JEE Main 2021 Mathematics February 25 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


JEE (MAIN)-2021 Phase-1 (25-02-2021)-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. If 
?
?
?
?
2
n
n
4
Icotxdx , then :
(1)
?? ?
24 3 5 4 6
111
,,
II I I I I
 are in G.P.
(2) ?? ?
243 54 6
II,I I,I I are in A.P.
(3)
?? ?
2
24 3 5 4 6
II,(II ) ,I I
are in G.P.
(4)
?? ?
24 3 5 4 6
111
,,
II I I I I
are in A.P.
Answer (4)
Sol.
??
??
?
??
??
??
22
nn2 2
n
44
I cot xdx cot x cot x dx
??
?
?
?
??
?
2
n2 2
4
cot x cosec x 1 dx
??
??
??
??
??
22
n2 2 n 2
44
cot xcosec xdx cot xdx
?
?
??
?
?
?? ? ? ?
?
?? ?
n1
2
n2 n2
4
cot x 1
II
n1 n1
?
?
?? ? ??
??
nn2
n2 n
11
II n1
n1 I I
 = a linear
expression in n.
? Sequence 
?
?
n2 n
1
II
 is an A.P.
PART–C : MATHEMATICS
2. Let ? and ? be the roots of ?? ?
2
x6x2 0 . If
?? ? ?
nn
n
a for ? n1, then the value of
10 8
9
a2a
3a
?
 is :
(1) 2 (2) 4
(3) 3 (4) 1
Answer (1)
Sol. ?, ? are roots of x
2
 – 6x – 2 = 0
??
2
 – 6 ? – 2 = 0
? ? ? ? ? ? ? ? ? ?
2
 – 2 = 6 ?
Similarly ?
2
 – 2 = 6 ?
??
??
? ??? ? ? ??
?
???
10 10 8 8
10 8
99
9
a2a 2
3a 3
??
??
??
?? ? ? ? ? ?
?
???
10 8 10 8
99
22
3
??
??
??
??
??
??
???? ? ?? ? ?? ? ?
??
??? ? ??
88 82 8 2
99 99
66 22
33
??
??
???
??
???
99
99
6
2
3
3. A hyperbola passes through the foci of the
ellipse ??
22
xy
1
25 16
 and its transverse and
conjugate axes coincide with major and minor
axes of the ellipse, respectively. If the product
of their eccentricities is one, then the
equation of the hyperbola is :
(1) ??
22
xy
1
916
(2) ??
22
x y 9
(3) ??
22
xy
1
94
(4) ??
22
xy
1
925
Answer (1)
Sol. Eccentricity of Ellipse ?? ?
1
16 3
e1
25 5
Foci = ( ?ae, 0) = ( ?3, 0)
For Hyperbola
Eccentricity ?
2
5
e
3
Page 2


JEE (MAIN)-2021 Phase-1 (25-02-2021)-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. If 
?
?
?
?
2
n
n
4
Icotxdx , then :
(1)
?? ?
24 3 5 4 6
111
,,
II I I I I
 are in G.P.
(2) ?? ?
243 54 6
II,I I,I I are in A.P.
(3)
?? ?
2
24 3 5 4 6
II,(II ) ,I I
are in G.P.
(4)
?? ?
24 3 5 4 6
111
,,
II I I I I
are in A.P.
Answer (4)
Sol.
??
??
?
??
??
??
22
nn2 2
n
44
I cot xdx cot x cot x dx
??
?
?
?
??
?
2
n2 2
4
cot x cosec x 1 dx
??
??
??
??
??
22
n2 2 n 2
44
cot xcosec xdx cot xdx
?
?
??
?
?
?? ? ? ?
?
?? ?
n1
2
n2 n2
4
cot x 1
II
n1 n1
?
?
?? ? ??
??
nn2
n2 n
11
II n1
n1 I I
 = a linear
expression in n.
? Sequence 
?
?
n2 n
1
II
 is an A.P.
PART–C : MATHEMATICS
2. Let ? and ? be the roots of ?? ?
2
x6x2 0 . If
?? ? ?
nn
n
a for ? n1, then the value of
10 8
9
a2a
3a
?
 is :
(1) 2 (2) 4
(3) 3 (4) 1
Answer (1)
Sol. ?, ? are roots of x
2
 – 6x – 2 = 0
??
2
 – 6 ? – 2 = 0
? ? ? ? ? ? ? ? ? ?
2
 – 2 = 6 ?
Similarly ?
2
 – 2 = 6 ?
??
??
? ??? ? ? ??
?
???
10 10 8 8
10 8
99
9
a2a 2
3a 3
??
??
??
?? ? ? ? ? ?
?
???
10 8 10 8
99
22
3
??
??
??
??
??
??
???? ? ?? ? ?? ? ?
??
??? ? ??
88 82 8 2
99 99
66 22
33
??
??
???
??
???
99
99
6
2
3
3. A hyperbola passes through the foci of the
ellipse ??
22
xy
1
25 16
 and its transverse and
conjugate axes coincide with major and minor
axes of the ellipse, respectively. If the product
of their eccentricities is one, then the
equation of the hyperbola is :
(1) ??
22
xy
1
916
(2) ??
22
x y 9
(3) ??
22
xy
1
94
(4) ??
22
xy
1
925
Answer (1)
Sol. Eccentricity of Ellipse ?? ?
1
16 3
e1
25 5
Foci = ( ?ae, 0) = ( ?3, 0)
For Hyperbola
Eccentricity ?
2
5
e
3
JEE (MAIN)-2021 Phase-1 (25-02-2021)-E
Semi-transverse axis ? a = 3
??
??? ??
??
??
22 2
25
ba(e 1)9 1 16
9
Equation of Hyperbola
??
22
xy
1
916
4. Let x denote the total number of one-one
functions from a set A with 3 elements to a set
B with 5 elements and y denote the total
number of one-one functions from the set A to
the set A × B. Then :
(1) ? 2y 273x (2) ? 2y 91x
(3) ? y 273x (4) ? y 91x
Answer (2)
Sol. n(A) = 3, n(B) = 5
x = 
5
C
3
 × 3! = 5 × 4 × 3
n(A × B) = 15
y = 
15
C
3
 × 3! = 15 × 14 × 13
??
??
??
y151413 91
x543 2
2y = 91x
5. A function f(x) is given by ?
?
x
x
5
f(x)
55
, then
the sum of the series
?? ?? ?? ??
????
?? ?? ?? ??
?? ?? ?? ??
12 3 39
fff ....f
20 20 20 20
 is equal to:
(1)
19
2
(2)
29
2
(3)
49
2
(4)
39
2
Answer (4)
Sol.
??
?
?
?? ?
??
2x
2x x
55
f2 x
5555
So f(x) + f(2 – x) = 1
??
?? ?? ?? ? ?
??? ?
?? ?? ? ? ??
?? ?? ? ? ??
??
39 19
r1 r1
rr r
ff f2 f(1)
20 20 20
?? ?
139
19
22
6. The minimum value of 
?
??
xx
a1a
f(x) a a ,
where a, ? xR and a > 0, is equal to :
(1) a + 1 (2) 2a
(3) ?
1
a
a
(4) 2a
Answer (2)
Sol. ??
x
x
a
a
a
f(x) a
a
?
?
?
x
x
x
x
a
a
a
a
a
a
a
a
a.
2
a
? ? f(x) 2a
?
min
f(x) 2 a
7. If the curve ??
22
x2y 2 intersects the line
?? x y 1 at two points P and Q, then the angle
subtended by the line segment PQ at the
origin is :
(1)
?
? ??
?
??
??
1
1
tan
24
(2)
?
? ??
?
??
??
1
1
tan
23
(3)
?
? ??
?
??
??
1
1
tan
23
(4)
?
? ??
?
??
??
1
1
tan
24
Answer (4)
Sol. y = 1 – x ...(i)
x
2
 + 2y
2
 = 2 ...(ii)
? x
2
 + 2 (1 – x)
2
 =2
3x
2
 – 4x = 0
B(0, 1)
(0, 0)
0
?
A
?
4
x0,
3
?
?
1
y1,
3
Page 3


