JEE Main 2021 Mathematics March 16 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


 
 
16
th
March. 2021 | Shift 1
SECTION –A 
 
1. Consider three observations a, b and c such that b = a+c. If the standard deviation of a+2, 
b+2, c+2 is d, then which of the following is true? 
 (1)
2 2 2 2
b a c 3d ? ? ? 
 (2)
? ?
2 2 2 2
b 3 a c 9d ? ? ? 
 (3)
? ?
2 2 2 2
b 3 a c 9d ? ? ? 
 (4)
? ?
2 2 2 2
b 3 a c d ? ? ?
 
Ans. (2) 
Sol. for a, b, c 
 mean = 
a b c
x
3
? ?
? 
 
2b  
x
3
? 
 S.D. of a, b, c = d 
 
2 2 2 2
2
a b c 4b
d
3 9
? ?
? ? 
 b
2
 = 3a
2
 + 3c
2
 – 9d
2
 
 
2. Let a vector 
ˆ ˆ
i j ? ? ? be obtained by rotating the vector 
ˆ ˆ
3i j ? by an angle 45° about the origin in 
counter clockwise direction in the first quadrant. Then the area of triangle having vertices ( , ) ? ? , 
(0, ) ? and (0, 0) is equal to : 
 (1) 1 
 (2)
1
2
 
 (3)
1
2
 
 (4)2 2
 
Ans. (2) 
Sol.
  
 
  ? ? , (2cos75 ,2 sin75 ) ? ? ? ? ? 
 Area = 
1
2
 (2 cos75°)(2 sin 75°) 
 = sin(150°) =
1
2
 square unit 
( ?, ?) 
( ?, ?) 
(0,0) 
r=2 
30° 
45° 
? ?
3,1
Page 2


 
 
16
th
March. 2021 | Shift 1
SECTION –A 
 
1. Consider three observations a, b and c such that b = a+c. If the standard deviation of a+2, 
b+2, c+2 is d, then which of the following is true? 
 (1)
2 2 2 2
b a c 3d ? ? ? 
 (2)
? ?
2 2 2 2
b 3 a c 9d ? ? ? 
 (3)
? ?
2 2 2 2
b 3 a c 9d ? ? ? 
 (4)
? ?
2 2 2 2
b 3 a c d ? ? ?
 
Ans. (2) 
Sol. for a, b, c 
 mean = 
a b c
x
3
? ?
? 
 
2b  
x
3
? 
 S.D. of a, b, c = d 
 
2 2 2 2
2
a b c 4b
d
3 9
? ?
? ? 
 b
2
 = 3a
2
 + 3c
2
 – 9d
2
 
 
2. Let a vector 
ˆ ˆ
i j ? ? ? be obtained by rotating the vector 
ˆ ˆ
3i j ? by an angle 45° about the origin in 
counter clockwise direction in the first quadrant. Then the area of triangle having vertices ( , ) ? ? , 
(0, ) ? and (0, 0) is equal to : 
 (1) 1 
 (2)
1
2
 
 (3)
1
2
 
 (4)2 2
 
Ans. (2) 
Sol.
  
 
  ? ? , (2cos75 ,2 sin75 ) ? ? ? ? ? 
 Area = 
1
2
 (2 cos75°)(2 sin 75°) 
 = sin(150°) =
1
2
 square unit 
( ?, ?) 
( ?, ?) 
(0,0) 
r=2 
30° 
45° 
? ?
3,1
 
 
3. If for a>0, the feet of perpendiculars from the points A(a, –2a, 3) and B(0, 4, 5) on the plane lx 
+ my + nz = 0 are points C(0, –a, –1) and D respectively, then the length of line segment CD is 
equal to : 
 (1) 41 
 (2) 55 
 (3) 31 
 (4) 66
 
Ans. (4) 
Sol.
 
 
A 
B 
R 
C 
D 
?
n
 
 
CD =  AR = |AB|sin ?  
CD = |AB| 
2
1 cos ? ?
 
2
AB. n
| AB| 1
| AB|
? ?
? ? ?
? ?
? ?
? ? ? ?
?
? ? ? ? 
= ?
? ? ? ?
?
2 2
(AB) (AB·n) 
 Cos ? = 
? ? ? ?
?
? ? ? ?
?
AB·n
|n|| AB|
 
 
? ? ? ?
| AB| = 
ˆ ˆ ˆ
ai –(2a 4) j 2k ? ? 
? ? ? ?
AB·
?
n = ?a – (2a + 4)– 2n 
C on plane   
0 ? – am – n = 0 …. (1) 
? ? ? ?
AC || 
?
n 
Page 3


 
 
16
th
March. 2021 | Shift 1
SECTION –A 
 
1. Consider three observations a, b and c such that b = a+c. If the standard deviation of a+2, 
b+2, c+2 is d, then which of the following is true? 
 (1)
2 2 2 2
b a c 3d ? ? ? 
 (2)
? ?
2 2 2 2
b 3 a c 9d ? ? ? 
 (3)
? ?
2 2 2 2
b 3 a c 9d ? ? ? 
 (4)
? ?
2 2 2 2
b 3 a c d ? ? ?
 
