JEE Main 2021 Mathematics February 24 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


JEE (MAIN)-2021 Phase-1 (24-02-2021)-E
SECTION - I
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4
choices (1), (2), (3) and (4), out of which ONLY ONE
is correct.
Choose the correct answer :
1. 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) 3600 3 m (2) 2400 3 m
(3) 1800 3 m (4) 1200 3 m
Answer (4)
Sol.
d x
60° 30°
d
h
A
Given ??
?
h
tan30
xd
 and ??
h
tan60
x
? ?? xd 3h and ?
h
x
3
?
??
??
??
??
12h
d3– h
33
Given 
?
?
d 432 1000
20 3600
? d = 2400 m
? ? ? ?
2h
2400 h 1200 3 m
3
2. The probability that two randomly selected
subsets of the set {1, 2, 3, 4, 5} have exactly
two elements in their intersection, is :
(1)
7
65
2
(2)
9
135
2
(3)
8
65
2
(4)
7
35
2
Answer (2)
PART–C : MATHEMATICS
Sol. Number of ways of selecting elements common
to both A and B = 
5
C
2
? Required probability 
?
??
53
2
59
C3 135
42
3. If P is a point on the parabola y = x
2
 + 4 which
is closest to the straight line y = 4x – 1, then the
co-ordinates of P are :
(1) (–2, 8) (2) (1, 5)
(3) (3, 13) (4) (2, 8)
Answer (4)
Sol. Closest point will be point of tangency of
tangent of same slope i.e. 4
Let equation of tangent y = 4x + c
? 4x + c = x
2
 + 4 have D = 0
i.e. x
2
 – 4x + (4 – c) = 0
D = 0 ???  16 – 4(4 – c) = 0 ???  c = 0
Tangent is y = 4x gives x = 2 and y = 8 as point
of tangency
? Nearest point (2, 8)
4. If ? n2 is a positive integer, then the sum of
the series
??
?
?? ? ? ?
n1 2 3 4 n
22 2 2 2
C 2 C C C ..... C
 is :
(1)
?? n(2n 1)(3n 1)
6
(2)
? n(n–1)(2n 1)
6
(3)
??
2
n(n 1) (n 2)
12
(4)
?? n(n 1)(2n 1)
6
Answer (4)
Sol. Sum of 
2
C
2
 + 
3
C
2
 + .... + 
n
C
2
 is coefficient of
x
2
 in (1 + x)
2
 + (1 + x)
3
 + .... + (1 + x)
n
i.e. coefficient of x
2
 in
??
??
? ?
?
?
??
?
n–1
2
n1
3
1x –1
1x C
(1 x–1)
Hence required sum = 
n+1
C
2
 + 2 ?
n+1
C
3
??
??
(n 1)(n) 2(n 1)n(n–1)
26
?? ? ?
??
n(n 1)(3 2n–2) n(n 1)(2n 1)
23 6
Page 2


