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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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FAQs on Class 12 Mathematics: CBSE Marking Scheme (2024-25) - Sample Papers for Class 12 Medical and Non-Medical

1. What is the CBSE marking scheme for Class 12 Mathematics in 2024-25?
Ans. The CBSE marking scheme for Class 12 Mathematics includes a total of 100 marks, which are divided into various sections: 20 marks for internal assessment, 80 marks for the theory exam. The theory exam consists of different types of questions, including very short answer, short answer, and long answer questions.
2. How are internal assessment marks calculated in Class 12 Mathematics?
Ans. Internal assessment marks in Class 12 Mathematics are calculated based on various components such as periodic tests, notebook submission, and practical work. Schools typically evaluate students continuously throughout the year, contributing a total of 20 marks to the final assessment.
3. What types of questions can we expect in the Class 12 Mathematics exam?
Ans. In the Class 12 Mathematics exam, students can expect a variety of question types, including multiple-choice questions (MCQs), very short answer questions, short answer questions, and long answer questions. The distribution of marks is designed to cover all topics in the syllabus comprehensively.
4. How can students prepare effectively for the Class 12 Mathematics exam?
Ans. Students can prepare effectively for the Class 12 Mathematics exam by understanding the syllabus thoroughly, practicing previous years' question papers, and solving sample papers. Regular revision and attending doubt-clearing sessions with teachers are also crucial for better understanding and retention of concepts.
5. What is the importance of the Class 12 Mathematics exam in a student's academic career?
Ans. The Class 12 Mathematics exam is crucial as it contributes significantly to a student's overall academic performance and is often a deciding factor for college admissions, especially for courses in science, engineering, and other mathematics-related fields. Good performance can open up various opportunities for higher education.
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