(IIT)(JEE)(Engineering)

FINAL JEE–MAIN EXAMINATION – JANUARY, 2024
(Held On Saturday 27th January, 2024) TIME : 9 : 00 AM to 12 : 00 NOON

MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A Sol. B = (2 +7, 3 – 2, 6 + 11)
1. n 1
Cr   k 2  8 n Cr 1 if and only if :

(1) 2 2  k  3 (2) 2 3  k  3 2

(3) 2 3  k  3 3 (4) 2 2  k  2 3
Ans. (1)
n-1
Sol. Cr = (k2 – 8) nCr+1
r  1  0, r  0
x  6 y  4 z 8
r 0 Point B lies on  
n 1
1 0 3
Cr
n
 k2  8 2  7  6 3  2  4 6  11  8
Cr 1  
1 0 3
r 1
 k2  8 3 – 6 = 0
n
 k2  8  0  = –2

 k  2 2  k  2 2   0 B  (3, 4, –1)


k   ,  2 2    2 2,   AB   7  3   4  2   11  1
2 2 2
…(I)

r 1  16  36  144
 n  r  1, 1
n
 196  14
 k2 – 8 1
3. Let x = x(t) and y = y(t) be solutions of the
k2 – 9 0
–3 k  3 ….(II) dx
differential equations  ax  0 and
From equation (I) and (II) we get dt

  
k   3,  2 2  2 2, 3 dy
dt
 by  0 respectively, a, b  R. Given that
2. The distance, of the point (7, –2, 11) from the line
x(0) = 2; y(0) = 1 and 3y(1) = 2x(1), the value of t,
x  6 y  4 z 8
  along the line for which x(t) = y(t), is :
1 0 3
x  5 y 1 z  5 (1) log 2 2 (2) log 4 3
  , is : 3
2 3 6
(1) 12 (2) 14 (3) log 3 4 (4) log 4 2
3
(3) 18 (4) 21
Ans. (2) Ans. (4)

, dx Sol. Equation of CE
Sol.  ax  0
dt y – 1 = (x – 3)
dx x+y=4
 adt
x
dx
 x  a  dt
ln | x | at  c
at t = 0, x = 2
ln 2  0  c orthocentre lies on the line x + y = 4
ln x  at  ln 2 so, a + b = 4
x
 eat b
2 I1   x sin  x(4  x)  dx …(i)
x  2e  at ….(i) a

dy Using king rule
 by  0 b
dt
I1    4  x  sin  x(4  x)  dx …(ii)
dy
 bdt a
y (i) + (ii)
ln | y | bt   b

t = 0, y = 1 2I1   4sin  x(4  x)  dx
0=0+ a

y = e–bt ….(ii) 2I1 = 4I2
According to question I1 = 2I2
3y(1) = 2x(1) I1
3e–b = 2(2 e–a) 2
I2
4
ea  b  36I1
3  72
For x(t) = y(t) I2
 2e–at = e–bt 5. If A denotes the sum of all the coefficients in the
2 = e(a – b)t expansion of (1 – 3x + 10x2)n and B denotes the
t
4 sum of all the coefficients in the expansion of
2 
3 (1 + x2)n, then :
log 4 2  t (1) A = B3 (2) 3A = B
3
3 (3) B = A (4) A = 3B
4. If (a, b) be the orthocentre of the triangle whose Ans. (1)
vertices are (1, 2), (2, 3) and (3, 1), and
b b
Sol. Sum of coefficients in the expansion of
I1   x sin  4x  x 2  dx , I2   sin  4x  x 2  dx (1 – 3x + 10x2)n = A
a a then A = (1 – 3 + 10)n  8n (put x = 1)
I1 and sum of coefficients in the expansion of
, then 36 is equal to :
I2 (1 + x2)n = B
(1) 72 (2) 88 then B = (1 + 1)n = 2n
(3) 80 (4) 66 A = B3
Ans. (1)

, 6. The number of common terms in the progressions x  4 y 1 z
th Sol.  
4, 9, 14, 19, ...... , up to 25 term and 3, 6, 9, 12, 1 2 3
......., up to 37th term is : x   y 1 z  2
 
(1) 9 (2) 5 2 4 5
(3) 7 (4) 8 the shortest distance between the lines
Ans. (3)
Sol. 4, 9, 14, 19, …., up to 25th term 
a  b  d  d 
1 2

T25 = 4 + (25 – 1) 5 = 4 + 120 = 124 d1  d 2
3, 6, 9, 12, …, up to 37th term
T37 = 3 + (37 – 1)3 = 3 + 108 = 111 4 0 2
st
Common difference of I series d1 = 5 1 2 3
nd
Common difference of II series d2 = 3 2 4 5

First common term = 9, and i j kˆ
their common difference = 15 (LCM of d1 and d2) 1 2 3
then common terms are
2 4 5
9, 24, 39, 54, 69, 84, 99
7. If the shortest distance of the parabola y2 = 4x from

   4  10  12   0  2  4  4 
the centre of the circle x2 + y2 – 4x – 16y + 64 = 0
2i  1j  0kˆ
is d, then d2 is equal to :
(1) 16 (2) 24
6 2    4
(3) 20 (4) 36 
5 5
Ans. (3)
3 = | – 4|
Sol. Equation of normal to parabola
y = mx – 2m – m3  – 4 = ±3
this normal passing through center of circle (2, 8)  = 7, 1
8 = 2m – 2m – m 3
Sum of all possible values of  is = 8
m = –2 1
1
So point P on parabola  (am2, –2am) = (4, 4)
9. If 
0 3  x  1 x
dx  a  b 2  c 3 , where

And C = (2, 8)
a, b, c are rational numbers, then 2a + 3b – 4c is
PC = 4  16  20 equal to :
2
d = 20 (1) 4 (2) 10
8. If the shortest distance between the lines (3) 7 (4) 8
x  4 y 1 z x   y 1 z  2 Ans. (4)
  and   is
1 2 3 2 4 5
3  x  1 x
1 1
1
6
, then the sum of all possible values of  is :
Sol. 
0 3  x  1 x
dx  
0  3  x   1  x 
dx
5
1 
 
1 1


2  0
(1) 5 (2) 8  3  x dx   1  x dx 
(3) 7 (4) 10 0 
Ans. (2)