Ans. (2) 
Sol. for a, b, c 
 mean = 
a b c
x
3
? ?
? 
 
2b  
x
3
? 
 S.D. of a, b, c = d 
 
2 2 2 2
2
a b c 4b
d
3 9
? ?
? ? 
 b
2
 = 3a
2
 + 3c
2
 – 9d
2
 
 
2. Let a vector 
ˆ ˆ
i j ? ? ? be obtained by rotating the vector 
ˆ ˆ
3i j ? by an angle 45° about the origin in 
counter clockwise direction in the first quadrant. Then the area of triangle having vertices ( , ) ? ? , 
(0, ) ? and (0, 0) is equal to : 
 (1) 1 
 (2)
1
2
 
 (3)
1
2
 
 (4)2 2
 
Ans. (2) 
Sol.
  
 
  ? ? , (2cos75 ,2 sin75 ) ? ? ? ? ? 
 Area = 
1
2
 (2 cos75°)(2 sin 75°) 
 = sin(150°) =
1
2
 square unit 
( ?, ?) 
( ?, ?) 
(0,0) 
r=2 
30° 
45° 
? ?
3,1
 
 
3. If for a>0, the feet of perpendiculars from the points A(a, –2a, 3) and B(0, 4, 5) on the plane lx 
+ my + nz = 0 are points C(0, –a, –1) and D respectively, then the length of line segment CD is 
equal to : 
 (1) 41 
 (2) 55 
 (3) 31 
 (4) 66
 
Ans. (4) 
Sol.
 
 
A 
B 
R 
C 
D 
?
n
 
 
CD =  AR = |AB|sin ?  
CD = |AB| 
2
1 cos ? ?
 
2
AB. n
| AB| 1
| AB|
? ?
? ? ?
? ?
? ?
? ? ? ?
?
? ? ? ? 
= ?
? ? ? ?
?
2 2
(AB) (AB·n) 
 Cos ? = 
? ? ? ?
?
? ? ? ?
?
AB·n
|n|| AB|
 
 
? ? ? ?
| AB| = 
ˆ ˆ ˆ
ai –(2a 4) j 2k ? ? 
? ? ? ?
AB·
?
n = ?a – (2a + 4)– 2n 
C on plane   
0 ? – am – n = 0 …. (1) 
? ? ? ?
AC || 
?
n 
 
 
16
th
March. 2021 | Shift 1
?
? ?
?
a a 4
m n
 
m = – ? & an + 4m = 0 … (2)  
From (1) and (2) 
a
2
m + an = 0 
4m + an = 0 
 (a
2
 – 4)m = 0 ? ? a = 2  .  
2m + n = 0  … (1) 
m + ? = 0 
?
2 
+ m
2 
+ n
2
 = 1 
m
2 
+ m
2 
+ 4m
2
 = 1 
m
2
 = 
1
6
 
m = 
1
6
 
n= 
2
6
?
 
? =  
1
6
?
 
Now   
? ? ? ?
?
AB.n = 2
1
6
? ? ?
? ?
? ?
– 8
1
6
? ?
? ?
? ?
–2
2
6
? ? ?
? ?
? ?
 
  = 
2 8 4
6
? ? ?
= – 6 
 
? ? ? ?
| AB| = 4 64 4 ? ? = 72 
 CD =  72 6 ? 
 CD = 66 
 
4. The range of a R ? for which the function 
 f(x) = (4a–3) (x + log
e
5) + 2(a–7) cot 
x
2
? ?
? ?
? ?
sin
2
x
2
? ?
? ?
? ?
, x 2n ,n N ? ? ? has critical points, is : 
 (1)
4
,2
3
? ?
?
? ?
? ?
 
 (2) ? 1, ? ?
?
 
Page 4


 
 
16
th
March. 2021 | Shift 1
SECTION –A 
 
1. Consider three observations a, b and c such that b = a+c. If the standard deviation of a+2, 
b+2, c+2 is d, then which of the following is true? 
 (1)
2 2 2 2
b a c 3d ? ? ? 
 (2)
? ?
2 2 2 2
b 3 a c 9d ? ? ? 
 (3)
? ?
2 2 2 2
b 3 a c 9d ? ? ? 
 (4)
? ?
2 2 2 2
b 3 a c d ? ? ?
 
