JEE Main 2024 January 29 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


  
            
 
  
 
 
 
 
 
 
 
 
SECTION-A 
1. Let A = 
2 1 2
6 2 11
3 3 2
??
??
??
??
??
 and P = 
1 2 0
5 0 2
7 1 5
??
??
??
??
??
. The sum 
of the prime factors of 
1
P AP 2I
?
? is equal to 
 
 (1) 26 (2) 27 (3) 66 (4) 23 
Ans. (1) 
Sol. 
1 1 1
P AP 2I P AP 2P P
? ? ?
? ? ? 
   = 
1
P (A 2I)P
?
?  
   = 
1
P A 2I P
?
? 
   = |A-2I| 
   
0 1 2
6 0 11
3 3 0
? = 69 
 So, Prime factor of 69 is 3 & 23 
 So, sum = 26 
2. Number of ways of arranging 8 identical books 
into 4 identical shelves where any number of 
shelves may remain empty is equal to  
 
 (1) 18 (2) 16 (3) 12 (4) 15 
Ans. (4) 
Sol. 3 Shelf empty : (8, 0, 0, 0) ? 1way 
 2 shelf empty : 
(7,1,0,0)
(6,2,0,0)
4ways
(5,3,0,0)
(4,4,0,0)
?
?
?
?
?
?
?
 
 1 shelf empty : 
(6,1,1,0) (3,3,2,0)
(5,2,1,0) (4,2,2,0) 5ways
(4,3,1,0)
?
?
?
?
?
?
 
 0 Shelf empty : 
(1,2,3,2) (5,1,1,1)
(2,2,2,2)
5ways
(3,3,1,1)
(4,2,1,1)
?
?
?
?
?
?
?
 
 Total = 15 ways 
3. Let P(3, 2, 3), Q (4, 6, 2) and R (7, 3, 2) be the 
vertices of ?PQR. Then, the angle ?QPR is  
 (1) 
6
?
 (2) 
1
7
cos
18
?
??
??
??
 
 (3) 
1
1
cos
18
?
??
??
??
 (4) 
3
?
 
Ans. (4) 
Sol. 
?
P(3, 2, 3)
R(7, 3, 2) Q(4, 6, 2)
 
 Direction ratio of PR = (4, 1, -1) 
 Direction ratio of PQ = (1, 4, -1) 
 Now, 
4 4 1
cos
.
18 18
??
?? 
 
3
?
?? 
4. If the mean and variance of five observations are 
24
5
 and 
194
25
 respectively and the mean of first 
four observations is 
7
,
2
 then the variance of the 
first four observations in equal to  
 (1) 
4
5
 (2) 
77
12
 (3) 
5
4
 (4) 
105
4
 
Ans. (3) 
Sol. 
2
24 194
X;
5 25
? ? ? 
 Let first four observation be x
1
, x
2
, x
3
, x
4
 
 Here, 
1 2 3 4 5
x x x x x 24
......(1)
55
? ? ? ?
? 
 Also, 
1 2 3 4
x x x x 7
42
? ? ?
? 
 
1 2 3 4
x x x x 14 ? ? ? ? ? 
Page 2


  
            
 
  
 
 
 
 
 
 
 
 
SECTION-A 
1. Let A = 
2 1 2
6 2 11
3 3 2
??
??
??
??
??
 and P = 
1 2 0
5 0 2
7 1 5
??
??
??
??
??
. The sum 
of the prime factors of 
1
P AP 2I
?
? is equal to 
 
 (1) 26 (2) 27 (3) 66 (4) 23 
Ans. (1) 
Sol. 
1 1 1
P AP 2I P AP 2P P
? ? ?
? ? ? 
   = 
1
P (A 2I)P
?
?  
   = 
1
P A 2I P
?
? 
   = |A-2I| 
   
0 1 2
6 0 11
3 3 0
? = 69 
 So, Prime factor of 69 is 3 & 23 
 So, sum = 26 
2. Number of ways of arranging 8 identical books 
into 4 identical shelves where any number of 
shelves may remain empty is equal to  
 
 (1) 18 (2) 16 (3) 12 (4) 15 
Ans. (4) 
Sol. 3 Shelf empty : (8, 0, 0, 0) ? 1way 
 2 shelf empty : 
(7,1,0,0)
(6,2,0,0)
4ways
(5,3,0,0)
(4,4,0,0)
?
?
?
?
?
?
?
 
 1 shelf empty : 
(6,1,1,0) (3,3,2,0)
(5,2,1,0) (4,2,2,0) 5ways
(4,3,1,0)
?
?
?
?
?
?
 
 0 Shelf empty : 
(1,2,3,2) (5,1,1,1)
(2,2,2,2)
5ways
(3,3,1,1)
(4,2,1,1)
?
?
?
?
?
?
?
 
 Total = 15 ways 
3. Let P(3, 2, 3), Q (4, 6, 2) and R (7, 3, 2) be the 
vertices of ?PQR. Then, the angle ?QPR is  
 (1) 
6
?
 (2) 
1
7
cos
18
?
??
??
??
 
 (3) 
1
1
cos
18
?
??
??
??
 (4) 
3
?
 
Ans. (4) 
Sol. 
?
P(3, 2, 3)
R(7, 3, 2) Q(4, 6, 2)
 
 Direction ratio of PR = (4, 1, -1) 
 Direction ratio of PQ = (1, 4, -1) 
 Now, 
4 4 1
cos
.
18 18
??
?? 
 
