# CBSE Class 12 Maths Question Paper 2022 (Term-II) with Solutions

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)