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Page 1 of 15
MARKING SCHEME
CLASS XII
MATHEMATICS (CODE-041)
SECTION: A (Solution of MCQs of 1 Mark each)
Q no. ANS HINTS/SOLUTION
1. (D)
For a square matrix A of order nn ? , we have ? ? .,
n
A adj A A I ? where
n
I is the identity matrix of
order . nn ?
So, ? ?
3
2025 0 0
. 0 2025 0 2025
0 0 2025
A adj A I
??
??
??
??
??
??
2025 A ?? & ? ?
2 31
2025 adj A A
?
??
? ?
2
2025 2025 . A adj A ? ? ? ?
2. (A)
3. (C)
?? = ?? ?? = >
???? ???? = ?? ??
In the domain (R) of the function,
????
????
> 0 , hence the function is strictly increasing in ( -8, 8)
4. (B)
5, A ?
? ?
2
2
2
1 1 2
5. B AB B A B A
??
? ? ?
5. (B)
A differential equation of the form ? ? ,
dy
f x y
dx
? is said to be homogeneous, if ? ? , f x y is a
homogeneous function of degree 0.
Now, log log
n
ee
dy y
x y e
dx x
??
??
??
??
? ? ? ? log . , ;
e n
dy y y
e f x y Let
dx x x
?? ??
? ? ?
?? ??
?? ??
. ? ? , f x y will be a
homogeneous function of degree 0, if 1. n ?
6. (A) Method 1: ( Short cut)
When the points ? ? ? ?
1 1 2 2
, , , x y x y and ? ?
1 2 1 2
, x x y y ?? are collinear in the Cartesian plane then
? ? ? ?
? ?
1 2 1 2 1 2 1 2
1 2 2 2 2 1 2 2
1 1 2 1 1 2 22
00
x x y y x x y y
x y x y x y x y
x x x y y y xy
?? ??
? ? ? ? ? ? ? ?
? ? ? ? ??
2 1 1 2
. x y x y ??
Page 2
Page 1 of 15
MARKING SCHEME
CLASS XII
MATHEMATICS (CODE-041)
SECTION: A (Solution of MCQs of 1 Mark each)
Q no. ANS HINTS/SOLUTION
1. (D)
For a square matrix A of order nn ? , we have ? ? .,
n
A adj A A I ? where
n
I is the identity matrix of
order . nn ?
So, ? ?
3
2025 0 0
. 0 2025 0 2025
0 0 2025
A adj A I
??
??
??
??
??
??
2025 A ?? & ? ?
2 31
2025 adj A A
?
??
? ?
2
2025 2025 . A adj A ? ? ? ?
2. (A)
3. (C)
?? = ?? ?? = >
???? ???? = ?? ??
In the domain (R) of the function,
????
????
> 0 , hence the function is strictly increasing in ( -8, 8)
4. (B)
5, A ?
? ?
2
2
2
1 1 2
5. B AB B A B A
??
? ? ?
5. (B)
A differential equation of the form ? ? ,
dy
f x y
dx
? is said to be homogeneous, if ? ? , f x y is a
homogeneous function of degree 0.
Now, log log
n
ee
dy y
x y e
dx x
??
??
??
??
? ? ? ? log . , ;
e n
dy y y
e f x y Let
dx x x
?? ??
? ? ?
?? ??
?? ??
. ? ? , f x y will be a
homogeneous function of degree 0, if 1. n ?
6. (A) Method 1: ( Short cut)
When the points ? ? ? ?
1 1 2 2
, , , x y x y and ? ?
1 2 1 2
, x x y y ?? are collinear in the Cartesian plane then
? ? ? ?
? ?
1 2 1 2 1 2 1 2
1 2 2 2 2 1 2 2
1 1 2 1 1 2 22
00
x x y y x x y y
x y x y x y x y
x x x y y y xy
?? ??
? ? ? ? ? ? ? ?
? ? ? ? ??
2 1 1 2
. x y x y ??
Page 2 of 15
Method 2:
When the points ? ? ? ?
1 1 2 2
, , , x y x y and ? ?
