JEE Main 2021 Answer Key Mathematics - Evening Shift (17-03-2021) JEE Notes | EduRev

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JEE : JEE Main 2021 Answer Key Mathematics - Evening Shift (17-03-2021) JEE Notes | EduRev

 Page 1


 
 
17
th
 March. 2021 | Shift 2
SECTION – A 
 
1. If the Boolean expression ?? pq ? ?? pq ?  is a tautology, then  and  ?  are respectively given by :  
 (1) ?,  ?   
 
(2) ?,  ?  
 (3) ?,  ? 
 
(4) ?,  ? 
Ans. (2)  
Sol. ?? pq ? ??? pq ?  
 ?? pq ? ??? ~p q ?  
  ??? ? ~p ~q ~p q ?? ?  
  ?? ~p ~q q ?? ?Tautology  
  ?   ?   ? 
 ? ?   ? 
 
2. Let the tangent to the circle x
2
 + y
2
 = 25 at the point R(3,4) meet x-axis and y-axis at points P 
and Q, respectively. If r is the radius of the circle passing through the origin O and having 
centre at the incentre of the triangle OPQ, then r
2
 is equal to : 
 (1) 
625
72
 
 
(2) 
585
66
 
 (3) 
125
72
 
 
(4) 
529
64
 
Ans. (1)  
Sol. Given equation of circle  
 x
2
 + y
2
 = 25 
 ?? ??Tangent equation at (3, 4) 
 T : 3x + 4y = 25 
   
 
Y 
 (0, 0) O
X
25
4
25
3
P 
25
,0
3
??
??
??
25
0,
4
??
??
??
Q 
125
12
Page 2


 
 
17
th
 March. 2021 | Shift 2
SECTION – A 
 
1. If the Boolean expression ?? pq ? ?? pq ?  is a tautology, then  and  ?  are respectively given by :  
 (1) ?,  ?   
 
(2) ?,  ?  
 (3) ?,  ? 
 
(4) ?,  ? 
Ans. (2)  
Sol. ?? pq ? ??? pq ?  
 ?? pq ? ??? ~p q ?  
  ??? ? ~p ~q ~p q ?? ?  
  ?? ~p ~q q ?? ?Tautology  
  ?   ?   ? 
 ? ?   ? 
 
2. Let the tangent to the circle x
2
 + y
2
 = 25 at the point R(3,4) meet x-axis and y-axis at points P 
and Q, respectively. If r is the radius of the circle passing through the origin O and having 
centre at the incentre of the triangle OPQ, then r
2
 is equal to : 
 (1) 
625
72
 
 
(2) 
585
66
 
 (3) 
125
72
 
 
(4) 
529
64
 
Ans. (1)  
Sol. Given equation of circle  
 x
2
 + y
2
 = 25 
 ?? ??Tangent equation at (3, 4) 
 T : 3x + 4y = 25 
   
 
Y 
 (0, 0) O
X
25
4
25
3
P 
25
,0
3
??
??
??
25
0,
4
??
??
??
Q 
125
12
 
 
 Incentre of ?OPQ. 
 I =
25 25 25 25
43 3 4
,
25 25 125 25 25 125
34 12 34 12
??
??
??
??
??
?? ??
??
??
. 
 ? I =
625 625
,
75 100 125 75 100 125
??
??
?? ??
??
=
25 25
,
12 12
??
??
??
 
 ? Distance from origin to incentre is r. 
 ? r
2
 =
2
25
12
??
??
??
+
2
25
12
??
??
??
=
625
72
 
 Therefore, the correct answer is (1) 
 
 
3. Let a computer program generate only the digits 0 and 1 to form a string of binary numbers 
with probability of occurrence of 0 at even places be  
1
2
 and probability of occurrence of 0 at the 
odd place be 
1
3
. Then the probability that ‘10’ is followed by ‘01’ is equal to :  
 (1) 
1
6
  
 
(2) 
1
18
 
 (3) 
1
9
 
 
(4) 
1
3
 
Ans. (3)  
Sol. P(0 at even place) = 
1
2
,   P(0 at odd place) = 
1
3
 
 P(1 at even place) = 
1
2
,   P(1 at odd place) = 
2
3
 
 P(10 is followed by 01) 
 = 
21 11 111 2
3 23 2 232 3
??? ?
?? ? ? ? ??
??? ?
??? ?
 