3
?
?? 
4. If the mean and variance of five observations are 
24
5
 and 
194
25
 respectively and the mean of first 
four observations is 
7
,
2
 then the variance of the 
first four observations in equal to  
 (1) 
4
5
 (2) 
77
12
 (3) 
5
4
 (4) 
105
4
 
Ans. (3) 
Sol. 
2
24 194
X;
5 25
? ? ? 
 Let first four observation be x
1
, x
2
, x
3
, x
4
 
 Here, 
1 2 3 4 5
x x x x x 24
......(1)
55
? ? ? ?
? 
 Also, 
1 2 3 4
x x x x 7
42
? ? ?
? 
 
1 2 3 4
x x x x 14 ? ? ? ? ? 
 
 
 
 Now from eqn -1 
 x
5
 = 10 
 Now, 
2
194
25
?? 
 
22222
1 2 3 4 5
xxxxx 576 194
5 25 25
????
?? 
 
2222
1 2 3 4
x x x x 54 ? ? ? ? ? 
 Now, variance of first 4 observations 
 Var = 
2
44
2
ii
i 1 i 1
xx
44
??
??
??
?? ?
??
??
??
??
 
  
54 49
44
?? = 
5
4
 
5. The function f(x) = 
2
3
2x 3(x) ,x , ?? has  
 (1)  exactly one point of local minima and no   
  point of local maxima 
 (2)  exactly one point of local maxima and no   
       point of local minima  
 (3)  exactly one point of local maxima and 
 exactly one point of local minima  
 (4)  exactly two points of local maxima and 
 exactly one point of local minima  
Ans. (3) 
Sol. 
2
3
f (x) 2x 3(x) ?? 
 
1
3
f '(x) 2 2x
?
?? 
 = 
1
3
1
21
x
??
??
?
??
??
 
 = 
1
3
1
3
x1
2
x
??
?
??
??
??
??
 
 
+
–1
M
0
m
– +
 
 So, maxima (M) at x = -1 & minima(m) at x = 0 
 
6. Let r and ? respectively be the modulus and 
amplitude of the complex number 
5
z 2 i 2tan
8
? ??
??
??
??
, then (r, ) ? is equal to  
 (1) 
33
2sec ,
88
?? ??
??
??
 
 (2) 
35
2sec ,
88
?? ??
??
??
 
 (3) 
53
2sec ,
88
?? ??
??
??
 
 (4) 
11 11
2sec ,
88
?? ??
??
??
 
Ans. (1) 
Sol. z = 
5
2 i 2tan
8
? ??
?
??
??
 = x + iy (let) 
 r = 
2 2 1
y
x y & tan
x
?
? ? ? 
 r = 
2
2
5
(2) 2tan
8
? ??
?
??
??
 
 
= 
53
2sec 2sec
88
?? ??
? ? ?
??
??
 
 = 2 sec 
3
8
?
 
 & 
1
5
2tan
8
tan
2
?
? ??
?
??
??
??
??
??
??
 
 =
1
5
tan tan
8
?
?? ? ??
??
?? ??
?? ??
 
 = 
3
8
?
 
7. The sum of the solutions x ? of the equation 
3
66
3cos2x cos 2x
cos x sin x
?
?
 = x
3
 – x
2
 + 6  is  
 (1) 0 (2) 1 
 (3) –1  (4) 3 
Page 3


  
            
 
  
 
 
 
 
 
 
 
 
SECTION-A 
1. Let A = 
2 1 2
6 2 11
3 3 2
??
??
??
??
??
 and P = 
1 2 0
5 0 2
7 1 5
??
??
??
??
??
. The sum 
of the prime factors of 
1
P AP 2I
?
? is equal to 
 
 (1) 26 (2) 27 (3) 66 (4) 23 
Ans. (1) 
Sol. 
1 1 1
P AP 2I P AP 2P P
? ? ?
? ? ? 
   = 
1
P (A 2I)P
?
?  
   = 
1
P A 2I P
?
? 
   = |A-2I| 
   
0 1 2
6 0 11
3 3 0
? = 69 
 So, Prime factor of 69 is 3 & 23 
 So, sum = 26 
2. Number of ways of arranging 8 identical books 
into 4 identical shelves where any number of 
shelves may remain empty is equal to  
 
 (1) 18 (2) 16 (3) 12 (4) 15 
Ans. (4) 
Sol. 3 Shelf empty : (8, 0, 0, 0) ? 1way 
 2 shelf empty : 
(7,1,0,0)
(6,2,0,0)
4ways
(5,3,0,0)
(4,4,0,0)
?
?
?
?
?
?
?
 
 1 shelf empty : 
(6,1,1,0) (3,3,2,0)
(5,2,1,0) (4,2,2,0) 5ways
(4,3,1,0)
?
?
?
?
?
?
 
 0 Shelf empty : 
(1,2,3,2) (5,1,1,1)
(2,2,2,2)
5ways
(3,3,1,1)
(4,2,1,1)
?
?
?
?
?
?
?
 
 Total = 15 ways 
3. Let P(3, 2, 3), Q (4, 6, 2) and R (7, 3, 2) be the 
vertices of ?PQR. Then, the angle ?QPR is  
 (1) 
6
?
 (2) 
1
7
cos
18
?
??
??
??
 
 (3) 
1
1
cos
18
?
??
??
??
 (4) 
3
?
 
Ans. (4) 
Sol. 
?
P(3, 2, 3)
R(7, 3, 2) Q(4, 6, 2)
 
 Direction ratio of PR = (4, 1, -1) 
 Direction ratio of PQ = (1, 4, -1) 
 Now, 
4 4 1
cos
.
18 18
??
?? 
 
3
?
?? 
4. If the mean and variance of five observations are 
24
5
 and 
194
25
 respectively and the mean of first 
four observations is 
7
,
2
 then the variance of the 
first four observations in equal to  
 (1) 
4
5
 (2) 
77
12
 (3) 
5
4
 (4) 
105
4
 
Ans. (3) 
Sol. 
2
24 194
X;
5 25
? ? ? 
 Let first four observation be x
1
, x
2
, x
3
, x
4
 
 Here, 
1 2 3 4 5
x x x x x 24
......(1)
55
? ? ? ?
? 
 Also, 
1 2 3 4
x x x x 7
42
? ? ?
? 
 
