# Get Answers to all your Questions

#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 9 Differentiability Exercise 9.1 Question 7 Sub Question 2 Maths Textbook Solution.

Answer: f(x) is continuous but not differentiable at x = 0; if 0 < 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:

Now we have to check for continuity at x = 2.For continuity,

$(LHL at \: x = 2) = (RHL at \: x = 2)$

$f\left ( 0 \right )=0$

$LHL=\lim_{x\rightarrow 0^{-}}f\left ( x \right )$

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

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

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

As we know $\left ( 0 \right )^{1}=0$ and $\sin \frac{1}{0}$ $=\sin (\infty)=\mathrm{K}, \quad\{\sin (-\theta) \Rightarrow-\sin \theta\},\{\text { when }-1 \leq k \leq 1\}$

$\Rightarrow 0\: x\: k$

$\Rightarrow 0$

Now consider,

$RHL=\lim_{x\rightarrow 0^{+}}f\left ( x \right )$

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

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

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

As we know $\left ( a \right )^{0}=1$  and $\sin \frac{1}{0}=\sin (\infty),\{\text { when }-1 \leq k \leq 1\}$

$\Rightarrow 0\; X \: k$

$\Rightarrow 0$

$LHL = RHL = f (0)$

Since f(x) is continuous at x = 0, we have to find its differentiability using the formula,

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

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

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

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

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

$\Rightarrow Not \: Defined$                                                                                            $\left [ 0< m< 1 \right ]$

\begin{aligned} &\mathrm{RHD} \text { at } \mathrm{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} \\ &=\lim _{h \rightarrow 0}(h)^{m-1} \sin \left(\frac{1}{h}\right) \end{aligned}

$\Rightarrow$Not defined

(LHD at x = 0) $\neq$ (RHD at x = 0)

Hence, f(x) is continuous but not differentiable at x = 0.