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Contents
S.no.
Chapter Name
Page no.
Questions
Solutions
1.
SETS
1-2
75-77
2.
RELATIONS & FUNCTIONS-1
3-4
77-81
3.
TRIGONOMETRIC FUNCTIONS
5-6
81-84
4.
PRINCIPLE OF MATHEMATICAL INDUCTION
7-8
84-88
5.
COMPLEX NUMBERS AND QUADRATIC EQUATIONS
9-11
88-92
6.
LINEAR INEQUALITIES
12-13
92-96
7.
PERMUTATIONS AND COMBINATIONS
14-15
96-98
8.
BINOMIAL THEOREM
16-17
99-101
9.
SEQUENCES AND SERIES
18-19
102-106
10.
STRAIGHT LINES
20-22
107-112
11.
CONIC SECTIONS
23-25
113-118
12.
LIMITS & DERIVATIVES
26-27
118-123
13.
MATHEMATICAL REASONING
28-29
123-124
14.
STATISTICS
30-31
124-127
15.
PROBABILITY-1
32-33
128-131
16.
RELATIONS & FUNCTIONS-2
34-35
132-135
17.
INVERSE TRIGONOMETRIC FUNCTIONS
36-38
136-140
18.
MATRICES
39-41
141-145
19.
DETERMINANTS
42-44
146-152
20.
CONTINUITY AND DIFFERENTIABILITY
45-47
153-158
21.
APPLICATION OF DERIVATIVES
48-50
159-165
22.
INDEFINITE INTEGRATION
51-53
165-170
23.
DEFINITE INTEGRATION
54-56
171-175
24.
APPLICATION OF INTEGRALS
57-59
176-182
25.
DIFFERENTIAL EQUATION AND ITS APPLICATIONS
60-62
182-187
26.
VECTOR ALGEBRA
63-65
188-193
27.
THREE DIMENSIONAL GEOMETRY
66-68
194-200
28.
PROBABILITY-2
69-71
201-207
29.
PROPERTIES OF TRIANGLE
72-74
208-212
1
SETS
MCQs with One Correct Answer
1.
2.
3.
4.
5.
6.
7.
If A and B are non-empty sets such that A É B,
then
(a) B' – A' = A – B
(b) B' – A' = B – A
(c) A' – B' = A – B
(d) A' Ç B' = B – A
Let A = {q : sin(q) = tan(q)} and B = {q : cos(q) = 1}
be two sets. Then :
(a) A = B
(b) A Ë B
(c) B Ë A
(d) A Ì B and B - A ¹ f
A set S contains 3 elements, the number of
subsets of which of following sets is 256.
(a) S
(b) P(S)
(c) P(P(S))
(d) None of these
In a college of 300 students every student reads
5 newspapers and every newspaper is read by
60 students. The number of newpapers is
(a) at least 30
(b) at most 20
(c) exactly 25
(d) None of these
Let A and B be two sets then
(A È B)'È (A 'Ç B) is equal to
(a) A¢
(b) A
(c) B¢
(d) None of these
The set (A \ B) È (B \ A) is equal to
(a) [ A \ ( A Ç B)] Ç [ B \ ( A Ç B)]
(b) ( A È B) \ ( A Ç B)
(c) A \ ( A Ç B )
(d) A Ç B \ A È B
Let A = { x : x ÎR, | x | < 1}
B = { x : x ÎR, | x – 1| ³1} and
AÈB = R – D, then the set D is :
8.
9.
10.
11.
12.
13.
14.
(a) { x : 1< x £ 2}
(b) {x : 1£ x < 2}
(c) {x : – 2 £ x £ 2}
(d) None of these
If aN = {ax : x Î N} and bN Ç cN = dN, where b,
c Î N are relatively prime, then
(a) d = bc
(b) c = bd
(c) b = cd
(d) None of these
The set (A È B È C) È (A Ç B' Ç C')' Ç C' is
equal to
(a) B Ç C'
(b) A Ç C
(c) B' Ç C'
(d) None of these
A dinner party is to be fixed for a group of 100
persons. In this party, 50 persons do not prefer
fish, 60 prefer chicken and 10 do not prefer either
chicken or fish. The number of persons who
prefer both fish and chicken is
(a) 20
(b) 22
(c) 25
(d) None of these
Let U be the universal set and A È B È C = U.
Then {(A – B) È (B – C) È (C – A)}¢ is equal to:
(a) A È B È C
(b) A È (B Ç C)
(c) A Ç B Ç C
(d) A Ç (B È C)
If n(A) = 1000, n(B) = 500 and if n(A Ç B) ³ 1 and
n(A È B) = p, then
(a) 500 £ p £ 1000
(b) 1001 £ p £ 1498
(c) 1000 £ p £ 1498
(d) 1000 £ p £ 1499
In a battle 70% of the combatants lost one eye,
80% an ear, 75% an arm, 85% a leg, x % lost all
the four limbs. The minimum value of x is
(a) 10
(b) 12
(c) 15
(d) None of these
The number of students who take both the
subjects mathematics and chemistry is 30. This
represents 10% of the enrolment in mathematics
and 12% of the enrolment in chemistry. How
MATHEMATICS
2
15.
16.
17.
18.
19.
20.
1
2
3
many students take at least one of these two
subjects?
(a) 520
(b) 490
(c) 560
(d) 480
At a certain conference of 100 people, there are
29 Indian women and 23 Indian men. Of these
Indian people 4 are doctors and 24 are either
men or doctors. There are no foreign doctors.
How many foreigners and women doctors are
attending the conference?
(a) 48, 1
(b) 34, 3
(c) 46, 4
(d) 42, 2
Each student in a class of 40, studies at least
one of the subjects English, Mathematics and
Economics. 16 study English, 22 Economics and
26 Mathematics, 5 study English and Economics,
14 Mathematics and Economics and 2 study all
the three subjects. The number of students who
study English an d Math ematics but not
Economics is
(a) 7
(b) 5
(c) 10
(d) 4
In a market research project, 20% opted for
'Nirma' detergent whereas 60% opted for 'Surf
blue' detergent. The remaining individuals were
not certain. If the difference between those who
opted for 'Surf blue' and those who were
uncertain was 720, how many respondents were
covered in the survey
(a) 1100
(b) 1150
(c) 1800
(d) None of these
In a school 80 students like chocolate, 40 like
coffee if the number of students doesn’t like any
of them is equal to the number of students who
like both of them then what is the total number
of students in the school?
(a) 115
(b) 90
(c) 120
(d) None of these
In a school there are 100 students 60 of them
don’t like Chocolate and 50 don’t like Biscuit
and 10 of them like none then how many of them
like both?
(a) 20
(b) 30
(c) 40
(d) None of these
A survey was conducted of 100 people whether
they have read recent issues of ‘Golmal', a
(a)
(b)
(c)
4
5
6
(c)
(a)
(b)
7
8
9
(b)
(a)
(a)
10
11
12
monthly magazine. Summarized information is
presented below:
Only September: 18
September but not August: 23
September and July: 8
September: 28
July: 48
July and August:10
None of the three months: 24
What is the number of surveyed people who
have read exactly for two consecutive months?
(a) 7
(b) 9
(c) 12
(d) 14
Numeric Value Answer
21.
22.
23.
24.
25.
An investigator interviewed 100 students to
determine their preferences for the three drinks :
milk (M), coffee (C) and tea (T). He reported the
following : 10 students had all the three drinks
M, C and T; 20 had M and C; 30 had C and T; 25
had M and T; 12 had M only; 5 had C only; and
8 had T only. If number of students who did not
n
take any of the three drinks is n, then is
5
In a class of 80 students numbered 1 to 80, all
odd numbered students opt of Cricket, students
whose numbers are divisible by 5 opt for
Football and those whose numbers are divisible
by 7 opt for Hockey. If the number of students
who do not opt any of the three games is n, then
n
is equal to
4
A survey shows that 61%, 46% and 29% of the
people watched “3 idiots”, “Rajneeti” and
“Avatar” respectively. 25% people watched
exactly two of the three movies and 3% watched
none. What percentage of people watched all
the three movies?
If n(A) = 4 and n(B) = 7, then the difference
between maximum and minimum value of
n(A È B) is
Two finite sets have m and n elements. The
number of subsets of the first set is 112 more
than that of the second set. The value of m – n is
ANSW ER KEY
(a) 13 (a) 16
(c) 14 (a) 17
(d) 15 (a) 18
(b)
(c)
(c)
19
20
21
(d)
(b)
(4)
22
23
24
(7)
(7)
(4)
25
(3)
2
RELATIONS &
FUNCTIONS-1
MCQs with One Correct Answer
1.
2.
x, y Î R
If f (x) . f (y) = f (x) + f (y) + f (xy) – 2
and if f (x) is not a constant function, then the
value of f (a) is
(a) 1
(b) 2
(c) 0
(d) – 1
The domain of the function
f ( x ) = x14 - x11 + x 6 - x3 + x 2 + 1 is
(a)
(-¥, ¥)
(a)
(c)
7.
(b) [0, ¥)
3.
4.
üï
1 ìï æ x ö
f (x)f (y) – í f ç ÷ + f ( xy ) ý is equal to :
2 ïî è y ø
ïþ
(a) 0
(b) 1
(c) –1
(d) None of these
The domain of the function
(d)
8.
5.
6.
a + 2b
a 2 - b2
a - b2
(d) None of these
(a)
[1, p)
(b)
(c)
æ pö
çè 0, ÷ø - {1}
2
(d) (0, 1)
( 0, 2p ) - [1, p)
1
æp x ö
Let f ( x) = - tan ç
÷ , -1< x <1 &
2
è 2 ø
(a)
é - 21, 21 ù
ë
û
9.
æ x + 59 ö
3f(x) + 2f ç
÷ = 10x + 30 for all real x ¹ 1.
è x -1 ø
The value of f(7) is
(a) 8
(b) 4
(c) –8
(d) 11
10.
é1 ù
êë 2 ,1úû
(b)
ö
é1
ê 2 , -1 ÷
ë
ø
é 1
ù
é 1 ö
(d) ê - , -1ú
ê - 2 ,1÷
2
ë
û
ë
ø
The domain of function
f(x) = log4[log5{log3 (18x – x2 – 77)}] is:
(a) (7, 11)
(b) (8, 10)
(c) (8, 11)
(d) (7, 10)
Let f(x) = ax(a > 0) be written as f(x) = g(x) + h(x),
where g(x) is an even function and h(x) is an
odd function. Then the value of g(x + y) + g(x – y)
is
(a) 2g(x) g(y)
(b) 2g(x + y) g(x – y)
(c) 2g(x)
(d) None of these
(c)
(c) éë -5, - 21 ùû È éë 21, 5 ùû È {0}
(d) (–¥, –5)
The function f satisfies the functional equation
æ 1 ö
If af(x + 1) + bf ç
÷ = x , x ¹ –1, a ¹ b, then
è x +1 ø
f(2) is equal to
)
a
2
g ( x) = 3 + 4 x - 4 x 2 , then dom (f + g) is given
by
f ( x ) = 10 - x 4 - 21x 2 is
(a) [5, ¥)
(b)
2 a2 - b 2
(b)
The domain of f(x) = cos(sin x ) + log x {x} ;
{.} denote the fractional part, is
R - [0,1]
(c) (- ¥, 0]
If f(x) = cos (log x) then
(
2a + b
MATHEMATICS
4
11.
12.
13.
14.
15.
16.
17.
1
2
3
Let f be a real valued function such that for any
real x, f (l + x) = f (l – x) and f (2l + x) = –f (2l – x)
for some l > 0. Then
(a) f is even and non-periodic
(b) f is odd and periodic
(c) f is odd and non-periodic
(d) f is even and periodic
Let f(x) = ([a]2 – 5[a] + 4)x3 – (6{a}2 – 5{a} + 1) x
– (tan x) sgn x, be an even function for all x Î R,
then sum of all possible values of 'a' is (where
[ ] and { } denote greatest integer function and
fractional part functions respectively)
17
53
31
35
(b)
(c)
(d)
(a)
6
6
3
3
The set of all integer values of n for which the
5x
function f(x) = cos nx . sin
is periodic with
n
period 2p is equal to
(a) {1, 5, 10}
(b) {1, 5}
(c) {±1, ±5}
(d) None of these
If f : ¡ ® ¡ & g: ¡ ® ¡ be two given
functions, then 2 min {f (x) – g(x), 0} equals
(a) f(x) + g(x) – |g(x) – f(x)|
(b) f(x) + g(x) + |g(x) – f(x)|
(c) f(x) – g(x) + |g(x) – f(x)|
(d) f(x) – g(x) – |g(x) – f(x)|
The domain of f(x) is (0, 1), therefore the domain
of y = f(ex) + f(ln | x |) is :
æ1 ö
(a) ç , 1÷
(b) (–e, –1)
èe ø
1ö
æ
(d) (–e, –1) È (1, e)
(c) ç -1, - ÷
eø
è
Suppose that f is a periodic function with period
1
æ 9ö
and that f(2) = 5 and f ç ÷ = 2 then
2
è 4ø
f(–3) – f æç 1 ö÷ has the value equal to
è 4ø
(a) 2
(b) 3
(c) 5
(d) 7
2
x
+
3
;
x
£
1
é
Let f ( x ) = ê 2
. If the range of f(x)
êë a x + 1 ; x > 1
= R (set of real numbers) then number of integral
value(s), which a may take is
(a) 2
(b) 3
(c) 4
(d) 5
(b)
(a)
(a)
4
5
6
(c)
(b)
(a)
7
8
9
(d)
(c)
(b)
10
11
12
18.
19.
20.
Let f(x) = sin x – cos x and g(x) = log 5x ; then
the range of g ( 2 f ( x) + 3) is
(a) [0, 1]
(b) [0, 2]
3ù
é
(c)
(d) None of these
êë 0, 2 úû
1
Let f(x) = 1 +
and g(x, y) = log y, then the
4 x
æ1
ö
domain of g ç , - g (2, f ( x)) - 1÷ is
è2
ø
(a) 0 < x < 1
(b) 0 < x £ 1
(c) x ³ 1
(d) Null set
Let f (x) be defined as
|x|
0 £ x <1
ì
ï
f ( x) = n í| x - 1| + | x - 2 | 1 £ x < 2
ï
| x -3|
2£ x<3
î
The range of function g(x) = sin (7(f (x)) is :
(a) [0, 1]
(b) [–1, 0]
1
1
é
ù
(c) ê - , ú
(d) [–1, 1]
ë 2 2û
Numeric Value Answer
21.
22.
23.
24.
25.
If f ( x ) =
1
and S = f(5) + f(4) + f(3) +....+
1 + e- x
f(–3)+ f(–4) + f(–5), then the value of S is
If the period of f(x) satisfying the condition:
f(x + p) = 1 + {1 – 3f(x) + 3f 2(x) – f 3(x)}1/3 is lp,
then evaluate l.
If f(x) is an odd function, f(1) = 3, and f(x + 2)
= f(x) + f(2), then the value of f(3) is
Let f(x, y) be a function satisfying the functional
equation: f(x, y) = f(2x + 2y, 2y – 2x) for all real
numbers x, y. Define g(x) by g(x) = f(2x, 0). Also
given that g(x) is a periodic function with period
k
k, then find value of .
2
Number of elements in the range set of
é x ù é 15 ù
f ( x) = ê ú ê - ú " x Î (0, 90) ; (where [.]
ë15 û ë x û
denotes greatest integer function)
ANSW ER KEY
(a) 13 (c) 16 (b) 19 (d) 22 (2) 25 (6)
(b) 14 (d) 17 (c) 20 (d) 23 (9)
(d) 15 (b) 18 (b) 21 (5.5 24 (6)
3
TRIGONOMETRIC
FUNCTIONS
6.
MCQs with One Correct Answer
1.
2.
1
, then the value of
5
1
1
2
4
+
+
+
cos2 a 1 + sin 2 a 1 + sin 4 a 1 + sin8 a
is equal to
(a) 2
(b) 4
(c) 6
(d) 10
If a, b, g, d are the smallest positive angles in
ascending order of magnitude which have their
sines equal to the positive quantity k, then the
a
b
g
d
value of 4 sin + 3 sin + 2 sin + sin
is
2
2
2
2
equal to
(a) 2 1 - k
(b) 2 1 + k
16
If sin a =
(c)
3.
2 k
(d) none of these
If cos a =
2cos b - 1
(0 < a < b < p) , then
2 - cos b
tan a / 2
is equal to
tan b / 2
(a) 1
4.
5.
(b)
2
a=
(a)
3
(d)
1
3
Let 0 £ a, b, g, d £ p whrere b and g are not
complementary such that
2 cos a + 6 cos b + 7 cos g + 9 cos d = 0
and 2 sin a – 6 sin b + 7 sin g – 9 sin d = 0
cos (a + d ) m
where m and n are
If
=
cos (b + g ) n
relatively prime positive numbers, then the
value of (m + n) is equal to:
(a) 11
(b) 10
(c) 9
(d) 7
If sin q + sin 2q + sin 3q = sin a and cos q
+ cos 2q + cos 3q = cos a, then q is equal to
(a) a/2 (b) a
(c) 2a
(c) a/6
p
is equal to:
5
5 -1
4
(b)
5 +1
4
5 +1
5 -1
(d)
2
2
sin q
sin (3q) sin(9q)
sin(27q)
+
+
+
=
cos (3q) cos(9q) cos(27q) cos(81q)
(c)
7.
8.
(c)
The value of the expression
sin 7a + 6sin 5a + 17 sin 3a + 12 sin a
where
sin 6a + 5 sin 4a + 12 sin 2a
(a)
sin(81 q)
(b)
2 cos(80 q) cos q
sin(80 q)
2 cos(81 q) cos q
(c)
sin(81 q)
cos(80 q) cos q
sin(80 q)
cos(81 q) cos q
If cosec q =
(d)
p+q
æp qö
, then cot ç + ÷ =
p-q
è4 2ø
p
q
(b)
(c)
pq (d) pq
q
p
9.
If sin 2q + sin 2f = 1/2 and cos 2q + cos 2f = 3/2,
then cos2 (q – f) =
(a) 3/8 (b) 5/8
(c) 3/4
(d) 5/4
10. If u = (1 + cos q) (1 + cos 2q) – sin q × sin 2q,
v = sin q (1 + cos 2q) + sin 2q (1 + cos q), then
u2 + v2 =
(a) 4(1 + cos q) (1 + cos 2q)
(b) 4(1 + sin q) (1 + sin 2q)
(c) 4(1 – cos q) (1 – cos 2q)
(d) 4(1 – sin q) (1 – sin 2q)
(a)
MATHEMATICS
6
11.
The product
19.
ìï 2 p üï
ìï 2 p üï
ì 2p ü
cos í 64 ý cos í 64 ý ...cos í 64 ý
î 2 - 1þ
ïî 2 - 1ïþ
îï 2 - 1 ïþ
2
64
16.
1
(d)
32
16
8
32
6464
16
8
Let x, y Î R satisfy the condition such that
sin x sin y + 3 cos y + 4 sin y cos x = 26.
The value of tan2 x + cot2 y is equal to
(a) 9 × 17
(b) 205
1
9
+
(c)
(d) None of these
16 17
The product
xö æ
xö æ
xö
x ö
æ
æ
ç cos ÷ × ç cos ÷ × ç cos ÷ ........ × ç cos
÷
2ø è
4ø è
8ø
256 ø
è
è
is equal to:
sin x
sin x
(a)
(b)
x
x
128 sin
256 sin
256
256
sin x
sin x
(c)
(d)
x
x
128 sin
512 sin
128
512
The maximum value of log20 (3 sin x – 4 cos x + 15)
is equal to:
(a) 1
(b) 2
(c) 3
(d) 4
The value of
cos 25°
cos 70°
cos 85°
+
+
sin 70° sin 85° sin 25° sin 85° sin 25° sin 70°
is
(a) 1/2 (b) 1
(c) 2
(d) 3/2
The number of solutions of the equation
17.
32sec x + 1 = 10.3tan x in the interval [0, 2p] is
(a) 8
(b) 6
(c) 4
(d) 2
Number of solutions of the equation
18.
4 (cos2 2x + cos 2x + 1) + tan x (tan x – 2 3 ) = 0
in [0, 2p] is
(a) 0
(b) 1
(c) 2
(d) 3
The number of solutions of the equation
1
(a)
12.
13.
14.
15.
1
(b)
2
(a) 2
(d)
(b)
(c)
20.
2cos 2 x - 3cos x + 1
(b) 3
4
5
6
(b)
(a)
(c)
= 1 in [0, p] is
(c) 4
7
8
9
(b)
(b)
(b)
solutions of the equation
æ
æ sin x ö ö
çè cos çè 2 ÷ø ÷ø
æ
x
ö
æ
2 xö
çè sin çè 2 tan 2 cos 2 ÷ø - 3÷ø + 2 = 0 in [0, 2p] is :
(a) 0
(b) 1
(c) 2
(d) 4
If the equation x2 + 12 + 3 sin (a + bx) + 6x = 0 has
atleast one real solution, where a, b Î [0, 2p], then
the value of a – 3b is (n Î Z)
(a) 2np
(b) (2n + 1) p
p
p
(c) (4n – 1)
(d) (4n + 1)
2
2
Numeric Value Answer
21.
The number of solutions of the equation
22.
+ log15 cos x)
for
51/2 + 51/2+ log5 (sinx) = 15 2
x Î [0, 100 p] is
Number of solution(s) of th e equation
sin x
sin 3x
sin 9x
+
= 0 in the interval
+
cos 3x cos 9x cos 27x
1
23.
24.
æ pö
ç 0, 4 ÷ is _______.
è
ø
Number of integral value(s) of m for which the
4m - 6
equation sin x – 3 cos x =
has
4-m
solutions, x Î [0, 2p], is _______.
The number of solutions of the equation
pö
æ
cos 2 ç x + ÷ + cos2 x – 2
è
6ø
pö
p
p
æ
cos ç x + ÷ × cos = sin 2 in interval
6ø
6
6
è
2
2 sin x - 3
1
2
3
1
(c)
The number of
æ
æ sin x ö ö
çè 2 sin çè 2 ÷ø ÷ø
25.
æ -p p ö
ç 2 , 2 ÷ is _______.
è
ø
The number of solutions of the equation
1 + cos x + cos 2x + sin x + sin 2x + sin 3x = 0,
which satisfy the condition
___________.
(d) 5
ANSW ER KEY
10 (a) 13 (b) 16 (c) 19
11 (a) 14 (a) 17 (c) 20
12 (c) 15 (c) 18 (b) 21
(a)
(c)
(50)
22
23
24
p
p
< 3x - £ p is
2
2
(6)
(4)
(2)
25
(2)
4
PRINCIPLE OF
MATHEMATICAL INDUCTION
MCQs with One Correct Answer
1.
5.
1
1
1
1
+
+
+ ...... +
n n +1 n + 2
2n - 1
If P(n) = 2 + 4 + 6 + .....+ 2n, n Î N , then
P(k) = k (k + 1) + 2
2.
= 1-
we can conclude that P(n ) = n(n + 1) + 2 for
(a) all n Î N
(b) for even values of n
(c) for odd values of n
(d) not true for any n
For all n ³ 1, find
(a) all n Î N
(b) n > 1
(c) n > 2
(d) nothing can be said
The greatest positive integer, which divides
(a) 2
6.
1
1
1
1
+
+
+ ..... +
1.2 2.3 3.4
n( n + 1)
(b) 6
(c) 24
(d) 120
(a)
n
n +1
(b)
(c)
1
n(n + 1)
(d) None of these
For any n Î N , the value of the expression
2 + 2 + ..... + 2 is
n - roots
(a)
(c)
4.
1 1 1
1
+ - + ...... +
holds for
2 3 4
2n - 1
Þ P(k + 1) = (k + 1)(k + 2) + 2 for all k Î N . So
n (n + 1)(n + 2)(n + 3) for all n Î N , is
3.
If n Î N , then the result
7.
æ p ö
2 cos çç
÷÷
è 2 n +1 ø
(b)
æ p ö
2 sin çç
÷÷
è 2 n +1 ø
Let a(n) = 1 +
1
1 1 1
+ + +…+ n
. Then
2 3 4
(2 ) - 1
(a) a(100) £ 100
(c) a(200) £ 100
(b) a(100) > 100
(d) a(200) < 100
For all natural numbers n, find
æ 3 ö æ 5 ö æ 7 ö æ 2n + 1 ö
ç1 + ÷ ç1 + ÷ ç1 + ÷ ..... ç1 + 2 ÷
è 1 øè 4 øè 9 ø è
n ø
2 cos( 2 n +1 p) (d) none of these
For a positive integer n,
1
n +1
8.
9.
(a) (n + 1)2
(b) (n – 1)2
(c) n(n + 1)
(d) None of these
2n > n2 when n Î N such that
(a) n > 2 (b) n > 3 (c) n < 5 (d) n ³ 5
If
4n
(2n)!
, then P(n) is true for
<
n + 1 (n!)2
(a) n ³ 1 (b) n > 0 (c) n < 0 (d) n ³ 2
MATHEMATICS
8
10.
11.
12.
13.
14.
15.
If P(n) : 3n < n!, n Î N, then P(n) is true
(a) for n ³ 6
(b) for n ³ 7
(c) for n ³ 3
(d) for all n
(a)
If n is a positive integer, then 2 . 42n + 1 + 33n + 1 is
divisible by :
(a) 2
(b) 7
(c) 11
(d) 27
For every natural number n, n(n2 – 1) is divisible
by
(a) 4
(b) 6
(c) 10
(d) None of these
If p is a prime number, then n p – n is divisible by p
when n is a
(a) Natural number greater than 1
(b) Irrational number
(c) Complex number
(d) Odd number
If 49n + 16n + l is divisible by 64 for all n Î N,
then the least negative value of l is
(a) –2
(b) –1
(c) –3
By mathematical induction,
(c)
16.
17.
1
2
3
4 ( n + 2 ) ( n + 3)
n (n + 2)
4 (n + 1) (n + 3)
(b)
is
n (n + 3)
4 (n + 1) (n + 2)
4
5
6
(a)
(a)
(a)
7
8
9
(a)
(d)
(d)
10
11
12
(2n + 1) 3n
4
+3
(d)
(2n – 1) 3n + 1 + 3
4
(2n – 1) 3n + 1 + 1
4
For all n Î N,
1+
1
1
1
+
+ ..... +
1+ 2 1+ 2 + 3
1 + 2 + 3 + ..... + n
is equal to
(a)
3n
(b)
n +1
n
(c)
n +1
2n
(d)
n –1
2n
n +1
19.
When 2301 is divided by 5, the least positive
remainder is
(a) 4
(b) 8
(c) 2
(d) 6
20.
By the principle of induction " n Î N, 32n
when divided by 8, leaves remainder
(a) 2
(b) 3
(c) 7
(d) 1
21.
If n Î N, then 11n + 2 + 122n+1 is divisible by
___________.
22.
For all n Î N, 41n – 14n is a multiple of
___________.
If m, n are any two odd positive integers with
n < m, then the largest positive integer which
divides all the numbers of the type m2 – n2 is
___________.
For every natural number n, 32n + 2 – 8n – 9 is
divisible by ___________.
The remainder when 599 is divided by 13, is
___________.
23.
(d) None of these
For all n Î N, 3.52n + 1 + 23n + 1 is divisible by
(a) 19
(b) 17
(c) 23
(d) 25
For all n Î N, 1.3 + 2.32 + 3.33 + ..... + n.3n is
equal to
(d)
(c)
(a)
4
(b)
Numeric Value Answer
equal to
n (n + 1)
18.
(d) – 4
1
1
1
+
+ ..... +
1× 2 ×3 2 × 3× 4
n (n + 1)(n + 2)
(a)
(c)
(2n + 1) 3n + 1 + 3
24.
25.
ANSW ER KEY
(b) 13 (a) 16
(c) 14 (b) 17
(b) 15 (b) 18
(b)
(b)
(d)
19 (c) 22
20 (d) 23
21 (133) 24
(27)
(8)
(16)
25
(8)
5
COMPLEX NUMBERS AND
QUADRATIC EQUATIONS
the point in the complex plane given by wk =
MCQs with One Correct Answer
1.
2.
(b) 16(x12 + y12)
(a) 32(x12 + y12)
(d) 32
(c) 4(x12 + y12)
Let z and w be two non-zero complex numbers
such that | z | = | w | and Arg z + Argg w = p, then
z equals
(a)
3.
cos 2a k + i sin 2a k
for k = 1, 2, 3. The origin, O
zk
Let x1 and y1 be real numbers. If z1 and z2 are
complex numbers such that |z1| = |z2| = 4, then
|x1z1 – y1z2|2 + |y1z1 + x1z2|2 =
w
(b) – w
(c)
w
7.
the line b z + b z = c, b ¹ 0 in the argand plane,
(d) – w
If | z - 1| + | z + 3 | £ 8, then the range of values
of | z - 4 | is
4.
(a) (0, 7) (b) (1, 8) (c) [1, 9] (d) [2, 5]
Which of the following is/are value of
2
is the
(a) incentre of DA1A2A3
(b) orthocentre of DA1A2A3
(c) circumcentre of DA1A2A3
(d) centroid of DA1A2A3
If a point z1 is the reflection of a point z2 through
8.
2
then b z2 + b z1 is equal to
(a) 4c
(b) 2c
(c) c
(d) None of these
z1 and z2 lie on a circle with centre at the origin.
The point of intersection z3 of the tangents at z1
and z2 is given by
sin ln (ii )i + cos ln (ii )i ?
(a) – 1
(b) 1
(c) 0
(d) None of these
(a)
1
( z1 + z2 )
2
(b)
2z1 z2
z1 + z2
5.
If n1, n2 are positive integers, then (1+ i ) n1 +
(c)
(d)
z1 + z2
z1 + z2
6.
(1 + i 3 )n1 + (1+ i 5 )n2 + (1+ i 7 )n2 is real number
if and only if:
(a) n1 = n2 + 1
(b) n1 + 1 = n2
(c) n1 = n2
(d) n1,n2 are any positive integers
Let rk > 0 and zk = rk (cos ak + i sin ak) for k = 1,
1æ 1 1 ö
+
2 çè z1 z2 ÷ø
2, 3 be such that
1
1
1
+
+
= 0 Let Ak be
z3
z2
z1
9.
If z1, z2 are two complex numbers such that
z1 – z2
= 1 and i z1 = Kz2, where K Î R, then
z1 + z2
the angle between z1 – z2 and z1 + z2 is
æ 2K ö
(a) tan– 1 ç 2 ÷
è K + 1ø
æ 2K ö
(b) tan– 1 ç
÷
è 1– K 2 ø
(c) – 2 tan– 1 K
(d) 2 tan– 1 K
MATHEMATICS
10
10.
Let 'z' be a complex number and 'a' be a real
parameter such that z2 + az + a2 = 0, then which
is of the following is not true?
(a) locus of z is a pair of staight lines
(b) |z| = |a|
16.
2p
3
(d) None of these
Let P denotes a complex number z = r(cos q + i
sin q) on the Argand's plane, and Q denotes a
17.
(c) arg (z) = ±
11.
æ
p
pö
complex number 2| z |2 çcosæçq+ ö÷ +i sinæçq+ ö÷÷ .
è 4øø
è è 4ø
12.
y = 2(a - x) ( x + x 2 + b2 ) is
18.
If 'O' is the origin, then DOPQ is
(a) isosceles but not right angled
(b) right angled but not isosceles
(c) right isosceles
(d) equilateral
If w ¹ 1 and w3 = 1, then
19.
1
1
= 1 and a = z2017 + 2017 and b is
z
z
n
2
the last digit of the number 2 – 1, when the
integer n > 1, the value of a2 + b2 is
(a) 23
(b) 24
(c) 26
(d) 27
If y1 = max ||z – w| – |z – w2||, where |z| = 2 and
20.
+
is equal to
a w2 + b w + c a + b w + c w2
(a) 2
(b) w
(c) 2w
(d) 2w2
13.
14.
If z +
1
y2 = max ||z – w| – |z – w2||, where |z| = and w and
2
w2 are complex cube roots of unity, then
15.
(a) y1 =
3 ; y2 =
3
(b) y1 <
3 ; y2 =
3
(c) y1 =
3 ; y2 <
3
(d) y1 > 3; y2 < 3
If a and b are the roots of the equation ax2 + bx
+ c = 0, (c ¹ 0) , then the equation whose roots
1
1
and
is
a
b
+b
aa + b
(a) acx2 – bx + 1 = 0 (b) x2 – acx + bc + 1 = 0
(c) acx2 + bx – 1 = 0 (d) x2 + acx – bc + 11 = 0
are
(a)
a 2 + b2
(b)
(c)
a 2 + 2b 2
(d) None of these
a2 - b2
If a, b are the roots of x 2 + px + q = 0, and
x 2 n + p n x n + q n = 0 and
a w2 + b + cw
a w + b + c w2
If the roots of the equation x2 + px + c = 0 are 2,
–2 and the roots of the equation x2 + bx + q = 0
are –1, –2, then the roots of the equation x2 + bx
+ c = 0 are
(a) –3, –2 (b) –3, 2 (c) 1, – 4 (d) –5, 1
If x Î R, then the maximum value of
a
is a root of
b
x n + 1 + ( x + 1)n = 0, a n ¹ bn , the n must be
(a) any integer
(b) an even integer
(c) an odd integer
(d) None of these
If 0 < a < b < g < p/2, then the equation (x – sin b)
(x – sin g) + (x – sin a) (x – sin g) + (x – sin a)
(x – sin b) = 0 has
(a) real and unequal roots.
(b) non-real roots.
(c) real and equal roots.
(d) real and unequal roots greater than 2.
The set of values of a for which inequation
(a – 1) x2 – (a + 1)x + a – 1 ³ 0 is true for all x ³ 2
(a)
é 7ù
ê1, 3 ú
ë û
(b) (– ¥, 1)
(c)
é7 ö
ê 3 , ¥÷ø
ë
(d) None of these
Numeric Value Answer
21.
a, b, c are integers, not all simultaneously equal
and w is cube root of unity (w ¹ 1), then minimum
value of |a + bw + cw2| is
22.
If 2 – i is a root of the equation ax2 + 12x + b = 0
(where a and b are real), then the value of ab is
equal to
23.
If the equations ax 2 + bx + c = 0 and
cx2 + bx + a = 0, a ¹ c have a negative
common root, then the value of a - b + c is
Complex Numbers and Quadratic Equations
24. If a, b are the roots the quadratic equation
x2 – (3 + 2 log2 3 –3 log 3 2 ) x –2(3log3 2 – 2log2 3 )
= 0, then the value of a2 + ab + b2 is equal to
11
28. If z and w are two complex numbers having
non-negative imaginary parts such that
æ z - 2ö
æ w - 1ö
= arg ç
= p / 2, then
arg ç
è z + 2 ÷ø
è w + 1÷ø
25. Let f (x) = x4 + ax3 + bx2 + cx + d be a polynomial
with real coefficients and real zeroes. If |f (i)| = 1,
(where i = -1 ) then a + b + c + d is equal to
26. If w and w2 be the non-real cube roots of unity
1
1
1
+
+
= 2w 2 and
a+w b+w c+w
and
1
a + w2
+
1
b + w2
+
1
30.
1
1
1
+
+
is
a + 1 b +1 c + 1
equal to :
27. For any real x, the maximum value of
2
k2
1
2
3
4
(
(x - k) x +
(a)
(d)
(c)
(b)
5
6
7
8
(d)
(d)
(c)
(b)
2
x +k
9
10
11
12
29. If z =
= 2w, where a, b, c
c + w2
are real then the value of
| w- z |< k; evaluate k. (Here k is least upper
bond)
2
)
æ 1 - pi
p
p -i ö
+
(1 + i )4 çç
÷÷ , where
4
è p + i 1 + pi ø
æ |z| ö
i = -1, then ç amp( z ) ÷ equals to
è
ø
If p, q, r are positive and are in A.P., then the
roots of the quadratic equation px2 + qx + r = 0
are real for
r
- 7 ³ k 3 then find the value
p
of k.
is equal to
(d)
(d)
(c)
(d)
13
14
15
16
ANSW ER KEY
(c) 17 (a) 21
(c) 18 (b) 22
(a) 19 (a) 23
(c) 20 (c) 24
(1)
(45)
(0)
(7)
25
26
27
28
(0)
(2)
(2)
(3)
29
30
(4)
(4)
MATHEMATICS
12
6
LINEAR INEQUALITIES
MCQs with One Correct Answer
1.
The set of all x satisfying the inequality
1
2x - 1
- 3
³0
x - x +1 x +1 x +1
(a) (–¥, 2]
(b) [1, 2]
(c) (–¥, –1) È (–1, 2] (d) (2, ¥]
The set of all x satisfying the inequality
(2x + 1) (x – 3) (x + 7) < 0
2
2
2.
-
(a) (–¥, 7)
3.
4.
5.
6.
æ 1 ö
(b) ç - ,3 ÷
è 2 ø
æ 1 ö
(c) (–¥, 7) È ç - ,3 ÷ (d) (–¥, 3)
è 2 ø
The set of real values of x satisfying
| x - 1| £ 3 and | x - 1| ³ 1 is
(a) [2, 4]
(b) (-¥, 2] È [4, + ¥)
(c) [-2, 0] È [2, 4]
(d) none of these
The system of equation
| x - 1 | +3y = 4, x - | y - 1 |= 2 has
(a) No solution
(b) A unique solution
(c) Two solutions
(d) More than two solutions
The number of real roots of the equation
| 2- | 1- | x |||= 1 is
(a) 1
(b) 3
(c) 5
(d) 6
The solution set of the inequality
| x + 2 | - | x - 1 |< x -
3
is
2
7.
8.
9.
(a)
ö
æ9
ç , ¥÷
2
ø
è
(c)
3ö
æ
ç - 2, - ÷
2ø
è
12x
³ 1 for all real values of x, the
4x 2 + 9
inequality being satisfied only if | x | is equal to
2
1
3
1
(a)
(c)
(d)
(b)
3
3
2
2
The equation | x - 1| +a = 4 can have real
solutions for x if ‘a’ belongs to the interval
(a) (-¥, + ¥)
(b) (-¥, 4]
If
(c) (4, + ¥)
(d) [–4, 4]
The interval(s) that satisfy the equation
x 2 - 8x + 12
x 2 - 10x + 21
10.
11.
3ö
æ
(b) ç - ¥, ÷
2ø
è
3ö
æ
(d) ç - 1, ÷
2ø
è
=-
x 2 - 8x + 12
x 2 - 10x + 21
is /are
(a) (-¥, 2]
(b) [2, ¥)
(c) [6, 7)
(d) [3, 6] È [7, ¥)
The set of all real x satisfying the inequality
3- | x |
³ 0 is
4- | x |
(a) [– 3, 3] È (– ¥, – 4) È (4, ¥)
(b) (– ¥, – 4) È (4, ¥)
(c) (– ¥, – 3) È (4, ¥)
(d) (– ¥, – 3) È (3, ¥)
If x, y, z be any three positive real number and
x +z
x+y
y+z
A= 2
+ 2
+ 21 2
2
2
x +y
y +z
x +z
and B =
1 1 1
+ + ;
x y z
Linear Inequalities
13
then which of the following is true?
(a) A ³ B (b) A £ B (c) A = B (d) A < B
12. If x, y, z are arbitrary real numbers satisfiying the
condition xy + yz + zx < 0 and if
x 2 + y2 + z 2
then only one of the
xy + yz + zx
following statements is always correct.
Which one is it?
(a) – 1 £ u < 0
(b) u takes all negative real values
(c) – 2 < u £ – 1
(d) u £ – 2
13. If a1,a2.....,an are positive real numbers whose
product is a fixed number c, then the minimum
value of a1 + a2 + .......+an-1 + 2an is
(a) n(2c)1/n
(b) (n + 1)c1/n
1/n
(c) 2nc
(d) (n + 1)(2c)1/n
14. The set of real values of x for which
u =
log 0.2
(a)
æ 3p ö æ 3p ö
(a) x Î (3, p) È ç p, ÷ È ç , 5 ÷
è 2 ø è 2 ø
(b) x Î (3, p) È (p, 5)
æ 5p ö
(c) x Î ç 3, ÷
è 2 ø
(d) None of these
20. If x is real, the maximum value of
3x 2 + 9 x + 17
3x 2 + 9 x + 7
x+2
£ 1 is
x
1
4
(a)
5ù
æ
ç - ¥, - ú È (0, ¥) (b)
2û
è
é5
ö
ê 2 , ¥÷
ë
ø
is
(b) 41
(c) 1
(d)
17
7
Numeric Value Answer
(-¥, - 2) È [0, ¥) (d) None of these
21. If ‘k’ any integer such that set of all real values
of x, satisfy the equation (sin3 x ) (cos3 x )
x 2 + 6x + 9
< - log 2 ( x + 1), then x lies
2(x + 1)
in the interval
3x + 3- x
; then find the minimum value k.
k
22. For any x, y Î R, xy > 0. Then the minimum value
(c)
15. If log1 / 2
(a)
(-1, - 1 + 2 2 )
(b)
1 + log 5 ( x 2 + 1) £ log 5 (ax 2 + 4x + a ) is true for
all x Î R is
(a) 6
(b) 7
(c) 10
(d) 1
17. The set of all real numbers x for which
x2 – [x + 2] + x > 0, is
(a) (- ¥,-2) È (2, ¥ )
(b)
(c)
(d)
(c)
(c)
(c)
(- ¥,- 2 )È (
2,¥
(- ¥,-1) È (1, ¥ )
(
2, ¥
4
5
6
)
7
8
9
of
2x
y3
10
11
12
x 3 y 4y 2
+ 4 is.
3
9x
x + 3 - x ³ 3 - x + 3 is
x+3 + x
> 1 and then find
x+2
modulus of smallest integral member of the
interval.
The number of integers satisfying the equation
24. Solve for x,
|x|+
(a)
(b)
(c)
+
23. Number of integeral solution satisfying
25.
)
(b)
(c)
(a)
£
(1 - 2 2 , 2)
(c) (-1, ¥)
(d) None of these
16. The least integer a, for which
1
2
3
18. Solution set of the in equality log102x
– 3 (log10 x) (log10 (x – 2)) + 2log102 (x – 2) < 0, is :
(a) (0, 4)
(b) (– ¥, 1 )
(c) (4, ¥)
(d) (2, 4)
19. The solution set of log|sin x| (x2 – 8x + 23)
3
>
contains
log 2 | sin x |
ANSW ER KEY
(a) 13 (a) 16
(b) 14 (a) 17
(d) 15 (a) 18
4 - x2
4 - x2
= x+
is
x
x
(b)
(b)
(c)
19
20
21
(a)
(b)
(4)
22
23
24
(2)
(1)
(4)
25
(4)
MATHEMATICS
14
PERMUTATIONS AND
COMBINATIONS
MCQs with One Correct Answer
1.
2.
3.
4.
Ten different letters of an alphabet are given
words with five letters are formed from three
given letters. Then the number of words which
have at least one letter repeated are
(a) 69760
(b) 30240
(c) 99748
(d) None of these
How many different nine digit numbers can be
formed from th e number 223355888 by
rearranging its digits so that the odd digits
occupy even positions ?
(a) 16
(b) 36
(c) 60
(d) 180
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
(a) 360 (b) 192 (c) 96
(d) 48
The total number of 5-digit numbers of different
digits in which the digit in the middle is the
largest is
(a)
9
å
n =4
5.
6.
n
p4
(b) 4563
(c) 2688
(d) 5292
All possible 120 permutations of WDSMC are
arranged in dictionary order, as if each were an
ordinary five-letter word. The last letter of the
86th word in the list, is :
(a) W
(b) D
(c) M
(d) C
If m be the number of different words that can be
formed with the letters of the word BHARAT in
which B and H are never together and n be
number of different words that can be formed
with the letters of the words BHARAT in which
7.
8.
9.
10.
11.
7
words always begin with B and end with T. Then
m/n is
(a) 10
(b) 20
(c) 1
(d) 2
Anil have tiled his square bathroom wall with
congruent square tiles. All the tiles are red,
except those along the two diagonals, which are
all blue. If he used 121 blue tiles, then the
number of red tiles used are
(a) 900 (b) 1800 (c) 3600 (d) 7200
The number of distinct natural numbers up to a
maximum of four digits and divisible by 5, which
can be formed with the digits 0, 1, 2, 3, 4, 5, 6, 7,
8, 9, each digit not occurring more than once in
each number, is
(a) 1246
(b) 952
(c) 1106
(d) None of these
6 white and 6 black balls are distributed among
ten identical urns, so that there is atleast one
ball in each urn. Balls are all alike except for the
colour and each box can hold any number of
balls. The number of different distributions of
the balls is:
(a) 26250 (b) 132 (c) 12
(d) 10
Number of ways in which two Americans, two
British, one Chinese, one Dutch and one Egyptian can sit on a round table so that persons of
the same nationality are separated is
(a) 48
(b) 240
(c) 336
(d) None of these
In an examination of 9 papers a candidate has to
pass in more papers than the number of papers
in which he fails in order to be successful. The
number of ways in which he can be unsuccessful
is
(a) 255 (b) 256 (c) 193 (d) 319
Permutations and Combinations
15
12. During a draw of lottery, tickets bearing numbers
1, 2, 3, ..., 40, 6 tickets are drawn out and then
arranged in the descending order of their
numbers. In how many ways, it is possible to
have 4th ticket bearing number 25?
(a) 15C3 × 24C2
(b) 12C3 × 20C2
(c) 15C3 + 24C2
(d) None of these
13. A teacher takes 3 children from her class to the
zoo at a time as often as she can, but she does
not take the same three children to the zoo more
than once he finds than she goes to the zoo 84
times more than a particular child goes to the
zoo. The number of children in her class is
(a) 12
(b) 10
(c) 60
(d) None of these
14. Messages are conveyed by arranging four white,
one blue, and three red flags on a pole. Flags of
the same colour are alike. If a message is
transmitted by the order in which the colours
are arranged, the total number of messages that
can be transmitted if exactly six flags are used is
(a) 45
(b) 65
(c) 125 (d) 185
15. There were two women participating in a chess
tournament. Every participant played two games
with the other participants. The number of games
that the men played between themselves proved
to exceed by 66 the number of games that the men
played with the women. The number of participants is
(a) 6
(b) 11
(c) 13
(d) None of these
16. In a class tournament, all participants were to
play different games with one another. Two
players fell ill after having played three games
each. If the total number of games played in the
tournament is equal to 84, the total number of
participants in the beginning was equal to
(a) 10
(b) 15
(c) 12
(d) 14
17. The number of ways in which 5 X's can be placed
in the squares of the figure so that no row
remains empty is
(a) 97
1
2
3
(a)
(c)
(c)
(b) 44
4
5
6
(d)
(b)
(b)
(c) 100
7
8
9
(c)
(c)
(d)
10
11
12
(d)
(c)
(b)
(a)
18. There are three coplanar parallel lines. If any p
points are taken on each the lines, the maximum
number of triangles with vertices at these points is
(a) 3p2(p – 1) + 1
(b) 3p2(p – 1)
2
(c) p (4p – 3)
(d) None of these
19. From the vertices of a regular polygon of 10 sides,
the number of ways of selecting three vertices
such that no two vertices are consecutive is
(a) 10
(b) 30
(c) 50
(d) 40
20. 5 different objects are to be distributed among 3
persons such that no two persons get the same
number of objects. Number of ways this can be done, is
(a) 60
(b) 90
(c) 120 (d) 150
Numeric Value Answer
21. If a, b, c are three natural numbers in AP such that
a + b + c = 21 and if possible number of ordered
triplet (a, b, c) is then the value of (l – 5) is
22. In a single correct match the column question,
column I contain 10 questions and Column II
contain 10 answers written in some arbitrary order.
If the number ways a student can answer this
question so that exactly 6 of his matching are
correct is k, then (sum of digits of k)/2 is equal to
23. There are 720 permutations of the digits 1, 2, 3, 4,
5, 6. Suppose these permutations are arranged
from smallest to largest numerical values,
beginning from 1 2 3 4 5 6 and ending with 6 5 4
3 2 1. Then the digit in unit place of number at
267th position is ...... .
24. If N is the number of ways in which a person can
walk up a stairway which has 7 steps if he can
take 1 or 2 steps up the stairs at a time, then the
value of N/3 is ...... .
25. In an international convention participants from
10 different countries were arranged in a row such
that all the participants from the same country
were together. Each country has different number
of participants with maximum 10 participants from
a country. If K is the number of ways that they
can be arranged in a row then find the highest
power of 10 in K.
126
ANSWER KEY
13 (b) 16
14 (d) 17
15 (c) 18
(b)
(b)
(c)
19
20
21
(c)
(b)
(8)
22
23
24
(9)
(6)
(7)
25
(9)
MATHEMATICS
16
8
BINOMIAL THEOREM
MCQs with One Correct Answer
1.
2 60 when divided by 7 leaves the remainder
(a) 1
(b) 6
(c) 5
(d) 2
2.
If 7 divides 32 32 , the remainder is
(a) 1
(b) 0
(c) 4
(d) 6
3.
4.
5.
8.
32
æ r -1
ö
n r
p÷
ç
C
C
2
is equal to
å
r
p
å ç
÷
r=1 è p =1
ø
(a) 4n – 3n + 1
(b) 4n – 3n – 1
n
n
(c) 4 – 3 + 2
(d) 4n – 3n
2 6
If (1 + x – 2x ) = 1+ a1x + a2x2 + ..... + a12 x 12,
then the value of a2 + a4 + ..... + a12 is :
(a) 1024
(b) 64
(c) 32
(d) 31
The last term in the binomial expansion of
n
1/ 3
(2
n
1/ 3
- 1/ 2) is (1/ 3.9
9.
10.
n
11.
) log 3 8. Then the
n
If (1 + ax) = 1+ 8x +24x2 + ......... ; than a – n is
a+n
equal to (n being a positive integer)
(a) 3
(b) – 3
(c) –
7.
1
3
(d)
1
3
The unit digit of 17 2009 + 112009 – 7 2009 is
(a) 1
(b) 2
(c) 3
(d) 0
1 ö
æ 1/3
ç 3 + 1/3 ÷ . If y = 12x then the value of n is :
4 ø
è
(a) 9
(b) 8
(c) 10
(d) 11
The coefficient of the term independent of x in
æ
ö
x +1
x –1 ÷
ç
–
the expansion of ç 2
1
1÷
çè 3
÷
x – x3 +1 x – x2 ø
5th term from the beginning is
(a) 10 C6
(b) 2.10C4
10
(c) 1/2. C4
(d) None
6.
Let N = 21224 – 1, a = 2153 + 277 + 1 and b = 2408
– 2204 + 1. Then which of the following
statement is correct ?
(a) a divides N but b does not
(b) b divides N but a does not
(c) a and b both divides N
(d) neither a nor b divides N
The number N = 20C7 – 20C8 + 20C9 – 20C10 +
..... – 20C20 is not divisible by :
(a) 3
(b) 7
(c) 11
(d) 19
Let x be the 7th term from the beginning and y be
the 7th term from the end in the expansion of
12.
10
is
(a) 70
(b) 112 (c) 105 (d) 210
If the middle term of (1 + x)2n (x > 0, n Î N) is the
greatest term of the expansion. Then the interval
in which x lies, is
(a)
én +1 n + 2ù
êë n , n úû
(b)
é n –1 n + 1ù
êë n , n úû
(c)
n + 1ù
é n
êë n + 1 , n úû
(d) None of these
Binomial Theorem
17
13. The number of irrational terms in the expansion
(
of
8
)
19. The value of
5 + 6 2 100 is
(a) 97
(b) 98
(c) 96
(d) 99
14. The greatest value of the term independent of x in
the expansion of ( x sin a + x
(a)
5
(b)
2
(c)
–1
1
10!
(5!)2
(d) None of these
.
2 5 (5!) 2
25
The largest term in the expansion of (2 + 3x )
where x = 2 is its.
(a) 13th term
(b) 19th term
th
(c) 20 term
(d) 26th term
If the middle term in the expansion of
15.
16.
10
æ1
ö
7
ç + x sin x ÷ equals to 7 then x is equal to
x
è
ø
8
(n Î I)
p
p
(b) n p +
(a) 2np ±
6
6
np
n 5p
(c) n p + (-1)
(d) n p + ( -1)
6
6
17. The sum of the coefficients of all the integral
40
powers of x in the expansion of (1 + 2 x ) is
(a) 3 40 + 1
(b) 3 40 – 1
1 40
1 40
(3 – 1)
(3 + 1)
(c)
(d)
2
2
18. If p (n) denotes product of all binomial coefficients
in (1+ x ) n , then ratio of p (2002) to p (2001) is
2001
(a) 2002
(c)
1
2
3
(a)
(c)
(d)
(b)
(2001) 2002
(2002)!
4
5
6
(d)
(a)
(c)
(a)
2n + 1
n +1
(b)
2n
n +1
(c)
2n + 1
n -1
(d)
2n - 1
n +1
cos a )10 , a Î R, is
10!
(2002)
(2001) !
C1 C 3 C 5
+
+
+ ....... is equal to
2
4
6
n
20. If I is integral part of (2 + 3) and f is its
fractional part. Then (I + f ) (1 – f ) is
(a) I + 1 (b) 1
(c) n
(d)
2n
Numeric Value Answer
21. The largest real value for x such that
æ 34 - k öæ x k ö 32
÷ç ÷ =
å çç
÷ç ÷ 3 is
k = 0 è ( 4 - k )! øè k! ø
4
22. Find the coefficient of x13 in the expansion of
(1 – x)5 (1 + x + x2 + x3 )4 is
23. If (1 + x)n = C0 + C1x + C2x2 + ..... + Cnxn, then
find the value of 3C0 – 5C1 + 7C2 + ...+ (–1)n
(2n+3)Cn
24. If
1
1
1
1
2k
+
+
+ ... +
=
1!10! 3!8! 5!6!
11!1! 11!
then
k
is equal to
2
25. (1 – 2x + 5x2 – 10x3) (1 + x)n = 1 + a1x + a2x2 + ...
and that a21 = 2a2, then the value of n is____
(d) 2001
7
8
9
(a)
(c)
(c)
10
11
12
(a)
(d)
(c)
ANSWER KEY
13 (a) 16
14 (c) 17
15 (c) 18
(c)
(d)
(b)
19
20
21
(d)
(b)
(1)
22
23
24
(4)
(0)
(5)
25
(6)
MATHEMATICS
18
9
SEQUENCES AND SERIES
1.
MCQs with One Correct Answer
Sum to n terms of the series
(a) x, y, z are in A.P.
3
13 + 3. 23 + 33 + 3. 43 + 5 + ................ is
(n is even)
(a)
n(n 2 + 1) (2 n + 1)
3
(b)
n(n3 + 4n2 + 10n + 8)
8
(c)
n(n3 + 1)
8
7.
(d)
2.
3.
(a)
n (4n2 –1) c 2
6
(b)
n (4n2 +1) c 2
3
n (4n2 +1) c 2
n (4n2 –1) c 2
(d)
6
3
If a1, a2, ..., an are in A.P. with common difference
d ¹ 0, then (sin d) [sec a1 seca2 + sec a2 sec a3 +
... + sec an–1 sec an] is equal to
(a) cot an – cot a1
(b) cot a1 – cot an
(c) tan an – tan a1
(d) tan an – tan an–1
1 1 1
1 3 1 3
If x = 2 + 2 + 2 + ..., y = 2 + 2 + 2 + 2
1 3 5
1 2 3 4
1 1 1 1
+ ... and z = 2 – 2 + 2 – 2 +..., then
1 2 3 4
8.
9.
(c)
4.
5.
y x z
, , are in A.P..
6 3 2
y x z
, , are in A.P.. (d) 6y, 3x, 2z are in A.P.
6 3 2
a +b
b+c
, b,
For a, b, c Î R – {0}, let
are in
1– ab 1– bc
A.P. If a, b are the roots of the quadratic
equation 2ac x2 + 2abc x + (a + c) = 0, then the
value of (1 + a)(1 + b) is
(a) 0
(b) 1
(c) – 1
(d) 2
An A.P. consist of even number of terms 2n
having middle terms equal to 1 and 7 respectively. If n is the maximum value which satisfy
t1t2n + 713 ³ 0, then the value of the first term
of the series is
(a) 17
(b) – 15 (c) 21
(d) – 23
If a, b, c are in G. P., x and y be the A. M.’s between
(c)
6.
n2 (2n 2 + 6 n + 5)
4
If 1, log9 (31–x + 2), log3 (4.3x – 1) are in A.P., then
x equals
(a) log3 4
(b) 1 – log3 4
(c) 1 – log4 3
(d) log4 3
In the sum of first n terms of an A.P. is cn2, then
the sum of squares of these n terms is
(b)
10.
11.
æ a cö æ b bö
a, b and b, c respectively, then ç + ÷ ç + ÷
è x yø è x yø
is equal to
(a) – 2 (b) – 4
(c) 2
(d) 4
Suppose a, b, c are in A.P. and a2, b2, c2 are in G.P.
3
if a < b < c and a + b + c = , then the value of a is
2
1
1
(a)
(b)
2
3
2 2
1 1
1 1
(c) 2 (d) 2 3
2
An infinite G.P. has first term ‘x’ and sum ‘5’,
then x belongs to
(a) x < – 10
(b) – 10 < x < 0
(c) 0 < x < 10
(d) x > 10
In the quadratic equation ax2 + bx + c = 0, D = b2
– 4ac and a + b, a2 + b2, a3 + b3, are in G.P. where
a, b are the root of ax2 + bx + c = 0, then
(a) D ¹ 0 (b) bD = 0 (c) cD = 0 (d) D = 0
Sequences and Series
19
12. The sum of an infinite geometric series is 2 and the
sum of the geometric series made from the cubes
of this infinite sereis is 24. Then the series is
3 3 3
3 3 3
(a) 3 + - + - .... (b) 3 + + + + ....
2 4 8
2 4 8
3 3 3
(c) 3 - + - + ... (d) None of these
2 4 8
13. If a, b, c are in G. P. and log a – log 2b, log 2b –
log 3c and log 3c – log a are in A. P., then a, b, c
are the sides of a triangle which is
(a) Acute angled
(b) Obtuse angled
(c) Right angled
(d) None of these
14. Ar ; r = 1, 2, 3, ........... , n are n points on the
parabola y 2 = 4x in the first quadrant.
If Ar = ( xr , yr ) , where x1 , x2 , x3 , ...............,
xn are in G. P. and x1 = 1, x2 = 2, then yn is equal to
n +1
–2 2
n
22
n +1
n +1
(a)
(b) 2
(c) ( 2)
(d)
15. If three successive terms of a G..P. with common
ratio r (r > 1) form the sides of a D ABC and [r]
denotes greatest integer function, then [r] + [– r] =
(a) 0
(b) 1
(c) – 1
(d) None of these
16. If a, b, c, are in A.P. and p, p¢ are respectively
A.M. and G.M. between a and b while q, q¢ are
respectively AM.and G.M. between b and c, then
(a) p 2 + q 2 = p '2 + q '2
(b) pq = p ' q '
(c) p 2 - q 2 = p '2 - q '2
(d) p 2 + p¢2 = q 2 + q '2
17. The sum of the series
1 + 2.2 + 3.2 2 + 4.2 3 + 5.2 4 + ... + 100.2 99 is
(a) 99.2100 – 1
(b) 100.2100
(c) 99.2100
(d) 99.2100 + 1
18. If a, b, c, d are positive real number such that
a + b + c + d = 2, then M = (a + b) (c + d) satisfies
the relation:
(a) 0 < M £ 1
(b) 1 £ M £ 2
(c) 2 £ M £ 3
(d) 3 £ M £ 4
2
3 1
4 æ1ö
5 æ1ö
. +
.ç ÷ +
.ç ÷
1 .2 2
2.3 è 2 ø
3.4 è 2 ø
+ ......... to n terms is equal to
19. The sum of
1
2
3
(d)
(b)
(c)
4
5
6
(c)
(b)
(b)
7
8
9
(d)
(d)
(d)
10
11
12
(c)
(c)
(c)
13
14
15
3
(a) 1 –
(c)
1-
1
( n + 1)2 n
1
(b)
n -
1
2
n +1
(d) None of these
n.2n+1
b
³ c for all positive x, where a < 0
20. Let ax2 +
x
and b < 0. The value of the expression 27ab2
cannot be less than
(a) 4c3 (b) 4c2
(c) 8c3
(d) c3
Numeric Value Answer
21. Sum of infinite number of terms of GP is 20 and sum
of their square is 100. The common ratio of GP is
22. Three numbers a, b, c are in GP. If a, b, c – 64 are
in AP and a, b – 8, c – 64 are in GP, then the sum
of the numbers may be
23. a, b, c are positive integers forming an
increasing G.P. and b – a is a perfect cube and
log6 a + log6 b + log6 c = 6, then a + b + c =
24. The sum to infinite term of the series
25.
26.
27.
28.
2 6 10 14
1 + + 2 + 3 + 4 + ... is
3 3
3 3
The 20th term of the series 2 + 3 + 5 + 9 + 16
+.......is
Two consecutive numbers from 1, 2, 3,.........., n
are removed. If the arithmetic mean of the
n
remaining numbers is 105/4 then
is equal to
10
Let a, b, c, d be four distinct real numbers in
A.P. Then half of the smallest positive value of
k satisfying 2(a – b) + k(b – c)2 + (c – a)3 = 2(a
– d) + (b – d)2 + (c – d)3 is ..... .
Let x1, x2, ... Î (0, p) denote the of values of x
satisfying the equation
2
3
27(1 + |cos x| + cos x + |cos x| + ....upto ¥) = 93, find
1
the value of ( x1 + x2 + ...)
p
29. For a, b > 0, let 5a – b, 2a + b, a + 2b be
in A.P. and (b + 1)2, ab + 1, (a – 1)2 are in
G.P., then the value of (a–1 + b– 1) is ..... .
30. If one geomteric mean G and two Arithmetic
means P and q be inserted between two quantities, then G2 = (kp – q)(kq – p) then find k.
ANSWER KEY
(b) 16 (c)
(c) 17 (d)
(c) 18 (a)
19 (a)
22 (124) 25 (990) 28
20 (a)
23 (189) 26 (5) 29
21 (0.60) 24 (3) 27 (8) 30
(1)
(6)
(2)
MATHEMATICS
20
10
STRAIGHT LINES
MCQs with One Correct Answer
1.
5.
Through the point P (a, b), where ab > 0 , the
x y
+ = 1 is drawn so as to form
a b
with coordinate axes a triangle of area S. If
ab > 0, then least value of S is
straight line
(a)
(c)
2.
3.
2ab
ab
(b)
(c)
(d) None of these
1ö
1ö
æ
æ
çè bc, bc ÷ø and ç ca, ca ÷ where a, b, c are the
è
ø
roots of the equation x3 – 3x2 + 6x + 1 = 0. The
coordinates of its centroid are.
(a) (1, 2) (b) (2, – 1) (c) (1, – 1) (d) (2, 3)
Consider points A (3, 4) and B (7, 13). If P be a
point on the line y = x such that PA + PB is
minimum, then coordinates of P are
(a)
(a) (a, a)
1
ab
2
1ö
æ
The vertices of a triangle are ç ab, ÷ ,
è abø
æ 12 12 ö
ç ,
÷
è7 7ø
x
y
+ = 1 meets the axis of x and y at A
a b
and B respectively and the line y = x at C so that
area of the trinagle AOC is twice the area of the
triangle BOC, O being the origin, then one of the
positions of C is
The line
æ 13 13 ö
(b) ç , ÷
è7 7ø
æ 31 31 ö
ç , ÷
(d) (0, 0)
è 7 7ø
If the straight lines 2x + 3y – 1 = 0, x + 2y – 1 = 0
and ax + by – 1 = 0 form a triangle with origin as
orthocentre, then (a, b) is given by
(a) (6, 4) (b) (– 3, 3) (c) (– 8, 8) (d) (0, 7)
6.
(a) 5 < b £ 7
(b)
1
£ b£1
2
5
7
£ b £
(d) None of these
3
2
The intercepts on the straight line y = mx by the
lines y = 2 and y = 6 is less than 5, then m belongs
to
(c)
7.
2a ö
3 ÷ø
2b ö
3 ÷ø
The range of values of b such that (0, b ) lie
on or inside the triangle formed by the lines
y + 3x + 2 = 0, 3y – 2x – 5 = 0, 4y + x – 14 = 0 is
(a)
æ 4 4ö
çè - , ÷ø
3 3
(b)
æ 4 3ö
çè , ÷ø
3 8
(c)
4ö æ4 ö
æ
ç -¥, - ÷ È ç , ¥ ÷
3ø è3 ø
è
(d)
æ4 ö
çè , ¥÷ø
3
(c)
4.
æ b bö
çè 3 , 3 ÷ø
æ 2a
,
3
æ 2b
(d) ç ,
è 3
(b) ç
è
Straight Lines
8.
21
If three distinct points A, B, C are given in the
2-dimensional coordinate plane such that the ratio
of the distance of each one of them from the
point (1, 0) to the distance from (– 1, 0) is equal
1
to , then the circumcentre of the triangle ABC
2
is at the point
(a)
æ5 ö
ç ,0÷
è3 ø
(a) ax + by + 2a = 0
(c) bx + ay – 2b = 0 (d) ay – bx + 2b = 0
13. If the point (a, 2) lies between the lines x – y – 1
= 0 and 2 (x – y) + 5 = 0, then the set of values of
‘a’ is
æ9
ö
(a) (– ¥ , 3) È ç , ¥ ÷
è2
ø
(b) (0, 0)
æ1 ö
(d) (3, 0)
ç ,0÷
è3 ø
9.
Let A (–3, 2) and B (–2, 1) be the vertices of a
triangle ABC. If the centroid of this triangle lies
on the line 3x + 4y + 2 = 0, then the vertex C lies
on the line :
(a) 4x + 3y + 5 = 0
(b) 3x + 4y + 3 = 0
(c) 4x + 3y + 3 = 0
(d) 3x + 4y + 5 = 0
10. The circumcentre of a triangle lies at the origin
and its centroid is the mid point of the line
segment joining the points (a2 + 1, a2 + 1) and
(2a, – 2a), a ¹ 0. Then for any a, the orthocentre
of this triangle lies on the line:
(a) y – 2ax = 0
(b) y – (a2 + 1)x = 0
(c) y + x = 0
(d) (a – 1)2x – (a + 1)2y = 0
11. The line parallel to the x- axis and passing through
the intersection of the lines ax + 2by + 3b = 0 and
bx – 2ay – 3a = 0, where (a, b) ¹ (0, 0) is
(c)
(a) below the x - axis at a distance of
3
from it
2
2
(b) below the x - axis at a distance of from it
3
(c) above the x - axis at a distance of
3
from it
2
2
from it
3
The straight line y = x – 2 rotates about a point
where it cuts the x-axis and becomes
perpendicular to the straight line
ax + by + c = 0. Then its equation is
(b)
æ 9ö
ç 3, ÷
è 2ø
(c) (– ¥ , 3)
(d)
æ 1 ö
çè - , 3÷ø
2
14. If two vertices of a triangle are (5, –1) and (–2, 3)
and its orthocentre is at (0, 0), then the third
vertex is
(a) (4, – 7)
(b) (– 4, – 7)
(c) (– 4, 7)
(d) (4, 7)
15. The base of an equilateral triangle is along the
line given by 3x + 4y = 9. If a vertex of the triangle
is (1, 2), then the length of a side of the triangle
is:
(a)
2 3
15
(b)
4 3
15
(c)
4 3
5
(d)
2 3
5
16. The equation of bisector of that angle between
the lines x + y + 1 = 0 and 2x – 3y – 5 = 0 which
contains the point (10, – 20) is
(a)
x ( 13 + 2 2 ) + y ( 13 – 3 2 )
+ ( 13 – 5 2 ) = 0
(b) x ( 13 – 2 2 ) + y ( 13 + 3 2 )
+ ( 13 + 5 2 ) = 0
(d) above the x - axis at a distance of
12.
(b) ax – by – 2a = 0
(c) x ( 13 + 2 2 ) + y ( 13 + 3 2 )
+ ( 13 + 5 2 ) = 0
(d) None of these
MATHEMATICS
22
17.
The bisector of the acute angle formed
always pass through a fixed point P for all
possible values of q. If the maximum value of the
difference of distances of P and B (3, 4) from a
between the lines 4 x - 3 y + 7 = 0 and
3x - 4 y + 14 = 0 has the equation :
(a)
x+ y+3= 0
(b)
x - y -3 = 0
(c)
x- y+3 = 0
(d) 3x + y - 7 = 0
22.
Numeric Value Answer
18.
A straight line through the origin O meets the
parallel lines 4x + 2y = 9 and 2x + y + 6 = 0 at
points P and Q respectively. If the point
O divides the segment PQ in the ratio
19.
20.
m
, then
n
m + n is ________.
The vertex of an equilateral triangle is (2, –1),
and the equation of its base is x + 2y = 1. If the
23.
24.
length of its sides is 2 / K , then value of K is
____.
If (sin q, cos q), q Î [0, 2p] and (1, 4) lie on the
same side or on the line
3x – y + 1 = 0, then the
maximum value of sin q will be ______.
21.
point on the line x - y + 3 = 0 is k then
The straight lines (3sec q + 5cosec q)x
+ (7 sec q - 3cosec q) y + 11(sec q - cosec q) = 0
25.
k2
is
10
equal to .
The straight line L º x + y + 1 = 0 and L1 º x + 2y
+ 3 = 0 are intersecting. m is the slope of the
straight line L2 such that L is the bisector of the
anlge between L1 and L2. The unit digit of
812m2 + 3 is equal to
If tana, tanb, tanl are the roots of the equation
t3 – 12t2 + 15t – 1 = 0; then the centroid of
triangle having vertices (tana, cota); (tanb,
cotbb); (tanl, cotl) is given by G(h, k); then
evaluate (h + k)/(k – h).
Consider a DABC whose sides AB, BC, and CA
are represented by the straight lines 2x + y = 0,
x + py = q, and x – y = 3, respectively. The point
P(2, 3) is the orthocenter. The value of (p + q)/10
is ..... .
In DABC, the vertex A = (1, 2), y = x is the perpendicular bisector of the side AB and x – 2y + 1 = 0
is the equation of the internal angle bisector of
L . If the equation of the side BC is ax + by – 5
= 0, then the value of a – b is ..... .
1
2
3
(a)
(b)
(c)
4
5
6
(c)
(d)
(c)
7
8
9
(c)
(a)
(b)
10
11
12
ANSW ER KEY
(d) 13 (d) 16
(a) 14 (b) 17
(d) 15 (b) 18
(a)
(c)
(7)
19
20
21
(15)
(0)
(4)
22
23
24
(1)
(9)
(5)
25
(4)
11
CONIC
SECTIONS
MCQs with One Correct Answer
1.
2.
3.
The line 4 x + 3 y - 4 = 0 divides the
circumference of the circle centered at (5, 3), in
the ratio 1 : 2. Then the equation of the circle is
(a)
x 2 + y 2 - 10 x - 6 y - 66 = 0
(b)
x 2 + y 2 - 10 x - 6 y + 100 = 0
(c)
x 2 + y 2 - 10 x - 6 y + 66 = 0
(d)
x 2 + y 2 - 10 x - 6 y –100 = 0
Let A(– 4, 0) and B(4, 0). Then the number of
points C = (x, y) on the circle x2 + y2 = 16 lying in
first quadrant such that the area of the triangle
whose vertices are A, B and C is a integer is
(a) 14
(b) 15
(c) 16
(d) None of these
If (a, b) is a point on the circle whose centre is
on the x-axis and which touches the line x + y = 0
at (2, –2), then the greatest value of a is
(a) 4 –
4.
5.
2
2
y = | x | + c and x + y 2 – 8| x | – 9 = 0 have no
solution is
(a) (– ¥ , – 3) È (3, ¥ )
(b) (– 3, 3)
(c) (–¥, – 2 ) È (5 2 , ¥)
(d) (5 2 –4, ¥)
(a) 2x2 + 2 y 2 + gx + fy = 0
2
(b) x + y 2 + gx + fy = 0
6.
(c) x 2 + y 2 + 2gx + 2fy = 0
(d) None of these
A ray of light incident at the point (– 2, – 1) gets
reflected from the tangent at (0, –1) to the circle
x 2 + y 2 = 1. The reflected ray touches the circle.
The equation the line along which the incident
ray moved, is
7.
(b) 6
(c) 4 + 2 2
(d) 4 + 2
The set of values of ‘c’ so that the equations
Tangents are drawn from O (origin) to touch the
circle x2 + y2 + 2gx + 2fy + c = 0 at points P and Q.
The equation of the circle circumscribing triangle
OPQ is
8.
(a)
4 x - 3 y + 11 = 0 (b)
4 x + 3 y + 11 = 0
(c)
3 x + 4 y + 11 = 0 (d)
4x + 3 y + 7 = 0
If the line y = mx + 1 meets the circle x2 + y2 + 3x
= 0 in two points equidistant from and on
opposite sides of x-axis, then
(a) 3m + 2 = 0
(b) 3m – 2 = 0
(c) 2m + 3 = 0
(d) 2m – 3 = 0
If a circle passes through the point (a, b) and
cuts the circle x 2 + y 2 = 4 orthogonally, then
the locus of its centre is
(a)
2ax - 2by - (a 2 + b2 + 4) = 0
(b)
2ax + 2by - (a 2 + b2 + 4) = 0
(c)
2ax - 2by + (a 2 + b2 + 4) = 0
(d)
2ax + 2by + (a 2 + b2 + 4) = 0
MATHEMATICS
24
9.
10.
11.
12.
13.
14.
15.
The set of all real values of l for which exactly
two common tangents can be drawn to the circles
x2 + y2 – 4x – 4y + 6 = 0 and
x2 + y2 – 10x – 10y + l = 0 is the interval:
(a) (12, 32)
(b) (18, 42)
(c) (12, 24)
(d) (18, 48)
A circle bisects the circumference of the circle
x2 + y2 – 2y – 3 = 0 and touches the line x = y
and the point (1, 1). Its radius is :
3
9
(a)
(b)
(c) 4 2 (d) 3 2
2
2
Let L1 be the length of the common chord of the
curves x2 + y2 = 9 and y2 = 8x, and L2 be the
length of the latus rectum of y2 = 8x, then:
(a) L1 > L2
(b) L1 = L2
L1
= 2
(c) L1 < L2
(d)
L2
If the tangent at the point P (x1, y1) to the parabola
y2 = 4ax meets the parabola y2 = 4a (x + b) at Q
and R, then the mid-point of QR is
(a) (x1 + b, y1 + b)
(b) (x1 – b, y1 – b)
(c) (x1, y1)
(d) (x1 + b, y1 – b)
Tangent to the curve y = x2 + 6 at a point (1, 7)
touches the circle x2 + y2 + 16x + 12y + c = 0 at a
point Q. Then the coordinates of Q are
(a) (–6, –11)
(b) (–9, –13)
(c) (–10, –15)
(d) (–6, –7)
A circle is drawn with centre at the focus S of the
parabola y2 = 4x so that a common chord of the
parabola and the circle is equidistant from the
focus and the vertex. Then the equation of the
circle is
9
9
(a) (x – 1)2 + y2 =
(b) (x – 1)2 = – y2
16
4
9
9
(c) (x – 1)2 + x2 =
(d) (y – 1)2 + x2 =
16
4
Locus of all such points so that sum of its
distances from (2, – 3) and (2, 5) is always 10, is
(a)
( x - 2)2
( y - 1)2
+
=1
25
9
(b)
( x - 2)2
( y - 1)2
+
=1
25
16
(c)
( x - 2)2
( y - 1)2
+
=1
16
25
( x - 2)2
( y - 1)2
+
=1
9
25
The radius of the circle passing through the foci
(d)
16.
2
2
of the ellipse x + y = 1 , and having its
16
9
centre at (0, 3) is
7
1
(d)
2
2
Equation of the line passing through the points
of intersection of the parabola x2 = 8y and the
(a) 4
17.
(b) 3
(c)
x2
+ y 2 = 1 is :
3
(a) y – 3 = 0
(b) y + 3 = 0
(c) 3y + 1 = 0
(d) 3y – 1 = 0
Equation of the largest circle with centre (1, 0)
that can be inscribed in the ellipse x2 + 4y2 = 16,
is
ellipse
18.
2
2
(a) 2 x + 2 y – 4x + 7 = 0
(b)
x 2 + y 2 – 2x + 5 = 0
2
(c) 3 x + 3 y 2 – 6x – 8 = 0
(d) None of these
æ 3ö
ç 2, ÷ to the ellipse,
è 2ø
19.
The normal at
20.
x2 y 2
+
= 1 touch es a par abola, whose
16 3
equation is
(a) y2 = – 104 x
(b) y2 = 14 x
2
(c) y = 26x
(d) y2 = – 14x
The angle subtended by the common tangent of
the two ellipse
( x - 4)2 y 2
+
25
4
= 1 and
( x + 1)2 y 2
+
= 1 at the origin is
1
4
(a)
p
2
(b)
p
4
(c)
p
3
(d)
p
6
Conic Sections
25
Numeric Value Answer
21. Two equal chords AB and AC of the circle x2 + y2
– 6x – 8y – 24 = 0 are drawn from the point
A
(
)
33 + 3, 0 . Another chord PQ is drawn
intersecting AB and AC at points R and S,
respectively given that AR = SC = 7 and RB = AS
= 3. The value of PR/QS is
22. If p and q be the longest and the shortest distance
respectively of the point (–7, 2) from any point
(a, b) on the curve whose equation
27. S1 and S2 be the foci of the hyperbola whose
transverse axis length is 4 and conjugate axis
length is 6, S3 and S4 be the foci of the conjugate
hyperbola. If the area of the quadrilateral S1 S3
A
.
13
28. If the ratio of the area of equilateral triangles
made of the common chord of the circles x2 + y2
= 4 and x2 + y2 – 8x + 4 = 0 and their respective
pairs of tangents drawn from points on the
S2 S4 is A, then find
positive x- axis is 57 + 24 3 : k then k is
________.
is x 2 + y 2 - 10 x - 14 y - 51 = 0 and G.M. of p
and q is 2 k , then value k is _______.
23. The straight line y = mx + c (m > 0) touches the
29.
are drawn through P touching the coordinate
axes, such that the length of common chord of
these circle is maximum. If possible values of
2 = 8 (x + 2) then the minimum value
parabolas y
taken by c is
24. Two tangents are drawn from a point (–2, –1) to
the curve, y2 = 4x. If a is the angle between them,
then |tan a| is equal to:
2
y
x2
25. Tangents are drawn to the ellipse
+
=1
5
9
at ends of latus rectum. The area of quadrilateral
so formed is
26. A trapezium is inscribed in the parabola y2 = 4x
such that its diagonal pass through the point
25
. If the area of
(1, 0) and each has length
4
éPù
trapezium be P then ê ú is equal to
ë4û
P(a , b) is a points in the first quadrant. Circles
a/b is k1 ± k2 2 then k1 + k2 is equal
to______.
30. C is the centre of the hyperbola
and ' A ' is any point on it. The tangent at A to
the hyperbola meets the line x - 2 y = 0 and
x + 2 y = 0 at Q and R respectively. The value
of CQ.CR is equal to
ANSW ER KEY
1
2
3
4
(a)
(b)
(c)
(d)
5
6
7
8
(b)
(b)
(b)
(b)
9
10
11
12
(b)
(b)
(c)
(c)
13
14
15
16
(d)
(a)
(d)
(a)
17
18
19
20
x2 y 2
= 1,
4
1
(d)
(c)
(a)
(a)
21
22
23
24
(1)
(11)
(4)
(3)
25
26
27
28
(27)
(4)
(2)
(9)
29
30
(5)
(5)
MATHEMATICS
26
12
LIMITS & DERIVATIVES
5.
MCQs with One Correct Answer
1.
Given a real valued function 'f ' such that
lim
(d)
3.
6.
7.
lim f ( x ) exist
lim f ( x ) does not exists
8.
x®0
4.
n
lim
n®¥
å
(c) –3
log ( n + r ) - log n
n
r =1
1
k
k
k 1/ n
then lim k [(n + 1) (n + 2) ...(n + n) ] is
n®¥ n
equal to
(a)
4k
e
(b)
æ 4ö
çè ÷ø
e
1/ k
æ 4ö
(c) ç ÷
è eø
k
(d)
æ eö
çè ÷ø
4
k
(
(d) 4
1/ x
The limit lim æç 2 - ÷
xø
x ®a è
a
p (b)
-
2a
p (c)
)
+ (1/ x)sin x ,
(d) 2
æ px ö
tan
a ö èç 2a ø÷
is equal to
-
2
p
(d) 1
(a) e
e
e
If a and b are the roots of the quadratic
lim
x®
(d) –1/3
1ö
æ
= 2 ç log 2 - ÷ ,
è
2ø
(c) 3
equation ax 2 + bx + c = 0 , then
æ x 2 + 5x + 3 ö
lim ç
÷
x ®¥ çè x 2 + x + 2 ÷ø
(a) e4
(b) e2
(c) e3
(d) 1
2
x
x3
£ f ( x) £ 3 +
for all x ¹ 0, then the
If 3 –
12
9
value of lim f ( x ) is equal to
h®0
is a finite non-zero
The value of lim (sin x)
x ®0
where x > 0 is
(a) 0
(b) –1
(c) 1
-
x®0
(a) 1/3 (b) 3
Given that
xn
number is
(a) 1
(b) 2
x
2.
(cos x - 1)(cos x - e x )
x®0
ì tan 2 {x}
for x > 0
ï 2
2
ï x - [ x]
ï
f ( x) = í
1
for x = 0
ï
ï {x} cot{x} for x < 0
ï
î
then
(a) LHL = 1
(b) RHL = cot 1
(c)
The integer n for which
9.
1
a
1 - cos(cx 2 + bx + a )
2(1 - ax )2
=
(a)
c æ 1 1ö
2a èç a bø÷
(b)
(c)
c æ 1 1ö
ab èç a bø÷
(d) None of these
c æ 1 1ö
2b èç a bø÷
cos2 (1- cos2 (1- cos2(............ - cos2(x))))..........)
x®0
ìï æ x + 4 - 2ö üï
sin ípç
÷ý
x
ø þï
ïî è
is equal to
p
p
4
2
(a)
(b)
(d)
(c)
p
2
4
p
lim
Limits & Derivatives
27
10. The value of
1/ cos2 x
lim [1
x®p / 2
1/ cos 2 x
+2
1/ cos 2 x cos 2 x
+ ..... + n
]
is
(a) 0
11.
(c) ¥
(b) n
(d)
lim {log n -1 (n) log n (n + 1)log n +1( n + 2)...
n(n+1)
2
n®¥
...log
n k -1
Numeric Value Answer
(n x )} is equal to
(a) n
(b) k
(c) ¥
(d) None of these
12. If [x] denotes the greatest integer £ x , then
lim
1
n®¥ n 3
(a)
{[12 x] + [22 x] + [32 x] + ... + [n 2 x]} equals
x / 2 (b)
(c)
x/3
(d) 0
x/6
If lim f ( x ) exist, then l =
x ®3
9
(b)
2
14. The value of
(a)
2
9
(c)
sec 2
ì
æ p öü
lim ísin 2 ç
÷ý
x ®0 î
è 2 - ax ø þ
e-a / b
p
2 -bx
(b)
2
3
(d) None
19.
é
ù
20. The value of lim ê x + x + x - x ú is
x ®¥ ë
û
f (x)
=1
21. Let f (x) be a function such that lim
x®0 x
x (1 + a cos x ) - b sin x
and lim
= 1, then b – 3a
x ®0
{ f ( x )3}
is equal to
22. The largest value of non-negative integer a for
is equal to
e-a
2
g (1 - cos 2 x)
x4
=
1
is
4
x
ü is
23. The value of lim ì
ï
ï
3
x ®¥ ï
x
ï
íx+
ý
3
x
ï
x+
....¥ ï
ïî
ïþ
x+3 x
24. If a, b are two distinct real roots of the
/ b2
equation ax3 + x - 1 - a = 0, (a ¹ -1, 0), none
of which is equal to unity, then the value of
is
4a
a
a
(b)
(c)
(d) None
b
4b
b
Let f (x) be a polynomial function of second
degree. If f (1) = f ( -1) and a1, a2 , a3 are in A.P..
then f '(a1 ), f '( a2 ), f '( a3 ) are in
(a) A. P.
(b) G. P.
(c) H. P.
(d) None of these
(a)
16.
p
2
1- x
x
4a / b
(d) e
a 2a / b
f (1 - cos x )
f ( x)
=b
If lim 2 = a and lim
x ®0 g ( x ) sin 2 x
x ®0 x
(where b ¹ 0 ), then lim
x ®0
g ( x) f ( a ) - g ( a ) f ( x )
is
x-a
x®a
The value of lim (sinx)tanx is
then the value of lim
ìï - ax + sin ( x - 1) + a üï1which lim í
ý
x ®1 î
ï x + sin ( x - 1) - 1 þï
(c)
15.
18. If f (a ) = 2, f ' (a) = 1 , g (a) = -1 , g ' (a) = 2 ,
x®
x
ì
ï (e( x+ 3) ln 27 ) 27 - 9
; x<3
ï
3x - 27
13. If f ( x) = í
ï 1 - cos( x - 3)
; x>3
ïl
î ( x - 3) tan( x - 3)
(a)
17. Let f(x) be a polynomial function satisfying
æ1ö
æ1ö
f ( x ) × f ç ÷ = f ( x ) + f ç ÷ . If f(4) = 65 and
è xø
èxø
l1, l2, l3 are in GP, then f '(l1 ), f '(l2 ), f '(l3 )
are in
(a) AP
(b) GP
(c) HP
(d) None of these
lim
(1 + a) x 3 - x 2 - a
(e1-ax - 1)( x - 1)
the value of kl.
x ®(1/ a )
25.
is
al ( k a - b)
. Find
a
3x + 33- x - 12
is equal to _______.
x ®2 3- x /2 - 31- x
lim
ANSW ER KEY
1
2
3
(d)
(a)
(b)
4
5
6
(c)
(c)
(c)
7
8
9
(c)
(a)
(b)
10
11
12
(b)
(b)
(b)
13
14
15
(c)
(a)
(c)
16
17
18
(a)
(b)
(5)
19
20
21
(1)
(0.50)
(6)
22
23
24
(2)
(1)
(1)
25
(36)
MATHEMATICS
28
13
MATHEMATICAL
REASONING
MCQs with One Correct Answer
1.
2.
For the statement “17 is a real number or a
positive integer”, the “or” is
(a) Inclusive
(b) Exclusive
(c) Only (a)
(d) None of these
Let p and q be any two logical statements and
7.
8.
r : p ® (: p Ú q) . If r has a truth value F, then
the truth values of p and q are respectively :
(a) F, F (b) T, T (c) T, F (d) F, T
3.
If
p : Ashok works hard
q : Ashok gets good grade
9.
The verbal form for (~ p ® q) is
5.
(a) If Ashok works hard then gets good grade
(b) If Ashok does not work hard then he gets
good grade
(c) If Ashok does not work hard then he does
not get good grade
(d) Ashok works hard if and only if he gets
grade
If p is false and q is true, then
(a) p Ù q is true
(b) p Ú ~ q is true
(d) p Þ q is true
(c) q Ù p is true
~ p Ù q is logically equivalent to
6.
