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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 9 Differentiability Exercise 9.2 Question 11 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 9 Differentiability Exercise 9.2 Question 11 Maths Textbook Solution.

Answers (1)

Answer: Function is continuous but not differentiable at x = c

Hint: To check whether the function f(x) is differentiable at point x=c, we need to check that f'\left ( c^{-} \right )=f'\left ( c^{+} \right )

Given: \mathrm{f}(\mathrm{x})=\left\{\begin{array}{c} (x-c) \cos \left(\frac{1}{x-c}\right), \mathrm{x} \neq \mathrm{c} \\ 0, x=c \end{array}\right.

Solution:

It is given that, f(c) = 0.

Consider,

\lim _{x \rightarrow c}(x-c) \cos \left(\frac{1}{x-c}\right) 

Substituting x – c = y, we get:

\begin{aligned} \lim _{y \rightarrow 0} y \cos \left(\frac{1}{y}\right) &=\lim _{y \rightarrow 0} y \cdot \lim _{y \rightarrow 0} \cos \left(\frac{1}{y}\right) \\ &=0 \cdot \lim _{y \rightarrow 0} \cos \left(\frac{1}{y}\right) \end{aligned}

                            =0

So,\lim_{x\rightarrow c}f\left ( x \right )=f\left ( c \right )=0

Thus, f(x) is continuous at x = c.

Now, we need to check differentiability at x=c.

Using the formula, we have:

f'\left ( c \right )=\lim_{x\rightarrow c}\frac{f\left ( x \right )-f(c)}{x-c}

So, LHD is given by,

f'\left ( c \right )=\lim_{x\rightarrow c}\frac{f\left ( x \right )-f(c)}{x-c}

            =\lim_{x\rightarrow c}\frac{\left ( x-c \right )\cos \left ( \frac{1}{x-c} \right )-0}{x-c}

            =\lim_{x\rightarrow c}\cos \left ( \frac{1}{x-c} \right )

Put x = c – h

f'\left ( c \right )=\lim_{x\rightarrow c}\cos \left ( \frac{1}{c-h-c} \right )

            =\lim_{x\rightarrow c}\cos \frac{1}{h}                                                                                                                                            \left [ \cos \left ( \Theta \right )=\cos \Theta \right ]

Since the value of cos is going to infinity, its limit will oscillate between -1 and 1.

Now, R.H.D is given by,

\begin{aligned} \mathrm{f}^{\prime}\left(\mathrm{c}^{+}\right) &=\lim _{x \rightarrow c^{+}} \frac{f(x)-f(c)}{x-c} \\ &=\lim _{x \rightarrow c^{+}} \frac{(x-c) \cos \left(\frac{1}{x-c}\right)-0}{x-c} \\ &=\lim _{h \rightarrow 0} \cos \left(\frac{1}{c-h-c}\right) \\ &=\lim _{h \rightarrow 0} \cos \left(\frac{1}{h}\right) \end{aligned}

Since the value of cos is going to infinity, its limit will oscillate between -1 and 1. As the value of limit is not a finite value, the function is not differentiable.

 

 

Posted by

infoexpert21

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