1 2 1 2
, x x y y ?? are collinear in the Cartesian plane then
11
22
1 2 1 2
1
10
1
xy
xy
x x y y
?
??
? ? ? ? ? ?
2 1 2 2 1 2 2 2 1 1 1 2 1 1 2 1 1 2 2 1
1. 1 0 x y x y x y x y x y x y x y x y x y x y ? ? ? ? ? ? ? ? ? ? ?
2 1 1 2
. x y x y ??
7. (A)
01
1
2 3 0
c
A a b
??
??
? ? ?
??
??
??
When the matrix A is skew symmetric then ;
T
ij ji
A A a a ? ? ? ? ?
2; 0 and 3 c a b ? ? ? ? ?
So , 0 3 2 1. a b c ? ? ? ? ? ?
8. (C)
? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ? ? ?
1 2 1
;;
2 3 4
11
;
23
1 1 1 7
Wehave,
2 3 4 12
P A P B P A B
P A P B
P A B P A P B P A B
? ? ? ?
? ? ?
? ? ? ? ? ? ? ? ?
? ?
? ?
? ?
? ?
? ?
? ?
7
1
1
5
12
.
2
8
3
P A B
P A B P A B
A
P
B
P B P B P B
?
?
?? ? ? ?
? ? ? ? ?
??
??
9. (B) For obtuse angle, cos ?? < 0 => ?? ?. ?? ? < 0
?? ?? ?? - ???? + ?? < ?? => ?? ?? ?? - ???? < ?? => ?? ? ( ?? , ?? )
10. (C)
3, 4, 5 a b a b ? ? ? ?
We have ,
? ?
? ?
2 2 2 2
2 2 9 16 50 5. a b a b a b a b ? ? ? ? ? ? ? ? ? ? ?
11. (B) Corner point Value of the objective function 43 Z x y ??
1. ? ? 0,0 O
0 z ?
2. ? ? 40,0 R
160 z ?
3. ? ? 30,20 Q
120 60 180 z ? ? ?
4. ? ? 0,40 P
120 z ?
Since , the feasible region is bounded so the maximum value of the objective function 180 z ? is at
? ? 30,20 . Q
Page 3
Page 1 of 15
MARKING SCHEME
CLASS XII
MATHEMATICS (CODE-041)
SECTION: A (Solution of MCQs of 1 Mark each)
Q no. ANS HINTS/SOLUTION
1. (D)
For a square matrix A of order nn ? , we have ? ? .,
n
A adj A A I ? where
n
I is the identity matrix of
order . nn ?
So, ? ?
3
2025 0 0
. 0 2025 0 2025
0 0 2025
A adj A I
??
??
??
??
??
??
2025 A ?? & ? ?
2 31
2025 adj A A
?
??
? ?
2
2025 2025 . A adj A ? ? ? ?
2. (A)
3. (C)
?? = ?? ?? = >
???? ???? = ?? ??
In the domain (R) of the function,
????
????
> 0 , hence the function is strictly increasing in ( -8, 8)
4. (B)
5, A ?
? ?
2
2
2
1 1 2
5. B AB B A B A
??
? ? ?
5. (B)
A differential equation of the form ? ? ,
dy
f x y
dx
? is said to be homogeneous, if ? ? , f x y is a
homogeneous function of degree 0.
Now, log log
n
ee
dy y
x y e
dx x
??
??
??
??
? ? ? ? log . , ;
e n
dy y y
e f x y Let
dx x x
?? ??
? ? ?
?? ??
?? ??
. ? ? , f x y will be a
homogeneous function of degree 0, if 1. n ?
6. (A) Method 1: ( Short cut)
When the points ? ? ? ?
1 1 2 2
, , , x y x y and ? ?
1 2 1 2
, x x y y ?? are collinear in the Cartesian plane then
? ? ? ?
? ?
1 2 1 2 1 2 1 2
1 2 2 2 2 1 2 2
1 1 2 1 1 2 22
00
x x y y x x y y
x y x y x y x y
x x x y y y xy
?? ??
? ? ? ? ? ? ? ?
? ? ? ? ??
2 1 1 2
. x y x y ??
Page 2 of 15
Method 2:
When the points ? ? ? ?
