MATHS | 31st Jan 2023 _ Shift-1

SECTION - A
𝑥2 𝑦2
61. If the maximum distance of normal to the ellipse + 𝑏2 = 1, 𝑏 < 2, from the origin is 1 , then the
4
eccentricity of the ellipse is :
1 √3 √3 1
(1) 2 (2) (3) (4)
4 2 √2
Sol.
x 2 y2 ax by
Normal to the ellipse  2  1 at point (a cos, b sin) is –  a 2 – b2
a 2
b cos  sin 
Its distance from origin is
| a 2 – b2 |
d
a 2 sec2   b2 cosec2 
| a 2 – b2 |
d
a 2  b 2  2ab  (a tan  – b cot) 2
| (a– b)(a  b) |
d
a 2  b2  2ab  (a tan  – b tan ) 2
| (a– b)(a  b) |
d max  | a– b |
ab
 dmax = 1

|2 – b| = 1
2 – b = 1 [ b < 2]

b 1

b2 1 3
Eccentricity = 1– 2
 1– 
a 4 2

3
 e
2
x
f t
62. Let a differentiable function 𝑓 satisfy f  x    dt  x  1, x  3 . Then 12𝑓(8) is equal to :
3
t
(1) 34 (2) 1 (3) 17 (4) 19
Sol.
x
f(t)
f(x)   dt  x  1, x  3
3
t
Differentiate both side w.r.t. x
f (x) 1
f 1 (x)  
x 2 x 1
 Above eqn. is linear differential equation
1
 x dx
I.f. = e  eln x  x
Solution is
x
f (x)·x   dx  C
2 x 1
1  x 1 1 
f (x)·x   
2  x 1
  dx  C
x 1 
1  1 
f (x)·x  
2  x 1 –  dx  C
x 1 
1 2 3 
f (x)·x   (x  1) 2 – 2 x  1   C
2 3 
 f(3) = 2
than
1 2 
2.3    8 – 2  2  C
2 3 
1 16 
6  – 4  C
2 3 
2
6 C
3
16
C
3
1 2 3  16
f(x). x   (x  1) 2
– 2 x  1 
2 3  3
Put x = 8
1 2  16
f(8)·8 =   27 – 2  3 
2 3  3
1 16
f(8)·8 = 12 
2 3
16 34
f(8)·8 = 6  
3 3
12f (8)  17

1
63. For all 𝑧 ∈ 𝐶 on the curve 𝐶1 : |𝑧| = 4, let the locus of the point 𝑧 + 𝑧 be the curve C2 . Then :
(1) the curve 𝐶1 lies inside 𝐶2 (2) the curve 𝐶2 lies inside 𝐶1
(3) the curves 𝐶1 and 𝐶2 intersect at 4 points (4) the curves 𝐶1 and 𝐶2 intersect at 2 points
Sol.
C1 : |z| = 4 then zz  16
1 z
z  z
z 16
x – iy
 x  iy 
16
1 17x 15y
z  i
z 16 16
17x 15
Let X  , Y y
16 16
X Y
 x, y
 17   15 
   
 16   16 
 x2 + y2 = 16

X2 Y2
2
 2
 16
 17   15 
   
 16   16 
x2 y2
 C2 : 2
 2
1 (Ellipse)
 17   15 
   
 4 4

x 2  y 2  16
x2 y2
2
 2
1
 17   15 
   
 4  4
O

Curve C1 and C2 intersect at 4 point.

3
𝜋 𝜋
64. 𝑦 = 𝑓(𝑥) = sin3 ⁡( 3 (cos⁡(3√2 (−4𝑥 3 + 5𝑥 2 + 1)2 ))). Then, at 𝑥 = 1,

(1) √2𝑦 ′ − 3𝜋 2 𝑦 = 0 (2) 𝑦 ′ + 3𝜋 2 𝑦 = 0 (3) 2𝑦 ′ + 3𝜋 2 𝑦 = 0 (4) 2𝑦 ′ + √3𝜋 2 𝑦 = 0

