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Paper on JEE exam which was held on 29th of feb 2024
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FINAL JEE–MAIN EXAMINATION – JANUARY, 2024(Held On Monday 29th January, 2024) TIME : 3 : 00 PM to 6 : 00 PM
MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A 3. Let P(3, 2, 3), Q (4, 6, 2) and R (7, 3, 2) be the
2 1 2 1 2 0 vertices of PQR. Then, the angle QPR is
1. Let A = 6 2 11 and P = 5 0 2 . The sum
7
3 3 2 7 1 5 (1) (2) cos1
6 18
of the prime factors of P 1AP 2I is equal to 1
(3) cos1 (4)
(1) 26 (2) 27 (3) 66 (4) 23 18 3
Ans. (1) Ans. (4)
1 1 1
Sol. P AP 2I P AP 2P P Sol. P(3, 2, 3)
= P 1 (A 2I)P
= P 1 A 2I P Q(4, 6, 2) R(7, 3, 2)
Direction ratio of PR = (4, 1, -1)
= |A-2I|
01 2 Direction ratio of PQ = (1, 4, -1)
6 0 11 = 69 4 4 1
Now, cos
3 3 0 18. 18
So, Prime factor of 69 is 3 & 23
So, sum = 26 3
2. Number of ways of arranging 8 identical books 4. If the mean and variance of five observations are
into 4 identical shelves where any number of 24 194
shelves may remain empty is equal to and respectively and the mean of first
5 25
(1) 18 (2) 16 (3) 12 (4) 15
7
Ans. (4) four observations is , then the variance of the
2
Sol. 3 Shelf empty : (8, 0, 0, 0) 1way
first four observations in equal to
(7,1,0,0)
4 77 5 105
(6, 2,0,0) (1) (2) (3) (4)
2 shelf empty : 4ways 5 12 4 4
(5,3,0,0)
Ans. (3)
(4, 4,0,0)
24 2 194
(6,1,1,0) (3,3, 2,0) Sol. X ;
5 25
1 shelf empty : (5, 2,1,0) (4, 2, 2,0) 5ways
Let first four observation be x1, x2, x3, x4
(4,3,1,0)
x1 x 2 x 3 x 4 x 5 24
(1, 2,3, 2) (5,1,1,1) Here, ......(1)
5 5
(2, 2, 2, 2)
0 Shelf empty : 5ways x1 x 2 x 3 x 4 7
(3,3,1,1) Also,
4 2
(4, 2,1,1)
x1 x 2 x 3 x 4 14
Total = 15 ways
, Now from eqn -1 6. Let r and respectively be the modulus and
x5 = 10 amplitude of the complex number
194 5
Now, 2 z 2 i 2 tan , then (r, ) is equal to
25 8
x12 x 22 x 32 x 24 x 52 576 194
3 3
5 25 25 (1) 2sec ,
8 8
x12 x22 x32 x42 54
3 5
Now, variance of first 4 observations (2) 2sec ,
8 8
2
4
4
x i2 xi 5 3
(3) 2sec ,
Var = i 1
i 1 8 8
4 4
11 11
(4) 2sec ,
8 8
54 49 5
=
4 4 4 Ans. (1)
2
5
5. The function f(x) = 2x 3(x) 3 ,x , has Sol. z = 2 i 2 tan = x + iy (let)
8
(1) exactly one point of local minima and no
y
point of local maxima r= x 2 y 2 & tan 1
x
(2) exactly one point of local maxima and no
2
point of local minima 5
r = (2)2 2 tan
8
(3) exactly one point of local maxima and
exactly one point of local minima 5 3
= 2sec 2sec
(4) exactly two points of local maxima and 8 8
exactly one point of local minima 3
Ans. (3) = 2 sec
8
2
Sol. f (x) 2x 3(x) 3 5
2 tan
& tan 1 8
1
f '(x) 2 2x 3 2
1
= 2 1 1 5
= tan 1 tan
x3 8
13 3
x 1 =
= 2 1 8
x3
7. The sum of the solutions x of the equation
+ – +
3cos2x cos3 2x
–1 0 = x3 – x2 + 6 is
cos6 x sin 6 x
M m
So, maxima (M) at x = -1 & minima(m) at x = 0 (1) 0 (2) 1
(3) –1 (4) 3
, Ans. (3) Ans. (1)
3cos2x cos3 2x Sol. logea, logeb, logec are in A.P.
Sol. = x3 – x2 + 6 b2 = ac …..(i)
cos x sin x
6 6
Also
cos2x (3 cos2 2x)
= x 3 – x2 + 6 a 2b 3c
cos2x (1 sin 2 x cos2 x) loge ,loge ,loge are in A.P.
2b 3c a
4(3 cos2 2x)
= x3 – x2 + 6 2
(4 sin 2 2x) 2b a 3c
3c 2b a
4(3 cos2 2x)
= x3 – x2 + 6 b 3
(3 cos2 2x)
c 2
x3 – x2 + 2 = 0 (x + 1)(x2 – 2x + 2) = 0
2b
so, sum of real solutions = –1 Putting in eq. (i) b2 = a ×
3
8. Let OA a,OB 12a 4b and OC b, where O
a 3
is the origin. If S is the parallelogram with adjacent
b 2
sides OA and OC, then a :b:c=9:6:4
area of the quadrilateral OABC 10. If
is equal to ___
area of S 3 3
sin 2 x cos 2 x
(1) 6 (2) 10 sin 3 x cos3 x sin(x )
dx A cos tan x sin B cos sin cot x C,
(3) 7 (4) 8 where C is the integration constant, then AB is
Ans. (4) equal to
B (1) 4 cosec(2) (2) 4sec
A
Sol. (3) 2sec (4) 8cosec(2)
12a + 4b
a Ans. (4)
3 3
sin 2 x cos 2 x
Sol. sin 3 x cos3 x sin(x )
dx
O C 3 3
b
sin 2 x cos 2 x
Area of parallelogram, S a b I= sin 3 x cos3 x (sin x cos cos x sin )
dx
Area of quadrilateral =Area(OAB)+Area (OBC) 3 3
= sin 2 x cos 2 x
dx =
=
1
2
a (12a 4b) b (12a 4b) 3
sin x cos x tan x cos sin
2 2
dx 3
sin x cos x cos cot xsin
2 2
sec2 x cosec2 x
= 8 (a b) tan x cos sin
dx
cos cot xsin
dx
8 (a b) I = I1 + I2 …… {Let}
Ratio = =8
(a b) For I1, let tan x cos – sin = t2
2t dt
9. If loge a, loge b, loge c are in an A.P. and loge a – sec 2 x dx
cos
loge2b, loge2b – loge3c, loge3c – loge a are also in
For I2 , let cos cot xsin z2
an A.P, then a : b : c is equal to
2z dz
(1) 9 : 6 : 4 (2) 16 : 4 : 1 cosec 2 x dx
sin
(3) 25 : 10 : 4 (4) 6 : 3 : 2
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