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Page 1 of 19 
 
                                                              SAMPLE QUESTION PAPER 
MARKING SCHEME 
CLASS XII 
MATHEMATICS (CODE-041) 
                                                       SECTION: A (Solution of MCQs of 1 Mark each) 
Q no.    ANS                            HINTS/SOLUTION 
1 (d) 
2
0 1 1 0
, .
1 0 0 1
A A
? ? ? ?
? ?
? ? ? ?
? ? ? ?
 
2 (d) 
? ?
1
1 1
. A B B A
?
? ?
? ? ? 
3 (b) 
3 0 1
1
Area 3 0 1 ,
2
0 1 k
?
? given that the area 9 . sq unit ? 
3 0 1
1
9 3 0 1 ;
2
0 1 k
?
? ? ? expanding along 
2
, C we get 3. k ? ? ? 
4 (a) Since, f is continuous at 0 x ? , 
therefore,  ? ? . . . . 0 . L H L R H L f a finite quantity ? ? ? 
? ? ? ? ? ?
0 0
lim lim 0
x x
f x f x f
? ?
? ?
? ? 
0 0
lim lim 3 3 3.
x x
kx
k
x
? ?
? ?
?
? ? ? ? ? ? 
5 (d) 
Vectors
 
? ?
2 3 6 &6 9 18 i j k i j k ? ? ? ?
? ? ? ?
 
are parallel and the fixed point 
?
i j k ? ?
? ?
 on the 
line 
? ?
? ?
2 3 6 r i j k i j k ? ? ? ? ? ? ?
?
? ? ? ?
 does not satisfy the other line 
? ?
? ?
2 6 9 18 ; r i j k i j k ? ? ? ? ? ? ?
?
? ? ? ?
 where & ? ? are scalars.
 
6 (c) 
The degree of the differential equation 
3
2
2
2
2
1 2
? ?
? ?
? ?
? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
dy d y
is
dx dx
 
7 (b) 
? ?   Z px qy i ? ? ? ? ? 
At  ? ? 3,0 , ? ? 3 Z p ii ? ? ? ? and  at ? ? 1,1 , ? ? Z p q iii ? ? ? ? ? ? 
From ? ? ? ?
 
 3 2 . & , ii iii p p q p q ? ? ? ? 
Page 2


Page 1 of 19 
 
                                                              SAMPLE QUESTION PAPER 
MARKING SCHEME 
CLASS XII 
MATHEMATICS (CODE-041) 
                                                       SECTION: A (Solution of MCQs of 1 Mark each) 
Q no.    ANS                            HINTS/SOLUTION 
1 (d) 
2
0 1 1 0
, .
1 0 0 1
A A
? ? ? ?
? ?
? ? ? ?
? ? ? ?
 
2 (d) 
? ?
1
1 1
. A B B A
?
? ?
? ? ? 
3 (b) 
3 0 1
1
Area 3 0 1 ,
2
0 1 k
?
? given that the area 9 . sq unit ? 
3 0 1
1
9 3 0 1 ;
2
0 1 k
?
? ? ? expanding along 
2
, C we get 3. k ? ? ? 
4 (a) Since, f is continuous at 0 x ? , 
therefore,  ? ? . . . . 0 . L H L R H L f a finite quantity ? ? ? 
? ? ? ? ? ?
0 0
lim lim 0
x x
f x f x f
? ?
? ?
? ? 
0 0
lim lim 3 3 3.
x x
kx
k
x
? ?
? ?
?
? ? ? ? ? ? 
5 (d) 
Vectors
 
? ?
2 3 6 &6 9 18 i j k i j k ? ? ? ?
? ? ? ?
 
are parallel and the fixed point 
?
i j k ? ?
? ?
 on the 
line 
? ?
? ?
2 3 6 r i j k i j k ? ? ? ? ? ? ?
?
? ? ? ?
 does not satisfy the other line 
? ?
? ?
2 6 9 18 ; r i j k i j k ? ? ? ? ? ? ?
?
? ? ? ?
 where & ? ? are scalars.
 
6 (c) 
The degree of the differential equation 
3
2
2
2
2
1 2
? ?
? ?
? ?
? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
dy d y
is
dx dx
 
7 (b) 
? ?   Z px qy i ? ? ? ? ? 
At  ? ? 3,0 , ? ? 3 Z p ii ? ? ? ? and  at ? ? 1,1 , ? ? Z p q iii ? ? ? ? ? ? 
From ? ? ? ?
 
