CBSE Class 12 Question Paper 2024 Solution Maths

Marking Scheme
Strictly Confidential
(For Internal and Restricted use only) Senior School Certificate Examination, 2024
MATHEMATICS PAPER CODE 65/2/1

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MS_XII_Mathematics_041_65/2/1 1

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MS_XII_Mathematics_041_65/2/1 2

Q. No. EXPECTED OUTCOMES/VALUE POINTS Marks
SECTION A
Questions no. 1 to 18 are multiple choice questions (MCQs) and
questions number 19 and 20 are Assertion-Reason based questions of 1
mark each
1.

Sol. (A) 0 1
2.

Sol. (C) Bijective 1
3.

Sol. (D) 4 1

4.

Sol. (B) 1 1

MS_XII_Mathematics_041_65/2/1 3

5.

Sol. (B) Strictly increasing on R 1
6.

Sol. (B) Zero (0) 1
7.

Sol. (A) 𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛 ( )
𝑦 1
𝑥

8.

Sol. (C) 𝑎⃗. 𝑏⃗⃗ ⩽ |𝑎⃗||𝑏⃗⃗| 1

MS_XII_Mathematics_041_65/2/1 4

9.

Sol. (D) (0, 0, 0) 1
10.

Sol. (D) a feasible region 1
11.

Sol. (C) 1 1
12.

Sol. (C) 𝑎13 > 𝑎31 1

13.

Sol. 2x 1
(B)
1 + x4

MS_XII_Mathematics_041_65/2/1 5

14.

Sol. (D) not defined 1
15.

Sol. (B) ĵ 1
16.

Sol. (D) 2, -1, 3 1
17.

Sol. (B) 2 1

18.

Sol. (A) 900 1

MS_XII_Mathematics_041_65/2/1 6

19.

Sol. (D) Assertion (A) is false, but Reason (R) is true 1
20.

Sol. (C) Assertion (A) is true, but Reason (R) is false. 1

SECTION B
21.

−𝜋 𝜋 𝜋
Sol. The given expression = +3−4 1
6 1
−𝜋 2
= 1
12
2

OR

Sol. −1 ⩽ 𝑥 2 − 4 ⩽ 1 1
⟹ 3 ⩽ 𝑥2 ⩽ 5 2
Domain = [−√5, −√3] ∪ [√3, √5] 1
2

MS_XII_Mathematics_041_65/2/1 7

𝜋 𝜋
Range = [− , ] 1
2 2

22.

Sol.
𝜋 𝜋 1
𝑓(𝑥) = −𝑡𝑎𝑛2𝑥, <𝑥<
4 𝜋 2 𝜋 1
2
𝑓′(𝑥) = −2𝑠𝑒𝑐 2𝑥, 4 < 𝑥 < 2 2
𝜋 1
𝑓′( ) = −2(−2)2 = −8
3 2

OR

Sol. 𝑦 = √1 + 𝑐𝑜𝑡 2 (𝑐𝑜𝑡 −1 𝑥) = √1 + 𝑥 2 1
2
𝑑𝑦 𝑥
⟹ = 1
𝑑𝑥 √1 + 𝑥 2
1
𝑑𝑦
⟹ √1 + 𝑥 2 −𝑥 =0 2
𝑑𝑥

23.

Sol.
1 𝑥 2 − 1 (𝑥 + 1)(𝑥 − 1)
𝑓′(𝑥) = 1 − = = 1
𝑥2 𝑥2 𝑥2
𝑓′(𝑥) = 0 ⟹ 𝑥 = −1, 1 2
2
𝑓′′(𝑥) = 𝑥 3 ⟹ 𝑓′′(−1) = −2 < 0
⸫ -1 is a point of local maximum 1
The local maximum value = 𝑓(−1) = −2 = 𝑀
2
𝑓′′(1) = 2 > 0
⸫ 1 is point of local minimum 1
The local minimum value = 𝑓(1) = 2 = 𝑚 2
1
𝑀 − 𝑚 = −4 2

MS_XII_Mathematics_041_65/2/1 8

24.