JEE (MAIN)-2021 Phase-1 (25-02-2021)-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. If 
?
?
?
?
2
n
n
4
Icotxdx , then :
(1)
?? ?
24 3 5 4 6
111
,,
II I I I I
 are in G.P.
(2) ?? ?
243 54 6
II,I I,I I are in A.P.
(3)
?? ?
2
24 3 5 4 6
II,(II ) ,I I
are in G.P.
(4)
?? ?
24 3 5 4 6
111
,,
II I I I I
are in A.P.
Answer (4)
Sol.
??
??
?
??
??
??
22
nn2 2
n
44
I cot xdx cot x cot x dx
??
?
?
?
??
?
2
n2 2
4
cot x cosec x 1 dx
??
??
??
??
??
22
n2 2 n 2
44
cot xcosec xdx cot xdx
?
?
??
?
?
?? ? ? ?
?
?? ?
n1
2
n2 n2
4
cot x 1
II
n1 n1
?
?
?? ? ??
??
nn2
n2 n
11
II n1
n1 I I
 = a linear
expression in n.
? Sequence 
?
?
n2 n
1
II
 is an A.P.
PART–C : MATHEMATICS
2. Let ? and ? be the roots of ?? ?
2
x6x2 0 . If
?? ? ?
nn
n
a for ? n1, then the value of
10 8
9
a2a
3a
?
 is :
(1) 2 (2) 4
(3) 3 (4) 1
Answer (1)
Sol. ?, ? are roots of x
2
 – 6x – 2 = 0
??
2
 – 6 ? – 2 = 0
? ? ? ? ? ? ? ? ? ?
2
 – 2 = 6 ?
Similarly ?
2
 – 2 = 6 ?
??
??
? ??? ? ? ??
?
???
10 10 8 8
10 8
99
9
a2a 2
3a 3
??
??
??
?? ? ? ? ? ?
?
???
10 8 10 8
99
22
3
??
??
??
??
??
??
???? ? ?? ? ?? ? ?
??
??? ? ??
88 82 8 2
99 99
66 22
33
??
??
???
??
???
99
99
6
2
3
3. A hyperbola passes through the foci of the
ellipse ??
22
xy
1
25 16
 and its transverse and
conjugate axes coincide with major and minor
axes of the ellipse, respectively. If the product
of their eccentricities is one, then the
equation of the hyperbola is :
(1) ??
22
xy
1
916
(2) ??
22
x y 9
(3) ??
22
xy
1
94
(4) ??
22
xy
1
925
Answer (1)
Sol. Eccentricity of Ellipse ?? ?
1
16 3
e1
25 5
Foci = ( ?ae, 0) = ( ?3, 0)
For Hyperbola
Eccentricity ?
2
5
e
3
JEE (MAIN)-2021 Phase-1 (25-02-2021)-E
Semi-transverse axis ? a = 3
??
??? ??
??
??
22 2
25
ba(e 1)9 1 16
9
Equation of Hyperbola
??
22
xy
1
916
4. Let x denote the total number of one-one
functions from a set A with 3 elements to a set
B with 5 elements and y denote the total
number of one-one functions from the set A to
the set A × B. Then :
(1) ? 2y 273x (2) ? 2y 91x
(3) ? y 273x (4) ? y 91x
Answer (2)
Sol. n(A) = 3, n(B) = 5
x = 
5
C
3
 × 3! = 5 × 4 × 3
n(A × B) = 15
y = 
15
C
3
 × 3! = 15 × 14 × 13
??
??
??
y151413 91
x543 2
2y = 91x
5. A function f(x) is given by ?
?
x
x
5
f(x)
55
, then
the sum of the series
?? ?? ?? ??
????
?? ?? ?? ??
?? ?? ?? ??
12 3 39
fff ....f
20 20 20 20
 is equal to:
(1)
19
2
(2)
29
2
(3)
49
2
(4)
39
2
Answer (4)
Sol.
??
?
?
?? ?
??
2x
2x x
55
f2 x
5555
So f(x) + f(2 – x) = 1
??
?? ?? ?? ? ?
??? ?
?? ?? ? ? ??
?? ?? ? ? ??
??
39 19
r1 r1
rr r
ff f2 f(1)
20 20 20
?? ?
139
19
22
6. The minimum value of 
?
??
xx
a1a
f(x) a a ,
where a, ? xR and a > 0, is equal to :
(1) a + 1 (2) 2a
(3) ?
1
a
a
(4) 2a
Answer (2)
Sol. ??
x
x
a
a
a
f(x) a
a
?
?
?
x
x
x
x
a
a
a
a
a
a
a
a
a.
2
a
? ? f(x) 2a
?
min
f(x) 2 a
7. If the curve ??
22
x2y 2 intersects the line
?? x y 1 at two points P and Q, then the angle
subtended by the line segment PQ at the
origin is :
(1)
?
? ??
?
??
??
1
1
tan
24
(2)
?
? ??
?
??
??
1
1
tan
23
(3)
?
? ??
?
??
??
1
1
tan
23
(4)
?
? ??
?
??
??
1
1
tan
24
Answer (4)
Sol. y = 1 – x ...(i)
x
2
 + 2y
2
 = 2 ...(ii)
? x
2
 + 2 (1 – x)
2
 =2
3x
2
 – 4x = 0
B(0, 1)
(0, 0)
0
?
A
?
4
x0,
3
?
?
1
y1,
3
JEE (MAIN)-2021 Phase-1 (25-02-2021)-E
? ??
??
??
41
B(0,1), A ,
33
?
?? ? ? ??
1
1
11
3
tan tan
4
44
3
?
?
???
1
1
AOB tan
24
8.
??
??
?? ??
??
?? ? ??
??
22 2
n
1n n n
lim ...
n
(n1)(n2)(2n 1)
is equal to :
(1)
1
2
(2)
1
3
(3) 1 (4)
1
4
Answer (1)
Sol.
??
?
??
? ?
?
n1
2
n
r0
n
lim
nr
= 
?
??
???
?
??
??
?
n1
2
n
r0
11
lim
n
r
1
n
?? ? ? ?
?
?
?
1
1
2
0
0
dx 1 1 1
–1
1x 2 2
(1 x)
9. Let A be a set of all 4-digit natural numbers
whose exactly one digit is 7. Then the
probability that a randomly chosen element of
A leaves remainder 2 when divided by 5 is :
(1)
2
9
(2)
97
297
(3)
122
297
(4)
1
5
Answer (2)
Sol. Number having exactly one 7 can be
(i) Having 7 at thousand’s place = 9
3
 = 729
(ii) Not 7 at thousand’s place = 3 × 8 × 4
2
= 1944
n(s) = 729 + 1944 = 2673
Favourable cases = having 7 at unit place or
having 2 at unit place.
i.e. = (9 × 9) + (8 × 9 × 2) + (8 × 9 × 9) = 873
Required probability = 
873 97
2673 297
?
10. A plane passes through the points A(1,2,3),
B(2, 3, 1) and C(2, 4, 2). If O is the origin and
P is (2, –1, 1), then the projection of 