1 2 3 4
x x x x 14 ? ? ? ? ? 
 
 
 
 Now from eqn -1 
 x
5
 = 10 
 Now, 
2
194
25
?? 
 
22222
1 2 3 4 5
xxxxx 576 194
5 25 25
????
?? 
 
2222
1 2 3 4
x x x x 54 ? ? ? ? ? 
 Now, variance of first 4 observations 
 Var = 
2
44
2
ii
i 1 i 1
xx
44
??
??
??
?? ?
??
??
??
??
 
  
54 49
44
?? = 
5
4
 
5. The function f(x) = 
2
3
2x 3(x) ,x , ?? has  
 (1)  exactly one point of local minima and no   
  point of local maxima 
 (2)  exactly one point of local maxima and no   
       point of local minima  
 (3)  exactly one point of local maxima and 
 exactly one point of local minima  
 (4)  exactly two points of local maxima and 
 exactly one point of local minima  
Ans. (3) 
Sol. 
2
3
f (x) 2x 3(x) ?? 
 
1
3
f '(x) 2 2x
?
?? 
 = 
1
3
1
21
x
??
??
?
??
??
 
 = 
1
3
1
3
x1
2
x
??
?
??
??
??
??
 
 
+
–1
M
0
m
– +
 
 So, maxima (M) at x = -1 & minima(m) at x = 0 
 
6. Let r and ? respectively be the modulus and 
amplitude of the complex number 
5
z 2 i 2tan
8
? ??
??
??
??
, then (r, ) ? is equal to  
 (1) 
33
2sec ,
88
?? ??
??
??
 
 (2) 
35
2sec ,
88
?? ??
??
??
 
 (3) 
53
2sec ,
88
?? ??
??
??
 
 (4) 
11 11
2sec ,
88
?? ??
??
??
 
Ans. (1) 
Sol. z = 
5
2 i 2tan
8
? ??
?
??
??
 = x + iy (let) 
 r = 
2 2 1
y
x y & tan
x
?
? ? ? 
 r = 
2
2
5
(2) 2tan
8
? ??
?
??
??
 
 
= 
53
2sec 2sec
88
?? ??
? ? ?
??
??
 
 = 2 sec 
3
8
?
 
 & 
1
5
2tan
8
tan
2
?
? ??
?
??
??
??
??
??
??
 
 =
1
5
tan tan
8
?
?? ? ??
??
?? ??
?? ??
 
 = 
3
8
?
 
7. The sum of the solutions x ? of the equation 
3
66
3cos2x cos 2x
cos x sin x
?
?
 = x
3
 – x
2
 + 6  is  
 (1) 0 (2) 1 
 (3) –1  (4) 3 
 
 
 
Ans. (3) 
Sol. 
3
66
3cos2x cos 2x
cos x sin x
?
?
= x
3
 – x
2
 + 6 
 
2
22
cos2x (3 cos 2x)
cos2x (1 sin xcos x)
?
?
?
= x
3
 – x
2
 + 6 
 
2
2
4(3 cos 2x)
(4 sin 2x)
?
?
?
= x
3
 – x
2
 + 6 
 
2
2
4(3 cos 2x)
(3 cos 2x)
?
?
?
 = x
3
 – x
2
 + 6 
 x
3
 – x
2
 + 2 = 0 ? (x + 1)(x
2
 – 2x + 2) = 0  
 so, sum of real solutions = –1 
8. Let OA a,OB 12a 4b ? ? ? and OC b, ? where O 
is the origin. If S is the parallelogram with adjacent 
sides OA and OC, then  
 
area of the quadrilateral OABC
area of S
 is equal to ___ 
 (1) 6 (2) 10 
 (3) 7  (4) 8 
Ans. (4) 
Sol. 
A
B
C O
a
b
12a + 4b
 
 Area of parallelogram,  S a b ?? 
 Area of quadrilateral =Area( ?OAB)+Area ( ?OBC) 
 = 
? ?
1
a (12a 4b) b (12a 4b)
2
? ? ? ? ? 
 = 8(a b) ? 
 Ratio = 
8(a b)
(a b)
?
?
 = 8    
9. If log
e
 a, log
e
 b, log
e
 c are in an A.P. and log
e
 a – 
log
e
2b, log
e
2b – log
e
3c, log
e
3c – log
e
 a are also in 
an A.P, then a : b : c is equal to   
 (1) 9 : 6 : 4 (2) 16 : 4 : 1 
 (3) 25 : 10 : 4 (4) 6 : 3 : 2 
Ans. (1) 
Sol. log
e
a, log
e
b, log
e
c   are in A.P. 
 ? b
2
 = ac   …..(i) 
 Also 
 
eee
a 2b 3c
log ,log ,log
2b 3c a
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
   are in A.P. 
 
2
2b a 3c
3c 2b a
??
??
??
??
 
 
b3
c2
? 
 Putting in eq. (i)   b
2
 = a × 
2b
3
 
  
a3
b2
? 
  a  : b : c = 9 : 6 : 4 
10. If 
33
22
33
sin x cos x
dx A cos tanx sin B cos sin cotx C,
sin xcos x sin(x )
?
? ? ? ? ? ? ? ? ?
??
?
 
where C is the integration constant, then AB is 
equal to  
 (1) 4cosec(2 ) ? (2) 4sec ? 
 (3) 2sec ? (4) 8cosec(2 ) ? 
Ans. (4) 
Sol. 
33
22
33
sin x cos x
dx
sin xcos x sin(x )
?
??
?
 
