General Instructions:

1. All questions are compulsory.

1. This question paper contains 29 questions divided into four sections A, B, C and D. Section A comprises of

4 questions of one mark each, Section B comprises of 8 questions of two marks each, Section C comprises of 11 questions of four marks each and Section D comprises of 6 questions of six marks each.

1. All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.

1. There is no overall choice. However, internal choice has been provided in 1 question of Section A, 3

questions of Section B, 3 questions of Section C and 3 questions of Section D. You have to attempt only one of the alternatives in all such questions.

1. Use of calculators is not permitted. You may ask logarithmic tables, if required.

Question 1

Find the acute angle between the planes <math>\overrightarrow{r}</math>. <math>\left(\hat{i}-2\hat{j}-2\hat{k}\right)=1</math> and <math>\overrightarrow{r}</math>. <math>\left(3\hat{i}-6\hat{j}+2\hat{k}\right)=0</math> .

OR Find the length of the intercept, cut off by the plane <math>2x + y - z = 5</math> on the x-axis

Solution

The vector equation of the planes is <math>\overrightarrow{r}\cdot\left(\hat{i}-2\hat{j}-2\hat{k}\right)=1</math> and <math>\overrightarrow{r}\cdot\left(3\hat{i}-6\hat{j}+2\hat{k}\right)=0</math>.

It is known that if <math>\vec{n}_1</math> and <math>\vec{n}_2</math> are normal to the planes, <math>\vec{r} \cdot \vec{n}_1 = d_1</math> and <math>\vec{r} \cdot \vec{n}_2 = d_2</math>, then the angle between them, is given by,

<math>\cos \theta = \left| \begin{array}{c} \frac{\rightarrow \rightarrow \rightarrow \rightarrow \rightarrow \rightarrow \rho}{\left| \frac{\rightarrow \rightarrow \rho}{n_1} \right| \frac{\rightarrow \rho}{n_2}} \right|</math>

So, the angle between the given planes will be

<math display="block">\cos\theta = \left| \frac{\left(\hat{i} - 2\hat{j} - 2\hat{k}\right) \cdot \left(3\hat{i} - 6\hat{j} + 2\hat{k}\right)}{\left(\sqrt{1^2 + (-2)^2 + (-2)^2}\right) \left(\sqrt{3^2 + (-6)^2 + (2)^2}\right)} \right|</math> <math>=\left|\frac{3+12-4}{3\times7}\right|</math> <math>= \left| \frac{11}{21} \right|</math> <math>\Rightarrow \theta = \cos^{-1} \left| \frac{11}{21} \right|</math> OR

The given plane is <math>2x + y - z = 5</math>.

Dividing both sides of equation by 5, we obtain

<math>\frac{2}{5}x + \frac{y}{5} - \frac{z}{5} = 1</math> <math display="block">\Rightarrow \frac{x}{5} + \frac{y}{5} + \frac{z}{-5} = 1</math>

<math display="block">\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1,</math>

It is known that the equation of a plane in intercept form is where a, b, c are the intercepts

cut off by the plane at x, y, and z axes respectively.

1. axis is <math>\frac{5}{2}</math>.

Therefore, for the given equation, the intercept made with the

Question 2

If <math>y = \log(\cos e^x)</math> then find <math>\frac{dy}{dx}</math>.

Solution

Let <math>y = \log(\cos e^x)</math>

By using the chain rule, we obtain

<math display="block">\frac{dy}{dx} = \frac{d}{dx} \Big[ \log \Big( \cos e^x \Big) \Big]</math> <math display="block">= \frac{1}{\cos e^x} \cdot \frac{d}{dx} \left(\cos e^x\right)</math> <math display="block">= \frac{1}{\cos e^x} \cdot \left(-\sin e^x\right) \cdot \frac{d}{dx} \left(e^x\right)</math> <math display="block">= \frac{-\sin e^x}{\cos e^x} \cdot e^x</math> <math>=-e^x \tan e^x, e^x \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{N}</math>

Question 3

A is a square matrix with <math>|A| = 4</math>. then find the value of <math>|A|</math>. (adj A).