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

  • question_answer
    Let \[f(x)\] satisfy all the conditions of mean value theorem in [0, 2]. If f (0) = 0 and \[|f'(x)|\,\le \frac{1}{2}\] for all x, in [0, 2] then

    A)            \[f(x)\le 2\]

    B)            \[|f(x)|\le 1\]

    C)            \[f(x)=2x\]

    D)            \[f(x)=3\]for at least one x in [0, 2]

    Correct Answer: B

    Solution :

               \[\frac{f(2)-f(0)}{2-0}=f'(x)\Rightarrow \frac{f(2)-0}{2}=f'(x)\]                    \[\Rightarrow \frac{df(x)}{dx}=\frac{f(2)}{2}\Rightarrow f(x)=\frac{f(2)}{2}x+c\]                    \[\therefore f(0)=0\Rightarrow c=0\]; \[\therefore f(x)=\frac{f(2)}{2}x\]                           .....(i)                    Given \[|f'(x)|\le \frac{1}{2}\Rightarrow \left| \frac{f(2)}{2} \right|\le \frac{1}{2}\]                      ?..(ii)                    (i) Þ \[|f(x)|=\left| \frac{f(2)}{2}x \right|=\left| \frac{f(2)}{2} \right||x|\le \frac{1}{2}|x|\][from (ii)]                    In [0, 2], for maximum \[x(x=2)\]                    \[|f(x)|\le \frac{1}{2}.\,\,\,2\Rightarrow |f(x)|\le 1\] .


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