# NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.2

## Chapter 6 Ex.6.2 Question 1

Show that the function given by $$f\left( x \right) = 3x + 17$$ is increasing on R.

### Solution

Let $${x_1}$$ and $${x_2}$$ be any two numbers in R.

Then,

$${x_1} < {x_2} \Rightarrow 3{x_1} + 17 < 3{x_2} + 17 = f\left( {{x_1}} \right) < f\left( {{x_2}} \right)$$

Thus, $$f$$ is strictly increasing on R.

## Chapter 6 Ex.6.2 Question 2

Show that the function given by $$f\left( x \right) = {e^{2x}}$$ is increasing on R.

### Solution

Let $${x_1}$$ and $${x_2}$$ be any two numbers in R.

Then,

${x_1} < {x_2} \Rightarrow 2{x_1} < 2{x_2} \\ \Rightarrow {e^{2{x_1}}} < {e^{2{x_2}}} = f\left( {x{ \ _1}} \right) < f\left( {{x_2}} \right)$

Thus, $$f$$ is strictly increasing on R.

## Chapter 6 Ex.6.2 Question 3

Show that the function given by $$f\left( x \right) = \sin x$$ is

(a) increasing in $$\left( {0,\frac{\pi }{2}} \right)$$

(b) decreasing in $$\left( {\frac{\pi }{2},\pi } \right)$$

(c) neither increasing nor decreasing in $$\left( {0,\pi } \right)$$

### Solution

It is given that $$f\left( x \right) = \sin x$$

Hence, $$f'\left( x \right) = \cos x$$

(a) Here, $$x \in \left( {0,\frac{\pi }{2}} \right)$$

\begin{align}&\Rightarrow \; \cos x > 0\\&\Rightarrow \; f'\left( x \right) > 0\end{align}

Thus, $$f$$ is strictly increasing in $$\left( {0,\frac{\pi }{2}} \right)$$.

(b) Here, $$x \in \left( {\frac{\pi }{2},\pi } \right)$$

\begin{align}&\Rightarrow \; \cos x < 0\\&\Rightarrow \; f'\left( x \right) < 0\end{align}

Thus, $$f$$ is strictly decreasing in $$\left( {\frac{\pi }{2},\pi } \right)$$.

(c) Here, $$x \in \left( {0,\pi } \right)$$

The results obtained in (a) and (b) are sufficient to state that $$f$$ is neither increasing nor decreasing in $$\left( {0,\pi } \right)$$.

## Chapter 6 Ex.6.2 Question 4

Find the intervals in which the function $$f$$ given by $$f\left( x \right) = 2{x^2} - 3x$$ is

(a) increasing

(b) decreasing

### Solution

The given function is $$f\left( x \right) = 2{x^2} - 3x$$

Hence,

$$f'\left( x \right) = 4x - 3$$

Therefore,

\begin{align}&f'\left( x \right) = 0\\&\Rightarrow \; x = \frac{3}{4}\end{align} In $$\left( { - \infty, \frac{3}{4}} \right)$$, $$f'\left( x \right) = 4x - 3 < 0$$

Hence, $$f$$ is strictly decreasing in $$\left( { - \infty, \frac{3}{4}} \right)$$.

In $$\left( { \frac{3}{4}}, \infty \right)$$, $$f'\left( x \right) = 4x - 3 > 0$$

Hence, $$f$$ is strictly increasing in $$\left( { \frac{3}{4}}, \infty \right)$$.

## Chapter 6 Ex.6.2 Question 5

Find the intervals in which the function $$f$$ given $$f\left( x \right) = 2{x^3} - 3{x^2} - 36x + 7$$ is

(a) increasing

(b) decreasing

### Solution

The given function is $$f\left( x \right) = 2{x^3} - 3{x^2} - 36x + 7$$

Hence,

\begin{align}f'\left( x \right) &= 6{x^2} - 6x - 36\\&= 6\left( {{x^2} - x - 6} \right)\\&= 6\left( {x + 2} \right)\left( {x - 3} \right)\end{align}

Therefore,

\begin{align}f'\left( x \right) &= 0\\ \Rightarrow \; x &= - 2,3\end{align} In $$\left( { - \infty, - 2} \right)$$ and $$\left( {3, \infty } \right), f'\left( x \right) > 0$$

Hence, $$f$$ is strictly increasing in $$\left( { - \infty, - 2} \right)$$ and $$\left( {3, \infty } \right)$$.

