41 Years’
CHAPTERWISE TOPICWISE
SOLVED PAPERS
2019-1979
IITJEE
(JEE Main & Advanced)
Mathematics
Amit M Agarwal
Arihant Prakashan (Series), Meerut
Arihant Prakashan (Series), Meerut
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© Author
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CONTENTS
1. Complex Numbers
1-26
2. Theory of Equations
27-47
3. Sequences and Series
48-71
4. Permutations and Combinations
72-83
5. Binomial Theorem
84-96
6. Probability
97-126
7. Matrices and Determinants
127-159
8. Functions
160-175
9. Limit, Continuity and Differentiability
176-224
10. Application of Derivatives
225-262
11. Indefinite Integration
263-277
12. Definite Integration
278-310
13. Area
311-334
14. Differential Equations
335-357
15. Straight Line and Pair of Straight Lines
358-382
16. Circle
383-416
17. Parabola
417-434
18. Ellipse
435-449
19. Hyperbola
450-460
20. Trigonometrical Ratios and Identities
461-473
21. Trigonometrical Equations
474-487
22. Inverse Circular Functions
488-497
23. Properties of Triangles
498-518
24. Vectors
519-548
25. 3D Geometry
549-572
26. Miscellaneous
573-598
Ÿ JEE Advanced Solved Paper 2019
1-18
SYLLABUS
JEE MAIN
UNIT I Sets, Relations and Functions
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.
UNIT II Complex Numbers and
Quadratic Equations
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 coefficients, 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 IV Permutations and
Combinations
Fundamental principle of counting, permutation
as an arrangement and combination as
selection, Meaning of P(n,r) and C (n,r), simple
applications.
UNIT V Mathematical Induction
Principle of Mathematical Induction and its
simple applications.
UNIT VI Binomial Theorem and its
Simple Applications
Binomial theorem for a positive integral index,
general term and middle term, properties of
Binomial coefficients and simple applications.
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.
,
form of the equation of a circle, its radius and
centre, equation of a 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
UNIT IX Integral Calculus
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
,
x2 ± a2
dx
,
ax + bx + c
2
(px + q) dx
,
ax 2 + bx + c
dx
a2 – x2
(px + q) dx
,
2
2
ax + bx + c
ax + bx + c
a 2 ± x 2 dx
dx
dx
,
a2 – x2
and
x 2 – a 2 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 dy + p(x)y = q(x)
dx
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 perpendicular lines, intercepts
of a line on the coordinate 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
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 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.
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)
1 Let z ∈ C with Im (z ) = 10 and it satisfies
for some natural number n, then
2z − n
= 2i − 1
2z + n
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.
5 + 3z
, then
5(1 − z )
(b) 5 Re (ω) > 1
(d) 5 Re(ω) > 4
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)
3π
(a)
4
5π
(b)
6
(c) π
6 i –3 i
4
3i
20
3
 3
−1  1 
(c) sin −1 
 (d) sin 

 3
 4 
π
6
1
–1 = x + iy, then
i
(1998, 2M)
(a) x = 3, y = 1 (b) x = 1, y = 1 (c) x = 0, y = 3 (d) x = 0, y = 0
13
∑ (i n + i n + 1 ), where i =
(b) i − 1
(a) i
4 Let  −2 − i =
(b) 85
7. If
(b)
2 + 3i sin θ
is purely imaginary, is
1 − 2i sin θ
(2016 Main)
−1, equals
n =1
3
(a) 91
π
3
8. The value of sum
(2019 Main, 9 April II)
(a) 4 Im(ω) > 5
(c) 5 Im (ω) < 1
(a)
(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
3 Let z ∈ C be such that|z|< 1. If ω =
6. A value of θ for which
2π
(d)
3
(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?
(a) 1 − 1 + y
(c) 1 + 1 + y
(2017 Adv.)
2
(b) − 1 − 1 − y
2
(d) − 1 +
2
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.
If z1 and z2 are complex numbers such that
z1 – 2z2
is
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
1
(d)
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
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
(b) 32
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| =  +
z
z
z
1
2
3
(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
(b)
 1 
1
2
(c)
(d)
⋅
2
z
+
1
z
1
|
z
+
1|2
|
+
|


1
|z + 1|2
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
(a) − 1
(b)
(a) x = nπ
(c) x = (n + 1/2) π
(b) lies in the interval (1, 2)
(a)
(a) 0
z −1
(where, z ≠ − 1), then Re (w) is
z+1
(2003, 1M)
17. The
(2014 Main)
(a) is equal to 5/2
1
(b)
2
13. If|z| = 1 and w =
(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
1
2
(b)| z | = 1 and z ≠ 1
(d) None of these
16. For positive integers n1 , n2 the value of expression
(b) straight line parallel toY -axis
minimum value of z +
(a)| z | = 1, z ≠ 2
(c) z = z
(d) 2
5 Let z be a complex number such that | z | + z = 3 + i
(a)
w = α + iβ, where β ≠ 0 and z ≠ 1, satisfies the
 w − wz 
condition that 
 is purely real, then the set of
 1−z 
values of z is
(2006, 3M)
12. If
(c) 40
(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
z
lie on
1 − z2
(2007, 3M)
(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
+  +
−  , then
 2 2
 2 2
21. If z = 
(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 − 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
Passage II
Objective Questions II
(One or more than one correct option)
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)
23. Let s, t, r be non-zero complex numbers and L be the set of
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
24. Let z1 and z2 be complex numbers such that z1 ≠ z2 and
|z1| = |z2|. If z1 has positive real part and z2 has negative
z + z2
imaginary part, then 1
may be
(1986, 2M)
z1 − z2
29. 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
30. The number of elements in the set A ∩ B ∩ C is
(a) 0
(c) 2
31. 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
25. If z1 = a + ib and z2 = c + id are complex numbers such
that |z1| = |z2| = 1 and Re (z1z2) = 0, then the pair of
complex numbers w1 = a + ic and w2 = b + id satisfies
(b)|w2| = 1
(d) None of these
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
(1985, 2M)
Read the following passages and answer the questions
that follow.
Passage I
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 }
z ∈s
32. If α , β, γ are the cube roots of p, p < 0, then for any x, y
2+ 3
2
3+ 3
(d)
2
(b)
and z then
20 π
(b)
3
16 π
(c)
3
32 π
(d)
3
28. Let z be any point in A ∩ B ∩ C and let w be any point
satisfying|w − 2 − i| < 3. Then,|z | − |w| + 3 lies between
(a) − 6 and 3
(c) − 6 and 6
(b) − 3 and 6
(d) − 3 and 9
xα + yβ + zγ
= ... .
xβ + yγ + zα
(1990, 2M)
33. For any two complex numbers z1 , z2 and any real
27. Area of S is equal to
10 π
(a)
3
the set of points
satisfying|Re( z)|≤ 2
the set of points z
satisfying| z| ≤ 3
Fill in the Blanks
26. min|1 − 3i − z|is equal to
2− 3
2
3− 3
(c)
2
Column II
Column I
Passage Based Problems
(a)
(b) 1
(d) ∞
Match the Columns
(a) zero
(b) real and positive
(c) real and negative
(d) purely imaginary
(a)|w1| = 1
(c) Re (w1 w2 ) = 0
(b) 30 and 34
(d) 40 and 44
numbers a and b,|az1 − bz2|2+ |bz1 + az2|2 = K .
(1988, 2M)


 x
 x
sin  2 + cos  2 − i tan (x)

34. If the expression 

 x 
1 + 2 i sin  2 


is real, then the set of all possible values of x is… .
(1987, 2M)
4 Complex Numbers
True/False
41. For complex numbers z and w, prove that
35. If three complex numbers are in AP. Then, they lie on a
circle in the complex plane
| z |2 w − |w|2 z = z − w, if and only if z = w or z w = 1.
(1999, 10M)
(1985 M)
36. 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)
37. 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
42. Find all non-zero complex numbers z satisfying
z = iz 2.
43. If
(1996, 2M)
iz + z − z + i = 0,
3
2
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.
n
that
|z| = 1.
(1995, 5M)
44. 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)
equation is satisfied
(1 + i ) x − 2i (2 − 3i ) y + i
+
= i.
3+ i
3−i
46. Express
47. If x + iy =
a rz r = 1, where|a r|< 2.
( 1980, 2M)
1
in the form A + iB.
(1 − cos θ ) + 2i sin θ
(1979, 3M)
(2004, 2M)
39. Prove that there exists no complex number z such that
∑
show
45. Find the real values of x and y for which the following
38. Find the centre and radius of the circle formed by all the
|z| < 1 /3 and
then
a + ib
a 2 + b2
, prove that (x2 + y2)2 = 2
c + id
c + d2
(1978, 2M)
r =1
(2003, 2M)
40. If z1 and z2 are two complex numbers such that
1 − z1z2
< 1.
|z1| < 1 < |z2|, then prove that
z1 − z2
(2003, 2M)
Integer Answer Type Question
48. If z is any complex number satisfying | z − 3 − 2i | ≤ 2,
then the maximum value of|2z − 6 + 5i |is …… (2011)
Topic 3 Argument of a Complex Number
Objective Questions I (Only one correct option)
1. 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
 1 + z
θ, then arg 
 is equal to
1 + z
(b)
π
−θ
2
(c) θ
(2013 Main)
(d) π − θ
3. If arg (z ) < 0 , then arg (−z ) − arg (z ) equals
(a) π
(c) − π/2
|w| ≤ 1 and|z + i w|= |z − iw| = 2 , then z equals
(1995, 2M)
(b) i or −i
(d) i or −1
(b) −w
(c) w
(d) −w
6. If z1 and z2 are two non-zero complex numbers such
(b) −
π
2
(c) 0
(d)
π
2
7. If a , b, c and u , v, w are the complex numbers
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
(a) have the same area
(c) are congruent
(b) are similar
(d) None of these
(2000, 2M)
(b) − π
(d) π /2
4. Let z and w be two complex numbers such that |z| ≤ 1,
(a) 1 or i
(c) 1 or −1
(a) w
(a) − π
2. If z is a complex number of unit modulus and argument
(a) − θ
such that |z| = |w| and arg (z ) + arg (w) = π, then z
equals
(1995, 2M)
that|z1 + z2| = |z1| + |z2|, then arg (z1 ) − arg (z2) is equal
(1987, 2M)
to
(b) Im(z ) = 0
(c) Re(z) = 0
5. Let z and w be two non-zero complex numbers
Objective Questions II
(One or more than one correct option)
8. For a non-zero complex number z, let arg(z ) denote the
principal argument with − π < arg(z ) ≤ π . Then, which of
the following statement(s) is (are) FALSE ? (2018 Adv.)
(a) arg (−1 − i ) =
π
, where i =
4
−1
Complex Numbers 5
(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 
2π.
10. Match the conditions/expressions in Column I with
statement in Column II (z ≠ 0 is a complex number)
9. Let z1 and z2 be two distinct complex numbers and let
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
complex number w, then
(2010)
(a) |z − z1| + |z − z2|= |z1 − z2|(b) arg (z − z1 ) = arg (z − z2 )
z − z1
z − z1
z2 − z1
z2 − z1
=0
Column II
Column I
Re ( z) = 0
π
arg ( z) =
4
A.
B.
(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 ) 
(c)
Match the Columns
p.
Re ( z2 ) = 0
q.
Im ( z2 ) = 0
r.
Re ( z2 ) = Im ( z2 )
Analytical & Descriptive Questions
11. |z| ≤ 1,|w|≤ 1, then show that
|z − w|2 ≤ (|z| − |w|)2 + (arg z − arg w)2
(1995, 5M)
12. 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.
(d) arg (z − z1 ) = arg (z2 − z1 )
Topic 4 Rotation of a Complex Number
Objective Questions I (Only one correct option)
5
(d)|z − 1|< 2,| arg (z + 1)|>
5
 3 i
 3 i
+  +
−  . If R(z ) and I (z )
2
2
 2
 2
1. Let z = 
(b) I (z ) = 0
(d) R (z ) = − 3
2. 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
3. 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
4. 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
Y
(a)|z + 1|> 2,| arg (z + 1)|<
Q
X′
P
S
O
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
5. If 0 < α <
respectively denote the real and imaginary parts of z,
(2019 Main, 10 Jan II)
then
(a) R (z ) > 0 and I (z ) > 0
(c) R (z ) < 0 and I (z ) > 0
π
2
X
(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. 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
(b) right angled isosceles
(c) equilateral
(d) obtuse angled isosceles
Objective Questions II
(One or more than one correct option)
7. Let a , b ∈ R and a 2 + b2 ≠ 0.


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
(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 a <
 2a 
2a
0, b ≠ 0
(c) the X-axis for a ≠ 0, b = 0
6 Complex Numbers
(d) the Y-axis for a = 0, b ≠ 0
13. Let bz + bz = c, b ≠ 0, be a line in the complex plane,
3+i
8. Let W =
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
(a)
(b)
(c)
(d)
(2013 JEE Adv.)
π
2
π
6
2π
3
5π
6
where b is the complex conjugate of b. If a point z1 is the
reflection of the point z2 through the line, then show
that c = z1b + z2b.
(1997C, 5M)
14. Let z1 and z2 be the roots of the equation z + pz + q = 0,
2
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)
15. 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
16. 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
Fill in the Blanks
9. Suppose z1 , z2, z3 are the vertices of an equilateral
triangle inscribed in the circle |z| = 2. If z1 = 1 + i 3,
(1994, 2M)
then z2 = K, z3 = … .
10. 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)
11. 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)
17. Prove that the complex numbers z1 , z2 and the origin
form an equilateral triangle only if z12 + z22 − z1z2 = 0.
(1983, 2M)
18. 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 Answer Type Question
kπ 
 kπ 
 + i sin   , where
 7
 7
i = −1. The value of the expression
19. For any integer k, let α k = cos 
12
∑|α k + 1 − α k|
k =1
Analytical & Descriptive Questions
is
3
12. If one of the vertices of the square circumscribing the
circle|z − 1| = 2 is 2 + 3i. Find the other vertices of
square.
(2005, 4M)
∑|α 4k − 1 − α 4k − 2|
k =1
(2016 Adv.)
Topic 5 De-Moivre’s Theorem, Cube Roots and nth Roots of Unity
Objective Questions I (Only one correct option)
1. If z and w are two complex numbers such that| zw| = 1
and arg(z ) − arg(w) =
(a) zw = − i
(c) zw = i
π
, then
2
(2019 Main, 10 April II)
1− i
2
− 1+ i
(d) zw =
2
(b) zw =
3 i
2. If z =
+ (i = −1 ), then (1 + iz + z5 + iz 8 )9 is equal
2
2
to
(2019 Main, 8 April II)
(a) 1
(b) (−1 + 2i )9 (c) −1
(d) 0
3. Let z0 be a root of the quadratic equation, x2 + x + 1 = 0,
If z = 3 + 6iz081 − 3iz093 , then arg z is equal to
(a)
π
4
(b)
π
6
(2019 Main, 9 Jan II)
(c) 0
(d)
π
3
15
4. Let z = cos θ + i sin θ . Then, the value of
∑ Im (z2m −1 ) at
m =1
θ = 2° is
1
sin 2°
1
(c)
2 sin 2°
(a)
(2009)
1
3 sin 2°
1
(d)
4 sin 2°
(b)
Complex Numbers 7
5. The minimum value of |a + bω + cω 2|, where a, b and c
are all not equal integers and ω (≠ 1) is a cube root of
unity, is
(2005, 1M)
1
(b)
2
(a) 3
(c) 1
(d) 0
6. 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
(2004, 1M)
(c) 5
(d) 6
1
3
, then value of the determinant
+i
2
2
1
1
1
2
1 −1 − ω ω 2 is
(2002, 1M)
1
ω2
ω
7. Let ω = −
(a) 3 ω
(b) 3 ω (ω − 1) (c) 3 ω
(d) 3 ω (1 − ω)
2
angled at the origin, then n must be of the form (where, k
is an integer)
(2001, 1M)
(b) 4k + 2
(c) 4k + 3
 1
 2
9. If i = −1, then 4 + 5  − +
i 3

2 
334
 1 i 3
+ 3 − +

2 
 2
365
is
(1999, 2M)
3
(b) −1 + i 3
(c) i
3
(d) −i
is equal to
(1998, 2M)
(c) 128 ω2
(d) −128 ω2
11. If ω (≠ 1) is a cube root of unity and (1 + ω )7 = A + Bω,
then A and B are respectively
(a) 0, 1
(c) 1, 0
6

∑ sin
k =1
(a) – 1
(1995, 2M)
(b) 1, 1
(d) –1, 1
12. The value of
2π
2π
+ i sin . Then
3
3
the number of distinct complex number z satisfying
14. Let ω be the complex number cos
ω
ω2
z+1
2
ω
1 = 0 is equal to ... .
z+ω
ω2
1
z+ω
(c) – i
(2010)
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….
(1996, 2M)
True/False
16. The cube roots of unity when represented on Argand
diagram form the vertices of an equilateral triangle.
Analytical & Descriptive Questions
17. 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.
2 πk
2 πk 
– i cos
 is
7
7 
(b) 0
Fill in the Blanks
(1988, 1M)
10. If ω is an imaginary cube root of unity, then (1 + ω − ω )
(b) −128 ω
S
(iii)
(iv)
(iv)
(iii)
3
27
(a) 128 ω
R
(iv)
(iii)
(iii)
(iv)
(d) 4k
equal to
(a) 1 − i
Q
(ii)
(i)
(ii)
(i)
15. The value of the expression
8. Let z1 and z2 be nth roots of unity which subtend a right
(a) 4k + 1
Codes
P
(a) (i)
(b) (ii)
(c) (i)
(d) (ii)
(1998, 2M)
18. If 1, a1 , a 2, ... , a n − 1 are the n roots of unity, then show
that
(d) i
(2002, 5M)
(1 − a1 ) (1 − a 2) (1 − a3 ) K (1 − a n − 1 ) = n
(1984, 2M)
Match the Columns
2 kπ 
13. Let zk = cos 
 + i sin
 10 
19. It is given that n is an odd integer greater than 3, but n
is not a multiple of 3. Prove that x3 + x2 + x is a factor
of
.
(x + 1)n − xn − 1
 2 kπ 

 ; k = 1, 2, …9.
 10 
Column I
(1980, 3M)
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
|1 − z1||1 − z2| … |1 − z9|
equal
10
(iii)
1
(iv)
2
R.
S.
9
2 kπ 
1 − ∑ cos 
 equals
 10 
k =1
20. If x = a + b, y = aα + bβ, z = aβ + bα , where α , β are
complex cube roots of unity, then show that
(1979, 3M)
xyz = a3 + b3 .
Integer Answer Type Question
21. 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
is ……
Then, the value of
(2011)
| a |2 + | b|2 + | c|2
(2011)
8 Complex Numbers
Answers
Topic 1
1. (c)
5. (d)
9. (d)
2. (c)
6. (d)
10. (b, d)
3. (b)
7. (d)
4. (a)
8. (b)
46. A + iB =
1

2 1 + 3 cos2

θ

2
−i
cot (θ / 2 )
1 + 3 cos2 (θ / 2 )
Topic 3
Topic 2
1. (*)
2. (c)
3. (a)
1. (b)
4. (d)
4. (c)
5. (d)
6. (c)
8. (a, b, d)
9. (a, c, d)
2. (b)
3. (d)
5. (b)
6. (c)
7. (b)
8. (c)
9. (d)
10. (a)
11. (d)
12. (b)
14. (b)
15. (a)
16. (d)
1. (b)
2. (d)
3. (d)
17. (d)
18. (b)
19. (b)
20. (d)
5. (d)
6. (c)
7. (d)
21. (b)
22. (a)
23. (a, c, d)
24. (a,d)
25. (a, b, c)
26. (c)
27. (b)
28. (d)
9. z 2 = − 2, z 3 = 1 − i 3
29. (c)
30. (b)
32. ω
2
33. (a + b )(| z1| + | z 2| )
2
2
4. (a)
8. (c, d)
i
3i
10. 3 − or 1 −
2
2
12. z 2 = − 3 i , z 3 = (1 − 3 ) + i and z 4 = (1 + 3 ) − i
19. (4)
−1
34. x = 2nπ + 2α , α = tan k, where k ∈(1, 2 ) or x = 2nπ
35. False
Topic 4
11. a = b = 2 ± 3
31. A → q, r ; B → p; C → p, s, t ; D → q, r, s, t
2
7. (b)
10. A → q ; B → p
13. (a)
2
48. 5
36. True
37. True
k (α − β )
α − k 2β

38. Centre =
, Radius = 
2
1 − k2
 1 −k 

3 i
42. z = i , ±
– 
2
2

Topic 5
1. (a)
2. (c)
3. (a)
4. (d)
5. (c)
6. (b)
7. (b)
11. (b)
8. (d)
12. (d)
9. (c)
13. (c)
10. (d)
14. (1)
16. True
21. (3)
2
 n (n + 1 )
15. 
 −n

2 
45. (x = 3 and y = −1)
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
On comparing real and imaginary parts, we get
x=
2α
α2 −1
and y = 2
α +1
α2 + 1
2
2
 α 2 − 1
 2α 
Now, x2 + y2 =  2
 + 2

 α + 1
 α + 1
=
α 4 + 1 − 2α 2 + 4α 2 (α 2 + 1)2
= 2
=1
(α 2 + 1)2
(α + 1)2
⇒
x2 + y 2 = 1
Which is an equation of circle with centre (0, 0) and
radius 1 unit.

α + i
So, S = 
; α ∈ R lies on a circle with radius 1.
α
i
−


3. Given complex number
ω=
⇒
⇒
⇒
5 + 3z
5(1 − z )
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|
Complex Numbers 9
ω + 3 > |ω − 1|
5

2
3

Let ω = x + iy, then  x +  + y2 > (x − 1)2 + y2

5
9
6
2
2
+ x > x + 1 − 2x
⇒ x +
25 5
16x 16
1
>
⇒ x > ⇒ 5x > 1
⇒
5
25
5
⇒
5 Re( ω ) > 1
⇒
3
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
 3 + 2i sin θ   1 + 2 i sin θ 
 ×

 1 − 2i sin θ   1 + 2 i sin θ 
∴
⇒
⇒
⇒
7. Given,
5. Let z = 
(rationalising the denominator)
=
[Q a 2 − b2 = (a + b)(a − b) and i 2 = − 1]
2
 3 − 4 sin θ   8 sin θ 
=
 +
i
 1 + 4 sin 2 θ   1 + 4 sin 2 θ 
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
3
⇒
sin 2 θ = ⇒ sin θ = ±
2
4
=
2 + 4i sin θ + 3i sin θ + 6i 2 sin 2 θ
12 − (2i sin θ) 2
=
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
−3 i
3i
3
1
−1  = x + iy
i
1
−1
i
1
−1  = x + i y
i
⇒
x + iy = 0 [Q C 2 and C3 are identical]
⇒
8.
x = 0, y = 0
13
13
13
n=1
n=1
n =1
∑ (i n + i n + 1 ) = ∑ i n (1 + i ) = (1 + i ) ∑
y=sin θ
 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 
Since, sum of any four consecutive powers of iota is zero.
∴
–π/2 –π/3
O π/3
2π/3
π
13
∑ (i n + i n + 1 ) = (i + i 2 + K + i13 )
n=1
X
+ (i 2 + i3 + K + i14 ) = i + i 2 = i − 1
n
−1
–√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
⇒
in
Alternate Solution
Y
1
√3/2
X′
 6i
4
20
 6i
−3 i  4
 20
⇒
3 − 4 sin 2 θ + 8i sin θ
1 + 4 sin 2 θ
(2 + 3i sin θ) (1 + 2i sin θ )
(1 − 2i sin θ ) (1 + 2i sin 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
4. We have,
=
9. Since,
 1 + i

 =1
1 − i
1 + i 1 +
×
⇒ 
1 − i 1 +
n
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
10 Complex Numbers
⇒
(a − b) y
=y
(x + 1)2 + y2
4. Since, the complex number
Q
∴
a − b =1
(x + 1) + y2 = 1
imaginary number, therefore
z −α
z −α
+
=0
z+α z+α
2
x = − 1 ± 1 − y2
∴
Topic 2 Conjugate and Modulus of
Complex Number
1. Let the complex number z = x + iy
z −α
(α ∈ R) is purely
z+α
[Qα ∈ R]
⇒
zz − αz + αz − α 2 + zz − αz + αz − α 2 = 0
⇒
2 z
⇒
α2 = z
⇒
α=±2
2
2
− 2 α2 = 0
2
[Qzz = z ]
[| z | = 2 given]
=4
5. We have,|z | + z = 3 + i
Also given,| z − i | = | z − 1|
⇒| x + iy − i | = | x + iy − 1|
z = x + iy
Let
∴
x2 + y2 + x + iy = 3 + i
⇒
(x + x2 + y2 ) + iy = 3 + i
On squaring both sides, we get
⇒
x + x2 + y2 = 3 and y = 1
x2 + y2 − 2 y + 1 = x2 + y2 − 2x + 1
Now,
⇒ y = x, which represents a line through the origin with
slope 1.
⇒
x + 1 = 9 − 6 x + x2
⇒
6x = 8 ⇒ x =
⇒ x + ( y − 1) = (x − 1) + y
2
2
2
2
[Q| z | = (Re(z ))2 + (Im(z ))2 ]
(1 + i )
a−i
(1 − 1 + 2i ) (a + i )
=
a2 + 1
2i (a + i ) −2 + 2ai
=
=
a2 + 1
a2 + 1
2. The given complex number z =
⇒
[Q i 2 = − 1]
…(i)
[given]
4 + 4a 2
2
2
2
⇒
=
=
2
5
5
(a 2 + 1)2
1+ a
4
2
= ⇒ a 2 + 1 = 10
1 + a2 5
⇒
∴
a2 = 9 ⇒ a = 3
–2 + 6i
z=
10
PLAN If z is unimodular, then| z| = 1. Also, use property of modulus
i.e. z z =| z|2
Given, z2 is not unimodular i.e.|z2|≠ 1
z − 2 z2
is unimodular.
and 1
2 − z1z2
z1 − 2z2
= 1 ⇒ |z1 − 2z2|2 =|2 − z1z2|2
2 − z1z2
⇒
[Qa > 0]
[From Eq. (i)]
1 3
 −2 + 6 i   1 3 
So, z = 
 =  − + i ⇒ z = − − i
 10   5 5 
5 5
[Qif z = x + iy, then z = x − iy]
3. Clearly|z1|= 9, represents a circle having centre C1 (0, 0)
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.
QC1C 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
4
+i
3
5
16
25
|z | =
+1=
⇒ |z | =
9
9
3
⇒
6.
4
3
z=
∴
z = 2 /5
Q
⇒
2
x2 + 1 = 3 − x
2
⇒ (z1 − 2z2)(z1 − 2z2) = (2 − z1z2) (2 − z1z2)
[zz = |z|2 ]
⇒ |z1| +4|z2| −2z1z2 − 2z1z2
2
2
= 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.
7. |z| ≥ 2 is the region on or outside circle whose centre is
(0, 0) and radius is 2.
1
Minimum z +
is distance of z, which lie on circle
2
| z | = 2 from (−1 / 2, 0).
1
 1 
∴ Minimum z +
= Distance of  − , 0 from (−2, 0)
 2 
2
2
2
1
3
3


 −1
=  −2 +  + 0 = = 
+ 2 + 0 =


 2
2
2
2
Complex Numbers 11
Y
 − r 2
(|α|2 − 1) ⋅ 
 = r 2(1 − 4|α|2 )
 2 
⇒
⇒
X′
D
(–2,0)
A
1 , (0,0)
–
(—
)
2 0
(2,0)
⇒
X
PLAN If ax + bx + c = 0 has roots α, β, then
α, β =
1
Geometrically Min z +
= AD
2
1
Hence, minimum value of z +
lies in the interval
2
(1, 2).
PLAN Intersection of circles, the basic concept is to solve the
equations simultaneously and using properties of modulus of
complex numbers.
|z|2 = z ⋅ z
(x − x0 )2 + ( y − y0 )2 = 4r 2 can be written as,
|z − z0|2 = r 2 and |z − z0|2 = 4r 2
∴
1
lies on first and second respectively.
α
2
1
|α − z0|2 = r 2 and
− z0 = 4 r 2
α
⇒
(α − z0 ) (α − z0 ) = r 2
⇒
|α|2 − z0α − z0α + |z0|2 = r 2
…(i)
⇒
1
 1

2
 − z0   − z0  = 4 r
α
 α

⇒
1
z
z
− 0 − 0 + |z0|2 = 4r 2
α
|α|2 α
⇒
2
⇒

r 2 + 2
(|α|2 − 1) 1 −
 = r 2(1 − 4|α|2 )
2 

Given,
|z0|2 =
2
r2 + 2
2
Method II
Here, z 2 + z + (1 − a ) = 0
∴
z=
− 1 ± 1 − 4 (1 − a )
2 ×1
⇒
z=
− 1 ± 4a − 3
2
a<
3
4
zz (z 2 + z 2) = 350
2 (x + y2) (x2 − y2) = 350
(x2 + y2) (x2 − y2) = 175
2
... (ii)
x2 = 16, y2 = 9 ⇒ x = ± 4, y = ± 3
∴ Area of rectangle = 8 × 6 = 48
2
(|α| − 1) (1 − |z0| ) = r (1 − 4|α| )
2
…(i)
2x + 1 = 0 or x = − 1 / 2
1
1
From Eq. (i),
− y2 − + 1 − a = 0
4
2
3
3
a = − y2 +
⇒ a<
⇒
4
4
…(ii)
⇒
2
(x2 − y2 + x + 1 − a ) + iy (2x + 1) = 0,
From Eqs. (i) and (ii),
(|α| − 1) + |z0| (1 − |α| ) = r (1 − 4|α| )
2
⇒
x2 − y2 = 7
On subtracting Eqs. (i) and (ii), we get
2
(x2 − y2 + x + 1 − a ) + i (2xy + y) = 0
Since, x, y ∈ I, the only possible case which gives
integral solution, is
... (i)
x2 + y2 = 25
1 − z0α − z0α + |α|2|z0|2 = 4r 2|α|2
2
a = (x + iy)2 + (x + iy) + 1
⇒
⇒
⇒
1
z ⋅α
z
− 0
− 0 ⋅ α + |z0|2 = 4r 2
|α|2 |α|2 |α|2
⇒
Method I Let z = x + iy
∴
10. Since,
|α|2 = α ⋅ α
Since,
As imaginary part of
For z do not have real roots, 4 a − 3 < 0 ⇒
2
1
− z0 = 4 r 2
α
and
Description of Situation
z = x + iy is non-zero.
⇒
y ≠0
⇒
Here, (x − x0 )2 + ( y − y0 )2 = r 2
Since, α and
2a
For roots to be real b 2 − 4ac ≥ 0.
but imaginary part of z, i.e. y is non-zero.
= |z1|2 − z1z2 − z2z1 + |z2|2
and
− b ± b 2 − 4ac
It is purely real, if y (2x + 1) = 0
|z1 − z2|2 = (z1 − z2) (z1 − z2)
and
|α| = 1 / 7
2
Y′
Formula used
7|α|2 = 1
∴
9.
8.
|α|2 − 1 = − 2 + 8|α|2
2
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
⇒
12 Complex Numbers
Alternate Solution
z
z
1
which is an imaginary.
Let E =
=
=
1 − z 2 zz − z 2 z − z
w − wz
be purely real ⇒ z1 = z1
12. Let 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| =
⇒ z − 1 = wz + w ⇒ z =
|1 − w|
1−w
⇒
|1 − w| = |1 + w|
[ Q|z| = 1]
1 + |w| − 2|w| Re (w) = 1 + |w| + 2|w| Re (w)
2
2
[using|z1 ± z2|2 = |z1|2 + |z2|2 ± 2|z1||z2| Re (z1z2)]
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|
≥ 12 − 5 − 5
∴
= [n1 C 0 +
n1
C1i +
+ [ C0 −
+ [n2 C 0 +
+ [n2 C 0 –
n1
= 2 [n1 C 0 +
n1
n1
C3 i3 + K ]
n1
C2 i2 +
+ 2 [ C0 +
n1
n1
C1 i + C 2i 2 − n1 C3 i3 + ... ]
n2
C1i + n2C 2i 2 + n2C3 i3 + K ]
n2
C1 i + n2C 2i 2 – n2C3 i3 + .. ]
n2
= 2 [n1 C 0 −
C 2i 2 +
n1
C2 +
n1
n2
n1
C 4i 4 + K ]
C 2i 2 +
n2
C 4i 4 + K ]
C 4 − K ] + 2 [n2 C 0 −
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]
17. Since, (sin x + i cos 2x) = cos x − i sin 2x
On squaring both sides, we get
⇒
⇒
16. (1 + i )n1 + (1 − i )n1 + (1 + i )n2 + (1 − i )n2
[using|z1 − z2| ≥ |z1| − |z2|]
|z1 − z2| ≥ 2
⇒
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.
Hence, no solution exists.
18. Since, z1 , z2, z3 , z4 are the vertices of parallelogram.
D(z4)
C(z3)
Alternate Solution
Clearly from the figure|z1 − z2|is minimum when z1 , z2
lie along the diameter.
Y
A
(3
X′
)
,4
C2
B
Z1
A(z1)
Z2
C1
12
X
|z1 − z2| ≥ C 2B − C 2A ≥ 12 − 10 = 2
15. Given,
Now,
⇒
Similarly,
Again now,
∴ Mid-point of AC = mid-point of BD
z1 + z3 z2 + z4
⇒
=
2
2
⇒
z1 + z3 = z2 + z4
 1 − iz 
= 1
z−i 
19. Since,|w| = 1 ⇒ 
Y′
∴
B(z2)
|z1| = |z2| = |z3| = 1
|z1| = 1
|z1|2 = 1 ⇒ z1z1 = 1
z2z2 = 1, z3 z3 = 1

1 + 1 + 1 
= 1
z1 z2 z3
⇒ | z1 + z2 + z3 |= 1 ⇒ |z1 + z2 + z3|= 1
⇒
|z1 + z2 + z3| = 1
⇒
⇒
|z − i| = |1 − iz|
|z − i | = |z + i |
[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.
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
Complex Numbers 13
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.
Y
X′
O
(2, 0) (3, 0) (4, 0)
X
24. Given,|z1| = |z2|
Y′
5
 3 i
 3 i
+  +
− 
2
2
 2
 2
5
21. Given, z = 
Now,
z1 + z2 z1 − z2 z1z1 − z1z2 + z2z1 − z2 z2
×
=
z1 − z2 z1 − z2
|z1 − z2|2

−1 + i 3
−1 − i 3 
and ω 2 =
Q ω =

2
2


 −1 + i 3 

 = − iω
2




−1 − i 3
3−i
=i
 = iω 2
2
2


3+i
= −i
2
Now,
and
∴
∴
z is purely real. i.e. Im (z ) = 0
z2z1 − z1z2
|z1 − z2|2
[Q|z1|2 = |z2|2 ]
which is purely imaginary or zero.
25. Since, z1 = a + ib and
z2 = c + id
⇒ |z1| = a + b = 1 and |z2|2 = c2 + d 2 = 1
2
2
2
…(i)
[Q|z1|=|z2| = 1]
z − 5i 

22. Given, 
= 1 ⇒ |z − 5i| = |z + 5i|
z + 5i 
[Q if|z − z1| = |z − z2|, then it is a perpendicular
bisector of z1 and z2 ]
Y
Also, Re (z1z2) = 0 ⇒ ac + bd = 0
a
d
⇒
=− =λ
b
c
[say]…(ii)
From Eqs. (i) and (ii), b2λ2 + b2 = c2 + λ2c2
⇒
b2 = c2 and a 2 = d 2
Also, given w1 = a + ic and w2 = b + id
(0, 5)
X′
=
z2z1 − z1z2 = 2i Im (z2z1 )
z1 + z2 2i Im (z2 z1 )
=
z1 − z2
|z1 − z2|2
∴
Alternate Solution
We know that,
z + z = 2 Re(z )
5
5
 3 i
 3 i
If
z=
+  +
−  , then
2
2
 2
 2
O
|z1|2 + (z2 z1 − z1 z2) − |z2|2
|z1 − z2|2
As, we know z − z = 2i Im (z )
z = (− iω )5 + (iω 2)5 = − iω 2 + iω
= i(ω − ω 2) = i (i 3 ) = − 3
Re(z ) < 0 and lm (z ) = 0
⇒
=
|w1| = a 2 + c2 = a 2 + b2 = 1
Now,
X
|w2| = b2 + d 2 = a 2 + b2 = 1
(0, –5)
Y′
∴ Perpendicular bisector of (0, 5) and (0, – 5) is X-axis.
23. 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
(a) 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.
and Re(w1 w2) = ab + cd = (bλ )b + c(− λc) [from Eq. (i)]
= λ (b2 − c2) = 0
26. min|1 − 3 i − z|= perpendicular distance of point (1, − 3)
Z ∈S
…(i)
…(ii)
from the line
3x + y = 0 ⇒
| 3 − 3| 3 − 3
=
2
3+1
27. Since, S = S1 ∩ S 2 ∩ S3
Y
X′
150°
O
X
(4, 0)
y = –3√x
Y′
14 Complex Numbers
Clearly, the shaded region represents the area of sector
∴
S=
= 2 (cos θ + i sin θ ) −
1 2
1
5π 20π
r θ = × 42 ×
=
2
2
3
6
=
|w − (2 + i )| < 3 ⇒ |w| − |2 + i| < 3
28. Since,
⇒
−3 + 5 < |w| < 3 + 5
⇒
−3 − 5 < − |w| < 3 − 5
…(i)
⇒
…(ii)
⇒
 2 y
 2x
  +   =1
5
 3
⇒
x2
y2
+
=1
9 / 4 25 / 4
2
⇒
−3 + 5 ≤ |z| ≤ 3 + 5
∴
−3 < |z| − |w| + 3 < 9
29. |z + 1 − i| + |z − 5 − i|
2
3
5
cos θ + i sin θ
2
2
z = x + iy
3
5
x = cos θ and y = sin θ
2
2
Let
|z − (2 + i )| = 3
Also,
2
= (x + 1)2 + ( y − 1)2 + (x − 5)2 + ( y − 1)2
= 2(4) + 28 = 36
e= 1−
∴
= 2(x2 + y2 − 4x − 2 y) + 28
[Q x2 + y2 − 4x − 2 y = 4]
…(i)
Set B consists of points lying on the circle, centre at
(2, 1) and radius 3.
i.e.
…(ii)
x2 + y2 − 4x − 2 y = 4
Set C consists of points lying on the x + y = 2
…(iii)
9 /4
4
=
25 / 4 5
z = x + iy = cos θ + i sin θ +
Then,
Set A corresponds to the region y ≥ 1
⇒
x = 2 cos θ , y = 0
xα + yβ + z γ
x( p)1/3 + y( p)1/3 ω + z ( p)1/3 ω 2
32.
=
xβ + yγ + zα x( p)1/3 ω 2 + y( p)1/3 ω3 + z ( p)1/3 ω
ω 2 (x + yω + z ω 2)
ω 2 (xω + yω 2 + z )
P
=
X′
(2,1)
y=1
Clearly, there is only one point of intersection of the line
x + y = 2 and circle x2 + y2 − 4x − 2 y = 4.
z = x + iy
31. A. Let
y x + y =0
2
2
⇒
y=0
⇒
Im (z ) = 0
B. We have
2ae = 8, 2a = 10
⇒
⇒
⇒
∴
C. Let
∴
= [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 )
Y′
⇒ we get
ω 2 (x + yω + zω 2)
= ω2
x + yω + zω 2
33. |az1 − bz2|2 + |bz1 + az2|2
X
(√2,0)
1
cos θ + i sin θ
= 2 cos θ
Y
(0,√2)
2
w = cos θ + i sin θ
D. Let
30. Let z = x + iy
1
(cos θ − i sin θ )
2
10e = 8
4
e=
5
16

b2 = 25 1 −  = 9

25
x2 y2
+
=1
25 9
w = 2 (cos θ + i sin θ )
1
z = 2 (cos θ + i sin θ ) −
2 (cos θ + i sin θ )
x
x

sin + cos  − i tan x

2
2
34.
∈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
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
x
x
x
sin + cos  cos x + 2 sin cos = 0


2
2
2
2
2
⇒
2 sin
⇒ sin
x 
x
x 
x
2x
2 x
sin + cos   cos − sin  + cos  = 0
2 
2
2 
2
2
2
∴
⇒
or
sin
x
=0
2
x = 2 nπ
... (i)
x
x 
x

2x
2 x
sin + cos   cos − sin  + cos = 0

2
2 
2
2
2
Complex Numbers 15
+ (|α|2 − k2|β|2 ) = 0
x
On dividing by cos3 , we get
2
x

 

2 x
2 x
 tan + 1 1 − tan  + 1 + tan  = 0

 
2
2 
2
⇒
tan3
Let
x
x
− tan − 2 = 0
2
2
x
tan = t
2
On comparing with equation of circle,
|z|2 + az + az + b = 0
whose centre is (− a ) and radius = |a|2 − b
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

35. 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.
36. Since, z1 , z2, z3 are vertices of equilateral triangle and
|z1| = |z2| = |z3|
⇒ z1 , z2, z3 lie on a circle with centre at origin.
k(α − β ) 

=
2
1−k 
39. Given, a1z + a 2z 2 + K + a nz n = 1
|z| <
and
⇒
|a1z| + |a 2z 2| + |a3 z3| + K + |a nz n| ≥ 1
⇒
2{(|z| + |z| + |z| + K + |z|n } > 1
[using|z1 + z2| ≤ |z1| + |z2|]
⇒
⇒
⇒
⇒
2
2|z|(1 − |z| )
>1
1 − |z|
n
⇒
⇒
1−x − y
2iy
−
∩ 0 + 0i
(1 + x)2 + y2 (1 + x)2 + y2
⇒
x2 + y2 ≥ 1
2
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
|z − α|
= k2
|z − β|2
2
(z − α )(z − α ) = k2(z − β ) (z − β )
⇒ |z|2 − αz − αz + |α|2 = k2(|z|2 − βz − β z + |β|2 )
⇒ |z| (1 − k ) − (α − k β )z − (α − β k ) z
2
2
…(i)
40. Given,|z1| < 1 and |z2| > 1
Then, to prove

1 − z1z2
< 1
 z1 − z2 
|1 − z1z2| < |z1 − z2|
⇒
On squaring both sides, we get,
…(i)



z1
 |z1|
using z = |z |
 2
2 

…(ii)
(1 − z1z2)(1 − z1z2) < (z1 − z2)(z1 − z2) [using|z|2 = zz ]
⇒
1 − z1z2 − z1z2 + z1z1z2z2 < z1z1 − z1z2 − z2z1 + z2z2
1 + |z1|2|z2|2 <|z1|2 + |z2|2
1 − |z1|2 − |z2|2+ | z1|2|z2|2 < 0
⇒
(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
38. As we know,|z|2 = z ⋅ z
2
r =1
⇒
or x2 + y2 ≥ 1 and y ≥ 0 which is true by Eq. (i).
⇒
…(ii)
|z| < 1 / 3 and ∑ a rz r = 1
⇒
−2 y ≤ 0
Given,
[using sum of n terms of GP]
n
(1 − x − iy) (1 + x − iy)
∩ 0 + 0i
(1 + x + iy) (1 + x − iy)
and
[using|a r| < 2]
3
∴ There exists no complex number z such that
1 ≤ x and 0 ≤ y
1−z
1 − x − iy
∩0 ⇒
∩0
1+ z
1 + x + iy
2
…(i)
|a1z + a 2z 2 + a3 z3 + K + a nz n| = 1
37. Let z = x + iy ⇒ 1 ∩ z gives 1 ∩ x + iy
Given,
1
3
∴
⇒
⇒ Circumcentre = Centroid
z + z2 + z3
⇒
0= 1
3
∴
z1 + z2 + z3 = 0
or
α − k2β
1 − k2
 α − k2β  α − k2β  αα − k2ββ
and radius = 
 −
 
1 − k2
 1 − k2   1 − k2 
f (1) = − 2 < 0
f (2) = 4 > 0
Then,
and
|α|2 − k2|β|2
(α − k β )
(α − β k )
−
z
z
= 0 …(i)
+
(1 − k2)
(1 − k2)
(1 − k2)
2
∴ Centre for Eq. (i) =
f (t ) = t3 − t − 2
and
⇒ |z|2 −
2
2
…(iii)
∴ Eq. (iii) is true whenever Eq. (ii) is true.
⇒

1 − z1z2
< 1
 z1 − z2 
Hence proved.
16 Complex Numbers
41. Given,|z|2 w − |w|2 z = z − w
⇒ zz w − ww z = z − w
[Q |z|2 = zz ] …(i)
Taking modulus of both sides, we get
|zw||z − w| = |z − w|
[∴ |z| = |z| ]
⇒
|zw||z − w| = |z − w|
⇒
| zw|| z − w | = |z − w |
⇒
|z − w|(|zw| − 1) = 0
⇒
|z − w| = 0
or
|zw| − 1 = 0
⇒
|z − w| = 0
or
|zw| = 1
z − w=0
or
|z w|= 1
⇒
z=w
or
|zw| = 1
Now, suppose z ≠ w
NOTE
‘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| ⋅z − |w| ⋅z
2
2
If
2
Hence proved.
∴
⇔
(|z| + 1)w = (|w| + 1)z
z |z|2 + 1
=
w |w|2 + 1
z
z
=
w w
⇒
iz3 − i 2z 2 − z + i = 0
⇒
iz (z − i ) − 1(z − i ) = 0
⇒
(iz 2 − 1)(z − i ) = 0
2
zw = zw
[Q i 2 = − 1]
2
z − i = 0 or iz 2 − 1 = 0
1
z = i or z 2 = = − i
i
z = i, then|z| = |i| = 1
|z|2 = 1 ⇒ |z| = 1
z − z2
44. Here, z1Rz2 ⇔ 1
is real
z1 + z2
z − z1
(i) Reflexive z1Rz1 ⇔ 1
=0
z1 + z2
∴
(|z|2 + 1)w − (|w|2 + 1)z = 0
z
is purely real.
w
⇒
⇒
Alternate Solution
We have,
|z|2 w − |w|2 z = z − w
2
⇔
|z| w − |w|2 z − z + w = 0
2
and − y = x2 − y2
If z 2 = − i, then |z 2| = |− i| = 1
= z − w = z − w = 0 = RHS
⇔
It is a compound equation, therefore we can generate
from it more than one primary equations.
x = − 2xy
If
zw = 1, then zw = 1 and
⇔
z = iz 2
(x + iy) = i (x + i y)2
x − iy = i (x2 − y2 + 2i xy)
x − iy = − 2xy + i (x2 − y2)
⇒
⇒
⇒
⇒
RHS = z − w = 0
LHS = zz w − ww z = z ⋅ 1 − w ⋅ 1
⇔
z = x + iy.
Given,
⇒
= |z| ⋅z − |z| ⋅z = 0
and
Therefore, |z|2 w − |w|2 z = z − w if and only if z = w or
zw = 1.
⇒
x + 2xy = 0 and x2 − y2 + y = 0
⇒
x(1 + 2 y) = 0
⇒
x = 0 or y = − 1 / 2
When x = 0, x2 − y2 + y = 0 ⇒ 0 − y2 + y = 0
⇒
y(1 − y) = 0 ⇒ y = 0 or y = 1
When, y = − 1 / 2 , x2 − y2 + y = 0
1 1
3
⇒
x2 − − = 0 ⇒ x2 =
4 2
4
3
x=±
⇒
2
3 i
Therefore, z = 0 + i 0 , 0 + i ; ±
−
2
2
3 i
⇒
[Q z ≠ 0]
z = i, ±
−
2
2
43. Given, iz3 + z 2 − z + i = 0
1 iφ
e
r
eiφ = eiθ ⇒ φ = θ
1
z = reiθ and w = eiθ
r
1
zw = reiθ . e−iθ = 1
r
2
[from Eq. (i)]
On equating real and imaginary parts, we get
On putting these values in Eq. (i), we get
1
 1
1
r 2 ei φ  − 2 (reiθ ) = reiθ − eiφ
 r
r
r
1
1
⇒
reiφ − eiθ = reiθ − eiφ
r
r
1  iφ 
1  iθ

⇒
r +  e = r +  e


r
r
⇒
z = w or zw = 1
[say]
z = reiθ and w =
Therefore,
(z − w)(zw − 1) = 0
NOTE
Then,|zw| = 1 or|z||w| = 1
1
⇒
|z| =
=r
|w|
⇒
⇔
⇔
42. Let
⇒
Let
|z|2 w − |w|2 z = z − w
z ⋅ zw − w ⋅ wz = z − w
z (zw − 1) − w (zw − 1) = 0
Again,
⇔
⇔
…(i)
z1Rz1 is reflexive.
z − z2
(ii) Symmetric z1Rz2 ⇔ 1
is real
z1 + z2
− (z2 − z1 )
is real ⇒ z2Rz1
⇒
z1 + z2
∴
z1Rz2 ⇒ z2Rz1
Therefore, it is symmetric.
[purely real]
Complex Numbers 17
(iii) Transitive
⇒
and
z1Rz2
z1 − z2
is real
z1 + z2
47. Since, (x + iy)2 =
z2Rz3
z2 − z3
is real
z2 + z3
⇒
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)
⇒
{(x1 − x2) + i ( y1 − y2)}{(x1 + x2) − i ( y1 + y2)}
(x1 + x2)2 + ( y1 + y2)2
2 x2y1 − 2 y2x1 = 0
x1 x2
=
y1 y2
⇒
Similarly,
z2Rz3
x2 x3
=
y2 y3
⇒
From Eqs. (i) and (ii), we have
|x + iy|2 =
⇒
(x2 + y2) =
⇒
(x2 + y2)2 =
a 2 + b2
c2 + d 2
a 2 + b2
c2 + d 2
…(i)
... (i)
i.e. ||z1| − |z2|| ≤ |z1 + z2|
5
5
∴
z −3 + i = z −3 −2i + 2i + i
2
2
x1 x3
=
⇒ z1Rz3
y1 y3
[transitive]
= (z − 3 − 2 i ) +
9
i
2
≥ |z − 3 − 2 i| −
9
9
5
≥ 2−
≥
2
2
2
z −3 +
⇒
Hence, R is an equivalence relation.
45.
1. (*) Given, 3|z1| = 4|z2|⇒
⇒ (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
46. Now,
=
(1 − cos θ ) + 2i sin θ 2 sin 2 θ + 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
θ
θ
sin − 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
⇒ A + iB =
5
5
i ≥ or|2z − 6 + 5 i| ≥ 5
2
2
Topic 3 Argument of a Complex Number
(1 + i ) x − 2i (2 − 3i ) y + i
+
=i
3+ i
3−i
⇒
Hence proved.
48. Given,|z − 3 − 2 i| ≤ 2
... (ii)
Thus, z1Rz2 and z2Rz3 ⇒ z1Rz3 .
 z1 |z1|
Q z = |z |
2
2

|a + ib|
|c + id|
To find minimum of|2z − 6 + 5 i|
5
or 2 z − 3 + i , using triangle inequality
2
⇒ ( y1 − y2) (x1 + x2) − (x1 − x2) ( y1 + y2) = 0
⇒
⇒
a + ib
c + id
∴
[Q z =|z|(cos θ + i sin θ) = |z| eiθ ]
z
3
eiθ and 2 = e−iθ
z1 4
2 z2 1 −iθ
z1
= e
= 2eiθ and
3 z1 2
z2
On adding these two, we get
1
3 z1 2 z2
z=
= 2eiθ + e−iθ
+
2
2 z2 3 z1
1
1
cos θ − i sin θ
2
2
[Q e± iθ = (cos θ ± i sin θ)]
= 2 cos θ + 2i sin θ +
=
5
3
cos θ + i sin θ
2
2
2
⇒
cot
θ
2
1
−i
θ
θ

1 + 3 cos 2
2 1 + 3 cos 2 

2
2
[Q z2 ≠ 0 ⇒|z2| ≠ 0]
z1 z1 iθ
z
z
e and 2 = 2 e−iθ
=
z2 z2
z1 z1
z1 4
=
z2 3
3
⇒
2
⇒
|z1| 4
=
|z2| 3
2
34
17
 5
 3
|z| =   +   =
=
 2
 2
4
2
Note that z is neither purely imaginary and nor purely
real.
‘*’ None of the options is correct.
18 Complex Numbers
2. Given,|z| = 1 , arg z = θ∴z = eiθ
Similarly, when
1
z
∴
z = e–iθ ⇒ z =
∴


 1 + z
 1 + z
arg 
 = arg (z ) = θ
 = arg 
1 + z
1 + 1

z
3. Since, arg (z ) < 0 ⇒
arg (z ) = − θ
Y
(–z)
π−θ
r
X′
X
–θ
O
r
Y′
⇒
(z)
z = r cos (−θ ) + i sin (−θ )
= r (cos θ − i sin θ )
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
Reason
 − z
arg (− z ) − arg z = arg 
 = arg (− 1) = π
 z 
 z 
and also arg z − arg (− z ) = arg 
 = arg (− 1) = π
 − z
4. Given,
|z + iw| = |z − iw |= 2
⇒
|z − (− iw)| = |z − (iw )| = 2
⇒
|z − (− iw)| = |z − (− iw )|
∴ 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 may take values given in option (c).
Alternate Solution
|z + i w| ≤ |z| + |iw|= |z| + |w|
≤1 +1 =2
∴
⇒
⇒
|z + iw| ≤ 2
|z + i w| = 2 holds when
arg z − arg i w = 0
z
arg
=0
iw
z
is purely real.
iw
z
is purely imaginary.
⇒
w
⇒
|z − i w| = 2 , then
z
is purely
w
imaginary
Now, given relation
|z + iw| = |z − iw| = 2
Put w = i, we get
|z + i 2| = |z + i 2| = 2
⇒
|z − 1| = 2
⇒
z = −1
Put w = − i , we get
|z − i 2| = |z − i 2| = 2
⇒
|z + 1| = 2 ⇒ z = 1
[Q|z| ≤ 1]
[Q|z| ≤ 1]
∴ z = 1 or − 1 is the correct option.
5. Since,|z|=|w|and arg (z ) = π − arg (w)
Let
w = re iθ , then w = re–iθ
∴
z = rei ( π − θ ) = re iπ ⋅ e−iθ = − re−iθ = − w
6. Given,|z1 + z2| = |z1| + |z2|
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|
⇒
2|z1||z2|cos (arg z1 − arg z2) = 2|z1||z2|
⇒
cos (arg z1 − arg z2) = 1
⇒
arg (z1 ) − arg (z2) = 0
7. Since a , b, c and u , v, w are the vertices of two triangles.
Also,
c = (1 − r ) a + rb
…(i)
w = (1 − r ) u + rv
 a u 1
Consider  b v 1 
c w 1 
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
and
a
= b
0
8. (a) Let
u
v
0
1
1 = 0
0 
[from Eq. (i)]
z = − 1 − i and arg(z) = θ
 im (z ) 
=  − 1  = 1
tan θ = 
 Re(z )   −1 
π
⇒
θ=
4
Since, x < 0, y < 0
π
3π

arg (z ) = −  π −  = −
∴

4
4
Now,
(b) We have, f (t ) = arg (−1 + it )
 π − tan −1 t , t ≥ 0
arg (−1 + it ) = 
−1
− (π + tan t ), t < 0
This function is discontinuous at t = 0.
Complex Numbers 19
10. Let z = a + ib.
(c) We have,
z 
arg  1  − arg (z1 ) + arg (z2)
 z2 
Now,
z 
arg  1  = arg (z1 ) − arg (z2) + 2nπ
 z2 
∴
z 
arg  1  − arg (z1 ) + arg (z2)
 z2 
= 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 )
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

z 2 = r 2 cos 2 − sin 2  + 2 ir 2 cos sin

4
4
4
4
= ir 2 sin π /2 = ir 2
Therefore, Re (z 2) = 0 ⇒ B → p.
⇒
a = b =2− 3
[Q a , b ← (0, 1)]
Then,
11. Let z = r1 (cos θ1 + i sin θ1 ) and w = r2(cos θ 2 + i sin θ 2)
 z − z1   z2 − z3 
⇒

 is purely real
 z − z3   z2 − z1 
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.
C(z3)
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|2 = (r1 cos θ1 − r2 cos θ 2)2
D(z)
=
B(z2)
r12 cos 2 θ1
+
r22 cos 2 θ 2
+ (r1 sin θ1 − r2 sin θ 2)2
− 2r1r2 cos θ1 cos θ 2
+ r12 sin 2 θ1 + r22 sin 2 θ 2 − 2r1r2 sin θ1 sin θ 2
= r12(cos 2 θ1 + sin 2 θ1 ) + r22(cos 2 θ 2 + sin 2 θ 2)
A(z1)
−2r1r2(cos θ1 cos θ 2 + sin θ1 sin θ 2)
∴(a), (b), (d) are false statement.
(1 − t ) z1 + t z2
9. Given, z =
(1 − t ) + t
A
z1
P
z
t : (1 - t)
=
2
 θ − θ 2 


≤ |r1 − r2|2 + 4sin  1

 2 

and |sin θ| ≤ |θ|, ∀ θ ∈ R
or
z2 − z1
z2 − z1
Option (c) is correct.
=0
[Q r1 , r2 ≤ 1]
2
θ − θ 2
Therefore, |z − w |2 ≤ |r1 − r2|2 + 4 1
 2 
≤ |r1 − r2|2 + |θ1 − θ 2|2
⇒
|z − w|2 ≤ (|z| − |w|)2 + (arg z − arg w)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|
z − z1
is purely real.
z2 − z1
z − z1
z − z1
=
⇒
z2 − z1 z2 − z1
z − z1
− 2r1r2 cos (θ1 − θ 2)
 θ − θ 2
= (r1 − r2)2 + 4r1r2 sin 2  1

 2 
− 2|z||w|cos (arg z − arg w)
 arg z − arg w
= (|z| − |w|) + 2|z||w|⋅ 2 sin 2
 …(i)


2
∴
z − z1
+
r22
= (r1 − r2)2 + 2r1r2[1 − cos (θ1 − θ 2)]
B
z2
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 
r12
2
∴
 arg z − arg w
|z − w|2 ≤ (|z| − |w|)2 + 4 ⋅ 1⋅ 1 



2
⇒
|z − w| ≤ (|z| − |w|) + (arg z − arg w)
2
[Q sin θ ≤ θ ]
2
2
2
20 Complex Numbers
12. Since, z1 = 10 + 6i , z2 = 4 + 6i
3. Let OA = 3,so that the complex
 z − z1  π
represents locus of z is a circle
and arg 
 =
 z − z2  4
shown as from the figure whose centre is (7, y) and
∠ AOB = 90°, clearly OC = 9 ⇒ OD = 6 + 3 = 9
6
∴ Centre = (7, 9) and radius =
=3 2
2
number associated with A is
3e iπ / 4 . If z is the complex
number associated with P,
then
4i
z − 3eiπ / 4 4 − iπ / 2
= e
=−
3
0 − 3eiπ / 4 3
⇒ 3z − 9eiπ / 4 = 12 ieiπ / 4
⇒
z = (3 + 4 i ) eiπ / 4
z
C
Y
4
3e iπ/4
A
3
X′
π/4
X
O
Y′
45
°
Y
P
6
O
(7, y)
(4, 6) A
4. 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
B (10, 6)
z2
z1
X
D (7, 0) (10, 0)
(4, 0)
⇒ Equation of circle is| z − (7 + 9i )| = 3 2
Also, argument of z + 1 with respect to positive direction
of X-axis is π/4.
π
…(i)
∴
arg (z + 1) ≤
4
Topic 4 Rotation of a Complex Number
5
 3 i
 3 i
− 
+  +
 2 2
 2 2
5
1. Given, z = 
Q Euler’s form of
π
3 i 
+ =  cos + i sin

2 2
6
and argument of z + 1 in anticlockwise direction is −π /4.
∴
π
i( π / 6 )
 =e
6
3 i
 −π 
 π
and
− = cos   + i sin  −  = e−iπ / 6
 6
 6
2 2
So,
i
5π
−i
= 2 cos
[Q eiθ = cos θ + i sin θ]
5π
6
5. In the Argand plane, P is represented by e i0 and Q is
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.
6.
⇒
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
π
 = (7 + 6i ) (i ) = − 6 + 7i
2
1 − i 23
2 (1 + i 3 )
z3
4
=
2 (1 + i 3 )
2
=
(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
=
π
= − 3 <0
6
5π
π
π


Q cos 6 = cos  π − 6  = − cos 6 


Imaginary axis
z2
z1 − z3 1 − i 3 (1 − i 3 )(1 + i 3 )
=
=
z2 − z3
2
2 (1 + i 3 )
∴ I (z ) = 0 and R(z ) = −2 cos
2.
…(ii)
From Eqs. (i) and (ii),
|arg (z + 1)|≤ π / 4
5π
z = (eiπ / 6 )5 + (e−iπ / 6 )5 = e 6 + e 6
5π
5π  
5π
5π 

− i sin 
=  cos
+ i sin  +  cos

6
6 
6
6
− π / 4 ≤ arg (z + 1)
⇒
π/3
Hence, the triangle is an equilateral.
Alternate Solution
z1 − z3 1 − i 3
∴
=
z2 − z3
2
π
π
z2 − z3
2
1 +i 3
= 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.
z1
Complex Numbers 21
a − ibt
1
×
a + ibt a − ibt
a − ibt
x + iy = 2
a + b2t 2
∴ z2 = cos π + i sin π , cos
7. Here, x + iy =
∴
+ i sin
− 3
+
2
2π
Thus, ∠z1Oz2 =
3
⇒ z2 = − 1 ,
a ≠ 0, b ≠ 0
− bt
a
and y = 2
∴
x= 2
2 2
a +bt
a + b2t 2
y − bt
ay
=
⇒t =
⇒
x
a
bx
a
, we get
On putting x = 2
a + b2t 2
Let

a 2 y2 
x  a 2 + b2 ⋅ 2 2  = a ⇒
bx 

x2 + y2 −
or
7π
6
i − 3 i
,
−
2
2
2
5π
,
,π
6
9. z1 = 1 + i 3 = r (cos θ + i sin θ )
⇒
⇒
So,
a 2(x2 + y2) = ax
Since, |z2| = |z3| = 2
… (i)
z1
1
1

2
x −
 + y =

2a 
4a 2
z2
P (2, 0)
P (–1, 0)
n
π
π
nπ
nπ

Here, P = W =  cos + i sin  = cos
+ i sin

6
6
6
6
n
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
Similarly, z2 = P ∩ H 2 i.e. those points for which
nπ
cos
<0
6
–√3 , 1
—– — Z2
2 2
√3 , 1
Z1 —–
—
2 2
…(i)

1
5π
5π 
3
 =2  − i
and z3 = 2  cos
+ i sin
 =1 − i 3
3
3
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
10. Given, D = (1 + i ), M = (2 − i )
and diagonals of a rhombus bisect each other.
Let B ≡ (a + ib), therefore
O
–√3 , –1 Z
—– — 2
2 2
Z1
X-axis
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, 0) Z2
√3 –1
—– , —
2 2
O
z3
PLAN It is the simple representation of points on Argand plane and
to find the angle between the points.
∴
[given]
Y-axis
x
=0
a
∴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.
8.
[let]
r cos θ = 1, r sin θ = 3
r = 2 and θ = π /3
z1 = 2 (cos π / 3 + sin π / 3)
2
or
5π
5π cos 7π
+ i sin
,
6
6
6
⇒
⇒
a+1
b+1
= 2,
= −1
2
2
a + 1 = 4, b + 1 = − 2 ⇒ a = 3, b = − 3
B ≡ (3 − 3i )
22 Complex Numbers
A
M
⇒ z3 = (2 − i ) ±
(2−i )
C
B
Again,
1
1
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
i
3 


∴ A is either 1 − i or 3 −  .


2
2 
D (1+i )
11. Since, z1 , z2 and z3 form an equilateral triangle.
DM = (2 − 1)2 + (− 1 − 1)2 = 1 + 4 = 5
BD = 2DM ⇒ BD = 2 5
2 AC = BD
⇒ 2 AC = 2 5
AC = 5 and
AC = 2 AM
5
5 = 2 AM ⇒ AM =
⇒
2
Now, let coordinate of A be (x + iy).
But in a rhombus AD = AB, therefore we have
AD 2 = AB2
⇒
(x − 1)2 + ( y − 1)2 = (x − 3)2 + ( y + 3)2
⇒ x2 + 1 − 2x + y2 + 1 − 2 y = x2 + 9 − 6x+ y2 + 9 + 6 y
⇒
4x − 8 y = 16
But
and
⇒
⇒
x − 2y = 4
⇒
x = 2y + 4
…(i)
5
5
Again,
AM =
⇒ AM 2 =
2
4
5
(x − 2)2 + ( y + 1)2 =
⇒
4
5
2
2
[from Eq. (i)]
⇒
(2 y + 2) + ( y + 1) =
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
⇒
⇒
⇒
⇒
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.
C
⇒
a 2 − b2 = a − b
⇒
If a = b ,
2 (2a ) = a 2 + 1
⇒
a − 4a + 1 = 0
4 ± 16 − 4
a=
=2 ± 3
⇒
2
2
If a + b = 1, 2 = a (1 − a ) + 1 ⇒ a − 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
12. 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
⇒
⇒
and
X
z4
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
and z4 = (1 + 3 ) − i
A (z3)
⇒
(a = b or a + b = 1)
2 (a + b) = ab + 1
and
= 1 ± i (1 + 3i ) = (1 + 3 ) ± i = (1 − 3 ) + i
M
(2– i) z2
z3 − (2 − i ) 1
= (± i )
−1 + 2 i
2
2 (a + b) = ab + 1
and
⇒
⇒
a 2 − 1 + 2ia + 1 − b2 + 2ib = a + i (ab + 1) − b
(a 2 − b2) + 2i (a + b) = (a − b) + i (ab + 1)
B
D z1(1+i )
(a + i ) + (1 + ib)2 + (0)2 = (a + i ) (1 + ib) + 0 + 0
⇒
x = 3, x = 1
i
3i 


Therefore, A is either 3 −  or 1 −  .



2
2
z12 + z22 + z32 = z1z2 + z2z3 + z3 z1
2
z3 − (2 − i ) 1
= (± i )
−1 + 2 i
2
13. 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|
Complex Numbers 23
⇒
⇒
|z − z1|2 = |z − z2|2
(z − z1 ) (z − z1 ) = (z − z2) (z − z2)
⇒
⇒
(z1 − z2)2 = 2{(z1z3 − z32 ) + (z2z3 − z1z2)}
⇒
(z1 − z2)2 = 2(z1 − z3 )(z3 − z2)
z (z1 − z2) + z (z1 − z2) = z1z1 − z2z2
16. We have, iz = zei π/2. This implies that iz is the vector
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
obtained by rotating vector z in anti-clockwise direction
through 90°. Therefore, OA ⊥ AB. So,
 z − z2 
 z − z2 
∴ z1b + z2b = z1  1

 + z2  1
 λ 
 λ 
zz − z2z2
= 1
=c
λ
…(i)
1
1
1
OA × OB = |z ||iz | = |z |2
2
2
2
Area of ∆ OAB =
A (z1)
⇒
⇒
⇒
p2
= − cot2(α /2)
p − 4q
z12 + z22 = z1z2
z12 + z22 − z1z2 = 0
18. Since, z1 , z2, z3 are the vertices of an equilateral
triangle.
 z + z2 + z3 
∴ Circumcentre (z0 ) = Centroid  1



3
...(i)
Also, for equilateral triangle
z12 + z22 + z32 = z1z2 + z2z3 + z3 z1
2
p
= − cot2(α /2)
(z1 + z2)2 − 4z1z2
z12 + z22 + 02 = z1z2 + z2 ⋅ 0 + 0 ⋅ z1
⇒
p2
= − cot2(α /2)
(z1 − z2)2
⇒
17. Since, z1 , z2 and origin form an equilateral triangle.
Q if z1 , z2, z3 from an equilateral triangle, then 



z12 + z22 + z32 = z1z2 + z2z3 + z3 z1
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)]
cot (α /2)
−p
=
= − i cot α /2 ⇒
= − i cot (α /2)
i
z1 − z2
... (ii)
On squaring Eq. (i), we get
9z02 = z12 + z22 + z32 + 2 (z1z2 + z2z3 + z3 z1 )
⇒ 9z02 = z12 + z22 + z32 + 2 (z12 + z22 + z32 )
2
p2 = − p2 cot2(α /2) + 4q cot2(α /2)
p (1 + cot α /2) = 4q cot2(α /2)
p2 cosec2(α /2) = 4q cot2(α /2)
p2 = 4q cos 2 α /2
2
X
Y′
Applying componendo and dividendo, we get
z1 + z2 cos α + i sin α + 1
=
z1 − z2 cos α + i sin α − 1
⇒
⇒
⇒
⇒
A
O
B (z2)
z1 cos α + i sin α
=
O
1
z2
[Q|z1|=|z2|]
On squaring both sides, we get
iz
[from Eq. (i)]
z1 |z1|
=
(cos α + i sin α )
z2 |z2|
⇒
B
z
X′
14. Since, z1 + z2 = − p and z1z2 = q
Now,
Y
z + iz
[say]
2
15. Since, triangle is a right angled isosceles triangle.
[from Eq. (ii)]
⇒ 3z02 = z12 + z22 + z32
kπ 
 kπ 
 + i sin  
 7
 7
19. Given, α k = cos 
 2 kπ 
 2 kπ 
= cos 

 + i sin 
 14 
 14 
∴ Rotating z2 about z3 in anti-clockwise direction
through an angle of π / 2 , we get
∴ 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
where,|z2 − z3|= |z1 − z3|
⇒
(z2 − z3 ) = i (z1 − z3 )
…(i)
=a
and α 4k−1 − α 4k− 2 = length of a side of the regular
polygon
…(ii)
=a
z2 − z 3 | z 2 − z 3 | iπ / 2 A (z1)
e
=
z1 − z3 | z1 − z 3 |
B (z3)
C (z2)
12
On squaring both sides, we get
(z2 − z3 ) = − (z1 − z3 )
2
2
⇒
z22 + z32 − 2z2z3 = − z12 − z32 + 2z1z3
⇒
z12 + z22 − 2z1z2 = 2z1z3 + 2z2z3 − 2z32 − 2z1z2
∴
∑
k =1
3
∑
k =1
αk+1 −αk
α 4k−1 − α 4k− 2
=
12 (a )
=4
3 (a )
24 Complex Numbers
Topic 5 De-Moivre’s Theorem, Cube Roots
and nth Roots of Unity
1. 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
= ei( − π / 2 − θ + θ ) = e− i( π / 2) = − i
[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
and
3. 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 =
−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
4. 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
i
π
π
3  1
+   i = cos + i sin = e 6


6
6
2
2
so, (1 + iz + z5 + iz 8 )9
π
5π
8π

i
i
i 
= 1 + ie 6 + e 6 + ie 6 




=
9
9
π

i 
Q i = e 2 


9
2π
2π  
5π
5π 


+ i sin 
= 1 +  cos
+ i sin  +  cos



3
3
6
6

11π
11π  

+  cos
+ i sin


6
6  

1 i 3
3 1
3 i
= 1 − +
−
+ i+
− 
2
2
2
2
2
2

9
9
9
9
1
3i 
π
π

= +
 =  cos + i sin 

2 
3
3
2
[Q for any natural number ‘n’
= cos 3π + i sin 3π
(cos θ + i sin θ )n = cos(nθ ) + i sin(nθ )]
= −1
sin (15 θ )sin (15 θ )
1
=
sin θ
4 sin 2°
5. Let z = |a + bω + cω 2|
π
π
π
4π
5π

i
i
i
i
i 
= 1 + e 2 ⋅ e 6 + e 6 + e 2 ⋅ e 3 




2π
5π
11 π 

i
i
i
= 1 + e 3 + e 6 + e 6 




ε ι ( 2 µ −1 ) θ
= sin θ + sin 3 θ + sin 5 θ + ... + sin 29 θ
 15 × 2 θ 
 θ + 29 θ 
sin 

 sin 



2
2 
=
 2 θ
sin 

 2 
π
Given, z =
−1 + 3 i
and
2
b2 − 4ac
]
2a
ω2 =
Key Idea Use, e i θ = cos θ + i sin θ
2.
−b±
⇒ z 2 = |a + bω + cω 2|2 = (a 2 + b2 + c2 − ab − bc − ca )
1
or z 2 = {(a − b)2 + (b − c)2 + (c − a )2}
…(i)
2
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)
2
⇒
z2 ≥ 1
Hence, minimum value of|z | is 1 .
6. Given, (1 + ω 2)n = (1 + ω 4 )n
⇒
⇒
⇒
(− ω )n = (− ω 2)n
[Qω3 = 1 and 1 + ω + ω 2 = 0]
ω =1
n = 3 is the least positive value of n.
n
Complex Numbers 25
1
1
11. (1 + ω )7 = (1 + ω ) (1 + ω )6
1
7. Let ∆ = 1 −1 − ω 2 ω 2
1
ω
= (1 + ω ) (−ω 2)6 = 1 + ω
ω
2
Applying R2 → R2 − R1 ; R3 → R3 − R1
1
1 
1
2
2

=  0 −2 − ω
ω −1 

2
0
ω −1 
ω −1

⇒
A + Bω = 1 + ω
⇒
A = 1, B = 1
6
12.

∑ sin
k =1
=
= (−2 − ω )(ω − 1) − (ω − 1)
= − 2ω + 2 − ω3 + ω 2 − (ω 4 − 2ω 2 + 1)
= 3 ω 2 − 3 ω = 3ω (ω − 1)
[Q ω 4 = ω ]
z
π
8. Since, arg 1 =
z2 2
z1
π
π
⇒
= cos + i sin = i
z2
2
2
2
∴
⇒
2
2
z1n
= (i )n ⇒ i n = 1
z2n
z2
=
z1 z1
∴
z2
=i
z1
⇒
A (z1)
[Q|z1| = |z2| = 1]
 1 i 3
+ 3 − +

2 
 2
 2π 
i
 ( k + j)
 10 
⇒ zk ⋅ z j = e
⇒ zk will also be 10th root of unity.
e iθ
= e i( θ − α )
2kπ
e iα
i 
z = zk / z1 = e  10
365
= 4 + 5 ⋅ (ω3 )111 ⋅ ω + 3 ⋅ (ω3 )121 ⋅ ω 2
= 4 + 5 ω + 3 ω2
[Q ω3 = 1]
= 1 + 3 + 2 ω + 3 ω + 3 ω2
= 1 + 2 ω + 3 (1 + ω + ω 2) = 1 + 2 ω + 3 × 0
[Q 1 + ω + ω 2 = 0]
[Q 1 + ω + ω 2 = 0]
= (−2ω 2)7 = (−2)7ω14= − 128 ω 2
π
5
−
2π 

10 
=e
i
π
( k − 1)
5
which is in the given set (False)
(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 

πk 
πk
πk 
πk

sin
− i cos
= 2 sin
= 2 sin
10
10
10
10



= 4 + 5 ω334 + 3 ω365
= 1 + (−1 + 3i ) = 3i
2 kπ
10
i( θ + α )
(R) PLAN
1
3
+
i
2
2
10. (1 + ω − ω 2)7 = (− ω 2 − ω 2)7
i
zk = e
i
9. We know that,
334
=e
For k = 2; z = e
in = 1
n = 4k, where k is an integer.
 1 i 3
∴ 4 + 5 − +

2 
 2
iα
zk is 10th root of unity.
(Q) PLAN
∴ z1 and z2 are nth roots of unity.
z1n = z2n = 1
n
 z2 
⇒
  =1
 z1 
ω=−
Given
e iθ⋅ e
Taking, z j as zk, we have zk ⋅ z j = 1 (True)
 z2 
n
  =i
 z1 
⇒
[Q e i14π /7= 1]
 e i2π / 7 − 1 
= −i 
=i
i2π / 7 
1 − e

n
⇒
⇒
 ei2π / 7 − ei14π / 7 
= −i

i2π / 7
 1−e

13. (P) PLAN
O
2 kπ
2 kπ 
+ i sin

7
7 

(1− ei12 π / 7 )
= − i  ei 2 π / 7

1− ei 2 π / 7 

B (z2)
π
2

 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 }
n = 4k
π
z2 i 2
e
6
∑ − i  cos
k =1
[Q|z2| = |z1| = 1]
Alternate Solution
z
π
Since,
arg 2 =
z1 2
2 kπ
2 kπ 
− i cos

7
7 
Now, required product is
π
2π
3π
8π
9π
⋅ sin
⋅ sin
K sin
⋅ sin
29 sin
10
10
10
10
10
10
2
2π
3π
4π 
5π
π

29 sin
sin
sin
sin
 sin

10
10
10
10 
10
=
10
2
2π
2π 
π
π

29 sin
cos
cos
⋅ sin
 ⋅1

10
10
10
10 
=
10
26 Complex Numbers
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  
⇒ (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


9
1+
9
∑
αk = 0
k =1
9
∑
1+
k =1
2 kπ
2 kπ 

+ i sin
 cos
 =0

10
10 
1+
1−
So,
9
∑
cos
2 kπ
=0
10
∑
cos
2 kπ
=2
10
k =1
9
k =1
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
(P) → (i), (Q) → (ii), (R) → (iii), (S) → (iv)
 1 ω ω 2


14. Let
A =  ω ω2 1 
2
ω
1 ω 

0 0 0
2
Now,
A = 0 0 0 and Tr ( A ) = 0, A = 0


0 0 0
A3 = 0
⇒
z+1
ω
ω2
ω
19. 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 .
ω2
z + ω2
1 = [ A + zl] = 0
1
z+ω
20. 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 )
⇒
z =0
⇒ z = 0, the number of z satisfying the given equation
is 1.
3
= (a + b)[a 2αβ + ab(α 2 + β 2) + b2αβ ]
15. Here, Tr = (r − 1) (r − ω ) (r − ω ) ] = (r − 1)
2
∴
Sn =
n
∑ (r3 − 1)=
r =1
q
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.
1 + α + α + α + K+ α =0
3
…(i)
⇒
(z − 1)(z − 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
p
18. Since, 1, a1 , a 2, ... , a n − 1 are nth roots of unity.
(S) Sum of nth roots of unity = 0
2
z p + q − z p − zq + 1 = 0
17. Given,
3
= (a + b)[a 2(ω ⋅ ω 2) + ab(ω 2 + ω 4 ) + b2(ω ⋅ ω 2)]
= (a + b)(a 2 − ab + b2)
2
 n (n + 1) 
−n


2
16. Since, cube root of unity are 1, ω , ω 2 given by
= a3 + b3
i
21. Priniting error = e
 1
3  1
3
A (1, 0), B − , −
 ,C  − , −

2  2
2
 2
Then,
Hence, cube roots of unity form an equilateral triangle.
Download Chapter Test
http://tinyurl.com/y3avd3wd
or
2π
3
|x|2|+ | y|2 + |z|2
=3
(a )2 + (b)2 + |c|2
NOTE Here, w = e
⇒ AB = BC = CA = 3
[Q1 + ω + ω 2 = 0 and ω3 = 1]
i
2π
3
, then only integer solution exists.
2
Theory of Equations
Topic 1
Quadratic Equations
Objective Questions I (Only one correct option)
1. If α and β are the roots of the quadratic equation,
 π
x2 + x sin θ − 2 sin θ = 0, θ ∈ 0,  , then
 2
12
12
α +β
is equal to
(α −12 + β −12 )(α − β )24
(2019 Main, 10 April I)
(a)
212
(sin θ + 8)12
(b)
26
(sin θ + 8)12
12
(c)
2
(sin θ − 4)
12
(d)
(sin θ − 8)6
2. Let p, q ∈R. If 2 − 3 is a root of the quadratic equation,
(a)
(b)
(c)
(d)
(2019 Main, 9 April I)
q 2 − 4 p − 16 = 0
p2 − 4q − 12 = 0
p2 − 4q + 12 = 0
q 2 + 4 p + 14 = 0
(b) 8 5
(d) 4 3
n
α
then the least value of n for which   = 1 is
 β
(b) 5
(d) 3
(2019 Main, 8 April I)
5. 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
6. 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
(2019 Main, 12 Jan II)
(a) 6
(b) 8
8. 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
(b) 144
(d) −300
9. If 5, 5r , 5r 2 are the lengths of the sides of a triangle, then
r cannot be equal to
5
(a)
4
7
(b)
4
(2019 Main, 10 Jan I)
3
(c)
2
(d)
3
4
(c) 7
(a)
4
9
(b) 1
(c)
15
8
(d) 2
11. 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)
If α and β are the roots of the equation x2 − 2x + 2 = 0,
(a) 2
(c) 4
(b) 4 − 2 3
(d) 2 − 3
of the quadratic equation, x2 + (3 − λ )x + 2 = λ has the
(2019 Main, 10 Jan II)
least value is
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)
4.
(a) − 2 + 2
(c) 4 − 3 2
10. The value of λ such that sum of the squares of the roots
3. If
(a) 10 5
(c) 8 3
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) 100
(c) −81
212
x2 + px + q = 0, then
7. If λ be the ratio of the roots of the quadratic equation in
(d) 3
(a) 5
(c) 4
(b) 2
(d) 3
12. Let α and β be two roots of the equation x2 + 2x + 2 = 0,
then α 15 + β15 is equal to
(a) 256
(c) −256
(2019 Main, 9 Jan I)
t(b) 512
(d) −512
13. 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
14. If α , β ∈C are the distinct roots of the equation
x2 − x + 1 = 0, then α 101 + β107 is equal to
(a) −1
(c) 1
(b) 0
(d) 2
(2018 Main)
28 Theory of Equations
15. For a positive integer n, if the quadratic equation,
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)
(a) 12
(b) 9
(c) 10
(d) 11
16. 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
(d) 5
π
π
17. 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 − 2a 8
is
a n = α − β , for n ≥ 1, then the value of 10
2a 9
n
(a) 6
(c) 3
(b) – 6
(d) – 3
(2015 Main)
19. 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
61
9
(b)
2 17
9
(2014 Main)
(c)
34
9
(d)
a n = α n − β n for n ≥ 1 , then the value of
(b) 2
(c) 3
a10 − 2a 8
is
2a 9
(d) 4
(2011)
22. Let p and q be real numbers such that p ≠ 0, p ≠ q and
p ≠ − 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
3
( 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
23. Let α, β be the roots of the equation x2 − px + r = 0 and
α
, 2 β be the roots of the equation x2 − qx + r = 0 . Then,
2
the value of r is
(2007, 3M)
2
( p − q) (2q − p )
9
2
(c) (q − 2 p ) (2q − p )
9
(a)
2
(q − p ) (2 p − q)
9
2
(d) (2 p − q) (2q − p )
9
(b)
4 5
1 5
(c) λ ∈  ,  (d) λ ∈  , 
 3 3
 3 3
5
3
25. 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)
26. The set of all real numbers x for which x2 − |x + 2| + x > 0
is
(2002, 1M)
(b) (− ∞ , − 2 ) ∪ ( 2 , ∞ )
(d) ( 2 , ∞ )
27. The number of solutions of log 4 (x − 1) = log 2(x − 3) is
(a) 3
(c) 2
(b) 1
(d) 0
(2001, 2M)
28. For the equation 3x2 + px + 3 = 0, p > 0, if one of the root
is square of the other, then p is equal to
(a) 1 /3
(b) 1
(c) 3
(2000, 1M)
(d) 2 /3
29. If α and β (α < β) are the roots of the equation
x2 + bx + c = 0, where c < 0 < b, then
(2000, 1M)
(b) α < 0 < β < |α|
(d) α < 0 < |α|< β
(1997C, 2M)
(a) no solution
(b) one solution
(c) two solutions
(d) more than two solutions
31. The equation x 4
3
(a)
(b)
(c)
(d)
(b) λ >
3
2 13
9
21. Let α and β be the roots of x2 − 6x − 2 = 0, with α > β. If
(a) 1
4
3
30. The equation x + 1 − x − 1 = 4x − 1 has
1 1
+ = 4, then the
α β
value of|α − β|is
(a)
(a) λ <
(a) 0 < α < β
(c) α < β < 0
20. Let α and β be the roots of equation px2 + qx + r = 0,
p ≠ 0. If p, q and r are in AP and
x2 − 2 (a + b + c) x + 3λ (ab + bc + ca ) = 0 has real roots,
(2006, 3M)
then
(a) (− ∞ , − 2) ∪ (2, ∞ )
(c) (− ∞ , − 1) ∪ (1, ∞ )
(b) 2secθ
(d) 0
18. Let α and β be the roots of equation x2 − 6x − 2 = 0. If
n
24. If a , b, c are the sides of a triangle ABC such that
(log 2 x )2 + log 2 x −
5
4
= 2 has
(1989; 2M)
(a) atleast one real solution
(b) exactly three real solutions
(c) exactly one irrational solution
(d) complex roots
32. If α and β are the roots of x2 + px + q = 0 and α 4 , β 4are
the roots of x2 − rx + s = 0, then
x2 − 4qx + 2q2 − r = 0 has always
the
equation
(1989, 2M)
(a) two real roots
(b) two positive roots
(c) two negative roots
(d) one positive and one negative root
33. The equation x −
2
2
has
=1 −
x−1
x−1
(a) no root
(c) two equal roots
34. For real x, the function
(x − a )(x − b)
will assume all real
(x − c)
values provided
(a) a > b > c
(c) a > c < b
(1984, 2M)
(b) one root
(d) infinitely many roots
(1984, 3M)
(b) a < b < c
(d) a ≤ c ≤ b
Theory of Equations 29
35. 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
(x − b) (x − c) + (x − a ) (x − c) + (x − a ) (x − b) = 0
(1980, 1M)
(a) positive
(b) negative
(c) real
(d) None of the above
37. Let a > 0, b > 0 and c > 0. Then, both the roots of the
equation ax2 + bx + c = 0
45. 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.
36. Both the roots of the equation
are always
Analytical & Descriptive Questions
(1979, 1M)
(a) are real and negative
(b) have negative real parts
(c) have positive real parts
(d) None of the above
46. If α , β are the roots of ax + bx + c = 0, (a ≠ 0) and
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
47. Let f (x) = Ax2 + Bx + C where, A , B, C are real
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
whenever x is an integer.
(1998, 3M)
48. Find the set of all solutions of the equation
2|y| − |2y − 1 − 1| = 2y − 1 + 1
Assertion and Reason
(1997 C, 3M)
49. Find the set of all x for which
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
38. 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
2x
1
>
x
+1
2x + 5x + 2
(1987, 3M)
50. Solve| x2 + 4x + 3 | + 2x + 5 = 0
(1987, 5M)
2
51. For a ≤ 0, determine all real roots of the equation
x2 − 2a|x − a|− 3a 2 = 0
52. Solve for x : (5 + 2 6 )x
2
−3
+ (5 − 2 6 )x
2
(1986, 5M)
−3
= 10 (1985, 5M)
53. If one root of the quadratic equation ax + bx + c = 0 is
2
equal to the nth power of the other, then show that
1
n n +1
(ac )
1
+ (a c)
n
n +1
+ b =0
(1983, 2M)
54. If α and β are the roots of x + px + q = 0 and γ , δ are the
2
where β 2 ∉ { −1, 0, 1}.
Statement I ( p − q) (b − ac) ≥ 0
Statement II b ∉ pa or c ∉ qa .
2
(2006, 6M)
2
2
(2008, 3M)
55. Solve 2 log x a + log ax a + 3 logb a = 0 ,
Fill in the Blanks
where a > 0, b = a 2 x
39. The sum of all the real roots of the equation
|x − 2|2 + |x − 2| − 2 = 0 is…… .
(1997, 2M)
40. 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 .
41. If 2 + i 3 is a root of the equation x2 + px + q = 0, where
(1982, 2M)
p and q are real, then ( p, q) = (…,…).
42. The coefficient of x99 in the polynomial
(x − 1)(x − 2)... (x − 100) is....
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.
(1982, 2M)
True/False
(1978 , 3M )
56. If α
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 )
Passage Type Questions
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.
57. a12 =
43. 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)
44. The equation 2x + 3x + 1 = 0 has an irrational root.
2
(1983, 1M)
(a) a11 + 2a10
(c) a11 − a10
(b) 2a11 + a10
(d) a11 + a10
58. If a 4 = 28, then p + 2q =
(a) 14
(b) 7
(c) 21
(d) 12
30 Theory of Equations
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 andx2 + 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
x2 + bx + a = 0 (a ≠ b) have a common root, then the
(1986, 2M)
numerical value of a + b is… .
(a) 0
(b) αβ
(c) αγ
(d) βγ
4. If
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,
y = 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
2
125
2
625
(c)
4
(a)
75
2
125
(d)
4
(b)
= 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)
(b) 10
(d) 18
(b) (1, 2)
(d) (−∞ , − 2) ∪ (2 , ∞ )
(a) a < − 5
(2004, 1M)
(b) −5 < a < 2 (c) a > 5
(d) 2 < a < 5
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)
7. If the roots of the equation x2 − 2ax + a 2 + a − 3 = 0 are
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) (−1, 0) ∪ (0, 1)
(c) (−2 , − 1)
which ‘a’ lies is
2. Consider the quadratic equation, (c − 5) x − 2cx + (c − 4)
(a) (4, 5)
(c) (5, 6)
+ a 2 = 0 (where, [x] denotes the greatest integer ≤ x) has
no integral solution, then all possible values of a lie in
(2014 Main)
the interval
5 For all ‘x’, x2 + 2ax + (10 − 3a ) > 0, then the interval in
2
(a) 11
(c) 12
4. If a ∈ R and the equation −3 (x − [x])2 + 2 (x − [x])
real and less than 3, then
(a) a < 2
(b) 2 ≤ a ≤ 3
(1999, 2M)
(c) 3 < a ≤ 4
(d) a> 4
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
(b) g (x) > 0
(c) g (x) = 0
(d) g (x) ≥ 0
Theory of Equations 31
11. Find all real values of x which satisfy x2 − 3x + 2 > 0 and
Analytical & Descriptive Questions
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 roots
(2003, 4M)
for all values of b.
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,
c
b
1 + +  < 0
a a 
then show that
x2 − 2x − 4 ≤ 0.
(1983, 2M)
Integer 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
and have values atleast 4, is …… .
(2009)
(1995, 5M)
Topic 5 Some Special Forms
Objective Questions I (Only one correct option)
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
sin 2 x − 2 sin x + 5
2
⋅
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
(2019 Main, 8 April I)
(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
α+β
2
(c) γ = α
β
2
(d) α < γ < β
(b) γ = α +
3ax + 2bx + c = 0 has
2
(1983, 1M)
x12 − x9 + x4 − x + 1 > 0 is
2
= ∫ (1 + cos 8 x)(ax2 + bx + c)dx
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
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|.
1
(a)  − , 0
 4 
3
3
1
(b)  − 11, −  (c)  − , − 



4
4
2
(d)  0,

1

4
21 11
(b)  , 
 64 16 
(c) (9, 10)
21
(d)  0, 
 64 
13. The function f ′ (x) is
(1982, 2M)
x)(ax2 + bx + c)dx
(a)
(b)
(c)
(d)
x = 0, y = 0 and x = t , lies in the interval
(b) 0 < x < 1
(d) −∞ < x < ∞
0
1 
 1 , 1
(c)  0,

 (d) 
 5 2

5
10. Let a ∈ R and f : R → R be given by f (x) = x5 − 5x + a.
3
(a)  , 3
4 
8. Let a , b, c be non-zero real numbers such that
8
1
(b)  –
, 0

5 
12. The area bounded by the curve y = f (x) and the lines
7. The largest interval for which
1
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?
11. The real numbers s lies in the interval
(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
∫0 (1 + cos
9. Let S be the set of all non-zero real numbers α such that
(2010)
6. If a + b + c = 0, then the quadratic equation
(a) −4 < x ≤ 0
(c) −100 < x < 100
Objective Questions II
(One or more than one correct option)
Then,
4. The
(a) γ =
(b) atleast one root in (1, 2)
(d) two imaginary roots
(2015 Adv.)
| x − 2| + x ( x − 4) + 2 = 0 (x > 0) is equal to
(a) lies between 1 and 2
(c) lies between − 1and 0
(a) no root in (0, 2)
(c) a double root in (0, 2)
1
1 
(a)  – , –

 2
5
3. The sum of the solutions of the equation
(a) 9
(c) 4
Then, the quadratic equation ax2 + bx + c = 0 has
(1981, 2M)
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 )
32 Theory of Equations
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)
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 )
14. The line y = x meets y = ke for k ≤ 0 at
x
(a) no point
(c) two points
16. For k > 0, the set of all values of k for which kex − x = 0
(b) one point
(d) more than two points
15. The positive value of k for which ke − x = 0 has only one
x
root is
(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
1
(a)
e
(b) 1
(c) e
(d) log e 2
has a unique root in the interval [1/2, 1] and identify it.
(2001, 4M)
Answers
Topic 1
1. (a)
5. (a)
9. (b)
2. (b)
6. (c)
10. (d)
3. (b)
7. (c)
11. (d)
4. (c)
8. (d)
12. (c)
1. (d)
14. (c)
15. (d)
17. (c)
18. (c)
19. (d)
20. (d)
Topic 3
21. (c)
22. (b)
23. (d)
24. (a)
1. (c)
25. (a)
26. (b)
27. (b)
28. (c)
29. (b)
30. (a)
31. (b)
32. (a)
33. (a)
34. (d)
35. (a)
36. (c)
37. (b)
38. (b)
39. 4
40. k = 2
41. ( −4, 7 )
42. –5050
43. True
44. False
45. 1210
48. y ∈ { −1 } ∪ [1, ∞ )
51. x = {a (1 − 2 ), a ( 6 − 1 )}
54. (q − s ) − rqp − rsp + sp + qr
2
56. q − p
2
2
2
57. (d)
2
2. (a)
3. (b)
4. (–1)
2. (a)
3. (a)
4. (a)
6. (d)
7. (a)
8. (b)
5. False
13. (c)
1
 2
49. x ∈ ( −2, − 1 ) ∪  − , − 
 3
2
16. (a)
Topic 2
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
50. −4 and ( − 1 − 3 )
1. (a)
2. (b)
3. (d)
4. (d)
5. (d)
6. (a)
7. (d)
8. (b)
52. ±2, ± 2
9. (a,d)
10. (b, d)
11. (c)
12. (a)
15. (a)
1

18. x = cos cos−1 p
3

16. (a)
55. x = a
58. (d)
−1/ 2
or a
− 4 /3
13. (b)
17. True
14. (b)
Hints & Solutions
Topic 1 Quadratic Equations
⇒
(x − 1)2 = − 1
⇒
x−1 = ± i
⇒
x = (1 + i ) or (1 − i )
Clearly, if α = 1 + i, then β = 1 − i
1. Given quadratic equation is
 π
x2 + x sin θ − 2 sin θ = 0, θ ∈ 0, 
 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
=
n
12
12
12

 αβ 

αβ
=
=

2
2


(
α
β
)
(
α
β
)
αβ
−
+
−
4


…(i)
n
⇒
12
12


2
=

 − sin θ − 8
5.
12
[Q α + β = − sin θ]
2. 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
[Q i 4 = 1]
Key Idea
(i) First convert the given equation in quadratic equation.
(ii) Use, Discriminant, D = b 2 − 4 ac < 0
(m2 + 1)x2 − 3x + (m2 + 1)2 = 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
…(i)
According to the question there is no solution of the
quadratic Eq. (i), then
D <0
∴ −8m(2m − 1)2 < 0 ⇒ m > 0
So, there are infinitely many values of ‘m’ for which,
there is no solution of the given quadratic equation.
3. Given quadratic equation is
…(i)
Let the roots of quadratic Eq. (i) are α and β, so
3
and αβ = m2 + 1
α+β= 2
m +1
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
4. Given, α and β are the roots of the quadratic equation,
⇒
n
 2i 
n
  =1⇒ i =1
 2
[from Eq. (i)]
212
(sin θ + 8)12
x2 − 2x + 2 = 0
(x − 1)2 + 1 = 0
 1 + i 2 + 2i 

 =1⇒
 1 − i2 
[by rationalization]
So, minimum value of n is 4.
12


2
=

 (α + β ) − 8
 (1 + i )(1 + i )

 =1
 (1 − i )(1 + i ) 
n
⇒
12


2(α + β)
=

2
(
α
+
β
)
−
8
(
α
+
β)


=
α
According to the question   = 1
 β
n
 1 + i
⇒

 =1
1 − i
+β
α +β
= 12
 1
 β + α 12
1
24
 12 + 12 (α − β)
 12 12  (α − β)24
α
β 
 α β 
α
12
[where i = −1]
6. 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
[Q m2 − 6m − 3 = 0 ⇒ m =
= 3 ± 12]
2
34 Theory of Equations
⇒
3 − 12 < m < 3 + 12
…(iv)
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.
1 − 5 1 + 5
r ∈
,

2 
 2
⇒
+
⇒
⇒
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
⇒
⇒
⇒
⇒
4
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
a+ b>c
⇒ 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
− b ± b2 − 4ac
and r 2 − r − 1 = 0
2a
⇒r =
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
and
9. Let a = 5, b = 5r and c = 5r 2
x=
c+ a >b
5r 2 + 5 > 5r
2
r − r + 1 >0
2
4
 4
α4 =   ⇒α = ±
 3
3
k = − 81 (α + α 3 )
= − 81 α (1 + α 2)
16
 4 
= − 81  ±  1 +
 = ± 300
 3 
9
⇒ 5 + 5r > 5r 2
–1+ √ 5
2
2
b
a
c
and product of roots = ]
a
Now,
+
–
–1– √ 5
2
…(ii)
1
 1
 1
⇒ r2 − 2 ⋅ r +   + 1 −   > 0
 2
 2
2
[Q for ax2 + bx + c = 0, sum of roots = −
∴
+
and
⇒
⇒
81x2 + kx + 256 = 0
Let one root be α, then other is α 3 .
k
256
Now, α + α 3 = −
and α ⋅ α 3 =
81
81
⇒

 −1 + 5

−1 − 5 
r ∈  − ∞,
, ∞
 ∪
2
2




[Qm ≠ 0]
8. Given quadratic equation is
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


⇒
The least value of m = 4 − 3 2
+
1– √ 5
2
7. Let the given quadratic equation in x,
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
(given)
Then,
λ + =1 ⇒ + =1
λ
β α
–
...(i)
1± 1+4 1± 5
]
=
2
2
1– √ 5
2
–1+ √ 5
2
… (iii)
1+ √ 5
2
∞
7
is the only value that does not satisfy.
4
10. Given quadratic equation is
x2 + (3 − λ )x + 2 = λ
x2 + (3 − λ )x + (2 − λ ) = 0
… (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.
Theory of Equations 35
11. For the roots of quadratic equation ax2 + bx + c = 0 to be
rational D = (b − 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.
2
12. We have, x2 + 2x + 2 = 0
⇒ x=
−2 ± 4 −8
2
[Q roots of ax2 + bx + c = 0 are
given by x =
−b±
b2 − 4ac
]
2a
⇒
x = −1 ± i
Let α = − 1 + i and β = − 1 − i.
Then, α 15 + β15 = (−1 + i )15 + (− 1 − i )15

= − 



= − 

+

= − [(1 − i )15 + (1 + i )15 ]
15
  1
i 
i  
 1
2
+
−
 
  +  2
 2
2 
2   
  2
15

π
π 

2  cos − i sin  


4
4 

15
π
π  
 
 2  cos + i sin   
4
4  

15
15π
15π  
15π
15π  

= − ( 2 )15  cos
− i sin
+ i sin
 +  cos




4
4
4
4  

[using De’ Moivre’s theorem
(cos θ ± i sin θ )n = cos nθ ± i sin nθ, n ∈ Z ]
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
 [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
)


13. 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| + y 2 − 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
⇒
14. 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
15. 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
16. Given, (x2 − 5x + 5)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
36 Theory of Equations
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 − 5x + 4 = 0
⇒
x2 − 4 x − 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, consider
a10 − 2a 8 α 10 − β10 − 2(α 8 − β 8 )
=
2a 9
2(α 9 − α 9 )
x2 − 6x − 2 = 0.
x2 = 6 x + 2
or
∴
α 2 = 6α + 2
⇒
α 10 = 6 α 9 + 2 α 8
Similarly,
⇒ a10
α 1 , β1 =
18. 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
…(ii)
(Q a n = α n − β n)
a10 = 6a 9 + 2a 8
a − 2a 8
− 2a 8 = 6a 9 ⇒ 10
=3
2a 9
coefficient of x must be equal to zero.
Let p(x) = ax2 + b with a, b of same sign and a , b ∈ R.
17. Here, x − 2x secθ + 1 = 0 has roots α 1 and β1.
∴
and
Thus,
=6β + 2β
8
19. If quadratic equation has purely imaginary roots, then
2
−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
...(i)
9
α 10 − β10 = 6(α 9 − β 9 ) + 2(α 8 − β 8 )
Hence, the sum of all real values of
α 2, β 2 =
β
10
On subtracting Eq. (ii) from Eq. (i), we get
Thus, in this case, we get x = 2.
i.e.
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
Since, α and β are the roots of the equation
⇒
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 .
=
Alternate Solution
When x = 3, x2 + 4x − 60 = 9 + 12 − 60 = − 39, which is not
an even integer.
∴
α 8 (α 2 − 2) − β 8 (β 2 − 2)
2(α 9 − β 9 )

Q α and β are the roots of
x2 − 6x − 2 = 0 or x2 = 6x + 2


⇒
α2 = 6 α + 2 ⇒ α2 − 2 = 6 α
2
2
and β = 6 β + 2 ⇒ β − 2 = 6 β 
Now, when x = 2, x2 + 4x − 60 = 4 + 8 − 60 = − 48, which is
an even integer.
x = − 10 + 6 + 4 + 1 + 2 = 3
=
Then,
p[ p(x)] = a (ax2 + b)2 + b
p(x) has imaginary roots say ix.
Then, also ax2 + b ∈ R and (ax2 + b)2 > 0
∴
Thus,
20.
PLAN
a (ax2 + b)2 + b ≠ 0, ∀ x
p [ p(x)] ≠ 0, ∀ x
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 px2 + qx + r = 0, p ≠ 0.
−q
r
, αβ =
α+β=
∴
p
p
…(i)
Since, p, q and r are in AP.
∴
Also,
⇒
⇒
2q = p + r
1 1
α+β
+ =4 ⇒
=4
α β
αβ
−q 4r
α + β = 4 αβ ⇒
=
p
p
q = − 4r
...(ii)
[from Eq. (i)]
On putting the value of q in Eq. (ii), we get
⇒
2 (−4r ) = p + r ⇒
p = − 9r
Theory of Equations 37
−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
2
(α − β ) =
81
2
|α − β| =
13
9
α+β=
Now,
and
∴
⇒
⇒
21.
a10 − 2a 8 (α 10 − β10 ) − 2 (α 8 − β 8 )
=
2a 9
2 (α 9 − β 9 )
=
2
8
α 8 (6 α ) − β 8 (6 β ) 6 (α 9 − β 9 )
=3
=
2 (α 9 − β 9 )
2 (α 9 − β 9 )
a 2 + b2 + c2 < 2 (ab + bc + ca )
a 2 + b2 + c2
<2
ab + bc + ca
⇒
…(ii)
From Eqs. (i) and (ii), we get
3λ − 2 < 2
α + α2 = − p
⇒ λ<
4
3
and α 3 = q
α (α + 1) = − p
(α + β ) (α 2 − αβ + β 2) = q
−q
α 2 + β 2 − αβ =
p
q (q + 1 − 3 p) = − p
p3 − (3 p − 1)q + q2 = 0
x2 − |x + 2| + x > 0
…(i)
x+ 2 ≥0
∴
x − x − 2 + x > 0 ⇒ x2 − 2 > 0
⇒
x < − 2 or x > 2
2
⇒
x ∈ [−2, − 2 ) ∪ ( 2 , ∞ )
…(ii)
Case II When x + 2 < 0
...(i)
(α + β )2 = p2
α 2 + β 2 + 2αβ = p2
⇒
[cubing both sides]
3
Case I When
Given, α + β = − p and α 3 + β3 = q
⇒
⇒
26. Given,
α 2 + β2
and product = 1
22. Sum of roots =
αβ
and
a 2 + b2 − c2 < 2ab
⇒
[and β is root of x2 − 6 x − 2 = 0 ⇒ β 2 − 2 = 6 β]
∴
and
⇒ α 3 {α 3 + 1 + 3α (α + 1)} = − p3
Qα is root of x − 6 x − 2 = 0 ⇒ α − 2 = 6α
⇒
c2+ a 2 − b2 < 2ca
⇒
2
=
Similarly,
⇒
2
2
b2 + c2 − a 2< 2bc
25. Let the roots of x2 + px + q = 0 be α and α 2.
α (α − 2) − β (β − 2)
2(α 9 − β 9 )
8
⇒
...(ii)
∴
x2 + x + 2 + x > 0
⇒
x2 + 2 x + 2 > 0
⇒
(x + 1)2 + 1 > 0
From Eqs. (i) and (ii), we get
p3 − 2q
3p
α 2 + β2 =
and αβ =
∴ Required equation is, x2 −
which is true for all x.
p3 + q
3p
∴
( p − 2q) x
+1 =0
( p3 + q)
2
3
27. Given, log 4 (x − 1) = log 2(x − 3) = log 41/ 2 (x − 3)
3
23. The equation x − px + r = 0 has roots α, β and the
2
α
equation x − qx + r = 0 has roots , 2 β .
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
2
4 (a + b + c)2 − 12λ (ab + bc + ca ) ≥ 0
⇒
(a + b + c)2 ≥ 3λ (ab + bc + ca )
⇒
⇒
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
⇒
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 − 6 x = x − 1
x − 7x + 10 = 0
⇒
⇒
2
(x − 2)(x − 5) = 0
x = 2, or x = 5
x = 5 [Q x = 2 makes log (x − 3) undefined].
⇒
Hence, one solution exists.
24. Since, roots are real, therefore D ≥ 0
⇒
…(iii)
x ∈ (−∞ , − 2 ) ∪ ( 2 , ∞ )
⇒ ( p + q) x − ( p − 2q) x + ( p + q) = 0
3
x ≤ − 2 or x ∈ (−∞ , − 2)
From Eqs. (ii) and (iii), we get
3
28. Let α , α 2 be the roots of 3x2 + px + 3 = 0
Now,
…(i)
⇒
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
38 Theory of Equations
⇒
⇒
− 1 = − p/3
p=3
c<0 < b
29. Given,
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
30. Since,
⇒
1 − 2x = 2 x2 − 1 ⇒ 1 + 4x2 − 4x = 4x2 − 4
5
⇒
4x = 5 ⇒ x =
4
But it does not satisfy the given equation.
⇒
⇒
⇒
(log 2 x )2 + log 2 x −
5
4
= 2
3
5
(log 2 x)2 + log 2 x − = log x 2
4
4
3
5
1
(log 2 x)2 + log 2 x − =
4
4 2 log 2 x
⇒
3(log 2 x) + 4(log 2 x) − 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
⇒
3
36. (x − a ) (x − b) + (x − b) (x − c) + (x − c) (x − a ) = 0
37. Since, a , b, c > 0 and ax2 + bx + c = 0
⇒
x=
−b
±
2a
b2 − 4ac
2a
Case I When b2 − 4ac > 0
⇒
4
4
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
Hence, no solution exist.
Hence, four real solutions exist.
Hence, both roots are real.
32. Since, α , β are the roots of x + px + q = 0 and α , β are
⇒ Roots are real and of opposite sign.
2
2
33. Given, x −
=1 −
⇒ x=1
x−1
x−1
But at x = 1, the given equation is not defined.
(|x| − 1) (|x| − 2) = 0
|x| = 1, 2
x = 1, − 1, 2, − 2
which is always positive.
2
2
⇒
⇒
∴
| x|2 − 3|x| + 2 = 0
⇒
3x2 − 2 (a + b + c) x + (ab + bc + ca ) = 0
Now, discriminant = 4 (a + b + c)2 − 12 (ab + bc + ca )
= 4 { a 2 + b2 + c2 − ab − bc − ca }
= 2 {(a − b)2 + (b − c)2 + (c − a )2}
Hence, no solution exists.
3
i.e. a ≤ c ≤ b or b ≤ c ≤ a
35. Since,
(x + 1) + (x − 1) − 2 x2 − 1 = 4x − 1
31. Given, x 4
x2 − (a + b)x + ab
x−c
⇒ yx − cy = x2 − (a + b)x + ab
⇒ x2 − (a + b + y)x + (ab + cy) = 0
For real roots, D ≥ 0
⇒
(a + b + y)2 − 4(ab + cy) ≥ 0
⇒ (a + b)2 + y2 + 2(a + b) y − 4ab − 4cy ≥ 0
⇒
y2 + 2(a + b − 2c) y + (a − b)2 ≥ 0
which is true for all real values of y.
∴
D ≤0
2
4 (a + b − 2c) − 4 (a − b)2 ≤ 0
⇒ 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
34. Let y =
and
x=
−b
+
2a
−b
−
2a
b2 − 4ac
2a
b2 − 4ac
both roots, are negative.
2a
Case II When b2 − 4ac = 0
−b
, i.e. both roots are equal and negative
⇒ x=
2a
Case III When b2 − 4ac < 0
⇒
x=
4ac − b2
−b
±i
2a
2a
have negative real part.
∴ From above discussion, both roots have negative real
parts.
Theory of Equations 39
38. Given,
x2 + 2 px + q = 0
∴
And
∴
and
α + β = −2 p
... (i)
αβ = q
... (ii)
ax + 2bx + c = 0
1
2b
α+ =−
β
a
α c
=
β a
... (iii)
... (iv)
Now, ( p2 − q) (b2 − ac)
Given,| x − 2|2 + | x − 2| − 2 = 0
⇒ (| x − 2| + 2) (| x − 2| − 1) = 0
∴
| x − 2| = − 2, 1
⇒
| x − 2| = 1
Sum of the roots = 4
2e2 log k − 1 = 7
e2 log e k = 4
k2 = 4
k =2
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)
42. The coefficient of x99 in (x − 1)(x − 2)... (x − 100)
= − (1 + 2 + 3 + ... + 100)
100
=−
(1 + 100) = −50(101) = −5050
2
1
⇒ α + ≠α +β
β
Since,
pa ≠ b
⇒
β 2 ≠ 1, β ≠ { −1, 0, 1}, which is correct.
Similarly, if c ≠ qa
α
1

⇒
a ≠ a α β ⇒ α β −  ≠ 0

β
β
α ≠ 0 and β −
⇒
β ≠ { −1 , 0 , 1 }
Hence, statement is true.
44. Given, 2x2 + 3x + 1 = 0
x + 4 − 4x + x − 2 − 2 = 0
2
x = 0, 3
⇒
x=3
∴ Roots are rational.
Hence, statement is false.
a + b = 10c and c + d = 10a
⇒ (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
⇒
Here, D = (3)2 − 4 ⋅ 2 ⋅ 1 = 1 which is a perfect square.
45. Here,
Case I When x ≥ 2
⇒
(x − 2)2 + (x − 2) − 2 = 0
x (x − 3) = 0
D1 + D2 = 2b2 ≥ 0
∴ Atleast one of D1 and D2 is (+ ve). Hence, atleast two
real roots.
1
≠0
β
39. Given,|x − 2|2 + |x − 2| − 2 = 0
⇒
D1 = b2 − 4ac and D2 = b2 + 4ac
Clearly,
Both Statement I and Statement II are true. But
Statement II does not explain Statement I.
x − 3x = 0
43. P (x) ⋅ Q (x) = (ax2 + bx + c) (−ax2 + bx + c)
Now,
Statement II is true.
⇒
[neglecting −2]
41. If 2 + i 3 is one of the root of x + px + q = 0. Then,
1

α + 

β
⇒
[neglecting –2]
⇒ x = 3, 1
2
∴ Statement I is true.
a
 α + β
Again, now
pa = − 
 a = − (α + β )
 2 
2
a
2
…(ii)
Hence, the sum of the roots is 3 + 1 = 4.
⇒
⇒
⇒
⇒
1
(α − β )2 
2
α −  . a ≥ 0

16
β
b=−
[4 is rejected]
40. Since, x2 − 3kx + 2e2 log k − 1 = 0 has product of roots 7.
2
⇒
x=1
⇒
2


1
α
+
2



 α + β

α 2
β
 − a
= 
 − αβ  


β
 −2
  2 





and
x = 1, 4
⇒
Alternate Solution
2
=
⇒
[0 is rejected]
…(i)
…(i)
…(ii)
Similarly, ‘a’ is the root of
x2 − 10cx − 11d = 0
Case II When x < 2
⇒
⇒
{ − (x − 2)}2 − (x − 2) − 2 = 0
On subtracting Eq. (iv) from Eq. (ii), we get
⇒
(x − 2) − x + 2 − 2 = 0
⇒
x2 + 4 − 4x − x = 0
⇒
x − 4x − (x − 4) = 0
⇒
x(x − 4) − 1 (x − 4) = 0
⇒
(x − 1) (x − 4) = 0
(c2 − a 2) = 11 (b − d )
2
2
a 2 − 10ca − 11d = 0
∴
⇒
∴
(c + a ) (c − a ) = 11 × 11 (c − a )
c + a = 121
a + b + c + d = 10c + 10a
= 10 (c + a ) = 1210
…(iii)
…(iv)
[from Eq. (i)]
40 Theory of Equations
b
c
, αβ =
a
a
46. Since, α + β = −
B
C
, (α + δ ) (β + δ ) =
A
A
α − β = (α + δ ) − (β + δ )
(α − β )2 = [(α + δ ) − (β + δ )]2
Now,
⇒
–
2
2
2
2
b
4c B
4C
−
= 2−
A
a2 a
A
b2 − 4ac B2 − 4 AC
=
a2
A2
⇒
⇒
47. Suppose f (x) = Ax2 + Bx + C is an integer, whenever x
is an integer.
∴ f (0), f (1), f (−1) are integers.
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.
Conversely, suppose 2 A , A + B and C are integers.
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.
48. Given, 2
− |2
y −1
− 1| = 2
y −1
∴
2
+1
+ (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
− 1) = 2
y −1
∴
x = − 1 − 3 is the only solution.
x = − 4 and (−1 − 3) are the only solutions.
a ≤0
Given,
x2 − 2 a | x − a | − 3 a 2 = 0
Case I When x ≥ a
51. Here,
⇒
x2 − 2a (x − a ) − 3a 2 = 0
⇒
x2 − 2ax − a 2 = 0
x = a ± 2a
⇒
…(i)
x = a (1 − 2 )
Case II
⇒
x2 + 2 ax − 5a 2 = 0 ⇒ x = − a ± 6a
[as a ( 6 − 1) < a and a (−1 − 6 ) > a]
y
∴ Neglecting x = a (−1 − 6 ) ⇒ x = a ( 6 − 1)
y = 1 ∈ (0, 1]
From Eqs. (i) and (ii), we get
2 =2
…(ii)
∴
2 −2
⇒
y −1
2 − 2 ⋅2
y
⇒
+ 1 =2
y −1
y −1
+1
From Eqs. (i), (ii) and (iii), we get
⇒
⇒
52. Given, (5 + 2 6 )
=0
2y − 2y = 0 true for all y > 1
49. Given,
...(ii)
x = { a (1 − 2 ), a ( 6 − 1)}
Case III When y ∈ (1, ∞ )
y
…(i)
When x < a ⇒ x2 + 2a (x − a ) − 3a 2 = 0
2y + (2y − 1 − 1) = 2y − 1 + 1
⇒
…(ii)
∴ Neglecting x = a (1 + 2 ) as x ≥ a
+1
y = − 1 ∈ (−∞ , 0]
⇒
[but x ∈ (−3, − 1)]
[as a (1 + 2 ) < a and a (1 − 2 ) > a]
y −1
Case II When y ∈(0, 1]
∴
– 1
2
1
 2
x ∈ (−2, − 1) ∪  − , − 
 3
2
∴
⇒
2−y = 2
⇒
–
– 2
3
–1
Case I When y ∈ (−∞ , 0]
−y
+
From Eqs. (i) and (ii), we get
Let n be any integer. We have,
 n (n − 1) 
+ ( A + B) n + C
f (n ) = An 2 + Bn + C = 2 A


2
|y|
–
50. Given,| x2 + 4x + 3|+ 2x + 5 = 0
4c  B 
4C
 b
= −  −
−  −
 a
a  A
A
2
+
–2
(α + β ) − 4αβ = (α + δ + β + δ ) − 4(α + δ ) (β + δ )
2
⇒
⇒
⇒
⇒
− (3x + 2)
> 0 ; using number line rule
(2x + 1) (x + 1) (x + 2)
α+δ+β+δ=−
and
⇒
⇒
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)
x 2 −3
Put y = (5 + 2 6 )x
From Eq. (i), y +
⇒
−3
x −3 =1
= 10
2
−3
…(i)
=
or
y=5±2 6
x 2 −3
=5 + 2 6
2
=5 −2 6
(5 + 2 6 )x
2
−3
1
= 10
y
(5 + 2 6 )
or
2
⇒ (5 − 2 6 )x
y2 − 10 y + 1 = 0 ⇒
⇒
⇒
2
+ (5 − 2 6 )x
−3
x − 3 = −1
2
⇒
x = ± 2 or x = ± 2
⇒
x = ± 2, ± 2
1
y
Theory of Equations 41
= (1 + γ 2 + γp)(1 − δp + δ 2) = (− qγ + γp)(−δp − δq)
53. Let α , β are roots of ax2 + bx + c = 0
Given,
⇒
α = βn
c
c
αβ =
⇒ βn + 1 =
a
a
 c
β= 
 a
⇒
[Q γ 2 + qγ + 1 = 0 and δ 2 + qδ + 1 = 0]
= (q2 − p2)(γδ ) = q2 − p2
α =α + 1
57.
1/( n + 1 )
β2 = β + 1
a n = pα n + qβ n
It must satisfy ax2 + bx + c = 0
 c
a 
 a
i.e.
⇒
c1/( n + 1)
a1/( n + 1)
⇒
2 /( n + 1 )
 c
+ b 
 a
1 /( n + 1 )
= p(α n − 1 + α n − 2) + q(β n − 1 + β n − 2)
+ c=0
a . c2/( n + 1) b. c1/( n + 1)
+ 1/( n + 1) + c = 0
a 2/( n + 1)
a
 a . c1/( n + 1)
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
= an − 1 + an − 2
∴ a12 = a11 + a10
α=
58.
= 2a 2 + a1
= 3a1 + 2a 0
28 = p(3α + 2) + q(3β + 2)
3 5
3

28 = ( p + q) + 2 + ( p − q)

2

 2 
and γ , δ are the roots of x + rx + s = 0
2
and
α + β = − p, αβ = q
∴
γ + δ = − r, γδ = s
and
Now, (α − γ )(α − δ )(β − γ )(β − δ )
= [α 2 − (γ + δ ) α + γδ ][β 2 − (γ + δ ) β + γδ ]
= (α 2 + rα + s)(β 2 + rβ + s)
= (αβ )2 + r (α + β )αβ + s(α 2 + β 2) + αβr 2 + rs(α + β ) + s2
= q2 − rqp + s( p2 − 2q) + qr 2 − rsp + s2
= (q − s) − rqp − rsp + sp + qr
2
2
2
55. 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
⇒
⇒
2
1
3
+
+
=0
log a x 1 + log a x 2 + log a x
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
⇒
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
56. Since, α + β = − p, αβ = 1 and γ + δ = − q, γδ = 1
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+ 5
1− 5
,β =
2
2
a 4 = a3 + a 2
54. Since, α , β are the roots of x2 + px + q = 0
∴
[Q γ δ = 1]
2
⇒
p− q =0
7
( p + q) × = 28
2
p+ q =8
⇒
p= q =4
∴
p + 2q = 12
Topic 2 Common Roots
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)
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. Given equations are x2 + 2x + 3 = 0
…(i)
ax + bx + c = 0
…(ii)
and
2
Since, Eq. (i) has imaginary roots, so Eq. (ii) will also
have both roots same as Eq. (i).
42 Theory of Equations
a b c
= =
1 2 3
Thus,
Hence, a : b : c is 1 : 2 : 3.
⇒
x = α β α , α β β are the roots
⇒
x = α 2β , αβ 2 are the roots
Divide the Eq. (i) by a3 , we get
3. If a1x + b1x + c1 = 0
2
x2 +
b c
⋅ ⋅ x+
a a
3
 c
  =0
 a
and a 2x2 + b2x + c2 = 0
have a common real root, then
⇒
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
⇒
(x − α 2β )(x − αβ 2) = 0
(a1c2 − a 2c1 ) = (b1c2 − b2c1 ) (a1b2 − a 2b1 )
2
x + bx − 1 = 0
 have a common root.
x2 + x + b = 0 
2
∴
⇒
(1 + b)2 = (b2 + 1) (1 − b)
⇒
b2 + 2b + 1 = b2 − b3 + 1 − b
⇒
b + 3b = 0
∴
b (b2 + 3) = 0
⇒ x = α 2β , αβ 2 which is the required answer.
Alternate Solution
⇒ x=
b = 0, ± 3 i
⇒
4. Given equations are x2 + ax + b = 0 and
⇒ x=
x2 + bx + a = 0 have common root
On subtracting above equations, we get
⇒ x=
(a − b) x + (b − a ) = 0
x=1
⇒
1 + a + b =0
⇒
a + b = −1
5. Since, (x − r ) is a factor of the polynomial
f (x) = a nxn + a n − 1xn − 1 + ... + a 0
Then, x = r is root of f ′ (x) = 0 repeated (m − 1) times.
Hence, statement is false.
1. Given, α , β are the roots of (x − a )(x − b) − c = 0
⇒
− (b/a )(c/a ) ±
(α + β ) (α β ) ±
(b/a )2(c/a )2 − 4(c/a )3
2
(α + β )2 (α β )2 − 4(α β )3
2
α + β + α − β α + β − α + β 
,


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
(x − a )(x − b) = (x − α )(x − β ) + c
1. Given parabola is y2 = 4x,
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. Since, ax2 + bx + c = 0 has roots α and β.
and
2 a3
⇒ x = αβ
⇒
(x − a )(x − b) − c = (x − α ) (x − β )
⇒ a , b are the roots of equation (x − α )(x − β ) + c = 0
⇒
(abc)2 − 4 . a3 ⋅ c3
(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


⇒
Topic 3 Transformation of Roots
⇒
− abc ±
⇒ x=
∴ x = 1 is the common root.
⇒
a3 x2 + abcx + c3 = 0
Since,
3
α + β = − b /a
αβ = c / a
Y
Now, a3 x2 + abcx + c3 = 0
…(i)
Q(9,6)
X (t 2,2t)
2
y 2=4x
On dividing the equation by c , we get
a3 2 abcx c3
x + 2 + 2 =0
c
c
c2
X′
X
O
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
P(4,–4)
Y′
∴Area of ∆PXQ =
1
2
4 −4 1
t 2 2t 1
9
6
1
Theory of Equations 43
1
[4(2t − 6) + 4(t 2 − 9) + 1(6t 2 − 18t ]
2
1
= |[8t − 24 + 4t 2 − 36 + 6t 2 − 18t ]|
2
= |5t 2 − 5t − 30| = |5(t + 2) (t − 3)|
3. According to given information, we have the following
=
graph
Y
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. 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.
O
2
3
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)
From the values of m obtained in (i), (ii), (iii) and (iv), we
get m ∈ (4, 5).
2
3
–∞
–4
In both cases f (0). f (2) < 0 and f (2) f (3) < 0
⇒
(c − 4) [4 (c − 5) − 4c + (c − 4)] < 0
⇒
(c − 4) (c − 24) < 0
⇒
c ∈ (4, 24)
+
–
4
49/4
⇒
5 29/5
∞
Given, a ∈ R and equation is
… (i)
+
−3 { x − [x]}2 + 2 { x − [x]} + a 2 = 0
Let t = x − [x], then equation is
−3 t 2 + 2 t + a 2 = 0
24
–
4
and lie between 0 ≤ { X } < 1 and then solve it.
Similarly, f (2) ⋅ f (3) < 0
⇒ [4 (c − 5) − 4c + (c − 4)]
[9(c − 5) − 6c + (c − 4)] < 0
⇒
(c − 24) (4c − 49) < 0
+
2
4. Put t = x − [x] = { X }, which is a fractional part function
f (0) f (2) < 0
Now, consider
X
5
1
29