Ans. (2) 
Sol. for a, b, c 
 mean = 
a b c
x
3
? ?
? 
 
2b  
x
3
? 
 S.D. of a, b, c = d 
 
2 2 2 2
2
a b c 4b
d
3 9
? ?
? ? 
 b
2
 = 3a
2
 + 3c
2
 – 9d
2
 
 
2. Let a vector 
ˆ ˆ
i j ? ? ? be obtained by rotating the vector 
ˆ ˆ
3i j ? by an angle 45° about the origin in 
counter clockwise direction in the first quadrant. Then the area of triangle having vertices ( , ) ? ? , 
(0, ) ? and (0, 0) is equal to : 
 (1) 1 
 (2)
1
2
 
 (3)
1
2
 
 (4)2 2
 
Ans. (2) 
Sol.
  
 
  ? ? , (2cos75 ,2 sin75 ) ? ? ? ? ? 
 Area = 
1
2
 (2 cos75°)(2 sin 75°) 
 = sin(150°) =
1
2
 square unit 
( ?, ?) 
( ?, ?) 
(0,0) 
r=2 
30° 
45° 
? ?
3,1
 
 
3. If for a>0, the feet of perpendiculars from the points A(a, –2a, 3) and B(0, 4, 5) on the plane lx 
+ my + nz = 0 are points C(0, –a, –1) and D respectively, then the length of line segment CD is 
equal to : 
 (1) 41 
 (2) 55 
 (3) 31 
 (4) 66
 
Ans. (4) 
Sol.
 
 
A 
B 
R 
C 
D 
?
n
 
 
CD =  AR = |AB|sin ?  
CD = |AB| 
2
1 cos ? ?
 
2
AB. n
| AB| 1
| AB|
? ?
? ? ?
? ?
? ?
? ? ? ?
?
? ? ? ? 
= ?
? ? ? ?
?
2 2
(AB) (AB·n) 
 Cos ? = 
? ? ? ?
?
? ? ? ?
?
AB·n
|n|| AB|
 
 
? ? ? ?
| AB| = 
ˆ ˆ ˆ
ai –(2a 4) j 2k ? ? 
? ? ? ?
AB·
?
n = ?a – (2a + 4)– 2n 
C on plane   
0 ? – am – n = 0 …. (1) 
? ? ? ?
AC || 
?
n 
 
 
16
th
March. 2021 | Shift 1
?
? ?
?
a a 4
m n
 
m = – ? & an + 4m = 0 … (2)  
From (1) and (2) 
a
2
m + an = 0 
4m + an = 0 
 (a
2
 – 4)m = 0 ? ? a = 2  .  
2m + n = 0  … (1) 
m + ? = 0 
?
2 
+ m
2 
+ n
2
 = 1 
m
2 
+ m
2 
+ 4m
2
 = 1 
m
2
 = 
1
6
 
m = 
1
6
 
n= 
2
6
?
 
? =  
1
6
?
 
Now   
? ? ? ?
?
AB.n = 2
1
6
? ? ?
? ?
? ?
– 8
1
6
? ?
? ?
? ?
–2
2
6
? ? ?
? ?
? ?
 
  = 
2 8 4
6
? ? ?
= – 6 
 
? ? ? ?
| AB| = 4 64 4 ? ? = 72 
 CD =  72 6 ? 
 CD = 66 
 
4. The range of a R ? for which the function 
 f(x) = (4a–3) (x + log
e
5) + 2(a–7) cot 
x
2
? ?
? ?
? ?
sin
2
x
2
? ?
? ?
? ?
, x 2n ,n N ? ? ? has critical points, is : 
 (1)
4
,2
3
? ?
?
? ?
? ?
 
 (2) ? 1, ? ?
?
 
 
 
 (3) ? ,–1 ? ? ?
?
 
 (4) (–3, 1) 
Ans. (1) 
Sol. f(x) = (4a – 3) (x + ln5) + 2(a – 7) 
2
x
cos
x
2
sin
x 2
sin
2
? ?
? ?
? ? ?
? ?
? ?
? ?
 
 f(x) = (4a – 3) (x + ln5) + (a – 7) sinx 
f'(x) = (4a – 3) + (a – 7) cosx = 0 
(4a 3)
cos x
a 7
? ?
?
?
 
4a 3
–1 1
a 7
?
? ? ?
?
 
4a 3
1 1
a 7
?
? ? ?
?
 
