JEE Main 2021 Mathematics March 16 Shift 2 Paper & Solutions

# JEE Main 2021 Mathematics March 16 Shift 2 Paper & Solutions | JEE Main & Advanced Mock Test Series PDF Download

``` Page 1

16
th
March. 2021 | Shift 2
SECTION – A

1. The least value of |z| where z is complex number which satisfies the inequality exp
? ? ? ?
e
2
z 3 z 1
log 2 log 5 7 9i ,i 1
z 1
? ?
? ?
? ? ? ? ? ?
? ? ?
? ?
is equal to :
(1) 2

(2) 3
(3) 8

(4) 5
Ans. (2)
Sol.
? ? ? ?
? ?
? ? ? ?
z 3 z 1
z 1
3
z 3 z 1
2 2 3
z 1
? ?
?
? ?
? ? ?
?

2
z 2 z 3 3 z 3 ? ? ? ? ?

2
z z 6 0 ? ? ? ?

? ? ? ?
z 3 z 2 0 ? ? ?

min
z 3 ?

2. Let f : S ? ?S where S = (0, ?) be a twice differentiable function such that f(x+1) = xf(x). If g :
S ? R be defined as g(x) = logef(x), then the value of |g”(5) – g”(1)| is equal to :
(1)
197
144

(2)
187
144

(3)
205
144

(4) 1
Ans. (3)
Sol. f(x+1) = xf(x)
g(x+1) = loge(f(x+1))
g(x+1) logex + logf(x)
g(x+1) – g(x) = logex
g’’(x+1) – g’’(x) = –
2
1
x

Page 2

16
th
March. 2021 | Shift 2
SECTION – A

1. The least value of |z| where z is complex number which satisfies the inequality exp
? ? ? ?
e
2
z 3 z 1
log 2 log 5 7 9i ,i 1
z 1
? ?
? ?
? ? ? ? ? ?
? ? ?
? ?
is equal to :
(1) 2

(2) 3
(3) 8

(4) 5
Ans. (2)
Sol.
? ? ? ?
? ?
? ? ? ?
z 3 z 1
z 1
3
z 3 z 1
2 2 3
z 1
? ?
?
? ?
? ? ?
?

2
z 2 z 3 3 z 3 ? ? ? ? ?

2
z z 6 0 ? ? ? ?

? ? ? ?
z 3 z 2 0 ? ? ?

min
z 3 ?

2. Let f : S ? ?S where S = (0, ?) be a twice differentiable function such that f(x+1) = xf(x). If g :
S ? R be defined as g(x) = logef(x), then the value of |g”(5) – g”(1)| is equal to :
(1)
197
144

(2)
187
144

(3)
205
144

(4) 1
Ans. (3)
Sol. f(x+1) = xf(x)
g(x+1) = loge(f(x+1))
g(x+1) logex + logf(x)
g(x+1) – g(x) = logex
g’’(x+1) – g’’(x) = –
2
1
x

g’’(2) – g’’(1) = –1
g’’(3) – g’’(2) = –
1
4

g’’(4) – g’’(3) = –
1
9

g’’(5) – g’’(4) = –
1
16

g’’(5) – g’’(1) = –
1 1 1
1
4 9 16
? ?
? ? ?
? ?
? ?

? ? ? ?
144 36 16 9
g'' 5 g'' 1
16 9
? ? ? ? ?
? ?
? ?
?
? ?
=
205
16 9
? ?
? ?
?
? ?

3. If y = y(x) is the solution of the differential equation ? ?
dy
tanx y sinx,0 x
dx 3
?
? ? ? ? , with y(0) =
0, then y
4
? ? ?
? ?
? ?
equal to :
(1) loge2

(2)
e
1
log 2
2

(3)
1
2 2
? ?
? ?
? ?
loge2

(4)
e
1
log 2
4

Ans. (3)
Sol. I.f. =
tanxdx
e
?

=
ln sec x
e
? ?
? ?

