A) continuous and differentiable
B) continuous but not differentiable
C) differentiable and not continuous
D) not continuous and not differentiable
Correct Answer: B
Solution :
\[f(x)=\left\{ \begin{matrix} {{x}^{3}}-1, & 1\le x\le \infty \\ x-1, & -\infty <x<1 \\ \end{matrix} \right.\] LHL\[=\underset{x\to {{1}^{-1}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,(1-h-1)\] \[=\underset{x\to 0}{\mathop{\lim }}\,-h=0\] RHL\[=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,[{{(1+h)}^{3}}-1]=0\] Hence,\[f(x)\]is continuous at\[x=1\] Now, \[Rf'(1)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(1+h)-f(1)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{(1+h)}^{3}}-1}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{3}}{{(1/h+1)}^{3}}-1}{h}=-1\] \[Lf'(1)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(1-h)-f(1)}{-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{1-h-1}{-h}=1\] So, \[Rf'(1)\ne Lf'(1)\] Hence, \[f(x)\]is not differentiable at\[x=1\]
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