A) f is not continuous on [ ?1, 1]
B) f is not differentiable on (?1,1)
C) \[f(-1)\ne f(1)\]
D) \[f(-1)=f(1)\ne 0\]
Correct Answer: B
Solution :
\[f(x)=\left\{ \begin{align} & -x,\,\text{when --1}\le x<0 \\ & \text{ }x,\ \text{when}\ \text{0}\le x\le \text{1} \\ \end{align} \right.\] Clearly \[f(-1)=|-1|=1=f(1)\] But \[Rf'(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{|h|}{h}\]\[=\underset{h\to 0}{\mathop{\lim }}\,\frac{h}{h}=1\] \[Lf'(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0-h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{|-h|}{-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{h}{-h}=-1\] \[\therefore Rf'(0)\ne Lf'(0)\] Hence it is not differentiable on \[(-1,\,\,1)\].
You need to login to perform this action.
You will be redirected in
3 sec