Sol.
    3 
y  f (x)  sin 3   cos   –4x 3  5x 2  1 2   
3 3 2 

 –4x 3  5x 2  1 2
3
Let g(x) 
3 2
2
g(1) =
3
 
y  sin 3  cos(g(x)) 
3 
Differentiate w.r.t. x
    
y '  3sin 2  cos(g(x))   cos  cos(g(x))     sin g  x   g '  x 
3  3  3

 –4x 3  5x 2  1 2  –12x 2  10x 
1
 g1 (x) 
3 2

g1 (1) 
2 2
 2   –2  –
3 3  – 3  32
y (1)       – 
1

4 2 3  2  16

32
y1 (1) 
16

 2  –1
y(1)  sin 3.  cos  
3 3  8

2y1 (1)  32 y(1)  0 

65. A wire of length 20 m is to be cut into two pieces. A piece of length 𝑙1 is bent to make a square of area
𝐴1 and the other piece of length 𝑙2 is made into a circle of area 𝐴2 . If 2𝐴1 + 3𝐴2 is minimum then
(𝜋𝑙1 ): 𝑙2 is equal to :
(1) 1:6 (2) 6: 1 (3) 3: 1 (4) 4: 1
Sol.
Total length of wire = 20 m
2
 
area of square (A1) =  1 
4
2
 
area of circle (A2) =   2 
 2 
Let S = 2A1 + 3A2
 2
3 22
S= 1

8 4
1  2  20 then
d
1 2
=0
d 1

d 2
= –1
d 1

ds 6 d
= 1  2. 2
=0
d 1 4 4 d 1

6 2
= 1

4 4
 6
= 1

2 1

66. Let a circle 𝐶1 be obtained on rolling the circle 𝑥 2 + 𝑦 2 − 4𝑥 − 6𝑦 + 11 = 0 upwards 4 units on the
tangent 𝑇 to it at the point (3,2). Let 𝐶2 be the image of 𝐶1 in 𝑇. Let 𝐴 and 𝐵 be the centers of circles
𝐶1 and 𝐶2 respectively, and 𝑀 and 𝑁 be respectively the feet of perpendiculars drawn from 𝐴 and 𝐵
on the 𝑥-axis. Then the area of the trapezium AMNB is :
(1) 4(1 + √2) (2) 3 + 2√2 (3) 2(1 + √2) (4) 2(2 + √2)
Sol.

(x2,y2)
 C1
A 2 (x1,y1)
(2,3)

2
2
B 
4
(3,2) C2
y = x–1 (x1,y1)

M N

(x', y') point lies on line y = x – 1 have distance 4 unit from (3, 2).
4
x' = 3 = 2 2 3
2
4
y' = 2 = 2 22
2
1 1
Slope of line AB is –1. i.e. = tan = –1 then sin = , cos = 
2 2
for point A and B
  1 
x =  2
 2

 2 2 3 
y =  2
 1 
 2

 2 2 2 
for point A we take +ve sign


(x2, y2) = 2 2  2, 2 2  3 
for point B we take –ve sign


(x1, y1) = 2 2  4, 2 2  1 
MN = x 2  x1 = 2
AM + BN = 2 2  3  2 2  1  4  4 2

area of trapezium =
1
2

 2 4  4 2 

= 4 1 2 

67. A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The
probability that the bag contains at least 5 black balls is
3 5 5 2
(1) 7 (2) 7 (3) 6 (4) 7
C2  6 C2
3
Sol. Pr obability  2
C2  C2  4 C2  5 C2  6 C2
3

10  15
=
1  3  6  10  15
5
=
7

1 1
68. Let 𝑦 = 𝑓(𝑥) represent a parabola with focus (− 2 , 0) and directrix 𝑦 = − 2.
𝜋
Then 𝑆 = {𝑥 ∈ ℝ: tan−1 ⁡(√𝑓(𝑥)) + sin−1 ⁡(√𝑓(𝑥) + 1) = 2 }:
(1) contains exactly two elements (2) contains exactly one element
(3) is an empty set (4) is an infinite set

equation of parabola which have focus   , 0  and directrix y =  is

1 1
Sol.
 2  2
2
 1  1
x   y 
 2  4
y = f(x) = (x2 + x)
 
S =  x  R : tan 1 



 
f  x   sin 1  
f x  1 
 

2 

tan 1  
f  x   sin 1  f x 1  
2
f  x   0 & f  x   1 can not greater then 1, so f(x) must be 0

i.e. f(x) = 0
 x 2  x  0
x  x  1  0
x  0, x= –1 
S contain 2 element.