 I = 
33
22
33
sin x cos x
dx
sin xcos x (sinxcos cosxsin )
?
? ? ?
?
 
= 
33
22
33
22
22
sin x cos x
dx dx
sin xcos x tanxcos sin sin xcos x cos cotxsin
?
? ? ? ? ? ?
??
= 
22
sec x cosec x
dx dx
tanxcos sin cos cotxsin
?
? ? ? ? ? ?
??
  
 I = I
1
 + I
2 ……
 {Let} 
 For I
1
, let    tan x cos ?  – sin ? = t
2
 
  
2
2t dt
sec xdx
cos
?
?
 
 For I
2
 ,   let 
2
cos cotxsin z ? ? ? ? 
   
2
2z dz
cosec x dx
sin
?
?
 
Page 4


  
            
 
  
 
 
 
 
 
 
 
 
SECTION-A 
1. Let A = 
2 1 2
6 2 11
3 3 2
??
??
??
??
??
 and P = 
1 2 0
5 0 2
7 1 5
??
??
??
??
??
. The sum 
of the prime factors of 
1
P AP 2I
?
? is equal to 
 
 (1) 26 (2) 27 (3) 66 (4) 23 
Ans. (1) 
Sol. 
1 1 1
P AP 2I P AP 2P P
? ? ?
? ? ? 
   = 
1
P (A 2I)P
?
?  
   = 
1
P A 2I P
?
? 
   = |A-2I| 
   
0 1 2
6 0 11
3 3 0
? = 69 
 So, Prime factor of 69 is 3 & 23 
 So, sum = 26 
2. Number of ways of arranging 8 identical books 
into 4 identical shelves where any number of 
shelves may remain empty is equal to  
 
 (1) 18 (2) 16 (3) 12 (4) 15 
Ans. (4) 
Sol. 3 Shelf empty : (8, 0, 0, 0) ? 1way 
 2 shelf empty : 
(7,1,0,0)
(6,2,0,0)
4ways
(5,3,0,0)
(4,4,0,0)
?
?
?
?
?
?
?
 
 1 shelf empty : 
(6,1,1,0) (3,3,2,0)
(5,2,1,0) (4,2,2,0) 5ways
(4,3,1,0)
?
?
?
?
?
?
 
 0 Shelf empty : 
(1,2,3,2) (5,1,1,1)
(2,2,2,2)
5ways
(3,3,1,1)
(4,2,1,1)
?
?
?
?
?
?
?
 
 Total = 15 ways 
3. Let P(3, 2, 3), Q (4, 6, 2) and R (7, 3, 2) be the 
vertices of ?PQR. Then, the angle ?QPR is  
 (1) 
6
?
 (2) 
1
7
cos
18
?
??
??
??
 
 (3) 
1
1
cos
18
?
??
??
??
 (4) 
3
?
 
Ans. (4) 
Sol. 
?
P(3, 2, 3)
R(7, 3, 2) Q(4, 6, 2)
 
 Direction ratio of PR = (4, 1, -1) 
 Direction ratio of PQ = (1, 4, -1) 
 Now, 
4 4 1
cos
.
18 18
??
?? 
 
3
?
?? 
4. If the mean and variance of five observations are 
24
5
 and 
194
25
 respectively and the mean of first 
four observations is 
7
,
2
 then the variance of the 
first four observations in equal to  
 (1) 
4
5
 (2) 
77
12
 (3) 
5
4
 (4) 
105
4
 
Ans. (3) 
Sol. 
2
24 194
X;
5 25
? ? ? 
 Let first four observation be x
1
, x
2
, x
3
, x
4
 
 Here, 
1 2 3 4 5
x x x x x 24
......(1)
55
? ? ? ?
? 
 Also, 
1 2 3 4
x x x x 7
42
? ? ?
? 
 
1 2 3 4
x x x x 14 ? ? ? ? ? 
 
 
 
 Now from eqn -1 
 x
5
 = 10 
 Now, 
2
194
25
?? 
 
22222
1 2 3 4 5
xxxxx 576 194
5 25 25
????
?? 
 
2222
1 2 3 4
x x x x 54 ? ? ? ? ? 
 Now, variance of first 4 observations 
 Var = 
2
44
2
ii
i 1 i 1
xx
44
??
??
??
?? ?
??
??
??
??
 
  
54 49
44
?? = 
5
4
 
5. The function f(x) = 
2
3
2x 3(x) ,x , ?? has  
 (1)  exactly one point of local minima and no   
  point of local maxima 
 (2)  exactly one point of local maxima and no   
       point of local minima  
 (3)  exactly one point of local maxima and 
 exactly one point of local minima  
 (4)  exactly two points of local maxima and 
 exactly one point of local minima  
Ans. (3) 
Sol. 
2
3
f (x) 2x 3(x) ?? 
 
1
3
f '(x) 2 2x
?
?? 
 = 
1
3
1
21
x
??
??
?
??
??
 
 = 
1
3
1
3
x1
2
x
??
?
??
??
??
??
 
 
+
–1
M
0
m
– +
 
 So, maxima (M) at x = -1 & minima(m) at x = 0 
 
6. Let r and ? respectively be the modulus and 
amplitude of the complex number 
5
z 2 i 2tan
8
? ??
??
??
??
, then (r, ) ? is equal to  
 (1) 
33
2sec ,
88
?? ??
??
??
 
 (2) 
35
2sec ,
88
?? ??
??
??
 
 (3) 
53
2sec ,
88
?? ??
??
??
 
 (4) 
11 11
2sec ,
88
?? ??
??
??
 
Ans. (1) 
Sol. z = 
5
2 i 2tan
8
? ??
?
??
??
 = x + iy (let) 
 r = 
2 2 1
y
x y & tan
x
?
? ? ? 
 r = 
2
2
5
(2) 2tan
8
? ??
?
??
??
 
