CBSE Class 12 Question Paper 2022 Solution Maths

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Senior Secondary School Term II Examination, 2022
Marking Scheme – MATHEMATICS (SUBJECT CODE – 041)
(PAPER CODE – 65/2/1)
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XII_127_041_65/2/1_Mathematics # Page-1

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XII_127_041_65/2/1_Mathematics # Page-2

MARKING SCHEME
Senior Secondary School Examination TERM–II, 2022
MATHEMATICS (Subject Code–041)
[Paper Code: 65/2/1]
Maximum Marks: 40
Q. No. EXPECTED ANSWER / VALUE POINTS Marks
SECTION – A

1.
2

Sol. Given differential equation can be written as
2
 dy  dy
2 xy   + y 2 + y = 0 Order = 1, Degree = 2 1
 dx  dx
Order × degree = 1 × 2 = 2 1

2

2. (a)
2

sin 3x 3sin x − 4sin 3 x ½
Sol.  sin x dx =  sin x
dx
½
(1−cos 2𝑥)
= ∫[3 − 4 ]𝑑𝑥
2

=  (1 + 2 cos 2 x)dx

= x + sin 2 x + C 1

2
Or

2. (b)
2

Sol. Let e x = t , e x dx = dt ½

XII_127_041_65/2/1_Mathematics # Page-3

1 log 3
ex 3 dt ½
 2 dx = 
2x 2
0 e +1 1 t +1
3
= tan −1 t ½
1

  
= − = ½
3 4 12
2
3.
2

Sol. |2𝑎⃗ + 3𝑏⃗⃗| = |3𝑎⃗ − 2𝑏⃗⃗|
2 2
⟹ |2𝑎⃗ + 3𝑏⃗⃗| = |3𝑎⃗ − 2𝑏⃗⃗|
2 2
⟹ 4|𝑎⃗|2 + 12 𝑎⃗. 𝑏⃗⃗ + 9|𝑏⃗⃗| = 9|𝑎⃗|2 − 12 𝑎⃗. 𝑏⃗⃗ + 4|𝑏⃗⃗| 1
As |𝑎⃗| = |𝑏⃗⃗| = 1
2 ½
∴ 24 𝑎⃗. 𝑏⃗⃗ = 5|𝑎⃗|2 − 5|𝑏⃗⃗| = 0 ⟹ 𝑎⃗. 𝑏⃗⃗ = 0

So, 𝑎⃗ ⊥ 𝑏⃗⃗ or Angle between them is ½
2
2
4.

2
Sol. A: number appearing on at least one die is 3
B: sum of numbers appearing on both dice is even
Clearly, 𝐴 ∩ 𝐵 = {(3,1), (3,5), (1,3), (5,3), (3,3)} ½
𝐴 𝑃(𝐴∩𝐵) 5⁄
𝑃 (𝐵) = = 18⁄36 1
𝑃(𝐵) 36
5 ½
=
18

2
5.

2

Sol. P (Problem is solved) =1 – P (Problem not solved) 1
= 1 − P ( A) P ( B )

XII_127_041_65/2/1_Mathematics # Page-4

1 2
= 1− 
3 5
13 1
=
15
2
6.

2

Sol. Required equation of line is given by
𝑥−2 𝑦−2 𝑧−1
= = 1
3 −1 −3

y − 2 −3 ½
putting z = –2, we get = =1
−1 −3
½
y − 2 = −1  y = 1

2
SECTION – B

7.
3

D a C
Sol. Let ⃗⃗⃗⃗⃗⃗ 𝐴𝐷 = 𝑏⃗⃗
𝐴𝐵 = 𝑎⃗ and ⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗ = 𝐴𝐵
𝐴𝐶 ⃗⃗⃗⃗⃗⃗ = 𝑎⃗ + 𝑏⃗⃗ = 𝑖̂ + 𝑗̂
⃗⃗⃗⃗⃗⃗ + 𝐵𝐶
⃗⃗⃗⃗⃗⃗⃗ = 𝐵𝐶
𝐵𝐷 ⃗⃗⃗⃗⃗⃗ = 𝑏⃗⃗ − 𝑎⃗ = 2𝑖̂ + 𝑗̂ + 𝑘̂
⃗⃗⃗⃗⃗⃗ + 𝐶𝐷 b b

⃗⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗
Adding we get, 2𝐴𝐷 𝐵𝐷 = 3𝑖̂ + 2𝑗̂ + 𝑘̂ A
𝐴𝐶 + ⃗⃗⃗⃗⃗⃗⃗ a B
3 1 1
𝐴𝐷 = 𝑖̂ + 𝑗̂ + 𝑘̂
⟹ ⃗⃗⃗⃗⃗⃗ 2 2

Subtracting, we get
̂ ⟹ ⃗⃗⃗⃗⃗⃗⃗ 1 1̂
⃗⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗
2𝐴𝐵 𝐴𝐶 − 𝐵𝐷⃗⃗⃗⃗⃗⃗⃗ = −𝑖̂ − 𝑘 𝐴𝐵 = − 2 𝑖̂ − 2 𝑘 1

𝑖̂ 𝑗̂ 𝑘̂
⃗⃗⃗⃗⃗⃗⃗ | = |1 1 0| = 𝑖̂ − 𝑗̂ − 𝑘̂
⃗⃗⃗⃗⃗⃗ × 𝐵𝐷
|𝐴𝐶
2 1 1
1
⃗⃗⃗⃗⃗⃗
Area = 2 |𝐴𝐶 × ⃗⃗⃗⃗⃗⃗⃗ 𝐵𝐷 |
3
= 1
2
XII_127_041_65/2/1_Mathematics # Page-5

3
8.(a)

3

Sol. 1 1
Consider  (tan −1 x) dx = tan −1 x  x −   x dx
1 + x2
1 1
= 𝑥 𝑡𝑎𝑛−1 𝑥 − 2 log(1 + 𝑥 2 )
1 1 1
∫0 (𝑡𝑎𝑛−1 𝑥)𝑑𝑥 = 𝑥 𝑡𝑎𝑛−1 𝑥 − 2 𝑙𝑜𝑔(1 + 𝑥 2 )]
0
 1
= − log 2
4 2
1
3
Or

8.(b)

3

Sol. 2 x  dx 2x
I = = dx ½
x 2 + 3x + 2 ( x + 1)( x + 2)
 −2 4 
=  +  dx using partial fraction 1½
 x +1 x + 2 
= −2 log | x + 1| + 4 log | x + 2 | + C 1
3

9.

