JEE Main 2024 January 27 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


  
        
 
  
 
 
 
 
 
 
 
 
SECTION-A 
1. 
? ?
n 1 2 n
r r 1
C k 8 C
?
?
?? 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 
 
r0
r 1 0, r 0
?
? ? ?
 
 
n1
2 r
n
r1
C
k8
C
?
?
?? 
 
2
r1
k8
n
?
?? 
 ? 
2
k 8 0 ?? 
 
? ? ? ?
k 2 2 k 2 2 0 ? ? ? 
 
? ? ? ?
k , 2 2 2 2, ? ? ? ? ? ?    …(I) 
 ? ? n r 1 ?? , 
r1
1
n
?
? 
? k
2
 – 8 ??1 
  k
2
 – 9 ??0 
  –3 ??k ? 3  ….(II) 
 From equation (I) and (II) we get ?
?
? ?
k 3, 2 2 2 2, 3
??
? ? ? ?
??
 
2. The distance, of the point (7, –2, 11) from the line 
x 6 y 4 z 8
1 0 3
? ? ?
?? along the line 
x 5 y 1 z 5
2 3 6
? ? ?
??
?
, is : 
 (1) 12 (2) 14 
 (3) 18  (4) 21 
 Ans. (2) 
Sol. B = (2 ? +7, ?3 ? – 2, 6 ? + 11) 
 
 Point B lies on  
x 6 y 4 z 8
1 0 3
? ? ?
?? 
  
2 7 6 3 2 4 6 11 8
1 0 3
? ? ? ? ? ? ? ? ? ?
?? 
 ?3 ? – 6 = 0 
 ? = –2 
 B ? (3, 4, –1) 
 
? ? ? ? ? ?
2 2 2
AB 7 3 4 2 11 1 ? ? ? ? ? ? 
  16 36 144 ? ? ? 
  196 14 ?? 
3. Let x = x(t) and y = y(t) be solutions of the 
differential equations 
dx
ax 0
dt
?? and 
dy
by 0
dt
?? 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


  
        
 
  
 
 
 
 
 
 
 
 
SECTION-A 
1. 
? ?
n 1 2 n
r r 1
C k 8 C
?
?
?? 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 
 
r0
r 1 0, r 0
?
? ? ?
 
 
n1
2 r
n
r1
C
k8
C
?
?
?? 
 
2
r1
k8
n
?
?? 
 ? 
2
k 8 0 ?? 
 
? ? ? ?
k 2 2 k 2 2 0 ? ? ? 
 
? ? ? ?
k , 2 2 2 2, ? ? ? ? ? ?    …(I) 
 ? ? n r 1 ?? , 
r1
1
n
?
? 
? k
2
 – 8 ??1 
  k
2
 – 9 ??0 
  –3 ??k ? 3  ….(II) 
 From equation (I) and (II) we get ?
?
? ?
k 3, 2 2 2 2, 3
??
? ? ? ?
??
 
2. The distance, of the point (7, –2, 11) from the line 
x 6 y 4 z 8
1 0 3
? ? ?
?? along the line 
x 5 y 1 z 5
2 3 6
? ? ?
??
?
, is : 
 (1) 12 (2) 14 
 (3) 18  (4) 21 
 Ans. (2) 
Sol. B = (2 ? +7, ?3 ? – 2, 6 ? + 11) 
 
 Point B lies on  
x 6 y 4 z 8
1 0 3
? ? ?
?? 
  
2 7 6 3 2 4 6 11 8
1 0 3
? ? ? ? ? ? ? ? ? ?
?? 
 ?3 ? – 6 = 0 
 ? = –2 
 B ? (3, 4, –1) 
 
? ? ? ? ? ?
2 2 2
AB 7 3 4 2 11 1 ? ? ? ? ? ? 
  16 36 144 ? ? ? 
  196 14 ?? 
3. Let x = x(t) and y = y(t) be solutions of the 
differential equations 
dx
ax 0
dt
?? and 
dy
by 0
dt
?? 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
?? 
 
dx
adt
x
?? 
 
dx
a dt
x
??
??
 
 ln | x | at c ? ? ? 
 at t = 0, x = 2 
 ln 2 0 c ?? 
 ln x at ln 2 ? ? ? 
 
at
x
e
2
?
? 
 
at
x 2e
?
?  ….(i) 
 
dy
by 0
dt
?? 
 
dy
bdt
y
?? 
 ln | y | bt ? ? ? ? 
 t = 0, y = 1 
 0 = 0 + ? 
 y = e
–bt
  ….(ii) 
 According to question 
  3y(1) = 2x(1) 
  3e
–b
 = 2(2 e
–a
) 
  
ab
4
e
3
?
?
 