(iv) f (5) > 0 ⇒ 25 − 5m + 4 > 0 ⇒ 5m < 29 ⇒m ∈  − ∞, 

5
Or
O
O
t=
⇒
Q
t = x − [x] = { X }
∴
0 ≤ t ≤1
+
0≤
24
 49

c ∈
, 24
4

From Eqs. (i) and (ii), we get

 49
, 24
c ∈

4
∴Integral values of c are 13, 14, ……, 23.
Thus, 11 integral values of c are possible.
…(ii)
1 ± 1 + 3a 2
3
[fractional part]
1 ± 1 + 3a 2
≤1
3
Taking positive sign, we get
0≤
⇒
⇒
1 + 1 + 3a 2
<1
3
[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)
44 Theory of Equations
∴
⇒
⇒
5. As we know, ax2 + bx + c > 0 for all x ∈ R, iff
a > 0 and D < 0.
Given equation is x2 + 2ax + (10 − 3a ) > 0, ∀ x ∈ R
D <0
Now,
⇒
4a − 4(10 − 3a ) < 0
⇒
4(a 2 + 3a − 10) < 0
⇒
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 )
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.
2
(a + 5)(a − 2) < 0 ⇒ a ∈ (−5, 2)
Thus, g (x) > 0, ∀ x ∈ R.
y = (x – a)(x – b) –1
6.
α a
9. Given,
x2 + (a − b) x + (1 − a − b) = 0 has real and unequal roots.
⇒
D >0
2
⇒
(a − b) − 4(1) (1 − a − b) > 0
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, ∞ ).
⇒ 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
7. Let f (x) = x − 2ax + a + a − 3
2
2
Since, both root are less than 3.
⇒
α < 3, β < 3
⇒
Sum, S = α + β < 6
α+β
<3
2
2a
<3
2
⇒
⇒
⇒
⇒
(4 − 2a )2 − 4 (a 2 + 4a − 4) < 0
⇒
16 − 16a + 4a 2 − 4a 2 − 16a + 16 < 0
⇒
−32a + 32 < 0
10.
⇒
a >1
Y
a<0
a <3
…(i)
y = ax2 + bx + c
Again, product, P = αβ
⇒
P <9
⇒ αβ < 9
⇒
a2 + a − 3 < 9
⇒
a + a − 12 < 0
⇒
–1
α
1
0
a>0
(a − 3) (a + 4) < 0
−4 < a < 3 …(ii)
Again, D = B − 4 AC ≥ 0
2
α
y = ax2 + bx + c
β
3
α
⇒ (−2a ) − 4 ⋅ 1 (a + a − 3) ≥ 0
2
⇒
9 − 6a + a + a − 3 > 0
⇒
a 2 − 5a + 6 > 0
⇒
(a − 2) (a − 3) > 0
…(iv)
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
a >0
b2 − 4ac < 0
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.
2
a ∈ (−∞ , 2) ∪ (3, ∞ )
0 1
…(iii)
From Eqs. (i), (ii), (iii) and (iv), we get
⇒
and
–1
2
⇒
4a 2 − 4a 2 − 4a + 12 ≥ 0
⇒
−4a + 12 ≥ 0 ⇒ a ≤ 3
Again,
a f (3) > 0
⇒ 1 [(3)2 − 2a (3) + a 2 + a − 3] > 0
∴
Y
2
⇒
⇒
X
β
…(i)
a (a − b + c) < 0
and
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
a (a + b + c) < 0
1+
b
c
+ <0
a a
Theory of Equations 45
∴The above inequality holds, iff
11. Since, x2 − 3x + 2 > 0 and x2 − 2x − 4 ≤ 0
⇒
(x − 1) (x − 2) > 0
⇒
x − 2x + 1 ≤ 5
(sin x − 1)2 + 4 = 2 = 2 sin 2 y
(x < 1 or x > 2) and (1 − 5 ≤ x ≤ 1 + 5 )
⇒ sin x = 1 and sin 2 y = 1
⇒ sin x = |sin y|
x ∈ [1 − 5 , 1) ∪ [1 + 5 , 2)
∴
12.
and
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
⇒
16 − 32k + 16 (k2 − k + 1) ≥ 0
⇒
k − 3k + 2 ≥ 0
⇒
(k − 2) (k − 1) ≥ 0
⇒
k ≤1
Given equation is
| x − 2| + x ( x − 4) + 2 = 0
⇒ | x − 2| + x − 4 x + 4 = 2
⇒ (| x − 2|)2 + | x − 2| − 2 = 0
2
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.
sin 2 x − 2sin x + 5
⋅
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
On differentiating w.r.t. x, we get
f ′ (x) = 6x2 + 3 > 0, ∀ x ∈ R
⇒ f (x) is strictly increasing function.
⇒ f (x) = 0 has only one real root, so two roots are not
possible.
5. Since, α is a root of a 2x2 + bx + c = 0
⇒
1
sin 2 y
a 2x2 − bx − c = 0
⇒
a 2β 2 − bβ − c = 0
Let
f (x) = a 2x2 + 2bx + 2c
∴
f (α ) = a 2α 2 + 2bα + 2c
⇒2
⋅2
⇒2
(sin x − 1 )2 + 4
≤ 22sin
−2sin 2 y
2
y
(sin x − 1)2 + 4 ≤ 2 sin 2 y
[if a > 1 and a ≤ a ⇒ m ≤ n]
m
Q Range of (sin x − 1) + 4 is [2, 2 2 ]
and range of 2 sin 2 y is [0, 2].
... (ii)
= a 2α 2 − 2a 2α 2 = − a 2α 2
[from Eq. (i)]
and
f (β ) = α 2β 2 + 2bβ + 2c
= a 2β 2 + 2a 2β 2 = 3a 2β 2 [from Eq. (ii)]
6. Let
≤1
2
... (i)
f (α ) f (β ) < 0
f (x) must have a root lying in the open interval (α , β ).
∴
α < γ <β
≤1
4
(sin x − 1 )2 + 4
a 2α 2 + bα + c = 0
and β is a root of
⇒
2. Given, inequality is
⇒
Key Idea Reduce the given equation into quadratic equation.
⇒ | x − 2| + ( x − 2)2 = 2
(iii) f (4) ≥ 0
2
3.
[from the options]
n
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).
∴ By Rolle’s theorem, f ′ (α ) = 0 for 0 < α < 1
46 Theory of Equations
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
Case I When x ≤ 0 ⇒ x12 > 0, − x9 > 0, x4 > 0, − x > 0
…(i)
∴
x12 − x9 + x4 − x + 1 > 0, ∀ x ≤ 0
Case II When 0 < x ≤ 1
x9 < x4 and x < 1 ⇒ − x9 + x4 > 0
∴
4
9
(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
…(iii)
8. Consider,
1
f (x) = ∫ (1 + cos 8 x)(ax2 + bx + c) dx
0
Obviously, f (x) is continuous and differentiable in the
interval [1, 2].
f (1) = f (2)
[given]
∴ By Rolle’s theorem, there exist atleast one point
k ∈ (1, 2), such that f ′ (k) = 0.
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, 4)
f ′ (k) = 0
–1
⇒ (1 + cos k)(ak + bk + c) = 0
8
⇒
2
(1, – 4)
ak2 + bk + c = 0
[as (1 + cos 8 k) ≠ 0]
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.
∴ x = k is root of ax2 + bx + c = 0,
where
k ∈ (1, 2)
9. Given, x1 and x2 are roots of αx2 − x + α = 0.
∴
Also,
⇒
or
⇒
⇒
x1 + x2 =
Also,
⇒
11. Given, f (x) = 4x3 + 3x2 + 2x + 1
f ′ (x) = 2 (6x2 + 3x + 1)
⇒
5α 2 − 1 > 0 or ( 5 α − 1) ( 5 α + 1) > 0
–1/√5
∴
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
+
–
+
1/√5
1  1


α ∈  −∞ , −
, ∞
 ∪

5  5 
D >0
 1 1
1 − 4α > 0 or α ∈  − , 
 2 2
2
1
Now,
f ′ (x) = (1 + cos 8 x)(ax2 + bx + c)
Now,
+
–
+
4
From Eqs. (i), (ii) and (iii), the above equation holds for
all x ∈ R.
Also,
(ii) Number of roots are taken out from the curve traced.
…(ii)
Case III When x > 1 ⇒ x > x and x > x
∴
x12 − x9 + x4 − x + 1 > 0, ∀ x > 1
12
PLAN
(i) Concepts of curve tracing are used in this question.
and 1 − x > 0
x − x + x − x + 1 > 0, ∀ 0 < x ≤ 1
9
10.
Let y = x5 − 5x
Here, three cases arises:
12
From Eqs. (i) and (ii), we get
 1 −1   1 1 
, 
α ∈ − ,  ∪ 
 2 5   5 2
…(i)
…(ii)
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
1
 3
f (x) changes its sign in  − , − 
 4
2
 3 1
Hence, f (x) = 0 has a root in  − , −  .
 4 2
Theory of Equations 47
12.
1/ 2
∫0
t
3/ 4
0
0
f (x) dx < ∫ f (x) dx < ∫
16. For two distinct roots, 1 + ln k < 0 (k > 0)
f (x) dx
ln k < −1 ⇒
Now, ∫ f (x) dx = ∫ (1 + 2x + 3x2 + 4x3 ) dx
= x + x2 + x3 + x4
Hence,
k<
1
e
 1
k ∈ 0, 
 e
17. Let f (x) = (x − a ) (x − c) + 2 (x − b) (x − d )
f (a ) = + ve
Here,
1
f (b) = − ve
f (c) = − ve
1 t 3
2
4
1/ 2
∫0
⇒
15 3
f (x) dx =
> ,
16 4
3/ 4
∫0
f (d ) = + ve
530
f (x) dx =
<3
256
∴ There exists two real and distinct roots one in the
interval (a , b) and other in (c, d ).
Hence, statement is true.
13. As, f ′′ (x) = 2 (12x + 3)
18. Let f (x) = 4x3 − 3x − p
1
and
4
1
f ′ (x) < 0, when x < − .
4
f ′ (x) > 0, when x > −
Now,
4 3
 1
 1
 1
f   =4   −3   − p= − − p
 2
 2
 2
8 2
= − (1 + p)
∴ It could be shown as
f (1) = 4(1)3 − 3(1) − p = 1 − p
1
 1
f   . f (1) = − (1 + p)(1 − p)
 2
⇒
= ( p + 1)( p − 1) = p2 − 1
S
–3
4
…(i)
3
–1
2
1
2
Which is ≤ 0 , ∀ p ∈ [−1, 1].
3
4
∴ f (x) has atleast one root in
14. Let y = x intersect the curve y = kex at exactly one point
when k ≤ 0 .
Y
1 
,1 .
2 
Now, f ′ (x) = 12x2 − 3 = 3 (2x − 1) (2x + 1)
3
1 
1
1 
=  x −   x +  > 0 in
,1
2 
4
2 
2
⇒ f (x) is an increasing function in [1 /2, 1]
X
X′
Therefore, f (x) has exactly one root in [1 /2, 1] for any
p ∈ [−1, 1].
Now, let x = cos θ
15. Let
f (x) = kex − x
f ′ (x) = kex − 1 = 0
⇒
⇒
f ′ ′ (x) = ke
⇒
[ f ′ ′ (x)] x = − ln k = 1 > 0
f (− ln k) = 1 + ln k
⇒
For one root of given equation
⇒
1


 π
⇒ θ ∈ 0,
 3 
4 cos3 θ − 3 cos θ = p ⇒ cos 3θ = p
x = − ln k
Hence,
1
,
2
From Eq. (i),
x
∴
x∈
∴
Y′
1 + ln k = 0
1
k=
e
⇒
Download Chapter Test
http://tinyurl.com/y67a3jgc
or
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)
(a)
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
2. If 19th term of a non-zero AP is zero, then its (49th
term) : (29th term) is
(b)
1
1
+
m n
(d) 0
Analytical and Descriptive Question
(2019 Main, 11 Jan II)
ai > 0, ∀ i, then show that
1
1
+
+ ...
a1 + a 2
a 2 + a3
(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
4. If Tr is the r th term of an AP, for r = 1, 2, 3, .... . If for
some positive integers m and n, we have Tm =
Tn =
(c) 1
5. If a1 , a 2 ..... , a n are in arithmetic progression, where
(b) 76
(d) 38
(a) 1 : 3
(c) 2 : 1
1
mn
1
, then Tmn equals
m
1
and
n
(1998, 2M)
even number, then the number of odd integers among
them is odd.
(1985, 1M)
Integer Answer Type Question
7. 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
(b) 280
(c) 120
(d) 150
2. Let S n denote the sum of the first n terms of an AP. If
S 4 = 16 and S 6 = − 48, then S10 is equal to
(2019 Main, 12 April I)
(a) − 260
(b) − 410
(c) − 320
(d) − 380
3. For x ∈ R, let [x] denote the greatest integer ≤ x, then the
sum of the series
1   1
2 
 1  1
 1 99 
− 3  + − 3 − 100  + − 3 − 100  + … + − 3 − 100  is
(2019 Main, 12 April I)
(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
(c) –25
(b) –36
(d) –35
5. Let the sum of the first n terms of a non-constant AP
n (n − 7)
A, where A is a constant.
2
If d is the common difference of this AP, then the
ordered pair (d , a50 ) is equal to
(2019 Main, 9 April I)
a1 , a 2, a3 .....be 50n +
(a) (A, 50 + 46A)
(c) (50, 50 + 46A)
(b) (50, 50 + 45A)
(d) (A, 50 + 45A)
Sequences and Series 49
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
30
7. Let a1 , a 2, ..... a30 be an AP, S = ∑ ai and
i =1
T=
∑ a( 2 i − 1). If a5 = 27 and S − 2T = 75,
15. Let p and q be the roots of the equation x2 − 2x + A = 0
i =1
(2019 Main, 9 Jan I)
(a) 42
(c) 52
(b) 57
(d) 47
8. Let bi > 1 for i = 1, 2, ... , 101 . Suppose log e b1 , log e b2,
... , 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
of squares of these n terms is
n (4n 2 − 1) c2
6
n (4n 2 − 1) c2
(c)
3
(2009)
n (4n 2 + 1) c2
3
n (4n 2 + 1) c2
(d)
6
(b)
is equal to the sum of the first n terms of the AP series
57, 59, 61,..., then n equals
(2001, 1M)
(b) 12
(d) 13
11. If S n = ∑
k( k + 1 )
(−1) 2 k2. Then, S
n
(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, . . .
12. The sum V1 + V 2 + ... + V n is
1
n (n + 1) (3n 2 − n + 1)
12
1
(c) n (2n 2 − n + 1)
2
(a)
1
n (n + 1) (3n 2 + n + 2)
12
1
(d) (2n3 − 2n + 3)
3
(b)
13. Tr is always
(a) an odd number
(c) a prime number
or 5 is ……
(1984, 2M)
18. The fourth power of the common difference of an
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)
x3 − x2 + βx + γ = 0 are in AP. Find the intervals in
(1996, 3M)
which β and γ lie.
20. The interior angles of a polygon are in arithmetic
progression. The smallest angle is 120° and the common
difference is 5°. Find the number of sides of the polygon.
21. Suppose that all the terms of an arithmetic progression
(2013 Adv.)
(a) 1056
(c) 1120
17. The sum of integers from 1 to 100 that are divisible by 2
(1980, 3M)
can take value(s)
2
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
Integer Answer Type Questions
Objective Question II
(One or more than one correct option)
4n
16. The sum of the first n terms of the series
19. The real numbers x1 , x2, x3 satisfying the equation
10. If the sum of the first 2n terms of the AP series 2,5,8,...,
(a) 10
(c) 11
and let r and s be the roots of the equation
x2 − 18x + B = 0. If p < q < r < s are in arithmetic
progression, then A = … and B = … .
(1997, 2M)
Analytical & Descriptive Questions
9. If the sum of first n terms of an AP is cn 2, then the sum
(a)
(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 = ...
Fill in the Blanks
15
then a10 is equal to
14. Which one of the following is a correct statement ?
(b) an even number
(d) a composite number
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.)
23. Let a1 , a 2, a3 , K , a100 be an arithmetic progression with
p
a1 = 3 and S p =
∑
ai , 1 ≤ p ≤ 100. For any integer n with
i =1
1 ≤ n ≤ 20, let m = 5n. If
is equal to ……
Sm
does not depend on n, then a 2
Sn
(2011)
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
50 Sequences and Series
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)
2
(b) a
3
(a) 5a
7
(d) a
3
(c) a
2. If three distinct numbers a , b and c are in GP and the
equations ax2 + 2bx + c = 0 and dx2 + 2ex + f = 0 have a
common root, then which one of the following
statements is correct?
(2019 Main, 8 April II)
(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
(d) 24
(b) 2(52 )
(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
(2019 Main, 9 Jan I)
(c) −2
(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
(2016 Main)
7
(d)
4
(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)
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
values of p and q respectively are
(2001, 1M)
(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)
(a) 53
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
consecutive terms of a GP.
(1983, 2M)
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)
Topic 4 Sum of n Terms & Infinite Terms of a GP
Objective Questions I (Only one correct option)
20
1. The sum
∑k
k =1
(a) 2 −
(c) 2 −
11
19
2
3
217
2
1
is equal to
2k
(b) 1 −
(d) 2 −
2. Let S n = 1 + q + q2 + K + qn and
(2019 Main, 8 April II)
11
220
21
220
n
 q + 1  q + 1
 q + 1
Tn = 1 + 
 +
 +K+
 , where q is a
 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
(b) 202
(d) 299
Sequences and Series 51
3. The sum of an infinite geometric series with positive
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)
(a)
4
9
(b)
2
3
(c)
2
9
(d)
1
3
4. Three positive numbers form an increasing GP. If the
middle term in this GP is doubled, then new numbers
are in AP. Then, the common ratio of the GP is
(a) 2 + 3
(c) 2 − 3
(b) 3 +
(d) 2 +
2
3
(2014 Main)
1
8
2
7
9
then k is equal to
(a)
121
10
(b)
441
100
9
(2014 Main)
(c) 100
(d) 110
6. The sum of first 20 terms of the sequence
0.7, 0.77, 0.777,… , is
7
(a)
(179 − 10− 20 )
81
7
(c)
(179 + 10− 20 )
81
(2013 Main)
belongs to
(b) −10 < x < 0 (c) 0 < x < 10
(d) x > 10
8. Consider an infinite geometric series with first term a
and common ratio r. If its sum is 4 and the second term
(2000, 2M)
is 3 /4, then
(b) a = 2, r = 3 /8
(d) a = 3, r = 1/4
1
2
9. Sum of the first n terms of the series +
(b) 8
(c) 9
(d) 10
Analytical & Descriptive Questions
3
n
3
3
3
3
11. 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.
12. 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
respectively, then
, , ,... ,
2 3 4
n+1
find the values of S12 + S 22 + S32 + ... + S 22n − 1. (1991, 4M)
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)
Integer Answer Type Questions
14. Let S k , where k = 1, 2, , K , 100, denotes the sum of the
infinite geometric series whose first term is
3 7 15
+ +
+ ...
4 8 16
is equal to
(1988, 2M)
(a) 2n − n − 1 (b) 1 − 2−n
(a) 7
13. The sum of the squares of three distinct real numbers,
(2004, 1M)
(a) a = 4/7, r = 3/7
(c) a = 3/2, r = 1/2
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
1 sq cm?
(1999, 3M)
common ratios are
7
(b)
(99 − 10− 20 )
9
7
(d)
(99 + 10− 20 )
9
7. An infinite GP has first term x and sum 5, then x
(a) x < − 10
10. Let S1 , S 2, ... be squares such that for each n ≥ 1 the
2
5. If (10) + 2(11) (10) + 3(11) (10) + ... + 10(11) = k(10) ,
9
Objective Question II
(One or more than one correct option)
(c) n + 2 − n − 1 (d) 2n + 1
the
common
1002
+
100 !
ratio
is
1
.
k
Then,
the
k −1
and
k!
value
of
100
∑|(k2 − 3k + 1) Sk |is ……
k =1
(2010)
Topic 5 Harmonic Progression (HP)
Objective Questions I (Only one correct option)
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
(b) 23
(c) 24
(d) 25
2. If the positive numbers a , b, c,d are in AP. Then,
abc, abd , acd , bcd are
(a) not in AP / GP / HP
(c) in GP
(2001, 1M)
(b) in AP
(d) in HP
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
4. If x > 1, y > 1, z > 1 are in GP, then
1
are in
1 + ln z
(a) AP
(b) HP
(d) 6
1
1
,
,
1 + ln x 1 + ln y
(1998, 2M)
(c) GP (d) None of these
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
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.
Statement II The numbers b1 , b2, b3 , b4 are in HP.
(2008, 3M)
52 Sequences and Series
8. Let a and b be positive real numbers. If a , A1 , A2, b are
Fill in the Blank
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
=
H 1H 2 H 1 + H 2
9ab
(2002, 5M)
6. If cos (x − y), cos x and cos (x + y)’ are in HP. Then
 y
cos x ⋅ sec   = K .
 2
(1997C, 2M)
Analytical & Descriptive Questions
9.
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, −
c
form a GP.
2
(2003, 4M)
(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)
7. The sum of the following series
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 )
2. The sum of series
+
12
12 + 22
7 × (13 + 23 + 33 )
+ .......... + upto 10th term, is
+
12 + 22 + 32
(2019 Main, 10 April I)
(a) 680
(c) 660
3.
(b) 600
(d) 620
(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
. If S12 + S 22 + ... + S10
A,
=
k
12
(2019 Main, 12 Jan I)
then A is equal to
5. Let S k =
(a) 156
(c) 283
(b) 301
(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)
1
(a)
2
1
(c)
4
(b) 1
(d)
(2019 Main, 9 Jan II)
(a) 7510
(c) 7830
m+ n
6mn
(b) 7820
(d) 7520
12
8. Let a1 , a 2, a3 , …, a 49 be in AP such that
∑ a 4k + 1 = 416
k=0
2
and a 9 + a 43 = 66. If a12 + a 22 + … + a17
= 140 m, then m
(2018 Main)
is equal to
(a) 66
The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 +... upto
(2019 Main, 9 April II)
11th term is
(a) 915
9 (12 + 22 + 32) 12 (12 + 22 + 32 + 42)
+
7
9
2
2
2
15 (1 + 2 + ... + 5 )
+
+ ... up to 15 terms is
11
1+6+
(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
 3
 2
 1
2
1  + 2  + 3  + 4 +
 5
 5
 5
m is equal to
(a) 102
(b) 101
2
16
 4
m, then
4  + K, is
 5
5
(2016 Main)
(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
(2015)
between l and n, then G14 + 2G24 + G34 equals
(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 +
tan 2 α
than or equal to
(a) 2 tan α
(b) 1
x2 + x
(d) l2m2n 2
of
the
series
(2015)
(d) 192
is always greater
(2003, 2M)
(c) 2
(d) sec2 α
Sequences and Series 53
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
relation
(2000, 2M)
(5 + 2 ) x2 − (4 + 5 ) x + 8 + 2 5 = 0 is
(b) 4
(c) 6
(1999, 2M)
(d) 8
17. The product of n positive numbers is unity, then their
sum is
(d) never less than n
expression (b + c − a ) (c + a − b) (a + b − c) − abc is
(a) positive
(c) non-positive
(b) negative
(d) non-negative
(1991, 2M)
integer, then
xi2
i =1
(c) n ∑
i =1
are equal and their nth terms are a , b and c
(1988, 2M)
respectively, then
(a) a = b = c
(b) a ≥ b ≥ c
(c) a + c = b
(d) ac − b2 = 0
xi2
 n 
<  ∑ xi 


i =1 
(1982, 1M)
2
 n 
≥ n  ∑ xi 


i =1 
n
(b) n ∑
i =1
xi2
 n 
≥  ∑ xi 


i =1 
means between any two positive numbers, then
y3 + z3
= ...
xyz
(1997C, 2M)
positive numbers are in the ratio 4 : 5. Then, the two
numbers are in the ratio… .
(1992, 2M)
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 >...
25. If x be is the arithmetic mean and y, z be two geometric
26. If the harmonic mean and geometric mean of two
19. If x1 , x2,... , xn are any real numbers and n is any positive
n
(1999, 3M)
(b) a (100) > 100
(d) a (200) > 100
Fill in the Blanks
(b) divisible by n
18. If a , b and c are distinct positive numbers, then the
(a) n ∑
(a) a (100) ≤ 100
(c) a (200)≤ 100
(1991, 2M)
(a) a positive integer
1
(c) equal to n +
n
n
1 1 1
1
+ + + ... + n
, then
2 3 4
(2 ) − 1
24. If the first and the (2n − 1)th term of an AP, GP and HP
(b) 1 ≤ M ≤ 2
(d) 3 ≤ M ≤ 4
16. The harmonic mean of the roots of the equation
(a) 2
23. For a positive integer n let
a (n ) = 1 +
15. If a , b, c are positive real numbers such that
(a) 0 < M ≤ 1
(c) 2 ≤ M ≤ 3
Objective Question II
(One or more than one correct option)
True/False
27. If x and y are positive real numbers and m, n are any
positive integers, then
1
xn ym
> .
(1 + x )(1 + y2m ) 4 (1989, 1M)
2n
28. For 0 < a < x, the minimum value of function
log a x + log x a is 2.
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)
30. Let a1 , a 2,.. be positive real numbers in geometric
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.
31. If p is the first of the n arithmetic means between two
numbers and q be the first on n harmonic means
between the same numbers. Then, show that q does not
2
 n + 1
(1991, 4M)
lie between p and 
 p.
 n − 1
32. If a > 0, b > 0 and c > 0, then prove that
(a + b + c)
1 1 1
+ + ≥9
a b c
(1984, 2m)
54 Sequences and Series
Integer Answer Type Question
33. Let a,b,c be positive integers such that b /a is an integer. If a,b,c are in geometric progression and the arithmetic mean of
a,b,c is b + 2, then the value of
a 2 + a − 14
is
a+1
(2014 Adv.)
−5
−4
−3
34. The minimum value of the sum of real numbers a , a , 3a , 1, a and a with a > 0 is ……
8
10
(2011)
Answers
Topic 4
Topic 1
1. (b)
6. False
2. (d)
7. (6)
3. (b)
4. (c)
1. (a)
5. (a)
9. (c)
2. (c)
6. (c)
10. (c)
3. (b)
7. (c)
11. (a, d)
13. (d)
14. (b)
15. (A = – 3 , B = 77)
n 2(n + 1 ) 
16. 
 17. (3050)
2


1
 1 

19. β ∈ – ∞,  and γ ∈ – , ∞  20. (9)
3
 27 

22. (5)
23. (9)
24. (0)
4. (c)
8. (b)
12. (b)
3. (b)
6. (c)
2.
6.
10.
14.
11. (7)
7. (c)
9. (c)
10. (b, c, d)
1
12. (2n )(2n + 1 )( 4n + 1 ) – 1
6
14. (4)
Topic 5
1. (d)
5. (c)
2. (d)
6. ± 2
3. (d)
4. (b)
9.(i) a = 1, b = 9
12. 29
21. (9)
Topic 3
(c)
(c)
(d)
(b)
2. (a)
5. (c)
8. (d)
Topic 2
1.
5.
9.
13.
1. (a)
4. (d)
(b)
3. (b)
(b)
7. (b)
(a)
11. (b)
(a = 5 ) (b = 8 ) (c = 12 )
4.
8.
12.
15.
(d)
(c)
(a)
Yes, infinite
Topic 6
1.
5.
9.
13.
17.
21.
25.
34.
(a)
(d)
(b)
(a)
(d)
(a)
2
(8)
2.
6.
10.
14.
18.
22.
26.
(c)
(c)
(b)
(a)
(b)
(b)
4:1
3.
7.
11.
15.
19.
23.
27.
(b)
(b)
(b)
(a)
(b)
(a, d)
False
4.
8.
12.
16.
20.
24.
28.
(b)
(c)
(b)
(b)
(c)
(a, b, d)
False
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 )
= 2 × 38 = 76
[from Eq. (i)]
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)]
3. We have,
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
Sequences and Series 55
⇒ 15a = 3b, 3b = 5c and 5c = 15a
∴ 15a = 3b = 5c
a b c
⇒
= = =λ
1 5 3
⇒
(say)
a = λ , b = 5λ , c = 3λ
∴b, c, a are in AP.
1
n
1
Tn = a + (n − 1) d =
m
Tm = a + (m − 1) d =
4. Let
and
…(i)
…(ii)
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
= a + (n − 1) d + (mn − n ) d
(m − 1)
1
1
= Tn + n (m − 1)
=
+
=1
mn m
m
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 a n + a1
d
 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. 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
Topic 2 Sum of n Terms of an AP
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
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.
∴
S 4 = 2[2a + 3d ] = 16
(given)
n


Q S n = 2 [2a + (n − 1)d ]
… (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
3. Given series is
 1
− 3  +
1 
 1
− 3 − 100  +
2 
 1
− 3 − 100  + K ... +
 1 99 
− 3 − 100 
[where, [x] denotes the greatest integer ≤ x]
Now,
1 
 1  1
− 3 , − 3 − 100 ,
2 
 1
 1 66 
− 3 − 100 , …+ − 3 − 100 
all the term have value − 1
 1 67   1 68 
 1 99 
, − −
, …, − −
all the term
and − −


 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 68 
− 3 − 100  + K +
= − 2 − 2 − 2 − 2 K 33 times
= (−2) × 33 = −66
 1 99 
− 3 − 100 
56 Sequences and Series
 1
∴ −  +
 3
1 
 1
− 3 − 100  +
2 
 1
− 3 − 100  + K +
 1 99 
− 3 − 100 
= (− 67) + (− 66) = − 133.
Alternate Solution
Q [− x] = − [x] − 1, if x ∉Integer,
1 
2

and [x] + x +  + x +  + K +
n 
n

n ∈N.
So given series
1 
 1  1
− 3  + − 3 − 100  +
n − 1

x + n  = [nx],
2 
 1
− 3 − 100  + … K +
 − 1 99 
 3 − 100 
 1
2  
99 

 1
− 1 + K +  −  +
+ −  +
− 1


 3 100  

 3 100 

1
= (− 1) × 100 −  × 100 = − 100 − 33 = − 133.

3
4. Let first three terms of an AP as a − d, a, a + d.
So,
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.
a1 −
d
7
= 50 − A
2
2
A
7
= 50 − A
2
2
⇒
a1 = 50 − 3 A
So
a50 = a1 + 49d
[Qd = A]
= (50 − 3 A ) + 49 A
[Q d = A]
= 50 + 46 A
Therefore, (d , a50 ) = ( A , 50 + 46 A )
6. Clearly, the two digit number which leaves remainder 2
when divided by 7 is of the form N = 7k + 2 [by Division
Algorithm]
For,
k = 2, N = 16
k = 3, N = 23
M
M
k = 13, N = 93
∴ 12 such numbers are possible and these numbers
forms an AP.
Now,
S=
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
Key Idea Use the formula of sum of first n terms of AP, i.e
∴13 such numbers are possible and these numbers also
forms an AP.
Sn =
Now,
n
[2 a + ( n − 1) d ]
2
Given AP, is
a1 , a 2, a3 ,… having sum of first n-terms
=
⇒
d = A and a1 −
⇒
  1
 1 
1 

=  −   − 1 +  −  +
− 1
  3 100 
 3 

5
On comparing corresponding term, we get
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
S′ =
13
[12 + 96] = 702
2
n

QS n = ( a +

2
Total sum = S + S′ = 654 + 702 = 1356

l )

7. We have, S = a1 + a 2 + … + a30
…(i)
= 15[2a1 + 29d ]
(where d is the common difference)
n


Q S n = 2 [2a + (n − 1)d ]
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
and
Sequences and Series 57
⇒
Now,
d =5
a10 = a5 + 5d
= 27 + 25 = 52
⇒
∴
11.
6n + 1 = 57 + n − 1 ⇒ 5n = 55
n = 11
Convert it into differences and use sum ofn terms of an AP,
n
S n = [2a + ( n − 1 )d ]
2
PLAN
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
Also, a1 , a 2, ... , a101 are in AP.
Given,
a1 = b1 and a51 = b51
⇒
i.e.
Now,
⇒
Now,
a1 + 50 D = 2 a1
t = b1 + b2 + K + b51
(251 − 1)
t = b1
2 −1
⇒
∴
or
= (32 − 12) + (42 − 22) + (72 − 52) + (82 − 62) + K
…(iv)
[Q a1 = b1 ]
…(v)
9. Let S n = cn 2
S n − 1 = c (n − 1)2 = cn 2 + c − 2cn
[Q Tn = S n − S n − 1 ]
Tn2 = (2 cn − c)2 = 4c2n 2 + c2 − 4c2n
∴ Sum = Σ Tn2 =
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)]
a101 = a1 + 251 a1 − 2a1 = 251 a1 − a1
a101 < 251 a1 and b101 > 251 a1
b101 > a101
Tn = 2cn − c
= 2{(4 + 6 + 12 + K ) + (6 + 14 + 22 + K )}
144424443 144424443
…(iii)
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 
 and b101 = 2100 a1
a101 = a1 + 100 
∴
50


∴
⋅ k2
n terms
t = a1 (251 − 1)
⇒
⇒
⇒
(−1)
= − (1)2 − 22 + 32 + 42 − 52 − 62 + 72 + 82 + K
[Q a1 = b1 ] …(ii)
s = a1 + a 2 + K + a51
51
=
(2a1 + 50 D )
2
and
∑
k( k + 1)
2
k =1
a1 + 50 D = 250 b1
50
Sn =
4n
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
= 2 [4n 2 + 4n 2 + 2n ] = 4n (4n + 1)
Here, 1056 = 32 × 33, 1088 = 32 × 34,
1120 = 32 × 35, 1332 = 36 × 37
1056 and 1332 are possible answers.
r
1
12. Here, V r = [ 2r + (r − 1) ( 2r − 1)] = ( 2r3 − r 2 + r )
2
2
1
3
2
∴ ΣV r = [ 2 Σr − Σr + Σr ]
2
2
1   n (n + 1)  n (n + 1) ( 2n + 1) n (n + 1) 
= 2 
+

 −
2  
2
6
2


n (n + 1)
=
[ 3n (n + 1) − ( 2n + 1) + 3]
⇒
12
1
=
n (n + 1) ( 3n 2 + n + 2)
12
1
1
13. V r + 1 − V r = (r + 1)3 − r3 − [(r + 1)2 − r 2] +
2
2
= 3r 2 + 2r − 1
∴
Tr = 3r 2 + 2r − 1 = (r + 1) ( 3r − 1)
which is a composite number.
14. Since, Tr = 3r 2 + 2r − 1
and Tr + 1 = 3 (r + 1)2 + 2 (r + 1) − 1
∴
Qr = Tr+1 − Tr = 3 [ 2r + 1] + 2 [1]
⇒
Qr = 6 r + 5
⇒
Qr+ 1 = 6(r + 1) + 5
Common difference = Qr+ 1 − Qr = 6
15. Given, p + q = 2, pq = A
and
r + s = 18, rs = B
and it is given that p, q, r , s are in an AP.
Therefore, let p = a − 3d , q = a − d , r = a + d
and
Since,
s = a + 3d
p<q<r<s
58 Sequences and Series
We have,
d >0
Now,
2 = p + q = a − 3d + a − d = 2a − 4d
⇒
Now, a 2 − 5d 2 = a 2 − 9d 2 + 4d 2
= (a − 3d )(a + 3d ) + (2d )2
a − 2d = 1
= I ⋅ I + I2
…(i)
18 = r + s = a + d + a + 3d
Again,
18 = 2a + 4d
⇒
=I
9 = a + 2d
…(ii)
8 = 4d ⇒ d = 2
x3 = a + d and x1 , x2, x3 be the roots of x3 − x2 + βx + γ = 0
On putting in Eq. (ii), we get a = 5
∴
p = a − 3d = 5 − 6 = − 1
Σα = a − d + a + a + d = 1
⇒
q = a − d =5 −2 =3
and
a = 1 /3
and
s = a + 3d = 5 + 6 = 11
From Eq. (i),
αβγ = (a − d ) a (a + d ) = − γ
2
2
2
n (n + 1)2
=
2
[when n is even] … (i)
When n is odd, 1 + 2 ⋅ 2 + 3 + 2 ⋅ 4 + 5 ... + n
2
2
2
2
…(iii)
3a = 1 ⇒ a = 1 / 3
16. Here, 1 + 2 ⋅ 2 + 3 + 2 ⋅ 4 + 5 + ... upto n terms
2
…(i)
Σαβ = (a − d ) a + a (a + d ) + (a − d ) (a + d ) = β …(ii)
r = a + d =5 + 2 = 7
Therefore, A = pq = − 3 and B = rs = 77
2
[where, I is any integer]
Therefore, P = (I )2 = Integer
19. Since, x1 , x2, x3 are in an AP. Let x1 = a − d , x2 = a and
On subtracting Eq. (i) from Eq. (ii), we get
∴
[given]
= I2 + I2 = I2
2
2
From Eq. (ii), 3a − d 2 = β
2
⇒
3 (1 / 3)2 − d 2 = β
⇒
1 /3 − β = d
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.
= {12 + 2⋅22 + 32 + 2 ⋅ 42 + ... + 2 (n − 1)2} + n 2
 (n − 1) (n )2 
2
[from Eq. (i)]
=
+ n
2


⇒
(n + 1)

n −1
= n2 
+ 1 = n 2

 2
2
⇒
∴ 12 + 2⋅ 22 + 32 + 2 ⋅ 42 + ... upto n terms, when n is odd
n 2 (n + 1)
=
2
17. Integers divisible by 2 are {2,4,6,8,10, ...,100}.
Integers divisible by 5 are {5,10,15, ...,100}.
[from Eq. (i)]
2
1
−β ≥0
[Q d 2 ≥ 0]
3
1
β≤
⇒ β ∈ [− ∞ , 1 / 3]
3
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
1
γ+
= d ⇒γ +
≥0
⇒
27 3
27
Thus, sum of integers divisible by 2
50
=
(2 + 100) = 50 × 51 = 2550
2
⇒
Sum of integers divisible by 5
20
=
(5 + 100) = 10 × 105 = 1050
2
Hence, β ∈ (− ∞ , 1 / 3] and γ ∈ [−1 / 27, ∞ )
Sum of integers divisible by 10
10
=
(10 + 100) = 5 × 110 = 550
2
∴ Sum of integers from 1 to 100 divisible by 2 or 5
= 2550 + 1050 − 550
= 2550 + 500 = 3050
18. Let four consecutive terms of the AP are a − 3d , a − d ,
a + d , a + 3d, which are integers.
Again, required product
P = (a − 3d )(a − d )(a + d )(a + 3d ) + (2d )4
[by given condition]
= (a 2 − 9d 2)(a 2 − d 2) + 16d 4
= a 4 − 10a 2d 2 + 9d 4 + 16d 4 = (a 2 − 5d 2)2
γ ≥ − 1 / 27

 1
, ∞
γ ∈ −

27

⇒
20. Since, angles of polygon are in an AP.
∴ Sum of all angles
= (n − 2) × 180° =
⇒
5n 2 − 125n + 720 = 0
⇒
n 2 − 25n + 144 = 0
⇒
(n − 9) (n − 16) = 0
⇒
n
{2 (120° ) + (n − 1) 5° }
2
n = 9, 16
If n = 9, then largest angle = a + 8d = 160°
Again, if n = 16, the n largest angle
= a + 15d = 120° + 75 = 195°
which is not possible.
[since, any angle of polygon cannot be > 180°]
Hence, n = 9
[neglecting n = 16]
Sequences and Series 59
21. Given,
S7
6
and 130 < t7 < 140
=
S11 11
Topic 3 Geometric Progression (GP)
7
[2a + 6d ]
6
7 (2a + 6d )
2
=
⇒
=6
11
(2a + 10d )
[2a + 10d ] 11
2
⇒
⇒
1.
a = 9d
…(i)
130 < t7 < 140
Also,
⇒
130 < a + 6d < 140
⇒
130 < 9d + 6d < 140
⇒
∴
[from Eq. (i)]
d =9
22. Let number of removed cards be k and (k + 1).
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
∴
23. Given, a1 = 3,m = 5n and a1 , a 2, …, is an AP.
∴
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
∴
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
⇒ a1 , a 2, . . . , a11 are in an AP.
2
11a 2 + 35 × 11d 2 + 10ad
a12 + a 22 + . . . + a11
∴
=
= 90
11
11
⇒ 225 + 35 d 2 + 150 d = 90
⇒ 35 d 2 + 150 d + 135 = 0 ⇒ d = − 3, −
Given,
a2 <
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
130 < 15d < 140
26
28
[since, d is a natural number]
<d<
3
3
⇒
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.
9
7
27
2
9
7
a1 + a 2 + . . . + a11 11
=
[ 30 − 10 × 3] = 0
11
2
Now, the terms 3a, 7b and 15c are the first three terms
of an AP, then
2( 7b) = 3a + 15 c
⇒
14ar = 3a + 15ar 2
⇒
14r = 3 + 15r 2
⇒
15r 2 − 14r + 3 = 0
⇒ 15r 2 − 5r − 9r + 3 = 0
⇒ 5r (3r − 1) − 3(3r − 1) = 0
⇒
(3r − 1) (5r − 3) = 0
3
1
r = or
⇒
5
3
1
 1
as, r ∈ 0, , so r =
 2
3
Now, the common difference of AP = 7b − 3a
2a

7
= 7ar − 3a = a  − 3 = −

3
3
 −2a 
So, 4th term of AP = 3a + 3
 =a
 3 
2. (b) Given, three distinct numbers a , b and c are in GP.
∴
b2 = ac
…(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
2
 b
 b
d  −  + 2e  −  + f = 0
 a
 a
⇒
⇒
⇒
⇒
⇒
∴ d = − 3 and d ≠ −
⇒
[as b = ar, c = ar 2]
[as a ≠ 0]
⇒
db2 − 2eba + a 2f = 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
60 Sequences and Series
11th term = b = A + 10d
13th term = c = A + 12d
Q a, b, c are also in GP
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
∴
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
[Q c ≠ 0]
⇒
+ =
a
c b
a
3. Let the three consecutive terms of a GP are , a and ar.
r
Now, according to the question, we have
a
⋅ a ⋅ ar = 512
r
⇒
a3 = 512
⇒
a=8
… (i)
Also, after adding 4 to first two terms, we get
8
+ 4, 8 + 4, 8r are in AP
r
⇒
2 (12) =
⇒
24 =
⇒
5=
8
+ 4 + 8r
8r
+ 8r + 4
r
2
+ 2r
r
⇒
2r 2 − 4r − r + 2 = 0
⇒ 2r (r − 2) − 1(r − 2) = 0
⇒
(r − 2) (2r − 1) = 0
⇒

2
20 = 4  + 2r

r
⇒
2r 2 − 5r + 2 = 0
6. Let b = ar and c = ar 2, where r is the common ratio.
a + b + c = xb
a + ar + ar 2 = xar
… (i) [Q a ≠ 0]
1 + r + r 2 = xr
1 + r + r2
1
=1+r +
x=
⇒
r
r
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
Then,
⇒
⇒
following sequence a1 , a 2 = a1r , a3 = a1r , ... , a10 = a1r
Now,
a3 = 25 a1
⇒
a1r 2 = 25 a1
2
r 2 + (1 − x )r + 1 = 0
For real solution of r , D ≥ 0.
⇒
(1 − x )2 − 4 ≥ 0
r = 2,
4. Let r be the common ratio of given GP, then we have the
⇒
⇒ ( A + 10d )2 = ( A + 6d ) ( A + 12d )
⇒ A 2 + 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
∴ a = A + 6d = − 14d + 6d = − 8d
and
c = A + 12d = − 14d + 12d = − 2d
a − 8d
=
=4
⇒
c − 2d
⇒
1
2
Thus, the terms are either 16, 8, 4 or 4, 8, 16. Hence,
required sum = 28.
⇒
∴ b2 = ac
9
r 2 = 25
a
a r8
Consider, 9 = 1 4 = r 4 = (25)2 = 54
a5 a1r
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]
⇒
x 2 − 2x − 3 ≥ 0
⇒
( x − 3)( x + 1) ≥ 0
+
–
–1
+
3
⇒ x ∈ ( −∞ , − 1] ∪ [3, ∞ )
∴ x cannot be 2.
7. Let a be the first term and d be the common difference.
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
Sequences and Series 61
11. Here, (a 2 + b2 + c2) p2 − 2 (ab + bc + cd ) p
Now, common ratio,
a + 4d 8 d + 4d 12 d 4
r=
=
=
=
a+d
9d 3
8d + d
+ (b2 + c2 + d 2) ≤ 0
⇒ (a p − 2abp + b ) + (b p − 2bcp + c2)
2 2
8. Since, (α + β), (α + β ), (α + β ) are in GP.
2
⇒
⇒
⇒
⇒
⇒
⇒
⇒
2
3
3
2 2
3
3
⇒ (ap − b) + (bp − c) + (cp − d ) ≤ 0
2
∴
∴
2
1
 1 1

Also,  − D , ,  + D are in GP.
2
 4 2

2
2
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,
αr 2 + αr3 = 4 ⇒ αr 2(1 + r ) = 4
From Eqs. (i) and (ii), r 2 = 4 ⇒ r = ± 2
Now,
On putting
α ⋅ αr = p and αr 2 ⋅ αr3 = q
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.
a , b, c,d are in GP.
12. Since, a , b, c are in GP.
⇒
b2 = ac
Given,
ax + 2bx + c = 0
2
ax + 2 ac x + c = 0
⇒
2
( a x+
⇒
c )2 = 0 ⇒ x = −
c
a
Since, ax2 + 2bx + c = 0 and dx2 + 2ex + f = 0 have
common root.
∴ x = − c/a must satisfy.
dx2 + 2ex + f = 0
1 1

 1
1

 1
⇒
=  − D 2
∴   =  − D  + D

 2
2

 4
16  4
1
1
1
1
⇒
⇒ D=±
− D2 = ±
⇒ D2 =
2
4
4
2
1
1
∴
a= ±
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
and
(ap − b)2 = (bp − c)2 = (cp − d )2 = 0
b c d
p= = =
a b c
⇒
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
2
2
[since, sum of squares is never less than zero]
Let
2
2
Since, each of the squares is zero.
9. Since, a, b and c are in an AP.
2
2 2
+ (c2p2 − 2cdp + d 2) ≤ 0
(α + β ) = (α + β ) (α + β )
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 or α = β
c
=0
or ∆ = 0
a
c∆ = 0
2
2
…(i)
…(ii)
⇒
c
c
d
f
2e
+ =0
− 2e
+ f =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
d⋅
⇒
Hence,
13. Here, t3 = 4 ⇒ ar 2 = 4
∴ Product of first five terms = a ⋅ ar ⋅ ar 2 ⋅ ar3 ⋅ ar 4
= a5 r10 = (a r 2)5 = 45
14. If a , b, c ∈ (2, 18), then
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
⇒
b = 8, a = 5
and
c = 12
.... (ii)
... (iii)
62 Sequences and Series
11
1
10
1
20
S
= 1 − 20 − 21 = 1 − 20 − 20 = 1 − 20
2
2
2
2
2
2
15. Let 27, 8, 12 be three terms of a GP.
⇒
tm = 27, tn = 8 and t p = 12
⇒
AR m − 1 = 27, ARn − 1 = 8
and AR p − 1 = 12
 27
R= 
8
∴
1/(m − n)
 27
 
8
⇒
and
1/(m − n)
8
R= 
 12
 2
= 
 3
3
1
+
m− n n− p
3
⇒
1
3
+
n− p m− n
2
 q + 1  q + 1
Tn = 1 + 
 +
 +…+
 2   2 
1/( n − p)
101
C1 +
⇒
101
=1
⇒ 3 (n − p) = n − m and
⇒
2n = 3 p – m
z = a + ( p − 1)d and z = br p−1
Now, x − y = (m − n )d, y − z = (n − p)d
C2(1 + q ) +
101
C1 +
=b
⋅r
101
C2
(1 − q 2 )
+
1− q
 1 − q4 
 +…+
C4 
 1−q
101
⇒
 1
∑ k 2k 
1. Let S =
k =1
S=
1 2
3
4
20
+
+
+
+ … + 20
2 22 23 24
2
S 1
2
3
19
20
=
+
+
+ … + 20 + 21
2 22 23 24
2
2
On subtracting Eq. (ii) from Eq. (i), we get
S−
1
1
20
S 1 1
= +
+
+ … + 20 − 21
2 2 22 23
2
2
1 
1
1 − 20 

S 2
20
2
=
− 21
1
2
2
1−
2
101
2
− ( q + 1)
1− q
101
[Q nC 0 + nC1 + … + nC n = 2n]
=α
2101 − ( q + 1)101
1− q
101

 q + 1 
 1− 
 
 2  
= α 1 ⋅

q+1 
1−


2


[Q q ≠ 1 ⇒ q + 1 ≠ 2 ⇒
=
…(ii)


a (1 − r n )
, r < 1
Q sum of GP =
r
1
−


 1 − q101 

C101 
 1−q 
101
 1 − rn
,r ≠1]
[Qfor a GP, S n = a 
1−r
…(i)
 1
On multiplying by   both sides, we get
 2
 1 − q3 
C3 

 1− q 
101
2
100

q + 1  q + 1
 q + 1 
+
1 +
 
 +…+ 
 2  
 2 
2


= b0 ⋅ r 0 = 1
20
C3 (1 + q + q 2 )
101
1
[{ 101C1 + 101C2 + … + 101C101 }
1− q
− { 101C1q + 101C2q 2 + … +101 C101q101 } = α ⋅ T100
1
⇒
[( 2101 − 1) − ((1 + q )101 − 1)] = αT100
(1 − q )
[(m − 1 )( n − p) + ( n − 1 )( p − m)+ ( p − 1 )(m − n)]d
Topic 4 Sum of n Terms and Infinite
Terms of a GP
C101S100 = αT100
101
⇒
⇒
= [br m − 1 ]( n − p)d ⋅ [br n − 1 ]( p − m) d ⋅ [br p − 1 ](m − n)d
[ n − p + p − m + m − n] d
n
C101 (1 + q + q2 + … + q100)
z − x = ( p − m)d
Again now, xy − z ⋅ yz − x ⋅ z x − y
⇒
+…+
C3S 2 + … +
101
= α ⋅ T100
16. Let a , d be the first term and common difference of an
y = a + (n − 1)d and y = br n−1
C1 +
101
101
+
AP and b, r be the first term and common ratio of a GP.
Then,
x = a + (m − 1) d and x = brm−1
C2S1 +
101
= α ⋅ T100
Hence, there exists infinite GP for which 27, 8 and 12 as
three of its terms.
and
 q + 1


 2 
Also, we have
2
3
1
1
3
+
= 0 and
+
=0
m−n n− p
n− p m−n
∴
11
219
2. (a) We have, S n = 1 + q + q2 + … + qn and
1/( n − p)
33/(m − n) ⋅ 31/( n − p) = 21/( n − p) ⋅ 23/(m − n)
⇒
S =2 −
q+1
≠ 1]
2
α [2101 − (q + 1)101 ]
⇒ α = 2100
(1 − q) ⋅ 2100
3. 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

a 
a3
27
⇒
=
Q S ∞ =

− r
1
19 1 − r3

Sequences and Series 63

a
27 (3 (1 − r ))3 
⇒ a = 3 (1 − r )
=
Q 3 =
1−r
19
1 − r3 

27 27 (1 − r ) (1 + r 2 − 2r )
=
19
(1 − r ) (1 + r + r 2)
⇒
⇒
⇒
⇒
⇒
⇒
∴
=
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 )
=
<
S
1
GP,
when
(
r
)


n
−
1
r


[Q (1 − r )3 = (1 − r ) (1 − r )2]
r + r + 1 = 19 (r − 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
2
2
4. 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
k ⋅ 10 = 10 + 2 (11) (10) + 3(11) (10) + ... + 10(11)
9
1
8
2
7
2
 11
 11
 11
k = 1 + 2   + 3  + ... + 10  
 10
 10
 10
⇒
9
2
2
∴
10
…(ii)
or
9
10
 11 10

1   − 1
10


10


 10 − 11
 11
− 10  
⇒ k
 =
 10 
 10
 11

− 1

 10



a (r n − 1)
, when r > 1
Q In GP,sum of n terms =
r −1


10
  11 10
 11 
− k = 10 10   − 10 − 10   
 10 
  10

k = 100
⇒
i.e.
or
⇒
20

 1  
1 −   
 10 


179 1
+

9
 9
1
 
 10
20 
7
[179 + (10)− 20 ]
=
 81
7
77
777
+
+
+ … upto 20 terms
10 102 103
11 111

1
=7
+
+ 3 + … upto 20 terms 
2
10
10
10


=
79
99
999

+
+
+… upto 20 terms 
9 10 100 1000

1 
7 
1 
1  
 1 −  +  1 − 2 +  1 − 3 
9 
10 
10 
10  
+…+ upto 20 terms]
 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
5
1−r=
−10 < − x < 0
0 < x < 10
3
4
8. Since, sum = 4 and second term = .
It is given first term a and common ratio r.
3
a
= 4, ar =
⇒
4
1−r
r=
⇒
6. Let S = 0.7 + 0.77 + 0.777 +…
=
7
9
...(i)
11
11  11
 11

 11
k 1 −  = 1 +
+   + ... +   − 10  
 10

 10
10
10  10
=
=
9
On subtracting Eq. (ii) from Eq. (i), we get
∴
7 
1
20 −
9 
9

9
 11
 11
 11
 11
 11
  k = 1   + 2   + ... + 9   + 10  
 10
 10
 10
 10
 10
⇒
=
7. We know that, the sum of infinite terms of GP is
5. Given,
9
7
[(1 + 1 +…+ upto 20 terms)
9
1
1
1

−  + 2 + 3 + … + upto 20 terms 
 10 10

10
⇒
a
1−
3
4a
3
4a
=4
⇒
4a 2
=4
4a − 3
⇒
(a − 1) (a − 3) = 0
⇒
a = 1 or
When a = 1, r = 3 / 4
and when a = 3, r = 1 / 4
3
64 Sequences and Series
9. 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
12. Consider an infinite GP with first term 1, 2, 3, ..., n and
common ratios
∴

1 1 1
= n −  + + + ... + n terms

2 4 8
1
1
1 − n 
2
2 
= n + 2−n − 1
=n−
1
1−
2
10. 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.
⇒
an = 2 an + 1
a
⇒
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
⇒
∴
and
⇒
S12 + S 22 + S32 + ... + S 22n − 1
a 2 + a 2r 2 + a 2r 4 = S 2
Put
1
 3
−  > −
 4
6
∴
⇒
(1 + r 2 − r ) 1
=
(1 + r 2 + r ) a 2
1
r+ +1
2
r
a =
1
r + −1
r
1
r+ =y
r
y+ 1
= a2
y−1
y + 1 = a 2y − a 2
y=
⇒
a2 + 1
>2
a2 −1
⇒
n = 1, −
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.
a2 + 1
a2 − 1
⇒
Obviously, it is true for all even values of n.
But for
...(ii)
(1 + r 2)2 − r 2 1
=
(1 + r + r 2)2 a 2
n
⇒
...(i)
a + ar + ar 2 = a S
a 2 (1 + r 2 + r 4 )
S2
= 2 2
2
2 2
a (1 + r + r )
aS
Hence, (b), (c), (d) are the correct answers.
n

1 −  − 3 

 4  1
3
1
<
An < ⇒
3
2
4
2
1+
4
M
M
2n − 1
=
= 2n
1 − 1 / 2n
On dividing Eq. (i) by Eq. (ii) after squaring it, we get
⇒
11. Bn = 1 − An > An
2
=3
1 − 1 /3
13. Let three numbers in GP be a , ar , ar 2.
[Q a n2 ≤ 1, given]
This is possible for n ≥ 8.
S2 =
= 22 + 32 + 42 + ... + (2n )2
1
= (2n ) (2n + 1) (4n + 1) − 1
6
⇒
100 ≤ 2n − 1
⇒
∴
⇒
100
≤1
2n − 1
⇒
1
=2
1 − 1 /2
S 2n − 1
[ Q a1 = 10, given]
 1
a n2 = 100  
 2
⇒
S1 =
M
n −1
2( n − 1)
1 1 1
1
, , , ... ,
.
2 3 4
n+1


1
Q | y | = r + r > 2


[where , (a 2 − 1) ≠ 0]
|a 2 + 1| > 2|a 2 − 1|
⇒ (a 2 + 1)2 − {2 (a 2 − 1)}2 > 0
⇒ {(a 2 + 1) − 2 (a 2 − 1)}{(a 2 + 1) + 2 (a 2 − 1)} > 0
⇒
∴
∴
(− a 2 + 3) (3a 2 − 1) > 0
1
< a2 < 3
3
1 
[Q a 2 ≠ 1]
a 2 ∈  , 1 ∪ (1, 3)
3 
k −1
14. We have, S k = k ! =
1
1−
k
1
(k − 1)!
Sequences and Series 65
1
Now, (k2 − 3k + 1) S k = {(k − 2) (k − 1) − 1} ×
(k − 1)!
1
1
=
−
(k − 3)! (k − 1)!
100
⇒
2
1 
100
 1
+
⇒ ∑|(k2 − 3k + 1) S k|= 1 + 1 + 2 − 
 =4−


99
!
98
!
100 !
k =1
100
⇒
1002
+ |(k2 − 3k + 1) S k|= 4
100 ! k∑
=1
Topic 5 Harmonic Progression (HP)
1.
1
nth term of HP, t n =
a + ( n − 1) n
PLAN
Here,
∴
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
⇒
∴
an < 0
1
⇒
+ (n − 1) d < 0
5
95
1
4
−
(n − 1) < 0 ⇒ (n − 1) >
⇒
4
5 19 × 25
Since,
95
or n > 24.75
n >1 +
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,
∴
⇒
⇒
1 1 6 ×1
= +
h7 2 − 54
1 1 1
18
= − ⇒ h7 =
h7 2 9
7
7 18
a 4h7 = ×
=6
3
7
⇒
h1 = 2, h10 = 3
1
1
1 1
=
+ 9d1 ⇒ = + 9d1
h10 h1
3 2
1
− = 9d1
6
1
1
1
and
=
+ 6d1
d1 = −
h7 h1
54
∴
4. Let the common ratio of the GP be r. Then,
y = xr and z = xr 2
⇒ ln y = ln x + ln r and ln z = ln x + 2 ln r
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
Let
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
cos 2 x ⋅ sec2 = 2
2
y
cos x ⋅ sec = ± 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
66 Sequences and Series
⇒
4b2 = –2ac ⇒ b2 = –
ac
2
Topic 6 Relation between AM, GM, HM
and Some Special Series
c
are in GP.
2
c
∴ Either a = b = c or a , b, − are in GP.
2
Hence,
a , b, –
1. Given series,
8. Since, a , A1 , A2, b are in AP.
⇒
A1 + A2 = a + b
a , G1 , G2, b are in GP
⇒ G1G2 = ab
a , H 1 , H 2, b are in HP.
3ab
3ab
H1 =
, H2 =
b + 2a
2b + a
and
⇒
∴
1
1
1 1
+
= +
H1 H 2 a b
⇒
H 1 + H 2 A1 + A2 1 1
=
= +
H 1H 2
G1G2
a b
…(i)
G1G2
ab
=
H 1H 2  3ab   3ab 

 

 2b + a   b + 2a 
(2a + b) (a + 2b)
=
9ab
Now,
…(ii)
From Eqs. (i) and (ii), we get
G1G2
A + A2 ( 2a + b) (a + 2b)
= 1
=
H 1H 2 H 1 + H 2
9ab
9. (i) Now,
=
 n
n (n + 1) (2n + 1) 

Q ∑ r 2 =
6

 r = 1
5
(a + b) − 15
2
[since, a , x, y, z are in AP]
5
Sum = (a + b) ⇒ a + b = 10
…(i)
∴
2
1 1 1 1 1
Since, a , x, y, z , b are in HP, then , , , ,
a x y z b
are in AP.
1 1  1 1 1 1 1   1 1 1
+ = + + + +  − + + 
a b  a x y z b  x y z 
5  1 1 5
 +  −
2  a b 3
a + b 10
9 × 10
=
⇒ ab =
ab
9
10
=
⇒
⇒
1 15 × 16 × 31 15 × 16 
+
2 
6
2 
a + b = (a + x + y + z + b) − (x + y + z )
=
Now,
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
+K+
+
S1 = 1 +
1+ 2
1+ 2+ 3
13 + 23 + 33 + K + 153
1 + 2 + 3 + K + 15
2
 n( n + 1)


15
15


13 + 23 + K + n3
2
= ∑
= ∑
+
1
n
n
(
)
+
+
+
1
2
K
n
n =1
n =1
2
2
n
 n
(
n
n
+ 1)
(
)
n
n
+
1


Q ∑ r3 = 

 and ∑ r =


2
2
 r = 1

r =1
15
15
n( n + 1) 1
= ∑
=
∑ (n 2 + n )
2
2 n =1
n =1
S =1+
ab = 9
Therefore, S = S1 − S 2 = 680 − 60 = 620.
2. Given series is
[from Eq. (i)]
…(ii)
a = 1, b = 9
= log (x + z ) + log
(x − z )2
(x + z )
= 2 log (x − z ) = RHS
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
On solving Eqs. (i) and (ii), we get
(ii) LHS = log (x + z ) + log (x + z − 2 y)

 2 xz  
= log (x + z ) + log x + z − 2 

 x + z 

1
[(5 × 8 × 31) + (15 × 8)]
2
= (5 × 4 × 31) + (15 × 4)
= 620 + 60 = 680
1
and S 2 = (1 + 2 + 3 + K + 15)
2
1 15 × 16
= ×
= 60
2
2
=
2xz 

Q y = x + z 
 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 Σ r = 


r =1
1
r
=
6
2




Sequences and Series 67
3n (n + 1) 3 2
= (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
Tn =
So,
S10 =
∴
=
2
3 10 × 11 × 21 10 × 11 
+
2 
6
2 
⇒
=
3
[(5 × 11 × 7) + (5 × 11)]
2
=
3
3
× 55(7 + 1) = × 55 × 8 = 3 × 55 × 4
2
2
11
r =1
= (11 × 4 × 23) − (11 × 6) = 11(92 − 6) = 11 × 86 = 946
4. Given series is
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)
Q1 + 23 + 33 + ... + n3 = 
 , n ∈N


2


27 225 × 256
=
×
64
4
= 27 × 225
⇒ S = 27 × 225 = 225 k
⇒ k = 27.
[Q ∑ n 2 =
6. Consider,
11 × (11 + 1)(2 × 11 + 1) 11 × (11 + 1)
=2
−
6
2
n
 n
n (n + 1)(2n + 1)
n (n + 1) 
Q ∑ r 2 =

and ∑ r =
6
2
 r = 1

r =1
 11 × 12 × 23  11 × 12
=

 −
  2 

3
3
10
1
∑ (k + 1)2 = 4 [22 + 32 + 42 + ... 112]
k =1
1 11 × (11 + 1) (2 × 11 + 1) 2
= 
−1 
4
6

11
r =1
r =1
5
1
A=
12
4
n ( n + 1) ( 2n + 1)
]
6
1 11 × 12 × 23
 1
− 1 = [(22 × 23) − 1]
4 
6
 4
1
1
= [506 − 1] = [505]
4
4
5
505
A=
⇒ A = 303
⇒
12
4
∑ r (2r − 1) = ∑ (2r 2 − r ) = 2 ∑ r 2 − ∑ r
r =1
10
5
2
A = S 12 + S 22 + S 23 + ... S 10
= ∑ S 2k
12
k =1
=
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
11
1
 k + 1
2
S 2k = 
 = (k + 1)
 2 
4
Now,
3. (b) Given series is
11
k ( k + 1) k + 1
=
2k
2
So,
= 12 × 55 = 660
S11 =
1 + 2 + 3 + ... + k
k
5. Since, S k =
[given]
=
xm yn
(1 + x )(1 + y2n )
2m
1
( x + x )( y n + y − n )
By using AM ≥ GM (because x , y ∈ R + ), we get
( x m + x − m ) ≥ 2 and ( y n + y − n ) ≥ 2
1
[Q If x > 0, then x + ≥ 2]
x
⇒ ( x m + x − m )( y n + y − n ) ≥ 4
1
1
≤
⇒ m
−m
n
−n
( x + x )( y + y ) 4
1
∴ Maximum value = .
4
m
−m
7. General term of the given series is
3r (12 + 22 + K + r 2) 3r [r (r + 1) (2r + 1)]
=
6(2r + 1)
2r + 1
1 3
2
= (r + r )
2
15
1 15
Now, required sum = ∑ Tr = ∑ (r3 + r 2)
2 r =1
r =1
Tr =
=
1
2
=
1
2
1
2
1
=
2
=
 n (n + 1)  2 n (n + 1) (2n + 1) 


 +
2
6
n = 15

 n (n + 1)  n 2 + n 2n + 1 
+



2
3 
 2

n = 15
 n (n + 1) (3n 2 + 7n + 2) 


2
6

n = 15
15 × 16 (3 × 225 + 105 + 2)
×
×
= 7820
2
6
… (i)
68 Sequences and Series
2
12
∑ a4k + 1 = 416 and a9 + a43 = 66
k =0
Let a1 = a and d = common difference
Q
a1 + a 5 + a 9 + L + a 49 = 416
∴ a + (a + 4d) + (a + 8d) + …(a + 48d) = 416
13
(2a + 48d) = 416
⇒
2
⇒
a + 24d = 32
Also
, a 9 + a 43 = 66
∴
a + 8d + a + 42d = 66
⇒
2a + 50d = 66
⇒
a + 25d = 33
Solving Eqs. (i) and (ii), we get
a = 8 and d = 1
2
Now, a12 + a 22 + a 23 + L + a17
= 140m
…(i)
…(ii)
2
2
∴
2
l + n = 2m
l, G1 , G2, G3 , n are in GP.
Let r be the common ratio of this GP.
1
∴ G1 = lr, G2 = lr , G3 = lr , n − lr
2
we have
2
 3
S10 = 1  +
 5
2
2
 1
 2
2
2  + 3  + 4 +
 5
 5
= l4 ×
12.
n
l
⇒
 n 4
r= 
 l
2
 n + l
2
2

 = ln × 4 m = 4lm n
 l 
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
S9 =
=
(n + 1)2 1 2
∑ 4 = 4 (2 + 32 + ... + 102) + 12 – 12]
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
≥
x2 + x ⋅
tan 2 α
x2 + x
[using AM ≥ GM]
2
 4
4 
 5
+ ... to 10 terms
4
= l4 × r 4 (1 + 2r 4 + r 6 ) = l4 × r 4 (r 4 + 1)2
12 + 2⋅ 22 + 32 + 2⋅ 42 + 52 + 2⋅ 62 + …
10. Let S10 be the sum of first ten terms of the series. Then,
3
Now, G14 + 2G24 + G34 = (lr )4 + 2(lr 2)4 + (lr3 )4
9. We have,
A = (12 + 22 + 32 + … + 202) + (22 + 42 + 62 + … + 202)
A = (12 + 22 + 32 + … + 202) + 41
( 2 + 22 + 32 + …+102)
20 × 21 × 41 4 × 10 × 11 × 21
A=
+
6
6
20 × 21
20 × 41 × 63
(41 + 22) =
A=
6
6
Similarly
B = (12 + 22 + 32 + … + 402) + 41
( 2 + 22 +…+ 202)
40 × 41 × 81 4 × 20 × 21 × 41
B=
+
6
6
40 × 41
40 × 41 × 123
(81 + 42) =
B=
6
6
Now, B − 2A = 100λ
40 × 41 × 123 2 × 20 × 21 × 63
−
= 100λ
∴
6
6
40
40
(5043 − 1323) = 100λ ⇒
× 3720 = 100λ
⇒
6
6
40 × 620
⇒ 40 × 620 = 100λ
⇒λ =
= 248
100
... (i)
and G1 , G2, G3 are geometric means between l and n.
⇒ (12 + 22 + 32 + … + 242) − (12 + 22 + 32 + … + 72) = 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
A = sum of first 20 terms
B = sum of first 40 terms
∴A = 12 + 2⋅ 22 + 32 + 2⋅ 42 + 52 + 2⋅ 62 + … + 2⋅ 202
2
2
11. Given, m is the AM of l and n.
8 + 9 + 10 + … + 24 = 140m
2
2
 8
 24
 12
 16
=   +   +   + 42 +   + ... to 10 terms
 5
5
5
5
1 2
2
2
2
= 2 (8 + 12 + 16 + 20 + 242 + ... to 10 terms)
5
42 2
= 2 (2 + 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
=
× 505 ⇒
(506 − 1) =
m=
× 505 = 101
25
25
5
25
8. We have, a1 , a 2, a3 , … a 49 are in AP.
⇒
x2 + x +
tan 2 α
x2 + x
≥ 2 tan α
Sequences and Series 69
Gn =
14. Given, a1 a 2 a3 ... a n = c
⇒
…(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]
∴
⇒ a1 + a 2 + a3 + ... + 2a n ≥ n (2c)1/ n
[from Eq. (i)]
⇒ Minimum value of
a1 + a 2 + a3 + ... + 2a n = n (2c)1/ n
Hn =
An − 1 H n − 1 ,
2 An − 1 H n − 1
An − 1 + H n − 1
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
15. Since, AM ≥ GM, then
(a + b) + (c + d )
≥ (a + b)(c + d ) ⇒ M ≤ 1
2
(a + b) + (c + d ) > 0
[Q a , b, c, d > 0]
Also,
∴
0 < M ≤1
16. Let α , β be the roots of given quadratic equation. Then,
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
H =
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
Using AM ≥ GM,
≥ (x1 ⋅ x2 ... xn )1/ n
n
⇒
x1 + x2 + ... + xn ≥ n (1)1/ n
[from Eq. (i)]
Hence, sum of n positive numbers is never less than n.
18. Since, AM > GM
∴
(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
1/ 2
b > [(a + b − c)(b + c − a )]
…(ii)
and
a > [(a + b − c)(c + a − b)]1/ 2
…(iii)
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 − 1
2
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 +
=1 +
= 1 + 1 + 1 + 1 + ... + 1 = n
1444
424444
3
(n ) times
Thus, a(100) ≤ 100
Therefore, (a) is the answer.
1  1 1  1
1
Again, a (n ) = 1 + +  +  +  + K + 
2  3 4  5
8
1
1 1

+ K +  n −1
+ K + n − n
2
2  2
+1
On multiplying Eqs. (i), (ii) and (iii), we get
>1 +
abc > (a + b − c)(b + c − a )(c + a − b)
2 4 8
2n −1
+ + + K + n −1
2 4 8
2
Hence, (a + b − c)(b + c − a )(c + a − b) − abc < 0
1  1 1  1 1
1
+  +  +  + +K+ 



2
4 4
8 8
8
1 1
 1
+ K +  n + K + n − n
2
2  2
19. Since, x1 , x2 ,…, xn are positive real numbers.
∴ Using nth power mean inequality
x12 +
x22 + ... +
xn2
n
 x + x2 + ... + xn 
≥ 1



n
2
⇒
 n   n 
n 2  n 2  n 
xi ≥ ∑ xi ⇒ n  ∑ xi2 ≥  ∑ xi 
∑

 

n  i = 1   i = 1 
 i =1   i =1 
20. Let a and b are two numbers. Then,
a+b
2ab
; G1 = ab ; H 1 =
2
a+b
An − 1 + H n − 1
,
An =
2
A1 =
1 2 4
2n − 1
1
+ + +K+ n − n
2 4 8
2
2
1 n
1 1 1
1 1 
= 1 + + + + K + − n = 1 − n  +
 2

2
2
2
2
2
2
14442444
3
n times
=1 +
2
2
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.
70 Sequences and Series
x+ y
=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
Thus, in arithmetic progression, tn =
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.
∴
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=
and
∴
y2
z
a=
and b =
log a x +
28. Since,
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]
⇒
y +z
= 2x
yz
⇒
y +z
=2
xyz
1 + a + b + c + ca + abc > 7(a 4b4c4 )1/7
{(1 + a )(1 + b)(1 + c)}7 > 77 (a 4b4c4 )
30. Let Gm be the geometric mean of G1 , G2, ... , Gn.
⇒
= [(a1 ) ⋅ (a1 ⋅ a1r )1/ 2 ⋅ (a1 ⋅ a1r ⋅ a1r 2)1/3
K (a1 ⋅ a1r ⋅ a1r 2 K a1r n − 1 )1/ n ]1/ n
where, r is the common ratio of GP a1 , a 2, K , a n.
Again,
⇒
H 4
=
G 5
2ka
k+1 4
=
ka 5
1
= [a1n ⋅ r 2
2 k 4
⇒
=
k+1 5
+1+
k=
Now,
An =
and
Hn =
5 ± 25 − 16 5 ± 3
1
=
= 2,
4
4
2
=
⇒ k = 4, 1 / 4.
Hence, the required ratio is 4 : 1.
=
27. Using AM ≥ GM,
⇒
⇒
∴
1 + x2 n
≥ 1 ⋅ x2 n
2
∴
1 + x2 n
≥ xn
2
⇒
n
1
x
≤
2n
2
1+x
1
xn ⋅ ym
≤
2n
(1 + x ) (1 + y2m ) 4
Hence, it is false statement.
⋅r
6/ 4
Kr
( n − 1 )n
2n
)]1/ n
n −1
3
+ L+
2
2 ]1/ n
1/ n
2k − 5 k + 2 = 0
⇒
⋅r
3 /3
 1  ( n − 1 )n  
 n −1 


= a1 r 2  2   = a1 r 4 






[given]
5 k = 2k + 2
⇒
⇒
= [(a1 ⋅ a1. K n times ) (r
1/ 2
G = ka ⋅ a = k ⋅ a
2(ka )a
2ka
H =
=
ka + a k + 1
and
Gm = (G1 ⋅ G2 K Gn )1/ n
3
26. Let the two positive numbers be ka and a , a > 0.
Then,
…(ii)
(1 + a )(1 + b)(1 + c) > 7(a 4b4c4 )1/7
or
3
3
a + b + c + ab + bc + ca + abc ≥ 7(a 4b4c4 )1/7
From Eqs. (i) and (ii), we get
y2 z 2
2x =
+
z
y
⇒
> 1, using AM > GM
Hence, it is a false statement.
On substituting the values of a and b in Eq. (i), we get
3
2
Here, equality holds only when x = a which is not
possible. So, log a x + log x a is greater than 2.
⇒
z2
y
1
log a x
An ⋅ H n =
n
∏
k =1
…(i)
a1 + a 2 + ... + a n a1 (1 − r n )
=
n
n (1 − r )
n
1
1
1
+
+K+


an
 a1 a 2
1
a1
n
1
1 

1 + + K + n − 1 


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
Ak H k =
∏ (a12r n − 1 )
k =1
= (a12 ⋅ a12 ⋅ a12K n times ) × r 0 ⋅ r1 ⋅ r 2K r n − 1
= a12n ⋅ r1 + 2 + K + ( n − 1)
= a12nr
n ( n − 1)
2
= [a1r
n −1
4 ]2 n
Sequences and Series 71
= [Gm ]2n
2
[from Eq. (i)]
 n

Gm =  ∏ AkH k 
k = 1

⇒
2
 n + 1
Hence, q cannot lie between p and 
 p.
 n − 1
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
d=
n+1
b−a
p = A1 = a + d = a +
n+1
na + b
p=
n+1
⇒

 n + 1 2
 p > p


 n − 1 
Q
Gm = ( A1 A2 K AnH 1H 2 K H n )1/ 2n
∴
 n + 1
q < p or q > 
 p
 n − 1
⇒
1/ 2 n
32. Since a , b, c > 0
(a + b + c)
> (abc)1/3
3
⇒
…(i)
Also,
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 1
1 1
(a − b)
= +D ⇒
= +
q a
q a (n + 1) ab
⇒
1
nb + a
=
q (n + 1) ab
⇒
q=
[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)  + + 
1
 a b c
≥ (abc)1/3
9
(abc)1/3
 1 1 1
(a + b + c)  + +  ≥ 9
 a b c
∴
33.
( n + 1) ab
nb + a
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
... (ii)
From Eq. (i),
Let a , b, c be a , ar , ar 2, where r ∈ N
a+ b+ c
Also,
=b+2
3
b = (n + 1) p − na.
Putting it in Eq. (ii), we get
q { n (n + 1) p − n 2a + a } = (n + 1) a {(n + 1) p − na }
⇒
⇒
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 .
⇒ n (n + 1) a 2 − {(n + 1)2 p + (n 2 − 1)q}a
+ n (n + 1) pq = 0
⇒ na 2 − {(n + 1) p + (n − 1)q} a + npq = 0
Since, a is real, therefore
{(n + 1) p + (n − 1)q}2 − 4n 2pq > 0
So, a can be 6 only.
⇒ (n + 1) p + (n − 1) q + 2 (n − 1) pq − 4n pq > 0
⇒ r −1 = ± 1 ⇒ r =2
⇒
and
2 2
2 2
2
2
(n + 1)2 p2 + (n − 1)2q2 − 2 (n 2 + 1) pq > 0
2
⇒
⇒
q2 −
 n + 1
2 (n 2 + 1)
pq + 
 p2 > 0
2
(n − 1)
 n − 1

 n + 1
q − 1 + 

 n − 1

2
2

 pq +

a 2 + a − 14 36 + 6 − 14
=
=4
a+1
7
34. Using AM ≥ GM,
2
 n + 1) 2

 p >0
 n −1

 n + 1 
(q − p) q − 
 p > 0
 n − 1 


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
2
⇒
…(i)
Download Chapter Test
http://tinyurl.com/y6r7ns95
⇒ a −5 + a −4 + 3a −3 + 1 + a 8 + a10 ≥ 8 ⋅ 1
Hence, minimum value is 8.
or
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 73
Match the Column
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)
1. The number of ways of choosing 10 objects out of 31
objects of which 10 are identical and the remaining 21
are distinct, is
(2019 Main, 12 April I)
20
(a) 2
−1
21
20
(b) 2
(c) 2
20
(d) 2
+1
2. 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
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)
(b) 190
(c) 225
(a) 100
(d) 157
4. There are m men and two women participating in a
(b) 400
(c) 200
(d) 50
8. 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
(2017 Main)
are in this party, is
(a) 485
3. Some identical balls are arranged in rows to form an
(a) 262
3
20


Ci − 1
k

 = , then k equals
∑
20
20
Ci + Ci − 1 
21
i=1
(2019 Main, 10 Jan I)
20
7. If
(b) 468
(c) 469
(d) 484
9. 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
10. 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
(2016 Adv.)
number of ways of selecting the team is
(a) 380
(b) 320
(c) 260
(d) 95
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)
11. Let Tn be the number of all possible triangles formed by
(a) 12
12. If r , s, t are prime numbers and p, q are the positive
(b) 11
(c) 9
(d) 7
5. If n C 4 , n C 5 and n C 6 are in AP, then n can be
(2019 Main, 12 Jan II)
(a) 9
(b) 11
25
6. If
∑{
50
Cr ⋅
(c) 14
50 − r
(d) 12
C 25 − r } = K ( C 25 ),
(b) 5
(c) 10
(d) 8
integers such that LCM of p, q is r 2s4 t 2 ,then the number
of ordered pairs ( p, q) is
(2006, 3M)
(b) 254
(c) 225
13. The value of the expression C 4 +
47
(d) 224
5
∑
52− j
C 3 is
j =1
then, K is equal to
(a) 2
(a) 7
(a) 252
50
r = 0
24
joining vertices of an n-sided regular polygon. If
(2013 Main)
Tn + 1 − Tn = 10, then the value of n is
25
(b) 2
(2019 Main, 10 Jan II)
−1
25
(c) 2
(d) (25)
2
(a) 47 C5
(c) 52C4
(b) 52C5
(d) None of these
(1980, 2M)
74 Permutations and Combinations
17. In a certain test, a i students gave wrong answers to at
Match Type Question
14. 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 Answer Type Question
15. Let n ≥ 2 be an integer. Take n distinct points on a circle
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)
Fill in the Blanks
16. 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)
least i questions, where i = 1, 2, K , k. No student gave
more that k wrong answers. The total number of wrong
(1982, 2M)
answers given is … .
True/False
18. The product of any r consecutive natural numbers is
always divisible by r !.
(1985, 1M)
Analytical & Descriptive Questions
19. 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)
20. 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
(1987, 3M)
value of n.
21. 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)
22. 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
(1985, 5M)
relatives?
23. 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)
24. 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
equal.
(1982, 5M)
25. 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)
Topic 3 Multinomial, Repeated Arrangement and Selection
Objective Question I (Only one correct option)
1. The number of 6 digits numbers that can be formed
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)
(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
Permutations and Combinations 75
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
(b) 120
(c) 240
(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
(b) 375
(c) 372
(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
(2019 Main, 9 Jan I)
team, is
(a) 350
(b) 500
(c) 200
(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
(b) 192
(c) 96
(d) 48
Numerical Value
Integer Answer Type Questions
9. 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
10. 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
(2015 Adv.)
n
11. Let n1 < n2 < n3 < n4 < n5 be positive integers such that
n1 + n2 + n3 + n4 + n5 = 20. The number of such distinct
arrangements (n1 , n2 , n3 , n4 , n5 ) is
(2014 Adv.)
Fill in the Blanks
12. 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)
13. Total number of ways in which six ‘+’ and four ‘–’ signs
can be arranged in a line such that no two ‘–’signs occur
(1988, 2M)
together is… .
Analytical & Descriptive Question
8. The number of 5 digit numbers which are divisible by 4,
14. Five balls of different colours are to be placed in three
with digits from the set {1, 2, 3, 4, 5} and the repetition of
(2018 Adv.)
digits is allowed, is ..................... .
boxes of different sizes. Each box can hold all five. In
how many different ways can we place the balls so that
(1981, 4M)
no box remains empty?
Topic 4 Distribution of Object into Group
Objective Questions I (Only one correct option)
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
2. Let S be the set of all triangles in the xy-plane, each
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)
(a) 36
(c) 18
(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
4. The total number of ways in which 5 balls of different
colours can be distributed among 3 persons so that each
person gets at least one ball, is
(2012)
(a) 75
(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
(b) 60
(c) 80
(d) 100
76 Permutations and Combinations
7. In how many ways can a pack of 52 cards be
Analytical & Descriptive Questions
2
6. Using permutation or otherwise, prove that
integer, where n is a positive integer.
n !
is an
(n !)n
(2004, 2M)
(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)
Topic 5 Dearrangement and Number of Divisors
Fill in the Blank
Objective Question I (Only one correct option)
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
25. (n = 9 and r = 3 )
22. (485)
1. (b)
2. (a)
3. (b)
4. (c)
5. (c)
6. (b)
7. (c)
8. (a)
9. (d)
10. (a)
11.
9
P4 × 9 P3 (11 )!
12. ( A → p; B → s; C → q ; D → q )
Topic 2
1. (c)
2. (c)
3. (b)
4. (a)
5. (c)
6. (c)
7. (a)
8. (a)
9. (c)
10. (a)
11. (b)
12. (c)
13. (c)
14. (c)
15. (5)
Topic 3
1. (a)
2. (c)
3. (b)
4. (a)
5. (d)
6. (d)
7. (c)
8. (625)
9. (5)
10. (5)
1
2
12. (2n − k + k − 2 )
2
11. (7)
17. 2n − 1
18. (True)
19. 6062, (i) 2702 (ii) 1008
20. n = 3
21. (64)
r =1
14. (300)
Topic 4
1. (c)
2. (a)
3. (a)
4. (b)
(52 )!
(52 )!
(52 )!
(ii)
(iii)
7. (i)
4
4
(13 !)
4 ! (13 !)
3 ! (17 ) 3
5. (a)
n
16. nn , ∑ ( −1 )n − r n Cr (r )n
13. (35 ways)
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-III
4
Case-I
4
3
2
0/1/2/3/4/5
6 ways
2×6×6=72 numbers
Case-IV
2/3/4/5
4 ways
4 numbers
Case-II
4
4/5
2 ways
5
6×6×6=216 numbers
0/1/2/3/4/5
6 ways
3
3/4/5 0/1/2/3/4/5
3 ways
6 ways
3×6=18 numbers
So, required total numbers = 4 + 18 + 72 + 216 = 310
Permutations and Combinations 77
2. Sum of diagonal entries of M T M is ∑ a i2.
9
∑a
2
i
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
=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
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
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
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)!.
4. There are two possible cases
Case I Five 1’s, one 2’s, one 3’s
7!
Number of numbers =
= 42
5!
Case II Four 1’s, three 2’s
7!
Number of numbers =
= 35
4!3!
∴ Total number of numbers = 42 + 35 = 77
5. X  X  X  X  X . The four digits 3, 3, 5, 5 can be
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
6. Distinct n-digit numbers which can be formed using
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
the students.
Then,
⇒
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.
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. Given that, out of 31 objects 10 are identical and
remaining 21 are distinct, so in following ways, we can
choose 10 objects.
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 ×
. . . . . .
. . . . . .
C10
21
C8
78 Permutations and Combinations
So, total number of ways in which we can choose 10
objects is
21
C10 +
21
C9 +
⇒ C11 +
21
21
21
C8 + K +
C12 +
21
21
C 0 = x (let)
C13 + K +
21
C 21 = x
… (i)
… (ii)
[Q C r = C n − r ]
n
n
On adding both Eqs. (i) and (ii), we get
2x = 21C 0 + 21C1 + 21C 2 + K + 21C10
+ 21C11 + 21C12 + K +
⇒
2x = 2
21
⇒
x=2
21
C 21
20
2. It is given that, there are 20 pillars of the same height
have been erected along the boundary of a circular
stadium.
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.
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
2
= 190 − 20 = 170
3. Let there are n balls used to form the sides of equilateral
triangle.
⇒ m2 − 5m − 84 = 0
⇒ m2 − 12m + 7m − 84 = 0
⇒ m(m − 12) + 7 (m − 12) = 0
⇒
m = 12
n
⇒
n = 19, − 10
⇒
n = 19
5. 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 )! 


2
⇒
5 ⋅ 4 !(n − 5) (n − 6)!
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
25
6. Given, Σ { 50C r .50− r C 25− r } = K 50C 25
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.]
⇒
[Qnumber of balls n > 0]
two
games
with
every
and the number of games played between the men and
the women =2 × mC1 × 2C1
Now, according to the question,
2 mC 2 = 2 mC1 2C1 + 84
m!
= m × 2 + 42
2 !(m − 2)!
m(m − 1) = 4m + 84
m2 − m = 4m + 84
25
C 25 Σ
r = 0
Cr = K
25
 50
50 ! 
Q C 25 = 25 ! 25 ! 


50
C 25
25
K= Σ
C r = 225
25
r = 0
[Q nC 0 + nC1 + n C 2 + ....+ nC n = 2n ]
⇒
K = 225
7. Given,
20
∑
i=1
other
∴ Number of games played by the men between
themselves = 2 × mC 2
50
⇒
4. Since, there are m-men and 2-women and each
⇒
⇒
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)
12 (n − 4) = 30 + n 2 − 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
=
25
Now, number of balls used to form an equilateral
n (n + 1)
triangle is
2
19 × 20
=
= 190.
2
⇒
[Q m > 0]
n
r=0
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
participant plays
participant.
n
⇒
3
20


Ci − 1
k
 =
 20
20
21
 Ci + Ci − 1 
20
∑
i=1
3
 20C i − 1 
k
 =
 21
21
Ci 

(Q nC r + nC r − 1 =
n +1
Cr )
3
⇒
20
∑
i=1
⇒
⇒

 20

Ci − 1 
k
n
 n
 =

Q C r =

r
21
 21 20C i − 1 

 i
3
20
k
 i
=


∑
 21
21
i=1
1
(21)3
20
∑i
i=1
3
=
k
21
n−1

Cr − 1 

Permutations and Combinations 79
Similarly, the number of ways to select t = 5
2
k
1  n (n + 1) 
=

(21)3 
2
21
n = 20
2
 3
 n (n + 1)  
3
3
Q
1
+
2
+
K
+
=
n


 
2


2
21  20 × 21
k=
 = 100

(21)3  2 
⇒
⇒
∴
∴ Total number of ways = 5 × 9 × 5 = 225
5
13. Here, 47 C 4 + ∑
= 47C 4 +
= ( C4 +
47
k = 100
N 2 = 10 × 4
N4 = 5 × 4
N5 = 1
N 1 + N 2 + N 3 + N 4 + N 5 = 126
(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
n +1
[Q nC r + nC r
+1
=
n +1
Cr
+1
]
C 2 = 10
Number of ways
1 way
1 way
3 ways
∴ Total number of ways to select, r = 5
Selection of s as under
s
s1
s2
s3
s4
s
s4
s4
s4
51
51
51
C3
C3
C3
C 3 = 52C 4
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.
∴ Selection of p and q are as under
4
50
= 40 + 30 + 4 = 74
M 1 is included → 4C 2 ⋅ 5C1 + 4C 3 = 30 + 4 =34
G1 and M 1 both are not included
4
C 4 + 4 C 3 ⋅ 5C1 + 4C 2 ⋅ 5C 2
[given]
12. Since, r , s, t are prime numbers.
0
49
51
(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
C3
C 3 − nC 3 = 10
q
r2
r2
r 0 , r1 , r 2
49
α 3 = 381
n =5
p
r0
r1
r2
C3
n
(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
10. We have, 6 girls and 4 boys. To select 4 members
⇒
51
(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
N 3 = 10 × 6
⇒
C3
(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
N1 = 5 × 1
n
50
5 girls G1 , G2 , G3 , G4 , G5
9. N i = 5C k × 4C 5 − k
⇒ n C 2 + nC 3 − nC 3 = 10
49
14. Given 6 boys M1 , M 2 , M 3 , M 4 , M 5 , M 6 and
= 1 + 144 + 324 + 16 = 485
n +1
48
= ( 50C 4 + 50C 3 ) +
= 51C 4 +
+ 3C1 × 4C 2 × 4C 2 × 3C1 + 3C 0 × 4C 3 × 4C 3 × 3C 0
− Tn =
47
C3 ) + C3 + C3 + C3 +
48
= ( C4 + C3 ) + C3 +
49
= 3C 3 × 4C 0 × 4C 0 × 3C 3 + 3C 2 × 4C1 × 4C1 × 3C 2
+1
C 3 + 50C 3 + 49C 3 + 48C 3 +
47
= ( C 4 + C 3 ) + C 3 + 50C 3 +
∴ Total number of required ways
Tn
51
[using C r + C r − 1 = n +1C r ]
48
men while Y has 7 friends, 3 of them are ladies and 4 are
men.
∴
C3
n
8. Given, X has 7 friends, 4 of them are ladies and 3 are
11. Given, Tn = nC 3 ⇒ Tn + 1 =
52 − j
j =1
1 way
1 way
1 way
1 way
5 ways
∴ Total number of ways to select s = 9
15.
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
∴
⇒
⇒
C2 − n = n
n (n − 1)
= 2n
2
n
n =5
80 Permutations and Combinations
16. Let A = { x1 , x2 , ... , xn }
∴ Number of functions from A to A is n n and out of these
n
∑ (−1)n − r nC r (r )n are onto functions.
r =1
17. The
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
=2
n−1
+2
n− 2
+2
n− 3
+ K+ 2 + 2 =2 −1
1
0
n
(x + r )! r !
⋅ =
(x)! r !
=
Number of ways = 3C 3 = 1
Total number of ways = 45 + 18 + 1 = 64
22. 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 = (4 C 2 ⋅3 C1 ) ⋅ (3C1 ⋅4 C 2 ) = 324
Case III A man invites (1 lady, 2 gentlemen) and
women invites (2 ladies, 1 gentleman).
18. Let r consecutive integers be x + 1, x + 2, K , x + r.
∴ (x + 1) (x + 2) K (x + r ) =
Case III When all three black balls are drawn
⇒
∴
(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.
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
19. 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 )
23. 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 Pn
(m)! ⋅ (m + 1)!
=
(m + 1 − n )!
+ (9C 8 . 8C 4 ) + (9C 9. 8C 3 )
= 1008 + 2352 + 2016 + 630 + 56 = 6062
(i) The women are in majority = 2016 + 630 + 56
24. Let mn squares of equal size are arrange to form a
rectangle of dimension m by n. Shown as, from figure.
= 2702
(ii) The man are in majority = 1008 ways
20. 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
2n +1
C 2 + ... +
C 2 + ... +
2n +1
C n = 63
... (i)
C 2 n + 1 = 22 n + 1
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
21. 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
⇒
n
Number of ways = 3C 2 ⋅6 C1 = 18
x6 x2
x5 x1 x3
x7 x4
m
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
x4 + x5
and
x7 =
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
Permutations and Combinations 81
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.
C
n−r+1
25. We know that, n r =
C r −1
r
84 7 n − r + 1
= =
⇒
36 3
r
⇒
3n − 10r + 3 = 0
n
Cr
84
Also given,
=
n
C r +1 126
r+1 2
⇒
=
n−r 3
n
⇒
[given]
…(i)
2n − 5r − 3 = 0
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,
number of two digit natural numbers that can be
formed = 20,
…(ii)
4×5
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
(Q 0 can not come in Ist box)
number of three digit natural numbers that can be
formed = 100
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
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.
members committee of 11 members is to be formed.
Alternate Method
According to the question, m = number of ways when
there is at least 6 males
Number of ways when A is selected and B is not
= ( C 6 × C 5 ) + ( C7 × C 4 ) + ( C 8 × C 3 )
= (28 × 1) + (8 × 5)+ (1 × 10)
= 28 + 40 + 10 = 78
and n = number of ways when there is at least 3 females
Number of ways when B is selected and A is not
8
5
8
5
8
5
= ( 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
= 5C 2 ⋅5 C 2 = 100
= 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.
82 Permutations and Combinations
6. Clearly, number of words start with A =
4!
= 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 ! .
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.
8. 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
10. Here,
B1
10 !
y 10
10 !
=
=5
= 10 × 9 ×
⇒
2!
9x 2
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
positions between boys = 6C 2
…(i)
4 girls are to be selected out of 5 = 5C 4
…(ii)
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)
6
5
Now, total number of ways
= C × C × 2! × 4! × 5!
2
∴
and
⇒
11.
4
[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.
12. 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
1
k(k + 1)
2
1
1
= (2k + 2n − 2 + k2 − k) = (2n − k2 + k − 2)
2
2
where, r = k + n − p − 1 = k + n − 1 −
9. x = 10 !
y = 10C1 × 9C 8 ×
x1 ≤ x2 ≤ x3 ≤ x4 ≤ x5
Now,
13. Since, six ‘+’ signs are + + + + + +
∴ 4 negative sign has seven places to be arranged in
7
⇒
C 4 ways = 35 ways
14. 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
Topic 4 Distribution of Object into Group
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


× n = 1750
⇒ 5 ×
 +


  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
[Q n ∈ N ]
Permutations and Combinations 83
2. According
to
given
information, we have the
following figure.
(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.
Y
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 !
(0, b)
B
X
O
A (a, 0)
= 60 + 90 = 150
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|
| b|
0
2
1
1
2
0
5. Total number of arrangements of word BANANA
=
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
2
2
2
2
= n C n . n − nC n . n − 2n C n . n − 3nC n .
=
⇒ 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.
7.
4. Objects
Distinct
Distinct
Groups
Distinct
Identical
Objects
Identical
Identical
Groups
Identical
Distinct
C n K nC n
(n 2 )!
(n 2 − n )!
n!
(n 2 )!
.
K
=
n ! (n 2 − n )! n ! (n 2 − 2n )!
n ! ⋅ 1 (n !)n
(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
Number of ways of selecting 4 novels from 6 novels is
6!
6
C4 =
= 15
2 !4 !
∴ Total number of arrangement of 4 novels and 1
dictionary where dictionary is always in the middle, is
15 × 3 × 4 ! = 45 × 24 = 1080
n 2 − 2n
⇒ Integer (as number of arrangements has to be integer).
3. Given 6 different novels and 3 different dictionaries.
Number of ways of selecting 1 dictionary is from 3
3!
dictionaries is 3C1 =
=3
1 !2 !
6!
= 60
3!2!
=
52
C17 × 35C18 × 18C17 × 1C1
(52)!
=
3!
3 !(17)3
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
Description of Situation Here, 5 distinct balls are
distributed amongst 3 persons so that each gets at least
one ball. i.e. Distinct → Distinct
Download Chapter Test
http://tinyurl.com/y32hjn72
1
1
1
1

own colour box = 4 ! 1 − +
−
+ 

1 ! 2 ! 3 ! 4 !
1
1 1
 12 − 4 + 1
= 4!  − +
 = 24 
 =9
 2 6 24


24
or
5
Binomial Theorem
Topic 1 Binomial Expansion and General Term
Objective Questions I (Only one correct option)
1. The coefficient of x in the product
(1 + x)(1 − x) (1 + x + x ) is
10
2 9
(2019 Main, 12 April I)
(b) − 126
(d) 126
(a) 84
(c) − 84
expansion of the expression (1 + ax + bx2) (1 − 3x)15 in
powers of x, then the ordered pair (a , b) is equal to
(2019 Main, 10 April I)
(b) (− 21, 714)
(d) (− 54, 315)
3. The term independent of x in the expansion of
6
1
x8  
3
−  . 2x2 − 2 is equal to

x 
 60 81 
(a) − 72
(c) − 36
(b) 36
(2019 Main, 12 April II)
(d) − 108
4. 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
5. If some three consecutive coefficients in the binomial
expansion of (x + 1)n in powers of x are in the ratio 2 : 15 :
70, then the average of these three coefficients is
(2019 Main, 9 April II)
(a) 964
(b) 227
(c) 232
(d) 625
6. If the fourth term in the binomial expansion of
6
log x 
2
7
 + x 8  (x > 0) is 20 × 8 , then the value of x is

x
(2019 Main, 9 April I)
(a) 8−2
(c) 8
(b) 83
(d) 82
(2019 Main, 8 April I)
x − 1 ) + (x − x − 1 ) , (x > 1) is equal to
6
3
(b) 32
6
(c) 26
(d) 24
9. The total number of irrational terms in the binomial
expansion of (71/5 − 31/10 )60 is
(a) 49
(b) 48
(2019 Main, 12 Jan II)
(c) 54
(d) 55
10. The ratio of the 5th term from the beginning to the 5th
term from the end in the binomial expansion of
10
 1

 3
1 
is
2 +
(2019 Main, 12 Jan I)
1



2(3)3 
1
1
(a) 1 : 2(6)3
(b) 1 : 4(16)3
1
1
(c) 4(36)3 : 1
(d) 2(36)3 : 1
11. The sum of the real values of x for which the middle
8
 x3 3
term in the binomial expansion of 
+  equals
x
3
(2019 Main, 11 Jan I)
5670 is
(a) 4
(b) 0
(c) 6
(d) 8
12. 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
13. 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
(d) 2 2
3
7. If the fourth term in the binomial expansion of
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
(a) 100
(c) 10
(x +
3
(a) 29
2. If the coefficients of x2 and x3 are both zero, in the
(a) (28, 315)
(c) (28, 861)
8. The sum of the coefficients of all even degree terms is x
in the expansion of
18
4
(b) 10
(d) 103
 1 − t6 
14. The coefficient of t in the expansion of 
 is
1−t
4
(2019 Main, 9 Jan II)
(a) 12
(b) 10
(c) 15
(d) 14
15. The sum of the coefficients of all odd degree terms in the
expansion of  x +
(a) −1
(b) 0
5
5
x3 − 1 +  x − x3 − 1 , (x > 1) is
(2018 Main)
(c) 1
(d) 2
Binomial Theorem 85
value
of
(21C1 − 10C1 ) + (21C 2 − 10C 2)
10
21
+ ( C3 − C3 ) + ( C 4 − 10C 4 ) + ... + (21C10 − 10C10 ) is
27. If the coefficients of x3 and x4 in the expansion of
16. The
21
(2017 Main)
(a) 221 − 211
(c) 220 − 29
(b) 221 − 210
(d) 220 − 210
251
(a)  16,


3 
272 
(c)  14,


3 
17. If the number of terms in the expansion of
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
18. The sum of coefficients of integral powers of x in the
binomial expansion (1 − 2 x )
1
1
(a) (350 + 1) (b) (350 )
2
2
is
1
(d) (250 + 1)
2
(1 + x2)4 (1 + x3 )7 (1 + x4 )12 is
(a) 1051
term
(b) 1106
(c) 1113

x+1
x−1 
−
 2/ 3

− x 1 /3 + 1 x − x 1 / 2 
x
(a) 4
(b) 120
of
x
30. For any odd integer n ≥ 1, n − (n − 1) + K
expansion
of
10
(2013 Main)
is
(c) 210
(d) 12C6 + 2
(c) 12C6
22. In the binomial expansion of (a − b) , n ≥ 5 the sum of
(b)
(2001, 1M)
n−4
5
r =1
(c)
5
n−4
(d)
6
n−5
23. If in the expansion of (1 + x) (1 − x) , the coefficients of
m
2n
2n
r=0
r=0
2n + 1
∑ a r (x − 2)r = ∑ br (x − 3)r
show that bn =
n
x and x are 3 and −6 respectively, then m is euqal to
(1999, 2M)
(b) 9
(c) 12
24. The expression [x + (x − 1)
3
] + [x − (x − 1) ]
3
1/ 2 5
polynomial of degree
(a) 5
(c) 7
25. The coefficient of x4 in  −
x
2
405
256
450
(c)
263
(a)
is a
(1992, 2M)
(b) 6
3

x2 
(b)
(1992, 6M)
(1985, 5M)
35. Given, sn = 1 + q + q + K + q
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)
(d) 8
10
is
(1983, 1M)
504
259
(d) None of these
26. 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
Cn+ 1
r

 1 3r
7r 15
r n

C
(
−
1
)
+
+
+
... upto m terms .
r
∑
r
2r
3r
4r
2
2
2
2
r=0

(d) 24
1/ 2 5
and a k = 1 , ∀ k ≥ n, then
34. Find the sum of the series
2
(a) 6
(1993, 5M)
n
the 5th and 6th terms is zero. Then, a / b equals
n−5
6
k
32. Prove that ∑ (−3)r − 1 3 nC 2r − 1 = 0, where k = (3n )/ 2 and n
33. If
n
(a)
Analytical & Descriptive Questions
(d) 310
(2003, 1M)
(b) 12C6 + 1
31. The larger of 9950 + 10050 and 10150 is ... .
is an even positive integer.
21. Coefficient of t 24 in (1 + t 2)12 (1 + t12) (1 + t 24 ) is
(a) 12C6 + 3
3
+ (− 1)n−113 = K
(d) 1120
in
(1983, 2M)
3
(2014 Adv.)
independent
29. If (1 + a x)n = 1 + 8x + 24x2 + … , then a = … and n = K .
(2015 Main)
1
(c) (350 − 1)
2
19. Coefficient of x11 in the expansion of
20. The
28. Let n be a positive integer. If the coefficients of 2nd, 3rd,
and 4th terms in the expansion of (1 + x)n are in AP,
then the value of n is… .
(1994, 2M)
(b) 2187
(d) 729
50
251
(b)  14,


3 
272 
(d)  16,


3 
Fill in the Blanks
(2016 Main)
(a) 64
(c) 243
(1 + ax + bx2)(1 − 2x)18 in powers of x are both zero, then
(a , b) is equal to
(b) n = 2r + 1
(d) None of these
Integer Answer Type Question
36. 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.)
37. The coefficient of x9 in the expansion of
(1 + x) (1 + x2) (1 + x3 ) ... (1 + x100 ) is
(2015 Adv.)
38. 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.)
86 Binomial Theorem
Topic 2 Properties of Binomial Coefficient
Objective Questions I (Only one correct option)
1. Let (x + 10) + (x − 10)
(c) (−1)n/ 2 (n + 1)
= a 0 + a1x + a 2x + K + a50x ,
a2
for all x ∈ R; then
is equal to
(2019 Main, 11 Jan II)
a0
50
(a) 12.25
50
2
(b) 12.50
(c) 12.00
50
Cr
C0 +
20
C1 +
20
20
Cr−1
Cr− 2 20C2 + .... + 20C 020C r
20
is maximum, is
(a) 15
(c) 11
(d) 20
403
2
k
3. If the fractional part of the number
is , then k is
15
15
equal to
(2019 Main, 9 Jan I)
(a) 14
(b) 6
10
C r,
(c) 4
(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
∑ Ar (B10Br − C10 Ar )
(2010)
 30 30 30 30 30
30 30
5. 30
0  10 −  1  11 +  2  12 + K + 20 30 is equal
to
(2005, 1M)
(a) 30 C11
(c) 30 C10
6. If
n −1
(b)
(d)
C10
65
C55
C r = (k2 − 3) nC r + 1, then k belongs to
(a) (− ∞ , − 2]
(c) [ − 3 , 3 ]
sum
(2004, 1M)
(b) [2, ∞ )
(d) ( 3 , 2]
m
7. The
60
∑
i=0
10 20  , where  p = 0 if p > q,
 i  m − i
 q
maximum when m is equal to
(a) 5
(b) 10
(c) 15
13. Prove that
 n  n
 n  n − 1
k − 2  n  n − 2
2k     − 2k −1   
 +2   
 − ...
 0  k
 1  k − 1
 2  k − 2
 n  n − k  n
+ (−1)k   
 =   (2003, 4 M)
 k  0   k
 n
 r
n
n
1
9. If a n = ∑ n , then
C
r
r=0
(a) (n − 1) an
(c)
1
n an
2
 n + 2
(c) 2 

 r 
 n + 2
(d) 

 r 
r
∑ nC r equals
r=0
(b) n an
(1998, 2M)
(1 + x + x2)n = a 0 + a1x + ... + a 2n x2n.
Then, show that, a 02 − a12 + ... + a 22n = a n.
(1994, 5M)
17. Prove that C 0 − 2 ⋅ C1 + 3 ⋅ C 2 − ... + (−1) (n + 1) 2⋅ C n
2
2
n
= 0 , n > 2, where C r = nC r.
(1989, 5M)
2
n
that the sum of the products of the Ci’s taken two at a
time represented by Σ Σ CiC j is equal to
(2n !)
.
0 ≤ i < j ≤ n 22 n −1 −
(1983, 3M)
2 (n !)2
19. Prove that C12 − 2 ⋅ C 22 + 3 ⋅ C32 − ...−2n ⋅ C 22n = (−1)n n ⋅ C n
 n  n
2  !  !
 2  2
[C 02 − 2 C12 + 3 C 22 − ... + (−1)n (n + 1) C n2 ],
n!
(b) (−1)n (n + 1)
(1997C, 5M)
18. If (1 + x) = C 0 + C1x + C 2 x + ... + C nx , then show
10. If C r stands for nC r, then the sum of the series
(a) (−1)n/ 2 (n + 2)
n
 nC 
3!
= ∑ (−1)r  r + 3 r  .
2(n + 3) r = 0
Cr 

n
(d) None of these
where n is an even positive integer, is
15. Prove that
16. If n is a positive integer and
 n   n 
+
 is equal to
 r − 1  r − 2
(2000, 2M)
 n + 1
(b) 2 

 r + 1
 n
If   = nCm. Prove that
m
 n   n − 1  n − 2
m  n + 1
 +
+ 
 + ... +   = 

m  m   m 
m m + 1
or
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
(2002, 1M)
8. For 2 ≤ r ≤ n,   + 2 
 n + 1
(a) 

 r − 1
14. For any positive integers m, n (with n ≥ m),
is
(d) 20
(1982, 2M)
Analytical & Descriptive Questions
is equal to
2
(b) A10 (B10
− C10 A10 )
(d) C10 − B10
1
X is .......... .
(2018 Adv.)
1430
(1 + x − 3x2)2163 is …. .
r =1
(a) B10 − C10
(c) 0
r ∈{1, 2,... , 10}
12. The sum of the coefficients of the polynomial
10
and (1 + x)30. Then,
+ ... + 10(10C10 )2,
denote
binomial
Fill in the Blank
(2019 Main, 11 Jan I)
(b) 10
11. Let X = (10C1 )2 + 2(10C 2)2 + 3(10C3 )2
coefficients. Then, the value of
2. The value of r for which
20
Numerical Value
where
(d) 12.75
(d) None of these
(1986, 2M)
(1979, 4M)
20. Prove that ( C 0 ) − ( C1 ) + ( C 2) − ... + (2 nC 2 n )2
2n
= (−1)n ⋅2n C n.
2
2n
2
2n
2
(1978, 4M)
Binomial Theorem 87
Answers
Topic 1
1. (a)
5.
9.
13.
17.
21.
25.
2. (a)
3. (c)
4. (d)
6.
10.
14.
18.
22.
26.
(d)
7. (c)
(c)
11. (b)
(c)
15. (d)
(a)
19. (c)
(b)
23. (c)
(a)
27. (d)
1
2
29. (a = 2, n = 4) 30. (n + 1 ) (2n − 1 )
4
(c)
(c)
(b)
(d)
(d)
(a)
8.
12.
16.
20.
24.
28.
(d)
(d)
(d)
(c)
(c)
(n = 7)
31. (101 ) 50
 2mn − 1 
34.  mn n

2 (2 − 1 ) 
36. (5)
37. (8)
38. (n = 6 )
2. (d)
6. (d)
10. (a)
3. (d)
7. (c)
11. (646)
Topic 2
1. (a)
5. (c)
9. (c)
4. (d)
8. (d)
12. ( − 1 )
Hints & Solutions
Topic 1 Binomial Expansion and General Term
⇒
1. Given expression is
(1 + x) (1 − x) (1 + x + x )
10
2 9
21a − b = 273
24a = 672 ⇒ a = 28
= (1 + x) (1 − x) [(1 − x) (1 + x + x2)]9
= (1 − x ) (1 − x )
2
3 9
(1 + x) (1 − x)10 (1 + x + x2)9
3.
= coefficient of x18 in the product (1 − x2) (1 − x3 )9
in (1 − x )
− coefficient of x16 in (1 − x3 )9
term in the expansion of
= coefficient of x
Since, (r + 1)th
Key Idea Use the general term (or (r + 1)th term) in the
expansion of binomial (a + b) n
3 9
(1 − x3 )9 is 9C r (− x3 )r = 9C r (− 1)r x3 r
Now, for x18, 3r = 18 ⇒ r = 6
and for x16, 3r = 16
16
⇒
r=
∉N.
3
i.e.
6
3

Let a binomial 2x2 − 2 , it’s (r + 1)th term

x 
r
 3
= Tr + 1 = 6C r (2x2)6 − r  − 2
 x 
= 6C r (−3)r (2)6 − r x12 − 4r
9!
9 ×8 × 7
∴Required coefficient is C 6 =
=
= 84
6 !3 !
3 ×2
2. 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 = 15 C r (−3)r xr
Now, coefficient of x2,
(1 + ax + bx2)(1 − 3x)15 is
in
the
expansion
of
C 2(−3)2 + a15C1 (−3)1 + b 15C 0 (−3)0 = 0 (given)
⇒ (105 × 9) − 45 a + b = 0
⇒ 45a − b = 945
Now, the
1
x8 
− 

 60 81
C3 (−3)3 + a 15C 2(−3)2 + b 15C1 (−3)1 = 0
⇒ − 12285 + 945a − 45b = 0
15
=
…(i)
63a − 3b = 819
(given)
…(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 
Similarly, the coefficient of x3 , in the expansion of
(1 + ax + bx2)(1 − 3x)15 is
⇒
T r + 1 = nC r a n − r b r
= 6C r (− 3)r (2)6 − r x12 − 2r − 2r
9
15
b = 315
(a , b) = (28, 315)
So,
⇒
Now, coefficient of x18 in the product
18
…(ii)
From Eqs. (i) and (ii), we get
=
6
C3
(− 3)3 (2)6 − 3 x12 − 4 (3 )
60
 1
+  −  6C5 (−3)5 (2)6 − 5 x12 − 4 (5 ) x8
 81
1
35 × 2(6)
(− 3)3 23 +
3
81
= 36 − 72 = − 36
[put r = 3]
[put r = 5]
88 Binomial Theorem
 2
T4 = T3 + 1 = 6 C3  
 x
3
 2 3 log 8 x
⇒ 20   x
= 20 × 87
 x
∴
n
1
4. Given binomial is  x2 + 3  , its (r + 1)th term, is

x 
r
1
 1
Tr + 1 = nC r (x2)n − r  3  = nC rx2n − 2r 3 r
x 
x
= nC rx2n − 2r − 3 r = nC rx2n − 5 r
For the coefficient of x ,
2n − 5r = 1 ⇒ 2n = 5r + 1 …(i)
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.
5.
6 −3
(xlog 8 x )3 = 20 × 87 (given)
[Q 6C3 = 20]

3
 log 2 x −3 

3
⇒ 23 x [3(log 8 x )−3 ]= (23 )7 ⇒ x
1


Q log an (x) = n log a x for x > 0; a > 0, ≠ 1
log x − 3 )
= 218
⇒ x( 2
On taking log 2 both sides, we get
(log 2 x − 3) log 2 x = 18
⇒
(log 2 x)2 − 3 log 2 x − 18 = 0
If n − r = 23, then from Eq. (i) on replacing the value of ‘
r’, we get 2n = 5(n − 23) + 1
⇒ (log 2 x)2 − 6 log 2 x + 3 log 2 x − 18 = 0
⇒ 2n = 5n − 115 + 1 ⇒ 3n = 114 ⇒ n = 38
So, the required smallest natural number n = 38.
⇒
⇒ log 2 x(log 2 x − 6) + 3 (log 2 x − 6) = 0
Key Idea Use general term of Binomial expansion ( x + a) i.e.
Tr + 1 = C r 1 x
n−r
a
(log 2 x − 6) (log 2 x + 3) = 0
⇒
n
n
= (23 )6
log 2 x = −3, 6
⇒ x = 2 −3 , 2 6 ⇒ x =
r
1 2
,8
8
6
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
n
Cr − 1
Now,
⇒
⇒
⇒
n
Cr
2n − 17r + 2 = 0
C
Similarly, n r
Cr + 1
⇒
3
1

 2 1 3
6
1 + log10 x   12

= 200
x
∴ C3 x

  


2
=
15
n!
(r − 1)!(n − r + 1)! 2
=
n!
15
r !(n − r )!
r
2
=
⇒ 15r = 2n − 2r + 2
n − r + 1 15
n
⇒



 

1
1
7. Given binomial is   1 + log10 x 

12
+x 
 x


Since, the fourth term in the given expansion is 200.
⇒
⇒
⇒
…(i)
n!
3
15
r !(n − r )!
=
⇒
=
n!
14
70
(r + 1)!(n − r − 1)!
⇒
⇒
Since,
…(ii)
n − 16 = 0 ⇒ n = 16 and r = 2
C1 +
(7 + log10 x) log10 x = 4 + 4 log10 x
t 2 + 7t = 4 + 4t
t + 3t − 4 = 0
t = 1 , −4 = log10 x
2
[let log10 x = t]
x = 10, 10−4
x = 10
x>1
8.
Key Idea Use formula :
C 2 + 16C3
Now, the average =
3
16 + 120 + 560
696
=
=
= 232
3
3
16
2
log x
6. Given binomial is  + x 8 

x
= 200
x
= 10

3
1
2(1 + log x) + 4  log10 x = 1


10
[applying log10 both sides]
⇒
⇒
⇒
On solving Eqs. (i) and (ii), we get
16
3
1
+
2(1 + log10 x ) 4
⇒ [6 + (1 + log10 x)] log10 x = 4(1 + log10 x)
r+1
3
=
⇒ 14r + 14 = 3n − 3r
n − r 14
3n − 17r − 14 = 0

3
1
 2 (1 + log x ) + 4 


10
20 × x
6
Since, general term in the expansion of (x + a )n is
Tr+ 1 = n C r xn− ra r
( a + b) n + ( a − b) n =
2 [ n C 0 a n + nC 2a n − 2b 2 + nC 4 a n − 4 b 4 + ...... ]
Given expression is (x +
x3 − 1 )6 + (x − x3 − 1 )6
= 2 [6C 0x6 + 6C 2x4 ( x3 − 1 )2
+ 6C 4x2( x3 − 1 )4 + 6C 6 ( x3 − 1 )6 ]
{Q (a + b)n + (a − b)n
= 2 [ C 0a + C 2a
n
n
n
n−2 2
b + nC 4a n − 4b4 + …]}
Binomial Theorem 89
= 2 [6C 0x6 + 6C 2x4 (x3 − 1) + 6C 4x2(x3 − 1)2 + 6C 6 (x3 − 1)3 ]
12. 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 
10 − r
r

 λ 
Tr+ 1 = x2  10C r ( x )10− r  2  =10C r x2 ⋅ x 2 λr x−2r
x  

The sum of the terms with even power of x
= 2 [6C 0x6 + 6C 2(− x4 ) + 6C 4x8 + 6C 4x2 + 6C 6 (−1 − 3x6 )]
= 2 [6C 0x6 − 6C 2x4 + 6C 4x8 + 6C 4x2 − 1 − 3x6 ]
Now, the required sum of the coefficients of even powers
of x in
(x +
x3 − 1 )6 + (x − x3 − 1 )6
=10C r λr x
= 2 [6C 0 − 6C 2 + 6 C 4 + 6C 4 − 1 − 3]
2+
10 − r
− 2r
2
Now, for the coefficient of x2, put 2 +
= 2 [1 − 15 + 15 + 15 − 1 − 3] = 2(15 − 3) = 24
10 − r
− 2r = 0
2
10 − r = 4r ⇒ r = 2
⇒
9. The general term in the binomial expansion of (a + b)n
10 − r
− 2r = 2
2
is Tr + 1 = nC r a n − rbr.
⇒
So, the general term in the binomial expansion of
(71/5 − 31/10 )60 is
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]
C r (71/5 )60 − r (−31/10 )r
Tr + 1 =
60
=
60
Cr 7
60 − r
5 (−1 )r
r
310 = (−1)r
60
12 −
Cr 7
r
5
r
310
The possible non-negative integral values of ‘r’ for
r
r
which
and
are integer, where r ≤ 60, are
5
10
r = 0, 10, 20, 30, 40, 50, 60.
13. 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
∴There are 7 rational terms in the binomial expansion
and remaining 61 − 7 = 54 terms are irrational terms.
5
⇒ 5 C 2(1)5 − 2(xlog 2 x )2 = 2560 (given)
10. Since, rth term from the end in the expansion of a
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
∴Required ratio =
= 5 =
T10 − 5 + 2 T7 T6 + 1
10
⇒
T5
=
T10 − 5 + 2
=
10
 1 
C 4 (21/3 )10− 4 

 2(3)1/3 
4
 1 
C 6 (21/3 )10− 6 

 2(3)1/3 
6
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
= 4(6)23
= 4(36)1/3
So, the required ratio is 4(36)1/3 : 1 .
8
3
+  , the middle term is T4 + 1.
x
3
 x3
11. In the expansion of 
[Q Here, n = 8, which is even, therefore middle term
 n + 2
=
 th term]
 2 
4
 x3   3 4 8 ⋅ 7 ⋅ 6 ⋅ 5 8
x
∴ 5670 = C4     =
1⋅ 2⋅ 3⋅ 4
 3   x
8− r
r

 3
 3 
QTr + 1 = 8Cr  x 
 
 x 

 3


⇒ x 8 = 34 ⇒ x = ± 3
8
So, sum of all values of x i.e + 3 and − 3 = 0
C 2(1)5 − 2(xlog 2 x )2
⇒
⇒
10 (xlog 2 x )2 = 2560
x( 2log 2 x ) = 256
⇒
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
2
(log 2 x) = 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 
14. 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)
15.
Key Idea Use formula :
= ( a + b) n + ( a − b) n
= 2 ( n C 0 a n + nC 2a n − 2b 2 + nC 4 a n − 4 b 4 + ...)
We have, (x +
x3 − 1 )5 + (x − x3 − 1 )5 , x > 1
= 2(5 C 0x5 + 5C 2x3 ( x3 − 1 )2 + 5C 4x( x3 − 1 )4 )
= 2(x5 + 10x3 (x3 − 1) + 5x(x3 − 1)2)
= 2(x5 + 10x6 − 10x3 + 5x7 − 10x4 + 5x)
Sum of coefficients of all odd degree terms is
2 (1 − 10 + 5 + 5) = 2
90 Binomial Theorem
16. ( 21C1 − 10C1 ) + (21C 2 − 10C 2) + (21C3 − 10C3 )
19. Coefficient of xr in (1 + x)n is nC r.
+ ... + (21C10 − 10C10 )
= ( C1 + C 2 + ... + C10 ) − ( C1 + 10C 2 + ... + 10C10 )
1
= (21C1 + 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
21
21
21
10
17. Clearly, number of terms in the expansion of
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
Hence, sum of coefficients = (1 − 2 + 4)6 = 36 = 729
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.
Note As
18. Let Tt +1 be the general term in the expension of
(1 − 2 x )50
∴
Tr+ 1 =
50 − r
50
C r (1)
(−2x ) =
1/ 2 r
C r2 x (−1)
50
r r/ 2
r
For the integral power of x, r should be even integer.
25
∴ Sum of coefficients = ∑ 50C 2r (2)2r
r= 0
1
1
= [(1 + 2)50 + (1 − 2)50 ] = (350 + 1)
2
2
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)
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.
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

20. 
x+1
x
2/3
10
 (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 = 10C r (x1/3 )10 − r (− x− 1/ 2)r = 10C r (− 1)r x
21. Here, Coefficient of t 24 in {(1 + t 2)12(1 + t12)(1 + t 24 )}
= Coefficient of t 24 in {(1 + t 2)12 ⋅ (1 + t12 + t 24 + t36 )}
= 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]
= Coefficient of t 24 = (12C12 +
C 4a n − 4b4 − nC5 a n − 5 b5 = 0
⇒
n
= C 0 + C 2(2 x )2 + ... + C50 (2 x )50
On putting x = 1, we get
(1 − 2 1 )50 + (1 + 2 1 )50
= C 0 + C 2 + ... + C50 (2)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
C6 +
12
C0 ) = 2 +
12
12
C6
22. Given, T5 + T6 = 0
(1 − 2 x )50 + (1 + 2 x )50
(1 − 2 x ) + (1 + 2 x )50
2
10 − r r
−
3
2
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
n
⇒
10
10
⇒
= 2 [C 0 + C 2(2 x )2 + ... + C50 (2 x )50 ] ...(iii)
( x − 1) 

x − x1/ 2 
 (x1/ 3 )3 + 13
{( x )2 − 1} 
−
=  23

1/3
/
x ( x − 1) 
x − x + 1
On adding Eqs. (i) and (ii), we get
50
− x1/ 3 + 1
−
C 4a n − 4b4 = nC5 a n − 5 b5 ⇒
a nC5 n − 4
=
=
b nC 4
5
m (m − 1) 2

x + K
2

n (n − 1) 2


x − ...
1 − nx +
2

m(m − 1) n (n − 1)
 2
= 1 + (m − n ) x + 
+
− mn  x + K
2
2


23. (1 + x)m (1 − x)n = 1 + mx +

term containing power of x ≥ 3.
Now,
and
m − n =3
…(i)
[Q coefficient of x = 3, given]
1
1
m(m − 1) + n (n − 1) − mn = − 6
2
2
Binomial Theorem 91
⇒
⇒
⇒
⇒
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
Similarly, coefficient of x4 = 0
⇒
…(ii)
On solving Eqs. (i) and (ii), we get m = 12
18
C 4 ⋅ 24 − a ⋅18 C3 23 + b ⋅18 C 2 ⋅ 22 = 0
∴
32a − 3b = 240
a = 16, b =
24. We know that,
(a + b)5 + (a − b)5 = 5C 0a5 + 5C1a 4b + 5C 2a3 b2
expansion of (1 + x)n is nC1, nC 2, nC3 .
According to given condition,
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
= 2 [a5 + 10a3 b2 + 5ab4 ]
∴ [x + (x − 1) ] + [x − (x − 1) ]
1/ 2 5
3
1/ 2 5
= 2 [x5 + 10x3 (x3 − 1) + 5x (x3 − 1)2]
Therefore, the given expression is a polynomial of
degree 7.
25. The general term in  −
x
2
⇒
⇒
⇒
10
3
 is
x2 
r
 3
r
 2 = (−1)
x 
For coefficient of x4, we put 10 − 3r = 4
⇒
⇒
⇒
⇒
⇒
3r = 6
or
⇒
r =2
tr + 1 = (−1)
 x
Cr  
 2
10 − r
r 10
3r
10
C r.
210 − r
10 − 3 r
⋅x
But C3 is true for n ≥ 3, therefore n = 7 is the answer.
10
32
= (−1)2. 10C 2. 8
2
45 × 9 405
=
=
256
256
29. Given,
⇒
26. In the expansion (1 + x)2n, t3 r = 2nC3 r − 1 (x)3 r − 1
tr + 2 =
2n
C r + 1 (x)
∴
r+1
Since, binomial coefficients of t3 r and tr + 2 are equal.
∴
⇒
⇒
⇒
But
C3 r −1 = 2nC r + 1
2n
= n3 + (n − 1)3 + (n − 2)3 + K + 13
rn
n
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
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
3
[Q n − 1, n − 3, ... , are even integers]
  n − 1 3 
= Σ n3 − 16 Σ 
 
  2  
2
18
17a − b =
− 2 [(n − 1)3 + (n − 3)3 + K + 23 )]

  n − 1 3  n − 3 3
3
= Σ n3 − 2 × 23 
 +K+1 
 +




2

 2
1
 n (n + 1) 
− 16 
=

2


2
Given, coefficient of x3 = 0
⇒
n =4
n3 − (n − 1)3 + (n − 2)3 − (n − 3)3 + K + (− 1)n−1 ⋅ 13
coefficient of x in (1 − x) is (−1) Cr and then simplify.
In expansion of (1 + ax + bx2)(1 − 2x)18.
⇒
and
and n − 1, n − 3, n − 5, etc., are even integers, then
27. To find the coefficient of x3 and x4, use the formula of
=
a =2
30. Since, n is an odd integer, (− 1)n−1 = 1
∴ We take, n = 2r
18
⇒
⇒
(1 + ax)n = 1 + 8x + 24x2 + ...
n (n − 1) 2 2
1 + anx +
a x + ... = 1 + 8x + 24x2 + ...
2!
n (n − 1)
an = 8 and a 2
= 24
2
8 (8 − a ) = 48
8 − a =6 ⇒ a =2
Hence,
3r − 1 = r + 1 or 2n = (3r − 1) + (r + 1)
or 2n = 4r
2r = 2
or
r =1
n = 2r
r >1
r
n=7
n
x 3
∴ Coefficient of x4 in  − 2
2 x 
and
272
3
28. Let the coefficients of 2nd, 3rd and 4th terms in the
+ 5C3 a 2b3 + 5C 4ab4 + 5C5 b5 + 5C 0a5 − 5C1a 4b
+ 5C 2a3 b2 − 5C3 a 2b3 + 5C 4ab4 − 5C5 b5
3
…(ii)
On solving Eqs. (i) and (ii), we get
..(i)

 n − 1  n − 1
+ 1 
 


 2   2
2
=
1 2
16 (n − 1)2(n + 1)2
n (n + 1)2 −
4
4 ×4 ×4
=
1
1
(n + 1)2 [n 2 − (n − 1)2] = (n + 1)2(2n − 1)
4
4
92 Binomial Theorem
31. Consider, (101)50 − (99)50 − (100)50
33. Let y = (x − a )m, where m is a positive integer, r ≤ m
= (100 + 1)50 − (100 − 1)50 − (100)50
= {(100)
(1 + 0.01)
50
− (1 − 0.01)
50
= (100) {2 ⋅ [ C1 (0.01) +
50
= (100) {2 [ C3 (0.01) +
50
50
50
∴
C3 (0.01)3 + K ] − 1}
50
50
⇒
C5 (0.01)5 + ... ]}
3
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
…………………………………
…………………………………
Now,
− 1)}
50
(101)50 − {(99)50 + (100)50 } > 0
⇒
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 )!
(101)50 > (99)50 + (100)50
32. Since, n is an even positive integer, we can write
n = 2 m ,m = 1, 2, 3, K
3m
3n 3(2m)
Also, k =
=
= 3m ∴ S = ∑ (−3)r − 1 ⋅6mC 2r − 1
2
2
r =1
S = (−3)
i.e.
C1 + (−3)
C3 + K
3 m − 1 6m
0 6m
and for r > m,
6m
+ (−3)
⋅
2n
…(i)
C3m − 1
C0 +
6m
C1x +
6m
6m
6m
C2
6m − 1
C 6 m − 1x
r=0
(1 − x)6 m =
C0 +
6m
+
C1 (− x) +
6m
C 6 m − 1 (− x)
6m
x2 + K
+
C6mx
6m
C 6 m (− x)6 m
6m
r=n
+
⇒
(1 + x)
− (1 − x)
2x
6m
6m
C3 x3
(1 +
⇒
[since, all the terms except first on RHS become zero]
C5 x5 + K + 6 m C 6 m − 1x6 m − 1 ]
C3 x2 +
⇒
y )6 m − (1 −
2 y
y )6 m
+
=
S=
=
⇒
and
Now,
= (n + 2C n + 1 +
C1 +
6m
6m
C3 y
C5 y2 + K + 6 m C 6 m − 1 y 3 m − 1
6m
⇒
S=
n+ 2
Cn + K +
z 6 m − z 6 m r 6 m (2i sin 6 m θ )
=
2i 3
2i 3
26 m sin 6 m θ
=
3
= 0 as m ∈ z , and θ = π /3
2n
Cn
n+ 2
C n ) + K + 2n C n
Cn + 1 + K +
Cn + 1 +
Cn =
2n
2n
Cn
2n + 1
= ....
Cn + 1

1
3r
7r
15r
 r + 2r + 3 r + 4r + ... upto m terms 
2
2
2

2
n
r
 1
∑ (−1)r nC r  2 +
r=0
r
n
 3
∑ (−1)r nC r  4 +
r=0
n
 7
∑ (−1)r nC r  8 
r
+ ... upto m terms
r=0
…(iv)
n
n
n
1
3
7



= 1 −  + 1 −  + 1 −  + ... upto m terms



2
4
8

using

n
n
n

r= 0

∑ (−1)r nC rxr = (1 − x)n 
n
 1
 1
 1
=   +   +   + ... upto m terms
 8
 4
 2
z = r (cos θ − i sin θ )
From Eq. (i),
=
∑ (−1)r nC r
=
m

 1 
1 −  n  
2  
2mn − 1
 1
=  
= mn n
1
 2 
 2 (2 − 1)
 1 − 2n 


(z )6 m = r 6 m (cos 6m θ − i sin 6m θ )
z 6 m − z 6 m = r 6 m (2i sin 6m θ )
n+3
2n
r=0
−3 )6m − (1 − −3 )6m
2 −3
(1 + i 3 )6m − (1 − i 3 )6m
2i 3
=
n
34.
r = |z| = 1 + 3 = 2
θ = π /3
z 6 m = [r (cos θ + i sin θ )]6 m
= r 6 m (cos 6m θ + i sin 6m θ )
Again,
Cn +
[Q a r = 1, ∀ r ≥ n ]
C5 x4 + ...
z = 1 + i 3 = r (cos θ + i sin θ )
Let
and
(1 +
n+1
6m
For the required sum we have to put y = − 3 in RHS.
∴
bn = nC n +
+ 6 mC 6 m − 1 x6 m − 2
x2 = y
Let
r=n
6m
6m
∑ a r (n !)rC n = (bn )n !
On putting x = 3, we get
6m
= 6mC1 +
r=n
2n
…(iii)
On subtracting Eq. (iii) from Eq. (ii), we get
(1 + x)6 m − (1 − x)6 m = 2 [6 mC1x +
2n
2n
∑ a r (n !)rC n (x − 2)r − n = ∑ br (n !)rC n (x − 3)r − n
…(ii)
C 2(− x)2 + K
+
[given]
r=0
On differentiating both sides n times w.r.t. x, we get
6m
6m
6m − 1
2n
∑ a r (x − 2)r = ∑ br (x − 3)r
Now,
From the binomial expansion, we write
(1 + x)6 m =
dr y
=0
dxr
n
…(v)
35.
n+ 1
C1 +
n+ 1
C 2 s1 +
n+ 1
=
n+ 1
C3 s2 + ... +
∑ n + 1C rsr −1,
r =1
n +1
C n+ 1sn
Binomial Theorem 93
where sn = 1 + q + q2 + ... + qn =
n+ 1
 1 − qr 
1
1 − qn + 1
1−q
Topic 2 Properties of Binomial Coefficient
 n +1
n+ 1

 r =1
r =1

∑ n + 1C r  1 − q  = 1 − q  ∑ n + 1C r − ∑ n + 1C r qr
∴
r =1
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 
(x + 10)50 + (x − 10)50 = a 0 + a1x + a 2x2 + … + a50x50
∴a 0 + a1x + a 2x2 + … + a50x50
= [(50C 0x50 +
C1 +
n+ 1
C 2 s1 +
2n + 1 − (q + 1)n + 1
=
2n (1 − q)
C3 s2 + ... +
n+ 1
a 2 = 2 50C 48 (10)48 ; a 0 = 2 50C50 (10)50 = 2(10)50
... (ii)
a 2 2(50C 2)(10)48
50 ⋅ 49 (10)48
=
=
2
a0
1 ⋅ 2 2 ⋅ (10)50
2 (10)50
∴
=
C n + 1sn = 2 S n
n
⇒
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.
(1 + x)(1 + x )(1 + x ) K (1 + x ) =Terms having x
= [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 + 5 − (r − 1)
=2
r
and
n−r+5 7
=
r+1
5
n − r + 6 = 2 r 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
20
(1 + x)20 =
20
C0 +
C r − 1xr − 1 +
20
C1x +
C 2x2 + ... +
20
C rxr + ... +
20
20
C 20x20
∴ (1 + x)20 ⋅ (1 + x)20 = (20C 0 +
20
C 2x + ... +
2
× (20C 0 +
20
C r − 1x
r −1
+
C1x +
20
C rx + ... +
20
20
r
C r − 1xr − 1 +
C1x + ...+
20
20
c20x20 )
20
C r xr
+ ....+ 20C 20x20 )
⇒ (1 + x)
40
= ( C0 .
20
Cr +
20
20
C1
20
C r − 1 ...
20
C r20C 0 ) xr + ...
On comparing the coefficient of xr of both sides, we get
20
C 020C r +
C120C r − 1 + ... +
20
9
38. Let the three consecutive terms in (1 + x)n + 5 be
tr , tr + 1 , tr + 2 having coefficients
n+ 5
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
10
5
Cr
C r −1
50 × 49
5 × 49 245
=
= 12 .25
=
20
2 ⋅ (10 × 10)
20
20
Cr
20
C0 =
40
Cr
The maximum value of 40C r is possible only when r = 20
37. Coefficient of x9 in the expansion of
⇒
C 48 = 50C 2]
50
2. We know that,
{(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
50
2
51
⇒
C3 + C 2m = (3n + 1) ⋅ 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
100
C50 ⋅ 1050 ]
50
By comparing coefficients, we get
36. Coefficient of x in the expansion of
3
C50 1050 )
C50 1050 )]
[Q
n+ 1
50
50
+…+
2
2
C 2 x48 ⋅ 102 + … +
50
+ (50C 0x50 − 50C1x49 10 + 50C 2 x48102− … +
= 2 [50C 0 x50 + 50C 2x48 ⋅ 102 + 50C 4x46 ⋅ 104
From Eqs. (i) and (ii),
n+ 1
C1x49 10 +
50
n
n +1
 q + 1
1−

 2 
=
 q + 1
1−

 2 
1. We have,
[Q nC n/2 is maximum when n is even]
Thus, required value of r is 20.
3. Consider,
2403 = 2400 + 3 = 8 ⋅ 2400 = 8 ⋅ (24 )100 = 8 (16)100= 8(1 + 15)100
= 8 (1 +
C1 (15) +
100
C 2(15)2 + … +
100
100
C100 (15)100 )
[By binomial theorem,
(1 + x)n = nC 0 + nC1x + nC 2x2 + … nC nxn , n ∈ N ]
= 8 + 8 (100C1 (15) +
C 2(15)2 + … +
100
100
C100 (15)100 )
= 8 + 8 × 15λ
where λ =100 C1 +......+ 100C100 (15)99 ∈ N
∴
⇒
2403 8 + 8 × 15λ
8
=
= 8λ +
15
15
15
2403  8

=
 15  15
(where {⋅} is the fractional part function)
∴
k =8
94 Binomial Theorem
m
Alternate Method
2
= 8 ⋅2
403
400
⇒
= 8(16)
100
10  20 
m
 is the coefficient of x in the
 
 i  m − i
∑
i=0
expansion of (1 + x)30
10  20  30
30
…(i)
 
 = Cm =  
 i  m − i
 m
Note that, when 16 is divided by 15, gives remainder 1.
∴ When (16)100 is divided by 15, gives remainder 1100 = 1
and when 8(16)100 is divided by 15, gives remainder 8.
(where {⋅} is the fractional part function)
⇒
k =8
4. A r = Coefficient of xr in (1 + x)10 = 10C r
B r = Coefficient of xr in (1 + x)20 =
20
C r = Coefficient of x in
30
(1 + x)
30
r
∴
=
Cr
Cr
10
10
10
r =1
r =1
r =1
=
10
10
r =1
r =1
=
 n   n    n + 1  n + 1  n + 2
+ 
+
 =

= 
+
 r − 1  r − 2   r   r − 1   r 
n
10
9. Let b =
10
∑ 10C10 − r 20C10 20C r − ∑ 10C10 − r 30C10 10C r
20
C10
10
10
r =1
r =1
C10 ( 30C10 − 1) − 30C10 (20C10 − 1)
30 30
 2  12
30 30
20 30
5. Let A =     −     +     − ... +    
A = 30C 0 ⋅30 C10 − 30C1 ⋅30 C11 +
= [C 02 − C12 + C 22 − C32 + ... + (−1)nC n2 ]
C 2 ⋅30 C12
−K +
− [ 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
C 20 ⋅ C30
30
= Coefficient of x20 in (1 − x2)30
30
∑ (−1)r
30
C r (x2)r
= (−1)n/ 2
r=0
= (−1)
[for coefficient of x20, put r = 10]
10 30
C10
= C10
30
6. Given ,
⇒
⇒
n −1
C r = (k2 − 3) nC r + 1
n −1
C r = (k2 − 3)
k2 − 3 =
n
r+1
n −1
r+1
n
Cr
r+1
≤ 1 and n , r > 0]
n
2
2
⇒
0 < k −3 ≤1 ⇒ 3 < k ≤4
k ∈ [−2, − 3 ) ∪ ( 3 , 2]
⇒
m
10  20 
m
7. ∑   
 is the coefficient of x in the expansion of
i
m
i
−




i=0
[since, n ≥ r ⇒
(1 + x) (x + 1) ,
10
20
n
an
2
C 02 − 2C12 + 3C 22 − 4C32 + ... + (−1)n (n + 1) C n2
= Coefficient of x20 in (1 + x)30 (1 − x)30
= Coefficient of x20 in
[Q nC r = nC n − r ]
10. We have,
30
30
Cr ]
n
= na n − b ⇒ 2b = na n ⇒ b =
C10 = C10 − B10
20
n+1
n
n−r
1
− ∑ n
C
r
r=0
r = 0 Cr
n
n−r
= na n − ∑ n
r = 0 Cn − r
20
30 30
 1  11
[Q nC r + nC r − 1 =
n
r
n − (n − r )
= ∑
n
n
Cr
Cr
r=0
∑
=n
∑ 10C10 − r ⋅ 20C r − 30C10 ∑ 10C10 − r 10C r
30 30
 0  10
∑
r=0
n
r =1
= 30C10 −
∴
 n   n   n  n  
+
 =   + 

 r − 1  r − 2  r   r − 1 
 n
 r
∑ 10C r 20C10 20C r − ∑ 10C r 30C10 10C r
r =1
=
∑
i=0
8.   + 2 
∑ Ar (B10 Br − C10 Ar ) = ∑ Ar B10 Br − ∑ Ar C10 Ar
=
i.e.
 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
2403  8

= .
 15  15
∴
m
n!
n

1 + 

n
n
2
   
 !  !
 2  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
11. We have,
X = (10C1 )2 + 2(10C 2)2+ 3(10C3 )2 + ... + 10 (10C10 )2
⇒
⇒
X=
10
∑ r (10C r )2 ⇒
X=
r =1
r =1
10
10 9
X = ∑r×
Cr − 1
r
r =1
10
∑ r 10C r 10C r
10
Cr
n
 n
Q C r = r

Cr − 1 

n −1
Binomial Theorem 95
10
r =1
10
⇒ X = 10
∑ 9C r − 1 10C10 − r
[Q nC r = nC n − r ]
r =1
⇒ X = 10 × 19C 9
[Q
n−1
C r − 1nC n − r =
m
 n
 n − 1
 n − 2
Let S1 =   + 2 
 +3
 + ... + (n − m + 1)  
m
m
 m 
 m 
2n − 1
Cn − 1 ]
 n   n − 1  n − 2
m 
=  + 
+
 + ... +   
m  m   m 
m

m 
 n − 1  n − 2
+
+
 + ... +   
 m   m 
m

 n − 2
m  n − m + 1 rows
+
 + ... +   
 m 
m

......
m 
+  
m 
1
10 × C 9
C9
C9
X=
=
=
1430
1430
143 11 × 13
19 × 17 × 16
=
= 19 × 34 = 646
8
19
Now,
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
∑ 9C r − 1 10C r
⇒ X = 10
19
19
12. 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.
 n + 1
Now, sum of the first row is 
.
m + 1
13. 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
 n 
Sum of the second row is 
.
m + 1
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
2
r=0
k
=
n!
∑ (−1)r. 2k − r.
r=0
Ck − r
(n − r )!
n!
k!
⋅
(n − k)!. k ! r !(k − r )!
 k
k.n
r. 1 .k
r. k − r n
k
.
C
Cr
(
1
)
2
C
C
=
2
−
k  ∑ (−1 )
k
r
∑
2r
r=0
r = 0
m + 1  n + 1 + 1  n + 2
+K+
= 
 =

m + 1  m + 2  m + 2



[from Eq. (i) replacing n by n + 1 and m by m + 1]
k
=2
k.n
1

C k 1 −  = n C k = RHS

2
 n
m
n
15.
 n − 1  n − 2
m  n + 1
+
 + ... +   = 
 …(i)
 m   m 
m m + 1
14. Let S =   + 
It is obvious that, n ≥ m.
NOTE
∑ (−1)r
r=0
n
= ∑ (−1)r
r=0
[given]
This question is based upon additive loop.
=
 n
m m + 1 m + 2
Now, S =   + 
 + ... +  
+
m
m  m   m 
m + 1 m + 1  m + 2
 n
= 
 + K 
+
+ 
m
1
m
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 =
m + 3
 n
=
 + ... +  
m + 1
m
= ...............................
 n   n   n + 1
=
 +   =
 , which is true.
m + 1 m m + 1
 n + 1  n   n − 1 
S=
+
+

m + 1 m + 1 m + 1
Thus,
k
=
 n − 1
Sum of the third row is 
,
m + 1
…………………………
m m + 1
Sum of the last row is   = 
.
m m + 1
∑ (−1)r 2k − r. r !(n − r )! . (k − r )!(n − k)!
r=0
k
=
Cr
.n − r
n
Cr
Cr
r +3
n
n!
n !⋅ 3 !
= 3 ! ∑ (−1)r
n
r
(
)
! (r + 3) !
−
(n − r ) ! (r + 3) !
r=0
3!
(n + 1)(n + 2)(n + 3)
(−1)r. (n + 3)!
r=0
=
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+1
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

Cr + 1 ]
…(ii)
n
∑ (n − r )!(r + 3)!
C1 − n + 3C 2
(n + 3)(n + 2) 

0 − 1 + (n + 3) −

2!


