JEE MAIN 2021

17th March Shift II Maths Question Paper

SECTION – A

1. If the Boolean expression p  q p  q is a tautology, then and  are

respectively given by :
(1)  ,  (2) , 

(3)  ,  (4)  , 
Ans. (2)
Sol. p  q  p  q
p  q  ~ p  q
~ p ~ q  ~ p  q
~ p  ~ q  q  Tautology

  
  
2. Let the tangent to the circle x2 + y2 = 25 at the point R(3,4) meet x-axis and y-axis
at points P and Q, respectively. If r is the radius of the circle passing through the
origin O and having centre at the incentre of the triangle OPQ, then r2 is equal to :
625 585
(1) (2)
72 66
125 529
(3) (4)
72 64
Ans. (1)
Sol. Given equation of circle

x2 + y2 = 25

Tangent equation at (3, 4)

T : 3x + 4y = 25

X
(0, 0) P
O

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Incentre of OPQ.

 25 25 25 25 
   
I = 4 3 3 4
, .
 25  25  125 25  25  125 
 3 4 12 3 4 12 

625 625  =  25 25 
I =  ,   , 
 75  100  125 75  100  125   12 12 

 Distance from origin to incentre is r.

2 2
 25   25  625
 r =  +  =
2
 12   12  72

Therefore, the correct answer is (1)

3. Let a computer program generate only the digits 0 and 1 to form a string of binary
1
numbers with probability of occurrence of 0 at even places be and probability of
2
1
occurrence of 0 at the odd place be . Then the probability that ‘10’ is followed by
3
‘01’ is equal to :

1 1
(1) (2)
6 18

1 1
(3) (4)
9 3
Ans. (3)
1 1
Sol. P(0 at even place) = , P(0 at odd place) =
2 3
1 2
P(1 at even place) = , P(1 at odd place) =
2 3
P(10 is followed by 01)
=  2  1  1  1    1  1  1  2 
3 2 3 2 2 3 2 3
1 1
= 
18 18
1
=
9

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
4. The number of solutions of the equation x + 2tanx = in the interval [0, 2π] is :
2
(1) 5 (2) 2
(3) 4 (4) 3
Ans. (4)
Sol.

– 0  2
–


x + 2 tan x = in [0, 2]
2


2 tan x = –x
2


2 tan x = –x
2

 x
tan x = –
4 2

–x 
y = tan x and y = +
2 4

3 intersection points

 3 solutions

option (4)

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5. If the equation of plane passing through the mirror image of a point (2, 3, 1) with
x 1 y –3 z  2 x –2 1– y z 1
respect to line   and containing the line   is
2 1 –1 3 2 1
x  y  z  24 , then      is equal to :
(1) 21 (2) 19

(3) 18 (4) 20

Ans. (2)
Sol.

A(2, 3, 1)

= = ...(L1)
M

B (image)

Let point M is (2 – 1,  + 3, – –2)

D.R.’s of AM line are 2 – 1 – 2,  + 3 – 3, ––2–1

2 – 3, , –– 3

AM  line L1

 2(2 – 3) + 1 () – 1 (– –3) = 0

1
6 = 3,  =  M   0, 7 , –5 
2  2 2 

M is mid-point of A & B

A B
M=
2

B=2M–A

B  (–2, 4, –6)

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Now we have to find equation of plane passing through B (–2, 4, –6) & also
containing the line

x–2 1– y z 1
= = ...(1)
3 2 1

x–2 y –1 z 1
= =
3 –2 1

B

L2
P

Point P on line is (2, 1, –1)


b 2 of line L2 is 3, –2, 1

  
n || ( b 2 × PB )


b 2 = 3 î – 2 ĵ + k̂


PB = – 4 î + 3 ĵ – 5 k̂


n = 7 î + 11 ĵ + k̂

   
 equation of plane is r . n = a . n

r .(7 î + 11 ĵ + k̂ ) = (–2 î + 4 ĵ – 6 k̂ ).(7 î + 11 ĵ + k̂ )

