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

Mock Test Series for JEE Main & Advanced 2021

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

 Page 1


 
 
17
th
 March. 2021 | Shift 1
SECTION – A 
 
1. Which of the following is true for y(x) that satisfies the differential equation  
dy
dx
= xy-1+x-y;y(0) = 0 : 
 (1) y(1) = 1 
(2)  y(1) = 
1
2
e 1 ? 
(3) y(1) = 
1 1
2 2
e e
?
? 
(4) y(1) = 
?
?
1
2
e 1
 
Ans. (4) 
 
Sol. ? ? ? ?
dy
x 1 y x 1
dx
? ? ? ? 
 ? ? ? ?
dy
x 1 y 1
dx
? ? ? 
 
? ?
dy
x 1 dx
y 1
? ?
?
 
 
? ?
? ? ? ? ?
2
x
n y 1 x c
2
 
 X = 0, y = 0 
 c 0 ? ? 
 
? ?
2
x
n y 1 x
2
? ? ? ? ? 
 putting x = 1, ? n(y + 1) =
1 1
1
2 2
? ? ? 
 
1
2
y 1 e
?
? ? 
 
1
2
y e 1
?
? ? 
 
? ?
?
? ? ?
1
2
y 1 e 1 
  
 
 
Page 2


 
 
17
th
 March. 2021 | Shift 1
SECTION – A 
 
1. Which of the following is true for y(x) that satisfies the differential equation  
dy
dx
= xy-1+x-y;y(0) = 0 : 
 (1) y(1) = 1 
(2)  y(1) = 
1
2
e 1 ? 
(3) y(1) = 
1 1
2 2
e e
?
? 
(4) y(1) = 
?
?
1
2
e 1
 
Ans. (4) 
 
Sol. ? ? ? ?
dy
x 1 y x 1
dx
? ? ? ? 
 ? ? ? ?
dy
x 1 y 1
dx
? ? ? 
 
? ?
dy
x 1 dx
y 1
? ?
?
 
 
? ?
? ? ? ? ?
2
x
n y 1 x c
2
 
 X = 0, y = 0 
 c 0 ? ? 
 
? ?
2
x
n y 1 x
2
? ? ? ? ? 
 putting x = 1, ? n(y + 1) =
1 1
1
2 2
? ? ? 
 
1
2
y 1 e
?
? ? 
 
1
2
y e 1
?
? ? 
 
? ?
?
? ? ?
1
2
y 1 e 1 
  
 
 
 
 
2. The system of equations kx + y + z=1, x + ky + z=k and x + y + zk =k
2
 has no solution if k is 
equal to: 
 (1) -2 
 
(2)-1 
 (3)1 
 
(4) 0 
Ans. (1)  
Sol. ? ?
k 1 1
D 1 k 1 0
1 1 k
 
 ? k(k
2
 – 1) – (k – 1) + (1 – k) = 0 
? (k – 1) (k
2
 + k – 1 – 1) = 0 
? (k – 1) (k
2
 + k – 2) = 0 
? (k –1) (k –1) (k + 2) = 0 
?k = 1, k = 2 
for k = 1 equation identical so k = –2 for no solution. 
   
3. The value of 
1
4
1
5
1
4
1
5
4 .......
?
?
?
?
? ?
     is: 
 (1) 
4
2 30
5
?  
 
(2) 
4
4 30
5
? 
 (3) 
2
2 30
5
? 
 
(4) 
2
5 0
5
3 ? 
Ans. (3)  
Sol. 
1
y 4
5
y
? ?
?
?
 
 ? 
y
y 4
5y 1
? ?
?
 
Page 3


 
 
17
th
 March. 2021 | Shift 1
SECTION – A 
 
1. Which of the following is true for y(x) that satisfies the differential equation  
dy
dx
= xy-1+x-y;y(0) = 0 : 
 (1) y(1) = 1 
(2)  y(1) = 
1
2
e 1 ? 
(3) y(1) = 
1 1
2 2
e e
?
? 
(4) y(1) = 
?
?
1
2
e 1
 
Ans. (4) 
 
Sol. ? ? ? ?
dy
x 1 y x 1
dx
? ? ? ? 
 ? ? ? ?
dy
x 1 y 1
dx
? ? ? 
 