1 1 2 2
, , , x y x y and ? ?
1 2 1 2
, x x y y ?? are collinear in the Cartesian plane then
11
22
1 2 1 2
1
10
1
xy
xy
x x y y
?
??
? ? ? ? ? ?
2 1 2 2 1 2 2 2 1 1 1 2 1 1 2 1 1 2 2 1
1. 1 0 x y x y x y x y x y x y x y x y x y x y ? ? ? ? ? ? ? ? ? ? ?
2 1 1 2
. x y x y ??
7. (A)
01
1
2 3 0
c
A a b
??
??
? ? ?
??
??
??
When the matrix A is skew symmetric then ;
T
ij ji
A A a a ? ? ? ? ?
2; 0 and 3 c a b ? ? ? ? ?
So , 0 3 2 1. a b c ? ? ? ? ? ?
8. (C)
? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ? ? ?
1 2 1
;;
2 3 4
11
;
23
1 1 1 7
Wehave,
2 3 4 12
P A P B P A B
P A P B
P A B P A P B P A B
? ? ? ?
? ? ?
? ? ? ? ? ? ? ? ?
? ?
? ?
? ?
? ?
? ?
? ?
7
1
1
5
12
.
2
8
3
P A B
P A B P A B
A
P
B
P B P B P B
?
?
?? ? ? ?
? ? ? ? ?
??
??
9. (B) For obtuse angle, cos ?? < 0 => ?? ?. ?? ? < 0
?? ?? ?? - ???? + ?? < ?? => ?? ?? ?? - ???? < ?? => ?? ? ( ?? , ?? )
10. (C)
3, 4, 5 a b a b ? ? ? ?
We have ,
? ?
? ?
2 2 2 2
2 2 9 16 50 5. a b a b a b a b ? ? ? ? ? ? ? ? ? ? ?
11. (B) Corner point Value of the objective function 43 Z x y ??
1. ? ? 0,0 O
0 z ?
2. ? ? 40,0 R
160 z ?
3. ? ? 30,20 Q
120 60 180 z ? ? ?
4. ? ? 0,40 P
120 z ?
Since , the feasible region is bounded so the maximum value of the objective function 180 z ? is at
? ? 30,20 . Q
Page 3 of 15
Section –B
[This section comprises of solution of very short answer type questions (VSA) of 2 marks each]
12. (A)
?
????
?? 3
( 1 + ?? 4
)
1
2
= ?
????
?? 5
( 1 +
1
?? 4
)
1
2
( Let 1 + ?? -4
= 1 +
1
?? 4
= ?? , ???? = -4?? -5
???? = -
4
?? 5
???? ?
????
?? 5
= -
1
4
???? )
= -
1
4
?
????
?? 1
2
= -
1
4
× 2 × v ?? + ?? , where '' c denotes any arbitrary constant of integration.
= -
1
2
v1 +
1
?? 4
+ ?? = -
1
2?? 2
v1 + ?? 4
+ ??
13. (A)
We know, ? ? ? ? ? ?
2
0
0, if 2
a
f x dx f a x f x ? ? ? ?
?
Let
? ?
7
cos f x ec x ? .
Now,
? ? ? ? ? ?
77
2 cos 2 cos f x ec x ec x f x ?? ? ? ? ? ? ? ?
2
7
0
cos 0; ec x dx
?
??
?
Using the property
? ? ? ? ? ?
2
0
0, if 2 .
a
f x dx f a x f x ? ? ? ?
?
14. (B)
The given differential equation ?? ?? '
= ?? =>
????
????
= log ??
???? = log ?? ???? => ????? = ?log ?? ????
?? = ?? log ?? - ?? + ??
hence the correct option is (B).
15. (B)
The graph represents
1
cos yx
?
? whose domain is ? ? 1,1 ? and range is ? ? 0, . ?
16. (D) Since the inequality 18 10 134 Z x y ? ? ? has no point in common with the feasible region hence
the minimum value of the objective function 18 10 Z x y ?? is 134 at ? ? 3,8 P .