 3 2 . & , ii iii p p q p q ? ? ? ? 
Page 2 of 19 
 
8 (a) 
Given, ABCD is a rhombus whose diagonals bisect each other. EA EC ?
? ? ? ? ? ? ? ?
and 
EB ED ?
? ? ? ? ? ? ? ?
but since they are opposite to each other so they are of opposite signs 
EA EC ? ? ?
? ? ? ? ? ? ? ?
and . EB ED ? ?
? ?? ? ? ?? ?
 
 
 
 
 
 
 
 
 
? ? ..... EA EC O i ? ? ?
? ? ? ? ? ? ? ? ?
and ? ? .... EB ED O ii ? ?
? ? ? ? ? ? ? ? ?
 
Adding (i) and (ii), we get . EA EB EC ED O ? ? ? ?
? ? ? ? ? ?? ? ? ? ? ? ?? ? ? ?
 
9 (b) 
? ? ? ?
2
cos 3
sin 2 1
x
f x e n x ? ? 
? ?
? ?
? ? ? ?
2
cos 3
sin 2 1
x
f x e n x
?
? ? ? ? 
2
cos 3
( ) sin (2 1) ? ? ? ?
x
f x e n x
  
2
cos 3
( ) ( )
, sin (2 1) 0
?
? ? ?
? ?
?
?
x
f x f x
So e n x dx
?
?
  
10 (b) Matrix A is a skew symmetric matrix of odd order. 0. A ? ? 
11 (c) 
We observe, ? ? 0,0 does not satisfy the inequality 1 x y ? ?
 
So, the half plane represented by the above inequality will not contain origin 
therefore, it will not contain the shaded feasible region. 
12 (b) 
Vector component of a
?
along b
?
?
? ?
2
. 18
3 4 .
25
a b
b j k
b
? ?
? ?
? ? ?
? ?
? ?
? ?
? ?
?
?
?
 
13 (d) 
? ? ? ? ? ? ? ?
2 2 2 2
3 6 6 8
2 2 2 2 2 2 2 . adj A A A A ? ? ? ? ? ? ? 
14 (d) Method 1:  
Let , , A B C be the respective events of solving the problem. Then, ? ? ? ?
1 1
,
2 3
P A P B ? ? 
and ? ?
1
.
4
P C ? Here, , , A B C are independent events. 
Problem is solved if at least one of them solves the problem.  
Required probability is ? ? ? ? ? ? ? ?
1 P A B C P A P B P C ? ? ? ? ? 
 
 
 
 
 
Page 3


Page 1 of 19 
 
                                                              SAMPLE QUESTION PAPER 
MARKING SCHEME 
CLASS XII 
MATHEMATICS (CODE-041) 
                                                       SECTION: A (Solution of MCQs of 1 Mark each) 
Q no.    ANS                            HINTS/SOLUTION 
1 (d) 
2
0 1 1 0
, .
1 0 0 1
A A
? ? ? ?
? ?
? ? ? ?
? ? ? ?
 
2 (d) 
? ?
1
1 1
. A B B A
?
? ?
? ? ? 
3 (b) 
3 0 1
1
Area 3 0 1 ,
2
0 1 k
?
? given that the area 9 . sq unit ? 
3 0 1
1
9 3 0 1 ;
2
0 1 k
?
? ? ? expanding along 
2
, C we get 3. k ? ? ? 
4 (a) Since, f is continuous at 0 x ? , 
therefore,  ? ? . . . . 0 . L H L R H L f a finite quantity ? ? ? 
? ? ? ? ? ?
0 0
lim lim 0
x x
f x f x f
? ?
? ?
? ? 
0 0
lim lim 3 3 3.
x x
kx
k
x
? ?
? ?
?
? ? ? ? ? ? 
5 (d) 
Vectors
 
? ?
2 3 6 &6 9 18 i j k i j k ? ? ? ?
? ? ? ?
 
are parallel and the fixed point 
?
i j k ? ?
? ?
 on the 
line 
? ?
? ?
2 3 6 r i j k i j k ? ? ? ? ? ? ?
?
? ? ? ?
 does not satisfy the other line 
? ?
? ?
2 6 9 18 ; r i j k i j k ? ? ? ? ? ? ?
?
? ? ? ?
 where & ? ? are scalars.
 
6 (c) 
The degree of the differential equation 
3
2
2
2
2
1 2
? ?
? ?
? ?
? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
dy d y
is
dx dx
 
7 (b) 
? ?   Z px qy i ? ? ? ? ? 
At  ? ? 3,0 , ? ? 3 Z p ii ? ? ? ? and  at ? ? 1,1 , ? ? Z p q iii ? ? ? ? ? ? 
From ? ? ? ?
 
