JEE Main 2021 Answer Key Mathematics - Evening Shift (18-03-2021) JEE Notes | EduRev

Mock Test Series for JEE Main & Advanced 2021

JEE : JEE Main 2021 Answer Key Mathematics - Evening Shift (18-03-2021) JEE Notes | EduRev

 Page 1


 
 
18
th
 March. 2021 | Shift 2
SECTION – A 
 
 
1. Let the system of linear equations 
 4x y 2z 0 ? ? ? ? 
 2x y z 0 ? ? ? 
 x 2y 3z 0, , R ? ? ? ? ? ? ? 
 Has a non-trivial solution. Then which of the following is true? 
(1)  6, R ? ? ? ? 
(2) 2, R ? ? ? ? 
(3) ? ? ? ? 3, R 
(4) 6, R ? ? ? ? ? 
Ans. (1) 
Sol. For non trivial solution 
  ? ? 0 
  
?
? ?
?
4 2
2 1 1 0
2 3
 
  4( 3 2) (6 ) 2(4 ) ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? 20 6 8 2 
  = ? ? ? ? ? ? ? 12 6 2 
 ? ? ? ? ? ? ? ? ? ? 12 6 2 
  ? ? ? ? 6, R 
 
2. A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from 
each corner of the park be 
?
3
. If the radius of the circumcircle of ?ABC is 2, then the height of the pole is 
equal to: 
 (1) 
1
3
 
 (2) 3 
 (3) 2 3 
Page 2


 
 
18
th
 March. 2021 | Shift 2
SECTION – A 
 
 
1. Let the system of linear equations 
 4x y 2z 0 ? ? ? ? 
 2x y z 0 ? ? ? 
 x 2y 3z 0, , R ? ? ? ? ? ? ? 
 Has a non-trivial solution. Then which of the following is true? 
(1)  6, R ? ? ? ? 
(2) 2, R ? ? ? ? 
(3) ? ? ? ? 3, R 
(4) 6, R ? ? ? ? ? 
Ans. (1) 
Sol. For non trivial solution 
  ? ? 0 
  
?
? ?
?
4 2
2 1 1 0
2 3
 
  4( 3 2) (6 ) 2(4 ) ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? 20 6 8 2 
  = ? ? ? ? ? ? ? 12 6 2 
 ? ? ? ? ? ? ? ? ? ? 12 6 2 
  ? ? ? ? 6, R 
 
2. A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from 
each corner of the park be 
?
3
. If the radius of the circumcircle of ?ABC is 2, then the height of the pole is 
equal to: 
 (1) 
1
3
 
 (2) 3 
 (3) 2 3 
 
 
 (4) 
2 3
3
 
 
Ans. (3) 
Sol. 
   
 ? ? ? ?
h
tan60 h 2 3
2
 
 
3.  Let in a series of 2n observations, half of them are equal to a and remaining half are equal to –a. Also by 
adding a constant b in each of these observations, the mean and standard deviation of new set become 5 
and 20, respectively. Then the value of a
2
 + b
2
 is equal to: 
 (1) 250 
 (2) 925 
 (3) 650 
 (4) 425 
Ans. (4) 
Sol. Given series 
 (a,a,a…….n times), (–a, –a, –a,…… n times) 
 Now 
i
x
x 0
2n
? ?
?
 
 as x i ? x i + b 
 then x x b ? ? 
 So , x + b = 5 ? b = 5 
 No change in S.D. due to change in origin 
 ? = 
2
2
i 2
x
2na
– (x) – 0
2n 2n
?
?
 
 20 = 
2
a ? a = 20 
 a
2
 + b
2
 = 425 
 
 
 
A 
B C 
2 
• 
Page 3


 
 
18
th
 March. 2021 | Shift 2
SECTION – A 
 
 
1. Let the system of linear equations 
 4x y 2z 0 ? ? ? ? 
 2x y z 0 ? ? ? 
 x 2y 3z 0, , R ? ? ? ? ? ? ? 
 Has a non-trivial solution. Then which of the following is true? 
(1)  6, R ? ? ? ? 
(2) 2, R ? ? ? ? 
(3) ? ? ? ? 3, R 
(4) 6, R ? ? ? ? ? 
Ans. (1) 
Sol. For non trivial solution 
  ? ? 0 
  