? ?
dy
x 1 dx
y 1
? ?
?
 
 
? ?
? ? ? ? ?
2
x
n y 1 x c
2
 
 X = 0, y = 0 
 c 0 ? ? 
 
? ?
2
x
n y 1 x
2
? ? ? ? ? 
 putting x = 1, ? n(y + 1) =
1 1
1
2 2
? ? ? 
 
1
2
y 1 e
?
? ? 
 
1
2
y e 1
?
? ? 
 
? ?
?
? ? ?
1
2
y 1 e 1 
  
 
 
 
 
2. The system of equations kx + y + z=1, x + ky + z=k and x + y + zk =k
2
 has no solution if k is 
equal to: 
 (1) -2 
 
(2)-1 
 (3)1 
 
(4) 0 
Ans. (1)  
Sol. ? ?
k 1 1
D 1 k 1 0
1 1 k
 
 ? k(k
2
 – 1) – (k – 1) + (1 – k) = 0 
? (k – 1) (k
2
 + k – 1 – 1) = 0 
? (k – 1) (k
2
 + k – 2) = 0 
? (k –1) (k –1) (k + 2) = 0 
?k = 1, k = 2 
for k = 1 equation identical so k = –2 for no solution. 
   
3. The value of 
1
4
1
5
1
4
1
5
4 .......
?
?
?
?
? ?
     is: 
 (1) 
4
2 30
5
?  
 
(2) 
4
4 30
5
? 
 (3) 
2
2 30
5
? 
 
(4) 
2
5 0
5
3 ? 
Ans. (3)  
Sol. 
1
y 4
5
y
? ?
?
?
 
 ? 
y
y 4
5y 1
? ?
?
 
 
 
17
th
 March. 2021 | Shift 1
 ? 5y
2
-20y-4 = 0 
 ? 
20 400 80
y
10
? ?
? 
 ?
20 4 30
y ,y 0
10
?
? ? 
 
10 2 30
y
5
?
? 
4. If the Boolean expression (p
?
q) ? (q * ( ? P)) is a tautology, then the Boolean expression p * 
( ~q) is equivalent to: 
 (1) ? ? p q 
 (2) ? p q 
 (3) ? q p 
 (4) ? ? q p 
Ans. (3)  
Sol. (p ?q) ? (q* ? P) 
 P q p ?q        q* ? P         ? q       ? q ? p    ? ( ? q ? p) 
 T T T  T  F  F  T 
 T F F  F  T  T  F  
 F T T  T  F  F  T 
 F F T  T  T  F  T 
 ? ? ( ? q ? p)  = q ? ? p 
   = ? p ? q 
 ? * is equivalent to ? 
 ? p * ? q = p ? ? q 
  = ? q ? p  
  = q ? ?p    
 
5. Choose the incorrect statement about the two circles whose equations are given below: 
 X
2
+y
2
-10x-10y+41=0 and  
 X
2
+y
2
-16x-10y+80=0  
(1) Distance between two centres is the average of radii of both the circles. 
(2) Circles have two intersection points. 
(3) Both circles’ centres lie inside region of one another. 
(4) Both circles pass through the centre of each other. 
Ans. (3) 
Sol. C1 (5, 5), C2 (8, 5) 
position of C1(5, 5) in S2 = 0 
= 25 + 25 – 80 – 50 + 80 
= 0 
position of C2(8, 5) in S1 = 0 
? 64 + 25 – 80 – 50 + 41 
= 0 
Page 4


 
 
17
th
 March. 2021 | Shift 1
SECTION – A 
 
1. Which of the following is true for y(x) that satisfies the differential equation  
dy
dx
= xy-1+x-y;y(0) = 0 : 
 (1) y(1) = 1 
(2)  y(1) = 
1
2
e 1 ? 
(3) y(1) = 
1 1
2 2
e e
?
? 
(4) y(1) = 
?
?
1
2
e 1
 
Ans. (4) 
 
Sol. ? ? ? ?
dy
x 1 y x 1
dx
? ? ? ? 
 ? ? ? ?
dy
x 1 y 1
dx
? ? ? 
 