= secx
Solution of the equation
y(secx) = ? ? ? ? sinx sec x dx
?

? ?
y
n sec x c
cos x
? ? ? ?
Put x = 0, c = 0
? ?y = cosx ?n(sec x)
put x = ?/4
Page 3

16
th
March. 2021 | Shift 2
SECTION – A

1. The least value of |z| where z is complex number which satisfies the inequality exp
? ? ? ?
e
2
z 3 z 1
log 2 log 5 7 9i ,i 1
z 1
? ?
? ?
? ? ? ? ? ?
? ? ?
? ?
is equal to :
(1) 2

(2) 3
(3) 8

(4) 5
Ans. (2)
Sol.
? ? ? ?
? ?
? ? ? ?
z 3 z 1
z 1
3
z 3 z 1
2 2 3
z 1
? ?
?
? ?
? ? ?
?

2
z 2 z 3 3 z 3 ? ? ? ? ?

2
z z 6 0 ? ? ? ?

? ? ? ?
z 3 z 2 0 ? ? ?

min
z 3 ?

2. Let f : S ? ?S where S = (0, ?) be a twice differentiable function such that f(x+1) = xf(x). If g :
S ? R be defined as g(x) = logef(x), then the value of |g”(5) – g”(1)| is equal to :
(1)
197
144

(2)
187
144

(3)
205
144

(4) 1
Ans. (3)
Sol. f(x+1) = xf(x)
g(x+1) = loge(f(x+1))
g(x+1) logex + logf(x)
g(x+1) – g(x) = logex
g’’(x+1) – g’’(x) = –
2
1
x

g’’(2) – g’’(1) = –1
g’’(3) – g’’(2) = –
1
4

g’’(4) – g’’(3) = –
1
9

g’’(5) – g’’(4) = –
1
16

g’’(5) – g’’(1) = –
1 1 1
1
4 9 16
? ?
? ? ?
? ?
? ?

? ? ? ?
144 36 16 9
g'' 5 g'' 1
16 9
? ? ? ? ?
? ?
? ?
?
? ?
=
205
16 9
? ?
? ?
?
? ?

3. If y = y(x) is the solution of the differential equation ? ?
dy
tanx y sinx,0 x
dx 3
?
? ? ? ? , with y(0) =
0, then y
4
? ? ?
? ?
? ?
equal to :
(1) loge2

(2)
e
1
log 2
2

(3)
1
2 2
? ?
? ?
? ?
loge2

(4)
e
1
log 2
4

Ans. (3)
Sol. I.f. =
tanxdx
e
?

=
ln sec x
e
? ?
? ?

= secx
Solution of the equation
y(secx) = ? ? ? ? sinx sec x dx
?

? ?
y
n sec x c
cos x
? ? ? ?
Put x = 0, c = 0
? ?y = cosx ?n(sec x)
put x = ?/4

16
th
March. 2021 | Shift 2
y =
1 1
ln 2 n2
2 2 2
? ?
y =
ln2
2 2

4. If the foot of the perpendicular from point (4, 3, 8) on the line L1 :
x a y 2 z b
3 4
? ? ?
? ?
?
, ? ? 0 is
(3, 5, 7), then the shortest distance between the line L1 and line L2 :
x 2 y 4 z 5
3 4 5
? ? ?
? ? is
equal to :
(1)
2
3

(2)
1
3

(3)
1
2

(4)
1
6

Ans. (4)
Sol. (3, 5, 7) lie on given line L1

3 a 3 7 b
3 4
? ?
? ?
?

7 b
1 b 3
4
?
? ? ?

3 a
1 3 a
?
? ? ? ? ?
?

A (4, 3, 8)
B (3, 5, 7)
DR’S of AB = (1, –2, 1)
AB ? line L1
(1)( ?) + (–2)(3) + 4(1) = 0
? ? ?? = 2
a = 1
a = 1, b = 3, ? = 2
Page 4

16
th
March. 2021 | Shift 2
SECTION – A

1. The least value of |z| where z is complex number which satisfies the inequality exp
? ? ? ?
e
2
z 3 z 1
log 2 log 5 7 9i ,i 1
z 1
? ?
