CBSE Class 12 Question Paper 2023 Solution Maths

e s t i o n P a p er
Qu
Solu t i o n
2023

Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Senior School Certificate Examination, 2023

MATHEMATICS PAPER 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
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,
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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. 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.

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
answer is correct, and no marks are awarded. This is 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 totalled 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 In Q1-Q20, if a candidate attempts the question more than once (without canceling the
previous attempt), marks shall be awarded for the first attempt only and the other

1 MS_Mathematics_041_65/1/1_2022-23

answer scored out with a note “Extra Question”.

10 In Q21-Q38, if a student has attempted an extra question, answer of the question
deserving more marks should be retained and the other answer scored out with a note
“Extra Question”.

11 No marks to be deducted for the cumulative effect of an error. It should be penalized only
once.

12 A full scale of marks __________ (example 0 to 80/70/60/50/40/30 marks as given in
Question Paper) must be used. Please do not hesitate to award full marks if the answer
deserves it.

13 Every examiner has to necessarily do evaluation work for full working hours i.e., 8 hours
every day and evaluate 20 answer books per day (Details are given in Spot Guidelines). This
is in view of the reduced syllabus and number of questions in question paper.

14 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 totalling 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 totalling on the title page.
● Wrong totalling 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 answer marked correct and the rest as wrong, but no marks awarded.
15 While 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.

16 Any unassessed portion, non-carrying over of marks to the title page, or totalling 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.

17 The Examiners should acquaint themselves with the guidelines given in the “Guidelines for
spot Evaluation” before starting the actual evaluation.

18 Every Examiner shall also ensure that all the answers are evaluated, marks carried over to the
title page, correctly totalled and written in figures and words.

19 The candidates are entitled to obtain 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.

2 MS_Mathematics_041_65/1/1_2022-23

EXPECTED ANSWER/VALUE POINTS
SECTION A
Q.No. EXPECTED ANSWER / VALUE POINTS Marks
SECTION-A
(Question nos. 1 to 18 are Multiple Choice Questions carrying 1 mark each)

1

Ans (b) 2 1

2.

Ans (c) 8 1

3.

Ans (d) 8 or – 8 1
4.

Ans −6 −8
(b)[ ] 1
−10 −4

5.

Ans (b) x(logx – 1) + C 1

3 MS_Mathematics_041_65/1/1_2022-23

6.

1
Ans (a) 1
√3

7.

Ans (c) 3 1

8.

Ans (a) 3 1

9.

Ans (d) -1 1

10.

Ans (b) 3 1

11.

2 −3 6
Ans (d) 7 , 7 , 7 1

4 MS_Mathematics_041_65/1/1_2022-23

12.

Ans (a) 0.6 1

13.

11
Ans (d) 4 1

14.

Ans (c) ± 5 1

15.

Ans 𝑥2
(d) y = 2 logx + 2 + C 1

16.

Ans (c) (-∞, 0) 1

5 MS_Mathematics_041_65/1/1_2022-23

17.

Ans (c) z is maximum at (40, 15) and minimum at (15, 20) 1

18.

Ans (a) 2 1
(Question Nos. 19 & 20 are Assertion-Reason based questions of 1 mark each)

19.

Ans (c) Assertion is True, Reason is False 1

20.

Ans (d) Assertion is False, Reason is True 1
SECTION-B
(Question nos. 21 to 25 are very short Answer type questions carrying 2 marks each)

21.

Ans 𝟏
(a) f(1) = 2, f(2) = 4, f(3) = 6, f(4) = 8 𝟏
𝟐
 B = {2, 4, 6, 8} ½
OR
𝟏
 3  𝟏
(b) Required value = + + 𝟐
4 4 4

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𝟏
5
= 𝟐
4

22.
Ans ^ ^ ^
^ ^ ^ i j k
Unit vector along i + j + k is + + 1
3 3 3
^ ^ ^ ^ ^ ^ 𝟏 𝟏
Required vectors are 3 i + 3 j + 3 k and – 3 i – 3 j – 3 k +
𝟐 𝟐

23.

Ans → → 5 → → 11
(a) According to question, c – a = ( b – a ) 2
4
→ →
→ 5b a 1
 c = – 2
4 4
OR
(b) D.r.s. of lines are < 2, 7, – 3 > and < – 1, 2, 4 > 1
Now 2. – 1 + 7·2 + – 3·4 = 0 1
 given lines are perpendicular

24.

Ans 2
 2 
dy    x  2  x + x – 1 
= 2  x + x – 1  1 + =
2
1½
dx    2 
x –1 x2 – 1
dy
x2 – 1 = 2y
dx ½
2
 dy 
(x2 – 1)   = 4y2
 dx 

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25.
Ans 16[4 + cos x] cos x + 16 sin 2 x
f (x ) = –1 1
(4 + cos x)2
cos x (56 – cos x)
= 1
(4 + cos x) 2 2


in ( , ), cos x < 0  f  (x) < 0 1
2
2

 f(x) in strictly decreasing in ( , )
2
SECTION-C
(Question nos. 26 to 31 are short Answer type questions carrying 3 marks each)

26.

