(IIT)(JEE)(Engineering)

FINAL JEE–MAIN EXAMINATION – JANUARY, 2024
(Held On Monday 29th January, 2024) TIME : 9 : 00 AM to 12 : 00 NOON

MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A 2  2x ,  1  x  0
 x ,  3  x  0
3. If f  x    x ; g x   ,
1. If in a G.P. of 64 terms, the sum of all the terms is 1 , 0  x  3 x, 0  x  1
 3
7 times the sum of the odd terms of the G.P, then then range of (fog(x)) is
the common ratio of the G.P. is equal to (1) (0, 1] (2) [0, 3)
(1) 7 (2) 4 (3) [0, 1] (4) [0, 1)
Ans. (3)
(3) 5 (4) 6
2  2g(x) , 1  g(x)  0 .....(1)
Ans. (4) 
Sol. f(g(x))   g(x)
 1  3 , 0  g(x)  3 .....(2)
Sol. a  ar  ar2  ar3  ....  ar63
By (1) x  
= 7(a  ar 2  ar 4 .....  ar 62 )
And by (2) x  [ 3,0] and x  [0,1]
64 64
a(1  r ) 7a(1  r ) y=f(x)
  (–3,3)
1 r 1  r2
(1,1)
r=6
2. In an A.P., the sixth terms a6 = 2. If the a1a 4a 5 is

the greatest, then the common difference of the 1
y=f(g(x))
A.P., is equal to 2/3

3 8 2 5 –3 O 1
(1) (2) (3) (4)
2 5 3 8 Range of f(g(x)) is [0, 1]
Ans. (2) 4. A fair die is thrown until 2 appears. Then the
probability, that 2 appears in even number of
Sol. a6  2  a  5d  2
throws , is
a1a 4a5  a(a  3d)(a  4d) 5 1 5 6
(1) (2) (3) (4)
6 6 11 11
= (2  5d)(2  2d)(2  d)
Ans. (3)
2 2 3
f(d)  8  32d  34d  20d  30d  10d Sol. Required probability =
3 5
f '(d)  2(5d  8)(3d  2) 5 1 5 1 5 1
      .....
6 6  6  6  6  6
– + –
2/3 8/5 5
1 5
8 =  6 =
d 6 1 25 11
5
36

,5.
1
If z   2i , is such that 7. In a ABC, suppose y = x is the equation of the
2
bisector of the angle B and the equation of the side
| z  1|  z  1  i  ,i  1 and ,   R , then
AC is 2x –y =2. If 2AB = BC and the point A and
   is equal to B are respectively (4, 6) and  , , then   2 is
(1) –4 (2) 3
equal to
(3) 2 (4) –1
(1) 42 (2) 39
Ans. (2)
(3) 48 (4) 45
1
Sol. z   2i Ans. (1)
2
Sol.
z  1  z  (1  i) A(4,6)

3 
 2i   2i    i D
y=x
2 2
(2,2)
3  
 2i          2  i B(,) A’(6,4) C(–2,–6)
2 2 
AD : DC = 1 : 2
 9
  2 and    4 4   10
2 4 
6 8
   3
=
 
 
3
   14 and  = 14
 2  cos 
1 1 
lim 
 
6. 2  x3   is equal to
dt
x    t   8. Let a, b and c be three non-zero vectors such that
3
2
 x   
 2 
b and c are non-collinear .if a  5b is collinear
3 32
(1) (2) with c,b  6c is collinear with a and
8 4
32 a  b c  0 , then    is equal to
3
(3) (4)
8 4 (1) 35 (2) 30
Ans. (3) (3) – 30 (4)–25
Sol. Using L’hopital rule Ans. (1)
0  cos x  3x 2 Sol. a  5b  c
= lim
   
x 2x   b  6c  a
2  2
Eliminating a
 
sin  x   6 1
 2  3 2 c  5b  c b
= lim   
   4
x 2x  
2  2 1
  ,   30
3 2 5
=
8   5,   30

,  a  
9. Let  5,  , be the circumcenter of a triangle with 11. If ,     is the solution of 4cos  5sin   1 ,
 4 2 2

  a then the value of tan  is
vertices A a, 2 , Ba, 6 and C  , –2  . Let 
4  10  10 10  10
(1) (2)
denote the circumradius,  denote the area and  6 12
denote the perimeter of the triangle. Then      is 10  10 10  10
(3) (4)
(1) 60 (2) 53 12 6

(3) 62 (4) 30 Ans. (3)
Ans. (2) Sol. 4  5 tan   sec 
a   a Squaring : 24tan 2   40tan   15  0
Sol. A(a, –2), B(a, 6), C  , 2  , O  5, 
4   4
10  10
AO = BO tan  
12
2 2
a  a 
(a  5)2    2   (a  5)2    6   10  10 
4  4  and tan      is Rejected.
 12 
a=8
(3) is correct.
AB = 8, AC = 6, BC = 10
12. A function y = f(x) satisfies
  5,   24,   24
f  x  sin 2x  sin x  1  cos2 x  f '  x   0 with condition
  
10. For x    ,  , if
 2 2   
f(0) = 0 . Then f   is equal to
cosec x  sin x 2
y x   dx and
cosec x sec x  tan x sin 2 x (1) 1 (2) 0 (3) –1 (4) 2
 Ans. (1)
lim  y  x   0 then y   is equal to

x   4 dy  sin 2x 
2
Sol.   y  sin x
 1  1  1  dx  1  cos 2 x 
(1) tan 1   (2) tan 1  
 2 2  2 I.F. = 1 + cos2x

(3) 
1  1 
tan 1  (4)
1  1
tan 1    y  1  cos2 x     sin x  dx

2  2 2  2
= – cosx + C
Ans. (4)
x = 0, C = 1
(1  sin2 x)cos x
Sol. y(x)   dx  
1  sin 4 x y   1
2
Put sinx = t
13. Let O be the origin and the position vector of A
 1
1 t 2
1 t  t  and B be 2iˆ  2jˆ  kˆ and 2iˆ  4jˆ  4kˆ respectively. If
1   C
=  dt = tan
4 the internal bisector of AOB meets the line AB
t 1 2 2
 at C, then the length of OC is
x  ,t 1 C=0
2 (1)
2
31 (2)
2
34
3 3
 1  1
y   tan 1    3 3
4 2  2 (3) 34 (4) 31
4 2