JEE Main 2021 24 Feb Shift 2 Maths Paper

JEE-Main-2021-Question-Paper-Maths-Feb-24-Shift-2JEE Main 2021 24Feb Shift 2 Page | 1
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1. Let , ∈ . If the mirror image of the point ( ,6,9)P a with respect to the line −3
7 =
−1
−9 is (20, , − − 9), then | a b |+ is equal to :
(1) 86 (2) 88 (3)84 (4)90
Ans. (2)
Mid point of = ( +20
2 ,
+20
2 −3
| + | = |−56 − 32| = 88
2. Let be a twice differentiable function defined on R such that (0) = 1, ′(0) = 2 and
′() ≠ 0 for all ∈ . If | () ′()
′() ′′() | = 0, for all ∈ , then the value of (1) lies in the
interval :
(1) (9, 12) (2) (6, 9) (3)(3, 6) (4)(0, 3)
Ans. (2)
= 0
⇒ ()
⇒ (0) = ′(0)
JEE Main 2021 24Feb Shift 2 Page | 2
⇒ ∫ 2 = ∫ ′()
⇒ 2 = ln |()|
∴ (1) = 2 ≈ 7.38
tan sin 4 8
Ans. (2)
4 =
1
√7
4. The probability that two randomly selected subsets of the set {1, 2, 3, 4, 5} have exactly two
elements in their intersection, is :
(1) 65
Ans. (2)
Sol. Let and be two subsets.
For each ∈ {1, 2, 3, 4, 5} , there are four possibilities :
∈ ∩ , ∈ ′ ∩ , ∈ ∩ ′, ∈ ′ ∩ ′
So, the number of elements in sample space = 45
Required probability
= 52×33
29
5. The vector equation of the plane passing through the intersection of the planes
ˆˆ ˆr ( ) 1 + + =i j k and ˆ ˆr ( 2 ) 2, − = −i j and the point (1, 0, 2) is :
(1) ⋅ ( − 7 + 3) = 7
3 (2) ⋅ ( + 7 + 3) = 7
(3) ⋅ (3 + 7 + 3) = 7 (4) ⋅ ( + 7 + 3) = 7
3
Sol. Family of planes passing through intersection of planes is
{ ⋅ ( + + ) − 1} + { ⋅ ( − 2) + 2} = 0
The above curve passes through + 2,
(3 − 1) + (1 + 2) = 0 ⇒ = − 2
3
Hence, equation of plane is
3{ ⋅ ( + + ) − 1} − 2{ ⋅ ( − 2) + 2} = 0
⇒ ⋅ ( + 7 + 3) = 7
TRICK: Only option (2) satisfies the point (1, 0, 2)
6. If P is a point on the parabola 2 4= +y x which is closest to the straight line 4 1,= −y x then
the co-ordinates of P are :
(1) (–2, 8) (2) (1, 5) (3) (3, 13) (4) (2, 8)
Ans. (4)
Sol. Tangent at P is parallel to the given line.

⇒ 1 = 2
Required point is (2, 8)
7. Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices
( , ),(2, )a c b and ( , )a b be 10 7
, . 3 3

If , are the roots of the equation 2 + + 1 = 0,
then the value of 2 2+ − is :
(1) 71
256 (2) −
3 =
10
}, solving
4 + 1 = 0
256 −
3
4 = −
71
256
2
1 2 2 d , − − x x x where [x] denotes the greatest integer less than
or equal to x , is :
(1) –4 (2) –5 (3) −√2 − √3 − 1 (4) −√2 − √3 + 1
Ans. (3)
I = (–6) +∫ [2] 2
0
I = −6 + (√2 − 1) + 2√3 − 2√2 + 6 − 3√3
I = −1 − √2 − √3
3 2
3 2
− −
= − − − − − −
x x x x
Let = { ∈ is increasing}. Then is equal to :
(1) (−5, −4) ∪ (4, ∞) (2) (−5, ∞)
(3) (−∞, −5) ∪ (4, ∞) (4) (−∞, −5) ∪ (−4, ∞)
Ans. (1)
6(2 − − 6) ; > 4
⇒ ′() = { – 55 ; <– 5
6( − 5)( + 4) ; – 5 < < 4 6( − 3)( + 2) ; > 4
Hence, () is monotonically increasing in (– 5, – 4) ∪ (4, ∞)
10. If the curve = 2 + + , ∈ passes through the point (1, 2) and the tangent line to
this curve at origin is = , then the possible values of , , are :
(1) = 1, = 1, = 0 (2) = −1, = 1, = 1
(3) = 1, = 0, = 1 (4) = 1
2 , =