 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 
? ?
b
2
1
a
I x sin 4x x dx ??
?
, 
? ?
b
2
2
a
I sin 4x x dx ??
?
, 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 = ?(x – 3) 
 x + y = 4 
 
 orthocentre lies on the line x + y = 4 
 so, a + b = 4 
 
? ?
b
1
a
I x sin x(4 x) dx ??
?
 …(i) 
 Using king rule 
 ? ? ? ?
b
1
a
I 4 x sin x(4 x) dx ? ? ?
?
 …(ii) 
 (i) + (ii) 
 ? ?
b
1
a
2I 4sin x(4 x) dx ??
?
 
 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


  
        
 
  
 
 
 
 
 
 
 
 
SECTION-A 
1. 
? ?
n 1 2 n
r r 1
C k 8 C
?
?
?? 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 
 
r0
r 1 0, r 0
?
? ? ?
 
 
n1
2 r
n
r1
C
k8
C
?
?
?? 
 
2
r1
k8
n
?
?? 
 ? 
2
k 8 0 ?? 
 
? ? ? ?
k 2 2 k 2 2 0 ? ? ? 
 
? ? ? ?
k , 2 2 2 2, ? ? ? ? ? ?    …(I) 
 ? ? n r 1 ?? , 
r1
1
n
?
? 
? k
2
 – 8 ??1 
  k
2
 – 9 ??0 
  –3 ??k ? 3  ….(II) 
 From equation (I) and (II) we get ?
?
? ?
k 3, 2 2 2 2, 3
??
? ? ? ?
??
 
2. The distance, of the point (7, –2, 11) from the line 
x 6 y 4 z 8
1 0 3
? ? ?
?? along the line 
x 5 y 1 z 5
2 3 6
? ? ?
??
?
, is : 
 (1) 12 (2) 14 
 (3) 18  (4) 21 
 Ans. (2) 
Sol. B = (2 ? +7, ?3 ? – 2, 6 ? + 11) 
 
 Point B lies on  
x 6 y 4 z 8
1 0 3
? ? ?
?? 
  
2 7 6 3 2 4 6 11 8
1 0 3
? ? ? ? ? ? ? ? ? ?
?? 
 ?3 ? – 6 = 0 
 ? = –2 
 B ? (3, 4, –1) 
 
? ? ? ? ? ?
2 2 2
AB 7 3 4 2 11 1 ? ? ? ? ? ? 
  16 36 144 ? ? ? 
  196 14 ?? 
3. Let x = x(t) and y = y(t) be solutions of the 
differential equations 
dx
ax 0
dt
?? and 
dy
by 0
dt
?? 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
?? 
 
dx
adt
x
?? 
 
dx
a dt
x
??
??
 
 ln | x | at c ? ? ? 
 at t = 0, x = 2 
 ln 2 0 c ?? 
 ln x at ln 2 ? ? ? 
 
at
x
e
2
?
? 
 
at
x 2e
?
?  ….(i) 
 
dy
by 0
dt
?? 
 
dy
bdt
y
?? 
 ln | y | bt ? ? ? ? 
 t = 0, y = 1 
 0 = 0 + ? 
 y = e
–bt
  ….(ii) 
 According to question 
  3y(1) = 2x(1) 
  3e
–b
 = 2(2 e
–a
) 
  
ab
4
e
3
?
?
 
 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 
? ?
b
2
1
a
I x sin 4x x dx ??
?
, 
? ?
b
2
2
a
I sin 4x x dx ??
?
, 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 = ?(x – 3) 
 x + y = 4 
 
 orthocentre lies on the line x + y = 4 
 so, a + b = 4 
 
? ?
b
1
a
I x sin x(4 x) dx ??
?
 …(i) 
 Using king rule 
 ? ? ? ?
b
1
a
I 4 x sin x(4 x) dx ? ? ?
?
 …(ii) 
 (i) + (ii) 
 ? ?
b
1
a
2I 4sin x(4 x) dx ??
?
 
