 7

x>3
,
log e (x − 2 )

−
+
At x = 1, f′′ (1 ) = 2 > 0 and f′′ (1 ) = 4 − 8 = −4 < 0
∴f ′ (x) is not differentiable at x = 1 and
f ′ (x) has a local maximum at x = 1.
For x ∈ (−∞ , 0)
f ′ (x) = 5x4 + 20x3 + 30x2 + 20x + 3
and since f′ (−1) = 5 − 20 + 30 − 20 + 3 = −2 < 0
So, f (x) is not increasing on x ∈ (−∞ , 0).
Now, as the range of function f (x) is R, so f is onto
function.
Hence, options (b), (c) and (d) are correct.
Application of Derivatives 265
Here, f (x) = 2|x|+ |x + 2| − ||x + 2| − 2|x||
− 2x − (x + 2) + (x − 2), if when x ≤ − 2
− 2 x + x + 2 + 3 x + 2 ,
if when − 2 < x ≤ −2 /3

2

=
if when − < x ≤ 0
− 4x,
3

if when 0 < x ≤ 2
4x,


if when x > 2
2x + 4,
x ≤ −2
− 2x − 4, if
 2x + 4, if −2 < x ≤ − 2 /3

2

=  − 4x,
if
− < x≤0
3

if
0 < x≤2
 4x,
 2x + 4, if
x>2
cos 2x sin 2x 
cos x − sin x

sin x
cos x 
 sin x
cos 2x(cos 2 x + sin 2 x) − cos 2x
(− cos 2 x + sin 2 x) + sin 2x(− sin 2x)
= cos 2x + cos 4x
f ′ (x) = − 2 sin 2x − 4 sin 4x = −2 sin 2x(1 + 4 cos 2x)
At
x = 0 ⇒ f ′ (x) = 0 and f (x) = 2
Also, f ′ (x) = 0 ⇒ sin 2x = 0
nπ
1
−1
or cos 2x = −
or
cos 2x =
⇒ x=
4
2
4
f (x) ⋅ g (x)
34. Here,
lim
=1
x → 2 f ′ (x) ⋅ g′ (x)
f (x) g′ (x) + f ′ (x) g (x)
⇒ lim
=1
x → 2 f ′ ′ (x) g′ (x) + f ′ (x) g′′ (x)
 cos 2x
33. f (x) = 
 − cos x
[using L’ Hospital’s rule]
f (2) g′ (2) + f ′ (2) g (2)
=1
f ′′ (2) g′ (2) + f ′ (2) g′′ (2)
f (2) g′ (2)
[Q f ′ (2) = g (2) = 0]
=1
f ′′ (2) g′ (2)
⇒
⇒
⇒
f (2) = f ′′ (2)
∴ f (x) − f ′′ (x) = 0, for atleast one x ∈ R.
⇒ Option (d) is correct.
Also, f : R → (0, ∞ ) ⇒ f (2) > 0
∴
f ′′ (2) = f (2) > 0
Since, f′ (2) = 0 and f ′′ (2) > 0
∴ f (x) attains local minimum at x = 2.
⇒ Option (a) is correct.
35.
Graph for y = f (x) is shown as
… (i)
36.
[from Eq. (i)]
PLAN
The problem is based on the concept to maximise volume of
cuboid, i.e. to form a function of volume, say f( x )find f ′( x )and f ′ ′( x ).
Put f ′ ( x ) = 0 and check f ′ ′( x ) to be + ve or − ve for minimum and
maximum, respectively.
Here,
a
PLAN
x , if x ≥
We know that,| x| = 
−
 x , if x <
 x − a,
⇒
| x − a| = 
 − ( x − a ),
l = 15x − 2a , b = 8x − 2a and h = a
a
0
0
8x–2a
if x ≥ a
if x < a
and for non-differentiable continuous function, the maximum or
minimum can be checked with graph as
Y
8x
15x – 2a
15x
∴ Volume = (8x − 2a ) (15x − 2a ) a
Y
V = 2a ⋅ (4x − a ) (15x − 2a )
…(i)
a
O
x=a
Minimum at x = a
X
x=a
Maximum at x = a
Y
X
8x – 2 a
15x – 2a
On differntiating Eq. (i) w.r.t a, we get
dv
= 6a 2 − 46ax + 60x2
da
Again, differentiating,
O
x=a
X
Neither maximum
nor minimum at x = a
Here,
d 2v
= 12a − 46x
da 2
 dv 
2
  = 0 ⇒ 6x − 23x + 15 = 0
 da 
266 Application of Derivatives
a = 5 ⇒ x = 3,
At
5
6
 d 2v 
 2 = 2 (30 − 23x)
 da 
⇒
 d 2v 
At x = 3,  2 = 2(30 − 69) < 0
 da 
∴ Maximum when x = 3, also at x =
 d 2v 
5
⇒  2 > 0
6
 da 
For maxima or minima, put f ′ (x) = 0 ⇒ x = 1, − 1
1
Now
f ′ ′ (x) = (114x)
4
At
x = 1 , f ′ ′ (x) > 0, minima
At
x = − 1, f ′ ′ (x) < 0 , maxima
∴ f (x) is increasing for [1, 2 5 ] .
∴ f (x) has local maxima at x = − 1 and f (x) has local
minima at x = 1.
Also,
f (0) = 34 /4
Hence, (b) and (c) are the correct answers.
∴ At x = 5 /6, volume is minimum.
39. f (x) = ∫
Thus, sides are 8x = 24 and 15x = 45
x
−1
37. Given,
 ex
,

f (x) = 2 − ex − 1 ,
 x−e ,

g (x) = ∫
and
⇒
x
0
–
–
–∞ 0
if 0 ≤ x ≤ 1
if 1 < x ≤ 2
if 2 < x ≤ 3
f ′ (x) =
Also,
 d
Q
 dx
g′ (x) = 0 ⇒ x = 1 + log e 2 and x = e.
 ex ,
if 0 ≤ x ≤ 1

g′′ (x) =  − ex − 1 , if 1 < x ≤ 2
 1
, if 2 < x ≤ 3
⇒
Let
g′′ (e) = 1 > 0, g (x) has a local minima.
Q f (x) is discontinuous at x = 1, then we get local
maxima at x = 1 and local minima at x = 2.
Hence, (a) and (b) are correct answers.
38. Since, f (x) has local maxima at x = − 1 and f ′ (x) has
local minima at x = 0.
∴
f ′ ′ (x) = λx
t (et − 1)(t − 1)(t − 2)3 (t − 3)5 dt

f (t ) dt = f { ψ (x)}ψ′ (x) − f { φ (x)} φ′ (x)

x = 0, 1, 2, 3
f ′ (x) = g (x) = x(ex − 1)(x − 1)(x − 2)3 (x − 3)5
+
−
1
+
−
2
3
This shows that f (x) has a local minimum at x = 1 and
x = 3 and maximum at x = 2 .
Therefore, (b) and (d) are the correct answers.
40. For −1 ≤ x ≤ 2 , we have
f (x) = 3x2 + 12x − 1
⇒
f ′ (x) = 6x + 12 > 0, ∀ − 1 ≤ x ≤ 2
2
x
+c
2
[Q f ′ (− 1) = 0]
λ
+ c = 0 ⇒ λ = − 2c
2
Again, function is an algebraic polynomial, therefore it
is continuous at x ∈ (−1, 2) and (2, 3).
For continuity at x = 2,
lim f (x) = lim (3x2 + 12x − 1)
x → 2−
…(i)
x3
+ cx + d
6
 8
f (2) = λ   + 2c + d = 18
 6
λ
f (1) = + c + d = − 1
6
= lim [3 (4 + h 2 − 4h ) + 24 − 12h − 1]
h→ 0
f (x) = λ
From Eqs. (i), (ii) and (iii),
1
f (x) = (19x3 − 57x + 34)
4
1
57
2
(x − 1) (x + 1)
∴ f ′ (x) = (57x − 57) =
4
4
x → 2−
= lim [3 (2 − h )2 + 12(2 − h ) − 1]
h→ 0
Again, integrating on both sides, we get
and
∞
Hence, f (x) is increasing in [–1, 2].
On integrating, we get
⇒
3
Using sign rule,
Also, at x = e,
⇒
ψ( x )
∫ φ ( x)
g′′ (1 + log e 2) = − elog e 2 < 0, g (x) has a local maximum.
f ′ (x) = λ
x
∫ −1
2
For local minimum, f ′ (x) = 0
x = 1 + log e 2,
At
d
dx
+
–
+
1
= x(ex − 1)(x − 1)(x − 2)3 (x − 3)5 × 1
f (t ) dt
g′ (x) = f (x)
Put
t (et − 1) (t − 1)(t − 2)3 (t − 3)5 dt
= lim (12 + 3h 2 − 12h + 24 − 12h − 1)
h→ 0
= lim (3h 2 − 24h + 35) = 35
…(ii)
h→ 0
and
…(iii)
lim f (x) = lim (37 − x)
x → 2+
x → 2+
= lim [37 − (2 + h )] = 35
h→ 0
and
f (2) = 3 ⋅ 22 + 12 ⋅ 2 − 1
= 12 + 24 − 1 = 35
Therefore, LHL = RHL = f (2) ⇒ function is continuous
at x = 2 ⇒ function is continuous in −1 ≤ x ≤ 3.
Application of Derivatives 267
Now, Rf ′ (2) = lim
x → 2+
f (x) − f (2)
x−2
⇒
(x0 , y0 ) = (4t , 8t ) = (1, 4)
f (2 + h ) − f (2)
= lim
h→ 0
h
y0 = 4
37 − (2 + h ) − (3 × 22 + 12 × 2 − 1)
= lim
h→ 0
h
−h
= lim
= −1
h→ 0 h
f (x) − f (2)
f (2 − h ) − f (2)
and Lf ′ (2) = lim
= lim
h→ 0
x−2
−h
x → 2−
f (x) =
42.
 (x2 − 1) 
 −2ax2 + 2a 
= 2a  2
= 2
2
2
 (x + ax + 1) 
 (x + ax + a ) 
3h 2 − 24h + 35 − 35
h→ 0
−h
3h − 24
= lim
= 24
h→ 0
−1
2x (x2 + ax + 1) − 2 (x2 − 1) (2x + a ) 
= 2a 

(x2 + ax + 1)3


4a (a + 2)
4a
Now, f ′′ (1) =
=
(a + 2)3
(a + 2)2
Since, Rf ′ (2) ≠ Lf ′ (2), f ′ (2) does not exist.
Again, f (x) is an increasing in [–1, 2] and is decreasing
in (2, 3), it shows that f (x) has a maximum value at x = 2.
Therefore, options (a), (b), (c), (d) are all correct.
and
f ′′ (−1) =
x2 < 1 ⇒ x2 − 1 < 0
(4t 2, 8t)
∴ f ′ (x) < 0, f (x) is decreasing.
4a
Also, at
>0
x = 1, f ′′ (1) =
(a + 2)2
E (0, 3)
X
O
fy
x = 0, y = at = 4 t
g′ (x) =
44.
Also, (4 t 2, 8 t ) satisfy y = mx + c .
4mt 2 − 8t + 3 = 0
0 3 1
1
1
Area of ∆ =
0 4t 1 = ⋅ 4t 2(3 − 4t )
2 2
2
4t 8t 1
A = 2[3t 2 − 4t3 ]
dA
= 2[6t − 12t 2] = − 12t (12t − 1)
dt
+
0
∴ Maximum at t =
–
1
2
1
and 4mt 2 − 8t + 3 = 0
2
⇒
m −4 + 3 =0
⇒
m =1
G (0, 4t ) ⇒ G (0,2)
f ′ (ex ) x
⋅e
1 + (ex )2
  ex 

e2x − 1
= 2a  2x


x
2
2x
 (e + ae + 1)   1 + e 
8 t = 4mt 2 + 3
–
[Q0 < a < 2]
So, f (x) has local minimum at x = 1.
Tangent at F , yt = x + at 2
∴
4a (a − 2)
− 4a
=
3
(2 − a )
(a − 2)2
43. When x ∈ (−1, 1),
Y
∴
…(ii)
∴ (2 + a )2 f ′ ′ (1) + (2 − a )2 f ′ ′ (−1) = 4a − 4a = 0
41. Here, y2 = 16x,0 ≤ y ≤ 6
⇒
...(i)
 (x2 + ax + 1)2 (2x) − 2(x2 − 1)


(x2 + ax + 1) (2x + a ) 
f ′ ′ (x) = 2a 

(x2 + ax + 1)4




= lim
⇒
(x2 + ax + 1) − 2ax
2ax
=1 − 2
x + ax + 1
x2 + ax + 1
 (x2 + ax + 1) ⋅ 2a − 2ax(2x + a ) 
f ′ (x) = − 

(x2 + ax + a )2


[12 + 3h 2 − 12h + 24 − 12h − 1] − 35
= lim
h→ 0
−h
At
 3 1 1
Area = 2 −  =
 4 2 2
∴

[3(2 − h )2 + 12(2 − h ) − 1]

− (3 × 22 + 12 × 2 − 1)
= lim
h→ 0
−h
G
y1 = 2
2
g′ (x) = 0, if e2x − 1 = 0, i.e. x = 0
x < 0, e2x < 1 ⇒ g′ (x) < 0
If
45. Let g (x) =
d
[ f (x) ⋅ f ′ (x)]
dx
To get the zero of g (x), we take function
h (x) = f (x) ⋅ f ′ (x)
between any two roots of h (x), there lies atleast one root
of h′ (x) = 0.
⇒ g (x) = 0 ⇒ h (x) = 0
⇒ f (x) = 0 or f ′ (x) = 0
If
f (x) = 0 has 4 minimum solutions.
f ′ (x) = 0 has 3 minimum solutions.
h (x) = 0 has 7 minimum solutions.
⇒ h′ (x) = g (x) = 0 has 6 minimum solutions.
268 Application of Derivatives
46. To maximise area of ∆ APB, we know that, OP = 10 and
sin θ = r /10 , where θ ∈ (0, π / 2)
… (i)
b
2(ax + b)
Y
P(6,8)
θ
A
r
θ
Q
X′
X
O
2ax − 1
b+1
2ax + b + 1
−1
2ax + 2b + 1
2ax + b
2ax
48. Given, f ′ (x) =
Applying R3 → R3 − R1 − 2R2, we get
2ax 2ax − 1 2ax + b + 1
f ′ (x) = b
b+1
−1
1
0
0
⇒
f ′ (x) = 2ax + b
On integrating both sides, we get
f (x) = ax2 + bx + c
B
Since, maximum at x = 5 /2 ⇒ f ′ (5 /2) = 0
Y′
⇒
1
∴ Area = (2 AQ ) (PQ )
2
= AQ . PQ = (r cos θ ) (10 − OQ )
= (r cos θ ) (10 − r sin θ )
= 10 sin θ cos θ (10 − 10 sin 2 θ )
[from Eq. (i)]
⇒
A = 100 cos3 θ sin θ
dA
⇒
= 100 cos 4 θ − 300 cos 2 θ ⋅ sin 2 θ
dθ
dA
Put
=0
dθ
⇒ cos 2 θ = 3 sin 2 θ
⇒
5a + b = 0
…(i)
Also,
f (0) = 2
⇒
c=2
and
f (1) = 1
⇒
a + b + c=1
On solving Eqs. (i), (ii) and (iii), we get
1
5
a = ,b = − ,c=2
4
4
1 2 5
Thus,
f (x) = x − x + 2
4
4
49. Let coordinates of P be (t , t 2 + 1)
Reflection of P in y = x is P1 (t 2 + 1, t )
which clearly lies on y2 = x − 1
Similarly, let coordinates of Q be (s2 + 1, s)
tan θ = 1 / 3
⇒ θ = π /6
dA
At which
< 0, thus when θ = π /6, area is maximum
dθ
From Eq. (i), r = 10 sin
π
= 5 units
6
x2 y2
47. Let us take a point P( 6 cos θ , 3 sin θ ) on +
=1.
6
3
Now, to minimise the distance from P to given straight
line x + y = 7, shortest distance exists along the
common normal.
Its reflection in y = x is
Q1 (s, s2 + 1), which lies on x2 = y − 1.
We have,
PQ12 = (t − s)2 + (t 2 − s2)2 = P1Q 2
⇒
PQ1 = P1Q
[Q both perpendicular to y = x]
Also PP1 || QQ1
Y
C1
x 2 = y −1
y=x
P1
Q1
(0, 1)
Q
O
X′
Y
(1, 0)
y
P
y2− = 1
X
N
P
X′
…(ii)
…(iii)
O
C2
x+y=7
Y′
X
Thus, PP1QQ1 is an isosceles trapezium.
Also, P lies on PQ1 and Q lies on P1Q , then
Y′
Slope of normal at P =
So,
and
6 sec θ
a 2 / x1
=
= 2 tan θ = 1
6 cosec θ
b2 / y1
2
3
1
sin θ =
3
cos θ =
Hence, required point is P(2, 1).
PQ ≥ min { PP1QQ1 }
Let us take min { PP1QQ1 } = PP1
∴
PQ 2 ≥ PP12 = (t 2 + 1 − t )2 + (t 2 + 1 − t 2)
= 2(t 2 + 1 − t 2) = f (t )
we have,
f ′ (t ) = 4(t 2 + 1 − t )(2t − 1)
= 4[(t − 1 / 2)2 + 3 / 4][2t − 1]
Now,
⇒
f ′ (t ) = 0
t = 1 /2
[say]
Application of Derivatives 269
Also, f ′ (t ) < 0 for t < 1 / 2
On adding Eqs. (i), (ii), (iii) and (iv), we get
and f ′ (t ) > 0 for t > 1 / 2
a 2 + b2 + c2 + d 2 = { x12 + (1 − x1 )2} + { y12 + (1 − y1 )2}
+ { x22 + (1 − x2)2} + { y22 + (1 − y2)2}
Thus, f (t ) is least when t = 1 / 2.
Corresponding to t = 1 / 2, point P0 on C1 is (1/2, 5/4) and
P1 (which we take as Q0) on C 2 are (5 / 4, 1 / 2). Note that
P0Q0 ≤ PQ for all pairs of (P , Q ) with P on C 2.
50. Let the square S is to be bounded by the lines x = ±1 / 2
and y = ±1 / 2.
2
1

1

a 2 =  x1 −  +  − y1

2

2
We have,
2
⇒ 4x = 2
X'
Again,
1/2
−1/2
X
B ( −1/2, y)
O
c
b
C(x2 −1/2)
and maximum value is 1 + 1 + 1 + 1 = 4
−1/2
Y′
51. f (x) is a differentiable function for x > 0.
1
=
− y1 +
2
1
2
2
2
b = x2 − y1 − x2 + y1 +
2
1
c2 = x22 − y22 + x2 + y2 +
2
1
d 2 = x12 − y22 + x1 − y2 +
2
x12 − y12 − x1
Similarly,
Therefore, for maxima or minima, f ′ (x) = 0 must satisfy.
1
Given,
f (x) = ln x − bx + x2, x > 0
8
1 1
⇒
f ′ (x) = . − b + 2x
8 x
⇒
a 2 + b2 + c2 + d 2 = 2(x12 + y12 + x22 + y22) + 2
1
Therefore, 0 ≤ x12, x22, y12, y22 ≤
4
⇒
0 ≤ x12 + x22 + y12 + y22 ≤ 1
Alternate Solution
c2 = x22 + y22
16x2 − 8bx + 1 = 0
⇒
(4x − b)2 = b2 − 1
⇒
(4x − b) = (b − 1) (b + 1)
... (i)
f ′ (x) =
Y
A(x , 1)
=
a
d
(0, y2)D
X
b
c
X′
O
X
C(x2 , 0)
Y′
b = (1 − x2) +
2
2
y12
[b ≥ 0, given]
1
− 1 + 2x
8x
2 2 1
1 2
=
x − x +
x
2
16 x
1

x − 

4
2
We have to check the sign of f ′ (x) at x = 1/4.
B (1, y1)
1/2
…(i)
2
Case I
0 ≤ b < 1 , has no solution. Since, RHS is
negative in this domain and LHS is positive.
1
Case II When b = 1, then x = is the only solution.
4
When b = 1,
0 ≤ 2(x12 + x22 + y12 + y22) ≤ 2
2 ≤ 2(x12 + x22 + y12 + y22) + 2 ≤ 4
But
f ′ (x) = 0
1
− b + 2x = 0
8x
For
∴
⇒
⇒ x = 1 /2
d 2y
=2+2 =4
dx2
1
Hence, y is minimum at x = and its minimum value is
2
1/4. Clearly, value is maximum when x = 1.
1 1 1 1
∴Minimum value of a 2 + b2 + c2 + d 2 = + + + = 2
2 2 2 2
1/2
d
D( −1/2, 1/2)
Now, consider the function y = x2 + (1 − x)2, 0 ≤ x ≤ 1
dy
differentiating ⇒
= 2x − 2(1 − x). For maximum or
dx
dy
minimum
= 0.
dx
⇒ 2x − 2(1 − x) = 0 ⇒ 2x − 2 + 2x = 0
Y
A(x , 1/2)
where x1 , y1 , x2, y2 all vary in the interval [0, 1].
...(ii)
a 2 = (1 − y1 )2 + (1 − x1 )2
...(iii)
d 2 = x12 + (1 − y2)2
...(iv)
Interval
Sign of f′(x)
Nature of f(x)
−∞, 0
−ve
↓
 0,


1

4
+ ve
↑
 1, ∞


4 
+ ve
↑
From sign chart, it is clear that f ′ (x) has no change of
sign in left and right of x = 1/4.
270 Application of Derivatives
Case III When b > 1, then
2
1
1
1
f ′ (x) =
− b + 2x =  x2 − bx +

x
2
16
8x
dA 1
= [− k2(− cosec2θ ) − h 2 sec2θ ]
dθ 2
1
= [k2 cosec2θ − h 2 sec2θ ]
2
dA
To obtain minimum value of A,
=0
dθ
⇒
=
2

2 
1 2
b
−
−
x
(b − 1)



4
16
x

=
b 1
b 1
2 
 

b 2 − 1  x − +
b 2 − 1 
x − −





x
4 4
4 4
⇒
2
(x − α ) (x − β )
x
1
where, α < β and α = (b − b2 − 1 ) and
4
1
β = (b + b2 − 1 ). From sign scheme, it is clear that
4
> 0, for 0 < x < α

f ′ (x) < 0, for α < x < β
> 0, for x > β

k2 cosec2 θ − h 2 sec2 θ = 0
⇒
k2
h2
=
sin 2 θ cos 2 θ
⇒
tan θ = ±
=
By the first derivative test, f (x) has a maxima at x = α
1
= (b − b2 − 1 )
4
1
and f (x) has a minima at x = β = (b + b2 − 1 )
4
= − [k2(1 + cot2 θ ) cot θ + h 2(1 + tan 2 θ ) tan θ ]
 
 d 2A 
h 2  − h 
= − k21 + 2   
 2
k   k
 dθ  tan θ = − k/ h
 

k 2   − k 
+ h 2 1 + 2   
h   h 

2
2
2
2
  k + h   h
 h + k   k 
= k2
  
   + h 2
2


 h2   h 
 k
  k
⇒
… (i)
 (k2 + h 2) h (h 2 + k2)(k) 
=
+

k
h


h
k


= (k2 + h 2)
+
>0
 k h 
Y
X′
O
(h, k)
P
1
 − k 
 −h 
2hk − k2  − h 2  
 h 
 k
2 
1
= [2hk + kh + kh ] = 2hk
2
53. Since x2 + y2 = 1 a circle S1 has centre (0, 0) and cuts
X-axis at P(–1, 0) and Q(1, 0) . Now, suppose the circle
S 2, with centre at Q(1, 0) has radius r. Since, the circle
has to meet the first circle, 0 < r < 2 .
Again, equation of the circle with centre at Q(1, 0) and
radius r is
A=
Y′
k 

The line (i) meets X-axis at P  h − , 0 and Y-axis at

m 
Q (0, k − mh ).
1
Let
A = area of ∆ OPQ = OP . OQ
2
1
k
=  h −  (k − mh )
2
m
Y
=
=−
=
R
S1
X′
[Qm = tan θ]
1
(k2 + h 2 tan 2 θ − 2hk tan θ )
2 tan θ
1
(2kh − k2 cot θ − h 2 tan θ )
2
[Q h, k > 0]
Therefore, A is least when tan θ = − k / h. Also, the least
value of A is
X
1  mh − k
 (k − mh )

2 m 
1
=−
(k − mh )2
2m
1
=−
(k − h tan θ )2
2 tan θ
k
h
[given]
tan θ < 0 , k > 0, h > 0
k
Therefore, tan θ = − (only possible value).
h
d 2A 1
2
Now,
= [−2k cosec2 θ cot θ − 2h 2 sec2 θ tan θ ]
dθ 2 2
For this line to intersect the positive direction of two
axes, m = tan θ < 0 , since the angle in anti-clockwise
direction from X-axis becomes obtuse.
Q
k2
= tan 2 θ
h2
Q
52. Let equation of any line through the point (h , k) is
y − k = m(x − h )
⇒
(–1,0)P
S2
r
O S
Q(1,0)
X
Y′
(x − 1)2 + y2 = r 2
…(i)
Application of Derivatives 271
To find the coordinates of point R, we have to solve it
with
… (ii)
x2 + y2 = 1
⇒
On subtracting Eq. (ii) from Eq. (i), we get
(x − 1)2 − x2 = r 2 − 1
⇒
x2 + 1 − 2 x − x 2 = r 2 − 1
⇒
1 − 2x = r 2 − 1
∴
2 − r2
x=
2
2
2 2
 4 × 2
 4 × 2
f ′ ′
 = 48 
 − 30 

 3 
 3 
 3
10 × 64
640
256
= 16 × 8 −
= 128 −
=−
<0
3
3
2
Therefore, f (r ) is maximum when, r =
Hence, maximum value of A
2
 2 − r 2
 + y2 = 1

 2 
2
 2 − r 2
(2 − r 2)2
y =1 − 
 =1 −
4
 2 
2
=1 −
⇒
r 4 − 4r 2 + 4
4
=
4 − r 4 + 4r 2 − 4
4
=
4r 2 − r 4
4
=
r 2(4 − r 2)
4
y=
=
2 . 12 − 8 2 . 2
4
4 3
=
=
=
3
9
3
3 3 3 3
∴
b ∈ (−2, − 1) ∪ [1, ∞ ]
SQ = 1 − (1 − r ) = r
Y
Let A denotes the area of ∆ QSR, then
2
1  4−r 
A = r r

2 
2 


1 2
2
= r 4−r
4
1 4
2
⇒
A =
r (4 − r 2)
16
⇒
55.
X′
f ′ (r ) = 16r − 6r = 2r (8 − 3r )
3
2
2r3 (8 − 3r 2) = 0
⇒
r = 0, 8 − 3r 2 = 0
⇒
r = 0, 3r 2 = 8
⇒
r = 0, r 2 = 8/3
⇒
r = 0, r =
2 2
3
[ Q0 < r < 2 , so r = 2 2 / 3]
Again,
f ′ ′ (r ) = 48r 2 − 30r 4
A
X
Y′
f (r ) = r 4 (4 − r 2) = 4r 4 − r 6
5
y = x2
B
For maxima and minima, put f ′ (r ) = 0
⇒
2 2
1  8
8
4−
 =   . 4−
4  3
3
 3
In order this value is not less than –1, we must have
b3 − b2 + b − 1
(b2 + 1) (b − 1)
≥
⇒
0
≥0
(b + 1) (b + 2)
b2 + 3b + 2
r 4 − r2
2
3
2
1 2 2


4 3
 3 (b3 − b2 + b − 1)
− x +
, if 0 ≤ x ≤ 1
54. Given, f (x) = 
(b2 + 3b + 2)

2x − 3
, if 1 ≤ x ≤ 3

is smallest at x = 1 .
So, f (x) is decreasing on [0, 1] and increasing on [1, 3].
Here, f (1) = − 1 is the smallest value at x = 1.
∴ Its smallest value occur as
(b3 − b2 + b − 1)
lim f (x) = lim (− x3 ) +
−
–
b2 + 3b + 2
x→1
x→1
Again, we know that, coordinates of S are (1 − r , 0),
therefore
Let
2
=
On putting the value of x in Eq. (i), we get
⇒
2 2
3
Any point on the parabola y = x2 is of the form (t , t 2).
dy
 dy 
Now,
= 2t
= 2x
⇒
dx
 dx 
x=t
Which is the slope of the tangent. So, the slope of the
normal to y = x2 at A (t , t 2) is –1/2t.
Therefore, the equation of the normal to
y = x2 at A (t , t 2) is
 1
y − t 2 =  −  (x − t )
 2t 
Suppose Eq. (i) meets the curve again at B(t1 , t12).
…(i)
272 Application of Derivatives
and when t = − 1 / 2 , the equation of AB is
y −2
x− 2
=
1
1

− 2 −   − 2
 2
2
1
(t1 − t )
2t
1
(t1 − t )
⇒ (t1 − t ) (t1 + t ) = −
2t
1
⇒
(t1 + t ) = −
2t
1
t1 = − t −
⇒
2t
Then,
t12 − t 2 = −

 1
1 
⇒ ( y − 2)  −
− 2 = (x − 2 )  − 2
2 


2
Therefore, length of chord,
L = AB2 = (t − t1 )2 + (t 2 − t12)2
= (t − t1 ) + (t − t1 ) (t + t1 )
2
2
the lengths of the other sides of the rectangle.
Total perimeter = 2a + 4b + πb = K
2
1 
1 
1 


L =  2t +   1 + 2 = 4t 2 1 + 2





2t
4t
4t 
3
Coloured
glass
2b
On differentiating w.r.t. t, we get
3
[say] …(i)
Now, let the light transmission rate (per square metre)
of the coloured glass be L and Q be the total amount of
transmitted light.
2
⇒
2x − 2 y + 2 = 0
56. Let 2b be the diameter of the circular portion and a be
2
= (t − t1 ) [1 + (t + t1 ) ]
2
2
1 
1 


=  t + t +  1 +  t − t −  


2t  
2t  
2
2 y − 4 = 2 (x − 2 )
⇒
⇒
2
1   2 
1 
dL


= 8t 1 + 2 + 12t 2 1 + 2  − 3 



dt
4t   4t 
4t
a
Clear glass
a
2
1  3
1   

= 2 1 + 2 4t 1 + 2  − 

4t  t 
4t   
2
2
1
1  

+ 4  1 + 2  2 + 2



4t
t 
⇒
 d 2L 
 2
 dt 
t = ± 1/ 2

= 0 + 4 1 +

2
1
 (2 + 2) > 0
2
Therefore, L is minimum, when t = ± 1 / 2.For t = 1 / 2 ,
point A is (1 / 2 , 1 / 2) and point B is (− 2 , 2). When
t = − 1 / 2 , A is (−1 / 2 , 1 / 2), B is ( 2 , 2).
Again, when t = 1 / 2 , the equation of AB is
y−2
x+ 2
=
1
1
+ 2
−2
2
2
 1

1

+ 2  = (x + 2)  − 2
⇒ ( y − 2 ) 




2
2


⇒
⇒
− 2 y + 4 = 2x + 2
2x + 2 y − 2 = 0
2b
2
1
2
1  
1  


= 2  1 + 2  4t −  = 4  1 + 2  2t − 


t
t
4t  
4t  
dL
For maxima or minima, we must have
=0
dt
1
1
⇒
2t − = 0 ⇒ t 2 =
t
2
1
⇒
t=±
2
1
1  1 
d 2L

Now,
= 8  1 + 2  − 3   2t − 

t
4t   2t  
dt 2
Then,
Q = 2ab (3L ) +
⇒
Q=
⇒
⇒
1
πb2(L )
2
L
(πb2 + 12ab)
2
L
Q = [πb2 + 6b (K − 4b − πb)]
2
L
Q = (6Kb − 24b2 − 5πb2)
2
[from Eq. (i)]
On differentiating w.r.t. b, we get
dQ L
= (6K − 48b − 10πb)
db 2
dQ
For maximum, put
=0
db
6K
b=
⇒
48 + 10 π
Now,
…(ii)
d 2Q L
= (− 48 − 10π ) < 0
db2 2
Thus, Q is maximum and from Eqs. (i) and (ii), we get
(48 + 10π ) b = 6 {2a + 4b + πb}
2b
6
Ratio =
∴
=
= 6 :6 + π
a 6+ π
57. Since, the chord QR is parallel to the tangent at P.
∴
ON ⊥ QR
Consequently, N is the mid-point of chord QR.
∴
QR = 2QN = 2r sin θ
Also,
ON = r cos θ
Application of Derivatives 273
∴
PN = r + r cos θ
Now,
P
d2
π
, we have
( AP )2 = − (16 − 2a 2) < 0
2
dθ 2
π
Thus, AP 2 i.e. AP is maximum when θ = .The point on
2
the curve 4x2 + a 2y2 = 4a 2 that is farthest from the point
π
π

A(0, − 2) is  a cos , 2 sin  = (0, 2)

2
2
r
For θ =
O
r
r
Q
N
R
M
59. Let AF = x and AE = y, ∆ABC and ∆EDC are similar.
Let A denotes the area of ∆PQR.
1
Then,
A = ⋅ 2r sin θ (r + r cos θ )
2
⇒
⇒
⇒
and
d2
( AP )2 = − {(8 − 2a 2) sin θ + 8} sin θ
dθ 2
+ (8 − 2a 2) ⋅ cos 2 θ
A = r 2(sin θ + sin θ cos θ )
1
A = r 2(sin θ + sin 2θ )
2
dA
2
= r (cos θ + cos 2θ )
dθ
C
dA
=0
dθ
Hence, the area of ∆PQR is maximum when θ =
A 2
x
(p , – p )
3 3 2
r sq units
4
58. Let P (a cos θ , 2 sin θ ) be a point on the ellipse
x2
y2
+
=1
4
a2
Let A(0, − 2) be the given point.
Then,
( AP )2 = a 2 cos 2 θ + 4 (1 + sin θ )2
d
⇒
( AP )2 = − a 2 sin 2θ + 8 (1 + sin θ ) ⋅ cos θ
dθ
d
⇒
( AP )2 = [(8 − 2a 2) sin θ + 8 ] cos θ
dθ
d
For maximum or minimum, we put
( AP )2 = 0
dθ
⇒ [(8 − 2a 2) sin θ + 8] cos θ = 0
4
⇒
cos θ = 0 or sin θ = 2
a −4
4x2 + a 2y2 = 4a 2, i.e.
4
> 1 ⇒ sin θ > 1, which is
a2 − 4
impossible]
B
(q 2, –q)
c
b
=
x b− y
⇒
π
.
3
F
c
⇒
The maximum area of ∆ PQR is given by
 3
2π 
3
π 1

+
A = r 2 sin + sin  = r 2 


3 2
3
4
 2
[Q 4 < a 2 < 8 ⇒
a
D
y
⇒ cos θ + cos 2θ = 0 ⇒ cos 2θ = − cos θ
π
⇒
cos θ = cos (π − 2θ ) ⇒ θ =
3
d 2A
π
Clearly,
< 0 for θ =
3
dθ 2
=
2
( r , – r)
b E
d 2A
= r 2(− sin θ − 2 sin 2θ )
dθ 2
For maximum and minimum values of θ,we put
AB AC
=
ED CE
∴
bx = c (b − y)
c
x = (b − y)
b
⇒
Let z denotes the area of par
allelogram AFDE.
z = xy sin A
c
z = (b − y) y ⋅ sin A
b
Then,
⇒
…(i)
On differentiating w.r.t. y we get
dz c
= (b − 2 y) sin A
dy b
d 2z −2c
=
sin A
b
dy2
and
For maximum or minimum values of z, we must have
dz
=0
dy
c
⇒
(b − 2 y) = 0
b
b
y=
⇒
2
Clearly,
d 2z
2c
=−
< 0, ∀ y
2
b
dy
Hence, z is maximum, when y =
On putting y =
b
.
2
b
in Eq. (i), we get
2
274 Application of Derivatives
the maximum value of z is
1
c
b b
z =  b −  ⋅ ⋅ sin A = bc sin A
2 2
4
b
=
g′ (x) = 0 ⇒ x = 0, x = ± 3
Now,
1
area of ∆ABC
2
g′ ′ (x) =
p2 − p 1
1 1 2
= ×
q
q 1
2 2 2
−r 1
r
At
∴
AC = x km
−p 1
1
( p + q) (r − p) q − p 1 0
4
r + p −1 0
1
( p + q) (r − p) (− q − r )
4
1
= ( p + q) (q + r ) ( p − r )
4
π
π
3
60. Let y = f (x) = sin x + λ sin 2 x, − < x <
2
2
Let
sin x = t
S
∴
∴
⇒
Let
∴
1−x
dy (1 + x ) ⋅ 1 − x (2x)
=
=
dx
(1 + x2)2
(1 + x2)2
dy
= g (x)
dx
B
SC = x2 + d 2
CB = (L − x)
and
Distance
Time =
Speed
Time from S to B = Time from S to C + Time from C to B
x
1 + x2
2
(L – x)
C
L
For exactly one minima and exactly one maxima dy/dt
must have two distinct roots ∈ (−1, 1).
2λ
⇒
t = 0 and t = −
∈ (−1, 1)
3
2λ
−1 < −
<1
⇒
3
3
3
⇒
− <λ<
2
2
 3 3
λ ∈− , 
⇒
 2 2
y=
x
A
∴
61. Given,
x 2 + d2
d
y = t3 + λt 2, − 1 < t < 1
dy
= 3t 2 + 2tλ = t (3 t + 2λ )
dt
⇒
AB = L km
Let the swimmer land at C on the shore and let
=
∴
x = 0, g′ ′ (x) = − 6 < 0
62. Let the house of the swimmer be at B.
−p
1
p2
1
2
=
q − p2 q + p 0
4
r 2 − p2 − r + p 0
p2
(1 + x2)3 (6x2 − 6) − (2x3 − 6x) ⋅ 3 (1 + x2)2 ⋅ 2x
(1 + x2)6
∴ g′ (x) has a maximum value at x = 0.
⇒ (x = 0, y = 0) is the required point at which tangent to
the curve has the greatest slope.
Applying R3 → R3 − R1 and R2 → R2 − R1
=
For greatest or least values of m, we should have
Let
⇒
x2 + d 2 L − x
+
u
v
L x
1
2
f (x) = T =
x + d2 +
−
u
v v
1
1 ⋅ 2x
1
+0−
f ′ (x) = ⋅
v
u 2 x2 + d 2
T=
For maximum or minimum, put f ′ (x) = 0
⇒
⇒
∴
v2x2 = u 2 (x2 + d 2)
x2 =
u 2d 2
v2 − u 2
ud
f ′ (x) = 0 at x = ±
v − u2
2
2
− ud
x≠
But
v2 − u 2
[i.e. slope of tangent]
1 − x2
g (x) =
(1 + x2)2
(1 + x2)2 ⋅ (−2x) − (1 − x2) ⋅ 2 (1 + x2) ⋅ 2x
⇒ g ′ (x) =
(1 + x2)4
−2x (1 + x2) [(1 + x2) + 2 (1 − x2)] −2x (3 − x2)
=
=
(1 + x2)4
(1 + x2)3
ud
∴ We consider, x =
Now,
f ′ ′ (x) =
, (v > u )
v − u2
2
1
u
d2
x2 + d 2 (x2 + d 2)
Hence, f has minimum at x =
ud
v − u2
2
> 0, ∀ x
.
Application of Derivatives 275
b
≥ c, ∀ x > 0 ; a , b > 0
x
b
f (x) = ax2 + − c
x
63. Given, ax2 +
Let
b 2ax3 − b
=
x2
x2
2b
⇒ f ′ ′ (x) = 2a + 3 > 0 [since, a , b are all positive]
x
∴
f ′ (x) = 2ax −
Now, put
At
 b
f ′ (x) = 0 ⇒ x =  
 2a 
 b
x= 
 2a 
and
⇒
1/3
>0
[Q a , b > 0]
1/3
, f ′ ′ (x) = + ve
 b
⇒ f (x) has minimum at x =  
 2a 
1/3
.
2 /3
  b  1/3 
b
 b
+
− c≥ 0
f     = a  




2
2
a
a
(
/
2
b
a )1/3


 2a 
= 
 b
1/3
 2a 
 
 b
1/3
π 5π 13π
,
,
,…
12 12 12
but θ ∈ { λ 1π , λ 2π , … λ rπ }, where 0 < λ 1 < … < λ r < 1.
π 5π
θ=
,
∴
12 12
1
5
So, λ 1 + … + λ r =
+
= 0.50
12 12
⇒
⋅
3b
− c≥0
2
⋅
3b
≥c
2
66. Given set S of polynomials with real coefficients
S = {(x2 − 1)2(a 0 + a1x + a 2x2 + a3 x3 ) :
a 0 , a1 , a 2, a3 ∈ R}
and for a polynomial f ∈ S, Let
f (x) = (x2 − 1)2 (a 0 + a1x + a 2x2 + a3 x3 )
it have − 1 and 1 as repeated roots twice, so graph of f (x)
touches the X-axis at x = − 1 and x = 1, so f ′ (x) having at
least three roots x = − 1, 1 and α . Where α ∈ (− 1, 1) and
f ′ ′ (x) having at least two roots in interval (− 1, 1)
So,
mf ′ = 3 and mf ′ ′ = 2
∴Minimum possible value of (mf ′ + mf ′ ′ ) = 5
67. On drawing the diagram of given situation
MD 2 + MC 2 = 64 + x2 + 121 + (10 − x)2
= f (x)
⇒
f (x) = 2x − 20x + 285
D
27ab ≥ 4c
2
3
11
64. Let f (x) = x + y, where xy = 1
⇒
[say]
2
C
On cubing both sides, we get
2a 27b3
⋅
≥ c3
8
b
⇒
θ=
f (x) = x +
8
1
x
⇒
1 x2 − 1
f ′ (x) = 1 − 2 =
x
x2
Also,
f ′ ′ (x) = 2 / x3
On putting f ′ (x) = 0, we get
x = ± 1, but x > 0 [neglecting x = − 1]
f ′ ′ (x) > 0, for x = 1
Hence, f (x) attains minimum at x = 1, y = 1
A
M
x
B
10 – x
For minima, f ′ (x) = 0 ⇒ 4x − 20 = 0 ⇒ x = 5
As
f ′ ′ (x) = 4 > 0
∴
AM = 5 m
68. Here, volume of cylindrical container, V = πr 2h …(i)
and let volume of the material used be T.
r+2
⇒ (x + y) has minimum value 2.
O r
65. The given function f : R → R be defined by
f ( θ ) = (sin θ + cos θ )2 + (sin θ − cos θ )4
= 1 + sin 2 θ + (1 − sin 2 θ )2
= 1 + sin 2 θ + 1 + sin 2 2 θ − 2 sin 2 θ
= sin 2 2 θ − sin 2 θ + 2
2
h
2
1
7

= sin 2 θ −  +

2
4
The local minimum of function ‘ f ’ occurs when
1
sin 2 θ =
2
π 5π 13π
2θ = ,
,
…
⇒
6 6
6
∴
⇒
T = π [(r + 2)2 − r 2] h + π (r + 2)2 × 2
V
T = π [(r + 2)2 − r 2] ⋅ 2 + 2π (r + 2)2
πr
[Q V = πr 2h ⇒ h =
V
]
πr 2
276 Application of Derivatives
2
 r + 2
2
T =V 
 + 2π (r + 2) − V
 r 
⇒
∴
On differentiating w.r.t. r, we get
or
69.
70.
1 9
=
2 2
PLAN
(i) Local maximum and local minimum are those points at which
f ′ ( x ) = 0, when defined for all real numbers.
(ii) Local maximum and local minimum for piecewise functions
are also been checked at sharp edges.
(h,0)
Also,
(x2 − 1),
y =|x2 − 1|= 
2
 (1 − x ),
R
(2 sec θ, 0)
∆ = Area of ∆PQR
1
= (2 3 sin θ ) (2 sec θ − 2 cos θ )
2
= 2 3 ⋅ sin3 θ/cos θ
1
≤ h ≤1
2
1
≤ 2 cos θ ≤ 1
2
1
1
≤ cos θ ≤
4
2
∴
⇒
∴
d∆ 2 3 {cos θ ⋅ 3 sin θ cos θ − sin θ (− sin θ )}
=
dθ
cos 2 θ
2
=
=
When
2 3 ⋅ sin θ
[3 cos 2 θ + sin 2 θ ]
cos 2 θ
2
2 3 sin 2 θ
⋅ [2 cos 2 θ + 1]
cos 2 θ
= 2 3 tan 2 θ (2 cos 2 θ + 1) > 0
1
1
≤ cos θ ≤ ,
4
2
if x ≤ − 1
if − 1 ≤ x ≤ 0
if 0 ≤ x ≤ 1
if x ≥ 1
 − x2 − x + 1 ,
if x ≤ − 1
 2
 − x − x + 1, if − 1 ≤ x ≤ 0
= 2
− x + x + 1, if 0 ≤ x ≤ 1
 x2 + x − 1, if
x≥1
x
y
cos θ +
sin θ = 1
2
3
R(2 sec θ , 0)
Since,
if x ≤ − 1 or x ≥ 1
if − 1 ≤ x ≤ 1
 − x + 1 − x2 ,

2
− x + 1 − x ,
y =|x|+ |x2 − 1|= 
2
x + 1 − x ,
 x + x2 − 1 ,
(2 sec θ, –√3 sin θ)
⇒
x, if x ≥ 0
y =|x|= 
 − x, if x < 0
Description of Situation
(2 cos θ, √3 sin θ)
Q
∴
1
2
8
∆1 − 8∆ 2 = 45 − 36 = 9
5
∴
Here, tangent at P(2 cos θ , 3 sin θ ) is
O
cos θ =
When
PLAN As to maximise or minimise area of triangle, we should find
area in terms of parametric coordinates and use second
derivative test.
P
1 45 5
=
8
4
 2 3 sin3 θ 
=

 cos θ 
r = 10
V
=π
1000
V
=4
250π
⇒
1  2 3 ⋅ sin3 θ 
=

4 
cos θ 
∆ 2 = ∆ min occurs at cos θ =
V
=π
r3
where
cos θ =
When
dT
 r + 2   −2 
= 2V ⋅ 
 ⋅   + 4π (r + 2)
 r   r2 
dr
dT
At r = 10,
=0
dr
V

Now,
0 = (r + 2) ⋅ 4  π − 3 

r 
⇒
∆1 = ∆ max occurs at cos θ =
which could be graphically shown as
Y
…(i)
– x 2– x + 1
– x 2– x + 1 – x 2+ x +1
x 2+x –1
1
–1 –1/2 O
X
1/2 1
Thus, f (x) attains maximum at x =
…(ii)
attains minimum at x = − 1, 0, 1.
⇒ Total number of points = 5
3
71.
1 −1
and f (x)
,
2 2
PLAN If f( x ) is least degree polynomial having local maximum and
local minimum at α and β.
Then,
f ′ (x) = λ (x − α ) (x − β )
Here,
p′ (x) = λ (x − 1) (x − 3) = λ (x2 − 4x + 3)
On integrating both sides between 1 to 3, we get
3
3
∫1 p′ (x) dx = ∫1 λ (x
2
− 4x + 3) dx
3
⇒
 x3

( p(x))31 = λ  − 2x2 + 3x
3
1
Application of Derivatives 277
⇒


1
p(3) − p(1) = λ  (9 − 18 + 9) −  − 2 + 3 

3

− 4
2 −6 = λ  
3 
⇒
⇒
λ =3
⇒
p′ (x) = 3 (x − 1) (x − 3)
∴
72.
p′ (0) = 9
f (x) = x − 4x3 + 12x2 + x − 1
4
f ′ (x) = 4x3 − 12x2 + 24x + 1
f ′′ (x) = 12x2 − 24x + 24 = 12 (x2 − 2x + 2)
= 12 {(x − 1)2 + 1} > 0 ∀x
⇒
f ′ (x) is increasing.
Since, f ′ (x) is cubic and increasing.
74. Let
f (θ ) =
1
sin 2 θ + 3 sin θ cos θ + 5 cos 2 θ
Again let, g (θ ) = sin 2 θ + 3 sin θ cos θ + 5 cos 2 θ
1 − cos 2 θ
 1 + cos 2 θ  3
=
+5
 + sin 2 θ
 2