In $$\left( { - 2,3} \right), f'\left( x \right) < 0$$

Hence, $$f$$ is strictly decreasing in $$\left( { - 2,3} \right)$$.

## Chapter 6 Ex.6.2 Question 6

Find the intervals in which the following functions are strictly increasing or decreasing.

(a)$${x^2} + 2x - 5$$

(b)$$10 - 6x - 2{x^2}$$

(c)$$- 2{x^3} - 9{x^2} - 12x + 1$$

(d)$$6 - 9x - 9{x^2}$$

(e)$${\left( {x + 1} \right)^3}{\left( {x - 3} \right)^3}$$

### Solution

(a) $$f\left( x \right) = {x^2} + 2x - 5$$

Hence,

$$f'\left( x \right) = 2x + 2$$

Therefore,

\begin{align}&\Rightarrow \; f'\left( x \right) = 0\\&\Rightarrow \; x = - 1\end{align}

$$x = - 1$$ divides the number line into intervals $$\left( { - \infty ,1} \right)$$ and $$\left( { - 1,\infty } \right)$$

In $$\left( { - \infty ,1} \right),f'\left( x \right) = 2x + 2 < 0$$

Thus, $$f$$ is strictly decreasing in $$\left( { - \infty ,1} \right)$$

In $$\left( { - 1,\infty } \right),f'\left( x \right) = 2x + 2 > 0$$

Thus, $$f$$ is strictly increasing in $$\left( { - 1,\infty } \right)$$

(b) $$f\left( x \right) = 10 - 6x - 2{x^2}$$

Hence,

$$f'\left( x \right) = - 6 - 4x$$

Therefore,

\begin{align}&\Rightarrow \; f'\left( x \right) = 0\\&\Rightarrow \; x = - \frac{3}{2}\end{align}

$$x = - \frac{3}{2}$$ , divides the number line into two intervals $$\left( { - \infty , - \frac{3}{2}} \right)$$ and $$\left( { - \frac{3}{2},\infty } \right)$$

In $$\left( { - \infty , - \frac{3}{2}} \right),f'\left( x \right) = - 6 - 4x < 0$$

Hence, $$f$$ is strictly increasing for $$x < - \frac{3}{2}$$

In $$\left( { - \frac{3}{2},\infty } \right),f'\left( x \right) = - 6 - 4x > 0$$

Hence, $$f$$ is strictly increasing for $$x > - \frac{3}{2}$$

(c) $$f\left( x \right) = - 2{x^3} - 9{x^2} - 12x + 1$$

Hence,

\begin{align}f'\left( x \right) = - 6{x^2} - 18x - 12\\ = - 6\left( {{x^2} + 3x + 2} \right)\\ = - 6\left( {x + 1} \right)\left( {x + 2} \right)\end{align}

Therefore,

\begin{align}&\Rightarrow \; f'\left( x \right) = 0\\&\Rightarrow \; x = - 1,2\end{align}

$$x = - 1$$ and $$x = - 2$$ divide the number line into intervals $$\left( { - \infty , - 2} \right)$$, $$\left( { - 2, - 1} \right)$$ and $$\left( { - 1,\infty } \right)$$.

In $$\left( { - \infty , - 2} \right)$$ and $$\left( { - 1,\infty } \right)$$, $$f'\left( x \right) = - 6\left( {x + 1} \right)\left( {x + 2} \right) < 0$$

Hence, $$f$$ is strictly decreasing for $$x < - 2$$ and $$x > - 1$$

In $$\left( { - 2, - 1} \right),f'\left( x \right) = - 6\left( {x + 1} \right)\left( {x + 2} \right) > 0$$

Hence, $$f$$ is strictly increasing in $$- 2 < x < - 1$$

(d)$$f\left( x \right) = 6 - 9x - {x^2}$$

Hence,

$$f'\left( x \right) = - 9 - 2x$$

Therefore,

\begin{align}&\Rightarrow \; f'\left( x \right) = 0\\&\Rightarrow \; x = - \frac{9}{2}\end{align}