  = 
11
18 18
?  
  = 
1
9
 
 
 
Page 3


 
 
17
th
 March. 2021 | Shift 2
SECTION – A 
 
1. If the Boolean expression ?? pq ? ?? pq ?  is a tautology, then  and  ?  are respectively given by :  
 (1) ?,  ?   
 
(2) ?,  ?  
 (3) ?,  ? 
 
(4) ?,  ? 
Ans. (2)  
Sol. ?? pq ? ??? pq ?  
 ?? pq ? ??? ~p q ?  
  ??? ? ~p ~q ~p q ?? ?  
  ?? ~p ~q q ?? ?Tautology  
  ?   ?   ? 
 ? ?   ? 
 
2. Let the tangent to the circle x
2
 + y
2
 = 25 at the point R(3,4) meet x-axis and y-axis at points P 
and Q, respectively. If r is the radius of the circle passing through the origin O and having 
centre at the incentre of the triangle OPQ, then r
2
 is equal to : 
 (1) 
625
72
 
 
(2) 
585
66
 
 (3) 
125
72
 
 
(4) 
529
64
 
Ans. (1)  
Sol. Given equation of circle  
 x
2
 + y
2
 = 25 
 ?? ??Tangent equation at (3, 4) 
 T : 3x + 4y = 25 
   
 
Y 
 (0, 0) O
X
25
4
25
3
P 
25
,0
3
??
??
??
25
0,
4
??
??
??
Q 
125
12
 
 
 Incentre of ?OPQ. 
 I =
25 25 25 25
43 3 4
,
25 25 125 25 25 125
34 12 34 12
??
??
??
??
??
?? ??
??
??
. 
 ? I =
625 625
,
75 100 125 75 100 125
??
??
?? ??
??
=
25 25
,
12 12
??
??
??
 
 ? Distance from origin to incentre is r. 
 ? r
2
 =
2
25
12
??
??
??
+
2
25
12
??
??
??
=
625
72
 
 Therefore, the correct answer is (1) 
 
 
3. Let a computer program generate only the digits 0 and 1 to form a string of binary numbers 
with probability of occurrence of 0 at even places be  
1
2
 and probability of occurrence of 0 at the 
odd place be 
1
3
. Then the probability that ‘10’ is followed by ‘01’ is equal to :  
 (1) 
1
6
  
 
(2) 
1
18
 
 (3) 
1
9
 
 
(4) 
1
3
 
Ans. (3)  
Sol. P(0 at even place) = 
1
2
,   P(0 at odd place) = 
1
3
 
 P(1 at even place) = 
1
2
,   P(1 at odd place) = 
2
3
 
 P(10 is followed by 01) 
 = 
21 11 111 2
3 23 2 232 3
??? ?
?? ? ? ? ??
??? ?
??? ?
 
  = 
11
18 18
?  
  = 
1
9
 
 
 
 
 
17
th
 March. 2021 | Shift 2
4. The number of solutions of the equation x + 2tanx = 
2
?
 in the interval [0, 2p] is : 
 (1) 5     
 
(2) 2 
 (3) 4     
 
(4) 3 
Ans. (4)  
Sol.  
   
x + 2 tan x = 
2
?
 in [0, 2 ?] 
2 tan x = 
2
?
– x 
2 tan x = 
2
?
 – x 
tan x = 
4
?
 – 
x
2
 
y = tan x and y = 
–x
2
 + 
4
?
 
3 intersection points 
? 3 solutions  
option (4)  
 
5. If the equation of plane passing through the mirror image of a point (2, 3, 1) with respect to 
line 
x1 y–3 z 2
21 –1
??
??  and containing the line  
x–2 1–y z 1
32 1
?
??  is xy z 24 ??? ?? ? , then ?? ?? ?  
is equal to : 
 (1) 21 
 
(2) 19  
 (3) 18 
 
(4) 20 
– ? 
– 
0
?
2 ?
  2
?
3
2
?
2
?
Page 4


 
 
17
th
 March. 2021 | Shift 2
SECTION – A 
 
1. If the Boolean expression ?? pq ? ?? pq ?  is a tautology, then  and  ?  are respectively given by :  
 (1) ?,  ?   
 
(2) ?,  ?  
 (3) ?,  ? 
 
