A) f is continuous but not differentiable
B) f is differentiable but not continuous
C) \[{f}'\] is continuous but not differentiable
D) \[{f}'\] is continuous and differentiable
Correct Answer: C
Solution :
\[f(x)=\left\{ \begin{matrix} \ \,0, & x<0 \\ {{x}^{2}}, & x\ge 0 \\ \end{matrix} \right.\]; \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(0-h)=0\] and \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(0+h)=\underset{h\to 0}{\mathop{\lim }}\,{{(0+h)}^{2}}=0\] Þ \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=f(0)\] Hence \[f(x)\] is continuous function at \[x=0\]. \[L\,{f}'(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{f(x)-f(0)}{x-0}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0-h)-0}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{0-0}{-h}=0\] \[R{f}'(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{f(x)-f(0)}{x-0}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{(0+h)}^{2}}-0}{h}=0\] Þ \[L{f}'(x)=R{f}'(x)\] Hence \[f(x)\] is differentiable at \[x=0\]. Now \[{f}'(x)=\left\{ \begin{matrix} 0\,\,\,\,, & x<0 \\ 2x\,\,, & x\ge 0 \\ \end{matrix} \right.\]; \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{f}'(x)=\underset{h\to 0}{\mathop{\lim }}\,{f}'(0-h)=0\] and \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{f}'(x)=\underset{h\to 0}{\mathop{\lim }}\,{f}'(0+h)=\underset{h\to 0}{\mathop{\lim }}\,2(0+h)=0\] Þ \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{f}'(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{f}'(x)=0\] Hence \[{f}'(x)\] is continuous function at \[x=0\]. Now \[L\,{f}''(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{f(x)-f(0)}{x-0}\]\[=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0-h)-f(0)}{-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{0-0}{-h}=0\] \[R\,{f}''(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{f(x)-f(0)}{x-0}\]\[=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0+h)-f(0)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{2(0+h)-0}{h}\]\[=\underset{h\to 0}{\mathop{\lim }}\,\frac{2h}{h}=2\]Þ \[L{f}''(x)\ne R{f}''(x)\] Hence \[{f}'(x)\] is not differentiable at \[x=0\].
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