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Paper of JEE exam which was held on 31st Jan 2024 2nd session
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FINAL JEE–MAIN EXAMINATION – JANUARY, 2024(Held On Wednesday 31st January, 2024) TIME : 3 : 00 PM to 6 : 00 PM
MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A 1
Where tan
1. The number of ways in which 21 identical apples 2
can be distributed among three children such that r 2cos 3sin 17
each child gets at least 2 apples, is
(1) 406 17 5 17 5
r
(2) 130 7 7
(3) 142 3. Let z1 and z2 be two complex number such that z1
(4) 136 + z2 = 5 and z13 z 32 20 15i . Then z14 z 24
Ans. (4)
equals-
Sol. After giving 2 apples to each child 15 apples left
now 15 apples can be distributed in 1531 C2 17C2 (1) 30 3
ways (2) 75
17 16 (3) 15 15
136
2
(4) 25 3
2. Let A (a, b), B(3, 4) and (–6, –8) respectively Ans. (2)
denote the centroid, circumcentre and orthocentre Sol.- z1 z 2 5
of a triangle. Then, the distance of the point
z13 z32 20 15i
P(2a + 3, 7b + 5) from the line 2x + 3y – 4 = 0
measured parallel to the line x – 2y – 1 = 0 is z13 z32 z1 z2 3z1z2 z1 z2
3
(1)
15 5 z13 z32 125 3z1. z 2 5
7
20 15i 125 15z1z2
17 5
(2) 3z1z2 25 4 3i
6
3z1z2 21 3i
17 5
(3) z1.z2 7 i
7
z1 z2 25
2
5
(4)
17 z12 z 22 25 2 7 i
Ans. (3)
11 2i
Sol. A(a,b), B(3,4), C(-6, -8)
z12 z22
2
2:1 121 4 44i
B
z14 z24 2 7 i 117 44i
C A 2
(-6, -8) (a, b) (3, 4)
z14 z 24 117 44i 2 49 1 14i
a 0, b = 0 P 3,5
Distance from P measured along x – 2y – 1 = 0 z14 z24 75
x 3 r cos , y = 5+rsin
,4. Let a variable line passing through the centre of the Sol.-
circle x2 + y2 – 16x – 4y = 0, meet the positive x
f x t t e
2 t2
dt
co-ordinate axes at the point A and B. Then the x
f x 2. x x 2 e x ............ 1
minimum value of OA + OB, where O is the 2
origin, is equal to
x2 1
(1) 12
g x t 2 e t dt
(2) 18 0
(3) 20 g x xe x 2x 0
2
(4) 24
f x g x 2xe x 2x 2e x 2x 2e x
2 2 2
Ans. (2)
Integrating both sides w.r.t.x
Sol.- y 2 m x 8
f x g x 2xe x dx
2
x-intercept
0
2
8 x t
2
m
y-intercept
e dt e t
t
8m 2
0
0
2 e
log 9 1
e
1
OA OB 8 8m 2
m 1
9 f (x) g(x) 1 9 8
f ' m
2
8 0 9
Let , , be mirror image of the point (2, 3, 5)
m2
6.
1
m2 x 1 y 2 z 3
4 in the line .
2 3 4
1
m Then 2 3 4 is equal to
2
(1) 32
1 (2) 33
f 18
2 (3) 31
Minimum = 18 (4) 34
Ans. (2)
Sol.
5. Let f ,g : (0, ) R be two functions defined by P(2,3,5)
t t e
2
x
dt and g(x) t 2 e t dt .
x
f (x)
1
2 t2
x 0
Then the value of f
loge 9 g loge 9 is
equal to R , ,
(1) 6
PR 2,3, 4
(2) 9
(3) 8 PR. 2,3, 4 0
(4) 10 2, 3, 5 . 2,3, 4 0
Ans. (3) 2 3 4 4 9 20 33
, 7. Let P be a parabola with vertex (2, 3) and directrix a 2 2b2
x 2 y2
2x + y = 6. Let an ellipse E : 2 2 1, a b of Put in (1) b
2 328
a b 25
1
eccentricity pass through the focus of the 2b 2
2
4b 2 1 328 656
2 2
b2 4
parabola P. Then the square of the length of the a a 2 25 25
latus rectum of E, is
385 8. The temperature T(t) of a body at time t = 0 is 160o
(1)
8 F and it decreases continuously as per the
347 dT
(2) K(T 80) , where K
8 differential equation
dt
512
(3) is positive constant. If T(15) = 120oF, then T(45) is
25
equal to
656
(4) (1) 85o F
25
(2) 95o F
Ans. (4)
Sol.- (3) 90o F
(4) 80o F
(1.6, 2.8)
Ans. (3)
( , )
axis Sol.-
(2, 3) Focur
dT
k T 80
dt
T t
dT
Slope of axis
1
T 80 0 Kdt
160
2
T
1
y 3 x 2
ln T 80 160 kt
2
ln T 80 ln 80 kt
2y 6 x 2
T 80
2y x 4 0 ln kt
80
2x y 6 0
T 80 80e kt
4x 2y 12 0
120 80 80e k.15
1.6 4 2.4
40 1
2.8 6 3.2 e k15
80 2
Ellipse passes through (2.4, 3.2)
T 45 80 80e k.45
2 2
24 32
80 80 e k.15
3
1
10 10
……….(1)
a2 b2 1
80 80
b2 1 b2 1 8
Also 1
a2 2 a 2 2 90
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