17. (D)
The graph of the function ?? : ?? ? ?? defined by ? ? ? ?; f x x ? ? ? ? ?
where . denotes . . G I F is a straight
line
? ? 2.5 ,2.5 x h h ? ? ? ? , '' h is an infinitesimally small positive quantity. Hence, the function is
continuous and differentiable at 2.5 . x ?
18. (B) The required region is symmetric about the y ? axis.
So, required area (in sq units ) is
4
3
4
2
0
0
64
2 2 4 .
3
3
2
y
ydy
??
??
? ? ?
??
??
??
?
19. (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
20. (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
Page 4
Page 1 of 15
MARKING SCHEME
CLASS XII
MATHEMATICS (CODE-041)
SECTION: A (Solution of MCQs of 1 Mark each)
Q no. ANS HINTS/SOLUTION
1. (D)
For a square matrix A of order nn ? , we have ? ? .,
n
A adj A A I ? where
n
I is the identity matrix of
order . nn ?
So, ? ?
3
2025 0 0
. 0 2025 0 2025
0 0 2025
A adj A I
??
??
??
??
??
??
2025 A ?? & ? ?
2 31
2025 adj A A
?
??
? ?
2
2025 2025 . A adj A ? ? ? ?
2. (A)
3. (C)
?? = ?? ?? = >
???? ???? = ?? ??
In the domain (R) of the function,
????
????
> 0 , hence the function is strictly increasing in ( -8, 8)
4. (B)
5, A ?
? ?
2
2
2
1 1 2
5. B AB B A B A
??
? ? ?
5. (B)
A differential equation of the form ? ? ,
dy
f x y
dx
? is said to be homogeneous, if ? ? , f x y is a
homogeneous function of degree 0.
Now, log log
n
ee
dy y
x y e
dx x
??
??
??
??
? ? ? ? log . , ;
e n
dy y y
e f x y Let
dx x x
?? ??
? ? ?
?? ??
?? ??
. ? ? , f x y will be a
homogeneous function of degree 0, if 1. n ?
6. (A) Method 1: ( Short cut)
When the points ? ? ? ?
1 1 2 2
, , , x y x y and ? ?
1 2 1 2
, x x y y ?? are collinear in the Cartesian plane then
? ? ? ?
? ?
1 2 1 2 1 2 1 2
1 2 2 2 2 1 2 2
1 1 2 1 1 2 22
00
x x y y x x y y
x y x y x y x y
x x x y y y xy
?? ??
? ? ? ? ? ? ? ?
? ? ? ? ??
2 1 1 2
. x y x y ??
Page 2 of 15
Method 2:
When the points ? ? ? ?
1 1 2 2
, , , x y x y and ? ?
1 2 1 2
, x x y y ?? are collinear in the Cartesian plane then
11
22
1 2 1 2
1
10
1
xy
xy
x x y y
?
??
? ? ? ? ? ?
2 1 2 2 1 2 2 2 1 1 1 2 1 1 2 1 1 2 2 1
1. 1 0 x y x y x y x y x y x y x y x y x y x y ? ? ? ? ? ? ? ? ? ? ?
2 1 1 2
. x y x y ??
7. (A)
01
1
2 3 0
c
A a b
??
??
? ? ?
??
??
??
When the matrix A is skew symmetric then ;
T
ij ji
A A a a ? ? ? ? ?
2; 0 and 3 c a b ? ? ? ? ?
So , 0 3 2 1. a b c ? ? ? ? ? ?
8. (C)
? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ? ? ?
1 2 1
;;
2 3 4
11
;
23
1 1 1 7
Wehave,
2 3 4 12
P A P B P A B
P A P B
P A B P A P B P A B
? ? ? ?
? ? ?
? ? ? ? ? ? ? ? ?
? ?
? ?
? ?
? ?
? ?
? ?
7
1
1
5
12
.
2
8
3
P A B
P A B P A B
A
P
B
P B P B P B
?
?
?? ? ? ?
? ? ? ? ?
??
??
9. (B) For obtuse angle, cos ?? < 0 => ?? ?. ?? ? < 0
?? ?? ?? - ???? + ?? < ?? => ?? ?? ?? - ???? < ?? => ?? ? ( ?? , ?? )
10. (C)
3, 4, 5 a b a b ? ? ? ?