OP on
this plane is of length:
(1)
2
5
(2)
2
7
(3)
2
3
(4)
2
11
Answer (4)
Sol. ??

ˆ ˆˆ
AB i j–2k
??

ˆ ˆˆ
AC i 2j–k
Normal to plane 
??
 

nABAC
= 
?? ? ?
?
ˆ ˆˆ
ij k
ˆ ˆˆ
11 2 3i j k
12 1
?? ?

ˆ ˆˆ
OP 2i j k
??
?? ? ?
?




OP.n 6 1 1 8
sin
11. 6 66 OP n
?? ? ?
64 1
cos 1
66
33
Projection = ?? ? ?

12
OPcos 6
11
33
11. The integral 
?
??
?
ee
ee e
3log 2x 2log 2x
4log x 3log x 2log x
e5e
e5e 7e
dx,
x > 0, is equal to :
(1) ?? ?
2
e
4logx5x7 c
(2) ?? ?
2
e
logx5x7 c
(3) ?? ?
2
e
1
log x 5x 7 c
4
(4) ?? ?
2
e
log x5x7c
Answer (1)
Sol.
?
?
??
?
3ln2x 2ln2x
4lnx 3lnx 2lnx
e5e
Idx
e5e 7e
?? ?? ?
?
??
?? ? ?
??
32
43 2 2
2x 5 2x
8x 20
Idx dx
x5x 7x x 5x7
Let x
2
 + 5x – 7 = t
(2x + 5)dx = dt
?? ?
?
dt
I4 4lnt c
t
????
2
I4lnx 5x 7 c
Page 4


