JEE Main 2024 February 1 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


  
        
  
 
 
 
 
SECTION-A 
1. Let f(x) = |2x
2
+5|x|–3|,x ?R. If m and n denote the 
number of points where f is not continuous and not 
differentiable respectively, then m + n is equal to : 
 (1) 5 (2) 2 
 (3) 0  (4) 3 
Ans.  (4) 
Sol. f(x) = |2x
2
+5|x|–3| 
 Graph of y = |2x
2 
+ 5x – 3| 
 
 Graph of f(x) 
  
  
0 
 
 Number of points of discontinuity = 0 = m 
 Number of points of non-differentiability = 3 = n 
2. Let ? and ? be the roots of the equation px
2
 + qx – 
r = 0, where p ? 0. If p, q and r be the consecutive 
terms of a non-constant G.P and , then 
the value of ( ??– ??
? ?
is : ? 
 (1)  (2) 9 
 (3)   (4) 8 
Ans.  (1) 
Sol. px
2
 + qx – r = 0  
 p = A, q = AR, r = AR
2
 
 Ax
2
 + ARx – AR
2
 = 0 
 x
2
 + Rx – R
2
 = 0  
   
 ?? ?
 ( ? ?? ? ?)
2
 = ( ? ??)
2
 – 4 ?? ? = R
2
 – 4(–R
2
) = 5 
  = 80/9 
3. The number of solutions of the equation 4 sin
2
x – 4 
cos
3
x + 9 – 4 cosx = 0; x ? [ –2 ?, 2 ?] is : 
 (1) 1 
 (2) 3 
 (3) 2 
 (4) 0 
Ans.  (4) 
Sol. 4sin
2
x – 4cos
3
x + 9 – 4 cosx = 0 ; x ? [ – 2 ??, 2 ?] 
 4 – 4cos
2
x – 4cos
3
x + 9 – 4 cosx = 0 
 4cos
3
x + 4cos
2
x + 4 cosx – 13 = 0 
 4cos
3
x + 4cos
2
x + 4cosx = 13 
 L.H.S. ? 12 can’t be equal to 13. 
4. The value of dx is equal to: 
 (1) 0 
 (2) 1 
 (3) 2 
 (4) –1 
Ans.  (1) 
Sol.  
 Using  where f(2a–x) = –f(x) 
 Here f(1–x) = f(x) 
 ? I = 0 
–3
1/2 – 5 / 4
1 1 3
4
??
??
80
9
20
3
?
?
?
?
1 1 3
4
??
??
2
3 R 4
R
4 R 4 3
? ? ? ? ?
? ? ? ? ?
? ? ?
16
9
??
??
??
1
1 32
3
0
(2x 3x x 1) ? ? ?
?
? ?
1
1
3
32
0
I 2x 3x x 1 dx ? ? ? ?
?
? ?
?
2a
0
f x dx
Page 2


  
        
  
 
 
 
 
SECTION-A 
1. Let f(x) = |2x
2
+5|x|–3|,x ?R. If m and n denote the 
number of points where f is not continuous and not 
differentiable respectively, then m + n is equal to : 
 (1) 5 (2) 2 
 (3) 0  (4) 3 
Ans.  (4) 
Sol. f(x) = |2x
2
+5|x|–3| 
 Graph of y = |2x
2 
+ 5x – 3| 
 
 Graph of f(x) 
  
  
0 
 
 Number of points of discontinuity = 0 = m 
 Number of points of non-differentiability = 3 = n 
2. Let ? and ? be the roots of the equation px
2
 + qx – 
r = 0, where p ? 0. If p, q and r be the consecutive 
terms of a non-constant G.P and , then 
the value of ( ??– ??
? ?
is : ? 
 (1)  (2) 9 
 (3)   (4) 8 
Ans.  (1) 
Sol. px
2
 + qx – r = 0  
 p = A, q = AR, r = AR
2
 
 Ax
2
 + ARx – AR
2
 = 0 
 x
2
 + Rx – R
2
 = 0  
   
 ?? ?
 ( ? ?? ? ?)
2
 = ( ? ??)
2
 – 4 ?? ? = R
2
 – 4(–R
2
) = 5 
  = 80/9 
3. The number of solutions of the equation 4 sin
2
x – 4 
cos
3
x + 9 – 4 cosx = 0; x ? [ –2 ?, 2 ?] is : 
 (1) 1 
 (2) 3 
 (3) 2 
 (4) 0 
Ans.  (4) 
Sol. 4sin
2
x – 4cos
3
x + 9 – 4 cosx = 0 ; x ? [ – 2 ??, 2 ?] 
 4 – 4cos
2
x – 4cos
3
x + 9 – 4 cosx = 0 
 4cos
3
x + 4cos
2
x + 4 cosx – 13 = 0 
 4cos
3
x + 4cos
2
x + 4cosx = 13 
 L.H.S. ? 12 can’t be equal to 13. 
4. The value of dx is equal to: 
 (1) 0 
 (2) 1 
 (3) 2 
 (4) –1 
Ans.  (1) 
Sol.  
 Using  where f(2a–x) = –f(x) 
 Here f(1–x) = f(x) 
 ? I = 0 
–3
1/2 – 5 / 4
1 1 3
4
??
??
80
9
20
3
?
?
?
?
1 1 3
4
??
??
2
3 R 4
R
4 R 4 3
? ? ? ? ?
? ? ? ? ?
? ? ?
16
9
??
??
??
1
1 32
3
0
(2x 3x x 1) ? ? ?
?
? ?
1
1
3
32
0
I 2x 3x x 1 dx ? ? ? ?
?
? ?
?
2a
0
f x dx
 
 
5. Let P be a point on the ellipse . Let the 
line passing through P and parallel to y-axis meet 
the circle x
2
 + y
2
 = 9 at point Q such that P and Q 
are on the same side of the x-axis. Then, the 
eccentricity of the locus of the point R on PQ such 
that PR : RQ = 4 : 3 as P moves on the ellipse, is : 
 (1)  (2)  
 (3)   (4)  
Ans.  (4) 
Sol.  
  
 h = 3cos ?; 
  
 ? locus =  
  
6. Let m and n be the coefficients of seventh and 
thirteenth terms respectively in the expansion of  
. Then is : 
 (1)  (2)  
 (3)   (4)  
Ans.  (4) 
Sol.  
  