? ?
? ? ? ? ? ?
? ? ?
? ?
is equal to :
(1) 2

(2) 3
(3) 8

(4) 5
Ans. (2)
Sol.
? ? ? ?
? ?
? ? ? ?
z 3 z 1
z 1
3
z 3 z 1
2 2 3
z 1
? ?
?
? ?
? ? ?
?

2
z 2 z 3 3 z 3 ? ? ? ? ?

2
z z 6 0 ? ? ? ?

? ? ? ?
z 3 z 2 0 ? ? ?

min
z 3 ?

2. Let f : S ? ?S where S = (0, ?) be a twice differentiable function such that f(x+1) = xf(x). If g :
S ? R be defined as g(x) = logef(x), then the value of |g”(5) – g”(1)| is equal to :
(1)
197
144

(2)
187
144

(3)
205
144

(4) 1
Ans. (3)
Sol. f(x+1) = xf(x)
g(x+1) = loge(f(x+1))
g(x+1) logex + logf(x)
g(x+1) – g(x) = logex
g’’(x+1) – g’’(x) = –
2
1
x

g’’(2) – g’’(1) = –1
g’’(3) – g’’(2) = –
1
4

g’’(4) – g’’(3) = –
1
9

g’’(5) – g’’(4) = –
1
16

g’’(5) – g’’(1) = –
1 1 1
1
4 9 16
? ?
? ? ?
? ?
? ?

? ? ? ?
144 36 16 9
g'' 5 g'' 1
16 9
? ? ? ? ?
? ?
? ?
?
? ?
=
205
16 9
? ?
? ?
?
? ?

3. If y = y(x) is the solution of the differential equation ? ?
dy
tanx y sinx,0 x
dx 3
?
? ? ? ? , with y(0) =
0, then y
4
? ? ?
? ?
? ?
equal to :
(1) loge2

(2)
e
1
log 2
2

(3)
1
2 2
? ?
? ?
? ?
loge2

(4)
e
1
log 2
4

Ans. (3)
Sol. I.f. =
tanxdx
e
?

=
ln sec x
e
? ?
? ?

= secx
Solution of the equation
y(secx) = ? ? ? ? sinx sec x dx
?

? ?
y
n sec x c
cos x
? ? ? ?
Put x = 0, c = 0
? ?y = cosx ?n(sec x)
put x = ?/4

16
th
March. 2021 | Shift 2
y =
1 1
ln 2 n2
2 2 2
? ?
y =
ln2
2 2

4. If the foot of the perpendicular from point (4, 3, 8) on the line L1 :
x a y 2 z b
3 4
? ? ?
? ?
?
, ? ? 0 is
(3, 5, 7), then the shortest distance between the line L1 and line L2 :
x 2 y 4 z 5
3 4 5
? ? ?
? ? is
equal to :
(1)
2
3

(2)
1
3

(3)
1
2

(4)
1
6

Ans. (4)
Sol. (3, 5, 7) lie on given line L1

3 a 3 7 b
3 4
? ?
? ?
?

7 b
1 b 3
4
?
? ? ?

3 a
1 3 a
?
? ? ? ? ?
?

A (4, 3, 8)
B (3, 5, 7)
DR’S of AB = (1, –2, 1)
AB ? line L1
(1)( ?) + (–2)(3) + 4(1) = 0
? ? ?? = 2
a = 1
a = 1, b = 3, ? = 2

x 1 y 2 z 3
2 3 4
? ? ?
? ?

x 2 y 4 z 5
3 4 5
? ? ?
? ?
S.D. =
1 2 2
2 3 4
3 4 5
1
ˆ ˆ ˆ 6
i j k
2 3 4
3 4 5
?

