CBSE Class 12 Question Paper 2025 Solution Mathematics

Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Senior Secondary Examination, 2025
SUBJECT NAME MATHEMATICS (Q.P. CODE – 65/1/1)
General Instructions: -

1 You are aware that evaluation is the most important process in the actual and correct
assessment of the candidates. A small mistake in evaluation may lead to serious problems
which may affect the future of the candidates, education system and teaching profession.
To avoid mistakes, it is requested that before starting evaluation, you must read and
understand the spot evaluation guidelines carefully.
2 “Evaluation policy is a confidential policy as it is related to the confidentiality of the
examinations conducted, Evaluation done and several other aspects. Its leakage to
the public in any manner could lead to derailment of the examination system and
affect the life and future of millions of candidates. Sharing this policy/document to
anyone, publishing in any magazine and printing in Newspaper/Website, etc. may
invite action under various rules of the Board and IPC.”
3 Evaluation is to be done as per instructions provided in the Marking Scheme. It should not
be done according to one’s own interpretation or any other consideration. The Marking
Scheme should be strictly adhered to and religiously followed. However, while evaluating,
answers which are based on latest information or knowledge and/or are innovative,
they may be assessed for their correctness otherwise and due marks be awarded to
them. In class-XII, while evaluating the competency-based questions, please try to
understand the given answer and even if reply is not from a marking scheme but
correct competency is enumerated by the candidate, due marks should be awarded.
4 The Marking Scheme carries only suggested value points for the answers.
These are Guidelines only and do not constitute the complete answer. The students can
have their own expression and if the expression is correct, the due marks should be
awarded accordingly.
5 The Head-Examiner must go through the first five answer books evaluated by each
evaluator on the first day, to ensure that evaluation has been carried out as per the
instructions given in the Marking Scheme. If there is any variation, the same should be zero
after deliberation and discussion. The remaining answer books meant for evaluation shall
be given only after ensuring that there is no significant variation in the marking of individual
evaluators.
6 Evaluators will mark (√) wherever answer is correct. For wrong answer CROSS ‘X’ be
marked. Evaluators will not put right (✓) while evaluating which gives the impression
that the answer is correct, and no marks are awarded. This is the most common
mistake which evaluators are committing.
7 If a question has parts, please award marks on the right-hand side for each part. Marks
awarded for different parts of the question should then be totaled up and written in the left-
hand margin and encircled. This may be followed strictly.
8 If a question does not have any parts, marks must be awarded in the left-hand margin and
encircled. This may also be followed strictly.
9 If a student has attempted an extra question, answer to the question deserving more marks
should be retained and the other answer scored out with a note “Extra Question”.

MS_XII_Mathematics_041_65/1/1_2024-25 Page 1

10 No marks to be deducted for the cumulative effect of an error. It should be penalized only
once.
11 A full scale of marks _ (example 0 to 80/70/60/50/40/30 marks as given in
Question Paper) has to be used. Please do not hesitate to award full marks if the answer
deserves it.
12 Every examiner must necessarily do evaluation work for full working hours, i.e., 8 hours
every day and evaluate 20 answer books per day in main subjects and 25 answer books
per day in other subjects (Details are given in Spot Guidelines). This is in view of the reduced
syllabus and number of questions in question paper.
13 Ensure that you do not make the following common types of errors committed by the
Examiner in the past: -
● Leaving answer or part thereof unassessed in an answer book.
● Giving more marks for an answer than assigned to it.
● Wrong totaling of marks awarded on an answer.
● Wrong transfer of marks from the inside pages of the answer book to the title page.
● Wrong question wise totaling on the title page.
● Wrong totaling of marks of the two columns on the title page.
● Wrong grand total.
● Marks in words and figures not tallying/not same.
● Wrong transfer of marks from the answer book to online award list.
● Answers marked as correct, but marks not awarded. (Ensure that the right tick mark is
correctly and clearly indicated. It should merely be a line. Same is with the X for
incorrect answer.)
● Half or a part of the answer marked correct and the rest as wrong, but no marks
14 Whileawarded.
evaluating the answer books if the answer is found to be totally incorrect, it should be
marked as cross (X) and awarded zero (0) Marks.
15 Any unassessed portion, non-carrying over of marks to the title page, or total error
detected by the candidate shall damage the prestige of all the personnel engaged in the
evaluation work as also of the Board. Hence, to uphold the prestige of all concerned, it is again
reiterated that the instructions be followed meticulously and judiciously.
16 The Examiners should acquaint themselves with the guidelines given in the “Guidelines for
Spot Evaluation” before starting the actual evaluation.
17 Every Examiner shall also ensure that all the answers are evaluated, marks carried over to
the title page, correctly totaled and written in figures and words.
18 The candidates are entitled to obtain a photocopy of the Answer Book on request on payment
of the prescribed processing fee. All Examiners/Additional Head Examiners/Head
Examiners are once again reminded that they must ensure that evaluation is carried out
strictly as per value points for each answer as given in the Marking Scheme.