7x + 11y + z = – 14 + 44 – 6

7x + 11 y + z = 24

 =7

 = 11

=1

  +  +  = 19

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option (2)

  1 
 2 – sin    x , x  0
6. Consider the function f : R  R defined by f(x) =   x  . Then f is

 0 , x0
(1) monotonic on (0, ∞) only
(2) Not monotonic on (–∞, 0) and (0, ∞)
(3) monotonic on (–∞, 0) only
(4) monotonic on (–∞, 0)  (0, ∞)
Ans. (2)
Sol.
  1
–  2 – sin x  x , x  0
  
f (x) =  0 , x0
 1
  2 – sin  x , x  0
  x

  1 1   1
– x  – cos x   – 2  –  2 – sin x  , x  0
   x   
f’(x) = 
 x  – cos 1   – 1    2 – sin 1  , x  0
  x   x2   x 

 1 1 1
– x cos x  sin x – 2, x  0

 1 cos 1 – sin 1  2, x  0
 x x x

Then f is not monotonic on (–∞, 0) and (0, ∞).

7. Let O be the origin. Let OP  xˆ ˆ and OQ  –i
ˆ– k
i  yj ˆ  2j ˆ , x, y  R, x > 0, be
ˆ  3xk

such that PQ  20 and the vector OP is perpendicular to OQ . If OR  3iˆ  zj ˆ, z

ˆ – 7k

 R, is coplanar with OP and OQ , then the value of x2+y2+z2 is equal to :

(1) 2 (2) 9
(3) 1 (4) 7
Ans. (2)
Sol. OP  xˆ ˆ OP  OQ
ˆ– k
i  yj

ˆ  2j
OQ  –i ˆ
ˆ  3xk

PQ  –1 – x  ˆi  2 – y  ˆj  3x  1 k
ˆ

PQ  –1 – x 2  2 – y 2  3x  12

 OP. OQ  0

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Page
4x = 2y
 JEE MAIN 2021
20  –1 – x 2  2 – y 2  3x  12
20  1  x2  2x  4  y2 – 4y  9x2  1  6x

20  10x2  y2  8x  6 – 4y

20  10x2  4x2  8x  6 – 8x

14 = 14x2  x2  1
 y2  4x2  y2  4

x = 1 as x > 0 and y = 2
x y –1
 –1 2 3x  0
3 z –7

1 2 –1
–1 2 3
3 z –7
1(–14 – 3z) – 2(7 – 9) –1(–z – 6)
–14 – 3z + 4 + z + 6 = 0

2z = –4 z  –2
x2 + y2 + z2 = 9

8. Let L be a tangent line to the parabola y2 = 4x – 20 at (6, 2). If L is also a tangent to

x2 y2
the ellipse   1 , then the value of b is equal to :
2 b
(1) 20 (2) 14
(3) 16 (4) 11
Ans. (2)
Sol. Parabola y2 = 4x – 20

Tangent at P(6, 2) will be

2y = 4  x  6  – 20
 2 

2y = 2x + 12 – 20

2y = 2x – 8

y=x–4

x–y–4=0 ....(1)

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x2 y2
This is also tangent to ellipse  1
2 b

Apply c2 = a2m2 + b2

(–4)2 = (2)(1) + b

b = 14

Option (2)

9. Let f : R  R be defined as f(x) = e–xsinx. If F : [0,1] R  is a differentiable

x 1
  F '  x   f  x  e dx lies in the
x
function such that F(x) = f(t)dt , then the value of
0 0

interval

(1)  330 , 331  (2)  327 , 329 

 360 360   360 360 

(3)  331 , 334  (4)  335 , 336 

 360 360   360 360 
Ans. (1)
Sol. F'(x) = f(x) by Leibnitz theorem

1 1


0

(F '(x)  f(x)) ex dx = 2f(x) ex dx
0


I = 2 sinx dx
0

I = 2(1 – cos 1)


  12 14 16 
= 1  1     ... 
 2! 4! 6! 