?
? ?
?
4 2
2 1 1 0
2 3
 
  4( 3 2) (6 ) 2(4 ) ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? 20 6 8 2 
  = ? ? ? ? ? ? ? 12 6 2 
 ? ? ? ? ? ? ? ? ? ? 12 6 2 
  ? ? ? ? 6, R 
 
2. A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from 
each corner of the park be 
?
3
. If the radius of the circumcircle of ?ABC is 2, then the height of the pole is 
equal to: 
 (1) 
1
3
 
 (2) 3 
 (3) 2 3 
 
 
 (4) 
2 3
3
 
 
Ans. (3) 
Sol. 
   
 ? ? ? ?
h
tan60 h 2 3
2
 
 
3.  Let in a series of 2n observations, half of them are equal to a and remaining half are equal to –a. Also by 
adding a constant b in each of these observations, the mean and standard deviation of new set become 5 
and 20, respectively. Then the value of a
2
 + b
2
 is equal to: 
 (1) 250 
 (2) 925 
 (3) 650 
 (4) 425 
Ans. (4) 
Sol. Given series 
 (a,a,a…….n times), (–a, –a, –a,…… n times) 
 Now 
i
x
x 0
2n
? ?
?
 
 as x i ? x i + b 
 then x x b ? ? 
 So , x + b = 5 ? b = 5 
 No change in S.D. due to change in origin 
 ? = 
2
2
i 2
x
2na
– (x) – 0
2n 2n
?
?
 
 20 = 
2
a ? a = 20 
 a
2
 + b
2
 = 425 
 
 
 
A 
B C 
2 
• 
 
 
18
th
 March. 2021 | Shift 2
4. Let g(x) =
?
x
0
f(t)dt, where f is continuous function in [0, 3] such that  ? ?
1
f(t) 1
3
for all ? ? ?
? ?
t 0,1 and 
? ?
1
0 f(t)
2
 for all t (1,3] ? . The largest possible interval in which g(3) lies is: 
 (1) [1,3]   (2) 
? ?
? ?
? ?
? ?
1
1,
2
  (3) 
? ?
? ?
? ?
? ?
3
, 1
2
  (4) 
? ?
? ?
? ?
1
,2
3
 
 
 
Ans. (4) 
Sol. 
1 3 1 3
0 1 0 1
1 1
dt 0.dt g(3) 1.dt dt
3 2
? ? ? ?
? ? ? ?
 
1
3
? g(3) ? 2 
 
5. If ? ? ? ?
4 4
15sin 10cos 6 , for some ? ? R , then the value of ? ? ?
6 6
27sec 8cosec is equal to: 
(1) 250 
(2) 500 
(3) 400 
(4) 350 
Ans. (1) 
Sol. 15 sin
4
 ? + 10 cos
4
 ? = 6 
 ? 15 sin
4
? + 10(1–sin
2
?)
2
 = 6 
 ? 25 sin
4
? – 20sin
2
? + 4 = 0 
 ? 
? ?
? ? ? ? ? ? ? ?
2
2 2 2
2 3
5sin 2 0 sin ,cos
5 5
 
 Now 
6 6
125 125
27cosec 8sec 27 8 250
27 8
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
 
 
6. Let ? ? ? ? ? ? ? f :R 3 R 1 be defind by 
?
?
?
x 2
f(x)
x 3
. 
 Let ? g:R R be given as ? ? g(x) 2x 3 . The, the sum of all the values of x for which 
? ?
? ?
1 1
13
f (x) g (x)
2
 is 
equal to  
(1) 7 
(2) 5 
(3) 2 
(4) 3 
 
 
 
 
 
Page 4


 
 
18
th
 March. 2021 | Shift 2
SECTION – A 
 
 
1. Let the system of linear equations 
 4x y 2z 0 ? ? ? ? 
 2x y z 0 ? ? ? 
 x 2y 3z 0, , R ? ? ? ? ? ? ? 
 Has a non-trivial solution. Then which of the following is true? 
(1)  6, R ? ? ? ? 
(2) 2, R ? ? ? ? 
(3) ? ? ? ? 3, R 
(4) 6, R ? ? ? ? ? 
Ans. (1) 
Sol. For non trivial solution 
  ? ? 0 
  