In $$\left( { - \frac{9}{2},\infty } \right),f'\left( x \right) < 0$$

Hence, $$f$$ is strictly decreasing for $$x > - \frac{9}{2}$$

In $$\left( { - \infty , - \frac{9}{2}} \right),f'\left( x \right) > 0$$

Hence, $$f$$ is strictly decreasing in $$x>-\frac{9}{2}$$

(e) $$f\left( x \right) = {\left( {x + 1} \right)^3}{\left( {x - 3} \right)^3}$$

Hence,

\begin{align}f'\left( x \right)& = 3{\left( {x + 1} \right)^2}{\left( {x - 3} \right)^3} + 3{\left( {x - 3} \right)^2}{\left( {x + 1} \right)^3}\\&= 3{\left( {x + 1} \right)^2}{\left( {x - 3} \right)^2}\left[ {x - 3 + x + 1} \right]\\&= 3{\left( {x + 1} \right)^2}{\left( {x - 3} \right)^2}\left( {2x - 2} \right)\\&= 6{\left( {x + 1} \right)^2}{\left( {x - 3} \right)^2}\left( {x - 1} \right)\end{align}

Therefore,

\begin{align}&f'\left( x \right) = 0\\&\Rightarrow \; x = - 1,3,1\end{align}

$$x = - 1,3,1$$ divides the number line into four intervals $$\left( { - \infty , - 1} \right),\left( { - 1,1} \right),\left( {1,3} \right)$$ and $$\left( {3,\infty } \right)$$

In $$\left( { - \infty , - 1} \right)$$ and $$\left( { - 1,1} \right)$$, $$f'\left( x \right) = 6{\left( {x + 1} \right)^2}{\left( {x - 3} \right)^2}\left( {x - 1} \right) < 0$$

Hence, $$f$$ is strictly decreasing in $$\left( { - \infty , - 1} \right)$$ and $$\left( { - 1,1} \right)$$

In $$\left( {1,3} \right)$$ and $$\left( {3,\infty } \right)$$, $$f'\left( x \right) = 6{\left( {x + 1} \right)^2}{\left( {x - 3} \right)^2}\left( {x - 1} \right) > 0$$

Hence, $$f$$ is strictly increasing in $$\left( {1,3} \right)$$ and $$\left( {3,\infty } \right)$$

## Chapter 6 Ex.6.2 Question 7

Show that $$y = \log \left( {1 + x} \right) - \frac{{2x}}{{2 + x}},x > - 1$$, is an increasing function of $$x$$ throughout its domain.

### Solution

It is given that $$y = \log \left( {1 + x} \right) - \frac{{2x}}{{2 + x}}$$

Therefore,

\begin{align}\frac{{dy}}{{dx}} &= \frac{1}{{1 + x}} - \frac{{\left( {2 + x} \right)\left( 2 \right) - 2x\left( 1 \right)}}{{{{\left( {2 + x} \right)}^2}}}\\&= \frac{1}{{1 + x}} - \frac{4}{{{{\left( {2 + x} \right)}^2}}}\\&= \frac{{{x^2}}}{{\left( {1 + x} \right){{\left( {2 + x} \right)}^2}}}\end{align}

Now, $$\frac{{dy}}{{dx}} = 0$$

Hence,

\begin{align}&\Rightarrow \; \frac{{{x^2}}}{{{{\left( {2 + x} \right)}^2}}} = 0\\&\Rightarrow \; {x^2} = 0\\&\Rightarrow \; x = 0\end{align}

Since, $$x > - 1,x = 0$$ divides domain $$\left( { - 1,\infty } \right)$$ in two intervals $$- 1 < x < 0$$ and $$x > 0$$

When, $$- 1 < x < 0$$

Then,

\begin{align}&x < 0 \Rightarrow {x^2} > 0\\&x > - 1 \Rightarrow \left( {2 + x} \right) > 0\\&\Rightarrow \; {\left( {2 + x} \right)^2} > 0\end{align}

Hence,

$$y = \frac{{{x^2}}}{{{{\left( {2 + x} \right)}^2}}} > 0$$

When, $$x > 0$$

Then,

\begin{align}&x > 0 \Rightarrow {x^2} > 0\\&\Rightarrow \; {\left( {2 + x} \right)^2} > 0\end{align}

Hence,

$$y = \frac{{{x^2}}}{{{{\left( {2 + x} \right)}^2}}} > 0$$

Thus, $$f$$ is increasing throughout the domain.