69. Let 𝑎⃗ = 2𝑖ˆ + 𝑗ˆ + 𝑘ˆ, and 𝑏⃗⃗ and 𝑐⃗ be two nonzero vectors such that |𝑎⃗ + 𝑏⃗⃗ + 𝑐⃗| = |𝑎⃗ + 𝑏⃗⃗ − 𝑐⃗| and
𝑏⃗⃗ ⋅ 𝑐⃗ = 0. Consider the following two statements:
(A) |𝑎⃗ + 𝜆𝑐⃗| ≥ |𝑎⃗| for all 𝜆 ∈ ℝ.
(B) 𝑎⃗ and 𝑐⃗ are always parallel.
Then.
(1) both (A) and (B) are correct (2) only (A) is correct
(3) neither (A) nor (B) is correct (4) only (B) is correct
Sol.
a  b  c  a  b  c , b.c  0
2 2
abc  abc
2
a  b  c  2a.b  2b.c  2a.c
2 2

2
 a  b  c  2a.b  2b.c  2a.c
2 2

2a.b  2b.c  2a.c  2a.b  2b.c  2a.c

a.b  a.c  a.b  a.c
a.c  0 (B is incorrect)

a  c  a
2 2

2
a  2 c  2a  c  a
2 2

= 2c2  0
True   R (A is correct)
 
2
 2  3sin x 
70. The value of  sin x 1  cos x  dx is equal to

3
10 7
(1) − √3 − log 𝑒 ⁡√3 (2) 2 − √3 − log 𝑒 ⁡√3
3
10
(3) −2 + 3√3 + log 𝑒 ⁡√3 (4) − √3 + log 𝑒 ⁡√3
3
Sol. (4)

2
 2  3sin x 
 sin x 1  cos x dx

3

 
2 2
2 3
=  dx +  dx
 sin x  sin x cos x  1  cos x
3 3

= I1 + I2
   2 x
1  tan  dx
 2
2 2
2dx
I1 =  sin x 1  cos x  = 2
 x
 
1  tan 2
3 3 x  2
2 tan  1 
2  1  tan 2 x

 2
 2 x  2 x
1  tan 1  tan  dx
= 2
2
2  2
x
 2 tan  2
3 2
 2 x 2 x
2 sec 1  tan 
2 2
= 2 dx

x
4 tan
3 2
x x 1
Let, tan = t then sec2 × dx = dt
2 2 2
1 t2
1
= 2 1 2t dt
3
1
 t2 
=  nt  
 2 1
3

1 1 1
=   n  
2 3 6

I1 =  n 3  
1
 3
 

dx 2 2
1  cos x
I 2  3 = 3 dx
 1  cos x
2
 sin x
3 3
 
I2  3cos ecx  cot x  2  3  3
3

1
I1  I2  ln 3   3  3
3
10
=  ln 3  3
3

𝑥−5 𝑦−𝜆 𝑧+𝜆

71. Let the shortest distance between the lines 𝐿: = = , 𝜆 ≥ 0 and
−2 0 1
𝐿1 : 𝑥 + 1 = 𝑦 − 1 = 4 − 𝑧 be 2√6. If (𝛼, 𝛽, 𝛾) lies on L, then which of the following is
NOT possible ?
(1) 𝛼 − 2𝛾 = 19 (2) 2𝛼 + 𝛾 = 7 (3) 2𝛼 − 𝛾 = 9 (4) 𝛼 + 2𝛾 = 24
Sol. (4)
Let b1  –2,0,1  a1  (5, , – )
b 2  1,1, –1  a 2  (–1,1, 4)
ˆi ˆj kˆ
Normal vector of both line is b1  b 2  –2 0 1
1 1 –1

ˆi(–1) – ˆj(1)  k(–2)