Ans /2 /2

 
 tan x 
Let I = [log sin x – log (2cos x)]dx = log   dx 1
 2  2
0 0
a a
Using property

0
f(x) dx =

0
f(a – x)dx

/2 1

 cot x 
We get, I = log   dx
 2 
0
/2 /2

 
 2I =
 tan x cot x 
log    dx = log 
1 
 dx
½
 2 2  4
0 0
 1  𝜋/2  1
2I = log  x = log ½
4 0 2 4
𝜋 1
I = 4 𝑙𝑜𝑔 4 OR − 2 𝑙𝑜𝑔 2
𝜋 ½
27.


Ans dx
Let I =
x ( x + 1) ( x + 2)

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1 ½
Let x = t, dx = dt
2 x


dt
I= 2
(t + 1) (t + 2)


 1 1 
=2  –  dt 1
t +1 t + 2
= 2[log  t + 1 – log  t + 2] + C 1
 x +1 
= 2[log ( x + 1) – log ( x + 2)] + C or 2 log  
 x +2 +C ½
 

28.

Let (a) Integrating factor = e 
Ans sec x dx 2
= etan x ½
Solution is yetan x =
 tan x sec2 x etan x dx + C ½

Let tan x = t sec2 x dx = dt ½

 etan x tan x sec2 x dx =
 et t dt= et (t – 1) ½

 yetan x = etan x (tan x – 1) + C ½
y(0) = 0 gives C = 1
Particular solution is yetan x = etan x (tan x – 1) + 1or y = tanx -1 + e-tanx ½
OR
(b)Given differential equation can be written as
2
dy y y
= + 1 +   --------- (i)
dx x x 1
2
dv dv
Let y = vx  =v+x substituting in (i)
dx dx 1
2
dv
We get v + x = v + 1 + v2
dx
dv dx
 =
1 + v2 x 1
2
Integrating both sides, we get
2
log | 1 + v + v| = log  x + log C
1
y+ x 2 + y 2 = C x2 1
2

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29.

Ans
Correct Graph 2

Corner points Value of Z
(2, 72) (12 + 216 = 228)
1
(15, 20) (90 + 60 = 150)
(40, 15) (240 + 45 = 285) Maximum

30.

𝑘 𝑘 𝑘
Ans (a) (i) 2 + 3 + 6 = 1 1
Gives k = 1 ½
5𝑘 5
(ii) P(1 X< 3) = = ½
6 6
k 2k k 5k
(iii)E(X) = pi.xi = + + = ½
2 3 2 3

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5 1
2
E(X) =
3
OR
– 1 – 1
(b)P(A) P( B ) = P( A ) P(B) = ½
4 6
Let P(A) = x P(B) = y
1 1 1
x(1 – y) = , (1 – x)y = x–y=
4 6 12
2 1
eliminating y, we get 12x – 13x + 3 = 0
1 3 1
gives x = ,
3 4
1 1
P(A) =  P(B) =
3 4
3 2 ½
P(A) =  P(B) =
4 3

31.

 e sin x dx
Ans x
(a)Let I =

= ex sin x –
 cos x e dx x
1

= ex sin x – cos x ex – I ½
1 x
I= e (sin x – cos x) ½
2
/2

 ex sin x dx = 2 e/2 + 2 or 2 (e/2 + 1)
1 1 1
 1
0
OR


1
(b)Let I = dx
cos (x – a) cos(x – b)
1 𝑠𝑖𝑛[(𝑥 – b) – (x – a)]
= 𝑠𝑖𝑛(a – b) ∫ 𝑐𝑜𝑠(𝑥 – a) cos(x –b) dx 1


1  sin (x – b) cos(x – a) cos(x – b) sin(x – a)  ½
= – dx
sin (a – b)  cos( x – a) cos(x – b) cos( x – a) cos(x – b) 


1  
=  [tan( x – b) – tan(x – a)]dx  ½
sin (a – b)  

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1 1
= [log  sec(x – b) – log  sec(x – a)] + C
sin (a – b)
SECTION-D
(Question nos. 32 to 35 are Long Answer type questions carrying 5 marks each)

32.

Ans For reflexive
(1, 1) R as 12 is rational (or any other counter example) 1½
R is not reflexive 1
For symmetric
Let (x, y) R  x.y is an irrational number
 (y.x) is an irrational number
 (y, x) R
1½
 R is symmetric 2
For Transitive
(1, √2) R, (√2, 2) R (or any other counter example)
but (1, 2) R 2
 R is not transitive

33.