 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 ?? 
 d
2
 = 20 
8. If the shortest distance between the lines 
x 4 y 1 z
1 2 3
??
??
?
 and 
x y 1 z 2
2 4 5
? ? ? ?
??
?
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
??
??
?
 
 
x y 1 z 2
2 4 5
? ? ? ?
??
?
 
 
the shortest distance between the lines 
 
? ? ? ?
12
12
a b d d
dd
? ? ?
?
?
 
 
4 0 2
1 2 3
2 4 5
ˆ
i j k
1 2 3
2 4 5
??
?
?
?
?
?
 
 
? ? ? ? ? ? 4 10 12 0 2 4 4
ˆ
2i 1j 0k
? ? ? ? ? ? ?
?
??
 
 
? ? 24
6
55
??
? 
 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
? ? ?
? ? ?
?
, 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 ? ? ?
?
? ? ? ?
1
0
3 x 1 x
dx
3 x 1 x
? ? ?
?
? ? ?
?
 
 
? ?
11
00
1
3 x dx 1 x dx
2
??
? ? ?
??
??
??
 
Page 4


  
        
 
  
 
 
 
 
 
 
 
 
SECTION-A 
1. 
? ?
n 1 2 n
r r 1
C k 8 C
?
?
?? 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 
 
r0
r 1 0, r 0
?
? ? ?
 
 
n1
2 r
n
r1
C
k8
C
?
?
?? 
 
2
r1
k8
n
?
?? 
 ? 
2
k 8 0 ?? 
 
? ? ? ?
k 2 2 k 2 2 0 ? ? ? 
 
? ? ? ?
k , 2 2 2 2, ? ? ? ? ? ?    …(I) 
 ? ? n r 1 ?? , 
r1
1
n
?
? 
? k
2
 – 8 ??1 
  k
2
 – 9 ??0 
  –3 ??k ? 3  ….(II) 
 From equation (I) and (II) we get ?
?
? ?
k 3, 2 2 2 2, 3
??
? ? ? ?
??
 
2. The distance, of the point (7, –2, 11) from the line 
x 6 y 4 z 8
1 0 3
? ? ?
?? along the line 
x 5 y 1 z 5
2 3 6
? ? ?
??
?
, is : 
 (1) 12 (2) 14 
 (3) 18  (4) 21 
 Ans. (2) 
Sol. B = (2 ? +7, ?3 ? – 2, 6 ? + 11) 
 
 Point B lies on  
x 6 y 4 z 8
1 0 3
? ? ?
?? 
  
2 7 6 3 2 4 6 11 8
1 0 3
? ? ? ? ? ? ? ? ? ?
?? 
 ?3 ? – 6 = 0 
 ? = –2 
 B ? (3, 4, –1) 
 
? ? ? ? ? ?
2 2 2
AB 7 3 4 2 11 1 ? ? ? ? ? ? 
  16 36 144 ? ? ? 
  196 14 ?? 
3. Let x = x(t) and y = y(t) be solutions of the 
differential equations 
dx
ax 0
dt
?? and 
dy
by 0
dt
?? 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
?? 
 
dx
adt
x
?? 
 
dx
a dt
x
??
??
 
 ln | x | at c ? ? ? 
 at t = 0, x = 2 
 ln 2 0 c ?? 
 ln x at ln 2 ? ? ? 
 
at
x
e
2
?
? 
 
at
x 2e
?
?  ….(i) 
 
dy
by 0
dt
?? 
 
dy
bdt
y
?? 
 ln | y | bt ? ? ? ? 
 t = 0, y = 1 
 0 = 0 + ? 
 y = e
–bt
  ….(ii) 
 According to question 
  3y(1) = 2x(1) 
  3e
–b
 = 2(2 e
–a
) 
  
ab
4
e
3
?
?
 
 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 
? ?
b
2
1
a
I x sin 4x x dx ??
?
, 
? ?
b
2
2
a
I sin 4x x dx ??
?
, 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 = ?(x – 3) 
 x + y = 4 
 
 orthocentre lies on the line x + y = 4 
 so, a + b = 4 
 
? ?
b
1
a
I x sin x(4 x) dx ??
?
 …(i) 
 Using king rule 
 ? ? ? ?
b
1
a
I 4 x sin x(4 x) dx ? ? ?
?
 …(ii) 
 (i) + (ii) 
 ? ?
b
1
a
2I 4sin x(4 x) dx ??
?
 