## Chapter 6 Ex.6.2 Question 8

Find the values of $$x$$ for which $$y = {\left[ {x\left( {x - 2} \right)} \right]^2}$$ is an increasing function.

### Solution

We have,

\begin{align}y = {\left[ {x\left( {x - 2} \right)} \right]^2}\\ = {\left[ {{x^2} - 2x} \right]^2}\end{align}

Therefore,

\begin{align}\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\left[ {{x^2} - 2x} \right]^2}\\ = 2\left( {{x^2} - 2x} \right)\left( {2x - 2} \right)\\ = 4x\left( {x - 2} \right)\left( {x - 1} \right)\end{align}

Now, $$\frac{{dy}}{{dx}} = 0$$

Hence,

\begin{align}&\Rightarrow \; 4x\left( {x - 2} \right)\left( {x - 1} \right)\\&\Rightarrow \; x = 0,x = 2,x = 1\end{align}

$$x = 0$$, $$x = 1$$ and $$x = 2$$ divide the number line intervals $$\left( { - \infty ,0} \right),\left( {0,1} \right),\left( {1,2} \right)$$ and $$\left( {2,\infty } \right)$$

In $$\left( { - \infty ,0} \right)$$ and $$\left( {1,2} \right)$$ , $$\frac{{dy}}{{dx}} < 0$$

Hence, $$y$$ is strictly decreasing in intervals $$\left( { - \infty ,0} \right)$$ and $$\left( {1,2} \right)$$

In $$\left( {0,1} \right)$$ and $$\left( {2,\infty } \right)$$, $$\frac{{dy}}{{dx}} > 0$$

Hence, $$y$$ is strictly increasing in intervals $$\left( {0,1} \right)$$ and $$\left( {2,\infty } \right)$$

## Chapter 6 Ex.6.2 Question 9

Prove that $$y = \frac{{4\sin \theta }}{{\left( {2 + \cos \theta } \right)}} - \theta$$ is an increment function of $$\theta$$ in $$\left[ {0,\frac{\pi }{2}} \right]$$.

### Solution

We have, $$y = \frac{{4\sin \theta }}{{\left( {2 + \cos \theta } \right)}} - \theta$$

Therefore,

\begin{align}\frac{{dy}}{{d\theta }}& = \frac{{\left( {2 + \cos \theta } \right)\left( {4\cos \theta } \right) - 4\sin \theta \left( { - \sin \theta } \right)}}{{{{\left( {2 + \cos \theta } \right)}^2}}} - 1\\&= \frac{{8\cos \theta + 4{{\cos }^2}\theta + 4{{\sin }^2}\theta }}{{{{\left( {2 + \cos \theta } \right)}^2}}} - 1\\&= \frac{{8\cos \theta + 4}}{{{{\left( {2 + \cos \theta } \right)}^2}}} - 1\end{align}

Now, $$\frac{{dy}}{{d\theta }} = 0$$

Hence,

\begin{align}&\Rightarrow \; \frac{{8\cos \theta + 4 }}{{{{\left( {2 + \cos \theta } \right)}^2}}} = 1\\&\Rightarrow \; 8\cos \theta + 4 = 4 + {\cos ^2}\theta + 4\cos \theta \\&\Rightarrow \; {\cos ^2}\theta - 4\cos \theta = 0\\&\Rightarrow \; \cos \theta \left( {\cos \theta - 4} \right) = 0\\&\Rightarrow \cos \theta=0 \text { or } \cos \theta=4\end{align}

Since, $$\cos \theta \ne 4$$

Therefore,

\begin{align}&\cos \theta = 0\\&\Rightarrow \; \theta = \frac{\pi }{2}\end{align}