Sol. 𝑒 4𝑥 −1
Given integral =∫ 𝑒 4𝑥 +1 𝑑𝑥
1
𝑒 2𝑥 −𝑒 −2𝑥
=∫ 𝑒 2𝑥 +𝑒 −2𝑥 𝑑𝑥 2

1 2(𝑒 2𝑥 −𝑒 −2𝑥 ) 1
=2 ∫ 𝑒 2𝑥 +𝑒 −2𝑥 𝑑𝑥
2
1
= 𝑙𝑜𝑔|𝑒 2𝑥 + 𝑒 −2𝑥 | + 𝐶
2 1

25.

Sol. 1 1
𝑓′(𝑥) = 𝑒 𝑥 + 𝑒 −𝑥 + 1 −
1 + 𝑥2
1 𝑥 2
=𝑒 𝑥 + 𝑒 𝑥 + 1+𝑥 2 > 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ 𝑅
1
⸫ f is strictly increasing over its domain R

Section C
26.

Sol 𝑑𝑥 1
= 𝑒 𝑐𝑜𝑠3𝑡 × (−𝑠𝑖𝑛3𝑡) × 3
𝑑𝑡
𝑑𝑦
= 𝑒 𝑠𝑖𝑛3𝑡 × (𝑐𝑜𝑠3𝑡) × 3
𝑑𝑡 1

𝑑𝑦
𝑑𝑦 𝑒 𝑠𝑖𝑛3𝑡 × (𝑐𝑜𝑠3𝑡)
= 𝑑𝑡 =
𝑑𝑥 𝑑𝑥 −𝑒 𝑐𝑜𝑠3𝑡 × (𝑠𝑖𝑛3𝑡)
𝑑𝑡

𝑥 = 𝑒 𝑐𝑜𝑠3𝑡 ⟹ 𝑐𝑜𝑠3𝑡 = 𝑙𝑜𝑔𝑥
𝑦 = 𝑒 𝑠𝑖𝑛3𝑡 ⟹ 𝑠𝑖𝑛3𝑡 = 𝑙𝑜𝑔𝑦
𝑑𝑦 −𝑦𝑙𝑜𝑔𝑥
∴ = 1
𝑑𝑥 𝑥𝑙𝑜𝑔𝑦

MS_XII_Mathematics_041_65/2/1 9

OR

Sol. 𝑑(|𝑥|) 𝑑(√𝑥 2 )
= ,𝑥 ≠ 0 1
𝑑𝑥 𝑑𝑥
1 1 𝑑(𝑥 2 )
= (𝑥 2 )−2 × 1
12
2 𝑑𝑥
1 𝑥 1
= 2𝑥 =
2√𝑥 2 |𝑥| 2

27.

Sol. 2 2−𝑥
(a) ∫−2 √2+𝑥 𝑑𝑥
2 2−𝑥 1
= ∫−2 √4−𝑥 2 𝑑𝑥
2
1
2 2 2 𝑥
= ∫−2 √4−𝑥 2 𝑑𝑥 − ∫−2 √4−𝑥 2 𝑑𝑥 2

2 2 2 𝑥 1
= 2∫0 √4−𝑥2 𝑑𝑥 − 0 [√4−𝑥 2is even, √4−𝑥 2 is odd]
2 1
= 4∫0 √4−𝑥2 𝑑𝑥
𝑥 2
1
= 4 𝑠𝑖𝑛−1 2| 2
0
1
= 2𝜋 2
OR

Sol. 1 1
Let 𝑙𝑜𝑔𝑥 = 𝑡 ⟹ 𝑥 𝑑𝑥 = 𝑑𝑡
2
1
1
The given integral becomes = ∫ 𝑡 2 −3𝑡−4 𝑑𝑡 2
1
=∫ 𝑑𝑥 1
3 2 5 2
(𝑡 − 2) − (2)

MS_XII_Mathematics_041_65/2/1 10

1 𝑡−4 1
= 𝑙𝑜𝑔 | |+𝐶
5 𝑡+1 2

1 𝑙𝑜𝑔𝑥 − 4 1
= 𝑙𝑜𝑔 | |+𝐶
5 𝑙𝑜𝑔𝑥 + 1 2

28.