  
  :  
  
7. Let ? be a non-zero real number. Suppose f : R ?  
R is a differentiable function such that f (0) = 2 and 
? ?
x
lim f x 1
? ? ?
? . If f '(x) = ?f(x) +3, for all x ? R, 
then f (–log
e
2) is equal to____. 
 (1) 3 (2) 5 
 (3) 9  (4) 7 
Ans.  (3 OR BONUS) 
Sol. f(0) = 2, 
? ?
x
lim f x 1
? ? ?
? 
 f’(x) – x.f(x) = 3 
 I.F = e
– ?x 
 
y(e
– ?x
) =  
 f(x). (e
– ?x
) =  
 x = 0 ?  ?  (1) 
 f(x) =  
 x ? – ? ?? 1 = 
 ? = –3 ? c = 1 
  
 = 1 + e
3ln2 
= 9 
 (But ? should be greater than 0 for finite 
value of c) 
22
xy
1
94
??
11
19
13
21
139
23
13
7
Q
P(3cos , 2sin ) ? ?
Q(3cos , 3sin ) ? ?
P
2 2
x y
1
9 4
? ?
4
P
(3C, 2S)
R
(h, k)
Q
(3C, 3S)
3
18
k sin
7
??
22
x 49y
1
9 324
??
324 117 13
e1
49 9 21 7
? ? ? ?
?
18
1
3
2
3
11
x
3
2x
??
??
?
??
??
1
3
n
m
??
??
??
4
9
1
9
1
4
9
4
18
12
33
xx
3
?
??
??
?
??
??
??
? ?
12 6
12
33
18 18
7 6 6
12 6
x x 1 1
t c c .
32 2
3
?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
? ?
6 12
12
33
18 18 –6
13 12 12
6 12
x x 1 1
t c c . .x
32 2
3
?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
18 –12 –6
6
m c .3 .2 ?
18 –12 –6
12
n c .2 .3 ?
1
1
2
–12 –6
3
3
12 6
n 2 .3 3 9
m 2 4 3 .2
??
??
? ? ? ?
? ? ?
??
? ? ? ?
? ? ? ?
??
x
3.e dx
??
?
x
3e
c
??
?
??
3
2c
?
??
?
3
c2 ??
?
x
3
c.e
?
?
?
?
3
c(0)
?
?
?
x
3
f ( ln 2) c.e
?
?
? ? ?
?
Page 3


  
        
  
 
 
 
 
SECTION-A 
1. Let f(x) = |2x
2
+5|x|–3|,x ?R. If m and n denote the 
number of points where f is not continuous and not 
differentiable respectively, then m + n is equal to : 
 (1) 5 (2) 2 
 (3) 0  (4) 3 
Ans.  (4) 
Sol. f(x) = |2x
2
+5|x|–3| 
 Graph of y = |2x
2 
+ 5x – 3| 
 
 Graph of f(x) 
  
  
0 
 
 Number of points of discontinuity = 0 = m 
 Number of points of non-differentiability = 3 = n 
2. Let ? and ? be the roots of the equation px
2
 + qx – 
r = 0, where p ? 0. If p, q and r be the consecutive 
terms of a non-constant G.P and , then 
the value of ( ??– ??
? ?
is : ? 
 (1)  (2) 9 
 (3)   (4) 8 
Ans.  (1) 
Sol. px
2
 + qx – r = 0  
 p = A, q = AR, r = AR
2
 
 Ax
2
 + ARx – AR
2
 = 0 
 x
2
 + Rx – R
2
 = 0  
   
 ?? ?
 ( ? ?? ? ?)
2
 = ( ? ??)
2
 – 4 ?? ? = R
2
 – 4(–R
2
) = 5 
  = 80/9 
3. The number of solutions of the equation 4 sin
2
x – 4 
cos
3
x + 9 – 4 cosx = 0; x ? [ –2 ?, 2 ?] is : 
 (1) 1 
 (2) 3 
 (3) 2 
 (4) 0 
Ans.  (4) 
Sol. 4sin
2
x – 4cos
3
x + 9 – 4 cosx = 0 ; x ? [ – 2 ??, 2 ?] 
 4 – 4cos
2
x – 4cos
3
x + 9 – 4 cosx = 0 
 4cos
3
x + 4cos
2
x + 4 cosx – 13 = 0 
 4cos
3
x + 4cos
2
x + 4cosx = 13 
 L.H.S. ? 12 can’t be equal to 13. 
4. The value of dx is equal to: 
 (1) 0 
 (2) 1 
 (3) 2 
 (4) –1 
Ans.  (1) 
Sol.  
 Using  where f(2a–x) = –f(x) 
 Here f(1–x) = f(x) 
 ? I = 0 
–3
1/2 – 5 / 4
1 1 3
4
??
??
80
9
20
3
?
?
?
?
1 1 3
4
??
??
2
3 R 4
R
4 R 4 3
? ? ? ? ?
? ? ? ? ?
? ? ?
16
9
??
??
??
1
1 32
3
0
(2x 3x x 1) ? ? ?
?
? ?
1
1
3
32
0
I 2x 3x x 1 dx ? ? ? ?
?
? ?
?
2a
0
f x dx
 