  

   
=2 1  1  1  1   < 2 (1 – cos 1) < 2 1  1  1  1  1  
  2 24    2 24 720 

330 331
< 2(1 – cos 1) <
360 360

330 331
<I<
360 360

(1) is correct

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10. If x, y, z are in arithmetic progression with common difference d, x  3d, and the
3 4 2 x
 
determinant of the matrix  4 5 2 y  is zero, then the value of k2 is :
 
5 k z

(1) 6 (2) 36
(3) 72 (4) 12
Ans. (3)

3 4 2 x
Sol. 4 5 2 y =0
5 k z

R1  R1 + R3 – 2R2

0 4 2  k  10 2 0
4 5 2 y =0 { 2y = x + z}
5 k z

 (k – 6 2 )(4z – 5y) = 0

k = 6 2 or 4z = 5y (Not possible  x, y, z in A.P.)

So, k2 = 72

 Option (3)
1
10 [sin2x] –

–1
11. If the integral dx  e  e 2   , where , ,  are integers and [x]
0 x– x 
e
denotes the greatest integer less than or equal to x, then the value of      is

equal to :
(1) 20 (2) 0
(3) 25 (4) 10
Ans. (2)
Sol. Given integral
10 1
sin2x  sin2x 
 dx  10 
 dx (using property of definite in.)
e 
x–  x  x
0 e 0

 12 1 
 –1 

= 10  0.dx 
ex
dx 
 0 1
2


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1
 e–x   –1 –1 
= –10    10 e – e 2 
 –1 1  
2

–1 –1
= 10e –10e 2
comparing with the given relation,
 = 10,  = –10,  = 0
 +  + = 0.
therefore, the correct answer is (2).

12. Let y = y(x) be the solution of the differential equation


cosx(3sinx + cosx + 3) dy = (1+y sinx(3sinx + cosx + 3))dx, 0≤x≤ , y(0) = 0.
2
Then, y    is equal to :
3
 2 3  10   3  7
(1) 2loge   (2) 2loge 
 11  2 
   
3 3 –8 2 3  9
(3) 2loge   (4) 2loge  
 4  6
   
Ans. (1)
Sol. cosx (3 sin x + cos x + 3) dy = (1 + y sin x (3 sin x + cos x + 3)) dx ...(1)

(3 sin x + cos x + 3) (cos x dy – y sin x dx) = dx

dx
 d(y.cos x) =  3 sin x  cos x  3
1
y cos x =
  x   2 x
 2  tan 2   1 – tan 23
3 
 1  tan2 x   1  tan2 x 
 2  2

x
sec2
2
y cos x =
 x
6 tan  1 – tan2 x
 3  3 tan2
x
2 2 2

x 1 x
sec2 sec2 dx
2 2 2
y cos x =
 2 tan 2 x x
 6 tan  4
=
 x x
tan2  3 tan  2
2 2 2 2

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x
tan
1
y cos x = ln 2 +c
x
tan  2
2

Put x =0 & y = 0

C = – ln  1  = ln (2)
2

1 3
y    = 2 ln + ln 2
3 12 3

5 3
= 2 ln + ln 2
11

2 3  10
= 2 ln
11

13. The value of the limit lim


tan  cos2   is equal to :
0 sin 2 sin 
2

1 1
(1) – (2) –
2 4
1
(3) 0 (4)
4
Ans. (1)
Sol. Given,

lim

tan  cos2  
0 sin 2 sin 2

tan   –  sin  2

= lim
0 sin 2 sin  2  cos   1 – sin 
2 2

– tan   sin 2

= lim  tan   –   – tan 

0 sin 2 sin  2

 tan( sin2 )   2 sin2  

1 1
= lim –     = –
 0 2 2 2
  sin    sin(2 sin )  2
Therefore, the correct answer is (1).