JEE (MAIN)-2021 Phase-1 (25-02-2021)-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. If 
?
?
?
?
2
n
n
4
Icotxdx , then :
(1)
?? ?
24 3 5 4 6
111
,,
II I I I I
 are in G.P.
(2) ?? ?
243 54 6
II,I I,I I are in A.P.
(3)
?? ?
2
24 3 5 4 6
II,(II ) ,I I
are in G.P.
(4)
?? ?
24 3 5 4 6
111
,,
II I I I I
are in A.P.
Answer (4)
Sol.
??
??
?
??
??
??
22
nn2 2
n
44
I cot xdx cot x cot x dx
??
?
?
?
??
?
2
n2 2
4
cot x cosec x 1 dx
??
??
??
??
??
22
n2 2 n 2
44
cot xcosec xdx cot xdx
?
?
??
?
?
?? ? ? ?
?
?? ?
n1
2
n2 n2
4
cot x 1
II
n1 n1
?
?
?? ? ??
??
nn2
n2 n
11
II n1
n1 I I
 = a linear
expression in n.
? Sequence 
?
?
n2 n
1
II
 is an A.P.
PART–C : MATHEMATICS
2. Let ? and ? be the roots of ?? ?
2
x6x2 0 . If
?? ? ?
nn
n
a for ? n1, then the value of
10 8
9
a2a
3a
?
 is :
(1) 2 (2) 4
(3) 3 (4) 1
Answer (1)
Sol. ?, ? are roots of x
2
 – 6x – 2 = 0
??
2
 – 6 ? – 2 = 0
? ? ? ? ? ? ? ? ? ?
2
 – 2 = 6 ?
Similarly ?
2
 – 2 = 6 ?
??
??
? ??? ? ? ??
?
???
10 10 8 8
10 8
99
9
a2a 2
3a 3
??
??
??
?? ? ? ? ? ?
?
???
10 8 10 8
99
22
3
??
??
??
??
??
??
???? ? ?? ? ?? ? ?
??
??? ? ??
88 82 8 2
99 99
66 22
33
??
??
???
??
???
99
99
6
2
3
3. A hyperbola passes through the foci of the
ellipse ??
22
xy
1
25 16
 and its transverse and
conjugate axes coincide with major and minor
axes of the ellipse, respectively. If the product
of their eccentricities is one, then the
equation of the hyperbola is :
(1) ??
22
xy
1
916
(2) ??
22
x y 9
(3) ??
22
xy
1
94
(4) ??
22
xy
1
925
Answer (1)
Sol. Eccentricity of Ellipse ?? ?
1
16 3
e1
25 5
Foci = ( ?ae, 0) = ( ?3, 0)
For Hyperbola
Eccentricity ?
2
5
e
3
JEE (MAIN)-2021 Phase-1 (25-02-2021)-E
Semi-transverse axis ? a = 3
??
??? ??
??
??
22 2
25
ba(e 1)9 1 16
9
Equation of Hyperbola
??
22
xy
1
916
4. Let x denote the total number of one-one
functions from a set A with 3 elements to a set
B with 5 elements and y denote the total
number of one-one functions from the set A to
the set A × B. Then :
(1) ? 2y 273x (2) ? 2y 91x
(3) ? y 273x (4) ? y 91x
Answer (2)
Sol. n(A) = 3, n(B) = 5
x = 
5
C
3
 × 3! = 5 × 4 × 3
n(A × B) = 15
y = 
15
C
3
 × 3! = 15 × 14 × 13
??
??
??
y151413 91
x543 2
2y = 91x
5. A function f(x) is given by ?
?
x
x
5
f(x)
55
, then
the sum of the series
?? ?? ?? ??
????
?? ?? ?? ??
?? ?? ?? ??
12 3 39
fff ....f
20 20 20 20
 is equal to:
(1)
19
2
(2)
29
2
(3)
49
2
(4)
39
2
Answer (4)
Sol.
??
?
?
?? ?
??
2x
2x x
55
f2 x
5555
So f(x) + f(2 – x) = 1
??
?? ?? ?? ? ?
??? ?
?? ?? ? ? ??
?? ?? ? ? ??
??
39 19
r1 r1
rr r
ff f2 f(1)
20 20 20
?? ?
139
19
22
6. The minimum value of 
?
??
xx
a1a
f(x) a a ,
where a, ? xR and a > 0, is equal to :
(1) a + 1 (2) 2a
(3) ?
1
a
a
(4) 2a
Answer (2)
Sol. ??
x
x
a
a
a
f(x) a
a
?
?
?
x
x
x
x
a
a
a
a
a
a
a
a
a.
2
a
? ? f(x) 2a
?
min
f(x) 2 a
7. If the curve ??
22
x2y 2 intersects the line
?? x y 1 at two points P and Q, then the angle
subtended by the line segment PQ at the
origin is :
(1)
?
? ??
?
??
??
1
1
tan
24
(2)
?
? ??
?
??
??
1
1
tan
23
(3)
?
? ??
?
??
??
1
1
tan
23
(4)
?
? ??
?
??
??
1
1
tan
24
Answer (4)
Sol. y = 1 – x ...(i)
x
2
 + 2y
2
 = 2 ...(ii)
? x
2
 + 2 (1 – x)
2
 =2
3x
2
 – 4x = 0
B(0, 1)
(0, 0)
0
?
A
?
4
x0,
3
?
?
1
y1,
3
JEE (MAIN)-2021 Phase-1 (25-02-2021)-E
? ??
??
??
41
B(0,1), A ,
33
?
?? ? ? ??
1
1
11
3
tan tan
4
44
3
?
?
???
1
1
AOB tan
24
8.
??
??
?? ??
??
?? ? ??
??
22 2
n
1n n n
lim ...
n
(n1)(n2)(2n 1)
is equal to :
(1)
1
2
(2)
1
3
(3) 1 (4)
1
4
Answer (1)
Sol.
??
?
??
? ?
?
n1
2
n
r0
n
lim
nr
= 
?
??
???
?
??
??
?
n1
2
n
r0
11
lim
n
r
1
n
?? ? ? ?
?
?
?
1
1
2
0
0
dx 1 1 1
–1
1x 2 2
(1 x)
9. Let A be a set of all 4-digit natural numbers
whose exactly one digit is 7. Then the
probability that a randomly chosen element of
A leaves remainder 2 when divided by 5 is :
(1)
2
9
(2)
97
297
(3)
122
297
(4)
1
5
Answer (2)
Sol. Number having exactly one 7 can be
(i) Having 7 at thousand’s place = 9
3
 = 729
(ii) Not 7 at thousand’s place = 3 × 8 × 4
2
= 1944
n(s) = 729 + 1944 = 2673
Favourable cases = having 7 at unit place or
having 2 at unit place.
i.e. = (9 × 9) + (8 × 9 × 2) + (8 × 9 × 9) = 873
Required probability = 
873 97
2673 297
?
10. A plane passes through the points A(1,2,3),
B(2, 3, 1) and C(2, 4, 2). If O is the origin and
P is (2, –1, 1), then the projection of 