 
5. Let P be a point on the ellipse . Let the 
line passing through P and parallel to y-axis meet 
the circle x
2
 + y
2
 = 9 at point Q such that P and Q 
are on the same side of the x-axis. Then, the 
eccentricity of the locus of the point R on PQ such 
that PR : RQ = 4 : 3 as P moves on the ellipse, is : 
 (1)  (2)  
 (3)   (4)  
Ans.  (4) 
Sol.  
  
 h = 3cos ?; 
  
 ? locus =  
  
6. Let m and n be the coefficients of seventh and 
thirteenth terms respectively in the expansion of  
. Then is : 
 (1)  (2)  
 (3)   (4)  
Ans.  (4) 
Sol.  
  
  
  :  
  
7. Let ? be a non-zero real number. Suppose f : R ?  
R is a differentiable function such that f (0) = 2 and 
? ?
x
lim f x 1
? ? ?
? . If f '(x) = ?f(x) +3, for all x ? R, 
then f (–log
e
2) is equal to____. 
 (1) 3 (2) 5 
 (3) 9  (4) 7 
Ans.  (3 OR BONUS) 
Sol. f(0) = 2, 
? ?
x
lim f x 1
? ? ?
? 
 f’(x) – x.f(x) = 3 
 I.F = e
– ?x 
 
y(e
– ?x
) =  
 f(x). (e
– ?x
) =  
 x = 0 ?  ?  (1) 
 f(x) =  
 x ? – ? ?? 1 = 
 ? = –3 ? c = 1 
  
 = 1 + e
3ln2 
= 9 
 (But ? should be greater than 0 for finite 
value of c) 
22
xy
1
94
??
11
19
13
21
139
23
13
7
Q
P(3cos , 2sin ) ? ?
Q(3cos , 3sin ) ? ?
P
2 2
x y
1
9 4
? ?
4
P
(3C, 2S)
R
(h, k)
Q
(3C, 3S)
3
18
k sin
7
??
22
x 49y
1
9 324
??
324 117 13
e1
49 9 21 7
? ? ? ?
?
18
1
3
2
3
11
x
3
2x
??
??
?
??
??
1
3
n
m
??
??
??
4
9
1
9
1
4
9
4
18
12
33
xx
3
?
??
??
?
??
??
??
? ?
12 6
12
33
18 18
7 6 6
12 6
x x 1 1
t c c .
32 2
3
?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
? ?
6 12
12
33
18 18 –6
13 12 12
6 12
x x 1 1
t c c . .x
32 2
3
?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
18 –12 –6
6
m c .3 .2 ?
18 –12 –6
12
n c .2 .3 ?
1
1
2
–12 –6
3
3
12 6
n 2 .3 3 9
m 2 4 3 .2
??
??
? ? ? ?
? ? ?
??
? ? ? ?
? ? ? ?
??
x
3.e dx
??
?
x
3e
c
??
?
??
3
2c
?
??
?
3
c2 ??
?
x
3
c.e
?
?
?
?
3
c(0)
?
?
?
x
3
f ( ln 2) c.e
?
?
? ? ?
?
 
 
 
 
8. Let P and Q be the points on the line 
 which are at a distance of 6 
units from the point R (1,2,3). If the centroid of the 
triangle PQR is ( ?, ? ?, ?), then ?
2
+ ? ?
2 
+ ??
2 
is: 
 (1) 26 
 (2) 36 
 (3) 18 
 (4) 24 
Ans.  (3) 
Sol.  
  
 P(8 ?? – 3, 2 ? + 4, 2 ? – 1) 
 PR = 6 
 (8 ?? – 4)
2 
+ ( 2 ? + 2)
2
 + (2 ? – 4)
2 
= 36 
 ? = 0, 1 
 Hence  P(–3, 4, –1) & Q(5, 6, 1) 
 Centroid of ?PQR = (1, 4, 1) ? ( ?, ? ?, ?)  
 ?
??
+ ?
?
 + ?
?
 = 18 
9. Consider a ?ABC where A(1,2,3,), B(–2,8,0) and 
C(3,6,7). If the angle bisector of ?BAC meets the 
line BC at D, then the length of the projection of 
the vector on the vector is: 
 (1)  
 (2)  
 (3)  
 (4)  
Ans.  (1) 
Sol.  
 A(1, 3, 2); B(–2, 8, 0); C(3, 6, 7); 
  
 AB =  
 AC =  
  
 Length of projection of  on  
 =   
10. Let S
n
 denote the sum of the first n terms of an 
arithmetic progression. If S
10
 = 390 and the ratio of 
the tenth and the fifth terms is 15 : 7, then S
15
 –S
5
 
is equal to: 
 (1) 800 
 (2) 890 
 (3) 790 
 (4) 690 
Ans.  (3) 
Sol. S
10
 = 390 
  
 ? 2a + 9d = 78  (1) 
  (2) 
 From (1) & (2) a = 3 & d = 8 
  
 =790 
x 3 y 4 z 1
8 2 2
? ? ?
??
P Q
R(1,2,3)
AD
?
AC
?
37
2 38
38
2
39
2 38
19
A (1,3,2)
C (3, 6, 7) B
(–2, 8, 0)
D
1 :   1
1 7
, 7,
2 2
? ?
? ?
? ?
ˆ ˆ ˆ
AC 2i 3j 5k ? ? ?
9 25 4 38 ? ? ?
4 9 25 38 ? ? ?
1 3 1
ˆ ˆ ˆ ˆ ˆ ˆ
AD i 4j k (i 8j 3k)
2 2 2
? ? ? ? ? ?
AD AC
AD.AC 37
| AC | 2 38
?
? ?
10
2a 10 1 d 390
2
?? ? ? ?
??
10
5
t 15 a 9d 15
8a 3d
t 7 a 4d 7
?
? ? ? ? ?
?
? ? ? ?
15 5
15 5
S – S 6 14 8 6 4 8
22
? ? ? ? ? ?
15 118 5 38
2
? ? ?
?
Page 4