(a) p ® q
(b) q ® p
(c) ~ (p ® q)
(d) ~ (q ® p)
Which of the following is a contradiction?
4.
(a)
(p Ù q)Ù ~ (p Ú q) (b) p Ú (~ p Ù q)
(c) (p Þ q) Þ p
(d) None of these
10.
(p Ù ~ q) Ù (~ p Ù q) is
(a) A tautology
(b) A contradiction
(c) Both a tautology and a contradiction
(d) Neither a tautology nor a contradiction
The false statement in the following is
(a) p Ù (~ p) is contradiction
(b) (p Þ q) Û (~ q Þ ~ p) is a contradiction
(c) ~ (~ p) Û p is a tautology
(d) p Ú (~ p) Û p is a tautology
The conditional ( p Ù q) Þ p is
(a) A tautology
(b) A fallacy i.e., contradiction
(c) Neither tautology nor fallacy
(d) None of these
If p and q are two statements, then
( p Þ q) Û (-q Þ~ p) is a
11.
12.
(a) contradiction
(b) tautology
(c) neither (a) nor (b) (d) None of these
Which of the following is false?
(a) p Ú ~ p is a tautology
(b) ~ (~p) « p is a tautology
(c) p Ù ~ p is a contradiction
(d) ((p Ù q) ® q) ® p is a tautology
In the truth table for the statement ( p ® q) «
(~ p Ú q), the last column has the truth value in
the following order is
(a) TTFF
(b) FFFF
(c) TTTT
(d) FTFT
Mathematical Reasoning
29
13. If p Þ (~ p Ú q) is false, then truth values of p
and q are respectively
(a) F. T
(b) F, F
(c) T, T
(d) T, F
14. Negation of “2 + 3 = 5 and 8 < 10” is
(a) 2 + 3 ¹ 5 and < 10 (b) 2 + 3 = 5 and 8 </ 10
21. The negation of (p Ú q)Ù (p Ú ~ r) is
(a) (~ p Ù ~ q) Ú (q Ù ~ r)
(b) (~ p Ù ~ q) Ú (~ q Ù r)
(c) (~ p Ù ~ q) Ú (~ q Ù r)
(d) (p Ù q) Ú (~ q Ù ~ r)
22. Let p, q and r be any three logical statements.
Which of the following is true?
(c) 2 + 3 ¹ 5 or 8 </ 10(d) None of these
15. If the compound statement p ® (~ p Ú q) is false
then the truth value of p and q are respectively
(a) T, T (b) T, F
(c) F, T
(d) F, F
16. The contrapositive of p ® (~q ® ~r) is
(a) (~ q Ù r) ® ~ p
(b) (q ® r) ® ~p
(c) (q Ú ~r) ® ~ p
(d) None of these
(c)
( p Ú ~ q ) Ú ~ p (d) None of these
( p Ù ~ q) Ú ~ p
18. The inverse of the statement (p Ù ~ q) ® r is
(a) ~ (p Ú ~q) ® ~ r (b) (~p Ù q) ® ~ r
(c) (~p Ú q) ® ~ r (d) None of these
19. ~ ( (~ p) Ù q ) is equal to
(a) p Ú (~ q)
(b) p Ú q
(c) p Ù (~ q)
(d) ~ p Ù ~ q
20. Which of the following is true?
(a) p Þ q º ~ p Þ ~ q
(b)
(c)
(d)
(b)
~ [( p Ú q ) Ù (~ r ) º (~ p) Ú (~ q) Ú (~ r )
(c)
~ [ p Ú (~ q)] º (~ p) Ù q
(d)
~ [ p Ú (~ q)] º (~ p)Ù ~ q
(b) [p Ú q] Ú (~ p) is a tautology
(c) [p Ù q) Ù (~ p) is a contradiction
p Ú (~ p Ú q) is
( p Ù ~ q ) Ù ~ p (b)
~ [ p Ù (~ q)] º (~ p) Ù q
23. Identify the false statements
(a) ~ [p Ú (~ q)] º (~ p) Ú q
17. The negation of the compound proposition
(a)
(a)
(d) ~ [p Ú q] º (~ p) Ú (~ q)
24. Negation of the statement (p Ù r) ® (r Ú q) is
(a) ~ (p Ù r) ® ~ (r Ú q)
(b) (~p Ú ~r) Ú (r Ú q)
(c) (p Ù r) Ù (r Ù q)
(d) (p Ù r) Ù (~ r Ù ~q)
25. Let A, B, C and D be four non-empty sets. The
contrapositive statement of “If A Í B and B Í D,
then A Í C ” is:
(a) If A Í C, then A Í B and B Í D
(b) If A Í C, then B Ì A or D Ì B
~ ( p Þ ~ q) º ~ p Ù q
~ (~ p Þ ~ q) º ~ p Ù q
~ (~ p Û q) º [~ ( p Þ q)Ù ~ (q Þ p)]
(c) If A Í C, then A Í B and B Í D
(d) If A Í C, then A Í B or B Í D
ANSW ER KEY
1
2
3
(a)
(c)
(b)
4
5
6
(d)
(d)
(a)
7
8
9
(b)
(b)
(a)
10
11
12
(b)
(b)
(c)
13
14
15
(d)
(c)
(b)
16
17
18
(a)
(a)
(c)
19
20
21
(a)
(c)
(c)
22
23
24
(c)
(d)
(d)
25
(d)
MATHEMATICS
30
14
STATISTICS
MCQs with One Correct Answer
1.
n.2n-1
2n - 1
(b)
6.
The median of 100 observations grouped in
classes of equal width is 25. If the median class
interval is 20 - 30 and the number of observations
less than 20 is 45, then the frequency of median
class is
(a) 10
(b) 20
(c) 15
(d) 12
If the mean deviation of the numbers 1, 1 + d,
1 + 2d, .... 1 + 100d from their mean is 255, then d
is equal to :
(a) 20.0 (b) 10.1 (c) 20.2 (d) 10.0
In a series of 2 n observations, half of them equal
a and remaining half equal –a. If the standard
deviation of the observations is 2, then | a |
equals
3n (n + 1)
2 (2n + 1)
(n + 1) (2 n + 1)
n (n + 1)
(c)
(d)
6
2
The A.M. of n observations is M. If the sum of
n – 4 observation is a, then the mean of remaining
4 observation is
(a)
nM - a
4
(b)
nM + a
2
nM – a
(d) nM + a
2
The mean income of a group of persons is ` 400.
Another group of persons has mean income
` 480. If the mean income of all the persons in
the two groups together is ` 430, then ratio of
the number of persons in the groups is
(c)
3.
The mean of n items is X . If the first item is
increased by 1, second by 2 and so on, the new
mean is :
x
(a) X +
(b) X + x
2
n +1
(d) none of these
(c) X +
2
The weighted mean of first n natural numbers
whose weights are equal to the number of
selections out of n n atural numbers of
corresponding numbers is
(a)
2.
5.
(a)
n1 5
=
n2 3
(b)
n1 2
=
n2 5
n1 7
=
(d) None of these
n2 4
The mean of six numbers is 30. If one number is
excluded, the mean of the remaining numbers is
29. The excluded number is
(a) 29
(b) 30
(c) 35
(d) 45
7.
8.
1
2
(b)
2 (c) 2
(d)
n
n
The standard deviations of two sets containing
10 and 20 members are 2 and 3 respectively
measured from their common mean 5. The SD for
the whole set of 30 members is
(a)
9.
(c)
4.
(a)
(c)
2
3
æ 22 ö
ç 3 ÷
è ø
(b)
6
(d)
3
Statistics
31
10. If M. D. is 12, the value of S.D. will be
(a) 15
(b) 12
(c) 24
(d) None of these
11. Suppose values taken by a variable x are such
that a £ xi £ b, where xi denotes the value of x in
ith case for i = 1, 2, ... n. Then
(a) a £ Var(x) £ b
(b) a2 £ Var(x) £ b2
a2
£ Var( x )
(d) (b – a)2 ³ Var(x)
4
12. If the median and the range of four numbers
{x, y, 2x + y, x – y}, where 0 < y < x < 2y, are 10
and 28 respectively, then the mean of the
numbers is :
(a) 18
(b) 10
(c) 5
(d) 14
(c)
13. Let X and M.D. be the mean and the mean
deviation about X of n observations xi, i = 1, 2,
........, n. If each of the observations is increased
by 5, then the new mean and the mean deviation
about the new mean, respectively, are :
(a)
(b)
X, M.D.
the following statistical measures will not change
even after the grace marks were given ?
(a) mean
(b) median
(c) mode
(d) variance
17. Coefficient of variation of two distribution are
60 and 70, and their standard deviations are 21
and 16, respectively. What are their arithmetic
means?
(a) 35, 22.85
(b) 22.85, 35.28
(c) 36, 22.85
(d) 35.28, 23.85
Numeric Value Answer
18. Variance of the data 2, 4, 5, 6, 8, 17 is 23.33.
Then, variance of 4, 8, 10, 12, 16, 34 will be
19. Coefficient of variation of two distributions are
50 and 60 and their arithmetic means are 30 and
25, respectively. Then, difference of their
standard deviations is
20. In an experiment with 15 observations on x, the
following results were available:
å x2 = 2830, å x = 170
X + 5, M.D.
(c) X, M.D. + 5
(d) X + 5, M.D. + 5
14. The mean and variance for first n natural
numbers are respectively
21.
2
n +1
n -1
, variance =
2
12
2
n -1
n +1
(b) mean =
, variance =
2
12
(a) mean =
22.
n +1
n2 - 1
, variance =
2
12
2
n -1
n +1
(d) mean =
, variance =
2
2
15. The variance of 20 observations is 5. If each
observation is multiplied by 2, then the new
variance of the resulting observation is
(a) 23 × 5 (b) 22 × 5 (c) 2 × 5 (d) 24 × 5
16. All the students of a class performed poorly in
Mathematics. The teacher decided to give grace
marks of 10 to each of the students. Which of
(c) mean =
23.
24.
25.
One observation that was 20 was found to be
wrong and was replaced by the correct value
30. The corrected variance is
Consider the following data
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
If 1 is added to each number, then variance of
the numbers so obtained is
Coefficient of variation of two distribution are
50% and 60% and their standard deviation are
10 and 15, respectively. Then, difference of their
arithmetic means is
The mean of 5 observation is 4.4 and their
variance is 8.24. If three of the observations are
1, 2 and 6, then difference of the other two
observations is
If the variance of the first n natural numbers is
10 and the variance of the first m even natural
numbers is 16, then m + n is equal to _______.
If the mean and variance of eight numbers 3, 7, 9,
12, 13, 20, x and y be 10 and 25 respectively, then
x × y is equal to ____.
ANSW ER KEY
1
2
3
(a)
(a)
(a)
4
5
6
(c)
(c)
(a)
7
8
9
(b)
(c)
(c)
10
11
12
(a)
(d)
(d)
13
14
15
(b)
(a)
(b)
16 (d) 19
17 (a) 20
18 (93.32)
(0)
(78)
21 (8.25) 24
22 (5) 25
23 (5)
(18)
(52)
MATHEMATICS
32
15
PROBABILITY-1
1.
MCQs with One Correct Answer
If two numbers p and q are choosen randomly
from the set {1, 2, 3, 4} with replacement, then
6.
the probability that p 2 ³ 4q is equal to
1
3
1
7
(b)
(c)
(d)
4
16
2
16
A natural number x is chosen at random from the
first 100 natural numbers. Then the probability,
100
for the equation x +
> 50 is
x
1
11
1
3
(b)
(c)
(d)
(a)
20
20
3
20
A person throws two dice, one the common cube
and the other regular tetrahedron with numbers
1, 2, 3, 4 on its faces, the number on the lowest
face being taken in the case of a tetrahedron.The
chance that the sum of numbers thrown is not
less than 5 is
5
1
3
4
(c)
(d)
(b)
(a)
6
4
5
4
A is a set containing n element. A subset P of A is
chosen at random, and the set A is reconstruced
by replacing the elements of P. Another subset
Q of A is now chosen at random. The probability
that P È Q contains exactly r elements, 1 £ r £ n,
is
(a)
2.
3.
4.
n
(a)
4n
n
(c)
5.
Cr 3r
Cr 3n - r
n
(b)
3n
4n
n
(d)
C r 3n
4
4n
10 persons sit around a circular table with 10
numbered chairs. The probability that the two
particular persons A and B are always together is
2
1
1
2
(b)
(c)
(d)
9
5
9
5
5 different games are to be distributed among 4
children randomly. The probability that each child
gets atleast one game is
15
21
1
1
(a)
(b)
(c)
(d)
64
64
4
4
If three squares are chosen at random on a chess
board, then find the probability that they should
be in a diagonal line
(a) 9/743
(b) 11/743
(c) 13/743
(d) None of these
If the letters of the word MATHEMATICS are
arranged arbitrarily, the probability that C comes
before E, E before H, H before I and I before S is
1
1
1
1
(a)
(b)
(c)
(d)
75
120
720
24
In a game of chance a player throws a pair of dice
and scores points equal to the difference between
the numbers on the two dice. Winner is the person
who scores exactly 5 points more than his opponent.
If two players are playing this game only one time,
then the probability that neither of them wins is
1
1
53
107
(a)
(b)
(c)
(d)
54
108
54
108
A box contains 2 fifty paise coins, 5 twenty five
paise coins and 15 ten paise coins. Five coins
are taken out of the box at random. Probability
that the value of these five coins is less than
one rupee and fifity paise is
150
170
(a) 22
(b) 1 - 22
C5
C5
(a)
7.
8.
9.
10.
(c)
1-
170
22
C5
(d) None of these
Probability-1
11.
17. From a well-shuffled pack of 52 cards, four cards
are selected at random. The probability of drawing exactly 2 spades and exactly 2 aces is
1494
1594
(a)
(b)
270725
270725
1296
1396
(c)
(d)
270725
270725
Numeric Value Answer
18. The probability that in the random arrangement
of the letters of the word ‘UNIVERSITY’, the
two I’s does not come together is
19. A four digit number is formed by the digits 1, 2,
3, 4 with no repetition. The probability that the
number is odd, is
20. Two numbers x and y are chosen at random
without replacement from amongst the numbers
1, 2, 3, .........,3n. The probability that x3 + y3 is
divisible by 3 is
21. 5 girls and 10 boys sit at random in a row having
15 chairs numbered as 1 to 15. If the probability
that the end seats are occupied by the girls and
between any two girls odd number of boys take
2
2
3p + 2p
p + 3p
(a)
(b)
n
20
, then
seat is
2
4
is equal to
1001
n
p + 3 p2
3 p + 2 p2
22. A quadratic equation is chosen from the set of
(c)
(d)
all quadratic equations which are unchanged by
2
4
squaring their roots. If the chance that the chosen
There are three events E1, E2 and E3. one of which
must, and only one can happen. The odds are 7
p
equation has equal roots, is , then p + q =
to 4 against E1 and 5 to 3 against E2. The odds
q
against E3 is
(a) 4 : 11 (b) 3 : 8 (c) 23 : 88 (d) 65 : 23 23. In a multiple choice question, there are five
alternative answers of which one or more than
The chance of an event happening is the square
one are correct. A candidate will get marks on the
of the chance of a second event but the odds
question, if he ticks all the correct answers. If he
against the first are the cube of the odds against
decides to tick answers at random, then the least
the second. The chances of the events are
number of choice should he be allowed so that
1 1
1 1
the probability of his getting marks on the
,
,
(a)
(b)
1
9 3
16 4
question exceeds
is
8
1 1
,
24.
A
die
is
rolled
three
times,
if p be the probability
(c)
(d) None of these
4 2
of getting a large number than the previous
Ashmit and Bishmit are two students who
number, then the value of 54p is
appeared in Class X exam, Probability that 25. A class contains 20 boys and 20 girls of which
Ashmit will pass is 0.5 while probability that
half the boys and half the girls have cat eyes. If
both of them will pass is at least 0.1 and at most
one student is selected from the class & if the
0.3, and probability that Bishmit will pass is P(B)
probability that either the student is a boy or
then find the range of P(B)
a
(a) (0.2, 0.8)
(b) (0.15, 0.25)
has cat eyes is , then a 2 + b2 =
(c) (0.1, 0.8)
(d) None of these
b
ANSWER KEY
(d)
(b)
(d) 10 (c) 13 (a) 16
(c)
4
7
19 (0.50) 22 (3) 25 (5)
(b)
(a)
(c) 11 (b) 14 (d) 17
(a)
5
8
20 (0.33) 23 (4)
(b)
(b)
(c) 12 (d) 15 (a) 18 (0.80) 21 (3) 24 (5)
6
9
Two distinct numbers a and b are chosen
randomly from the set {2, 22, 23, .... 225}. Then the
probability that logab is an integer is
31
131
(a)
(b)
300
300
21
62
(c)
(d)
200
300
12. There are two vans each having numbered seats,
3 in the front and 4 at the back. There are 3 girls
and 9 boys to be seated in the vans. The probability of 3 girls sitting together in a back row on
adjacent seats, is
1
1
1
1
(a)
(b)
(c)
(d)
13
39
65
91
13. For the three events A, B, and C, P (exactly one
of the events A or B occurs) = P (exactly one of
the two events B or C occurs) = P(exactly one of
the events C or A occurs) = p and P (all the three
events occur simultaneously) = p2, where
0 < p < 1/2. Then the probability of at least one
of the three events A, B and C occurring is
14.
15.
16.
1
2
3
33
16
RELATIONS
& FUNCTIONS-2
MCQs with One Correct Answer
1.
2.
Let P be the relation defined on the set of all real
numbers such that P = {(a, b) : sec2a – tan2b
= 1}. Then P is:
(a) reflexive and symmetric but not transitive.
(b) reflexive and transitive but not symmetric.
(c) symmetric and transitive but not reflexive.
(d) an equivalence relation.
Consider the following relations. R = {(x, y) | x, y
are real numbers and x = wy for some rational
3.
æm pö
number w} S = {ç , ÷ | m, n, p and q are
èn qø
integers such that, q ¹ 0, n ¹ 0 and qm = pn}.
Then
(a) Neither R nor S is an equivalence relation
(b) S is an equivalence relation but R is not an
equivalence relation
(c) R and S both are equivalence relations
(d) R is an equivalence relation but S is not an
equivalence relation
For real numbers x and y, we define xRy iff
4.
x - y + 5 is an irrational number. Then,
relation R is
(a) reflexive
(b) symmetric
(c) transitive
(d) None of these
If f : R ® S , defined by
5.
f ( x) = sin x - 3 cos x + 1, is onto, then the
interval of S is
(a) [ –1, 3]
(b) [–1, 1]
(c) [ 0, 1]
(d) [0, 3]
If the function f : (– ¥ ,¥) ® B defined by
f (x) = – x2 + 6x – 8 is bijective, then B =
(a) [1,¥)
(b) (– ¥, 1]
(c) (– ¥, ¥)
(d) None of these
6.
7.
8.
Let f : f : {x, y, z} ® {1, 2, 3} be a one-one
mapping such that only one of the following
three statements is true and remaining two are
false : f (x) ¹ 2, f (y) = 3, f (z) ¹ 1, then –
(a) f (x) > f (y) > f (z) (b) f (x) < f (y) < f (z)
(c) f (y) < f (x) < f (z) (d) f (y) < f (z) < f (x)
Let A º {1,2,3,4} B º {a, b, c}, then number of
functions from A ® B, which are not onto is
(a) 8
(b) 24
(c) 45
(d) 6
A quadratic polynomial maps from [–2, 3] onto
[0, 3] and touches X-axis at x = 3, then the
polynomial is
3 2
3 2
( x - 6 x + 16) (b)
( x - 6 x + 9)
(a)
16
25
3 2
3 2
( x - 6 x + 16) (d)
( x - 6 x + 9)
(c)
25
16
9.
Let f : R + ® {-1, 0, 1} defined by
10.
f ( x ) = sgn( x - x 4 + x 7 - x8 - 1) where sgn
denotes signum function. Then f(x) is
(a) many-one and onto
(b) many-one and into
(c) one-one and onto
(d) one-one and into
Let f : R ® R be a function defined by
f ( x) = x + x2 , then f is
(a) injective
(b) surjective
(c) bijective
(d) None of these
11. If f (x) = sin x + cos x, g (x) = x2 – 1, then g (f (x))
is invertible in the domain
(a)
(c)
é pù
ê0 , 2 ú
ë
û
é p pù
ê- 2 , 2 ú
ë
û
(b)
é p pù
ê- 4 , 4 ú
ë
û
(d) [0, p]
Relations & Functions-2
35
12. Let g : R ® R be given by g (x) = 3 + 4x.
If g n ( x) = gogo....... og (x), then g-n (x) =
(where g–n (x) denotes inverse of g n ( x) )
(a)
(4n - 1) + 4n x
(b)
( x + 1)4- n - 1
(c)
( x + 1)4n - 1
(d)
(4- n - 1) x + 4n
13. If f (x) = 3 | x | - x - 2 and g (x) = sin x, then
domain of definition of fog (x) is
pü
ì
(a) í2np + ý
2 þ n ÎI
î
(b)
(c)
U
n ÎI
7p
11p ö
æ
, 2np +
çè 2np +
÷
6
6 ø
7p ü
ì
í 2 np + ý
6 þ n ÎI
î
p
7p
11p ù
é
(d) {(4m + 1) : m ÎI} U ê2np + , 2np +
2
6
6 úû
nÎI ë
pö
pö
æ
æ
14. If f ( x) = sin 2 x + sin 2 ç x + ÷ + cos x cos ç x + ÷
è
è
3ø
3ø
æ5ö
and g ç ÷ = 1, then graph of y = g [f (x)] is
è4ø
± np , n Î{1, 2,....}
p
+ 2np, n Î{... - 2, -1, 0,1, 2....}
2
(d) 2np, n Î{... - 2, -1, 0,1, 2,....}
Let f : [0, 1] ® [–1, 1] and g : [–1, 1] ® [0, 2] be
two functions such that g is injective and
gof : [0, 1] ® [0, 2] is surjective. Then,
(a) f must be injective but need not be
surjective
(b) f must be surjective but need not be
injective
(c) f must be bijective
(d) f must be a constant function
(c)
16.
Numeric Value Answer
18. Let E = {1, 2, 3, 4} and F = {1, 2}. Then the
number of onto functions from E to F is
19. If for x > 0, f ( x ) = (a - x n )1/ n ,
g ( x) = x 2 + px + q; p, q Î R
and the equation g (x) – x = 0 has imaginary
roots, then number of real roots of equation
g (g (x)) – f (f (x)) = 0 is
ì -1, x < 0
ï
20. Let g(x) = 1 + x - [x] and f ( x ) = í 0, x = 0 .
ï 1, x > 0
î
Then for all x, f (g(x)) is equal to
21. Let g ( x) =
(a) a circle
(b) a straight line
(c) a parabola
(d) None of these
15. Let f (x) = x2 and g(x) = sin x for all x Î R. Then
the set of all x satisfying (f o g o g o f) (x) =
(g o g o f) (x), where (f o g) (x) = f (g(x)), is
(a) ± np , n Î{0,1, 2,....}
(b)
17. Let f : [0, 1] ® R be an injective continuous
function that satisfies the condition
–1 < f (0) < f (1) < 1.
Then, the number of functions g : [–1, 1] ® [0, 1]
such that (gof ) (x) = x for all x Î [0, 1] is
(a) 0
(b) 1
(c) more than 1, but finite
(d) infinite
e x - e- x
and g(f(x)) = x, then
2
æ e22 -1ö
f ç 11 ÷ - 5 =.
ç 2e ÷
è
ø
x£0
ìï - x + 1,
, then the number
22. If f ( x) = í
2
ïî-( x - 1) , x ³ 1
of solutions of f ( x) - f -1 ( x) = 0 is
æp ö
æp ö
23. If f (x) = cos2 x + cos2 ç + x ÷ - cos x cos ç + x ÷
è3 ø
è3 ø
and g(3/4) = 2, then find the value of (gof) (1).
24. If a non-zero function f(x) is symmetrical about
y = x, then the value of p (constant) such that
f 2 ( x) = ( f -1 (x))2 - px × f ( x) × f -1( x) + 2x2 f ( x)
for all x Î R + is ....... .
25. Let f be defined on the natural numbers as follows :
f (1) = 1 and for n > 1, f (n) = f [f (n – 1)]
+ f [n – f (n – 1)], the value of
1 20
å f (r ) is
30 r =1
ANSWER KEY
1
2
3
(d)
(b)
(a)
4
5
6
(a)
(b)
(c)
7
8
9
(c)
(b)
(b)
10
11
12
(d)
(b)
(b)
13
14
15
(d)
(b)
(a)
16
17
18
(b)
(d)
(14)
19
20
21
(0)
(1)
(6)
22
23
24
(4)
(2)
(2)
25
(7)
17
INVERSE TRIGONOMETRIC
FUNCTIONS
4.
MCQs with One Correct Answer
1.
ac
(a)
a 2 + c2
2 ac
(c)
2.
f (x) = sin -1 (log[ x ]) + log(sin -1[ x]); (where [.]
denotes the greatest integer function) is
(a) R
(b) [1, 2)
If ax + b (sec (tan–1 x)) = c and
ay + b (sec (tan–1 y)) = c, then
2
2
x+ y
=
1 - xy
(b)
2 ac
a-c
(d)
a+c
1 - ac
a -c
The range of the function
5.
4
å tan -1 xi
(a)
ìp ü
í , pý
î2 þ
(b)
(c)
{p}
æ pö
(d) ç 0, ÷
è 2ø
If
å cos-1 br =
r =1
lim
(1 + x )
(a)
1
2
r =1
1/ 4
- (1 - 2 x )
(b) 0
7.
å (b r ) r . Then
(c)
p
-b
(d)
2
The maximum value of f(x)
æ 12 - 2 x 2 ö
÷ is
–1
= tan çç 4
x + 2x2 + 3 ÷
è
ø
(
A
2
(d)
p
2
p
- 2b
2
)
(a) 18º
(b) 36º
(c) 22.5º
(d) 15º
The number of solutions of the equation
tan -1 x =
=
(b) p – 2b
(c)
6.
k
x + x2
x® A
(a) p – b
kp
for any k ³ 1 where
2
br ³ 0"r and A =
2 1/ 3
1ü
ì
í0, - ý
2þ
î
is equal to
i =1
-1 é 2
k
3.
pü
ì
(d) {– sin 1}
ílog ý
2þ
î
If x1, x2, x3, x4 are roots of the equation
x4 – x3 sin 2b + x2 cos 2b – x cos b – sin b = 0, then
(c)
1ù
-1 é 2 1 ù
êë x + 2 úû + cos êë x - 2 úû , where
[.] is the greatest integer function, is
f (x) = sin
The range of the function
(a) 2
(c) 4
( x2 + 1)
2
- 4 x 2 is
(b) 3
(d) none of these
Inverse Trigonometric Functions
8.
9.
37
æ
ö
1
1
ö
-1 æ
S = tan-1 ç
÷ + tan ç 2
÷ + ...
2
è n + n +1ø
è n + 3n + 3 ø
æ
ö
1
+ tan -1 ç
÷ , then tan S is
1
+
(
n
+
19)(
n
+
20)
è
ø
equal to :
20
n
(a)
(b)
2
401 + 20 n
n + 20n + 1
20
n
(c)
(d)
2
401 + 20 n
n + 20n + 1
(a) a = –3 & b = 1
(b) a = 1 & b = -
1
1
&b=
6
2
(d) None of these
(c)
-1
x-
1
= cos
-1
x-
1
is
cos -1 x
(a) 0
(b) 1
(c) 2
(d) 3
10. The minimum integral value of a for which the
quadratic equation (cot–1 a)x2 – (tan–1 a)3/2 x +
2(cot–1 a)2 = 0 has both positive roots
(a) 1
(b) 2
(c) 3
(d) 4
11. The sum of series
sin -1 x
sec -1 2 + sec-1
sec -1
a=
cot–1
7
19
39
+ cot –1 + cot –1 + ..... then
4
4
4
(a)
S n = tan –1
n
2n + 5
(b)
Sn = cot –1
n+5
2n
(c)
S n = cot –1
4n
2n + 5
1
2
–1
15. Find : tan (1/3) + tan–1(1/7) + tan–1 (1/13)
(d)
S¥ = cot –1
10
50
+ sec -1
+ . . . . +
3
7
( n2 + 1)( n2 - 2n + 2) is
2
( n2 - n + 1)
1
3
14. If Sn denotes the sum to n terms of the series
The number of roots of the equation
sin
æ
ö
÷ +....¥
è n + n +1ø
+..... + tan -1 ç
1
2
p
p
p
p
(b)
(c)
(d)
2
3
4
6
16. If the equation x3 + bx2 + cx + 1 = 0 (b < c) has
only one real root a.
Then the value of 2 tan –1 (cosec a) + tan –1
(2 sin a sec2 a) is:
(a)
(a) tan–1 1
(b) tan–1 n
(c) tan–1 (n + 1)
(d) tan–1 (n – 1)
12. The value of x satisfying the equation
(sin -1 x)3 - (cos-1 x)3 +
(sin -1 x)(sin -1 x - cos -1 x ) =
p3
16
is:
13.
when
(a)
p
cos
5
(b)
p
cos
4
(c)
cos
p
8
(d)
cos
-
(c)
p
2
p
2
(b) –p
(d) p
17. Solution of equation cos -1 x 3 + cos -1 x =
is
p
12
æ
ö
a 2 a3
sin -1 ç a +
+ .... ÷
ç
÷
3
9
è
ø
+ cos -1 (1 + b + b2 + ...) =
(a)
p
2
1
2
(a)
x=
(c)
x= ±
(b)
1
2
x= -
1
2
(d) None of these
p
2
MATHEMATICS
38
18.
The sum of infinite series
Numeric Value Answer
æ 1 ö
-1 æ 2 - 1 ö
-1 æ 3 - 2 ö
sin -1 ç
÷ + sin çç
÷ + sin çç
÷
÷÷
è 2ø
è 6 ø
è 2 2 ø
æ n - n -1 ö
÷ + ...
+... + sin -1 ç
is
ç n ( n + 1) ÷
è
ø
p
(b) p
2
The value of
(a)
19.
p
3
(c)
21.
lim æ π - tan -1 x ö
÷ø
x ®¥ çè 2
22.
Find the value of
3
a
aö b
æ1
æ1
æ b öö
cosec2 ç tan-1 ÷ + sec2 ç tan -1 ç ÷ ÷
b
2
2
2
2
è a øø
è
ø
è
is equal to
20.
(a)
( a - b ) ( a 2 + b2 ) (b) ( a + b ) ( a2 - b2 )
(c)
( a + b ) ( a2 + b2 ) (d)
None of these
cot–1(2.12) + cot–1(2.22) + cot–1(2.32) +... is equal
p2
+ (sec–1 y)2 + (tan–1 z)2 is
4
Find the number of solution of
cos (2 sin–1(cot(tan–1(sec (6 cosec–1x))))) + 1 = 0;
where x > 0.
If the domain of the function
(sin–1 x)2 =
24.
25.
to
(a)
p
p
< A<
4
2
The number of ordered triplets (x, y, z) that satisfy
the equation
when
23.
3
f(x) =
p
4
(b)
p
3
(c)
p
2
p
5
(d)
equals
æ1
ö
tan -1 ç (tan 2 A) + tan -1 (cot A) + tan -1 (cot 3 A) ÷ ,
2
è
ø
p
4
(d)
1/ x
3cos -1 ( 4x ) - p is [a, b], then the value
of (4a + 64b) is _______.
ANSWER KEY
1
2
3
(c)
(c)
(a)
4
5
6
(c)
(c)
(d)
7
8
9
(c)
(c)
(c)
10
11
12
(b)
(b)
(c)
13
14
15
(b)
(d)
(c)
16
17
18
(b)
(a)
(a)
19
20
21
(c)
(a)
(1)
22
23
24
(0)
(2)
(3)
25
(7)
Matrices
39
18
MATRICES
MCQs with One Correct Answer
1.
The product of the matrices
é cos 2 q
A=ê
êë cos q sin q
cos q sin q ù
ú and
sin 2 q úû
(c) I + P
(d) None of these
For each real x : – 1 < x < 1. Let A (x) be the matrix
x+ y
é 1 – xù
then
(1 – x)–1 ê
and z =
ú
1 + xy
ë–x 1 û
(a) A (z) = A (x) + A(y)
q-f =
(c) A (z) = A (x) A (y)
(2n + 1)
p
2
(c) 2np
(b) A (z) = A (x) [A (y)]–1
(d) A (z) = A (x) – A (y)
(b) np
(d)
n
p
2
5.
é - 3 2 ù é1 0 ù
é2 1ù
A ê
If ê
ú= ê
ú , then the
ú
ë 5 - 3û ë0 1 û
ë 3 2û
matrix A equals
(a)
(c)
3.
(b) 2I + P
é cos2 f
cos f sin f ù
B=ê
ú is a null matrix if
êëcos f sin f
sin 2 f úû
(a)
2.
4.
(a) I + 2p
é1 1 ù
ê1 0ú
ë
û
(b)
é1 0ù
ê1 1 ú
ë
û
(d)
é1 1ù
ê
ú
ë0 1û
é0 1ù
ê1 1ú
ë
û
- tan (q / 2) ù
é 0
If matrix P = ê
ú , then find
tan
q
/
2
0
ë
û
é cos q - sin q ù
(I – P) ê
ú
ë sin q cos q û
6.
é1 0 ù
If A = ê1/ 2 1ú , then A50 is
ë
û
é1 0 ù
ê50 1 ú
ë
û
(a)
é1 0 ù
ê0 50 ú
ë
û
(b)
(c)
é1 25 ù
ê0 1 ú
ë
û
(d) None of these
If A = [aij]m × n and aij = (i2 + j2 – ij) (j – i), n is odd,
then which of the following is not the value of
Tr(A)
(a) 0
(b) | A |
(c) 2| A |
(d) None of these
MATHEMATICS
40
7.
é 3
ê
2
If P = ê
ê 1
êë 2
1 ù
ú
é1 1ù
2 ú
and A = ê
ú and Q =
3ú
ë 0 1û
ú
2 û
10.
11.
é u1
ê
P = ê u2
êë u 3
PAPT and x = PTQ2005P then x is equal to
8.
(a)
é1 2005ù
ê0
1 úû
ë
(b)
é 4 + 2005 3
6015 ù
ê
ú
4 - 2005 3 úû
ëê 2005
(c)
1 é2 + 3
1 ù
ú
ê
4 êë - 1
2 - 3 úû
(d)
1 é 2005 2 - 3 ù
ú
ê
4 ëê2 + 3 2005 ûú
(a)
9.
±
(b)
±
(c)
±
(d)
±
1
2
1
3
1
2
1
2
,±
12.
13.
,±
,±
,±
1
6
1
2
1
6
1
2
,±
,±
,±
,±
1
v3
é1ù
é u1 v1 w1 ù é x ù
ê ú
ú êyú
ê
the equation ê u 2 v 2 w 2 ú ê ú = ê 1 ú ,
êë 5 úû
êë u 3 v 3 w 3 úû êë z úû
the value of y comes out to be –3. Then the
value of m is equal to
(a) 27
(b) 7
(c) – 27 (d) – 7
If A, B are two square matrices such that AB = A
and BA = B, then
(a) only B is idempotent
(b) A, B are idempotent
(c) only A is idempotent
(d) None of these
If A1, A3, ..., A2n–1 are n skew – symmetric matrices
of same order, then
(a)
(b)
(c)
(d)
1
6
1
3
2
v2
2 1ù
w1 ù
é 2
1 ê
ú
ú
w 2 ú ; Q = ê 13 – 5 m ú
9
êë – 8 1 5 úû
w 3 úû
X = A1 + 3 A33 + ...(2 n - 1)(A 2n -1 )2 n -1 will be
2
1
v1
are such that PQ = I, an identity matrix. Solving
é 0 2b c ù
The value of a, b, c when êê a b - c úú is
êë a - b c úû
orthogonal, are
Matrix A is such that A2 = 2A – I when I is unit
matrix then An is equal to
(a) nA – (n – 1) I
(b) nA – I
(c) 2n–1 A – I
(d) none of these
The matrices
14.
symmetric
skew-symmetric
neither symmetric nor skew symmetric
depends on ‘n’ is even or odd
éa b c ù
ê
ú
For A = ê d e f ú , a, b, c, d, e, f, g, h, i Î C, we
ëê g h i úû
é 0 1 0ù
If A = ê 0 0 1ú then A3 – rA2 – qA =
ê
ú
êë p q r úû
éa d g ù
ê
ú
say A = êb e h ú and we say that A is the
êc f i ú
ë
û
(a) p I
(c) r I
Hermitian matrix if A = Aq. Suppose A is the
Hermitian matrix such that A2 = O then
(a) A = – A'
(b) A = A'
(c) A = O
(d) A = I
(b) q I
(d) None of these
q
Matrices
41
15. A, B, C are three matrices of the same order such
that any two are symmetric and the 3rd one is
skew symmetric. If X = ABC + CBA and Y =
ABC – CBA, then (XY)T is
(a) symmetric
(b) skew symmetric
(c) I – XY
(d) – YX
écos q sin q ù
é 1 0ù
T
16. If A = ê
ú , B = ê –1 1ú , C = ABA ,
sin
q
–
cos
q
ë
û
ë
û
then ATCnA equals to (n Î I+)
(a)
é – n 1ù
ê 1 0ú
ë
û
(b)
é1 – n ù
ê0 1 ú
ë
û
(c)
é0 1 ù
ê1 – n ú
ë
û
(d)
é 1 0ù
ê – n 1ú
ë
û
éa b c ù
21. If matrix A = ê b c a ú where a, b, c are real
ê
ú
êë c a b úû
positive numbers, abc = 1 and ATA = I, then find
the value of a3 + b3 + c3.
22. Let z =
-1 + 3i
, where i =
2
-1 , and r, s Î
é (- z)r
{1, 2, 3}. Let P = ê 2s
êë z
z 2s ù
ú and I be the
z r úû
identity matrix of order 2. Then the total number
of ordered pairs (r, s) for which P2 = –I is
17. Let A be a 3 × 3 matrix given by A = (aij)3 × 3. If for
every column vector X satisfies X 'AX = 0 and a12
= 2008, a13 = 1010 and a23 = – 2012. Then the
value of a21 + a31 + a32 =
(a) – 6
(b) 2006
(c) – 2006
(d) 0
Numeric Value Answer
é 1 0 0ù
ê
ú
23. Let P = ê 4 1 0 ú and I be the identity matrix
êë16 4 1 úû
of order 3. If Q = [qij] is a matrix such that P50 – Q
=I, then
q 31 + q32
equals
103 q 21
24. Let X be the solution set of the equation Ax = I;
18. If B, C are square matrices of order n and if
A = B + C, BC = CB, C2 = 0, then for any positive
integer N, A N + 1 = BK[B + ( N + 1) C], then K/N is
19. A and B are two square matrices such that A2B =
BA and if (AB)10 = AkB10, then k is
20. If the product of n matrices
é1 1 ù é1 2 ù é1 3 ù é1 n ù
ê 0 1ú ê 0 1ú ê 0 1ú ... ê 0 1ú is equal to the
ë
ûë
ûë
û ë
û
é1 378ù
matrix ê
ú , then the value of n is equal to
ë0 1 û
é2
where A = ê
ë3
–1 ù
and I is the corresponding
– 2 úû
unit matrix and x Î ¥ , then find the minimum
ì np
ü
value of S(cos x q + sin x q), qΡ – í ; n ΢ ý .
î2
þ
é0 a ù
50
25. Let A = ê
ú and (A + I) – 50A =
ë0 0 û
find a + b + c + d.
éa b ù
êc d ú ,
ë
û
ANSWER KEY
1
2
3
(a)
(a)
(c)
4
5
6
(c)
(d)
(d)
7
8
9
(a)
(c)
(a)
10
11
12
(a)
(d)
(b)
13
14
15
(b)
(c)
(d)
16
17
18
(d)
(c)
(1)
19
20
21
(1023) 22
(27) 23
(4)
24
(1)
(1)
(2)
25
(2)
19
DETERMINANTS
1.
MCQs with One Correct Answer
The value of the determinant
5.
1
a
a2
cos (n –1) x cos nx cos (n + 1) x
is zero, if
sin (n –1) x sin nx sin (n + 1) x
(a) sin x = 0
2.
Q = (cos(b - a ),sin b) and
R = (cos(b - a + q),sin(b - q)) , where
(b) cos x = 0
p
0 < a, b, q < . Then,
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-collinear
1 + a2
(c) a = 0
(d) cos x =
2a
If a, b, c, are the sides of a triangle ABC such
a2
b2
(a) 0
(b) positive
(c) negative
(d) can not be determined
Consider three points
P = ( - sin(b - a ), - cos b),
c2
that (a + 1) 2 (b + 1)2 (c + 1) 2 = 0 , then
(a - 1)2
3.
(c - 1)2
(a) DABC is non-isosceles right-angled triangle
(b) D ABC is an equilateral
(c) D ABC is an acute angled triangle with no
two angles being equal
(d) D ABC is an isosceles
If a, b, c, are the sides of a triangle A B C and A,
B, C are angles opposite to a, b, c, and
a
2
D = b sin A
c sin A
4.
(b - 1)2
6.
3r - 2
n -1
a
1
(n - 1)(3n - 4)
2
n -1
then the value of
å Dr
r =1
cos A
then,
1
(a) D = area of triangle
(b) D = perimeter of the triangle
(c) D = Sa2
(d) None of these
If P, Q and R represent the angles of an acute
angled triangle, no two of them being equal then
1 1 + sin P sin P(1 + sin P)
1 1 + sin Q sin Q(1 + sin Q)
the value of
is
1 1 + sin R sin R(1 + sin R)
If D r =
2r - 1
1
n(n - 1) (n - 1)2
2
b sin A c sin A
1
cos A
r
n
2
7.
(a) depends only on a
(b) depends only on n
(c) depends both on a and n
(d) is independent of both a and n
If a2 + b2 + c2 + ab + bc + ca £ 0 " a, b,
c Î R, then value of the determinant,
( a + b + 2)2
a 2 + b2
1
(b + c + 2)
c2 + a2
1
equals
(a) 65
(c) 4(a2 + b2 + c2)
1
2
2
b + c2
(c + a + 2) 2
(b) a2 + b2 + c2 + 31
(d) 0
Determinants
8.
43
If A is matrix of order 3 such that |A| = 5 and B =
adj A, then the value of |A
9.
–1
T
| ( AB)
is equal
to (where |A| denotes determinant of matrix A.AT
denotes transpose of matrix A, A– 1 denotes inverse
of matrix A. adj A denotes adjoint of matrix A)
1
(a) 5
(b) 1
(c) 25
(d)
25
If f(x), h(x) are polynomials of degree 4 and
f ( x ) g ( x ) h( x )
a
b
c = mx4 + nx3 + rx2 + sx + t be
p
q
r
an identity in x, then
f "'(0) – f "(0) g "'(0) – g "(0) h "'(0) – h "(0)
a
b
c
is
p
q
r
(a) 2(3n – r)
(b) 2(2n – 3r)
(c) 3(n – 2r)
(d) None of these
1 1 1
10. If a b c = (a – b) (b – c) (c – a) (a + b + c),
a 3 b3 c 3
where a, b, c are all different, then the determinant
1
1
1
( x – a)2
( x – b) 2
( x – c) 2
( x – b)( x – c ) ( x – c )( x – a) ( x – a)( x – b)
vanishes when
(a) a + b + c = 0
(b) x =
1
(a + b + c)
3
1
(a + b + c) (d) x = a + b + c
2
2
If t is real and l = t - 3t + 4 , then the number
t 2 + 3t + 4
(c) x =
11.
of solutions of the system of equations
3x – y + 4z = 3, x + 2y – 3z = – 2, 6x + 5y +lz = –3 is :
(a) one (b) two (c) zero (d) infinite
12. The set of homoegeneous equations :
tx + (t + 1) y + (t - 1) z = 0;
(t + 1) x + ty + (t + 2) z = 0;
(t - 1) x + (t + 2) y + tz = 0
has non-trivial solution for
(a) three values of t (b) two values of t
(c) one value of t
(d) no value of t
13. If a, b, c are non-zeros, then the system of
equations :
(a + a) x + ay + az = 0; ax + (a + b) y + az = 0;
ax + ay + (a + c ) z = 0 has a non-trivial
solution if :
(a) a -1 = -(a -1 + b-1 + c -1 )
(b) a -1 = a + b + c
(c) a + a + b + c = 1
(d) None of these
14. If the system of equations ax + y + z = 0,
x + by + z = 0 and x + y + cz = 0 (a, b, c ¹ 1) has a
non-trivial solution, then the value of
1
1
1
+
+
is
1– a 1– b 1– c
(a) –1
(b) 0
(c) 1
(d) None of these
15. If a, b, c are non-zeros, then the system of equations (a + a)x + ay + az = 0, ax + (a + b)y + az
= 0, ax + ay + (a + c)z = 0 has a non-trivial
solution if
(a) a– 1 = – (a– 1 + b– 1 + c– 1)
(b) a– 1 = a + b + c
(c) a + a + b + c = 1
(d) None of these
16. If c < 1 and the system of equations x + y – 1 = 0,
2x – y – c = 0, and –bx + 3by – c = 0 is consistent,
then the possible real values of b are
3ö
æ
(a) b Î ç – 3, ÷
è
4ø
3
(b) b Î æç – , 4 ö÷
è 2 ø
3
(c) b Î æç – ,3ö÷
è 4 ø
(d) None of these
1
x+3
np ( -1)n n ! then
cos
2
3n +1
3
a
a5
x n -1 cos x
17. If f(x) =
dn
0
a
[f ( x )]x =0 =
dx n
(a) 1
(c) 0
a1
18. Suppose D = a2
a3
(b) –1
(d) None of these
b1 c1
b2
b3
c2 and
c3
a1 + pb1
b1 + qc1
c1 + ra1
D¢ = a2 + pb2
b2 + qc2
c2 + ra2 . Then
a3 + pb3
b3 + qc3
c3 + ra3
MATHEMATICS
44
19.
20.
(a) D¢ = D
(b) D¢ = D(1 – pqr)
(c) D¢ = D(1+p+q+r) (d) D¢ = D(1 + pqr)
é x 2 –1ù
–11ö
ê
ú æ
If A = ê –1 1 2 ú ; çè x ¹
÷ and det (adj(adj
3 ø
êë 2 –1 1úû
A)) = (14)4. Then the value of x is
(a) 2
(b) – 2
(c) 0
(d) None of these
3
2
x
é
ù
ê1 y 4 ú
Matrix A = ê
ú , if x y z = 60 and 8x + 4y + 3z
êë 2 2 z úû
= 20, then A(adj A) is equal to
é88 0 0ù
é 64 0 0ù
ê
ú
ê
ú
(a)
(b) ê 0 88 0ú
0 64 0
ê
ú
êë 0 0 88úû
ëê 0 0 64ûú
(c)
é 68 0 0 ù
ê 0 68 0 ú
ê
ú
êë 0 0 68úû
(d)
24.
D = log an + 6 log an + 8 log an + 10 is equal to:
25.
value of (1 + aq)2 (1 + bq)2 (1+ cq)2
x
22.
23.
1
2
3
4
The value of the determinant
æ 3p ö
æ 5p ö
æ 7p ö
sin2 ç x + ÷ sin2 ç x + ÷ sin2 ç x + ÷
è
è
è
2ø
2ø
2ø
æ 5p ö
sin ç x + ÷
è
2ø
7p ö
æ
sin ç x + ÷ is
è
2ø
æ 3p ö
sin ç x - ÷
è
2ø
æ 5p ö
sin ç x - ÷
è
2ø
7p ö
æ
sin ç x - ÷
è
2ø
ìæ a
ü
a ö
Let S = íç 11 12 ÷ : aij Î{0,1, 2}, a11 = a22 ý
îè a21 a22 ø
þ
Then the number of non-singular matrices in the
set S is :
é3; when $i = $j
Let A = [aij]3 × 3 be such that aij = ê
$i ¹ $j ,
ëê 0;
ì det (adj(adj A)) ü
then í
ý equals: (where {×}
5
î
þ
denotes fractional part function)
(a)
(d)
(d)
(d)
5
6
7
8
(d)
(d)
(a)
(b)
9
10
11
12
(a)
(b)
(a)
(c)
13
14
15
16
(a)
(c)
(a)
(c)
is
(1 + ar )2 (1 + br )2 (1 + cr )2
27.
æ 3p ö
sin ç x + ÷
è
2ø
log an + 12 log an + 14 log an + 16
Area of triangle whose vertices are (a, a2) (b, b2)
1
(c, c2) is , and area of another triangle whose
2
vertices are (p, p2) , (q, q2) and (r, r2) is 4, then the
(1 + ap)2 (1 + bp)2 (1+ cp)2
Numeric Value Answer
21.
log an + 2 log an + 4
log a n
26.
é34 0 0 ù
ê 0 34 0 ú
ê
ú
êë 0 0 34 úû
If a1, a2, a3, ..., an are in G.P. and ai > 0 for each i,
then the determinant
28.
29.
30.
sin x
cos x
2
If f ( x) = x - tan x
2 x sin 2 x
- x3 then lim
x ®0
5x
f ¢( x)
x
is equal to
If a, b are the roots of the equation x2 – 2x + 4 = 0,
find the value of
a+b
a2 + b2
a 3 + b3
a 2 + b2
a 3 + b3
a4 + b4
a 3 + b3 a 4 + b 4 a 5 + b5
Find the co-efficient of x in the expansion of the
determinant
(1 + x) 2
(1 + x )4
(1 + x)6
(1 + x )3
(1+ x)6
(1 + x)9 .
(1 + x) 4
(1 + x )8
(1 + x)12
éa b ù
Suppose A = ê
ú is a real matrix with nonëc d û
zero entries, ad – bc = 0 and A2 = A. Then, a + d
equal to
The number of integers x satisfying
é1 x
-3x 4 + det ê1 x 2
ê1 x3
ë
ANSWER KEY
17 (c) 21 (0) 25
18 (d) 22 (20) 26
19 (d) 23 (0.20) 27
20 (c) 24 (0) 28
x2 ù
x 4 ú = 0 is equal to
x 6 úû
(16)
(4)
(0)
(0)
29
30
(1)
(2)
20
CONTINUITY AND
DIFFERENTIABILITY
MCQs with One Correct Answer
1.
The function f ( x ) = (sin 2 x ) tan
p
defined at x = . The value of f
4
is continuous at x =
(a)
(c) 2
2.
3.
4.
e
2
2x
is not
5.
æ pö
ç ÷ so that f
è 4ø
æ
ç
ç
f ( x) = ç
ç
ç
ç
è
p
is
4
(b)
1
e
(d) None of these
ìï g ( x), x £ 0
. If f (x) is
be defined by f ( x) = í sin x
,x>0
ïî x
(cos x - sin x)cosec x , a
,
1/ x
e
ae
+e
2/ x
2/ x
+e
+ be
3/ x
3/ x
,
p
<x<0
2
x=0
0< x<
p
2
If f (x) is continuous at x = 0, then (a, b) =
ì1 - [ x]
ï
, x ¹ -1
If f (x) = í 1 + x
, then the value of
ïî1
, x = -1
f ( 2 k ) will be (where [] shows the greatest
integer function]
(a) Continuous at x = –1
(b) Continuous at x = 0 1
(c) Discontinuous at x =
2
(d) All of these
Suppose 'f ' is continuous function from R to R
and f ( f (a)) = a for some a Î R then the equation
f (x) = x has
(a) no solution
(b) exactly one solution
(c) at most one solution
(d) atleast three solutions
Let g (x) be a polynomial of degree one and f (x)
continuous satisfying f '(1) = f (–1), then g (x) is
(a) (1 + sin 1) x +1
(b) (1– sin1) x + 1
(c) (1– sin 1) x –1
(d) (1 + sin 1) x – 1
Let f (x) be defined as follows :
(a)
6.
æ 1ö
çè e, ÷ø
e
æ1 ö
e
(b) ç , e÷
è
ø
(c) (e, e)
(d)
The function f defined by
( e -1 , e -1 )
ìï (1 + sin p x )t - 1 üï
f ( x) = lim í
ý is
t
t ®¥ ï
x
(1
+
sin
p
)
+
1
î
þï
(a) every where continuous
(b) discontinuous at all integer values of x
(c) continuous at x = 0
(d) None of these
7.
Let f : R ® R be a function defined by
f (x) = min {x + 1, x + 1} ,then which of the
following is true?
(a) f (x) is differentiable everywhere
(b) f (x) is not differentiable at x = 0
(c) f (x) ³ 1 for all x Î R
(d) f (x) is not differentiable at x = 1
MATHEMATICS
46
8.
Let g ( x) =
( x – 1) n
log cos m ( x – 1)
; 0 < x < 2, m and n
14.
æ
x ®1+
9.
10.
11.
12.
15.
Let f: R ® R be a function such that f ( x ) £ x 2 ,
for all x Î R . Then, at x = 0, f is:
(a) continuous but not differentiable.
(b) continuous as well as differentiable.
(c) neither continuous nor differentiable.
(d) differentiable but not continuous.
æ xx – x – x ö
If f (x) = cot–1 ç
÷ , then f ¢ (1) is
2
è
ø
(a) – 1
(b) 1
(c) log 2
(d) – log 2
1
If g is the inverse of f and f ' (x) =
, then
1 + x3
find g' (x).
(a) 1 + [g(x)]2
(b) 1 – [g(x)]2
3
(c) 1 + [g(x)]
(d) 1 – [g(x)]3
æp
2ö
If f ( x) = sin ç [ x] - x ÷ , where 2 < x < 3 and
3
è
ø
[ . ] represetns greatest integer function, then
5p
3
(c)
13.
-2
5p
3
(b)
2
5p
3
(d)
-
5p
3
16.
(a) 1
(c)
x n -1
y n -1
(b) x/y
(d) None of these
f ''(0)
f n (0)
+ .... +
is
2!
n!
(a) n
(b) 2n
(c) 2n– 1
(d) None of these
If a and b are any two roots of equation ex cos
x = 1, then the equation ex sin x – 1 = 0 has
(a ) exactly one root in (a, b)
(b) exactly two roots in (a, b)
(c) at least one root in (a, b)
17.
18.
(d) no root in (a, b)
Let f(x) = x |sin x|, x Î R. Then
(a) f is differentiable for all x, except at
x = np, n = 1, 2, 3, ....,
(b) f is differentiable for all x, except at
x = np, n = ±1, ±2, ±3, ...
(c) f is differentiable for all x, except at
x = np, n = 0, 1, 2, 3
(d) f is differentiable for all x, except at
x = np, n = 0, ± 1, ± 2, ± 3, ...
f : [–1, 1] ® R be a function defined by
ì 2
æp ö
ï x cos ç ÷ for x ¹ 0
f ( x) = í
è xø
ï
0
for x = 0
î
The set of points where f is not differentiable is
(a) {x Î [–1, 1] : x ¹ 0}
(b)
If 1 - x 2n + 1 - y 2n = a( x n - y n ) , then
1 - x 2 n dy
is equal to
1 - y 2 n dx
(a) f (x, y) = y /x
(b) f (x, y) = y2 / x2
2
2
(c) f (x, y) = 2y / x (d) f (x, y) = x2 / y2
If f (x) = (1 + x)n, then the value of
f (0) + f '(0) +
æ 5p ö
f ¢ç
is equal to
ç 3 ÷÷
è
ø
(a)
6ö
è1- x ø
hand derivative of |x – 1| at x = 1. If lim g(x) = p,
(b) n = 1, m = – 1
(d) n > 2, m = n
dy
3
3
(1 - x 6 ) + (1 - y 6 ) = a(x – y ), and dx
= f (x, y) ç 1 - y6 ÷ , then
are integers, m ¹ 0, n > 0 , and let p be the left
then
(a) n = 1, m = 1
(c) n = 2, m = 2
If
2
ì
ü
, n ÎZ ý
í x Î[ -1,1] : x =
2n + 1
î
þ
(d) [–1, 1]
Let f : R ® R be a continuous function such that
f (x2) = f (x3) for all x Î R. Consider the following
statements.
(c)
19.
2
ì
ü
, n ÎZý
í x Î[-1, 1] : x = 0or x =
2
n
+
1
î
þ
Continuity and Differentiability
47
I. f is an odd function.
II. f is an even function.
III. f is differentiable everywhere
Then
(a) I is true and III is false
(b) II is true and III is false
(c) both I and III are true
(d) both II and III are true
20. Let f : R ® R be function defined by
26. If y = (1 + 1/ x ) x then find the value of
2 y2 (2) + 1/ 8
log 3/ 2 - 1/ 3
27. If y = ( x - sin x ) + ( x - sin x ) + ..., then
( )
ì sin x 2
ï
if x ¹ 0,
f (x) = í x
ï
if x = 0,
î 0
Then, at x = 0, f is
(a) not continuous
(b) continuous but not differentiable
(c) differentiable and the derivative is not
continuous
(d) differentiable and the derivative is
continuous
Numeric Value Answer
dx
dy
28.
æ pö
2
25.
1
2
3
4
5
6
7
8
(b)
(b)
(a)
(c)
- 2p = .........
ì
ï3 - cos x ï
=í
ï 2 + cos x +
ï
î
9
10
11
12
(b)
(a)
(c)
(b)
13
14
15
16
(c)
(d)
(b)
(c)
1
2
1
2
, | sin x |<
, | sin x |³
1
2
1
is
2
non-differentiable is
30. If the derivative of the function
ì 1
ü
f ( x ) = cos-1 í
(2 cos x - 3sin x) ý
î 13
þ
ì 1
ü
+ sin -1 í
(2 cos x + 3sin x) ý
î 13
þ
2
æ æ x öö æ æ x öö
F ( x) = ç f ç ÷ ÷ + ç g ç ÷ ÷ and given that
è è 2 øø è è 2 øø
F (5) = 5, then F (10) is equal to
Number of functions f : [0, 1] ® [0, 1] satisfying |
f (x) – f (y) | = |x – y| for all x, y in [0, 1] is
(b)
(d)
(d)
(b)
p
2
29. Number of points where function f (x) defined
as f :[0, 2p ] ® R, f ( x)
2
cos 2 n–1 x and n > 1, then f ' çè 2 ÷ø is
24. If f '' (x) = – f (x) and g (x) = f ' (x) and
x=
ìa cot x b
ïï x + x 2 , 0 <| x |£ 1
Let f ( x) = í
ï1
x=0
ïî 3
If f (x) is continuous at x = 0, then the value of
a2 + b2 is
21. If f is a real valued differentiable function
satisfying | f (x) – f (y) | £ ( x - y )2 , x, y Î R and
f (0) = 0, then f (1) is equal to
22. If x = cosec q - sin q ; y = cosecn q- sin n q ,
æ dy ö
and ( x2 + 4) ç ÷ - n 2 y 2 = kn2, then value of
è dx ø
k is
23. If f (x) = cos x cos 2x cos 22 x cos 23 x .....
2
3 10
w.r.t. 1 + x 2 at x = is , then b =
4
b
ANSWER KEY
17 (b) 21
18 (c) 22
19 (d) 23
20 (d) 24
(0)
(4)
(1)
(5)
25
26
27
28
(2)
(3)
(3)
(2)
29
30
(4)
(3)
21
APPLICATION OF
DERIVATIVES
MCQs with One Correct Answer
1.
2.
tan–1(sin
The function f (x) =
increasing function in
x + cos x) is an
(a)
æ pö
çè 0, ÷ø
2
æ p pö
(b) ç - , ÷
è 2 2ø
(c)
æ p pö
çè , ÷ø
4 2
p p
(d) æç - , ö÷
è 2 4ø
5.
6.
The greatest of the numbers
1, 21 / 2 , 31 / 3 , 41 / 4 , 51 / 5 , 61 / 6 and 71 / 7 is
(a)
3.
21 / 2
(b) 31 / 3
(c) 71 / 4
(d) All but 1 are equal
The interval of decrease of the function
7.
f ( x) = x 2 log 27 - 6 x log 27
2
+ (3x - 18 x + 24) log( x - 6 x + 8) is
4.
(3 - 1 + 1/ 3e , 2) È (4,3 + 1 + 1/ 3e )
(b)
(3 - 1 + 1/ 3e , 3 + 1 + 1/ 3e )
(c) (-¥, 3 - 1 + 1/ 3e ) È (3, 4 + 1 + 1/ 3e )
(d) None of these.
The set of positive values of the parameter ‘a’
for each of which the function
f ( x) = sin 2 x - 8(a + 1)sin x - (4a 2 + 8a - 14) x
is monotonic increasing in R and has no
critical points are
(a)
(0, 6 - 2)
(b)
(-2 - 6, 6 - 2)
(c)
(-2 - 6, 0)
(d) None of these
æ p pö
Let the function g : ( -¥, ¥) ® ç - , ÷ be
è 2 2ø
p
-1 u
given by g (u ) = 2 tan (e ) - . Then, g is
2
(a) even and is strictly increasing in (0, ¥ )
(b) odd and is strictly decreasing in ( -¥, ¥)
(c) odd and is strictly increasing in ( -¥, ¥)
(d) neither even nor odd, but is strictly
increasing in ( -¥, ¥)
The largest term in the sequence,
an =
2
(a)
x
x
and g(x) =
, where 0 < x £ 1,
sin x
tan x
then in this interval
(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.
If f (x ) =
n2
n3 + 200
(a) a 6
(c) a 8
8.
9.
is
(b) a 7
(d) None of these
x2 + 1
[ x]
for all x Î [1, 4], where [.] denotes the greatest
integer function. then, f (x) is monotonically.
(a) increasing on [1, 4)
(b) decreasing on [1, 4)
(c) increasing on [1, 2)
(d) decreasing on [2, 3)
If A > 0, B > 0 and A + B = p/3, then the maximum
value of tan A tan B is
1
1
(a)
(c) 3
(d)
(b)
3
3
3
Let f (x) be a function given by f ( x) =
Application of Derivatives
49
10. The set of values for which the function
x
x
f ( x) = (4a - 3)( x + ln 5) +2( a - 7) cot sin 2
2
2
does not possess critical points is
(a)
4ö
æ
ç -¥, - ÷ È (2, ¥)
3ø
è
(b)
(-¥, 2)
(c)
[1, ¥)
(d) (1, ¥)
11. A given right circular cone has a volume p, and
the largest right circular cylinder that can be
inscribed in the cone has a volume q. then p : q is
(a) 9 : 4
(b) 8 : 3
(c) 7 : 2
(d) None of these
12.
ìï 2 - | x 2 + 5 x + 6 |;
x ¹ -2
f ( x) = í
2
;
x = -2
ïî a + 1
then the range of a is so, that f(x) has maxima
at x = – 2 is
(a) | a |³ 1
(b) | a |< 1
(c)
a >1
(d)
a <1
p
p
<x< .
2
2
In order that f ( x) has exactly one minimum,
15. Two men P and Q start with velocities v at the
same time from the junction of two roads inclined
at 45o to each other. If they travel by different
roads, the rate at which they are being separated
is
(a)
v 2
(b)
v 2+ 2
(c)
v 2- 2
(d)
v/ 2
16. The general value of a such that the line x cos a+
y sin a = p is a normal to the curve (x + a) y = c2
is
p
3p
(a) æç 2np + ,(2n +1)p ö÷ È æç 2np + ,(2n + 2)p ö÷
2
2
è
ø è
ø
(b)
3p ö
æ
ç 2np + p , 2np + ÷
2 ø
è
(c)
3p
æ
ö
ç 2np + ,(2n + 2)p ÷
2
è
ø
(d)
p ö
æ
ç 2np , 2np + , ÷ È
2 ø
è
13. Let f ( x) = sin 3 x + l sin 2 x , -
3p
æ
ç (2n + 1)p , 2np +
2
è
l should belong to
(a) (–1, 1)
æ 3 ö æ 3ö
(b) ç - , 0 ÷ È ç 0, ÷
è 2 ø è 2ø
æ 3 1 ö æ 1 3ö
(d) ç - , - ÷ Èç , ÷
è 2 2 ø è 2 2ø
p
14. The function f (x) = 1 + x (sin x) [cos x], 0 < x £
2
(where [ . ] is G.I.F.)
(c)
(0, ¥)
pö
÷
2ø
æ pö
(b) is strictly increasing in ç 0, ÷
è 2ø
(a) is continuous on æç 0,
è
(c) is strictly decreasing in æç 0, p ö÷
è 2ø
(d) has global maximum value 2
ö
, ÷ (n Î I )
ø
17. Let P be a point on the hyperbola x 2 - y 2 = a 2 ,
where a is a parameter, such that P is nearest to
the line y = 2x. Then the locus of P is
(a) y = 2x
(b) y = x
(c) 2y = x
(d) x + y = 0
18. For the curve y = 3 sinq cosq, x = eq sin q,
0 £ q £ p, the tangent is parallel to x-axis when q is:
(a)
3p
4
(b)
p
2
(c)
p
4
(d)
p
6
19. Let C be the curve y3 – 3xy + 2 = 0. If H is the set
of points on the curve C, where the tangent is
parallel to x-axis and V is the set of points on
C where the tangent is parallel to y-axis, then
MATHEMATICS
50
20.
(a) H = {(x, y): y = 0, x Î R}, V = {(1, 1)}
(b) H = {(x, y) : x = 0, y Î R}, V = {(1, 1)}
(c) H = f, V = {(1, 1)}
(d) H = {(1, 1)} , V = {(x, y) : y = 0, x Î R}
The number of polynomials p : R ® R satisfying
p(0) = 0, p(x) > x2 for all x ¹ 0 and p¢¢(0) =
25.
26.
f ( x ) = (b2 - 3b + 2)(cos 2 x - sin 2 x)
1
is
2
27.
(a) 0
(b) 1
(c) more than 1, but finite
(d) infinite
28.
Numeric Value Answer
21.
22.
The fuel charges for running a train are
proportional to th e square of the speed
generated at 16 miles per hour and costs ` 48 per
hour. The most economical speed if the fixed
charges i.e. salaries etc. amount to ` 300 per hour
is
The curve x + y – ln (x + y) = 2 x +5 has a vertical
tangent at the point (a, b). Then a + b is equal to
23.
24.
1
2
3
Two ships A and B are sailing straightaway from
a fixed point O along routes such that ÐAOB is
always 120°. At a certain instance, OA = 8 km,
OB = 6 km and the ship A is sailing at the rate of
20 km/hr while the ship B sailing at the rate of 30
km/hr. If the distance between A and B is
k
changing at the rate
km/hr, then value of k
37
is _______.
Let P be a non-zero polynomial such that
P(1 + x) = P(1- x) for all real x, and P(l) = 0. Let m
be the largest integer such that (x - l)m divides
P(x) for all such P(x). Then m is equal to
The integral value of b for which the function
+(b - 1) x + sin(b + 1)
does not possess stationary point is
If the curves ax2 + by2 = 1 and a1x2 + b1y2 = 1
may cut each other orthogonally such that
æ1 1 ö
1 1
- = l ç - ÷ then l is equal to
a a1
è b b1 ø
Tangent at P1 (2,3) on the curve 3 y = x3 + 1
meets the curve again at P2 . The tangent at P2
meets the curve at P3 and so on. If the sum of
the ordinates for P1, P2 , P3 ,.....P60 be S then
æ 2183 - 8 ö
S +ç
÷ is equal to 5k, when k is equal to
è 27 ø
29. A conical vessel is to be prepared out of a
circular sheet of metal of unit radius. In order
that the vessel has maximum volume, the
sectorial area that must be removed from the
sheet is A1 and the area of the given sheet is A. If
A
= m + n , then m + n is equal to
A1
30.
An arch way is in the shape of semi-ellipse, the
road level being the major axis. If the breadth of
the road is 30 m and the height of the arch is 6 m
at a distance of 2 m from the side, the greatest
height of the arch in metres, is
A rectangle with its sides parallel to the X-axis
and Y-axis is inscribed in the region bounded
by the curves y = x2 – 4 and 2y = 4 – x2. The
maximum possible area of such a rectangle is
(d)
(b)
(a)
4
5
6
(a)
(c)
(c)
7
8
9
(b)
(c)
(b)
10
11
12
(a)
(a)
(a)
ANSWER KEY
13 (b) 16 (a)
14 (a) 17 (c)
15 (c) 18 (c)
19
20
21
(c) 22
(1)
25
(a) 23 (260) 26
(40) 24 (9.22) 27
(2)
(2)
(1)
28
29
30
(4)
(9)
(6)
22
INDEFINITE INTEGRATION
MCQs with One Correct Answer
1.
If
then value of (A, B) is
(a) (- cos a, sin a)
(b)
(cos a, sin a)
(- sin a, cos a)
(sin a, cos a)
(c)
2.
sin x
ò sin( x - a) dx = Ax + B log sin( x - a) + C ,
(d)
If f ( x ) = lim [2 x + 4 x3 + ......... + 2nx 2 n -1 ] ;
n®¥
(0 < x < 1), then ò ( f ( x )) dx is equal to
(a)
(c)
3.
4.
- 1 - x2 + c
1
x2 - 1
(b)
+c
(d)
1
1 - x2
1
1 - x2
+c
(a)
1 + nx n -1 - x 2 n
(1– x n ) 1 - x 2 n
(a)
æ 1 - xn ö
ex ç
÷ +C
è 1 + xn ø
(c)
æ 1+ xn ö
ex ç
÷ + C (d)
çè 1 - x n ÷ø
(b)
1 æ g ( x) ö
ç
÷ +C
2 è f ( x) ø
(c)
æ g ( x) ö ö
1æ
ç log ç
÷÷ +C
2è
è f ( x) ø ø
(d)
æ æ g ( x ) ö2 ö
log ç ç
÷ ÷+C
ç è f ( x) ø ÷
è
ø
2
(b)
æ 1 + xn ö
ex ç
÷ +C
è 1 - xn ø
æ 1 - xn ö
ex ç
÷ +C
çè 1 + x n ÷ø
If f ( x ) = lim n 2 ( x1/ n - x1/( n +1) ) , x > 0 then
n ®¥
6.
(a)
x2
+ ln x + C
2
(b) -
x2
x2
ln x +
+C
4
2
(c)
x3
+ xl n x + C
3
(d)
x2
x2
ln x +C
2
4
If f ( x ) = lim
xn - x -n
n®¥ x n
ò (sin
2
dx =
ò xf ( x) dx is equal to
+c
æ f ( x ) g '( x ) - f '( x ) g ( x ) ö
÷ (log ( g ( x ))
ò çè
f ( x) g ( x)
ø
- log ( f ( x))) dx is equal to
æ g ( x) ö
log ç
÷+C
è f ( x) ø
5.
x
òe
-1
+ x -n
, 0 < x < 1 , n Î N then
x ) f ( x)dx is equal to
(a)
-[ x sin -1 x + 1 - x 2 ] + C
(b)
x sin -1 x + 1 - x 2 + C
(c)
x2
+C
2
(d)
1
(sin -1 x )2 + C
2
MATHEMATICS
52
7.
x
æ ln a a x / 2
ln bb
ç
x
+
ò ç 3a5 x / 2b3 x 2a 2 x b 4 x
è
ö
÷ dx (where
÷
ø
(c)
+ log{log e e 3 x 3 } + C
+
a, b Î R ) is equal to
(a)
(b)
(c)
(d)
8.
(d)
a 2 x b3 x
2 x 3x
a
b
+k
ln
e
6 ln a 2b3
1
1
1
2 3
2 x 3x
6 ln a b a b
1
1
ln
1
2x 3x
ea b
+k
11.
ln(a 2 x b 3 x ) + k
6 ln a 2b 3 a 2 x b3 x
1
1
ln(a 2 x b3 x ) + k
2 3 2 x 3x
6 ln a b a b
n
If I n = ò (sin x + cos x ) dx
(n - 1)
(sin x + cos x )n -1
I n -2 ,
(sin x – cos x) +2
=
n
n
then I5 equals (t = (sin x + cos x))
(sin x - cos x) 5
[t - 3t 3 + 8t - 32] + C
(a)
15
(b)
12.
(sin x - cos x) 4
[3t + 8t 2 + 32] + C
15
(d) None of these
sin 2q
æ cos q + sin q ö 1
´ loge ç
÷ - f ( x) + c
2
è cos q - sin q ø 2
where f (x) is
(a) sec 2q
(b) log e (sec 2q)
(c) 2 tan–1q
(d) tan 2q
1
ò x{logex e × log e2 x e × loge3 x e} dx is equal to
1
(a)
log{log e e 2 x} - log{log e e3 x}
2
+ log{loge e 4 x} + C
(b)
1
log{loge x} - log{log e x}
2
+ log{loge x} + C
dx
( x - a )(b - x )
is
(a)
2
x -a
+C
a -b b - x
(b)
2
( x - a) (b - x) + C
a -b
If
æp
ö
tan ç - x ÷
è4
ø
ò cos2 x
tan 3 x + tan 2 x + tan x
= –2 tan–1 u + c then u is equal to
( a)
1 + tan x + cot x (b)
(c)
tan x + cot x
dx.
1 + tan x + tan 2 x
(d) tan -1 (tan x + cot x)
dx
ò (a + bx2 )
= K tan–1 ( L tan q) + M ,
b - ax 2
M being constant of integration then KL is equal to
1
1
(a) 1
(b)
(c)
(d)
a
a
a
a
13.
If
14.
If ò f ( x)sin x cos xdx =
=
10.
ò ( x - b)
a-b
( x - a) b - x + C
2
(d) None of these
(sin x - cos x ) 5
[t + 3t 4 + 8t 3 + 32t + 1] + C
15
æ cos q + sin q ö
ò cos 2q loge çè cos q - sin q ÷ø d q
1
log{loge ex} - log{log e e 2 x}
2
1
+ log{log e e 3 x} + C
2
(c)
(c)
9.
1
log{log e ex} - log{log e e 2 x 2 }
2
1
2(b - a 2 )
then f (x) is
(a)
(b)
(c)
(d)
1
a sin x + b cos x
1
a 2 sin 2 x + b2 cos2 x
1
a 2 sin x + b2 cos x
1
a sin 2 x + b cos2 x
2
l n f ( x) + c,
Indefinite Integration
53
15. The integral
æ sec 6α
sec18α
sec54α ö
ò çè cosec 2α + cosec 6α + cosec18α ÷ø dα
is equal to
ln | sec 54a | ln | sec 2a |
+c
(a)
108
4
ln | sec 6a | ln | sec18a | ln | sec54a |
+
+
(b)
+c
6
18
54
1
sec 6α
1
sec18α
ln
+ ln
(c)
6
cosec 2α 18 cosec 6α
1
sec 54α
+ ln
+c
54 cosec18α
cos x
+ ax + b l n | 2 cos x - sin x | + c,
2 cos x - sin x
then value of |a + b| is ______.
=
20. If
21.
16. If I n = ò (sin x + cos x)n dx (n ³ 2), then
22.
n
(a)
(sin x + cos x) (sin x - cos x )
(b)
(sin x + cos x) n -1 (sin x - cos x)
(c)
(sin x + cos x)n +1 (sin x - cos x)
x ®0
f ( x)
x2
exists finitely
1/ x
f ( x) ö
æ
lim ç1 + x +
x ÷ø
x ®0 è
1
2
3
= e , and
ò f ( x) log e x dx
cos 2 x + sin 2 x
ò (2 cos x - sin x)2 dx
(b)
(d)
(c)
4
5
6
(c)
(d)
(a)
7
8
9
= tan -1 f (q) + c
then the least value of f(q) for allowable values
of q is equal to
Let f(x) be a function such that f ( xy ) = f ( x) × f ( x )
" x, y Î R. If f (1 + x) = 1 + x(1 + g(x)) where
f ( x)
xk
+ C,
k
t ®0
ò f '( x) dx =
99
ò {sin (101x) × sin x}dx =
sin(100 x)(sin x)l
,
m
then evaluate the numerical quantity k.
23.
then
24. If
l
is equal to ..........
m
dx
ò cos3 x - sin3 x
= A tan -1 f ( x) + B ln
2 + f ( x)
+C
2 - f ( x)
where f (x) = sin x + cos x find the value of
(12 A + 9 2 B) - 3.
3
is equal to ax 3 ( log e x - b ) + c , then value of
a + b is _______.
19. If
ò
lim g ( x) = 0. Further if
(d) (sin x - cos x )n -1 (sin x + cos x )
17. Let ln x denote the logarithm of x with respect to
the base e. Let S Ì R be the set of all points
where the function ln (x2 –1) is well-defined.
Then, the number of functions f : S ® R that
are differentiable, satisfy f ¢ (x) = ln (x 2 – 1)
for all x Î S and f (2) = 0, is
(a) 0
(b) 1
(c) 2
(d) infinite
Numeric Value Answer
18. If lim
1
log e | (sin x ) k + (cos x ) k | + c
k
then k is equal to
q
sin3 dq
2
If
q
3
cos
cos q + cos 2 q + cos q
2
=
(d) None of these
nI n - 2(n - 1) I n- 2 is equal to
1 - (cot x )2010
ò tan x + (cot x)2011 dx
(b)
(c)
(b)
10
11
12
25. If
sin8 x
ò cos x dx = -
sin 7 x sin 5 x
+
a
b
sin 3 x sin x
æp xö
+
+ ln tan ç + ÷ + C ,
c
d
è 4 2ø
then evaluate (a + b + c + d).
ANSWER KEY
(d) 13 (c) 16
(a) 14 (b) 17
(a) 15 (a) 18
-
(b)
(d)
(1)
19 (0.60) 22
20 (2012) 23
(3)
21
24
(2)
(1)
(8)
25
(4)
23
DEFINITE INTEGRATION
MCQs with One Correct Answer
1.
Let p (x) be a polynomial of least degree whose
graph has three points of inflection (–1, –1),
(1, 1) and a point with abscissa 0 at which the
curve is inclined to the axis of abscissa at an
±1
(c)
±2
p/2
5.
If
1
angle of 60°. Then
ò
3 3 +4
14
(c)
3+ 7
14
p
2.
ò (cos px - sin qx)
(b)
(d)
2
3.
2p
ò
Let I =
(a)
(c) p
(d) 2p
the definite integral
4.
(a) 0
(b) p
(c) 2p
(d) ep
If p, q, r, s are in arithmetic progression and
p + sin x q + sin x
f ( x) = q + sin x r + sin x
r + sin x
s + sin x
2
such that
p - r + sin x
-1 + sin x
s - q + sin x
ò f ( x)dx = -4 , then the common
0
difference of the progession is
2
=
p
. Then
16
ò
sin x
dx , then I belongs to
x
æ 3 2ö
ç
÷
,
ç 8 6 ÷
è
ø
æ 2
(b) ç
,
ç 2
è
3ö
÷
2 ÷ø
æ1
2ö
(d) None of these
çç ,
÷
2 ÷ø
è2
Let f(x) be positive, continuous and
differentiable on the interval (a, b) and
(c)
7.
0
2
a cos x + b sin x
p/4
dx; where p, q are integers;
ecos q cos q(sin q)d q is
2
p/3
6.
-p
is equal to
(a) – p (b) 0
The value of
dx
2
minimum value of a cos x + b sin x is
(a) – 4
(b) – 8
(c) – 2
(d) -2 2
3 3
7
3+2
7
ò
0
p ( x)dx is equal to
0
(a)
1
2
(d) None of these
(b)
(a)
lim f ( x ) = 1 , lim f ( x ) = 31/ 4 . If
x®a +
x ®b –
1
then the greatest value
f '( x ) ³ f 3 ( x ) +
f ( x)
of b – a is
(b) 31 / 4
(a) 1
(c)
(31/ 4 - 1 )
p
24
(d)
p
24
Definite Integration
8.
55
Let a, b, c be non-zero real numbers such that
13. Suppose the limit L = lim
1
n®¥
8
2
ò (1 + cos x)(ax + bx + c ) dx
0
exists and is larger than
2
= ò (1 + cos8 x)(ax2 + bx + c) dx .
ax 2 + bx + c = 0 has
(a) no root in (0, 2)
(b) at least one root in (0, 2)
(c) a double root in (0, 2)
(d) two imaginary roots
x
x
0
Then the value of
x
| f '(w) | where w is a complex cube root of unity
is
(a)
(b)
3
3 3
2 3 (c)
(d)
4 3
sec 2 x
ò
10.
lim
p
x®
4
(a)
(c)
p2
x 16
equals
2
8
f (2)
p
2 æ 1ö
fç ÷
p è 2ø
2
f (2)
p
(b)
(d) 4 f (2)
11. The value of the integral
ò
p
0
(1- | sin8 x |)dx is
(a) 0
(b) p –1
(c) p –2
(d) p –3
12. Let S be the set of real numbers p such that there
is no non-zero continuous function f : R ® R
satisfying
S is
(a)
(b)
(c)
(d)
Then f (l) equals
(a) e
(b) e 2
(c) e 4
(d) e 6
1
x
dx.
8
0 1+ x
Consider the following assertions:
15. Let J = ò
I.
J>
1
4
II.
J<
p
8
Then
(a) only I is true
(b) only II is true
(c) both I and II are true
(d) neither I nor II is true
f (t )dt
2
1
. Then,
2
f (x) = 2ò tf (t )dt + 1 for all x ³ 0.
2
ò (t - 1)dt .
dx
1
<L<2
(b) 2 < L < 3
2
(c) 3 < L < 4
(d) L > 4
14. Suppose a continuous function f : [0, ¥) ® R
satisfies
Then the quadratic equation
Let f ( x ) =
1
1
0 (1 + x 2 ) n
(a)
0
9.
nò
ò f ( t )dt = p f (x) for all x Î R. Then,
x
0
the empty set
the set of all rational numbers
the set of all irrational numbers
the whole set R.
16. For x ÎR, let f (x) = | sin x | and g (x) =
x
ò f (t ) dt.
0
2
Let p(x) = g(x) - x. Then
p
(a) p(x + p) = p(x) for all x
(b) p(x + p) ¹ p(x) for at least one but finitely
many x
(c) p(x + p) ¹ p(x) for infinitely many x
(d) p is a one-one function
17. Let f: [0, 1] ® [0, 1] be a continuous function
such that
1
x2 + (f(x))2 £ 1 for all x Î [0, 1] and ò f ( x ) dx =
1
2
Then
0
f ( x)
ò 1 - x2 dx equals
1
2
p
.
4
MATHEMATICS
56
(a)
(c)
p
12
2 -1
p
2
(b)
p
15
(d)
p
10
22.
x
x
ò
ò
x (1 - t ) f (t )dt = t f (t )dt "x Î (0, ¥) and
0
Numeric Value Answer
18.
Let f : (0, ¥) ® R be a differentiable function
such that
Let f : R ® R be a differentiable function and
23.
f (x)
ò
f (1) = 4. Then the value of lim
x ®1
0
f(1) = 1. The value of the limit lim f ( x ) is
x ®¥
equal to
If p and q are different roots of the equation
2t dt
4
1
tan x = x then
, if
x -1
0
f ' (1) = 2 is
3
sin x
19.
f ( x ) = cosec x
If
tan x
24.
2
sec x
x -1
x sin x
cos x
x tan x
x2 + 1
f (x)dx=
-p / 3
20.
é1 x 2
1
1 ù
lim ê 5 ò e-t dt - 4 + 2 ú =
x®0 ê x
3x úû
x
ë 0
If I (n) =
ò
4
5
6
7
8
9
(d)
(b)
(c)
10
11
12
–1
ò t | f ( f (t)) | dt for
all
F ( x) 1
= , then the value of
x ®1 G ( x ) 14
xÎ[–1, 2]. If lim
q sin n qd q, n Î N , n > 3, then
(a)
(a)
(a)
greatest integral value of x.
Let f : ¡ ® ¡ be a continuous odd function,
which vanishes exactly at one point and
x
1
f (1) = . Suppose that F(x) = ò f (t )dt for all
2
–1
1
-1
[2010 I (2010) -2009 I (2008)] is
1005
equal to
(a)
(d)
(c)
25.
é x ù dx , when [x] is the
ë û
x Î [–1, 2] and G(x) =
0
1
2
3
ò
x
p/2
21.
Find the value
0
then
p/3
ò
ò 2 sin( pt )sin(qt )dt is equal to
æ 1ö
f ç ÷ is
è 2ø
ANSWER KEY
(a) 13 (a) 16
(c) 14 (a) 17
(d) 15 (a) 18
(a) 19
(0)
22
(a) 20 (0.30) 23
(16) 21
(2)
24
0
0
(2)
25
(7)
24
APPLICATION OF
INTEGRALS
5.
MCQs with One Correct Answer
1.
points ( x1, y1 ) and (x2, y2) where (x1 < x2), then
The area bounded by the x-axis, the curve
y = f(x) and the lines x = 1, x = b, is equal to
the value of m for which
b 2 + 1 - 2 for all b > 1, then f(x) is
(a)
x -1
(b)
(c)
2
(d)
x +1
x
1+ x 2
6.
2
Area of region bounded by [ x ] = [ y ]
3.
if x Î [1, 5] (where [.] represents the greatest
integer function), is
(a) 10 sq. unit
(b) 8 sq. unit
(c) 6 sq. unit
(d) 5 sq. unit
If a point P moves such that its distance from
7.
(a)
ln 2
(c) ln 4
(d)
The
area
2ln 2
8
1
(c)
(d) 0
3
3
bounded by the curve
(a) 4p square units (b) 8p square units
(c) 4 square units
(d) 2 square units
The area bounded by the curve y2 (2a – x) = x3
and the line x = 2a is
(c)
8.
3pa 2
sq. unit
4
(b)
3pa 2
sq. unit
2
(d)
6pa 2
sq. unit
5
The area bounded by the curves y = xe x ,
y = xe - x and the line x = 1 is
2
2
1
1
(b) 1 (c)
(d) 1 e
e
e
e
Area enclosed by the curve
| x + y – 1| + |2x + y – 1| = 1 is
(a) 2 sq. units
(b) 3 sq. units
(c) 6 sq. units
(d) 7 sq. units
(a)
(b) 2 ln 2
(b)
(a) 3pa2 sq. unit
(a)
p
and the x-axis, is
2
2
between the ordinates x = 0 to x = 2p is
(2 3, -1), then area of the curve, described by
P, enclosed between the coordinate axes is
3
(b) 2 3 (c) 6
(d)
3
2
The area bounded by the curve f (x) = ||tan x +
cot x| – |tan x – cot x|| between the lines x = 0,
(a)
f ( x) = x + sin x and its inverse function
line y = 3x - 7 is same as its distance from
x=
ò
æ
x2 ö
mx
+
2
ç
÷ dx is
2ø
è
minimum, is
2.
4.
x2
x1
x +1
2
If the line y = mx + 2 cuts the parabola 2y = x2 at
9.
MATHEMATICS
58
10.
The area bounded by the curve
(a)
f ( x) = max.{4 - x 2 ,| x - 2|, ( x - 2)1/ 3}
for x Î[-2, 4] and x-axis is
(a)
45
sq. units
4
(b)
23
sq. units
4
15.
