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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 39

  • question_answer
    The value of c in Rolle's theorem for the function \[f(x)=\left\{ \begin{matrix}    {{x}^{2}}\cos \left( \frac{1}{x} \right), & x\ne 0  \\    0, & x=0  \\ \end{matrix} \right.\]in the interval \[[-1,1]\] is

    A) \[\frac{-1}{2}\]                    

    B)        \[\frac{1}{4}\]                    

    C) \[0\]    

    D)        None of these

    Correct Answer: C

    Solution :

    [c] Clearly, Rolle?s theorem is applicable for the function \[f(x)\] in \[[-1,\,1]\]. Also , \[f'(0)=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{{{h}^{2}}\cos \left( \frac{1}{h} \right)-0}{h}=0\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,c=0\]               


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