(IIT)(JEE)(Engineering)

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) cos1  
6  18 
of the prime factors of P 1AP  2I is equal to  1 
(3) cos1   (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