  
        
  
 
 
 
 
SECTION-A 
1. Let f(x) = |2x
2
+5|x|–3|,x ?R. If m and n denote the 
number of points where f is not continuous and not 
differentiable respectively, then m + n is equal to : 
 (1) 5 (2) 2 
 (3) 0  (4) 3 
Ans.  (4) 
Sol. f(x) = |2x
2
+5|x|–3| 
 Graph of y = |2x
2 
+ 5x – 3| 
 
 Graph of f(x) 
  
  
0 
 
 Number of points of discontinuity = 0 = m 
 Number of points of non-differentiability = 3 = n 
2. Let ? and ? be the roots of the equation px
2
 + qx – 
r = 0, where p ? 0. If p, q and r be the consecutive 
terms of a non-constant G.P and , then 
the value of ( ??– ??
? ?
is : ? 
 (1)  (2) 9 
 (3)   (4) 8 
Ans.  (1) 
Sol. px
2
 + qx – r = 0  
 p = A, q = AR, r = AR
2
 
 Ax
2
 + ARx – AR
2
 = 0 
 x
2
 + Rx – R
2
 = 0  
   
 ?? ?
 ( ? ?? ? ?)
2
 = ( ? ??)
2
 – 4 ?? ? = R
2
 – 4(–R
2
) = 5 
  = 80/9 
3. The number of solutions of the equation 4 sin
2
x – 4 
cos
3
x + 9 – 4 cosx = 0; x ? [ –2 ?, 2 ?] is : 
 (1) 1 
 (2) 3 
 (3) 2 
 (4) 0 
Ans.  (4) 
Sol. 4sin
2
x – 4cos
3
x + 9 – 4 cosx = 0 ; x ? [ – 2 ??, 2 ?] 
 4 – 4cos
2
x – 4cos
3
x + 9 – 4 cosx = 0 
 4cos
3
x + 4cos
2
x + 4 cosx – 13 = 0 
 4cos
3
x + 4cos
2
x + 4cosx = 13 
 L.H.S. ? 12 can’t be equal to 13. 
4. The value of dx is equal to: 
 (1) 0 
 (2) 1 
 (3) 2 
 (4) –1 
Ans.  (1) 
Sol.  
 Using  where f(2a–x) = –f(x) 
 Here f(1–x) = f(x) 
 ? I = 0 
–3
1/2 – 5 / 4
1 1 3
4
??
??
80
9
20
3
?
?
?
?
1 1 3
4
??
??
2
3 R 4
R
4 R 4 3
? ? ? ? ?
? ? ? ? ?
? ? ?
16
9
??
??
??
1
1 32
3
0
(2x 3x x 1) ? ? ?
?
? ?
1
1
3
32
0
I 2x 3x x 1 dx ? ? ? ?
?
? ?
?
2a
0
f x dx
 
 
5. Let P be a point on the ellipse . Let the 
line passing through P and parallel to y-axis meet 
the circle x
2
 + y
2
 = 9 at point Q such that P and Q 
are on the same side of the x-axis. Then, the 
eccentricity of the locus of the point R on PQ such 
that PR : RQ = 4 : 3 as P moves on the ellipse, is : 
 (1)  (2)  
 (3)   (4)  
Ans.  (4) 
Sol.  
  
 h = 3cos ?; 
  
 ? locus =  
  
6. Let m and n be the coefficients of seventh and 
thirteenth terms respectively in the expansion of  
. Then is : 
 (1)  (2)  
 (3)   (4)  
Ans.  (4) 
Sol.  
  
  
  :  
  
7. Let ? be a non-zero real number. Suppose f : R ?  
R is a differentiable function such that f (0) = 2 and 
? ?
x
lim f x 1
? ? ?
? . If f '(x) = ?f(x) +3, for all x ? R, 
then f (–log
e
2) is equal to____. 
 (1) 3 (2) 5 
 (3) 9  (4) 7 
Ans.  (3 OR BONUS) 
Sol. f(0) = 2, 
? ?
x
lim f x 1
? ? ?
? 
 f’(x) – x.f(x) = 3 
 I.F = e
– ?x 
 
y(e
– ?x
) =  
 f(x). (e
– ?x
) =  
 x = 0 ?  ?  (1) 
 f(x) =  
 x ? – ? ?? 1 = 
 ? = –3 ? c = 1 
  
 = 1 + e
3ln2 
= 9 
 (But ? should be greater than 0 for finite 
value of c) 
22
xy
1
94
??
11
19
13
21
139
23
13
7
Q
P(3cos , 2sin ) ? ?
Q(3cos , 3sin ) ? ?
P
2 2
x y
1
9 4
? ?
4
P
(3C, 2S)
R
(h, k)
Q
(3C, 3S)
3
18
k sin
7
??
22
x 49y
1
9 324
??
324 117 13
e1
49 9 21 7
? ? ? ?
?
18
1
3
2
3
11
x
3
2x
??
??
?
??
??
1
3
n
m
??
??
??
4
9
1
9
1
4
9
4
18
12
33
xx
3
?
??
??
?
??
??
??
? ?
12 6
12
33
18 18
7 6 6
12 6
x x 1 1
t c c .
32 2
3
?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
? ?
6 12
12
33
18 18 –6
13 12 12
6 12
x x 1 1
t c c . .x
32 2
3
?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
18 –12 –6
6
m c .3 .2 ?
18 –12 –6
12
n c .2 .3 ?
1
1
2
–12 –6
3
3
12 6
n 2 .3 3 9
m 2 4 3 .2
??
??
? ? ? ?
? ? ?
??
? ? ? ?
? ? ? ?
??
x
3.e dx
??
?
x
3e
c
??
?
??
3
2c
?
??
?
3
c2 ??
?
x
3
c.e
?
?
?
?
3
c(0)
?
?
?
x
3
f ( ln 2) c.e
?
?
? ? ?
?
 