2
2
3
= 3 + 2 cos 2 θ + sin 2 θ
2
9
g (θ )min = 3 − 4 +
∴
4
5 1
=3− =
2 2
1
∴ Maximum value of f (θ ) =
=2
12
75. Given, A = { x|x2 + 20 ≤ 9x} = { x|x ∈ [4, 5]}
⇒ f ′ (x) has only one real root and two imaginary roots.
Y
∴ f (x) cannot have all distinct roots.
⇒ Atmost 2 real roots.
O
Now, f (− 1) = 15, f (0) = − 1, f (1) = 9
2
3
4
5
∴ f (x) must have one root in (− 1, 0) and other in (0, 1).
⇒ 2 real roots.
X
–16
–20
–21
73. Let g (x) = e f ( x ), ∀ x ∈ R
⇒ g′ (x) = ef ( x ) ⋅ f ′ (x)
⇒ f ′ (x) changes its sign from positive to negative in the
neighbourhood of x = 2009
⇒ f (x) has local maxima at x = 2009.
So, the number of local maximum is one.
Now,
f ′ (x) = 6(x2 − 5x + 6)
Put
f ′ (x) = 0 ⇒ x = 2, 3
f (2) = − 20, f (3) = − 21, f (4) = − 16, f (5) = 7
From graph, maximum value of f (x) on set A is f (5) = 7.
11
Indefinite Integration
Topic 1 Some Standard Results
Objective Questions I (Only one correct option)
cos θ
dθ = A log e|B(θ )| + C , where C is a
5 + 7 sin θ − 2 cos 2 θ
B(θ )
can be
constant of integration, then
A
1. If ∫
(2020 Main, 5 Sep II)
2 sin θ + 1
sin θ + 3
5(sin θ + 3)
(c)
2 sin θ + 1
(a) −
(c)
1
and f (x) = 9 (x − 1)
27
1
(b) A =
and f (x) = 3 (x − 1)
81
1
(c) A =
and f (x) = 3 (x − 1)
54
1
(d) A =
and f (x) = 9 (x − 1)2
54
x
dx (x ≥ 0). Then f (3) − f (1) is equal to
(1 + x)2
(2020 Main, 4 Sep I)
π 1
3
+ +
6 2
4
π 1
3
+ −
6 2 4
(b) −
(d)
π 1
3
+ +
12 2
4
π 1
3
+ −
12 2 4
1
7. If ∫
∫
sin 2 α + C, where C is a constant of integration, then the
functions A (x) and B (x) are respectively
(2019 Main, 12 April II)
(a) x + α and log e|sin(x + α )| (b) x − α and log e|sin(x − α )|
(c) x − α and log e|cos(x − α )| (d) x + α and log e|sin(x − α )|
2x3 − 1
dx is equal to
x4 + x
|x + 1|
1
log e
+C
2
x2
x3 + 1
(c) log e
+C
x
(a)
(2019 Main, 12 April I)
(x + 1)2
1
+C
log e
2
|x3|
|x3 + 1|
(d) log e
+C
x2
3
(b)
∫
1
(b) −
6x3
1
(d)
2x2
1
2x3
3
x2
5x
2 dx is equal to
x
sin
2
sin
(where, C is a constant of integration)
(2019 Main, 8 April I)
(a)
(b)
(c)
(d)
2x + sin x + 2 sin 2x + C
x + 2 sin x + 2 sin 2x + C
x + 2 sin x + sin 2x + C
2x + sin x + sin 2x + C
3x13 + 2x11
dx is equal to (where C
(2x4 + 3x2 + 1)4
(2019 Main, 12 Jan II)
is a constant of integration)
9. The integral ∫
(here C is a constant of integration)
3
(c) −
8.
tan x + tan α
dx = A (x) cos 2α + B (x)
tan x − tan α
5. The integral ∫
(a) −
(b) (1, 1 − tan θ)
(d) (−1, 1 − tan θ)
4. Let α ∈ (0, π / 2) be fixed. If the integral
dx
= xf (x)(1 + x6 )3 + C
/
x (1 + x6 )23
3
where, C is a constant of integration, then the
function f (x) is equal to
(2019 Main, 8 April II)
dθ
3. If ∫
=
λ tan θ + 2 log e| f (θ )| + C
2
cos θ (tan 2 θ + sec 2 θ )
where C is a constant of integration, then the ordered pair
(2020 Main, 9 Jan II)
(λ , f (θ )) is equal to
(a) (1, 1 + tan θ)
(c) (−1, 1 + tan θ)
2
(a) A =
2 sin θ + 1
5(sin θ + 3)
5(2 sin θ + 1)
(d)
sin θ + 3
(b)
(a)
2. Let f (x) = ∫
dx
(x − 2x + 10)2


f (x)
 x − 1
= A  tan −1 
 + C , where, C is a
 +
 3  x2 − 2x + 10

(2019 Main, 10 April I)
constant of integration, then
6. If ∫
(a)
(c)
x4
6(2x4 + 3x2 + 1)3
x4
(2x + 3x + 1)
4
2
3
+C
+C
(b)
(d)
x12
6(2x4 + 3x2 + 1)3
x12
(2x + 3x2 + 1)3
4
+C
+C
Indefinite Integration 279
x+1
dx = f (x) 2x − 1 + C, where C is a constant
2x − 1
of integration, then f (x) is equal to (2019 Main, 11 Jan II)
10. If ∫
2
(x + 2)
3
2
(c) (x − 4)
3
is equal to
(c)
1
−1
(b)
9x4
−1
3 (1 + tan3 x)
1
(c)
+C
1 + cot3 x
(2019 Main, 11 Jan I)
27x6
17.
(sin θ
cos θ
dθ is equal to
sin n + 1 θ
(where C is a constant of integration)
∫
(a)
1
n 

1 −

sin n + 1 θ 
n 2 − 1
(b)
1
1 +



n 2 − 1
sin n − 1 θ 
(c)
1
1 −



sin n − 1 θ 
n 2 − 1
(d)
n
n
1
n 

1 −

sin n − 1 θ 
n 2 + 1
(2019 Main, 10 Jan I)
n+1
n
+C
n+1
n
+C
n+1
n
+C
n
+C
5x8 + 7x6
13. If f (x) = ∫ 2
dx, (x ≥ 0), and f (0) = 0, then
(x + 1 + 2x7 )2
the value of f (1) is
(2019 Main, 9 Jan I)
(a) −
(c)
1
2
1
4
(b) −
(d)
1
4
1
2
14. For x ≠ nπ + 1, n ∈ N (the set of natural numbers), the
2
integral
2 sin(x2 − 1) − sin 2(x2 − 1)
∫ x 2 sin(x2 − 1) + sin 2(x2 − 1) dx is equal to
(where C is a constant of integration ) (2019 Main, 9Jan I)
1
log e|sec(x2 − 1)| + C
2
 x2 − 1
 + C
(b) log e sec 
 2 
1
(c) log e sec2 (x2 − 1) + C
2
 x2 − 1
1
 + C
(d) log e sec2 
2
 2 
(a)
3 (1 + tan3 x)
−1
(d)
+C
1 + cot3 x
(2015 Main)
1
(b) (x4 + 1) 4 + c
(c) − (x +
 x4 + 1 4
(d) −  4  + c
 x 
1
1) 4
1
+c
sec2 x
∫ (sec x + tan x)9/ 2 dx equals to
(for some arbitrary constant K)
−1
1 −

(sec x + tan x)11/ 2 11
1
1 −
(b)

(sec x + tan x)11/ 2 11
−1
1 +
(c)

(sec x + tan x)11/ 2 11
1
1 +
(d)

(sec x + tan x)11/ 2 11
(a)
18. If I = ∫
n+1
+C
 x4 + 1 4
(a)  4  + c
 x 
4
π
2
1
− sin θ ) n
(2018 Main)
−1
(b)
1
12. Let n ≥ 2 be a natural number and 0 < θ < . Then,
n
+C
(where C is a constant of integration)
dx
16. The value of ∫ 2 4
is
x (x + 1)3/ 4
3x3
1
(d)
27x9
1
(a)
dx = A (x)( 1 − x2 )m + C,
x4
for a suitable chosen integer m and a function A (x),
where C is a constant of integration, then ( A (x))m equals
dx
+ cos5 x)2
(b)
1 − x2
11. If ∫
sin 2 x cos 2 x
∫ (sin5 x + cos3 x sin 2 x + sin3 x cos2 x
1
(x + 4)
3
1
(d) (x + 1)
3
(a)
(a)
15. The integral
4x
e
(2012)
1
(sec x + tan x)2  + K
7

1
2
(sec x + tan x)  + K
7

1
2
(sec x + tan x)  + K
7

1
2
(sec x + tan x)  + K
7

ex
e− x
dx, J = ∫ − 4x
dx.
2x
+e +1
+ e−2x + 1
e
Then, for an arbitrary constant c, the value of J − I
(2008, 3M)
equals
(a)
(c)
e4 x − e2x + 1
1
+c
log 4 x
2
e + e2x + 1
e2x − ex + 1
1
+c
log 2x
2
e + ex + 1
19. If f (x) =
(b)
e2x + ex + 1
1
+c
log 2x
2
e − ex + 1
(d)
e4 x + e2x + 1
1
log 4 x
+c
2
e − e2x + 1
x
for n ≥ 2 and g (x) = ( fofo ... of ) (x).
14243
(1 + xn )1/ n
f occurs n times
Then, ∫ xn − 2g (x) dx equals
1
1−
nxn ) n
(2007, 3M)
(a)
1
(1 +
n (n − 1)
(c)
1+
1+
1
1
(1 + nxn ) n + c (d)
(1 + nxn ) n + c
n (n + 1)
n+1
+ c (b)
1
(1 +
n −1
1
1−
nxn ) n
1
20. The value of ∫
(a) 2 2 −
(c)
2
x2
+
1
(x2 − 1) dx
3
x
1
x4
2 x4 − 2 x2 + 1
+c
2
1
1
2− 2 + 4 + c
2
x
x
+c
is
(b) 2 2 +
(2006, 3M)
2
x2
+
1
x4
(d) None of these
+ c
280 Indefinite Integration
Analytical & Descriptive Questions
Objective Questions II
(One or more than one correct option)
24. For any natural number m, evaluate
21. Let f : R → R and g : R → R be two non-constant
differentiable functions. If f ′ (x) = (e( f ( x ) − g( x )) ) g′ (x) for
all x ∈ R and f (1) = g (2) = 1, then which of the following
statement(s) is (are) TRUE?
(2018 Adv.)
(a) f (2) < 1 − log e 2
(b) f (2) > 1 − log e 2
(c) g (1) > 1 − log e 2
(d) g (1) < 1 − log e 2
22. Let f : R → R be a differentiable function with f (0) = 1
and satisfying the equation
+ f ′ (x) f ( y) for all x, y ∈ R.
f (x + y) = f (x) f ′ ( y)
Then, the value of log e ( f (4)) is ....... .
(2018 Adv.)
+ x2m + xm ) (2 x2m + 3 xm + 6)1/m dx, x > 0. (2002, 5M)
25. Evaluate
 1 − x


 1 + x
∫
1/ 2
⋅
1− x
dx.
1+ x
dx
27. Evaluate ∫ 2 4
.
x (x + 1)3/ 4
28. Evaluate the following:
1 
(i) ∫ 1 + sin  x dx
2 
dx
x
(1997C, 3M)
29. Integrate
(1985, 2 1 M)
2
(1984, 2M)
(1980, 4M)
(ii)
∫
x2
dx
1−x
x2
.
(a + bx)2
(1979, 2M)
30. Integrate
sin x ⋅ sin 2 x ⋅ sin 3x + sec2 x ⋅ cos 2 2x + sin 4 x ⋅ cos 4 x.
Fill in the Blank
(1979, 1M)
−x
4e + 6e
dx = Ax + B log (9e2x − 4) + C, then A = K,
9ex − 4e− x
(1989, 2M)
B = ... and C = K .
23. If ∫
3m
26. Evaluate ∫
Integer & Numerical Answer Type Question
x
∫ (x
x
.
31. Integrate the curve
1 + x4
1
sin x
32. Integrate
or
.
1 − cot x sin x − cos x
(1978, 1M)
(1978, 2M)
Topic 2 Some Special Integrals
Objective Question I (Only one correct option)
1. The integral ∫ sec
23
/
4/3
x cosec x dx is equal to (here C is a
constant of integration)
(a) 3 tan −1/3 x + C
(2017 Main)
1
1
(c)  , − 1 (d)  − , 0
5

 5 

1
3
 x+
∫(
4
x
tan x +
+
ln (1 + 6 x )
 dx.
3
x+ x 
(1992, 4M)
cot x ) dx.
(1988, 3M)
(cos 2x)1/ 2
(1987, 6M)
dx.
sin x
2 sin x − sin 2 x
6. If f (x) is the integral of
, where x ≠ 0, then
x3
(1979, 3M)
find lim f ′ (x).
5. Evaluate
I 4 + I 6 = a tan5 x + bx5 + C, where C is a constant of
integration, then the ordered pair (a , b) is equal to
1
(b)  , 0
5 
∫
4. Evaluate
2. Let I n = ∫ tan n x dx (n > 1). If
1
(a)  − , 1
 5 
3. Find the indefinite integral
(2019 Main, 9 April I)
(b) −3 tan −1/3 x + C
3
(d) − tan −4 /3 x + C
4
(c) −3 cot −1/3 x + C
Analytical & Descriptive Questions
∫
x→ 0
Topic 3 Integration by Parts
Objective Questions (Only one correct option)
1. If
5 −x
∫xe
2
2
dx = g (x)e− x + C, where C is a constant of
integration, then g (− 1) is equal to (2019 Main, 10 April II)
(a) − 1
1
(c) −
2
(b) 1
(d) −
5
2
2. If ∫ esec x (sec x tan x f (x) + (sec x tan x + sec2 x))
dx = esec x f (x) + C, then a possible choice of f (x) is
(2019 Main, 9 April II)
(a) x sec x + tan x +
1
2
(b) sec x + tan x +
1
2
(c) sec x + x tan x −
1
2
(d) sec x − tan x −
1
2
Indefinite Integration 281
∫ cos (log e x) dx is equal to (where C is a
3. The integral
constant of integration)
6. If ∫ f (x) dx = ψ (x), then ∫ x5 f (x3 ) dx is equal to
1 3
[x ψ(x3 ) − ∫ x2ψ(x3 )dx] + c
3
1 3
(b)
x ψ(x3 ) − 3 ∫ x3 ψ(x3 ) dx + c
3
1
(c) x3 ψ(x3 ) − ∫ x2ψ(x3 ) dx + c
3
1
(d) [x3 ψ(x3 ) − ∫ x3 ψ(x3 ) dx] + c
3
(2019 Main, 12 Jan I)
(a)
x
(a) [cos(log e x) + sin(log e x)] + C
2
(b) x [cos(log e x) + sin(log e x)] + C
(c) x [cos(log e x) − sin(log e x)] + C
x
(d) [sin(log e x) − cos(log e x)] + C
2
1 −4 x 3
e
f (x) + C ,
48
where C is a constant of integration, then f (x) is equal to
(2013 Main)
3
4. If ∫ x5 e−4x dx =
(2019 Main, 10 Jan II)
5.
(a) − 4x3 − 1
(b) 4x3 + 1
(c) − 2x3 − 1
(d) − 2x3 + 1
∫
1
1 x + x
e
dx is

1 + x − 

x
x+
(a) (x − 1) e
x+
(b) x e
1
x

2x + 2
 dx.
 4x2 + 8 x + 13 



7. Evaluate ∫ sin −1 
(c) (x + 1) e
1
x+
−x e x
1
x
(2000, 5M)
8. Find the indefinite integral
 cos θ + sin θ 
∫ cos 2θ log cos θ − sin θ  dθ.
(2014 Main)
+ c
9. Evaluate ∫
+ c
x+
(d)
1
x
equal to
Analytical & Descriptive Questions
10. Evaluate ∫
+ c
sin −1 x − cos −1 x
dx.
sin −1 x + cos −1 x
(x − 1) ex
dx.
(x + 1)3
11. Evaluate ∫ (elog x + sin x) cos x dx.
+ c
(1994, 5M)
(1986, 2½ M)
(1983, 2M)
(1981, 2M)
Topic 4 Integration, Irrational Function and Partial Fraction
Objective Questions (Only one correct option)
1. The integral ∫
(a)
2x12 + 5x9
dx is equal to
(x5 + x3 + 1)3
− x5
(x + x + 1)
5
3
2
(c)
x
2(x5 + x3 + 1)2
x5
2(x + x + 1)
5
3
2
−x
10
(d)
2(x5 + x3 + 1)2
(c) sin x − 2 (sin x)−1 − 6 tan −1 (sin x) + c
(d) sin x − 2 (sin x)−1 + 5 tan −1 (sin x) + c
Analytical & Descriptive Questions
3.
+C
where, C is an arbitrary constant.
(1995, 2M)
(b) sin x − 2 (sin x)−1 + c
+C
+C
cos3 x + cos5 x
dx is
sin 2 x + sin 4 x
(a) sin x − 6 tan −1 (sin x) + c
(2016 Main)
+C
10
(b)
2. The value of ∫
∫
x3 + 3x + 2
dx.
(x + 1)2 (x + 1)
2
4. Evaluate ∫
(x + 1)
dx.
x (1 + xex )2
(1999, 5M)
(1996, 2M)
Answers
Topic 1
1.
5.
9.
13.
17.
(d)
(c)
(b)
(c)
(c)
21. (b,c)
3.
2.
6.
10.
14.
18.
(d)
(c)
(b)
(b)
(c)
22. (2)
3.
7.
11.
15.
19.
4. (b)
8. (c)
12. (c)
16. (d)
20. (c)
3
35
23. A = − , B = and C ∈R
2
36
(c)
(b)
(c)
(b)
(a)
1
⋅(2 x 3m + 3 x 2m + 6 xm ) (m + 1)/m + c
6 (m + 1 )
1
25. 2 [cos−1 x − log| 1 + 1 − x | − log | x| ] + c
2
−1
26. −2 1 − x + cos
x + x (1 − x ) + c
24.
29.
30.
31.
4.
 tan x − cot x 
2 tan −1 
 +c
2


5. − log| cot x +
cot 2 x − 1 | +
1
log
2
2 + 1 − tan 2 x
+c
2 − 1 − tan 2 x
Topic 3
1. (d)
( x 4 + 1 )1/4
+c
x
x
x
(i) 4 sin − 4 cos + c
4
4
2
1


(ii) − 2  1 − x − (1 − x ) 3/2 + (1 − x ) 5/2  + c
3
5



1 
a2
+ c
a + bx − 2a log (a + bx ) −
a + bx
b3 

cos 4 x cos 2 x cos 6 x
−
−
+
+ sin 2 x + tan x − 2 x
16
8
24
3 x sin 4 x sin 8 x
+
−
+
128
128
1024
1
1
x
−1
2
32. log (sin x − cos x ) + + c
tan ( x ) + c
2
2
2
Topic 2
1. (b)
+ 12 ln (1 + x1/2 ) − 3 [ ln(1 + x1/6 )]2 + c
6. (1)
27. −
28.
3 2/3 12 7/12 4 1/2 12 5/12 1 1/3
x −
x
+ x −
x
+ x − 4 x 1/ 4 − 7 x 1/ 6
2
7
3
5
2
− 12 x1/12 + (2 x1/2 − 3 x1/3 + 6 x1/6 + 11 ) ln (1 + x1/6 )
7. ( x + 1 ) tan −1
8.
3. (a)
4. (a)
6. (c)
2
 x + 2 3
2
 − log( 4 x + 8 x + 13 ) + c

 3  4
 cosθ + sin θ  1
1
sin 2 θ ln 
 + ln(cos2 θ ) + c
 cosθ − sin θ  2
2
2
[ x − x 2 − (1 − 2 x ) sin −1 x ] − x + c
π
cos2 x
ex
10.
+c
+ c 11. x sin x + cos x −
4
(x + 1)2
9.
Topic 4
1. (b)
2. (c)
1
1
3
x
3. − log| x + 1 | + log| x 2 + 1 | + tan −1 x + 2
+c
2
4
2
x +1
4. log
2. (b)
2. (b)
5. (b)
1
xe x
+
+c
1 + xe x
1 + xe x
Hints & Solutions
Topic 1 Some Standard Results
cos θ
1. Since, I = ∫
dθ
5 + 7 sin θ – 2 cos 2 θ
cos θ
=∫
dθ
3 + 7 sin θ + 2 sin 2 θ
Let,
sin θ = t ⇒ cos θ dθ = dt
dt
dt
1
= ∫
∴ I=∫ 2
7
3
2
2t + 7t + 3
t2 + t +
2
2
dt
1
1
= ∫
= ∫
2
2 
7
 3 49 2 
t +  +  – 
t +

 2 16

4
7 5
t+ –
1
1
4
4 + C'
log e
= ×
7 5
2 2×5
t+ +
4
4 4
dt
2
7
 5
 – 
 4
4
2
∴
=
1
2t + 1
+ C'
log e
5
2t + 6
=
1
1
2 sin θ + 1
+ C, where C = C′– log e 2
log e
5
sin θ + 3
5
= A log e|B(θ )|+ C
B(θ ) 5(2 sin θ + 1)
.
=
A
(sin θ + 3)
[given]
Indefinite Integration 283
Now, put x − α = t ⇒ dx = dt, so
2. It is given that,
f (x) = ∫
x
dx, (x ≥ 0)
(1 + x)2
I=∫
Put x = tan 2 t ⇒ dx = 2 tan t sec2 tdt
2 tan 2 t sec2 t
∴ f (x) = ∫
dt = ∫ 2 sin 2 t dt
sec4 t
[Q2 sin 2 x = 1 − cos 2x]
= ∫ (1 − cos 2t )dt
sin 2t
+C
2
tan t
x
So,
=t−
+ C = tan −1 x −
+C
2
1+ x
1 + tan t


3
∴f (3) − f (1) =  tan −1 3 −
+ C
1+3


sin t cos 2 α + sin 2α cos t
dt
sin t
cos t 

= ∫  cos 2 α + sin 2 α
 dt

sin t 
=∫
= t (cos 2 α ) + (sin 2 α ) log e |sin t |+ C
= (x − α ) cos 2 α + (sin 2 α ) log e |sin (x − α )| + C
=t−


1
−  tan −1 (1) −
+ C


1+1
π
3 π 1 π
1
3
−
− + =
+ −
3
4
4 2 12 2
4
Hence, option (d) is correct.
dθ
3. Given integral, I = ∫
cos 2 θ (tan 2 θ + sec 2 θ )
=
=∫
sec2 θ cos 2 θ
dθ
(sin 2 θ + 1)
=∫
sec2 θ (1 − tan 2 θ )
dθ
(2 tan θ + 1 + tan 2 θ )
=∫
sec2 θ (1 − tan θ )
dθ
1 + tan θ
= A (x) cos 2 α + B(x) sin 2 α + C (given)
Now on comparing, we get
A (x) = x − α and B(x) = log e |sin (x − α )|
5.
Key Idea
(i) Divide each term of numerator and denominator by x 2.
1
(ii) Let x 2 +
= t
x
2x3 − 1
2 x − 1 / x2
dx = ∫
dx
4
1
x +x
x2 +
x
[dividing each term of numerator and
Let integral is I = ∫
denominator by x2]
1

Put x2 + = t ⇒ 2x +

x
I=∫
∴
= − t + 2 log e|1 + t| + C
= − tan θ + 2 log e |1 + tan θ| + C
= λ tan θ + 2 log e| f (θ )| + C (given)
(on comparing)
∴
(λ , f (θ )) = (−1, 1 + tan θ )
Hence, option (c) is correct.
tan x + tan α
 π
4. Let
I=∫
dx, α ∈ 0, 
 2
tan x − tan α
sin x sin α
+
cos x cos α
dx
=∫
sin x sin α
−
cos x cos α
sin x cos α + sin α cos x
dx
=∫
sin x cos α − sin α cos x
sin (x + α )
=∫
dx
sin (x − α )
= log e
6. Let I = ∫
 1 
 − 2  dx = dt
 x 
dt
= log e|(t )| + C
t

= log e  x2 +

Let,
tan θ = t
⇒ sec2 θdθ = dt
1 −t
Then,
I=∫
dt
1+t

2 
= ∫  −1 +
 dt

1 + t
sin (t + 2α )
dt
sin t
1
 +C
x
x3 + 1
+C
x
dx
dx
=∫
2
((x − 1)2 + 32)2
(x − 2x + 10)
2
Now, put x − 1 = 3 tan θ ⇒ dx = 3 sec2θ dθ
3sec2θ dθ
3sec2θ dθ
=∫ 4
2
2 2
(3 tan θ + 3 )
3 sec4θ
1
1 1 + cos 2θ
=
dθ
cos 2θ dθ =
27 ∫
27 ∫
2
1 + cos 2θ 

Q cos 2 θ =


2
1 
sin 2θ 
1
=
(1 + cos 2θ ) dθ =
 +C
θ +
54 
2 
54 ∫
So,I = ∫
=
2
1
1  2 tan θ 
 x − 1
tan −1 

 +C
 +


54
3
108  1 + tan 2 θ 

2 tan θ 
Q sin 2θ =

+ tan 2 θ 
1

284 Indefinite Integration
 x − 1


 3 
1
1
 x − 1
tan −1 
=
 +
 3  54
54
+C
2
 x − 1
1+ 

 3 


x−1

 +C
2
2
 (x − 1) + 3 


x−1
 2
 +C
 x − 2x + 10
=
1
1
 x − 1
tan −1 
 +
 3  18
54
=
1
1
 x − 1
tan −1 
 +
 3  18
54
=
1 
3(x − 1) 
 x − 1
tan −1 
+C
 +
 3  x2 − 2x + 10 
54 
It is given, that


f (x)
 x − 1
+C
I = A tan −1 
 + 2
 3  x − 2x + 10 

On comparing, we get A =
1
and f (x) = 3(x − 1).
54
dx
x (1 + x6 )2/ 3
7. Let I = ∫
3
dx
=∫
1

x3 ⋅ x4  6 + 1
x

2/ 3
=∫
dx
1

x7  6 + 1
x

2/ 3
1
+ 1 = t3
x6
Now, put
6
dx
t2
dx = 3t 2dt ⇒ 7 = − dt
7
2
x
x
1 2
− t dt
1
2
= − ∫ dt
2
t2
−
⇒
So, I = ∫
1/3
1 1
1
1
 3


Qt = 6 +1
t + C = −  6 + 1 + C



2 x
2
x
1 1
(1 + x6 )1/3 + C
=−
2 x2
[given]
= x ⋅ f (x) ⋅ (1 + x6 )1/3 + C
On comparing both sides, we get
1
f (x) = − 3
2x
5x
5x
x
sin
2 sin
cos
2 dx =
2
2 dx
8. Let I = ∫
∫
x
x
x
sin
2 sin cos
2
2
2
x
[multiplying by 2 cos in numerator and
2
denominator]
sin 3x + sin 2x
=∫
dx
sin x
[Q2 sin A cos B = sin( A + B) + sin( A − B) and
sin 2 A = 2 sin A cos A]
(3 sin x − 4 sin3 x) + 2 sin x cos x
=∫
dx
sin x
[Q sin 3x = 3 sin x − 4 sin3 x]
2
= ∫ (3 − 4 sin x + 2 cos x)dx
=−
= ∫ [3 − 2(1 − cos 2x) + 2 cos x]dx
= ∫ [3 − 2 + 2 cos 2x + 2 cos x]dx
[Q2 sin 2 x = 1 − cos 2x]
= ∫ [1 + 2 cos 2x + 2 cos x]dx
= x + 2 sin x + sin 2x + C
9. Let
I=∫
3
2
+ 5
3
3x13 + 2x11
x
x
dx = ∫
dx
4
(2x4 + 3x2 + 1)4
3
1

2 + 2 + 4 

x
x 
[on dividing numerator and denominator by x16]
3
1
Now, put 2 + 2 + 4 = t
x
x
⇒
−6 4 
 3 − 5  dx = dt
x
x 
⇒
2
dt
3
 3 + 5  dx = −

x
2
x
1
1 t− 4 + 1
− dt
=− ×
+C= 3 +C
4
2 −4 + 1
2t
6t
3
1
1

=
+C
Qt =2 + 2 + 4
3


x
x
3
1


6 2 + 2 + 4 

x
x 
So, I = ∫
=
x12
+C
6 (2x + 3x2 + 1)3
4
10. We have,
∫
x+1
dx = f (x) 2x − 1 + C
2x − 1
Let I = ∫
...(i)
x+1
dx
2x − 1
Put 2x − 1 = t 2
⇒
2dx = 2tdt ⇒ dx = tdt
2
t +1
+1
1
I=∫ 2
tdt = ∫ (t 2 + 3) dt
t
2

t2 + 1 
2
Q 2x − 1 = t ⇒ x = 2 



1  t3
t 2
=  + 3t + C = (t + 9) + C
23
6

2x − 1
=
[Q t = 2x − 1 ]
(2x − 1 + 9) + C
6
2x − 1
=
(2x + 8) + C
6
x+4
=
2x − 1 + C
3
On comparing it with Eq. (i), we get
x+4
f (x) =
3
Indefinite Integration 285
11. We have,
∫
1 − x2
x4
dx = A (x) ( 1 − x2 )m + C
… (i)
1

x2 2 − 1
x

1 − x2
LetI = ∫
13. We have, f (x) = ∫
=∫
dx = ∫
dx
x4
x4
1
x 2 −1
1 1
x
=∫
− 1 dx
dx = ∫ 3
x4
x x2
1
−2
1
Put 2 − 1 = t 2 ⇒ 3 dx = 2t dt ⇒ 3 dx = − t dt
x
x
x
t3
2
∴
I = − ∫ t dt = −
+C
3
=−
1  1 − x2 
.

3  x2 
3/ 2
1 1
( 1 − x2 )3 + C
3 x3
On comparing Eqs. (i) and (ii), we get
1
A (x) = − 3 and m = 3
3x
1
m
( A (x)) = ( A (x))3 = −
∴
27 x9
(sin θ − sin θ ) cos θ
dθ
sin n + 1 θ
sin θ = t ⇒ cos θ dθ = dt
1/ n
t 
 n
−
t
1


 
(t n − t )1/ n
t n  
dt = ∫ 
I=∫
dt
n+1
n+1
t
t
t (1 − 1 / t n−1 )1/ n
(1 − 1 / t n − 1 )1/ n
dt = ∫
dt
=∫
n+1
t
tn
1
1 − n −1 = u
t
(n − 1)
1 − t −( n − 1) = u ⇒
dt = du
tn
dt
du
=
n
n −1
t
12. Let I = ∫
Put
∴
Put
or
⇒
1
⇒
…(ii)
1/ n
n
I=∫
u1/ ndu
=
n −1
+1
un
+C

1
(n − 1)  + 1

n
n+1
1  n

n 1 − n−1 

t 
=
+C
(n − 1) (n + 1)
=
1


n 1 −


sin n − 1 θ 
n2 − 1
 x6 
 x8 
5 14  + 7 14 
x 
x 
 x2 1 2x7 
 7 + 7 + 7
x
x 
x
2
dx
(dividing both numerator and denominator by x14)
5 x − 6 + 7 x− 8
= ∫ −5
dx
(x + x− 7 + 2)2
1/ 2

 
1
+ C Q t =  2 − 1 

x


=−
5 x8 + 7 x 6
dx
(x2 + 1 + 2x7 )2
n+1
n
Let
x− 5 + x − 7 + 2 = t
⇒ (− 5x− 6 − 7x− 8 )dx = dt
⇒
(5x− 6 + 7x− 8 )dx = − dt
dt
∴ f (x) = ∫ − 2 = − ∫ t −2dt
t
t− 2 + 1
t− 1
1
=−
+ C =−
+C= +C
−2 + 1
−1
t
=
x
−5
1
x7
+C= 7
+C
−7
+x +2
2 x + x2 + 1
Q f (0) = 0
∴
0
+ C ⇒C = 0
0+0+1
0=
∴ f (x) =
x7
2 x + x2 + 1
⇒
f (1) =
7
14. Let I = ∫ x
1
1
=
2
2(1) + 1 + 1 4
7
2 sin(x2 − 1) − sin 2(x2 − 1)
dx
2 sin(x2 − 1) + sin 2(x2 − 1)
x2 − 1
=θ
2
⇒ x2 − 1 = 2θ
⇒ 2x dx = 2 dθ
⇒
x dx = dθ
2 sin 2 θ − sin 4 θ
Now, I = ∫
dθ
2 sin 2 θ + sin 4 θ
Put
=∫
2 sin 2 θ − 2 sin 2 θ cos 2 θ
dθ
2 sin 2 θ + 2 sin 2 θ cos 2 θ
=∫
2 sin 2 θ (1 − cos 2 θ )
dθ
2 sin 2 θ (1 + cos 2 θ )
=∫
1 − cos 2 θ
dθ = ∫
1 + cos 2 θ
(Qsin 2 A = 2 sin A cos A)
2 sin 2 θ
dθ
2 cos 2 θ
[Q1 − cos 2 A = 2 sin 2 A and 1 + cos 2 A = 2 cos 2 A]
+C
1


Q u = 1 − n − 1 and t = sin θ


t
=∫
tan 2 θ d θ = ∫ tan θd θ
 x2 − 1 
= log e|sec θ| + C = log e sec 
 +C
 2 

x2 − 1 
Q θ = 2 


286 Indefinite Integration
15. We have,
sin 2 x ⋅ cos 2 x
I=∫
dx
5
3
(sin x + cos x ⋅ sin 2 x + sin3 x ⋅ cos 2 x + cos5 x)2
sin 2 x cos 2 x
dx
=∫
{sin3 x(sin 2 x + cos 2 x) + cos3 x(sin 2 x + cos 2 x)}2
=∫
sin 2 x cos 2 x
sin 2 x cos 2 x
=
dx
dx
∫
(sin3 x + cos3 x)2
cos 6 x(1 + tan3 x)2
=∫
tan 2 x sec2 x
dx
(1 + tan3 x)2
Put tan3 x = t ⇒ 3 tan 2 x sec2 xdx = dt
1
dt
I= ∫
∴
3 (1 + t )2
−1
⇒
I=
+C
3 (1 + t )
−1
+C
⇒
I=
3 (1 + tan3 x)
16.
dx
1+
Put
18. Since,
∴
Put
1

x5 1 + 4 

x 
3/ 4
1
= t4
x4
−4
dx = 4t3 dt
x5
dx
= − t3 dt
x5
⇒
⇒
Hence, the integral becomes
∫
17.
1
− t3 dt

= − ∫ dt = − t + c = − 1 + 4 

t3
x 
I=∫
J −I=∫
4x
e
ex
e3 x
dx and J = ∫
dx
2x
+ e +1
1 + e2x + e4x
(e3 x − ex )
dx
1 + e2x + e4x
ex = u ⇒ ex dx = du
1

 1 − 2

(u 2 − 1)
u  du
∴
du = ∫
J −I = ∫
2
4
1
1+ u + u
1 + 2 + u2
u
1

 1 − 2

u 
du
=∫
2
1

u +  −1

u
1
Put
u+ =t
u
1


⇒
1 − 2 du = dt

u 
dx
∫ x2(x4 + 1)3/ 4 = ∫
1
1 dt
t +  ⋅
1 
t t 1  1
2
= ∫  9/ 2 + 13/ 2 dt
⇒ I=∫
2 t
t 9/ 2
t 
1 2
2 
= −  7/ 2 +
+ K
2 7 t
11 t11/ 2 


1
1
= −
+
+K
7/ 2
11/ 2 
7
(sec
x
tan
x
)
11
(sec
x
tan
x
)
+
+


1
−1
1

+ (sec x + tan x)2 + K
=
11/ 2 
11
7
(sec x + tan x)


1/ 4
+c
=∫
PLAN Integration by Substitution
i.e.
Put
∴
I = ∫ f { g ( x )} ⋅ g ′ ( x )dx
=
u2 − u + 1
1
log 2
+c
2
u +u+1
I = ∫ f(t )dt
=
e2x − ex + 1
1
log 2x
+c
2
e + ex + 1
g ( x ) = t ⇒ g ′ ( x )dx = dt
Description of Situation Generally, students gets
confused after substitution, i.e. sec x + tan x = t.
Now, for sec x, we should use
sec2 x − tan 2 x = 1
⇒
⇒
Here,
Put
(sec x − tan x) (sec x + tan x) = 1
1
sec x − tan x =
t
I=∫
sec2 dx
(sec x + tan x)9/ 2
∴
∴
x
for n ≥ 2
(1 + xn )1/ n
f (x)
x
∴
=
ff (x) =
n 1/ n
[1 + f (x) ]
(1 + 2 xn )1/ n
x
and fff (x) =
(1 + 3xn )1/ n
x
g (x) = ( fofo ... of ) (x) =
∴
14243
(1 + n xn )1/ n
19. Given, f (x) =
n times
sec x + tan x = t
⇒ (sec x tan x + sec2 x) dx = dt
⇒
dt
1
t −1
= log
+c
t+1
t −1 2
2
dt
t
1
1
1
sec x − tan x =
⇒ sec x =  t + 
2
t
t
sec x ⋅ sec x dx
I =∫
(sec x + tan x)9/ 2
Let
xn − 1 dx
(1 + nxn )1/ n
d
(1 + nxn )
n 2 xn − 1 dx
1 dx
=
dx
(1 + nxn )1/ n n 2 ∫ (1 + nxn )1/ n
I = ∫ xn − 2g (x) dx = ∫
sec x ⋅ t dx = dt ⇒ sec x dx =
=
1
n2 ∫
1
I=
1−
1
(1 + nxn ) n + c
n (n − 1)
Indefinite Integration 287
20. Let I = ∫
Put
⇒
∴ I=
(x2 − 1) dx
2x − 2x + 1
[dividing numerator and enominator by x5 ]
1
1
 3 − 5  dx
x
x 
=∫
2
1
2− 2+ 4
x
x
2
1
2− 2+ 4 =t
x
x
4
4
 3 − 5  dx = dt
x
x 
1
4
∫
3
4
x
ef ( x )
g′ (x)
eg( x )
f ′ (x) g′ (x)
= g( x )
⇒
ef ( x )
e
⇒
e− f ( x ) f ′ (x) = e− g( x ) g′ (x)
On integrating both side, we get
f (0) = 1, then A = 1
1
x
f (x) = e2
1
1
log e f (x) = x ⇒ log e f (4) = × 4 = 2
2
2
Hence,
∴
4ex + 6e− x
dx = Ax + B log (9e2x − 4) + c
9ex − 4e− x
4e2x + 6
dx
9e2x − 4
4e2x + 6 = A (9e2x − 4) + B (18 e2x )
9 A + 18B = 4 and − 4 A = 6
3
35
and
A=−
B=
2
36
∫
1
A (9e2x − 4) + B (18e2x )
dx = A ∫ 1 dx + B ∫
dt
t
9e2x − 4
where t = 9e2x − 4
e− f (1) = e− g(1) + C
[Q f (1) = 1]
=e
+C
= A x + B log (9e2x − 4) + c
3
35
=− x+
log (9e2x − 4) + c
2
36
3
35
A =− ,B=
2
36
…(i)
[Q g(2) = 1]
…(ii)
∴
c = any real number
and
…(iii)
I = ∫ (x3 m + x2m + xm )
− f (2) < log e 2 − 1
⇒
e− g(1) + e− f (2) = 2e−1
⇒
− g(1 )
24. For any natural number m , the given integral can be
written as
(2x3 m + 3x2m + 6xm )1/ m
dx
x
⇒ I = ∫ (2 x3 m + 3x2m + 6 xm )
f (2) > 1 − log e 2
[from Eq. (iii)]
−1
<2e
− g (1) < log e 2 − 1
⇒
If
⇒
− g(1 )
e
x
f (x) = Ae2 , where eC = A
⇒
e− f ( 2) = 2e−1 − e− g(1)
⇒
e− f ( 2) > 2e−1
We know that, e− x is decreasing
∴
1
⇒
Let
e− f ( x ) = e− g( x ) + C
e
At x = 2
e− f ( 2) = e− g( 2) + C
⇒
e− f ( 2) = e−1 + C
From Eqs. (i) and (ii)
On integrating, we get
1
log f (x) = x + C
2
LHS = ∫
f ′ (x) =
−1
⇒
23. Given, ∫
21. We have, f ′ (x) = e( f ( x ) − g( x )) g′ (x) ∀ x ∈ R
At x = 1
f (x) =
⇒
dt 1 t1/ 2
= ⋅
+c
t 4 1 /2
1
2
1
=
2− 2+ 4 + c
2
x
x
⇒
1
f (x) + f ′ (x)
2
1
f ′ (x) 1
f ′ (x) = f (x) ⇒
=
2
f (x) 2
⇒
2
g (1) > 1 − log e 2
22. Given,
f (x + y) = f (x) f ′ ( y) + f ′ (x) f ( y), ∀x, y ∈ R
and
f (0) = 1
Put x = y = 0, we get
f (0) = f (0) f ′ (0) + f ′ (0) f (0)
1
⇒
1 = 2 f ′ (0) ⇒ f′ (0) =
2
Put x = x and y = 0, we get
f (x) = f (x) f ′ (0) + f ′ (x) f (0)
1/ m
(x3 m − 1 + x2m − 1 + xm − 1 ) dx
Put 2 x3 m + 3x2m + 6xm = t
(6 mx3 m−1 + 6 mx2m−1 + 6 mxm −1 ) dx = dt
⇒
1
∴ I=∫
=
25. Let
Put
+1
1
tm
dt
t1/m
=
⋅
1
6m 6m 

 + 1

m
1
⋅ (2 x3m + 3x2m + 6 xm )(m + 1)/m + c
6 (m + 1)
I=∫
 1 − x


 1 + x
1/ 2
⋅
dx
x
x = cos 2 θ ⇒ dx = − 2 cos θ sin θ dθ
288 Indefinite Integration
∴
 1 − cos θ 
I=∫ 

 1 + cos θ 
1/ 2
⋅
− 2 cos θ ⋅ sin θ
dθ
cos 2 θ
θ
− 2 sin θ
2
=∫
⋅
dθ
θ cos θ
cos
2
θ
θ
θ
θ
2 sin ⋅ 2 sin ⋅ cos
2 sin 2
2
2
2 d θ −2
=−∫
∫ cos θ 2 d θ
θ
cos ⋅ cos θ
2
1 − cos θ
= −2∫
dθ
cos θ
= 2∫ (1 − sec θ ) dθ = 2 [θ − log|sec θ + tan θ|] + c

1
−1 + c
x

1

x − log|1 + 1 − x|− log|x| + c

2

1
+
I = 2 cos −1 x − log
x

⇒

I = 2 cos −1

26. Let
I=∫
x2
dx
1−x
Put 1 − x = t 2 ⇒ − dx = 2 t dt
sin
⇒
I=∫
(ii) Let
(1 − t 2)2 ⋅ (−2t )
dt
t
2
4
= − 2 ∫ (1 − 2t + t ) dt
I=∫
∴

2t3
t5 
= − 2 t −
+  +c
3
5

2
1


= − 2  1 − x − (1 − x)3/ 2 + (1 − x)5/ 2 + c
3
5


I=
29. Let
x2
(a + bx)2
Put a + bx = t ⇒ b dx = dt
2
 t − a


 b  dt
∴
⋅
I=∫
b
t2
 t 2 − 2 at + a 2
1
= 3 ∫ 
 dt
b
t2


1− x
dx
1+ x

2 a a 2
+ 2  dt
1 −
t
t 

Put x = cos 2 θ ⇒ dx = − 2sin θ cos θ dθ
=
1 − cos θ
⋅ (− 2 sin θ cos θ ) dθ
1 + cos θ
1
b3
=
1 
a 2
 t − 2 a log t −  + c
3
t
b 
=
1
b3
∴
I=∫
θ
θ
= − ∫ 2 tan ⋅ sin θ cos θ dθ = − 2 ∫ 2 sin 2 ⋅ cos θ dθ
2
2
= − 2∫ (1 − cos θ ) cos θ dθ = − 2 ∫ ( cos θ − cos θ ) dθ
2
= − 2∫ cos θ dθ +
= − 2 sin θ + θ +
sin 2θ
+c
2
dx
=
x2 (x4 + 1)3/ 4 ∫
Put 1 + x− 4 = t ⇒ −
∴
I=−
1
4
1
(sin 4x + sin 2 x − sin 6x) dx
4∫
cos 4x cos 2 x cos 6 x
=−
−
+
16
8
24
=
x (1 − x) + c
dx
1

x ⋅ x 1 + 4 

x 
2
I 2 = ∫ sec2 x ⋅ cos 2 2 x dx
3/ 4
3
= ∫ sec2 x (2 cos 2 x − 1)2dx
4
dx = dt
x5
dt
1 t1/ 4
= ∫ (4 cos 2 x + sec2 x − 4) dx

1
∫ t3/ 4 = − 4 ⋅ 1 / 4 + c = − 1 + x4 
(x4 + 1)1/ 4
=−
+c
x
28. (i) Let
I=∫
= ∫ (2 cos 2 x + sec2 x − 2) dx
1/ 4
+c
= sin 2 x + tan x − 2 x
and I3 = ∫ sin 4 x cos 4 x dx
1
(3 − 4 cos 4x + cos 8x) dx
128 ∫
3x sin 4x sin 8x
=
−
+
128 128
1024
=
x
1 + sin dx
2
x
x
x
x
+ sin 2 + 2 sin cos dx
4
4
4
4
x
x

= ∫  cos + sin  dx

4
4
x
x
= 4 sin − 4 cos + c
4
4
=∫
cos 2


a2
+ c
 a + bx − 2 a log (a + bx) −
a + bx 

30. Let I1 = ∫ sin x sin 2 x sin 3x dx
∫ (1 + cos 2 θ ) dθ
= − 2 1 − x + cos −1 x +
27. Let I = ∫
∫
∴
I = I1 + I 2 + I3
cos 4x cos 2 x cos 6 x
=−
−
+
+ sin 2 x + tan x − 2 x
16
8
24
3x sin 4x sin 8 x
+
−
+
128 128
1024
Indefinite Integration 289
31. Let
I=∫
x dx 1
2x
=
dx
1 + x4 2 ∫ 1 + (x2)2
3. Let
Put x = u ⇒ 2 x dx = du
2
∴
du
1
1
1
I= ∫
= tan −1 (u ) + c = tan −1 (x2) + c
2
2
2
2 1+ u
∴

1
where, I1 = ∫  3
 x+
sin x
dx
sin x − cos x
Again, let sin x = A (cos x + sin x) + B(sin x − cos x),
A=
∴
I2 = ∫
Now,
1
1
, B=
2
2
1
1
(cos x + sin x) + (sin x − cos x)
2
2
I=∫
dx
(sin x − cos x)
1 cos x + sin x
1
= ∫
dx + ∫ 1 dx + c
2 sin x − cos x
2
1
1
= log (sin x − cos x) + x + c
2
2
⇒
Put
∴
2
4
∫
dx
4
x 3
 sin


 cos x
4
cos3
2
x cos3
dx
4
tan3
x cos 2 x
sec2 x dx
1
t3
−3
1
dt
t+1
+ 12 ln (t + 1)
∴
x = u6 ⇒
I2 = ∫
dx = 6 u5 du
ln (1 + u ) 5
ln (1 + u )
6u du = ∫ 2
. 6 u5 du
u (1 + u )
u 2 + u3
u3
ln (1 + u ) du
(u + 1)

1 
= 6 ∫  u2 − u + 1 −
 ln (1 + u ) du

u + 1
+ C = −3 tan
−
1
3
x+C
(tan x)3
∴ I n + I n + 2 = ∫ tan n x dx +
∫ tan
n+ 2
x dx
= ∫ tan n x(1 + tan 2 x) dx
= ∫ tan n x sec2 x dx =
n+1
tan
x
+C
n+1
tan5 x
+C
5
1
a = and b = 0
5
Put n = 4, we get I 4 + I 6 =
= 6 ∫ (u 2 − u + 1) ln (1 + u ) du − 6 ∫
II
I
ln (1 + u )
du
(u + 1)
 u3 u 2

=6 
−
+ u ln (1 + u )
2
 3

2. We have, I n = ∫ tan n x dx
∴
t 8 dt
t+1
 u3 − 1 + 1
=6 ∫ 
 ln(1 + u ) du
 u+1 
dt
+C
=
−4
t 4/3
+1
3
+C =
t11
dt
t 4 + t3
 ln (1 + 6 x ) 
 dx
 3
 x+ x 
=6 ∫
4
x)3
−4
+1
t3
1
I1 = 12 ∫
+ 12∫
Put
Now, put tan x = t ⇒ sec x dx = dt
= −3

 dx
x
4
x
(tan
4
x = t12 ⇒ dx = 12 t11dt
x sin3 x
2
∴I=∫

1
I1 = ∫  3
 x+
and I 2 = ∫
=∫

 dx,
x
 t8
t7 t 6 t5 t 4 t3 t 2 
= 12  −
+
− +
− + − t
7
6 5
4 3 2
8

[dividing and multiplying by cos 4/3 x in denominator]
=∫
4
= 12∫ (t7 − t 6 + t5 − t 4 + t3 − t 2 + t − 1) dt
dx
2
cos3
x
ln (1 + 6 x )
 dx
3
x+ x 
ln (1 + 6 x )
dx
3
x+ x
= 12 ∫
Topic 2 Some Special Integrals
1. Let I = ∫ sec3 x cos ec3 x dx = ∫
4
+
I = I1 + I 2
32. Let I = ∫
then A + B = 1 and A − B = 0

1
I = ∫ 3
 x+
−∫
2 u3 − 3u 2 + 6 u
1
du − 6 [ln (1 + u )]2
2
u+1
= (2 u3 − 3 u 2 + 6 u ) ln (1 + u )

11 u 
2
−∫ 2 u 2 − 5 u +
 du − 3 [ln (1 + u )]

u + 1
= (2 u3 − 3 u 2 + 6 u ) ln (1 + u )
 2 u3 5 2

− 
− u + 11u − 11 ln (u + 1) − 3 [ln (1 + u )]2
2
 3

290 Indefinite Integration
∴
I=
3 23
12 7/12
12 5/12
x/ −
x + 2x1/ 2 −
x
+ 3x1/3 − 4x1/ 4
2
7
5
− 6 x1/ 6 − 12 x1/12 + 12 ln (x1/12 + 1)
+ (2 x1/ 2 − 3x1/3 + 6 x1/ 6 ) ln (1 + x1/ 6 )
2 1/ 2 5 1/3