−
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
96 Binomial Theorem
16. (1 + x + x2)n = a 0 + a1x + K + a 2nx2n
=
…(i)
Replacing x by −1 / x, we get
…(ii)
Now, a 02 − a12 + a 22 − a32 + K + a 22n = coefficient of the
term independent of x in
∴
=
=
= − (−1)n n ⋅ C n
 1
Again, the coefficient of   on the RHS
 x
r=0
∑ (−1) r
r=0
n
⋅ C r + 2 ∑ (−1) r ⋅ C r +
r
n
r=0
n
∑ (−1)
r. n
Cr
r=0
= − (C12 − 2 ⋅ C 22 + 3 ⋅ C32 − ... − 2n C 22n )
n
∑ (−1)r. r (r − 1) ⋅n C r + 3 ⋅ ∑ (−1)r. r ⋅ nC r
r=0
r=0
+
=
∑ (−1) n (n − 1)
r
n
C r − 2 + 3 ∑ (−1) n .
n− 2
r=2
r
r =1
+
∑ (−1)r nC r
1
20. (1 + x)2 n 1 − 

n −1
C r −1
n
∑ (−1)
+ { C 0 − C1 + C 2 + ... + (−1)
= n (n − 1) . 0 + 3n . 0 + 0, ∀n > 2 = 0, ∀n > 2
18. We know that,
n
2 ∑ ∑ Ci C j = ∑
0 ≤i < j ≤ n
i=0
n
n
n
r n
Cr
n n
Cn}
n
∑ Ci C j − ∑ ∑ CiC j
j=0
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
= 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}
n
…(iv)
C12 − 2 ⋅ C 22 + 3 ⋅ C32 − ... − 2n ⋅ C 22n = (−1)n n ⋅ C n
r=0
n
…(iii)
From Eqs. (iii) and (iv),
n
r=0
n
1
 1 
in 2n  2 n  (1 + x)2 n−1 (x − 1)2 n
x 
x
= Coefficient of x2 n−1 in 2n (1 − x2)2 n−1 (1 − 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
2
n
r 2 n
2n
= Coefficient of
4 n
n
(2n )!
22 n − 2 nC n
= 22 n−1 −
2
2 (n !)2
 1
Coefficient of   on the LHS
 x
∑ (−1) r(r + 1)2 nC r = ∑ (−1)r (r 2 + 2r + 1) nC r
r=0
n
CiC j =
= [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

17. C 0 − 2 ⋅ C1 + 3 ⋅ C 2 − ... + (−1) (n + 1) ⋅ C n
n
i=0
1

2n (1 + x)2 n−1 1 − 

x
= Coefficient of x in (1 + x + x )
= Coefficient of t n in (1 + t + t 2)n = a n
=
j=0
On multiplying Eqs. (i) and (ii), we get
= Coefficient of the term independent of x in
1
(1 + x2 + x4 )n
x2n
n
i=0
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
n
2
n
19. We know that, (1 + x)2 n = C 0 + C1x + C 2 x2 + ... + C 2 n x 2 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 )
=
=
x2 n
x2n
2
2
2
2
2
Thus, a 0 − a1 + a 2 − a3 + K+ a 2n
2
∑∑
0 ≤i< j≤n
[a 0 + a1x + a 2x2 + K + a 2nx2n ]
a 2n 

a
a
× a 0 − 1 + 22 − K + 2n 
x
x
x 

= Coefficient of the term independent of x in
n
1 1

(1 + x + x2)n 1 − + 2

x x 
2
n
= 2n 2n − (2 nC n ) = 22 n − 2 nC n
n
a 2n
a1 a 2 a3
1 1

+ 2 − 3 + K + 2n
 1 − + 2 = a 0 −

x x 
x
x
x
x
2n
n
∑ Ci ∑ C j − ∑ Ci2
Independent terms of x on RHS
= (2nC 0 )2 − (2nC1 )2 + (2nC 2)2 − ...+ (2nC 2n )2
 x − 1
LHS = (1 + x)2n 

 x 
=
1
(1 − x2)2n
x2n
Independent term of x on the LHS = (−1)n ⋅2n C n.
i = 0 j=0
Download Chapter Test
http://tinyurl.com/yxkarvhw
2n
or
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
equal to
(2019 Main, 12 Jan I)
(a)
(c)
175
(b)
5
6
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
(2019 Main, 11 Jan II)
‘‘nice’’, is
(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
3 balls, is
(2015 Main)
(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.
Then the probability that z is even, is
(2017 Adv.)
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
98 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
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
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
(2014 Adv.)
card drawn from the i th box i = 1, 2, 3.
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
(a)
9
105
(b)
10
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
For the following questions, choose the correct answer
from the codes (a), (b), (c) and (d) defined as follows.
(a)
Fill in the Blanks
(c)
11
105
(d)
7
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
this list?
(2001, 5M)
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
seats?
(1996, 5M)
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 99
Topic 2 Addition and Subtraction Law of Probability
Objective Questions I (Only one correct option)
1. 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 one
1
and P (all the three events occur
4
1
simultaneously) = , then the probability that atleast
16
(2017 Main)
one of the events occurs, is
of C or A occurs) =
(a)
7
32
(b)
7
16
3
4
then P (B ∩ C ) is equal to
(c)
2. If P (B) = , P ( A ∩ B ∩ C ) =
1
(a)
12
1
(b)
6
7
64
(d)
3
16
1
1
and P ( A ∩ B ∩ C ) = ,
3
3
(2002, 3M)
1
(c)
15
1
(d)
9
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
(1996, 2M)
p + 3p2
(b)
4
3p + 2p2
(d)
4
+ P (B) − P ( A ) P (B), then
(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 )
(1984, 3M)
(b) P (M ) + P (N ) − P (M ∪ N )
(c) P (M ) + P (N ) − 2 P (M ∩ N )
Fill in the Blanks
10. Three numbers are chosen at random without
P ( A ) and P (B) is… .
(1985, 2M)
True/False
12. 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
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)
14. A, B, C are events such that
6. The probability that at least one of the events A and 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)
(d) 1.4
7. Two events A and B have probabilities 0.25 and 0.50,
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)
(a) 0.39
(c) 0.11
exactly one of them occurs is
(a) P (M ) + P (N ) − 2 P (M ∩ N )
13. In a certain city only two newspapers A and B are
5. If 0 < P ( A ) < 1, 0 < P (B) < 1 and P ( A ∪ B) = P ( A )
(c) 1.2
9. If M and N are any two events, then the probability that
11. P ( A ∪ B) = P ( A ∩ B) if and only if the relation between
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 simultaneously)
1
= p2, where 0 < p < . Then, the probability of atleast
2
one of the three events A, B and C occurring is
(b) 0.8
(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 )
replacement from {1, 2,…, 10}. The probability that the
minimum of the chosen number is 3, or their maximum
is 7, is … .
(1997C, 2M)
4. For the three events A, B and C, P(exactly one of the
(a) 0.4
8. For two given events A and B, P ( A ∩ B) is
(d) P (M ∩ N ) − P (M ∩ N )
3. If
3p + 2p2
(a)
2
p + 3p2
(c)
2
Objective Questions II
(One or more than one correct option)
(b) 0.25
(d) None of these
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
If Pr ( A ∪ B ∪ C ) ≥ 0.75, then show that Pr (BC ) lies in
(1983, 2M)
the interval [ 0.23, 0.48 ].
15. 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
(1982, 2M)
0.9?
100 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.)
16. 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
17. 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
probabilities , , 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)
(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 )
…… , 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)
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)
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
(b) P (A/B ) ≥ P (A )
(d) P (A/B ) = 1
4. Two integers are selected at random from the set { 1, 2,
(a)
(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
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)
235
256
(b)
21
256
(c)
3
256
(d)
253
256
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
(2007, 3M)
seated adjacent to his wife, is
(a)
1
2
(b)
1
3
(c)
2
5
(d)
1
5
Probability 101
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)
13
(a)
32
1
(b)
4
1
(c)
32
3
(d)
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
(a) 16/81
(c) 80/81
(b) 1/81
(d) 65/81
(1993, 1M)
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
(b) 1 − P (A / B )
1 − P (A ∪ B )
(c)
P (B )
P (A )
(d)
P (B )
(1982, 2M)
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
(1980, 1M)
event A happens atleast once, is
(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
4
(a) P (Y ) =
15
(c) P (X ∪Y ) =
22. If
1
2
(2017 Adv.)
1
(b) P (X ′/Y ) =
2
2
5
(d) P (X ∩ Y ) =
1
5
and Y
are
two
events
such
that
1
1
1
P (X / Y ) = , P (Y /X ) = and P (X ∩ Y ) . Then, which of
2
3
6
(2012)
the following is/are correct?
X
(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
that both E and F happen is 1/12 and the probability that
neither E nor F happen is 1/2. Then,
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)
(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
102 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.
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)
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, then
(1990, 2M)
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
(1988, 2M)
found to be red, is….
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
probability… .
(1985, 2M)
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.
(ii) A, B, C are independent.
(1992, 6M)
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 marks in the question.
(1985, 5M)
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
(1984, 2M)
occurrence of A.
42. Cards are drawn one by one at random from a well
34. If A and B are two independent events, prove that
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)
35. A is targeting to B, B and C are targeting to A.
43. An anti-aircraft gun can take a maximum of four shots
Analytical and Descriptive Questions
P ( A ∪ B) ⋅ P ( A′ ∩ B ′ ) ≤ P (C ), where C is an event
defined that exactly one of A and B occurs.
(2004, 2M)
2 1
Probability of hitting the target by A, B and C are , and
3 2
1
, respectively. If A is hit, then find the probability that B
3
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 that
he will qualify.
(2003, 2M)
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
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 Answer Type Question
45. 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
Probability 103
(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,
probability of occurrence of E1
is equal to
probability of occurrence of E3
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.
46. P (X > Y ) is
Passage Type Questions
(a)
Passage
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
4
(b)
5
12
(c)
1
2
(d)
7
12
(b)
1
3
(c)
13
36
(d)
1
2
20
23
(d)
9
20
47. P (X = Y ) is
(a)
11
36
Topic 4 Law of Total Probability and Baye’s Theorem
Objective Question I (Only one correct option)
(a)
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
3
5
6
7
(b)
(c)
Objective Question II
(One or more than one correct option)
5. A ship is fitted with three engines E1 , E 2 and E3 . The
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.
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
(2018 Main)
is red, is
(a) P
3
(a)
10
(b) P [exactly two engines of the ship are functioning] =
2
(b)
5
1
(c)
5
3
(d)
4
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
4
5
1
respectively, is received by station A and then
5
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
and
Which of the following is/are true?
[X1c| X ] =
(2012)
3 / 16
7
8
5
16
7
(d) P [X | X1 ] =
16
(c) P [X | X 2 ] =
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
(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
6. Let H 1 , H 2,... , H n be mutually exclusive events with
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
n
Statement II
i = 1, 2, . . . , n
∑ P (Hi ) = 1
i =1
(2007, 3M)
104 Probability
14. If P (ui ) ∝ i, where i = 1, 2, 3,... , n , then lim P (W ) is
Passage Based Problems
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,
(2015 Adv.)
respectively in box II.
7. 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
(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
(2011)
random from U 2.
9. The probability of the drawn ball fromU 2 being white, is
13
30
(b)
23
30
(c)
19
30
(d)
11
30
10. Given that the drawn ball from U 2 is white, the
probability that head appeared on the coin is
(a)
17
23
(b)
11
23
(c)
15
23
(d)
12
23
A fair die is tossed repeatedly until a six is obtained. Let X
denote the number of tosses required.
(2009)
11. The probability that X = 3 equals
25
216
(b)
25
36
(c)
5
36
(d)
125
216
12. The probability that X ≥ 3 equals
(a)
125
216
(b)
25
36
(c)
5
36
(d)
25
216
13. The conditional probability that X ≥ 6 given X > 3 equals
(a)
125
216
(b)
25
216
(c)
5
36
1
4
15. If P (ui ) = c , where c is a constant, then P (un / W ) is
equal to
(a)
2
n+1
(b)
1
n+1
(c)
n
n+1
(d)
1
2
16. 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 
(b)
n+ 2
2 (n + 1)
(c)
n
n+1
(d)
1
n+1
17. 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)
18. 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)
19. 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)
20. An urn contains m white and n black balls. A ball is
Passage III
(a)
(c)
Analytical and Descriptive Questions
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)
2
(b)
3
3
(d)
4
(a) 1
(a) n + 2
2n + 1
8. A ball is drawn at random from box I and transferred to
(a) n1 = 4 and n2 = 6
(c) n1 = 10 and n2 = 20
n→ ∞
equal to
Passage I
(d)
25
36
Passage IV
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 denotes the
event of getting a white balls.
(2006, 5M)
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)
21. 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 that the player P4 reaches the final?
(1999, 10M)
22. 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 .
(1998, 8M)
Determine α, β and γ (in terms of p).
Probability 105
23. 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
(1997C, 5M)
S1 and S2 is among the eight winners.
24. In a test an examinee either guesses or copies of knows
the answer to a multiple choice question with four
1
choices. The probability that he make a guess is and
3
1
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
(1991, 4M)
it.
25. 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)
26. 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. 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 is
4
, then the probability that he is unable to solve less
5
than two problem is
(2019 Main, 12 April II)
(a)
201  1 
 
5  5
49
(b)
316  4 
 
25  5 
48
(c)
54  4 
 
5  5
49
(d)
164  1 
 
25  5 
48
2. 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
that the probability of getting atleast one head is more
than 99% is
(2019 Main 10 April II)
(b) 6
(c) 7
(d) 5
4. The minimum number of times one has to toss a fair
coin so that the probability of observing atleast one head
is atleast 90% is
(2019 Main, 8 April II)
(a) 2
(b) 3
(c) 5
(d) 4
5. 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)
400
400
loss (b)
loss
(a)
3
9
(c) 0
(a) 6
(b) 3
(c) 5
(d) 4
7. 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)
(a)
25
169
(b)
52
169
(c)
49
169
(d)
400
(d)
gain
3
6. 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
24
169
8. 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)
(d) 137
3. Minimum number of times a fair coin must be tossed so
(a) 8
probability of hitting the target at least once is greater
5
than , is
(2019 Main, 10 Jan II)
6
(a)
12
5
(b) 6
(c) 4
(d)
6
25
9. 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
35
(b)
13
35
(c)
11
35
(d)
10
35
10. 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
(1992, 2M)
points, is
(a) 0.8750
(b) 0.0875
(c) 0.0625
(d) 0.0250
11. 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
106 Probability
15.
2 and 1 respectively, then the probability that X takes a
value greater than one is equal to… .
(1991, 2M)
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.
13. For a biased die the probabilities for the different faces
16. Suppose the probability for A to win a game against B is
Fill in the Blanks
12. If the mean and the variance of a binomial variate X are
to turn up are given below
Face
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)
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
draw).
(1989, 5M)
17. 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)
Analytical & Descriptive Questions
14. 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
3 times.
(1993, 5M)
Integer Type Question
18. The minimum number of times a fair coin needs to be
tossed, so that the probability of getting atleast two
heads is atleast 0.96, is
(2015 Adv.)
Answers
Topic 1
1. (c)
39. (i) A, B and C are pairwise independent
2. (a)
5.
9.
13.
17.
(a)
6. (c)
(a)
10. (c)
(a)
14. (b)
(d)
18. (b)
1
1
1
22. ≤ p ≤
21.
36
3
2
(3n − 3. 2n + 3 ) × 6C 3
25.
6n
10 (n + 2 )
28. 1 − n + 7
C5
3. (a)
4. (d)
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
Topic 2
1. (b)
5. (c)
9. (a, c)
13. 13.9%
2. (a)
6. (c)
11
10.
40
15. No
40.
3. (c)
7. (a)
4. (a)
8. (a, b, c)
11. P ( A ∩ B ) 12. False
16. (a)
17. (c)
Topic 3
1. (d)
5. (c)
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
41.
1 1
or
3 2
46. (b)
43. 0.6976
44.
1
1260
1
5
45. 6
47. (c)
Topic 4
1. (d)
5. (b, d)
9. (b)
2. (b)
3. (c)
4. (c)
6. (d)
7. (b)
8. (d)
10. (d)
11. (a)
12. (b)`
13. (d)
14. (b)
15. (a)
16.
1
17.
7
12
C 2 ⋅6 C 4 10C1 ⋅2 C1 12C1 ⋅6 C 5 11C1 ⋅1 C1
19.
18.
⋅ 12
+ 18
⋅ 12
18
C6
C2
C6
C2
m
4
21.
20.
m+n
35
p
p (1 − p )
p − 2p 2 + p 3
22. α =
,γ =
,β =
3
3
1 − (1 − p )
1 − (1 − p )
1(1 − p ) 3
1
8
24
23
24.
25.
26.
23. (i) (ii)
2
15
29
30
(b)
9m
8N + m
99
1900
Topic 5
1. (c)
5. (c)
2. (d)
6. (c)
3. (c)
7. (a)
9. (c)
10. (b)
11. (d)
13.
5
21
14.
16. Best of 3 games
97
25 4
 3
15.  
 4
17.
11
4. (d)
8. (a)
11
12.
16
n
C 6( 0 . 24 ) 5 18. (8)
Hints & Solutions
Topic 1 Classical Probability
S − { 1, 2, 4 }
∴Number of favourables cases = 5
5
Hence, required probability = 20
2
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
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
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 fours
4 4
(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].
5 × 35 175
Hence, the required probability =
= 5
65
6
A6
4, 8
2
6, 10
4
0, 8
6
2, 10
8
0, 4
10
2, 6
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
7. We have mentioned that boxes are different and one
particular box has 3 balls.
A1
A2
A5
y
0
∴Required probability =
3. Since, there is a regular hexagon, then the number of
ways of choosing three vertices is 6C3 . 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.
…(i)
…(ii)
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}
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 =
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
9.
PLAN As one of the dice shows a number appearing on one of P1, P2
and P3.
108 Probability
13. 71 = 7, 72 = 49, 73 = 343, 74 = 2401, …
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
Therefore, for 7r, r ∈ N the number ends at unit place
7, 9, 3, 1, 7, …
∴ 7m + 7n will be divisible by 5 if it end at 5 or 0.
But it cannot end at 5.
Also for end at 0.
For this m and n should be as follows
Case I First we should select one number for D4
which appears on all i.e. 6C1 × 1.
1
2
3
4
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.
r2
r3
Favorable events ω r 1 + ω + ω = 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
11. Since, three distinct numbers are to be selected from
first 100 natural numbers.
⇒
n (S ) = 100C3
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
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
m
n
4r
4r − 1
4r − 2
4r − 3
4r − 2
4r − 3
4r
4r − 1
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
NOTE
Power of prime numbers have cyclic numbers in their unit
place.
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.
—W—W—W—W —W—W—W—
This can be done in 8C3 ways.
∴
Required probability =
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
Only two equilateral
triangles can be formed
∆AEC and ∆BFD.
∴ Favourable ways = 2
So, required probability
2
1
=
=
20 10
16. Since, three dice are rolled.
D
E
C
F
B
A
∴ 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.
Probability 109
Thus, the favourable number of cases = 97 − 87.
97 − 87
∴ Required probability =
157
18. The number of all possible determinants of the form
a
c
b
= 24 = 16
d
⇒
0 1
0 0
,
0 0
1 0
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.
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.
∴ 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
=
20. Since, x1 , x2, x3 are in AP.
∴
x1 + x3 = 2x2
So, x1 + x3 should be even number.
Either both x1 and x3 are odd or both are even.
∴ Required probability =
=
C1 × 4C1 + 1C1 × 3C1
3
C1 × 5C1 × 7C1
2
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
P ( yellow at the first toss) =
4 + 12 p + 3 − 3 p + 6 − 12 p ≤ 12
⇒
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
19.
1 + 3p 1 − p
1 − 2p
are the probability of
,
and
3
4
2
mutually exclusive events.
1 + 3p 1 − p 1 −2p
+
+
≤1
∴
3
4
2
22. Since,
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 =
3
16
24. Total number of ways to arrange ‘ASSASSIN’ is
8!
.
4 !⋅ 2 !
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
Hence, it is a false statement.
25. Let us define a onto function F from A : [ r1 , r2, K , rn] to
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)]
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
⇒
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).
110 Probability
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
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
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!
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.
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. 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 ) =
8 16 16
16 

3
4
2. Given, P (B) = , P ( A ∩ B ∩ C ) =
(i) When one 50 paise coin and four 25 paise coins are
chosen.
B
A
(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)
n+7
C5 − 10(n + 2)
n+7
C5
10 (n + 2)
=1 − n+7
C5
∴ Required probability =
Number of pairs of p, q
1
10
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)
1
3
(A ∩ B ∩ C)
(A ∩ B ∩ C)
(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 ))}
Probability 111
=
3  1 1 3 2 1
− +  = − =
4  3 3 4 3 12
3. It is given that, P (E ) ≤ P (F ) ⇒ E ⊆ F
…(i)
P (E ∩ F ) > 0 ⇒ E ⊂ F
and
(a) occurrence of E ⇒ occurrence of F
(b) occurrence of F ⇒ occurrence of E
(c) non-occurrence of E ⇒ occurrence of
Hence, option (c) is not correct.
…(ii)
[from Eq. (i)]
[from Eq. (ii)]
F
[from Eq. (i)]
4. We know that,
P (exactly one of A or B occurs)
P ( A ∩ B)min. , when P ( A ∪ B)max = 1
⇒
P ( A ∩ B) ≥ P ( A ) + P (B) − 1
∴ 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.
∴
= 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)
9. P(exactly one of M, N occurs)
On adding Eqs. (i), (ii) and (iii), we get
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)
− P (B ∩ C ) − P (C ∩ A ) + P ( A ∩ B ∩ C )
3p
[from Eqs. (iv) and (v)]
=
+ p2
2
=
∴
3 p + 2 p2
2
5. Since, P ( A ∩ B) = P ( A ) ⋅ P (B)
= 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 (M ∪ N ) − P (M ∩ N ) = P (M ) + P (N ) − 2P (M ∩ N )
Hence, (a) and (c) are correct answers.
10. 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
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)′
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)′
6. 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
7. 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
8. We know that,
P ( A ∩ B) = P ( A ) + P (B) − P ( A ∪ B)
Also,
P ( A ∪ B) ≤ 1
P (E1 ∪ E 2) = P (E1 ) + P (E 2) − P (E1 ∩ E 2)
7
1 1 11
=
+ −
=
40 8 40 40
11. 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).
12. 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.
13. Let P ( A ) and P (B) denote respectively the percentage of
city population that reads newspapers A and B.
112 Probability
= 5 ! − [4C1 4 ! 2 ! − (3 C1 3 ! 2 ! + 3C1 3 ! 2 ! 2 !)
Then,
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.
+ (2C1 2 ! 2 ! + 4C1 2 ⋅ 2 !) − 2]
P ( A) =
∴ P (C ) = 30% of P ( A ∩ B) + 40% of P ( A ∩ B)
⇒
= 120 − [192 − (36 + 72) + (8 + 16) − 2]
∴
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 ) =
[since, A ∩ B, A ∩ B and A ∩ B are all mutually
exclusive]
3
17
2 3
1 2
139
×
+ ×
+ ×
=
= 13 . 9%
⇒ P (C ) =
10 100 5 25 2 25 1000
14. We know that,
15. 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.
16. 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
9
3
=
∴ Required probability =
120 40
17. Here, n (T1 ∩ T2 ∩ T3 ∩ T4 )
Total = − n (T1 ∪ T2 ∪ T3 ∪ T4 )
⇒ n (T1 ∩ T2 ∩ T3 ∩ T4 )
= 120 − [192 − 108 + 24 − 2] = 14
14
7
Required probability =
=
120 60
Topic 3 Independent and Conditional
Probability
+ 50% of P ( A ∩ B)
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
n (T1 ∩ T2 ∩ T3 ∩ T4 )
=
2.
 1
 
 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)
Probability 113
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
=
=
5
C 2 + C 2 15 + 10 5
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
=
∴
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.
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.
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
dice is thrown =
36
Also, let P (E 2) = Probability of getting 7 or 8 when a
2
card is picked from cards numbered 1, 2, ...., 9 =
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
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)
and
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
P ( A ∩ B) ≤
⇒
5
1 2 3 4
P ( A ∩ B) =
,
,
,
⇒
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
9. Since, P ( A ) =
114 Probability
3
2
3
⇒
× P (B) =
10
5
10
3
P (B) = , not possible
4
4
4
P ( A ∩ B) =
⇒ P ( A ) . p(B) =
10
10
⇒
P ( A )P (B) =
(iv) Now,
⇒
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
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 
15. P (2 white and 1 black) = P (W1W 2B3 or W1B2W3 or
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
5
2
 A  n ( A ∩ E ) (4 !) × (2 !)
=
=
P  =
∴
 E
n (E )
(5 !) × (2 !)4 5
Alternate Solution
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
12. Let E be the event of getting 1 on a die.
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 ) =
= { 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
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
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
⇒
A ={2, 3, 4, 5}
⇒ P ( A) =
4
6
But die is rolled four times, therefore the probability in
getting four throws
 4  4  4  4 16
=     =
 6  6  6  6 81
18. Let A, B and C denote the events of passing the tests I,
II and III, respectively.
Evidently A, B and C are independent events.
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
⇒ 1 = 2 pq + p − pq
⇒ 1 = p(q + 1)
…(i)
The values of option (c) satisfy Eq. (i).
[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.]
19. Since, P ( A / B ) + P ( A / B ) = 1
∴
P(A / B) = 1 − P(A / B)
20. Given that, P ( A ) = 0.4, P ( A ) = 0.6
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
Probability 115
1
3
 X  P (X ∩ Y ) 1
P  =
=
Y 
P (Y )
2
 Y  P (X ∩ Y ) 2
P  =
=
 X
P (X )
5
2
P (X ∩ Y ) =
15
4
P (Y ) =
15
21. P (X ) =
 X ′  P (Y ) − P (X
P  =
Y 
P (Y )
P (X ∪ Y ) =
22.
1
4
2
+
−
3 15 15
E
⇒
4
2
−
∩ Y ) 15 15 1
=
=
4
2
15
7
7
=
=
15 15
(i) Conditional probability, i.e. P( A / B) =
11
25
2
From Eq. (ii),
( 1 − P (E )) ( 1 − P (F )) =
25
2
1 − P (E ) − P (F ) + P (E ) ⋅ P (F ) =
⇒
25
P ( A ∩ B)
P( B)
…(iv)
7
12
and P (E ) ⋅ P (F ) =
5
25
7
 12
P (E ) ⋅
− P (E ) =
 5
 25
7
12
(P (E ))2 − P (E ) +
=0
5
25
3 
4

P (E ) −
P (E ) −
=0

5  
5 
4
3
3
4
⇒ P (F ) = or
P (E ) = or
5
5
5
5
P (E ) + P (F ) =
1
Y  1
,P  =
 X 3
2
and P (X ∩ Y ) = 6
 X  P (X ∩ Y )
P  =
∴
Y 
P (Y )
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 )
independent events
∴
⇒
⇒
∴
24. Let A, B and C respectively denote the events that the
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
⇒
Q
1 − P ( A′ ∩ B ′ ∩ C ′ ) = 0.75
[P ( A ) = 1 − P ( A )
[P ( A ∪ B ∪ C ] = P ( A′ ∩ B′ ∩ C′ )]
⇒ 1 − P ( A′ ) . P (B ′ ) . P (C′ ) = 0.75
and
Q A, B and C are independent events, therefore A′, B′
and C ′ are independent events.
P (X c ∩ Y ) = P (Y ) − P (X ∩ Y )
1 1 1
= − =
3 6 6
F
E
…(iii)
From Eqs. (iii) and (iv),
Here, P (X /Y ) =
23.
…(ii)
From Eq. (i), P (E ) + P (F ) − 2 P (E ∩ F ) =
(ii) P ( A ∪ B) = P( A ) + P( B) − P ( A ∩ B)
(iii) Independent event, then P ( A ∩ B) = P( A ) ⋅ P( B)
i.e.
2
25
2
P (E ∩ F ) =
25
Neither of them occurs =
PLAN
∴
F
⇒
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
⇒ 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 (E ∪ F ) − P (E ∩ F ) =
11
25
+ P ( A ) P (B) P (C ) = 0.4
…(i)
⇒
pm (1 − c) + p(1 − m) c + (1 − p) mc = 0.4
[i.e. only E or only F]
⇒
pm − pmc + pc − pmc + mc − pmc = 0.4
…(ii)
116 Probability
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
⇒
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),
pm + pc + mc − 3 pcm = 0.4 …(iv)
From Eq. (i),
0.25 = 1 − (m + p + c) + ( pm + pc + cm) − pcm
…(v)
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 ) =
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 )
P (E ∩ F ) P (E ∩ F )
+
≠1
P (F )
1 − P (F )
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.
Now, P (E ∩ F ) = P (E ) − P (E ∩ F ) = P (E ) − P (E ) ⋅ P (F )
Therefore, option (b) is not correct.
P (E ∩ F ) P (E ∩ F )
(c) P (E / F ) + P (E / F ) =
+
P (F )
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 )
= 1 − [1 − P (E ) ⋅ P (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 )
P (F )
=
=1
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 )
Therefore, option (d) is correct.
29. P ( A c ) = 0.3
1
12
and neither E nor F happens ⇒ P (E ∩ F ) =
and
1
12
[given]
⇒
1
2
But for independent events, we have
P (E ∩ F ) = P (E ) P (F ) =
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)
⇒
Therefore, option (a) is correct.
P (E ∩ F ) P (E ∩ F )
(b) P (E / F ) + P (E / F ) =
+
P (F )
P (F )
26. Both E and F happen ⇒ P (E ∩ F ) =
 A  P ( A ∩ B) P ( A ) + P (B) − P ( A ∪ B)
=
P  =
 B
P (B)
P (B)
⇒
P (E ∩ F ) + P (E ∩ F )
P (F )
P (F )
=
=1
P (F )
=
27. We know that,
Since,
P (E ∩ F ) P (E ∩ F )
+
P (F )
P (F )
…(ii)
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
⇒
⇒
=
=
1
1
= 1 − { P (E ) + P (F )} +
2
12
1
1
7
P (E ) + P (F ) = 1 − +
=
2 12 12
⇒
…(i)
P (E ∩ F ) = P (E ) P (F )
= {1 − P (E )}{(1 − P (F )}
= 1 − P (E ) − P (F ) + P (E )P (F )
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
Probability 117
1
[P ( A ) − P ( A ∩ Bc )]
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
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
35. Given, P ( A ) = probability that A will hit B =
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
P (B) = probability that B will hit A =
1.2
P (B) . P (C ) 2 3 1
= P (B ∩ C / E ) =
=
=
2
P (E )
2
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.
∴
Then
34. Here, P ( A ∪ B) . P ( A′ ∩ B ′ )
⇒ { 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′ )
[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′ )]
+ [P (E1′ ) × P (E 2 / E ′1 ) × P (E3 / E 2)]
p
p
= p + p(1 − p) . + (1 − p) . . p
2
2
2 p2 + p2 − p3 + p2 − p3
= 2 p2 − p3
P (E ) =
2
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).
For n = 2, p2 = 1 − p (two heads simultaneously occur).
= 1 − p(HH ) = 1 − pp = 1 − p2
For n ≥ 3, pn = pn − 1 (1 − p) + pn − 2(1 − p) p
⇒
pn = (1 − p) pn − 1 + p(1 − p) pn − 2
Hence proved.
38. Let, E1 = the event noted number is 7
E 2 = the event noted number is 8
H = getting head on coin
T = be getting tail on coin
C
P ( A ) = 100 2
C2
 B  P ( A ∩ B)
P  =
 A
P ( A)
5
C1 1
= 10
=
C2 9
P (E ) = [P (E1 ) × P (E 2 / E1 )]
+ [P (E1 ) × P (E 2′ /E1 ) × P (E3 / E 2′ )]
10
and
2
3
∴ By law of total probability,
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
118 Probability
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  91 
91
 =

2  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.
50 1
Therefore, P (N 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,
A ∩ B = { D1N 2}, B ∩ C = { N 1N 2}.
C ∩ A = { D1D2} and A ∩ B ∩ C = φ
Also,
P ( A ) = P{ D1D2} + P{ D1N 2}
= 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
= 4C1 + 4C 2 + 4C3 + 4C 4 = 24 − 1 = 15
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 ) =
∴
[1 − P ( A )] [1 − P (B)] =
⇒
⇒
⇒
⇒
⇒
⇒
∴
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
Let
⇒
1
3
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
=
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.
Probability 119
Let the corresponding probabilities be
2
,
9
1
p6 = ,
4
p1 =
Then,
p1 , p2,... , p9
1
4
3
2
p2 = , p3 = , p4 = , p5 =
8
7
6
5
3
2
p7 = , p8 = , p9 = 1
3
2
∴ Required probabilitie
p1 . p2 . p3 ⋅ K ⋅ p9
 2
= 
 9
45.
1
 1  4   3   2  1   3  2 
              (1) =
 8  7  6   5  4   3  2 
1260
PLAN
…(v)
From above equations, x = 2 y and y = 3z
x = 6z
x
=6
z
46. 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
47. 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.
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
Key idea Use the theorem of total probability
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
6
4
, P  =
10  E 2 12
By law of total probability
 A
 A
P ( A ) = P (E1 ) × P   + P (E 2) × P  
 E1 
 E 2
=
4
6
6
4 24 + 24 48 2
×
+
×
=
=
=
10 12 10 12
120
120 5
3. Let x = P (computer turns out to be defective, given that
 D
x= P  
 T2
…(i)
where, D = Defective computer
= 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 )
⇒
P (E 2) =
⇒
P( E1 ∩ E 2 ∩ E 3 ) = P( E1 ) ⋅ P( E 2 ) ⋅ P( E 3 )
P( E1 ∩ E 2 ∩ E 3 ) = P(only E1 occurs)
∴
⇒
 A
4
6
, P  =
10  E1  12
it is produced in plant T2)
Forthe events to be independent,
Given, (α − 2 β ) p = αβ and ( β − 3γ ) p = 2 βγ
P (E1 ) =
∴ 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
Also,
P (T1 ) =
Using law of total probability,
 D
 D
P (D ) = 9(T1 ) ⋅ P   + P (T2) ⋅ P  
 T1 
 T2
∴
⇒
∴
7
 20 
=
 ⋅ 10x +
100  100
 80 
⋅x

 100
1
40
 D  10
and P   =
 T1  40
7 = (280)x ⇒ x =
 D
1
P  =
 T2 40
…(iii)
 D
 D
1 39
10 30
and P   = 1 −
…(iv)
=
=
⇒ P   =1−
40 40
40 40
 T2
 T1 
Using Baye’s theorem,
P (T2 ∩ D )
T 
P  2 =
 D  P (T1 ∩ D ) + P (T2 ∩ D )
=
 D
P (T2) ⋅ P  
 T2
 D
P (T1 ) ⋅ P   + P (T2) ⋅ P
 T1 
80 39
⋅
78
100
40
=
=
20 30
80 39 93
⋅
+
⋅
100 40 100 40
 D
 
 T2
120 Probability
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
1 1 1 1 3 1 1 1 3
⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅
P (X ∩ X1 ) 2 4 4 2 4 4 2 4 4
(d) P (X / X1 ) =
=
P (X1 )
1 /2
7
=
16
4. From the tree diagram, it follows that
=
S
1
5
4
5
G
3
4
R
1
4
AR
AG
3
4
1 3
4 4
11
44
BG
BR
BG
3
4
1
4
AR
1 AG
3 4
3
4
4
BR
BG
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.
6. 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]
PLAN It is based on law of total probability and Bay’s Law.
Hence, Statement I may not always be true.
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.
Statement II Clearly, H 1 ∪ H 2 ∪ . . . ∪ H n = S
[sample space]
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 ) 
⇒
P (H 1 ) + P (H 2) + . . . + P (H n ) = 1
Hence, Statement II is ture.
Passage I
7.
 1 1 3 1 3 1 1 1 1  1 1 1
= ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ 
 2 4 4 2 4 4 2 4 4  2 4 4
= 1 /4
Now, (a) P (X1c / X )
1 1 1
 X1c ∩ X  P (E1 ∩ E 2 ∩ E3 ) 2 ⋅ 4 ⋅ 4 1
=
=P
=
=
1
P (X )
8
 P (X ) 
4
(b) P (exactly two engines of the ship are functioning)
=
P (ship is operating with E 2 function )
=
P (X 2)
Red
n3
Red
n2
Black
n4
Black
Box I
Let
∴
Given,
⇒
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)
n1
⇒
⇒
Box II
A = Drawing red ball
P ( A ) = P (B1 ) ⋅ P ( A / B1 ) + P (B2) ⋅ P ( A / B2)
1  n1  1  n3 
= 
+ 

2  n1 + n2 2  n3 + n4 
1
P (B2 / A ) =
3
P (B2) ⋅ P (B2 ∩ A ) 1
=
P ( A)
3
1  n3 


2  n3 + n4 
1
=
1  n1  1  n3  3

 + 

2  n1 + n2 2  n3 + n4 
n3 (n1 + n2)
1
=
n1 (n3 + n4 ) + n3 (n1 + n2) 3
Now, check options, then clearly options (a) and (b)
satisfy.
Probability 121
8.
10. P (Head appeared/white from U 2)
1 Red
(n1 – 1)
Red
n2
(n3 + 1)
n4
Black
Box I
1
2
 3C1 2C1
C
C 
× 2
+ 5 1 × 2 1
5
C
C
C
C1 
1
1
= P (H ) .  1
23 / 30
2 1
3
 ×1 + × 
1 5
5 2
= 
2
20 / 30
12
=
23
Red
Black
Box II
1 Black
Red
n1
or
Red
n3
(n2 – 1) Black
(n4 + 1) Black
Box I
Passage III
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
Initial
2R
Passage IV
U1
14. Here, P (ui ) = ki, Σ P (ui ) = 1
U2
Head appears
or
1W
1R
U1
3W
2R
U2
U1
1R
U2
3W
1W
2W
2W
0R
2R
1R
1R
U1
U2
U1
U2
3 Cases
9. Now, probability of the drawn ball from U 2 being white
is
i=1
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
×
n
2
 un  n + 1
15. P   =
=
Σi
W 
n+1
n+1
W  2 + 4 + 6 + ...
n+2
=
 =
n (n + 1)
 E
2 (n + 1)
2
16. P 
17. As, the statement shows problem is to be related to
Baye’s law.
2
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
3
Again, L be the event of the person reaching office late.
1
2
2
 3C
C
C
C 
P (white / U 2) = P (H ) ⋅  5 1 × 2 1 + 5 1 × 2 1 
C
C
C
C1 
1
1
 1
+
2i 2
2n (n + 1)(2n + 1)
= 2 /3
= lim
n→∞
6n (n + 1)2
Tail appears
2W
n
∑ n (n + 1)2
n→ ∞
n→ ∞
1W
1R
2
n (n + 1)
lim P (W ) = lim
2 Cases
U2
U1
3W
1W
k=
⇒
2W
2W
2R
5 5 1 25
6 6 6 216
5 5
25
12. P (X ≥ 3) = ⋅ ⋅ 1 =
6 6
36
P {(X > 3) / (X ≥ 6)} ⋅ P (X ≥ 6)
13. P {(X ≥ 6) / (X > 3)} =
P (X > 3)
5
  5  1  5 6  1

1 ⋅   ⋅   +   ⋅   + ... ∞ 
 6  6  6  6
 = 25
= 
  5 3 1  5 4 1

36
  ⋅ +   ⋅ + ... ∞ 
6
6
6
 6

11. P (X = 3) = ⋅ ⋅ =
C1 

C2 
13
1 1
6 2  23
× +
× =
 ×1 +
2 10
10 3 10 3  30
∴ L be the event of the person reaching office in time.
 L 5
 L 8
 L 7
Then, P   = , P   = , P   =
 B 9
S 9
C 9
and
 L 8
P  =
T 9
122 Probability
 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
18. 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]
 B
 B
P (B) = P   P ( A1 ) + P   P ( A2)
 A1 
 A2 
∴
12
=
C 2.6 C 4
18
C6
10
.
C1.2 C1
+
12
C2
C1.6 C5 . 11C1.1 C1
12
18
C6
C2
12
19. 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
20. 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
21. 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 8 × 7 ×6 ×5
=
×
=
= 105
×
×
2 .2 .2 .2
4 ! 2 ! ×1 2 ! × 1 2 ! × 1
=
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 .
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
22. Let q = 1 − p = probability of getting the tail. We have,
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
=
=
1 − P (T )3 1 − q3
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
Again, we have
α + β + γ =1
⇒
γ = 1 − (α + β ) = 1 −
=1 −
=
γ=
p + p(1 − p)
1 − q3
p + p(1 − p)
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
=
1 − (1 − p)3
1 − (1 − p)3
Probability 123
Also,
α=
p(1 − p)
p
, β=
3
1 − (1 − p)3
1 − (1 − p)
23. (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.
(14)!
(16)!
and n (S ) =
∴ n (E ) =
7
(2 !) ⋅ 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
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.
(14)! 
(14)! 



(2 !)7 ⋅ 7 !  1 1
(2 !)7 ⋅ 7 !  1 1 
= 1 −
× × + 1 −
× ×
(16)!  2 2 
(16)!  2 2




(2 !)8 ⋅ 8 ! 
(2 !)8 ⋅ 8 ! 
1 14 × (14)! 7
= ×
=
2 15 × (14)! 15
∴ Required probability =
E1 = the examinee guesses the answer
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
− =
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 ) =
=
 29
 1 1  1 1  1
 ×  +  ×  +  × 1

 3 4  6 8  2
25. Let
and
Bi = ith ball drawn is black.
Wi = ith ball drawn is white, where i = 1, 2
A = third ball drawn is black.
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 }
1
7
8
+
=
15 15 15
24. Let E1 , E 2, E3 and A be the events defined as
∴ P (E1 ) + P (E 2) + P (E3 ) = 1
P (E3 ) = 1 −
⇒
+ 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 1
 2 3 4
=  ×  ×1 +  ×  ×
 4 3
 4 5 6
 2 2 3  2 2 3
+ ×  × + ×  ×
 4 3 4  4 5 4
1 1 1 3 23
= + + +
=
6 5 4 20 30
124 Probability
26. The testing procedure may terminate at the twelfth
testing in two mutually exclusive ways.
∴
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
= 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. 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
50 − r
r
 1  4
, r = 0, 1, ... , 50
P (X = r ) = 50C r    
 5  5
 4 50 54
 +  =
5 5 
5
n×
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
 1
=16 C 0  
 2
∴ Required probability
∴Required probability
= P (X < 2) = P (X = 0) + P (X = 1)
50
49
449
 4
 4
= 50C 0   + 50C1
=


 5
(5)50  5
1
1
⇒ p =1−q =
2
2
1
= 8 ⇒ n = 16
2
16
 1
P (X = r ) =16 C r  
 2
Now,
I : When lot contains 2 defective articles.
=
q=
 4
 
 5
49
2. 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
=
⇒
16
 1
+ 16 C1  
 2
16
 1
+ 16 C 2 
 2
k
1 + 16 + 120 137
= 16 = 16
216
2
2
16
(given)
k = 137
3. As we know probability of getting a head on a toss of a
fair coin is P (H ) =
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 
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
4. The required probability of observing atleast one head
= 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
n
⇒ n ≤
⇒ 2 ≥ 10 ⇒ n ≥ 4
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.
5. Let p and q represents the probability of success and
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)
2
3
1
  2 1
 2  1
 2
=  × 100 +  ×  (50) +     (0) +   (− 150)
3
  3 3
 3  3
 3
Probability 125
=
100 100
1200
+
+0−
3
9
27
10. 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.
900 + 300 − 1200 1200 − 1200
=
=
=0
27
27
Therefore, the probability of the required event
= 4C3 (0.05) (0.5)3 + 4C 4 (0.5)4
6. The probability of hitting a target at least once
= 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.
[by using binomial distribution]
1 2
=
3 3
n
0
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.
Here,
1
p=
3
and
q =1− p =1−
7. 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
4
1
Then,
p=
=
52 13
1 12
and
=
q =1 − p=1 −
13 13
Here, number of trials, n = 2
Clearly, X follows binomial distribution with parameter
1
n = 2 and p = .
13
2− x
x
 1   12
Now, P (X = x) = 2C x    
, x = 0, 1, 2
 13  13
∴
P (X = 1) + P (X = 2)
1
2
 1   12
 1   12
= 2C1     + 2C 2   
 13  13
 13  13
0
1
 12 
=2
 +
 169 169
24
1
25
=
+
=
169 169 169
8. Given box contains 15 green and 10 yellow balls.
∴Total number of balls = 15 + 10 = 25
15 3
= = p = Probability of success
P(green balls) =
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
9. 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
 1 2
answers = 5C 4   ⋅ + 5C5   = 5 ⋅ 5 + 5 = 5
 3
 3 3
3
3
3
= [4(0.05) + 0.5 ] (0.5)3 = 0.0875
11. Let X be the number of coins showing heads. Let X be a
binomial variate with parameters n = 100 and p.
Since,
P (X = 50) = P (X = 51)
⇒
⇒
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
p=
⇒
51
101
12. For Binomial distribution, mean = np
and
variance = npq
∴
⇒
⇒
∴
Now,
np = 2 and
npq = 1
q = 1 / 2 and p + q = 1
p = 1 /2
n = 4, p = q = 1 / 2
P (X > 1) = 1 − { P (X = 0) + P (X = 1)}
0
4
 1  1
= 1 − 4C 0     − 4C1
 2  2
1
4 11
=1 −
−
=
16 16 16
0 .1
0 .1
5
13. Probability (face 1) =
=
=
0 .1 + 0 .32 0 .42 21
[given]
1
 1  1
   
 2  2
3
14. Let E be the event that product of the two digits is 18,
therefore required numbers are 29 , 36, 63 and 92.
4
Hence, p = P (E ) =
100
and probability of non-occurrence of E is
4
96
q = 1 − P (E ) = 1 −
=
100 100
Out of the four numbers selected, the probability that
the event E occurs atleast 3 times, is given as
P = 4C3 p3 q + 4C 4 p4
3
4
97
 4   96   4 
=4 
 
 +
 = 4
 100  100  100
25
15. 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.
126 Probability
Clearly, P1 > P2. Therefore, first option i.e. ‘best of 3
games’ has higher probability of winning the match.
But, r can vary from 0 to n.
∴ Total number of disjoint sets P and Q
=
n
∑ nC r2n − r = (1 + 2)n = 3n
17. The man will be one step away from the starting point,
if
r=0
Hence, required probability =
3 n  3
= 
4 n  4
(i) either he is one step ahead or (ii) one step behind the
starting point.
n
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 .
16. 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,
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
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)
= 11C 6 (0.4)6 (0.6)5 +
= 3C 2(0.4)2(0.6)1 + 3C3 (0.4)3 (0.6)0
18. Using Binomial distribution,
= 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
C 6 (0.6)6 (0.4)5 = 11C 6 (0.24)5
11
q = 0.6
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
Download Chapter Test
http://tinyurl.com/y4kwqyyl
P (X ≥ 2) = 1 − P (X = 0) − P (X = 1)
n
n −1

 1
 1  1
= 1 −   −  nC1 ⋅   ⋅  

 2
 2  2


n
1
1
1
+


= 1 − n − nC1 ⋅ n = 1 −  n 
 2 
2
2
Given, P (X ≥ 2) ≥ 0. 96
(n + 1) 24
1−
≥
∴
25
2n
n+1 1
⇒
≤
25
2n
∴
n =8
or
7
Matrices and Determinants
Topic 1 Types of Matrices, Addition, Subtraction,
Multiplication and Transpose of a Matrix
Objective Question 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 
2x − y 1 
(2019 Main, 9 April II)
(a) 2
(b) 4
(c) 3
(d) 6
cos α − sin α 
such
that
3. Let
A=
, (α ∈ R)
sin α cos α 
0 −1
A32 = 
. Then, a value of α is
1 0 
(2019 Main, 8 April I)
(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
9 3 1
q + q31
such that Q − P = I3 . Then, 21
is equal to
q32 (2019 Main, 12 Jan I)
5
(a) 10
(b) 135
(c) 9
(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
(b) 103
(d) 205
1 2
2. The total number of matrices A = 2x y −1 ,
(x, y ∈ R, x ≠ y) for which AT A = 3I3 is
0 0
1 0 and I be the identity matrix of order 3.
16 4 1
q + q32
If Q = [qij ] is a matrix, such that P50 − Q = I, then 31
q21
(2016 Adv.)
equals
1
6. Let P =  4
(d)
1
6
2
7. If A = 2 1 −2 is a matrix satisfying the equation
 c 2 b 
AA = 9 I, where, I is 3 × 3 identity matrix, then the
ordered pair (a, b) is equal to
(2015 Main)
T
(a) (2, − 1)
(c) (2, 1)
(b) (−2, 1)
(d) (−2, − 1)
 3 /2
 −1 /2
T 2005
P Q
P is
8. If P = 
1 1
1 /2 
T
, A = 
 and Q = PAP , then
3 / 2
0 1
 1 2005
(a) 
1 
0
1
0

(c) 

 2005 1
9. If A = α
1

A 2 = B, is
0
1
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)
128 Matrices and Determinants
16. Let ω be a solution of x3 − 1 = 0 with Im (ω ) > 0. If a = 2
Objective Question II
One or more than one correct option)
with b and c satisfying Eq. (i) then the value of
3
1
3
is
+
+
ω a ωb ω c
11. 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.)
(a)Y Z − Z Y
(c) X 4 Z 3 − Z 3 X 4
3
4
(b) X + Y
(d) X 23 + Y 23
4 3
44
44
12. For 3 × 3 matrices M and N , which of the following
statement(s) is/are not correct ?
(2013 Adv.)
T
(a) N 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
(a) − 2
(c) 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)
17. The number of A in T p such that det (A) is not divisible
by p, is
(a) 57
(b) 55
(c) 58
18. 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
skew-symmetric or both and det (A) is divisible by p is
(d) 56
(a) ( p − 1)2
(c) ( p − 1)2 + 1
NOTE
Passage I

 , 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 .
20. If matrix A =  b
(2011)
the plane 2x + y + z = 1, then the value of 7a + b + c is
(c) 7
(d) 6
15. Let b = 6, with a and c satisfying Eq. (i). If α and β are
the roots of the quadratic equation ax + bx + c = 0, then
n
∞
 1 1
∑  α + β is equal to
n= 0
(b) 7
(c)
6
7
b
c
a
c
a
b
Integer Type Question
−1 + 3 i
, where i = −1, and r , s ∈ {1, 2, 3}. Let
2
2
r
s
(− z ) z 
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
21. Let z =
2
(a) 6
The trace of a matrix is the sum of its diagonal entries.
a
...(i)
14. If the point P (a , b, c), with reference to Eq. (i), lies on
(b) 12
(b) 2 ( p − 1)
(d) 2 p − 1
Analytical and Descriptive Questions
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) p3 − ( p − 1)2
(d) ( p − 1) ( p 2 − 2)
19. The number of A in T p such that A is either symmetric or
Passage Based Problems
(a) 0
(b) p3 − 5 p
(d) p3 − p 2
(a) 2 p 2
(c) p3 − 3 p
13. 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
(2013 Adv.)
when n is equal to
(b) 2
(d) − 3
(d) ∞
Topic 2 Properties of Determinants
Objective Questions I (Only one correct option)
−6
−1
x
2 − 3x x − 3 = 0, is equal to
− 3 2x x + 2
1. A value of θ ∈ (0, π / 3), for which
sin 2 θ
1 + cos 2 θ
4 cos 6θ
2
cos θ
1 + sin 2 θ
4 cos 6θ = 0, is
cos 2 θ
sin 2 θ
1 + 4 cos 6θ
π
(a)
9
π
(b)
18
7π
(c)
24
2. The sum of the real roots of the equation
(2019 Main, 12 April II)
(d)
7π
36
(a) 0
(b) − 4
(c) 6
(d) 1
(2019 Main, 10 April II)
Matrices and Determinants 129
sin θ cos θ
x
3. If ∆1 = − sin θ −x
1
cos θ
1
x
Then, the number of elements in S, is (2019 Main, 10 Jan II)
(a) 4
(b) 2
(c) 10
(d) infinitely many
1
b
2

2
10. Let A = b b + 1 b, where b > 0. Then, the minimum
1
2 
b
det ( A )
is
value of
(2019 Main, 10 Jan II)
b
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)
(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
 1 0
(a) 

12 1
1 n − 1 1 78
,then the
=
1  0 1 
(2019 Main, 9 April I)
 1 0
 1 −13
(b) 
 (c) 13 1
0
1




 1 −12
(d) 

0 1 
Then, for y ≠ 0 in R,
α
β
y+1
α
1 is equal to
y+β
1
(2019 Main, 9 April I)
(d) y3
1
b c .
4 b2 c2
1
6. Let the numbers 2, b, c be in an AP and A = 2
1
If det( A ) ∈ [2, 16], then c lies in the interval
(2019 Main, 8 April II)
(a) [3, 2 + 23 / 4 ] (b) (2 + 23 / 4 , 4) (c) [4, 6]
(d) [2, 3)
sin θ
1 
 1
 3π 5π 
7. If A = − sin θ
1
sin θ ; then for all θ ∈ 
, ,
 4 4
 − 1
1 
− sin θ
(2019 Main, 12 Jan II)
det( A ) lies in the interval
3
(c)  0, 
 2 
5
(d)  1, 
 2 
2a
2a
a−b−c
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
(b) − 2(a + b + c)
(a) − (a + b + c)
(d) abc
(c) 2(a + b + c)
8. If
9. 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
log e a 4r
log e a7r
a 2k
a5k
a 8k
log e a 2ra3k
log e a5r a 6k
log e a 8r a 9k
3
(c) 2( 2 + 1)
(d) 2( 2 + 2)
2x 
x − 4 2x

12. If
2x x − 4 2x  = ( A + Bx)(x − A )2,


2x x − 4
 2x
ordered pair ( A , B) is equal to
(b) (−4, 3)
(c) (−4, 5)
then
the
(2018 Main)
(d) (4, 5)
13. Let ω be a complex number such that 2ω + 1 = z, where
y+α
5
(b)  , 4
 2 
(d)
(2019 Main, 10 Jan I)
(b) −7
(a) (−4, − 5)
(a) y( y2 − 1) (b) y ( y2 − 3) (c) y3 − 1
3
(a)  , 3
 2 
(c) 2 3
4+ d
(sin θ ) − 2

 −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
(a) −5
5. Let α and β be the roots of the equation x2 + x + 1 = 0.
β
(b) −2 3
11. Let d ∈ R, and
4. If 
 ... 
.
.
0 1 0 1 0 1 0
1 n 
inverse of 
 is
0 1 
(a) − 3
log e a3r
log e a 6r
log e a 9r
a 4k
a7k
k
a10
=0
1
1
1
z = − 3. If 1 −ω 2 − 1 ω 2 = 3 k, then k is equal to
ω2
ω7
1
(2017 Main)
(a) − z
(c) − 1
(b) z
(d) 1
14. If α, β ≠ 0 and f (n ) = α + β and
n
n
3
1 + f (1) 1 + f (2)
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)
(a) αβ
(b)
1
αβ
(c) 1
(d) −1
15. 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
16. If A = α
and| A3| = 125, then the value of α is
2 α


(a) ± 1
(b) ± 2
(2004, 1M)
(c) ± 3
17. The number of distinct real roots of
 sin x
 cos x
 cos x
(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
x
x+1
1

x (x − 1)
(x + 1) x
2x
,
 3x (x − 1) x (x − 1) (x − 2) (x + 1) x (x − 1) 

18. If f (x) = 
then f (100) is equal to
(a) 0
(b) 1
(1999, 2M)
(c) 100
(d) –100
130 Matrices and Determinants
19. The parameter on which the value of the determinant


a2
a
1


−
p
+
d
x
px
cos
(
p
d
)
x
cos
cos
(
)


sin ( p − d ) x sin px sin ( p + d ) x 
26. Let P be a matrix of order 3 × 3 such that all the entries in
P are from the set { − 1, 0, 1}. Then, the maximum possible
value of the determinant of P is ......... .
does not depend upon, is
(a) a
(b) p
(1997, 2M)
(c) d
xp + y
x
20. The determinant yp + z
y
0
xp + y
(a) x, y, z are in AP
(c) x, y, z are in HP
(d) x
y
21. 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)
22. Which of the following is(are) NOT the square of a
3 × 3 matrix with real entries?
(2017 Adv.)
1 0 0 
(b)  0 − 1 0 


 0 0 −1
(2 + α )2 (2 + 2α )2 (2 + 3 α )2 = − 648 α ?
(3 + α )2 (3 + 2α )2 (3 + 3 α )2
(2015 Adv.)
(d) 4
MN = NM . Further, if M ≠ N 2 and M 2 = N 4, then
(2014 Adv.)
(a) determinant of (M + MN ) is 0
(b) there is a 3 × 3 non-zero matrix U 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
 a
25. The determinant  b
aα + b
2
b
c
bα + c
is equal to zero, then
(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
(1988, 2M)
x
29. Given that x = − 9 is a root of 2
7
roots are... and... .
7
2  = 0, the other two
x
3
x
6
(1983, 2M)
1
4
1
2x
20 
5 
 = 0 is… .
5x2 (1981, 2M)
 λ2 + 3λ
31. Let pλ + qλ + rλ + sλ + t = 
 λ+1
 λ −3
4
3
2
λ − 1 λ + 3
− 2λ λ − 4 

λ + 4 3λ 
1 a a2
32. The determinants 1 b ca and 1 b b2 are not
1 c ab
1 c c2
bc
identically equal.
24. Let M and N be two 3 × 3 matrices such that
2
1

1 a
(1 + α )2 (1 + 2α )2 (1 + 3 α )2
(c) −9
a − bc
b2 − ca is … .

c2 − ab 
a
b
c
True/False
 1 0 0
(d)  0 1 0


 0 0 1
(b) 9
1
28. The value of the determinant 1
(1993, 2M)
2
be an identity in λ , where p,q,r,s and t are constants.
(1981, 2M)
Then, the value of t is…. .
23. Which of the following values of α satisfy the equation
(a) −4
log x y log x z
1
determinant log y x
log y z is…… .
1
log z x log z y
1
30. The solution set of the equation
1 − 2
Objective Question II
(One or more than one correct option)
− 1 0 0 
(c)  0 − 1 0 


 0 0 −1
Fill in the Blanks
27. For positive numbers x, y and z, the numerical value of the
= 0, if
z
yp + z (1997C, 2M)
(b) x, y, z are in GP
(d) xy, yz , zx are in AP
1 0 0 
(a)  0 1 0 


 0 0 −1
Numerical Value
aα + b
bα + c 
0 
(1986, 2M)
(1983, 1M)
Analytical and Descriptive Questions
33. 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)
34. Let a , b, c be real numbers with a + b + c = 1 . Show
2
2
2
that the equation
bx + ay
cx + a
 ax − by − c

cy + b  = 0 represents a
− ax + by − c
 bx + ay
cy + b
− ax − by + c 
 cx + a
(2001, 6M)
straight line.
35. Prove that for all values of θ
sin θ
cos θ
2π 

sin θ +


3
2π 

sin θ −


3
2π 

cos θ +


3
2π 

cos θ −


3
sin 2θ
4π 

sin 2θ +
 =0

3
4π 

sin 2 θ −


3
(2000, 3M)
Matrices and Determinants 131
36. Suppose, f (x) is a function satisfying the following
conditions
(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
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
37. Find the value of the determinant p q
ab
r , where
1 1 1
a , b and c are respectively the pth , qth and rth terms of a
(1997C, 2M)
harmonic progression.
38. Let a > 0, d > 0. Find the value of the determinant
1
a
1
(a + d )
1
(a + 2d )
1
a (a + d )
1
(a + d ) (a + 2d )
1
(a + 2d ) (a + 3d )
1
(a + d ) (a + 2d )
1
(a + 2d ) (a + 3d )
1
(a + 3d ) (a + 4d )
 A 3 6
 8 9 C is divisible by k.
2 B 2 
 a −1
n
2n 2
3n3
2
43. Let ∆ a = 
 (a − 1)3
 (a − 1)
(1990, 4M)

6
4n − 2 

3n 2 − 3n 
n
Show that
∑ ∆ a = c ∈ constant.
(1989, 5M)
a =1
44. Show that
 xC
r
 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)
45. If α be a repeated root of a quadratic equation f (x) = 0
and A (x), B (x) and C (x) be polynomials of degree 3, 4
and 5 respectively, then show that
 A (x)
 A (α )
 A′ (α )
39. For all values of A , B, C and P , Q , R, show that
is divisible by
derivatives.
B (x)
B (α )
B′ (α )
C (x) 
C (α ) 
C′ (α ) 
f (x), where prime denotes the
(1984, 3M)
(1994, 4M)
 cos ( A − P )

 cos (B − P )
 cos (C − P )
cos ( A − Q )
cos (B − Q )
cos (C − Q )
cos ( A − R) 
cos (B − R) 
= 0
cos (C − R) 
40. For a fixed positive integer n, if
 n!
D =  (n + 1)!
 (n + 2)!
(n + 1)!
(n + 2)!
(n + 3)!
(n + 2)! 
(n + 3)! ,
(n + 4)! 

 D
then show that 
− 4 is divisible by n.
3

 (n !)
p b
41. If a ≠ p, b ≠ q, c ≠ r and a q
a b
p
Then, find the value of
p−a
46. 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)
(1992, 4M)
c
c = 0
r
q
r
. (1991, 4M)
+
+
q−b r−c
42. 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
47. 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)
Integer Type Question
48. The total number of distincts x ∈ R for which
x x2
1 + x3
2
2x 4x
1 + 8x3 = 10 is
3x 9x2 1 + 27x3
(2016 Adv.)
Topic 3 Adjoint and Inverse of a Matrix
Objective Questions I (Only one correct option)
5 2α 1 
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
1. If B = 0
(2019 Main, 12 April I)
(a) 0
(b) −1
(c) 1
(d) 2
et
e− t cos t
 t
−t
2. If A = e −e cos t − e−t sin t
et
2e− t sin t


e− t sin t

−t
− e sin t + e cos t  then A is
−t

−2e cos t

−t
(2019 Main, 9 Jan II)
132 Matrices and Determinants
each of a , b and c is either ω or ω 2. Then, the number of
distinct matrices in the set S is
(2011)
(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
(a) 2
det( ABAT ) = 8 and det( AB− 1 ) = 8, then det(BA − 1BT ) is
(2019 Main, 11 Jan II)
equal to
1
1
(a) 1
(b)
(c)
(d) 16
4
16
cos θ − sin θ 
4. If A = 
 , then the matrix
sin θ cos θ 
π
A −50 when θ = , is equal to
(2019 Main, 9 Jan I)
12
3
2 

1

2 
1 
2 

3

2 


(b) 




(d) 


3
2
1
2
1
2
3
2
1
− 
2

3

2 
3
−
2 

1

2 
 2 −3 
2
 , then adj (3 A + 12 A ) is equal to
 −4 1 
(2017 Main)
5a
3
6. If A = 
 51 63
(b) 

 84 72
72
−
63

(d) 

 − 84 51 
(c) 4
(d) 13
7. 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
(b) I
(c) B −1
(2014 Main)
(d) (B −1 )T
1 α 3
8. If P = 1 3 3 is the adjoint of a 3 × 3 matrix A and
2 4 4
(2013 Main)
| A | = 4 , then α is equal to
(a) 4
(b) 11
(c) 5
(d) 0
9. If P is a 3 × 3 matrix such that P = 2P + I, where PT is
T
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 (b) PX = X
 
 0
(c) PX = 2X
(a) M 2
(b) −N 2
(c) −M 2
(d) MN (2011)
1 0 0
12. 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)
Objective Questions II
(One or more than one correct option)
3 −1 −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
k2
k
identity matrix of order 3. If q23 = − and det (Q ) = ,
2
8
(2016 Adv.)
then
(a) α = 0, k = 8
(c) det (P adj (Q )) = 29
(b) 4α − k + 8 = 0
(d) det (Q adj (P )) = 213
Then, M is invertible, if
(2016 Main)
(b) 5
matrices such that MN = NM . If PT denotes the
transpose of P, then M 2N 2(M T N )−1 (MN −1 )T is equal to
14. Let M be a 2 × 2 symmetric matrix with integer entries.
− b
and A adj A = AAT , then 5a + b is equal
2 
to
(a) − 1
(d) 8
13. Let P = 2
5. If A = 
 72 − 84
(a) 

 − 63 51 
51
84


(c) 

 63 72
(c) 4
11. Let M and N be two 3 × 3 non-singular skew-symmetric
3. Let A and B be two invertible matrices of order 3 × 3. If
 1

(a)  2
3
−
 2
 3

(c)  2
1
−
 2
(b) 6
(d) PX = − X
10. Let ω ≠ 1 be a cube root of unity and S be the set of all
 1 a b
non-singular matrices of the form  ω 1 c , where
ω 2 ω 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
1 4 4 
15. 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
(c) 1
(b) − 1
(d) 2
Integer Answer Type Question
16. Let k be a positive real number and let
 2k − 1 2 k 2 k 


1
A= 2 k
− 2k  and
 − 2 k 2k
− 1 

 0
2k − 1
k

0
2 k
B =  1 − 2k
 − k −2 k
0






If det (adj A ) + det (adj B) = 106, then [k] is equal to……
(2010)
Matrices and Determinants 133
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 
θ∈ ,  ∪
2 3 
 7π 
π, 

6
 π 2π 
(c) has a unique solution if θ ∈  , 
2 3 
has a non-trivial solution, is
(a) −1
(a)
(b)
(c)
(d)
(d) λ2 + λ − 6 = 0
x+ y+ z =5
x + 2 y + 2z = 6
has
infinitely
x + 3 y + λz = µ,(λ , µ ∈ R),
solutions, then the value of λ + µ is
many
(a) (2, 4)
(b) 12
(d) 9
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 ),
x y z
then + + + k is equal to
y z x
(2019 Main, 9 April II)
1
4
(d)
3
4
5. If 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
(2019 Main, 8 April II)
straight line whose equation is
(a) 3x − 4 y − 4 = 0
(c) 4x − 3 y − 4 = 0
(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
(b) (− 4, 2)
(c) (1, − 3)
(d) (−3, 1)
9. If the system of linear equations
2x + 2 y + 3z = a
3x − y + 5z = b
x − 3 y + 2z = c
where a , b, c are non-zero real numbers, has more than
one solution, then
(2019 Main, 11 Jan I)
(a) b − c − a = 0
(c) b − c + a = 0
(b) a + b + c = 0
(d) b + c − a = 0
10. The number of values of θ ∈ (0, π ) for which the system of
x + 3 y + 7z = 0,
− x + 4 y + 7z = 0,
(sin 3θ )x + (cos 2θ ) y + 2z = 0
has a non-trivial solution, is
(a) two
(2019 Main, 10 April I)
(c) −
(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
linear equations
3. If the system of linear equations
1
2
(2019 Main, 12 Jan II)
contains exactly two elements.
contains more than two elements.
is a singleton.
is an empty set.
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
(b)
(d) 0
equations x − 2 y − 2z = λx,x + 2 y + z = λyand − x − y = λz
has a non-trivial solution
equations
(a) −4
(c) 2
7. The set of all values of λ for which the system of linear
2. Let λ be a real number for which the system of linear
(a) 7
(c) 10
(2019 Main, 8 April I)
8. An ordered pair (α , β) for which the system of linear
 7π 
and have infinitely many solutions if θ ∈  π , 

6
(d) have infinitely many solutions if
 π 2π   7π 
θ ∈  ,  ∪ π, 
2 3  
6
(c) λ2 − λ − 6 = 0
1
(b)
2
(b) three
(2019 Main, 10 Jan II)
(c) four
(d) one
11. If the system of equations
x+ y+z=5
x + 2 y + 3z = 9
x + 3 y + αz = β
has infinitely many solutions, then β − α equals
(2019 Main, 10 Jan I)
(a) 8
(b) 18
(c) 21
(d) 5
12. If 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)
(a) 2 g + h + k = 0
(c) g + h + k = 0
(b) g + 2h + k = 0
(d) g + h + 2k = 0
13. The system of linear equations
x + y + z = 2,
2x + 3 y + 2z = 5
2x + 3 y + ( a 2 − 1)z = a + 1
(a)
(b)
(c)
(d)
has infinitely many solutions for a = 4
is inconsistent when a = 4
has a unique solution for|a| = 3
is inconsistent when|a| = 3
(2019 Main, 9 Jan I)
134 Matrices and Determinants
14. If the system of linear equations
x + ky + 3z = 0, 3x + ky − 2z = 0
2x + 4 y − 3z = 0
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
16. The set of all values of λ for which the system of linear
equations 2x1 − 2x2 + x3 = λx1, 2x1 − 3x2 + 2x3 = λx2 and
(2015 Main)
− x1 + 2x2 = λx3 has a non-trivial solution
(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.
23. Consider the system of equations
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
1
(2008, 3M)
Objective Questions II (Only or More Than One)
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
equation
(k + 1) x + 8 y = 4 y ⇒ kx + (k + 3) y = 3k − 1
(2013 Main)
has no solution, is
(c) 2
(d) 3
18. The number of 3 × 3 matrices A whose entries are either
 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,
then the value of λ such that the given system of
(2004, 1M)
equations has no solution, is
(a) 3
(b) 1
(c) 0
(d) –3
20. The number of values of k for which the system of
equations (k + 1) x + 8 y = 4k and kx + (k + 3) y = 3k − 1
(2002, 1M)
has infinitely many solutions, is/are
(a) 0
(b) 1
(c) 2
(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
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 of k are
(2000, 2M)
(a) –1, 2
(b) 1, 2
(c) 0, 1
1
24. Let S be the set of all column matrices  b2  such that b1,
17. The number of value of k, for which the system of
(b) 1
4
k ≠ 0.
 b1 
(a) is an empty set
(b) is a singleton set
(c) contains two elements
(d) contains more than two elements
(a) infinite
x − 2 y + 3 z = −1 ,
(d) –1, 1
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
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 
(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
25. The system of equations λx + y + z = 0 , − x + λy + z = 0
and − x − y + λz = 0 will have a non-zero solution, if real
(1982, 2M)
values of λ are given by ...
Analytical and Descriptive Questions
a 2
1
f 
c , U =  g , V = 0 
0 
1
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
(2004, 4M)
that BX = V has no solution.
a
26. A = 1
0
c
d
1
a
b, B = 0
 f
b
1
d
g
27. Let λ and α be real. Find the set of all values of λ for
which the system of linear equations
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 a non-trivial solution.
For λ = 1, find all values of α.
(1993, 5M)
Matrices and Determinants 135
28. Let α 1 , α 2, β1 , β 2 be the roots of ax2 + bx + c = 0 and
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
33. 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)
Integer Answer Type Question
34. For a real number α , if the system
29. Consider the system of linear equations in x, y, z
 1 α α 2  x   1 
   

 α 1 α   y = −1
α 2 α 1   z   1 
   

(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
(1986, 5M)
solution.
of linear equations, has infinitely many solutions,
then 1 + α + α 2 =
(2017 Adv.)
30. 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)
0
 −1 
0
 3 
35. Let M be a 3 × 3 matrix satisfying M 1 =  2 ,
1 1
M −1 =  1 , and M
 0  −1
31. For what values of m, does the system of equations
3x + my = m and 2x − 5 y = 20 has a solution satisfying the
(1979, 3M )
conditions x > 0, y > 0?
32. For what value of k, does the following system of equations
1  0 
1 =  0  ,
   
1 12
Then, the sum of the diagonal entries of M is …
possess a non-trivial solution over the set of rationals
(2011)
x + y − 2z = 0, 2x − 3 y + z = 0, and x − 5 y + 4z = k
Find all the solutions.
(1979, 3M )
Answers
Topic 3
Topic 1
1. (b)
5. (b)
9.
13.
17.
21.
(d)
(b, c, d)
(d)
(1)
2. (b)
6. (b)
3. (c)
7. (d)
4. (a)
8. (a)
10. (a)
14. (d)
18. (c)
11. (c, d)
15. (b)
19. (d)
12. (c, d)
16. (a)
20. (4)
Topic 2
1. (a)
2. (a)
3. (c)
4. (b)
5. (d)
9. (d)
6. (c)
10. (c)
7. (a)
11. (a)
8. (b)
12. (c)
13.
17.
21.
25.
29.
(a)
14. (c)
15. (d)
(c)
18. (a)
19. (b)
(b)
22. (a, c)
23. (b, c)
(b,d)
26. (4)
27. (0)
(2 and 7)
30. {–1,2}
31. (0)
1
5
1 2 5


36. a = , b = − and f ( x ) = x − x + 2


4
4
4
4
4


4d

38. 
 a (a + d ) 2(a + 2d ) 3(a + 3d ) 2(a + 4d )
48. (2)
16.
20.
24.
28.
32.
(c)
(b)
(a, b)
(0)
False
1. (c)
2. (b)
3. (c)
4. (c)
5. (b)
9. (d)
13. (b,c)
6. (b)
10. (a)
14. (c, d)
7. (b)
11. (c)
15. (a,d)
8. (b)
12. (a)
16. (4)
2.
6.
10.
14.
18.
22.
3.
7.
11.
15.
19.
23.
Topic 4
1.
5.
9.
13.
17.
21.
25.
29.
30.
37. (0)
31.
41. (2)
4.
8.
12.
16.
20.
24.
π
27. − 2 ≤ λ ≤ 2 , α = nπ , nπ +
λ=0
4
n π
θ = nπ , nπ + ( −1 ) , n ∈ Z
6
4 − 5k
13k − 9
x=
,y =
,z = k
7
7
15
m < − or m > 30
2
(a)
(c)
(a)
(d)
(d)
(a)
(c)
(b)
(a)
(b)
(a)
(d)
(c)
(c)
(a)
(d)
(b)
(a)
(b)
(a)
(a)
(c)
(b)
(a,d)
32. (k = 0, the given system has infinitely many solutions)
34. (1)
35. (9)
136 Matrices and Determinants
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)
On taking transpose both sides, we get
T
2 3 
( A + B)T = 

5 − 1
2 5 
⇒
A +B =

3 − 1
Given, AT = A and BT = − B
2 5 
⇒
A−B=

3 − 1
T
T
… (ii)
On solving Eqs. (i) and (ii), we get
0 − 1
2 4 
A=
 and B = 1 0 
4
1
−




2 4  0 − 1  4 − 2
So, AB = 

=

4 − 1 1 0  − 1 − 4
2. Given matrix
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


⇒  0 6 y2 0 = 0 3 0
 0
0 3 0 0 3

Here, two matrices are equal, therefore equating the
corresponding elements, we get
8x2 = 3 and 6 y2 = 3
⇒
and
cos α
∴ A2 = 
sin α

cos2 α − sin 2 α
=
 sin α cosα + cosα sin α
− cosα sin α − sin α cosα 

− sin 2 α + cos2 α

cos 2 α − sin 2 α 
=

sin 2 α cos 2 α 
Similarly,
cos(nα ) − sin(nα )
An = 
, n ∈ N
sin(nα ) cos(nα ) 
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
4. Given matrix
1 0 0 0 0 0
P = 3 1 0 = 3 0 0 +
9 3 1 9 3 0
⇒
P = X + I (let)
Now, P5 = (I + X )5
1 0 0
0 1 0


0 0 1
= I + 5C1 (X ) + 5C 2(X 2) + 5C3 (X 3 ) + …
[Q I = 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
2
Here,
X = 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
and X 3 = X 2 ⋅ X = 0 0 0 3 0 0 = 0 0
9 0 0 9 3 0 0 0
n
 0 2y 1 
A = 2x y −1 , (x, y ∈ R, x ≠ y)
2x − y 1 
⇒
cos α − sin α 

sin α cos α 
− sin α cos α − sin α 
cos α sin α cos α 
3. Given, matrix A = 
3
8
1
y=±
2
x=±
Q There are 2 different values of x and y each.
So, 4 matrices are possible such that AT A = 3I3 .
0 0 0
⇒ X = X = 0 0 0
0 0 0
0 0 0
0 0 0


5
So,
P = I + 5 3 0 0 + 10 0 0 0
9 3 0
9 0 0
0 0
 1
=  15 1 0
135 15 1
4
5
0 0
 2
and 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
0
0
0
Matrices and Determinants 137
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
0 − 2q2 + r 2 p2 − q2 − r 2 p2 + q2 + r 2 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
Thus,
1
7. Given, A = 2
a
1
AAT = 2
a
0 0
1 0
4 1
0
0
0 0  1 0 0  1
1
0
1 0  4 1 0 =  4 + 4
4 1 16 4 1 16 + 32 4 + 4 1
1
0
0


…(i)
1
0
=  4 ×2
16 (1 + 2) 4 × 2 1
1
0
0  1 0 0


3
and P =  4 × 2
1
0  4 1 0
16 (1 + 2) 4 × 2 1 16 4 1
1
6. Here, P =  4
16
1
∴
P 2 =  4
16
1
0
0

...(ii)
4 ×3
1
0
= 
16 (1 + 2 + 3) 4 × 3 1
From symmetry,
1
0
0

P50 = 
4 × 50
1
0
16 (1 + 2 + 3 + ... + 50) 4 × 50 1
[given]
Q
P50 − Q = I


1 − q11
− q12
− q13  1 0 0


∴
200 − q21
1 − q22
− q23  = 0 1 0


50
16 ×
(51) − q31 200 − q32 1 − q33  0 0 1


2
16 × 50 × 51
⇒
200 − q21 = 0,
− q31 = 0,
2
200 − q32 = 0
∴
q21 = 200, q32 = 200, q31 = 20400
2 2
1 −2, AT
2 b 
2 2  1
1 −2 2
2 b  2
1 2 a 
= 2 1 2  and
2 −2 b 
2 a
1 2 
−2 b 
a + 4 + 2b 
9
0


0
9
2a + 2 − 2b
=
a + 4 + 2b 2a + 2 − 2b a 2 + 4 + b2 
…(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
q31 + q32 20400 + 200 20600
=
= 103
=
200
q21
200
It is given that, AAT = 9I
9
0
a + 4 + 2b 

1 0 0
⇒ 
0
9
2a + 2 − 2b = 9 0 1 0
a + 4 + 2b 2a + 2 − 2b a 2 + 4 + b2 
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
…(ii)
and a 2 + 4 + b2 = 9
…(iii)
On solving Eqs. (i) and (ii), we get
a = − 2, b = − 1
This satisfies Eq. (iii)
Hence, (a , b) ≡ (−2,−1)
 3 /2 −1 /2   3 /2
PT P = 

3 /2  −1 /2
 1 /2
1 0
PT P = 

0 1
8. Now,
⇒
⇒
1 /2 

3 /2
PT P = I
PT = P −1
⇒
Q = PAPT
Since,
∴ PTQ 2005 P = PT [(PAPT )(PAPT ) K 2005 times ] P
= (PT P ) A (PT P ) A (PT P ) K (PT P ) A (PT P )
144444444
42444444444
3
∴
= IA 2005 = A 2005
1 1
A=

0 1
1
A2 = 
0
1
A3 = 
0
2005 times
1 1 1 1
=
1 0 1 0
2 1 1 1
=
1 0 1 0
2
1
3
1
……………………
……………………
138 Matrices and Determinants
1 2005
A 2005 = 
1 
0
1 2005
PTQ 2005 P = 
1 
0
∴
9. Given, A = α
1
⇒
0
1
, B = 
1
5
α
A 2 = 
1
0  α
1 1
Also, given, A = B
 2
0 1
⇒ α
=
α + 1 1 5
0
1
0  α 2
=
1 α + 1
0 
1
 p11 p12  ω 2 ω3  ω 2 1 
=
P = [ pij ]2× 2 = 

= 3
4
 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
2
P = [ pij ]3 × 3
0
1
ω 2 1 ω  ω 2 1 ω  0 0 0



P =  1 ω ω 2  1 ω ω 2 = 0 0 0 = 0
 ω ω 2 1   ω ω 2 1  0 0 0

 


⇒ α 2 = 1 and α + 1 = 5
2
Which is not possible at the same time.
∴ No real values of α exists.
10. If A and B are square matrices of equal degree, then
∴
11. Given, X T = − X , Y T = − Y , Z T = Z
P =Y Z − Z Y
P = (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,
4
4 3
T
12. (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.
13. Here,
P = [ pij ]n × n with pij = wi + j
∴ When n = 1
P = [ pij ]1 × 1 = [ω ]
P 2 = [ω 4 ] ≠ 0
2
⇒
∴ When
n =2
P 2 = 0, when n is a multiple of 3.
P 2 ≠ 0, when n is not a multiple of 3.
A+ B=B+ A
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 

 

⇒
n = 57 is not possible.
∴
n = 55, 58, 56 is possible.
14. 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
15. If b = 6 ⇒ a = 1 and c = − 7
∴ ax2 + bx + c = 0 ⇒ x2 + 6x − 7 = 0
⇒ (x + 7) (x − 1) = 0
∴ x = 1, − 7
∞
1
6  6
 1 1
⇒∑  −  =1 + +   + L+ ∞ =
6




1
7
7
7
n= 0
1−
7
1
=
=7
1 /7
2
n
16. If a = 2, b = 12, c = − 14
3
1
3
+
+
ω a ωb ω c
3
3
1
3
⇒
+ 12 + −14 = 2 + 1 + 3ω 2 = 3ω + 1 + 3ω 2
2
ω
ω
ω
ω
∴
= 1 + 3 (ω + ω 2) = 1 − 3 = − 2
17. 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.
∴
Required number = ( p − 1) p2 − ( p − 1)2 + ( p − 1)2
= p3 − p2
18. Trace of A = 2a, will be divisible by p, iff a = 0.
| 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.
Matrices and Determinants 139
b
, a, b, c ∈ {0, 1, 2 ,... , p − 1}
a 
a
c
19. 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
...(ii)
p divides (a + b).
⇒ p choices, including a = b = 0 included in Eq. (i).
∴ Total number of choices are ( p + p − 1) = 2 p − 1
a
b
c
a
20. Given, A = b
c
Now, AT A = I
a b c  a
⇒ b c a  b
c a b  c
c
a , abc = 1 and AT A = I
b 
b
c
a
c  1
a  = 0
b  0
ab + bc + ca
a 2 + b2 + c2
ab + bc + ca
 a 2 + b2 + c2
⇒ ab + bc + ca
ab + bc + ca

1
= 0
0
⇒ a 2 + b2 + c2 = 1
0
1
0
…(i)
0
0
1
0 − sin 2 θ − 2 − 4 cos 6 θ
∆ = 0 1 − sin 2 θ − 2 − 4 cos 6 θ = 0
1 + 4 cos 6 θ
sin 2 θ
1
0
0
1
…(ii)
= (a + b + c)(a 2 + b2 + c2 − ab − bc − ca )
⇒ a + b + c = (a + b + c) (1 − 0) + 3
[from Eqs. (i) and (ii)]
3
3
∴ a3 + b3 + c3 = (a + b + c) + 3
Now,
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


2. Given equation
…(iii)
(a + b + c)2 = a 2 + b2 + c2 + 2(ab + bc + ca )
=1
…(iv)
From Eq. (iii), a + b + c = 1 + 3 ⇒ a + b + c = 4
3
3
3
3
3
3
−1 + i 3
=ω
2
(−ω )r ω 2s 
Q
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

r
ω 4s + ω 2r
[(− 1) + 1]
ω

Given, P 2 = − 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}
21. Here,
Applying C1 → C1 + C 2, we get
Applying R1 → R1 − 2R3 and R2 → R2 − 2R3 , we get
We know, a3 + b3 + c3 − 3abc
3
sin 2 θ
1 + cos 2 θ
4 cos 6 θ
2
1. Let ∆ = cos θ
1 + sin 2 θ
4 cos 6 θ
=0
1 + 4 cos 6 θ
cos 2 θ
sin 2 θ
4 cos 6 θ
sin 2 θ
∆ = 2 1 + sin 2 θ
4 cos 6 θ
=0
2
1 + 4 cos 6 θ
1
sin θ
ab + bc + ca = 0
and
Topic 2 Properties of Determinants
2
ab + bc + ca 
ab + bc + ca 

a 2 + b2 + c2 
0
1
0
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.
z=
−6
−1
x
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
3. Given determinants are
x
∆1 = − sin θ
cos θ
sin θ cos θ
−x
1
1
x
=0
140 Matrices and Determinants
= − x3 + sin θ cos θ − sin θ cos θ + x cos 2 θ − x + x sin 2 θ
= − x3
∆= α
β
sin 2θ cos 2θ
x
and ∆ 2 = − sin 2θ
cos 2θ
−x
1
1
x
On applying C 2 → C 2 − C1 and C3 → C3 − C1, we get
0
0
y
,x≠0
y + β −α
1 −β
1 −α
y + α −β
= y[( y + (β − α )) ( y − (β − α )) − (1 − α ) (1 − β )]
= − x (similarly as ∆1)
3
[expanding along R1]
So, according to options, we get ∆1 + ∆ 2 = − 2x3
= y [ y − (β − α ) − (1 − α − β + αβ )]
2
4. Given
= y [ y2 − β 2 − α 2 + 2αβ − 1 + (α + β ) − αβ ]
1 1 1 2 1 3 1 n − 1 1 78
=
0 1 0 1 0 1 ... 0
1  0 1 
 



= y [ y2 − (α + β )2 + 2αβ + 2αβ − 1 + (α + β ) − αβ ]
= y[ y2 − 1 + 3 − 1 − 1] = y3 [Qα + β = −1 and αβ = 1]
1 1 1 
6. Given, matrix A = 2 b c , so
4 b2 c2
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
det( A ) = 2 b c


2
2
4 b c 
:
:
:
:
:
:
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

1
=
0

n (n − 1)  1 78
=
2

 0 1 
1

Since, both matrices are
corresponding element, we get
n (n − 1)
= 78 ⇒ n (n − 1) = 156
2
equal,
so
equating
= 13 × 12 = 13(13 − 1)
⇒
n = 13
1 −13
1 13
−1
A=
 = A = 0 1 
0
1




So,
 d − b 
a b
−1
[Q if|A|= 1and A = 
 , then A =  − c a  


 c d
5. 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
2
y+α
[Qα + β = −1]
On applying, C 2 → C 2 − C1 and C3 → C3 − C1,
0
0 
1

we get det( A ) = 2 b − 2 c − 2


2
2
4 b − 4 c − 4
b − 2 c − 2

= 2
2
b − 4 c − 4
b −2
c−2
=
(b − 2)(b + 2) (c − 2)(c + 2)
1 
 1

= (b − 2)(c − 2)
b + 2 c + 2
[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
So,
[given]
| A| = d (2d )d = 2d3 ∈ [2, 16]
3
⇒
d ∈ [1, 8] ⇒ d ∈ [1, 2]
∴ 2 + 2d ∈ [2 + 2, 2 + 4]
= [4, 6] ⇒ c ∈ [4, 6]
1 
sin θ
 1
7. Given matrix A = − sin θ
1
sin θ 
 −1
− sin θ 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)
⇒| A| = 2 (1 + sin 2 θ )
 3π 5π 
As we know that, for θ ∈ 
, 
 4 4
1
 1
sin θ ∈  −
,


2
2
…(i)
Matrices and Determinants 141
1
 1


⇒ sin 2 θ ∈ 0,  ⇒ 1 + sin 2 θ ∈ 0 + 1, + 1


2
 2

3


⇒ 1 + sin 2 θ ∈ 1, 
 2
3
⇒ 2(1 + sin 2θ ) ∈ [2, 3) ⇒| A| ∈ [2, 3) ⊂  ,
2
 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

log e a r + kRk
⇒ log e a r + kR3 r + 4k

3

log e a r + kR6r + 7k
2a
2a
a−b−c
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
∆=
 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

8. Let ∆ =
2c
2c
1
1
c−a −b
1
2b
= (a + b + c) 2b b − c − a
2c
2c
c−a −b
(taking common (a + b + c) fromR1)
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]
9. Given,
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 a3r a 4k
log e a 6r a7k
k
log e a 9r a10
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.
=0
⇒ log e a 4r a5k
log e a7ra 8k
1
1
2
b
b
2 
1
b
1
b2 + 1
b
2
= 2 [2(b + 1) − b ] − b(2b − b)
+1(b2 − b2 − 1)
2
2
2
= 2[2b + 2 − b ] − b − 1
2
log e a 4r a5k log e a5r a 6k − log e a 4r a5k
log e a7ra 8k log e a 8r a 9k − log e a7ra 8k
log e a1ra 2k
b
So, det ( A ) =| A| = b
On applying elementary operations
C 2 → C 2 − C1 and C3 → C3 − C1, we get
log e a1ra 2k log e a 2ra3k − log e a1ra 2k
log e a3r a 4k − log e a1ra 2k
log e a 6r a7k − log e a 4r a5k
k
log e a 9r a10
− log e a7ra 8k
2
10. Given matrix, A = b b2 + 1 b, b > 0
2
= 2b2 + 4 − b2 − 1 = b2 + 3
=0
 a ra k 
 arak 
log e  2r 3k  log e  3r k4 
 a1 a 2 
 a1 a 2 
 arak 
 a5r a 6k 
log e  r k  log e  6r 7k  = 0
 a 4a5 
 a 4a5 
k 
 a 9r a10
 a 8r a 9k 
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 ]
det( A ) b2 + 3
3
=
= b+
b
b
b
Now, by AM ≥ GM, we get
3
1/ 2
b+
b ≥  b × 3



2
b
3
b+ ≥2 3
⇒
b
⇒
So, minimum value of
det (A )
=2 3
b
11. Given,
(sin θ ) − 2 
4+ d
 −2


A =  1 (sin θ ) + 2
d

 5 (2 sin θ ) − d (− sin θ ) + 2 + 2d 
{Q b > 0 }
142 Matrices and Determinants
4+ d
−2
(sin θ ) − 2
∴ | A| = 1
(sin θ ) + 2
d
5 (2 sin θ ) − d (− sin θ ) + 2 + 2d
⇒
[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
(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.
0
1
therefore
⇒
⇒
∴
(d + 2) = 9
2
⇒
14.
d+2=±3
⇒
d = 1, − 5
12. 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
ω2
ω7
ω ω 2 = 3k
1 ω2 ω
1
3(ω 2 − ω 4 ) = 3k
(ω 2 − ω ) = k
 − 1 − 3i   − 1 + 3i 
 −
 = − 3i = − z
k=
2
2

 

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.
3
⇒
∆= 1+α +β
1 + α 2 + β2
=
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)
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
1 ⋅1 + 1 ⋅1 + 1 ⋅1
1 ⋅1 + 1 ⋅α + 1 ⋅β
1 ⋅ 1 + 1 ⋅ α 2 + 1 ⋅ β2
1 ⋅1 + 1 ⋅α + 1 ⋅β
1 ⋅1 + α ⋅α + β ⋅β
1 ⋅ 1 + α 2 ⋅ α + β2 ⋅ β
1 ⋅ 1 + 1 ⋅ α 2 + 1 ⋅ β2
1 ⋅ 1 + α ⋅ α 2 + β ⋅ β2
1 ⋅ 1 + α 2 ⋅ α 2 + β2 ⋅ β2
1
= 1
(5x − 4)(x + 4)2 = ( A + Bx)(x − A )2
Equating, we get, A = − 4 and B = 5
1
0
1 + f (2) 1 + f (3) 1 + f (4)
Apply R2 → R2 − R1 and R3 → R3 − R1
2x
0 
1
0  = ( A + Bx)(x − A )2
∴ (5x − 4)0 − x − 4


−x − 4
0
0
Expanding along C1, we get
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
Now, 1 − ω 2 − 1 ω 2 = 3k
0
Given, f (n ) = α n + β n,
f (3) = α 3 + β3 ,
3
1+
Let ∆ = 1 + f (1) 1 +
Taking common (5x − 4) from C1, we get
2x 
1 2x
(5x − 4)1 x − 4 2x  = ( A + Bx)(x − A )2


1 2x x − 4
13. Given,
⇒
ω
⇒
(d + 2)2 − 1 = 8
⇒
ω2
3
| A|min = 8,
Q
1 1 1
1 ω ω 2 = 3k
1 ω2 ω
1
α
1
β
1 α 2 β2
1 1 1
1 1 1
1 α β = 1 α β
1 α 2 β2
1 α 2 β2
2
On expanding, we get ∆ = (1 − α )2(1 − β )2(α − β )2
But given,
[Q z = − 3]
∆ = 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
15.
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.
 a11 a12 a13 
if a = [aij ]3 × 3 =  a21 a22 a23 
 a31 a32 a33 
Matrices and Determinants 143
 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 
Here,
19. Let
Applying C1 → C1 + C3
a
1 + a2
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
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
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
a13
Applying C1 → C1 − 2 cos dx C 2
= 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

1 + a 2 − 2a cos dx
a
a2

⇒ ∆ =
+
0
px
p
d
x
cos
cos
(
)


0
sin px sin ( p + d ) x 

⇒ ∆ = (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
a13
a 23
a33
= 212 ⋅ P = 212 ⋅ 2 = 213
16. We know,
Since,
⇒
| A n| = | A|n
| A3| = 125 ⇒ | A|3 = 125
α 2 = 5
⇒ α2 − 4 = 5 ⇒ α = ± 3
2 α 
 sin x cos x cos x 
17. 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
cos x 
cos x 
sin x 
cos x
sin x
cos x
cos x 
cos x  = 0
sin x 
cos x
cos x
1

0
⇒ (2 cos x + sin x) 0 sin x − cos x
= 0
0
sin x − cos x 
0
⇒
(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
1
x

f (x) =
2x
x (x − 1)
 3x (x − 1) x (x − 1) (x − 2)
Applying C3 → C3 − (C1 + C 2)
x
1

x (x − 1)
2x
=
 3x (x − 1) x (x − 1)(x − 2)
∴
which is independent of p.
xp + y
20. Given, yp + z
0
x
y
xp + y
y
=0
z
yp + z
Applying C1 → C1 − ( p C 2 + C3 )
x
0
⇒
y
0
Applying R2 → R2 − R1 , R3 → R3 − R1
18. Given,


a2
a
1
p
px
cos
(
p
d
)
x
cos
cos
(
∆ =
−
+ d) x 


sin ( p − d ) x sin px sin ( p + d) x 
x+1

(x + 1) x

(x + 1) x (x − 1) 
0
0 = 0
0
f (x) = 0 ⇒ f (100) = 0
− (xp2 + yp + yp + z ) xp + y
⇒
y
z
=0
yp + z
− (xp2 + 2 yp + z ) (xz − y2) = 0
∴
Either xp2 + 2 yp + z = 0 or y2 = xz
⇒ x, y, z are in GP.
21. 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.
[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.
22. For a matrix to be square of other matrix its
determinant should be positive.
(a) and (c) → Correct
(b) and (d) → Incorrect
23. Given determinant could be expressed as product of two
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
144 Matrices and Determinants
1 + 2α + α2
⇒ 4 + 4α + α2
1 + 4α + 4α2
4 + 8α + 4α2
1 + 6α + 9 α2
4 + 12 α + 9 α 2
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
9 + 6 α + α 2 9 + 12 α + 4 α 2 9 + 18 α + 9 α 2
= − 648 α
α α2 1 1 1
4 2 α α 2 ⋅ 2 4 6 = − 648 α
9 3α α2 1 4 9
1
⇒
⇒
⇒
∴
24.
−1 1
1
1 1 1 1 1 1
⇒ α 3 4 2 1 ⋅ 2 4 6 = − 648 α
9 3 1 1 4 9
−8 α 3 = − 648 α
α − 81α = 0 ⇒ α (α 2 − 81) = 0
α = 0, ± 9
3
Hence, maximum value is 4.
log x y log x z
1
27. Let
log y z
∆ = log y x
1
log z x log z y
1
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
⇒
⇒
=
M2 − N 4 = 0
(M − N 2) (M + N 2) = 0
[as MN = NM ]
⇒
M + N2 =0
⇒
det (M + N ) = 0
2
Also, det (M + MN ) = (det M) (det M + N )
2
2
2
= (det M) (0) = 0
As,
det (M + MN ) = 0
2
2
Thus, there exists a non-zero matrix U such that
(M 2 + MN 2) U = 0
aα + b
b
 a
25. Given,
bα + c  = 0
c
 b
0 
 aα + b bα + c
Applying C3 → C3 − (α C1 + C 2)




a
b
0

b
c
0
= 0
2
a
b
α
+
b
α
+
c
−
a
α
+
b
α
+
c
2
(
)


⇒
⇒
log x
log y
log x
log z
28.
1 a
1 b
Now,
1
b1
c1
b3
c3
= 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
by log x,
log y
log z
log y log z = 0
log y log z
c
a a 2 abc
1
b b2 abc
ca =
abc
c c2 abc
ab
bc
1 a a2
a a2 1
1
=
⋅ abc b b2 1 = 1 b b2
abc
c c2 1
1 c c2
∴
1 a a 2 − bc
1 b b2 − ca = 0
1 c c2 − ab
26. Let Det (P ) = a 2 b2 c2
a3
1
Applying R1 → aR1 , R2 → bR2, R3 → cR3
aα 2 + 2bα + c = 0 or b2 = ac
a1
log y
log z
1 a a 2 − bc
1 a a2
1 a bc
2
1 b b − ca = 1 b b2 − 1 b ca
1 c c2 − ab
1 c c2
1 c ab
− (aα 2 + 2bα + c) (ac − b2) = 0
⇒ x − α is a factor of ax2 + 2bx + c or a , b, c are in GP.
1
log z
log x
log z
log y
On dividing and multiplying R1 , R2, R3
log y, log z, respectively.
log x
1
=
log x
log x log y log z
log x
M ≠ N2
Also,
log y
log x
1
29. Given,
x 3 7
2 x 2 =0
7 6 x
Applying R1 → R1 + R2 + R3
x+9 x+9 x+9
1 1 1
⇒
2
x
2 = 0 ⇒ (x + 9) 2 x 2 = 0
7
6
x
Applying C 2 → C 2 − C1
7 6 x
and C3 → C3 − C1
Matrices and Determinants 145
1
0
⇒ (x + 9) 2 x − 2
−1
7
0
0
= 0 ⇒ (x + 9) (x − 2) (x − 7) = 0
x−7
⇒
x = − 9, 2, 7 are the roots.
1
⇒
a
Applying C1 → C1 + bC 2 + cC3
∴ Other two roots are 2 and 7.
1
4
30. Given, 1 −2
20
5
a 2x − aby − ac
cx + a
bx + ay
− ax + by − c
cy + b
abx + a 2y
=0
cy + b
− ax − by + c
acx + a 2
⇒
 (a 2 + b2 + c2) x
cx + a 
ay + bx
1 2
(a + b2 + c2) y by − c − ax
b + cy  = 0

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 
=0
1 2 x 5 x2
⇒ 1 (− 10 x − 10x) − 4 (5x − 5) + 20 (2x + 2) = 0
2
2
⇒
−30x2 + 30x + 60 = 0
⇒
(x − 2) (x + 1) = 0
⇒
[Q a 2 + b2 + c2 = 1]
Applying C 2 → C 2 − bC1 and C3 → C3 − cC1
ay
a

1x
b
⇒
 y − c − ax
= 0
a 1
cy
− ax − by 
x = 2, − 1
Hence, the solution set is { −1, 2}.
λ 2 + 3λ
31. Given, λ + 1
λ −3
λ −1
−2 λ
λ+3
λ −4
λ+4
3λ
⇒
= pλ 4 + qλ 3 + rλ 2 + sλ + t
x2
axy
ax
1
y − c − ax
b
=0
ax
1
cy
− ax − by
Thus, the value of t is obtained by putting λ = 0.
0 −1 3
1
0 −4 = t
⇒
−3 4
0
Applying R1 → R1 + yR2 + R3
⇒
2
 2

0
0
1 x + y +1


y
b
− c − ax

= 0
ax
1
cy
ax
by
−
−


⇒
t =0
[Q determinants of odd order skew-symmetric matrix
is zero]
⇒
1
[(x2 + y2 + 1) {(− c − ax)(− ax − by) − b(cy)}] = 0
ax
a a 2 abc
 1 a bc 
1
32. Let ∆ =  1 b ca  =
b b2 abc
 abc

c c2 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
Hence, statement is false.
|M − I | = |IM − M T M |
[Q IM = M ]
⇒ |M − I | = |(I − M )M |= |(I − M ) ||M |= | I − M |
T
T
= (− 1)3 |M − I |[Q I − M is a 3 × 3 matrix]
= − |M − I |
⇒
2|M − I| = 0
⇒
|M − I | = 0
4π 

sin 2θ −


3
Applying R2 → R2 + R3
33. Since, M T M = I and|M | = 1
∴
1
[(x2 + y2 + 1) (acx + bcy + a 2x2 + abxy − bcy)] = 0
ax
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 θ
cos θ
sin 2θ
2π 
2π 
4π 



sin θ +
 cos θ +
 sin 2θ +






3
3
3
35. Let ∆ =
⇒
bx + ay
cx + a
 ax − by − c
34. Given,  bx + ay −ax + by − c
cy + b
cy + b
− ax − by +
 cx + a

= 0
c
sin θ
2π 

sin θ +


3
=
2π 

+ sin θ −


3
2π 

sin θ −


3
cos θ
2π 

cos θ +


3
2π 

+ cos θ −


3
2π 

cos θ −


3
2π 
2π 


Now, sin θ +
 + sin θ − 


3
3
sin 2θ
4π 

sin 2θ +


3
4π 

+ sin 2θ −


3
4π 

sin 2θ −


3
146 Matrices and Determinants
2π
2π 
2π
2π 


+θ−
−θ +


θ +
θ +
3
3  cos 
3
3 
= 2 sin 
2
2








 2π 
 1
= 2 cos θ cos   = 2 cos θ  −  = − cos θ
 3
 2
4π 

2θ −


3
[since, R1 and R2 are proportional]
36. Given, f ′ (x) =
2ax − 1
2ax + b + 1
b
b+1
2 (ax + b) 2ax + 2b + 1
2ax − 1   2ax −1 
=
b+1   b
1
⇒
1
c
r
1
[from Eq. (i)]
A + ( p − 1) D
p
A + (q − 1) D
q
A + (r − 1) D
r
1
1
1
Applying R1 → R1 − ( A − D ) R3 − DR2
bc ca ab
0 0 0
p q
r =0
= abc p q r = 0 ⇒
1 1 1
1 1 1
38. Given, a > 0, d > 0 and let
1
a
1
∆=
(a + d )
1
(a + 2d )
Taking
⇒ ∆=
1
1
a (a + d )
(a + d ) (a + 2d )
1
1
(a + d ) (a + 2d ) (a + 2d ) (a + 3d )
1
1
(a + 2d ) (a + 3d ) (a + 3d ) (a + 4d )
1
common from R1 ,
a (a + d ) (a + 2d )
f ′ (x) = 2ax + b
f ′ (5 / 2) = 0 ⇒ 5a + b = 0
⇒ ∆=
…(i)
f (0) = 2 ⇒ c = 2 and f (1) = 1
a + b + c=1
On solving Eqs. (i) and (ii) for a , b, we get
1
5
a = ,b = −
4
4
…(ii)
1
a (a + d )2(a + 2d )3 (a + 3d )2(a + 4d )
a
 (a + d )(a + 2d ) (a + 2d )

 (a + 2d )(a + 3d ) (a + 3d ) (a + d ) 
 (a + 3d )(a + 4d ) (a + 4d ) (a + 2d ) 
[C 2 → C 2 − C1]
where c is an arbitrary constant.
⇒
1
b
q
1
1
from R3
(a + 2d ) (a + 3d )(a + 4d )
Since, f has maximum at x = 5 / 2.
⇒
q
1
1
a
r = abc p
1
1
ab
…(i)
1
from R2,
(a + d )(a + 2d )(a + 3d )
On integrating, we get f (x) = ax2 + bx + c,
Also,
Let ∆ = p
1
−1
2ax + b
Applying R3 → R3 − R1 − 2R2, we get
 2ax 2ax − 1 2ax + b + 1 
f ′ (x) =  b
b+1
−1 
0
1
 0
2ax
=
 b
bc ca
= abc
4π
4π 
4π
4π 


− 2θ +
+ 2θ −
 2θ +


 2θ +
3
3
3
3 
 cos 
= 2 sin 
2
2








4π
π

= 2 sin 2θ cos
= 2 sin 2θ cos  π + 

3
3
π
= − 2 sin 2θ cos = − sin 2θ
3




sin 2 θ
cos θ
sin θ
− sin 2θ = 0
∴ ∆ =  − sin θ
− cos θ


sin θ − 2π  cos θ − 2π  sin 2θ − 4π  






3 
3
3

2ax
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
⇒
2π
2π 
2π
2π 


−θ +
+θ−

θ +

θ +
3
3
3
3 
 cos 
= 2 cos 
2
2








1 2 5
x − x+2
4
4
37. Since, a , b, c are pth , qth and rth terms of HP.
⇒
2π
π

= 2 sin θ cos  π − 

3
3
π
= − 2 sin θ cos = − sin θ
3
2π 
2π 


and cos θ +
 + cos θ −




3
3
= 2 sin θ cos
4π 

and sin 2θ +
 + sin

3
f (x) =
Thus,
1
∆′
a (a + d )2(a + 2d )3 (a + 3d )2(a + 4d )
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 R2 → R2 − R1 , R3 → R3 − R2
(a + d ) (a + 2d ) (a + 2d ) a
d
d
∆′ = (a + 2d ) (2d )
⇒
d
d
(a + 3d ) (2d )
Matrices and Determinants 147
Applying R3 → R3 − R2
(a + d )(a + 2d ) (a + 2d ) a
∆′ =
(a + 2d )2d
d
d
2
2d
0
0
∴
⇒
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)
⇒
⇒
 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) 
 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)!
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) = 2n (n 2 + 4n + 5) + 4
⇒
(n !)3
D
− 4 = 2n (n 2 + 4n + 5)
⇒
(n !)3

 D
which shows that 
− 4 is divisible by n.
3

 (n !)
p b c
41. Let ∆ = a q c
a
b r
Applying R1 → R2 − R1 and R3 → R3 − R1 , we get
p
b
c
0
∆= a− p q−b
a−p
0
r−c
=c
p
b
a− p q−b
+ (r − c)
0
a− p q−b
a−p
= − 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
42. We know, A 28 = A × 100 + 2 × 10 + 8
∆ =0 + 0 =0
40. Given, D =  (n + 1)!
(n + 1) (n + 2) 
(n + 2) (n + 3) 
(n + 3) (n + 4) 
Expanding along C1 , we get
∆′ = (2d 2)(d )(a + 2d − a ) = 4d 4
39. Let
(n + 1)
(n + 2)
(n + 3)
Applying R2 → R2 − R1 and R3 → R3 − R2, we get
 1 (n + 1) (n + 1) (n + 2) 
D = n !(n + 1)!(n + 2)! 0
1
2n + 4

1
2n + 6
0

Expanding along R3 , we get
a + 2d a 
∆′ = 2d 2
d
 d
∴
1
D = n ! (n + 1)! (n + 2) ! 1
1
(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.
and
3B9 = 3 × 100 + B × 10 + 9
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)
148 Matrices and Determinants
∴
A 3 6
∆= 8 9 C
2 B 2
Applying R2 → 100R1 + 10R3 + R2
A
3
⇒
∆ = 100 A + 2 × 10 + 8 100 × 3 + 10 × B + 9
2
 xC
r
44. Let ∆ =  y C r
z
C
 r
x
y
y
Cr + 1
Cr + 1
z
Cr + 1
Applying C3 → C3 + C 2
 xC
r
∆ = y C r
z
C
 r
B
6
100 × 6 + 10 × 2 + C
2
3
6
A
= A28 3B9 62C
2
2
B
A
3
A
3
6
= m1k m2k m3 k = k m1 m2 m3
2
B
2
2
B
2
∴
∆ = mk
Hence, determinant is divisible by k.
n

 a −1
6
2
2

43. Given, ∆ a = 
(
)
4
2
1
2
−
n
n
−
a


3
3
2
 (a − 1) 3n 3n − 3n 

 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
∴
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
1
1
6
n (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
2
Applying C3 → C3 − 6 C1
1
1 0
n3 (n − 1)
=
2n − 1 6n 0 = 0
12
n − 1 6n 0
⇒
n
∑ ∆a = c
a=1
[c = 0, i.e. constant]
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 =
n+1
Cr ]
Applying C 2 → C 2 + C1
 xC
r
∆ = y C r
z
C
 r
[from Eq. (i)]
6
Cr + 2 
Cr + 2 

z
Cr + 2 
x
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
 xC
r
⇒ ∆ = y C r
z
C
 r
x+1
x+ 2
y+1
y+ 2
Cr + 1
Cr + 1
z +1
Cr + 1
Cr + 2 
Cr + 2 

z+ 2
Cr + 2 
Hence proved.
45. Since, α is repeated root of f (x) = 0.
∴ f (x) = a (x − α )2, a ∈ constant (≠ 0)
C (x) 
B(x)
 A (x)
Let
B(α ) C (α ) 
φ (x) =  A (α )
 A ′ (α ) B ′ (α ) C ′ (α ) 
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 ) 
=0
[Q R1 and R2 are identical]
B
x
A
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 ) 
=0
[Q R1 and R3 are identical]
Thus, α is a repeated root of φ (x) = 0.
Hence, φ (x) is divisible by f (x).
 x2 + x
x+1
x−2
46. Let ∆ =  2x2 + 3x − 1
3x
3x − 3
 2
2x − 1 2x − 1
 x + 2x + 3
Applying R2 → R2 − (R1 + R3 ), we get
 x2 + x
x+ 1 x−2 
0
4
0 
∆ =
−
 2

 x + 2x + 3 2x − 1 2x − 1 
Applying R1 → R1 +
and R3 → R3 +
x2
R2
4
x2
R2, we get
4
Matrices and Determinants 149
x+ 1 x−2
 x
∆ =  −4
0
0

−
−1
2
2
x
1
x
 2x + 3
 x + 0 x + 1 x − 2
Applying R3 → R3 − 2R1 =  − 4
0
0 
−3
3 
 3
x
x  0
1 −2 
 x
0
0
=  −4
0
0  +  −4
3
3   3 −3
 3 −3
1 1  0
1 −2 
 1
= x − 4
0 0  +  −4
0
0
3
 3 −3 3   3 −3
⇒
∆ = Ax + B
1 1
 1
where, A =  −4
0 0
 3 −3 3 
and
 0
B =  −4
 3
1
0
−3
a
b
c
c
a
b
⇒
⇒
⇒
x3 (48x3 − 4 − 36x3 + 6) = 10
12x6 + 2x3 = 10
6x6 + x3 − 5 = 0
5
x3 = , − 1
6
1/3
 5
x=   ,−1
 6
⇒
Hence, the number of real solutions is 2.
Topic 3 Adjoint and Inverse of a Matrix
1. Given matrix B is the inverse matrix of 3 × 3 matrix A,
5 2α
where B = 0 2
α 3
We know that,
−2 
0
3
det( 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.
1 + x3
1 + 8x3
= 10
3x 9x2 1 + 27x3
⇒
x⋅ x
2
1 1 1 + x3
2 4 1 + 8x3 = 10
3 9 1 + 27x3
Apply R2 → R2 − 2R1 and R3 → R3 − 3R1, we get
1 1
1 + x3
3
x 0 2 − 1 + 6x3 = 10
0 6 − 2 + 24x3
⇒
x3 ⋅
2 6x3 − 1
= 10
6 24x3 − 2

1 
−1
Q det( A ) = det( A ) 


