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Page 1
FINAL JEE –MAIN EXAMINATION – JANUARY, 2024
(Held On Saturday 27
th
January, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A
1.
?H ?I n 1 2 n
r r 1
C k 8 C
?M ?K ?]?M 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)
Sol.
n-1
C
r
= (k
2
– 8)
n
C
r+1
r 0
r 1 0, r 0
?? ?K ?? ??
n 1
2 r
n
r 1
C
k 8
C
?M ?K ?]?M
2
r 1
k 8
n
?K ?]?M
??
2
k 8 0 ?M?^
?H ?I ?H ?I k 2 2 k 2 2 0 ?M ?K ?^
?H ?I ?H ?I k , 2 2 2 2, ?? ?M ?? ?M ?? ?? …(I)
?| ?@ n r 1 ???K ,
r 1
1
n
?K ??
?? k
2
– 8 ???@ 1
k
2
– 9 ???@ 0
–3 ???@ k ?? 3 ….(II)
From equation (I) and (II) we get ?@ ?@ ?I ?H k 3, 2 2 2 2, 3
????
?? ?M ?M ??????
2. The distance, of the point (7, –2, 11) from the line
x 6 y 4 z 8
1 0 3
?M ?M ?M ?]?] along the line
x 5 y 1 z 5
2 3 6
?M ?M ?M ?]?]
?M , is :
(1) 12 (2) 14
(3) 18 (4) 21
Ans. (2)
Sol. B = (2 ?? +7, ?M 3 ?? – 2, 6 ?? + 11)
Point B lies on
x 6 y 4 z 8
1 0 3
?M ?M ?M ?]?]
2 7 6 3 2 4 6 11 8
1 0 3
?? ?K ?M ?M ?? ?M ?M ?? ?K ?M ?]?]
?M 3 ?? – 6 = 0
?? = –2
B ?? (3, 4, –1)
?H ?I ?H ?I ?H ?I 2 2 2
AB 7 3 4 2 11 1 ?] ?M ?K ?K ?K ?K
16 36 144 ?] ?K ?K
196 14 ?]?]
3. Let x = x(t) and y = y(t) be solutions of the
differential equations
dx
ax 0
dt
?K?] and
dy
by 0
dt
?K?] respectively, a, b ?? R. Given that
x(0) = 2; y(0) = 1 and 3y(1) = 2x(1), the value of t,
for which x(t) = y(t), is :
(1)
2
3
log 2 (2)
4
log 3
(3)
3
log 4
(4)
4
3
log 2
Ans. (4)
Page 2
FINAL JEE –MAIN EXAMINATION – JANUARY, 2024
(Held On Saturday 27
th
January, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A
1.
?H ?I n 1 2 n
r r 1
C k 8 C
?M ?K ?]?M 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)
Sol.
n-1
C
r
= (k
2
– 8)
n
C
r+1
r 0
r 1 0, r 0
?? ?K ?? ??
n 1
2 r
n
r 1
C
k 8
C
?M ?K ?]?M
2
r 1
k 8
n
?K ?]?M
??
2
k 8 0 ?M?^
?H ?I ?H ?I k 2 2 k 2 2 0 ?M ?K ?^
?H ?I ?H ?I k , 2 2 2 2, ?? ?M ?? ?M ?? ?? …(I)
?| ?@ n r 1 ???K ,
r 1
1
n
?K ??
?? k
2
– 8 ???@ 1
k
2
– 9 ???@ 0
–3 ???@ k ?? 3 ….(II)
From equation (I) and (II) we get ?@ ?@ ?I ?H k 3, 2 2 2 2, 3
????
?? ?M ?M ??????
2. The distance, of the point (7, –2, 11) from the line
x 6 y 4 z 8
1 0 3
?M ?M ?M ?]?] along the line
x 5 y 1 z 5
2 3 6
?M ?M ?M ?]?]
?M , is :
(1) 12 (2) 14
(3) 18 (4) 21
Ans. (2)
Sol. B = (2 ?? +7, ?M 3 ?? – 2, 6 ?? + 11)
Point B lies on
x 6 y 4 z 8
1 0 3
?M ?M ?M ?]?]
2 7 6 3 2 4 6 11 8
1 0 3
?? ?K ?M ?M ?? ?M ?M ?? ?K ?M ?]?]
?M 3 ?? – 6 = 0
?? = –2
B ?? (3, 4, –1)
?H ?I ?H ?I ?H ?I 2 2 2
AB 7 3 4 2 11 1 ?] ?M ?K ?K ?K ?K
16 36 144 ?] ?K ?K
196 14 ?]?]
