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1. Marking Scheme
Class – XII (MATHEMATICS)
Section: A (Multiple Choice Questions -1 Mark each)
Question No. Answer Hints / Solution
1 A 𝜋
3
2 D 216
3 B K=6
4 C 𝜋
4
5 C log x
6 A 43
2
sq. units
7 D 1
16
8 A |𝐴|3
9 C x cos x
10 A 𝑥4
+
1
𝑥3
−
129
8
11 C sin−1
(𝑥 − 1)+C
12 B sin (
𝑦
𝑥
)=Cx

2. 13 D −
3
2
14 C 𝜆 = −
2
3
15 A −
2
3
16 D P (
𝐴
𝐵
) = 0
17 A 1
√14
, -
3
√14
,
2
√14
18 D the vertex which is maximum distance from (0,0)
19 A
20 B
Section: B (VSA questions of 2 marks each)
21 cos−1
(
1
2
) = cos−1
(cos
𝜋
3
) =
𝜋
3
sin−1
(
1
2
)
= sin−1
(sin
𝜋
6
) =
𝜋
6
cos−1
(
1
2
) + 2sin−1
(
1
2
)=
𝜋
3
+ 2(
𝜋
6
)=
𝜋
3
+
𝜋
3
=
2𝜋
3
½
½
1

3. OR
For injectivity, let f(x1) = f(x2)
𝑥1−2
𝑥1−3
=
𝑥2−2
𝑥2−3
x1 = x2, so f(x) is an injective function.
For surjectivity, let y=
𝑥−2
𝑥−3
x-2=xy-3y
x(1-y) =2-3y
x=
2−3y
1−y
or x =
3y−2
y−1
∈A,∀ y∈B, so f(x) is surjective
function.
Hence, f(x) is a bijective function.
1
1
22 Given:
𝑑𝑉
𝑑𝑡
=12, let height = y and radius = x, then y=x/6 or x=6y
V=
1
3
𝜋𝑥2
y, putting x=6y,
V=
1
3
𝜋(6𝑦)2
y=12𝜋𝑦3
𝑑𝑉
𝑑𝑦
=36𝜋𝑦2
, ∴
𝑑𝑉
𝑑𝑦
×
𝑑𝑦
𝑑𝑡
=12
36𝜋𝑦2
×
𝑑𝑦
𝑑𝑡
=12, so
𝑑𝑦
𝑑𝑡
=
1
36𝜋𝑦2
½
1/2
½

4. When y= 4,
𝑑𝑦
𝑑𝑡
=
1
48𝜋
cm/sec. 1/2
23 x √1 + 𝑦 +y √1 + 𝑥 =0
x √1 + 𝑦 = -y √1 + 𝑥
squaring both sides, y = −
𝑥
1+𝑥
𝑑𝑦
𝑑𝑥
= −
1
(1+𝑥)2
1
1
24 𝑎
⃗ + 𝑏
⃗⃗⃗⃗=4 𝑖̂ + 𝑗̂-𝑘
̂ , 𝑎
⃗ − 𝑏
⃗⃗⃗⃗= -2 𝑖̂ + 3𝑗̂-5𝑘
̂
(𝑎
⃗⃗⃗⃗⃗ + 𝑏
⃗⃗⃗⃗).( 𝑎
⃗ + 𝑏
⃗⃗⃗⃗) = -8+3+5=0
OR
Let P divides QR in the ratio 𝜆: 1
Then coordinates of are (
5𝜆+2
𝜆+1
,
𝜆+2
𝜆+1
,
−2𝜆+1
𝜆+1
)
∴
5𝜆+2
𝜆+1
=4, ∴ 𝜆=2
So, z-coordinate of P = -1
1
1
1
½
1/2