(4) ?,  ? 
Ans. (2)  
Sol. ?? pq ? ??? pq ?  
 ?? pq ? ??? ~p q ?  
  ??? ? ~p ~q ~p q ?? ?  
  ?? ~p ~q q ?? ?Tautology  
  ?   ?   ? 
 ? ?   ? 
 
2. Let the tangent to the circle x
2
 + y
2
 = 25 at the point R(3,4) meet x-axis and y-axis at points P 
and Q, respectively. If r is the radius of the circle passing through the origin O and having 
centre at the incentre of the triangle OPQ, then r
2
 is equal to : 
 (1) 
625
72
 
 
(2) 
585
66
 
 (3) 
125
72
 
 
(4) 
529
64
 
Ans. (1)  
Sol. Given equation of circle  
 x
2
 + y
2
 = 25 
 ?? ??Tangent equation at (3, 4) 
 T : 3x + 4y = 25 
   
 
Y 
 (0, 0) O
X
25
4
25
3
P 
25
,0
3
??
??
??
25
0,
4
??
??
??
Q 
125
12
 
 
 Incentre of ?OPQ. 
 I =
25 25 25 25
43 3 4
,
25 25 125 25 25 125
34 12 34 12
??
??
??
??
??
?? ??
??
??
. 
 ? I =
625 625
,
75 100 125 75 100 125
??
??
?? ??
??
=
25 25
,
12 12
??
??
??
 
 ? Distance from origin to incentre is r. 
 ? r
2
 =
2
25
12
??
??
??
+
2
25
12
??
??
??
=
625
72
 
 Therefore, the correct answer is (1) 
 
 
3. Let a computer program generate only the digits 0 and 1 to form a string of binary numbers 
with probability of occurrence of 0 at even places be  
1
2
 and probability of occurrence of 0 at the 
odd place be 
1
3
. Then the probability that ‘10’ is followed by ‘01’ is equal to :  
 (1) 
1
6
  
 
(2) 
1
18
 
 (3) 
1
9
 
 
(4) 
1
3
 
Ans. (3)  
Sol. P(0 at even place) = 
1
2
,   P(0 at odd place) = 
1
3
 
 P(1 at even place) = 
1
2
,   P(1 at odd place) = 
2
3
 
 P(10 is followed by 01) 
 = 
21 11 111 2
3 23 2 232 3
??? ?
?? ? ? ? ??
??? ?
??? ?
 
  = 
11
18 18
?  
  = 
1
9
 
 
 
 
 
17
th
 March. 2021 | Shift 2
4. The number of solutions of the equation x + 2tanx = 
2
?
 in the interval [0, 2p] is : 
 (1) 5     
 
(2) 2 
 (3) 4     
 
(4) 3 
Ans. (4)  
Sol.  
   
x + 2 tan x = 
2
?
 in [0, 2 ?] 
2 tan x = 
2
?
– x 
2 tan x = 
2
?
 – x 
tan x = 
4
?
 – 
x
2
 
y = tan x and y = 
–x
2
 + 
4
?
 
3 intersection points 
? 3 solutions  
option (4)  
 
5. If the equation of plane passing through the mirror image of a point (2, 3, 1) with respect to 
line 
x1 y–3 z 2
21 –1
??
??  and containing the line  
x–2 1–y z 1
32 1
?
??  is xy z 24 ??? ?? ? , then ?? ?? ?  
is equal to : 
 (1) 21 
 
(2) 19  
 (3) 18 
 
(4) 20 
– ? 
– 
0
?
2 ?
  2
?
3
2
?
2
?
 
 
 
Ans. (2)  
Sol.   
  