Now,

\begin{align}\frac{{dy}}{{d\theta }} &= \frac{{8\cos \theta + 4 - \left( {4 + {{\cos }^2}\theta + 4\cos \theta } \right)}}{{{{\left( {2 + \cos \theta } \right)}^2}}}\\&= \frac{{4\cos \theta - {{\cos }^2}\theta }}{{{{\left( {2 + \cos \theta } \right)}^2}}}\\&= \frac{{\cos \left( {4 - \cos \theta } \right)}}{{{{\left( {2 + \cos \theta } \right)}^2}}}\end{align}

In interval $$\left[ {0,\frac{\pi }{2}} \right]$$ , we have $$\cos \theta > 0$$

Also,

\begin{align}&4 > \cos \theta \\&\Rightarrow \; 4 - \cos \theta > 0\end{align}

Hence, $$\cos \theta \left( {4 - \cos \theta } \right) > 0$$ and also $${\left( {2 + \cos \theta } \right)^2} > 0$$

Therefore,

$$\frac{{\cos \theta \left( {4 - \cos \theta } \right)}}{{{{\left( {2 + \cos \theta } \right)}^2}}} > 0$$

Hence, $$\frac{{dy}}{{d\theta }} > 0$$

So, $$y$$ is strictly increasing in $$\left( {0,\frac{\pi }{2}} \right)$$ and the given function is continuous at $$x = 0$$ and $$x = \frac{\pi }{2}$$

Thus, $$y$$ is increasing in interval $$\left[ {0,\frac{\pi }{2}} \right]$$.

## Chapter 6 Ex.6.2 Question 10

Prove that the logarithmic function is strictly increasing on $$\left( {0,\infty } \right)$$ .

### Solution

The given function is $$f\left( x \right) = \log x$$

Therefore, $$f'\left( x \right) = \frac{1}{x}$$

For, $$x > 0,f'\left( x \right) = \frac{1}{x} > 0$$

Thus, the logarithmic function is strictly increasing in interval $$\left( {0,\infty } \right)$$.

## Chapter 6 Ex.6.2 Question 11

Prove that the function $$f$$ given by $$f\left( x \right) = {x^2} - x + 1$$ is neither strictly increasing nor decreasing on $$\left( { - 1,1} \right)$$.

### Solution

The given function is $$f\left( x \right) = {x^2} - x + 1$$

Therefore,

$$f'\left( x \right) = 2x - 1$$

Now,

\begin{align}&f'\left( x \right) = 0\\&\Rightarrow \; x = \frac{1}{2}\end{align}

$$x = \frac{1}{2}$$ divides the interval $$v$$ into $$\left( { - 1,\frac{1}{2}} \right)$$ and $$\left( {\frac{1}{2},1} \right)$$

In interval $$\left( {\frac{1}{2},1} \right),f'\left( x \right) = 2x - 1 > 0$$

Hence, $$f$$ is strictly decreasing in $$\left( { - 1,\frac{1}{2}} \right)$$

In interval $$\left(\frac{1}{2}, 1\right), f^{\prime}(x)=2 x-1>0$$

Hence, $$f$$ is strictly increasing in $$\left( {\frac{1}{2},1} \right)$$

Thus, $$f$$ is strictly increasing nor strictly decreasing in interval $$\left( { - 1,1} \right)$$

## Chapter 6 Ex.6.2 Question 12

Which of the following principles are decreasing on $$\left( {0,\frac{\pi }{2}} \right)$$?

(A)$$\cos x$$

(B)$$\cos 2x$$

(C)$$\cos 3x$$

(D)$$\tan x$$

### Solution

(A) Let $${f_1}\left( x \right) = \cos x$$

Therefore, $${f_1}^\prime \left( x \right) = - \sin x$$

In interval $$\left( {0,\frac{\pi }{2}} \right),{f_1}^\prime \left( x \right) = - \sin x < 0$$

Thus, $$\cos x$$ is strictly decreasing in $$\left( {0,\frac{\pi }{2}} \right)$$.