5. 25
|
𝑖̂ 𝑗̂ 𝑘
̂
2 6 27
1 𝜆 𝜇
| = 0
⃗⃗
(6𝜇-27𝜆)𝑖̂ – (2𝜇-27)𝑗̂ +(2𝜆-6)𝑘
̂=0
6𝜇-27𝜆=0, 2𝜇-27=0, 2𝜆-6=0
𝜇 =
27
2
and 𝜆 =3
½
½
½
½
Section: C (SA questions of 3 marks each)
26
∫
𝑥+3
√5−4𝑥−𝑥2
𝑑𝑥 = −√5 − 4𝑥 − 𝑥2 + sin−1 (𝑥+2)
3
+C 2+1
27 Correct Figure
Corner points (0,0), (20,0), (12,6), (0,10)
Z has maximum value 1680 at (12,6)
1
1
1
28 Let I=∫ log(1 + tan x)
π
4
0
dx …… (i)
Changing x to (
𝜋
4
-x) [ Using ∫ 𝑓(𝑥)𝑑𝑥
𝑎
0
= ∫ 𝑓(𝑎 − 𝑥)𝑑𝑥
𝑎
0
]
I= ∫ 𝑙𝑜𝑔 [1 + tan(
𝜋
4
− x)]
𝜋
4
0
dx = ∫
𝜋
4
0
log(
2
1+𝑙𝑜𝑔𝑥
) dx (ii)
½
1

6. Adding equations (i) and (ii), we have
2I =∫ log(1 + tan x)
π
4
0
dx +∫
𝜋
4
0
log(
2
1+𝑙𝑜𝑔𝑥
) dx
= ∫
𝜋
4
0
log 2 dx =
𝜋
8
log 2
OR
I = ∫
𝜋
0
𝑥 𝑡𝑎𝑛𝑥
𝑠𝑒𝑐𝑥+𝑡𝑎𝑛𝑥
dx
I = ∫
𝑥 sin 𝑥
1+sin 𝑥
𝜋
0
dx (i)
[ Using ∫ 𝑓(𝑥)𝑑𝑥
𝑎
0
= ∫ 𝑓(𝑎 − 𝑥)𝑑𝑥
𝑎
0
]
I = ∫
(𝜋−𝑥) sin(𝜋−𝑥)
1+sin(𝜋−𝑥)
𝜋
0
dx = ∫
(𝜋−𝑥) sin 𝑥
1+sin𝑥
𝜋
0
dx (ii)
Adding equations (i) and (ii), we have
2I = 𝜋 ∫
sin 𝑥
1+sin 𝑥
𝜋
0
dx
2I = 𝜋 ∫ (1 −
1
1+sin 𝑥
𝜋
0
) dx
2I = 𝜋2
− 2𝜋 = 𝜋(𝜋 −2)
∴ I =
𝜋
2
(𝜋 − 2)
½
1
½
1
½
1

7. 29 2𝑥2 𝑑𝑦
𝑑𝑥
−2xy+𝑦2
=0
𝑑𝑦
𝑑𝑥
=
2xy−𝑦2
𝑥2
It is homogeneous differential equation. Put y= v x
𝑑𝑦
𝑑𝑥
=v+x
𝑑𝑦
𝑑𝑥
Its solution is log |x|+C =
2𝑥
𝑦
OR
𝑑𝑦
𝑑𝑥
=1+x+y+xy
𝑑𝑦
𝑑𝑥
= (1+x) (1+y)
Its general solution is log|1+y| = x+
𝑥2
2
+C
C = -
3
2
Hence, Particular solution is log|1+y| = x+
𝑥2
2
-
3
2
½
½
1+1
1+1
½
1/2

8. 30 I= ∫
2 𝑑𝑥
(1−𝑥)(1+𝑥2)
Let
2 𝑑𝑥
(1−𝑥)(1+𝑥2)
=
𝐴
1−𝑥
+
𝐵𝑥+𝐶
1+𝑥2
A=B=C=1
I = ∫
2 𝑑𝑥
(1−𝑥)(1+𝑥2)
=∫
1
1−𝑥
dx +
1
2
∫
2𝑥
1+𝑥2dx +∫
1
1+𝑥2dx
I= - log|1-x| +
1
2
log (1+𝑥2
) +tan−1
𝑥 +C
½
1
½
1
31 S = {(B, B), (G, B), (B, G), (G, G)}
E: both the children are boys
F: at least one of the child is a boy
Then E = {(B, B)} and F = {(B, B), (G, B), (B, G)}
E∩F = {(B, B)}, Thus P(F) =
3
4
and P(E∩F) =
1
4
Therefore P (
𝐸
𝐹
)=
P(E∩F)
𝑃(𝐹)
=
1
3
OR
S = {1,2,3,4,5,6}
E= {3,6}, F= {2,4,6} and E∩ F = {6}
Then P(E) =
1
3
, P(F) =
1
2
and P(E∩F) =
1
6
Clearly P(E∩F) = P(E). P(F)
Hence E and F are independent events.
½
1
½
1
½
1
½
1

