(IIT)(JEE)(Engineering)

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 1531 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