MATHS | 30th Jan 2023 _ Shift-1

SECTION - A
61. A straight line cuts off the intercepts OA = a and OB = b on the positive directions of x-axis and y
π
axis respectively. If the perpendicular from origin O to this line makes an angle of 6 with positive
98
direction of y-axis and the area of △ OAB is 3
√3, then a2 − b2 is equal to:
392 196
(1) (2) (3) 98 (4) 196
3 3
Sol. 1

In AOB
 OB b
tan  
6 OA a
1 b
 
3 a

 a  3b

1 98
area of triangle OAB   ab   3
2 3
3b2 98
 
2 3
98
 b2  2
3
196
 b
3

a  196

196 588  196

a 2  b 2  196  
3 3
392
 a 2  b2 
3

62. The minimum number of elements that must be added to the relation R={(a, b), (b, c)} on the set
{a, b, c} so that is becomes symmetric and transitive is :
(1) 3 (2) 4 (3) 5 (4) 7
Sol. 4
R  {(a,b),(b,c)}
For symmetric relation (b, a), (c, b) must be added in R
For transitive relation (a, c), (a, a), (b, b), (c, c), (c, a) must be added in R
So, minimum number of element = 7

63. If an unbiased die, marked with −2, −1,0,1,2,3 on its faces, is thrown five times, then the probability
that the product of the outcomes is positive, is :
881 27 440 521
(1) 2592 (2) 288 (3) 2592 (4) 2592
Sol. 4
Unbiased die. Marked with –2, –1, 0, 1, 2, 3
Product of outcomes is positive if
All time get positive number, 3 time positive and 2 time negative, 1 time positive and 4 time negative.
5 3 2 4
P (Product of the outcomes is positive) = 5 C5  3   5 C3  3   2   5 C1  3  2
 
6   6 6     6 6   
All positive 3positive, 2 negative 1positive, 4 negative

35 10  33  22 5  3  24
=  
65 65 65
1563 521
= 
65 2592

64. If a⃗, ⃗b, c are three non-zero vectors and n̂ is a unit vector perpendicular to c such that a⃗ = αb
⃗ − n̂, (α ≠
0) and ⃗b ⋅ c = 12, then |c × (a⃗ × ⃗b)| is equal to :
(1) 9 (2) 15 (3) 6 (4) 12
Sol. 4
a  b  n,
ˆ b.c  12

c  (a  b)  (c.b)a  (c.a)b
c  (a  b)  12a  (c.a)b ….(1)

a  b  n
c.a  c.b  c.n

c.a  12 ….(2)

Equation (2) put in equation (1)
c  (a  b)  12a  12b
c  (a  b)  12 a   b  a  b  n then | a  b |  1
 

 c  (a  b)  12
65. Among the statements :
(S1) ((p ∨ q) ⇒ r) ⇔ (p ⇒ r)
(S2) ((p ∨ q) ⇒ r) ⇔ ((p ⇒ r)V(q ⇒ r))
(1) only (S2) is a tautology (2) only (S1) is a tautology
(3) neither ( S1) nor (S2) is a tautology (4) both (S1) and (S2) are tautologies
Sol. 3
S1 :  (p  q)  r   (p  r)
(p q)  r
p q r ~ pr  (p  q)  r   (p  r)
(~ p  ~ q)  r
T T T T T T
T T F F F T
T F T T T T
T F F F F T
F T T T T T
F F F T T T
F T F F T F
F F T T T T
S1 is not a tautology
S2 =  (p  q)  r    (p  r)  (q  r) 
p q r (p  q)  r (p  r)  (q  r)  (p  q)  r    (p  r)  (q  r) 
T T T T T T
T T F F F T
T F T T T T
T F F F T F
F T T T T T
F F F T T T
F T F F T F
F F T T T T
S2 is not a tautology
So, neither S1 nor S2 is a tautology.

66. If P(h, k) be a point on the parabola x = 4y 2 , which is nearest to the point Q(0,33), then the distance
of P from the directrix of the parabola y 2 = 4(x + y) is equal to :
(1) 2 (2) 6 (3) 8 (4) 4
Sol. 2
Equation of normal of the parabola x  4y 2
 t 2 2t 
At a point P  ,  is
 16 16 
2t 1 3
y  tx   t
16 16
 Normal pass through Q(0,33) then
t t3
33  
8 16
 t 3  2t  528  0
 (t  8)(t 2  8  166)  0

 t 8
Point P is (4, 1)
Given parabola is y 2  4(x  y)
y 2  4y  4x
(y  2) 2  4(x  1)
directrix is x + 1 = – 1
x  2
Distance of P(4, 1) from the directrix x = –2 is 6.