We have ,
? ?
? ?
2 2 2 2
2 2 9 16 50 5. a b a b a b a b ? ? ? ? ? ? ? ? ? ? ?
11. (B) Corner point Value of the objective function 43 Z x y ??
1. ? ? 0,0 O
0 z ?
2. ? ? 40,0 R
160 z ?
3. ? ? 30,20 Q
120 60 180 z ? ? ?
4. ? ? 0,40 P
120 z ?
Since , the feasible region is bounded so the maximum value of the objective function 180 z ? is at
? ? 30,20 . Q
Page 3 of 15
Section –B
[This section comprises of solution of very short answer type questions (VSA) of 2 marks each]
12. (A)
?
????
?? 3
( 1 + ?? 4
)
1
2
= ?
????
?? 5
( 1 +
1
?? 4
)
1
2
( Let 1 + ?? -4
= 1 +
1
?? 4
= ?? , ???? = -4?? -5
???? = -
4
?? 5
???? ?
????
?? 5
= -
1
4
???? )
= -
1
4
?
????
?? 1
2
= -
1
4
× 2 × v ?? + ?? , where '' c denotes any arbitrary constant of integration.
= -
1
2
v1 +
1
?? 4
+ ?? = -
1
2?? 2
v1 + ?? 4
+ ??
13. (A)
We know, ? ? ? ? ? ?
2
0
0, if 2
a
f x dx f a x f x ? ? ? ?
?
Let
? ?
7
cos f x ec x ? .
Now,
? ? ? ? ? ?
77
2 cos 2 cos f x ec x ec x f x ?? ? ? ? ? ? ? ?
2
7
0
cos 0; ec x dx
?
??
?
Using the property
? ? ? ? ? ?
2
0
0, if 2 .
a
f x dx f a x f x ? ? ? ?
?
14. (B)
The given differential equation ?? ?? '
= ?? =>
????
????
= log ??
???? = log ?? ???? => ????? = ?log ?? ????
?? = ?? log ?? - ?? + ??
hence the correct option is (B).
15. (B)
The graph represents
1
cos yx
?
? whose domain is ? ? 1,1 ? and range is ? ? 0, . ?
16. (D) Since the inequality 18 10 134 Z x y ? ? ? has no point in common with the feasible region hence
the minimum value of the objective function 18 10 Z x y ?? is 134 at ? ? 3,8 P .
17. (D)
The graph of the function ?? : ?? ? ?? defined by ? ? ? ?; f x x ? ? ? ? ?
where . denotes . . G I F is a straight
line
? ? 2.5 ,2.5 x h h ? ? ? ? , '' h is an infinitesimally small positive quantity. Hence, the function is
continuous and differentiable at 2.5 . x ?
18. (B) The required region is symmetric about the y ? axis.
So, required area (in sq units ) is
4
3
4
2
0
0
64
2 2 4 .
3
3
2
y
ydy
??
??
? ? ?
??
??
??
?
19. (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
20. (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
Page 4 of 15
21
cot
-1
( 3?? + 5)>
?? 4
= cot
-1
1
=>3x + 5 < 1 ( as cot
-1
?? is strictly decreasing function in its domain)
=> 3x < – 4
=> ?? < -
4
3
? ?? ? ( -8, -
4
3
)
1
2
1
2
1
22.
The marginal cost function is ? ?
2
' 0.00039 0.004 5 C x x x ? ? ? .
? ? ' 150 C ? ? 14.375 .
1
1
23.(a)
1
tan yx
?
?
and log
e
zx ?
.
Then
2
1
1
dy
dx x
?
?
and
1 dz
dx x
?
So,
2
2
1
1
.
1
1
dy
dy
dx
dz
dz
dx
x
x
x
x
?
?
??
?
1
2
1
2
1
2
1
2
OR
23.(b)
Let
x
yx (cos ) . ? Then,
log cosx x
e
ye ?
On differentiating both sides with respect to
, x
we get
log cos
( log cos )
xx
e
e
dy d
e x x
dx dx
?
(cos ) log cos ( ) (log cos )
x
ee
dy d d
x x x x x
dx dx dx
??