 
= 
53
2sec 2sec
88
?? ??
? ? ?
??
??
 
 = 2 sec 
3
8
?
 
 & 
1
5
2tan
8
tan
2
?
? ??
?
??
??
??
??
??
??
 
 =
1
5
tan tan
8
?
?? ? ??
??
?? ??
?? ??
 
 = 
3
8
?
 
7. The sum of the solutions x ? of the equation 
3
66
3cos2x cos 2x
cos x sin x
?
?
 = x
3
 – x
2
 + 6  is  
 (1) 0 (2) 1 
 (3) –1  (4) 3 
 
 
 
Ans. (3) 
Sol. 
3
66
3cos2x cos 2x
cos x sin x
?
?
= x
3
 – x
2
 + 6 
 
2
22
cos2x (3 cos 2x)
cos2x (1 sin xcos x)
?
?
?
= x
3
 – x
2
 + 6 
 
2
2
4(3 cos 2x)
(4 sin 2x)
?
?
?
= x
3
 – x
2
 + 6 
 
2
2
4(3 cos 2x)
(3 cos 2x)
?
?
?
 = x
3
 – x
2
 + 6 
 x
3
 – x
2
 + 2 = 0 ? (x + 1)(x
2
 – 2x + 2) = 0  
 so, sum of real solutions = –1 
8. Let OA a,OB 12a 4b ? ? ? and OC b, ? where O 
is the origin. If S is the parallelogram with adjacent 
sides OA and OC, then  
 
area of the quadrilateral OABC
area of S
 is equal to ___ 
 (1) 6 (2) 10 
 (3) 7  (4) 8 
Ans. (4) 
Sol. 
A
B
C O
a
b
12a + 4b
 
 Area of parallelogram,  S a b ?? 
 Area of quadrilateral =Area( ?OAB)+Area ( ?OBC) 
 = 
? ?
1
a (12a 4b) b (12a 4b)
2
? ? ? ? ? 
 = 8(a b) ? 
 Ratio = 
8(a b)
(a b)
?
?
 = 8    
9. If log
e
 a, log
e
 b, log
e
 c are in an A.P. and log
e
 a – 
log
e
2b, log
e
2b – log
e
3c, log
e
3c – log
e
 a are also in 
an A.P, then a : b : c is equal to   
 (1) 9 : 6 : 4 (2) 16 : 4 : 1 
 (3) 25 : 10 : 4 (4) 6 : 3 : 2 
Ans. (1) 
Sol. log
e
a, log
e
b, log
e
c   are in A.P. 
 ? b
2
 = ac   …..(i) 
 Also 
 
eee
a 2b 3c
log ,log ,log
2b 3c a
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
   are in A.P. 
 
2
2b a 3c
3c 2b a
??
??
??
??
 
 
b3
c2
? 
 Putting in eq. (i)   b
2
 = a × 
2b
3
 
  
a3
b2
? 
  a  : b : c = 9 : 6 : 4 
10. If 
33
22
33
sin x cos x
dx A cos tanx sin B cos sin cotx C,
sin xcos x sin(x )
?
? ? ? ? ? ? ? ? ?
??
?
 
where C is the integration constant, then AB is 
equal to  
 (1) 4cosec(2 ) ? (2) 4sec ? 
 (3) 2sec ? (4) 8cosec(2 ) ? 
Ans. (4) 
Sol. 
33
22
33
sin x cos x
dx
sin xcos x sin(x )
?
??
?
 
 I = 
33
22
33
sin x cos x
dx
sin xcos x (sinxcos cosxsin )
?
? ? ?
?
 
= 
33
22
33
22
22
sin x cos x
dx dx
sin xcos x tanxcos sin sin xcos x cos cotxsin
?
? ? ? ? ? ?
??
= 
22
sec x cosec x
dx dx
tanxcos sin cos cotxsin
?
? ? ? ? ? ?
??
  
 I = I
1
 + I
2 ……
 {Let} 
 For I
1
, let    tan x cos ?  – sin ? = t
2
 
  
2
2t dt
sec xdx
cos
?
?
 
 For I
2
 ,   let 
2
cos cotxsin z ? ? ? ? 
   
2
2z dz
cosec x dx
sin
?
?
 
 
 
 
 
  I = I
1
 + I
2
 
  = 
2t dt 2z dz
cos t sin z
?
??
??
 
  = 
2t 2z
cos sin
?
??
 
 = 2sec tan xcos sin 2cosec cos cotxsin ? ? ? ? ? ? ? ? ? 
 Comparing  
  AB = 8cosec2 ? 
11. The distance of the point (2, 3) from the line 2x – 
3y + 28 = 0, measured parallel to the line 
3x y 1 0, ? ? ? is equal to  
 (1) 42 (2) 63 
 (3) 3 4 2 ?  (4) 4 6 3 ? 
Ans. (4) 
Sol. 
P
2x – 3y + 28 = 0
r
A
(2, 3)
 
 Writing P in terms of parametric co-ordinates 2 + r 
cos ?, 3 + r sin ? as tan ??= 3 
 
r 3r
P(2 ,3 )
22
?? 
 P must satisfy 2x – 3y + 28 = 0 
 So, 
r 3r
2(2 ) 3(3 ) 28 0
22
? ? ? ? ? 
 We find r = 4 6 3 ? 
12. If sin
e
y
log |x|
x2
? ??
??
??
??
 is the solution of the 
differential equation x cos 
y dy y
ycos x
x dx x
? ? ? ?
??
? ? ? ?
? ? ? ?
 
and y(1) = ,
3
?
 then 
2
? is equal to  
 (1) 3 (2) 12 
 (3) 4  (4) 9 
Ans. (1) 
Sol. Differential equation :– 
  
y dy y
xcos ycos x
x dx x
?? 
  
y dy
cos x y x
x dx
??
??
??
??
 