−
x − x 11 x1/ 6 − 11 ln (1 + x1/ 6 )
2
3

−3 [ln (1 + x1/ 6 )]2 + c
12 7/12 4 1/ 2 12 5/12
−
x + x −
x
7
3
5
1
+ x1/3 − 4x1/ 4 − 7x1/ 6 − 12 x1/12
2
3
= x2 / 3
2
= − ∫ secθ dθ + 2 ∫
cos θ
dθ
1 + cos 2 θ
= − log|sec θ + tan θ | + 2 ∫
= − log|sec θ + tan θ | +
I = ∫ ( tan x +
4. Let
cot x ) dx = ∫
+
tan x + 1
dx
tan x
I=∫
t2 + 1
t2
=2 ∫
t−
Put
⋅
t2 +
1
−2 + 2
t2
x

2 sin 2 
 sin x 
2
lim f ′ (x) = lim 2 


 x   x2 
x→ 0
x→ 0


dt = 2 ∫
1+
1
t2
2
1

2
t −  + ( 2)

t
du
u 2 + ( 2 )2
2
 u
tan −1   + c
I=
 2
2
⇒
 tan x − cot x 
= 2 tan −1 
 +c
2


5. Let I = ∫
cos 2 x
dx = ∫
sin x
=∫
cos 2 x − sin 2 x
dx
sin 2 x
cot2 x − 1 dx
Put cot x = sec θ
∴
I=∫

x 
 sin 2 
2 =1
= 4 ⋅ 1 ⋅ lim 
2
x→ 0 
 x 
4 ×   
2 

dt
1
1
= u ⇒ 1 + dt = du
t
t2
I =2 ∫
∴
1
t2
⇒ − cosec2x dx = sec θ tan θ dθ
sec θ ⋅ tan θ
sec2 θ − 1 ⋅
dθ
− (1 + sec2 θ )
=−∫
sec θ ⋅ tan 2 θ
dθ
1 + sec2 θ
=−∫
sin 2 θ
dθ
cos θ + cos3 θ
=−∫
1 − cos 2 θ
dθ
cos θ (1 + cos 2 θ )
=−∫
(1 + cos 2 θ ) − 2 cos 2 θ
dθ
cos θ (1 + cos 2 θ )

x
On differentiating w.r.t. x, we get
2 sin x − sin 2 x 2 sin x  1 − cos x
f ′ (x) =
=


x  x2 
x3
t2 + 1
2t
dt
=
2
∫ t 4 + 1 dt
t4 + 1
1+
 2 + 1 − tan 2 x 
1
log 
+ c
2
2
 2 − 1 − tan x 
2 sin x − sin 2 x
6. Given, f (x) = ∫ 
 dx
3

Put tan x = t 2 ⇒ sec2x dx = 2t dt
2t
⇒ dx =
dt
1 + t4
∴
 2 + sin θ 
1
+ c
log 
2 2
 2 − sin θ 
cot2 x − 1 |
+ (2 x1/ 2 − 3x1/3 + 6 x1/ 6 + 11) ln (1 + x1/ 6 )
+ 12 ln (1 + x1/12) − 3 [ln (1 + x1/ 6 )]2 + c
dt
∫ 2 − t 2 , where sin θ = t
= − log|sec θ + tan θ |+ 2 ⋅
= − log|cot x +
cos θ
dθ
2 − sin 2 θ
Topic 3 Integration by Parts
2
1. Let given integral, I = ∫ x5 e− x dx
Put x2 = t ⇒ 2xdx = dt
1
So,
I = ∫ t 2e− t dt
2
1
= [(− t 2e− t ) + ∫ e− t (2t ) dt ] [Integration by parts]
2
1
= [− t 2e− t + 2t (− e− t ) + ∫ 2e− t dt ]
2
1
= [− t 2e− t − 2te− t − 2e− t ] + C
2
e− t 2
(t + 2t + 2) + C
=−
2
2
e− x
(x4 + 2x2 + 2) + C [Q t = x2]
2
Q It is given that,
=−
2
2
I = ∫ x5 e− x dx = g (x) ⋅ e− x + C
By Eq. (i), comparing both sides, we get
1
g (x) = − (x4 + 2x2 + 2)
2
1
5
So,
g(− 1) = − (1 + 2 + 2) = −
2
2
…(i)
Indefinite Integration 291
2. Given, ∫ esec x [(sec x tan x) f (x) + (sec x tan x + sec2 x)]dx
= e sec x ⋅ f (x) + C
On differentiating both sides w.r.t. x, we get
esec x [(sec x tan x) f (x) + (sec x tan x + sec2 x)]
= esec x f ′ (x) + esec x (sec x tan x) f (x)
⇒ esec x (sec x tan x + sec2 x) = esec x f ′ (x)
⇒
f ′ (x) = sec x tan x + sec2 x
So, f (x) = ∫ f ′ (x)dx = ∫ (sec x tan x + sec2 x)dx
x+
=∫e
x+
=∫e
So, possible value of f (x) from options, is
1
f (x) = sec x + tan x + .
2
Put
∴
1
= x cos(log e x) − ∫ x(− sin(log e x)) ⋅ dx
x
[using integration by parts]
= x cos(log e x) + ∫ sin(log e x) dx
1
[using integration by parts]
1  te−4t e−4t 
+C
= 
+
3  −4
−16 
e−4x
[4x3 + 1] + C
48
5.
[Q t = x3 ]
f (x) = −1 − 4x3 (comparing with given equation)

1
x+
∫ 1 + x − x e
1
x dx
x+
1
x dx
+
x+
1
x dx
+ xe
=∫e
=∫e
1
∫
1 x+

x 1 − 2 e x dx

x 
x+
1
x
1
−∫
1
x
dx
x+
dx + xe
1
x
− ∫ ex
x+
1
x
x+
dx = xe
1
x
+c
I = ∫ x5 f (x3 ) dx
x3 = t
dt
x2dx =
3
…(i)
1
t f (t ) dt
3∫
1
d
 
= t ⋅ ∫ f (t ) dt − ∫  (t ) ∫ f (t ) dt  dt 
3
 dt
 
I=


2x + 2
 dx
I = ∫ sin −1 
 4x2 + 8 x + 13 


7. Let
=
∫ sin


 dx
 (2x + 2)2 + 9 


−1 
2x + 2
Put 2x + 2 = 3 tan θ ⇒ 2 dx = 3 sec2θ dθ

3 tan θ  3
∴ I = ∫ sin −1 
⋅ sec2 θ dθ
 9 tan 2 θ + 9  2


 3 tan θ  3 2
= ∫ sin −1 
 ⋅ sec θ dθ
 3 secθ  2
 sin θ  3 2
= ∫ sin −1 
 ⋅ sec θ dθ
 cos θ ⋅ sec θ  2
3
∴
x+
−∫e
[integration by parts]
1
= [t ψ (t ) − ∫ ψ (t ) dt ]
3
1 3
[from Eq. (i)]
= [x ψ (x3 ) – 3 ∫ x2ψ (x3 ) dx] + c
3
1
= x3 ψ (x3 ) − ∫ x2ψ (x3 ) dx + c
3
∫ x(cos(log e x)) x dx
[again, using integration by parts]
⇒
I = x cos(log e x) + x sin((log e x) − I
x
I = [cos(log e x) + sin(log e x)] + C.
⇒
2
1 −4 x 3
5 −4 x 3
4. Given, ∫ x e
dx =
e
f (x) + C
48
In LHS, put x3 = t
⇒
3x2dx = dt
3
1
So, ∫ x5 e−4x dx = ∫ t e−4t dt
3
1  e−4t
e−4t 
dt 
= t
−∫
3  −4
−4

=−
1
x
⇒
3. Let I = ∫ cos(log e x)dx
1 −4 t
e [4t + 1] + C
48
1
x
6. Given, ∫ f ( x ) dx = ψ( x )
Let
=−
x+
+ xe
1
1

x+ 
1 x+

Q ∫ 1 − 2 e x dx = e x 

x 


= sec x + tan x + C
= x cos(log e x ) + x sin(log e x ) −
1
x dx
x+
d
(x)e x dx
dx
3
sin −1 (sin θ ) ⋅ sec2θ dθ
2∫
3
3
= ∫ θ ⋅ sec2θ dθ = [θ ⋅ tan θ − ∫ 1 ⋅ tan θdθ]
2
2
3
= [θ tan θ − log sec θ ] + c
2
3
 2 x + 2  2 x + 2
= tan −1 

 ⋅
 3   3 
2
=
2 
 2 x + 2
− log 1 + 
  + c1
 3  

292 Indefinite Integration
θ
1
cos 2θ + sin 2θ
2
4
1
1
= − θ (1 − 2 sin 2 θ ) + sin θ 1 − sin 2 θ
2
2
1
1
−1
= − sin
x (1 − 2 x) +
x 1−x
2
2
2

 2x + 2 3
 2 x + 2 
= (x + 1) tan −1 
 − log 1 + 
  + c1
 3  4
 3 


=−
 2 x + 2 3
2
= (x + 1) tan −1 
 − log (4x + 8 x + 13) + c
 3  4
3


let log 3 + c1 = c


2
 cos θ + sin θ 
cos 2 θ ln 
 dθ
 cos θ − sin θ 
8. I = ∫
[given]
We integrate it by taking parts
 cos θ + sin θ 
ln 
 as first function
 cos θ − sin θ 
=
−
But
From Eqs. (i) and (ii),
4  1
1

I=
− (1 − 2 x) sin −1 x +
x − x2 − x + c
π  2
2

2
= [ x − x2 − (1 − 2 x) sin −1 x ] − x + c
π
10. Let I = ∫
 cos θ + sin θ 
sin 2 θ
ln 

 cos θ − sin θ 
2
1 d   cos θ + sin θ  
ln 
 sin 2 θdθ
2 ∫ dθ   cos θ − sin θ  
d
[ln (cos θ + sin θ ) − ln (cos θ − sin θ )]
dθ
1
(− sin θ − cos θ )
=
. (− sin θ + cos θ ) −
(cos θ + sin θ )
cos θ − sin θ
(cos θ − sin θ ) (cos θ − sin θ ) − (cos θ + sin θ )
(− sin θ − cos θ )
=
(cos θ + sin θ ) (cos θ − sin θ )
=
Applying integration by parts,

 1
−2
=
⋅ ex − ∫ ex ⋅
dx
2
3
+
+
(
x
1
)
(
x
1
)


− 2 ∫ ex ⋅
= ∫ (x + sin x) cos x dx
=
2
=
= x sin x + cos x −
 cos θ + sin θ  1
1
sin 2 θ ln 
 + ln (cos 2 θ ) + c
 cos θ − sin θ  2
2
9. Let I = ∫
=∫
2
π
∫
sin −1 x − cos −1 x
dx
sin −1 x + cos −1 x

π
sin −1 x −  − sin −1 x

2
dx
π
2
π
4

−1
x −  dx = ∫ sin −1 x dx − x + c …(i)
2 sin

2
π
Now, ∫ sin −1 x dx
Put x = sin 2 θ ⇒ dx = sin 2θ
θ cos 2 θ
= ∫ θ ⋅ sin 2 θ dθ = −
+
2
1
2
∫
cos 2 x
+c
4
cos 2 x
+c
4
Topic 4 Integration, Irrational Function
and Partial Fraction
1. Let I = ∫
=∫
2x12 + 5x9
2x12 + 5x9
dx
=
∫ x15 (1 + x− 2 + x− 5 ) 3 dx
(x5 + x3 + 1) 3
2x− 3 + 5x− 6
dx
(1 + x− 2 + x− 5 ) 3
Now, put 1 + x− 2 + x− 5 = t
⇒
(− 2x− 3 − 5x− 6 ) dx = dt
(2x− 3 + 5x− 6 ) dx = − dt
dt
∴ I = − ∫ 3 = − ∫ t − 3 dt
t
⇒
=−
1
cos 2θ dθ
2
∫ (sin 2x) dx
= (x ⋅ sin x − ∫ 1 ⋅ sin x dx) −
2
Therefore, on putting this value in Eq.(i), we get
 cos θ + sin θ  1
1
2
I = sin 2θ ln 
dθ
sin 2 θ
 −
 cos θ − sin θ  2 ∫
2
cos 2 θ
=
= ∫ x cos x dx +
2
2 (cos θ + sin θ )
=
cos 2 θ
cos 2θ
2
1
ex
dx
=
+c
(x + 1)3
(x + 1)2
11. Let I = ∫ (elog x + sin x) cos x dx
cos θ − cos θ sin θ − sin θ cos θ + sin θ + cos θ sin θ
+ cos 2 θ + sin 2 θ + cos θ ⋅ sin θ
=
cos 2 θ − sin 2 θ
2
(x − 1) ex
dx
(x + 1)3
x + 1 − 2 x
 x
 1
2
e dx = ∫ 
I=∫
−
e dx
3 
2
3
+
x
+
+
(
1
)
(
x
1
)
(
x
1
)




1
1
= ∫ ex ⋅
dx − 2 ∫ ex ⋅
dx
(x + 1)2
(x + 1)3
…(i)
d   cos θ + sin θ  
ln 

dθ   cos θ − sin θ  
…(ii)
=
t− 3 + 1
1
+C= 2+C
−3 + 1
2t
x10
+C
2 (x + x3 + 1) 2
5
Indefinite Integration 293
I=∫
2. Let
=∫
cos3 x + cos5 x
dx
sin 2 x + sin 4 x
On putting x = 0, we get
(cos 2 x + cos 4 x) ⋅ cos x dx
(sin 2 x + sin 4 x)
⇒
0=B+C
x3 + 3x + 2
x+1
1
2
−
+
=
(x + 1)2 (x + 1) 2 (x2 + 1) 2 (x + 1) (x2 + 1)2
2
Put sin x = t ⇒ cos x dx = dt
∴
I=∫
[(1 − t 2) + (1 − t 2)2]
dt
t2 + t4
⇒
I=∫
1 − t 2 + 1 − 2t 2 + t 4
dt
t2 + t4
⇒
I=∫
2 − 3t 2 + t 4
dt
t 2 (t 2 + 1)
⇒
∴
x3 + 3x + 2
dx
(x + 1)2 (x + 1)
∴ I=∫
2
=−
…(i)
Using partial fraction for
y2 − 3 y + 2
A
B
=1 +
+
y ( y + 1)
y
y+1
B = − C = 1 /2
[where, y = t 2]
A = 2, B = − 6
y − 3y + 2
2
6
=1 + −
y ( y + 1)
y y+1
2

2
6 
Now, Eq. (i) reduces to, I = ∫ 1 + 2 −
 dt

1 + t 2
t
1
2
3.
Again,
=
x(x + 1) + 2(x + 1)
(x2 + 1)2(x + 1)
=
x
2
+ 2
2
(x + 1)(x + 1) (x + 1)2
2
x
Ax + B
C
+
=
(x + 1) (x + 1) (x2 + 1) (x + 1)
2
⇒
On putting
x = ( Ax + B) (x + 1) + C (x2 + 1)
x = − 1, we get
−1 = 2 C ⇒ C = − 1 / 2
On equating coefficients of x2, we get
0= A+C
⇒
A = − C = 1 /2
x+1
1
dx
(x2 + 1)2
1
1
log|x + 1| + log|x2 + 1|
2
4
1
+ tan −1 x + 2 I1
2
dx
where, I1 = ∫
(x2 + 1)2
⇒I = −
…(i)
x = tan θ
Put
⇒
dx = sec2θ dθ
∴ I1 = ∫
1
sec2θdθ
= ∫ cos 2 θ dθ = ∫ (1 + cos 2 θ )dθ
2
2
2
(tan θ + 1)
1
1

θ + sin 2θ

2 
2
1
1
tan θ
= θ+ ⋅
2
2 (1 + tan 2 θ)
=
2
= t − − 6 tan −1 (t ) + c
t
2
= sin x −
− 6 tan −1 (sin x) + c
sin x
x3 + 3x + 2
x3 + 2x + x + 2
= 2
2
2
(x + 1) (x + 1) (x + 1)2(x + 1)
dx
∫ x + 1 + 2 ∫ x2 + 1 dx + 2 ∫
=
1
1
x
tan −1 x + ⋅
2
2 (1 + x2)
From Eq. (i),
1
3
1
x
I = − log|x + 1| + log|x2 + 1|+ tan −1 x + 2
+c
2
2
4
x +1
4. Let
I=∫
(x + 1)
ex (x + 1)
dx
dx = ∫
x 2
x (1 + xe )
x ex (1 + xex )2
Put 1 + xex = t ⇒ (ex + x ex ) dx = dt
 1
dt
1 1
∴
I=∫
=
− −
dt
(t − 1)t 2 ∫  t − 1 t t 2 
1
= log|t − 1| − log|t| + + c
t
t − 1 1

= log
+ +c
 t  t
 xex 
1
+
+c
= log 
x
x
1 + xe  1 + xe
12
Definite Integration
Topic 1 Properties of Definite Integral
Objective Questions I (Only one correct option)
1.
π
∫−π|π −|x||dx is equal to
(2020 Main, 3 Sep I)
2π 2
(a)
(b) 2 π 2
π2
(d)
2
(c) π 2
8. The value of ∫
π −1
(a)
2
∫π / 6
2
2
If f (x + 5) = g (x), then ∫ f (t )dt equals
2
7
(b)
18
9
(d)
2
1
(c) −
18
3. A value of α such that
α
dx
 9
= log e   is
 8
(x + α ) (x + α + 1)
(a) − 2
4. If ∫
π /2
0
(b)
1
2
(c) −
1
(a) −
2
1
(c)
2
(2019 Main, 12 April II)
1
2
(d) 2
(b) 1
(d)
π−2
4
∫ g (t )dt
(b)
x+5
10. If f (x) =
(d) −1
π /3
π /6
(b) 37/ 6 − 35 / 6
(d) 34/3 − 31/3
2π
∫ [sin 2x (1 + cos 3x)] dx, where [t ] denotes
0
the greatest integer function, is
(2019 Main, 10 April I)
(c) − 2π
(b) 2π
1
−1
(d) π
7. The value of the integral ∫ x cot (1 − x + x )dx is
2
4
0
π 1
− log e 2
4 2
π
(c)
− log e 2
4
(2019 Main, 9 April II)
π 1
− log e 2
2 2
π
(d)
− log e 2
2
(b)
(c) 2
5
∫ g (t )dt
5
(d)
5
∫ g (t )dt
x+5
2 − x cos x
and g (x) = log e x, (x > 0) then the
2 + x cos x
π/ 4
−π/ 4
g ( f (x))dx is
(2019 Main, 8 April I)
(a) log e 3
(c) log e 2
(b) log e e
(d) log e 1
11. Let f and g be continuous functions on
[0, a] such that f (x) = f (a − x) and g (x) + g (a − x) = 4,
a
then ∫ f (x) g (x) dx is equal to
(2019 Main, 12 Jan I)
a
(a) 4∫ f (x) dx
(b) ∫ f (x) dx
(c) 2∫ f (x) dx
(d) − 3∫ f (x) dx
0
a
sec 2/ 3 x cosec4/3 x dx is equal to
(a) 35 / 6 − 32/ 3
(c) 35 /3 − 31/3
(2019 Main, 8 April II)
x+5
0
(2019 Main, 10 April II)
(a)
∫ g (t )dt
(a) 5
a
5. The integral ∫
(a) − π
0
x+5
5
value of the integral ∫
cot x
dx = m(π + n ), then m ⋅ n is equal to
cot x + cosec x
(2019 Main, 12 April I)
6. The value of
(2019 Main, 9 April I)
x
3
(2020 Main, 4 Sep II)
∫
π −1
(c)
4
0
+ 3 tan x ⋅ sin 6x) dx is equal to
α+1
π−2
(b)
8
x
tan x ⋅ sin 3x(2 sec x ⋅ sin 3x
1
(a) −
9
0
sin3 x
dx is
sin x + cos x
9. Let f (x) = ∫ g (t )dt, where g is a non-zero even function.
2. The integral
π /3
π/ 2
0
a
0
0
e
12. The integral ∫  
1  e

x
2x
x
 e 
−    log e x dx is
 x

equal to
3
1
(a)
−e− 2
2
2e
1
1
(c)
−e− 2
2
e
13. The integral ∫
(2019 Main, 12 Jan II)
1 1
1
+ −
2 e 2e2
3 1
1
(d)
− −
2 e 2e2
(b) −
π /4
π /6
dx
equals
sin 2x(tan5 x + cot5 x)
1 π
1 
(a)  − tan − 1 



5 4
3 3
(c)
1π
−1 1 
 − tan 

 9 3
10  4
(2019 Main, 11 Jan II)
1
1 
(b)
tan − 1 


20
9 3
(d)
π
40
Definite Integration 295
14. The value of the integral ∫
sin 2 x
dx
−2  x 
1
+
 π  2
2
24. The value of the integral ∫
is
(where, [x] denotes the greatest integer less than or equal
to x) is
(2019 Main, 11 Jan I)
(a) 4 − sin 4
(b) 4
(c) sin 4
(d) 0
(2012)
π2
(b)
−4
2
(a) 0
25. The value of
b
15. Let I = ∫ (x4 − 2x2) dx. If I is minimum, then the ordered
(a)
a
pair (a , b) is
(2019 Main, 10 Jan I)
(a) (− 2 , 0)
(c) ( 2 , − 2 )
16. If ∫
π /3
0
(b) (0, 2 )
(d) (− 2 , 2 )
0
(2019 Main, 9 Jan II)
1
(b)
2
(d) 4
(a) 1
(c) 2
π
(2019 Main, 9 Jan I)
0
(a)
2
3
(b) −
18. The value of ∫
π
(a)
8
19.
3 π/ 4
∫π / 4
4
3
(d)
sin 2 x
dx is
+ 2x
π /2
π
(d)
4
(c) 4π
(b) 2
x2 cos x
dxis equal to
−π / 2 1 + ex
π
−2
4
(c) π 2 − e− π / 2
2
22. The integral ∫
∫0
π /2
π /4
(c) 1
log( 1 + 2)
(eu − e−u )17 du
∫0
(c)
∫0
log( 1 + 2)
∫0
π
0
(c) 4 3 − 4
1
0
(d) 1
1−x
dx is
1+ x
(c) −1
(2004, 1M)
(d) 1
1
(d) log  
 2
(c) 1
cos 2 x
dx, a > 0,is
−π 1 + a x
π
(2001, 1M)
(b) aπ
(d) 2π
ecos x sin x , for | x|≤ 2
2 ,
otherwise

then ∫
3
−2
(2000, 2M)
f (x) dx is equal to
(b) 1
(c) 2
(d) 3
log x
31. The value of the integral ∫ −1  e  dx is
e 
x 
(a) 3 / 2
(b) 5 / 2
2π
− 4− 4 3
3
(d) 4 3 − 4 − π /3
(d) 5
than or equal to y, then the value of the integral
3π/2
(1999, 2M)
∫ π / 2 [2 sin x] dx is
3 π /4
∫ π /4
(a) 2
x
dx is equal to
2
(2014 Main)
(c) 3
(2000, 2M)
32. If for a real number y, [ y] is the greatest integer less
(2014 Adv.)
−u 16
(b)
1
3
log
6
2
30. If f (x) = 
33.
x
2
(d)

 1 + x 
 [x] + log 
  dx equals (2002, 1M)
 1 − x 

(a) − π
π
(c) −
2
23. The integral ∫ 1 + 4 sin 2 − 4 sin
(a) π − 4
(d) 6
2(e − e ) du
u
1/ 2
−1/ 2
(a) π
π
2
−u 16
(eu + e−u )17 du
(b)
3
2
(2011)
e2
(2 cosec x)17dx is equal to
log( 1 + 2)
π
−1
2
(b) 0
29. The value of ∫
2(e + e ) du
u
(b)
1
2
(a) 0
log x2
dx is equal to
log x2 + log(36 − 12x + x2)
(2015, Main)
(b) 4
log( 1 + 2)
(a) −
(b)
21. The integral ∫
π
+1
2
28. The integral ∫
(2016 Adv.)
π
+ 2
4
(d) π 2 + eπ / 2
(c) log
(c) 4
27. The value of the integral ∫
2
4
1
3
log
2
2
(b)
(b) 3
π /2
(a)
x sin x2
dx is
log 2 sin x2 + sin (log 6 − x2 )
+ 3x2 + 3x + 3 + (x + 1) cos (x + 1)] dx is (2005, 1M)
3
(a) 0
(2017 Main)
(d) − 1
π2
(d)
2
log 3
(c)
(c) 4
2
(a) 2
4
3
(2018 Main)
dx
is equal to
1 + cos x
20. The value of ∫
(d)
(c) 0
−π / 2 1
π
(b)
2
(a) − 2
(a)
∫ −2 [x
(a)
17. The value of ∫ |cos x|3 dx is
1
3
log
4
2
∫
π2
(c)
+ 4
2
26. The value of
tan θ
1
, (k > 0), then the value of k
dθ = 1 −
2k sec θ
2
is
 2
π − x
 x + log
 cos x dx

π + x
π /2
− π /2
(b) 0
π
(d)
2
dx
is equal to
1 + cos x
(b) −2
(1999, 2M)
1
(d) −
2
1
(c)
2
34. Let f (x) = x − [x] , for every real number x, where [x] is
the integral part of x. Then,
1
∫ −1 f (x) dx is
(a) 1
(b) 2
(c) 0
(d) −
1
2
(1998, 2M)
296 Definite Integration
x
35. If g (x) = ∫ cos 4 t dt , then g (x + π ) equals
(1997, 2M)
0
(a) g (x) + g ( π )
(b) g (x) − g ( π )
g (x )
g( π)
(c) g (x) g ( π )
(d)
36. Let f be a positive function.
If I1 = ∫
I2 = ∫
k
1−k
(a) Statement I is correct; Statement II is correct;
Statement II is a correct explanation for Statement I
(b) Statement I is correct; Statement II is correct;
Statement II is not a correct explanation for
Statement I
(c) Statement I is correct; Statement II is false
(d) Statement I is incorrect; Statement II is correct
x f [x (1 − x)] dx and
k
Passage Based Questions
f [x (1 − x)] dx, where 2k − 1 > 0.
1−k
I
Then, 1 is
I2
Passage I
(1997C, 2M)
(a) 2
(c) 1/2
(b) k
(d) 1
37. The value of
2π
∫0
44. The correct statement(s) is/are
[ 2 sin x ] dx, where [.] represents the
greatest integral functions, is
5π
(a) −
3
(b) − π
Let F : R → R be a thrice differentiable function. Suppose that
F (1) = 0, F (3) = − 4 and F ′ (x) < 0 for all x ∈ (1, 3). Let
(2015 Adv.)
f (x) = xF (x) for all x ∈ R.
(1995, 2M)
5π
(c)
3
(d) −2π
πx
38. If f (x) = A sin   + B,
 2 
1
2A
1
f ′   = 2 and ∫ f (x) dx =
, then constants
0
 2
π
A and B are
(a) f ′ (1) < 0
(b) f (2) < 0
(c) f ′ (x) =/ 0 for any x ∈ (1, 3) (d) f ′ (x) = 0 for some x ∈ (1, 3)
3
45. If ∫ x2 F ′ (x) dx = − 12 and
1
39. The value of ∫
π /2
0
4
(c) 0 and −
π
4
(d) and 0
π
dx
is
1 + tan3 x
(a) 0
(c) π / 2
(b) ∫ f (x) dx = 12
(c) 9f ′ (3) – f ′ (1) + 32 = 0
(d) ∫ f (x) dx = − 12
Then, the value of the integral
(b) 1
F (c) =
(c) −1
cos 2 x
∫0 e
(1990, 2M)
(d) 0
a+b
2
b
b−a
∫ a f (x) dx = 4 { f (a ) + f (b) + 2 f (c)} dx
t
cos (2n + 1) x dx has the value
3
(a) π
(c) 0
(1985, 2M)
(a) π / 4
(c) π
π /2
0
cot x
dx is
cot x + tan x
(b) π / 2
(d) None of these
46. If lim
∫ a f (x) dx −
t→ a
(t − a )
{ f (t ) + f (a )}
2
= 0,
(t − a )3
(1983, 1M)
(a) 0
(c) 3
(b) 1
(d) 2
47. If f ′ ′ (x) < 0, ∀x ∈ (a , b), and (c, f (c)) is point of maxima,
where c ∈ (a , b), then f ′ (c) is
f (b) − f (a )
b− a
f (b) − f (a ) 

(c) 2
 b − a 
(b) 3 

(a)
Assertion and Reason
43. Statement I The value of the integral
π /3
∫π / 6
1+
Statement II
dx
is equal to π /6 .
tan x
b
b
∫a f (x) dx = ∫a f (a + b − x) dx
(2006, 6M)
then degree of polynomial function f (x) atmost is
(b) 1
(d) None of these
42. The value of the integral ∫
c−a
b−c
[ f (a ) − f (c)] +
[ f (b) − f (c)]
2
2
When c =
41. For any integer n, the integral
π
 b − a
 { f (a ) + f (b)},
2 
∫ a f (x) dx = 
for more acurate results for c ∈ (a , b),
(b) 1
(d) π / 4
(a) π
3
1
Passage II
b
f (− x)] [ g (x) − g (− x)] dx is
1
For every function f (x) which is twice differentiable, these
will be good approximation of
(1993, 1M)
∫ −π / 2 [ f (x) +
3
(a) 9f ′ (3) + f ′ (1) − 32 = 0
40. Let f : R → R and g : R → R be continuous functions.
π/ 2
F ′ ′ (x) dx = 40, then the
correct expression(s) is/are
(1995, 2M)
π
π
2
3
(b) and
(a) and
2
2
π
π
3 3
∫1 x
(2013 Main)
48. Good approximation of ∫
(a) π / 4
(c) π ( 2 + 1) / 8
f (b) − f (a ) 
b − a 
(d) 0
π /2
0
sin x dx, is
(b) π ( 2 + 1) / 4
π
(d)
8
Definite Integration 297
Integer & Numerical Answer Type Questions
Objective Questions II
(One or more than one correct option)
56. Let [t ] denote the greatest integer less than or equal to t.
49. Let f : R → (0, 1) be a continuous function. Then, which
of the following function(s) has (have) the value zero at
(2017 Adv.)
some point in the interval (0, 1) ?
x
(a) ex − ∫ f (t ) sin t dt
(c) x −
0
π
−x
2
∫ f (t ) cos t dt
(b) f (x) +
π
2
2
Then the value of ∫ |2x − [3x]|dx is ……… .
1
(2020 Main, 2 Sep II)
57. The value of the integral
π /2
∫ f (t ) sin t dt
∫
0
0
(d) x9 − f (x)
3 cos θ
d θ equals …………
( cos θ + sin θ )5
58. If I =
0
k+1
50. If I = ∑ k = 1
98
∫k
k+1
dx , then
x(x + 1)
(2017 Adv.)
(a) I > log e 99
(b) I < log e 99
49
(c) I <
50
(d) I >
 π π
x ∈  – ,  . Then, the correct expression(s) is/are
 2 2
π/4
1
12
π/4
1
(c) ∫ x f (x) dx =
0
6
0
x f (x) dx =
1
1/ 2
π/ 4
0
(d) ∫
f (x) dx = 0
π/ 4
0
(2015 Adv.)
f (x) dx = 1
3
192x
 1
for all x ∈ R with f   = 0. If
 2
2 + sin 4 πx
52. Let f ′ (x) =
m≤∫
(b) ∫
f (x) dx ≤ M , then the possible values of m and M
are
(1 + e
sin x
− π/ 4
4π
equation
4
e4 π − 1
eπ − 1
(b) a = 2, L =
e4 π + 1
eπ + 1
(c) a = 4, L =
e4 π − 1
eπ − 1
(d) a = 4, L =
e4 π + 1
eπ + 1
0
22
−π
7
55. If I n = ∫
π
−π
(b)
10
∑ I2 m = 0
m =1
x4 (1 − x)4
dx is (are)
1 + x2
2
105
(c) 0
(2010)
(d)
71 3 π
−
15 2
sin nx
dx , n = 0 , 1 , 2 ,... , then
(1 + π x ) sin x
(a) I n = I n + 2
1+ 3
dx is
((x + 1)2 (1 − x)6 )1/ 4
(2018 Adv.)
where tan −1 x takes only principal values, then the
3π 

value of  log e|1 + α | −
(2015 Adv.)
 is

4
 d2

(1 − x2)5  dx is
2
dx


(2014 Adv.)
Fill in the Blanks
(a) a = 2, L =
1
1/ 2
0
[x], x ≤ 2
60. Let f : R → R be a function defined by f (x) = 
,
0 , x > 2
where [x] denotes the greatest integer less than or equal
2
xf (x2)
to x. If I = ∫
dx, then the value of (4I − 1)
−1 2 + f (x + 1 )
is
(2015 Adv.)
2
 12 + 9x 
1
−1
61. If α = ∫ (e9x + 3 tan x ) 
 dx,
0
 1 + x2 
0
∫0 e (sin at + cos at )dt = L, is/are
π t
6
4
∫0 e (sin at + cos at )dt
(2015 Adv.)
54. The value(s) of ∫
(2019 Adv.)
59. The value of the integral ∫
1
6
t
dx
then 27I 2 equals .........
) (2 − cos 2x)
62. The value of ∫ 4x3 
53. The option(s) with the values of a and L that satisfy the
(c)
∫
(2015 Adv.)
(a) m = 13, M = 24
1
1
(b) m = , M =
4
2
(c) m = − 11, M = 0
(d) m = 1, M = 12
(a)
π/ 4
.......
49
50
51. Let f (x) = 7 tan 8 x + 7 tan 6 x – 3 tan 4 x – 3 tan 2 x for all
(a) ∫
2
π
(2019 Adv.)
10
(b)
∑ I2 m + 1 = 10 π
m =1
(d) I n = I n + 1
(2009)
63. Let f : [1, ∞ ] → [2, ∞ ] be differentiable function such
x
that f (1) = 2. If 6∫ f (t ) = dt = 3x f (x) − x3 , ∀ x ≥ 1 then
1
the value of f (2) is ....
(2011)
d
e sin x
64. Let
F (x) =
, x > 0.
dx
x
2
4 2 e sin x
If ∫
dx = F (k) − F (1), then one of the possible
1
x
values of k is ..… .
(1997, 2M)
37 π π sin (π log x)
65. The value of ∫
dx is …… .
1
x
(1997, 2M)
66. For n > 0, ∫
x sin 2n x
dx = …… .
sin x + cos 2n x
2π
2n
0
(1996, 2M)
1
1
67. If for non-zero x, af (x) + bf   = − 5, where a ≠ b,
 x x
2
then ∫ f (x) dx = …… .
1
68. The value of ∫
3
2
x
5−x+
(1996, 2M)
x
dx is …… .
(1994, 2M)
298 Definite Integration
69. The value of ∫
3 π /4
x
dx …… .
1 + sin x
π /4
76. Evaluate
(1993, 2M)
π + 4x3
dx.
π

2 − cos |x|+ 

3
π/ 3
∫ −π / 3
2
70. The value of ∫ |1 − x2| dx is … .
−2
1.5
∫0
71. The integral
(1989, 2M)
2
[x ] dx, where [.] denotes the greatest
function, equals …… .
(1988, 2M)
77. If
is
f
π /2
∫0
an
Column I
1
dx
∫ −1 1 + x 2
A.
1
∫0
B.
3
∫2
C.
2
∫1
D.
P.
Q.
2
2 log  
 3
dx
1− x 2
dx
1− x
dx
R.
π
3
x x2 −1
S.
π
2
2
8
1
≤ 2, satisfying f( 0) = 0 and ∫ f( x ) dx = 1, is
Q. The number of points in the interval
(ii)
[− 13, 13 ] at which f( x ) = sin( x 2 ) + cos( x 2 )
attains its maximum value, is
∫−2 1 + e x dx equals
(iii)
 1/ 2
1 + x  
 ∫ cos 2 x log 
 dx 
− 1/ 2
1− x  

equals
 1/ 2
1 + x  
dx 
 ∫ cos 2 x log 

1− x  
 0
(iv)
2
4
0
(1998, 8M)
log (1 + tan x) dx.
π
∫ −π
2x (1 + sin x)
dx. (1995, 5M)
1 + cos 2 x
 x4 
 2x 
 cos −1 

 dx.
4
 1 + x2 
1 − x 
(1995, 5M)
2x5 + x4 − 2x3 + 2x2 + 1
dx.
(x2 + 1) (x4 − 1)
(1993, 5M)
1/ 3
3
∫2
83. Evaluate
3
has relative
minimum / maximum at x = − 1 and x = 1/3.
1
If ∫ f (x) dx = 14 / 3, find the cubic f (x).
(1992, 4M)
85. Evaluate ∫
π

π
x sin (2x) sin  cos x

2
0
2x − π
86. Show that ,
π /2
∫0
f (sin 2x) sin x dx = 2
π /4
∫0
dx .
(1991, 4M)
f (cos 2x) cos x dx.
(1990, 4M)
sin 2kx
= 2 [cos x + cos 3x + K + cos(2k − 1) x]
sin x
Hence, prove that
π /2
∫0
sin 2kx ⋅ cot x dx = π /2.
(1990, 4M)
88. If f and g are continuous functions on [0, a ] satisfying
f (x) = f (a − x) and g (x) + g (a − x) = 2, then show that
a
π |cos x |
1

1

 dx = 2 ∫ tan −1 x dx.
0
 1 − x + x2 
(1997C, 2M)
∫ −1/
(5050) ∫
1
∫0
1
0
(1 − x50 )100 dx
(1 − x50 )101 dx
a
∫ 0 f (x) g(x) dx = ∫ 0 f (x) dx.
Analytical & Descriptive Questions
∫ 0e
(1999, 3M)
87. Prove that for any positive integer k,
P Q R S
(a) (iii) (ii) (iv) (i)
(b) (ii) (iii) (iv) (i)
(c) (iii) (ii) (i) (iv)
(d) (ii) (iii) (i) (iv)
75. Evaluate
dx.
−1
Codes
74. The value of
(2003, 2M)
84. A cubic f (x) vanishes at x = − 2 and
0
S.
π /4
0
List II
3x 2
f (sin 2x) cos x dx.
82. Evaluate the definite integral
(2014)
(i)
that
−1 
81. Determine the value of
List I
2
prove
0
using codes given below the lists.
R.
1
80. Integrate ∫
The number of polynomials f( x ) with
non-negative integer coefficients of degree
0
then
Hence or otherwise, evaluate the integral
1
−1
2
∫ tan (1 − x + x ) dx.
73. Match List I with List II and select the correct answer
P.
+ e− cos x
∫ 0 tan
Column II
1
2
log  
 3
2
e
0 ecos x
79. Prove that
π /4
cos x
π
72. Match the conditions/expressions in Column I with
statement in Column II.
function,
f (cos 2x) cos x dx = 2 ∫
78. Evaluate ∫
Match the Columns
even
(2004, 4M)
89. Prove
is
(2006, 6M)

1

1

2 sin  2 cos x + 3 cos  2 cos x  sin x dx.


(2005, 2M)
2a
∫0
that
the
value
of
the
[ f (x) /{ f (x) + f (2a − x)}] dx is equal to a.
90. Evaluate
1
∫ 0 log[
91. Evaluate ∫
π
01
(1 − x) +
(1 + x)] dx.
x dx
, 0 < α < π.
+ cos α sin x
(1989, 4M)
integral,
(1988, 4M)
(1988, 5M)
1
(1986, 2 M)
2
Definite Integration 299
92. Evaluate ∫
93. Evaluate
π /2
0
1/ 2
∫0
94. Evaluate ∫
π /4
0
x sin x cos x
dx.
cos 4 x + sin 4 x
x sin
−1
1−x
x
2
1
(1985, 2 M)
2
dx.
(1984, 2M)
sin x + cos x
dx.
9 + 16 sin 2x
π
95. (i) Show that ∫ x f (sin x) dx =
0
π
2
(1983, 3M)
π
∫ 0 f (sin x) dx.(1982, 2M)
(ii) Find the value of
96. Evaluate
1
3/ 2
∫ −1 | x sin πx| dx.
(1982, 3M)
∫ 0 (tx + 1 − x) dx,
n
where n is a positive integer and t is a parameter
independent of x. Hence, show that
1 k
1
n−k
.
∫ 0 x (1 − x) dx = nC k (n + 1), for k = 0, 1,... , n(1981,
4M)
Topic 2 Periodicity of Integral Functions
Objective Questions I (Only one correct option)
(c)
π/ 2
dx
1. The value of ∫
, where [t ] denotes
− π / 2 [x] + [sin x] + 4
the greatest integer less than or equal to t, is
(2019 Main, 10 Jan II)
1
(a)
(7 π − 5)
12
3
(c)
(4 π − 3)
10
1
(b)
(7 π + 5)
12
3
(d)
(4 π − 3)
20
3
I
2
(c) 3I
f (2x) dx is
(2002,1M)
(d) 6I
x
3. Let g (x) = ∫ f (t ) dt, where f is such that
0
t ∈ [0,1] and 0 ≤ f (t ) ≤
nπ + v
0
|sin x|dx = 2n + 1 − cos v, where n is a
(1994, 4M)
every interval on the real line and f (t + x) = f (x), for
every x and a real t, then show that the integral
a+ t
(1984, 4M)
∫ f (x) dx is independent of a.
Integer & Numerical Answer Type Question
6. For any real number x, let [x] denotes the largest
(b) I
(a)
4. Show that ∫
a
0
3
Analytical & Descriptive Questions
5. Given a function f (x) such that it is integrable over
continuous function such that for all
T
x ∈ R. f (x + T ) = f (x). If I = ∫ f (x) dx,
3 + 3T
(d) 2 < g(2) < 4
positive integer and 0 ≤ v < π .
2. Let T > 0 be a fixed real number. Suppose, f is a
then the value of ∫
3
< g(2) ≤ 5 / 2
2
1
≤ f (t ) ≤ 1 for
2
1
for t ∈ [1, 2]. Then, g(2) satisfies
2
the inequality
integer less than or equal to x. Let f be a real valued
function defined on the interval [− 10, 10] by
if f (x) is odd
 x − [x[,
f(x) = 
1
[
[
if
+
x
−
x,
f (x) is even

π 2 10
(2010)
Then, the value of
f (x) cos πx dx is……
10 ∫− 10
(2000, 2M)
1
3
(a) − ≤ g(2) <
2
2
(b) 0 ≤ g(2) < 2
Topic 3 Estimation, Gamma Function and
Derivative of Definite Integration
Objective Questions I (Only one correct option)
x
1. If ∫ f (t ) dt = x2 +
0
 1
∫x t f (t )dt , then f′  2 is
1 2
(2019 Main, 10 Jan II)
24
25
6
(c)
25
(a)
18
25
4
(d)
5
(b)
2. Let f : [0, 2] → R be a function which is continuous on
[0, 2] and is differentiable on (0, 2) with f (0) = 1.
Let F (x) = ∫
x2
0
f ( t ) dt, for x ∈ [0, 2]. If F ′ (x) = f ′ (x), ∀
x ∈ (0, 2), then F (2) equals
(a) e2 − 1
(c) e − 1
(2014 Adv.)
(b) e4 − 1
(d) e4
3. The intercepts on X-axis made by tangents to the curve,
x
y = ∫ | t | dt , x ∈ R, which are parallel to the line y = 2x,
0
are equal to
(a) ± 1
(c) ± 3
(2013 Main)
(b) ± 2
(d) ± 4
300 Definite Integration
4. Let f be a non-negative function defined on the interval
[0, 1].
x
x
∫0
If
1 − ( f ′ (t ))2 dt = ∫ f (t ) dt , 0 ≤ x ≤ 1
0
f (0) = 0 , then
1
1
(a) f   <
 2 2
1
1
(b) f   >
 2 2
1
1
(c) f   <
 2 2
1
1
(d) f   >
 2 2
5. If
1
∫ sin x
and
(2009)
1
1
and f   >
 3 3
1
1
and f   >
 3 3
1
1
and f   <
 3 3
1
1
and f   <
 3 3
(1998, 2M)
1
(a)
2
(b) 0
(c) 1
(d) −
Then, the value of lim ∫
x→1
f ( x)
4
(a) 8 f ′(1)
(c) 2 f ′(1)
1
2
2t
dt is
x −1
(b) 3
(d) None of these
6. If f (x) is differentiable and
4
f   equals
 25
t2
(2005, 1M)
2
∫ 0 x f (x) dx = 5 t
5
,
then
(2004, 1M)
(1990, 2M)
(b) 4 f ′(1)
(d) f ′(1)
1
is
2
13. The value of the definite integral ∫ (1 + e−x ) dx is
0
(a) −1
(b) 2
(c) 1+ e−1
(d) None of the above
(1981, 2M)
Objective Question II
(One or more than one correct option)
14. Which of the following inequalities is/are TRUE?
(2020 Adv.)
5
(b) −
2
5
(d)
2
2
(a)
5
(c) 1
7. If f (x) = ∫
1
x
∫ 0 f (t ) dt = x + ∫ x t f (t ) dt, then the value of f (1) is
12. Let f : R → R be a differentiable function and f (1) = 4.
 1
t 2 f (t ) dt = 1 − sin x, ∀ x ∈ (0, π / 2), then f  
 3
(a) 3
(c) 1/3
11.
x 2 + 1 −t 2
e
x2
3
(a) ∫ x cos x dx ≥
0
8
1 2
1
(c) ∫ x cos x dx ≥
0
2
1
dt, then f (x) increases in
(a) (2, 2 )
(c) (0, ∞ )
(2003, 1M)
(b) no value of x
(d) (−∞ ,0)
1
8. If I (m, n ) = ∫ tm (1 + t )n dt, then the expression for
15. If g (x) = ∫
sin( 2x )
sin x
3
(b) ∫ x sin x dx ≥
0
10
1
2
(d) ∫ x2 sin x dx ≥
0
9
1
sin −1 (t ) dt, then
π
(a) g′  −  = 2 π
 2
π
(c) g′   = 2 π
 2
(2017 Adv.)
π
(b) g′  −  = − 2 π
 2
π
(d) g′   = − 2 π
 2
0
I (m, n ) in terms of I (m + 1, n − 1) is
(2003, 1M)
2n
n
(a)
−
I (m + 1, n − 1)
m+1 m+1
n
(b)
I (m + 1, n − 1)
m+1
Let f (x) = (1 − x)2 sin 2 x + x2, ∀ x ∈ R and
x  2 (t − 1 )

g (x) = ∫ 
− ln t f (t ) dt ∀ x ∈ (1, ∞).
1  t+1

16. Consider the statements
2n
n
+
I (m + 1, n − 1)
m+1 m+1
m
(d)
I (m + 1, n − 1)
m+1
(c)
9. Let f (x) = ∫
x
1
2 − t 2 dt. Then, the real roots of the
equation x2 − f '(x) = 0 are
(a) ±1
(c) ±
1
2
(2002, 1M)
1
(b) ±
2
(d) 0 and 1
P : There exists some x ∈ R such that,
f (x) + 2x = 2 (1 + x2).
Q : There exists some x ∈ R such that,
2 f (x) + 1 = 2x (1 + x).
Then,
(a)
(b)
(c)
(d)
both P and Q are true
P is true and Q is false
P is false and Q is true
both P and Q are false
17. Which of the following is true?
x
10. Let f : (0, ∞ ) → R and F (x) = ∫ f (t ) dt.
0
If F (x2) = x2 (1 + x), then f (4) equals
5
(a)
4
Passage Based Questions
(b) 7
(c) 4
(2001, 1M)
(d) 2
(a)
(b)
(c)
(d)
g is increasing on (1, ∞ )
g is decreasing on (1, ∞ )
g is increasing on (1, 2) and decreasing on (2, ∞ )
g is decreasing on (1, 2) and increasing on (2, ∞ )
Definite Integration 301
Fill in the Blank
sec x cos x sec2x + cot x cosec x
18. f (x) = cos 2 x cos 2 x
cosec2x
.
2
2
1
cos x
cos x
Then, ∫
π /2
0
|x|→ ∞ , then show that every line y = mx intersects the
curve y2 +
(1987, 2M)
Analytical & Descriptive Questions
1
(1988, 5M)
Integer & Numerical Answer Type Question
24. Letf : R → R be a differentiable function such that its
derivative f′ is continuous and f (π) = − 6.
x
dg
differentiable function. If
> 0, ∀ x prove that
dx
a
b
∫ g(x) dx + ∫ g(x) dx increases as (b − a ) increases.
0
If F : [0, π ] → R is defined by F (x) = ∫ f (t )dt, and if
0
π
∫ ( f ′ (x) + F (x)) cos x dx = 2
0
(1997, 5M)
then the value of f (0) is ……… .
21. Determine a positive integer n ≤ 5, such that
∫ 0e
x
f (x) = ∫ [2(t − 1)(t − 2)3 + 3(t − 1)2(t − 2)2] dt.
x
20. Let a + b = 4, where a < 2 and let g (x) be a
1 x
(1991, 2M)
1
ln t
dt. Find the function
1+ t
f (x) + f (1 / x) and show that f (e) + f (1 / e) = 1 / 2, where
(2000, 5M)
ln t = log e t.
0
x
∫ 0 f (t ) dt = 2
23. Investigate for maxima and minima the function,
f (x) dx = K .
19. For x > 0, let f (x) = ∫
x
∫ 0 f (t ) dt → ∞ as
22. If ‘ f ’ is a continuous function with
(x − 1)n dx = 16 − 6e
(2020 Adv.)
(1992, 4M)
Topic 4 Limits as the Sum
Objective Question I (Only one correct option)
 (n + 1)1/3
(n + 2)1/3
(2n )1/3 
+
+ ..... +
 is equal to
4/3
4/3
n
n 4/3 
 n
1. lim 
n→ ∞
(2019 Main, 10 April I)
3 4/3 4
(2) −
3
4
4 3/ 4
(d)
(2)
3
4 4/3
(2)
3
3
3
(c)
(2)4/3 −
4
4
(a)
(b)

n
n
n
1
+ 2
+ 2
+ ... +
 is equal to
2
2
2
n→ ∞ n + 1
5n 
n +2
n +3
2. lim 
(a)
(b)
(c)
(d)
2
(2019 Main, 12 Jan II)
tan − 1 (3)
tan − 1 (2)
π/4
π/2
(a)
(b)
(c)

