JEE Main 2021 Mathematics March 16 Shift 1 Paper & Solutions

# JEE Main 2021 Mathematics March 16 Shift 1 Paper & Solutions | JEE Main & Advanced Mock Test Series 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
```

## JEE Main & Advanced Mock Test Series

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## JEE Main & Advanced Mock Test Series

366 docs|219 tests

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