Sol. Given differential equation can be written as
𝑑𝑦 2𝑥𝑦 + 𝑦 2 𝑦 𝑦 2 1
= = + 2
𝑑𝑥 2𝑥 2 𝑥 2𝑥 2
𝑑𝑦 𝑑𝑣
Let 𝑦 = 𝑣𝑥 ⟹ 𝑑𝑥 = 𝑣 + 𝑥 𝑑𝑥
1
The equation becomes
𝑑𝑣 1 2 2
𝑥 = 𝑣
𝑑𝑥 2
𝑑𝑣 1 𝑑𝑥 1
⟹ 2= ×
𝑣 2 𝑥 2
Integrating both sides, we get

−1 1
= 𝑙𝑜𝑔|𝑥| + 𝐶
𝑣 2
𝑥 1
⟹ − = 𝑙𝑜𝑔|𝑥| + 𝐶 1
𝑦 2
1
𝑦 = 2, 𝑥 = 1 gives 𝐶 = − 2
The particular solution is
𝑥 1 1 2𝑥 1
− = 𝑙𝑜𝑔|𝑥| − 𝑜𝑟, 𝑦 = 2
𝑦 2 2 1 − 𝑙𝑜𝑔|𝑥|

OR

Sol. Given differential equation can be written as
𝑑𝑥 𝑥 1
− = 2𝑦
𝑑𝑦 𝑦
∫
−1
𝑑𝑦 1 1
Integrating Factor = 𝑒 𝑦 =𝑦
1
Solution is 𝑥 𝑦 = ∫ 2𝑑𝑦 1
𝑥 2
⟹ = 2𝑦 + 𝐶
𝑦 1
⟹ 𝑥 = 2𝑦 2 + 𝐶𝑦
2

MS_XII_Mathematics_041_65/2/1 11

29.

Sol.
⃗⃗⃗⃗⃗⃗
𝐴𝐵 . ⃗⃗⃗⃗⃗⃗
𝐴𝐶 (−𝑖̂ − 2𝑗̂ − 6𝑘̂). (𝑖̂ − 3𝑗̂ − 5𝑘̂) 35 √35
𝑐𝑜𝑠𝐴 = = = =
⃗⃗⃗⃗⃗⃗ ||𝐴𝐶
|𝐴𝐵 ⃗⃗⃗⃗⃗⃗ | √41√35 √41√35 √41
√35 1
𝐴 = 𝑐𝑜𝑠 −1 ( )
√41
⃗⃗⃗⃗⃗⃗. 𝐵𝐶
𝐵𝐴 ⃗⃗⃗⃗⃗⃗ (𝑖̂ + 2𝑗̂ + 6𝑘̂). (2𝑖̂ − 𝑗̂ + 𝑘̂ ) 6 √6
𝑐𝑜𝑠𝐵 = = = =
⃗⃗⃗⃗⃗⃗||𝐵𝐶
|𝐵𝐴 ⃗⃗⃗⃗⃗⃗ | √41√6 √41√6 √41

√6 1
𝐵 = 𝑐𝑜𝑠 −1 ( )
√41
⃗⃗⃗⃗⃗⃗. ⃗⃗⃗⃗⃗⃗
𝐶𝐵 𝐶𝐴 (−2𝑖̂ + 𝑗̂ − 𝑘̂). (−𝑖̂ + 3𝑗̂ + 5𝑘̂)
𝑐𝑜𝑠𝐶 = = =0
⃗⃗⃗⃗⃗⃗ ||𝐶𝐴
|𝐶𝐵 ⃗⃗⃗⃗⃗⃗| ⃗⃗⃗⃗⃗⃗ ||𝐶𝐴
|𝐶𝐵 ⃗⃗⃗⃗⃗⃗| 1
𝜋
cos C = 0 ⟹𝐶=2

30.