ˆ

b1  b 2  –1, –1, –2 
a1 – a 2  6,  –1, – – 4 
 a 2 – a1    b1  b2 
Shortest distance d =
b1  b2

 6,  – 1, –  – 4    –1, –1, –2 
2 6
(1)2  (1)2  (2)2
12  –6 –   1  2  8
  3  12
  9, -15
  9    0
 (, , ) lies on line L then
 –5 –9 9
  K
–2 0 1
= 5 –2K,  = 9K,  = –9 + K
+ 2, = 5 – 2K – 18 + 2K = –13  24
Therefore  + 24 is not possible.
72. For the system of linear equations
𝑥+𝑦+𝑧 =6
𝛼𝑥 + 𝛽𝑦 + 7𝑧 = 3
𝑥 + 2𝑦 + 3𝑧 = 14
which of the following is NOT true ?
(1) If 𝛼 = 𝛽 and 𝛼 ≠ 7, then the system has a unique solution
(2) If 𝛼 = 𝛽 = 7, then the system has no solution
(3) For every point (𝛼, 𝛽) ≠ (7,7) on the line 𝑥 − 2𝑦 + 7 = 0 , the system has infinitely many
solutions
(4) There is a unique point (𝛼, 𝛽) on the line 𝑥 + 2𝑦 + 18 = 0 for which the system has infinitely
many solutions
Sol. (3)
x+y+z=6 … (1)
x + y + 7z = 3 … (2)
x + 2y + 3z = 14 … (3)
equation (3) – equation (1)
y + 2z = 8
y = 8 – 2z
From (1) x = –2 + z
Value of x and y put in equation (2)
(–2 + z) + (8 – 2z) + 7z = 3
–2+ z + 8 – 2z + 7z = 3
(– 2 + 7) z = 2 – 8 + 3
if  – 2 + 7 0 then system has unique solution
if ( – 2 + 7 = 0) and 2 – 8 + 3 0 then system has no solution
if ( – 2 + 7 = 0) and 2 – 8 + 3 = 0 then system has infinite solution

[𝑥]
73. If the domain of the function 𝑓(𝑥) = 1+𝑥 2, where [𝑥] is greatest integer ≤ 𝑥, is [2,6), then its range is
5 2 5 2 9 27 18 9
(1) (26 , 5] (2) (37 , 5] − {29 , 109 , 89 , 53}
5 2 5 2 9 27 18 9
(3) (37 , 5] (4) (26 , 5] − {29 , 109 , 89 , 53}

Sol. (3)
[x]
f (x)  , x 
1  x2
  2
1  x 2 x  [2,3)

 3 x  [3, 4)

f (x)  1  x
2

 4 x  [4,5)
1  x 2
 5
 x  [5,6)
1  x 2
 ƒ(x) is  in x[2, 6)
2
5

5
37
0 2 3 4 5 6

 5 2
range is  , 
 37 5 

74. Let R be a relation on N×N defined by (𝑎, 𝑏)R(𝑐, 𝑑) if and only if 𝑎𝑑(𝑏 − 𝑐) = 𝑏𝑐(𝑎 − 𝑑). Then R is
(1) transitive but neither reflexive nor symmetric
(2) symmetric but neither reflexive nor transitive
(3) symmetric and transitive but not reflexive
(4) reflexive and symmetric but not transitive
Sol. (2)
(a, b) R (c, d) ad(b – c) = bc(a – d)
For reflexive
(a, b) R (a, b)
 ab (b – a)  ba(a – b)
R is not reflexive
For symmetric:
(a,b) R(c,d) ad(b –c) = bc (a –d)
then we check
(c, d) R (a, b)  cb(d – a) = ad(c – b)
cb(a – d) = ad(b – c)
R is symmetric :
For transitive:
 (2,3) R (3,2) and (3,2) R (5,30)
But (2,3) is not related to (5,30)
R is not transitive.
75. (S1)(𝑝 ⇒ 𝑞) ∨ (𝑝 ∧ (∼ 𝑞)) is a tautology
(S2)((∼ 𝑝) ⇒ (∼ 𝑞)) ∧ ((∼ 𝑝) ∨ 𝑞) is a contradiction.
Then
(1) both (S1) and (S2) are correct (2) only ( S1) is correct
(3) only (S2) is correct (4) both (S1) and (S2) are wrong
Sol. (2)
S1 : (P  q) V (P  (~q))
P q Pq ~q P~q (P  q) V (P  ~q)
T T T F F T
T F F T T T
F T T F F T
F F T T F T
S1 is a tautology