 3 2 . & , ii iii p p q p q ? ? ? ? 
Page 2 of 19 
 
8 (a) 
Given, ABCD is a rhombus whose diagonals bisect each other. EA EC ?
? ? ? ? ? ? ? ?
and 
EB ED ?
? ? ? ? ? ? ? ?
but since they are opposite to each other so they are of opposite signs 
EA EC ? ? ?
? ? ? ? ? ? ? ?
and . EB ED ? ?
? ?? ? ? ?? ?
 
 
 
 
 
 
 
 
 
? ? ..... EA EC O i ? ? ?
? ? ? ? ? ? ? ? ?
and ? ? .... EB ED O ii ? ?
? ? ? ? ? ? ? ? ?
 
Adding (i) and (ii), we get . EA EB EC ED O ? ? ? ?
? ? ? ? ? ?? ? ? ? ? ? ?? ? ? ?
 
9 (b) 
? ? ? ?
2
cos 3
sin 2 1
x
f x e n x ? ? 
? ?
? ?
? ? ? ?
2
cos 3
sin 2 1
x
f x e n x
?
? ? ? ? 
2
cos 3
( ) sin (2 1) ? ? ? ?
x
f x e n x
  
2
cos 3
( ) ( )
, sin (2 1) 0
?
? ? ?
? ?
?
?
x
f x f x
So e n x dx
?
?
  
10 (b) Matrix A is a skew symmetric matrix of odd order. 0. A ? ? 
11 (c) 
We observe, ? ? 0,0 does not satisfy the inequality 1 x y ? ?
 
So, the half plane represented by the above inequality will not contain origin 
therefore, it will not contain the shaded feasible region. 
12 (b) 
Vector component of a
?
along b
?
?
? ?
2
. 18
3 4 .
25
a b
b j k
b
? ?
? ?
? ? ?
? ?
? ?
? ?
? ?
?
?
?
 
13 (d) 
? ? ? ? ? ? ? ?
2 2 2 2
3 6 6 8
2 2 2 2 2 2 2 . adj A A A A ? ? ? ? ? ? ? 
14 (d) Method 1:  
Let , , A B C be the respective events of solving the problem. Then, ? ? ? ?
1 1
,
2 3
P A P B ? ? 
and ? ?
1
.
4
P C ? Here, , , A B C are independent events. 
Problem is solved if at least one of them solves the problem.  
Required probability is ? ? ? ? ? ? ? ?
1 P A B C P A P B P C ? ? ? ? ? 
 
 
 
 
 
Page 3 of 19 
 
1 1 1 1 3
1 1 1 1 1 .
2 3 4 4 4
? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
 
Method 2: 
The problem will be solved if one or more of them can solve the problem. The probability is 
? ? ? ? ? ? ? ? ? ? ? ? ? ? P ABC P ABC P ABC P ABC P ABC P ABC P ABC ? ? ? ? ? ?
1 2 3 1 1 3 1 2 1 1 1 3 1 2 1 1 1 1 1 1 1 3
. . . . . . . . . . . . . . .
2 3 4 2 3 4 2 3 4 2 3 4 2 3 4 2 3 4 2 3 4 4
? ? ? ? ? ? ? ? 
Method 3: 
Let us think quantitively. Let us assume that there are 100 questions given to . A  A
solves 
1
100 50
2
? ? questions then remaining 50 questions is given to B and B solves 
1
50 16.67
3
? ? questions . Remaining 
2
50
3
? questions is given to C and C solves 
2 1
50 8.33
3 4
? ? ? questions.  
Therefore, number of questions solved is 50 16.67 8.33 75 ? ? ? . 
So, required probability is 
75 3
.
100 4
? 
15 (c) Method 1:  
2
1
0 0 0 .
ydx xdy x
ydx xdy d x y y cx
y y c
? ? ?
? ? ? ? ? ? ? ? ? ?
? ?
? ?
 
Method 2:  
0 ydx xdy ydx xdy ? ? ? ?
dy dx
y x
? ? ; on integrating 
dy dx
y x
?
? ?
 
log log log
e e e
y x c ? ?  
since , , 0 x y c ? , we write  log log log
e e e
y x c ? ?  . y cx ? ? 
16 (d) Dot product of two mutually perpendicular vectors is zero.
? ? 2 3 1 2 1 0 8. ? ? ? ? ? ? ? ? ? ? ? 
17 (c) Method 1:  
? ?
2 , 0
0 , 0
x x
f x x x
x
? ?
? ? ?
?
?
?
 
