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JEE Main & Advanced Mathematics Limits, Continuity and Differentiability Question Bank Self Evaluation Test - Continuity and Differentiability

  • question_answer
    Which of the following is correct for \[f(x)=\left\{ \begin{matrix}    (x-e){{2}^{-{{2}^{\left( \frac{1}{(e-x)} \right)}},}} & x\ne e\,\,at\,\,x=e  \\    0, & x=e  \\ \end{matrix} \right.\]

    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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