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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 2

  • question_answer
    If \[f(x)\] is differentiable and strictly increasing function, then the value of \[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{f({{x}^{2}})-f(x)}{f(x)-f(0)}\] is

    A) 1                     

    B) 0    

    C) -1                         

    D) 2

    Correct Answer: C

    Solution :

    Let\[L=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{f({{x}^{2}})-f(x)}{f(x)-f(0)}\]
    Using L.H. Rule, we get \[L=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{f'({{x}^{2}}).2x-f'(x)}{f'(x)}\]
    \[=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{f'({{x}^{2}}).2x}{f'(x)}-1\] \[\left[ \begin{align}   & \because \,\,f'\,(a)>0,f\,\,being \\  & strictly\,\,\text{increasing} \\ \end{align} \right]\]
    \[=0-1=-1\]
     


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