14. If the curve y = y(x) is the solution of the different equation

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   
2 x2  x5 /4 dy – y x  x1/4 dx  2x9 /4dx , x > 0 which passes through the point

 4 
1,1 – 3 loge 2  , then the value of y(16) is equal to :
 

(1)  31 – 8 loge 3  (2) 4  31  8 loge 3 

 3 3   3 3 

(3)  31  8 loge 3  (4) 4  31 – 8 loge 3 

 3 3   3 3 
Ans. (4)

Sol. dy y x9/4
  5/4 3/4
dx 2x x (x  1)

dx 1

  ln x
e 2x  e 2
1
 1/2
If =
x

x9/4.x 1/2
y.x–1/2 =
x 5/4
(x3/4  1)
dx

x1/2
 (x 3/4
 1)
dx

x = t4  dx = 4t3dt

t2.4t3dt
 (t3  1)

t2 (t3  1  1)
4  (t3  1)
dt

t2

4 t2dt  4 t 3
1
dt

4t3 4
 ln(t3  1)  C
3 3

4x3/4 4
yx–1/2 = – ln(x3/4 + 1) + C
3 3

4 4 4
1– loge 2= – loge 2 + C
3 3 3

1
C = –
3

4 5/4 4 x
y= x – x ln(x3/4 + 1) –
3 3 3

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4 4 4
y(16) = × 32 – × 4ln9 –
3 3 3

124 32
= – ln 3 = 4  31  8 ln3 
3 3  3 3 

15. Let S1, S2 and S3 be three sets defined as


S1  z  C : z – 1  2 
S2  z  C :Re((1 – i)z)  1

S3  z  C : Im  z   1
Then the set S1  S2  S3

(1) has infinitely many elements (2) has exactly two elements
(3) has exactly three elements (4) is a singleton
Ans. (1)
Sol.

S1S2S3

y=1

O (1, 0)
(x – 1)2 + y2 = 2
x+y=1

Let, z = x + iy

S1  (x – 1)2 + y2  2 ...(1)

S2  x + y  1 ...(2)

S3  y  1 ...(3)

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 S1  S2  S3 has infinitely many elements.

16. If the sides AB, BC, and CA of a triangle ABC have, 3, 5 and 6 interior points
respectively, then the total number of triangles that can be constructed using these
points as vertices, is equal to:
(1) 360 (2) 240
(3) 333 (4) 364
Ans. (3)
Sol.

6
3

B C
5

Total number of triangles

= 3C1 × 5C1 × 6C1

+ 3C1 × 5C2 + 5C1 × 3C2

+ 3C1 × 6C2 + 6C1 × 3C2

+ 5C1 × 6C2 + 6C1 × 5C2

= 90 + 30 + 15 + 45 + 18 + 75 + 60

= 333

17. The value of

r   2r   ...  nr 
lim     2 ,
x  n
Where r is a non-zero real number and [r] denotes the greatest integer less than or
equal to r, is equal to :

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(1) 0 (2) r

r (4) 2r
(3)
2

Ans. (3)
Sol. We know that
r  [r] < r + 1
and 2r  [2r] < 2r + 1
3r  [3r] < 3r + 1
  
nr  [nr] < nr + 1
______________________

r + 2r + ..... + nr
 [r] + [2r] + .....+ [nr] < (r + 2r + .....+ nr) + n
n(n  1) n(n  1)
.r [r]  [2r]  .....  [nr] r n
2  < 2
2
n2 n n2
Now,
n(n  1).r r
lim 2

n 2.n 2
n(n  1)r
n
2 r
and lim 
2 2
n n
So, by Sandwich Theorem, we can conclude that
[r]  [2r]  .....  [nr] r
lim 2

n n 2
6
18. The value of 
r 0
6
Cr  6 C6–r  is equal to :

(1) 1124 (2) 924

(3) 1324 (4) 1024
Ans. (2)
6
Sol. 
r 0
6
Cr6C6 r

=6C0.6C6 + 6C1.6C5 + ....... + 6C6.6C0

Now,

(1 + x)6 (1 + x)6 = (6C0 + 6C1x + 6C2x2 + .... + 6C6x6) (6C0 + 6C1x + 6C2x2 + .... +
6
C6x6)

Comparing coefficient of x6 both sides

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6
C0.6C6 + 6C1 +6C5 + ..... + 6C6.6C0= C6
12

= 924

Therefore, the correct answer is (2).

