A) Continuous as well as differentiable at \[x=1\]
B) Differentiable but not continuous at \[x=1\]
C) Continuous but not differentiable at \[x=1\]
D) Neither continuous nor differentiable at\[x=1\]
Correct Answer: D
Solution :
| we have \[f(x)=\left\{ \begin{align} & \begin{matrix} \frac{-1}{x-1}, & 0<x<1 \\ \frac{1-1}{x-1}, & 1<x<21 \\ \end{matrix} \\ & \begin{matrix} 0, & x=1 \\ \end{matrix} \\ \end{align} \right.\] |
| \[{}^{\underset{h\to 0}{\mathop{\lim }}\,f(1-h)}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{(1-h)-1}\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{h}=\infty \] |
| \[\therefore \]\[f\left( x \right)\] is not continuous and hence not differentiable at x=1. |
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