(B) Let $${f_2}\left( x \right) = \cos 2x$$

Therefore, $${f_2}^\prime \left( x \right) = - 2\sin 2x$$

Now,

\begin{align}&\Rightarrow \; 0 < x < \frac{\pi }{2}\\&\Rightarrow \; 0 < 2x < \pi \\&\Rightarrow \; \sin 2x > 0\\&\Rightarrow \; - 2\sin 2x < 0\end{align}

Hence, $${f_2}^\prime \left( x \right) = - 2\sin 2x < 0$$ in $$\left( {0,\frac{\pi }{2}} \right)$$

Thus, $$\cos 2x$$ is strictly decreasing in $$\left( {0,\frac{\pi }{2}} \right)$$

(C) Let $${f_3}\left( x \right) = \cos 3x$$

Therefore, $${f_3}^\prime \left( x \right) = - 3\sin 3x$$

Now,

\begin{align} &{f_3}^\prime \left( x \right) = 0 \hfill \\ &\Rightarrow \sin 3x = 0 \hfill \\& \Rightarrow 3x = \pi \;\;\;\;\;\;\;\;\;\;\left[ {\because x \in \left( {0,\frac{\pi }{2}} \right)} \right] \hfill \\& \Rightarrow x = \frac{\pi }{3} \hfill \\ \end{align}

The point $$x = \frac{\pi }{3}$$, divides $$\left( {0,\frac{\pi }{2}} \right)$$ into $$\left( {0,\frac{\pi }{3}} \right)$$ and $$\left( {\frac{\pi }{3},\frac{\pi }{2}} \right)$$

In interval $$\left( {0,\frac{\pi }{3}} \right),{f_3}\left( x \right) = - 3\sin 3x < 0\;\;\;\;\;\left[ \begin{array}{l}0 < x < \frac{\pi }{3}\\0 < 3x < \pi \end{array} \right]$$

Hence, $${f_3}$$ is strictly decreasing in $$\left( {0,\frac{\pi }{3}} \right)$$

In interval $$\left( {\frac{\pi }{3},\frac{\pi }{2}} \right),{f_3}\left( x \right) = - 3\sin 3x > 0\;\;\;\;\left[ {\frac{\pi }{3} < x < \frac{\pi }{2} < \pi < 3x < \frac{{3\pi }}{2}} \right]$$

Hence, $${f_3}$$ is strictly increasing in $$\left( {\frac{\pi }{3},\frac{\pi }{2}} \right)$$

Thus, $$\cos 3x$$ is neither increasing nor decreasing in interval $$\left( {0,\frac{\pi }{2}} \right)$$

(D) Let $${f_4}\left( x \right) = \tan x$$

Therefore, $${f_4}^\prime \left( x \right) = {\sec ^2}x$$

In interval $$\left( {0,\frac{\pi }{2}} \right),{f_4}^\prime \left( x \right) = {\sec ^2}x > 0$$

Thus, $$\tan x$$ is strictly increasing in $$\left( {0,\frac{\pi }{2}} \right)$$

Thus, the correct options are A and B.

## Chapter 6 Ex.6.2 Question 13

On which of the following intervals is the function $$f$$ is given by $$f\left( x \right) = {x^{100}} + \sin x - 1$$ decreasing?

(A)$$\left( {0,1} \right)$$

(B)$$\left( {\frac{\pi }{2},\pi } \right)$$

(C)$$\left( {0,\frac{\pi }{2}} \right)$$

(D) None of these

### Solution

We have,

$$f\left( x \right) = {x^{100}} + \sin x - 1$$

Therefore,

$$f'\left( x \right) = 100{x^{99}} + \cos x$$

In interval $$\left( {0,1} \right),\cos x > 0$$ and $$100{x^{99}} > 0$$

Hence, $$f'\left( x \right) > 0$$

Thus, $$f$$ is strictly increasing in $$\left( {0,1} \right)$$

In interval $$\left( {\frac{\pi }{2},\pi } \right),\cos x < 0$$ and $$100{x^{99}} > 0$$

Hence, $$f'\left( x \right) > 0$$

Thus, $$f$$ is strictly increasing in interval $$\left( {\frac{\pi }{2},\pi } \right)$$

Now, in interval $$\left( {0,\frac{\pi }{2}} \right),\cos x > 0$$ and $$100{x^{99}} > 0$$

Hence, $$f'\left( x \right) > 0$$

Thus, $$f$$ is strictly increasing in interval $$\left( {0,\frac{\pi }{2}} \right)$$

Hence, $$f$$ is strictly decreasing in none of the intervals.