19. Two tangents are drawn from a point P to the circle x2 + y2 – 2x – 4y + 4 = 0, such

that the angle between these tangents is tan–1  12  , where tan–1  12   0,   . If the
 15   5 
centre of the circle is denoted by C and these tangents touch the circle at points A
and B, then the ratio of the areas of P and CAB is :
(1) 11 : 4 (2) 9 : 4
(3) 2 : 1 (4) 3 : 1
Ans. (2)
Sol.

r=1 L

 /2
C P
M

 12 
Let  = tan–1  5  
 

12
tan  =
5


2 tan 12
 2 =
 5
1  tan2
2

 2  2  3
 tan = sin = and cos =
2 3 2 3 2 13

In CAP,

 1
tan =
2 AP

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3
 AP =
2

 AM  PM
In APM, sin = , cos =
2 AP 2 AP

3 9
 AM =  PM =
13 2 13

6
 AB =
13

1
 Area of PAB = × AB × PM
2

1 6 9 27
= × × =
2 13 2 13 26


Now,  = 90º – .
2

In CAM,

CM
cos  =
CA

 CM = 1.cos     
2 2

 2
= 1.sin =
2 13

1
 Area of CAB = × AB × CM
2

1 6 2 6
= × × =
2 13 13 13

Area of PAB 27 / 26 9
 = =
Area of CAB 6 / 13 4

Therefore, the correct answer is (2).

Aliter

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 JEE MAIN 2021
A

L
R
 /2
C P
M
R L

B
RL3 LR 3
Area of PAB = and Area of CAB =
R 2  L2 R2  L
RL3
2
2 L  2
Therefore, required ratio = R 3L   
LR R 
2
R L

2 cos2
2  2
= cot 
2 
2 sin2
2
1  cos 

1  cos 
5
1
 13
5
1
13
9

4
20. The number of solutions of the equation sin–1  x2  1   cos–1  x2 – 2   x2 , for
 3  3

x  –1,1 , and [x] denotes the greatest integer less than or equal to x, is :

(1) 0 (2) 2
(3) 4 (4) Infinite
Ans. (1)
Sol. There are three cases possible for x  [–1, 1]
 2
Case I : x  –1,– 
 3 

 sin–1 1  cos–1 0   x2

 
 x2     x   (Reject)
2 2
 2 2
Case II : x   – ,
 3 3 
 

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 JEE MAIN 2021
 sin–1 0   cos–1 –1  x2
 0    x2  x   (Reject)
 2 
Case III : x   , 1
 3 
 

 sin–1 0   cos–1 0   x2
 x2    x  (Reject)
 No solution. There, the correct answer is (1)

SECTION - B

1. Let the coefficients of third, fourth and fifth terms in the expansion of
n
 a
 x  2  , x  0, be in the ration 12:8:3. Then the term independent of x in the
 x 
expansion, is equal to ………
Ans. (4)
r
 a
Sol. Tr 1  ncr xn–r .  2 
x 
n
= crar xn–3r
T3  n c2a2xn–6 , T4  n c3a3xn–9
T5  n c 4a4 xn–12
coefficient of T3 n c2.a2 3 3
Now,  n  
coefficient of T4 c3.a3 a n – 2  2
 a(n– 2) = 2 ……… (i)
coefficient of T4 n c3.a3 4 8
and  n  
coefficient of T5 c4.a4 a n – 3 3
3
 a(n– 3) = ……… (ii)
2
1
by (i) and (ii) n = 6, a =
2
for term independent of ‘x’
n 6
n – 3r = 0  r=  r= =2
3 3
2
1 15
T3  6 c2   x0  = 3.75  4
2 4