 
 
 
8. Let P and Q be the points on the line 
 which are at a distance of 6 
units from the point R (1,2,3). If the centroid of the 
triangle PQR is ( ?, ? ?, ?), then ?
2
+ ? ?
2 
+ ??
2 
is: 
 (1) 26 
 (2) 36 
 (3) 18 
 (4) 24 
Ans.  (3) 
Sol.  
  
 P(8 ?? – 3, 2 ? + 4, 2 ? – 1) 
 PR = 6 
 (8 ?? – 4)
2 
+ ( 2 ? + 2)
2
 + (2 ? – 4)
2 
= 36 
 ? = 0, 1 
 Hence  P(–3, 4, –1) & Q(5, 6, 1) 
 Centroid of ?PQR = (1, 4, 1) ? ( ?, ? ?, ?)  
 ?
??
+ ?
?
 + ?
?
 = 18 
9. Consider a ?ABC where A(1,2,3,), B(–2,8,0) and 
C(3,6,7). If the angle bisector of ?BAC meets the 
line BC at D, then the length of the projection of 
the vector on the vector is: 
 (1)  
 (2)  
 (3)  
 (4)  
Ans.  (1) 
Sol.  
 A(1, 3, 2); B(–2, 8, 0); C(3, 6, 7); 
  
 AB =  
 AC =  
  
 Length of projection of  on  
 =   
10. Let S
n
 denote the sum of the first n terms of an 
arithmetic progression. If S
10
 = 390 and the ratio of 
the tenth and the fifth terms is 15 : 7, then S
15
 –S
5
 
is equal to: 
 (1) 800 
 (2) 890 
 (3) 790 
 (4) 690 
Ans.  (3) 
Sol. S
10
 = 390 
  
 ? 2a + 9d = 78  (1) 
  (2) 
 From (1) & (2) a = 3 & d = 8 
  
 =790 
x 3 y 4 z 1
8 2 2
? ? ?
??
P Q
R(1,2,3)
AD
?
AC
?
37
2 38
38
2
39
2 38
19
A (1,3,2)
C (3, 6, 7) B
(–2, 8, 0)
D
1 :   1
1 7
, 7,
2 2
? ?
? ?
? ?
ˆ ˆ ˆ
AC 2i 3j 5k ? ? ?
9 25 4 38 ? ? ?
4 9 25 38 ? ? ?
1 3 1
ˆ ˆ ˆ ˆ ˆ ˆ
AD i 4j k (i 8j 3k)
2 2 2
? ? ? ? ? ?
AD AC
AD.AC 37
| AC | 2 38
?
? ?
10
2a 10 1 d 390
2
?? ? ? ?
??
10
5
t 15 a 9d 15
8a 3d
t 7 a 4d 7
?
? ? ? ? ?
?
? ? ? ?
15 5
15 5
S – S 6 14 8 6 4 8
22
? ? ? ? ? ?
15 118 5 38
2
? ? ?
?
 
 
11. If , where a and b are 
rational numbers, then 9a + 8b is equal to : 
 (1) 2 (2) 1 
 (3) 3  (4)  
Ans.  (1) 
Sol.
  
  
  
  
  
  
  
 =  
 ? 
 ? 9a + 8b =  
12. If z is a complex number such that |z| ?1, then the 
minimum value of is: 
 (1)  (2) 2 
 (3) 3  (4)  
 Ans. (Bonus) 
Sol. |z| ? 1 
  
 Min. value of  is actually zero. 
13. If the domain of the function f(x) = 
+log
10
 (x
2
 + 2x – 15) is (– ?, ?) U [ ?, ?), then  
?
2
 + ?
3
 is equal to :  
 (1) 140 (2) 175 
 (3) 150  (4) 125 
 Ans. (3) 
Sol. ƒ(x) = + log
10
(x
2
 + 2x - 15) 
 Domain : x
2
 – 25 ? 0  ? x ? (– ?, -5] ? [5, ?) 
 4 – x
2
 ? 0 ? x ?{–2, 2} 
 x
2
 + 2x – 15 > 0 ?  (x + 5) (x – 3) > 0  
 ? x ? (– ?, –5) ? (3, ?) 
 ??x ? (– ?, –5) ? [5, ??) 
 ??= –5; ??= 5 
 ? ??
?
?? ? ?
3
?= 150 
14. Consider the relations R
1
 and R
2
 defined as aR
1
b 
? a
2
 + b
2
 = 1 for all a , b, ? R and (a, b) R
2
(c, d) 
? a + d = b + c for all (a,b), (c,d) ? N × N. Then 
 (1) Only R
1
 is an equivalence relation 
 (2) Only R
2
 is an equivalence relation 
 (3) R
1
 and R
2
 both are equivalence relations 
 (4) Neither R
1
 nor R
2
 is an equivalence relation 
Ans.  (2) 
Sol. aR
1
 b ? a
2
 + b
2
 = 1; a, b ? R 
 (a, b) R
2
 (c, d) ? a + d = b + c; (a, b), (c, d) ? N 
 for R
1
 : Not reflexive symmetric not transitive 
 for R
2
 : R
2
 is reflexive, symmetric and transitive  
 Hence only R
2
 is equivalence relation. 
3
4
0
cos x dx a b 3
?
? ? ?
?
3
2
/3
4
0
cos xdx
?
?
2
/3
0
1 cos2x
dx
2
?
???
?
??
??
?
/3
2
0
1
(1 2cos2x cos 2x)dx
4
?
? ? ?
?
/3 /3 /3
0 0 0
1 1 cos4x
dx 2 cos2x dx dx
42
? ? ?
??
?
? ? ?
??
??
? ? ?
/3 /3
00
1 1 1
(sin 2x) (sin 4x)
4 3 2 3 8
??
?? ?? ??
? ? ? ?
?? ??
?? ??
/3 /3
00
1 1 1
(sin 2x) (sin 4x)
4 3 2 3 8
??
?? ?? ??
? ? ? ?
?? ??
?? ??
1 3 1 3
4 2 2 8 2
?? ??
?
? ? ? ? ?
?? ??
??
??
?? ??
73
2 64
?
?
17
a ; b
8 64
??
97
2
88
??
1
z (3 4
2
?? i)
5
2
3
2
O
P
3
, 2
2
? ? ?
?
? ?
? ?
3
z 2i
2
??
2
2
x 25
(4 x )
?
?
2
2
x 25
4x
?
?
Page 5