Since (0, 0) lies on curve,
∴ = 0, = 1
: (0,0) lies on the curve. Only option (1) has = 0
11. The negation of the statement ~ ∧ ( ∨ ) is :
(1) ~p ∧ (2) ∧ ~q (3) ~p ∨ (4) ∨∼
JEE Main 2021 24 Feb Shift 2
Ans. (4)
∼ [~ ∧ ( ∨ )]
≡ ∨ ∼ ( ∨ )
≡ ∨ (∼ ∧∼ )
≡ ∨∼
12. For the system of linear equations :
2 1, 2, 4 6,− = − + = − + = Rx y x y kz ky z k
consider the following statements :
(A) The system has unique solution if 2, 2 −k k .
(B) The system has unique solution if 2= −k .
(C) The system has unique solution if 2=k .
(D) The system has no-solution if 2=k .
(E) The system has infinite number of solutions if 2 −k .
Which of the following statements are correct ?
(1) (B) and (E) only (2) (C) and (D) only
(3) (A) and (D) only (4) (A) and (E) only
Ans. (3)
− + = −2
= | 1 −2 0 1 −1 0 4
| = 4 − 2
≠ ±2
− + 2 = −2
0. + 2 + 4 = 6
= | 1 −2 0
| = (−8) + 2[−20]
⇒ = −48 ≠ 0
So, for = 2, the system has no solution.
13. For which of the following curves, the line 3 2 3+ =x y is the tangent at the point
3 3 1 , ?
(3) 2 = 1
Ans. (1)
Sol. Tangent to 2 + 92 = 9 at point ( 3√3
2 ,
1
⇒ Option (1) is true.
14. The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 20
seconds at the speed of 432 km/hour, the angle of elevation changes to 30°. If the jet plane is
flying at a constant height, then its height is :
(1) 1200√3 m (2) 1800√3 m (3) 3600√3 m (4) 2400√3 m
Ans. (1)
Sol.
D
P
Distance PQ = v × 20 = 2400 m
In ΔPAC
⇒ h = 1200√3 m
15. For the statements and , consider the following compound statements :
(a) (~ ∧ ( → )) → ~
(b) (( ∨ )) ∧ ~) →
Then which of the following statements is correct ?
(1) (a) is a tautology but not (b) (2) (a) and (b) both are not tautologies.
(3) (a) and (b) both are tautologies. (4) (b) is a tautology but not (a).
Ans. (3)
Sol. (a)
~ → ~ ∧ ( → ) ~ (~) ∧ ( → ) → ~
(a) is tautology.
(b)
∨ ~ ( ∨ ) ∧ ~ (( ∨ ) ∧ ~) →

(b) is tautology.
∴ (a) and (b) both are tautologies.
16. Let A and B be 3 3 real matrices such that A is symmetric matrix and B is skew-symmetric
matrix. Then the system of linear equations ( )2 2 2 2A B B A X O,− = where X is a 3 1
column matrix of unknown variables and O is a 3 1 null matrix, has :
(1) a unique solution (2) exactly two solutions
(3) infinitely many solutions (4) no solution
Ans. (3)
Let A2B2 – B2A2 = P
= (B2)T (A2)T– (A2)T (B2)T
[ 0
] [
] = [ 0 0 0
Δ = 0 and Δ1 = Δ2 = Δ3 = 0
∴ System of equations has infinite number of solutions.
17. If ≥ 2 is a positive integer, then the sum of the series
( )n 1 2 3 4 n
2 2 2 2 2C 2 C C C . C+ + + + + + is :
(1) (+1)2(+2)
Sol. 2C2 = 3C3
Let S = 3C3 + 3C2 + ……. + nC2 = n+1C3 ( nCr + nCr–1 = n+1Cr)
∴ n+1C2 + n+1C3 + n+1C3
= n+2C3 + n+1C3
TRICK : Put = 2 and verify the options.
18. If a curve ( )=y f x passes through the point (1, 2) and satisfies 4d b ,
d + =
1 () =
2 =
5 +
Also, ∫ ( 4