67. Let y = x + 2,4y = 3x + 6 and 3y = 4x + 1 be three tangent lines to the circle (x − h)2 + (y −
k)2 = r 2 .
Then h + k is equal to :
(1) 5(1 + √2) (2) 5√2 (3) 6 (4) 5
Sol. 4

In centre of triangle is (h, k)

 5(2)  2  7 2  5  5 3  (7 2)  0  5  7  5 
  , 
 557 2 557 2 
 14 2  15 21 2  35 
  , 
 10  7 2 10  7 2 
14 2  15 21 2  35
So, h  k  
10  7 2 10  7 2
35 2  50 5(7 2  10)
hk   5
7 2  10 7 2  10
 hk 5
68. The number of points on the curve y = 54x 5 − 135x 4 − 70x 3 + 180x 2 + 210x at which the normal
lines are parallel to x + 90y + 2 = 0 is :
(1) 4 (2) 2 (3) 0 (4) 3
Sol. 1
Given curve is y  54x 5  135x 4  70x 3  180x 2  210x
dy
 270x 4  540x 3  210x 2  360x  210
dx
Normal is parallel to x + 90y + 2 = 0
Then tangent is  r to x + 90 y + 2 = 0
Then (270x 4  540x 3  210x 2  360x  210)  1   1
 90 
270x  540x  210x  360x 120  0
4 3 2

 9x4 18x3  7x 2  12x  4  0

 (x 1)(x  2)(3x  1)(3x  2)  0
1 2
 x  1, 2,  , 
3 3
Number of points are 4

−2
69. If an = 4n2 −16n+15, then a1 + a2 + ⋯ … + a25 is equal to:
52 49 50 51
(1) 147 (2) 138 (3) 141 (4) 144
Sol. 3
2
given that a n 
4n  16n  15
2

25
2
a1  a 2  a 3  ....a 25  
n 1 (2n  3)(2n  5)
25
(2n  5)  (2n  3)

n 1 (2n  3)(2n  5)

25
 1 1 
   
n 1  2n  3 (2n  5) 
1 1
 
1 3
1 1
 
1 1
1 1
 
3 1

1 1

47 45
1 1
 
47 3
3  47 50
 
141 141
 1 1 1
70. If tan⁡15∘ + tan⁡75∘ + tan⁡105∘ + tan⁡195∘ = 2a, then the value of (a + a) is :
3
(1) 2 (2) 4 − 2√3 (3) 5 − 2 √3 (4) 4
Sol. 4
1 1
tan15º    tan195º  2a
tan 75º tan105º
1 1
 tan15º    tan15º  2a
cot15º cot15º
 tan15º  tan15º  tan15º  tan15º  2a
 2tan15º  2a
 a  tan15º
1 1
a  tan15º 
a tan15º
 tan15º  cot15º
2 3 2 3

1
a  4
a

   , is 1
, where 
71. If the solution of the equation log cos x cotx 4 logsin x tan x 1, x  0,  sin
 2 2

and  are integers, then  +  is equal to :

(1) 5 (2) 6 (3) 4 (4) 3
Sol. 3
 
log cos x cot x  4logsin x tan x  1, x   0, 
 2
cos x sin x
 log cos x  4logsin x 1
sin x cos x
1  logcos x sin x  4  4logsin x cos x  1
 4  logcos x sin x  4logsin x cos x
Let logcos x sin x  t
4
4  t 
t
 t 2  4t  4  0