3. Let x = x(t) and y = y(t) be solutions of the
differential equations
dx
ax 0
dt
?K?] and
dy
by 0
dt
?K?] respectively, a, b ?? R. Given that
x(0) = 2; y(0) = 1 and 3y(1) = 2x(1), the value of t,
for which x(t) = y(t), is :
(1)
2
3
log 2 (2)
4
log 3
(3)
3
log 4
(4)
4
3
log 2
Ans. (4)
Sol.
dx
ax 0
dt
?K?]
dx
adt
x
?]?M
dx
a dt
x
?]?M
????
ln | x | at c ?] ?M ?K
at t = 0, x = 2
ln 2 0 c ?]?K
ln x at ln 2 ?] ?M ?K
at
x
e
2
?M ?]
at
x 2e
?M ?] ….(i)
dy
by 0
dt
?K?]
dy
bdt
y
?]?M
ln | y | bt ?] ?M ?K ??
t = 0, y = 1
0 = 0 + ??
y = e
–bt
….(ii)
According to question
3y(1) = 2x(1)
3e
–b
= 2(2 e
–a
)
a b
4
e
3
?M ?]
For x(t) = y(t)
?? 2e
–at
= e
–bt
2 = e
(a – b)t
t
4
2
3
????
?] ????
????
4
3
log 2 t ?]
4. If (a, b) be the orthocentre of the triangle whose
vertices are (1, 2), (2, 3) and (3, 1), and
?H ?I b
2
1
a
I x sin 4x x dx ?]?M
?? ,
?H ?I b
2
2
a
I sin 4x x dx ?]?M
?? , then
1
2
I
36
I
is equal to :
(1) 72 (2) 88
(3) 80 (4) 66
Ans. (1)
Sol. Equation of CE
y – 1 = ?M (x – 3)
x + y = 4
orthocentre lies on the line x + y = 4
so, a + b = 4
?H ?I b
1
a
I x sin x(4 x) dx ?]?M
?? …(i)
Using king rule
?H ?I ?H ?I b
1
a
I 4 x sin x(4 x) dx ?] ?M ?M ?? …(ii)
(i) + (ii)
?H ?I b
1
a
2I 4sin x(4 x) dx ?]?M
??
2I
1
= 4I
2
I
1
= 2I
2
1
2
I
2
I
?]
1
2
36I
72
I
?]
5. If A denotes the sum of all the coefficients in the
expansion of (1 – 3x + 10x
2
)
n
and B denotes the
sum of all the coefficients in the expansion of
(1 + x
2
)
n
, then :
(1) A = B
3
(2) 3A = B
(3) B = A
3
(4) A = 3B
Ans. (1)
Sol. Sum of coefficients in the expansion of
(1 – 3x + 10x
2
)
n
= A
then A = (1 – 3 + 10)
n
?] 8
n
(put x = 1)
and sum of coefficients in the expansion of
(1 + x
2
)
n
= B
then B = (1 + 1)
n
= 2
n
A = B
3
Page 3
FINAL JEE –MAIN EXAMINATION – JANUARY, 2024
(Held On Saturday 27
th
January, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A
1.
?H ?I n 1 2 n
r r 1
C k 8 C
?M ?K ?]?M 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)
Sol.
n-1
C
r
= (k
2
– 8)
n
C
r+1
r 0
r 1 0, r 0
?? ?K ?? ??
n 1
2 r
n
r 1
C
k 8
C
?M ?K ?]?M
2
r 1
k 8
n
?K ?]?M
??
2
k 8 0 ?M?^
?H ?I ?H ?I k 2 2 k 2 2 0 ?M ?K ?^
?H ?I ?H ?I k , 2 2 2 2, ?? ?M ?? ?M ?? ?? …(I)
?| ?@ n r 1 ???K ,
r 1
1
n
?K ??
?? k
2
– 8 ???@ 1
k
2
– 9 ???@ 0
–3 ???@ k ?? 3 ….(II)
From equation (I) and (II) we get ?@ ?@ ?I ?H k 3, 2 2 2 2, 3
????
?? ?M ?M ??????
2. The distance, of the point (7, –2, 11) from the line
x 6 y 4 z 8
1 0 3
?M ?M ?M ?]?] along the line
x 5 y 1 z 5
2 3 6
?M ?M ?M ?]?]
?M , is :
(1) 12 (2) 14
(3) 18 (4) 21
Ans. (2)
Sol. B = (2 ?? +7, ?M 3 ?? – 2, 6 ?? + 11)
Point B lies on
x 6 y 4 z 8
1 0 3
?M ?M ?M ?]?]
2 7 6 3 2 4 6 11 8
1 0 3
?? ?K ?M ?M ?? ?M ?M ?? ?K ?M ?]?]
?M 3 ?? – 6 = 0
?? = –2
B ?? (3, 4, –1)
?H ?I ?H ?I ?H ?I 2 2 2
AB 7 3 4 2 11 1 ?] ?M ?K ?K ?K ?K
16 36 144 ?] ?K ?K
196 14 ?]?]