18
e4
27
e2
9
e2
(d) 3 log 3 − 2
4. For a ∈ R,|a| > 1, let




3
3
1 + 2 +…+ n


= 54
lim
n→∞ 


1
1
1

 n7/3 
+
+
…
+


 (an + 1)2 (an + 2)2
(an + n )2 

Then the possible value(s) of a is/are
(2020 Adv.)
(a) −6
(b) 7
n

1/ n
(d) −9
(c) 8
5. For a ∈ R (the set of all real numbers), a ≠ − 1,
(1a + 2a + K + n a )
1
=
.
n→∞ (n + 1 )
[(na + 1) + (na + 2) + K + (na + n )] 60
Then, a is equal to
lim
a−1
(a) 5
(n + 1)(n + 2) K 3n 
3. lim 

2n
n→ ∞
Objective Questions II
(One or more than one correct option)
(b) 7
(c)
−15
2
(d)
−17
2
n−1
n
n
and Tn = ∑ 2
, for
2
+
+
+
+ k2
n
kn
k
n
kn
k=0
k=0
(2008, 4M)
n = 1, 2, 3, ... , then
n
is equal to
(2016 Main)
6. Let S n = ∑
(a) Sn <
2
π
π
(b) Sn >
3 3
3 3
(c) Tn <
π
3 3
(d) Tn >
π
3 3
Analytical & Descriptive Question
 1
1
1
+
+K+
 = log 6.
n + 1 n + 2
6n 
(1981, 2M)
7. Show that, lim 
n→ ∞
Answers
Topic 1
π
(log 2 )
8
1
1
83. log 6 −
2
10
π
[ π + 3 log (2 + 3 ) − 4 3 ]
12
8
84. f ( x ) = x 3 + x 2 − x + 2
85. 2
π
81. π 2
80.
1. (c)
5. (b)
9. (d)
13. (c)
17. (d)
21. (c)
25. (a)
29. (c)
33. (a)
37. (a)
41. (c)
45. (c, d)
49. (c, d)
53. (a, c)
57. (0.5)
61. (9)
2.
6.
10.
14.
18.
22.
26.
30.
34.
38.
42.
46.
50.
54.
58.
62.
65. (2)
66. π 2
 1
68.  
 2
69. π ( 2 − 1 ) 70. (4)
(c)
(a)
(d)
(d)
(d)
(a)
(c)
(c)
(a)
(d)
(a)
(b)
(b, d)
(a)
(4.00)
(2)
3.
7.
11.
15.
19.
23.
27.
31.
35.
39.
43.
47.
51.
55.
59.
63.
(a)
(a)
(c)
(d)
(b)
(d)
(b)
(b)
(a)
(d)
(d)
(a)
(a, b)
(a, b, c)
(2)
(8/3)
67.
1
a − b2
2
4.
8.
12.
16.
20.
24.
28.
32.
36.
40.
44.
48.
52.
56.
60.
64.
(d)
(c)
(a)
(c)
(a)
(b)
(a)
(c)
(c)
(d)
(a, b, c)
(c)
(*)
(1)
(0)
(k = 16)
7 

a log 2 − 5a + b

2 
71. (2 − 2 )
72. A → S; B → S; C → P; D → R 73. (d)
24
75.
5
74. 5051
78.
π
2
1
1 π
log 2 − +
2
2 4
91.
απ
sin α
92.
 3
1
3π + 1 
1

93.  −
π +  94.
(log 3 ) 95. ( ii )
 π 2 
20

2
 12
Topic 2
1. (d)
2. (c)
3. (b)
6. (4)
2. (b)
6. (a)
10. (c)
3. (a)
7. (d)
11. (a)
4. (c)
8. (a)
12. (a)
14. (a,b,d)
15. (*)
15
π
32
+


18. − 

 60 
16. (c)
1
19.
2
Topic 3
1. (a)
5. (a)
9. (a)
13. (d)
17. (b)
21. (n = 3 )
7
23. At x = 1 and , f ( x ) is maximum and minimum respectively.
5
Topic 4
79. log 2
1. (c)
2. (b)
5. (b, d)
6. (a, d)
3. (b)
4. (c,d)
Hints & Solutions
Topic 1 Properties of Definite Integral
π
1. Given integral ∫ |π −|x|| dx
=∫
π /3
π/ 6
−π
π
π
0
π
0
= 2 ∫ |π −|x|| dx = 2∫ |π − x | dx
= 2 ∫ (π − x) dx
π


 π 2
x2 
π2
= 2  πx −  = 2  π 2 −
= 2   = π2

2 0
2
 2


Hence, option (c) is correct.
=
1
1
1  
(9 × 0) –  × 1  = −
9  
2 
18
α+1
3. Let I =
=
π /3
π /3
π/ 6
=
3
2
∫
dx
(x + α ) (x + α + 1)
∫
(x + α + 1) − (x + α )
dx
(x + α ) (x + α + 1)
∫
 1

1
−

 dx
 x + α x + α + 1
α
α+1
tan3 x ⋅ sin 2 3x(2 sec2 x sin 2 3x + 3 tan x ⋅ sin 6x)dx
π/ 6
=∫
π /3

1
tan 4 x sin 4 3x

2
π/ 6
α
α+1
2. Given integral
∫
 d  tan 4 x sin 4 3x 
 dx
 
2

 dx 
=
[Q x ∈ (0, π)]
0
I=
4
α
[2 tan x sec x sin 3x
+ 3 tan x(2 sin 3x cos 3x)]dx
4
π2
16
24. (4)


 1
 1 e
e cos  2 + 2 sin  2 − 1 


 4π
 1 
76.  tan −1   
 2 
 3
90.
82.
3
α +1
= [ log e (x + α ) − log e (x + α + 1)] α
Definite Integration 303
π /2
α+1
π
1
x

= x − tan
= − 1 = (π − 2)


2 0
2
2

 x+ α 
= log e 

 x + α + 1  α

2α + 1
2α
= log e
− log e
2α + 2
2α + 1
 2α + 1 2α + 1
 9
= log e 
×
 = log e  
 8
 2α + 2
2α 
(given)
⇒
(2 α + 1)2 9
= ⇒ 8 [4α 2 + 4α + 1] = 36 (α 2 + α )
4α (α + 1) 8
⇒
⇒
8α 2 + 8α + 2 = 9α 2 + 9α ⇒ α 2 + α − 2 = 0
(α + 2) (α − 1) = 0 ⇒ α = 1, − 2
π /3
5. Let
From the options we get α = − 2
π /2
cot x
4. Let I = ∫
dx
0
cot x + cosec x
cos x
π / 2 cos x
π /2
sin
x
dx
=∫
dx = ∫
0
0
cos x
1
1 + cos x
+
sin x sin x
π / 2 cos x(1 − cos x)
=∫
dx
0
1 − cos 2 x
=∫
=∫
=∫
0
π /2
0
=
cos 2/ 3
1
dx =
x sin 4/3 x
π /3
sec2 x
∫ (tan x)4/ 3 dx
π/6
and sec2 x dx = dt
I=
So,
3
dt  t − 1/3 
=

t 4/3  − 1 / 3 1/
∫
1/ 3
3
 1

= − 3  1/ 6 − 31/ 6 = 3 ⋅ 31/ 6 − 3 ⋅ 3− 1/ 6
3

(cosec x cot x − cosec x + 1) dx
2
1
2 x
1 − sec  dx

2
2
x cosec4/3 x dx
[multiplying and dividing the denominator by cos 4/3 x]
Put, tan x = t, upper limit, at x = π / 3 ⇒ t = 3 and
lower limit, at x = π / 6 ⇒ t = 1 / 3
(cosec x cot x − cot2 x) dx
π /2
∫
π /6
= 37/ 6 − 35/ 6
6. Given integral
2π
I = ∫ [sin 2x ⋅ (1 + cos 3x)]dx
0
π
= ∫ [sin 2x ⋅ (1 + cos 3x)]dx
0
2π
+ ∫ [sin 2x ⋅ (1 + cos 3x)]dx
π
= I1 + I 2 (let)
[given]
cos x
π / 2 cos x
π /2
sin x
dx
=∫
dx = ∫
0
0
cos x
1
cos x + 1
+
sin x sin x
x
2 cos 2 − 1
π /2
2
=∫
dx
0
x
2 cos 2
2
θ
θ

Q cos θ = 2 cos 2 − 1 and cos θ + 1 = 2 cos 2

2
2 
0
2/3
3
= [− cosec x + cot x + x]π0 / 2
π /2


2 x 
2
sin
−


π /2


cos x − 1 
2 

= x+
= x +

x
x
sin x  0


2 sin cos 
2
2  0

π /2
x
π
1


= x − tan
= − 1 = [π − 2]

2  0
2
2
= m[π + n ]
1
On comparing, we get m = and n = − 2
2
∴
m⋅ n = − 1
Alternate Solution
π /2
cot x
Let I = ∫
dx
0
cot x + cosec x
=∫
∫ sec
π /6
cos 2 x
dx
sin 2 x
π /2
I=
π /3
π / 2 cos x −
0
I = m(π − n )
1
∴ m(π − n ) = (π − 2)
2
On comparing both sides, we get
1
m = and n = − 2
2
1
Now, mn = × − 2 = − 1
2
Since,
Now,
... (i)
2π
I 2 = ∫ [sin 2x ⋅ (1 + cos 3x)]dx
π
let 2π − x = t, upper limit t = 0 and lower limit t = π
and
dx = − dt
0
I 2 = − ∫ [− sin 2x ⋅ (1 + cos 3x)]dx
So,
π
π
= ∫ [− sin 2x ⋅ (1 + cos 3x)]dx
0
π
…(ii)
I = ∫ [sin 2x ⋅ (1 + cos 3x)]dx
∴
0
+
π
∫0 [− sin 2x ⋅ (1 + cos 3x)]dx
[from Eqs. (i) and (ii)]
π
= ∫ (−1)dx] [Q [x] + [− x] = − 1, x ∉Integer]
0
= −π
1
7. Let I = ∫ x cot−1 (1 − x2 + x4 )dx
0
Now, put x2 = t ⇒ 2xdx = dt
304 Definite Integration
Lower limit at x = 0, t = 0
Upper limit at x = 1, t = 1
x
9. Given, f (x) = ∫ g (t )dt
0
1
∴I=
1
cot−1 (1 − t + t 2)dt
2 ∫0
On replacing x by (−x), we get
−x


1
1
= ∫ tan −1 
 dt
2
20
1 − t + t 
1
1

Q cot−1 x = tan −1

x 
1
=
 t − (t − 1) 
1
tan −1 
 dt
∫
 1 + t (t − 1)
20
=
1

1
−1
−1
∫ ( tan t − tan (t − 1)dt 
2 0


y

1
Q ∫ tan −1 (t − 1)dt = ∫ tan −1 (1 − t − 1)dt = − ∫ tan −1 (t )dt
0
a
a
0
So, I =
⇒
f (− x) = − f (x)
⇒ f is an odd function.
Now, it is given that f (x + 5) = g (x)
∴
f (5 − x) = g (− x) = g (x) = f (x + 5)
f (5 − x) = f (x + 5)
0
Put t = u + 5 ⇒ t − 5 = u ⇒ dt = du
t
dt
t2
1
+
0
0
∴
[by integration by parts method]
π 1
π 1
= − [log e (1 + t 2)]10 = − log e 2
4 2
4 2
π /2
0
5 −x
2I = ∫
0
=∫
π
2
0
=∫
π
2
0
∫
5
g (− t )dt =
b
∫ g(t )dt
5 −x
a
[Q − ∫ f (x)dx = ∫ f (x)dx and g is an even function]
a
3
sin x
dx
sin x + cos x
…(i)
I=
b
5
∫ f ′ (t )dt
[by Leibnitz rule f ′ (x) = g (x)]
5 −x
b
b
∫a f (x)dx = ∫a f (a + b − x) dx,
= f (5) − f (5 − x) = f (5) − f (5 + x)
5
5
5+ x
5+ x
=
cos3 x
dx
cos x + sin x
…(ii)
On adding integrals (i) and (ii), we get
π/ 2
−5
5
we get
I=∫
−5
u = −t
du = − dt, we get
Put
⇒
b
On applying the property,
x −5
∫ f (u + 5)du = ∫ g(u )du
I=−
∫a f ( x) dx = ∫a f ( a + b − x) dx
π
2
0
x −5
I=
Key Idea Use property of definite integral.
I=∫
10. The given functions are
g (x) = log e x, x > 0
2 − x cos x
f (x) =
2 + x cos x
nd
(sin x + cos x) (sin 2 x + cos 2 x − sin x cos x)
dx
sin x + cos x
Let
I=∫
Then,
I=∫
π
1


 1

1 − (2 sin x cos x) dx = ∫ 2 1 − sin 2x dx

 2
0 

2
π /4
−π / 4
π /4
−π / 4
g ( f (x))dx
 2 − x cos x 
log e 
 dx
 2 + x cos x
π/ 2
π 1 π −1
I= − =
4 4
4
[from Eq. (i)]
∫ f ′ (t )dt = ∫ g(t )dt
sin3 x + cos3 x
dx
sin x + cos x
π 1
1
π
 1


= x + cos 2x
=  − 0 + (−1 − 1) = −




4
2
2 2
4
0
⇒
…(i)
Let I = ∫ f (t )dt
1
1
Let
0
x
1
( tan −1 t + tan −1 t )dt
2 ∫0
b
0
⇒
0
= ∫ tan −1 tdt = [t tan −1 t ]10 − ∫
8.
x
[Q g is an even function]
because ∫ f (x)dx = ∫ f (a − x)dx
0
1
x
[Q g is an even function]
1
0
0
Now, put t = − u, so
f (− x) = − ∫ g (− u )du = − ∫ g (u )du = − f (x)

−1 x − y
−1
−1
Q tan 1 + xy = tan x − tan

1
∫ g(t )dt
f (− x) =
Now, by using the property
b
∫
a
b
f (x)dx = ∫ f (a + b − x)dx, we get
a
…(i)
Definite Integration 305
I=∫
 2 + x cos x
log e 
 dx
−π / 4
 2 − x cos x 
π /4
⇒ 5 tan 4 x sec 2x dx = dt
…(ii)

 2 − x cos x 
 2 + cos x  
log e 
 + log e 
 dx
−π / 4 
 2 + x cos x
 2 x cos x 

π /4
 2 − x cos x 2 + x cos x
=∫
log e 
×
 dx
−π / 4
 2 + x cos x 2 − x cos x 
2I = ∫
π /4
⇒ 2I = ∫
∴ I=
log e (1)dx = 0 ⇒ I = 0 = log e (1)
−π / 4
a
11. Let I = ∫ f (x) g (x) dx
… (i)
0
a
= ∫ f (a − x) g (a − x) dx
0
Q a f (x) dx =
 ∫0
a
a
I = ∫ f (x) [4 − g (x)] dx
⇒
0

∫0 f (a − x) dx
[Q f (x) = f (a − x) and g (x) + g (a − x) = 4]
a
a
0
0
1 1
⋅
2 5
1
∫(1/
3 )5
a
I = 4 ∫ f (x) dx − I
[from Eq. (i)]
0
a
⇒ 2I = 4 ∫ f (x) dx ⇒
0
e
12. Let I = ∫  
1  e
x
2x

⇒
1 π
−1  1  
 − tan 

9 3
10  4
14. Let I = ∫
sin 2 x
dx
−2 1
x
+
2  π 
sin 2 x
1 x
+
2  π 
Also, let f (x) =
sin 2(− x)
(replacing x by − x)
1  x
+ −
2  π 
Then, f (− x) =
a
I = 2 ∫ f (x) dx.
0
=
sin 2 x
1 
 x 
+ − 1 −
 π 
2 
if

 − [x],
Q [− x] = − 1 − [x], if


 x
  = log t
 e
x (log e x − log e e) = log t
1
  1

x  x + (log e x − log e e) dx = t dt


1
1
⇒ (1 + log e x − 1) dx = dt ⇒ (log e x) dx = dt
t
t
Also, upper limit x = e ⇒ t = 1 and lower limit x = 1 ⇒
1
t=
e
I=∫
1
⇒ I=∫
1
 2 1 1
 t −  ⋅ dt
t t
f (− x) = −
⇒
x ∈ I
x ∉ I 
sin 2 x
= − f (x)
1 x
+
2  π 
i.e. f (x) is odd function
∴ I =0

a
Q ∫ f (x) dx =
−a

0, if f (x) is odd function



 a f (x)dx, f (
if
x)
is
even
function
2
 ∫0

b
15. We have, I = ∫ (x4 − 2x2)dx
1/ e 
1/ e
3 )5
2
⇒
∴
1
dt
(tan −1 (t ))(11/
=
t 2 + 1 10
=
 e
−    log e x dx
 x

x
1
1 
−1
−1  1  
 tan (1) − tan 

9 3
10 
x
 x
Now, put   = t ⇒ x log e
 e
5
=
= ∫ 4 f (x) dx − ∫ f (x) g (x) dx
⇒
π
4
 1 


 3
t
[Q log e A + log e B = log e AB]
π /4
π
6
x
On adding Eqs. (i) and (ii), we get
a
(t − t −2) dt
Let
1
  t 2 1 
1
 1
  1
 3
I =  +   =  + 1 −  2 + e  = − e − 2




t
2
2
2
2
e
2
e


1




f (x) = x4 − 2x2 = x2(x2 − 2) = x2(x − 2 ) (x + 2 )
Graph of y = f (x) = x4 − 2x2 is
Y
y=f(x)
e
13. Let I = ∫
=∫
π /4
π /6
π /4
π /6
dx
sin 2x(tan5 x + cot5 x)
(1 + tan x) tan x
dx
2 tan x (tan10 x + 1)
2
5
1 π / 4 tan 4 x sec 2x
= ∫
dx
2 π / 6 (tan10 x + 1)
Put
tan5 x = t

2 tan x 
Q sin 2x =

1 + tan 2 x 

– √2
X
O
√2 ∴
f(x) < 0 for – √2 < x < √2
–
–
+
+
– √2
0
√2
b
[Qsec2 x = 1 + tan 2 x]
Note that the definite integral ∫ (x4 − 2x2)dx represent
a
the area bounded byy = f (x) , x = a, b and the X -axis.
306 Definite Integration
But between x = − 2 and x = 2 , f (x) lies below the
X-axis and so value definite integral will be negative.
Also, as long as f (x) lie below the X-axis, the value of
definite integral will be minimum.
∴(a , b) = (− 2 , 2 ) for minimum of I.
π /3
1
tan θ
, (k > 0)
16. We have, ∫
dθ = 1 −
0
2
2k sec θ
π /3 tan θ
π /3
tan θ
1
Let I = ∫
dθ =
dθ
∫
0
0
sec θ
2k sec θ
2k
π /3
π /3 sin θ
(sin θ)
1
1
=
dθ =
dθ
∫
∫
cos θ
2k 0
1
2k 0
(cos θ)
cos θ
Let cos θ = t ⇒ − sin θ dθ = dt ⇒ sin θ dθ = − dt
for lower limit, θ = 0 ⇒ t = cos 0 = 1
π
π 1
for upper limit, θ = ⇒ t = cos =
3
3 2
⇒
I=
1
2k
1/ 2 −
∫1
dt
−1
=
2k
t
1/ 2 −
∫1
t

1
1 
1
=−
[2 t ]12
 =−

2k  − 1 + 1
2k
1
 2

2  1
 2 − 1 =
2k 

1
I =1 −
2
2 
1
1
⇒
1 −
 =1 −
2k 
2
2
Q
∴
⇒
π
∴I=
1
2 
1 −


2
2k
(given)
2 2
( cos 3x + 3 cos x) dx
4 ∫0
π
=
1 sin 3x
2
+ 3 sin x
2  3
 0
=
1  1
3π
π  1

− sin 0 + 3 sin 0 
+ 3 sin
 sin

2 3
2
2  3
=
1  1


 (−1) + 3 − [0 + 0]
2 3


3π
π
π



Q sin 2 = sin  π + 2  = − sin 2 = − 1


1
2dt
1
1
− +1 2
t 2 
=−
Now, as cos 3x = 4 cos3 x − 3 cos x
1
cos3 x = (cos 3x + 3 cos x)
∴
4
=
1 1
 4
− +3 =
 3
2  3
π /2
Let
I=
⇒
I=
∫
−π/ 2
a
π π

 − + − x

 2 2
2
1+2
−
π
π
+
−x
2
2
dx
b
 b

Q ∫ f (x)dx = ∫ f (a + b − x)dx
 a

a
π /2
⇒
I=
⇒
I=
⇒
2I =
⇒
2I =
Y
O
a
sin x
dx
1
+ 2x
−π/ 2
∫
π / 2 sin
2 = 2k ⇒ 2k = 4 ⇒ k = 2
X′
b
2
2
=1
2k
17. We know, graph of y = cos x is
b
18. Key idea Use property = ∫ f (x)dx = ∫ f (a + b − x)dx
sin 2 x
∫ 1 + 2−x dx
−π / 2
π /2
π/2
π
X
2x sin 2 x
dx
2x + 1
−π / 2
∫
π /2
 2 x + 1
sin 2 x x
 dx
 2 + 1
−π / 2
∫
π /2
Y′
∴ The graph of y =|cos x|is
Y
∫ sin
2
x dx
−π / 2
π /2
⇒
y=|cos x|
2I = 2 ∫ sin 2 x dx [Qsin 2 x is an even function]
0
π /2
⇒
X′
π
O
Y′
I = ∫ |cos x|3 = 2 ∫
0
π/2
π
I=
π
(Q y = |cos x|is symmetric about x = )
2
π
π




= 2∫ 2 cos3 x dx
Q cos x ≥ 0 for x ∈ 0, 2 
0


2
0
π /2
X
π
2|cos x|3 dx
0
∫ sin xdx
⇒
I=
∫
cos 2xdx
0
π /2
∫ dx
⇒
2I =
⇒
2I = [x]0π / 2
π
I=
4
0
⇒
a
 a

Q ∫ f (x)dx = ∫ f (a − x)dx
 0

0
Definite Integration 307
19. Let I = ∫
π /4
=∫
=∫
3 π / 4 1 − cos x
dx
=∫
dx
π / 4 1 − cos 2 x
1 + cos x
3 π /4
3 π /4
π /4
22.
Given, I = ∫
1 − cos x
dx
sin 2 x
3 π /4
π /4
=∫
(cosec2 x − cosec x cot x)dx
= [− cot x + cosec
x]3π π/ 4/ 4
2
π /2
x cos x
…(i)
dx
+ ex
Q b f (x) dx = b f (a + b − x) dx
∫a
 ∫a

− π /2 1
x2 cos (− x)
⇒ I=∫
dx
−π / 2 1 + e− x
π /2
…(ii)
On adding Eqs. (i) and (ii), we get
π /2 2
 1
1 
x cos x 
+
2I = ∫
 dx
x
−π / 2
+
e
+
e− x 
1
1

=∫
π /2 2
x cos x ⋅ (1) dx
−π / 2
⇒ 2I = 2∫


a
a
Q ∫ f (x) dx = 2 ∫ f (x) dx, when f (− x) = f (x)
0

 − a
π /2 2
0
2I = 2 [x2(sin x) − (2x) (− cos x) + (2) (− sin x)] π0 / 2

π2
− 2
⇒ 2I =2
4


21.
⇒
(− cosec x ⋅ cot x − cosec2x) dx = dt
and
cosec x − cot x = 1 /t
⇒
2 cosec x = t +
I=−∫
∴
⇒
1
217
2+1
⇒
u = ln ( 2 + 1 )
⇒
I=−∫
=2 ∫
log x2
dx
I=∫
2 log x2 + log(36 − 12 x + x2 )
4
2 log x
dx
=∫
2 2 log x + log(6 − x)2
4
2 log x dx
=∫
2 2 [log x + log(6 − x)]
4
log x dx
…(i)
I=∫
2 [log x + log(6 − x)]
4
log(6 − x)
…(ii)
I=∫
dx
2 log(6 − x) + log x
Q b f (x)dx = b f (a + b − x)dx
∫a
 ∫a

4
On adding Eqs. (i) and (ii), we get
4 log x + log(6 − x)
2I = ∫
dx
2 log x + log(6 − x)
4
⇒
2I = 2 ⇒ I = 1
2
dt
t
0
ln ( 2 + 1 )
ln ( 2 + 1 )
0
2(eu + e− u )16
eudu
eu
(eu + e− u )16 du
x − a,
x ≥a
 − ( x − a ), x < a
to break given integral in two parts and then integrate separately.
π
∫0
2
π
x
x

1 − 2 sin  dx = ∫ |1 − 2 sin |dx
0

2
2
π
π 
x
x

= ∫ 3 1 − 2 sin  dx − ∫ π 1 − 2 sin  dx
0 


2
2
3
⇒ 2I = 2
π
x 3 
x
π

=  x + 4 cos  −  x + 4 cos  = 4 3 − 4 −

2 0 
2 π
3
b
2I = ∫ dx = [x]42
16
and when t = 2 + 1, eu = 2 + 1
π
⇒
1

t + 
t

2




When t = 1, eu = 1 ⇒ u = 0
∫a f(x )dx = ∫a f(a + b − x )dx and then add.
⇒
1
t
Let t = eu ⇒ dt = eudu.
PLAN Apply the property
Let
π /4
217 (cosec x)16 cosec x (cosec x + cot x)
dx
(cosec x + cot x)
cosec x + cot x = t
π2
I=
−2
4
b
π /2
(2 cosec x)17 dx
23. PLAN Use the formula,| x − a | = 
x cos x dx
Using integration by parts, we get
∴
π /2
π /4
Let
= [(1 + 2 ) − (− 1 + 2 )] = 2
20. Let I = ∫
PLAN This type of question can be done using appropriate
substitution.
3
24. I = ∫
As,
π /2
− π /2
 2
 π − x 
x + log  π + x  cos x dx


a
∫− a f (x) dx = 0, when f (− x) = − f (x)
∴ I=∫
π /2
− π /2
x2 cos x dx + 0 = 2∫
= 2 {(x2 sin x)π0 / 2 − ∫
π /2
0
π /2
0
(x2 cos x) dx
2x ⋅ sin x dx}
π2
π /2
=2 
− 2 {(− x ⋅ cos x)0π / 2 − ∫
1 ⋅ (− cos x) dx}
0

4



  π2
π2
π2
− 4
=2 
− 2 (sin x)0π / 2 = 2 
− 2 = 


  2
4
4
25. Put x2 = t
⇒ x dx = dt /2
∴
dt
2
I=∫
log 2 sin t + sin (log 6 − t )
log 3
sin t ⋅
...(i)
308 Definite Integration
b
b
∫a f (x) dx = ∫a f (a + b − x) dx
Using,
sin (log 2 + log 3 − t )
∫log 2 sin (log 2 + log 3 − t ) + sin dt
(log 6 − (log 2 + log 3 − t ))
1 log 3
sin (log 6 − t )
= ∫
dt
2 log 2 sin (log 6 − t ) + sin (t )
1
=
2
sin (log 6 − t )
dt
sin (log 6 − t ) + sin t
log 3
log 2
2I = ∫
=∫
…(ii)
0
−2
=∫
−2
0
⇒
2I = π
30. Given,
[x3 + 3x2 + 3x + 3 + (x + 1) cos (x + 1)] dx
∴
I=∫
=∫
1
−1
t dt + 2 ∫
3
−1
1
−1
dt +
1
∫ −1 t cos t dt
=∫
1 1−x
1−x
dx = ∫
dx
0
1+ x
1 − x2
1
1
1
0
1−x
2
dx − ∫
1
x
0
1−x
2
= (sin
1 − sin
−1
0
dx = [sin −1 x] 10+
∫1
31.
e2
∫e
−1
−1/ 2
=∫
−1/ 2
=∫
1/ 2
0
−1/ 2
[x] dx +
0) + [t ]10 = π /2 − 1
∫ −1/ 2log 1 − x dx


 1 + x
Q log  1 − x  is an odd function 


[x]dx + 0
[x] dx +
1/ 2
∫0

= − [x]−01/ 2 = − 0 +

29. Let I = ∫
=∫
π
−π
−π
π
[x] dx = ∫
0
−1/ 2
(−1) dx +
1/ 2
∫0
(0) dx
1
1
 =−
2
2
cos 2(− x)
d (− x)
1 + a −x
⇒ I = π /2
3
2
−2
ecos x sin x dx +
3
∫ 2 2 dx
= 0 + 2 [x]32
e 2 log x
log e x
e 
dx + ∫ 
dx
1
x
 x 
2
1
1
= − [(log e x)2]1e−1 + [(log e x)2] 1e
2
2
1
1
5
= − {0 − (−1)2} + (22 − 0) =
2
2
2
1
e −1
figure. From the graph, it is clear that
x = π /2
 2, if
 1, if
π / 2 < x ≤ 5π /6

[2 sin x] =  0, if
5π / 6 < x ≤ π
−1, if π < x ≤ 7π / 6

−2, if 7π / 6 < x ≤ 3π /2
Y
2
1
π/2
2
cos x
dx
1 + ax
cos 2x dx = π + 0
32. The graph of y = 2 sin x for π /2 ≤ x ≤ 3π / 2 is given in
 1 + x
1/ 2
π /2
0
π
∫ 0 cos 2x dx
2
log e x  dx = 1  log e x  dx − e  log e x  dx
−
1
∫ e  x  ∫1  x 
 x 

since, 1 is turning point for 

 log e x

 for + ve and − ve values


 x 
t
dt
t

 1 + x 
28. ∫ [x] + log 
 dx
−1/ 2
 1 − x  

1/ 2
π
0
= −∫
1/ 2
=∫
π
2
[where, t 2 = 1 − x2 ⇒ t dt = − x dx]
−1
1 + cos 2x
dx
2
[Q ecos x sin x is an odd function]
Q 3 f (x) dx = 2
= 2 [3 − 2] = 2
 ∫ −2

[since, t3 and t cos t are odd functions]
0
π
0
0
=∫
= 0 + 2 ⋅ 2 [x]10 + 0 = 4
27. I = ∫
cos 2 x dx = 2 ∫
3
(t3 + 2 + t cos t ) dt
1
π
−π
ecos x sin x, for | x| ≤ 2

f (x) =  2
otherwise
,


Put x + 1 = t
dx = dt
1 + ax 

 cos 2x dx
−π  1 + a x 
∫ −2 f (x) dx = ∫ −2 f (x) dx + ∫ 2 f (x) dx
∴
[(x + 1)3 + 2 + (x + 1) cos (x + 1)] dx
⇒
…(ii)
π
= [x]π0 + 2∫
1 log 3 1
(t )log 2 = (log 3 − log 2)
2
2
1
 3
I = log  
 2
4
26. Let I = ∫
cos 2 x
dx
1 + ax
= ∫ (1 + cos 2x) dx = ∫ 1 dx +
2I =
∴
−π
ax
On adding Eqs. (i) and (ii), we get
On adding Eqs. (i) and (ii), we get
1 log 3 sin t + sin (log 6 − t )
2I =
dt
2 ∫log 2 sin (log 6 − t ) + sin t
⇒
π
log 3
I=∫
∴
I=∫
⇒
…(i)
–1
–2
5π/6
π
7π/6
3π/2
X
Definite Integration 309
Therefore,
=∫
5 π /6
π /2
3 π /2
∫ π /2
dx + ∫
⇒
[2 sin x]dx
x
π
5π / 6
0 dx +
7 π /6
∫π
(−1) dx + ∫
3 π /2
7π / 6
∴
3 π /4
π /4
3 π /4
I=∫
π /4
k
⇒
2I = ∫
I=∫
⇒
dx
1 + cos (π − x)
π /4
dx
1 − cos x
…(ii)
π /4
1
1
−1
[x] dx
0
∫ −1 [x] dx = ∫ −1 [x]
0
−1
1
[Q x is an odd function]
I1 1
=
I2 2
subdivide the interval π to 2π as under keeping in view
that we have to evaluate [2 sin x ]
Y
1,π/2
O
30°
30°
∴
⇒
∫ 0 [x] dx
1
∫0
11π
π
π
1

= sin 2π −  = − sin = −


6
6
6
2
9π
3π
sin
= sin
= −1
6
6
sin
Hence, we divide the interval π to 2π as
7π   7π 11 π   11π


,
, 2π
π ,
, 
, 


6  6
6   6
0 dx
1
∫ −1 f (x) dx = 1
1 
1  1 

sin x = 0, −  ,  − 1, −  ,  − , 0

2 
2  2 
x
⇒
0
⇒
4
cos t dt
π
0
π+x
π
2 sin x = (0, − 1), (−2, − 1), (−1, 0)
[2 sin x] = − 1
=∫
cos 4 t dt = I1 + I 2
7π / 6
π
[2 sin x] dx +
4
0
and
Put
I2 = ∫
π+x
π
t=π+ y
4
cos t dt
=∫
7π / 6
π
=−
(−1) dx +
11 π / 6
∫ 7π / 6
11 π / 6
∫ 7π / 6
+
π
I1 = ∫ cos t dt = g (π )
–1/2,11π/6
Y'
π 1
=
6 2
7π
1
π

sin  π +  = sin
=−

6
6
2
1
(−1) dx +
= ∫ cos 4 t dt + ∫
where,
X
(0,2π)
We know that, sin
x=1
dx +
–1,3π/2
–1/2,7π/6
⇒
g (x) = ∫ cos 4 t dt
0
⇒
xf (1 − x)]dx
37. It is a question of greatest integer function. We have,
−1 ≤ x < 0
0 ≤ x<1
= − [x]−01 + 0 = −1; ∴
π+x
k
1 −k
I1 = I 2 − I1
X'
(0,π)
1
−1, if

[x] =  0, if
 1, if

g (x + π ) = ∫
f [(1 − x)]dx] − ∫
cosec2x dx = [− cot x] 3π π/ 4/ 4
1
=∫
⇒
k
1−k

2

 dx
2
 1 − cos x 
3 π /4
=0−∫
35. Given,
∫ (1 − x) f [(1 − x) x] dx
=∫
⇒
∫ −1 f (x) dx = ∫ −1 (x − [x]) dx = ∫ −1 x dx − ∫ −1 [x] dx
∴
∫ x f [x (1 − x)] dx
3 π /4 
1
But
I1 =
…(i)
3π
π

= − cot
+ cot
= − (−1) + 1 = 2

4
4 
34. Let
I1 =
1−k
On adding Eqs. (i) and (ii), we get
3 π /4 

1
1
2I = ∫
+

 dx
π / 4  1 + cos x
1 − cos x
⇒
x
0
1−k
k
dx
1 + cos x
3 π /4
I=∫
⇒
x
0
g (x + π ) = g (π ) + g (x)
36. Given,
 7 − 9
 1
 5 − 3
=π
 = −π / 2
 + π −  + π 
 3 
 6
 6 
π /4
0
= ∫ (− cos y)4 dy = ∫ cos 4 y dy = g (x)
7

7

 5 1
= π  −  + π 1 −  + π  − 3

3

 6 2
6
I=∫
I 2 = ∫ cos 4 ( y + π ) dy
(−2) dx
/6
= [x] 5π π/ 2/ 6 + [− x] 7π
+ [−2x] 37ππ //62
π
 5π π   7π
  −2 ⋅ 3 π 2 ⋅ 7 π 
=
−  + −
+ π + 
+

 6


  2
2
6
6 
33. Let
dt = dy
[2 sin x] dx
2π
∫ 11π / 6 [2 sin x] dx
(−2) dx +
π
10π
5π
 4π  π
−2  − =−
=−


6
6
6
6
3
2π
∫ 11π / 6 (−1) dx
310 Definite Integration
πx
 1
 + B, f ′   = 2
2
 2
1
2A
and
∫ 0 f (x) dx = π
πx
π Aπ
Aπ
 1  Aπ
cos
cos =
f ′ (x) =
⇒ f′  =
 2
2
2
2
4 2 2
Aπ
4
 1
But
f′   = 2 ∴
= 2 ⇒ A=
 2
π
2 2
1
1 
π
x
2A
2A



Now, ∫ f (x) dx =
⇒ ∫  A sin 
 + B dx =
0
0


π
π
2


38. Given, f (x) = A sin 
On adding Eqs. (i) and (ii), we get
2I = ∫
π /2
0
1 dx
π
4
π /3
dx
43. Let I = ∫
π / 6 1 + tan x
I=
∴
I=∫
∴
1
πx
 2A
 2A
⇒ −
cos
+ Bx =

 π
π
2
0
2A 2A
B+
=
⇒
π
π
⇒
I=∫
⇒
⇒
π /2
0
π /2
1
dx = ∫
3
0
1 + tan x
1
dx
sin3 x
1+
cos3 x
cos3 x
dx
0
cos x + sin3 x
π

cos3  − x
π /2
2

I=∫
dx
0


3π
3π
cos  − x + sin  − x
2

2

I=∫
I=∫
π /2
3
π /2
0
sin3 x
dx
sin3 x + cos3 x
40. Let I = ∫
π /2
0
π /2
−π / 2
…(i)
⇒
φ (− x) = [ f (− x) + f (x)] [ g (− x) − g (x)]
⇒
φ (− x) = − φ (x)
But
…(ii)
41. Let I = ∫ e
cos x
0
f (π − x) = (ecos
∴
2
b
I=∫
π /2
0
π /2
0
b
f (x) = x F (x)
We have,
⇒
f ′ (x) = F (x) + x F ′ (x)
⇒
f ′ (1) = F (1) + F ′ (1 ) < 0
…(i)
f (2) = 2F (2) < 0
[using F (x) < 0,∀x ∈ (1, 3)]
[using F (x) < 0, ∀ x ∈ (1, 3)]
45. Given,
f ′ (x) < 0
3
∫1 x F ′ (x) dx = − 12
2
3
⇒ [x2F (x)]31 − ∫ 2x ⋅ F (x) dx = − 12
1
3
⇒ 9 F (3) − F ( 1 ) − 2∫ f (x) dx = − 12
1
[QxF (x) = f (x), given]
⋅ cos3 {(2n + 1) x}
x
1 π π  π
−
=
2  3 6  12
∫ a f (x)dx = ∫ a f (a + b − x)dx is a true statement by
⇒
⇒
){ − cos3 (2n + 1) x} = − f (x)
∴
I =0
42. Let I = ∫
⇒
x
dx
property of definite integrals.
3
2
π /3
π /6
Now, f ′ (x) = F (x) + x F ′ (x) < 0
⋅ cos {(2n + 1) x} dx
Again, let f (x) = ecos
…(ii)
tan x
[given F (1) = 0 and F ′ (x) < 0]
f (a − x) = − f (x)
0,

a
a /2
Using∫ f (x) dx = 
0
2 ∫ 0 f (x) dx, f (a − x) = f (x)
∴
I=
Also,
∫ −π / 2 φ (x) dx = 0
2
tan xdx
π /6 1 +
F ′ (x) < 0, ∀x ∈ (1, 3)
π /2
π
π /3
dx
cot x
44. According to the given data,
⇒ φ(x) is an odd function.
∴
π /3
π /6 1 +
Statement I is false.
[f (x) + f (− x) ] [ g (x) − g (− x)] dx
⇒
π

1 + tan  − x
2

2I = [x]ππ //36 dx
⇒
φ (x) = [ f (x) + f (− x)] [ g (x) − g (− x)]
Let
dx
π /6
2I = ∫
⇒ 2I = [x]0π / 2 = π / 2 ⇒ I = π /4
1 dx
π /3
On adding Eqs. (i) and (ii), we get
On adding Eqs. (i) and (ii), we get
2I = ∫
I=∫
⇒
B =0
39. Let
⇒
=∫
…(i)
cot x
dx
cot x + tan x
…(i)
tan x
dx
cot x + tan x
…(ii)
3
− 36 − 0 − 2 ∫ f (x) dx = −12
1
3
∫1 f (x) dx = − 12
3
and ∫ x3 F ′ ′ (x)dx = 40
1
3
⇒
[x3 F ′ (x)] 31 − ∫ 3x2F ′ (x) dx = 40
⇒
[x2(xF ′ (x)] 31 − 3 × (−12) = 40
⇒
1
{ x2 ⋅ [ f ′ (x) − F (x)]} 31 = 4
Definite Integration 311
⇒
9 [ f ′ (3) − F (3)] − [ f ′ (1) − F (1)] = 4
⇒
9 [ f ′ (3) + 4] − [ f ′ (1) − 0] = 4
⇒
t
46. Given, lim
∫a
t→ a
9 f ′ (3) − f ′ (1) = − 32
(t − a )
f (x) dx −
{ f (t ) + f (a )}
2
=0
3
(t − a )
Using L’Hospital’s rule, put t − a = h
a+ h
h
∫ a f (x) dx − 2 { f (a + h ) + f (a )}
=0
lim
⇒
h→ 0
h3
h
1
f (a + h ) − { f (a + h ) + f (a )} − { f ′ (a + h )}
2
2
=0
⇒ lim
h→ 0
3h 2
Again, using L’ Hospital’s rule,
1
1
h
f ′ (a + h ) − f ′ (a + h ) − f ′ (a + h ) − f ′ ′ (a + h )
2
2
2
=0
lim
h→ 0
6h
h
− f ′′ (a + h )
2
⇒
lim
= 0 ⇒ f ′′ (a ) = 0, ∀ a ∈ R
h→ 0
6h
⇒ f (x) must have maximum degree 1.
47. F ′ (c) = (b − a ) f ′ (c) + f (a ) − f (b)
F ′ ′ (c) = f ′ ′ (c)(b − a ) < 0
f (b) − f (a )
⇒ F ′ (c) = 0 ⇒ f ′ (c) =
b−a
48.
π /2
∫0
π
−0
2
sin x dx =
4
=
49.
π


0 +  

 π
2

sin 0 + sin   + 2 sin 
2
 2 




(d) Let g (x) = x9 − f (x)
g (0) = − f (0) < 0
g (1) = 1 − f (1) > 0
∴ Option (d) is correct.
50. I =
98 k + 1
∑ ∫
k =1 k
(k + 1)
dx
x(x + 1)
I>
Clearly,
⇒
I>
I>
⇒
⇒
I>
Also,
I<
∫
∴
(c) Let h (x) = x −
98 k + 1
0
h (1) = 1 −
π
−1
2
∫ f (t ) cos t dt > 0
0
∴ Option (c) is correct.
1
k =1
98
k+1
dx = ∑ [log e (k + 1) − log e k]
x(k + 1)
k =1
∑ ∫
= (7 tan 6 x − 3 tan 2 x) sec2 x
Now, ∫
π /4
π /4
0
0
x f (x)dx = ∫
x (7 tan 6 x − 3 tan 2 x) sec2 x dx
I
II
= [x (tan x − tan x)]
7
3
−∫
=0 − ∫
= –∫
π /4
π /4
1 (tan7 x − tan3 x) dx
tan3 x (tan 4 x − 1)dx
0
0
π /4
0
π /4
0
tan 3 x (tan 2 x − 1) sec2 x dx
Put tan x = t ⇒ sec2 x dx = dt
∴
π /4
∫0
1
x f (x)dx = − ∫ t3 (t 2 − 1) dt
0
1
 t 4 t5 
1 1 1
= ∫ (t − t )dt =  −  = − =
0
4
5
 0 4 6 12

1
Also,
π /4
∫0
3
5
f (x) dx = ∫
π /4
0
(7 tan 6 x − 3 tan 2 x) sec2 x dx
1
= ∫ (7t 6 − 3t 2)dt
0
0
h (0) = − ∫ f (t ) cos t dt < 0
98
∑k+2
1
1
98
49
+…+
>
⇒ I>
3
100 100
50
∫ f (t ) cos t dt,
π
2
I>
f (x) = 7 tan 6 x sec2 x − 3 tan 2 x sec2 x
0
π
−x
2
1 
 1
∑ (−(k + 1)) k + 2 − k + 1  ⇒
 −π π 
for all x ∈ 
, 
 2 2
f (t )sin t dt always positive
∴Option (b) is incorrect.
k
98
51. Here, f (x) = 7 tan 8 x + 7 tan 6 x − 3 tan 4 x − 3 tan 2 x
x
(b) f (x) +
k =1
1
dx
(x + 1)2
I < log e 99
0
π
2
k+1
∑ (k + 1) ∫
k =1 k
x
So, option (a) is incorrect.
98
k =1
Q ex ∈ (1, e) in (0, 1) and ∫ f (t )sin t dt ∈ (0, 1) in (0, 1)
0
(k + 1)
dx
(x + 1)2
k =1 k
π
(1 + 2 )
8
∴ ex − ∫ f (t )sin t dt cannot be zero.
98 k + 1
∑ ∫
= [t7 − t3 ]10 = 0
192 x3
192x3
192x3
∴
≤ f ′ (x) ≤
4
3
2
2 + sin πx
1
On integrating between the limits to x, we get
2
52. Here,
f ′ (x) =
x
∫1/ 2
x
x 192 x3
192x3
dx ≤ ∫ f ′ (x)dx ≤ ∫
dx
1
2
1
/
/2
3
2
312 Definite Integration
1
4x2 
= ∫ (x2 − 1) (1 − x)4 + (1 + x2) − 4x +
 dx
0
(1 + x2) 

1
4 
= ∫  (x2 − 1) (1 − x)4 + (1 + x2) − 4x + 4 −
 dx
0
1 + x2 
192  4 1 
4 3
 x −  ≤ f (x) − f (0) ≤ 24x −
12 
16
2
3
⇒ 16x4 − 1 ≤ f (x) ≤ 24x4 −
2
1
Again integrating between the limits to 1, we get
2
1
1
1 
4 3
4
∫1/ 2 (16x − 1) dx ≤ ∫1/ 2 f (x) dx ≤∫1/ 2 24x − 2 dx
⇒
=∫
1
1
24x5 3 

16x5
⇒ 
− x
− x ≤ ∫ f (x)dx ≤ 
2 1/ 2
 5
1/ 2 1/ 2
 5
=
1
6
 11 2
 33
⇒ 
+  ≤ ∫ f (x)dx ≤ 
+

1/ 2
5
 10 10
5
2 .6 ≤ ∫
⇒
1
1 4 5 4
π
 22
− + − + 4 − 4  − 0 =
−π
4

7 6 5 3
7
In = ∫
55. Given
f (x) dx ≤ 3 . 9
1/ 2
4π
0
π t
b
In = ∫
= ∫ e (sin 6 at + cos 4 at ) dt
+∫
2π
π
et (sin 6 at + cos 4 at ) dt
+∫
3π t
2π
e (sin 6 at + cos 4 at ) dt
+∫
∴
I3 = ∫
π
π
e (sin at + cos at ) dt
π +x
0
I4 = ∫
Now,
…(i)
6
t
I3 = ∫ e
∴
4
dt = dx
⋅ (sin 6 at + cos 4 at ) dt = eπ ⋅ I 2 …(ii)
0
and I5 = ∫
…(iii)
e (sin 6 at + cos4 at ) dt
I5 = ∫ e
0
3π
(sin at + cos at ) dt = e
6
4
⋅ I2
…(iv)
From Eqs. (i), (ii), (iii) and (iv), we get
I1 = I 2 + eπ ⋅ I 2 + e2π ⋅ I 2 + e3 π ⋅ I 2 = (1 + eπ + e2π + e3 π ) I 2
4π
∴
L=
∫0 e (sin at + cos at )dt
π t
6
4
∫0 e (sin at + cos at )dt
6
t
π
2π
= (1 + e + e
54. Let I = ∫
1
0
In = ∫
Since,
4
1⋅ (e4π − 1)
for a ∈ R
+ e )=
eπ − 1
3π
1 (x4 − 1 ) (1 − x)4 + (1 − x)4
x4 (1 − x)4
dx
dx = ∫
2
0
(1 + x2)
1+ x
1
1 (1 +
0
0
= ∫ (x2 − 1) (1 − x)4 dx + ∫
π
0
…(iii)
sin nx
dx ⇒ I1 = π and I 2 = 0
sin x
I 2 = I 4 = I 6 = ... = 0
and
t = 3π + x
π 3 π +x
…(ii)
From Eq. (iii) I1 = I3 = I5 = .... = π
4π t
3π
π x sin nx
dx
(1 + π x ) sin x
π
∴ In + 2 = In
I 4 = ∫ ex + 2π (sin 6 at + cos 4 at ) dt = e2π ⋅ I 2
π
−π
π
sin (n + 1) x 
= 2 ∫ cos (n + 1) x dx = 2 
 =0
0
 (n + 1)  0
e (sin 6 at + cos 4 at ) dt
π
…(i)
On adding Eqs. (i) and (ii), we have
π sin nx
π sin nx
2I n = ∫
dx = 2 ∫
dx
0 sin x
− π sin x
sin nx
is an even function]
[Q f (x) =
sin x
π sin nx
⇒
In = ∫
dx
0 sin x
π sin (n + 2 )x − sin nx
Now, I n + 2 − I n = ∫
dx
0
sin x
π 2 cos (n + 1 ) x ⋅ sin x
dx
=∫
0
sin x
3π t
2π
t = 2π + x ⇒ dt = dx
Put
∴
2π
t = π + x⇒
Put
Put
e (sin 6 at + cos 4 at ) dt
I1 = I 2 + I3 + I 4 + I5
Now,
∴
4π t
3π
sin nx
dx
(1 + π x ) sin x
b
et (sin 6 at + cos 4 at )dt
0
π
−π
∫ a f (x) dx = ∫ a f (b + a − x) dx, we get
Using
(*) None of the option is correct.
53. Let I1 = ∫
4 
 6
5
4
2
 x − 4x + 5x − 4x + 4 −
 dx

