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JEE Main & Advanced Mathematics Functions Question Bank Critical Thinking

  • question_answer
    If \[f(x)=sgn ({{x}^{3}})\], then [DCE 2001]

    A) f is continuous but not derivable at \[x=0\]

    B) \[f'({{0}^{+}})=2\]

    C) \[f'({{0}^{-}})=1\]

    D) f is not derivable at \[x=0\]

    Correct Answer: D

    Solution :

    • Here, \[f(x)=sgn {{x}^{3}}=\left\{ \begin{align}   & \left\{ \begin{matrix}    \frac{{{x}^{3}}}{|{{x}^{3}}|}, & \text{for} & {{x}^{3}}\ne 0  \\    0\text{    }, & \text{for} & {{x}^{3}}=0  \\ \end{matrix} \right. \\  & \left\{ \begin{matrix}    \frac{x}{|x|}, & \text{for} & x\ne 0  \\    0\text{   ,} & \text{for} & x=0  \\ \end{matrix} \right. \\  & \left\{ \begin{matrix}    -1, & x0  \\ \end{matrix} \right. \\ \end{align} \right.\]           
    • Thus, \[f(x)=sgn {{x}^{3}}=sgn x,\] which is neither continuous nor derivable at 0.
    • Note that \[{f}'({{0}^{+}})=\underset{h\to {{0}^{+}}}{\mathop{\text{lim}}}\,\,\frac{f(0+h)-f(0)}{h}\]\[=\underset{h\to {{0}^{+}}}{\mathop{\text{lim}}}\,\,\frac{1-0}{h}\to \infty \]  and \[{f}'({{0}^{-}})=\underset{h\to {{0}^{-}}}{\mathop{\text{lim}}}\,\,\frac{f(0-h)-f(0)}{h}\]\[=\underset{h\to {{0}^{-}}}{\mathop{\text{lim}}}\,\,\frac{-1-0}{h}\to \infty \].  \[{f}'({{0}^{+}})\ne {f}'({{0}^{-}})\],   f is not derivable at \[x=0\].


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