3. Let x = x(t) and y = y(t) be solutions of the
differential equations
dx
ax 0
dt
?K?] and
dy
by 0
dt
?K?] respectively, a, b ?? R. Given that
x(0) = 2; y(0) = 1 and 3y(1) = 2x(1), the value of t,
for which x(t) = y(t), is :
(1)
2
3
log 2 (2)
4
log 3
(3)
3
log 4
(4)
4
3
log 2
Ans. (4)
Sol.
dx
ax 0
dt
?K?]
dx
adt
x
?]?M
dx
a dt
x
?]?M
????
ln | x | at c ?] ?M ?K
at t = 0, x = 2
ln 2 0 c ?]?K
ln x at ln 2 ?] ?M ?K
at
x
e
2
?M ?]
at
x 2e
?M ?] ….(i)
dy
by 0
dt
?K?]
dy
bdt
y
?]?M
ln | y | bt ?] ?M ?K ??
t = 0, y = 1
0 = 0 + ??
y = e
–bt
….(ii)
According to question
3y(1) = 2x(1)
3e
–b
= 2(2 e
–a
)
a b
4
e
3
?M ?]
For x(t) = y(t)
?? 2e
–at
= e
–bt
2 = e
(a – b)t
t
4
2
3
????
?] ????
????
4
3
log 2 t ?]
4. If (a, b) be the orthocentre of the triangle whose
vertices are (1, 2), (2, 3) and (3, 1), and
?H ?I b
2
1
a
I x sin 4x x dx ?]?M
?? ,
?H ?I b
2
2
a
I sin 4x x dx ?]?M
?? , then
1
2
I
36
I
is equal to :
(1) 72 (2) 88
(3) 80 (4) 66
Ans. (1)
Sol. Equation of CE
y – 1 = ?M (x – 3)
x + y = 4
orthocentre lies on the line x + y = 4
so, a + b = 4
?H ?I b
1
a
I x sin x(4 x) dx ?]?M
?? …(i)
Using king rule
?H ?I ?H ?I b
1
a
I 4 x sin x(4 x) dx ?] ?M ?M ?? …(ii)
(i) + (ii)
?H ?I b
1
a
2I 4sin x(4 x) dx ?]?M
??
2I
1
= 4I
2
I
1
= 2I
2
1
2
I
2
I
?]
1
2
36I
72
I
?]
5. If A denotes the sum of all the coefficients in the
expansion of (1 – 3x + 10x
2
)
n
and B denotes the
sum of all the coefficients in the expansion of
(1 + x
2
)
n
, then :
(1) A = B
3
(2) 3A = B
(3) B = A
3
(4) A = 3B
Ans. (1)
Sol. Sum of coefficients in the expansion of
(1 – 3x + 10x
2
)
n
= A
then A = (1 – 3 + 10)
n
?] 8
n
(put x = 1)
and sum of coefficients in the expansion of
(1 + x
2
)
n
= B
then B = (1 + 1)
n
= 2
n
A = B
3
6. The number of common terms in the progressions
4, 9, 14, 19, ...... , up to 25
th
term and 3, 6, 9, 12,
......., up to 37
th
term is :
(1) 9 (2) 5
(3) 7 (4) 8
Ans. (3)
Sol. 4, 9, 14, 19, …., up to 25
th
term
T
25
= 4 + (25 – 1) 5 = 4 + 120 = 124
3, 6, 9, 12, …, up to 37
th
term
T
37
= 3 + (37 – 1)3 = 3 + 108 = 111
Common difference of I
st
series d
1
= 5
Common difference of II
nd
series d
2
= 3
First common term = 9, and
their common difference = 15 (LCM of d
1
and d
2
)
then common terms are
9, 24, 39, 54, 69, 84, 99
7. If the shortest distance of the parabola y
2
= 4x from
the centre of the circle x
2
+ y
2
– 4x – 16y + 64 = 0
is d, then d
2
is equal to :
(1) 16 (2) 24
(3) 20 (4) 36
Ans. (3)
Sol. Equation of normal to parabola
y = mx – 2m – m
3
this normal passing through center of circle (2, 8)
8 = 2m – 2m – m
3
m = –2
So point P on parabola ?? (am
2
, –2am) = (4, 4)
And C = (2, 8)
PC = 4 16 20 ?K?]
d
2
= 20
8. If the shortest distance between the lines
x 4 y 1 z
1 2 3
?M?K
?]?]
?M and
x y 1 z 2
2 4 5
?M ?? ?K ?M ?]?]
?M is
6
5
, then the sum of all possible values of ?? is :
(1) 5 (2) 8
(3) 7 (4) 10
Ans. (2)
Sol.
x 4 y 1 z
1 2 3
?M?K
?]?]
?M
x y 1 z 2
2 4 5
?M ?? ?K ?M ?]?]
?M
the shortest distance between the lines
?H ?I ?H ?I 1 2
1 2
a b d d
d d
?M ?? ?? ?] ??
4 0 2
1 2 3
2 4 5
ˆ
i j k
1 2 3
2 4 5
???M
?M ?M ?] ?M ?M
?H ?I ?H ?I ?H ?I 4 10 12 0 2 4 4
ˆ
2i 1j 0k
?? ?M ?M ?K ?M ?K ?M ?] ?M?K
?H ?I 2 4
6
5 5
???M
?]
3 = | ?? – 4|
?? – 4 = ±3
?? = 7, 1
Sum of all possible values of ?? is = 8
9. If
1
0
1
dx a b 2 c 3
3 x 1 x
?] ?K ?K ?K ?K ?K ?? , where
a, b, c are rational numbers, then 2a + 3b – 4c is
equal to :
(1) 4 (2) 10
(3) 7 (4) 8
Ans. (4)
Sol.