 Divide both sides by x
2
 
  
2
dy
xy
y1
dx
cos
xx x
??
?
??
?
??
??
??
??
 
 Let 
y
t
x
? 
  
dt 1
cos t
dx x
??
?
??
??
 
  
1
cost dt dx
x
? 
 Integrating both sides  
  sin t = ln | x | + c  
  
y
sin ln x c
x
?? 
 Using y(1) = ,
3
?
 we get c = 
3
2
 
 So, 
2
33 ? ? ? ? ? 
13. If each term of a geometric progression a
1
, a
2
, a
3
,… 
with 
1
1
a
8
? and 
21
a a , ? is the arithmetic mean of 
the next two terms and S
n
 = a
1
 + a
2
 + …+a
n
, then 
S
20
 – S
18
 is equal to  
 (1) 2
15 
(2) –2
18 
 (3) 2
18  
(4) –2
15 
Ans. (4) 
Sol. Let r’th term of the GP be ar
n–1
. Given,  
  2a
r
 = a
r+1
 + a
r+2
 
  2ar
n–1
 = ar
n
 + ar
n+1
 
  
2
1r
r
?? 
  r
2
 + r – 2 = 0 
Page 5


  
            
 
  
 
 
 
 
 
 
 
 
SECTION-A 
1. Let A = 
2 1 2
6 2 11
3 3 2
??
??
??
??
??
 and P = 
1 2 0
5 0 2
7 1 5
??
??
??
??
??
. The sum 
of the prime factors of 
1
P AP 2I
?
? is equal to 
 
 (1) 26 (2) 27 (3) 66 (4) 23 
Ans. (1) 
Sol. 
1 1 1
P AP 2I P AP 2P P
? ? ?
? ? ? 
   = 
1
P (A 2I)P
?
?  
   = 
1
P A 2I P
?
? 
   = |A-2I| 
   
0 1 2
6 0 11
3 3 0
? = 69 
 So, Prime factor of 69 is 3 & 23 
 So, sum = 26 
2. Number of ways of arranging 8 identical books 
into 4 identical shelves where any number of 
shelves may remain empty is equal to  
 
 (1) 18 (2) 16 (3) 12 (4) 15 
Ans. (4) 
Sol. 3 Shelf empty : (8, 0, 0, 0) ? 1way 
 2 shelf empty : 
(7,1,0,0)
(6,2,0,0)
4ways
(5,3,0,0)
(4,4,0,0)
?
?
?
?
?
?
?
 
 1 shelf empty : 
(6,1,1,0) (3,3,2,0)
(5,2,1,0) (4,2,2,0) 5ways
(4,3,1,0)
?
?
?
?
?
?
 
 0 Shelf empty : 
(1,2,3,2) (5,1,1,1)
(2,2,2,2)
5ways
(3,3,1,1)
(4,2,1,1)
?
?
?
?
?
?
?
 
 Total = 15 ways 
3. Let P(3, 2, 3), Q (4, 6, 2) and R (7, 3, 2) be the 
vertices of ?PQR. Then, the angle ?QPR is  
 (1) 
6
?
 (2) 
1
7
cos
18
?
??
??
??
 
 (3) 
1
1
cos
18
?
??
??
??
 (4) 
3
?
 
Ans. (4) 
Sol. 
?
P(3, 2, 3)
R(7, 3, 2) Q(4, 6, 2)
 
 Direction ratio of PR = (4, 1, -1) 
 Direction ratio of PQ = (1, 4, -1) 
 Now, 
4 4 1
cos
.
18 18
??
?? 
 
3
?
?? 
4. If the mean and variance of five observations are 
24
5
 and 
194
25
 respectively and the mean of first 
four observations is 
7
,
2
 then the variance of the 
first four observations in equal to  
 (1) 
4
5
 (2) 
77
12
 (3) 
5
4
 (4) 
105
4
 
Ans. (3) 
Sol. 
2
24 194
X;
5 25
? ? ? 
 Let first four observation be x
1
, x
2
, x
3
, x
4
 
 Here, 
1 2 3 4 5
x x x x x 24
......(1)
55
? ? ? ?
? 
 Also, 
1 2 3 4
x x x x 7
42
? ? ?
? 
 
1 2 3 4
x x x x 14 ? ? ? ? ? 
 
 
 
 Now from eqn -1 
 x
5
 = 10 
 Now, 
2
194
25
?? 
 
22222
1 2 3 4 5
xxxxx 576 194
5 25 25
????
?? 
 
2222
1 2 3 4
x x x x 54 ? ? ? ? ? 
 Now, variance of first 4 observations 
 Var = 
2
44
2
ii
i 1 i 1
xx
44
??
??
??
?? ?
??
??
??
??
 
  
54 49
44
?? = 
5
4
 
5. The function f(x) = 
2
3
2x 3(x) ,x , ?? has  
 (1)  exactly one point of local minima and no   
  point of local maxima 
 (2)  exactly one point of local maxima and no   
       point of local minima  
 (3)  exactly one point of local maxima and 
 exactly one point of local minima  
 (4)  exactly two points of local maxima and 
 exactly one point of local minima  
Ans. (3) 
Sol. 
2
3
f (x) 2x 3(x) ?? 
 
1
3
f '(x) 2 2x
?
?? 
 = 
1
3
1
21
x
??
??
?
??
??
 
 = 
1
3
1
3
x1
2
x
??
?
??
??
??
??
 