1 + x2 
 x7 4x6 5x5

4x3
= −
+
−
+ 4x − 4 tan −1 x
7
6
5
3

0
1
1
1
0
x2 − 2x)2
dx
(1 + x2)
10
10
∑ I 2m + 1 = 10 π
⇒
m=1
and
∑ I 2m = 0
m =1
∴ Correct options are (a), (b), (c).
2
56. The integral, I = ∫ |2x − [3x]| dx
4/3
1
5 /3
2
= ∫ |2x − 3| dx +
∫4/3 |2x − 4|dx + ∫1|2x − 5|dx
=∫
∫4/3 (4 − 2x) dx + ∫5/3 (5 − 2x) dx
1
4/3
1
(3 − 2x) dx +
5 /3
2
= [3x − x2]14/3 + [4x − x2]54//33 + [5x − x2]52/3
16
  20 25 16 16

−
−
+
= 4 −
− 3 + 1 + 

 3

9
9
3
9
25 25

+ 10 − 4 −
+


3
9
Definite Integration 313
 20 16 25  16 25 16 25
= (2 + 6) + 
−
−  + −
−
+
+ 
3
3
3  9
9
9
9
=8 −
Put 3 tan x = t ⇒ 3 sec2 dx = dt , and at x = 0, t = 0 and
at x = 3, t = 3
2 3 1 dt
2
So,
=
I= ∫
[tan −1 t ]03
π 0
3 1 + t2
3π
2  π
2
=
⇒ 27I 2 = 4.00
  =
3π  3  3 3
21
=8 − 7 =1
3
Hence, answer 1.00 is correct.
57.
Key Idea Use property ∫ f ( x) dx = ∫ f ( a − x) dx
a
a
0
0
I=∫
59. Let
The given integral
π/ 2
∫(
I=
0
π/ 2
⇒
I=
∫(
0
3 cos θ
dθ
cos θ + sin θ )5
… (i)
3 sin θ
dθ
sin θ + cos θ )5
…(ii)
I=∫
⇒
1/ 2
0
1+ 3
dx
[(x + 1)2 (1 − x)6 ]1/ 4
1+ 3
1/ 2
1/ 4
 1 − x  6 
(1 − x)2 
 
 1 + x 
1−x
− 2 dx
= dt
=t ⇒
1+ x
(1 + x)2
0
a
a
Put
0
0
when x = 0, t = 1, x =
[Using the property ∫ f (x) dx = ∫ f (a − x) dx]
Now, on adding integrals (i) and (ii), we get
π/ 2
2I =
∫(
0
π/ 2
=
∫
0
⇒
Now, let tan θ = t 2 ⇒ sec2 θ dθ = 2 t dt
π
and at θ = , t → ∞
2
and at θ = 0, t → 0
∞ 6 t dt
∞ t + 1 −1
So, 2 I = ∫
=6 ∫
dt
0 (1 + t )4
0 (t + 1 )4
⇒
∞
∞
 ∞ dt


dt 
1
1
−
I = 3 ∫
3
=
−
+



∫0 (t + 1)4   2(t + 1)2 3(t + 1)3 
3
 0 (t + 1)
0
1 1 
 1 1
⇒ I =3
−
= 3   = ⇒ I = 0.5
2 3 
 6 2
dx
2 π/4
58. Given, I = ∫
π −π/4 (1 + esin x ) (2 − cos 2x)
− (1 + 3 )  −2 
 t 
2
1
I=
⇒
I = (1 + 3 ) ( 3 − 1)
⇒
I =3 −1 =2
[x], x ≤ 2
 0, x > 2
60. Here, f (x) = 
I=∫
∴
=∫
0
−1
x f (x2)
dx
2 + f (x + 1)
2
−1
x f (x2)
dx +
2 + f (x + 1)
+
∫1
2
π/4
…(i)
=∫
b
a
∫−π/4
esin x dx
sin x
(1 + e
) (2 − cos 2x)
∫0
x f (x2)
dx
2 + f (x + 1)
x f (x2)
dx +
2 + f (x + 1)
0
−1
0 dx +
1
∫0 0 dx + ∫1
we get
2
I=
π
1
+
On applying property ∫ f (x)dx = ∫ f (a + b − x) dx,
a
1/3
1
1
1
,t =
2
3
(1 + 3 ) dt
−2(t )6/ 4
1/3
3 sec2 θ
dθ
(1 + tan θ )4
b
I=∫
∴
3
dθ
sin θ + cos θ )4
dx
2
On adding integrals (i) and (ii), we get
2 π/4
dx
2I = ∫
π − π/4 2 − cos 2x

1 π/4
1 − tan 2 x 
dx
=
x
2
⇒
I= ∫
as
cos


π − π/4
1 − tan 2 x
1 + tan 2 x 

2−
2
1 + tan x
2 π/4 sec2 x
dx
= ∫
π 0 1 + 3 tan 2 x


sec2 x
is even function 
Q
2
1
+
3
x
tan


3
2
2
∫3
x f (x2)
dx
2 + f (x + 1)
x f (x2)
dx
2 + f (x + 1)
x⋅1
dx
2+0
+
....(ii)
∫
3
∫
2
0 dx +
2
∫3
0 dx
Q − 1 < x < 0 ⇒ 0 < x2 < 1 ⇒ [x2] = 0,
0 < x < 1 ⇒ 0 < x2 < 1 ⇒ [x2] = 0,

⇒ [x2] = 1
1 < x2 < 2
1 < x< 2 ⇒
2 < x + 1 < 1 + 2 ⇒ f (x+ 1) = 0,
2 < x < 3 ⇒ 2 < x2 < 3 ⇒ f (x2) = 0,
and 3 < x < 2 ⇒ 3 < x2 < 4 ⇒ f (x2) = 0
⇒
∴
I=∫
1
2
2
 x2 
1
1
x
dx =   = (2 − 1) =
2
4
4
4
 1
4I = 1 ⇒ 4I − 1 = 0
314 Definite Integration
1
61. Here, α = ∫ e( 9x + 3 tan
−1
 12 + 9x2

 dx
 1 + x2 
x)
0
Put
⇒
9x + 3 tan −1 x = t

3 
9 +
 dx = dt

1 + x2 
∴ α=∫
⇒
62.
⇒
16
∫1
2
e sin t
dt = F (k) − F (1)
t
16
∫1
⇒
⇒
[F (t )]16
1 = F (k ) − F (1 )

 d
e sin x
Q dx { F (x)} = x , given 


9+3π /4 t
e dt = [et ]90 + 3 π / 4 = e9 + 3 π / 4 − 1
0
log e 1 + α = 9 +
3π
3π
=9
⇒ log e α + 1 −
4
4
d

∫ f(x ) g (x ) dx = f(x ) ∫ g (x ) dx − ∫  dx [f(x )] ∫ g (x ) dx  dx
Given, I = ∫
1
0
4x3
I
⇒
F (16) − F (1) = F (k) − F (1)
∴
PLAN Integration by parts
1
1
d
d


(1 − x2)5 − ∫ 12 x2
(1 − x2)5 dx
= 4x3
0
dx
dx
 0

k = 16
π sin (π log x)
dx
x
37 π
65. Let I = ∫
Put
d2
(1 − x2)5 dx
dx2
II
e sin t dt
⋅
= F (k) − F (1)
2
t
1
π log x = t
π
dx = dt
⇒
x
∴ I=∫
37π
0
π
= − [cos 37π − cos 0]
sin (t ) dt = − [cos t ]37
0
1
= 4x3 × 5 (1 − x2)4 (− 2x)

0
1

2
25 1
− 12 [x (1 − x ) ]0 − ∫ 2x (1 − x2)5 dx


0
= − [(−1) − 1] = 2
66. Let I = ∫
0
I=∫
0
1
= 0 − 0 − 12 (0 − 0) + 12 ∫ 2x (1 − x2)5 dx
0
2n
x sin x
dx
sin 2n x + cos 2n x
2π
(2π − x)[sin (2π − x)]2n
dx
[sin (2π − x)]2n + [cos (2π − x)]2n
2π
2 6 1
 (1 − x )
1

= 12 × −
 = 12 0 + 6  = 2
6



0
1
and 6 ∫ f (t )dt = 3x f (x) − x3 , ∀ x ≥ 1
1
3
Using Newton-Leibnitz formula.
Differentiating both sides
⇒
6 f (x) ⋅ 1 − 0 = 3 f (x) + 3xf ′ (x) − 3x2
1
f ’ (x) − f (x) = x
⇒ 3xf ′ (x) − 3 f (x) = 3x2 ⇒
x
xf ’ (x) − f ’ (x)
⇒
=1
x2
d x
⇒
 =1
dx  x 
On integrating both sides, we get
⇒
∴
For that f (x) = x2 −
64. Given,
Put
⇒
I=∫
4
∫1
2
2
4 8
x and f (2) = 4 − =
3
3 3
2e sin x
dx = F (k) − F (1)
x
x2 = t
2x dx = dt
a
0
2π
0
2π
(2π − x) ⋅ sin 2n x
dx
sin 2n x + cos 2n x
2π
x sin 2n x
2π sin 2n x
dx − ∫
dx
2
2n
2n
0 sin n x + cos 2n x
sin x + cos x
⇒
I=∫
⇒
I=∫
0
⇒
I=∫
0
⇒
 π
π sin 2n x
dx
I = π ∫
2
2n
n
 0 sin x + cos x
0
2π
2π
2π sin 2n x
dx − I
sin 2n x + cos 2n x
[from Eq. (i)]
π sin 2n x
dx
sin x + cos 2n x
2n

sin 2n (2π − x)
dx
0 sin (2 π − x) + cos 2n (2 π − x)


using property
 2a f (x) dx = a [f (x) + f (2a − x)]dx
∫0

∫ 0
+∫
f (x)
1

= x+ c
Q f (1) =

3 
x`
1
2
2
= 1 + c ⇒ c = and f (x) = x2 − x
3
3
3
4 8
f (2) = 4 − =
3 3
NOTE Here, f(1) = 2, does not satisfy given function.
1
∴ f(1 ) =
3
a
0
[Q ∫ f (x) dx = ∫ f (a − x) dx]
x
63. Given, f (1) =
…(i)
π
2n
 π
sin 2n x dx
dx
I = π ∫
2n
2n
 0 sin x + cos x
+∫
sin 2n x dx
dx
0 sin 2n x + cos 2n x
π
π
0

sin 2n x
dx
2n
sin x + cos x 
⇒
I = 2π ∫
⇒
 π /2

sin 2n x dx
dx
I = 4π ∫
2n
2n
0
+
x
x
sin
cos


⇒
I = 4π ∫
π /2
0
2n
sin 2n (π / 2 − x)
dx
sin (π / 2 − x) + cos 2n (π / 2 − x)
2n
…(ii)
Definite Integration 315
⇒
I = 4π ∫
π /2
0
cos 2n x
dx
cos 2n x + sin 2n x
…(iii)
On adding Eqs. (ii) and (iii), we get
2I = 4π ∫
2I = 4π ∫
⇒
π / 2 sin 2n
0
sin
π /2
0
2n
x + cos 2n x
dx
x + cos 2n x
1 dx = 4π
[x]π0 / 2
1
−5
x
Replacing x by 1 / x in Eq. (i), we get
67. Given, af (x) + bf (1 / x) =
…(i)
af (1 / x) + bf (x) = x − 5
…(ii)
On multiplying Eq. (i) by a and Eq. (ii) by b, we get

1
…(iii)
a 2f (x) + abf (1 / x) = a  − 5

x
abf (1 / x) + b2f (x) = b (x − 5)
…(iv)
2 a
1
2

2
=
=
68. Let
I=∫
3
2
x
b

+ 5 (a − b)

2
2+3−x
2
(2 + 3) − (5 − x) + 2 + 3 − x
⇒
I=∫
3
5−x
2
x + 5 −x
x+ 5−x
2
5−x+
69. Let I = ∫
3 π /4
π /4
=
π
2
∫ π /4
=
π
2
∫ π /4
=
π
2
∫ π /4
3 π /4
dx
(1 + sin x)
3 π /4
(1 − sin x)
dx
(1 + sin x) (1 − sin x)
3 π /4
(1 − sin x)
dx
1 − sin 2 x
3 π /4 
1
sin x 
−

 dx
 cos 2 x cos 2 x
2
2
−1
−2
(x2 − 1) dx + ∫
1
2
(1 − x2) dx + ∫ (x2 − 1) dx
−1
1
1
2
71.
1.5
∫0
1
[x2] dx = ∫ 0 dx + ∫
0
2
1
1 dx +
72. (A) Let I = ∫
dx
x2
1
−1 1 +
⇒
dx = sec2 θ dθ
I = 2∫
∴
…(i)
(B) Let
 π 3π

− x
 +
3 π /4
4

4
⇒ I=∫
dx
π /4
 π 3π

1 + sin  +
− x
4

4
Put
⇒
b
b
a
a
1.5
2
= ( 2 − 1) + 2 (1.5 − 2 )
= 2 −1 + 3 −2 2 =2− 2
Put x = tan θ
1
dx ⇒ 2I = ∫ 1 dx = 1 ⇒ I =
2
2
x
3
x
dx
1 + sin x
∫
= 0 + [x]1 2 + 2 [x]1.52
On adding Eqs. (i) and (ii), we get
3
∫ π /4
dx
…(ii)
dx
π
2
[from Eq. (i)]
=4
…(i)
3
=
x
dx
1 + sin x
dx
−I
1 + sin x
−1
dx
I=∫
3 π /4
π /4
3 π /4
∫ π /4



 x3
 x3
x3 
=
− x + x −
+
− x

3
3
3
 −1 
 −2 
1

1
8
1
1
1

 

8

=  − + 1 + − 2 + 1 − + 1 −  +  − 2 − + 1
 3





3
3
3
3
3
1
7 

a log 2 − 5a + b
2 
2 
(a − b ) 
x
5−x+
π
dx −
1 + sin x
3 π /4
∫ −2 |1 − x |dx
=∫
2
⇒
2I = ∫
70.
1
[a log 2 − 2b − 10 (a − b)
(a 2 − b2)
− a log 1 +
π /4
π 3 π /4
(sec2 x − sec x ⋅ tan x) dx
2 ∫ π /4
π
= [tan x − sec x]3π π/ 4/ 4
2
π
= [− 1 − 1 − (− 2 − 2 )]
2
π
= (− 2 + 2 2 ) = π ( 2 − 1)
2
∫ 1 f (x) dx = (a 2 − b2) ∫ 1  x − bx − 5a + 5b dx
b
1


a log|x| − x2 − 5(a − b)x
= 2

2
(a − b2) 
1
=∫
=
On subtracting Eq. (iv) from Eq. (iii), we get
a
(a 2 − b2) f (x) = − bx − 5a + 5b
x
1
a

f (x) = 2
⇒
 − bx − 5a + 5b

(a − b2)  x
⇒
π /4
= π∫
π
= 4π ⋅ ⇒ I = π 2
2
π−x
dx
1 + sin (π − x)
3 π /4
=∫
[Q ∫ f (x) dx = ∫ f (a + b − x) dx]
∴
I=
dθ =
dx
1
∫0
π /4
0
1 − x2
x = sinθ
dx = cosθ d θ
I=
π/ 2
∫0
1 dθ =
π
2
π
2
2 dx
316 Definite Integration
3
1
 1 + x 
∫ 2 1 − x 2 2 log  1 − x  
2
1
 2 
 3  1 
 4
= log   − log    = log   
 3 
 −1  2 
 −2
2
3
(C)
(D) ∫
=
dx
2
1
dx
x x −1
2
= [sec−1 x]12 =
f ( x ) = ax 2 + bx + c
f (0) = 0 ⇒ c = 0
Hence, f (x) is an odd function.
(P) → (ii); (Q) → (iii); (R) → (i); (S) → (iv)
0
= [(1 − x50 )101 ⋅ x]10 +
1
 ax3
α β
bx2
+ =1
+
 =1 ⇒

3 2
2 0
 3
=0 − ∫
As a , b are non-negative integers.
1
0
(Q) PLAN Such type of questions are converted into only sine or
f ( x ) = sin ( x 2 ) + cos ( x 2 )
1
 1

cos (x2) +
sin (x2)
= 2
 2

2
π
π


= 2 cos x2 cos + sin sin (x2)


4
4
1
0
1
+ (50) (101) ∫
f (x) = 2x or f (x) = 3x2
cosine expression and then the number of points of
maxima in given interval are obtained.
)
50 ⋅ x49⋅x dx
0
(1 − x50 )101 dx
(1 − x50 )100 dx = 5050I 2 + 5050I1
∴ I 2 + 5050I 2 = 5050I1 ⇒
75. Let
π

I = ∫ e |cos x | 2 sin
0

(5050)I1
= 5051
I2

1

1
 cos x + 3 cos  cos x  sin x dx

2

2
π

1
⇒ I = ∫ e |cos x | ⋅ sin x ⋅ 2 sin  cos x dx
0

2
π

= 2 cos  x2 − 

4
+
π
π
= 2 n π ⇒ x2 = 2 n π +
4
4
50 100
(50) (101) (1 − x50 )100 (− x50 ) dx
= − (50) (101) ∫
a = 0, b = 2 or a = 3, b = 0
∴
1
∫ 0 (1 − x
[using integration by parts]
2a + 3b = 6
So,
f (x) dx = 0
1
∫0 f (x) dx = 1
⇒
1/ 2
∫−1/ 2
So,
74. Let I 2 = ∫ (1 − x50)101 dx,
1
⇒
 1 + x
f ( x ) = cos 2x log 

 1 − x
Let
 1 − x
f (− x) = cos 2x log 
 = − f (x)
 1 + x
(ii) The other conditions are also applied and the number
of polynomial is taken out.
Now,
f( x ) dx = 0
If f( − x ) = − f( x ), i.e. f( x ) is an odd function.
π
π
−0 =
3
3
73. (P) PLAN (i) A polynomial satisfying the given conditions is taken.
Let
a
∫− a
(S) PLAN
π |cos x |
∫ 0e
1

⋅ 3 cos  cos x ⋅ sin x dx
2

So, f (x) attains maximum at 4 points in [− 13 , 13].
…(i)
⇒ I = I1 + I 2
2
a

using
∫ 0 f (x) dx


0,
f (2a − x) = − f (x) 


a


=

2 ∫ 0 f (x) dx, f (2a − x) = + f (x)
(R) PLAN
where,
For maximum value, x2 −
⇒
π
,for n = 0 ⇒ x = ±
4
x=±
(i)
(ii)
a
f( x ) dx =
a
f( x ) dx = 2
∫− a
∫− a
a
∫− a
9π
, for n = 1
4
f( − x ) dx
a
∫0
I=
2
3x 2
∫−2 1 + ex
and
3 x2
dx
I=∫
−2 1 + e− x
⇒
2I = ∫
dx
−2
2I = ∫
2
−2
I = [x3 ]20 = 8
I 2 = 6∫
Now,
1
 t
I 2 = 6 ∫ et ⋅ cos   dt
0
 2
0
e
1
t 

 t 1
= 6 et cos   + ∫ et sin dt 


2
2
2 0

 3x2
3x2(ex )
+ x
 dx

x
e +1
1 + e
3x2 dx
[Q f (π − x) = − f (x)] …(ii)
1

⋅ sin x ⋅ cos  cos x dx
2

[put cos x = t ⇒ − sin x dx = dt]
2
2
π / 2 cos x
and
f( x ) dx , if f( − x ) = f( x ) , i.e. f is an even
function.
I1 = 0
2
⇒ 2I = 2 ∫ 3x2 dx
0
1

t
et
t 
 t 1 
cos dt 
= 6 et cos   +  et sin − ∫
 2 2 
2
2
2 

0
1
t 1
t
I

= 6 et cos + et sin
− 2

2 2
2  0 4
Definite Integration 317
=


 1
 1 e
e cos  2 + 2 sin  2 − 1


24
5
…(iii)
From Eq. (i), we get
24 

 1 e
 1
I=
e cos   + sin   − 1
 2 2
 2
5 

76. Let I = ∫
π /3
x dx
π dx
+ 4∫
− π /3
π
π


2 − cos |x|+ 
2 − cos |x|+ 


3
3
∴
2I = − 2
⇒
2I = 2
∴
I= 2
3
π /3
− π /3
f (− x) = − f (x)
0,

a
Using ∫ f (x) dx = 
f (x) dx, f (− x) = f (x)
2
−a
 ∫ 0
=∫
I =2 ∫
∴
I = 2π
x+
Put
π /3
0
π /3
∫0
π dx
+0
π

2 − cos |x| + 

3


x3 dx
is odd
Q
 2 − cos |x|+ π 

3

Put
⇒





dx
2 − cos (x + π / 3)
=∫
π /2
4π
4π
 1
(tan −1 3 − tan −1 1) =
tan −1  
 2
3
3
⇒ I=∫
0
π /2
f (cos 2x) cos x dx
I=∫
1
∫ 0 tan
π /2
f (cos 2x) sin x dx
On adding Eqs. (i) and (ii), we get
2I = ∫
π /2
0
= 2
Put − x +
f (cos 2x) (sin x + cos x) dx
π /2
∫0
e
a
0
a
f (x) dx = ∫ f (a − x) dx]
0
e− cos x
dx
0 e− cos x + ecos x
π
…(ii)
π
ecos x + e− cos x
dx = ∫ 1 dx = [x] π0 = π
0
ecos x + e− cos x
π
0
−1
1


 1−x+ x 
1

 dx = ∫ tan −1 
 dx
2
0
1 − x + x 
1 − x(1 − x) 
0
1
= ∫ tan −1 [1 − (1 − x) ] dx +
0
−1
x dx
1


1
Now, ∫ tan −1 
 dx
2
0
1 − x + x 
1 π


1
= ∫  − cot−1 
 dx
2 
0 2
1 − x + x  

π
−
2
1
∫0
1
∫0
tan −1 (1 − x + x2) dx
1
π
1
tan −1
−
dx
2 ∫0
(1 − x + x2)
π
= − 2I1
2
tan −1 (1 − x + x2) dx =
…(i)
where, I1 =
…(ii)
1
∫ 0 tan
1
a
a
= 2∫ tan −1x dx Q ∫ f (x) dx = ∫ f (a − x) dx …(i)
0
0
0


1
∫ 0 tan
−1
x dx = [x tan−1 x ]10 −
1
x dx
∫ 0 1 + x2
π 1
π 1
− [log(1 + x 2 )]10 = − log 2
4 2
4 2
1
π
π 1

∴ ∫ tan−1(1 − x + x 2 ) dx = − 2  − log 2 = log 2
0


2
4 2
=
π /4
80. Let I = ∫
0
I=∫
0
f (cos 2x) [cos (x − π / 4)] dx
π
= t ⇒ − dx = dt
4
0
= ∫ [ tan −1 (1 − x) − tan −1 x] dx
∴



π
π
f  cos 2  − x  ⋅ cos  − x dx


2
2

0
cos ( π − x )
=
using a f (x) dx = a f (a − x) dx
∫0
∫0


⇒
ecos ( π − x )
dx
+ e− cos ( π − x )
π
1
π + 4x3
4π
 1
dx =
tan −1  
∫ −π /3
 2
π


3
2 − cos |x|+ 


3
0
…(i)
⇒ I = π /2
π /3
77. Let I = ∫
f (sin 2t ) cos t dt
On adding Eqs. (i) and (ii), we get
79.
t
sec dt
2 π /3
2 π /3
dt
2
I = 2π ∫
= 2π ∫
π /3
π / 3 2 − cos t
t
1 + 3 tan 2
2
t
t
tan = u ⇒ sec2 dt = 2 du
2
2
3
2 du
4π
=
I = 2π ∫
[ 3 tan −1 3u ] 13
1/ 3 1 + 3u 2
3
3
∴
π /4
∫0
ecos x
dx
0 ecos x + e− cos x
⇒ I=∫
π
= t ⇒ dx = dt
3
=
π /4
∫ −π / 4 f (sin 2t ) cos t dt
[Q∫
2
∴


π
f cos  − 2t  cos t dt


2

π
78. Let I = ∫
a
− π /4
∫ π /4
∴
π /4
I=∫
π /4
0
log (1 + tan x) dx
π
− x)) dx
4

1 − tan x 
log 1 +
 dx

1 + tan x
log (1 + tan (
…(i)
318 Definite Integration
0
I=∫
0
⇒ 2I =
81. Let
π /4


2
log 2 dx − I
log 
 dx ⇒ I = ∫
0
 1 + tan x
π /4
π
π
2x
2x sin x
dx
dx + ∫
2
− π 1 + cos x
− π 1 + cos 2 x
I = I1 + I 2
∴
I1 = 0
Again, let
⇒ g (− x) =
⇒ g (− x) = g (x) which shows that g (x) is an even
function.
π 2 x sin x
π
x sin x
∴ I2 = ∫
dx = 2 ⋅ 2 ∫
dx
2
− π 1 + cos x
0 1 + cos 2 x
=4 ∫
0
(π − x)sin (π − x)
(π − x)sin x
dx = 4 ∫
dx
0 1 + cos 2x
1 + [cos (π − x)]2
π
π x sin x
π sin x
=4 ∫
dx
dx − 4 ∫
0 1 + cos 2x
0 1 + cos 2x
π
π
sin x
dx
⇒ 2I 2 = 4π ∫
0 1 + cos 2x
cos x = t
1
dt
1
I = I1 + I 2 = 0 + π 2 = π 2
1/ 3
− 1/ 3
 x4 
 2x 
 cos −1 

 dx
4
 1 + x2
1 − x 
Put x = − y ⇒ dx = − dy
∴
I=∫
−1/ 3
1/ 3
1/ 3
x4
dx = π ∫
4
1/ 3
−
1−x
1/ 3
−1/ 3
=−π∫
[from Eq. (i)]
1/ 3
−1/ 3
1 dx + π

1 
 −1 +
 dx
1 − x4 

dx
1 − x4
1/ 3
∫ −1/
3
= − π [x]1−/1/3 3 + π I1, where I1 = ∫
dx
1 − x4
1/ 3
−1/ 3
1
2π
 1
2I = − π 
+ π I1
+
 + πI1 = −
 3
3
3
⇒
1/ 3
∫ −1/
3
1/
dx
=2 ∫
4
0
1−x
3
1/ 3
dx
1 − x4
=
∫0
=
∫0
=
∫0
=
1
2
1 + 1 + x2 − x2
dx
(1 − x2) (1 + x2)
1/ 3
1/
1
dx + ∫
2
0
1−x
1/ 3
1/
1
dx + ∫
0
(1 − x) (1 + x)
1/ 3
∫0
3
1
1 1/
dx + ∫
1−x
2 0
1
dx
1 + x2
3
3
1
dx
(1 + x2)
1/
1
dx + ∫
0
1+ x
1/ 3
  1 + x 


ln 1 − x 
 0
 
=
1
2
=
1 
1 + 1 / 3
 + tan −1 1
ln 
2 1 − 1 / 3
3
=
1 
3 + 1
+ π
ln 
2  3 − 1 6
+ [tan −1 x]01/
1 
( 3 + 1 )2 
+ π
ln 
2  3 −1  6
1
π
= ln (2 + 3 ) +
2
6
=
= 4π (π / 4 − 0) = π 2
82. Let I = ∫
⇒ 2I = π ∫
dt
∫ −1 1 + t 2 = 4 π ∫ 0 1 + t 2
= 4π [tan −1 t ] 01 = 4π [tan −1 1 − tan −1 0]
∴
x4
dx − I
1 − x4
1/ 3
−1/ 3
3
1
dx
1 + x2
1/ 3
sin x
⇒ I 2 = 4π ∫
dx − I 2
0 1 + cos 2x
− sin x dx = dt
−1
dt
I2 = − 2 π ∫
=2 π
1 1 + t2
−1/ 3
1/ 3
 2x 
x4
x4
cos −1 
dx − ∫
 dx
4
−1/ 3 1 − x4
 1 + x2
1−x
1
 1

= − ln|1 − x| + ln|1 + x| + tan −1 x
2
 2
 0
π
Put
1/ 3
[since, the integral is an even function]
2 (− x)sin (− x)
2x sin x
=
= g (x)
1 + cos 2(− x) 1 + cos 2 x
π
−1/ 3
Now, I1 =
2x sin x
g (x) =
1 + cos 2 x
1/ 3
 2y 
y4
y4
dy − ∫
cos −1 
 dy
4
−1/ 3 1 − y4
 1 + y2 
1− y
1/ 3
I = π∫
⇒
2x
Now, I1 = ∫
dx
− π 1 + cos 2 x
2x
Let f (x) =
1 + cos 2 x
−2 x
− 2x
⇒ f (− x) =
= − f (x)
=
2
1 + cos (− x) 1 + cos 2 x
⇒ f (− x) = − f (x) which shows that f (x) is an odd
function.
∴
= π∫
=π∫
π
⇒
Now, cos −1 (− x) = π − cos −1 x for −1 ≤ x ≤ 1.

1/ 3
y4 
−1  2 y 
∴I=∫
 dy
π − cos 
2 
−1/ 3 1 − y4


+
y
1


π
π
log 2 ⇒
I = (log 2)
4
8
π 2 x (1 + sin x)
dx
I=∫
− π 1 + cos 2 x
I=∫
⇒
 1 + tan x + 1 − tan x
log 
 dx


1 + tan x
π /4
=∫
…(i)
 −2 y 
y4
cos −1 
 (−1) dy
4
 1 + y2 
1− y
− 2π π
π2
+ ln (2 + 3 ) +
2
6
3
π
= [π + 3 ln (2 + 3 ) − 4 3 ]
6
∴ 2I =
3
Definite Integration 319
⇒ I=
π
[π + 3 ln (2 + 3 ) − 4 3 ]
12
3
Alternate Solution
π
Since,
cos −1 y = − sin −1 y
2
π
π
2x
−1  2 x 
−1
∴ cos 
= − 2 tan −1 x
 = − sin
 1 + x2  2
1 + x2 2
−1/ 3
∴
83. Let I = ∫
∫03

1
1 
+
 −2 +
 dx
2

1 − x 1 + x2 
π
2
∫0
=
π
2

1
1+ x
−1 
−2x + 2 ⋅ 1 log 1 − x + tan x

0
=
π
2
 2
1
3 + 1 π
+ log
+ 
−
6
3 −1
 3 2
=
π
[π + 3 log (2 + 3 ) − 4 3 ]
12
2
3
=∫
2
3
1/ 3
2x + x − 2x + 2x + 1
dx
(x2 + 1) (x4 − 1)
5
4
3
2
2x5 − 2x3 + x4 + 1 + 2x2
dx
(x2 + 1) (x2 − 1)(x2 + 1)
2
=∫
2 (x2
=∫
2x3
dx +
2 (x + 1 )2
2x3 (x2 − 1)
dx +
+ 1)2(x2 − 1)
3
3
2
I1 = ∫
3
(x2 + 1)2
3
1
∫ 2 (x2 − 1) dx
2x
dx
+ 1 )2
3
2 (x2
∴
I1 = ∫
10
5
10 1
10 1
(t − 1)
dt = ∫
dt − ∫
dt
2
5
5
t
t
t2
10
= [log t ]10
5 +
Now, integrating w.r. t. x, we get
 x3
x2 x
f (x) = a 
+
− +c
3 3
3
where, c is constant of integration.
Again, f (−2) = 0
 8 4 2
∴
f (−2) = a  − + +  + c
 3 3 3
 −8 + 4 + 2 
0=a
+c


3
2a
−2 a
+ c ⇒ c=
3
3
 x3
x2 x 2a a 3
f (x) = a 
+
− +
= (x + x2 − x + 2)
3 3
3
3
3
0=
∴
Again, ∫
⇒
Put x 2 + 1 = t ⇒ 2x dx = dt
1 
 t 
5
=
1
2
1
1
3
1
∫ 2 (x − 1) dx − 2 ∫ 2 (x + 1) dx
f (x) dx =
14
3
[given]
14
3
1 a
14
2
∫ −1 3 (0 + x + 0 + 2) dx = 3
1
a
3
+ x2 − x + 2) dx =
[Q y = x3 and y = − x are odd functions]
1
a 1 2
14
2 x dx + 4∫ 1 dx =
⇒
0
 3
3  ∫ 0
= log 10 − log 5 +
3
1
−1
∫ −1 3 (x
⇒
1 1
1
− = log 2 −
10 5
10
3
3
1
1
Again, I 2 = ∫
dx = ∫
dx
2 (x2 − 1 )
2 (x − 1 )(x + 1 )
1
1 1
= log 6 −
10 2
10
2
1

= a  x2 + x − 

3
3
⇒
3
]−
quadratic polynomial and f (x) has relative maximum
1
and x = − 1 respectively,
and minimum at x =
3
therefore, –1 and 1/3 are the roots of f ′ (x) = 0.
1
1
1


f ′ (x) = a (x + 1)  x −  = a  x2 − x + x − 
∴



3
3
3
⇒
∫ 2 (x2 + 1)2(x2 − 1) dx
−1/ 2
84. Since, f (x) is a cubic polynomial. Therefore, f ′ (x) is a
⇒
2x3 (x2 − 1) + (x2 + 1)2
dx
(x2 + 1)2 (x2 − 1)
=∫
1
1
1
4
+ log 2 − log
10 2
2
3
 4
= log [2 ⋅ 21/ 2 
 3
1/ 3
⇒ I = I1 + I 2
Now,

1 
 −1 +
 dx + 0

1 − x4 
1
=
3
= log 2 −
4


x4
2 tan −1 x is an odd function 
Q
4
1
−
x


π
I =2⋅
2
From Eq. (i), I = I1 + I 2

π
x
x
−
2 tan −1 x dx
2 ⋅
4
4
1−x

 1−x
4
1/ 3
I=∫
3

 1 
1
log (x − 1) −  log (x + 1)
=
 2
2
2
 2

1
2 1
4
= log − log
2
1 2
3
⇒
⇒
Hence,
a
3
1
 2x3
  14
+ 4x  =

3

 0 3

a
3
2
 14
 + 4 =
3

3
a
3
 14 14
⇒ a =3
  =
3
3
f (x) = x3 + x2 − x + 2
320 Definite Integration
π

x sin (2x) ⋅ sin  cos x
2

I=∫
dx
0
(2x − π )
85. Let
Then
⇒
⇒
87. We know that,
π
…(i)
π

(π − x) ⋅ sin 2 (π − x) ⋅ sin
cos(π − x)
 2

π
I=∫
dx
0
2 (π − x) − π
…(ii)
π

(π − x) ⋅ sin 2x ⋅ sin  cos x
π
2

I=∫
dx
0
π − 2x
π
I=∫
0
π

(x − π ) sin 2x ⋅ sin  cos x
2

dx
(2x − π )
… (iii)
2 sin x [cos x + cos 3x + cos 5x + K + cos (2k − 1) x]
= 2 sin x cos x + 2 sin x cos 3x + 2 sin x cos 5x
+ K + 2 sin x cos (2k − 1) x
= sin 2x + (sin 4x − sin 2x) + (sin 6x − sin 4x)
+ K + {sin 2kx − sin (2k − 2) x}
= sin 2kx
∴ 2 [cos x + cos 3x + cos 5x + K + cos (2k − 1) x]
sin 2kx
=
sin x
sin 2kx
Now, sin 2kx ⋅ cot x =
⋅ cos x
sin x
= 2 cos x [cos x + cos 3x + cos 5x + K + cos (2k − 1) x]
[from Eq. (i)]
On adding Eqs. (i) and (iii), we get
π

π
2I = ∫ sin 2x ⋅ sin  cos x dx
0

2
= [2 cos 2 x + 2 cos x cos 3x + 2 cos x cos 5x +
K + 2 cos x cos (2k − 1) x ]
π

π
2I = 2 ∫ sin x cos x ⋅ sin  cos x dx
0

2
⇒
= (1 + cos 2x) + (cos 4x + cos 2x)
+ (cos 6x + cos 4x) + ... + {cos 2kx + cos (2k − 2) x}
= 1 + 2 [cos 2x + cos 4x + cos 6x + K + cos (2k − 2) x]
+ cos 2kx

π
⇒ I = ∫ sin x cos x ⋅ sin  cos x dx
0

2
π
π
π
2 

put cos x = t ⇒ − sin x dx = dt ⇒ sin x dx = − dt

2
2
π 
/
π
2
−
2
2t
I=− ∫
⋅ sin t dt
∴
π π /2 π
4 π /2
= 2∫
t sin t dt
π −π / 2
4
I = 2 [− t cos t + sin t ]π− π/ 2/ 2
⇒
π
4
8
= 2 ×2= 2
π
π
86. Let
I=∫
π /2
0
=∫


π

π
f sin 2  − x  sin  − x dx




2
2


π /2
Then, I = ∫
0
π /2
0
…(i)
f (sin 2x) sin x dx
f [sin 2x] ⋅ cos x dx
∴∫
π /2
0
=∫
(sin 2kx) ⋅ cot x dx
π /2
0
1 ⋅ dx + 2 ∫
π /2
0
=2 ∫
=2 2 ∫
∴
I= 2
Hence, ∫
π /2
0
π /2
∫0
cos (2k) x dx
π /2
=
sin 2x sin 4x
π
sin (2k − 2) x 
+2
+
+K+
2
2
4
(2k − 2)  0

+
π /2
sin (2k) x 


2k
0
=
π
2
a
88. Let I = ∫ f (x) ⋅ g (x) dx
0
a
I = ∫ f (a − x) ⋅ g (a − x) dx
0
…(ii)
a
= ∫ f (x) ⋅ {2 − g (x)} dx
[Q f (a − x) = f (x) and g (x) + g (a − x) = 2]
f (sin 2x)(sin x + cos x) dx
π /4
0
π /4
∫0
π /4
∫0
π /4
∫0
π

f (sin 2x) sin  x +  dx

4
π

π

π
f sin 2  − x  sin  − x +  dx
4

4

4
f (cos 2x) cos x dx
f (cos 2x) cos x dx
f (sin 2x) ⋅ sin x dx = 2 ∫
a
a
0
0
= 2 ∫ f (x) dx − ∫ f (x) g (x) dx
f (sin 2x)(sin x + cos x) dx
0
=2 2
( cos 2x + cos 4x K cos (2k − 2) x) dx
0
π /4
=2 2
π/ 2
0
+
On adding Eqs. (i) and (ii), we get
2I = ∫
…(i)
⇒
a
0
a
a
∫ 0 f (x) g(x) dx = ∫ 0 f (x) dx
∴
89. Let
a
0
I = 2 ∫ f (x) dx − I ⇒ 2I = 2 ∫ f (x) dx
2a
I=∫
0
I=∫
0
2a
f (x)
dx
f (x) + f (2a − x)
…(i)
f (2a − x)
dx
f (2a − x) + f (x)
…(ii)
On adding Eqs. (i) and (ii), we get
π /4
0
f (cos 2x) cos x dx
2I = ∫
2a
0
1 dx = 2a ⇒ I = a
Definite Integration 321
π
sin α
π
=
sin α
1
90. Let I = ∫ log ( 1 − x + 1 + x ) dx
Put x = cos 2θ
⇒ dx = − 2 sin 2θ dθ
∴ I = −2 ∫
0
π /4
= −2 ∫
log [ 1 − cos 2θ + 1 + cos 2θ ] (sin 2θ ) dθ
0
= −2 ∫
π /4
0
π /4
log [ 2 (sin θ + cos θ )] sin 2θ dθ
[(log 2 ) sin 2θ
+ log (sin θ + cos θ ) ⋅ sin 2θ ] dθ
0
 − cos 2θ 

 2
π /4
= − 2 log 2
−2 ∫
0
π /4
log (sin θ + cos θ ) ⋅ sin 2θ dθ
I
II
0
 
cos 2θ 
= log 2 − 2 − log (sin θ + cos θ ) ⋅

2 π / 4
 
−∫
⇒
(π − x)
I=∫
dx
0 1 + cos α sin (π − x)
⇒
I=∫
…(i)
π
(π − x)
dx
1 + cos α sin x
… (ii)
On adding Eqs. (i) and (ii), we get
π
dx
2 I = π∫
0 1 + cos α sin x
x
sec2 dx
π
2
2I=π ∫
⇒
0
x
x
(1 + tan 2 ) + 2 cos α tan
2
2
x
x
Put tan = t ⇒ sec2 dx = 2 dt
2
2
∞
2 dt
∴
2I = π ∫
0 1 + t 2 + 2 t cos α
I=
∞
∫0
dt
(t + cos α )2 + sin 2 α
π
sin α

−1  t + cos α  
tan  sin α  

0
2I = 2π
∞
0

π
 − x ⋅ sin x cos x

π /2  2
cos 4 x + sin 4 x
0
⇒ I=
π
2
∫0
=
π
2
∫0
π /2
π / 2 x sin x ⋅ cos x
sin x cos x
dx − ∫
dx
0
sin 4 x + cos 4 x
sin 4 x + cos 4 x
π /2
π
2
dx
sin x ⋅ cos x
dx − I
sin 4 x + cos 4 x
x ⋅ sec2x
dx
tan 4 x + 1
π / 2 tan
∫0
1
π 1 π /2
⋅
d (tan 2 x)
2 2 ∫ 0 1 + (tan 2 x)2
π
π
⇒ 2 I = ⋅ [tan −1 t ]0∞ = (tan −1 ∞ − tan –1 0)
4
4
[where, t = tan 2 x]
2
π
⇒ I=
16
=
⇒
⇒ I=∫

π

π

π
 − x sin  − x ⋅ cos  − x

2

2

dx
π
π




sin 4  − x + cos 4  − x
2

2

π /2  2
⇒ 2I=
0
π
⇒ I=∫
⇒ 2I=
1
cos 2θ 

log 2 − θ +

2
2  π / 4
0
απ
sin α
π / 2 x sin x ⋅ cos x
92. Let I = ∫
dx
0
cos 4 x + sin 4 x

 cos θ − sin θ − cos 2θ 
×

 dθ 
π / 4  cos θ + sin θ
2 

1
1 π
1 π 1
log 2 −  −  = log 2 − +
2 4 2
2
2 4
π
x
91. Let I = ∫
dx
0 1 + cos α sin x
απ
π π

 −  − α   =
2
 sin α
2
I=
∴
0
1 0


= log ( 2 ) − 2 0 + ∫
( cos θ − sin θ )2 dθ


2 π /4
0
1
= log 2 − ∫
(1 − sin 2θ ) dθ
π /4
2
=
[tan −1 (∞ ) − tan −1 (cot α )]
=
0
93. Let I = ∫
1/ 2
x sin −1 x
1−x
0
2
dx Put sin −1 x = θ ⇒ x = sin θ
⇒ dx = cos θ dθ
π/6
π /6
θ sin θ
⋅ cos θ d θ = ∫
θ sin θ d θ
∴ I=∫
2
0
0
1 − sin θ
= [− θ cos θ ]π0 / 6 +
π/6
∫0
cos θ d θ
π
π
3π 1

 
 π
=  − cos + 0 + sin − sin 0 = −
+

 
 6
6
6
12
2
94. Let
π /4
I=∫
0
I=∫
0
π /4
(sin x + cos x)
dx
9 + 16 sin 2x
sin x + cos x
dx
25 − 16 (sin x − cos x)2
Put
4 (sin x − cos x) = t ⇒ 4 (cos x + sin x) dx = dt
∴
I=
1
4
0
5 + t 
dt
1 1
∫ −4 25 − t 2 = 4 ⋅ 2 (5) log 5 − t 
−4
0

 5 + 0 



5 − 4 

log 5 − 0 − log 5 + 4 



 
1 
1
1
1
=
log 9 =
(log 3)
 log 1 − log  =
40 
9
40
20
I=
1
40
322 Definite Integration
π
95. (i) Let I = ∫ x f (sin x) dx
…(i)
I=∫
…(ii)
0
⇒
π
(π − x) f (sin x) dx
0
Topic 2 Periodicity of Integral Functions
1. Let
I=∫
On adding Eqs. (i) and (ii), we get
2I=∫
∴
π
∫0
π
0
π
x f (sin x) dx =
2
I=∫
(ii) Let
3/ 2
−1
=∫
π f (sin x) dx
1
−1
+∫
|x sin πx| dx
x sin πx dx +
3/ 2
∫1

 − x cos πx 3/ 2
3 / 2  − cos πx
−∫ 
−
 dx


1 

π
π
1


1
1
1
0
So,
π
2
1
+∫
−1
= ∫ −π
2
I = ∫ (t x + 1 − x) dx = ∫
0
=∫
1
0
0
[(1 − x) + t x] dx
[ C 0 (1 − x) + C1 (1 − x)
n
n
n
n
n −1
(t x)
C 2(1 − x)n − 2(tx)2+ ... + nC n (tx)n + ] dx
n
=∫
=
1
0

 n n
C r (1 − x)n − r (t x)r  dx
rΣ
=0


n
Σ
r=0
n
r=0
1
C r ∫ (1 − x)n − r ⋅ xr dx t r
 0

n
1
∫ 0 (1 − x)
n−k k
x dx =
∫0
4
π
2
1
+∫
dx
5
−20 + 10 + 5 − 4 5π + π
+
20
10
9
3π
3
=−
+
=
(4π − 3)
20
5 20
1
(n + 1)nC k
2.
3 + 3T
∫3
f (2x) dx Put 2x = y ⇒ dx =
∴
1
2
6 + 6T
∫6
f ( y) dy =
1
dy
2
6I
= 3I
2
x
3. Given, g (x) = ∫ f (t ) dt
0
1
0
0
2
1
1
≤ f (t ) ≤ 1 for t ∈ [0,1]
2
Now,
We get
2
g (2) = ∫ f (t ) dt = ∫ f (t ) dt + ∫ f (t )dt
…(ii)
1
1
C r ∫ (1 − x)n − r ⋅ xr dx t r=
(1 + t + K + t n )
 0

n+1
n
On equating coefficient of t k on both sides, we get
1
1
n
C k ∫ (1 − x)n − k ⋅ xk dx =
 0
 n + 1
⇒
1 dx
=
⇒
From Eqs. (i) and (ii), we get
Σ
dx
∫ −1 2 +
0
1 1 1  π π 

=  −1 + + −  +  + 

2 4 5  2 10 
1
1
dx
+
1
dx
1+0+4
1
1 π
π 1


=  −1 +  + (0 + 1) + (1 − 0) +  − 1


2 2
4
5 2
 ((t − 1) x + 1)n + 1 
1  t n + 1 − 1
=
=



 (n + 1) (t − 1)  0 n + 1  t − 1 
1
…(i)
(1 + t + t 2 + K + t n )
=
n+1
n
− π / 2 < x < −1
−1 ≤ x < 0
0 ≤ x<1
1 ≤ x < π /2
3/ 2
sin πx 
 π 
1
I = ∫ (t x + 1 − x)n dx = ∫ {(t − 1) x + 1 }n dx
1
π
dx
dx
+∫2
0 [x] + [sin x] + 4
1 [x] + [sin x] + 4
1
[Q For x < 0,−1 ≤ sin x < 0 and for x > 0, 0 < sin x ≤ 1]
−1
0
1
dx
dx
dx
I = ∫ −π
+∫
+∫
−1 − 1 − 1 + 4
00+0+4
2
1
4
−
−
+
2
2 2
1
1
= + 2 (0 − 0) + + 2 (+ 1 − 0)
π π
π π
3 1  3π + 1
= + 2=

 π2 
π π
0
dx
0
∫−1 [x] + [sin x] + 4
 − 1 , − π / 2 < x < −1

and [sin x] = −01, , − 01 << xx << 10
 0, 1 < x < π / 2
− x sin πx dx
 1  2 sin πx   1  1
=2   + ⋅
− −  −
 π  π  π  0  π  π
− 2 ,

[x] =  0−,1,
 1,
Q
1
Again,
dx
+
[x] + [sin x] + 4
π
1
 − cos πx
 x cos πx 
− 2 ∫ 1⋅ 
=2 −
 dx


0

π
π 
0
96. Let
dx
[x] + [sin x] + 4
∫ 0 f (sin x) dx
 x sin πx, −1 < x ≤ 1
3
Since, |x sin πx| = 
− x sin πx, 1 < x < 2
∴ I=∫
π
2
−π
2
−1
−π
2
11
1
1
∫ 0 2 dt ≤ ∫ 0 f (t ) dt ≤ ∫ 0 1 dt
1
1
≤ ∫ f (t ) dt ≤ 1
0
2
1
for t ∈ [1, 2]
0 ≤ f (t ) ≤
2
⇒
Again,
2
2
2
⇒
∫ 10 dt ≤ ∫ 1 f (t ) dt ≤ ∫ 1 dt
⇒
0 ≤ ∫ f (t ) dt ≤
2
1
1
2
…(i)
…(ii)
Definite Integration 323
2
From Eqs. (i) and (ii), we get
1
2
1
3
≤
f (t ) dt + ∫ f (t ) dt ≤
1
2 ∫0
2
1
3
⇒
≤ g (2) ≤ ⇒ 0 ≤ g (2) < 2
2
2
4.
nπ + v
∫0
π
|sin x|dx = ∫
|sin x|dx + ∫
0
+∫
n
=∑∫
r =1
nπ
( n − 1 )π
rπ
( r − 1 )π
2π
π
0
1
Now,
0
nπ + v
nπ + v
∫ nπ
rπ
r −1
0
10
π
π
0
0
sin t dt
| sin x|dx, putting x = nπ + t
nπ + v
∫ nπ
v
v
0
0
| sin x|dx = ∫ |(−1)n sin t|dt = ∫
n
nπ + v
|sin x|dx = ∑ ∫
0
r =1
=
n
φ (a ) = ∫
a+ t
a
| sin x|dx +
( r − 1 )π
∑2+
r =1
5. Let
rπ
2 10
π
f (x) cos π x dx = 4
10 −∫10
|sin x|dx
|sin x|dx = 2n + 1 − cos v
f (x) dx is independent of a.
if [x] is odd.
 x − [x],
1
if
+
[
x
]
−
x
,
[x] is even.

