#### Need Solution for R.D.Sharma Maths Class 12 Chapter 9 Differentiability Exercise 9.1 Question 7 Sub Question 1 Maths Textbook Solution.

Answer: f(x) is differentiable at x = 0, if m > 1.

Hint: The function y = f(x) is said to be differentiable in the closed interval [a, b] if R f  (a) and L f  (b) exist and f `(x) exist for every point of (a, b).

If the left hand limit, right hand limit and the value of the function at x = c exist and are equal to each other, then f is said to be continuous at x = c.

Given: $f\left ( x \right )=$$\left\{\begin{array}{c} x^{m} \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0, x=0 \end{array}\right.$

Solution:

Differentiability at x = 0:

$LHD\: at x=0: \lim_{x\rightarrow 0^{-}}\frac{f\left ( x \right )-f\left ( 0 \right )}{x-0}$

$= \lim_{h\rightarrow 0}\frac{f\left ( 0-h \right )-f\left ( 0 \right )}{\left ( 0-h \right )-0}$

$= \lim_{h\rightarrow 0} \frac{\left ( 0-h \right )^{m}\sin\left ( \frac{1}{-h} \right )}{-h}$

$= \lim_{h\rightarrow 0} \left ( -h \right )^{m-1}\sin \left ( \frac{1}{-h} \right )$                                                        $\left \{ \frac{a^{m}}{a^{n}}\Rightarrow a^{m-n} \right \}$

$= \lim_{h\rightarrow 0} \left ( -h \right )^{m-1}\sin \left ( \frac{1}{h} \right )$                                                         $\left \{ \sin \left ( -\theta \right )\Rightarrow -\sin \theta \right \}$

\begin{aligned} &\text { RHD at } x=0: \lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} \\ &=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{(0+h)-0} \\ &=\lim _{h \rightarrow 0} \frac{h^{m} \sin \left(\frac{1}{h}\right)}{h} \\ &\Rightarrow\{\text { when }-1 \leq k \leq 1\} \end{aligned}                                    $\left\{\begin{array}{c} (0)^{m-1}=0 \\ \sin \left(\frac{1}{0}\right)=\sin (\infty) \end{array}\right.$

$\Rightarrow 0\; x\; k$

$\Rightarrow 0$

$(LHD \: at \: x = 0)= (RHD\: at\: x = 0)$

Hence, f(x) is differentiable at x = 0.