Date: 07/06/2022

SET No. 1

Question Paper Code

65/5/1

Time: 2 hrs.

Class-XII

Max. Marks: 40

MATHEMATICS

Term-II (CBSE-2022)

GENERAL INSTRUCTIONS

Read the following instructions carefully and strictly follow them:

1. This question paper contains THREE Sections - A, B and 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 subparts of 2 marks each.

SECTION-A

Question numbers 1 to 6 carry 2 marks each.

1. Find: <math display="block">\int \frac{dx}{\sqrt{4x-x^2}}</math> [2]

<math display="block">Sol. \quad I = \int \frac{dx}{\sqrt{4x - y^2}}</math> <math display="block">=\int \frac{dx}{\sqrt{4-4+4x-x^2}}</math> <math display="block">=\int \frac{dx}{\sqrt{4-(x-2)^2}}</math> <math>=\sin^{-1}\left(\frac{x-2}{2}\right)+C</math>, where C is integration constant. <math display="block">\left\{\because \int \frac{1}{\sqrt{a^2-y^2}}dx = \sin^{-1}\frac{x}{a}+C\right\}</math>

1. Find the general solution of the following differential equation:

[2] <math display="block">\frac{dy}{dx} = e^{x-y} + x^2 e^{-y}</math>

Sol. <math>\frac{dy}{dx} = e^{x-y} + x^2 e^{-y}</math> <math>\Rightarrow</math> <math>e^y dy = (e^x + x^2) dx</math>. <math>\Rightarrow</math> <math>e^y = e^x + \frac{x^3}{3} + C</math>, where C is constant of integration.

1. Let X be a random variable which assumes values <math>x_1</math>, <math>x_2</math>, <math>x_3</math>, <math>x_4</math> such that <math>2P(X = x_1) = 3P(X = x_2) = P(X = x_3) = 5P(X = x_4)</math> Find the probability distribution of X.

[2] Sol. : <math>2P(X = x_1) = 3P(X = x_2) = P(X = x_3) = 5P(X = x_4) = \lambda \text{ (say)}</math>

So, <math>P(X = x_1) = \frac{\lambda}{2}</math>, <math>P(X = x_2) = \frac{\lambda}{3}</math>

<math>P(X = x_3) = \lambda</math> and <math>P(X = x_4) = \frac{\lambda}{5}</math>

<math display="block">\therefore \sum_{i=1}^{4} P(X = x_i) = 1 \Rightarrow \frac{\lambda}{2} + \frac{\lambda}{3} + \lambda + \frac{\lambda}{5} = 1 \Rightarrow \lambda = \frac{30}{61}</math>

Probability distribution of X

X_{1_} <math>\boldsymbol{x}_2</math> <math>x_3</math> <math>X_4</math> <math>P(X = x_i)</math> 15<br>61 <u>10</u> 30<br>61 6<br>61