 (t  2) 2  0
t  2
 logcos x sin x  2

 sin x = cos2 x
 sin x = 1 – sin2x
 sin2 x + sin x – 1 = 0
  sin x  1  1  4
2

 sin x  1  5 , 1  5  
x   0,  then
1  5
not possible
2 2  2 2

 1  5 
 x  sin 1  
 2 
  1,   5 then

   4

72. Let the system of linear equations

x + y + kz = 2
2x + 3y − z = 1
3x + 4y + 2z = k
have infinitely many solutions. Then the system
(k + 1)x + (2k − 1)y = 7
(2k + 1)x + (k + 5)y = 10
has:
(1) infinitely many solutions (2) unique solution satisfying x − y = 1
(3) unique solution satisfying x + y = 1 (4) no solution
Sol. 3
x + y + kz = 2
2x + 3y – z = 1
3x + 4y + 2z = k
Have Infinitely many solution then
1 1 k
2 3 1  0
3 4 2
1(10)  1(7)  k(8  9)  0
 10 – 7 – k = 0

 k 3
For k = 3
4x + 5y = 7
7x + 8y = 10
has unique solution and solution is (–2, 3).
Hence solution is unique and satisfying x + y = 1
73. The line l1 passes through the point (2,6,2) and is perpendicular to the plane 2x + y − 2z = 10.
x+1 y+4 z
Then the shortest distance between the line l1 and the line = = 2 is :
2 −3
13 19
(1) (2) (3) 7 (4) 9
3 3
Sol. 9
x 2 y6 z2
equation of l1 is  
2 1 2
x 1 y  4 z
Let l2 is  
2 3 2
Point on l1 is a = (2, 6, 2), direction p  2,1, 2 

Point on l2 is b = (–1, –4, 0) direction q  2, –3,2 

(a  b).(p  q)
Shortest distance between l1 and l2 =
| pq |

i j k
p  q  2 1 2  i(4)  j(8)  k(8)
2 3 2

3,10, 2 · 4, 8, 8


16  64  64
12  80  16

144
108
=
12
=9
Shortest distance between the lines is 9.

m n
74. Let A = ( p q) , d = |A| ≠ 0 and |A − d(AdjA)| = 0. Then
(1) 1 + d2 = m2 + q2 (2) 1 + d2 = (m + q)2
(3) (1 + d)2 = m2 + q2 (4) (1 + d)2 = (m + q)2
Sol. 4
m n 
A=   , d =|A| = mq – np
 p q
m n  q –n 
A – d(Adj. A) =   – d 
 p q  –p m 
 m – dq n + dn 
=  
 p + pd q – dm 
|A – d(Adj A)| = (m – dq)(q – dm) – (n + dn)(p + pd) = 0
 mq – m2d – dq2 + d2qm = np(1+d)2
 (mq – m2d – dq2 + d2qm) = (mq – d)(1 + d)2
  mq – m2d – dq2 + d2qm = mq + mqd2 + 2mqd – d (1 + d)2
 d(1+d) 2 = m2d + dq2 + 2mqd
(1+d) 2 = (m+q) 2

3(e−1) 2 3
75. If [t] denotes the greatest integer ≤ t, then the value of ∫1 x 2 e[x]+[x ] dx is :
e
(1) e8 − 1 (2) e7 − 1 (3) e8 − e (4) e9 − e
Sol. 3
2
3(e – 1) 2
x e
2 [x]+[x 3 ]
dx
e 1

x e
2 [x]+[x 3 ]
Let I = dx
1

x e
2 1+[x 3 ]
I= dx dx
1

I = e  x e dx
3
2 [x ]

1
3
Let x = t
3x2 dx = dt
8
e [t]
I =  e dt
31
e 
2 3 4 8
I =  e dt +  e 2
dt +  e 3
dt + ..... +  e 7dt 
3 1 2 3 7 
e
I = e + e2 + e3 + ..... + e7 
3
e  e(e7 – 1) 
 I=  
3 e –1 
2
3(e – 1) 3(e – 1) e 2 (e 7 – 1)

2 [x]+[x 3 ]
Therefore x e dx = 
e 1 e 3 e –1
2
3(e – 1) 2 [x]+[x3 ]

e 1 x e dx = e8 – e

76. ̂ make angles α, β, γ with the positive directions of the co-ordinate axes OX, OY, OZ
Let a unit vector OP
π
̂ is perpendicular to the plane through points (1,2,3), (2,3,4) and
respectively, where β ∈ (0, 2 ). If OP
(1,5,7), then which one of the following is true ?
π π π π
(1) α ∈ (0, 2 ) and γ ∈ (0, 2 ) (2) α ∈ (0, 2 ) and γ ∈ ( 2 , π)
π π π π
(3) α ∈ (2 , π) and γ ∈ (2 , π) (4) α ∈ ( 2 , π) and γ ∈ (0, 2 )
Sol. 3
 OP makes angle , ,  with positive directions of the co-ordinate axes then cos2  + cos2  + cos2
= 1.
Point on planes are a(1, 2, 3), b(2, 3, 4) and c(1, 5, 7).
 ab = <1, 1, 1>

