Provide solution for RD Sharma maths class 12 chapter 9 Differentiability exercise Very short answer type question  10

Answer: $x=0,1$

Hint: $|x|= \begin{cases}x & x>0 \\ -x & x<0\end{cases}$

Given:$f(x)=|x|+|x-1|$

Explanation:

$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc} x+x-1 & x>1 \\ x-(x-1) & 0 \leq x \leq 1 \\ -x-(x-1) & x<0 \end{array}\right.$

$=\left\{\begin{array}{lr} 2 x-1 & x>1 \\ 1 & 0 \leq x \leq 1 \\ -2 x+1 & x<0 \end{array}\right.$

$f(x)$ is a polynomial function for $x > 1, 0\leq x\leq 1$ and $x < 0$ . so, $f(x)$ is differentiable for all  $x>1, 0\leq x\leq 1$ & $x < 0$

We only have to check at $x= 0, 1$

$LHD\; at\; x = 0$

\begin{aligned} \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=& \lim _{x \rightarrow 0} \frac{-2 x+1-1}{x-0} \\ &=-2 \end{aligned} \quad \text { [L-Hospital rule ] }

$\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0} \frac{1-1}{x-0}$

$=0$

$\text { As } \mathrm{LHD} \neq \mathrm{RHD} \text { at } \mathrm{x}=0$

$f(x)$ is not differentiable at $x=0$

\begin{aligned} &\text { At } x=1 \\ &\text { LHD at } x=1 \end{aligned}

\begin{aligned} &\text { At } x=1 \\ &\text { LHD at } x=1 \\ &\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 0} \frac{1-1}{x-1}=0 \end{aligned}

\begin{aligned} &\mathrm{RHD} \text { at } \mathrm{x}=1 \\ &\begin{aligned} \lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} &=\lim _{x \rightarrow 0} \frac{2 x-1-1}{x-1} \\ &=2 \end{aligned} \quad \text { [L-Hospital rule] } \\ & \end{aligned}

$\text { As } \mathrm{LHD} \neq \mathrm{RHD} \text { at } \mathrm{x}=1$

$f(x)$ is not differentiable at $x=1$ so that point of non-differentiability

are $0, 1$