5. If (x, y, z) be an arbitrary point lying on a plane P which passes through the points (42, 0, 0),
(0, 42, 0) and (0, 0, 42), then the value of the expression

? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
2 2 2 2 2 2
x 11 y 19 z 12 x y z
3
14 x 11 y 19 z 12
y 19 z 12 x 11 z 12 z 11 y 19
? ? ? ? ?
? ? ? ?
? ? ?
? ? ? ? ? ?
is equal
to :
(1) 3

(2) 0
(3) 39

(4) –45
Ans. (1)
Sol. equation of plane x + y + z = 42
Let pt. on plane x = 10, y = 21, z = 11

? ?
? ? ? ?
? ?
? ? ? ?
? ?
? ? ? ? ? ? ? ? ? ?
1 2 1
42
3
4 1 1 1 1 4 14 1 2 1
? ?
? ? ? ?
? ?

1 1 3
3 2
4 4 2
? ? ? ? = 3

6. Consider the integral

x
10
x 1
0
x e
I dx
e
? ?
? ?
?
? ?
? ?
?
?

Where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to :
(1) 45 (e – 1)

(2) 45 (e + 1)
(3) 9(e – 1)

(4) 9(e + 1)
Page 5

16
th
March. 2021 | Shift 2
SECTION – A

1. The least value of |z| where z is complex number which satisfies the inequality exp
? ? ? ?
e
2
z 3 z 1
log 2 log 5 7 9i ,i 1
z 1
? ?
? ?
? ? ? ? ? ?
? ? ?
? ?
is equal to :
(1) 2

(2) 3
(3) 8

(4) 5
Ans. (2)
Sol.
? ? ? ?
? ?
? ? ? ?
z 3 z 1
z 1
3
z 3 z 1
2 2 3
z 1
? ?
?
? ?
? ? ?
?

2
z 2 z 3 3 z 3 ? ? ? ? ?

2
z z 6 0 ? ? ? ?

? ? ? ?
z 3 z 2 0 ? ? ?

min
z 3 ?

2. Let f : S ? ?S where S = (0, ?) be a twice differentiable function such that f(x+1) = xf(x). If g :
S ? R be defined as g(x) = logef(x), then the value of |g”(5) – g”(1)| is equal to :
(1)
197
144

(2)
187
144

(3)
205
144

(4) 1
Ans. (3)
Sol. f(x+1) = xf(x)
g(x+1) = loge(f(x+1))
g(x+1) logex + logf(x)
g(x+1) – g(x) = logex
g’’(x+1) – g’’(x) = –
2
1
x

g’’(2) – g’’(1) = –1
g’’(3) – g’’(2) = –
1
4

g’’(4) – g’’(3) = –
1
9

g’’(5) – g’’(4) = –
1
16

g’’(5) – g’’(1) = –
1 1 1
1
4 9 16
? ?
? ? ?
? ?
? ?

? ? ? ?
144 36 16 9
g'' 5 g'' 1
16 9
? ? ? ? ?
? ?
? ?
?
? ?
=
205
16 9
? ?
? ?
?
? ?

3. If y = y(x) is the solution of the differential equation ? ?
dy
tanx y sinx,0 x
dx 3
?
? ? ? ? , with y(0) =
0, then y
4
? ? ?
? ?
? ?
equal to :
(1) loge2

(2)
e
1
log 2
2

(3)
1
2 2
? ?
? ?
? ?
loge2

(4)
e
1
log 2
4

Ans. (3)
Sol. I.f. =
tanxdx
e
?

=
ln sec x
e
? ?
? ?

= secx
Solution of the equation
y(secx) = ? ? ? ? sinx sec x dx
?

? ?
y
n sec x c
cos x
? ? ? ?
Put x = 0, c = 0
? ?y = cosx ?n(sec x)
put x = ?/4

16
th
March. 2021 | Shift 2
y =
1 1
ln 2 n2
2 2 2
? ?
y =
ln2
2 2

4. If the foot of the perpendicular from point (4, 3, 8) on the line L1 :
x a y 2 z b
3 4
? ? ?
