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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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Document Description: Class 12 Mathematics: CBSE Marking Scheme (2023-24) for Class 12 2024 is part of Sample Papers for Class 12 Medical and Non-Medical preparation.
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