1
0
1
dx
3 x 1 x ?K ?K ?K ?? ?H ?I ?H ?I 1
0
3 x 1 x
dx
3 x 1 x
?K ?M ?K ?] ?K ?M ?K ??
?H ?I 1 1
0 0
1
3 x dx 1 x dx
2
????
?K ?M ?K ????
????
????
Page 4
FINAL JEE –MAIN EXAMINATION – JANUARY, 2024
(Held On Saturday 27
th
January, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A
1.
?H ?I n 1 2 n
r r 1
C k 8 C
?M ?K ?]?M 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)
Sol.
n-1
C
r
= (k
2
– 8)
n
C
r+1
r 0
r 1 0, r 0
?? ?K ?? ??
n 1
2 r
n
r 1
C
k 8
C
?M ?K ?]?M
2
r 1
k 8
n
?K ?]?M
??
2
k 8 0 ?M?^
?H ?I ?H ?I k 2 2 k 2 2 0 ?M ?K ?^
?H ?I ?H ?I k , 2 2 2 2, ?? ?M ?? ?M ?? ?? …(I)
?| ?@ n r 1 ???K ,
r 1
1
n
?K ??
?? k
2
– 8 ???@ 1
k
2
– 9 ???@ 0
–3 ???@ k ?? 3 ….(II)
From equation (I) and (II) we get ?@ ?@ ?I ?H k 3, 2 2 2 2, 3
????
?? ?M ?M ??????
2. The distance, of the point (7, –2, 11) from the line
x 6 y 4 z 8
1 0 3
?M ?M ?M ?]?] along the line
x 5 y 1 z 5
2 3 6
?M ?M ?M ?]?]
?M , is :
(1) 12 (2) 14
(3) 18 (4) 21
Ans. (2)
Sol. B = (2 ?? +7, ?M 3 ?? – 2, 6 ?? + 11)
Point B lies on
x 6 y 4 z 8
1 0 3
?M ?M ?M ?]?]
2 7 6 3 2 4 6 11 8
1 0 3
?? ?K ?M ?M ?? ?M ?M ?? ?K ?M ?]?]
?M 3 ?? – 6 = 0
?? = –2
B ?? (3, 4, –1)
?H ?I ?H ?I ?H ?I 2 2 2
AB 7 3 4 2 11 1 ?] ?M ?K ?K ?K ?K
16 36 144 ?] ?K ?K
196 14 ?]?]
3. Let x = x(t) and y = y(t) be solutions of the
differential equations
dx
ax 0
dt
?K?] and
dy
by 0
dt
?K?] respectively, a, b ?? R. Given that
x(0) = 2; y(0) = 1 and 3y(1) = 2x(1), the value of t,
for which x(t) = y(t), is :
(1)
2
3
log 2 (2)
4
log 3
(3)
3
log 4
(4)
4
3
log 2
Ans. (4)
Sol.
dx
ax 0
dt
?K?]
dx
adt
x
?]?M
dx
a dt
x
?]?M
????
ln | x | at c ?] ?M ?K
at t = 0, x = 2
ln 2 0 c ?]?K
ln x at ln 2 ?] ?M ?K
at
x
e
2
?M ?]
at
x 2e
?M ?] ….(i)
dy
by 0
dt
?K?]
dy
bdt
y
?]?M
ln | y | bt ?] ?M ?K ??
t = 0, y = 1
0 = 0 + ??
y = e
–bt
….(ii)
According to question
3y(1) = 2x(1)
3e
–b
= 2(2 e
–a
)
a b
4
e
3
?M ?]
For x(t) = y(t)
?? 2e
–at
= e
–bt
2 = e
(a – b)t
t
4
2
3
????
?] ????
????
4
3
log 2 t ?]
4. If (a, b) be the orthocentre of the triangle whose
vertices are (1, 2), (2, 3) and (3, 1), and
?H ?I b
2
1
a
I x sin 4x x dx ?]?M
?? ,
?H ?I b
2
2
a
I sin 4x x dx ?]?M
?? , then
1
2
I
36
I
is equal to :
(1) 72 (2) 88
(3) 80 (4) 66
Ans. (1)
Sol. Equation of CE
y – 1 = ?M (x – 3)
x + y = 4
orthocentre lies on the line x + y = 4
so, a + b = 4
?H ?I b
1
a
I x sin x(4 x) dx ?]?M
?? …(i)
Using king rule
?H ?I ?H ?I b
1
a
I 4 x sin x(4 x) dx ?] ?M ?M ?? …(ii)
(i) + (ii)
?H ?I b
1
a
2I 4sin x(4 x) dx ?]?M
??
2I
1
= 4I
2
I
1
= 2I
2
1
2
I
2
I
?]
1
2
36I
72
I
?]