(a)
ò
0
t
(1 + t ) 1 - t 2
2 -1
(b)
ò
0
4t
(1 + t 2 ) 1 - t 2
2 +1
(c)
ò
0
4t
(1 + t 2 ) 1 - t 2
2 +1
(d)
ò
0
12.
t
(1 + t 2 ) 1 - t 2
dt
1
(1 + log 2 )
2
}
x + 3 , 5y £ x + 9 £ 15
16.
1
4
5
3
(b)
(c)
(d)
6
3
3
2
The maximum possible area bounded by the
parabola y = x2 + x + 10 and a chord of the
parabola of length 1 is
17.
1
1
1
1
(b)
(c)
(d)
12
6
3
2
The area of the region bounded by the lines
x = l, x = 2 and the curves x(y - ex) = sin x and
2xy = 2sin x + x3 is
dt
dt
(a)
e2 - e -
1
6
(b)
e2 - e -
7
6
(c)
e2 - e +
1
6
(d)
e2 - e +
7
6
Numeric Value Answer
18.
dt
If [x] is the greatest integrer £ x , then
2
ò min{x - [ x], - x - [ - x ]} dx =
3
13.
The area enclosed by the curves y = x , y = x ,
x = 0 and x = p, where p > 1, is 1/6. Then p equals
(a) 8/3 (b) 16/3 (c) 2
(d) 4/3
If a straight line y – x = 2 divides the region
14.
x2 + y 2 £ 4 into two parts, then the ratio of the
area of the smaller part to the area of the greater
part is
(a) 3p – 8 : p + 8
(b) p – 3 : 3p + 3
(c) 3p – 4 : p + 4
(d) p – 2 : 3p + 2
The area of the region above the x-axis bounded
p
by the curve y = tan x, 0 £ x £ and the tangent to
2
p
is:
4
(d)
(a)
2
the curve at x =
1æ
1ö
ç log 2 + ÷
2è
2ø
(a)
p
is
4
2
(b)
is equal to
1 - sin x
1 + sin x
y=
and y =
bounded by
cos x
cos x
2 -1
{( x, y ) Î ¡ : y ³
2
59
53
sq. units
(d)
sq. units
4
4
The area of the region between the curves
the lines x = 0 and x =
1
(1 - log 2 )
2
Area of the region
(c)
(c)
11.
1æ
1ö
ç log 2 - ÷
2è
2ø
-2
19.
Let for all n Î N , n - 2 ³ 2 log e n then the
minimum value of the area bounded by the curve
| y|+
20.
1
£ e -| x| is
n
ïì {x},
If f ( x ) = í
ïî1,
x ÏI
x ÎI
2
and g ( x ) = {x} ,
(where {x} is fractional part of x), then area
bounded between f (x) and g(x) "x Î[0,10] is .
Application of Integrals
59
21. P (x, y) is a point, which moves in the xy plane
such that 2 [ y ] = 3 [ x ] where [ . ] denotes the
greatest integer function and -2 £ x £ 5 and
-3 £ y £ 6 . The area of the region containing
the point P (x, y) is equal to
22. If the value of a ( a ³ 1) for which the area of the
figure bounded by pair of straight lines
2
y 2 - 3 y + 2 = 0 and the curves y = [ a ] x ,
1
[ a ] x2 is greatest, is [a, b), then
2
(b – a) = where [.] denotes the greatest integer
function.
If the area bounded by y = f (x) and the curve
y=
23.
y=
2
1 + x2
24. Let f ( x) be a polynomial of degree 3 such that
the curve y = f ( x) has relative extremes at
x = ±2 / 3 and passes through (0, 0) and
(1, –2) dividing the circle x2 + y2 = 4 in two parts.
Find out the difference of areas of these two
parts.
25. Area of the region bounded by the curve
y = {x 2 }, where {.} denotes fractional part
function
7ö
æ
x Î [–2, 2] is p ç q + r - ÷ . Find
è
3ø
the value of p + q + r.
where f is a continuous function
satisfying the conditions f (x) · f (y) = f (xy). "
aö
æ
x, y, Î R and f ¢ (1) = 2, f (1) = 1is ç p - ÷ , then
3ø
è
a=
1
2
3
(d)
(b)
(a)
4
5
6
(c)
(d)
(d)
7
8
9
(b)
(a)
(a)
10
11
12
ANSWER KEY
(c) 13 (d) 16
(b) 14 (a) 17
(d) 15 (c) 18
(b)
(b)
(1)
(2)
19
22
20 (3.33) 23
(4)
21
24
(1)
(2)
(0)
25
(7)
25
DIFFERENTIAL EQUATION
AND ITS APPLICATIONS
MCQs with One Correct Answer
1.
2.
Through any point (x, y) of a curve which passes
through the origin, lines are drawn parallel to
the co-ordinate axes. The curve, given that it
divides the rectangle formed by the two lines
and the axes into two areas, one of which is twice
the other, represents a family of
(a) circles
(b) pair of straight lines
(c) parabolas
(d) rectangular hyperbolas
Lef f (x) be a positive, continuous and
differentiable function on the interval (a, b). If
lim f ( x ) = 1 an d
x®a
+
4.
5.
1
then
f ¢ (x) ³ (x) +
f ( x)
p
p
(a) b – a ³
(b) b – a £
4
4
p
(c) b – a £
(d) None of these
24
Let f be a non-negative function defined on the
interval [0, 1].
6.
f3
3.
If
x
ò0
x
1 - ( f '(t )) 2 dt = ò f (t ) dt , 0 £ x £ 1 and
f (0) = 0, then
(a) f (1/ 2) < 1/ 2
(b) f (1/ 2) > 1/ 2
(c) f (1/ 2) < 1/ 2
(d) f (1/ 2) > 1/ 2
0
and
and
and
and
f (1/ 3) > 1/ 3
f (1/ 3) > 1/ 3
f (1/ 3) < 1/ 3
f (1/ 3) < 1/ 3
(b)
y - y 2 - x 2 = cxy
(c)
y y 2 - x2 = cx + y
(d) x y 2 - x 2 = cx + y
The solution of the differential equation,
2x2 y
lim f ( x) = 31/ 4 . Also
x ®b -
The solution of the differential equation
dy
x3
= y 3 + y 2 y 2 - x 2 is
dx
(a) y + y 2 - x2 = cxy
dy
= tan( x 2 y 2 ) – 2 xy 2 given y (1) =
dx
is
(a) sinx2y2 = ex–1
(b) sin(x2y2) = x
2
2
(c) cosx y + x = 0 (d) sin(x2y2) = e.ex
Solution of the differental equation
æ yö
æ yö
x cos ç ÷ ( ydx + xdy ) = y sin ç ÷ ( xdy - ydx)
x
è ø
èxø
is
(a)
æxö
y = cx cos ç ÷
è yø
y
(b) sec æç ö÷ = cxy
èxø
æ yö æ yö
ç ÷ sec ç ÷ = c (d) None of these
è xø èxø
The family of curves whose tangents form an
p
with the hyperbolas xy = C are
angle of
4
(a) pair of straight lines
(c)
7.
p
2
(b)
y 2 - xy - x 2 = C
(c)
y 2 - 2 xy - x 2 = C
(d)
y 2 + 2 xy - x 2 = C
Differential Equation and its Applications
8.
9.
Solution of the differential equation
ìï 1
ìï x 2
1 üï
y 2 üï
- ý dy = 0 is
dx + í
í 2ý
2
y þï
îï x ( x - y ) þï
îï ( x - y )
x
xy
+
=c
(a) ln
y x- y
xy
=c
(b) ln | xy | +
( x - y)
xy
= ce x / y
(c)
( x - y)
xy
= c e xy
(d)
( x - y)
(where c is arbitrary constant)
The equation of the curve passing through the
points (3a, a) (a > 0) in the form x = f(y) which
satisfy the differential equation;
a 2 dx x y
× = + - 2, is
xy dy y x
æ 1 + e y -k ö
x
=
y
+
a
çç
(a)
y-k ÷
÷
è 1 - 2e
ø
y
k
æ1+ e
ö
(b) x = y + a çç
y-k ÷
÷
è 1- e
ø
y
k
æ1+ e
ö
(c) y = x + a çç
y-k ÷
÷
è 1- e
ø
(d) None of these
10. The solution of the differential equation
( y + x xy ( x + y ))dx + ( y xy ( x + y) - x)dy = 0,
is
(a)
x2 + y 2
x
+ 2 tan -1
=C
2
2y
(b)
x2 + y 2
+ 2 tan -1
2
x
=C
y
x2 + y 2
x
=C
y
+ 2 tan -1
2
(d) None of these
(c)
11.
dy
y3
Solution of dx = 2 x
is
e + y2
(a)
e
-2 x 2
y + 2 ln | y |= c
61
(b)
e2 x y 2 - 2 ln | y |= c
(c) e x + ln | y |= c
(d) None of these
12. The orthogonal trajectory of the curve
an–1 y = xn is
(a) n2 + x2 = constant (b) ny2 + x2 = constant
(c) x2 + y2 = constant (d) y2 + nx2 = constant
13. A curve f (x) passes through the point P (1, 1).
The normal to the curve at point P is
a( y - 1) + ( x - 1) = 0 . If the slope of the tangent
at any point on the curve is proportional to the
ordinate at that point, then the equation of the
curve is
(a)
y = e ax - 1
(b)
y - 1 = eax
(d) y - a = e ax
(c) y = e a ( x -1)
14. A tangent and a normal to a curve at any point P
meet the x and y axes at A, B and C, D respectively.
If the centre of circle through O, C, P and B lies
on the line y = x (O is the origin) then the
differential equation of all such curves is :
(a)
(c)
dy y - x
=
dx y + x
dy x - y
=
dx
xy
(b)
dy y 2 - x 2
=
dx y 2 + x 2
(d) None of these
dy
- y log e 2 = 2sin x (cos x - 1) log e 2, then y =
dx
(a) 2sin x + c 2 x
(b) 2cos x + c 2 x
15. If
(c) 2sin x + c 2- x
(d) 2cos x + c 2- x
16. Solution of differential equation
æ d
ö
t ç ( g ( x )) ÷ - t 2
dt
dx
ø
is
= è
dx
g ( x)
(a)
t=
g ( x)
+c
x
(b) t =
g ( x)
(c)
t=
g ( x)
x+c
(d) t = g ( x) + x + c
x2
+c
MATHEMATICS
62
17.
(a)
1 2 x2
+
3x
3
(b)
1 4 x2
- +
3x
3
(c)
1 2
- + 2
x x
(d)
1
x
22.
æ 1 + x 2 - y2 ö
x dx - y dy
= ç
÷ be
x dy - y dx
è x2 - y 2 ø
æ
f ( x, y ) + 1 + f ( x, y ) = c ç
è
Find the sum of the order ‘O’ and degree D of
the differential equation
n®¥
23.
24.
g ( x) dx equals to.
1/ e
25.
If the solution of the differential equation
where c is an arbitrary constant then f (3, 2) is
equal to
A curve y = f (x) is such that f ( x) ³ 0 and
equal to 10 k, where k equals
Let y = f(x) is a polynomial function satisfying
æ1ö
æ1ö
æ1ö
x2 f '( x) f ç ÷ - f ( x) f ' ç ÷ = x2 f '( x) - f ' ç ÷ .
x
x
è ø
è ø
èxø
If f(1) = 2 and f(5) = 26, then find f(6) – 30.
e
ò
x+ y ö
÷
f ( x, y ) ø
f (0) = 0 and bounds a curvilinear trapezoid
with the base [0, x] whose area is proportional to
(n + 1)th power of f (x). If f (1) = 1, then {f(10)}n is
n
2
xn æ dy ö
æ dy ö x æ dy ö
y = 1 + x ç ÷ + ç ÷ + ..... + ç ÷
n! è dx ø
è dx ø 2! è dx ø
+........¥ .
Let y = f (x) be a curve passing through (e, ee),
which satisfy the differential equation (2ny + xy
logex) dx – x logex dy = 0, x > 0, y > 0. If
g ( x) = lim f ( x), then
dy
1
=
is
2
dx xy[ x sin y 2 + 1]
If the ar ea bounded by y = f ( x), x =
1
,
2
3
and the X-axis is A sq. units where
2
2
2 4
2 4 6
f ( x ) = x + x3 + × x 5 + × × x 7 +
3
3 5
3 5 7
....¥, | x |< 1, then the value of [4A] is (where
[.] is G..I.F)
x=
2
21.
0
æ 1ö
then y ç ÷ is equal to
è 2ø
If the solution of the differential equation
2
20.
0
t 2 f ( x ) - x 2 f (t )
lim
= 1 for each x > 0, then f(x)
t®x
t-x
Numeric Value Answer
19.
x
x ò y (t )dt = ( x + 1) ò ty (t )dt , x > 0 , and y(1) = e,
is
18.
x
Let f(x) be continuously differentiable on the
interval (0, ¥ ) such that f(1) = 1, and
x 2 (cos y 2 - sin y 2 - 2Ce - y ) = k , then value
of k is ______.
If the equation of a curve y = y(x) satisfies the
differential equation
ANSWER KEY
1
2
3
(c)
(c)
(c)
4
5
6
(a)
(a)
(b)
7
8
9
(c)
(a)
(b)
10
11
12
(b)
(a)
(b)
13
14
15
(c)
(a)
(a)
16
17
18
(c)
(a)
(2)
19
20
21
(0)
(2)
(8)
22
23
24
(5)
(1)
(7)
25
(1)
26
VECTOR ALGEBRA
1.
2.
3.
4.
MCQs with One Correct Answer
r
The vectors a ( x) = cos xiˆ + sin xjˆ and
r
b ( x) = xiˆ + sin xjˆ are collinear for
p
(a) Unique value of x, 0 < x <
6
p
p
(b) Unique value of < x <
6
3
(c) No value of x
p
(d) Infinitely many values of x, 0 < x <
2
If cos a ¹ 1, cos b ¹ 1 and cos g ¹ 1, then the
r
r
vector a = iˆ cos a + ˆj + kˆ, b = iˆ + ˆj cos b + kˆ ,
r
c = iˆ + ˆj + kˆ cos g are
(a) coplanar vectors
(b) coplanar vectors if cos a = cos b = cos g ¹ 1
(c) coplanar vectors if os a ¹ cos b ¹ cos g
(d) never coplanar
Let a, b, c be distinct non-negative numbers. If
the vectors aiˆ + ajˆ + ckˆ, iˆ + kˆ and ciˆ + cjˆ + bkˆ
lie in a plane, then c is
(a) the Arithmetic Mean of a and b
(b) the Geometric Mean of a and b
(c) the Harmonic Mean of a and b
(d) equal tor zero
r
Let a and b are two vectors making angle q
with each other, then which of the following
r
represents unit vectors along bisector of a and
r
b is:
aˆ + bˆ
aˆ + bˆ
(a) ±
(b) ±
2 cos q
2
ˆ
aˆ + b
(aˆ + bˆ)
(c) ±
(d)
2cos q / 2
| aˆ + bˆ |
5.
uuur
uuur
In a parallelogram ABCD, | AB |= a,| AD |= b and
uuur
uuuur uuur
| AC |= c , the value of DB . AB is
(a)
3a 2 + b2 - c 2
2
(b)
a 2 + 3b2 - c 2
2
a 2 + 3b2 + c 2
a 2 - b2 + 3c 2
(d)
2
2
r
r
r
a and c are unit vectors and | b | = 4 . The angle
(c)
6.
r
r
r
æ 1ö
r
r
between a and c is cos -1 ç ÷ . If b - 2c = la ,
è 4ø
then l is equal to
1 3
,
4 4
1 3
(c) - 3,4
(d) - ,
4 4
The angles of a triangle, two of whose side are
r
represented by the vectors 3(aˆ ´ b) and
r
r
r
ˆ ˆ where b is a non-zero vector and â
b - (a.b).a
is a unit vector, are
(a) 3, – 4
7.
(b)
(a)
æ 3+2ö
æ 1 ö
æ 1ö
tan -1 ç
; tan -1 ç ÷ ; tan -1 ç
÷
÷
è 2ø
è 3ø
è1- 2 3ø
(b)
æ 1ö
tan-1( 3); tan -1 ç ÷ ;cot -1(0)
è 3ø
(c)
æ 3+2ö
tan -1 ( 3); tan -1 (2); tan -1 ç
÷
è 2 3 - 1ø
(d)
æ 2 +3ö
tan -1 ( 3); tan -1 (2); tan -1 ç
÷
è 3 2 - 1ø
MATHEMATICS
64
8.
uur
If the vectors a = (c log 2 x)iˆ - 6 ˆj + 3kˆ and
uur
b = (log 2 x)iˆ + 2 ˆj + (2c log 2 x)kˆ make an
12.
vectors iˆ + ˆj - kˆ and 2iˆ + 2 ˆj - kˆ is
obtuse angle for any x Î(0, ¥) , then the
interval to which ‘c’ belongs is
(0, ¥)
(a)
9.
(a)
(-¥, 0)
4
(c) (- , 0)
(d) ( -1, 0) È (0, 2 )
3
3
ur
r
Let p = ai$+ b$j + ck$ and q = bi$+ c$j + ak$,
ur
where a, b, c Î R. If 'q' be the angle between p
r
and q then,
q Î (0, p / 2)
(a)
10.
(b)
13.
(b) q Î[0, 2p / 3]
(c) q Î (2p / 3, p]
(d) q Î[p / 2, p]
The value of ‘a’ so that th e volume of
parallelopiped formed by iˆ + ajˆ + kˆ, ˆj + akˆ and
aiˆ + kˆ becomes minimum is
(a) –3
11.
(b) 3
(c) 1
3 (d)
® ® ®
3
14.
If a , b , c are three non-zero, non-coplanar
vectors and
®
b1
®
c1
®
c2
®
c3
®
® ®
® ®
b× a ® ® ® b× a ®
a , b2 = b +
a,
= b®
®
2
2
|a|
|a|
®
® ®
® ®
® ®
® ®
c× a ® b× c ®
= ca+
b1 ,
®
®
2
2
|a|
|c|
®
c × a ® b1× c ®
= c- ® a- ®
b1 ,
2
2
|a|
| b1 |
®
® ®
® ®
® ® ®
(a)
( a , b1, c3 )
(b) ( a , b , c )
1 2
(c)
( a , b1, c1)
(d) ( a , b , c )
2 2
® ® ®
® ® ®
5
ˆ
3i + 2 ˆj + 2kˆ
(b)
3iˆ + 2 ˆj - 2 kˆ
r iˆ ˆj kˆ
v = + + , then
a b c
r
r
(a) u and v are parallel vectors
r
r
(b) u and v are orthogonal vectors
r r
(c) u × v = 1
r r
(d) u ´ v = iˆ + ˆj + kˆ
If the two adjacent sides of two rectangles are
represented by vectors
r
r
r r r
r
r
r r
p = 5a - 3b ; q = -a - 2b an d r = -4a - b ;
r
r r
s = - a + b , r espectively, then the angle
r 1 r r r
between the vectors x = ( p + r + s ) and
3
r 1 r r
y = ( r + s ) is
5
æ 19 ö
(a) - cos -1 ç
÷
è 5 43 ø
æ 19 ö
cos -1 ç
÷
è 5 43 ø
æ 19 ö
p cos -1 ç
÷
è 5 43 ø
(d) Cannot be evaluated
r r
r
a, b and c be three non-coplanar vectors and
r
d be a non-zero vector, which is perpendicular
r r r
to (a + b + c ). Now if
r
r r
r r
r r
d = sin x(a ´ b ) + cos y (b ´ c ) + 2(c ´ a ), then
2
2
minimum value of x + y is equal to
(a) p2
(b) 0
(c) p2/4 (d) 5p2/4
(c)
15.
2 ˆj + kˆ
17
ˆ
2i + 2 ˆj - 2kˆ
(c)
(d)
3
17
If a, b and c are pth, qth, rth terms of HP and
r
u = (q - r )iˆ + (r - p) ˆj + ( p - q)kˆ,
(b)
c× a ® b× c ®
a+
b1 ,
= c®
®
2
2
|c|
|c|
uur uur
uur uur
uur uur c . a uur b . c uur
c4 = c - uur a = uur b1 ,
| c |2
| b |2
then the set of orthogonal vectors is
® ® ®
A unit vector which is perpendicular to the
vector 2iˆ - ˆj + 2kˆ and is coplanar with the
Vector Algebra
65
r
16. The vectors a = 2l 2iˆ + 4lˆj + kˆ and
r
b = 7iˆ - 2 ˆj + lkˆ make an obtuse angle
r
whereas the angle between b and k̂ is acute
and less than p/6,
1
(a) 0 < l <
(b) l > 159
2
1
(c) - < l < 0
(d) null set
2
r
17. The angle q between two non-zero vectors a
r
and b satisfies the relation
r
r
r
r
cos q = (a ´ iˆ) × (b ´ iˆ) + (a ´ ˆj ) × (b ´ ˆj )
r
r
+(a ´ kˆ) × (b ´ kˆ),
then the least value of |a| + |b| is equal to (where
q ¹ 90°)
(a)
(c)
1
2
(b) 2
a,
b,
where x1, x2, x3 Î {– 3, – 2, – 1, 0, 1, 2}.
r
r
If the number of possible vectors b such that a
r
and b are mutually perpendicular is t, then t /5 =
22. Find the absolute value of parameter t for which
the area of the triangle whose vertices are
A(–1, 1, 2); B(1, 2, 3) and C(t, 1, 1) is minimum.
r
23. Given a vector A defined as
r
r r
r r
r r
A = (a ´ b ) ´ (c ´ d ) + (a ´ c )
r r
r r
r r
´(b ´ d ) + (a ´ d ) ´ (b ´ c ) , then find the value
r r
of | A ´ a | .
r é1ù
r
24. Let v0 be a fixed vector and v0 = ê ú . Then for
ë0û
n ³ 0 a sequence is defined
(d) 4
2
g are scalars. If
a= k1 ( Fˆ × aˆ ) - k2 ( Fˆ × bˆ), then the value of
2(k1 + k2) is
r
r
k , b = x1$i + x2 $j + x3 $
k,
21. Let a = $i + $j + $
where
Numeric Value Answer
18. If the sum of two unit vectors is a unit vector,
and the magnitude of their difference is k ,
then value of k is
ˆ yˆ and ẑ are three unit vectors in three19. If x,
dimensional space, then the minimum value of
2
2
2
xˆ + yˆ + yˆ + zˆ + zˆ + xˆ
r
r æ1ö
vn +1 = vn + ç ÷
è2ø
25.
20. Let â and b̂ be two unit vectors such that
r
1
aˆ × bˆ = and aˆ ´ bˆ = cˆ. Also F = aaˆ + bbˆ + gcˆ,
3
n +1
é0 -1ù
ê
ú
ë1 0 û
n +1
r
v0 then
éa ù
r
a
lim vn = ê ú . Find .
n ®¥
b
b
ë û
If A1, A2, ...., Ag are vertices of a regular octagon,
7 uuur
uuur
uuur uuur
then å (OA j ´ OA j +1 ) = K (OA1 ´ OA2 )
j =1
where the value of K is ........
ANSWER KEY
1
2
3
(b)
(d)
(b)
4
5
6
(c)
(a)
(a)
7
8
9
(b)
(c)
(b)
10
11
12
(c)
(b)
(d)
13
14
15
(b)
(b)
(d)
16
17
18
(d)
(c)
(3)
19
20
21
(3)
(3)
(5)
22
23
24
(2)
(0)
(2)
25
(7)
27
THREE DIMENSIONAL
GEOMETRY
MCQs with One Correct Answer
1.
The angle between the lines whose direction
cosines are given by the equations
3l + m + 5n = 0 , 6nm - 2nl + 5lm = 0 is
(a)
(c)
2.
æ1ö
cos -1 ç ÷
è6ø
æ2ö
cos -1 ç ÷
è3ø
(b)
(d)
æ 1ö
cos -1 ç - ÷
è 6ø
4.
æ pù
ç 0, ú
è 4û
(b)
5.
(b)
l1 + l2
m + m2
n +n
, 1
, 1 2
2 cos q / 2 2 cos q / 2 2 cos q / 2
1
(d) –2
2
L1 and L2 are two lines whose vector equations
r
are L1 : r = l[(cos q + 3)ˆi +
(b) –1
(c)
-
r
and L2 : r = m(aˆi + bˆj + ckˆ ) where, l and m are
scalars and a is the acute angle between L1 and
L2. If the angle a is independent of q, then the
value of a is
é p pù
æ p pù
(d)
ç , ú
ê 4 , 2ú
ë
û
è 3 2û
If l1, m1, n1 and l2, m2, n2 are DCs of the two
lines inclined to each other at an angle q, then
the DCs of the internal bisector of the angle
between these lines are
l1 + l2
m + m2
n +n
, 1
, 1 2
2 sin q / 2 2 sin q / 2 2 sin q / 2
If the straight lines
x = 1 + s , y = -3 - l s , z = 1 + l s
( 2 sin q)ˆj + (cos q - 3)kˆ ]
é p pù
ê6 , 3ú
ë
û
(a)
l1 - l2
m - m2
n -n
, 1
, 1 2
2 cos q / 2 2 cos q / 2 2 cos q / 2
(a) 0
(c)
3.
(d)
t
, y = 1 + t , z = 2 - t , with parameters
2
s and t respectively, are co-planar, then l equals.
A line in the 3-dimensional space makes an angle
(a)
l1 - l2
m - m2
n -n
, 1
, 1 2
2 sin q / 2 2 sin q / 2 2 sin q / 2
and x =
æ 5ö
cos -1 ç - ÷
è 6ø
pö
æ
q çè 0 < q £ ÷ø with both the x and y axes. Then
2
the set of all values of q is the interval:
(c)
(a)
6.
p
6
(b)
p
4
(c)
p
3
(d)
p
2
x + 6 y + 10 z + 14
=
=
is the
5
3
8
hypotenuse of an isosceles right angled triangle
whose opposite vertex is (7, 2, 4). Then which of
the following is not the side of the triangle?
The line
Three Dimensional Geometry
(a)
x-7 y-2 z-4
=
=
2
6
-3
(b)
x-7 y-2 z-4
=
=
3
6
2
ABC satisfies the relation 1 + 1 + 1 = k ,
x2 y 2 z 2
then the value k is
(a) 3
x-7 y-2 z-4
=
=
3
5
-1
(d) None of these
The vertex A of the triangle ABC is on the line
r
r = iˆ + ˆj + lkˆ and the vertices B and C have
(c)
7.
67
12. The
8.
9.
(c) [–2, 2]
(d) [-4, - 2] È [2, 4]
A line with positive direction cosines passes
through the point P(2, –1, 2) and makes equal
angles with the coordinate axes. The line meets
the plane 2x + y + z = 9 at point Q. The length of
the line segment PQ equals
(a) 1
(b)
(d) 2
2 (c)
3
Through a point P (h, k, l) a plane is drawn at
right angles to OP to meet the co-ordinate axes
in A, B and C. If OP = p, then the area of DABC
is :
(a)
p 2 hk
l2
(b)
p 3l
3hk
p5
p 2t 2
(d)
2hkl
2hk
10. Projection of the line x + y + z – 3 = 0 = 2x + 3y +
4z – 6 on the plane z = 0 is
(c)
(a)
x
y-6 z
=
=
-2
1
0
(b)
x y-6 z
=
=
1
-2
0
x y-6 z
=
=
(d) none of these
1
2
0
11. A variable plane at a distance of the one unit
from the origin cuts the coordinates axes at
A, B and C. If the centroid D (x, y, z) of triangle
(c)
plane
(c)
1
3
containing
(d) 9
the
line
x -1 y - 2 z - 3
=
=
and parallel to the line
1
2
3
respective position vectors iˆ and ĵ . Let D be
é 3 33 ù
the area of the triangle and D Î ê 2 , 2 ú , then
ë
û
the range of values l corresponding to 'A' is
(a) [-8, - 4] È [4, 8] (b) [–4, 4]
(b) 1
13.
x
=
1
(a)
(c)
The
y z
= passes through the point:
1 4
(1, – 2, 5)
(b) (1, 0, 5)
(0, 3, –5)
(d) (– 1, – 3, 0)
position vectors of points a and b are
iˆ - ˆj + 3kˆ and 3iˆ + 3 ˆj + 3kˆ respectively. The
equation of a plane is r × (5iˆ + 2 ˆj - 7 kˆ) + 9 = 0.
The points a and b
(a) lie on the plane
(b) are on the same side of the plane
(c) are on the opposite side of the plane
(d) None of the above
14. The angle between the pair of planes represented by equation
2 x 2 - 2 y 2 + 4 z 2 + 6 xz + 2 yz + 3xy = 0 is
(a)
æ1ö
cos -1 ç ÷
è 3ø
(b)
æ 4ö
cos -1 ç ÷
è 21 ø
(c)
æ4ö
cos -1 ç ÷
è9ø
(d)
æ 7 ö
cos -1 ç
÷
è 84 ø
15. Let P = (–3, 1, 1) and Q = (3, 4, 2). R divides PQ
in the ratio PR : PQ = 1 : 3. Then, the equation of
uuur
the plane perpendicular to PQ at R is
(a) 18x + 9y + 3z = 8 (b) 18x + 9y + 3z = 4
(c) 9x + 18y + 3z = 4 (d) 3x + 9y + 18z = 8
16. Let s1, s2, s3 be planes passing through the
origin. Assume that s1 is perpendicular to the
vector (1, 1, 1), s2 is perpendicular to a vector
(a, b, c), and s3 is perpendicular to the vector
(a2, b2, c2).
MATHEMATICS
68
17.
What are all the positive values of a, b and
c so that s1 Ç s2 Ç s3 is a single point?
(a) Any positive value of a, b, and c other than 1
(b) Any positive values of a, b and c where
either a ¹ b, b ¹ c or a ¹ c
(c) Any three distinct positive values of a, b,
and c
(d) There exist no such positive real numbers
a, b and c
r ˆ ˆ ˆ r
r
Let a = i + j + k , b = 2iˆ + 2 ˆj + kˆ , and c = 5iˆ + ˆj - kˆ
be three vectors. The area of the region formed
by the set of points whose position vectors
r
r r
r s art i sfy t h e eq u a t i on s r . a = 5 an d
r
r r
| r - b | + | r - c | = 4 is closest to the integer
(a) 4
(c) 14
(b) 9
(d) 19
21.
22.
f (-2) = 1, f (1) ¹ 1, f (0) ¹ 2 and the remaining
two are false. If the area of the triangle formed
by (–2, 1, 0) and (f(–2), f(1), f(0)) and the origin is
19.
20.
A line with direction ratios (2, 1, 2) intersects
r
the lines r = - ˆj + l(iˆ + ˆj + kˆ) and
r
r = -iˆ + m(2iˆ + ˆj + kˆ) at A and B, respectively,,
then length of AB is equal to
The shortest distance between the z-axis and the
line, x + y + 2z - 3 = 0 , 2x + 3y + 4z - 4 = 0 is
The distance of the point (1, 0, –3) from the plane
x - y - z = 9 measured parallel to the line
k
; then sum of digits of k is.
2
If the equation of the plane through the
intersection of the planes x + 2y + 3z – 4 = 0 and
2x + y – z + 5 = 0 and perpendicular to the plane
5x + 3y + 6z + 8 = 0 is ax + by + cz + 173 = 0, then
b – 9(a + c) is equal to ........
If a line is passing through (a, b, c) and
intersecting y = 0, z2 = 4ax lies on the surface
given by
23.
Numeric Value Answer
18.
Let the equation of the plane containing line
x – y – z – 4 = 0 = x + y + 2z – 4 and parallel to the
line of intersection of the planes 2x + 3y + z = 1
and x + 3y + 2z = 2 be x + Ay + Bz + C = 0. Then
the values of |A + B + C – 4| is ........ .
Let f be a one-one function with domain {–2, 1, 0}
and range {1, 2, 3} such that exactly one of the
following statements is true :
24.
25.
(bz - cy)2 = k a(b - y)(bx - ay); then find the
value of k.
If the volume enclosed by the equation
| x |£ 8, | y |£ 8, | z |£ 8 and | x + y + z |£ 8 is t,
then
t
=
512
x-2 y+2 z -6
is
=
=
2
3
-6
ANSW ER KEY
1
2
3
(b)
(c)
(b)
4
5
6
(d)
(a)
(c)
7
8
9
(d)
(c)
(d)
10
11
12
(b)
(d)
(b)
13
14
15
(c)
(c)
(b)
16
17
18
(c)
(a)
(3)
19
20
21
(2)
(b)
(7)
22
23
24
(7)
(6)
(4)
25
(4)
28
PROBABILITY-2
1.
Let A and B be the sets {1, 2, …10} and {1, 2,
…20} respectively. A function is selected
randomly from A to B the probability that the
function is non-decreasing is
29
(a)
(20)
29
3.
(b)
10
(c)
2.
29
C10
C19
(20)10
Raj and Sanchita are playing game in which they
throw two dice alternately till one of them gets 9.
Which one of the following could be the
probability that Sanchita win the game?
(a) 7/15 or 8/15
(b) 6/11 or 5/11
(c) 8/17 or 9/17
(d) None of these
A boy whose hobby is tossing a fair coin is to
score one point for every tail and 2 points for
every head. The boy goes on tossing the coin,
till his score reaches n or exceeds n where n > 2.
If pn is the probability that his score attains
exactly n, then pn is equal to
(b)
The odds in favour of a book reviewed by three
independent critics are, respectively, 5 : 2, 4 : 3
and 3 : 4. The probability that majority of the
critics give favourable remark is
(a)
(20)10
1
p n -1 + pn -2
2
n
4.
C20
(d) None of these
(a) pn–1 + pn–2
5.
n
2n +1 + ( -1)
2 ( -1)
+
(c)
(d)
3
2n
3.2 n
Entries of a 2 × 2 determinant are chosen from
the set {–1, 1}. The probability that determinant
has zero value is
1
3
(a)
1
4
(b)
(c)
1
2
(d) None of these
6.
(c)
211
343
(d)
205
343
1
1
+ 3
6
7
7
(b)
1
1
1
+ 3- 6
7
7 7
9
1
1
1
1
+ 9
(d)
+
3
6
7
7
7
7
79
If a Î[– 20, 0], then probability that the graph of
the function y =16x2 + 8(a + 5) x – 7a – 5 is
strictly above the x-axis is
(c)
1
3
13
17
7
3
(c)
(d)
(b)
20
20
20
20
In a Competitive test, a candidate guesses,
copies or knows the answer to a multiple choice
question with four choices. The probability that
he makes a guess is 1/3 and the probability that
he copies the answer is 1/6. The probability that
his answer is correct given that he copied it is 1/
8. Find the probability that he knew the answer
to the question, given that he answered it
correctly.
(a) 24/29
(b) 26/29
(c) 22/29
(d) None of these
(a)
8.
209
343
Matrices of order 3 × 3 are formed by using the
elements of the set A = {–3, –2, –1, 0, 1, 2, 3},
then probability that matrix is either symmetric
or skew symmetric is
(a)
7.
210
(b)
343
MATHEMATICS
70
9.
A wire of length l is cut into three pieces. Then
the probability that the three pieces form a triangle
is
(a)
1
2
(b)
(a)
14.
2
(d) None of these
3
Given that the sum of two non-negative
quantities is 200, the probability that their
product is not less than
2
1
, otherwise it is . If he writes
3
4
2 books, the probability that at least one book
will be published is
3
times their greatest
4
7
101
9
10
(a)
(b)
(c)
(d)
16
201
16
16
On each evening a boy either watches
DOORDARSHAN channel or TEN SPORTS. The
3
1
that he will be asleep, while it is
4
4
when he watches TEN SPORTS. On one day, the
boy is found to be asleep. The probability that
the boy watched DOORDARSHAN is
15.
parameters n = 8 and p =
(a)
(c)
13.
( 2 - p )k +1
2a . p
(2 - p)
(b)
pk
( 2 - p )k +1
k
k +1
(d) None of these
A die is thrown 2n + 1 times, n Î N. The
probability that faces with even numbers show
odd numbers of times, is
121
119
117
115
(b)
(c)
(d)
128
128
128
128
Ravi and Rashmi are each holding 2 red cards
and 2 black cards (all four red and all four black
cards are identical). Ravi picks a card at random
from Rashmi, and then Rashmi picks a card at
random from Ravi. This process is repeated a
second time. Let p be the probability that both
have all 4 cards of the same colour. Then p
satisfies
(a) p £ 5%
(b) 5% < p £ 10%
(c) 10% < p £ 15%
(d) 15% < p
The probability of men getting a certain disease
1
is
and that of women getting the same
2
1
disease is . The blood test that identifies the
5
4
disease gives the correct result with probability .
5
(a)
16.
(a)
of n children have the same probability. If k ³ 1 ,
then the probability that a family contains exactly
k boys is
1
, then p(|x – 4| < 2) is
2
equal to
chance of
5
2
3
4
(b)
(c)
(d)
7
7
7
7
Let the probability Pn that a family has exactly n
children be apn, when n ³ 1 and P0 = 1 – ap
(1 + p + p2 + ...). Suppose that all sex distributions
407
411
405
307
(b)
(c)
(d)
576
576
576
576
If X follows a binomial distribution with
(a)
4
.
5
If he watches DOORDARSHAN, there is a
2a
1
(d) None of these
2
A book writer writes a good book with probability
be published is
probability that he watches TEN SPORTS is
12.
1
2
1
. If it is a good book, the probability that it will
2
product value is
11.
(b) less than
(c) greater than
1
4
(c)
10.
2n + 1
4n + 3
17.
Suppose a person is chosen at random from a
group of 30 males and 20 females, and the blood
test of the person is found to be positive. What
is the probability that the chosen person is a
man?
(a)
75
107
(b)
3
5
(c)
15
19
(d)
3
10
Probability-2
71
Numeric Value Answer
18. A and B play a game of tennis. The situation of
the game is as follows : if one scores two
consecutive points after a deuce, he wins, if loss
of a point is followed by win of a point, it is
deuce. The chance of a server to win a point is
2
. The game is at deuce and A is serving.
3
Probability that A will win the match is, (serves
are changed after each game).
22. A number x is selected from the set of first 9
natural numbers (i.e., x = 1, 2, 3, ......., 9). If the
probability that f(f(x)) = x where f(x) = x2 – 3x + 3
is
23. A special die is so constructed that the
probabilities of throwing 1, 2, 3, 4, 5 and 6 are
(1 – k) / 6, (1 + 2k) / 6, (1 – k) / 6, (1 + k) / 6,
(1 – 2k) / 6 and (1 + k) / 6, respectively. If two
such dice are thrown and the probability of
19. The probablilty that the length of a randomly
chosen chord of a circle lies between
getting a sum equal to 9 lies between
2
5
and
3
6
of its diameter is
20. In a hurdle race, a runner has probability p of
jumping over a specific hurdle. Given that in 5
trials, the runner succeeded 3 times, the
conditional probability that the runner had
succeeded in the first trial, is
24. A sportsman's chance of shooting an animal at a
distance r > a ("a" is constant) is given to be
a2
r2
. He fires at r = 2a and if he misses, then
again fires at r = 3a. He repeats the same process
at r = 4a, 5a and 6a. When he misses at r = 6a, the
animal escapes into the jungle. If the odds
against the sportsman are p : q, then q – p is
_______.,
8
or at the point (II) with probability
9
1
. There are 21 shells each of which can be
9
1
2
and ,
9
9
then the integral value of k is.
21. An artillery target may be either at point I with
probability
m
, then m is equal to _________.
9
25. A determinant of the second order is made with
m
be the probability
n
that the determinant made is non-negative, where
m and n are relative primes, then the value of
n – m is
the elements 0 and 1. If
fired either at point I or II. Each shell may hit the
target independently of the other shell with
1
. Minimum number of shells that
2
must be fired at point I to hit the target with
maximum probability is equal to 2k. Then value
of k is.
probability
ANSW ER KEY
1
2
3
(c)
(c)
(d)
4
5
6
(c)
(b)
(d)
7
8
9
(b)
(a)
(b)
10
11
12
(b)
(c)
(c)
13
14
15
(d)
(a)
(b)
16
(a)
19
17
(a)
20
18 (0.50) 21
(0.25)
(0.60)
(6)
22
23
24
(2)
(0)
(5)
25
(3)
29
PROPERTIES OF
TRIANGLE
5.
MCQs with One Correct Answer
1.
2.
If the sines of the angles A and B of a triangle
ABC satisfy the equation c2x2 – c(a + b) x +
ab = 0, then the triangle is (a,b,c, are sides of D)
(a) acute angled
(b) obtuse angled
(c) right angled
(d) no such triangle is possible
sin 3 A + sin 3 B + sin 3 C
(a) 7
3.
4.
6.
= 7, then the
maximum possible value of a is
(b) 49
(c)
3
7
7.
(d)
1
7
If in a triangle ABC, sin A, sin B, sin C are in
A.P., then
(a) the altitudes are in A.P.
(b) the reciprocals of altitudes are in A.P.
(c) the altitudes are in G.P.
(d) the medians are in A.P.
Let a, b and A are given and c1 and c2 are two
values of c, then the value of c12 + c22 – 2c1c2
cos 2A =
(a) a2sin2A
(b) a2 cos2 A
2
2
(c) 2(a + b )
(d) 4a2cos2 A
(a + b + c )2
, then
ab + bc + ca
(a) P Î [1, 2]
(b) P Î [3, 4)
(c) P Î (2, 4]
(d) None of these
If a, b, c are the sides of a triangle such that
b. c = l2, for some positive l, then
P=
In DABC, a ³ b ³ c . If
a 3 + b 3 + c3
If a, b, c be the sides of a triangle and
(a)
a ³ 2l sin
A
2
(b) b ³ 2l sin
B
2
(c)
c ³ 2l sin
C
2
(d) all are correct
In DABC, a2 (s – a) + b2 (s – b) + c2 (s – c) =
(a) 4R(cos A + cos B + cos C)
(b) 4RD(sin A + sin B + sin C)
A
B Cö
æ
4RD ç 1 + 4 sin sin sin ÷
2
2
2ø
è
(d) None of these
In an obtuse angled triangle, the obtuse angle is
(c)
8.
3p
and the other two angles are equal to two
4
values of q satisfying a tan q + b sec q = c,
2
2
where | b | £ a + c , then a2 – c2 is equal to
(a) ac
(c)
a
c
(b) 2ac
(d) None of these
Properties of Triangle
9.
73
1
In a triangle ABC, 2ac sin ( A - B + C ) =
2
(a) a2 + b2 – c2
(c) b2 – c2 – a2
(b) c2 + a2 – b2
(d) c2 – a2 – b2
10. The sides of a triangle are sin a, cos a and
1 + sin a cos a for some 0 < a <
p
. Then the
2
greatest angle of the triangle is
(a) 150° (b) 90°
(c) 120° (d) 60°
11. Given a triangle DABC such that sin2 A + sin2 C
=1001 × sin 2 B. Then the value of
2 (tan A + tan C) × tan 2 B
tan A + tan B + tan C
1
1
1
1
(a)
(b)
(c)
(d)
2000
1000
500
250
12. In DABC, if
sin A sin B sin C c
b
a
+
+
= + + , then
c sin B
c
b
ab ac bc
the value of angle A is
(a) 120° (b) 90°
(c) 60°
(d) 30°
13. If the hypotenuse of a right-angled triangle is
four times the length of the perpendicular drawn
from the opposite vertex to it, then the difference of the two acute angles will be
(a) 60° (b) 15°
(c) 75°
(d) 30°
14. In a DABC, ÐB =
p
p
and ÐC =
let D divide
3
4
BC internally in the ratio 1 : 3, then
sin (ÐBAD)
sin (ÐCAD)
is equal to:
1
1
1
2
(b)
(c)
(d)
3
3
6
3
15. In a triangle ABC, a = 5, b = 4 and cos (A – B)
(a)
31
, then the third side is equal to:
=
32
(where symbols used have usual meanings)
(b) 6 6
(a)
6
(c) 6
(d) (216)1/4
16. In triangle ABC if A : B : C = 1 : 2 : 4, then (a2 – b2)
(b2 – c2) (c2 – a2) = l a2b2c2, where l =
(where notations have their usual meaning)
(a) 1
(b) 2
(c) 4
(d) 9
17. If in a right angle triangle ABC, 4 sinA cosB – 1 = 0
and tan A is real, then
(a) angles are in A.P. (b) B2 = AC
2 AC
(d) None of these
A+ C
18. A tower of height b subtends an angle at a point
O on the level of the foot of the tower and at a
distance a from the foot of the tower. If a pole
mounted on the tower also subtends an equal
angle at O, the height of the pole is
(c)
B=
(a)
æ a 2 - b2
bç
ç a 2 + b2
è
ö
÷
÷
ø
æ a2 + b2 ö
(b) b ç
÷
ç a 2 - b2 ÷
è
ø
(c)
æ a 2 - b2
aç
ç a2 + b2
è
ö
÷
÷
ø
(d)
æ a2 + b2
aç
ç a 2 - b2
è
ö
÷
÷
ø
19. An observer on the top of a tree, finds the angle
of depression of a car moving towards the tree
to be 30°. After 3 minutes this angle becomes
60°. After how much more time, the car will reach
the tree?
(a) 4 min.
(b) 4.5 m
(c) 1.5 min
(d) 2 min.
20. Two flagstaffs stand on a horizontal plane. A
and B are two points on the line joining their feet
and between them. The angles of elevation of
the tops of the flagstaffs as seen from A are 30°
and 60° and as seen from B are 60º and 45°. If AB
is 30m, then the distance between the flagstaffs
in metres is
(a)
30 + 15 3
(b)
45 + 15 3
(c)
60 - 15 3
(d)
60 + 15 3
MATHEMATICS
74
24.
Numeric Value Answer
21.
Points D, E are taken on the side BC of an acute
angled triangle ABC, such that BD = DE = EC. If
ÐBAD = x, ÐDAE = y and ÐEAC = z then the
value of
22.
sin (x + y) sin (y +z)
is _________.
sin x sin z
In triangle ABC, sin A sin B + sin B sin C + sin C
sin A = 9/4 and a = 2, then the value of 3D,
where D is the area of triangle, is _________.
23.
Triangle ABC is right angled at A. The points P
and Q are on hypotenuse BC such that BP = PQ
= QC. If AP = 3 and AQ = 4, if BC = a b then
AD, BE, CF are internal angular bisectors of
A
DABC and I is the incentre. If a (b + c) sec
2
B
C
ID + b(a + c) sec
IE + c(a + b) sec
2
2
IF = kabc, then the value of k is
25.
A man from the top of a 100 metres high tower
see a car moving towards the tower at an angle
of depression of 30°. After some time, the angle
of depression becomes 60°. If the distance (in
metres) travelled by the car during this time is
a b
æ a
ö
, then ç - b ÷ =
3
è 20
ø
| a - b |=
1
2
3
(c)
(c)
(b)
4
5
6
(d)
(b)
(a)
7
8
9
(c)
(b)
(b)
10
11
12
(c)
(d)
(b)
ANSWER KEY
13 (a) 16
14 (a) 17
15 (c) 18
(a)
(a)
(b)
19
20
21
(c)
(d)
(4)
22
23
24
(3)
(2)
(2)
25
(7)
Hints & Solutions
CHAPTER
Sets
1
1.
B
A
B¢ – A¢ and A – B
2.
3.
4.
5.
(b) Let A = {q : sin q = tan q}
and B = {q : cos q = 1}
sin q ü
ì
\ A = íq : sin q =
ý
cos
qþ
î
= {q : sin q (cos q – 1) = 0} = {q = 0, p, 2p, 3p,.....}
For B : cos q = 1 Þ q = p, 2p, 4p,......
This shows that A is not contained in B. i.e. A Ë B.
but B Ì A.
(c) n(P(S)) = 23 = 8 elements.
n(P(P(S))) = 28 = 256 elements.
(c) Let n be the number of newspapers which
are read. Then 60 n = (300) (5) Þ n = 25
(a) From Venn-Euler’s Diagram.
(A È B) '
A
7.
Total 100
Chicken (60)
Fish (50)
c
a
b
n
U
(A 'Ç B)
6.
(a) bN = {bx : x ÎN}; cN = {cx : x ÎN}
\ bN Ç cN = {x : x is multiple of b and c both}
= { x: x is multiple of l.c.m. of b and c }
= { x : x is multiple of b c}
[given b and c are relatively prime \ l.c.m. of b
and c = bc]
\ bN Ç cN = {bc x : x ÎN} = dN (Given)
\ d = bc.
9. (a) (A È B È C) È (A Ç B' Ç C' )' Ç C'
= (A È B È C) Ç (A' È B È C ) Ç C'
= [(A Ç A') È (B È C)] Ç C' = (f È B È C) Ç C'
= (B È C) Ç C' = (B Ç C') È (C Ç C’)
= (B Ç C') È f = B Ç C'
10. (a) Total number of persons = a + b + c + n
= 100
8.
(a)
B
\ (A È B) ' È (A 'Ç B) = A '
(b) Given set can be written as
(A – B) È (B – A) = (A È B) – (A Ç B)
(By definition of symmetric difference)
Hence, (A \ B) È (B \ A) = (A È B) \ (A Ç B)
(b) A = {x : |x| < 1} = (–1, 1)
Since, |x| < 1 Þ –1< x < 1
B = {x : |x–1| ³ 1} = (– ¥, 0] È [2, ¥)
Since, |x –1| ³ 1 Þ x–1£ –1 or x – 1³ 1
Þ x £ 0 or x ³ 2
\ A È B = (– ¥ , 0] È [2, ¥ ) È (–1, 1)
= (– ¥, 1) È [2, ¥ ) = R – [1, 2)
\ D = [1, 2) = { x : 1 £ x < 2}
Do not prefer fish b + n = 50
60 prefer chicken hence b + c = 60
Do not like fish and chicken is n = 10
On solving these equations we will get a = 30,
b = 40, c = 20
The number of persons who prefer both fish
and chicken is = c = 20
11. (c) {(A – B) È (B – C) È (C – A)}¢
= (A – B)¢ Ç (B – C)¢ (C – A)¢
= [(U – (A – B)) Ç (U – (B – C) Ç (U – (C – A))]
A
B
A– B
C –A C
By Venn - diagram = (A Ç B Ç C)
MATHEMATICS
76
12. (d) n(A) = 1000, n(B) = 500,
n(A Ç B) ³ 1 & n(A È B) = p
Also, n(A È B) = n(A) + n(B) – n(A Ç B)
Þ p = 1000 + 500 – n (A Ç B)
Q 1 £ n(A Ç B) £ 500
Hence p £ 1499 and p ³ 1000 Þ 1000 £ p £ 1499
13. (a) Minimum value of n
= 100 – (30 + 20 + 25 +15) = 100 – 90 = 10
14. (a) Let the number of students who take only
Math be x and only Chemistry be y.
C
M
x
30
y
So, from the Venn diagram, we have total number
of students who take Math = x + 30 and take
Chemistry = y + 30.
According to question, we have
10
( x + 30)
30 =
100
12
( 30 + y )
Þ x = 270 and 30 =
100
Þ y = 220
x + y + 30= 270 + 220 + 30 = 520.
15. (a) See the following Venn diagram
M
I
23
29
Hence a = 20k and b = 60k, then
n = 100k – 20k – 60k = 20k
The difference between those who opted for
‘Surf blue’ and those who were uncertain
= 60k – 20k = 40k = 720 hence, k = 18,
Hence total number of persons covered in
survey
= 100k = 1800
18. (c) From the given information (C) = 80,
(F) = 40, and (C Ç F) = (n)
Hence (U) = (C) + (F) – (C Ç F) + (n)
Or 80 + 40 – x + x = 120
19. (d) If number of students who like chocolate
=a+c
Number of students who like Biscuit = b + c
Number of students who like Both = c
Number of students who like none = n = 10
From the given condition 100 = a + b + c + n
Since 60 of them don’t like Chocolate, hence
b + n = 60 or b = 50
And 50 of them don’t like Biscuit hence a + n
= 50, a = 40
Hence 100 = 40 + 50 + c + 10 or c = 0
20. (b) The given information can be represented
as follows.
U = 100
F
4–x
x
W
n (I) = 29 + 23 = 52; n (F) = 100 – 52 = 48
n (m È D ) = n ( m ) + n ( D ) - n ( m Ç D )
24 = 23 + 4 - n (m Ç D) \ n (m Ç D) = 3
\ n( W Ç D) = 4 - 3 = 1
16. (b) C stands for set of students taking
economics
E
b
g
a
c
C
e
d
f
M
a + b + c + d + e + f + g = 40; a + b + d + g = 16
b + c + e + g = 22; d + e + f + g = 26
b + g = 5; e + g = 14; g = 2
Go by backward substitution
e = 12, b = 3, d + f = 12, c + e = 17 Þ c = 5;
a + d = 11
a +d + f = 18 Þ f = 7
\ d = 12 – 7 = 5
17. (c) Let those who opted for Nirma = a and
those who opted Surf Blue = b and those who
opted for none is n.
21.
Now from the given information we can frame
following equationsc = 18, f + c = 23, f + g = 8,
c + f + g + e = 28, a + d + f + g = 48,
d + g = 10, n = 24
People who read exactly two consecutive
months is represented by d and e.
f + c = 23 and c = 18
\ f= 5
f + g = 8 and f = 5
\ g=3
d + g = 10 and g = 3
\ d=7
c + f + g + e = 28, c = 18, f = 5 and g = 3
\ e = 2 or \ d + e = 9
(4) We have
n (U) = 100, where U stands for universal set
n (M Ç C Ç T) = 10; n (M Ç C) = 20;
n (C Ç T) = 30; n (M Ç T) = 25;
n (only M ) = 12; n (only C) = 5; n (only T) = 8
Filling all the entries we obtain the Venn diagram
as shown below :
Solutions
77
M
C
12
15
10
10
8
= Total students – n(C È F È H)
U
n 28
=
=7
4 4
23. (7) The given condition is as follows-
5
n(C¢ Ç F¢ Ç H¢)= 80 – 52 = 28, So
20
T
\ n (M È C È T)
= 12 + 10 + 5 + 15 + 10 + 20 + 8 = 80
\ n (M È C È T)' = 100 – 80 = 20 = n.
n 20
=4
So, =
5 5
22. (7) Numbers which are divisible by 5 are 5,
10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70,
75, 80 they are 16 in numbers. Now, Numbers
which are divisible by 7 are 7, 14, 21, 28, 35, 42,
49, 56, 63, 70, 77 they are 11 in numbers.
Also, total odd numbers = 40
Let C represents the students who opt. for
cricket, F for football and H for hockey.
\ we have n(C) = 40, n(F) = 16, n(H) = 11
Now, C Ç F = Odd numbers which are divisible
by 5.
CÇ H = Odd numbers which are divisible by 7.
F Ç H = Numbers which are divisible by both 5
and 7.
n(C Ç F), 8, n(C Ç H) = 6,
n(FÇ H) = 2, n (C Ç F Ç H) = 1
We Know
n(CÈFÈH) = n(C) + n(F) + n(H)
– n(C Ç F) – n(C Ç H)
– n(F Ç H) + n(C Ç H Ç F)
n(CÈFÈH) = 67 – 16 + 1 = 52
\ n(C¢ Ç F¢ Ç H¢)
CHAPTER
2
1.
(b) Put x = y = 1, (f (1))2 = 3 f (1) – 2
Þ f (1) = 1 or 2
Let f (1) = 1, then put y = 1
f (x) . f (1) = f (x) + f (1) + f (x) – 2
Þ f (x) = 1 constant function \ f (1) ¹ 1,
2.
hence f (1) = 2
(a) Given
f ( x) = x14 - x11 + x 6 - x3 + x 2 + 1
for f (x) to be defined,
x14 - x11 + x6 - x 3 + x 2 + 1 ³ 0
Case 1 : x ³ 1
x14 - x11 + x 6 - x3 + x 2 + 1
We know that {(a + d + e + g) + ( b + d + f + g)
+ (c + e + f + g)} – ( d + e + f) – 2g
=a+b+c+d+e+f+g
or 61x + 46x + 29x – 25x – 2g = 97x
or 2g = 14x or g = 7x. So 7% of people
watched all the three movies)
24. (4) n (A È B) is minimum when A Í B. In this
case, (A È B) = B and hence minimum value of
n (A È B) = n(B) = 7.
n (A È B) is maximum when A and B are disjoint.
\ Maximum value of n (A È B) = 4 + 7 = 11.
So 11 – 7 = 4
25. (3)
2m - 2n = 112 Þ 2 n (2 m- n - 1) = 16.7
\ 2n (2m -n - 1) = 24 (23 - 1)
Comparing, we get n = 4 and m – n = 3
Þ n = 4 and m = 7
So m – n = 7 – 4 = 3
Relations & Functions-1
= ( x14 - x11 ) + ( x 6 - x3 ) + ( x 2 + 1) > 0
Case 2 : 0 £ x £ 1
x14 - x11 + x 6 - x3 + x 2 + 1
= x14 - {( x11 - x6 ) + ( x3 - x 2 )} + 1 > 0
{Q x11 - x6 £ 0, x 3 - x 2 £ 0 }
Case 3 : x < 0
x14 - x11 + x 6 - x3 + x 2 + 1 > 0
(Q x11 < 0, x3 < 0, x14 , x 6 , x 2 > 0)
Thus, for all real x,
x14 - x11 + x6 - x 3 + x 2 + 1 ³ 0
Hence, the domain of f (x) = R = (-¥, ¥)
MATHEMATICS
78
3.
(a) Given f (x) = cos (log x)
\ f (xy) = cos (log xy)
f (xy) = cos [log x + log y]
æ xö
æ
Replacing x + 1 by
....(1)
1
æ 1 ö
af ç
-1
÷ + bf ( x + 1) =
x +1
è x +1ø
Eq. (1) × a – Eq. (2) × b
ö
And f ç ÷ = cos log x
çè
è yø
y ÷ø
æ xö
f ç ÷ = cos (log x – log y)
è yø
Adding (1) and (2), we get
....(2)
Þ ( a 2 - b 2 ) f ( x + 1) = a ( x + 1) - a -
7.
Then the value of f (x)f (y)
ü
1 ì æ xö
– í f ç ÷ + f ( xy)ý
2 î è yø
þ
1
= f ( x) f ( y) - .2 { f ( x). f ( y)} = 0
2
(c) We must x 4 - 21x 2 ³ 0 and
8.
10 - x 4 - 21x 2 ³ 0
(
)
x 2 x 2 - 21 ³ 0
(
)(
)
Þ (2x – 3 ) (2x +1) £ 0 Þ -
æ x + 59 ö
3 f ( x) + 2 f ç
÷ = 10 x + 30
è x -1 ø
For x = 7, 3f (7) + 2f (11) = 70 + 30 = 100
For x = 11, 3f (11) + 2f (7) = 140
f (7) f (11)
-1
=
=
or f(7) = 4
-20 -220 9 - 4
(b)
6.
(a)
æ 1 ö
af ( x + 1) + bf ç
÷ = ( x + 1) - 1 ...(1)
è x +1 ø
1
3
£x£
2
2
1 3
Domain of g(x) = éê- , ùú
ë 2 2û
Hence, domain of (f + g) = intersection of their
(as x2 + 4 > 0 always)
\
é -5, - 21 ù È é 21,5 ù È {0}
ë
û ë
û
5.
b 2a + b
=
2
2
(d) f (x) = cos(sin x ) + log x {x}
Domain cos(sin x) ³ 0 {x} > 0, x > 0, x ¹ 1,
logx {x} ³ 0
(i) cos(sin x) ³ 0 for all x, x Î R [–1, 1]
(ii) {x} > 0, x Ï Int. (iii) x > 0, x Î(0, ¥)
(iv) x ¹ 1
(v) log x {x} ³ 0 Þ 1 > f ( x ) ³ 0 ,
so 1 > x ³ 0 logx f(x) is positive x Î[0, 1)
Þ x Î(0, 1)
(c) Given that,
g ( x ) = 3 + 4 x - 4 x 2 = -(2 x - 3)(2 x + 1)
x 2 - 25 x 2 + 4 £ 0
Þ x 2 - 25 £ 0
Þ -5 £ x £ 5
Domain is given by
b
+b
2
1
æ px ö
f ( x ) = - tan ç ÷ , -1 < x < 1
è 2ø
2
Here, domain of f(x) = (– 1, 1) and
...(1)
4
2
...(2)
and 100 ³ x - 21x
Eq. (1) gives x = 0 or x £ - 21 or x ³ 21
Eq. (2) Þ x 4 - 21x 2 - 100 £ 0
Þ
b
+b
x +1
=a+
æ xö
Þ f (xy) + f ç ÷ = 2 f (x). f (y)
è yø
Þ
...(2)
2
2
Putting x = 1, (a - b ) f (2) = 2a - a -
æ xö
f (xy) + f ç ÷
è yø
= cos (log x + log y) + cos (log x – logy)
= 2 cos (log x). cos (log y)
4.
1
, we get
x +1
9.
é 1 ö
domains = ê - ,1÷ .
ë 2 ø
(b) We have f(x)
= log4[log5{log3 (18x – x2 – 77)}]
Since, loga x is defined for all x > 0, f(x) is defined if
log5 {log3 (18x – x2 – 77)} > 0 and 18x – x2 – 77 > 0
or log3 (18x – x2 – 77) > 50 and x2 – 18x + 77 < 0
or log3 (18x – x2 – 77) > 1 and (x – 11)(x – 7) < 0
or 18x – x2 – 77 > 31 and 7 < x < 11
or 18x – x2 – 80 > 0 and 7 < x < 11
or x2 – 18x + 80 < 0 and 7 < x < 11
Solutions
79
or (x – 10) (x – 8) < 0 and 7 < x < 11
or 8 < x < 10 and 7 < x < 11 or 8 < x < 10
or x Î (8, 10)
Hence, the domain of f(x) is (8, 10).
1 x
-x
10. (a) Clearly, g(x) = (a + a ) and
2
1
h( x ) = (a x - a - x ) . Now g(x + y) + g(x – y)
2
1 x+ y
1
+ a - ( x + y ) ) + (a x - y + a - x + y )
= (a
2
2
1 x y
x -y
-x y
-x -y
= (a a + a a + a a + a a )
2
1 x y
-y
-x y
-y
= (a (a + a ) + a (a + a ))
2
æ1 x
öæ1
ö
= 2 ç ( a + a - x )÷ ç ( a y + a - y ) ÷ = 2 g ( x ) g ( y )
è2
øè2
ø
11. (b) Given f (l + x) = f (l – x)
...(1)
f (2l + x) = – f (2l – x)
...(2)
for l > 0
Replacing x by l – x in l – x in (1), we get
f(2l – x) = f(x)
...(3)
\ From (2) and (3), f(x) = –f(2l + x)
Þ f(x) = –[–f (2l + 2l + x)]
Þ f(x) = f(x + 4l)
...(4)
Þ f(x) is periodic with period 4l.
Further from (3), replacing x by –x, we get
f (2l + x) = f (–x)
...(5)
From (2), (3) and (5), we have
f (–x) = f (2l + x) = –f (2l – x) = –f (x)
i.e. f (–x) = –f (x)
Þ f(x) is odd function
Thus, f is odd and periodic function.
12. (d)
f ( x) = ax3 - bx - (tan x )sgn x
Since f (- x) = f ( x)
Þ
-ax3 + bx - tan x sgn x
= ax 3 - bx - (tan x ) (sgn x )
(
)
Þ
2 ax 2 - b x = 0 " x Î R Þa = 0 and b = 0
\
[ a ]2 - 5 [ a ] + 4 = 0
and 6 {a}2 - 5{a} + 1 = 0
Þ ([a] – 1)([a] – 4) = 0 and
(3{x} – 1) (2{x} – 1) = 0
1 1
Þ [a] = 1, 4 and {a} = ,
3 2
\
1
1
1
1
a = 1+ , 1+ , 4 + , 4 +
3
2
3
2
Sum of all possible values of a =
35
3
æ 5x ö
13. (c) f (x) = cos nx . sin ç ÷ ;
è nø
Period of cos nx =
Period of sin
2p
|n|
5x 2p 2 | n | p
=
=
5
n
5
n
æ 2p 2 | n | p ö
\ Period of f (x) = L.C.M. ç
,
= 2p
è|n|
5 ÷ø
(given)
æ 1 | n |ö
L.C.M. (1, | n |)
,
=1=
=1
Þ L.C.M. ç
H.C.F. (| n |, 5)
è | n | 5 ÷ø
|n|
= 1 Þ H.C.F. (|n|, 5) = |n|
H.C.F. (| n |, 5)
If g.c.d. (|n|, 5) = 1 Þ |n| = 1 Þ n = 1
If g.c.d. (|n|, 5) ¹ 1 Þ |n| = 5m; m Î ¥
Þ g.c.d. (5m, 5) = 1
Þ |n| = 5 Þ n = ±5 \ n Î {±1, ±5}
14. (d) f : ¡ ® ¡ , g; ¡ ® ¡
We know that min. {f1(x), f2(x)}
( f1 ( x) + f 2 ( x )) - | f1 ( x) - f 2 ( x) |
=
2
\ min {f(x) – g(x), 0}
( f ( x ) - g ( x ) + 0) - | f ( x) - g ( x) - 0 |
=
2
( f ( x ) - g ( x)) - | f ( x ) - g ( x) |
=
2
15. (b) y = f(ex) + f(|ln| |x|) Domain f(x) = (0, 1)
Þ 0 < ex < 1 Þ x < 0
...(1)
and 0 < ln |x| < 1 Þ 1 < |x| < e
Þ x Î {–e, –1} È (1, e)
... (2)
Taking intersection x Î (–e, –1)
Þ
16. (b)
1ö
æ
f ç x + ÷ = f ( x ); f (2) =5;
2ø
è
æ9ö
f ç ÷ = 2, f (-3) è4ø
æ1ö
f ç ÷ =?
è4ø
MATHEMATICS
80
Q f(x) is periodic with period
1
2
Þ
nö
æ
Þ f(x) = f ç x + ÷ " x Î ¥
è
2ø
10 ö
æ
Þ f(–3) = f çè - 3 + ÷ø = f (2) = 5
2
Þ
20.
(d)
< 2- 1
4 x
1
1
-1 <
<Þ x Î f (null set)
2
4 x
y
1
æ1
æ 1ö
æ 1ö ö
æ 9ö
and f ç ÷ = f ç + 4 ç ÷ ÷ = f ç ÷ = 2
è 4ø
è 2ø ø
è 4ø
è4
0
æ 1ö
\ f (-3) - f çè ÷ø = 5 - 2 = 3
4
17.
18.
é 2x + 3 x £ 1
(c) f(x) = ê 2
êë a x + 1 x > 1
For x £ 1 ;
f(x) £ 5
So for range of f(x) to be R.
Þ a2 + 1 £ 5 and a ¹ 0 Þ a Î [–2, 2]
Hence, a = {–2, – 1, 1, 2}
(b) g ( 2 f ( x) + 3)
= log 5 ( 2 (sin x - cos x) + 3)
We know that
- 2 £ (sin x - cos x) £ 2 " x Î ¡
éQ - a 2 + b2 £ a sin x + b cos x £ a 2 + b 2 ù
ëê
ûú
Þ
- 2 £ 2(sin x - cos x) £ 2
Þ
1 £ 2(sin x - cos x) + 3 £ 5
Þ 0 £ log
( 2 (sin x - cos x ) + 3) £ 2
(Q logax is increasing for a > 1)
5
1 ö
æ
(d) –g(2, f(x)) = - log 2 ç 1 +
÷
è
4 xø
Þ
1 ö
æ
–g(2, f(x)) – 1 = - log 2 ç1 +
÷ -1
è
4 xø
æ1
ö
\ g ç , - g (2, f ( x )) - 1÷
è2
ø
æ
1 ö ö
æ
= log1/2 ç - log 2 ç 1 +
÷ - 1÷
4 xø ø
è
è
Þ
1 ö
æ
log 2 ç1 +
÷ +1< 0
è
4 xø
1
2
x
3
0 £ f (x) £ 1 Þ 0 £ 7 f(x) £ 7
Þ – 1 £ sin (7 f(x)) £ 1
21.
(5.5) Here f ( - x ) =
1
1 + ex
1
1
+
So, f ( x ) + f ( - x ) =
-x
1+ e
1 + ex
=
ex
+
1
ex + 1 1+ ex
=
ex +1
ex +1
=1
\ S = {f ( 5 ) + f ( -5 )} + ... + {f (1) + f ( -1)} + f ( 0 )
1 11
=
2 2
1+ e
(2) f(x + p) = 1 + {1 – 3f (x) + 3f 2(x) – f 3(x)}1/3
Þ f(x + p) = 1 + (1 – f(x)) = 2 – f(x)
Þ f(x + p) = 2 – [2 – f(x – p)]
Þ f(x + p) = f(x – p)
Þ f(x) = f(x + 2p)
Þ Period of f(x) = 2p = lp (given)
Þ l=2
(9) Given f(x + 2) = f(x) + f(2)
Put x = –1. Then f(1) = f(–1) + f(2)
or f(1) = –f(1) + f(2) [as f(x) is an odd function]
or f(2) = 2f(1) = 6
Now, put x = 1
We have f(3) = f(1) + f(2) = 3 + 6 = 9
(6) Given f(x, y) = f(2x + 2y, 2y – 2x) " x, y Î ¡ ,
f(x) = f(2x, 0) and f(x) is periodic with period k.
Þ f(x) = f(2x, 0) = f(2.2x + 2(0), 2(0) – 2.2x)
= f(2x + 1, 2x +1)
= f(2.2x + 1 – 2.2x + 1, – 2.2x + 1 – 2.2x + 1)
= f (0, – 2x + 3)
= f(2.(–2x + 3), –2.2x + 3) = f(– 2x + 4, – 2x + 4)
= f(–2x + 6, 0)
= f(–2x +7, 2x + 7) = f(0, 2x + 9)
= f(2x + 10, 20x + 10) = f(2x + 12, 0) = f(x + 12)
Þ f(x) is periodic with period 12 Þ k = 12.
= 1 + 1 + 1 + 1 + 1 + f(0) = 5 +
22.
23.
Hence, range of g ( 2 f ( x) + 3) is [0, 2].
19.
1
0 <1+
24.
1
0
= 5+
Solutions
81
é x ù é 15 ù
25. (6) f(x) = ê ú ê - ú
ë15 û ë x û
0 £ x < 15
15 £ x < 30
30 £ x < 45
45 £ x < 60
f(x) = – 3
60 £ x < 75
f(x) = – 4
75 £ x < 90
f(x) = – 5
Total integers in range f(x) = {0, –1, –2, –3, –4, –5}
x Î (0, 90)
f(x) = 0
f(x) = – 1
f(x) = – 2
CHAPTER
Trigonometric Functions
3
1.
(d)
2
+
1
2
cos a 1 + sin a
2
=
2.
1
2
+
2
4
+
4
a
Þ sin 2 =
2
8
1 + sin a 1 + sin a
4.
Þ b = p - a , g = 2p + a, d = 3p - a .
So that the given expression is equal to
2p + a
3p - a
a
æ p-a ö
+ sin
+ 3 sinç
÷ + 2 sin
2
2
2
è 2 ø
= 4sin
a
a
a
a
+ 3cos - 2sin - cos
2
2
2
2
aö
a a
æ a
= 2çsin + cos ÷ = 2 1+ 2sin cos = 2 1+ k
è 2
2ø
2
2
3.
(c)
1 + cos a = 1 +
=
… (i)
b
b
b
b
- cos 2
1 + 2 sin 2 - [1 - sin 2 ]
2
2=
2
2
2b
2b
1 + 2 sin
1 + 2 sin
2
2
1 + 2sin 2
5.
2 cos b - 1 2 - cos b + 2cos b - 1
=
2 - cos b
2 - cos b
b
cos 2
1 + cos b
2 a
2
=
Þ cos
=
b
2
2 - cos b
1 + 2 sin 2
2
b
cos 2
2a
2
=1Þ 1 - cos
b
2
1 + 2sin 2
2
1 + 2 sin 2
6.
…(ii)
a
2 = 3
b
tan
2
(b) Gives equations can be written as
2 cos a + 9 cos d = – 6 cos b – 7 cos g ..(i)
2 sin a – 9 sin d = 6 sin b – 7 sin g ...(ii)
Square and add equations (i) and (ii),
Þ 4 + 36 + 36 [cos a cos d – sin d sin a]
= 36 + 49 + 84 [cos b cos g – sin b sin g]
Þ 36 [cos(a + d)] = 84 [cos (b + g)]
a
b
tan 2 = 3 tan 2 Þ
2
2
sin a = sin b = sin g = sin d = k
4 sin
b
2
b
2
Divide equation (ii) by (i)
4
+
+
1 - sin 4 a 1 + sin 4 a 1 + sin8 a
4
4
8
=
+
=
1 - sin 8 a 1 + sin 8 a 1 - sin16 a
40
8
1ö
= 10 æ
=
=
16
çQ sin a = ÷
1
4
5ø
1è
5
(b) a < b < g < d and
3 sin 2
tan
cos (a + d) 84 7 m
=
= = ; m + n = 10
cos (b + g ) 36 3 n
(a) sin q + sin 3q + sin 2q = sin a
Þ 2 sin 2q cos q + sin 2q = sin a
Þ sin 2q(2 cos q + 1) = sin a
...(i)
Now cos q + cos 3q + cos 2q = cos a
2 cos 2q cos q + cos 2q = cos a
cos 2q (2 cos q + 1) = cos a
...(ii)
From (i) and (ii),
tan 2q = tan a Þ 2q = a Þ q = a/2
(c) (sin 7a + sin 5a ) + 5(sin 5a + sin 3a)
+12(sin 3a + sin a )
sin 6a + 5 sin 4a + 12 sin 2a
2sin 6a cos a + 5(2sin 4a cos a)
+ 12(2sin 2a cos a )
=
sin 6a + 5sin 4a + 12sin 2a
= 2 cos a =
5 +1
2
MATHEMATICS
82
7.
(b)
2
é 2sin q cos q 2sin 3q cos3q 2sin9q cos9q ù
+
+
1 êê cos q cos3q cos9q cos3q cos9q cos 27q úú
2ê
2sin 27q cos 27q ú
+
ê
cos 27q cos81q úû
ë
é sin(3q - q) sin(9q - 3q) sin(27q - 9q) ù
+
+
ê
ú
1 ê cos q cos3q cos3q cos9q cos9q cos 27 q ú
= 2ê
sin(81q - 27q) ú
+
ê
cos
27q cos81q úû
ë
=
8.
1
1 é sin 80 q ù
[tan 81q - tan q] = ê
2
2 ë cos q cos 81q úû
11.
qö
q
æ
= 4 ç 2 cos q× cos ÷ = 16 cos 2 q× cos 2
è
ø
2
2
= 4 (1 + cos q) (1 + cos 2q)
(a) Let
ìï 22 p üï
ìï 264 p üï
ì 2p ü
P = cos í
cos
...cos
ý
í 64 ý
í 64 ý
î 264 - 1 þ
îï 2 - 1 þï
îï 2 - 1 þï
Þ P=
æ 22 p ö æ 22 p ö
sin ç
÷ cos ç
÷
æ 2p ö çè 264 -1 ÷ø çè 264 -1 ÷ø
2sin ç
÷
è 264 -1 ø
1
p+q
1
p+q
Þ
=
p-q
sin q p - q
Apply componendo and dividendo
æ 264 p ö
...cos ç
÷
ç 264 - 1 ÷
è
ø
(b) cosec q =
1 + sin q p + q + p - q
=
1 - sin q p + q - p + q
Þ P=
2
2
q
qü
ì
qü
ì
cos
sin
+
1
tan
+
ï
ï
2
2ï = p
2ï = p
Þ í
Þí
ý
ý
q
q
q
q
ï cos - sin ï
ï1 - tan q ï
î
2
2þ
î
2þ
qö q
2æp
æp qö p
Þ tan 2 ç + ÷ =
Þ cot ç + ÷ =
4
2ø p
è
è 4 2ø q
9.
(b) Given, sin 2q + sin 2f = 1/2
...(i)
and cos 2q + cos 2f = 3/2
...(ii)
Square and adding,
\ (sin2 2q + cos2 2q) + (sin2 2f + cos2 2f)
+ 2[sin 2q sin 2f + cos 2q cos 2f] = 1/4 + 9/4
Þ cos 2q cos 2f + sin 2q sin 2f = 1/4
Þ cos (2q – 2f) = 1/4 Þ cos2 (q – f) = 5/8
10. (a) u = 1 + cos q + cos 2q + cos (q + 2q)
= (1 + cos 3q) + (cos q + cos 2q)
3q
3q
3q
+ 2 cos
+ cos
2
2
2
3q é
3q
qù
cos
+ cos ú
= 2 cos
2 êë
2
2û
= 2 cos 2
Similarly v = 2 sin
\
3q é
3q
qù
cos
+ cos ú
ê
2 ë
2
2û
3q
qö
æ
u2 + v2 = ç cos + cos ÷
2
2ø
è
2
Þ P=
æ 23 p ö
æ 264 p ö
sin ç
...cos ç
÷
÷
ç 264 -1÷
æ 2p ö çè 264 -1÷ø
è
ø
2sin ç
÷
è 264 -1ø
1
æ 265 p ö
1
1
sin ç
÷ = 64
64
ç
÷
æ
p
ö
2
è 2 -1 ø 2
264 sin ç
÷
è 264 - 1 ø
12. (c) sin x sin y + 3 cos y + 4 sin y cos x =
\
\
26
3 cos y + (sin x + 4 cos x) sin y =
3 cos y + (sin x + 4 cos x) sin y
26
£ 9 + (sin x + 4 cos x) 2 £ 9 + 1 + 16 =
26
\
sin x sin y + 3 cos y + 4 sin y cos x = 26
cos y sin y cos x
=
Þ sin x sin y =
3
4
Þ 3 tan y = cosec x and tan x = 1/4
Þ 9 tan2 y = cosec2 x = (1 + cot2 x) = 17
1
9
+
Þ tan2 x + cot2 y =
16 17
x
x
x
x
x
× cos ×
× cos × ×.....cos × × cos ×
13. (b) cos ×
256
128
64
4
2
sin x
=
æ x ö
256 sin ç
è 256 ÷ø
14. (a) – 5 £ 3 sin x – 4 cos x £ 5
10 £ 3 sin x – 4 cos x + 15 £ 20
log20 10 £ log20 (3 sin x – 4 cos x + 15) £ log20 20
Solutions
83
15. (c) Given expression
sin 50° + sin 140° + sin 170°
=
2 sin 25° sin 70° sin 85°
4
(Using sin 2A + sin 2B + sin 2C = 4 sin A
2
sin B sin C, where A + B + C = 180°)
=2
=
16. (c)
Þ
32(1 + tan
2
x)
+ 1 - 10 × 3tan
2
2
x
2
32 (3tan x )2 + 1 - 10.3tan
Put 3tan
2
x
=t;
x
=0
=0.
9t2 – 10t + 1 = 0
1
and t = 1
Þ t=
9
Þ 3tan 2 x = 3–2 and 3tan 2 x = 1
Þ tan2x = –2 gives no solution and tan2 x = 1
gives four solutions.
17. (c) (4cos2 2x + 4 cos2x + 1)
2
Þ
Þ
Þ
Þ
Þ
\
18. (b)
+ (tan x – 2 3 tan x + 3) = 0
(2 cos 2x + 1)2 + (tan x – 3 )2 = 0
1
cos 2x = - and tan x = 3
2
1
2cos2 x – 1 = - and tan x = 3
2
1
cos x = ± - and tanx = 3
2
p 4p
x= ,
3 3
Number of solutions of equation = 2.
2 sin x - 3
2cos2 x - 3cos x + 1
=1
2
Case I: 2 cos x – 3 cos x + 1 = 0
1
p
cos x = ,1 ; x = 0,
2
3
p
But at x =
L.H.S. = 0°
3
p
\ x=
(rejected)
\ x = 0 is a
3
solution
Case II: 2 sin x – 3 = 1, – 1
\
\
\
3 +1 3 -1
3 -1
,
sin x =
2
2
2
x has 2 values in [0, p].
Total number of solutions = 2 + 1 = 3.
sin x =
19. (a) sin2 (sin x) – 3 sin (sin x) + 2 = 0
{sin (sin x) – 2} {sin(sin x) – 1} = 0
Equation has no solution.
20. (c) x2 + 12 + 3 sin (a + bx) + 6x = 0
Þ (x + 3)2 + 3 + 3 sin (a + bx) = 0
Þ (x + 3)2 + 3 = – 3 sin (a + bx)
L.H.S. ³ 3 but R.H.S. £ 3
L.H.S. = R.H.S. = 3
\ x = – 3 and sin (a + bx) = – 1
Þ sin (a – 3b) = – 1
p
or a – 3b = (4n – 1)
n Î Z.
2
1
1
21. (50) 5 2 + 5 2
Þ
Þ
Þ
1
52
+ log5 (sin x)
=
1
+ log5 cos x
15 2
1
52
+ × 5log5 (sin x) = 151/ 2 ×15 log 15 cos x
5 + 51/2 × sin x = 151/2 × cos x
1 + sin x = 3 cos x
1/2
3
sin x 1
cos x =
2
2
2
pö
p
æ
Þ cos ç x + ÷ = cos
6
3
è
ø
p
p
Þ x + = 2np ± , n Î Z
6
3
p
p
Þ x = 2np - , 2np + , n Î Z .
2
6
But we must have sin x, cos x > 0
\ x = 2np + p/6, n Î Z.
sin x
sin 3x
sin 9x
+
+
=0
22. (6)
cos 3x cos 9x cos 27x
2sin x cos x 2sin 3x cos3x
+
or
2cos3x cos x 2cos9x cos3x
2sin 9x cos9x
+
=0
2cos 27x cos9x
sin(3x - x)
sin(9x - 3x)
+
or
2cos3x cos x 2cos9x cos3x
sin(27x - 9x)
+
=0
2cos 27x cos9x
or (tan3x – tan x) + (tan9x – tan3x)
+ (tan27x – tan9x) = 0
or tan 27x – tan x = 0 or tan x = tan 27 x
np
, nÎI
Þ 27x = np + x, n Î I or x =
26
p 2p 3p 4p 5p 6p
or
, , ,
, ,
26 26 26 26 26 26
Hence, there are six solutions.
Þ
MATHEMATICS
84
23. (4) Since – 2 £ sin x –
3 cos x £ 2, we have
2m - 3
4m - 6
£1
-2 £
£ 2 or -1 £
4-m
4-m
If
2m - 3
£ 1, we have
4-m
(2m - 3) - (4 - m)
3m - 7
£ 0 or
³0.
4-m
m-4
...(i)
m +1
2m - 3
£0
Þ
Also, - 1 £
...(ii)
m-4
4-m
x
p æ x pö
x = 2mp, mÎ Zor - ç + ÷ = np + , n ÎI
2 è 2 6ø
2
p
.
\ x = 0 or x =
3
p
p
25. (2) Given < 3x - £ p
2
2
Þ
p æ
pö
p ö -p
æ
< ç 3x - ÷ £ p or - p £ ç 3x - ÷ <
2 è
2ø
2ø 2
è
3p -p
Þ p< 3x = or £ x < 0
2
6
é -p ö æ p p ù
\ xÎê ,0 ÷ È ç , ú
ë 6 ø è 3 2û
Now, 1 + cos x + cos 2x + sin x + sin 2x + sin
3x = 0
Þ 2 cos2x + cos x + sin 2x + 2 sin 2x cos x = 0
Þ cos x (2 cos x + 1) + sin 2x (2 cos x + 1) = 0
Þ (cos x + sin 2x) (2 cos x + 1) = 0
Þ cos x (1 + 2 sin x) (2 cos x + 1) = 0
-1
Þ cos x = 0 or sin x =
2
(as for given interval, cos x > 0)
p
-p
.
Þ x = or x =
2
6
Hence, there are 2 solutions.
Þ
7ù
é
From Eqs. (i) and (ii), we get m Î ê -1, ú .
3û
ë
Therefore, the possible integers are – 1, 0, 1, 2.
24. (2) We have,
2
æ
pö
æ
æ p öö
2
ç cos ç x + ÷ - cos ç ÷ ÷ = sin x
6
6
è
ø
è
ø
è
ø
Þ
æxö
æ x pö
æxö
æxö
4sin2 ç ÷ × sin2 ç + ÷ = 4sin2 ç ÷ × cos2 ç ÷
è 2ø
è 2 6ø
è 2ø
è2ø
\
sin
x
x
æ x pö
= 0 or sin 2 ç + ÷ = cos 2
2
2
è2 6ø
CHAPTER
Principle of Mathematical Induction
4
1.
2.
3.
(d) We note that P(1) = 2 and hence,
P(n) = n (n + 1) + 2 is not true for n = 1.
So the principle of mathematical induction is not
applicable and nothing can be said about the
validity of the statement P(n) = n (n + 1) + 2.
(c) The product of r consecutive integers is
divisible by r ! . Thus n (n + 1 ) (n + 2) (n + 3) is
divisible by 4 ! = 24.
2 + 2 + 2 + ...... 2 = 2 + 2 + 2 + ...... 2
144
42444
3
( k +1) terms
æ p ö
÷÷
= 2 + 2 cosçç
è 2 k +1 ø
(a) Let f (n ) = 2 + 2 + ..... + 2 (number of
roots is n)
Then f(1) = 2 = 2 cos p or 2 sin p
4
4
\ f (1) may be true for (a) as well as for (b)
Again f(2) =
2 + 2 = 2´
\ f (2) is true for (a).
We check it for any integer.
æ p ö
÷÷ for some
è 2 k +1 ø
Let 2 + 2 + ...... 2 = 2 cosçç
k - roots
2+ 2
p
= 2 cos
2
8
...(i)
k ³1
Now ,
k terms
[From (i)]
p
é
p ù
2 p
= 2ê1+ cos k+1 ú = 2.2 cos k+2 = 2 cos k+2
2 û
2
2
ë
\ The result is true for n = k + 1. Hence, by
the principle of mathematical induction
4.
æ p ö
÷÷ for all n Î N .
2 + 2 + ....+ 2 = 2 cosçç
è 2n +1 ø
n -roots
(a) It can be proved with the help of
n
mathematical induction that
< a(n) £ n.
2
Solutions
85
200
< a(200)
2
Þ a(200) > 100 and a(100) £ 100.
1
1
1
1
+
+ .... +
(a) Let Un = +
n n +1 n + 2
2n - 1
1 1 1
1
Vn = 1 - + - + ..... +
2 3 4
2n - 1
Let T(n) be the statement Un = Vn
Then T(1) is true. For U1 = 1 and V1 = 1, so that
U1 = V1
Let T(k) be true for some positive integer K. Now,
\
5.
1
1 ù
é 1
U k +1 - U k = ê
+
+ .... +
k
1
k
2
+
+
2
k
+ 1 úû
ë
7.
1
1
1 1
=
Þ = is true
1.2 1 + 1
2 2
thus P(n) is true for n = 1
Suppose that P(k) is true for some natural number
‘k’
1
1
1
1
k
+
+
+ ..... +
=
......(1)
1.2 2.3 3.4
k ( k + 1) k + 1
Now,
1
1
1
1
1
+
+
+ ..... +
+
1.2 2.3 3.4
k (k + 1) (k + 1)(k + 2)
=
k (k + 2) + 1
(k + 1)(k + 2)
[From (1)]
(k + 1)2
k +1
k +1
=
=
(k + 1)(k + 2) k + 2 (k + 1) + 1
Thus P(k + 1) is true whenever P(k) is true. Hence.
by the principle of mathematical induction P(n)
is true for all natural numbers.