?
? ?
?
4 2
2 1 1 0
2 3
 
  4( 3 2) (6 ) 2(4 ) ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? 20 6 8 2 
  = ? ? ? ? ? ? ? 12 6 2 
 ? ? ? ? ? ? ? ? ? ? 12 6 2 
  ? ? ? ? 6, R 
 
2. A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from 
each corner of the park be 
?
3
. If the radius of the circumcircle of ?ABC is 2, then the height of the pole is 
equal to: 
 (1) 
1
3
 
 (2) 3 
 (3) 2 3 
 
 
 (4) 
2 3
3
 
 
Ans. (3) 
Sol. 
   
 ? ? ? ?
h
tan60 h 2 3
2
 
 
3.  Let in a series of 2n observations, half of them are equal to a and remaining half are equal to –a. Also by 
adding a constant b in each of these observations, the mean and standard deviation of new set become 5 
and 20, respectively. Then the value of a
2
 + b
2
 is equal to: 
 (1) 250 
 (2) 925 
 (3) 650 
 (4) 425 
Ans. (4) 
Sol. Given series 
 (a,a,a…….n times), (–a, –a, –a,…… n times) 
 Now 
i
x
x 0
2n
? ?
?
 
 as x i ? x i + b 
 then x x b ? ? 
 So , x + b = 5 ? b = 5 
 No change in S.D. due to change in origin 
 ? = 
2
2
i 2
x
2na
– (x) – 0
2n 2n
?
?
 
 20 = 
2
a ? a = 20 
 a
2
 + b
2
 = 425 
 
 
 
A 
B C 
2 
• 
 
 
18
th
 March. 2021 | Shift 2
4. Let g(x) =
?
x
0
f(t)dt, where f is continuous function in [0, 3] such that  ? ?
1
f(t) 1
3
for all ? ? ?
? ?
t 0,1 and 
? ?
1
0 f(t)
2
 for all t (1,3] ? . The largest possible interval in which g(3) lies is: 
 (1) [1,3]   (2) 
? ?
? ?
? ?
? ?
1
1,
2
  (3) 
? ?
? ?
? ?
? ?
3
, 1
2
  (4) 
? ?
? ?
? ?
1
,2
3
 
 
 
Ans. (4) 
Sol. 
1 3 1 3
0 1 0 1
1 1
dt 0.dt g(3) 1.dt dt
3 2
? ? ? ?
? ? ? ?
 
1
3
? g(3) ? 2 
 
5. If ? ? ? ?
4 4
15sin 10cos 6 , for some ? ? R , then the value of ? ? ?
6 6
27sec 8cosec is equal to: 
(1) 250 
(2) 500 
(3) 400 
(4) 350 
Ans. (1) 
Sol. 15 sin
4
 ? + 10 cos
4
 ? = 6 
 ? 15 sin
4
? + 10(1–sin
2
?)
2
 = 6 
 ? 25 sin
4
? – 20sin
2
? + 4 = 0 
 ? 
? ?
? ? ? ? ? ? ? ?
2
2 2 2
2 3
5sin 2 0 sin ,cos
5 5
 
 Now 
6 6
125 125
27cosec 8sec 27 8 250
27 8
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
 
 
6. Let ? ? ? ? ? ? ? f :R 3 R 1 be defind by 
?
?
?
x 2
f(x)
x 3
. 
 Let ? g:R R be given as ? ? g(x) 2x 3 . The, the sum of all the values of x for which 
? ?
? ?
1 1
13
f (x) g (x)
2
 is 
equal to  
(1) 7 
(2) 5 
(3) 2 
(4) 3 
 
 
 
 
 
 
 
Ans. (2) 
Sol. 
? ?
? ?
1 1
13
f (x) g (x)
2
 
 ? 
? ?
? ?
?
3x 2 x 3 13
x 1 2 2
 
 ? ? ? ? ? ? ? 2(3x 2) (x 1)(x 3) 13(x 1) 
 ? ? ? ?
2
x 5x 6 0 
 ? ? x 2 or 3 
 
7. Let S 1 be the sum of frist 2n terms of an arithmetic progression. Let S 2 be the sum of first 4n terms of the 
same arithmetic progression. If (S 2–S 1) is 1000, then the sum of the first 6n terms of the arithmetic 
progression is equal to : 
 (1) 3000  (2) 7000  (3) 5000  (4) 1000 
 