6. Given, f (x) = 

d
Q
 dx

(1 + x2) f (x) = 2x ⇒ f (x) =
f (x) cos π x dx = 2
− 10
∫
f (x) cos π x dx
d
d
d  ψ (x )
φ ( x )
ψ ( x ) − f { φ ( x )} 
f(t ) dt  = f { ψ ( x )} 

dx  ∫ φ ( x )
 dx
 dx


⇒
X
1 2
9 10
2
 1
3
2 −2   2 − 1
 4
24
 1
2
∴f′   =
= 2 =
=
=
2
2
25 25
 2 
2 2
1
5




1
1 + 
 
1 +   
16

 4
4

2 

∴
Also,
0
–2 –1 0
(1 + x2)(2) − (2x) (0 + 2x) 2 + 2x2 − 4x2 2 − 2x2
=
=
(1 + x2)2
(1 + x2)2
(1 + x2)2
 1
2 −2  
 2
Given,
Y
–10 – 9
2x
1 + x2
2. PLAN Newton-Leibnitz’s formula
10
10
∫
x


ψ( x )
d
d
ψ (x) − f (φ (x))
φ(x)
 ∫ f (t )dt  = f (ψ (x))
dx
dx


φ ( x )

⇒
f (x) and cos π x both are periodic with period 2 and both
are even.
∴
0
f (x) = 2x + 0 − x2f (x)
f ′ (x) =
∴ φ (a ) is constant.
a+ t
1
On differentiating w.r.t. ‘x’ we get
f (x) dx
φ ′ (a ) = f (a + t ) ⋅ 1 − f (a ) ⋅ 1 = 0 [given, f (x + t ) = f (x)]
∫a
x
1. Given, ∫ f (t ) dt = x2 + ∫ t 2f (t )dt
sin t dt
nπ + v
∫ nπ
Topic 3 Estimation, Gamma Function and
Derivative of Definite Integration
On differentiating w.r.t. a, we get
⇒
0
40
π2
On differentiating both sides, w.r.t. ‘x’, we get
nπ + v
∫ nπ
1
f (x) cos πx dx = − 20 ∫ u cos π u du =
− 10
= [− cos t ]v0 = − cos v + cos 0 = 1 − cos v
∴∫
∫
∴
= [− cos t ] π0 = − cos π + cos 0 = 2
nπ + v
cos πx dx
sin t| dt
= ∫ |sin t| dt = ∫
Then,
1
1
⇒
π
nπ
1
= − ∫ u cos π u du
when x = (r − 1) π, t = 0 and when
Again, ∫
2
∫ ( r −1) π| sin x |dx , we have
∫ ( r − 1)π|sin x| dx = ∫ 0 |(−1)
0
2
∫ f (x) cos π x dx = ∫ (x − 1)
and
|sin x| dx
x = r π, t = π
∴
1
0
|sin x|dx
nπ
x = ( r − 1)π + t
sin x = sin [(r − 1) π + t ] = (−1)r − 1 sin t
⇒
1
= ∫ (1 − x) cos π x dx = − ∫ u cos π u du
rπ
Now to solve,
and
∫ f (x) cos π x dx
|sin x|dx + ...
|sin x|dx + ∫
|sin x| dx +
= 10 ∫ f (x) cos π x dx
F (x) = ∫
x2
0
f ( t ) dt
F ′ (x) = 2x f (x)
F ′ (x) = f ′ (x) ⇒
f ′ (x)
= 2x
⇒
f (x)
⇒
In f (x) = x + c
⇒
f (x) = ex
⇒
f (x) = K ex
∫
2x f (x) = f ′ (x)
f ′ (x)
dx = ∫ 2x dx
f (x)
2
2
+ c
2
[K = ec ]
324 Definite Integration
Now,
∴
f (0) = 1
1=K
Hence,
f (x) = ex
2
4
F (2) = ∫ et dt = [et ]40 = e4 − 1
0
x
3. Given, y = ∫ |t| dt
0
∴
dy
= | x|⋅ 1 − 0 = |x|
dx
t2
2 5
t
5
Using Newton Leibnitz's formula, differentiating both
sides, we get
d

d

t 2{ f (t 2)}  (t 2) − 0 ⋅ f (0)  (0) = 2t 4
 dt

 dt

6. Here,
[by Leibnitz’s rule]
∫0
x f (x) dx =
⇒
t 2f (t 2)2t = 2t 4
∴
2
4
f   =−
 25
5
⇒
4 2
f   =
 25 5
x
Q Tangent to the curve y = ∫ |t| dt , x ∈ R are parallel
0
to the line y = 2x
∴ Slope of both are equal ⇒ x = ± 2
y=∫
Points,
± 2
0
7. Given,
| t | dt = ± 2
y − 2 = 2 (x − 2) and y + 2 = 2 (x + 2)
0 − 2 = 2 (x − 2) and 0 + 2 = 2 (x + 2) ⇒ x = ± 1
4. Given ∫
0
= e− ( x
x
1 − { f ′ (t )} dt = ∫ f (t ) dt , 0 ≤ x ≤ 1
Differentiating both sides w.r.t. x by using Leibnitz’s
rule, we get
1 − { f ′ (x)} = f (x) ⇒
⇒
∫
f ′ (x)
1 − { f (x)}
2
dx = ±
f ′ (x) = ± 1 − { f (x)}
2
[Q f (0) = 0]
sin x < x, ∀ x > 0
 1 1
 1 1
sin   < and sin   <
 2 2
 3 3
 1 1
 1 1
f   < and f   <
 2 2
 3 3
1
sin x
4
⋅ 2x − e− ( x
2
+ 2x + 1 )
2
∴
)
⋅ 2x
(1 − e2x
2x + 1
[where, e
2 2
2
+1
)
> 1, ∀ x and e− ( x
4
+ 2x 2 + 1 )
> 0, ∀ x]
f ′ (x) > 0
1
As we know that,
5. Since ∫
+ 1 )2
0
x = 0 ⇒ sin { f (0)} = c
c = sin −1 (0) = 0
f (x) = ± sin x
f (x) ≥ 0, ∀ x ∈ [0, 1]
f (x) = sin x
⇒
2
8. Here, I (m, n ) = ∫ t m(1 + t )n dt reduce into I (m + 1, n − 1)
−1
∴
dt
which shows 2x < 0 or x < 0 ⇒ x ∈ (−∞ , 0)
∫ dx
⇒ sin −1 { f (x)} = ± x + c
Put
⇒
∴
but
∴
+ 1 )2 
= 2xe− ( x
0
2
e
x2
d 2


− ( x 2 )2  d
(x2)

 (x + 1) − e
 dx

 dx

2
f ′ (x) = e− ( x
For x intercept put y = 0, we get
2
2

putting t =

5 
x 2 + 1 −t 2
f (x) = ∫
f (t 2) = t
On differentiating both sides using Newton’s Leibnitz’s
formula, we get
Equation of tangent is
x
⇒
t 2 f (t ) dt = 1 − sin x, thus to find f (x).
On differentiating both sides using Newton Leibnitz
formula
d 1
d
i.e.
t 2 f (t ) dt =
(1 − sin x)
dx ∫ sin x
dx
⇒ {12 f (1)} ⋅ (0) − (sin 2 x) ⋅ f (sin x) ⋅ cos x = − cos x
1
f (sin x) =
⇒
sin 2 x
 1
For f   is obtained when sin x = 1 / 3
 3
 1
i.e.
f   = ( 3 )2 = 3
 3
[we apply integration by parts taking (1 + t )n as first
and t m as second function]
1
1

tm + 1 
tm + 1
∴ I (m, n ) = (1 + t )n ⋅
− ∫ n (1 + t )( n − 1) ⋅
dt

0
m + 1 0
m+1

1
n
2n
(1 + t )( n − 1) ⋅ tm + 1 dt
=
−
∫
m+1 m+1 0
n
2n
∴ I (m, n ) =
−
⋅ I (m + 1, n − 1)
m+1 m+1
9. Given,
f (x) = ∫
x
1
2 − t 2 dt ⇒
f ′ (x) = 2 − x2
Also, x2 − f ′ (x) = 0
∴
⇒
x2 = 2 − x 2
⇒ x4 = 2 − x 2
x4 + x2 − 2 = 0 ⇒ x = ± 1
10. Given, F (x) = ∫
x
0
f (t ) dt
By Leibnitz’s rule,
F ′ (x) = f (x)
But
…(i)
F (x2) = x2 (1 + x) = x2 + x3
⇒
F (x) = x + x3/ 2 ⇒
⇒
f (x) = F ′ (x) = 1 +
⇒
f (4) = 1 +
F ′ (x) = 1 +
3 1/ 2
x
2
[given]
3 1/ 2
x
2
[from Eq. (i)]
3 1/ 2
3
(4) ⇒ f (4) = 1 + × 2 = 4
2
2
Definite Integration 325
1
x
∫ 0 f (t )dt = x + ∫ x t f (t ) dt
11. Given,
On differentiating both sides w.r.t. x, we get
f (x) 1 = 1 − x f (x) ⋅ 1 ⇒ (1 + x) f (x) = 1
1
1
1
⇒
f (x) =
⇒ f (1) =
=
1+ x
1+1 2
f ( x)
f ( x)
12. lim ∫
x→1
2t
dt = lim
x→1
x−1
∫4
2t dt
x−1
[using L’ Hospital’s rule]
2 f (x) ⋅ f ′ (x)
= 2 f (1) ⋅ f ′ (1)
= lim
x→1
1
[Q f (1) = 4]
= 8 f ′ (1)
4
13. If f (x) is a continuous function defined on [a , b], then
b
m (b − a ) ≤ ∫ f (x) dx ≤ M (b − a )
where, M and m are maximum and minimum values
respectively of f (x) in [a , b].
−x 2
Here, f (x) = 1 + e
is continuous in [0, 1].
x2
0 < x<1 ⇒x < x ⇒ e
2
Now,
x2
Again, 0 < x < 1 ⇒ x > 0 ⇒ e
2
< e ⇒e
−x 2
> e ⇒e
0
<1
2
e− x < e− x < 1, ∀ x ∈ [0, 1]
⇒
1 + e− x < 1 + e− x < 2, ∀ x ∈ [0, 1]
2
1
−x
∫ 0 (1 + e
1
1
2
) dx < ∫ (1 + e− x ) dx < ∫ 2 dx
2−
⇒
0
0
1
2
1
< ∫ (1 + e− x ) dx < 2
0
e
x2 x4 x6
+
−
+ ...
2! 4! 6!
x3
x5 x7
sin x = x −
+
−
+ ...
3! 5! 7!
cos x = 1 −
14. Q
and
∴
−x
>e
∴
⇒
1
1

3
x
x
2
∫ 0 x cos x dx ≥ ∫ 0  x − 2 ! dx =  2
⇒
and,
−
sin 2x
sin x
sin −1 (t )dt
g′ (x) = 2 cos 2x sin −1 (sin 2x) − cos x sin −1 (sin x)
 π
 π
g′   = − 2 sin −1 (0) = 0 ⇒ g′  −  = − 2 sin −1 (0) = 0
 2
 2
No option is matching.
16. Here, f (x) + 2x = (1 − x)2 ⋅ sin 2 x + x2 + 2x
where, P : f (x) + 2x = 2 (1 + x)2
and,
4 1
1 1 3
x
= − =

8 0 2 8 8
∫ 0 x cos x dx ≥ 8
1 2
x4 
∫ 0 x sin dx ≥ ∫ 0  x − 6  dx
1
 x3 x5 
1 1
9
3
= −
 = 3 − 30 = 30 = 10
3
30
0

1
3
∫ 0 x sin x dx ≥ 10
1 2
1 3
x5 
∫ 0 x cos x dx ≥ ∫ 0  x − 2  dx
∴
6 1
x
x
1 1
2 1
= −
 = 4 − 12 = 12 = 6
4
12
0

1 2
1
∫ 0 x cos x dx ≥ 6
4
…(ii)
2 (1 + x2) = (1 − x)2 sin 2 x + x2 + 2x
(1 − x)2 sin 2 x = x2 − 2x + 2
⇒
(1 − x)2 sin 2 x = (1 − x)2 + 1 ⇒ (1 − x)2 cos 2 x = − 1
∴ P is false.
Again, let Q : h (x) = 2 f (x) + 1 − 2x (1 + x)
where,
h (0) = 2 f (0) + 1 − 0 = 1
h (1) = 2 f (1) + 1 − 4 = − 3, as h (0) h (1) < 0
⇒ h (x) must have a solution.
∴ Q is true.
17. Here, f (x) = (1 − x)2 ⋅ sin 2 x + x2 ≥ 0, ∀ x.
x  2 (t − 1 )

and g (x) = ∫ 
− log t f (t ) dt
1 t + 1


2 (x − 1)
g′ (x) = 
(x)
− log x ⋅ f {
x
(
)
1
+
 + ve

…(i)
For g′ (x) to be increasing or decreasing,
2 (x − 1)
let φ(x) =
− log x
(x + 1)
φ′ (x) =
4
1 − (x − 1)2
− =
2
x x (x + 1)2
(x + 1)
φ′ (x) < 0, for x > 1 ⇒ φ (x) < φ (1)
⇒
φ(x) < 0
…(ii)
From Eqs. (i) and (ii), we get
1
⇒
…(i)
⇒
⇒
3
1
15. g (x) = ∫
which is never possible.
−x 2
x
∴
∴
a
1
1 3
 x4 x6 
x5 
∫ 0 x sin x dx ≥ ∫ 0  x − 6  dx =  4 − 36 
0
1 1
8 2
= −
=
=
4 36 36 9
1 2
2
∫ 0 x sin x dx ≥ 9
1 2
and,
g′ (x) < 0 for x ∈ (1, ∞ )
∴g (x) is decreasing for x ∈ (1, ∞ ).
sec x
cos x
18. Given, f (x) = cos 2 x cos 2 x
1
2
cos x
1
R3 ,
cos x
sec x cos x
cosec x ⋅ cotx + sec2x
cosec2x
cos 2 x
Applying R3 →
f (x) = cos x cos x cos x
sec x cos x
2
2
cosec x ⋅ cot x + sec2 x
cosec2 x
cos x
326 Definite Integration
Applying R1 → R1 − R3 ⇒ f (x)
cosec x ⋅ cot x + sec2x − cos x
0
0
= cos x cos 2 x cos 2 x
sec x cos x
⇒
cosec2x
a
∴
∫0
3
2
f (x) dx = −


Q


π /2
∫0
2
=∫
∫0


π /2
∫ 0 f (x) dx = − 


5
n +2
2
2
sinm x ⋅ cos n x dx =
m +n+2
2
2
3
1
6
1 
⋅
⋅
2
2
2
2 
+

2 2
7 
2
2 







2 π
8

1 /2 ⋅ π
 15π + 32
π
=−
+

 = − 
= −  +
5 3 1
2
60 
4
15



2⋅ ⋅ ⋅ ⋅ π 


2 2 2
19. f (x) = ∫
x
1
ln t
dt for x > 0
1+ t
f (1 / x) = ∫
Now,
1/ x
1
[given]
ln t
dt
1+ t
Now,
=∫
 1
f (x) + f   = ∫
 x
x
ln t
dt +
1 (1 + t )
⇒
⇒
⇒
t =4 − a − a
t = 4 − 2a
t
a =2 −
2
and
t = b − (4 − b)
⇒
t = 2b − 4
t
= b −2
2
t
b =2 +
2
⇒
⇒
2 + t/ 2
g (x) dx
1 
t 1 
g 2 −  + g 2 +

2
2 2 
t

2
[given]
We get g (2 + t / 2) − g (2 − t / 2) > 0, ∀ t > 0
F ′ (t ) > 0, ∀ t > 0
So,
Hence, F (t ) increases with t, therefore F (t ) increases as
(b − a ) increases.
21. Let I n = ∫
1
0
ex (x − 1)n dx
Put x − 1 = t
In = ∫
⇒
dx = dt
0 t+1
−1
e
⋅ t ndt = e ∫
= e 0 − (−1)n e−1 − n

ln t
∫ 1 t (1 + t ) dt
x
= (−1)n + 1 − ne ∫
⇒
0 n t
−1
t e dt
0
−1
0
∫ −1 t
e dt

n −1 t
t n − 1et dt
I n = (−1)n + 1 − nI n − 1
…(i)
1 x
For n = 1, I1 = ∫ e (x − 1) dx = [e (x
x
0
− 1)]10
1
− ∫ ex dx
0
= e1 (1 − 1) − e0 (0 − 1) − [ex ] 10 = 1 − (e − 1) = 2 − e
Therefore, from Eq. (i), we get
I 2 = (−1)2 + 1 − 2I1 = − 1 − 2(2 − e) = 2e − 5
Put x = e,
20. Let t = b − a and a + b = 4
∫0
0
= e  [t net ] −01 − n ∫ t n − 1et dt


−1
x ln t
1
1
t ) ln t
dt = ∫
dt = [(ln t )2]1x = (ln x)2
1
2
2
t (1 + t )
t
1
 1 1
f (e) + f   = (ln e)2 =
 e 2
2
g (x) dx +
g (x) dx
Now, 2 − t / 2 < 2 + t / 2 ∴ t > 0
x (1 +
1
∫0
dg
> 0, ∀ x ∈ R
dx
Again,
∴
x ln t
ln u
du = ∫
dt
1 u (u + 1 )
1 t (1 + t )
x
0
2 + t/ 2
g (x) dx +
[using Leibnitz’s rule]
1 
t 1 
t
= g 2 +  − g 2 − 
2 
2 2 
2
Put t = 1 / u ⇒ dt = (−1 / u 2) du
x ln (1 / u ) (−1 )
⋅ 2 du
∴
f (1 / x) = ∫
1 1 + 1 /u
u
=∫
0
2 − t/ 2
For t > 0, F ′ (t ) = −
(sin x + cos x) dx
2
F (x) = ∫
Let
2 − t/ 2
5
m+ 1
π /2
0
3
= − sin x − cos x (1 − sin x) = − sin x − cos x
π /2
b
0
= (cosecx ⋅ cot x + sec x − cos x) ⋅ (cos x − cos x) ⋅ cos x
sin 2 x + cos3 x − cos3 x ⋅ sin 2 x 
2
2
=−
 ⋅ cos x ⋅ sin x
sin 2 x ⋅ cos 2 x


2
[given]
Now, ∫ g (x) dx + ∫ g (x) dx
cos x
2
a <2
π
π
2 − < 2 ⇒ >0 ⇒ t >0
2
2
Again,
Hence proved.
[given]
and
I3 = (−1)3 + 1 − 3I 2 = 1 − 3(2e − 5) = 16 − 6e
Hence, n = 3 is the answer.
22. Since, f is continuous function and ∫
x
0
f (t ) dt → ∞ ,
as| x|→ ∞. To show that every line y = mx intersects the
x
curve y2 + ∫ f (t ) dt = 2
0
Y
(0,√2)
A
O
B (0,–√2)
(Xp,0)
X
Definite Integration 327
At x = 0, y = ± 2
Hence, (0, 2 ), (0, − 2 ) are the point of intersection of
the curve with the Y-axis.
As x → ∞, ∫
x
∫0
x
0
f (t ) dt → ∞ for a particular x (say xn ), then
x
0
23. Given, f (x) = ∫ [2(t − 1) (t − 2) + 3 (t − 1) (t − 2) ] dt
2
1
=
lim 2n
n
∑
n → ∞ r=1 n 2 + r 2
0
F ′ (π ) = f (π ) = − 6
π
0
= [cos x f (x)]0π +
… (i)
{by integration by parts}
= (− 1) (− 6) − f (0) + [(sin x) F (x)]π0
⇒
π
∫ 0 F (x) cos x dx
= 6 − f (0) = 2
f (0) = 4
(given)
1
2
2n  




log 1 + n  + log 1 + n  + ... + log 1 + n  


 (n + 1)1/3
(n + 2)1/3
(2n )1/3 
1. Let p = lim 
+
+…+

4
3
4
3
/
/
n→ ∞ 
n
n
n 4/3 
n→ ∞
n
(n + r )1/3
n 4/3
r =1
∑
⇒
⇒
log l = lim
n→ ∞
1
n
2n

r
∑ log 1 + n 
r =1
2
log l = ∫ log (1 + x) dx
0
2
⇒
1/3
r

1/3
 n
n 1 +

1
n
= lim ∑
= lim
4/3
n→ ∞
n→ ∞ n
n
r =1
pn


p
1  r
Q lim ∑ f   = ∫ 0 f (x)dx
n→∞
n n
r =1


Taking log on both sides, we get
1 
1 
2
2 n  


log l = lim log 1 +  1 +  ... 1 +


n→ ∞ n 
n 
n
n  

1
⇒ log l = lim
n→ ∞ n
Topic 4 Limits as the Sum
= lim
2 dx
1
=∫
0
n
1 + x2
  n + 1  n + 2
 n + 2n   n
= lim  

 K
 
 n  
n→ ∞   n   n 
π
π
⋅
1
∫ 0 f (x) sin x dx + ∫ 0 F (x) cos x dx
0
 r
1+  
 n
2
1
π
− ∫ F (x) cos x dx +
r =1
1
 (n + 1) ⋅ (n + 2) ... (n + 2n ) n
= lim 


n→ ∞ 
n 2n
∫ 0 F (x) cos xdx
π
n→ ∞
2n
∑
(n + 1) ⋅ (n + 2) K (3n ) n
3. Let l = lim 


n→ ∞ 
n 2n
( f ′ (x) + F (x)) cos x dx
= ∫ f ′ (x) cos xdx +
lim
1
x
F (x) = ∫ f (t )dt , where f (π ) = − 6
∫0
0
= [tan −1 x]20 = tan −1 2
and F : [0, π ] → R ,
Now,

3
(1 + x)4/3

4
lim  n

n
n
n
+
+
+ ....+ 2


n → ∞  n 2 + 12 n 2 + 22 n 2 + 32
n + (2n )2
=
7/5
24. It is given that, for functions f : R → R
π

r =1
=
+
∴ f (x) attains maximum at x = 1 and f (x) attains
7
minimum at x = .
5
⇒
n
r

∑ 1 + n 
r =1
Now, as per integration as limit of sum.
r
1
Let = x and = dx
n
n

lim  n
1
n
n
+ 2
+ 2
+ ...+

 2
2
2
2

n→ ∞ n +1
5n 
n +2
n +3
= (x − 1) (x − 2)2 (5x − 7)
1
1
2. Clearly,
= (x − 1) (x − 2)2 [2 (x − 2) + 3 (x − 1)]
–
 r
∑ f  n  = ∫0 f (x) dx
3
3
3
= (24/3 − 1) = (2)4/3 −
4
4
4
f ′ (x) = [2 (x − 1) (x − 2)3 + 3 (x − 1)2(x − 2)2] ⋅ 1 − 0
+
n
1
=
2
3

1
Q lim
 n → ∞ n
1
So, p = ∫ (1 + x)1/3 dx
f (t ) dt = 2 and for this value of x, y = 0
The curve is symmetrical about X-axis.
Thus, we have that there must be some x, such that
f (xn ) = 2.
Thus, y = mx intersects this closed curve for all values of
m.
∴
Then, upper limit of integral is 1 and lower limit of
integral is 0.
1/3
[Q n → ∞]
⇒


1
log l = log (1 + x) ⋅ x − ∫
⋅ x dx
+
x
1

0
2 x+ 1 −1
2
log l = [log (1 + x) ⋅ x]0 − ∫
dx
0
1+ x
⇒
2
1 
log l = 2 ⋅ log 3 − ∫ 1 −
 dx
0 
1 + x
⇒
log l = 2 ⋅ log 3 − [x − log 1 + x ]20
328 Definite Integration
⇒
⇒
⇒
log l = 2 ⋅ log 3 − [2 − log 3]
log l = 3 ⋅ log 3 − 2
log l = log 27 − 2
∴
l = elog 27 − 2 = 27 ⋅ e− 2 =
n
n




3
3
3
1 + 2 + 3 + ... + n
,
lim 
n→ ∞ 

1
1
1
7/3 
+
+ ... +

n 
 (an + 1)2 (an + 2)2
(an + n)2 

a ∈ R,|a| > 1
1/3
n
n
 r 1
Σ  
Σ (r1/3 )
r = 1  n
n
r =1
= lim
= lim n
1
1
1
n→ ∞
n → ∞ 7/3 n
Σ
Σ
n
2
r =1 
r = 1 (an + r )2
r n
a + 

n
1
1/3
=
∫
0
dx
(a + x)2
= 54,
(given)
3 4/3 1
[x ]0
3 /4
4
= 54
= 54 ⇒
1
1
1

1 
−
+
a+1 a
− x + a 

0
3
1
=
4 × 54 a (a + 1)
⇒
⇒
⇒
a 2 + a = 72
2
⇒
a + 9a − 8a − 72 = 0
⇒
a (a + 9) − 8(a + 9) = 0
⇒
(a − 8) (a + 9) = 0
⇒
a = 8 or− 9
Hence, options (c) and (d) are correct.
5. PLAN
 r
2∑  
 n
Converting Infinite series into definite Integral
h( n)
i.e.
lim
n→ ∞ n
h ( n)
1
 r
lim
∑ f  n  = ∫ f(x )dx
n→ ∞ n
r = g ( n)
g ( n)
n→ ∞ n
lim
r
is replaced with x.
n
Σ is replaced with integral.
where,
n→ ∞
1
1a + 2a + K + n a
=
a −1
(n + 1) {(na + 1) + (na + 2) + K + (na + n )} 60
1
60
a
n→ ∞
2 ⋅ [xa+ 1 ] 10
1
=
(2a + 1) ⋅ (a + 1) 60
2
1
=
⇒ (2a + 1) (a + 1) = 120
(2a + 1) (a + 1) 60
⇒
⇒ 2a 2 + 3a + 1 − 120 = 0 ⇒ 2a 2 + 3a − 119 = 0
−17
⇒ (2a + 17) (a − 7) = 0 ⇒ a = 7,
2
n
n
2
+
+ k2
n
kn
k=0
6. Given, S n = ∑




1 
1
 < lim
=∑ ⋅
n 
k k2  n → ∞
k=0
1
+
+



n n2 
n
=∫
1
0




1
1
∑ n  k k  2

k=0
1 + +   

n  n 
n
1
dx
1 + x + x2
1
 2
1  
 2 
tan −1 
=
x +  


2   0
3
3

2  π π
π
⋅ −  =
3  3 6 3 3
π
Similarly,
Tn >
3 3
=
i.e. S n <
π
3 3
 1
1
1  5n 1
+
+K+
 =∑
n→ ∞  n + 1
n+2
6n  r=1 n + r
7. lim 
= lim
n→ ∞
Here,
lim
=
1
r =1
=
a −1
60
1


⋅ (2na + n + 1)
1 + 

n
a
1
1  n  r 
1
=
⇒ lim 2 ∑    ⋅ lim
a −1
 n→ ∞


n→ ∞ n 
60
n
1
1





 r =1
⋅ 2a + 1 + 
1 + 


n
n
1 a
1
1
⇒ 2∫ (x ) dx ⋅
=
0
1 ⋅ (2a + 1) 60
⇒ lim
∴
dx
0
1
n (n + 1) 

(n + 1)a − 1 ⋅ n 2a +


2
n→ ∞
27
e2
ra
r =1
⇒ lim
4. Since,
∫x
∑
=∫
5
0
1
n
5n
∑
r =1
1
r
1 + 

n
dx
= [log (1 + x)] 50 = log 6 − log 1 = log 6
1+ x
13
Area
Topic 1 Area Based on Geometrical Figures
Without Using Integration
Objective Questions I (Only one correct option)
1. The area (in sq. units) of the largest rectangle ABCD
whose vertices A and B lie on the X-axis and vertices
C and D lie on the parabola, y = x2 − 1 below the
(2020 Main, 4 Sep II)
X-axis, is
(a)
4
3 3
(b)
2
3 3
(c)
1
(d)
3 3
4
3
2. If the area enclosed between the curves y = kx2 and
x = ky2, (k > 0), is 1 square unit. Then, k is
(2019 Main, 10 Jan I)
(a)
3
(b)
1
3
(c)
2
3
(d)
3
2
3. The area (in sq units) of the region {(x, y) : y2 ≥ 2x
and x2 + y2 ≤ 4x, x ≥ 0, y ≥ 0} is
4
3
4 2
(c) π −
3
(2016 Main)
8
3
π 2 2
(d) −
2
3
(a) π −
(b) π −
the parabola y2 = 8x touch the circle at the points
P,Q and the parabola at the points R, S. Then, the
area (in sq units) of the quadrilateral PQRS is
(2014 Adv.)
(b) 6
(c) 9
end points of latusrectum to the ellipse
(a) 27/4 sq units
(c) 27/2 sq units
(d)15
5. The area of the equilateral triangle, in which three
coins of radius 1 cm are placed, as shown in the
figure, is
x2 y2
+
= 1, is
9
5 (2003, 1M)
(b) 9 sq units
(d) 27 sq units
7. The area (in sq units) bounded by the curves y = | x | − 1 and
(2002, 2M)
y = − | x | + 1 is
(a) 1
(c) 2 2
(b) 2
(d) 4
8. The triangle formed by the tangent to the curve
f (x) = x2 + bx − b at the point (1,1) and the coordinate axes,
lies in the first quadrant. If its area is 2 sq units, then the
value of b is
(2001, 2M)
(a) – 1
4. The common tangents to the circle x2 + y2 = 2 and
(a) 3
6. The area of the quadrilateral formed by the tangents at the
(b) 3
(c) – 3
(d) 1
Objective Question II
(One or more than one correct option)
9. Let P and Q be distinct points on the parabola y2 = 2x such
that a circle with PQ as diameter passes through the vertex
O of the parabola. If P lies in the first quadrant and the
area of ∆OPQ is 3 2, then which of the following is/are the
(2015 Adv.)
coordinates of P ?
(a) (4 , 2 2 ) (b) (9 , 3 2 )
1 1 
(c)  ,

4
2
(d) (1, 2 )
Integer & Numerical Answer Type Questions
10. A farmer F1 has a land in the shape of a triangle with
(2005, 1M)
(a) (6 + 4 3 ) sq cm
(c) (7 + 4 3 ) sq cm
(b) (4 3 − 6) sq cm
(d) 4 3 sq cm
vertices at P(0, 0), Q(1, 1) and R(2, 0). From this land, a
neighbouring farmer F2 takes away the region which lies
between the sides PQ and a curve of the form y = xn (n > 1).
If the area of the region taken away by the farmer F2 is
exactly 30% of the area of ∆PQR, then the value of n is
.................... .
(2018 Adv.)
330 Area
y = f (x). If x ∈ (− 2, 2), the equation implicitly defines a
unique real-valued differentiable function y = g (x),
(2008, M)
satisfying g (0) = 0.
Fill in the Blanks
11. The area of the triangle formed by the positive X-axis
and the normal and the tangent to the circle x2 + y2 = 4
(1989, 2M)
at (1, 3 ) is … .
14. If f (− 10 2 ) = 2 2 , then f′′ (− 10 2 ) is equal to
12. The area enclosed within the curve|x| + | y| = 1 is ....... .
(1981, 2M)
Analytical & Descriptive Question
(a)
4 2
4 2
73 32
(c)
4 2
(d) −
73 3
4 2
73 3
15. The area of the region bounded by the curve y = f (x),
the X-axis and the lines x = a and x = b, where
− ∞ < a < b < − 2, is
1
) be the vertices of a
3
triangle. Let R be the region consisting of all those
points P inside ∆ OAB which satisfy d (P , OA ) ≥ min
{ d (P , OB), d (P , AB)}, where d denotes the distance
from the point to the corresponding line. Sketch the
(1997C, 5M)
region R and find its area.
13. Let O (0, 0), A (2, 0) and B (1,
(a) ∫
x
b
(b) −
(c) ∫
x
b
∫a 3[{f (x)}2 − 1] dx + bf (b) − af (a)
x
b
− 1]
a 3[{f (x)}2
(d) −
16.
dx + bf (b) − af (a )
− 1]
a 3[{f (x)}2
Passage Based Questions
Consider the functions defined implicity by the
equation y3 − 3 y + x = 0 on various intervals in the real
line. If x ∈ (−∞ , − 2) ∪ (2, ∞ ), the equation implicitly
defines a unique real-valued differentiable function
(b) −
73 32
dx − bf (b) + af (a )
x
b
∫a 3[{f (x)}2 − 1] dx − bf (b) + af (a)
1
∫− 1 g ′ (x) dx is equal to
(a) 2 g(− 1)
(c) − 2 g(1)
(b) 0
(d) 2 g(1)
Topic 2 Area Using Integration
Objective Questions I (Only one correct option)
1. Let the functions f : R → R and g : R → R be defined by
1 x −1
(e
+ e1 − x ).
2
Then the area of the region in the first quadrant
bounded by the curves y = f (x), y = g (x) and x = 0 is
f (x) = ex − 1 − e−|x − 1|and g (x) =
(2020 Adv.)
1
(a) (2 − 3 ) + (e − e−1 )
2
1
(c) (2 − 3 ) + (e + e−1 )
2
(b) (2 +
(d) (2 +
1
3 ) + (e − e−1 )
2
1
3 ) + (e + e−1 )
2
2. Consider the region R = {(x, y) ∈ R : x ≤ y ≤ 2x }. If a
2
2
line y = α divides the area of region R into two equal
parts, then which of the following is true?
(2020 Main, 2 Sep II)
(a) α3 − 6α 2 + 16 = 0
(c) 3α 2 − 8α + 8 = 0
(b) 3α 2 − 8α3 / 2 + 8 = 0
(d) α3 − 6α3 / 2 − 16 = 0
3. The
area
(in
sq.
units)
of
the
region
A = {(x, y) : (x − 1)[x] ≤ y ≤ 2 x, 0 ≤ x ≤ 2}, where [t ]
denotes the greatest integer function, is
(2020 Main, 5 Sep II)
1
8
(a)
2−
2
3
8
(c)
2−1
3
4
(b)
2+1
3
1
4
(d)
2−
2
3
(2020 Adv.)
7
(b) 8 log e 2 −
3
(d) 16 log e 2 − 6
1
y2 = 4λx and the line y = λx, λ > 0, is , then λ is equal to
9
(2019 Main, 12 April II)
(a) 2 6
(b) 48
(c) 24
(d) 4 3
6. If
the area (in sq units) of the region
{(x, y): y2 ≤ 4x, x + y ≤ 1, x ≥ 0, y ≥ 0} is a 2 + b, then
(2019 Main, 12 April I)
a − b is equal to
(a)
10
3
(b) 6
(c)
8
3
(d) −
2
3
7. The area (in sq units) of the region bounded by the
curves y = 2x and y = | x + 1|, in the first quadrant is
(2019 Main, 10 April II)
(a)
3
2
(b) log e 2 +
3
1
(c)
2
2
(d)
3
1
−
2 log e 2
8. The area (in sq units) of the region


y2
A = (x, y) :
≤ x ≤ y + 4 is
2


53
(b)
3
(a) 30
(2019 Main, 9 April II)
(c) 16
(d) 18
9. The
area
(in
sq
units)
A = {(x, y) : x2 ≤ y ≤ x + 2} is
(a)
4. The area of the region {(x, y) : xy ≤ 8,1 ≤ y ≤ x2} is
14
(a) 8 log e 2 −
3
14
(c) 16 log e 2 −
3
5. If the area (in sq units) bounded by the parabola
13
6
(b)
9
2
(c)
of
the
region
(2019 Main, 9 April I)
31
6
(d)
10
3
10. Let S (α ) = {(x, y) : y2 ≤ x, 0 ≤ x ≤ α} and A(α ) is area of the
region S(α ). If for λ, 0 < λ < 4, A (λ ) : A (4) = 2 : 5, then λ
(2019 Main, 8 April II)
equals
1
4 3
(a) 2 
 25 
1
2 3
(b) 4 
 5
1
4 3
(c) 4 
 25 
1
2 3
(d) 2 
 5
Area 331
11. The tangent to the parabola y2 = 4x at the point where it
intersects the circle x2 + y2 = 5 in the first quadrant,
passes through the point
1
(a)  ,
4
3
(b)  ,
4
3

4
7

4
1
(c)  − ,
 3
1
(d)  − ,
 4
4

3
1

2
53
(a)
6
(2019 Main, 8 April I)
59
(c)
6
(b) 8
(d)
26
3
13. The area (in sq units) of the region bounded by the
parabola, y = x + 2 and the lines, y = x + 1, x = 0 and
(2019 Main, 12 Jan I)
x = 3, is
2
(a)
15
2
(b)
17
4
(c)
21
2
(d)
15
4
14. The area (in sq units) in the first quadrant bounded by
the parabola, y = x + 1, the tangent to it at the point
(2, 5) and the coordinate axes is
(2019 Main, 11 Jan II)
2
(a)
14
3
(b)
187
24
(c)
8
3
(d)
37
24
15. The area (in sq units) of the region bounded by the curve
x2 = 4 y and the straight line x = 4 y − 2 is
(2019 Main, 11 Jan I)
7
(a)
8
9
(b)
8
5
(c)
4
3
(d)
4
16. The area of the region A = {(x, y); 0 ≤ y ≤ x| x| + 1 and
− 1 ≤ x ≤ 1} in sq. units, is
4
(b)
3
(a) 2
(2019 Main, 9 Jan II)
1
(c)
3
2
(d)
3
y = x2 − 1, the tangent at the point (2, 3) to it and the
(2019 Main, 9 Jan I)
Y -axis is
8
3
(b)
56
3
(c)
32
3
(d)
14
3
18. Let g (x) = cos x2, f (x) = x and α , β (α < β) be the roots of
the quadratic equation 18x2 − 9πx + π 2 = 0. Then, the
area (in sq units) bounded by the curve y = ( gof )(x) and
the lines x = α, x = β and y = 0, is
(2018 Main)
1
( 3 − 1)
2
1
(c) ( 3 − 2 )
2
{(x, y) : x ≥ 0, x + y ≤ 3, x2 ≤ 4 y and y ≤ 1+
(b)
3
2
(c)
7
3
x } is
(d)
5
2
(a)
1
6
of
the
(b)
4
3
(c)
3
2
(d)
5
3
21. The area (in sq units) of region described by (x, y) y ≤ 2x
2
and y ≥ 4x − 1 is
(a)
7
32
(b)
(2015 JEE Main)
5
64
(c)
15
64
(c)
π 2
−
2 3
(2014 Main)
(d)
π 2
+
2 3
(2014 Adv.)
(b) 2 2 ( 2 − 1)
(d) 2 2 ( 2 + 1)
24. The area (in sq units) bounded by the curves
y = x, 2 y − x + 3 = 0, X-axis and lying in the first
(2013 Main, 03)
quadrant, is
(a) 9
(b) 6
(c) 18
(d)
27
4
25. Letf : [−1, 2] → [0, ∞ ) be a continuous function such that
f (x) = f (1 − x), ∀x ∈ [−1, 2]. If R1 = ∫
2
−1
xf (x) dx and R2 are
the area of the region bounded by y = f (x), x = − 1, x = 2
and the X-axis. Then,
(2011)
(a) R1 = 2R2
(c) 2R1 = R2
(b) R1 = 3R2
(d) 3R1 = R2
26. If the straight line x = b divide the area enclosed by
y = (1 − x)2, y = 0 and x = 0 into two parts R1 (0 ≤ x ≤ b)
1
and R2(b ≤ x ≤ 1) such that R1 − R2 = . Then, b equals
4
3
4
(a)
(b)
1
2
(c)
1
3
(d)
1
4
(2011)
area of the region between the curves
1 + sin x
1 − sin x
and bounded by the
y=
and y =
cos x
cos x
π
lines x = 0 and x = is
4
(2008, 3M)
2 −1
(a) ∫
0
(c) ∫
0
t
(1 + t ) 1 − t
2+1
4t
2
2
(1 + t 2 ) 1 − t 2
2 −1
4t
dt
(b) ∫
0
dt
(d) ∫
(1 + t ) 1 − t 2
2+1
t
0
(1 + t 2 ) 1 − t 2
2
dt
dt
(d)
9
32
1
is
4
(2005, 1M)
1
2
sq unit (b) sq unit
3
3
(c)
1
sq unit
4
(d)
1
sq unit
5
29. The area enclosed between the curves y = ax2 and
x = ay2 (a > 0) is 1 sq unit. Then, the value of a is
(2004, 1M)
region {(x, y)} ∈ R2 : y ≥ |x + 3|,
(2016 Adv.)
5 y ≤ (x + 9) ≤ 15} is equal to
20. Area
π 4
−
2 3
(a) 4( 2 − 1)
(c) 2 ( 2 + 1)
(a)
(2017 Main)
59
12
(b)
23. The area enclosed by the curves y = sin x + cos x and
and y =
19. The area (in sq units) of the region
(a)
π 4
+
2 3
28. The area bounded by the curves y = (x − 1)2, y = (x + 1)2
1
( 3 + 1)
2
1
(d) ( 2 − 1)
2
(b)
(a)
(a)
27. The
17. The area (in sq units) bounded by the parabola
(a)
A = {(x, y) : x2 + y2 ≤ 1 and y2 ≤ 1 − x } is
 π
is
y = | cos x − sin x|over the interval 0,
 2 
12. The area (in sq units) of the region
A = {(x, y) ∈ R × R|0 ≤ x ≤ 3,
0 ≤ y ≤ 4, y ≤ x2 + 3x} is
22. The area (in sq units) of the region described by
1
(a)
3
1
(b)
2
(c) 1
1
(d)
3
30. The area bounded by the curves y = f (x),the X-axis and
the ordinates x = 1 and x = b is (b − 1) sin (3b + 4). Then,
(1982, 2M)
f (x) is equal to
(a) (x − 1) cos (3x + 4)
(b) 8sin (3x + 4)
(c) sin (3x + 4) + 3(x − 1) cos (3x + 4)
(d) None of the above
332 Area
31. The slope of tanget to a curve y = f (x) at [x, f (x)] is 2x + 1.
If the curve passes through the point (1, 2), then the
area bounded by the curve, the X-axis and the line x = 1
is
(a)
3
2
(b)
4
3
(c)
5
6
(d)
1
12
Objective Questions II
(One or more than one correct option)
divides the area of region
R = {(x, y) ∈ R2 : x3 ≤ y ≤ x, 0 ≤ x ≤ 1} into two equal
(2017 Adv.)
parts, then
(b) α 4 + 4α 2 − 1 = 0
1
(d) 0 < α ≤
2
2
y = e− x , y = 0, x = 0 and x = 1. Then,
1
e
1
1
(c) S ≤  1 +

e
4
(b) S ≥ 1 −
(d) S ≤
bounded by the Y-axis and the curve
jπ
( j + 1)π
. Then, show that
xe = sin by,
≤ y≤
b
(b)
S 0 , S1 , S 2,... , S n are in geometric progression. Also,
find their sum for a = − 1 and b = π .
(2001, 5M)
ay
2x,
|x| ≤ 1
f (x) =  2
x + ax + b, |x| > 1
Then, find the area of the region in the third quadrant
bounded by the curves x = −2 y2 and y = f (x) lying on the
left on the line 8x + 1 = 0.
(1999, 5M)
42. Let C1 and C 2 be the graphs of functions y = x2 and
33. If S be the area of the region enclosed by
(a) S ≥
region
41. If f (x) is a continuous function given by
32. If the line x = α
(a) 2α 4 − 4α 2 + 1 = 0
1
(c) < α < 1
2
40. Let b ≠ 0 and for j = 0,1,2..., n. If S j is the area of the
(2012)
1
e
1
1 
1 
+
1 −

e 
2
2
y = 2x, 0 ≤ x ≤ 1, respectively. Let C3 be the graph of a
function y = f (x), 0 ≤ x ≤ 1, f (0) = 0. For a point P on C1,
let the lines through P, parallel to the axes, meet C 2 and
C3 at Q and R respectively (see figure). If for every
position of P(on C1) the areas of the shaded regions OPQ
and ORP are equal, then determine f (x).
(1998, 8M)
Y
34. Area of the region bounded by the curve y = e and lines
x
x = 0 and y = e is
(a) e − 1
1
(c) e − ∫ ex dx
0
(d)
35. For which of the following values of m, is the area of the
region bounded by the curve y = x − x2 and the line
9
(1999, 3M)
y = mx equals ?
2
(a) – 4
(b) –2
(c) 2
(d) 4
4a 2 4a 1  f (−1)  2
3a + 3a 
36. If 4b2 4b 1  f (1)  = 3b2 + 3b ,
  2


 4c2 4c 1  f (2)  3c + 3c 


f (x) is a quadratic function and its maximum value
occurs at a point V. A is a point of intersection of y = f (x)
with X-axis and point B is such that chord AB subtends
a right angle at V. Find the area enclosed by f (x) and
chord AB.
(2005, 5M)
37. Find the area bounded by the curves x = y, x = − y
and y2 = 4x − 3.
2
(2005, 4M)
38. A curve passes through (2,0) and the slope of tangent at
(x + 1) + y − 3
.
(x + 1)
2
point P (x, y) equals
(0,0) O C
3
(1,0)
X
R
43. Letf (x) = max { x , (1 − x) , 2x (1 − x)}, where 0 ≤ x ≤ 1.
2
2
Determine the area of the region bounded by the curves
(1997, 5M)
y = f (x), X-axis, x = 0 and x = 1.
the bounded region enclosed between the parabolas
x2
is maximum.
y = x − bx2 and y =
(1997C, 5M)
b
45. If An is the area bounded by the curve y = (tan x)n and
the lines x = 0, y = 0 and x =
π
.
4
Then, prove that for n > 2 , An + An + 2 =
and deduce
1
1
.
< An <
2n + 2
2n − 2
1
n+1
(1996, 3M)
46. Consider a square with vertices at (1,1), (–1, 1), (–1, –1)
and (1, –1). If S is the region consisting of all points
inside the square which are nearer to the origin than to
any edge. Then, sketch the region S and find its area.
(1995, 5M)
Find the equation of the curve and area enclosed by the
curve and the X-axis in the fourth quadrant. (2004, 5M)
39. Find the area of the region bounded by the curves
y = x2, y = | 2 − x2| and y = 2 ,
which lies to the right of the line x = 1.
P
44. Find all the possible values of b > 0, so that the area of
Analytical & Descriptive Questions
2
C1
Q
e
∫1 ln (e + 1 − y) dy
e
∫1 ln y dy
(1,1)
C2
(2009)
(b)
(1/2,1)
(0,1)
47. In what ratio, does the X-axis divide the area of the
region bounded by the parabolas y = 4x − x2 and
(1994, 5M)
y = x2 − x ?
48. Sketch the region bounded by the curves y = x2 and
(2002, 5M)
y = 2 / (1 + x2). Find its area.
(1992, 4M)
Area 333
49. Sketch the curves and identify the region bounded by
x = 1 /2, x = 2, y = log x and y = 2x . Find the area of this
(1991, 4M)
region.
55. Sketch the region bounded by the curves y = 5 − x2 and
y = |x − 1|and find its area.
56. Find the area of the region bounded by the X-axis
50. Compute the area of the region bounded by the curves
log x
, where log e = 1.
y = ex log x and y =
(1990, 4M)
ex
51. Find all maxima and minima of the function
y = x (x − 1)2, 0 ≤ x ≤ 2 .
Also, determine the area bounded by the curve
(1989, 5M)
y = x (x − 1)2, the Y-axis and the line x = 2 .
and the curves defined by y = tan x, −
y = cot x,
53. Find the area bounded by the curves
x2 + y2 = 25, 4 y = |4 − x2|and x = 0 above the X-axis.
(1987, 6M)
54. Find the area bounded by the curves x2 + y2 = 4,
x2 = − 2 y and x = y.
(1986, 5M)
π
π
≤x≤ .
6
3
π
π
and
≤x≤
3
3
(1984, 4M)
57. Find the area bounded by the X-axis, part of the curve
8

y = 1 + 2 and the ordinates at x = 2 and x = 4. If the

x 
ordinate at x = a divides the area into two equal parts,
then find a.
(1983, 3M)
52. Find the area of the region bounded by the curve
C: y = tan x, tangent drawn to C at x = π / 4 and the
X-axis.
(1988, 5M)
( 1985, 5M)
58. Find the area bounded by the curve x2 = 4 y and the
straight line x = 4 y − 2.
et + e– t
et – e– t
is a point on the
59. For any real t, x =
,y=
2
2
hyperbola x2 – y2 = 1. Find the area bounded by this
hyperbola and the lines joining its centre to the points
corresponding to t1 and – t1.
(1982, 3M)
Answers
Topic 1
1. (a)
5. (a)
9. (a,d)
2. (b)
6. (d)
10. (4)
3. (b)
4. (d)
7. (b)
8. (c)
11. 2 3 sq units
12. 2 sq units
13. (2 − 3 ) sq unit
14. (b)
15. (a)
16. (d)
17
sq unit
27
47. 121 : 4
1. (a)
2. (b)
3. (a)
4. (c)
5. (c)
6. (b)
7. (d)
8. (d)
9. (b)
10. (c)
11. (b)
12. (c)
13. (a)
14. (d)
15. (b)
16. (a)
17. (a)
18. (a)
19. (d)
20. (c)
21. (d)
22. (a)
23. (b)
24. (a)
25. (c)
26. (b)
27. (b)
28. (a)
29. (a)
30. (c)
31. (c)
33. (b, d)
34. (b, c, d)
35. (b, d)
32. (a, c)
125
sq units
36.
3
38. y = x 2 – 2 x,
4
sq units
3
1
sq unit
3
 761
41. 
 sq units
 192
43.
Topic 2
37.
 20 − 12 2 
39. 
 sq units
3


 π (1 + e ) (en + 1 − 1 ) 
40. 
⋅
2
e − 1 
 (1 + π )
42. f ( x ) = x 3 − x 2, 0 ≤ x ≤ 1
44. b = 1
46.
2

48.  π −  sq units

3
4 − 2 5
49. 
− log 2 +
2
 log 2
1

(16 2 − 20 ) sq units
3

 e 2 − 5
3
 sq units
 sq units 50. 
2
 4e 
4
10


51. y max = , y min = 0, sq units


27
3
1


52.  log 2 −  sq units


4


1

54.  − π  sq units
3

1


56.  loge 3 sq units

2
59.

