A) continuous and differentiable at\[x=3\]
B) continuous but not differentiable at\[x=3\]
C) differentiable but not continuous at\[x=3\]
D) neither differentiable nor continuous at \[x=3\]
Correct Answer: B
Solution :
\[f'({{3}^{+}})=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(3+h)-f(3)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{(2-{{e}^{h}})-1}{h}=-\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{e}^{h}}-1}{h} \right)=-1\]\[f'({{3}^{-}})=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(3-h)-f(3)}{-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{10-{{(3-h)}^{2}}}-1}{-h}=-\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+(6h-{{h}^{2}})}-1}{-h}\]\[=\underset{h\to 0}{\mathop{\lim }}\,\frac{6h-{{h}^{2}}}{-h(\sqrt{1+6h-{{h}^{2}}+1})}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{h(h-6)}{h(\sqrt{1+6h-{{h}^{2}}+1})}=\frac{-6}{2}=-3\] Hence, \[f'({{3}^{+}})\ne f({{3}^{-}})\]
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