5. If A denotes the sum of all the coefficients in the
expansion of (1 – 3x + 10x
2
)
n
and B denotes the
sum of all the coefficients in the expansion of
(1 + x
2
)
n
, then :
(1) A = B
3
(2) 3A = B
(3) B = A
3
(4) A = 3B
Ans. (1)
Sol. Sum of coefficients in the expansion of
(1 – 3x + 10x
2
)
n
= A
then A = (1 – 3 + 10)
n
?] 8
n
(put x = 1)
and sum of coefficients in the expansion of
(1 + x
2
)
n
= B
then B = (1 + 1)
n
= 2
n
A = B
3
6. The number of common terms in the progressions
4, 9, 14, 19, ...... , up to 25
th
term and 3, 6, 9, 12,
......., up to 37
th
term is :
(1) 9 (2) 5
(3) 7 (4) 8
Ans. (3)
Sol. 4, 9, 14, 19, …., up to 25
th
term
T
25
= 4 + (25 – 1) 5 = 4 + 120 = 124
3, 6, 9, 12, …, up to 37
th
term
T
37
= 3 + (37 – 1)3 = 3 + 108 = 111
Common difference of I
st
series d
1
= 5
Common difference of II
nd
series d
2
= 3
First common term = 9, and
their common difference = 15 (LCM of d
1
and d
2
)
then common terms are
9, 24, 39, 54, 69, 84, 99
7. If the shortest distance of the parabola y
2
= 4x from
the centre of the circle x
2
+ y
2
– 4x – 16y + 64 = 0
is d, then d
2
is equal to :
(1) 16 (2) 24
(3) 20 (4) 36
Ans. (3)
Sol. Equation of normal to parabola
y = mx – 2m – m
3
this normal passing through center of circle (2, 8)
8 = 2m – 2m – m
3
m = –2
So point P on parabola ?? (am
2
, –2am) = (4, 4)
And C = (2, 8)
PC = 4 16 20 ?K?]
d
2
= 20
8. If the shortest distance between the lines
x 4 y 1 z
1 2 3
?M?K
?]?]
?M and
x y 1 z 2
2 4 5
?M ?? ?K ?M ?]?]
?M is
6
5
, then the sum of all possible values of ?? is :
(1) 5 (2) 8
(3) 7 (4) 10
Ans. (2)
Sol.
x 4 y 1 z
1 2 3
?M?K
?]?]
?M
x y 1 z 2
2 4 5
?M ?? ?K ?M ?]?]
?M
the shortest distance between the lines
?H ?I ?H ?I 1 2
1 2
a b d d
d d
?M ?? ?? ?] ??
4 0 2
1 2 3
2 4 5
ˆ
i j k
1 2 3
2 4 5
???M
?M ?M ?] ?M ?M
?H ?I ?H ?I ?H ?I 4 10 12 0 2 4 4
ˆ
2i 1j 0k
?? ?M ?M ?K ?M ?K ?M ?] ?M?K
?H ?I 2 4
6
5 5
???M
?]
3 = | ?? – 4|
?? – 4 = ±3
?? = 7, 1
Sum of all possible values of ?? is = 8
9. If
1
0
1
dx a b 2 c 3
3 x 1 x
?] ?K ?K ?K ?K ?K ?? , where
a, b, c are rational numbers, then 2a + 3b – 4c is
equal to :
(1) 4 (2) 10
(3) 7 (4) 8
Ans. (4)
Sol.
1
0
1
dx
3 x 1 x ?K ?K ?K ?? ?H ?I ?H ?I 1
0
3 x 1 x
dx
3 x 1 x
?K ?M ?K ?] ?K ?M ?K ??
?H ?I 1 1
0 0
1
3 x dx 1 x dx
2
????
?K ?M ?K ????
????
????
?H ?I ?H ?I 1
3 3
2 2
0
3 x 2 1 x
1
2
2 3 3
????
?K?K
????
?M ????
????
????
?H ?I 3
2
1 2 2
8 3 3 2 1
2 3 3
???? ????
?M ?M ?M ???? ????
????
???? ????
1
8 3 3 2 2 1
3
????
?M ?M ?K ????
2
3 3 2 a b 2 c 3
3
?] ?M ?M ?] ?K ?K
2
a 3, b , c 1
3
?] ?] ?M ?] ?M
2a + 3b – 4c = 6 – 2 + 4 = 8
10. Let S = {l, 2, 3, ... , 10}. Suppose M is the set of all
the subsets of S, then the relation
R = {(A, B): A ?? B ?? ?? ; A, B ?? M} is :
(1) symmetric and reflexive only
(2) reflexive only
(3) symmetric and transitive only
(4) symmetric only
Ans. (4)
Sol. Let S = {1, 2, 3, …, 10}
R = {(A, B): A ?? B ?? ?? ; A, B ?? M}
For Reflexive,
M is subset of ‘S’
So ?? ?@?? M
for ?? ?? ?? = ??
?? but relation is A ?? B ?? ??
So it is not reflexive.
For symmetric,
ARB A ?? B ?? ?? ,
?? BRA ?? B ?? A ?? ?? ,
So it is symmetric.