(a) Let us write the statement
æ 2n +1ö
3öæ 5ö æ 7ö
P(n) : çæ1+ ÷ç
1+ 1+ .......ç1+
= (n +1)2
2 ÷ø
è 1øè 4÷ø çè 9÷ø
è
n
We note that
æ 3ö
P : ç1 + ÷ = 4 = (1 + 1) 2 , which is true
è 1ø
Thus P (n) is true for n = 1
Suppose that P(k) is true for some natural number
‘k’ i.e.,
æ 3ö æ 5 ö æ 7 ö æ 2k + 1ö
2
çè1 + ÷ø çè1 + ÷ø çè1 + ÷ø ..... ç1 + 2 ÷ = ( k + 1)
è
1
4
9
k ø
1 ù
é 1 1
- ê1 - + - ..... +
2
3
2
k
- 1 úû
ë
6.
k
1
+
k + 1 (k + 1)(k + 2)
=
1
1 ù
é1
+ .... +
– ê +
2k - 1 úû
ë k k +1
1
1
1
1
1
+
- =+
=
...(i)
2k 2k + 1 k
2k 2k + 1
Also,
1 ù
é 1 1
Vk +1 - Vk = ê1 - + - ...... +
2
3
2
k
+ 1úû
ë
1
1
+
=...(ii)
2k 2k + 1
From (i) and (ii), we find that
U k +1 - U k = Vk +1 - Vk
Since, Uk = Vk, therefore, it follows that
Uk+1 = Vk+1
\ T (n ) is true for all n Î N .
(a) Let us write the statement
1
1
1
1
n
P(n) : 1.2 + 2.3 + 3.4 + ..... + n(n + 1) = n + 1
=
....(1)
Now,
æ 3ö æ 5 ö æ 7 ö æ 2k + 1ö
çè1 + ÷ø çè1 + ÷ø çè1 + ÷ø ..... çè1 + 2 ÷ø
1
4
9
k
ì
2k + 3 ïü
2 ï
= ( k +1) í1 +
2ý
îï (k + 1) þï
ìï
2k + 3 üï
í1 +
2ý
îï ( k + 1) þï
[Using (1)]
ï (k + 1)2 + 2k + 3üï 2
2ì
= ( k +1) í
ý = k + 2k + 1 + 2k + 3
(k +1)2
îï
þï
we note that P(1) :
= k 2 + 4 k + 4 = ( k + 2) 2 = {( k + 1) + 1}
2
8.
9.
Thus , P(k + 1) is true whenever P(k) is true.
Hence by the principle of mathematical induction
P(n) is true for all natural numbers.
(d) Q 21 > 12. 22 = 22, 23 < 32, 24 = 42
But 25 > 52, 26 > 62 and so on.
4n
(2n)!
<
(d) Let P(n) :
n + 1 (n !)2
For n = 2,
MATHEMATICS
86
16 24
42
4!
<
<
Þ
2
3
4
2 + 1 (2)
which is true.
Let for n = m ³ 2, P(m) is true.
P(2) :
4m
(2m)!
<
m + 1 ( m !) 2
i.e.
Now,
=
(2m)! 4(m + 1)
4m+1
4m 4(m + 1)
·
·
=
<
(m !) 2 (m + 2)
m+2
m +1 m + 2
(2m)!(2m + 1)(2m + 2)4(m + 1)(m + 1) 2
(2m + 1)(2m + 2)(m !) 2 ( m + 1) 2 ( m + 2)
[2(m + 1)]!
[2(m + 1)]!
2(m + 1) 2
·
<
2
[(m + 1)!] (2m + 1)(m + 2) [(m + 1)!]2
Hence, for n ³ 2, P(n) is true.
10. (b) Given that, P(n) : 3n < n!
Now, P(7) : 37 < 7! is true
Let P(k) : 3k < k!
Þ P(k + 1) : 3k+1 = 3.3k < 3.k! < (k + 1)!
(Q k + 1 > 3)
11. (c) Let P(n) = 2 . 42n + 1 + 33n + 1
Then P(1) = 2 . 43 + 34 = 209, which is divisible
by 11 but not divisible by 2, 7 or 27.
Further, let P(k) = 2 . 42k + 1 + 33k + 1 is divisible by
11, that is,
=
2.4
2 k +1
3k +1
+3
= 11q for some integer q.
Now P (k + 1) = 2 . 42 k +3 + 33k + 4
2k +1
=2 . 4
2
3k +1 3
.4 + 3
2k +1
.3 =16 . 2.4
13. (a) n p - n is divisible by p for any natural
number greater than 1.
Trick : Let n = 4 and p = 2
Then (4)2 – 4 = 16 – 4 = 12, it is divisible by 2. So,
it is true for any natural number greater than 1
14. (b) For n = 1, we have
49n + 16n + l = 49 + 16 + l = 65 + l
= 64 + ( l + 1), which is divisible by 64 if l = – 1
For n = 2, we have
49n + 16n + l = 492 + 16 × 2 + l = 2433 + l
= 64 × 38 + ( l + 1), which is divisible by 64
if l = – 1
Hence, l = – 1
15. (b) Let P(n) :
1
1
1
+
+ ..... +
1× 2 × 3 2 × 3 × 4
n ( n + 1) (n + 2 )
=
n (n + 3)
4 (n + 1) (n + 2)
For n = 1,
+ 27.3
= 16 . 2 . 42 k +1 + (16 + 11) . 33k +1
= 16 [2. 42 k +1 + 33k +1 ] + 11 . 33k +1
1(1 + 3)
1
=
4 (1 + 1) (1 + 2) 6
\ P(1) is true. Let P(k) is true, then P(k) :
1
1
1
+
+ ..... +
1× 2 × 3 2 × 3 × 4
k ( k + 1) (k + 2 )
k (k + 3)
4 (k + 1) (k + 2)
For n = k + 1,
1
1
+
+ .....
P(k + 1) :
1× 2 × 3 2 × 3 × 4
= 16 . 11q + 11 . 33k +1
= 11(16q + 33k +1 ) = 11m
where m = 16q + 33k + 1 is another integer.
\ P(k +1) is divisible by 11.
\ P(n ) = 2 . 42 n +1 + 33n +1 is divisible by 11 for
all n Î N .
+
=
(k + 1) ( k + 4)
4 ( k + 2 ) ( k + 3)
+
=
... (i)
1
1
+
k (k + 1) (k + 2 ) (k + 1) (k + 2 ) (k + 3)
L.H.S. =
2
12. (b) n(n - 1) = (n - 1)(n)( n + 1)
It is product of three consecutive natural
numbers, so according to Langrange’s theorem
it is divisible by 3! i.e., 6
1
1
= and
1× 2 ×3 6
R.H.S.=
=
3k +1
L.H.S.=
1
1
+
+ .....
1× 2 × 3 2 × 3 × 4
1
1
+
k (k + 1) (k + 2 ) (k + 1) (k + 2 ) (k + 3)
k ( k + 3)
4 (k + 1) (k + 2 )
+
1
(k + 1) (k + 2) (k + 3)
[from (i)]
Solutions
=
87
(k + 1)2 (k + 4)
(k +1)(k + 4)
=
4 (k + 1)(k + 2)(k + 3) 4 (k + 2)(k + 3)
= R.H.S.
Hence, P(k + 1) is true.
Hence, by principle of mathematical induction
for all n Î N, P(n) is true.
16. (b) 3.52n + 1 + 23n + 1
Put n = 1, we get
(3 × 53) + 24 = 391, which is divisible by 17.
Put n = 2, we get
(3 × 55) + 27 = 9503, which is divisible by 17
only.
17. (b) Let the statement P(n) be defined as
P(n) = 1.3 + 2.32 + 3.33 + ..... + n.3n
=
(2n – 1) 3n + 1 + 3
4
Step I : For n = 1,
P(1) : 1.3 =
(2.1 – 1) 31 + 1 + 3
=
32 + 3
4
4
9 + 3 12
=
= 3 = 1.3, which is true.
=
4
4
Step II : Let it is true for n = k,
i.e. 1.3 + 2.32 + 3.33 + ..... + k.3k
=
(2k – 1) 3k + 1 + 3
... (i)
4
Step III : For n = k + 1,
(1.3 + 2.32 + 3.33 + ..... + k.3k) + (k + 1)3k+1
=
(2k – 1) 3k + 1 + 3
4
[Using equation (i)]
=
(2k – 1) 3k + 1 + 3 + 4 (k + 1) 3k + 1
4
k +1
=
+ (k + 1)3
k+1
3
(2k – 1 + 4k + 4) + 3
4
[taking 3k + 1 common in first and last term of
numerator part]
=
3k + 1 (6k + 3) + 3
=
3k + 1 × 3 (2k + 1) + 3
4
4
[taking 3 common in first term of numerator part]
k + 1) + 1
3(
[ 2k + 2 – 1] + 3
=
4
k + 1) + 1
éë 2 ( k + 1) – 1ûù 3(
+3
=
4
Therefore, P(k + 1) is true when P(k) is true.
Hence, from the principle of mathematical
induction, the statement is true for all natural
numbers n.
18. (d) Let the statement P(n) be defined as
1
1
+
+ .....
P(n) : 1 +
1+ 2 1+ 2 + 3
+
i.e. P(n) : 1+
1
2n
=
1 + 2 + 3 + ..... + n n + 1
1
1
2
2n
+
+ ..... +
=
1+ 2 1+ 2 + 3
n (n +1) n +1
é
n (n + 1) ù
êQ sum of natural numbers =
ú
2 û
ë
Step I : For n = 1,
2 ´1 2
= = 1, which is true.
P(1) : 1 =
1+1 2
Step II : Let it is true for n = k,
1
1
2
2k
+
+ ..... +
=
i.e. 1 +
1+ 2 1+ 2 + 3
k (k + 1) k + 1
...(i)
Step III : For n = k + 1,
æ
1
1
2 ö
2
ç1+ 1+ 2 + 1+ 2 + 3 + ..... + k (k +1) ÷ + (k +1)(k + 2)
è
ø
=
2k
2
+
[using equation (i)]
k + 1 (k + 1) (k + 2)
é 2
ù
2k (k + 2 ) + 2 2 ë k + 2k + 1û
=
=
(k + 1) (k + 2) (k + 1) (k + 2)
[taking 2 common in numerator part]
2 (k + 1)
2 (k + 1)
2 (k + 1)
=
=
=
(k + 1) (k + 2)
k+2
( k + 1) + 1
Therefore, P(k + 1) is true, when P(k) is true.
Hence, from the principle of mathematical
induction, the statement is true for all natural
numbers n.
19. (c) 24 º 1 (mod 5) Þ (24)75 º (1)75 (mod 5)
i.e. 2300 º 1 (mod 5) Þ 2300 × 2 º (1.2) (mod 5)
Þ 2301 º 2 (mod 5)
\ Least positive remainder is 2.
20. (d) Let P(n) be the statement given by
P(n) : 32n when divided by 8, the remainder is 1.
or P(n) : 32n = 8l + 1 for some l Î N
For n = 1,
P(1) : 32 = (8 × 1) + 1 = 8l + 1, where l = 1
2
MATHEMATICS
88
\ P(1) is true.
Let P(k) be true.
Then, 32k = 8l + 1 for some l Î N
... (i)
We shall now show that P(k + 1) is true, for
which we have to show that 32(k + 1) when
divided by 8, the remainder is 1.
Now, 32(k + 1) = 32k . 32 = (8l + 1) × 9
[Using (i)]
= 72l + 9 = 72l + 8 + 1 = 8(9l + 1) + 1
= 8m + 1, where m = 9l + 1 Î N
Þ P(k + 1) is true.
Thus, P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical
induction P(n) is true for all n Î N.
21. (133) Putting n = 1 in 11n + 2 + 122n + 1
We get, 111+2 + 122×1+1 = 113 + 123 = 3059,
which is divisible by 133.
22. (27) Let P(n) be the statement given by
P(n) : 41n – 14n is a multiple of 27
For n = 1,
i.e. P(1) = 411 – 141 = 27 = 1 × 27,
which is a multiple of 27.
\ P(1) is true.
Let P(k) be true, i.e. 41k – 14k = 27l
... (i)
For n = k + 1,
CHAPTER
2.
25. (8)
599 = (5)(52 )49 = 5(25) 49 = 5(26 - 1) 49
= 5 ´ (26) ´ (Positive terms) –5 , So when it is
divided
by 13 it gives the remainder –5 or (13 – 5) i.e., 8.
Complex Numbers and Quadratic Equations
5
1.
41k + 1 – 14k + 1 = 41k 41 – 14k 14
= (27l + 14k) 41 – 14k 14 [using (i)]
= (27l × 41) + (14k × 41) – (14k × 14)
= (27l × 41) + 14k (41 – 14) = (27l × 41)
+ (14k × 27)
k
= 27(41l + 14 ),
which is a multiple of 27.
Therefore, P(k + 1) is true when P(k) is true.
Hence,from the principle of mathematical
induction, the statement is true for all natural
numbers n.
23. (8) Let m = 2k + 1, n = 2k – 1 (k Î N)
\ m2 – n 2
= 4k2 + 1 + 4k – 4k2 + 4k – 1 = 8k
Hence, all the numbers of the form m2 – n 2 are
always divisible by 8.
24. (16) 32n+ 2 – 8n – 9, " n Î N
Putting n = 2
Þ 32 × 2 + 2 – 8 × 2 – 9 = 729 – 16 – 9 = 704
It is divisible by 16.
(a) | x1z1 – y1z2 |2 + | y1z1 + x1z2 |2
= | x1z1 |2 + | y1z2|2 – 2Re(x1y1z1z2)
+ | y1z1 |2 + | x1z2 |2 + 2Re(x1y1z1z2)
2
2
= x1 | z1 | + y12 | z2 |2 + y12 | z1 |2 + x12 | z2 |2
= x12 | z1 |2 + y12 | z2 |2 + y12 | z1 |2 + x12 | z2 |2
= 2(x12 + y12) (42) = 32(x12 + y12)
(d) Q | z | = | w| and arg z = p - arg w
Let w = reiq , then z = rei (p-q)
4.
5.
2
+ (1– i ) n2
z = reip . e-iq = (re -iq ) (cos p + i sin p)
= w (-1) = -w
(c) Q | z - 1| + | z + 3 | £ 8
\ z lies inside or on the ellipse whose foci are (1, 0)
and (– 3, 0) and vertices are (– 5, 0) and (3, 0).
Clearly the minimum and maximum values of
| z - 4 | are 1 and 9 respectively, representing
Þ
3.
S'
(–3, 0)
S
(1, 0)
A (4, 0)
(3, 0) P
Observe that (1+ i ) n1 + (1– i )n1
n1p ü
ì
ý,
= 2 í 2cos
4 þ
î
always real " n1 Î ¥
Similarly (1 + i )n2 + (1– i ) n2 is always real
" n2 Î ¥
Þ n1, n2 may be any positive integers
(cos a k + i sin a k )2
(d) wk =
zk
( )
the distances PA and PA¢. Thus, 1 £ | z - 4 | £ 9 .
A'
(– 5, 0)
2
i i
i i
(b) Let z = sin ln (i ) + cos ln (i )
ì æ p ö –1 ü
ì æ p ö –1 ü
–
ï
ï
ï
ï
= sin íln ç e 2 ÷ ý + cos íln ç e 2 ÷ ý
ç
÷
ç
÷
ï è ø ï
ï è
ø ï
î
þ
î
þ
p
p
= sin + cos = 1
2
2
(d) Let z = (1+ i ) n1 + (1– i )n1 + (1 + i )n2
6.
Solutions
=
89
zk 2
rk2 zk
=
1
zk2
= z We have
zk zk zk
k
1
1
1
+
+
z3
z1
z2
w1 + w2 + w3 =
7.
æ1 1 1ö
= ç + + ÷ =0
è z1 z2 z3 ø
\ The origin O is the centroid of DA1A2A3.
(c) If P(z1) is the reflection of Q(z2) through
the line b z + b z = c in the argand plane. Then,
Þ
O
9.
(d)
P(z)1
Þ
æ z +z ö
Rç 1 2÷
è 2 ø
bz + bz =c
=
Q(z)2
æ z1 + z2 ö
æ z1 + z2 ö
+
=c
bç
è 2 ÷ø b çè 2 ÷ø
b z1 + b z1 + b z2 + b z2 = 2c
Þ
... (i)
\
Since, PQ is perpendicular to the line b z + b z = c.
Therefore
Slope of PQ + Slope of the line = 0
Þ
æ z2 – z1 ö æ – b ö
Þ ç z – z ÷ + çè
÷ =0
b ø
è 2 1ø
Þ b (z2 – z1) – b ( z2 – z1 ) = 0
8.
Þ b z2 – b z1 – b z2 + b z1 = 0
Adding Eqs. (i) and (ii), we get
2 ( b z1 + b z2) = 2c Þ b z1 + b z2 = c
(b) As DOAC is a right triangle with right angle
at A, |z1|2 + |z3 – z1|2 = |z3|2
Þ 2|z1|2 – z3 z1 – z1 z3 = 0
Þ
2z1 – z3 –
z1
z =0
z1 3
z
Similarly, 2z2 – z3 – 2 z3 = 0
z2
Subtracting (2) from (1) we get
æz z ö
2 ( z2 – z1 ) = z3 ç 1 – 2 ÷
è z1 z2 ø
z1 – z2
= cos a + i sin a
z1 + z2
C(z3)
B(z2)
cos a + i sin a + 1
2 z1
=
cos a –1 + i sin a
– 2 z2
2 cos 2 a / 2 + 2i sin a / 2 cos a / 2
2i sin a / 2 cos a / 2 – 2sin 2 a / 2
2cos a / 2[cos a / 2 + i sin a / 2]
= 2i sin a / 2[cos a / 2 + i sin a / 2]
æ z1 + z2 ö
Rç
lies on the line.
è 2 ÷ø
Þ
æ 2 2 ö
2r 2 ( z1 – z2 )
2 z2 – z1 )
= z3 r ç 2 2 ÷
z1 z2
è z1 z2 ø
[\ |z1|2 = |z2|2 = r2]
A(z1)
2z1 z2
z3 =
r
z1 + z2
Þ
... (1)
z1
a
z
K
= i cot
Þ Given 1 =
3
1
z2
z2
tan a/2 = – 1/K
–2 / K
1–1/ K
2
Þ
Þ tan a =
2 tan a / 2
1– tan 2 a / 2
–2 K
K 2 –1
æ 2K ö
÷ Þ 2 tan– 1 (K)
a = tan– 1 ç
è 1– K 2 ø
10. (d) z2 + az + a2 = 0
Þ z = aw, aw2
(where 'w' is non-real root of cube unity)
Þ Locus of z is a pair of straight lines
and arg(z) = arg(a) + arg(w) or arg(a)
+ arg(w2)
2p
Þ arg(z) = ±
3
Also, |z| = |a||w| or |a||w2| Þ |z| = |a|.
11. (c) Zp = r(cos q + i sin q)
y
Q
... (2)
2r
p/4
O
a
q
x
r
P(z)
x
MATHEMATICS
90
ZQ =
æ
pö
pöö
æ
æ
2 | z |2 ç cos ç q + ÷ + i sin ç q + ÷ ÷
è
ø
è
è
4
4øø
æ -b ö
a ´ ç ÷ + 2b
b
è a ø
=
=
c
b
ac
æ
ö
a 2 . + ab . ç ÷ + b2
a
è a ø
é æ
pö
pö ù
æ
2r êcos ç q + ÷ + i sin ç q + ÷ ú
è
ø
è
4
4ø û
ë
From the figure,
=
2
2
2
2
2
2
2
p 2r + r – x
3r – x
3r – x
=
=
2 \1 =
2
4
2 × 2r × r
2r
2r2
Þ r2 = x2 Þ x = r Þ Triangle is right isosceles.
cos
12. (d)
=
=
aw + b + cw 2
aw2 + bw 2 + c
aw 2 + bw + cw 3
w (aw 2 + bw + c)
+
+
aw 2 + b + cw
a + bw + cw 2
a w 3 + bw + c w 2
17. (a) Let t = x + x2 + b2
w (a + bw + cw 2 )
1 (aw 2 + bw + c) (a + bw + cw 2 )
+
w (aw 2 + bw + c) w (a + bw + cw 2 )
(Q w3 = 1)
2
1 1 2 2w
+ = =
= 2w 2
w w w
w3
1
13. (c) Q z + = 1 Þ z2 – z + 1 = 0
z
=
\
z=
– (–1) ± (1– 4)
2
n
n–4
= – w, – w2
Þa+b= =
n–4
1
.
2
1
1
b
c
+
and ab = =
a
a
+
b
a
b
+b
a
a
ab + b + aa + b
(aa + b) (ab + b)
2
2
t
t
b
Þ t - b = 2x & t +
= 2 x2 + b2
Þ a 2 + b2 - (a 2 - 2at + t 2 )
= a 2 + b 2 - (a - t )2 £ a 2 + b2
18. (b) a, b are the roots of
Again, a, b are roots of x 2 n + p n x n + q n = 0
and 22 = 24.2
has last digit 6.
=16 2
\ b=6–1=5
Hence, a2 + b2 = 12 + 52 = 26
14. (c) We have ||z1| – |z2|| £ | z1 – z2|| and equality
holds only when arg z1 = arg z2
Þ ||z – w| – |z – w2|| £ |w2 – w| £ 3 and equality
15. (a) a and b are roots of ax2 + bx + c = 0
b2
x 2 + px + q = 0 Þ a + b = - p and ab = q ...(i)
1
can hold only when |z| = 2 and not when |z| =
x2 + b2 - x
1
1
=
=
t x + x2 + b2
æ
b2 ö
= ç 2a - t + ÷ (t ) = 2at - t 2 + b2
t ø
è
1ö
æ
= – çw + ÷
2017
è
ø
w
z
= – (w + w2) = 1
a = z2017 +
Þ
\ 2(a - x)( x + x2 + b2 )
[w is cube root of unity]
and z2017 = (– w)2017 = – w,
z2017 = (– w2)2017 = – w2
\
1
1
=
(aa + b) (ab + b) ac
\ Required quadratic equation
x2 – (sum of roots) x + product of roots = 0
Þ acx2 – bx + 1 = 0
16. (c) c = – 4, – b = –3
So, x2 + bx + c = 0 becomes x2 + 3x – 4 = 0
or (x + 4) (x – 1) = 0
This gives 1 and – 4 as its roots.
Þ a 2 n + pn a n + q n = 0 and
b 2 n + p nb n + q n = 0
Þ ( a 2 n - b 2 n ) + p n (a n - b n ) = 0
Þ a n + bn = - p n
Now
...(ii)
a
is a root of ( x n + 1) + ( x + 1) n = 0
b
æ an
ö æ a ön
Þ ç n + 1 ÷ + ç + 1÷ = 0
çb
÷ èb ø
è
ø
Þ (a n + b n ) + (a + b ) n = 0
Þ - p n + (- p)n = 0,
Which holds only if n is an even integer.
Solutions
91
19. (a) Let f (x) = (x – sin b) (x – sin g) + (x – sin a)
(x – sin g) + (x – sin a) (x – sin b)
Now, f (sin a) = (sin a – sin b) (sin a – sin g)
= (–) (–) = positive
f (sin b) = (sin b – sin a) (sin b – sin g) = (+)(–)
= negative
f (sin g) = (sin g – sin a) (sin g – sin b) = (+)(+)
= positive
Þ Roots of f (x) = 0 are real and distinct.
20. (c) (a – 1) x2 – (a + 1)x + (a – 1) ³ 0 " x ³ 2
For a = 1, – 2x ³ 0 " x £ 2 Þ a ¹ 1
Þ a > 1 and f ' (x) ³ 0 " x ³ 2, f (2) ³ 0
Þ 2(a – 1)x – (a + 1) ³ 0 " x ³ 2, f (2) ³ 0
Þ 2(a – 1)x ³ (a + 1) " x ³ 2, 3a – 7 ³ 0
Þ
Þ
a +1
7
"x ³ 2, x ³
2(a –1)
3
a +1
7
£ 2; a ³
2(a - 1)
3
x³
7
Þ a + 1 £ 4a – 4; a ³ ; a > 1
3
5
7
é7 ö
Þ a ³ ; a ³ ; a > 1 Þ a Î ê , ¥÷
3
3
ë3 ø
21. (1) Given that a, b, c are integers not all equal,
w is cube root of unity ¹ 1, then
| a + bw + cw2 |
æ -1 + i 3 ö æ -1 - i 3 ö
= a + bç
÷ + cç
÷
è
2
ø è
2
ø
æ 2a - b - c ö æ b 3 - c 3 ö
= ç
÷
÷ +iç
è
2
ø è
2
ø
=
1
(2a - b - c)2 + 3(b - c) 2
2
=
1
4a2 + b2 + c2 - 4ab + 2bc - 4ac + 3b2 + 3c2 - 6bc
2
22. (45) If 2 – i is the root then 2 + i is also the root
sum of roots = 4 Þ a = – 3
b
Product of roots =
= (2 – i)(2 + i)
a
Þ b = – 15 \ ab = 45
23. (0) As the coefficients of two equations are in
reverse order, if the roots of ax2 + bx + c = 0 are
a and b then the roots of second equation are
1 1
, . Given that one negative root is common,
a b
two possibilities may arise.
1
1
Either a = < 0 Þ a = -1 or a = < 0
b
a
c
Þ ab = 1 Þ = 1 Þ c = a (not possible)
a
\ a = -1 is the common root. Put a = -1 in
any of the equations, we get a - b + c = 0.
24. (7)
x 2 - (3 + 2
log 2 3
–3
log 3 2
)x
log 3 2
– 2(3
– 2log 2 3 ) = 0
Þ x2 – 3(x) – 2(2 – 3) = 0 Þ x2 – 3x + 2 = 0
Þ a = 2, b = 1 Þ a2 + ab + b2 = 4 + 2 + 1 = 7
25. (0) Let f(x) = (x – x1) (x – x2)(x – x3)(x – x4)
Þ |f(i)| = 1 + x12 1 + x22 1 + x33 1 + x44 = 1
Þ x1 = x2 = x3 = x4 = 0
Þ All four roots are zero.
Þ f (x) = x4 Þ a + b + c + d = 0
26. (2) The given relation can be rewritten as
2
1
1
1
+
=
+
a+w b+w c+w w
1
1
1
2
+
= 2
and
2 +
2
2
b+w
a+w
c+w
w
2
Þ w and w are roots of
1
1
1
2
+
+
=
a+ x b+ x c+ x x
2
3x 2 + 2(a + b + c) x + bc + ca + ab
=
x
(a + x )(b + x)(c + x )
= a 2 + b2 + c 2 - ab - bc - ca
Þ
1
[(a - b) 2 + (b - c ) 2 + (c - a) 2 ]
2
R.H.S. will be min. when a = b = c, but we cannot
take a = b = c as per question.
\ The min value is obtained when any two are
zero and third is a minimum magnitude integer
i.e. 1.
Thus b = c = 0, a = 1 gives us the minimum value 1.
Þ x3 - (bc + ca + ab) x - 2abc = 0
=
Two roots of the equation (i) are w and w
the third root be a , then
a + w + w2 = 0 Þ a = – w – w 2 = 1.
\ a = 1 will satisfy equation (i)
1
1
1
+
+
=2
Þ
a +1 b +1 c +1
...(i)
2 . Let
MATHEMATICS
92
k2
=
Þ
y
x2 + k 2 - x
k2
y
\
2x
=
x +k -x
y
2
2
(
29. (4) Q (1 + i )4 = [(1 + i )2 ]2
k2
27. (2) Let y = x + x 2 + k 2 =
Let z = 2 ( k - x ) x + x 2 + k 2
= (1 + i 2 + 2i )2 = (1 - 1 + 2i) 2
= 4i2 = –4
...(i)
&
)
æ
k2 ö
÷y
= çç 2k - y +
y ÷ø
è
= k 2 - y 2 + 2ky = 2k 2 - ( k - y )2 £ 2k 2
\
z
k
2
Consider z = 2- + 0i+ and w = -1+ + 0 + i
Y
O
or
or
-2pi - 2i
= -2i
p +1
...(ii)
p 2 + r 2 + 2 pr
- 4 pr ³ 0
4
Þ p 2 + r 2 - 14 pr > 0
X
Þ
p r
r
+ > 14 Þ - 7 ³ 4 3 Þ k = 4
r p
p
Linear Inequalities
6
or
p - i - pi - p + p - pi - i - p
p +1
Since D ³ 0 Þ
CHAPTER
(c)
=
æ | z | ö 2p
= 4.
÷=
Now, ç
è amp( z ) ø p / 2
30. (4) If p, q, r are in A.P., p + r = 2q
Þ| z - w |< 3 so k = 3 (as |z – w| < k and k is least
upper bond).
1.
(1- pi)( p - i) ( p - i)(1- pi)
+
p +1
1+ p
p
= ( - 4)(- 2i ) = 2pi
4
[from Eqs. (i) and (ii)]:
w
z
p -i
1 + pi
æ 1- pi
p
p -i ö
Given, z = (1 + i )4 ç
ç p + i + 1+ pi ÷÷
4
è
ø
æ z-2ö p
28. (3) Arg ç
gives semi-circle with
÷=
è z+2ø 2
radius r = 2.
æ w -1 ö p
Arg ç
÷ = gives semi circle with radius
è w +1 ø 2
r = 1.
p +i
+
=
=
£2
1 - pi
...(i)
1
2x - 1
- 3
³0
x
+
1
x - x +1
x +1
2
-
2
(
)
2 ( x + 1) - x 2 - x + 1 - ( 2 x - 1)
( x + 1) ( x2 - x + 1)
(
- x2 - x - 2
)
)
³0
( x + 1) ( x 2 - x + 1)
³0
( x + 1) ( x
2
- x +1
- ( x - 2 )( x + 1)
³0
2.
3.
2- x
³ 0 , where x ¹ – 1
x - x +1
or 2 - x ³ 0, x ¹ -1 (as x2 – x+ 1 > 0 for " x ÎR)
Þ x £ 2, x ¹ - 1 Þ x Î (– ¥, – 1) È (–1, 2]
(c) (2x + 1) (x – 3) (x + 7) < 0
Sign scheme of (2x + 1) (x – 3) (x + 7) is as
follows :
–
+
–
+
or
2
–7
–1/2
3
Hence, solution is (–¥, 7) È (–1/2, 3)
(c) | x - 1| £ 3 Þ -3 £ x - 1 £ 3 Þ -2 £ x £ 4
and | x - 1 | ³ 1Þ x - 1 £ -1
Solutions
4.
93
or x - 1 ³ 1 Þ x £ 0 or x ³ 2
Taking the common values of x, we get
x Î [-2, 0] È [2, 4]
(b) The given equations are
(iii) If x ³ 1 , then
ì x + 3y = 5, x ³ 1
| x - 1| +3y = 4 Þ í
î - x + 3y = 3, x < 1
and
Q
( x + 2) - ( x - 1) < x -
L (i)
L (ii)
L (iii)
ì x - y = 1, y ³ 1
x - | y - 1| = 2 Þ í
L (iv)
î x + y = 3, y < 1
Solving (i) and (iii), we get x = 2, y = 1
Solving (i) and (iv), we get x = 2, y = 1
no solution (Q x ³ 1 and y < 1)
Solving (ii) and (iii), we get x = 3, y = 2
no solution (Q x < 1, y ³ 1)
5.
Þ x Îf
5
3
solving (ii) and (iv) we get, x = , y =
2
2
no solution (Q x < 1, y < 1)
Hence solution is x = 2, y = 1(a unique solution)
(c) 2- |1- | x || = 1 Þ 2- |1- | x ||= ±1
9
> 1. Þ common solution is
2
x>
9
æ9 ö
Þ x Îç ,¥÷
2
è2 ø
ö
æ9
\ Solution set is x Î ç , ¥ ÷ .
2
ø
è
7.
12x
(a)
4x 2 + 9
³1Þ
12 | x |
³1
4x 2 + 9
Q 4x 2 + 9 > 0
Þ 4x 2 - 12 | x | +9 £ 0 Þ 4| x |2 -12| x | +9 £ 0
Þ (2 | x | -3) 2 = 0 Þ | x | =
8.
3
2
| x - 1| + a = 4 Þ | x - 1| + a = ±4
(b)
Þ| x - 1| = - a ± 4
Þ|1- | x ||= 1 or 3
or 2 Þ x = 0 or ± 2
The above equation holds if -a + 4 ³ 0
or -a - 4 ³ 0
Þ a £ 4 or a £ -4 Þ a Î (-¥, 4] È (-¥, - 4]
If 1- | x | = 3 Þ 1- | x |= ±3 Þ| x |= -2 or 4
Þ a Î (-¥, 4]
If 1- | x | = 1 Þ 1- | x |= ±1 Þ| x |= 0
Þ| x |= 4 Þ x = ±4
[Q | x |¹ -2]
\ Solution set is {– 4, – 2, 0, 2, 4}, hence 5 real
roots in all.
6.
9
3
Þx>
2
2
3
(a) The inequality is | x + 2 | - | x - 2 |< x - .
2
Dividing the problem into three intervals :
(i) If x < –2, then
- ( x + 2) + ( x - 1) < x -
3
3
Þx > 2
2
3
> -2 , hence no common values
2
Þ x Îf
(ii) If -2 £ x < 1, then
But -
( x + 2) + ( x - 1) < x -
But -
3
5
Þx<2
2
5
< -2 , hence no common values
2
9.
(c) We know that |f(x)| = – f(x) if f ( x ) £ 0
\
Þ
x 2 - 8x + 12
x 2 - 10x + 21
=-
x 2 - 8x + 12
x 2 - 10x + 21
x 2 - 8 x + 12
£0
x 2 - 10 x + 21
( x - 2)( x - 6)
Þ
£ 0, x ¹ 3, 7
( x - 3)(x - 7)
Þ ( x - 2)( x - 3)(x - 6)(x - 7) £ 0 , x ¹ 3, 7
+
+
2
–
3
6
Þ 2 £ x < 3 or 6 £ x < 7
Þ x Î [2, 3) È [6, 7)
10. (a) Case (i) : When x ³ 0
\ |x| = x
–
+
7
MATHEMATICS
94
3- x
(3 - x)(4 - x)
³0Þ
³0
4-x
(4 - x) 2
Þ (x – 3) (x – 4) ³ 0 and x ¹ 4 Þ x £ 3 or x > 4
but x ³ 0 so x Î [0, 3] È (4, ¥)
...(i)
Case (ii) : When x < 0
3+ x
(x + 3)(x + 4)
³0Þ
³0
4+x
(x + 4) 2
Þ x < – 4 or x ³ – 3
but x < 0 so x Î (– ¥, – 4) È [–3, 0)
...(ii)
So union of (i) and (ii) gives
[ -3,3] È ( -¥, - 4) È (4, ¥)
\
11.
(b)
|x| = –x \
x+y
2
x +y
x+y
2
x +y
2
-
2
-
1 æ1 1ö
ç + ÷
2 èx yø
æ
x+y
1
1 ö
= ( x + y) ç 2
ç x + y2 2xy ÷÷
2xy
è
ø
æ 2xy - (x 2 + y 2 ) ö
= ( x + y ) çç 2
2
÷÷
è (x + y ) (2 xy) ø
æ - (x - y) 2 ö
and ( x + y ) çç 2
÷÷ £ 0
2
è (x + y ) (2 xy) ø
Similarly,
y+z
y 2 + z2
£
1 æ1 1ö
ç + ÷ and
2 èy zø
x+z
1 æ 1 1ö
£ ç + ÷
2
2
2 èx zø
x +z
\ Adding; we get
x+y
y +z
z+x
1 1 1
+ 2
+ 2
£ + +
2
2
2
2
x y z
x +y
y +z
z +x
and Hence A £ B
12. (d) xy + yz + zx < 0 and m =
\
\
Þ
Þ
x 2 + y2 + z 2
xy + yz + zx
(x + y + z)2 = x2 + y2 + z2 + 2 (xy + yz + zx)
(x + y + z)2 ³ 0
x2 + y2 + z2 + 2 (xy + yz + zx) ³ 0
x2 + y2 + z2 ³ – 2 (xy + yz + zx)
x 2 + y2 + z 2
£–2
xy + yz + zx
13. (a) We have
(a1 + a2 + ... + an -1 + 2an ) (a a ...a 2a )1/n
³ 1 2 n–1 n
n
[Using A.M. ³ G.M.]
1/n
Þ a1+ a2 + a3 + ..... + an– 1+ 2an ³ n(2c)
Þ
14. (a) The inequality is log 0.2
x+2
£ 1 . The
x
x+2
>0
x
Þ x( x + 2) > 0 Þ x < -2 or x > 0 .
Solving the inequality, we get (note that base < 1)
x+2 1
x+2
4 x + 10
1
- ³0Þ
³0
³ 0. 2 = Þ
x
5
x
5x
5
L.H.S is valid if
5
or x ³ 0 .
2
Taking the intersection, we get
5
x£or x > 0
2
x (2x + 5) ³ 0 Þ x £ -
æ
5ù
Þ x Î çç - ¥, - ú È (0, ¥)
2û
è
15. (a) The log functions are defined if
(x + 3) 2
x 2 + 6x + 9
>0
> 0 and x + 1 > 0 Þ
2( x + 1)
2(x + 1)
and x + 1 > 0 Þ x > –1
Now the inequality is
log
2 -1
x 2 + 6x + 9
< - log 2 ( x + 1)
2(x + 1)
Þ - log 2
Þ log 2
Þ
x 2 + 6x + 9
< - log 2 (x + 1)
2( x + 1)
x 2 + 6x + 9
> log 2 ( x + 1)
2( x + 1)
2
x 2 + 6x + 9
> ( x + 1) Þ - x + 2x + 7 > 0
2( x + 1)
2( x + 1)
Þ ( x + 1)(x 2 - 2x - 7) < 0
Þ x 2 - 2x - 7 < 0
[Q x + 1 > 0]
Þ -1 - 2 2 < x < -1 + 2 2 ,
but x > -1 Þ -1 < x < -1 + 2 2
16. (b) For the validity of inequality ax2 + 4x + a > 0,
which is possible if a > 0 and 16 – 4a2 < 0
Þ a >2
...(i)
Further, the inequality can be rewritten as
log 5 5 + log 5 ( x 2 + 1) £ log 5 (ax 2 + 4x + a )
Þ 5( x 2 + 1) £ ax 2 + 4x + a
Solutions
Þ (a - 5) x 2 + 4 x + (a - 5) ³ 0 .
It holds if a - 5 > 0 and 16 - 4(a - 5) 2 £ 0
...(ii)
Þ a > 5 and a £ 3 or a ³ 7 Þ a ³ 7
Combining the results of (i) and (ii) for common
values, we get a Î [7, ¥)
17. (b) For x ³ -2 , x2 – x – 2 + x > 0
2
Þ x > 2 Þ x Î (-¥, - 2 ) È ( 2 , ¥)
Þ x Î [-2,- 2 ) È ( 2 , ¥)
For x < – 2
x2 + x + 2 + x > 0 or x2 + 2x + 2 > 0
which is true for all x.
Hence x Î (-¥, - 2 ) È ( 2 , ¥)
18. (c) log10 x = A
x >0
log10 (x – 2) = B, x – 2 > 0 Þ x > 2
Þ A2 – 3AB + 2B2 < 0 Þ(A – 2B) (A – B) < 0
Þ (log x – 2 log (x – 2)(log x – log (x – 2)) < 0
Case - I : log x – 2 log (x – 2) < 0
and
log x – log (x – 2) > 0
Case - II :log x – 2 log (x – 2) > 0
and
log x – log (x – 2) < 0
From (1) & (2), x Î (4, ¥)
19. (a) x ¹ (2n + 1) p/2, npnÎ I. The given
inequality can be written as
log 2 (x 2 - 8x + 23)
3
>
log 2 | sin x |
log 2 |sin x |
As log2 |sin x| < 0, we get; log (x2 – 8x + 23) < 3
Þ x2 – 8x + 23 < 23 = 8 Þ x2 – 8x + 15 < 0
Þ (x – 5) (x – 3) < 0 Þ 3 < x < 5
p 3p
For x Î (3, 5), x ¹ p, ,
. Hence
2 2
æ 3p ö æ 3p ö
x Î (3, p) È ç p, ÷ È ç , 5 ÷ .
è 2 ø è 2 ø
3 x 2 + 9 x + 17
20. (b) y =
3 x2 + 9 x + 7
3x 2 ( y - 1) + 9 x ( y - 1) + 7 y - 17 = 0
D ³ 0 Q x is real
81( y - 1)2 - 4 ´ 3( y - 1)(7 y - 17) ³ 0
Þ ( y - 1)( y - 41) £ 0 Þ 1 £ y £ 41
\ Max value of y is 41
21. (4) Consider the L.H.S of the equation,
(sin 3x) (cos 3x)
95
2sin 3x . cos 3x sin (2 . 3x ) 1
=
£
2x
2
2
x
3 +3
1
x
–x
Þ
£ but (3 + 3 ) ³ 2
k
2
Þ Minimum value of k = 4
22. (2) As x, y Î R, and xy > 0, so x and y will
be of same sign.
=
Therefore, all the quantities
2x x 3 y 4y 2
,
,
y3 3 9x 4
are positive.
Now, A.M. ³ G.M.
1/3
æ æ 2x ö æ x3y ö æ 4y2 ö ö
2x x3 y 4y2
÷÷ çç 4 ÷÷ ÷÷
Þ 3+
+ 4 ³ 3 çç çç 3 ÷÷ çç
3
y
9x
è è y ø è 3 ø è 9x ø ø
=3×
2
=2
3
23. (1) x + 3 - x ³ 3 - x + 3 Þ x ³ 3
But 3 – x ³ 0 Þ x £ 3
Hence, x = 3 is the only integral solution.
x+3 + x
>1
24. (4) We have
x+2
x+3 -2
x+3 + x
-1 > 0 Þ
>0
Þ
x+2
x+2
Now two cases arise :
Case I : When x + 3 ³ 0, i.e., x ³ – 3. Then
x+3 -2
>0 Þ
x+3- 2
>0
x+2
x+2
x +1
> 0 Þ {(x + 1) > 0 and x + 2 > 0}
Þ
x+2
or {x + 1 < 0 and x + 2 < 0}
Þ {x > – 1 and x > – 2}
or {x < – 1 and x < –2} Þ x > – 1 or x < – 2
Þ x Î (-1, ¥) or x Î ( - ¥, - 2)
Þ x Î ( -3, -2) È ( -1, ¥) [Since x ³ -3 ]…(i)
Case II : When x + 3 < 0, i.e., x < – 3
x+3 -2
-x - 3 - 2
>0
>0 Þ
x+2
x+2
-( x + 5)
x+5
>0 Þ
<0
Þ
x+2
x+2
Þ (x + 5 < 0 and x + 2 > 0 )
or (x + 5 > 0 and x + 2 < 0)
Þ (x < – 5 and x > – 2)
or (x > – 5 and x < – 2). It is not possible.
...(ii)
Þ x Î (– 5, – 2)
MATHEMATICS
96
Þ ab ³ 0
Combining (i) and (ii), the required solution is
x Î (– 5, – 2) È (– 1, ¥ ),
smallest integral member = – 4
\ Modulus of smallest integral member =|–4| = 4
25. (4)
4 - x2
4 - x2
= x+
x
x
As | a | + | b | = | a + b |
|x|+
CHAPTER
7
1.
2.
3.
4.
æ 4 - x2 ö
\ x çç
÷÷ ³ 0, x ¹ 0
è x ø
Þ x2 – 4 £ 0
Þ x Î [– 2, 2] – {0}
Number of integral values = 4.
Permutations and Combinations
and 3, and a number cannot begin with 0)
Similarly, the number of numbers with 5 in the
middle
= 5P4 – 4P3, etc.
\ The required number of numbers
= (4P4 – 3P3) + (5P4 – 4P3) + (6P4 – 5P3) +
(a) Total number of words that can be formed
using 5 letters out of 10 given different letters
= 10 × 10 × 10 × 10 × 10 (as letters can repeat)
= 1, 00, 000
Number of words that can be formed using 5
different letters out of 10 different letters
= 10P5 (none can repeat)
10!
= 30, 240
=
5!
\ Number of words in which at least one letter
is repeated
= total words-words with none of the letters
repeated
= 1,00,000 – 30,240 = 69760
(c) X – X – X – X – X. The four digits 3, 3, 5, 5
can be arranged at (–) places in 4! = 6 ways.
2!2!
The five digits 2, 2, 8, 8, 8 can be arranged at
5!
= 10 ways.
(X) places in
2!3!
Total no. of arrangements = 6 × 10 = 60 ways.
(c) The letter of word COCHIN in alphabetic
order are C, C, H, I, N, O.
Fixing first letter C and keeping C at second place,
rest 4 can be arranged in 4! ways.
Similarly the words starting with CH, CI, CN are
4! in each case.
Then fixing first two letters as CO next four places
when filled in alphabetic order give the word
COCHIN.
\ Numbers of words coming before COCHIN
are 4 × 4! = 4 × 24 = 96
(d) The smallest number, which can occur in
the middle is 4.
The number of numbers with 4 in the middle
= 4 P4 – 3 P3
(Q The other four places are to be filled by 0, 1, 2
..................... + (9P4 – 8P3) =
5.
9
å ( n P4 - n -1 P3 )
n=4
(b) C ...... = 4! = 24
D ...... = 4! = 24
M ...... = 4! = 24
S C ...... = 3! = 6
S D ...... = 3! = 6
S M C DW=1
6.
S M C WD=1
(b) m = Number of permutations of letters of
the word BHARAT, when B and H are never
together
7.
5!
6!
–
× 2! (keeping (BH) together and
2!
2!
permuting with A, R, A, T)
= 360 – 120 = 240 and n = number of permutations
of the letters of the word BHARAT when each
word begins with B an d ends with
4!
m 240
T=
= 12 \ =
= 20
2!
n 12
(c) Let the square has dimensions n × n.
If n is even, total no. of blue (diagonal) tiles
= n + n = 2n
If n is odd, then total no. of blue tiles
= n + n – 1 = 2n – 1
(the middle tile is be counted twice).
Hence, 2n – 1 = 121
Þ n = 61
So, number of red tiles = 61 × 61 – 121 = 3600
=
Solutions
8.
97
(c)
1 9
8
. 2 = 2 = 256.
2
12. (a) Arrangement does not matter because of
descending order there should be only one
arrangement
\ required number of ways = 15C3 × 24C2
1 ................ 24 25 26 ................ 40
=
Number of
digits
Numbers
ending
with 0
Numbers
ending
with 5
Total
1
0
1
1
2
8
9
17
3
9.8 = 72
8.8 = 64
136
4
9.8.7 = 504 8.8.7 = 448
Total
9.
952
1106
(d) 10 identical boxes can contain 12 balls such
that each box contains at least one ball in the
following cases:
Case (i) : 8 boxes contains single ball each + 2
boxes contains two balls each
Single ball boxes Boxes with doubleball
6
474
8
2W,6B;
64
4744
8
(WW)(WW)
3W,5B;
(WW)(WB)
4W, 4B;
5W,3B;
(WW)(BB);
(WB)(BB)
(WB)(WB)
6W, 2B;
(BB)(BB)
Case (ii): 9 boxes single ball each + 1 box
containing 3 balls
Singleballboxes Triple ball boxes
6474
8
3W + 6 B;
4W + 5B;
6
474
8
(WWW )
(WWB )
5W + 4 B;
(WBB )
6W + 3B;
( BBB )
\ Total 10 ways are there
10. (c) Required number of ways = (Total number
of ways of arranging 7 persons) – (Number of
ways where two American are together + Number
of ways when two Britishian are together –
Number of ways when both American and both
Britishian are together)
= 6! – [2 × 5! + 2 × 5! – 4! × 2! × 2!)]
= 720 – [480 – 96]= 240 + 96 = 336
11. (b) The candidate is unsuccessful if he fails in
9 or 8 or 7 or 6 or 5 papers.
\ The number of ways to be unsuccessful
= 9 C9 + 9 C8 + 9 C7 + 9 C6 + 9 C5
= 9 C0 + 9 C1 + 9 C2 +
=
9
C3 + 9 C 4
1 9
( C0 + 9 C1 + ......... + 9 C9 )
2
Selecting any 2 out of
Selecting any 3 from
24
15
26 to 40 = C3
1 to 24 = C2
13. (b) Let the number of children in the class = n.
Number of groups of 3 children = nC3
Each child would go to zoo n– 1C2 times.
\ According the question: nC3 = 84 + n – 1C2
n(n - 1)(n - 2)
(n - 1)(n - 2)
= 84 +
Þ
6
2
Þ n(n – 1) (n – 2) = 504 + 3(n – 1) (n – 2)
Þ (n – 1) (n – 2) (n – 3) = 504
Þ (n – 1) (n – 2) (n – 3) = 7 × 8 × 9 Þ x = 10
14. (d) We will consider the following cases:
Case
Flags
4 alike and 2
others alike
4 alike and 2
others different
3alike and 3
others alike
3alike and 2
others alike and1
different
4 white and 2 red
4 white,1red and1blue
3 white,3red
3 white,1blue,
2 red or 3 red, 1
blue, 2 white
Total
No.of signals
6!
=15
4!2!
6!
= 30
4!
6!
= 20
3!3!
2
C1 ×
6!
=120
3!2!
185
15. (c) Let the number of men = n
Number of women = 2
\ Total participants = (n + 2)
Now number of games played by men among
the m selves = 2(nC2)
Number of games men played with women = 4n
According to question 2(nC2) = 4n + 66
Þ n(n – 1) = 4n + 66 Þ n2 – 5n – 66 = 0
Þ (n – 11)(n + 6) = 0 Þ n = 11 or n = – 6
\ Total number of participants = 11 + 2 = 13
16. (b) Suppose there n players in the beginning.
The total number of games to be played was
equal to nC2 and each player would have played
n – 1 games.
Let us assume that A and B fell ill. Now the total
number of games of A and B is (n – 1) + (n – 1) – 1
= 2n – 3. But they have played 3 games each.
Then their remaining number of games is
MATHEMATICS
98
17.
\
Number of ways
é 1 1 1 1ù
= 10C6 × 4! ê1– + – + ú
ë 1! 2! 3! 4!û
2n – 3 – 6 = 2n – 9. Given,
nC – (2n – 9) = 84
2
Þ n2 – 5n – 150 = 0 or n = 15
(b) Required ways = 8C5 – (Number of ways
when R2 or R3 are unused)
=
R1
R2
R3
18.
23.
= 8C5 – (6C5 + 6C5) = 8C5 – (12) = 56 – 12 = 44
(c) The number of triangles with vertices on
p
C1 × p C1 = p3.
The number of triangles with 2 vertices on one
line and the third vertex on any one of the other
two lines
different lines =
20.
22.
{
p
C2 ´
2p
C1
} = 6p
5!
5!
´ 3!+
´ 3! = 30 + 60 = 90 ways
0!.1!.4!
0!.2!.3!
(8) a + b + c = 21 Þ 3b = 21 Þ b =7
[Q a + c = 2b]
Þ a + b + c = 21 Þ a + c = 14
Þ l = 14 – 1C2 – 1 = 13C1 = 13
Hence l – 5 = 13 – 5 = 8
(9) If exactly 6 of his matching are correct, then
the remaining 4 statements from column I must
be incorrectly matched with 4 statements from
column II
=
21.
p
p( p - 1)
2
\ The required number of triangles
= p3 + 3p2(p – 1).
(c) Let A1, A2, ..., A10 be vertices of a regular
polygon of 10 sides.
The number of ways of selecting 3 vertices
is 10C3.
The number of ways of selecting 3 consecutive
vertices is (i.e., A1 A2 A3, A2 A3 A4, ..., A10 A1 A2)
= 10.
The number of ways of selecting three vertices
such that two vertices are consecutive = (First
select 2 consecutive vertices, leave their
neighbouring two vertices and select one more
from the remaining 6 vertices) is
10 × 6C1 = 60 The total number of required
selections is
10C – 10 – 60 = 120 – 70 = 50
3
(b) Division of objects can be (0, 1, 4) or (0, 2, 3)
So total number of distribution ways
3
= C1
19.
C1 ×
24.
25.
10!
é1 1 1 ù
× 4! – +
6!4! êë 2 6 24 úû
é12 – 4 + 1ù
= (10 × 9 × 8 × 7) ê
= 1890
ë 24 úû
(6) Number of numbers beginning with 1 = 120
1
Number of numbers beginning with 2 = 120
2
Starting with 31 ................................. = 24
3 1
Starting with 3214 ................................. = 2
3 2 1 4
Finally = 1 3 2 1 5 4 6
Hence, unit place digit of 267th number is 6.
(7) If x denotes the number of times he can
take unit step and y denotes the number of times
he can take 2 steps, then x + 2y = 7.
We must have x = 1, 3, 5
If x = 1, the steps will be 1 2 2 2
4!
Þ Number of ways =
=4
3!
If x = 3, the steps will be 1 1 1 2 2
5!
Þ Number of ways = 2!.3! = 10
If x = 5, the steps will be 1 1 1 1 1 2
Þ Number of ways = 6C1 = 6
If x = 7, the steps will be 1 1 1 1 1 1 1
Þ 7C0 = 1
Hence total number of ways = N = 21
Þ N/3 = 7
(9)
In this case number of participants from
10 different countries are 1, 2, 3… 10. number
of ways that they can be arranged in a row
such that all the participants from the same
country were together is (10!)(1!)(2!)(3!)(4!)….
(10!)
Hence K = (10!)(1!)(2!)(3!)(4!)…. (10!) highest
power of 10 or that of 5 can be given by only
(5!)(6!)(7!)(8!) (9!)(10!)(10!) which is
1+1+1+1+1+2+2=9
Solutions
99
CHAPTER
Binomial Theorem
8
1.
2 60 = (1 + 7) 20
(a)
=
20
C0 . +
20
C1. 7 +
6.
20
2
C2 . 7 +
............ +
2.
20
20
\ the remainder = C0 . = 1
(c) 32 = 25 Þ (32) 32 = (25 ) 32
\ (32)
n
, remainder = 4
When 7 divides (32)
n æ r –1
ö
(d) å ç å n Cr .r C p .2 p ÷
÷
r =1 çè p = 0
ø
n
–
1
r
æ
ö
= å n C r ç å r C p .2 p ÷
ç p=0
÷
r =1
è
ø
=
n
å nCr { (1 + 2)r – 2r }= å
r =1
r=1
= n
4.
n
n
7.
\
n/2
(– 1) .(1/2)
5
- . 3log3 2
3
= 3
æ 1 ö
= ç 5/3 ÷
è3
ø
1/3 6
... (iv)
8
in equation (iv), from equation (iii)
a
(a)
17 2009 + 112009 – 7 2009 =
2009
2009
–7
is 0.
Cleary last digit in 17
Also, last digit in 112009 is 1.
Last digit of 17 2009 + 112009 – 72009 is
0 + 1 = 1.
… (i)
8.
…(ii)
9.
log3 8.
= 3-5 log 3 2 = 2 -5
So, t5 = C4 (2 ) (–1/ 2 )4 = 10C4. =
10
n(n – 1) 2
a = 24
2
...(iii)
2009
(17 2008 + 17 2007.71 + ..... + 7 2008 ) + 11
n
(-2)6 = 1 - a1 + a2 – .... + a12
Adding (i) and (ii),
2 + 2a2 + 2a4 + .... + 2a12 = 64
64 - 2
= 31
Þ a2 + a4 + .... + a12 =
2
(a) The last term = nCn. (–1 / 2 )n
n
C1 a = 8 Þ na = 8
(17 2009 – 72009 ) + 112009 = (17 – 7)
n
(4 –1) – (3 –1) = 4 – 3
2 6
(d) Given expression is (1 + x - 2 x )
= (1/3. 3 9 )log3 8.
...(ii)
éæ 8 ö 2 8 ù
2
Þ ê çè ÷ø – ú a = 48 Þ 64 – 8a = 48
a
a
úû
ëê
Þ a=2
\ n=4
2–4
a–n
1
=
=–
\
2+ 4
a+n
3
Cr 3r – n Cr 2r
= 1 + a1 x + a2 x 2 + ..... + a12 x12
Put x = 1 both side;
0 = 1 + a1 + a2 + .... + a12
Put x = – 1 both sides;
5.
Put n =
3232
n
C0 = 1 Þ 1 = 1
and n C2 . a 2 = 24 Þ
= 23(5 m + 1) 2 2 = 4. 85m + 1
4(7 + 1)5 m + 1 = 4(7n + 1), n Î N = 28n + 4
3.
C0 + n C1 ( ax )1 + n C2. (ax)2 + ........
= 1 + 8x + 24x2 +
... (i)
n
3m+1
= (32)
= 25(3m + 1)
\
n
Compairing coefficients of x 0 , x1 , x 2 in
equation (i),
C20 . 7 20
= 2160 = (3 – 1)160 = 3m + 1, m Î N
3232
(c)
10
C10–4.
153
2
\ a = x2 + 2 x + 1
Let y = 2204 \ b = y2 – y + 1
Þ N = x16 – 1 Þ N = (x4 – 1) (x4 + 1)(x8 + 1)
(c) Let x = 2
Þ N = (x4 – 1) (x2 + 2 x + 1 ) (x2 – 2 x + 1 )(x8 + 1)
Þ N = y6 – 1 = (y3 – 1) (y3 + 1)
Þ (y3 – 1)(y + 1)(y2 – y + 1)
(c) N = 20C7 – 20C8 + 20C9 – 20C10 + ..... – 20C20
= (20C7 + 20C9 + 20C11 + ..... + 20C19) –
(20C8 – 20C10 + ..... + 20C20)
= (20C0 + 20C2 + 20C4 + 20C6) –
(20C1 + 20C3 + 20C5)
= (1 + 190 + 4845 + 38760) – (20 + 1140 + 15504)
= 43796 – 16664 = 27132 = 3 × 4 × 7 × 19 × 17
MATHEMATICS
100
10.
11.
(a) x = T7 = nC6(31/3)n – 6. (4–1/3)6
y = Tn – 5 = nCn – 6(31/3)6. (4–1/3)n – 6
Þ y = 12x
n
Cn – 6(31/3)6 (4–1/3)n – 6
= 12. nC 6(31/3)n – 6.(4–1/3)6
Þ 12 = (121/3)12 – n Þ n = 9
(d) Given expression
=
=
( x1/ 3 )3 + (1)3
x
2/3
1/ 3
–x
+1
–
n + 1ù
é n
So, from (iii) and (iv), x Î ê
,
n úû
ën +1
x
( x 2 / 3 - x1/ 3 + 1)
( x1/ 2 + 1) ( x1/ 2 - 1)
x1/ 2 ( x1/ 2 - 1)
1/ 3
æ
x +1
x -1 ö
Þ çè 2 / 3 1/ 3 ÷
x - x + 1 x - x1/ 2 ø
1/ 3
= (x
-x
–1/ 2
– x
10
= ( -1)r
)
cos 5 a =
Cr ( x1/ 3 )10 - r . ( -1)r . ( x -1/ 2 )r
10
Cr x
independent of x if
Hence required coefficient =
12.
(c) Middle term =
2n
=
which is
æ 10 - r r ö
- ÷
çè
3
2ø
10
=0 Þ r=4
4
C4 (-1) = 210.
n
Cn . x = t n+1
If tn + 1 is the greatest term also, then
t n+1 ³ tn
....(i)
t n+1 ³ t n+ 2
.....(ii)
2n
Cn . x n ³ 2 n C . x n –1
n –1
(2n)!
(2n)!
Þ
Þ (n + 1) x ³ n
³
(n – 1)!(n + 1)!
n! n!
From (i),
.
(6 2)
\ The term independent of x =
–1/ 2 10
æ 10 - r r ö
- ÷
çè
3
2ø
Cr
r
. As 2 and
It is independent of x if r = 5
Tr +1 in ( x1/ 3 - x –1/ 2 )10 is
10
100 – r
14.
( x1/ 3 + 1) ( x 2 / 3 - x1/ 3 + 1)
1/ 3
= ( x + 1) – (1 + x -1/ 2 ) = x
(8 5)
5 are coprime, t r +1 will be rational if 100 – r is a
multiple of 8 and r is a multiple of 6.
Also 0 £ r £ 100.
\ r = 0, 6, 12, ............. , 96
Þ 100 – r = 4, 10, 16, ............, 100
But 100 – r is to be a multiple of 8.
So, 100 – r = 0, 8, 16, 24, ... 96.
The common terms in (1) and (2) are 16, 40, 64
and 88.
\ r = 84, 60, 36, 12 give rational terms.
\ The number of irrational terms
= 101 – 4 = 97.
r
æ cos a ö
(c) t r +1 = 10 Cr (x sin a )10 – r . çè x ÷ø .
( x1/ 2 - 1)
–
100
(a)
( x - 1)
1/ 2
t r +1 =
13.
n
Þ x³
n +1
From (ii),
2n
Cn . x n ³ 2 n Cn +1 . x n +1
(2n)!
(2n)!
x
Þ
³
(n –1)!(n + 1)!
n!n !
n +1
Þ x£
n
...(iii)
...(iv)
15.
10
1
C5 . 5 (sin 2a )5 £
2
10
C5 . sin 5 a .
10
1
C5 5
2
1
. 5
(5!) 2
(10)!
2
T
(c) If Tr + 1 ³ Tr , then r +1 ³ 1
Tr
25 – r + 1 . 3x
³1
r
2
26 – r .
Þ
3 ³1
r
Þ
(Q x = 2)
1
Þ 78 – 3r ³ r \ r £ 78 Þ r £ 19
2
4
Þ r = 19
Largest term = (r + 1)th term = 20th term
16. (c) Middle term in the expansion is
æ 10 ö
çè + 1÷ø
2
th
i.e., 6th term.
7
1
63
Þ 10C5 5 . x5 sin5 x =
8
8
x
63
1
Þ 252.sin 5 x =
Þ sin 5 x =
8
32
1
π
Þ sin x = \ x = nπ + (-1)n
2
6
Thus T6 = 7
Solutions
101
17. (d) The cofficients of the integral powers of x
are
40
Þ
C0 ,
40
C2 . 22 ,
(1 + 2) 40 =
40
40
C4 . 24 , ..... , 40 C40 . 2 40
C0 + 40 C1 2+ 40 C2 . 22 +......
+
Þ
(1 – 2) 40 =
Adding, 3 40
18. (b)
= 2[ n C0 2n + n C2 2n - 2.3 + n C4 .2n - 4.32 +.............]
= 2 × Integer = Integer
Q I + f + F is integer Þ f + F must be integer..
\ 0 £ f < 1 and 0 < F < 1 Þ 0 < f + F < 2
Þ f + F = 1 Þ F = 1 – f \ ( I + f ) (1 – f )
= (I + f ) F = (2 + 3)n (2 – 3) n = 1
40
40
C40 . 2 40
C0 – 40 C1 . 2
4
+ 40 C2 . 22 –......+ 40 C40 . 2 40
+ 1 = 2 × (required sum)
21. (1)
n
n
=
n
C 0 . C1. C 2 ..... C n
p( n )
=
n
+
1
p(n + 1)
C 0 .n +1 C1.n +1 C 2 ......n +1 C n +1
æ n Cn ö
1 æç n C 0 ö÷ æç n C1 ö÷
÷
.......ç
= n +1
÷
ç n +1 C
C 0 çè n +1 C1 ÷ø çè n +1 C 2 ÷ø
n +1 ø
è
n
é
C
r +1 ù
êQ n +1 r =
ú
C r +1 n + 1 úû
êë
19.
n+1
=
n!
\
p (2002) (2002) 2001
=
p (2001)
(2001)!
(n + 1)
(n + 1) n
(d)
n n (n - 1)(n - 2)
C1 C 3 C 5
+
+
+ ..... = +
2
4!
2
4
6
+
n (n - 1)(n - 2)(n - 3)(n - 4)
+ ......
6!
=
1 é (n + 1)n (n + 1)n (n - 1)(n - 2)
+
+
ê
n + 1 ë 2!
4!
or
and 0 £ f < 1. We note that (2 + 3 ) (2 – 3 )
n
= 1. So let us assume that F = (2 – 3) .
Clearly 0 < F < 1. Now,
I + f + F = (2 + 3) n + (2 – 3)n
k =0 è
4 4
øè
ø
4- k k
å
(3 + x)4 32
=
4!
3
4
(3 + x) = 256 or
x + 3 = 4 or
x=1
1
1
1 ù
é 1
11! ê
+
+
+ ... +
11!1!úû
ë1!10! 3!8! 5!6!
11
11
11
11
+
+
+ ... +
1!10! 3!8! 5!6!
11!
= 11C1 + 11C3 + 11C5 + 11C7 + 11C9 + 11C11
= sum of the even coefficient in the expansion of
(1 + x)11 = 210
Therefore K = 10
(6) (1 – 2x + 5x2 – 10x3) [C0 + C1x + C2x2 + ...]
= 1 + a1x + a2x2 + ...
=
=
20. (b) Given (2 + 3)n = I + f, where I is integer
æ 34- k ö æ x k ö 4!
å çç (4 - k )! ÷÷ çç k ! ÷÷ 4!
22. (4) Expression = (1 – x)5 .(1 + x)4 (1 + x2)4
= (1 – x) (1 – x2)4 (1 + x2)4 = (1 – x) (1 – x4)4
\ so the coefficient of x13 = – 4C3 (–1)3 = 4
23. (0) We have 3C0 – 5C1 + 7C2 + ...
+ (–1)n (2n+3)Cn
= 3C0 – 3C1 + 3C2 + ...+ (–1)n 3Cn – 2C1 + 4C2+...
+ (–1)n 2n Cn
n
= 3 [C0 – C1 + C2 + ....+ (–1) Cn]
–2[C1 – 2C2 + .... (–1)nnCn]
=3×0–2×0=0
24. (5) We have
ù
(n + 1)n (n - 1)(n - 2)(n - 3)(n - 4)
+ .....ú
6!
û
1 n +1
[ C 2 + n +1C 4 + n +1C6 + .....]
n +1
1
2n - 1
éë 2n +1-1 -n +1 C0 ùû =
=
n +1
n +1
ö æ xk ö
÷
÷
øè k! ø
Ck .3
x
(3 + x )4
=
4!
4!
k =0
According to the question.
=
1æ 1 öæ 2 ö
æ n +1ö
÷
÷........ç
֍
= ç
1 è n + 1ø è n +1 ø
è n -1ø
(n +1)!
k =0 è
4
n
=
æ 34 - k
å çç (4 - k )! ÷÷ çç
25.
Þ a1 = n – 2
and
a2 =
n(n - 1)
– 2n + 5
2
a21 = 2a2. Therefore, we have
(n – 2)2 = n (n – 1) – 4n + 10
Þ n2 – 4n + 4 = n2 – 5n + 10
\ n=6
MATHEMATICS
102
CHAPTER
Sequences and Series
9
1.
(d) We have
å n2 - 4 å n + nùû
= c2 é 4
ë
Sn = 13 + 3. 23 + 33 + 3. 43 + 53 + ............
Let n = 2 m
3
3
é 4n(n + 1)(2n + 1) 4n(n + 1)
ù
= c2 ê
+ nú
6
2
ë
û
3
Then S2m = ( 1 + 3 + 5 + ...... to m terms)
é 2(2n2 + 3n + 1) - 6(n + 1) + 3 ù
= c2 n ê
ú
3
ëê
ûú
+ 3(23 + 43 + 63 + ...to m terms)
= {13 + 23 + 33 + 43 + ...... + (2m – 1)3 + (2m)3}
– {23 + 43 + ....+(2m)3} + 3{23 + 43 + 63 + ..... +
(2m)3}
2
é 2m(2m + 1) ù
3
= ê
+ 8 × 2{ 13 + 23 + 3
ú
2
ë
û
4.
\
3
+ ... + m }
m2 (m + 1)2
= m (2m + 1) + 16.
4
2
é 4n2 - 1ù n(4n 2 - 1)c 2
= c2 n ê
ú=
3
êë 3 úû
(c) d = a2 – a1 = a3 – a2 = ... = an – an – 1
=
2
sin d [sec a1 sec a2 + sec a2 sec a3 + ...
+ sec an–1 sec an]
sin(a 2 - a1 ) sin(a 3 - a 2 )
+
+ ...
cos a1 cos a 2 cos a 2 cos a 3
+
n2 (2n 2 + 6 n + 5)
n
[Put m = ]
4
2
(b) 1, log9 (31 – x + 2), log3 (4.3x – 1) are in A.P.
Þ 2 log9 (31– x+2) = 1 + log3 (4.3x – 1)
Þ log3 (31 – x + 2) = log33 + log3 (4.3x – 1)
Þ log3 (31– x + 2) = log3 [3(4 × 3x – 1)]
Þ 31– x + 2 = 3 (4.3x – 1)
=
2.
Put 3x = t Þ
sin a 3 cos a 2 - cos a 3 sin a 2
+ ...
cos a 2 cos a 3
= tan a2 – tan a1 + tan a3 – tan a2 + ... +
tan an – tan an– 1 = tan an – tan a1
5.
3
(as 3 x ¹ - ve )
4
æ 3ö
è 4ø
3.
Then, Tn = S n - S n -1 = cn 2 - c (n - 1)2
= (2n - 1)c
\ Sum of squares of n terms of this A.P
=
åTn2 = å (2n - 1)2.c2
1
1
æ 1
ö
(b) y – x = 3 çè 2 + 2 + 2 + ...÷ø
2
4
6
1
1
æ 1
ö
x – z = çè 2 + 2 + 2 + ...÷ø
2
4
6
\ (y – x) = 3(x – z) Þ 4x = y + 3z
Þ x = log3 ç ÷ or x = log3 3 – log3 4
Þ x = 1 – log3 4
(c) Given that for an A.P, Sn = cn2
sin a 2 cos a1 - cosa 2 sin a1
cosa1 cos a 2
+
3
+ 2 = 12t - 3 or 12t2 – 5t – 3 = 0;
t
1 3
Hence t = - ,
3 4
Þ 3x =
=
sin(a n - a n -1 )
cos a n cos a n -1
x y z
Þ2 = +
3 6 2
6.
(b) Given
Þb–
a +b
b+c
, b,
are in A.P..
1– ab 1– bc
a+b
b+ c
=
–b
1– ab 1– bc
Solutions
7.
8.
103
– a(b 2 + 1) c(b 2 + 1)
=
Þ
1– ab
1– bc
Þ a + c = 2abc
Now, given quadratic equation is
2ac x2 + 2abc x + 2abc = 0
(Substituing a + c
= 2abc and then cancelling 2ac)
Þ x2 + bx + b = 0
(d) Given mid terms tn = 1 and tn + 1 = 7
\ tn + tn + 1= 8 = t1 + t2n
and tn + 1 – tn = 6 = d (common difference of
A.P.)
tn + tn + 1 = 8
\ a + (n – 1)d + a + nd = 8 \ a + 6n = 7
Now 4t1t2n = [(t1 + t2n)2 – (t2n – t1)2]
= 64 – 36(2n – 1)2 [as t2n – t1 = (2n – 1) × 6]
\ t1t2n = 16 – 9(2n – 1)2
\ 16 – 9(2n – 1)2 + 713 ³ 0
\–4£n£5 \ n=5
Hence, from a + 6n = 7, a = – 23
(d) Given, b 2 = ac, x =
Now,
=
a
c
2a
2c
+
=
+
x
y
a+b
b+c
2( ab + ac + ac + bc)
ab + ac + b 2 + bc
Again,
=
9.
a+b
b+c
,y=
2
2
=2
[Q b 2 = ac]
b b
1 ù
é 1
+ = 2b ê
+
ú
x y
ëa +b b + cû
2b(b + c + a + b)
ab + ac + b2 + bc
Þ (a – c)2 = 1 + 1= 2 Þ a – c = ± 2
but a < c Þ a - c = - 2
....(ii)
Solving (i) and (ii) we get a =
10. (c)
1 1
2
2
x
x
= 5 Þ r = 11- r
5
Since G.P. contains infinite terms
\ – 1< r<1
x
5
x
5
Þ -1 < 1 - < 1 Þ - 2 < - < 0
11.
Þ – 10 < x < 0. Þ 0 < x < 2 Þ 0 < x < 10.
5
(c) In the quadratic equation ax2 + bx + c = 0
b
c
D = b 2 - 4 ac and a + b = - , ab =
a
a
a 2 + b2 = (a + b)2 - 2ab
b 2 2c b2 - 2ac
=
= 2- =
a
a
a2
3
3
and a + b =-
b3
a
3
-
3c æ b ö
ç- ÷
a è aø
æ b - 3abc ö
÷
= -ç
a3
è
ø
ATQ a + b, a2 + b2, a3 + b3 are in G.P.
3
2
3
Þ - b , - b - 2ac , - (b - 3 abc) are in G.P.
a
a2
a3
=2
æ a cö æ b bö
\ ç + ÷ ç + ÷ = 4.
è x yø è x yø
(d) Given that a, b, c are in A.P.
Þ 2b = a + c
but given a + b + c = 3/2 Þ 3b = 3/2
Þ b = 1/2 and then a + c = 1
Again a2, b2, c2, are in G.P. Þ b4 = a2 c2
1
1
Þ b = ± ac Þ ac = or 4
4
and a + c = 1
Considering a + c = 1 and ac = 1/4
Þ (a – c)2 = 1 – 1 = 0 Þ a = c but
a ¹ c as given that a < b < c
\ We consider a + c = 1 and ac = – 1/4
2
æ b2 - 2 ac ö
b æ b3 - 3 abc ö
= ç
÷
÷
Þ ç
aè
a3
è a2
ø
ø
4
2
2
2
4
Þ b + 4a c – 4ab c = b – 3ab2c
Þ 4a2c2 – ab2c = 0 Þ ac D = 0
Þ c D = 0 ( Q In quadratic a ¹ 0)
12. (c) Let first term = a, common ratio = r, where
–1 < r < 1
Then,
2
....(i)
a3
a
= 24
= 2 and
1- r
1 - r3
MATHEMATICS
104
\
1- r3
(1 - r ) 3
(Q r > 1)
1
3
=
\
i.e., 1 – 2r + r2 = 3 (1 + r + r2 ) or 2r 2 + 5r + 2 = 0
\ r = -2 or
-1
2
16.
As – 1 < r < 1 \ we have r = -
1
2
13.
3 3 3
+ - + ...
2 4 8
(b) Given b2 = ac (Q a, b, c are in G. P.) and
2(log 2b – log 3c) = log a – log 2b
+ log 3c – log a
[Q given terms are in A. P.]
2
3c
æ 3c ö
æ 2b ö
Þ log ç ÷ = log ç ÷ Þ b =
è 2b ø
è 3c ø
2
\ p2 - q2 =
14.
17.
b2 + c2 – a 2
2bc
18.
\ S = 100.2100 - 2100 + 1 = 99.2100 + 1
(a) Consider two real numbers (a + b) and
(c + d). Using G. M. £ A. M.
M = (a + b) (c + d) > 0. \ 0 < M £ 1
19.
(a)
Since r > 1 \ ar 2 is greatest side
1- 5
1+ 5
1+ 5
<r<
Þ 1<r<
2
2
2
( a + b) + ( c + d )
2
2
M £ 2 Þ M£ 1
Also a, b, c, d are positive. Therefore
yn = 2( 2)n –1 = ( 2)n + 1
2
\ a + ar > ar Þ r2 – r – 1 < 0
( a + b)(c + d ) £
Þ
2.
(c) Let sides of triangle be a, ar, ar 2
1(2100 - 1)
- 100.2100 = 2100 - 1 - 100.2100
2 -1
We get
2, 2 2 , 2 4 , 2 8 , ................ is a G.P. where
15.
(a - c )(a + c + 2b)
= (a - c)b = p '2 - q '2
4
(d) Let S = 1 + 2.2 + 3. 22 + 4 .23 + 5.24 + ....
+ 100 . 299
\ 2S = 1.2 + 2.22 + 3.23 + ... + 99.299 + 100 . 2100
Subtracting, we get
–S = 1 + 1.2 + 1.22 + .... + 1.299 – 100. 2100
= (1 + 2 + 22 + ... + 299) – 100 . 2100
=
(c) y = 2 x , being in the first quadrant. The
sequence of x-coordinate 1, 2, 4, 8, ................
\ the sequence of y-coordinate
the common ratio is
(a + b)2 - (b + c )2
4
=
3b
9c
b2
=
=
2
4
c
\ a is the largest side.
9c 2
81
+ c2 – c2
4
16
=
= negative
3c
2´ ´c
2
\ A > 90° \ triangle is obtuse.
a+b
b+c
,q=
2
2
Again, p ' = ab and q ' = bc
Now, a =
Now, cos A =
\ [– r] = – 2
[r] + [– r] = 1 – 2 = – 1.
(c) We have 2b = a + c and a, p, b, q, c are in
A.P.
Þ p=
\ The series is 3 -
1+ 5
<–r<–1
2
[r] = 1. Also –
=
tn =
n+2
n(n + 1)
æ 1ö
.ç ÷
è 2ø
2(n + 1) - n æ 1 ö
.ç ÷
n(n + 1) è 2 ø
n
n
Solutions
105
1
=
n
æ1ö
ç ÷
è 2ø
n-1
n
1 æ1ö
–
.
.
n + 1 çè 2 ÷ø
ìï1 æ 1 ö° 1 æ 1 ö1 üï
Sn = å t n = í ç ÷ - ç ÷ ý
n=1
ïî1 è 2 ø 2 è 2 ø ïþ
n
ìï 1 æ 1 ö1 1 æ 1 ö 2 üï
+ í ç ÷ - ç ÷ ý + ..........
ïî 2 è 2 ø 3 è 2 ø ïþ
n
ìï 1 æ 1 ö n -1
1
1 æ 1 ö üï
+ í ç ÷
ý =1–
ç
÷
n è 2ø
n + 1 è 2ø ï
( n + 1)2 n
îï
þ
20. (a) As A.M. ³ G.M., we get
1æ 2 b b ö æ 2 b b ö
ç ax + + ÷ø ³ çè ax × × ÷ø
3è
2x 2 x
2 x 2x
Þ
1 æ 2 b ö æ ab ö
ç ax + ÷ø ³ çè ÷ø
x
3è
4
1/ 3
But the least value of ax2 +
æ ab2 ö
3ç
÷
è 4 ø
1/ 3
b
is c, therefore
x
³ c Þ 27ab2 ³ 4c3
21. (0.60) Let a = first term of G.P. and r = common
ratio of G.P.; Then G.P. is a, ar, ar2
a
= 20
1- r
Þ a = 20(1 – r)
Also a 2 + a 2 r 2 +
Þ
a2r4
b
36
and c = br Þ b = 36 and a =
r
r
r = 2, 3, 4, 6, 9, 12, 18
Let a =
Þ
æ 1ö
Also 36 ç1– ÷ is a perfect cube. Þ r = 4
è rø
Þ a + b + c = 9 + 36 + 144 = 189.
24. (3) We have
2 6 10 14
S = 1 + + 2 + 3 + 4 + .......¥
3 3
3 3
Multiplying both sides by
1/ 3
Given S ¥ = 20 Þ
From Eqs. (i) and (ii) , we get
r2 – 2r + 1 = 16r – 64 Þ r2 – 18r + 65 = 0
(r – 5) (r – 13) = 0
r = 5 or 13
If r = 5, then a(80 – 64) = 64 and hence a = 4. In
this case the numbers are 4,20,100 and their sum
is 124.
23. (189) log6 (abc) = 6 Þ (abc) = 66
... (i)
+ ... to ¥ =100
1
, we get
3
1
1 2 6 10
S = + 2 + 3 + 4 + ........¥
....(ii)
3
3 3
3 3
Subtracting eqn. (ii) from eqn. (i), we get
2
1 4 4
4
S = 1 + + 2 + 3 + 4 + ........¥
3
3 3
3 3
2
4 4
4 4
S = + 2 + 3 + 4 + ........¥
Þ
3
3 3
3 3
2
Þ
S=
3
a2
=100
1 - r2
... (ii)
Þ a2 = 100(1 – r)(1 + r)
2
2
From (i), a = 400(1 – r) ;
From (ii), we get 100(1 – r)(1 + r) = 400(1 – r)2
Þ 1 + r = 4 – 4r Þ 5r = 3 Þ r = 3/5 = 0.60
22. (124) Let b = ar and c = ar2. Given that a, ar, ar2
– 4 are in AP. Therefore
a + (ar2 – 64) = 2(ar)
...(i)
2
Þ a(r – 2r + 1) = 64
Again a, ar – 8, ar2 – 64 are in GP. Therefore
a(ar2 – 64) = (ar – 8)2
...(ii)
Þ a(16r – 64) = 54
....(i)
4
3 = 4´3 Þ S =3
1 3 2
13
25. (990) Let S1= 2 + 3 + 5 +9 + 16 +............+ xn
S1 =
2 + 3 + 5 + 9 + ..........x n -1 + x n
0 = 2 + [1 + 2 + 4 + 7 + ......
+ to (n - 1) term] - xn
\ x n = 2 + [1 + 2 + 4 + 7 + .....to (n - 1) terms]
Again let
S2 = 1 + 2 + 4 + 7 + ........+ tn–1
S2 =
1 + 2 + 4 + 7..... + t n - 2 + t n -1
0 = 1 + [1 + 2 + 3 + ....... + (n - 2) term] - t n -1
MATHEMATICS
106
(n - 2)(n - 1) n 2 - 3n + 4
t n -1 = 1 +
=
2
2
n -1
\ S2 =
1
3
å t n -1 = 2 Sn 2 - 2 Sn + 2S1
n =1
=
1 ( n - 1) n ( 2n - 1) 3 n ( n - 1)
+ 2( n - 1)
2
6
2
2
28.
é 2n 2 - n 3n
ù
= (n - 1) ê
- + 2ú
4
úû
ëê 12
=
1
1– | cos x |
=
n -1 é 2
2n - n - 9n + 24 ù
û
12 ë
\
=
n -1 é 2
n 3 - 6n 2 + 17n - 12
n - 5n + 12ù =
û
6 ë
6
Þ |cos x| =
\ x n = 2 + S2 = 2 +
n 3 - 6n 2 + 17n - 12
6
n 3 - 6n 2 + 17n
6
So,
x20 = 990
(5) Let k and k + 1 be removed. Then,
29.
n
(n + 1) - 2k - 1
105
2
=
n-2
4
4m 2 + 103(1 - m)
. Clearly
4
(1 – m) must be divisible by 4.
Let m = 1 + 4t, then we get k = 16t2 – 95t + 1 and
1£ k < n
27.
1
1
Þ cos x = ±
2
2
p 2p
1
,
Þ ( x1 + x2 ) = 1.
p
3 3
(6) For the given A.P., we have
2(2a + b) = (5a – b) + (a + 2b)
Þ b = 2a
...(1)
Also for the given G.P., we have (ab + 1)2 =
(a – 1)2 (b + 1)2
...(2)
Putting b = 2a from (1) in (2), we get a = 0, – 2, or
1
1
and b = 2a =
4
2
Hence, (a– 1 + b– 1) = 2 + 4 = 6.
(2) Let the two quantities be a and b
But a > 0, so a =
30.
G = ab ; p = a +
(kp – q) = ka +
Þ 1 £ 16t 2 - 95t + 1 < 8t + 2
Þ t = 6 and so, n = 50
(8) Since a, b, c, d are in A.P., we have
b – a = c – b = d – c = D (let common difference)
or d = a + 3D
Þ a – d = – 3d and d = b + 2D
or b – d = – 2D
Also c = a + 2D or c – a = 2D
\ Given equation 2(a – b) + k(b – c)2 +
3
=6
1– | cos x |
1
4
Þ 2n 2 - 103n - 8k + 206 = 0
Since n and k are integers, so n must be even,
say n = 2m, then k =
33/(1 – |cos x|) = 36 Þ
Þ x=
=
26.
(c – a)3 = 2(a – d) + (b – d)2 + (c – d)3
becomes – 2D + kD2 + (2D)3 = – 6D + 4D2 – D3
Þ 9D2 + (k – 4)D + 4 = 0
Since D is real, we have (k – 4)2 – 4(4) (9) ³ 0
or k2 – 8k – 128 ³ 0 or (k – 16) (k + 8) ³ 0
\ k Î(– ¥, – 8] È [16, ¥)
Hence, the smallest positive value of k = 16.
(1) x Î (0, p), implies |cos x| < 1. Thus,
S¥ = 1 + |cos x| + cos2 x + |cos3 x| + ... up to ¥
= 1 + |cos x| + |cos x|2 + |cos x|3 + ... up to ¥
–2
b–a
2(b – a )
;q=a+
3
3
k (b – a )
–a
3
(b – a )
b–a
= a( k –1) +
( k – 2)
3
3
kq – p = ak +
2k (b – a)
(b – a )
–a–
3
3
b–a
(2k – 1)
3
\ (kp – q) (kq – p) = ab
Þ a(k – 1) +
\k=2
Solutions
107
CHAPTER
Straight Lines
10
1.
(a) Given P º (a, b)
13 - 3
(x – 4)
7-4
Þ 3y – 10x + 31 = 0.
Now, AB is y – 3 =
x y
...(i)
+ =1
a b
If line (i) cuts x and y axes at A and B respectively,
then A = (a, 0) and B = (0,b).
Given line is
Also the area of D OAB = S i.e.
(7,13)
Y
1
ab = S
2
Ry = x
A (3,4)
Þ ab = 2S
Since line (i) passes through P( a , b )
P
a b
ab
a
+
=1 Þ
+
=1
a
2
S
b
a
A’
(4,3)
X
2
Þ a b – (2S)a + 2 a S = 0
Since a is real, 4S 2 – 8abS ³ 0 Þ S ³ 2ab
2.
Hence the least value of S = 2ab .
(b) a, b, c are the roots of equation
x 3 - 3 x 2 + 6x + 1 = 0.
So, a + b + c = 3, ab + bc + ca = 6
abc = – 1
Now, the centroid of the triangle is
4.
and
æ aö
b
Since AO isperpendicular to BC, (– 1) ç - ÷ = – 1
è ø
\ a =–b
A
(c) Consider a point A’ , the image of A in y = x
\ Coordinates of A’ = (4, 3)
[Notice that A and B lie to the same side with
respect to y = x].
Then PA = PA’. Thus, PA + PB is minimum,
if PA’+ PB is minimum, i.e. if P, A’, B are collinear.
O
=0
3.
B
+
0
-1
æ6 3 ö
=ç ,
÷ or (2, – 1)
è 3 -3ø
2x
1=
2y
æ ab + bc + ca a + b + c ö
,
3
3abc ÷ø
i.e. ç
è
3y
x+
1
1 1ö
æ
+ +
ç ab + bc + ca ab bc ca ÷
,
ç
÷
3
3
ç
÷
è
ø
æ 31 31 ö
It intersects y = x at ç , ÷ , which is the
è 7 7ø
required point P.
(c) Equation of AO is 2x + 3y – 1 + l (x + 2y – 1) = 0
Where l = – 1, since the line passes through
the origin. So, x + y = 0
ax + by - 1 = 0
C
Similarly, (2x + 3y – 1) + m (ax – ay – 1) = 0
will be the equation of BO for m = – 1.