Sol.
X 0 1 2 3 4 5
P(X) 6 1 10 5 8 2 6 1 4 1 2 1 1
= = = = = = ×6
36 6 36 18 36 9 36 6 36 9 36 18 2
=3
31.

Sol. 3
Let 𝑥 2 = 𝑡
3 1 1
⟹ 𝑥 2 𝑑𝑥 = 𝑑𝑡 2
2
2 1
The given integral becomes 3 ∫ 𝑡 𝑠𝑖𝑛−1 𝑡𝑑𝑡
2
2 𝑡2 1 𝑡2 1
= 3 [𝑠𝑖𝑛−1 𝑡 × 2 − ∫ √1−𝑡 2 × 2 𝑑𝑡]
1 1−𝑡 2 −1
= 3 [𝑠𝑖𝑛−1 𝑡 × 𝑡 2 + ∫ √1−𝑡 2 𝑑𝑡]
1 1
= 3 [𝑠𝑖𝑛−1 𝑡 × 𝑡 2 + ∫ √1 − 𝑡 2 𝑑𝑡 − ∫ √1−𝑡 2 𝑑𝑡]
1 𝑡 1
= 3 [𝑡 2 𝑠𝑖𝑛−1 𝑡 + 2 √1 − 𝑡 2 + 2 𝑠𝑖𝑛−1 𝑡 − 𝑠𝑖𝑛−1 𝑡]+C

MS_XII_Mathematics_041_65/2/1 12

1 𝑡 1
= [𝑡 2 𝑠𝑖𝑛−1 𝑡 + √1 − 𝑡 2 − 𝑠𝑖𝑛−1 𝑡] + 𝐶
3 2 2
3
1 3 −1 3 𝑥2 1 3
= [𝑥 𝑠𝑖𝑛 (𝑥 2 ) + √1 − 𝑥 3 − 𝑠𝑖𝑛−1 (𝑥 2 )] + 𝐶 1
3 2 2

Section D
32.

Sol. Let 𝑓(𝑥1 ) = 𝑓(𝑥2 ) 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑥1 , 𝑥2 ∈ 𝑅
2𝑥 2𝑥
Then 1+𝑥1 2 = 1+𝑥2 2
1 2

⟹ 𝑥1 + 𝑥1 𝑥2 2 = 𝑥2 + 𝑥1 2 𝑥2
⟹ (𝑥1 − 𝑥2 ) − 𝑥1 𝑥2 (𝑥1 − 𝑥2 ) = 0
⟹ (𝑥1 − 𝑥2 )(1 − 𝑥1 𝑥2 ) = 0
⟹ 𝑥1 − 𝑥2 = 0 or 1 − 𝑥1 𝑥2 = 0
⟹ 𝑥1 = 𝑥2 or 𝑥1 𝑥2 = 1, so if 𝑥1 𝑥2 = 1, 𝑥1 ≠ 𝑥2
Hence f is not one -one 2
Let y = f(x) where 𝑥 ∈ 𝑅
2𝑥
Then 𝑦 = 1+𝑥 2 . Here, for x = 0, y = 0

2𝑥
If 𝑦 ≠ 0, then 𝑦 = 1+𝑥 2
⟹ 𝑦𝑥 2 − 2𝑥 + 𝑦 = 0
2 ± √4 − 4𝑦 2
⟹𝑥=
2𝑦
For x to be real, 4 − 4𝑦 2 ≥ 0
⟹ 𝑦2 ≤ 1
⟹ −1 ≤ 𝑦 ≤ 1
Hence, range = [−1,1] ≠ 𝑐𝑜𝑑𝑜𝑚𝑎𝑖𝑛
Hence, f is not onto. 2

