# JEE Main 2021 Answer Key Mathematics - Morning Shift (24-02-2021) Notes | Study Mock Test Series for JEE Main & Advanced - JEE

## JEE: JEE Main 2021 Answer Key Mathematics - Morning Shift (24-02-2021) Notes | Study Mock Test Series for JEE Main & Advanced - JEE

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JEE (MAIN)-2021 Phase-1 (24-02-2021)-M
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.
1. If
?
? ? ? ?
? ?
? ?
? ? ?
?
1
cosx sinx sinx cosx
dx asin c,
b
8 sin2x
where c is a constant of integration, then the
ordered pair (a, b) is equal to :
(1) (–1, 3) (2) (3, 1)
(3) (1, –3) (4) (1, 3)
Sol.
? ?
2
cos x sinx cos x sinx
dx dx
8 sin2x
9 sinx cosx
? ?
?
?
? ?
? ?
let sinx + cosx = t
(cosx – sinx)dx = dt
2
dt
9 t
?
?
?
1
t
sin c
3
?
? ?
? ?
? ?
? ?
1
sinx cosx
sin c
3
?
? ? ?
? ?
? ?
? ?
Hence (a, b) = (1, 3)
2. The area (in sq. units) of the part of the circle
x
2
+ y
2
= 36, which is outside the parabola
y
2
= 9x, is :
(1) ? ? 12 3 3 (2) ? ? 24 3 3
(3) ? ? 12 3 3 (4) ? ? 24 3 3
Sol. Area of the shaded region
3 6
2
0 3
2 3 xdx 36 x dx
? ?
? ? ?
? ?
? ?
? ?
y = 9x
2
(3, 3 3)
3
x + y = 36
2 2
PART–C : MATHEMATICS
3
6
3/2 2 1
3 0
1 x
2 2x x 36 x 18sin
2 6
?
? ?
? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ?
? ?
9 3
2 6 3 9 3 3 3 12
2
? ?
? ? ? ? ? ? ? ? ?
? ?
? ?
? ?
Required area ? ?
36 3 3 12 ? ? ? ? ?
24 3 3 ? ? ?
3. If
? ?
? ? ? ?
2 4 6
e
cos x cos x cos x ... log 2
e satisfies the
equation t
2
– 9t + 8 = 0, then the value of
? ? ?
? ?
? ?
? ? ?
2sinx
0 x
2
sinx 3 cosx
is :
(1)
3
2
(2) 3
(3) 2 3 (4)
1
2
Sol.
2
2 2
cos x
ln2
1 cos x cot x
e 2
? ?
? ?
? ?
?
? ?
?
? t = 1 or 8
So,
2
cot x 0
2 2 ?
or 2
3
? cot
2
x = 0 or 3
? x 0,
2
? ? ?
?
? ?
? ?
then cot x 3 x
6
?
? ? ?
1
2
2sinx 1 2
2
sinx 3 cos x 1 3
3
2 2
? ?
? ?
? ?
? ?
? ? ?
? ?
? ?
? ?
? ?
4. The population P = P(t) at time ‘t’ of a certain
species follows the differential equation
? ?
dP
0.5P 450.
dt
If P(0) = 850, then the time at t
which population becomes zero is :
(1) log
e
18 (2)
e
1
log 18
2
(3) log
e
9 (4) 2log
e
18
Sol. ? ? ?
dP 1
P 900
dt 2
? ?
Page 2

JEE (MAIN)-2021 Phase-1 (24-02-2021)-M
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.
1. If
?
? ? ? ?
? ?
? ?
? ? ?
?
1
cosx sinx sinx cosx
dx asin c,
b
8 sin2x
where c is a constant of integration, then the
ordered pair (a, b) is equal to :
(1) (–1, 3) (2) (3, 1)
(3) (1, –3) (4) (1, 3)
Sol.
? ?
2
cos x sinx cos x sinx
dx dx
8 sin2x
9 sinx cosx
? ?
?
?
? ?
? ?
let sinx + cosx = t
(cosx – sinx)dx = dt
2
dt
9 t
?
?
?
1
t
sin c
3
?
? ?
? ?
? ?
? ?
1
sinx cosx
sin c
3
?
? ? ?
? ?
? ?
? ?
Hence (a, b) = (1, 3)
2. The area (in sq. units) of the part of the circle
x
2
+ y
2
= 36, which is outside the parabola
y
2
= 9x, is :
(1) ? ? 12 3 3 (2) ? ? 24 3 3
(3) ? ? 12 3 3 (4) ? ? 24 3 3
Sol. Area of the shaded region
3 6
2
0 3
2 3 xdx 36 x dx
? ?
? ? ?
? ?
? ?
? ?
y = 9x
2
(3, 3 3)
3
x + y = 36
2 2
PART–C : MATHEMATICS
3
6
3/2 2 1
3 0
1 x
2 2x x 36 x 18sin
2 6
?
? ?
? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ?
? ?
9 3
2 6 3 9 3 3 3 12
2
? ?
? ? ? ? ? ? ? ? ?
? ?
? ?
? ?
Required area ? ?
36 3 3 12 ? ? ? ? ?
24 3 3 ? ? ?
3. If
? ?
? ? ? ?
2 4 6
e
cos x cos x cos x ... log 2
e satisfies the
equation t
2
– 9t + 8 = 0, then the value of
? ? ?
? ?
? ?
? ? ?
2sinx
0 x
2
sinx 3 cosx
is :
(1)
3
2
(2) 3
(3) 2 3 (4)
1
2
Sol.
2
2 2
cos x
ln2
1 cos x cot x
e 2
? ?
? ?
? ?
?
? ?
?
? t = 1 or 8
So,
2
cot x 0
2 2 ?
or 2
3
? cot
2
x = 0 or 3
? x 0,
2
? ? ?
?
? ?
? ?
then cot x 3 x
6
?
? ? ?
1
2
2sinx 1 2
2
sinx 3 cos x 1 3
3
2 2
? ?
? ?
? ?
? ?
? ? ?
? ?