25 =
62
5
19. The area of the region : 2 2( , ) : 5 2 9= +R x y x y x is :
(1) 9√3 square units (2) 12√3 square units
(3) 11√3 square units (4) 6√3 square units
Ans. (2)
0
= 2|9– 3|0 √3 = 12√3
20. Let () be a differentiable function defined on [0, 2] such that ′() = ′(2 − ) for all ∈
(0, 2), (0) = 1 and (2) = 2. Then the value of 2
0 ( )d f x x is :
(1) 1 + e2 (2) 1 – e2 (3) 2(1– e2) (4) 2(1+e2)
Ans. (1)
() = −(2 − ) +
= ∫ () 2
1
Section B
1. The number of the real roots of the equation 2 27 ( 1) | 5 |
4 + + − =x x is _______.
4
4
∴ 2 real roots.
2. The students 1 2 10, , ,S S S are to be divided into 3 groups A, B and C such that each group
has at least one student and the group C has at most 3 students. Then the total number of
possibilities of forming such groups is _______.
JEE Main 2021 24Feb Shift 2
Ans. 31650
= 128 [40 + 90 + 120 ] – 350
= (128 × 250) – 350
= 10[3165] = 31650
3. If + = 1, + = 2 and 1
( ) , 0,
then the value of the expression
1 ( )
1
)] = ( +
4. If the variance of 10 natural numbers 1, 1, 1, … , 1, is less than 10, then the maximum
possible value of k is ________.
Ans. 11
⇒ 92 − 18 + 9 < 1000
Maximum possible integral value of is 11.
5. Let be an integer. If the shortest distance between the lines − = 2 − 1 = −2 and
2= + = −x y z is 7
, 2 2
Ans. 1
1
|1×2|
|
|
⇒ |10 + 3| = 7 ⇒ = −1 as is an integer.
⇒ || = 1
6. Let 1= −i . If 21 21
24 24
i i k
i i and = [||] be the greatest integral part of
|k|. Then 5 5
Ans. 310
Sol. (2
∴ = 0
5 =0
= [52 + 62 + 72 + 82 + 92 + 102] − [5 + 6 + 7 + 8 + 9 + 10]
= [(12 + 22+. . . +102) − (12 + 22 + 32 + 42)] − [(1 + 2 + 3+. . . +10) − (1 + 2 + 3 + 4)]
= (385 − 30) − [55 − 10]
= 355 – 45 = 310
7. Let a point P be such that its distance from the point (5, 0) is thrice the distance of P from the
point ( 5,0).− If the locus of the point P is a circle of radius r, then 4r2 is equal to ________.
Ans. 56.25
Sol. Let P be (h,k), A(5, 0) and B(-5, 0)
Given PA = 3PB
⇒ 82 + 82 + 100 + 200 = 0
∴ Locus of is 2 + 2 + ( 25
2 ) + 25 = 0
= 625
0, otherwise
r
1
k k
i ii k i i k i exists, is equal to________.
Ans. *
(1+x)10 = 10C0 + 10C1x + 10C2x2 + …… + 10C10x10
(1+x)15 = 15C0 + 15C1x + ….15Ck–1xk–1 + 15Ckxk + 15Ck+1xk+1+……15C15x15
∑ (10)(15−) =0 =10C0. 15Ck+ 10C1.15Ck–1+…..+10Ck.15C0
Coefficient of xk in (1+x)25
= 25Ck
∑ (12)(13+1−)+1 =0 =12C0.13Ck+1 + 12C1. 13Ck +……+ 12Ck+1.13C0
Coefficient of xk+1 in (1+x)25
= 25Ck+1
By the given definition of (
), can be as large as possible.
9. The sum of first four terms of a geometric progression (G.P.) is 65
12 and the sum of their
respective reciprocals is 65
. 18
If the product of first three terms of the G.P. is 1, and the third
term is , then 2 is__________.
Ans. 3
12 …….(1)
⇒ 1
(
2 ) = 1 ⇒ =
∴ 2 = 3
10. If the area of the triangle formed by the positive x-axis, the normal and the tangent to the
circle 2 2( 2) ( 3) 25− + − =x y at the point (5, 7) is A, then 24A is equal to ______.
Ans.
Sol.
( − 7) = ( 7−3
5−2 ) ( − 5)
⇒ 3y – 21 = 4x – 20
⇒ 4x – 3y + 1 = 0
( − 7) = − 3
4 ( − 5)
3 , 0)
The question is wrong. The normal cuts at a point on the negative axis.
P(5,7)
A(2,3)
M
y

JEE Main 2021 24 Feb Shift 2 Maths Paper

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