JEE (MAIN)-2021 Phase-1 (24-02-2021)-E
SECTION - I
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4
choices (1), (2), (3) and (4), out of which ONLY ONE
is correct.
Choose the correct answer :
1. 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) 3600 3 m (2) 2400 3 m
(3) 1800 3 m (4) 1200 3 m
Answer (4)
Sol.
d x
60° 30°
d
h
A
Given ??
?
h
tan30
xd
 and ??
h
tan60
x
? ?? xd 3h and ?
h
x
3
?
??
??
??
??
12h
d3– h
33
Given 
?
?
d 432 1000
20 3600
? d = 2400 m
? ? ? ?
2h
2400 h 1200 3 m
3
2. The probability that two randomly selected
subsets of the set {1, 2, 3, 4, 5} have exactly
two elements in their intersection, is :
(1)
7
65
2
(2)
9
135
2
(3)
8
65
2
(4)
7
35
2
Answer (2)
PART–C : MATHEMATICS
Sol. Number of ways of selecting elements common
to both A and B = 
5
C
2
? Required probability 
?
??
53
2
59
C3 135
42
3. If P is a point on the parabola y = x
2
 + 4 which
is closest to the straight line y = 4x – 1, then the
co-ordinates of P are :
(1) (–2, 8) (2) (1, 5)
(3) (3, 13) (4) (2, 8)
Answer (4)
Sol. Closest point will be point of tangency of
tangent of same slope i.e. 4
Let equation of tangent y = 4x + c
? 4x + c = x
2
 + 4 have D = 0
i.e. x
2
 – 4x + (4 – c) = 0
D = 0 ???  16 – 4(4 – c) = 0 ???  c = 0
Tangent is y = 4x gives x = 2 and y = 8 as point
of tangency
? Nearest point (2, 8)
4. If ? n2 is a positive integer, then the sum of
the series
??
?
?? ? ? ?
n1 2 3 4 n
22 2 2 2
C 2 C C C ..... C
 is :
(1)
?? n(2n 1)(3n 1)
6
(2)
? n(n–1)(2n 1)
6
(3)
??
2
n(n 1) (n 2)
12
(4)
?? n(n 1)(2n 1)
6
Answer (4)
Sol. Sum of 
2
C
2
 + 
3
C
2
 + .... + 
n
C
2
 is coefficient of
x
2
 in (1 + x)
2
 + (1 + x)
3
 + .... + (1 + x)
n
i.e. coefficient of x
2
 in
??
??
? ?
?
?
??
?
n–1
2
n1
3
1x –1
1x C
(1 x–1)
Hence required sum = 
n+1
C
2
 + 2 ?
n+1
C
3
??
??
(n 1)(n) 2(n 1)n(n–1)
26
?? ? ?
??
n(n 1)(3 2n–2) n(n 1)(2n 1)
23 6
JEE (MAIN)-2021 Phase-1 (24-02-2021)-E
5. Let f(x) be a differentiable function defined on
[0, 2] such that f ?(x) = f ?(2 – x) for all ? x (0, 2) ,
f(0) = 1 and f(2) = e
2
. Then the value of 
?
2
0
f(x)dx
is :
(1) 2(1 + e
2
) (2) 1 + e
2
(3) 2(1 – e
2
) (4) 1 – e
2
Answer (2)
Sol. Given f ?(x) = f ?(2 – x)
? f ?(x) – f ?(2 – x) = 0
Integrating both sides, we get
f(x) + f(2 – x) = c ...(i)
Put x = 0, we get
c = f(0) + f(2) = 1 + e
2
Integrating 0 to 2 equation (i) both sides, we get
????
??
22
2
2
0
00
f(x)dx f(2–x)dx (1e ) 8
Also ?
??
22
00
f(x)dx f(2–x)dx
Hence ??
?
2
2
0
2f(x)dx 2(1e )
? ??
?
2
2
0
f(x)dx 1 e
6. For the statements p and q, consider the
following compound statements:
(a)?? ?? ? ~q (p q) ~p
(b) ?? ? ((p q) ~p) q
Then which of the following statements is
correct?
(1) (a) and (b) both are tautologies.
(2) (a) is a tautology but not (b).
(3) (b) is a tautology bot not (a).
(4) (a) and (b) both are not tautologies.
Answer (1)
Sol. Truth table for required statements
??
?
?? ? ?
?? ?
??
~
~
~~ ~
~
~
(q ((p q)
(q) (p q)
p q p q p q (p q)) p q p)
(p q) p
pq
TT F F T F T T F T
TF F T F F T T F T
FT T F T F T T T T
F F T T T T T (a) is tautology F F T (b) is tautology
7. For the system of linear equations :
x – 2y = 1, x – y + kz = –2, ky + 4z = 6, k ? R,
consider the following statements :
(A)The system has unique solution if
??? k2,k 2.
(B) The system has unique solution if k = – 2.
(C) The system has unique solution if k = 2
(D) The system has no solution if k = 2.
(E) The system has infinite number of solutions
if k ? – 2.
Which of the following statements are correct?
(1) (A) and (E) only (2) (A) and (D) only
(3) (B) and (E) only (4) (C) and (D) only
Answer (2)
Sol. Using Cramer’s Rule, we have
?
?? ? ? ? ? ? ? ?
22
120
11k 1(4k ) 2(4) 4k
0k 4
?
?? ? ? ? ? ? ? ? ? ? ? ?
22
x
120
21k 1(4k ) 2( 2k 6) 84k k
6k 4
Now, ? = 0 if k = ±2
if k = –2, ? = 0 and  ?
x
 ? 0
Hence no solution
Also if k = 2, ? = 0 and  ?
x
 ? 0
Now
?? ? ? ? ? ? ?? ? ?
y
11 0
12k 1(86k) 1(4) 6k 12 0
06 4
Hence, the system has no solution if k = ±2
and unique solution if k ? ±2
8. Let a, b, c be in arithmetic progression. Let the
centroid of the triangle with vertices
(a, c), (2, b) and (a, b) be 
??
??
??
10 7
,
33
. If ?, ? are
the roots of the equation ax
2
 + bx + 1 = 0, then
the value of ?
2
 + ?
2
 – ?? is :
(1)
69
256
(2) –
71
256
(3) –
69
256
(4)
71
256
Answer (2)
Page 3