OP on
this plane is of length:
(1)
2
5
(2)
2
7
(3)
2
3
(4)
2
11
Answer (4)
Sol. ??

ˆ ˆˆ
AB i j–2k
??

ˆ ˆˆ
AC i 2j–k
Normal to plane 
??
 

nABAC
= 
?? ? ?
?
ˆ ˆˆ
ij k
ˆ ˆˆ
11 2 3i j k
12 1
?? ?

ˆ ˆˆ
OP 2i j k
??
?? ? ?
?




OP.n 6 1 1 8
sin
11. 6 66 OP n
?? ? ?
64 1
cos 1
66
33
Projection = ?? ? ?

12
OPcos 6
11
33
11. The integral 
?
??
?
ee
ee e
3log 2x 2log 2x
4log x 3log x 2log x
e5e
e5e 7e
dx,
x > 0, is equal to :
(1) ?? ?
2
e
4logx5x7 c
(2) ?? ?
2
e
logx5x7 c
(3) ?? ?
2
e
1
log x 5x 7 c
4
(4) ?? ?
2
e
log x5x7c
Answer (1)
Sol.
?
?
??
?
3ln2x 2ln2x
4lnx 3lnx 2lnx
e5e
Idx
e5e 7e
?? ?? ?
?
??
?? ? ?
??
32
43 2 2
2x 5 2x
8x 20
Idx dx
x5x 7x x 5x7
Let x
2
 + 5x – 7 = t
(2x + 5)dx = dt
?? ?
?
dt
I4 4lnt c
t
????
2
I4lnx 5x 7 c
JEE (MAIN)-2021 Phase-1 (25-02-2021)-E
12. If for the matrix, A = 
?? ??
??
??
??
1
, AA
T
 = I
2
, then
the value of ???
44
 is :
(1) 1 (2) 2
(3) 4 (4) 3
Answer (1)
Sol.
?? ??
?
??
??
??
1
A
? AA
T
 = I
So, I + ?
2
 = 1 ??
2
 = 0
and ?
2
 + ?
2
 = 1 ??
2
 = 1
then ?
4
 + ?
4
 = 1
13. The following system of linear equations
2x + 3y + 2z = 9
3x + 2y + 2z = 9
x – y + 4z = 8
(1) has a solution ( ?, ?, ?) satisfying
?? ?? ?
2
 + ?
3 
= 12
(2) has a unique solution
(3) does not have any solution
(4) has infinitely many solutions
Answer (2)
Sol. Determinant of coefficients of given equations
is
?? ? ? ? ??
?
23 2
32 2 2 (8 2) 3 (12 2) 2 ( 3 2)
114
= 20 – 30 – 10 = –20 ? 0
? Hence the system of equation have unique
solution
14. cosec
??
??
?? ??
?
?? ??
?? ??
11
4
2cot 5 cos
5
 is equal to :
(1)
65
56
(2)
65
33
(3)
75
56
(4)
56
33
Answer (1)
Sol.
??
?? ??
?
?? ??
?? ??
11
4
cosec 2cot 5 cos
5
?
??
?? ?? ? ?
?
?? ?? ? ?
?? ? ? ??
11
53
cosec tan tan
12 4
?
?
?? ??
?? ??
?? ??
1
56
cosec tan
33
?
?
?? ??
?
?? ??
?? ??
1
65 65
cosec cosec
56 56
15. If ??? ,R are such that 1 – 2i (here i
2
 = –1) is
a root of z
2
 + ???? z0, then ?? ??? is equal
to:
(1) –3 (2) –7
(3) 7 (4) 3
Answer (2)
Sol. As ?, ? ? R roots are 1 – 2i and 1 + 2i
– ? = 2 ? ? = –2
and ? = (1)
2
 – (2i)
2
 = 5 ? ? – ? = –7
16. The contrapositive of the statement “If you will
work, you will earn money” is :
(1) If you will earn money, you will work
(2) You will earn money, if you will not work
(3) If you will not earn money, you will not
work
(4) To earn money, you need to work
Answer (3)
Sol. Contrapositive of A ? B is ~B ? ~A
? Contrapositive of the given statement will
be ~(you will earn money) ? ~(you will
work)
i.e., if you will not earn money, you will not
work
17. In a group of 400 people, 160 are smokers and
non-vegetarian; 100 are smokers and
vegetarian and the remaining 140 are non-
smokers and vegetarian. Their chances of
getting a particular chest disorder are 35%,
20% and 10% respectively. A person is chosen
from the group at random and is found to be
suffering from the chest disorder. The
probability that the selected person is a
smoker and non-vegetarian is :
(1)
7
45
(2)
28
45
(3)
14
45
(4)
8
45
Answer (2)
Sol. n(smokers + Non vegetarian) = 160 = n(A
1
)(Let)
? P(A
1
) = 0.4
n(smokers + vegetarian) = 100 = n(A
2
)
similarly P(A
2
) = 0.25
n(Non-smokers + vegetarian) = 140 = n(A
3
) and
P(A
3
) = 0.35
Let event E of getting chest disorder i.e.,
Page 5