Ans (1) 
Sol. S 4n – S 2n = 1000 
 ? 
4n
2
(2a + (4n–1)d)–
2n
2
(2a+(2n–1)d)=1000 
 ? ?2an + 6n
2
d–nd = 1000 
 ? ?
6n
2
?2a + (6n–1)d) = 3000 
 ? S 6n = 3000 
 
 
8. Let ? ?
2 2
1
S : x y 9 and ? ? ?
2 2
2
S :(x 2) y 1. Then the locus of center of a variable circle S which touches 
S 1 internally and S 2 externally always passes through the points: 
 (1) 
? ?
? ? ?
? ?
? ?
1 5
,
2 2
 
 (2) 
? ?
?
? ?
? ?
3
2,
2
 
 (3) ? ? ? 1, 2 
 (4) 
? ?
? 0, 3 
Ans. (2) 
Sol. C 1 : (0,0) , r 1 = 3 
 C 2 : (2, 0), r 2 = 1 
 Let centre of variable circle be C 3(h,k) and radius be r. 
Page 5


 
 
18
th
 March. 2021 | Shift 2
SECTION – A 
 
 
1. Let the system of linear equations 
 4x y 2z 0 ? ? ? ? 
 2x y z 0 ? ? ? 
 x 2y 3z 0, , R ? ? ? ? ? ? ? 
 Has a non-trivial solution. Then which of the following is true? 
(1)  6, R ? ? ? ? 
(2) 2, R ? ? ? ? 
(3) ? ? ? ? 3, R 
(4) 6, R ? ? ? ? ? 
Ans. (1) 
Sol. For non trivial solution 
  ? ? 0 
  
?
? ?
?
4 2
2 1 1 0
2 3
 
  4( 3 2) (6 ) 2(4 ) ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? 20 6 8 2 
  = ? ? ? ? ? ? ? 12 6 2 
 ? ? ? ? ? ? ? ? ? ? 12 6 2 
  ? ? ? ? 6, R 
 
2. A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from 
each corner of the park be 
?
3
. If the radius of the circumcircle of ?ABC is 2, then the height of the pole is 
equal to: 
 (1) 
1
3
 
 (2) 3 
 (3) 2 3 
 
 
 (4) 
2 3
3
 
 
Ans. (3) 
Sol. 
   
 ? ? ? ?
h
tan60 h 2 3
2
 
 
3.  Let in a series of 2n observations, half of them are equal to a and remaining half are equal to –a. Also by 
adding a constant b in each of these observations, the mean and standard deviation of new set become 5 
and 20, respectively. Then the value of a
2
 + b
2
 is equal to: 
 (1) 250 
 (2) 925 
 (3) 650 
 (4) 425 
Ans. (4) 
Sol. Given series 
 (a,a,a…….n times), (–a, –a, –a,…… n times) 
 Now 
i
x
x 0
2n
? ?
?
 
 as x i ? x i + b 
 then x x b ? ? 
 So , x + b = 5 ? b = 5 
 No change in S.D. due to change in origin 
 ? = 
2
2
i 2
x
2na
– (x) – 0
2n 2n
?
?
 
 20 = 
2
a ? a = 20 
 a
2
 + b
2
 = 425 
 
 
 
A 
B C 
2 
• 
 
 
18
th
 March. 2021 | Shift 2
4. Let g(x) =
?
x
0
f(t)dt, where f is continuous function in [0, 3] such that  ? ?
1
f(t) 1
3
for all ? ? ?
? ?
t 0,1 and 
? ?
1
0 f(t)
2
 for all t (1,3] ? . The largest possible interval in which g(3) lies is: 
 (1) [1,3]   (2) 
? ?
? ?
? ?
? ?
1
1,
2
  (3) 
? ?
? ?
? ?
? ?
3
, 1
2
  (4) 
? ?
? ?
? ?
1
,2
3
 
 
 
Ans. (4) 
Sol. 
1 3 1 3
0 1 0 1
1 1
dt 0.dt g(3) 1.dt dt
3 2
? ? ? ?
? ? ? ?
 
