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

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

Answers (1)

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

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