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 JEE MAIN 2021
a b  0
2. Let A =  and B      such that AB = B and a + d = 2021, then the value
c d 
  0 
of ad-bc is equal to ……………. .
Ans. (2020)
a b 
Sol. A =   , B =  
c d  
AB = B
a b      
c d    =   
     
a  b     a  b   ....(1)
 c  d      
    and c  d   ....(2)
  a –1  –b and c   1 – d

 –b  1– d
 & 
 a –1  c
–b 1– d
 
a –1 c
–bc = (a – 1) (1 – d)
–bc = a – ad – 1 + d
ad – bc = a + d – 1
= 2021–1
= 2020

3. Let f : [–1,1] → R be defined as f(x) = ax2+bx+c for all x  [–1, 1], where a, b, c  R
1
such that f(–1) = 2, f’(–1) = 1 and for x  [–1, 1] the maximum value of f’’ (x) is .
2
If f(x)   , x  [–1, 1], then the least value of  is equal to …………… .
Ans. (5)
Sol. f(x) = ax2 + bx + c
f'(x) = 2ax + b,

f''(x) = 2a

1 1
Given f''(–1) = a=
2 4

3
f'(–1) = 1  b – 2a = 1 b=
2

13
f(–1) = a – b + c = 2 c=
4

1 2
Now f(x) = (x + 6x + 13), x[–1, 1]
4

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 JEE MAIN 2021
1
f'(x) = (2x + 6) = 0  x = –3  [–1, 1]
4

f(1) = 5, f(–1) = 2

f(x)  5

So minimum = 5

e
x19 log x  dx, where n  N. If 20 I10  I9  I8 , for natural numbers
n
4. Let In  1

and , then – equal to ………………… .

Ans. (1)
e

x  nx  .dx
19 n
Sol. In =
1

e e
x20  nx 
n 1
x20
 nx  
n
 . n dx
20 x 20
1 1

e20 n
In   I 
20 20 n1

20In  e20  n In1


20I10  e20  10I9  ....(1)

20I9  e20  9I8 ....(2)

– – –
__________________

20I10  10I9  9I8

= 10, = 9   –  = 1

5. Let f : [–3, 1]→ R be given as


min  x  6  , x2 , –3,  x  0

f x  

 max x, x ,

2

0  x  1. 
If the area bounded by y = f(x) and x-axis is A, then the value of 6A is equal to
…………… .
Ans. (41)
Sol.

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 JEE MAIN 2021

x2

ξ𝑥
4
3

–6
–3 –2 0 1

–2 0 1
Area is

–3
(x  6)dx +
–2
 x2dx + 
0
xdx  A

0 1
7  3 2 
= + x  +  x /2 
2  3 –2  3 0

7 8 2 41
= + + =
2 3 3 6
So, 6A = 41

6. Let x be a vector in the plane containing vectors ˆ–ˆ

a  2i ˆ and b  ˆ
j k ˆ . If the
ˆ– k
i  2j

vector x is perpendicular to 3iˆ  2jˆ – kˆ and its projection on a is 17 6 , then the
2
2
value of x is equal to ……………….. .
Ans. (486)
Sol. Let, x = k( a + b )

I. x is perpendicular to 3iˆ  2j ˆ
ˆ k

k{(2 + )3 + (2 – 1)2 + (1 – )(–1) = 0

 8 + 3 = 0

3
 =
8

II. Also projection of x on a is therefore

x.a = 17 6
| a| 2

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 JEE MAIN 2021
 (a  b).a  17 6
 k =
 6  2

 
 k 6   3    17  6
  8  2

51
 k= 8
51

k=8

13 ˆ 14 ˆ 11 ˆ 
x = 8  i j k
 8 8 8 

= 13iˆ  14j ˆ
ˆ  11k

| x |2 = 169 + 196 + 121 = 486

7. Consider a set of 3n numbers having variance 4. In this set, the mean of first 2n
numbers is 6 and the mean of the remaining n numbers is 3. A new set is
constructed by adding 1 into each of first 2n numbers, and subtracting 1 from each
of the remaining n numbers. If the variance of the new set is k, then 9k is equal to
………………. .
Ans. (68)