Ans 1 2 –2 3 –1 1
(a)A = [ – 1 3 0], B–1 = [–15 6 –5 ]
0 –2 1 5 –2 2 ½
(AB)–1 = B–1A–1
 A = 1(3) – 2(– 1) – 2(2) = 3 + 2 – 4 = 1  0 1
3 2 6

adj(A) = 1 1 2 2
 
2 2 5 
3 2 6
A–1 =
1 1 1 2
1   ½
2 2 5 
3 –1 1 3 2 6
 B–1A–1 = [–15 6 –5 ] [1 1 2]
5 –2 2 2 2 5

12 MS_Mathematics_041_65/1/1_2022-23

 10 7 21 

= – 49 – 34 – 103 
  1
 17 12 36 
OR
(b)Given system is
1 2 3  x  6 
2 –1 1   y  2 
    
3 2 – 2   z  = 3 
A . X = B  X = A–1B
 A = 35  0 ½
A11 = 0 A12 = 7 A13 = 7 1
A21 = 10 A22 = –11 A23 = 4
A31 = 5 A32 = 5 A33 = – 5 1½
0 10 5
 A–1 =
1
[7 –11 5] 1
35
7 4 –5
1 0 10 5 6 1 35 1
X= [7 –11 5 2
] [ ] = [35]= [1] 1
35 7 4 –5 3 35 35 1
x=1 y=1 z=1

34.

Ans (a)
Vector equation of required line through (1, 2, – 4) is
^ ^ ^ ^ ^ ^ 1
𝑟⃗⃗ = i + 2 j – 4 k + 𝜆(2 i + 3 j + 6 k )
𝑥–1 y–2 z+4
and cartesian equation: 2
= 3 = 6 1
Equation of line through A(3, 3, – 5) and B(1, 0, – 11) is
^ ^ ^ ^ ^ ^ ½
𝑟⃗⃗ = 3 i + 3 j – 5 k + 𝜇(2 i + 3 j + 6 k )
⃗|
⃗⃗⃗⃗ − 𝑎⃗ )× 𝑏
|(𝑎
Distance between parallel lines is given by d = 2 ⃗ 1
|𝑏|
→ ^ ^ ^ ^ ^ ^ → ^ ^ ^
Here b = 2 i + 3 j + 6 k , 𝑎1 = i + 2 j – 4 k , a 2 = 3 i + 3 j – 5 k
→ ^ ^ ^
( a 2 – 𝑎1 ) = 2 i + j – k
→ → → ½
^ ^ ^
( a 2 – a 1 ) × b = 9 i – 14 j + 4 k 11

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d=
√293 1
7

OR
𝑥–1 y–2 z–3
(b)Equation of line AB is 2
= 3 = 6
1
Let coordinates of required point on AB be (2 + 1, 3 + 2, 6 + 3) for some 
1
According to Question
(2 - 2)2 + (3 - 3)2 + (6 - 6 )2 = 142 gives 2 -2 -3 = 0 1
Solving we get = 3 and -1 1
 required points are (7, 11, 21) and (– 1, – 1, – 3) 1

35.
Ans Let Correct Graph : 1½

x coordinates of point of intersection are – 1, 2
–1 0 ½
Required area =

–2
(x + 2) dx +

–1
x2 dx 1½

2 –1 3 0
(x + 2) x
= +
2 3 1
–2 –1
1 1 5
= + = ½
2 3 6
SECTION-E
(Question nos. 36 to 38 are source based/case based/passage based/integrated units of assessment
questions carrying 4 marks each)

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36.

Ans (i) x + 0·21 = 0·44  x = 0·23 1 1
(ii) 0·41 + y + 0·44 + 0.11= 1  y = 0·04 1 1
 C  P (C  B)
(iii) (a) P   =
B P (B )
P(B) = 0·09 + 0·04 + 0·23= 0·36 1 1
 C  0·23 23 1
P  = =
 B  0·36 36
OR
(iii) (b) P(A or B but not C)
= 0·32 + 0.09 + 0.04 1½
= 0·45 ½

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37.

Ans 1 2 1
((i) v = r h =  r3 [ as  = 45 gives r = h] 1
3 3
𝑑𝑣 𝑑𝑟
(ii) = r2 ½
𝑑𝑡 𝑑𝑡
 dr  1
  =– cm/sec ½
 dt r = 2 2 4𝜋
(iii)(a) C = rl = r 2 r = 2 r2 1
𝑑𝐶 𝑑𝑟
= 2  2r ½
𝑑𝑡 𝑑𝑡
𝑑𝐶
( 𝑑𝑡 ) = – 2 cm2/sec
𝑟 = 2√2 ½
OR
(iii)(b) l2 = h2 + r2
l=4 r=h= 2 2
1
h = r  𝑑ℎ
𝑑𝑡
𝑑𝑟 1
= 𝑑𝑡 = – 4𝜋 cm/sec
1

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38.

Ans 1 1+1
(i)– 1 = a(– 27)  a =
27
1
(ii)f(x) = (x + 9)(x + 1)(x – 3) ½
27
1
= (x3 + 7x2 – 21x – 27)
27
1
f(x) = (3x2 + 14x – 21) ½
27
f(x) =
6𝑥 + 14 ½
27
f(1) =
20
27 ½

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