 4 
53. 4 + 25 sin −1    sq units
 5 

 5π 1
55. 
−  sq units
 4 2
9
58. sq units
57. 2 2
8
e 2t1 − e −2t1 1 2t1
− (e − e −2t1 − 4t1 )
4
4
Hints & Solutions
Topic 1 Area Based on Geometrical Figures
Without Using Integration
Now, the required area is the area of shaded region, i.e.
Y
1. Equation of given parabola y = x2 – 1 According to
A (2, 2)
symmetry let, the coordinate of A (– a , 0), B(a , 0)
C (a , a 2 – 1) and D (– a , a 2 – 1).
∴Area of rectangle P (a ) = 2a (a 2 – 1)
x 2 + y 2 = 4x
X′
(0, 0)
y2 =
Y
B(a, 0)
A(–a, 0)
(1, 0)
O
D
(–a, a2–1)
2
Area of a circle
Required area =
− ∫ 2x dx
0
4
2
2
 x3/ 2 
π (2)2
=
− 2 ∫ x1/ 2dx = π − 2 

0
4
 3 /2  0
8
2 2

=π−
[2 2 − 0] =  π −  sq unit

3
3
X
C(a, a2–1)
(0, –1)
4.
Now, for maxima P ' (a ) = 0 ⇒ 2(a – 1) + 4a = 0
1
units
⇒
3a 2 = 1 ⇒ a =
3
2
2
∴ Area of largest rectangle is
2 1
4

sq. units
 – 1 =
 3 3
3 3
2. We know that, area of region bounded by the parabolas
16
(ab) sq units.
3
On comparing y = kx2 and x = ky2 with above equations,
1
1
we get 4a = and 4b =
k
k
1
1
and b =
⇒
a=
4k
4k
∴ Area enclosed between y = kx2 and x = ky2 is
16  1   1 
1
   = 2




3 4k 4k
3k
1
= 1 [given, area = 1 sq.unit]
⇒
3k2
1
1
⇒
k2 =
⇒ k=±
3
3
1
[Q k > 0]
k=
⇒
3
x2 = 4ay and y2 = 4bx is
3. Given equations of curves are y2 = 2x
which is a parabola with vertex (0, 0) and axis parallel
to X-axis.
.
..(i)
And
x2 + y 2 = 4 x
which is a circle with centre (2, 0) and radius = 2 ...(ii)
On substituting y2 = 2x in Eq. (ii), we get
x2 + 2x = 4x ⇒ x2 = 2x ⇒ x = 0 or x = 2
[using Eq. (i)]
⇒
y = 0 or y = ± 2
2x
Y′
y=x2–1
(–1, 0)
X
B (2,0)
PLAN (i) y = mx + a / m is an equation of tangent to the parabola
y 2 = 4ax .
(ii) A line is a tangent to circle, if distance of line from centre is
equal to the radius of circle.
(iii) Equation of chord drawn from exterior point ( x 1, y1 ) to a
circle/parabola is given by T = 0.
1
(iv) Area of trapezium = (Sum of parallel sides)
2
Let equation of tangent to parabola be y = mx +
2
m
It also touches the circle x2 + y2 = 2.


2

= 2
2
m 1 + m 
m4 + m2 = 2
4
m + m2 − 2 = 0
2
(m − 1) (m2 + 2) = 0
[rejected m2 = − 2]
m = ± 1,m2 = − 2
∴
⇒
⇒
⇒
⇒
So, tangents are y = x + 2, y = − x − 2.
They, intersect at (−2, 0).
R
Y
P
T(–2,0)
X′
X
O
Q
Y′
S
Equation of chord PQ is −2x = 2 ⇒ x = − 1
Equation of chord RS is O = 4(x − 2) ⇒ x = 2
∴ Coordinates of P, Q, R, S are
P (−1, 1), Q (−1, − 1), R(2, 4), S (2, − 4)
(2 + 8) × 3
∴ Area of quadrilateral =
= 15 sq units
2
Area 335
5. Since, tangents drawn from external points to the circle
7. The region is clearly square with vertices at the points
subtends equal angle at the centre.
(1, 0), (0, 1), (– 1, 0) and (0, – 1).
Y
A
y = –IxI + 1
y = IxI –1
(0,1)
O3
X′
X
(1,0)
(–1,0)
O1
B
(0,–1)
O2
1cm 1cm
30°
30°
2 cm E
3 cm D
Y′
C
∴ Area of square = 2 × 2 = 2 sq units
3 cm
8. Let y = f (x) = x2 + bx − b
∴
∠ O1BD = 30°
OD
In ∆O1BD, tan 30° = 1
BD
Y
B
⇒
BD = 3 cm
Also,
DE = O1O2 = 2 cm and EC = 3 cm
Now,
BC = BD + DE + EC = 2 + 2 3
⇒
Area of ∆ABC =
=
(1,1)
P
O
3
(BC )2
4
The equation of the tangent at P (1, 1)
to the curve 2 y = 2x2 + 2bx − 2b is
y + 1 = 2x ⋅ 1 + b (x + 1) − 2b ⇒ y = (2 + b) x − (1 + b)
Its meet the coordinate axes at
1+ b
xA =
and yB = − (1 + b)
2+ b
1
∴ Area of ∆ OAB = OA × OB
2
1 (1 + b)2
[given]
=− ×
=2
2 (2 + b)
3
⋅ 4 (1 + 3 )2 = (6 + 4 3 ) sq cm
4
x2 y2
+
=1
9
5
To find tangents at the end points of latusrectum, we
find ae.
6. Given,
ae = a 2 − b2 = 4 = 2
i.e.
4 5

b2(1 − e2) = 5 1 −  =

9 3
and
⇒
⇒
By symmetry, the quadrilateral is a rhombus.
(1 + b)2 + 4(2 + b) = 0
(b + 3)2 = 0
Y
A
t 12
P 2 , t1
L(ae, √ b2(1 – e2))
D
O
B
⇒ b2 + 6b + 9 = 0
⇒ b = −3
9. Since, ∠ POQ = 90°
Y
X′
X
A
X
L′
C
X′
X
O
(0, 0)
Y′
So, area is four times the area of the right angled triangle
formed by the tangent and axes in the Ist quadrant.
 5
∴ Equation of tangent at 2,  is
 3
2
5 y
x
y
x+ ⋅ =1 ⇒
+ =1
9
3 5
9 /2 3
∴ Area of quadrilateral ABCD
[area of ∆ AOB]
=4
1 9 
= 4  ⋅ ⋅ 3 = 27 sq units
2 2 
Y′
⇒
Q
∴
⇒
t 22
Q 2 , t2
t1 − 0 t2 − 0
⋅ 2
= − 1 ⇒ t1t2 = − 4
t12
t
−0 2 −0
2
2
ar (∆OPQ ) = 3 2
0
0 1
1 2
t1 / 2 t1 1 = ± 3 2
2 2
t2 / 2 t2 1
1  t12t2 t1t22
−

 =±3 2
2 2
2 
…(i)
336 Area
⇒
4
1
( −4t1 + 4t2) = ± 3 2 ⇒ t1 + = 3 2 [Q t1 > 0 for P]
t1
4
13. Let the coordinates of P be (x, y).
Y
⇒ t12 − 3 2t 1 + 4 = 0 ⇒ (t1 − 2 2 ) (t1 − 2 ) = 0
t1 = 2 or 2 2
⇒
∴
P (1, 2 ) or P (4, 2 2 )
B
P
10. We have, y = xn , n > 1
Q P (0, 0) Q (1, 1) and R(2, 0) are vertices of ∆ PQR.
X′
O (0, 0)
A (2, 0)
y
Y′
y=
x
Q(1,1)
F2
F1
y = xn
x′
P(0,0) (1,0) R(2,0)
x
y′
∴ Area of shaded region = 30% of area of ∆ PQR
1
30 1
n
⇒
∫0 (x − x ) dx = 100 × 2 × 2 × 1
1
 x2 xn + 1 
1
3
1 
3
 −
 =
 2 − n + 1  = 10 ⇒


2
+
1
10
n
0

2 1
1
1 3
=
= ⇒ n + 1 =5 ⇒ n =4
= −
⇒
n + 1 2 10 10 5
⇒
11. Equation of tangent at the point (1, 3 ) to the curve
x2 + y2 = 4 is x +
whose X-axis intercept (4, 0).
3y = 4
Y
Equation of line OA be y = 0.
Equation of line OB be 3 y = x.
Equation of line AB be 3 y = 2 − x.
d (P , OA ) = Distance of P from line OA = y
| 3 y − x|
d (P , OB) = Distance of P from line OB =
2
| 3 y + x − 2|
d (P , AB) = Distance of P from lineAB =
2
Given, d (P , OA ) ≤ min { d (P , OB), d (P , AB)}
 | 3 y − x| | 3 y + x − 2|
,
y ≤ min 

2
2


| 3 y − x|
| 3 y + x − 2|
and y ≤
2
2
| 3 y − x|
Case I When y ≤
[since, 3 y − x < 0]
2
x − 3y
⇒ (2 + 3 ) y ≤ x ⇒ y ≤ x tan 15°
y≤
2
| 3 y + x − 2|
Case II When y ≤
,
2
2 y ≤ 2 − x − 3 y [since, 3 y + x − 2 < 0]
⇒
(2 + 3 ) y ≤ 2 − x ⇒ y ≤ tan 15°⋅ (2 − x)
⇒
y≤
Y
P (1,√3)
X′
(0,0) O
A (4,0)
B (1, 1/ 3)
X
P
X′
Y′
Thus, area of ∆ formed by (0, 0) (1, 3 ) and (4, 0)
0 0 1
1
1
=
1
3 1 = |(0 − 4 3 )|= 2 3 sq units
2
2
4 0 1
12. The area formed by | x| + | y| = 1 is square shown as
below :
Y
−x + y = 1
X'
−1
x+y=1
O
1
x+y=1
x−y=1
Y'
∴ Area of square = ( 2 )2 = 2 sq units
X
X
O (0, 0)
C
(1, 0)
A
(2, 0)
X
Y′
From above discussion, P moves inside the triangle as
shown below :
⇒ Area of shaded region
= Area of ∆OQA
1
= (Base) × (Height)
2
1
= (2) (tan 15° ) = tan 15° = (2 − 3 ) sq unit
2
14. Given,
⇒
y3 − 3 y + x = 0
dy
dy
3 y2
−3
+ 1 =0
dx
dx
2
 d 2y
d 2y
 dy
⇒ 3 y2 2  + 6 y   − 3
=0
 dx
dx2
 dx 
At x = − 10 2 , y = 2 2
…(i)
…(ii)
Area 337
On substituting in Eq. (i) we get
dy
dy
3(2 2 )2 ⋅
− 3⋅
+ 1 =0
dx
dx
dy
1
=−
⇒
dx
21
Again, substituting in Eq. (ii), we get
So, required area is
1/ 2 log 3e + 1
∫0
=∫
2
2
=
2
d y
d y
 1
+ 6 (2 2 ) ⋅  −  − 3 ⋅ 2 = 0
3(2 2 )2
2


21
dx
dx
⇒
21 ⋅
⇒
1/ 2 log 3e + 1
0
b
b
a
a
[ f (x) ⋅ x]ba
− ∫ f ′ (x)x dx
a
a
b
xdx
∫a 3[{ f (x)}2 − 1]
y3 − 3 y + x = 0
y = g (x)
{ g (x)}3 − 3 g (x) + x = 0
y
…(iii)
g (1) + g (− 1) = 0
g (1) = − g ( − 1)
I = g (1) − g (− 1)
= g (1) − { − g (1)} = 2 g (1)
Topic 2 Area Using Integration
1. The given functions f : R → R and g : R → R be defined
, x≥1
0,
x<1
1 x −1
1−x
and g (x) = (e
+e
)
2
For point of intersection of curves f (x) and g (x) put
f (x) = g (x)
1
for x ≥ 1, ex − 1 − e1 − x = (ex − 1 + e1 − x )
2
⇒
ex − 1 = 3e1 − x
1
⇒
e2x = 3e2 ⇒ x = log3e + 1
2
y=2x
y=α
x
O
α
∫0 
⇒ [ g (1) + g (− 1)][{ g (1)}2 + { g (− 1)}2 − g (1) g (− 1) − 3] = 0
−e
3
(2, 4)
{ g (1)}3 + { g (− 1)}3 − 3 { g (1) + g (− 1)} = 0
e
y=x2
[from Eq. (i)]
At x = − 1, { g (− 1)}3 − 3 g (− 1) − 1 = 0
On adding Eqs. (i) and (ii), we get
f (x) = ex − 1 − e−|x − 1| =
1
3
log e + 1
log e + 1
1 x −1
[e
− e1 − x ]02
− [ex − 1 + e1 − x ]12
2
1
1
1
 

=
− e−1 + e − 3 +
− 1 −1
3−
2 
3
3
 

1
1
4
1
=
+ (e − e−1 ) −
+ 2 = (2 − 3 ) + (e − e−1 )
2
3 2
3
2. According to the question,
…(ii)
1 −x
X
½ loge3 +1
…(i)
{ g (1)} − 3 g (1) + 1 = 0
x −1
1
1
g′ (x) dx = [ g (x)]1− 1 = g (1) − g (− 1)
by
(ex − 1 − e1 − x ) dx
=
3
⇒
⇒
∴
1/ 2 log 3e + 1
∫1
y=f(x)
O


dy
−1
−1
=
Q f ′ (x) = dx =

2
2
−
3
(
1
)
3
[{
(
)}
1
]
−
y
f
x


Since,
and
∴
At x = 1,
f (x)dx
y=g(x)
b
= bf (b) − af (a ) +
1
1
d 2y
12 2
=−
2
dx
(21)2
2
d y − 12 2 − 4 2
=
= 3 2
(21)3
dx2
7 ⋅3
= bf (b) − af (a ) − ∫ f ′ (x)x dx
−1
1/ 2 log 3e + 1
Y
b
16. Let I = ∫
g (x)dx − ∫
1 1/ 2log 3e + 1 x − 1
(e
+ e1 − x )dx −
2 ∫0
15. Required area = ∫ y dx = ∫ f (x) dx
=
( g (x) − f (x))dx
4
y
y −  dy = ∫ ( y − y / 2) dy
α
2
α α α 2 4(2) 16 α α α 2
−
=
−
−
+
3 /2
4
3 /2
4
3 /2
4
2
4
64 − 48 16 4
α
⇒
α α −
+
=
=
3
2
12
12 3
⇒
3 α 2 − 8 α3/ 2 + 8 = 0
Hence, option (b) is correct.
⇒
3. As we know that,
 0, 0 ≤ x < 1

y = (x – 1)[x] =  x – 1, 1 ≤ x < 2
2(x – 1), x = 2

Now, on drawing the graph of given region with the
help of equation of curves y = (x – 1)[x] and y = 2 x
y
y=2√x
1
x–
y=
O
1
2
x
338 Area
∴Area of given region
1
Now, required area is
2
= ∫ 2 x dx + ∫ (2 x – x + 1)dx
0
=
1
2
4

x
4 3/ 2
+  x3/ 2 –
+ x
x
3

3
2
0

1

 8 2 1
4
8 2
4 1
= +
– sq. units.
– 2 + 2 – + – 1 =
3
2
3  3
3 2

1
2
=
xy=8
y=x2
Q (2,4)
R(8,1)
P
X¢
4/ λ
 x2 
−λ 
 2 0
2
4
4 4 λ  4
λ
−  
3
λ λ 2  λ
32 8 32 − 24
8
=
− =
=
3λ λ
3λ
3λ
1
It is given that area =
9
8
1
⇒
=
3λ 9
⇒
λ = 24
6. Given region is {(x, y) : y2 ≤ 4x, x + y ≤ 1, x ≥ 0, y ≥ 0}
y=1
X
O
4/ λ
=
Y
A
λx − λx) dx


 x3 / 2 
=2 λ 
3 


 2 0
From the figure, region A and B satisfy the given region,
but only A is bounded region, so area of bounded region
(1,1)
∫ (2
0
4. The given region {(x, y) : xy ≤ 8, 1 ≤ y ≤ x2}.
B
4/ λ
B(0,1)
P
y2=4x
x+y=1
xy=8
2
A = ∫ (x − 1) dx + ∫
2
1
Y¢
O
8 8
2

 − 1 dx
x 
R(8, 1)]
2

 x3
8
=  − x + [8 log|x|− x]2
3
1

1
8

=  − 2 − + 1 + 8 log 8 − 8 − 8 log 2 + 2
3

3
14
14
=−
+ 16 log 2 = 16 log 2 −
3
3
y2 = 4λx
y = λx
λ> 0
x=
6 ± 36 − 4
2
= 3 ± 2 2.
Since, x-coordinate of P less than x-coordinate of point
A(1, 0).
Area bounded by above two curve is, as per figure
∴ x=3 −2 2
Now, required area
=∫
y2=4λx
X
y=λx
the intersection point A we will get on the solving Eqs. (i)
and (ii), we get
4
λ2x2 = 4λx ⇒ x = , so y = 4.
λ
4
So,
A  , 4
λ 
3 −2 2
0
=2
=
O
x2 − 6 x + 1 = 0
⇒
…(i)
…(ii)
A
(1 − x)2 = 4x ⇒ x2 + 1 − 2x = 4x
⇒
5. Given, equation of curves are
Y
X
Now, for point P, put value of y = 1 − x to y2 = 4x, we get
[Q Points P (1, 1), Q (2, 4)and
and
A(1,0)
x3/ 2
3 /2
2 x dx +
3 −2 2
0
1
∫3 − 2 2 (1 − x) dx
1

x2 
+ x − 
2 3 − 2

2
(3 − 2 2 ) 2
4
1
(3 − 2 2 ) 3/ 2 +  1 −  − (3 − 2 2 ) +


3
2
2
4
1
1
[( 2 − 1)2]3/ 2 + − 3 + 2 2 + (9 + 8 − 12 2 )
3
2
2
4
5
17
= ( 2 − 1)3 − + 2 2 +
−6 2
3
2
2
4
= (2 2 − 3(2) + 3( 2 ) − 1) − 4 2 + 6
3
4
8 2 10
= (5 2 − 7) − 4 2 + 6 =
−
3
3
3
=
=a 2+b
(given)
Area 339
8
10
and b = −
3
3
8 10
a−b= +
=6
3
3
Now, the area enclosed by the region A
So, on comparing a =
∴
7. Given, equations of curves
 x + 1 ,x ≥ − 1
y = 2x and y = | x + 1| = 
− x − 1 , x < − 1
4
9. Given region is A = {(x, y) : x2 ≤ y ≤ x + 2}
Q The figure of above given curves is
Y
4

 y2
y2 
y3 
∫  ( y + 4) − 2 dy =  2 + 4 y − 6 
−2
−2
16
64   4
8
8
−
−
+
= 
+ 16 −

 
 2
6  2
6
4
32
= 8 + 16 −
− 2 + 8 − = 30 − 12 = 18 sq unit.
3
3
=
Now, the region is shown in the following graph
y=x+1
Y
(1,2)
y=–x–1
y=x+2
x2=y
y=2x
B(2,4)
(0,1)
X′
O
(–1,0)
A
X
X'
In first quadrant, the above given curves intersect each
other at (1, 2).
1
So, the required area = ∫ ((x + 1) − 2 ) dx
x
0
1
x
2 
= + x−

2
log
e 2 0

2
x


ax
x
Q ∫ a dx = log a + C 
e


1
2
1 
= +1−
+

2
log
2
log

e
e 2
3
1
= −
2 log e 2

8. Given region A = (x, y) :

y2
=x
2
⇒
y2 = 2x
and x = y + 4 ⇒ y = x − 4
Y'
2
] dx
2
=8−
…(i)
…(ii)
1 9
1
1 9
− = 8 − − 3 = 5 − = sq units
2 3
2
2 2
10. Given, S (α ) = {(x, y) : y2 ≤ x, 0 ≤ x ≤ α } and
A(α ) is area of the region S(α )
Y
Graphical representation of A is
y 2 =x
y2
=x
2
X
O
X
O
2
∫ [(x + 2) − x
 x2
x3 
1
8
1
4
=  + 2x −  =  + 4 −  −  − 2 + 



3
3
2
2
2
3
 −1

∴
X'
X'
2
−1

y
≤ x ≤ y + 4
2

Q
O
For intersecting points A and B
Taking,
x2 = x + 2 ⇒ x2 − x − 2 = 0
⇒
x2 − 2x + x − 2 = 0
⇒
x(x − 2) + 1(x − 2) = 0
⇒
x = −1, 2 ⇒ y = 1, 4
So, A(−1, 1) and B (2, 4).
Now, shaded area =
2
Y
(–2,0)
–1
(0,2)
A(λ)
x=
y+
4
P
Y'
On substituting y = x − 4 from Eq. (ii) to Eq. (i), we get
(x − 4)2 = 2x
2
⇒
x − 8x + 16 = 2x
⇒
x2 − 10x + 16 = 0
⇒
(x − 2)(x − 8) = 0
⇒
x = 2, 8
[from Eq. (ii)]
∴
y = − 2, 4
So, the point of intersection of Eqs. (i) and
(ii) are P (2, − 2) and Q(8, 4).
x=λ
λ
Clearly, A (λ ) = 2∫
0
Since,
⇒
⇒
λ
 x3/ 2 
4
x dx = 2 
= λ3/ 2

3 / 2 0 3
A (λ ) 2
= , (0 < λ < 4)
A (4) 5
3
λ3/ 2 2
=
43/ 2 5
λ 4
= 
4  25
 2
 λ
⇒   = 
 5
 4
1/3
⇒
2
4
λ =4  
 25
1/3
340 Area
11. Given equations of the parabola y2 = 4x
…(i)
14. Given, equation of parabola is y = x2 + 1, which can be
and circle
…(ii)
x + y =5
So, for point of intersection of curves (i) and (ii), put
y2 = 4x in Eq. (ii), we get
x2 + 4 x − 5 = 0
2
⇒
x + 5x − x − 5 = 0
⇒
(x − 1)(x + 5) = 0
⇒
x = 1, − 5
For first quadrant x = 1 , so y = 2 .
Now, equation of tangent of parabola (i) at point (1, 2)
is T = 0
⇒
2 y = 2(x + 1) ⇒ x − y + 1 = 0
 3 7
The point  ,  satisfies, the equation of line
 4 4
written as x2 = ( y − 1). Clearly, vertex of parabola is
(0, 1) and it will open upward.
y+5
Now, equation of tangent at (2, 5) is
= 2x + 1
2
2
2
[QEquation of the tangent at (x1 , y1 ) is given by
1
T = 0. Here, ( y + y1 ) = xx1 + 1]
2
y = 4x − 3
y= 4x–3
Y
P (2, 5)
x− y+ 1 =0
(0, 1)
12. Given, y ≤ x + 3x
2
2
2
3
9
3
9



⇒ x +  ≥  y + 
y ≤ x +  −



2
4
2
4
⇒
O
Since,
0 ≤ y ≤ 4 and 0 ≤ x ≤ 3
∴The diagram for the given inequalities is
Y
R
Q (2, 0)
3,
0
4
X
Required area = Area of shaded region
y=x2+3x
2
= ∫ y(parabola) dx − (Area of ∆PQR)
0
2
= ∫ (x2 + 1) dx − (Area of ∆PQR)
y=4
0
2
–3/2
O
9
–
4
(–3, 0)
1

 x3
1
3
=
+ x − 2 −  ⋅ 5

3
2
4
0

X
3
x=3
[QArea of a triangle =
and points of intersection of curves y = x2 + 3x and y = 4
are (1, 4) and (−4, 4)
Now required area
1  5

8
=  + 2 − 0 −   5

3
2  4
1
 x3 3x2 
3
4
dx
=
 3 + 2  + [4x]1
∫


1
0
0
1 3
2+9
11
59
sq units
= + + 4(3 − 1) =
+8 =
+8=
3 2
6
6
6
1
= ∫ (x2 + 3x)dx +
3
=
14 25 112 − 75 37
−
=
=
3
8
24
24
15. Given equation of curve is x2 = 4 y, which represent a
parabola with vertex (0, 0) and it open upward.
y
x2
4
x+2
y=
4
Y
13. Given equation of parabola is y = x2 + 2, and the line is
y = x+1
y=
y=x2 +2
y=x+1
(0,2)
1
× base × height]
2
X′
B
A
–1 O
X
2
1
1
O
(3,0)
x
The required area = area of shaded region
=∫
3
0
((x2 + 2) − (x + 1)) dx = ∫
3
3
0
(x2 − x + 1) dx
 x3 x2

 27 9

=  − + x = 
− + 3 − 0


3
2
3
2

0
9
9 15
sq units
= 9 − + 3 = 12 − =
2
2 2
Y′
Now, let us find the points of intersection of x2 = 4 y and
4y = x + 2
For this consider, x2 = x + 2 ⇒ x2 − x − 2 = 0
⇒
(x − 2) (x + 1) = 0 ⇒ x = − 1, x = 2
1
When x = − 1, then y =
4
and when x = 2, then y = 1
1
Thus, the points of intersection are A  − 1,  and B (2, 1).

4
Area 341
Now, required area = area of shaded region
=
∫−1 {y (line) − y (parabola)} dx
2
 x + 2 x2 
x3 
1  x2
∫−1  4 − 4  dx = 4  2 + 2x − 3 
−1
1 
8  1
1 
=
2 + 4 −  −  − 2 + 
3 
3  2
4  
=
1
4
1

 1
8 − −3 =

 4
2
1 9

= sq units.
5−

2  8
16. We have,
A = {(x, y) : 0 ≤ y ≤ x| x|+ 1and − 1 ≤ x ≤ 1}
When x ≥ 0, then 0 ≤ y ≤ x2 + 1
and when x < 0, then 0 ≤ y ≤ − x2 + 1
Now, the required region is the shaded region.
y
2
y=–x2+1
y=x2+1
1
–1
x
1
y=0
[Q y = x2 + 1⇒ x2 = ( y − 1), parabola with vertex (0, 1) and
y = − x2 + 1⇒ x2 = − ( y − 1) ,
parabola with vertex (0,1) but open downward]
We need to calculate the shaded area, which is equal to
0
∫−1 (− x
2
y+3
= 2x − 1 ⇒ y = 4x − 5
2
Now, required area = area of shaded region
⇒
2
=
y1 = 3.
and
2
1
2
= ∫ (y(parabola) − y(tangent)) dx
0
2
= ∫ [(x2 − 1) − (4x − 5)] dx
0
2
2
0
0
= ∫ (x2 − 4x + 4) dx = ∫ (x − 2)2 dx
=
(x − 2)3
3
2
(2 − 2)3 (0 − 2)3 8
−
= sq units.
3
3
3
=
0
18. We have,
⇒
⇒
18x2 − 9πx + π 2 = 0
18x − 6πx − 3πx + π 2 = 0
(6x − π )(3x − π ) = 0
π π
x= ,
⇒
6 3
π
Now, α < β
α= ,
6
π
β=
3
2
Given, g(x) = cos x and f (x) = x
2
y = gof (x)
∴
y = g ( f (x)) = cos x
Area of region bounded by x = α,x = β, y = 0 and curve
y = g ( f (x)) is
π /3
+ 1)dx + ∫ (x2 + 1) dx
A=
0
0
1
3
∫ cos x dx
π /6

x

 x
+ x
= −
+ x + 
3
3
0
− 1 

3
A = [sin x]ππ //36
π
π
3 1
A = sin − sin =
−
3
6
2
2
 3 − 1
A=

 2 

 (− 1)3
  1

=  0 − −
+ (− 1)  +   + 1 − 0
   3



3



1
 4 2 4
= −  − 1 + = + = 2
3
 3 3 3
19. Required area
17. Given, equation of parabola is y = x − 1, which can be
2
rewritten as x2 = y + 1 or x2 = ( y − (−1)).
⇒ Vertex of parabola is (0, − 1) and it is open upward.
=∫
1
0
(1 +
x )dx +
2
∫1
(3 − x)dx − ∫
2
0
x2
dx
4
Y
y=1+√x
(0, 3) (1, 2)
4y=x 2
y=x2–1
(2, 3)
(0, 1)
(2, 1)
x+y=3
2
(0, –1)
y=4x–5
Equation of tangent at (2, 3) is given by T = 0
y + y1
⇒
= x x1 − 1, where, x1 = 2
2
X′
X
(0, 0) (1, 0)(2, 0) (3, 0)
Y′
3/ 2  1
2
2
 x3 


x
x2 
+ 3x −  −  
= x +

3 /2  0 
2  1 12  0

2

= 1 +  +

3
5 3 2
= + −
3 2 3
1  8 

6 − 2 − 3 +  −  

2  12
3 5
= 1 + = sq units
2 2
342 Area
and y ≥ 4x − 1 represents a region to the left of the line
…(ii)
y = 4x − 1
The point of intersection of the curves (i) and (ii) is
(4x − 1)2 = 2x
2
⇒
16x + 1 − 8x = 2x
⇒
16x2 − 10x + 11 = 0
1 1
⇒
x= ,
2 8
1 
So, the points where these curves intersect are  , 1
2 
 1 1
and  ,  .
 8 2
20. Here, {(x, y) ∈ R2 : y ≥ |x + 3|, 5 y ≤ (x + 9) ≤ 15}
∴
y ≥
x+3
 x + 3 , when x ≥ − 3
y≥ 
 − x − 3 , when x ≤ − 3
⇒
 x + 3 , when x ≥ − 3
y2 ≥ 
− 3 − x, when x ≤ − 3
or
Shown as
Y
y2=–x–3
y2=x+3
4x
−1
Y
1,1
2
X
0
–3
Also,
5 y ≤ (x + 9) ≤ 15
⇒
Shown as
(x + 9) ≥ 5 y and x ≤ 6
X′
1
2
−1
2
−1
−1
X
0
∴ Required area = ∫
x=6
–9
(–4,0) A
6
1  1
1  1 
1
 1
 + 1 −  −   − 1 + 
4 2
8
2  6
8
1 3 3  1 9 
=  + −  
4 2 8  6 8 
1 15 3
9
sq units
= ×
−
=
4 8 16 32
X
22. Given, A = {(x, y) : x2 + y2 ≤ 1 and y2 ≤ 1 − x}
Y′
∴ Required area = Area of trapezium ABCD
− Area of ABE under parabola
− Area of CDE under parabola
1
−3
1
= (1 + 2) (5) − ∫
− (x + 3) dx − ∫
(x + 3) dx
4
−
−
3
2
−3
 y + 1 y2 
−  dy

−1/ 2  4
2
1
=
(1,2)
(1,0)
0
E (–3,0) D
Y′
−1
Y
C
Y
X′
(0,1)
(–1,0)
X
1




15  (− 3 − x)3/ 2 
 (x + 3)3/ 2 
=
−
 −

3
3
2 
−




 –3
− 4 

2
2
15 2
15 2 16 15 18 3
2
=
+ [0 − 1] − [8 − 0] =
− −
=
−
=
2
3
2 3 3
2
3 2
3
21. Given region is {(x, y) : y2 ≤ 2x and y ≥ 4x − 1}
y2 ≤ 2x repressents a region inside the parabola
y2 = 2x
X

1  y2
1
− ( y3 )1−1/ 2
= 
+ y
42
 −1/ 2 6
∴ {(x, y) ∈ R2 : y ≥ |x + 3|, 5 y ≤ (x + 9) ≤ 15}
)
1
1 −1
,
8 2
−1
2
(0, 9/5)
4,1
B (–
1
2
O
Y
X′
y 2 = 2x
1
Y′
(–9, 0)
y=
X′
…(i)
Y′
Required area =
1
1 2
πr + 2∫ (1 − y2)dy
0
2
1
=

1
y3 
π (1)2 + 2  y − 
2
3 0

 π 4
=  +  sq units
 2 3
Area 343
23.
PLAN To find the bounded area between y = f( x ) and y = g ( x )
between x = a to x = b.
Y
Hence, required area
f(x)
a
O
3
g(x)
c
∴ Area bounded =
=
2
25. R1 = ∫ x f (x) dx
b
b
∫a[g (x ) − f(x )]dx + ∫c [f(x ) − g (x )]dx
∫a | f(x ) − g (x )|dx
g (x) = y = | cos x − sin x|
π

cos x − sin x, 0 ≤ x ≤

4
=
π
π
sin x − cos x,
≤x≤

4
2
could be shown as
π
2
Y
π/4
O
∴ Area bounded = ∫
+
π /4
2 sin x dx +
π /2
∫π / 4 2 cos x dx
…(i)
2y − x + 3 = 0
…(ii)
Y
x
2y –
x+
3
0
3=
X
–3
2
Y'
On solving Eqs. (i) and (ii), we get
2 x − ( x )2 + 3 = 0
∴
b
∫0
∴
X
{(sin x + cos x) − (cos x − sin x)} dx
X'
…(iv)
f (x) dx
2R1 = R2
24. Given curves are y = x
⇒
⇒
2
−1
(1 − x)2 dx −
1
∫b
(1 − x)2 dx =
b
y=
…(iii)
f (x) dx
From Eqs. (iii) and (iv), we get
= − 2 [cos x]π0 / 4 + 2 [sin x ⋅ n ]ππ // 24
= 4 − 2 2 = 2 2( 2 − 1) sq units
and
−1
2R1 = ∫
∫π / 4 {(sin x + cos x) − (sin x − cos x)} dx
0
2
R2 = ∫
On adding Eqs. (i) and (ii), we get
0
π /2
=∫
…(ii)
26. Here, area between 0 to b is R1 and b to 1 is R2.
π/2
π /4
2
−1
∴
g(x)
g(x)
(1 − x) f (1 − x) dx
[f (x) = f (1 − x), given]
Given, R2 is area bounded by f (x), x = − 1 and x = 2.
= √2 sin x + π
4
1
2
−1
R1 = ∫ (1 − x) f (x) dx
∴
y = sin x + cos x
f(x)
b
R1 = ∫
b
and
√2
b
∫ a f (x) dx = ∫ a f (a + b − x) dx
Using
c
…(i)
−1
X
f (x) = y = sin x + cos x, when 0 ≤ x ≤
Here,
3
0

y3 
= 9 + 9 − 9 = 9 sq units
=  y2 + 3 y −
3  0

g(x)
f(x)
3
0
= ∫ (x2 − x1 ) dy = ∫ {(2 y + 3) − y2} dy
1
 (1 − x)3 
 (1 − x)3 
1
 −3  −  −3  = 4
b
0 

⇒
1
1
1
[(1 − b)3 − 1] + [0 − (1 − b)3 ] =
3
4
3
2
1
1 1
1
3
⇒ − (1 − b) = − + = −
⇒ (1 − b)3 =
3
8
3 4
12
1
1
(1 − b) =
⇒ b=
⇒
2
2
π / 4  1 + sin x
1 − sin x 
27. Required area = ∫
−
 dx

0
cos x
cos x 

⇒−

x
2 tan

2
 1+
2x

1
+
tan
π /4 
2 −
=∫

0
2x
1 − tan

2

x

1 + tan 2

2
( x )2 − 2 x − 3 = 0
( x − 3) ( x + 1) = 0 ⇒
x =3
[since, x = − 1 is not possible]
y=3
1
4
=∫
π /4
0
 1 + sin x
Q

cos x
x
2 tan
2
1−
1 − sin x

>0

cos x



x
1 + tan 2 
2  dx

2x
1 − tan

2 
x 
1 + tan 2

2 

x
x
1 − tan 
 1 + tan
2
2

 dx
−
x
 1 − tan x
1 + tan 


2
2
>
344 Area
=∫
π /4
0
Put tan
π
8
tan
x
1
x
= t ⇒ sec2 dx = dt = ∫
0
2
2
2
∫0
As
∴ Required area OABCO = Area of curve OCBDO
– Area of curve OABDO
1/a  x

2
[given]
⇒
∫ 0  a − ax  dx = 1
x
x
x
− 1 + tan
2 tan
2
2 dx = π / 4
2 dx
∫0
x
x
1 − tan 2
1 − tan 2
2
2
1 + tan
2 −1
4t dt
4t dt
⇒
π
= 2 − 1]
8
⇒
[Q tan
(1 + t 2) 1 − t 2
1/ a
(1 + t 2) 1 − t 2
28. The curves y = (x − 1)2, y = (x + 1)2 and y = 1 /4 are
 1 x3/ 2
2
1
ax3 
−
=1
⋅
−
=1 ⇒


2
3 0
3a
3a 2
 a 3 /2
1
1
[Q a > 0]
⇒ a=
a2 =
3
3
b
30. Since, ∫ f (x) dx = (b − 1) sin (3b + 4)
1
shown as
Y y = (x + 1)2
1/4
y = (x – 1)
2
P
R
Q
O 1/2
–1 –1/2
y = 1/4
X
1
On differentiating both sides w.r.t. b, we get
f (b) = 3(b − 1) ⋅ cos (3b + 4) + sin (3b + 4)
∴ f (x) = sin (3x + 4) + 3(x − 1) cos (3x + 4)
dy
31. Given,
= 2x + 1
dx
On integrating both sides
∫ dy = ∫ (2x + 1) dx
⇒
∴
⇒
∴
where, points of intersection are
1
1
⇒ x=
(x − 1)2 =
4
2
1
1
2
and
(x + 1) = ⇒ x = −
4
2
 1 1
 1 1
i.e.
Q  ,  and R  − , 
 2 4
 2 4
∴ Required area = 2 ∫
1/ 2
0
y = x2 + x + C which passes through (1, 2)
2 =1+1+C
C =0
y = x2 + x
Y
y = x (x + 1)
1

(x − 1)2 − dx

4 
X'
1/ 2
 (x − 1)3 1 
=2 
− x
4 0
 3
1  1
 1
 8 1
= 2 −
− −  − − 0  =
= sq unit
  24 3
 8 ⋅3 8  3
B
C
1
1
 x3 x2 
1 1 5
= ∫ (x2 + x)dx =  +  = + = sq unit
0
3
2

0 3 2 6
32.
O
1
∫0 (x − x
3
α
)dx = 2∫ (x − x3 )dx
0
α2 α4
1
= 2
−

4
4
 2
x
y 2= a
⇒
(1/a,1/a)
A
X'
X
x=1
Y'
y = ax2
Y
O
Thus, the required area bounded by X-axis, the curve
and x = 1
29. As from the figure, area enclosed between the curves is
OABCO.
Thus, the point of intersection of
y = ax2 and x = ay2
−1
X
D
33.
2α 4 − 4α 2 + 1 = 0
4 − 16 − 8
α2 =
4
1
2
α =1 −
2
(Q α ∈ (0, 1))
PLAN (i) Area of region f( x ) bounded between x = a to x = b is
Y'
⇒
⇒
x = a (ax2)2
1
x = 0,
⇒
a
y = 0,
y = f (x)
1
a
 1 1
So, the points of intersection are (0, 0) and  ,  ⋅
 a a
a a 1 a 2 a3
b
D
∫a f(x )dx = Sum of areas of rectangle shown in shaded part.
Area 345
(ii) If f( x )≥ g ( x ) when defined in [a, b ], then
b
∫a
OABCO and points of intersection are (0,0) and
{1 − m, m(1 − m)}.
b
f( x )dx ≥ ∫ g ( x )dx
a
2
Description of Situation As the given curve y = e − x
cannot be integrated, thus we have to bound this function by
using above mentioned concept.
Graph for y = e− x
∴ Area of curve OABCO = ∫
1 −m
[x − x2 − mx] dx
0
Y
2
Y
A
1
B
(0, 0) O
1
—
√e
O
1
—
√2
C
X
1
2
− x2 ≥ − x or e− x ≥ e− x
1 − x2
∫0 e
∴
Also,
∴
1
dx ≥ ∫ e− x dx
0
S ≥ − (e− x )10 = 1 −
⇒
1 − x2
∫0 e
1
e
…(i)
dx ≤ Area of two rectangles
1 
1 1

≤ 1 ×
×
 + 1 −

e
2 
2
1
1 
1
…(ii)
≤
+
1 −

2
e
2
1
1 
1
1
[from Eqs. (i) and (ii)]
+
1 −
 ≥S ≥1 −
e
2
e
2
Also,
∫1 ln (e + 1 − y) dy
=∫
1
∴
⇒
⇒
(1 − m)3 = 27
1 −m =3
Y
y = mx
[put e + 1 − y = t ⇒ − dy = dt]
e
e
1
1
1– m
A
35. Case I When m = 0
In this case,
…(i)
y = x − x2
and
…(ii)
y=0
are two given curves, y > 0 is total region above X-axis.
Therefore, area between y = x − x2 and y = 0
is area between y = x − x2 and above the X-axis
1
⇒
2
Hence, no solution exists.
Case II When m < 0
In this case, area between y = x − x2 and y = mx is
0
1 −m
(x − x2 − mx) dx
0
X
 x2 x3 
1 1 1 9
∴ A = ∫ (x − x ) dx =  −  = − = ≠
0
3 0 2 3 6 2
2
1
y = x − x2
∴ Area of shaded region = ∫
A
B
X
O (0,0)
B
Y
O
y = mx
⇒
m = −2
Case III When m > 0
In this case, y = mx and y = x − x2 intersect in (0,0) and
{(1 – m), m( 1 – m)} as shown in figure
ln t (− dt ) = ∫ ln t dt = ∫ ln y dy = 1
e
x – x2

x2 x3 
= (1 − m) − 
2
3 0

1
1
1
= (1 − m)3 − (1 − m)3 = (1 − m)3
2
6
3
1
9
3
[given]
(1 − m) =
6
2
1
34. Shaded area = e −  ∫ ex dx = 1
0
e
y=
1 −m
Since, x2 ≤ x when x ∈ [0, 1]
⇒
X
{1 – m, m (1 – m)}

x2 x3 
= (1 − m)
− 
2
3 1−m

1
1
= − (1 − m) (1 − m)2 + (1 − m)3
2
3
1
3
= − (1 − m)
6
9
1
[given]
= − (1 − m)3
2
6
⇒
(1 − m)3 = − 27
⇒
(1 − m) = − 3
⇒
m =3 + 1 =4
Therefore, (b) and (d) are the answers.
346 Area
4a 2 4a 1
f (−1) 3a 2 + 3a 

 2
36. Given, 4b 4b 1  f (1)  = 3b2 + 3b 
 2

 4c2 4c 1  f (2)  3c + 3c 


⇒
4a 2 f (−1) + 4a f (1) + f (2) = 3a 2 + 3a ,
4b f (−1) + 4b f (1) + f (2) = 3b + 3b
2
2
4c2 f (−1) + 4cf (1) + f (2) = 3c2 + 3c
and
…(i)
…(ii)
…(iii)
where, f (x) is quadratic expression given by,
f (x) = ax2 + bx + c and Eqs. (i), (ii) and (iii).
⇒
4x2 f (−1) + 4x f (1) + f (2) = 3x2 + 3x
or {4 f (−1) − 3} x2 + {4 f (1) − 3} x + f (2) = 0
As above equation has 3 roots a, b and c.
So, above equation is identity in x.
i.e. coefficients must be zero.
⇒
f (−1) = 3 / 4, f (1) = 3 / 4, f (2) = 0
Q
f (x) = ax2 + bx + c
∴ a = − 1 / 4, b = 0 and c = 1, using Eq. (v)
4 − x2
Thus,
shown as,
f (x) =
4
Let A (−2, 0), B = (2t , − t 2 + 1)
 x3  1  (4x − 3)3/ 2 1 
= 2   − 
 
 3  0  3 ⋅ 4 / 2  3/ 4 
1 1
 1 1
= 2  −  = 1 ⋅ = sq unit
 3 6
6 3
…(iv)
…(v)
Since, AB subtends right angle at vertex V (0, 1).
1 −t 2
⇒
⋅
= −1 ⇒ t =4
2 2t
∴
B (8, − 15)
− (3x + 6)
So, equation of chord AB is y =
.
2
8  4 − x2
3x + 6
∴ Required area = ∫
+

 dx
−2 
4
2 
38. Here, slope of tangent,
dy (x + 1)2 + y − 3
dy
( y − 3)
= (x + 1) +
,
⇒
=
dx
dx
(x + 1)
(x + 1)
Put x + 1 = X and y − 3 = Y
dy dY
⇒
=
dx dX
dY
Y
∴
=X +
dX
X
dY
1
⇒
− Y =X
dX X
∫
IF = e
2
 128


= 8 −
+ 48 + 24 −  −2 + + 3 − 6 


3
3

=
125
sq units
3
37. The region bounded by the curves y = x2, y = − x2 and
y2 = 4 x − 3 is symmetrical about X-axis, where y = 4x − 3
meets at (1, 1).
∴ Area of curve (OABCO )
1
1
( 4x − 3 ) dx
= 2 ∫ x2 dx − ∫
3/ 4
 0

y=x
2
1
dX
X
= e− log X =
∴ Solution is, Y ⋅
1
X
1
1
Y
= X ⋅ dX + c ⇒
=X +c
X ∫
X
X
Y
y = x 2 _ 2x
X′
y − 3 = (x + 1)2 + c(x + 1), which passes through (2, 0).
⇒
− 3 = (3)2 + 3c ⇒ c = − 4
∴ Required curve
y = (x + 1)2 − 4(x + 1) + 3 ⇒ y = x2 − 2x
2
  x3
 
2
− x2 
∴ Required area =  ∫ (x2 − 2x)dx=  
 0
 3
 0

8
4
= − 4 = sq units
3
3
39. The points in the graph are
A (1, 1), B ( 2, 0), C (2, 2), D ( 2, 2)
Y
y = |2 – x 2 |
y = x2
D (√2,2)
C (2,2)
y=2
Y
(1,1) A
B (3/4, 0)
X'
O
C
y = –x 2
Y′
X
2
O
Y′
8


x3 3x2
= x −
+
+ 3x
12
4

 −2
−
X
y 2 = 4x – 3
A
X′
x = 1 B (√2,0)
Y′
X
Area 347
∴ Required area
2
=∫
1
=∫
{ x2 − (2 − x2)} dx +
2
1
(2x2 − 2) dx +
2x3

=
− 2x
3

1
∫
2
2
∫
2
2
{2 − (x2 − 2)} dx
∴
(4 − x2) dx
Sj
Sj −1
3 2
2

x
+ 4x − 
3

a 2 + b2
a
− π
b
=e
= constant
⇒ S 0 , S1 , S 2, K , S j form a GP.
For a = − 1 and b = π
1
. πj
π ⋅ eπ
Sj =
(1 + π 2)
− ay
40. Given, x = (sin by) e
Now,
−1 ≤ sin by ≤ 1
⇒
− e−ay ≤ e− ay sin by ≤ e− ay ⇒ − e− ay ≤ x ≤ e− ay
n
⇒
Y
∑ Sj =
j=0
j
 1 ⋅π

 eπ + 1 = π ⋅ e (1 + e)

 (1 + π 2)


π ⋅ (1 + e)
(1 + π )2
n
π (1 + e)
∑ e j = (1 + π 2) (e0 + e1 + ...+ en )
j=0
π (1 + e) (en + 1 − 1)
⋅
=
e−1
(1 + π 2)
x = e – ay
S3
|x| ≤ 1
 x + ax + b,|x| > 1
41. Given, f (x) =  2x2 ,
S2
S1
Y
S0
X′
a
− jπ
e b
a 2 + b2
=
= a
a
− aπ


− ( j − 1 )π
− ( j − 1 )π
 e b + 1
e b
be b




2
4 2
  8
2
2 2
=
− 2 2 − + 2 + 8 − − 4 2 +
3
3
3
3 

 
 20 − 12 2 
=
 sq units
3


x = e – ay
 − aπ

 e b + 1




a
− jπ
b
be
X
O
x = – 2y2
1
x=–
8
y = f (x)
–2
X′
Y′
–1
O
− ay
In this case, if we take a and b positive, the values − e
and e− ay become left bond and right bond of the curve
and due to oscillating nature of sin by, it will oscillate
between x = e− ay and x = − e− ay
Now,
Sj = ∫
( j + 1 ) π /b
jπ /b
∴
Y′
sin by ⋅ e− ay dy

since, I = sin by ⋅ e− ay dy
∫


− ay
e
−

I =
a
by
b
by
(
sin
+
cos
)


a 2 + b2
⇒
−
a
Sj = −
− ( j + 1 )π
1
[e b
2
2
a +b
a

−
b (− 1) j e b

=
 a 2 + b2

lim f (x) = lim f (x) = f (−1)
x → −1 −
{a sin ( j + 1)π + b cos ( j + 1) π}
− ajπ
e b
jπ