Thus, the correct option is D.

## Chapter 6 Ex.6.2 Question 14

For what values of $$a$$ the function $$f$$ given $$f\left( x \right) = {x^2} + ax + 1$$ is increasing on $$\left[ {1,2} \right]$$?

### Solution

We have

$$f\left( x \right) = {x^2} + ax + 1$$

Therefore,

$$f'\left( x \right) = 2x + a$$

Now, the function $$f$$ is strictly increasing on $$\left[ {1,2} \right]$$

Therefore,

\begin{align}&\Rightarrow \; f'\left( x \right) > 0\\&\Rightarrow \; 2x + a > 0\\&\Rightarrow \; 2x > - a\\&\Rightarrow \; x > \frac{{ - a}}{2}\end{align}

Here, we have $$1 \le x \le 2$$

Thus,

\begin{align}&\frac{{ - a}}{2} > 1\\&a > - 2\end{align}

## Chapter 6 Ex.6.2 Question 15

Let I be any interval disjoint from $$\left[ { - 1,1} \right]$$. Prove that the function $$f$$ given by $$f\left( x \right) = x + \frac{1}{x}$$ is increasing on I.

### Solution

We have

$$f\left( x \right) = x + \frac{1}{x}$$

Therefore,

$$f'\left( x \right) = 1 - \frac{1}{{{x^2}}}$$

Now,

\begin{align}&f'\left( x \right) = 0\\&\Rightarrow \; 1 - \frac{1}{{{x^2}}} = 0\\&\Rightarrow \; {x^2} = 1\\&\Rightarrow \; x = \pm 1\end{align}

The points $$x = 1$$ and $$x = - 1$$ divide the real line intervals $$\left( { - \infty ,1} \right),\left( { - 1,1} \right)$$ and $$\left( {1,\infty } \right)$$

In interval $$\left( { - 1,1} \right)$$, $$- 1 < x < 1$$

\begin{align}&\Rightarrow \; {x^2} < 1\\&\Rightarrow \; 1 < \frac{1}{{{x^2}}}\;\;\;\;\;\;\;\;\;\;\;x \ne 0\\&\Rightarrow \; 1 - \frac{1}{{{x^2}}} < 0\;\;\;\;\;\;x \ne 0\end{align}

Therefore, $$f'\left( x \right) = 1 - \frac{1}{{{x^2}}} < 0$$ on $$\left( { - 1,1} \right) \sim \left\{ 0 \right\}$$

Hence, $$f$$ is strictly decreasing on $$\left( { - 1,1} \right) \sim \left\{ 0 \right\}$$

Now, in interval $$\left( { - \infty , - 1} \right)$$ and $$\left( {1,\infty } \right)$$, $$x < - 1$$ or $$1 < x$$

\begin{align}&\Rightarrow \; {x^2} > 1\\&\Rightarrow \; 1 > \frac{1}{{{x^2}}}\\&\Rightarrow \; 1 - \frac{1}{{{x^2}}} > 0\end{align}

Therefore, $$f'\left( x \right) = 1 - \frac{1}{{{x^2}}} > 0$$ on $$\left( { - \infty , - 1} \right)$$ and $$\left( {1,\infty } \right)$$

Hence, $$f$$ is strictly increasing on $$\left( { - \infty , - 1} \right)$$ and $$\left( {1,\infty } \right)$$

Thus, $$f$$ is strictly increasing in I in $$\left[ { - 1,1} \right]$$

## Chapter 6 Ex.6.2 Question 16

Prove that the function $$f$$ given by $$f\left( x \right) = \log \sin x$$ is increasing on $$\left( {0,\frac{\pi }{2}} \right)$$ and decreasing on $$\left( {\frac{\pi }{2},\pi } \right)$$.