? ? ?
??
??
1
(cos ) log cos . ( sin )
cos
x
e
dy
x x x x
dx x
??
? ? ? ?
??
??
(cos ) (log cos tan ) .
x
e
dy
x x x x
dx
? ? ?
1
2
1
2
1
24.(a)
We have b
? ?
+ ?c ? = ( -1 + 3?) i^ + ( 2 + ? ) j^ + k
^
( b
? ?
+ ?c ?) . a ? ? = 0 => 2( -1 + 3? )+ 2 ( 2 + ? )+ 3 = 0
? = -
5
8
1
2
1
1
2
OR
24.(b)
????
? ?? ? ??
= ????
???? ??
- ????
??????
= ( 4?? ^ + 3?? ^
)- ?? ^
= 4?? ^ + 2?? ^
1
2
Page 5
Page 1 of 15
MARKING SCHEME
CLASS XII
MATHEMATICS (CODE-041)
SECTION: A (Solution of MCQs of 1 Mark each)
Q no. ANS HINTS/SOLUTION
1. (D)
For a square matrix A of order nn ? , we have ? ? .,
n
A adj A A I ? where
n
I is the identity matrix of
order . nn ?
So, ? ?
3
2025 0 0
. 0 2025 0 2025
0 0 2025
A adj A I
??
??
??
??
??
??
2025 A ?? & ? ?
2 31
2025 adj A A
?
??
? ?
2
2025 2025 . A adj A ? ? ? ?
2. (A)
3. (C)
?? = ?? ?? = >
???? ???? = ?? ??
In the domain (R) of the function,
????
????
> 0 , hence the function is strictly increasing in ( -8, 8)
4. (B)
5, A ?
? ?
2
2
2
1 1 2
5. B AB B A B A
??
? ? ?
5. (B)
A differential equation of the form ? ? ,
dy
f x y
dx
? is said to be homogeneous, if ? ? , f x y is a
homogeneous function of degree 0.
Now, log log
n
ee
dy y
x y e
dx x
??
??
??
??
? ? ? ? log . , ;
e n
dy y y
e f x y Let
dx x x
?? ??
? ? ?
?? ??
?? ??
. ? ? , f x y will be a
homogeneous function of degree 0, if 1. n ?
6. (A) Method 1: ( Short cut)
When the points ? ? ? ?
1 1 2 2
, , , x y x y and ? ?
1 2 1 2
, x x y y ?? are collinear in the Cartesian plane then
? ? ? ?
? ?
1 2 1 2 1 2 1 2
1 2 2 2 2 1 2 2
1 1 2 1 1 2 22
00
x x y y x x y y
x y x y x y x y
x x x y y y xy
?? ??
? ? ? ? ? ? ? ?
? ? ? ? ??
2 1 1 2
. x y x y ??
Page 2 of 15
Method 2:
When the points ? ? ? ?
1 1 2 2
, , , x y x y and ? ?
1 2 1 2
, x x y y ?? are collinear in the Cartesian plane then
11
22
1 2 1 2
1
10
1
xy
xy
x x y y
?
??
? ? ? ? ? ?
2 1 2 2 1 2 2 2 1 1 1 2 1 1 2 1 1 2 2 1
1. 1 0 x y x y x y x y x y x y x y x y x y x y ? ? ? ? ? ? ? ? ? ? ?
2 1 1 2
. x y x y ??
7. (A)
01
1
2 3 0
c
A a b
??
??
? ? ?
??
??
??
When the matrix A is skew symmetric then ;
T
ij ji
A A a a ? ? ? ? ?
2; 0 and 3 c a b ? ? ? ? ?
So , 0 3 2 1. a b c ? ? ? ? ? ?
8. (C)
? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ? ? ?
1 2 1
;;
2 3 4
11
;
23
1 1 1 7
Wehave,
2 3 4 12
P A P B P A B
P A P B
P A B P A P B P A B
? ? ? ?
? ? ?
? ? ? ? ? ? ? ? ?
? ?
? ?
? ?
? ?
? ?
? ?
7
1
1
5
12
.
