(IIT)(JEE)(Engineering)

FINAL JEE–MAIN EXAMINATION – JANUARY, 2024
(Held On Thursday 01st February, 2024) TIME : 3 : 00 PM to 06 : 00 PM

MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A 
Sol. px2 + qx – r = 0 
2
1. Let f(x) = |2x +5|x|–3|,xR. If m and n denote the
p = A, q = AR, r = AR2
number of points where f is not continuous and not
Ax2 + ARx – AR2 = 0
differentiable respectively, then m + n is equal to : 
(1) 5 (2) 2 x2 + Rx – R2 = 0 
(3) 0 (4) 3 1 1 3
 
Ans. (4)   4
Sol. f(x) = |2x2+5|x|–3|   3 R  4
    R 
Graph of y = |2x2 + 5x – 3|  4 R 2 4 3
 16 
()2 = ()2 – 4 = R2 – 4(–R2) = 5  
 9 
= 80/9
3. The number of solutions of the equation 4 sin2x – 4
cos3x + 9 – 4 cosx = 0; x  [ –2, 2] is :
–3 –5/4 1/2 (1) 1
(2) 3
Graph of f(x) (3) 2
(4) 0
Ans. (4)
Sol. 4sin2x – 4cos3x + 9 – 4 cosx = 0 ; x  [ – 2, 2]
4 – 4cos2x – 4cos3x + 9 – 4 cosx = 0
4cos3x + 4cos2x + 4 cosx – 13 = 0
4cos3x + 4cos2x + 4cosx = 13
0
L.H.S.  12 can’t be equal to 13.
1
1
4. The value of  0
(2x 3  3x  x 2
 1) 3 dx is equal to:
Number of points of discontinuity = 0 = m
(1) 0
Number of points of non-differentiability = 3 = n
(2) 1
2. Let  and  be the roots of the equation px2 + qx – (3) 2
r = 0, where p  0. If p, q and r be the consecutive (4) –1
1 1 3 Ans. (1)
terms of a non-constant G.P and   , then
  4 1

 
1 3
3 2

the value of (–  is : Sol. I   2x  3x  x  1 dx
0
80
(1) (2) 9 2a
9 Using  f  x dx where f(2a–x) = –f(x)
0
20
(3) (4) 8 Here f(1–x) = f(x)
3
 I=0
Ans. (1)

, 18
x 2 y2  1 2

5. Let P be a point on the ellipse   1 . Let the
9 4  x3 x 3 
Sol.  3  
line passing through P and parallel to y-axis meet  
 
the circle x2 + y2 = 9 at point Q such that P and Q 12 6
are on the same side of the x-axis. Then, the  1   2 
 x3  x3  18 1 1
eccentricity of the locus of the point R on PQ such t 7  18c6    2   c6 12 . 26
 3     3
that PR : RQ = 4 : 3 as P moves on the ellipse, is :    
11 13 6 12
(1) (2)  1   2 
19 21  x3  x3  1 1
t13  18c12    2
18
  c12 . 12 .x –6
 3 2
6
139 13  3   
(3) (4)    
23 7
Ans. (4) m  18c6 .3–12.2 –6 : n  18c12 .2–12.3–6
1 1
2
 n  3  2 .3  3  3 
–12 –6
9
Q x2 y2    12 6    
 
P  1  m   3 .2   2  4
9 4
P(3cos , 2sin ) 7. Let  be a non-zero real number. Suppose f : R 
Q(3cos , 3sin) R is a differentiable function such that f (0) = 2 and
lim f  x   1 . If f '(x) = f(x) +3, for all x  R,
x 
Sol.
4 3 then f (–loge2) is equal to____.
(1) 3 (2) 5
P R Q (3) 9 (4) 7
(3C, 2S) (h, k) (3C, 3S) Ans. (3 OR BONUS)
h = 3cos;
Sol. f(0) = 2, lim f  x   1
x 
18
k sin  f’(x) – x.f(x) = 3
7
I.F = e–x
x 2 49y 2
 locus =  1 y(e–x) =  3.e x dx
9 324
324 117 13 3e x
e  1   f(x). (e–x) = c
49  9 21 7 
6. Let m and n be the coefficients of seventh and 3 3
x=0 2 c  c2 (1)
thirteenth terms respectively in the expansion of  
18 3
 1  1
f(x) =  c.ex
 1 x3  1  . Then  n  3 
3 2    is :
m 3
 2x 3  x  –  1 =  c(0)

4 1
(1) (2)  = –3  c = 1
9 9
3
1 9 f ( ln 2)   c.ex
(3) (4) 
4 4
= 1 + e3ln2 =9
Ans. (4)
(But  should be greater than 0 for finite
value of c)

, 8. Let P and Q be the points on the line A (1,3,2)

x  3 y  4 z 1
  which are at a distance of 6
8 2 2
units from the point R (1,2,3). If the centroid of the
1: 1
triangle PQR is (,,), then 2+2 +2 is:
B D C (3, 6, 7)
(1) 26 (–2, 8, 0) 1 7
 2 , 7, 2 
(2) 36 Sol.  
(3) 18 A(1, 3, 2); B(–2, 8, 0); C(3, 6, 7);
(4) 24 AC  2iˆ  3jˆ  5kˆ
Ans. (3)
AB = 9  25  4  38
Sol.
R(1,2,3) AC = 4  9  25  38
1 3 1
AD  ˆi  4jˆ  kˆ  (iˆ  8jˆ  3k)
ˆ
2 2 2
Length of projection of AD on AC
P Q
AD.AC 37
P(8 – 3, 2 + 4, 2 – 1) = 
| AC | 2 38
PR = 6
10. Let Sn denote the sum of the first n terms of an
(8 – 4)2 + ( 2 + 2)2 + (2 – 4)2 = 36
arithmetic progression. If S10 = 390 and the ratio of
 = 0, 1 the tenth and the fifth terms is 15 : 7, then S15 –S5
Hence P(–3, 4, –1) & Q(5, 6, 1) is equal to:
Centroid of PQR = (1, 4, 1)  (,, ) (1) 800
  
 +  +  = 18 (2) 890
9. Consider a ABC where A(1,2,3,), B(–2,8,0) and (3) 790
C(3,6,7). If the angle bisector of BAC meets the (4) 690

line BC at D, then the length of the projection of Ans. (3)
  Sol. S10 = 390
the vector AD on the vector AC is:
10
2a  10  1 d   390
(1)
37 2 
2 38
 2a + 9d = 78 (1)
38 t10 15 a  9d 15
(2)     8a  3d (2)
2 t5 7 a  4d 7
39
(3) From (1) & (2) a=3&d=8
2 38
15 5
(4) 19
S15 – S5   6  14  8    6  4  8 
2 2
Ans. (1) 15  118  5  38
 =790
2