? ?
? ?
? ?
4. The population P = P(t) at time ‘t’ of a certain
species follows the differential equation
? ?
dP
0.5P 450.
dt
If P(0) = 850, then the time at t
which population becomes zero is :
(1) log
e
18 (2)
e
1
log 18
2
(3) log
e
9 (4) 2log
e
18
Sol. ? ? ?
dP 1
P 900
dt 2
? ?
JEE (MAIN)-2021 Phase-1 (24-02-2021)-M
?
dP 1 1
dt ln P 900 t c
P 900 2 2
? ? ? ? ?
?
When t = 0, P = 850 ? c = ln50
When P = 0, t = 2(ln900 – ln50) = 2ln18
5. The statement among the following that is a
tautology is :
(1) A ? (A ? B) (2) B ? [A ? (A ? B)]
(3) [A ? (A ? B)] ? B (4) A ? (A ? B)
Sol. (1) A ? (A ? B) = A
(2) ? A ? (A ? B) = A ??( ? A ??B) = A ? B
So, B ? (A ? B) = ? B ??(A ? B) = ? B ? A
(3) (A ? (A ? B)) ? B = (A ? B) ? B = ? (A ? B)
? B = ? A ? ? B ? B (Tautology)
(4) A ? (A ? B) = A
6. Let p and q be two positive numbers such that
p + q = 2 and p
4
+ q
4
= 272. Then p and q are
roots of the equation :
(1) x
2
– 2x + 2 = 0 (2) x
2
– 2x + 8 = 0
(3) x
2
– 2x + 136 = 0 (4) x
2
– 2x + 16 = 0
Sol. ? p
4
+ q
4
= (p + 4)
4
– 4pq(p
2
+ q
2
) – 6p
2
q
2
? 272 = 16 – 4pq(4 – 2pq) – 6p
2
q
2
? 2p
2
q
2
– 16pq – 256 = 0
? pq = –8 or 16
? p, q > 0, so pq = 16
x
2
– 2x + 16 = 0
7. The system of linear equations
3x – 2y – kz = 10
2x – 4y – 2z = 6
x + 2y – z = 5m
is inconsistent if :
(1)
4
k 3, m
5
? ? (2) k 3, m R ? ?
(3)
4
k 3, m
5
? ? (4)
4
k 3, m
5
? ?
Sol. Here
3 2 k
2 4 2 0
1 2 1
? ?
? ? ? ? ?
?
? 3(4 + 4) + 2(–2 + 2) – k(4 + 4) = 0
? 24 + 0 – 8k = 0 ? k 3 ?
Now,
1
10 2 3
6 4 2 10(4 4) 2( 6 10m) 3(12 20m)
5m 2 1
? ?
? ? ? ? ? ? ? ? ? ? ?
?
= 80 – 12 + 20 m – 36 – 60 m
= 32 – 40 m
2
3 10 3
2 6 2 3( 6 10 m) 10( 2 2) 3(10 m 6)
1 5m 1
?
? ? ? ? ? ? ? ? ? ? ?
?
= –18 + 30 m + 0 – 30 m + 18 = 0
3
3 2 10
2 4 6 3( 20m 12) 2(10m 6) 10(4 4)
1 2 5 m
?
? ? ? ? ? ? ? ? ? ?
= –60 m – 36 + 20 m – 12 + 80
= –40 m + 32
For inconsistent we have k = 3, &
32 – 40 m ? 0 ?
4
m
5
?
8. A scientific committee is to be formed from 6
Indians and 8 foreigners, which includes at
least 2 Indians and double the number of
foreigners as Indians. Then the number of
ways, the committee can be formed, is :
(1) 560 (2) 1050
(3) 1625 (4) 575
Sol. Indians = 6, Foreigners = 8
According to questions
The no. of ways to form the committee are
(2I, 4F) or (3I, 6F) or (4I, 8F)
?
6
C
2
×
8
C
4
+
6
C
3
×
8
C
6
+
6
C
4
×
8
C
8
= 15 × 70 + 20 × 28 + 15 × 1
= 1625
9. The equation of the plane passing through the
point (1, 2, –3) and perpendicular to the planes
3x + y – 2z = 5 and 2x – 5y – z = 7, is:
(1) 3x – 10y – 2z + 11 = 0
(2) 6x – 5y – 2z – 2 = 0
(3) 6x – 5y + 2z + 10 = 0
(4) 11x + y  + 17z + 38 = 0
Page 3

JEE (MAIN)-2021 Phase-1 (24-02-2021)-M
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.
1. If
?
? ? ? ?
? ?
? ?
? ? ?
?
1
cosx sinx sinx cosx
dx asin c,
b
8 sin2x
where c is a constant of integration, then the
ordered pair (a, b) is equal to :
(1) (–1, 3) (2) (3, 1)
(3) (1, –3) (4) (1, 3)
Sol.
? ?
2
cos x sinx cos x sinx
dx dx
8 sin2x
9 sinx cosx
? ?
?
?
? ?
? ?
let sinx + cosx = t
(cosx – sinx)dx = dt
2
dt
9 t
?
?
?
1
t
sin c
3
?
? ?
? ?
? ?
? ?
1
sinx cosx
sin c
3
?
? ? ?
? ?
? ?
? ?
Hence (a, b) = (1, 3)
2. The area (in sq. units) of the part of the circle
x
2
+ y
2
= 36, which is outside the parabola
y
2
= 9x, is :
(1) ? ? 12 3 3 (2) ? ? 24 3 3
(3) ? ? 12 3 3 (4) ? ? 24 3 3
Sol. Area of the shaded region
3 6
2
0 3
2 3 xdx 36 x dx
? ?
? ? ?
? ?
? ?
? ?
y = 9x
2
(3, 3 3)
3
x + y = 36
2 2
PART–C : MATHEMATICS
3
6
3/2 2 1
3 0
1 x
2 2x x 36 x 18sin
2 6
?
? ?
? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ?
? ?
9 3
2 6 3 9 3 3 3 12
2
? ?