? ?
dy
x 1 dx
y 1
? ?
?
 
 
? ?
? ? ? ? ?
2
x
n y 1 x c
2
 
 X = 0, y = 0 
 c 0 ? ? 
 
? ?
2
x
n y 1 x
2
? ? ? ? ? 
 putting x = 1, ? n(y + 1) =
1 1
1
2 2
? ? ? 
 
1
2
y 1 e
?
? ? 
 
1
2
y e 1
?
? ? 
 
? ?
?
? ? ?
1
2
y 1 e 1 
  
 
 
 
 
2. The system of equations kx + y + z=1, x + ky + z=k and x + y + zk =k
2
 has no solution if k is 
equal to: 
 (1) -2 
 
(2)-1 
 (3)1 
 
(4) 0 
Ans. (1)  
Sol. ? ?
k 1 1
D 1 k 1 0
1 1 k
 
 ? k(k
2
 – 1) – (k – 1) + (1 – k) = 0 
? (k – 1) (k
2
 + k – 1 – 1) = 0 
? (k – 1) (k
2
 + k – 2) = 0 
? (k –1) (k –1) (k + 2) = 0 
?k = 1, k = 2 
for k = 1 equation identical so k = –2 for no solution. 
   
3. The value of 
1
4
1
5
1
4
1
5
4 .......
?
?
?
?
? ?
     is: 
 (1) 
4
2 30
5
?  
 
(2) 
4
4 30
5
? 
 (3) 
2
2 30
5
? 
 
(4) 
2
5 0
5
3 ? 
Ans. (3)  
Sol. 
1
y 4
5
y
? ?
?
?
 
 ? 
y
y 4
5y 1
? ?
?
 
 
 
17
th
 March. 2021 | Shift 1
 ? 5y
2
-20y-4 = 0 
 ? 
20 400 80
y
10
? ?
? 
 ?
20 4 30
y ,y 0
10
?
? ? 
 
10 2 30
y
5
?
? 
4. If the Boolean expression (p
?
q) ? (q * ( ? P)) is a tautology, then the Boolean expression p * 
( ~q) is equivalent to: 
 (1) ? ? p q 
 (2) ? p q 
 (3) ? q p 
 (4) ? ? q p 
Ans. (3)  
Sol. (p ?q) ? (q* ? P) 
 P q p ?q        q* ? P         ? q       ? q ? p    ? ( ? q ? p) 
 T T T  T  F  F  T 
 T F F  F  T  T  F  
 F T T  T  F  F  T 
 F F T  T  T  F  T 
 ? ? ( ? q ? p)  = q ? ? p 
   = ? p ? q 
 ? * is equivalent to ? 
 ? p * ? q = p ? ? q 
  = ? q ? p  
  = q ? ?p    
 
5. Choose the incorrect statement about the two circles whose equations are given below: 
 X
2
+y
2
-10x-10y+41=0 and  
 X
2
+y
2
-16x-10y+80=0  
(1) Distance between two centres is the average of radii of both the circles. 
(2) Circles have two intersection points. 
(3) Both circles’ centres lie inside region of one another. 
(4) Both circles pass through the centre of each other. 
Ans. (3) 
Sol. C1 (5, 5), C2 (8, 5) 
position of C1(5, 5) in S2 = 0 
= 25 + 25 – 80 – 50 + 80 
= 0 
position of C2(8, 5) in S1 = 0 
? 64 + 25 – 80 – 50 + 41 
= 0 
 
 
 
6. The sum of possible values of x for 
? ?
1 1 1
1 8
tan 1 cot tan
1 31
? ? ?
? ? ? ?
? ? ?
? ? ? ?
?
? ? ? ?
x
x
  is: 
 (1) 
33
4
? 
 (2) 
32
4
? 
 (3) 
31
4
? 
 (4) 
30
4
? 
Ans. (2) 
Sol. Taking tan both sides 
 
? ? ? ?
? ? ? ?
1 x x 1
8
1 1 x x 1 31
? ? ?
?
? ? ?
 
 
2
2x 8
31 2 x
? ?
?
 