For the given function to become onto, 𝐴 = [−1,1] 1

MS_XII_Mathematics_041_65/2/1 13

OR

Sol. Let (𝑎, 𝑏) ∈ 𝑁 × 𝑁

We have
𝑎−𝑎 =𝑏−𝑏
This implies that (a, b) R (a, b) ∀(𝑎, 𝑏) ∈ 𝑁 × 𝑁
1
Hence R is reflexive 12

Let (a, b) R (c, d) for some (𝑎, 𝑏), (𝑐, 𝑑) ∈ 𝑁 × 𝑁
Then 𝑎 − 𝑐 = 𝑏 − 𝑑
⟹𝑐−𝑎 =𝑑−𝑏
⟹(c, d) R (a, b) 1
12
Hence, R is symmetric.

Let (a, b) R (c, d), (c, d) R (e, f) for some (𝑎, 𝑏), (𝑐, 𝑑), (𝑒, 𝑓) ∈ 𝑁 × 𝑁
Then 𝑎 − 𝑐 = 𝑏 − 𝑑, 𝑐 − 𝑒 = 𝑑 − 𝑓
⟹𝑎−𝑐+𝑐−𝑒 =𝑏−𝑑+𝑑−𝑓
⟹𝑎−𝑒 =𝑏−𝑓
⟹(a, b) R (e, f)
Hence, R is transitive 2

Thus, R is an equivalence relation.
33.

Sol Let direction ratios of the required line be a, b, c.
∵ the required line is perpendicular to both the given lines

⸫ 3𝑎 − 16𝑏 + 7𝑐 = 0 1
and 3𝑎 + 8𝑏 − 5𝑐 = 0 1
𝑎 𝑏 𝑐 2
⟹ = =
24 36 72
𝑎 𝑏 𝑐 1
12
⟹ = =
2 3 6
The mid-point of the line-segment AB is (3, 4, 6) 1
Hence, the required equation of the line is
𝑥−3 𝑦−4 𝑧−6
= = 1
2 3 6

MS_XII_Mathematics_041_65/2/1 14

34.

Sol. Given system of linear equations is equivalent to AX = B, where
1

2 3 10 𝑥 4 1
1
𝐴 = [4 −6 5 ], 𝑋 = 𝑦 , 𝐵 = [1] 2
6 9 −20 1 2
[𝑧 ] 1
|𝐴| = 1200 ≠ 0 2
Cofactors of the elements of A are
𝐴11 = 75, 𝐴12 = 110, 𝐴13 = 72
𝐴21 = 150, 𝐴22 = −100, 𝐴23 = 0
𝐴31 = 75, 𝐴32 = 30, 𝐴33 = −24

75 150 75 2
𝑎𝑑𝑗𝐴 = [110 −100 30 ]
72 0 −24

𝑎𝑑𝑗𝐴 1 75 150 75 1
−1
𝐴 = = [110 −100 30 ]
|𝐴| 1200 2
72 0 −24

75 150 75 4
−1 1
𝑋 = 𝐴 𝐵 = 1200 [110 −100 30 ] [1]
72 0 −24 2
1
600 2
1 1
=1200 [400] = 3 1
240 1
[5] 1
⸫x = 2, y = 3, z = 5 2