? ? ? ? ? ? ? ? ?
? ?
? ?
? ?
Required area ? ?
36 3 3 12 ? ? ? ? ?
24 3 3 ? ? ?
3. If
? ?
? ? ? ?
2 4 6
e
cos x cos x cos x ... log 2
e satisfies the
equation t
2
– 9t + 8 = 0, then the value of
? ? ?
? ?
? ?
? ? ?
2sinx
0 x
2
sinx 3 cosx
is :
(1)
3
2
(2) 3
(3) 2 3 (4)
1
2
Sol.
2
2 2
cos x
ln2
1 cos x cot x
e 2
? ?
? ?
? ?
?
? ?
?
? t = 1 or 8
So,
2
cot x 0
2 2 ?
or 2
3
? cot
2
x = 0 or 3
? x 0,
2
? ? ?
?
? ?
? ?
then cot x 3 x
6
?
? ? ?
1
2
2sinx 1 2
2
sinx 3 cos x 1 3
3
2 2
? ?
? ?
? ?
? ?
? ? ?
? ?
? ?
? ?
? ?
4. The population P = P(t) at time ‘t’ of a certain
species follows the differential equation
? ?
dP
0.5P 450.
dt
If P(0) = 850, then the time at t
which population becomes zero is :
(1) log
e
18 (2)
e
1
log 18
2
(3) log
e
9 (4) 2log
e
18
Sol. ? ? ?
dP 1
P 900
dt 2
? ?
JEE (MAIN)-2021 Phase-1 (24-02-2021)-M
?
dP 1 1
dt ln P 900 t c
P 900 2 2
? ? ? ? ?
?
When t = 0, P = 850 ? c = ln50
When P = 0, t = 2(ln900 – ln50) = 2ln18
5. The statement among the following that is a
tautology is :
(1) A ? (A ? B) (2) B ? [A ? (A ? B)]
(3) [A ? (A ? B)] ? B (4) A ? (A ? B)
Sol. (1) A ? (A ? B) = A
(2) ? A ? (A ? B) = A ??( ? A ??B) = A ? B
So, B ? (A ? B) = ? B ??(A ? B) = ? B ? A
(3) (A ? (A ? B)) ? B = (A ? B) ? B = ? (A ? B)
? B = ? A ? ? B ? B (Tautology)
(4) A ? (A ? B) = A
6. Let p and q be two positive numbers such that
p + q = 2 and p
4
+ q
4
= 272. Then p and q are
roots of the equation :
(1) x
2
– 2x + 2 = 0 (2) x
2
– 2x + 8 = 0
(3) x
2
– 2x + 136 = 0 (4) x
2
– 2x + 16 = 0
Sol. ? p
4
+ q
4
= (p + 4)
4
– 4pq(p
2
+ q
2
) – 6p
2
q
2
? 272 = 16 – 4pq(4 – 2pq) – 6p
2
q
2
? 2p
2
q
2
– 16pq – 256 = 0
? pq = –8 or 16
? p, q > 0, so pq = 16
x
2
– 2x + 16 = 0
7. The system of linear equations
3x – 2y – kz = 10
2x – 4y – 2z = 6
x + 2y – z = 5m
is inconsistent if :
(1)
4
k 3, m
5
? ? (2) k 3, m R ? ?
(3)
4
k 3, m
5
? ? (4)
4
k 3, m
5
? ?
Sol. Here
3 2 k
2 4 2 0
1 2 1
? ?
? ? ? ? ?
?
? 3(4 + 4) + 2(–2 + 2) – k(4 + 4) = 0
? 24 + 0 – 8k = 0 ? k 3 ?
Now,
1
10 2 3
6 4 2 10(4 4) 2( 6 10m) 3(12 20m)
5m 2 1
? ?
? ? ? ? ? ? ? ? ? ? ?
?
= 80 – 12 + 20 m – 36 – 60 m
= 32 – 40 m
2
3 10 3
2 6 2 3( 6 10 m) 10( 2 2) 3(10 m 6)
1 5m 1
?
? ? ? ? ? ? ? ? ? ? ?
?
= –18 + 30 m + 0 – 30 m + 18 = 0
3
3 2 10
2 4 6 3( 20m 12) 2(10m 6) 10(4 4)
1 2 5 m
?
? ? ? ? ? ? ? ? ? ?
= –60 m – 36 + 20 m – 12 + 80
= –40 m + 32
For inconsistent we have k = 3, &
32 – 40 m ? 0 ?
4
m
5
?
8. A scientific committee is to be formed from 6
Indians and 8 foreigners, which includes at
least 2 Indians and double the number of
foreigners as Indians. Then the number of
ways, the committee can be formed, is :
(1) 560 (2) 1050
(3) 1625 (4) 575
Sol. Indians = 6, Foreigners = 8
According to questions
The no. of ways to form the committee are
(2I, 4F) or (3I, 6F) or (4I, 8F)
?
6
C
2
×
8
C
4
+
6
C
3
×
8
C
6
+
6
C
4
×
8
C
8
= 15 × 70 + 20 × 28 + 15 × 1
= 1625
9. The equation of the plane passing through the
point (1, 2, –3) and perpendicular to the planes
3x + y – 2z = 5 and 2x – 5y – z = 7, is:
(1) 3x – 10y – 2z + 11 = 0
(2) 6x – 5y – 2z – 2 = 0
(3) 6x – 5y + 2z + 10 = 0
(4) 11x + y  + 17z + 38 = 0
JEE (MAIN)-2021 Phase-1 (24-02-2021)-M
Sol. The given planes are 3x + y – 2z = 5 ...(1)
2x – 5y – z = 7 ...(2)
Since the required plane passes through (1, 2, –3)
So equation of this plane is
a(x – 1) + b(y – 2) + c(z + 3) = 0 ...(3)
Now this plane (3) is ? to the planes (1) & (2)
So 3a + b – 2c = 0
& 2a – 5b – c = 0
?
a b c
11 1 17
? ?
? ? ?
