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Paper of JEE exam which was held on 29th feb 2024 2nd session
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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 2i 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 32
(1) (2) with c,b 6c is collinear with a and
8 4
32 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 2x b 6c a
2 2
Eliminating a
sin x 6 1
2 3 2 c 5b c b
= lim
4
x 2x
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 , Ba, 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
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