2
8
3
P A B
P A B P A B
A
P
B
P B P B P B
?
?
?? ? ? ?
? ? ? ? ?
??
??
9. (B) For obtuse angle, cos ?? < 0 => ?? ?. ?? ? < 0
?? ?? ?? - ???? + ?? < ?? => ?? ?? ?? - ???? < ?? => ?? ? ( ?? , ?? )
10. (C)
3, 4, 5 a b a b ? ? ? ?
We have ,
? ?
? ?
2 2 2 2
2 2 9 16 50 5. a b a b a b a b ? ? ? ? ? ? ? ? ? ? ?
11. (B) Corner point Value of the objective function 43 Z x y ??
1. ? ? 0,0 O
0 z ?
2. ? ? 40,0 R
160 z ?
3. ? ? 30,20 Q
120 60 180 z ? ? ?
4. ? ? 0,40 P
120 z ?
Since , the feasible region is bounded so the maximum value of the objective function 180 z ? is at
? ? 30,20 . Q
Page 3 of 15
Section –B
[This section comprises of solution of very short answer type questions (VSA) of 2 marks each]
12. (A)
?
????
?? 3
( 1 + ?? 4
)
1
2
= ?
????
?? 5
( 1 +
1
?? 4
)
1
2
( Let 1 + ?? -4
= 1 +
1
?? 4
= ?? , ???? = -4?? -5
???? = -
4
?? 5
???? ?
????
?? 5
= -
1
4
???? )
= -
1
4
?
????
?? 1
2
= -
1
4
× 2 × v ?? + ?? , where '' c denotes any arbitrary constant of integration.
= -
1
2
v1 +
1
?? 4
+ ?? = -
1
2?? 2
v1 + ?? 4
+ ??
13. (A)
We know, ? ? ? ? ? ?
2
0
0, if 2
a
f x dx f a x f x ? ? ? ?
?
Let
? ?
7
cos f x ec x ? .
Now,
? ? ? ? ? ?
77
2 cos 2 cos f x ec x ec x f x ?? ? ? ? ? ? ? ?
2
7
0
cos 0; ec x dx
?
??
?
Using the property
? ? ? ? ? ?
2
0
0, if 2 .
a
f x dx f a x f x ? ? ? ?
?
14. (B)
The given differential equation ?? ?? '
= ?? =>
????
????
= log ??
???? = log ?? ???? => ????? = ?log ?? ????
?? = ?? log ?? - ?? + ??
hence the correct option is (B).
15. (B)
The graph represents
1
cos yx
?
? whose domain is ? ? 1,1 ? and range is ? ? 0, . ?
16. (D) Since the inequality 18 10 134 Z x y ? ? ? has no point in common with the feasible region hence
the minimum value of the objective function 18 10 Z x y ?? is 134 at ? ? 3,8 P .
17. (D)
The graph of the function ?? : ?? ? ?? defined by ? ? ? ?; f x x ? ? ? ? ?
where . denotes . . G I F is a straight
line
? ? 2.5 ,2.5 x h h ? ? ? ? , '' h is an infinitesimally small positive quantity. Hence, the function is
continuous and differentiable at 2.5 . x ?
18. (B) The required region is symmetric about the y ? axis.
So, required area (in sq units ) is
4
3
4
2
0
0
64
2 2 4 .
3
3
2
y
ydy
??
??
? ? ?
??
??
??
?
19. (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
20. (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
Page 4 of 15
21
cot
-1
( 3?? + 5)>
?? 4
= cot
-1
1
=>3x + 5 < 1 ( as cot
-1
?? is strictly decreasing function in its domain)
=> 3x < – 4
=> ?? < -
4
3
? ?? ? ( -8, -
4
3
)
1
2
1
2
1
22.
The marginal cost function is ? ?
2
' 0.00039 0.004 5 C x x x ? ? ? .
? ? ' 150 C ? ? 14.375 .
1
1
23.(a)
1
tan yx
?
?
and log
e
zx ?
.
Then
2
1
1
dy
dx x
?
?
and
1 dz
dx x
?
So,
2
2
1
1
.
1
1
dy
dy
dx
dz
dz
dx
x
x
x
x
?