 
+
–1
M
0
m
– +
 
 So, maxima (M) at x = -1 & minima(m) at x = 0 
 
6. Let r and ? respectively be the modulus and 
amplitude of the complex number 
5
z 2 i 2tan
8
? ??
??
??
??
, then (r, ) ? is equal to  
 (1) 
33
2sec ,
88
?? ??
??
??
 
 (2) 
35
2sec ,
88
?? ??
??
??
 
 (3) 
53
2sec ,
88
?? ??
??
??
 
 (4) 
11 11
2sec ,
88
?? ??
??
??
 
Ans. (1) 
Sol. z = 
5
2 i 2tan
8
? ??
?
??
??
 = x + iy (let) 
 r = 
2 2 1
y
x y & tan
x
?
? ? ? 
 r = 
2
2
5
(2) 2tan
8
? ??
?
??
??
 
 
= 
53
2sec 2sec
88
?? ??
? ? ?
??
??
 
 = 2 sec 
3
8
?
 
 & 
1
5
2tan
8
tan
2
?
? ??
?
??
??
??
??
??
??
 
 =
1
5
tan tan
8
?
?? ? ??
??
?? ??
?? ??
 
 = 
3
8
?
 
7. The sum of the solutions x ? of the equation 
3
66
3cos2x cos 2x
cos x sin x
?
?
 = x
3
 – x
2
 + 6  is  
 (1) 0 (2) 1 
 (3) –1  (4) 3 
 
 
 
Ans. (3) 
Sol. 
3
66
3cos2x cos 2x
cos x sin x
?
?
= x
3
 – x
2
 + 6 
 
2
22
cos2x (3 cos 2x)
cos2x (1 sin xcos x)
?
?
?
= x
3
 – x
2
 + 6 
 
2
2
4(3 cos 2x)
(4 sin 2x)
?
?
?
= x
3
 – x
2
 + 6 
 
2
2
4(3 cos 2x)
(3 cos 2x)
?
?
?
 = x
3
 – x
2
 + 6 
 x
3
 – x
2
 + 2 = 0 ? (x + 1)(x
2
 – 2x + 2) = 0  
 so, sum of real solutions = –1 
8. Let OA a,OB 12a 4b ? ? ? and OC b, ? where O 
is the origin. If S is the parallelogram with adjacent 
sides OA and OC, then  
 
area of the quadrilateral OABC
area of S
 is equal to ___ 
 (1) 6 (2) 10 
 (3) 7  (4) 8 
Ans. (4) 
Sol. 
A
B
C O
a
b
12a + 4b
 
 Area of parallelogram,  S a b ?? 
 Area of quadrilateral =Area( ?OAB)+Area ( ?OBC) 
 = 
? ?
1
a (12a 4b) b (12a 4b)
2
? ? ? ? ? 
 = 8(a b) ? 
 Ratio = 
8(a b)
(a b)
?
?
 = 8    
9. If log
e
 a, log
e
 b, log
e
 c are in an A.P. and log
e
 a – 
log
e
2b, log
e
2b – log
e
3c, log
e
3c – log
e
 a are also in 
an A.P, then a : b : c is equal to   
 (1) 9 : 6 : 4 (2) 16 : 4 : 1 
 (3) 25 : 10 : 4 (4) 6 : 3 : 2 
Ans. (1) 
Sol. log
e
a, log
e
b, log
e
c   are in A.P. 
 ? b
2
 = ac   …..(i) 
 Also 
 
eee
a 2b 3c
log ,log ,log
2b 3c a
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
   are in A.P. 
 
2
2b a 3c
3c 2b a
??
??
??
??
 
 
b3
c2
? 
 Putting in eq. (i)   b
2
 = a × 
2b
3
 
  
a3
b2
? 
  a  : b : c = 9 : 6 : 4 
10. If 
33
22
33
sin x cos x
dx A cos tanx sin B cos sin cotx C,
sin xcos x sin(x )
?
? ? ? ? ? ? ? ? ?
??
?
 
where C is the integration constant, then AB is 
equal to  
 (1) 4cosec(2 ) ? (2) 4sec ? 
 (3) 2sec ? (4) 8cosec(2 ) ? 
Ans. (4) 
Sol. 
33
22
33
sin x cos x
dx
sin xcos x sin(x )
?
??
?
 
 I = 
33
22
33
sin x cos x
dx
sin xcos x (sinxcos cosxsin )
?
? ? ?
?
 
= 
33
22
33
22
22
sin x cos x
dx dx
sin xcos x tanxcos sin sin xcos x cos cotxsin
?
? ? ? ? ? ?
??
= 
22
sec x cosec x
dx dx
tanxcos sin cos cotxsin
?
? ? ? ? ? ?
??
  
 I = I
1
 + I
2 ……
 {Let} 
 For I
1
, let    tan x cos ?  – sin ? = t
2
 
  
2
2t dt
sec xdx
cos
?
?
 
 For I
2
 ,   let 
2
cos cotxsin z ? ? ? ? 
   
2
2z dz
cosec x dx
sin
?
?
 
 
 
 
 
  I = I
1
 + I
2
 
  = 
2t dt 2z dz
cos t sin z
?
??
??
 
  = 
2t 2z
cos sin
?
??
 