MS_XII_Mathematics_041_65/1/1_2024-25 Page 2

MARKING SCHEME
SENIOR SECONDARY EXAMINATION 2024-25
MATHEMATICS (Code–041)
[ Paper Code: 65/1/1]

Q. No. EXPECTED ANSWER / VALUE POINTS Marks
SECTION - A
Questions no. 1 to 18 are multiple choice questions (MCQs) of 1 mark each.

Q1.

Ans
1

Q2.

Ans 1

Q3.

Ans 1

MS_XII_Mathematics_041_65/1/1_2024-25 Page 3

Q4.

Ans 1

Q5.

1
Ans
Q6.

1
Ans

Q7.

Ans 1

Q8.

Ans 1

MS_XII_Mathematics_041_65/1/1_2024-25 Page 4

Q9.

1
Ans

Q10.

Ans 1

Q11.

Ans
1

Q12.

Ans 1

MS_XII_Mathematics_041_65/1/1_2024-25 Page 5

Q13.

Ans 1

Q14.

Ans 1

Q15.

Ans 1
Q16.

Ans 1

Q17.

Ans
1

MS_XII_Mathematics_041_65/1/1_2024-25 Page 6

Q18.

Ans

1

Q19.

Ans 1

MS_XII_Mathematics_041_65/1/1_2024-25 Page 7

Q20.

Ans 1

SECTION B
This section comprises very short answer (VSA) type questions of 2 marks each.

Q21.

Ans(a) du cos2 x
Let u = 2cos x  =2 ( −2cos x sin x ) log 2
2
1
dx
dv
Let v = cos 2 x  = − 2cos x sin x ½
dx
 du 
du  dx  cos2 x
Now = =2 log 2 ½
dv  dv 
 dx 
 

OR
Ans(b) tan −1 ( x 2 + y 2 ) = a 2  x 2 + y 2 = tan a 2 ½

Differentiatebothsides wrt x ,
dy
2x + 2 y = 0 1
dx
dy x
 =− ½
dx y

MS_XII_Mathematics_041_65/1/1_2024-25 Page 8

Q22.

Ans   3 
tan −1  2sin  2cos −1 
  2  
     
= tan −1  2sin  2    = tan −1  2sin  1
  6   3
 3 
= tan −1  2   = tan
−1
3= 1
 2  3

Q23.

Ans
iˆ ˆj kˆ
a  b = 2 −1 1 = − 2iˆ + 3 ˆj + 7kˆ 1
1 3 −1
1
Areaof parallelogram = a b
2
1 62
( −2 ) + 3 2 + 7 2 =
2
= 1
2 2
Q24.

Ans 15
f ( x ) = 5 x 3/2 − 3 x 5/2  f  ( x ) =
x (1 − x ) 1
2
For increasing / decreasing,put f  ( x ) = 0
 x = 0,1
( i ) When x 0,1 , f  ( x )  0.So, f isincreasing when x  0,1 ½

( Theintervals ( 0,1) ,  0,1) or ( 0,1 canalsobeconsidered.)
( ii ) When x1,  ) , f  ( x )  0.So, f isdecreasing when x 1,  ) ½

( The interval ( 1,  ) canalsobeconsidered.)

MS_XII_Mathematics_041_65/1/1_2024-25 Page 9

Q25.