1
det(B)
det( A ) + 1 = 0
1
+ 1 =0
det(B)
Since,
47. Let ∆ =  b c a
x x2
48. Given, 2x 4x2
1 
1 
− 1
(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
2. | 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]
1
cos t
sin t
= 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)
−t
2
2
= e (5 cos t + 5 sin 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]
150 Matrices and Determinants
3. Given,| ABAT| = 8
[Q|XY | = |X ||Y |]
⇒
| A||B|| AT| = 8
…(i) [Q| AT| = | A|]
∴
| A|2|B| = 8
Also, we have
| AB−1| = 8 ⇒| A||B−1| = 8
1 
| A|

…(ii) Q| A −1|=| A|−1 =
=8
⇒
|
A|
|B|

On multiplying Eqs. (i) and (ii), we get
| A|3 = 8 ⋅ 8 = 43
⇒
| A| = 4
| A| 4 1
⇒
|B| =
= =
8
8 2
1
 1 1
Now, |BA −1BT| = |B| |B| =  
 2 4
| A|
5. We have,
∴
Now,
 2 − 3
A=

− 4 1 
 2 − 3  2 − 3
A2 = A ⋅ A = 


− 4 1  − 4 1 
 4 + 12 − 6 − 3
=

− 8 − 4 12 + 1 
− 9
13 
− 9
 2 − 3
+ 12 

13 
− 4 1 
 48 − 27  24 − 36
=

+ 
− 36 39  − 48 12 
 72 − 63
=

− 84 51 
 16
=
− 12
 16
3 A 2 + 12 A = 3 
− 12
1
 1
  =
 2 16
cos θ − sin θ 

sin θ cos θ 
4. We have, A = 
∴
and
| A| = cos 2 θ + sin 2 θ = 1
 cos θ sin θ 
adj A = 

− sin θ cos θ 
b
 d − b
, then adj A = 
]

d
−c a 

 cos θ sin θ 
adj A 
A −1 = 
Q A −1 =


| A| 

− sin θ cos θ 
a
[Q If A = 
c
⇒
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 
=

− sin 2 θ + cos 2 θ 
− cos θ sin θ − cos θ sin θ
 cos 2 θ sin 2 θ 
=

− sin 2 θ cos 2 θ 
−3
−2
−1
Also, A = ( A )( A )
 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 
 cos 50 θ sin 50 θ 
Similarly, A −50 = 

− sin 50 θ cos 50 θ 
25
25 

 cos 6 π sin 6 π 
π

=
 when θ = 
25
25 

12
− sin
π cos
π
6
6 

π
π   Q cos  25π  = cos 4π + π  = cos π 





 cos 6 sin 6  
 6 

6
6 

=

π
π 
25π 
π
π


− sin
cos  and sin 
 = sin 4π +  = sin 
 6 


6
6  

6
6
 3 1 


2 
= 2
3
 −1
 2
2 
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
=
I 2 = (10a + 3b) I 2
3
2
5a
3
6. Given, A = 
and
1 0
= (10a + 3b) 

0 1
0

10a + 3b
=
0
10a + 3b

5a − b  5a 3
AAT = 
2  − b 2
3
25a 2 + b2 15a − 2b
=

13 
 15a − 2b
Q
A (adj A ) = AAT
0
 25a 2 + b2 15a − 2b
10a + 3b
=
∴

0
10a + 3b  15a − 2b
13 

...(i)
...(ii)
[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
15
⇒
b =3
⇒
Matrices and Determinants 151
Now, substituting the value of b in Eq. (iii), we get
5a = 2
Hence, 5a + b = 2 + 3 = 5
7.
12. Every square matrix satisfied its characteristic
equation,
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
i.e. | A − λI | = 0 ⇒
λ3 − 6λ2 + 11λ − 6 = 0
∴
∴
2 α − 6 = 16
⇒
2 α = 22
⇒
α = 11
∴
[Q|adj A| = | A|n −1 ]
…(i)
T
⇒
P =2P + I
⇒
P = 2 (2 P + I ) + I
⇒
P = 4P + 3I
⇒
PX = − IX = –X
As,
⇒
⇒
3P = − 3I
⇒
T
or
1
10. | A| ≠ 0, as non-singular ω
ω
⇒
2
a
b
1 c ≠0
ω 1
1 (1 − cω ) − a (ω − cω 2) + b (ω 2 − ω 2) ≠ 0
⇒ 1 − cω − aω + acω 2 ≠ 0
⇒
(1 − cω ) (1 − aω ) ≠ 0 ⇒ a ≠
1
1
,c≠
ω
ω
⇒ a = ω , c = ω and b ∈{ω , ω 2} ⇒ 2 solutions
11. 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 2 N (NN −1 )(− M )−1 (N T )−1 (− M )
⇒ M 2 N I (− M −1 ) ( − N )−1 (− M )
⇒ − M 2 NM −1N −1 M
⇒ − M ⋅ (MN )M −1N −1 M = − M (NM )M −1N −1 M
⇒ − MN (NM −1 )N −1 M = − M (N N −1 )M ⇒ − M 2
NOTE
∴
T
(P ) = (2 P + I ) = 2 P + I
T T
Now,
[given]
9. Given, PT = 2 P + I
…(i)
= A + cA + dI , multiplying both sides by
2
…(ii)
c = − 6 and d = 11
| P | = 1(12 − 12) − α (4 − 6) + 3 (4 − 6) = 2 α − 6
|P| = |adj A|= | A|2 = 16
A3 − 6 A 2 + 11 A − 6I = O
−1
On comparing Eqs. (i) and (ii), we get
13. Here,
P = adj ( A )
1
=0
4−λ
6I = A3 + cA 2 + dA ⇒ A3 + cA 2 + dA − 6I = O
1 α 3
8. Given, P = 1 3 3
2 4 4
Q
0
(1 − λ ) {(1 − λ ) (4 − λ ) + 2} = 0
= IT = I
∴
1−λ
−2
⇒
Given, 6 A
A, we get
= A −1 AAT ( A −1 )T [Q AAT = AT A]
= IAT ( A −1 )T
[Q A −1 A = I ]
= AT ( A −1 )T = ( A −1 A )T
[Q ( AB)T = BT AT ]
0
0
0
⇒
⇒
= A −1 AT A ( A −1 )T [Q ( AB)T = BT AT ]
1−λ
Here, non-singular word should not be used, since there
is no non-singular 3 × 3 skew-symmetric matrix.
⇒
Q
∴
3 − 1 − 2
P = 2 0
α 
3 − 5 0 
|P| = 3(5α ) + 1(− 3α ) − 2(− 10)
= 12α + 20
T
2α
− 10
 5α


6
12 
adj (P ) = − 10
2 
 − α − (3α + 4)
…(i)
−α 
 5α − 10

…(ii)
=  2α
− 3α − 4
6
− 10 12

2
PQ = kI
|P||Q| = |kI|
|P||Q| = k3
 k2

k2 
|P|  = k3
given,|Q| = 
2
 2

…(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 
6
=
= 2α
− 3α − 4
2
2

− 10 12
2
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.
152 Matrices and Determinants
⇒ ∆ = [sin θ ] − [− cos θ ] [cot θ ]
 π 2π 
When
θ ∈ , 
2 3 
Now, |P adj (Q )| = |P||adj Q|
2
 k2
k5 210
=
= 29
= 2k  =
2
2
 2
 3 
sin θ ∈ 
, 1
 2

∴ Option (c) is correct.
14.
PLAN A square matrix M is invertible, iff dem (M) or| M| ≠ 0.
a b
M = 
 b c
a  b
[let]
(a) Given,   =   ⇒ a = b = c = α
 b  c
α α 
⇒ M =
 ⇒ |M |= 0 ⇒ M is non-invertible.
α α 
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.
| adj P | = 2 1 7 = 1 (3 − 7) − 4 (6 − 7) + 4 (2 − 1)
1 1 3
= − 4 + 4 + 4 = 4 ⇒ |P |= ± 2
16. | A | = (2k + 1)3 ,| B| = 0
But det (adj A) + det (adj B) = 106
(2k + 1)6 = 106
9
k=
⇒
2
⇒
⇒
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

 2 
6
⇒
[sin θ ] = − 1
 3 
, 1 ⇒ [cos θ ] = 0
− cos θ ∈ 

 2
cot θ ∈ ( 3 , ∞ ) ⇒ [cot θ ] = n , n ∈ N .
2. Given, system of linear equations
|adj P | = | P |2
1 4 4
⇒
and
…(iv)
∆ = − 1 − (0 × n ) = − 1
 7π 
Thus, for θ ∈  π,
 , the given system has a unique

6
solution.
1 4 4 
Here, adj P3 × 3 = 2 1 7
1 1 3 
∴
[− cos θ ] = 0
 1

cot θ ∈  −
, 0

3 
…(iii)
So,
PLAN If| A n × n| = ∆, then|adj A| = ∆A − 1
⇒
[sin θ ] = 0
 1
− cos θ ∈ 0, 
 2
⇒
and
M is invertible.
15.
⇒
x+ y+ z =6
4 x + λy − λz = λ − 2
and
3x + 2 y − 4z = − 5
has infinitely many solutions, then ∆ = 0
1 1
⇒ 4 λ
3 2
… (i)
…(ii)
…(iii)
1
− λ =0
−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.
[k] = 4
3. Given system of linear equations
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
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)
Matrices and Determinants 153
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 − 1
c
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
4. Given system of linear equations
2x + 3 y − z = 0,
x + ky − 2z = 0
and 2x − y + z = 0 has a non-trivial solution (x, y, z ).
2 3 −1
∴ ∆ = 0 ⇒1 k −2 = 0


2 −1 1 
2(k − 2) − 3(1 + 4) − 1(−1 − 2k) = 0
⇒
⇒
⇒
2k − 4 − 15 + 1 + 2k = 0
9
4 k = 18 ⇒ k =
2
So, system of linear equations is
2x + 3 y − z = 0
2x + 9 y − 4z = 0
and
2x − y + z = 0
⇒ 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
7. The given system of linear equations is
…(i)
…(ii)
…(iii)
From Eqs. (i) and (iii), we get
x
1
4x + 2 y = 0 ⇒ = −
y
2
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
1
 y 1
1
x
Q z = 2 and y = − 2 


x y z
1 1
9 1
+ + + k= − + −4 + = .
∴
y z x
2 2
2 2
…(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.
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
2−λ
1
1
λ
=0
[Q If a1x + b1 y + c1z = 0; a 2x + b2y + c2z = 0
a3 x + b3 y + c3 z = 0]
x x y
z
1
= × = − ⇒ = −4
z y z
x
4
5. Given system of linear equations
6.
1
2
1
Clearly, the greatest value of c is .
2
From Eqs. (i) and (ii), we get
y 1
6 y − 3z = 0, =
z 2
So,
c = − 1 or

c1

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
has a non-trivial solution, then a 2
⇒
⇒
⇒
⇒
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
(1 + β ) 1 ≠ 0
α
β
2
Apply R1 → R1 − R3 and R2 → R2 − R3
1 0 −1
0 1 −1 ≠0
α β
2
b1
b2
154 Matrices and Determinants
⇒ 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
has more than one solution, then
D1 = D2 = D3 = 0. In the given problem,
a 2 3
D1 = 0 ⇒ b − 1 5 = 0
c
D =0
−3 2
10. We know that,
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
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
[Q sin θ ≠ 0, θ ∈ (0, π )]
11. Since, the system of equations has infinitely many
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
a3
In given interval (0, π ),
1
sin θ =
2
π 5π
θ= ,
⇒
6 6
Hence, 2 solutions in (0, π )
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]
⇒ 28 sin3 θ + 28 sin 2 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 ≠ − 2 as − 1 ≤ sin θ ≤ 1


solution, therefore D = D1 = D2 = D3 = 0
Here,
1 1 1
D = 1 2 3 = 1 (2α − 9) − 1 (α − 3) + 1(3 − 2)
1 3 α
=α −5
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
⇒
∴
D =0
α −5 =0 ⇒ α =5
D3 = 0 ⇒ β − 13 = 0
β = 13
β − α = 13 − 5 = 8
1 −4 7
12. (a) Here, D = 0
−2
3
5
−5
−9
= 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
2
2 3 a −1
2 1 a −3
(Applying C 2 → C 2 − C1 and C3 → C3 − C1)
(Expanding along R1)
2
1
0
2
1
1
and D1 = 5
3
−1
3
2 = 5
a + 1 3 a2 − 1 a + 1 3 a2 − 1 −3
= a2 −3
(Applying C3 → C3 − C 2)
Matrices and Determinants 155
2
=
0
Hence, given system of linear equation has a non-trivial
solution for exactly three values of λ.
0
5
−1
2
a + 1 3 − (a + 1) a 2 − 1 − 3
2
5
3−
16. Given system of linear equations
2x1 − 2x2 + x3 = λx1
⇒
1
(Applying C 2 → C 2 − C1)
2
2
5
0
1
2
−1
a+1
5 a
−
2 2
a2 − 4
=
0
2x1 − 3x2 + 2x3 = λx2
⇒ (−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
⇒ (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
⇒
17. Given equations can be written in matrix form
AX = B
k+1
where, A = 
 k
…(i)
…(ii)
…(iii)
x
4k
8 
, X =   and B =
k + 3 
3k − 1
 y
For no solution, A = 0 and (adj A) B ≠ 0
k+1
8 
Now, A = 
=0
k + 3 
 k
⇒
(k2 + 1)(k + 3) − 8k = 0
k2 + 4k + 3 − 8k = 0
⇒
k2 − 4k × 3 = 0
⇒
(k − 1)(k − 3) = 0
⇒
k = 1, k = 3,
k+3
adj A = 
 −k
k+3
(adj A)B = 
 −k
Now
Now,
−8 
k + 1 
− 8   4k

k + 1   3 k + 1 
 (k + 3)(4k) − 8 (3k − 1)
=
2

 − 4k + (k + 1)(3k − 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
λ = 1, 1, − 3
Hence, λ contains two elements.
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,
⇒
⇒
⇒
⇒
− x1 + 2x2 − λx3 = 0
Since, the system has non-trivial solution.
−2
1
2 − λ
 =0
 2
(
λ
)
∴
−
3
+
2


 −1
λ 
2
15. Given, system of linear equation is
1
…(iii)
− x1 + 2x2 = λx3
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
1
λ
…(ii)
⇒ 2x1 − ( 3 + λ )x2 + 2x3 = 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.
i.e.
…(i)
(2 − λ )x1 − 2x2 + x3 = 0
 4k2 − 12k + 8 
=

2
 − k + 2k − 1 
Put k = 1
4 − 12 + 8 0
(adj A) B = 
not true
=
−1 + 2 − 1  0
156 Matrices and Determinants
Put k = 3
36 − 36 + 8  8 
(adj A) B = 
=
≠ 0 true
−9 + 6 − 1 −4
Hence, required value of k is 3.
Alternate 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
∴
Take
8
k+1
=
k
k+3
⇒
k2 + 4k + 3 = 8k
⇒
k2 − 4k + 3
k = 1, 3
8
4.1
, false
If k –1, then
≠
1+3 2
8
4.3
, true
≠
6
9 −1
Therefore, k = 3
Hence, only one value of k exist.
 x  1
18. Since, A  y = 0 is linear equation in three variables
 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 
 2 −1
Let 1 −2 1  = 0 and ∆ z =  1 −2 −4  = 6 ≠ 0
4
1
1 λ
1
1
⇒
λ =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 
1 + a3 = 0 or a = − 1
22. Since, the given system has non-zero solution.
∴
1
k
1
−k
−1
1
⇒
NOTE
2(k + 1) − (k + 1)2 = 0
(k + 1)(2 − k − 1) = 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 −2 3
~ 0 −1 1
0 −1 1
 −1 
 1
 
 k
 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. 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 −4 3
1
20. For infinitely many solutions, we must have
⇒
⇒
− C 2, C 2 → C 2 + C3
−1 
−1  = 0
−1 
Applying R2 → R2 − R1 , R3 → R3 + R1
⇒ (k –1) (k –3) = 0
And, if k = 3, then
Applying C1 → C1
 1 + k −k − 1
−2
⇒ 1 + k
0
 0
−1 
−1  = 0
−1 
and
⇒
−2 2
D1 = D2 = D3 = 0
b1 2 5
D1 = b2 − 4 3
b3 − 2 2
= − 2b1 − 14b2 + 26b3 = 0
⇒ b1 + 7b2 = 13b3
1 2 3
(a) D = 0 4 5 = 1(24 − 10) + 1(10 − 12)
1 2 6
= 14 − 2 = 12 ≠ 0
Here, D ≠ 0 ⇒ unique solution for any b1, b2, b3 .
...(i)
Matrices and Determinants 157
1
1
3
(b) D = 5
2
6
−2 −1 −3
a
Again, | A3|= 1
1
0
c
d
f
g =0 ⇒
h
g=h
= 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
⇒
a
| A2| = 1
1
f
g
h
1
b =0 ⇒
b
g=h
and
f
| A1| = g
h
0
c
d
1
b =0 ⇒
b
g=h
−1 −3
b3
∴
= b1 (− 6 + 6) − b2(− 3 + 3) + b3 (6 − 6)
=0
1 b1 3
D2 =
λ (λ 2 + 3) = 0 ⇒
⇒
∴
0
c
d
|B2| ≠ 0
BX = V has no solution.
∆ =0
λ
sin α
cos α
1
cos α
sin α
=0
− cos α
λ (− cos α − sin α ) − sin α (− cos α + sin α )
2
2
− 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 


∴
λ =0
∴
a (bc − bd ) + 1(d − c) = 0 ⇒ (d − c)(ab − 1) = 0
− 2≤λ≤ 2
…(i)
Again, when λ = 1, cos 2α + sin 2α = 1
1
1
1
⇒
cos 2α +
sin 2α =
2
2
2
⇒
d=c
and
x + (cos α ) y + (sin α ) z = 0
⇒
1
b =0
b
ab = 1 or
adf ≠ 0 ⇒ |B2| ≠ 0
27. Given, λx + (sin α ) y + (cos α ) z = 0
⇒
26. Since, AX = U has infinitely many solutions.
a
1
1
[since, c = d and g = h]
1
[Q c = d ]
c = a 2cf = a 2df
h
∴
⇒ λ (λ 2 + 1) − 1 (− λ + 1) + 1(1 + λ ) = 0
λ 3 + 3λ = 0
a2
0
0
+ cos α (sin α + cos α ) = 0
−1 −1 λ
⇒
a
|B2| = 0
f
[from Eq. (i)]
−1 sin α
1
1 =0
λ3 + λ + λ − 1 + 1 + λ = 0
1
c =0
h
|B| = 0
⇒
⇒
1
d
g
⇒
− x − y + λz = 0
will have non-zero solution, if
λ
1
−1 λ
a2
|B1|= 0
0
∴
25. Given system λx + y + z = 0, − x + λy + z = 0
⇒| A| = 0 ⇒
[from Eq. (i)]
and − x + (sin α ) y − (cos α ) z = 0 has non-trivial
solution.
1 4 −5
⇒
1
c =0
h
BX = V has no solution.
Since,
= 1(0 − 12) − 2 (− 10 − 3) + 5 (8 − 0)
3
= 54
D ≠0
It has unique solution for any b1, b2, b3 .
and
∴
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
1
d
g
…(i)
[since, C 2 and C3 are equal]
= − 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
−2
BX = V
a
|B| = 0
f
Now,
5 b2 6
− 2 b3 − 3
1
g = h , c = d and ab = 1
⇒
∴
cos (2α − π / 4) = cos π / 4
2α − π / 4 = 2n π ± π / 4
2α = 2nπ − π / 4 + π / 4 or 2α = 2nπ + π / 4 + π / 4
α = nπ or nπ + π /4
158 Matrices and Determinants
28. Since, α 1 , α 2 are the roots of ax2 + bx + c = 0.
⇒
α1 + α 2 = −
b
a
and α 1α 2 =
c
a
Also, β1 , β 2 are the roots of px2 + qx + r = 0.
q
r
and β1β 2 =
β1 + β 2 = −
⇒
p
p
...(i)
Let z = − k, then equations become
...(ii)
Given system of equations
α 1 y + α 2z = 0
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,
⇒ 3 (2λ + 15) + 1 (λ + 18) + 4 (5 − 12) ≠ 0
⇒
7 (λ + 5) ≠ 0
⇒
λ ≠ −5
α 1 + α 2 β1 + β 2
=
α 1 − α 2 β1 − β 2
⇒ (α 1 + α 2) (β1 − β 2) = (α 1 − α 2) (β1 + β 2)
3x − y = 3 − 4k
and
x + 2 y = 3k − 2
On solving, we get
4 − 5k
13k − 9
x=
,y=
,z=k
7
7
31. Given system of equations are
Here,
3x + my = m and 2x − 5 y = 20
3 m
= −15 − 2m
∆=
2 −5
and
∆x =
m m
= −25m
20 −5
∆y =
3 m
= 60 − 2m
2 20
⇒ (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
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
29. The system of equations has non-trivial solution, if ∆ = 0.
⇒
sin 3θ
−1 1
cos 2θ
4
3 =0
2
7
7
Expanding along C1 , we get
sin 3θ ⋅ (28 − 21) − cos 2 θ (−7 − 7) + 2 (−3 − 4) = 0
⇒
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
⇒
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
30. The given system of equations
3x − y + 4z = 3
x + 2 y − 3z = − 2
6 x + 5 y + λz = − 3
has atleast one solution, if ∆ ≠ 0.
3 −1 4
∴
∆= 1
2
−3 ≠ 0
6
5
λ
y=
and
For x > 0 ,
∆y
∆
=
60 − 2m
2m − 60
=
− 15 − 2m 15 + 2m
25m
>0
15 + 2m
⇒ m >0
15
2
15
2m − 60
and y > 0,
> 0 ⇒ m > 30 or m < −
2
2m + 15
or
m<−
From Eqs. (i) and (ii), we get m < −
…(i)
…(ii)
15
or m > 30
2
32. Since, the given system of equations posses non-trivial
0 1 −2
solution, if 0 −3 1 = 0 ⇒ k = 0
k −5 4
On solving the equations x = y = z = λ
[say]
∴ For k = 0, the system has infinite solutions of λ ∈R.
33. 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.
Matrices and Determinants 159
Therefore, the coefficient of determinant must be zero.
−1 c
b
∴
c
b
0 − 1
M 1 =  2 ⇒ M
0  3
∴
−1 a = 0
a −1
1  0
M ⋅ 1 =  0
1 12
⇒ − 1 (1 − a ) − c (− c − ab) + b (ca + b) = 0
2
⇒
a 2 + b2 + c2 + 2abc − 1 = 0
⇒
a 2 + b2 + c2 + 2abc = 1
1
34.
α
α2
⇒
α α
1 α =0
α 1
2
⇒ α 4 − 2α 2 + 1 = 0
⇒
α2 = 1
⇒
α = ±1
But α = 1 not possible
∴
α = −1
Hence, 1 + α + α 2 = 1
a1 a 2 a3 
35. Let
M =  b1 b2 b3 
 c1 c2 c3 
 1  1
− 1 =  1
   
 0 − 1
⇒
[Not satisfying equation]
a 2 − 1 a1 − a 2  1
 b  =  2,  b − b  =  1,
2
 2    1
 
 c2   3  c1 − c2  − 1
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
⇒
Download Chapter Test
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or
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 the function y (x) is given by
the equation 2x + 2y = 2 , is
1. The domain of the definition of the function
1
f (x) =
+ log10 (x3 − x) is
4 − x2
(a) (−1, 0) ∪ (1, 2) ∪ (3, ∞ )
(c) (−1, 0) ∪ (1, 2) ∪ (2, ∞ )
(a) 0 < x ≤ 1
(c) − ∞ < x ≤ 0
(2019 Main, 9 April II)
(b) (−2, − 1) ∪ (−1, 0) ∪ (2, ∞ )
(d) (1, 2) ∪ (2, ∞ )
8. 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
2. Let f (x) = a x (a > 0) be written as f (x) = f1 (x) + f2(x),
where f1 (x) is an even function and f2(x) is an odd
function. Then f1 (x + y) + f1 (x − y) equals
y=
(b) 2f1 (x + y) ⋅ f1 (x − y)
(d) 2f1 (x) ⋅ f1 ( y)
1
(a)  − ,
 4
1
(c)  − ,
 2
π
for real valued x, is
6
1
(b)  − ,
 2
1
(d)  − ,
 4
1
2 
1

9
4. Range of the function f (x) =
(a) (1, ∞)
(c) (1, 7/3]
(2003, 2M)
x + x+2
; x ∈ R is
x2 + x + 1
(2003, 2M)
2
(2001, 1M)
1
(b)  0, 
 2 
(2007, 6M)
Column I
Column II
A.
If −1 < x < 1, then f ( x ) satisfies
p.
0 < f (x ) < 1
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.
If x > 5, then f ( x ) satisfies
s.
f (x ) < 1
Objective Question II
(One or more than one correct option)
(d) (0, 1]
6. The domain of definition of f (x) =
log 2(x + 3)
is
x2 + 3x + 2
(2001, 1M)
(a) R / {− 1, − 2}
(b) (− 2, ∞ )
(c) R / {− 1, − 2, − 3}
(d) (− 3, ∞ ) / {− 1, − 2}
x2 − 6x + 5
.
x2 − 5x + 6
2
minimum value of f (x). As b varies, the range of m (b) is
1
(c)  , 1
 2 
(1983, 1M)
Match the Columns
10. Let f (x) =
5. Let f (x) = (1 + b ) x + 2bx + 1 and let m (b) be the
(a) [0, 1]
x + 2 is
Match the conditions / expressions in Column I with
statement in Column II.
1
2 
1
4 
(b) (1, 11/7)
(d) (1, 7/5)
2
1
+
log10 (1 − x)
(a) (− 3, − 2) excluding − 2. 5 (b) [0, 1] excluding 0.5
(c) (−2, 1) excluding 0
(d) None of these
3. Domain of definition of the function
f (x) = sin −1 (2x) +
(2000, 1M)
(b) ≤ 0, for all real θ
(d) ≤ 0, only when θ ≤ 0
9. The domain of definition of the function
(2019 Main, 8 April II)
(a) 2f1 (x + y) ⋅ f2 (x − y)
(c) 2f1 (x) ⋅ f2 ( y)
(2000, 1M)
(b) 0 ≤ x ≤ 1
(d) −∞ < x < 1
11. If S is the set of all real x such that
positive, then S contains
2x − 1
is
2x3 + 3x2 + x
(1986, 2M)
3
(a)  − ∞ , − 

2
3
1
(b)  − , − 
 2
4
1 1
(c)  − , 
 4 2
1
(d)  , 3
2 
Functions 161
Fill in the Blanks
 4 − x2 
 , then the domain of
12. If f (x) = sin log 
 1−x 


(1985, 2M)
f (x) is… .

13. The domain of the function f (x) = sin −1  log 2

given by ...
True/False
15. 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
16. Find the range of values of t for which
x2 
 is
2
2 sin t =
(1984, 2M)
 π2

14. The values of f (x) = 3 sin 
− x2 lie in the

 16

interval…
(1983, 2M)
y=
17. Let
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)
Topic 2 Composite of Functions
Objective Questions I (Only one correct option)
6. Let a , b, c ∈ R. If f (x) = ax2+ bx + c be such that
 3
let
x ∈ 0,  ,
f (x) = x , g (x) = tan x
 2
1 − x2
 π
. If φ(x) = ((hof )og )(x), then φ   is equal to
h (x) =
2
 
1+ x
(20193Main, 12 April I)
1. For
π
(a) tan
12
7π
(c) tan
12
11π
(b) tan
12
5π
(d) tan
12
(2019 Main, 10 April I)
(b) g (f (S)) ≠ S
(d) f(g(S)) ≠ f (S)
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
(c) 3
(b) 4
(d) 16
 1 − x
 ,|x| < 1, then
 1 + x
4. If f (x) = log e 
5. For
x ∈ R − {0, 1},
 2x 
f
 is equal to
 1 + x2 
(b) 2f (x2 )
(d) −2f (x)
(a) 2f (x)
(c) (f (x))2
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)
(c) f1 (x)
(b) f3 (x)
1
(d) f3 (x)
x
(2017 Main)
(b) 165
(c) 190
(d) 255
(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 ?
8. Let f (x) =
10
∑ f (a + k) = 16(210 − 1),
(a) 330
all x satisfying
( fog )(x) = f ( g (x)), is
For
any
define
A ⊆ R,
f (x) = x2, 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
n =1
7. Let f (x) = x2 and g (x) = sin x for all x ∈ R. Then, the set of
2. Let
(a) f ( g (S)) = S
(c) g (f (S)) = g (S)
a + b + c = 3 and f (x + y) = f (x) + f ( y) + xy, ∀ x, y ∈ R,
10
and
then ∑ f (n ) is equal to
(a) 2
(b) −
2
(c) 1
(d) −1
− 1 , x < 0

9. 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)
10. 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

11. If f (x) = cos(log x), then f (x) ⋅ f ( y) −  f   + f (xy)
2   y

has the value
(a) −1
(c) − 2
(1983, 1M)
1
(b)
2
(d) None of these
162 Functions
12. Let f (x) = | x − 1|. Then,
(1983, 1M)
(a) f (x2 ) = {f (x) }2
(b) f (x + y) = f (x) + f ( y)
(c) f (| x|) = | f (x)|
(d) None of the above
1 − x2
(b) g (x) = 1 − x2
(c) g (x) = − 1 − x2
(d) g (x) = 1 + x2
16. 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
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.)
13. 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
14. If f (x) = cos [π 2] x + cos [− π 2] x, where [x] stands for the
greatest integer function, then
(a) f ( π / 2) = − 1
(c) f (− π ) = 0
(a) g (x) = ±
(1991, 2M)
(b) f ( π ) = 1
(d) f ( π / 4) = 1
Fill in the Blanks
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)
17. If
True/False
18. If f (x) = (a − xn )1/ n, where a > 0 and n is a positive
integer, then f [ f (x)] = x.
Analytical & Descriptive Questions
19. Find the natural number a for which
n
15. Let g (x) be a function defined on [− 1, 1]. If the area of
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)
(1983, 1M)
∑
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
(1992, 6M)
further f (1) = 2.
Topic 3 Types of Functions
Objective Questions I (Only one correct option)
defined by
f : R − {1, − 1} → A
x2
, is surjective, then A is equal to
f (x) =
1 − x2
(2019 Main, 9 April I)
1. If the function
(a) R − {−1}
(b) [0, ∞ )
(c) R − [−1, 0)
(d) R − (−1, 0)
(a)
(b)
(c)
(d)
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
3. The number of functions f from {1, 2, 3, … , 20} onto {1,
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
(a)
(b)
(c)
(d)
(15)! × 6!
56 × 15
5! × 6!
65 × (15)!
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]
5. Let N be the set of natural numbers and two functions f
2. Let a function f : (0, ∞ ) → (0, ∞ ) be defined by
f (x) = 1 −
4. Let f : R → R be defined by f (x) =
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
(2019 Main, 10 Jan II)
and g (n ) = n − (−1)n. Then, fog is
(a) one-one but not onto
(c) both one-one and onto
(b) onto but not one-one
(d) neither one-one nor onto
6. Let A = { x ∈ R : x is not a positive integer}. Define a
function f : A → R as f (x) =
2x
, then f is
x−1
(2019 Main, 9 Jan II)
(a) injective but not surjective
(b) not injective
(c) surjective but not injective
(d) neither injective nor surjective
Functions 163
1 1
7. The function f : R → − ,  defined as f (x) =
 2 2
x
is
1 + x2
(2017 Main)
(a) invertible
(b) injective but not surjective
(c) surjective but not injective
(d) neither injective nor surjective
14. Let the functions defined in Column I have domain
(−π /2, π /2) and range (−∞ , ∞ )
8. The function f: [0, 3] → [1, 29], defined by
f (x) = 2x − 15 x + 36x + 1, is
3
Codes
A B C D
(a) r p s q
(b) p r s q
(c) r p q s
(d) p r q s
2
Column I
(2012)
(a) one-one and onto
(b) onto but not one-one
(c) one-one but not onto (d) neither one-one nor onto
1+ 2x
p.
onto but not one-one
B.
tan x
q.
one-one but not onto
r.
one-one and onto
s.
neither one-one nor onto
(2005, 1M)
(a) one-one and into
(c) many one and onto
(b) neither one-one nor onto
(d) one-one and onto
10. If f : [0, ∞ ) → [0, ∞ ) and f (x) =
(a) one-one and onto
(c) onto but not one-one
x
, then f is
1+ x
(2003, 2M)
(b) one-one but not onto
(d) neither one-one nor onto
11. 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
12. Let E = {1, 2, 3, 4} and F = {1, 2}. Then, the number of
onto functions from E to F is
(a) 14
(c) 12
(2001, 1M)
(b) 16
(d) 8
Objective Question II
(One or more than one correct option)
π π
15. Let f :  − ,  → R be given by


2 2
f (x) = [log(sec x + tan x)]3 . Then,
(a)
(b)
(c)
(d)
f (x) is an odd function
f (x) is a one-one function
f (x) is an onto function
f (x) is an even function
Fill in the Blanks
16. There are exactly two distinct linear functions, …,
and… which map {– 1, 1} onto {0, 2}.
17. The function f (x) =
Match the conditions/expressions in Column I with
statement in Column II.
(2014 Adv.)
sin x, if x < 0
|x|, if x < 0
; f2(x) = x2; f3 (x) = 
f1 (x) =  x
 x, if x ≥ 0
 e , if x ≥ 0
 f [ f (x)], if x < 0
f4 (x) =  2 1
 f2[ f1 (x)] − 1, if x ≥ 0
Column I
x2 + 4x + 30
is not one-to-one.
x2 − 8x + 18
(1983, 1M)
Analytical & Descriptive Question
13. Let f1 : R → R, f2 : [0, ∞ ] → R, f3 : R → R and
f4 : R → [0, ∞ ) be defined by
(1989, 2M)
True/False
Match the Columns
and
Column II
A.
0, if x is rational
x, if x is rational
, g (x) = 
9. f (x) = 
x, if x is irrational
0, if x is irrational
Then, f − g is
(1992, 2M)
Column II
A.
f4 is
p.
onto but not one-one
B.
f3 is
q.
neither continuous nor one-one
C.
f2of1 is
r.
differentiable but not one-one
D.
f2 is
s.
continuous and one-one
18. A function f : IR → IR, where IR, is the set of real
numbers, is defined by f (x) =
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.
(1996, 5M)
19. 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)
164 Functions
Topic 4 Inverse and Periodic Functions
Objective Questions I (Only one correct option)
7. Which of the following functions is periodic?
1. 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 and
f
−1
(D ) = { x : f (x) ∈ D } for D ⊆ Y ,
for any A ⊆ Y and B ⊆ Y , then
(2005, 1M)
−1
(a) f {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
(2004, 1M)
π
(a)  0, 
 2 
π π
(b)  − , 
 4 4 
π π
(c)  − , 
 2 2 
(d) [0 , π ]
3. Suppose f (x) = (x + 1)2 for x ≥ − 1. If g (x) is the function
whose graph is reflection of the graph of f (x) with
respect to the line y = x, then g (x) equals
(2002, 1M)
(a) −
x − 1, x ≥ 0
(c) x + 1 , x ≥ − 1
(b)
1
(x + 1)2
,x> −1
(d) x − 1, x ≥ 0
1
x
4. If f : [1, ∞ ) → [2, ∞ ) is given by f (x) = x + , then f −1(x)
equals
(a)
(c)
(2001, 1M)
x+
x −4
2
2
x−
x2 − 4
(b)
x
x2 − 4
f : [1, ∞ ) → [1, ∞ ) is
, then f −1 (x) is
5. If the function
f (x) = 2
x ( x − 1)
x ( x − 1)
1
(a)  
 2
1
(c) (1 − 1 + 4 log2 x )
2
(b)
1
(1 +
2
constant such that 0 < b < 1. Then,
b−x
, where b is a
1 − bx
(2011)
(a) f is not invertible on (0, 1)
1
f ′ (0)
1
(c) f = f −1 on (0, 1) and f ′ (b) =
f ′ (0)
(d) f −1 is differentiable on (0, 1)
(b) f ≠ f −1 on (0, 1) and f ′ (b) =
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.
(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.
9. Let F (x) be an indefinite integral of sin 2 x.
1 + x2
(d) 1 +
2
Objective Question II
(One or more than one correct option)
8. Let f : (0, 1) → R be defined by f (x) =
2. If f (x) = sin x + cos x, g (x) = x2 − 1, then g { f (x) } is
(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
defined
by
Analytical & Descriptive Question
10. Let f be a one-one function with domain { x, y, z } and
(d) not defined
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
(2007, 3M)
(1999, 2M)
1 + 4 log2 x )
6. If f (x) = 3x − 5, then f −1 (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.
(1998, 2M)
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) .
(1982, 2M)
11. 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)
Functions 165
Answers
Topic 3
Topic 1
1. (c)
5. (d)
9. (c)
1.
5.
9.
13.
2. (d)
3. (a)
4. (c)
6. (d)
7. (d)
8. (c)
10. A → p; B → q; C → q; D → p
11. (a,d)
12. (–2,1)
13. Domain ∈ [ −2,−1 ] ∪ [1, 2 ]
 3 
14. 0,

2 
 π π
16. t ∈ − ,
∪
 2 10 
17. x ∈ [ −1, 2 ) ∪ [3, ∞ )
3π π 
,
 10 2 
15. True
(b)
(b)
(b)
(a,b,c)
1
2.
6.
10.
14.
18.
4. (a)
3. (c)
7. (b)
11. (d)
15. (b, c)
19. (a = 3)
(c)
(a)
(a)
(a, c)
True
(d)
3. (a)
(a)
7. (c)
(b)
11. (a)
A → q; B → r
4. (a)
8. (b)
12. (a)
15. (a, b, c)
16. y = x + 1 and y = − x + 1
17. True
18. 2 ≤ α ≤ 14, No
Topic 4
Topic 2
1.
5.
9.
13.
17.
2.
6.
10.
14.
(c)
(b)
(d)
(d)
1. (c)
2. (b)
3. (d)
4. (a)
5. (b)
6. (b)
7. (a)
8. (b)
9. (d)
8. (d)
12. (d)
16. (a, d)
10. f
−1
(1 ) = y
 ± 3 ± 5
11. 

2


Hints & Solutions
Topic 1 Classification of Functions,
Domain and Range
1. Given function f (x) =
=
1
+ log10 (x3 − x)
4 − x2
 a x + a −x   a y + a −y 
=2

 = 2 f1 (x) ⋅ f1 ( y)
2
2



For domain of f (x)
4 − x2 ≠ 0 ⇒ x ≠ ± 2
and
x3 − x > 0
⇒
x(x − 1)(x + 1) > 0
…(i)
–1
+
0
–
+1
x ∈ (−1, 0) ∪ (1, ∞ )
+∞
…(ii)
From Eqs. (i) and (ii), we get the domain of f (x) as
(−1, 0) ∪ (1, 2) ∪ (2, ∞ ).
2. 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
So, f1 (x + y) + f1 (x − y)
1
1
= [a x + y + a − ( x + y ) ] + [a x − y + a − ( x − y ) ]
2
2
1 x y
1
ax ay 
= a a + x y + y + x 
2
a a
a
a 
=
1  x y
1
1
a a + y + x
2  
a  a
π
, to find domain we must
6
π
≥0
6
π
π
− ≤ sin −1 (2x) ≤
6
2
π
 – π
⇒
sin 
 ≤ 2x ≤ sin
 6 
2
–1
1
≤x≤
4
2
– 1 1
Q
x∈
,
 4 2 
sin −1 (2x) +
+
–
3. Here, f (x) = sin −1 (2x) +
have,
From Wavy curve method,
–∞
1 x
1  y
1
a + x a + y
2
a 
a 
 1
y 
 y + a 

a
4. Let y = f (x) =
∴
π
π

but − ≤ sin −1 θ ≤

2
2 
–1
1
≤ 2x ≤
2
2
x2 + x + 2
, x ∈R
x2 + x + 1
x2 + x + 2
x2 + x + 1
1
y=1 + 2
x + x+1
y=
[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
…(i)
166 Functions
⇒
( y − 1) (− 3 y + 7) ≥ 0
7
⇒
1≤ y≤
3
 7
From Eqs. (i) and (ii), Range ∈ 1 ,
 3 
A. If − 1 < x < 1 ⇒ 0 < f (x) < 1
…(ii)
B. If 1 < x < 2 ⇒ f (x) < 0
C. If 3 < x < 5 ⇒ f (x) < 0
D. If x > 5 ⇒ 0 < f (x) < 1
2x − 1
11. Since,
>0
3
2 x + 3 x2 + x
5. Given, f (x) = (1 + b2) x2 + 2bx + 1
2

b 
b2
= (1 + b )  x +
+
1
−


1 + b2
1 + b2
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)
6. Given, f (x) = 2 2
=
(x + 3x + 2) (x + 1) (x + 2)
m (b) = minimum value of f (x) =
⇒
(2x − 1)
>0
x(2x2 + 3x + 1)
⇒
(2x − 1)
>0
x (2x + 1) (x + 1)
…(i)
and for denominator, (x + 1) (x + 2) ≠ 0
x≠ −1 ,−2
…(ii)
From Eqs. (i) and (ii),
Domain is (− 3 , ∞ ) /{ − 1, − 2}
2x , 2y > 0, ∀ x , y ∈ R
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
f (θ ) = sin θ (sin θ + sin 3 θ )
= (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.
4 − x2
> 0 , 4 − x2 > 0 and 1 − x ≠ 0
1−x
⇒
(1 − x) > 0
and
⇒
x<1
and
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)
10. Given, f (x) =
(x − 2) (x − 3)
x+ 2 >0
x > −2

13. Given, f (x) = sin −1  log 2

−1 ≤ log 2
x2
≤ 1
2
1 x2
≤
≤2
2 2
⇒
1 ≤ x2 ≤ 4
⇒
1 ≤ |x| ≤ 2
Domain ∈ [−2, − 1] ∪ [1, 2]
14. Given, f (x) = 3 sin
π2
− x2
16
 π2

− x2  ⋅
∴ For range, f ′ (x) = 3 cos 
 16



Where,
5
 π2

cos 
− x2  = 0
 16



or
1(−2x)
2
π2
− x2
16
=0
x=0

 π2

π2
π2
− x2  = 0 ⇒
− x2 =

neglecting cos 

16
4

 16



3π 2
2
, never possible

⇒ x = −


16
y=1
Y'
x2

2
⇒
Y
3
| x| < 2 ⇒ −2 < x < 1
 π π
⇒ Domain ∈ − ,
 4 4 
The graph of f (x) is shown as :
2
4 − x2 > 0
Thus, domain ∈ (−2, 1).
⇒
9. For domain of y,
1
∞
 4 − x2 

 1−x 


For domain,
= (sin θ + 3 sin θ − 4 sin3 θ ) sin θ
0
+
1/2
12. Given, f (x) = sin log 
8. It is given,
X'
0
Hence, (a) and (d) are the correct options.
For domain,
7. Given, 2x + 2y = 2, ∀ x , y ∈ R
But
−1/2
x ∈ (−∞ , − 1) ∪ (−1/2, 0) ∪ (1 / 2, ∞ )
x> −3
⇒
−
+
−1
Hence, the solution set is,
For numerator, x + 3 > 0
⇒
−
+
−∞
X
⇒
x =0
Functions 167
f (0) = 3 sin
Thus,
π
3
=
4
2
Topic 2 Composite of Functions and
Even, Odd Functions
 π
 π
f −  = f   =0
 4
 4
and
1. Given, for x ∈ (0, 3 / 2), functions
and
15. Since, domains of f1 (x) and f2(x) are D1 and D2 .
Thus, domain of [ f1 (x) + f2(x)] is D1 ∩ D2 .
Hence, given statement is true.
16. Given, 2 sin t =
Put
f (x) = x
g (x) = tan x
1 − x2
h (x) =
1 + x2
 3 
range ∈ 0,

2 
Hence,
y=
1 − 2 x + 5 x2
 π π
,t ∈ − ,
2
 2 2 
3x − 2x − 1
1 − 2 x + 5 x2
3 x2 − 2 x − 1
⇒ (3 y − 5)x2 − 2x( y − 1) − ( y + 1) = 0
 π
 π π
φ   = tan  − 
 3
 4 3
 3π − 4π 
 π
= tan 
 = tan  − 
 12 
 12
π
π

= − tan   = tan  π − 
 12

12
 11π 
= tan 

 12 
Now,
Since, x ∈ R − {1, − 1 / 3}
[as, 3x2 − 2x − 1 ≠ 0 ⇒ (x − 1)(x + 1 / 3) ≠ 0]
∴
D ≥0
⇒
⇒
4( y − 1)2 + 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. 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]
1− 5
2
1+ 5
y≥
2
1− 5
2 sin t ≤
2
1+ 5
2 sin t ≥
2
 π
sin t ≤ sin  − 
 10
⇒
y≤
or
⇒
or
⇒
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 ).
 3π 
sin t ≥ sin  
 10 
or
3. Given,
π
3π
t≤−
t≥
or
⇒
10
10
π  3π π 
 π
.
Hence, range of t is − , −
∪
,
10   10 2 
 2
17. Since, y =
−∞
⇒
∴
Let
Q
∴ λ =2
−
+
−1
−1 ≤ x < 2
2
or
f (x) = λx
f (1) = 2
[where λ > 0]
(given)
So,
10
 10 
Σ f (a + k) = Σ λa+ k = λa  Σ λk
 k=1 
k =1
k =1
= 2a [21 + 22 + 23 + ......+210 ]
2(2 − 1) 
= 2a 

 2 −1 
10
(x + 1) (x − 3)
≥0
(x − 2)
−
f (x + y) = f (x) ⋅ f ( y)
10
(x + 1) (x − 3)
takes all real values only
(x − 2)
when
… (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
2 sin t = y ⇒ − 2 ≤ y ≤ 2
∴
… (i)
… (ii)
+
3
x≥3
x ∈ [−1, 2) ∪ [3, ∞ ).
∞
[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

168 Functions
⇒
⇒
⇒
2ax2 − x2 = 0
1
⇒
a=
2
Also, a + b + c = 3
1
5
+ b + 0 =3⇒ b =
⇒
2
2
2
x + 5x
f (x) =
∴
2
2a+ 1 (210 − 1) = 16 (210 − 1) (given)
2a+ 1 = 16 = 24 ⇒ a + 1 = 4 ⇒ a = 3
 1 − x
 ,|x| < 1, then
 1 + x
4. Given, f (x) = log e 
2x 

1−


 2x 
1 + x2 

=
f
log

e
 1 + 2x 
 1 + x2



1 + x2 
  2x  
 < 1
Q
2
 1 + x  
 1 − x
= 2 log e 

 1 + x
f (n ) =
Now,
 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
10

 1 − x 
Q f (x) = log e  1 + x 


n =1
1
[Q f2(x) = 1 − x and f3 (x) =
]
1−x
1
 1
1 − J  =
 x 1 − x
⇒
1
[Q f1 (x) = ]
x
1
 1
J  = 1 −
 x
1−x
⇒
⇒
1
= X , then
x
−1
J (X ) = X
1
1−
X
−1
1
=
=
X −1 1 − X
Again, put y = − x
∴
f (0) = f (x) + f (− x) − x2
⇒
0 = ax2 + bx + ax2 − bx − x2
n =1
n =1
f (x) = x2, g (x) = sin x
7.
( gof )(x) = sin x2
go( gof ) (x) = sin (sin x2)
( fogogof ) (x) = (sin (sin x2))2
Again, ( gof ) (x) = sin x
Given,
⇒
…(i)
2
( gogof ) (x) = sin (sin x2)
…(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 =
π
is not possible as − 1 ≤ sin θ ≤ 1]
2
x = ± nπ
αx
8. Given, f (x) =
x+1
1

Qx=
X 

 αx 
f [ f (x) ] = f 
 =
 x + 1
J (X ) = f3 (X ) or J (x) = f3 (x)
6. 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
10
5
1 10 × 11 × 21 5 10 × 11
⋅
+ ×
2
6
2
2
385 275 660
=
+
=
= 330
2
2
2
∴
1 − x−1
−x
=
=
1−x
1−x
Now, put
10
=
5. We have,
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
1
∑ f (n ) = 2 ∑ n 2 + 2 ∑ n
∴
[Q log e| A|m = m log e| A|]
= 2 f (x)
n 2 + 5n 1 2 5
= n + n
2
2
2
⇒
⇒
⇒
⇒
⇒
 αx 
α

 x + 1
αx
+1
x+1
α 2x
α 2x
x+1
=
=
= x [given] …(i)
α x + (x + 1) (α + 1) x + 1
x+1
α 2 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).
Functions 169
9. 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
10. 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|
11. 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) −
12.
Given,
∴
and
⇒
π
π

sin  sin  sin x 


6
2

π
(sin x)
2
π
π
 π
π

sin  sin  sin x 
sin  sin x
2
 6
2

6

= 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
...(ii)
sin { f (x)} ∈ − ,
∴
 2 2 
2
[from Eqs. (i) and (ii)]
⇒ sin { f (x)} ≠ ,
π
i.e. No solution.
∴ Option (d) is not correct.
f (x)
(c) lim
= lim
x → 0 g (x)
x→ 0
14. Since, f (x) = cos [π 2] x + cos [−π 2] x
⇒
f (x) = |x − 1|
f (x2) = |x2 − 1|
∴
{ f (x)}2 = (x − 1)2
⇒
f (− π ) = cos 9π + cos 10π = − 1 + 1 = 0
1
9π
10π
1
 π
+ cos
=
+0=
f   = cos
 4
4
4
2
2
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
π

6
2


π

π
 π π
= sin  sin θ , θ ∈ − , , where θ = sin x
 2 2 

6
2
π
 π π
= sin α, α ∈ − ,
,where α = sin θ
 6 6 
6
 1 1
∴
f (x) ∈ − ,
 2 2 
 1 1
Hence, range of f (x) ∈ − ,
 2 2 
Hence, (a) and (c) are correct options.
15. Since, area of equilateral triangle =
⇒
13. (a) f (x) = sin  sin  sin x  , x ∈ R


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.
[using [π 2] = 9 and [− π 2] = − 10]
9
π
π
 
+ cos 5π = − 1
f   = cos
 2
2
f (π ) = cos 9π + cos 10π = − 1 + 1 = 0
f (x2) ≠ ( f (x))2, hence (a) is false.
Also, f (x + y) = |x + y − 1|
and
f (x) = cos (9) x + cos (−10) x
3
(BC )2
4
3
3
=
⋅ [x2 + g 2(x)] ⇒ g 2(x) = 1 − x2
4
4
A
B
(0,0)
⇒
C
(x,g(x))
g (x) = 1 − x2 or − 1 − x2
Hence, (b) and (c) are the correct options.
x+2
16. Given , y = f (x) =
x−1
⇒
⇒
yx − y = x + 2 ⇒ x( y − 1) = y + 2
y+2
x=
⇒ x = f ( y)
y−1
170 Functions
Here, f (1) does not exist, so domain ∈ R − {1}
dy (x − 1) ⋅ 1 − (x + 2) ⋅ 1
=
dx
(x − 1)2
Again, let f (k) is true.
⇒
f (k) = 2k, for some k ∈ N .
Again, f (k + 1) = f (k) ⋅ f (1)
3
=−
(x − 1)2
= 2k ⋅ 2
= 2k + 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,
f (n ) = 2n , ∀ n ∈ N
17. f (x) = sin 2 x + sin 2(x + π / 3) + cos x cos (x + π / 3)
⇒
⇒
n
n
n
k =1
k =1
k =1
∑ f (a + k) = ∑ f (a ) f (k) = f (a ) ∑ 2k
2 (2n − 1)
= f (a ) ⋅
2 −1
+ cos x cos (x + π / 3)
2
 sin x ⋅ 1 cos x . 3 
f (x) = sin 2 x + 
+

2
 2

+ cos x (cos x cos π / 3 − sin x sin π / 3)
⇒
Now,
f (x) = sin 2 x + (sin x cos π / 3 + cos x sin π / 3)2
f (x) = sin 2 x +
sin x 3 cos x 2 ⋅ 3
+
+
sin x cos x
4
4
4
2
= 2a ⋅ 2 (2n − 1) = 2a + 1 (2n − 1)
n
But
2
∑ f (a + k) = 16 (2n − 1) = 24 (2n − 1)
k =1
a + 1 =4 ⇒ a =3
Therefore,
2
3
cos 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
= sin 2 x + cos 2 x =
4
4
4
+
gof (x) = g { f (x)} = g (5 / 4) = 1
and
Alternate Solution
f (x) = sin 2 x + sin 2(x + π / 3) + cos x cos (x + π / 3)
⇒ 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)
1
π


= 2 sin (2x + π / 3) ⋅
− sin 2x +  = 0



2
3
⇒ f (x) = c, where c is a constant.
But f (0) = sin 2 0 + sin 2(π / 3) + cos 0 cos π / 3
2
 3
1 3 1 5
=  + = + =
2
2 4 2 4
 
Therefore, ( gof ) (x) = g [ f (x)] = g(5 / 4) = 1
Topic 3 Types of Functions
1. Given, function f : R – {1, − 1} → A defined as
x2
=y
1 − x2
x2 = y(1 − x2)
2
x (1 + y) = y
y
x2 =
1+ y
f (x) =
⇒
⇒
⇒
Q
⇒
⇒
f [ f (x)] = [a − {(a − xn )1/ n }n ]1/ n = (xn )1/ n = x
∴
f [ f (x)] = x
(let)
[Q x2 ≠ 1]
[provided y ≠ −1]
x2 ≥ 0
y
≥ 0 ⇒ y ∈ (−∞ , − 1) ∪ [0, ∞ )
1+ y
Since, for surjective function, range of f = codomain
∴Set A should be R − [−1, 0).
 (x − 1)
|x − 1| − x , if 0 < x ≤ 1
2. 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
f (x) = (a − xn )1/ n
18. Given,
[by definition]
[from induction assumption]
x=0
Hence, given statement is true.
y=1
19. Let f (n ) = 2 for all positive integers n.
n
Now, for n = 1, f (1) = 2 = 2 !
⇒ It is true for n = 1.
O
1
y=0
X
Functions 171
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.
∴f (x) is neither injective nor surjective
3. 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}]
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
4. We have, f (x) =
, x ∈R
1 + x2
Ist Method f (x) is an odd function and maximum
occur at x = 1
Y (1, 1/2)
y= 1
2
–1
X
O 1
y =– 1
2
(–1, 1/2)
From the graph it is clear that range of f (x) is
 1 1
− ,
 2 2 
1
IInd Method f (x) =
1
x+
x
1
≥2
x
1
⇒ 0 < f (x) ≤
2
If x > 0, then by AM ≥ GM, we get x +
1
≤
1 2
x+
x
1
If x < 0, then by AM ≥ GM, we get x + ≤ −2
x
1
1
1
⇒
≥−
⇒ – ≤ f (x) < 0
1
2
2
x+
x
0
If x = 0, then f (x) =
=0
1+0
1
1
Thus,
− ≤ f (x) ≤
2
2
 1 1
Hence, f (x) ∈ − ,
 2 2 
⇒
1
IIIrd Method
x
Let y =
⇒ yx2 − x + y = 0
1 + x2
Q x ∈ R, so D ≥ 0
⇒
1 − 4 y2 ≥ 0
 1 1
⇒ (1 − 2 y) (1 + 2 y) ≥ 0 ⇒ y ∈ − ,
 2 2 
+
–
–
–1/2
1/2
 1 1
So, range is − , .
 2 2 
n + 1
, if n is odd

5. Given, f (n ) =  n 2
 , if n is even,
2
+ 1 , if n is odd
and
g (n ) = n − (−1)n = n
n − 1, if n is even
+ 1), if n is odd
Now, f ( g (n )) = ff ((n
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
6. We have a function f : A → R defined as, f (x) =
x −1
One-one Let x1, x2 ∈ A such that
f (x1 ) = f (x2)
2x1
2x2
⇒
=
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
7. We have, f (x) =
1 + x2
1
x
 1
∴
= f (x)
f  = x =
1 1 + x2
 x
1+ 2
x
172 Functions
 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
∴
⇒
As,
∴
⇒
y + x y = x ⇒ yx − x + y = 0
x ∈R
(− 1)2 − 4 ( y)( y) ≥ 0
1 − 4 y2 ≥ 0
− 1 1
⇒
,
y∈
 2 2 
− 1 1
,
∴ Range = Codomain =
 2 2 
So, f (x) is surjective.
Hence, f (x) is surjective but not injective.
8.
2
2
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)
−
+
+
For given domain [0, 3], f (x) is increasing as well as
decreasing ⇒ many-one
f ′ (x) = 0
⇒
x = 2, 3
⇒
10. 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)
x
Again,
y = f (x) =
1+ x
⇒
∴ f (x) is one-one but not onto.
11. Given,
⇒
f ' (x) > 0 , ∀x ∈ R
Hence, f (x) is one-to-one and onto.
12. The number of onto functions from
E = {1, 2, 3, 4} to F = {1, 2}
= Total number of functions which map E to F
− Number of functions for which map f (x) = 1 and
f (x) = 2 for all x ∈ E = 24 − 2 = 14
13. 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.
− x, x < 0
f1 (x) =  x
e , x ≥ 0
f2(x) = x2, x ≥ 0
sin x, x < 0
f3 (x) = 
x≥0
x,
x<0
 f2( f1 (x)),
f4 (x) = 
f
(
f
(
x
))
−
1
,
x≥0
 2 1
Now,
To check onto
y = − x for rational and irrational values
⇒
Range of f (x) ∈ R.
Now, to check one-one.
Take any straight line parallel to X-axis which will
intersect φ(x) only at one point.
y = x and
f ' (x) = 2 + cos x
∴ f (x) takes all intermediate values between (−∞ , ∞ ).
 x, x ∈ Q
− x, x ∉ Q
⇒ y ∈ real numbers.
f (x) = 2x + sin x
which shows f (x) is one-one, as f (x) is strictly increasing.
Since, f (x) is increasing for every x ∈ R,
9. Let φ (x) = f (x) − g (x) = 
 x, x ∈ Q
, which shows
f (x) = 
− x, x ∉ Q
y
y
⇒
≥0
1− y
1− y
Since, x ≥ 0, therefore 0 ≤ y < 1
∴ Onto but not one-one.
As
x=
i.e. Range ≠ Codomain
Range ∈[1, 29]
⇒ φ(x) is one-one.
y + yx = x
⇒
f (0) = 1, f (2) = 29, f (3) = 28
Thus, for range
Range = Codomain ⇒ onto
Thus, f − g is one-one and onto.
3
2
Now, put
∴
x2, x < 0
f2( f1 (x)) =  2x
e , x ≥ 0
 x2 ,
x<0
f4 =  2x
e − 1 , x ≥ 0
x<0
2x,
As f4 (x) is continuous, f ′ 4 (x) =  2x
,
2
e
x>0

⇒
Functions 173
f (− x) = − [ln (sec x + tan x)]3
f4(x)
f (x) + f (− x) = 0
⇒ f (x) is an odd function.
x
O
16. Let y = ax + b and y = cx + d be two linear functions.
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.
14. 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.
When
x = − 1, y = 0 and x = 1, y = 2 , then
∴
0 = −a + b
y=x+1
− c + d = 2 and c + d = 0
⇒
d = 1 and c = − 1
…(ii)
∴
y= −x+ 1
Hence, two linear functions are y = x + 1 and y = − x + 1
y = 1 + 2x
X′
−π
2
π
2
O
⇒
X
It is clear from the graph that y = tan x is one-one and
onto, therefore (B) → (r).
15. 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
f (x) = [ln (sec x + tan x)]
3
f ′ (x) =
3 [ln (sec x + tan x)] (sec x tan x + sec x)
(sec x + tan x)
2
2
 −π π 
, 
f ′ (x) = 3 sec x [ln (sec x + tan x)]2 > 0, ∀x ∈ 
 2 2
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
⇒
∴ f (x) is many one.
Hence, given statement is true.
18. Let y =
αx2 + 6x − 8
α + 6 x − 8 x2
⇒
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)2 + 4 (α + 8 y) (8 + αy) ≥ 0
⇒ 9 (1 − 2 y + y2) + [8α + (64 + α 2) y + 8 αy2] ≥ 0
⇒
⇒
and
y2 (9 + 8α ) + y (46 + α 2) + 9 + 8α ≥ 0
α > − 9 /8
and [46 + α 2 − 2 (9 + 8α )][46 + α 2 + 2 (9 + 8α )] ≤ 0
α > − 9 /8
(α 2 − 16α + 28) (α 2 + 16α + 64) ≤ 0
⇒
α > − 9 /8
− ∞ < ln (sec x + tan x) < ∞
and [(α − 2) (α − 14)] (α + 8)2 ≤ 0
− ∞ < [ln (sec x + tan x)]3 < ∞
⇒
−∞ < f (x) < ∞
and
Range of f (x) is R and thus f (x) is an ont function.
 

1
f (− x) = [ln (sec x − tan x)]3 = ln 

  sec x + tan x 
3
…(i)
A > 0, D ≤ 0, ⇒ 9 + 8 α > 0
(46 + α 2)2 − 4 (9 + 8α )2 ≤ 0
⇒
and
0 < sec x + tan x < ∞
⇒
2 (−6x2 − 12x + 156) −12 (x2 + 2x − 26)
=
(x2 − 8x + 18)2
(x2 − 8x + 18)2
which shows f ′ (x) is positive and negative both.
⇒
 π x
0 < tan  +  < ∞
 4 2
x2 + 4x + 30
x2 − 8x + 18
 (x2 − 8x + 18) (2x + 4) 


− (x2 + 4x + 30) (2x − 8)
f ′ (x) = 
(x2 − 8x + 18)2
=
Y′
...(i)
Again, when x = − 1, y = 2 and x = 1, y = 0, then
17. Given, f (x) =
Y
and a + b = 2 ⇒ a = b = 1
⇒
α > − 9 /8
(α − 2) (α − 14) ≤ 0
α > − 9 /8
and
2 ≤ α ≤ 14
⇒
2 ≤ α ≤ 14
[Q (α + 8)2 ≥ 0]
174 Functions
Thus, f (x) =
αx2 + 6x − 8
will be onto, if 2 ≤ α ≤ 14
α + 6 x − 8 x2
⇒
3x + 6x − 8
, in this case f (x) = 0
3 + 6 x − 8 x2
2
3 x2 + 6 x − 8 = 0
⇒
x=
x±
x2 − 4
2
Since, the range of the inverse function is [1, ∞), then
Again, when α = 3
f (x) =
f −1 (x) =
⇒
− 6 ± 36 + 96 − 6 ± 132 1
=
= (− 3 ± 33 )
6
6
3
we take
f −1 (x) =
If we consider
f −1 (x) =
x+
x2 − 4
2
x − x2 − 4
, then f −1 (x) > 1
2
This is possible only if (x − 2)2 > x2 − 4
⇒
x2 + 4 − 4 x > x 2 − 4
This shows that
1

1

f
(− 3 + 33) = f
(− 3 − 33) = 0


3
3
⇒
8 > 4x
⇒
x < 2, where x > 2
Therefore, f is not one-to-one.
Therefore, (a) is the answer.
19. 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.
Topic 4 Inverse and Periodic Functions
1. Since, only (c) satisfy given definition
i.e.
f {f
−1
Taking log 2 on both sides, we get
⇒
⇒
x2 − x − log 2 y = 0
x=
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
⇒
2. By definition of composition of function,
log 2 y = log 2 2x ( x − 1)
log 2 y = x (x − 1)
⇒
⇒
(B)} = B
B ⊆ f (x)
Only, if
5. Let y = 2x ( x − 1), where y ≥ 1 as x ≥ 1
⇒
1 + 4 log 2 y ≥ 1
− 1 + 4 log 2 y ≤ − 1
1 − 1 + 4 log 2 y ≤ 0
x≥1
g ( f (x) ) = (sin x + cos x)2 − 1, is invertible
(i.e. bijective)
But
⇒
Therefore, we take x =
1
(1 + 1 + 4 log 2y )
2
⇒
f −1 ( y) =
1
(1 + 1 + 4 log 2 y )
2
⇒
f −1 (x) =
1
(1 + 1 + 4 log 2 x )
2
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
3. It is only to find the inverse.
Let
±
Let
y = x + 1,
⇒
⇒
y = x+1
x=
y −1
⇒
f −1 ( y) =
y −1
⇒
f −1 (x) = x − 1
1
x
⇒ xy = x2 + 1
⇒ y=
f −1 ( y) =
y±
x≥ −1
⇒
y ≥ 0, x + 1 ≥ 0
⇒
x≥0
y±
y2 − 4
2
⇒
y = f (x) = 3x − 5
y+5
x=
3
y+5
−1
f ( y) =
3
x+5
−1
f (x) =
3
[given]
⇒
y + 5 = 3x
7. Clearly, f (x) = x − [x] = { x}
x2 + 1
x
⇒ x2 − xy + 1 = 0 ⇒ x =
⇒
6. Given, f (x) = 3x − 5
y = f (x) = (x + 1)2, for x ≥ − 1
⇒
4. Let y = x +
So, x = 1 − 1 + 4 log 2 y is not possible.
y2 − 4
2
which has period 1.
1
And sin , x cos x are non-periodic functions.
x
b−x
, where 0 < b < 1, 0 < x < 1
8. Here, f (x) =
1 − bx
For function to be invertible, it should be one-one onto.
∴ Check Range :
Functions 175
f (x) = y
Let
⇒
y=
b−x
1 − bx
Case III When f (z ) ≠ 2 is true.
If f (z ) ≠ 2 is true, then remaining statements are false.
y − bxy = b − x ⇒ x (1 − by) = b − y
b− y
, where 0 < x < 1
⇒ x=
1 − by
∴ If
b− y
<1 ⇒
∴ 0<
1 − by
Hence,
⇒
⇒
Thus, we have
b− y
b− y
> 0 and
<1
1 − by
1 − by
1
b
1
(b − 1) ( y + 1)
<0−1 < y<
b
1 − by
y>
y < b or
…(i)
…(ii)
1 − cos 2x
dx
2
1
(2x − sin 2x) + C
4
Hence, Statement I is false.
But Statement II is true as sin 2 x is periodic with
period π.
10. It gives three cases
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.
Case II When f ( y) ≠ 1 is true.
If f ( y) ≠ 1 is true, then the remaining statements are
false.
∴
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.
Download Chapter Test
http://tinyurl.com/y47c8lwl
 − x + 1
f ( x) = f 

 − x + 2
[Q f (− x) = f (x)]
Taking f −1 on both sides, we get
−x+1
x=
−x+2
⇒
Again,
Case I When f (x) = 1 is true.
In this case, remaining two are false.
…(i)
 − x + 1
f (− x) = f 

 − x + 2
⇒
⇒
⇒
Since, F (x + π ) ≠ F (x)
∴
then f (− x) = f (x), ∀ x ∈ (− 5, 5)
 x + 1
Given ,
f (x) = f 

 x + 2
⇒
1

y ∈  − 1,  ⊂ Codomain

b
F (x) =
f (x) = 2, f ( y) = 1 and f (z ) = 3
f −1 (1) = y
11. Since, f is an even function,
From Eqs. (i) and (ii), we get
9. Given, F (x) = ∫ sin 2 x dx = ∫
f (x) ≠ 1 and f ( y) = 1
But f is injective.
⇒
− x2 + 2 x = − x + 1
x − 3x + 1 = 0
2
3 ± 9 −4 3 ± 5
=
2
2
 x + 1
f (x) = f 

 x + 2
x=
 x + 1
f (− x) = f 

 x + 2
[Q f (− x) = f (x)]
Taking f −1 on both sides, we get
x+1
−x=
x+2
⇒
⇒
x2 + 3 x + 1 = 0
−3 ± 9 −4 −3 ± 5
x=
=
2
2
Therefore, four values of x are
or
±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
x + 2 sin x + 1 − sin 2 x − x + 1
2
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
(d) 5
x4 − 1
x3 − k3
, then k is
= lim 2
x − 1 x → k x − k2
(2019 Main, 10 April I)
4
3
(a)
(c) − 7
(b) 1
(b)
3
8
(c)
3
2
(d)
8
3
2
sin x
4. lim
equals
x→ 0 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
(d) 8 2
(2019 Main, 11 Jan II)
(d) 2
(2019 Main, 9 Jan I)
y4
(d) exists and equals
1
16
1
(d)
4
(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
(2013 Main)
(c) 2
(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
given
that
f ′ (2) = 6 and
(2003, 2M)
(b) is equal to −3/2
(d) is equal to 3
{(a − n ) nx − tan x} sin nx
= 0, where n is non-zero
x2
real number, then a is equal to
(2003, 2M)
13. If lim
x→ 0
(a) 0
(b)
n+1
n
(c) n
(d) n +
1
n
(cos x − 1) (cos x − ex )
is a
x→ 0
xn
1
4 2
finite non-zero number, is
(a) 1
(c) 3
(b) does not exist
(c) exists and equals
(2017 Main)
14. The integer n for which lim
1 + 1 + y4 − 2
(a) exists and equals
1
24
1
(c)
8
(a)
f (2h + 2 + h 2) − f (2)
,
h → 0 f (h − h 2 + 1 ) − f (1 )
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)
15. lim
x→ 0
(a) 2
1
(c)
2
(2002, 2M)
(b) 2
(d) 4
x tan 2x − 2x tan x
is
(1 − cos 2x)2
(1999, 2M)
(b) −2
1
(d) −
2
Limit, Continuity and Differentiability 177
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
1
2
n 
+
+ ... +
 is equal to (1984, 2M)
2
n → ∞  1 − n2
1−n
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)
j =1
(b) ∑
j =1
10
−1
(1997C, 2M)
 x2 + 1 , x ≠ 0 , 2 


and g (x) =  4,
x = 0 , then lim g [f(x)] is ………
x→0
 5,
x = 2 

(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
lim g (x) exist.

1
fn (x) = Σ tan 
 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?
5
log (1 + 2h ) − 2 log (1 + h )
=K .
h2


25. If f (x) = sin x, x ≠ nπ , n = 0, ± 1, ± 2, ... 

 2,
other wise
24. lim
x→ a
22. For any positive integer n, define fn : (0, ∞ ) → R as
(a) ∑
(2009)
Fill in the Blanks
30. If
Objective Question II
(One or more than one correct option)
n
(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 −
(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)
tan 2 (f j (0)) = 55
(a + h )2 sin (a + h ) − a 2 sin a
32. Evaluate lim
. (1980, 3M)
h→ 0
h
(1+ f ′ j (0)) sec2 (f j (0)) = 10
33. Evaluate lim
x − sin x
.
x + cos 2 x
x→ 0
1
n
(d) For any fixed positive integer n, lim sec2 (fn (x)) = 1
(c) For any fixed positive integer n, lim tan(fn (x)) =
x →∞
x →∞
x −1

.
 2 x2 − 7 x + 5 
34. Evaluate lim 
x→1
(1979, 3M)
(1978, 3M)
178 Limit, Continuity and Differentiability
36. Let m and n be two positive integers greater than 1. If
Integer Type Questions
x sin (βx)
35. Let α , β ∈ R be such that lim
= 1 . Then,
x → 0 αx − sin x
(2016 Adv)
6 (α + β ) equals
2
 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
x
 1 + f (3 + x) − f (3)
f ′ (3) + f ′ (2) = 0. Then lim
 is equal
x→ 0  1 + f (2 − x) − f (2 ) 
(2019 Main, 8 April II)
to
(b) e−1
(a) e
2.
π − 2 sin
1−x
lim
x→1 −
(a)
π
2
(b)
−1
(c) e2
x
(d) 1
is equal to
(c)
π
(d)
1
2π
3. Let [x] denote the greatest integer less than or equal to
x. Then,
tan(π sin 2 x) + (|x| − sin(x[x]))2
lim
x→ 0
x2
(a) equals π
(c) equals 0
(2019 Main, 11 Jan I)
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
(2019 Main, 9 Jan II)
15
 x 
1
x→ 0
(a) ±
π
4
 1 
1 − x (1 + |1 − x|)
cos 

|1 − x|
 1 − x
(c) limx→ 1 − f (x) = 0
(d) limx→ 1 + f (x) does not exist
π
3
(b) ±
π
6
(c) ±
11. For x > 0, lim (sin x)1/ x +  
 x
x→0

1
(b) – 1
(d) ±
sin x 
 is

π
2
(2011)
(2006, 3M)
(c) 1
(d) 2
12. Let f : R → R be such that f (1) = 3 and f ′ (1) = 6. Then,
1/ x
 f (1 + x) 
lim

x→ 0
 f (1) 
equals
(2002, 2M)
(2018 Main)
(c) e2
(b) e 2
(d) e3
x
x−3 
 is equal to
x+ 2 
13. For x ∈ R , lim 
(b) e−1
(2000, 2M)
(c) e−5
(d) e5
Fill in the Blanks
 1 + 5 x2 
14. lim 

x → 0  1 + 3 x2 
 x + 6

x → ∞  x + 1
15. lim 
(b) is equal to 15
(d) does not exist (in R)
(b) limx→ 1 − f (x) does not exist
1
and −1
2
9
(d) − and 3
2
(b) −
10. If lim [1 + x log (1 + b2)] x = 2b sin 2 θ, b > 0
(a) e
equal to t. Then,
for x ≠ 1. Then
(a) limx→ 1 + f (x) = 0
1
4
a→ 0 +
5
(a) − and 1
2
7
(c) − and 2
2
x→ ∞
6. For each t ∈R, let [t ] be the greatest integer less than or
7. Let f (x) =
(d)
1
(b) sin 1
(d) 1
(a) is equal to 0
(c) is equal to 120
1
2
( 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
(a) 1
5. For each x ∈ R, let [x] be the greatest integer less than or
  1  2
lim x   +   + … +
+  x
x→ 0
   x
(c)
9. Let α (a ) and β (a ) be the roots of the equation
(a) 0
equal to t. Then,
equal to x. Then,
x([x] + |x|) sin [x]
is equal to
lim
|x|
x → 0−
(2016 Main)
(b) 1

4. For each t ∈ R, let [t ] be the greatest integer less than or
(a) 0
(c) − sin 1
(a) 2
and θ ∈ (− π , π ], then the value of θ is
(b) equals π + 1
(d) does not exist
(a) equals 0
(c) equals − 1
x→ 0
a→ 0+
(2019 Main, 12 Jan II)
2
π
8. Let p = lim+ (1 + tan 2 x )1/ 2x , then log p is equal to
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)
Integer Answer Type Question
17. The largest value of the non-negative integer a for
1−x
 − ax + sin(x − 1) + a 1−
which lim 

x→1  x + sin(x − 1 ) − 1 
x
=
1
is
4
(2014 Adv)
Limit, Continuity and Differentiability 179
Topic 3 Squeeze, Newton-Leibnitz’s Theorem and Limit Based on
Converting infinite Series into Definite Integrals
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
to
r =1
(2019 Main, 12 April I)
21
346
1
(c)
12
29
358
7
(d)
116
(a)
(b)
sec 2 x
∫2
2. lim
π
x→
4
1
(a) f   ≥ f (1)
 2
equals
π2
16
(2007, 3M)
n→ ∞
2n
∑
r =1
(2016 Adv.)
Numerical Value
5. For each positive integer n, let
yn =
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.)
................. .
(d) 4f (2)
1
n
x
n
n
n


 ...  x + 

2
n
 ,
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
8
(a) f (2)
π
2
(b) f (2)
π
2  1
(c) f  
π  2
3. lim

 n n (x + n )  x +

4. Letf (x) = lim 
n→ ∞

 n ! (x2 + n 2)  x2 +


for all x = 0. Then
f (t ) dt
x2 −
Objective Questions II
(One more than one correct option)
r
n +r
2
2
equals
(a) 1 + 5
(c) −1 + 2
(1999, 2M)
(b) 5 − 1
(d) 1 + 2
Fill in the Blank
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
(a)
1
2
(c) 1
x→ 0
2x − 1
 π, [⋅] denotes the
 2 
greatest integer function, is discontinuous at
4. The function f (x) = [x] cos 
(2019 Main, 9 April I)
(b) 2
1
2
2. The function f (x) = [x]2 − [x2] (where, [x] is the greatest
integer less than or equal to x), is discontinuous at
(a) all integers
(b) all integers except 0 and 1
(c) all integers except 0
(d) all integers except 1
(1993, 1M)
(b) f (x) is continuous at x = 0
(c) f (x) is not differentiable at x = 0
(d) f ′ (0) = 1
π
4 is continuous,
π
4
(d)
f (x) = [tan 2 x], then
(a) lim f (x) does not exist
 6 3
 2 cos x − 1
, x≠

f (x) =  cot x − 1

k,
x=

then k is equal to
3. Let [.] denotes the greatest integer function and
(1999, 2M)
(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
180 Limit, Continuity and Differentiability
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
(c) log a + log b
(b) a + b
(d) None of these
is continuous for 0 ≤ x ≤ π.
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.
Then, at which of the following point(s) the function
(2017 Adv.)
f (x) = x cos (π (x + [x])) is discontinuous ?
(b) x = 1
(d) x = 2
such that max { f (x): x ∈ [0, 1]} = max { g (x): x ∈ [0,1]}.
The correct statement(s) is (are)
(2014 Adv.)
[f (c)]2 + 3f (c) = [ g (c)]2 + 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]
9. For every integer n, let a n and bn be real numbers. Let
a n + sin πx, for x ∈ [2n , 2n + 1]
,
f (x) = 
bn + cos πx, for x ∈ (2n − 1, 2n )
function
f (x)
satisfying
(1987, 6M)
 sin (a + 1) x + sin x
, for x < 0

x

f (x) =  c,
for x = 0
 (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
2
(i) f1 (x) = sin( 1− e− x ),
for all integers n.
If f is continuous, then which of the following hold(s) for
all n ?
(2012)
(b) an − bn = 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) = .... .
Find the continuous
f ′ (1) = f (− 1).
16. Let f1 : R → R, f2 :  −
function f : R → R be given by
(a) an−1 − bn−1 = 0
(c) an − bn + 1 = 1
x≤0
 g (x),
1/ x

defined by f (x) =  (1 + x) 
 (2 + x)  , x > 0


15. Determine the values a, b, c, for which the function
8. For every pair of continuous function f , g : [0, 1] → R
(a)
(b)
(c)
(d)
(1989)
 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

(1983, 1M)
Objective Questions II
(One or more than one correct option)
(a) x = − 1
(c) x = 0
13. Find the values of a and b so that the function
(1982, 2M)
Analytical & Descriptive Questions
 |sin x|
if x ≠ 0
(ii) f2 (x) =  tan − 1 x
, where the inverse

1
if x = 0
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,
1
 2
x sin   if x ≠ 0
(iv) f4 (x) = 
 x

if x = 0
0
List-I
π

a/|sin x |
< x<0
,
{1 + |sin x|}
6

11. Let f (x) = 
b,
x=0
π

tan 2 x / tan 3 x
0<x<
,
e

6
Determine a and b such that f (x) is continuous at x = 0.
(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.
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.
S.
The function f4
is
4.
The correct option is
(a) P → 2; Q → 3; R → 1; S → 4
(b) P → 4; Q → 1; R → 2; S → 3
(c) P → 4; Q → 2; R → 1; S → 3
(d) P → 2; Q → 1; R → 4; S → 3
differentiable at x = 0
and its derivative is
NOT continuous at x = 0
differentiable at x = 0
and its derivative is
continuous at x = 0
Limit, Continuity and Differentiability 181
Topic 5 Continuity in a Domain
Objective Question I (Only one correct option)
1. Let f : R → R be a continuously differentiable
1
function such that f (2) = 6 and f′ (2) = . If
48
f ( x) 3
∫ 4t dt = (x − 2) g(x), then lim g (x) is equal to
x→ 2
6
(a) 18
(c) 12
(2019 Main, 12 April I)
(a)
(b)
(c)
(d)
sin ( p + 1) x + sin x
,x<0

x

2. If f (x) = 
q,
x=0

x>0
x + x2 − x

,
3/ 2

x
1 3
(b)  − , 
 2 2
3 1
(d)  − , 
 2 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
x
4. If f (x) = [x] −  , x ∈ R where [x] denotes the greatest
(2019 Main, 9 April II)
(a) lim f (x) exists but lim f (x) does not exist
x→ 4 +
x→ 4 −
(c) Both lim f (x) and lim f (x) exist but are not equal
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, 9 Jan I)
(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
Objective Questions II
(One or more than one correct option)
8. The following functions are continuous on (0, π)
1
dt
t
x sin x, 0 < x ≤
(d)  π
sin ( π + x ),
 2
(b) ∫
(a) tan x
x
0
t sin
(1991, 2M)
π /2
π
< x< π
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
(1986, 2M)
(b) continuous in (−1, 0)
(d) differentiable in (−1, 1)
π 
 , where [⋅] denotes the
 [x + 1]
greatest integer function. The domain of f is…… and
the points of discontinuity of f in the domain are…… .

10. Let f (x) = [x] sin 
(1996, 2M)
Analytical & Descriptive Question
(2019 Main, 8 April II)
where, [t ] denotes the greatest integer less than or
equal to t. Then, f is discontinuous at
(a) four or more points
(c) only three points
x≥5
if
Fill in the Blank
(b) f is continuous at x = 4
x→ 4 −
if 1 < x < 3
if 3 ≤ x < 5
 1, 0 ≤ x≤ 3 π /4
(c) 
2
3π
2 sin x,
< x< π

9
4
(b)
4 
integer function, then
x≤1
if
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) =
is continuous at x = 0 , then the ordered pair ( p, q) is
equal to
(2019 Main, 10 April I)
(a)
 5,
a + bx,

f (x) = 
 b + 5x,
 30,
Then, f is
(b) 24
(d) 36
3
1
(a)  − , − 
 2
2
5
1
(c)  , 
 2 2
6. Let f: R → R be a function defined as
(b) only two points
(d) only one point

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].
(1983, 2M)
182 Limit, Continuity and Differentiability
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
function. Prove that f (x) + g (x) is a discontinuous
function.
(1987, 2M)
x→ ∞
1 + x, 0 ≤ x ≤ 2
3 − x, 2 < x ≤ 3
(c) for all x in the interval [1, ∞ ), f (x + 2) − f (x) > 2
(d) f ′(x) is strictly decreasing in the interval [1, ∞ )
4. Let f (x) = 
Determine the form of g (x) = f [ f (x)] and hence find the
(1983, 2M)
points of discontinuity of g, if any
Analytical & Descriptive Questions
 x + a , if x < 0
f (x) = 
|x − 1|, if x ≥ 0
if x < 0
 x + 1,
g (x) = 
2
(x − 1) + b, if x ≥ 0
2. Let
and
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 continuous
at all x.
(1981, 2M)
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)
(2019 Main, 10 April I)
2. If f : R → R is a differentiable function and
f ( x)
∫
6
(a) 12f ′ (2)
(c) 24f ′ (2)
2t dt
is
(x − 2)
(2019 Main, 9 April II)
of x, at which the function, g (x) = f ( f (x)) is not
differentiable, is
(2019 Main, 9 April I)
(b) {5, 10, 15}
(d) {10, 15}
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?
π
π
(a) − , 0, 
4
 4
3π
π 3π π 

(c) −
,− ,
, 
4 4 4
 4
(2019 Main, 12 Jan I)
π
π π π
(b) − , − , , 
4 4 2
 2
3π
π π 3π 
(d) −
,− , ,

2 2 4
 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.
Then, the set K is equal to
(2019 Main, 11 Jan II)
(a) {0}
(c) { π }
(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)
(b) 0
(d) 2f ′ (2)
3. Let f (x) = 15 − x − 10 ; x ∈ R. Then, the set of all values
(a) {5, 10, 15, 20}
(c) {10}
g (x) = | f (x)| + f (|x|). Then, in the interval (−2, 2), g is
(2019 Main, 11 Jan I)
not differentiable
differentiable if f ′ (c) ≠ 0
not differentiable if f ′ (c) = 0
differentiable if f ′ (c) = 0
lim
f (2) = 6, then
x→ 2
−2 ≤ x < 0
 −1 ,
and
2
x − 1 , 0 ≤ x ≤ 2
6. Let f (x) = 
(b) φ (an empty set)
(d) {0, π }
(a) three elements
(c) two elements
(b) five elements
(d) one element
max {|x|, x2},
8. Let f (x) = 
|x| ≤ 2
2 < |x| ≤ 4
 8 − 2|x|,
Let S be the set of points in the interval (−4, 4) at which f
(2019 Main, 10 Jan I)
is not differentiable. Then, S
(a) equals {−2, − 1, 0, 1, 2}
(c) is an empty set
(b) equals {−2, 2}
(d) equals {−2,−1, 1, 2}
9. Let f be a differentiable function from R to R such that
3
| f (x) − f ( y)| ≤ 2|x − y|2 , for all x, y ∈ R. If f (0) = 1, then
1
∫f
2
0
(x) dx is equal to
(a) 2
1
(b)
2
(2019 Main, 9 Jan II)
(c) 1
(d) 0
S = (t ∈ R : f (x) = |x − π |(
⋅ e|x| − 1)sin| x| is not
differentiable at t}.Then, the set S is equal to (2018 Main)
10. Let
(a) φ (an empty set)
(c) { π }
(b) {0}
(d) {0, π }
Limit, Continuity and Differentiability 183
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) 2f ′ (c) = g ′(c)
(c) f ′(c) = g ′(c)
(b) 2 f ′(c) = 3 g ′(c)
(d) f ′(c) = 2 g ′(c)
π
 2
13. Let f (x) = x cos x , x ≠ 0, x ∈ R, then f is
 0,
x =0
(a)
(b)
(c)
(d)
(2012)
n
x→1 +
(a) n = 1, m = 1
(c) n = 2 , m = 2
(b) n = 1, m = −1
(d) n > 2, m = n
(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
(a) R − {0}
(c) R − {−1}
(2002, 2M)
(2001, 2M)
(b) cos (| x|) − | x|
(d) sin (| x|) − | x|
is an integer, is
(2001, 2M)
(b)(−1)k − 1 (k − 1) π
(d) (−1)k − 1 kπ
The set of all points, where f (x) is not differentiable, is
(b) {−1, 0 }
(d) {−1, 0,1 }
(d) 2
x
is
1 + | x|
(1987, 2M)
(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. For every twice differentiable function f : R → [−2, 2]
(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
27. Let f : (0, π)→ R be a twice differentiable function such
that lim
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
20. Let f : R → R be a function defined by f (x) = max { x, x3 }.
(a) {−1,1 }
(c) {0,1 }
(1999, 2M)
(c) 1
differentiable, is
(a) (− ∞ , ∞ )
(c) (− ∞ , 0) ∪ (0, ∞ )
t→ x
19. The left hand derivative of f (x) = [x] sin (π x) at x = k, k
(a) (−1)k (k − 1) π
(c) (−1)k kπ
(b) 0
(d) There exists α ∈ (−4, 4) such that f (α ) + f ′′(α ) = 0 and
f ′ (α ) ≠ 0
18. Which of the following functions is differentiable
(a) cos (| x|) + | x|
(c) sin (| x|) + | x|
not differentiable at
x→ ∞
(b) R − {1}
(d) R − {−1, 1}
at x = 0 ?
22. The function f (x) = (x2 − 1)| x2 − 3x + 2| + cos (| x|) is
with ( f (0))2 + ( f ′ (0))2 = 85, which of the following
statement(s) is (are) TRUE ?
(2018 Adv.)
17. The domain of the derivative of the functions
 tan −1 x ,
if | x| ≤ 1

is
f (x) = 1
2 (| x| − 1), if | x| > 1
(2000, 2M)
23. The set of all points, where the function f (x) =
(x − 1)
; 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
g (x) =
g (x) =| f (x)|, ∀ x. Then, g is
(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
(a) −1
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
14. Let
21. Let f : R → R be any function. Define g : R → R by
(2001, 2M)
(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
184 Limit, Continuity and Differentiability
28. Let f : R → R, g : R → R and h : R → R be differentiable
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,
1
15
(b) h ′(1) = 666
(c) h(0) = 16
(d) h ( g (3)) = 36
(a) g ′(2) =
(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)
33. Let f : [a , b] → [1, ∞ ) be a continuous function and
a , b ∈ R and
f : R → R be defined by
f (x) = a cos (|x3 − x|) + b|x|sin (|x3 + x|). Then, f is
29. Let
(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
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 
31. Let g : R → R be a differentiable function with
g (0) = 0, g′ (0) = 0 and g′ (1) =/ 0.
x< a
a ≤ x ≤ b.
x> b
Then,
30. Let f : − , 2 → R and g : − , 2 → R be functions
 2 
 2 
1

, if
0
 x
g : R → R be defined as g (x) = ∫ f (t )dt , if
a
 b
 f (t )dt , if
∫a
(a)
(b)
(c)
(d)
π
π

 −x− 2 , x≤− 2

π
34. If f (x) = − cos x, − < x ≤ 0, then
2

0 < x≤1
 x − 1,
 ln x,
x>1
(a)
(b)
(c)
(d)
(2015 Adv.)
 x
g (x), x =/ 0

Let f (x) = |x|
 0 ,
x=0
(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
(2011)
π
f (x) is continuous at x = −
2
f (x) is not differentiable at x = 0
f (x) is differentiable at x = 1
3
f (x) is differentiable at x = −
2
35. 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
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
(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
36. If f (x) = min { 1, x2, x3 }, then
(c) foh is differentiable at x = 0
(d) hof is differentiable at x = 0
32. 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.)
(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
37. 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
38. The function f (x) =  x2
(a) continuous at x = 1
(c) discontinuous at x = 1
(1998, 2M)
x≥1
x<1
is (1988, 2M)
(b) differentiable at x = 1
(d) differentiable at x = 3
Limit, Continuity and Differentiability 185
39. 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
40. If x + | y| = 2 y, then y as a function of x is
(1984, 2M)
(b) continuous at x = 0
dy 1
(d) such that
= for x < 0
dx 3
(a) defined for all real x
(c) differentiable for all x
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
(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 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)
Match the Columns
42. In the following, [x] denotes the greatest integer less
than or equal to x.
Column I
Column II
A.
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)

45. For the function f (x) = 1 + e1/ x , x ≠ 0 ;
x

0
, x=0
the derivative from the right, f ′ (0+ ) = … and the
(1983, 2M)
derivative from the left, f ′ (0− ) = … .

2
46. Let f (x) = (x − 1) sin (x − 1) − |x|, if x ≠ 1 be a real
1
− 1,
if x = 1
valued function. Then, the set of points, where f (x) is
(1981, 2M)
not differentiable, is …. .
True/False
47. The derivative of an even function is always an odd
function.
(1983, 1M)
Analytical & Descriptive Questions
1

−1  x + c
b sin  2  , − 2 < x < 0

1
48. 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
(2004, 4M)
value of a and prove that 64b2 = (4 − c2).
1
49. If f : [−1, 1] → R and f ′ (0) = lim nf   and f (0) = 0.
 n
n→ ∞
2
 1
Find the value of lim
(n + 1) cos −1   − n, given that
 n
n→ ∞π

 1  π
0 <  lim cos −1    < .
 n  2
n → ∞
50. 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)
43. Match the conditions/expressions in Column I with
statement in Column II
Column I
(1992, 2M)
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
Fill in the Blanks
44. 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 )
(1997C, 2M)
and h ′ (x0 ) = kh (x0 ), then k = K .
51. Determine the values of x for which the following
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)
 −  1 + 1 
52. Let f (x) = x e  |x | x  , x ≠ 0

, x=0
0

Test whether
(i) f (x) is continuous at x = 0.
(ii) f (x) is differentiable at x = 0.
(1997C, 5M)
53. 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)
186 Limit, Continuity and Differentiability
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)
54. A
55. 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)
56. 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
that f (x) is a constant.
(1988, 2M)
57. Let f (x) be a function satisfying the condition
f (− x) = f (x), ∀ x. If f ′ (0) exists, find its value. (1987, 2M)
58. Let f (x) be defined in the interval [− 2, 2] such that
(1986, 5M)
59. Let f (x) = x − x − x + 1
3
(1978, 3M)
Integer Answer Type Questions
62. 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.)
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 .
g (x) = f (| x|) + | f (x)|.
Test the differentiability of g (x) in (− 2, 2).
61. If f (x) = x tan −1 x , find f ′ (1) from first principle.
63. Let f : R → R and g : R → R be respectively given by
 −1 , −2 ≤ x ≤ 0
f (x) = 
x − 1 , 0 < x ≤ 2
and
x −1

2x2 − 7x + 5 , when x ≠ 1
.
60. Find f ′ (1) , if f (x) = 
1

when x = 1
− ,

3
(1979, 3M)
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
The number of points at which h (x) is not differentiable
(2014 Adv.)
is
64. 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 
(2010)
is ……… .
Topic 8 Differentiation
Objective Questions I (Only one correct option)
(a)
C1 + (22) 20C 2 + (32) 20C3 + ..... + (202)20C 20 = A (2β )
, then the ordered pair ( A , β) is equal to
1. If
20
(2019 Main, 12 April II)
(a) (420, 19) (b) (420, 18) (c) (380, 18) (d) (380, 19)
 sin x − cos x 
x
 , with respect to ,
x
x
sin
cos
+
2


2. The derivative of tan −1 

 π
where  x ∈ 0,   is
 2

(a) 1
(b)
2
3
(2019 Main, 12 April II)
(c)
1
2
 dy d y
, 2  at x = 0 is equal
 dx dx 
2
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
4. If
(a) 12
(b) 9

derivative
of
(2019 Main, 8 April II)
(c) 15
2
(d) 33
 3 cos x + sin x 

  , x ∈ 0,

 cos x − 3 sin x 
5. If 2 y =  cot−1 

equal to
1
1
(d)  − , − 2 
 e
e 
the
(b) x −
dy
π
is
 then

dx
2
(2019 Main, 8 April I)
π
6
(c)
π
−x
3
(d) 2x −
6. For x > 1, if (2x)2y = 4e2x − 2y , then (1 + log e 2x)2
equal to
dy
is
dx
x log e 2x − log e 2
(b)
x
(d) log e 2x
7. If x log e (log e x) − x2 + y2 = 4( y > 0), then
(a)
π
3
(2019 Main, 12 Jan I)
x log e 2x + log e 2
(a)
x
(c) x log e 2x
equal to
(d) 2
3. If ey + xy = e, the ordered pair 
π
−x
6
e
4+e
2
(b)
(2e − 1)
2 4+e
2
(c)
(1+ 2e )
4+e
2
dy
at x = e is
dx
(2019 Main, 11 Jan I)
(d)
(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
8. Let
(a) 30
(b) − 4
(c) − 2
(d) 8
9. If x = 3 tan t and y = 3sec t, then the value of
is
1
(a)
6
1
(c)
3 2
d 2y
π
at t = ,
4
dx2
(2019 Main, 9 Jan II)
(b)
1
6 2
3
(d)
2 2
Limit, Continuity and Differentiability 187
1
10. For x ∈ 0,  , if the derivative of

18. If x2 + y2 = 1 , then
4
 6x x 
tan −1 
 is x ⋅ g (x), then g (x) equals
 1 − 9x3 
(2017 Main)
(a)
9
1 + 9x
3
(b)
3x x
1 − 9x
3
(c)
3x
(d)
1 − 9x
3
11. If g is the inverse of a function f and f ′ (x) =
3
(2015)
(b) 5 x4
(a)
1
2
(b)
1
2
dy
at x = 1 is equal to
dx
(c) 1
(d)
(2013)
2
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 + +
+ ... +
2
9
25
−
(
2
N
1
)




1 1
1
(b) 4 1 + +
+ ... +

9 25
(2 N − 1)2 

(2008, 3M)
d
f (x) at x = 0 is
dx3
(a) p
(c) p + p3
(1988, 2M)
(a) P ′ ′ ′ (x) + P ′ (x)
(c) P (x) P ′ ′ ′ (x)
(b) P ′ ′ (x) ⋅ P ′ ′ ′ (x)
(d) a constant
Fill in the Blanks
21. If x exy = y + sin 2 x, then at x = 0,
dy
= ...… .
dx
(1996, 2M)
22. Let f (x) = x|x|. The set of points, where f (x) is twice
differentiable, is … .
(1992, 2M)
(1990, 2M)
1 
24. The derivative of sec  − 2  with respect to
 2x − 1
1
(1986, 2M)
1 − x2at x = is …… .
2
25. If f (x) = log x (log x), then f ′ (x) at x = e is ...... . (1985, 2M)
(2007, 3M)
 d y
(b) −  2 
 dx 
2
−1
 dy 
 
 dx 
−3
−1 
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) ,
h1 (x) h2(x) h3 (x)
 d 2 y  dy − 3
(d) −  2   
 dx   dx 
 d 2 y  dy − 2
(c)  2   
 dx   dx 
then F ′ (x) at x = a is …… .
15. If f ′′ (x) = − f (x), where f (x) is a continuous double
2
2
  x 
  x 
If F (x) =  f    +  g    and F (5) = 5,
  2 
  2 
then F (10) is
(b) 5
(c) 10
 2x − 1
dy
= ....... .
 and f ′ (x) = sin 2 x, then
dx
 x2 + 1 
(1982, 2M)
Analytical & Descriptive Questions
(2006, 3M)
16. Let f be twice differentiable function satisfying
17. If y is a function of x and log (x + y) = 2xy, then the value
of y′ (0) is
(2004, 1M)
(c) 2
c
ax2
bx
+
+ 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
28. If y =
(d) 25
f (1) = 1, f (2) = 4, f (3) = 9, then
(d) 0
(1985, 2M)
27. If y = f 
differentiable function and g (x) = f ′ (x).
(a) 1
(b) p + p 2
(d) independent of p
26. If fr (x), gr (x), hr (x), r = 1, 2, 3 are polynomials in x such
−1
(a) 0
(1997, 2M)
d  3 d 2y
 equals
y
dx 
dx2 
x > 2.
d 2x
equals
dy2
 d y
(a)  2 
 dx 
Then,
23. If f (x) = |x − 2|and g (x) = f [ f (x)] , then g ′ (x) = K… for


1 1
1
(c) − 4 1 + +
+ ... +
2
9 25
(2 N + 1) 



1 1
1
(d) 4 1 + +
+ ... +
2
9 25
(2 N + 1) 

2
sin x cos x
−1
, where p is constant.
0
p2
p3
3
2
13. Let g (x) = log f (x), where f (x) is a twice differentiable
14.
p
(b) yy′ ′ + ( y ′ )2 + 1 = 0
(d) yy′ ′ + 2 ( y ′ )2 + 1 = 0
20. If y2 = P (x) is a polynomial of degree 3, then
(d) 1 + { g (x)}5
12. If y = sec (tan − 1 x), then
19. Let f (x) = 6
1 + 9x
g′ (x) is equal to
(a) 1 + x5
1
(c)
1 + { g (x)}5
x3
3
1
, then
1 + x5
(2000, 1M)
(a) yy′ ′ − 2 ( y ′ )2 + 1 = 0
(c) yy′ ′ + ( y ′ )2 − 1 = 0
29. 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)
30. 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)
188 Limit, Continuity and Differentiability
31. 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.
(1984, 4M)
32. Find the derivative with respect to x of the function

y = (log cos x sin x) (log sin x cos x)−1 + sin −1

π
at x = .
4
 2x  

 
 1 + x2  
(1984, 4M)
33. If (a + bx) ey/ x = x, then prove that
d y  dy

x
= x
− y .
2


dx
dx
3
(1983, 3M)
f ′ ′ (x) = − f (x) , f ′ (x) = g (x) and
h (x) = [ f (x) ]2 + [ g (x)]2
Find h (10), if h (5) = 11.
3
dy
35. Let y = ex sin x + (tan x)x , find
.
(1981, 2M)
dx
5x
dy
2
36. Given, y =
+ cos (2x + 1), find .
dx
3 (1 − x)2
(1980)
Integer Type Questions
37. Let f : R → R be a continuous odd function, which
1
vanishes exactly at one point and f (1) = .
2
Suppose that F (x) = ∫
2
2
34. Let f be a twice differentiable function such that
x
(1983, 3M)
G (x) = ∫ t| f { f (t )}| dt
–1
x
–1
f (t ) dt for all x ∈ [−1 , 2] and
for
all
x ∈ [−1 , 2].
F (x) 1
 1
lim
= , then the value of f   is
 2
x → 1 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
31. loge 4
(d)
(c)
(b)
(c)
a =2
2.
6.
10.
14.
(b)
(c)
(d)
e2
3.
7.
11.
15.
(d)
(d)
(c)
e5
4.
8.
12.
16.
(a)
(c)
(c)
e2
Topic 3
1. (c)
5. (1)
2. (a)
6. 1
3. (b)
4. (b,c)
2. (b)
6. (b)
3. (b)
7. (a,b,d)
4. (c)
8. (a, d)
Topic 4
1. (a)
5. (c)
9. (b,d)
10. f ( x ) = 4 − x 2
Topic 5
2. (d)
3. (d)
4. (b)
6. (d)
7. (b)
8. (b, c)
10. x ∈ (– ∞,– 1 ) ∪ [ 0, ∞ ),[ − 1, 0 )
11. f and f ′′ are continuous and f ′ is discontinuous at x = {1, 2 }.
Topic 2
1.
5.
9.
13.
17.
2
π
–π
13. a = , b =
11. a = , b = e 2/3 12. a = 8
3
6
12
2 
2
1

 
3  log  3 – 6 x, x ≤ 0

14. f ( x ) = 
1/ x
  1 + x  ,
x>0
  2 + x 
–3
1
16. (d)
15. a = , c = and b ∈ R
2
2
1. (a)
5. (c)
9. (a,b, d)
Topic 6
1. (b, c, d)
x + a + 1,
( x + a − 1 ) 2,

2. g{ f ( x )} =  2
x + b ,
2
( x − 2 ) + b,
If
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 }
Limit, Continuity and Differentiability 189
Topic 7
1.
5.
9.
13.
17.
21.
25.
29.
2.
6.
10.
14.
18.
22.
26.
30.
(b)
(b)
(c)
(b)
(d)
(c)
(d)
(a,b)
33. (b, c)
37. (a, c, d)
3.
7.
11.
15.
19.
23.
27.
31.
(a)
(a)
(a)
(c)
(d)
(d)
(a,b,d)
(b,c)
34. (a, b, c, d)
38. (a, b)
(b)
(a)
(b)
(b)
(a)
(a)
(b,c,d)
(a,d)
35. (b, c)
39. (b, d)
4.
8.
12.
16.
20.
24.
28.
32.
(c)
(a)
(d)
(a)
(d)
(a)
(b,c)
(b,c)
36. (a, d)
40. (a, b, d)
41. (b)
42. (A) → p, q, r, s; (B) → p, s; (C) → r, s; (D) → p, s
43. (A) → p; (B) → r
45. f ′( 0 + ) = 0, f ′( 0 − ) = 1
44. (24)
46. x = 0
48. (a = 1 )
47. True
51. (1, 2 )
52. (i) Yes (ii) No
53. ( − 1 )
54. e 2x
55. { 0, 1, 2 )
2

49. 1 − 

π
57. f ′( 0 ) = 0
58. g ( x ) is differentiable for all x ∈ ( − 2, 2 ) − { 0, 1 )
59. g ( x ) is continuous for all x ∈ ( 0, 2 ) − {1 ] and g ( x ) is
differentiable for all x ∈ ( 0, 2 ) − {1 }
 2
1 π
61.  + 
62. (2)
63. (3)
60.  − 
 9
2 4
64. p (2 ) = 0
Topic 8
1.
5.
9.
13.
17.
21.
2.
6.
10.
14.
18.
22.
(b)
(b)
(b)
(a)
(a)
1
1
25.
e
29.
3
π π2 –3
(b)
4. (d)
(b)
8. (c)
(d)
12. (a)
(b)
16. (c)
(d)
20. (c)
1
24. –4
–2 ( x 2 – x – 1 )
 2x – 1 
26. 0
27.
⋅ sin 2  2

 x + 1
(x 2 + 1)2
−8
32
32.
33. 11
+
2
loge
16 + π 2
(d)
(b)
(a)
(d)
(b)
x ∈ R − { 0}
3.
7.
11.
15.
19.
23.
3
35. 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

3 (1 – x ) 2 – 2 sin ( 4 x + 2 ), x < 1
37. (7)
36.  –5
+
>
2
4
2
1
x
x
–
sin
(
),

2
3 ( x – 1 )
Hints & Solutions
Topic 1
0
∞
and Form
0
∞
x2 − ax + b
=5
x→1
x−1
2. It is given that lim
1. Let
P = lim
x→ 0
x + 2 sin x
x2 + 2 sin x + 1 − sin 2 x − x + 1
0

0 form
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 )
= 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
0 form
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
2
x→ 0
x + 2 sin x + 1 − sin 2 x − x + 1
…(i)
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
⇒
1 − b =5
…(iii)
⇒
b = −4
On putting value of ‘b’ from Eq. (iii) to Eq. (ii), we get
a = −3
So,
a+ b=−7
3. Given,
lim
x→1
⇒
x4 − 1
x3 − k3
= lim 2
x − 1 x → k x − k2
(x − 1)(x + 1)(x2 + 1)
x→1
x −1
(x − k)(x2 + k2 + xk)
= lim
x→ k
(x − k)(x + k)
lim
190 Limit, Continuity and Differentiability
3k2
2k
8
k=
3
⇒
7. Clearly,
2 ×2 =
⇒
4. Given limit is lim
x→ 0
x
2 − 2 cos
2
sin 2 x
= lim
x→ 0
x

2 1 − cos 

2
x→ 0
= lim
x→ 0
x2
 x
2 2 
 4
2
=
x→ 0
1 + 1 + y4 + 2
y4 ( 1 + 1 + y4 + 2 )
[Q (a + b) (a − b) = a 2 − b2]
×
1 + y4 + 1
1 + y4 + 1
[again, rationalising the numerator]
y4
y4 ( 1 + 1 + y 4 + 2 ) ( 1 + y 4 + 1 )
1
2 2 ×2
(by cancelling y4 and then by direct substitution).
=
1 − tan 4 x
1
×
3
1
tan
x
(cos x − sin x)
2
2 (1 + tan 2 x)
(1 − tan 2 x)
= lim
×
x → π / 4 cos x − sin x
tan3 x
x → π /4
1 

Q cot x =
tan x 

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]
(cos x − sin x) (cos x + sin x)
2 sec4 x
= lim
×
x → π /4
(cos x − sin x)
tan3 x
[Q (a 2 − b2) = (a − b) (a + b)]
4
2 sec x
= lim
(cos x + sin x)
x → π / 4 tan3 x
2 ( 2 )4  1
1
[on applying limit]
=
+


3

2
2
(1)
 2
= 4 2   = 8.
 2
= lim
2
x cot 4x
x
1 tan 2x
= lim
.
2
2
x
→0
tan 4x sin 2 x
1
sin x. cot 2x
2
1 4x
x
tan 2 2x
.
= lim
x→0 4 (tan 4 x) sin 2 x
x2
6. lim
x→0
2
4
1 + y4 − 1
y→ 0
=
2
1 4x
 x   tan 2x 4
 .
 .

 2x  1
x→ 0 4 (tan 4 x)  sin x
= lim
1
4
⋅ 1 ⋅1 ⋅1 ⋅ = 1
4
1
y ( 1 + 1 + y + 2)
= lim
[lim sin x = lim x]
cot3 x − tan x
x → π /4
π

cos  x + 

4
=
(1 + 1 + y ) − 2
4
y→ 0
x

2 x
Q 1 − cos 2 = 2 sin 4 
x→ 0
1 + 1 + y4 + 2
[rationalising the numerator]
= lim
16
=4 2
2 2
×
4
5. Given, limit = lim
= lim
y4
y→ 0
y→ 0
 x
2 × 2 sin 2 
 4
x→ 0
1 + 1 + y4 − 2
= lim
= lim
sin 2 x
= lim
0

0 form

2 x
Q 1 + cos x = 2 cos 2 
sin x
= lim
y4
y→ 0
sin 2 x
2 − 1 + cos x
2
1 + 1 + y4 − 2
lim
x
tan x 

= 1 = lim
Q lim
x→ 0 sin x
x→ 0
x 
8.
1
4 2
lim
x→ π / 2
.
cot x − cos x
1 cos x(1 − sin x)
= lim ⋅
3
x → π/ 2 8
(π − 2x)3
π

sin x − x
2

1
= lim ⋅
h→ 0 8
=

π

π
cos  − h 1 − sin  − h 

2

2
3
π
 π π

sin  − h  − + h
2
 2 2

sin h (1 − cos h )
1
lim
8 h→ 0
cos h ⋅ h3
1
= lim
8 h→ 0
1
= lim
4 h→ 0
h

sin h 2 sin 2 

2
cos h ⋅ h3
 h
sin h ⋅ sin 2 
 2
h3 cos h
2
h

sin 
1
1
1
1 1 1
 sin h  
2
= lim 
⋅ = × =
  h  ⋅


→
h
0
4
h

 cos h 4 4 4 16
 2 
sin(π cos 2 x)
sin π (1 − sin 2 x)
=
lim
x→0
x→ 0
x2
x2
2
sin(π − π sin x)
= lim
x→ 0
x2
9. lim
sin(π sin 2 x)
x→ 0
x2
= lim
[Q sin (π − θ ) = sin θ ]
 sin 2 x
sin π sin 2 x
× (π )  2  = π
2
x→ 0
π sin x
 x 
= lim
sin θ


= 1
Q lim
θ→ 0 θ

Limit, Continuity and Differentiability 191
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
= 1 and lim
= 1
Q θlim
θ→ 0 θ
→0 θ
4




x2 x4
− ...
 1 − +

x
!
!
2
4






2
 −2 sin  
3
2



x
x
2 
+
+ ... 
− 1 + x +

2! 3!
 

10. We have, lim
= lim
= lim
PLAN
lim
x→∞
 ∞  form
 
∞
n
n−1
b 0x + b1x
m−1
a0x + a1x
m
⇒
lim
x→ ∞
+ K + an
+ K + bm
 0,
 a0
 ,
=  b0
 + ∞,

 − ∞,
if n< m
= lim
if n = m
if n>m and a0b 0 < 0
1−a −b =4
⇒
b = − 4 [Q (1 − a ) = 0]
h→ 0
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
⇒
⇒
{(a − n ) n − 1} n = 0
(a − n ) n = 1
⇒
(cos x − 1) (cos x − ex )
x→ 0
xn
14. lim
2
a=n+
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
NOTE
In trigonometry try to make all trigonometric functions in
same angle. It is called 3rd Golden rule of trigonometry.
x
= lim
x→ 0
Here, we make degree of numerator
= degree of denominator
∴
1−a =0
⇒ a =1
x (1 − a − b) + (1 − b)
and
lim
=4
x→ ∞
x+1
1
n


x2
1 + x +
+ K
3!


 x
2   xn − 3
 2
if n> mand a0b 0 > 0
x2 + x + 1 − ax2 − ax − bx − b
=4
x+1
⇒
x
2
x→ 0
x2(1 − a ) + x(1 − a − b) + (1 − b)
lim
=4
x→ ∞
x+1
12. Here, lim
2
 x
4   xn − 2
 2
sin 2
Description of Situation As to make degree of
numerator equal to degree of denominator.

 x2 + x + 1
∴
lim 
− ax − b = 4
x→ ∞ 
x+1

⇒

2x2 x3

2 x 
−
− ...
 − 2 sin   − x −

2 
2! 3!

x→ 0
=2
11.
xn
x→ 0
2 tan x
− 2x tan x
1 − tan 2 x
(2 sin 2 x)2
1


−1
2x tan x 
1 − tan 2 x 

= lim
x→ 0
4 sin 4 x
1 − 1 + tan 2 x 
2x tan x 

2
 1 − tan x 
= lim
x→ 0
4 sin 4 x
3
 tan x
3
x
 ⋅x
 x 
1
2x tan3 x
= lim
x → 0 2 sin 4 x (1 − tan 2 x)
x → 0 2 sin 4 x (1 − tan 2 x)
= lim
=
 tan x


 x 
3
1
1
1 ⋅ (1)3
=
lim
=
4
4
→
x
0
2
2(1) (1 − 0) 2
 sin x
2

 (1 − tan x)
 x 
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 )|
= 2 lim
h→ 0
−h
sin h
= 2 lim
=− 2
h → 0 −h
Again, RHL = lim
x →1+
= lim
x →1+
1 − cos 2 (x − 1)
x−1
|sin (x − 1)|
2
x−1
192 Limit, Continuity and Differentiability
x = 1 + h, h > 0
Put
= G′ (1) =
+
For x → 1 , h → 0
= lim 2
h→ 0
∴
|sin h |
sin h
= lim 2
= 2
h→ 0
h
h
LHL ≠ RHL.
Hence,
17. lim
x→ 0
At
and
lim
x→1
f (x) does not exist.
1
(1 − cos 2 x)
1 .|sin x|
2
= lim
x
→
0
x
x
2
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]
 0,
[x] = 0
sin [x]
, x ∈ R − [0, 1)

⇒
f (x) =  [x]
 0,
0 ≤ x<1
At x = 0,
RHL = lim 0 = 0
x→ 0 +
sin [x]
sin [0 − h ]
= lim
h→ 0
[x]
[0 − h ]
sin (−1)
= lim
= sin 1
h→ 0
−1
Since, RHL ≠ LHL
⇒
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
2
n 
lim 
+
+K+

2
2
1
1
1
−
n
−
n
−
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
n→∞
j=1
fn (x) = (tan − 1 (x + 1) − tan − 1 x)
+ (tan − 1 (x + 2) − tan − 1 (x + 1))
⇒
+ (tan − 1 (x + 3) − tan − 1 (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, ∞ )
(c) fn (x) = tan − 1 (x + n ) − tan − 1 x
lim tan( fn (x)) = lim tan(tan − 1 (x + n ) − tan − 1 x)
x→ ∞
lim
x→ a
g (x) f (a ) − g (a ) f (x)
x−a
g ′ (x) f (a ) − g (a ) f ′ (x)
,
1 −0
[using L’ Hospital’s rule]
= g ′ (a ) f (a ) − g (a ) f ′ (a )
= lim
x→ a
= 2 (2) − (−1) (1) = 5
21. Given,
∴ lim
x→1
G (x) = − 25 − x2
G (x) − G (1)
G ′ (x) − 0
= lim
x→1
x−1
1 −0
[using L’ Hospital’s rule]
x→ ∞


n
lim tan( fn (x)) = lim tan  tan − 1

2
x→ ∞
1 + nx + x 

n
=0
= lim
x → ∞ 1 + nx + x2
⇒
x→ ∞
∴ (c) statement is false.
(d) lim sec2( fn (x)) = lim (1 + tan 2 fn (x))
x→ ∞
x→ ∞
= 1 + lim tan 2( fn (x)) = 1 + 0 = 1
x→ ∞
∴ (d) statement is true.
23. L = lim
a − a 2 − x2 −
x2
4 , a >0
x4
1 1 


 x2
 1 x2 2  2 − 1 x4
a − a ⋅ 1 − ⋅ 2 +
⋅ 4 − ... −
2
a
 4
 2 a


= lim
x→ 0
x4
x→ 0
20. Given, f (a ) = 2, f ′ (a ) = 1, g (a ) = − 1, g ′ (a ) = 2
∴
n
x→ 0 −
∴ Limit does not exist.
19.


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
LHL = lim
and
1
24
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
24. lim
h→ 0
0

0 form
Limit, Continuity and Differentiability 193
= lim
2 + 2h − 2 − 4h
2h (1 + 2h ) (1 + h )
= lim
−1
= −1
(1 + 2h ) (1 + h )
h→ 0
h→ 0
n = 0, ± 1, ± 2, K
sin x, x ≠ nπ ,
25. Given, f (x) = 
,
2
otherwise

{ f (x)}2 + 1 , f (x) ≠ 0, 2

g [ f (x)] = 
4
, f (x) = 0

5
, f (x) = 2

∴
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
Put
π 

 2 2
 y
2
= lim 
⋅ =
π  π π
y→ 0 
tan y

2 
g [ f (x)] = (sin x) + 1, x ≠ , nπ = 0, ± 1, K
5 , x = nπ

2
Now,
lim g [ f (x)] = lim (sin 2 x) + 1 = 1
x→ 0
x→ 0
30. If lim [ f (x) g (x)] exists, then both lim f (x) and lim g (x)
x→ a
26. Given, P = 2 ( 2hr − h 2 + 2hr )
x→ a
x→ a
may or may not exist. Hence, it is a false statement.
A
(2x − 1)( 1 + x + 1)
2x − 1
1+ x+1
×
= lim
x
1 + x −1
1+ x + 1 x→ 0
31. lim
x→ 0
= log e (2) ⋅ (2)
= 2 log e 2 = log e 4
h
B
D
C
Here, BD = r 2 − (h − r )2 = 2hr − h 2
1
∴
A = . 2BD. h = ( 2hr − h 2) h
2
∴ lim
h→ 0
h 2hr − h 2
A
lim
=
P 3 h→ 0 8 ( 2hr − h 2 + 2hr )3
= lim
h→ 0
=
(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
h
h


a 2 ⋅ 2 cos  a +  ⋅ sin

2
2
= lim
+ (2a + h ) sin (a + h )
h
h→ 0
2⋅
2
32. Here, lim
h–r
r
h3/ 2 ( 2r − h )
8 h3/ 2 ( 2r − h + 2r )3
1
1
2r
=
⋅
3
128 r
8 ( 2r + 2r )
= a 2 cos a + 2a sin a
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
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 + |x |3  x → −∞
1 − x3
x → −∞ 


On dividing by x3 , we get
sin (1 / x) 1
+
1
x
1+0
x
=
= −1
lim
1
x → −∞
0 −1
−1
3
x
 x3 + x2 − 16x + 20
28. f (x) = 
(x − 2)2

k
, if x ≠ 2
x→ 0
0 .0
=
=0
1
1
x −1

 = lim
 (x − 1)(2x − 5)  x→1 (2x − 5)
1
=−
3

34. lim 
x→1
x2 sin (βx)
=1
x → 0 αx − sin x
35. Here, lim
, if x = 2
Since, continuous at x = 2.
x3 + x2 − 16x + 20
,[using L’Hospital’s rule]
⇒ f (2) = lim
x→ 2
(x − 2)2
⇒


(βx)3
(βx)5
x2  β x −
+
− K
3!
5!


lim
=1
3
5
x→ 0


x
x
+
− K
αx −  x −
3! 5!


194 Limit, Continuity and Differentiability


β3 x2 β 5 x4
+
− K
x3 β −
3!
5!


1
 1 + f (3 + x) − f (3) x
So, lim
 =1
x→ 0  1 + f (2 − x) − f (2 ) 
=1
x3
x5
+
−K
3! 5!
Limit exists only, when α − 1 = 0
⇒
α =1
3 2


β
β5 x4
x
x3 β −
+
− K
3!
5!