JEE (MAIN)-2021 Phase-1 (25-02-2021)-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. If 
?
?
?
?
2
n
n
4
Icotxdx , then :
(1)
?? ?
24 3 5 4 6
111
,,
II I I I I
 are in G.P.
(2) ?? ?
243 54 6
II,I I,I I are in A.P.
(3)
?? ?
2
24 3 5 4 6
II,(II ) ,I I
are in G.P.
(4)
?? ?
24 3 5 4 6
111
,,
II I I I I
are in A.P.
Answer (4)
Sol.
??
??
?
??
??
??
22
nn2 2
n
44
I cot xdx cot x cot x dx
??
?
?
?
??
?
2
n2 2
4
cot x cosec x 1 dx
??
??
??
??
??
22
n2 2 n 2
44
cot xcosec xdx cot xdx
?
?
??
?
?
?? ? ? ?
?
?? ?
n1
2
n2 n2
4
cot x 1
II
n1 n1
?
?
?? ? ??
??
nn2
n2 n
11
II n1
n1 I I
 = a linear
expression in n.
? Sequence 
?
?
n2 n
1
II
 is an A.P.
PART–C : MATHEMATICS
2. Let ? and ? be the roots of ?? ?
2
x6x2 0 . If
?? ? ?
nn
n
a for ? n1, then the value of
10 8
9
a2a
3a
?
 is :
(1) 2 (2) 4
(3) 3 (4) 1
Answer (1)
Sol. ?, ? are roots of x
2
 – 6x – 2 = 0
??
2
 – 6 ? – 2 = 0
? ? ? ? ? ? ? ? ? ?
2
 – 2 = 6 ?
Similarly ?
2
 – 2 = 6 ?
??
??
? ??? ? ? ??
?
???
10 10 8 8
10 8
99
9
a2a 2
3a 3
??
??
??
?? ? ? ? ? ?
?
???
10 8 10 8
99
22
3
??
??
??
??
??
??
???? ? ?? ? ?? ? ?
??
??? ? ??
88 82 8 2
99 99
66 22
33
??
??
???
??
???
99
99
6
2
3
3. A hyperbola passes through the foci of the
ellipse ??
22
xy
1
25 16
 and its transverse and
conjugate axes coincide with major and minor
axes of the ellipse, respectively. If the product
of their eccentricities is one, then the
equation of the hyperbola is :
(1) ??
22
xy
1
916
(2) ??
22
x y 9
(3) ??
22
xy
1
94
(4) ??
22
xy
1
925
Answer (1)
Sol. Eccentricity of Ellipse ?? ?
1
16 3
e1
25 5
Foci = ( ?ae, 0) = ( ?3, 0)
For Hyperbola
Eccentricity ?
2
5
e
3
JEE (MAIN)-2021 Phase-1 (25-02-2021)-E
Semi-transverse axis ? a = 3
??
??? ??
??
??
22 2
25
ba(e 1)9 1 16
9
Equation of Hyperbola
??
22
xy
1
916
4. Let x denote the total number of one-one
functions from a set A with 3 elements to a set
B with 5 elements and y denote the total
number of one-one functions from the set A to
the set A × B. Then :
(1) ? 2y 273x (2) ? 2y 91x
(3) ? y 273x (4) ? y 91x
Answer (2)
Sol. n(A) = 3, n(B) = 5
x = 
5
C
3
 × 3! = 5 × 4 × 3
n(A × B) = 15
y = 
15
C
3
 × 3! = 15 × 14 × 13
??
??
??
y151413 91
x543 2
2y = 91x
5. A function f(x) is given by ?
?
x
x
5
f(x)
55
, then
the sum of the series
?? ?? ?? ??
????
?? ?? ?? ??
?? ?? ?? ??
12 3 39
fff ....f
20 20 20 20
 is equal to:
(1)
19
2
(2)
29
2
(3)
49
2
(4)
39
2
Answer (4)
Sol.
??
?
?
?? ?
??
2x
2x x
55
f2 x
5555
So f(x) + f(2 – x) = 1
??
?? ?? ?? ? ?
??? ?
?? ?? ? ? ??
?? ?? ? ? ??
??
39 19
r1 r1
rr r
ff f2 f(1)
20 20 20
?? ?
139
19
22
6. The minimum value of 
?
??
xx
a1a
f(x) a a ,
where a, ? xR and a > 0, is equal to :
(1) a + 1 (2) 2a
(3) ?
1
a
a
(4) 2a
Answer (2)
Sol. ??
x
x
a
a
a
f(x) a
a
?
?
?
x
x
x
x
a
a
a
a
a
a
a
a
a.
2
a
? ? f(x) 2a
?
min
f(x) 2 a
7. If the curve ??
22
x2y 2 intersects the line
?? x y 1 at two points P and Q, then the angle
subtended by the line segment PQ at the
origin is :
(1)
?
? ??
?
??
??
1
1
tan
24
(2)
?
? ??
?
??
??
1
1
tan
23
(3)
?
? ??
?
??
??
1
1
tan
23
(4)
?
? ??
?
??
??
1
1
tan
24
Answer (4)
Sol. y = 1 – x ...(i)
x
2
 + 2y
2
 = 2 ...(ii)
? x
2
 + 2 (1 – x)
2
 =2
3x
2
 – 4x = 0
B(0, 1)
(0, 0)
0
?
A
?
4
x0,
3
?
?
1
y1,
3
JEE (MAIN)-2021 Phase-1 (25-02-2021)-E
? ??
??
??
41
B(0,1), A ,
33
?
?? ? ? ??
1
1
11
3
tan tan
4
44
3
?
?
???
1
1
AOB tan
24
8.
??
??
?? ??
??
?? ? ??
??
22 2
n
1n n n
lim ...
n
(n1)(n2)(2n 1)
is equal to :
(1)
1
2
(2)
1
3
(3) 1 (4)
1
4
Answer (1)
Sol.
??
?
??
? ?
?
n1
2
n
r0
n
lim
nr
= 
?
??
???
?
??
??
?
n1
2
n
r0
11
lim
n
r
1
n
?? ? ? ?
?
?
?
1
1
2
0
0
dx 1 1 1
–1
1x 2 2
(1 x)
9. Let A be a set of all 4-digit natural numbers
whose exactly one digit is 7. Then the
probability that a randomly chosen element of
A leaves remainder 2 when divided by 5 is :
(1)
2
9
(2)
97
297
(3)
122
297
(4)
1
5
Answer (2)
Sol. Number having exactly one 7 can be
(i) Having 7 at thousand’s place = 9
3
 = 729
(ii) Not 7 at thousand’s place = 3 × 8 × 4
2
= 1944
n(s) = 729 + 1944 = 2673
Favourable cases = having 7 at unit place or
having 2 at unit place.
i.e. = (9 × 9) + (8 × 9 × 2) + (8 × 9 × 9) = 873
Required probability = 
873 97
2673 297
?
10. A plane passes through the points A(1,2,3),
B(2, 3, 1) and C(2, 4, 2). If O is the origin and
P is (2, –1, 1), then the projection of 