For transitive,
If A = {(1, 2), (2, 3)}
B = {(2, 3), (3, 4)}
C = {(3, 4), (5, 6)}
ARB & BRC but A does not relate to C
So it not transitive
11. If S = {z ?? C : |z – i| = |z + i| = |z–1|}, then, n(S) is:
(1) 1 (2) 0
(3) 3 (4) 2
Ans. (1)
Sol. |z – i| = |z + i| = |z – 1|
ABC is a triangle. Hence its circum-centre will be
the only point whose distance from A, B, C will be
same.
So n(S) = 1
12. Four distinct points (2k, 3k), (1, 0), (0, 1) and
(0, 0) lie on a circle for k equal to :
(1)
2
13
(2)
3
13
(3)
5
13
(4)
1
13
Ans. (3)
Sol. (2k, 3k) will lie on circle whose diameter is AB.
(x – 1) (x) + (y – 1) (y) = 0
x
2
+ y
2
– x – y = 0 …(i)
Satisfy (2k, 3k) in (i)
(2k)
2
+ (3k)
2
– 2k – 3k = 0
13k
2
– 5k = 0
k = 0,
5
k
13
?]
hence
5
k
13
?]
Page 5
FINAL JEE –MAIN EXAMINATION – JANUARY, 2024
(Held On Saturday 27
th
January, 2024) TIME : 9 : 00 AM to 12 : 00 NOON
MATHEMATICS TEST PAPER WITH SOLUTION
SECTION-A
1.
?H ?I n 1 2 n
r r 1
C k 8 C
?M ?K ?]?M 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)
Sol.
n-1
C
r
= (k
2
– 8)
n
C
r+1
r 0
r 1 0, r 0
?? ?K ?? ??
n 1
2 r
n
r 1
C
k 8
C
?M ?K ?]?M
2
r 1
k 8
n
?K ?]?M
??
2
k 8 0 ?M?^
?H ?I ?H ?I k 2 2 k 2 2 0 ?M ?K ?^
?H ?I ?H ?I k , 2 2 2 2, ?? ?M ?? ?M ?? ?? …(I)
?| ?@ n r 1 ???K ,
r 1
1
n
?K ??
?? k
2
– 8 ???@ 1
k
2
– 9 ???@ 0
–3 ???@ k ?? 3 ….(II)
From equation (I) and (II) we get ?@ ?@ ?I ?H k 3, 2 2 2 2, 3
????
?? ?M ?M ??????
2. The distance, of the point (7, –2, 11) from the line
x 6 y 4 z 8
1 0 3
?M ?M ?M ?]?] along the line
x 5 y 1 z 5
2 3 6
?M ?M ?M ?]?]
?M , is :
(1) 12 (2) 14
(3) 18 (4) 21
Ans. (2)
Sol. B = (2 ?? +7, ?M 3 ?? – 2, 6 ?? + 11)
Point B lies on
x 6 y 4 z 8
1 0 3
?M ?M ?M ?]?]
2 7 6 3 2 4 6 11 8
1 0 3
?? ?K ?M ?M ?? ?M ?M ?? ?K ?M ?]?]
?M 3 ?? – 6 = 0
?? = –2
B ?? (3, 4, –1)
?H ?I ?H ?I ?H ?I 2 2 2
AB 7 3 4 2 11 1 ?] ?M ?K ?K ?K ?K
16 36 144 ?] ?K ?K
196 14 ?]?]
3. Let x = x(t) and y = y(t) be solutions of the
differential equations
dx
ax 0
dt
?K?] and
dy
by 0
dt
?K?] respectively, a, b ?? R. Given that
x(0) = 2; y(0) = 1 and 3y(1) = 2x(1), the value of t,
for which x(t) = y(t), is :
(1)
2
3
log 2 (2)
4
log 3
(3)
3
log 4
(4)
4
3
log 2
Ans. (4)
Sol.
dx
ax 0
dt
?K?]
dx
adt
x
?]?M
dx
a dt
x
?]?M
????
ln | x | at c ?] ?M ?K
at t = 0, x = 2
ln 2 0 c ?]?K
ln x at ln 2 ?] ?M ?K
at
x
e
2
?M ?]
at
x 2e
?M ?] ….(i)
dy
by 0
dt
?K?]
dy
bdt
y
?]?M
ln | y | bt ?] ?M ?K ??
t = 0, y = 1
0 = 0 + ??
y = e
–bt
….(ii)
According to question
3y(1) = 2x(1)
3e
–b
= 2(2 e
–a
)
a b
4
e
3
?M ?]
For x(t) = y(t)
?? 2e
–at
= e
–bt
2 = e
(a – b)t
t
4
2
3
????
?] ????
????
4
3
log 2 t ?]
