#### please solve rd sharma class 12 chapter 9 Differentiability exercise Multiple choice question, question 3 maths textbook solution

(a)

HINTS: Understand the  definition of differentiability and modulus function/

GIVEN: $f(x)=x|x|$

SOLUTION:

$f(x)=\left\{\begin{matrix} x(-x) & & x<0\\ x(x)& & x\geq 0 \end{matrix}\right.\\$

$=\left\{\begin{matrix} -x^{2} & & x<0\\ x^{2}& & x\geq 0 \end{matrix}\right.$

For $x<0$ and $x>0$ function is differentiable as it is a polynomial function.

Now at x=0

LHD=

\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)^{2}-0}{-h} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-\left(-h^{2}\right)-0}{-h} \\ &=\lim _{h \rightarrow 0}-h \\ &=0 \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{h^{2}-0}{h} \\ &=\lim _{h \rightarrow 0} \mathrm{~h} \\ &=0 \end{aligned}

As LHD at x=0 and RHD at x=0

Hence, f(x) is  differentiable at $(-\infty ,\infty )$