∴
=1
lim
2
x→ 0


1
x
3
− K
x  −

3! 5!
⇒
lim
x→ 0
(α − 1)x +
2. Let L =
…(i)
L=
π − 2 sin −1 x
, then
1−x
lim
x → 1−
π − 2 sin −1 x
π + 2 sin −1 x
×
1−x
π + 2 sin −1 x
lim
x → 1−
[on rationalization]
⇒
6β = 1
From Eqs. (i) and (ii), we get
6(α + β ) = 6α + 6 β
=6 + 1 = 7
lim π − 2 sin −1 x
=
×
x → 1−
1−x
…(ii)
lim
=
x → 1−
 ecos ( α n ) − e 
e
=−
m
α→ 0 
2
α


1
π + 2 sin −1 x
π

π − 2 − cos −1 x
2

×
1−x
1
π + 2 sin −1 x
π

−1
−1
Q sin x + cos x = 2 
36. Given, lim 
⇒
cos( α n ) −1
=
e {e
− 1} cos(α ) − 1 − e
⋅
=
lim
α→ 0
2
cos(α n ) − 1
αm
n
2α
=
n
−2 sin
 ecos( α n ) −1 − 1 
2 = − e /2
lim
⋅
⇒ lim e 

n
α→ 0
αm
 cos(α ) − 1  α→ 0
 α n
sin 2 
 2  α 2n
−e
⇒
⋅ m=
e × 1 × (−2) lim
α→ 0
2
4α
α 2n
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
lim
l = ex→ 0
=
1 + f (3 + x ) − f (3 ) 
1
1 −

1 + f ( 2 − x ) − f ( 2) 
x
 1 + f ( 2 − x ) − f ( 2) − 1 − f (3 + x ) + f (3 ) 
lim 

x(1 + f ( 2 − x ) − f ( 2))
x→ 0 

e
 f ( 2 − x ) − f (3 + x ) + f (3 ) − f ( 2) 
lim 

x(1 + f ( 2 − x ) − f ( 2))

x→ 0 
=e
lim 2 cos x
1
−
2 π x→ 1
1−x
1
=
⋅ 2 lim
2 π
θ→ 0 +
=
3.
1
⇒
−1
1
2 2 =
2 π

lim 

− f ′( 2 − x )− f ′(3 + x )
1 − xf ′( 2 − x ) + f ( 2 − x ) − f ( 2) 
x→a
[1 form]
 − f ′( 2) − f ′(3 ) 


 1 − 0 + f ( 2) − f ( 2) 
l=e
= e0 = 1
 θ
2⋅  
 2
 θ
sin  
 2
lim


θ
Q x → 0+ sin θ = 1


2
π
lim f ( x) = lim f ( x)
x → a+
x → a−
At x = 0,
RHL = lim
x → 0+
= lim
x→ 0
+
= lim
x → 0+
On applying limit, we get
π + 2 sin −1 x
lim

π
−1
Q x → 1− sin x = 2 


Key Idea lim f ( x) exist iff
∞
On applying L’Hopital rule, we get
l = ex→ 0 
1
Put x = cos θ, then as x → 1− , therefore θ → 0+
lim
1
2θ
Now, L =
+
2 π θ→0
1 − cos θ
lim
1
2θ

2 θ
=
+
Q 1 − cos θ = 2 sin 2 
 θ
2 π θ→0
2 sin  
 2
Topic 2 1∞ Form, RHL and LHL
 1 + f (3 + x) − f (3) x
1. Let l = lim

x→ 0  1 + f (2 − x) − f (2 ) 
lim 2 cos −1 x
lim
×
−
x → 1−
x→ 1
1−x
tan(π sin 2 x) + (|x| − sin(x [x]))2
x2
tan(π sin 2 x) + (x − sin(x ⋅ 0))2
x2
 Q|x| = x for x > 0 
and [x] = 0 for 0 < x < 1


tan (π sin 2 x) + x2
x2

 tan(π sin 2 x) π sin 2 x
.
= lim 
+ 1
2
2
+
π sin x
x→ 0 
x

Limit, Continuity and Differentiability 195
= π lim
x→ 0
+
π

( − h + sin h ) sin  (− 1)
2

= lim
+
h (− 1)
h→ 0
sin 2 x
tan (π sin 2 x)
. lim
+1
2
+
x→ 0
x2
π sin x
tan x

=1
Q xlim
→0
x

sin x
=1
and xlim
→0
x

=π+1





(Q [x] = − 1 for − 1 < x < 0 and h → 0+ ⇒ − h → 0− )
= lim
h→ 0 +
and LHL
= lim
x→ 0
−
= lim
x→ 0
−
tan (π sin 2 x) + (|x| − sin (x [x])2
x2
= lim
x → 0−
 sin h 
 h
= lim 
 − lim+  
h→ 0 +  h 
h → 0  h


sin h
 sin h 
 h
= lim 
= 1
 − lim+   = 1 − 1 = 0 Q lim+
h
h→ 0 +  h 
h → 0  h

 h→ 0
tan (π sin 2 x) + (− x − sin(x (− 1))2
x2
 Q|x| = − x for x < 0

and [x] = − 1 for − 1 < x < 0


5.
lim
x → 0−
tan(π sin 2 x) + (x + sin(− x))2
x2
x([x] − x) sin [x]
x([x] + | x|) sin [x]
= lim
| x|
−x
x → 0−
(Q| x| = − x, if x < 0)
x(− 1 − x) sin (− 1)
(Q lim [x] = − 1)
= lim
−x
x → 0−
x → 0−
− x(x + 1) sin(− 1)
= lim
= lim (x + 1)sin(− 1)
−
−x
x→ 0
x → 0−
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 
(− h + sinh)
 − π
sin 

 2 
−h
= (0 + 1) sin (− 1) (by direct substitution)
(Qsin(− θ) = − sin θ)
= − sin 1
6.
Key Idea Use property of greatest integer function [ x ] = x − { x }.
 tan (π sin 2 x) π sin 2 x
sin 2 x
sin x
.
= lim 
−2
+ 1+

2
2
−
x 
π sin x
x→ 0 
x
x2
 1  2 
15 
lim x   +   + …+  
  x  x
 x 
x→ 0 +
We know, [x] = x − { x}
1  1 1 
∴
 x  = x −  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
−
 n  n n 
 x  = x −  x 
15 15 
 1 1  2 2 
∴Given limit = lim x −   + −   + … −  
x  x 
x x
x→ 0 +  x  x 
Similarly,
Q RHL ≠ LHL
∴ Limit does not exist
 1  2 
15 
= lim (1 + 2 + 3+ ...+15) − x   +   + ... +  
 x x
x→ 0 +
 x 
4. Given,
π

(1 − |x| + sin|1 − x|) sin  [1 − x]
2

lim
|1 − x|[1 − x]
x →1+
= 120 − 0 = 120
Put x = 1 + h , then
x → 1+ ⇒ h → 0+
π

(1 − |x| + sin|1 − x|) sin  [1 − x]
2

∴ lim
|1 − x|[1 − x]
x →1+
π

(1 − |h + 1| + sin|− h|) sin  [− h ]
2

= lim
|− h|[− h ]
h→ 0 +
π

(1 − (h + 1) + sin h ) sin  [− h ]
2

= lim
h [− h ]
h→ 0 +
(Q|− h| = h and|h + 1| = h + 1 as h > 0)


n 

Q 0 ≤  x  < 1, therefore


n 
n 
0 ≤ x  < x ⇒ lim x  = 0


x→ 0 +  x 
x
7.
f (x) =
 1 
1 − x( 1 + |1 − x|)
cos 

|1 − x|
 1 − x
Now, lim f (x) = lim
x→1 −
x→1 −
 1 
1 − x(1 + 1 − x)
cos 

1−x
 1 − x
 1 
= lim (1 − x) cos 
 =0
−
 1 − x
x→1
and
lim f (x) = lim
x→1 +
x→1 +
 1 
1 − x(1 − 1 + x)
cos 

x−1
 1 − x
 1 
= lim − (x + 1) ⋅ cos 
 , which does not exist.
 x + 1
x→1 +
196 Limit, Continuity and Differentiability
8. Given,
p = lim (1 + tan 2 x )
x → 0+
tan 2 x
+
2x
x→ 0
=e
9.
PLAN
1
e2
log p = log
=
Applying L’Hospital’s rule, we get
(1∞ form)
1
 tan x 
lim 

2x → 0+ 
x 
lim
∴
1
2x
lim
2
=e
=
ex → 0
1
e2
 1 
x −

 x2 
⇒


1
lim log y = lim 
⋅ f ′ (1 + x)
x → 0  f (1 + x)

x→ 0
[using L’ Hospital’s rule]
f (1) 6
=
=
f (1) 3
i.e. ax2 + bx + c = 0, ∀ x ∈ R or a = b = c → 0


⇒ log  lim y = 2 ⇒
 x→ 0 
Thus, first we should make above equation independent
from coefficients as 0.
Let a + 1 = t . Thus, when a → 0, t → 1.
6
(t − 1) x + (t − 1)x + (t − 1) = 0
x→ ∞
2x + 3x + 1 = 0
2
2 x2 + 2 x + x + 1 = 0
⇒
(2x + 1) (x + 1) = 0
 x − 3
(1 − 3 / x)x e−3
= 2 = e−5
 = lim
x→∞ (1 + 2 / x)x
 x + 2
e
 1 + 5 x2 
lim 

x → 0  1 + 3 x2 
 x + 6

x → ∞  x + 1
15. lim 
lim α (a ) = − 1 / 2
2
1/ x 2
x+4
=
lim β (a ) = − 1
a→ 0
+
10. Here, lim {1 + x log (1 + b )}
[1 form]
x→ 0
lim {x log (1 + b 2 )} ⋅
= ex → 0
= elog (1 + b
2
)
= (1 + b2)
Given, lim {1 + x log (1 + b )}
2 1/ x
x→ 0
…(i)
= 2b sin θ
2
(1 + b ) = 2b sin θ
2
∴
2
sin 2 θ =
By AM ≥ GM,
1 + b2
2b
…(ii)
1
1/ 2
b ≥  b ⋅ 1


 b
2
17.
b+
PLAN lim
x→0
1/ x
11. Here, lim (sin x)
x→ 0
1
log  
 x
= 0 + lim e
x→ 0
sin x
…(iii)
sin x
=1
x
x ) (1 − x )
1− x
1+

 sin (x − 1)
 (x − 1) − a 
lim 
sin(x − 1) 
x→1

1 +

(x − 1) 
π
, as θ ∈ (− π , π ]
2
 1
+ lim  
x → 0  x
sin x
log (1/ x )
x → 0 cosec x
=e
= e5
sin (x − 1) + a (1 − x) 
Given, lim 

x → 1  (x − 1 ) + sin (x − 1 ) 
sin 2 θ = 1
θ=±
5 ( x + 4)
x +1
[1∞ form]


π
lim tan  + x 

4
x→ 0
1/ x
π


1/ x
tan
tan
x
+
1 + tan x 


4
= lim 
= lim 


π
x→ 0
x → 0  1 − tan x 
1 − tan tan x 


4
e1
[(1 + tan x)1/tan x ]tan x/ x
= −1 = e2
= lim
x
/
x
1
x
/tan
tan
−
−
x → 0 [(1 − tan x)
e
]
From Eqs. (ii) and (iii),
⇒
lim
x→ ∞
x+ 4
(1 +
b2 + 1
≥1
2b
⇒
2
lim [(1 + 3x2)1/3 x ]3
e5
= e2
e3
=
1/ x
16.
1
x
x→ 0

5 
= lim 1 +

x→ ∞ 
x + 1
=e
∞
2 1/ x
lim [(1 + 5x2)1/5 x ]5
x→ 0
a→ 0 +
and
⇒
14.
x = − 1, − 1 / 2
Thus,
x→ 0
13. For x ∈ R , lim 
3
⇒
lim y = e2
x
⇒ (t − 1) {(t + 1) x2 + (t 2 + t + 1) x + 1} = 0, as t → 1
or
= e0 = 1
1/ x
Description of Situation As for given equation,
when a → 0 the equation reduces to identity in x.
∴
sin x
tan x
x
1
 f (1 + x) 
⇒ log y = [log f (1 + x) − log f (1)]
x
f (1) 
1
2
2
= ex → 0
12. Let y = 

To make the quadratic into simple form we should
eliminate radical sign.
2
lim
− cosec x cot x
lim
 lim (sin x)1/ x → 0 

x → 0
∞
as, (decimal) → 0
=
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.
Limit, Continuity and Differentiability 197
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
Taking log on both sides, we get
25
1
α+β=
=
375 15
2
αβ = −
375
∴
and
n
Now, lim
n→ ∞
… (i)
… (ii)
n
∑ α r + nlim
∑ βr
→∞
r =1

 n n (x + n )  x +

log e { f (x)} = lim log 

n →∞

 n !(x2 + n 2)  x2 +


n

1 

Π x +


r =1 
x
/ n
r
= lim ⋅ log 
n
n
n→ ∞ n

 2
1 
 Π (r / n )
x +
 rΠ
=1 
(
/
r
n )2  r = 1

r =1
= (α + α + α + K + upto infinite terms)+
2
3
(β + β 2 + β 3 + K + upto infinite terms)


a
for GP
Q S ∞ =
1−r


α (1 − β) + β (1 − α ) α − αβ + β − αβ
=
=
(1 − α ) (1 − β )
1 − α − β + αβ
=
=
α
β
+
1 −α 1 −β
= x lim
n→ ∞
(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
sec 2 x
∫ f (t ) dt
0

0 form
π2
x −
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. lim
π
x→
4
2
2
1
3. Let I = lim
n→ ∞ n
= 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
dx = [ 1 +
4. Here,

 n n (x + n )  x +

f (x) = lim 
n→ ∞ 

 n ! (x2 + n 2)  x2 +


x2 ]20
1
n
n
∑
r =1


n


x+

r
log 
2


n
r

  x2 + 2 
r  n 
 
tx=z
xdt = dz
Put,
⇒
∴
⇒
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 
Here, at x = 1 ,
f ′ (1)
= log (1) = 0
f (1)
∴
f′ (1) = 0
Now, sign scheme of f ′ (x) is shown below
= 5 −1
+
–
x=1
x
n
n
n

 … x +  


n 
2
,x>0
2

n
n 2 
 K  x2 + 2 
4
n  







 r


⋅x + 1 
∑ log  rn2 2 
r =1
 2 ⋅ x + 1
n

Converting summation into definite integration, we get
1
 xt + 1 
log e{ f (x)} = x ∫ log  2 2
 dt
0
 x t + 1
1
= x lim
n→ ∞ n
r /n
1 + (r / n )2
x
n
n
n 

 K x +  

2
n 
 2 n 2 
n 2
 K  x + 2 
4
n  

∴ 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.
… (i)
198 Limit, Continuity and Differentiability
Also,
f ′ (x) < 0, when x > 1
⇒
f′ (2) < 0
∴ Option (c) is correct.
 1 + x
f ′ (x)
Also,
= log 

f (x)
 1 + x2
∴
Topic 4 Continuity at a Point
1. Given function is
 2 cos x − 1

f (x) =  cot x − 1

k

f ′ (3) f ′ (2)
4
 3
−
= log   − log  
 10
 5
f (3)
f (2)
Q Function f (x) is continuous, so it is continuous at
π
x= .
4
 π
∴
f   = lim f (x)
 4  x→ π
= log (2 / 3) < 0
f ′ (3) f ′ (2)
<
f (3)
f (2)
⇒
∴ Option (d) is incorrect.
4
5. We have,
⇒ log L = lim
n→ ∞
⇒ log L =
Put x =
1
2 
3 
n  n
 1 +  ... 1 +  
n 
n 
n 
1
2
n 


 + log 1 +  … log 1 +  


n
n
n 
1
(1 + x) dx
∫0 1II × log
I
⇒ log L = [x log(1 +
⇒ log L = log 2 −
x)]10
1
d
−∫
(log(1 + x) ∫ dx  dx

0
 dx
[by using integration by parts]
x
−∫
dx
01 + x
=
1
1

∫0  x + 1 − x + 1  dx
=
⇒ log L = log 2 − [x]10 + [log(x + 1)]10
4 
[L ] =   = 1
 e
x2
6. lim
x→ 0

lim  (1 − cos h ) + sin h
(sin h + cos h )

h→0 
2 sin h

lim
4
4
⇒ L=
e
e
cos 2tdt
x sin x
⇒
0

0 form 
Applying L’Hospital’s rule, we get
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
lim cos h − sin h − 1
−2
h→0
1 + cot h
h
h


2h

 2 sin 2 + 2 sin 2 cos 2
=
(sin h + cos h )

h
h
h→ 0

4 sin cos
2
2


⇒ log L = log 2 − 1 + log 2 − 0
⇒ log L = log 4 − log e = log

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
1
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
cos h −
sin
2
2

=
cot h − 1
h→0
−1
cot h + 1
1
r
∑ log 1 + n 
n r =1
x))10
2 cos x − 1
cot x − 1
lim
n
⇒ log L = (x ⋅ log (1 +
∫0
lim
π
k=
x→
4
⇒
1
[(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
yn =
1

L = lim  1 +  1 +
⇒
n→ ∞  
n 
1
⇒ log L = lim  log 1 +

n→ ∞ n 
π
4
π
, x=
4
,x≠
h
h


lim sin 2 + cos 2

1
=
× (sin h + cos h ) ⇒ k =
h
h → 0 
2

2 cos


2
2.
NOTE All integers are critical point for greatest integer function.
Case I When x ∈ I
f (x) = [x]2 − [x2] = x2 − x2 = 0
Case II When x ∉ I
If
0 < x < 1, then [x] = 0
Limit, Continuity and Differentiability 199
and
0 < x2 < 1, then [x2] = 0
Next, if
1 ≤ x <2 ⇒ 1 ≤ x< 2
a log (1 + ax) b log (1 − bx)
+
− bx
ax
log (1 + x)
[using lim
= 1]
= a ⋅1 + b ⋅1
x→ 0
x
= lim
x→ 0
2
⇒
[x] = 1
and
[x2] = 1
Therefore, f (x) = [x]2 − [x2] = 0, if 1 ≤ x < 2
Therefore,
=a+b
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
3. Given, f (x) = [tan 2 x]
x→ 0
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
⇒ f (c) − g (c) = 0 for some c ∈ [0, 1] as f and g are
continuous functions.
, 0 ≤ x<1
, 1 ≤ x<2
⇒ f (c) − g (c) = 0 for some c ∈ [0, 1] for all cases.
, 2 ≤ x<3
Option (a) ⇒ f (c) − g (c) + 3 [ f (c) − g (c)] = 0
2
Also,
f (1) = 0
∴ Continuous at x = 1.
Similarly, when x = 2,
lim f (x) = lim f (x) = 0
x → 2−
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
5. Given, f (x) = x ( x + x + 1 )
f {(2n + 1)+ } = bn + 1 − 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,
or
because LHL does not exist.
Hence, option (c) is correct.
6. For f (x) to be continuous, we must have
f (0) = lim f (x)
x→ 0
log (1 + ax) − log (1 − bx)
= lim
x→ 0
x
...(i)
2
which is true from Eq. (i).
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 

−
2
1
x



2 cos  2  π

x →0
f (x) = 0
∴It is continuous at x = 0 and clearly discontinuous at
other integer points.
and
Now, − 45° < x < 45°
⇒
⇒
⇒
⇒
⇒
f (0) = (a + b)
10. Given,
or
a n = bn + 1 − 1 or a n − bn + 1 = − 1
a n − 1 − bn = − 1
x + y =4
2
2
⇒
y = 4 − x2
f (x) = 4 − x2
 {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

Since, f (x) is continuous at x = 0.
200 Limit, Continuity and Differentiability
∴
⇒
⇒
∴
and
 1 + x 
x
− log 

+
+
x
x
(
)(
)
1
2
 2 + x 

∴


x2




2 1
2

 
⇒
f ′ (1) =  − log   
 3 
3 6
and f (−1) = − a + b = − a
[from Eq. (i)]
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

RHL (at x = 0) = LHL (at x = 0) = f (0)
tan 2 h/tan 3 h
lim e
h→ 0
e
23
/
= lim {1 + |sin h |}
a /|sin h|
h→ 0
a
1 +
f ′ (x) = 
2 +
=b
=e =b
a = 2 /3
/
b = e 23
12. Since, f (x) is continuous at x = 0.
∴
⇒
f (0) = LHL
1 − cos 4h
a = lim
h→ 0
h2
2 sin 2 2h 4
×
h→ 0
4
h2
⇒
a = lim
⇒
a =8
⇒
⇒
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
 1 + x
RHL = lim 

x → 0  2 + x
…(i)
x→ 0
…(ii)
⇒
(LHL at x = 0) = (RHL at x = 0) = f (0)
 sin (a + 1) x sin x 
+
⇒ lim 
x→ 0 
x
x 
(1 + bx)1/ 2 − 1
=c
x→ 0
bx
x≤0
 ax + b,
1/ x

f (x) =  1 + x
 2 + x , x > 0

Since, f (x) is continuous and f ′ (1) = f (−1)
(LHL at x = 0) = (RHL at x = 0)
⇒
Also,
⇒
⇒
 x + 1
lim (ax + b) = lim 

x→ 0
x→ 0  x + 2
⇒
(a + 1) + 1 = lim
⇒
a+2=
∴
1/ x
b =0
and

 1
 1 + x 
1 
−
x

 − 1 log 
 2 + x 
f ′ (x)
1 + x 2 + x 

=
f (x)
x2
x→ 0
bx
1
⋅
=c
bx 1 + bx + 1
1
=c
2
3
1
a =− ,c=
2
2
b ∈R
2
…(i)
f ′ (1) = f (−1)
1/ x
 1 + x
f (x) = 
 ,x>0
 2 + x
1
log f (x) = [log (1 + x) − log (2 + x)]
x
On differentiating both sides, we get

 2 1 
log  3 − 6  x = 0


= lim
∴
⇒
=0
Hence, f (x) is continuous for all x.
14. Let g (x) = ax + b be a polynomial of degree one.
⇒
2
3
1/ x
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.
−a−b=b
⇒
a + 2b = 0
On solving Eqs. (i) and (ii), we get
π
π
and b = −
a=
6
12
LHL = lim
∴
2π
π  π
π


− b sin  = 2 ⋅ ⋅ cot + b
⇒  a cos


2
2  2
2
⇒
1/ x 
Now, to check continuity of f (x) (at x = 0).
13. Since, f (x) is continuous for 0 ≤ x ≤ π
∴
x

x
16. (i) Given, f1 : R → R and f1 (x) = sin ( 1 − e−x )
∴ f1 (x) is continuous at x = 0
2
Now, f1 ′ (x) = cos 1 − e− x .
1
2
−x 2
2 1− e
At x = 0
f1 ′ (x) does not exists.
∴ f1 (x) is not differential at x = 0
Hence, option (2) for P.
 |sin x|
, if x ≠ 0
(ii) Given, f2 (x) =  tan −1 x

if x = 0
1,
(2xe− x )
Limit, Continuity and Differentiability 201
⇒
 − 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
x=0
Clearly, f2 (x) is not continuous at x = 0.
(iii) Given, f3 (x) = [sin (log e (x + 2))] , where [ ] is G.I.F.
4t3 dt = (x − 2) g (x)
f ( x)
∫
g (x) = 6
⇒
∴ Option (1)for Q.
f ( x)
6
4t3 dt
f ( x)
∫
So, lim g (x) = lim 6
x→ 2
x→ 2
4t3 dt
x−2

 0
Q 0 form as x → 2 ⇒ f (2) = 6
and f3 : (−1, eπ / 2 − 2) → R
− 1 < x < eπ / 2 − 2
It is given
− 1 + 2 < x + 2 < eπ/ 2 − 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
⇒
⇒
⇒
1
 2
x sin   , 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
Thus,

2x sin
f4 ′ (x) = 

 1  − cos  1  , x ≠ 0
 
 
 x
 x
0,
x=0

Again, lim f ′ (x) = lim  2x sin
x→ 0
x→ 0 
 1  − cos  1  
 
 
 x
 x 
does not exists.
1
Since, lim cos   does not exists.
 x
x→ 0
Hence, f ′ (x) is not continuous at x = 0.
∴Option (3) for S.
[provided x ≠ 2]
(x − 2)
4( f (x))3 f ′ (x)
x→ 2
1
lim g (x) = lim
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
=
4 × 216
= 18
48
1
,
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)


= a
= p + 1 + 1 = p + 2 Q lim
x

 x→ 0
lim f (x) = lim
x→ 0 −
x→ 0 −
and lim f (x) = lim
x + x2 − x
x3/ 2
x [(1 + x)1/ 2 − 1]
= lim
+
x x
x→ 0
1 1



 − 1




1
2
2
1 + x +
x2 + .... − 1


2
2!




= lim
+
x
x→ 0
[Q (1 + x)n
n (n − 1) 2 n (n − 1(n − 2)) 3
= 1 + nx +
x +
x + ... ,|x|< 1]
1 ⋅2
1 ⋅2 ⋅3
x→ 0 +
x→ 0 +
202 Limit, Continuity and Differentiability
− x − 1 , − 1 ≤ x < 0
 x,
0 ≤ x<1

1 ≤ x<2
=  2x,
 x + 2, 2 ≤ x < 3

 6,
x=3
1 1



 − 1




1
1
2
2
= lim  +
x + ... =
 2
2!
x→ 0 +  2




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
So,
( p, q ) =  − , 
 2 2
Q
Q
x = 5.
…(i)
lim f (x) = lim[a|π − (5 − h )| + 1]
h→ 0
= a(5 − π ) + 1
lim f (x) = 1 ≠ f (1)
…(ii)
lim f (x) = 4 = f (2) = lim f (x) = 4
[Q f (2) = 4]
lim f (x) = 5 ≠ f (3)
[Q f (3) = 6]
x→ 2−
∴ Function f (x) is discontinuous at points 0, 1 and 3.
6.
Key Idea A function is said to be continuous if it is continuous at
each point of the domain.
We have,
 5
a + bx

f (x) = 
 b + 5x
 30
h→ 0
= b(5 − π ) + 3
Q Function f (x) is continuous at x = 5.
…(iii)
∴f (5) = lim f (x) = lim f (x)
x→5 +
x→5 −
⇒ a (5 − π ) + 1 = b(5 − π ) + 3
⇒
(a − b)(5 − π ) = 2
2
a−b=
⇒
5−π
x→ 2 +
x→ 3 −
and lim f (x) = lim[b| (5 + h ) − π| + 3]
x→5 +
x →1+
[Q lim f (x) = lim f (x) = f (1)]
x→1 −
4 
h→ 0

4 − h 
and lim f (x) = lim [4 − h ] − 

h→ 0 
x→ 4 −
 4 
[Q put x = 4 − h, when x → 4−
then h → 0]
= lim(3 − 0) = 3
h→ 0
4 
and f (4) = [4] −   = 4 − 1 = 3
4 
Q lim f (x) = f (4) = lim f (x) = 3
x→ 4 −
x→ 4 +
So, function f (x) is continuous at x = 4.
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 +
lim (b + 5x) = b + 25 = 30
[Q lim f (x) = lim

4 + h 
Now, lim f (x) = lim  [4 + h ] − 

h→ 0 
x→ 4 +
 4 
= lim(4 − 1) = 3
if
x≤1
if 1 < x < 3
if 3 ≤ x < 5
if
x≥5
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
x
4. Given function f (x) = [x] −  , x ∈ R
[Q put x = 4 + h, when x → 4+ , then h → 0]
[Q f (1) = 2]
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)
x →5−
x →5−
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
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,
⇒
∴
lim tan [ f (x)] = − tan 1
x→ 2−
Limit, Continuity and Differentiability 203
lim tan [ f (x)] = 0
and
10.
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 
So, tan [ f (x)] and 
 are both discontinuous at x = 2 .
 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
3π
 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
 2
 h→ 0 2
 2
h→ 0
lim
x→ ( π / 2)
+
π

f (x) = lim f  + h
2

h→ 0
π
 3π
 π
+ h =
sin 
 2
 2
2
So, f (x) is not continuous at x = π / 2.
= lim
h→ 0
⇒
−1 < x < 1
0 ≤ x sin π x ≤ 1 / 2
∴
[x sin πx] = 0
9. We have, for
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).
At x = 1, f (x) is discontinuous, since lim (1 − h ) = 0
and
lim (1 + h ) = − 1
h→ 0
h→ 0
∴ f (x) is not differentiable at x = 1.
Hence, (a), (b) and (d) are correct answers.
 π 
f (x) = [x] sin 

 [x + 1]
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
Thus, the function is defined in the interval.

x2

2
11. Given, f (x) = 
2x2 − 3x + 3

2
, 0 ≤ x<1
…(i)
, 1 ≤ x≤2
Clearly, RHL (at x = 1) = 1 / 2 and LHL (at x = 1) = 1 / 2
f (x) = 1 / 2
Also,
∴ f (x) is continuous for all x ∈[0, 2].
On differentiating Eq. (i), we get
, 0 ≤ x<1
 x
f ′ (x) = 
4x − 3 , 1 ≤ x ≤ 2
…(ii)
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}.
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  
 x
x3
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.
As f ′ (1) = sin 1 + cos 1 > 1
f ′ (x) is strictly decreasing and lim f ′ (x) = 1
Now,
x→ ∞
x→ ∞
So, graph of f ′ (x) is shown as below.
Y
(1, sin1+cos1)
1
0 1
X
204 Limit, Continuity and Differentiability
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
1
−
+
{
(
)
}
,
f
x
b
f (x) ≥ 0

if
1
,
x
a
x<−a
+
+


2
if − a ≤ x < 0
=  (x + a − 1) + b,
(| x − 1| − 1)2 + b,
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
⇒
gof (0) = gof (0+ ) = gof (0− )
⇒
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
LHL = lim f (k − h ) = lim [ f (k) + f (− h )]
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
⇒ lim φ (x) = lim { f (x) + g (x)} = does not exist.
= f ′ (c)
x→ k
Case II g (x) is discontinuous as, lim g (x) ≠ g (k).
x→ k
quantity
but
x→ k
[Q f is differentiable at x = c]
2.
Key Idea (i) First use L’ Hopital rule
(ii) Now, use formula
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.
1 + x, 0 ≤ x ≤ 2
4. Given, f (x) = 
3 − x, 2 < x ≤ 3
∴
[Qh > 0]
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.
∴ φ (x) is discontinuous.
x→ k
[as f (c) = 0 (given)]
f (c + h ) − f (c)
= lim
h→ 0
h
f (c + h ) − f (c)
= lim
h→ 0
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
h→ 0
= f (k) + f (0− ) = f (k) + f (0)
lim f (x) = f (k)
⇒ f (x) is continuous for all x ∈ R.
3. As, f (x) is continuous and g (x) is discontinuous.
φ (x) = f (x) + g (x).
h→ 0
= f (k) + f (0)
In the neighbourhood of x = 0 , gof (x) = x2, which is
differentiable at x = 0.
∴
…(i)
RHL = lim f (k + h ) = lim [ f (k) + f (h )] = f (k) + f (0+ )
b = b = (a − 1) + b ⇒ a = 1
if x < −1
x + 2,


if − 1 ≤ x < 0
gof (x) = 
x2 ,
(| x − 1| − 1)2,
if x ≥ 0

x→ k
x→ 0
⇒
f (0 ) = f (0− ) = f (0) = 0
To show, continuous at x = k
2
Hence,
lim f (x) = f (0)
+
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


⇒
4 − x, 2 < x ≤ 3

( fof ) (x) = 2 + x, 0 ≤ x ≤ 1
2 − x, 1 < x ≤ 2

d
dx
φ2 ( x )
∫ f (t ) dt = f [ φ2( x)] ⋅ φ′2 ( x) − f [ φ1( x)] ⋅ φ′1( x)
φ1 ( x )
f ( x)
f ( x)
Let l = lim
x→ 2
∫
6
∫ 2tdt
2tdt
= lim 6
(x − 2) x→ 2 (x − 2)
0

0 form, as f (2) = 6
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 φ ( x )
1
− f (φ1 (x)) ⋅ φ1′ (x)]
So, l = 2 f (2) ⋅ f ′ (2) = 12 f ′ (2)
[Q f (2) = 6]
l = lim
f ( x)
∴ lim
x→ 2
∫
6
2tdt
= 12 f ′ (2)
x−2
Limit, Continuity and Differentiability 205
function is f (x) = 15 − |x − 10|, x ∈ R
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
3. Given
 − 1, − 2 ≤ x < 0
We have, f (x) =  2
x − 1 , 0 ≤ x ≤ 2
and
and g (x) = | f (x)| + f (|x|)
−2 ≤ x<0
 1,
Clearly, | f (x)| =  2
|x − 1|, 0 ≤ x ≤ 2
1,
−2 ≤ x<0


= − (x2 − 1), 0 ≤ x < 1
 x2 − 1 ,
1 ≤ x≤2

and
x<5
 15 + (x − 5) = 10 + x ,
 15 − (x − 5) = 20 − x , 5 ≤ x < 10

=
 15 + (x − 15) = x , 10 ≤ x < 15
15 − (x − 15) = 30 − x ,
x ≥ 15
[Q f (|x|) = − 1 is not possible as|x| </ 0]
= x2 − 1 , |x| ≤ 2

−2 ≤ x<0
1 + x2 − 1 ,

2
2
= − (x − 1) + x − 1, 0 ≤ x < 1
 x2 − 1 + x 2 − 1 ,
1 ≤ x≤2

4. Let us draw the graph of y = f (x), as shown below
–p
–3p
4
p
2
y= cos x
y= sin x
p
O
p/4
–1
 x2 ,
−2 ≤ x<0

0,
0 ≤ x<1
=
2 (x2 − 1), 1 ≤ x ≤ 2

X
y= min {sin x, cos x}
Now, let us draw the graph of y = g (x), as shown in the
figure.
Clearly, the function f (x) = min {sin x, cos x} is not
− 3π
π
[these are point of
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
differentiable at x =
(–2,4)
X′
–2
X
–1 O
1
1
( y + 2) represent a parabola
2
with vertex (0, − 2) and it open upward]
[ Here, y = 2 (x2 − 1) or x2 =
[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
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
=3 −1 −0 =2
and
Lf ′ (0) = lim (cos x + 1 − 2(x − π )sin x)
x → 0−
Y
Rf ′ (0) = Lf ′ (0)
So, f (x) is differentiable at all values of x.
⇒
K =φ
6.
Key Idea This type of problem can be solved graphically.
{−|x|, −
O
–1 , –1 ,
√2 √2
–1
y=–|x|
1
y=f(x)
1 , –1 ,
√2 √2
}
1 − x2
On drawing the graph, we get the follwong figure.
=1 + 1 −0 =2
Q
2
Y′
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
x→ 0 +
(2,6)
y=x2 y=0
y=2 (x2 –1)
5. We have,
Clearly, f (x) is differentiable everywhere except
possibly at x = 0
[Q f ′ (x) exist for x < 0 and x > 0]
Here, Rf ′ (0) = lim (3 cos x − 1 − 2(x − π )sin x)
[Q |x|2 = x2]
= x2 − 1 , − 2 ≤ x ≤ 2
∴ g (x) = | f (x)| + f (|x|)
From the above definition it is clear that g (x) is not
differentiable at x = 5, 10, 15.
1
f (|x|) = |x|2 − 1, 0 ≤ |x| ≤ 2
206 Limit, Continuity and Differentiability
For| x| ∈ (2, 4]
[ Q graph of y = −| x|is
x ∈ (2, 4]
8 − 2x,
f (x) = 8 − 2|x|= 
8 + 2x, x ∈ [− 4, − 2)
Y
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
X′
–5 –4 –3 –2 –1 O
2
3
4
5
X
Hence, the graph of y = f (x) is
4
8–
1
–2
–1
2x
–4
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.
1
Y′
y=
8+
2x
\[Q x2 + y2 = 1 represent a complete circle]
1

2
−1 < x ≤ −
 − 1−x ,
2

1
1
⇒ f (x) =  −|x|,
−
<x≤
2
2

1
 − 1 − x2 ,
< x<1
2

y=8–2x
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|,
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
Let us first draw the graph of y =|x|and y = x2 as shown
in the following figure.
y=x2
y=|x|
2
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.
1
| 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
f (x + h ) − f (x)
≤ lim 2|(x + h ) − x|2
h→ 0
(x + h ) − x
⇒ lim
1
h→ 0
1
–2
0
–1
1
2
–1
–2
Clearly, y =|x|and y = x2 intersect at x = − 1, 0, 1
Now, the graph of y = max {|x|, x2} for|x| ≤ 2 is
f (x + h ) − f (x)
≤ lim 2| h |2
h→ 0
h→ 0
h
f (x + h ) − f (x)
≤0
⇒ lim
h→ 0
h
⇒ lim
[substituting limit directly on right hand
side and using lim | f (x)| = lim f (x) ]
x→ a
x→ a
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
Now, ∫ ( f (x))2dx = ∫ dx = [x]10 = (1 − 0) = 1
Limit, Continuity and Differentiability 207
10.
We have,
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
π
h 2 cos
−0
π
h
= lim
= lim h ⋅ cos
=0
h→ 0
h→ 0
h
h
f (x) = |x − π|(
⋅ e|x| − 1)sin|x|
−
x
 (x − π )(e − 1)sin x,
x<0

x
f (x) = − (x − π )(e − 1)sin x, 0 ≤ x < π
 (x − π )(ex − 1)sin x,
x≥π

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
π
)(
e
)
cos
x + (ex − 1) sin x
1

+ (x − π ) sin xex , x > π

Clearly,
lim f ′ (x) = 0 = lim f ′ (x)
 π 
(2 + h )2 cos 
 −0
 2 + h
= lim
x→ π +
h→ 0
∴ f is differentiable at x = 0 and x = π
Hence, f is differentiable for all x.
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
and
L { f ′ (2)} = lim
h→ 0
= lim
h→ 0
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)
⇒
13.
PLAN
f ′ (c) = 2 g′ (c)
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.
Description of Situation
As,
and
f (x + h ) − f (x)
h
f (x − h ) − f (x)
L{ f ′ (x)} = lim
h→ 0
−h
R{ f ′ (x)} = lim
h→ 0
π
π
− 22 ⋅ cos
2−h
2
−h

π 
(2 − h )2 −  − cos
 −0
2 − h

= lim
h→ 0
−h
π
π 
− (2 − h )2 ⋅ sin  −

 2 2 − h
= lim
h→ 0
h

πh 
2
(2 − h ) ⋅ sin  −

 2 (2 − h )
−π
= lim
×
=−π
−π
h→ 0
2
2
− h)
(
h×
2 (2 − h )
…(i)
Differentiating Eq. (i) at c , we get
⇒
f ′ (c) − 2 g′ (c) = 0
 π 
(2 + h )2 ⋅ cos 

 2 + h
= lim
h→ 0
h
f (2 − h ) − f (2)
−h
(2 − h )2 ⋅ cos
f and g are differentiable in (0, 1).
h (x) = f (x) − 2 g (x)
h (0) = f (0) − 2 g (0) = 2 − 0 = 2
=0
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
2
2
+ h)
(
h⋅
2 (2 + h )
12. Given, f (0) = 2 = g (1), g(0) = 0 and f (1) = 6
Let
−h
To check differentiability at x = 2,
f (2 + h ) − f (2)
R{ f ′ (2)} = lim
h→ 0
h
x→ 0 +
x→ π −
π
h 2 cos  −  − 0
 h
So, f (x) is differentiable at x = 0.
lim f ′ (x) = 0 = lim f ′ (x)
x→ 0 −
and
f (0 − h ) − f (0)
L {f ′ (0)} = lim
= lim
h→ 0
h→ 0
−h
Thus, f (x) is differentiable at x = 0 but not at x = 2.
(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
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)
are
208 Limit, Continuity and Differentiability
⇒
lim
h→ 0
hn
= −1
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
 n h
⇒ lim  −  ⋅
=1
= − 1 ⇒   lim
 m h → 0  tan h 
h → 0  m  tan h 




 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
n
=1 ⇒ m = n =2
n = 2 and
m
⇒
∴ Domain of f ′ (x) ∈ R − { −1, 1}
sin h − h
18. RHD of sin (| x|) − | x| = hlim
= 1 −1 = 0
→0
h
[Q f (0) = 0]
LHD of sin (| x|) − | x|
sin| − h | − | − h | sin h − h
= lim
=
=0
h→ 0
−h
−h
15. Given, f (1) = f   = f   = K = lim f   = 0
 2
 3
 n
n→ ∞
1
as
⇒
1
1
many
points
f (x) = max { x, x3 } considering
separately, y = x3 and y = x
20. Given,
Now,
Y
⇒
Y
y = |x| – 1
O
X
–1
–1
1
x
 3
f (x) = x
x
x3

1
 2
f ′ (x) = 3x
1
3x2

Y
X
(ii) y = |x| – 1
X′
(–1, –1) B
Y
–1
O 1
graph
in (− ∞ , − 1]
in (−1, 0]
in (0, 1]
in (1, ∞ )
in (− ∞ , − 1]
in (−1, 0]
in (0, 1]
in (1, ∞ )
–1
(i) y = x – 1
the
NOTE y = x 3 is odd order parabola and y = x is always
intersect at (1, 1) and ( −1, − 1).
As, we know that, the function is not differentiable at
sharp edges.
1
(k − 1) sin π (k − h )
−h
[where x = k − h ]
(k − 1) (−1)k − 1 ⋅ sin h π
k
= (−1) (k − 1) π
= lim
h→ 0
−h
in
16. Using graphical transformation.
O
= lim
h→ 0
f (0) = f ′ (0) = 0
Hence,
(k − 1) sin π x
,
x−k
x→ k
 1
f   = 0; n ∈ integers and n ≥ 1.
 n
 1
lim f   = 0 ⇒ f (0) = 0
 n
n→ ∞
Since, there are infinitely
neighbourhood of x = 0.
∴
f (x) = 0
⇒
f ′ (x) = 0
⇒
f ′ (0) = 0
= lim
X
(iii) y = ||x| – 1|
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
2 (– x − 1),

−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.
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.
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.
Limit, Continuity and Differentiability 209
(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,
⇒
 cos x,
cos| x| = 
cos x,
if x < 0
if x ≥ 0
f (x) − f (2)
R f ′ (2) = lim
x → 2+
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

cos x − cos 1 
= lim (x2 − 1) (x − 2) +

−
x −1
x→1 

= 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
RHD at x = 0
⇒ lim
1
=1
(1 + x)2
and
LHD at x = 0
⇒
1
=1
(1 − x)2
x→ 0
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−
and
∴
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
1 + [x]2 ≠ 0
Q
∴
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. We have,
[same approach]
Q Lf ′ (1) = Rf ′ (1). Therefore, function is differentiable
at x = 1.
f (x) − f (2)
Again, Lf ′ (2) = lim
−
x−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
lim
x→ 0
and
( f (0))2 + ( f ′ (0))2 = 85
f : R → [− 2, 2]
(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.
210 Limit, Continuity and Differentiability
(b) By L.M.V.T
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   = −
6  6
12
π
π
π
 π
f   = − sin + C sin
∴
 6
6
6
6
f (b) − f (a )
f (b) − f (a )
f ′ (c) =
⇒| f ′ (c)| =
b−a
b−a
⇒
| f ′ (x0 )| =
f (0) − f (− 4)
f (0) − f (− 4)
=
4
0+4
Range of f is [− 2, 2]
=−
f (0) − f (− 4)
∴
− 4 ≤ f (0) − f (−4) ≤ 4 ⇒ 0 ≤
≤1
4
⇒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
Hence,| f ′ (x0 )| ≤ 1.
Hence, statement is true.
(c) As no function is given, then we assume
 85 x
f (x) = 2 sin 

 2 
(b) f (x) = − x sin x
Now, ( f (0)) + ( f ′ (0)) = (2 sin 0) + ( 85 cos 0)
2
2
2
⇒
2
( f (0))2 + ( f ′ (0))2 = 85
and lim f (x) does not exists.
x→ ∞
Hence, statement is false.
(d) From option b,| f ′ (x0 )| ≤ 1 and x0 ∈ (− 4, 0)
∴
( f ′ (x0 ))2 ≤ 1
Hence, g (x0 ) = ( f (x0 ))2 + ( f ′ (x0 )2 ≤ 4 + 1
[Q f (x0 ) ∈ [−2, 2]]
⇒
x3
, ∀ x ∈ (0, π )
6
x4
− x sin x < − x2 +
6
x4
2
f (x ) <
− x , ∀ x ∈ (0, π )
6
sin x > x −
 85 x
f ′ (x) = 85 cos 

 2 
∴
π
π
1
=−
+ C
12
12 2
⇒
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
Y
g (x0 ) ≤ 5
Now, let p ∈ (−4, 0) for which g ( p) = 5
X′
Similarly, let q be smallest positive number q ∈ (0, 4)
o
π2
X
3π/2
π1
such that g (q) = 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
and f (x))2 ≤ 4 ⇒ ( f ′ (x))2 ≥ 1in ( p, q)
Thus,
⇒
g′ (c) = 0
Y
⇒ Their exists α ∈(0, π) for which f′(α) = 0
It is true
(d) f(x) = − x sin x
f ′(x) = − sin x − x cos x
f′ f + f′ f′ ′ = 0
f ′ ′ (x) = − 2 cos x + x sin x
π
π
π
π
f ′ ′   = , f   = −
 2 2
 2
2
So, f (α ) + f ′ ′ (α ) = 0 and f′ (α ) ≠ 0
Hence, statement is true.
f (x) sin t − f (t ) sin x
27. Given, lim
= sin 2 x
t→ x
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
y=–x
∴
π
π
f ′ ′   + f   = 0
 2
 2
It is true.
28. As, g ( f (x)) = x
Thus, g (x) is inverse of f (x).
⇒
⇒
∴
g ( f (x)) = x
g′ ( f (x)) ⋅ f ′ (x) = 1
g′ ( f (x)) =
1
f ′ (x)
…(i)
[where, f ′ (x) = 3x2 + 3]
Limit, Continuity and Differentiability 211
f (x) = 2, then
x3 + 3x + 2 = 2
⇒
x=0
i.e. when x = 0, then f (x) = 2
1
at (0, 2)
∴
g′ ( f (x)) = 2
3x + 3
1
g′ (2) =
⇒
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.
When
and
= (| x| + |4x − 7|) 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

∴
∴
|x|sin (|x3 + x|) = x sin (x3 + x), ∀ x ∈ R
⇒
3 ≤ x<2
x=2
Y
1
X′
–1/2
X
0
1
√2
√3
2
–1
–2
Y′
…(i)
Clearly, f (x) is discontinuous at 4 points.
∴ Option (b) is correct.
Graph for g ( x)
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)
Hence, f (x) is always continuous and differentiable.
− 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

x=2
 1,
− 1 /2 ≤ x < 0
0 ≤ x<1
1≤x< 2
2 ≤x< 3
–3
∴
which is clearly sum and composition of differential
functions.
30. Here,
15x + 21,
9x − 21,

6x − 14,
g (x) = 
3x − 7,
0,

5x − 7,
Now, the graphs of f (x)and g (x)are shown below.
Graph for f ( x)
29. 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 )}
g (x) = | x| f (x) + |4x − 7| f (x)
X′
–1/2
0
X
3 √ 3–71
3 √ 2–7
√2
√3
2
–8
–12
–21
Y′
Clearly, g (x) is not differentiable at 4 points, when
x ∈ (− 1 / 2, 2).
∴ Option (c) is correct.
212 Limit, Continuity and Differentiability
31.
 g (x), x > 0

Here, f (x) =  0,
x=0
− g (x), x < 0

 g′ (x), x ≥ 0
f ′ (x) = 
− g′ (x), x < 0
Given that, f : [a , b] → [1, ∞ )
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
(
),

+
Now, (hof )′ (0) = lim
e
= 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
h
→
→
0
0
g (x)
|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
32. 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
33. 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.
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
but
g′ (a + ) = f (a ) ≥ 1
−
⇒
( 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
−

,
x<a
 0
 x
g (x) = ∫ f (t )dt , a ≤ x ≤ b
a
 b
∫ f (t )dt , x > b
 a
f (x) − f (a )
f (x) − f (a )
= lim
+
x−a
x−a
x→ a
−
+
f ′ (a ) = f ′ (a )
[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
−
cos
x
,
−
<
x
≤
0
34. f (x) = 
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 213
π

x≤−
 − 1,
2

π
sin x, − < x ≤ 0
f ′ (x) = 
2

1,
0 < x≤1
 1
,
x>1

 x
37. 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
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, LHD = 1, RHD = 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|
38. Here, f (x) =  x2 3x 13
 4 − 2 + 4 , x < 1
∴ RHL at x = 1, lim |1 + h − 3| = 2
3
2
35. f (x + y) = f (x) + f ( y), as f (x) is differentiable at x = 0.
⇒
f ′ (0) = k
Now,

0
0 form
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 ∴
lim
…(i)
f (x + h ) − f (x)
f ′ (x) = lim
h→ 0
h
f (x) + f (h ) − f (x)
= lim
h→ 0
h
f (h )
= lim
h→ 0 h
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,
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

 differentiable at x = 1.
1 3
∴
LHD at x = 1 ⇒ − = − 1

2 2
RHD at x = 3 ⇒ 1 
not differentiable at x = 3.
Again,
LHD at x = 3 ⇒ − 1
39. We know that, f (x) = 1 + |sin x|could be plotted as,
(1) y = sin x
…(i)
Y
Hence, (b) and (c) are correct.
1
36. Here, f (x) = min { 1, x2, x3 } which could be graphically
shown as
X′
y = x3
Y
y=
−π
O
1
X
Hence, (a) and (d) are correct answers.
3π
y = sin x
X
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.
2π
−1
y=1
X′
π
O
x2
X′
–2 π
–π
O
Y′
y=|sin x|
π
2π
3π
X
214 Limit, Continuity and Differentiability
y = 1+|sin x|
(3)
…(iii)
Y
43. We know, [x] ∈ I , ∀ x ∈ R.
–2
y=1 +|sin x|
–1
X′
–π
2π
π
O
X
3π
Y′
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
40. Since, x + | y| = 2 y ⇒ 
 x − y = 2 y, when y < 0
⇒
D. |x − 1| + |x + 1| = 2 in (−1, 1)
⇒ The function is continuous and differentiable in
(−1, 1).
 y = x, when y > 0 ⇒ x > 0

 y = x /3, when y < 0 ⇒ x < 0
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.
44. Given, F (x) = f (x) ⋅ g (x) ⋅ h (x)
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 )
where,
…(i)
F ′ (x0 ) = 21 F (x0 ), f ′ (x0 ) = 4 f (x0 )
g′ (x0 ) = − 7 g (x0 ) and h ′ (x0 ) = k h (x0 )
which could be plotted as,
On substituting in Eq. (i), we get
Y
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°
X′
O
y =x , x < 0
3
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

41. We have, lim
0 form
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)
⇒
21 = 4 − 7 + k , [using F (x0 ) = f (x0) g (x0 ) h (x0 )]
∴
k = 24

x

0,
45. Given, f (x) = 1 + e1/ x
, x ≠0
x=0
h
−0
1
1 + e1/ h
= lim
=0
∴ Rf ′ (0) = f ′ (0+ ) = lim
h→ 0
h → 0 1 + e1/ h
h
−h
−0
1
+
e−1/ h
−
and Lf ′ (0) = f ′ (0 ) = lim
h→ 0
−h
1
1
=
=1
= lim
1
h→ 0
1+0
1 + 1/ h
e
∴
f ′ (0+ ) = 0 and f ′ (0− ) = 1
1

(x − 1)2 sin
− | x|, if x ≠ 1
46. Given, f (x) = 
(x − 1)
if x = 1
−1 ,

Statement II is not a correct explanation of Statement I.
1

2
 (x − 1) sin (x − 1) − x, 0 ≤ x − {1}

1

f (x) = (x − 1)2 sin
+ x,
x<0
(x − 1)

,
x=1
−1


42. A. x|x| is continuous, differentiable and strictly
Here, f (x) is not differentiable at x = 0 due to|x|.
x→ 0
⇒ Statement I is true.
Statement II f ′ (x) = g′ (x) sin x + g (x) cos x
⇒
f ′ (0) = g (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.
As,
Thus, f (x) is not differentiable at x = 0.
47. It is always true that differential of even function is and
odd function.
48. Since, f (x) is differentiable at x = 0.
⇒ It is continuous at x = 0.
Limit, Continuity and Differentiability 215
i.e.
lim f (x) = lim f (x) = f (0)
x → 0+
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
 2 
h→ 0
2
x → 0−
Here, lim f (x) = lim
+
∴
b sin −1
⇒
c a 1
= =
2 2 2
a =1
R f ′ (0+ ) = L f ′ (0– )
eh/ 2 − 1 1
−
2
h
Rf ′ (0+ ) = lim
h→ 0
h
[Q a = 1]
4−c
2
⇒
=
⇒
f (x) is differentiable at x = α.
Clearly,
1−
c
4
lim g (x) = f ′ (α )
x→ α
g (x) is continuous at x = α.
is
continuous and differentiable at all points except
possibly at x = 1 and 2.
Continuity at x = 1,
and
LHL = lim f (x) = lim (1 − x)
64b2 = (4 − c2)
x → 1−
x → 1−
= lim [1 − (1 − h )] = lim h = 0
h→ 0
and
1
 1 2 
 1
where, f   = 1 +  cos −1   − 1 = f ′ (0)
 n π 
 n
n
2 
π
2
− 1 +  = 1 −
2
π 
π
…(ii)
∴ From Eqs. (i) and (ii), we get
2
2
 1
lim
(n + 1) cos −1   − n = 1 −
 n
n→ ∞ π
π
x →1+
h→ 0
h→ 0
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
f (1 + h ) − f (1)
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
h→ 0
Since, L [ f ′ (1)] = Rf ′ (1), therefore f is differentiable at
x = 1.
Continuity at x = 2,
50. 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
x →1+
= lim − h ⋅ (1 − h ) = 0
∴

 1 
nf   
given, f ′ (0) = nlim
 n 
→∞

2
1
…(i)
lim (n + 1) cos −1 − n = f ′ (0)
∴
n→ ∞ π
n
2
where, f (x) = (1 + x) cos −1 x − 1, f (0) = 0
π

2 
−1
⇒
f ′ (x) = (1 + x)
+ cos −1 x
π 
1 − x2


h→ 0
RHL = lim f (x) = lim (1 − x) (2 − x)
= lim [1 − (1 + h )] [2 − (1 + h )]
1

2 
 1
 1
= lim n  1 +  cos −1   − 1 = lim nf  
 n
 n
n→ ∞ π 
n
 n→ ∞
f ′ (0) =
f ′ (α ) = g (α )
x<1
(1 − x),

f (x) = (1 − x)(2 − x), 1 ≤ x ≤ 2
(3 − x),
x>2

2
2
1
49. Here, lim (n + 1) cos −1   − n
 n
n→ ∞ π
⇒
x→ α
51. It is clear that the given function
1
8
a =1
f ′ (α ) = lim g (x) ⇒
b /2
64b2 = (4 − c2)
⇒
f (x) − f (α )
= lim g (x)
x→ α
x−α
Hence, f (x) is differentiable at x = α, iff g (x)
continuous at x = α.
 c − h 1
b sin −1 
 −
 2  2
Lf ′ (0– ) = lim
=
h→ 0
−h
b
lim
x→ α
⇒
⇒
2eh/ 2 − 2 − h 1
= lim
=
h→ 0
8
2h 2
∴
⇒
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
x=α
 f ′ (α ) ,
Also, it is differentiable at x = 0
and
f (x) − f (α )
= g (x)
(x − α )
⇒
x → 0−
[given]
LHL = lim f (x) = lim (1 − x) (2 − x)
x → 2−
x → 2−
= lim [1 − (2 − h )] [(2 − (2 − h )]
h→ 0
216 Limit, Continuity and Differentiability
= lim (− 1 + h ) h = 0
f (x + h ) − f (x)
h
 2x + 2h 
f
 − f (x)
 2 
= lim
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
Again, f ′ (x) = lim
h→ 0
h→ 0
RHL = lim f (x) = lim (3 − x)
and
x → 2+
x → 2+
= lim [3 − (2 + h )] = lim (1 − h ) = 1
h→ 0
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
− − + 
52. Given, f (x) = xe  x x  , x < 0

, x=0
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
⇒
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

h
h→ 0

[Q f (0) = 1]
∴
⇒
 x
2 f   = f (x) + 1
 2
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 (0) = 1,
54. 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)
53. Given, f 
, ∀x, y ∈ R
 =
 2 
2
 x
f (x) = 2 f   − 1, ∀ x, y ∈ R
 2
f (x) = − x + k, where, k is a constant.
But
therefore f (0) = −0 + k
∴ f (x) is continuous at x = 0
⇒
∫
⇒
⇒
⇒
f (0) = 0
On putting y = 0, we get
 x f (x) + f (0) 1
f   =
= [1 + f (x)]
 2
2
2
f ′ (x) = −1, ∀x ∈ R
f ′ (x)dx = ∫ − 1 dx
…(i)
[from Eq. (i)]
On integrating both sides between 0 to x, we get
x f ′ (x)
∫ 0 f (x) dx = 2x
⇒
…(ii)
f ′ (x) = 2 f (x)
f ′ (x)
=2
f (x)
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 217
−2 ≤ x ≤ 0
 −1 ,
(x − 1), 0 < x ≤ 2
Since, x ∈ [−2, 2]. Therefore,|x|∈ [0, 2]
55. y = [x] + |1 − x|, − 1 ≤ x ≤ 3
58. 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,

0 ≤ x<1
g (x) =  0,
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} .
56. 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 
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

59. Given,
f ( y) − f (x)
 ≤ lim (x − y)
lim 
y−x  y→ x
y → x
⇒
| f ′ (x)|2 ≤ 0
which is only possible, if| f ′ (x)| = 0
∴ f ′ (x) = 0
or f ′ (x) = Constant
57. 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}
…(i)
2
⇒
 − 1 , −2 ≤ x ≤ 0

g ′ (x) =  0,
0 ≤ x<1
 2,
1 ≤ x≤2

X
⇒
 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,
RHL = lim (3 − x) = 2
[Q f (− h ) = f (h )]
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}.
218 Limit, Continuity and Differentiability
x −1

2x2 − 7x + 5 , when x ≠ 1
60. 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 }




2h 2
2
= lim 
 =−
h→ 0  3 h (−3 h + 2 h 2 )
9
f (1 − h ) − f (1)
−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]
 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
63.
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
LHD = lim
h→ 0
= lim
h→ 0
2h 2
2
= − ∴ LHD = RHD
9
−3h (2h 2 + 3h )
2
Hence, required value of f ′ (1) = − .
9
61. Given, f (x) = x tan −1 x
Using first principle,
 f (1 + h ) − f (1) 
f ′ (1) = lim 

h→ 0 
h
Hence, number of points of non-differentiability of h (x)
is 3.
64. 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 
 (1 + h ) tan −1 (1 + h ) − tan −1 (1) 
= lim 

h→ 0
h


∴ lim
 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  4
h→ 0 
(2 + h ) ⋅

2 + h 

−1  h  

 tan 
 2 + h  π 1 π
1 
+ = +
= lim
h
h→ 0 2 + h 
 4 2 4


(2 + h ) 

π
62. g (x) = ∫ 2 (f ′ (t ) cosec t − cot t cosec t f (t ))dt
x
π
 π
g (x) = f   cosec − f (x) cosec x
 2
2
f (x)
⇒ g (x) = 3 −
sin x
∴
h (x) is not differentiable at x = ± 1 and 0.
As, h (x) take sharp turns at x = ± 1 and 0.
ax4 + bx3 + (c + 1) x2 + dx + e
=2
x→ 0
x2
⇒
c + 1 = 2, d = 0, e = 0
c=1
From Eqs. (i) and (ii), we get
4a + 3b = − 2
and
⇒
∴
⇒
⇒
32a + 12b = − 4
1
a = and b = − 1.
4
x4
− x3 + x2
4
16
p(2) =
−8 + 4
4
p(x) =
p(2) = 0
Topic 8 Differentiation
1. 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
... (i)
... (ii)
[given]
Limit, Continuity and Differentiability 219
On multiplying both sides by x, we get
On differentiating both sides w.r.t. x, we get
dy
dy
ey
+x
+ y=0
dx
dx
 y 
dy
⇒
=− y

dx
 e + x
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
Again differentiating Eq. (ii) w.r.t. ‘x’, we get
2
d 2y
d 2y dy dy
y  dy
ey
+
=0
e
+
+x 2+


2
 dx
dx dx
dx
dx
C1 + (22) nC 2 + (32) nC3 + ... + (n 2) nC n
n
n −1
= n (2
For n = 20, we get
+ (n − 1) 2
n−2
)
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
20
C 20
Now, on putting x = 0, y = 1 and
 sin x − cos x 
−1  tan x − 1 

 = tan 
 tan x + 1
 sin x + cos x

π 

 tan x − tan
4 

π



 1 +  tan  (tan x)


4
π
π


 tan  x −   = x −


4
4
d 
π
x −  = 2
4
dx 
1
d 2y
= 2
2
dx ( 0, 1) e
⇒
 dy d 2y
 1 1
, 2  at (0, 1) is  − , 2

 e e
 dx dx 
4. 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
∴
= f ′ ( f ( f (1))) ⋅ f ′ ( f (1)) ⋅ f ′ (1) + 2 f (1) f ′ (1)
x=1
dy
= f ′ ( f (1)) ⋅ f ′ (1) ⋅ (3) + 2(1)(3)
dx x = 1
[Q f (1) = 1 and f′ (1) = 3]
2


 3 cot x + 1 
 3 cos x + sin x 
2 y =  cot−1 

  =  cot−1 
 cot x − 3  
 cos x − 3 sin x  


Given equation is
…(i)
2
[dividing each term of numerator and
denominator by sin x]
π




 cot cot x + 1 
−1
6
=  cot 


 cot x − cot π  


6 

 π

=  cot−1  cot − x  

 6

Key Idea Differentiating the given equation twice w.r.t. ‘x’.
ey + xy = e
2
 d 2y  1  1
d 2y
1  1
e
+
−
+ 0 2  +  −  +  −  = 0


2
 e
dx
 dx   e  e
5. Given expression is

 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)
3.
dy
1
= − in
dx
e
= f ′ (1) ⋅ (3) ⋅ (3) + 6
= (3 × 9) + 6 = 27 + 6 = 33
Then,
=2 ×
e1
So,
π 


= tan −1 tan  x −  

4 

 tan A − tan B

= tan ( A − B)
Q
tan
tan
A
B
+
1


 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
f (x) = tan
…(iv)
Eq. (iv), we get
[dividing numerator and denominator
 π
by cos x > 0, x ∈ 0,  ]
 2
−1
…(iii)
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
2. Let f (x) = tan −1 
= tan −1
…(ii)
2
2
π


Q cot 6 = 3 

cot A cot B + 1 
Q cot( A − B) =

cot B − cot A 

220 Limit, Continuity and Differentiability



=
 π

= 4 + e2 − e log e (1) = 4 + e2 − 0
= e2 + 4
2
π

 − x ,
6

π
6
2
π
π
π

+  − x  ,
<x<
6

6
2
0<x<
2
 π
π

  − x , 0 < x <

 6
6
⇒2y = 
2
7
π
π
π


<x<
− x ,
 6

6
2
π
 π

2 − x (−1), 0 < x <
dy   6

6
⇒2
=
π
π

dx   7π
<x<
− x (−1),
2

  6
2
6
π
π

dy  x − 6 , 0 < x < 6
=
⇒
dx x − 7π , π < x < π

6
6
2
6. Given equation is
(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)
m
[Q log e n = 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)
... (i)
2
2x
[using Eq. (ii)]
8. We have,
f (x) = x3 + x2f ′ (1) + xf ′ ′ (2) + f ′ ′ ′ (3)
… (i)
⇒ f ′ (x) = 3x2 + 2xf ′ (1) + f ′ ′ (2)
… (ii)
⇒ f ′ ′ (x) = 6x + 2 f ′ (1)
… (iii)
⇒ f ′ ′ ′ (x) = 6
⇒ f′ ′ ′ (3) = 6
Putting x = 1 in Eq. (i), we get
… (iv)
f ′ (1) = 3 + 2 f ′ (1) + f ′ ′ (2)
and putting x = 2 in Eq. (ii), we get
…(v)
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
⇒
f′ (1) = − 5
[using Eq. (v)]
⇒
f′ ′ (2) = 12 + 2 (− 5) = 2
∴
f (x) = x3 + x2f ′ (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
d 2y d  dy
d  dy dt
=
  =
 ⋅
dx2 dx  dx dt  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
d 2y 
π
4 = 1
Now,
=
 at t =  =
3
4
3(2 2 ) 6 2
dx2 
and
dy  x log e (2x) − log e 2
=


dx 
x
7. We have, x log e (log e x) − x2 + y2 = 4, which can be
written as
y2 = 4 + x2 − x log e (log e x)
dy 2e − 1 − 0
2e − 1
=
=
dx 2 e2 + 4 2 e2 + 4
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
(1 + log e (2x))2
1
1 + log e (2x) − 1 − log e 2
x
=
(1 + log e (2x))2
So, (1 + log e (2x))2
∴ At x = e and y = e + 4,
9. We have, x = 3 tan t and y = 3 sec t
On differentiating ‘y’ w.r.t. ‘x’, we get
(1 + log e (2x))1 − (x + log e 2)
[Q y > 0]
2

π + θ , − π < θ < 0 
Q cot−1 (cot θ ) =  θ ,
0 <θ < π 



θ − π , π < θ < 2π 



dy
=
dx
y = e2 + 4
⇒
… (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)
dy
log
 
ex
… (ii)
⇒
  =
 dx
2y
Now, at x = e, y2 = 4 + e2 − e log e (log e e)
[using Eq. (i)]
3/ 2
 6x x 
− 1  2 ⋅ (3 x ) 
tan
=


3/ 2 2 
 1 − 9x3 
1 − (3x ) 
10. Let y = tan − 1 
∴
∴

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 =
3/ 2 2
dx
2
1 + 9x3
1 + (3x )
9
g (x) =
1 + 9x3
Limit, Continuity and Differentiability 221
  x 
 x 1
+ 2  g    ⋅ g′   ⋅ = 0 [from Eq.(i)]
 2 2
  2 
11. Here, g is the inverse of f (x).
⇒
fog (x) = x
∴
On differentiating w.r.t. x, we get
1
f ′{ g (x)} × g′ (x) = 1 ⇒ g′ (x) =
f ′ ( g (x))
⇒
16. Let, g (x) = f (x) − x2

1 
Q f ′ (x) =

1 + x5 

1
=
1
1 + { g (x)}5
g′ (x) = 1 + { g (x)}5
12. Given, y = sec (tan− 1 x )
x
1+
x
θ
y = sec θ = 1 + x2
∴
1
On differentiating w.r.t. x, we get
dy
1
=
⋅ 2x
dx 2 1 + x2
At x = 1,
dy
1
=
dx
2
On differentiating Eq. (i) w.r.t. x, we get
⇒
g (x + 1) = log x + g (x)
... (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
1
1
−4


... (ii)
∴ g′′  x +  − g′′  x −  =


2
2 (2x − 1)2
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
(2N − 1) 

−1
On differentiating w.r.t. x, we get
2x + 2 yy′ = 0
⇒
⇒
g′ (x) = − f (x)
Also,

F (x) =  f

x3
19. Given,
⇒
0
p3
p2
f ′ (x) = 6
−1
0
p
2
3
p
x3
+ 0
p
p
sin x cos x
0
p
x3
+ 6
0
3x2 cos x − sin x
f ′ (x) = 6
−1
0
p
2
  x 
 x 1
F ′ (x) = 2  f    ⋅ f ′   ⋅
 2 2
  2 
−1
3x2 cos x − sin x
[Q g (x) = f ′ (x), given]…(i)
2
sin x cos x
f (x) = 6
p
On differentiating w.r.t. x, we get
⇒
  x 
 x 
   + g   
 2 
  2 
x + yy′ = 0.
Again, differentiating w.r.t. x, we get
1 + y ′ y ′ + yy ′′ = 0
⇒
1 + ( y′ )2 + yy′′ = 0
f ′′ (x) = − f (x)
d
{ f ′ (x)} = − f (x)
dx
 dy
=1
 
 dx ( 0, 1)
18. Given, x2 + y2 = 1
1
dx
 dy
14. Since,
=
= 
dy dy / dx  dx
−1
d  dx
d  dy dx
⇒
 
  =
dy  dy dx  dx dy
 d 2y  dy −3
 d 2y  dy − 2 dx
d 2x
⇒
=
−
=
−
 2  






2
dy2
 dx   dx
 dx   dx  dy
⇒
dy 2 y (x + y) − 1
=
dx 1 − 2 (x + y) x
⇒
i.e.
15. Since,
…(i)
∴ At x = 0, ⇒ log ( y) = 0 ⇒ y = 1
dy
∴ To find
at (0, 1)
dx
dy
1 
dy
+ 2y ⋅ 1
 = 2x
1 +
x+ y 
dy
dx
13. Since, f (x) = e g ( x ) ⇒ e g ( x + 1)= f (x + 1) = xf (x) = xe g ( x )
and
⇒ 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)
17. Given that, log (x + y) = 2xy
2
tan − 1 x = θ
x = tan θ
Let
⇒
F (x) is constant ⇒ F (10) = F (5) = 5
⇒
p2
p3
6x − sin x − cos x
f ′ ′ (x) = 6
−1
0
+0+0
p
p2
p3
2
0
p3
sin x cos x
−1
0
0
0
222 Limit, Continuity and Differentiability
− cos x sin x
−1
0 +0+0
6
and f ′ ′ ′ (x) = 6
p2
p
−1
6
∴
f ′ ′ ′ (0) = 6 −1
p p2
∴
p3
g (x) = f [ f (x)] =||x − 2|− 2|
x>2
When,
0
g (x) = |(x − 2) − 2| = | x − 4| = x − 4
0 = 0 = independent of p
p3
∴
g′ (x) = 1 when x > 2
1
24. Let u = sec−1  − 2  and v = 1 − x2
 2x − 1
Put
x = cos θ
∴
u = sec−1 (− sec 2 θ ) and v = sin θ
20. Since, y2 = P (x)
On differentiating both sides, we get
2 yy1 = P ′ (x),
⇒
Again, differentiating, we get
v = sin θ
du
= −2
dθ
dv
d
= cos θ
dθ
du
2
 du 
,  
=−
= −4
dv
cos θ  dv  θ = π /3
⇒
2 y3 y2 + 2 y2y12 = y2P ′ ′ (x)
⇒
2 y3 y2 = y2P ′ ′ (x) − 2 ( yy1 )2
⇒
{ P ′ (x)}2
2 y y2 = P (x) ⋅ P ′ ′ (x) −
2
an
3
⇒
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 
21. Given, xexy = y + sin 2 x
⇒
f (x) = log x (log x)
log (log x)
f (x) =
log x
25. Given,
∴
On differentiating both sides, we get
1
 1 1
(log x) 
⋅  − log (log x) ⋅
 log x x
x
f ′ (x) =
(log x)2
…(i)
f ′ (e) =
∴
On putting x = 0, we get
0 . e0 = y + 0
⇒ f ′ (e) =
y=0
1
 1 1
1 ⋅  ⋅  − log (1) ⋅
 1 e
e
(1)2
1
e
f1 (x)
On differentiating Eq. (i) both sides w.r.t. x, we get
 dy
 dy
1 . exy + x ⋅ exy  x ⋅
+ y =
+ 2 sin x cos x
 dx
 dx
On putting x = 0, y = 0, we get
 dy 
e0 + 0(0 + 0) =  
 dx 
F ′ (x) = g1 (x)
+ 2 sin 0 cos 0
f1 (x)
 dy 
=1
 dx 
0, 0
⇒
f ′ (x) = 2x,
 − 2x,
f ′ ′ (x) = 2,
− 2 ,
x>0
x<0
x>0
x<0
Therefore, f (x) is twice differentiable for all x ∈ R − {0}.
h1 (x) h2 (x)
f2 (x)
f3 (x)
g3 (x)
h3 (x)
f1 (x)
f2 (x)
f3 (x)
g2 (x)
g3 (x)
g2 ′ (x)
g3 ′ (x) + g1 (x)
h1 (x)
h2 (x)
h3 (x)
⇒
Therefore,
g2 (x)
+ g1 ′ (x)
⇒
 2
if x ≥ 0
f (x) = x , 2
if x < 0
− x ,
f (x) is not differentiable at x = 0 but all R − {0} .
f3 (x)
h1 (x) h2 (x) h3 (x)
f1 ′ (x) f2 ′ (x) f3 ′ (x)
∴
22. Given, f (x) = x|x|
f2 (x)
26. Given, F (x) = g1 (x) g2 (x) g3 (x)
( 0, 0)
⇒
[Q sec–1 (− x) = π − sec−1 x]
u = π − 2θ
and
2 yy2 + 2 y12 = P ′ ′ (x)
⇒
f (x) = | x − 2|
23. Given,
h1 ′ (x) h2 ′ (x) h3 ′ (x)
F ′ (a ) = 0 + 0 + 0 = 0
[Q fr (a ) = gr (a ) = hr (a ) ; r = 1, 2, 3]
 2x − 1
y= f  2

 x + 1
f ′ (x) = sin 2 x
 2x − 1 d
dy
= f′  2
⋅
dx
 x + 1 dx
27. Given,
and
∴
 2x − 1

 2
 x + 1
 2x − 1  (x2 + 1) ⋅ 2 − (2x − 1) (2x) 
= sin 2 2
 ⋅

(x2 + 1)2
 x + 1 

Limit, Continuity and Differentiability 223
 2x − 1 −2x2 + 2x + 2
= sin 2  2
⋅
(x2 + 1)2
 x + 1
=
28.
y=
 2x − 1
−2 (x2 − x − 1)
sin 2  2

(x2 + 1)2
 x + 1
c
ax2
bx
+
+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)

 a
x2
x3
=
+ 1 ⇒ y =

(x − c) (x − b)  x − 1
(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  a
b
c 
= 
+
+

y x  a − x b − x c − x
π
sin x
y) 2
3
29. Here, (sin
+
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

π 
x  −1
2 
dy
⋅ cos y ⋅
dx
2
x
3
2
2 ⋅ sec {log (x + 2)}
+
⋅
+
(x + 2)
2 (2|x|) 4x2 − 1
+ 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
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
=
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)
31. Let φ (x) = A (α )
B(x)
B (α )
C (x)
C (α )
…(i)
A ′ (α ) B ′ (α ) C ′ (α )
∴ 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).
⇒
(x − α ) is a factor of φ′ (x) also.
or we can say (x − α )2 is a factor of f (x).
⇒
φ (x) is divisible by f (x).
2x  

 1 + x2 

32. Given,y = (log cos xsin x) ⋅ (logsin x cos x)−1 + sin −1 
∴
⇒

2
 log e (sin x)
−1  2 x 
y=

 + sin 
x
log
(cos
)


 1 + x2
e
(log e (cos x) ⋅ cot x
 log e (sin x) + loge ( in x) ⋅ tan x )
dy
=2 
⋅
dx
{log e (cos x)}2
 log e (cos x)
 dy
⇒  
 dx  x = π 

π −3
2
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
30. Given,
⇒
Given that, α is repeated root of quadratic equation
f (x) = 0.
1  1
1  1
1 
y ′ 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)
 sin
π 


+ sin x (sin y)

2 
dy
= 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
and
4

 1 
 2 ⋅ log   
 2 
2

+
= 2 1 ⋅
2
2 
  log 1   1 + π


 
16
2  

8
32
=−
+
log e 2 16 + π 2


2
+
x2
1
+


x 

 a + bx

33. Given, (a + bx) ey/ x = x ⇒ y = x log 
⇒
y = x [log (x) − log (a + bx)]
…(i)
224 Limit, Continuity and Differentiability
On differentiating both sides, we get
1
dy
b 
=x −
 + 1 [log (x) − log (a + bx)]
dx
 x a + bx
⇒x
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)]


dy
a
= x2 
 + y
dx
 x (a + bx)
⇒ x y1 =
36. Given, y =
ax
+ y
a + bx
…(ii)
 (a + bx) ⋅ 1 − x ⋅ b 
x y2 + y1 = a 
 + y1
(a + bx)2


a x
(a + bx)2
x3 y2 =
⇒
 ax 
x3 y2 = 

 (a + bx)
⇒
x3 y2 = (xy1 − y)2
⇒
x3
d 2y  dy

= x
− y

dx2  dx
2
[from Eq. (ii)]
2
⇒
34. Given, h (x) = [ f (x)]2 + [ g (x)]2
⇒
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 

2 2
⇒
[Q f ′ (x) = g (x) and g′ (x) = − f (x)]
∴ h (x) is constant.
and
+ (tan x) , then
x
⇒
3
y = u + v, where u = ex sin x and v = (tan x)x
dy  du dv
=
+

dx  dx dx
⇒
⇒
As
⇒ h (10) = h (5) = 11
35. Since, y = e
F (x) 1
=
G (x) 14
F ′ (x) 1
lim
=
x→1 G′ (x)
14
x→1
= 2 [ f (x) ⋅ g (x) − g (x) ⋅ f (x)]
x sin x 3
5

− 2 sin (4x + 2), x < 1
dy  3 (1 − x)2
=
5
dx −
− 2 sin (4x + 2), x > 1
 3 (x − 1)2
37. Here, lim
h ′ x = 2 f (x) ⋅ f ′ (x) + 2 g (x) ⋅ g ′ (x)
=0
 5x
2
3(1 − x) + cos (2x + 1), x < 1
y=
5x

+ cos 2(2x + 1), x > 1
3(x − 1)
⇒
Again, differentiating both sides, we get
∴
…(i)
3
Here, u = ex sin x and log v = x log (tan x)
On differentiating both sides w.r.t. x, we get
3
du
= ex sin x ⋅ (3x3 cos x3 + sin x3 )
dx
and
5x
+ cos 2(2x + 1)
3|1 − x|
Given,
…(ii)
1 dv x ⋅ sec2x
⋅
=
+ log (tan x)
v dx
tan x
dv
= (tan x)x [2x ⋅ cosec (2x) + log (tan x)] …(iii)
dx
Download Chapter Test
http://tinyurl.com/y6dl84lx
∴
or
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
(b) | 6α + 2β | = 9
(c) | 2α + 6β | = 19
(d) | 2α + 6β | = 11
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)
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)
1
(a)  , − 1

3
1 

(c) − , 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 ?
(c) 4
8. The normal to the curve y(x − 2) (x − 3) = x + 6 at the
point, where the curve intersects the Y -axis passes
through the point
(2017 Main)
1
1
(a)  − , − 
 2
2
1
1

(c)  , − 
2
3
1
(b)  ,
2
1
(d)  ,
2
 1 + sin x 
 π
 , x ∈ 0,  .
 2
 1 − sin x 
A normal to y = f (x) at x =
(b) (2, − 2)
(d) (2, − 1)
4. The tangent to the curve y = x2 − 5x + 5, parallel to the
line 2 y = 4x + 1, also passes through the point
(2019 Main, 12 Jan II)
1 7
(a)  , 
 4 2
1
(c)  − , 7
 8 
7 1
(b)  , 
 2 4
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)
1 7
(c)
6 3
(d)
1
2
1

2
1

3
9. Consider f (x) = tan −1 
(2019 Main, 9 April I)
(a) (−2, 2)
(c) (−2, 1)
(2018 Main)
7
(b)
2
9
(d)
2
2
1
(b)  , 1
3 
1
(d) − , − 1

 3
8
17
7. If the curves y2 = 6x, 9x2 + by2 = 16 intersect each other
(a) 6
3
5
(b)
6
(d)
at right angles, then the value of b is
2. Let S be the set of all values of x for which the tangent to
1 7
(a)
3 3
4
9
point
(a) (0, 0)
π
also passes through the
6
(2016 Main)
2π 
(b)  0,


3
π
(c)  , 0
6 
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)
4
(a)  ±
, − 2


3
(c) (0, 0)
 11 
(b)  ±
, 0
3 

4
(d)  ±
, 2

3 
(2002, 2M)
226 Application of Derivatives
12. If the normal to the curve y = f (x) at the point
(3, 4) makes an angle
f ′ (3) is equal to
3π
with the positive X-axis, then
4
(2000, 1M)
(a) –1
(c) 4/3
(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
(1983, 1M)
(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
Analytical & Descriptive Questions
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).
19. Tangent at a point P1 {other than (0, 0)} on the curve
2
tangents are parallel to the line 8x = 9 y, are (1999, 2M)
2
(b)  − ,
 5
2
(d)  , −
5
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)
P(−2, 0) and cuts the Y-axis at a point Q, where its
gradient is 3. Find a, b, c.
(1994, 5M)
14. On the ellipse 4x + 9 y = 1, the point at which the
2 1
(a)  , 
 5 5
2
1
(c)  − , − 
 5
5
16. Let C be the curve y3 − 3xy + 2 = 0. If H is the set of
18. The curve y = ax3 + bx2 + cx + 5, touches the X-axis at
Objective Questions II
(One or more than one correct option)
2
Fill in the Blank
1

5
1

5
y = x3 meets the curve again at P2. The tangent at P2
meets the curve at P3 and so on.
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.
15. If the line ax + by + c = 0 is a normal to the curve xy = 1,
then
(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.
(a) a > 0, b > 0
(b) a > 0, b < 0
(c) a < 0, b > 0
(d) a < 0, b < 0
(1986, 2M)
(1985, 5M)
Integer 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
(2019 Main, 10 April II)
(a)
1
9π
(b)
1
18 π
(c)
1
36 π
(d)
5
6π
2. Let f (x) = e − x and g (x) = x − x, ∀ x ∈ R. Then, the set
x
2
of all x ∈ R, where the function h (x) = ( fog ) (x) is
(2019 Main, 10 April II)
increasing, is
1
(a)  0,  ∪ [1, ∞ )
 2 
(c) [0, ∞ )
−1
1
(b)  −1,  ∪  , ∞ 

2   2 
−1
(d)  , 0 ∪ [1, ∞ )
 2 
3. A water tank has the shape of an inverted right circular
 1
cone, whose semi-vertical angle is tan −1   . 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
(2019 Main, 9 April II)
is
2
π
1
(c)
15π
1
5π
1
(d)
10π
(a)
(b)
4. Let f : [0, 2] → R be a twice differentiable function such
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
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
(d) 5
Application of Derivatives 227
x
d−x
, x ∈ R, where a, b
a +x
b + (d − x)2
and d are non-zero real constants. Then,
6. Let f (x) =
2
2
−
2
(2019 Main, 11 Jan II)
(a)
(b)
(c)
(d)
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 −1 (eu ) − . Then, g is
2
(a)
(b)
(c)
(d)
(2008, 3M)
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
(−∞ , ∞ )
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
(2004, 2M)
3 sin x − 4 sin3 x is increasing, is
π
3
3π
(c)
2
(b)
(2002, 2M)
π
2
(d) π
(2001, 2M)
(b) decreasing in R
(d) decreasing in [−1 / 2, 1]
11. For all x ∈ (0, 1)
(2000, 1M)
(a) ex < 1 + x
(c) sin x > x
(b) log e (1+ x) < x
(d) log ε x > x
12. Let f (x) = ∫ ex (x − 1) (x − 2) dx. Then, f decreases in the
interval
(2000, 2M)
(a) (−∞ , − 2) (b) (−2, − 1)
(c) (1, 2)
(d) (2, ∞ )
(a) 0 < x <
(c)
(b)
3π
5π
< x<
8
8
Objective Questions II
(One or more than one correct option)
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) > e2x in (0, ∞ )
(b) f ′ (x) < e2x in (0, ∞ )
(c) f (x) is increasing in (0, ∞ )
(d) f (x) is decreasing in (0 ∞ )
18. If f : (0, ∞ ) → R be given by
f (x) = ∫ e
1/ x
(2014 Adv.)
dt
.
t
Then,
(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
19. If h (x) = f (x) − f (x)2 + f (x)3 for every real number x.
Then,
(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
Fill in the Blanks
20. The set of all x for which log (1 + x) ≤ x is equal to ..... .
4
π
3π
< x<
4
8
5π
3π
(d)
< x<
8
4
π
8
(a) always negative
(b) always positive
(c) strictly increasing
(d) None of these
(1987, 2M)
13. The function f (x) = sin x + cos x increases, if
4
respectively from [0, ∞ ) to [0, ∞ ) and h (x) = f { g (x)}. If
(1987, 2M)
h(0) = 0, then h (x) − h (1) is
(d) f (2x ) is an odd function of x on R
10. If f (x) = xex (1 − x ) , then f (x) is
(a) increasing in [−1 / 2, 1]
(c) increasing in R
16. Let f and g be increasing and decreasing functions
 1
x −  t + 
t
9. The length of a longest interval in which the function
(a)
(c) increasing on (0, π / e), decreasing on ( π / e, ∞ )
(d) decreasing on (0, π / e), increasing on ( π / e, ∞ )
(1999, 2M)
x
x
and g (x) =
, where 0 < x ≤ 1, then in
sin x
tan x
this interval
(1997, 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)
21. The
2
14. If f (x) =
Directions (Q.Nos. 22-24) by appropriately matching
the information given in the three columns of the
following table.
(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
15. The function f (x) =
log (π + x)
is
log (e + x)
(a) increasing on (0, ∞ )
(b) decreasing on (0, ∞ )
Match the Columns
(1995, 1M)
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.
228 Application of Derivatives
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)
(III) f ′ ( x ) = 0 for some
x ∈ ( 0, 1)
(iii) lim f ′ ( x ) = − ∞
(R) f′ is increasing
in (0, 1)
(IV) f ′′( x ) = 0 for some
x ∈ (1, e )
(iv) lim f ′′( x ) = 0
(S) f′ is decreasing
in (e, e 2)
x→ ∞
x→ ∞
x→ ∞
22. Which of the following options is the only INCORRECT
combination?
(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)
24. Which of the following options is the only CORRECT
Column II
A. x + sin x
p. increasing
B. sec x
q. decreasing
r. neither increasing nor decreasing
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)
27. Using the relation 2 (1 − cos x) < x , x ≠ 0 or prove that
2
sin (tan x) ≥ x , ∀ x ∈ [0, π / 4].
(2003, 4M)
−1 ≤ p ≤ 1, then show that the equation
4x3 − 3x − p = 0 has a unique root in the interval [1 / 2, 1]
and identify it.
28. If
(2001, 5M)

xeax ,
x≤0
29. Let f (x) = 
2
3
+
−
x
ax
x
x >0
,

where, a is a positive constant. Find the interval in
which f ′ (x) is increasing.
(1996, 3M)
30. Show that 2 sin x + 2 tan x ≥ 3x, where 0 ≤ x < π /2.
combination?
(a) (III) (iii) (R)
(c) (II) (ii) (Q)
Column I
(b) (IV) (iv) (S)
(d) (I) (i) (P)
25. Match the conditions/expressions in Column I with
statements in Column II.
Let the functions defined in Column I have domain
(−π / 2, π / 2).
(1990, 4M)
31. Show that 1 + x log (x + x2 + 1 ) ≥ 1 + x2 ∀x ≥ 0.
32. Given A = x :

(1983, 2M)
π
π
≤ x ≤  and f (x) = cos x – x (1 + x). Find
6
3
(1980, 2M)
f ( A ).
Topic 3 Rolle’s and Lagrange’s Theorem
Objective Questions 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
1
(c) 0 < f ′ (1) ≤
2
(b) f ′ (1) > 1
1
(d) < f ′ (1) ≤ 1
2
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)
(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
3. If f (x) and g (x) are differentiable functions for 0 ≤ x ≤ 1,
f (0) = 2, g (0) = 0
f (1) = 6, g (1) = 2
Then, show that there exists c satisfying 0 < c < 1 and
(1982, 2M)
f ′ (c) = 2 g′ (c).
such that
Topic 4 Maxima and Minima
Objective Questions I (Only one correct option)
1. 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
2. If the volume of parallelopiped formed by the vectors
$ , $j + λk
$ and λi$ + k
$ is minimum, then λ is equal
i$ + λ$j + k
to
(2019 Main, 12 April I)
(a) −
1
3
(b)
1
3
(c)
3
(d) −
3
Application of Derivatives 229
3. If m is the minimum value of k for which the function
f (x) = x kx − x 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
2
(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
5. 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
(c)
3
(b) 2 3
2
(d)
3
3
minimum and local maximum points of the function,
f (x) = 9x4 + 12x3 − 36x2 + 25, x ∈ R, then
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)
curve y = x − 2 is
(2019 Main, 8 April I)
(c)
7
4 2
(d)
11
4 2
8. 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
(c) 32
(c) − 222
(d) 222
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)
1
4
3
(c) 31
4
(b) 32
(d) 30
1
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
minimum, then
(2016 Main)
(a) 2x = ( π + 4)r
(c) x = 2r
(b) (4 − π )x = πr
(d) 2x = r
x > 0, is
1
≥ 1 , for all
x
(2016 Adv.)
1
(a)
64
1
(b)
32
1
(c)
27
1
(d)
25
16. Let f (x) be a polynomial of degree four having extreme
cone having slant height 3m is
(b) 2 3π
(d) 6π
(a) −8
(c) 0
lim 
x→ 0 1+

f (x) 
= 3, then f (2) is
x2 
(b) −4
(d) 4
x = − 1 and x = 2 are
f (x) = α log|x| + βx2 + x, then
17. If
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
(2019 Main, 9 Jan I)
(2013 Adv.)
(b) 4
(c) 2
(d) 0
19. 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
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
20. The total number of local maxima and local minima of
(2 + x)3 , − 3 < x ≤ −1
the function f (x) =  2
is
x 3 ,
−1 < x < 2
(a) 0
11. The maximum volume (in cu.m) of the right circular
4
(a) π
3
(c) 3 3π
14. A wire of length 2 units is cut into two parts which
(a) 6
10. Let A (4, − 4) and B(9, 6) be points on the parabola,
(a) 31
(b) 10
(d) 30
x2 − x sin x − cos x = 0, is
f (x) = 3x3 − 18x2 + 27x − 40
on the set S = { x ∈ R : x2 + 30 ≤ 11x} is (2019 Main, 11 Jan I)
(b) − 122
(a) 12.5
(c) 25
18. The number of points in (− ∞ , ∞ ) for which
(b) 20 2
(d) 18 3
9. The maximum value of the function
(a) 122
the form of a circular sector, then the maximum area (in
sq. m) of the flower-bed is
(2017 Main)
equal to
2
(a) 2
(b) −3
(d) 2 2
(a) 3
(c) −2 2
values at x = 1 and x = 2. If
7. The shortest distance between the line y = x and the
7
(b)
8
f (x)
, then the local minimum value of h (x) is
g (x)
(2018 Main)
15. The least value of α ∈ R for which 4αx2 +
6. If S1 and S 2 are respectively the sets of local
(a)
(b)
(c)
(d)
h (x) =
1
1
and g (x) = x − , x ∈ R − { −1, 0, 1}. If
2
x
x
13. If 20 m of wire is available for fencing off a flower-bed in
4. If f (x) is a non-zero polynomial of degree four, having
(a)
(b)
(c)
(d)
12. Let f (x) = x2 +
(b) 1
(c) 2
(2008, 3M)
(d) 3
21. If f (x) = x + 2bx + 2c and g (x) = − x − 2cx + b2, such
2
2
2
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
(c)| c|< | b| 2
(d)| c|> | b| 2
230 Application of Derivatives
22. If f (x) = | x|, for 0 < | x| ≤ 2 . Then, at x = 0, f has
1
, for
x=0
(a) a local maximum
(c) a local minimum
(2000, 1M)
(b) no local maximum
(d) no extremum
x2 − 1
, for every real number x, then the
x2 + 1
minimum value of f
(1998, 2M)
23. 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
24. 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
(d) infinite
25. On the interval [0,1], the function x
25
(1 − x)75 takes its
maximum value at the point
(a) 0
(b) 1/4
(c) 1/2
(d) 1/3
x2 y2
+
= 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 

(b) 
,

 a 2 – b2 a 2 – b2 
 ± a2
± b2 

(d) 
,

 a 2 – b2 a 2 + b2 
27. If P (x) = a 0 + a1x + a 2x +... + a nx is a polynomial in
2
4
2n
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
28. 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
31. 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
= 1, then
x → 2 f ′ (x) g′ (x)
(2016 Adv.)
(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
32. 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
(1995, 1M)
26. Find the coordinates of all the points P on the ellipse
 ± a2
± b2 

(a) 
,

 a 2 + b2 a 2 + b2 
 ± a2
± b2 

(c) 
,

 a 2 + b2 a 2 – b2 
(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 )
(1983, 1M)
(b) a = 2, b = −
1
2
(d) None of the above
29. If p, q and r are any real numbers, then
(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
(a) −2
(2013 Adv.)
(c) 2
(d) 2/3
33. A rectangular sheet of fixed perimeter with sides
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.)
(a) 24
(b) 32
 ex
,

34. If f (x) = 2 − ex − 1 ,
 x−e ,

(c) 45
0 ≤ x≤1
1 < x≤2
2 < x≤3
(d) 60
and g (x) = ∫
x
0
x ∈ [1, 3], then
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
35. 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)
(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
36. The function
f (x) = ∫
x
−1
t (et − 1) (t − 1) (t − 2)3 (t − 3)5 dt has a local
minimum at x equals
(a) 0
Objective Questions II
(One or more than one correct option)
cos(2x) cos(2x) sin(2x)

30. If f (x) = 
− cos x cos x − sin x , then
sin x
cos x 
 sin x
−2
(b)
3
(b) 1
(1999, 3M)
(c) 2
2
37. If f (x) = 3x + 12x − 1, − 1 ≤ x ≤ 2, then
37 − x,
(2017 Adv.)
2 < x≤3
(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
(d) 3
(1993, 3M)
Application of Derivatives 231
43. For the circle x2 + y2 = r 2, find the value of r for which
Match the Columns
38. A line L : y = mx + 3 meets Y -axis at E (0, 3) and the
arc of the parabola y2 = 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
Match List I with List II and select the correct answer
using the codes given below the list.
Column I
Column II
P.
m=
1.
1/2
Q.
Maximum area
of ∆EFG is
2.
4
R.
y0 =
3.
2
S.
y1 =
4.
1
the area enclosed by the tangents drawn from the point
P(6, 8) to the circle and the chord of contact is
(2003, 2M)
maximum.
44. Find a point on the curve x2 + 2 y2 = 6 whose distance
from the line x + y = 7, is minimum.
(2003, 2M)
45. 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
(iii) For all x, f ′ (x) =
b
b+ 1
2 (ax + b) 2 ax + 2b + 1
46.
Codes
(a)
(b)
(c)
(d)
P
4
3
1
1
Q
1
4
3
3
R
2
1
2
4
S
3
2
4
2
47.
Passage Based Problems
Consider the function f : (−∞ , ∞ ) → (−∞ , ∞ ) defined by
x2 − ax + 1
; 0 < a < 2.
f (x) = 2
(2008, 12M)
x + ax + 1
48.
39. Which of the following is true ?
(a) (2 + a )2 f ′ ′ (1) + (2 − a )2 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
40. 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
41. Let g (x) = ∫
ex
0
f ′ (t )
dt. Which of the following is true?
1 + t2
(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 (− ∞ , ∞ )
Analytical & Descriptive Questions
42. 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
49.
2 ax + b + 1
−1
2 ax + b
where, a and b are some constants. Determine the
constants a , b and the function f (x).
(1998, 8M)
Let C1 and C 2 be respectively, the parabolas x2 = y – 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.
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)
Determine the points of maxima and minima of the
1
functionf (x) = In x − bx + x2, x > 0, where b ≥ 0 is a
8
(1996, 5M)
constant.
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)
50. 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)
 3 (b3 − b2 + b − 1)
− x +
, 0 ≤ x≤1
51. 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)
52. What normal to the curve y = x2 forms the shortest
chord?
(1992, 6M)
53. 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)
232 Application of Derivatives
54. A point P is given on the circumference of a circle of
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.
radius r. Chord QR is parallel to the tangent at P.
Determine the maximum possible area of the ∆PQR.
(1990, 4M)
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
(2015 Adv.)
is
250 π
55. Find the point on the curve 4x + a y = 4a , 4 < a < 8
2
2 2
that is farthest from the point (0, –2).
2
2
(1987, 4M)
56. Let A ( p , − p) B(q , q), C (r , − r ) be the vertices of the
2
2
2
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
1
area of such a parallelogram is ( p + q)(q + r )( p − r ).
4
(1986, 5M)
π
π
3
2
57. Let f (x) = sin x + λ sin 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.
63. 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.
If ∆(h ) = area of the ∆ PQR, ∆1 = max ∆ (h ) and
1/ 2 ≤ h ≤ 1
8
∆ 2 = min ∆ (h ), then
∆1 − 8∆ 2 is equal to (2013 Adv)
1/ 2 ≤ h ≤ 1
5
intersects the ellipse
64. Let f : R → R be defined as f (x) = | x| + | x2 − 1|. The
total number of points at which f attains either a local
maximum or a local minimum is
(2012)
58. Find the coordinates of the point on the curve y =
65. Let p(x) be a real polynomial of least degree which has a
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
(1984, 4M)
59. A swimmer S is in the sea at a distance d km from the
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
(1983, 2M)
possible time?
66. The number of distinct real roots of
x4 − 4x3 + 12x2 + x − 1 = 0 is……
67. 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)
60. If ax2 + b / x ≥ c for all positive x where a > 0 and b > 0,
then show that 27ab2 ≥ 4c3 .
(1982, 2M)
61. If x and y be two real variables such that x > 0 and xy = 1.
Then, find the minimum value of x + y.
68. The maximum value of the expression
1
is ……
sin 2 θ + 3 sin θ cos θ + 5 cos 2 θ
(1981, 2M)
Integer Answer Type Questions
(2010)
69. The maximum value of the function
f (x) = 2x3 − 15x2 + 36x − 48 on the set
A = { x | x2 + 20 ≤ 9x} is ……… .
62. 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
(2009)
Answers
Topic 2
Topic 1
1. (a)
5. (c)
2. (c)
6. (b)
3. (b)
7. (d)
4. (d)
8. (b)
9. (b)
13. (c)
10. (d)
14. (b, d)
11. (d)
15. (b, c)
12. (d)
16. H = φ, V = {1, 1 }
1
3
18. a = – , b = – , c = 3
2
4
19. 1 : 16
20. y + x − 1 = 0
π
− 3π
21. x + 2y = and x + 2y =
2
2
22. (8)
17. y − 2 = 0
1.
5.
9.
13.
17.
(b)
(c)
(a)
(c)
(c, d)
2.
6.
10.
14.
18.
(a)
(c)
(b)
(b)
(a, c)
3.
7.
11.
15.
19.
(b)
(a)
(c)
(d)
x > −1
4.
8.
12.
16.
1
 1 
1 

 1
20. x ∈  − , 0 ∪  , ∞ , x ∈  − ∞, −  ∪  0, 
 2 
2 

 2
2
21. (d)
22. (c)
23. (c)
 2 a
24. A → p, B → r
28. – ,
 a 3 
1 π 
π 3 π 
π 
31.  − 1 +  ,
− 1 +  



3 2
6
6
2 3
(c)
(a)
(b)
(a, c)
Application of Derivatives 233
Topic 3
44. (2, 1)
1. (b)
2. (b)
1
46. P0 =  ,
2
Topic 4
1. (a)
5. (b)
9. (a)
13.
17.
21.
25.
29.
33.
37.
(c)
(c)
(d)
(b)
(b)
(a, c)
(a, b, c, d)
2. (b)
6. (c)
10. (a)
3. (b)
7. (c)
11. (b)
4. (d)
8. (c)
12. (d)
14.
18.
22.
26.
30.
34.
15.
19.
23.
27.
31.
35.
16.
20.
24.
28.
32.
36.
(c)
(c)
(a)
(a)
(b, c)
(a, b)
(c)
(d)
(d)
(c)
(a, d)
(b, c)
49. 2hk
(c)
(c)
(b)
(b)
(a, b)
(b, d)
39. (a)
40. (a)
43. 5 units
42. 6
(b − b 2 − 1 )
1
and minima at x = (b + b 2 − 1 )
4
4
4 3
51. b ∈ ( −2, − 1 ) ∪ [1, ∞ ]
50.
9
48. Maxima at x =
38. (a) P → 4 Q → 1 R → 2 S → 3
41. (a)
5

4
1
−5
1
5
45. a = , b = ; f ( x ) = x 2 − x + 2
4
4
4
4
5
1


and Q0 =  , 
 4 2
52.
2 x − 2y + 2 = 0, 2 x + 2y − 2 = 0
53. 6 : 6 + π
54.
3 3 2
r sq. units
4
 3 3
57. λ ∈  − ,  58. x = 0, y = 0 59.
 2 2
62. (4)
66. (2)
63. (9)
67. (1)
55. (0, 2)
ud
v − u2
64. (5)
68. (2)
2
61. (2)
65. (9)
69. (7)
Hints & Solutions
Topic 1 Equations of Tangent and Normal
(−1, f (−1)) is m =
1. Equation of given curve is
y=
x
, x ∈ R, (x ≠ ± 3 )
x −3
…(i)
2
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
∴ Slope of this line = − =
6 dx ( α , β )
⇒
−
1
α2 + 3
=−
2
2
3
(α − 3)
⇒
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
3
α
1
1
or
⇒ β=
β= 2
= or −
9 −3 2
9 −3
2
α −3
Now,
f (1) − f (−1) −2 − 0
=
= −1
1 − (−1)
1+1
dy
= 3 x2 − 2 x − 2
dx
[differentiating Eq. (i), w.r.t. ‘x’]
According to the question,
dy
=m
dx
2
⇒
3 x − 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 = x3 + ax − b
…(i)
passes through point P(1, − 5).
[Q α ≠ 0]
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.
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
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
…(ii)
∴
−5 =1+ a −b⇒b−a =6
and slope of tangent at point P(1, − 5) to the curve (i), is
dy
= [3x2 + a ](1, −5 ) = a + 3
m1 =
dx (1, −5 )
...(i)
∴
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
…(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]
234 Application of Derivatives
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
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).
5. The helicopter is nearest to the soldier, if the tangent to
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
3 1/ 2 x3/ 2
= −1
x
1
2
x−
2
3 2
1
x = −x +
2
2
3 x2 + 2 x − 1 = 0
3 x2 + 3 x − x − 1 = 0
3x(x + 1) − 1(x + 1) = 0
1
x = , −1
3
…(i)
x≥0
1
∴
x=
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

Q
Now, the nearest distance
2
2
3
2 
3/ 2

 1
 1 1
 1
 1
=  −  +  7 −   − 7 =   +  
 3
 6
 2 3
 3


=
6.
1
1
3+4
7
1
+
=
=
=
36 27
108
108 6
Key Idea Angle between two curves is the angle between the
tangents to the curves at the point of intersection.
Given equation of curves are
y = 10 − x2
and
y = 2 + x2
[from Eqs. (i) and (ii)]
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
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
⇒
[Q y = x3/ 2 + 7]
…(i)
…(ii)
For point of intersection, consider
|tan θ| =
…(ii)
7
3
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.
⇒
⇒
Application of Derivatives 235
∴ m1m2 = − 1 ⇒
27x1
⇒
27x1
=1
6bx1
9
b=
2
⇒
8. Given curve is
by12
 2π 
which passes through 0,  .
 3
=1
[Q y12 = 6x1]
2x + 2xy′ + 2 y − 6 yy′ = 0 ⇒ y′ =
…(i)
y(− 2) (− 3) = 6 ⇒ y = 1
dy 1 (x − 2)(x − 3) − (x + 6)(x − 3 + x − 2)
=
dx
(x − 2)2(x − 3)2
6 + 30 36
 dy
=
=
=1
 
 dx ( 0, 1)
4 ×9
36
∴ Equation of normal at (0, 1) is given by
−1
y−1 =
(x − 0)
1
⇒ x+ y−1 =0
 1 1
which passes through the point  ,  .
 2 2
⇒
f (x) = tan − 1
1 + sin x
 π
, x ∈ 0, 
 2
1 − sin x
x

2
2
x
x

 cos − sin 

2
2
2
x

 cos + sin

2
x
x

 cos + sin 
2
2
= tan 

 cos x − sin x 

2
2
−1
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

⇒
Equation of normal at (1, 1) is
1
y − 1 = − (x − 1) ⇒ y − 1 = − (x − 1)
1
⇒
x+ y=2
…(ii)
On solving Eqs. (i) and (ii) simultaneously, we get
⇒
x2 + 2x(2 − x) − 33(2 − x)2 = 0
2
⇒ x + 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
(1,1)
X'
X
O
(3,–1)
1
 π 1
f ′ (x) = ⇒ f′   =
 6 2
2
Now, equation of normal at x =
x+ y
3y − x
 dy
=1
 
 dx (1, 1)
i.e.
So, point of intersection is (0, 1).
x+6
Now, y =
(x − 2)(x − 3)
9. We have, f (x) = tan − 1
…(i)
x = 1, y = 1, y′ = 1
At
Put x = 0 in Eq. (i), we get
⇒
x2 + 2xy − 3 y2 = 0
On differentiating w.r.t x, we get
y(x − 2)(x − 3) = x + 6
⇒
10. Given equation of curve is
Y'
π
is given by
6
π

 π

 y − f    = − 2  x − 

6 
6
π
π 
π 4π π 


 π π
⇒  y −  = − 2  x −  Q f   = +
=
=






3
6 
6
4 12 12 3 
11. Given,
y + 3x = 12 y
3
2
On differentiating w.r.t. x, we get
dy
dy
⇒
3 y2
+ 6x = 12
dx
dx
dy
6x
⇒
=
dx 12 − 3 y2
...(i)
236 Application of Derivatives
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)
[given]
f ′ (3) = 1
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.
13. Given, x = a (cos θ + θ sin θ )
16. Given,
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
y3 − 3xy + 2 = 0
On differentiating w.r.t. x, we get
dy
dy
3 y2
− 3x
− 3y = 0
dx
dx
dy
dy
3y
⇒
= 2
(3 y2 − 3x) = 3 y ⇒
dx
dx 3 y − 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
⇒ − 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
On putting value of h in Eq. (ii), we get
4(− 2k)2 + 9 k2 = 1
⇒
16k2 + 9k2 = 1
⇒
25k2 = 1
1
⇒
k2 =
25
1
⇒
k=±
5
Thus, the point, where the tangents are parallel to
1
 2 1
2
8x = 9 y are  − ,  and  , −  .
 5 5
5
5
…(i)
On differentiating w.r.t. x, we get
dy
8x + 18 y
=0
dx
dy
8x
4x
=−
=−
⇒
dx
18 y
9y
The tangent at point (h , k) will be parallel to 8x = 9 y,
then
4h 8
−
=
9k 9
⇒
h = − 2k
Point (h , k) also lies on the ellipse.
...(ii)
∴
4h 2 + 9k2 = 1
⇒ 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
3
⇒
y − 1 = 0 ⇒ y3 = 1
⇒
y=1, x=1
Then,
V = {1, 1}
17. As | f (x1) − f (x2)| ≤ (x1 − x2)2, ∀x1 , x2 ∈ R
⇒ | f (x1 ) − f (x2)| ≤ |x1 − x2|2
[as x2 =|x|2 ]
Application of Derivatives 237
∴
⇒
f (x1 ) − f (x2)
≤ |x1 − x2|
x1 − x2
19. Let any point P1 on y = x3 be (h , h3 ).
Then, tangent at P1 is
 f (x1 ) − f (x2) 
 ≤ lim |x1 − x2|
lim 
x1 → x2 
x1 − x2  x1 → x2
y − h3 = 3h 2(x − h )
It meets y = x at P2.
⇒ | f ′ (x1 )| ≤ 0, ∀x1 ∈ R
On putting the value of y in Eq. (i), we get
x3 − h3 = 3h 2(x − h )
∴ | f ′ (x)| ≤ 0, which shows| f ′ (x)| = 0
[as modulus is non negative or| f ′ (x)| ≥ 0]
⇒
(x − h ) (x + xh + h 2) = 3h 2(x − h )
2
∴ f ′ (x) = 0 or f (x) is constant function.
⇒
⇒ Equation of tangent at (1, 2) is
y−2
= f ′ (x)
x−1
⇒
x2 + xh − 2h 2 = 0
⇒
(x − h ) (x + 2h ) = 0
y−2 =0
or
...(i)
3
x2 + xh + h 2 = 3h 2 or x = h
⇒
[Q as f ′ (x) = 0]
⇒ y − 2 = 0 is required equation of tangent.
x = h or x = − 2h
Therefore, x = − 2h is the point P2,
which implies y = − 8h3
18. Given, y = ax3 + bx2 + cx + 5 touches X-axis at P(−2, 0)
Hence, point P2 ≡ (−2h , − 8h3 )
which implies that X-axis is tangent at (–2, 0) and the
curve is also passes through (–2, 0).
Again, tangent at P2 is y + 8h3 = 3 (−2h )2(x + 2h ).
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
⇒
0 = 12a − 4b + c
3 = 3a (0) + 2b (0) + c
⇒
3=c
x + 8h = 12h 2(x + 2h )
3
⇒
(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
…(i)
D ′ ∆ P1P2 P3
=
=
D ′ ′ ∆ P2 P3 P4
2
…(ii)
Since, the curve passes through (–2, 0).
0 = a (−2)3 + b(−2)2 + c(−2) + 5
⇒
0 = − 8a + 4b − 2c + 5
… (iii)
=
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
3
x2 − 2hx − 8h 2 = 0
⇒
Again, slope of tangent at (0, 5) is 3.
dy
∴
=3
dx ( 0, 5 )
⇒
y = x3 at P3
It meets
h
h3
1
−2h −8h3
2
4h 64h3
−2 h
−8h3
1
4h
64h3
2
−8h −512h3
h
h3
1
−2h −8h3
2
4h 64h3
1
1
1
1
1
1
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
Now,
u = (1 + x)y
On taking logarithm both sides, we get
log e u = y log e (1 + x)
238 Application of Derivatives
⇒
1 du
y
dy
=
+
{log e (1 + x)}
u dx 1 + x dx
⇒
 y

du
dy
= (1 + x)y 
+
log e (1 + x)
dx
1 + x dx

…(ii)
sin v = sin 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
2
⇒
⇒
⇒
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
⇒
dy
=1
dx
Again, the slope of the normal is
1
m=−
= −1
dy / dx
Hence, the required equation of the normal is
y − 1 = (−1) (x − 0)
i.e.
y+ x−1 =0
21. Given, y = cos (x + y)
dy

 dy
⇒   = − sin (x + y) ⋅ 1 +


 dx
dx
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
⇒
∴
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 
∴ Equation of tangents are
y−0
1
=−
x − π /2
2
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

⇒ 2 ( y − x5 ) 
− 5x4 = (1 + x2)2 + 2x (1 + x2) ⋅ 2x
 dx

Put x = 1 and y = 3, dy /dx = 8
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 )
y = (1 + 0)y + sin −1sin (0) = 1
⇒
are the required equations of tangents.
1. Let the thickness of layer of ice is x cm, the volume of
At x = 0,
1 −1
+ 2 sin 0 ⋅ cos 0 / (1 − sin 4 0)
dy 1 (1 + 0)
=
dx
1 − (1 + 0)1 log e (1 + 0)
and
π
y−0
1
= − ⇒ 2y = − x +
2
x + 3π / 2
2
3π
2y = − x −
2
π
x + 2y =
2
3π
x + 2y = −
2
Topic 2 Increasing and Decreasing
Functions
y −1
+ 2 sin x cos x / 1 − sin 4 x
dy y(1 + x)
=
dx
1 − (1 + x)y log e (1 + x)
∴
and
⇒
v = sin −1 (sin 2 x)
Again,
⇒
and
…(i)
So, the thickness of the ice decreases at the rate of
1
cm /min.
18π
2. The given functions are
f (x) = ex − x,
and
g (x) = x2 − x, ∀ x ∈ R
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, ∞ )
Application of Derivatives 239
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 
So,
 1
x ∈ 0,
 2 
 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.
Given, semi-vertical angle of right circular cone
 1
= tan −1  
 2
 1
Let
α = tan −1  
 2
1
tan α =
⇒
2
r
r 1
[from fig. tan α = ]
⇒
=
h
h 2
1
…(i)
⇒
r= h
2
r
h
l
α
1 2
πr h
3
2
1
1 1 
V = π  h (h ) =
πh3
3 2 
12
Q Volume of cone is (V ) =
∴
[from Eq. (i)]
On differentiating both sides w.r.t. ‘t’, we get
dV
dh
1
=
π (3h 2)
dt 12
dt
dh
4 dV
⇒
=
dt πh 2 dt
dh
dV
4
[Q given
= 5 m3 /min]
=
×5
⇒
dt πh 2
dt
Now, at h = 10 m, the rate at which height of water
dh
level is rising =
dt h = 10
4
1
m/min
=
×5=
5π
π (10)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).
5. Given that function,
f (x) = x3 − 3 (a − 2) x2 + 3ax + 7, for some
increasing in (0, 1] and decreasing in [1, 5).
f′ (1) = 0
[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)
− 2
2 1/ 2
(a + x )
(b + (d − x)2)1/ 2
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]
=
a 2 + x2 − x2 b2 + (d − x)2 − (d − x)2
+
(a 2 + x2)3/ 2
(b2 + (d − x)2)3/ 2
=
a2
b2
+
> 0,
(a 2 + x2)3/ 2 (b2 + (d − x)2)3/ 2
4. Given, φ (x) = f (x) + f (2 − x), ∀ x ∈ (0, 2)
⇒
φ′ (x) = f ′ (x) − f ′ (2 − x)
Also, we have f ′ ′ (x) > 0 ∀ x ∈ (0, 2)
⇒ f ′ (x) is a strictly increasing function
…(i)
∀ x ∈ (0, 2).
a ∈ R is
∀ x ∈R
Hence, f (x) is an increasing function of x.
240 Application of Derivatives
π
for u ∈ (−∞ , ∞ )
2
π
π
g (− u ) = 2 tan −1 (e− u ) − = 2 (cot−1 (eu )) −
2
2
π
π
−1 u 
= 2  − tan (e ) −
 2
2
7. Given, g (u ) = 2 tan −1 (eu ) −
= π / 2 − 2 tan −1 (eu ) = − g (u )
∴
for 0 < x < 1
⇒ log e (1 + x) − x < 0
for 0 < x < 1
log e (1 + x) < x
for 0 < x < 1
Let h (x) = sin x − x
2 eu
1 + e 2u
h′ (x) = cos x − 1
g′ (u ) > 0, ∀ x ∈ R
[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]
∴
for 0 < x < 1
g (x) < g (0)
Option sin x > x
We have, g (u ) = 2 tan −1 (eu ) − π 2
⇒
g (x) decreases
⇒
Therefore, option (b) is the answer.
⇒ g (u ) is an odd function.
8. Given,
⇒
or
g (− u ) = − g (u )
g′ (u ) =
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
D = (–ve) or D < 0
For x ∈(0, 1), cos x − 1 < 0
⇒ 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
⇒ f ′ (x) = 3x2 + 2bx + c > 0 ∀x ∈ (− ∞ , ∞ )
[as D < 0 and a > 0]
⇒
log x − x < 0
Hence, f (x) is strictly increasing function.
⇒
log x < x
9. Let f (x) = 3 sin x − 4 sin x = sin 3x
3
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 )
⇒
+ xe
x (1 − x )
[1 + x (1 − 2x)]
f ′ (x) = e
x (1 − x )
(1 − 2x)
= ex (1 − x ) (1 + x − 2x2)
x (1 − x )
= −e
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
x (1 − x )
=e
⇒ 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
(2x − x − 1)
2
x (1 − x )
= 4 sin x cos x (sin 2 x − cos 2 x)
= −e
(x − 1) (2x + 1)
 1 
which is positive in  − , 1 .
 2 
 1 
Therefore, f (x) is increasing in − , 1 .
 2 
11.
p(x) = log x − x
1
p′ (x) = − 1 > 0, ∀x ∈ (0, 1)
x
= 2 sin 2x (− cos 2x)
= − sin 4x
Now,
⇒
f ′ (x) > 0 , if sin 4x < 0
π < 4x < 2π
π
π
<x<
4
2
PLAN Inequation based upon uncompatible function. This type
of inequation can be solved by calculus only.
⇒
Option (a) Let f (x) = ex − 1 − x.
⇒ Option (a) is not proper subset of Eq. (i), so it is not
correct.
π
3π
Now,
<x<
4
8
then
f ′ (x) = e − 1 > 0, ∀x ∈ (0, 1)
x
⇒
f (x) increase in (0, 1)
⇒
f (x) > f (0) for 0 < x < 1
⇒ ex − 1 − x > 0 or ex > 1 + x for 0 < x < 1
…(i)
Since, option (b) is the proper subset of Eq. (i), so it is
correct.
Application of Derivatives 241
x
, where 0 < x ≤ 1
tan x
Now, g (x) is continuous in [0, 1] and differentiable in
]0, 1[.
14. Given, g (x) =
g′ (x) =
= (+ ) (− ) = − ve
tan x − x sec x
tan 2 x
[since, f (x) is an increasing function f ′ ( g (x)) is + ve and
g (x) is decreasing function g ′ ( f (x)) is −ve ]
Now, H (x) is continuous in [0, 1] and differentiable in
]0, 1[.
0 < x < 1, H (x) = tan x − x sec x, 0 ≤ x ≤ 1
2
For
⇒
H ′ (x) = sec x − sec x − 2x sec x tan x
2
2
2
= − 2x sec x tan x < 0
2
H (x) < H (0)
for 0 < x < 1
⇒
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
∴ f (x) is decreasing function.
0 ≤ x<1
When
⇒
h (x) − h (1) = + ve
x ≥ 1,
⇒
h (x) − h (1) = − ve
Hence, for
x > 0,
h (x) − h (1) is neither always positive nor always
negative, so it is not strictly increasing throughout.
⇒
Also,
f ′ (x) is –ve.
Since,
When
Hence, H (x) is decreasing function in [0, 1].
Thus,
F ′ (x) = f ′ { g (x)} ⋅ g ′ (x)
2
H (x) = tan x − x sec2 x, 0 ≤ x ≤ 1
Again,
16. Let F (x) = h (x) − h (1) = f { g (x)} − f { g (1)}
∴
0 < x < 1,
For
From Eqs. (i) and (iv), f ′ (x) < 0
∴ f (x) is decreasing for x ∈ (0, ∞ ).
Therefore, option (d) is the answer.
dy
17. f ′ (x) > 2 f (x) ⇒
> 2dx
y
f ( x)
∫
⇒
1
0 < x≤1
for
0 < x≤1
for
x=0
Now, f is continuous in [0, 1] and differentiable in ]0, 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
ln( f (x)) > 2x
∴
f (x) > e2x
Also, as f ′ (x) > 2 f (x)
∴
f ′ (x) > 2c2x > 0
18. Given, f (x) = ∫
log (e + x) ⋅
f ′ (x) =
=
As
x > 0, π + x > e + x
∴
log (π + x) > log (e + x)
1
1
>
e+ x π + x
and
On multiplying Eqs. (ii) and (iii), we get
log (π + x) log (e + x)
>
π+x
e+ x
 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
1
− log (π + x) ⋅
e+ x
π+x
…(i)
[log (e + x)]2
When
x
1
x
f ′ (x) = 1 ⋅
⇒ f (x) increases in [0, 1].
x
increases in 0 < x ≤ 1.
Thus, f (x) =
sin x
Therefore, option (c) is the answer.
log (π + x)
15. Given, f (x) =
log(e + x)
x
dy
> 2∫ dx
y
0
 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, ∞ )
... (ii)
Now, let
... (iii)
g (x) = f (2x ) = ∫
−x
2x
∴ f (2x ) is an odd function.
e
2− x
g (− x) = f (2 ) = ∫
…(iv)
 1
− t + 
 t
2− x
2x
dt
t
 1
− t + 
 t
e
t
dt = − g (x)
242 Application of Derivatives
⇒ f (x) = 0 for some x ∈ (1, e2)
19. Given, h (x) = f (x) − f (x)2 + f (x)3
∴ I is correct
On differentiating w.r.t. x, we get
h ′ (x) = f ′ (x) − 2 f (x) ⋅ f ′ (x) + 3 f 2(x) ⋅ f ′ (x)
f ′ (x) = 1 +
= f ′ (x)[1 − 2 f (x) + 3 f (x)]
2
1

= 3 f ′ (x) ( f (x))2 − f (x) +

3
3 
2
f ′ (x) > 0 for (0, 1)
f ′ (x) < 0 for (e, ∞ )
∴ P and Q are correct, II is correct, III is incorrect.
−1 1
f ′′ (x) = 2 −
x
x
2

1
1 1
= 3 f ′ (x)  f (x) −  + − 


3
3 9

2

1
3 − 1
= 3 f ′ (x)  f (x) −  +


3
9 

f ′′ (x) < 0 for (0, ∞ )
∴ S, is correct, R is incorrect.
2

1
2
= 3 f ′ (x)  f (x) −  + 


3
9

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
and
when
x
1
−1 = −
1+ x
1+ x
f ′ (x) > 0
−1 < x<0
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
2
log
(
),
x
−
−
x
x
<0

1

dy 4x − x , x > 0
=
dx 4x − 1 , x < 0

x
(2x − 1) (2x + 1)
4x2 − 1
, x ∈ R − {0} =
x
x
 1  1 
Increasing when x ∈  − , 0 ∪  , ∞
 2  2 
=
∴
IV is incorrect.
lim f (x) = − ∞
x→∞
h ′ ( x ) < 0, if f ′ ( x ) < 0 and h ′ ( x ) > 0, if f ′ ( x ) > 0
NOTE
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
1
1
− ln x − 1 = − ln x
x
x
lim f ′ (x) = − ∞
x→∞
lim f ′′ (x) = 0
x→∞
∴ ii, iii, iv are correct.
22. (d)
23. (c)
24.(c)
π
π
d
x
(x + sin x) = 1 + cos x = 2 cos 2 > 0 for − < x < .
2
2
dx
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,
⇒
π
 2 
 π
.
∴
f ′ (x) is decreasing for all x ∈ 0,
 2 
⇒
f ′ (x) > 0
⇒
f ′ (x) > f ′ (π / 2)
∴ f (x) is increasing.
Thus, when x ≥ 0, f (x) ≥ f (0)
3x(x + 1)
⇒ sin x + 2x −
≥0
π
3x (x + 1)
⇒ sin x + 2x ≥
π
27. Let
f (x) = sin (tan x) − x
f ′ (x) = cos (tan x) ⋅ sec2x − 1
= cos (tan x) (1 + tan 2 x) − 1
[ Q x < π /2]
Application of Derivatives 243
= tan 2 x {cos (tan x)} + cos (tan x) − 1
2
tan x
> tan 2 x cos (tan x) −
2

x2 
2
Q2 (1 − cos x) < x , x ≠ 0 ⇒ cos x > 1 − 2 

tan 2 x 

 ⇒
cos (tan x) > 1 −


2
1

f ′ (x) > tan 2 x cos (tan x) −

2 
2
> tan x [cos (tan x) − cos (π / 3)] > 0
⇒
(ax + 1) eax ,
f ′ (x) = 
1,
1 + 2ax − 3x2,
Hence,
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
if x > 0
2a − 6x,
Lf ′ ′ (0) = lim
and
x → 0−
= lim
f (x) is increasing function, for all x ∈ [0, π / 4]
As
28. Given,
Let
Now,
Also,
x→ 0
sin (tan x) ≥ x
−1 ≤ p ≤ 1
x→ 0
and
[Q p ≤ 1]
∴ f (x) has atleast one real root between [1 /2, 1].
⇒
f ′ (x) increasing on [1 /2, 1]
⇒
To find a root, we observe f (x) contains 4x − 3x, which is
multiple angle formula for cos 3θ.
3
x = cos θ
⇒ 4 cos3 θ − 3 cos θ − p = 0
p = cos 3 θ ⇒ θ = (1 / 3) cos −1 ( p)

1
∴ Root is cos  cos −1 ( p) .

3
⇒
29.
NOTE This type is asked in 1983 and repeat after 13 years.
At x = 0, LHL = lim f (x) = lim xeax = 0
x→ 0
and
−
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
ax
x
+
−
1
2
3
, if x > 0

and
Lf ′ (0) = lim
x → 0−
= lim
x → 0−
and
Rf ′ (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
= lim
x
x →0 +
= lim 1 + ax − x2 = 1
x → 0+
Therefore, Lf ′ (0) = Rf ′ (0) = 1 ⇒ f ′ (0) = 1
x→ 0
−
eax − 1
ax
= lim
(1 + 2ax − 3x2) − 1
x
= lim
2ax − 3x2
= lim 2a − 3x = 2a
x
x → 0+
x → 0+
x→ 0
f has only one real root between [1 /2, 1].
∴ Put
−
= ae0 + a (1) = 2a
f ′ (x) − f ′ (0)
Rf ′ ′ (0) = lim
x+0
x → 0+
[Q p ≥ − 1]
f (1) = 4 − 3 − p = 1 − p ≥ 0
f ′ (x) = 12x2 − 3 > 0 on [1 / 2, 1]
(ax + 1)eax − 1
x
= lim aeax + a ⋅ lim
f (x) = 4x3 − 3x − p = 0
1 3
f (1 / 2) = − − p = − 1 − p ≤ 0
2 2
Also,
−
f ′ (x) − f ′ (0)
x−0

eax − 1 
= lim aeax +
x 
x → 0− 
f (0) = 0 ⇒ f (x) ≥ 0, for all x ∈ [0, π / 4]
⇒
if x < 0
if x = 0
if 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
Now, for x < 0, f ′ ′ (x) > 0, if ax + 2 > 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,
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
⇒
For
f ′ (x) = 2 cos x + 2 sec2 x − 3
0 ≤ x < π / 2, f ′ (x) > 0
Thus, f (x) is increasing.
x ≥ 0, f (x) ≥ f (0)
When
⇒
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 ⋅

1 +


x+


x + 1 
x
2
x +1
2
+ log (x +
x2 + 1 )
244 Application of Derivatives
−
⇒
x
x +1
2
=
x
x +1
2
f ′ (x) = log (x +
⇒
+ log (x +
x
x2 + 1 ) −
x +1
2
Topic 4 Maxima and Minima
1. Given functions are f (x) = 5 − |x − 2|
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
x + 1)
2
f ′ (x) ≥ 0
[Q log (x +
x + 1 ) ≥ 0]
2
∴ 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 :

π
π
≤x≤ 
6
3
and f (x) = cos x − x − x2
Key Idea Volume of parallelopiped formed by the vectors a,
b and c is V = [ a b c].
2.
⇒
f ' (x) = − sin x − 1 − 2x = − (sin x + 1 + 2x)
π π 
which is negative for x ∈
,
 6 3 
$ and λi$ + k
$ , which
$ , $j + λk
Given vectors are i$ + λ$j + k
forms a parallelopiped.
∴
∴Volume of the parallelopiped is
f ' (x) < 0
or f (x) is decreasing.
  π  π 
Hence, f ( A ) =  f   , f   
  3  6 
1 π 
π 3 π 
π 
− 1 +  
=  − 1 +  ,



2
3
3
2
6
6 

⇒
Topic 3 Rolle’s and Lagrange’s Theorem
1. f ′ (x) is increasing
1
For some x in  ,
2
∴

1

f ′ (x) = 1
f′ (1) > 1
[LMVT]
2. Given, f (x) = 2 + cos x, ∀x ∈ R
Statement I
f ′ (c) = 0
There exists a point ∈ [t , t + r ], where
Hence, Statement I is true.
Statement II f (t ) = f (t + 2π ) is true. But statement II
is not correct explanation for statement I.
3. Since, f (x) and g (x) are differentiable functions for
0 ≤ x ≤ 1.
∴
f ′ (c) =
f (1) − f (0)
1 −0
Using Lagrange’s Mean Value theorem,
6 −2
=4
1 −0
and
⇒
g (1) − g (0)
1 −0
2 −0
=
=2
1 −0
g ′ (c) =
f ′ (c) = 2 g′ (c)
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
1

2 3 > 0 , for λ =

d 2V
3
and
= 6λ = 
1
dλ2
2 3 < 0 , for λ = −
3

1
d 2V
, so volume ‘V ’ is minimum
is positive for λ =
3
dλ2
1
for λ =
3
Q
3. Given function f (x) = x kx − x2
… (i)
the function f (x) is defined if kx − x ≥ 0
2
⇒
x2 − kx ≤ 0
⇒
x ∈ [0, k]
… (ii)
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 − x2) + kx − 2x2
2 kx − x2
=
3kx − 4x2
2 kx − x2
Application of Derivatives 245
as f (x) is increasing in interval x ∈ [0, 3], so
f ′ (x) ≥ 0 ∀ x ∈ (0, 3)
⇒
3kx − 4x2 ≥ 0
⇒
4x2 − 3kx ≤ 0
3k
 3k 

(as k is positive)
⇒ 4x x −  ≤ 0 ⇒ x ∈ 0,
 4 

4
3k
So,
3≤
⇒k ≥4
4
⇒ Minimum value of k = m = 4
and the maximum value of f in [0, 3] is f (3).
Q f is increasing function in interval x ∈ [0, 3]
⇒
tan 2 θ = 2
Therefore, ordered pair (m, M ) = (4, 3 3 )
4. 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.
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
⇒
tan θ = 2
2
1
and cos θ =
⇒
sin θ =
3
3
From Eqs. (i) and (ii), we get
1
h =6
=2 3
3
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
–
+
1
0
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.
∴S1 = { −2, 1} and S 2 = {0}.
Since,
y2 = x − 2
and the equation of line is
y=x
Y
h
r
Q Volume of cylinder V = πr 2h
= π (3 sin θ )2(6 cos θ ) = 54π sin 2 θ cos θ .
…(ii)
y2=x–2
M
P(t2+2, t)
O
3
…(i)
y=x
r
θ
…(ii)
7. Given equation of curve is
Let a sphere of radius 3, which inscribed a right circular
cylinder having radius r and height is h, so
h
From the figure, = 3 cos θ
2
⇒
h = 6 cos θ
and
…(i)
r = 3 sin θ
h/2

 π 
Q θ ∈ 0, 2  


6. Given function is
Q M = f (3) = 3 4 × 3 − 32 = 3 3
5.
⇒
⇒
dV
=0
dθ
2
3
54π [2 sin θ cos θ − sin θ ] = 0
sin θ [2 cos 2 θ − sin 2 θ ] = 0
For maxima or minima,
(2, 0)
X
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.
246 Application of Derivatives
∴
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 =
2t
2
dy
=1]
dx
[Q P (x, y) = P (t + 2, t )]
2
⇒
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]
[Q f ′ (x) > 0 for x ∈[5, 6]
∴f (6) is maximum, where
f (6) = 3(6)3 − 18(6)2 + 27(6) − 40 = 122
10. According to given information, we have the following
figure.
 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
|ax1 + by1 + c |
ax + by + c = 0 is

a 2 + b2

7
units
=
4 2
For y2 = 4ax, parametric coordinates of a point is (at 2,
2at).
∴For y2 = 4x, let coordinates of C be (t 2, 2t).
t 2 2t 1
1
Then, area of ∆ABC =
9 6 1
2
4 −4 1
8. 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)
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
Then, area of rectangle PQRS
= 2 × (Area of rectangle PQMO)
Thus, for A (t ), critical point is at t =
[due to symmetry aboutY -axis]
1
2
2
= 2 × [a (12 − a 2)] = 24a − 2a3 = ∆(let).
The area function ∆ a will be maximum, when
d∆
=0
da
⇒
24 − 6a 2 = 0
⇒
a2 = 4 ⇒ a = 2
[Q a > 0]
So, maximum area of rectangle
PQRS = (24 × 2) − 2 (2)3
= 48 − 16 = 32 sq units
125
1
1
 1
 1
Now,A   = 5   − − 6 =
= 31 [Using Eq. (i)]
 2
 2
2
4
4
11. Let h = height of the cone,
r = radius of circular base
= (3)2 − h 2
= 9 − h2
h
l=
…(i)
[Q l2 = h 2 + r 2]
…(i)
3
9. We have,
f (x) = 3x3 − 18x2 + 27x − 40
⇒ 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
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
...(i)
Let,
A (t ) = 5| t 2 − t − 6|
Clearly, A (4, − 4) ≡ A (t12, 2t1 ) ⇒ 2t1 = − 4
⇒
t1 = − 2
r
Application of Derivatives 247
Now, volume (V ) of cone =
1 2
πr h
3
14. According to given information, we have Perimeter of
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,
A = x2 + π r 2
(1 − 2x) 2
= x2 + π
π2
(1 − 2x)2
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
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
V (h ) =
⇒
12. We have,
f(x) = x2 +
1
x2
and g( x ) = x −
1
f(x)
⇒ h( x ) =
g( x )
x
Now, from Eq. (i), we get
2
1 −2⋅
1
π + 4 π + 4 −4
r=
=
=
π(π + 4) π + 4
π
2
1

x2 + 2  x −  + 2

x
x =
∴ h( x ) =
1
1
x−
x−
x
x
1
2

⇒
h( x ) =  x −  +
1


x
x−
x
1
1
2

∈ [2 2 , ∞ )
x − > 0,  x −  +

x
x x − 1
x
1
2
1

∈ ( −∞ , 2 2 ]
x − < 0,  x −  +

x x − 1
x
x
∴ Local minimum value is 2 2.
1
From Eqs. (ii) and (iii), we get
θ=
4 αx2 +
⇒
∴
⇒
∴
r
rθ
 1
∴ y attains minimum when x =  
 8α 
A = 10r − r 2
dA
= 10 − 2r
dr
For maxima or minima, put
to
2
r
Now, area of flower-bed,
1
A = r 2θ
2
1 2 20 − 2r 
⇒
A= r 

r 
2 
1
≥ 1, for all x > 0.
x
find the minimum value of α when
1
y = 4αx + ; x > 0 attains minimum value of α.
x
dy
1
…(i)
∴
= 8αx − 2
dx
x
2
d 2y
…(ii)
Now,
= 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 
i.e.
θ
20 − 2r
r
x = 2r
15. Here, to find the least value of α ∈ R, for which
13. Total length = 2r + r θ = 20
⇒
...(iii)
i.e.
 1
4α  
 8α 
1/3
.
/
23
+ (8α )1/3 ≥ 1
⇒
α 1/3 + 2α 1/3 ≥ 1
1
⇒
3α 1/3 ≥ 1 ⇒ α ≥
27
1
Hence, the least value of α is
.
27
dA
= 0.
dr
10 − 2r = 0 ⇒ r = 5
1
20 − 2 (5) 
Amax = (5)2


2
5
1
= × 25 × 2 = 25 sq. m
2
16.
PLAN Any function have extreme values (maximum or minimum) at
its critical points, where f ′ ( x ) = 2.
Since, the function have extreme values at x = 1 and
x = 2.
248 Application of Derivatives
∴
⇒
f ′ (x) = 0 at x = 1 and x = 2
f′ (1) = 0 and f′ (2) = 0
g′ (x) = 0 ⇒ x cos x = 0
π 3π 5π 7π
x = 0, ,
,
,
2 2 2 2
Put
∴
Also, it is given that,
f (x) 
lim 
=3
x→ 0 1 +
x2 

f (x)
⇒
1 + lim
x→ 0 2 = 3
x
lim f (x)
=2
⇒
x→ 0
x2
⇒ f (x) will be of the form ax4 + bx3 + 2x2.
g (x)
/2
–5 π/ 2
–π –π/2 O
[Q f (x) is of four degree polynomial]
Let
⇒
⇒
and
⇒
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
Y′
3π 7π
,
,... , f ′′ (x) > 0, so minimum
2 2
π 5π 9π
At x = ,
,
,... , f ′ (x) < 0, so maximum
2 2 2
...(i)
...(ii)
So, graph of f (x) and g (x) are shown as
Y
f (x)
x4
f (x) =
− 2x3 + 2x2
2
f (2) = 8 − 16 + 8 = 0
⇒
X′
f (x) = α log|x| + βx2 + x, then
α
f ′ (x) = + 2 βx + 1
x
f ′ (−1) = − α − 2 β + 1 = 0
Y′
So, number of solutions are 2.
…(i)
[at extreme point, f ′ (x) = 0]
α
…(ii)
f ′ (2) = + 4 β + 1 = 0
2
On solving Eqs. (i) and (ii), we get
1
α = 2, β = −
2
Here, x2 = x sin x + cos x
X′
2
2
2
2
⇒
(2 + x)3 , if − 3 < x ≤ −1
f (x) = 
x 2 / 3 , if − 1 < x < 2

3 (x + 2)2 , if − 3 < x ≤ −1

f ′ (x) = 2 − 1
3
, if − 1 < x < 2
3 x
Y
X′
(–3,0)
X
Y′
g (x) = x sin x + cos x
g′ (x) = x cos x + sin x − sin x
g′ (x) = x cos x
g′′ (x) = − x sin x + cos x
X
(–1,0)
Y′
g (x) = x sin x + cos x
We know that, the graph for f (x) = x2
To plot,
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
x2
O
Let f (x) = x2 and
2
19. Given function, f (x) = ex + e− x , g (x) = xex + e− x and
20. Given,
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.
Y
g (x)
X
–π/2 O π/2
17. Here, x = − 1 and x = 2 are extreme points of
18.
5π/2
At x = 0,
On solving Eqs. (i) and (ii),
1
we get
a = , b = −2
2
∴
π/2 π
…(i)
…(ii)
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 .
Application of Derivatives 249
21. Given f (x) = x2 + 2bx + 2c2 and g (x) = − x2 − 2cx + b2
Then, f (x) is minimum and g (x) is maximum at

−b
−D 
and f (x) =
x =
 , respectively.