9. Section: D (LA questions of 5 marks each)
32 For Reflexive
For Symmetric
For Transitive
OR
For Reflexive
For Symmetric
For Transitive
1
2
2
1
2
2

10. 33 AX=B
Where A=[
3 −2 3
2 1 −1
4 −3 2
], X=[
𝑥
𝑦
𝑧
], and B=[
8
1
4
]
|A|= -17≠0
𝐴−1
= −
1
17
[
−1 −5 −1
−8 −6 9
−10 1 7
]
X=𝐴−1
B =−
1
17
[
−1 −5 −1
−8 −6 9
−10 1 7
] [
8
1
4
]
[
𝑥
𝑦
𝑧
] = −
1
17
[
−17
−34
−51
] =[
1
2
3
]
Hence x=1, y=2 and z=3
1
½
2
½
½
1/2
34 Correct Figure
The points of intersection of the parabola
y = 𝑥2
and the line y = x are (0,0) and (1,1).
Required Area=∫ 𝑥2
1
0
dx +∫ 𝑥 𝑑𝑥
2
1
=
1
3
+
3
2
=
11
6
1
1
1
1+1
35 Comparing the given equations with
equations 𝑟
⃗ = 𝑎1
⃗⃗⃗⃗⃗ +𝜆 𝑏1
⃗⃗⃗⃗ and 𝑟
⃗ = 𝑎2
⃗⃗⃗⃗⃗ +𝜆 𝑏2
⃗⃗⃗⃗⃗

11. 𝑎1
⃗⃗⃗⃗⃗ = 𝑖̂ + 𝑗̂, 𝑏1
⃗⃗⃗⃗ = 2𝑖̂ − 𝑗̂+𝑘
̂ 𝑎𝑛𝑑
𝑎2
⃗⃗⃗⃗⃗ = 2𝑖̂ + 𝑗̂-𝑘
̂, 𝑏2
⃗⃗⃗⃗⃗ = 3𝑖̂ − 5𝑗̂+2𝑘
̂
Therefore, 𝑎2
⃗⃗⃗⃗⃗ - 𝑎1
⃗⃗⃗⃗⃗ = 𝑖̂ - 𝑘
̂
𝑏1
⃗⃗⃗⃗ × 𝑏2
⃗⃗⃗⃗⃗ =|
𝑖̂ 𝑗̂ 𝑘
̂
2 −1 1
3 −5 2
|
= 3𝑖̂ − 𝑗̂ − 7𝑘
̂
| 𝑏1
⃗⃗⃗⃗ × 𝑏2
⃗⃗⃗⃗⃗ | =√9 + 1 + 49
=√59
S.D.= d = |
(𝑏1
⃗⃗⃗⃗⃗ ×𝑏2
⃗⃗⃗⃗⃗ ).( 𝑎2
⃗⃗⃗⃗⃗⃗ − 𝑎1
⃗⃗⃗⃗⃗⃗)
| 𝑏1
⃗⃗⃗⃗⃗ ×𝑏2
⃗⃗⃗⃗⃗ |
| = |
3−0+7
√59
|
d=
10
√59
units
1
1/2
1
½
1
1

12. Section: E (Case Studies/Passage based questions of 4 Marks each)
Q36. CASE STUDY:1 Answer
1 4-x
2 4
3 8
4 6
OR
1
Q37. CASE STUDY:2
1 12.5
2 Rs 38281.25
3 (0, 12.5)
4 37730
OR
15

13. Q38. CASE STUDY:3
1
7
10
2
1
10
3
7
25
4
3
4
OR
1
4