JEE (MAIN)-2021 Phase-1 (24-02-2021)-E
SECTION - I
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4
choices (1), (2), (3) and (4), out of which ONLY ONE
is correct.
Choose the correct answer :
1. 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) 3600 3 m (2) 2400 3 m
(3) 1800 3 m (4) 1200 3 m
Answer (4)
Sol.
d x
60° 30°
d
h
A
Given ??
?
h
tan30
xd
 and ??
h
tan60
x
? ?? xd 3h and ?
h
x
3
?
??
??
??
??
12h
d3– h
33
Given 
?
?
d 432 1000
20 3600
? d = 2400 m
? ? ? ?
2h
2400 h 1200 3 m
3
2. The probability that two randomly selected
subsets of the set {1, 2, 3, 4, 5} have exactly
two elements in their intersection, is :
(1)
7
65
2
(2)
9
135
2
(3)
8
65
2
(4)
7
35
2
Answer (2)
PART–C : MATHEMATICS
Sol. Number of ways of selecting elements common
to both A and B = 
5
C
2
? Required probability 
?
??
53
2
59
C3 135
42
3. If P is a point on the parabola y = x
2
 + 4 which
is closest to the straight line y = 4x – 1, then the
co-ordinates of P are :
(1) (–2, 8) (2) (1, 5)
(3) (3, 13) (4) (2, 8)
Answer (4)
Sol. Closest point will be point of tangency of
tangent of same slope i.e. 4
Let equation of tangent y = 4x + c
? 4x + c = x
2
 + 4 have D = 0
i.e. x
2
 – 4x + (4 – c) = 0
D = 0 ???  16 – 4(4 – c) = 0 ???  c = 0
Tangent is y = 4x gives x = 2 and y = 8 as point
of tangency
? Nearest point (2, 8)
4. If ? n2 is a positive integer, then the sum of
the series
??
?
?? ? ? ?
n1 2 3 4 n
22 2 2 2
C 2 C C C ..... C
 is :
(1)
?? n(2n 1)(3n 1)
6
(2)
? n(n–1)(2n 1)
6
(3)
??
2
n(n 1) (n 2)
12
(4)
?? n(n 1)(2n 1)
6
Answer (4)
Sol. Sum of 
2
C
2
 + 
3
C
2
 + .... + 
n
C
2
 is coefficient of
x
2
 in (1 + x)
2
 + (1 + x)
3
 + .... + (1 + x)
n
i.e. coefficient of x
2
 in
??
??
? ?
?
?
??
?
n–1
2
n1
3
1x –1
1x C
(1 x–1)
Hence required sum = 
n+1
C
2
 + 2 ?
n+1
C
3
??
??
(n 1)(n) 2(n 1)n(n–1)
26
?? ? ?
??
n(n 1)(3 2n–2) n(n 1)(2n 1)
23 6
JEE (MAIN)-2021 Phase-1 (24-02-2021)-E
5. Let f(x) be a differentiable function defined on
[0, 2] such that f ?(x) = f ?(2 – x) for all ? x (0, 2) ,
f(0) = 1 and f(2) = e
2
. Then the value of 
?
2
0
f(x)dx
is :
(1) 2(1 + e
2
) (2) 1 + e
2
(3) 2(1 – e
2
) (4) 1 – e
2
Answer (2)
Sol. Given f ?(x) = f ?(2 – x)
? f ?(x) – f ?(2 – x) = 0
Integrating both sides, we get
f(x) + f(2 – x) = c ...(i)
Put x = 0, we get
c = f(0) + f(2) = 1 + e
2
Integrating 0 to 2 equation (i) both sides, we get
????
??
22
2
2
0
00
f(x)dx f(2–x)dx (1e ) 8
Also ?
??
22
00
f(x)dx f(2–x)dx
Hence ??
?
2
2
0
2f(x)dx 2(1e )
? ??
?
2
2
0
f(x)dx 1 e
6. For the statements p and q, consider the
following compound statements:
(a)?? ?? ? ~q (p q) ~p
(b) ?? ? ((p q) ~p) q
Then which of the following statements is
correct?
(1) (a) and (b) both are tautologies.
(2) (a) is a tautology but not (b).
(3) (b) is a tautology bot not (a).
(4) (a) and (b) both are not tautologies.
Answer (1)
Sol. Truth table for required statements
??
?
?? ? ?
?? ?
??
~
~
~~ ~
~
~
(q ((p q)
(q) (p q)
p q p q p q (p q)) p q p)
(p q) p
pq
TT F F T F T T F T
TF F T F F T T F T
FT T F T F T T T T
F F T T T T T (a) is tautology F F T (b) is tautology
7. For the system of linear equations :
x – 2y = 1, x – y + kz = –2, ky + 4z = 6, k ? R,
consider the following statements :
(A)The system has unique solution if
??? k2,k 2.
(B) The system has unique solution if k = – 2.
(C) The system has unique solution if k = 2
(D) The system has no solution if k = 2.
(E) The system has infinite number of solutions
if k ? – 2.
Which of the following statements are correct?
(1) (A) and (E) only (2) (A) and (D) only
(3) (B) and (E) only (4) (C) and (D) only
Answer (2)
Sol. Using Cramer’s Rule, we have
?
?? ? ? ? ? ? ? ?
22
120
11k 1(4k ) 2(4) 4k
0k 4
?
?? ? ? ? ? ? ? ? ? ? ? ?
22
x
120
21k 1(4k ) 2( 2k 6) 84k k
6k 4
Now, ? = 0 if k = ±2
if k = –2, ? = 0 and  ?
x
 ? 0
Hence no solution
Also if k = 2, ? = 0 and  ?
x
 ? 0
Now
?? ? ? ? ? ? ?? ? ?
y
11 0
12k 1(86k) 1(4) 6k 12 0
06 4
Hence, the system has no solution if k = ±2
and unique solution if k ? ±2
8. Let a, b, c be in arithmetic progression. Let the
centroid of the triangle with vertices
(a, c), (2, b) and (a, b) be 
??
??
??
10 7
,
33
. If ?, ? are
the roots of the equation ax
2
 + bx + 1 = 0, then
the value of ?
2
 + ?
2
 – ?? is :
(1)
69
256
(2) –
71
256
(3) –
69
256
(4)
71
256
Answer (2)
JEE (MAIN)-2021 Phase-1 (24-02-2021)-E
Sol. Here, 2b = a + c ...(i)
and centroid of ? is 
??
??
??
10 7
,
33
?
??
? ? ?
a2 a 10
a4
33
and 
?? ? ?
? ? ?
cbb 7 c (ac) 7
33 3 3
 ? 2c + a = 7
? 2c + 4 = 7
? c = 
3
2
So from (i) 
? ? ?
11 11
2b b
24
So the Q.E. if 
???
2
11
4x x 1 0
4
? 16x
2
 + 11x + 4 = 0 ??
?
?? ? ? ?? ?
11 1
,
16 4
Now, ??? ???
22
 = ??? ? ??? ??
22
23
= 
?? ?? ? ? ??
2
3
= 
???
?
??
??
2
11 3
16 4
= 
???
121 3 71
256 4 256
9. A possible value of 
?
??
??
??
1
163
tan sin
48
 is :
(1)
? 22 1
(2)
? 71
(3)
1
7
(4)
1
22
Answer (3)
Sol. 
?
??
? ??
? ? ?? ??
?? ??
1
163
tan sin tan ?
48 4
Let
?
?? ?? ? ?? ? ??
1
63 1 63
sin and cos sin tan 63 and 6
88 8
?
?? ?
?? ? ? ? ?
22 2
19 9
2cos 1 2cos cos
28 28 216
?
?
?
3
cos
24
2   2
7
1
?
?
??
2
3
2cos 1
44
? 
?
?
2
7
2cos
44
? 
??
? ? ?
2
77
cos cos
48 4 22
? 
?
?
1
tan
4 7
10. Let ? f:R R be defined as
?
???
?
?? ? ??? ?
?
?? ? ? ?
?
32
32
55x, if x 5
f(x) 2x 3x 120x, if 5 x 4
2x 3x 36x 336, ifx 4,
Let A = {x ? R : f is increasing}. Then A is equal
to:
(1) (–5, ? ) (2) ?? ? ? ? ( ,5)(4, )
(3) ?? ? ? ( 5, 4)(4, ) (4) ?? ? ? ? ? ( ,5)( 4, )
Answer (3)
Sol.
?
???
?
?
? ? ? ??? ?
?
?? ? ? ?
?
32
32
55x , x 5
f(x) 2x 3x 120x , 5 x 4
2x 3x 36x 336 , x 4
Now, 
?
???
?
?
?? ? ? ? ? ? ?
?
?? ? ?
?
2
2
55 , x 5
f(x) 6x 6x 120 , 5 x 4
6x 6x 36 , x 4
?
??? ?
?
?? ? ? ? ? ?
?
?
?? ?
?
55 , x 5
f(x) 6(x5)(x4) ,5 x 4
6(x 3)(x 2) , x 4
For increasing f ?(x) > 0
So clearly f(x) is increasing for ?? ? ? ? x ( 5, 4)(4, )
11. The value of the integral,  ??
?
3
2
1
[x 2x 2]dx,
where [x] denotes the greatest integer less
than or equal to x, is:
(1) –5 (2)
?? ? 231
(3)
?? ? 231
(4) –4
Answer (3)
Sol. I = 
??
?
3
2
1
[x 2x 2]dx
 