4a
4a 
∴
and
min f (x) =
26. Let the coordinates of P be (a cos θ , b sin θ )
Equations of tangents at P is
Y
N
− (4b2 − 8c2)
= (2c2 − b2)
4
max g (x) = −
(4c2 + 4b2)
= (b2 + c2)
4(−1)
X'
O
Now, min f (x) > max g (x)
⇒
2
2
2
⇒
c2 > 2b2
⇒
| c| > 2| b|
Y'
x
y
cos θ + sin θ = 1
a
b
22. It is clear from figure that at x = 0, f (x) is not continuous.
Again, equation of normal at point P is
ax secθ − by cosec θ = a 2 − b2
Y
y=x
y = –x
X′
X
O
Y′
Here, f (0) > RHL at x = 0 and f (0) > LHL at x = 0
So, local maximum at x = 0.
23. Given, f (x) =
2
x2 − 1
=1 − 2
x2 + 1
x +1
2
is maximum,
f (x) will be minimum, when 2
x +1
i.e.
when x2 + 1 is minimum,
i.e.
at x = 0.
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 =
=
2
2
2
2
cos θ sin θ
b cos θ + a 2 sin 2 θ
+
2
2
a
b
[perpendicular from O, to line NP]
and OM =
24. 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.
⇒
Let f (θ ) =
f ' (θ ) =
f ′ (x) = 25 x24 (1 − x)75 − 75x25 (1 − x)74
= 25x24 (1 − x)74 [(1 − x) − 3x]
For maximum value of f (x), put f ′ (x) = 0
⇒
Also, at
=
(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 )
ab(a 2 − b2) tan θ
2(a 2 tan 2 θ + b2)
tan θ
α 2 tan 2 θ + b2
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
x = 0, y = 0
[sec2 θ ≠ 0, a tan θ + b ≠ 0, 0 < θ < π / 2 ]
At
x = 1, y = 0
⇒
and at
x = 1 / 4, y > 0
−1

Also, f ′ (θ )> 0, if 0 < θ−1< tan (b / a )
b
a
<
< θ < π /2
,
tan
(
/
)
0
if

∴ f (x) attains maximum at x = 1 / 4.
[0 < θ < π / 2]
sec2 θ( a 2 tan 2 θ + b2) − tanθ (2a 2 tanθ sec2θ)
(a 2 tan 2 θ + b2)2
=
= 25x24 (1 − x)74 (1 − 4x)
25x24 (1 − x)74 (1 − 4x) = 0
1
x = 0, 1,
4
a 2 sec2 θ + b2cosec2θ
=
f (x) = x25 (1 − x)75 , x ∈[0, 1]
⇒
a 2 − b2
Thus, area of ∆OPN =
∴ Minimum value of f (x) is f (0) = − 1
25. Let
X
M
K
2c − b > b + c
2
P
tan θ = b / a
250 Application of Derivatives
Therefore, f (θ ) has maximum, when
b
 b
θ = tan −1   ⇒ tan θ =
 a
a
Again,
b
sin θ =
a 2 + b2
, cos θ =
a
a 2 + b2
By using symmetry, we get the required points
 ±a2
± b2 
,


 a 2 + b2
a 2 + b2 
27. Given, P (x) = a 0 + a1x2 + a 2 x4 + K+ a nx2n
where, a n > a n − 1 > a n − 2 > K > a 2 > a1 > a 0 > 0
⇒ P ′ (x) = 2a1x + 4a 2 x3 + ... + 2na nx2n − 1
= 2x (a1 + 2a 2 x2 + K + na nx2n − 2)
…(i)
2n − 2
where, (a1 + 2a 2 x + 3a3 x + K + na nx
P ′ (x) > 0, when x > 0
Thus,

P ′ (x) < 0, when x < 0
2
4
) > 0, ∀ x ∈ R.
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 .
cos 2x cos 2x sin 2x

30. f (x) = 
− cos x cos x − sin x
sin x
cos x 
 sin x
2
2
cos 2x(cos x + sin 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
−1
or
cos 2x =
4
nπ
1
or cos 2x = −
x=
⇒
2
4
f (x) ⋅ g (x)
31. 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)
28. y = a log x + bx2 + x has extremum at x = − 1 and x = 2.
∴
and
and
∴
and
⇒
⇒
dy
= 0, at x = − 1
dx
a
x=2 ⇒
+ 2bx + 1 = 0, at x = − 1
x
⇒
⇒
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.
x=2
−a − 2b + 1 = 0
a
+ 4b + 1 = 0
2
a = 2 and b = −
1
2
 p, if p > q
 q, if q > p
29. 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
∴
⇒
1
(p +
1
( p + q −| p − q|) = 12
2
 (p +
2
 q, if
=
 p, if
[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)
32.
… (i)
[from Eq. (i)]
PLAN
x , if x ≥
We know that,| x| = 
 − x , if x <
x − a,
⇒
| x − a| = 
(
−
 x − a ),
0
0
if x ≥ a
if x < a
and for non-differentiable continuous function, the maximum or
minimum can be checked with graph as
q − p + q), if p ≥ q
Y
Y
q + p − q), if p < q
p≥q
p<q
1
{ p + q − | p − q|} = min ( p, q)
2
O
x=a
Minimum at x = a
X
x=a
Maximum at x = a
X
Application of Derivatives 251
Y
On differntiating Eq. (i) w.r.t a, we get
dv
= 6a 2 − 46ax + 60x2
da
Again, differentiating,
O
d 2v
= 12a − 46x
da 2
 dv 
2
  = 0 ⇒ 6x − 23x + 15 = 0
 da 
5
a = 5 ⇒ x = 3,
6
2
 d v
 2 = 2 (30 − 23x)
 da 
X
x=a
Here,
Neither maximum
nor minimum at x = a
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,
− < x≤0
if
3

if
0 < x≤2
 4x,
 2x + 4, if
x>2
At
⇒
 d 2v 
At x = 3,  2 = 2(30 − 69) < 0
 da 
∴ Maximum when x = 3, also at x =
∴ At x = 5 /6, volume is minimum.
Thus, sides are 8x = 24 and 15x = 45
34. Given,
 ex
,

x −1
,
f (x) = 2 − e
 x−e ,

Graph for y = f (x) is shown as
g (x) = ∫
and
⇒
Also,
Put f ′ ( x ) = 0 and check f ′ ′( x ) to be + ve or − ve for minimum and
maximum, respectively.
Here,
l = 15x − 2a , b = 8x − 2a and h = a
f (t ) dt
g′ (x) = 0 ⇒ x = 1 + log e 2 and x = e.
 ex ,
if 0 ≤ x ≤ 1

g′′ (x) =  − ex − 1 , if 1 < x ≤ 2
, if 2 < x ≤ 3
 1
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 ).
x
0
if 0 ≤ x ≤ 1
if 1 < x ≤ 2
if 2 < x ≤ 3
g′ (x) = f (x)
Put
33.
 d 2v 
5
⇒  2 > 0
6
 da 
x = 1 + log e 2,
At
g′′ (1 + log e 2) = − elog e 2 < 0, g (x) has a local maximum.
Also, at x = e,
g′′ (e) = 1 > 0, g (x) has a local minima.
a
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.
a
8x–2a
8x
35. Since, f (x) has local maxima at x = − 1 and f ′ (x) has
15x – 2a
local minima at x = 0.
∴
15x
f ′ ′ (x) = λx
On integrating, we get
∴ Volume = (8x − 2a ) (15x − 2a ) a
V = 2a ⋅ (4x − a ) (15x − 2a )
f ′ (x) = λ
…(i)
⇒
a
x2
+c
2
λ
+ c = 0 ⇒ λ = − 2c
2
Again, integrating on both sides, we get
f (x) = λ
8x – 2a
15x – 2a
[Q f ′ (− 1) = 0]
x3
+ cx + d
6
…(i)
252 Application of Derivatives
 8
f (2) = λ   + 2c + d = 18
 6
⇒
λ
+ c+ d = −1
6
f (1) =
and
…(ii)
…(iii)
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
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
and
lim f (x) = lim (37 − x)
x → 2+
h→ 0
f (2) = 3 ⋅ 22 + 12 ⋅ 2 − 1 = 12 + 24 − 1 = 35
and
Therefore, LHL = RHL = f (2) ⇒ function is continuous
at x = 2 ⇒ function is continuous in −1 ≤ x ≤ 3.
f (x) − f (2)
Now, Rf ′ (2) = lim
+
x−2
x→ 2
= lim
h→ 0
36. f (x) = ∫
37 − (2 + h ) − (3 × 22 + 12 × 2 − 1)
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
= lim
−1
–
–
–∞ 0
f ′ (x) =
d
dx
x
∫ −1
5
2
3
 d
Q
 dx
ψ( x )
∫ φ ( x)
= lim
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)

For local minimum, f ′ (x) = 0
⇒
Let
3h 2 − 24h + 35 − 35
h→ 0
−h
3h − 24
= lim
= 24
h→ 0
−1
∞
= x(ex − 1)(x − 1)(x − 2)3 (x − 3)5 × 1
x = 0, 1, 2, 3
f ′ (x) = g (x) = x(ex − 1)(x − 1)(x − 2)3 (x − 3)5
−h
[12 + 3h 2 − 12h + 24 − 12h − 1] − 35
= lim
h→ 0
−h
+
–
+
1
[3(2 − h )2 + 12(2 − h ) − 1]


− (3 × 22 + 12 × 2 − 1)
h→ 0
t (e − 1) (t − 1)(t − 2) (t − 3) dt
3
t
f (2 + h ) − f (2)
h
= lim
Hence, (b) and (c) are the correct answers.
x
x → 2+
= lim [37 − (2 + h )] = 35
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.
38. Here, y2 = 16x,0 ≤ y ≤ 6
Y
Using sign rule,
+
−
1
+
−
2
37. For −1 ≤ x ≤ 2 , we have
f (x) = 3x2 + 12x − 1
f ′ (x) = 6x + 12 > 0, ∀ − 1 ≤ x ≤ 2
Hence, f (x) is increasing in [–1, 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−
E (0, 3)
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.
⇒
(4t 2, 8t)
G
fy
Tangent at F , yt = x + at 2
⇒
⇒
∴
x → 2−
h→ 0
h→ 0
= lim (12 + 3h 2 − 12h + 24 − 12h − 1)
h→ 0
= lim (3h 2 − 24h + 35) = 35
h→ 0
x = 0, y = at = 4 t
At
Also, (4 t 2, 8 t ) satisfy y = mx + c .
= lim [3 (2 − h )2 + 12(2 − h ) − 1]
= lim [3 (4 + h 2 − 4h ) + 24 − 12h − 1]
X
O
∴
8 t = 4mt 2 + 3
4mt − 8t + 3 = 0
2
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
–
1
2
Application of Derivatives 253
∴ Maximum at t =
⇒ g (x) = 0 ⇒ h (x) = 0
1
and 4mt 2 − 8t + 3 = 0
2
⇒
m −4 + 3 =0
⇒
m =1
⇒ f (x) = 0 or f ′ (x) = 0
If
G (0, 4t ) ⇒ G (0,2)
⇒
h (x) = 0 has 7 minimum solutions.
y1 = 2
⇒ h′ (x) = g (x) = 0 has 6 minimum solutions.
(x0 , y0 ) = (4t 2, 8t ) = (1, 4)
43. To maximise area of ∆ APB, we know that, OP = 10 and
y0 = 4
sin θ = r /10 , where θ ∈ (0, π / 2)
 3 1 1
Area = 2 −  =
 4 2 2
∴
f (x) =
39.
P(6,8)
2ax
(x + ax + 1) − 2ax
=1 − 2
x + ax + 1
x2 + ax + 1
θ
2
 −2ax + 2a 
 (x − 1) 
= 2
= 2a  2
2
2
+
+
(
)
x
ax
a


 (x + ax + 1) 
2
A
r
...(i)
4a (a − 2)
− 4a
=
(2 − a )3
(a − 2)2
∴ (2 + a )2 f ′ ′ (1) + (2 − a )2 f ′ ′ (−1) = 4a − 4a = 0
40. When x ∈ (−1, 1),
∴ f ′ (x) < 0, f (x) is decreasing.
4a
Also, at
>0
x = 1, f ′′ (1) =
(a + 2)2
[Q0 < a < 2]
So, f (x) has local minimum at x = 1.
f ′ (e ) x
⋅e
1 + (ex )2
g′ (x) = 0, if e2x − 1 = 0, i.e. x = 0
42. Let g (x) =
= (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
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
x
  ex 

e2x − 1
= 2a  2x


x
2
2x
 (e + ae + 1)   1 + e 
If
Y′
1
∴ Area = (2 AQ ) (PQ )
2
= AQ . PQ = (r cos θ ) (10 − OQ )
⇒
2
g′ (x) =
B
⇒ cos 2 θ = 3 sin 2 θ
x <1 ⇒x −1 <0
41.
X
O
2
2x (x2 + ax + 1) − 2 (x2 − 1) (2x + a ) 
= 2a 
 …(ii)
(x2 + ax + 1)3


4a (a + 2)
4a
Now, f ′′ (1) =
=
(a + 2)3
(a + 2)2
2
θ
Q
X′

 (x2 + ax + 1)2 (2x) − 2(x2 − 1)

(x2 + ax + 1) (2x + a ) 
f ′ ′ (x) = 2a 

(x2 + ax + 1)4




f ′′ (−1) =
… (i)
Y
 (x2 + ax + 1) ⋅ 2a − 2ax(2x + a ) 
f ′ (x) = − 

(x2 + ax + a )2


and
f (x) = 0 has 4 minimum solutions.
f ′ (x) = 0 has 3 minimum solutions.
x2 y2
+
=1.
6
3
Now, to minimise the distance from P to given straight
line x + y = 7, shortest distance exists along the
common normal.
44. Let us take a point P( 6 cos θ , 3 sin θ ) on
x < 0, e2x < 1 ⇒ g′ (x) < 0
d
[ f (x) ⋅ f ′ (x)]
dx
To get the zero of g (x), we take function
Y
y
N
X′
P
O
h (x) = f (x) ⋅ f ′ (x)
between any two roots of h (x), there lies atleast one root
of h′ (x) = 0.
π
= 5 units
6
Y′
x+y=7
X
254 Application of Derivatives
Slope of normal at P =
cos θ =
So,
6 sec θ
a 2 / x1
=
= 2 tan θ = 1
6 cosec θ
b2 / y1
we have,
1
2
and sin θ =
3
3
Now,
2ax
2ax − 1
b
2(ax + b)
b+1
2ax + 2b + 1
2ax + b + 1
−1
2ax + b
Applying R3 → R3 − R1 − 2R2, we get
2ax 2ax − 1 2ax + b + 1
f ′ (x) = b
b+1
−1
1
0
0
⇒
= 4[(t − 1 / 2)2 + 3 / 4][2t − 1]
f ′ (t ) = 0
⇒
Hence, required point is P(2, 1).
45. Given, f ′ (x) =
f ′ (t ) = 4(t 2 + 1 − t )(2t − 1)
t = 1 /2
Also, f ′ (t ) < 0 for t < 1 / 2
and f ′ (t ) > 0 for t > 1 / 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.
47. Let the square S is to be bounded by the lines x = ±1 / 2
and y = ±1 / 2.
f ′ (x) = 2ax + b
2
1

1

a 2 =  x1 −  +  − y1

2

2
We have,
On integrating both sides, we get
Y
f (x) = ax2 + bx + c
A(x , 1/2)
Since, maximum at x = 5 /2 ⇒ f ′ (5 /2) = 0
⇒
5a + b = 0
f (0) = 2
⇒
c=2
and
f (1) = 1
⇒
a + b + c = 1 …(iii)
…(ii)
X'
1/2
−1/2
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
Similarly,
Its reflection in y = x is
Q1 (s, s2 + 1), which lies on x2 = y − 1.
∴
= (t − s) + (t − s ) = P1Q
PQ1 = P1Q
2 2
a 2 + b2 + c2 + d 2 = 2(x12 + y12 + x22 + y22) + 2
1
Therefore, 0 ≤ x12, x22, y12, y22 ≤
4
2
[Q both perpendicular to y = x]
Also PP1 || QQ1
Y
C1
x 2 = y −1
y=x
P1
Q1
(0, 1)
Q
O
X′
(1, 0)
0 ≤ x12 + x22 + y12 + y22 ≤ 1
⇒
0 ≤ 2(x12 + x22 + y12 + y22) ≤ 2
But
2 ≤ 2(x12 + x22 + y12 + y22) + 2 ≤ 4
Alternate Solution
P
y2− = 1
X
c2 = x22 + y22
A(x , 1)
Y′
a
d
Thus, PP1QQ1 is an isosceles trapezium.
Also, P lies on PQ1 and Q lies on P1Q , then
B (1, y1)
1/2
(0, y2)D
PQ ≥ min { PP1QQ1 }
X
b
c
Let us take min { PP1QQ1 } = PP1
X′
PQ 2 ≥ PP12 = (t 2 + 1 − t )2 + (t 2 + 1 − t 2)
= 2(t 2 + 1 − t 2) = f (t )
... (i)
Y
C2
∴
−1/2
1
2
1
2
2
2
b = x2 − y1 − x2 + y1 +
2
1
2
2
2
c = x2 − y2 + x2 + y2 +
2
1
2
2
2
d = x1 − y2 + x1 − y2 +
2
Similarly, let coordinates of Q be (s2 + 1, s)
⇒
b
= x12 − y12 − x1 − y1 +
which clearly lies on y2 = x − 1
We have,
c
Y′
Reflection of P in y = x is P1 (t 2 + 1, t )
2
X
B ( −1/2, y)
O
C(x2 −1/2)
46. Let coordinates of P be (t , t 2 + 1)
2
1/2
d
D( −1/2, 1/2)
…(i)
Also,
PQ12
2
[say]
O
Y′
C(x2 , 0)
X
Application of Derivatives 255
b2 = (1 − x2)2 + y12
...(ii)
a 2 = (1 − y1 )2 + (1 − x1 )2
...(iii)
d 2 = x12 + (1 − y2)2
...(iv)
On adding Eqs. (i), (ii), (iii) and (iv), we get
a 2 + b2 + c2 + d 2 = { x12 + (1 − x1 )2} + { y12 + (1 − y1 )2}
+ { x22 + (1 − x2)2} + { y22 + (1 − y2)2}
From sign chart, it is clear that f ′ (x) has no change of
sign in left and right of x = 1/4.
Case III When b > 1, then
2
1
1
1
f ′ (x) =
− b + 2x =  x2 − bx +

x
2
16
8x
=
2 
 x −
x 
=
b 1
b 1
2 
 

b 2 − 1  x − +
b 2 − 1 
x − −
 

x 
4 4
4 4
where x1 , y1 , x2, y2 all vary in the interval [0, 1].
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
⇒ 4x = 2
⇒ 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
Again,
and maximum value is 1 + 1 + 1 + 1 = 4
48. f (x) is a differentiable function for x > 0.
2

1 2
b
(b − 1)
 −

4
16

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

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
=
49. Let equation of any line through the point (h , k) is
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
y − k = m(x − h )
Y
f ′ (x) = 0
For
⇒
1
− b + 2x = 0
8x
⇒
16x2 − 8bx + 1 = 0
Q
⇒
(4x − b)2 = b2 − 1
⇒
(4x − b)2 = (b − 1) (b + 1)
f ′ (x) =
X′
…(i)
[b ≥ 0, given]
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,
1
2
1
1 2
1
− 1 + 2 x =  x2 − x +
 = x − 
4
8x
x
2
16 x 
We have to check the sign of f ′ (x) at x = 1/4.
Interval
Sign of f′(x)
Nature of f(x)
−∞, 0
−ve
↓
 0, 1 


 4
+ ve
↑
 1, ∞


4 
+ ve
↑
2
… (i)
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.
O
(h, k)
P
X
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
1  mh − k
1
(k − mh )2

 (k − mh ) = −
2 m 
2m
1
[Qm = tan θ]
=−
(k − h tan θ )2
2 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
256 Application of Derivatives
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
To find the coordinates of point R, we have to solve it
with
… (ii)
x2 + y 2 = 1
⇒
⇒
On subtracting Eq. (ii) from Eq. (i), we get
(x − 1)2 − x2 = r 2 − 1
k cosec θ − h sec θ = 0
2
2
2
2
2
2
⇒
k
h
=
sin 2 θ cos 2 θ
⇒
tan θ = ±
⇒
⇒
x + 1 − 2 x − x2 = r 2 − 1
⇒
1 − 2x = r 2 − 1
2
k
= tan 2 θ
h2
k
h
2
 2 − r 2
 + y2 = 1

 2 
2
⇒
  k2 + h 2  h 
 h 2 + k2  k  
= k2
  
   + h 2
2
 h2   h 
  k
  k
 (k2 + h 2) h (h 2 + k2)(k) 
=
+

k
h


k
2
2 h
= (k + h )
+
>0
 k h 
⇒
[Q h, k > 0]
Therefore, A is least when tan θ = − k / h. Also, the least
value of A is
1
 − k 
 −h 
2hk − k2  − h 2  
 h 
 k
2 
1
= [2hk + kh + kh ] = 2hk
2
2
50. Since x + 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
X′
(–1,0)P
r
O S
Q(1,0)
X
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=
r 4 − r2
2
Again, we know that, coordinates of S are (1 − r , 0),
therefore
SQ = 1 − (1 − r ) = r
Let A denotes the area of ∆ QSR, then
2
1  4−r 
A = r r

2 
2 


1 2
= r 4 − r2
4
1 4
⇒
A2 =
r (4 − r 2)
16
Let
⇒
f (r ) = r 4 (4 − r 2) = 4r 4 − r 6
f ′ (r ) = 16r3 − 6r5 = 2r3 (8 − 3r 2)
For maxima and minima, put f ′ (r ) = 0
⇒
S2
 2 − r 2
(2 − r 2)2
y2 = 1 − 
 =1 −
4
 2 
=1 −
= − [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 

R
2 − r2
2
On putting the value of x in Eq. (i), we get
[given]
tan θ < 0 , k > 0, h > 0
k
Therefore, tan θ = − (only possible value).
h
d 2A 1
2
2
Now,
=
[
−
2
cosec
θ cot θ − 2h 2 sec2 θ tan θ ]
k
dθ 2 2
S1
x=
∴
Q
⇒
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]
Y′
Again,
(x − 1) + y = r
2
2
2
…(i)
f ′ ′ (r ) = 48r 2 − 30r 4
Application of Derivatives 257
⇒
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 =
2 2
3
Hence, maximum value of A
=
1 2 2


4 3
2
2
2 2
1  8
8
4−
 =   . 4−


4 3
3
 3
1
(t1 − t )
2t
1
(t1 − t )
⇒ (t1 − t ) (t1 + t ) = −
2t
1
⇒
(t1 + t ) = −
2t
1
t1 = − t −
⇒
2t
Then,
Therefore, length of chord,
L = AB2 = (t − t1 )2 + (t 2 − t12)2
= (t − t1 )2 + (t − t1 )2(t + t1 )2
2 12 − 8 2 . 2
4
4 3
= .
=
=
=
3
9
3
3 3 3 3
 3 (b3 − b2 + b − 1)
− x +
, if 0 ≤ x ≤ 1
(b2 + 3b + 2)

2x − 3
, if 1 ≤ x ≤ 3

= (t − t1 )2[1 + (t + t1 )2]
2
2
1 
1 


=  t + t +  1 +  t − t −  


2t  
2t  
51. Given, f (x) = 
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–
In order this value is not less than –1, we must have
b3 − b2 + b − 1
≥0
b2 + 3b + 2
⇒
(b2 + 1) (b − 1)
≥0
(b + 1) (b + 2)
∴
b ∈ (−2, − 1) ∪ [1, ∞ ]
Y
52.
y = x2
B
X′
2
⇒
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
…(i)
Suppose Eq. (i) meets the curve again at B(t1 , t12).
3
3
2
1   2 
1 
dL


= 8t 1 + 2 + 12t 2 1 + 2  − 3 


dt
4t   4t 
4t 
2
1  3
1   

= 2 1 + 2 4t 1 + 2  − 

4t  t 
4t   
2
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 ⇒
t2 =
t
2
1
⇒
t=±
2
2
1
1
dL
 1 

Now,
= 8  1 + 2  − 3   2t − 
2





t
4t
2t
dt
2
1
1  

+ 4  1 + 2  2 + 2

4t  
t 
X
Y′
1 
1 
1 


L =  2t +   1 + 2 = 4t 2 1 + 2


2t  
4t 
4t 
On differentiating w.r.t. t, we get
⇒
A
y = x2 at A (t , t 2) is
 1
y − t 2 =  −  (x − t )
 2t 
t12 − t 2 = −
 d 2L 
 2
 dt  t = ± 1 /
2
2
1

= 0 + 4 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
258 Application of Derivatives
and when t = − 1 / 2 , the equation of AB is
y −2
x− 2
=
1
1

− 2 −   − 2
 2
2
P
r
O
r

 1
1 
⇒ ( y − 2)  −
− 2 = (x − 2 )  − 2
2 


2
Q
R
N
2 y − 4 = 2 (x − 2 )
⇒
⇒
r
M
2x − 2 y + 2 = 0
53. Let 2b be the diameter of the circular portion and a be
the lengths of the other sides of the rectangle.
Total perimeter = 2a + 4b + πb = K
[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.
Let A denotes the area of ∆PQR.
1
Then,
A = ⋅ 2r sin θ (r + r cos θ )
2
⇒
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
⇒
⇒
Coloured
glass
2b
d 2A
= r 2(− sin θ − 2 sin 2θ )
dθ 2
and
For maximum and minimum values of θ,we put
a
Clear glass
a
⇒ 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
2b
Then,
⇒
⇒
⇒
1
Q = 2ab (3L ) + πb2(L )
2
L
2
Q = (πb + 12ab)
2
L
Q = [πb2 + 6b (K − 4b − πb)]
2
L
Q = (6Kb − 24b2 − 5πb2)
2
Hence, the area of ∆PQR is maximum when θ =
[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,
∴
QR = 2QN = 2r sin θ
Also,
ON = r cos θ
∴
PN = r + r cos θ
 3
2π 
3
π 1

A = r 2 sin + sin  = r 2 
+


3 2
3
4
 2
3 3 2
r sq units
4
4x2 + a 2y2 = 4a 2, i.e.
…(ii)
54. Since, the chord QR is parallel to the tangent at P.
Consequently, N is the mid-point of chord QR.
π
.
3
55. Let P (a cos θ , 2 sin θ ) be a point on the ellipse
d 2Q L
= (− 48 − 10π ) < 0
db2 2
ON ⊥ QR
The maximum area of ∆ PQR is given by
=
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
∴
dA
=0
dθ
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
cos θ = 0
[Q 4 < a 2 < 8 ⇒
or
sin θ =
4
a −4
2
4
> 1 ⇒ sin θ > 1, which is
a2 − 4
impossible]
Application of Derivatives 259
Now,
d2
( AP )2 = − {(8 − 2a 2) sin θ + 8} sin θ
dθ 2
+ (8 − 2a 2) ⋅ cos 2 θ
d
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
=
p2 − p 1
1 1 2
q
q 1
= ×
2 2 2
r
−r 1
2
For θ =
Applying R3 → R3 − R1 and R2 → R2 − R1
p2
−p
1
1
2
2
q −p
q+ p 0
=
4
r 2 − p2 − r + p 0
56. Let AF = x and AE = y, ∆ABC and ∆EDC are similar.
C
1
( p + q) (r − p) (− q − r )
4
1
= ( p + q) (q + r ) ( p − r )
4
π
π
57. Let y = f (x) = sin3 x + λ sin 2 x, − < x <
2
2
Let
sin x = t
=
2
( r , – r)
a
b E
D
y
A 2
x
(p , – p )
B
(q 2, –q)
F
∴
c
⇒
⇒
c
b
=
x b− y
bx = c (b − y)
⇒
⇒ x=
c
(b − y)
b
z = xy sin A
c
z = (b − y) y ⋅ sin A
b
…(i)
On differentiating w.r.t. y we get
dz c
d 2z −2c
= (b − 2 y) sin A and
=
sin A
dy b
b
dy2
For maximum or minimum values of z, we must have
dz
=0
dy
c
b
(b − 2 y) = 0 ⇒ y =
⇒
b
2
Clearly,
y = t3 + λt 2, − 1 < t < 1
dy
= 3t 2 + 2tλ = t (3 t + 2λ )
dt
⇒
Let z denotes the area of par
allelogram AFDE.
Then,
p2
−p 1
1
= ( p + q) (r − p) q − p 1 0
4
r + p −1 0
AB AC
=
ED CE
∴
1
area of ∆ABC
2
d 2z
2c
=−
< 0, ∀ y
2
b
dy
b
Hence, z is maximum, when y = .
2
b
On putting y =
in Eq. (i), we get
2
the maximum value of z is
c
1
b b
z =  b −  ⋅ ⋅ sin A = bc sin A
b
2 2
4
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=
58. Given,
x
1 + x2
⇒
1 − x2
dy (1 + x2) ⋅ 1 − x (2x)
=
=
dx
(1 + x2)2
(1 + x2)2
Let
dy
= g (x)
dx
∴
g (x) =
⇒ g ′ (x) =
=
[i.e. slope of tangent]
1 − x2
(1 + x2)2
(1 + x2)2 ⋅ (−2x) − (1 − x2) ⋅ 2 (1 + x2) ⋅ 2x
(1 + x2)4
−2x (1 + x2) [(1 + x2) + 2 (1 − x2)] −2x (3 − x2)
=
(1 + x2)4
(1 + x2)3
For greatest or least values of m, we should have
g′ (x) = 0 ⇒ x = 0, x = ± 3
260 Application of Derivatives
Now,
g′ ′ (x) =
At
(1 + x ) (6x − 6) − (2x − 6x) ⋅ 3 (1 + x ) ⋅ 2x
(1 + x2)6
23
2
3
2 2
x = 0, g′ ′ (x) = − 6 < 0
∴ 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.
59. Let the house of the swimmer be at B.
∴
AB = L km
Let the swimmer land at C on the shore and let
b 2ax3 − b
=
x2
x2
2b
⇒ f ′ ′ (x) = 2a + 3 > 0 [since, a , b are all positive]
x
∴
f ′ (x) = 2ax −
Now, put
 b
f ′ (x) = 0 ⇒ x =  
 2a 
and
S
 b 
f   
  2a 
1/3 
b
 = a  

2
a

2
2
x +d
d
⇒
x
A
(L – x)
C
B
L
∴
SC = x2 + d 2
Time =
∴
CB = (L − x)
and
∴
Let
⇒
x +d
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
2
f ′ (x) = 0 at x = ±
x≠
∴ We consider, x =
f ′ ′ (x) =
27ab2 ≥ 4c3
1
x
⇒
f ′ (x) = 1 −
Also,
f ′ ′ (x) = 2 / x3
1 x2 − 1
=
x2
x2
x = ± 1, but x > 0 [neglecting x = − 1]
f ′ ′ (x) > 0, for x = 1
Hence, f (x) attains minimum at x = 1, y = 1
62. Here, volume of cylindrical container, V = πr 2h …(i)
ud
v − u2
− ud
2
, (v > u )
and let volume of the material used be T.
r+2
v2 − u 2
ud
O r
2
h
d2
x2 + d 2 (x2 + d 2)
Hence, f has minimum at x =
ud
v − u2
2
b
≥ c, ∀ x > 0 ; a , b > 0
x
b
f (x) = ax2 + − c
x
60. Given, ax2 +
3b
≥c
2
f (x) = x +
v2 − u 2
1
u
⋅
⇒ (x + y) has minimum value 2.
∴
But
3b
− c≥0
2
On putting f ′ (x) = 0, we get
v2x2 = u 2 (x2 + d 2)
⇒
Let
1/3
T=
u 2d 2
x2 = 2
v − u2
Now,
 2a 
 
 b
b
− c≥ 0
(b / 2a )1/3
⋅
⇒
For maximum or minimum, put f ′ (x) = 0
⇒
1/3
+
.
61. Let f (x) = x + y, where xy = 1
Time from S to B = Time from S to C + Time from C to B
2
 2a 
= 
 b
2 /3
1/3
On cubing both sides, we get
2a 27b3
⋅
≥ c3
8
b
⇒
Distance
Speed
[Q a , b > 0]
, f ′ ′ (x) = + ve
 b
⇒ f (x) has minimum at x =  
 2a 
AC = x km
>0
1/3
 b
x= 
 2a 
At
1/3
> 0, ∀ x
.
∴
⇒
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
Application of Derivatives 261
2
 r + 2
2
T =V 
 + 2π (r + 2) − V
 r 
⇒
On differentiating w.r.t. r, we get
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 
or
cos θ =
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.
PLAN As to maximise or minimise area of triangle, we should find
area in terms of parametric coordinates and use second
derivative test.
Also,
(2 cos θ, √3 sin θ)
(h,0)
Q
(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
…(i)
= 2 3 ⋅ sin3 θ/cos θ
1
≤ h ≤1
2
1
≤ 2 cos θ ≤ 1
2
1
1
…(ii)
≤ cos θ ≤
4
2
Since,
∴
⇒
∴
d∆ 2 3 {cos θ ⋅ 3 sin 2 θ cos θ − sin3 θ (− sin θ )}
=
dθ
cos 2 θ
=
2 3 ⋅ sin 2 θ
[3 cos 2 θ + sin 2 θ ]
cos 2 θ
=
2 3 sin θ
⋅ [2 cos 2 θ + 1]
cos 2 θ
2
= 2 3 tan θ (2 cos θ + 1) > 0
1
1
≤ cos θ ≤ ,
4
2
2
When
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)
⇒
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
Here, tangent at P(2 cos θ , 3 sin θ ) is
O
1 9
=
2 2
8
∆1 − 8∆ 2 = 45 − 36 = 9
5
∴
64.
P
1
2
 2 3 sin3 θ 
=

 cos θ 
When
V
=π
1000
V
=4
250π
⇒
1  2 3 ⋅ sin3 θ 
=

4 
cos θ 
1 45 5
=
8
4
∆ 2 = ∆ min occurs at cos θ =
r = 10
where
63.
When cos θ =
V
=π
r3
⇒
∆1 = ∆ max occurs at cos θ =
∴
which could be graphically shown as
Y
– 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 =
attains minimum at x = − 1, 0, 1.
⇒ Total number of points = 5
65.
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
2
3
3
∫1 p′ (x) dx = ∫1 λ (x
⇒
2
− 4x + 3) dx
3
x

( p(x))31 = λ  − 2x2 + 3x
3
1
3
262 Application of Derivatives
⇒


1
p(3) − p(1) = λ  (9 − 18 + 9) −  − 2 + 3 

3

⇒
λ =3
⇒
f (θ ) =
p′ (x) = 3 (x − 1) (x − 3)
∴
= 3 + 2 cos 2 θ +
p′ (0) = 9
9
4
5 1
=3− =
2 2
f ′ (x) = 4x3 − 12x2 + 24x + 1
f ′′ (x) = 12x2 − 24x + 24 = 12 (x2 − 2x + 2)
∴ Maximum value of f (θ ) =
= 12 {(x − 1)2 + 1} > 0 ∀x
f ′ (x) is increasing.
Since, f ′ (x) is cubic and increasing.
3
sin 2 θ
2
g (θ )min = 3 − 4 +
∴
f (x) = x − 4x3 + 12x2 + x − 1
4
⇒
1
sin θ + 3 sin θ cos θ + 5 cos 2 θ
2
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
− 4
2 −6 = λ  
3 
⇒
66.
68. Let
1
=2
12
69. 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
So, the number of local maximum is one.
Download Chapter Test
http://tinyurl.com/yxhc5me3
4
X
–16
–20
–21
67. Let g (x) = e f ( x ), ∀ x ∈ R
⇒ f (x) has local maxima at x = 2009.
3
5
∴ f (x) must have one root in (− 1, 0) and other in (0, 1).
⇒ 2 real roots.
⇒ g′ (x) = ef ( x ) ⋅ f ′ (x)
⇒ f ′ (x) changes its sign from positive to negative in the
neighbourhood of x = 2009
2
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.
or
11
Indefinite Integration
Topic 1 Some Standard Results
Objective Questions (Only one correct option)
1. Let α ∈ (0, π / 2) be fixed. If the integral
∫
tan x + tan α
dx = A (x) cos 2α + B (x)
tan x − tan α
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 )|
2. The integral ∫
x3 + 1
+C
x
(c) log e
(a)
(2019 Main, 12 April I)
(x3 + 1)2
1
+C
log e
2
|x3|
|x3 + 1|
(d) log e
+C
x2
(b)


f (x)
 x − 1
= A  tan −1 
 + C,
 + 2


3
x
x
2
10
−
+


where, C is a constant of integration, then
3. If
dx
∫ (x2 − 2x + 10)2
(2019 Main, 10 April I)
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
(a) A =
4. If ∫
dx
= xf (x)(1 +
/
x3 (1 + x6 )23
(c) −
1
3
6x
1
2x2
x4
6(2x + 3x + 1)
4
2
3
x4
(2x + 3x + 1)
4
2
3
+C
+C
(b)
(d)
x12
6(2x + 3x 2 + 1)3
4
x12
(2x + 3x 2 + 1)3
4
+C
+C
x+1
dx = f (x) 2x − 1 + C, where C is a
2x − 1
constant of integration, then f (x) is equal to
7. If
∫
(2019 Main, 11 Jan II)
2
(x + 2)
3
2
(c) (x − 4)
3
1
(x + 4)
3
1
(d) (x + 1)
3
(a)
1 − x2
x4
(b)
dx = A (x)( 1 − x2 )m + C,
for a suitable chosen integer m and a function A (x),
where C is a constant of integration, then ( A (x))m
(2019 Main, 11 Jan I)
equals
1
x6 )3
(a)
+C
(b) −
(d)
(c)
8. If ∫
where, C is a constant of integration, then the function
f (x) is equal to
(2019 Main, 8 April II)
(a) −
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
is a constant of integration)
(2019 Main, 12 Jan II)
2x3 − 1
dx is equal to
x4 + x
|x3 + 1|
1
+C
log e
2
x2
(2019 Main, 8 April I)
(a)
(b)
(c)
(d)
6. The integral ∫
(here C is a constant of integration)
(a)
5x
2 dx is equal to
5. ∫
x
sin
2
(where, C is a constant of integration )
sin
1
3
2x
3
x2
(c)
1
(b)
9x4
−1
(d)
27x9
−1
3x3
1
27x6
π
2
9. Let n ≥ 2 be a natural number and 0 < θ < . Then,
1
(sin n θ − sin θ ) n cos θ
dθ is equal to
∫
sin n + 1 θ
(where C is a constant of integration)
(2019 Main, 10 Jan I)
264 Indefinite Integration
(a)
1
1 −



sin n + 1 θ 
n 2 − 1
1
1 +

(b) 2

n −1 

sin
θ
n −1
n
1
1 −



sin n − 1 θ 
n 2 − 1
(d)
1
1 −



sin n − 1 θ 
n 2 + 1
(a)
+C
n+1
n
(c)
−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
n+1
n
n
+C
n+1
n
n
+C
n+1
n
n
+C
15. If I = ∫
5x + 7x
dx, (x ≥ 0), and f (0) = 0, then
(x2 + 1 + 2x7 )2
(2019 Main, 9 Jan I)
the value of f (1) is
8
10. If f (x) = ∫
6
1
(a) −
2
1
(c)
4
1
(b) −
4
1
(d)
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
(a) log e|sec(x2 − 1)| + C
2
 x2 − 1
(b) log e sec 
 +C
 2 
(a)
is equal to
dx
5
f occurs n times
Then, ∫ xn − 2g (x) dx equals
1
1 4
x +
(a)  4  + c
 x 
(c) − (x +
4
1
1) 4
+c
1
n
1
(1 + nxn )
n (n − 1)
(b)
1−
1
(1 + nxn ) n + c
n −1
(c)
1+
1
(1 + nxn ) n + c
n (n + 1)
(d)
1+
1
(1 + nxn ) n + c
n+1
(a) 2 2 −
(2018 Main)
−1
+C
3 (1 + tan3 x)
−1
(d)
+C
1 + cot3 x
(b)
(2007, 3M)
+c
(b) (x4 +
1
1) 4
+
x3
1
x4
2 x4 − 2 x2 + 1
+c
2
1
1
2− 2 + 4 + c
2
x
x
is
(b) 2 2 +
(2006, 3M)
2
x2
+
1
x4
+ c
(d) None of these
18. 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.)
(2015 Main)
+c
(a) f (2) < 1 − log e 2
(c) g (1) > 1 − log e 2
(b) f (2) > 1 − log e 2
(d) g (1) < 1 − log e 2
Numerical Value
1
 x4 + 1 4
(d) −  4  + c
 x 
sec x
∫ (sec x + tan x)9/ 2 dx equals to
(for some arbitrary constant K)
(c)
2
x2
(x2 − 1) dx
One or More Than One
19. Let f : R → R be a differentiable function with f (0) = 1
2
14.
1−
(a)
2
(where C is a constant of integration)
dx
13. The value of ∫ 2 4
is
x (x + 1)3/ 4
4
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
17. The value of ∫
+ cos x)
+C
(d)
1
sin 2 x cos 2 x
3 (1 + tan3 x)
1
(c)
+C
1 + cot3 x
e2x + ex + 1
1
+c
log 2x
2
e − ex + 1
1
∫ (sin5 x + cos3 x sin 2 x + sin3 x cos2 x
(a)
e2x − ex + 1
1
+c
log 2x
2
e + ex + 1
(b)
1
1
sec2 (x2 − 1) + C
2
 x2 − 1
1
(d) log e sec2 
 +C
2
 2 
1
e4 x − e2x + 1
1
+c
log 4 x
2
e + e2x + 1
16. If f (x) =
(c) log e
12. The integral
ex
e− x
,
=
dx
J
∫ e− 4x + e−2x + 1 dx.
+ e2x + 1
Then, for an arbitrary constant c, the value of J − I
(2008, 3M)
equals
(c)
11. For x2 ≠ nπ + 1, n ∈ N (the set of natural numbers), the
e4x
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

(2012)
and satisfying the equation f (x + y) = f (x) f ′ ( y)
+ f ′ (x) f ( y) for all x, y ∈ R.
Then, the value of log e ( f (4)) is ....... .
(2018 Adv.)
Indefinite Integration 265
Fill in the Blank
dx
∫ x2(x4 + 1)3/ 4 .
24. Evaluate
4ex + 6e− x
20. If ∫ x
dx = Ax + B log (9e2x − 4) + C, then A = K,
9e − 4e− x
(1989, 2M)
B = ... and C = K .
25. Evaluate the following:
(i)
∫
1 
1 + sin  x dx
2 
Analytical & Descriptive Questions
26. Integrate
21. For any natural number m, evaluate
∫ (x
3m
+ x2m + xm ) (2 x2m + 3 xm + 6)1/m dx, x > 0. (2002, 5M)
22. Evaluate
∫
23. Evaluate ∫
 1 − x


 1 + x
1/ 2
⋅
dx
x
(1997C, 3M)
1− x
dx.
1+ x
(1985, 2 1 M)
2
(1984, 2M)
(1980, 4M)
(ii)
∫
2
x
dx
1−x
x2
.
(a + bx)2
(1979, 2M)
27. Integrate
sin x ⋅ sin 2 x ⋅ sin 3x + sec2 x ⋅ cos 2 2x + sin 4 x ⋅ cos 4 x
(1979, 1M)
.
x
(1978, 1M)
28. Integrate the curve
.
1 + x4
1
sin x
(1978, 2M)
29. Integrate
or
.
1 − cot x sin x − cos x
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
3. Find the indefinite integral
∫
(2019 Main, 9 April I)
−1/3
x+C
(b) −3 tan
x+C
3
−4 /3
(d) − tan
x+C
4
(c) −3 cot −1/3 x + C
4
x
tan x +
+
ln (1 + 6 x )
 dx.
3
x+ x 
cot x ) dx.
(1992, 4M)
(1988, 3M)
(cos 2x)1/ 2
(1987, 6M)
∫ sin x dx.
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
(2017 Main)
1
(b)  , 0
5 

1
3
 x+
∫(
4. Evaluate
2. Let I n = ∫ tan n x dx (n > 1). If
1
(a)  − , 1
 5 
Analytical & Descriptive Questions
1
1
(c)  , − 1 (d)  − , 0
5

 5 
x→ 0
Topic 3 Integration by Parts
Objective Questions (Only one correct option)
1. If
5 − x2
∫xe
− x2
dx = g (x)e
+ C, where C is a constant of
3. The integral ∫ cos (log e x) dx is equal to
constant of integration)
x
[cos(log e x) + sin(log e x)] + C
2
integration, then g (− 1) is equal to (2019 Main, 10 April II)
(a)
(a) − 1
1
(c) −
2
(b) x [cos(log e x) + sin(log e x)] + C
(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)
1
(a) x sec x + tan x +
2
(b) sec x + tan x +
1
2
1
2
(d) sec x − tan x −
1
2
(c) sec x + x tan x −
(where C is a
(2019 Main, 12 Jan I)
(c) x [cos(log e x) − sin(log e x)] + C
x
(d) [sin(log e x) − cos(log e x)] + C
2
3
4. If ∫ x5 e−4x dx =
1 −4 x 3
e
f (x) + C ,
48
where C is a constant of integration, then f (x) is equal to
(2019 Main, 10 Jan II)
(a) − 4x3 − 1
(b) 4x3 + 1
(c) − 2x3 − 1
(d) − 2x3 + 1
266 Indefinite Integration
1
5.
∫
1 x +

1 + x −  e x dx is equal to

x
x+
1
x
x+
1
x
(a) (x − 1) e
(c) (x + 1) e
(2014 Main)
x+
+ c
(b) x e
+ c
(d) − x e
1
x
x+
Analytical & Descriptive Questions

 dx.
 4x2 + 8 x + 13 



7. Evaluate ∫ sin −1 
+ c
1
x
+ c
(a)
(b)
(c)
(d)
(2000, 5M)
8. Find the indefinite integral
 cos θ + sin θ 
6. If ∫ f (x) dx = ψ (x), then ∫ x f (x ) dx is equal to
5
2x + 2
3
1 3
[x ψ(x3 ) − ∫ x2ψ(x3 )dx] + c
3
1 3
x ψ(x3 ) − 3 ∫ x3 ψ(x3 ) dx + c
3
1 3
x ψ(x3 ) − ∫ x2ψ(x3 ) dx + c
3
1 3
[x ψ(x3 ) − ∫ x3 ψ(x3 ) dx] + c
3
∫ cos 2θ log cos θ − sin θ  dθ.
(2013 Main)
9. Evaluate ∫
10. Evaluate ∫
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.
(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 ∫
2x12 + 5x9
dx is equal to
(x5 + x3 + 1)3
−x
5
(a)
(b)
(c)
(d)
(x + x + 1)
5
3
2
2(x5 + x3 + 1)2
x5
2(x + x + 1)
3
2
− x10
2(x5 + x3 + 1)2
cos3 x + cos5 x
dx is
sin 2 x + sin 4 x
(1995, 2M)
(a) sin x − 6 tan −1 (sin x) + c
(2016 Main)
(b) sin x − 2 (sin x)−1 + c
+C
x10
5
2. The value of ∫
(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
+C
Analytical & Descriptive Questions
+C
3.
+C
∫
x3 + 3x + 2
dx.
(x + 1)2 (x + 1)
2
4. Evaluate ∫
where, C is an arbitrary constant.
(x + 1)
dx.
x (1 + xex )2
(1999, 5M)
(1996, 2M)
Answers
Topic 1
1. (b)
( x 4 + 1 )1/4
+c
x
x
x
25. (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


24. −
2. (c)
3. (c)
4. (b)
5. (c)
6. (b)
7. (b)
8. (c)
9. (c)
10. (c)
11. (b)
12. (b)
13. (d)
14. (c)
15. (c)
16. (a)
17. (c)
18. (b,c)
19. (2)
3
35
20. A = − , B = and C ∈R
2
36
1
3m
21.
⋅(2 x + 3 x 2m + 6 xm ) (m + 1)/m + c
6 (m + 1 )
1
22. 2 [cos−1 x − log| 1 + 1 − x | − log | x| ] + c
2
23. −2 1 − x + cos−1 x + x (1 − x ) + c
26.

1 
a2
log
(
)
2
a
bx
a
a
bx
+
−
+
−
+ c

3
a + bx
b 

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
29. log (sin x − cos x ) + + c
28. tan ( x ) + c
2
2
2
27. −
Indefinite Integration 267
Topic 2
8.
1. (b)
2. (b)
3 2/3 12 7/12 4 1/2 12 5/12 1 1/3
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 )
+ 12 ln (1 + x1/2 ) − 3 [ ln(1 + x1/6 )]2 + c
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
2 − 1 − tan x
2
6. (1)
Topic 3
1. (d)
2. (b)
5. (b)
7. ( x + 1 ) tan
−1
3. (a)
4. (a)
+c
 cosθ + sin θ  1
1
sin 2 θ ln 
 + ln(cos2 θ ) + c
2
 cosθ − sin θ  2
2
[ x − x 2 − (1 − 2 x ) sin −1 x ] − x + c
π
ex
10.
+c
(x + 1)2
cos2 x
11. x sin x + cos x −
+c
4
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
6. (c)
 2x + 2 3
2
 − log( 4 x + 8 x + 13 ) + c

 3  4
1
xe x
+
+c
1 + xe x
1 + xe x
Hints & Solutions
Topic 1 Some Standard Results
1. Let I = ∫
tan x + tan α
 π
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 )
Now, put x − α = t ⇒ dx = dt, so
I=∫
sin (t + 2α )
dt
sin t
sin t cos 2 α + sin 2α cos t
dt
sin t
cos t 

= ∫  cos 2 α + sin 2 α
 dt

sin t 
2.
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
=
∫ 2 1 dx
x4 + x
x +
x
[dividing each term of numerator and
Let integral is I = ∫
denominator by x2]
1

 1 
Put x2 + = t ⇒ 2x +  − 2  dx = dt
 x 

x
∴
I=∫
=∫
= 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
= 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 )|
dt
= log e|(t )| + C
t
1

= log e  x2 +  + C

x
= log e
3. Let I = ∫
x3 + 1
+C
x
dx
dx
=
(x2 − 2x + 10)2 ∫ ((x − 1)2 + 32)2
Now, put x − 1 = 3 tan θ ⇒ dx = 3 sec2θ dθ
So,I = ∫
3sec2θ dθ
3sec2θ dθ
=∫ 4
2
2 2
(3 tan θ + 3 )
3 sec4θ
2
268 Indefinite Integration
1
1 1 + cos 2θ
dθ
cos 2θ dθ =
27 ∫
27 ∫
2
1 + cos 2θ 

2

Q cos θ =
2
1 
sin 2θ 
1
=
(1 + cos 2θ ) dθ =
 +C
θ +
54 
2 
54 ∫
=
=
1
1  2 tan θ 
 x − 1
tan −1 

 +C
 +
 3  108  1 + tan 2 θ 
54

2 tan θ 
Q sin 2θ =

1 + tan 2 θ 

 x − 1


 3 
1
1
 x − 1
tan −1 
=
 +
 3  54
54
+C
2
 x − 1
1+ 

 3 

1
1 
x−1
 x − 1
tan −1 
+C
=

 +
2
2


54
3
18  (x − 1) + 3 
= ∫ [3 − 2(1 − cos 2x) + 2 cos x]dx
= ∫ [3 − 2 + 2 cos 2x + 2 cos x]dx

x−1
1
1 
 x − 1
tan −1 
=
 +C
 2
 +


54
3
18  x − 2x + 10
= x + 2 sin x + sin 2x + C
6. Let
It is given, that


f (x)
 x − 1
I = A tan −1 
 + 2
+C


3
x − 2x + 10 

4. Let I = ∫
I=∫
1
and f (x) = 3(x − 1).
54
dx
x3 (1 + x6 )2/ 3
dx
=∫
2/ 3
=∫

1
x3 ⋅ x4  6 + 1

x
1
Now, put
+ 1 = t3
x6
6
− 7 dx = 3t 2dt
⇒
x
So, I = ∫
=
dx
t
= − dt
7
2
x
So, I = ∫
−
∫
1/3
x12
+C
6 (2x4 + 3x2 + 1)3
7. We have,
1 2
t dt
1
2
= − ∫ dt
2
t2
1 1
1

t + C = −  6 + 1 + C

2 x
2
1 1
(1 + x6 )1/3 + C
=−
2 x2
= x ⋅ f (x) ⋅ (1 + x6 )1/3 + C
On comparing both sides, we get
1
f (x) = − 3
2x
=−
[on dividing numerator and denominator by x16]
3
1
Now, put 2 + 2 + 4 = t
x
x
− dt
1 t− 4 + 1
1
=− ×
+C= 3 +C
4
2 −4 + 1
6t
2t
1
3
1

+C
=
3
Q t = 2 + x2 + x4 
3
1

6 2 + 2 + 4 

x
x 
2/ 3
2
⇒
3
2
+ 5
3
3x13 + 2x11
x
x
dx
dx = ∫
4
(2x4 + 3x2 + 1)4
3
1

2 + 2 + 4 

x
x 
2
dt
−6 4 
3
⇒  3 − 5  dx = dt ⇒  3 + 5  dx = −
x
x
2
x 
x 
dx

1
x7  6 + 1

x
[Q2 sin 2 x = 1 − cos 2x]
= ∫ [1 + 2 cos 2x + 2 cos x]dx
1 
3(x − 1) 
−1  x − 1 
=
 + 2
tan 
+C
54 
3  x − 2x + 10 
On comparing, we get A =
x
5x
5x
cos
2 sin
2
2
2
5. Let I = ∫
dx
dx = ∫
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
sin
1

 3
Q t = x6 + 1
[given]
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
t2 + 1
+1
1
I=∫ 2
tdt = ∫ (t 2 + 3) dt
t
2

t2 + 1 
2

Q 2x − 1 = t ⇒ x =
2 

Indefinite Integration 269

1  t3
t
 + 3t + C = (t 2 + 9) + C
23
6

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
1
=
⇒
I=∫
u1/ ndu
=
n −1
[Q t = 2x − 1 ]
n+1
1  n

n 1 − n−1 

t 
+C
=
(n − 1) (n + 1)
1


n 1 −


sin n − 1 θ 
=
2
n −1
1 − x2
x4
dx = A (x) ( 1 − x2 )m + C
LetI = ∫
… (i)

1
x2 2 − 1

x
1 − x2
dx = ∫
dx
x4
x4
1
x 2 −1
1 1
x
dx = ∫ 3
=∫
− 1 dx
x4
x x2
1
−2
1
Put 2 − 1 = t 2 ⇒ 3 dx = 2t dt ⇒ 3 dx = − t dt
x
x
x
t3
I = − ∫ t 2dt = −
+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
( A (x))m = ( A (x))3 = −
∴
27 x9
(sin θ − sin θ ) cos θ
dθ
sin n + 1 θ
sin θ = t ⇒ cos θ dθ = dt
(t n − t )1/ n
dt
I=∫
tn + 1
1/ n
t 
 n
−
1
t


 
t n  
=∫ 
dt
n+1
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)
dt = du
1 − t −( n − 1) = u ⇒
tn
dt
du
=
tn n − 1
9. Let I = ∫
Put
∴
Put
or
⇒
n
1/ n
10. We have, f (x) = ∫
+C
5 x8 + 7 x 6
dx
(x + 1 + 2x7 )2
2
=∫
 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

=−
n+1
n
1


Q u = 1 − t n − 1 and t = sin θ 
8. We have,
∫
+1
un
+C

1
(n − 1)  + 1

n
…(ii)
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
1
t− 2 + 1
t− 1
=−
+ C =−
+C= +C
t
−2 + 1
−1
=
x
−5
1
x7
+C= 7
+C
−7
2 x + x2 + 1
+x +2
Q f (0) = 0
∴
0=
0
+ C ⇒C = 0
0+0+1
∴ f (x) =
x7
2 x + x2 + 1
⇒
f (1) =
7
11. 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 ⇒ x2 − 1 = 2θ ⇒ 2x dx = 2 dθ
2
⇒
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
(Qsin 2 A = 2 sin A cos A)
270 Indefinite Integration
=∫
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
Put sec x + tan x = t
⇒ (sec x tan x + sec2 x) dx = dt
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]
∴
tan θ d θ = ∫ tan θd θ
∴
=∫
2
 x2 − 1 
= log e|sec θ| + C = log e sec 
 +C
 2 
12. We have,

x2 − 1 

Q θ =
2 

=∫
sin x ⋅ cos x
dx
(sin5 x + cos3 x ⋅ sin 2 x + sin3 x ⋅ cos 2 x + cos5 x)2
sin 2 x cos 2 x
dx
3
2
{sin x(sin 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
3
3
2
(sin x + cos x)
cos 6 x(1 + tan3 x)2
=∫
tan 2 x sec2 x
dx
(1 + tan3 x)2
2
I=∫
dx
1+
Put
1

x5 1 + 4 

x 
3/ 4
1
= t4
x4
−4
dx = 4t3 dt
x5
dx
= − t3 dt
x5
⇒
⇒
Hence, the integral becomes
∫
14.
=−
1
− t3 dt

= − ∫ dt = − t + c = − 1 + 4 

t3
x 
1/ 4
1 2
2 
+

+ K
2  7 t7/ 2 11 t11/ 2 


1
1
= −
+
+K
7/ 2
11/ 2 
11 (sec x + tan x) 
 7 (sec x + tan x)
=
15. Since,
dx
∫ x2(x4 + 1)3/ 4 = ∫
⇒ I=∫
2
Put tan3 x = t ⇒ 3 tan 2 x sec2 xdx = dt
1
dt
I= ∫
∴
3 (1 + t )2
−1
−1
⇒
+C
I=
+C ⇒ I=
3 (1 + t )
3 (1 + tan3 x)
13.
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
1
1 dt
t +  ⋅
1 
t t 1  1
2
= ∫  9/ 2 + 13/ 2 dt
2 t
t 
t 9/ 2
sec x ⋅ t dx = dt ⇒ sec x dx =
⇒
∴
Put
−1
1
1

+ (sec x + tan x)2 + K
11/ 2 
(sec x + tan x)
11 7

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 
+c
=∫
PLAN Integration by Substitution
i.e.
I = ∫ f { g ( x )} ⋅ g ′ ( x )dx
Put
g ( x ) = t ⇒ g ′ ( x )dx = dt
∴
I = ∫ f(t )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,
(sec x − tan x) (sec x + tan x) = 1
1
sec x − tan x =
t
I=∫
sec2 dx
(sec x + tan x)9/ 2
1
t −1
dt
= log
+c
t+1
t2 − 1 2
=
u2 − u + 1
1
log 2
+c
2
u +u+1
=
e2x − ex + 1
1
log 2x
+c
2
e + ex + 1
x
for n ≥ 2
(1 + xn )1/ n
f (x)
x
=
ff (x) =
∴
[1 + f (x)n ]1/ n (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
16. Given, f (x) =
n times
Indefinite Integration 271
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 = ∫
Let
=
1
n2 ∫
1
I=
17. Let I = ∫
Put
⇒
∴ I=
1−
1
(1 + nxn ) n + c
n (n − 1)
(x2 − 1) dx
2 x4 − 2 x2 + 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
∫
x3
dt 1 t1/ 2
= ⋅
+c
t 4 1 /2
=
1
2
1
2− 2+ 4 + c
2
x
x
−f ( x )
1
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
2x
t
9e − 4
where t = 9e2x − 4
…(i)
…(ii)
…(iii)
e− f ( 2) = 2e−1 − e− g(1)
⇒ e− f ( 2) > 2e−1
We know that, e− x is decreasing
∴
− f (2) < log e 2 − 1
f (2) > 1 − log e 2
[from Eq. (iii)]
⇒
e− g(1) + e− f (2) = 2e−1
⇒
e− g(1) < 2 e−1
− g (1) < log e 2 − 1
⇒
g (1) > 1 − log e 2
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)
4e2x + 6
dx
9e2x − 4
4e2x + 6 = A (9e2x − 4) + B (18 e2x )
= A x + B log (9e2x − 4) + c
3
35
=− x+
log (9e2x − 4) + c
2
36
3
35
A =− ,B=
2
36
∴
c = any real number
and
[Q g(2) = 1]
x
4ex + 6e− x
dx = Ax + B log (9e2x − 4) + c
9ex − 4e− x
20. Given, ∫
− g( x )
[Q f (1) = 1]
x
f (x) = e2
1
⇒
log e f (x) = x
2
1
⇒ log e f (4) = × 4 = 2
2
Hence,
⇒
f ( x)
e
=e
+C
At x = 1
e− f (1) = e− g(1) + C
−1
e = e− g(1) + C
At x = 2
e− f ( 2) = e− g( 2) + C
⇒
e− f ( 2) = e−1 + C
From Eqs. (i) and (ii)
19. Given,
1
⇒
f (x) = Ae2 , where eC = A
If f (0) = 1, then A = 1
Let
e
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 ′ (x) =
1 = 2 f ′ (0) ⇒ f′ (0) =
LHS = ∫
18. We have, f ′ (x) = e( f ( x ) − g( x )) g′ (x) ∀ x ∈ R
⇒
1
2
Put x = x and y = 0, we get
f (x) = f (x) f ′ (0) + f ′ (x) f (0)
1
f (x) = f (x) + f ′ (x)
⇒
2
f ′ (x) 1
1
=
⇒
f ′ (x) = f (x) ⇒
f (x) 2
2
On integrating, we get
1
log f (x) = x + C
2
⇒
21. For any natural number m , the given integral can be
written as
I = ∫ (x3 m + x2m + xm )
(2x3 m + 3x2m + 6xm )1/ m
dx
x
⇒ I = ∫ (2 x3 m + 3x2m + 6 xm )
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=∫t
1 /m
=
+1
1
tm
dt
=
⋅
6m 6m  1

 + 1

m
1
⋅ (2 x3m + 3x2m + 6 xm )(m + 1)/m + c
6 (m + 1)
272 Indefinite Integration
22. Let
 1 − x
I=∫ 

 1 + x
1/ 2
⋅
x

= ∫  cos + sin

4
dx
x
Put
x = cos 2 θ ⇒ dx = − 2 cos θ sin θ dθ
∴
 1 − cos θ 
I=∫ 

 1 + cos θ 
1/ 2
⋅
− 2 cos θ ⋅ sin θ
dθ
cos 2 θ
Put 1 − x = t 2 ⇒ − dx = 2 t dt
θ
sin
− 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
+
I = 2 cos −1 x − log
x

⇒
1


I = 2 cos −1 x − log|1 + 1 − x|− log|x| + c
2


23. Let
I=∫

1
−1 + c
x

(1 − t 2)2 ⋅ (−2t )
dt
t
= − 2 ∫ (1 − 2t 2 + t 4 ) 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=
26. Let
I=∫
1− x
dx
1+ x
1 − cos θ
⋅ (− 2 sin θ cos θ ) dθ
1 + cos θ
θ
θ
= − ∫ 2 tan ⋅ sin θ cos θ dθ = − 2 ∫ 2 sin 2 ⋅ cos θ dθ
2
2
= − 2 sin θ + θ +
24.
1
b3
=
1 
a 2
 t − 2 a log t −  + c
3
t
b 
=

1 
a2
+ c
 a + bx − 2 a log (a + bx) −
3
a + bx 
b 
1
(sin 4x + sin 2 x − sin 6x) dx
4∫
cos 4x cos 2 x cos 6 x
=−
−
+
16
8
24
∫ (1 + cos 2 θ ) dθ
I 2 = ∫ sec2 x ⋅ cos 2 2 x dx
= ∫ sec2 x (2 cos 2 x − 1)2dx
= − 2 1 − x + cos −1 x + x (1 − x) + c
dx
dx
Let I = ∫ 2 4
=
3/ 4
x (x + 1)3/ 4 ∫ 2 3 
1
x ⋅ x 1 + 4 

x 
∴
1
I=−
4
=−
25. (i) Let
I=∫
=∫
∫
= ∫ (4 cos 2 x + sec2 x − 4) dx
= ∫ (2 cos 2 x + sec2 x − 2) dx
4
dx = dt
x5
dt
1
1 t1/ 4

=− ⋅
+ c = − 1 + 4 
3/ 4

4 1 /4
x 
t
= sin 2 x + tan x − 2 x
1/ 4
+c
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
=
(x4 + 1)1/ 4
+c
x
1 + sin
cos 2
x
dx
2
x
x
x
x
+ sin 2 + 2 sin cos dx
4
4
4
4
∫
=
sin 2θ
+c
2
Put 1 + x− 4 = t ⇒ −

2 a a 2
+ 2  dt
1 −
t
t 

=
27. Let I1 = ∫ sin x sin 2 x sin 3x dx
= − 2∫ (1 − cos θ ) cos θ dθ = − 2 ∫ ( cos θ − cos 2 θ ) dθ
= − 2∫ cos θ dθ +
x2
(a + bx)2
Put a + bx = t ⇒ b dx = dt
2
 t − a


 t 2 − 2 at + a 2
 b  dt 1
∴
I=∫
= 3 ∫ 
⋅
 dt
2
b b
t
t2


Put x = cos 2 θ ⇒ dx = − 2sin θ cos θ dθ
∴
x2
dx
1−x
I=∫
(ii) Let
x
x
x
 dx = 4 sin − 4 cos + c

4
4
4
∴
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 273
28. Let
I=∫
x dx 1
2x
= ∫
dx
4
2
1+ x
1 + (x2)2
Put x2 = u ⇒ 2 x dx = du
∴
du
1
1
1
I= ∫
= tan −1 (u ) + c = tan −1 (x2) + c
2
2 1 + u2 2
∴

1
I = ∫ 3
 x+
3. Let

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
=∫
I1 = 12 ∫
= 12 ∫
∴I=∫
= −3
4/3
x in denominator]
sec2 x dx
1
t3
+C =
1
dt
t+1
+ 12 ln (t + 1)
and I 2 = ∫
∴
 ln (1 + 6 x ) 
 3
 dx
 x+ x 
x = u6 ⇒
I2 = ∫
dx = 6 u5 du
ln (1 + u ) 5
ln (1 + u )
. 6 u5 du
6u du = ∫ 2
2
3
u (1 + u )
u +u
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
∴ 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 =
tan n + 1 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 )
3
2


(tan x)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 
+1
−3
t11
dt
t + t3
4
+ 12∫
=6 ∫
4
x)3
dt
t3
+C
=
4/3
−4
t
+1
3
1

 dx
x
4
x sin3 x
Now, put tan x = t ⇒ sec2 x dx = dt
−4
4
= 12∫ (t7 − t 6 + t5 − t 4 + t3 − t 2 + t − 1) dt
Put
(tan

 dx,
x
 t8
t7 t 6 t5 t 4 t3 t 2 
= 12  −
+
− +
− + − t
7
6 5
4 3 2

8
x
[dividing and multiplying by cos
4
x = t12 ⇒ dx = 12 t11dt
dx
2
cos3
ln (1 + 6 x )
 dx
3
x+ x 
ln (1 + 6 x )
dx
3
x+ x

1
I1 = ∫  3
 x+
Topic 2 Some Special Integrals
1. Let I = ∫ sec3 x cos ec3 x dx = ∫
x
+
I = I1 + I 2
29. Let I = ∫
then A + B = 1 and A − B = 0
4
−∫
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
274 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 )
5

2
−  x1/ 2 − x1/3 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
+ (2 x
1/ 2
I=∫
4. Let
− 3x
1/3
+ 6x
1/ 6
I=∫
t2
=2 ∫
t−
Put
⋅
t2 +
1
−2 + 2
t2
I=
⇒
dt = 2 ∫
1+
1
t2
2
1

2
t −  + ( 2)

t
2
 u
tan −1   + c
 2
2
=∫
cos 2 x − sin 2 x
dx
sin 2 x
cot2 x − 1 dx
Put cot x = sec θ
∴
I=∫

⇒ − 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 )
 2 + 1 − tan 2 x 
1
log 
+ c
2
2
 2 − 1 − tan x 

x

x 
 sin 2 
2 =1
= 4 ⋅ 1 ⋅ lim 
2
x→ 0 
 x 
4 ×   
2 

dt
du
u 2 + ( 2 )2
cos 2 x
dx = ∫
sin x
cot2 x − 1 |
x

2 sin 2 
 sin x 
2
lim f ′ (x) = lim 2 

 x   x2 
x→ 0
x→ 0


 tan x − cot x 
= 2 tan −1 
 +c
2


5. Let I = ∫
 2 + sin θ 
1
+ c
log 
2 2
 2 − sin θ 
2 sin x − sin 2 x
6. Given, f (x) = ∫ 
 dx
3
1
1
= u ⇒ 1 + dt = du
t
t2
I =2 ∫
∴
1
t2
dt
∫ 2 − t 2 , where sin θ = t
= − log|sec θ + tan θ |+ 2 ⋅
= − log|cot x +
cos θ
dθ
2 − sin 2 θ
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
2t
t2 + 1
dt
=
2
∫ t 4 + 1 dt
t4 + 1
1+
= − log|sec θ + tan θ | +
+ 11) ln (1 + x )
2
t2 + 1
= − log|sec θ + tan θ | + 2 ∫
+
Put tan x = t ⇒ sec x dx = 2t dt
2t
dt
⇒ dx =
1 + t4
∴
cos θ
dθ
1 + cos 2 θ
1/ 6
+ 12 ln (1 + x1/12) − 3 [ln (1 + x1/ 6 )]2 + c
tan x + 1
( tan x + cot x ) dx = ∫
dx
tan x
2
= − ∫ secθ dθ + 2 ∫
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 275
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
∫ 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
3
1 −4 x 3
4. Given, ∫ x5 e−4x 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

[using integration by parts]
1  te
e 
= 
+
+C
3  −4
−16 
−4 t
=−
e−4x
[4x3 + 1] + C
48
[Q t = x3 ]
f (x) = −1 − 4x3 (comparing with given equation)
1
5.
∫
1 x +

1 + x −  e x dx

x
x+
1
x dx
+
x+
1
x dx
+ xe
=∫e
=∫e
1

x+
∫ x 1 − x2 e
x+
1
x
1
x dx
1
−∫
1
x
− ∫ ex
x+
dx
1
1
x+ 
1 x+ x

x
=
1
−
e
dx
e



x2

1
x
x+
dx = xe
1
x
+c
I = ∫ x5 f (x3 ) dx
x3 = t
dt
x2dx =
3
1
t f (t ) dt
3∫
1
= t ⋅ ∫ f (t ) dt − ∫
3
…(i)
I=
d
 
 (t ) ∫ f (t ) dt  dt 
dt

 


2x + 2
 dx
I = ∫ sin −1 
 4x2 + 8 x + 13 


7. Let
=
∫ sin

2x + 2
 dx
 (2x + 2)2 + 9 



−1 
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+
dx + xe
1
x
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
⇒
1 −4 t
e [4t + 1] + C
48
1
x
⇒
3. Let I = ∫ cos(log e x)dx
=−
1
x
6. Given, ∫ f ( x ) dx = ψ( x )
Let
−4 t
x+
+ xe

Q ∫

= 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 
+ c1
− log 1 + 

 3  

276 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 2 log 3 + c1 = c
 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
1 d   cos θ + sin θ  
  sin 2 θdθ
ln 
2 ∫ dθ   cos θ − sin θ  
d
dθ
From Eqs. (i) and (ii),
4  1
1

I = − (1 − 2 x) sin −1 x +
x − x2  − x + c
π  2
2

2
2
−1
x] − x + c
= [ x − x − (1 − 2 x) sin
π
10. Let I = ∫
 cos θ + sin θ 
sin 2 θ
ln 

2
 cos θ − sin θ 
(x − 1) ex
dx
(x + 1)3
x + 1 − 2 x
 1
 x
2
e dx = ∫ 
I=∫
−
e dx
3 
2
3
(x + 1) 
 (x + 1) 
 (x + 1)
1
1
= ∫ ex ⋅
dx − 2 ∫ ex ⋅
dx
2
(x + 1)
(x + 1)3
…(i)
  cos θ + sin θ  

ln 
  cos θ − sin θ  
Applying integration by parts,
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 )
 1

−2
=
⋅ ex − ∫ ex ⋅
dx
2
3
(x + 1)
 (x + 1)

=
− 2 ∫ ex ⋅
= ∫ (x + sin x) cos x dx
= ∫ x cos x dx +
= x sin x + cos x −
Therefore, on putting this value in Eq.(i), we get
=
 cos θ + sin θ  1
1
sin 2 θ ln 
 + ln (cos 2 θ ) + c
2
 cos θ − sin θ  2
9. Let I = ∫
sin −1 x − cos −1 x
dx
sin −1 x + cos −1 x
π
sin −1 x −  − sin −1
2
=∫
π
2
π
2 
= ∫ 2 sin −1 x −  dx

π
2
1
2

x

∫ (sin 2x) dx
= (x ⋅ sin x − ∫ 1 ⋅ sin x dx) −
2
2 (cos 2 θ + sin 2 θ )
=
=
cos 2 θ
cos 2θ
 cos θ + sin θ  1
1
2
dθ
sin 2θ ln 
sin 2 θ
 −
2
cos 2 θ
 cos θ − sin θ  2 ∫
1
ex
dx =
+c
3
(x + 1)
(x + 1)2
11. Let I = ∫ (elog x + sin x) cos x dx
cos 2 θ − cos θ sin θ − sin θ cos θ + sin 2 θ + cos θ sin θ
+ cos 2 θ + sin 2 θ + cos θ ⋅ sin θ
=
cos 2 θ − sin 2 θ
I=
…(ii)
cos 2 x
+c
4
Topic 4 Integration, Irrational Function
and Partial Fraction
1. Let I = ∫
=∫
2x12 + 5x9
2x12 + 5x9
dx
dx = ∫ 15
3
3
x (1 + x− 2 + x− 5 ) 3
(x + x + 1)
5
2x− 3 + 5x− 6
dx
(1 + x− 2 + x− 5 ) 3
Now, put 1 + x− 2 + x− 5 = t
dx
⇒
(− 2x− 3 − 5x− 6 ) dx = dt
(2x− 3 + 5x− 6 ) dx = − dt
dt
∴ I = − ∫ 3 = − ∫ t − 3 dt
t
⇒
4
=
π
∫ sin
−1
x dx − x + c …(i)
Now, ∫ sin −1 x dx
Put x = sin 2 θ ⇒ dx = sin 2θ
θ cos 2 θ
= ∫ θ ⋅ sin 2 θ dθ = −
+
2
=−
∫
1
cos 2θ dθ
2
cos 2 x
+c
4
=
t− 3 + 1
1
+C= 2+C
−3 + 1
2t
x10
+C
2 (x + x3 + 1) 2
5
Indefinite Integration 277
I=∫
2. Let
=∫
cos3 x + cos5 x
dx
sin 2 x + sin 4 x
On putting x = 0, we get
(cos x + cos x) ⋅ cos x dx
(sin 2 x + sin 4 x)
⇒
2
0=B+C
4
B = − C = 1 /2
x3 + 3x + 2
x+1
1
2
−
+ 2
=
2
2
2
(
x
+
)
2
1
(x + 1)2
(x + 1) (x + 1) 2 (x + 1)
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)
y − 3y + 2
A
B
=1 +
+
y ( y + 1)
y
y+1
⇒
[where, y = t 2]
⇒
∴ I1 = ∫

2
6 
1 + 2 −
 dt
1 + t 2
t

1
sec θdθ
= cos 2 θ dθ = ∫ (1 + cos 2 θ )dθ
2
(tan 2θ + 1)2 ∫
1
1

θ + sin 2θ 
2 
2

1
1
tan θ
= θ+ ⋅
2
2 (1 + tan 2 θ)
=
=
x + 3x + 2
x + 2x + x + 2
=
(x2 + 1)2 (x + 1) (x2 + 1)2(x + 1)
x
2
+
(x2 + 1)(x + 1) (x2 + 1)2
4. Let
x
Ax + B
C
+
=
(x + 1) (x + 1) (x2 + 1) (x + 1)
On putting
x = ( Ax + B) (x + 1) + C (x + 1)
2
(x + 1)
ex (x + 1)
dx
=
∫ x ex (1 + xex )2 dx
x (1 + xex )2
 1
dt
1 1
=∫ 
− − 2  dt
2
−
t
t t 
1
(t − 1)t

1
= log|t − 1| − log|t| + + c
t
+ 1 + c
t − 1
= log
 t  t
I=∫
∴
x = − 1, we get
−1 = 2 C ⇒ C = − 1 / 2
On equating coefficients of x2, we get
0= A+C
⇒
I=∫
Put 1 + xex = t ⇒ (ex + x ex ) dx = dt
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
x(x2 + 1) + 2(x + 1)
=
(x2 + 1)2(x + 1)
Again,
…(i)
dx = sec2θ dθ
3
=
dx
(x + 1)2
2
2
2
− 6 tan −1 (t ) + c
t
2
= sin x −
− 6 tan −1 (sin x) + c
sin x
3.
x+1
1
x = tan θ
Put
=t−
3
dx
∫ x + 1 + 2 ∫ x2 + 1 dx + 2 ∫
1
1
log|x + 1| + log|x2 + 1|
2
4
1
+ tan −1 x + 2 I1
2
dx
where, I1 = ∫
(x2 + 1)2
A = 2, B = − 6
Now, Eq. (i) reduces to, I = ∫
1
2
⇒I = −
…(i)
y2 − 3 y + 2
2
6
=1 + −
y ( y + 1)
y y+1
∴
2
=−
Using partial fraction for
2
x3 + 3x + 2
dx
(x + 1)2 (x + 1)
∴ I=∫
 xex 
1
+
+c
= log 
x
1
1
xe
xex
+
+


A = − C = 1 /2
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12
Definite Integration
Topic 1 Properties of Definite Integral
Objective Questions I (Only one correct option)
1. A value of α such that
α+1
∫
α
dx
 9
= log e   is
 8
(x + α ) (x + α + 1)
(a) − 2
2. If ∫
π /2
0
(a) −
(b)
1
2
(c) −
(b) 1
3. The integral ∫
π /3
π /6
(c)
1
2
(d) 2
1
2
(d) −1
(b) 37/ 6 − 35 / 6
(d) 34/3 − 31/3
∫ [sin 2x (1 + cos 3x)] dx, where [t ] denotes
0
the greatest integer function, is
(2019 Main, 10 April I)
(c) − 2π
(b) 2π
(d) π
1
5. The value of the integral ∫ x cot−1 (1 − x2 + x4 )dx is
0
(2019 Main, 9 April II)
π 1
− log e 2
4 2
π
(c)
− log e 2
4
π 1
− log e 2
2 2
π
(d)
− log e 2
2
(b)
(a)
(a)
π −1
2
π/ 2
0
(b)
sin3 x
dx is
sin x + cos x
π−2
8
(c)
π −1
4
(2019 Main, 9 April I)
(d)
π−2
4
x
x
If f (x + 5) = g (x), then ∫ f (t )dt equals
0
x+5
5
(a) 5
∫ g (t )dt
(b)
x+5
x+5
(c) 2
∫ g (t )dt
5
(b) log e e
(d) log e 1
9. 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)
0
a
a
(a) 4∫ f (x) dx
(b) ∫ f (x) dx
(c) 2∫ f (x) dx
(d) − 3∫ f (x) dx
0
a
0
a
(d)
(2019 Main, 8 April II)
∫ g (t )dt
e
11. The integral ∫
2x
x
 e 
−    log e x dx is
 x

(2019 Main, 12 Jan II)
1 1
1
(b) − + − 2
2 e 2e
3 1
1
(d)
− −
2 e 2e2
π /4
π /6
dx
equals
sin 2x(tan5 x + cot5 x)
1 π
1 
(a)  − tan − 1 

 3 3
5 4
1π
−1 1 
 − tan 

 9 3

10 4
12. The value of the integral ∫
(2019 Main, 11 Jan II)
1
1 
(b)
tan − 1 

 9 3
20
(d)
π
40
sin 2 x
dx
−2  x 
1
+
 π  2
2
(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
b
13. Let I = ∫ (x4 − 2x2) dx. If I is minimum, then the ordered
a
pair (a , b) is
5
(a) (− 2 , 0)
(c) ( 2 , − 2 )
∫ g (t )dt
x
equal to
3
1
(a)
−e− 2
2
2e
1
1
(c)
−e− 2
2
e
5
x+5
0
10. The integral ∫  
1  e
(c)
7. Let f (x) = ∫ g (t )dt, where g is a non-zero even function.
0
(a) log e 3
(c) log e 2

2π
6. The value of ∫
g ( f (x))dx is
0
(2019 Main, 10 April II)
(a) − π
π/ 4
−π/ 4
(2019 Main, 8 April I)
(2019 Main, 12 April II)
sec 2/ 3 x cosec4/3 x dx is equal to
(a) 35 / 6 − 32/ 3
(c) 35 /3 − 31/3
4. The value of
2 − x cos x
and g (x) = log e x, (x > 0) then the
2 + x cos x
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)
1
2
8. If f (x) =
(2019 Main, 10 Jan I)
(b) (0, 2 )
(d) (− 2 , 2 )
Definite Integration 279
14. If ∫
π /3
0
1
tan θ
dθ = 1 −
, (k > 0), then the value of k
2
2k sec θ
is
(2019 Main, 9 Jan II)
1
2
(d) 4
(a) 1
(a)
(b)
(c) 2
π
4
(b) −
3
16. The value of ∫
π
(a)
8
(c) 0
sin x
dx is
−π / 2 1 + 2x
π
(b)
2
3 π/ 4
4
(d)
3
(c) 4π
(a) − 2
x2 cos x
18. The value of ∫
dxis equal to
−π / 2 1 + ex
19. The integral ∫
(a) 2
π /2
π /4
(c) 1
2(eu + e−u )16 du
log( 1 + 2)
(eu + e−u )17 du
∫0
(b)
∫0
log( 1 + 2)
∫0
∫0
x
dx is equal to
2
(2014 Main)
2π
− 4− 4 3
3
(d) 4 3 − 4 − π /3
(c) 4 3 − 4
22. The value of the integral ∫
π /2
− π /2
 2
π − x
 x + log
 cos x dx

π + x
(2012)
(a)
1
3
log
4
2
∫
(c)
π2
+ 4
2
(d)
π2
2
x sin x2
dx is
2
log 2 sin x + sin (log 6 − x2 )
(b)
(c) log
3
2
(d)
(2011)
1
3
log
6
2
24. The value of
0
∫ −2 [x
(a) 0
3
3 π /4
∫ π /4
(d) 5
(b) 0
π
(d)
2
dx
is equal to
1 + cos x
(b) −2
(1999, 2M)
1
(d) −
2
1
(c)
2
the integral part of x. Then,
+ 3x2 + 3x + 3 + (x + 1) cos (x + 1)] dx is (2005, 1M)
(c) 4
1
∫ −1 f (x) dx is
(a) 1
(b) 2
(c) 0
(d) −
(1998, 2M)
1
2
33. If g (x) = ∫ cos 4 t dt , then g (x + π ) equals
0
(a) g (x) + g ( π )
(c) g (x) g ( π )
(d) 1
k
If I1 = ∫
1−k
I2 = ∫
1−k
Then,
(b) 3
(c) 3
(2000, 2M)
(1997, 2M)
(b) g (x) − g ( π )
g (x )
g( π)
(d)
34. Let f be a positive function.
log 3
1
3
log
2
2
(d) 3
x
is
23. The value of
(b) 5 / 2
(a) − π
π
(c) −
2
(2014 Adv.)
(b)
π2
−4
2
(c) 2
32. Let f (x) = x − [x] , for every real number x, where [x] is
x
2
(b)
(b) 1
(a) 2
(a) π − 4
(a) 0
(2000, 2M)
f (x) dx is equal to
than or equal to y, then the value of the integral
3π/2
(1999, 2M)
∫ π / 2 [2 sin x] dx is
31.
21. The integral ∫ 1 + 4 sin 2 − 4 sin
0
3
−2
30. If for a real number y, [ y] is the greatest integer less
(d) 6
2(eu − e−u )16 du
π
(2001, 1M)
(d) 2π
(a) 3 / 2
(e − e ) du
log( 1 + 2)
cos 2 x
dx, a > 0,is
−π 1 + a x
π
log x
29. The value of the integral ∫ −1  e  dx is
e 
x 
−u 17
u
1
(d) log  
 2
(c) 1
e2
(2 cosec x)17dx is equal to
log( 1 + 2)
(a)
(d)
2
(d) 1

 1 + x 
 [x] + log 
  dx equals (2002, 1M)
 1 − x 

(b) 0
(a) 0
(2016 Adv.)
log x2
dx is equal to
log x2 + log(36 − 12x + x2)
(2015, Main)
(b) 4
20. The integral ∫
(c)
4
(c) −1
(2004, 1M)
(b) aπ
then ∫
π2
(b)
+ 2
4
(d) π 2 + eπ / 2
1−x
dx is
1+ x
ecos x sin x , for | x|≤ 2
2 ,
otherwise

π /2
π2
(a)
−2
4
(c) π 2 − e− π / 2
1
0
28. If f (x) = 
(d) − 1
(c) 4
1/ 2
−1/ 2
(c)
(2017 Main)
(b) 2
1
2
π
−1
2
(a) π
π
2
(2018 Main)
π
(d)
4
dx
is equal to
1 + cos x
∫π / 4
(b)
27. The value of ∫
2
π /2
(a) −
(2019 Main, 9 Jan I)
0
2
(a)
3
π
+1
2
26. The integral ∫
15. The value of ∫ |cos x|3 dx is
17.
25. The value of the integral ∫
(a) 2
(c) 1/2
k
x f [x (1 − x)] dx and
f [x (1 − x)] dx, where 2k − 1 > 0.
I1
is
I2
(1997C, 2M)
(b) k
(d) 1
280 Definite Integration
35. The value of
2π
∫0
[ 2 sin x ] dx, where [.] represents the
greatest integral functions, is
5π
(a) −
3
(b) − π
(1995, 2M)
5π
(c)
3
(d) −2π
π x
 + B,
 2 
36. If f (x) = A sin 
A and B are
(1995, 2M)
37. The value of ∫
π /2
0
4
(c) 0 and −
π
(1993, 1M)
38. Let f : R → R and g : R → R be continuous functions.
Then, the value of the integral
π/ 2
f (− x)] [ g (x) − g (− x)] dx is
(a) π
(b) 1
(c) −1
(1990, 2M)
(d) 0
39. For any integer n, the integral
π
cos 2 x
∫0 e
cos3 (2n + 1) x dx has the value
(a) π
(c) 0
(1985, 2M)
(b) 1
(d) None of these
40. The value of the integral ∫
(a) π / 4
(c) π
π /2
0
(1983, 1M)
Assertion and Reason
41. Statement I The value of the integral
π /3
∫π / 6
(2013 Main)
dx
is equal to π /6 .
1 + tan x
Statement II
b
3
(b) ∫ f (x) dx = 12
(c) 9f ′ (3) – f ′ (1) + 32 = 0
(d) ∫ f (x) dx = − 12
1
3
1
For every function f (x) which is twice differentiable, these
will be good approximation of
 b − a
 { f (a ) + f (b)},
2 
b
∫ a f (x) dx = 
for more acurate results for c ∈ (a , b),
c−a
b−c
[ f (a ) − f (c)] +
[ f (b) − f (c)]
2
2
a+b
2
b
b−a
∫ a f (x) dx = 4 { f (a ) + f (b) + 2 f (c)} dx
t
(t − a )
∫ a f (x) dx − 2 { f (t ) + f (a )}
44. If lim
= 0,
t→ a
(t − a )3
When c =
(2006, 6M)
then degree of polynomial function f (x) atmost is
(a) 0
(c) 3
(b) 1
(d) 2
45. If f ′ ′ (x) < 0, ∀x ∈ (a , b), and (c, f (c)) is point of maxima,
cot x
dx is
cot x + tan x
(b) π / 2
(d) None of these
F ′ ′ (x) dx = 40, then the
(a) 9f ′ (3) + f ′ (1) − 32 = 0
F (c) =
(b) 1
(d) π / 4
∫ −π / 2 [ f (x) +
3 3
∫1 x
correct expression(s) is/are
4
(d) and 0
π
dx
is
1 + tan3 x
(a) 0
(c) π / 2
1
Passage II
1
2A
1
f ′   = 2 and ∫ f (x) dx =
, then constants
0
 2
π
π
π
2
3
(b) and
(a) and
2
2
π
π
3
43. If ∫ x2 F ′ (x) dx = − 12 and
b
∫a f (x) dx = ∫a f (a + b − x) dx
(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
Passage Based Questions
Passage I
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.
42. The correct statement(s) is/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)
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)
f (b) − f (a ) 
b − a 
(d) 0
46. Good approximation of ∫
(a) π / 4
π /2
0
sin x dx, is
(b) π ( 2 + 1) / 4
π
(d)
8
(c) π ( 2 + 1) / 8
Objective Questions II
(One or more than one correct option)
47. Let f : R → (0, 1) be a continuous function. Then, which
of the following function(s) has (have) the value zero at
some point in the interval (0, 1) ?
(2017 Adv.)
x
(a) e − ∫ f (t ) sin t dt
x
(c) x −
0
π
−x
2
∫ f (t ) cos t dt
(b) f (x) +
π
2
∫ f (t ) sin t dt
0
(d) x9 − f (x)
0
48. If I = ∑ k = 1
98
k+1
∫k
k+1
dx , then
x(x + 1)
(a) I > log e 99
(b) I < log e 99
49
(c) I <
50
(d) I >
49
50
(2017 Adv.)
Definite Integration 281
49. Let f (x) = 7 tan 8 x + 7 tan 6 x – 3 tan 4 x – 3 tan 2 x for all
 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
(a) ∫
0
x f (x) dx =
(b) ∫
π/ 4
0
(d) ∫
f (x) dx = 0
π/ 4
0
(2015 Adv.)
f (x) dx = 1
192x3
 1
for all x ∈ R with f   = 0. If
50. Let f ′ (x) =
4
 2
2 + sin πx
m≤∫
1
1/ 2
f (x) dx ≤ M , then the possible values of m and M
are
(2015 Adv.)
(a) m = 13, M = 24
1
1
(b) m = , M =
4
2
(c) m = − 11, M = 0
(d) m = 1, M = 12
equation
= L, is/are
e4 π − 1
(c) a = 4, L = π
e −1
e4 π + 1
(d) a = 4, L = π
e +1
x4 (1 − x)4
dx is (are)
1 + x2
22
−π
7
53. If I n = ∫
π
−π
2
105
(b)
(d)
2
then ∫ f (x) dx = …… .
1
60. The value of ∫
61. The value of ∫
2
3 π /4
x
dx is …… .
(1994, 2M)
x
dx …… .
1 + sin x
π /4
(1993, 2M)
2
1.5
∫0
(b)
(1989, 2M)
2
[x ] dx, where [.] denotes the greatest
(1988, 2M)
64. Match the conditions/expressions in Column I with
statement in Column II.
Column I
1
(2009)
10
∑ I2 m + 1 = 10 π
dx
A.
∫ −1 1 + x 2
B.
∫0
Column II
P.
1
2
log  
 3
2
Q.
2
2 log  
 3
1
dx
3
1− x
dx
R.
π
3
2
1− x 2
dx
x x2 −1
S.
π
2
C.
∫2
D.
∫1
2
65. Match List I with List II and select the correct answer
using codes given below the lists.
(2014)
m =1
List I
10
∑ I2 m = 0
(d) I n = I n + 1
m =1
P.
List II
The number of polynomials f( x ) with
non-negative integer coefficients of degree
(i)
8
(ii)
Q. The number of points in the interval
[− 13, 13 ] at which f( x ) = sin( x 2 ) + cos( x 2 )
attains its maximum value, is
2
1
≤ 2, satisfying f( 0) = 0 and ∫ f( x ) dx = 1, is
Numerical Value
0
54. The value of the integral ∫
1/ 2
0
1+ 3
dx is
((x + 1)2 (1 − x)6 )1/ 4
.......
(2018 Adv.)
55. Let f : [1, ∞ ] → [2, ∞ ] be differentiable function such
x
that f (1) = 2. If 6∫ f (t ) = dt = 3x f (x) − x , ∀ x ≥ 1 then
3
1
the value of f (2) is ....
sin x
(2011)
d
e
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
(1997, 2M)
values of k is ..… .
3x 2
R.
∫−2 1 + e x dx equals
(iii)
4
S.
 1/ 2
1 + x  
 ∫ cos 2 x log 
 dx 
1− x  
 − 1/ 2
equals
 1/ 2
1 + x  
dx 
 ∫ cos 2 x log 

1− x  
 0
(iv)
0
Fill in the Blanks
56. Let
(1996, 2M)
x
5−x+
3
function, equals …… .
71 3 π
−
15 2
sin nx
dx , n = 0 , 1 , 2 ,... , then
(1 + π x ) sin x
(a) I n = I n + 2
(c)
(1996, 2M)
1
1
59. If for non-zero x, af (x) + bf   = − 5, where a ≠ b,
 x x
(2010)
(c) 0
(1997, 2M)
Match the Columns
(2015 Adv.)
e4 π + 1
(b) a = 2, L = π
e +1
(a)
0
63. The integral
e4 π − 1
(a) a = 2, L = π
e −1
0
x sin 2n x
dx = …… .
2n
sin x + cos 2n x
2π
−2
∫0
π t
6
4
∫0 e (sin at + cos at )dt
1
π sin (π log x)
dx is …… .
x
62. The value of ∫ |1 − x | dx is … .
et (sin 6 at + cos 4 at )dt
52. The value(s) of ∫
58. For n > 0, ∫
37 π
1
2
51. The option(s) with the values of a and L that satisfy the
4π
57. The value of ∫
2
Codes
P Q R S
(a) (iii) (ii) (iv) (i)
(c) (iii) (ii) (i) (iv)
P Q R S
(b) (ii) (iii) (iv) (i)
(d) (ii) (iii) (i) (iv)
282 Definite Integration
80. If f and g are continuous functions on [0, a ] satisfying
Analytical & Descriptive Questions
(5050) ∫
66. The value of
67. Evaluate
π |cos x |
∫ 0e
f
π /2
∫0
1
∫0
0
f (x) = f (a − x) and g (x) + g (a − x) = 2, then show that
(1 − x50 )100 dx
a
is
(1 − x )
50 101
dx
π + 4x
dx.
π

2 − cos |x|+ 

3
is
an
even
function,
f (cos 2x) cos x dx = 2 ∫
π /4
0
(2004, 4M)
then
prove
1
∫ 0 tan
(1999, 3M)
−1 
1

1

 dx = 2 ∫ tan −1 x dx.
2
0
1 − x + x 
Hence or otherwise, evaluate the integral
1
2
−1
∫ tan (1 − x + x ) dx.
0
72. Integrate ∫
π /4
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
∫ −1/
3
∫2
3
76. A cubic f (x) vanishes at x = − 2 and
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)
−1
π

x sin (2x) sin  cos x
2

77. Evaluate ∫
dx .
0
2x − π
π
78. Show that ,
∫0
01
84. Evaluate ∫
0
85. Evaluate
π /2
x sin x cos x
dx.
cos 4 x + sin 4 x
x sin −1 x
1/ 2
∫0
86. Evaluate ∫
(1 + x)] dx.
x dx
, 0 < α < π.
+ cos α sin x
1 − x2
π /4
0
dx.
87. (i) Show that ∫ x f (sin x) dx =
0
3/ 2
(1991, 4M)
1
π /4
∫0
(1984, 2M)
π
2
(1983, 3M)
π
∫ 0 f (sin x) dx.(1982, 2M)
∫ −1 | x sin πx| dx.
79. Prove that for any positive integer k,
sin 2kx
= 2 [cos x + cos 3x + K + cos(2k − 1) x]
sin x
π /2
∫0
Integer Answer Type Questions
[x], x ≤ 2
,
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.)
89. Let f : R → R be a function defined by f (x) = 
−1
x
(1990, 4M)
 12 + 9x2
)
 dx,
 1 + x2 
where tan −1 x takes only principal values, then the
3π 

value of  log e|1 + α | −
 is

4
(2015 Adv.)
1
sin 2kx ⋅ cot x dx = π /2.
(1982, 3M)
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)
1
(1990, 4M)
1
(1986, 2 M)
2
n
90. If α = ∫ (e9x + 3 tan
f (cos 2x) cos x dx.
(1988, 5M)
∫ 0 (tx + 1 − x) dx,
0
f (sin 2x) sin x dx = 2
Hence, prove that
88. Evaluate
integral,
(1988, 4M)
1
(1985, 2 M)
2
sin x + cos x
dx.
9 + 16 sin 2x
(ii) Find the value of
74. Evaluate the definite integral
π /2
π
83. Evaluate ∫
(1 − x) +
(1997C, 2M)
73. Determine the value of
75. Evaluate
1
∫ 0 log[
π
(1998, 8M)
log (1 + tan x) dx.
0
82. Evaluate
(2003, 2M)
0 ecos x
71. Prove that
that
f (sin 2x) cos x dx.
ecos x
dx.
+ e− cos x
π
70. Evaluate ∫
∫0
that
the
value
of
the
[ f (x) /{ f (x) + f (2a − x)}] dx is equal to a.
3
∫ −π / 3
(1989, 4M)
81. Prove


1

1
2 sin  2 cos x + 3 cos  2 cos x  sin x dx.


(2005, 2M)
π/ 3
a
∫ 0 f (x) g(x) dx = ∫ 0 f (x) dx.
(2006, 6M)
2a
68. Evaluate
69. If
1
 d2

(1 − x2)5  dx is
2
dx


91. The value of ∫ 4x3 
0
(2014 Adv.)
Definite Integration 283
Topic 2 Periodicity of Integral Functions
Objective Questions I (Only one correct option)
π/ 2
dx
, where [t ] denotes
1. The value of ∫
− π / 2 [x] + [sin x] + 4
the greatest integer less than or equal to t, is
(2019 Main, 10 Jan II)
1
(7 π − 5)
12
3
(c)
(4 π − 3)
10
1
(7 π + 5)
12
3
(d)
(4 π − 3)
20
(a)
(b)
f (2x) dx is
3
I
2
(c) 3I
(a)
(2002,1M)
(d) 2 < g(2) < 4
Analytical & Descriptive Questions
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.
a
Integer Answer Type Question
(b) I
6. For any real number x, let [x] denotes the largest
(d) 6I
x
3. Let g (x) = ∫ f (t ) dt, where f is such that
0
t ∈ [0,1] and 0 ≤ f (t ) ≤
3
< g(2) ≤ 5 / 2
2
5. Given a function f (x) such that it is integrable over
0
3
(c)
positive integer and 0 ≤ v < π .
continuous function such that for all
T
x ∈ R. f (x + T ) = f (x). If I = ∫ f (x) dx,
then the value of ∫
(b) 0 ≤ g(2) < 2
4.. Show that ∫
2. Let T > 0 be a fixed real number. Suppose, f is a
3 + 3T
1
3
≤ g(2) <
2
2
(a) −
1
≤ f (t ) ≤ 1 for
2
1
for t ∈ [1, 2]. Then, g(2) satisfies
2
the inequality
(2000, 2M)
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 + [x[− x, if f (x) is even
π 2 10
Then, the value of
(2010)
f (x) cos πx dx is……
10 ∫− 10
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 2
∫x t
 1
f (t )dt , then f′   is
 2
(2019 Main, 10 Jan II)
24
(a)
25
18
(b)
25
6
(c)
25
(d)
4
5
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
(b) e4 − 1
(2014 Adv)
(c) e − 1
(d) e4
3. The intercepts on X-axis made by tangents to the curve,
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
1
1
and f   >
 3 3
1
1
and f   >
 3 3
1
1
and f   <
 3 3
1
1
and f   <
 3 3
 1
t 2 f (t ) dt = 1 − sin x, ∀ x ∈ (0, π / 2), then f  
 3
is
(a) 3
(c) 1/3
(b) 3
(d) None of these
x
y = ∫ | t | dt , x ∈ R, which are parallel to the line y = 2x,
0
are equal to
(a) ± 1
(2013 Main)
(b) ± 2
(c) ± 3
(d) ± 4
4. Let f be a non-negative function defined on the interval
[0, 1].
If
x
∫0
f (0) = 0 , then
x
1 − ( f ′ (t ))2 dt = ∫ f (t ) dt , 0 ≤ x ≤ 1
0
and
(2009)
6. If f (x) is differentiable and
4
f   equals
 25
(a)
2
5
(c) 1
t2
(2005, 1M)
2
∫ 0 x f (x) dx = 5 t
5
,
then
(2004, 1M)
(b) −
(d)
5
2
5
2
284 Definite Integration
x 2 + 1 −t 2
7. If f (x) = ∫
e
x2
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
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
2n
n
+
I (m + 1, n − 1)
m+1 m+1
m
(d)
I (m + 1, n − 1)
m+1
9. Let f (x) = ∫
1
2 − t dt. Then, the real roots of the
2
1
(b) ±
2
(a) ±1
(2002, 1M)
1
(c) ±
2
(d) 0 and 1
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
11.
x
(b) 7
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,
16. Which of the following is true?
2
equation x − f '(x) = 0 are
15. Consider the statements
(a) both P and Q are true (b) P is true and Q is false
(c) P is false and Q is true (d) both P and Q are false
(c)
x
Passage Based Questions
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

(c) 4
(2001, 1M)
(d) 2
1
∫0
f (t ) dt = x + ∫ t f (t ) dt, then the value of f (1) is
(a)
1
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, ∞ )
Fill in the Blank
sec x cos x sec2x + cot x cosec x
17. f (x) = cos 2 x cos 2 x
cosec2x
.
2
2
1
cos x
cos x
Then, ∫
π /2
0
f (x) dx = K .
(1987, 2M)
x
(1998, 2M)
(b) 0
(d) −
(c) 1
1
2
12. Let f : R → R be a differentiable function and f (1) = 4.
Then, the value of lim ∫
x→1
(a) 8 f ′(1)
(c) 2 f ′(1)
f ( x)
4
2t
dt is
x −1
(1990, 2M)
(b) 4 f ′(1)
(d) f ′(1)
1
0
) dx is
(1981, 2M)
14. If g (x) = ∫
sin x
a
1
dg
> 0, ∀ x prove that
dx
g (x) dx increases as (b − a ) increases.
(2017 Adv.)
(1997, 5M)
20. Determine a positive integer n ≤ 5, such that
(x − 1)n dx = 16 − 6e
21. If ‘ f ’ is a continuous function with
−1
π
(b) g′  −  = − 2 π
 2
π
(d) g′   = − 2 π
 2
b
∫ 0 g(x) dx + ∫ 0
1 x
sin (t ) dt, then
π
(a) g′  −  = 2 π
 2
π
(c) g′   = 2 π
 2
x
19. Let a + b = 4, where a < 2 and let g (x) be a
∫ 0e
Objective Question II
(One or more than one correct option)
sin( 2x )
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.
18. For x > 0, let f (x) = ∫
differentiable function. If
−x 2
13. The value of the definite integral ∫ (1 + e
(a) −1
(b) 2
(c) 1+ e−1
(d) None of the above
Analytical & Descriptive Questions
(1992, 4M)
x
∫ 0 f (t ) dt → ∞ as
|x|→ ∞ , then show that every line y = mx intersects the
x
curve y2 + ∫ f (t ) dt = 2
0
(1991, 2M)
22. Investigate for maxima and minima the function,
x
f (x) = ∫ [2(t − 1)(t − 2)3 + 3(t − 1)2(t − 2)2] dt.
1
(1988, 5M)
Definite Integration 285
Topic 4 Limits as the Sum
Objective Question I (Only one correct option)
 (n + 1)1/3
(n + 2)1/3
(2n )1/3 
1. lim 
+
+ ..... +
 is equal to
4
3
4
3
/
/
n→ ∞ 
n
n
n 4/3 
Objective Questions II
(One or more than one correct option)
4. For a ∈ R (the set of all real numbers), a ≠ − 1,
(1a + 2a + K + n a )
1
=
.
n→∞ (n + 1 )a−1 [(na + 1 ) + (na + 2 ) + K + (na + n )]
60
Then, a is equal to
(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)

2. lim 
n → ∞  n2
(b)
(a)
(c)

18
π
3 3
π
(c) Tn <
3 3
is equal to
(2016 Main)
−17
2
π
3 3
π
(d) Tn >
3 3
(b) Sn >
Analytical & Descriptive Question
e2
(d) 3 log 3 − 2
e2
(d)
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
1/ n
(b)
e4
9
−15
2
2
(a) Sn <
27
(c)
n
(2019 Main, 12 Jan II)
n
(b) 7
5. Let S n = ∑
(b) tan − 1 (2)
(d) π /2
(n + 1)(n + 2) K 3n 
3. lim 

2n

(a) 5
n
n
n
1
+ 2
+ 2
+ ... +
 is equal to
2
2
2
n
5
+1
n +2
n +3
(a) tan − 1 (3)
(c) π / 4
n→ ∞
lim
 1
1
1
+
+K+
 = log 6.
n + 1 n + 2
6n 
(1981, 2M)
6. Show that, lim 
n→ ∞
Answers
Topic 1
1.
5.
9.
13.
17.
21.
25.
29.
33.
37.
41.
45.
49.
53.
(a)
(a)
(c)
(d)
(b)
(d)
(b)
(b)
(a)
(d)
(d)
(a)
(a, b)
(a, b, c)
75.
2.
6.
10.
14.
18.
22.
26.
30.
34.
38.
42.
46.
50.
54.
(d)
(c)
(a)
(c)
(a)
(b)
(a)
(c)
(c)
(d)
(a, b, c)
(c)
(*)
(2)
3.
7.
11.
15.
19.
23.
27.
31.
35.
39.
43.
47.
51.
55.
(b)
(d)
(c)
(d)
(c)
(a)
(c)
(a)
(a)
(c)
(c, d)
(c, d)
(a, c)
(8/3)
59.
1
a − b2
57. (2)
58. π 2
 1
60.  
 2
61. π ( 2 − 1 ) 62. (4)
2
4.
8.
12.
16.
20.
24.
28.
32.
36.
40.
44.
48.
52.
56.
(a)
(d)
(d)
(d)
(a)
(c)
(c)
(a)
(d)
(a)
(b)
(b, d)
(a)
(k = 16)
7 

a log 2 − 5a + b

2 
63. (2 − 2 )
64. A → S; B → S; C → P; D → R 65.
24 
 1
66. 5051
67.
e cos   +
 2
5 
 4π
 1 
68.  tan −1   
 2 
 3
π
72.
73. π 2
(log 2 )
8
(d)
e

 1
sin   − 1 
2  2

π
70.
71. log 2
2
74.
π
[ π + 3 log (2 + 3 ) − 4 3 ]
12
1
1
log 6 −
2
10
76. f ( x ) = x 3 + x 2 − x + 2
8
π2
π2
84.
16
77.
1
απ
1 π
83.
log 2 − +
2
sin α
2 4
 3
1
1

3π + 1 
85.  −
(log 3 ) 87. ( ii )
π +  86.
20
 π 2 
2
 12

82.
 tn+ 1 − 1 
88. 

