JEE Main 2021 Answer Key Mathematics - Evening Shift (16-03-2021) JEE Notes | EduRev

Mock Test Series for JEE Main & Advanced 2021

JEE : JEE Main 2021 Answer Key Mathematics - Evening Shift (16-03-2021) JEE Notes | EduRev

 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
 
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