OP on
this plane is of length:
(1)
2
5
(2)
2
7
(3)
2
3
(4)
2
11
Answer (4)
Sol. ??

ˆ ˆˆ
AB i j–2k
??

ˆ ˆˆ
AC i 2j–k
Normal to plane 
??
 

nABAC
= 
?? ? ?
?
ˆ ˆˆ
ij k
ˆ ˆˆ
11 2 3i j k
12 1
?? ?

ˆ ˆˆ
OP 2i j k
??
?? ? ?
?




OP.n 6 1 1 8
sin
11. 6 66 OP n
?? ? ?
64 1
cos 1
66
33
Projection = ?? ? ?

12
OPcos 6
11
33
11. The integral 
?
??
?
ee
ee e
3log 2x 2log 2x
4log x 3log x 2log x
e5e
e5e 7e
dx,
x > 0, is equal to :
(1) ?? ?
2
e
4logx5x7 c
(2) ?? ?
2
e
logx5x7 c
(3) ?? ?
2
e
1
log x 5x 7 c
4
(4) ?? ?
2
e
log x5x7c
Answer (1)
Sol.
?
?
??
?
3ln2x 2ln2x
4lnx 3lnx 2lnx
e5e
Idx
e5e 7e
?? ?? ?
?
??
?? ? ?
??
32
43 2 2
2x 5 2x
8x 20
Idx dx
x5x 7x x 5x7
Let x
2
 + 5x – 7 = t
(2x + 5)dx = dt
?? ?
?
dt
I4 4lnt c
t
????
2
I4lnx 5x 7 c
JEE (MAIN)-2021 Phase-1 (25-02-2021)-E
12. If for the matrix, A = 
?? ??
??
??
??
1
, AA
T
 = I
2
, then
the value of ???
44
 is :
(1) 1 (2) 2
(3) 4 (4) 3
Answer (1)
Sol.
?? ??
?
??
??
??
1
A
? AA
T
 = I
So, I + ?
2
 = 1 ??
2
 = 0
and ?
2
 + ?
2
 = 1 ??
2
 = 1
then ?
4
 + ?
4
 = 1
13. The following system of linear equations
2x + 3y + 2z = 9
3x + 2y + 2z = 9
x – y + 4z = 8
(1) has a solution ( ?, ?, ?) satisfying
?? ?? ?
2
 + ?
3 
= 12
(2) has a unique solution
(3) does not have any solution
(4) has infinitely many solutions
Answer (2)
Sol. Determinant of coefficients of given equations
is
?? ? ? ? ??
?
23 2
32 2 2 (8 2) 3 (12 2) 2 ( 3 2)
114
= 20 – 30 – 10 = –20 ? 0
? Hence the system of equation have unique
solution
14. cosec
??
??
?? ??
?
?? ??
?? ??
11
4
2cot 5 cos
5
 is equal to :
(1)
65
56
(2)
65
33
(3)
75
56
(4)
56
33
Answer (1)
Sol.
??
?? ??
?
?? ??
?? ??
11
4
cosec 2cot 5 cos
5
?
??
?? ?? ? ?
?
?? ?? ? ?
?? ? ? ??
11
53
cosec tan tan
12 4
?
?
?? ??
?? ??
?? ??
1
56
cosec tan
33
?
?
?? ??
?
?? ??
?? ??
1
65 65
cosec cosec
56 56
15. If ??? ,R are such that 1 – 2i (here i
2
 = –1) is
a root of z
2
 + ???? z0, then ?? ??? is equal
to:
(1) –3 (2) –7
(3) 7 (4) 3
Answer (2)
Sol. As ?, ? ? R roots are 1 – 2i and 1 + 2i
– ? = 2 ? ? = –2
and ? = (1)
2
 – (2i)
2
 = 5 ? ? – ? = –7
16. The contrapositive of the statement “If you will
work, you will earn money” is :
(1) If you will earn money, you will work
(2) You will earn money, if you will not work
(3) If you will not earn money, you will not
work
(4) To earn money, you need to work
Answer (3)
Sol. Contrapositive of A ? B is ~B ? ~A
? Contrapositive of the given statement will
be ~(you will earn money) ? ~(you will
work)
i.e., if you will not earn money, you will not
work
17. In a group of 400 people, 160 are smokers and
non-vegetarian; 100 are smokers and
vegetarian and the remaining 140 are non-
smokers and vegetarian. Their chances of
getting a particular chest disorder are 35%,
20% and 10% respectively. A person is chosen
from the group at random and is found to be