Ans(a) Let the required angle between the kite strings be  .
a .b
Then,cos =
a b

 cos = =
( 3iˆ + ˆj + 2kˆ )( 2iˆ − 2 ˆj + 4kˆ )
=
12
=
3
1½
9 + 1 + 4 4 + 4 + 16 336 21
 12  −1  3 
  = cos −1   or cos   ½
 336   21 
OR
Ans(b) BA = − 6iˆ + 2 ˆj + 3kˆ 1
Requiredunit vector of magnitude 21
 −6iˆ + 2 ˆj + 3kˆ 
= 21    ½
 36 + 4 + 9 
( )
= 3 −6iˆ + 2 ˆj + 3kˆ or − 18iˆ + 6 ˆj + 9kˆ ½

SECTION C
This section comprises short answer (SA) type questions of 3 marks each.
Q26.

Ans da
Let' a 'bethesideof thetriangle,so = 3cm/s ½
dt
3 2
Now areaof anequilateral triangle, A = a
4
dA 3a da
 =  1½
dt 2 dt
dA  3  15 45 3
  = 3= cm 2 /s 1
dt  a = 15cm 2 2
MS_XII_Mathematics_041_65/1/1_2024-25 Page 10

Q27.

Ans

For
correct
graph
and
shading
1½

Corner Point Value of Z = x + 2 y
For

O ( 0,0 )
correct
0 table

A ( 2, 2 ) 6 1

Since feasible region is unbounded. Plot x + 2y > 6 which has common region
½
with feasible region, thus Z has no maximum value.

Ans

MS_XII_Mathematics_041_65/1/1_2024-25 Page 11

Ans(a) x + sin x
 1 + cos x dx
x x
x + 2sin cos
= 2 2 dx 1
x
2cos 2
2
1 x x
=  x  sec2  dx +  tan dx ½
2 2 2
x x x
= x tan −  tan dx +  tan dx
2 2 2 1
x
= x tan + C
2 ½

OR
Ans(b)  /4
dx
0 cos3 x 2sin 2 x
 /4
1 dx
=  4
½
2 0 cos x tan x
1 ( 1 + tan x ) sec x
 /4 2 2

=  dx
2 0 tan x
Put tan x = t  sec 2 x dx = dt ½
1 1 + t2
1
I =  dt ½
20 t
1  1 
1
=  + t 3/2  dt
2 0 t 
1
1 2 
=  2 t + t 5/2  1
2 5 0
6
= ½
5

MS_XII_Mathematics_041_65/1/1_2024-25 Page 12

Q29.

Ans(a) Rewriting the lines, we get
( ) ( ) ( ) (
r = iˆ − 2 ˆj + 3kˆ +  − iˆ + ˆj − 2kˆ and r = iˆ − ˆj − kˆ +  iˆ + 2 ˆj − 2kˆ ) ½

Let a1 = iˆ − 2 ˆj + 3kˆ , a2 = iˆ − ˆj − kˆ , b1 = − iˆ + ˆj − 2kˆ , b2 = iˆ + 2 ˆj − 2kˆ
Note that the dr'sof given lines are not proportional so, they are not parallel lines.
The lines will be skew if they do not intersect each other also.
iˆ ˆj kˆ
Here a − a = ˆj − 4kˆ , b  b = −1 1 −2 = 2iˆ − 4 ˆj − 3kˆ
2 1 1 2
½+½

1 2 −2

(
Consider ( a2 − a1 ) . b1  b2 )
( )(
= ˆj − 4kˆ . 2iˆ − 4 ˆj − 3kˆ = 8  0 )
Hence lines will not intersect. So the lines are skew. ½

( a2 − a1 ) . ( b1  b2 )
Shortest Distance =
b1  b2
8 8
= = 1
4 + 16 + 9 29
OR
Ans(b) Let the wicket keeper divides the line segment in ratio k :1
k F + 1.B 1
W =
k +1
 12k + 2  ˆ  18k + 8  ˆ
 6iˆ + 12 ˆj =  i + j 1
 k +1   k +1 
2
k=
3
Hence, the required ratio is 2 : 3 1

MS_XII_Mathematics_041_65/1/1_2024-25 Page 13

Q30.

