Let $f$ be defined on $D=[R-\{-1,1\}$ by $f(x)=\frac{|x|}{1-|x|}$, then

Let $f$ be defined on $D=[R-\{-1,1\}$ by $f(x)=\frac{|x|}{1-|x|}$, then
  1. f is differentiable on D
  2. f is differentiable on D except at x = 0
  3. f is continuous but not differentiable on D
  4. f is differentiable but not continuous on D

Solution

$ \begin{aligned} & \text {We have, } f(x)=\frac{|x|}{1-|x|}=\left\{\begin{array}{l} \frac{-x}{1+x}, x < 0 \\ \frac{x}{1-x}, x \geq 0 \end{array}\right. \\ & \text { LHD }(\text { at } x=0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{f(-h)-0}{-h}=\lim _{h \rightarrow 0} \frac{h}{\frac{1-h}{-h}} \\ & =\lim _{h \rightarrow 0} \frac{-1}{1-h}=-1 \\ & \text { RHD }(\text { at } x=0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\ & =\lim _{h \rightarrow 0} \frac{f(h)-0}{h}=\lim _{h \rightarrow 0} \frac{h}{1-h} \end{aligned} $ So, $f(x)$ is not differentiable at $x=0$. Hence, $f(x)$ is differentiable on $D$ except at $x=0$

Asked in: AP EAMCET 2018 (23 Apr Shift 1)

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