  Let point M is (2 ? – 1,  ? + 3, –  ??–2) 
D.R.’s of AM line are   2 ? – 1 – 2,   ? + 3 – 3,   –  ? – 2 – 1 
      2 ? – 3,    ?,     – ??– 3 
  AM  ? line L1 
  ? 2(2 ? – 3) + 1 ( ?) – 1 (– ? –3) = 0 
  6 ? = 3,  ? = 
1
2
  ? M  ? 
7–5
0, ,
22
??
??
??
 
  M is mid-point of A & B 
  M = 
AB
2
?
 
  B = 2 M – A 
  B  ? (–2, 4, –6)  
  Now we have to find equation of plane passing through B (–2, 4, –6) & also containing the line 
x–2
3
 = 
1–y
2
 = 
z1
1
?
     ...(1) 
x–2
3
 = 
y–1
–2
 = 
z1
1
?
 
 
Point P on line is (2, 1, –1) 
2 b
?
 of line L2 is 3, –2, 1 
n
?
 || ( 2 b
?
 × PB
?
) 
?B
P
L2
M
A(2, 3, 1)
B (image) 
x1
2
?
 = 
y–3
1
 = 
z2
–1
?
 ...(L1) 
Page 5


 
 
17
th
 March. 2021 | Shift 2
SECTION – A 
 
1. If the Boolean expression ?? pq ? ?? pq ?  is a tautology, then  and  ?  are respectively given by :  
 (1) ?,  ?   
 
(2) ?,  ?  
 (3) ?,  ? 
 
(4) ?,  ? 
Ans. (2)  
Sol. ?? pq ? ??? pq ?  
 ?? pq ? ??? ~p q ?  
  ??? ? ~p ~q ~p q ?? ?  
  ?? ~p ~q q ?? ?Tautology  
  ?   ?   ? 
 ? ?   ? 
 
2. Let the tangent to the circle x
2
 + y
2
 = 25 at the point R(3,4) meet x-axis and y-axis at points P 
and Q, respectively. If r is the radius of the circle passing through the origin O and having 
centre at the incentre of the triangle OPQ, then r
2
 is equal to : 
 (1) 
625
72
 
 
(2) 
585
66
 
 (3) 
125
72
 
 
(4) 
529
64
 
Ans. (1)  
Sol. Given equation of circle  
 x
2
 + y
2
 = 25 
 ?? ??Tangent equation at (3, 4) 
 T : 3x + 4y = 25 
   
 
Y 
 (0, 0) O
X
25
4
25
3
P 
25
,0
3
??
??
??
25
0,
4
??
??
??
Q 
125
12
 
 
 Incentre of ?OPQ. 
 I =
25 25 25 25
43 3 4
,
25 25 125 25 25 125
34 12 34 12
??
??
??
??
??
?? ??
??
??
. 
 ? I =
625 625
,
75 100 125 75 100 125
??
??
?? ??
??
=
25 25
,
12 12
??
??
??
 
 ? Distance from origin to incentre is r. 
 ? r
2
 =
2
25
12
??
??
??
+
2
25
12
??
??
??
=
625
72
 
 Therefore, the correct answer is (1) 
 
 
3. Let a computer program generate only the digits 0 and 1 to form a string of binary numbers 
with probability of occurrence of 0 at even places be  
1
2
 and probability of occurrence of 0 at the 
odd place be 
1
3
. Then the probability that ‘10’ is followed by ‘01’ is equal to :  
 (1) 
1
6
  
 
(2) 
1
18
 
 (3) 
1
9
 
 
(4) 
1
3
 
Ans. (3)  
Sol. P(0 at even place) = 
1
2
,   P(0 at odd place) = 
1
3
 
 P(1 at even place) = 
1
2
,   P(1 at odd place) = 
2
3
 
 P(10 is followed by 01) 
 = 
21 11 111 2
3 23 2 232 3
??? ?
?? ? ? ? ??
??? ?
??? ?
 
  = 
11
18 18
?  
  = 
1
9
 
 
 
 
 
17
th
 March. 2021 | Shift 2
4. The number of solutions of the equation x + 2tanx = 
2
?
 in the interval [0, 2p] is : 
 (1) 5     
 
(2) 2 
 (3) 4     
 
(4) 3 
Ans. (4)  
Sol.  
   
x + 2 tan x = 
2
?
 in [0, 2 ?] 
2 tan x = 
2
?
– x 
2 tan x = 
2
?
 – x 
tan x = 
4
?
 – 
x
2
 
y = tan x and y = 
–x
2
 + 
4
?
 
