A) f(x) is discontinuous at x = e
B) f(x) is differentiable at x = e
C) f(x) is non-differentiable at x = e
D) None of these
Correct Answer: C
Solution :
[c] \[f({{e}^{+}})=\underset{h\to 0}{\mathop{\lim }}\,(e+h-e){{2}^{-{{2}^{\frac{1}{e-(e+h)}}}}}\] |
\[=\underset{h\to 0}{\mathop{\lim }}\,(h){{2}^{-{{2}^{-\frac{1}{h}}}}}=0\times 1=0\] |
\[\left( \text{As}\,\,\text{for}\,\,\text{h}\to 0,-\frac{1}{h}\to -\infty \Rightarrow {{2}^{-\,\frac{1}{h}}}\to 0 \right)\] |
\[f({{e}^{-}})=\underset{h\to 0}{\mathop{\lim }}\,(-h){{2}^{-{{2}^{\frac{1}{h}}}}}=0\times 0=0\] |
Hence, \[f(x)\] is continuous at \[x=e\]. |
\[f'({{e}^{+}})=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(e+h)-f(e)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{h\times {{2}^{-{{2}^{\frac{1}{h}}}}}-0}{h}\] |
\[=\underset{h\to 0}{\mathop{\lim }}\,{{2}^{-{{2}^{\frac{1}{h}}}}}=1\] |
\[f'({{e}^{-}})=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(e-h)-f(0)}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{(-h){{2}^{-{{2}^{\frac{1}{h}}}}}-0}{-h}\] |
\[=\underset{h\to 0}{\mathop{\lim }}\,{{2}^{-{{2}^{\frac{1}{h}}}}}=0\]. |
Hence, \[f(x)\] is non-differentiable at \[x=e\]. |
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