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JEE Main & Advanced Mathematics Functions Question Bank Critical Thinking

  • question_answer
    If f is strictly increasing function, then \[\underset{x\to 0}{\mathop{\lim }}\,\frac{f({{x}^{2}})-f(x)}{f(x)-f(0)}\] is equal to                   [IIT Screening 2004]

    A) 0

    B) 1

    C) -1

    D) 2

    Correct Answer: C

    Solution :

    • \[\underset{x\to 0}{\mathop{\lim }}\,\frac{f({{x}^{2}})-f(x)}{f(x)-f(0)}\], \[\left( \frac{0}{0}\text{form} \right)\]                   
    • \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2xf'({{x}^{2}})-f'(x)}{f'(x)}\], (using L' Hospital's rule)                   
    • \[=-1+\underset{x\to 0}{\mathop{\lim }}\,\frac{2xf'({{x}^{2}})}{f'(x)}=-1,f'(0)\ne 0,\,\]                   
    • as f is strictly increasing.


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