 
 
 
 
 
 
There is a sharp corner at 0 x ? , so ? ? f x is not differentiable at 0 x ? . 
Method 2:  
 
 
 
 
 
 
 
Page 4


Page 1 of 19 
 
                                                              SAMPLE QUESTION PAPER 
MARKING SCHEME 
CLASS XII 
MATHEMATICS (CODE-041) 
                                                       SECTION: A (Solution of MCQs of 1 Mark each) 
Q no.    ANS                            HINTS/SOLUTION 
1 (d) 
2
0 1 1 0
, .
1 0 0 1
A A
? ? ? ?
? ?
? ? ? ?
? ? ? ?
 
2 (d) 
? ?
1
1 1
. A B B A
?
? ?
? ? ? 
3 (b) 
3 0 1
1
Area 3 0 1 ,
2
0 1 k
?
? given that the area 9 . sq unit ? 
3 0 1
1
9 3 0 1 ;
2
0 1 k
?
? ? ? expanding along 
2
, C we get 3. k ? ? ? 
4 (a) Since, f is continuous at 0 x ? , 
therefore,  ? ? . . . . 0 . L H L R H L f a finite quantity ? ? ? 
? ? ? ? ? ?
0 0
lim lim 0
x x
f x f x f
? ?
? ?
? ? 
0 0
lim lim 3 3 3.
x x
kx
k
x
? ?
? ?
?
? ? ? ? ? ? 
5 (d) 
Vectors
 
? ?
2 3 6 &6 9 18 i j k i j k ? ? ? ?
? ? ? ?
 
are parallel and the fixed point 
?
i j k ? ?
? ?
 on the 
line 
? ?
? ?
2 3 6 r i j k i j k ? ? ? ? ? ? ?
?
? ? ? ?
 does not satisfy the other line 
? ?
? ?
2 6 9 18 ; r i j k i j k ? ? ? ? ? ? ?
?
? ? ? ?
 where & ? ? are scalars.
 
6 (c) 
The degree of the differential equation 
3
2
2
2
2
1 2
? ?
? ?
? ?
? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
dy d y
is
dx dx
 
7 (b) 
? ?   Z px qy i ? ? ? ? ? 
At  ? ? 3,0 , ? ? 3 Z p ii ? ? ? ? and  at ? ? 1,1 , ? ? Z p q iii ? ? ? ? ? ? 
From ? ? ? ?
 
 3 2 . & , ii iii p p q p q ? ? ? ? 
Page 2 of 19 
 
8 (a) 
Given, ABCD is a rhombus whose diagonals bisect each other. EA EC ?
? ? ? ? ? ? ? ?
and 
EB ED ?
? ? ? ? ? ? ? ?
but since they are opposite to each other so they are of opposite signs 
EA EC ? ? ?
? ? ? ? ? ? ? ?
and . EB ED ? ?
? ?? ? ? ?? ?
 
 
 
 
 
 
 
 
 
? ? ..... EA EC O i ? ? ?
? ? ? ? ? ? ? ? ?
and ? ? .... EB ED O ii ? ?
? ? ? ? ? ? ? ? ?
 
Adding (i) and (ii), we get . EA EB EC ED O ? ? ? ?
? ? ? ? ? ?? ? ? ? ? ? ?? ? ? ?
 
9 (b) 
? ? ? ?
2
cos 3
sin 2 1
x
f x e n x ? ? 
? ?
? ?
? ? ? ?
2
cos 3
sin 2 1
x
f x e n x
?
? ? ? ? 
2
cos 3
( ) sin (2 1) ? ? ? ?
x
f x e n x
  
2
cos 3
( ) ( )
, sin (2 1) 0
?
? ? ?
? ?
?
?
x
f x f x
So e n x dx
?
?
  
10 (b) Matrix A is a skew symmetric matrix of odd order. 0. A ? ? 
11 (c) 
We observe, ? ? 0,0 does not satisfy the inequality 1 x y ? ?
 
So, the half plane represented by the above inequality will not contain origin 
therefore, it will not contain the shaded feasible region. 
12 (b) 
Vector component of a
?
along b
?
?
? ?
2
. 18
3 4 .
25
a b
b j k
b
? ?
? ?
? ? ?
? ?
? ?
? ?
? ?
?
?
?
 
13 (d) 
? ? ? ? ? ? ? ?
2 2 2 2
3 6 6 8
2 2 2 2 2 2 2 . adj A A A A ? ? ? ? ? ? ? 
14 (d) Method 1:  
Let , , A B C be the respective events of solving the problem. Then, ? ? ? ?
1 1
,
2 3
P A P B ? ? 
and ? ?
1
.
4
P C ? Here, , , A B C are independent events. 
Problem is solved if at least one of them solves the problem.  
Required probability is ? ? ? ? ? ? ? ?
1 P A B C P A P B P C ? ? ? ? ? 
 
