A) \[f(x)\] is differentiable at \[x=0\]
B) \[f(x)\] is continuous at \[x=0\]
C) \[f(x)\] is differentiable at \[x=1\]
D) \[f(x)\] is continuous at \[x=1\]
Correct Answer: B
Solution :
\[f(x)=\left\{ \begin{align} & {{e}^{x}}\,\,;\,\,\,x\le 0 \\ & 1-x;\,\,0<x\le 1 \\ & x-1\,;\,\,x>1 \\ \end{align} \right.\] \[Rf'(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1-h-1}{h}=-1\] \[Lf'(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0-h)-f(0)}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{e}^{-h}}-1}{-h}=1\] So, it is not differentiable at \[x=0\]. Similarly, it is not differentiable at \[x=1\]. But it is continous at \[x=0\], 1.
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