BO is perpendicular to AC
ì (2 - a) ü æ 1 ö
Þ íý ç - ÷ = – 1. \ a = – 8, b = 8.
î 3+ a þ è 2ø
MATHEMATICS
108
5.
(d) The coordinates of A are (a, 0) and of B are
(0, b). If the coordinates of C are (x1, x1) then
area of the
D AOC =
æ6 ö
y = 6 intersects y = mx at Q ç , 6 ÷
èm ø
1
OA × x1 and the area of the
2
2
6 2
Thus PQ = æç - ö÷ + ( 6 - 2 )2 < 5
èm mø
1
OB × x1 .
2
According to the given condition,
D BOC =
2
æ4ö
Þ ç ÷ + 16 < 25
èmø
1
1
OA × x1 = 2 × OB × x1 Þ OA = 2OB
2
2
Þ a = 2b.
=
Ry
B ( 0, b)
2
4
3 1 3
æ 4ö
Þ ç ÷ < 9 Þ -3 < < 3 Þ - < <
è mø
m
4 m 4
\
x
8.
C
O
Þ
Þ
Þ
Þ
Þ
2b a
=
3
3
Hence the coordinates of C are
Þ x1 =
6.
ry+3x+2=0
AP BP CP 1
=
=
=
AQ BQ CQ 2
2AP = AQ Þ 4(AP)2 = AQ2
4[(x – 1)2 + y2] = (x + 1)2 + y2
4(x2 + 1 – 2x) + 4y2 = x2 + 1 + 2x + y2
3x2 + 3y2 – 8x – 2x + 4 – 1 = 0
3x2 + 3y2 – 10x + 3 = 0
10
x +1=0
...(1)
3
\ A lies on the circle given by (1). As B and C
also follow the same condition.
\ Centre of circumcircle of DABC = centre of
circle given by (1)
Þ
aö
.
3÷ø
ry
4y+x-14=0
7/2
9.
5/3
(a) Let P(1, 0) and Q(–1, 0), A(x, y)
Given:
A (a, 0)
x y
Equation of AB is therefore
+ =1
2b b
or x + 2y = 2b
Since C lies on it, x1 + 2 x1 = 2b
æ 2b 2b ö
æa
çè 3 , 3 ÷ø or çè 3 ,
(c)
4ö æ 4 ö
æ
m Îç - ¥, - ÷ È ç , ¥÷
è
3ø è 3 ø
3y-2x-5=0
x2 + y2 –
æ5 ö
= ç ,0÷ .
è3 ø
(b) Let C = (x1, y1)
A(– 3, 2)
1
–5/2 –2 –1 –2/3 O
1 2
–1
–2
4
5
7
£ b £ .
3
2
æ2 ö
(c) y = 2 intersects y = mx at P ç , 2 ÷
èm ø
From diagram it is clear that
7.
3
X
2
E
1
B
(– 2, 1)
D
æ x1 - 2 y1 + 1 ö
,
ç
÷
è 2
2 ø
æ x - 5 y1 + 3 ö
Centroid, E = ç 1
,
÷
3 ø
è 3
C
(x1, y1)
Solutions
109
Since centroid lies on the line
3x + 4y + 2 = 0
æ x -5ö
æ y1 + 3 ö
3ç 1
÷ + 4ç
÷+2 = 0
è 3 ø
è 3 ø
Þ 3x1 + 4y1 + 3 = 0
Hence vertex (x1, y1) lies on the line 3x + 4y + 3 = 0
10. (d) Circumcentre = (0, 0)
\
æ (a + 1) (a - 1)
Centroid = ç 2 , 2 ÷
è
ø
2
2ö
We know the circumcentre (O),
Centroid (G) and orthocentre (H) of a triangle
lie on the line joining the O and G.
HG
2
Also,
= Þ Coordinate of orthocentre
GO
1
3(a + 1) 2 3(a - 1)2
,
2
2
Now, these coordinates satisfies eqn given in
option (d)
Hence, required eqn of line is
(a – 1)2 x – (a + 1)2 y = 0
(a) The line passing through the intersection
of lines ax + 2by + 3b = 0 and
bx - 2ay - 3a = 0 is
b
,
a
since it is ^ to the line ax + by + c = 0 and it cuts
the x-axis at (2,0). Hence, the required line passes
12. (d) Slope of the line in the new position is
through (2, 0) and its slope is
y-0 =
b
( x - 2)
a
Þ ay = bx - 2b Þ ay - bx + 2b = 0
13. (d) Let L1 (x, y) = x – y – 1 and L 2 (x, y)
= 2 (x – y ) + 5
Then, from figure P(a, 2) lies below L 2 ,
So
2a - 4 + 5
1
<0 Þ a > -2
2
=
11.
ax + 2by + 3b + l (bx – 2ay – 3a) = 0
Þ (a + b l ) x + (2b – 2a l )y + 3b – 3 l a = 0
As this line is parallel to x-axis.
\ a + b l = 0 Þ l = – a/b
a
Þ ax + 2by + 3b – (bx – 2ay – 3a) = 0
b
2a 2
3a 2
=0
y+
Þ ax + 2by + 3b – ax +
b
b
æ
2a 2 ö
3a 2
=0
y ç 2b +
÷ + 3b +
ç
b ÷ø
b
è
2
æ 2
y ç 2b + 2a
ç
b
è
2
y=
2
-3(a + b )
2(b2 + a 2 )
=
ö
æ 3b2 + 3a 2 ö
÷ = -ç
÷
÷
ç
÷
b
ø
è
ø
-3
2
So it is 3/2 units below x-axis.
b
. Required eq. is
a
L 2 : 2(x – y) + 5 = 0
P (a, 2)
L1 : x - y - 1 = 0
Also P(a, 2) lies above L1 ,
So
a - 2 -1
>0 Þ a<3
-1
Taking intersection, we get -
1
<a<3
2
æ 1 ö
Þ a Î ç - , 3÷ .
è 2 ø
14. (b)
A(5, – 1)
H
B (– 2, 3)
(0, 0)
C (a, b)
MATHEMATICS
110
Let the third vertex of DABC be (a, b).
Orthocentre = H(0, 0)
Let A (5, – 1) and B (– 2, 3) be other two vertices
of DABC.
Now, (Slope of AH) × (Slope of BC) = – 1
Þ
æ -1 - 0 ö æ b - 3 ö
çè
֍
÷ = -1
5 - 0 ø è a + 2ø
2
a2
æ 2ö
Þ a2 = ç ÷ +
è 5ø
4
Þ a2 -
Þ b – 3 = 5 (a + 2)
Similarly,
(Slope of BH) × (Slope of AC) = – 1
Þ
Now, In DPCB, (PB)2 = (PC)2 + (CB)2
(By Pythagoras theoresm)
...(1)
a2 =
æ 3ö æ b + 1ö
-ç ÷ ´ç
= -1
è 2 ø è a - 5 ÷ø
a4 4
3a 2
4
Þ
=
=
4 25
4
25
16
Þa=
75
\ Length of Equilateral triangle ( a) =
Þ 3b + 3 = 2a – 10
Þ 3b – 2a + 13 = 0
On solving equations (1) and (2) we get
a = – 4, b = – 7
Hence, third vertex is (– 4, – 7).
15. (b)
...(2)
and L 2 (x, y) = 2x – 3y – 5
\ L1 (10, – 20) = 10 – 20 + 1 = – 9 Þ ‘–’ve
and L 2 ( 10, – 20) = 20 + 60 – 5 = 75 Þ ‘+’ve
Equation of the bisector will be
x + y +1
2
C
3x + 4y = 9
Shortest distance of a point (x1, y1) from line
ax1 + by1 - c
a2 + b2
Now shortest distance of P (1, 2) from 3x + 4y
= 9 is
PC = d =
3(1) + 4(2) - 9
2
3 +4
2
=
2
5
Given that DAPB is an equilateral triangle
Let 'a' be its side
then PB = a, CB =
a
2
æ 2 x - 3 y - 5ö
÷
13 ø
= -ç
è
Þ x ( 13 + 2 2 ) + y ( 13 – 3 2 ) + ( 13
B
ax + by = c is d =
4 3
15
16. (a) Let, L1 (x, y) = x + y + 1
\
P (1, 2)
A
16
4
3 4 3
=
´
=
75 5 3
15
3
–5 2 )=0
17. (c) If a point is equidistant from the two
intersecting lines, then the locus of this point is
the angle bisector of those lines.
Now, let (h, k) be the point which is equidistant
from the lines 4x – 3y + 7 = 0 and 3x – 4y + 14 = 0
Then 4h - 3k + 7 = ± 3h - 4k + 14
4 2 + ( -3) 2
32 + ( -4) 2
Þ 4h - 3k + 7 = ± (3h - 4k + 14)
Þ h + k – 7 = 0 and 7h – 7k + 21 = 0
Hence locus of (h, k) is x + y – 7 = 0
and x – y + 3 = 0
Solutions
111
18. (7) The given lines are
2x + y = 9/2
… (i)
and 2x + y = – 6
… (ii)
Signs of constants on R.H.S. show that two
lines lie on opp. sides of origin. Let any line
through origin meets these lines in P and Q
respectively then req. ratio is OP : OQ
Y
(0, 9/2)
B
P
C (–3, 0)
A X
O
x'
(9/4, 0)
Q
Ð POA = Ð QOC (ver. opp. Ð' s)
Ð PAO = Ð OCQ (alt. int. Ð' s)
\
DOPA ~ DOQC (by AA similarity)
OP OA 9 / 4 3
=
=
=
OQ OC
3
4
Req. ratio is 3 : 4.
= OP : OQ =
9
20
:
-6
5
1
5
æQ Perpendicular distance of
ç
ç ax + by = c from (x1 , y1 )is
ç
ax + by1 + c
ç
p= 1
ç
a 2 + b2
è
m
n
Þ m + n = 7.
19. (15) Let ABC be an equilateral triangle with base
BC.
So, AD ^ BC
A (2,–1)
ö
÷
÷
÷
÷
÷
ø
Let AB = BC = AC = x
So, in DABD , AD 2 + BD 2 = AB 2
BC ö
æ
÷
çQ BD =
2 ø
è
4
2
2
Þx=
=
15
K
15
K = 15
Þ x2 =
Þ
20. (0) Since
so
= 3: 4 =
5
=
2
Now in DOPA and DOQC,
\
2 - 2 -1
æ 1 ö
x2
÷÷ +
Þ çç
= x2
4
è 5ø
D (0, –6)
y'
\
=
Þ
3 .1 – 4 + 1 < 0 ,
3 sin q – cos q + 1 £ 0
1
3
1
sin q – cos q £ –
2
2
2
1
pö
æ
Þ sin ç q - ÷ £ –
2
6ø
è
Þ
11p
p
4p
7p
£ q £ 2p
Þ
£ q–
£
6
6
3
6
Þ maximum value of sin q is 0.
21. (4) Family of lines can be written as
(3x + 7 y + 11)sec q + cosec q (5 x - 3 y - 11) = 0
B
D
C
x+2y =1
Now, length AD =
2 + 2(-1) - 1
(1)2 + (2)2
Þ
(5 x - 3 y - 11) + tan q (3 x + 7 y + 11) = 0
So, for any value of q this line passes through
the point of intersection of lines
5 x - 3 y - 11 = 0and 3x + 7 y + 11 = 0 i.e from
P(1, – 2).
MATHEMATICS
112
Let A( x, y ) be a point on the line
x- y+3=0
Þ
k + h 5+ 4
=
=9
k –h 5–4
Applying cosines rule in DPAB
PB2 = PA2 + AB2 - 2PA. AB cos q ³ PA2 + AB2
A(1, – 2)
24. (5)
2x
3
( PA - AB) 2 £ PB 2
y=
+y
x–
Þ
=0
-2PA. AB = ( PA - AB )2
So, maximum value of
| PA - PB | is PB = 4 + 36 = 2 10.
22.
Þ k 2 = 40
(1) Let the equation of L2 be L1 + lL = 0
Þ
(1 + l)x + (2 + l)y + 3 + l = 0
Slopes of L2, L and L1 are –
1+ l
, – 1, – 1/2
2+l
Since L is the bisector of the angle between L1
and L2
1+ l
+1
2+ l
–1 + 1/ 2
Þl=–3
=
1+ l
1 + 1/ 2
1+
2+l
–
\
So the equation of L2 is y + 2x = 0
Þ m = – 2 and 812m2 + 3 = 812 × 4 + 3 = 3251
23. (9) tana, tanb, tanl are the roots of equation
t3 – 12t2 + 15t – 1 = 0 (given).
So, tana, tanb, tanl = 12 and tana, tanb,
tanl = 1. Also Stana tanb = 15
When A(tana, cota), B(tanb, cotb) and C(tanl,
cotl)
The centroid G(h, k)
æ tan a + tan b + tan l cot a + cot b+ cot l ö
,
=ç
÷
3
3
è
ø
æ 12
D tan a.tan b ö
=ç ,
÷ = (4, 5)
3
3(tan
a tan b tan l) ø
è
B
x + py = q
æ
q
2q ö
çè 2 p 1 , 2 p 1÷ø
C
æ 3 p + q q 3ö
çè p + 1 , p + 1÷ø
P is the orthocenter. Therefore, AP ^ BC
æ 1 ö æ 3 + 2ö
5
or ç – ÷ ç
= – 1 or
= 1 or p = 5
÷
è p ø è 2 –1 ø
p
Since BP ^ AC, we have
27 – 2q
=–1
18 + q
or q = 27 + 18 or
q = 45
\ p + q = 5 + 45 = 50
25. (4) Since y = x is the perpendicular bisector of
the side AB and A = (1, 2), we have B = (2, 1).
Since the image A'(x, y) of A in the angular bisector x – 2y + 1 = 0 lies on the line BC, we have
x –1 y – 2 – 2(1– 2(2) + 1) 4
=
=
=
1
–2
5
12 + 2 2
æ 9 2ö
Therefore, A' = ç , ÷
è 5 5ø
Since equation of BC is the equation of BA', we
have the equation of BC as
y–1=
1– (2 / 5)
( x – 2)
2 – (9 / 5)
Þ y – 1 = 3(x – 2)
Þ 3x – y – 5 = 0
So that a = 3, b = – 1. Hence, a – b = 4.
Solutions
113
CHAPTER
Conic Sections
11
1.
2.
(a) Since 4x + 3y – 4 = 0 is dividing the
circumference in the ratio 1 : 2, angle subtended
at the centre = 2 p / 3.
Also the perpendicular distance from the centre
of the given line is 5
Þ Radius = 10 Þ Equation of the circle is
Þ y = x + (5 2 – 4)
For no solution c > 5 2 – 4
5.
\ c Î (5 2 – 4, ¥ ).
(b) Equation of the given circle is
x 2 + y 2 + 2gx + 2fy + c = 0
x 2 + y 2 - 10 x - 6 y - 66 = 0.
(b)
y
C(4 cos q , 4 sin q )
......(i)
Y
given circle
P
R
A
B
O
(– 4, 0)
x
Q
O
3.
1
A = . 8 . 4 sinq = |16 sin q|
2
1 2 15
Now, sinq can be equal to
, ,...
16 16 16
i.e. there are 15 points in each quadrant.
(c) If (a, 0) is the centre C and P is (2, – 2),
then Ð COP = 45°.
Since the equation of OP is x + y = 0.
\
Equation of the chord of contact PQ, drawn from
the origin (0, 0) to the given circle will be
gx + fy + c = 0
.....(ii)
Eq. of any circle passing through the intersection
points of the given circle and the chord PQ can
be written as
2
2
( x + y + 2gx + 2fy + c ) + l (gx + fy + c) = 0
....(iii)
If this circle passes through the origin, then we
have,
c + l c = 0 gives l = – 1
Putting the above value of l in equation (iii)
gives the equation of the required circle as
OP = 2 2 = CP. Hence OC = 4
Y
O
C
A
B
circumcircle X
of DOPQ
X
x 2 + y 2 + gx + fy = 0
P
,
(2
–2
6.
(b)
)
The point on the circle with the greatest x
coordinate is B.
a = OB = OC + CB = 4 + 2 2 .
4.
2
(d) Since y = | x | + c and x + y 2 – 8 | x | – 9 = 0
both are symmetrical about y-axis we consider
the case x > 0, when the equations become
2
y = x + c and x + y 2 – 8x – 9 = 0 . Equation of
2
P (–2, –1) (0, –1)
Any line through ( – 2, – 1) is y + 1 = m ( x + 2 )
It touches the circle if
tangent to circle x 2 + y 2 – 8x – 9 = 0 parallel to
y = x + c is y = (x – 4) + 5 1 + 1
2
x + y =1
B
Þ m = 0,
4
3
2m - 1
1 + m2
=1
MATHEMATICS
114
4
\ Equation of PB is y + 1 = ( x + 2)
3
Þ 4 x– 3y+ 5=0
A point on PB is (– 5, – 5), (we can choose some
other point as well)
Its image by the line y = – 1 is P¢ ( – 5 , 3).
Hence equation of incident ray PP¢ is
3 +1
( x + 5) Þ
4 x + 3 y + 11 = 0
-5 + 2
(b) Circle : x2 + y2 + 3x = 0,
y -3 =
7.
æ 3 ö
Centre, B = ç – , 0÷
è 2 ø
Radius =
3
units.
2
B
æ 3 ö
çè – , 0÷ø
2
8.
O
X
y = mx + 1
y-intercept of the line = 1
\ A = (0, 1)
OA
Slope of line, m = tan q =
OB
1 2
Þ m= =
Þ 3m – 2 = 0
3 3
2
(b) Let the variable circle is
x 2 + y 2 + 2 gx + 2 fy + c = 0
It passes through (a, b)
\ a 2 + b2 + 2 ga + 2 fb + c = 0
2
2
(i) cuts x + y = 4 orthogonally
\ 2( g ´ 0 + f ´ 0) = c - 4 Þ c = 4
2
2
\ from (ii) a + b + 2 ga + 2 fb + 4 = 0
\ Locus of centre (–g,–f ) is
a 2 + b2 - 2ax - 2by + 4 = 0
or
2ax + 2by = a 2 + b 2 + 4
...(i)
...(ii)
r1 = 50 - l , r2 = 2
For exactly two common tangents we have
r1 - r2 < C1C2 < r1 + r2
Þ
50 - l - 2 < 3 2 < 50 - l + 2
Þ 50 - l - 2 < 3 2 or 3 2 < 50 - l + 2
50 - l < 4 2 or 2 2 < 50 - l
Þ 50 - l < 32 or 8 < 50 - l
Þ l > 18 or l < 42
Required interval is (18, 42)
10. (b) Equation of required circle :
S : (x – 1)2 + (y – 1)2 + l (x – y) = 0
S' : x2 + y2 – 2y – 3 = 0
Common chord of S = 0 and S¢ = 0 is S – S ' = 0
(l – 2) x – (l + 4) y + 5 = 0
Centre of S ' : (0, – 1) lies on common chord
Þ l =–9
9
S : (x – 1)2 + (y – 1)2 – 9 (x – y) = 0 Þ r =
2
11. (c)
2
y = 8x
Y
Y¢
Line :
(b) The equations of the circles are
x 2 + y 2 - 10 x - 10 y + l = 0
and x 2 + y 2 - 4 x - 4 y + 6 = 0
C1 = centre of (i) = (5, 5)
C2 = centre of (ii) = (2, 2)
d = distance between centres = C1C2
= 9 + 9 = 18
Þ
A(0, 1)
X¢
9.
A
2
2
2
x +y =3
B
...(i)
...(ii)
We
Þ
Þ
Þ
2
have : x + (8x) = 9
x2 + 9x – x – 9 = 0
x (x + 9) – 1 (x + 9) = 0
(x + 9) (x – 1) = 0 Þ x = –9, 1
for x = 1, y = ± 2 2 x = ± 2 2
L1 = Length of AB
(2 2 + 2 2) 2 + (1 - 1) 2 = 4 2
L2 = Length of latus rectum = 4a = 4 × 2 = 8
L1 < L2
12. (c) Equation of the tangent at P (x1, y1) to
y2 = 4ax is yy1 – 2ax – 2ax1 = 0
...(i)
Equation of the chord of y2 = 4a(x + b) whose
mid-point is (x', y') is
=
Solutions
115
yy' – 2ax – 2ax' – 4ab
= y'2 – 4a x' – 4ab or yy' – 2ax – (y'2 – 2a x') = 0
...(ii)
Equation (i) and (ii) represent the same line.
y1
2 ax
2a
=
=–
\
2
y'
2a
y ' - 2ax '
This gives y' = y1 and then 2ax1 = y'2 – 2ax'
= y12 – 2ax' = 4ax1 – 2ax' \ x' = x1
\ mid-point (x', y') = (x1, y1).
13. (d) The given curve is y = x2 + 6
Equation of tangent at (1, 7) is
1
...(i)
( y + 7) = x .1 + 6 Þ 2x - y + 5 = 0
2
As given this tangent (1) touches the circle
x2 + y2 +16x + 12y + c = 0 at Q
Centre of circle = (– 8, – 6).
=0
+5
y
–
2x
P(1,7)
Q
C
(–8,–6)
Then equation of CQ which is perpendicular to
(1) and passes through (– 8, – 6) is
1
y + 6 = – ( x + 8) Þ x + 2y + 20 = 0
...(ii)
2
Now Q is pt. of intersection of (i) and (ii)
\ Solving eqs (i) & (ii) we get ; x = – 6, y = – 7
\ Req. pt. is (– 6, – 7).
14. (a) Since S = (a, 0) = (1, 0), the circle is of the
form (x – 1)2 + y2 = r2
Suppose AB is a common chord. Since this is
equidistant from the focus and the vertex.
M(1/2,0) lies on AB and AB is double ordinate of
the parabola, let A = (1/2, y) so that
æ1ö
y2 = 4 ç ÷ Þ y = ± 2
è2ø
æ1
ö
æ1
ö
Þ A = ç , 2 ÷ and B = ç , - 2 ÷
2
2
è
ø
è
ø
Since DAMS is right-angled triangle, we have
1
9
SA2 = SM2 + MA2 = + 2 = = (Radius)2
4
4
Hence, the equation of the circle is
(x – 1)2 + y2 =
9
4
15. (d) As per the definition, the locus must be an
ellipse, with given points as foci and 10 as its
major axis. Since the line segment joining (2, –3)
and (2, 5) is parallel to y-axis, therefore, ellipse is
vertical.
4
\ 2 be = 8 and 2b = 10 Þ b = 5 and e =
5
2
\ a = b 2 (1 - e 2 ) = 9 and centre of the
ellipse is (2, 1)
\ Equation of the required ellipse is
( x - 2) 2
( y - 1)2
=1
+
9
25
x2
16. (a) For ellipse
2
a = 4, b = 3
4
æ3ö
+
2
y2
32
7
Þ
e = 1–ç ÷ =
4
è4ø
\
Foci are
(
=1
)
(
7, 0 and – 7,0
)
Centre of circle is at (0, 3) and it passes through
(±
7, 0 , therefore radius of circle
)
=
( 7)
2
+ ( 3) = 4
2
17. (d) x2 = 8y
...(i)
2
x
+ y2 = 1
3
...(ii)
8y
1
+ y 2 = 1 Þ y = – 3,
3
3
When y = – 3, then x2 = – 24, which is not possible.
From (i) and (ii),
2 6
1
, then x = ±
3
3
Point of intersection are
æ 2 6 1ö
æ 2 6 1ö
, ÷ and ç , ÷
ç
3 3ø
è 3 3ø
è
When y =
Required equation of the line, y Þ 3y – 1 = 0
1
=0
3
y2
x2
+
=1
... (i)
4
16
Equation of a circle centered at (1, 0) can be
written as (x –1)2 + y2 = r2
... (ii)
The abscissae of the intersection points of the
circle and the ellipse is given by the equation
18. (c) Given ellipse is
MATHEMATICS
116
( x - 1)2 +
See figure B (4, 2) is one end of the minor axis of
16 - x 2
2
=r
4
( x - 4)2 y 2
+
= 1 and (– 1, 2) is one
25
4
end of the major axis of the second ellipse.
Therefore,
AB = 5, OB = 16 + 4 = 20 , OA = 1 + 4 = 5
We have (OA)2 + (OB)2 = 25 = (AB)2
p
Therefore Ð AOB =
2
21. (1)
C
the ellipse
(4, 0)
(1, 0)
2
2
i.e. 4 ( x – 2x + 1) + 16 – x = 4 r 2
2
2
i.e. 3 x – 8x + 20 – 4 r = 0
If the circle lies inside the ellipse, then the roots
of the above equation must be imaginary or equal
Q
i.e. D £ 0 i.e. 64 + 12 (4 r 2 – 20) £ 0
Þ r= £
S
11
3
B
R
Hence, greatest value of r =
equation of required circle is
11
( x - 1)2 + y 2 =
3
11
and the
3
2
i.e. 3 ( x + y 2 ) – 6x – 8 = 0.
x2 y 2
+
=1
16 3
Now, equation of normal at (2, 3/2) is
16 x 3 y
= 16 - 3
2
3/ 2
13
Þ 8x – 2y = 13 Þ y = 4 x 2
13
Let y = 4 x - touches a parabola
2
y2 = 4ax.
We know, a straight line y = mx + c touches a
parabola y2 = 4ax if a – mc = 0
æ 13ö
\ a - ( 4) ç - ÷ = 0 Þ a = – 26
è 2ø
Hence, required equation of parabola is
y2 = 4 (– 26)x = – 104 x
y
20. (a)
19. (a) Ellipse is
A
B
O
A
P
Point A ( 33 + 3, 0) lies on the given circle,
x2 + y2 – 6x – 8y – 24 = 0
PQ and AB intersect inside the circle.
Let PR = a, RS = b, QS = c
Since PR × RQ = AR × RB Þ a(b + c) = 3 × 7
Also, QS × SP = 3 × 7 Þ c(a + b) = 3 × 7
Þ a = c \ PR/QS = 1
22. (11) The centre C of the circle = ( 5,7 ) and the
radius
= 52 + 7 2 + 51 = 5 5
PC = 122 + 52 = 13 Þ q = PA = 13 - 5 5
and p = PB = 13 + 5 3
\ G.M. of p and q
= pq = (13 - 5 5(13 + 5 5))
= 169 - 125 = 2 11 = 2 k Þ k = 11.
A
C (5, 7)
B
P (–7, 2)
(0, –1)
2
23. (4) Tangent to y = 8 (x + 2) is
x
y = m (x + 2) +
c = 2m +
2
m
1ö
2
c æ
= çè m + ÷ø
Þ
m
m
2
Solutions
117
1
c
³2 Þ
³ 2 Þ c³4
m
2
Þ The minimum value of c = 4.
24. (3) The locus of the point of intersection of
tangents to the parabola y2 = 4 ax inclined at
an angle a to each other is tan2a (x + a)2
= y2 – 4ax
Given equation of Parabola y2 = 4x {a = 1}
Point of intersection (–2, –1)
tan2a (–2 + 1)2 = (–1)2 – 4 × 1 × (–2)
Þ tan2a = 9 Þ tan a = ± 3 Þ |tan a| = 3
5 4
x2
y2
25. (27)
+
= 1 Þ e2 = 1 - =
9 9
9
5
2
Þ e=
3
æ 5ö
One end of latusrectum is ç 2, ÷
è 3ø
y
2x
æ 5ö
=1
+
Equation of tangent at ç 2, ÷ is
3
9
è 3ø
Q
m+
Q
2ö
æ
ç - ae, b ÷
ç
a ÷ø
è
R
F¢
2ö
æ
ç- ae, - b ÷
ç
a ÷ø
è
2ö
æ
ç ae, b ÷
ç
a ÷ø
è
C
F
O
P
1
9
27
Area of D CPQ =
×
×3=
2
2
4
\
Area of quadrilateral PQRS = 4 ×
27
= 27.
4
26. (4) Focus of the parabola y 2 = 4x is (1, 0)
So diagonals are focal chord.
AS = 1 + t 2 = c (say)
1
1
=1
Q + 25
c
-c
4
For c = 5, 1 + t 2 = 5 Þ t = ± 2
æ1 ö
A º ç , 1÷ , B º (4 , 4), C º (4, – 4) and
è4 ø
ö
æ1
D º ç , -1÷
ø
è4
AD = 2 and BC = 8, distance between AD and BC
15
=
4
\ Area of trapezium ABCD
1
15
75
= ( 2 + 8) ×
=
sq. units.
2
4
4
27. (2) Due to symmetry the desired area
1
= 4 ´ area of DS1OS3 = 4 ´ ae ´ be1
2
S
3
2ö
æ
ç ae, - b ÷
ç
÷
a
è
ø
S
25
25
c – c 2 Þ 4c 2 – 25 c + 25 = 0
=
4
4
5
Þ c= ,5
4
5
5
1
1
Þ t2 =
For c = , 1 + t 2 =
Þt =±
4
4
4
2
1
1ù
é 1
êëQ AS + CS = a úû
S2
S4
Where e1 is eccentricity of conjugate hyperbola
= 2 ´ 2e ´ 3e1 = 12ee1
Now b 2 = a 2 (e 2 - 1) Þ e 2 = 13/ 4
1
1
13
and 2 + 2 = 1 Þ e12 =
9
e
e1
\ Required area = 12 ´
28. (9)
P
A
2
O
D
C
13
13
´
= 26
2
3
C
y
B
S
S1
L
A Ö12
30°
N(4, 0)
B M(1, 0)
D
x=1
30°
Q(4 + 4Ö3, 0)
x
MATHEMATICS
118
Common chord of both the circles is x = 1.
Now, we have to find the ratio of areas of
equilateral triangles ANB and CQD.
Now in triangle OPN,
ON = OP cosec 30° = 2 × 2 = 4.
Area of triangle NAB
1
MN . AB = MN . AM = MN. MN tan 30°
2
= (ON – OM)2 tan 30°
1
9
sq. units.
= (4 - 1) 2
=
3
3
Now in triangle NLQ,
NQ = NL cosec 30° = 4 3.
1
Since area of tr iangle CQD = QM .CD
2
= QM .CM
For maximum length of the common chord, it
must pass through the centre of the smaller circle
( r2 , r2 ) , so
4r2 = r1 + r2 Þ
Þ
Þ
30.
1 57 + 24 3
=
sq. units.
3
3
57 + 24 3
So, ratio of area of trianlges =
.
9
(5) Let r be the radius of the circle. Its equation
is x 2 + y 2 - 2r ( x + y ) + r 2 = 0. Since it passes
through P(a , b)
a 2 + b 2 - 2r ( a + b ) + r 2 = 0
r1 = a + b + 2ab
...(1)
r2 = a + b - 2ab
Now, the equations of two circles are
x 2 + y 2 - 2r1 ( x + y ) + r12 = 0 and
x 2 + y 2 - 2r2 ( x + y ) + r2 2 = 0
Þ 2 ( r2 - r1 ) ( x + y ) + r12 - r2 2 = 0
CHAPTER
a 2 - 6 ab + b 2 = 0
6b ± 36b2 - 4b2
= (3 ± 2 2 ) b
2
a
= 3± 2 2
b
(5) The tangent at any point A(2sec q, tan q)
x sec q y tan q
= 1.
is given by
2
1
It meets the line x - 2 y = 0
2
x sec q x tan q
Þ
=1 Þ x =
sec q - tan q
2
2
2
1
æ
ö
Þ Q ºç
,
...(1)
÷
è sec q - tan q sec q - tan q ø
Also, the tangent meets the line x + 2 y = 0 at
R, so
x
x
sec q + tan q = 1
Þ
2
2
2
Þ x=
sec q + tan q
-1
2
æ
ö
...(2)
Þ Rºç
,
÷
è sec q + tan q sec q + tan q ø
Now,
2 2 + 12
2 2 + 12
(sec q - tan q) 2
(sec q + tan q) 2
Limits & Derivatives
12
lim f ( x) = lim {- h}cot{-h}
x ®0 -
\ lim f ( x ) does not exist.
h ®0
x ®0
= lim (1 - h) cot(1 - h) = cot 1
h ®0
lim f ( x ) = lim
x ®0+
= 3 Þ 2 ( a + b ) = 4 2 ab
= 22 + 12
Þ CQ.CR=5
Þ 2 ( x + y ) = r1 + r2
(d)
a=
CQ.CR =
The common chord is S1 - S2 = 0
1.
a + b - 2 ab
Þ
= (3 + 4 3) 2
Solving
a + b + 2ab
Þ ( a + b ) 2 = 8ab Þ
QM. QM tan 30° = (MN + NQ) 2 tan 30°
29.
r1
=3
r2
tan 2 {h}
h ®0 h 2
- [h]2
= lim
h®0
tan 2 h
h2
2.
=1
(a)
x
æ
4x +1 ö
æ x 2 + 5x + 3 ö
lim ç 2
÷ = xlim
çè1+ 2
÷
®¥
x + x + 2ø
x ®¥ çè x + x + 2 ÷ø
x
Solutions
119
( 4 x +1) x
é
x2 + x + 2 ù x2 + x + 2
êæ
4 x + 1 ö 4 x+1 ú
ú
= lim êç1 + 2
÷
x ®¥ ê è
x + x + 2ø
ú
êë
úû
4+
lim
2
4x + x
lim
3.
1
x
6.
x ®¥ 1+ 1 + 2
x x2
1
e x - sin x
[using L¢ Hospital's rule]
= 2 lim
x® 0 ( x - 2) x n - 3
For this limit to be finite, n – 3 = 0
Þ n=3
(c) lim [(sin x)1/x + (1/x)sin x]
x® 0
æ 1ö
1/ x
= lim (sin x ) + lim ç ÷
x®0
x®0 è x ø
2
= e4
= e x®¥ x + x + 2 = e
(b) According to the question
æ
æ
x2 ö
x3 ö
lim ç 3 –
÷ £ lim f ( x) £ lim ç 3 + ÷
12 ÷ø x®0
9 ÷ø
x ®0 ç
x ®0 ç
è
è
Þ (3 – 0) £ lim f ( x ) £ (3 + 0)
=0+
x ®0
Hence lim f ( x) = 3 (from Sandwitch Theorem)
=e
x®0
4.
(c)
Let y
1
k
k
k 1/ n
= lim k ((n + 1) (n + 2) ...(n + n) )
n®¥ n
1
Þ ln y = lim
n®¥ n
k
k
æ æ n + 1 ök
æn+2ö
æn+nö ö
ç ln ç
÷
+ ln ç
+ ... + ln ç
÷
÷
÷
ç è n ø
è n ø
è n ø ÷ø
è
k æ æ n + 1ö
æ n + 2ö
æ n + nö ö
= lim ç ln ç
÷ + ln çè
÷ + ... + ln çè
÷
n ø
n ø ÷ø
n ®¥ n è è n ø
n
å
= lim k .
n®¥
r =1
n
r =1
7.
xn
number
= lim
n
x (1 + cos x )
x®0
æ sin x ö æ e x - cos x ö æ
1 ö
= lim ç 2 ÷ . ç
÷ . çè
÷
n
2
x®0 è x
ø è x
ø 1 + cos x ø
2
2
= lim 1 .
x ®0
e x - cos x 1
.
2
xn - 2
(c) Limit is of the form 1¥ , so
tan
px
2a
ì æ aö ü
= lim í1 + ç1 - ÷ ý
è xø þ
x®a î
= lim
h®0
h
æ p ph ö
tan ç + ÷
a + h è 2 2a ø
= lim
h®0
h æ
æ ph ö ö
- cot ç ÷ ÷
ç
è 2a ø ø
a+hè
e
e
= lim
h® 0
e
= finite non-zero
(1 - cos x ) (1 + cos x )(e x - cos x )
= e0 = 1
tan
px
2a
p
æ h ö
tan ( a + h)
ç
÷
= lim
2a
e h® 0 è a + h ø
x
x®0
[Using L' Hospital rule]
sin x
lim
. tan x
x
x
0
®
e
aö
px
px
æ x - aö
= lim ç
tan
÷
x
2a ex®a è x ø
2a
e
Let x – a = h we get the limit
ln (n + r ) - ln n ö
÷ø
n
(cos x - 1) (cos x - e )
=e
-1/ x
lim
x ® 0 - cosec x cot x
æ
k
æ æ
1ö ö
æ 4ö
æ 4ö
= k . ç 2 ç ln 2 - ÷ ÷ = ln ç ÷ . Þ y = ç ÷ .
è eø
è è
è eø
2ø ø
(c) Given that,
lim
[Q | sin x | < 1 when x ® 0]
- log x
lim
x ® 0 cosec x
= lim ç1 - ÷ tan
x®a è
ø
k
5.
æ 1ö
lim sin x log ç ÷
è xø
e x®0
aö
æ
lim ç 2 - ÷
xø
x®a è
æ ln (n + r ) - ln n ö
çè
÷ø
n
æ
çè
å
n®¥
= k . lim
=
sin x
-h
p
1
. .
æ ph ö 2 a p / 2 a
(a + h) tan ç ÷
è 2a ø
2a
2
-2a
ph / 2a
.
= e pa = e p
= lim
eh®0 p(a + h) tan(ph / 2a)
8.
(a) Q ax 2 + bx + c = 0 has roots a and b
a b
then 2 + + c = 0
x
x
1
1
i.e., cx 2 + bx + a = 0 has roots and .
a
b
MATHEMATICS
120
1ö
æ
Þ cx 2 + bx + a = c ç x - ÷
è
aø
Now lim
x®
1
a
æ
1ö
x- ÷
èç
bø
cos2 (sin2(sin2).........(sin2(x)).....) ( x + 4 + 2)
.
p
x®0
ìï æ (x + 4) - 2ö üï
sin ípç
ý
÷
x
ø ïþ
îï è
æ (x + 4) - 2ö
çp
÷
x
è
ø
= lim
æ 1 - cos(cx 2 + bx + a ) ö
ç
÷
2(1 - ax ) 2
è
ø
1
ì 2 æ cx 2 + bx + a ö ü 2
ï sin ç
֕
2
è
øï
ï
= lim í
ý
2
x ®1/ a ï
(1 - ax )
ï
ï
ï
î
þ
=
10.
æ cx + bx + a ö
sin ç
÷
2
è
ø
= lim
(1 - ax)
x®1/ a
2
=
lim
x®1/ a
c æ 1 ö æ 1ö
çx- ÷ x2 è aø èç bø÷
cos2 x
é
ù
1
= t ³ 1ú
ê Putting
2
cos x
ë
û
1/ t
c æ 1ö
x2 çè bø÷
. lim
-a
x®1/ a
t 1/ t
= lim ( n )
t ®¥
1/ t
11.
(b)
ì log e n
lim í
n®¥ î log e (n - 1)
´
(b) Let P
cos2 (1 - cos 2 (1 - cos 2 (......cos 2 ( x))......
x ®0
ìï æ ( x + 4) - 2 ö üï
sin í p ç
÷ø ý
è
x
ïî
ïþ
2
2
2
2
cos sin (1 - cos (......cos ( x))......
= lim
x ®0
ïì æ ( x + 4) - 2 ö ïü
sin í p ç
÷ý
x
ø þï
ïî è
2
2
2
cos (sin (sin (......(sin 2 ( x))......)
= lim
x®0
ïì æ x + 4 - 2 ö ïü
sin íp ç
÷ýì
x
ø þï ï æ x + 4 - 2 ö üï
îï è
íp
÷ý
x
ìï æ x + 4 - 2 ö üï îï çè
ø þï
íp ç
ý
÷
x
ø þï
ïî è
t
éæ 1 öt æ 2 öt
ænö ù
êç ÷ + ç ÷ + ...... + ç ÷ ú
è n ø úû
êëè n ø è n ø
t
éæ 1 öt æ 2 öt
ænö ù
= n lim êç ÷ + ç ÷ + ...... + ç ÷ ú
t ®¥ êè n ø
ènø
è n ø úû
ë
= n (0 + 0 + ...... + 1)0 = n
c æ 1 1ö
2 çè a bø÷
c æ 1 1ö
= 1.
=
2a èç a bø÷
-a
9.
2
2
2
(b) lim éê11/ cos x + 12 / cos x + ..... + n1/ cos x ùú
û
x ®p / 2 ë
(1t + 2t + .......nt )1/ t
= tlim
®¥
ìc æ
1ö æ
1ö ü
sin í ç x - ÷ ç x - ÷ ý
è
ø
2
a
è
bø
î
þï
= lim
1ö
x®1/ a
æ
-a ç x - ÷
è
aø
æ c æ 1 ö æ 1ö ö
sin ç ç x - ÷ ç x - ÷ ÷
è 2 è aø è bø ø
cos 2 0 (2 + 2) 4
.
=
1
p
p
= lim
12.
´
log e ( n + 1)
log e n
log e (n + 2)
log e n k ïü
´ .... ´
ý
log e ( n + 1)
log e (n k - 1) ïþ
ìï log e n k üï
log e n
lim í
ý = k lim
log
(
n
1)
log
n®¥ î
n
®¥
e ( n - 1)
ï e
þï
1
Let n = , then
h
æ 1ö
log e ç ÷
è hø
- log e ( h)
= k lim
= k lim
æ 1ö
h® 0
h® 0 log e (1 - h) - log e ( h )
log e ç 1 - ÷
è hø
1
1
= k lim
=k×
=k
(1 - 0)
h®0 ì
log e (1 - h) ü
í1 ý
loge ( h) þ
î
(b) Q lim 1 {[12 x] + [22 x] + [32 x] + ... + [ n2 x]}
n®¥ n3
ì n 2 ü
æ n 2
ö
2
ï å [r x ] ï
ç å r x - {r x}÷
í r =1
ý
ç r =1
÷
= lim ï
ï = lim ç
÷ø
n ®¥ î
n 3 þ n®¥ è
n3
Solutions
121
æ n (n + 1)(2 n + 1)
ö
x.
n
{r 2 x}÷
ç
6
-å 3 ÷
= lim ç
n®¥
n3
r =1 n ÷
çè
ø
(1)(1)(2)
x
= x.
-0=
6
3
2
æxö
2 x
ç ÷ tan
a
2ø
è
2 x
2
= a lim
´
= lim
´ tan
2
x®0 g ( x )
x ®0 g ( x )
2
æxö
ç ÷
è2ø
x2
x2
4b
\ lim
=
x ®0 g ( x )
x ®0 4 g ( x )
a
lim f ( x) = lim f ( x )
13. (c)
x ®3-
Þ lim
= a lim
x ®3+
( x +3) x
(27) 27
-9
= lim l
3x - 27
x ®3+
2
æ x + 3 x -18 ö
ç
÷
9
- 1÷
32 ç 3
ç
÷ l
è
ø=
Þ lim
2
33 (3x -3 - 1)
x ®3x ®3-
Þ lim
1 - cos( x - 3)
( x - 3) 2
2
1 x + 3x -18 l
1
2
l
= Þ ×9 = Þ l = .
9( x - 3)
2 27
2
3
x®3 3
æ
æ p öö
14. (a) Here, lim ç sin 2 ç
÷÷
x ®0 è
è 2 - ax ø ø
æ p ö
sec 2 ç
÷
è 2-bx ø
æ p ö
sec 2
ö ü çè 2 -bx ÷ø
ì
æ p
= lim í1 - cos 2 ç
÷ý
x ®0 î
è 2 - ax ø þ
=e
ì
ü
1
ïï 2 æ p ö
ïï
lim - ícos ç
ý
÷×
è 2 - ax ø cos 2 æ p ö ï
x®0 ï
ç
÷
è 2 -bx ø þï
îï
=e
ì
pa ü
æ p ö æ p ö
ï 2sin ç 2 - ax ÷ cos ç 2- ax ÷´
ï
è
ø è
ø (2- ax )2 ï
ï
lim - í
ý
pb ï
æ p ö æ p ö
x ®0 ï
2sin ç
÷ cosç
÷´
2
ï
è 2 -bx ø è 2 -bx ø (2 -bx ) ïþ
î
[using L'Hospital's rule]
æ 2p ö
sin ç
÷
2
è 2 - ax ø × a × (2 - bx )
- lim
p
2
ö b (2 - ax )2
x ®0 æ
sin ç
÷
2 - bx ø
è
=e
=e
15. (c)
a (2 -bx )3
- lim ×
x ®0 b (2 - ax )3
lim
x ®0
=e
f (1 - cos x)
-
a
b.
g ( x ) sin 2 x
xö
æ
æ
2 xö
f ç 2 sin 2 ÷
ç 2sin ÷
2ø
2ø
è
è
= lim
´
x®0
x
x
xö
æ
ö
æ
öæ
g ( x ) ç 2 sin 2 ÷ 4 ç sin 2 ÷ç cos 2 ÷
2ø
2 øè
2ø
è
è
Now, lim
g (1 - cos 2 x)
x ®0
x
4
= lim
g (2sin 2 x)
x®0
x4
g (2sin 2 x) (2sin 2 x) 2
a
a
´
=
´4 = .
x ® 0 (2sin 2 x ) 2
x4
4b
b
= lim
16. (a) Let f ( x ) = lx 2 + mx + n
Þ f '( x) = 2lx + m
Now , f (1) = f ( -1) Þ l + m + n = l - m + n
Þ m=0
\ f '(x) = 2lx
\ f '(a1) = 2la, f '(a2 ) = 2la2 , f '(a3 ) = 2la3
As a1, a2 , a3 are in A.P..
\ f '(a1 ), f '(a2 ), f '(a3 ) are in A. P..
17. (b) Since, f(x) is a polynomial function
satisfying
æ1ö
f ( x) × f ç ÷ = f ( x ) +
è xø
æ1ö
f ç ÷,
è xø
\ f ( x) = x n + 1 or f ( x ) = - x n + 1
If f ( x) = - x n + 1, then f (4) = -4n + 1 ¹ 65
So, f ( x ) = x n + 1
Since, f(4) = 65
\ 4 n + 1 = 65
Þ n = 3 \ f ( x) = x3 + 1 Þ f '( x) = 3 x 2
\ f '(l1 ) = 3l12 , f '(l2 ) = 3l22 , f '(l3 ) = 3l32
Since, l1, l2, l3 are in GP.
\ f '(l1 ), f '(l2 ), f '(l3 ) are also in GP..
g ( x) f (a) - g (a) f ( x)
x-a
x®a
g (a + h) f (a) - g ( a) f (a + h)
= lim
h
h®0
[For x = a + h]
18. (5)
= lim
h®0
lim
g (a + h) f (a) - g (a) f (a) + g (a) f (a) - g (a) f (a + h)
h
MATHEMATICS
122
é g (a + h ) - g ( a ) ù
é f (a + h) - f (a) ù
lim f (a ) ê
g (a ) ê
ú - hlim
ú
= h ®0
h
h
ë
û ®0
ë
û
1- x
ïì - ax + sin ( x - 1) + a ïü1- x 1
lim í
=
ý
x ®1 ï
4
î x + sin ( x - 1) - 1 ïþ
22. (2)
= f (a) g' (a) – g (a) f ' (a) = 2 × 2 – (– 1) × 1 = 5
lim (sin x)
19. (1)
x®
=
e
tan x
p
2
2
lim
=
20. (0.50)
ì
sin( x - 1) ü
ïï - a +
ï
x -1 ï
Þ lim í
ý
sin ( x - 1) ï
x ®1 ï
1+
ïî
x - 1 ïþ
æ0
ö
çè form÷ø
0
2sin x cos x -cos x
- sin x
2
= e0 =1
Required limit,
(x +
l = lim
x ®¥
)
x+ x -x
x+ x+ x + x
Dividing numerator & denominator by
get
x ®¥
1+
x+ x
x
1+
x ®¥
21. (6) lim
x , we
=
2
\y=
1
x
=
{ f ( x )}3
1
1
= = 0.50
1+1 2
=1
ìï x 2 x 4
üï ìï
üï
x3 x5
x + ax í1 +
- ......ý - b í x - +
- ......ý
3! 5!
ïî 2! 4!
þï îï
þï = 1
Þ lim
3
x ®0
{ f ( x)}
æ a bö
1+ a - b æ a b ö
+ ç - + ÷ + x 2 ç - ÷ + .......
è 2! 3!ø
è 4! 5!ø
x2
x ®0
3
ì f ( x) ü
í
ý
î x þ
Þ 1 + a - b = 0 and -
Þa=-
x 5/ 3
x
5/ 3
+y
x
=
x+
or y 2 + ( x 5/ 3 ) y - x 5/ 3
+1
1
1
1+
+ 3/2 +1
x x
x (1 + a cos x ) - b sin x
x®0
Þ lim
1
æ -a + 1 ö
Þç
÷ = Þ a = 0 or 2
4
è 2 ø
\ Largest value of a is 2.
23. (1) Let
x
x
=
y=
3
1
x
x
x + 2/3
x+
3
3
x
x
x ....¥
+
x + x ....¥
x
1+
x
l = lim
= lim
1+ x
ìï a (1 - x ) + sin ( x - 1) üï
ý
Þ lim í
x ®1 ï
î ( x - 1) + sin ( x - 1) ïþ
1+ x
sin 2 x - sin x
lim
cos x
x®p
x® p
e 2
lim ( tan x (sin x -1))
x® p
=e 2
a b
+ =1
2! 3!
5
3
and b = - . Thus b – 3a = 6
2
2
=1
y
x2 /3
=0
- x5 / 3 ± x10 / 3 + 4 x5 / 3
2
- x 5 / 3 + x10 / 3 + 4 x5 / 3
2
4 x5 /3
=
2( ( x10/ 3 + 4 x 5/ 3 ) + x5 / 3 )
2
=
4 ö
æ
ç1 + 5/ 3 ÷ + 1
è x ø
2
2
\ lim y =
= = 1.
x ®¥
1+ 0 + 1 2
=
(Q y > 0)
a( x3 - 1) + ( x - 1) = 0
24. (1)
or ( x - 1)(ax 2 + ax + a + 1) = 0
a, b ¹ 1 so, a, b are roots of
ax 2 + ax + a + 1 = 0
a +1
a + b = -1, ab =
a
lim
x®
1
a
(1 + a) x3 - x2 - a
(e1-ax -1)( x -1)
= lim
x®
1
a
( x3 - x2 ) + a( x3 -1)
(e1-ax -1)( x -1)
Solutions
= lim
123
[x2 + a(x2 + x +1)]
(e1-ax -1)
1
x®
a
(1 + a)x2 + ax + a
1-ax
1
-1 ö
x® æ e
aç
ç 1 - ax ÷÷ (1 - ax)
è
ø
= lim
25. (36) Let 3x = t2
t2 +
lim
t ®3
éæ1+ a ö 2
ù
êç a ÷ x + (1)x +1ú
2
è
ø
û = lim a (abx -(a+b)x +1)
= lim a ë
1
1
(1- ax)
(1-ax)
x®
x®
a
1
x®
a
2.
3.
4.
5.
6.
7.
8.
t 4 - 12t 2 + 27
t -3
t ®3
= lim
1 3
t t2
Mathematical Reasoning
13
1.
- 12
(t 2 - 3)(t + 3)(t - 3)
t -3
t ®3
2
= (3 – 3) (3 + 3) = 36.
(1 - (a ) x )(1 - (b) x) a (a - b)
=
.
a
(1 - ax )
CHAPTER
t2
= lim
a
= lim a
27
(a) Inclusive “or”. 17 is a real number or a
positive integer or both.
(c) p ® (~ p Ú q) has truth value F.
It means p ® (~ p Ú q) is false.
It means p is true and ~ p Ú q is false.
Þ p is true and both ~ p and q are false.
Þ p is true and q is false.
(b) ~p : Ashok does not work hard
Use '®' symbol for then
(~p ® q) mean = If Ashok does not work hard
then he gets good grade.
(d) When p is false and q is true, then p Ù q is
false, pÚ ~ q is false.
(Q both p and ~q are false)
and q Þ p is also false,
only p Þ q is true.
(d) ~ p Ù q = ~ (q ® p)
(a)
p Ù q p Ú q ~(p Ú q) (p Ù q) Ù ~ (p Ú q)
p
q
T
T
T
T
F
F
T
F
F
T
F
F
F
T
F
T
F
F
F
F
F
F
T
F
\ (p Ù q) Ù (~ (p Ú q)) is a contradiction.
(b) (p Ù ~ q) Ù (~ p Ù q) = (p Ù ~ q) Ù (~ q Ù q)
= f Ùf = f
(By using associative laws and commutative laws)
\ (pÙ ~ q) Ù (~ p Ù q) is a contradiction.
(b) p Þ q is logically equivalent to ~ q Þ ~ p
\ (p Þ q) Û (~ q Þ ~ p) is a tautology but
not a contradiction.
9.
p Ù q (p Ù q)Þ p
T
T
p q
T T
(a)
T
F
F
F
T
F
F
F
F
T
T
T
\ ( p Ù q) Þ p is a tautology..
10. (b)
11.
pÞq
~q Þ ~p
p Þ q Û ~q Þ ~p
T
T
T
F
F
T
T
T
T
T
T
T
(b) The truth value of ~(~p) « p as follow
~(~p ) ~(~p) ®p p ®~(~p ) ~(~p )«p
p
~p
T
F
T
T
T
T
F
T
F
T
T
T
Since last column of above truth table contains
only T.
Hence ~ (~p) ® p is a tautology.
12. (c)
p
T
T
F
F
q p®q ~p ~pÚq (p®q)«~(pÚq)
T
T
F
T
T
F
F
F
F
T
T
T
T
T
T
F
T
T
T
T
MATHEMATICS
124
13.
14.
(d) p Þ (~ p Ú q) is false means p is true and
~ p Ú q is false.
Þ p is true and both ~p and q are false. Þ p is true
and q is false
(c) Let p : 2 + 3 = 5, q : 8 < 10
Given proposition is : p Ù q .
18.
(c) The inverse of the proposition
(p Ù ~ q) ® r is ~ (p Ù ~ q) ® ~ r
º ~ p Ú ~ (~q) ® ~ r º ~ p Ú q ® ~ r
19.
(a)
~ ((~ p) Ù q) º ~ (~ p) Ú ~ q º p Ú (~ q)
20.
(c)
~ ( p Þ q) º p Ù ~ q
\ ~ (~ p Þ ~ q) º ~ p Ù ~ (~ q) º ~ p Ù q .
Its negation is ~ ( p Ù q) = ~ p Ú ~ q
15.
16.
17.
\ we have 2 + 3 ¹ 5 or 8 </ 10.
(b) We know that p ® q is false only when p is
true and q is false.
So p ® (~ p Ú q) is false only when p is true and
(~ p Ú q) is false.
But (~ p Ú q) is false if q is false because ~ p is
false.
Hence p ® (~ p Ú q) is false when truth value of
p and q are T and F respectively.
(a) We know that the contrapositive of p ® q
is ~ q ® ~ p.
So contrapositive of p ® (~q ® ~r) is
~ (~q ® ~r) ® ~p º ~ q Ù [~ (~r)] ~p
Q ~ (p ® q) º p Ù ~q º ~ q Ù r ® ~p
(a) ~ [ p Ú (~ p Ú q)] º ~ p Ù ~ (~ p Ú q)
21.
22.
23.
24.
25.
º ~ p Ù (~ (~ p)Ù ~ q) º ~ p Ù ( p Ù ~ q ) .
Thus ~ (~ p Þ ~ q) º ~ p Ù q
(c) ~ [ (p Ú q) Ù (q Ú ~ r)]
º ~ ( p Ú q) Ú ~ (q Ú ~ r)
º (~ p Ù ~ q) Ú (~ q Ù r)
(c) Statement given in option (c) is correct.
~ [p Ú (~ q) ] = (~ p) Ù ~ (~ q) = (~ p) Ù q.
(d) Since ~ (p Ú q) º ~ pÙ ~ q
(By De-Morgans’ law)
\ ~ (p Ú q) ¹ ~ p Ú ~ q
\ (d) is the false statement.
(d) We know that ~ (p ® q) º p Ù ~q
\ ~((p Ù r) ® (r Ú q)) º (p Ù r) Ù [~(r Ú q)]
º (p Ù r) Ù (~r Ù ~q)
(d) Let P = A Í B, Q = B Í D, R = A Í C
Contrapositive of (P Ù Q) ® R is (® ~ R ® ~
(P Ù Q)).
~R®~ PÚ~Q
CHAPTER
Statistics
14
1.
(a) X =
n
n
X=
å r. nCr
r =1
n
å
n
r =1
2.
Cr
C1 + nC2 + .... + n Cn
n
=
n å n -1 Cr -1
r =1
n
å
r =1
n
Cr
n.2n -1
.
= n
2 -1
(a) Let the mean of the remaining 4
observations be x1 .
Then, M =
3.
(a)
Q
\
1. n C1 + 2. n C2 + 3. n C3 + .... + n. nCn
x=
a + 4 x1
nM - a
Þ x1 =
.
(n - 4) + 4
4
n1 x1 + n 2 x 2
n1 + n 2
x1 = 400, x 2 = 480, x = 430
5.
n1 (400) + n 2 (480)
Þ 30n1= 50n2
n1 + n 2
n1 5
=
n2 3
(c) Sum of 6 numbers = 30 × 6 = 180
Sum of remaining 5 numbers = 29 × 5 = 145
\ Excluded number = 180 – 145 = 35.
(c) Let the items be a1, a2, ........, an.
a + a + ........ + a n
then X = 1 2
.
n
Now, according to the given condition:
(a + 1) + (a 2 + 2) + ........ + (a n + n)
X= 1
n
1 + 2 + 3 + ........ + n
n(n + 1)
=X+
=X+
n
2n
(using sum of n natural nos.)
n +1
=X+
.
2
Þ
4.
430 =
Solutions
6.
7.
125
(a) Median is given as
N
-F
M = l+ 2
´C
f
where
l
= lower limit of the median - class
f
= frequency of the median class
N = total frequency
F = cumulative frequency of the class just
before the median class
C = length of median class.
Now, given, M = 25, N = 100, F = 45,
C = 20 – 30 = 10, l = 20.
\ By using formula, we have
50 – 45
25 = 20 +
´ 10
f
50
50
25 – 20 =
Þ5=
Þ f = 10.
f
f
101 + d (1 + 2 + 3 + ......+100)
(b) Mean =
101
d × 100 × 101
=1+
= 1 + 50 d
101 × 2
Q Mean deviation from the mean = 255
1
[|1 - (1 + 50d ) | + | (1 + d ) - (1 + 50d ) | +
Þ
101
| (1 + 2d )|
-(1 + 50d ) | +....+ | (1 + 100d ) - (1 + 50 d ) |] = 255
Þ
Þ
Þ
8.
2d [1 + 2 + 3 + ... + 50] = 101´ 255
50 ´ 51
= 101´ 255
2
101´ 255 25755
d=
=
= 10.1
50 ´ 51
2550
2d ´
(c) Clearly mean A = 0.
Standard deviation s =
2=
2
å ( x - A)2
2n
2
( a - 0) + ( a - 0) + ...(0 - a) 2 + ...
2n
a 2 .2n
= | a | . Hence | a | = 2.
2n
(c) Using,
=
9.
s=
22
n1 (s12 + d12 ) + n 2 (s22 + d 22 )
=
.
n1 + n 2
3
10. (a) We know that Q.D
11.
5
5
= ´ M .D. = ´12 = 10
6
6
3
3
\ S.D = ´ Q.D. = ´ 10 Þ S .D. = 15.
2
2
(d) Since S.D. £ Range = b – a
Var ( x ) £ (b – a )2 or (b - a)2 ³ Var ( x) .
12. (d) Since 0 < y < x < 2y
x
x
\ y>
Þ x- y <
2
2
\ x – y < y < x < 2x + y.
y+x
= 10
Hence median =
2
Þ x + y = 20.
...(i)
And range = (2x + y) – (x – y) = x + 2y.
But range = 28 \ x + 2y = 28
...(ii)
From equations (i) and (ii), x = 12, y = 8
( x - y ) + y + x + (2 x + y ) 4 x + y
=
\ Mean =
4
4
y
8
= x + = 12 + = 14.
4
4
13. (b) Let xi be n observations, i = 1, 2, ...n
Let X be the mean and M.D be the mean
\
deviation about X .
If each observation is increased by 5 then new
mean will be X + 5 and new M.D. about new
n
æ
xi ö
=
Q
Mean
ç
å
÷
mean will be M.D.
nø
è
i =1
14. (a) The first n natural numbers are 1, 2, 3,
................n
1 + 2 + 3 + 4 + ... + n
Their mean, x =
n
n(n + 1) n + 1
=
.
=
2n
2
[Q The sum of Ist n natural numbers is n (n + 1) ]
2
2
(
x
x
)
å i
Now, Variance = s 2 =
n
1é
2
2 ù
= å ( xi - 2 xx i + x )
û
në
=
å xi2 - 2 x å xi + x 2 .n
n
n
n
MATHEMATICS
126
=
å xi2 - 2 x 2 + x 2 = å xi2 - æ å xi ö
2
ç
÷
n
n
è n ø
[Since frequency of each variate is one]
n(n + 1)(2n + 1)
Q å xi2 =
.
6
\ Variance =
n(n + 1)(2n + 1) æ (n + 1) ö
-ç
è 2 ÷ø
6n
2
(n + 1)(2n + 1) (n + 1)2
=
6
4
æ 2n + 1 n + 1ö (n + 1)(n - 1)
= (n + 1) ç
÷ =
è 6
4 ø
12
n2 - 1
.
12
15. (b) Let the observations be x1, x2, ...., x20 and x
be their mean. Given that, variance = 5 and n = 20.
We know that,
=
1 20
2
å ( xi - x )
n i=1
( )
Variance s 2 =
i.e. 5 =
or
20
1 20
( x i - x )2
å
20 i =1
å ( xi - x )
2
= 100
...(i)
i =1
If each observation is multiplied by 2 and the
new resulting observations are yi, then
1
yi = 2xi i.e., xi = yi .
2
Therefore, y =
= 2.
1 20
å xi
20 i =1
1 20
1 20
yi =
å
å 2x i
n i =1
20 i=1
1
y.
2
On substituting the values of xi and x in eq. (i),
we get
i.e., y = 2x or x =
2
20
1 ö
æ1
å çè 2 yi - 2 y ÷ø = 100
i =1
i.e.,
20
å ( yi - y )
i =1
2
= 400 .
Thus, the variance of new observations
1
´ 400 = 20 = 22 ´ 5 .
20
(d) If initially all marks were xi then
=
16.
si2 =
å ( xi - x )2
i
.
N
Now each is increased by 10
2
å ëé( xi +10)-( x +10)ûù
i
å ( xi - x )2
i
=
= si2
N
N
Hence, variance will not change even after the
grace marks were given.
17. (a) C.V. (1st distribution) = 60, s1 = 21
C.V. (2nd distribution) = 70, s2 = 16
Let x1 and x2 be the means of 1st and 2nd
si2 =
distribution, respectively, Then
s
C.V. (1st distribution) = 1 ´ 100
x1
21
21
´ 100 = 35
´100 or x1 =
\ 60 =
60
x1
s2
´100
and C.V. (2nd distribution) =
x2
16
16
´ 100 or, x =
´ 100 = 22.85
i.e., 70 =
2
x2
70
18. (93.32) When each observation is multiplied
by 2, then variance is also multiplied by 2.
We are given, 2, 4, 5, 6, 8, 17.
When each observation multiplied by 2, we get
4, 8, 10, 12, 16, 34.
\ Variance of new series = 22 × Variance of given
data = 4 × 23.33 = 93.32.
19. (0) We know that,
s
Coefficient of variation = ´ 100
x
s1
´ 100
\ CV of 1st distribution =
30
s1
Þ 50 = ´ 100 [CV of 1st distribution = 50
30
(given)]
Þ s1 = 15
s2
´ 100
Also, CV of 2nd distribution =
25
s2
´100
Þ 60 =
25
60 ´ 25
Þ s2 =
Þ s2 = 15
100
Thus, s1 – s2 = 15 – 15 = 0.
Solutions
127
å x = 170, å x 2 = 2830 increase in
å x = 10 , then å x¢ = 170 + 10 = 180
Increase in å x 2 = 900 - 400 = 500 , then
å x¢2 = 2830 + 500 = 3330
1
i.e., 8.24 = [(3.4)2 + (2.4)2 + (1.6)2 + x2 + y2
5
– 2 × 4.4 (x + y) + 2 × (4.4)2 ]
or 41.20 = 11.56 + 5.76 + 2.56 + x2 + y2
– 8.8 ×13 + 38.72.
Therefore, x2 + y2 = 97
… (ii)
But from eq. (i), we have
x2 + y2 + 2xy = 169
… (iii)
From eqs. (ii) and (iii), we have
2xy = 7 2
… (iv)
On subtracting eq, (iv) from eq. (ii), we get
x2 + y2 – 2xy = 97 – 72
i.e. (x – y)2 = 25 or x – y = ± 5
… (v)
So, from eqs. (i) and (v), we get
x = 9, y = 4 when x – y = 5
or x = 4, y = 9 when x – y = –5.
Thus, the remaining observations are 4 and 9.
Required difference = 5.
24. (18) Var (1, 2, ....., n) = 10
20. (78)
Variance =
1
æ1
ö
x '2 - ç å x ' ÷
å
n
èn
ø
2
2
=
1
æ1
ö
´ 3330 - ç ´ 180 ÷ = 222 - 144 = 78.
15
è 15
ø
21. (8.25)
2
11 ×12.23
æ 11.12 ö
-1 ç
-1 ÷
2
6
s =
-ç 2
÷ = 8.25
10
è 10 ø
22. (5) We have,
CV of 1st distribution (CV1) = 50.
CV of 2nd distribution (CV2) = 60
s1 = 10 and s2 = 15.
s
We know that, CV = ´ 100
x
s
10
\ CV1 = 1 ´ 100 Þ 50 = ´ 100
x1
x1
2
Þ
2
(n + 1)(2n + 1) æ n + 1 ö
-ç
÷ = 10
6
è 2 ø
Þ n2 – 1 = 120
Þ n = 11
Var (2, 4, 6, ....., 2m) = 16 Þ Var (1, 2, ....., m) = 4
Þ m2 – 1 = 48 Þ m = 7
Þ m + n =18
25. (52) Mean
Þ
10 ´ 100
= 20 .
50
s
Also, CV2 = 2 ´ 100
x2
Þ x1 =
Þ 60 =
15 ´100
x2
15 ´100
Þ x 2 = 25.
60
Thus, x 2 - x1 = 25 - 20 = 5.
23. (5) Let the other two observations be x and y.
Therefore, the series is 1, 2, 6, x, y.
2
Also, variance (s ) = 8.24 =
1
n
5
å ( xi - x )2
i =1
Þ
Variance = s2 =
s2 =
… (i)
3 + 7 + 9 + 12 + 13 + 20 + x + y
= 10
8
x + y = 16
=x =
Þ x2 =
1+ 2 + 6 + x + y
Now, mean ( x ) = 4.4 =
5
or 22 = 9 + x + y.
Therefore, x + y = 13
12 + 2 2 + ..... + n2 æ 1 + 2 + ..... + n ö
-ç
÷ = 10
n
n
è
ø
...(i)
S ( xi )2
- ( x )2 = 25
8
9 + 49 + 81+144 +169 + 400 + x 2 + y 2
- 100 = 25
8
Þ x2 + y2 = 148
...(ii)
From eqn. (i), (x + y)2 = (16)2
Þ x2 + y2 + 2xy = 256
Using eqn. (ii), 148 + 2xy = 256
Þ xy = 52
MATHEMATICS
128
CHAPTER
Probability-1
15
1.
2.
(d) Total number of outcomes
S = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3),
(2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}
n(S) = 16
Number of favourable outcomes
E = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (4, 4)}
n(E) = 7
n(E) 7
=
Required probability =
n(S) 16
(b) Given equation
100
x+
> 50 Þ x 2 - 50x + 100 > 0
x
So, for each ai Î A , there are four possibilities.
\ Total no. of cases = 4 × 4 × ............ × 4 (n times) = 4n
Further out of above four possibilities, first three
satisfy ai Î P È Q
So, the number of cases, when exactly r element
of A belong to P È Q = n Cr (3)r
\ Required probability =
5.
Þ ( x - 25) 2 > 525
Þ x - 25 < - (525) or x - 25 > (525)
Þ x < 25 - (525) or x > 25 + (525)
As x is positive integer and
(525) = 22.91 , we
must have x £ 2 or x ³ 48
Let E be the event for favourable cases and S be
the sample space.
\ E = {1, 2, 48, 49, ......100}
3.
\ n(E ) = 55 and n(S) = 100
Hence the required probability
n( E) 55 11
P(E) =
=
=
.
n( S ) 100 20
(b) Total number of combinations of numbers
on the cube and the tetrahedron = 6 × 4 = 24
Favourable number of ways of getting a sum
not less than 5
= sum of coefficients of x6, x7, .... x10 in the product
2
3
4
5
6
= (x + x + x + x + x + x )
( x + x 2 + x3 + x 4 )
= ( x 2 + 2 x 3 + 3 x 4 + 4 x5 + 4 x 6
4.
+ 4 x7 + 3x8 + 2 x9 + x10 )
= 4 + 4 + 4 + 3 + 2 + 1 = 18
18
3
\ Required probability =
=
6´ 4
4
(b) Let A = {a1, a2, a3, ........... an}. For any
ai Î A , 1 £ i £ n , we have following choices
(1) ai Î P and ai Î Q (2) ai Î P and ai Ï Q
(3) ai Ï P and ai Î Q (4) ai Ï P and ai Ï Q
n
Cr (3)r
4n
(a) Since the chairs are numbered, so for
counting of total number of cases it is equivalent
to linear permutation. Hence, total number of
cases = 10!
If two particular persons A and B sit together
then the total number of linear arrangements =2!
9!. Consider one of such arrangements in which
the arrangement started at chair 1 (C1) and ends
at chair 10 (C10).
C1 – C2 –C3 – ......... –C9 –C10
If two persons sit at C1 and C10 then it will lead
to 2! 8! new arrangements. So the favourable
number of cases = 2! 9! + 2! 8! = 2!8! (10)
2!8!(10) 2
=
10!
9
(b) Total ways of distribution = 45 Þ n(S) = 45
Total ways of distribution so that each child get
\ Probability =
6.
7.
atleast one game = 45 - 4 C1 35 + 4C2 25 - 4 C3
= 1024 - 4 ´ 243 + 6 ´ 32 - 4 = 240
n(E) = 240
n( E ) 240 15
=
= .
required probablity =
64
n(S )
45
(d) A chess board is a square divided into 64
equal squares.
In 1st diagonal we have only 1 square
In 2nd diagonal we have only 2 squares
In 3rd diagonal we have 3 squares so selection
can be done in 3C3 ways
In 4 diagonal we have 4 squares so selection
can be done in C34 ways
And so on
Hence the total number of ways in which 3
squares can be chosen
2(3C3 + 4C3 + 5C3 + 6C3 + 6C3 + 7C3) + 8C3
[Note that we do not have 2 × 8C3]
Hence the total number of favourable ways m
Solutions
129
= 4(3C3 + 4C3 + 5C3 + 6C3 + 7C3) + 2 × 8C3 = 392.
64 × 63 × 62 64
And total number of ways =
C3
1× 2 × 3
8.
9.
= 32 × 21 × 64
Hence the required probability
m
392
7
=
= =
n 32 × 21 × 62 744
(c) The total numbers of arrangements is
11!
11!
=
2!2!2! 8
The number of arrangements in which C, E, H,
6!
11!
I, S appear in that order 11 C5
=
2!2!2! 8 × 5!
11!
1
\ Required Probability = 8 × 5! =
11! 120
8
(c) Player A can win if A throws (1, 6) or (6, 1)
and B throws ((1, 1), (2, 2), (3, 3), (4, 4), (5, 5) or
(6, 6)). Thus the number of ways is 12.
Similarly the number of ways in which B can win
is 12.
Total number of ways in which either A wins or B
wins = 24.
Thus the number of ways in which none of the
two wins = 64 – 24.
\ The required probability =
6 4 - 24
6
4
=
53
.
54
10. (c) Value 50p. 25p. 10p.
No.
2
5
15
5 coins out of 22 can be chosen in 22C2 ways.
Let x be the no. of 50 P coins selected. y be the
no. of 25p. coins selected and z be the no. of 10
p. coins selected. We desire that 50x + 25y + 10z
< 150 or 10x + 5y + 2z < 30 where 0 £ x £ 2; or
0 £ y £ 5, 0 £ z £ 14 and x + y + z = 5. Thus
following cases are impossible.
Case (i) : x = 2, y = 2, z = 1
No. of ways of such selections
= 2C2 ´ 5C2 ´ 15C1 = 1´10 ´15 = 150.
Case (ii) : x = 1, y = 4, z = 0
No. of ways of such selections = 2C1 ´ 5C4 = 10.
Case (iii) : x = 2, y = 3, z = 0. No. of ways of such
selection = 2C2 ´ 5C3 ´ 15C0 = 1´10 = 10.
\ Total = 150 + 10 + 10 = 170 ways are to be rejected.
\ Required probability = 1 11.
170
22
C5
.
log b
log a
Let b = 2m and a = 2n where m and n denote the
exponents on the base 2 in the given set then
(b)
log a b =
m
n
Therefore, logab is an integer only if n divides m.
Now, total no. of ways m and n can be chosen
= 25 × 24 = 600.
For favourable cases
Let n = 1, So m can take values 1, 2, 3, 4,
5, 6, ..., 24
= 24
if n = 2 m = 4, 6, 8, 10, 12, 14, 16, 18,
20, 22, 24
= 11
n = 3m = 6, 9, 12, 15, 18, 21, 24
=7
n = 4m = 8, 12, 16, 20, 24
=5
n = 5m = 10, 15, 20, 25
=4
n = 6m = 12, 18, 24
=3
n = 7m = 14, 21
=2
n = 8m = 16, 24
=2
n = 9, 10, 11, 12 m = 1 for each
=4
62
log a b =
Hence, required probability = 62 = 31 .
600 300
12. (d)
n(S) = Number of total ways
14!
= 14 P12 =
= 7 ´ 13!
2!
The girls can be seated together in the back seats
leaving a corner seat in 4 × 3! = 24 ways and the
boys can be seated in the remaining 11 seats in
11! 1
11
= ´ 11! ways
P9 =
2! 2
\ n(E) = Number of favourable ways
1
= 24 ´ ´11! = 12!
2
n( E )
12!
1
=
=
The required probability =
n(S ) 7 ´13! 91
13. (a) We know that P (exactly one of A or B occurs)
= P (A) + P (B) – 2P (A Ç B).
Therefore, P (A) + P (B) – 2P (A Ç B) = p … (i)
Similarly, P (B) + P (C) – 2P (B Ç C) = p … (ii)
and P (C) + P (A) – 2P (C Ç A) = p
…(iii)
MATHEMATICS
130
Adding (i), (ii) and (iii) we get
2 [P (A) + P (B) + P (C) – P (A Ç B)
– P (B Ç C) – P (C Ç A)] = 3p
Þ P (A) + P (B) + P (C) – P (A Ç B)
– P (B Ç C) – P (C Ç A) = 3p/2 … (iv)
We are also given that,
P (A Ç B Ç C) = p2
… (v)
Now, P (at least one of A, B and C)
= P (A) + P (B) + P (C) – P (A Ç B) – P (B Ç C)
– P (C Ç A) + P (A Ç B Ç C)
3p
3 p + 2 p2
+ p2 =
[using (iv) and (v)]
2
2
14. (d) Since, one and only one of the three events
E1, E2 and E3 can happen, therefore P (E1) + P
(E2) + P (E3) = 1
.....(i)
Q Odds against E1 are 7 : 4
4
4
Þ P(E1 ) =
=
.....(ii)
4 + 7 11
Q Odds against E2 are 5 : 3
3
3
Þ P (E 2 ) =
=
......(iii)
3+5 8
From (i), (ii) and (iii), we have,
4 3
+ + P(E 3 ) = 1.
11 8
4 3 88 - 32 - 33
Þ P(E3 ) = 1 - - =
11 8
88
23
23
=
=
88 23 + 65
Hence odds against E3 are 65 : 23.
15. (a) Let p1, p2 be the chances of happening of
the first and second event, respectively; then,
according to the given conditions, we have
=
p1 =
Þ
p22
1 - p22
p22
1 - p1 æ 1 - p2 ö
=ç
and
÷
p1
è p2 ø
3
æ 1 - p2 ö
2
=ç
÷ Þ p2 (1 + p2 ) = (1 - p2 )
p
è 2 ø
1
1
and so p1 = .
3
9
(c) Let P(A) = probability that Ashmit will pass,
and P(B) = probability that Bishmit will pass,
then from the given information
P(A) = 0.5, 0.1 £ P(A Ç B) £ 0.3.
We have to find the value of P(B).
Since P(A È B) = P(A) + P(B) – P(A Ç B)
Or P(B) = P(A È B) + P(A Ç B) – P(A)
Þ 3 p2 = 1 Þ p2 =
16.
3
For maximum value of P(B) both P(A È B) and
P(A Ç B) should be maximum, maximum value of
P(A È B) is 1 and maximum value of P(A Ç B)
= is 0.3.
Hence P(B) = 1 + 0.3 – 0.5 = 0.8.
Similarly For minimum value of P(B) both P(A È B)
and P(A ÇB) should be minimum, minimum value
of P(AÈB) = P(A) = 0.5 and minimum value of
P(A Ç B) = is 0.1.
Hence P(B) = 0.5 + 0.1 – 0.5 = 0.1.
So required range is (0.1, 0.8).
17. (a) Total number of outcomes = 52C4 = 270725
E = Event of exactly two spade cards and exists
two aces.
A = Event of 1 spade ace, 1 non-spade ace and 1
spade card and 1 non-spade card
B = Event of 2 non-spade aces and 2 spade cards.
P ( A)
1( 3C1 )( 12 C1 )( 36C1 )
P( B) =
52
3
C4
C2 ´ 12 C2
52
C4
=
=
3 ´ 12 ´ 36
1296
=
270725
270725
3 ´ 66
198
=
270725 270725
Now, E = A È B and A Ç B = f. This implies
1494
.
P ( E ) = P ( A È B) = P ( A) + P ( B ) =
270725
18. (0.80)
Total no. of arrangements of the letters
10!
of the word UNIVERSITY is
.
2!
No. of arrangements when both I's are together
= 9!
So. the no. of ways in which 2 I’s do not together
10!
- 9!
=
2!
10!
- 9!
10!- 9! 2!
2!
=
\ Required probability =
10!
10!
2!
10 ´ 9!- 9!.2! 9![10 - 2]
8 4
=
=
=
= = 0.80
10!
10 ´ 9!
10 5
19. (0.50)
Total number of numbers = 4! = 24
For odd nos. 1 or 3 has to be at unit's place
If 1 is at unit place, then total number of numbers
= 3! = 6
And if 3 is at unit place, then total number of
numbers = 3! = 6
\ Total number of odd number = 6 + 6 = 12
12 1
= = 0.50
\ Required probability =
24 2
Solutions
20. (0.33)
=
131
Total number of cases =
3n
C2
3n(3n - 1)
2
Now x3 + y3 = ( x + y ) ( x 2 - xy + y 2 )
x3 + y3 is divisible by 3 if 3 divdes x + y or divides
x2 – xy + y2 we arrange the 3n numbers in 3
sequences.
A : {1, 4, 7, ............., 3n – 2}
B : {2, 5, 8, .............., 3n – 1}
C : {3, 6, 9, ............., 3n}
Clearly we must choose either one number from
the first sequence and other number from the
second sequence or both numbers from the third
sequence only.
\ Number of favourable cases
n (n - 1) n
= (3n - 1)
= n × n + n C2 = n 2 +
2
2
n
(3n - 1)
1
2
= = 0.33
\ Required probability =
3n(3n - 1) 3
2
21. (3) Total no. of arrangements = 15!
Extreme chairs are occupied by girls, thus there
are four gaps among 5 girls where boys can be
seated. Let the number of boys in these four gaps
be 2x + 1, 2y + 1, 2z + 1 and 2t + 1, then
2x + 1 + 2y + 1 + 2z + 1 + 2t + 1 = 10
Þ x+y+z+t=3
Where x, y, z, t are integers and
0 £ x £ 3, 0 £ y £ 3, 0 £ z £ 3, 0 £ t £ 3
\ The number of ways of selecting positions
for boys
= coefficient of x3 in (1 + x + x2 + x3)4
4
æ 1 - x4 ö
= coefficient of x3 in ç
÷
è 1- x ø
= coefficent of x3 in (1 – x4)4 (1 – x)–4= 6C3 = 20
\ Number of arrangements of boys and girls
with given condition = 20 × 10 ! × 5 !
\ Required probability = 20 ´ 10!´ 5! = 20
15!
3003
n
3003
Þ
=
=3
1001 1001
22. (3) Let a and b be the roots of the quadratic
equation. According to question,
a + b = a 2 + b2 and ab = a 2b2
Þ ab(ab - 1) = 0 Þ ab = 1 or ab = 0
Þ a = 1, b = 1; a = w, b = w2
[Cube roots and unity]
a = 1, b = 0, a = 0, b = 0
\ n( S ) = Number of quadratic equations which
are unchanged by squaring their roots = 4
and n(E) = Number of quadratic equations have
equal roots = 2.
n( E ) 2 1 p
= = = .
\ Required probability =
n(S ) 4 2 q
Þ p + q = 3.
23. (4) The number of ways to answer a question
= 25 – 1 = 31.
i.e. In 31 ways only one correct.
Let number of choices = n
n 1
>
Now, according to the question
31 8
31
Þn>
Þ n > 3.8 \ Least value of n = 4.
8
24. (5) Let S be the sample space and E be the event
of getting a large number than the previous number.
\ n( S ) = 6 ´ 6 ´ 6 = 216
Now, we count the number of favourable ways.
Obviously, the second number has to be greater
than 1. If the second number is i(i > 1), then the
number of favourable ways = (i – 1) × (6 – i)
\ n(E) = Total number of favourable ways
6
= å (i - 1) ´ (6 - i )
i =1
=0+1× 4+2×3+3×2+4×1+5×0
= 4 + 6 + 6 + 4 = 20.
Therefore, the required probability,
n( E ) 20
5
P( E ) =
=
=
=p
(given)
n(S ) 216 54
\ 54 p = 5
25. (5) Let A be the event of a boy and B the event
of having cat eyes. So
20 1
20 1
P ( A) =
= and P ( B ) =
=
40 2
40 2
10 1
=
Now, P ( A Ç B) =
40 4
\ P ( A È B ) = P( A) + P ( B ) - P ( A Ç B )
1 1 1 3 a
= + - = =
2 2 4 4 b
=
a 2 + b 2 = 32 + 4 2 = 5
MATHEMATICS
132
CHAPTER
Relations & Functions-2
16
1.
(d) P = {( a, b) : sec 2 a - tan 2 b = 1}
For reflexive :
Þ x = w1 y and y = w2z for some w1, w2 ÎQ
sec 2 a - tan 2 a = 1 (true " a)
For symmetric :
sec2 b – tan2 a = 1
Þ xRz as w1 × w2 ÎQ
Þ x = w1 ( w2 z ) = ( w1 × w2 ) z
m p
p r
Now let n S q and q S t ; where n, q, t ¹ 0 .
L.H.S. 1 + tan 2 b - (sec 2 a - 1)
2.
= 1 + tan 2 b - sec 2 a + 1
= – (sec2 a – tan2b) + 2 = – 1+2 = 1
So, Relation is symmetric.
For transitive :
if sec2 a – tan2 b = 1 and sec2 b – tan2 c = 1
sec2 a – tan2 c = (1 + tan2 b) – (sec2 b – 1)
= –sec2b + tan2b + 2 = – 1 + 2 = 1
So, relation is transitive.
Hence, relation P is an equivalence relation.
(b) Reflexivity : xRx as x = 1 . x and 1 is a
rational number.
3.
\ ( x, x) Î R
So, R is reflexive.
( 5, 1) Î R because 5 - 1 + 5 = 2 5 - 1
which is an irrational number.
But (1, 5) Ï R
\ R is not symmetric.
We have, ( 5, 1), (1, 2 5) Î R because
m m
S iff m . n = m . n, which is true.
n n
Thus, R and S are reflexive relations.
Symmetricity : Let xRy
Þ x = wy for some rational number w
Þy=
5 -1+ 5 = 2 5 -1
If 1 - 2 5 + 5 = 1 - 5 are irrational numbers.
Also, ( 5, 2 5) Î R and 5 - 2 5 + 5 = 0
which is not an irrational numbers.
\ ( 5, 2 5) Ï R
So, R is not transitive.
1
x
w
Þ yRx provided w ¹ 0, but if x = 0 and y is any
non-zero integer, then x = 0y and there exist no
rational number w for which y = wx.
Þ y R x thus R is not symmetric.
Now, let
m p
S
n q
Þ mq = pn Þ pn = mq Þ
(a) x Î R Þ x - x + 5 = 5 is an irrational
number.
p m
S
q n
Þ S is a symmetric relation.
Transitivity : Let xRy and yRz.
4.
(a) f ( x) is onto
\ S = range of f (x).
Now f (x) = sin x - 3 cos x + 1
pö
æ
= 2sin ç x - ÷ + 1
è
3ø
pö
æ
Q -1 £ sin ç x - ÷ £ 1
è
3ø
pö
æ
-1 £ 2sin ç x - ÷ + 1 £ 3
è
3ø
\ f ( x) Î[ -1, 3] = S .
Solutions
5.
6.
7.
133
(b)
(c) Let f (x) ¹ 2 be true and f (y) = 2, f (z) ¹ 1 are
false
Þ f (x) ¹ 2, f (y) ¹ 2, f (z) = 1
Þ f (x) = 3, f (y) = 3, f (z) = 1
but then function is many one, similarly two other
cases.
(c) Total number of functions from A ® B
= 34 = 81 and number of onto mappings
= coefficient of x4 in 4! (ex – 1)3.
= coefficient of x4 in 4! (e3x – 3e2x + 3e2x + 3ex – 1)
æ 34 3.24 3.14
ö
+
– 0 ÷÷
= 4! çç –
4!
è 4! 4!
ø
= 81 – 48 + 3 = 81 – 45 = 36
\ Number of functions from A ® B, which are
not onto is 81 – 36 = 45.
8.
2
(b) Let f ( x ) = ax + bx + c as it touches
X-axis at x = 3
-b
=3
2a
Þ
Þ
Also,
9.
(b )
...(1)
...(2)
...(3)
3
18
27
,b=- ,c=
25
25
25
Þ f ( x) =
\ Range of f = (0, ¥) Ì R.
11.
(b) f (x) = sin x + cos x, g (x) = x2 – 1
Þ
g (f (x)) = (sin x + cos x)2 – 1 = sin 2x
p
p
£ 2x £
2
2
[Q sin q is invertible when – p/2 £ q £ p/2]
Clearly g (f (x)) is invertible in –
p
p
£x£ .