(a sin jπ + b cos jπ ) 

{0 + b(−1) j + 1 }
− e− ajπ / b{0 + b(− 1) j }]

 −aπ

 e b + 1 





[Q (−1) j + 2 = (−1)2 (−1) j = (−1) j ]
=
a
− jπ
be b
x2 + ax + b, if x < − 1

if − 1 ≤ x < 1
f (x) = 2x,
x2 + ax + b, if x ≥ 1

f is continuous on R, so f is continuous at –1 and 1.
 − a ( j + 1)π
e b
 −1
S j = 2
2
a + b 
X
 − aπ

 e b + 1
2
2 

a +b 

and
⇒
⇒
∴
x → −1 +
lim f (x) = lim f (x) = f (1)
x → 1−
x →1+
1 − a + b = − 2 and 2 = 1 + a + b
a − b = 3 and a + b = 1
a =2, b = −1
 x2 + 2 x − 1 ,

Hence, f (x) =  2x,
 x2 + 2x − 1,
if
if
if
x < −1
−1 ≤ x < 1
x≥1
Next, we have to find the points x = − 2 y2 and y = f (x).
The point of intersection is (–2, –1).
−1/ 8  − x

Required area = ∫
− f (x) dx
∴
−2  2


−1/ 8
−1
−1/ 8
−x
dx − ∫
(x2 + 2x − 1)dx − ∫
2x dx
=∫
−2
−2
−1
2
348 Area
−1
 x3

=−
− 
+ x2 − x  − [x2] −−11/ 8
[(−
3
3 2

  −2

3/ 2
  1
2   1

3/ 2
=−
  − 2  −  − + 1 + 1
3
3 2 8


 1
 8
−1
+  − + 4 + 2 −

 64
 3
2
5
63
=
[2 2 − 2−9/ 2] + +
3
3 64
63
509
761
sq units
=
+
=
16 × 3 64 × 3 192
2
x)3/ 2]−−12/ 8
42. Refer to the figure given in the question. Let the
coordinates of P be (x, x2), where 0 ≤ x ≤ 1.
For the area (OPRO ),
Upper boundary: y = x2 and
lower boundary : y = f (x)
Lower limit of x : 0
Upper limit of x : x
∴ Area (OPRO ) =
x
∫0
t 2 dt − ∫
x
0
Now, to get the point of intersection of y = x2 and
y = 2x (1 − x), we get
x2 = 2x (1 − x)
⇒
3 x2 = 2 x
⇒
x (3x − 2) = 0
⇒
x = 0, 2 / 3
Similarly, we can find the coordinate of the points of
intersection of
y = (1 − x2) and y = 2x (1 − x) are x = 1 / 3 and x = 1
From the figure, it is clear that,
 (1 − x)2,
if 0 ≤ x ≤ 1 / 3

if 1 / 3 ≤ x ≤ 2 / 3
f (x) = 2x (1 − x),
2
if

2 /3 ≤ x ≤ 1
x ,
∴ The required area
1
A=∫
0
=∫
0
f (x) dx
1/3
(1 − x)2 dx +
x
x
 t3 
=   − ∫ f (t ) dt
0
 3 0
x
x3
=
− ∫ f (t ) dt
0
3
For the area (OPQO ),
The upper curve : x = y
and the lower curve : x = y /2
Lower limit of y : 0
and upper limit of y : x2
x2
x2 t
∴ Area (OPQO ) = ∫
t dt − ∫
dt
0
0 2
2
2
2
1
2
x4
= [t3/ 2]x0 − [t 2]x0 = x3 −
3
4
3
4
According to the given condition,
x
2
x3
x4
− ∫ f (t ) dt = x3 −
0
3
3
4
On differentiating both sides w.r.t. x, we get
x2 − f (x) ⋅ 1 = 2x2 − x3
⇒
f (x) = x3 − x2, 0 ≤ x ≤ 1
2x (1 − x) dx +
1
∫ 2/ 3
x
 1

= − (1 − x)3
 3

0
1

2x
1 3 
x
+  x2 −
+

3 1/ 3 3  2 / 3

 1  2 3 1 
= −   +  +
 
3
 3 3
  2 2 2
  −
3
 3
2
3
2
 1
 2
  −  +
 3
 3
3
19 13 19 17
+
+
=
sq unit
81 81 81 27
x2
44. Eliminating y from y = and y = x − bx2, we get
b
x2 = bx − b2x2
b
x = 0,
⇒
1 + b2
=
Y
y=
x2
b
y = x – bx 2
X′
–1
O
X
b
2
1+ b
y = 2x(1 − x) in following figure
Y′
(1, 1)
(0, 1)
y = (1 – x)
Thus, the area enclosed between the parabolas
b/(1 + b )2 
x2 
A=∫
 x − bx2 −  dx
0
b

2
2
Q
y=x
(1/2, 1/2)
A
B
y = 2x (1 – x)
X′
O
1/3
Y′
2/3
1
X
dx
3
 1 
  
 3

1
1
+  (1) −
3
3
43. We can draw the graph of y = x2, y = (1 − x2) and
Y
2
3 2 / 3
1/ 3
f (t ) dt
2 /3
∫ 1/ 3
b/(1 + b )2
 x2 x3 1 + b2 
= −
⋅
3
b  0
2
=
b2
1
⋅
6 (1 + b2)2
3
 2 
  
 3

Area 349
On differentiating w.r.t. b, we get
dA 1 (1 + b2)2 ⋅ 2b − 2b2 ⋅ (1 + b2) ⋅ 2b
= ⋅
db 6
(1 + b2)4
1 b (1 − b2)
= ⋅
3 (1 + b2)3
dA
For maximum value of A, put
=0
db
⇒
b = − 1, 0, 1, since b > 0
∴ We consider only b = 1.
dA
Sign scheme for
around b = 1 is as shown below :
db
–
+
0
1
n −1
1
An <
⇒
2n − 2
1
1
From Eqs. (i) and (ii),
< An <
2n + 2
2n − 2
⇒
AB : y = 1, BC : x = − 1, CD : y = − 1, DA : x = 1
Y
B(–1,1)
–
(0,1/2)
X′
π /4
∫0
(tan x)n+ 1 dx < ∫
(tan x)n dx
Y′
Let the region be S and (x, y) is any point inside it.
Then, according to given conditions,
An + An + 2 = ∫
=∫
π /4
0
π /4
0
x2 + y2 < |1 − x|,|1 + x|,|1 − y|,|1 + y|
[( tan x)n + (tan x)n + 2] dx
⇒
⇒
x2 + y2 < (1 − x)2, (1 + x)2, (1 − y)2, (1 + y)2
x2 + y 2 < x 2 − 2 x + 1 , x 2 + 2 x + 1 ,
y2 − 2 y + 1 , y 2 + 2 y + 1
2
2
2
⇒
y < 1 − 2x, y < 1 + 2x, x < 1 − 2 y and x2 < 2 y + 1
Now, in y2 = 1 − 2x and y2 = 1 + 2x, the first equation
represents a parabola with vertex at ( 1/2,0) and second
equation represents a parabola with vertex ( –1/2, 0)
and in x2 = 1 − 2 y and x2 = 1 + 2 y, the first equation
represents a parabola with vertex at (0, 1/2) and second
equation represents a parabola with vertex at (0, –1 /2) .
Therefore, the region S is lying inside the four parabolas
y2 = 1 − 2x, y2 = 1 + 2x, x2 = 1 + 2 y, x2 = 1 − 2 y
(tan x) (1 + tan x) dx
2
n
Y
y = (tan x)
n
B
X
O
Y
2
(tan x) sec x dx
2x
An + 2 < An + 1 < An,
An + An + 2 < 2 An
1
< 2 An
n+1
1
< An
2n + 2
Also, for n > 2 An + An < An + An − 2 =
G
I
O
–1
π /4
 1

=
(tan x)n + 1 
 (n + 1)
0
1
1
=
(1 − 0) =
(n + 1)
n+1
⇒
A(1,1)
=
0
n
2
π /4
(0,1)
y
=∫
⇒
D(1,–1)
C(–1,–1)
π /4
0
X
(1/2,0)
O
(0,–1/2)
⇒
An+ 1 < An
Now, for n > 2
Since,
then
(–1/2,0)
(tan x)n dx
Since, 0 < tan x < 1, when 0 < x < π /4
We have, 0 < (tan x)n+ 1 < (tan x)n for each n ∈N
⇒
A(1,1)
1
π /4
0
…(ii)
46. The equations of the sides of the square are as follow :
From sign scheme, it is clear that A is maximum at
b = 1.
45. We have, An = ∫
2 An <
E x2
H F
=
2y
–1
(1,0)
X
where, S is the shaded region.
Now, S is symmetrical in all four quadrants, therefore
S = 4 × Area lying in the first quadrant.
…(i)
1
n −1
Now, y2 = 1 − 2x and x2 = 1 − 2 y intersect on the line
y = x. The point of intersection is E ( 2 − 1, 2 − 1).
Area of the region OEFO
= Area of ∆ OEH + Area of HEFH
350 Area
1/ 2
1
( 2 − 1 )2 + ∫
1 − 2x dx
2 −1
2
1/ 2
1
2 1


= ( 2 − 1)2 + (1 − 2x)3/ 2 ⋅ (− 1)


2
3 2
2−1
1
1
3/ 2
= (2 + 1 − 2 2 ) + (1 + 2 − 2 2 )
2
3
1
1
= (3 − 2 2 ) + (3 − 2 2 )3/ 2
2
3
1
1
= (3 − 2 2) + [( 2 − 1)2]3/ 2
2
3
1
1
= (3 − 2 2 ) + ( 2 − 1)3
2
3
1
1
= (3 − 2 ) + [2 2 − 1 − 3 2 ( 2 − 1)]
2
3
1
1
= (3 − 2 2 ) + [5 2 − 7]
2
3
1
1
= [9 − 6 2 + 10 2 − 14] = [4 2 − 5] sq units
6
6
1
Similarly, area OEGO = (4 2 − 5) sq units
6
Therefore, area of S lying in first quadrant
2
1
= (4 2 − 5) = (4 2 − 5) sq units
6
3
4
1
Hence, S = (4 2 − 5) = (16 2 − 20) sq units
3
3
=
47. Given parabolas are y = 4x − x2
and
y = − (x − 2)2 + 4 or (x − 2)2 = − ( y − 4)
Therefore, it is a vertically downward parabola with
vertex at (2,4) and its axis is x = 2
2
1
1

and
y = x2 − x ⇒ y =  x −  −


2
4
Also, y = x2 − x meets X-axis at (0,0) and (1, 0).
5/ 2
∴ Area, A1 = ∫
0
=∫
0
5/ 2
[(4x − x2) − (x2 − x)] dx
(5x − 2x2) dx
2
5/ 2
=
5  5
2  5
5 2 2 3 
x − x
=   − . 
2
2  2
3  2
3  0
=
5 25 2 125
⋅
− ⋅
2 4 3 8
=
125 
2 125
sq units
1 −  =
8 
3
24
3
This area is considering above and below X-axis both.
Now, for area below X-axis separately, we consider
1
 x2 x3 
1
1 1 1
A2 = − ∫ (x2 − x) dx =  −  = − = sq units
0
3 0 2 3 6
2
Therefore, net area above the X-axis is
125 − 4 121
sq units
A1 − A2 =
=
24
24
Hence, ratio of area above the X-axis and area below
X-axis
121 1
=
: = 121 : 4
24 6
48. The curve y = x2 is a parabola. It is symmetric about
Y-axis and has its vertex at (0, 0) and the curve
2
is a bell shaped curve. X-axis is its asymptote
y=
1 + x2
and it is symmetric about Y-axis and its vertex is (0, 2).
Y
y = x2
A 2
B (1,1)
y= 2 2
1+ x
(–1, 1) C
X′
Since,
and
⇒
2
⇒
1
1

x −  = y +

2
4
1
1
This is a parabola having its vertex at  , −  ⋅
2
4
1
Its axis is at x = and opening upwards.
2
The points of intersection of given curves are
⇒
4 x − x 2 = x2 − x ⇒ 2 x 2 = 5 x
5
x (2 − 5x) = 0
⇒ x = 0,
2
⇒
But
O
M
X
y = x2
2
y=
1 + x2
2
y=
⇒ y2 + y − 2 = 0
1+ y
…(i)
…(ii)
( y − 1) ( y + 2) = 0 ⇒ y = − 2, 1
y ≥ 0, so y = 1 ⇒ x = ± 1
Therefore, coordinates of C are (–1, 1) and coordinates
of B are (1,1).
∴ Required area OBACO = 2 × Area of curve OBAO
1
 1 2

= 2 ∫
dx − ∫ x2 dx
0 1 + x2
0


1

3
x 
2
2π 1  
= 2 [2 tan −1 x]10 −    = 2
−
=  π −  sq unit



3
3
4

 3  0 
Area 351
49. The required area is the shaded portion in following
figure
y = 2x
Y
∴ The required area
1  (log x)

=∫ 
− ex log x dx
1/ e 

ex
2
1
1/2
O
 x2

 e2 − 5
1  (log x)2 
(
log
)
e
x
−
−
=
2
1
 sq units

4

e  2 1/ e

1/ e  4e 
51. Given,
y = loge x
X′
1
1
=
y = x (x − 1)2 ⇒
dy
= x ⋅ 2 (x − 1) + (x − 1)2
dx
Y
X
max
4
27
X′
Y′
y = x(x – 1)2
O 1/3 1 min
X
∴ The required area
=∫
2

 2x
(2x − log x) dx = 
− (x log x − x)
1/ 2
 1/ 2
 log 2
2
Y′
4 − 2 5
3
=
− log 2 +  sq units
2
2
 log 2
50. Both the curves are defined for x > 0.
∴
Both are positive when x > 1 and negative when 0 < x < 1.
We know that, lim (log x) → −∞
x→ 0 +
log x
Hence, lim
→ −∞. Thus, Y-axis is asymptote of
ex
x→ 0 +
second curve.
[(0) × ∞ form]
And lim ex log x
x→ 0 +
e log x
 ∞

= lim
− form
 ∞

x→ 0 + 1 / x
 1
e 
 x
[using L’Hospital’s rule]
=0
= lim
+ 
1
x→ 0
 − 2
 x 
Thus, the first curve starts from (0, 0) but does not
include (0, 0).
Now, the given curves intersect, therefore
log x
ex log x =
ex
i.e.
(e2x2 − 1) log x = 0
1
[Qx > 0]
x = 1,
⇒
e
Y
y=
log x
ex
y = ex log x
X′
1/e
O
Y′
1
X
= (x − 1) ⋅ (2x + x − 1)
= (x − 1) (3x − 1)
+•
− • +
1 /3
1
Maximum at x = 1 / 3
2
1  2
4
ymax =  −  =


3
3
27
Minimum at x = 1
ymin = 0
Now, to find the area bounded by the curve y = x (x − 1)2,
the Y-axis and line x = 2 .
Y
B
C
2
4
27
X′
O
1
A
x=2
X
Y′
∴ Required area = Area of square OABC − ∫
2
= 2 × 2 − ∫ x (x − 1)2 dx
0

 x (x − 1)3  2 1 2
= 4 − 
− ∫ (x − 1)3 ⋅ 1 dx

3


0 3 0
2
x
(x − 1)4 
= 4 −  (x − 1)3 −
12  0
3
1  10
2 1
sq units
=4− −
+
=
3 12 12  3
dy
52. Given,
= sec2x
y = tan x ⇒
dx
 dy
∴
=2
 
 dx x = π
4
2
0
y dx
352 Area
π 
Hence, equation of tangent at A  , 1 is
4 
y−1
=2
x − π /4
π
⇒
y − 1 = 2x −
2

25
 4  1  8 
sin −1    − 8 −
= 2 6 +
 5  4  3 
2

−
25

 4 4 4 4 
= 2 6 +
sin −1   − − − 
 5 3 3 3 
2

y = tan x
Y

 4 
= 4 + 25 sin −1    sq units
 5 

A
O
X′
1  64
  8  
− 16 −  − 8 

  3  
4  3
1
X
B L
2
54. Given curves are x2 + y2 = 4, x2 = − 2 y and x = y.
−1
Y
Y′
y=x
2

π
(2x − y) =  − 1

2
⇒
x + y2 = 4
∴ Required area is OABO
=∫
π /4
0
X′
( tan x) dx − area of ∆ ALB
1
⋅ BL ⋅ AL
2
1  π π − 2
= log 2 −  −
 ⋅1
2 4
4 
1

=  log 2 −  sq unit

4
−2
−√ 2
√2
O
X
2
= [log|sec x|]π0 / 4 −
−2
Thus, the required area
curves, x2 + y2 = 25, 4 y = |4 − x2| could be
sketched as below, whose points of intersection are
(4 − x2)2
x2 +
= 25
16
53. Given
=
X′
4y = 4 – x 2
4y =x 2 – 4
5
O
–5 –4
2
–2
X
4
x 2 + y 2 = 25
∫−
=2 ∫
Y
4y = x 2 – 4
x 2 − √ 2y
Y′
2
2
2
0
4 − x2 dx −
0
∫−
2
x dx −
∫0
2
− x2
dx
2
  x2 0   x3  2

4 − x2 dx −   
 −
2− 2
3 2 0




2
4
x
2
x
=2 
4 − x2 − sin −1  − 1 −
2
2 0
3
2
5 1

= (2 − π ) − =  − π  sq units

3 3
55. Given curves y = 5 − x2 and y = |x − 1|could be sketched
as shown, whose point of intersection are
5 − x2 = (x − 1)2
Y
–5
Y′
(x2 + 24) (x2 − 16) = 0
x=±4
 4
2  4 − x2 
∴ Required area = 2 ∫ 25 − x2 dx − ∫ 
 dx
0 
0
4 


4  x2 − 4
−∫ 
 dx
2 
4 

y = –x + 1
⇒
⇒
4
 x
25
 x 
sin −1   
25 − x2 +
= 2 
 5 
2
2
0
−
1
4
4
2

 
x3 
1  x3
x
−
−
−
x
4
4

 
3  0 4  3

 2 
X′
y=x–1
–1
1
X
2
Y′
⇒ 5 − x = x − 2 x + 1 ⇒ 2 x2 − 2 x − 4 = 0 ⇒ x = 2 , − 1
∴ Required area
2
=∫
2
−1
2
5 − x2 dx − ∫
1
−1
2
( − x + 1) dx − ∫ (x − 1) dx
1
2
1
2
 − x2

 x2

x
5
 x 
5 − x2 + sin −1    − 
=
+ x −  − x


2
2
2
2
5
 −1 

 −1 
1
Area 353
5
5
2 

−1  − 1  
= 1 + sin −1
 − −1 + sin   
 5 

2
2
5  
1
1

 
 1
−  − + 1 + + 1 − 2 − 2 − + 1

 
 2
2
2
5  −1 2
1
1

= sin
+ sin −1
 −
2
5
5 2
 2
5
= sin −1 
2
 5
=
58. The point of intersection of the curves x2 = 4 y and
x = 4 y − 2 could be sketched are x = − 1 and x = 2.
∴ Required area
 x2 
2  x + 2 
= ∫ 
 −    dx
−1  4 
 4

2
4 1
1−  −
5 2
1
1
1− +
5
5
5
1  5π 1
−  sq units
sin −1 (1) − = 
2
2  4
2
π
π

tan x, − ≤ x ≤

3
3
56. Given, y = 
π
π
 cot x,
≤x≤

6
2
which could be plotted as Y-axis.
1  x2
x3 
2
+
−
x
4  2
3  −1
=
1
4
=
1 10  −7  1 9 9
−   = ⋅ = sq units
4  3  6   4 2 8
8  1
1 

2 + 4 − 3 −  2 − 2 + 3 


 et1 + e– t1 et1 – e– t1 
,

2
2


59. Let P = 
 e– t + et1 e– t1 – et 
and Q = 
,

2
2


Y
y = cot x
y = tan x
X′
=
X
O
We have to find the area of the region bounded by the
curve x2 – y2 = 1 and the lines joining the centre x = 0,
y = 0 to the points (t1 ) and (– t1 ).
Y
π/3
P (t1)
X′
Y′
∴ Required area = ∫
π /4
0
( tan x) dx +
π /3
3
1
= log
− 2 log
2
2
3
1 1

= log
− log =  log e 3 sq units

2
2 2
57. Here,
∫2
4
8
1 + 2 dx = ∫
a

x 
a
⇒
⇒
⇒
⇒
⇒
∴
A
1 N
C
∫ π / 4 ( cot x) dx
= [− log|cos x|]π0 / 4 + [log sin x] ππ //34
3
1
1

 
= −  log
− 0 +  log
− log


 
2
2
2
a
−1
8

1 + 2 dx

x 
4
8
8


x−
= x−

x  2 
x  a
8
8


 a −  − (2 − 4) = (4 − 2) −  a − 


a
a
8
8
a − + 2 =2 − a +
a
a
16
2a −
= 0 ⇒ 2 (a 2 − 8) = 0
a
[neglecting –ve sign]
a = ±2 2
a =2 2
X
Q (−t1)
Y′
Required area
e t1 + e – t1


2
= 2 area of ∆PCN – ∫
ydx
1





1  et1 + e– t1   et1 – e– t1 
t1
dy
=2  
⋅ dt 

 –∫ y
1
dt
2
2



2 
 e2t1 – e–2t1 t1  et – e– t  
–∫
=2
 dt 
8
 2  

0
=
e2t1 – e–2t1 1 t1 2t
– ∫ (e + e–2t – 2)dt
4
2 0
=

e2t1 – e–2t1 1  e2t e–2t
– 
–
– 2t 
4
22
2

=
e2t1 – e–2t1 1 2t1
– (e – e–2t1 – 4t1 )
4
4
14
Differential Equations
Topic 1 Solution of Differential Equations by Variable
Separation Method
Objective Questions I (Only one correct option)
1. The solution curve of the differential equation,
(1 + e− x )(1 + y2)
dy
= y2, which passes through the point
dx
(0, 1) is
(2020 Main, 3 Sep I)

1+ e
(a) y2 + 1 = y  log e 
 2

−x
(c) 2e
2
(d) 2e
dy
= (x − y)2,
dx
(2019 Main, 11 Jan II)
1+ x − y
=x+ y−2
1− x + y
differential

x

−1
dx,
x>0
(2017 Adv.)
(c) 9
(d) 80
(2016 Adv.)
8. The differential equation
dy
1 − y2
determines a
=
dx
y
family of circles with
(2007, 3M)
(a) variable radii and a fixed centre at (0, 1)
(b) variable radii and a fixed centre at (0, – 1)
(c) fixed radius 1 and variable centres along the X-axis
(d) fixed radius 1 and variable centres along the Y-axis
(a) 1/3
2 + sin x
y+1
(b) 2/3
 dy
  = − cos x , y (0) = 1, then
 dx
(2004, 1M)
(c) − 1 / 3
(d) 1
10. A solution of the differential equation
2
x, y ∈ [0, 1] and
f (0) ≠ 0. If y = y (x) satisfies the
dy
differential equation,
= f (x) with y(0) = 1, then
dx
 3
 1
(2019 Main, 9 Jan II)
y   + y   is equal to
 4
 4
(c) 2
and
(b) 2(3 − 3 )
(d) 2(2 + 3 )
 π
y   equals
 2
1− x + y
= 2(x − 1)
1+ x − y
equation
1
is equal
1
π
(
−
)
π
k

 π kπ 
sin  +
 sin  +

4

4
6
6
(a) 3 − 3
(c) 2( 3 − 1)
9. If y = y (x) and
2− x
=x− y
2− y
(b) 3
4
(d)
3
to
4. Let f : [0, 1] → R be such that f (xy) = f (x). f ( y), for all
(a) 5
∑
k =1
2− y
= 2( y − 1)
2− x
(d) − log e
(b) 3
13
when y(1) = 1, is
(c) log e
the

x ) dy =  4 + 9 +

7. The value of
3. The solution of the differential equation,
(b) − log e
satisfies
1
(c) −
3
y(0) = 7, then y(256) =
f ′ (x) = f (x) for all x ∈ R. If h (x) = f ( f (x)), then h′ (1) is
equal to
(2019 Main, 12 Jan II)
(a) log e
y = y(x)
(a) 16
2. Let f be a differentiable function such that f (1) = 2 and
(b) 4e
(2017 Main)
2
(b) −
3
1
(a)
3
8 x( 9+
 1 + ex 
(c) y2 = 1 + y log e 

 2 
 1 + e− x 
(d) y2 = 1 + y log e 

 2 
(a) 4e
dy
 π
+ ( y + 1) cos x = 0 and y(0) = 1, then y 
 2
dx
is equal to
6. If


 + 2




 1 + ex 
(b) y2 + 1 = y  log e 
 + 2
 2 


2
5. If (2 + sin x)
(d) 4
dy
 dy
+ y = 0 is
  −x
 dx
dx
(a) y = 2
(b) y = 2x
(1999, 2M)
(c)y = 2x − 4
(d) y = 2x2 − 4
11. The order of the differential equation whose general
solution is given by y = (c1 + c2) cos (x + c3 ) − c4 ex + c5 ,
where c1 , c2, c3 , c4 , c5 are arbitrary constants, is
(1998, 2M)
(a) 5
(b) 4
(c) 3
(d) 2
Differential Equations 355
Integer & Numerical Answer Type Questions
Objective Questions II
(One or more than one correct option)
17. Let f : R → R be a differentiable function with f (0) = 0.
12. Let b be a nonzero real number. Suppose f : R → R is a
differentiable function such that f (0) = 1. If the
derivative f′ of f satisfies the equation
f (x)
f ′ (x) = 2
b + x2
for all x ∈R , then which of the following statements
(2020 Adv.)
is/are TRUE?
(a) If b > 0, then f is an increasing function
(b) If b < 0, then f is a decreasing function
(c) f (x)f (− x) =1for all x∈ R
(d) f (x) − f (− x) = 0 for all x ∈ R
x x− t
∫0 e
f (t ) dt for all x ∈ [0, ∞ ). Then,
which of the following statement(s) is (are) TRUE?
(2018 Adv.)
(a) The curve y = f (x) passes through the point (1, 2)
(b) The curve y = f (x) passes through the point (2, − 1)
(c) The area of the region
π−2
{(x, y) ∈ [0, 1] × R : f (x) ≤ y ≤ 1 − x2 } is
4
(d) The area of the region
π −1
{(x, y) ∈ [0, 1] × R : f (x) ≤ y ≤ 1 − x2 } is
4
14. Let y(x) be a solution of the differential equation
(1 + ex ) y′ + yex = 1. If y(0) = 2, then which of the
following statement(s) is/are true?
(2015 Adv.)
(a) y (−4) = 0
(b) y (−2) = 0
(c) y(x) has a critical point in the interval (−1 , 0)
(d) y(x) has no critical point in the interval (−1 , 0)
15. Consider the family of all circles whose centres lie on the
straight line y = x.If this family of circles is represented by
the differential equation Py′ ′+ Qy′ + 1 = 0, where P , Q are
dy
d 2y
), then
the functions of x, y and y′ (here, y′ =
, y′ ′ =
dx
dx2
which of the following statement(s) is/are true? (2015 Adv.)
f (t ) dt . Then, the value of f (ln 5) is … .
(2009)
Assertion and Reason
(a) Statement I is true, Statement II is also true;
Statement II is the correct explanation of Statement I.
(b) Statement I is true, Statement II is also true;
Statement II is not the correct explanation of
Statement I.
(c) Statement I is true; Statement II is false.
(d) Statement I is false; Statement II is true.
19. Let a solution y = y(x) of the differential equation
x x2 − 1 dy − y y2 − 1 dx = 0 satisfy y(2) =
2
3
π

Statement I y(x) = sec sec−1 x −  and

6
1 2 3
1
Statement II y(x) is given by =
− 1− 2
y
x
x
(2008, 3M)
Analytical & Descriptive Questions
20. If P(1) = 0 and
P (x) > 0, ∀ x > 1.
dP (x)
> P (x), ∀ x ≥ 1 , then prove that
dx
(2003, 4M)
21. Let y = f (x) be a curve passing through (1, 1) such that
the triangle formed by the coordinate axes and the
tangent at any point of the curve lies in the first
quadrant and has area 2 unit. Form the differential
equation and determine all such possible curves.
Passage
c), where c is a positive parameter,
(1999, 3M)
(b) order 2
(d) degree 4
x
0
Passage Based Problems
16. The differential equation representing the family of
(a) order 1
(c) degree 3
f (x) = ∫
(1995, 5M)
(a) P = y + x
(b) P = y − x
(c) P + Q = 1 − x + y + y′ + ( y′ )2
(d) P – Q = x + y – y′ – ( y′ )2
curves y2 = 2c (x +
is of
18. Let f : R → R be a continuous function, which satisfies
For the following question, choose the correct answer
from the codes (a), (b), (c) and (d) defined as follows.
13. Let f : [0, ∞ ) → R be a continuous function such that
f (x) = 1 − 2x +
If y = f (x) satisfies the differential equation
dy
= (2 + 5 y) (5 y − 2), then the value of lim f (x) is ...... .
x→− ∞
dx
(2018 Adv.)
Let f : [0, 1] → R (the set of all real numbers) be a function.
Suppose the function f is twice differentiable,
f ( 0) = f (1) = 0 and satisfies
f ′ ′ (x) − 2 f ′ (x) + f (x) ≥ ex , x ∈ [0, 1]
(2013 Adv.)
356 Differential Equations
22. If the function e−x f (x) assumes its minimum in the
interval [0, 1] at x = 1 / 4, then which of the following is
true?
1
3
< x<
4
4
1
(c) f ′ (x) < f (x), 0 < x <
4
(a) f ′ (x) < f (x),
1
4
(b) f ′ (x) > f (x), 0 < x <
(d) f ′ (x) < f (x),
3
< x<1
4
(c) −
(b) −
1
< f (x ) < 1
4
(a) g is increasing on (1, ∞ )
(b) g is decreasing on (1, ∞ )
(c) g is increasing on (1, 2) and decreasing on (2, ∞ )
(d) g is decreasing on (1, 2) and increasing on (2, ∞ )
25. Consider the statements.
I. There exists some x ∈ R such that, f (x) + 2x = 2(1 + x2)
II. There exists some x ∈ R such that,
23. Which of the following is true?
(a) 0 < f (x) < ∞
24. Which of the following is true?
1
1
< f (x ) <
2
2
2 f (x) + 1 = 2x (1 + x)
(d) − ∞ < f (x) < 0
(a) Both I and II are true (b) I is true and II is false
(c) I is false and II is true (d) Both I and II are false
Topic 2 Linear Differential Equation and
Exact Differential Equation
Objective Questions I (Only one correct option)
6. The
1. The general solution of the differential equation
( y2 − x3 )dx − xydy = 0 (x ≠ 0) is (where, C is a constant of
integration)
(2019 Main, 12 April II)
(a) y2 − 2x2 + Cx3 = 0
(c) y2 + 2x2 + Cx3 = 0
(b) y2 + 2x3 + Cx2 = 0
(d) y2 − 2x3 + Cx2 = 0

1
y
If value of y is 1 when x = 1, then the value of x for which
(2019 Main, 12 April I)
y = 2, is
2. Consider the differential equation, y2dx +  x −  dy = 0.


(a)
5
1
+
2
e
(b)
3
1
−
2
e
(c)
1
1
+
2
e
(d)
3
−
2
π
π
(a) y ′   − y ′  −  = π −
 4
 4
π
π
y ′   + y ′  −  = − 2
 4
 4
 π π
x ∈− , ,
 2 2
such
that
(2019 Main, 10 April II)
1
(c) 2 +
e
π2
2 3
(b) −
2
(d) e − 2
(2019 Main, 9 April II)
π2
2 3
(a)
1
4
(b)
1
2
y(0) = 0.
that
If
(2019 Main, 8 April I)
(c) 1
(d)
1
16
8. If a curve passes through the point (1, − 2) and has slope
of the tangent at any point (x, y) on it as
x2 − 2 y
, then
x
(c) −
π2
4 3
(d) −
π2
2
(b) (− 1, 2)
(c) (−
2 , 1)
(d) (3, 0)
9. Let y = y(x) be the solution of the differential equation,
dy
x

 π
− y sin x = 6x, 0 < x <  and y  = 0, then

 3
2
dx
 π
y  is equal to
 6
(a)
dy
+ 2x(x2 + 1) y = 1 such
dx
π
a y(1) = , then the value of ‘a’ is
32
(x2 + 1)2
(2019 Main, 12 Jan II)
dy
 π π
= (tan x − y) sec2 x, x ∈  − ,  , such that y (0) = 0,
 2 2
dx
 π
(2019 Main, 10 April I)
then y  −  is equal to
 4
5. If cos x
7. Let y = y(x) be the solution of the differential equation,
(a) ( 3 , 0)
4. If y = y(x) is the solution of the differential equation
1
(b)
−e
2
1
x3
+
5
5x2
4
1
(d) y = x3 +
5
5x2
(b) y =
the curve also passes through the point
2 (b)
π
π
π2
π
π
(c) y   + y −  =
+ 2 (d) y   − y −  =
 4
 4
 4
 4
2
1
(a) − 2
e
3
x2
+
4
4x2
3
1
(c) y = x2 +
4
4x2
(a) y =
e
3. Let y = y(x) be the solution of the differential equation,
dy
+ y tan x = 2x + x2 tan x,
dx
y(0) = 1. Then
solution
of
the
differential
equation
dy
2
(2019 Main, 9 April I)
x
+ 2 y = x (x ≠ 0) with y(1) = 1, is
dx
dy
+ y = x log e x, (x > 1). If 2 y(2) = log e 4 − 1, then y(e)
dx
is equal to
(2019 Main, 12 Jan I)
x
(a) −
e
2
(b) −
e2
2
(c)
e
4
(d)
e2
4
10. If y(x) is the solution of the differential equation
dy  2x + 1
−2x
+
 y = e , x > 0,
dx  x 
1
(2019 Main, 11 Jan I)
where y (1) = e−2, then
2
1
(a) y(x) is decreasing in  , 1
2 
(b) y(x) is decreasing in (0, 1)
(c) y(log e 2) = log e 4
log e 2
(d) y(log e 2) =
4
Differential Equations 357
11. Let f
be a
differentiable function such
3 f (x)
f ′ (x) = 7 −
, (x > 0) and f (1) ≠ 4. Then, lim x
4 x
x→ 0 +
that
 1
f 
 x
(2019 Main, 10 Jan II)
(a) does not exist
4
7
(d) exists and equals 4
(b) exists and equals
(c) exists and equals 0
dy
3
1
 −π π 
 π 4
,  and y  = , then
+
y=
,x ∈ 
2
2


 4 3
3 3
dx cos x
cos x
π


y −  equals
 4
(2019 Main, 10 Jan I)
12. If
1
(a)
+ e6
3
4
(b) −
3
1
(c)
+ e3
3
1
(d)
3
13. If y = y(x) is the solution of the differential equation,
x
dy
 1
+ 2 y = x2 satisfying y(1) = 1, then y  is equal to
 2
dx
(2019 Main, 9 Jan I)
(a)
13
16
(b)
1
4
(c)
49
16
(d)
7
64
14. Let y = y(x) be the solution of the differential equation
dy
+ y cos x = 4x, x ∈ (0, π ).
dx
 π
 π
If y  = 0, then y  is equal to
 6
 2
sin x
(a)
4 2
π
9 3
(b)
−8 2
π
9 3
(c) −
8 2
π
9
(2018 Main)
4 2
π
9
(d) −
satisfies the differential equation, y(1 + xy)dx = x dy,
 1
then f  −  is equal to
 2
(2016 Main)
2
5
(b) −
4
5
(c)
2
5
(d)
4
5
dy
+ y = 2x log x, (x ≥ 1). Then, y(e) is equal to
dx
(2015 Main)
(a) e
(b) 0
(c) 2
(d) 2e
17. The function y = f (x) is the solution of the differential
equation
dy
xy
x4 + 2x
+ 2
=
dx x − 1
1 − x2
f (0) = 0. Then, ∫
(a)
π
3
−
3
2
3 /2
−
(b)
3
2
in
(−1, 1) satisfying
f (x) dx is
π
3
−
3
4
(c)
(2014 Adv.)
π
3
−
6
4
(d)
π
3
−
6
2
18. Let f : [1 /2, 1] → R (the set of all real numbers) be a
positive, non-constant and differentiable function such
that f ′ (x) < 2 f (x) and f (1 / 2) = 1 . Then, the value of
1
∫ f (x) dx lies in the interval
(2013 Adv.)
1/ 2
(a) (2e − 1, 2e)
e−1
(c) 
, e − 1
 2

t→ x
t 2f (x) − x2f (t )
= 1 for each x > 0 . Then,
t−x
f (x) is
1
(a)
+
3x
1
(c) − +
x
(2007, 3M)
2x2
3
2
1 4x2
(b) −
+
3x
3
1
(d)
x
x2
20. If x dy = y (dx + y dy), y (1) = 1 and y (x) > 0. Then, y (−3)
is equal to
(2005, 1M)
(a) 3
(c) 1
(b) 2
(d) 0
21. If y (t ) is a solution of (1 + t )
dy
− ty = 1 and y (0) = − 1,
dt
then y (1) is equal to
(a) −1 / 2
(c) e − 1 / 2
(2003, 1M)
(b) e + 1 / 2
(d) 1 / 2
Objective Questions II
(One or more than one correct option)
(b) (e − 1, 2e − 1)
e − 1
(d)  0,


2 
f ′ (x) = 2 −
f (x)
for all x ∈ (0, ∞ ) and f (1) ≠ 1. Then
x
(2016 Adv.)
1
(a) lim f ′  = 1
x → 0+  x 
1
(b) lim x f   = 2
x → 0+
 x
(c) lim x2f ′(x) = 0
x → 0+
(d)|f (x)|≤ 2 for all x ∈ (0, 2)
23. If
16. Let y(x) be the solution of the differential equation
(x log x)
f (1) = 1, and lim
22. Let f : (0, ∞ ) → R be a differentiable function such that
15. If a curve y = f (x) passes through the point (1, − 1) and
(a) −
19. Let f (x) be differentiable on the interval (0, ∞) such that
satisfies
the
differential
y(x)
y′ − y tan x = 2 x sec x and y(0), then
π
(a) y   =
 4
π
(c) y   =
 3
π2
8 2
π2
9
equation
(2012)
π
π2
(b) y′   =
 4  18
π
4π
2π 2

(d) y′   =
+
 3
3
3 3
Analytical & Descriptive Question
24. Let u (x) and v (x) satisfy the differential equations
du
dv
+ p (x) u = f (x) and
+ p (x) v = g (x), where
dx
dx
p (x), f (x) and g (x) are continuous functions. If
u (x1 ) > v (x1 ) for some x1 and f (x) > g (x) for all x > x1,
prove that any point (x, y) where x > x1 does not satisfy
(1997, 5M)
the equations y = u (x) and y = v (x).
Integer & Numerical Answer Type Question
25. Let y′ (x) + y(x) g′ (x) = g (x) g′ (x), y(0) = 0, x ∈ R, where
d f (x)
and g (x) is a given non-constant
dx
differentiable function on R with g (0) = g (2) = 0. Then,
the value of y(2) is ……
(2011)
f ′ (x) denotes
358 Differential Equations
Topic 3 Applications of Homogeneous Differential Equations
Objective Questions I (Only one correct option)
1. If a curve y = f (x), passing through the point (1, 2), is the
solution
of
equation,
7. Let Γ denote a curve y = y(x) which is in the first
(2020 Main, 2 Sep II)
quadrant and let the point (1, 0) lie on it. Let the
tangent to Γ at a point P intersect the y-axis at YP . If
PYP has length 1 for each point P on Γ, then which of the
following options is/are correct?
(2019 Adv.)
differential
 1
2x dy = (2xy + y )dx, then f   is equal to
 2
2
(a)
the
2
1
1
(b)
(c) 1 + log e 2
1 + log e 2
1 − log e 2
(d)
−1
1 + log e 2
2. Given that the slope of the tangent to a curve y = y(x) at
2y
. If the curve passes through the
x2
centre of the circle x2 + y2 − 2x − 2 y = 0, then its
(2019 Main, 8 April II)
equation is
any point (x, y) is
(a) x2 log e|y| = − 2(x − 1)
(b) x log e|y|= x − 1
(c) x log e|y| = 2(x − 1)
(d) x log e|y| = − 2(x − 1)
3. The curve amongst the family of curves represented by
the differential equation, (x2 − y2)dx + 2xydy = 0, which
passes through (1, 1), is
(2019 Main, 10 Jan II)
(a) a circle with centre on the Y-axis
(b) a circle with centre on the X-axis
(c) an ellipse with major axis along the Y-axis
(d) a hyperbola with transverse axis along the X-axis.
4. Let the population of rabbits surviving at a time
t be governed by the differential equation
dp(t ) 1
= p(t ) − 200. If p(0) = 100, then p(t ) is equal to
dt
2
(2014 Main)
(a) 400 −
t
300 e 2
(b) 300 − 200 e
(c) 600 −
t
500 e 2
(d) 400 − 300 e
−
t
2
−
t
2
y
(b) cosec   = log x + 2
 x
2y
1
(d) cos   = log x +
 x 
2
6. At present, a firm is manufacturing 2000 items. It is
estimated that the rate of change of production P with
respect to additional number of workers x is given by
dP
= 100 − 12 x. If the firm employees 25 more
dx
workers, then the new level of production of items is
(2013 Main)
(a) 2500
(b) 3000
(c) 3500
(a) xy′ +
1 − x2 = 0
(b) xy′ − 1 − x2 = 0
 1 + 1 − x2 
 − 1 − x2
(c) y = log e 


x


 1 + 1 − x2 
 +
(d) y = − log e 


x


8. A
solution
1 − x2
curve
of the differential equation
dy
(x + xy + 4x + 2 y + 4)
− y2 = 0, x > 0, passes through
dx
(2016 Adv.)
the point (1, 3). Then, the solution curve
2
(a) intersects y = x + 2 exactly at one point
(b) intersects y = x + 2 exactly at two points
(c) intersects y = (x + 2)2
(d) does not intersect y = (x + 3)2
9. Tangent is drawn at any point P of a curve which passes
through (1, 1) cutting X-axis and Y-axis at A and B,
respectively. If BP : AP = 3 : 1, then
(2006, 3M)
dy
+ y =0
dx
dy
(b) differential equation of the curve is 3x
− y=0
dx
1
(c) curve is passing through  , 2
8 
(a) differential equation of the curve is 3x
(d) normal at (1, 1) is x + 3 y = 4.
π
5. A curve passes through the point 1,  . Let the slope of
 6
y
 y
the curve at each point (x, y) be + sec   , x > 0.
 x
x
Then, the equation of the curve is
(2013 Adv.)
1
y
(a) sin   = log x +
 x
2
2y 

(c) sec   = log x + 2
 x 
Objective Questions II
(One or more than one correct option)
(d) 4500
Fill in the Blank
10. A spherical rain drop evaporates at a rate proportional
to its surface area at any instant t. The differential
equation giving the rate of change of the rains of the
rain drop is …. .
(1997C, 2M)
Analytical & Descriptive Questions
11. If length of tangent at any point on the curve y = f (x)
intercepted between the point and the X-axis is of
length 1. Find the equation of the curve.
(2005, 4M)
12. A right circular cone with radius R and height H
contains a liquid
which evaporates at a rate
proportional to its surface area in contact with air
(proportionality constant = k > 0). Find the time after
(2003, 4M)
which the cone is empty.
Differential Equations 359
13. A hemispherical tank of radius 2 m is initially full of
15. A curve passing through the point (1, 1) has the property
water and has an outlet of 12 cm2 cross-sectional area
at the bottom. The outlet is opened at some instant.
The flow through the outlet is according to the law
v (t ) = 0.6 2 gh (t ), where v (t ) and h (t) are respectively
the velocity of the flow through the outlet and the
height of water level above the outlet at time t and g is
the acceleration due to gravity. Find the time it takes
to empty the tank.
(2001, 10M)
that the perpendicular distance of the origin from the
normal at any point P of the curve is equal to the distance
of P from the X-axis. Determine the equation of the
(1999, 10M)
curve.
16. A and B are two separate reservoirs of water. Capacity of
reservoir A is double the capacity of reservoir B. Both the
reservoirs are filled completely with water, their inlets
are closed and then the water is released simultaneously
from both the reservoirs. The rate of flow of water out of
each reservoir at any instant of time is proportional to
the quantity of water in the reservoir at the time.
Hint Form a differential equation by relating the
decreases of water level to the outflow.
14. A country has food deficit of 10%. Its population grows
continuously at a rate of 3% per year. Its annual food
production every year is 4% more than that of the last
year. Assuming that the average food requirement per
person remains constant, prove that the country
will become self- sufficient in food after n years, where
n is the smallest integer bigger than or equal to
ln 10 − ln 9
.
ln (1.04) − (0.03)
(2000, 10M)
One hour after the water is released, the quantity of
1
water in reservoir A is 1 times the quantity of water in
2
reservoir B. After how many hours do both the reservoirs
have the same quantity of water?
(1997, 7M)
17. Determine the equation of the curve passing through the
origin in the form y = f (x), which satisfies the
dy
differential equation
= sin (10x + 6 y)
(1996, 5M)
dx
Match the Columns
18. Match the conditions/expressions in Column I with statements in Column II.
Column I
π /2
(2006, 6M)
Column II
A.
∫0
B.
Area bounded by − 4 y 2 = x and x − 1 = − 5 y 2
q.
0
C.
The angle of intersection of curves y = 3 x − 1 log x and y = x x − 1 is
r.
3e y / 2
D.
If
s.
4
3
cos x
(sin x )
{cos x cot x − log (sin x )
sin x
p.
}dx
dy
2
passing through (1, 0), then ( x + y + 2 ) is
=
dx
x + y
1
Answers
21. (a)
Topic 1
22. (a)
23. (a, d)
25. (0)
2. (c)
3. (b)
4. (a)
6. (c)
7. (a,c)
1. (c)
2. (b)
3. (d)
4. (b)
5. (a)
6. (b)
7. (c)
8. (c)
1. (a)
9. (a)
10. (c)
11. (c)
12. (a,c)
5. (a)
14. (a, c)
15. (b, c)
8. (a, d)
9. (a, c)
13. (b, c)
16. (a, c)
17. (0.40)
18. (0)
19. (b)
dy
d 2y
21. Differential Equation: 2 = 0, x 2
+1 = 0
dx
dx
Curves : x + y = 2, xy = 1
22. (c)
23. (d)
24. (b)
25. (c)
Topic 2
1. (b)
2. (b)
3. (a)
4. (d)
5. (b)
6. (a)
7. (d)
8. (a)
9. (c)
13. (c)
10. (a)
14. (c)
11. (d)
15. (d)
12. (a)
16. (c)
17. (b)
18. (d)
19. (a)
20. (a)
Topic 3

 dr
10.  = − λ 

 dt


1 + 1 −y2
11.  1 − y 2 − log
= ± x + c
2


1 − 1 −y


H

12. T = 

k

 14 π × 10 5
13. 
unit

 27 g

1 
15. ( x 2 + y 2 = 2 x ) 16.  log 3
2 

4
17.
1
4

tan −1  tan  4 x + tan −1

3
5
3  3  5x
−
 −
4 5  3
18. A → p; B → s; C → q; D → r
Hints & Solutions
Topic 1 Solution of Differential Equations
by Variable Separation Method
⇒
1. Given differential equation
dy
(1 + e ) (1 + y )
= y2
dx
ex
1 + y2
⇒
=
dy
∫ y2
∫ 1 + ex dx
−x
2
which passes through (0, 1),
so
− 1 + 1 = log e 2 + C ⇒ C = − log e 2
So, equation of required curve is
 1 + ex 
y2 = 1 + y log e 

 2 
Hence, option (c) is correct.
2. Given that,
⇒
∫
⇒
⇒
∴
⇒
⇒
Q
So,
⇒
⇒
f ′ (x) = f (x)
f ′ (x)
=1
f (x)
f ′ (x)
dx = ∫ 1 ⋅ dx
f (x)
⇒
[by integrating both sides w.r.t. x]
Put f (x) = t ⇒ f ′ (x)dx = dt
dt
∫ t = ∫ 1 dx
dx


Q
= ln|x|+ C
ln|t|= x + C

 ∫ x
…(i)
ln| f (x)|= x + C
[Q t = f (x)]
f (1) = 2
[using Eq. (i)]
ln (2) = 1 + C
C = ln 2 − ln e
[Q ln e = 1]
 2
 A
[Q ln A − ln B = ln   ]
C = ln  
 e
 B
From Eq. (i), we get
⇒
⇒
Now,
ef (x)
=x
2
e
f (x) = ex
2
| f (x)|= 2ex −1


a+x
1
dx
log e
=
+ C
∫ 2
2
a
a
x
2
−

 a −x
 1 + x − y
1
[Q t = x − y]
log e 
 =x+C
 1 − x + y
2
Since, y = 1 when x = 1, therefore
 1 + 0
1
log e 
 =1 + C
 1 + 0
2
⇒
∴
⇒
C = −1
1
 1 + x − y
log e 
 = x −1
 1 − x + y
2
− log e
[Q log 1 = 0]
1−x+ y
= 2(x − 1)
1+ x− y
1
= log x−1 = − log x]
x
4. Given, f (xy) = f (x) ⋅ f ( y), ∀ x, y ∈ [0, 1]
 2
⇒ ln| f (x)|− ln   = x
 e
ln
⇒