NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.2

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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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