So equation of plane is 11(x – 1) + (y – 2)
+ 17(2 + 3) = 0
? 11x + y + 17z + 38 = 0
10. If the tangent to the curve y = x
3
at the point
P(t, t
3
) meets the curve again at Q, then the
ordinate of the point which divides PQ
internally in the ratio 1 : 2 is :
(1) –2t
3
(2) 2t
3
(3) 0 (4) –t
3
Sol. Curve is y = x
3
...(1)
So equation of tangent at (t, t
3
)
(y – t
3
) = 3t
2
(x – t) ...(2)
? It meets again the curve at Q
So solving (1) & (2) we get
x = –2t ? Q = (–2t, – 8t
3
)
Now by section formula
Ordinate
3 3
2t 8t
1 2
?
?
?
3
6t
3
?
?
= –2t
3
11. Let ? f : R R ? be defined as f(x) = 2x – 1 and
g : R – {1} ? R be defined as ? ?
?
?
?
1
x
2
g x .
x 1
Then the composition function f(g(x)) is :
(1) neither one-one nor onto
(2) onto but not one-one
(3) both one-one and onto
(4) one-one but not onto
Sol. Here f : R ? R, f(x) = 2x – 1
and g : R – {1} ? R g(x)
?
?
?
1
x
2
x 1
So, f(g(x)) = 2 g(x) – 1
? ?
?
? ?
? ?
? ?
?
? ?
? ?
? ?
1
x
2
2 1
x 1
? ? ? ? ?
? ?
? ?
2x 1 x 1 x 1 1
x 1 x 1
? ?
?
1
1
x 1
So Clearly it is one-one but not onto
0
(1, 1)
12.
? ?
?
?
2
X
0
3
x 0
sin t dt
lim
x
is equal to :
(1) 0
(2)
1
15
(3)
2
3
(4)
3
2
Sol.
? ?
?
?
?
2
x
0
3
x 0
sin t dt
0
lim (form)
0
x
? By D, L Hospital rule
? ?
? ?
?
? ?
? ?
2
x 0 x 0
2x sinx 2 sinx
lim lim
3 x
3x
? ? ?
2 2
1
3 3
Page 4

JEE (MAIN)-2021 Phase-1 (24-02-2021)-M
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.
1. If
?
? ? ? ?
? ?
? ?
? ? ?
?
1
cosx sinx sinx cosx
dx asin c,
b
8 sin2x
where c is a constant of integration, then the
ordered pair (a, b) is equal to :
(1) (–1, 3) (2) (3, 1)
(3) (1, –3) (4) (1, 3)
Sol.
? ?
2
cos x sinx cos x sinx
dx dx
8 sin2x
9 sinx cosx
? ?
?
?
? ?
? ?
let sinx + cosx = t
(cosx – sinx)dx = dt
2
dt
9 t
?
?
?
1
t
sin c
3
?
? ?
? ?
? ?
? ?
1
sinx cosx
sin c
3
?
? ? ?
? ?
? ?
? ?
Hence (a, b) = (1, 3)
2. The area (in sq. units) of the part of the circle
x
2
+ y
2
= 36, which is outside the parabola
y
2
= 9x, is :
(1) ? ? 12 3 3 (2) ? ? 24 3 3
(3) ? ? 12 3 3 (4) ? ? 24 3 3
Sol. Area of the shaded region
3 6
2
0 3
2 3 xdx 36 x dx
? ?
? ? ?
? ?
? ?
? ?
y = 9x
2
(3, 3 3)
3
x + y = 36
2 2
PART–C : MATHEMATICS
3
6
3/2 2 1
3 0
1 x
2 2x x 36 x 18sin
2 6
?
? ?
? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ?
? ?
9 3
2 6 3 9 3 3 3 12
2
? ?
? ? ? ? ? ? ? ? ?
? ?
? ?
? ?
Required area ? ?
36 3 3 12 ? ? ? ? ?
24 3 3 ? ? ?
3. If
? ?
? ? ? ?
2 4 6
e
cos x cos x cos x ... log 2
e satisfies the
equation t
2
– 9t + 8 = 0, then the value of
? ? ?
? ?
? ?
? ? ?
2sinx
0 x
2
sinx 3 cosx
is :
(1)
3
2
(2) 3
(3) 2 3 (4)
1
2
Sol.
2
2 2
cos x
ln2
1 cos x cot x
e 2
? ?
? ?
? ?
?
? ?
?
? t = 1 or 8
So,
2
cot x 0
2 2 ?
or 2
3
? cot
2
x = 0 or 3
? x 0,
2
? ? ?
?
? ?
? ?
then cot x 3 x
6
?
? ? ?
1
2
2sinx 1 2
2
sinx 3 cos x 1 3
3
2 2
? ?
? ?
? ?
? ?
? ? ?
? ?
? ?
? ?
? ?
4. The population P = P(t) at time ‘t’ of a certain
species follows the differential equation
? ?
dP
0.5P 450.
dt
If P(0) = 850, then the time at t
which population becomes zero is :
(1) log
e
18 (2)
e
1
log 18
2
(3) log
e
9 (4) 2log
e
18
Sol. ? ? ?
dP 1
P 900
dt 2
? ?
JEE (MAIN)-2021 Phase-1 (24-02-2021)-M
?
dP 1 1
dt ln P 900 t c
P 900 2 2
? ? ? ? ?
?