3 intersection points 
? 3 solutions  
option (4)  
 
5. If the equation of plane passing through the mirror image of a point (2, 3, 1) with respect to 
line 
x1 y–3 z 2
21 –1
??
??  and containing the line  
x–2 1–y z 1
32 1
?
??  is xy z 24 ??? ?? ? , then ?? ?? ?  
is equal to : 
 (1) 21 
 
(2) 19  
 (3) 18 
 
(4) 20 
– ? 
– 
0
?
2 ?
  2
?
3
2
?
2
?
 
 
 
Ans. (2)  
Sol.   
  
  Let point M is (2 ? – 1,  ? + 3, –  ??–2) 
D.R.’s of AM line are   2 ? – 1 – 2,   ? + 3 – 3,   –  ? – 2 – 1 
      2 ? – 3,    ?,     – ??– 3 
  AM  ? line L1 
  ? 2(2 ? – 3) + 1 ( ?) – 1 (– ? –3) = 0 
  6 ? = 3,  ? = 
1
2
  ? M  ? 
7–5
0, ,
22
??
??
??
 
  M is mid-point of A & B 
  M = 
AB
2
?
 
  B = 2 M – A 
  B  ? (–2, 4, –6)  
  Now we have to find equation of plane passing through B (–2, 4, –6) & also containing the line 
x–2
3
 = 
1–y
2
 = 
z1
1
?
     ...(1) 
x–2
3
 = 
y–1
–2
 = 
z1
1
?
 
 
Point P on line is (2, 1, –1) 
2 b
?
 of line L2 is 3, –2, 1 
n
?
 || ( 2 b
?
 × PB
?
) 
?B
P
L2
M
A(2, 3, 1)
B (image) 
x1
2
?
 = 
y–3
1
 = 
z2
–1
?
 ...(L1) 
 
 
17
th
 March. 2021 | Shift 2
2 b
?
 = 3
ˆ
i  – 2
ˆ
j  + 
ˆ
k  
PB
?
 = – 4
ˆ
i  + 3
ˆ
j  – 5
ˆ
k  
n
?
 = 7
ˆ
i  + 11
ˆ
j  + 
ˆ
k  
? equation of plane is  r
?
. n
?
 =  a
?
. n
?
 
r
?
.(7
ˆ
i  + 11
ˆ
j  + 
ˆ
k) = (–2
ˆ
i  + 4
ˆ
j  – 6
ˆ
k).(7
ˆ
i  + 11
ˆ
j  + 
ˆ
k) 
7x + 11y + z = – 14 + 44 – 6 
7x + 11 y + z = 24 
?   ? = 7 
? = 11 
? = 1 
?  ? +  ? +  ? = 19 
option (2) 
 
6. Consider the function f : R  ?R defined by f(x) =  
1
2– sin x ,  x 0 
x
         0         ,   x 0
??? ??
?
??? ??
? ?? ??
?
?
?
. Then f is :  
 (1) monotonic on (0, 8) only 
 
(2) Not monotonic on (– 8, 0) and (0, 8) 
 (3) monotonic on (– 8, 0) only 
 
(4) monotonic on (– 8, 0) ?  (0, 8) 
Ans. (2)  
Sol.  
 f (x) = 
1
–2–sin x , x 0
x
0,x0
1
2–sin x , x 0
x
???
?
???
??
?
?
?
?
?
??
? ?
??
???
?
 
f’(x) = 
2
2
11 1
–x –cos – – 2–sin ,x 0
xx
x
11 1
x–cos – 2–sin ,x 0
xx
x
???? ?? ?
?
???? ?? ?
???? ?? ?
?
??? ?? ?
?
??
??? ?? ?
?
??? ?? ? ?
 
11 1
–cos sin –2, x 0
xx x
11 1
cos – sin 2, x 0
xx x
?
??
?
?
?
?
??
?
?
 
 
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