 
 
 
 
Page 3 of 19 
 
1 1 1 1 3
1 1 1 1 1 .
2 3 4 4 4
? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
 
Method 2: 
The problem will be solved if one or more of them can solve the problem. The probability is 
? ? ? ? ? ? ? ? ? ? ? ? ? ? P ABC P ABC P ABC P ABC P ABC P ABC P ABC ? ? ? ? ? ?
1 2 3 1 1 3 1 2 1 1 1 3 1 2 1 1 1 1 1 1 1 3
. . . . . . . . . . . . . . .
2 3 4 2 3 4 2 3 4 2 3 4 2 3 4 2 3 4 2 3 4 4
? ? ? ? ? ? ? ? 
Method 3: 
Let us think quantitively. Let us assume that there are 100 questions given to . A  A
solves 
1
100 50
2
? ? questions then remaining 50 questions is given to B and B solves 
1
50 16.67
3
? ? questions . Remaining 
2
50
3
? questions is given to C and C solves 
2 1
50 8.33
3 4
? ? ? questions.  
Therefore, number of questions solved is 50 16.67 8.33 75 ? ? ? . 
So, required probability is 
75 3
.
100 4
? 
15 (c) Method 1:  
2
1
0 0 0 .
ydx xdy x
ydx xdy d x y y cx
y y c
? ? ?
? ? ? ? ? ? ? ? ? ?
? ?
? ?
 
Method 2:  
0 ydx xdy ydx xdy ? ? ? ?
dy dx
y x
? ? ; on integrating 
dy dx
y x
?
? ?
 
log log log
e e e
y x c ? ?  
since , , 0 x y c ? , we write  log log log
e e e
y x c ? ?  . y cx ? ? 
16 (d) Dot product of two mutually perpendicular vectors is zero.
? ? 2 3 1 2 1 0 8. ? ? ? ? ? ? ? ? ? ? ? 
17 (c) Method 1:  
? ?
2 , 0
0 , 0
x x
f x x x
x
? ?
? ? ?
?
?
?
 
 
 
 
 
 
 
There is a sharp corner at 0 x ? , so ? ? f x is not differentiable at 0 x ? . 
Method 2:  
 
 
 
 
 
 
 
Page 4 of 19 
 
 
  Section –B 
   [This section comprises of solution of very short answer type questions (VSA) of 2 marks each] 
? ? ? ?
' '
0 0 & 0 2 Lf Rf ? ? ; so, the function is not differentiable at 0 x ? 
For ? ? ? ? 0, 2 x f x x
  
(linear function) & when ? ? ? ? 0, 0 x f x
 
(constant function) 
Hence ? ? f x is differentiable when ? ? ? ? ,0 0, . x ? ? ? ? ? 
18 (d) 
We know, 
2 2 2 2
2 2 2
1 1 1 1
1 1 3 1 l m n
c c c c
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
3. c ? ? ? 
19 (a) 
? ? ? ? ? ? ? ?
3 2
1 3
d
f x x x
dx
? ? ?  
Assertion : ? ? f x has a minimum at 1 x ? is true as  
? ? ? ? ? ? 0, 1 ,1
d
f x x h
dx
? ? ? ? and ? ? ? ? ? ? 0, 1,1 ;
d
f x x h
dx
? ? ? ? where,  
' ' h is an infinitesimally small positive quantity , which is in accordance with  
the Reason statement.       
20 (d) Assertion is false. As element 4 has no image under , f so relation f is not a function. 
Reason is true. The given function 
? ? ? ? : 1,2,3 , , , f x y z p ?
 
is one – one, as for each 
? ? 1,2,3 , a ? there is different image in ? ? , , , x y z p under . f 
21 
1 1 1 1
33 3 3 3
sin cos sin cos 6 sin cos sin sin
5 5 5 2 5
? ? ? ?
? ? ? ? ? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ?
? ? ? ? ?
? 
3
.
2 5 10
? ? ?
? ? ? ? 
1 
 
 
1 
21 OR 
? ?
2 2
1 4 1 3 5 3 5 x x x ? ? ? ? ? ? ? ? ? ? 
5, 3 3, 5 . x
? ? ? ?
? ?
?
? ?
? ? ?
? So, required domain is 5, 3 3, 5 .
? ? ? ?
? ? ?
? ? ? ?
 