Sol. Let first 2n observations ae x1, x2 ………… , x2n0

and last n observations are y1, y2 ………… , yn

Now, x i
 6,
y i
3
2n n

 x i  12n , y i  3n  x  y
i i

15n
5
3n 3n

Now, x  y 2
i
2
i
– 52  4
3n
 x  y 2
i
2
i  29  3n  87n

Now, mean is  x i  1    y – 1  15n  2n – n  16

i
3n 3n 3

  x  1    y –1
2 2 2
i i  16 
Now, variance is – 
3n  3 

=
x  y2
i
2
i 2   x –  y   3n –  16 
i i
2

3n  3 
 
87n  2  9n  3n  16 
2
= – 
3n  3 

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2
 16 
29  6  1 –  
 3 
324 – 256 68
=  k
9 9
 9k  68
Therefore, the correct answer is 68.

8. If 1, log10(4x–2) and log10  4x  18  are in arithmetic progression for a real number

 5 

 1
2  x –  x –1 x2
 2
x, then the value of the determinant 1 0 x is equal to :
x 1 0

Ans. (2)
Sol.  
1, Log10 4x – 2 , Log10  4x  18  in AP.
 5 

 
2. Log10 4x – 2 = 1 + Log10  4x  18 
 5 

    18  
2
Log10 4x – 2  log10 10.  4x  
  5  

4   18 
2
x
–2  10.  4x 
 5 

4 
2
x
 4 – 4.4x  10.4x  36

4 
2
x
– 14.4x – 32  0

4 
2
x
 2.4x – 16.4x – 32  0

  
4x 4x  2 –16. 4x  2  0
4  24 –16  0
x x

4x  –2 4x  16
rejected x=2
2(x  1 / 2) x  1 x2
Therefore 1 0 x
x 1 0
3 1 4
= 1 0 2
2 1 0
= 3(–2) – 1(0 – 4) + 4(1 – 0)
= –6 + 4 + 4

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 JEE MAIN 2021
=2

9. Let P be an arbitrary point having sum of the squares of the distances from the
planes x + y + z = 0, lx – nz = 0 and x – 2y + z = 0, equal to 9. If the locus of the
point P is x2 + y2 +z2 = 9, then the value of l–n is equal to …………….. .
Ans. (0)
Sol. Let point P is (, , )
2 2 2
    – n    – 2   
  +  2  +   =9
 3    n2   6 
(x  y  z)2 2 (x – 2 y  z)2
Locus is + ( n – nz) + =9
3 2
 n2 6

1 2  1 n2  1 n 
x2   + y2 + z 2   2
2 2 2
 n  2  n  + 2zx  2 –
2 2 – 9 = 0
 n2 
  
Since its given that x2 + y +z =9
2 2

After solving  = n,
then  – n = 0

10. Let tan , tan and tan ;   2n – 1  ,n  N be the slopes of three line
2
segment OA, OB and OC, respectively, where O is origin. If circumcentre of C
coincides with origin and its orthocentre lies on y-aixs, then the value of
2
 cos 3  cos 3  cos 3 
  is equal to
 cos  cos  cos  
Ans. (144)
Sol. Since orthocentre and circumcentre both lies on y-axis
 Centroid also lies on y-axis
 cos  = 0
cosα + cos  + cos = 0

 cos3   cos3   cos3   3cos  cos  cos 

cos 3  cos 3  cos 3

cos  cos  cos 

=
 
4 cos3   cos3   cos3  – 3  cos   cos   cos  
= 12
cos  cos  cos 
2
 cos 3  cos 3  cos 3 
then,   = 144
 cos  cos  cos  

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