  
        
  
 
 
 
 
SECTION-A 
1. Let f(x) = |2x
2
+5|x|–3|,x ?R. If m and n denote the 
number of points where f is not continuous and not 
differentiable respectively, then m + n is equal to : 
 (1) 5 (2) 2 
 (3) 0  (4) 3 
Ans.  (4) 
Sol. f(x) = |2x
2
+5|x|–3| 
 Graph of y = |2x
2 
+ 5x – 3| 
 
 Graph of f(x) 
  
  
0 
 
 Number of points of discontinuity = 0 = m 
 Number of points of non-differentiability = 3 = n 
2. Let ? and ? be the roots of the equation px
2
 + qx – 
r = 0, where p ? 0. If p, q and r be the consecutive 
terms of a non-constant G.P and , then 
the value of ( ??– ??
? ?
is : ? 
 (1)  (2) 9 
 (3)   (4) 8 
Ans.  (1) 
Sol. px
2
 + qx – r = 0  
 p = A, q = AR, r = AR
2
 
 Ax
2
 + ARx – AR
2
 = 0 
 x
2
 + Rx – R
2
 = 0  
   
 ?? ?
 ( ? ?? ? ?)
2
 = ( ? ??)
2
 – 4 ?? ? = R
2
 – 4(–R
2
) = 5 
  = 80/9 
3. The number of solutions of the equation 4 sin
2
x – 4 
cos
3
x + 9 – 4 cosx = 0; x ? [ –2 ?, 2 ?] is : 
 (1) 1 
 (2) 3 
 (3) 2 
 (4) 0 
Ans.  (4) 
Sol. 4sin
2
x – 4cos
3
x + 9 – 4 cosx = 0 ; x ? [ – 2 ??, 2 ?] 
 4 – 4cos
2
x – 4cos
3
x + 9 – 4 cosx = 0 
 4cos
3
x + 4cos
2
x + 4 cosx – 13 = 0 
 4cos
3
x + 4cos
2
x + 4cosx = 13 
 L.H.S. ? 12 can’t be equal to 13. 
4. The value of dx is equal to: 
 (1) 0 
 (2) 1 
 (3) 2 
 (4) –1 
Ans.  (1) 
Sol.  
 Using  where f(2a–x) = –f(x) 
 Here f(1–x) = f(x) 
 ? I = 0 
–3
1/2 – 5 / 4
1 1 3
4
??
??
80
9
20
3
?
?
?
?
1 1 3
4
??
??
2
3 R 4
R
4 R 4 3
? ? ? ? ?
? ? ? ? ?
? ? ?
16
9
??
??
??
1
1 32
3
0
(2x 3x x 1) ? ? ?
?
? ?
1
1
3
32
0
I 2x 3x x 1 dx ? ? ? ?
?
? ?
?
2a
0
f x dx
 
 
5. Let P be a point on the ellipse . Let the 
line passing through P and parallel to y-axis meet 
the circle x
2
 + y
2
 = 9 at point Q such that P and Q 
are on the same side of the x-axis. Then, the 
eccentricity of the locus of the point R on PQ such 
that PR : RQ = 4 : 3 as P moves on the ellipse, is : 
 (1)  (2)  
 (3)   (4)  
Ans.  (4) 
Sol.  
  
 h = 3cos ?; 
  
 ? locus =  
  
6. Let m and n be the coefficients of seventh and 
thirteenth terms respectively in the expansion of  
. Then is : 
 (1)  (2)  
 (3)   (4)  
Ans.  (4) 
Sol.  
  
