#### Provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 2

The correct option is (a) and (b)

Hint:

(i) A function f(x) is said to be continuous at a point x = a of its domain, if

\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)\\ &\text { (ii) } \lim _{h \rightarrow 0}\left\{\frac{f(a+h)-f(a)}{h}\right\}=f^{\prime}\left(a^{+}\right) \text {for right hand derivative }\\ &\lim _{h \rightarrow 0}\left\{\frac{f(a-h)-f(a)}{-h}\right\}=f^{\prime}\left(a^{-}\right) \text {for left hand derivative } \end{aligned}

Given:

$f(x)=|x-a| \phi(x)$

Solution:

Using formula (ii)

\begin{aligned} &=\lim _{h \rightarrow 0}\left\{\frac{f(a+h)-f(a)}{h}\right\} \\ &=\lim _{h \rightarrow 0}\left\{\frac{|h+a-a| \phi(a+h)-|a-a| \phi(a)}{h}\right\} \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{h \phi(a+h)}{h}\\ &=\lim _{h \rightarrow 0} \phi(a+h)=\phi(a)=f^{\prime}\left(a^{+}\right)\; \; \; ....(i) \end{aligned}

For Left Hand Derivative

\begin{aligned} &=\lim _{h \rightarrow 0}\left\{\frac{f(a-h)-f(a)}{-h}\right\} \\ &=\lim _{h \rightarrow 0}\left\{\frac{|a-h-a| \phi(a-h)-|a-a| \phi(a)}{h}\right\} \\ &=\lim _{h \rightarrow 0} \frac{|-h| \phi(a-h)}{-h} \\ &=\lim _{h \rightarrow 0}-\phi(a-h) \end{aligned}

Therefore

\begin{aligned} &=\lim _{h \rightarrow 0}-\phi(a-h)\\ &=-\phi(a)\\ &=f^{\prime}\left(a^{-}\right)\; \; \; \; \; \; .....(ii) \end{aligned}

From (i) and (ii), we get

$f^{\prime}\left(a^{+}\right) \neq f^{\prime}\left(a^{-}\right)$

So, correct option is (a) and (b)