ac = <0,3, 4>
ˆi ˆj ˆ
k
 
normal vector of plane = 1 1 1
0 3 4
 

= ˆi(1) – ˆj(4) + ˆ
k(3)
= <1, –4, 3>
1 4 3
direction cosine of normal is =  , ,
26 26 26
1 4 3
then direction cosine of op is  , ,
26 26 26
  π 
 β   0,  
  2 

π  π 
Hence    ,   and    ,  
2  2 

77. The coefficient of x 301 in (1 + x)500 + x(1 + x)499 + x 2 (1 + x)498 + ⋯ … . . x 500 is :

(1) ⁡500 C300 (2) ⁡501 C200 (3) ⁡501 C302 (4) ⁡500 C301
Sol. 2
x0(1+x)500 + x(1+ x)499 + x2(1+x)498 + … + x500
 x 501 
  – 1
 1 + x  
= (1+x)500
x
–1
1+ x
(1 + x)500 (x 501 – (1 + x)501 )
=
 –1 
(1 + x)501  
x + x
= (1+x)501 – x501
Coefficient of x301 in above expression is 501C301 or 501C200.
78. Let the solution curve y = y(x) of the differential equation
dy 3x5 tan−1 (x3 ) x3 −tan−1 ⁡x3
− 3 y = 2x⁡exp⁡{ } pass through the origin. Then y(1) is equal to :
dx (1+x6 )2 √(1+x6 )

4+π 1−π π−4 4−π

(1) exp⁡( 4√2 ) (2) exp⁡( 4√2 ) (3) exp⁡( 4√2 ) (4) exp⁡( 4√2 )
Sol. 4
 dy  3x 5 tan –1 (x 3 )  x 3 – tan –1x 3 
 dx  – y = 2x exp  
  (1 + x 6 )3/2  (1 + x 6 ) 
 
above equation is linear differential equation.
–3x5 tan –1 (x3 )
 (1+x 6 )3/ 2
dx
I.F. = e
3x 2 ·x 3 tan –1 (x3 )
–  (1+x 6 )3/ 2
dx
=e
Let tan–1(x3) = t then
3x 2 ·dx
= dt
1 + x6
t tan t
–  dt
=e 1+tan 2 t

t tan t
–  sec t
dt
=e

– t sin t dt
=e
= e– t cost sint
= et cost t–sint
tan –1 x 3 x3
–
I.F. = e 1+x
6
1+x 6

Solution is
 x 3 – tan –1 x 3  tan –1 x 3 – x 3
 tan x 
–1 3
x3  
–  
y  e 1+x  =  2x e
6
1+x 6  1+x 6  1+x 6
·e dx
 
 
 tan x – x 
–1 3 3

y  e 1+x =  2x dx = x2 + c
6

 
 
above eq. is passing through (0, 0) then c = 0
x 3 – tan –1 x 3

y = x2 e 1+x
6

Put x = 1 then
π
1–
4 4– π

y(1) = e 2
= e4 2

4 – π
y(1) = exp  
 4 2 
 1 15
79. If the coefficient of x15 in the expansion of (ax 3 + ) is equal to the coefficient of x −15 in the
bx1/3
1 15
expansion of (ax1/3 − bx3 ) , where a and b are positive real numbers, then for each such ordered pair
(a, b) :
(1) ab = 3 (2) ab = 1 (3) a = b (4) a = 3b
Sol. 2
15
 3 1 
 ax + 1/3 
 bx 
r
15 3 15–r  1 
general term is Tr+1 = Cr (ax )  1/3 
 bx 
15 – r
r
Tr+1 = 15Cr a r x45 – 3r –
b 3
r
For coefficient of x15 45 – 3r – = 15
3
10r
30 =
3
r=9
Coefficient of x is = C9 a b–9
15 15 6
… (1)
15
 1 
 general term of  ax1/3 –  is
 bx 3 
r
 –1 
Tr+1 = 15Cr (ax1/3)15–r  3 
 bx 
15 – r
For coefficient of x–15  – 3r = –15
3
 15 – r – 9r = – 45
60 = 10 r
r=6
Coefficient of x–15 is = 15C6 a9 b–b … (2)
 both coefficient are equal then
15
C9 a6 b–9 = 15C6 a9 b–6
a6 b‫–ؘ‬9 = a9 b–6
a3 b3 = 1
ab = 1