4. If (a, b) be the orthocentre of the triangle whose
vertices are (1, 2), (2, 3) and (3, 1), and
?H ?I b
2
1
a
I x sin 4x x dx ?]?M
?? ,
?H ?I b
2
2
a
I sin 4x x dx ?]?M
?? , then
1
2
I
36
I
is equal to :
(1) 72 (2) 88
(3) 80 (4) 66
Ans. (1)
Sol. Equation of CE
y – 1 = ?M (x – 3)
x + y = 4
orthocentre lies on the line x + y = 4
so, a + b = 4
?H ?I b
1
a
I x sin x(4 x) dx ?]?M
?? …(i)
Using king rule
?H ?I ?H ?I b
1
a
I 4 x sin x(4 x) dx ?] ?M ?M ?? …(ii)
(i) + (ii)
?H ?I b
1
a
2I 4sin x(4 x) dx ?]?M
??
2I
1
= 4I
2
I
1
= 2I
2
1
2
I
2
I
?]
1
2
36I
72
I
?]
5. If A denotes the sum of all the coefficients in the
expansion of (1 – 3x + 10x
2
)
n
and B denotes the
sum of all the coefficients in the expansion of
(1 + x
2
)
n
, then :
(1) A = B
3
(2) 3A = B
(3) B = A
3
(4) A = 3B
Ans. (1)
Sol. Sum of coefficients in the expansion of
(1 – 3x + 10x
2
)
n
= A
then A = (1 – 3 + 10)
n
?] 8
n
(put x = 1)
and sum of coefficients in the expansion of
(1 + x
2
)
n
= B
then B = (1 + 1)
n
= 2
n
A = B
3
6. The number of common terms in the progressions
4, 9, 14, 19, ...... , up to 25
th
term and 3, 6, 9, 12,
......., up to 37
th
term is :
(1) 9 (2) 5
(3) 7 (4) 8
Ans. (3)
Sol. 4, 9, 14, 19, …., up to 25
th
term
T
25
= 4 + (25 – 1) 5 = 4 + 120 = 124
3, 6, 9, 12, …, up to 37
th
term
T
37
= 3 + (37 – 1)3 = 3 + 108 = 111
Common difference of I
st
series d
1
= 5
Common difference of II
nd
series d
2
= 3
First common term = 9, and
their common difference = 15 (LCM of d
1
and d
2
)
then common terms are
9, 24, 39, 54, 69, 84, 99
7. If the shortest distance of the parabola y
2
= 4x from
the centre of the circle x
2
+ y
2
– 4x – 16y + 64 = 0
is d, then d
2
is equal to :
(1) 16 (2) 24
(3) 20 (4) 36
Ans. (3)
Sol. Equation of normal to parabola
y = mx – 2m – m
3
this normal passing through center of circle (2, 8)
8 = 2m – 2m – m
3
m = –2
So point P on parabola ?? (am
2
, –2am) = (4, 4)
And C = (2, 8)
PC = 4 16 20 ?K?]
d
2
= 20
8. If the shortest distance between the lines
x 4 y 1 z
1 2 3
?M?K
?]?]
?M and
x y 1 z 2
2 4 5
?M ?? ?K ?M ?]?]
?M is
6
5
, then the sum of all possible values of ?? is :
(1) 5 (2) 8
(3) 7 (4) 10
Ans. (2)
Sol.
x 4 y 1 z
1 2 3
?M?K
?]?]
?M
x y 1 z 2
2 4 5
?M ?? ?K ?M ?]?]
?M
the shortest distance between the lines
?H ?I ?H ?I 1 2
1 2
a b d d
d d
?M ?? ?? ?] ??
4 0 2
1 2 3
2 4 5
ˆ
i j k
1 2 3
2 4 5
???M
?M ?M ?] ?M ?M
?H ?I ?H ?I ?H ?I 4 10 12 0 2 4 4
ˆ
2i 1j 0k
?? ?M ?M ?K ?M ?K ?M ?] ?M?K
?H ?I 2 4
6
5 5
???M
?]
3 = | ?? – 4|
?? – 4 = ±3
?? = 7, 1
Sum of all possible values of ?? is = 8
9. If
1
0
1
dx a b 2 c 3
3 x 1 x
?] ?K ?K ?K ?K ?K ?? , where
a, b, c are rational numbers, then 2a + 3b – 4c is
equal to :
(1) 4 (2) 10
(3) 7 (4) 8
Ans. (4)
Sol.
1
0
1
dx
3 x 1 x ?K ?K ?K ?? ?H ?I ?H ?I 1
0
3 x 1 x
dx
3 x 1 x
?K ?M ?K ?] ?K ?M ?K ??
?H ?I 1 1
0 0
1
3 x dx 1 x dx
2
????
?K ?M ?K ????
????
????
?H ?I ?H ?I 1
3 3
2 2
0
3 x 2 1 x
1
2
2 3 3
????
?K?K
????
?M ????
????
????
?H ?I 3
2
1 2 2
8 3 3 2 1
2 3 3
???? ????
?M ?M ?M ???? ????
????
???? ????
1
8 3 3 2 2 1
3
????
?M ?M ?K ????