??
??
?? ??
???? ? ?
?? ??
?? ??
?? ?
33 3
2
2
11 1
x2x1 3dx x1 dx–3dx
Now when ? ? ? x1 , 3
Page 4


JEE (MAIN)-2021 Phase-1 (24-02-2021)-E
SECTION - I
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4
choices (1), (2), (3) and (4), out of which ONLY ONE
is correct.
Choose the correct answer :
1. 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) 3600 3 m (2) 2400 3 m
(3) 1800 3 m (4) 1200 3 m
Answer (4)
Sol.
d x
60° 30°
d
h
A
Given ??
?
h
tan30
xd
 and ??
h
tan60
x
? ?? xd 3h and ?
h
x
3
?
??
??
??
??
12h
d3– h
33
Given 
?
?
d 432 1000
20 3600
? d = 2400 m
? ? ? ?
2h
2400 h 1200 3 m
3
2. The probability that two randomly selected
subsets of the set {1, 2, 3, 4, 5} have exactly
two elements in their intersection, is :
(1)
7
65
2
(2)
9
135
2
(3)
8
65
2
(4)
7
35
2
Answer (2)
PART–C : MATHEMATICS
Sol. Number of ways of selecting elements common
to both A and B = 
5
C
2
? Required probability 
?
??
53
2
59
C3 135
42
3. If P is a point on the parabola y = x
2
 + 4 which
is closest to the straight line y = 4x – 1, then the
co-ordinates of P are :
(1) (–2, 8) (2) (1, 5)
(3) (3, 13) (4) (2, 8)
Answer (4)
Sol. Closest point will be point of tangency of
tangent of same slope i.e. 4
Let equation of tangent y = 4x + c
? 4x + c = x
2
 + 4 have D = 0
i.e. x
2
 – 4x + (4 – c) = 0
D = 0 ???  16 – 4(4 – c) = 0 ???  c = 0
Tangent is y = 4x gives x = 2 and y = 8 as point
of tangency
? Nearest point (2, 8)
4. If ? n2 is a positive integer, then the sum of
the series
??
?
?? ? ? ?
n1 2 3 4 n
22 2 2 2
C 2 C C C ..... C
 is :
(1)
?? n(2n 1)(3n 1)
6
(2)
? n(n–1)(2n 1)
6
(3)
??
2
n(n 1) (n 2)
12
(4)
?? n(n 1)(2n 1)
6
Answer (4)
Sol. Sum of 
2
C
2
 + 
3
C
2
 + .... + 
n
C
2
 is coefficient of
x
2
 in (1 + x)
2
 + (1 + x)
3
 + .... + (1 + x)
n
i.e. coefficient of x
2
 in
??
??
? ?
?
?
??
?
n–1
2
n1
3
1x –1
1x C
(1 x–1)
Hence required sum = 
n+1
C
2
 + 2 ?
n+1
C
3
??
??
(n 1)(n) 2(n 1)n(n–1)
26
?? ? ?
??
n(n 1)(3 2n–2) n(n 1)(2n 1)
23 6
JEE (MAIN)-2021 Phase-1 (24-02-2021)-E
5. Let f(x) be a differentiable function defined on
[0, 2] such that f ?(x) = f ?(2 – x) for all ? x (0, 2) ,
f(0) = 1 and f(2) = e
2
. Then the value of 
?
2
0
f(x)dx
is :
(1) 2(1 + e
2
) (2) 1 + e
2
(3) 2(1 – e
2
) (4) 1 – e
2
Answer (2)
Sol. Given f ?(x) = f ?(2 – x)
? f ?(x) – f ?(2 – x) = 0
Integrating both sides, we get
f(x) + f(2 – x) = c ...(i)
Put x = 0, we get
c = f(0) + f(2) = 1 + e
2
Integrating 0 to 2 equation (i) both sides, we get
????
??
22
2
2
0
00
f(x)dx f(2–x)dx (1e ) 8
Also ?
??
22
00
f(x)dx f(2–x)dx
Hence ??
?
2
2
0
2f(x)dx 2(1e )
? ??
?
2
2
0
f(x)dx 1 e
6. For the statements p and q, consider the
following compound statements:
(a)?? ?? ? ~q (p q) ~p
(b) ?? ? ((p q) ~p) q
Then which of the following statements is
correct?
(1) (a) and (b) both are tautologies.
(2) (a) is a tautology but not (b).
(3) (b) is a tautology bot not (a).
(4) (a) and (b) both are not tautologies.
Answer (1)
Sol. Truth table for required statements
??
?
?? ? ?
?? ?
??
~
~
~~ ~
~
~
(q ((p q)
(q) (p q)
p q p q p q (p q)) p q p)
(p q) p
pq
TT F F T F T T F T
TF F T F F T T F T
FT T F T F T T T T
F F T T T T T (a) is tautology F F T (b) is tautology
7. For the system of linear equations :
x – 2y = 1, x – y + kz = –2, ky + 4z = 6, k ? R,
consider the following statements :
(A)The system has unique solution if
??? k2,k 2.
(B) The system has unique solution if k = – 2.
(C) The system has unique solution if k = 2
(D) The system has no solution if k = 2.
(E) The system has infinite number of solutions
if k ? – 2.
Which of the following statements are correct?
(1) (A) and (E) only (2) (A) and (D) only
(3) (B) and (E) only (4) (C) and (D) only
Answer (2)
Sol. Using Cramer’s Rule, we have
?
?? ? ? ? ? ? ? ?
22
120
11k 1(4k ) 2(4) 4k
0k 4
?
?? ? ? ? ? ? ? ? ? ? ? ?
22
x
120
21k 1(4k ) 2( 2k 6) 84k k
6k 4
Now, ? = 0 if k = ±2
if k = –2, ? = 0 and  ?
x
 ? 0
Hence no solution
Also if k = 2, ? = 0 and  ?
x
 ? 0
Now
?? ? ? ? ? ? ?? ? ?
y
11 0
12k 1(86k) 1(4) 6k 12 0
06 4
Hence, the system has no solution if k = ±2
and unique solution if k ? ±2
8. Let a, b, c be in arithmetic progression. Let the
centroid of the triangle with vertices
(a, c), (2, b) and (a, b) be 
??
??
??
10 7
,
33
. If ?, ? are
the roots of the equation ax
2
 + bx + 1 = 0, then
the value of ?
2
 + ?
2
 – ?? is :
(1)
69
256
(2) –
71
256
(3) –
69
256
(4)
71
256
Answer (2)
JEE (MAIN)-2021 Phase-1 (24-02-2021)-E
Sol. Here, 2b = a + c ...(i)
and centroid of ? is 
??
??
??
10 7
,
33
?
??
? ? ?
a2 a 10
a4
33
and 
?? ? ?
? ? ?
cbb 7 c (ac) 7
33 3 3
 ? 2c + a = 7
? 2c + 4 = 7
? c = 
3
2
So from (i) 
? ? ?
11 11
2b b
24
So the Q.E. if 
???
2
11
4x x 1 0
4
? 16x
2
 + 11x + 4 = 0 ??
?
?? ? ? ?? ?
11 1
,
16 4
Now, ??? ???
22
 = ??? ? ??? ??
22
23
= 
?? ?? ? ? ??
2
3
= 
???
?
??
??
2
11 3
16 4
= 
???
121 3 71
256 4 256
9. A possible value of 
?
??
??
??
1
163
tan sin
48
 is :
(1)
? 22 1
(2)
? 71
(3)
1
7
(4)
1
22
Answer (3)
Sol. 
?
??
? ??
? ? ?? ??
?? ??
1
163
tan sin tan ?
48 4
Let
?
?? ?? ? ?? ? ??
1
63 1 63
sin and cos sin tan 63 and 6
88 8
?
?? ?
?? ? ? ? ?
22 2
19 9
2cos 1 2cos cos
28 28 216
?
?
?
3
cos
24
2   2
7
1
?
?
??
2
3
2cos 1
44
? 
?
?
2
7
2cos
44
? 
??
? ? ?
2
77
cos cos
48 4 22
? 
?
?
1
tan
4 7
10. Let ? f:R R be defined as
?
???
?
?? ? ??? ?
?
?? ? ? ?
?
32
32
55x, if x 5
f(x) 2x 3x 120x, if 5 x 4
2x 3x 36x 336, ifx 4,
Let A = {x ? R : f is increasing}. Then A is equal
to:
(1) (–5, ? ) (2) ?? ? ? ? ( ,5)(4, )
(3) ?? ? ? ( 5, 4)(4, ) (4) ?? ? ? ? ? ( ,5)( 4, )
Answer (3)
Sol.
?
???
?
?
? ? ? ??? ?
?
?? ? ? ?
?
32
32
55x , x 5
f(x) 2x 3x 120x , 5 x 4
2x 3x 36x 336 , x 4
Now, 
?
???
?
?
?? ? ? ? ? ? ?
?
?? ? ?
?
2
2
55 , x 5
f(x) 6x 6x 120 , 5 x 4
6x 6x 36 , x 4
?
??? ?
?
?? ? ? ? ? ?
?
?
?? ?
?
55 , x 5
f(x) 6(x5)(x4) ,5 x 4
6(x 3)(x 2) , x 4
For increasing f ?(x) > 0
So clearly f(x) is increasing for ?? ? ? ? x ( 5, 4)(4, )
11. The value of the integral,  ??
?
3
2
1
[x 2x 2]dx,
where [x] denotes the greatest integer less
than or equal to x, is:
(1) –5 (2)
?? ? 231
(3)
?? ? 231
(4) –4
Answer (3)
Sol. I = 
??
?
3
2
1
[x 2x 2]dx
 