Ans(a) ( i ) Since  P ( X ) = 1 p + 2 p + 3 p + p = 1 ½
1
p= ½
7
( ii ) Mean =  X . P ( X ) = 0 ( p ) + 2 ( 2 p ) + 4 ( 3 p ) + 5 ( p ) 1
1
= 21 p = 21   = 3 1
7

OR
Ans(b) Let E1 :Theapplicant isamale
½
E2 :Theapplicant isafemale
A :The candidate chosen will have distinction in the written test.
1 2
P ( E1 ) = , P ( E2 ) = , P ( A | E1 ) = 0.4, P ( A | E2 ) = 0.35 1
3 3
 P ( A ) = P ( E1 ) P ( A | E1 ) + P ( E 2 ) P ( A | E 2 )
1 2
=  0.4 +  0.35 1
3 3
11
= ½
30

MS_XII_Mathematics_041_65/1/1_2024-25 Page 14

Q31.

For
Ans correct
graph:
1 mark

Required Area
0
½
=  y dx
−6
0
= 2  ( x + 3 ) dx ½
−3
0
 ( x + 3)2 
=2  ½
 2  −3
=9 ½
SECTION D
This section comprises long answer (LA) type questions of 5 marks each.
Q32.

Ans(a) Let x = sin A, y = sin B  A = sin −1 x , B = sin −1 y 1

 1 − x2 + 1 − y2 = a ( x − y )
 cos A + cos B = a ( sin A − sin B )
 A+ B  A− B  A+ B  A− B
 2cos   cos   = 2a cos   sin  
1
 2   2   2   2 
 A− B
 cot   = a  A − B = 2cot −1 a 1
 2 
 sin −1 x − sin −1 y = 2cot −1 a ½
differentiatebothsides w rt x ,
1 1 dy
− =0 1½
1 − x2 1 − y 2 dx
dy 1 − y2
 =
dx 1 − x2
OR

MS_XII_Mathematics_041_65/1/1_2024-25 Page 15

Ans(b)  
x = a  cos  + log tan 
 2
 
dx  1  1
 = a  − sin  +  sec 2   ½
d   2 2
tan
 2 
 1   1 − sin  
2
= a  − sin  + =a 
sin    sin  
½

dx
= a cot  cos  ½
d
dy
Also, y = sin   = cos 
d ½
dy tan 
 = 1
dx a
Differetiating wrt x ,
d 2 y sec 2  d
= 
dx 2 a dx
sec  tan 
3
= 1
a2
d2 y 2 2
2 
= 2 1
dx  at =  a
4

Q33.

Ans f ( x ) = 2 x 3 − 15 x 2 + 36 x + 1
 f  ( x ) = 6 ( x 2 − 5 x + 6 ) = 6 ( x − 2 )( x − 3 ) 1

f  ( x ) = 0  x = 2 , 3 1, 5 1

Now f ( 1) = 24, f ( 2 ) = 29, f ( 3 ) = 28, f ( 5 ) = 56 2

Hence, theabsolute maximum value is 56 and the absolute minimum value is 24. 1

Q34.

MS_XII_Mathematics_041_65/1/1_2024-25 Page 16

Ans(a) x y −1 z − 2
Theequation of given line is = = =
1 2 3
Any arbitrary point on the line is M (  , 2 + 1, 3 + 2 ) 1
dr's of AM are   − 1, 2 − 5, 3 − 1 
Here1 (  − 1) + 2 ( 2 − 5 ) + 3 ( 3 − 1) = 0 1
 = 1 ½
 M ( 1, 3, 5 ) is thefoot perpendicular of the point A to the given line.
Let image of point A in the line be A ( ,  ,  )
 1+ 6+  3+  
Since M is the mid-point of AA,so M  = M (1, 3, 5 ) ½
2 
, ,
 2 2
 A ( 1, 0, 7 ) is the image of A. 1
x −1 y − 6 z − 3
Also, Equation of AA is = = 1
0 −3 2

OR
Ans(b) x+5 y+3 z−6
The given line is = = =  and Q ( 2, 4, −1)
1 4 −9
Any random point on the line will be given by P (  − 5, 4 − 3, −9 + 6 ) 1

(  − 7 ) + ( 4 − 7 ) + ( − 9 + 7 ) = 7
2 2 2
Since PQ = 7  1

 98 (  2 − 2 + 1) = 0   = 1 1
Hence, the required point is P ( −4,1, −3 ) 1
x + 4 y −1 z + 3 x −2 y−4 z +1
The equation of line PQ is = = or = = 1
6 3 2 6 3 2

Q35.

