Students can use CBSE Previous Year Question Papers Class 12 Maths with Solutions and CBSE Class 12 Maths Question Paper 2022 (Term-II) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Maths Question Paper 2022 (Term-II) with Solutions
Time Allowed: 2 hours
Maximum Marks: 40
General Instructions:
Read the following instructions very carefully and strictly follow them:
- This question paper contains three Sections – Section A, Section B and Section C.
- Each section is compulsory.
- Section—A has 6 short-answer type-I questions of 2 marks each.
- Section—B has 4 short-answer type-II questions of 3 marks each.
- Section — C has 4 long-answer type questions of 4 marks each.
- There is an internal choice in some questions.
- Question 14 is a Case Study Based question with two sub-parts of 2 marks each.
Section – A
Question numbers 1 to 6 carry 2 marks each.
Question 1.
Find: \(\int \frac{d x}{\sqrt{4 x-x^2}}\)
Solution:
Question 2.
Find the general solution of the following differential equation:
\(\frac{d y}{d x}\) = \(e^{x-y}+x^2 e^{-y}\)
Solution:
Question 3.
Let X be a random variable which assumes values x
1
, x
2
, x
3
, x
4
such that 2P(X = x
1
) = 3P(X = x
2
) = P(X = x
3
) = 5P(X = x
4
).
Find the probability distribution of X.
Solution:
We have, 2P(X = x
1
) = 3P(X = x
2
)
P(X = x
3
) = 5P(X = x
4
) = k (let)
Here, 2P(X = x
1
) = k ⇒ P(X = x
1
) = \(\frac{k}{2}\)
3P(X = x
2
) = k ⇒ P(X = x
2
) = \(\frac{k}{3}\)
P(X = x
3
) = k
5P(X = x
4
) = k ⇒ P(X = x
4
) = \(\frac{k}{5}\)
As we know,
P(X = x
1
) + P(X = x
2
) + P(X = x
3
) + P(X = x
4
) = 1
∴ \(\frac{k}{2}\) + \(\frac{k}{3}\) + k + \(\frac{k}{5}\) = 1
⇒ \(\frac{15 k+10 k+30 k+6 k}{30}\) = 1
⇒ 61k = 30
⇒ k = \(\frac{30}{61}\)
P(X = x
1
) = \(\frac{1}{2}\left(\frac{30}{61}\right)\) = \(\frac{15}{61}\)
P(X = x
2
) = \(\frac{1}{3}\left(\frac{30}{61}\right)\) = \(\frac{10}{61}\)
P(X = x
3
) = \(\frac{30}{61}\)
P(X = x
4
) = \(\frac{1}{5}\left(\frac{30}{61}\right)\) = \(\frac{6}{61}\)
∴ Probability distribution is
Question 4.
If \(\vec{a}\) = i + j + k, \(\vec{a}, \vec{b}\) = 1 and \(\vec{a} \times \vec{b}\) = \(\hat{j}-\hat{k}\), then find \(|\vec{b}|\).
Solution:
Question 5.
If a line makes an angle α, β, γ with the coordinate axes, then find the value of cos 2α + cos 2 β + cos 2 γ.
Solution:
Let, l = cos α, m = cos β, h = cos γ
We know that, l
2
+ m
2
+ h
2
= 1
cos
2
α + cos
2
β + cos
2
γ = 1 ………. (1)
Now, cos 2α + cos 2β + cos 2γ
= 2cos
2
α – 1 + 2cos
2
β – 1 + 2cos
2
γ – 1
= 2(cos
2
α + cos
2
β + cos
2
γ) – 3
= 2(1) – 3 ………… (From (i)
= -1
Question 6.
(a) Events A and B are such that P(A) = \(\frac{1}{2}\), P(B) = \(\frac{7}{12}\) and \(\mathrm{P}(\overline{\mathrm{A}} \cup \overline{\mathrm{B}})\) = \(\frac{1}{4}\)
Find whether the events A and B are independent or not.
Solution:
Given. \(\mathrm{P}(\overline{\mathrm{A}} \cup \overline{\mathrm{B}})\) = \(\frac{1}{4}\)
1 – P(A ∩ B) = \(\frac{1}{4}\) …….. [By De Morgan’s law
1 – \(\frac{1}{4}\) = P(A ∩ B)
∴ P(A ∩ B) = \(\frac{3}{4}\) ….. (i)
Now, P(A) × P(B) = \(\frac{1}{2} \times \frac{7}{12}\) = \(\frac{7}{24}\) ……. (ii)
Therefore, Events A ‘and B are not independent.
Or,
(b) A box B
1
contains 1 white ball and 3 red balls. Another box B
2
contains 2 white balls and 3 red balls. If one ball is drawn at random from each of the boxes B
1
and B
2
, then find the probability that the two balls drawn are of the same colour.