 ? 4x
2
+31x-8 = 0 
 
1
x 8,
4
? ? ? 
 but at 
1
x
4
? 
 LHS>
2
?
and RHS<
2
?
 
So, only solution is x = –8 
 
7. Lt 2 3
ˆ ˆ ˆ
4 ? ? ?
?
a i j k and 
ˆ ˆ ˆ
7 6 ?
?
b i + j - k  
 
 If 
? ?
ˆ ˆ ˆ
r a r b,r. i 2j k 3 ? ? ? ? ? ? ?
?
? ? ? ?
, then 
? ?
ˆ ˆ ˆ
r. 2i 3j k ? ?
?
 is equal to : 
(1) 10 
(2) 13 
(3) 12 
(4) 8 
Ans. (3) 
Sol. ? ? 2, 3, 4 ? ?
?
a  , ? ? 7,1, 6 ? ?
?
b 
Page 5


 
 
17
th
 March. 2021 | Shift 1
SECTION – A 
 
1. Which of the following is true for y(x) that satisfies the differential equation  
dy
dx
= xy-1+x-y;y(0) = 0 : 
 (1) y(1) = 1 
(2)  y(1) = 
1
2
e 1 ? 
(3) y(1) = 
1 1
2 2
e e
?
? 
(4) y(1) = 
?
?
1
2
e 1
 
Ans. (4) 
 
Sol. ? ? ? ?
dy
x 1 y x 1
dx
? ? ? ? 
 ? ? ? ?
dy
x 1 y 1
dx
? ? ? 
 
? ?
dy
x 1 dx
y 1
? ?
?
 
 
? ?
? ? ? ? ?
2
x
n y 1 x c
2
 
 X = 0, y = 0 
 c 0 ? ? 
 
? ?
2
x
n y 1 x
2
? ? ? ? ? 
 putting x = 1, ? n(y + 1) =
1 1
1
2 2
? ? ? 
 
1
2
y 1 e
?
? ? 
 
1
2
y e 1
?
? ? 
 
? ?
?
? ? ?
1
2
y 1 e 1 
  
 
 
 
 
2. The system of equations kx + y + z=1, x + ky + z=k and x + y + zk =k
2
 has no solution if k is 
equal to: 
 (1) -2 
 
(2)-1 
 (3)1 
 
(4) 0 
Ans. (1)  
Sol. ? ?
k 1 1
D 1 k 1 0
1 1 k
 
 ? k(k
2
 – 1) – (k – 1) + (1 – k) = 0 
? (k – 1) (k
2
 + k – 1 – 1) = 0 
? (k – 1) (k
2
 + k – 2) = 0 
? (k –1) (k –1) (k + 2) = 0 
?k = 1, k = 2 
for k = 1 equation identical so k = –2 for no solution. 
   
3. The value of 
1
4
1
5
1
4
1
5
4 .......
?
?
?
?
? ?
     is: 
 (1) 
4
2 30
5
?  
 
(2) 
4
4 30
5
? 
 (3) 
2
2 30
5
? 
 
(4) 
2
5 0
5
3 ? 
Ans. (3)  
Sol. 
1
y 4
5
y
? ?
?
?
 
 ? 
y
y 4
5y 1
? ?
?
 
 
 
17
th
 March. 2021 | Shift 1
 ? 5y
2
-20y-4 = 0 
 ? 
20 400 80
y
10
? ?
? 
 ?
20 4 30
y ,y 0
10
?
? ? 
 
10 2 30
y
5
?
? 
4. If the Boolean expression (p
?
q) ? (q * ( ? P)) is a tautology, then the Boolean expression p * 
( ~q) is equivalent to: 
 (1) ? ? p q 
 (2) ? p q 
 (3) ? q p 
 (4) ? ? q p 
Ans. (3)  
Sol. (p ?q) ? (q* ? P) 
 P q p ?q        q* ? P         ? q       ? q ? p    ? ( ? q ? p) 
 T T T  T  F  F  T 
 T F F  F  T  T  F  
 F T T  T  F  F  T 
 F F T  T  T  F  T 
 ? ? ( ? q ? p)  = q ? ? p 
   = ? p ? q 
 ? * is equivalent to ? 
 ? p * ? q = p ? ? q 
  = ? q ? p  
  = q ? ?p    
 