??
??
?? ??
???? ? ?
?? ??
?? ??
?? ?
33 3
2
2
11 1
x2x1 3dx x1 dx–3dx
Now when ? ? ? x1 , 3
JEE (MAIN)-2021 Phase-1 (24-02-2021)-E
we see that (x – 1)
2
 ? [0, 4]
So, I = 
?? ??
?? ??
?? ?? ?
?? ??
?? ??
?? ?
12 3
22
01 2
0dxx1dxx1dx
??
??
??
??
??
??
23
2
1 3
x1 dx 3dx
= 
?? ? ?
?? ?
23 2
1 23
0 1dx 2dx 3dx 6
= ?? ?? ?? ?? ?
23 2
12 3
0x 2x 3x
= 
?? ? ? ? ? 2 1 23 22 6 3 3 6
= 
?? ? 321
= 
?? ? 231
12. The negation of the statement
~ p ? (p ? q) is :
(1) p ? ~ q (2) ~ p ??q
(3) ~ p ? q (4) p ??~ q
Answer (4)
Sol. ?? ??
?? ~~pp q = ?? p ~(p q)
= ?? p(~p ~q)
= ?? ? (p ~p) (p ~q)
= ? p ~q
13. If the curve y = ax
2
 + bx + c, x ? R, passes
through the point (1, 2) and the tangent line to
this curve at origin is y = x, then the possible
values of a, b, c are :
(1) ??
11
a, b, c = 1
22
(2) a = 1, b = 0, c = 1
(3) a = 1, b = 1, c = 0
(4) a = –1, b = 1, c = 1
Answer (3)
Sol. y = ax
2
 + bx + c passes through (1, 2)
So a + b + c = 2 ...(1)
also (0, 0) satisfies ? c = 0 ...(2)
also slope of tangent at origin is 1 i.e.
y ? = 2ax + b ? b = 1 and a = 1
? a = 1 = b, c = 0
14. For which of the following curves, the line
?? x3y 23 is the tangent at the point
??
??
??
??
33 1
,
22
?
(1) x
2
 + 9y
2
 = 9 (2)
?
2
1
yx
63
(3) 2x
2
 – 18y
2
 = 9 (4) x
2
 + y
2
 = 7
Answer (1)
Sol. ?? x3y23
?
?
t
1
m
3
 and point of tangency 
??
??
??
??
33 1
,
22
Option (1) ??
22
x9y 9 ? 2x + 18yy’
?
= 0
?
??
??
??
??
t
33 1
m,
22
i.e. y ? = 
?
??
??
??
??
??
3
3
x1
2
1 9y
3
9
2
Option (2)
?
x
y'
63
?
?
1
y'
12 3y
i.e. 
?
t
1
m
63
Option (3) 
??
22
2x 18y 9
? 4x – 36 yy ? = 0
i.e. ?? ?
t
x33 1
m
1
9y
3
2.9
2
Option (4) ?
22
x y 7 ?
??
x
y'
y
i.e. ??
t
m33
Hence only option (1) is correct
15. If a curve y = f(x) passes through the point
(1, 2) and satisfies ??
4
dy
xybx
dx
, then for what
value of b, ?? ?
?
2
1
62
fxdx ?
5
(1)
31
5
(2)
62
5
(3) 10 (4) 5
Answer (3)
Page 5