1 
1 
22 
                ? ? ? ? ? ?
'
1
x x
f x x e f x e x ? ? ? ? 
When ? ? ? ? ? ?
'
1, , 1 0 & 0 0
x
f x x e x ? ? ? ? ? ? ? ?  ? ? ? f x increases in this interval. 
or, we can write  ? ? ? ? ? ?
'
1
x x
f x x e f x e x ? ? ? ? 
For ? ? f x to be increasing, we have ? ? ? ?
'
1 0 1
x
f x e x x ? ? ? ? ? ?  as 0,
x
e x ? ? ? ? 
Hence, the required interval where ? ? f x increases  is ? ? 1, . ? ? 
1 
1 
1
2
 
1 
1
2
 
23 
Method 1 : ? ?
2
1
4 2 1
f x
x x
?
? ?
 ,
  
 
 
Page 5


Page 1 of 19 
 
                                                              SAMPLE QUESTION PAPER 
MARKING SCHEME 
CLASS XII 
MATHEMATICS (CODE-041) 
                                                       SECTION: A (Solution of MCQs of 1 Mark each) 
Q no.    ANS                            HINTS/SOLUTION 
1 (d) 
2
0 1 1 0
, .
1 0 0 1
A A
? ? ? ?
? ?
? ? ? ?
? ? ? ?
 
2 (d) 
? ?
1
1 1
. A B B A
?
? ?
? ? ? 
3 (b) 
3 0 1
1
Area 3 0 1 ,
2
0 1 k
?
? given that the area 9 . sq unit ? 
3 0 1
1
9 3 0 1 ;
2
0 1 k
?
? ? ? expanding along 
2
, C we get 3. k ? ? ? 
4 (a) Since, f is continuous at 0 x ? , 
therefore,  ? ? . . . . 0 . L H L R H L f a finite quantity ? ? ? 
? ? ? ? ? ?
0 0
lim lim 0
x x
f x f x f
? ?
? ?
? ? 
0 0
lim lim 3 3 3.
x x
kx
k
x
? ?
? ?
?
? ? ? ? ? ? 
5 (d) 
Vectors
 
? ?
2 3 6 &6 9 18 i j k i j k ? ? ? ?
? ? ? ?
 
are parallel and the fixed point 
?
i j k ? ?
? ?
 on the 
line 
? ?
? ?
2 3 6 r i j k i j k ? ? ? ? ? ? ?
?
? ? ? ?
 does not satisfy the other line 
? ?
? ?
2 6 9 18 ; r i j k i j k ? ? ? ? ? ? ?
?
? ? ? ?
 where & ? ? are scalars.
 
6 (c) 
The degree of the differential equation 
3
2
2
2
2
1 2
? ?
? ?
? ?
? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
dy d y
is
dx dx
 
7 (b) 
? ?   Z px qy i ? ? ? ? ? 
At  ? ? 3,0 , ? ? 3 Z p ii ? ? ? ? and  at ? ? 1,1 , ? ? Z p q iii ? ? ? ? ? ? 
From ? ? ? ?
 
 3 2 . & , ii iii p p q p q ? ? ? ? 
Page 2 of 19 
 
8 (a) 
Given, ABCD is a rhombus whose diagonals bisect each other. EA EC ?
? ? ? ? ? ? ? ?
and 
EB ED ?
? ? ? ? ? ? ? ?
but since they are opposite to each other so they are of opposite signs 
EA EC ? ? ?
? ? ? ? ? ? ? ?
and . EB ED ? ?
? ?? ? ? ?? ?
 
 
 
 
 
 
 
 
 
? ? ..... EA EC O i ? ? ?
? ? ? ? ? ? ? ? ?
and ? ? .... EB ED O ii ? ?
? ? ? ? ? ? ? ? ?
 
Adding (i) and (ii), we get . EA EB EC ED O ? ? ? ?
? ? ? ? ? ?? ? ? ? ? ? ?? ? ? ?
 
9 (b) 
? ? ? ?
2
cos 3
sin 2 1
x
f x e n x ? ? 
? ?
? ?
? ? ? ?
2
cos 3
sin 2 1
x
f x e n x
?
? ? ? ? 
2
cos 3
( ) sin (2 1) ? ? ? ?
x
f x e n x
  
2
cos 3
( ) ( )
, sin (2 1) 0
?
? ? ?
? ?
?
?
x
f x f x
So e n x dx
?
?
  
10 (b) Matrix A is a skew symmetric matrix of odd order. 0. A ? ? 
11 (c) 
We observe, ? ? 0,0 does not satisfy the inequality 1 x y ? ?
 