? ?
?
, ? ? 0 is
(3, 5, 7), then the shortest distance between the line L1 and line L2 :
x 2 y 4 z 5
3 4 5
? ? ?
? ? is
equal to :
(1)
2
3

(2)
1
3

(3)
1
2

(4)
1
6

Ans. (4)
Sol. (3, 5, 7) lie on given line L1

3 a 3 7 b
3 4
? ?
? ?
?

7 b
1 b 3
4
?
? ? ?

3 a
1 3 a
?
? ? ? ? ?
?

A (4, 3, 8)
B (3, 5, 7)
DR’S of AB = (1, –2, 1)
AB ? line L1
(1)( ?) + (–2)(3) + 4(1) = 0
? ? ?? = 2
a = 1
a = 1, b = 3, ? = 2

x 1 y 2 z 3
2 3 4
? ? ?
? ?

x 2 y 4 z 5
3 4 5
? ? ?
? ?
S.D. =
1 2 2
2 3 4
3 4 5
1
ˆ ˆ ˆ 6
i j k
2 3 4
3 4 5
?

5. If (x, y, z) be an arbitrary point lying on a plane P which passes through the points (42, 0, 0),
(0, 42, 0) and (0, 0, 42), then the value of the expression

? ? ? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
2 2 2 2 2 2
x 11 y 19 z 12 x y z
3
14 x 11 y 19 z 12
y 19 z 12 x 11 z 12 z 11 y 19
? ? ? ? ?
? ? ? ?
? ? ?
? ? ? ? ? ?
is equal
to :
(1) 3

(2) 0
(3) 39

(4) –45
Ans. (1)
Sol. equation of plane x + y + z = 42
Let pt. on plane x = 10, y = 21, z = 11

? ?
? ? ? ?
? ?
? ? ? ?
? ?
? ? ? ? ? ? ? ? ? ?
1 2 1
42
3
4 1 1 1 1 4 14 1 2 1
? ?
? ? ? ?
? ?

1 1 3
3 2
4 4 2
? ? ? ? = 3

6. Consider the integral

x
10
x 1
0
x e
I dx
e
? ?
? ?
?
? ?
? ?
?
?

Where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to :
(1) 45 (e – 1)

(2) 45 (e + 1)
(3) 9(e – 1)

(4) 9(e + 1)

16
th
March. 2021 | Shift 2
Ans. (1)
Sol. I =
10
x 1 x
0
x e dx
? ? ? ?
? ?
? ? ?
? ?
?

=
2 3 4 10
2 x 3 x 4 x 10 x
1 2 3 9
e dx 2 e dx 3 e dx ...... 9e dx
? ? ? ?
? ? ? ? ? ?
? ? ? ?

= –{(1 – e) + 2(1 – e) + 3(1 – e)+……+9(1 – e)}
= 45(e – 1)

7. Let A (–1, 1), B (3, 4) and C(2, 0) be given three points. A line y = mx, m > 0, intersects lines
AC and BC at point P and Q respectively. Let A1 and A2 be the areas of ?ABC and ?PQC
respectively, such that A1 = 3A2, then the value of m is equal to :
(1)
4
15

(2) 1
(3) 2

(4) 3
Ans. (2)
Sol.

1
1 1 1
1
A ABC 2 0 1
2
3 4 1
?
? ? ?

1
13
A
2
?
Equation of line AC is y – 1 =
1
3
? (x + 1)
solve it with line y = mx, we get P
2 2m
,
3m 1 3m 1
? ?
? ?
? ?
? ?

Equation of line BC is y – 0 = 4(x –2)
A(–1, 1)

P

Q

C(2, 0)
B(3, 4)

y = mx

```

## JEE Main & Advanced Mock Test Series

366 docs|219 tests

## JEE Main & Advanced Mock Test Series

366 docs|219 tests

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