#### please solve rd sharma class 12 chapter 9 Differentiability exercise Multiple choice question, question 1 maths textbook solution

(a)

HINTS: Learn the  definition of continuity and differentiability

GIVEN: $f(x)=|x|, g(x)=\left|x^{3}\right|$

SOLUTION:

\begin{aligned} &f(x)=|x|=\left\{\begin{array}{cc} x & x>0 \\ -x & x<0 \end{array}\right. \\ &g(x)=\left|x^{3}\right|=\left\{\begin{array}{cc} x^{3} & x>0 \\ -x^{3} & x<0 \end{array}\right. \end{aligned}

Now for the continuity of f(x),

Check at x=0

$\lim _{x \rightarrow 0} f(x)=\lim _{h \rightarrow 0} f(0-h)$

\begin{aligned} \lim _{h \rightarrow 0} f(-h) &=\lim _{h \rightarrow 0} f(-(-h)) \\ &=0 \\ \lim _{x \rightarrow 0} f(x) &=\lim _{h \rightarrow 0} f(0+h) \\ &=\lim _{h \rightarrow 0} h \\ &=0 \\ \end{aligned}

As \begin{aligned} \lim _{x \rightarrow 0^{-}} f(x) &=\lim _{x \rightarrow 0^{+}} f(x) \end{aligned}
therefore f(x) continuous for x=0

For differentiability of f(x) at x=0

LHD  at x=0

\begin{aligned} \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))-0}{-h} \\ &=\lim _{h \rightarrow 0}-1 \\ &=-1 \end{aligned}

RHD at x=0

\begin{aligned} \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{f(h)-f(0)}{h} \\ &=\lim _{h \rightarrow 0} \frac{h-0}{a} \\ &=\lim _{h \rightarrow 0} 1 \\ &=1 \end{aligned}

As LHD and RHD at x=0

f(x) is not differentiable at x=0

Now, continuity of g(x)

Check at x=0

\begin{aligned} \lim _{x \rightarrow 0^{-}} g(x) &=\lim _{h \rightarrow 0} g(0-h) \\ &=\lim _{h \rightarrow 0}-(-h)^{3} \\ &=0 \\ \end{aligned}

\begin{aligned} \lim _{x \rightarrow 0^{+}} g(x) &=\lim _{h \rightarrow 0} g(0+h) \\ &=\lim _{h \rightarrow 0} h^{3} \\ &=0 \end{aligned}

Differentiability of g(x) at x=0

LHD of x=0

\begin{aligned} \lim _{x \rightarrow 0^{-}} \frac{g(x)-g(0)}{x-0} &=\lim _{h \rightarrow 0} \frac{g(0-h)-g(0)}{0-h-0} \\ &=\lim _{h \rightarrow 0} \frac{-(-h)^{3}-0}{-h} \\ &=\lim _{h \rightarrow 0}-h^{2} \\ &=0 \end{aligned}

RHD at x=0

\begin{aligned} \lim _{x \rightarrow 0^{+}} \frac{g(x)-g(0)}{x-0} &=\lim _{h \rightarrow 0} \frac{g(0+h)-g(0)}{0+h-0} \\ &=\lim _{h \rightarrow 0} \frac{h^{3}-0}{h} \\ &=\lim _{h \rightarrow 0} h^{2} \\ &=0 \end{aligned}

As LHD and RHD at x=0

g(x) is differentiable at x=0