4a 3
1 0
a 7
?
? ?
?
and
4a 3
1 0
a 7
?
? ?
?
 
4
a 2
3
?
? ? ? 
 
5. Let the functions f: R ?R and g: R ?R be defined as : 
 
2
x 2, x 0
f(x)
x , x 0
? ? ?
?
?
?
? ?
?
and
3
x , x 1
g(x)
3x 2, x 1
?
? ?
?
?
? ? ?
?
 
Then, the number of points in R where (fog)(x) is NOT differentiable is equal to :  
 (1) 1 
 (2) 2 
 (3) 3 
 (4) 0 
Ans. (1) 
Sol. 
3
6
2
x 2, x 0
fog(x) x 0 x 1
(3 2 x
,
, x ) 1
?
? ?
?
?
? ? ?
?
?
? ?
?
?
 
 ?fog(x) is discontinuous at x = 0 then non-differentiable at x = 0 
 Now, 
at x = 1 
2
h 0 h 0
f(1 h) f(1) (3(1 h) 2) 1
RHD lim lim 6
h h   ? ?
? ? ? ? ?
? ? ? 
6
h 0 h 0
f(1 h) f(1) (1 h) 1
LHD lim lim 6
h h ? ?
? ? ? ?
? ? ?
? ?
 
 Number of points of non-differentiability = 1 
 
Page 5


 
 
16
th
March. 2021 | Shift 1
SECTION –A 
 
1. Consider three observations a, b and c such that b = a+c. If the standard deviation of a+2, 
b+2, c+2 is d, then which of the following is true? 
 (1)
2 2 2 2
b a c 3d ? ? ? 
 (2)
? ?
2 2 2 2
b 3 a c 9d ? ? ? 
 (3)
? ?
2 2 2 2
b 3 a c 9d ? ? ? 
 (4)
? ?
2 2 2 2
b 3 a c d ? ? ?
 
Ans. (2) 
Sol. for a, b, c 
 mean = 
a b c
x
3
? ?
? 
 
2b  
x
3
? 
 S.D. of a, b, c = d 
 
2 2 2 2
2
a b c 4b
d
3 9
? ?
? ? 
 b
2
 = 3a
2
 + 3c
2
 – 9d
2
 
 
2. Let a vector 
ˆ ˆ
i j ? ? ? be obtained by rotating the vector 
ˆ ˆ
3i j ? by an angle 45° about the origin in 
counter clockwise direction in the first quadrant. Then the area of triangle having vertices ( , ) ? ? , 
(0, ) ? and (0, 0) is equal to : 
 (1) 1 
 (2)
1
2
 
 (3)
1
2
 
 (4)2 2
 
Ans. (2) 
Sol.
  
 
  ? ? , (2cos75 ,2 sin75 ) ? ? ? ? ? 
 Area = 
1
2
 (2 cos75°)(2 sin 75°) 
 = sin(150°) =
1
2
 square unit 
( ?, ?) 
( ?, ?) 
(0,0) 
r=2 
30° 
45° 
? ?
3,1
 
 
3. If for a>0, the feet of perpendiculars from the points A(a, –2a, 3) and B(0, 4, 5) on the plane lx 
+ my + nz = 0 are points C(0, –a, –1) and D respectively, then the length of line segment CD is 
equal to : 
 (1) 41 
 (2) 55 
 (3) 31 
 (4) 66
 
Ans. (4) 
Sol.
 
 
A 
B 
R 
C 
D 
?
n
 
 
CD =  AR = |AB|sin ?  
CD = |AB| 
2
1 cos ? ?
 
2
AB. n
| AB| 1
| AB|
? ?
? ? ?
? ?
? ?
? ? ? ?
?
? ? ? ? 
= ?
? ? ? ?
?
2 2
(AB) (AB·n) 
 Cos ? = 
? ? ? ?
?
? ? ? ?
?
AB·n
|n|| AB|
 
 
? ? ? ?
| AB| = 
ˆ ˆ ˆ
ai –(2a 4) j 2k ? ? 
? ? ? ?
AB·
?
n = ?a – (2a + 4)– 2n 
C on plane   
0 ? – am – n = 0 …. (1) 
? ? ? ?
AC || 
?
n 
 
 
16
th
March. 2021 | Shift 1
?
? ?
?
a a 4
m n
 
m = – ? & an + 4m = 0 … (2)  
From (1) and (2) 
a
2
m + an = 0 
4m + an = 0 
 (a
2
 – 4)m = 0 ? ? a = 2  .  
2m + n = 0  … (1) 
m + ? = 0 
?
2 
+ m
2 
+ n
2
 = 1 
m
2 
+ m
2 
+ 4m
2
 = 1 
m
2
 = 
1
6
 
m = 
1
6
 
n= 
2
6
?
 