2
3 3 2 a b 2 c 3
3
?] ?M ?M ?] ?K ?K
2
a 3, b , c 1
3
?] ?] ?M ?] ?M
2a + 3b – 4c = 6 – 2 + 4 = 8
10. Let S = {l, 2, 3, ... , 10}. Suppose M is the set of all
the subsets of S, then the relation
R = {(A, B): A ?? B ?? ?? ; A, B ?? M} is :
(1) symmetric and reflexive only
(2) reflexive only
(3) symmetric and transitive only
(4) symmetric only
Ans. (4)
Sol. Let S = {1, 2, 3, …, 10}
R = {(A, B): A ?? B ?? ?? ; A, B ?? M}
For Reflexive,
M is subset of ‘S’
So ?? ?@?? M
for ?? ?? ?? = ??
?? but relation is A ?? B ?? ??
So it is not reflexive.
For symmetric,
ARB A ?? B ?? ?? ,
?? BRA ?? B ?? A ?? ?? ,
So it is symmetric.
For transitive,
If A = {(1, 2), (2, 3)}
B = {(2, 3), (3, 4)}
C = {(3, 4), (5, 6)}
ARB & BRC but A does not relate to C
So it not transitive
11. If S = {z ?? C : |z – i| = |z + i| = |z–1|}, then, n(S) is:
(1) 1 (2) 0
(3) 3 (4) 2
Ans. (1)
Sol. |z – i| = |z + i| = |z – 1|
ABC is a triangle. Hence its circum-centre will be
the only point whose distance from A, B, C will be
same.
So n(S) = 1
12. Four distinct points (2k, 3k), (1, 0), (0, 1) and
(0, 0) lie on a circle for k equal to :
(1)
2
13
(2)
3
13
(3)
5
13
(4)
1
13
Ans. (3)
Sol. (2k, 3k) will lie on circle whose diameter is AB.
(x – 1) (x) + (y – 1) (y) = 0
x
2
+ y
2
– x – y = 0 …(i)
Satisfy (2k, 3k) in (i)
(2k)
2
+ (3k)
2
– 2k – 3k = 0
13k
2
– 5k = 0
k = 0,
5
k
13
?]
hence
5
k
13
?]
13. Consider the function.
?H ?I ?H ?I 2
2
sin x 3
x [x]
a 7x 12 x
, x 3
b x 7x 12
f (x) 2 , x 3
b , x 3
?M ?M ?? ?M?M
?? ?\ ?M?K
?? ?? ???]?^
?? ?? ?] ?? ?? ?? ??
Where [x] denotes the greatest integer less than or
equal to x. If S denotes the set of all ordered pairs
(a, b) such that f(x) is continuous at x = 3, then the
number of elements in S is :
(1) 2 (2) Infinitely many
(3) 4 (4) 1
Ans. (4)
Sol.
?H ?I ?H ?I 2
2
7x 12 x
a
f 3
b x 7x 12
?M ?M?M
?] ?M?K
(for f(x) to be cont.)
?? f(3
–
) =
?H ?I ?H ?I ?H ?I ?H ?I x 3 x 4
a
;x 3
b x 3 x 4
?M?M
?M ?\ ?M?M
??
a
b
?M
Hence
?H ?I a
f 3
b
?M ?M ?]
Then
?H ?I ?H ?I x 3
sin x 3
lim
x 3
f 3 2 2
?K ?? ?M ????
????
????
?M ?K ????
?]?] and
f(3) = b.
Hence f(3) = f(3
+
) = f(3
–
)
?? b = 2 =
a
b
?M
b = 2, a = –4
Hence only 1 ordered pair (–4, 2).
14. Let a
1
, a
2
, ….. a
10
be 10 observations such that
10
k
k 1
a 50
?] ?] ?? and
k j
k j
a a 1100
?B?\
???]
?? . Then the
standard deviation of a
1
, a
2
, .., a
10
is equal to :
(1) 5 (2) 5
(3) 10 (4) 115
Ans. (2)
Sol.
10
k
k 1
a 50
?] ?] ??
a
1
+ a
2
+ … + a
10
= 50 ….(i)
k j
k j
a a 1100
?B?\
?] ?? ....(ii)
If a
1
+ a
2
+ … + a
10
= 50.
(a
1
+ a
2
+ … + a
10
)
2
= 2500
?? 10
2
i k j
i 1 k j
a 2 a a 2500
?]?\
?K?]
????
?? ?H ?I 10
2
i
i 1
a 2500 2 1100
?] ?]?M
??
10
2
i
i 1
a 300
?] ?] ?? , Standard deviation ‘ ?? ’
2
2
2
i i
a a
300 50
10 10 10 10
????
????
????
???? ?] ?M ?] ?M ????
???? ????
????
????
????
30 25 5 ?] ?M ?]
15. The length of the chord of the ellipse
2 2
x y
1
25 16
?K?] ,
whose mid point is
2
1,
5
????
????
????
, is equal to :
(1)
1691
5
(2)
2009
5
(3)
1741
5
(4)
1541
5
Ans. (1)
Sol. Equation of chord with given middle point.
T = S
1
x y 1 1
25 40 25 100
?K ?] ?K
8x 5y 8 2
200 200
?K?K
?]
10 8x
y
5
?M ?] …(i)
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students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of JEE Main 2024 January 27 Shift 1 Paper & Solutions is prepared as per the latest JEE syllabus.
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