4
4
12. (b) Here
Þ
–
g 2 ( x ) = gog ( x) = g{g ( x)} = g (3 + 4 x)
= 15 + 42 x = (42 - 1) + 42 x
7
4
8
\ x - x + x - x -1 < 0
For x Î (1, ¥ ); x < x 4 , x 7 < x8
\ x - x 4 + x 7 - x8 - 1 < 0
7
g n ( x) = (4n - 1) + 4n x = y (say)
\ g - n ( x) = ( x + 1)4- n - 1
For x Î (0, 1); x - 1 < 0, x - x < 0
4
= 63 + 43 x = (43 - 1) + 43 x
Generalizing, we get
Þ g - n ( y ) = ( y + 1)4- n - 1
3 2
( x - 6 x + 9)
25
7
Also, f is not onto as f ( x) ³ 0, "x Î R.
then x = ( y + 1 - 4n )4 - n
f ( x ) = sgn( x - x 4 + x 7 - x8 - 1)
4
f (-1) = f (-2) = 0 but -1 ¹ -2
g3 ( x) = gogog(x) = g(15 + 42 x) = 3 + 4(15 + 42 x)
b = -6 a
9a + 3b + c = 0
4a – 2b + c = 3
Þa=
10. (d) We have, f ( x) = x + x2 = x+ | x |
Clearly, f is not onto as
8
Also for x = 1; x - x + x - x - 1 = -1
Thus x - x 4 + x7 - x8 - 1 < 0 for all x Î R +
Hence sgn( x - x 4 + x 7 - x8 - 1) = -1"x Î R + .
Therefore f (x) is many-one and into.
13. (d)
f ( x) = 3 | x | - x - 2 and g (x) = sin x
3 | sin x | - sin x - 2 which is
defined if
3 |sinx | – sin x – 2 ³ 0
If sin x > 0 then 2 sin x – 2 ³ 0 Þ sin x ³ 1
for fog (x) =
p
2
If sin x < 0 then – 4 sin x – 2 ³ 0
Þ sin x = 1 Þ x = 2np +
Þ – 1 £ sin x £ –
1
2
7p
11p ù
Þ x Î é2np + , 2np +
êë
6
6 úû
MATHEMATICS
134
7p
11p ù ì
pü
é
x Î ê 2np +
È í 2 m p + ý , n, m Î I
, 2np +
6
6 ûú î
2þ
ë
14. (b)
æ 3
ö
1
f ( x) = sin 2 x + ç
cos x + sin x ÷
ç 2
÷
2
è
ø
.
2
æ1
ö
3
+ cos x ç cos x sin x ÷
ç2
÷
2
è
ø
3
1
3
= sin 2 x + cos 2 x + sin 2 x +
cos x sin x
4
4
2
1
3
+ cos 2 x cos x sin x
2
2
=
5
5
(sin 2 x + cos 2 x ) =
4
4
æ5ö
Now, y = g (f (x)) = g ç ÷ = 1.
è4ø
Clearly y = 1 is a straight line.
2
15. (a) Given that f (x) = x and g(x) = sin x, " x Î R.
Then (gof) (x) = sin x2
Þ (gogof) (x) = sin (sin x2)
Þ (fogogof) (x) = sin 2 (sin x2)
As given that (fogogof) (x) = (gogof) (x)
Þ sin2 (sin x2) = sin (sin x2)
Þ sin (sin x2) = 0, 1
p
Þ sin x = np or ( (4n + 1) where n Î Z
2
2
2
2
Þ sin x = 0 Q sin x Î[ -1,1]
Þ x2 = np Þ x = ± np where n Î W
16. (b) Let h(x) = g( f (x)) where, g(x) is injective
and h(x) is surjective.
\ Codomain of h(x) = Range of h(x).
Range of h(x) = [0, 2]
Þ codomain of g is [0, 2]. So, g(x) must be
surjective.
Now, domain of g = [–1, 1] which must be range
of f.
But codomain of f = [–1, 1]
Þ f must be surjective.
17. (d) We have, –1< f (0) < f (1) <1
g = [–1,1] ®[0,1]
and gof ( x) = x, " xÎ[0,1]
Only condition that g(x) should satisfy for
gof ( x) = x, " xÎ[0,1] is that g(x) should attain
all values in [0, 1] when rangte of f (x) a subset of
(–1, 1) is used as image for g(x). Thus, there can
be infinite such functions g(x) with domain
[–1, 1] and range [0, 1]
18. (14) If set A has m elements and set B has n
elements then number of onto functions from A
to B is
n
å (-1)n-r
n
Cr r m where 1 £ n £ m
r =1
Here E = {1, 2,3, 4}, F = {1, 2}
m = 4, n = 2
\ No. of onto functions from E to F
2
=
å (-1)2-r 2Cr (r )4
r =1
= (-1) 2C1 + 2C2 (2) 4 = – 2 +16 = 14.
19. (0)
f ( x) = (a - xn )1/n , x > 0
Þ f ( f (x)) = [a -{(a - xn )1/ n}n ]1/ n = x
2
Also g (x) = x + px + q
\ g (x) – x = 0 is quadratic equation
x 2 + ( p - 1) x + q = 0
Given that this equation has imaginary roots
\ x 2 + ( p - 1) x + q > 0 for all real x
[\ coefficient of x2 = 1 > 0]
\ g ( x) - x > 0 for all real x
Þ g (g(x)) – g (x) > 0 " x Î R
Now g (g (x)) – f (f (x)) = g (g (x)) – x
= [g (g(x))– g (x)] + [g(x) – x] > 0 " x Î R
\ The equation g (g (x)) – f (f (x)) = 0
has no real root.
20. (1) g (x) = 1 + x – [x];
ì –1, x < 0
ï
f (x) = í 0, x = 0
ï 1, x > 0
î
Solutions
135
For integral values of x; g (x) = 1.
For x < 0; (but not integral value) x – [x] > 0 Þ g (x) > 1.
For x > 0;(but not integral value) x – [x] > 0
Þ g (x) >1
\ g (x) ³ 1, " x \ f (g (x)) = 1, " x.
21. (6) Let y = g ( x) =
e x - e- x
2
Þ e2 x - 1 = 2 ye x
Substituting ex = t, we have t2 – 2yt – 1 = 0
Þt=
2 y ± 4 y2 + 4
= y ± y2 +1
2
(Q e x > 0)
Þ e x = y + y2 + 1
ù
æp ö é æp ö
Þ f (1) = cos2 1 + cos ç + 1÷ × êcos ç + 1÷ - cos1ú
è3 ø ë è3 ø
û
æp ö é
æ p ö pù
Þ f (1) = cos2 1+ cos ç +1÷ × ê-2sin ç +1÷ sin ú
è3 ø ë
è 6 ø 6û
æp ö æp ö
Þ f (1) = cos 2 1 - cos ç + 1÷ sin ç + 1÷
è3 ø è6 ø
Þ f (1) = cos 2 1 -
1é
1ù
Þ f (1) = cos 2 1 - êcos 2 - ú
2ë
2û
1
1 3
= cos 2 1 - (2 cos 2 1 - 1) + =
2
4 4
2
Þ x = ln( y + y + 1)
\ gof (1) = f ( f (1)) = f (3/ 4) = 2.
Þ g -1 ( y ) = ln( y + y 2 + 1)
-1
24. (2) Since, f (x) is symmetric about y = x.
2
\ g ( x ) = f ( x) = ln( x + x + 1)
(Q g ( f ( x)) = x Þ g -1 ( x) = f ( x))
æ 22
æ e 22 - 1 ö
ç e -1 +
\fç
=
ln
÷
ç 2e11 ÷
ç 2e11
ç
è
ø
è
æ 22
e -1
= ln ç
+
ç 2e11
ç
è
22. (4)
æ e 22 + 1 ö
çç
11 ÷
÷
è 2e ø
ìï - x + 1,
f -1 ( x ) = í
ïî1 + - x ,
2
2
ö
æ e22 - 1 ö
÷
+
1
çç
÷
11 ÷
÷
2
e
÷
è
ø
ø
ö
÷ = ln[e11 ] = 11.
÷
÷
ø
x ³1
x£0
the solution of
f ( x) - f -1 ( x) = 0 are x = –1, 1, 0, 2.
23. (2)
1é æp
ö
æ p öù
sin ç + 2 ÷ + sin ç - ÷ ú
2 êë è 2
ø
è 6 øû
æp
ö
f ( x ) = cos 2 x + cos 2 ç + x ÷
è3
ø
æp
ö æ3ö
- cos x × cos ç + x ÷ ; g ç ÷ = 2;
è3
ø è4ø
Þ f -1 ( x) = f ( x )
\ f 2 ( x) = ( f -1(x))2 - px × f (x) f -1(x) + 2x2 f (x).
Þ f 2 ( x) = f 2 ( x ) - px × f ( x ) × f ( x) + 2 x 2 f ( x )
Þ f ( x ) × {2 x 2 - px × f ( x )} = 0
Þ f ( x) =
2x2
px
Þ f ( x) =
2x
Þ p = 2.
p
25. (7) Here, f(1) = 1
f(2) = f [f (1)] + f [2 – f (1)], using f(1) = 1
f(2) = f(1) + f (1) = 2.
f(3) = f [f(3)] + f[3 – f(2)] = f (2) + f(1) = 2 + 1 = 3.
Thus, f (n) = n
\
1 20
1
å f (r ) = 30 [1 + 2 + 3 + .... + 20]
30 r =1
=
1 20(20 + 1)
´
= 7.
30
2
gof(x) = ?
æp ö
æp ö
f (1) = cos 2 1 + cos 2 ç + 1÷ - cos1cos ç + 1÷
3
è
ø
è3 ø
[as f ( x ) ¹ 0 ]
MATHEMATICS
136
CHAPTER
Inverse Trigonometric Functions
17
1.
(c) Let tan–1 x = a and tan –1y = b
Þ tan a = x, tan b = y
The given system of equations is :
a tan a + b sec a = c and a tan b + b sec b = c
\ a and b are roots of a tan q + b sec q = c
2
Þ (b sec q ) = (c - a tan q)
3.
cos -1 b1 + cos-1 b2 + .... + cos -1 bk = k
-1
We know that cos x £
2
-1
\ cos br £
Þ (a 2 - b 2 ) tan 2 q - 2ac tan q + c 2 - b 2 = 0
2ac
and
\ tan a + tan b = 2
a - b2
\ x+ y =
Þ 1 - xy =
2.
k
Þ
c 2 - b2
tan a tan b =
2ac
a -b
2
and xy =
2
c -b
k
Thus A=
a -b
x+ y
2ac
\
=
2
1 - xy a - c 2
1ù
1ù
é
é
(c) f ( x) = sin-1 ê x2 + ú + cos-1 ê x2 - ú
2û
2û
ë
ë
1ù
1 ù
é
é
= sin -1 ê x 2 + ú + cos-1 ê x2 + - 1ú
2
2 û
ë
û
ë
æé
1ù
1ù ö
é
= sin -1 ê x2 + ú + cos-1 ç ê x2 + ú - 1÷
2
2û ø
ë
û
èë
1 1 é
1ù
Since x 2 + ³ , ê x2 + ú = 0 or 1 as
2 2 ë
2û
1ù
é
sin -1 ê x 2 + ú is defined only for these two
2û
ë
values.
1ù
é
Hence ê x2 + ú = 0
2û
ë
é 2 1ù
Þ f ( x) = sin -1 0 + cos -1 (-1) = p ê x + ú = 1
2û
ë
-1 ( )
-1
Þ f ( x) = sin 1 + cos 0 = p .
p
2
Þ b1 = b2 = .... = b k = 0 .
a2 - b2
2
Therefore range of f ( x) = {p}
kp
2
cos -1 b1 = cos -1 b2 = ..... = cos -1 b k =
2
a 2 - c2
2
p
"x ³ 0
2
p
"r = 1, 2,3,...., k
2
å cos-1b r £
r =1
p
2
So the given equality holds only if
a 2 - b2
2
(a) Given
å ( br ) r = 0 .
r =1
Required limit = lim
(1 + x 2 )1/ 3 - (1 - 2 x)1/ 4
x ®0
4.
x + x2
1
1
(1 + x 2 ) -2 / 3 (2 x ) - (1 - 2 x) -3 / 4 ( -2)
4
= lim 3
1+ 2x
x®0
(L’ Hospital Rule)
1
=
2
(c) sin–1 (log[x]) is defined if -1 £ log[ x ] £ 1
and [ x] > 0
Þ
1
£ [ x ] £ e Þ [x] = 1, 2 Þ x Î [1, 3)
e
Again, log(sin -1[ x ]) is defined if
sin -1[ x ] > 0 and -1 £[ x ] £ 1
Þ [ x ] > 0 and - 1 £ [ x] £ 1 Þ 0 < [ x] £ 1
Þ x Î[1, 2)
\ Domain of f ( x) = [1, 2)
For 1 £ x < 2, [x] = 1
\ f ( x) = sin -1 0 + log
p
p
= log , "x Î [1, 2)
2
2
Solutions
5.
137
pü
\ Range of f ( x) = ì
ílog ý .
î 2þ
(c) We have
S1 = S x1 = sin 2b
S2 = S x1x2 = cos 2b
S3 = S x1x2x3 = cos b
S4 = x1x2x3x4 = – sin b
4
S - S3
2 + S4
å tan -1 xi = tan -1 1 - 1S
So that
i =1
sin 2b - cos b
1 - cos 2b - sin b
cos b ( 2sin b - 1)
= tan -1
sin b ( 2sin b - 1)
= tan -1
= tan cot b = tan -1 ( tan ( p / 2 - b ) ) = p / 2 - b
6.
(d)
f ( x ) = tan
æ
(
)
12 - 2 x 2 ö
÷
ç x4 + 2 x 2 + 3 ÷
è
ø
-1 ç
æ
ö
ç 2 3 -1 ÷
= tan ç
÷
ç x2 + 3 + 2 ÷
ç
÷
x2
è
ø
3
2
x
+
³
2
3
As
[using AM > GM]
x2
3
Þ x2 + 2 + 2 ³ 2 + 2 3
x
æ 2 3 -1 ö
-1
÷= p
\ ( f ( x ) ) max = tan çç
÷
12
2 3 +1
è
ø
(
-1
)
(
(
7.
(c)
( x + 1)
2
2
)
)
( x - 1)
- 4 x2 =
2
2
= | x2 - 1 |
Þ |tan–1 | x || = | x2 – 1 |
Draw the graphs of y = |tan –1 |x|| and y = |x2 – 1|
y
2
y = |x – 1|
p/2
–1
y = |tan – |x||
–3
8.
–2
–1
0
1
1
1
+ tan -1
+ ...... +
1 + (n - 1)n
1 + n(n + 1)
1
n + 19
tan -1
= tan -1
1 + (n + 19) (n + 20)
n + 21
1
-1 n - 1
+ tan -1
Þ tan
n +1
1 + n ( n + 1)
1
+ tan -1
1 + (n + 1) (n + 2)
1
n + 19
+ ...... +
= tan -1
n + 21
1 + (n + 19) (n + 20)
1
1
+ tan-1
+ ...... +
Þ tan-1
1+ n(n +1)
1+ (n +1)(n + 2)
1
n -1
-1 n + 19
- tan -1
= tan
1 + ( n + 19) ( n + 20)
n + 21
n +1
1 ö
1
æ
ö
-1 æ
Þ tan-1 ç 2
÷ + tan ç 2
÷ +...... +
è n + n +1 ø
è n + 3n + 3 ø
1
-1
tan
1 + ( n + 19) ( n + 20)
æ n + 19 n - 1 ö
ç
÷
= tan -1 ç n + 21 n + 1 ÷
çç 1 + n + 19 ´ n - 1 ÷÷
n + 21 n + 1 ø
è
20
-1
= tan
=S
2
n + 20n + 1
20
\ tan S = 2
n + 20n + 1
1
1
-1
-1
9. (c) sin x - -1 = cos x sin x
cos -1 x
-1
(sin x . cos-1 x + 1)
Þ (sin-1 x - cos-1 x)
=0
sin -1 x . cos-1 x
-1
+ tan
2
3
From the graph, it is clear that equation has four
roots.
(c) We know that,
1
1
1
tan-1
+ tan -1
+ tan -1
+ ......
1+ 2
1+ 2 ´ 3
1+ 3´ 4
Þ sin-1 x = cos-1 x or sin -1 x cos-1 x +1 = 0
1
æp
ö
Þ x=
or sin -1 x ç - sin -1 x ÷ + 1 = 0
2
2
è
ø
2
æ
ö
p
p
± ç
+ 4÷
ç 4
÷
2
1
è
ø
or sin -1 x =
Þ x=
2
2
æ p2 ö
p
- ç1 +
÷
ç 16 ÷
4
2
è
ø
10. (b) (cot–1 a)x2 – (tan –1 a)3/2 x + 2(cot–1 a)2 = 0
Equation has real roots
Þ x=
1
or sin -1 x =
MATHEMATICS
138
(
\ D ³ 0 Þ tan -1 a
) - 8 ( cot -1 a )
3
3
13. (b) The given relation is possible when
³0
a 2 a3
+
+ ... = 1 + b + b 2 + ...
3
9
Also
a-
\ tan -1 a ³ 2cot -1 a = p - 2 tan -1 a
Þ tan -1 a ³
p
Þa³ 3
3
( tan-1 a )
3/ 2
Sum of roots > 0 Þ
2 cot -1 a
>0
tan–1 a > 0 Þ a > 0
Product of roots > 0 Þ 2 cot–1 a > 0
Þ
aÎR Þ a ³ 3
11. (b) Tn = sec
Þ
( n2 + 1)( n2 - 2n + 2)
2
( n2 - n + 1)
-1
sec 2 Tn =
(n 2 + 1)2 ( n 2 - 2n + 2)
(n 2 - n + 1) 2
(n + 1) 2 + ( n 2 + 1) - 2 n( n 2 + 1)
2
Þ
sec 2 Tn =
Þ
sec2 Tn =
Þ
tan Tn = 2
n - n +1
Þ
tan Tn =
Þ
\
\
12. (c)
(n 2 - n + 1) 2
1 + (n2 + 1 - n)2
(n 2 - n + 1) 2
1
1ö
1ö
æ
æ
= tan –1 ç r + ÷ – tan –1 ç r – ÷
è
è
2ø
2ø
n - ( n - 1)
1 + n ( n - 1)
tan–1 n – tan–1 (n – 1)
S = T1 + T2 + T3 +....+ Tn
S = tan –1 n
(sin–1 x)3 – (cos–1 x)3 + (sin –1 x) (cos–1 x)
p3
(sin–1 x – cos–1 x) =
16
(sin–1 x – cos–1 x) {(sin –1 x)2 + (cos–1 x)2
+ (2 cos–1 x sin –1 x)} =
(sin–1 x – cos–1 x) (sin –1 x + cos–1 x)2 =
(sin–1 x – cos–1 x)
p2 p3
=
4 16
p p
3p
= ; 2 sin –1 x =
2 4
4
3
p
3
x
x
or cos
Þ sin–1 x =
Þ x = sin
8
8
8
2sin -1 x -
a2 a3
+ + ... £ 1& -1 £ 1+ b + b2 + ... £ 1
3 9
a
1
Þ | b |< 1 Þ| a |< 3 and
=
a 1- b
1+
3
3a
1
Þ
=
, there are infinitely many
a + 3 1- b
solution. But in given options it is satisfied only
1
when a = 1 and b = - .
3
2
æ
ö
1
–1 4 r + 3
–1
÷ = tan
14. (d) Tr = cot ç
1
è 4 ø
1+ r2 –
4
1ö æ
1ö
æ
çè r + ÷ø – çè r – ÷ø
2
2
= tan –1
1ö æ
1ö
æ
1+ ç r + ÷ ç r – ÷
è
2ø è
2ø
-1 £ a -
p3
16
p3
16
1ö
1
æ
Sn = å Tr = tan -1 ç n + ÷ – tan –1
2
2
è
ø
æ 4n ö
= tan –1 ç
÷
è 2n + 5 ø
1
S ¥ = lim S n = tan –1 2 = cot –1 .
2
n®¥
æ (n + 1) - n ö
1
æ
ö
15. (c) Tn = tan -1 ç 2
= tan-1 ç
è n + n + 1÷ø
è 1 + (n + 1) n ÷ø
–1
= tan (n + 1) – tan –1 n
–1
Þ T1 = tan (1/3) = tan –1 2 – tan–1 1
T2 = tan–1 (1/7) = tan –1 3 – tan–1 2
..... ..........................................
Tn = tan–1 (n + 1) – tan–1 n
On adding
T1 + T2 + T3 +.....+Tn = tan–1 (n + 1) – tan–1 1
æ n ö
= tan -1 ç
è n + 2 ÷ø
\ lim (T1 + T2 + T3 + ..... + Tn )
n®¥
æ 1 ö
= lim tan -1 ç
÷
n®¥
è 1+ 2 / n ø
Solutions
139
p
4
16. (b) x3 + bx2 + cx + 1 = 0 ; f(–1) = b – c < 0
f(0) = 1 > 0 Þ –1 < a < 0
a = –B
B Î (0, 1)
æ 1 ö ì -1 æ
= sin -1 ç
÷ + ísin ç
è 2ø î
è
ì
æ
+ ísin -1 ç
è
î
= tan–1 1 =
æ 2 sin B ö
y = -2 tan -1 ( cosec B ) - tan -1 ç
÷
è cos 2 B ø
2cosec B ö
2sin B
æ
= - ç p + tan -1
- tan -1
= -p
2 ÷
è
1 - cosec B ø
cos 2 B
p
17. (a) cos -1 x 3 + cos -1 x = is given equation
2
...(i)
Now cos -1 x 3, cos-1 x ³ 0 and their sum is
Þ
é
ê 0,
ë
Þ
cos-1 x 3, cos-1 x both must belong to
pù
2 úû
x 3, x Î [0,1]
Now from (i) cos -1 x 3 + cos -1 x =
p
- cos-1 x
2
Þ
cos -1 x 3 =
Þ
cos -1 x 3 = sin -1 x
Þ
cos -1 x 3 = cos -1 1 - x 2
Þ x 3 = 1- x
function
2
as
p
2
S = sin
19. (c)
æ1
a3
a ö b3
bö
æ1
cosec 2 ç tan -1 ÷ + sec 2 ç tan -1 ÷
bø 2
aø
2
è2
è2
1
1
+ b3
æ -1 æ a ö ö
æ -1 b ö
1 + cos ç tan
1 - cos ç tan ç ÷ ÷
÷
aø
è
è b øø
è
1
= a3
æ
æ
öö
b
1 - cos ç cos -1 ç
÷÷
çè a 2 + b2 ÷ø ÷
çè
ø
1
+b3
æ
ö
a
÷
1 + cos ç cos -1
çè
a 2 + b2 ÷ø
= a3
1
cos –1
1-
x is one-one
1
2
1
But x Î[ 0, 1] Þ x = is the only possible
2
solution
Þ (a) is the correct option.
é n - n -1 ù
-1
ú
18. (a) Let Tn = sin ê
ê n ( n + 1) ú
ë
û
é 1
1
1
1ù
= sin -1 ê
11- ú
1
n
+
nû
n +1
ë n
æ 1 ö
1 ö
-1 æ
Tn = sin -1 ç
÷ - sin ç
÷
n
n
+1 ø
è
ø
è
-1 æ
ì
æ 1 ö
-1 æ 1 ö ü
+ ísin -1 ç
÷ ý + ...
÷ - sin ç
è 4ø
è 5 øþ
î
p p
æ 1 ö
= 2 sin -1 ç
÷ =2. =
4 2
è 2ø
= a3
3x 2 = 1 - x 2 Þ 4 x2 = 1 Þ x = ±
\
p
2
1 ö
ç
÷ + T2 + T3 + T4 + ...
è 2ø
1 ö
-1 æ 1 ö ü
÷ - sin ç
÷ý
2ø
è 3 øþ
1 ö
-1 æ 1 ö ü
÷ - sin ç
÷ý
3ø
è 4 øþ
1
+ b3
b
2
a +b
a
1+
2
2
a + b2
æ
ö
a3
b3
÷
= a 2 + b2 ç +
çè
a 2 + b2 - b
a 2 + b2 + a ÷ø
æ
2
2ç 3
= a +b ç a
ç
è
(
) (
a2 +b2 +b
a2
3
a2 +b2 -a
+b
b2
)
ö
÷
÷
÷
ø
é
ù
= a 2 + b2 êa æç a 2 + b2 + b ö÷ ú
øû
ë è
+b æç a 2 + b2 - a ö÷
è
ø
= a 2 + b2 ( a + b ) a 2 + b2
= (a + b)(a 2 + b 2 )
20. (a) cot–1(2.12) + cot–1(2.22) + cot–1(2.32) +...
=
¥
( )
å cot -1 2.r 2 =
r =1
¥
æ 1 ö
å tan -1 çè 2r 2 ÷ø
r =1
MATHEMATICS
140
=
é (1 + 2r ) + (1 - 2r ) ù
¥
å tan -1 êê1 - (1 + 2r )(1 - 2r ) úú
ë
r=1
=
û
¥
å éë tan -1 (1 + 2r ) + tan -1 (1 - 2r ) ùû
r =1
= tan–1 3 – tan–1 1 + tan–1 5 – tan–1 3 + tan–1 7
– tan–1 5 +....+ tan –1 ¥
p p p
=- + =
4 2 4
21.
æp
-1 ö
(1) Let y = lim ç - tan x ÷
ø
x®¥ è 2
–1
1/x
= lim (cot x)
x®¥
\ log y = lim
x®¥
1/ x
ö
log cot -1 x æ ¥
çè form÷ø
¥
x
–1
log y = lim
(1 + x ) cot -1 x
(1 + x 2 )-1
log y = – lim
x®¥ cot -1 x
-2 x
2
x®¥
lim
= – x®¥
( 0. ¥ form)
æ0
ö
çè form÷ø
0
1 + x2
x
1
lim
= – 2 x®¥
= 2 lim
=0
1 + x2
x® ¥ 2 x
\ y = e0 =1
22. (0) We know that cot A > 1 if 0 < A <
p
p
<A<
4
2
p
4
tan -1 (cot A) + tan -1 (cot 3 A)
= p + tan -1
cot A + cot 3 A
1 - cot 4 A
,
p
4
and tan–1(cotA) + tan–1 (cot3A)
p
p
cot A + cot 3 A
if < A <
= tan -1
4
4
2
1 - cot A
If 0 < A <
Also,
cot A + cot 3 A
4
1 - cot A
=
2
(sin A + cos 2 A)(sin 2 A - cos 2 A)
sin 2A
1
== - tan 2A
2 cos 2 A
2
Hence,
æ1
ö
tan -1 ç tan 2 A÷ + tan -1 (cot A) + tan -1 (cot 3 A)
è2
ø
p
ì
if 0 < A <
ïï p
4 [Q tan -1 (- x) = - tan -1 x]
=í
p
p
ï0
if < A <
ïî
4
2
é p pù
23. (2) (sin–1x) Î ê - , ú
ë 2 2û
p2
Þ (sec–1 y)2, (tan–1 z)2 > 0
4
p2
p2
\ (sin–1 x)2 =
\ RHS ³
4
4
\ (sec–1 y)2 + (tan–1 z)2 = 0
or sec–1 y = tan –1 z = 0
p
-1
\ sin x = ± , y = 1, z = 0
2
\ x = ±1, y = 1, z = 0
24. (3) cos (2 sin–1 (cot (tan –1 (sec (6 cosec–1x)))))
= –1
p
sin–1 (cot (tan–1 (sec (6 cosec–1 x)))) = ±
2
cot (tan–1 (sec (6 cosec–1x))) = ± 1
p
tan–1 (sec (6 cosec–1x)) = ±
4
sec (6 cosec–1 x) = ±1
–1
6 cosec x = ± 3p, ± 2p, ± p
2
p p p
,2
cosec–1x = ± , ± , ± Þ x = 1,
3
2
3
6
(
\ sin -1 x
(1 + x 2 )2
-1
and cot A < 1 if
sin A cos A
=
cot A cosec2 A.sin 4 A
sin 4 A - cos 4 A
)
2
£
(Q
25. (7)
x > 0)
f ( x ) = 3cos -1 ( 4 x ) - p is defined
1
p
1
Þ 4 x £ or x £
...(i)
3
2
8
-1
1
£x£
Also, -1 £ 4 x £ 1 or
...(ii)
4
4
Therefore, from eqs. (i) and (ii), we have domain:
é -1 1 ù
x Î ê , ú Þ 4a + 64b = 7
ë 4 8û
-1
If cos 4 x ³
Solutions
141
CHAPTER
Matrices
18
1.
é 2 - 1ù é3 2ù é1 1ù
\ (B -1 )(C -1 ) = ê
úê
ú=ê
ú
ë- 3 2 û ë5 3û ë1 0û
é cos 2 q
cos q sin q ù
(a) AB = ê
ú
êëcos q sin q
sin 2 q úû
é1 1 ù
\A = ê
ú
ë1 0û
é cos2 f
cos f sin f ù
ê
ú
êë cos f sin f
sin 2 f úû
écos q cos f + cos q cos f sin q sin f
=ê
2
2
êë cos f cos q sin q + sin q sin f sin f
2
2
3.
cos 2 q cos f sin f + sin 2 f sin q cos qù
ú
cos q cos f sin q sin f + sin 2 q sin 2 f úû
écos q - sin q ù
\ (I – P) ê sin q cos q ú
ë
û
écos q cos f(cos q cos f + sin q sin f)
=ê
ë sin q cos f(cos q cos f + sin f sin f
cos q sin f(cos q cos f + sin q sin f ù
sin q cos f(cos q cos f + sin q sin f)úû
écos q cos f cos( q - f) cos q sin f cos(q - f)ù
=ê
ú
ë sin q cos f cos( q - f) sin q sin f cos(q - f) û
Clearly AB is the zero matrix if cos( q - f) = 0
i.e. q - f is an odd multiple of
2.
p
.
2
1
tan (q / 2) ù
é
= ê - tan q / 2
ú.
1
ë
û
Þ I(AC) = B -1 Þ AC = B-1
Þ ACC -1 = B -1C -1 Þ AI = B -1C -1
é
ù
-2sin(q / 2)cos(q / 2)
1- 2sin2 (q / 2)
ê
2
+ tan(q / 2)(2cos2 (q / 2) -1) ú
+
q
2sin
(
/
2)
ú
=ê
ê- tan (q / 2)(2cos2 q / 2 -1) tan(q / 2)(2sin(q / 2) cos(q / 2))ú
ê
ú
+(1 - 2sin2 (q / 2))
ë+2sin (q / 2) cos(q / 2)
û
- tan (q / 2) ù
é 1
= ê
ú = I+P
1
ë tan q / 2
û
4.
é 1 – xù é 1 – y ù
(c) (1 – x)–1 (1 – y)–1 ê
úê
ú
ë – x 1 û ë– y 1 û
–( x + y ) ù
é 1 + xy
= (1 + xy – (x + y))–1 ê –( x + y ) 1 + xy ú
ë
û
\ A = (B -1 )(C -1 )
1 é 2 - 1ù
-1
Now B =
ê
ú=
4 - 3 ë- 3 2 û
1 é - 3 - 2 ù é3
C -1 =
ê
ú=ê
9 - 10 ë- 5 - 3 û ë5
é 2 - 1ù
ê- 3 2 ú
ë
û
2ù
3 úû
écos q - sin q ù
ê sin q cos q ú
ë
û
é cos q + tan( q / 2) sin q - sin q + tan(q / 2) cos q ù
= ê - tan(q / 2) cos q + sin q tan(q / 2) sin q + cos q ú
ë
û
é2 1ù
é- 3 2 ù
(a) Let B = ê
and C = ê
ú
ú
ë 3 2û
ë 5 - 3û
-1
-1
Given BAC = I Þ B (BAC) = B I
0
- tan (q / 2) ù
é1 0 ù é
(c) I– P = ê
–ê
ú
ú
0
û
ë0 1 û ë tan (q / 2)
1
tan (q / 2) ù
é
= ê - tan (q / 2)
ú
1
ë
û
é
æ
x+ y ö
= ç1 –
÷
+ xy ø
1
è
–1 ê
1
ê
ê x+ y
ê – 1 + xy
ë
Þ A( x) A( y) = A( z)
–
x+ yù
1 + xy ú
ú = A( z)
ú
1 ú
û
MATHEMATICS
142
5.
(d) We have
0 ù é1
0ù
é1
A2 = ê
ú ê1/ 2 1ú =
1/
2
1
ë
ûë
û
0 ù é1
é1
ê 2(1/ 2) 1ú = ê1
ë
û ë
0ù
1úû
8.
é0
(c) Let A = ê a
ê
ëê a
é1 0 ù é1 0 ù é1 0 ù
A4 = ê
úê
ú=ê
ú;
ë1 1û ë1 1û ë 2 1û
b
-c
aù
-b ú
ú
c úû
a ù é1 0 0 ù
é 0 2b c ù é 0 a
ê
ú
ê
Þ a b - c 2b b -b ú = ê 0 1 0 ú
ê
úê
ú ê
ú
ëê a -b c ûú êë c -c c ûú ëê 0 0 1 ûú
é 4b2 + c 2
ê
Þ ê 2b 2 - c 2
ê
ê -2b2 + c 2
ë
é 3 1 ù
ê
ú
2
2 ú
(a) Given that, P = ê
ê 1
3ú
ê–
ú
ë 2 2 û
é1 1ù
T
T 2005
A=ê
P
ú and Q = P A P and X = P Q
ë0 1û
We observe that Q = P A PT
Þ Q2 = (P A PT) (P A PT) = P A (PT P) A PT
= PA (I A) PT
\ P A2 PT
Proceeding in the same way, we get
Q 2005 = P A2005 PT
é1 2005 ù
= IA2005 I = A2005 = ê
1 úû
ë0
a
\ AAT = I
é1 0ù
A50 = ê
ú
ë 25 1û
(d) As aij = (i2 + j2 – ij) (j – i)
aji = (j2 + i2 – ji) (i – j) = – aij
Þ A is skew symmetric Þ Tr(A) = 0.
Also |A| = 0.
é1 2005 ù
A2005 = ê
1 úû
ë0
Now, X = PTQ2005 P
= PT(PA2005 PT)P = (PT P) A2005 (PT P)
cù
-c ú .
ú
c ûú
Q A is orthogonal.
Þ
é1 1ù
é1 2 ù
Þ A2 = ê
Also A = ê
ú
ú
ë0 1û
ë0 1 û
and proceeding in the same way
-b
é0
Now A = ê 2b
ê
êë c
0ù
é1
In general, by induction An = ê
ú
ë n / 2 1û
7.
b
T
é1 0 ù é1 0 ù é1 0 ù
A8 = ê 2 1ú ê 2 1ú = ê4 1ú
ë
ûë
û ë
û
6.
2b
-2b 2 + c 2 ù
ú
a 2 - b2 - c 2 ú
ú
a 2 + b2 + c 2 ú
û
2b 2 - c 2
a 2 + b2 + c 2
a 2 - b2 - c 2
é1 0 0ù
= ê 0 1 0ú
ê
ú
ëê 0 0 1ûú
Equating the corresponding elements, we get
4b2 + c2 = 1
....(i)
2b2 – c2 = 0
....(ii)
a2 + b2 + c 2 = 1
....(iii)
a2 – b2 – c 2 = 0
....(iv)
From solving (i), (ii) and (iii) we get,
a=±
9.
(a)
1
2
;b=±
1
6
; c=±
1
3
;
é 0 1 0 ù é 0 1 0ù
A2 = AA = ê 0 0 1 ú ´ ê 0 0 1 ú
ê
ú ê
ú
êë p q r úû êë p q r úû
é0
ê
=ê p
ê
ë pr
ù
ú
q
r ú
ú
p + qr q + r 2 û
0
1
Solutions
143
Again
é0
0
1 ù é 0 1 0ù
ê
ú
A3 = A2 A = ê p
q
r ú ´ êê 0 0 1 úú
ê
2 ú ê p q r úû
ë pr p + qr q + r û ë
p
q
r
é
ù
ê
ú
2
p + qr
q+r
= ê pr
ú
ê
2
2
2
3ú
p + 2qr + r û
ë pq + r p pr + q + qr
ép
= êê 0
êë 0
0
p
0
0ù é 0
q
ê
ú
0ú + ê 0
0
ê
2
ú
pû
ë pq q
é 0
ê
+ ê pr
ê 2
ë pr
0ù
ú
qú
ú
qr û
0
ù
ú
r
ú
3ú
pr + r û
r
2
qr
pr + qr 2
é1 0 0 ù
é 0 1 0ù
= p êê 0 1 0 úú + q êê 0 0 1úú
êë 0 0 1 úû
êë p q r úû
é0
ê
+r ê p
ê
ë pr
= pI + qA + rA2
2
ù
ú
q
r ú
ú
p + qr q + r 2 û
XT = - [A1 + 3 A33 + ... ( 2 n – 1) ( A2n-1 )
2n -1
]
= –X, so skew-symmetric
éa b c ù
ê
ú
14. (c) A = ê d e f ú A2 = A × A = A Aq
ëê g h i úû
1
\ A3 - rA2 - qA = pI .
10. (a) Given A = 2A – I Þ A A = (2A – I) A
Þ A3 = 2A2 – A = 2 (2A – I) – A
= 4A – 2I – A = 3A – 2I
Again A3 A = (3A –2 I) A = 3 A2 – 2A
Þ A4 = 3 (2A – I) – 2A = 4A – 3I
........
........
In general An = nA – (n –1) I
11. (d) PQ = I Þ P–1 = Q
Now the system in matrix notation is PX = B
\ X = P–1 B = QB
2 1 ù é1 ù
é2
éxù
úê ú
ê ú 1ê
\ ê yú = ê 13 - 5 m ú ê1 ú
9
êë- 8 1 5 úû êë5úû
êëz úû
X = A1 + 3 A 33 + ... ( 2 n – 1 ) ( A 2n -1 )2n -1
We know that if A is a skew-symmetric matrix
then AT = –A
[Q Given that A = Aq]
0
2
1
(13 - 5 + 5m)
9
Þ – 27 = 8 + 5m (Given y = – 3)
\m=–7
12. (b) We have, AB = A and BA=B.
Now, AB = A Þ (AB) A = A . A Þ A (BA) = A2
Þ AB = A2 (Q BA = B) Þ A = A2 (Q AB = A)
Again, BA = B Þ (BA)B = B2 Þ B(AB) = B2
Þ BA=B2 (Q AB = A) Þ B = B2 ( Q BA = B)
\ A and B are idempotent matrices.
13. (b) Given that
\ y=
Þ
éa b c ù éa d g ù
ú
ê
ú ê
A = êd e f ú êd e h ú
êë g h i úû ê c f i ú
ë
û
2
é| a |2 + | b |2 + | c |2 ad + be + cf
ù
ag + bh + ci
ê
ú
2
2
2
ú
Þ ê da + eb + fc
| d | + | e | + | f | dg + eh + fi
ê
ú
2
2
2
ê ga + hb + ic
gd + he + if
| g | +| h | +| i | ú
ë
û
é0 0 0ù
ê
ú
= ê 0 0 0 ú (Q Given A2 = O)
êë 0 0 0 úû
Þ |a|2 + |b|2 + |c|2 = 0
Þ Re(a) = Re(b) = Re(c) = Im(a) = Im(b)
= Im(c) = 0
Similarly
Re(d) = Re(e) = Re(f) = Im(d) = Im(e) = Im(f) = 0
and Re(g) = Re(h) = Re(i) = Im(g) = Im(h)
= Im(i) = 0
MATHEMATICS
144
é 0 + 0i 0 + 0i 0 + 0i ù
ê
ú
\ A = ê 0 + 0i 0 + 0i 0 + 0i ú
êë 0 + 0i 0 + 0i 0 + 0i úû
= Null matrix of order 3 × 3
15. (d) (XY)T = YTXT
Y T = (ABC – CBA)T = CTBTAT – ATBTCT
= – CBA + ABC = Y
XT = (ABC + CBA)T = CTBTAT + ATBTCT
= – CBA – ABC = – X
= BN+1 + ( N + 1) BNC
= BN[ B + (N + 1) C]
Thus K = N Þ K/N = 1.
19. (1023) (AB)×(AB) = A(BA)B = A3B2
(AB)(AB)(AB) = A7B3
n
s o (AB)n = A2 - 1 Bn
s o k = 210 – 1 = 1023
20. (27) Observe A1A2
é1 1 ù é1 2 ù é1 3 ù é1 (2 + 1) ù
=ê
úê
ú=ê
ú=ê
1 úû
ë0 1û ë0 1û ë0 1û ë0
é cos q sin q ù
16. (d) A = ê
ú
ësin q – cos q û
\ AAT = I
...(i)
Now, C = ABAT
Þ ATC = BAT
...(ii)
Now ATCnA = ATCCn – 1 A = BATCn – 1 A
(from (ii))
= BATCCn – 2 A = B2ATCn – 2 A = ...
é 1 0ù
= Bn – 1ATCA = Bn – 1BATA = Bn = ê
ú
ë – n 1û
é1 n + ( n –1) + ( n – 2) + ... + 3 + 2 + 1ù
=ê
ú
1
ë0
û
n(n + 1)
= 378 Þ n = 27
2
(4) Given that AT A = I
So
21.
é x1 ù
ê ú
17. (c) Let X = ê x2 ú , (x1 x2 x3)
êë x3 úû
éa b cù
where A = êê b c a úú
êë c a b úû
æ a11 a12 a13 ö æ x1 ö
ça
a
a ÷ çx ÷ =0
ç 21 22 23 ÷ ç 2 ÷
çè a
÷ç ÷
31 a32 a33 ø è x3 ø
éa b c ù
\A = êb c a ú
ê
ú
êë c a b úû
T
a11 x12 + a22 x22 + a33 x32 + (a 12 + a 21 )x 1 x 2 +
(a13 + a31)x1x3 + (a23 + a32)x2x3 = 0
It is true for every x1, x2, x3,
then a11 = a22 = a33 = 0, a12 + a21 = 0, a13 + a31
= 0, a23 + a32 = 0
\ A is a skew symmetric matrix
a21 = – 2008; a31 = – 2010
a32 = 2012
18. (1) We have, BC = CB, and
AN + 1 = (B + C) N + 1
= N+1C0 B N+1 + N+1C1 BNC + N+1C2BN–1C2 +
... +
é1 3 ù é1 3 ù é1 (3 + 2 + 1) ù
and A1A2A3 = ê
úê
ú=ê
ú
1
ë 0 1û ë 0 1û ë0
û
So, in general A1A2A3 ... An
CrB N+1–rC r + . . . .
\ AT A = I
éa
Þ êb
ê
êë c
b
c
a
b
c
a
c ù é1
a ú = ê0
ú ê
b úû êë 0
0
1
0
0ù
0ú
ú
1 úû
é a2 + b2 + c2 ab + bc + ca ab + bc + ca ù
ê
ú
Þ ê ab + bc + ca a2 + b2 + c2 ab + bc + ca ú
ê
ú
ê ab + bc + ca ab + bc + ca a2 + b2 + c 2 ú
ë
û
N+1
But given that C2 = 0 Þ C3 = C4 = .... = Cr = 0
Hence, AN+1 = N+1CNBN+1 + N+1C1BNC
c ù éa
a ú êb
úê
b úû êë c
é1 0 0 ù
= êê0 1 0 úú
êë0 0 1 úû
Solutions
145
Þ a2 + b2 + c 2 = 1
...(i)
and ab + bc + ca = 0
...(ii)
abc = 1 (given)
...(iii)
Now a3 + b3 + c3 – 3abc
= (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
Þ a3 + b3 + c3 = (a + b + c) [1 – 0] + 3 × 1
[Using (i), (ii) and (iii)]
3
3
3
Þa +b +c =a+b+c+3
Now (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)
Þ (a + b + c)2 = 1
Þ (a + b + c) = 1
[Q (a + b + c) ¹ – 1 as a, b, c all are + ve
numbers.]
\ We get
a3 + b3 + c 3 = 4
22. (1) z =
-1 + i 3
Þ z3 = 1 and 1 + z + z2 = 0
2
é (- z)r
P = ê 2s
êë z
2
z 2s ù é( - z) r
úê
z r úû êë z 2s
é z 2r + z 4s
ê
=ê
2s
r
r
êë z (-z) + z
(
)
z 2s ù
ú
z r úû
(
)
z 2s (- z)5 + z r ù
ú
ú
4s
2r
z +z
úû
For P2 = – I we should have
z2r + z4s = 1 and z2s ((–z)r + zr) = 0
Þ z2r + zs + 1 = 0
Þ r is odd and s = r but not a multiple of 3.
Which is possible when s = r = 1
\ only one pair is there.
23. (1)
é 1 0 0ù
é 0 0 0ù
ê
ú
P = ê 4 1 0ú = I + êê 4 0 0úú = I + A
êë16 4 0úû
êë16 4 1úû
é 0 0 0ù
é0 0 0ù
ê
ú
3 ê
A = ê 0 0 0ú and A = ê 0 0 0 úú
ëê16 0 0úû
ëê 0 0 0 ûú
2
\ An = O, n ³ 3
Now P50 = (I + A)50 = 50C0 I50 + 50C1 I49 A
+ 50C2 + I48 A2 + O
= I + 50A + 25 × 49 A2.
50
\ Q = P – I = 50A + 25 × 49 A2.
Þ q21 = 50 × 4 = 200
Þ q31 = 50 × 16 + 25 × 49 × 16 = 20400
Þ q32 = 50 × 4 = 200
\
q31 + q32 20600
q + q 32
=
= 103 Þ 31
=1
q 21
200
103 q 21
–1 ù é 2
–1 ù é1 0 ù
é2
24. (2) A2 = ê
ú
ê
ú= ê
ú = I;
ë 3 – 2 û ë 3 – 2 û ë0 1 û
x = 2, 4, 6, 8....
Þ S(cosx q + sinx q) = (cos2 q + sin2 q)
+ (cos4 q + sin4 q) + (cos6 q + sin6 q) + ... + ...
=
cos 2 q
+
sin 2 q
1 – cos 2 q 1– sin 2 q
= cot2 q + tan2 q
Using AM ³ GM,
1
(cot 2 q + tan 2 q) ³ cot 2 q tan 2 q
2
Þ (cot2 q + tan2 q) ³ 2
Þ So, minimum value = 2
25. (2) As A2 = O, Ak = O
k
Thus, (A + I)50 = I + 50A
Þ (A + I)50 – 50A = I
\ a = 1, b = 0, c = 0, d = 1
2.
MATHEMATICS
146
CHAPTER
Determinants
19
1.
(a) Given determinant
2
1
a
a
cos(n - 1) x cos nx cos(n + 1) x
sin (n - 1) x
sin (n + 1) x
sin nx
=0
sin x = 0 or cos x =
æ 1 + a2 ö
a
2.
( a + 1)
(d)
b
2
2
c
(b + 1)
2
2
( b - 1) 2
( c - 1) 2
a2
b2
c2
a
b
c
(a - 1)2
(b - 1)2
(c - 1)2
Þ
b2
(using R 3 ® R3 + 2 R2 )
b-a
0
æ C ® C2 - C1 ö
c-a = 0 ç 2
÷
è C3 ® C3 - C1 ø
0
Þ a = b or b = c or c = a
Þ D ABC is an isosceles.
a2
(d)
D = b sin A
c sin A
b sin A c sin A
1
cos A
cos A
1
sin A sin B sin C
=
=
=k
a
b
c
Þ sin A = ak , sin B = bk , sin C = ck So,
a2
abk
ack
ack
1
cos A .
cos A
1
Take a common from C1 and R1 both, we get
1
D = a bk
bk
1
ck
cos A
ck
cos A
1
1
= a 2 sin B
sin B
1
sin C
cos A
2
=0
(taking R2 ® R2 - R3 )
a2
c =0
1
D = abk
( c + 1) 2 = 0
( a - 1) 2
Þ4
b
1
Using,
As a ¹ 1 \ç
÷ >1
è 2a ø
Þ cos x > 1 It is not possible. \ sin x = 0
2
Þ a
1
Þ (b2 - a 2 )(c - a) - (b - a)(c 2 - a 2 ) = 0
Þ (b - a)(c - a )(b - c ) = 0
3.
1 + a2
2a
c2
Þ 1
1
By applying C1 ® C1 + C3 – 2 cos x C2
By expanding
(1 + a2 – 2a cos x) [cos nx sin (n + 1) x
– sin nx cos (n + 1) x]= 0
Now, (1 + a2 – 2a cos x) sin (n + 1 – n) x = 0
Þ (1 + a - 2a cos x)sin x = 0
b2
a2 b2 - a 2 c 2 - a 2
1 + a 2 - 2a cos x
a
a2
0
cos nx cos(n + 1) x = 0
Þ
0
sin nx sin (n + 1) x
2
a2
c2
a
b
c =0
1 - 2a 1 - 2b 1 - 2c
(u sin g R3 ® R3 - R1 )
sin C cos A
1
Operate
C2 ® C2 - C1 sin B and C3 ® C3 - C1 sin C
Solutions
147
1
2
D = a sin B
0
0
2
1 - sin B
Where 0 < a , b, q <
cos A - sin B sin C
sin C cos A - sin B sin C
1 - sin 2 C
1
1
1
D
=
sin(
b
a
)
cos(
b
a
)
cos(
b
a + q)
\
- cos b
sin b
sin(b - q)
= a 2 [(1 - sin 2 C - sin 2 B + sin 2 B sin 2 C )
- (cos A - sin B sin C ) 2 ]
2
2
2
Operating C3 - C1 sin q - C2 cos q , we get
2
= a [sin A - sin B - sin C + 2sin B sin C cos A]
The above expression does not represent area
or perimeter of the triangle.
4.
1
1
1 - sin q - cos q
D = - sin(b - a ) cos(b - a )
0
- cos b
sin b
0
1 1 + sin P sin P (1 + sin P)
(d) D = 1 1 + sin Q sin Q (1 + sin Q)
1 1 + sin R sin R (1 + sin R)
= (1 - sin q - cos q)[cos b cos(b - a) - sin b sin(b - a)]
Þ D = [1 - (sin q + cos q)] cos(2b - a)
Q 0 < a , b, q <
1 sin P sin P + sin 2 P
n -1
6.
(C3 ® C3 – C2)
å
(d)
r =1
n -1
å
1 sin R sin 2 R
=
sin 2 P
sin P
0 sin Q - sin P sin 2 Q - sin 2 P
2
2
0 sin R - sin P sin R - sin P
1
sin P sin 2 P
0 1
0 1
sin Q + sin P
sin R + sin P
= (sinQ – sinP) (sinR – sinP) (sinR – sinQ)
Now
5.
D > 0 if P < Q < R
D < 0 if P > Q > R
Hence the sign of D cannot be determined.
(d) The given points are
P ( - sin(b - a), – cos b ) , Q (cos(b - a ),sin b)
R (cos(b - a + q),sin(b - q))
= (n – 1)2
n -1
å
r =1
(R2 ® R2– R1, R3 ® R3 – R1)
= (sinQ – sinP) (sinR – sinP)
r = 1 + 2 + 3 + ... + (n – 1) = n(n - 1)
2
(2r - 1) = 1 + 3 + 5 + ... + [2 (n – 1) – 2]
r =1
1
\ sin q + cos q ¹ 1
Also 2b - a <
1 sin R sin R + sin 2 R
= 1 sin Q sin 2 Q
p
4
p
Þ cos(2b - a ) ¹ 0
4
\ D ¹ 0 Þ the three points are non-collinear..
= 1 sin Q sin Q + sin 2 Q (C2 ® C2 – C1)
1 sin P sin 2 P
p
4
(3r - 2) = 1 + 4 + 7 + .. + (3n – 3 – 2)
=
n -1
\
å
r =1
Dr
Sr
n
=
2
n(n - 1)
2
n -1
å
r =1
(n - 1)(3n - 4)
2
S (2r - 1)
S (3r - 2)
n -1
a
(n - 1)2
(n - 1)(3n - 4)
2
D r consists of (n – 1) determinants in
L.H.S. and in R.H.S every constituent of first
MATHEMATICS
148
row consists of (n – 1) elements and hence it
can be splitted into sum of (n – 1) determinants.
n -1
å
\
r =1
f "( x ) g "( x ) h "( x )
a
(n - 1)2
(n - 1)(3n - 4)
2
n -1
a
(n - 1)2
(n - 1)(3n - 4)
2
n -1
Hence, value of
å
r= 1
a
(a + b + 2)
2
1
2
=0
a +b
(b + c + 2)
c2 + a 2
1
2
b +c
1
1
||A– 1|(AB)T| = | (AB)T| = 3 |AB| = 1
5
5
(a) Differentiating given equation w.r.t.x, we
get
b
c
r
a
p
= 2(3n – r)
(b) We have
1
1
a
b
3
...(1)
Again differentiating w.r.t.x, we get
c
...(5)
b
b
c
q
r
1
c = (a – b)(b – c)(c – a)(a + b + c)
3
c3
...(1)
1
1
Also, a
b
1
a
c = abc 1
1
1 1
b c
1 1
a 2 b2 c2
(taking a, b, c common from R1, R2, R3)
bc ac ab
= 1
1
= 4mx3 + 3nx2 + 2rx + 5
...(4)
f "'(0) – f "(0) g "'(0) – g "(0) h "'(0) – h "(0)
a
f '( x) g '( x ) h '( x)
q
b
a 3 b 3 c3
\
p
c
p
q
r
From (5) and (4), we get
–1
a
a
6n =
a
1
1
=
(b) |A | =
|A| 5
| (AB)T | = |AB| = |A×(adj A)| = |A|×|adj (A)|
= 5 × 52 =53
b
f "'(0) g "'(0) h "'(0)
10.
(c + a + 2)2
= 1 4 0 = 65
0 1 4
9.
... (3)
p
q
r
Putting x = 0 in (3), we get
2
4 0 1
8.
a
2r =
1
2
c
f "(0) g "(0) h "(0)
Dr is independent of
2
b
= 24mx + 6n
p
q
r
putting x = 0 in (2), we get
both 'a' and 'n'.
(a) We have a2 + b2 + c2 + ab + bc + ca £ 0
\ (a + b)2 + (b + c)2 + (c + a)2 £ 0
\ a + b = 0, b + c = 0 and c + a = 0
\ a=b=c=0
Þ
= 12mx2 + 6nx + 2r ...(2)
f "'( x ) g "'( x ) h "'( x )
(Q R1 and R3 are identical)
7.
c
p
q
r
Again differentiating w.r.t.x, we get
Dr
n(n - 1)
2
n
=
2
n(n - 1)
2
b
2
1
b
1
1
2
c
(Multiplying R1 by abc)
2
1
= a 2 b2 c 2 then,
bc ac ab
Solutions
149
1
D=
1
1
a+a a a
( x – a)2
( x – b) 2
( x – c) 2
( x – b)( x – c) ( x – c)( x – a) ( x – a)( x – b)
= (a – b)(b – c)(c – a)(3x – a – b – c)
Now given that a, b, c are all different, then
1
D = 0. Therefore, x = (a + b + c)
3
11.
Operate R2 ® R2 - R1 ; R3 ® R3 - R1 , then
(a)
t 2 + 3t + 4
Þ (l - 1)t 2 + 3(l + 1)t + 4(l - 1) = 0
Since, t is real
Þ (3l + 3 - 4l + 4)(3l + 3 + 4l - 4) ³ 0
Now, D = 1
6
1
£l£7
7
a 1 1
1 b 1 =0
14. (c)
1 1 c
-3
5
l
é 1
ù
êQ 7 £ l £ 7 ú
ë
û
Hence the given system of equations has a
unique solution.
(c) Determinant of coefficients
= 7 (l + 5) ¹ 0
t +1
t
= t +1
t
t -1
t
1
t + 2 = t + 1 -1
t -1 t + 2
t
t -1
3
-1
= 2t + 1 0
0
2t - 1 4
0
1
= -4(2t + 1)
1
.
2
13. (a) The given system of equations will have
a non-trivial solution if,
For non-trivial solution t = -
a+a
a
a
a+b
a
a
a
a+c
=0
1
c
–1
0
+1
0
–1
1
1
1– a
1
1– b
c
1– c
= 0,
æ 1 ö
æ 1 ö
C1 ® ç
÷ C1 , C 2 ® ç
÷ C2 ,
è1 – a ø
è1 – b ø
Þ (1 – a) (1 – b) (1 – c)
æ 1 ö
C3 ® ç
÷ C3
è1 – c ø
ì 1
ü
1
c
(0 + 1) –
(–1 + 0) +
(1– 0)ý = 0
í
(1–
a)
(1–
b)
1–
c
î
þ
1
1
c
+
+
= 0,
1– a 1– b 1– c
as a ¹ 1, b ¹ 1, c ¹ 1 (given)
Þ
1
1
c – 1+ 1
+
+
=0
1– a 1– b
1– c
1
1
1
Þ
+
+
=1
1– a 1– b 1– c
Þ
a
0 1–c
b – 1 1 – c = 0,
1
1 -1
t
0
R1 ® R1 – R3, R2 ® R2 – R3
Þ (1 – a) (1 – b) (1 – c)
[Determinant of coefficients of equations]
12.
a –1
Þ
1
4
2
c
1
æ 1 1 1ö
= -ç + + ÷
è a b cø
a
Þ 9(l + 1) 2 - 16(l - 1)2 ³ 0
3 -1
0 =0
0
Þ aab + c(ab + ab + aa ) = 0 since a, b, c ¹ 0
\
Þ ( 7 - l )(7l - 1) ³ 0 Þ
b
-a
Þ a (bc + ca + ab) + abc = 0
t 2 - 3t + 4
l=
-a
MATHEMATICS
150
15.
(a) The given system of equations will have a
non-trivial solution if
a+a
a
a
a
a+b
a
18.
=0
a
a a+c
Operating R2 ® R2 – R1 and R3 ® R3 – R1, we get
a+a a a
–a
16.
0 =0
–a 0 c
or aab + c(ab + ab + aa) = 0
or a(bc + ca + ab) + abc = 0
1
æ 1 1 1ö
= –ç + + ÷
(Q a, b, c ¹ 0)
or
è a b cø
a
(c) The given system is consistent.
Therefore,
1
D=
17.
b
1
–1
or
–c =0
– b 3b – c
c + bc – 6b + b + 2c + 3bc = 0
or
3c + 4bc – 5b = 0 or
2 –1
c=
5b
4b + 3
Now, c < 1
5b
5b
< 1 or
–1 <0
Þ
4b + 3
4b + 3
b–3
æ 3 ö
or
< 0 Þ b Î ç – ,3÷
è 4 ø
4b + 3
(c) We have,
np ö (-1)n n !
æ
0 cos ç x + ÷
è
2 ø ( x + 3)n +1
dn
np
(-1)n n !
=
f
(
x
)
0
cos
[
]
2
3n +1
dx n
3
a
a
a5
np ( -1)n n !
2
3n +1
np ( -1)n n !
= 0 cos
=0
2
3n +1
a a3
a5
19.
20.
dn
dx n
[ f ( x) ]x = 0
(Q R1 and R2 are identical)
b1 + qc1
c1 + ra1
D¢ = a2 + pb2
b2 + qc2
c2 + ra2
a3 + pb3
b3 + qc3
c3 + ra3
a1
b1 + qc1
c1 + ra1
= a2
b2 + qc2
c2 + ra2
a3
b3 + qc3
c3 + ra3
pb1
b1 + qc1
c1 + ra1
+ pb2
b2 + qc2
c2 + ra2
pb3
b3 + qc3
c3 + ra3
In the first determinant, apply C3 ® C3 – rC1 and
then C2 ® C2 – qC3.
In second determinant take p common from C1
and then apply
C2 ® C2 – C1. Then take q common from C2 and
apply
C3 ® C3 – C2. Finally taking r common from C3,
we have ultimately D¢ = (1 + pqr)D.
(d) det{Adj(Adj A)} = (14)4 = |A|(n – 1)2
Þ |A| = 14 = 3x + 11 Þ x = 1
(c) A(Adj A) = |A|In;
|A| = xyz – (8x + 4y + 3z) + 28
é 68 0 0 ù
ê
ú
= 60 – 20 + 28 = 68 Þ A(Adj A) = ê 0 68 0 ú
êë 0 0 68úû
21.
(0) Apply C3 ® C3 – C1
æ 3pö
æ 5p ö
sin2 ç x + ÷ sin2 ç x + ÷
è
è
2ø
2ø
0 cos
\
(d)
a1 + pb1
22.
sin(2x + 5p)sin(2p)
æ 3p ö
Þ sin çè x + ÷ø
2
æ 5pö
sin ç x + ÷
è
2ø
æ 5pö
2cos ç x + ÷ sin(p) = 0
è
2ø
æ 3pö
sin ç x - ÷
è
2ø
æ 5pö
sin ç x - ÷
è
2ø
æ 5pö
2cos ç x - ÷ sin(-p)
è
2ø
Q All elements of C3 are zero.
(20) The matrices in the form
é a11 a12 ù
êë a21 a22 úû , aij Î {0, 1, 2}, a11 = a12 are
é 0 0/1/ 2ù é 1 0/1/ 2ù é 2 0/1/ 2ù
êë0/1/ 2 0 úû , êë0/1/ 2 1 úû , êë0/1/ 2 2 úû
Solutions
151
At any place, 0/1/2 means 0, 1 or 2 will be the
element at that place.
Hence there are total 27 = 3 × 3 + 3 × 3 + 3 × 3)
matrices of the above form. Out of which the
matrices which are singular are
é 0 0 / 1/ 2 ù é 0 0 ù é1 1ù é 2 2ù
,
,
,
êë 0
0 úû êë1/ 2 0 úû êë1 1úû êë 2 2úû
Hence there are total 7(= 3 + 2 + 1 + 1) singular
matrices.
Therefore number of all non-singular matrices in
the given form = 27 – 7 = 20
cos x
x
sin x
cos x
2
2
2
- tan x
2cos 2 x
- x3
5
+ 2 x - sec x -3x + x
2 x sin 2 x
5x
2
Operate
1
1
1
R2 , R3 ® R3 , R2 ® R2 ,
x
x
x
respectively on three determinants
R2 ®
1
cos x
- sin x
- x2
5x
sin x
cos x
tan x
+ 2 x - sec 2 x -3x 2 + x - x2
x
sin 2 x
2 2cos 2 x
5
2
5
x
x
ìï 27 4 üï 1
í
ý = = 0.20
îï 5 þï 5
24. (0)
log a + (n –1)log r log a + (n +1)log r log a + (n + 3)log r
log a + (n + 9)log r
log a + (n +11)log r log a + (n +13)log r log a + (n +15log r
D = C3 ® C3 – C2
sin x
f '( x)
tan x
= x x
x
2 x sin 2 x
é3 0 0ù
ê
ú
23. (0.20)
A = ê0 3 0ú
êë0 0 3úû
det (adj(adj(A))) = |A|4 = 274
log a + (n + 5)log r log a + (n + 7)log
x
C2 ® C2 – C1
sin x
cos x
x
1 1 0
0 0 1
f '( x )
= 0 -1 0 + 0 -1 0
Lt
x ®0 x
0 0 0
2 2 5
0
log a + (n –1) log r 2 log r 2log r
log a + (n + 5) log r 2log r 2 log r = 0
log a + (n + 11) log r 2 log r 2log r
Þ
(1 + ap)2 (1 + bp)2 (1+ cp)2
25. (16) (1 + aq)2 (1 + bq)2 (1+ cq)2
(1 + ar )2 (1 + br )2 (1 + cr )2
1 2a a 2
1 p p2
= 1 2b b2 ´ 1 q q 2 = 2 × 2D1 × 2D2
1 2c c 2
= 8D1D2 = 8 ×
1 r r2
1
26. (4)
2
cos x
f '( x) = x - tan x
2 x sin 2 x
2
- x3
5x
2
5
27. (0) The two roots of the equation are 1 ± i 3 ,
so that we can take
æ1
3ö
a = 1 + i 3 = 2 çç 2 + i 2 ÷÷ = 2eip/3 and
è
ø
æ1
3ö
b = 1 – i 3 = 2 çç 2 - i 2 ÷÷ = 2e–ip/3
è
ø
(
- sin x
1
= 0 + 2(0 + 1) + 2(0 + 1) = 4
i2 p / 3
+ e-i2p / 3
a + b = 2, a2 + b2 = 22 e
1
× 4 = 16
2
0
+ 0 -1 0
= 22 . 2 cos
(
a4 + b4 = 24 ( e
2p
= – 22 ,
3
)
ip
-ip
a3 + b3 = 23 e + e
= – 24 ,
i4 p / 3
)
+ e-i4 p / 3 = – 24
)
MATHEMATICS
152
(
29. (1) We have,
éa b ù
A= ê
ú
ëc d û
)
i5 p / 3
+ e-i5p / 3 = 25
and a5 + b5 = 25 e
-22
-2 4
Determinant = -22
-24
-2 4
-2 4
-24
25
2
1
2
-2 -2
2
= – 28 1 2
-1 -1
28.
2
1
éa b ù éa b ù
Þ A2 = ê
úê
ú
ëc d û ëc d û
1 -2 -22
= – 28 0 6
0 -3
6
-3
(1 + x) 2
(1 + x )4
(1 + x)6
3
(0) Let f (x) = (1 + x )
(1+ x)6
(1 + x)9
(1 + x) 4
(1 + x )8
(1 + x)12
Coefficient of 'x' is f ' (0)
2(1 + x )2
= 0.
é a 2 + bc ab + bd ù
ú
=ê
êë ac + cd bc + d 2 úû
Given, A2 = A and ad – bc = 0
é a 2 + bc ab + bd ù é a b ù
\ ê
ú=ê
ú
êë ac + cd bc + d 2 úû ë c d û
Þ ab + bd = b Þ b(a + d) = b
Þ a+d=1
x
x2
2
30. (2) –3x4 + 1 x
1 x3
x4
1
x6
=0
4(1 + x )3
6(1 + x)5
3
f ' (x) = (1+ x )
(1 + x)6
(1 + x )9
(1 + x )4
(1 + x)8
(1 + x )12
Þ –3x4 + x3
(1 + x )2
(1 + x )4
(1 + x )6
Now, apply R1 ® R1 – R2, R2 ® R2 – R3
+ 3(1 + x)3
6(1 + x )5
9(1 + x)8
(1 + x )4
(1 + x )8
(1 + x )12
(1 + x )2
(1 + x)2
(1+ x)6
+ (1 + x )3
(1 + x)6
(1 + x )9
4(1+ x)3 8(1 + x )7 12(1 + x )11
Put x = 0, f ¢(0) = 0
1
1
1
1
x
x2
1 x2
x4
0
1- x
2
Þ –3x4 + x3 0 x - x
1
x2
=0
1 - x2
x 2 - x4
x4
=0
Þ –3x4 + x3[x2(1 – x2)(1 – x)
– x(1 – x)(1 – x2)] = 0
Þ x4[– 3 + (1 – x) (1 – x2) (x – 1)] = 0
x = 0, x = 2
Solutions
153
CHAPTER
Continuity and Differentiability
20
1.
(b) f is continuous at x = p / 4 ,
if lim f ( x ) = f (p / 4) .
3.
(d) If f (a) = a, then obviously x = a is the
solution.
Let f (a) > a and g(x) = f (x) – x then g (a) > 0 and
g( f (a)) = f ( f (a)) – f (a) = a – f (a) < 0
Since g(x) is continuous, so at least for one
c Î(a, f (a)) , g(c) = 0.
Similarly we can argue for f (a) < a.
4.
(b)
x ®p / 4
Now, L = lim (sin 2x ) tan
2
2x
x ®p / 4
Þ log L = lim tan 2 2x log sin 2x
x ®p / 4
log sin 2 x æ ¥ ö
= lim
ç ÷
x ®p /4 cot 2 2 x è ¥ ø
1
= lim
=2
x ®p / 4 - 2 cot 2x cos ec 2x.2
2
or L = e -1 / 2
\ f (p / 4) = e -1 / 2 = 1 / e
lim
=e
=e
æ sin x ö
- lim ç
tan x
è x ÷ø
+
x®0
=e
+
1x
x ® 0+ - cosec x cot x
= e -1´ 0 = 1
Let g (x) = ax + b Þ b = 1 Þ g ( x ) = ax + 1
For
sin x ù
é
x > 0, f '( x ) = esin x ln(| x|) êcos x ln(| x |) +
x úû
ë
,x < 0
ì1
ï
f (x) = í1 - x
,x ³ 0
îï1 + x
f '(1) = 1(0 + sin1) = sin1
f (-1) = - a + 1 Þ a = 1 - sin1
,x < 0
ì1
ï
f (2x) = í1 - [ 2 x ] , x > 0
ï1 + [ 2 x ]
î
5.
g ( x) = (1 - sin1) x + 1.
(b) We apply check for continuity at x = 0
LHL = lim f ( x) = lim f (0 - h)
x ® 0-
,x< 0
,0 £ x <
ln x
x ® 0+ cosec x
lim sin x ln x
= e x®0
\ f (0- ) = g (0) = 1
ì1 - [ x]
ï
, x ¹ -1
(d) f (x) = í 1 + x
and
ïî1
, x = -1
ì1
ï
ï1
ï
Þ f ( 2x) = í
ï0
ï 1
ïî 3
sin x
x ®0 +
lim
2 cot 2x
2.
f (0+ ) = lim x
¥
= lim (cos h + sin h) -cosec h ( 1 from )
1
2
1
, £ x £1
2
3
,1 £ x <
2
Þ f (x), for all values of x where x <
h ®0
= exp{ lim (cos h + sin h - 1) ´ - cosec h}
h® 0
1
2
1
and x = 1
2
f (x) be a discontinuous function.
continuous function and for x =
h® 0
ì
h
hö
æ
2h
= exp í lim ç -2sin 2 + 2sin 2 cos 2 ÷
h
®
0
è
ø
î
´-
1
ü
ï
h
hý
2sin cos ï
2
2þ
MATHEMATICS
154
h
höü
ì
æ
sin - cos ï
ç
ïï
2
2 ÷ ï = e-1
= exp í lim ç
÷ý
h
ïh®0 ç
֕
cos
è
ø ïþ
2
îï
7.
(a) f (x) = min {x + 1, | x | + 1}
Þ f (x) = x + 1 " x Î R
Y
h ®0
x ®0
= lim
1/ h
e
= lim
h® 0
e
(e
-2 / h
3/ h
e
(0, 1)
+ e 2/ h + e3/ h
ae 2/ h + be3/ h
h ®0
3/ h
{ae
+e
-1/ h
-1/ h
X'
+ 1)
+ b}
=
1
b
é
ù
e-1/ h = 0ú
êëQhlim
®0
û
8.
\ For continuity at x = 0
1
e -1 = a = b -1 Þ a = , b = e
e
6.
y=x+1
y=–x+1
RHL = lim+ f ( x ) = lim f (0 + h)
x® k
according as
k Î (2n, 2n + 1) or k Î (2n + 1, 2n + 2)
( x - 1) n
Where g (x) =
log cos m ( x - 1)
m, n are integers, m ¹ 0, n > 0
\ we get,
lim
x ®1+
( x - 1) n
log cos m ( x - 1)
Þ lim
hn
h® 0 log cos m
h
, 0 < x < 2,
= -1
= -1
hn
= -1
h®0 m(log cos h)
[Using L' Hospital's rule]
1
ì
1ï
t
ï lim (1- sin px) , 2n < x < 2n +1
1
ït®¥
1+
ï
(1+ sin px)t
f (x) = í
ï
(1+ sin px)t + 1 2n +1 < x < 2n + 2
ï lim
,
ï t®¥ (1+ sin px)t + 1
ï
x = 0,1,2,.....
0,
î
If k Î I, f (k) = 0, but lim f ( x) = 1 or – 1
Y'
Hence, f (x) is differentiable everywhere for all
x Î R.
(c) As per question,
p = left hand derivative of |x –1| at x = 1
Þ p = –1
Also lim g ( x ) = p
x ®1+
ìï (1 + sin px )t - 1 üï
(b) f (x) = lim í
ý
t ®¥ ï (1 + sin px )t - 1 ï
î
þ
Q sin p x > 0 (in I and II quadrants)
\ 2n p < p x < (2n + 1) p
Þ 2n < x < 2n + 1, n Î I
and sin px < 0
(in III and IV quadrrants)
\ (2n + 1) < p x < (2n + 2) p
Þ 2n + 1 < x < 2n + 2, n Î I and
sin p x = 0 if x = 0, 1, 2,.........
ì 1, 2n < x < 2n + 1
ï
= í -1 2n + 1 < x < 2n + 2
ï 0, x = 0,1, 2..........
î
X
(– 1, 0)
Þ lim
n h n-1 cos h
= -1
h®0 m( - sin h)
[Using L' Hospital's rule]
Þ lim
n hn - 2 cos h
= 1 Þ n = 2 and m = 2
sin h ö
h® 0
m æç
÷
è h ø
(b) Let | f (x) | £ x2, "x Î R
Now, at x = 0, | f (0) | £ 0
Þ f (0) = 0
Þ lim
9.
\ f ¢ (0) = lim
h® 0
f (h ) - f (0)
f (h)
= lim
...(1)
h-0
h® 0 h
Solutions
155
f (h)
£|h|
Now,
h
2
(Q | f ( x ) | £ x )
f (h)
f (h)
® 0 ...(2)
£ | h | Þ lim
h® 0 h
h
(using sandwich Theorem)
\ from (1) and (2), we get f ¢ (0) = 0,
i.e. - f (x) is differentiable, at x = 0
Since, differentiability Þ Continutity
\ | f (x) | £ x2, for all x Î R is continuous as
well as differentiable at x = 0.
Þ - | h |£
10. (a)
æ xx – x– x
f ( x) = cot –1 ç
ç
2
è
æ 2x x
= tan –1 çç
x 2
è –1 + ( x )
\
11.
f ¢( x ) = –
ö
æ ( x) 2 x – 1 ö
÷ = cot –1 ç
÷
÷
ç 2xx ÷
ø
è
ø
ö
–1 x
÷÷ = –2 tan x
ø
2
1 + ( x x )2
´ x x (1 + log e x )
–2
f ¢(1) =
´ 1 = –1
\
2
(c) We have, g = inverse of f = f –1
Þ g(x) = f –1 (x) Þ f [g(x)] = x
Differentiating w.r.t. x, we get f ' [g (x)]. g ' (x) = 1
\ g ' (x) =
1
= 1 + [g (x)]3
f '[ g (x)]
é
ù
1
1
\ f ¢[ g ( x)] =
êQ f ¢ ( x) =
3
3ú
1+ x
1 + [ g ( x)] úû
êë
12. (b) Since, 2 < x < 3, \ [x] = 2
æ 2p
ö
f ( x) = sin ç - x 2 ÷
\
è 3
ø
\ f ¢( x) = cos çæ 2p - x 2 ÷ö (-2 x)
è 3
ø
æ 5p ö
5p
\ f ¢ç
=2
ç 3 ÷÷
3
è
ø
13. (c) Put x n = cos a , y n = cos b
Þ
Þ
Þ
a -b
= -a
2
cos–1(xn) – cos–1( yn) = 2 tan–1(– a)
tan
y n -1
1 - y 2n
dy
x n -1
x n -1
=
= n -1
dx
y
1 - x2n
14. (d) Let x3 = cos p and y3 = cos q
3
3
(1 - x 6 ) + (1 - y 6 ) = a (x – y )
Given
Þ
Þ
Þ
(1 - cos 2 p ) + (1 - cos 2 q )
= a(cos p – cos q)
sin p + sin q = a(cos p–cos q)
æ p + qö
æ p - qö
2sin ç
cos ç
è 2 ÷ø
è 2 ÷ø
p - qö
æ p + qö
= –2a sin æç
÷ sin ç
÷
è
2 ø
è
2 ø
Þ
1
æ p - qö
tan ç
=- Þ
è 2 ÷ø
a
Þ
cos–1x3 – cos–1 y3 = tan–1 çè - a ÷ø
æ 1ö
p – q = tan–1 çè - ÷ø
a
æ 1ö
Differentiate w.r.t. x, we have
-
3x2
1 - x6
+
dy x 2 1 - y 6
dy
=
=0 Þ
.
dx y 2 1 - x 6
1 - y 6 dx
3 y2
Hence, f (x, y) = x2 / y2
15. (b) f ¢ (x) = n (1 + x)n –1, f ¢¢ (x) = n (n – 1) (1 + x)n – 2
\ f n (x) = n!, f n (0) = n!
Þ
n (n –1)
n!
+ ... +
2!
n!
= nC0 + nC1 + nC2 + ... + nCn = 2n
1+ n +
16. (c) Given ea cos a = 1
....(i)
b
and e cos b = 1
.....(ii)
–x
Let f(x) = e – cos x, then f(x) is continuous and
differentiable.
Also, f(a) = f(b) = 0
(from (i) and (ii))
Therefore by Rolle's MVT, f '(x) = 0 has at least
one root in (a, b).
Þ - e - x + sin x = 0 for at least one x Î ( a, b)
17. (b) We have, f(x) = x|sin x|, x Î R
x Î(2 np,(2 n + 1) p)
ì x sin x,
f ( x) = í
î – x sin x, x Î(2n + 1) p , 2pn)
f '(np ) = lim
x ®nx
f ( x ) – f ( np )
x – np
x | sin x |
x – np
Clearly, f(x) is differentiable for all x except
x = np, n = ±1, ±2, ±3,....
= lim
x ® nx
MATHEMATICS
156
18. (c) We have,
ì
æ pö
f ( x ) = í x 2 cos ç ÷
è xø
î
0
For x = 0, f (0) = 0
20. (d)
for x ¹ 0
R.H.D.
sin h 2
-0
f (0 + h) - 0
= lim
= lim h
=1
h
h® 0
h
h®0
æ pö
h 2 cos ç ÷ – 0
p
è hø
= lim
= lim – h cos = 0 .
h
x®0
–h
x ®0
Simillarly, R f ¢ (0) = 0
Þ differentiable at x = 0.
sin h2
-0
f (0 - h ) - 0
L.H.D. = lim
= lim -h
=1
-h
h® 0
-h
h® 0
R.H.D. = L.H.D.
p
=0
x
p
p
p
cos = 0 Þ = (2 n +1)
x
x
2
2
2 x +1
f (x) is not differentiable at
x=
2
x Î[–1,1]: x =
, n ÎZ
2n + 1
19. (d) Let f : R ® R be a continuous function
such that f (x2) = f (x3)
...(i) for all x Î R
Put x = –x, in eqn. (i), then f (x2) = f (–x3)
from (i) f (x3) = f (–x3)
Let x3 = t, then f (t) = f(–t)
Þ f (x) is an even function
Now take x3 = t, then from (i)
f (t2/3) = f (t)
Put t = t2/3 2
2/3
Þ f (t (2/3) ) = f (t )
(2/3) 2
)
Þ f (t) = f (t 2/3 ) = f (t
3
n
= f (t (2/3) ) ... = f (t (2/3) )
This is true for all t Î R and any n Î I.
n
x2
Now, to check the differentiability of f (x)
f (0 – h) – 0
Now Lf ¢ (0) = lim
–h
x ®0
Q
x ®0
x®0
Þ f (x) is continuous
p
=0.
n
x®0
Þ f (x) is continuous at x = 0
2
and lim x cos
Þ
sin x2
æ sin x 2 ö
ç
÷ x = (1)(0) = 0 = f (0)
= xlim
®0 è x 2 ø
for x = 0
Now, f (x) is not differentiable at cos
lim f ( x ) = lim
æ 2ö
When n ® ¥, ç ÷ ® 0
è 3ø
0
Then f (t) = f (t ) = 1
Hence, f (x) is a constant function and
therefore it is differentiable everywhere
Þ f (x) is differentiable
Now, differentiate f (x) w.r.t. x, we get
f ¢(x) =
x cos( x 2 )2 x - sin( x 2 )
x2
2
f ¢(x) = 2cos x -
sin x 2
x2
2
ì
2 sin x
, x¹ 0
ï 2cos x - 2
f ¢(x) = í
x
ï
1
, x =0
î
Now, to check the continuity of f ¢(x)
2
lim f ¢ ( x ) = lim 2cos x 2 - sin x = 2 – 1 = 1
x®0
x ®0
x2
Þ f ¢(x) is continuous
21. (0)
f '(x) = lim
h®0
f ( x + h ) - f ( x)
h
f ( x + h) - f ( x)
(h)2
£ lim
h
h®0
h®0 h
| f '( x) | = lim
Þ | f '( x) | £ 0
Þ
f '( x) = 0
Þ f (x) = constant
As f (0) = 0 Þ f (1) = 0.
Solutions
157
22. (4) Q x = cosec q - sin q
Þ
2
x + 4 = (cosec q - sin q) + 4
= (cosec q + sin q)2
....... (i)
and y 2 + 4 = (cosec n q - sin n q) 2 + 4
= (cosec n q + sin n q)2
é æ xö æ xö
æ x ö æ x öù
F ¢( x) = ê f ç ÷ . f ¢ ç ÷ + g ç ÷ g ¢ ç ÷ ú
2
2
è 2 ø è 2 øû
ë è ø è ø
Here, g (x) = f ' (x)
and g' (x) = f '' (x) = – f (x)
24. (5)
2
........(ii)
æ dy ö
ç ÷
dy è dq ø
=
Now ,
dx æ dx ö
ç ÷
è dq ø
x
x
So, F ¢( x) = f çæ ÷ö g çæ ÷ö 2
2
è ø è ø
Þ F (x) is constant function
So, F (10) = 5
25. (2) Given, | f (x) – f (y) | = | x – y | for all in [0, 1]
Þ
n(cosecn-1q)(-cosec q cot q) - n sin n-1 q cos q
=
-cosec q cot q - cos q
n -1
n
Þ
| f '( x) – f (y) |
=1
| x– y |
lim
x® y
| f (x) – f (y) |
= 1Þ | f '(x) | =1
x– y
Þ f '( x) = ±1Þ f ( x ) = ± x xÎ[0,1]
\ f (x) has exactly 2 function.
26. (3) Taking logarithm of both sides, we get
=
n(cosec q cot q + sin q cos q)
(cosec q cot q + cos q)
=
n cot q (cosecn q + sin n q)
cot q (cosec q + sin q)
é
æ 1 öù
log y = x êlog ç1 + ÷ ú
x øû
è
ë
=
2
n(cosecn q + sin n q) n y + 4
=
(cosec q + sin q)
x2 + 4
Þ
[From (i) and (ii)]
Squaring both side, we get
2
n 2 ( y 2 + 4)
æ dy ö
çè ÷ø =
dx
( x 2 + 4)
2
æ dy ö
( x 2 + 4) ç ÷ - n 2 y 2 = 4n 2 = kn 2
è dx ø
Þ k = 4.
23. (1) We have,
f (x) = cos x cos 2x cos22 x cos 23 x .....
cos 2n–1x
or
Þ f ( x) =
Þ f '( x) =
Þ
1
æ 1ö
+ log ç 1 + ÷
x +1
xø
è
Since y (2) = (1 + 1/2)2 = 9/4
2n sin 2 x
n
n-1
æ p ö 2 cos 2 p
f 'ç ÷ =
= cos 2n–1 p
è 2ø
2n
= (–1) 2 n – 1 = 1
......... (1)
3ö
æ 1
So, y1 (2) = (9/4) çè - + log ÷ø
3
2
Multiplying (1) by y and then differentiating, we
get
æ
1
æ 1 öö
y2 ( x ) = y1 ( x ) ç + log ç 1 + ÷ ÷
x øø
è
è x +1
æ 1
x æ 1 öö
+ y ( x ). ç
+
ç- ÷÷
è ( x + 1)2 x + 1 è x 2 ø ø
sin 2 x
2n cos 2n x sin x - sin 2n x cos x
1
x2 æ 1 ö
1ö
æ
y1 ( x ) =
ç÷ + log ç 1 + ÷
y
x + 1 è x2 ø
xø
è
=-
n
2n sin x
æ xö æ xö
f ç ÷ gç ÷=0
è2ø è2ø
So,
3ö
æ 1
æ1 1ö
y2 (2) = y1 (2) ç - + log ÷ + y(2) ç - ÷
2ø
è 3
è9 6ø
2
3ö
1
æ 9ö æ 1
= ç ÷ ç - + log ÷ è 4ø è 3
2ø
8
MATHEMATICS
158
27. (3) Here,
y=
So, y =
(
x - sin x ) + ( x - sin x) + ...¥
\ y = x – sin x + y
( x - sin x) + y
2
Differentiating, we get : 2 y
At x =
)
dy
dy
= 1 - cos x +
dx
dx
1 p 3
p 2
p
, y - y = -1 Þ y2 - y + = 2
2
4 2 4
2
1ö
2p - 3
æ
Þ (2 y - 1) = ± 2p - 3
Þ çy- ÷ =
è
ø
2
4
From Eq. (i), we get (2 y - 1)
\
28. (2)
Þ
dy
dx
x=
p
2
=
lim f ( x) =
x ®0
2p - 3 Þ
1
3
dy
= 1 - cos x
dx
dx
dy
2
x=
p
2
- 2p = 3
x × a cot x + b 1
=
x®0
3
x2
Þ lim
xa + b tan x 1
lim
=
x ® 0 x 2 tan x
3
æ
ö
x3
ax + b ç x + + ...¥ ÷
3
è
ø 1
=
Þ lim
x®0
3
3 æ tan x ö
x ç
è x ÷ø
æb ö
(a + b) x + ç ÷ x3 + ...¥
è3ø
1
=
Þ lim
3
x®0
3
x
So a + b = 0
Also, b = 1 Þ a = –1 Þ a2 + b2 = 2
ì
ï 3 - cos x ï
29. (4) f`(x) = í
ï 2 + cos x +
ïî
ì
ï 3 - cos x ï
=í
ï 2 + cos x +
ïî
1
1
, | sin x |<
2
2
1
1
, | sin x |³
2
2
1
1
, | cos x |>
2
2
1
1
, | cos x |£
2
2
1
1
ì
ïï 3 - cos x + 2 , | cos x |> 2
=í
ï 2 + cos x + 1 , | cos x |£ 1
ïî
2
2
Thus, f (x) is discontinuous at
p 3p 5 p 7 p
1
or x = , , ,
| cos x |=
4 4 4 4
2
30. (3) Put cos f =
2
3
3
;sin f =
; tan f =
2
13
13
-1
-1
y = cos {cos( x + f)} + sin {cos( x - f)}
-1
= cos {cos( x + f)} +
p
- cos -1{cos(f - x)}
2
p
-f+ x
2
p
y = 2 x + ; z = 1 + x2
2
dy
dy dx 2 1 + x 2
=
=
Now,
dz dz
x
dx
10
æ dy ö
\ç ÷
=
3
è dz ø x=
3
= x+f+
4
Solutions
159
CHAPTER
Application of Derivatives
21
1.
2.
(d) Given f (x) = tan –1 (sin x + cos x)
1
.(cos x - sin x )
f '(x) =
1 + (sin x + cos x) 2
1
æ 1
ö
2. ç
cos x sin x÷
è 2
ø
2
=
2
1 + (sin x + cos x )
p
p
æ
ö
çè cos .cos x - sin .sin x÷ø
4
4
=
1 + (sin x + cos x) 2
pö
æ
2 cos ç x + ÷
è
4ø
\ f '(x) =
1 + (sin x + cos x )2
if f ' (x) > 0 then f (x) is increasing function.
p
p p
Hence f (x) is increasing, if - < x + <
2
4 2
3p
p
Þ <x<
4
4
æ p pö
Hence, f (x) is increasing when n Î ç - , ÷
è 2 4ø
(b) If f ( x ) = x1/ x , then
1
f ' ( x ) = 2 é x1/ x (1 - ln x ) ù
û
x ë
f is decreasing if x > e and f is increasing if x < e.