  
  :  
  
7. Let ? be a non-zero real number. Suppose f : R ?  
R is a differentiable function such that f (0) = 2 and 
? ?
x
lim f x 1
? ? ?
? . If f '(x) = ?f(x) +3, for all x ? R, 
then f (–log
e
2) is equal to____. 
 (1) 3 (2) 5 
 (3) 9  (4) 7 
Ans.  (3 OR BONUS) 
Sol. f(0) = 2, 
? ?
x
lim f x 1
? ? ?
? 
 f’(x) – x.f(x) = 3 
 I.F = e
– ?x 
 
y(e
– ?x
) =  
 f(x). (e
– ?x
) =  
 x = 0 ?  ?  (1) 
 f(x) =  
 x ? – ? ?? 1 = 
 ? = –3 ? c = 1 
  
 = 1 + e
3ln2 
= 9 
 (But ? should be greater than 0 for finite 
value of c) 
22
xy
1
94
??
11
19
13
21
139
23
13
7
Q
P(3cos , 2sin ) ? ?
Q(3cos , 3sin ) ? ?
P
2 2
x y
1
9 4
? ?
4
P
(3C, 2S)
R
(h, k)
Q
(3C, 3S)
3
18
k sin
7
??
22
x 49y
1
9 324
??
324 117 13
e1
49 9 21 7
? ? ? ?
?
18
1
3
2
3
11
x
3
2x
??
??
?
??
??
1
3
n
m
??
??
??
4
9
1
9
1
4
9
4
18
12
33
xx
3
?
??
??
?
??
??
??
? ?
12 6
12
33
18 18
7 6 6
12 6
x x 1 1
t c c .
32 2
3
?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
? ?
6 12
12
33
18 18 –6
13 12 12
6 12
x x 1 1
t c c . .x
32 2
3
?
? ? ? ?
? ? ? ?
??
? ? ? ?
? ? ? ?
? ? ? ?
18 –12 –6
6
m c .3 .2 ?
18 –12 –6
12
n c .2 .3 ?
1
1
2
–12 –6
3
3
12 6
n 2 .3 3 9
m 2 4 3 .2
??
??
? ? ? ?
? ? ?
??
? ? ? ?
? ? ? ?
??
x
3.e dx
??
?
x
3e
c
??
?
??
3
2c
?
??
?
3
c2 ??
?
x
3
c.e
?
?
?
?
3
c(0)
?
?
?
x
3
f ( ln 2) c.e
?
?
? ? ?
?
 
 
 
 
8. Let P and Q be the points on the line 
 which are at a distance of 6 
units from the point R (1,2,3). If the centroid of the 
triangle PQR is ( ?, ? ?, ?), then ?
2
+ ? ?
2 
+ ??
2 
is: 
 (1) 26 
 (2) 36 
 (3) 18 
 (4) 24 
Ans.  (3) 
Sol.  
  
 P(8 ?? – 3, 2 ? + 4, 2 ? – 1) 
 PR = 6 
 (8 ?? – 4)
2 
+ ( 2 ? + 2)
2
 + (2 ? – 4)
2 
= 36 
 ? = 0, 1 
 Hence  P(–3, 4, –1) & Q(5, 6, 1) 
 Centroid of ?PQR = (1, 4, 1) ? ( ?, ? ?, ?)  
 ?
??
+ ?
?
 + ?
?
 = 18 
9. Consider a ?ABC where A(1,2,3,), B(–2,8,0) and 
C(3,6,7). If the angle bisector of ?BAC meets the 
line BC at D, then the length of the projection of 
the vector on the vector is: 
 (1)  
 (2)  
 (3)  
 (4)  
Ans.  (1) 
Sol.  
 A(1, 3, 2); B(–2, 8, 0); C(3, 6, 7); 
  
 AB =  
 AC =  
  
 Length of projection of  on  
 =   
10. Let S
n
 denote the sum of the first n terms of an 
arithmetic progression. If S
10
 = 390 and the ratio of 
the tenth and the fifth terms is 15 : 7, then S
15
 –S
5
 
is equal to: 
 (1) 800 
 (2) 890 
 (3) 790 
 (4) 690 
Ans.  (3) 
Sol. S
10
 = 390 
  
 ? 2a + 9d = 78  (1) 
  (2) 
 From (1) & (2) a = 3 & d = 8 
  
 =790 
x 3 y 4 z 1
8 2 2
? ? ?
??
P Q
R(1,2,3)
AD
?
AC
?
37
2 38
38
2
39
2 38
19
A (1,3,2)
C (3, 6, 7) B
(–2, 8, 0)
D
1 :   1
1 7
, 7,
2 2
? ?
? ?
? ?
ˆ ˆ ˆ
AC 2i 3j 5k ? ? ?
9 25 4 38 ? ? ?
4 9 25 38 ? ? ?
1 3 1
ˆ ˆ ˆ ˆ ˆ ˆ
AD i 4j k (i 8j 3k)
2 2 2
? ? ? ? ? ?
AD AC
AD.AC 37
| AC | 2 38
?
? ?
10
2a 10 1 d 390
2
?? ? ? ?
??
10
5
t 15 a 9d 15
8a 3d
t 7 a 4d 7
?
? ? ? ? ?
?
? ? ? ?
15 5
15 5
S – S 6 14 8 6 4 8
22
? ? ? ? ? ?
15 118 5 38
2
? ? ?
?
 
 
11. If , where a and b are 
rational numbers, then 9a + 8b is equal to : 
 (1) 2 (2) 1 
 (3) 3  (4)  
Ans.  (1) 
Sol.
  
  
  
  
  
  
  
 =  
 ? 
 ? 9a + 8b =  
12. If z is a complex number such that |z| ?1, then the 
minimum value of is: 
 (1)  (2) 2 
 (3) 3  (4)  
 Ans. (Bonus) 
Sol. |z| ? 1 
  