(t − 1 )(n + 1 ) 
91. (2)
89. (0)
90. (9)
2. (c)
3. (b)
6. (4)
2. (b)
6. (a)
10. (c)
3. (a)
7. (d)
11. (a)
4. (c)
8. (a)
12. (a)
Topic 2
1. (d)
Topic 3
1. (a)
5. (a)
9. (a)
13. (d)
14. (*)
15. (c)
16. (b)
1
 15 π + 32
2
20. (n = 3 )
18. (ln x )
17. − 

2

 60 
7
22. At x = 1 and , f ( x ) is maximum and minimum respectively.
5
Topic 4
1. (c)
5. (a, d)
2. (b)
3. (b)
4. (b, d)
Hints & Solutions
Topic 1 Properties of Definite Integral
α+1
1. Let I =
∫
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
=
α
α+1
=
α
Alternate Solution
π /2
cot x
Let I = ∫
dx
0
cot x + cosec x
cos x
sin
x
=∫
dx
0
cos x
1
+
sin x sin x
π / 2 cos x
=∫
dx
0
cos x + 1
π /2
α +1
= [ log e (x + α ) − log e (x + α + 1)] α
α+1

 x+ α 
= log e 

 x + α + 1  α

2α + 1
2α
= log e
− log e
2α + 2
2α + 1
 2α + 1 2α + 1
= log e 
×
 = log e
 2α + 2
2α 
⇒
⇒
⇒
⇒
⇒
=∫
 9
 
 8
(given)
=∫
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,
=∫
π /3
3. Let
I=
π /2
0
2/3
x cosec4/3 x dx
π /6
=
∫
π /6
cos 2 x
dx
2
sin x
cos
2/ 3
1
dx
x sin 4/3 x
π /3
=
(cosec x cot x − cot2 x) dx
sec2 x
dx
(tan x)4/ 3
π/6
∫
[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 − cosec2 x + 1) dx
= [− 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

= − 1 = [π − 2]
= x − tan

2  0
2
2
= m[π + n ]
1
On comparing, we get m = and n = − 2
2
∴
m⋅ n = − 1
∫ sec
π /3
π / 2 cos x −
0
π /2
1
2 x
1 − sec  dx

2
2
π /2
0
π /2
From the options we get α = − 2
π /2
cot x
2. 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
1
x

= x − tan
= − 1 = (π − 2)


2 0
2
2
8α 2 + 8α + 2 = 9α 2 + 9α
α2 + α − 2 = 0
(α + 2) (α − 1) = 0
α = 1, − 2
0
x
−1
2
dx
x
2 cos 2
2
2 cos 2
θ
θ

Q cos θ = 2 cos 2 − 1 and cos θ + 1 = 2 cos 2

2
2 
(2 α + 1)2 9
= ⇒ 8 [4α 2 + 4α + 1] = 36 (α 2 + α )
4α (α + 1) 8
=∫
π /2
and sec2 x dx = dt
3
So,
I=
∫
1/ 3
 1

= − 3  1/ 6 − 31/ 6
3

[given]
3
dt  t − 1/3 
=

t 4/3  − 1 / 3 1/
= 3 ⋅ 31/ 6 − 3 ⋅ 3− 1/ 6
= 37/ 6 − 35/ 6
3
Definite Integration 287
4. Given integral
6.
Key Idea Use property of definite integral.
2π
I = ∫ [sin 2x ⋅ (1 + cos 3x)]dx
b
b
∫a f ( x) dx = ∫a f ( a + b − x) dx
0
π
= ∫ [sin 2x ⋅ (1 + cos 3x)]dx
0
2π
I=∫
Let
+ ∫ [sin 2x ⋅ (1 + cos 3x)]dx
π
= I1 + I 2 (let)
... (i)
sin3 x
dx
sin x + cos x
On applying the property,
2π
I 2 = ∫ [sin 2x ⋅ (1 + cos 3x)]dx
Now,
0
I 2 = − ∫ [− sin 2x ⋅ (1 + cos 3x)]dx
So,
I=∫
π
= ∫ [− sin 2x ⋅ (1 + cos 3x)]dx
2I = ∫
…(ii)
0
π
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
⇒
−x
∫ g(t )dt
0
x
x
0
0
f (− x) = − ∫ g (− u )du = − ∫ g (u )du = − f (x)

y

1
1
0
a
0
Q ∫ tan −1 (t − 1)dt = ∫ tan −1 (1 − t − 1)dt = − ∫ tan −1 (t )dt
because ∫ f (x)dx = ∫ f (a − x)dx
[Q g is an even function]
⇒
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)
[Q g is an even function]
⇒
f (5 − x) = f (x + 5)
x
Let I = ∫ f (t )dt
0
1
So, I = ∫ ( tan −1 t + tan −1 t )dt
20
0
Put t = u + 5 ⇒ t − 5 = u ⇒ dt = du
1
t ]10
f (− x) =
Now, put t = − u, so
1

1
= ∫ ( tan −1 t − tan −1 (t − 1)dt 
2 0


−1 x − y
−1
−1
Q tan 1 + xy = tan x − tan

tdt = [t tan
π 1 π −1
− =
4 4
4
On replacing x by (−x), we get
1
−1
I=
1

Q cot−1 x = tan −1

x 
 t − (t − 1) 
1
= ∫ tan −1 
 dt
 1 + t (t − 1)
20
0
π
2
0
0
1


1
1
= ∫ tan −1 
 dt
2
20
1 − t + t 
= ∫ tan
(sin x + cos x) (sin 2 x + cos 2 x − sin x cos x)
dx
sin x + cos x
x
1
cot−1 (1 − t + t 2)dt
2 ∫0
−1
0
π
2
0
7. Given, f (x) = ∫ g (t )dt
1
1
sin3 x + cos3 x
dx
sin x + cos x
π/ 2
π/ 2
0
0
1
…(ii)
π 1
1
 1


π
= x + cos 2x
=  − 0 + (−1 − 1) = −




2
4
4
2 2
0
Now, put x2 = t ⇒ 2xdx = dt
Lower limit at x = 0, t = 0
Upper limit at x = 1, t = 1
a
0
cos3 x
dx
cos x + sin x

 1
1 − (2 sin x cos x) dx

 2
π
1


= ∫ 2 1 − sin 2x dx
0 

2
5. Let I = ∫ x cot−1 (1 − x2 + x4 )dx
0
π /2
=∫
1
1
b
∫a f (x)dx = ∫a f (a + b − x) dx,
On adding integrals (i) and (ii), we get
π
∴
b
…(i)
we get
π
let 2π − x = t, upper limit t = 0 and lower limit t = π
and
dx = − dt
∴I=
π
2
0
t
dt
−∫
1
t2
+
0
[by integration by parts method]
π 1
π 1
= − [log e (1 + t 2)]10 = − log e 2
4 2
4 2
x −5
x −5
−5
−5
∫ f (u + 5)du = ∫ g(u )du
∴
I=
Put
u = − t ⇒ du = − dt, we get
5 −x
I=−
∫
5
g (− t )dt =
5
∫ g(t )dt
5 −x
…(i)
288 Definite Integration
b
Also, upper limit x = e ⇒ t = 1 and lower limit x = 1 ⇒
1
e
1
1 
1 1
∴ I = ∫  t 2 −  ⋅ dt ⇒ I = ∫ (t − t −2) dt
1/ e 
1/ e
t t
a
[Q − ∫ f (x)dx = ∫ f (x)dx and g is an even function]
a
b
5
∫ f ′ (t )dt
I=
t=
[by Leibnitz rule f ′ (x) = g (x)]
5 −x
1
= f (5) − f (5 − x) = f (5) − f (5 + x)
5
∫
=
  t 2 1 
1
 1
 3
  1
I =  +   =  + 1 −  2 + e  = − e − 2




2
t
2
2
2
e
2
e
1 


[from Eq. (i)]
5
∫ g(t )dt
f ′ (t )dt =
5+ x
e
5+ x
11. Let I = ∫
8. The given functions are
g (x) = log e x, x > 0 and f (x) =
π /4
Let
I=∫
Then,
I=∫
−π / 4
π /4
−π / 4
2 − x cos x
2 + x cos x
=∫
g ( f (x))dx
π /4
π /6
π /4
dx
sin 2x(tan5 x + cot5 x)
π /6
(1 + tan 2 x) tan5 x
dx
2 tan x (tan10 x + 1)
1 π / 4 tan 4 x sec 2x
dx
2 ∫π / 6 (tan10 x + 1)
Put tan5 x = t [Q sec 2x = 1 + tan 2 x]
=
 2 − x cos x 
log e 
 dx
 2 + x cos x
…(i)
⇒ 5 tan 4 x sec 2x dx = dt
Now, by using the property
b
b
a
a
 2 + x cos x
log e 
 dx
−π / 4
 2 − x cos x 
π /4
…(ii)
∴I=
π /4
−π / 4
⇒ 2I = ∫
=
1 π
−1  1  
 − tan 

9 3
10  4
−π / 4
a
[Q log e A + log e B = log e AB]
log e (1)dx = 0 ⇒
I = 0 = log e (1)
… (i)
0
I=
∫0 f (a − x) g (a − x) dx
Q a f (x) dx =
 ∫0
a

∫0 f (a − x) dx
⇒
∫0
4f (x) dx −
a
∫0 f (x) g (x) dx
a
I = 4 ∫ f (x) dx − I
0
a
⇒ 2I = 4 ∫ f (x) dx ⇒
0
[from Eq. (i)]
a
I = 2 ∫ f (x) dx.
⇒
0
x
e 
10. Let I = ∫   −    log e x dx
1  e
 x


x
 x
 x
Now, put   = t ⇒
x log e   = log t
 e
 e
⇒ x (log e x − log e e) = log t
e  x
2x
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
sin 2 x
1 x
+
2  π 
sin 2 (− x)
(replacing x by − x)
1  x
+ −
2  π 
if

sin 2 x
 − [x],
=
Q [− x ] = 
− 1 − [x], if
x   
1 


+  − 1−
 π  
2 
∫0 f (x) [4 − g (x)] dx
a
sin 2 x
dx
−2 1
x
+
2  π 
Then, f (− x) =
[Q f (x) = f (a − x) and g (x) + g (a − x) = 4]
=
3 )5
2
Also, let f (x) =
a
a
12. Let I = ∫
1
1
dt
(tan −1 (t ))(11/
=
t 2 + 1 10
 2 − x cos x 2 + x cos x
log e 
×
 dx
 2 + x cos x 2 − x cos x 
9. Let I = ∫ f (x) g (x) dx
⇒
3 )5
1 
−1
−1  1  
 tan (1) − tan 

9 3
10 
π /4
=
1
∫(1/
5
=
−π / 4 
=∫
1 1
⋅
2 5
π
4
 2 − x cos x 
 2 + cos x  
log e 
 + log e 
 dx
 2 + x cos x
 2 x cos x 


π /4
 1 


 3
t
On adding Eqs. (i) and (ii), we get
2I = ∫
π
6
x
∫ f (x)dx = ∫ f (a + b − x)dx, we get
I=∫

2 tan x 
Q sin 2x =

1 + tan 2 x 

f (− x ) = −
x ∈ I
x ∉ I 
sin 2 x
= − f (x )
1 x
+
2  π 
i.e. f (x) is odd function
 a
∴ I = 0 Q ∫ f (x) dx =
 − a

 a 0, if f (x) is odd function
2 f (x)dx, if f (x) is even function 
 ∫0

b
13. We have, I = ∫ (x4 − 2x2)dx
a
Let
f (x) = x4 − 2x2 = x2(x2 − 2)
= x2(x − 2 ) (x + 2 )
Definite Integration 289
Y
Graph of y = f (x) = x4 − 2x2 is
y=|cos x|
Y
y=f(x)
– √2
X′
X
√2 ∴
f(x) < 0 for – √2 < x < √2
–
–
+
+
O
– √2
√2
0
X
π
π/2
O
Y′
π
π
I = ∫ |cos x|3 = 2 ∫ 2|cos x|3 dx
0
0
b
(Q y = |cos x|is symmetric about x =
Note that the definite integral ∫ (x4 − 2x2)dx represent
a
the area bounded byy = f (x) , x = a, b and the X -axis.
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
tan θ
1
, (k > 0)
14. We have, ∫
dθ = 1 −
0
2k sec θ
2
π
⇒
0
Now, as cos 3x = 4 cos3 x − 3 cos x
1
∴ cos3 x = (cos 3x + 3 cos x)
4
π
∴I=
1
I=
2k
1/ 2 −
∫1
−1
dt
=
2k
t
1/ 2 −
π
∫1
t
=
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 ]12
 =−

2k  − 1 + 1
2k
1
 2

1
2  1
2 
=−

1 −
 2 − 1 =

k
2
2k 
2

∴
2 2
( cos 3x + 3 cos x) dx
4 ∫0
π /3
∫0
1
1
− +1 2
t 2 
Q

 π 
Q cos x ≥ 0 for x ∈ 0, 2 


= 2∫ 2 cos3 x dx
tan θ
1 π /3 tan θ
dθ
dθ =
secθ
2k sec θ
2k ∫0
π /3
π /3 sin θ
(sin θ)
1
1
dθ =
dθ
=
cos θ
2k ∫0
1
2k ∫0
(cos θ)
cosθ
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
Let I =
1
2
1
1 
2 
2
⇒
=1
1 −
 = 1−
2
2
2k 
2k
I = 1−
=
1 1
 4
− +3 =

 3
2 3
π /2
Let
I=
a
a
(given)
sin x
dx
+ 2x
1
−π/ 2
∫
π / 2 sin
⇒
I=
∫
−π/ 2
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
Y
⇒
O
b
2
15. We know, graph of y = cos x is
X′
b
16. Key idea Use property = ∫ f (x)dx = ∫ f (a + b − x)dx
2 = 2k ⇒ 2k = 4 ⇒ k = 2
⇒
π/2
π
X
I=
sin 2 x
∫ 1 + 2−x dx
−π / 2
π /2
⇒
I=
⇒
2I =
⇒
2I =
2x sin 2 x
dx
2x + 1
−π / 2
∫
π /2
Y′
π
)
2
∴ The graph of y =|cos x|is
 2 x + 1
sin 2 x x
 dx
 2 + 1
−π / 2
∫
π /2
∫ sin
−π / 2
2
x dx
290 Definite Integration
π /2
0
π /2
⇒
I=
⇒
∫ sin xdx
2
0
π /2
⇒
I=
∫
a
 a

Q ∫ f (x)dx = ∫ f (a − x)dx
 0

0
cos 2xdx
0
π /2
0
π
⇒ I=
4
2I = [x]π0 / 2
⇒
17. 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
π /4
20.
2I = ∫ dx = [x]42
⇒
2I = 2 ⇒ I = 1
PLAN This type of question can be done using appropriate
substitution.
=∫
(cosec x − cosec x cot x)dx
2
π /2
π /4
π /2
π /4
(2 cosec x)17 dx
217 (cosec x)16 cosec x (cosec x + cot x)
dx
(cosec x + cot x)
Let
= [(1 + 2 ) − (− 1 + 2 )] = 2
⇒
(− cosec x ⋅ cot x − cosec2x) dx = dt
and
cosec x − cot x = 1 /t
x2 cos x
dx
− π / 2 1 + ex
π /2
…(i)
Q b f (x) dx = b f (a + b − x) dx
∫a
 ∫a

⇒ I=∫
π /2
x cos (− x)
dx
1 + e− x
2
…(ii)
−π / 2
On adding Eqs. (i) and (ii), we get
π /2 2
 1
1 
x cos x 
+
2I = ∫
 dx
x
−π / 2
1 + e− x 
1 + e
=∫
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)
−a
0


π /2 2
0
x cos x dx
Using integration by parts, we get
2I = 2 [x2(sin x) − (2x) (− cos x) + (2) (− sin x)] π0 / 2

π2
− 2
⇒ 2I =2

4
π2
∴ I=
−2
4
cosec x + cot x = t
⇒
2 cosec x = t +
I=−∫
∴
b
⇒
u = ln ( 2 + 1 )
⇒
I=−∫
=2 ∫
log x
dx
log x + log(36 − 12x + 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)]
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.
21. PLAN Use the formula,| x − a | = 
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
π
π
x 3 
x

=  x + 4 cos  −  x + 4 cos 



2 0
2 π
π
=4 3 −4 −
3
2
4
217
2+1
16
and when t = 2 + 1, eu = 2 + 1
b
2
1
1

t + 
t

 2 


When t = 1, eu = 1 ⇒ u = 0
∫a f(x )dx = ∫a f(a + b − x )dx and then add.
I=∫
1
t
Let t = eu ⇒ dt = eudu.
PLAN Apply the property
Let
⇒ 2I = 2
2
= [− cot x + cosec x]3π π/ 4/ 4
18. Let I = ∫
19.
4
⇒
Given, I = ∫
1 − cos x
dx
sin 2 x
3 π /4
4
2
On adding Eqs. (i) and (ii), we get
4 log x + log(6 − x)
2I = ∫
dx
2 log x + log(6 − x)
∫ dx
2I =
⇒
log x dx
…(i)
[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

I=∫
⇒
2I = 2 ∫ sin 2 x dx [Qsin 2 x is an even function]
⇒
2
22. I = ∫
As,
π /2
− π /2
a
3
 2
 π − x 
x + log  π + x  cos x dx


∫− a f (x) dx = 0, when f (− x) = − f (x)
Definite Integration 291
∴ I=∫
π /2
x2 cos x dx + 0 = 2∫
− π /2
= 2 {(x2 sin x)0π / 2 − ∫
π /2
0
π /2
0
(x2 cos x) dx
26.
2x ⋅ sin x dx}
1/ 2
π /2
π
1 ⋅ (− cos x) dx}
=2 
− 2 {(− x ⋅ cos x)0π / 2 − ∫
0

4
2


 π
π
π
− 4
=2 
− 2 (sin x)π0 / 2 = 2 
− 2 = 
4
4
2


 


2
2
2
23. Put x = t ⇒ x dx = dt /2
2
b
...(i)
b
∫a f (x) dx = ∫a f (a + b − x) dx
Using,
1
2
=
I=∫
−1/ 2
=∫
−1/ 2
1/ 2
0
sin (log 6 − t )
dt
sin (log 6 − t ) + sin t
…(ii)
24. Let I = ∫
−2
=∫
−2
0
⇒
Put x + 1 = t
−1
=∫
−1
1
1
dt +
1
=4
[since, t3 and t cos t are odd functions]
=∫
1
0
1
0
1 1−x
1−x
dx
dx = ∫
0
1+ x
1 − x2
1
1 − x2
dx − ∫
0
= [sin −1 x] 10+
∫1
−1
−1
x
1
0
1 − x2
dx
t
dt
t
[where, t 2 = 1 − x2 ⇒ t dt = − x dx]
= (sin
1 − sin
0) + [t ]10 = π /2 − 1
(0) dx
…(i)
cos 2 x
dx
1 + ax
…(ii)
π
cos 2 x dx = 2 ∫
−π
π
π
0
1 + cos 2x
dx
2
π
0
π /2
0
π
∫ 0 cos 2x dx
cos 2x dx = π + 0
⇒ I = π /2
ecos x sin x, for | x| ≤ 2

f (x) =  2
otherwise
,


2
3
2
−2
ecos x sin x dx +
3
∫ 2 2 dx
= 0 + 2 [x]32
[Q ecos x sin x is an odd function]
Q 3 f (x) dx = 2
= 2 [3 − 2] = 2
 ∫ −2

∫ −1 t cos t dt
= 0 + 2 ⋅ 2 [x]10 + 0
25. I = ∫
−π
ax
=∫
(t + 2 + t cos t ) dt
−1
π
3
3
t3 dt + 2 ∫
1/ 2
∫0
∫ −2 f (x) dx = ∫ −2 f (x) dx + ∫ 2 f (x) dx
dx = dt
I=∫
π
2I = π
28. Given,
∴
∴
(−1) dx +
cos 2(− x)
d (− x)
1 + a −x
−π
= [x]π0 + 2∫
2
1
0
−1/ 2
cos 2 x
dx
1 + ax
0
[(x + 1)3 + 2 + (x + 1) cos (x + 1)] dx
⇒
[x] dx = ∫
= ∫ (1 + cos 2x) dx = ∫ 1 dx +
[x + 3x + 3x + 3 + (x + 1) cos (x + 1)] dx
3
π
−π
=∫
1 log 3 1
(t )
= (log 3 − log 2)
2 log 2 2
1
 3
I = log  
 2
4
0
1/ 2
∫0
On adding Eqs. (i) and (ii), we get
π 1 + ax 
2I = ∫ 
 cos 2x dx
−π  1 + a x 
2I =
∴
[x] dx +
I=∫
⇒
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
⇒


 1 + x
Q log  1 − x  is an odd function 


[x]dx + 0
=∫
∫log 2 sin (log 2 + log 3 − t ) + sin dt
log 2
=∫
 1 + x
1/ 2
∫ −1/ 2log 1 − x dx
[x] dx +
27. Let I = ∫
(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 2 + log 3 − t )
log 3
log 3
1/ 2
=∫
1
1

= − [x]0−1/ 2 = − 0 +  = −

2
2
dt
sin t ⋅
log 3
2
I=∫
log 2 sin t + sin (log 6 − t )
∴
 1 + x 

∫ −1/ 2[x] + log 1 − x  dx
29.
e2
∫e
−1
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 
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
292 Definite Integration
30. The graph of y = 2 sin x for π /2 ≤ x ≤ 3π / 2 is given in
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
1
0
∫ −1 [x] dx = ∫ −1 [x]
∴
=∫
33. Given,
2
0
π
3π/2
7π/6
=
π /2
where,
3 π /2
∫ π /2
dx + ∫
π
5π / 6
⇒
7 π /6
∫π
0 dx +
(−1) dx + ∫
3 π /2
7π / 6
3 π /4
π /4
I=∫
I=∫
2I = ∫
⇒
I=∫
∴
π /4
3 π /4
π /4
⇒
π /4
But
I1 =
I1 =
∫ x f [x (1 − x)] dx
1−k
⇒
dx
1 − cos x
…(ii)
∫ (1 − x) f [(1 − x) x] dx
=∫
…(i)
dx
1 + cos (π − x)
f [(1 − x)]dx] − ∫
1
I1 = I 2 − I1
⇒
∫ −1 f (x) dx = ∫ −1 (x − [x]) dx = ∫ −1 x dx − ∫ −1 [x] dx
1
−1
[x] dx
−1, if

[x] =  0, if
 1, if

[Q x is an odd function]
xf (1 − x)]dx
I1 1
=
I2 2
35. It is a question of greatest integer function. We have,
subdivide the interval π to 2π as under keeping in view
that we have to evaluate [2 sin x ]
Y
1,π/2
O
X'
(0,π)
1
k
1 −k
30°
30°
cosec2x dx = [− cot x] 3π π/ 4/ 4
1
=0−∫
0
1−k
k

2

 dx
2
 1 − cos x 
3 π /4
0
k
3 π /4 
π /4
x
1−k
k
dx
1 + cos x
3 π /4
x
g (x + π ) = g (π ) + g (x)
34. Given,
3π
π

= − cot
+ cot
= − (−1) + 1 = 2

4
4 
32. Let
0
= ∫ (− cos y)4 dy = ∫ cos 4 y dy = g (x)
On adding Eqs. (i) and (ii), we get
3 π /4 

1
1
2I = ∫
+

 dx
π / 4  1 + cos x
1 − cos x
⇒
x
I 2 = ∫ cos 4 ( y + π ) dy
(−2) dx
 7 − 9
 5 − 3
 1
=π
 = −π / 2
 + π −  + π 
 3 
 6 
 6
I=∫
cos 4 t dt
dt = dy
[−2x] 37ππ //62
7π / 6
π
π+x
π
t=π+ y
Put
[2 sin x]dx
7
 5 1

7

= π  −  + π 1 −  + π  − 3
 6 2

3

6
1
I1 = ∫ cos 4 t dt = g (π )
I2 = ∫
and
+ [− x]
+
5
π
π
7
π

 
  −2 ⋅ 3 π 2 ⋅ 7 π 
=
−  + −
+ π + 
+

 6


  2
2
6
6 
⇒
cos 4 t dt = I1 + I 2
0
[x] 5π π/ 2/ 6
31. Let
π+x
π
π
X
–2
=∫
cos 4 t dt
0
–1
5 π /6
π+x
0
= ∫ cos 4 t dt + ∫
π
1
∫ −1 f (x) dx = 1
x
1
5π/6
dx
g (x) = ∫ cos 4 t dt
g (x + π ) = ∫
⇒
π/2
1
∫0 0
(−1) dx +
−1
= − [x]−01 + 0 = −1; ∴
Y
Therefore,
0
1
∫ 0 [x] dx
dx +
–1/2,7π/6
–1,3π/2
X
(0,2π)
–1/2,11π/6
Y'
π 1
=
6 2
7π
1
π

sin  π +  = sin
=−

6
6
2
We know that, sin
∴
−1 ≤ x < 0
0 ≤ x<1
⇒
x=1
⇒
11π
π
π
1

= sin 2π −  = − sin = −

6
6
6
2
9π
3π
sin
= sin
= −1
6
6
sin
Definite Integration 293
π
Hence, we divide the interval π to 2π as
7π   7π 11 π   11π


,
, 2π
π ,
, 
, 


6  6
6   6
39. Let I = ∫ ecos
∴
[2 sin x] = − 1
=∫
7π / 6
π
[2 sin x] dx +
+
=∫
7π / 6
π
=−
(−1) dx +
11 π / 6
∫ 7π / 6
11 π / 6
∫ 7π / 6
2π
∫ 11π / 6
(−2) dx +
[2 sin x] dx
[2 sin x] dx
2π
∫ 11π / 6 (−1) dx
π
10π
5π
 4π  π
−2  − =−
=−


6
6
6
6
3
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
= 2 ⇒ A=
f′   = 2 ∴
 2
π
2 2
1
1
2A
2A


 π x
Now, ∫ f (x) dx =
⇒ ∫  A sin 
 + B dx =
0
0 
 2 
π
π

1
⇒
⇒
⇒
B+
x
f (π − x) = (ecos
2
∴
I=∫
π /2
0
π /2
1
dx = ∫
0
1 + tan3 x
2A 2A
=
π
π
40. Let I = ∫
0
I=∫
0
⇒
π /2
I=∫
π /2
π /2
sin3 x
dx
sin x + cos3 x
0
2I = ∫
38. Let I = ∫
Let
0
π /2
−π / 2
1 dx
⇒ 2I =
[x]0π / 2 = π
φ (− x) = [ f (− x) + f (x)] [ g (− x) − g (x)]
⇒
φ (− x) = − φ (x)
π /2
π /2
0
1 dx
π
4
π /3
dx
41. Let I = ∫
π / 6 1 + tan x
I=∫
∴
=∫
I=∫
…(i)
π /3
dx
π /6

π
1 + tan  − x

2
π /3
π /6 1 +
π /3
dx
cot x
tan xdx
π /6 1 +
2I = ∫
…(i)
tan x
…(ii)
π /3
π /6
dx
2I = [x]ππ //36 dx
⇒
⇒
I=
1 π π  π
−
=
2  3 6  12
Statement I is false.
But
/ 2 ⇒ I = π /4
φ (x) = [ f (x) + f (− x)] [ g (x) − g (− x)]
∫ −π / 2 φ (x) dx = 0
…(ii)
b
b
∫ a f (x)dx = ∫ a f (a + b − x)dx is a true statement by
property of definite integrals.
F ′ (x) < 0, ∀x ∈ (1, 3)
⇒
∴
tan x
dx
cot x + tan x
42. According to the given data,
[f (x) + f (− x) ] [ g (x) − g (− x)] dx
⇒ φ(x) is an odd function.
…(i)
I=
∴
On adding Eqs. (i) and (ii), we get
π /2
){ − cos3 (2n + 1) x} = − f (x)
On adding Eqs. (i) and (ii), we get
…(ii)
3
x
cot x
dx
cot x + tan x
2I = ∫
⇒
1
dx
sin3 x
1+
cos3 x
cos3 x
dx
3
0
cos x + sin3 x

π
cos3  − x
π /2

2
I=∫
dx
0
π

π


cos3  − x + sin3  − x

2

2
I=∫
⋅ cos3 {(2n + 1) x}
I =0
π /2
B =0
37. Let
⇒
⇒
2
On adding Eqs. (i) and (ii), we get
36. Given, f (x) = A sin 
πx
 2A
 2A
⇒ −
cos
+ Bx =

 π
π
2
0
⋅ cos3 {(2n + 1) x} dx
Again, let f (x) = ecos
2 sin x = (0, − 1), (−2, − 1), (−1, 0)
⇒
x
0,
f (a − x) = − f (x)

a
a /2
Using∫ f (x) dx = 
0
2 ∫ 0 f (x) dx, f (a − x) = f (x)
1 
1  1 

sin x = 0, −  ,  − 1, −  ,  − , 0

2 
2  2 
⇒
2
0
We have,
f (x) = x F (x)
⇒
f ′ (x) = F (x) + x F ′ (x)
⇒
f ′ (1) = F (1) + F ′ (1 ) < 0
…(i)
[given F (1) = 0 and F ′ (x) < 0]
Also,
f (2) = 2F (2) < 0
[using F (x) < 0,∀x ∈ (1, 3)]
Now, f ′ (x) = F (x) + x F ′ (x) < 0
[using F (x) < 0, ∀ x ∈ (1, 3)]
⇒
f ′ (x) < 0
294 Definite Integration
43. Given,
3
∫1 x F ′ (x) dx = − 12
2
(b) f (x) +
3
⇒ [x2F (x)]31 − ∫ 2x ⋅ F (x) dx = − 12
π
2
∫ f (t )sin t dt always positive
0
1
∴Option (b) is incorrect.
3
⇒ 9 F (3) − F ( 1 ) − 2∫ f (x) dx = − 12
1
[QxF (x) = f (x), given]
(c) Let h (x) = x −
3
− 36 − 0 − 2 ∫ f (x) dx = −12
⇒
3
∫1 f (x) dx = − 12
and
0
3 3
∫1 x F ′ ′ (x)dx = 40
h (0) = − ∫ f (t ) cos t dt < 0
0
3
⇒
[x3 F ′ (x)] 31 − ∫ 3x2F ′ (x) dx = 40
⇒
[x2(xF ′ (x)] 31 − 3 × (−12) = 40
1
⇒
{ x2 ⋅ [ f ′ (x) − F (x)]} 31 = 4
⇒
9 [ f ′ (3) − F (3)] − [ f ′ (1) − F (1)] = 4
⇒
9 [ f ′ (3) + 4] − [ f ′ (1) − 0] = 4
⇒
t
44. Given, lim
t→ a
∫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 )}
lim
=0
⇒
h→ 0
h3
1
h
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
h→ 0
6h
⇒
f ′′ (a ) = 0, ∀ a ∈ R
⇒ f (x) must have maximum degree 1.
h (1) = 1 −
F ′ ′ (c) = f ′ ′ (c)(b − a ) < 0
f (b) − f (a )
⇒ F ′ (c) = 0 ⇒ f ′ (c) =
b−a
π


0 +  

 π
2

sin 0 + sin   + 2 sin 
2
 2 




π
= (1 + 2 )
8
48. I =
98 k + 1
∑ ∫
k =1 k
(k + 1)
dx
x(x + 1)
I>
Clearly,
98 k + 1
∑ ∫
k =1 k
I>
⇒
k+1
k =1
k
∑ (k + 1) ∫
98
1
dx
(x + 1)2
 1
1 
∑ (−(k + 1)) k + 2 − k + 1  ⇒
⇒
I>
⇒
1
1
98
49
I > +…+
>
⇒ I>
3
100 100
50
Also,
I<
I>
k =1
98 k + 1
∑ ∫
k =1 k
98
1
∑k+2
k =1
98
k+1
dx = ∑ [log e (k + 1) − log e k]
x(k + 1)
k =1
I < log e 99
49. Here, f (x) = 7 tan 8 x + 7 tan 6 x − 3 tan 4 x − 3 tan 2 x
 −π π 
for all x ∈ 
, 
 2 2
∴
f (x) = 7 tan 6 x sec2 x − 3 tan 2 x sec2 x
Now, ∫
π /4
π /4
0
0
= (7 tan 6 x − 3 tan 2 x) sec2 x
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 − ∫
0
= –∫
Q e ∈ (1, e) in (0, 1) and ∫ f (t )sin t dt ∈ (0, 1) in (0, 1)
x
So, option (a) is incorrect.
(k + 1)
dx
(x + 1)2
98
x
0
∫ f (t ) cos t dt > 0
∴ Option (c) is correct.
(d) Let g (x) = x9 − f (x)
g (0) = − f (0) < 0
g (1) = 1 − f (1) > 0
∴ Option (d) is correct.
x
∴ ex − ∫ f (t )sin t dt cannot be zero.
π
−1
2
0
45. F ′ (c) = (b − a ) f ′ (c) + f (a ) − f (b)
π
−0
π /2
2
46. ∫ sin x dx =
0
4
∫ f (t ) cos t dt,
π
2
1
∴
47.
π
−x
2
π /4
0
π /4
0
π /4
0
π /4
0
1 (tan7 x − tan3 x) dx
tan3 x (tan 4 x − 1)dx
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
Definite Integration 295
1
1
 t 4 t5 
1 1 1
= ∫ (t3 − t5 )dt =  −  = − =
0
5  0 4 6 12
4
π /4
∫0
Also,
f (x) dx = ∫
1
π /4
0
I1 = I 2 + eπ ⋅ I 2 + e2π ⋅ I 2 + e3 π ⋅ I 2 = (1 + eπ + e2π + e3 π ) I 2
4π
(7 tan 6 x − 3 tan 2 x) sec2 x dx
= ∫ (7t − 3t )dt = [t
6
From Eqs. (i), (ii), (iii) and (iv), we get
2
7
0
− t3 ]10
∴
=0
f ′ (x) =
=
x
x 192 x3
192x3
dx
f
x
dx
(
)
≤
′
≤
∫1/ 2 3
∫1/ 2
∫1/ 2 2 dx
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
52. Let I = ∫
x
=∫
1
1/ 2
4π
0
π t
+∫
2π
π
Using
3π t
2π
e (sin 6 at + cos 4 at ) dt
+∫
I3 = ∫
π
π
π +x
0
I4 = ∫
…(i)
e (sin at + cos at ) dt
6
t
I3 = ∫ e
4
dt = dx
⋅ (sin 6 at + cos 4 at ) dt = eπ ⋅ I 2 …(ii)
I 4 = ∫ ex + 2π (sin 6 at + cos 4 at ) dt = e2π ⋅ I 2
0
3π
…(iii)
I5 = ∫ e3 π + x (sin 6 at + cos4 at ) dt = e3 π ⋅ I 2
sin nx
dx
(1 + π x ) sin x
b
…(i)
b
∫ a f (x) dx = ∫ a f (b + a − x) dx, we get
π x sin nx
dx
(1 + π x ) sin x
π
−π
∴ In + 2 = In
Since,
e (sin 6 at + cos4 at ) dt
⇒
t = 3π + x
π
π
−π
…(ii)
π
4π t
0
In = ∫
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
and I5 = ∫
1 4 5 4
 22
π
− + − + 4 − 4  − 0 =
−π

4
7 6 5 3
7
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⇒
Now,
Put
e (sin 6 at + cos 4 at ) dt
I1 = I 2 + I3 + I 4 + I5
∴
∴
4π t
3π
x2 − 2x)2
dx
(1 + x2)
4 
 6
5
4
2
 dx
 x − 4x + 5x − 4x + 4 −

1 + x2 
In = ∫
et (sin 6 at + cos 4 at ) dt
+∫
Put
1
0
et (sin 6 at + cos 4 at )dt
0
Now,
1 (1 +
0
53. Given
f (x) dx ≤ 3 . 9
= ∫ e (sin 6 at + cos 4 at ) dt
∴
1
0
1
=
(*) None of the option is correct.
51. Let I1 = ∫
1 (x4 − 1 ) (1 − x)4 + (1 − x)4
x4 (1 − x)4
dx
=
dx
∫0
0
(1 + x2)
1 + x2
1
 x7 4x6 5x5

4x3
= −
+
−
+ 4x − 4 tan −1 x
6
5
3
7
0
1
6
 11 2
 33
+  ≤ ∫ f (x)dx ≤ 
+
⇒ 

1/ 2
5
 10 10
5
2 .6 ≤ ∫
1⋅ (e4π − 1)
for a ∈ R
eπ − 1
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 
24x5 3 

16x5
1
− x
− x ≤ ∫ f (x)dx ≤ 
⇒ 
2 1/ 2
 5
1/ 2 1/ 2
 5
⇒
4
= ∫ (x2 − 1) (1 − x)4 dx + ∫
1
1
6
t
= (1 + eπ + e2π + e3 π )
192 x3
192x3
192x3
∴
≤ f ′ (x) ≤
4
3
2
2 + sin πx
1
On integrating between the limits to x, we get
2
50. Here,
∫ e (sin at + cos at )dt
L = 0π
6
4
t
∫0 e (sin at + cos at )dt
From Eq. (iii)
…(iv)
and
…(iii)
In = ∫
π
0
sin nx
dx
sin x
I1 = π and I 2 = 0
I1 = I3 = I5 = .... = π
I 2 = I 4 = I 6 = ... = 0
296 Definite Integration
10
∑ I 2m + 1 = 10 π
⇒
10
and
m=1
∑ I 2m = 0
⇒
[F (t )]16
1 = F (k ) − F (1 )
 d

e sin x
Q dx { F (x)} = x , given 


m =1
∴ Correct options are (a), (b), (c).
I=∫
54. (2) Let
Put
1+ 3
dx
[(x + 1)2 (1 − x)6 ]1/ 4
1+ 3
1/ 2
∫0
I=
⇒
1/ 2
0
I=
1/ 3
∫1
dx
 1 − x 
(1 − x )2 
 
 1 + x  
1−x
− 2 dx
=t ⇒
= dt
1+ x
(1 + x )2
6
when x = 0, t = 1, x =
∴
1/ 4
⇒
F (16) − F (1) = F (k) − F (1)
∴
57. Let I = ∫
Put
k = 16
π sin (π log x)
dx
x
37 π
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) − 1] = 2
1
1
,t =
2
3
58. Let I = ∫
(1 + 3 ) dt
−2( t )
(2π − x)[sin (2π − x)]2n
dx
[sin (2π − x)]2n + [cos (2π − x)]2n
2π
0
I=
−(1 + 3 )  −2
 t 
2
1
I = (1 + 3 ) ( 3 − 1) ⇒ I = 3 − 1 = 2
x
1
55. Given, f (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)
d x
⇒
=1 ⇒
 =1
dx  x 
x2
On integrating both sides, we get
I=∫
⇒
⇒
∴
56. Given,
4
∫1
Put
⇒
⇒
⇒
16
∫1
2
16
∫1
e sin t dt
⋅
= F (k) − F (1)
2
t
e
sin t
t
dt = F (k) − F (1)
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=∫
⇒
 π
π sin 2n x
dx
I = π ∫
2n
2n
 0 sin x + cos x
0
2π
0
2π
0
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
+∫
π
2n
 π
sin 2n x dx
dx
I = π ∫
0 sin 2n x + cos 2n x

+∫
π
0

sin 2n x
dx
2n
sin x + cos x 
2n
sin 2n x dx
dx
0 sin 2n x + cos 2n x
π
⇒
I = 2π ∫
⇒

 π /2
sin 2n x dx
dx
I = 4π ∫
2n
2n
0
+
x
x
sin
cos


⇒
I = 4π ∫
⇒
I = 4π ∫
2
2e sin x
dx = F (k) − F (1)
x
x2 = t
2x dx = dt
2π
I=∫
NOTE Here, f(1) = 2, does not satisfy given function.
1
∴ f(1 ) =
3
2
4 8
x and f (2) = 4 − =
3
3 3
a
0
⇒
f (x)
1

= x+ c
Q f (1) =

x`
3 
1
2
2
= 1 + c ⇒ c = and f (x) = x2 − x
3
3
3
4 8
f (2) = 4 − =
3 3
For that f (x) = x2 −
a
0
[Q ∫ f (x) dx = ∫ f (a − x) dx]
1/ 3
⇒
…(i)
2n
0
I=∫
6/ 4
x sin 2n x
dx
sin x + cos 2n x
2π
π /2
0
π /2
0
…(ii)
sin 2n (π / 2 − x)
dx
sin 2n (π / 2 − x) + cos 2n (π / 2 − x)
cos 2n x
dx
cos x + sin 2n x
2n
…(iii)
Definite Integration 297
On adding Eqs. (ii) and (iii), we get
2I = 4π ∫
⇒
2I = 4π ∫
⇒
I=π
π / 2 sin
2n
0
2n
sin
π /2
0
x + cos x
dx
x + cos 2n x
2n
1 dx = 4π [x]π0 / 2 = 4π ⋅
π
2
1
−5
x
Replacing x by 1 / x in Eq. (i), we get
59. 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)
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
⇒
2 a
1
2
2
60. Let
I=∫
⇒
I=∫
⇒
I=∫
3
2
b
1


a log|x| − x2 − 5(a − b)x
2
2 

2
(a − b ) 
1
=
1
[a log 2 − 2b − 10 (a − b)
(a − b2)
2
(2 + 3) − (5 − x) + 2 + 3 − x
3
5−x
2
x + 5 −x
61. Let
3
x+ 5−x
2
5−x+
I=∫
3 π /4
π /4
x
dx ⇒ 2I = ∫ 1 dx = 1 ⇒ I =
x
dx
1 + sin x
∫ π /4
=
π
2
∫ π /4
=
π
2
∫ π /4
=
π
2
∫ π /4
dx
−I
1 + sin x
[from Eq. (i)]
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
2
 cos x cos 2 x
2
2
−1
−2
(x2 − 1) dx + ∫
1
−1
2
(1 − x2) dx + ∫ (x2 − 1) dx
1
1
2
1.5
∫0
1
[x2] dx = ∫ 0 dx + ∫
0
2
1
1 dx +
1
2
…(i)
 π 3π

− x
 +
3 π /4
4

4
⇒ I=∫
dx
π /4
 π 3π

− x
1 + sin  +
4

4
= ( 2 − 1) + 2 (1.5 − 2 )
= 2 −1 + 3 −2 2
=2− 2
1
dx
64. (A) Let I = ∫
−1 1 + x2
Put x = tan θ ⇒ dx = sec2 θ dθ
π /4
π
I = 2∫
dθ =
∴
0
2
(B) Let
1
I=
∫0
I=
∫0
dx
1 − x2
Put x = sinθ
⇒ dx = cosθ d θ
b
b
a
a
∫
= 0 + [x]1 2 + 2 [x]1.52
…(ii)
2
π
2
dx
dx
3
=
x
dx
1 + sin x
=4
63.
2+3−x
π /4
−1
…(i)
dx
3
3 π /4
3 π /4
∫ π /4



 x3
 x3
x3 
=
− x + x −
+
− x

3
3
3
 −1 
 −2 
1

1
1
8
1
1

 
8


=  − + 1 + − 2 + 1 − + 1 −  +  − 2 − + 1
 3





3
3
3
3
3
b

+ 5 (a − b)

2
1
7 

a log 2 − 5a + b
2 
(a 2 − b2) 
x
π
dx −
1 + sin x
3 π /4
∫ −2 |1 − x |dx
=∫
On adding Eqs. (i) and (ii), we get
2I = ∫
62.
2
x
5−x+
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
− a log 1 +
π /4
= π∫
2
π−x
dx
1 + sin (π − x)
3 π /4
=∫
[Q ∫ f (x) dx = ∫ f (a + b − x) dx]
∴
π/ 2
1 dθ =
π
2
1.5
2
2 dx
298 Definite Integration
3
1
 1 + x 
∫ 2 1 − x 2 2 log  1 − x  
2
1
 2 
 3  1 
 4
= log   − log    = log   

 3 



2
−1  2 
−2
3
(C)
(D) ∫
=
dx
2
1
dx
x x −1
2
= [sec−1 x]12 =
f ( x ) = ax + bx + c
2
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 +
2 1
 ax
α β
bx
+ =1
+
 =1 ⇒

3 2
2 0
 3
⇒
=0 − ∫
As a , b are non-negative integers.
(Q) PLAN Such type of questions are converted into only sine or
cosine expression and then the number of points of
maxima in given interval are obtained.
f ( x ) = sin ( x 2 ) + cos ( x 2 )
1
 1

cos (x2) +
sin (x2)
2
 2

π
π


= 2 cos x2 cos + sin sin (x2)


4
4
π

= 2 cos  x2 − 

4
π
π
For maximum value, x2 − = 2nπ ⇒ x2 = 2nπ +
4
4
⇒
π
,for n = 0 ⇒ x = ±
4
x=±
9π
, for n = 1
4
So, f (x) attains maximum at 4 points in [− 13 , 13].
(R) PLAN
(i)
(ii)
a
∫− a
a
∫− a
f( x ) dx =
a
∫− a
f( x ) dx = 2
f( − x ) dx
a
∫0
2
3x 2
∫−2 1 + ex
dx
3 x2
dx
−2 1 + e− x
2
and
I=∫
⇒
2I = ∫
2
−2
2I = ∫
2
−2
 3x2
3x2(ex )
+
 dx

ex + 1 
 1 + ex
3x2 dx
I = [x3 ]20 = 8
2
⇒ 2I = 2 ∫ 3x2 dx
0
)
50 ⋅ x49⋅x dx
(50) (101) (1 − x50 )100 (− x50 ) dx
1
0
(1 − x50 )101 dx
1
(1 − x50 )100 dx = 5050I 2 + 5050I1
0
67. Let
π

1

1

I = ∫ e |cos x | 2 sin  cos x + 3 cos  cos x  sin x dx
0
2

2


π
1

⇒ I = ∫ e |cos x | ⋅ sin x ⋅ 2 sin  cos x dx
0
2

+
π |cos x |
∫ 0e
1

⋅ 3 cos  cos x ⋅ sin x dx
2

…(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)
where,
f( x ) dx , if f( − x ) = f( x ) , i.e. f is an even
50 100
∴ I 2 + 5050I 2 = 5050I1
(5050)I1
⇒
= 5051
I2
I1 = 0
π / 2 cos x
[Q f (π − x) = − f (x)] …(ii)
1

⋅ sin x ⋅ cos  cos x dx
2

and
I 2 = 6∫
Now,
1
 t
I 2 = 6 ∫ et ⋅ cos   dt
0
 2
function.
I=
0
+ (50) (101) ∫
f (x) = 2x or f (x) = 3x2
= 2
1
= − (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,
66. Let I 2 = ∫ (1 − x50)101 dx,
1
3
 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
65. (P) PLAN (i) A polynomial satisfying the given conditions is taken.
Let
a
∫− a
(S) PLAN
0
e
[put cos x = t ⇒ − sin x dx = dt]
1
t 
 t
 t 1 t
= 6 e cos   + ∫ e sin dt 
 2 2
2 0

1

t
et
t 
 t 1 
= 6 et cos   +  et sin − ∫
cos dt 
 2 2 
2
2
2 

0
1
t 1
t
I

− 2
= 6 et cos + et sin


2 2
2 0 4
Definite Integration 299
=

 1 e
e cos  2 + 2 sin

24
5

 1
  − 1
 2

…(iii)
From Eq. (i), we get
24 

 1 e
 1
I=
e cos   + sin   − 1
 2 2
 2
5 

68. Let I = ∫
π /3
x3 dx
π dx
+ 4∫
− π /3
− π /3
π
π


2 − cos |x|+ 
2 − cos |x|+ 


3
3
∴
2I = − 2
⇒
2I = 2
∴
I= 2
π /3
f (− x) = − f (x)
0,

a
Using ∫ f (x) dx = 
−a
2 ∫ 0 f (x) dx, f (− x) = f (x)
π /3
π dx
+0
I =2 ∫
∴
0
π

2 − cos |x| + 

3
=∫
a
x+
∴
Put
I = 2π
tan
∫ π /3
t
=u
2
I = 2π
⇒
=
π /3
π /2
69. Let I = ∫
0
⇒ I=∫
0
π /2





∫ 1/
3
I=∫
2 π /3
∫ π /3
2I = ∫
0
= 2
f (cos 2x) (sin x + cos x) dx
∫0
0
…(ii)
π
ecos x + e− cos x
dx = ∫ 1 dx = [x] π0 = π
− cos x
cos x
0
+e
e
π
0
⇒ I = π /2
71.
1
∫ 0 tan
−1
1


 1−x+ x 
1

 dx = ∫ tan −1 
 dx
0
 1 − x + x2
1 − x(1 − x) 
1
…(i)
−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  

1
π
= − ∫ tan −1 (1 − x + x2) dx
0
2
1
1
π
1
∴ ∫ tan −1 (1 − x + x2) dx = − ∫ tan −1
dx
0
0
2
(1 − x + x2)
π
= − 2I1
2
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
4 2

2
=
π /4
72. Let I = ∫
0
I=∫
0
f (cos 2x) [cos (x − π / 4)] dx
π
Put − x + = t ⇒ − dx = dt
4
a
f (x) dx = ∫ f (a − x) dx]
0 e− cos x
0
π + 4x3
4π
 1
dx =
tan −1  
 2
π

3
2 − cos |x|+ 


3
f (cos 2x) sin x dx
π /2
a
0
e− cos x
dx
+ ecos x
π
= ∫ tan −1 [1 − (1 − x) ] dx +
t
dt = 2 du
2
2 du
4π
=
[ 3 tan −1 3u ] 13
1 + 3u 2 3
3
On adding Eqs. (i) and (ii), we get
π /2
=∫
t
dt
2
t
1 + 3 tan 2
2

π

π

f  cos 2  − x  ⋅ cos  − x dx
2

2


π /2
0
On adding Eqs. (i) and (ii), we get
sec2
f (cos 2x) cos x dx
0
ecos ( π − x )
dx
ecos ( π − x ) + e− cos ( π − x )
π
1
using a f (x) dx = a f (a − x) dx
∫0
∫0


⇒
…(i)
0
⇒ sec2
3
f (sin 2t ) cos t dt
= ∫ [ tan −1 (1 − x) − tan −1 x] dx
4π
4π
 1
(tan −1 3 − tan −1 1) =
tan −1  
 2
3
3
∫ −π /3
∴
dt
= 2π
2 − cos t
π /4
∫0
0 ecos x
⇒ I=∫
π
= t ⇒ dx = dt
3
2 π /3
π /4
∫ −π / 4 f (sin 2t ) cos t dt
[Q∫


x3 dx
is odd
Q
 2 − cos |x|+ π 


3
π /3
dx
I = 2π ∫
0
2 − cos (x + π / 3)
Put


π
f cos  − 2t  cos t dt

2

ecos x
dx
+ e− cos x
π
70. Let I = ∫
− 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)
300 Definite Integration
0
I=∫
0
⇒ 2I =
73. 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 + ∫
− π 1 + cos 2 x
− π 1 + cos 2 x
π
⇒ 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
π (π − x)sin x
π (π − x)sin (π − x)
=4 ∫
dx
dx = 4 ∫
0 1 + [cos (π − x)]2
0 1 + cos 2x
π π sin x
π x sin x
=4 ∫
dx
dx − 4 ∫
0 1 + cos 2x
0 1 + cos 2x
π
sin x
⇒ I 2 = 4π ∫
dx − I 2
0 1 + cos 2x
π
sin x
⇒ 2I 2 = 4π ∫
dx
0 1 + cos 2x
cos x = t ⇒ − sin x dx = dt
−1
1
dt
dt
I2 = − 2 π ∫
=2 π ∫
=4 π
−1 1 + t 2
1 1 + t2
= 4π [tan
t ] 01
= 4π [tan
−1
1 − tan
−1
1
dt
∫ 0 1 + t2
0]
= 4π (π / 4 − 0) = π 2
∴
I = I1 + I 2 = 0 + π 2 = π 2
74. Let I = ∫
1/ 3
− 1/ 3
 x4 
 2x 
 cos −1 

 dx
 1 + x2
 1 − x4 
Put x = − y ⇒ dx = − dy
∴
I=∫
−1/ 3
1/ 3
−1/ 3
…(i)
 −2 y 
y4
cos −1 
 (−1) dy
 1 + y2 
1 − y4
1/ 3
−1/ 3
1/ 3
 2y 
y4
y4
−
dy
cos −1 
 dy
∫
4
4
−
/
1
3
 1 + y2 
1− y
1− y
1/ 3
 2x 
x4
x4
dx − ∫
cos −1 
 dx
4
−1/ 3 1 − x4
 1 + x2 
1−x
x4
dx − I
1 − x4
1/ 3
−1/ 3
[from Eq. (i)]
1/ 3 
1 
x4
−1 +
dx = π ∫
 dx
−1/ 3 
−1/ 3 1 − x4
1
−
x4 

1/ 3
1/ 3
dx
=−π∫
1 dx + π ∫
−1/ 3
−1/ 3 1 − x4
1/ 3
dx
= − π [x]1−/1/3 3 + π I1, where I1 = ∫
−1/ 3 1 − x4
1/ 3
⇒ 2I = π ∫
−π
−1
1/ 3
I = π∫
⇒
2x
dx
1 + cos 2 x
2x
Let f (x) =
1 + cos 2 x
−2 x
− 2x
⇒ f (− x) =
= − f (x)
=
1 + cos 2(− x) 1 + cos 2 x
⇒ f (− x) = − f (x) which shows that f (x) is an odd
function.
∴
I1 = 0
2x sin x
Again, let
g (x) =
1 + cos 2 x
2 (− x)sin (− x)
2x sin x
=
= g (x)
⇒ g (− x) =
1 + cos 2(− x) 1 + cos 2 x
∴
= π∫
=π∫
I = I1 + I 2
Now, I1 = ∫
Put
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
1 + y  

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
=∫
2π
1
 1
2I = − π 
+ π I1
+
 + πI1 = −
 3
3
3
⇒
Now, I1 =
1/ 3
∫ −1/
3
1/
dx
=2 ∫
4
0
1−x
dx
1 − x4
3
[since, the integral is an even function]
1/ 3
=
∫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
3
1
dx
1 + x2
1/ 3
1
 1

= − ln|1 − x| + ln|1 + x| + tan −1 x
 2

2
0
1/ 3
  1 + x 


ln 1 − x 
 0
 
=
1
2
=
1 
1 + 1 / 3
 + tan −1 1
ln 
2 1 − 1 / 3
3
+ [tan −1 x]01/
3
2


1 
3 + 1
 + π = 1 ln  ( 3 + 1)  + π
ln 
2  3 − 1 6 2  3 − 1  6
1
π
= ln (2 + 3 ) +
2
6
=
− 2π π
π2
+ ln (2 + 3 ) +
2
6
3
π
= [π + 3 ln (2 + 3 ) − 4 3 ]
6
π
⇒ I=
[π + 3 ln (2 + 3 ) − 4 3 ]
12
∴ 2I =
Definite Integration 301
3
3

 1 
1
log (x − 1) −  log (x + 1)
 2 2 
2
 2
1
2 1
4
= log − log
2
1 2
3
Alternate Solution
π
Since,
cos −1 y = − sin −1 y
2
π
2x
π
−1  2 x 
−1
= − 2 tan −1 x
∴ cos 
 = − sin
 1 + x2  2
1 + x2 2

π
x4
x4
−
2 tan −1 x dx
2 ⋅
4
4
1−x

 1−x
1/ 3
I=∫
−1/ 3
=
From Eq. (i), I = I1 + I 2
= log 2 −


x4
2 tan −1 x is an odd function 
Q
4
 1−x

∴
I =2⋅
π
2
1
∫03

1 
 −1 +
 dx + 0

1 − x4 

1
1 
+
 −2 +
 dx

1 − x2 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
2
3
3
−1

3
2
1/ 3
3
2x5 + x4 − 2x3 + 2x2 + 1
dx
(x2 + 1) (x4 − 1)
2x5 − 2x3 + x4 + 1 + 2x2
dx
(x2 + 1) (x2 − 1)(x2 + 1)
=∫
2
=∫
2
=∫
2x3 (x2 − 1)
dx +
2 (x + 1 )2 (x2 − 1 )
=∫
2x3
dx +
2 (x2 + 1 )2
3
2
3
3
(x2 + 1)2
∫ 2 (x2 + 1)2(x2 − 1) dx
∫ 2 (x2 − 1) dx
⇒ I = I1 + I 2
Put x + 1 = t ⇒ 2x dx = dt
I1 = ∫
10 1
10 1
(t − 1)
dt = ∫
dt − ∫
dt
2
5 t2
5 t
t
10
5
10
= [log
t ]10
5
1 
+
 t 
5
= log 10 − log 5 +
= log 2 −
1 1
−
10 5
1
10
3
1
1
Again, I 2 = ∫
dx = ∫
dx
2 (x2 − 1 )
2 (x − 1 )(x + 1 )
3
1
=
2
1
1
∫ 2 (x − 1) dx − 2
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
3
2
x
x
x 2a a 3
f (x) = a 
+
− +
= (x + x2 − x + 2)
3 3
3
3
3
0=
⇒
∴
⇒
2
∴
Now, integrating w.r. t. x, we get
 x3
x2 x
f (x) = a 
+
− +c
3 3
3
Again, ∫
2x3
dx
I1 = ∫
2 (x2 + 1 )2
3
Now,
1
∫ 2 (x + 1) dx
3
1
1 1
= log 6 −
10 2
10
2
1

= a  x2 + x − 

3
3
⇒
1
3
]−
quadratic polynomial and f (x) has relative maximum
1
and x = − 1 respectively,
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
∴
2x3 (x2 − 1) + (x2 + 1)2
dx
(x2 + 1)2 (x2 − 1)
3
−1/ 2
and minimum at x =
π
=
[π + 3 log (2 + 3 ) − 4 3 ]
12
75. Let I = ∫
 4
= log [2 ⋅ 21/ 2 
 3
76. Since, f (x) is a cubic polynomial. Therefore, f ′ (x) is a
=
1/ 3
1
1
1
4
+ log 2 − log
10 2
2
3
⇒
1
−1
f (x) dx =
14
3
[given]
14
3
1 a
14
2
∫ −1 3 (0 + x + 0 + 2) dx = 3
1
a
∫ −1 3 (x
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
⇒
⇒
⇒
Hence,
a
3
1
 2x3
  14
+ 4x  =

 0 3
 3
a
3
 14
2
 + 4 =

3
3
⇒
a
3
 14 14
  =
3
3
a =3
f (x) = x3 + x2 − x + 2
302 Definite Integration
π

x sin (2x) ⋅ sin  cos x
2

I=∫
dx
0
(2x − π )
79. We know that,
π
77. Let
Then
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
π

(π − x) ⋅ sin 2 (π − x) ⋅ sin
cos(π − x)
 2

dx
2 (π − x) − π
π
I=∫
…(i)
0
+ 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}
…(ii)
⇒
⇒
π

(π − x) ⋅ sin 2x ⋅ sin  cos x
π
2

I=∫
dx
0
π − 2x
I=∫
π

π
(x − π ) sin 2x ⋅ sin  cos x

2
0
(2x − π )
dx
… (iii)
= 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 ]
= (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

2I = 2 ∫ sin x cos x ⋅ sin  cos x dx
0
2

π
⇒
π

π
⇒ I = ∫ sin x cos x ⋅ sin  cos x dx
0

2
2 
π
π

put cos x = t ⇒ − sin x dx = dt ⇒ sin x dx = − dt

π 
2
2
2 − π / 2 2t
∴
I=− ∫
⋅ sin t dt
π π /2 π
4 π /2
= 2∫
t sin t dt
π −π / 2
4
4
8
I = 2 [− t cos t + sin t ]π− π/ 2/ 2 = 2 × 2 = 2
⇒
π
π
π
78. Let
I=∫
π /2
0
=∫
0
π /2
0
π /2
0
=∫
2I = ∫
0
f [sin 2x] ⋅ cos x dx
=2 ∫
=2 2 ∫
=2 2
=2 2
∴
I= 2
Hence, ∫
π /2
0
π /4
0
π /4
∫0
π /4
∫0
π /4
∫0
( cos 2x + cos 4x K cos (2k − 2) x) dx
π /2
∫0
cos (2k) x dx
sin 2x sin 4x
π
sin (2k − 2) x 
+2
+
+K+
2
4
(2k − 2)  0
 2
+
π /2
sin (2k) x 


2k
0
=
π
2
0
a
a
0
0
I = ∫ f (a − x) ⋅ g (a − x) dx = ∫ f (x) ⋅ {2 − g (x)} dx
…(ii)
[Q f (a − x) = f (x) and g (x) + g (a − x) = 2]
a
a
0
0
= 2 ∫ f (x) dx − ∫ f (x) g (x) dx
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
a
I = 2 ∫ f (x) dx − I
⇒
0
a
2I = 2 ∫ f (x) dx
⇒
0
a
81. Let
π /4
0
a
∫ 0 f (x) g(x) dx = ∫ 0 f (x) dx
∴
2a
I=∫
0
I=∫
0
2a
f (cos 2x) cos x dx
f (sin 2x) ⋅ sin x dx = 2 ∫
π/ 2
0
80. Let I = ∫ f (x) ⋅ g (x) dx
f (sin 2x)(sin x + cos x) dx
0
1 ⋅ dx + 2 ∫
π /2
=
f (sin 2x)(sin x + cos x) dx
π /4
π /2
0
+
On adding Eqs. (i) and (ii), we get
π /2
(sin 2kx) ⋅ cot x dx
a


π

π
f sin 2  − x  sin  − x dx




2
2


π /2
Then, I = ∫
∴∫
…(i)
f (sin 2x) sin x dx
…(i)
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
f (cos 2x) cos x dx
2I = ∫
2a
0
1 dx = 2a ⇒ I = a
Definite Integration 303
1
82. Let I = ∫ log ( 1 − x + 1 + x ) dx
Put x = cos 2θ
π
sin α
π
=
sin α
0
π /4
= −2 ∫
= −2 ∫
log [ 1 − cos 2θ + 1 + cos 2θ ] (sin 2θ ) dθ
0
π /4
0
π /4
log [ 2 (sin θ + cos θ )] sin 2θ dθ
+ log (sin θ + cos θ ) ⋅ sin 2θ ] dθ
 − cos 2θ 

 2
π /4
−2 ∫
0
π /4
⇒ I=∫

π

π

π
 − x sin  − x ⋅ cos  − x

2

2

dx


4 π
4 π
sin  − x + cos  − x
2

2

⇒ I=∫
 
cos 2θ 
= log 2 − 2 − log (sin θ + cos θ ) ⋅

2 π / 4
 

 cos θ − sin θ − cos 2θ 
−∫
×

 dθ 
π / 4  cos θ + sin θ
2 

0
1


= log ( 2 ) − 2 0 + ∫
( cos θ − sin θ )2 dθ


2 π /4
0
1
= log 2 − ∫
(1 − sin 2θ ) dθ
π /4
2
0
0
1
cos 2θ 

= log 2 − θ +

2
2  π / 4
1
1 π
1 π 1
log 2 −  −  = log 2 − +


2
2
2 4
2 4
π
x
83. Let I = ∫
dx
0 1 + cos α sin x
=
⇒
(π − x)
I=∫
dx
0 1 + cos α sin (π − x)
⇒
I=∫
(π − x)
dx
1 + cos α sin x
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
2x
Put tan = t ⇒ sec dx = 2 dt
2
2
∞
2 dt
∴
2I = π ∫
0 1 + t 2 + 2 t cos α
⇒
…(i)
2I = 2π
∫0
dt
(t + cos α )2 + sin 2 α

π
 − x ⋅ sin x cos x

π /2  2
… (ii)
cos 4 x + sin 4 x
0
⇒ I=
π
2
∫0
=
π
2
∫0
⇒ 2I=
π /2
π
2
dx
π / 2 x sin x ⋅ cos x
sin x cos x
dx − ∫
dx
4
4
0
sin 4 x + cos 4 x
sin x + cos x
π /2
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)
0
2 2
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]
π2
⇒ I=
16
⇒ 2I=
85. Let I = ∫
π
∞
0
II
0
π
π /2  2
log (sin θ + cos θ ) ⋅ sin 2θ dθ
I
απ
π π

 −  − α   =
2
 sin α
2
απ
sin α
π / 2 x sin x ⋅ cos x
84. Let I = ∫
dx
0
cos 4 x + sin 4 x
0
0
[tan −1 (∞ ) − tan −1 (cot α )]
I=
∴
[(log 2 ) sin 2θ
= − 2 log 2
∞

−1  t + cos α  
tan  sin α  

0
=
⇒ dx = − 2 sin 2θ dθ
∴ I = −2 ∫
π
sin α
I=
0
1/ 2
x sin −1 x
1 − x2
0
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
86. Let
I=∫
I=∫
π /4
0
π /4
0
(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
304 Definite Integration

 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=
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
40
π
87. (i) Let I = ∫ x f (sin x) dx
…(i)
0
π
I=∫
⇒
(π − x) f (sin x) dx
0
…(ii)
∴
π
∫0
π
0
π
x f (sin x) dx =
2
I=∫
(ii) Let
3/ 2
−1
1
I=∫
∴
π f (sin x) dx
=∫
π
∫ 0 f (sin x) dx
−1
3/ 2
∫1
x f (x2)
dx
2 + f (x + 1)
2
−1
x f (x2)
dx +
2 + f (x + 1)
0
−1
+
|x sin πx| dx
x sin πx dx +
x dx =
[x], x ≤ 2
 0, x > 2
 x sin πx, −1 < x ≤ 1
3
Since, |x sin πx| = 
− x sin πx, 1 < x < 2
∴ I=∫
∫1
2
1
∫0
=∫
x f (x2)
dx +
2 + f (x + 1)
0
0 dx +
−1
1
∫0 0 dx + ∫1
2
1
1
 − cos πx
 x cos πx 
− 2 ∫ 1⋅ 
=2 −
 dx


0

π
π 
0
2 2
1
1
+
(0 − 0) + + 2 (+ 1 − 0)
π π2
π π
3 1  3π + 1
= + 2=

 π2 
π π
0
1
1
1
0
0
=∫
1
0
=
1
0
Put
⇒
Σ
⇒
C r ∫ (1 − x)n − r ⋅ xr dx t r
 0

n
91.
Σ
…(ii)
r=0
1
1
C r ∫ (1 − x)n − r ⋅ xr dx t r=
(1 + t + K + t n )
 0

n+1
n
0 dx
2
 12 + 9x2
 dx

 1 + x2 
9+3π /4 t
e dt = [et ]90 + 3 π / 4 = e9 + 3 π / 4 − 1
0
log e 1 + α = 9 +
log e α + 1 −
3π
=9
4
3π
4
PLAN Integration by parts
d
From Eqs. (i) and (ii), we get
n
2
∫3
9x + 3 tan −1 x = t

3 
9 +
 dx = dt

1 + x2 
∴ α=∫
C 2(1 − x)n − 2(tx)2+ ... + nC n (tx)n + ] dx
r=0
0 dx +
4I = 1 ⇒ 4I − 1 = 0
0
⇒
1
2
 x2 
1
1
x
dx =   = (2 − 1) =
2
4
4
4
 1
90. Here, α = ∫ e
[nC 0 (1 − x)n + nC1 (1 − x)n − 1 (t x)
n
2
1 ( 9 x + 3 tan −1 x )
[(1 − x) + t x]n dx
 n n

C r (1 − x)n − r (t x)r  dx
rΣ
=0


1
∴
n
=∫
I=∫
⇒
 ((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
I = ∫ (t x + 1 − x)n dx = ∫
3
∫
and 3 < x < 2 ⇒ 3 < x2 < 4 ⇒ f (x2) = 0
I = ∫ (t x + 1 − x)n dx = ∫ {(t − 1) x + 1 }n dx
Again,
x f (x2)
dx
2 + f (x + 1)
2 < x < 3 ⇒ 2 < x2 < 3 ⇒ f (x2) = 0,
=
0
2
∫3
x f (x2)
dx
2 + f (x + 1)

⇒ [x2] = 1
1 < x2 < 2
1 < x< 2 ⇒
2 < x + 1 < 1 + 2 ⇒ f (x+ 1) = 0,
 1  2 sin πx   1  1 sin πx 
=2   + ⋅
− −  −
 π  π  π  0  π  π  π 1
88. Let
3
2
Q − 1 < x < 0 ⇒ 0 < x2 < 1 ⇒ [x2] = 0,
0 < x < 1 ⇒ 0 < x2 < 1 ⇒ [x2] = 0,
3/ 2
1
∫
x⋅1
dx
2+0
+

 − x cos πx 3/ 2
3 / 2  − cos πx
−∫ 
−
 dx


1 

π
π
1


1
x f (x2)
dx
2 + f (x + 1)
+
− x sin πx dx
1
1
(n + 1)nC k
n−k k
89. Here, f (x) = 
On adding Eqs. (i) and (ii), we get
2I=∫
1
∫ 0 (1 − x)
⇒

∫ f(x ) g (x ) dx = f(x ) ∫ g (x ) dx − ∫  dx [f(x )] ∫ g (x ) dx  dx
Given, I = ∫
1
0
4x3
I
d2
(1 − x2)5 dx
dx2
II
1
1
d
d


= 4x3
(1 − x2)5 − ∫ 12 x2
(1 − x2)5 dx
0
dx
dx
 0

Definite Integration 305
1
= 4x3 × 5 (1 − x2)4 (− 2x)

0
1
− 12 [x2 (1 − x2)5 ]10 − ∫ 2x (1 − x2)5 dx


0
1
Again,
0
1
 (1 − x2)6 
1

= 12 × −
 = 12 0 + 6  = 2
6


0

Topic 2 Periodicity of Integral Functions
Let I = ∫
1.
=∫
+∫
− 2 ,

[x] =  0−,1,
 1,
Q
dx
0
∫−1 [x] + [sin x] + 4
2
dx
+
1
0 ≤ ∫ f (t ) dt ≤
2
1
4.
∫0
0
∫ −1
dx
+
2
1 dx
∫0
4
π
2
1
+∫
∴
rπ
( r − 1 )π
|sin x|dx + ∫
|sin x| dx +
nπ + v
|sin x|dx
nπ
nπ + v
∫ nπ
|sin x| dx
x = r π, t = π
∴
rπ
π
∫ ( r − 1)π|sin x| dx = ∫ 0 |(−1)
r −1
sin t| dt
π
π
0
0
= ∫ |sin t| dt = ∫
= [− cos t ] π0 = − cos π + cos 0 = 2
nπ + v
Again, ∫
| sin x|dx, putting x = nπ + t
nπ
nπ + v
∫ nπ
sin t dt
v
v
0
0
| sin x|dx = ∫ |(−1)n sin t|dt = ∫
sin t dt
= [− cos t ]v0 = − cos v + cos 0 = 1 − cos v
∴∫
1
f (2x) dx Put 2x = y ⇒ dx = dy
2
1 6 + 6T
6I
= 3I
f ( y) dy =
2∫6
2
3. Given, g (x) = ∫ f (t ) dt
n
nπ + v
|sin x|dx = ∑ ∫
0
r =1
=
rπ
| sin x|dx +
( r − 1 )π
n
nπ + v
∫ nπ
|sin x|dx
nπ + v
∑ 2 + ∫ nπ
|sin x|dx
r =1
= 2n + 1 − cos v
φ (a ) = ∫
5. Let
a+ t
a
f (x) dx
On differentiating w.r.t. a, we get
0
Now,
( n − 1 )π
|sin x|dx + ...
when x = (r − 1) π, t = 0 and when
and
x
⇒
nπ
2π
π
sin x = sin [(r − 1) π + t ] = (−1)r − 1 sin t
⇒
Then,
−20 + 10 + 5 − 4 5π + π
+
20
10
9
3π
3
=−
+
=
(4π − 3)
20
5 20
∫3
|sin x|dx + ∫
x = ( r − 1)π + t
=
2.
0
rπ
dx
5
1 1 1  π π 

=  −1 + + −  +  + 

2 4 5  2 10 
1
2
∫ ( r −1) π| sin x |dx , we have
Now to solve,
1
1 π
π 1


=  −1 +  + (0 + 1) + (1 − 0) +  − 1


2 2
4
5 2
3 + 3T
π
|sin x|dx = ∫
r =1
dx
1+0+4
…(ii)
0 ≤ g (2) < 2
nπ + v
n
+∫
−1
⇒
=∑∫
π
2
1
…(i)
2
+∫
[Q For x < 0,−1 ≤ sin x < 0 and for x > 0, 0 < sin x ≤ 1]
0
1
−1
dx
dx
dx
I = ∫ −π
+∫
+∫
1
0
−
−
−
+
−
−
+
+
2
1
4
1
1
4
0
0+4
2
= ∫ −π
2
∫ 10 dt ≤ ∫ 1 f (t ) dt ≤ ∫ 1 dt
− π / 2 < x < −1
−1 ≤ x < 0
0 ≤ x<1
1 ≤ x < π /2
 − 1 , − π / 2 < x < −1

and [sin x] = −01, , − 01 << xx << 10
 0, 1 < x < π / 2
So,
2
⇒
⇒
π
dx
dx
+∫2
0 [x] + [sin x] + 4
1 [x] + [sin x] + 4
1
1
From Eqs. (i) and (ii), we get
1
2
1
3
≤ ∫ f (t ) dt + ∫ f (t ) dt ≤
0
1
2
2
1
3
⇒
≤ g (2) ≤
2
2
dx
[x] + [sin x] + 4
dx
+
[x] + [sin x] + 4
1
1
1
≤ ∫ f (t ) dt ≤ 1
0
2
1
for t ∈ [1, 2]
0 ≤ f (t ) ≤
2
⇒
= 0 − 0 − 12 (0 − 0) + 12 ∫ 2x (1 − x2)5 dx
π
2
−π
2
−1
−π
2
11
∫ 0 2 dt ≤ ∫ 0 f (t ) dt ≤ ∫ 0 1 dt
We get
2
1
0
0
g (2) = ∫ f (t ) dt = ∫ f (t ) dt + ∫ f (t )dt
1
≤ f (t ) ≤ 1 for t ∈ [0,1]
2
φ ′ (a ) = f (a + t ) ⋅ 1 − f (a ) ⋅ 1 = 0 [given, f (x + t ) = f (x)]
2
1
∴ φ (a ) is constant.
⇒
a+ t
∫a
f (x) dx is independent of a.
306 Definite Integration
if [x] is odd.
 x − [x],
1 + [x] − x, if [x] is even.
f (x) and cos π x both are periodic with period 2 and both
are even.
10
10
− 10
0
∫ f (x) cos π x dx = 2 ∫ f (x) cos π x dx
∴
2. PLAN Newton-Leibnitz’s formula
d
d
d  ψ (x )
φ ( x )
ψ ( x ) − f { φ ( x )} 
f(t ) dt = f { ψ ( x )} 

dx  ∫ φ ( x )
 dx
 dx


Y
F (x) = ∫
Given,
–10 – 9
–2 –1 0
1 2
9 10
X
2
0
f (x) cos π x dx
0
1
1
= ∫ (1 − x) cos π x dx = − ∫ u cos π u du
0
and
0
2
2
1
1
∫ f (x) cos π x dx = ∫ (x − 1)
0
1
40
∫ f (x) cos πx dx = − 20 ∫ u cos π u du = π 2
− 10
0
f (x) = ex
Now,
∴
f (0) = 1
1=K
Hence,
f (x) = ex
π2
f (x) cos π x dx = 4
10 −∫10
Topic 3 Estimation, Gamma Function and
Derivative of Definite Integration
1
0
x
1. Given, ∫ f (t ) dt = x2 + ∫ t 2f (t )dt
f (x) = 2x + 0 − x2f (x)
⇒


ψ( x )
d
d
ψ (x) − f (φ (x))
φ(x)
 ∫ f (t )dt  = f (ψ (x))
dx
dx


φ ( x )

(1 + x2) f (x) = 2x ⇒ f (x) =
2
[K = ec ]
2
4
x
3. Given, y = ∫ |t| dt
0
∴
dy
= | x|⋅ 1 − 0 = |x|
dx
[by Leibnitz’s rule]
x
Q Tangent to the curve y = ∫ |t| dt , x ∈ R are parallel
0
y=∫
± 2
0
| t | dt = ± 2
Equation of tangent is
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 ∫
(1 + x2)(2) − (2x) (0 + 2x)
(1 + x2)2
2 + 2 x2 − 4 x2 2 − 2 x 2
=
=
(1 + x2)2
(1 + x2)2
x
0
x
1 − { f ′ (t )}2 dt = ∫ f (t ) dt , 0 ≤ x ≤ 1
0
Differentiating both sides w.r.t. x by using Leibnitz’s
rule, we get
2x
1 + x2
On differentiating w.r.t. ‘x’ we get
f ′ (x) =
⇒ f (x) = K ex
For x intercept put y = 0, we get
On differentiating both sides, w.r.t. ‘x’, we get

d
Q
 dx

+ c
F (2) = ∫ et dt = [et ]40 = e4 − 1
Points,
x
2
to the line y = 2x
∴ Slope of both are equal ⇒ x = ± 2
10
⇒
f ( t ) dt
0
1
∴
0
⇒
cos πx dx
= − ∫ u cos π u du
10
x2
F ′ (x) = 2x f (x)
F ′ (x) = f ′ (x)
2x f (x) = f ′ (x)
f ′ (x)
= 2x
f (x)
f ′ (x)
dx = ∫ 2x dx ⇒ In f (x) = x2 + c
f (x)
∫
⇒
1
∫
∴
Also,
⇒
⇒
= 10 ∫ f (x) cos π x dx
Now,
2
 1
3
2 −2   2 − 1
 4
 1
2 = 2 = 24
∴f′   =
=
=
2
2
25 25
 2 
2 2
1

 5
1

1
+




1 +   
16




4
4
 2 

 1
2 −2  
 2
6. Given, f (x) = 
1 − { f ′ (x)}2 = f (x) ⇒
⇒
∫
f ′ (x)
1 − { f (x)}2
dx = ±
∫ dx
f ′ (x) = ± 1 − { f (x)}2
⇒ sin −1 { f (x)} = ± x + c
Put
x = 0 ⇒ sin −1 { f (0)} = c
⇒
c = sin −1 (0) = 0
[Q f (0) = 0]
Definite Integration 307
∴
f (x) = ± sin x
but
f (x) ≥ 0, ∀ x ∈ [0, 1]
∴
f (x) = sin x
1
m+ 1
1

tm + 1 
( n − 1) t
1
dt
n
t
−
(
+
)
⋅
∴ I (m, n ) = (1 + t )n ⋅
m + 1  0 ∫ 0
m+1

1
n
2n
=
−
(1 + t )( n − 1) ⋅ tm + 1 dt
∫
0
m+1 m+1
n
2n
∴ I (m, n ) =
−
⋅ I (m + 1, n − 1)
m+1 m+1
As we know that,
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
⇒
5. Since ∫
1
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
2
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,
t
∫0
t 2f (t 2)2t = 2t 4
∴
2
4
f   =−
 25
5
⇒
4 2
f   =
 25 5
e
x2
f (t 2) = t
= e− ( x
⇒
x4 + x2 − 2 = 0 ⇒ x = ± 1
10. Given, F (x) = ∫
2
+ 1 )2 
d 2

− ( x 2 )2
 (x + 1) − e
dx


2
2
+ 1)
⋅ 2x − e− ( x
− ( x 4 + 2x 2 + 1 )
= 2xe
∴
)
f (t ) dt
F ′ (x) = f (x)
…(i)
F (x ) = x (1 + x) = x + x
2
But
2
2
⇒
F (x) = x + x
⇒
f (x) = F ′ (x) = 1 +
⇒
d 2
(x )


 dx
[given]
3
F ′ (x) = 1 + x1/ 2
2
⇒
3/ 2
f (4) = 1 +
3
3 1/ 2
x
2
[from Eq. (i)]
3 1/ 2
3
(4) ⇒ f (4) = 1 + × 2 = 4
2
2
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
12. lim ∫
x→1
f ( x)
2t
dt = lim
x→1
x−1
∫4
2t dt
x−1
[using L’ Hospital’s rule]
2 f (x) ⋅ f ′ (x)
= lim
= 2 f (1) ⋅ f ′ (1)
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 )
a
where, M and m are maximum and minimum values
respectively of f (x) in [a , b].
⋅ 2x
2x 2 + 1
(1 − e
2x 2 + 1
[where, e
2 2
x
0
By Leibnitz’s rule,
dt
On differentiating both sides using Newton’s Leibnitz’s
formula, we get
f ′ (x) = e− ( x
f ′ (x) = 2 − x2
⇒ x4 = 2 − x 2
f ( x)
2

putting t =

5 
x 2 + 1 −t 2
f (x) = ∫
7. Given,
⇒
2 − t 2 dt ⇒
x2 = 2 − x 2
∴
x f (x) dx =
⇒
x
1
Also, x2 − f ′ (x) = 0
t 2 f (t ) dt = 1 − sin x, thus to find f (x).
sin x
f (x) = ∫
9. Given,
2
Here, f (x) = 1 + e− x is continuous in [0, 1].
)
− ( x 4 + 2x 2 + 1 )
> 1, ∀ x and e
> 0, ∀ x]
f ′ (x) > 0
Now,
2
2
1
8. Here, I (m, n ) = ∫ t m(1 + t )n dt reduce into I (m + 1, n − 1)
0
[we apply integration by parts taking (1 + t ) as first
and t m as second function]
n
2
Again, 0 < x < 1 ⇒ x2 > 0 ⇒ ex > e0 ⇒ e− x < 1
2
e− x < e− x < 1, ∀ x ∈ [0, 1]
∴
which shows 2x < 0 or x < 0 ⇒ x ∈ (−∞ , 0)
2
0 < x < 1 ⇒ x2 < x ⇒ ex < ex ⇒ e− x > e− x
2
1 + e− x < 1 + e− x < 2, ∀ x ∈ [0, 1]
⇒
⇒
⇒
1
−x
∫ 0 (1 + e
1
2
1
) dx < ∫ (1 + e− x ) dx < ∫ 2 dx
0
1
2
1
2 − < ∫ (1 + e− x ) dx < 2
0
e
0
308 Definite Integration
14. g (x) = ∫
sin 2x
sin x
= (cosecx ⋅ cot x + sec2 x − cos x) ⋅ (cos3 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


sin −1 (t )dt
g′ (x) = 2 cos 2x sin −1 (sin 2x) − cos x sin −1 (sin x)
 π
g′   = − 2 sin −1 (0) = 0
 2
= − sin 2 x − cos3 x (1 − sin 2 x) = − sin 2 x − cos5 x
 π
g′  −  = − 2 sin −1 (0) = 0
 2
∴
π /2
∫0
f (x) dx = −


Q


No option is matching.
15. Here, f (x) + 2x = (1 − x)2 ⋅ sin 2 x + x2 + 2x
where, P : f (x) + 2x = 2 (1 + x)2
∴
…(i)
…(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
π /2
∫0
∴ P is false.
Again, let Q : h (x) = 2 f (x) + 1 − 2x (1 + x)
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.
16. 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

18. f (x) = ∫
=∫
φ(x) < 0
g′ (x) < 0 for x ∈ (1, ∞ )
sec x cos x cosec x ⋅ cotx + sec2x
17. Given, f (x) = cos 2 x cos 2 x
cosec2x
cos 2 x
1
R3 ,
cos x
sec x cos x cosec x ⋅ cot x + sec2 x
f (x) = cos x cos 2 x cos 2 x
cosec2 x
Applying R3 →
cos x
Applying R1 → R1 − R3 ⇒ f (x)
cosec x ⋅ cot x + sec2x − cos x
0
0
= cos x cos x cos x
sec x cos x
2
2
ln t
dt for x > 0
1+ t
1/ x
1
[given]
ln t
dt
1+ t
=∫
x
 1
f (x) + f   = ∫
 x
x
…(ii)
1
x ln t
ln u
du = ∫
dt
1 t (1 + t )
u (u + 1)
ln t
dt +
1 (1 + t )
ln t
Put x = e,
1
 1 1
f (e) + f   = (ln e)2 =
 e 2
2
⇒
⇒
⇒
and
⇒
⇒
⇒
Again,
⇒
2
cosec x
cos x
x
∫ 1 t (1 + t ) dt
x ln t
t ) ln t
1
1
dt = ∫
dt = [(ln t )2]1x = (ln x)2
1
t (1 + t )
t
2
2
19. Let t = b − a and a + b = 4
∴g (x) is decreasing for x ∈ (1, ∞ ).
cos x





x (1 +
1
From Eqs. (i) and (ii), we get
sec x
∫0
f (1 / x) = ∫
Now,
4
1 − (x − 1)2
φ′ (x) =
− =
2
x x (x + 1)2
(x + 1)
cos 2 x
x
1
…(i)
For g′ (x) to be increasing or decreasing,
2 (x − 1)
let φ(x) =
− log x
(x + 1)
1
π /2
Put t = 1 / u ⇒ dt = (−1 / u 2) du
x ln (1 / u ) (−1 )
∴
f (1 / x) = ∫
⋅ 2 du
1 1 + 1 /u
u

2 (x − 1)
− log x ⋅ f {
g′ (x) = 
(x)
+
x
(
)
1
 + ve

φ′ (x) < 0, for x > 1 ⇒ φ (x) < φ (1) ⇒
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 
m+ 1


f (x) dx = − 


Now,
∴ Q is true.
⇒
(sin 2 x + cos5 x) dx


8
1 /2 ⋅ π

2 π
 15π + 32
π
=−
+

 = − 
= −  +
5 3 1
60 
4
15
2



2⋅ ⋅ ⋅ ⋅ π 


2 2 2
which is never possible.
where,
π /2
∫0
⇒
t =4 − a − a
t = 4 − 2a
t
a =2 −
2
t = b − (4 − b)
t = 2b − 4
t
= b −2
2
t
b =2 +
2
a <2
π
2 − <2
2
π
>0 ⇒ t >0
2
Hence proved.
[given]
[given]
Definite Integration 309
a
b
0
0
Now, ∫ g (x) dx + ∫ g (x) dx
=∫
F (x) = ∫
Let
2 − t/ 2
0
2 − t/ 2
0
For t > 0, F ′ (t ) = −
As x → ∞, ∫
2 + t/ 2
g (x) dx +
∫0
g (x) dx +
∫0
x
∫0
g (x) dx
2 + t/ 2
g (x) dx
t 1 
t
1 
g 2 −  + g 2 + 
2 
2 2 
2
[using Leibnitz’s rule]
1 
t 1 
t
= g 2 +  − g 2 − 
2 
2 2 
2
dg
> 0, ∀ x ∈ R
dx
Again,
x
0
f (t ) dt → ∞ for a particular x (say xn ), then
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.
x
22. Given, f (x) = ∫ [2(t − 1) (t − 2)3 + 3 (t − 1)2(t − 2)2] dt
1
∴
f ′ (x) = [2 (x − 1) (x − 2)3 + 3 (x − 1)2(x − 2)2] ⋅ 1 − 0
= (x − 1) (x − 2)2 [2 (x − 2) + 3 (x − 1)]
[given]
= (x − 1) (x − 2)2 (5x − 7)
Now, 2 − t / 2 < 2 + t / 2 ∴ t > 0
+
We get g (2 + t / 2) − g (2 − t / 2) > 0, ∀ t > 0
F ′ (t ) > 0, ∀ t > 0
So,
1
Hence, F (t ) increases with t, therefore F (t ) increases as
(b − a ) increases.
20. Let I n = ∫
1
0
ex (x − 1)n dx
Put x − 1 = t
∴
In = ∫
⇒
−1
e
⋅ t ndt = e ∫
0 n t
= e 0 − (−1)n e−1 − n

⇒
−1
0
−1
0
∫ −1
n→ ∞
n→ ∞
t n − 1et dt

1/3
…(i)
For n = 1, I1 = ∫ e (x − 1) dx = [e (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
I3 = (−1)3 + 1 − 3I 2 = 1 − 3(2e − 5) = 16 − 6e
Hence, n = 3 is the answer.
21. Since, f is continuous function and ∫
r

1/3
1 +  n

n
= lim ∑
n→ ∞
n 4/3
r =1
x
0
f (t ) dt → ∞ ,
= lim
n→ ∞
n
r

∑ 1 + n 
r =1
1/3
1
So, p = ∫ (1 + x)1/3 dx
0

1
Q lim
n→ ∞ n

n
 r
1

∑ f  n  = ∫0 f (x) dx
r =1

1
=
0
Y
(0,√2)
A
1
n
Now, as per integration as limit of sum.
r
1
[Q n → ∞]
Let = x and = dx
n
n
Then, upper limit of integral is 1 and lower limit of
integral is 0.
as| x|→ ∞. To show that every line y = mx intersects the
x
curve y2 + ∫ f (t ) dt = 2
O
(n + r )1/3
n 4/3
r =1
∑
n
I n = (−1)n + 1 − nI n − 1
x
 (n + 1)1/3
(n + 2)1/3
(2n )1/3 
+
+…+

4/3
4/3
n
n 4/3 
 n
n
= lim
t n − 1et dt
1 x
and
7/5
∴ f (x) attains maximum at x = 1 and f (x) attains
7
minimum at x = .
5
1. Let p = lim 
t e dt
0
= e  [t net ] −01 − n ∫ t n − 1et dt


−1
= (−1)n + 1 − ne ∫
+
Topic 4 Limits as the Sum
dx = dt
0 t+1
–
3
3
3

3
(1 + x)4/3 = (24/3 − 1) = (2)4/3 −

4
4
4
4
0
2. Clearly,
(Xp,0)
X
B (0,–√2)
At x = 0, y = ± 2
Hence, (0, 2 ), (0, − 2 ) are the point of intersection of
the curve with the Y-axis.
lim  n
1
n
n
+
+
+ ...+


n → ∞  n 2 + 12 n 2 + 22 n 2 + 32
5n 
=
=
lim  n

n
n
n
+
+
+ ....+ 2


n → ∞  n 2 + 12 n 2 + 22 n 2 + 32
n + (2n )2
lim
2n
n
n → ∞ r=1 n 2 + r 2
∑
310 Definite Integration
=
lim
n→ ∞
2n
∑
r =1
1
 r
1+  
 n
2
⋅
2 dx
1
=∫
0
n
1 + x2
lim
n→ ∞
(n + 1)
a −1
1
1a + 2a + K + n a
=
{(na + 1) + (na + 2) + K + (na + n )} 60
n
pn


p
1  r
Q lim ∑ f   = ∫ 0 f (x)dx
n→∞
n n
r =1


= [tan −1 x]20 = tan −1 2
∑
⇒ lim
n→ ∞
(n + 1)
a −1
1
 (n + 1) ⋅ (n + 2) ... (n + 2n )
= lim 


n→ ∞ 
n 2n
n→ ∞
1

1 + 

n
1 n
⇒ lim 2 ∑
n→ ∞ n 
 r =1
1
⇒
1
n

∴
⇒
⇒
⇒
⇒
⇒
⇒
∴
⋅ (2na + n + 1)
r
∑ log 1 + n 
r =1
a
n
n
2
k = 0 n + kn + k
5. Given, S n = ∑
2
log l = ∫ log (1 + x) dx
0
2


1
log l = log (1 + x) ⋅ x − ∫
⋅ x dx
1+ x

0
2
x+ 1 −1
log l = [log (1 + x) ⋅ x]20 − ∫
dx
0
1+ x




1
1
 < lim
= ∑ ⋅
n 
k k2  n → ∞
k=0
1 + + 2

n n 
2
1 
log l = 2 ⋅ log 3 − ∫ 1 −
 dx
0 
1 + x
=∫
n
log l = 2 ⋅ log 3 − [x − log 1 + x ]20
log l = 2 ⋅ log 3 − [2 − log 3]
log l = 3 ⋅ log 3 − 2
log l = log 27 − 2
27
l = elog 27 − 2 = 27 ⋅ e− 2 = 2
e
4. PLAN
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)
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 +  
 3
2   0
 3
2  π π
π
π
i.e. S n <
⋅ −  =
3  3 6 3 3
3 3
π
Similarly,
Tn >
3 3
=
 1
1
1  5n 1
+
+K+
 =∑
n→ ∞  n + 1
n+2
6n  r=1 n + r
6. lim 
g ( n)
lim
n→ ∞ n
= lim
n→ ∞
r
is replaced with x.
n
Σ is replaced with integral.
where,
Here,
1
60
⇒ 2a 2 + 3a + 1 − 120 = 0 ⇒ 2a 2 + 3a − 119 = 0
−17
⇒ (2a + 17) (a − 7) = 0 ⇒ a = 7,
2
2
⇒
=
2 ⋅ [xa+ 1 ] 10
1
=
(2a + 1) ⋅ (a + 1) 60
2
1
=
⇒ (2a + 1) (a + 1) = 120
(2a + 1) (a + 1) 60
⇒
1
2
2n  




log 1 + n  + log 1 + n  + ... + log 1 + n  


n→ ∞
1
60
a
r =1
a −1
 r
 
 n
=
1
1
 ⋅ lim
=
a −1
 n→ ∞
1  60
1



⋅ 2a + 1 + 
1 + 


n
n
1
1
1
⇒ 2∫ (xa ) dx ⋅
=
0
1 ⋅ (2a + 1) 60
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
log l = lim
 r
2∑  
 n
⇒ lim
1
n
  n + 1  n + 2
 n + 2n   n
= lim  
 K

 
 n  
n→ ∞   n   n 
⇒
n (n + 1) 

⋅ n 2a +
2


n
(n + 1) ⋅ (n + 2) K (3n ) n
3. Let l = lim 


n→ ∞ 
n 2n
2n
ra
r =1
=∫
Download Chapter Test
http://tinyurl.com/y6h4j36z
or
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. 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
2. The area (in sq units) of the region {(x, y) : y2 ≥ 2x
and x2 + y2 ≤ 4x, x ≥ 0, y ≥ 0} is
4
(a) π −
3
4 2
(c) π −
3
(2016 Main)
8
(b) π −
3
π 2 2
(d) −
2
3
7. 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
(b) 3
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
Objective Questions II
(One or more than one correct option)
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 )
(2014 Adv.)
(b) 6
(c) 9
(d)15
4. The area of the equilateral triangle, in which three
coins of radius 1 cm are placed, as shown in the
figure, is
(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
5. The area of the quadrilateral formed by the
tangents at the end points of latusrectum to the
x2 y2
ellipse
+
= 1, is
9
5
(2003, 1M)
(a) 27/4 sq units
(c) 27/2 sq units
(b) 2
(c) 2 2
1 1 
(c)  ,

4
2
(d) (1, 2 )
Numerical Value
9. A farmer F1 has a land in the shape of a triangle with
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.)
Fill in the Blanks
10. The area of the triangle formed by the positive X-axis and
the normal and the tangent to the circle x2 + y2 = 4 at (1, 3 )
is … .
(1989, 2M)
11. The area enclosed within the curve|x| + | y| = 1 is ....... .
(1981, 2M)
Analytical & Descriptive Question
(b) 9 sq units
(d) 27 sq units
6. The area (in sq units) bounded by the curves
(2002, 2M)
y = | x | − 1 and y = − | x | + 1 is
(a) 1
(d) 1
8. Let P and Q be distinct points on the parabola y2 = 2x such
3. The common tangents to the circle x2 + y2 = 2 and
(a) 3
(c) – 3
(d) 4
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
12. Let O (0, 0), A (2, 0) and B (1,
312 Area
{ 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.
14. 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
(a) ∫
Passage Based Questions
(b) −
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
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.
4 2
(b) −
73 32
4 2
73 32
(c)
4 2
(d) −
73 3
(c) ∫
4 2
x
− 1]
b
dx − bf (b) + af (a )
x
∫a 3[{f (x)}2 − 1] dx − bf (b) + af (a)
1
∫− 1 g ′ (x) dx is equal to
(a) 2 g(− 1)
73 3
x
b
∫a 3[{f (x)}2 − 1] dx + bf (b) − af (a)
b
(d) −
dx + bf (b) − af (a )
− 1]
a 3[{f (x)}2
15.
13. If f (− 10 2 ) = 2 2 , then f′′ (− 10 2 ) is equal to
(a)
x
b
a 3[{f (x)}2
(b) 0
(c) − 2 g(1)
(d) 2 g(1)
Topic 2 Area Using Integration
Objective Questions I (Only one correct option)
8. The area (in sq units) of the region
1. If the area (in sq units) bounded by the parabola
1
y2 = 4λx and the line y = λx, λ > 0, is , then λ is equal to
9
A = {(x, y) ∈ R × R|0 ≤ x ≤ 3,
0 ≤ y ≤ 4, y ≤ x2 + 3x} is
(a)
(2019 Main, 12 April II)
(a) 2 6
(b) 48
(c) 24
(d) 4 3
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
8
(c)
3
(b) 6
2
(d) −
3
curves y = 2 and y = | x + 1|, in the first quadrant is
x
(2019 Main, 10 April II)
3
2
(b) log e 2 +
3
1
(c)
2
2
(d)
3
1
−
2 log e 2


y2
A = (x, y) :
≤ x ≤ y + 4 is
2


(b)
53
3
(2019 Main, 9 April II)
(c) 16
(b)
9
2
(c)
of
the
region
(2019 Main, 9 April I)
31
6
(d)
10
3
6. Let S (α ) = {(x, y) : y ≤ 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
1
2 3
(b) 4 
 5
1
4 3
(c) 4 
 25 
1
2 3
(d) 2 
 5
7. The tangent to the parabola y = 4x at the point where it
2
intersects the circle x2 + y2 = 5 in the first quadrant,
passes through the point
1
(a)  ,
4
3

4
3
(b)  ,
4
7

4
parabola, y = x2 + 2 and the lines, y = x + 1, x = 0 and
(2019 Main, 12 Jan I)
x = 3, is
(a)
15
2
(b)
17
4
(c)
21
2
(d)
15
4
the parabola, y = x2 + 1, the tangent to it at the point
(2, 5) and the coordinate axes is
(2019 Main, 11 Jan II)
(a)
14
3
(b)
187
24
(c)
8
3
(d)
37
24
11. The area (in sq units) of the region bounded by the curve
1
(c)  − ,
 3
4

3
(a)
7
8
(b)
9
8
(c)
5
4
(d)
3
4
− 1 ≤ x ≤ 1} in sq. units, is
(a) 2
4
(b)
3
(2019 Main, 9 Jan II)
1
(c)
3
(d)
2
3
13. The area (in sq units) bounded by the parabola
2
4 3
(a) 2 
 25 
26
3
12. The area of the region A = {(x, y); 0 ≤ y ≤ x| x| + 1 and
(d) 18
area
(in
sq
units)
A = {(x, y) : x2 ≤ y ≤ x + 2} is
13
6
(d)
(2019 Main, 11 Jan I)
5. The
(a)
59
6
x2 = 4 y and the straight line x = 4 y − 2 is
4. The area (in sq units) of the region
(a) 30
(2019 Main, 8 April I)
(c)
10. The area (in sq units) in the first quadrant bounded by
3. The area (in sq units) of the region bounded by the
(a)
(b) 8
9. The area (in sq units) of the region bounded by the
2. If
10
(a)
3
53
6
1
(d)  − ,
 4
1

2
y = x2 − 1, the tangent at the point (2, 3) to it and the
(2019 Main, 9 Jan I)
Y -axis is
(a)
8
3
(b)
56
3
(c)
32
3
(d)
14
3
14. 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
(a)
1
( 3 + 1)
2
1
(d) ( 2 − 1)
2
(b)
Area 313
25. The area enclosed between the curves y = ax2 and
15. The area (in sq units) of the region
{(x, y) : x ≥ 0, x + y ≤ 3, x2 ≤ 4 y and y ≤ 1+
x } is
(2017 Main)
(a)
59
12
(b)
3
2
(c)
7
3
(d)
5
2
(a)
of
1
6
the
(b)
4
3
(c)
3
2
(d)
5
3
17. The area (in sq units) of region described by (x, y) y2 ≤ 2x
and y ≥ 4x − 1 is
7
(a)
32
(2015 JEE Main)
5
(b)
64
15
(c)
64
(d)
9
32
(a)
π 4
+
2 3
(b)
π 4
−
2 3
(c)
π 2
−
2 3
(2014 Main)
(d)
π 2
+
2 3
19. The area enclosed by the curves y = sin x + cos x and
 π
is
y = | cos x − sin x|over the interval 0,
 2 
(a) 4( 2 − 1)
(c) 2 ( 2 + 1)
(2014 Adv.)
(b) 2 2 ( 2 − 1)
(d) 2 2 ( 2 + 1)
(c) 18
(d)
27
4
21. 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
(2011)
and the X-axis. Then,
(a) R1 = 2R2
(c) 2R1 = R2
(b) R1 = 3R2
(d) 3R1 = R2
22. 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
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
t
0
(c) ∫
(1 + t 2 ) 1 − t 2
2+1
4t
0
(1 + t 2 ) 1 − t 2
2 −1
4t
dt
(b) ∫
0
dt
(d) ∫
(1 + t 2 ) 1 − t 2
2+1
t
0
(1 + t 2 ) 1 − t 2
dt
dt
24. The area bounded by the curves y = (x − 1)2, y = (x + 1)2
and y =
(a)
1
is
4
1
2
sq unit (b) sq unit
3
3
(2005, 1M)
(c)
1
sq unit
4
(c) 1
1
3
(d)
26. 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
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)
(d)
divides the area of region
R = {(x, y) ∈ R : x ≤ y ≤ x, 0 ≤ x ≤ 1} into two equal
parts, then
(2017 Adv.)
3
(a) 2α 4 − 4α 2 + 1 = 0
1
(c) < α < 1
2
1
sq unit
5
(b) α 4 + 4α 2 − 1 = 0
1
(d) 0 < α ≤
2
29. If S be the area of the region enclosed by
2
y = e− x , y = 0, x = 0 and x = 1. Then,
1
e
1
1
(c) S ≤  1 +

e
4
(a) S ≥
(b) S ≥ 1 −
(d) S ≤
(2012)
1
e
1
1 
1 
+
1 −

e 
2
2
30. Area of the region bounded by the curve y = ex and lines
x = 0 and y = e is
(a) e − 1
1
(2009)
(b)
(c) e − ∫ e dx
x
0
(2011)
23. The
(a) ∫
1
2
2
y = x, 2 y − x + 3 = 0, X-axis and lying in the first
quadrant, is
(2013 Main, 03)
(b) 6
(b)
28. If the line x = α
20. The area (in sq units) bounded by the curves
(a) 9
1
3
27. The slope of tanget to a curve y = f (x) at [x, f (x)] is 2x + 1.
18. The area (in sq units) of the region described by
A = {(x, y) : x2 + y2 ≤ 1 and y2 ≤ 1 − x } is
(2004, 1M)
(a)
region {(x, y)} ∈ R2 : y ≥ |x + 3|,
(2016 Adv)
5 y ≤ (x + 9) ≤ 15} is equal to
16. Area
x = ay2 (a > 0) is 1 sq unit. Then, the value of a is
(d)
e
∫1
e
∫1 ln y dy
ln (e + 1 − y) dy
31. 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
Analytical & Descriptive Questions
4a 2 4a 1  f (−1)  2
3a + 3a 
32. 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)
314 Area
33. Find the area bounded by the curves x2 = y, x2 = − y
and y = 4x − 3.
2
and deduce
(2005, 4M)
34. A curve passes through (2,0) and the slope of tangent at
(x + 1)2 + y − 3
point P (x, y) equals
.
(x + 1)
Find the equation of the curve and area enclosed by the
curve and the X-axis in the fourth quadrant. (2004, 5M)
35. 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.
(2002, 5M)
36. Let b ≠ 0 and for j = 0,1,2..., n. If S j is the area of the
bounded by the Y-axis and the curve
jπ
( j + 1)π
ay
. Then, show that
xe = sin by,
≤ y≤
b
(b)
S 0 , S1 , S 2,... , S n are in geometric progression. Also,
(2001, 5M)
find their sum for a = − 1 and b = π .
region
1
1
.
< An <
2n + 2
2n − 2
(1996, 3M)
42. 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)
43. 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 ?
44. Sketch the region bounded by the curves y = x2 and
y = 2 / (1 + x2). Find its area.
(1992, 4M)
45. 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.
46. Compute the area of the region bounded by the curves
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
(1999, 5M)
left on the line 8x + 1 = 0.
log x
, where log e = 1.
(1990, 4M)
ex
47. 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 .
38. Let C1 and C 2 be the graphs of functions y = x2 and
48. Find the area of the region bounded by the curve
37. If f (x) is a continuous function given by
y = ex log x and y =
2x,
|x| ≤ 1
f (x) =  2
x + ax + b, |x| > 1
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
(1/2,1)
(0,1)
Q
49. Find the area bounded by the curves
x2 + y2 = 25, 4 y = |4 − x2|and x = 0 above the X-axis.
(1987, 6M)
50. Find the area bounded by the curves x2 + y2 = 4,
x2 = − 2 y and x = y.
(1,1)
C2
C: y = tan x, tangent drawn to C at x = π / 4 and the
X-axis.
(1988, 5M)
51. Sketch the region bounded by the curves y = 5 − x2 and
C1
y = |x − 1|and find its area.
P
( 1985, 5M)
52. Find the area of the region bounded by the X-axis
and the curves defined by y = tan x, −
(0,0) O C
3
(1,0)
X
y = cot x,
R
39. Letf (x) = max { x2, (1 − x)2, 2x (1 − x)}, where 0 ≤ x ≤ 1.
Determine the area of the region bounded by the curves
(1997, 5M)
y = f (x), X-axis, x = 0 and x = 1.
40. Find all the possible values of b > 0, so that the area of
the bounded region enclosed between the parabolas
x2
is maximum.
y = x − bx2 and y =
(1997C, 5M)
b
41. If An is the area bounded by the curve y = (tan x) and
n
the lines x = 0, y = 0 and x =
(1986, 5M)
π
.
4
Then, prove that for n > 2 , An + An + 2 =
1
n+1
π
π
≤x≤ .
6
3
π
π
and
≤x≤
3
3
(1984, 4M)
53. 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,
(1983, 3M)
then find a.
54. 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
55. For any real t, x =
,y=
2
2
hyperbola x2 – y2 = 1. Find the area bounded by this
hrperbola and the lines joining its centre to the points
corresponding to t1 and – t1.
(1982, 3M)
Area 315
Answers
2. (b)
3. (d)
4. (a)
 761
37. 
 sq units
 192
6. (b)
7. (c)
8. (a,d)
39.
Topic 1
1. (b)
5. (d)
9. (4)
10. 2 3 sq units
11. 2 sq units
12. (2 − 3 ) sq unit
14. (a)
15. (d)
13. (b)
Topic 2
1.
5.
9.
13.
15.
19.
23.
27.
(b)
3.
(c)
7.
(d)
11.
(a)
(c)
17.
(a)
21.
(a)
25.
(a, c)
29.
125
sq units
31. (b, d)
32.
3
4
2
34. y = x – 2 x, sq units
3
 20 − 12 2 
36.
35. 
 sq units
3


(c)
(b)
(a)
(a)
(d)
(b)
(b)
(c)
2.
6.
10.
14.
16.
20.
24.
28.
(d)
(b)
(b)
4. (d)
8. (c)
12. (a)
(d)
(c)
(a)
(b, d)
18.
22.
26.
30.
(a)
(b)
(c)
(b, c, d)
1
33. sq unit
3
17
sq unit
27
43. 121 : 4
38. f ( x ) = x 3 − x 2, 0 ≤ x ≤ 1
40. b = 1
42.
2

44.  π −  sq units

3
1

(16 2 − 20 ) sq units
3

 e 2 − 5
4 − 2 5
3
45. 
− log 2 +  sq units 46. 
 sq units
2
2
 log 2
 4e 
4
10


47. y max = , y min = 0, sq units


27
3

 4 
49. 4 + 25 sin −1    sq units
 5 

 5π 1
51. 
−  sq units
 4 2
9
53. 2 2
54. sq units
8
1


48.  log 2 −  sq units


4



1
50.  − π  sq units

3
1

52.  loge 3 sq units
2

55.
 π (1 + e ) (en + 1 − 1 ) 
⋅

2
e − 1 
 (1 + π )
e 2t1 − e −2t1 1 2t1
− (e − e −2t1 − 4t1 )
4
4
Hints & Solutions
Topic 1 Area Based on Geometrical Figures
Without Using Integration
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
Now, the required area is the area of shaded region, i.e.
1. 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
1
16  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
2. 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)
Y
A (2, 2)
x 2 + y 2 = 4x
X′
(0, 0)
B (2,0)
X
y 2 = 2x
Y′
2
Area of a circle
Required area =
− ∫ 2x dx
0
4
2
 x3/ 2 
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
3.
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
316 Area
Let equation of tangent to parabola be y = mx +
2
m
By symmetry, the quadrilateral is a rhombus.
Y
It also touches the circle x2 + y2 = 2.
A


2

= 2
2
m 1 + m 
⇒ m4 + m2 = 2 ⇒ m4 + m2 − 2 = 0
⇒
(m2 − 1) (m2 + 2) = 0
⇒
[rejected m2 = − 2]
m = ± 1,m2 = − 2
L(ae, √ b2(1 – e2))
∴
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
= 15 sq units
∴ Area of quadrilateral =
2
X′
O
D
X
B
L′
C
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 
6. The region is clearly square with vertices at the points
(1, 0), (0, 1), (– 1, 0) and (0, – 1).
Y
(0,1)
y = –IxI + 1
y = IxI –1
4. Since, tangents drawn from external points to the circle
subtends equal angle at the centre.
X′
A
X
(1,0)
(–1,0)
(0,–1)
Y′
O3
∴ Area of square = 2 × 2 = 2 sq units
O1
B
7. Let y = f (x) = x2 + bx − b
O2
1cm 1cm
30°
Y
30°
3 cm D
2 cm E
∠ O1BD = 30°
OD
In ∆O1BD, tan 30° = 1
BD
C
B
3 cm
(1,1)
∴
P
⇒ BD = 3 cm
Also,
DE = O1O2 = 2 cm and EC = 3 cm
Now,
BC = BD + DE + EC = 2 + 2 3
⇒
3
3
⋅ 4 (1 + 3 )2
(BC )2 =
4
4
= (6 + 4 3 ) sq cm
Area of ∆ABC =
x2 y2
+
=1
9
5
To find tangents at the end points of latusrectum, we
find ae.
5. Given,
i.e.
and
ae = a 2 − b2 = 4 = 2
4 5

b2(1 − e2) = 5 1 −  =

9 3
O
A
X
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
=− ×
=2
2 (2 + b)
⇒
⇒
(1 + b)2 + 4(2 + b) = 0 ⇒ b2 + 6b + 9 = 0
(b + 3)2 = 0 ⇒ b = − 3
[given]
Area 317
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
8. Since, ∠ POQ = 90°
Y
t 12
P 2 , t1
11. The area formed by| x| + | y| = 1 is square shown as below
X′
X
O
(0, 0)
Y
t 22
Q 2 , t2
Y′
⇒
:
t1 − 0 t2 − 0
⋅ 2
= − 1 ⇒ t1t2 = − 4
t12
t
−0 2 −0
2
2
−x + y = 1
X'
…(i)
ar (∆OPQ ) = 3 2
0
0 1
1 2
1  t12t2 t1t22
∴
t1 / 2 t1 1 = ± 3 2 ⇒
−

 =±3 2
2 2
2 2
2 
t2 / 2 t2 1
4
1
= 3 2 [Q t1 > 0 for P]
⇒ ( −4t1 + 4t2) = ± 3 2 ⇒ t1 +
t1
4
−1
x+y=1
O
1
x+y=1
X
x−y=1
Y'
Q
∴ Area of square = ( 2 )2 = 2 sq units
12. Let the coordinates of P be (x, 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 )
9. We have, y = xn , n > 1
Q P ( 0, 0 ) Q (1, 1) and R( 2, 0 ) are vertices of ∆ PQR.
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
× × 2×1
⇒ ∫ ( x − xn ) dx =
0
100 2
1
 x2 xn + 1 
3
⇒ −
⇒
 =
n + 1 0 10
2
⇒
1
1 
3
 =
 −
 2 n + 1 10
1
1 3
2 1
= −
=
= ⇒ n + 1= 5 ⇒ n = 4
n + 1 2 10 10 5
10. Equation of tangent at the point (1, 3 ) to the curve
x2 + y2 = 4 is x +
whose X-axis intercept (4, 0).
3y = 4
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 ≤
y≤
⇒
2
2
| 3 y − x|
Case I When y ≤
[since, 3 y − x < 0]
2
x − 3y
y≤
⇒ (2 + 3 ) y ≤ x ⇒ y ≤ x tan 15°
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
B (1, 1/ 3)
P (1,√3)
X′
(0,0) O
A (4,0)
P
X
X′
Y′
O (0, 0)
Y′
C
(1, 0)
A
(2, 0)
X
318 Area
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
⇒
2 d
⇒
A
y2=4λx
X
y=λx
the intersection point A we will get on the solving Eqs. (i)
and (ii), we get
λ2x2 = 4λx
4
x = , so y = 4.
⇒
λ
4
So,
A  , 4
λ 
21 ⋅
Now, required area is
2
=
=
d 2y − 12 2 − 4 2
= 3 2
=
7 ⋅3
(21)3
dx2


 x3 / 2 
=2 λ 
3 


 2 0
b
b
a
a
[ f (x) ⋅ x]ba
− ∫ f ′ (x)x dx
a
b
a
xdx
b
∫a 3[{ f (x)}2 − 1]


dy
−1
−1
=
Q f ′ (x) = dx =

2
2
−
−
y
f
x
3
(
1
)
3
[{
(
)}
1
]


1
4/ λ
y3 − 3 y + x = 0
y = g (x)
{ g (x)}3 − 3 g (x) + x = 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
B(0,1)
…(i)
P
y2=4x
[from Eq. (i)]
{ g (1)}3 − 3 g (1) + 1 = 0
4/ λ
 x2 
−λ 
 2 0
2. Given region is {(x, y) : y2 ≤ 4x, x + y ≤ 1, x ≥ 0, y ≥ 0}
g′ (x) dx = [ g (x)]1− 1 = g (1) − g (− 1)
x+y=1
…(ii)
At x = − 1, { g (− 1)}3 − 3 g (− 1) − 1 = 0
On adding Eqs. (i) and (ii), we get
…(iii)
{ g (1)}3 + { g (− 1)}3 − 3 { g (1) + g (− 1)} = 0
⇒ [ g (1) + g (− 1)][{ g (1)} + { g (− 1)} − g (1) g (− 1) − 3] = 0
2
λx − λx) dx
=
b
= bf (b) − af (a ) +
∫ (2
0
= bf (b) − af (a ) − ∫ f ′ (x)x dx
−1
4/ λ
d y
12 2
=−
dx2
(21)2
14. Required area = ∫ y dx = ∫ f (x) dx
∴
Area bounded by above two curve is, as per figure
…(ii)
d 2y
d 2y
 1
+ 6 (2 2 ) ⋅  −  − 3 ⋅ 2 = 0
2
 21
dx
dx
⇒
⇒
…(i)
…(ii)
O
2
3(2 2 )2
⇒
and
2
At x = − 10 2 , y = 2 2
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
Since,
and
∴
At x = 1,
y2 = 4λx
y = λx
λ> 0
…(i)
2
d 2y
y
 dy
=0
⇒ 3 y  2 + 6 y   − 3
 dx
dx2
 dx 
15. Let I = ∫
1. Given, equation of curves are
Y
y3 − 3 y + x = 0
dy
dy
3 y2
−3
+ 1 =0
dx
dx
13. Given,
Topic 2 Area Using Integration
2
g (1) + g (− 1) = 0
g (1) = − g ( − 1)
I = g (1) − g (− 1)
= g (1) − { − g (1)} = 2 g (1)
O
A(1,0)
X
Now, for point P, put value of y = 1 − x to y2 = 4x, we get
(1 − x)2 = 4x ⇒ x2 + 1 − 2x = 4x
⇒
⇒
x2 − 6 x + 1 = 0
x=
6 ± 36 − 4
2
= 3 ± 2 2.
Area 319
Since, x-coordinate of P less than x-coordinate of point
A(1, 0).
⇒
y2 = 2x
and x = y + 4 ⇒ y = x − 4
∴ x=3 −2 2
Now, required area
Graphical representation of A is
=∫
3 −2 2
0
x3/ 2
=2
3 /2
∫3 − 2 2 (1 − x) dx
Q
1

x2 
+ x − 
2 3 − 2

X'
X
2
P
y+
4
(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
3
= ( 2 − 1) − + 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)
8
10
So, on comparing a = and b = −
3
3
8 10
a−b= +
=6
∴
3
3
4

 y2
y2 
y3 
(
y
4
)
dy
4
+
−
=
+
−
y




∫
2
6  −2
2
−2 
8
64   4
16
= 
+ 16 −
 −  − 8+ 
 2
6
6  2
4
32
= 8 + 16 −
− 2+ 8−
3
3
= 30 − 12 = 18 sq unit.
 x + 1 ,x ≥ − 1
y = 2x and y = | x + 1| = 
− x − 1 , x < − 1
Q The figure of above given curves is
Y
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).
Now, the area enclosed by the region A
=
3. Given, equations of curves
4
5. Given region is A = {(x, y) : x2 ≤ y ≤ x + 2}
y=x+1
Now, the region is shown in the following graph
Y
(1,2)
y=–x–1
y=x+2
x2=y
y=2x
B(2,4)
(0,1)
X′
X
O
(–1,0)
A
In first quadrant, the above given curves intersect each
other at (1, 2).
X'
(–2,0)
–1
0
1


ax
x
Q
a
dx
=
+ C
 ∫
log e a


1
2
1 
= +1−
+

2
log
2
log

e
e 2
3
1
= −
2 log e 2

4. Given region A = (x, y) :

∴
y2
=x
2

y2
≤ x ≤ y + 4
2

(0,2)
O
X'
2
Y'
1
So, the required area = ∫ ((x + 1) − 2x ) dx
 x2
2x 
= + x−
log e 2  0
2
y2
=x
2
O
x=
=
0
Y
1
2 x dx +
3 −2 2
…(i)
…(ii)
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
∫ [(x + 2) − x
2
] dx
−1
2
 x2
1
8
1
4
x3 
=  + 2x −  =  + 4 −  −  − 2 + 



3
3
2
2
2
3
 −1

=8−
1 9
1
1 9
− = 8 − − 3 = 5 − = sq units
2 3
2
2 2
320 Area
and points of intersection of curves y = x2 + 3x and y = 4
are (1, 4) and (−4, 4)
Now required area
6. Given, S (α ) = {(x, y) : y2 ≤ x, 0 ≤ x ≤ α } and
A(α ) is area of the region S(α )
Y
= ∫ (x2 + 3x)dx +
X
O
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
y 2 =x
3
9. Given equation of parabola is y = x2 + 2, and the line is
A(λ)
y = x+1
y
y=x2 +2
x=λ
λ
x
4 3/ 2
x dx = 2 
 =3λ
3
2
/
0

Clearly, A (λ ) = 2∫
0
Since,
⇒
⇒
y=x+1
(0,2)
3/ 2 λ
1
A (λ ) 2
= , (0 < λ < 4)
A (4) 5
1
3
λ3/ 2 2
=
43/ 2 5
 λ
 2
⇒   = 
 4
 5
λ 4
= 
4  25
1/3
=∫
1/3
x
3
0
((x2 + 2) − (x + 1)) dx = ∫
3
0
(x2 − x + 1) dx
3
7. Given equations of the parabola y2 = 4x
…(i)
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
⇒
x2 + 5 x − 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
2
(3,0)
The required area = area of shaded region
2
4
λ =4  
 25
⇒
O
2

 x3 x2
 27 9

=  − + x = 
− + 3 − 0

0  3 2
3 2
9
9 15
sq units
= 9 − + 3 = 12 − =
2
2 2
10. Given, equation of parabola is y = x2 + 1, which can be
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
[Q Equation 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
8. Given, y ≤ x2 + 3x
2
2
⇒
3
9
3
9



⇒ x +  ≥  y + 
y ≤ x +  −



2
4
2
4
(0, 1)
O
Since,
0 ≤ y ≤ 4 and 0 ≤ x ≤ 3
∴The diagram for the given inequalities is
R
Q (2, 0)
3,
0
4
X
Y
y=x2+3x
Required area = Area of shaded region
2
= ∫ y(parabola) dx − (Area of ∆PQR)
y=4
–3/2
(–3, 0)
O
9
–
4
1
0
2
X
3
x=3
= ∫ (x2 + 1) dx − (Area of ∆PQR)
0
2

 x3
1
3
=
+ x − 2 −  ⋅ 5

3
2
4
0

Area 321
[Q Area of a triangle =
1
× base × height]
2
1  5

8
=  + 2 − 0 −   5

3
2  4
We need to calculate the shaded area, which is equal to
0
∫−1 (− x
2
0
parabola with vertex (0, 0) and it open upward.
X′
1
 4
= −  − 1 +
3
 3
2 4
= + =2
3 3
x2
4
x+2
y=
4
y=
B
13. Given, equation of parabola is y = x2 − 1, which can be
X
–1 O
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
Now, required area = area of shaded region
=
y+3
= 2x − 1
2
y = 4x − 5
⇒
⇒
y=x2–1
(2, 3)
2
2
(0, –1)
2
∫−1 
2
3
1
1
1 9
 1
8 − −3 =
5−
=
= sq units.
 4 
4 
2
2  8
12. 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
1
y=4x–5
2
x + 2 x
x 
1 x
−  dx =  + 2x −

42
3  −1
4
4
1 
8  1
1 
=
2 + 4 −  −  − 2 + 
4  
3  2
3 
=
rewritten as x2 = y + 1 or x2 = ( y − (−1)).
⇒ Vertex of parabola is (0, − 1) and it is open upward.
Equation of tangent at (2, 3) is given by T = 0
y + y1
= x x1 − 1, where, x1 = 2
⇒
2
and
y1 = 3.
∫−1 {y (line) − y (parabola)} dx
2
1

 (− 1)3
  1

=  0 − −
+ (− 1)  +   + 1 − 0





3


   3
11. Given equation of curve is x2 = 4 y, which represent a
A
0

 x3

 x3
+ x
= −
+ x + 
0
− 1  3
 3
14 25 112 − 75 37
=
−
=
=
3
8
24
24
Y
1
+ 1)dx + ∫ (x2 + 1) dx
x
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]
Now, required area = area of shaded region
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
=
0
14. We have,
⇒
⇒
⇒
Now, α < β
(2 − 2)3 (0 − 2)3 8
= sq units.
−
3
3
3
18x2 − 9πx + π 2 = 0
18x − 6πx − 3πx + π 2 = 0
(6x − π )(3x − π ) = 0
π π
x= ,
6 3
π
α= ,
6
π
β=
3
2
322 Area
Given, g(x) = cos x 2 and f(x) = x
Shown as
Y
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
A=
∫
(0, 9/5)
X
0
(–9, 0)
cos x dx
x=6
π /6
A = [sin x]ππ //36
π
π
3 1
A = sin − sin =
−
3
6
2
2
 3 − 1
A=

 2 
∴ {(x, y) ∈ R2 : y ≥ |x + 3|, 5 y ≤ (x + 9) ≤ 15}
Y
4,1
B (–
15. Required area
(1 +
0
x )dx +
2
∫1
(3 − x)dx − ∫
2
0
X′
x2
dx
4
Y
y=1+√x
(0, 3) (1, 2)
4y=x 2
(0, 1)
∴ 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
(2, 1)
X
(0, 0) (1, 0)(2, 0) (3, 0)
−3
2
2
 x3 


x3/ 2 
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
17. Given region is {(x, y) : y2 ≤ 2x and y ≥ 4x − 1}
y2 ≤ 2x repressents a region inside the parabola
…(i)
y2 = 2 x
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 ⇒ 16x2 + 1 − 8x = 2x
1 1
⇒
16x2 − 10x + 11 = 0 ⇒ x = ,
2 8
1 
So, the points where these curves intersect are  , 1
2 
 1 1
and  ,  .
 8 2
16. 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
1




 (x + 3)3/ 2 
15  (− 3 − x)3/ 2 
=
−
 −

3
3
2 
−





− 4 
 –3
2
2
15 2
15 2 16 15 18 3
2
=
+ [0 − 1] − [8 − 0] =
− −
=
−
=
2
3
2 3 3
2
3 2
3
Y′
1
X
6
Y′
x+y=3
X′
(1,0)
E (–3,0) 0 D
(–4,0) A
–9
y2=x+3
Y
4x
−1
=∫
1
(1,2)
C
)
X′
0
–3
Y′
Also,
⇒
5 y ≤ (x + 9) ≤ 15
(x + 9) ≥ 5 yand x ≤ 6
y=
1,1
2
y 2 = 2x
1
X
X′
−1
−1
2
1
2
1
2
O
1 −1
,
8 2
−1
2
−1
Y′
1
X
Area 323
∴ Required area = ∫
 y + 1 y2 
−  dy

−1/ 2  4
2
∴ Area bounded = ∫
1
+
−1
=

1  y2
1
− ( y3 )1−1/ 2
+ y

42
 −1/ 2 6
1  1
1  1
 1
 + 1 −  −   −
4  2
8
2  6
1 3 3  1 9 
=  + −  
4 2 8  6 8 
1 15 3
9
sq units
= ×
−
=
4 8 16 32
=
π /4
∫π / 4 {(sin x + cos x) − (sin x − cos x)} dx
=∫
1

1 + 
8

{(sin x + cos x) − (cos x − sin x)} dx
0
π /2
π /4
0
π /2
∫π / 4 2 cos x dx
2 sin x dx +
= − 2 [cos x]0π / 4 + 2 [sin x ⋅ n ]ππ // 24
= 4 − 2 2 = 2 2( 2 − 1) sq units
20. Given curves are y = x
…(i)
2y − x + 3 = 0
and
…(ii)
Y
18. Given, A = {(x, y) : x2 + y2 ≤ 1 and y2 ≤ 1 − x}
x
y=
x
2y –
Y
X'
+3
=0
X
3
X′
(0,1)
(–1,0)
–3
2
X
Y'
On solving Eqs. (i) and (ii), we get
2 x − ( x )2 + 3 = 0
Y′
Required area =
=
⇒
1
1 2
πr + 2∫ (1 − y2)dy
0
2
⇒
( x − 3) ( x + 1) = 0 ⇒
x =3
[since, x = − 1 is not possible]
∴
y=3
Hence, required area
3 1

y
1
π (1)2 + 2  y − 
2
3 0

 π 4
=  +  sq units
 2 3
19.
( x )2 − 2 x − 3 = 0
3
2
21. R1 = ∫ x f (x) dx
f(x)
a
O
∴ Area bounded =
=
X
b
c
∫a
b
| f( x ) − g ( x )|dx
f (x) = y = sin x + cos x, when 0 ≤ x ≤
Here,
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
and
Y
f(x)
√2
1
= √2 sin x + π
4
π/4
π/2
X
2
π
2
(1 − x) f (1 − x) dx
−1
2
R1 = ∫ (1 − x) f (x) dx
∴
−1
…(ii)
[f (x) = f (1 − x), given]
Given, R2 is area bounded by f (x), x = − 1 and x = 2.
2
R2 = ∫
∴
−1
…(iii)
f (x) dx
On adding Eqs. (i) and (ii), we get
2R1 = ∫
2
−1
…(iv)
f (x) dx
From Eqs. (iii) and (iv), we get
2R1 = R2
22. Here, area between 0 to b is R1 and b to 1 is R2.
∴
b
∫0
(1 − x)2 dx −
b
⇒
g(x)
g(x)
O
y = sin x + cos x
b
R1 = ∫
∫a[g (x ) − f(x )]dx + ∫c [f(x ) − g (x )]dx
b
b
∫ a f (x) dx = ∫ a f (a + b − x) dx
Using
g(x)
c
…(i)
−1
g(x)
f(x)
3
0

y3 
= 9 + 9 − 9 = 9 sq units
=  y2 + 3 y −
3  0

PLAN To find the bounded area between y = f( x ) and y = g ( x )
between x = a to x = b.
Y
3
0
= ∫ (x2 − x1 ) dy = ∫ {(2 y + 3) − y2} dy
1
∫b
(1 − x)2 dx =
1
1
4
 (1 − x)3 
 (1 − x)3 
1
 −3  −  −3  = 4

0 
b
324 Area
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
23. 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
2

1 + tan

2
 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
=∫
−
0
x
 1 − tan x
1 + tan 


2
2
x
x
x
1 + tan − 1 + tan
2 tan
π /4
π /4
2
2
2
=∫
dx = ∫
dx
0
0
x
x
1 − tan 2
1 − tan 2
2
2
tan
x
1
x
= t ⇒ sec2 dx = dt = ∫
0
2
2
2
∫0
As
2 −1
4t dt
(1 + t 2) 1 − t 2
π
8
–1 –1/2
2
Q
O 1/2
1
y = 1/4
X
where, points of intersection are
1
1
1
1
⇒ x = and (x + 1)2 = ⇒ x = −
(x − 1)2 =
4
2
4
2
 1 1
 1 1
i.e.
Q  ,  and R  − , 
 2 4
 2 4
1/ 2 
1
(x − 1)2 − dx
∴ Required area = 2 ∫

0
4 
(1/a,1/a)
X
D
O
Y'
⇒
x = a (ax2)2
1
x = 0,
⇒
a
⇒
y = 0,
1
a
 1 1
So, the points of intersection are (0, 0) and  ,  ⋅
 a a
∴ Required area OABCO = Area of curve OCBDO
– Area of curve OABDO
1/a  x
2
[given]
⇒
∫ 0  a − ax  dx = 1
1/ a
⇒
⇒
y = (x – 1)
x
y 2= a
A
X'
π
= 2 − 1]
8
P
R
C
⇒
shown as
1/4
B
(1 + t 2) 1 − t 2
[Q tan
y = ax2
Y
4t dt
24. The curves y = (x − 1)2, y = (x + 1)2 and y = 1 /4 are
Y y = (x + 1)2
OABCO.
Thus, the point of intersection of
y = ax2 and x = ay2
>
π /4
Put tan
25. As from the figure, area enclosed between the curves is
 1 x3/ 2
ax3 
=1
⋅
−

3  0
 a 3 /2
2
1
−
=1
2
3a
3a 2
1
1
a2 =
⇒ a=
3
3
b
26. Since, ∫ f (x) dx = (b − 1) sin (3b + 4)
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
27. Given,
= 2x + 1
dx
On integrating both sides
∫ dy = ∫ (2x + 1) dx
⇒
∴
⇒
∴
y = x2 + x + C which passes through (1, 2)
2 =1+1+C
C =0
y = x2 + x
Y
y = x (x + 1)
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
⋅

[Q a > 0]
X'
−1
O
Y'
x=1
X
Area 325
Thus, the required area bounded by X-axis, the curve
and x = 1
2 1
1
 x3 x
= ∫ (x2 + x)dx =  + 
0
2 0
3
1 1 5
= + = sq unit
3 2 6
28.
∫0
1
Also,
0
e
∫1 ln (e + 1 − y) dy
=∫
(x − x3 )dx = 2∫ (x − x3 )dx
⇒
[from Eqs. (i) and (ii)]
0
α
1
1
1 
1
1
+
1 −
 ≥S ≥1 −


e
2
e
2
30. Shaded area = e −  ∫ ex dx = 1
1
e
[put e + 1 − y = t ⇒ − dy = dt]
e
e
1
1
ln t (− dt ) = ∫ ln t dt = ∫ ln y dy = 1
31. Case I When m = 0
α2 α4
1
= 2
−

4
4
 2
29.
∴
2α 4 − 4α 2 + 1 = 0
4 − 16 − 8
α2 =
4
1
2
α =1 −
2
(Q α ∈ (0, 1))
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
Y
PLAN (i) Area of region f( x ) bounded between x = a to x = b is
A
O
y = f (x)
a a 1 a 2 a3
1
1
 x2 x3 
1 1 1 9
∴ A = ∫ (x − x2) dx =  −  = − = ≠
0
2
3
0 2 3 6 2

D
b
∫a f(x )dx = Sum of areas of rectangle shown in shaded part.
(ii) If f( x )≥ g ( x ) when defined in [a, b ], then
b
b
∫a f(x )dx ≥ ∫a g (x )dx
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
X
B
Hence, no solution exists.
Case II When m < 0
In this case, area between y = x − x2 and y = mx is
OABCO and points of intersection are (0,0) and
{1 − m, m(1 − m)}.
∴ Area of curve OABCO = ∫
2
1 −m
0
[x − x2 − mx] dx
Y
Y
1
A
1
—
√e
B
O
1
—
√2
1
(0, 0) O
X
C
Since, x2 ≤ x when x ∈ [0, 1]
2
− x2 ≥ − x or e− x ≥ e− x
⇒
∴
1 − x2
∫0 e
dx ≥ ∫ e
0
1 − x2
∫0 e
1
e
…(i)
dx ≤ Area of two rectangles
1 
1 1

≤ 1 ×
 + 1 −
×

2 
e
2
1
1 
1
≤
+
1 −

2
e
2
y=
x – x2
1 −m
dx
S ≥ − (e− x )10 = 1 −
⇒
Also,
1 −x
X
{1 – m, m (1 – m)}
∴
…(ii)
⇒
⇒
⇒
y = mx

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 − m)3 = 27
1 −m =3
m = −2
326 Area
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
⇒
∴
t =4
B (8, − 15)
− (3x + 6)
.
2
 4 − x2 3 x + 6 
+

 dx
2 
 4
So, equation of chord AB is y =
Y
∴ Required area =
y = mx
8
∫ −2
8
1– m
X
O (0,0)
A


x3 3x2
= x −
+
+ 3x
12
4

 −2
B
2
 128


= 8 −
+ 48 + 24 −  −2 + + 3 − 6 


3
3

y = x − x2
=
∴ Area of shaded region = ∫
0
1 −m
(x − x2 − mx) dx
33. The region bounded by the curves y = x2, y = − x2 and
3 0

x2 x
= (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.
4a 2 4a 1
2
 f (−1) 3a + 3a 
32. Given, 4b2 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 ,
4b2 f (−1) + 4b f (1) + f (2) = 3b2 + 3b
and
4c2 f (−1) + 4cf (1) + f (2) = 3c2 + 3c
or {4 f (−1) − 3} x + {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
shown as,
Thus,
f (x) =
4
Let A (−2, 0), B = (2t , − t 2 + 1)
Since, AB subtends right angle at vertex V (0, 1).
1 −t 2
⇒
⋅
= −1
2 2t
1
1
( 4x − 3 ) dx
= 2 ∫ x2 dx − ∫
 0

3/ 4
y = x2
Y
(1,1) A
B (3/4, 0)
X'
O
C
…(i)
…(iii)
y 2 = 4x – 3
Y′
34. Here, slope of tangent,
dy (x + 1)2 + y − 3
=
dx
(x + 1)
dy
( y − 3)
= (x + 1) +
,
dx
(x + 1)
…(iv)
⇒
…(v)
X
 x3  1  (4x − 3)3/ 2 1 
= 2   − 
 
 3  0  3 ⋅ 4 / 2  3/ 4 
 1 1
=2  − 
 3 6
1 1
= 1 ⋅ = sq unit
6 3
…(ii)
4x2 f (−1) + 4x f (1) + f (2) = 3x2 + 3x
2
y2 = 4 x − 3 is symmetrical about X-axis, where y = 4x − 3
meets at (1, 1).
∴ Area of curve (OABCO )
y = –x 2
where, f (x) is quadratic expression given by,
f (x) = ax2 + bx + c and Eqs. (i), (ii) and (iii).
⇒
125
sq units
3
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
−
1
dX
X
= e− log X =
1
X
Area 327
∴ Solution is, Y ⋅
⇒
36. Given, x = (sin by) e−ay
1
1
= X ⋅ dX + c
X ∫
X
Y
=X +c
X
−1 ≤ sin by ≤ 1
− e− ay ≤ e− ay sin by ≤ e− ay
− e− ay ≤ x ≤ e− ay
Now,
⇒
⇒
Y
Y
x = e – ay
y = x 2 _ 2x
x = e – ay
S3
S2
S1
X′
O
S0
X′
X
2
Y′
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 − 2 x
2
  x3
2
 
− x2 
∴ Required area =  ∫ (x2 − 2x)dx=  
 0
 3
 0

8
4
= − 4 = sq units
3
3
Y′
In this case, if we take a and b positive, the values − e− ay
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
sin by ⋅ e− ay dy

since, I = sin by ⋅ e−ay dy
∫


− ay
−
e

I =
a
by
b
by
(
sin
+
cos
)


a 2 + b2
∴
35. The points in the graph are
 − 1  − a ( j + 1)π
S j = 2
e b
2
+
a
b



{a sin ( j + 1)π + b cos ( j + 1) π}
A (1, 1), B ( 2, 0), C (2, 2), D ( 2, 2)
−e
Y
Sj = −
X
=
∴ Required area
=∫
2
1
2
1
{ x2 − (2 − x2)} dx +
(2x2 − 2) dx +
2x3

=
− 2x
3

1
2
∫
2
2
∫
2
2
− e−ajπ / b{0 + b(− 1) j }]


 −aπ
 e b + 1 





{2 − (x2 − 2)} dx
a
− jπ
be b
 − aπ

 e b + 1

a + b 

2
a
− jπ
b
∴
3 2
2
4 2
  8
2
2 2
=
− 2 2 − + 2 + 8 − − 4 2 +
3
3 
 3
  3
 20 − 12 2 
=
 sq units
3


2
be
(4 − x2) dx

x
+ 4x − 
3

jπ

(a sin jπ + b cos jπ ) 

{0 + b(−1) j + 1 }
[Q (−1) j + 2 = (−1)2 (−1) j = (−1) j ]
Y′
=∫
− ( j + 1 )π
1
[e b
a + b2
2
a

−
j
b
(
1
)
b
e
−
=
 a 2 + b2

A
x = 1 B (√2,0)
− ajπ
b
a
y = |2 – x 2 |
y = x2
D (√2,2)
C (2,2)
y=2
X′
X
O
Sj
Sj −1
 − aπ

 e b + 1




a
− jπ
e b
a 2 + b2
=
= a
a
a
−
π

− ( j − 1 )π
− ( j − 1 )π 
 e b + 1
e b
be b




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
328 Area
1
. πj
π ⋅ eπ
Sj =
(1 + π 2)
n
∑ 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| ≤ 1
 x + ax + b,|x| > 1
x=
1
8
y = f (x)
–2
X′
–1
X
O
Y′
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.
lim f (x) = lim f (x) = f (−1)
x → −1 −
x → −1 +
lim f (x) = lim f (x) = f (1)
and
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 + 2x − 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
2
dx − ∫
(x + 2x − 1)dx − ∫
2x dx
=∫
−2
−2
−1
2
−1
 x3

=−
+ x2 − x  − [x2] −−11/ 8
[(− x)3/ 2]−−12/ 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
∫0
t 2 dt − ∫
x
0
f (t ) dt
x
Y
x=–
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 ) =
37. Given, f (x) =  2x2 ,
– 2y2
38. Refer to the figure given in the question. Let the
x
 t3 
=   − ∫ f (t ) dt
0
3
 0
x
x3
=
−
f (t ) dt
3 ∫0
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
= [t3/ 2]0x − [t 2]0x
3
4
2 3 x4
= x −
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
39. We can draw the graph of y = x2, y = (1 − x2) and
y = 2x(1 − x) in following figure
Y
(1, 1)
(0, 1)
y = (1 – x)
Q
2
y=x
(1/2, 1/2)
A
2
B
y = 2x (1 – x)
X′
O
1/3
2/3
1
X
Y′
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
Area 329
∴ We consider only b = 1.
dA
Sign scheme for
around b = 1 is as shown below :
db
From the figure, it is clear that,
 (1 − x)2,
if 0 ≤ x ≤ 1 / 3

f (x) = 2x (1 − x),
if 1 / 3 ≤ x ≤ 2 / 3
if 2 / 3 ≤ x ≤ 1

x2 ,
–
∴ The required area
1
A=∫
0
=∫
0
(1 − x)2 dx +
2 /3
∫ 1/ 3
2x (1 − x) dx +
1
∫ 2/ 3
x
2
41. We have, An = ∫
dx
2 /3
 1

= − (1 − x)3
 3

0
1

2x3 
1 3 
x
+ x2 −
+

3 1/ 3 3  2 / 3

 1  2 3 1 
= −   +  +
 
3
 3 3
  2 2 2
  −
3
 3
1/ 3
2
3
2
 1
 2
  −  +
 3
 3
3
π /4
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
3
 1 
  
 3

1
1
+  (1) −
3
3

3
 2
  
 3

⇒
π /4
∫0
(tan x)n+ 1 dx < ∫
π /4
(tan x)n dx
0
⇒
An+ 1 < An
Now, for n > 2
19 13 19 17
+
+
=
sq unit
81 81 81 27
x2
40. Eliminating y from y = and y = x − bx2, we get
b
x2 = bx − b2x2
b
⇒
x = 0,
1 + b2
π /4
An + An + 2 = ∫
0
=∫
0
=
Y
–
1
From sign scheme, it is clear that A is maximum at
b = 1.
f (x) dx
1/3
+
0
π /4
[( tan x)n + (tan x)n + 2] dx
(tan x)n (1 + tan 2 x) dx
Y
y = (tan x)
n
(π/4, 1)
x2
y=
b
B
y = x – bx
X′
O
–1
X
b
2
1+ b
=∫
=
b2
1
⋅
6 (1 + b2)2
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
=
(tan x)n sec2 x dx
π /4
Thus, the area enclosed between the parabolas
b/(1 + b )2 
x2 
A=∫
 x − bx2 −  dx
0
b

b/(1 + b )2
π /4
0
 1

(tan x)n + 1 
=
n
(
)
+
1

0
1
1
=
(1 − 0) =
(n + 1)
n+1
Y′
 x2 x3 1 + b2 
= −
⋅
3
b  0
2
X
π/4
O
2
Since,
then
⇒
⇒
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 =
…(i)
1
n −1
1
n −1
1
An <
⇒
2n − 2
1
1
From Eqs. (i) and (ii),
< An <
2n + 2
2n − 2
⇒
2 An <
…(ii)
330 Area
42. The equations of the sides of the square are as follow :
AB : y = 1, BC : x = − 1, CD : y = − 1, DA : x = 1
Y
B(–1,1)
A(1,1)
(0,1/2)
X′
(–1/2,0)
X
(1/2,0)
O
(0,–1/2)
D(1,–1)
C(–1,–1)
Y′
Let the region be S and (x, y) is any point inside it.
Then, according to given conditions,
x2 + y2 < |1 − x|,|1 + x|,|1 − y|,|1 + y|
⇒
⇒
x2 + y2 < (1 − x)2, (1 + x)2, (1 − y)2, (1 + y)2
x2 + y2 < x2 − 2x + 1, x2 + 2x + 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
Y
43. 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
Y
y = x2 − x
(2,4)
A(1,1)
y = 4x − x2
2
y
(0,1)
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
=
=
2x
O
X'
–1
G
I
E x2
H F
=
2y
O
–1
(1,0)
1
(1/2,–1/4)
5/2
X
Y'
X
2
where, S is the shaded region.
Now, S is symmetrical in all four quadrants, therefore
S = 4 × Area lying in the first quadrant.
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
1/ 2
1
1 − 2x dx
= ( 2 − 1 )2 + ∫
2 −1
2
1/ 2
1
2 1


