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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:

  1. This question paper contains three Sections – Section A, Section B and Section C.
  2. Each section is compulsory.
  3. Section—A has 6 short-answer type-I questions of 2 marks each.
  4. Section—B has 4 short-answer type-II questions of 3 marks each.
  5. Section — C has 4 long-answer type questions of 4 marks each.
  6. There is an internal choice in some questions.
  7. 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 x1, x2, x3, x4 such that 2P(X = x1) = 3P(X = x2) = P(X = x3) = 5P(X = x4).
Find the probability distribution of X.
Solution:
We have, 2P(X = x1) = 3P(X = x2)
P(X = x3) = 5P(X = x4) = k (let)
Here, 2P(X = x1) = k ⇒ P(X = x1) = \(\frac{k}{2}\)
3P(X = x2) = k ⇒ P(X = x2) = \(\frac{k}{3}\)
P(X = x3) = k
5P(X = x4) = k ⇒ P(X = x4) = \(\frac{k}{5}\)
As we know,
P(X = x1) + P(X = x2) + P(X = x3) + P(X = x4) = 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 = x1) = \(\frac{1}{2}\left(\frac{30}{61}\right)\) = \(\frac{15}{61}\)
P(X = x2) = \(\frac{1}{3}\left(\frac{30}{61}\right)\) = \(\frac{10}{61}\)
P(X = x3) = \(\frac{30}{61}\)
P(X = x4) = \(\frac{1}{5}\left(\frac{30}{61}\right)\) = \(\frac{6}{61}\)
∴ 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, l2 + m2 + h2 = 1
cos2α + cos2β + cos2γ = 1 ………. (1)
Now, cos 2α + cos 2β + cos 2γ
= 2cos2α – 1 + 2cos2β – 1 + 2cos2γ – 1
= 2(cos2α + cos2β + cos2γ) – 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 B1 contains 1 white ball and 3 red balls. Another box B2 contains 2 white balls and 3 red balls. If one ball is drawn at random from each of the boxes B1 and B2, 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 B1) × P(white ball from box B2) + P(red balls from box B1) × P(red balls from box B2)
= \(\frac{1}{4} \times \frac{2}{5}\) + \(\frac{3}{4} \times \frac{3}{5}\) = \(\frac{2}{20}+\frac{9}{20}\)
= \(\frac{2+9}{20}\) = \(\frac{11}{20}\)

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 E1 : A is appointed
E2 : B is appointed
E3 : C is appointed
D: Increase in the profit does not take place
∴ P(E1) = \(\frac{1}{7}\), P(E2) = \(\frac{2}{7}\), P(E3) = \(\frac{4}{7}\)
and P(D | E1) = 1 – 0.8 = 0.2
P(D | E2) = 1 – 0.5 = 0.5
P(D | E3) = 1 – 0.3 = 0.7 [By Baye’s theorem
P(E1 | 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 – sin2x)dx + tan x dy = 0
Solution:
We have, (y – sin2x)dx + tan x dy = O
tan x dy = -(y – sin2x)dx

Or,

(b) Find the general solution of the differential equation:
(x3 + y3)dy = x2y 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:
L1 : \(\vec{r}\) = λ\((\hat{i}+2 \hat{j}-\hat{k})\) …….(i)
Let P(λ, 2λ, -λ) be any point on line (i)
L2: \(\vec{r}\) = \((3 \hat{i}+3 \hat{j})\) + \(\mu(2 \hat{i}+\hat{j}+\hat{k})\) ………(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)


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