JEE (MAIN)-2021 Phase-1 (24-02-2021)-E
SECTION - I
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4
choices (1), (2), (3) and (4), out of which ONLY ONE
is correct.
Choose the correct answer :
1. 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) 3600 3 m (2) 2400 3 m
(3) 1800 3 m (4) 1200 3 m
Answer (4)
Sol.
d x
60° 30°
d
h
A
Given ??
?
h
tan30
xd
 and ??
h
tan60
x
? ?? xd 3h and ?
h
x
3
?
??
??
??
??
12h
d3– h
33
Given 
?
?
d 432 1000
20 3600
? d = 2400 m
? ? ? ?
2h
2400 h 1200 3 m
3
2. The probability that two randomly selected
subsets of the set {1, 2, 3, 4, 5} have exactly
two elements in their intersection, is :
(1)
7
65
2
(2)
9
135
2
(3)
8
65
2
(4)
7
35
2
Answer (2)
PART–C : MATHEMATICS
Sol. Number of ways of selecting elements common
to both A and B = 
5
C
2
? Required probability 
?
??
53
2
59
C3 135
42
3. If P is a point on the parabola y = x
2
 + 4 which
is closest to the straight line y = 4x – 1, then the
co-ordinates of P are :
(1) (–2, 8) (2) (1, 5)
(3) (3, 13) (4) (2, 8)
Answer (4)
Sol. Closest point will be point of tangency of
tangent of same slope i.e. 4
Let equation of tangent y = 4x + c
? 4x + c = x
2
 + 4 have D = 0
i.e. x
2
 – 4x + (4 – c) = 0
D = 0 ???  16 – 4(4 – c) = 0 ???  c = 0
Tangent is y = 4x gives x = 2 and y = 8 as point
of tangency
? Nearest point (2, 8)
4. If ? n2 is a positive integer, then the sum of
the series
??
?
?? ? ? ?
n1 2 3 4 n
22 2 2 2
C 2 C C C ..... C
 is :
(1)
?? n(2n 1)(3n 1)
6
(2)
? n(n–1)(2n 1)
6
(3)
??
2
n(n 1) (n 2)
12
(4)
?? n(n 1)(2n 1)
6
Answer (4)
Sol. Sum of 
2
C
2
 + 
3
C
2
 + .... + 
n
C
2
 is coefficient of
x
2
 in (1 + x)
2
 + (1 + x)
3
 + .... + (1 + x)
n
i.e. coefficient of x
2
 in
??
??
? ?
?
?
??
?
n–1
2
n1
3
1x –1
1x C
(1 x–1)
Hence required sum = 
n+1
C
2
 + 2 ?
n+1
C
3
??
??
(n 1)(n) 2(n 1)n(n–1)
26
?? ? ?
??
n(n 1)(3 2n–2) n(n 1)(2n 1)
23 6
JEE (MAIN)-2021 Phase-1 (24-02-2021)-E
5. Let f(x) be a differentiable function defined on
[0, 2] such that f ?(x) = f ?(2 – x) for all ? x (0, 2) ,
f(0) = 1 and f(2) = e
2
. Then the value of 
?
2
0
f(x)dx
is :
(1) 2(1 + e
2
) (2) 1 + e
2
(3) 2(1 – e
2
) (4) 1 – e
2
Answer (2)
Sol. Given f ?(x) = f ?(2 – x)
? f ?(x) – f ?(2 – x) = 0
Integrating both sides, we get
f(x) + f(2 – x) = c ...(i)
Put x = 0, we get
c = f(0) + f(2) = 1 + e
2
Integrating 0 to 2 equation (i) both sides, we get
????
??
22
2
2
0
00
f(x)dx f(2–x)dx (1e ) 8
Also ?
??
22
00
f(x)dx f(2–x)dx
Hence ??
?
2
2
0
2f(x)dx 2(1e )
? ??
?
2
2
0
f(x)dx 1 e
6. For the statements p and q, consider the
following compound statements:
(a)?? ?? ? ~q (p q) ~p
(b) ?? ? ((p q) ~p) q
Then which of the following statements is
correct?
(1) (a) and (b) both are tautologies.
(2) (a) is a tautology but not (b).
(3) (b) is a tautology bot not (a).
(4) (a) and (b) both are not tautologies.
Answer (1)
Sol. Truth table for required statements
??
?
?? ? ?
?? ?
??
~
~
~~ ~
~
~
(q ((p q)
(q) (p q)
p q p q p q (p q)) p q p)
(p q) p
pq
TT F F T F T T F T
TF F T F F T T F T
FT T F T F T T T T
F F T T T T T (a) is tautology F F T (b) is tautology
7. For the system of linear equations :
x – 2y = 1, x – y + kz = –2, ky + 4z = 6, k ? R,
consider the following statements :
(A)The system has unique solution if
??? k2,k 2.
(B) The system has unique solution if k = – 2.
(C) The system has unique solution if k = 2
(D) The system has no solution if k = 2.
(E) The system has infinite number of solutions
if k ? – 2.
Which of the following statements are correct?
(1) (A) and (E) only (2) (A) and (D) only
(3) (B) and (E) only (4) (C) and (D) only
Answer (2)
Sol. Using Cramer’s Rule, we have
?
?? ? ? ? ? ? ? ?
22
120
11k 1(4k ) 2(4) 4k
0k 4
?
?? ? ? ? ? ? ? ? ? ? ? ?
22
x
120
21k 1(4k ) 2( 2k 6) 84k k
6k 4
Now, ? = 0 if k = ±2
if k = –2, ? = 0 and  ?
x
 ? 0
Hence no solution
Also if k = 2, ? = 0 and  ?
x
 ? 0
Now
?? ? ? ? ? ? ?? ? ?
y
11 0
12k 1(86k) 1(4) 6k 12 0
06 4
Hence, the system has no solution if k = ±2
and unique solution if k ? ±2
8. Let a, b, c be in arithmetic progression. Let the
centroid of the triangle with vertices
(a, c), (2, b) and (a, b) be 
??
??
??
10 7
,
33
. If ?, ? are
the roots of the equation ax
2
 + bx + 1 = 0, then
the value of ?
2
 + ?
2
 – ?? is :
(1)
69
256
(2) –
71
256
(3) –
69
256
(4)
71
256
Answer (2)
JEE (MAIN)-2021 Phase-1 (24-02-2021)-E
Sol. Here, 2b = a + c ...(i)
and centroid of ? is 
??
??
??
10 7
,
33
?
??
? ? ?
a2 a 10
a4
33
and 
?? ? ?
? ? ?
cbb 7 c (ac) 7
33 3 3
 ? 2c + a = 7
? 2c + 4 = 7
? c = 
3
2
So from (i) 
? ? ?
11 11
2b b
24
So the Q.E. if 
???
2
11
4x x 1 0
4
? 16x
2
 + 11x + 4 = 0 ??
?
?? ? ? ?? ?
11 1
,
16 4
Now, ??? ???
22
 = ??? ? ??? ??
22
23
= 
?? ?? ? ? ??
2
3
= 
???
?
??
??
2
11 3
16 4
= 
???
121 3 71
256 4 256
9. A possible value of 
?
??
??
??
1
163
tan sin
48
 is :
(1)
? 22 1
(2)
? 71
(3)
1
7
(4)
1
22
Answer (3)
Sol. 
?
??
? ??
? ? ?? ??
?? ??
1
163
tan sin tan ?
48 4
Let
?
?? ?? ? ?? ? ??
1
63 1 63
sin and cos sin tan 63 and 6
88 8
?
?? ?
?? ? ? ? ?
22 2
19 9
2cos 1 2cos cos
28 28 216
?
?
?
3
cos
24
2   2
7
1
?
?
??
2
3
2cos 1
44
? 
?
?
2
7
2cos
44
? 
??
? ? ?
2
77
cos cos
48 4 22
? 
?
?
1
tan
4 7
10. Let ? f:R R be defined as
?
???
?
?? ? ??? ?
?
?? ? ? ?
?
32
32
55x, if x 5
f(x) 2x 3x 120x, if 5 x 4
2x 3x 36x 336, ifx 4,
Let A = {x ? R : f is increasing}. Then A is equal
to:
(1) (–5, ? ) (2) ?? ? ? ? ( ,5)(4, )
(3) ?? ? ? ( 5, 4)(4, ) (4) ?? ? ? ? ? ( ,5)( 4, )
Answer (3)
Sol.
?
???
?
?
? ? ? ??? ?
?
?? ? ? ?
?
32
32
55x , x 5
f(x) 2x 3x 120x , 5 x 4
2x 3x 36x 336 , x 4
Now, 
?
???
?
?
?? ? ? ? ? ? ?
?
?? ? ?
?
2
2
55 , x 5
f(x) 6x 6x 120 , 5 x 4
6x 6x 36 , x 4
?
??? ?
?
?? ? ? ? ? ?
?
?
?? ?
?
55 , x 5
f(x) 6(x5)(x4) ,5 x 4
6(x 3)(x 2) , x 4
For increasing f ?(x) > 0
So clearly f(x) is increasing for ?? ? ? ? x ( 5, 4)(4, )
11. The value of the integral,  ??
?
3
2
1
[x 2x 2]dx,
where [x] denotes the greatest integer less
than or equal to x, is:
(1) –5 (2)
?? ? 231
(3)
?? ? 231
(4) –4
Answer (3)
Sol. I = 
??
?
3
2
1
[x 2x 2]dx
 