 Min. value of  is actually zero. 
13. If the domain of the function f(x) = 
+log
10
 (x
2
 + 2x – 15) is (– ?, ?) U [ ?, ?), then  
?
2
 + ?
3
 is equal to :  
 (1) 140 (2) 175 
 (3) 150  (4) 125 
 Ans. (3) 
Sol. ƒ(x) = + log
10
(x
2
 + 2x - 15) 
 Domain : x
2
 – 25 ? 0  ? x ? (– ?, -5] ? [5, ?) 
 4 – x
2
 ? 0 ? x ?{–2, 2} 
 x
2
 + 2x – 15 > 0 ?  (x + 5) (x – 3) > 0  
 ? x ? (– ?, –5) ? (3, ?) 
 ??x ? (– ?, –5) ? [5, ??) 
 ??= –5; ??= 5 
 ? ??
?
?? ? ?
3
?= 150 
14. Consider the relations R
1
 and R
2
 defined as aR
1
b 
? a
2
 + b
2
 = 1 for all a , b, ? R and (a, b) R
2
(c, d) 
? a + d = b + c for all (a,b), (c,d) ? N × N. Then 
 (1) Only R
1
 is an equivalence relation 
 (2) Only R
2
 is an equivalence relation 
 (3) R
1
 and R
2
 both are equivalence relations 
 (4) Neither R
1
 nor R
2
 is an equivalence relation 
Ans.  (2) 
Sol. aR
1
 b ? a
2
 + b
2
 = 1; a, b ? R 
 (a, b) R
2
 (c, d) ? a + d = b + c; (a, b), (c, d) ? N 
 for R
1
 : Not reflexive symmetric not transitive 
 for R
2
 : R
2
 is reflexive, symmetric and transitive  
 Hence only R
2
 is equivalence relation. 
3
4
0
cos x dx a b 3
?
? ? ?
?
3
2
/3
4
0
cos xdx
?
?
2
/3
0
1 cos2x
dx
2
?
???
?
??
??
?
/3
2
0
1
(1 2cos2x cos 2x)dx
4
?
? ? ?
?
/3 /3 /3
0 0 0
1 1 cos4x
dx 2 cos2x dx dx
42
? ? ?
??
?
? ? ?
??
??
? ? ?
/3 /3
00
1 1 1
(sin 2x) (sin 4x)
4 3 2 3 8
??
?? ?? ??
? ? ? ?
?? ??
?? ??
/3 /3
00
1 1 1
(sin 2x) (sin 4x)
4 3 2 3 8
??
?? ?? ??
? ? ? ?
?? ??
?? ??
1 3 1 3
4 2 2 8 2
?? ??
?
? ? ? ? ?
?? ??
??
??
?? ??
73
2 64
?
?
17
a ; b
8 64
??
97
2
88
??
1
z (3 4
2
?? i)
5
2
3
2
O
P
3
, 2
2
? ? ?
?
? ?
? ?
3
z 2i
2
??
2
2
x 25
(4 x )
?
?
2
2
x 25
4x
?
?
 
 
 
 
15. If the mirror image of the point P(3,4,9) in the line 
is ( ?, ? ?, ?), then 14 ( ??+ ? ? + ?) 
is : 
 (1) 102 (2) 138 
 (3) 108  (4) 132 
Ans.  (3) 
Sol.
  
 ? 
 3(3?? – 2) + 2 (2 ?? – 5) + ( ? – 7) = 0  
 14 ?? = 23 ? ?? ? ? ? ?
? ?
? ?? ?
? ?
? ?
 Ans. 14 ( ? ?? ? ? ? ? ?r) = 108 
16. Let f(x) =  x ? N. If for some  
 a ? ?N, f(f(f(a))) = 21, then  
where [t] denotes the greatest integer less than or 
equal to t, is equal to : 
 (1) 121 
 (2) 144 
 (3) 169 
 (4) 225 
 Ans. (2) 
Sol. ƒ(x) =   
 ƒ(ƒ(ƒ(a))) = 21 
 C –1: If a = even 
  ƒ(a) = a – 1 = odd 
 f(f(a)) = 2(a – 1) = even 
 ƒ(ƒ(ƒ(a))) = 2a – 3 = 21 ? a = 12 
 C –2: If a = odd 
  ƒ(a) = 2a = even 
  ƒ(ƒ(a)) = 2a – 1 = odd 
  ƒ(ƒ(ƒ(a))) = 4a – 2 = 21 (Not possible) 
 Hence a = 12 
 Now  
  
  
 = 144 – 0 = 144. 
17. Let the system of equations x + 2y +3z = 5, 2x + 
3y + z = 9, 4x + 3y + ?z = ? have infinite number 
of solutions. Then ? + 2??is equal to : 
 (1) 28 (2) 17 
 (3) 22  (4) 15 
Ans.  (2) 
Sol. x + 2y + 3z = 5 
 2x + 3y + z = 9 
 4x + 3y + ?z = µ 
 for infinite following ? = ?
1
 = ?
2
 = ?
3
 = 0 
 ? ?? ?= 0 ?? ? ?? –13 
 ?
?
?? ? = 0 ??? = 15 
 ?
?
?? ?= 0 
x 1 y 1 z 2
3 2 1
? ? ?
??
P(3, 4, 9)
A( , ) ? ?? ?
N
(3 + 1, 2 –1,  + 2) ? ? ?
PN.b 0 ?
23
14
83 32 51
N , ,
14 14 14
??
??
??
3 83 62
2 14 7
??
? ? ? ?
4 32 4
2 14 7
??
? ? ? ?
9 51 12
2 14 7
? ? ?
? ? ? ?
x 1,xis even,
2x, xis odd,
? ?
?
?
xa
lim
?
?
3
| x | x
,
aa
??
??
?
??
??
??
??
x 1; x even
2x; x odd
?? ?
?
?
?
3
x 12
| x | x
lim
2 12
?
?
??
??
?
??
??
??
??
3
x 12 x 12
| x | x
lim lim
12 12
??
??
??
??
??
??
1 2 3
2 3 1
43 ?
5 2 3
9 3 1
µ 3 13 ?
1 5 3
2 9 1
4 15 13 ?