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JEE Main & Advanced Mathematics Functions Question Bank Critical Thinking

  • question_answer
    If \[f(x)\,=\,\left\{ \begin{matrix} x{{e}^{-\,\left( \frac{1}{|\,x\,|}\,+\,\frac{1}{x} \right)}}, & x\ne 0  \\  0\,\,\,\,\,\,\,\,\,\,\,\,\,, & x=0  \\ \end{matrix} \right.\] , then \[f(x)\,\] is [AIEEE 2003]

    A) Continuous as well as differentiable for all x

    B) Continuous for all x but not differentiable at \[x=0\]

    C) Neither differentiable nor continuous at \[x=0\]

    D) Discontinuous every where

    Correct Answer: B

    Solution :

    • \[f(0)=0\] and \[f(x)=x{{e}^{-\left( \frac{1}{|x|}+\frac{1}{x} \right)}}\]           
    • R.H.L. = \[\underset{h\to 0}{\mathop{\lim }}\,(0+h){{e}^{-2/h}}=\underset{h\to 0}{\mathop{\lim }}\,\frac{h}{{{e}^{2/h}}}=0\]           
    • L.H.L. = \[\underset{h\to 0}{\mathop{\lim }}\,(0-h){{e}^{-\left( \frac{1}{h}\,-\,\frac{1}{h} \right)}}=0\]; \\[f(x)\] is continuous.           
    • \[R{f}'\,(x)=\underset{h\to 0}{\mathop{\lim }}\,\frac{(0+h){{e}^{-\left( \frac{1}{h}+\frac{1}{h} \right)}}-h{{e}^{-\left( \frac{1}{h}+\frac{1}{h} \right)}}}{h}=0\]           
    • \[L{f}'(x)=\underset{h\to 0}{\mathop{\lim }}\,\frac{(0-h){{e}^{-\left( \frac{1}{h}-\frac{1}{h} \right)}}-h{{e}^{-\left( \frac{1}{h}+\frac{1}{h} \right)}}}{-h}=1\]           
    • Þ \[L{f}'(x)\ne R{f}'(x)\]. \[f(x)\] is not differentiable at \[x=0.\]


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