? =  
1
6
?
 
Now   
? ? ? ?
?
AB.n = 2
1
6
? ? ?
? ?
? ?
– 8
1
6
? ?
? ?
? ?
–2
2
6
? ? ?
? ?
? ?
 
  = 
2 8 4
6
? ? ?
= – 6 
 
? ? ? ?
| AB| = 4 64 4 ? ? = 72 
 CD =  72 6 ? 
 CD = 66 
 
4. The range of a R ? for which the function 
 f(x) = (4a–3) (x + log
e
5) + 2(a–7) cot 
x
2
? ?
? ?
? ?
sin
2
x
2
? ?
? ?
? ?
, x 2n ,n N ? ? ? has critical points, is : 
 (1)
4
,2
3
? ?
?
? ?
? ?
 
 (2) ? 1, ? ?
?
 
 
 
 (3) ? ,–1 ? ? ?
?
 
 (4) (–3, 1) 
Ans. (1) 
Sol. f(x) = (4a – 3) (x + ln5) + 2(a – 7) 
2
x
cos
x
2
sin
x 2
sin
2
? ?
? ?
? ? ?
? ?
? ?
? ?
 
 f(x) = (4a – 3) (x + ln5) + (a – 7) sinx 
f'(x) = (4a – 3) + (a – 7) cosx = 0 
(4a 3)
cos x
a 7
? ?
?
?
 
4a 3
–1 1
a 7
?
? ? ?
?
 
4a 3
1 1
a 7
?
? ? ?
?
 
4a 3
1 0
a 7
?
? ?
?
and
4a 3
1 0
a 7
?
? ?
?
 
4
a 2
3
?
? ? ? 
 
5. Let the functions f: R ?R and g: R ?R be defined as : 
 
2
x 2, x 0
f(x)
x , x 0
? ? ?
?
?
?
? ?
?
and
3
x , x 1
g(x)
3x 2, x 1
?
? ?
?
?
? ? ?
?
 
Then, the number of points in R where (fog)(x) is NOT differentiable is equal to :  
 (1) 1 
 (2) 2 
 (3) 3 
 (4) 0 
Ans. (1) 
Sol. 
3
6
2
x 2, x 0
fog(x) x 0 x 1
(3 2 x
,
, x ) 1
?
? ?
?
?
? ? ?
?
?
? ?
?
?
 
 ?fog(x) is discontinuous at x = 0 then non-differentiable at x = 0 
 Now, 
at x = 1 
2
h 0 h 0
f(1 h) f(1) (3(1 h) 2) 1
RHD lim lim 6
h h   ? ?
? ? ? ? ?
? ? ? 
6
h 0 h 0
f(1 h) f(1) (1 h) 1
LHD lim lim 6
h h ? ?
? ? ? ?
? ? ?
? ?
 
 Number of points of non-differentiability = 1 
 
 
 
16
th
March. 2021 | Shift 1
6. Let a complex number z, |z| ? 1, satisfy 
1
2
2
| z| 11
log 2
(| z | 1)
? ?
?
?
? ?
? ?
?
? ?
. Then, the largest value of |z| is 
equal to ____ 
 (1) 5 
 (2) 8 
 (3) 6 
 (4) 7 
Ans. (4) 
Sol. 
? ?
2
| z | 11 1
2
| z | 1
?
?
?
 
2|z| + 22 ?  (|z| – 1)
2
 
2|z| + 22 ?|z|
2
 – 2|z| + 1 
|z|
2
 – 4|z| – 21 ?0 
(|z| – 7) (|z| + 3) ?0 
? |z| ?7 
?|z|
max
 = 7 
 
7. A pack of cards has one card missing. Two cards are drawn randomly and are found to be 
spades. The probability that the missing card is not a spade, is : 
 (1)
3
4
 
 (2)
52
867
 
 (3)
39
50
 
 (4)
22
425
 
Ans. (3) 
Sol. 
? ?
missing P S / both found spade = 
? ? m
P S BFS
P(BFS)
?
 
 
13 13 12
1
52 51 50
13 13 12 13 12 11
1
52 51 50 52 51 50
? ?
? ? ?
? ?
? ?
?
? ?
? ? ? ? ? ?
? ?
? ?
 
 
39
50
? 
8. If n is the number of irrational terms in the expansion of 
60
1 1
8 4
3 5
? ?
? ? ?
? ?
? ?
, then (n–1) is divisible by : 
 (1) 8 
 (2) 26 
 (3) 7 
 (4) 30