MS_XII_Mathematics_041_65/1/1_2024-25 Page 17

Ans Let x, y and z be the no. of students allocated to Sports, Music
and Drama clubs respectively.
x
Here, x = y + z , y = + 20 , x + y + z = 180
2
 x − y − z = 0, x − 2 y = −40, x + y + z = 180 1½
Given equations can be written as AX = B
1 −1 −1  0   x
where, A = 1 −2 0  , B =  −40 , X =  y 
    ½
1 1 1   180   z 
A = −4  0  A−1 exists. ½
 −2 0 −2 
adjA =  −1 2 −1 1
 3 −2 −1
 2 0 2
 adjA =  1 −2 1 
−1 1 1
A = ½
A 4
 −3 2 1 
−1
X=A B
 2 0 2   0   90 
=  1 −2 1   −40  =  65 
1 1
4
 −3 2 1   180   25 
 x = 90, y = 65, z = 25
Number of students allocated in sports, music and drama are
90 , 65 and 25 respectively .

SECTION E
This section comprises 3 case study-based questions of 4 marks each.
Q36.

MS_XII_Mathematics_041_65/1/1_2024-25 Page 18

Ans ( i ) 2 x + 3 y = 300 1
x
( ii ) A = xy = ( 300 − 2 x ) 1
3
( iii )( a ) A = ( 300 − 2 x ) = ( 300 x − 2 x 2 )
x 1
3 3
dA 1
 = ( 300 − 4 x ) ½
dx 3
dA
For critical points,put = 0  x = 75 ½
dx
d2A 4 ½
Also, 2 = −  0 . So, A is maximum at x = 75
dx 3
75
Also,maximum area is A = ( 300 − 150 ) = 3750m 2 ½
3
OR

( iii )( b ) A = ( 300 − 2 x ) = ( 300 x − 2 x 2 )
x 1
3 3
dA 1
 = ( 300 − 4 x ) ½
dx 3
dA
For critical points,put = 0  x = 75 ½
dx
dA
As changes its sign from positive to negative as x passes through
dx ½
x =75 from left to right, which means x = 75 is the point of maximum.
75
Also,maximum area is A = ( 300 − 150 ) = 3750m 2 ½
3

Note : Full credit to be given if the student takes equation as
2x + 2y = 300 or 2x + 4y = 300 or 4x + 4y = 300 or 4x + 3y = 300
The solutions of sub-parts will differ and marks may be given accordingly.

MS_XII_Mathematics_041_65/1/1_2024-25 Page 19

Q37.

Ans ( i ) R4 1
( ii ) R5 1

( iii )( a ) R1 and R3 1+1
OR
( iii )( b ) Required pairs to beadded to make therelation R2 as an equivalencerelation are:
( 1,1) , ( 2, 2 ) , ( 3, 3 ) , ( 2,1) , ( 3,1) and ( 2, 3 ) 2

MS_XII_Mathematics_041_65/1/1_2024-25 Page 20

Q38.

MS_XII_Mathematics_041_65/1/1_2024-25 Page 21

Ans E1 :customer avails loan on fixed rate
E2 :customer avails loan on floating rate
E3 :customer avails loan on variable rate
A:the person defaults on the loan
1 2 7
P ( E1 ) = , P ( E2 ) = , P ( E3 ) =
10 10 10
5 3 1
P ( A | E1 ) = , P ( A | E2 ) = , P ( A | E3 ) =
100 100 100
( i ) P ( A) = P ( E1 ) .P ( A | E1 ) + P ( E2 ) .P ( A | E2 ) + P ( E3 ) .P ( A | E3 )
1 5 2 3 7 1
=  +  +  1
10 100 10 100 10 100
18 9
= or 1
1000 500
P ( E3 ) .P ( A | E3 )
( ii ) P ( E3 | A) =
P ( E1 ) .P ( A | E1 ) + P ( E2 ) .P ( A | E 2 ) + P ( E 3 ) .P ( A | E 3 )
7 1

= 10 100 1
18
1000
7 1
=
18

MS_XII_Mathematics_041_65/1/1_2024-25 Page 22