 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 ?? 
 d
2
 = 20 
8. If the shortest distance between the lines 
x 4 y 1 z
1 2 3
??
??
?
 and 
x y 1 z 2
2 4 5
? ? ? ?
??
?
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
??
??
?
 
 
x y 1 z 2
2 4 5
? ? ? ?
??
?
 
 
the shortest distance between the lines 
 
? ? ? ?
12
12
a b d d
dd
? ? ?
?
?
 
 
4 0 2
1 2 3
2 4 5
ˆ
i j k
1 2 3
2 4 5
??
?
?
?
?
?
 
 
? ? ? ? ? ? 4 10 12 0 2 4 4
ˆ
2i 1j 0k
? ? ? ? ? ? ?
?
??
 
 
? ? 24
6
55
??
? 
 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
? ? ?
? ? ?
?
, 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 ? ? ?
?
? ? ? ?
1
0
3 x 1 x
dx
3 x 1 x
? ? ?
?
? ? ?
?
 
 
? ?
11
00
1
3 x dx 1 x dx
2
??
? ? ?
??
??
??
 
 
 
 
 
 
? ? ? ?
1
33
22
0
3 x 2 1 x
1
2
2 3 3
??
??
??
?
??
??
??
 
 
? ?
3
2
1 2 2
8 3 3 2 1
2 3 3
?? ??
? ? ?
?? ??
??
?? ??
 
 
1
8 3 3 2 2 1
3
??
? ? ?
??
 
 
2
3 3 2 a b 2 c 3
3
? ? ? ? ? ? 
 
2
a 3, b , c 1
3
? ? ? ? ? 
 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


  
        
 
  
 
 
 
 
 
 
 
 
SECTION-A 
1. 
? ?
n 1 2 n
r r 1
C k 8 C
?
?
?? 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 
 
r0
r 1 0, r 0
?
? ? ?
 
 
n1
2 r
n
r1
C
k8
C
?
?
?? 
 
2
r1
k8
n
?
?? 
 ? 
2
k 8 0 ?? 
 
? ? ? ?
k 2 2 k 2 2 0 ? ? ? 
 
? ? ? ?
k , 2 2 2 2, ? ? ? ? ? ?    …(I) 
 ? ? n r 1 ?? , 
r1
1
n
?
? 
? k
2
 – 8 ??1 
  k
2
 – 9 ??0 
  –3 ??k ? 3  ….(II) 
 From equation (I) and (II) we get ?
?
? ?
k 3, 2 2 2 2, 3
??
? ? ? ?
??
 
2. The distance, of the point (7, –2, 11) from the line 
x 6 y 4 z 8
1 0 3
? ? ?
?? along the line 
x 5 y 1 z 5
2 3 6
? ? ?
??
?
, is : 
 (1) 12 (2) 14 
 (3) 18  (4) 21 
 Ans. (2) 
Sol. B = (2 ? +7, ?3 ? – 2, 6 ? + 11) 
 
 Point B lies on  
x 6 y 4 z 8
1 0 3
? ? ?
?? 
  
2 7 6 3 2 4 6 11 8
1 0 3
? ? ? ? ? ? ? ? ? ?
?? 
 ?3 ? – 6 = 0 
 ? = –2 
 B ? (3, 4, –1) 
 
? ? ? ? ? ?
2 2 2
AB 7 3 4 2 11 1 ? ? ? ? ? ? 
  16 36 144 ? ? ? 
  196 14 ?? 
3. Let x = x(t) and y = y(t) be solutions of the 
differential equations 
dx
ax 0
dt
?? and 
dy
by 0
dt
?? 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
?? 
 
dx
adt
x
?? 
 
dx
a dt
x
??
??
 
 ln | x | at c ? ? ? 
 at t = 0, x = 2 
 ln 2 0 c ?? 
 ln x at ln 2 ? ? ? 
 
at
x
e
2
?
? 
 
at
x 2e
?
?  ….(i) 
 
dy
by 0
dt
?? 
 
dy
bdt
y
?? 
 ln | y | bt ? ? ? ? 
 t = 0, y = 1 
 0 = 0 + ? 
 y = e
–bt
  ….(ii) 
 According to question 
  3y(1) = 2x(1) 
  3e
–b
 = 2(2 e
–a
) 
  
ab
4
e
3
?
?
 