### Solution

We have

$$f\left( x \right) = \log \sin x$$

Therefore,

\begin{align}f'\left( x \right) &= \frac{1}{{\sin x}}\cos x\\&= \cot x\end{align}

In interval $$\left( {0,\frac{\pi }{2}} \right), f'\left( x \right) = \cot x > 0$$

Hence, $$f$$ is strictly increasing in $$\left( {0,\frac{\pi }{2}} \right)$$.

In interval $$\left( {\frac{\pi }{2},\pi } \right), f'\left( x \right) = \cot x < 0$$

Hence, $$f$$ is strictly decreasing in $$\left( {\frac{\pi }{2},\pi } \right)$$.

## Chapter 6 Ex.6.2 Question 17

Prove that the function $$f$$ given by $$f\left( x \right) = \log \left| {\cos x} \right|$$ is decreasing on $$\left( {0,\frac{\pi }{2}} \right)$$ and increasing on $$\left( {\frac{{3\pi }}{2},2\pi } \right)$$.

### Solution

We have $$f\left( x \right) = \log \left| {\cos x} \right|$$

Therefore,

\begin{align}f'\left( x \right) &= \frac{1}{{\cos x}}\left( { - \sin x} \right)\\&= - \tan x\end{align}

In interval $$\left( {0,\frac{\pi }{2}} \right),\, \tan x > 0 \Rightarrow - \tan x < 0$$

Hence, $$f'\left( x \right) < 0$$

Thus, $$f$$ is strictly decreasing on $$\left(0, \frac{\pi}{2}\right)$$ .

In interval $$\left( {\frac{{3\pi }}{2},2\pi } \right),\, \tan x < 0 \Rightarrow - \tan x > 0$$

Hence, $$f'\left( x \right) > 0$$

Thus, $$f$$ is strictly increasing on $$\left( {\frac{{3\pi }}{2},2\pi } \right)$$.

## Chapter 6 Ex.6.2 Question 18

Prove that the function given by $$f\left( x \right) = {x^3} - 3{x^2} + 3x - 100$$ is increasing in R

### Solution

We have

$$f\left( x \right) = {x^3} - 3{x^2} + 3x - 100$$

Therefore,

\begin{align}f'\left( x \right) &= 3{x^2} - 6x + 3\\&= 3\left( {{x^2} - 2x + 1} \right)\\&= 3{\left( {x - 1} \right)^2}\end{align}

For $$x \in {\bf{R}}$$, $${\left( {x - 1} \right)^2} \ge 0$$

So, $$f'\left( x \right)$$ is always positive in R.

Thus, the function is increasing in R.

## Chapter 6 Ex.6.2 Question 19

The interval in which $$y = {x^2}{e^{ - x}}$$ is increasing is

(A)$$\left( { - \infty ,\infty } \right)$$

(B)$$\left( { - 2,0} \right)$$

(C)$$\left( {2,\infty } \right)$$

(D)$$\left( {0,2} \right)$$

### Solution

We have $$y = {x^2}{e^{ - x}}$$

Therefore,

\begin{align}\frac{{dy}}{{dx}} &= 2x{e^{ - x}} - {x^2}{e^{ - x}}\\&= x{e^{ - x}}\left( {2 - x} \right)\end{align}

Now, $$\frac{{dy}}{{dx}} = 0$$

Hence, $$x = 0$$ and $$x = 2$$

The points $$x = 0$$ and $$x = 2$$ divide the real line into three disjoint intervals i.e., $$\left( { - \infty ,0} \right)$$, $$\left( {0,2} \right)$$ and $$\left( {2,\infty } \right)$$.

In intervals $$\left( { - \infty ,0} \right)$$ and $$\left( {2,\infty } \right)$$, $$f'\left( x \right) < 0$$ as $${e^{ - x}}$$ is always positive.

Hence, $$f$$ is decreasing on $$\left( { - \infty ,0} \right)$$ and $$\left( {2,\infty } \right)$$

In interval $$\left( {0,2} \right)$$, $$f'\left( x \right) > 0$$

Hence, $$f$$ is strictly increasing in $$\left( {0,2} \right)$$

Thus, the correct option is D.

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