#### Please Solve R.D.Sharma class 12 Chapter 9 Differentiability Exercise 9.2 Question 6 Maths textbook Solution.

Answer: f(x) is not differentiable at x = 1 but differentiable at x = 2

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:   $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{c} x, x \leq 1 \\ 2-x, 1 \leq \mathrm{x} \leq 2 \\ -2+3 x-x^{2}, x>2 \end{array}\right.$

Solution:

Differentiability at x = 1:

\begin{aligned} &\mathrm{LHD} \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{(0-h)^{m} \sin \left(\frac{1}{-h}\right)}{-h} \\ &=\lim _{h \rightarrow 0}(-h)^{m-1} \sin \left(\frac{1}{-h}\right) \end{aligned}                                                                                $\left \{ \frac{a^{m}}{a^{n}}\Rightarrow a^{m-n} \right \}$

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

$\Rightarrow \left \{ when-1\leq \frac{1}{h} \leq 1\right \}$

$\Rightarrow 0\: x\: k$

$\Rightarrow 0$

\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} \end{aligned}

$\Rightarrow \left \{ when-1\leq k\leq 1 \right \}$                                                                                                                    $\left\{\begin{array}{c} (0)^{m-1}=0 \\ \sin \left(\frac{1}{0}\right)=\sin (\infty) \end{array}\right.$

$\Rightarrow 0\; x\: k$

$\Rightarrow 0$

(LHD at x = 0)= (RHD at x = 0)

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