 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 
? ?
b
2
1
a
I x sin 4x x dx ??
?
, 
? ?
b
2
2
a
I sin 4x x dx ??
?
, 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 = ?(x – 3) 
 x + y = 4 
 
 orthocentre lies on the line x + y = 4 
 so, a + b = 4 
 
? ?
b
1
a
I x sin x(4 x) dx ??
?
 …(i) 
 Using king rule 
 ? ? ? ?
b
1
a
I 4 x sin x(4 x) dx ? ? ?
?
 …(ii) 
 (i) + (ii) 
 ? ?
b
1
a
2I 4sin x(4 x) dx ??
?
 
 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 ?? 
 d
2
 = 20 
8. If the shortest distance between the lines 
x 4 y 1 z
1 2 3
??
??
?
 and 
x y 1 z 2
2 4 5
? ? ? ?
??
?
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
??
??
?
 
 
x y 1 z 2
2 4 5
? ? ? ?
??
?
 
 
the shortest distance between the lines 
 
? ? ? ?
12
12
a b d d
dd
? ? ?
?
?
 
 
4 0 2
1 2 3
2 4 5
ˆ
i j k
1 2 3
2 4 5
??
?
?
?
?
?
 
 
? ? ? ? ? ? 4 10 12 0 2 4 4
ˆ
2i 1j 0k
? ? ? ? ? ? ?
?
??
 
 
? ? 24
6
55
??
? 
 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
? ? ?
? ? ?
?
, 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 ? ? ?
?
? ? ? ?
1
0
3 x 1 x
dx
3 x 1 x
? ? ?
?
? ? ?
?
 
 
? ?
11
00
1
3 x dx 1 x dx
2
??
? ? ?
??
??
??
 
 
 
 
 
 
? ? ? ?
1
33
22
0
3 x 2 1 x
1
2
2 3 3
??
??
??
?
??
??
??
 
 
? ?
3
2
1 2 2
8 3 3 2 1
2 3 3
?? ??
? ? ?
?? ??
??
?? ??
 
 
1
8 3 3 2 2 1
3
??
? ? ?
??
 
 
2
3 3 2 a b 2 c 3
3
? ? ? ? ? ? 
 
2
a 3, b , c 1
3
? ? ? ? ? 
 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. 
 
? ?
? ?
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
?
?
?
??
? ?
??
?
?
?
??
?
?
?
?
?
?
?
  
 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. 
? ?
? ?
2
2
7x 12 x
a
f3
b x 7x 12
?
??
?
??
  (for f(x) to be cont.) 
 ? f(3
–
) = 
? ? ? ?
? ? ? ?
x 3 x 4
a
;x 3
b x 3 x 4
??
?
?
??
 ?  
a
b
?
 
 Hence 
? ?
a
f3
b
?
?
? 
 Then 
? ?
? ?
x3
sin x 3
lim
x3
f 3 2 2
?
?
? ??
??
??
? ?
??
?? and 
 f(3) = b. 
 Hence f(3) = f(3
+
) = f(3
–
) 
 ? b = 2 = 
a
b
? 
 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
k1
a 50
?
?
?
 and 
kj
kj
a a 1100
??
??
?
. 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
k1
a 50
?
?
?
 
 a
1
 + a
2
 + … + a
10
 = 50 ….(i) 
 
kj
kj
a a 1100
??
?
?
 ....(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
??
??
??
 
 ? ? ?
10
2
i
i1
a 2500 2 1100
?
??
?
 
  
10
2
i
i1
a 300
?
?
?
, Standard deviation ‘ ?’ 
 
2
2
2
ii
aa
300 50
10 10 10 10
??
??
??
?? ? ? ? ?
??
?? ??
??
??
??
30 25 5 ? ? ? 
15. The length of the chord of the ellipse 
22
xy
1
25 16
?? , 
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
? ? ? 
 
8x 5y 8 2
200 200
??
? 
 
10 8x
y
5
?
?  …(i)