So, the half plane represented by the above inequality will not contain origin 
therefore, it will not contain the shaded feasible region. 
12 (b) 
Vector component of a
?
along b
?
?
? ?
2
. 18
3 4 .
25
a b
b j k
b
? ?
? ?
? ? ?
? ?
? ?
? ?
? ?
?
?
?
 
13 (d) 
? ? ? ? ? ? ? ?
2 2 2 2
3 6 6 8
2 2 2 2 2 2 2 . adj A A A A ? ? ? ? ? ? ? 
14 (d) Method 1:  
Let , , A B C be the respective events of solving the problem. Then, ? ? ? ?
1 1
,
2 3
P A P B ? ? 
and ? ?
1
.
4
P C ? Here, , , A B C are independent events. 
Problem is solved if at least one of them solves the problem.  
Required probability is ? ? ? ? ? ? ? ?
1 P A B C P A P B P C ? ? ? ? ? 
 
 
 
 
 
Page 3 of 19 
 
1 1 1 1 3
1 1 1 1 1 .
2 3 4 4 4
? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
 
Method 2: 
The problem will be solved if one or more of them can solve the problem. The probability is 
? ? ? ? ? ? ? ? ? ? ? ? ? ? P ABC P ABC P ABC P ABC P ABC P ABC P ABC ? ? ? ? ? ?
1 2 3 1 1 3 1 2 1 1 1 3 1 2 1 1 1 1 1 1 1 3
. . . . . . . . . . . . . . .
2 3 4 2 3 4 2 3 4 2 3 4 2 3 4 2 3 4 2 3 4 4
? ? ? ? ? ? ? ? 
Method 3: 
Let us think quantitively. Let us assume that there are 100 questions given to . A  A
solves 
1
100 50
2
? ? questions then remaining 50 questions is given to B and B solves 
1
50 16.67
3
? ? questions . Remaining 
2
50
3
? questions is given to C and C solves 
2 1
50 8.33
3 4
? ? ? questions.  
Therefore, number of questions solved is 50 16.67 8.33 75 ? ? ? . 
So, required probability is 
75 3
.
100 4
? 
15 (c) Method 1:  
2
1
0 0 0 .
ydx xdy x
ydx xdy d x y y cx
y y c
? ? ?
? ? ? ? ? ? ? ? ? ?
? ?
? ?
 
Method 2:  
0 ydx xdy ydx xdy ? ? ? ?
dy dx
y x
? ? ; on integrating 
dy dx
y x
?
? ?
 
log log log
e e e
y x c ? ?  
since , , 0 x y c ? , we write  log log log
e e e
y x c ? ?  . y cx ? ? 
16 (d) Dot product of two mutually perpendicular vectors is zero.
? ? 2 3 1 2 1 0 8. ? ? ? ? ? ? ? ? ? ? ? 
17 (c) Method 1:  
? ?
2 , 0
0 , 0
x x
f x x x
x
? ?
? ? ?
?
?
?
 
 
 
 
 
 
 
There is a sharp corner at 0 x ? , so ? ? f x is not differentiable at 0 x ? . 
Method 2:  
 
 
 
 
 
 
 
Page 4 of 19 
 
 
  Section –B 
   [This section comprises of solution of very short answer type questions (VSA) of 2 marks each] 
? ? ? ?
' '
0 0 & 0 2 Lf Rf ? ? ; so, the function is not differentiable at 0 x ? 
For ? ? ? ? 0, 2 x f x x
  
(linear function) & when ? ? ? ? 0, 0 x f x
 
(constant function) 
Hence ? ? f x is differentiable when ? ? ? ? ,0 0, . x ? ? ? ? ? 
18 (d) 
We know, 
2 2 2 2
2 2 2
1 1 1 1
1 1 3 1 l m n
c c c c
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
3. c ? ? ? 
19 (a) 
? ? ? ? ? ? ? ?
3 2
1 3
d
f x x x
dx
? ? ?  
Assertion : ? ? f x has a minimum at 1 x ? is true as  
? ? ? ? ? ? 0, 1 ,1
d
f x x h
dx
? ? ? ? and ? ? ? ? ? ? 0, 1,1 ;
d
f x x h
dx
? ? ? ? where,  
' ' h is an infinitesimally small positive quantity , which is in accordance with  
the Reason statement.       
20 (d) Assertion is false. As element 4 has no image under , f so relation f is not a function. 
Reason is true. The given function 
? ? ? ? : 1,2,3 , , , f x y z p ?
 
is one – one, as for each 
? ? 1,2,3 , a ? there is different image in ? ? , , , x y z p under . f 
21 
1 1 1 1
33 3 3 3
sin cos sin cos 6 sin cos sin sin
5 5 5 2 5
? ? ? ?
? ? ? ? ? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ? ?
? ? ? ? ?
? 
3
.
2 5 10
? ? ?
? ? ? ? 
1 
 
 
1 
21 OR 
? ?
2 2
1 4 1 3 5 3 5 x x x ? ? ? ? ? ? ? ? ? ? 
5, 3 3, 5 . x
? ? ? ?
? ?
?
? ?
? ? ?
? So, required domain is 5, 3 3, 5 .
? ? ? ?
? ? ?
? ? ? ?
 