Solution:
P(Two balls drawn are of the same colour) = P(both white balls) + P(both red balls)
= P(white ball from box B
1
) × P(white ball from box B
2
) + P(red balls from box B
1
) × P(red balls from box B
2
)
= \(\frac{1}{4} \times \frac{2}{5}\) + \(\frac{3}{4} \times \frac{3}{5}\) = \(\frac{2}{20}+\frac{9}{20}\)
= \(\frac{2+9}{20}\) = \(\frac{11}{20}\)
Section – B
Question numbers 7 to 10 carry 3 marks each.
Question 7.
Evaluate: \(\int_0^{\pi / 4} \frac{d x}{1+\tan x}\)
Solution:
Question 8.
(a) If \(\vec{a}\) and \(\vec{b}\) are two vectors such that \(|\vec{a}+\vec{b}|\) = \(|\vec{b}|\), then prove that \(|\vec{a}+2 \vec{b}|\) is prependicular to \(\vec{a}\).
Solution:
Or,
(b) If \(\vec{a}\) and \(\vec{b}\) are unit vectors and θ is the angle between them, then prove that
\(\sin \frac{\theta}{2}\) = \(\frac{1}{2}|\vec{a}-\vec{b}|\)
Solution:
Question 9.
Find the integrating factor of the differential equation \(\frac{d y}{d x}\) + y = \(\frac{1+y}{x}\)
Solution:
Question 10.
(a) Find: \(\int e^x \cdot \sin 2 x d x\).
Solution:
Or,
(b) Find : \(\int \frac{2 x}{\left(x^2+1\right)\left(x^2+2\right)} d x\)
Solution:
Section – C
Question numbers 11 to 14 carry 4 marks each.
Question 11.
Three persons A, B and C apply for a job of manager in a private company. Chances of their selection are in the ratio 1 : 2 : 4. The probability that A, B and C can introduce changes to increase the profits of a company are 0.8, 0.5 and 0.3 respectively. If increase in the profit does not take place, find the probability that it is due to the appointment of A.
Solution:
Let E
1
: A is appointed
E
2
: B is appointed
E
3
: C is appointed
D: Increase in the profit does not take place
∴ P(E
1
) = \(\frac{1}{7}\), P(E
2
) = \(\frac{2}{7}\), P(E
3
) = \(\frac{4}{7}\)
and P(D | E
1
) = 1 – 0.8 = 0.2
P(D | E
2
) = 1 – 0.5 = 0.5
P(D | E
3
) = 1 – 0.3 = 0.7 [By Baye’s theorem
P(E
1
| D) =
Question 12.
Find the area bounded by the curves y = |x – 1| and y = 1, using integration.
Solution:
Area of the shaded region
Question 13.
(a) Solve the following differential equation:
(y – sin
2
x)dx + tan x dy = 0
Solution:
We have, (y – sin
2
x)dx + tan x dy = O
tan x dy = -(y – sin
2
x)dx
Or,
(b) Find the general solution of the differential equation:
(x
3
+ y
3
)dy = x
2
y dx
Solution:
Case Study Based Question
Question 14.
Two motorcycles A and B are running at the speed more than the allowed speed on the roads represented by the lines \(\vec{r}\) = \(\lambda(\hat{i}+2 \hat{j}-\hat{k})\) and \(\vec{r}\) = \((3 \hat{i}+3 \hat{j})\) + \(\mu(2 \hat{i}+\hat{j}+\hat{k})\) respectively.
Based on the above information, answer the following questions:
(a) Find the shortest distance between the given lines.
Solution:
L
1
: \(\vec{r}\) = λ\((\hat{i}+2 \hat{j}-\hat{k})\) …….(i)
Let P(λ, 2λ, -λ) be any point on line (i)
L
2
: \(\vec{r}\) = \((3 \hat{i}+3 \hat{j})\) + \(\mu(2 \hat{i}+\hat{j}+\hat{k})\) ………(ii)
= (3 + 2µ)\(\hat{i}\) + (3 + µ)\(\hat{j}\) + µ\(\hat{k}\)
Let Q(3 + 2µ, 3 + µ, µ) be any point of line (ii).
If given two lines intersect,
P and Q must coinside for some λ and µ
λ = 3 + 2µ ………. (iii)
2λ = 3 + µ ……..(iv)
-λ = µ
Solving (iii) and (v), we get
µ = -1 and λ = 1
Putting the value λ and µ in (v),
-1 = -1 which is true.
Hence, the given two lines are intersecting.
So, the shortest distance between the given lines = 0
(b) Find the point at which the motorcycles may collide.
Solution:
Their point of intersection
= P(λ, 2λ, -λ)
∴ P(1, 2, -1) [∵ λ = 1 From (a)