3

Sol. Given differential equation can be written as
𝑑𝑦 dy 1
𝑥
𝑑𝑥
− 𝑦 = 3𝑥 2 or − y = 3x ½
dx x
1
1
I.F = 𝑒 ∫ −𝑥𝑑𝑥 = 𝑒 − 𝑙𝑜𝑔 𝑥 = −𝑥 −1 = 𝑥 1
1 1
Solution is y  =  3x dx + C
x x ½
y
= 3x + C
x ½

XII_127_041_65/2/1_Mathematics # Page-6

x = 1, y = 1 gives C = −2
y
Particular solution is = 3x − 2 or y = 3x 2 − 2 x
x ½

3

10.(a)
3

Sol. Equation of plane is given by
x−2 y −1 z
1 −3 −2 = 0 1½
−1 0 −7

21(𝑥 − 2) + 9(𝑦 − 1) − 3𝑧 = 0

i.e., 7 x + 3 y − z = 17 1

Distance of plane from origin is
|0+0−0−17| 17 ½
𝑑= =
√59 √59

3

Or

10.(b)

3

Sol. x y −1 z − 2 x + 1 y + 2 z −1
For lines = = and = =
1 2 3 1 2 3
𝑎1 = 𝑗̂ + 2𝑘̂ , 𝑏⃗⃗ = 𝑖̂ + 2𝑗̂ + 3𝑘̂
Let ⃗⃗⃗⃗⃗
𝑎2 = −𝑖̂ − 2𝑗̂ + 𝑘̂, 𝑏⃗⃗ = 𝑖̂ + 2𝑗̂ + 3𝑘̂
⃗⃗⃗⃗⃗
Clearly lines are parallel ½
Hence, Shortest distance or distance is given by
|(𝑎 𝑎1 × 𝑏⃗⃗|
⃗⃗⃗⃗⃗2 − ⃗⃗⃗⃗⃗)
|𝑏⃗⃗|
⃗⃗⃗⃗⃗ 𝑎1 = −𝑖̂ − 3𝑗̂ − 𝑘̂
𝑎2 − ⃗⃗⃗⃗⃗ ½
𝑖̂ 𝑗̂ 𝑘̂
(𝑎 𝑎1 × 𝑏⃗⃗ = |−1 −3 −1|
⃗⃗⃗⃗⃗2 − ⃗⃗⃗⃗⃗)
1 2 3

XII_127_041_65/2/1_Mathematics # Page-7

= −7iˆ + 2 ˆj + kˆ 1
49 + 4 + 1 27 3 21 1
Required distance = = or
1+ 4 + 9 7 7

3

SECTION – C

11.

4

Sol. General point of the line is
(3 + 2, 4 − 1, 12 + 2) 1

This will be the point of intersection with the plane, if
(3 + 2) − (4 − 1) + (12 + 2) = 5

 11 + 5 = 5,   = 0 1

Thus, the point of intersection is (2, –1, 2) 1

Distance = (2 + 1)2 + (−1 + 5)2 + (2 + 10)2

= 9 + 16 + 144

= 169 = 13 1

4

12.
4

XII_127_041_65/2/1_Mathematics # Page-8

Sol. 1
I =  x(1 − x)n dx
0
1
=  (1 − x)[1 − (1 − x)]n dx [using property] 1
0
1
=  x n (1 − x) dx
0
1 1 1
= ∫0 𝑥 𝑛 𝑑𝑥 − ∫0 𝑥 𝑛+1 𝑑𝑥
1 1
𝑥 𝑛+1 𝑥 𝑛+2
= [ 𝑛+1 ] − [ 𝑛+2 ] 1
0 0

1 1 1
= − Or
n +1 n + 2 (𝑛+1)(𝑛+2)
1

4

13.(a)
4

Sol. Clearly point of intersection are Y
3/2
3   3
 , 0  &  0,  ½
2   2
X’ X
O 3
2

Correct Figure 1
Y’

3/2 9 2 3/2  3 
Required area =  − x dx −   − x  dx 1
0 4 0 2 

1

9 9 ½
= −
16 8

4

Or

XII_127_041_65/2/1_Mathematics # Page-9

13.(b)

4

Sol.
Y y 2 = 4ax

256
Given area =
3
X’ X

Y’ x = 4a
1
Correct Figure
4a
Area of Shaded region = 2 4ax dx 1
0

4a
x3/ 2
=8 a
3 1
0

64a 2
= ½
3
64𝑎2 256
=
3 3

⟹ a 2 = 4 gives a = 2 (as a > 0) ½

4

XII_127_041_65/2/1_Mathematics # Page-10

14.

2

2

Sol. (a) Let X denote the number of tails:
½
X 0 1 2
1 1 1 1½
P(X)
4 2 4

(b) S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
P (at least one head) = 1–P (no head) 1
1
= 1−
8
7
= 1
8

2+2

***

XII_127_041_65/2/1_Mathematics # Page-11