S2 : ((~P)  (~q))  ((~P) Vq)

~P ~q ~Pq ~P v q ((~P)  (~q))  (~P) vq)
F F T T T
F T T F F
T F F T F
T T T T T
S2 is not a contradiction

76. If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296 , respectively,
then the sum of common ratios of all such GPs is
9
(1) 7 (2) 3 (3) 2 (4) 14
Sol. (1)
a a
Four term of G.P. , ,ar,ar 3
r3 r
a a
3
  ar  ar 3  126
r r
a a
  ar  ar 3  1296
r3 r
a4 = 1296
a=6
6 6
 + 6r + 6r3 = 126
r3 r
  1 3 1
 r    r  3  21
 r r
3
 1  1  1
 r     r    3  r    21
 r  r  r
1
Let r + =t
r
t3 – 2t = 21
t=3
1
r+ =3
r
r2 – 3r + 1= 0
3 9 4
r=
2
3 5
r
2
 5 3 5 9 5 3 5
Sum of common ratio =     
4 4 2 4 4 2
=7
1 0 0 
77. Let A   0 4 1  . Then the sum of the diagonal elements of the matrix (A + I)11 is equal to
 0 12 3 
 
(1) 6144 (2) 2050 (3) 4097 (4) 4094
Sol. (3)
1 0 0 
A =  0 4 1
 0 12 3

1 0 0   1 0 0  1 0 0 
2  
A  0 4 1  0 4 1 =  0 4 1
 
0 12 3  0 12 3  0 12 3

A3 = A4 = A5 … = A
(A + I)11 = 11 C0 A11  11 C1A10  11 C2 A9  11 C11I
=   11 C0  11 C1  11 C2  11 C10  A  I

= (211 – 1)A + I
= 2047 A + I
Sum of diagonal element = 2047(1 + 4– 3) +3
= 4097
78. The number of real roots of the equation √𝑥 2 − 4𝑥 + 3 + √𝑥 2 − 9 = √4𝑥 2 − 14𝑥 + 6, is :
(1) 3 (2) 1 (3) 2 (4) 0
Sol. (2)

x2  4x  3  x2  9  4x2  14x  6

 x  3 x  1   x  3 x  3  4x 2  12x  2x  6

 x 3  
x  1  x  3  4x  x  3  2  x  3

 x 3  x 1  x  3    4x  2 x  3

 x 3  
x  1  x  3  2  2x  1  0

x  3  0 or x  1  x  3  2  2x  1  0

x = 3 or x  1  x  3  2  2x  1

x – 1 + x + 3 + 2  x  1 x  3 = 4 x – 2

2  x  1 x  3  2x  4
  x  1 x  3   x  2 
2

 x2  2x  3  x 2  4  4x
 6x  7

7
x (not possible)
6
Number of real root = 1

𝛼 4 77
79. If sin−1 ⁡17 + cos−1 ⁡5 − tan−1 ⁡36 = 0,0 < 𝛼 < 13, then sin−1 ⁡(sin⁡𝛼) + cos −1 ⁡(cos⁡𝛼) is equal to
(1) 16 (2) 0 (3) 𝜋 (4) 16 − 5𝜋
Sol. (3)
 4 77
sin–1  cos1  tan 1  0 , 0 <  < 13
17 5 36
 7 
sin–1  tan–1 – tan–1
17 36 4
 77 3 
 36  4 
–1
= tan  
 1  77  3 
 36 4 
  8 8
sin–1 = tan–1   = sin
–1
 