80. Suppose f: ℝ → (0, ∞) be a differentiable function such that 5f(x + y) = f(x) ⋅ f(y), ∀x, y ∈ ℝ. If
f(3) = 320, then ∑5n=0 f(n) is equal to :
(1) 6875 (2) 6525 (3) 6825 (4) 6575
Sol. 3
f : ℝ (0, )
5 f(x+y) = f(x) · f(y)
Put x = 3, y = 0 then
5 f(3) = f(3) f(0)
f(0) = 5
Put x = 1, y = 1 then 5 f(2) = f2(1)
Put x = 1, y = 2 then 5 f(3) = f (1) f(2)
3
5 × 320 = f (1) = f(1) = 20
5
f(2) = 80
Put x = 2, y = 2 then 5 f(4) = f(2) f(2)
80 × 80
f(4) = =1280
5
Put x = 2, y = 3 then 5 f(5) = f(2) · f(2)
80 × 320
f(5) = = 5120
5
5

 F(n) = f(0) + f(1) + …. + f(5)

n =0

= 5 + 20 + 80 + 320 + 1280 + 5120

= 5(1 + 22 + 24 + 26 + 28 + 210)
= 6825

Section B
1+iz‾ 12
81. Let z = 1 + i and z1 = 1 . Then arg⁡(z1 ) is equal to
z‾(1−z)+ π
z

Sol. 9
z = 1 + i, z = 1 – i, i z =1+i

z1 = 1 + iz
1
z(1 – z) +
z
i+z
z1 =
1
z – zz) +
z
i+2
z1 =
1–i
1–i– 2+
2
i+2
z1 =
1 3i
– –
2 2
 z1 = –2(i + 2) × (1 – 3i)
(1 + 3i) (1 – 3i)
–2(5 – 5i)
z1 =
10
z1 = –1+i
1  3
arg. (z1) =  – tan–1   =   =
1 4 4
12 3
 arg (z1) = × =9
 4

82. If λ1 < λ2 are two values of λ such that the angle between the planes P1 : r ⋅ (3î − 5ĵ + k̂) = 7 and
2√6
P2 : r ⋅ (λî + ĵ − 3k̂) = 9 is sin−1 ⁡( ), then the square of the length of perpendicular from the point
5
(38λ1 , 10λ2 , 2) to the plane P1 is
Sol. 315s
Plane P1 : r .(3 î – 5ˆj + k̂ ) = 7
P2 : r. i  j  3k   9
angle between plane is same as angle between their normal.
angle between normal  then
3, –5,1 · λ,1, –3
Cos  =
9 + 25 + 1  2 + 1 + 9
3 – 5 – 3
Cos  = … (1)
35  2 + 10
 
  = sin–1  2 6  then
 
 5 

sin  = 2 6
5
1
cos  =
5
from equation (1)
3 – 8 1
=
35  + 10 5
2

(3 – 8) 2 1
 =
35( 2 + 10) 25
5 (3 – 8)2 = 7 ( + 10)
5 (9 – 48+64) = 7 + 70
38 – 240 + 250 = 0
19 – 120 + 125 = 0
25
  = 5,
19
25
= , 2 = 5
19
 Point (38, 10 , 2)  (50, 50, 2)
distance of (50, 50, 2) from plane P1 is
3 × 50 – 5 × 50 + 2 – 7
d=
9 + 25 + 1
150 – 250 + 2 – 7
d=
35
105
d=
35

d = 3 35
d2 = 315

83. Let α be the area of the larger region bounded by the curve y 2 = 8x and the lines y = x and x = 2,
which lies in the first quadrant. Then the value of 3α is equal to
Sol. 22
8

area () =  (2 2 x – x)dx

2

8
 2

=  2 2·2 x 3/ 2 – x 
 3 2 

= 4 2 [8 × 2 2 –2 2 ] – 30
3
28 × 4
= – 30
3
112
= – 30
3
22
=
3
3 = 22
 n3 ((2n)!)+(2n−1)(n!) b 1
84. Let ∑∞
n=0 = ae + e + c , where a, b, c ∈ ℤ and e = ∑∞ 2
n=0 n! Then a − b + c is
(n!)((2n)!)
equal to
Sol. 26

n 3 ((2n)!) + (2n – 1)n!
Let 
n 0 (n!)((2n)!)