suffering from the chest disorder. The
probability that the selected person is a
smoker and non-vegetarian is :
(1)
7
45
(2)
28
45
(3)
14
45
(4)
8
45
Answer (2)
Sol. n(smokers + Non vegetarian) = 160 = n(A
1
)(Let)
? P(A
1
) = 0.4
n(smokers + vegetarian) = 100 = n(A
2
)
similarly P(A
2
) = 0.25
n(Non-smokers + vegetarian) = 140 = n(A
3
) and
P(A
3
) = 0.35
Let event E of getting chest disorder i.e.,
JEE (MAIN)-2021 Phase-1 (25-02-2021)-E
P(E/A
1
) = 0.35, P(E/A
2
) = 0.2, P(E/A
3
) = 0.1
to find P(A
1
/E)
using Baye’s theorem we get
??
??? ?
??? ? ??? ? ??? ?
?
?
?? ? ?
11
1
11 2 2 3 3
PE/A PA
PA /E
PE/A PA PE/A PA PE/A PA
??? ?? ?
?
?
?? ? ? ?
0.35 0.4
0.35 0.4 0.2 0.25 0.1 0.35
???
??
140 140 28
1405035 225 45
18. If 0 < x, y < ??and cos x + cos y – cos (x + y) = 
3
2
,
then sin x + cos y is equal to :
(1)
? 13
2
(2)
3
2
(3)
1
2
(4)
? 13
2
Answer (1)
Sol. LHS = cosx + cosy – cos(x + y)
?? ? ?? ? ?? ?
?? ?
?? ? ?? ?
?? ? ?? ?
2
xy x y xy
2cos cos 2cos 1
22 2
??
?? ?
2
xy xy
2cos 2cos 1
22
?
xy
,
222
???? ??
??
?? ?
?? ?
 ? 
xy
0cos 1
2
? ? ??
??
?? ?
?? ?
?? ?? ?? ??
?? ?
?? ?? ??
?? ?? ??
2
xy xy
12cos cos
22
??
?? ? ??
??
?? ? ?
?? ??
?? ?? ??
??
2
x y 11
1 2 cos
22 4
?? ? ??
?? ? ?
?? ??
?? ??
2
3xy13
2cos
2222
But given that LHS ?
3
2
?
?? ??
??
??
??
x y x y 1
cos 1andcos
222
? x – y = 0 and 
?
??
2
xy
3
?
?
?? xy
3
?
?
??
31
sinx cosy
2
19. The shortest distance between the line
x – y = 1 and the curve x
2
 = 2y is :
(1)
1
2
(2)
1
2
(3) 0 (4)
1
22
Answer (4)
Sol. Equation of line parallel to x – y = 1 is
x – y = c ...(i)
If line x – y = c is tangent to parabola x
2
 = 2y
then x
2
 = 2 (x – c) has unique roots
x
2
 – 2x + 2c = 0
? D = 0 ? 4 – 4 × 1 × 2 c = 0
? ?
1
c
2
? Tangent of parabola is 
??
1
xy
2
? Shortest distance 
?
??
1
1
1 2
units
222
20. Let A be a 3 × 3 matrix with det(A) = 4. Let R
i
denote the i
th
 row of A. If a matrix B is
obtained by performing the operation R
2
? 2R
2
+ 5R
3 
on 2A, then det(B) is equal to :
(1) 64 (2) 128
(3) 80 (4) 16
Answer (1)
Sol. Given det (A) = 4
On application of R
2
 ? 2R
2
 + 5R
3
 on 2A we
have 2
3
.2 det (A) = 16 × 4 = 64
SECTION - II
Numerical Value Type Questions: This section
contains 10 questions. In Section II, attempt any five
questions out of 10. The answer to each question is
a NUMERICAL VALUE. For each question, enter the
correct numerical value (in decimal notation,
truncated/rounded-off to the second decimal place;
e.g. 06.25, 07.00, –00.33, –00.30, 30.27, –27.30)
using the mouse and the on-screen virtual numeric
keypad in the place designated to enter the answer.
1. If the curve, y = y(x) represented by the
solution of the differential equation (2xy
2
 – y) dx
+ xdy = 0, passes through the intersection of
the lines, 2x – 3y = 1 and 3x + 2y = 8, then ?? y 1
is equal to ___________.
Answer (01)