JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Rolle's theorem Lagrange's mean value theorem

  • question_answer
    If \[f(x)=\cos x,0\le x\le \frac{\pi }{2}\], then the real number ?c? of the mean value theorem is                                 [MP PET 1994]

    A)            \[\frac{\pi }{6}\]

    B)            \[\frac{\pi }{4}\]

    C)            \[{{\sin }^{-1}}\left( \frac{2}{\pi } \right)\]

    D)            \[{{\cos }^{-1}}\left( \frac{2}{\pi } \right)\]

    Correct Answer: C

    Solution :

               We know that \[f'(c)=\frac{f(b)-f(a)}{b-a}\]                    \[\Rightarrow f'(c)=\frac{0-1}{\pi /2}=-\frac{2}{\pi }\]                                  .....(i)                    But \[f'(x)=-\sin x\Rightarrow f'(c)=-\sin c\]                      ....(ii)                    From (i) and (ii), we get \[-\sin c=-\frac{2}{\pi }\Rightarrow c={{\sin }^{-1}}\left( \frac{2}{\pi } \right)\].

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