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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 9 Differentiability Exercise 9.2 Question 12 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 9 Differentiability Exercise 9.2 Question 12 Maths Textbook Solution.

Answers (1)

Answer: |\sin x| is not differentiable at x=n\pi and |\cos x| is differentiable everywhere.

Hint:

Check LHD at x=n\pi is equal to RHD at x=n\pi or not for n is even and n is odd respect for the function|\sin x|

Given:

Given functions are |\sin x| and |\cos x|.

Solution:

Let f\left ( x \right )=|\sin x|

                 = \begin{cases}-\sin x & , x<n \pi \\ \sin x & , x \geq n \pi\end{cases}

Case  1

            For x=n\pi (Where n is even)

LHD at x=n\pi

            \begin{aligned} &=\lim _{x \rightarrow n-1} \frac{f(x)-f(n \pi)}{x-n \pi} \\ &=\lim _{h \rightarrow 0} \frac{-\sin (n \pi-h)-\sin n \pi}{n \pi-h-n \pi} \\ &=\lim _{h \rightarrow 0} \frac{\sinh -0}{-h} \end{aligned}

LHD at x=n\pi =-1                                                                                                 … (i)

RHD at x=n\pi,

               \begin{aligned} &=\lim _{h \rightarrow 0} \frac{\sin (n \pi+h)-\sin n \pi}{h} \\ &=\lim _{h \rightarrow 0} \frac{\sinh }{h} \end{aligned}

RHD at x=n\pi =1                                                                                                ....(iii)

From(i) and (ii)

            LHD at x=n\pi \neq RHD \: at\: x=n\pi

Case II

For x=n\pi (Where n is odd)

LHD at x=n\pi

           \begin{aligned} &=\lim _{h \rightarrow 0} \frac{-\sin (n \pi-h)-\sin n \pi}{-h} \\ &=\lim _{h \rightarrow 0} \frac{-\sinh }{-h} \end{aligned}

RHD at x=n\pi =-1                                                                        .....(iv)

From (iii) and (iv)

LHD at x=n\pi \neq RHD \: at\: x=n\pi

Thus, f\left ( x \right )=|\sin x| is not differentiable at x=n\pi

Now we need to check \cos |x| is differentiable or not

Let,

        g\left ( x \right )=\cos |x|

        \cos \left ( -x \right )=\cos x

        g\left ( x \right )=\cos \: x

g\left ( x \right )=\cos |x|  is differentiable everywhere.

Posted by

infoexpert21

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