17  
15  17 
 8

17 17
=8
sin –1 (sin 8) + cos–1 (cos 8)
= 3 – 8 + 8 – 2
=

𝑥2 𝑥3 𝑥𝑛
80. Let 𝛼 ∈ (0,1) and 𝛽 = log 𝑒 ⁡(1 − 𝛼). Let 𝑃𝑛 (𝑥) = 𝑥 + + +⋯+ , 𝑥 ∈ (0,1).
2 3 𝑛
𝛼 𝑡 50
Then the integral ∫0 𝑑𝑡 is equal to
1−𝑡
(1) 𝛽 + 𝑃50 (𝛼) (2) 𝑃50 (𝛼) − 𝛽 (3) 𝛽 − 𝑃50 (𝛼) (4)−(𝛽 + 𝑃50 (𝛼))
Sol. 4
   0,1 ,   log e 1   

x 2 x3 xn
Pn  x   x    …… + , x  (0, 1)
2 3 n
 t 50  1  1
 0 1 t
dt

 1  t 50  1
 dt   dt
0 1 t 0 1 t

 1  t  t 2 t 49 dt  ln 1  t  0

 


 t 2 t3 t 50 
  t      ln 1   
 2 3 50  0

  2 3 50 
        ln 1   
 2 3 50 

– P50 () – ln (1 – )
–  + P50())

Section : Mathematics Section B

2 30
2
81. Let 𝛼 > 0, be the smallest number such that the expansion of (𝑥 3 + 𝑥 3 ) has a term

𝛽𝑥 −𝛼 , 𝛽 ∈ ℕ. Then 𝛼 is equal to
Sol. 2
30  r
 2
4
 2 
Tr 1  30
Cr  x 3   3
  x 
60 11r
30
= Cr 2r x 3

60  11r
0
3
11 r > 60
60
r>
11
r=6
T7 = 30 C6 26 x 2 then
  30 C6  26  N
2

82. Let for 𝑥 ∈ ℝ,

𝑥 + |𝑥| 𝑥, 𝑥 < 0
and 𝑔(𝑥) = { 2
𝑓(𝑥) = .
2 𝑥 , 𝑥≥0
Then area bounded by the curve 𝑦 = (𝑓 ∘ 𝑔)(𝑥) and the lines 𝑦 = 0,2𝑦 − 𝑥 = 15 is equal to
Sol. 72
x x x x  0
f x  = 
2 0 x  0
x 2 x0
g(x) = 
x x0
g(x) g  x   0

Fog(x) = f{g(x)} = 

 0 gx  0
x 2 x0
fog(x) = 
0 x0
given lines are 2y – x = 15 and y = 0

 15 
 0, 
 2 
  x  15 
3
1 15
Area =    x 2 dx +   15
0 
2 2 2
3
x 2 15x x 3  225
=    
4 2 3 0 4
9 45 225
=  9
4 2 4
Area = 72

83. Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11 , is
equal to
Sol. 710
4 digit number which are less then 2800 are 1000 – 2799
Number which are divisible by 3
2799 = 1002 + (n – 1) 3
n = 600
Number which are divisible by 11 in 1000 – 2799
= (Number which are divisible by 11 in 1 – 2799)
– (Number which are divisible by 11 in 1 – 999)

= 
2799   999 
 
 11   11 
= 254 – 90
= 164
Number which are divisible by 33 in 1000 – 2799
= (Number which are divisible by 33 in 1 – 2799) – (Number which are divisible by 33 in 1 – 999)

= 
2799   999 
 
 33   33 
= 84 – 30 = 54
total number = n(3) + n(11) – n(33)
= 600 + 164 – 54 = 710

84. If the variance of the frequency distribution

𝑥𝑖 2 3 4 5 6 7 8

Frequency 𝑓𝑖 3 6 16 𝛼 9 5 6

is 3 , then 𝛼 is equal to
Sol. 5
xi fi di = xi – 5 (fi di)2 fidi
2 3 –3 27 –9
3 6 –2 24 –12
4 16 –1 16 –16
5  0 0 0
6 9 1 9 9
7 5 2 20 10
8 6 3 54 18
f d   fi di
2
2