5. Choose the incorrect statement about the two circles whose equations are given below: 
 X
2
+y
2
-10x-10y+41=0 and  
 X
2
+y
2
-16x-10y+80=0  
(1) Distance between two centres is the average of radii of both the circles. 
(2) Circles have two intersection points. 
(3) Both circles’ centres lie inside region of one another. 
(4) Both circles pass through the centre of each other. 
Ans. (3) 
Sol. C1 (5, 5), C2 (8, 5) 
position of C1(5, 5) in S2 = 0 
= 25 + 25 – 80 – 50 + 80 
= 0 
position of C2(8, 5) in S1 = 0 
? 64 + 25 – 80 – 50 + 41 
= 0 
 
 
 
6. The sum of possible values of x for 
? ?
1 1 1
1 8
tan 1 cot tan
1 31
? ? ?
? ? ? ?
? ? ?
? ? ? ?
?
? ? ? ?
x
x
  is: 
 (1) 
33
4
? 
 (2) 
32
4
? 
 (3) 
31
4
? 
 (4) 
30
4
? 
Ans. (2) 
Sol. Taking tan both sides 
 
? ? ? ?
? ? ? ?
1 x x 1
8
1 1 x x 1 31
? ? ?
?
? ? ?
 
 
2
2x 8
31 2 x
? ?
?
 
 ? 4x
2
+31x-8 = 0 
 
1
x 8,
4
? ? ? 
 but at 
1
x
4
? 
 LHS>
2
?
and RHS<
2
?
 
So, only solution is x = –8 
 
7. Lt 2 3
ˆ ˆ ˆ
4 ? ? ?
?
a i j k and 
ˆ ˆ ˆ
7 6 ?
?
b i + j - k  
 
 If 
? ?
ˆ ˆ ˆ
r a r b,r. i 2j k 3 ? ? ? ? ? ? ?
?
? ? ? ?
, then 
? ?
ˆ ˆ ˆ
r. 2i 3j k ? ?
?
 is equal to : 
(1) 10 
(2) 13 
(3) 12 
(4) 8 
Ans. (3) 
Sol. ? ? 2, 3, 4 ? ?
?
a  , ? ? 7,1, 6 ? ?
?
b 
 
 
17
th
 March. 2021 | Shift 1
 0 ? ? ? ?
? ?
?
?
r a r b 
 
? ?
0 ? ? ?
?
?
r a b 
 
? ?
? ?
?
? ?
r a b ?  
 
? ?
? ?
5 4 10 ? ? ? ?
?
?
i j k r ? 
 ? ? 3 ? 2, ,1 ? ? ?
?
r  
if  ? ? 1, 2, 3 1 ? ? ?
?
r 
? ? 3 5 8 10 ? ? ? ? ? ? ? ? = 1 
 ? (-5, -4, 10) . (2, -3, 1) 
  ? –10 + 12 + 10 = 12 
 
8. The equation of the plane which contains the y-axis and passes through the point (1, 2, 3)is: 
 (1) 3x + z = 6 
 (2) 3x –z=0 
 (3) x + 3z=10 
 (4) x + 3z = 0 
Ans. (2) 
Sol. Let the equation of the plane is a(x – 1) + b(y – 2) + c (z – 3) = 0 
Y–axis lies on it D.R.’s of y-axis are 0, 1, 0 
?0.a + 1.b + 0.c = 0 ? b = 0 
? Equation of plane is a(x – 1) + c (z – 3) = 0 
x = 0, z = 0 also satisfy it –a – 3c = 0 ? a = –3c 
–3c (x – 1) + c (z – 3) = 0 
–3x + 3 + z – 3 = 0 
 3x – z = 0 
 
 
9. If 
0 sin
A
sin 0
? ? ?
?
? ?
?
? ?
 and det 
2
1
A I 0
2
? ?
? ?
? ?
? ?
, then  a possible value of ? is: 
 (1)
6
?
 
 (2) 
2
?
 
 (3)
3
?
 
 (4) 
4
?
 
Ans. (4) 
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