??
??
?? ??
???? ? ?
?? ??
?? ??
?? ?
33 3
2
2
11 1
x2x1 3dx x1 dx–3dx
Now when ? ? ? x1 , 3
JEE (MAIN)-2021 Phase-1 (24-02-2021)-E
we see that (x – 1)
2
 ? [0, 4]
So, I = 
?? ??
?? ??
?? ?? ?
?? ??
?? ??
?? ?
12 3
22
01 2
0dxx1dxx1dx
??
??
??
??
??
??
23
2
1 3
x1 dx 3dx
= 
?? ? ?
?? ?
23 2
1 23
0 1dx 2dx 3dx 6
= ?? ?? ?? ?? ?
23 2
12 3
0x 2x 3x
= 
?? ? ? ? ? 2 1 23 22 6 3 3 6
= 
?? ? 321
= 
?? ? 231
12. The negation of the statement
~ p ? (p ? q) is :
(1) p ? ~ q (2) ~ p ??q
(3) ~ p ? q (4) p ??~ q
Answer (4)
Sol. ?? ??
?? ~~pp q = ?? p ~(p q)
= ?? p(~p ~q)
= ?? ? (p ~p) (p ~q)
= ? p ~q
13. If the curve y = ax
2
 + bx + c, x ? R, passes
through the point (1, 2) and the tangent line to
this curve at origin is y = x, then the possible
values of a, b, c are :
(1) ??
11
a, b, c = 1
22
(2) a = 1, b = 0, c = 1
(3) a = 1, b = 1, c = 0
(4) a = –1, b = 1, c = 1
Answer (3)
Sol. y = ax
2
 + bx + c passes through (1, 2)
So a + b + c = 2 ...(1)
also (0, 0) satisfies ? c = 0 ...(2)
also slope of tangent at origin is 1 i.e.
y ? = 2ax + b ? b = 1 and a = 1
? a = 1 = b, c = 0
14. For which of the following curves, the line
?? x3y 23 is the tangent at the point
??
??
??
??
33 1
,
22
?
(1) x
2
 + 9y
2
 = 9 (2)
?
2
1
yx
63
(3) 2x
2
 – 18y
2
 = 9 (4) x
2
 + y
2
 = 7
Answer (1)
Sol. ?? x3y23
?
?
t
1
m
3
 and point of tangency 
??
??
??
??
33 1
,
22
Option (1) ??
22
x9y 9 ? 2x + 18yy’
?
= 0
?
??
??
??
??
t
33 1
m,
22
i.e. y ? = 
?
??
??
??
??
??
3
3
x1
2
1 9y
3
9
2
Option (2)
?
x
y'
63
?
?
1
y'
12 3y
i.e. 
?
t
1
m
63
Option (3) 
??
22
2x 18y 9
? 4x – 36 yy ? = 0
i.e. ?? ?
t
x33 1
m
1
9y
3
2.9
2
Option (4) ?
22
x y 7 ?
??
x
y'
y
i.e. ??
t
m33
Hence only option (1) is correct
15. If a curve y = f(x) passes through the point
(1, 2) and satisfies ??
4
dy
xybx
dx
, then for what
value of b, ?? ?
?
2
1
62
fxdx ?
5
(1)
31
5
(2)
62
5
(3) 10 (4) 5
Answer (3)
JEE (MAIN)-2021 Phase-1 (24-02-2021)-E
Sol.
??
4
dy
xybx
dx
?
??
3
dy y
bx
dx x
I.F = 
?
1
dx
x
e
 = e
ln x
 = x
? ?
?
4
yx bx dx
? ??
5
bx
xy c
5
   ? (1 , 2)
?
??
b
2c
5
 ??
??
b
c2
5
?
??
?? ?
??
??
4
bx 1 b
y2
5x 5
??
???
??
??
?
2
2
5
2
1
1 1
bx b
f(x)dx 2 ln x
25 5
??
?? ?
??
??
31b b 62
2ln2
25 5 5
?
?? ??
???
?? ??
?? ??
bb31
2ln2 2
555
?
??
b
20
5
 ??b = 10
16. The area of the region : R = {(x, y) : 5x
2
 ? y
? 2x
2
 + 9} is :
(1)
63
 square units
(2) 11 3 square units
(3) 12 3 square units
(4)
93
 square units
Answer (3)
Sol.
y
Required Area
y = 2x + 9
2 
??
? 3,15
??
3,15
y = 5x
2
x
Required Area = ??? ? ? ?
??
?
3
22
0
22x 9 5x dx
??
?? ??
??
??
3
3
0
3x
29x
3
??
?? ? 29 3 3 3 12 3sq.units
17. The vector equation of the plane passing
through the intersection of the planes
?? ??
?? ? ? ??