 2
 i i
 
f i
 f
 i 
150
= 0 = 3
45  
 150 = 135 + 3
 3 = 15
 =5

85. Let 𝜃 be the angle between the planes 𝑃1 : 𝑟⃗ ⋅ (𝑖ˆ + 𝑗ˆ + 2𝑘ˆ) = 9 and 𝑃2 : 𝑟⃗ ⋅ (2𝑖ˆ − 𝑗ˆ + 𝑘ˆ) = 15. Let L be
the line that meets 𝑃2 at the point (4, −2,5) and makes an angle 𝜃 with the normal of 𝑃2 . If 𝛼 is the
angle between L and 𝑃2 , then (tan2 ⁡𝜃)(cot 2 ⁡𝛼) is equal to
Sol. 9

   90 – 
3

1, 1, 2 . 2,  1, 1
cos  
6
2 1 2 1
= =
6 2


3
 
then   
2 3


6
 tan  cot   =  3    3 
2 2
2 2
=9
86. Let 5 digit numbers be constructed using the digits 0,2,3,4,7,9 with repetition allowed, and are
arranged in ascending order with serial numbers. Then the serial number of the number 42923 is
Sol. 2997
2 = 1296
6 6 6 6

3 = 1296
6 6 6 6

4 0 =216
6 6 6

4 2 0 = 36
6 6

4 2 2 = 36
6 6

4 2 3 = 36
6 6

4 2 4 = 36
6 6

4 2 7 = 36
6 6

4 2 9 0 =6
6
4 2 9 2 0 =1
4 2 9 2 2=1

4 2 9 2 3=
2997

87. Let 𝑎⃗ and 𝑏⃗⃗ be two vectors such that |𝑎⃗| = √14, |𝑏⃗⃗| = √6 and |𝑎⃗ × 𝑏⃗⃗| = √48.
Then (𝑎⃗ ⋅ 𝑏⃗⃗)2 is equal to
Sol. 36

 
2 2 2
ab  a b  ab
2

 
2
48 = 14 × 6 - a  b

a  b
2
 84  48

a  b
2
 36

𝑥−1 𝑦+1 𝑧−3

88. Let the line 𝐿: = = intersect the plane 2𝑥 + 𝑦 + 3𝑧 = 16 at the point
2 −1 1
𝑃. Let the point 𝑄 be the foot of perpendicular from the point 𝑅(1, −1, −3) on the line 𝐿. If 𝛼 is the
area of triangle 𝑃𝑄𝑅, then 𝛼 2 is equal to
Sol. 180
Point on line L is (2 + 1, – – 1,  + 3)
If above point is intersection point of line L and plane then
2 (2 + 1) + (–  – 1)+ 3 ( + 3) = 16
=1
Point P = (3, –2, 4)
Dr of QR = < 2, – ,  + 6 >
Dr of L = < 2 –1, 1 >
4 +  +  + 6 = 0
 = –1
Q = (–1, 0, 2)

QR  2iˆ  ˆj  5kˆ

QP  4iˆ  2ˆj  2kˆ

ˆi ˆj kˆ
QR  QP  2 1 5  12iˆ  24ˆj
4 2 2

1
   144  576
2
720
2   180
4
2  180

89. Let 𝑎1 , 𝑎2 , … , 𝑎𝑛 be in A.P. If 𝑎5 = 2𝑎7 and 𝑎11 = 18, then

1 1 1
12 ( + + ⋯+ ) is equal to
√𝑎10 +√𝑎11 √𝑎11 +√𝑎12 √𝑎17 +√𝑎18

Sol. 8
Given that
a5 = 2a7
a1 + 4d = 2(a1 + 6d)
a1 + 8d = 0
a1 + 10 d = 18
 a1 = –72, d = 9
a18 = a1 + 17d = –72 + 153 = 81
a10 = a1 + 9d = 9
 a11  a10 a  a11 a  a17 
12   12  ...... 18 
 d d d 
 
 a18  a10 
= 12  
 d 
 
12   9  3
= 8
9

90. The remainder on dividing 599 by 11 is :

Sol. 9
599 = 54 595
= 625 (55)19
= 625 (3125)19
= 625(3124 + 1)19
= 625(11 + 1)
= 11 × 625 + 625
= 11  × 625 + 616 + 9
= 11 × k + 9
Remainder = 9