OR

Sol. |𝐴| = 1 + 𝑐𝑜𝑡 2 𝑥 = 𝑐𝑜𝑠𝑒𝑐 2 𝑥 1
1 −𝑐𝑜𝑡𝑥 2
𝑎𝑑𝑗𝐴 = [ ] 1
𝑐𝑜𝑡𝑥 1 12

MS_XII_Mathematics_041_65/2/1 15

𝑎𝑑𝑗𝐴 1 1 −𝑐𝑜𝑡𝑥
𝐴−1 = = 2
[ ]
|𝐴| 𝑐𝑜𝑠𝑒𝑐 𝑥 𝑐𝑜𝑡𝑥 1
1
1 −𝑐𝑜𝑡𝑥
𝐴′ = [ ] 1
𝑐𝑜𝑡𝑥 1
2
1 1 − 𝑐𝑜𝑡 2 𝑥 −2𝑐𝑜𝑡𝑥 ]
𝐴′ 𝐴−1 = [
𝑐𝑜𝑠𝑒𝑐 2 𝑥 2𝑐𝑜𝑡𝑥 1 − 𝑐𝑜𝑡 2 𝑥
2 2
= [𝑠𝑖𝑛 𝑥 − 𝑐𝑜𝑠 𝑥 −2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 ]
2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 𝑠𝑖𝑛2 𝑥 − 𝑐𝑜𝑠 2 𝑥
−𝑐𝑜𝑠2𝑥 −𝑠𝑖𝑛2𝑥
=[ ]
𝑠𝑖𝑛2𝑥 −𝑐𝑜𝑠2𝑥 1
12

35.

Sol.

2

3 1
1 𝑥2 4 1
𝐴1 = Area (region OABO) = ∫0 2√𝑥 𝑑𝑥 = 2 [ 3 ]| = 3
2 0

4 2 64
𝐴2 = Area (region ODEO) = 2 ∫0 2√𝑥 𝑑𝑥 = 4 × 3 [23 ] = 3 1

4 64
𝐴1 : 𝐴2 = : = 1: 16
3 3 1

MS_XII_Mathematics_041_65/2/1 16

Section E
36.

Sol. (i) When V = 40 km/h, F = 36/5 ℓ/100km 1

(ii)
𝑑𝐹 𝑉 1 1
= −
𝑑𝑉 250 4

(iii)(a)
𝑑𝐹
=0 1
𝑑𝑉
⟹V = 62.5 km/h 1
2
𝑑2 𝐹 1
= 250 > 0 at V = 62.5 km/h
𝑑𝑉 2
1
Hence, F is minimum when V = 62.5 km/h
2

MS_XII_Mathematics_041_65/2/1 17

OR

(iii) (b)
𝑑𝐹
= −0.01
𝑑𝑉
𝑉 1 −1
⟹ − =
250 4 100 1
⟹ 𝑉 = 60 km/h

602 60 1
𝐹= − + 14 = 6.2 ℓ⁄100𝑘𝑚
500 4 2

Quantity of fuel required for 600 km 1
= 6.2 × 6 = 37.2 ℓ 2

37.

MS_XII_Mathematics_041_65/2/1 18

Sol. (i)Constraints are 𝑥 + 2𝑦 ≥ 10
1
𝑥+𝑦 ≥6 12
3𝑥 + 𝑦 ≥ 8 1
𝑥≥0
2
𝑦≥0

(ii)
Corner points Value of Z = 16x + 20y
A (10, 0) 160
B (2, 4) 112
1
C (1, 5) 116 12
D (0, 8) 160
1
The minimum cost is ₹112 2

38.

MS_XII_Mathematics_041_65/2/1 19

Sol.
(i) 𝑃(𝐸2 ) = 1 − 0.0000001

= 0.9999999 1
95 195
(ii) 𝑃(𝐴⁄𝐸1 ) + 𝑃(𝐴⁄𝐸2 ) = 100 + 1 = 100 1

(iii)(a) 𝑃(𝐴) = 𝑃(𝐸1 ) × 𝑃(𝐴⁄𝐸1 ) + 𝑃(𝐸2 ) × 𝑃(𝐴⁄𝐸2 ) 1

1 95 9999999
= 10000000 × 100 + 10000000 × 1

95+999999900 999999995
= 1000000000 = 1000000000 1

OR

(iii)(b)
𝑃(𝐸 )×𝑃(𝐴⁄𝐸2 )
𝑃(𝐸2 ⁄𝐴) = 𝑃(𝐸 )×𝑃(𝐴⁄2𝐸 )+𝑃(𝐸 )×𝑃(𝐴 1
1 1 2 ⁄𝐸 )
2

9999999
10000000 999999900
=
999999995
= 1
999999995
1000000000

MS_XII_Mathematics_041_65/2/1 20