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JEE MAIN QUESTION AND ANSWER
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I HAVE UPLOAD HERE JEE MAIN QUESTION PAPER OF 2022 YEAR WITH SOLN OF BATCH 1 AND 2
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JEE (Main)-2022 : Phase-2 (25-07-2022)-EveningMATHEMATICS
SECTION - A = 92 – 9 | | – 43
= 9 | |2 – 9 | | – 43
Multiple Choice Questions: This section contains 20
= 0 for 2 values of | | out of which one is –ve
multiple choice questions. Each question has 4 choices
and other is +ve
(1), (2), (3) and (4), out of which ONLY ONE is correct.
So, 2 values of satisfy the system of equations to
Choose the correct answer : obtain no solution
3. The number of bijective functions f : {1, 3, 5, 7, …,
1. For z if the minimum value of
99} → {2, 4, 6, 8, ….., 100} such that
( z − 3 2 + z − p 2i ) is 5 2, then a value of p
f ( 3 ) f ( 9 ) f (15 ) f ( 21) .... f ( 99 ) , is_____.
is _________.
50 50
7 (A) P17 (B) P33
(A) 3 (B)
2 50!
(C) 33! 17! (D)
9 2
(C) 4 (D)
2 Answer (B)
Answer (C) Sol. As function is one-one and onto, out of 50 elements
Sol. of domain set 17 elements are following restriction
f(3) > f(9) > f(15) ….. > f(99)
So number of ways =50 C17 · 1 · 33!
50
= P33
It is sum of distance of z from (3 2, 0 ) and
4. The remainder when (11)1011 + (1011)11 is divided
by 9 is
( 0, p 2 ) (A) 1 (B) 4
(C) 6 (D) 8
For minimising, z should lie on AB and AB = 5 2
Answer (D)
( AB ) 2
= 18 + 2 p 2
(11)1011 + (1011)11 1011 + 311
p = 4 Sol. Re = Re 2
9 9
2. The number of real values of , such that the
system of linear equations 21011
For Re
2x – 3y + 5z = 9 9
x + 3y – z = –18
21011 = ( 9 − 1) C0 9337 ( −1)
337 337 0
=
3x – y + (2 – | |)z = 16
C19336 ( −1)
337 1
has no solutions, is +
C2 9335 ( −1) + ......
337 2
(A) 0 (B) 1 +
(C) 2 (D) 4
C337 90 ( −1)
337 337
+
Answer (C)
2 −3 5 so, remainder is 8
Sol.
= 1 3 −1 (
= 2 3 2 − 3 | | −1) 311
and Re =0
3 −1 2 − | | +3 ( − | | +3 )
2 9
+5 ( −1 − 9 ) So, remainder is 8
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, JEE (Main)-2022 : Phase-2 (25-07-2022)-Evening
21
3
5. The sum ( 4n − 1)( 4n + 3 ) is equal to
1
1 1 1 1
n =1 Sol. I = lim n + + + ..... +
n → 2
1 2 3 2 n
− 1
7 7 1 − n 1− n 1− n 1 − n
(A) (B) 2 2 2 2
87 29
14 21 Let 2n = t and if n → then t →
(C) (D)
87 29
1 t −1 1
Answer (B) l = lim
n → t
r =1 r
21
3 3 21 1 1 1−
Sol. ( 4n − 1)( 4n + 3 ) = 4 4n − 1 − 4n + 3 t
n =1 n =1
1
dx
1
dx a a
3 1 1 1 1 1 1 l= = f ( x ) dx = f ( a − x ) dx
= − + − + ..... + − 0 1− x 0 x 0 0
4 3 7 7 11 83 87
1
3 1 1 3 84 7 1
= − = = = 2 x 2 = 2
4 3 87 4 3.87 29 0
8 2 − ( cos x + sin x )
7
8. If A and B are two events such that
6. lim is equal to
x→
2 − 2 sin2x 1 1 1
4
P ( A ) = , P ( B ) = and P ( A B ) = , then
3 5 2
(A) 14 (B) 7
P ( A B ) + P ( B A ) is equal to
(C) 14 2 (D) 7 2
Answer (A) 3 5
(A) (B)
4 8
8 2 − ( cos x + sin x )
7
0
Sol. lim 0 form
x→
4
2 − 2 sin2 x (C)
5
(D)
7
4 8
−7 ( cos x + sin x ) ( − sin x + cos x )
6
= lim using L–H Answer (B)
x→
−2 2 cos2x
4
1 1 1
Rule Sol. P ( A ) = , P ( B ) = and P ( A B ) =
3 5 2
56 ( cos x − sin x ) 0
= lim 0 P (A B) =
1 1 1 1
+ − =
x→
4
2 2 cos2x 3 5 2 30
−56 ( sin x + cos x )
= lim using L–H Rule
x→
4
−4 2 sin2x
= 7 2 2 = 14
1 1 1 1 1
7. lim + + + ..... +
n → 2n
1 2 3 2 n
− 1
1 − n 1− n 1− n 1 − n
2 2 2 2
P ( A B ) P ( B A )
is equal to Now, P ( A B ) + P ( B A ) = +
P ( B ) P ( A )
1
(A) (B) 1 9 5
2
5
(C) 2 (D) –2 = 30 + 30 =
4 2 8
Answer (C) 5 3
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