n 3 (2n)! (2n – 1)n!
= 
n0 n!(2n)!
+
n!(2n)!
= S1 + S2
  
n 3 (2n)! n3 n2
Let S1 = 
n0 n!(2n)!
=  =
n0 n!
 (n – 1)!
n1


n2 – 1 + 1
= 
n1 (n – 1)!

 
(n + 1) 1
= +  (n – 1)!
n2 (n – 2)! n1


(n  2)  3  1
 
n 2 (n  2)! n 1 (n  1)!
  
1 1 1
  3 
n 3 (n  3)! n  2 (n  2)! n 1 (n  1)!

S1  e  3e  e  5e

(2n  1)n!
S2  
n  0 n!(2n)!


2n  1

n  0 (2n)!

 
1 1
 
n 1 (2n  1)! n  0 (2n)!

=  1  1  1  ....   1  1  1  .....

 1! 3! 5!   2! 4! 
1 1 1 1 1
= 1       ....
1! 2! 3! 4! 5!

=  1  1  1  1  1  .... 
 1! 2! 3! 4! 
= – e–1
1 b
S1  S2  5e   ae   c
e e
Compare both side
a = 5, b = –1, c = 0
a2 – b + c = 25 + 1 + 0 = 26
85. If the equation of the plane passing through the point (1,1,2) and perpendicular to the line x − 3y +
2z − 1 = 0 = 4x − y + z is Ax + By + Cz = 1, then 140(C − B + A) is equal to
Sol. 15
give line is x – 3y + 2z – 1 = 0 = 4x – y + z
i j k
Direction of line a  1 3 2 = i(1)  j(7)  k(11)
4 1 1

 a   1, 7, 11 
Line is  r to the plane then direction of line is parallel to normal of plane.
n   1, 7, 11 
Equation of plane is
1(x  1)  7(y  1)  11(z  2)  0
x  7y  11z  1  7  22  0
 x  7y  11z  28
1 7 11
  x y  z 1
28 28 28
1 7 11
A   , B  ,C 
28 28 28
 11 7 1 
140(C  B  A)  140    
 28 28 28 
140  3
 = 15
28

86. Number of 4-digit numbers (the repeation of digits is allowed) which are made using the digits 1, 2, 3
and 5 , and are divisible by 15 , is equal to
Sol. 21
5
Last digit must be 5 and sum of digits is divisible by 3 for divisible by 15
Remaining 3 digits Arrange
(1, 1, 2) 3!
3
2
(1, 3, 3) 3!
3
2
(1, 5, 1) 3!
3
2
(2, 2, 3) 3!
3
2
(2, 3, 5) 3! = 6
(3, 5, 5) 3!
3
2

Total numbers = 21
 3x+2 −3
87. Let f 1 (x) = 2x+3 , x ∈ 𝐑 − { 2 }
For n ≥ 2, define f n (x) = f 1 of f n−1 (x)
ax+b
If f 5 (x) = bx+a , gcd⁡(a, b) = 1, then a + b is equal to
Sol. 3125
f1(x) = 3x  2 , x  R   3 
2x  3  2
 3x  2 
f 2 (x)  f 1 0 f 1 (x)  f 1  
 2x  3 
 3x  2) 
3 2
 2x  3) 

 3x  2 
2 3
 2x  3 
9x  6  4x  6

6x  4  6x  9
13x  12
f 2 (x) 
12x  13
f 3 (x)  f 1 o f 2 (x)

 13x  12 
 f1  
 12x  13 

 13x  12 
3 2
 12x  13 

 13x  12 
2 3
 12x  13 
39x  36  24x  26

26x  24  36x  39
63x  62
f 3 (x) 
62x  63
 63x  62 
f 4 (x)  f 1  
 62x  63 

 63x  62 
3 2
 62x  63 

 63x 62 
2  2
 62x 63 
313x  312
f 4 (x) 
312x  313
 313x  312 
f 5 (x)  f 1  
 312x  313 
  313x  312 
3 2
 312x  313 