When t = 0, P = 850 ? c = ln50
When P = 0, t = 2(ln900 – ln50) = 2ln18
5. The statement among the following that is a
tautology is :
(1) A ? (A ? B) (2) B ? [A ? (A ? B)]
(3) [A ? (A ? B)] ? B (4) A ? (A ? B)
Sol. (1) A ? (A ? B) = A
(2) ? A ? (A ? B) = A ??( ? A ??B) = A ? B
So, B ? (A ? B) = ? B ??(A ? B) = ? B ? A
(3) (A ? (A ? B)) ? B = (A ? B) ? B = ? (A ? B)
? B = ? A ? ? B ? B (Tautology)
(4) A ? (A ? B) = A
6. Let p and q be two positive numbers such that
p + q = 2 and p
4
+ q
4
= 272. Then p and q are
roots of the equation :
(1) x
2
– 2x + 2 = 0 (2) x
2
– 2x + 8 = 0
(3) x
2
– 2x + 136 = 0 (4) x
2
– 2x + 16 = 0
Sol. ? p
4
+ q
4
= (p + 4)
4
– 4pq(p
2
+ q
2
) – 6p
2
q
2
? 272 = 16 – 4pq(4 – 2pq) – 6p
2
q
2
? 2p
2
q
2
– 16pq – 256 = 0
? pq = –8 or 16
? p, q > 0, so pq = 16
x
2
– 2x + 16 = 0
7. The system of linear equations
3x – 2y – kz = 10
2x – 4y – 2z = 6
x + 2y – z = 5m
is inconsistent if :
(1)
4
k 3, m
5
? ? (2) k 3, m R ? ?
(3)
4
k 3, m
5
? ? (4)
4
k 3, m
5
? ?
Sol. Here
3 2 k
2 4 2 0
1 2 1
? ?
? ? ? ? ?
?
? 3(4 + 4) + 2(–2 + 2) – k(4 + 4) = 0
? 24 + 0 – 8k = 0 ? k 3 ?
Now,
1
10 2 3
6 4 2 10(4 4) 2( 6 10m) 3(12 20m)
5m 2 1
? ?
? ? ? ? ? ? ? ? ? ? ?
?
= 80 – 12 + 20 m – 36 – 60 m
= 32 – 40 m
2
3 10 3
2 6 2 3( 6 10 m) 10( 2 2) 3(10 m 6)
1 5m 1
?
? ? ? ? ? ? ? ? ? ? ?
?
= –18 + 30 m + 0 – 30 m + 18 = 0
3
3 2 10
2 4 6 3( 20m 12) 2(10m 6) 10(4 4)
1 2 5 m
?
? ? ? ? ? ? ? ? ? ?
= –60 m – 36 + 20 m – 12 + 80
= –40 m + 32
For inconsistent we have k = 3, &
32 – 40 m ? 0 ?
4
m
5
?
8. A scientific committee is to be formed from 6
Indians and 8 foreigners, which includes at
least 2 Indians and double the number of
foreigners as Indians. Then the number of
ways, the committee can be formed, is :
(1) 560 (2) 1050
(3) 1625 (4) 575
Sol. Indians = 6, Foreigners = 8
According to questions
The no. of ways to form the committee are
(2I, 4F) or (3I, 6F) or (4I, 8F)
?
6
C
2
×
8
C
4
+
6
C
3
×
8
C
6
+
6
C
4
×
8
C
8
= 15 × 70 + 20 × 28 + 15 × 1
= 1625
9. The equation of the plane passing through the
point (1, 2, –3) and perpendicular to the planes
3x + y – 2z = 5 and 2x – 5y – z = 7, is:
(1) 3x – 10y – 2z + 11 = 0
(2) 6x – 5y – 2z – 2 = 0
(3) 6x – 5y + 2z + 10 = 0
(4) 11x + y  + 17z + 38 = 0
JEE (MAIN)-2021 Phase-1 (24-02-2021)-M
Sol. The given planes are 3x + y – 2z = 5 ...(1)
2x – 5y – z = 7 ...(2)
Since the required plane passes through (1, 2, –3)
So equation of this plane is
a(x – 1) + b(y – 2) + c(z + 3) = 0 ...(3)
Now this plane (3) is ? to the planes (1) & (2)
So 3a + b – 2c = 0
& 2a – 5b – c = 0
?
a b c
11 1 17
? ?
? ? ?
So equation of plane is 11(x – 1) + (y – 2)
+ 17(2 + 3) = 0
? 11x + y + 17z + 38 = 0
10. If the tangent to the curve y = x
3
at the point
P(t, t
3
) meets the curve again at Q, then the
ordinate of the point which divides PQ
internally in the ratio 1 : 2 is :
(1) –2t
3
(2) 2t
3
(3) 0 (4) –t
3
Sol. Curve is y = x
3
...(1)
So equation of tangent at (t, t
3
)
(y – t
3
) = 3t
2
(x – t) ...(2)
? It meets again the curve at Q
So solving (1) & (2) we get
x = –2t ? Q = (–2t, – 8t
3
)
Now by section formula
Ordinate
3 3
2t 8t
1 2
?
?
?
3
6t
3
?
?
= –2t
3
11. Let ? f : R R ? be defined as f(x) = 2x – 1 and
g : R – {1} ? R be defined as ? ?
?
?
?
1
x
2
g x .
x 1
Then the composition function f(g(x)) is :
(1) neither one-one nor onto
(2) onto but not one-one
(3) both one-one and onto
(4) one-one but not onto
Sol. Here f : R ? R, f(x) = 2x – 1
and g : R – {1} ? R g(x)
?
?
?
1
x
2
x 1
So, f(g(x)) = 2 g(x) – 1
? ?
?
? ?
? ?
? ?
?
? ?
? ?
? ?
1
x
2
2 1
x 1
? ? ? ? ?
? ?
? ?
2x 1 x 1 x 1 1
x 1 x 1
? ?
?
1
1
x 1
So Clearly it is one-one but not onto
0
(1, 1)
12.
? ?
?
?
2
X
0
3
x 0
sin t dt
lim
x
is equal to :
(1) 0
(2)
1
15
(3)
2
3
(4)
3
2
Sol.
? ?
?
?
?
2
x
0
3
x 0
sin t dt
0
lim (form)
0
x
? By D, L Hospital rule
? ?
? ?
?
? ?
? ?
2
x 0 x 0
2x sinx 2 sinx
lim lim
3 x
3x
? ? ?
2 2
1
3 3
JEE (MAIN)-2021 Phase-1 (24-02-2021)-M
13. An ordinary dice is rolled for a certain number
of times. If the probability of getting an odd
number 2 times is equal to the probability of
getting an even number 3 times, then the
probability of getting an odd number for odd
number of times is :
(1)
3
16
(2)
1
2
(3)
1
32
(4)
5
16
Sol. Let number of trials be ‘n’
given
? ?