?
??
?
1
2
1
2
1
2
1
2
OR
23.(b)
Let
x
yx (cos ) . ? Then,
log cosx x
e
ye ?
On differentiating both sides with respect to
, x
we get
log cos
( log cos )
xx
e
e
dy d
e x x
dx dx
?
(cos ) log cos ( ) (log cos )
x
ee
dy d d
x x x x x
dx dx dx
??
? ? ?
??
??
1
(cos ) log cos . ( sin )
cos
x
e
dy
x x x x
dx x
??
? ? ? ?
??
??
(cos ) (log cos tan ) .
x
e
dy
x x x x
dx
? ? ?
1
2
1
2
1
24.(a)
We have b
? ?
+ ?c ? = ( -1 + 3?) i^ + ( 2 + ? ) j^ + k
^
( b
? ?
+ ?c ?) . a ? ? = 0 => 2( -1 + 3? )+ 2 ( 2 + ? )+ 3 = 0
? = -
5
8
1
2
1
1
2
OR
24.(b)
????
? ?? ? ??
= ????
???? ??
- ????
??????
= ( 4?? ^ + 3?? ^
)- ?? ^
= 4?? ^ + 2?? ^
1
2
Page 5 of 15
????
^
=
4
2v5
?? ^ +
2
2v5
?? ^
=
2
v5
?? ^ +
1
v5
?? ^
So, the angles made by the vector ????
? ?? ? ??
with the , xy and the z axes are respectively
??????
-1
(
2
v 5
),
?? 2
, ??????
-1
(
1
v 5
) .
1
2
1
25.
?? 1
? ? ? ??
= ?? ? + ?? ? ?
= 4?? ^ - 2?? ^ - 2?? ^
, ?? 2
? ? ? ??
= ?? ? - ?? ? ?
= -6?? ^ - 8?? ^
Area of the parallelogram =
1
2
|?? 1
? ? ? ??
× ?? 2
? ? ? ??
| =
1
2
||
?? ^ ?? ^ ?? ^
4 -2 -2
0 -6 -8
|| = 2|?? ^ + 8?? ^ - 6?? ^
|
Area of the parallelogram = 2v101 sq. units.
1
2
1
1
2
Section –C
[This section comprises of solution short answer type questions (SA) of 3 marks each]
26.
?? 2
+ 3
2
= ?? 2
?? h???? ?? = 5 ?? h???? ?? = 4, ?????? 2??
????
????
= 2??
????
????
4 ( 200 )= 5
????
????
=>
????
????
= 160 cm/s
1
2
1
2
1
1
27.
?? =
1
3
v ?? ?
????
????
=
1
6
?? -
1
2
=
1
6v ?? ; ??? ? ( 5,18)
????
????
=
1
6v ?? ?
?? 2
?? ?? ?? 2
= -
1
12?? v ??
So,
?? 2
?? ?? ?? 2
< 0, ??? ? ( 5,18)
This means that the rate of change of the ability to understand spatial concepts decreases
(slows down) with age.
1
1
1
2
1
2
28(a)
(i) ?? = ?????? -?? (
?? ?? ? ? ??
.?? ?? ? ? ??
|?? ?? ? ? ??
|.|?? ?? ? ? ??
|
) = ?????? -?? (
( i^ -2?^ +3k
^
) .( 3i^ -2?^ + ?? ^
)
|( i^ -2?^ +3k
^
) || ( 3i^ -2?^ + ?? ^
) |
)
= ?????? -?? (
?? +?? +?? v ?? +?? +?? v ?? +?? +?? )= ?????? -?? (
????
????
)= ?????? -?? (
?? ?? ) .
(ii) Scalar projection of ?? ?? ? ? ??
on ?? ?? ? ? ??
=
?? ?? ? ? ??
.?? ?? ? ? ??
|?? ?? ? ? ??
|
=
( i^ -2?^ +3k
^
) .( 3i^ -2?^ + ?? ^
)
| ( 3i^ -2?^ + ?? ^
) |
=
3+4+3
v 9+4+1
=
10
v 14
.
1
1
2
1
1
2
3
3
y
x
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