 313x  312 
2 3
 312x  313 
1563x  1562
f 5 (x) 
1562x  1563
a  1563,b  1562

a  b  3125

88. The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted
and a and b are respectively mean and variance of remaining 6 observation, then a + 3b − 5 is equal
to
Sol. 37
7

x
i 1
i
Mean of 7 observations =
7
7
 x
i 1
i  7  8  56

x i2
Variance =  (x) 2
n
x i2  7(16  64)  560

If 14 is removed then
7

x i  14
Mean = a  i 1
 6a  56  14
6
a 7
7

x 2
i  (14)2
Variance = b  i 1
 49
6
 6b = 560 – 196 – 294
 6b = 70
 3b = 35
 a + 3b – 5 = 7 + 35 – 5 = 37

89. Let S = {1,2,3,4,5,6}. Then the number of one-one functions f: S → P(S), where P(S) denote the
power set of S, such that f(n) ⊂ f(m) where n < m is
Sol. 3240
Case – I
f(6) = S i.e. 1 option
f(5) = any 5 element subset A of S i.e. 6 C5  6 options
f(4) = any 4 element subset B of A i.e. 5 C 4  5 options
f(3) = any 3 element subset C of B i.e. 4 C3  4 options
f(2) = any 2 element subset D of C i.e. 3 C 2  3 options
f(1) = any 1 element subset E of D or empty subset i.e. 3 options
Total function = 6×5×4×3×2×3=1080
Case – II
f(6) = S
f(5) = any 4 element subset A of S i.e. 6 C 4  15 options
f(4) = any 3 element subset B of A i.e. 4 C3  4 options
f(3) = any 2 element subset C of B i.e. 3 C 2  2 options
f(2) = any 1 element subset D of C i.e. 2 C1  2 options
f(1) = empty subset i.e. 1 options
Total function = 15×4×3×2×1=360
Case – III
f(6) = S
f(5) = any 5 element subset A of S i.e. 6 C5  6 options
f(4) = any 3 element subset B of A i.e. 5 C3  10 options
f(3) = any 2 element subset C of B i.e. 3 C 2  3 options
f(2) = any 1 element subset D of C i.e. 2 C1  2 options
f(1) = empty subset i.e. 1 options
Total function = 6×10×3×2×1=360
Case – IV
f(6) = S
f(5) = any 5 element subset A of S i.e. 6 C5  6 options
f(4) = any 4 element subset B of A i.e. 5 C 4  5 options
f(3) = any 2 element subset C of B i.e. 4 C 2  6 options
f(2) = any 1 element subset D of C i.e. 2 C1  2 options
f(1) = empty subset i.e. 1 options
Total function = 6×5×6×2×1=360
 Case – V
f(6) = S
f(5) = any 5 element subset A of S i.e. 6 C5  6 options
f(4) = any 4 element subset B of A i.e. 5 C 4  5 options
f(3) = any 3 element subset C of B i.e. 4 C3  4 options
f(2) = any 1 element subset D of C i.e. 3 C1  3 options
f(1) = empty subset i.e. 1 options
Total function = 6×5×4×3×1=360
Case – VI
f(6) = any 5 element subset A of S i.e. 6 C5  6 options
f(5) = any 4 element subset B of A i.e. 5 C 4  5 options
f(4) = any 3 element subset C of B i.e. 4 C3  4 options
f(3) = any 2 element subset D of C i.e. 3 C 2  3 options
f(2) = any 1 element subset E of D i.e. 2 C1  2 options
f(1) = empty subset i.e. 1 options
Total function = 6×5×4×3×2×1=720

Total number of such functions = 1080 + (4× 360) +720 = 3240

48 x t3
90. lim ∫0 dt is equal to
x 0 x4 t6 +1

Sol. 12
0
x
48 t 3
lim
x 0 x 4 
0
6
   form
t 1  0 
Using L Hopital Rule
x3
48 
 lim x6  1
x 0 4x 3
48
 lim
x 0 4(x 6  1)
48

4
= 12