? ? ? ? ? ? ? ?
?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
2 n 2 3 n 3
n n
2 3
1 1 1 1
C C
2 2 2 2
? n = 5
Probability of getting odd number for odd
number of times is
1 3 5
5
1
= (5C + 5C + 5C )
2
? ?
4
5
2 1
2
2
14. Two vertical poles are 150 m apart and the
height of one is three times that of the other. If
from the middle point of the line joining their
feet, an observer finds the angles of elevation
of their tops to be complementary, then the
height of the shorter pole (in meters) is :
(1) 25 3 (2) 30
(3) 25 (4) 20 3
Sol.
h
3h
75 75
90 – ? ?
Given
? ?
3h
tan
75
...(i)
and tan(90 – ?)
?
h
75
...(ii)
? Multiplying (i) and (ii) we get,
? ?
?
2
2
3h
1
75
?
? h 25 3
15. If f : R ? R is a function defined by f(x) = [x – 1]
? ? ?
?
? ?
? ?
2x 1
cos
2
, where [ ?] denotes the greatest
integer function, then f is :
(1) discontinuous only at x = 1
(2) continuous for every real x
(3) discontinuous at all integral values of x
except at x = 1
(4) continuous only at x = 1
Sol.
? ? ? ?
? ? ?
? ? ?
? ?
? ?
2x 1
f x x 1 cos
2
at x = 1
? ?
?
?
? ? ?
? ? ?
? ?
? ? x 1
2x 1
lim x 1 cos 0
2
? ?
?
?
? ? ?
? ? ?
? ?
? ? x 1
2x 1
lim x 1 cos 0
2
f(1) = 0
at any general integer x = k
? ?
?
?
? ? ?
? ? ?
? ?
? ? x K
2k 1
lim x 1 cos 0
2
? ?
?
?
? ? ?
? ? ?
? ?
? ? x K
2k 1
lim x 1 cos 0
2
f(k) = 0
? f(x) is continuous
? ? x R
16. The distance of the point (1, 1, 9) from the point
of intersection of the line
? ?
x – 3 y – 4 z – 5
1 2 2
and the plane x + y + z = 17 is:
(1) 38 (2) 19 2
(3) 2 19 (4) 38
Sol. Let a point P( ?) on the line
? ? ? ? ?
x – 3 y – 4 z – 5
1 2 2
? P( ? + 3, 2 ? + 4, 2 ? + 5)
as P also satisfies the given plane
Page 5

JEE (MAIN)-2021 Phase-1 (24-02-2021)-M
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.
1. If
?
? ? ? ?
? ?
? ?
? ? ?
?
1
cosx sinx sinx cosx
dx asin c,
b
8 sin2x
where c is a constant of integration, then the
ordered pair (a, b) is equal to :
(1) (–1, 3) (2) (3, 1)
(3) (1, –3) (4) (1, 3)
Sol.
? ?
2
cos x sinx cos x sinx
dx dx
8 sin2x
9 sinx cosx
? ?
?
?
? ?
? ?
let sinx + cosx = t
(cosx – sinx)dx = dt
2
dt
9 t
?
?
?
1
t
sin c
3
?
? ?
? ?
? ?
? ?
1
sinx cosx
sin c
3
?
? ? ?
? ?
? ?
? ?
Hence (a, b) = (1, 3)
2. The area (in sq. units) of the part of the circle
x
2
+ y
2
= 36, which is outside the parabola
y
2
= 9x, is :
(1) ? ? 12 3 3 (2) ? ? 24 3 3
(3) ? ? 12 3 3 (4) ? ? 24 3 3
Sol. Area of the shaded region
3 6
2
0 3
2 3 xdx 36 x dx
? ?
? ? ?
? ?
? ?
? ?
y = 9x
2
(3, 3 3)
3
x + y = 36
2 2
PART–C : MATHEMATICS
3
6
3/2 2 1
3 0
1 x
2 2x x 36 x 18sin
2 6
?
? ?
? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ?
? ?
9 3
2 6 3 9 3 3 3 12
2
? ?
? ? ? ? ? ? ? ? ?
? ?
? ?
? ?
Required area ? ?
36 3 3 12 ? ? ? ? ?
24 3 3 ? ? ?
3. If
? ?
? ? ? ?
2 4 6
e
cos x cos x cos x ... log 2
e satisfies the
equation t
2
– 9t + 8 = 0, then the value of
? ? ?
? ?
? ?
? ? ?
2sinx
0 x
2
sinx 3 cosx
is :
(1)
3
2
(2) 3
(3) 2 3 (4)
1
2
Sol.
2
2 2
cos x
ln2
1 cos x cot x
e 2
? ?
? ?
? ?
?
? ?
?
? t = 1 or 8
So,
2
cot x 0
2 2 ?
or 2
3
? cot
2
x = 0 or 3
? x 0,
2
? ? ?
?
? ?
? ?
then cot x 3 x
6
?
? ? ?
1
2
2sinx 1 2
2
sinx 3 cos x 1 3
3
2 2
? ?
? ?
? ?
? ?
? ? ?
? ?
? ?
? ?
? ?
4. The population P = P(t) at time ‘t’ of a certain
species follows the differential equation
? ?
dP
0.5P 450.
dt
If P(0) = 850, then the time at t
which population becomes zero is :
(1) log
e
18 (2)
e
1
log 18
2
(3) log
e
9 (4) 2log
e
18
Sol. ? ? ?
dP 1
P 900
dt 2
? ?
JEE (MAIN)-2021 Phase-1 (24-02-2021)-M
?
dP 1 1
dt ln P 900 t c
P 900 2 2
? ? ? ? ?
?
