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#### Please Solve R.D.Sharma class 12 Chapter 9 Differentiability Exercise 9.2 Question 5 Maths textbook Solution.

Answer: f(x) is continuous on (-1,2) but not differentiable at x = 0,1.

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: $f\left ( x \right )=|x|+|x-1|$

Solution:

The given function f(x) can be defined as:

$f\left ( x \right )=$ $\left\{\begin{array}{cc} x+x+1, & 1

$\Rightarrow \mathrm{f}(\mathrm{x})=\left\{\begin{array}{c} 2 x+1,-1

We know that a polynomial and a constant function is continuous and differentiable everywhere. So f(x) is continuous and differentiable for x $\epsilon$  (-1, 0) and x $\epsilon$ (0, 1) and (1, 2).

Continuity at  $x=0:$

\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(2 x+1)=1 \\ &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} 1=1 \\ &\mathrm{f}(0)=2(0)+1=1 \end{aligned}

Since, $\lim_{x\rightarrow 0^{-}}f\left ( x \right )=\lim_{x\rightarrow 0^{+}}f\left ( x \right )=f\left ( 0 \right ),f\left ( x \right )$is continuous at x = 0.

Continuity at x = 1:

\begin{aligned} &\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 1=1 \\ &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{-}} 1=1 \\ &\mathrm{f}(1)=1 \end{aligned}

Since, $\lim_{x\rightarrow 1^{-}}f\left ( x \right )=\lim_{x\rightarrow 1^{+}}f\left ( x \right )=f\left ( 0 \right ),f\left ( x \right )$ is continuous at x = 1.

Differentiability at x = 0:

\begin{aligned} &\text { LHD at } \mathrm{x}=0: \lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0} \\ &=\lim _{x \rightarrow 0^{-}} \frac{2 x+1-1}{x-0} \\ &=\lim _{x \rightarrow 0^{-}} \frac{2 x}{x} \\ &=2 \end{aligned}

\begin{aligned} &\text { RHD at } \mathrm{x}=0: \lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0} \\ &=\lim _{x \rightarrow 0^{+}} \frac{1-1}{x} \\ &=\lim _{x \rightarrow 0^{+}} \frac{0}{h} \\ &=0 \end{aligned}

LHD at x = 0 $\neq$ RHD at x = 0

Hence, f(x) is continuous but not differentiable at x = 0.

Differentiability at x = 1:

$LHD\: at\: x=1:\lim_{x\rightarrow 1^{-}}\frac{f\left ( x \right )-f\left ( 1 \right )}{x-1}$

$\lim_{x\rightarrow 1^{-}}\frac{1-1}{x-1}$

$=0$

$RHD\: at\: x=1:\lim_{x\rightarrow 1^{+}}\frac{f\left ( x \right )-f\left ( 1 \right )}{x-1}$

$=\lim_{x\rightarrow 1^{+}}\frac{2x+1-1}{x-1}$

$=\lim_{x\rightarrow 1^{+}}\frac{2x}{x-1}$

$=\frac{1}{0}$

$=\infty$

LHD at x = 1 $\neq$ RHD at x = 1

Hence, f(x) is continuous but not differentiable at x = 1.