ˆ ˆˆ ˆ ˆ
r· i j k1andr·i 2j 2, and the point
(1, 0, 2) is :
(1)??
?? ?

7
ˆ ˆˆ
r· i 7j 3k
3
(2)??
?? ?

ˆ ˆˆ
r· 3i 7j 3k 7
(3)??
?? ?

7
ˆ ˆˆ
r· i 7j 3k
3
(4)??
?? ?

ˆ ˆˆ
r· i 7j 3k 7
Answer (4)
Sol. P
1
 + ? P
2
 = 0
? ?? ?? ?? ??
?? ? ? ?? ?? ? ?

ˆ ˆˆ ˆ ˆ
ri jk1 r i 2j 20
...(1)
??
??
ˆ ˆ
i2k
? (1 + 2 – 1) + ? (1 + 2) = 0 
? ?? ?
2
3
...(2)
by (1) and (2)
??
??
?? ? ? ? ? ? ?
??
??

24
ˆ ˆˆ ˆ ˆ
ri j k i 2j 1 0
33
?
??
?? ? ?
??
??
??
 ˆ
i7 7
ˆ ˆ
rjk
33 3
???
?? ? ?

ˆ ˆˆ
ri 7j 3k 7
18. 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
(A
2
B
2
 – B
2
A
2
) X = O, where X is a 3 × 1 column
matrix of unknown variables and O is a 3 × 1
null matrix, has :
(1) exactly two solutions
(2) infinitely many solutions
(3) no solution
(4) a unique solution
Answer (2)
Sol. Let C = A
2
B
2
 – B
2
A
2
Then C
T
 = (A
2
B
2
 – B
2
A
2
)
T
= (B
T
)
2
.(A
T
)
2
 – (A
T
)
2
.(B
T
)
2
= (–B)
2
A
2
 – A
2
.(–B)
2
{? A
T
 = A and B
T
 = –B}
= (B)
2
A
2
 – A
2
B
2
? C + C
T
 = 0
? C is a skew symmetric odd order matrix
? |C| = |A
2
B
2
 – B
2
A
2
| = 0
? Equation (A
2
B
2
 – B
2
A
2
) X = 0 has
Infinite many solution