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Paper of JEE exam which was held on 31st Jan 2024 1st session
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FINAL JEE–MAIN EXAMINATION – JANUARY, 2024(Held On Wednesday 31st January, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A 2. Let a be the sum of all coefficients in the
1. For 0 < c < b < a, let (a + b – 2c)x2 + (b + c – 2a)x expansion of (1 – 2x + 2x2)2023 (3 – 4x2+2x3)2024
+ (c + a – 2b) = 0 and 1 be one of its root. x log 1 t
2024 dt
Then, among the two statements t 1 . If the equations
and b lim 0
x 0 x2
(I) If 1,0 , then b cannot be the geometric
mean of a and c 2 2
cx + dx + e = 0 and 2bx + ax + 4 = 0 have a
(II) If 0,1 , then b may be the geometric common root, where c, d, e R, then d : c : e
mean of a and c equals
(1) 2 : 1 : 4 (2) 4 : 1 : 4
(1) Both (I) and (II) are true
(3) 1 : 2 : 4 (4) 1 : 1 : 4
(2) Neither (I) nor (II) is true Ans. (4)
(3) Only (II) is true Sol. Put x = 1
a 1
(4) Only (I) is true
x
ln 1 t
Ans. (1) 1 t 2024
dt
2
Sol. f(x) = (a + b – 2c) x + (b + c – 2a) x + (c + a – 2b) b lim 0
x 0 x2
f(x) = a + b – 2c + b + c – 2a + c + a – 2b = 0 Using L’ HOPITAL Rule
f(1) = 0 ln 1 x 1 1
b lim
c a 2b
x 0
1 x 2024
2x 2
1
a b 2c Now, cx2 + dx + e = 0, x2 + x + 4 = 0
(D < 0)
c a 2b
c d e
a b 2c
1 1 4
If, –1 < < 0 3. If the foci of a hyperbola are same as that of the
c a 2b x 2 y2
1 0 ellipse 1 and the eccentricity of the
a b 2c 9 25
15
ac hyperbola is times the eccentricity of the
b + c < 2a and b 8
2
ellipse, then the smaller focal distance of the point
therefore, b cannot be G.M. between a and c.
14 2
If, 0 1 2, on the hyperbola, is equal to
3 5
c a 2b
0 1 2 8 2 4
a b 2c (1) 7 (2) 14
5 3 5 3
ac 2 16 2 8
b > c and b (3) 14 (4) 7
2 5 3 5 3
Therefore, b may be the G.M. between a and c. Ans. (1)
, x 2 y2 13 x = 39
Sol. 1
9 25 x = 3, y = 2
a = 3, b = 5 Center of given circle is (5, –2)
9 4 Radius 25 4 13 4
e 1 foci 0, be = (0, ± 4)
25 5 CM 4 16 5 2
4 15 3
eH CP 16 20 6
5 8 2
5. The area of the region
Let equation hyperbola
xy x 1 x 2
x 2 y2 x, y : y 4x, x 4, 0, x 3
2
1 x 3 x 4
A 2 B2
8 is
B eH 4 B
3 16 64
(1) (2)
64 9 80 3 3
A2 B2 e2H 1 1 A 9
2
9 4 8 32
(3) (4)
x 2 y2 3 3
1 Ans. (4)
80 64
9 9 Sol. y2 4x, x 4
B 16 xy x 1 x 2
Directrix : y 0
eH 9
x 3 x 4
3 14 2 16
PS = e PM Case – I : y 0
2 3 5 9
x x 1 x 2
2 8 0
7 x 3 x 4
5 3
x 0,1 2,3
4. If one of the diameters of the circle x2 + y2 – 10x +
4y + 13 = 0 is a chord of another circle C, whose Case – II : y < 0
center is the point of intersection of the lines 2x + x x 1 x 2
0, x 1, 2 3, 4
3y = 12 and 3x – 2y = 5, then the radius of the x 3 x 4
circle C is
(1) 20 (2) 4
(3) 6 (4) 3 2
Ans. (3)
P
4
M
(5, –2)
C
4
(3, 2)
Area 2 x dx
0
Sol.
2 4 32
2x + 3y = 12 2 x 3/2
3 0 3
3x – 2y = 5
, 4x 3 2 1 2 1
6. If f x , x and (fof) (x) = g(x), where
6x 4 3 Sol. D 2 3
2 2 3 1
g: , then (gogog) (4) is equal
3 3
= 1( + 3) + 2(2 – 9) + 1(–2 – 3)
to
= + 3 + 4 – 18 – 2 – 3
19 19
(1) (2) For infinite solutions D = 0, D1 = 0, D2 = 0 and
20 20
D3 = 0
(3) – 4 (4) 4
D=0
Ans. (4)
– 3 + 4 = 17 ….(1)
4x 3
Sol. f x
6x 4 4 2 1
4x 3 D1 5 3 0
4 3 34x
6x 4 3 1
gx x
4x 3 34
6 4 1 4 1
6x 4
D2 2 5 3 0
g x x g g g 4 4 3 3
e2 sin x 2 sin x 1
7. lim 15 9 4 2 9 1 6 15 0
x 0 x2
(1) is equal to – 1 (2) does not exist 13 9 36 9 0
(3) is equal to 1 (4) is equal to 2 54
13 54, put in (1)
Ans. (4) 13
e2 sin x 2 sin x 1 54 54
Sol. lim 3 4 17
x 0 x2 13 13
e2 sin x 2 sin x 1 sin 2 x 54 39 216 221
lim 2
2
x 0
sin x x 1
15 5
Let |sinx| = t 3
e 2t 2t 1 sin 2 x 1 54
lim 2
lim 2 Now, 12 13 12. 13.
t 0 t x 0 x 3 13
2e2t 2 = 4 + 54 = 58
lim 1 2 1 2
t 0 2t 9. The solution curve of the differential equation
8. If the system of linear equations
dx
x 2y z 4 y x log e x log e y 1 , x > 0, y > 0 passing
dy
2x y 3z 5
through the point (e, 1) is
3x y z 3
has infinitely many solutions, then 12 + 13 is y y
(1) loge x (2) log e y2
equal to x x
(1) 60 (2) 64 x x
(3) log e y (4) 2 log e y 1
(3) 54 (4) 58 y y
Ans. (4)
Ans. (3)
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