= ( 2 − 1)2 + (1 − 2x)3/ 2 ⋅ (− 1)
2
3 2


2−1
1
1
= (2 + 1 − 2 2 ) + (1 + 2 − 2 2 )3/ 2
2
3
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
4x − x2 = x2 − x ⇒ 2x2 = 5x
5
⇒
x (2 − 5x) = 0
⇒ x = 0,
2
Also, y = x2 − x meets X-axis at (0,0) and (1, 0).
∴ Area, A1 = ∫
5/ 2
0
[(4x − x2) − (x2 − x)] dx
Area 331
=∫
5/ 2
0
(5x − 2x2) dx
45. The required area is the shaded portion in following
2
5/ 2
=
5  5
2  5
5 2 2 3 
x − x
=   − . 
2  2
3  2
3  0
2
=
5 25 2 125
⋅
− ⋅
2 4 3 8
=
125
8
figure
3
y = 2x
Y
2 125

sq units
1 −  =

3
24
y = loge x
This area is considering above and below X-axis both.
Now, for area below X-axis separately, we consider
X′
2
1
1/2
O
X
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
∴ The required area
Hence, ratio of area above the X-axis and area below
X-axis
121 1
=
: = 121 : 4
24 6
44. 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
⇒
⇒
⇒
But
O
M
X
y = x2
2
1 + x2
2
y=
1+ y
y=
Y′
…(i)
…(ii)
y2 + y − 2 = 0
( y − 1) ( y + 2) = 0 ⇒ y = − 2, 1
y ≥ 0, so y = 1 ⇒ x = ± 1
=∫
2

 2x
(2x − log x) dx = 
− (x log x − x)
1/ 2
 1/ 2
 log 2
2
4 − 2 5
=
− log 2 +
2
 log 2
3
 sq units
2
46. 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


+
∞
/
1
x
x→ 0

 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=
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
2
0
 01 + x


3 1 

x
2
2π 1  
= 2 [2 tan −1 x]10 −    = 2
−
= π −  sq unit
 4
 
3
3
3


 0 
log x
ex
y = ex log x
X′
1/e
O
Y′
1
X
332 Area
∴ The required area
1  (log x)

=∫ 
− ex log x dx
1/ e 

ex
1
1
=
 x2

 e2 − 5
1  (log x)2 
− e  (2 log x − 1) = 
 sq units


e  2 1/ e
4
1/ e  4e 
47. Given,
⇒
π 
Hence, equation of tangent at A  , 1 is
4 
π
y−1
= 2 ⇒ y − 1 = 2x −
2
x − π /4
y = x (x − 1)
dy
= x ⋅ 2 (x − 1) + (x − 1)2
dx
Y
A
1
X
B L
2
y = x(x – 1)
2
−1
Y′
π

(2x − y) =  − 1
2

⇒
X′
O 1/3 1 min
X
∴ Required area is OABO
=∫
Y′
π /4
0
( 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
= [log|sec x|]π0 / 4 −
= (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
curves, x2 + y2 = 25, 4 y = |4 − x2| could be
sketched as below, whose points of intersection are
(4 − x2)2
x2 +
= 25
16
49. Given
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
Y
4y = 4 – x 2
4y =x 2 – 4
5
4y = x 2 – 4
B
C
2
4
27
X′
O
X′
max
4
27
∴
y = tan x
Y
2
X′
O
1
A
x=2
O
–5 –4
2
–2
X
X
4
x 2 + y 2 = 25
Y′
–5
∴ 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
48. Given,
= sec2x
y = tan x ⇒
dx
 dy
∴
=2
 
 dx x = π
4
2
0
Y′
y dx
⇒
⇒
(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 

4
 x
25
 x 
25 − x2 +
= 2 
sin −1   
 5 
2
2
0
−
1
4
2

x3 
1
x
−
4
−

3  0 4

4
 
 x3
x
−
4
 
3
 2 

Area 333

25
 4  1  8 
sin −1    − 8 −
= 2 6 +
 5  4  3 
2


−
⇒
∴ Required area
1  64
  8  
− 16 −  − 8 



  3  
4 3
25

 4 4 4 4 
= 2 6 +
sin −1   − − − 
 5 3 3 3 
2


 4 
= 4 + 25 sin −1    sq units
 5 

50. Given curves are x2 + y2 = 4, x2 = − 2 y and x = y.
Y
y=x
2
=∫
X′
−2
√2
O
X
2
1
2
−1
( − x + 1) dx − ∫ (x − 1) dx
1
2
1
2
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
2
1 1
= sin −1
+ sin −1
 −
2
5
5 2
=
 2
5
sin −1 
2
 5
=
5
1  5π 1
−  sq units
sin −1 (1) − = 
2
2  4
2
1−
4 1
1−  −
5 2
1
1
+
5
5
π
π

tan x, − ≤ x ≤

3
3
52. Given, y = 
π
π
 cot x,
≤x≤

6
2
which could be plotted as Y-axis.
−2
x 2 − √ 2y
Y′
5 − x2 dx − ∫
 − x2

 x2

x
5
 x 
5 − x2 + sin −1    − 
=
+ x −  − x


2
2
2
2
5  −1 

 −1 
1
x + y2 = 4
−√ 2
2
−1
x = 2, − 1
Y
Thus, the required area
=
∫−
=2 ∫
2
2
2
0
0
∫−
4 − x2 dx −
2
x dx −
∫0
2
− x2
dx
2
  x2 0   x3  2

4 − x2 dx −   
 −
2
3 2 0
  − 2  
y = cot x
y = tan x
X′
O π/4
–π/2 –π/4
π/2
π
X
π/3
2
4
x
2
x
=2 
4 − x2 − sin −1  − 1 −
2
2
2
3

0
5
= (2 − π ) −
3

1
=  − π  sq units

3
51. Given curves y = 5 − x
Y′
∴ Required area = ∫
2
5 − x2 = (x − 1)2
y = –x + 1
53. Here,
1
⇒
⇒
2 √5
Y′
5 − x = x2 − 2 x + 1
2
2x − 2x − 4 = 0
2
π /3
∫ π / 4 ( cot x) dx
3
1
− 2 log
2
2
3
1 1

= log
− log =  log e 3 sq units

2
2 2
y=x–1
1
( tan x) dx +
= log
Y
√5 –1
0
= [− log|cos x|]π0 / 4 + [log sin x] ππ //34
1
3
1
 

− log
= −  log
− 0 +  log



2
2
2

and y = |x − 1| could be
sketched as shown, whose point of intersection are
X′
π /4
X
a
8
4
∫ 2 1 + x2 dx = ∫ a
a
⇒
⇒
8

1 + 2 dx

x 
4
8
8


x−
= x−

x  2 
x  a
8
8


 a −  − (2 − 4) = (4 − 2) −  a − 


a
a
334 Area
⇒
⇒
⇒
∴
8
8
16
+ 2 =2 − a +
⇒ 2a −
=0
a
a
a
2
2 (a − 8) = 0
[neglecting –ve sign]
a = ±2 2
a =2 2
Y
a−
54. The point of intersection of the curves x2 = 4 y and
P (t1)
X′
−1
x = 4 y − 2 could be sketched are x = − 1 and x = 2.
∴ Required area
 x2  
2  x + 2 
= ∫ 
 −    dx
−1  4 
 4

A
1 N
C
X
Q (−t1)
Y′
Required area
3 2
=
1
4
 x2
x
 2 + 2x − 3 
 −1

=
1
4
8  1

2 + 4 − 3 −  2 − 2 +

=
1 10  −7  1 9 9
−   = ⋅ = sq units
4  3  6   4 2 8
1 

3 
 et1 + e– t1 et1 – e– t1 
55. Let P = 
,

2
2


 e– t + et1 e– t1 – et 
and Q = 
,

2
2


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 ).
Download Chapter Test
http://tinyurl.com/y5eq3q4r
e t1 + e – t1


2
= 2 area of ∆PCN – ∫
ydx
1




1  et1 + e– t1   et1 – e– t1 

t1
dy
=2  
⋅ dt 

 –∫ y
1
2
2
dt


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
or
14
Differential Equations
Topic 1 Solution of Differential Equations by Variable
Separation Method
Objective Questions I (Only one correct option)
1. Let f be a differentiable function such that f (1) = 2 and
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) 4e2
(b) 4e
(d) 2e2
(c) 2e
2. The solution of the differential equation,
when y(1) = 1, is
(2019 Main, 11 Jan II)
1+ x − y
2− y
= 2( y − 1) (b) − log e
=x+ y−2
1− x + y
2− x
(a) log e
(c) log e
dy
= (x − y)2,
dx
2− x
=x− y
2− y
(d) − log e
1− x + y
= 2(x − 1)
1+ x − y
3. Let f : [0, 1] → R be such that f (xy) = f (x). f ( y), for all
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
y   + y   is equal to
 4
 4
(2019 Main, 9 Jan II)
(a) 5
(b) 3
4. If (2 + sin x)
(c) 2
(d) 4
dy
 π
+ ( y + 1) cos x = 0 and y(0) = 1, then y 
 2
dx
is equal to
(a)
5. If
1
3
(2017 Main)
(b) −
y = y(x)
(
2
3
(c) −
satisfies
1
3
the
)

8 x 9 + x dy =  4 + 9 +

y(0) = 7, then y(256) =
(a) 16
(c) 9
(d)
differential

x

−1
equation
x>0
and
(2017 Adv.)
(b) 3
(d) 80
13
6. The value of
∑
k =1
1
is equal
 π (k − 1)π 
 π kπ 
sin  +
 sin  +

4

4
6
6
to
(a) 3 −
dx,
4
3
(2016 Adv.)
3
(b) 2(3 −
3)
(c) 2( 3 − 1)
(d) 2(2 +
3)
7. 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
8. If y = y (x) and
 π
y   equals
 2
2 + sin x  dy
  = − cos x , y (0) = 1, then
y + 1  dx
(2004, 1M)
(a) 1/3
(c) − 1 / 3
(b) 2/3
(d) 1
9. A solution of the differential equation
2
dy
 dy
+ y = 0 is
  −x
 dx
dx
(a) y = 2
(c)y = 2x − 4
(1999, 2M)
(b) y = 2x
(d) y = 2x2 − 4
10. 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
(a) 5
(c) 3
(b) 4
(d) 2
(1998, 2M)
Objective Questions II
(One or more than one correct option)
11. Let f : [0, ∞ ) → R be a continuous function such that
f (x) = 1 − 2x +
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
336 Differential Equations
12. Let y(x) be a solution of the differential equation
(1 + ex ) y′ + yex = 1. If y(0) = 2, then which of the
(2015 Adv.)
following statement(s) is/are true?
(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)
13. 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.)
(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
(1999, 3M)
(b) order 2
(d) degree 4
(2003, 4M)
18. 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.
(1995, 5M)
Integer Answer Type Question
19. Let f : R → R be a continuous function, which satisfies
x
0
f (t ) dt . Then, the value of f (ln 5) is … .
(2009)
Passage
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]
Numerical Value
(2013 Adv.)
−x
20. If the function e f (x) assumes its minimum in the
15. Let f : R → R be a differentiable function with f (0) = 0.
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.)
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.
16. Let a solution y = y(x) of the differential equation
x x2 − 1 dy − y y2 − 1 dx = 0 satisfy y(2) =
P (x) > 0, ∀ x > 1.
dP (x)
> P (x), ∀ x ≥ 1 , then prove that
dx
Passage Based Problems
c), where c is a positive parameter,
(a) order 1
(c) degree 3
17. If P(1) = 0 and
f (x) = ∫
14. The differential equation representing the family of
curves y2 = 2c (x +
is of
Analytical & Descriptive Questions
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)
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),
(b) f ′ (x) > f (x), 0 < x <
(d) f ′ (x) < f (x),
1
4
3
< x<1
4
21. Which of the following is true?
(a) 0 < f (x) < ∞
(c) −
1
< f (x ) < 1
4
(b) −
1
1
< f (x ) <
2
2
(d) − ∞ < f (x) < 0
22. Which of the following is true?
(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, ∞ )
23. 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,
2 f (x) + 1 = 2x (1 + x)
(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)
8. If a curve passes through the point (1, − 2) and has slope
1. The general solution of the differential equation
( y − x )dx − xydy = 0 (x ≠ 0) is (where, C is a constant of
integration)
(2019 Main, 12 April II)
2
3
(a) y2 − 2x2 + Cx3 = 0
(c) y2 + 2x2 + Cx3 = 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
5
1
+
2
e
(b)
3
1
−
2
e
(c)
1
1
+
2
e
(d)
3
−
2
e
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)
2 (b)
π
π
π2
π
π
(c) y   + y −  =
+ 2 (d) y   − y −  =
 4
 4
 4
 4
2
2
4. If y = y(x) is the solution of the differential equation
dy
 π π
= (tan x − y) sec2 x, x ∈  − ,  , such that y (0) = 0,
 2 2
dx
 π
then y  −  is equal to
(2019 Main, 10 April I)
 4
1
(a) − 2
e
1
(b)
−e
2
1
(c) 2 +
e
π2
2 3
(b) −
π2
2 3
(c) −
π2
4 3
(d) e − 2
(d) −
π2
2
solution
of
the
differential
equation
dy
(2019 Main, 9 April I)
+ 2 y = x2(x ≠ 0) with y(1) = 1, is
dx
3
x2
+
4
4x2
3
1
(c) y = x2 +
4
4x2
1
x3
+
5
5x2
4
1
(d) y = x3 +
5
5x2
(a) y =
dy
(x + 1)
+ 2x(x2 + 1) y = 1 such
dx
π
a y(1) = , then the value of ‘a’ is
32
(a)
2
1
4
(b)
dy
+ y = x log e x, (x > 1). If 2 y(2) = log e 4 − 1, then y(e)
dx
(2019 Main, 12 Jan I)
is equal to
x
1
2
(c) 1
that
(b) −
e2
2
(c)
e
4
(d)
e2
4
dy  2x + 1
−2x
+
 y = e , x > 0,
dx  x 
1
where y (1) = e−2, then
(2019 Main, 11 Jan I)
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
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)
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 ∈ 
 3 3
 4 3
dx cos 2 x
cos 2 x
 π
y −  equals
 4
(2019 Main, 10 Jan I)
12. If
1
+ e6
3
(b) −
4
3
(c)
1
+ e3
3
(d)
1
3
13. If y = y(x) is the solution of the differential equation,
x
dy
+ 2 y = x2 satisfying y(1) = 1, then
dx
 1
y  is equal to
 2
(2019 Main, 9 Jan I)
(a)
y(0) = 0.
If
(2019 Main, 8 April I)
(d)
e
2
10. If y(x) is the solution of the differential equation
(b) y =
7. Let y = y(x) be the solution of the differential equation,
2
9. Let y = y(x) be the solution of the differential equation,
(a)
6. The
x
(b) (− 1, 2)
(d) (3, 0)
(a) ( 3 , 0)
(c) (− 2 , 1)
(a) does not exist
x
dy
π
5. If cos x − y sin x = 6x, 0 < x <  and y  = 0, then

 3
2
dx
 π
(2019 Main, 9 April II)
y  is equal to
 6
(a)
the curve also passes through the point
(a) −
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
x2 − 2 y
, then
x
(2019 Main, 12 Jan II)
(b) y2 + 2x3 + Cx2 = 0
(d) y2 − 2x3 + Cx2 = 0
2. Consider the differential equation, y2dx +  x −  dy = 0.
(a)
of the tangent at any point (x, y) on it as
1
16
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
9 3
π
2
−8 2
(b)
π
9 3
8
(c) − π 2
9
(2018 Main)
(d) −
4 2
π
9
338 Differential Equations
15. If a curve y = f (x) passes through the point (1, − 1) and
satisfies the differential equation, y(1 + xy)dx = x dy,
 1
then f  −  is equal to
 2
(2016 Main)
(a) −
2
5
(b) −
4
5
(c)
2
5
(d)
4
5
16. Let y(x) be the solution of the differential equation
(x log x)
dy
+ y = 2x log x, (x ≥ 1). Then, y(e) is equal to
dx
(2015 Main)
(a) e
(b) 0
(c) 2
(d) 2e
equation
dy
xy
x + 2x
+
=
dx x2 − 1
1 − x2
f (0) = 0. Then, ∫
(a)
π
3
−
3
2
3 /2
−
(b)
3
2
in
π
3
−
3
4
(c)
π
3
−
6
4
(d)
π
3
−
6
2
(2013 Adv.)
(b) (e − 1, 2e − 1)
e − 1
(d)  0,


2 
t 2f (x) − x2f (t )
= 1 for each x > 0 . Then,
t−x
f (x) is
(2007, 3M)
2x2
3
2
x2
1 4x2
(b) −
+
3x
3
1
(d)
x
is equal to
(d)|f (x)|≤ 2 for all x ∈ (0, 2)
23. If
satisfies
the
differential
y(x)
y′ − y tan x = 2 x sec x and y(0), then
π
π2
(a) y   =
 4 8 2
π
π2

(c) y   =
 3
9
equation
(2012)
π
π2
(b) y′   =
 4  18
π
4π
2π 2

(d) y′   =
+
 3
3
3 3
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 Answer Type Question
20. If x dy = y (dx + y dy), y (1) = 1 and y (x) > 0. Then, y (−3)
(a) 3
(c) 1
f (x)
for all x ∈ (0, ∞ ) and f (1) ≠ 1. Then
(2016 Adv.)
x
Analytical & Descriptive Question
19. Let f (x) be differentiable on the interval (0, ∞) such that
1
(a)
+
3x
1
(c) − +
x
f ′ (x) = 2 −
x → 0+
1/ 2
t→ x
22. Let f : (0, ∞ ) → R be a differentiable function such that
(c) lim x2f ′(x) = 0
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
f (1) = 1, and lim
Objective Questions II
(One or more than one correct option)
(2014 Adv.)
18. Let f : [1 /2, 1] → R (the set of all real numbers) be a
(a) (2e − 1, 2e)
e−1
(c) 
, e − 1

 2
(2003, 1M)
(b) e + 1 / 2
(d) 1 / 2
(−1, 1) satisfying
f (x) dx is
dy
− ty = 1 and y (0) = − 1,
dt
then y (1) is equal to
(a) −1 / 2
(c) e − 1 / 2
1
(a) lim f ′  = 1
x → 0+  x 
1
(b) lim x f   = 2
x → 0+
 x
17. The function y = f (x) is the solution of the differential
4
21. If y (t ) is a solution of (1 + t )
(2005, 1M)
(b) 2
(d) 0
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
Topic 3 Applications of Homogeneous Differential Equations
Objective Questions I (Only one correct option)
1. 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)
2. 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.
Differential Equations 339
3. 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
(c) 600 −
t
500 e 2
(b) 300 −
t
−
200 e 2
−
(d) 400 − 300 e
 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.)
y
(b) cosec   = log x + 2
 x
2y
1
(d) cos   = log x +
 x 
2
5. 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
9. If length of tangent at any point on the curve y = f (x)
intercepted between the point and the X-axis is of
(2005, 4M)
length 1. Find the equation of the curve.
10. A right circular cone with radius R and height H
t
2
π
4. A curve passes through the point 1,  . Let the slope of
1
y
(a) sin   = log x +
 x
2
2y 

(c) sec   = log x + 2
 x 
Analytical & Descriptive Questions
(d) 4500
Objective Questions II
(One or more than one correct option)
6. A
solution curve of the differential equation
dy
(x + xy + 4x + 2 y + 4)
− y2 = 0, x > 0, passes through
dx
the point (1, 3). Then, the solution curve
(2016 Adv.)
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
7. 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.
Fill in the Blank
8. 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)
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
which the cone is empty.
(2003, 4M)
11. A hemispherical tank of radius 2 m is initially full of
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)
Hint Form a differential equation by relating the
decreases of water level to the outflow.
12. 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)
13. A curve passing through the point (1, 1) has the property
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 curve.
(1999, 10M)
14. 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.
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)
15. Determine the equation of the curve passing through
the origin in the form y = f (x), which satisfies the
dy
differential equation
(1996, 5M)
= sin (10x + 6 y)
dx
340 Differential Equations
Match the Columns
16. 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
Topic 1
1. (b)
5. (b)
9. (c)
2. (d)
6. (c)
10. (c)
3. (b)
7. (c)
11. (b, c)
4. (a)
8. (a)
12. (a, c)
13. (b, c)
14. (a, c)
15. (0.40)
16. (b)
dy
d 2y
18. Differential Equation: 2 = 0, x 2
+1 = 0
dx
dx
Curves : x + y = 2, xy = 1
19. (0)
20. (c)
21. (d)
22. (b)
23. (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)
21. (a)
18. (d)
22. (a)
19. (a)
23. (a, d)
20. (a)
25. (0)
1. (c)
2. (b)
3. (a)
4. (a)
5. (c)
6. (a, d)
7. (a, c)


1 + 1 −y2
9.  1 − y 2 − log
= ± x + c


1 − 1 −y2


Topic 3

 dr
8.  = − λ 

 dt
H

10. T = 

k
 14 π × 10 5

11. 
unit
 27 g


1 
13. ( x 2 + y 2 = 2 x ) 14.  log 3  

2 

4
15.
1
4

tan −1  tan  4 x + tan −1

3
5
3  3  5x
−
 −
4 5  3
16. A → p; B → s; C → q; D → r
Hints & Solutions
Topic 1 Solution of Differential Equations
by Variable Separation Method
1. Given that, 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


⇒
ln|t|= x + C
Q ∫ x = ln|x|+ C 
…(i)
⇒ ln| f (x)|= x + C
[Q t = f (x)]
⇒
Q
So,
⇒
⇒
f (1) = 2
ln (2) = 1 + C
C = ln 2 − ln e
 2
C = ln  
 e
[using Eq. (i)]
[Q ln e = 1]
 A
[Q ln A − ln B = ln   ]
 B
From Eq. (i), we get
 2
ln| f (x)|= x + ln  
 e
 2
⇒ ln| f (x)|− ln   = x
 e
⇒
⇒
ln
ef (x)
=x
2
[Q ln A − ln B = ln
e
f (x) = ex [Q ln a = b ⇒ a = eb , a > 0]
2
A
]
B
Differential Equations 341
⇒
Now,
⇒
⇒
e
 e

Q 2 f (x) = 2 | f (x)| 


| f (x)|= 2ex −1
⇒
⇒
f (x) = 2ex −1 or −2ex −1
h (x) = f ( f (x))
h′ (x) = f ′ ( f (x)) ⋅ f ′ (x)
[on differentiating both sides w.r.t. ‘x’]
h′ (1) = f ′ ( f (1)) ⋅ f ′ (1)
[Q f (1) = 2 (given)]
= f ′ (2) ⋅ f ′ (1)
= 2e2−1 ⋅ 2e1−1 [Q f ′ (x) = 2ex −1 or −2ex −1]
= 4e
dy
2. We have,
= (x − y)2 which is a differential equation of
dx
the form
dy
= f (ax + by + c)
dx
Put x − y = t
dy dt
dy
dt
1−
=
⇒
⇒
=1 −
dx dx
dx
dx
dt
dy
2
[ Q = (x − y)2]
1−
=t
⇒
dx
dx
dt
dt
= 1 − t2 ⇒ ∫
= dx
⇒
dx
1 − t2 ∫
⇒
Q
∴
⇒
∴
y= x + C
y(0) = 1
1 =0 + C
C =1
y=x+1
5
7
 1 1
 3 3
Now, y   = + 1 = and y   = + 1 =
 4 4
 4 4
4
4
 3 5 7
 1
⇒
y  + y  = + =3
 4 4 4
 4
⇒
which is a linear differential equation.
∫
∴ IF = e
⇒
∴
⇒
[Q log 1 = 0]
y(0) = 1
∴
1(2 + sin 0) = − sin 0 + C
⇒
C =2
π
2 − sin
2 − sin x
 π
2 =1
y=
⇒ y  =
π 3
 2
2 + sin x
2 + sin
2
1
∴
dy
=
dx 8 x 9 +
x 4+ 9+
x
y= 4 + 9+ x + c
⇒
Now, y(0) = 7 + c
1−x+ y
− log e
= 2(x − 1)
1+ x− y
[Q log
= elog ( 2 + sin x ) = 2 + sin x
y(2 + sin x) = − sin x + C
Also,
5.
C = −1
1
 1 + x − y
log e 
 = x −1
 1 − x + y
2
cos x
dx
2 + sin x
∴Required solution is given by
− cos x
y ⋅ (2 + sin x) = ∫
⋅ (2 + sin x)dx + C
2 + sin x


a+x
1
dx
=
+ C
log e
∫ 2
2
a
a
x
2
−
a
x
−


 1 + x − y
1
⇒
[Q t = x − y]
log e 
 =x+C
2
 1 − x + y
Since, y = 1 when x = 1, therefore
⇒
dy
+ ( y + 1) cos x = 0
dx
dy
cos x
− cos x
y=
+
dx 2 + sin x
2 + sin x
4. We have, (2 + sin x)
[separating the variables]


1
1+ t
log e 
 =x+C
2
 1 − t
 1 + 0
1
log e 
 =1 + C
2
 1 + 0
dy
dy
= f (x) ⇒
=1
dx
dx
∫ dy = ∫ dx
So,
⇒
1
= log x−1 = − log x]
x
3. Given, f (xy) = f (x) ⋅ f ( y), ∀ x, y ∈ [0, 1]
Putting x = y = 0 in Eq. (i), we get
f (0) = f (0) ⋅ f (0)
⇒
f (0) [ f (0) − 1] = 0
⇒
f (0) = 1 as f (0) ≠ 0
Now, put y = 0 in Eq. (i), we get
f (0) = f (x) ⋅ f (0)
⇒
f (x) = 1
...(i)
c=0
y(256) = 4 + 9 + 16 = 4 + 5 = 3
13
1
k
1
π
(
)
π
−

 π kπ 
k = 1 sin

 +
 sin  +
4
6
6
4

Converting into differences, by multiplying and dividing
 π k π   π (k − 1)π 
 π
by sin  +
 − +
, i.e. sin   .
 6
6  4
6

 4
6. Here,
∴
∑
13
∑
k =1
π  π
π 
 π
sin  + k  −  + (k − 1) 
6 4
6 
 4
π
π 
π 
π
π
sin sin  + (k − 1)  sin  + k  
4
6
6 
6 
4
342 Differential Equations
π 

π
 π kπ 
sin  4 + 6  cos  4 + (k − 1) 6  

π
π
 π kπ  
− sin  + (k − 1)  cos  +

4
6
6  

4
π
π
π
π
sin  + (k − 1)  sin  + k 
4
6
6
4
13
= 2∑
k =1
13
π
π 

π
π
= 2 ∑ cot  + (k − 1)  − cot  + k  

4
6
4
6 


k =1 

 π π 
 π
= 2 cot   − cot  +  
 4 6 
 4

(a) y = 2 ⇒
On putting in Eq. (i),
02 − x (0) + y = 0
⇒ y = 0 which is not satisfied.
dy
(b)
=2
y = 2x ⇒
dx
On putting in Eq. (i),
⇒ y = 2x which is not satisfied.
dy
(c)
=2
y = 2x − 4 ⇒
dx
On putting in Eq. (i)
(2)2 − x − 2 + y
π
π 

π
= 2 cot − cot  + 13  

4
4
6 

4 − 2x + 2x − 4 = 0
5 π 



 29 π  
= 2 1 − cot 

 = 2 1 − cot 2π +

 12  
12  


7. Given,
⇒
y = 2x2 − 4
dy
= 4x
dx
(d)
5π


Q cot 12 = (2 − 3 )
On putting in Eq. (i),
(4x)2 − x ⋅ 4x + y = 0
= 2 ( 3 − 1)
⇒ y = 0 which is not satisfied.
10. Given, y = (c1 + c2) cos (x + c3 ) − c4 ex + c5
dy
1 − y2
=
dx
y
∫
⇒
y
1 − y2
dy = ∫ dx
⇒
− 1 − y = x + c ⇒ (x + c) + y = 1
2
2
2
…(i)
y = (c1 + c2) cos (x + c3 ) − c4 e ⋅ e
x
Now, let
Here, centre (– c, 0) and radius = 1
dy − cos x ( y + 1)
8. Given,
=
dx
2 + sin x
c5
c1 + c2 = A , c3 = B, c4ec5 = c
y = A cos (x + B) − cex
…(ii)
On differentiating w.r.t. x, we get
dy
= − A sin ( A + B) − cex
dx
…(iii)
Again, on differentiating w.r.t. x, we get
d 2y
= − A cos (x + B) − cex
dx2
dy
− cos x
=
dx
y + 1 2 + sin x
…(v)
d 2y
+ y = − 2 cex
dx2
⇒
log ( y + 1) = − log (2 + sin x) + log c
…(iv)
d 2y
= − y − 2 cex
dx2
⇒
On integrating both sides
dy
cos x
∫ y + 1 = − ∫ 2 + sin x dx
⇒
[Q y = 2x − 4]
y = 2x − 4 is satisfied.
= 2 (1 − 2 + 3 )
⇒
⇒
(2)2 − x ⋅ 2 + y = 0
4 − 2x + y = 0
⇒

 π π
 π 2π  
+ cot  +  − cot  +

 4 6
4
6 

π
π  

π
π
+ K + cot  + 12  − cot  + 13  
4
4
6
6  

5π

= 2 1 − cot
12 

dy
=0
dx
Again, on differentiating w.r.t. x, we get
When x = 0, y = 1 ⇒ c = 4
4
⇒
y+1=
2 + sin x
d3 y dy
+
= − 2 cex
dx
dx3
d3 y dy d 2y
+
=
+ y
dx2 dx dx2
…(vi)
∴
 π 4
y  = −1
 2 3
⇒
⇒
 π 1
y  =
 2 3
which is a differential equation of order 3.
11. We have,
f ( x ) = 1 − 2x +
9. Given differential equation is
2
dy
 dy
+ y=0
  −x
 dx
dx
…(i)
x
∫0 e
x −t
f ( t ) dt
On multiplying e− x both sides, we get
x
e− x f ( x ) = e− x − 2xe− x + ∫ e− t f ( t ) dt
0
[from Eq. (v)]
Differential Equations 343
On differentiating both side w.r.t. x, we get
e− x f ′ ( x ) − e− x f ( x ) = − e− x − 2e− x + 2xe− x + e− x f ( x )
⇒
f ′ ( x ) − 2 f ( x ) = 2x − 3
[dividing both sides by e− x ]
Let
f(x) = y
dy
⇒
f ′ (x) =
dx
dy
− 2 y = 2x − 3
∴
dx
which is linear differential equation of the form
dy
+ Py = Q. Here, P = −2 and Q = 2x − 3.
dx
P dx
−2 dx
Now,
IF = e∫
= e∫
= e −2x
∴ Solution of the given differential equation is
y ⋅ e−2x = ∫ ( 2x − 3) e−2x dx + C
II
I
− ( 2x − 3) ⋅ e−2x
e −2x
+ 2∫
dx + C
2
2
[by using integration by parts]
− ( 2x − 3) e−2x e−2x
−2x
⇒ y⋅e
=
−
+C
2
2
⇒ y = (1 − x ) + Ce2x
On putting x = 0 and y = 1, we get
1=1+C ⇒ C = 0
∴ y=1− x
y = 1 − x passes through ( 2, − 1)
y ⋅ e −2x =
Now, area of region bounded by curve
y = 1 − x is shows as
y = 1 − x2 and
y= 1–x2
A
(1,0)
O
⇒
−4 + 4
=0
1 + e−4
Now at x = − 4, y =
∴
y(−4) = 0
…(i)
dy
=0
dx
For critical points,
dy (1 + ex ) ⋅ 1 − (x + 4)ex
=0
=
dx
(1 + ex )2
i.e.
⇒
ex (x + 3) − 1 = 0
e− x = (x + 3)
or
y = e–x
Y
y=x+3
(–1, e)
(–1, 2)
X′
X
O
–1
Y′
Clearly, the intersection point lies between (− 1, 0).
x2 + y2 − 2ax − 2ay + c = 0
x
y′
∴ Area of shaded region
= Area of 1st quadrant of a
circle − Area of ∆ OAB
π
1
= (1)2 − × 1 × 1
4
2
π 1 π −2
= − =
4 2
4
Hence, options b and c are correct.
12. Here, (1 + ex ) y′ + y ex = 1
dy
dy
+ ex ⋅
+ yex = 1
dx
dx
⇒ dy + ex dy + yex dx = dx
⇒
C =4
y (1 + ex ) = x + 4
x+4
y=
1 + ex
∴ Equation of circle is
y=1–x
⇒
⇒
∴
13. Since, centre lies on y = x.
B (0, 1)
–1,0
2 + e0 ⋅ 2 = 0 + C
∴ y(x) has a critical point in the interval (− 1, 0).
y
x′
y (0) = 2
Given,
⇒
dy + d (e y) = dx
x
On integrating both sides, we get
y + ex y = x + C
On differentiating, we get
2x + 2 yy′ − 2a − 2ay′ = 0
⇒
x + yy′ − a − ay′ = 0
x + yy′
a=
1 + y′
⇒
Again differentiating, we get
0=
(1 + y′ )[1 + yy′ ′+ ( y′ )2] − (x + yy′ ) ⋅ ( y′ ′ )
(1 + y′ )2
⇒ (1 + y′ ) [1 + ( y′ )2 + yy′ ′ ] − (x + yy′ ) ( y′′ ) = 0
⇒
1 + y′ [( y′ )2 + y′ + 1] + y′′ ( y − x) = 0
On comparing with Py′′ + Qy′ + 1 = 0, we get
P = y−x
and
14. Given,
Q = ( y′ )2 + y′ + 1
y2 = 2c (x +
c)
On differentiating w.r.t. x, we get
dy
dy
2y
= 2c ⇒ c = y
dx
dx
…(i)
344 Differential Equations
On putting this value of c in Eq. (i), we get
dy 
dy 
y2 = 2 y
x + y

dx 
dx 
y=2
⇒
⇒
y − 2x
dy
 dy
⋅ x + 2 y1/ 2  
 dx
dx
dy
 dy
=2 y  
 dx
dx
2
dy
 dy

 y − 2x  = 4 y  
 dx

dx
⇒
At x = 2 , y =
⇒ c=−
3/ 2

1
3
= cos cos −1 − cos −1

x
2 

3

= cos cos −1

Therefore, order of this differential equation is 1 and
degree is 3.
y=
15. We have,
dy
= ∫ dx
2
 2
y − 
 5
y − 2/ 5
1
1
⇒
log
= x+C
×
2
y + 2/ 5
25 2 ×
5
5y − 2
= 20( x + C )
⇒ log
5y + 2
⇒
∫
3 1
1
+
1− 2
2x 2
x
dP (x)
− P (x) > 0, ∀ x ≥ 1
dx
On multiplying Eq. (i) by e− x , we get
2
e− x ⋅
d
(P (x) ⋅ e− x ) > 0
dx
⇒
⇒
P (x) ⋅ e− x is an increasing function.
⇒
P (x) ⋅ e− x > P (1) ⋅ e−1 , ∀ x ≥ 1
⇒
P (x) > 0, ∀ x > 1
A (x, y) is Y − y =
dy
(X − x)
dx
y
(1, 1)
Q
x → −∞
⇒ lim
n → −∞
⇒
2
lim f (x) = = 0.4
n→ − ∞
5
2
dy y y − 1
=
dx x x2 − 1
∫
dy
y y −1
2
−1
sec
=∫
x
P
y′
dx 

whose, x-intercept  x − y ⋅
, 0

dy 
lim 5 f ( x ) −2 = 0
16. Given,
O
Given,
⇒
⇒
∆OPQ = 2
1 
dx 
dy
⋅ x − y   y − x  = 2
2 
dy 
dx
1

 x − y  ( y − xp) = 4, where

p
dx
⇒
p2x2 − 2 pxy + 4 p + y2 = 0
2
x x −1
⇒
( y − px)2 + 4 p = 0
x+ c
∴
−1
y = sec
y = f(x)
dy

y-intercept 0, y − x 

dx
n→−∞
⇒
⇒
x ′'
5 f(x) − 2
= 0
5 f(x) + 2
(x, y)
A
5 y −2
= e20x
5y + 2
5 f ( x ) −2
= lim e20x
x → −∞
5 f(x) + 2
[Q P (1) = 0 and e− x > 0]
18. Equation of tangent to the curve y = f (x) at point
5y − 2
= Ae20x [Q e20C = A]
5y + 2
lim
...(i)
d
d −x
P (x) ⋅
e >0
dx
dx
when x = 0 ⇒ y = 0, then A = 1
∴
3 
1 − 
4  
 3
1
+ 1− 2

2
x
x

17. Given, P(1) = 0 and


dy
1  dy 
 = dx

⇒
= dx ⇒
25  y 2 − 4 
25 y 2 − 4

25 
On integrating both sides, we get
1
25
π
6
π

Now, y = sec sec−1 x − 

6
3/ 2
dy
= ( 2 + 5 y ) ( 5 y − 2)
dx
2 π π
;
= +c
3 6 3
⇒
p=
dy
dx
y − px = 2 − p
y = px + 2 − p
…(i)
Differential Equations 345
On differentiating w.r.t. x, we get
p= p+
dp
dp
 1
⋅ x + 2 ⋅   ( − p)−1/ 2 ⋅ ( −1)
 2
dx
dx
dp
{ x − ( − p)−1/ 2 } = 0
dx
dp
= 0 or x = ( − p)−1/ 2
dx
dp
= 0 ⇒ p= c
dx
⇒
⇒
If
On putting this value in Eq. (i), we get y = cx + 2 − c
 1
⇒ e− x f ′ (x) − e− x f (x) < 0, x ∈ 0, 
 4
1
⇒ f ′ (x) < f (x) ,0 < x <
4
21. Here, f ′′ (x) − 2 f ′ (x) + f (x) ≥ ex
⇒
f ′′ (x)e−x − f ′ (x)e− x − f ′ (x)e− x + f (x)e− x ≥ 0
⇒
d
d
{ f ′ (x)e− x } −
{ f (x)e− x } ≥ 1
dx
dx
⇒
d
{ f ′ (x)e− x − f (x)e− x } ≥ 1
dx
⇒
d 2 −x
{ e f (x)} ≥ 1, ∀x ∈ [0, 1]
dx2
This curve passes through (1, 1).
⇒
⇒
∴
⇒
1 = c + 2 −c
c= −1
y=−x+2
x+ y=2
∴ φ (x) = e− x f (x) is concave function.
f (0) = f (1) = 0
Again, if
x = (− p)−1/ 2
1
⇒ − p = 2 putting in Eq. (i)
x
y=
−x
x
2
+ 2⋅
⇒
1
x
⇒ xy = 1
Thus, the two curves are xy = 1 and x + y = 2.
19. From given integral equation, f (0) = 0 .
Also, differentiating the given integral equation w.r.t. x
f ′ (x) = f (x)
f (x) ≠ 0
If
f ′ (x)
=1
f (x)
⇒
⇒
log f (x) = x + c
⇒
f (x) = ecex
Q
f (0) = 0 ⇒
∴
f (x) = 0, ∀ x ∈ R
⇒
ec = 0 , a contradiction
22. 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) = 
− log x ⋅ f{
(x)
x
(
+
)
1
 + ve

⇒
φ(x) < 0
From Eqs. (i) and (ii), g′ (x) < 0, x ∈ (1, ∞ )
f (0) = 0 and
f ′ (x) = f (x)
f ′ (x)
= 1 ⇒ ln f (x) = x + c
f (x)
⇒
f (x) = ec ⋅ ex
Q
f (0) = 0
…(i)
For g′ (x) to be increasing or decreasing.
2(x − 1)
Let φ(x) =
− log x
x+1
4
1 − (x − 1)
φ′ (x) =
− =
2
x x (x + 1)2
(x + 1)
x
⇒
⇒ ec = 0, a contradiction
f (x) = 0, ∀ x ∈ R
f (ln 5) = 0
20. Let φ(x) = e−x f (x)
and
f (x) < 0
0
If f (x) ≠ 0
Here,
∴
φ (x) < φ (1)
f (x) = ∫ f (t ) dt
⇒
φ (x) < 0
e− x f (x) < 0
⇒
Given,
∴
⇒
⇒
φ′ (x) < 0, ∀ x > 1
f (ln 5) = 0
Alternate Solution
⇒
φ(0) = 0 = f (1)
 1
φ′ (x) < 0, x ∈ 0, 
 4
1 
φ′ (x) > 0, x ∈  , 1
4 
…(ii)
∴g (x) is decreasing on x ∈ (1, ∞ ).
23. Here, f (x) + 2x = (1 − x)2 ⋅ sin 2 x + x2 + 2x
where, I: f (x) + 2x = 2(1 + x)
∴
2
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
which is never possible.
∴I is false.
Again, let h (x) = 2 f (x) + 1 − 2x(1 + x)
where,
h (0) = 2 f (0) + 1 − 0 = 1
h(1) = 2(1) + 1 − 4 = − 3 as [h (0)h (1) < 0]
⇒ h (x) must have a solution.
∴ II is true.
…(i)
…(ii)
346 Differential Equations
Topic 2 Linear Differential Equation and
Exact Differential Equation
1. Given differential equation is
( y2 − x3 ) dx − xy dy = 0, (x ≠ 0)
dy
xy
− y2 = − x3
⇒
dx
dy dt
dy 1 dt
Now, put y2 = t ⇒ 2 y
=
⇒y
=
dx dx
dx 2 dx
x dt
− t = − x3
∴
2 dx
dt 2
− t = − 2x2
⇒
dx x
which is the linear differential equation of the form
dt
+ Pt = Q.
dx
2
Here,
P = − and Q = − 2x2.
x
−∫
IF = e
Now,
2
dx
x
=
1
x2
Q Solution of the linear differential equation is
(IF) t = ∫ Q (IF)dx + λ
[where λ is integrating constant]
∴
⇒
1
1
t  2  = − 2 ∫  x2 × 2  dx + λ
x 

x 
t
= − 2x + λ
x2
⇒
y2
+ 2x = λ
x2
⇒
or
y2 + 2x3 − λx 2 = 0
y2 + 2x3 + Cx 2 = 0
[Q t = y2]
[let C = − λ]
dx 1
1
+ 2 x = 3 , which is the linear differential
dy y
y
dx
equation of the form
+ Px = Q.
dy
−
x ⋅ (IF) ∫ Q (IF)dy + C
= − tet +
∫ e dt + C
t
= − tet + et + C
3. Given differential equation is
dy
+ y tan x = 2x + x2 tan x , which is linear differential
dx
dy
equation in the form of
+ Py = Q .
dx
Here, P = tan x and Q = 2x + x2 tan x
∴IF = e∫ tan x dx = elog e (sec x ) = sec x
Now, solution of linear differential equation is given as
y × IF = ∫ (Q × IF)dx + C
∴ y(sec x) = ∫ (2x + x2 tan x) sec x dx + C
∫x
2
sec x tan x dx + C
Therefore, solution is
⇒ y sec x = x2 sec x + C …(i)
Q y(0) = 1 ⇒ 1(1) = 0(1) + C ⇒ C = 1
Now, y = x2 + cos x
and y′ = 2x − sin x
According to options,
1
 π
 − π   π
y′   − y′ 
 = 2  −

 4
 4    4
2
1
  π
− 2 −  +
 =π− 2
  4
2
[from Eq. (i)]
2
1
π2
1
 π
 π π
and y  − y −  =
+
−
−
=0
 4
 4  16
2 16
2
4. Given differential equation
1 − 1/ y
e dy + C
y3
∴ x e− 1/ y = ∫ (− t ) et dt + C
e1/ 2 3
1
1
+1−
= −
e
2
2
e
2
1
π2 1
π2
 π
 π π
and y  + y −  =
+
+
=
+ 2
+
 4
 4  16
4
2
2 16
1
y
∴The solution of linear differential equation is
⇒ x e− 1/ y = ∫
So, at y = 2, the value of x =
1
 π   π 1    π
 π
and y′   + y′  −  = 2  −
 + 2 −  +
 =0
 4   4
 4
2   4
2
1
1
and Q = 3
y2
y
=e
Now, at y = 1, the value of x = 1, so
1
1 ⋅ e− 1 = e− 1 + e− 1 + C ⇒ C = −
e
On putting the value of C, in Eq. (i), we get
1
e1/ y
x= +1−
y
e
y sec x = 2∫ x sec x dx + x2 sec x − 2∫ x sec x dx + C
⇒
1
∫ 2 dy
y
… (i)
Q ∫ x2 sec x tan x dx = x2 sec x − ∫ (2x sec x) dx
1

y2dx +  x −  dy = 0

y
Now, IF = e
1 − 1/ y
e
+ e− 1/ y + C
y
= ∫ (2x sec x) dx +
2. Given differential equation is
Here, P =
⇒ x e− 1/ y =
[Qlet −
1
1
= t ⇒ + 2 dy = dt]
y
y
[integration by parts]
dy
= (tan x − y)sec2 x
dx
dy
+ (sec2 x) y = sec2x tan x,
dx
which is linear differential equation of the form
dy
+ Py = Q,
dx
where P = sec2 x and Q = sec2 x tan x
⇒
Differential Equations 347
sec
IF = e∫
2
x dx
= etan x
So, solution of given differential equation is
y × IF = ∫ (Q × IF)dx + C
y(etan x ) = ∫ etan x ⋅ sec2 x tan x dx + C
Let
tan x = t ⇒ sec2x dx = dt
yetan x = ∫ et ⋅ t dt + C = tet − ∫ et dt + C
[using integration by parts method]
= et (t − 1) + C
⇒
y ⋅ etan x = etan x (tan x − 1) + C
[Q t = tan x]
Q
y(0) = 0
⇒
0 = 1(0 − 1) + C ⇒ C =1
∴
y ⋅ etan x = etan x (tan x − 1) + 1
π
Now, at x = −
4
ye−1 = e−1 (−1 − 1) + 1
⇒
ye−1 = − 2e−1 + 1⇒ y = e − 2
5. Key Idea (i)
First convert the given differential equation into
dy
linear differential equation of the form
+ Py = Q
dx
(ii) Find IF
(iii) Apply formula, y( IF) = ∫ Q( IF) dx + C
Given differential equation
dy
− (sin x) y = 6x
cos x
dx
6x
dy
, which is the linear
⇒
− (tan x) y =
cos x
dx
differential equation of the form
dy
+ Px = Q,
dx
6x
where P = − tan x and Q =
cos x
− tanx dx
So, IF = e ∫
= e− log(sec x ) = cos x
∴Required solution of differential equation is
x2
cos x
y(cos x) = ∫ (6x)
dx + C = 6 + C = 3x2 + C
cos x
2
π

Given,
y  = 0
 3
2
So,
∴
Now, at x =
π
6
π
π
0 = 3  + C ⇒C = −
 3
3
2
π
y(cos x) = 3x2 −
3
2
 3
π2 π2
π2
π2
−
=−
y
⇒ y=−
 =3
36
3
4
2 3
 2 
6. Given differential equation is
dy
+ 2 y = x2, (x ≠ 0)
dx
dy  2
+   y = x,
dx  x
which is a linear differential equation of the form
dy
+ Py = Q
dx
2
Here, P = and Q = x
x
∫
IF = e
∴
= e2log x = x2
Since, solution of the given differential equation is
y × IF = ∫ (Q × IF) dx + C
∴ y(x2) = ∫ (x × x2) dx + C ⇒ yx2 =
Q
x4
+C
4
1
3
+ C ⇒C =
4
4
2
4
3
3
x
x
+ 2
+ ⇒y=
yx2 =
4 4x
4
4
y(1) = 1, so 1 =
∴
7. Given differential equation is
dy
+ 2x(x2 + 1) y = 1
dx
1
dy  2x 
⇒
+ 
y =
dx  1 + x2 
(1 + x2 )2
[dividing each term by (1 + x2 )2] …(i)
This is a linear differential equation of the form
dy
+ P⋅y =Q
dx
2x
1
Here, P =
and Q =
(1 + x2 )
(1 + x2 )2
(x2 + 1)2
2x
∫
∴Integrating Factor (IF) = e 1 +
x2
= eln(1 +
dx
x2)
= (1 + x2 )
and required solution of differential Eq. (i) is given by
y ⋅ (IF) =
∫ Q (IF)dx + C
1
(1 + x2 )dx + C
(1 + x2 )2
dx
+C
⇒ y(1 + x2 ) = ∫
1 + x2
⇒ y(1 + x2 ) =
∫
⇒ y(1 + x2 ) = tan −1 (x) + C
Q
y(0) = 0
∴
C=0
∴
y(1 + x2 ) = tan −1 x
tan −1 x
⇒
y=
1 + x2
 tan −1 x 
ay = a 
⇒

 1 + x2 
[Q C = 0]
[multiplying both sides by a]
Now,
at x = 1
x
⇒
2
dx
x
a y (1) =
∴
a=
 tan −1 (1) 
a
 =
 1+ 1 
1
1
⇒a =
4
16
π
a 4 =
2
aπ
π
(given)
=
8
32
348 Differential Equations
8. We know that, slope of the tangent at any point (x, y) on
the curve is
dy x2 − 2 y
(given)
=
dx
x
dy 2
…(i)
+ y=x
⇒
dx x
which is a linear differential equation of the form
dy
+ P (x) ⋅ y = Q (x),
dx
2
where
P (x) = and Q (x) = x
x
Now, integrating factor
P ( x )dx
(IF) = e∫
=
= elog e x
2
∫ dx
e x
= e2log e x
2
[Q m log a = log am ]
= x2
[Q elog e f ( x ) = f (x)]
and the solution of differential Eq. (i) is
y(IF) = ∫ Q (x)(IF)dx + C ⇒ y(x ) = ∫ x ⋅ x dx + C
2
2
x4
…(ii)
+C
⇒
yx =
4
Q The curve (ii) passes through the point (1, − 2),
therefore
1
9
−2 = + C ⇒C = −
4
4
2
∴ Equation of required curve is 4 yx = x4 − 9.
2
Now, checking all the option, we get
only ( 3 , 0) satisfy the above equation.
9. Given differential equation is
dy
+ y = x log e x, (x > 1)
dx
dy 1
⇒
+ y = log e x
dx x
Which is a linear differential equation.
x
∫
So, if = e
1
dx
x
…(i)
= elog e x = x
Now, solution of differential Eq. (i), is
y × x = ∫ (log e x) x dx + C
⇒
yx =
x2
x2 1
× dx + C
log e x − ∫
2
2 x
[using integration by parts]
x2
x2
log e x −
⇒
yx =
+C
2
4
Given that, 2 y(2) = log e 4 − 1
On substituting, x = 2, in Eq. (ii),
we get
4
4
2 y(2) = log e 2 − + C,
2
4
… (ii)
… (iii)
[where, y(2) represents value of y at x = 2]
⇒
2 y(2) = log e 4 − 1 + C
… (iv)
[Q m log a = log am ]
From Eqs. (iii) and (iv), we get
C =0
So, required solution is
x2
x2
yx =
log e x −
2
4
e2
e2
Now, at
x = e, ey(e) = log e e −
2
4
[where, y(e) represents value of y at x = e]
⇒
y(e) =
e
4
[Q log e e = 1].
dy  2x + 1
− 2x
+
y=e
dx  x 
dy
which is of the form
+ Py = Q, where
dx
2x + 1
and Q = e−2x
P=
x
10. We have,
 1 + 2x 
 dx
x 
∫
Pdx
Now, IF = e∫
=e 

1
∫  + 2 dx
= e x 
= eln x + 2x = eln x . e2x = x. e2x
and the solution of the given equation is
y ⋅ (IF) = ∫ (IF ) Q dx + C
⇒
y(xe2x ) = ∫ (x e2x . e−2x ) dx + C
= ∫ x dx + C =
x2
+C
2
… (i)
1 −2
e when x = 1
2
1 −2 2 1
∴
e . e = + C ⇒ C = 0 (using Eq. (i))
2
2
x
x2
2x
⇒ y = e−2x
y (xe ) =
∴
2
2
dy 1 −2x x −2x
1

Now,
= e + e (− 2) = e−2x  − x < 0,
2
dx 2
2

1
if < x < 1
[by using product rule of derivative]
2
−2
log e 2 −2log e 2 1
and y(log e 2) =
e
= log e 2 elog e 2
2
2
1
1
−2
= . log e 2 ⋅ 2 = log e 2
2
8
3 f (x)
11. Given, f ′ (x) = 7 −
, (x > 0)
4 x
dy
On putting f (x) = y and f ′ (x) = , then we get
dx
dy
3 y
= 7−
dx
4 x
dy 3
…(i)
⇒
+
y=7
dx 4x
Since, y =
which is a linear differential equation of the form
3
dy
and Q = 7.
+ Py = Q, where P =
4x
dx
3
∫ dx
Now, integrating factor (IF) = e 4x
3
= e4
log x
= elog x
3/ 4
= x3/ 4
Differential Equations 349
and solution of differential Eq. (i) is given by
13. Given differential equation can be rewritten as
dy  2
+   ⋅ y = x, which is a linear differential equation of
dx  x
2
dy
the form
+ Py = Q, where P = and Q = x.
x
dx
Now, integrating factor
y(IF) = ∫ (Q ⋅ (IF))dx + C
yx3/ 4 = ∫ 7x3/ 4dx + C
3
⇒
3/ 4
yx
+1
x4
=7
+C
3
+1
4
7
y x3/ 4 = 4x 4 + C
⇒
⇒
So,
Now,
∴
 1
f 
 x
 1
lim x f  
 x
x→ 0 +
y = 4 x + C x− 3 / 4
y = f (x) = 4x + C ⋅ x−3/ 4
4
= + C ⋅ x3/ 4
x

4
= lim x + Cx3/ 4 = lim (4 + Cx7/ 4 ) = 4
 x→ 0 +
x→ 0 +  x
12. Given, differential equation is
1
dy  3 
, which is a linear differential
+
y=
dx  cos 2 x
cos 2 x
3
dy
equation of the form
and
+ Py = Q, where P =
dx
cos 2 x
1
.
Q=
cos 2 x
Now, Integrating factor
3
∫ 2 dx
3 sec 2 x dx
IF = e cos x = e∫
= e3 tan x and the solution of
differential equation is given by
y(IF) = ∫ (Q. (IF)) dx
⇒
Let
e3 tan x . y = ∫ e3 tan x sec2 x dx
… (i)
I = ∫ e3 tan x sec2 x dx
3 tan x = t
3 sec2 x dx = dt
et
et
e3 tan x
∴
I = ∫ dt =
+C=
+C
3
3
3
From Eq. (i)
e3 tan x
e3 tan x . y =
+C
3
It is given that when,
π
4
x = , y is
4
3
e3
3 4
⇒
e
=
+C
3 3
⇒
C = e3
e3 tan x
Thus, e3 tan x y =
+ e3
3
π
e−3
Now, when x = − , e−3 y =
+ e3
3
4
Put
⇒
⇒
y = e6 +
1
3
2
∫


 π
Q tan  − 4  = − 1


dx
2
(IF) = e x = e2log x = elog x = x2
[Q elog f ( x ) = f (x) ]
and the solution is given by
y(IF ) = ∫ (Q × IF ) dx + C
⇒
yx2 = ∫ x3 dx + C
x4
…(i)
+C
4
Since, it is given that y = 1 when x = 1
∴ From Eq. (i), we get
1
3
…(ii)
1 = + C ⇒C =
4
4
[using Eqs. (i) and (ii)]
∴
4 x2 y = x 4 + 3
x4 + 3
⇒
y=
4 x2
1
+3
49
 1 16
Now,
=
y  =
1
 2
16
4×
4
⇒
yx2 =
14. We have,
dy
dy
+ y cot x = 4xcosec x
sin x
+ y cos x = 4x ⇒
dx
dx
This is a linear differential equation of form
dy
+ Py = Q
dx
where P = cot x, Q = 4x cosec x
cotxdx
Now, IF = e∫ Pdx = e∫
= elogsin x = sin x
Solution of the differential equation is
y ⋅ sin x = ∫ 4x cosec x sin xdx + C
⇒
Put x =
y sin x = ∫ 4xdx + C = 2x2 + C
π
, y = 0, we get
2
C=−
Put
∴
⇒
x=
π
6
π2
π2
⇒ y sin x = 2x2 −
2
2
 π 2 π 2
 1
y  = 2  −
 2
2
 36 
y=
π2
8π 2
− π2 ⇒ y = −
9
9
Alternate Method
dy
We have, sin x
+ y cos x = 4x, which can be written as
dx
d
(sin x ⋅ y) = 4x
dx
350 Differential Equations
On integrating both sides, we get
d
∫ dx (sin x ⋅ y) ⋅ dx = ∫ 4x ⋅ dx
4 x2
⇒
y ⋅ sin x =
+ C ⇒ y ⋅ sin x = 2x2 + C
2
π
Now, as y = 0 when x =
2
π2
∴
C=−
2
2
2 π
⇒
y ⋅ sin x = 2x −
2
π
Now, putting x = , we get
6
2

1
π
π2
8π 2
π2
 
y  = 2  −
⇒ y=
− π2 = −
 2
9
9
 36  2
x=1 ⇒ c=2
y ⋅ log x = 2 [x log x − x] + 2
x = e, y = 2(e − e) + 2
y=2
At
⇒
At
⇒
where,
(ii)
a
∫− a
y ⋅ (IF) = ∫ Q ⋅ (IF) dx + c
P dx
IF = e ∫
f( x ) dx = 2
⇒
This is a linear differential equation.
∫
IF = e
y dx + xy2 dx = x dy
x dy − y dx
= x dx
y2
−
…(i)
=2 ∫
x x2 1
=
+
y 2 2
2x
x2 + 1 = −
y
2x
y=− 2
x +1
3 /2
0
π /3
=2 ∫
0
⇒
=2 ∫
0
 1 4
f −  =
 2 5
16. Given differential equation is
dy
+ y = 2x log x
dx
dy
y
+
=2
dx x log x
(x log x)
⇒
This is a linear differential equation.
1
∴
IF = e
∫ x log x dx
= elog(log x ) = log x
Now, the solution of given differential equation is given
by
y ⋅ log x = ∫ log x ⋅ 2dx
⇒
y ⋅ log x = 2∫ log xdx
⇒
y ⋅ log x = 2 [x log x − x] + c
ln | x 2 − 1 |
x(x3 + 2)
1−x
2
= 1 − x2
⋅ 1 − x2 dx
y 1 − x2 = ∫ (x4 + 2x) dx =
Now, from Eq. (i) −
∴
= e2
[using property]
Q It passes through (1, − 1).
1
1
∴
1= +C ⇒ C =
2
2
⇒
1
dx
x5
+ x2 + c
5
x5
+ x2
f (0) = 0 ⇒ c = 0 ⇒ f (x) 1 − x2 =
5
3 /2
3 /2
x2
Now, ∫
f (x) dx = ∫
dx
− 3 /2
− 3 /2
1 − x2
or
On integrating both sides, we get
x x2
=
+C
y 2
x
x2 − 1
⇒ Solution is y 1 − x2 = ∫
( y dx − x dy)
 x
= x dx ⇒ − d   = x dx
2
 y
y
−
f( x ) dx , if f( − x ) = f( x )
x4 + 2x
dy
x
+ 2
y=
dx x − 1
1 − x2
y(1 + xy) dx = x dy
⇒
a
∫0
Given differential equation
15. Given differential equation is
⇒
dy
+ Py = Q is
dx
17. PLAN (i) Solution of the differential equation
π /3
x2
1 − x2
sin 2 θ
cos θ dθ
cos θ
[taking x = sin θ ]
π /3
sin 2 θ dθ = ∫ (1 − cos 2θ ) dθ
0
π /3
sin 2θ 

= θ −


2 0
18. PLAN
dx
=
π sin 2π / 3 π
3
−
= −
3
2
3
4
Whenever we have linear differential equation containing
inequality, we should always check for increasing or
decreasing,
dy
dy
i.e. for
+ Py < 0 ⇒
+ Py > 0
dx
dx
Pdx
and convert into
Multiply by integrating factor, i.e. e ∫
total differential equation.
Here, f ′ (x) < 2 f (x), multiplying by e− ∫ 2dx
d
( f (x) ⋅ e−2x ) < 0
f ′ (x) ⋅ e−2x − 2e−2x f (x) < 0 ⇒
dx
1 
∴ φ (x) = f (x)e−2x is decreasing for x ∈  , 1
2 
1
Thus, when x >
2
 1
 1
φ (x) < φ   ⇒ e−2x f (x) < e−1 ⋅ f  
 2
 2
Differential Equations 351
⇒
 1
f (x) < e2x −1 ⋅ 1, given f   = 1
 2
1
0<∫
1
⇒
0<∫
1
⇒
e−1
0 < ∫ f (x) dx <
1/ 2
2
⇒
1/ 2
1/ 2
f (x) dx < ∫
1/ 2
19. Given,
c=0
∴
1
lim
t f (x) − x f (t )
=1
t−x
x2f ′ (x) − 2x f (x) + 1 = 0
⇒
x2f ′ (x) − 2x f (x) 1
+ 4 =0
(x2)2
x
y(x) = ∫ 2(x) dx + C
⇒
⇒
yx = x2 + C
C
y=x+
x
∴
∴ Option (a) is correct.
 1
(b) lim x f   = lim (1 + Cx2) = 1
 x x → 0 +
x → 0+
∴ Option (b) is incorrect.
On integrating both sides, we get
x
=− y+ c
y
(c) lim x2f ′ (x) = lim (x2 − C ) = − C ≠ 0
x → 0+
For C > 0,
…(i)
∴ Option (d) is incorrect.
23. PLAN
c=2
x
Now, Eq. (i) becomes, + y = 2
y
–3 + y2 = 2 y
⇒
( y + 1) ( y − 3) = 0
⇒
∴
As y > 0, take y = 3, neglecting y = − 1.
dy  t 
1
21. Given,
and y (0) = − 1
−
 y=
dt  1 + t 
(1 + t )
Which represents linear differential equation of first
order.

t

∫ −  1 + t  dt
= e− t + log (1 + t ) = e− t ⋅ (1 + t )
∴
IF = e 
Required solution is,
1
ye− t (1 + t ) = ∫
⋅ e− t (1 + t ) dt + c = ∫ e− t dt + c
1+ t
⇒
ye
−t
Linear differential equation under one variable.
Pdx
dy
+ Py = Q; IF = e ∫
dx
∴ Solution is, y (IF) = ∫ Q ⋅ (IF) dx + C
Again, for x = − 3
y2 − 2 y − 3 = 0
lim f (x) = ∞
x → 0+
∴ Function is not bounded in (0, 2).
⇒ x = 1, y = 1
⇒
x → 0+
∴ Option (c) is incorrect.
C
(d) f (x) = x + , C ≠ 0
x
∴
⇒
[QC ≠ 0, as f (1) ≠ 1]
 1
(a) lim f ′   = lim (1 − Cx2) = 1
+
 x x → 0 +
x→ 0
x dy − y dx = y2dy
x dy − y dx
 x
= dy ⇒ d   = − dy
 y
y2
y (1) = 1
f (x)
x
∴ Required solution is y ⋅ (IF) = ∫ Q (IF)dx + C
20. Given, x dy = y(dx + y dy), y > 0
Since,
−t
(1 + t ) = − e
1
2
1
On integrating both sides, we get
1
f (x) = cx2 +
3x
2
Also, f (1) = 1,
c=
3
2
1
Hence,
f (x) = x2 +
3
3x
⇒
y (1) = −
∫ dx
Integrating Factor, IF = e x = elog x = x
d  f (x)
1
 =− 4

dx  x2 
x
⇒
1
⇒
(1 + t )
dy y
+ = 2 [i.e. linear differential equation in y]
dx x
or
2
⇒
⇒
y=−
22. Here, f ′ (x) = 2 −
1
t→ x
−1 ⋅ e0 (1 + 0) = − e0 + c
e2x − 1 dx
 e2x − 1 
f (x) dx < 

 2  1/ 2
2
y (0) = − 1
Since,
⇒
+c
y ′ − y tan x = 2x sec x and y (0) = 0
dy
− y tan x = 2x sec x
dx
IF = ∫ e− tan x dx = elog|cos x| = cos x
Solution is y ⋅ cos x = ∫ 2x sec x ⋅ cos x dx + C
⇒
As
∴
Now,
⇒
y ⋅ cos x = x2 + C
y(0) = 0 ⇒ C = 0
y = x2 sec x
π2
 π
y  =
 4 8 2
π
π2
 π
+
y′   =
 4
2 8 2
2
 π  2π
y  =
 3
9
⇒
2
 π  4π 2π
+
y′   =
 3
3 3 3
352 Differential Equations
24. Let
w (x) = u (x) − v (x)
and
h (x) = f (x) − g (x)
…(i)
On differentiating Eq. (i) w.r.t. x
dw du dv
=
−
dx dx dx
[given]
= { f (x) − g (x)} − p (x) [u (x) − v (x)]
dw
…(ii)
⇒
= h (x) − p (x) ⋅ w (x)
dx
dw
+ p (x) w (x) = h (x) which is linear differential
⇒
dx
equation .
The integrating factor is given by
p ( x ) dx
IF = e∫
= r (x)
[let]
On multiplying both sides of Eq. (ii) of r (x), we get
dw
r (x) ⋅
+ p (x) (r (x)) w (x) = r (x) ⋅ h (x)
dx
d

 dr
[r (x) w (x)] = r (x) ⋅ h (x)
⇒
Q dx = p (x) ⋅ r (x)
dx
and
Thus,
r (x) = e∫
P ( x ) dx
> 0, ∀ x
h (x) = f (x) − g (x) > 0, for x > x1
d
[r (x) w (x)] > 0, ∀ x > x1
dx
r (x) w (x) increases on the interval [x, ∞ [
Therefore, for all x > x1
r (x) w (x) > r (x1 ) w (x1 ) > 0
[Q r (x1 ) > 0 and u (x1) > v (x1)]
⇒
w (x) > 0 ∀ x > x1
⇒
u (x) > v (x) ∀ x > x1
[Q r (x) > 0]
Hence, there cannot exist a point (x, y) such that x > x1
and y = u (x) and y = v (x).
25.
dy
+ y ⋅ g′ (x) = g (x) g′ (x)
dx
g ′( x ) dx
IF = e∫
= e g( x )
∴ Solution is y (e g( x ) ) = ∫ g (x) ⋅ g′ (x) ⋅ e g( x ) dx + C
Put
g (x) = t, g′ (x) dx = dt
y(eg( x ) ) = ∫ t ⋅ et dt + C
= t ⋅ et − ∫ 1 ⋅ et dt + C = t ⋅ et − et + C
…(i)
ye
= ( g (x) − 1) e
+C
Given,
y(0) = 0, g (0) = g (2) = 0
∴ Eq. (i) becomes,
y(0) ⋅ eg( 0) = ( g (0) − 1) ⋅ eg( 0) + C
⇒
0 = (− 1) ⋅ 1 + C ⇒ C = 1
∴
y(x) ⋅ eg( x ) = ( g (x) − 1) eg( x ) + 1
⇒
y(2) ⋅ eg( 2) = ( g (2) − 1) eg( 2) + 1, where g(2) = 0
⇒
y(2) ⋅ 1 = (− 1) ⋅ 1 + 1
y(2) = 0
g( x )
dy 2 y
=
dx x2
2
dy
[integrating both sides]
⇒
∫ y = ∫ x2dx
2
…(i)
⇒
log e| y| = − + C
x
Since, curve (i) passes through centre (1, 1) of the circle
x2 + y2 − 2x − 2 y = 0
2
log e (1) = − + C ⇒ C = 2
∴
1
∴ Equation required curve is
2
[put C = 2 in Eq. (i)]
log e| y| = − + 2
x
1. Given,
= { f (x) − p (x) ⋅ u (x)} − { g (x) − p (x) v (x)}
Now,
Topic 3 Applications of Homogeneous
Differential Equations
g( x )
⇒
x log e| y| = 2(x − 1)
2. Given differential equation is
(x2 − y2)dx + 2xy dy = 0, which can be written as
dy y2 − x2
=
2xy
dx
Put y = vx
[Q it is in homogeneous form]
dy
dv
⇒
= v+ x
dx
dx
Now, differential equation becomes
dv v2x2 − x2
dv (v2 − 1)x2
⇒ v+ x
=
v+ x
=
2x(vx)
dx
dx
2vx2
⇒
⇒
⇒
dv v2 − 1
v2 − 1 − 2v2
=
−v=
dx
2v
2v
dx
2v dv
dv
1 + v2
x
=−
⇒ ∫
= −∫
2
dx
2v
x
1+ v
ln (1 + v2) = − ln x − ln C
f ′ (x)


Q ∫ f (x) dx ⇒ ln| f (x)|+ C 


x
⇒ ln|(1 + v2)Cx|= 0
[Q ln A + ln B = ln AB]
2
⇒
(1 + v )Cx = 1
[log e x = 0 ⇒ x = e0 = 1]
y
Now, putting v = , we get
x

y2 
1 + 2  Cx = 1 ⇒ C (x2 + y2) = x
x 

Q The curve passes through (1, 1), so
1
C (1 + 1) = 1 ⇒ C =
2
Thus, required curve is x2 + y2 − 2x = 0, which represent
a circle having centre (1, 0)
∴ The solution of given differential equation represents
a circle with centre on the X-axis.
dp 1
3. Given, differential equation is
− p(t ) = −200 is a
dt 2
linear differential equation.
Differential Equations 353
p(t ) =
Here,
−1
, Q (t ) = −200
2
 1
⇒
[(x + 2)2 + y(x + 2)]
Put x + 2 = X and y = Y , then
t
−
∫ −   dt
IF = e  2 = e 2
(X 2 + XY )
Hence, solution is
p (t ) ⋅ IF = ∫ Q (t ) ⋅ IF dt
−
p(t )⋅ e
t
2
t
−
p(t )⋅ e 2
−
= ∫ −200 ⋅ e
t
−
= 400 e 2
t
2dt
+K
⇒
p(t ) = 400
4. PLAN
t
− 300 e2
To solve homogeneous differential equation, i.e. substitute
y
=v
x
dy
dv
∴ y = vx ⇒
=v +x
dx
dx
Here, slope of the curve at (x, y) is
dy y
 y
= + sec  
 x
dx x
y
=v
x
dv
v+ x
= v + sec (v)
dx
dv
dx
∫ sec v = ∫ x
Put
∴
⇒
log c =
5. Given,
⇒ x
⇒
dv
= sec (v)
dx
∴
∫ cos v dv = ∫
⇒
dx
x
dP
= (100 − 12 x ) ⇒dP = (100 − 12 x ) dx
dx
dP = ∫ (100 − 12 x ) dx
P = 100x − 8x
Since, it passes through the point (1, 3).
− log 3 = 1 + C
C = − 1 − log 3 = − (log e + log 3)
y
− log (3e) = 0
x+2
y
| y|
=0
log   +
 3e  x + 2
log| y| +
1
2
3/ 2
On integrating both sides, we get
Y
− log|Y| =
+ C, where x + 2 = X and y = Y
X
y
…(i)
− log| y| =
+C
⇒
x+ 2
= − log 3e
On integrating both sides, we get
∫
X 2dY + Y (XdY − YdX ) = 0
dY XdY − YdX
−
=
Y
X2
Y 
− d (log|Y |) = d  
 X
∴ Eq. (i) becomes
1
 y
sin   = log x +
 x
2
∴
⇒
⇒
 y
⇒
sin v = log x + log c ⇒ sin   = log(cx)
 x
 π
 π
As it passes through 1,  ⇒ sin   = log c
 6
 6
⇒
X 2dY + XYdY − Y 2dX = 0
⇒
If p(0) = 100, then k = − 300
+C
When x = 0, then P = 2000 ⇒ C = 2000
Now, when x = 25, then is
P = 100 × 25 − 8 × (25)3/ 2 + 2000
⇒
= 4500 − 1000 = 3500
dy
(x2 + xy + 4x + 2 y + 4)
− y2 = 0
dx
dy
[(x2 + 4x + 4) + y(x + 2)]
− y2 = 0
dx
…(ii)
Now, to check option (a), y = x + 2 intersects the curve.
|x + 2| x + 2
|x + 2|
= 0 ⇒ log 
⇒
log 
+
 = −1
 3e  x + 2
 3e 
⇒
|x + 2|
1
= e−1 =
3e
e
⇒
|x + 2| = 3 or x + 2 = ± 3
∴ x = 1, − 5 (rejected), as x > 0
[given]
∴ x = 1 only one solution.
Thus, (a) is the correct answer.
To check option (c), we have
y
| y|
=0
y = (x + 2)2 and log   +
 3e  x + 2
|x + 2|2  (x + 2 )2
|x + 2|2 
⇒ log 
= 0 ⇒ log 
+
 = − (x + 2)
x+2
 3e 
 3e 
⇒
3e
(x + 2)2
= e−( x + 2) or (x + 2)2 ⋅ ex + 2 = 3e ⇒ex+ 2 =
3e
(x + 2)2
Y
= 2500 − 8 × 125 + 2000
6. Given,
dY
−Y 2 = 0
dX
⇒
⇒
p(t ) = 400 + ke−1/ 2
⇒
dy
− y2 = 0
dx
e x +2
e2
3e /4
O
3e /( x + 2)2
X
354 Differential Equations
Clearly, they have no solution.
To check option (d), y = (x + 3)
|x + 3|2  (x + 3)2
i.e.
log 
=0
+
 3e  (x + 2)
2
To check the number of solutions.
(x + 3)
− log (3e)
(x + 2)
2
Let g (x) = 2 log (x + 3) +
∴
 (x + 2) ⋅ 2 (x + 3) − (x + 3)2 ⋅ 1
2
+
g′ (x) =
 −0
x+ 3 
(x + 2)2

(x + 3)(x + 1)
2
=
+
x+3
(x + 2)2
Clearly, when x > 0, then, g′ (x) > 0
∴
g (x) is increasing, when x > 0.
when x > 0, then g (x) > g (0)
 3 9
g (x) > log   + > 0
 e 4
Thus,
Hence, there is no solution. Thus, option (d) is true.
7. Since, BP : AP = 3 : 1. Then, equation of tangent is
Y − y = f ′ (x) (X − x)
The intercept on the coordinate axes are


y
A x −
, 0
f ′ (x) 

dr
= − λ is required differential equation.
dt
 dx

 dy
9. Since, the length of tangent = y 1 + 
Since, P is internally intercepts a line AB,

y 
3 x −
 + 1 ×0
f ′ (x)

x=
3+1
∴
Y
B
∴
dy
=±
dx
⇒
1 − y2
dy = ±
y
P
1 − y2
dy = ± x + C
y
∫
⇒
∫ x dx
Put y = sin θ ⇒ dy = cos θ dθ
cos θ
∴
∫ sin θ ⋅ cos θ dθ = ± x + C
⇒
∫
cos 2 θ
⋅ sin θ dθ = ± x + C
sin 2 θ
Again put cos θ = t ⇒ − sin θ dθ = dt
−∫
∴
⇒
∫
t2
dt = ± x + C
1 − t2

1 
1 −
 dt = ± x + C
1 − t 2

⇒
⇒ 1 − y2 − log
t − log
1+ t
=± x+C
1−t
1 + 1 − y2
=±x+C
1 − 1 − y2
surface area.
(x, y)
y = f (x)
1
dy
y
=
dx − 3x
dV
∝ −S
dt
⇒
X
A
⇒
∫
y
1 − y2
10. Given, liquid evaporates at a rate proportional to its
(1, 1)
3
=1
2

 dx 
y2  1 +    = 1

 dy 

⇒
B [0, y − x f ′ (x)]
and
2
⇒
dy
1
=−
dx
y
3x
We know that, volume of cone =
…(i)
1 2
πr h
3
R
On integrating both sides, we get
xy3 = c
r
Since, curve passes through (1, 1), then c = 1.
∴
1
⇒ y=2
8
Hence, (a) and (c) are correct answers.
At
x=
8. Since, rate of change of volume ∝ surface area
⇒
⇒
H
xy = 1
3
dV
∝ SA
dt
dr
4 πr 2 ⋅
= − λ 4 πr 2
dt
h
and surface area = πr 2
1
or
V = πr 2h and
S = πr 2
3
R
r
Where,
tan θ =
= tan θ
and
H
h
…(ii)
…(iii)
Differential Equations 355
From Eqs. (ii) and (iii), we get
1
V = πr3 cot θ and
3
S = πr
2
cot θ ∫
⇒
⇒
dr = − k ∫
R
T
R cot θ = kT
H
T=
k
⇒
−2πr3
A (0.6) 2 gr
∫1
⇒
T=
−2πr3
A (0.6) 2 gr
∫1
⇒
T=
2πr3
A (0.6) 2 gr
∫1
⇒
T=
2πr3
A (0.6) 2 gr
 t15 2t13 
 −

3 1
5
⇒
T=
⇒
T=
dt
0
cot θ (0 − R) = − k (T − 0)
⇒
T=
…(iv)
On substituting Eq. (iv) in Eq. (i), we get
1
dr
cot θ ⋅ 3r 2
= − kπ r 2
3
dt
0
⇒
⇒ H = kT [from Eq. (iii)]
∴ Required time after which the cone is empty, T =
H
k
11. Let O be the centre of hemispherical tank. Let at any
instant t, water level be BAB1 and at t + dt, water level
is B′ A′ B1. Let ∠ O1OB1 = θ.
O
A
B
B'
B1
B ′1
A′
[1 − (1 + t14 − 2t12)] dt1
0
[1 − 1 − t14 + 2t12] dt1
0
(t14 − 2t12) dt1
0
1
2

⋅ 0 − − 0 + 
5
3

2π ⋅ 25/ 2 (102) 5/ 2 2 1 
 − 
3
12 ⋅ ⋅ 2 ⋅ g 3 5 
5
=
2π × 105 ⋅ 4 ⋅ 5 10 − 3 
(12 × 3) g  15 
=
2π × 105 × 7 14π × 105
unit
=
3⋅3⋅ g ⋅3
27 g
O1
θ
2π ⋅ r5/ 2
6
A   2 gr
 10
0
12. Let X 0 be initial population of the country andY 0 be its
O2
⇒ AB1 = r cos θ and OA = r sin θ decrease in the water
volume in time dt = π AB12 ⋅ d (OA )
[πr 2 is surface area of water level and d (OA ) is depth
of water level]
initial food production. Let the average consumption be
a unit. Therefore, food required initially aX 0. It is given
 90 
…(i)
Y p = aX 0 
 = 0.9 aX 0
 100
Now, outflow rate Q = A ⋅ v (t ) = A ⋅ 0.6 2 gr (1 − sin θ )
Let X be the population of the country in year t.
dX
Then,
= Rate of change of population
dt
3
=
X = 0.03 X
100
dX
dX
⇒
= ∫ 0.03 dt
= 0.03 dt ⇒ ∫
X
X
Where, A is the area of the outlet.
⇒
Thus, volume flowing out in time dt.
⇒
= πr 2 ⋅ cos 2 θ ⋅ r cos θ dθ
= πr3 ⋅ cos3 θ dθ
Also, h (t ) = O2A = r − r sin θ = r (1 − sin θ )
Q dt = A ⋅ (0.6) ⋅ 2 gr ⋅ 1 − sin θ dt
⇒
We have, πr3 cos3 θ dθ = A (0.6) ⋅ 2 gr ⋅ 1 − sin θ dt
⇒
πr3
⋅
A (0.6) 2 gr
cos3 θ
dθ = dt
(1 − sin θ )
T =∫
π/ 2
0
cos3 θ
πr3
⋅
dθ
A (0.6) ⋅ 2 gr
1 − sin θ
− πr3
=
A (0.6) 2 gr
Let
⇒
∴
π/ 2 1
∫0
− sin 2 θ (− cos θ )
dθ
1 − sin θ
t1 = 1 − sin θ
− cos θ
dt1 =
dθ
1 − sin θ
T=
−2 πr
A (0.6) 2 gr
3
X = A ⋅ e0. 03 t , where A = ec
At t = 0, X = X 0, thus X 0 = A
∴
X = X 0 e0. 03 t
Let Y be the food production in year t.
t
Then,
4 

t
Y = Y 0 1 +
 = 0.9aX 0 (1.04)

100
Q
Y 0 = 0.9 aX 0
Let the time taken to empty the tank be T.
Then,
log X = 0.03 t + c
[from Eq. (i)]
0. 03 t
Food consumption in the year t is aX 0 e
Again,
⇒
⇒
Y − X ≥0
[given]
0.9 X 0 a (1.04) > a X 0 e
t
0. 03 t
(1.04)t
1
10
>
= .
9
e0. 03 t 0.9
Taking log on both sides, we get
0
∫1
[1 − (1 − t12)2] dt1
.
t[log (1.04) − 0.03] ≥ log 10 − log 9
log 10 − log 9
⇒
t≥
log (1.04) − 0.03
356 Differential Equations
Thus, the least integral values of the year n, when the
country becomes self-sufficient is the smallest integer
log 10 − log 9
greater than or equal to
.
log (1.04) − 0.03
14.
dV
∝ V for each reservoir.
dt
dV
∝ − VA ⇒
dx
13. Equation of normal at point (x, y) is
dx
Y − y=−
(X − x)
dy
…(i)
Distance of perpendicular from the origin to Eq. (i)
dx
y+
⋅x
dy
=
2
 dx
1+  
 dy
Also, distance between P and X-axis is |y|.
dx
y+
⋅x
dy
= | y|
∴
2
 dx
1+  
 dy
⇒
y2 +
2

dx 2
dx
 dx 
⋅ x + 2xy
= y2 1 +   
 dy 
dy
dy

2
⇒
⇒
dx
 dx
2
2
=0
  (x − y ) + 2xy
 dy
dy
dx
dy

 dx 2
2
 dy (x − y ) + 2xy = 0


dx
= 0 or
dy
dx
But
=0
dy
⇒
dy y2 − x2
=
2xy
dx
⇒x
⇒
⇒
⇒
dv v2 − 1
v2 − 1 − 2v2
v2 + 1
=
−v=
=−
dx
2v
2v
2v
dx
−2 v
dv =
x
v2 + 1
c1 − log (v + 1) = log|x|
2
log| x|(v2 + 1) = c1

 y2
⇒ | x|  2 + 1 = ec1

x
⇒ x2 + y2 = ± ec1 x or x2 + y2 = ± ecx is passing through
(1, 1).
∴
1 + 1 = ± ec ⋅ 1
⇒
± ec = 2
Hence, required curve is x2 + y2 = 2x .
∫V
⇒
⇒
log
A
[K 1 is the proportional constant]
t
dV A
= − K 1 ∫ dt
0
VA
V ′A
= − K 1t
VA
Similarly for B,
⇒ V A′ = V A ⋅ e−K1 t
VB′ = VB ⋅ e−K 2t
…(i)
…(ii)
On dividing Eq. (i) by Eq. (ii), we get
V A′ V A − (K1
=
⋅e
VB′ VB
− K2) t
It is given that at t = 0, V A = 2 VB and at
3
3
t = , V A′ = VB′
2
2
3
3
− (K1 − K 2 )t
Thus,
…(iii)
= 2⋅ e
⇒ e− (K1 − K 2 ) =
2
4
Now, let at t = t0 both the reservoirs have some quantity
of water. Then,
V A′ = VB′
−(K − K 2 )
From Eq. (iii), 2e
=1
 3
2⋅  
 4
⇒
t0
=1
t0 = log3/ 4 (1 / 2)
15. Given,
Let
⇒
⇒ x = c, where c is a constant.
Since, curve passes through (1, 1), we get the equation
of the curve as x = 1.
dy y2 − x2
is a homogeneous equation.
The equation
=
2xy
dx
dy
dv
Put
y = vx ⇒
=v+ x
dx
dx
dv v2x2 − x2
=
v+ x
dx
2x2v
V ′A
dV A
= − K 1V A
dt
dy
= sin (10x + 6 y)
dx
10x + 6 y = t
dy  dt 
10 + 6
= 
dx  dx
…(i)
dy 1  dt

= 
− 10

dx 6  dx
⇒
Now, the given differential equation becomes
1  dt

sin t = 
− 10

6  dx
dt
⇒
6 sin t =
− 10
dx
dt
⇒
= 6 sin t + 10
dx
dt
⇒
= dx
6 sin t + 10
On integrating both sides, we get
1
dt
=x+ c
2 ∫ 3 sin t + 5
Let
I1 = ∫
dt
=
3 sin t + 5 ∫
=∫
…(ii)
dt
 2 tan t / 2 
3
 +5
2
 1 + tan t / 2
(1 + tan 2 t / 2) dt
t

2 t
6 tan + 5 + 5tan 

2
2
Differential Equations 357
Put tan t /2 = u
1
2 du
sec2 t / 2 dt = du ⇒ dt =
⇒
2
sec2 t / 2
2 du
1 + tan 2 t / 2
⇒
dt =
∴
I1 = ∫
2
5
⇒ dt =
17. A. I = ∫
=∫
2 du
1 + u2
2 (1 + u 2)du
2
=
(1 + u 2) (5u 2 + 6u + 5) 5
∫
=
∫
du
6
u + u+1
5
(– 4, 1)
2
1
5u + 3  1
5 tan t / 2 + 3 
= tan −1 
tan −1 


2
4
2
4



t


5 tan + 3 
1
−1 
2
tan 
=x+ c
4
4




⇒
⇒
tan −1
X′
⇒
Then,
Y′
1
1

y3 
4
= 2 ∫ (1 − y2)dy = 2  y −
 = sq units
0
3
0 3

C. The point of intersection y = 3x −1 log x and y = xx − 1
is (1, 0).
Hence,
1
[5 tan (5x + 3 y) + 3] = tan (4x + 4c)
4
⇒
⇒
⇒
3 y = tan
y=
−1
 dy
=1
 
 dx (1, 0)
∴
If θ is angle between the curves, then tan θ = 0.

5 tan (5x + 3 y) + 3 = 4 tan 4x + tan −1

⇒
3

4
⇒

−1 3   3 
− 
4x + tan

4  5 
⇒
⇒
x + y + 2 = key/ 2
It passing through (1, 0).
⇒
k =3
4 
3

−1 3  
  −  − 5x
 tan 4x + tan

5
4
 5
 
4 
1

tan −1  tan 4x + tan −1

3
5 
θ = 0°
dy
2
dx x y
⇒
− =
=
dx x + y
dy 2 2
1
xe− y/ 2 = ⋅ ∫ y ⋅ e− y/ 2dy
2
 ye− y 2
1
e− y/ 2 
xe− y/ 2 = 
−
+ k
2  − 1 / 2 (1 / 2)2 
D.
4
3 3

tan 4x + tan  −

5
4 5

tan

y = xx − 1
dy
= xx (1 + log x)
dx
For
5 tan 0 + 3 = 4 tan (4c)
3
= tan 4c
4
3
4c = tan −1
4
dy 3x −1
=
+ 3x −1 ⋅ log 3 ⋅ log x
dx
x
 dy
=1
 
 dx (1, 0)
∴
5 tan (5x + 3 y) + 3 = 4 tan (4x + 4c)
4
⇒ 5x + 3 y = tan −1 
5
X
∴Required area
1
1
= 2 ∫ (1 − 5 y2) dy − ∫ − 4 y2 dy
0
0


t


5 tan 2 + 3 

 = 4x + 4c
4




⇒ tan (5x + 3 y) =
(1, 0)
(– 4, –1)
When x = 0, y = 0, we get
⇒
d
(sin x)cos x dx = 1
dx
Y
2
On putting this in Eq. (ii), we get
⇒
π /2
0
x
(sin x)cos {cos x ⋅ cot x − log (sin x)sin x } dx
B. The point of intersection of − 4 y2 = x and
x − 1 = − 5 y2 is (− 4, − 1) and (− 4, 1).
du
6
9
9
−
+1
u + u+
5
25 25
2 5
2
du
 u + 3 / 5
= ∫
= ⋅ tan −1 

2
2
 4 /5 
5
4
5 
3
 4
u +  +  
 5

5
=
π /2
0
3  3  5x
 −  −
4  5  3
∴
Download Chapter Test
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x + y + 2 = 3e y/ 2
or
15
Straight Line and
Pair of Straight Lines
Topic 1 Various Forms of Straight Line
Objective Questions I (Only one correct option)
1. A straight line L at a distance of 4 units from the origin
makes positive intercepts on the coordinate axes and the
perpendicular from the origin to this line makes an angle
of 60° with the line x + y = 0. Then, an equation of the line
L is
(2019 Main, 12 April II)
(a) x + 3 y = 8
(b) ( 3 + 1) x + ( 3 − 1) y = 8 2
(c) 3x + y = 8
(d) ( 3 − 1)x + ( 3 + 1) y = 8 2
(a)
(b)
(c)
(d)
(2019 Main, 12 April I)
second and third quadrants only
first, second and fourth quadrants
first, third and fourth quadrants
third and fourth quadrants only
3. Lines are drawn parallel to the line 4x − 3 y + 2 = 0, at a
3
from the origin. Then which one of the
5
following points lies on any of these lines?
distance
(2019 Main, 10 April I)
1
2
(a)  − , − 
 4
3
1
1

(c)  , − 
4
3
1 2
(b)  − , 
 4 3
1 1
(d)  , 
 4 3
(a) rhombus of side length 2 units
(2019 Main, 9 April II)
2
(a)
5
2
(c)
5
2
(b)
5
2
(d)
5
intersecting the line, x + y = 7 at a distance of 4 units
from P, is
(2019 Main, 9 April I)
(a)
1−
1+
5
5
(b)
7−1
7+1
(c)
1−
1+
7
7
(d)
5−1
5+1
7. If a point R(4, y, z ) lies on the line segment joining the
points P(2, − 3, 4) and Q(8, 0, 10), then the distance of R
from the origin is
(2019 Main, 8 April II)
(a) 2 21
(c) 2 14
8. Suppose that the points (h , k), (1, 2) and (−3, 4) lie on
(2019 Main, 8 April II)
(2019 Main, 10 April I)
1
(a) −
7
(c) 3
1
(b)
3
(d) 0
9. A point on the straight line, 3x + 5 y = 15 which is
equidistant from the coordinate axes will lie only in
(b) rhombus of area 8 2 sq units
(c) square of side length 2 2 units
(d) square of area 16 sq units
(b) 53
(d) 6
the line L1. If a line L 2 passing through the points (h , k)
and (4, 3) is perpendicular to L1, then k /h equals
4. The region represented by |x − y| ≤ 2 and |x + y| ≤ 2 is
bounded by a
2x + a 2y = 1, (a ∈ R − {0, 1}) are perpendicular, then the
distance of their point of intersection from the origin is
6. Slope of a line passing through P (2, 3) and
2. The equation y = sin x sin(x + 2) − sin 2(x + 1) represents a
straight line lying in
5. If the two lines x + (a − 1) y = 1 and
(2019 Main, 8 April I)
(a) IV quadrant
(c) I and II quadrants
(b) I quadrant
(d) I, II and IV quadrants
Straight Line and Pair of Straight Lines 359
10. If a straight line passing through the point P(− 3, 4) is
such that its intercepted portion between the
coordinate axes is bisected at P, then its equation is
(2019 Main, 12 Jan II)
(a)
(b)
(c)
(d)
x− y+ 7= 0
4x − 3 y + 24 = 0
3x − 4 y + 25 = 0
4x + 3 y = 0
(b) − 5
(c) −
35
3
(d) 5
12. If in a parallelogram ABDC, the coordinates of A , B and
C are respectively (1, 2), (3, 4) and (2, 5), then the
equation of the diagonal AD is
(2019 Main, 11 Jan II)
(a)
(b)
(c)
(d)
3x + 5 y − 13 = 0
3x − 5 y + 7 = 0
5x − 3 y + 1 = 0
5x + 3 y − 11 = 0
(2019 Main, 10 Jan II)
4
(a)  , 2e
3

(b) (3, 6e)
(c) (2, 3e)
5
(d)  , 2e
3

(b) (2, 6)
(d) (3, 5)
15. The shortest distance between the point  , 0 and the
3
2
curve y = x , (x > 0), is

(2019 Main, 10 Jan I)
(c)
3
2
(d)
5
2
16. If the line 3x + 4 y − 24 = 0 intersects the X-axis at the
point A and theY -axis at the point B, then the incentre
of the triangle OAB, where O is the origin, is
(2019 Main, 10 Jan I)
(b) (3, 4)
(d) (2, 2)
17. A point P moves on the line 2x − 3 y + 4 = 0. If Q(1, 4) and
R(3, − 2) are fixed points, then the locus of the centroid of
(2019 Main, 10 Jan I)
∆PQR is a line
2
3
(c) parallel to Y -axis
3
2
(d) parallel to X-axis
(b) with slope
18. A straight line through a fixed point (2, 3) intersects the
coordinate axes at distinct points P and Q. If O is the
origin and the rectangle OPRQ is completed, then the
(2018 Main)
locus of R is
(a) 3x + 2 y = 6
(c) 3x + 2 y = xy
(a) 2bc − 3ad = 0
(c) 2ad − 3bc = 0
(b) 2bc + 3ad = 0
(d) 3bc + 2ad = 0
21. If PS is the median of the triangle with vertices P(2, 2),
Q(6, – 1) and R(7, 3), then equation of the line passing
(2014 Main, 2000)
through (1, – 1) and parallel to PS is
(a) 4x − 7 y − 11 = 0
(c) 4x + 7 y + 3 = 0
(b) 2x + 9 y + 7 = 0
(d) 2x − 9 y − 11 = 0
(a) 2 +
2
(b) 2 −
2
(c) 1 +
2
(d) 1 −
2
an angle 60° to the line 3x + y = 1. If L also intersects
(2011)
the X-axis, then the equation of L is
(b) y − 3x + 2 + 3 3 = 0
(d) 3 y + x − 3 + 2 3 = 0
24. The locus of the orthocentre of the triangle formed by
x + y = 3 and x − y + 3 = 0. If its diagonals intersect at (2,
4), then one of its vertex is
(2019 Main, 10 Jan II)
(a) with slope
1

2
intersection of the lines 4ax + 2ay + c = 0 and
5bx + 2by + d = 0 lies in the fourth quadrant and is
equidistant from the two axes, then
(2014 Main)
(a) y + 3x + 2 − 3 3 = 0
(c) 3 y − x + 3 + 2 3 = 0
14. Two sides of a parallelogram are along the lines,
(a) (4, 3)
(c) (4, 4)
(d)  2,

23. A straight line L through the point (3, −2) is inclined at
the point (1, e) also passes through the point
5
(b)
4
3
(c)  1, − 

4
the coordinates of mid-points of its sides as (0, 1), (1, 1)
and (1, 0) is
(2013 Main)
2
3
(a)
2
3

4
22. The x-coordinate of the incentre of the triangle that has
13. The tangent to the curve, y = xex passing through
(a) (3, 6)
(c) (2, 1)
(b)  1,

20. Let a , b, c and d be non-zero numbers. If the point of
the line passing through the points (7, 17) and (15, β),
then β equals
(2019 Main, 12 Jan I)
35
3
(k, − 3k), (5, k) and (− k, 2) has area
28 sq units. Then, the orthocentre of this triangle is at
the point
(2017 Main)
1
(a)  2, − 

2
11. If the straight line, 2x − 3 y + 17 = 0 is perpendicular to
(a)
19. Let k be an integer such that the triangle with vertices
(b) 2x + 3 y = xy
(d) 3x + 2 y = 6xy
the lines (1 + p) x − py + p (1 + p) = 0 ,
(1 + q) x − qy + q (1 + q) = 0 and y = 0, where p ≠ q , is (2009)
(a) a hyperbola
(b) a parabola
(c) an ellipse
(d) a straight line
25. Let O(0, 0), P(3, 4) and Q(6, 0) be the vertices of a ∆OPQ.
The point R inside the ∆OPQ is such that the triangles
OPR, PQR and OQR are of equal area. The coordinates
of R are
(2007, 3M)
4
(a)  , 3
3 
(b)  3,

2

3
(c)  3,

4

3
4
(d)  ,
3
2

3
26. Orthocentre of triangle with vertices (0, 0), (3, 4) and
(4, 0) is
5
(a)  3, 
 4
(2003, 2M)
(b) (3, 12)
3
(c)  3, 
 4
(d) (3, 9)
27. The number of integer values of m, for which the
x-coordinate of the point of intersection of the lines
3x + 4 y = 9 and y = mx + 1 is also an integer, is (2001, 1M)
(a) 2
(b) 0
(c) 4
(d) 1
28. A straight line through the origin O meets the parallel
lines 4x + 2 y = 9 and 2x + y + 6 = 0 at points P and Q
respectively. Then, the point O divides the segment PQ
(2000, 1M)
in the ratio
(a) 1 : 2
(b) 3 : 4
(c) 2 : 1
(d) 4 : 3
29. The incentre of the triangle with vertices (1, 3 ), (0, 0)
and (2, 0) is

3
(a)  1,

 2 
(2000, 2M)
2 1 
(b)  ,

 3 3
 2 3
(c)  ,

3 2 
(d)  1,

1 

3
360 Straight Line and Pair of Straight Lines
30. If A0 , A1 , A2, A3 , A4 and A5 be a regular hexagon
39. The point (4, 1) undergoes the following three
inscribed in a circle of unit radius. Then, the product of
the lengths of the line segments A0 A1 , A0 A2 and A0 A4 is
transformations successively
I. Reflection about the line y = x.
II. Transformation through a distance 2 units along
the positive direction of X-axis.
π
III. Rotation through an angle about the origin in the
4
counter clockwise direction.
Then, the final position of the point is given by the
coordinates
(1980, 1M)
(a) 3 / 4
(b) 3 3
3 3
(d)
2
(c) 3
(1998, 2M)
31. If the vertices P , Q , R of a ∆PQR are rational points,
which of the following points of the ∆PQR is/are always
(1998, 2M)
rational point(s)
(a) centroid
(c) circumcentre
(b) incentre
(d) orthocentre
(A rational point is a point both of whose coordinates are
rational numbers)
32. If P (1, 2), Q (4, 6), R (5, 7) and S (a , b) are the vertices of
a parallelogram PQRS, then
(a) a = 2, b = 4
(c) a = 2, b = 3
(1998, 2M)
(b) a = 3, b = 4
(d) a = 3, b = 5
33. The diagonals of a parallelogram PQRS are along the
lines x + 3 y = 4 and 6x − 2 y = 7. Then, PQRS must be a
(a) rectangle
(c) cyclic quadrilateral
(b) square
(d) rhombus
(1998, 2M)
34. The graph of the function cos x cos (x + 2) − cos (x + 1)
2
(1997, 2M)
is
(a) a straight line passing through (0, − sin 1) with slope 2
(b) a straight line passing through (0, 0)
(c) a parabola with vertex (1, − sin 2 1)
π
(d) a straight line passing through the point  , − sin 2 1
2

2
and parallel to the X-axis
35. The orthocentre of the triangle formed by the lines
xy = 0 and x + y = 1,is
1
(a)  ,
2
1

2
(c) (0, 0)
(1995, 2M)
1 1
(b)  , 
 3 3
1 1
(d)  , 
 4 4
perpendicular lines in a plane is 1, then its locus is
(b) circle
(1992, 2M)
(d) two intersecting lines
37. Line L has intercepts a and b on the coordinate axes.
When, the axes are rotated through a given angle,
keeping the origin fixed, the same line L has intercepts
(1990, 2M)
p and q, then
(a) a 2 + b2 = p 2 + q2
(b)
(c) a 2 + p 2 = b2 + q2
(d)
1
a2
1
a
2
+
+
1
b2
1
p
(b) (−
2, 7 2)
(d) ( 2, 7 2)
40. The points (−a , − b), (0, 0), (a , b) and (a 2, a3 ) are
(a) collinear
(b) vertices of a rectangle
(c) vertices of a parallelogram
(d) None of the above
(1979, 2M)
Objective Questions II
(One or more than one correct option)
41. Let a, λ, µ ∈R. Consider the system of linear equations
ax + 2 y = λ and 3x − 2 y = µ.
Which of the following statement(s) is/are correct?
(2016 Adv.)
(a) If a = − 3, then the system has infinitely many
solutions for all values of λ and µ
(b) If a ≠ − 3, then the system has a unique solution for all
values of λ and µ
(c) If λ + µ = 0, then the system has infinitely many
solutions for a = − 3
(d) If λ + µ ≠ 0, then the system has no solution for a = − 3
42. For a > b > c > 0, the distance between (1, 1) and the
36. If the sum of the distance of a point from two
(a) square
(c) straight line
1 7 
(a) 
,

 2 2
1 7 
(c)  −
,


2 2
2
=
=
1
p2
1
2
b
+
+
1
q2
1
2
q
38. If P = (1, 0), Q = (−1, 0) and R = (2, 0) are three given
point of intersection of the lines ax + by + c = 0 and
(2014 Adv.)
bx + ay + c = 0 is less than 2 2. Then,
(a) a + b − c > 0
(c) a − b + c > 0
(b) a − b + c < 0
(d) a + b − c < 0
43. All points lying inside the triangle formed by the points
(1, 3), (5, 0) and (– 1, 2) satisfy
(a) 3x + 2 y ≥ 0
(c) 2x − 3 y − 12 ≤ 0
(1986, 2M)
(b) 2x + y − 13 ≥ 0
(d) −2x + y ≥ 0
Fill in the Blanks
44. Let the algebraic sum of the perpendicular distance
from the points (2, 0) , (0, 2) and (1, 1) to a variable
straight line be zero, then the line passes through a
fixed point whose coordinates are… .
(1991, 2M)
points, then locus of the points satisfying the relation
(1988, 2M)
SQ 2 + SR2 = 2 SP 2, is
45. The orthocentre of the triangle formed by the lines
(a) a straight line parallel to X-axis
(b) a circle passing through the origin
(c) a circle with the centre at the origin
(d) a straight line parallel to Y-axis
46. If a, b and c are in AP, then the straight line
x + y = 1, 2 x + 3 y = 6 and 4x − y + 4 = 0 lies in quadrant
number… .
(1985, 2M)
a x + by + c = 0 will always pass through a fixed point
whose coordinates are (…) .
(1984, 2M)
Straight Line and Pair of Straight Lines 361
47. y = 10x is the reflection of y = log10 x in the line whose
equation is .... .
(1984, 2M)
True/False
48. The lines 2x + 3 y + 19 = 0 and 9x + 6 y − 17 = 0 cut the
coordinate axes in concyclic points.
(1988, 1M)
49. No tangent can be drawn from the point (5/2, 1) to the
circumcircle of the triangle with vertices (1, 3 ),
(1985, 1M)
(1, − 3 ) and (3, 3 ).
50. The straight line 5x + 4 y = 0 passes through the point of
intersection of the straight lines x + 2 y − 10 = 0 and
(1983, 1M)
2x + y + 5 = 0.
B (0, − 5). A variable line PQ is drawn perpendicular to
AB cutting the X-axis in P and the Y-axis in Q. If AQ
(1990, 4M)
and BP inters at R, find the locus of R.
60. Let ABC be a triangle with AB = AC. If D is mid point of
BC, the foot of the perpendicular drawn from D to AC
and F the mid-point of DE. Prove that AF is
perpendicular to BE.
(1989, 5M)
61. The equations of the perpendicular bisectors of the
sides AB and AC of a ∆ABC are x − y + 5 = 0 and
x + 2 y = 0, respectively. If the point A is (1, – 2), find the
(1986, 5M)
equation of the line BC.
62. One of the diameters of the circle circumscribing the
Analytical & Descriptive Questions
51. A straight line L through the origin meets the line
x + y = 1 and x + y = 3 at P and Q respectively. Through
P and Q two straight lines L1 and L 2 are drawn, parallel
to 2x − y = 5 and 3x + y = 5, respectively. Lines L1 and L 2
intersect at R, show that the locus of R as L varies, is a
straight line.
(2002, 5M)
52. A straight line L with negative slope passes through the
point (8, 2) and cuts the positive coordinate axes at
points P and Q. Find the absolute minimum value of
OP + OQ, as L varies, where O is the origin. (2002, 5M)
53. For points P = (x1 , y1 ) and Q = (x2, y2) of the coordinate
plane, a new distance d (P , Q ) is defined by
d (P , Q ) = | x1 − x2| + | y1 − y2|.
Let O = (0, 0) and A = (3, 2). Prove that the set of points
in the first quadrant which are equidistant (with
respect to the new distance) from O and A consists of the
union of a line segment of finite length and an infinite
ray. Sketch this set in a labelled diagram. (2000, 10M)
54. A rectangle PQRS has its side PQ parallel to the line
y = mx and vertices PQ and S on the lines y = a , x = b
and x = − b, respectively. Find the locus of the vertex R.
(1996, 2M)
55. A line through A (−5, − 4) meets the line x + 3 y + 2 = 0,
2x + y + 4 = 0 and x − y − 5 = 0 at the points B, C and D
respectively. If (15 / AB)2 + (10 / AC )2 = (6 / AD )2, find
the equation of the line.
(1993, 5M)
56. Determine all values of α for which the point (α , α 2) lies
inside the triangles formed by the lines 2x + 3 y − 1 = 0,
x + 2 y − 3 = 0, 5x − 6 y − 1 = 0
59. A line cuts the X-axis at A (7, 0) and the Y-axis at
(1992, 6M)
57. Find the equations of the line passing through the point
(2, 3) and making intercept of length 3 unit between the
lines y + 2x = 2 and y + 2x = 5.
(1991, 4M)
58. Straight lines 3x + 4 y = 5 and 4x − 3 y = 15 intersect at
the point A. Points B and C are chosen on these two
lines such that AB = AC. Determine the possible
equations of the line BC passing through the point
(1, 2).
(1990, 4M)
rectangle ABCD 4 y = x + 7. If A and B are the points
(−3, 4) and (5, 4) respectively, then find the area of
(1985, 3M)
rectangle.
63. Two sides of a rhombus ABCD are parallel to the lines
y = x + 2 and y = 7x + 3. If the diagonals of the rhombus
intersect at the point (1, 2) and the vertex A is on the
Y-axis, find possible coordinates of A.
(1985, 5M)
64. Two equal sides of an isosceles triangle are given by the
equations 7x − y + 3 = 0 and x + y − 3 = 0 and its third
side passes through the point (1, −10). Determine the
equation of the third side.
(1984, 4M)
65. The vertices of a triangle are
[at1t2, a (t1 + t2)], [at2 t3 , a (t2 + t3 )],
[at3 t1 , a (t3 + t1 )].
(1983, 3M)
Find the orthocentre of the triangle.
66. The ends A , B of a straight line segment of constant
length c slide upon the fixed rectangular axes OX , OY
respectively. If the rectangle OAPB be completed, then
show that the locus of the foot of the perpendicular
drawn from P to AB is
x2 / 3 + y2 / 3 = c2 / 3
(1983, 2M)
67. The points (1, 3) and (5, 1) are two opposite vertices of a
rectangle. The other two vertices lie on the line
y = 2x + c. Find c and the remaining vertices. (1981, 4M)
68. Two vertices of a triangle are (5, − 1) and (−2, 3). If the
orthocentre of the triangle is the origin, find the
(1978, 3M)
coordinates of the third vertex.
69. One side of a rectangle lies along the line
4x + 7 y + 5 = 0. Two of its vertices are (−3, 1) and (1, 1).
Find the equations of the other three sides. (1978, 3M)
Integer Answer Type Question
70. For a point P in the plane, let d1 (P ) and d2(P ) be the
distances of the point P from the lines x − y = 0 and
x + y = 0, respectively. The area of the region R
consisting of all points P lying in the first quadrant of
the plane and satisfying 2 ≤ d1 (P ) + d2(P ) ≤ 4, is
(2014 Adv.)
Topic 2 Angle between Straight Lines and Equation of
Angle Bisector
Objective Questions II
(One or more than one correct option)
1. A ray of light along x + 3 y = 3 gets reflected upon
reaching X-axis, the equation of the reflected ray is
(2013 Main)
(a) y = x + 3
(c) y = 3x − 3
(b) 3 y = x − 3
(d) 3 y = x − 1
P = {– sin ( β – α ) – cos β}, Q = {cos( β – α ),sin β}
and R = {cos ( β – α + θ ) sin ( β – θ )},
π
where 0 < α , β , θ < . Then,
(2008, 4M)
4
(a) P lies on the line segment RQ
(b) Q lies on the line segment PR
(c) R lies on the line segment QP
(d) P, Q, R are non-colinear
(c) 3 x + y = 0
Statement II In any triangle, bisector of an angle
divides the triangle into two similar triangles. (2007, 3M)
5. The vertices of a triangle are A (− 1, − 7), B(5, 1) and
point. Then, the equation of the bisector of the angle
(2001, 1M)
PQR is
(b) x +
Statement I The ratio PR : RQ equals 2 2 : 5 .
Because
Fill in the Blank
3. Let P = (–1, 0), and Q (0, 0) and R = (3, 3 3 ) be three
3
x+ y = 0
2
4. Lines L1 : y − x = 0 and L 2 : 2x + y = 0 intersect the line
L3 : y + 2 = 0 at P and Q, respectively. The bisector of the
acute angle between L1 and L 2 intersects L3 at R.
2. Consider three points
(a)
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
C (1, 4). The equation of the bisector of the angle ABC
is… .
(1993, 2M)
Analytical & Descriptive Questions
6. The area of the triangle formed by the intersection of
3y = 0
line parallel to X-axis and passing through (h, k) with
the lines y = x and x + y = 2 is 4h 2. Find the locus of point
(2005)
P.
3
(d) x +
y=0
2
7. Find the equation of the line which bisects the obtuse
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
(b) Statement I is true, Statement II is also true;
Statement II is not the correct explanation of
Statement I
angle between the lines x − 2 y + 4 = 0 and 4x − 3 y + 2 = 0.
(1993, 2M)
L1 ≡ ax + by + c = 0 and L 2 ≡ lx + my + n = 0
intersect at the point P and makes an angle θ with each
other. Find the equation of a line L different from L 2
which passes through P and makes the same angle θ
(1988, 5M)
with L1.y
8. Lines
Topic 3 Area and Family of Concurrent Lines
Objective Questions I (Only one correct option)
3. Two sides of a rhombus are along the lines, x − y + 1 = 0
1. A triangle has a vertex at (1, 2) and the mid-points of
the two sides through it are (−1, 1) and (2, 3). Then, the
(2019 Main, 12 April II)
centroid of this triangle is
(a)  1,

7

3
1
(b)  , 2
3 
1
(c)  , 1
3 
1
(d)  ,
3
5

3
2. Consider the set of all lines px + qy + r = 0 such that
3 p + 2q + 4 r = 0. Which one of the following statements
(2019 Main, 9 Jan I)
is true?
(a) Each line passes through the origin.
 3 1
(b) The lines are concurrent at the point  , 
 4 2
(c) The lines are all parallel
(d) The lines are not concurrent
and 7x − y − 5 = 0. If its diagonals intersect at (− 1, − 2),
then which one of the following is a vertex of this
rhombus?
(2016 Main)
(a) (− 3, − 9)
1
8
(c)  , − 
3
3
(b) (− 3, − 8)
10
7
(d)  −
, − 
 3
3
4. Area of the parallelogram formed by the lines
y = mx, y = mx + 1, y = nx and y = nx + 1 equals
(a)
|m + n|
(m − n ) 2
1
(c)
|m + n|
(2001, 1M)
2
|m + n|
1
(d)
|m − n|
(b)
Straight Line and Pair of Straight Lines 363
8
5. The points 0,  , (1, 3) and (82, 30) are vertices of

(a)
(b)
(c)
(d)
3
(1986, 2M)
an obtuse angled triangle
an acute angled triangle
a right angled triangle
None of the above
6. The straight lines x + y = 0, 3x + y − 4 = 0, x + 3 y − 4 = 0
form a triangle which is
(A)
L1, L2, L3 are concurrent, if
(p)
k= −9
One of L1, L2, L3 is parallel to
at least one of the other two, if
(q)
k=−
(C)
L1, L2, L3 form a triangle, if
(r)
k=
(D)
L1, L2, L3 do not form a
triangle, if
(s)
k=5
6
5
5
6
Fill in the Blank
10. The set of lines ax + by + c = 0, where 3a + 2b + 4c = 0 is
concurrent at the point… .
7. Given the four lines with the equations
x + 2 y − 3 = 0, 3x + 4 y − 7 = 0,
2x + 3 y − 4 = 0,4x + 5 y − 6 = 0, then
(1980, 1M)
(a) they are all concurrent
(b) they are the sides of a quadrilateral
(c) only three lines are concurrent
(d) None of the above
x1
x2
x3
a1 b1 1
y1 1
y2 1 = a 2 b2 1
a3 b3 1
y3 1
,then the two triangles
with vertices (x1 , y1 ), (x2, y2), (x3 , y3 ) and (a1 , b1 ), (a 2, b2),
(1985, 1M)
(a3 , b3 ) must be congruent.
Analytical & Descriptive Questions
8. Three lines px + qy + r = 0, qx + ry + p = 0 and
rx + py + q = 0 are concurrent, if
(1982, 2M)
True/False
11. If
Objective Question II
(One or more than one correct option)
(1985, 2M)
(b) p 2 + q2 + r 2 = pr + rq
(d) None of these
Match the Columns
Match the conditions/expressions in Column I with
statement in Column II.
9. Consider the lines given by
L1 : x + 3 y − 5 = 0,
L3 : 5x + 2 y − 12 = 0
Column II
(B)
(1983, 1M)
(a) isosceles
(b) equilateral
(c) right angled
(d) None of the above
(a) p + q + r = 0
(c) p3 + q3 + r3 = 3 pqr
Column I
12. Using coordinate geometry, prove that the three
altitudes of any triangle are concurrent.
(1998, 8M)
13. The coordinates of A , B, C are (6, 3), (−3, 5), (4, − 2)
respectively and P is any point (x, y). Show that the
ratio of the areas of the triangles ∆ PBC and ∆ ABC is
x+ y−2
(1983, 2M)
.
7
14. A straight line L is perpendicular to the line in
5x − y = 1. The area of the triangle formed by the line L
and the coordinate axes is 5. Find the equation of the
line L.
(1980, 3M)
L 2 : 3x − ky − 1 = 0,
Topic 4 Homogeneous Equation of Pair of Straight Lines
Objective Questions I (Only one correct option)
1. Let a and b be non-zero and real numbers. Then, the
equation (ax2 + by2 + c) (x2 − 5xy + 6 y2) = 0 represents
2. Area of triangle formed by the lines x + y = 3 and angle
bisectors of the pair of straight lines x2 − y2 + 2 y = 1 is
(2008, 3M)
(a) four straight lines, when c = 0 and a , b are of the
same sign
(b) two straight lines and a circle, when a = b and c is of
sign opposite to that of a
(c) two straight lines and a hyperbola, when a and b are
of the same sign and c is of sign opposite to that of a
(d) a circle and an ellipse, when a and b are of the same
sign and c is of sign opposite to that of a
(2004, 2M)
(a) 2 sq units
(c) 6 sq units
(b) 4 sq units
(d) 8 sq units
Analytical & Descriptive Question
3. Show that all chords of curve 3x2 − y2 − 2x + 4 y = 0,
which subtend a right angle at the origin pass through
a fixed point. Find the coordinates of the point.
(1991, 4M)
364 Straight Line and Pair of Straight Lines
Topic 5 General Equation of Pair of Straight Lines
Objective Question I (Only one correct option)
1. Let PQR be a right angled isosceles triangle, right
angled at P (2, 1). If the equation of the line QR is
2x + y = 3, then the equation representing the pair of
(1999, 2M)
lines PQ and PR is
(a) 3x2 − 3 y2 + 8xy + 20x + 10 y + 25 = 0
(b) 3x2 − 3 y2 + 8xy − 20x − 10 y + 25 = 0
(c) 3x2 − 3 y2 + 8xy + 10 x + 15 y + 20 = 0
(d) 3x2 − 3 y2 − 8xy − 10 x − 15 y − 20 = 0
Answers
65. [( −a, a (t1 + t 2 + t 3 + t1t 2t 3 )]
Topic 1
67. c = − 4,( 4, 4 ),(2, 0 )
1. (d)
2. (d)
3. (b)
4. (c)
68. ( − 4, − 7 )
5. (d)
6. (c)
7. (c)
8. (c)
69. 7 x − 4y + 25 = 0, 4 x + 7y = 11 = 0,7 x − 4y − 3 = 0
9. (c)
10. (b)
11. (d)
12. (c)
13. (a)
14. (a)
15. (d)
16. (d)
17. (a)
18. (c)
19. (d)
20. (c)
21. (b)
22. (b)
23. (b)
24. (d)
25. (c)
26. (c)
5. 7y = x + 2
27. (a)
28. (b)
29. (d)
30. (c)
6. 2 x = ± (y − 1 )
31. (a)
32. (c)
33. (d)
34. (d)
7. ( 4 + 5 ) x − (2 5 + 3 ) y + ( 4 5 + 2 ) = 0
35. (c)
36. (a)
37. (b)
38. (d)
39. (c)
40. (a)
41. (b, c, d)
42. (a, c)
8. 2 (al + bm )(ax + by + c ) − (a 2 + b 2 ) (lx + my + n ) = 0
43. (a, c)
44. (1, 1)
45. Ist
46. (1, − 2 )
Topic 3
47. (y = x )
48. True
49. True
50. True
1. (b)
5. (d)
Topic 2
52. (OP + OQ = 18)
3
1
56. − < α < −1 ∪ < α < 1
2
2
55. 2 x + 3y + 22 = 0
57. x = 2 and 3 x + 4y = 18
2. (d)
2. (b)
6. (a)
3. (c)
3. (c)
7. (c)
4. (c)
4. (d)
8. (a, c)
 3 1
10.  , 
 4 2
11. False
14. x + 5y = ±5 2
Topic 4
1. (b)
58. x − 7y + 13 = 0 and 7 x + y − 9 = 0
59. x + y − 7 x + 5y = 0
61. 14 x + 23y − 40 = 0
2
62. 32 sq units
1. (b)
9. A → s; B → p, q; C → r; D → p, q, s
54. (m 2 − 1 ) x − my + b (m 2 + 1 ) + am = 0
2
70. 6 sq units
 5
63.  0,  or ( 0, 0 )
 2
2. (a)
3. (1, − 2 )
Topic 5
1. (b)
64. x − 3y − 31 = 0 or 3 x + y + 7 = 0
Hints & Solutions
Topic 1 Various Forms of Straight Line
1. According to the question, we have the following figure.
(0,b)
M
60°
O
(a,0)
α
x+
y
=0
x
y
— + — =1
a b
Let θ be the inclination of the line x + y = 0. Then,
tan θ = − 1 = tan (180° − 45° )
⇒
tan θ = tan 135°
⇒
θ = 135°
⇒
α + 60° = 135°
⇒
α = 75°
Since, line L having perpendicular distance OM = 4.
So, equation of the line ‘L’ is
x cos α + y sin α = 4
⇒
x cos 75° + y sin 75° = 4
⇒ x cos (45° + 30° ) + y sin (45° + 30° ) = 4
Straight Line and Pair of Straight Lines 365
 3
 3
1 
1 
+
⇒ x
−
=4
+ y
2
2
2
2
2
2
2
2



⇒
2.
5.
( 3 − 1) x + y ( 3 + 1) = 8 2
=
Key Idea Use formulae :
2sin A sin B = cos( A − B) − cos( A + B) and cos 2 θ = 1 − 2 sin2 θ
= − sin 2(1) < 0 ⇒ y < 0
and as we know that y < 0, is in third and fourth
quadrants only.
3. Since, equation of a line parallel to line ax + by + c = 0 is
ax + by + k = 0
…(i)
[as per question’s requirement]
⇒
| k| = 3
⇒
k=± 3
So, possible lines having equation, either 4x − 3 y + 3 = 0
or 4x − 3 y − 3 = 0
1 2
Now, from the given options the point  − ,  lies on the
 4 3
line 4x − 3 y + 3 = 0.
[Q If lines are perpendicular, then product of their
slopes is −1]
⇒
a 2(a − 1) = − 2 ⇒ a3 − a 2 + 2 = 0
⇒ (a + 1)(a 2 − 2a + 2) = 0 ⇒ a = − 1
∴Equation of lines are
…(i)
x − 2y = 1
and
…(ii)
2x + y = 1
On solving Eq. (i) and Eq. (ii), we get
3
1
x = and y = −
5
5
∴ Point of intersection of the lines (i) and (ii) is
1
3
P , −  .
5
5
1
3
Now, required distance of the point P  , −  from
5
5
9
1
10
2
origin =
+
=
=
25 25
25
5
6. Let the slope of line is m, which is passing through
P(2, 3).
Y
8
7
6
5
4. The given inequalities are
|x − y| ≤ 2 and|x + y| ≤ 2 .
On drawing, the above inequalities, we get a square
4
P(2,
3)
2
4
1
(0, 2)
X′
(–2, 0)
(2, 0)
O
X
(0, –2)
Y′
Now, the area of shaded region is equal to the area of a
square having side length (2 − 0)2 + (0 − 2)2 = 2 2
units.
R
d
3
Y
X′
(x 2 − x 1) 2 + (y 2 − y 1) 2
Given, lines x + (a − 1) y = 1
and 2x + a 2y = 1, where a ∈ R − {0, 1}
are perpendicular to each other

1   2
∴
−
 × −  = −1
 a − 1  a 2
Given equation is y = sin x sin(x + 2) − sin 2(x + 1)
1
1
= [cos 2 − cos(2x + 2)] − [1 − cos(2x + 2)]
2
2
[Q2 sin A sin B = cos( A − B) − cos( A + B) and
cos 2 θ = 1 − 2 sin 2 θ ⇒ 2 sin 2 θ = 1 − cos 2 θ]
1
1
1 1
= cos 2 − cos(2x + 2) − + cos(2x + 2)
2
2
2 2
1
1
= (cos(2) − 1) = − (2 sin 2(1))
2
2
∴Equation of line parallel to line
4x − 3 y + 2 = 0 is 4x − 3 y + k = 0
Now, distance of line (i) from the origin is
| k|
3
=
2
2
5
4 + 3
Key Idea
(i) If lines are perpendicular to each other, then
product of their slopes is −1, i.e. m1m2 = − 1
(ii) Distance between two points ( x 1, y 1) and (x 2, y 2)
1
Y′
2
3
4
θ
5
Q
6
7
8
9
X
x+y=7
Since, the distance of a point (x1 , y1 ) from the line
ax + by1 + c
.
ax + by + c = 0 is d = 1
a 2 + b2
∴The distance of a point P (2, 3) from the line
x + y − 7 = 0, is
|2 + 3 − 7| 2
d=
=
= 2
1+ 1
2
Now, in ∆PRQ,
QR = 16 − d 2 = 16 − 2 = 14
366 Straight Line and Pair of Straight Lines
d
2
1
m+ 1
=
=
=
QR
14
7 1− m
1
m+ 1
=±
1− m
7
1
1
m+ 1
m+ 1
or
=
=−
1− m
7
1− m
7
1− 7
−1 − 7
or m =
m=
1+ 7
7 −1
∴ tanθ =
⇒
⇒
⇒

m2 − m1 
Q tanθ =

1
+ m1 m2 

9. Given equation of line is 3x + 5 y = 15
…(i)
Clearly, a point on the line (i), which is equidistance
from X and Y -axes will lie on the line either y = x or
y = − x.
Y
B
y=x
(0, 3)
7. Given points are P(2, − 3, 4), Q(8, 0, 10) and R(4, y, z ).
Now, equation of line passing through points P and Q is
x − 8 y − 0 z − 10
=
=
6
3
6
[Since equation of a line passing through two points
A (x1 , y1 , z1 ) and B (x2 , y2 , z2 ) is given by
x − x1
y − y1
z − z1 
=
=

x2 − x1
y2 − y1 z2 − z1 
x − 8 y z − 10
…(i)
= =
⇒
2
1
2
Q Points P , Q and R are collinear, so
4 − 8 y z − 10
= =
2
1
2
z − 10
⇒
−2 = y =
2
⇒
y= −2
and
z=6
So, point R is (4, − 2, 6), therefore the distance of point R
from origin is
OR = 16 + 4 + 36
y=–x
A
(5, 0)
X
3x+5y=15
0
In the above figure, points A and B are
on the line (i) and are equidistance from the coordinate
axes.
 15 15
On solving line (i) and y = x, we get A 
, .
8 8
Similarly, on solving line (i) and y = − x, we get
 15 15
, .
B −
 2 2
So, the required points lie only in I and II quadrants.
10. Let the equation of required line having intercepts a
and b with the axes is
Y
x y
+ =1
a b
…(i)
B (0,b)
= 56 = 2 14
8. Given, points (1, 2), (−3, 4) and (h , k) are lies on line L1,
b
P (–3,4)
so slope of line L1 is
4− 2 k − 2
=
−3 − 1 h − 1
−1 k − 2
=
m1 =
2
h −1
m1 =
⇒
⇒
2(k − 2) = − 1(h − 1)
⇒
2k − 4 = − h + 1
⇒
h + 2k = 5
and slope of line L2 joining points (h , k ) and
3− k
(4, 3), is m2 =
4− h
A (a , 0 )
…(i)
…(ii)
…(iii)
Since, lines L1 and L2 are perpendicular to each other.
∴
m1 m2 = − 1


−
k
3
1
− 
= − 1[from Eqs. (i) and (iii)]
⇒


 2   4 − h 
⇒
⇒
3 − k = 8 − 2h
2h − k = 5
On solving Eqs. (ii) and (iv), we get
So,
(h , k ) = (3, 1)
k 3
= =3
h 1
…(iv)
O
a
X
Now, according to given information,
P is the mid-point of AB
a b
P =  ,  = (−3, 4)
∴
 2 2
[given]
⇒
(a , b) = (−6, 8)
On putting the value of a and b in Eq. (i), we get
x
y
+ = 1 ⇒ 8x − 6 y = −48
−6 8
⇒
4x − 3 y + 24 = 0
11. Slope of the line 2x − 3 y + 17 = 0 is
2
= m1, (let) and the slope of line joining the points (7, 17)
3
β − 17 β − 17
and (15, β ) is
=
= m2 (let)
15 − 7
8
According to the question, m1m2 = − 1
2 β − 17
= − 1 ⇒ β − 17 = − 12 ⇒ β = 5.
⇒ ×
3
8
Straight Line and Pair of Straight Lines 367
12. According to given information, we have the following
figure
C(2, 5)
D
P
A(1, 2)
B(3, 4)
We know that, diagonals of a parallelogram intersect at
mid-point.
5 9
∴ P = Mid-point of BC and so, P ≡  , 
 2 2
Now, equation of AD is.
⇒
⇒
⇒
9
−2
( y − 2) = 2
(x − 1)
5
−1
2
5
y − 2 = (x − 1)
3
3 y − 6 = 5x − 5
5x − 3 y + 1 = 0
x2
13. Given equation of curve is y = xe
0 + x2
3 + y2
;4=
2
2
⇒
x2 = 4 and y2 = 5
∴Thus,
C ≡ (4, 5)
Now, equation of line BC is given by
⇒
2=
( y − y1 ) = m (x − x1 )
y − 5 = 1(x − 4)
[li