As e < 3 <4 < 5 < 6 < 7
\
i.e. 3( x 2 - 6 x + 8)e < 1 i.e. x2 – 6x + (8 – 1/3e) < 0
i.e. ( x - (3 + 1 + 1/ 3e ))( x - (3 - 1 + 1/ 3e ) < 0
Û 3 - 1 + 1/ 3e < x < (3 + 1 + 1/ 3e )
Hence x Î (4, 3 + 1 + 1/ 3e )
Similarly if x < 2, then f ' (x) < 0,
If log 3(x2 – 6x + 8) e > 0
i.e., x < 3 - 1 + 1/ 3e or x > 3 + 1 + 1/ 3e
4.
-(4a2 + 8a - 14) > 0"x Î R
But 21 / 2 = 41 / 4 < 31 / 3
3.
5.
+ (6 x - 18) log( x 2 - 6 x + 8)
+
(3 x 2 - 18 x + 24)(2 x - 6)
x2 - 6x + 8
= 6 ( x - 3 ) éê log 3 + log x 2 - 6 x + 8 + 1ùú
ë
û
(
(
2
= 6 ( x - 3) log 3e x - 6 x + 8
)
)
For f (x) to be defined x 2 - 6 x + 8 > 0
Þ x < 2 or x > 4
If x > 4 then f '( x) < 0 if log 3( x 2 - 6 x + 8)e < 0
4cos2 x - 8(a + 1)cos x - (4a2 + 8a -12) > 0
Þ
cos 2 x - 2(a + 1) cos x - ( a 2 + 2a - 3) > 0
(cos x - a )(cos x - b) > 0
where
a = (a + 1) + 2a 2 + 4a - 2
and
b = (a + 1) - 2a 2 + 4a - 2
Þ
cos x - b < 0 since cos x - a < 0 "a > 0
Þ
(a + 1) - 2a 2 + 4a - 2 > cos x "x Î R
Þ
(a + 1) - 2a 2 + 4a - 2 > 1
Þ
2
2 a 2 + 4a - 2 < a Þ a + 4 a - 2 < 0
a Î (-2 - 6, 6 - 2)
Þ
\ Max {1, 21 / 2 } = 21 / 2
\ the greatest number is 31 / 3
(a) f '( x) = 2 x log 27 - 6 log 27
Þ
Þ
Max {31/ 3 , 41/ 4 ,51/ 5 , 61/ 6 , 71/ 7 } = 31/ 3
Also 1 < 2 < e
Hence x Î (3 - 1 + 1/ 3e , 2)
(a) Here f ¢ ( x) > 0"x Î R
Þ 2cos 2 x - 8(a + 1)cos x
Hence a Î (0, 6 - 2) since a > 0.
x
,0 < x £ 1
(c) We have f (x) =
sin x
sin x - x cos x
Þ f ¢ ( x) =
sin 2 x
where sin2 x is always +ve, when 0 < x £ 1 . But
to check Nr., we again let h (x) = sin x – x cos x
Þ h ¢( x ) = x sin x > 0 for 0 < x £ 1
Þ h (x) is increasing Þ h (0) < h (x),
when 0 < x £ 1
Þ 0 < sin x – x cos x, when 0 < x £ 1
Þ sin x – x cos x > 0, when 0 < x £ 1
Þ f ' (x) > 0, x Î (0,1] Þ f (x) is increasing on (0, 1]
x
Again g ( x) =
tan x
MATHEMATICS
160
Þ
g ¢( x) =
tan x - x sec 2 x
, when 0 < x £ 1
tan 2 x
Here tan2 x > 0 But to check Nr. we consider
p (x) = tan x – x sec2 x
p ¢ ( x ) = sec 2 x - sec2 x - x.2 sec x.sec x tan x
6.
7.
2
Þ p ¢( x ) = -2 x sec x tan x < 0 for 0 < x £ 1
Þ p (x) is decreasing, when 0 < x £ 1
Þ p (0) > p (x) Þ 0 > tan x – x sec2 x
\ g ¢( x) < 0
Hence g (x) is decreasing when 0 < x £ 1 .
p
-1 u
(c) Given that g(u) = 2 tan (e ) 2
p
-1
\ g(–u) = 2 tan ( e - u ) 2
1
p
æ
ö
-1
= 2 tan ç u ÷ èe ø 2
p
ép
ù p
-1 u
= 2cot ( e ) - = 2 ê - tan -1 ( eu ) ú 2
ë2
û 2
p
p
-1 u
-1 u
= p - 2 tan (e ) - = - 2 tan (e ) = – g(u)
2 2
\ g is an odd function.
2eu
Also g '(u) =
> 0 , " u Î (– ¥, ¥)
1 + e 2u
\ g is strictly increasing on (–¥, ¥) .
x2
in the
(b) Consider the function f (x) = 3
x + 200
interval [1, ¥).
Since the derivative f ¢(x) =
8.
x(400 – x3 )
3
2
is
( x + 200)
positive at 0 < x 3 400 and negative at x > 3 400 , the
function f(x) increases at 0 < x < 3 400 < 8 it follows
that the largest term in the sequence can be either a7
or a8. Since a7 = 49/534 > a8 = 8/89, the largest
49
term in the given sequence is a7 =
.
543
(c) We have,
x2 + 1
f ( x) =
, for all x Î [1, 4)
[ x]
ì
ï x 2 + 1, for all
x Î[1, 2)
ï 2
ï x +1
, for all
x Î[2, 3)
Þ f ( x) = í
ï 2
ï x2 +1
, for all x Î[3, 4)
ï
î 3
9.
ì
ï2 x, for all x Î[1, 2)
ï
for all x Î[2,3)
Þ f ¢ ( x ) = í x,
ï 2x
ï , for all x Î[3, 4)
î3
Clearly, f ¢ (x) > 0 in each of the intervals
(1, 2), (2, 3) and (3, 4). So, f (x) is increasing in each
of these intervals. It is to note here that f (x) is
not increasing on [1, 4] because values of f (x) in
(1.5, 2) are greater than the values of f (x) in (2, 3).
p
(b) We have, A + B =
3
3 - tan A
p
- A Þ tan B =
3
1 + 3 tan A
Let Z = tan A. tan B. Then,
\B =
Z = tan A.
Þ Z=
Þ
3 - tan A
1 + 3 tan A
3x - x
3 tan A - tan 2 A
=
1 + 3 tan A
2
1 + 3x
, where x = tan A
dZ
( x + 3 ) ( 3x - 1)
=dx
(1 + 3x ) 2
For max Z,
dZ
1
=0Þ x =
, - 3.
dx
3
x ¹ - 3 because A+B = p/3 which implies that
x = tan A > 0. It can be easily checked that
d 2Z
dx
2
< 0 for x =
for x =
1
3
1
3
. Hence, Z is maximum
i.e. tan A =
1
3
or A = p / 6.
1
.
3
f '( x) = (4a - 3) + ( a - 7) cos x
For this value of x, Z =
10. (a)
For non-existence of critical point f '( x) ¹ 0 for
any x Î R .
Þ
cos x ¹
3 - 4 a for any x Þ 3 - 4a > 1
a-7
a-7
Þ | 4a - 3 | - | a - 7 |> 0
(i)
a ³ 7 Þ 4a - 3 - a + 7 > 0 Þ a > -
Hence a ³ 7
4
.
3
Solutions
161
3
£ a < 7 Þ 4a - 3 + a - 7 > 0
4
Þ 5a > 10 Þ a > 2 Þ 2 < a < 7
3
(iii) If a < Þ 3 - 4a + a - 7 > 0
4
4
Þ a<3
4ö
æ
\ a Î ç -¥, - ÷ È (2, ¥)
3ø
è
11. (a) Let H be the height of the cone and a be
its semi vertical angle. Suppose that x is the
radius of the inscribed cylinder and h be height
then
h = QL = OL – OQ = H – x cot a
V= volume of cylinder = px 2 ( H - x cot a)
1
Also p = p(H tan a) 2 H
...(i)
3
dV
= p(2 Hx - 3 x 2 cot a )
dx
O
Y
(ii) If
Q
L
So,
a
x
P
M
2
dV
= 0 Þ x = 0 , x = H tan a;
3
dx
d 2V
dx 2 x = 2 H tan a
= -2pH < 0
3
So, V is maximum when x =
2
H tan a and
3
4
1
q = Vmax = p H 2 tan 2 a H
9
3
4 p3 p tan 2 a 4
=
= p [ from (i)]
27 p tan 2 a
9
Hence p : q = 9 : 4
2
12. (a) The graph of y = 2 - x + 5 x + 6 is drawn in
the adjacent figure. Clearly f (x) will have maxima at
x = –2 only if a 2 + 1 ³ 2 Þ | a | ³ 1
2
–3
X
O
–2
13. (b) f '( x ) = sin x cos x(3sin x + 2l)
For maximum or minimum
2l
3
æ 2l ö
p p
\ Critical pointsin æç - , ö÷ arex = 0, sin -1 ç ÷
2
2
è 3 ø
ø
è
One of these is point of minima and other is point
2l
<1
of maxima, provided - 1 < 3
3
3
Þ – <l<
2
2
But if l = 0 , then sinx = 0, which gives only one
critical points
f '(x) = 0 Þ sin x = 0 or cos x = 0 or sin x = -
æ 3 3ö
\ l Î ç - , ÷ - {0}
è 2 2ø
p
14. (a) For 0 < x £ ; [cos x] = 0
2
æ pù
Hence, f (x) = 1 for all çè 0, ú
2û
æ pö
Trivially f (x) is continuous on çè 0, ÷ø
2
This function is neither strictly increasing nor
strictly decreasing and its global maximum is 1.
15. (c) Let R and S be the positions of men P and
Q at any time t. Since velocities are same
\ OR = OS = x (say) and given dx = v and let SR = y
dt
S
x
Q 45°
O P
y
x
R
Now in triangle ORS, applying cosine rule, we get
y2 = x2 + x2 – 2x . x cos 45º = 2 x 2 - x 2 2
\ y = x (2 - 2)
MATHEMATICS
162
dy
dx
= { (2 - 2)}
= v (2 - 2)
dt
dt
Hence the required rate at which they are
Dividing (1) by (2)
being separated is v 2 - 2.
3 ( cos q + sin q ) (cos q - sin q)
dy
=
dx
eq ( sin q + cos q )
\
16. (a) Here y =
2
dy
3(cos 2 q - sin 2 q)
3cos 2q
= q
= q
dx
e (sin q + cos q)
e (sin q + cos q)
2
c
dy
c
Þ
=x+a
dx
+
( x a)2
Slope of normal is Þ
( x + a)
2
c2
> 0 (for all x)
\ x cos a + y sin a = p is normal if,
cos a
> 0 or cot a < 0
sin a
i.e., a lies in II or IV quadrant.
So,
p
3p
æ
ö æ
ö
a Î ç 2 np + , (2n + 1)p ÷ È ç 2np + , (2n + 2)p ÷
2
2
è
ø è
ø
17. (c) Any point on x2 – y2 = a2 is
(a sec q, a tan q)
This point is nearest to y = 2x if the tangent at
this point is parallel to y = 2x
dy x
Now, x 2 - y 2 = a 2 Þ
=
dx y
æ dy ö
= cosec q
ç ÷
è dx ø(a sec q ,a tan q )
cosecq = 2 Þ q =
p
6
p
pö
æ
Hence the point is ç a sec , a tan ÷ i.e.
6
6ø
è
18.
æ 2a a ö
,
ç
÷ , Clearly they lie on the line 2y = x
è 3 3ø
(c) Given, y = 3 sin q.cos q
dy
= 3[sin q(- sin q) + cos q(cos q)]
dq
dy
= 3[cos 2 q - sin 2 q] = 3 cos 2q
...(1)
dq
and x = eq sin q
dx
= eq cos q + sin q eq
dq
dx
= eq (sin q + cos q)
dq
Given tangent is parallel to x-axis then
0=
-
Þ
dy
3(cos q - sin q)
=
dx
eq
...(2)
dy
=0
dx
3(cos q - sin q)
eq
or cos q – sin q = 0 Þ cos q = sin q
p
tan p
Þ tan q = 1 Þ tan q =
Þ q=
4
4
19. (c) We have,
y3 – 3xy + 2 = 0
dy
æ dy
ö
Þ 3y2
- 3 ç x + y÷ = 0
è dx
ø
dx
dy
y
= 2
dx y - x
If the tangent is parallel to x-axis, then
dy
y
=0Þ 2
= 0 Þ y = 0.
dx
y -x
But, y = 0 does not satisfy equation (i). So, there
is no point on the curve where tangent is parallel
to x-axis. Therefore, H = f.
20. (a) We have,
1
p(x) > x2, p(0) = 0, p”(0) =
2
Let g(x) = p(x) –x2
g(x) > 0, " x ¹ 0 and g(0) = p(0) – 0 = 0
Þ x = 0 should be minima.
\ g¢¢(x) should be ³ 0 at x = 0
Now, g¢(x) = p¢(x) – 2x
g¢¢(x) = p¢¢(x) – 2
1
3
g¢¢(0) = p¢¢(0) – 2 = – 2 = –
2
2
But g¢¢(0) ³ 0 Þ No polynimial exists.
21. (40) Let the speed of the train be v and distance
to be covered be s so that total time taken is s/v
hours. Cost of fuel per hour = kv2 (k is constant)
3
Also 48 = k. 162 by given condition \ k =
16
Þ
Solutions
163
3 2
v .
16
Other charges per hour are 300.
Total running cost,
3s
300 s
ös
æ 3
C = ç v 2 + 300 ÷ = v +
16
v
øv
è 16
\ Cost to fuel per hour
24. (9.22) Equation of parabola
y = x2 – 4 and 2y = 4 – x2
dC 3s 300 s
= - 2 = 0 Þ v = 40
dv 16
v
2
d C
dy x + y + 1
=
dx x + y - 1
a+b
dt
=
= ¥ when a + b = 1.
dx (a,b ) a + b - 1
23. (260)
120°
O
B
Let OA = x km, OB = y km, AB = R
(AB)2 = (OA)2 + (OB)2 – 2 (OA) (OB) cos 120°
æ 1ö
R2 = x2 + y2 – 2 xy ç - ÷ = x 2 + y 2 + xy
è 2ø
...(1)
R at x = 6 km, and y = 8 km
R = 62 + 82 + 6 ´ 8 = 2 37
Differentiating equation (1) with respect to t
dx
dy æ dy
dx ö
dR
= 2 + 2y + ç x + y ÷
dt
dt è dt
dt ø
dt
1
[2 ´ 8 ´ 20 + 2 ´ 6 ´ 30 + (8 ´ 30 + 6 ´ 20)]
2R
dR
1
260
k
[1040] =
=
=
dt
2 ´ 2 37
37
37
=
A
B
y=
4 - x2
2
æ 4 – h2 ö
Let point B = ç h,
÷
è
2 ø
æ
4 – h2 ö
A = ç – h,
C = (h, h2 – 4)
÷
è
2 ø
D = (–h, h2 – 4)
\ Area of rectangle ABCD
æ 4 – h2
ö
= AB × BC = 2h × ç
– h 2 + 4÷
è 2
ø
Þ
A
2R
C
600 s
=
> 0 \ v = 40 results in minimum
dv 2
v3
running cost
22. (1) Given x + y – ln (x + y) = 2 x +5
dy
1 æ dy ö
Þ 1+
ç1 + ÷ = 2
dx x + y è dx ø
Þ
y = x2 – 4
D
A = 12h – 3h3 Þ
dA
= 12 – 9h2
dh
For maxima of minima
dA
= 0 Þ 12 – 9h2 = 0
dh
2
2
h= ±
\ maximum at h=
3
3
3
24 8
æ 2 ö æ 2 ö
–
\ A = 12 ç ÷ – 3 ç ÷ =
è 3ø è 3ø
3
3
16 ´173
16 16 3
= 9.22
=
ÞA=
3
3
3
25. (2) Q P(x) is non -zero polynomial and
P(1 + x) = P(1 – x) for all x
Now, differentiate w.r.t. x, we get
P¢(1 + x) = –P¢(1 – x)
Put x = 0, P¢(1) = –P¢(1) Þ P¢(1) = 0
and P(1) = 0 Þ P(x) touch x-axis at x = 1
\ P(x) = (x – 1)2 Q(x)
Since, m is the lergest integer such that
(x – 1)m divides polynomial P(x).
Þ m = 2 such that (x – 1)m divides P(x) for all
such P(x)
=
MATHEMATICS
164
26. (2) f '( x) = (b 2 - 3b + 2)(-2 sin 2 x) + b - 1 ¹ 0
for any x Î R
Þ (b - 1){1 - (b - 2)(2sin 2 x)} ¹ 0
Þ b ¹ 1 and
1
æ 3 ö æ 5ö
> 1 Þ b Î ç , 2 ÷ È ç 2, ÷
2(b - 2)
è 2 ø è 2ø
When b = 2, f (x) = x + sin 3 Þ f '( x) = 1 ¹ 0
æ 3 5ö
\ b Î ç , ÷ Þ integral value of b = 2
è 2 2ø
27. (1) Equation of given curves are
2
2
ax + by = 1
2
...(i)
2
from (i) and (ii), ax 2 + by 2 = a1x 2 + b1 y 2
2
2
or (a - a1 ) x = (b1 - b) y
b -b
= 1
2
a - a1
y
...(vi)
x2
y2
x22 + x12 + x1 x2 = 3 x12
2 x12 - x1 x2 - x22 = 0 Þ x2 = -2 x1
Similarly x3 = -2 x2 = 4 x1
Þ
x4 = -2 x3 = -8 x1
....................................
x2n = -22n -1 x1
= ( x13 + x23 + x33 + ......... + x2 n3 ) + 2n
x3
(( -8) 2 n - 1)
+ 2 = - 1 (82 n - 1) + 2n
-8 - 1
9
i.e. y1 + y2 + y3 + .... + y2n
= x13
3
æx ö
æ 2n ö
= - ç 1 ÷ (26 n - 1) + ç ÷ ,
è 3 ø
è 3ø
Now put n = 30 and x1 = 2 then
y1 + y2 + y3 + ...... + y60 = - æç 2 ö÷
è3ø
dy
= x 2 Þ Equation of tangent at
dx
P1 ( x1 , y1 ) is y - y1 = x12 ( x2 - x1 )
3
( 2180 - 1) + 20
æ 2183 - 8 ö
2183 - 8
+ 20 or S + ç
÷ = 20
è 27 ø
27
Þ 5k = 20 Þ k = 4
29. (9) Sectorial area AOB is removed and the
remaining part be folded into a cone of height
h and radius r.
ArcAB ArcAB
=
= Arc AB
Q q = Angle =
radius
1
Þ
S =-
A
I
O
I
q
q
I
in (v), we get
aa1 æ b1 - b ö
ç
÷ = -1
bb1 è a - a1 ø
a - a1
b - b b - b1
1 1 1 1
or
or
- = =- 1
=
a1 a b1 b
aa1
bb1
bb1
28. (4)
Þ
A
x2
Putting the value of
x23 - x13 = 3x12 ( x2 - x12 )
3( y1 + y2 + y 3 ........ + y2n )
and a1 x + b1 y = 1
...(ii)
dy
dy
a x
From (i) 2ax + 2by
=...(iii)
=0\
dx
b y
dx
dy
From (ii) 2a1 x + 2b1 y
= 0,
dx
a x
dy
\
=- 1
...(iv)
dx
b1 y
Curve (i) and (ii) will cut each other at right
dy
angles if the product of the values of
for
dx
the two curves is –1
æ a x ö æ a1 x ö
aa1 x 2
i.e., ç = -1
÷ = -1 or
֍ bb1 y 2
è b y øè b1 y ø
...(v)
\
i.e.
O
q
I
B
circular sheet
r
B
after removed part
2
2
2pr = 2p - q and r + h = 1
\ volume of cone
1
1
V = pr 2 h = pr 2 1 - r 2
3
3
p2 r 4 (1 - r 2 )
9
2 4
p
(r - r 6 )
Let, y = V 2 =
9
Þ V2 =
I
h
conical vessel
Solutions
165
The point P (-(a - 2), 6) lies on it.
dy p2
=
(4r 3 - 6r 5 ) and
dr
9
\
\
d2y
p2
(12r 2 - 30r 4 )
=
9
dr 2
dy
For max or min of y, = 0
dr
Then,
2
d y
=
or
2
4ö
16 2
p æ .2
çè 12 - 30. ÷ø = - p
9
3
9
27
A
(–a, 0)
or b2 =
\
2
3 = 3+ 6 = m + n
=
2
2
113
3
1+
a2
+
y2
b2
C (0,b)
D
O
9a2
1 ü
ì
= 9 ía + 1 +
ý
(a - 1)
- 1þ
a
î
d (b) 2
1 ïü
d 2 (b 2 )
18
ïì
= 9 í1 =
and
ý
2
2
da
(a - 1)3
da
îï ( a - 1) þï
db 2
=0
da
Indefinite Integration
sin( x - a + a)
dx
sin( x - a)
sin( x - a ) cos a + cos( x - a)sin a
= ò
dx
sin( x - a )
= ò {cos a + sin a cot( x - a )}dx
= (cos a) x + (sin a) logsin( x - a) + C
\ A = cos a, B = sin a
(d) Let gn ( x )
sin x
ò sin( x - a) dx = ò
2
X
B
(a, 0)
9(2)2
Þ b = 6m
(2 - 1)
Hence greatest height of the arch is 6 m.
=1
22
2.
a2
Þ we get a = 2 \ b2 =
m + n = 3 + 6 = 9.
x2
( a - 1)
For extremum value of b,
CHAPTER
(b)
=
6
1 2
.(1) .q
2
30. (6) Equation of the ellipse is
1.
b
2
2
Also, A = p
Þ
9
a
(– (a–2),6) P
æ
2ö
= p(1 - r ) = p çç1 ÷ [from (1)]
3 ÷ø
è
1
=1
Y
1
1
.q = (2p - 2pr )
2
2
A
\
=
A1
b2
a
....(1)
\ Removed sectorial area =
36
+
b
dr 2 r = 2 / 3
\ y is maximum and hence V is also maximum
A1 =
a2
2
or 36 = 1 - ( a - 2) = 4(a - 1)
2
2
2
æ2ö
è3ø
\ r= ç ÷
æ 2ö
at r = ç ÷
è 3ø
( a - 2) 2
4
= 1 + x + x + .... + x
2n
=
= hn ( x ) = g 'n ( x ) =
( x 2 - 1)2
Now f ( x) = lim hn ( x ) =
n®¥
Thus
2x
2
( x - 1) 2
as 0 < x <1
2x
ò f ( x)dx = ò ( x2 - 1)2 dx
1
1
=
+c
= - 2
x - 1 1 - x2
x2n+ 2 - 1
x2 - 1
So, 2 x + 4 x 3 + ..... + 2nx 2 n-1
2 x (nx 2 n + 2 - (n + 1) x 2 n + 1)
3.
(c)
æ f ( x) g '( x) - f '( x) g ( x ) ö æ g ( x ) ö
÷ ln ç
÷ dx
f ( x ) g ( x)
ø è f ( x) ø
ò çè
MATHEMATICS
166
Let
Þ
4.
g ( x)
=t
f ( x)
f ( x ) g '( x) - g ( x) f '( x )
dx = dt ;
( f ( x )) 2
ln t
(ln t )2
dt
=
+C
ò t
2
1 + nx n -1 - x 2 n
(c) We have ò e x
dx
(1 - x n ) 1 - x 2 n
é 1 + xn
ù
nx n -1
x
údx
+
= òe ê
ê 1 - x n (1 – x n ) 1 - x 2 n ú
ë
û
1 + xn
x
=e
+C
1 - xn
Here, f ( x) =
f '( x ) =
and
5.
=ò
(d)
òe
8.
1 + xn
, then
1 - xn
nx n -1
(1 - x n ) 1 - x 2 n
x
( f ( x) + f '( x ))dx = e x f ( x) + C
lim n 2 ( x1/ n - x1/(n +1) )
n®¥
= lim
1
h® 0 h 2
(
x
h
h
h
- x +1
)
æ æ h ö ö h
1 è èç h - h +1ø÷ ø h +1
= lim
-1 x
x
h® 0 h 2
2
æ h
ö
ç x h +1 - 1÷ æ 1 ö h
= lim ç
÷ .ç
÷ x h +1 = ln x.1.1
h®0 ç
h2 ÷ è h + 1ø
çè
÷
h +1 ø
\ f ( x) = ln x
So, I = ò xf ( x)dx = ò x ln xdx
6.
7.
x2
x2 1
x2
x2
ln x =
ln x - ò . dx =
+C
2
2 x
2
4
x 2n - 1
(a) f ( x ) = lim 2n
= -1 (Q 0 < x < 1) ,
n® ¥ x +1
so ò sin -1 x( f ( x))dx = - ò sin -1 x dx
= - [x sin -1 x + 1 - x 2 ] + C
x
æ ln aa x /2
ln bb
(b) I = ò x ç 5 x /2 3 x + 2 x 4 x
ç 3a b
2a b
è
ö
÷dx
÷
ø
ln a2 x b3 x
dx
6a 2 x b 3 x
Let a 2 x b3 x = t .
Then t ln a 2b3dx = dt .
Therefore,
1
ln t
I =ò
dt
2 3 2
6 ln a b t
1
-1 ö
æ - ln t
=
- ò 2 dt ÷
2 3ç
6ln a b è t
t
ø
1
æ ln et ö
=ç
÷+k
6 ln a 2 b3 è t ø
æ ln a 2 x b3 x e ö
1
=÷+k
2 3ç
6ln a b çè a 2 x b3 x ÷ø
(c) I n = (sin x + cos x)n dx
ò
(sin x + cos x) n -1
2( n - 1)
=
(sin x - cos x) +
I n -2
n
n
8
t4
t4
Þ I5 = (sin x - cos x) + I3 = (sin x - cos x )
5
5
5
2
æ
8 t
4 ö
+ ç (sin x - cos x) + I1 ÷
5 çè 3
3 ÷ø
t4
8
= (sin x - cos x ) + t 2 (sin x - cos x )
5
15
32
+ (sin x - cos x) + C
15
(sin x - cos x ) 4
=
(3t + 8t 2 + 32) + C ;
15
where t = (sin x + cos x).
9.
(b) Let I be the given integral. We can see that
cos q + sin q ö
the derivative of loge æç
÷ is 2 sec 2q,
è cos q - sin q ø
so that using integration by parts we have
I=
sin 2q
æ cos q + sin q ö
loge ç
÷
2
è cos q - sin q ø
-ò
=
sin 2q
× (2 sec 2q)d q
2
sin 2q
æ cos q + sin q ö
log e ç
÷ - ò tan 2q d q
2
è cos q - sin q ø
=
sin 2q
æ cos q + sin q ö 1
log e ç
÷ - log e (sec 2q) + c
2
è cos q - sin q ø 2
Therefore
f ( x) = log e (sec 2q).
Solutions
167
10. (d) We have,
1
ò x [log ex e × loge2 x × loge3 x e] dx
1
dx
=ò
x loge ex log ee 2 x.loge e 3 x
=
ò x(log
Put tan x = t
=-
1
e
e + loge x)(log e e2 + loge x)
dx
(loge e3 + loge x)
d (loge x)
=ò
(1 + loge x)(2 + log e x)(3 + log e x)
1
=ò
dt , where t = log x
e
(1 + t )(2 + t )(3 + t )
1
1 1 ö
æ1 1
= òç ×
+ ×
÷ dt
è 2 1+ t 2 + t 2 3 + t ø
1
= log | 1 + log e x | - log | 2 + log e x |
2
1
+ log | 3 + log e x | +C
2
1
= log{log e ex} - log{log e e 2 x}
2
1
+ log{log e e 3 x} + C
2
11. (a) I =
ò ( x - b)
=-
òæ
1
t3 + t 2 + t
t2
1ö
1
çt + 2 + ÷ t +1+
t
t
è
ø
2du
ò 1+ u2
dt
1
t
2
dt , put 1 + t + = u
= -2 tan -1 u + c,
1
tan x
b
sin q
a
2ax dx = 2b sin q cos qd q
b
2b sin q cos q
dq =
cos qd q
Þ dx =
a
2ax
b
cos qd q
b 2 ö
a æ
çè a + b. sin q÷ø . b cos q
a
ò
I=
= a
= a
( x - a)(b - x )
dq
ò (a 2 + b 2 sin 2 q)
sec 2 qd q
ò (a 2 sec2 q + b2 tan 2 q)
Putting tan q= z Þ sec 2 qd q = dz then
dz
I= a
2
a (1 + z 2 ) + b 2 z 2
ò
= a
( x - b) = ( a - b) sin 2 q
2(a - b) sin q cos q dq
ò (a - b) sin 2 q (b - a) sin q cos q
=
2
dq
2
=
cos ec 2 q dq
ò
2
b - a sin q b - a ò
2
2
(- cot q) + c =
cot q + C
=
b-a
a -b
Now x = a sin 2 q + b cos 2 q
=
=
Þ x(1 + cot 2 q) = a + b cot 2 q
dz
ò (a 2 + b 2 ) z 2 + a 2
a
2
(a + b
2
a
a 2 + b2
\I=
Þ x cos ec 2 q = a + b cot 2 q
12. (a) I = - ò
1-
dt
13. (c) Let ax 2 = b sin 2 q Þ x =
dx
x-a
2
; \I=
a -b
b-x
t3 + t2 + t
t 2 -1
where u = 1 + tan x +
Also, ( x - a) = (b - a ) cos 2 q
\cot q =
(t - 1)
ò (t 2 + 2t + 1)
\ I=-
Put x = a sin 2 q + b cos 2 q
dx = 2(a - b) sin q cos q dq
\I =
ò (t + 1)
I =-
)ò
dz
z2 +
a2
2
(a + b2 )
æ z a2 + b2
a 2 + b2
tan -1 ç
a
ç
a
.
è
æ z a 2 + b2 ö
tan -1 ç
÷ + c,
2
2
çè
a
÷ø
a(a + b )
1
where z = tan q .
x-a
+C
b- x
(tan x - 1) sec 2 xdx
(tan x + 1) tan 3 x + tan 2 x + tan x
dx
14.
ö
÷+c
÷
ø
(b) Given ò f ( x ) sin x cos xdx
1
=
ln f ( x) + c
2
2( b - a 2 )
MATHEMATICS
168
Differentiating both sides w.r.t.x then
1
f '( x )
f ( x) sin x cos x =
.
2
2
2(b - a ) f ( x )
f '( x)
Þ 2(b 2 - a 2 ) sin x cos x =
{ f ( x )}2
Þ 2b 2 sin x cos x - 2 a 2 sin x cos x =
+(n - 1) ò (sin x + cos x)n- 2 [2 - (sin x + cos x) 2 ] dx
f '( x)
{ f ( x )}2
Integrating both side w.r.t. x we get
1
-b cos x - a sin x = f ( x)
1
or f ( x) = 2 2
( a sin x + b 2 cos2 x )
sec6α
sec18α
sec54α
15. (a) We have
+
+
cosec2α cosec6α cosec18α
sin 2a sin 6a
sin18a
=
+
+
cos 6a cos18a cos 54a
sin 2a 1
sin 4a
=
Now,
cos 6a 2 cos 2a cos 6a
1 é sin 6a cos 2a - cos 6a sin 2a ù
= ê
ú
2ë
cos 2a cos 6a
û
1
= (tan 6a - tan 2a )
2
sin 6a
1
Similarly,
= (tan18a - tan 6a )
cos18a 2
and sin18a = 1 (tan 54a - tan18a )
cos 54a 2
1
Thus integral = ò (tan 54a - tan 2a ) dx
2
1 é l n | sec54a | l n | sec 2a | ù
= ê
úû + c
2ë
54
2
16. (b) Let
I n = ò (sin x + cos x ) n -1 (sin x + cos x ) dx
2
2
2
2
Take, u = (sin x + cos x )n -1
and dv = (sin x + cos x )dx
so that, v = sin x - cos x
Therefore, using integration by parts we have
I n = (sin x + cos x)n -1 (sin x - cos x)
-ò (n - 1)(sin x + cos x) n- 2
´ (cos x - sin x )(sin x - cos x ) dx
= (sin x + cos x )n -1 (sin x - cos x )
+(n - 1) ò (sin x + cos x)n -2 (cos x - sin x) 2 dx
= (sin x + cos x )n -1 (sin x - cos x )
= (sin x + cos x )n -1 (sin x - cos x )
+2(n - 1) I n -2 - (n - 1) I n
Therefore,
I n + (n - 1) I n - 2(n - 1) I n- 2
= (sin x + cos x )n -1 (sin x - cos x )
Þ nI n - 2( n - 1) I n- 2
= (sin x + cos x )n -1 (sin x - cos x ).
17. (d) We have,
f ' (x) = ln(x2 + 1)
f (x) = ò ln ( x 2 –1) dx
f (x) = xln(x2 – 1) –
2x2
ò x2 - 1dx
æ x 2 –1
1 ö
÷ dx
è x –1 x –1ø
f (x) = xln(x2 – 1) – 2ò ç
2
+
2
æ x –1ö
f (x) = xln(x2 – 1) – 2x – ln ç
+C
è x + 1ø÷
æ 1ö
f (2) = 2ln(3) – 4 –ln çè ÷ø + C = 0 [Q f (2) = 0]
3
Þ C = 4 – 3ln3
( x –1) + 4 – 3ln 3
\ f (x) = xln(x2– 1) – 2x – ln
( x +1)
\ the above function is defined for S
infinite C values possible in set S
such that f ¢ (x) = ln(x2 – 1)
1/ x
f ( x) ö
æ
18. (1) We have lim ç1 + x +
÷
x ø
x ®0 è
= e3
1/ x
æ
f ( x) ö ö
æ
Þ lim ç1 + x ç1 +
÷÷
x ®0 è
x2 ø ø
è
= e3
x
é
ù
ê æ x 2 + f ( x) ö x 2 + f ( x ) ú
Þ lim êç 1 +
÷
ú
÷
x ®0 ê ç
x
ú
è
ø
êë
úû
2
x + f ( x)
Þ
= 3 Þ f ( x) = 2 x2
x2
Therefore
2
ò f ( x)log e x dx = 2ò x loge x dx
x2 + f ( x) 1
x
x
= e3
é x3
x3 1 ù
= 2 ê loge x - ò × dx ú (By Parts)
3 x úû
êë 3
2
2
2 æ
1ö
= x3 log e x - x3 + c = x 3 ç log e x - ÷ + c
3
9
3 è
3ø
Þ a+b=1
Solutions
169
19. (0.60) Let I = ò
ò
(cos 2 x + sin 2 x )
(2 cos x - sin x) 2
dx
(cos x + 2sin x ) cos x
=ò
(sin x )2010 - (cos x )2010
(sin x )
=ò
2010
×
(sin x ) 2011 cos x
(sin x ) 2012 + (cos x )2012
[(sin x)2010 - (cos x) 2010 ]sin x cos x
(sin x )2012 + (cos x ) 2012
dx
dx
Put t = (sin x) 2012 + (cos x )2012 . Then
dt = [2012(sin x )2011 cos x
+2012(cos x )2011 (- sin x )]dx
= 2012[(sin x)2010 - (cos x )2010 ](sin x cos x) dx
Therefore
1æ 1 ö
1
I =ò ç
log e | t | + c
÷ dt =
t è 2012 ø
2012
1
=
log e | (sin x) 2012 + (cos x) 2012 | + c
2012
q
2
dq
q
cos3 q + cos2 q + cos q
2
q
qö
æ
2q
çè 2sin .cos ÷ø .sin
2
2
2
dq
=ò
2q
3
2
2 cos
cos q + cos q + cos q
2
q
2 sin 2 sin q d q
2
=ò
2q
3
2 cos
cos q + cos2 q + cos q
2
Put cos q = t Þ - sin q d q = dt
q
q
Also cos q = 2cos 2 - 1 = 1 - 2sin 2 = t
2
2
1- t
( - dt )
2
\ I= ò
(1 + t ) t 3 + t 2 + t
1
(t 2 - 1)dt
= ò
2 (t + 1)2 t 3 + t 2 + t
æ
1ö
çè1 - 2 ÷ø (dt )
1
t
= ò
2 æ 1 ö
1
çè t + + 2÷ø t + + 1
t
t
1
Put t + + 1 = u 2 Þ æç1 - 1 ö÷ dt = 2u du
t
è t2 ø
1
2udu
I= ò
2 1+ u2 u
21. (3)
dx
(2 cos x - sin x )2
Integrating by part , taking cos x as the first
(cos x + 2 sin x )
as the second function,
and
(2 cos x - sin x ) 2
we have
1
- sin xdx
ì
ü
= cos x í
ý-ò
(2cos x - sin x)
î 2 cos x - sin x þ
1
- sin x dx
ì
ü
= cos x í
ý+ò
(2cos x - sin x)
î 2 cos x - sin x þ
cos x
=
(2 cos x - sin x)
1
2
- (2 cos x - sin x ) - ( -2 sin x - cos x)
5
5
dx
+ò
(2 cos x - sin x )
d
é g
ù
êë N = l Dr + m dx Dr úû
cos x
1
=
(2 cos x - sin x) 5
2 ( -2 sin x - cos x)
ò dx - 5 ò 2 cos x - sin x dx
cos x
1
=
- x
(2cos x - sin x ) 5
2
- ln | 2 cos x - sin x | +c
5
Þ |a + b| = 0.60.
1 - (cot x)2010
20. (2012) Let I = ò
dx
tan x + (cot x) 2011
=
I =ò
sin 3
(
cos
)
1
u = tan -1 1 + + 1 + c
t
-1
1/ 2
= tan (cos q + sec q + 1) + c
So, f (q) = cosq + secq + 1 > 2 + 1 =3
f ( x + h) - f ( x )
22. (2) f '( x ) = lim
h
h® 0
æ æ h öö
f ç x ç 1 + ÷ ÷ - f ( x)
è x øø
= lim è
h
h ®0
æ hö
f ( x) × f ç 1 + ÷ - f ( x)
è xø
= lim
h
h ®0
(Q f ( xy ) = f ( x ) × f ( y ) " x, y Î R )
= tan
-1
æ hö
f ç1 + ÷ - 1
xø
= f ( x ) lim è
h
h ®0
MATHEMATICS
170
ì
hæ
æ h öö ü
1 + ç1 + g ç ÷ ÷ - 1 ï
ï
x
ï
è x øø ï
è
= f ( x ) í lim
ý
h
ï h ®0
ï
ïî
ïþ
ì
1æ
æ h ö öü
= f ( x ) × í lim ç1 + g ç ÷ ÷ ý
è x ø øþ
îh ®0 x è
h
As h ® 0; ® 0
x
æhö
Þ g ç ÷ ® 0 as lim g ( x ) = 0 (given)
x®0
èxø
f ( x)
1
\ f '( x ) = f ( x) × Þ
=x
x
f '( x)
Þò
23. (1)
f ( x)
x2
dx =
+ C Þ k = 2.
f '( x)
2
I = ò {sin(100 x + x) × (sin x)99 } dx
= ò {sin(100 x ) cos x + cos100 x sin x}(sin x )99 dx
= ò sin(100 x ) cos x × (sin x )99 dx
1424
3 1442443
I
II
+ ò cos(100 x ) × (sin x)100 dx
=
sin(100 x)(sin x )100
100
100
cos(100 x)(sin x )100 dx
100 ò
+ ò cos(100 x )(sin x )100 dx
sin(100 x)(sin x )100
+C
100
l 100
Þ l = 100, m = 100 Þ =
= 1.
m 100
dx
24. (8) ò
(cos x - sin x )(1 + sin x cos x)
(cos x - sin x )dx
= 2ò
(cos x - sin x) 2 (2 + (sin x + cos x) 2 - 1)
(cos x - sin x )dx
= 2ò
2
((sin x + cos2 x ) - 2sin x cos x )
(1 + (sin x + cos x ) 2 )
(cos x - sin x)dx
= 2ò
(2 - (sin x + cos x ) 2 )(1 + (sin x + cos x) 2 )
dt
= 2ò
where t = sin x + cos x
2
(2 - t )(1 + t 2 )
dt
2é 1
dt ù
= 2ò
= êò
dt + ò
ú
2
2
2
3
(2 - t )(1 + t )
2 - t2 û
ë 1+ t
=
æ 2 + t öù
2 é -1
1
ln çç
ê tan (t ) +
÷ú + C
3 ëê
2 2 è 2 - t ÷ø ûú
2
1
\A= , B =
3
3 2
2
1
\12 A + 9 2 B - 3 = 12 × + 9 2
- 3 = 8.
3
3 2
sin8 x
sin8 x × cos x
dx = ò
dx
25. (4) I = ò
cos x
cos2 x
=
=ò
sin8 x
(1 - sin 2 x )
× cos x dx
Put sin x = t , cos x dx = dt
t8
t8 -1 + 1
dt = ò
dt
\I = ò
1- t2
1- t2
(t 2 - 1)(t 2 + 1)(t 4 + 1)
1
=ò
dt + ò
dt
2
1- t
1 - t2
1
= ò (t 2 + 1)(-1 - t 4 ) dt + ò
dt
1- t2
1
dt
= ò (-t 6 - t 4 - t 2 - 1) dt + ò
1- t2
-t 7 t 5 t 3
1 1+ t
=
- - - t + ln
+C
7
5 3
2 1- t
- sin 7 x sin 5 x sin 3 x
=
7
5
3
1 1 + sin x
- sin x + ln
+C
2 1 - sin x
2
x
xö
æ
ç sin + cos ÷
2
2ø
Now, ln è
2
...(1)
x
xö
æ
ç cos - sin ÷
2
2ø
è
1 + tan x / 2
æp xö
= 2 ln
= 2 ln tan ç + ÷
...(2)
1 - tan x / 2
è4 2ø
\ From (1) and (2)
- sin 7 x sin5 x sin 3 x
\I =
- sin x
7
5
3
æp xö
+ ln tan ç + ÷ + C
è 4 2ø
7
5
3
- sin x sin x sin x sin x
=
+
+
a
b
c
d
æp xö
+ ln tan ç + ÷ + C (given)
è 4 2ø
Þ a = 7, b = -5, c = 3, d = -1
Þ (a + b + c + d ) = (7 - 5 + 3 - 1) = 4.
Solutions
171
CHAPTER
Definite Integration
23
1.
(a) Since the required function is a polynomial,
the absciassae of the points of inflection can
only be among the roots of the second derivative.
Consequently
p ''( x ) = ax( x - 1)( x + 1) = a ( x3 - x )
Since at the point x = 0, p '(0) = tan 60° = 3,
x
æ x 4 x2 ö
P '( x ) = P ''( x )dx + 3 = a ç
- ÷+ 3
4
2ø
è
0
hen, since P(1) =1, we get
P (x) =
x
æ x 5 x3 7 ö
- + ÷ + 3( x - 1) + 1
P '( x )dx + 1 =a ç
è 20 6 60 ø
ò
ò
1
Since, P (-1) = -1 so , a =
Thus
1
60( 3 - 1)
7
1
é 3 -1 5
ù
3
ò P( x)dx = ò êë 7 (3x - 10 x ) + x 3 úûdx
0
0
3 - 1 æ x6 5 4 ö x2
3
ç - x ÷+
7 è 2 2 ø 2
=
0
3 -1 æ 1 5 ö
3 3 3 2
=
+
ç - ÷+
7 è 2 2ø 2
14 7
=
p
2.
1
I=
(d)
ò
(cos2 px + sin2 qx – 2 cos px sin qx) dx
-p
sin 2 qx, cos2 px
are even functions of x and
Q
cos px . sin qx is an odd function.
p
ò
\
p
ò
-p
-p
p
cos 2 px dx = 2 ò cos 2 px dx
p
0
p
p
2
2
I = 2 ò cos px dx + 2 ò sin qx dx = 0
\
p
0
p
p
p
0
0
I=
(c)
òe
cos q
cos(sin q)d q = Real part of
0
2p
òe
cos q
{cos(sin q) + i sin(sin q) d q
0
2p
= Real part of
òe
cos q i sin q
e
d q = Real part of
0
2p
òe
cos q+i sin q
2p
d q = Real part of
0
òe
ei q
dq
0
= Real part of
2p é
2iq
ù
e3iq
iq e
1
+
e
+
+
+ ............ ú d q
ê
òê
2!
3!
ûú
0 ë
= Real part of
2p
é
ù
1
ò êë1 + (cos q + i sin q) + 2! (cos 2q + i sin 2q) + ...úû d q
0
=
2p
é
ù
1
ò êë1 + cos q + 2! cos 2q + .......úû
0
dq
2p
sin 2 q
é
ù
= ê q + sin q +
+ .......ú = 2p
2.2!
ë
û0
4. (a) Let q = p + d , r = p + 2d , s = p + 3d
p + sin x
p + d + sin x -2d + sin x
\ f (x) = p + d + sin x p + 2d + sin x -1+ sin x
p + 2d + sin x p + 3d + sin x 2d + sin x
ò (1 + cos 2 px) dx + ò (1 - cos 2qx) dx
0
2
p + d + sin x
p + 2d + sin x -1 + sin x
p + 2d + sin x
p + 3d + sin x 2d + sin x
= 2[( p + d + sin x)( p + 3d + sin x)
-( p + 2d + sin x )2 = -2d 2
0
æ 1 + cos 2 px ö
æ 1 - cos 2qx ö
= 2ò ç
÷ dx + 2 ò ç
÷ dx
2
2
è
ø
è
ø
0
0
=
3.
f ( x) =
ò cos px sin qx dx = 0
-p
2p
0
0
p
p
Applying R ® R1 + R3 - 2 R2 , we get
sin 2 qx dx = 2 ò sin 2 qx dx
and
p
sin 2qx ù
é sin 2 px ù é
= êx +
ú + ê x - 2q ú = 2p
p
2
ë
û0 ë
û0
2
ò f ( x)dx = -4
Given
Þ
2
0
ò ( -2d
0
2
)dx = -4 d 2 = 1 Þ d = ±1
MATHEMATICS
172
p /2
5.
dx
ò a2 cos2 x + b2 sin2 x
(a)
0
p /2
sec2 x dx
ò b2 tan 2 x + a 2
=
=
0
=
Þ
1
b2
¥
ò
0 t2
dt
æ aö
+ç ÷
è bø
1
b
2
p/2
ò
0
Þ b-a £
2
sec x dx
æ aö
tan 2 x + ç ÷
è bø
2
(Putting tan x = t )
2
=
8.
1
p
24
ò
ò
7.
p/4
3
1
f ( x ) f '( x )
Þ
³1
f ( x)
1 + f 4 ( x)
Integrating on the interval (a, b), we get
b
b
f ( x ) f '( x)
dx
³
ò 1 + f 4 ( x)
ò dx
a
a
(d)
f '( x ) ³ f ( x ) +
1
ò 0 (1 + cos
8
x )( ax 2 + bx + c ) dx
2
2
ò1 (1 + cos8 x)(ax 2 + bx + c) dx
+
2
ò 1 (1 + cos
Þ
- a 2 + b2 , which is least if a = b = 2 2
6.
³ b-a
ò 0 (1 + cos8 x )(ax 2 + bx + c) dx
1
= ò (1 + cos8 x )(ax 2 + bx + c ) dx
0
¥
Þ - a 2 + b 2 ³ - 8 + 8 = -4
sin x
(a) Let f ( x) =
x
x cos x - sin x
\ f '( x ) =
x2
( x - tan x) cos x
ép p ù
< 0" x Îê , ú
=
2
ë4 3 û
x
sin x
\ f ( x) =
is decreases on the interval
x
ép p ù
ê4, 3 ú
ë
û
Þ The least value of the function
æ p ö sin( p / 3) 3 3
=
m= f ç ÷ =
è 3ø
( p / 3)
2p
and the greastest value of the function
æ p ö sin(p / 4) 2 2
=
M = fç ÷=
è 4ø
( p / 4)
p
therefore
p/3
sin x
æ p pö 3 3
æ p pö 2 2
<
dx < ç - ÷
èç 3 4 ø÷ 2p
è 3 4ø p
x
p/4
[Mean Value Theorem of Integral Calculus]
p/3
3
sin x
2
Hence,
<
dx <
x
8
6
b
a
ù
1é
-1 2
-1 2
ê lim tan f ( x) - lim tan f ( x) ú
+
2 ëê x ® b
x®a
ûú
(b) Given
=
1 æ pö
b
bt
p
´ tan -1
= ç ÷=
2 a
è
ø
a
ab
2
2
ab
b
0
p
p
=
Thus
Þ ab = 8
2ab 16
Now minimum value of a cos x + b sin x is
=
1
tan -1 f 2 ( x )
2
8
x )( ax 2 + bx + c ) dx = 0
Now we know that if
b
òa f ( x)dx = 0
then it
means that f (x) is + ve on some part of (a, b) and
– ve on other part of (a, b).
But here 1 + cos8 x is always + ve,
\ ax2 + bx + c is + ve on some part of [1 , 2]
and – ve on other part [1, 2]
\ ax 2 + bx + c = 0 has at least one root in (1, 2).
Þ ax2 + bx + c = 0 has at least one root in (0,2).
x2
9.
f ( x) =
(c)
ò (t - 1)dt
x
2
Þ
f '(x) = (x -1)2x - (x -1) = (x -1)(2x2 + 2x -1)
\
f '(w ) = (w - 1)(2w 2 + 2w - 1)
= (w - 1){2( -1) - 1} = 3(1 - w )
| f '(w) |= 3 | 1 - w |
\
Now,
| 1 - w |2 = (1 - w )(1 - w) = (1 - w )(1 - w 2 )
10.
= 1 - w - w 2 + w3 = 3
\ | 1 - w |= 3 Þ | f '(w) |= 3 3
(a) We have
d é h ( x)
ù
f (t ) dt ú = f ( h( x)) h '( x) - f ( g ( x)).g '( x)
dx êë g ( x )
û
ò
=
f (2) ´ 2 ´ 2 ´ 1 8
= f (2)
p
p
2´
4
2
sec x
Let L = lim
p
x®
4
ò2
x2 -
f (t )dt
2
p
16
é0
ù
êë 0 form úû
Solutions
173
On applying L' Hospital's rule, we get
2
ù
d é sec x
f (t )dt ú
ê
dx êë 2
úû
L = lim
2
p
d æ 2 p ö
x®
x - ÷
4
dx çè
16 ø
If is possible when f (x) = 0.
Hence, " p ÎR. There is no non-zero continous
ò
L = lim
p
x®
4
function.
Hence, S ÎR.
13. (a) We have,
f (sec 2 x ).2sec 2 x tan x
2x
Q
1
1
<
(1+ x 2 )n 1+ nx 2
1
1 dx
dx
Þ ò
<ò
2
n
0 (1 + x )
0 1 + nx 2
1
1
dx
Þ ò
<
[tan –1 nx ]10
0 (1 + x 2 ) n
n
1
1
dx
Þ ò
tan –1 n
<
0 (1 + x 2 ) n
n
Þ L
n
1
1
(tan –1 n )
= lim n ò
dx < lim
n
2
0
x®¥ n
x ®¥
(1+ x )
Þ
pö
p
p
æ
f ç sec 2 ÷ .2.sec 2 . tan
8
4
4
4
è
ø
=
= . f (2)
p
p
2.
4
11. (c) Let I =
p
p
ò0 (1– | sin 8 x) |) dx
p
ò0 dx – ò0
Þ
I=
Þ
I= p–
| sin 8 x ) | dx
p 8´ 8
ò
| sin 8 x | dx
0
Þ
p
8
I = p –8ò sin8x dx
0
[Q sin 8x is periodic with period
p /8
p
]
8
é – cos8 x ù
I = p –8 ê
ë 8 úû 0
Þ I = p + (cos p – cos0) = p – 2
12. (d) Given,
Þ
0
x
ò0
f (t )dt = 0, " x Î R
L < tan –1 ¥ =
\
1
<L<2
2
14. (a) f (x) = 2ò t f (t )dt + 1
...(i)
0
...(1)
Differentiating, we get,
Þ f (x) = p f ¢(x)
...(2)
f '( x ) 1
=
Þ
f ( x) p
Integrating, we get,
x
Þ log f (x) = p + C
Þ f (x) = Aex/p
...(3)
Putting x = 0 in eqn. (i) we get,
0 = p f (x)
pf (0) = 0
Case I f (0) = 0, p ¹ 0
Þ A=0
(from 2)
Þ f (x) = 0
pf (0) = 0
Þ f ¢ (x) = 0
(from (2))
Þ there is no non-zero continuous to f (x).
Case II p = 0
\
p
2
é p ù
êëQ 2 < 2 úû
Þ
x
x
ò f ( x)dt = pf ( x)
1
1
dx
0
x ®¥
(1+ x 2 )n
(1 + x2)n > 1 + nx2
nò
L = lim
\ f (0) = 1
Now, differentiate (i) w.r.t. x
f ¢(x) = 2x f (x)
f ¢ ( x)
= 2x
f ( x)
Integrate both sides, we get
ln | f (x) | = x2 + c
Q f (0) = 1 Þ c = 0
Then from eqn. (ii), ln | f (x) | = x2
Þ
x
f (x) = e
1
15. (a) J =
2
Þ f (1) = e
x
ò 1 + x8 dx
0
Q 0 < x8 < 1
1
1
x
1 é x2 ù
1
Þ J > ò 2 dx Þ J > 2 ê 2 ú Þ J >
4
ëê úû0
0
1
1
é x2 ù
x
1
J < ò 1 + 0 dx Þ J < ê 2 ú Þ J <
2
ê
ú
ë û0
0
Hence, (I) is true and (II) is false.
...(ii)
MATHEMATICS
174
16. (a) It is given that f (x) = | sin x |,
( f ( x ))2 - 42
x -1
x ®1
2 f ( x). f ¢( x )
= lim
(L-Hospital rule)
1
x ®1
= 2f (1) . f ' (1) = 2 × 4 × 2 = 16
x
= lim
ò f (t ). dt
g(x) =
0
2
x
p
and p(x) = g(x) –
2
(x + p)
p
Now, p(x + p) = g(x + p) –
p+ x
p
0
- tan x
2
x-2
ò
=ò
p
p
0
Q f (x) is periodic function with period p
p+ x
\
ò
p
f (t )dt -
ò
1
sin x dx + g ( x ) -
1
1
2
1
2
f ( x)
ò 1- x
Þ
=
2
1
2
dx =
2
ò
1
2
1 - x2
2
(1 - x )
dx
2
2
0ö
æ
3 2
- e- x + 1
ç form ÷
= . lim
2
0ø
è
5 4 x®0
x
Again using L' Hospital’s rule we get
2
3
-e - x .(-2 x )
3
3
=
lim
1 = = 0.30
10 x® 0
(2 x)
10.
10
p/2
= ésin
ë
-1
p p
p
xù
= - =
û1/2
4 6 12
1/ 2
f ( x)
ò
18. (16) lim
x ®1
ò
q.sin n q d q .
ò
q.sin n - 2 q(1 - cos 2 q) d q
ò
(q.cos q).cos q.sin n - 2 qd q
0
p/2
1 - x2
1
2
4
x -1
Þ
I ( n) =
0
p/2
= I (n - 2) –
2t dt
= lim
x ®1
2
=
3
e - x .( -2 x ) - 0 + 2 x
lim
5 x® 0
4 x3
21. ( 2) I (n) =
dx
ò
3e - x - 3 + 3x 2
0ö
æ
ç form ÷
0ø
è
3
e- x - 1 + x 2
lim
5 x® 0
x® 0
5x4
x4
0ö
æ
ç form ÷
0ø
è
Again using L' Hospital’s rule, we get
lim
éx
ù
2 1
-1
ò f ( x)dx £ êë 2 1 - x + 2 sin ( x) úû0
0
1
p
ò f ( x)dx £ 4 Þ f (x) = 1 - x2
2
x5
Using L’ Hospital’s rule we get
0
0
1
2
x®0
1
0
1
Þ
f ( x) dx = 0.
0
20. (0.30) We have lim
2
x-2
p
f ( x )dx £ ò 1 - x 2 dx
ò
Þ
ò
0
f (x) £ 1 - x 2
Þ
x2 + 1
x
2
= 2 + g(x) – x – 2
p
Þ p(x + p) = p(x) for all x
(a) If is given that, x2 + (f (x))2 £ 1
Þ
x tan x
3ò e-t dt - 3x + x3
0
17.
cos x
-p /3
x
p
x sin x
p /3
Þ
f (t )dt = ò f (t )dt
Þ p(x + p) =
x2 -1
= - f ( x ) Þ f ( x) is an odd function
p+ x
f (t )dt +
f (- x ) = -cosec x
19. (0)
2
f (t )dt - x - 2
p
ò
=
sec x
- sin x
(t 2 ) |4f ( x )
x -1
0
= I (n - 2) - q.cos q.
sin n -1 q p / 2
n -1 0
Solutions
175
p/2
+
ò
0
n -1
[q.(- sin q) + cos q]. sin q d q
n -1
1
= I (n - 2) (n - 1)
p/2
ò
n
q.sin q d q
ò
cos q. sin
n -1
q dq
0
x
ò
ò
22. (0) We have x (1 - t ) f (t )dt = t f (t )dt
0
0
Differentating both sides with respect to x
we get,
x
2
2q
sin q cos p +
cos qt.cos ptdt
p
p
ò
0
1
ù
q
sin pt sin qtdt ú
p
ú
0
û
2
2 q æ sin p cos q q I ö
= - sin q cos p +
+
p
p çè
p
p 2 ÷ø
é q2 ù
2
2q
or, I ê1 ú = - sin q cos p + 2 sin p cos q
2
p
p
ëê p ûú
+
3
24. (2)
ò
0
Differentiating again with respect to x on both
sides, we get
x 2 f ¢( x ) + 2 xf ( x ) = (1 - x ) f ( x)
f ¢ ( x) 1 - 3 x
= 2
Þ
f ( x)
x
Integrating both the sides, we get
1
ln f ( x ) = - + 3lnx + l
x
1
3
Þ ln éë x f ( x) ùû + = l and f (1) = 1
x
Þ l =1
Þ
f ( x) =
æ 1ö
çè 1- ÷ø
x
e
x3
Thus lim f ( x) = 0
x ®¥
.
1
3
0
1
ò [ x ] dx = ò [ x ] dx + ò [ x ] dx
0
é
êQ
ë
0
x
x 2 f ( x ) = (1 - t ) f (t )dt
ò
é 2q tan p 2
ù
= cos p cos q ê
- tan q ú
2
p
êë p
úû
As p and q are different roots of equation,
tanx = x
Þ tan p = p and tan q = q
\ I = 0.
ò
1
0
1
2
2q éæ
sin pt ö
êç cos qt
= - sin q cos p +
p
p êè
p ÷ø 0
ë
x(1 - x ) f ( x) + (1 - t ) f (t )dt = xf ( x )
Þ
ò
1
p/2
1
1
.I (n) +
.sin n q
= I (n - 2) 0
(n - 1)
( n - 1)( n)
n
1
Þ
I (n) = I (n - 2) +
,
n -1
(n - 1)(n)
n -1 1
Þ I (n) - I (n - 2).
= 2.
n
n
1
Þ n I ( n ) - ( n - 1) I ( n - 2 ) =
n
Put n = 2010
1
Þ 2010 I ( 2010 ) - 2009 I ( 2008) =
2010
-1
2009
I
(2008)]
[2010
I
(2010)
= 2010
Þ
x
1
1
æ
cos pt ö
cos pt
I = ç -2 sin qt
+ 2q cos qt.
dt
÷
è
p ø0
p
=-
0
p/2
1
+
n -1
23. (0) Using integration by parts
1
3
0
1
ì0, if
éë x ùû = í
î1, if
0 £ x < 1ù
ú
1 £ x < 3û
= ò 0 dx + ò 1dx = [ x ]1 = 2
3
x
25. (7)
ò f (t )dt
F ( x) 1
lim
=
Þ lim x –1
x ®1 G ( x) 14
x®1
ò t f ( f (t )) dt
–1
1
1
Q ò-1 f (t )dt = 0 and ò-1t f ( f (t )) dt = 0
f(t) being odd function
\ Using L Hospital’s rule, we get
f ( x)
1
lim
=
14
x ®1 x f ( f ( x ))
f (1)
1
1/ 2
1
= Þ
=
Þ
f ( f (1)) 14
æ 1 ö 14
fç ÷
è 2ø
Þ
æ 1ö
æ 1ö
f ç ÷ =7 Þ fç ÷ =7
è 2ø
è 2ø
MATHEMATICS
176
CHAPTER
Application of Integrals
24
b
1.
(d) Given
ò f (x)dx =
b2 +1 - 2
= 2[ln | sec x ||0p / 4 +ln | sin x ||pp // 24 ]
1 ù
é
= êln 2 + ln 1 - ln
ú = 2[2ln 2] = ln 4.
2û
ë
y
1
2.
Differentiate with respect to b
x
b
Þ f ( x) =
f (b) =
2
x2 +1
b +1
(b) If 1 £ x < 2 Þ [x] = 1 Þ [ y] = ±1
Þ y Î[-1, 0) È [1, 2)
5.
p/4
O
p/2
x
(d) Eliminating y from two equations,
We get 2mx + 4 = x 2 Þ x 2 - 2m - 4 = 0
x1 and x2 are roots of this quadratic equation
3.
y = mx + 2
If 2 £ x < 3 Þ [x] = 2 Þ y = ±2
Þ y Î[-2, -1) È [2, 3)
If 3 £ x < 4 Þ [x] = 3 Þ y = ±3
Þ y Î[ -3, -2) È [3, 4)
If 4 £ x < 5 Þ [x] = 4 Þ y = ±4
Þ y Î[-4, -3) È [4, 5)
Clearly, the required area consists of eight
squares, each of area unity (see figure)
\ Required area = 8 sq. units
(a) As point lies on the line. Locus of the point
is straight line perpendicular to given line passing
through (2 3, -1) i.e. y +
x
3
2
x1
=1
=
y = 3x - 7
P
6.
4.
(c) Here f ( x ) = min{| tan x |,|cot x |}
p/4
Required area = 2
ò
0
p/2
tan xdx + 2
ò
p/4
cot xdx
x2
\ x1 + x2 = 2m and x1x2 = -4
x2
æ
x2 ö
2
mx
+
ç
÷ dx
Now, ò ç
2 ÷ø
x1 è
1 3
m 2
2
3
= ( x2 - x1 ) + 2( x2 - x1 ) - ( x2 - x1 )
2
6
1 2
ém
2 ù
= ( x2 - x1 ) ê ( x2 + x1 ) + 2 - ( x1 + x1x2 + x2 )ú
6
ë2
û
3 ´1
.
Þ required area =
2
O
O
( x2 + x1 ) 2 - 4 x1 x2
1
ém
ù
2
ê 2 ( x2 + x1 ) + 2 - 6 {( x2 + x1 ) - x2 x1}ú
ë
û
é m2 + 4 ù
2
= 4m + 16 ê
ú
êë 3 úû
Clearly above is minumum if m = 0
(d) Required area = 4A , where
p
p
0
0
A = ò ( x + sin x)dx - ò xdx
Solutions
177
Both the curves pass through origin.
1
y=x
ò ( xe x - xe- x )dx
\ Required area =
A
=
–1
f (x)
p
7.
2a
ò
2a
ydx =
0
2a - x
0
dx
Put x = 2a sin2q and dx = 4a sin q cosq dq
p/2
ò 8a
I=
2
1ö
æ
x
= çe + ÷ - e
è
eø
9.
0
=
Y
2
3/2
1
X'
–1
1/2
O 1/3
3
y = x ex y = xe–x
=
2y
y=
+2
Y
1
3
3pa 2
sq. unit
2
(a)
0
+ e- x
3x
8.
0
1
2
e
(a) If both x + y - 1 and 2 x + y - 1 are
positive, then x + y – 1 + 2x + y – 1 = 1
Þ 3x + 2y = 3
x
y
Þ
+
=1
1 3/ 2
If x + y - 1 is positive and 2 x + y - 1 is
negative, then (x + y – 1) – (2x + y – 1) = 1
Þ – x – 1 = 0 Þ x = –1
If first negative and second positive, then
–( x + y – 1) + (2x + y – 1) = 1
Þ x–1=0
Þ x=1
If both negative, then
– (x + y – 1) – (2x + y – 1) = 1 Þ– 3x – 2y + 1 = 0
x
y
+
=1
Þ – 3x – 2y + 1 = 0 Þ
1/ 3 1/ 2
ò
=
ò
+
3x
é3 1 pù
sin 4 q dq = 8a 2 ê . . ú
ë4 2 2û
0
m +1 n +1ö
p/2
G
G
æ
2
2 ÷
ç using
sin m x cos n x dx =
m+n+2 ÷
çè
2G
÷
0
ø
2
\
1
1
x
-x
x
-x
= x(e + e ) |0 - (e + e )dx
2p
x3 / 2
ò
ò x(e x - e- x )dx
0
p2
p2
A = - cos p + cos0 =2 sq. units.
2
2
(b) Let the equation of curve
y2(2a – x) = x3
...(i)
and equation of line x = 2a
...(ii)
The given curve is symmetrical about x-axis and
passes through origin.
x3
From (i) we have, y2 =
2a - x
x3
But
< 0 for x > 2a and x < 0
2a - x
So, curve does not lie in the portion x > 2a and
x < 0, therefore curve lies in 0 £ x £ 2a .
\ Area bounded by the curve and line
=
0
1
p y = sin x + y = f(x)
1
X
–1
1,e
A(
O
Y'
)
B(1, 1
e)
x=1
\
X
Given curves are y = xe x and y = xe - x
æ 1ö
Line x = 1 meets the curves at A(1, e) and B ç 1, ÷ .
è eø
Required area
2
ö 1
æ1
ö 1æ4
= ç ´ 2 ´ 3÷ - ç ´ 2÷ + ´ 1 ´
è2
ø 2è3
ø 2
3
= 2 sq units
10. (c) y = 4 - x 2 is the parabola EMA
y = |x – 2| is the pair of straight lines DCA and ABF,
y = (x – 2)1/3 is the curve IAHBG
MATHEMATICS
178
Y
D (–2, 4)
G
C(–1, 3)
y = (x–2)
H
E (–2, 0)
Also when x ® 0, t ® 0 and when x ®
y = x –2
M (0, 4)
O
1/3
B (3, 1)
p
8
tan
X
A (2, 0)
t ® tan
\ A=
ò
p
8
2
(1 + t ) 1 - t
0
I (0, –2)
Thus f (x) = max. {4 – x2, |x – 2|, (x – 2)1/3} is
2 -1
4t
-1
ò
2
(2 - x )dx +
ò
y = x3
y = x2
(p ,0)
(0.0)(1,0) x = p
0
4
1
p/4
=
ò
0
2 tan
x
2
1 - tan 2
1 x3 x4
=
6 3
4
Þ
1
0
+
p
0
(
)
1
1 æ 1 1 ö æ p4 p3 1 1ö
= ç - ÷ +ç
- + ÷
6 è 3 4ø è 4
3 4 3ø
1 1 1 1 1 3 p 4 – 4 p3
– + + – =
6 3 4 4 3
12
3
p ( 3 p - 4)
Þ
= 0 Þ p3 (3p – 4) = 0
12
4
Þ p = 0 or
3
Since, it is given that p > 1 \ p can not be zero.
4
Hence, p =
3
13. (d) Let I be the smaller portion and II be the
greater portion of the given figure then,
Þ
Y
x
2
1
dx
x
1
x
= t Þ sec2 dx = dt
2
2
2
2
dt
Þ dx =
1+ t2
Let tan
x 4 x3
4
3
p
2
0
æ
x
xö
1 - tan ÷
ç 1 + tan 2
2
ç
÷ dx
x
ç 1 - tan x
1 + tan ÷
è
2
2ø
)
x+
7
3 3 59
sq. units.
+9+ + =
2
4 2 4
11. (b) The given curves are
x
1 + tan
1 + sin x
2
y=
=
...(1)
x
cos x
1 - tan
2
x
1 - tan
1 - sin x
2
=
and y =
...(2)
cos x
x
1 + tan
2
\ The area bounded by the above curves, by
p
the lines x = 0 and x =
is given by
4
=
(
2
3
3
2
Required area = ò x - x dx + ò x - x dx
3
ò
dt
ò
ò
A=
(1 + t ) 1 - t 2
3
+ ( x - 2)dx
p/4
0
4t
2
(4 - x) dx + ( x - 2)1/3 dx
-1
2
ò
p
,
4
12. (d) Given curves are y = x2 and y = x3
Also, x = 0 and x = p, p > 1
Now, intersecting point is (1, 1)
ì 2 - x - 2 £ x < -1
ï
2
ï4 - x , - 1 £ x < 2
f ( x) = í
1/ 3
ï( x - 2) , 2 £ x < 3
ï x - 2, 3 £ x £ 4
î
Now, area bounded by the curve and x-axis.
=
2
dt =
y=
y=2–x
(0, 2)
I
II
X¢
(– 2, 0)
Y¢
X
Solutions
179
0
The graph of given region is as follows-
Area of I = ò éê 4 - x 2 - ( x + 2 ) ùú dx
û
-2 ë
0
0
é x2
ù
éx
4
æ xö ù
4 - x 2 + sin -1 ç ÷ ú - ê + 2 x ú
=ê
è 2ø û
2
ë2
-2 ëê 2
ûú -2
4
p
é
ù
1
= ëé 2sin ( -1) ûù - ê - + 4ú = 2 ´ - 2 = p - 2
2
ë 2 û
Now, area of II = Area of circle – area of I.
= 4p – (p – 2) = 3p + 2
area of I
p-2
Hence, required ratio =
=
area of II 3p + 2
14. (a) The given curve is y = tanx
...(1)
p
when x = , y = 1
4
Equation of tangent at P is
pö
æ 2 pö æ
y – 1 = ç sec ÷ ç x - ÷
è
4ø è
4ø
Y
y = 2x + 1 –
p
2
æp ö
P ç ,1÷
è4 ø
O
X¢
L
M
X
y = tan x
Y¢
p
or y = 2x + 1 –
2
Area of shaded region
= area of OPMO – ar (DPLM)
=
p
4 tan x
0
ò
...(2)
(1, 2)
(– 4, 1)
P
S
Q
R
T
(– 3, 0)
Required area
= Area (trap PQRS) – Area (PST + TQR)
1
1
é -3
ù
= ´ (1 + 2 ) ´ 5 - ê ò-4 -x - 3 dx + ò-3 x + 3 dx ú
2
ë
û
éæ
3/2 ö-3 æ
3/2 ö1 ù
2 ( x + 3)
15 ê 2 ( -x - 3)
÷ +ç
÷ ú
- êç
=
ç
÷
ç
÷ ú
2
-3
3
ø-4 è
ø-3 úû
êëè
15 é 2 16 ù 15
3
- +
- 6 = sq. units
=
=
2 êë 3 3 úû
2
2
16. (b) Since,
y = x 2 + x +10
1
1
Þ x2 + x + = y –10 +
4
4
2
æ 1 ö æ 39 ö
Þ ç x + ÷ =ç y - ÷
è 2ø è
4ø
Þ Latusrectum of the above parabola is 1.
\ length of chord is also 1.
Area is maximum when chord is latusrectum.
Area of shaded reion =
Area of rectangle OABC – 2 ò
1
dx - (OM - OL)PM
2
B
-1
,10
2
1 ì p p - 2ü
= [ log sec x ] - í ý ´1
2î4
4 þ
1é
1ù
= êlog 2 - ú sq. unit
2ë
2û
y³
0
x + 3 Þ y2 = x + 3
ìï ( x + 3 ) if x < -3
Þ y2 = í
ïî ( x + 3 ) if x ³ -3
O
–1/2
= 10 – 2 ò ( x 2 + x +10)dx
....(i)
x+9
and x £ 6
....(ii)
5
Solving (i) and (ii) we get intersection points as
(1, 2), (6, 3), (– 4, 1), (–39, –6)
Also y £
A
–1
2
parabola
C(10, 10)
p
4
0
15. (c)
0
–1/2
0
é x3 x2
ù
= 10 – 2 ê + +10 ú
ëê 3 2
ûú –1
2
é –1 1 10 ù 1
= 10 – 2 ê + – ú =
ë 24 8 2 û 6
MATHEMATICS
180
sin x x
sin x x 2
+ e and y =
+
x
x
2
Required area of the region
æ sin x x 2 ö üï
2ì
ï sin x
x
+
e
+ ÷ ý dx
í
= ò1
ç
2 øï
è x
îï x
þ
2
2
3
é x x ù
2æ x x ö
= ò1 ç e - 2 ÷ dx = ê e - 6 ú
è
ø
êë
úû1
1ö
7
2 8 æ
2
= e - -çe- ÷ = e -e6 è
6ø
6
18. (1) The graph of f (x) = min{x – [x], – x – [–x]}
is shown as in figure. Therefore
Y
17. (b) y =
O
1
21. (4) The desired region consists of four disjoint
squares. So the area = 1 × 4 = 4 square unit.
6
5
4
3
-2 -1
2
ò min
-2
1ö
æ1
= Area of shaded region = 4ç ´ 1 ´ ÷ = 1
2
2ø
è
Y
1
2
–2 –3/2 –1 –1/2
0 1/2
1
A=4
log e n
ò
0
3/2 2
X
22. (1) The curves y = [a] x2 and
1
y = [ a ] x 2 represent parabolas which are
2
symmetric about y-axis. The equation
y2 – 3y + 2 = 0 gives a pair of straight lines
y = 1, y = 2 which are parallel to x-axis.
Thus the area bounded is shown as;
Y
2
y = [a]x y =1/2 [a]x
æ n - 1 - loge n ö ...(1)
æ -x 1ö
çè e - ÷ø dx = 4 çè
÷ø
n
n
1–
y=1
1
n
logen
1
–1+
n
Now, n - 2 ³ 2loge n Þ 2n - 2 - 2log e n ³ n
n - 1 - log e n 1
Þ
³
n
2
1
=2
2
20. (3.33)
Here both functions f (x) and g(x)
are periodic. Thus required area
\ From (1), A ³ 4 ´
1
æ1
ö
é x 3/2 x 3 ù
2
ç
÷
= 10 ( x - x )dx = 10 ê
- ú
ç
÷
êë 3 / 2 3 úû 0
è0
ø
10
= = 3.33.
3
ò
2
y=2
O
– logen
l
1 2 3 45
-2
-3
{x - [ x ], - x - [- x ]}dx
19. (2) The area of the region is
X
3
2
X
From above figure the required area
y =2
=2
=
=
2
æ
ö
2y
y
ò ( x2 - x1 ) dy = 2 ò ç [ a ] - [ a ] ÷ dy
ø
y =1
1 è
2( 2 - 1)
[a]
2
ò
y dy =
1
2( 2 - 1) 2 3/2 2
- .( y )1
[a]
3
4 ( 2 - 1) 3/2
. (2 - 1)
3 [a]
4 (5 - 2 - 2 2 ) 4 (5 - 3 2 )
=
3
[a ]
3
[a]
\ Area is greatest, when [a] is least, i.e., 1.
\ Area is greatest, when [a] = 1
=
Þ
a Î éë1, 2 )
Solutions
181
23. (2) f ¢(x)= lim
h® 0
f ( x + h) - f ( x)
h
2x 2
( x - 4) . f ( x) meet the x-axis
3
at (0, 0) (–2, 0) and (2, 0).
Since f ( - x) = - f ( x ) , the curve y = f ( x ) is
symmetrical about the origin.
Thus f ( x) =
f ( x(1 + h / x )) - f ( x )
= lim
h
h® 0
=
f ( x)
f (1 + h / x ) - f (1)
lim
x h® 0
h/x
=
f ( x)
· f ' (1)
x
Also, as a = 2, f '( x) > 0 for x < -2 / 3,
x > 2 / 3 and f '( x) < 0 for
f '( x) 2
2 f ( x)
or
=
x
f ( x) x
Integrating both sides, we get, f (x) = cx2
Since f (1) = 1, \ c = 1
Y
f (x) = x2
\
f ¢ (x) =
-2 / 3 < x < 2 / 3 .
Thus, shape of y = f ( x ) is as shown in figure
Y
E
B(2, 0)
O
A
y=
C(2, 0)
X
D
2
x2 + y 2 = 4
1+ x2
(–1,0)
O
(1,0)
X
So, f (x) = x 2
2
= x2 Þ x 4 + x2 - 2 = 0
Now,
2
1+ x
Þ x2 = 1 Þ x = ± 1
é1 æ 2
ù
2ö
x
ú
dx
Required area = 2 ê ò ç
÷
ø ú
êë 0 è 1 + x 2
û
3 ù1
é
ép 1 ù
x
1
= 2 ê 2 tan x - ú = 2 ê - ú
ë2 3 û
ë
3 û0
F
Area of the region ACBDOEA
= 2p - ò
But
0
2
-2
f ( x ) dx + ò f ( x )dx
0
0
0
ò-2 | f ( x) | dx = - ò-2 f ( x)dx
0
2
2
0
= - ò f ( -t )( -1) dt = ò f (t )dt
\ Area (ACBDOEA) = 2p
Thus, the area of the other region
AEODBFA = 2p.
25. (7)
Y
2ö
æ
= çè p - ÷ø sq. units.
3
24. (0) Since y = f ( x ) has relative extremes at
4
3
x = ± 2 / 3, these points are critical points, and
hence they must be roots of f '( x) = 0 (clearly
is differentiable everywhere). Thus
2
f '( x) = a ( x - 2 / 3)( x + 2 / 3)
1
2
= a ( x - 4 / 3)
Þ f ( x ) = a ( x 3 / 3 - 4 x / 3) + b . This passes
through (0, 0) and (1, –2). So b = 0 and
a (1/ 3 - 4 / 3) = -2 Þ a = 2 .
–2
–Ö 2 –1
–Ö3
O
1 Ö2
2
Ö3
X
MATHEMATICS
182
\ Area of region bounded by y = {x} 2
between x-axis for the x Î [0, 2] is
As we know that fractional part of any thing must lie
between 0 and 1 thus
Y
1
= ò x 2 dx +
0
{x}2
2
ò
( x 2 - 1) dx +
1
3
ò
( x 2 - 2) dx
2
2
x 2 –1 x 2 – 2 x 2 – 3
+ ò ( x 2 - 3) dx
3
x2
A0 =
O
1
2
3
CHAPTER
7ö
æ
Required area = 2A0 = 2 ç 2 + 3 - ÷
è
3ø
\
X
(c) Let P (x, y) be the point on the curve passing
through the origin O (0, 0) and let PN and PM be
the lines parallel to the x-and y-axis, respectively. If
the equation of the curve is y = y (x), the area POM
x
x
0
0
³1
1 + f 4 ( x)
Integrating with respect to x, from x = a to x = b
b
1
tan –1 ( f 2 ( x)) ³ b – a
a
2
ü
1ì
–1 2
or (b – a) £ í lim (tan ( f ( x ))ý
2 î x ®b þ
(
Assuming that 2(POM) =PON, we therefore have
x
x
x
0
0
0
f ( x ). f '( x)
or
equals ò ydx and the area PON equals xy - ò ydx .
)
x ®a +
y
O
3.
x
Differentiating both sides of this gives
dy
dy
dx
dy
Þ
3y = x + y Þ 2 y = x
=2
y
x
dx
dx
2.
Þ log y = 2log x + C Þ y = Cx 2
With C being a constant. This solution
represents a parabola. We will get a similar result
if we have started instead with 2 (PON) = POM
1
3
(c) f '( x ) ³ f ( x ) +
f ( x)
4
or f '( x ). f ( x) ³ 1 + f ( x )
)
p
24
(c) Differentiating w.r.t. x the given equation,
Þ b–a £
P(x, y)
M
(
– lim tan –1 ( f 2 ( x))
2ò y dx = xy - ò y dx Þ 3ò y dx = xy
N
7
3
Differential Equation and its Applications
25
1.
2
2+ 3-
we have 1 - ( f '( x)) 2 = f ( x)
If y = f(x), then we have,
2
dy
æ dy ö
1 - ç ÷ = y2
= ± 1- y2
dx
è dx ø
dy
Þ
= ±dx
1 - y2
Þ sin -1 y = C ± x Þ y = sin(C ± x)
Since f(0) = 0 so 0 = sin C Þ C = 0
Thus f ( x ) = ± sin x
As f ( x ) ³ 0 for x Î [0, 1] so we have
f ( x ) = sin x.
Since sin x < x " x > 0, we get f ( x) < x for
0 < x £ 1. Thus f (1/ 2) < 1/ 2 and
f (1/ 3) < 1/ 3.
Solutions
4.
183
(a) Put y = x sec q
dy
dq
\
= x sec q tan q + sec q
dx
dx
From the given equation,
dq
æ
ö
x 3 ç x sec q tan q + sec q÷
dx
è
ø
= x3 sec3 q+ x3 sec2 q tan q
Dividing both sides by x 3 sec q , we get
dq
x tan q + 1 = sec 2 q+ sec q tan q
dx
dq
x
= (tan q+ sec q)
dx
dx
or (sec q- tan q)d q =
x
Integrating we get
ln(sec q+ tan q) - ln sec q = ln x + ln c
7.
where p =
y 2 - x2
= cx Þ y + y 2 - x 2 = cxy
y
where c is an arbitrary constant.
(a) Put x2y2 = z
dy
Given, x 2 × 2 y + y 2 × 2 x = tan( x 2 y 2 )
dx
d 2 2
( x y ) = tan( x 2 y 2 ) put x2y2 = z
dx
Now, given expression transform to
dz
= tan z \ ò dx = ò cot z dz
dx
x = ln (sin z) + C
p
p
Þ C=1
Þ z=
2
2
x = ln sin (x2y2) + 1 \ ln sin (x2y2) = x – 1
sin (x2y2) = ex – 1
dy y / x + 1
Putting y / x = v , the last
=
dx y / x - 1
equation reduces to
dv v + 1
v 2 - 2v - 1
x
=
-v= dx v - 1
v -1
dx
2(v - 1)
Þ 2 =dv
x
v 2 - 2v - 1
Þ
Þ 2ln x = - ln (v 2 - 2v - 1) + ln C
Þ ln x 2 ( y 2 / x 2 - 2 y / x - 1) = ln C
when x = 1, y =
\
6.
(b)
æ
x ç cos
è
yö
÷ ( y dx + xdy)
xø
æ yö
= y sin ç ÷ ( x dy - y dx )
èxø
2
Dividing both sides by x dx we get
y ö æ y dy ö y
æ
æ y ö æ dy
Þ ç cos ÷ ç + ÷ = sin ç ÷ ç è
ø
è
ø
è x ø è dx
x x dx
x
Which is homogeneous equation
dy
dv
Putting y = vx we get
= v+x
dx
dx
dv ö
æ
æ dv ö
or cos v ç v + v + x ÷ = v sin v ç x ÷
è
è dx ø
dx ø
8.
Þ y 2 - 2 xy - x 2 = C , which does not
represent a pair of straight lines if C ¹ 0 .
(a) The given equation can be written as
æ dx dy ö ( x2 dy – y 2 dx)
=0
ç - ÷+
y ø
( x – y) 2
è x
Þ
yö
÷
xø
dy
.
dx
Replacing p by p + tan p / 4 = p + 1 ,
1 - p tan p / 4 1 - p
p +1
we have y +
x=0
1- p
or 1 +
5.
Þ 2v cos v = x dv (v sin v - cos v)
dx
Separating the variables, we get
2dx æ v sin v - cos v ö
Þ
=ç
dv
è v cos v ÷ø
x
Integrating 2 ln x + ln c = ln sec v - ln v
sec v
Þ cx 2 =
Þ sec v = cx 2 v
v
y
Þ sec = cxy
x
where ‘c’ is an arbitrary constant.
(c) Differentiating xy = C , we get y + xp = 0
æ dy dx ö
– ÷
ç
æ dx dy ö è y 2 x 2 ø
=0
ç – ÷+
y ø æ 1 1 ö2
è x
ç – ÷
è y xø
dy dx
–
y 2 x2
æ dx dy ö
=0
ç - ÷+
y ø æ 1 1 ö2
è x
ç – ÷
èx yø
Integrating, we get: ln | x | – ln | y | –
Þ
1
æ1 1ö
ç x – y÷
è
ø
=c
MATHEMATICS
184
x
xy
x
xy
+
=c
–
= c Þ ln
y
x- y
y
y-x
a 2 dx x y
(b) Here,
× = + -2
xy dy y x
Þ
9.
ln
dy 2
( x + y 2 - 2 xy )
dx
dy
Þ ( x - y )2 ×
= a2 ,
dx
dy dv
=
put x - y = v \1 dx dx
æ dv ö
Þ v 2 ç1 - ÷ = a 2
è dx ø
Þ a2 =
dv
v 2 dv
Þ 2
= dx
dx
v - a2
æ
a2 ö
Þ ç1 + 2
dv = dx
ç v - a 2 ÷÷
è
ø
Integrating both the sides, we get
Þ v 2 - a 2 = v2
a
v-a
v + log
= x+C
2
v+a
Þ (x - y) +
æ x- y-aö
a
log ç
÷ = x+C
2
è x- y+aø
æ x- y-aö
a
log ç
÷
2
è x- y+aø
It passes through (3a, a)
Þ y+C =
Þ a+C =
a
æ1ö
log ç ÷
2
è 3ø
a
æ1ö
æ 2 + log 3 ö
Þ C = -a + log ç ÷ Þ C = - a ç
÷
2
2
è3ø
è
ø
æ x- y-aö
a
\ y = (2 + log 3) + log ç
÷
2
è x- y+aø
Þ
x- y-a
= e y - k , where k = a (2 + log 3)
x- y+a
2
y -k
x - y 1+ e
=
a
1 - e y-k
æ 1 + e y -k ö
Þ x = y + aç
,
ç 1 - e y -k ÷÷
è
ø
a
where k = (2 + log 3)
2
Which is required equation of curve.
Hence, (b) is the correct answer.
\
10. (b) The given equation can be written as
( ydx - xdy) + x xy (x + y)dx + y xy ( x + y)dy = 0
Þ ( y dx - x dy ) + ( x + y ) xy ( x dx + y dy ) = 0
y dx - x dy æ x ö x æ x 2 + y 2 ö
Þ
+ ç + 1÷ ×
dç
÷÷ = 0
ç
y2
èy ø y è 2 ø
æ x2 + y 2 ö æ x ö x
æxö
Þ dç ÷+d ç
+1
=0
ç 2 ÷÷ ç y ÷ y
è yø
ø
è
øè
æxö
dç ÷
æ x2 + y2 ö
è yø
Þ d çç
=0
÷÷ +
è 2 ø æ x + 1ö × x
ç
÷
èy ø y
æ -1 æ x ö ö
æ x2 + y 2 ö
or d çç
÷÷ + 2d çç tan çç
÷÷ ÷÷ = 0
è 2 ø
è y øø
è
Integrating both the sides, we get
x2 + y2
x
+ 2 tan -1
=C
2
y
Hence, (b) is the correct answer.
dy
y3
11. (a)
= 2x
dx e + y 2
Þ
Dividing by e2x : \
dy
y 3e -2 x
=
dx 1 + y 2 e-2 x
\ dy + y 2 e -2 x dy = y 3e -2 x dx
dy
\ ò 2 + ò 2( ye -2 x dy - y 2 e -2 x dx ) = 0
y
dy
\ ò 2 + ò d (e -2 x y 2 ) + c = 0
y
\ 2ln | y | + e-2 x y 2 = c
...(i)
12. (b) Given curve is a n -1 y = x n
Differentiating equation (i) w.r.t. x, we get
dy
a n -1
= nx n -1
...(ii)
dx
Eliminating a from equations using (i) & (ii), we get
x n dy
= nx n-1
y dx
Putting -
...(iii)
dy
dx
,
in equation (iii) at the place of
dx
dy
we get
- xdx
x n dx
= n,
- . = nx n -1 , Þ
ydy
y dy
Þ – xdy = ny dy,
x2
ny 2
Þ +c =
, Þ ny2 + x2 = constant.