1
3
? g(3) ? 2 
 
5. If ? ? ? ?
4 4
15sin 10cos 6 , for some ? ? R , then the value of ? ? ?
6 6
27sec 8cosec is equal to: 
(1) 250 
(2) 500 
(3) 400 
(4) 350 
Ans. (1) 
Sol. 15 sin
4
 ? + 10 cos
4
 ? = 6 
 ? 15 sin
4
? + 10(1–sin
2
?)
2
 = 6 
 ? 25 sin
4
? – 20sin
2
? + 4 = 0 
 ? 
? ?
? ? ? ? ? ? ? ?
2
2 2 2
2 3
5sin 2 0 sin ,cos
5 5
 
 Now 
6 6
125 125
27cosec 8sec 27 8 250
27 8
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
 
 
6. Let ? ? ? ? ? ? ? f :R 3 R 1 be defind by 
?
?
?
x 2
f(x)
x 3
. 
 Let ? g:R R be given as ? ? g(x) 2x 3 . The, the sum of all the values of x for which 
? ?
? ?
1 1
13
f (x) g (x)
2
 is 
equal to  
(1) 7 
(2) 5 
(3) 2 
(4) 3 
 
 
 
 
 
 
 
Ans. (2) 
Sol. 
? ?
? ?
1 1
13
f (x) g (x)
2
 
 ? 
? ?
? ?
?
3x 2 x 3 13
x 1 2 2
 
 ? ? ? ? ? ? ? 2(3x 2) (x 1)(x 3) 13(x 1) 
 ? ? ? ?
2
x 5x 6 0 
 ? ? x 2 or 3 
 
7. Let S 1 be the sum of frist 2n terms of an arithmetic progression. Let S 2 be the sum of first 4n terms of the 
same arithmetic progression. If (S 2–S 1) is 1000, then the sum of the first 6n terms of the arithmetic 
progression is equal to : 
 (1) 3000  (2) 7000  (3) 5000  (4) 1000 
 
Ans (1) 
Sol. S 4n – S 2n = 1000 
 ? 
4n
2
(2a + (4n–1)d)–
2n
2
(2a+(2n–1)d)=1000 
 ? ?2an + 6n
2
d–nd = 1000 
 ? ?
6n
2
?2a + (6n–1)d) = 3000 
 ? S 6n = 3000 
 
 
8. Let ? ?
2 2
1
S : x y 9 and ? ? ?
2 2
2
S :(x 2) y 1. Then the locus of center of a variable circle S which touches 
S 1 internally and S 2 externally always passes through the points: 
 (1) 
? ?
? ? ?
? ?
? ?
1 5
,
2 2
 
 (2) 
? ?
?
? ?
? ?
3
2,
2
 
 (3) ? ? ? 1, 2 
 (4) 
? ?
? 0, 3 
Ans. (2) 
Sol. C 1 : (0,0) , r 1 = 3 
 C 2 : (2, 0), r 2 = 1 
 Let centre of variable circle be C 3(h,k) and radius be r. 
 
 
18
th
 March. 2021 | Shift 2
 
 C 3C 1 = 3 – r  
 C 2C 1 = 1 + r 
 C 3C 1 + C 2C 1 = 4 
 So locus is ellipse whose focii are C 1 & C 2 
 And major axis is 2a = 4 and 2ae = C 1C 2 = 2 
 ? ? e = 
1
2
 
 ?  b
2
 = 4 
? ?
? ?
? ?
? ?
1
1 3
4
 
Centre of ellipse is midpoint of C 1 & C 2 is (1,0) 
Equation of ellipse is 
? ?
? ?
2
2
2 2
x 1
y
1
2
3
?
? ? 
Now by cross checking the option 
? ?
?
? ?
? ?
3
2,
2
 satisfied it.  
 
9. Let the centroid of an equilateral triangle ABC be at the origin. Let one of the sides of the equilateral triangle 
be along the straight line x + y = 3. If R and r be the radius of circumcircle and incircle respectively of ?ABC, 
then (R+ r ) is equal to  
 (1) 2 2 
 (2) 3 2 
 (3) 7 2 
 (4) 
9
2
 
 
 
C 1 
C
2
 
C
3(h,k)
 
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