(IIT)(JEE)(Engineering)

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
ac 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
ac 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 
gx   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  15  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)