 = 2sec tan xcos sin 2cosec cos cotxsin ? ? ? ? ? ? ? ? ? 
 Comparing  
  AB = 8cosec2 ? 
11. The distance of the point (2, 3) from the line 2x – 
3y + 28 = 0, measured parallel to the line 
3x y 1 0, ? ? ? is equal to  
 (1) 42 (2) 63 
 (3) 3 4 2 ?  (4) 4 6 3 ? 
Ans. (4) 
Sol. 
P
2x – 3y + 28 = 0
r
A
(2, 3)
 
 Writing P in terms of parametric co-ordinates 2 + r 
cos ?, 3 + r sin ? as tan ??= 3 
 
r 3r
P(2 ,3 )
22
?? 
 P must satisfy 2x – 3y + 28 = 0 
 So, 
r 3r
2(2 ) 3(3 ) 28 0
22
? ? ? ? ? 
 We find r = 4 6 3 ? 
12. If sin
e
y
log |x|
x2
? ??
??
??
??
 is the solution of the 
differential equation x cos 
y dy y
ycos x
x dx x
? ? ? ?
??
? ? ? ?
? ? ? ?
 
and y(1) = ,
3
?
 then 
2
? is equal to  
 (1) 3 (2) 12 
 (3) 4  (4) 9 
Ans. (1) 
Sol. Differential equation :– 
  
y dy y
xcos ycos x
x dx x
?? 
  
y dy
cos x y x
x dx
??
??
??
??
 
 Divide both sides by x
2
 
  
2
dy
xy
y1
dx
cos
xx x
??
?
??
?
??
??
??
??
 
 Let 
y
t
x
? 
  
dt 1
cos t
dx x
??
?
??
??
 
  
1
cost dt dx
x
? 
 Integrating both sides  
  sin t = ln | x | + c  
  
y
sin ln x c
x
?? 
 Using y(1) = ,
3
?
 we get c = 
3
2
 
 So, 
2
33 ? ? ? ? ? 
13. If each term of a geometric progression a
1
, a
2
, a
3
,… 
with 
1
1
a
8
? and 
21
a a , ? is the arithmetic mean of 
the next two terms and S
n
 = a
1
 + a
2
 + …+a
n
, then 
S
20
 – S
18
 is equal to  
 (1) 2
15 
(2) –2
18 
 (3) 2
18  
(4) –2
15 
Ans. (4) 
Sol. Let r’th term of the GP be ar
n–1
. Given,  
  2a
r
 = a
r+1
 + a
r+2
 
  2ar
n–1
 = ar
n
 + ar
n+1
 
  
2
1r
r
?? 
  r
2
 + r – 2 = 0 
 
 
 
 Hence, we get, r = – 2 (as r ? 1) 
 So, S
20
 – S
18
 = (Sum upto 20 terms) – (Sum upto 
18 terms) = T
19
 + T
20
 
  T
19
 + T
20
 = ar
18
 (1 + r) 
 Putting the values a = 
1
8
 and r = – 2;  
 we get T
19
 + T
20
 = –2
15 
14. Let A be the point of intersection of the lines 3x + 
2y = 14, 5x – y = 6 and B be the point of 
intersection of the lines 4x + 3y = 8, 6x + y = 5. 
The distance of the point P(5,  –2) from the line 
AB is  
 (1) 
13
2
 (2) 8 (3) 
5
2
 (4) 6 
Ans. (4) 
Sol. Solving lines L
1
 (3x + 2y = 14) and L
2
 (5x – y = 6) 
to get A(2, 4) and solving lines L
3
 (4x + 3y = 8) 
and L
4
 (6x + y = 5) to get B
1
,2 .
2
??
??
??
 
 Finding Eqn. of AB : 4x – 3y + 4 = 0 
 Calculate distance PM  
 ? 
4(5) 3( 2) 4
6
5
? ? ?
? 
15. Let x = 
m
n
 (m, n are co-prime natural numbers) be 
a solution of the equation 
? ?
1
1
cos 2sin x
9
?
? and let 
, ( ) ? ? ? ? ? be the roots of the equation mx
2
 – nx – 
m + n = 0. Then the point ( , ) ?? lies on the line  
 (1) 3x + 2y = 2 (2) 5x – 8y = –9 
 (3) 3x – 2y = –2 (4) 5x + 8y = 9 
Ans. (4) 
Sol. Assume sin
–1
 x = ? 
   
1
cos(2 )
9
?? 
   
2
sin
3
? ? ? 
 as m and n are co-prime natural numbers,  
   
2
x
3
? 
 i.e. m = 2, n = 3 
 So, the quadratic equation becomes 2x
2
 – 3x + 1 = 
0 whose roots are 
1
1,
2
? ? ? ? 
 
1
1,
2
??
??
??
 lies on 5x + 8y = 9 
16. The function f(x) = 
2
x
,x
x 6x 16
?
??
 –{–2, 8}  
 (1) decreases in (–2, 8) and increases in        
      ( , 2) (8, ) ? ? ? ? ?  
 (2) decreases in ( , 2) ( 2,8) (8, ) ? ? ? ? ? ? ? 
 (3) decreases in ( , 2) ? ? ? and increases in (8, ) ? 
 (4) increases in ( , 2) ( 2,8) (8, ) ? ? ? ? ? ? ? 
Ans. (2) 
Sol. f(x) = 
2
x
x 6x 16 ??
 
 Now,  
 
2
22
(x 16)
f'(x)
(x 6x 16)
??
?
??
 
 f'(x) 0 ? 
 Thus f(x) is decreasing in  
  ( , – 2) ( 2 , 8 ) ( 8 , ) ? ? ? ? ? ? 
17. Let y = 
2
e 2
1x
log ,
1x
?? ?
??
?
??
 –1 < x < 1. Then at x = 
1
,
2
the value of 225(y' y") ? is equal to  
 (1) 732 (2) 746 
 (3) 742  (4) 736 
Ans. (4) 
Sol. y = 
2
e 2
1x
log
1x
?? ?
??
?
??
 
 
4
dy 4x
y'
dx 1x
?
??
?
  
 Again,  
 
24
2 4 2
d y 4(1 3x )
y"
dx (1 x )
??
??
?
 
 Again 
 
4
4 4 2
4x 4(1 3x )
y' y"
1 x (1 x )
??
? ? ?
??
 
 at  
1
x,
2
?