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JEE Main & Advanced Sample Paper JEE Main Sample Paper-29

  • 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) non-existent in the interval

    Correct Answer: C

    Solution :

    Cleary, Rolle's Theorem is applicable for the function \[f(x)\] is\[[-1,\,\,1]\]. 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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