When t = 0, P = 850 ? c = ln50
When P = 0, t = 2(ln900 – ln50) = 2ln18
5. The statement among the following that is a
tautology is :
(1) A ? (A ? B) (2) B ? [A ? (A ? B)]
(3) [A ? (A ? B)] ? B (4) A ? (A ? B)
Sol. (1) A ? (A ? B) = A
(2) ? A ? (A ? B) = A ??( ? A ??B) = A ? B
So, B ? (A ? B) = ? B ??(A ? B) = ? B ? A
(3) (A ? (A ? B)) ? B = (A ? B) ? B = ? (A ? B)
? B = ? A ? ? B ? B (Tautology)
(4) A ? (A ? B) = A
6. Let p and q be two positive numbers such that
p + q = 2 and p
4
+ q
4
= 272. Then p and q are
roots of the equation :
(1) x
2
– 2x + 2 = 0 (2) x
2
– 2x + 8 = 0
(3) x
2
– 2x + 136 = 0 (4) x
2
– 2x + 16 = 0
Sol. ? p
4
+ q
4
= (p + 4)
4
– 4pq(p
2
+ q
2
) – 6p
2
q
2
? 272 = 16 – 4pq(4 – 2pq) – 6p
2
q
2
? 2p
2
q
2
– 16pq – 256 = 0
? pq = –8 or 16
? p, q > 0, so pq = 16
x
2
– 2x + 16 = 0
7. The system of linear equations
3x – 2y – kz = 10
2x – 4y – 2z = 6
x + 2y – z = 5m
is inconsistent if :
(1)
4
k 3, m
5
? ? (2) k 3, m R ? ?
(3)
4
k 3, m
5
? ? (4)
4
k 3, m
5
? ?
Sol. Here
3 2 k
2 4 2 0
1 2 1
? ?
? ? ? ? ?
?
? 3(4 + 4) + 2(–2 + 2) – k(4 + 4) = 0
? 24 + 0 – 8k = 0 ? k 3 ?
Now,
1
10 2 3
6 4 2 10(4 4) 2( 6 10m) 3(12 20m)
5m 2 1
? ?
? ? ? ? ? ? ? ? ? ? ?
?
= 80 – 12 + 20 m – 36 – 60 m
= 32 – 40 m
2
3 10 3
2 6 2 3( 6 10 m) 10( 2 2) 3(10 m 6)
1 5m 1
?
? ? ? ? ? ? ? ? ? ? ?
?
= –18 + 30 m + 0 – 30 m + 18 = 0
3
3 2 10
2 4 6 3( 20m 12) 2(10m 6) 10(4 4)
1 2 5 m
?
? ? ? ? ? ? ? ? ? ?
= –60 m – 36 + 20 m – 12 + 80
= –40 m + 32
For inconsistent we have k = 3, &
32 – 40 m ? 0 ?
4
m
5
?
8. A scientific committee is to be formed from 6
Indians and 8 foreigners, which includes at
least 2 Indians and double the number of
foreigners as Indians. Then the number of
ways, the committee can be formed, is :
(1) 560 (2) 1050
(3) 1625 (4) 575
Sol. Indians = 6, Foreigners = 8
According to questions
The no. of ways to form the committee are
(2I, 4F) or (3I, 6F) or (4I, 8F)
?
6
C
2
×
8
C
4
+
6
C
3
×
8
C
6
+
6
C
4
×
8
C
8
= 15 × 70 + 20 × 28 + 15 × 1
= 1625
9. The equation of the plane passing through the
point (1, 2, –3) and perpendicular to the planes
3x + y – 2z = 5 and 2x – 5y – z = 7, is:
(1) 3x – 10y – 2z + 11 = 0
(2) 6x – 5y – 2z – 2 = 0
(3) 6x – 5y + 2z + 10 = 0
(4) 11x + y  + 17z + 38 = 0
JEE (MAIN)-2021 Phase-1 (24-02-2021)-M
Sol. The given planes are 3x + y – 2z = 5 ...(1)
2x – 5y – z = 7 ...(2)
Since the required plane passes through (1, 2, –3)
So equation of this plane is
a(x – 1) + b(y – 2) + c(z + 3) = 0 ...(3)
Now this plane (3) is ? to the planes (1) & (2)
So 3a + b – 2c = 0
& 2a – 5b – c = 0
?
a b c
11 1 17
? ?
? ? ?
So equation of plane is 11(x – 1) + (y – 2)
+ 17(2 + 3) = 0
? 11x + y + 17z + 38 = 0
10. If the tangent to the curve y = x
3
at the point
P(t, t
3
) meets the curve again at Q, then the
ordinate of the point which divides PQ
internally in the ratio 1 : 2 is :
(1) –2t
3
(2) 2t
3
(3) 0 (4) –t
3
Sol. Curve is y = x
3
...(1)
So equation of tangent at (t, t
3
)
(y – t
3
) = 3t
2
(x – t) ...(2)
? It meets again the curve at Q
So solving (1) & (2) we get
x = –2t ? Q = (–2t, – 8t
3
)
Now by section formula
Ordinate
3 3
2t 8t
1 2
?
?
?
3
6t
3
?
?
= –2t
3
11. Let ? f : R R ? be defined as f(x) = 2x – 1 and
g : R – {1} ? R be defined as ? ?
?
?
?
1
x
2
g x .
x 1
Then the composition function f(g(x)) is :
(1) neither one-one nor onto
(2) onto but not one-one
(3) both one-one and onto
(4) one-one but not onto
Sol. Here f : R ? R, f(x) = 2x – 1
and g : R – {1} ? R g(x)
?
?
?
1
x
2
x 1
So, f(g(x)) = 2 g(x) – 1
? ?
?
? ?
? ?
? ?
?
? ?
? ?
? ?
1
x
2
2 1
x 1
? ? ? ? ?
? ?
? ?
2x 1 x 1 x 1 1
x 1 x 1
? ?
?
1
1
x 1
So Clearly it is one-one but not onto
0
(1, 1)
12.
? ?
?
?
2
X
0
3
x 0
sin t dt
lim
x
is equal to :
(1) 0
(2)
1
15
(3)
2
3
(4)
3
2
Sol.
? ?
?
?
?