1 
1 
22 
                ? ? ? ? ? ?
'
1
x x
f x x e f x e x ? ? ? ? 
When ? ? ? ? ? ?
'
1, , 1 0 & 0 0
x
f x x e x ? ? ? ? ? ? ? ?  ? ? ? f x increases in this interval. 
or, we can write  ? ? ? ? ? ?
'
1
x x
f x x e f x e x ? ? ? ? 
For ? ? f x to be increasing, we have ? ? ? ?
'
1 0 1
x
f x e x x ? ? ? ? ? ?  as 0,
x
e x ? ? ? ? 
Hence, the required interval where ? ? f x increases  is ? ? 1, . ? ? 
1 
1 
1
2
 
1 
1
2
 
23 
Method 1 : ? ?
2
1
4 2 1
f x
x x
?
? ?
 ,
  
 
 
Page 5 of 19 
 
Let
  
? ?
2
2 2
1 1 3 1 3 3
4 2 1 4 2 4
4 16 4 4 4 4
g x x x x x x
? ? ? ?
? ? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
 
?maximum value of ? ?
4
.
3
f x ? 
Method 2 : ? ? ?
? ?
2
1
,
4 2 1
f x
x x
 
 let
  
? ?
2
4 2 1 g x x x ? ? ? 
? ? ? ? ? ? ? ? ? ? ? ? ? ?
2
' ' "
2
1
8 2 and 0 at also 8 0
4
d d
g x g x x g x x g x g x
dx dx
? ? ? ? ? ? ? ? ? ? 
? ? g x ? is minimum when 
1
4
x ? ? so , ? ? f x is maximum at 
1
4
x ? ? 
?maximum value of  ? ?
? ?
? ? ? ?
? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
2
1 1 4
.
4 3
1 1
4 2 1
4 4
f x f 
Method 3 : ? ?
2
1
4 2 1
f x
x x
?
? ?
 
On differentiating w.r.t x ,we get ? ?
? ?
? ?
? ?
2
2
8 2
' ....
4 2 1
x
f x i
x x
? ?
?
? ?
 
For maxima or minima , we put ? ?
1
' 0 8 2 0
4
f x x x ? ? ? ? ? ? ? . 
Again, differentiating equation (i) w.r.t. x ,we get 
? ?
? ? ? ? ? ? ? ? ? ?
? ?
? ?
? ? ? ? ? ? ? ?
? ?
? ?
? ?
? ? ? ?
? ?
2
2 2
4
2
4 2 1 8 8 2 2 4 2 1 8 2
"
4 2 1
x x x x x x
f x
x x
 
At 
1
4
x ? ? , 
? ?
? ?
? ?
? ?
1
" 0
4
f 
? ? f x is maximum at 
1
.
4
x ? ? 
?maximum value of ? ? f x is 
2
1 1 4
.
4 3
1 1
4 2 1
4 4
f
? ?
? ? ?
? ?
? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
 
Method 4: ? ?
2
1
4 2 1
f x
x x
?
? ?
 
On differentiating w.r.t x ,we get ? ?
? ?
? ?
? ?
2
2
8 2
' ....
4 2 1
x
f x i
x x
? ?
?
? ?
 
For maxima or minima , we put ? ?
1
' 0 8 2 0
4
f x x x ? ? ? ? ? ? ? . 
When
1 1
,
4 4
x h
? ?
? ? ? ?
? ?
? ?
, where ' ' h is infinitesimally small positive quantity. 
? ? 4 1 8 2 8 2 0 8 2 0 x x x x ? ? ? ? ? ? ? ? ? ? ? ? and 
? ?
2
2
4 2 1 0 x x ? ? ? ? ?
'
0 f x ? ? 
1
1
2
 
1
2
 
 
 
1 
1
2
 
 
1
2
 
 
 
1
2
 
1
2
 
 
1
2
 
 
 
 
1
2
 
 
 
 
 
1
2
 
 
1
2
 
 
 
 
 
 
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