2
2
Solutions
185
13. (c) Given, equation of normal at P (1,1) is
ay + x = a + 1
æ dy ö
=a
\ Slope of tangent at P = a Þ ç ÷
è dx ø(1,1)
Given
æ dy ö
dy
dy
=k=a
µy Þ
= ky Þ ç ÷
è dx ø (1,1)
dx
dx
Y
Q
P(1, 1)
Normal
curve
O
P
X
dy
dy
= ay Þ
= adx
dx
y
(variable being separated)
Þ ln y = ax + c
It is passing through (1, 1) then c = – a
a ( x -1)
Þ equation of the curve is y = e
dy ö
æ
14. (a) Let P be (x, y). C is ç x + y , 0 ÷ and B is
dx ø
è
dy ö
æ
çè 0, y - x ÷ø . Centre of the circle through O,
dx
C, P and B has its centre at the mid-point of BC.
Let it be (a, b) then
dy
dy
2a = x + y
and 2b = y - x
dx
dx
B
P
O
C
A
D
Now, (a, b) lies on y = x
so, y - x dy = x + y dy Þ dy = y - x
dx
dx
dx x + y
dy
- y log e 2 = 2sin x (cos x - 1) log e 2
dx
This is linear differential equation
15. (a)
log 2
I.F. = e - e ò dx = e - x loge 2 = 2- x
(i) Solution is
y 2 - x = ò 2 - x 2sin x (cos x - 1) log e 2 dx
(ii) put sin x - x = t Þ (cos x - 1)dx = dt
\ y 2 - x = log e 2 ò 2t dt
\ y 2- x = 2t + c
\ y = 2 x + t + c2 x
\ y = 2sin x + c2 x
16. (c) Rearranging the terms of equation, we get
dt
g '( x )
t2
-t
=dx
g ( x)
g ( x)
1 dt 1 g '( x)
1
Þ- 2
+
=
...(i)
t dx t g ( x ) g ( x)
1
1 dt dz
=
Let z = Þ - 2
t
t dx dx
Thus, from (i) we obtain dz + g '( x ) z = 1
dx g ( x )
g ( x)
dz
which is clearly linear in z and
with
dx
g '( x )
dx
ò
I.F. = e g ( x ) = elog[ g ( x )] = g ( x)
Þ Thus, complete solution is
1
z × g ( x) = ò g ( x ) ×
dx + C
g ( x)
1
g ( x)
Þ g ( x) = x + c Þ
=t
t
x+c
2
2
17. (a) 1 = lim t f ( x) - x f (t ) æç 0 form ö÷
t®x
t-x
è0
ø
2
2tf ( x ) - x f '(t )
= lim
t ®x
1
Þ 2 xf ( x) - x2 f '( x) = 1
dy
Þ x2
- 2 xy = -1, y = f ( x )
dx
dy 2
1
Þ
- y=- 2
dx x
x
which is linear equation whose I.F.
1
= e -2log x = 2 .
x
Multiplying both sides of the last equation by
I.F., we get
d æ y ö
1
y
1
ç ÷ = - 4 Þ 2 = 3 + K.
dx è x 2 ø
x
x
3x
Since y = f(1) = 1.
2
1 2 2
+ x
so K = . Thus y =
3
3x 3
x
dy
dy
18. (2) The equation is y = e dx Þ x
= ln y
dx
MATHEMATICS
186
19. (0) (2ny + xy logex) dx = x logex dy
ö
dy æ 2n
Þ
=ç
+ 1÷ dx
y è x log e x ø
Again differentiating both sides w.r.t. x
y ( x) = x 2 y '( x) + y ( x )2 x + xy ( x )
or (1 - 3x ) y ( x ) = x 2 y '( x)
y '( x ) æ 1 3 ö
=
or
y ( x ) çè x 2 x ÷ø
Integrating, we get
1
l n y ( x ) = - - 3l n x + l n c
x
æ x3 y ( x ) ö
1
x3 y ( x)
or l n ç
= e -1/ x
÷ = - or
x
c
è c ø
Þ log( y ) = 2n log | log x | + x + c and c = 0
(Q curve passes through (e, ee))
\ y = e x + log(log x)
2n
= e x (log x )2n
Þ f ( x) = e x (log x)2 n
Now,
ì
ï® ¥ if
ï
ï
g ( x) = lim f ( x) = í 0
if
n ®¥
ï
ï ® ¥ if
ï
î
x<
1
e
1
<x<e
e
x>e
or y ( x) =
ò
g ( x ) dx = 0
2
1/ e
20. (2) The given differential equation can be
dx
written as
= xy[ x 2 sin y 2 + 1]
dy
1 dx 1
y = y sin y 2
3 dy
2
x
x
This equation is reducible to linear equation, so
1
putting = u, the last equation can be
2
x
written as
du
+ 2uy = 2 y sin y 2
dy
Þ
2
The integrating factor of this equation is e y .
So required solution is
2
2
ue y = ò 2 y sin y 2 .e y dy + C
(t = y 2 )
= ò (sin t )et dt + C
2
1 y2
2
2
Þ ue y = e (sin y - cos y ) + C
2
2
Þ 2u = (sin y 2 - cos y 2 ) + Ce - y
Þ 2 = x 2 [cos y 2 - sin y 2 - 2Ce - y ]
21. (8) Differentiating both sides w.r.t., x of the
given equation
x
x
x. y ( x ) + ò y (t )dt .1 = ( x + 1) x. y ( x ) + ò ty (t )dt
x
or
0
ò y(t )dt = x
0
2
x
y ( x ) + ò ty (t )dt
0
2
x -y = r
0
2
So,
... (1)
y
x
then differentiating (1) we get,
2 xdx - 2 ydy = 2rdr
and sin q =
...(2)
or xdx - ydy = rdr
and differentiaing (2) we get,
... (3)
xdy - ydx
x2
= cos qd q
2
or xdy - ydx = x cos qd q
= r 2 sec 2 q cos qd q = r 2 sec qd q
...(4)
Substituting values from (3) and (4) in the given
differetial equation, we get
æ 1+ r2
=
çç 2
r 2 sec 2 q d q
è r
dr
or
= sec q d q
(1 + r 2 )
Intergrating both sides,
rdr
2
x3
æ 1ö
So, y(1) = e Þ c = e2 \ y çè ÷ø = 8
2
22. (5) Put x = r sec q and y = r tan q
e
\
ce -1/ x
ö
1+ r2
÷÷ =
r
ø
ln(r + (1 + r 2 )) = ln(sec q + tan q) + ln c
Where c is an arbitrary constant.
or (r + (1 + r 2 )) = c(sec q+ tan q)
æ
ö
x+ y
or ( ( x2 - y2 ) + (1+ x2 - y2 ) = c ç
÷
çè (x2 - y2 ) ÷ø
23. (1) Area of curvilinear trapezoid
Solutions
187
x
OABCO = ò f ( x )dx according to question
0
x
n +1
ò f ( x)dx µ { f ( x)}
0
or
x
n +1
ò f ( x) dx = k{ f ( x)}
0
Y
y
x)
= f(
B
A
x
æ1ö
ln | f ( x) - 1 | + ln f ç ÷ - 1 = c
èxø
æ æ1ö ö
Þ f ( x ) - 1) ç f ç ÷ - 1 ÷ = ec = k
è èxø ø
æ1ö
æ1ö
Þ f ( x) f ç ÷ - f ( x) - f ç ÷ + 1 = k
è xø
èxø
Þ ( f (1))2 - f (1) - f (1) + 1 = k
(substituting x = 1)
(Q f (1) = 2, given)
Þ k =1
C
O
Integrating, we get
æ1ö
Þ f ( x) f ç ÷ = f ( x) +
è xø
X
Where k is constant of proportionality.
Differentiating both sides w.r.t. x,
f ( x ) = k (n + 1)( f ( x )) n f '( x )
1
or { f ( x)n-1} f '( x) =
k (n + 1)'
Integrating both sides w.r.t. x,
x
{ f ( x )}n
=
+c
n
k (n + 1)
{ f (0)}n
Putting x = 0
= 0+ c Þ 0= 0+c
n
(Q f (0) = 0)
n
{ f ( x )}
x
=
n
k (n + 1)
nx
n
Þ { f ( x)} = k (n + 1) ....... (1)
\
n
Again putting x = 1 then { f ( x )}n =
=1
k (n + 1)
(Q f (1) = 1)
From (1), ( f ( x ))n = x or f ( x ) = x1/ n
Þ (f(10))n = 10 Þ k =1
24. (7) Given
æ1ö
æ1ö
æ1ö
x2 f '( x) f ç ÷ - f ( x) f ' ç ÷ = x 2 f '( x) - f ' ç ÷
èxø
è xø
èxø
é
ù
1
1
æ
ö
æ
ö
Þ x 2 f '( x ) ê f ç ÷ - 1ú = f ' ç ÷ [ f ( x) - 1]
èxø
ë èxø û
ì æ1ö ü 1
æ1ö
Þ f '( x) í f ç ÷ - 1ý = 2 f ' ç ÷ { f ( x ) - 1}
è xø
î èxø þ x
æ 1 öæ 1 ö
f ' ç ÷ç - 2 ÷
f '( x)
x
x ø
Þ
+ è øè
=0
f ( x) - 1
æ1ö
f ç ÷ -1
èxø
æ1ö
fç ÷
è xø
(From functional equation)
Þ f ( x) = 1 ± x n
Since f(1) = 2, we must have f(x) = 1 + xn
Since f(5) = 26, 26 = 1 + 5n Þ n = 2
\ f ( x) = 1 + x 2
25. (1) Here,
\ f (6) = 1 + (6)2 = 37
2
2 4
f ' ( x ) = 1 + 2 x 2 + × 4 x 4 + × × 6 x 6 + ....¥
3
3 5
æd
ö
= 1 + x ç ( xf ( x)) ÷
dx
è
ø
Þ f '( x) = 1 + x | xf '( x) + f ( x) |
Þ (1 - x2 ) f '( x) = 1 + xf ( x)
Þ
dy
x
1
×y= 2,
dx 1 - x 2
x
-x
I.F. = e
ò 1- x2 dx
1
= e2
\ y × 1 - x2 = ò
log|1- x 2 |
1
1- x
2
= 1- x2
× 1- x2 + C
Þ y 1 - x = sin -1 x + C ,
2
as f (0) = 0 Þ C = 0
Þy=
sin -1 x
1- x2
Þ A=ò
3/2
1/ 2
sin -1 x
1 - x2
p
dx
æ t 2 ö 2 1 é p2 p2 ù
= ò t dt = ç ÷ = ê - ú
ç 2 ÷ p 2 4 36
p/ 6
è ø
ë
û
p/ 3
6
\[4 A] = 1.
MATHEMATICS
188
CHAPTER
26
1.
(b)
Vector Algebra
r
r
a ( x ) and b ( x) are collinear if and only if
a
b
g
+ cosec2 + cosec2 = 2
2
2
2
which is not possible as
Þ cosec 2
cos x = x . Now let f ( x ) = x - cos x , then
f '( x ) = 1 + sin x ³ 0
Þ f ( x) is increasing and hence f (x) = 0 for a
unique value of x.
p
p
For x ³
, f ( x ) > 0 and x <
, f ( x) < 0 .
6
3
Thus cos x = x , for a unique value of
cosec2
3.
a a c
or cos a(cos b - 1)(cos g - 1)
4.
cos a
1
1
+
+
=0
1 - cos a 1 - cos b 1 - cos g
-(1 - cos a) + 1
1
1
Þ
+
+
=0
(1 - cos a)
(1 - cos b) (1 - cos g )
1
1
1
+
+
=0
1 - cos a 1 - cos b 1 - cos g
1
1
1
Þ
+
+
=1
1 - cos a 1 - cos b 1 - cos g
Þ -1 +
0
c c b
c c b–c
Operating C3 ® C3 – C1
Expanding along R2, we get
a c–a
–
c b – c = c (c – a) – a (b – c) = 0
Þ c2 – ac – ab + ac = 0
Þ c2 = ab Þ a, c, b are in G.P.
\ c is the G.M. of a and b.
(c) In DABC, let AD is angle bisector of angle A.
A
-(1 - cos a)(cos g - 1) - (1 - cos a)(cos b - 1) = 0
Dividing through out by
(1 - cos a )(1 - cos ß)(1 - cos g ) ; we get
a a c–a
\ 1 0 1 =0Þ 1 0
r r r
(d) Suppose that a , b , c are coplanar..
cos a
1
1
Þ
1
cos b
1 =0
1
1
cos g
applying R2 ¾¾
® R1
® R2 - R1 and R3 ¾¾
cos a
1
1
or 1 - cos a cos b - 1
1
=0
1 - cos a
0
cos g - 1
So the vectors cannot be coplanar.
(b) a, b, c are distinct non-negative numbers
and the vectors ai$ + a $j + ck$ , $i + k$ and
r
ci$ + c $j + bk are coplanar..
æp p ö
x, x Î ç , ÷ .
è6 3 ø
2.
a
g
ß
³ 1, cosec2 ³ 1, cosec2 ³ 1.
2
2
2
a
q q
2 2
b
B ak D bk C
\ BD = ak , DC = bk \ BC = (a + b)k
Applying cosine formula, we have
cos q =
=
( AB)2 + ( AC )2 - ( BC ) 2
2( AB)( AC )
a 2 + b 2 - ( a + b) 2 k 2
2ab
...(1)
Solutions
189
Also in DADC and DABD
uur uur
uur uur
1
1
Þ a . b - 2. = l Þ a . b = l + .
2
4
Again,
uur uur
uur
uur uur
uur uur
uur uur
b - 2 c = l a Þ b.b - 2b .c = l b .a
q b 2 + ( AD)2 - b 2 k 2 a 2 + ( AD)2 - a 2 k 2
cos =
=
2
2b AD
2a AD
Þ ( AD) 2 = ab(1 - k 2 )
2
2
ïì a + b - 2ab cos q ïü
= ab í1 ý
(a + b)2
îï
þï
5.
4a 2 b2 cos 2 q / 2
2ab cos q / 2
( a + b)
( a + b)
r
r
r r
uuur
(ab + b a)
ab æ a b ö
\ AD = ±
=±
ç + ÷
(a + b)
(a + b) è a b ø
ab
=±
( aˆ + bˆ)
(a + b)
uuur
uuur AD
(aˆ + bˆ)
\ AD =
=±
AD
2cos q / 2
uur
uur
uur
uur
uur
uur
(a) Let A B = a and AD = b and AC = c
=
2
when a , b and c are non-collinear coplanar
vectors.
uuur uuur uuur r r
DB = AB - AC = a - b .
uur uuuur
uur uur uur
uur uur uur uur
Now, DB. AB = ( a - b ).( a ) = a . a - b . a
2
a 2 - ab cos q = a -
c 2 - a 2 - b 2 3a 2 + b 2 - c 2
=
2
2
é
a 2 + b2 - c 2 ù
ê\ In DABC , cos(p - q ) =
ú
2ab
êë
úû
D
C
®
b
6.
2
Þ 8 - l - l - 2(1) = l æç 1 ö÷
è 4ø
2 4
Þ l 2 + l - 12 = 0 Þ l = -4, 3.
7.
ur r r
(b) Let the triangle be PQR with sides p, q, r
ur
r
r r
r
Let p = 3(aˆ ´ b ) and q = b - (aˆ. b) aˆ
ur r
r r
r
\ p.q = 3(aˆ ´ b).{b - (aˆ.b) aˆ}
rr
r
uur
= 3[aˆ b b] - 3(aˆ.b)[aˆ b aˆ ] = 0
p
Þ ÐR = .
2
ur
r
r
Also | p |= 3(aˆ ´ b) = 3 | aˆ | | b | sin q
r
= 3 | b | sin q ,
r
q is the angle between â and b and | aˆ |= 1
r
r
r
r
r
and | q |= {b - ( aˆ.b)aˆ}.{b - ((aˆ.b ) aˆ}
r
r
r
= | b |2 -2(aˆ.b)2 + (aˆ.b)2 (aˆ )2
r
r
r
= | b |2 + | b |2 cos 2 q. - 2 | b |2 cos 2 q
r
= | b | sin q
®
c
q
A
2
uur uur
Þ b.c = 8- l - l .
2 4
Furthermore,
uur uur
uur
uur uur uur uur
uur uur
b - 2c = l a Þ b.c - 2 c.c = l a.c
Þ AD =
uur
uur uur
uur uur
1ö
æ
Þ 16 - 2 b . c = l ç l + ÷
è
2ø
[from (1)]
p–q
®
a
B
uur
uur
uur
(a) Given | a | = 1, | c | = 1 and | b | = 4
uur uur
1 1
Þ a . c = 1.1. = .
4 4
uur uur
uur
uur uur
uur uur
Again, b - 2 c = l a Þ a . b - 2 a . c = l a 2
R
®
q
®
p
Q
®
r
P
MATHEMATICS
190
\
8.
ur
| p|
r = 3 . But
|q|
ur
| p|
r = tan P Þ P = p
|q|
3
pù
é
êëQ ÐR = 2 úû
p
æ 1 ö
p
\ P = = tan -1( 3) , Q = = tan -1 ç ÷
è 3ø
6
3
p
and R = = cot -1 (0) .
2
p
[Note that as soon as we get R = , only one
2
p
option has the value of an angle = cot -1 (0) ,
2
hence (b) is the correct answer].
r
r
(c) For the vector a and b to be inclined at an
rr
obtuse angle, we must have a. b < 0 for all
x Î (0, ¥)
Þ c (log 2 x)2 - 12 + 6c log 2 x < 0 for all
x Î (0, ¥)
Þ cy 2 + 6cy - 12 < 0 for all y Î R ,
where y = log 2 x
Þ c < 0 and 36c 2 + 48c < 0 ,
(using ax 2 + bx + c < 0, "x Î R if a < 0 and D < 0 )
æ 4 ö
Þ c < 0 and c (3c + 4) < 0 Þ c Î ç - , 0÷ .
è 3 ø
ur r
9.
(b) p . q = ab + bc + ca
= a 2 + b 2 + c 2 b 2 + c 2 + a 2 cos q
Þ cos q =
ab + bc + ca
.
(a 2 + b2 + c 2 )
Þ Now (a - b)2 + (b - c )2 + (c - a) 2 ³ 0
Þ
Also
a 2 + b 2 + c 2 ³ ab + bc + ca
ab + bc + ca
£ 1.
a 2 + b2 + c 2
(a + b + c )2 = a 2 + b 2 + c 2 + 2(ab + bc + ca) ³ 0
Þ
Þ
ab + bc + ca
³ -1/ 2
a 2 + b2 + c 2
1
- £ cos q £ 1 Þ q Î[0, 2p / 3].
2
10. (c) Vol. of paralleopiped formed by
r
r
µ = ai$+ $
u = $i + a$j + $
k , v = $j + ak$, w
k is
1 a 1
r r ur
V = [u v w] = 0 1 a
a 0 1
= 1(1 - 0) - a(0 - a 2 ) + 1(0 - a) = 1 + a 3 - a
For V to be min
Þ 3a 2 - 1 = 0
11.
dV
=0
da
Þ a=±
1
3
.
(b) We observe that
rr
rr
r r æ b.a ö r r r r r r
a.b1 = a.b - ç r 2 ÷ a.a = a.b - a.b = 0
è|a| ø
r ur
rr
rr
r æ r c.a r c.b r ö
a.c 2 = a. ç c - r a - uur 1 b1 ÷
çè | a |2
| b1 |2 ÷ø
r ur
rr
r r r a.c r 2 c.b r r
= a.c - c. r | a | - uur 1 (a.b1 )
| a |2
| b1 |2
rr rr
rr
= a.c - a.c - 0 = 0
[ Q a.b1 = 0 ]
r ur
rr
r r
r æ r c.a r c.b r ö
and b1 .c 2 = b1. ç c - r a - uur 1 b1 ÷
çè
| a |2
| b1 |2 ÷ø
r r ur r
r ur
r r (c.a)(b .a ) c.b r r
1
= b1.c - uur 1 b1.b1
r
| a |2
| b1 |2
r r
r r
r
= b1.c - 0 - b1.c = 0
(Using b1.a = 0 )
rr
r uur r r
Hence a.b1 = a.c2 = b1.c 2 = 0 .
r uur r
Þ (a, b1, c 2 ) is a set of orthogonal vectors.
12. (d) Let xi$ + y $j + zk$ be the required unit
vector.
Since a$ is perpendicular to (2$i - $j + 2k$ ) .
\ 2x – y + 2z = 0
......... (i)
$
$
$
Since vector xi + y j + zk is coplanar with the
vector $i + $j - k$ and 2$i + 2 $j - k$ .
\ xi$ + y $j + zk$
= p ( $i + $j - k$ ) + q ( 2$i + 2 $j - k$ ),
Solutions
191
where p and q are some scalars.
Þ xi$ + y $j + zk$
= ( p + 2q)i$ + ( p + 2q ) $j - ( p + q )k$
Þ x = p + 2q, y = p + 2q, z = – p – q
Now from equation (i),
2p + 4q – p – 2q – 2p – 2q = 0
Þ –p=0Þp=0
\ x = 2q, y = 2q, z = –q.
Since vector xi$ + y $j + zk$ is a unit vector,,
therefore
2
2
2
| xi$ + y $j + zk$ | = 1 Þ x + y + z = 1
Þ x2 + y2 + z2 = 1 Þ 4q2 + 4q2 + q2 = 1
Þ
9q2 = 1 Þ q = ±
1
3
2
1
2
1
, then x = , y = , z = – .
3
3
3
3
2
1
2
1
When q = – , then x = – , y = – , z = .
3
3
3
3
2$ 2 $ 1 $
Here required unit vector is i + j - k
3 3
3
2$ 2 $ 1 $
or - i - j + k .
3 3
3
13. (b) Let A be the first term and D be the common
difference of the corresponding AP. Then,
1
1
= A + ( p - 1) D, = A + ( q - 1) D ,
a
b
1
= A + (r - 1) D
c
Þ a -1 (q - r ) + b-1 (r - p) + c -1 ( p - q) = 0
r r
r r
Þ v × u = 0 Þ u ^ v.
r
r
Hence u and v are orthogonal vectors.
r r
14. (b) We have p × q = 0
r
r r
r
Þ (5a - 3b ) × (-a - 2b ) = 0
r
r
r r
...(1)
Þ 6 | b |2 -5 | a |2 -7a × b = 0
r r
Also r × s = 0
r r
r r
Þ (- 4a - b ) (-a + b ) = 0
r
r
r r
...(2)
Þ 4 | a |2 - | b |2 -3a × b = 0
When q =
Now xr = 1 ( pr + rr + sr )
3
r
r
1 r
r r r r
= (5a - 3b - 4a - b - a + b ) = -b
3
r 1 r r 1
r
r
and y = ( r + s ) = ( -5a ) = - a
5
5
r
Angle between xr and y , i.e.,
r r
r r
x×y
a ×b
cos q = r r = r r
...(3)
| x || y | | a || b |
From (1) and (2),
r
r
25 r r
43 r r
| a |=
a × b and | b |=
a ×b
19
19
r r
25 ´ 43 r r
\ | a || b |=
× a ×b
19
æ 19 ö
q = cos -1 ç
÷
è 5 43 ø
r r
r r r
rrr
15. (d) Let d × a = (cos y )[a b c ] = -d × (b + c )
r r r r
[as d × (a + b + c ) = 0]
r r r
d × (b + c ) .
Þ cos y = - r r r
...(1)
[a b c ]
r r r
d × (a + b )
Similarly, sin x = - r r r
...(2)
[
a
b
c
]
r r r
d × (a + c )
2=- rrr .
...(3)
[a b c ]
Adding above equations, we get
sin x + cos y + 2 = 0
Þ sin x + cos y = -2
Þ sin x = -1, cos y = -1
Þ x = (4n - 1)p / 2, y = (2n + 1); n Î Z
x = (4n - 1)p / 2, y = (2n + 1) p .
Since we want minimum value of x2 + y2,
2
p
so x = - , y = ±p Þ x 2 + y 2 = 5p .
2
4
r
r
16. (d) As angle between a and b is obtuse,
r r
a ×b < 0
Þ (2l2iˆ + 4lˆj + kˆ) × (7iˆ - 2 ˆj + lkˆ) < 0
Þ 14l 2 - 8l + l < 0 Þ l(2l - 1) < 0
Þ0<l<
1
2
...(1)
MATHEMATICS
192
r
Angle between b and k̂ is acute and less than
p
.
6
r
r
b × kˆ = | b | × | kˆ | cos q
Þ cos q >
53 + l 2
B
bˆ
60°
A
a
uuur
uuur
From the figure, OA = aˆ, OC = bˆ
uuur uuur uuur
OA + OC = OB = aˆ + bˆ .
O
Þ l = 53 + l 1× cos q
l
aˆ
C
b̂
2
Þ cos q =
18. (3)
\ q<
p
p
Þ cos q > cos
6
6
3
3
l
Þ
>
2
2
2
53 + l
2
Þ 4l 2 - 3(53 + l 2 ) > 0 Þ l > 159
...(2)
Þ l < - 159
From Eqs. (1) and (2), l = f \ Domain of l is
null set.
17. (c) Given,
r
r
r
r
cos q = (a ´ iˆ) × (b ´ iˆ) + (a ´ ˆj ) × (b ´ ˆj )
r
r
...(1)
+(a ´ kˆ) × (b ´ kˆ)
Consider,
r
r
r
r
r
r
(a ´ iˆ) × (b ´ iˆ) = [(a ´ iˆ)b iˆ] = ((a ´ iˆ) ´ b ) × iˆ
r r
r r
((a × b )iˆ) - ((iˆ × b )a )iˆ = (a × b)(iˆ × iˆ) - (iˆ × bˆ)(a × iˆ)
r r rr
= a × b - a1b1 .
r
r
r r r r
Similarly, (a ´ ˆj ) × (b ´ ˆj ) = a × b - a2b2
r
r
r r r r
and (a ´ kˆ)(b ´ kˆ) = a × b - a3b3
\ From Eq. (i), we get
r r
cos q = 3a × b - (a1b1 + a2b2 + a3b3 )
r r r r
= 3a × b - a × b
r
r
a ×b
r r
r r 1
Þ r r = 2 a × b Þ | a || b |= .
| a || b |
2
r r
Now, use AM ³ GM on | a |, | b |
r
1
r
r
|a| +|b |
2
r r
r
\
³ (| a | × | b |) 2 Þ| a | + | b | ³
2
2
r
r
Þ| a|+|b |³ 2 .
Now | aˆ + bˆ |= 1 \ DOAB is equilateral.
\ In DAOC, cos120° =
OA2 + OC 2 - AC 2
2OA × OC
r r
1 1 + 1 - AC 2
\ AC = | a - b |= 3 = k .
=
2
2
Þ k = 3.
\-
19. (3)
Þ
( xˆ + yˆ + zˆ)2 ³ 0
3 + 2 S xˆ. yˆ ³ 0
2
Þ
2
2 S xˆ. yˆ ³ -3
2
Now, xˆ + yˆ + yˆ + zˆ + zˆ + xˆ
= 6 + 2 S xˆ. yˆ ³ 6 + ( -3)
Þ
20. (3)
2
2
2
xˆ + yˆ + yˆ + zˆ + zˆ + xˆ ³ 3 .
r
F = aaˆ + bbˆ + gcˆ
r
Fˆ × (bˆ ´ cˆ)
\ F × (bˆ ´ cˆ) = a[aˆ bˆ cˆ] \ a =
[aˆ bˆ cˆ]
Now, bˆ ´ cˆ = bˆ ´ (aˆ ´ bˆ) = (bˆ × bˆ)aˆ - (bˆ × aˆ )bˆ
1
= aˆ - bˆ
3
ˆ ˆ] = (aˆ ´ bˆ)2 = | aˆ |2 | bˆ |2 - (aˆ × bˆ)2
[aˆ bc
= 1-
1 8
=
9 9
9
3
9ì
1
ü
\a = í Fˆ × aˆ - Fˆ × bˆ ý \ k1 = , k2 =
8
8
8î
3
þ
Þ k1 + k2 =
12 3
= \ 2(k1 + k 2 ) = 3.
8 2
Solutions
193
r r
21. (5) a . b = 0 Þ x1 + x2 + x3 = 0
We have to obtain the number of integral solution
of this equation Þ Coefficient of
x 0 in ( x -3 + x -2 + x -1 + x 0 + x + x 2 )3
æ 1 + x + x 2 + x3 + x 4 + x5 ö
= Coeff. of x in ç
÷
x3
è
ø
3
0
= Coeff. of x 9 in (1 - x 6 )3 (1 - x ) -3
= 11C9 - 3. 5C3 = 25 .
uuur
uuur
22. (2) AB = 2iˆ + ˆj + kˆ, AC = (t + 1)iˆ + 0 ˆj - kˆ
iˆ
uuur uuur
AB ´ AC = 2
ˆj
kˆ
1
1
t + 1 0 -1
= -iˆ + (t + 3) ˆj - (t + 1)kˆ
= 1 + (t + 3)2 + (t + 1)2 = 2t 2 + 8t + 11 .
1 uuur uuur
Area of DABC = | AB ´ AC |
2
1
=
2t 2 + 8t + 1
2
1 2
(2t + 8t + 1)
4
f '(t ) = 0 Þ t = -2
At t = –2, f '' (t) > 0.
So D is minimum at t = –2.
r
r r
r r
r r r
23. (0 ) Let V1 = (a ´ b ) ´ (c ´ d ) = (a ´ b ) ´ p
2
Let f (t ) = D =
(say)
r r r r r r r r r r r r r r
= (a × p)b - (b × p)a = a × (c ´ d )b - b × (c ´ d )a
r r r r r r r r
...(1)
= [a c d ]b - [b c d ]a
r r r r r
r
r r
Similarly V2 = (a ´ c ) ´ (d ´ b ) = q ´ (d ´ b )
r r r r r r r r
...(2)
= [a c b ]d - [a c d ]b
r
r
r
r
r
r
r
r
And V3 = (a ´ d ) ´ (b ´ c ) = (a ´ d ) ´ r
r r r r rrr r
...(3)
= [a b c ]d - [d b c ]a
r r r r
rr r r r
r r r
\ A = V1 + V2 + V3 = -2[b c d ]a ¹ 0 as b , c , d
are non-coplanar and which is a vector parallel
to ar
r r
r r r r r r
Þ A ´ a = -2[b c d ](a ´ a ) = 0
r r
Þ| A ´ a |= 0 .
r
r æ é 0 -1/ 2ù ö
24. (2) vn +1 - vn = ç ê
÷
0 úû ø
è ë1/ 2
r r é 0 -1/ 2 ù r
v2 - v1 = ê
v0
0 úû
ë1/ 2
n +1
r
v0
3
r r é 0 -1/ 2 ù r
v3 - v2 = ê
v0
0 úû
ë1/ 2
n
é 0 -1/ 2 ù r
r r
vn - vn -1 = ê
v0
0 úû
ë1/ 2
Adding all the equations,
r r
r
vn - v0 = ( A + A2 + A3 + .....An )v0
where
é 0 -1/ 2 ù
r
r
A=ê
Þ vn = ( I + A + A2 + ....)v0 .
ú
0 û
ë1/ 2
25. (7) Each side subtends an angle of 2p/8 = p/4
at the centre of the octagon. Let 'O' be the centre
of the octagon and r the radius of the circumcircle of the octagon. Therefore
uuur uuur
pör
æ
OA1 ´ OA2 = ç r 2 sin ÷ n
4ø
è
r
where n is the vector perpendicular to the plane
of the polygon such that from the side of nr , the
points A1, A2, A3, ...., An are in counterclock sense.
Hence
uuur uuur æ
pö r
OA2 ´ OA3 = ç r 2 sin ÷ n
4ø
è
uuur uuur
pör
æ
OA3 ´ OA4 = ç r 2 sin ÷ n , ..... etc.
4ø
è
Therefore
7
uuur
uuur
æ
pör
å (OA j ´ OA j +1 ) = 7 çè r 2 sin 4 ÷ø n
j =1
uuur uuur
= 7(OA1 ´ OA2 ) .
MATHEMATICS
194
CHAPTER
Three Dimensional Geometry
27
1.
(b) The given equations are 3l + m + 5n = 0...(i)
Þ cos 2q = 0 or cos2q = – 1
and 6mn - 2nl + 5lm = 0
From (i), we have m = –3l – 5n. Putting
m = –3l – 5n in (ii), we get
Þ 2q = p/2 or 2q = p
...(ii)
Þ q = p/4 or q =
6(-3l - 5n )n - 2nl + 5l (-3l - 5n) = 0
3.
Þ (n + l )(2n + l ) = 0
Þ either l = –n or l = –2n.
If l = –n, then putting l = –n in (i),
we obtain m = –2n.
If l = –2n, then putting l = –2n in (i),
we obtain m = n.
Thus, the direction ratios of two lines are
–n, –2n, n and –2n, n, n i.e., 1, 2, –1 and –2, 1, 1.
Hence, the direction cosines are
1
6
,
2
6
,
-1
6
or
-2
6
,
1
6
,
1
6
ép pù
p
Þq= ê , ú
ë 4 2û
2
(b) Let OA and OB be two lines with DC's l1,
m1, n1 and l2, m2, n2. Let OA = OB = 1. Then
co-ordinates of A and B are (l1, m1, n1) and
(l2, m2, n2) respectively. Let OC Z be the
bisector of ÐAOB such that C is the midpoint of AB and so its co-ordinates are
æ l1 + l2 m1 + m2 n1 + n2 ö
,
ç 2 ,
2
2 ÷ø
è
\ DR's of OC are
. The angle q
l1 + l2 m1 + m2 n1 + n2
,
,
2
2
2
\ We have
between the lines is given by
cos q =
-2
2
1
-1
1
æ -1 ö
Þ q = cos -1 ç ÷ .
è 6 ø
2.
2
(c) It makes q with x and y-axes.
l = cosq, m = cosq, n = cos (p – 2q)
we have l2 + m2 + n2 = 1
Þ cos2q + cos2q + cos2 (p – 2q) = 1
Þ 2 cos2q + (–cos2q)2 = 1
Þ 2 cos2q – 1 + cos22q = 0
Þ cos2q – [1 + cos2q] = 0
2
æ l + l ö æ m + m2 ö æ n1 + n2 ö
OC = ç 1 2 ÷ + ç 1
÷ +ç
÷
2
è 2 ø è
ø è 2 ø
-1
´
+
´
+
´
=
6
6
6
6
6
6
6
1
=
2
1
(l12 + m12 + n12 ) + (l22 + m22 + n22 ) + 2(l1l2 + m1m2 + n1n2 )
2
=
1
2 + 2 cos q [Q cos q = l1l2 + m1m2 + n1n2 ]
2
=
1
æqö
2(1 + cos q) = cos ç ÷ .
2
è2ø
l1 + l2 m1 + m2 n1 + n2
\ DCs of OC are 2(OC ) , 2(OC ) , 2(OC )
i.e.,
l1 + l2
m + m2
n +n
, 1
, 1 2
2 cos q / 2 2 cos q / 2 2 cos q / 2
Solutions
195
Y
B
æ l1 + l2 m1 + m2 n1 + n2 ö
,
ç 2 ,
2
2 ÷ø
è
C A
A(l , m , n )
1
X'
O
1
=
1
a 2 + b 2 + c2
X
E æ l1 - l2 m1 - m2 n1 - n2 ö
,
ç 2 ,
2
2 ÷ø
è
D
(- l2 , - m2 , - n2 )
4.
a(cos q + 3) + (b 2)sin q + c(cos q - 3)
=
a 2 + b 2 + c2 2 + 6
(d) The given lines are
y + 3 z -1
=
=s
-l
l
and 2 x = y - 1 =
In order that cos a in independent of q
..........(i)
z-2
=t
-1
a + c = 0 and b = 0
.........(ii)
\ cos a =
The lines are coplanar, if
6.
0 - ( -1) -1 - 3 -2 - ( -1)
1
-l
l
=0
1
1
-1
2
+(cos q - 3)2
(a + c) cos q + b 2 sin q + (a - c) 3
Y'
x -1 =
(cos q + 3) 2 + 2sin 2 q
2a 3
a 2 ×2 2
=
3
p
Þa=
2
6
(c) Given one vertex A(7, 2, 4) and line
x + 6 y + 10 z + 14
=
=
5
3
8
General point on above line,
B º (5l - 6, 3l - 10, 8l - 14)
Direction ratios of line AB are
1
-5 -1
C2 ® C2 + C3 ; 1
0
l =0
1
2
0
-1
l
Þ 5( -1 - ) = 0 Þ l = -2
2
5.
(a) Both the lines pass through origin.
Line L1 is parallel to the vector
r
V1 = (cos q + 3)ˆi + ( 2 sin q)ˆj + (cos q - 3)kˆ
and L2 is parallel to the vector
r
V2 = aˆi + bˆj + ckˆ
r r
V1 × V2
r
r
\ cos a =
| V1 || V2 |
< 5l - 13, 3l - 12, 8l - 18 >
Direction ratios of line BC are <5, 3, 8>
Since, angle between AB and BC is
cos
p
=
4
p
.
4
(5l - 3)5 + 3(3l - 12) + 8(8l - 18)
52 + 32 + 8 2 ×
(5l - 13) 2 + (3l - 12) 2
+(8l - 18)2
Squaring and solving, we have l = 3, 2.
Hence, equation of lines are
x-7 y-2 z-4
=
=
and
-3
2
6
x-7 y-2 z-4
=
=
.
3
6
2
MATHEMATICS
196
7.
D=
(d)
1 ˆ
( j + l kˆ) ´ (iˆ + lkˆ)
2
z
C
1
1
= - kˆ + l iˆ + lˆj =
2l 2 + 1
2
2
P
9 1
33
Þ £ (2l 2 + 1) £
4 4
4
O
Þ 4 £ l 2 £ 16 Þ 2 £| l |£ 4.
8.
x
y
B
A
(c) The line has +ve and equal direction
1
cosines, these are
3
1
,
3
1
,
3
or direction
ratios are 1, 1, 1. Also the lines passes through P
(2, – 1, 2).
\ Equation of line is
x - 2 y +1 z - 2
=
=
= l (say)
1
1
1
h
k
l
x + y + z = p or hx + ky + lz = p 2
p
p
p
æ
ö
æ
æ p2
ö
p2
p2 ö
B
0,
,
0
C
0,
0,
ç
÷
ç
÷
,
,
A
,
0,
0
÷
\ çç
ç
÷
ç
÷
k
l ÷ø
è
ø
è
è h
ø
Now, Area of DABC , D =
Let Q (l + 2, l - 1, l + 2) be a point on this line
where it meets the plane
2
where, Axy
is area of projection of DABC on xy
plane = area of DAOB
2x + y + z = 9
Then Q must satisfy the eqn of plane
2(l + 2) + l - 1 + l + 2 = 9 Þ l = 1
i.e.
2
2
2
Axy
+ Ayz
+ Azx
\ Q has coordinates (3, 0, 3)
Hence the length of line segment
p2 / h
0
1
p4
2 | hk |
1
Now, Axy =
2
0
Similarly, Ayz =
p4
p4
and Azx =
2 | kl |
2 | lh |
0
2
p /k 1 =
0
1
= (2 - 3)2 + (-1 - 0)2 + (2 - 3)2 = 3
9.
2
2
2
(d) Here OP = h + k + l = p
\ DRs of OP are :
h
h2 + k 2 + l 2
or
,
k
h2 + k 2 + l2
,
l
h2 + k 2 + l2
h k l
, ,
p p p
Since OP is normal to the plane, therefore,
equation of plane is
p5
2
2 ,
+ A2yz + Azx
\ D 2 = Axy
D=
2hkl
10. (b) A plane containing the given lines is
2 x + 3 y + 4 z - 6 + l ( x + y + z - 3) = 0
… (i)
This plane is perpendicular to plane z = 0
if 4 + l = 0 Þ l = –4
So, the equation (i) becomes
–2 x - y + 6 = 0 Þ 2 x + y - 6 = 0
… (ii)
Solutions
197
Equation of the projection will be the line of
intersection of plane (2) and the plane z = 0. If
the line has d.c. proportional to l, m, n then
2l + m = 0 and n = 0
Þ l : m : n = 1 : –2 : 0. Obviously (0, 6, 0) is a
point on both the planes, hence lies on the line
as well.
x y -6 z
=
\ Equation of the line is =
1
-2
0
x y z
+ + = 1 which meets the axes at
a b c
A (a, 0, 0), B (0, b, 0) and C (0. 0, c).
æ a b cö
Centroid of D ABC is ç , , ÷
è 3 3 3ø
and it satisfies the relation
1
x
2
Þ
+
1
y
2
+
1
z
2
=k Þ
9
9
9
+ 2 + 2 =k
a
b
c
2
1
1 k
+ 2+ 2 =
9
a
b
z
1
...(i)
2
Also given that the distance of plane
Þ
1
a2
Þ
1
a2
+
1
1
b2
+
1
....... (iii)
\ a = 5k, b = –k, c = –k
On putting the value of a, b and c in equation (i),
5(x – 1) – (y – 2) – (z – 3) = 0
Þ 5x – y – z = 0
....... (iv)
when x = 1, y = 0 and z = 5; then
L.H.S. of equation (iv) = 5x – y – 2
=5×1–0–5
=0
= R.H.S. of equation (iv)
Hence coordinates of the point (1, 0, 5) satisfy
the equation plane represented by equations (iv),
Therefore the plane passes through the point
(1,0,5).
13. (c) The position vectors of two given points
the equation of the given plane is
r = (5iˆ + 2 ˆj - 7kˆ) + 9 = 0 or r × n + d = 0
=1
We have
c2
a × n + d = (iˆ - ˆj + 3kˆ) × (5iˆ + 2 ˆj - 7kˆ) + 9
1
+ 2 + 2 =1
b
c
From (i) and (ii), we get
\
a.1 + b.1 + c.4 = 0
i.e., a + b + 4c = 0
From (ii) and (iii),
are a = iˆ - ˆj + 3kˆ and b = 3iˆ + 3 ˆj + 3kˆ and
x y z
+ + = 1 from (0, 0, 0) is 1 unit.
a b c
1
x y z
= =
1 1 4
a
b
c
=
=
= k (let)
8 - 3 3 - 4 1- 2
11. (d) Let the eqn of variable plane be
\
where a.1 + b.2 + c.3 = 0
i.e.,
a + 2b + 3c = 0
....... (ii)
Since the plane (i) parallel to the line
...(ii)
= 5 – 2 – 21 + 9 < 0
and
b × n + d = (3iˆ + 3 ˆj + 3kˆ) × (5iˆ + 2 ˆj - 7kˆ) + 9
k
= 1 i.e. k = 9
9
12. (b) Equation of the plane containing the line
x -1 y - 2 z - 3
is
=
=
1
2
3
a (x – 1) + b (y – 2) + c (z – 3) = 0 ....... (i)
= 15 + 6 – 21 + 9 > 0
So, the points a and b are on the opposite
sides of the plane.
14. (c)
2 x 2 - 2 y 2 + 4 z 2 + 6 xz + 2 yz + 3xy = 0
or 2 x 2 + x(6 z + 3 y ) - 2 y 2 + 4 z 2 + 2 yz = 0
MATHEMATICS
198
\x =
-(6 z + 3 y) ± 36 z 2 + 9 y 2 + 36 yz - 8( -2 y 2 + 4 z 2 + 2 yz )
4
-(6 z + 3 y) ± (2 x + 5 y)2
\x =
4
\x =
-(6 z + 3 y ) ± (2 z + 5 y )
4
or 2 x - y + 2 z = 0, x + 2 y + 2 z = 0
\ Angle between planes
q = cos -1
(2)(1) + ( -1)(2) + (2)(2)
2
(2) + ( -1) 2 + (2) 2 (1) 2 + (2) 2 + (2) 2
æ4ö
= cos -1 ç ÷
è9ø
15. (b)
PR : PQ = 1: 3 Þ 3PR = PQ
1
P
R
(–3, 1, 1)
2
Q
(3, 4, 2)
s2 : ax + by + cz = 0
s3 : a2x + b2y + c2z = 0
D=
1
a
1
b
1
c
a2
b2
c2
So, for unique solution, D ¹ 0
Þ D = (a – b) (b – c) (c – a) ¹ 0
Þ a ¹ b, b ¹ c, c ¹ a
rr
17. (a) The equation of plane is r. a = 5
r r
r r
Q r -b + r-c =4
r
Þ sum of distances of a point (r ) from two
r
r
fixed points with position vector b and c is
constant.
Þ such points lies on ellipsoid.
r
r
Now points with position vector b and c
r r
satisfies the equation of plane r . a = 5, then
r r
r r
b . a = 5 and c . a = 5
Þ 3PR = PR + RQ Þ 2 PR = RQ
Therefore, PR : RQ = 1 : 2. Hence
B(b)
C(c)
Plane r × a = 5
4ö
æ -6 + 3 2 + 4 2 + 2 ö æ
R =ç
,
,
÷ = ç -1, 2, ÷
3
3 ø è
3ø
è 1+ 2
The normal to the required plane is
PQ = (6, 3, 1). Hence, the equation of the
Area in the plane constitutes an ellipse
r
r
Distance between b and c
required plane is
= 2 × (semi major axis) × e = 14
4ö
æ
6( x + 1) + 3( y - 2) + 1ç z - ÷ = 0
3ø
è
2ae = 14
Þ 18 x + 9 y + 3z - 4 = 0
16. (c) s1 is perpendicular to (iˆ + ˆj + kˆ)
s2 is perpendicular to (aiˆ + bjˆ + ckˆ)
and s3 is perpendicular to
(a 2 iˆ + b2 ˆj + c 2 kˆ)
Then, the planes are
s1 : x + y + z = 0
...(i)
Sum of distance = constant = major axis = 4
2a = 4
...(ii)
From eqn (i) and (ii)
e=
1
14
Þ b = a (1 - e2 ) =
(semi
2
4
minor axis)
Solutions
199
Area of ellipse = p.a.b.
= p.2.
18. (3)
L2 :
L1 :
20. (7)
21. (7) A plane containing line of intersection of
the given planes is
1
= 2 p » 4.443
2
x - y - z - 4 + l ( x + y + 2 z - 4) = 0
x - 0 y +1 z - 0
=
=
=l
1
1
1
i.e., (l + 1) x + (l - 1) y + (2l - 1) z - 4(l + 1) = 0
vector normal to it
V = (l + 1)iˆ + (l - 1) ˆj + (2l - 1) kˆ
x +1 y - 0 z - 0
=
=
=m
2
1
1
Hence any point on L1 and L2 can be
Now the vector along the line of intersection
of the planes
(l, l - 1, l) and (2m - 1, m, m) , respectively..
2x + 3 y + z - 1 = 0
According to the question,
and x + 3 y + 2 z - 2 = 0 is given by
2m - 1 - l m - l + 1 m - l
=
=
2
1
2
iˆ
On solving, we get m = 1 and l = 3
\ A = (3, 2, 3) B = (1, 1, 1) \ AB = 3.
19. (2) The equation of any plane passing through
given line is :
( x + y + 2z - 3) + l(2x + 3y + 4z - 4) =
Þ (1 + 2l)x + (1 + 3l) y + (2 + 4l)z - (3 + 4l) = 0
....(i)
If this plane is parallel to z-axis whose direction
cosines are 0, 0, 1; then the normal to the plane
will be perpendicular to z-axis
\ (1 + 2l)(0) + (1 + 3l)(0) + (2 + 4l)(1) = 0
Þl=-
1
2
n = 2 3 1 = 3(iˆ - ˆj + kˆ)
1 3 2
As n is parallel to the plane (i), therefore,
n ×V = 0
(l + 1) - (l - 1) + (2l - 1) = 0
Þ 2 + 2l - 1 = 0 Þ l =
-1
2
Hence, the required plane is
x 3y
- 2z - 2 = 0
2 2
Þ x - 3 y - 4z - 4 = 0
Hence, | A + B + C - 4 |= 7
Put in eq (i), the required plane is
1
(x + y + 2z - 3) - (2x + 3y + 4z - 4) = 0
2
...(ii)
Þ y+2 =0
\ S.D. = distance of any point say (0, 0, 0) on
z-axis from plane (ii) =
ˆj kˆ
2
(1)
2
=2
22. (7) Under the given conditions the possible
situation is f(–2) = 2; f(0) = 3; f(1) = 1.
{where f(–2) = 1 is false, f (0) ¹ 2 is true and
f (1) ¹ 1 is false}. The triangle formed is with
vertices
A(–2, 1, 0), B(2, 1, 3) and O(0, 0, 0)
iˆ ˆj kˆ
1
Area of DAOB =
2 1 3
2
-2 1 0
MATHEMATICS
200
1
1
| -3iˆ - 6 ˆj + 4 kˆ |=
61 square units
2
2
k
=
; so k = 61.
2
(6) The required plane is of the form
Þ [bc + bl, (c - 2at ) - cb - cbl ]2
( x + 2 y + 3z - 4) + l(2 x + y - z + 5) = 0
= -kbal(ba + abl - ablt 2 - ab - abl)
=
23.
whose normal is (1 + 2l, 2 + l, 3 - l). This
plane is perpendicular to the plane
5 x + 3 y + 6 z + 8 = 0. So we have
5(1 + 2l) + 3(2 + l) + 6(3 - l) = 0
= k a (-bl )(ba + bl)(a - at 2 ) - (ab - abl )
Þ [bc + blc - 2ablt - cb - cbl )2
Þ 4a 2 b2 l 2 t 2 = -kbal (-ablt 2 )
Þ 4a 2 b 2 l 2 t 2 = k 2b 2 a 2 l 2 t 2 Þ k = 4.
25. (4)
| x |£ 8 Þ x Î [-8, 8] similarly for y and z.
This represent a cube of side 16 units with
centre at origin.
-29
Þ 7 l = -29 Þ l =
7
Therefore, the required plane is
Now, -8 £ x + y + z £ 8 gives space between
two panes namely
Z
29
( x + 2 y + 3 z - 4) - (2 x + y - z + 5) = 0
7
(0,0,8)
Þ 51x + 15 y - 50 z + 173 = 0
Comparing this with ax + by + cz + 173 = 0 we
get a = 51, b = 15, c = –50
Y'
(0,–8,0)
24. (4) Any point on parabola z2 = 4ax, y = 0 is
given by Q(at2, 0, 2at)
Now equation of line joining P(a, b, c) and Q
(at2, 0, 2at) is given by
x-a
y -b
z-c
=
=
= l (say)
2
b
0
c
- 2at
a - at
Þ x = a + l(a - at 2 ); y = b + bl; z = c + l;
z = c + l(c - at ) by given condition point Q
lies on (bz - cy )2 = k a (b - y )(bx - ay )
(8,0,0)
X
Z'
x + y + z + 8 = 0 and x + y + z – 8 = 0 subject
to the limits of x, y, z Î [-8, 8]
æ1 1ö
This will take out ç + ÷ volume of the cube
è4 4ø
1
3
Þ The required volume = ´ (16) = 2018.
2
Þ t = 2048 Þ
B(0,b,0)
O
L(a,b,0)
Y
Y
(0,0,–8)
Z
2
(0,8,0)
O
so that, b - 9(a + c) = 15 - 9 = 6.
(at ,0,2at)
Q
0)
,
(a,0 A
X'
t
=4
512
Solutions
201
CHAPTER
Probability-2
28
1.
(c) Total number of functions from A to B is
(20)10.
The number of non-decreasing function is the
number of non-negative integral solutions of the
equation
x1 + x2 +¼+ x20 = 10 = 29 C10
1
( pn - 2 + pn -1 )
2
Now
=
1
1
p n + p n -1 = pn -1 + pn - 2
2
2
1
= p n - 2 + p n -3
2
(Here xi is the pre-image of number i, i = 1, 2, 3,
....,20)
Þ the probability =
2.
29
(20)10
(c) When two dice is thrown then sample
space has 6 × 6 = 36 elements so n(S) = 36.
Now consider the event of getting 9 is (3, 6),
(4, 5), (5, 4) and (6, 3).
So probability of getting 9 when two dice is
thrown is 4/36 = 1/9.
If Sanchita starts the game then the probability
that she wins is
æ 1 ö æ 8 öæ 8 öæ 1 ö
ç 9 ÷ + ç 9 ÷ç 9 ÷ç 9 ÷ +..... ¥ =
è ø è øè øè ø
3.
1
9 = 1 ´ 81 = 9
64 9 17 17
181
And if Raj starts the game then probability
that Sanchita wins the game is 1– 9/17 = 8/17
(d) As usual H denotes head and T denotes
tail so that
1
= P (T)
2
Let En denote the event that the score is n. One
can easily see that
En = (En–2 Ç H) È (En–1 Ç T)
Therefore
Pn = P(En)
= P(En–2 Ç H) + P(En–1 Ç T)
= P(En–2) P(H) + P(En–1)P(T)
P (H) =
1
= p n -3 + p n -4
2
.................................
.................................
C19
1
= p 2 + p1
2
Since
p1 = P ( T ) =
1
2
and p 2 = P ( ( T Ç T ) È H )
= P ( T ) P (T ) + P ( H)
1 1 1 3
´ + =
2 2 2 4
We have
=
1
1
3 1
pn + pn -1 = p2 + p1 = + = 1
2
2
4 4
Therefore
pn -
2 1 1
= - p n -1
3 3 2
2
1æ
2ö
1
2
= - ç p n -1 - ÷ = æç - ö÷ æç p n - 2 - ö÷
2è
3ø è 2ø è
3ø
3
2ö
æ 1ö æ
= ç - ÷ ç pn -3 - ÷ .....
3ø
è 2ø è
æ -1 ö
=ç ÷
è 2ø
n -1
2ö
æ
ç p1 - 3 ÷
è
ø
MATHEMATICS
202
æ 1ö
= ç- ÷
è 2ø
Hence
pn =
2
=
4.
n -1
2 æ 1ö
+ç- ÷
3 è 2ø
n +1
5 4 4 2 4 3 5 3 3 5 4 3
= ´ ´ + ´ ´ + ´ ´ + ´ +
7 7 7 7 7 7 7 7 7 7 7 7
æ1 2ö
ç - ÷
è2 3ø
+ ( -1)
n -1
æ 1 ö 2 1æ 1 ö
ç- ÷ = + ç- ÷
è 6ø 3 3è 2ø
6.
80 + 24 + 45 + 60 209
=
7´ 7´ 7
343
(d)
7.
(b) The graph of y = 16 x 2 + 8(a + 5)
=
n
n
x - 7a - 5 is strictly above the x-axis
Þ y>0 " x Î R
3.2n
(c) The number of determinants formed = 16.
Observe that the determinant is non-zero when
exactly once (–1) appears as shown
1
1
Þ 16 x 2 + 8(a + 5) x - 7a - 5 > 0 " x Î R
The above inequality holds if discriminant < 0
[Q coefficient of x 2 > 0 ]
= 2 Þ 4 ways
-1 1
Similarly the determinant is non-zero when (–1)
is used exactly three times as shown
-1 -1
-1
1
Þ 64(a + 5)2 -4.16( -7 a - 5) < 0
Þ a 2 + 17 a + 30 < 0
Þ ( a + 2 ) (a + 15) < 0
= -2 Þ 4 ways.
Þ - 15 < a < - 2
So non-zero determinant can be obtained in 8
ways. Similarly determinant will be zero in 8
determinants.
1
2
(b) Let E1, E2 and E3 be the events of the critics
giving favourable remarks. Then
Þ
5.
S
E = ( E1 Ç E 2 Ç E3 ) È ( E1 Ç E 2 Ç E3 )
È ( E1 Ç E 2 Ç E 3 ) È ( E1 Ç E 2 Ç E3 )
Hence P(E)
= P ( E1 ) P ( E 2 ) P ( E 3 ) + P ( E1 ) P ( E 2 ) P ( E 3 )
+ P ( E1 ) P ( E 2 ) P ( E 3 ) + P ( E1 ) P ( E 2 ) P ( E 3 )
5 4 æ 3ö æ 5ö 4 3 5
= ´ ´ ç1 - ÷ + ç1 - ÷ ´ ´ +
7 7 è 7ø è 7ø 7 7 7
æ 4ö 3 5 4 3
´ ç1 - ÷ ´ + ´ ´
è 7ø 7 7 7 7
–15
–
+
–2
Given -20 £ a £ 0 and favourable cases
-15 < a < -2
\ Required probability
P(E) =
5
4
3
P ( E1 ) = , P ( E 2 ) = and P ( E 3 ) =
7
7
7
Let E be the event that majority reviewed
favourably. Therefore
+
=
8.
length of interval (-15, - 2) - 2 - (-15) 13
=
=
length of interval (-20, 0) 0 - (-20) 20
(a) Let G, C, K, and A are events where
G = Guess answer, C = Copy answer, K = Know
the answer, and A = Answer correctly
P(G) = Probability that the Candidate Guess the
answer = 1/3 (given)
P(C) = Probability that the Candidate Copy the
answer = 1/6 (given)
P(K) = Probability that the Candidate Know the
answer = ?
Now, G,C and K are mutually exclusive and
exhaustive events
Therefore, P(G) + P(C) + P(K) = 1
P(K) = 1 – 1/3 – 1/6 = 1/2.
Say G(Candidate guesses) has occurred. As,
there are four choices out of which only one is
correct, then the probability that the candidate
made a guess is 1/4.
Solutions
9.
203
P(A/G) = 1/4
P( A/C) = 1/8 ( given) then
P( A/K) = Answer correctly that the candidate
know = 1
Applying Baye’s Theorem we have,
P(K/A) = [ P(K).P(A/K)]/[P(G) × P(A/G) + P(C) ×
P(A/C) + P(K) × P(A/K) = 24/29.
(b) Let the sides be x, y, l – (x + y)
Since the triangle will be formed when sum of
two sides is larger than the third.
Y
1
4
and P ( E 2 ) =
5
5
Let E be the event of the boy falls asleep. Again
by hypothesis
P ( E1 ) =
P ( E / E1 ) =
3
1
and P ( E / E 2 ) =
4
4
Now,
B
E = E Ç ( E1 È E2 ) = ( E1 Ç E ) È ( E 2 Ç E )
so that,
y
l /2
11. (c) Let E1 and E2 be the events of the boy
watching DOORDARSHAN and TEN SPORTS,
respectively. It is given that
Q
P ( E ) = P ( E1 ) P ( E / E1 ) + P ( E 2 ) P ( E / E 2 )
By Bayes' theorem
P
l
A
X
O
l /2
l
i.e. l – y > y, l – x > x and x + y > l – (x + y)
P ( E1 / E ) =
R
Þ 0< y<
l
l
l
, 0 < x < and < (x + y) < l
2
2
2
DPQR 1
=
DOAB 4
(as OQ = 1/2 OB)
10. (b) Let x and y be the two quantities. When the
sum of two non-negative quantities is fixed, the
product will be maximum when they are equal.
So, the greatest product = xy = 10000
where x = y = 100
Hence required probability =
3
Now, xy ³ ´10000
4
Þ
xy ³ 7500 Þ x ( 200 - x ) ³ 7500
Þ
x 2 - 200x + 7500 £ 0
Þ ( x - 50 )( x - 150 ) £ 0
Þ 50 £ x £ 150
So, favourable number of ways = 101
Total number of ways = 201
101
201
Hence, (b) is the correct answer.
So, required probability =
P ( E1 ) P ( E / E1 )
P ( E1 ) P ( E / E1 ) + P ( E 2 ) P ( E / E 2 )
=
(1/ 5 ) ´ ( 3/ 4 )
(1/ 5) ´ ( 3 / 4 ) + ( 4 / 5 ) ´ (1/ 4 )
=
3
7
12. (c) Here, Pn = apn, n > 1
and P0 = 1 – ap (1 + p + p2 + ...)
Consider the following events :
Ej = There are j children in the family;
j = 0, 1, 2,....., n
A = There are exactly k boys in the family
We have, P(Ej) = pj = apj; j = 0, 1, 2, ...., n
æ A ö jC
Pç ÷ = k , j ³ k
ç Ej ÷
2j
è ø
Þ
Now, A =
¥
U (A Ç E j )
j= k
æ ¥
P (A) = P ç U A Ç E j
ç j= k
è
(
Þ
\ P (A) =
=
ö
) ÷÷
ø
¥
¥
j= k
j= k
j
¥
æ j Ck ö
æpö j
j
a
=
a
p
ç
÷
å ç j ÷ å çè 2 ÷ø . Ck
j= k
j= k
è 2 ø
¥
æAö
÷
÷
è jø
å P ( A Ç E j ) = å P ( E j ) P çç E
MATHEMATICS
204
¥
=aå
k +r
j= k
æpö
Cr ç ÷
è2ø
æpö
P (A) = a ç ÷
è2ø
published) = P(X = 1) + P(X = 2)
k+r
k ¥
å
2
k +r
r=0
æpö
Cr ç ÷
è2ø
r
= 2´
We have, k+rCr = (–1)r–(k+1)Cr
k
\
¥
æpö
æ pö
- k +1
P ( A ) = a ç ÷ å ( ) Cr ç - ÷
è 2 ø r=0
è 2ø
k
æpö æ pö
= a ç ÷ ç1 - ÷
è2ø è 2ø
-( k +1)
k
k +1
2
æpö
= aç ÷ .
è 2 ø ( 2 - p )k +1
3 1
=
6 2
\ Required probability
=
2n
+
3
æ1ö
C3. ç ÷
è2ø
2n +1
(
2n +1
C1 +
2n +1
C3 + ..... +
æ1ö
.ç ÷
è2ø
2n - 2
æ1ö
C2n +1 ç ÷
è2ø
2n +1
)
æ 1ö
C2n +1 ç ÷
è 2ø
2n +1
8
æ1 1ö
\ The binomial distribution is ç + ÷
è2 2ø
Also, |x – 4| < 2 Þ –2 < x – 4 < 2 Þ 2 < x < 6
\ p(|x – 4| < 2) = p(x = 2) + p(x = 3)
+ p(x = 4) + p(x = 5) + p(x = 6)
6
3
2n +1
E = ( G È G ') Ç E = ( G Ç E ) È ( G 'Ç E )
1
2
; P ( E / G ') =
3
4
1
P ( G ) = = P (G ')
2
1 2 1 1 11
´ + ´ =
2 3 2 4 24
Further, X denotes the number of books
published. Then P (at least one book will be
5
4
æ1ö æ1ö
æ1ö æ1ö
æ1ö æ1ö
= 8C2 ç ÷ ç ÷ + 8C3 ç ÷ ç ÷ + 8C4 ç ÷ ç ÷
è2ø è2ø
è 2ø è 2ø
è 2ø è 2ø
3
6
æ1ö æ1ö
æ1ö æ1ö
+ 8C5 ç ÷ ç ÷ + 8C6 ç ÷ ç ÷
è2ø è2ø
è2ø è 2ø
2n +1
2n +1
Therefore, P ( E ) =
1 1
=
2 2
5
1
æ1ö
= 22n. ç ÷
=
2
è2ø
14. (a) Let G = Event of good book
G' = Event of not a good book
E = Event of publication
Then
Now, P ( E / G ) =
1
,n=8
2
q = 1 – p = 1-
\
2
+.... +
=
2
407
11 13 æ 11 ö
´ +ç ÷ =
576
24 24 è 24 ø
15. (b) Here, p =
r
13. (d) The probability of showing an even number
in a throw
1 æ1ö
= 2n +1C1. . ç ÷
2 è2ø
0
æ 11 öæ 13 ö
æ 11 ö æ 13 ö
= 2C1 ç ÷ç ÷ + 2C2 ç ÷ ç ÷
è 24 øè 24 ø
è 24 ø è 24 ø
=
8
2
C2 + 8C3 + 8C4 + 8C5 + 8C6
28
238 119
=
256 128
16. (a) If both have all 4 cards of the same color,
then there are two possibilities at the end.
Posibility 1: Ravi holds 4 red cards and
Rashmi 4 black cards.
Probability of this possibility = P(Ravi
picks red in 1st pick 'AND' Rashmi picks
black in 1st pick 'AND' Ravi picks red in 2nd
pick' AND 'Rashmi pick black in 2nd pick')
All 4 picks are independent
=
2 2 1 1
1
´ ´ ´ =
4 5 4 5 100
Posibility 2: Ravi holds 4 black cards and
Rashmi 4 red cards similarly probability
1
=
100
=
4
Solutions
205
[A wins his serve then B wins his serve or A
loses his serve then B also loses his serve]
So, probability that ‘A’ wins the game after n
deuces
Hence, required probability
P=
1
1
2
+
=
100 100 100
1
= 0.02 = 2%
50
17. (a) We have 30 males and 20 females.
n
=
P(males) =
30 3
=
50 5
P(females) =
20 2
=
50 5
¥
2 1
. =
5 9 2 = 0.50
19
19. (0.25)
Let l be the length of the chord AB of
the given circle of radius a and r be the distance
of the mid point D of the chord from the centre C,
then r = a cos q and l = 2a sin q . According to
given condition :
2/5
Male
1/2
Disease
+ve
4/5
(1)
Disease
No disease
not +ve
+ve
not
+ve
+ve
not +ve
+ve
not
+ve
1/5 1/5
4/5
4/5
1/5 1/5
4/5
(2)
(3)
(4)
(5)
Required probability =
1
Female
1/5
4/5
No disease
n
æ 5ö 2 1
= å çè 9 ÷ø . 3 . 3 =
n= 0
3/5
1/2
æ 5ö 2 1
= ç ÷ . . [After nth deuce, A serves and
è 9ø 3 3
wins then B serves and loses]
\ Required probability of ‘A’ Winning the
game
(6)
(7)
C
a
q
(8)
15
50 = 75
= 15 16 107
+
50 125
18. (0.50)
Let us assume that A wins after n
deuces, n = 0, 1, 2, 3 ...........
2 2 1 1 5
Probability of a deuce = . + . =
3 3 3 3 9
B
A
(1) + ( 3)
(1) + ( 3) + ( 5) + ( 7 )
3 1 4 3 1 1
´ ´ + ´ ´
5 2 5 5 2 5
=
3 1 4 3 1 1 2 1 4 2 4 1
´ ´ + ´ ´ + ´ ´ + ´ ´
5 2 5 5 2 5 5 5 5 5 5 5
r
D
2
5
(2a) < 2a sin q < (2a )
6
3
Þ
2
5
5
11
< sin q <
< cos q <
3
6 Þ 6
3
11
5
a <r <
a
6
3
\ The given condition is satisfied if the mid
point of the chord lies within the region
Þ
between the concentric circles of radii
and
5
a.
3
11
a
6
MATHEMATICS
206
Hence, the required probability
2
=
æ 5 ö
æ 11 ö
pç
a÷ - p ç
a÷
è 3 ø
è 6 ø
For maximum probability;
2
=
1
= 0.25
4
Þ-
pa2
20. (0.60)
Let A denotes the event that the
runner succeeds exactly 3 times out of five and
B denotes the event that the runner succeeds
on the first trial.
2
21. (6) Let x shell are fixed at point I. Define the
following events
E1 : The target is at point I Þ P(E1 ) =
8
9
E 2 : The target is at point II Þ P ( E2 ) =
1
9
A : The target is hit
The target will be hit if at least one shell hits the
target.
P ( A / E1 ) = 1 – None of the shells hit when the
x
æ 1ö
target is at point I = 1 - ç ÷ and
è 2ø
æ 1ö
P ( A / E2 ) = 1 - ç ÷
è 2ø
\
P (A) =
2
Also, d P ( A)
dx 2
x -3
21- x
1 é æ 1ö
æ 1ö
= 1- ê ç ÷
+ç ÷
è 2ø
9 êë è 2ø
2
. Therefore m = 2.
9
(0) Let E1, E2, E3, E4, E5 and E6 be the events
of occurrence of 1, 2, 3, 4, 5 and 6 on the dice
respectively and let E be the event of getting a
sum of numbers equal to 9.
f(f(x)) = x is
23.
\
ù
ú
úû
x -3
21- x
ù
1 é æ 1ö
æ 1ö
(ln2)2 + ç ÷
(ln2)2 ú < 0
êç ÷
è 2ø
9 ê è 2ø
ú
ë
û
\ P(A) is maximum when x = 12 Þ k = 6
22. (2) Clearly all the solutions of f(x) = x are also
solutions of f(f(x)) = x. First, we solve f(x) = x
f(x) = x Þ x2 – 3x + 3 = x
Þ x2 – 4x + 3 = 0
Þ (x – 1) (x – 3) = 0
Þ x = 1, 3
Therefore x = 1, 3 are also solutions of f(f(x)) = x.
We want to seek if there are any more solutions
of f(f(x)) = x other than 1 and 3
f(f(x)) = x Þ f(x2 – 3x + 3) = x
Þ (x2 – 3x + 3)2 – 3(x2 – 3x + 3) + 3 = 0
Þ x4 – 6x3 + 12x2 – 9x + 3 = 0
Þ (x2 – 4x + 3)(x2 – 2x + 1) = 0
Þ (x – 1) (x – 3) (x – 1)2 = 0 Þ x = 1, 3
In this case we have no additional solutions.
Therefore the probability that x satisfies equation
21- x
8 é æ 1 ö x ù 1 é æ 1 ö 21- x ù
ê 1- ç ÷ ú + ê 1 ú
9 ëê è 2ø ûú 9 êë çè 2 ÷ø
ûú
û
Þ x = 12
=-
2
3
æ B ö 6p (1 - p )
= = 0.60
Thus, P ç ÷ =
è A ø 10p3 (1 - p )2 5
x-3
21- x
ù
1 é æ 1ö
æ 1ö
ln 2 - ç ÷
ln2 ú = 0
êç ÷
9 ê è 2ø
è 2ø
ú
ë
æ B ö P ( B Ç A)
Pç ÷ =
P ( A)
èAø
But P ( B Ç A ) = P (clearing succeeding in the
first trial and exactly once in two other trials)
P (4C2 p2 (1 – p2)) = 6p2 (1 – p)2
and P(A) = 5C3 p3 (1 – p)2 = 10p3 (1 – p)2
dP( A)
=0
dx
P ( E1 ) =
1- k
1 + 2k
1- k
; P (E2 ) =
; P ( E3 ) =
6
6
6
P ( E4 ) =
1+ k
1 - 2k
1+ k
; P ( E5 ) =
; P ( E6 ) =
6
6
6
and
1
2
£ P (E) £
9
9
Solutions
207
Then, E º {(3, 6), (6, 3), (4, 5), (5, 4)}
Hence, P(E) = P(E3E6) + P(E6E3) + P(E4E5) +
P(E5E4)
= P(E3) P(E6) + P(E6) P(E3) + P(E4) P(E5) + P(E5)
P(E4)
6
æ
1ö æ
j = 2è
1
[2 - k - 3k 2 ]
18
1
2
£ P (E) £
9
9
Since,
Þ
1 1 é
2
£
2 - k - 3k 2 ù £
û 9
9 18 ë
Þ
2 £ 2 - k - 3k 2 £ 4
Þ
2
2 £ 2 - k - 3k 2 and 2 - k - 3k £ 4
Þ
1ö
æ
3k ç k + ÷ £ 0 and 3k2 + k + 2 > 0
3ø
è
Þ
1
- £ k £ 0 and k Î R
3
1
- £k £0
3
Hence, integral value of k is 0.
24. (5) The sportsman's chance of missing when
r = ja is
\
1-
a2
=1 -
1
( j = 2,3, 4,5, 6 )
2
j2 a 2
j
The animal escapes when the sportsman misses
in all the five shots. Therefore the probability of
animal escaping to jungle is
1 öæ
1 öæ
1 ö
éæ 1 ö æ 1 ö æ 1 ö æ 1 ö æ 1 ö ù
= êç 1 - ÷ ç 1 - ÷ ç 1 - ÷ ç 1 - ÷ ç 1 - ÷ ú
ëè 2 ø è 3 ø è 4 ø è 5 ø è 6 ø û
éæ 1 öæ 1 ö æ 1 ö æ 1 ö æ 1 ö ù
´ êç1 + ÷ç1 + ÷ ç1 + ÷ ç1 + ÷ ç1 + ÷ ú
ëè 2 øè 3 ø è 4 ø è 5 ø è 6 ø û
[since E1, E2, E3, E4, E5 and E6 are independent]
=
1 öæ
ø
= 2P(E3) P(E6) + 2P(E4) P(E5)
æ 1 - k öæ 1 + k ö æ 1 + k öæ 1 - 2k ö
= 2ç
֍
÷ + 2ç
֍
÷
è 6 øè 6 ø è 6 øè 6 ø
1 öæ
1 - ÷ç1 - ÷ç1 - ÷ç1 - ÷
Õ çç1 - j2 ÷÷ = çè1 - 22 ÷ç
øè 23 øè 2 4 øè 25 øè 26 ø
æ 1 2 3 4 5 öæ 3 4 5 6 7 ö
= ç ´ ´ ´ ´ ÷ç ´ ´ ´ ´ ÷
è 2 3 4 5 6 øè 2 3 4 5 6 ø
1 7 7 p
= ´ = =
6 2 12 q
Therefore, q – p = 12 – 7 = 5
25. (3) Let S be the sample space, then
n(S) = Total number of determinants that can be
made with 0 and 1 = 2 × 2 × 2 × 2 = 16
Q
a
b
c d
, each element can be replaced by two
types
i.e., 0 and 1
and let E be the event that the determinant made
is non-negative.
Also, E' be the event that the determinant is
negative.
\
\
ïì 1 1 0 1 0 1 ïü
E' = í
,
,
ý
îï 1 0 1 1 1 0 ïþ
P(E') = 3
n ( E ') 3
then P(E') = n S = 16
( )
Hence, the required probability,
3 13 m
[given]
= =
16 16 n
m = 13 and n = 16, then n – m = 3
P(E) = 1 – P(E') = 1 Þ
MATHEMATICS
208
CHAPTER
Properties of Triangle
29
1.
(c) Sum of the roots of the equation is given by
c(a + b)
sin A + sin B =
c2
= (c1 + c2 )2 - 2c1c2 (1 + cos 2 A)
a+b
=
c
= (2b cos A) 2 - 2(b2 - a 2 )(2cos 2 A)
sin A + sin B
sin C = 1
sin C
Þ the triangle is right angled.
(c) If D is the diameter of the circumscribed
circle of D ABC, then a = D sin A, b = D sin B,
c = D sin C
=
2.
\
a 3 + b3 + c 3
å sin3 A
=7Þ
D3
5.
( å sin A) = 7
3
å sin3 A
\ D=37.
Since no side of triangle can exceed the diameter
of the circle, the maximum possible value of a
3.
Þ
6.
b
p
C
a
= sin B sin C : sin C sin A : sin A sin B
B
1
1
1
:
:
sin A sin B sin C
\ sin A, sin B, sin C are in A.P.
=
4.
(d)
cos A =
b2 + c2 - a 2
2bc
Þ c 2 - (2b cos A)c + b2 - a 2 = 0 .
It is a quadratic in c, whose roots are c1 and c2,
so c1 + c2 = 2b cos A and c1c2 = b2 - a 2
\ c12 + c22 - 2c1c2 cos 2 A
(a + b + c )2
³3Þ P³3
ab + bc + ca
\ 3 £ P < 4 or P Î[3, 4)
(a) We have,
b+c
³ bc = l 2 Þ b + c ³ 2l
2
a
b
c
b+c
Now,
=
=
=
sin A sin B sin C sin B + sin C
Þ
\ p : q : r = b sin C : c sin A : a sin B
c
(a + b + c )2
<4ÞP<4
ab + bc + ca
Again (a - b)2 + (b - c )2 + (c - a) 2 ³ 0
is 3 7 .
(b) Let the altitudes from A, B, C be p, q, r
respectively. Then,
p = b sin C, q = c sin A, r = a sin B
A
= 4b 2 cos 2 A - 4b 2 cos 2 A + 4 a 2 cos 2 A
= 4a2 cos2A.
(b) a, b, c are sides of a triangle
\ a + b > c, b + c > a, c + a > b
\ a > | b – c |, b > | c – a |, c > |a – b | square and
add
a2 + b2 + c2 < 2 (ab + bc + ca)
Þ a2 + b2 + c2 + 2 (ab + bc + ac)
< 4 (ab + bc + ca)
Þ
a
b+c
=
A
A
B+C
B-C
2 sin cos
2 sin
cos
2
2
2
2
A
2
Þa=
æ B -Cö
cos ç
è 2 ÷ø
(b + c )sin
B-C
£ 1 and
2
A
b + c ³ 2l Þ a ³ 2l sin .
2
Q 0 £ cos
Solutions
7.
(c) L.H.S. =
209
Clearly a and b < 1 but c > 1 as sin a > 0 and
cos a > 0
\ c is the greatest side and greatest angle is C
1 2
(a (b + c – a) + b2 (c + a – b) + c2
2
(a + b – c)
1
(a (b2 + c2 – a2) + b (c2 + a2 – b2)
2
+ c (a2 + b2 – c2))
1
= (2abc cos A + 2abc cos B + 2abc cos C)
2
A
B
Cö
æ
= abc ç1 + 4 sin sin sin ÷
2
2
2ø
è
\ cos C =
=
A
B
Cö
æ
= 4RD ç1 + 4 sin sin sin ÷ .
2
2
2ø
è
8.
(b) a tan q + b sec q = c
Þ b2 sec 2 q = (c - a tan q)2
Þ b 2 (1 + tan 2 q) = c 2 - 2ca tan q + a 2 tan 2 q
Þ (a 2 - b2 ) tan 2 q - 2ca tan q + c 2 - b2 = 0 ...(i)
Roots of equation (i) are tan a and tan b , where
a and b are the two angles of the triangle.
2ca
We have tan a + tan b = 2
and
a - b2
tan a. tan b =
c 2 - b2
a 2 - b2
2ca
\ tan(a + b) =
9.
a 2 - b 2 = 2ca
c2 - b2 a 2 - c2
1- 2
a - b2
a2 + b 2 - c 2
2ab
1
sin 2 a + cos 2 a - 1 - sin a cos a
=2 sin a cos a
2
\ C = 120°
=
11.
(d) sin2 A + sin2 C = 1001 sin2 B
Þ a2 + c2 = 1001 b2 (using sine rule)
Now,
=
2 (tan A + tan C) × tan 2 B
tan A + tan B + tan C
2 (tan A + tan C) × tan 2 B
tan A × tan B × tan C
æ cot A + cot C ö
= 2ç
÷ø
cot B
è
=
2 (cos A sin C + sin A cos C)
sin B
sin A × sin C × cos B
=
2 sin (p - B) × sin B
2 sin 2 B
=
sin A sin C cos B
sin A sin C cos B
=
2 ´ 2b 2
2 ´ 2b 2
1
2 ´ 2b2
=
=
= 2 2
2
2
250
2ac × cos B a + c - b
1000b
12. (b)
sin A sin B sin C c
b
a
+
+
=
+ +
c sin B
c
b
ab ac bc
or
a sin B sin C
c
b
a
+
+
=
+
+
bc
c
b
ab ac bc
or
sin B sin C
c
b
+
=
+
c
b
ab ac
é1
ù
Now 2ac sin ê ( A - B + C ) ú
ë2
û
or
b sin B + c sin C c 2 + b2
=
bc
abc
= 2ac sin (90° - B )
= 2ac cos
or
a=
3p
2ca
\ tan æç p - ö÷ =
Þ a 2 - c 2 = 2ca
è
4 ø a 2 - c2
(b) We know that A + B + C = 180°
A + C – B = 180° – 2B
Þ
2ac (a 2 + c 2 - b2 )
= a 2 + c2 - b2
2ac
10. (c) Let a = sin a, b = cos a and
B=
c = 1 + sin a cos a
=
Þ
b2 + c2
b sin B + c sin C
b (2R sin B) + c (2R sin C)
=2R
b sin B + c sin C
ÐA =
p
2
MATHEMATICS
210
13. (a)
A
cos C =
4p
D
b
p
B
C
a
1
1
ab = p (4p) Þ ab = 4p2
2
2
Also, a2 + b2 = c2 = 16p2
\ (a – b)2 = a2 + b2 – 2ab = 8p2
Also, (a + b)2 = a2 + b2 + 2ab = 24p2
A-B a-b
C 1
=
cot =
´1
2
a+b
2
3
A-B
or
= 30° or A – B = 60°
2
BD
AD
14. (a) In DBAD,
=
sin q sin 60°
In DCAD,
Þ
Þ
p
2p
4p
, ÐB =
, ÐC =
7
7
7
æ
b2 öæ
c2 öæ
a2 ö
1 - 2 ÷ç 1 - 2 ÷
= a2b2c2 çç 1 - 2 ÷ç
a ֍
b ֍
c ÷ø
è
øè
øè
= a2b2c2
2p öæ
4p öæ
p ö
æ
sin 2
sin 2
sin 2 ÷
ç
֍
֍
7 17 17
ç1 ÷ç
֍
÷
2 p ֍
2 2p ֍
2 4p ÷
ç
sin
sin
sin
7 øè
7 øè
7 ø
è
= a2b2c2
17. (a) Since 4 sin A cos B = 1, so A and B can not
be
p
2
p
p
, then cos B = 0 and if A = , then
2
2
tan A is not defined]
CD
AD
=
sin (75° - q) sin 45°
[As if B =
BD
CD sin 45°
sin 60° =
sin q
sin (75° - q)
so, C =
sin q
BD sin 60°
1
=
=
sin (75° - q) CD sin 45°
6
A
æp
ö
Þ 4sin A cos ç - A ÷ = 1
è2
ø
q
B
16. (a) ÐA =
(a2 – b2) (b2 – c2) (c2 – a2)
D=
tan
a 2 + b 2 - c2 1
= Þ c= 6
2ab
8
p
p
ÞB= -A
2
2
Þ sin 2 A =
75°–q
60°+ q
60°
1 D
3
45°
p
p
1
1
Þ sin A = Þ A = Þ B =
4
2
6
3
so angles are
C
A - Bö
1 - tan ç
÷
è 2 ø = 31
15. (c) cos (A – B) =
æ A - B ö 32
1 + tan 2 ç
÷
è 2 ø
2æ
1
æA -Bö
Þ tan ç
=
÷
è 2 ø 3 7
C
1
æA -Bö a -b
tan ç
cot
Þ cos C =
÷=
2
8
è 2 ø a+b
18. (b)
Þ
Þ
p p p
, , which are in A.P..
6 3 2
2(b / a)
b
p+b
=
tan a = , tan 2a =
2
a
a
1 - (b / a)
2ba
2
a -b
2
=
p+b
a
2ba 2 - a 2 b + b3
Þ p=
a2 - b2
P
p
=p
b( a 2 + b 2 )
(a 2 - b 2 )
O
a
a
b
a
Solutions
211
19. (c) d = h cot 30° – h cot 60° and time = 3 min.
21. (4)
A
h(cot 30° - cot 60°)
\ Speed =
per minute
3
x y z
b
c
h
60°
30°
d
B
It will travel distance h cot 60° in
h cot 60° ´ 3
= 1.5minute
h(cot 30° - cot 60°)
20. (d) Let x and y be the heights of the flagstaffs
at P and Q respectively
Then,
AP = x cot 60° =
x
3
, AQ = y cot 30° = y 3
y
BP = x cot 45° = x, BQ = y cot 60° =
Þ AB = BP - AP = x -
x
3
[Q AB = 30 m]
3
Þ 30 3 = ( 3 - 1) x Þ x = 15(3 + 3)
S
R
x
y
60°
Q
45°
B
30°
30 m
60°
A
1 ö
æ
Similarly, 30 = y ç 3 ÷ Þ y = 15 3
è
3ø
so that PQ = BP + BQ = x +
y
3
= 15(3 + 3) + 15 = (60 + 15 3)m
P
a/3
D
a/3
E
a/3
a
AD
=
3sin x sin B
2a
AE
=
3sin(x + y) sin B
Dividing Eq. (1) by Eq. (2), we get
sin(x + y) AD
=
2sin x
AE
2a
AD
=
3sin (y + z) sin C
C
...(1)
...(2)
...(3)
...(4)
a
AE
=
3sin z sin C
Dividing Eq. (5) by Eq. (4), we get
...(5)
sin(y + z) AE
=
2sin z
AD
Multiplying Eq. (3) and Eq. (6), we get
...(6)
sin (x + y) sin(y + z)
\
=4
sin x × sin z
22. (3) The given equation is
4 sin A sin B + 4 sin B sin C + 4 sin C sin A = 9
Þ 2 cos (A – B) – 2 cos (A + B) + 2 cos (B – C)
–2 cos (B + C) + 2 cos (C – A) – 2 cos (C + A) = 9
or 2[cos(A – B) + cos(B – C) + cos(C – A)]
3
= 9 – 2 (cos A + cos B + cos C) ³ 9 – 2 × = 6
2
or cos (A – B) + cos (B – C) + cos (C – A) ³ 3
But cos (A – B) £ 1, cos (B – C) £ 1,
cos (C – A) £ 1
or cos (A – B) = 1, cos (B – C) = 1,
cos (C – A) = 1
Thus, A = B = C, i.e., triangle ABC is an
equilateral triangle.
Hence, D =
3.
MATHEMATICS
212
23. (2)
x
B
ID
ID
=
=
AD 2bc cos A
b+c
2
P
x
3
Q
x
c
A
9 = c2 + x2 – 2cx cos B
c
But cos B =
3x
c2
Þ
9= x +
...(1)
3
Similarly using cosine rule is DACQ, we get
2
16 = x2 +
b2
3
...(2)
Adding (1) and (2), we get 25 = 2x2 +
2abc
\
ID
ID
a
=
=
AD AI + ID a + b + c
\
ID
IE
IF a + b + c
+
+
=
=1
AD BE CF a + b + c
\
a(b + c) sec
\ k= 2
25. (7)
A
B
ID + b(a + c) sec IE
2
2
C
+ c(a + b) sec IF = 2abc
2
B
b 2 + c2
3
(3x)2
\ 25 = 2x2 + 3x2
3
\ x2 = 5 \ BC = 3x = 3 5
\
25
= 2x2 +
F
B
100 m
30°
A
24. (2)
A
I
D
A
× ID
2
Now AI = AB = c = b + c
ac
ID BD
a
b+c
4
C
b
Let BP = PQ = QC = x
In DABP, using cosine rule
a(b + c) sec
60°
d
x
D
C
100
100
In DCBD, tan 60° =
Þ x=
x
3
100
In DACB, tan30° =
x+d
x + d = 100 3 .
E
C
Þ
d = 100 3 -
100
3
=
200
3
=
200 3
m
3
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