2
x
0
3
x 0
sin t dt
0
lim (form)
0
x
? By D, L Hospital rule
? ?
? ?
?
? ?
? ?
2
x 0 x 0
2x sinx 2 sinx
lim lim
3 x
3x
? ? ?
2 2
1
3 3
JEE (MAIN)-2021 Phase-1 (24-02-2021)-M
13. An ordinary dice is rolled for a certain number
of times. If the probability of getting an odd
number 2 times is equal to the probability of
getting an even number 3 times, then the
probability of getting an odd number for odd
number of times is :
(1)
3
16
(2)
1
2
(3)
1
32
(4)
5
16
Sol. Let number of trials be ‘n’
given
? ?
? ? ? ? ? ? ? ?
?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
2 n 2 3 n 3
n n
2 3
1 1 1 1
C C
2 2 2 2
? n = 5
Probability of getting odd number for odd
number of times is
1 3 5
5
1
= (5C + 5C + 5C )
2
? ?
4
5
2 1
2
2
14. Two vertical poles are 150 m apart and the
height of one is three times that of the other. If
from the middle point of the line joining their
feet, an observer finds the angles of elevation
of their tops to be complementary, then the
height of the shorter pole (in meters) is :
(1) 25 3 (2) 30
(3) 25 (4) 20 3
Sol.
h
3h
75 75
90 – ? ?
Given
? ?
3h
tan
75
...(i)
and tan(90 – ?)
?
h
75
...(ii)
? Multiplying (i) and (ii) we get,
? ?
?
2
2
3h
1
75
?
? h 25 3
15. If f : R ? R is a function defined by f(x) = [x – 1]
? ? ?
?
? ?
? ?
2x 1
cos
2
, where [ ?] denotes the greatest
integer function, then f is :
(1) discontinuous only at x = 1
(2) continuous for every real x
(3) discontinuous at all integral values of x
except at x = 1
(4) continuous only at x = 1
Sol.
? ? ? ?
? ? ?
? ? ?
? ?
? ?
2x 1
f x x 1 cos
2
at x = 1
? ?
?
?
? ? ?
? ? ?
? ?
? ? x 1
2x 1
lim x 1 cos 0
2
? ?
?
?
? ? ?
? ? ?
? ?
? ? x 1
2x 1
lim x 1 cos 0
2
f(1) = 0
at any general integer x = k
? ?
?
?
? ? ?
? ? ?
? ?
? ? x K
2k 1
lim x 1 cos 0
2
? ?
?
?
? ? ?
? ? ?
? ?
? ? x K
2k 1
lim x 1 cos 0
2
f(k) = 0
? f(x) is continuous
? ? x R
16. The distance of the point (1, 1, 9) from the point
of intersection of the line
? ?
x – 3 y – 4 z – 5
1 2 2
and the plane x + y + z = 17 is:
(1) 38 (2) 19 2
(3) 2 19 (4) 38
Sol. Let a point P( ?) on the line
? ? ? ? ?
x – 3 y – 4 z – 5
1 2 2
? P( ? + 3, 2 ? + 4, 2 ? + 5)
as P also satisfies the given plane
JEE (MAIN)-2021 Phase-1 (24-02-2021)-M
? + 3 + 2 ? + 4 + 2 ? + 5 = 17
? 5 ? = 5 ? ? = 1
? P ? (4, 6, 7)
Distance from (1, 1, 9) is

? ?
2 2 2
(4 – 1 ) (6 – 1 ) (7 – 9)
? ? ? ? 9 25 4 38
17. The locus of the mid-point of the line segment
joining the focus of the parabola y
2
= 4ax to a
moving point of the parabola, is another
parabola whose directrix is
(1) x = a
(2) ?
a
x
2
(3) x = 0
(4) ?
a
x –
2
Sol. Let the moving point be P(at
2
, 2at)
Focus of given parabola is (a, 0)
Let point of required locus (h, k)
?
?
?
2
at a
h
2
...(i)
and
?
?
2at 0
k
2
...(ii)
? ? ?
2
a
(t 1 ) h
2
...(iii)
and ?
k
t
a
...(iv)
By (iii) and (iv) we have
? ?
? ? ? ?
? ?
? ?
2
2
a k
1 h
2
a
Locus is k
2
+ a
2
= 2ah
?
? ?
?
? ?
? ?
2
a
y 2a x –
2
Equation of directrix ? ?
a a
x – 0
2 2
? x = 0
18. A man is walking on a straight line. The
arithmetic mean of the reciprocals of the
intercepts of this line on the coordinate axes is
1
4
. Three stones A, B and C are placed at the
points (1, 1), (2, 2) and (4, 4) respectively. Then
which of these stones is/are on the path of the
man?
(1) C only (2) B only
(3) All the three (4) A only
Sol. Let line be ? ?
x y
1
a b
...(i)
given
?
?
1 1
1
a b
2 4
? ? ?
1 1 1
a b 2
...(ii)
By (i) and (ii), we get
? ?
? ?
? ?
? ?
x 1 1
– y 1
a 2 a
?
? ?
? ? ?
? ?
? ?
y
(x – y) – 1 0
2
? Represents family of line passing through
(2, 2)
19. The function
? ?
3 2
4x – 3x
f(x) – 2 sinx (2x – 1 )cosx :
6
(1) Decreases in
? ?
?
? ?
? ?
1
,
2
(2) Decreases in
? ?
?
? ?
? ?
1
– ,
2
(3) Increases in
? ?
?
? ?
? ?
1
– ,
2
(4) Increases in
? ?
?
? ?
? ?
1
,
2
Sol. ?
? ?
3 2
4x – 3x
f(x) – 2 sinx (2x – 1 )cos x
6
On differentiating both sides w.r .t. x we get
? ? ?
2
12x – 6x
f (x) – 2cosx – (2x – 1)sinx 2cos x
6
? ?
2
f (x) 2x – x – (2x – 1 )sinx
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## Mock Test Series for JEE Main & Advanced

2 videos|324 docs|160 tests

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