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BCECE Engineering BCECE Engineering Solved Paper-2015

  • question_answer
    Let \[f(x)\] be twice differentiable function such that \[f''(0)=2.\] Then,\[\underset{x\to 0}{\mathop{\lim }}\,\frac{2f(x)-3f(2x)+f(4x)}{{{x}^{2}}}\]is equal to

    A) 6           

    B) 3

    C) 12                                         

    D) None of these

    Correct Answer: B

    Solution :

    \[\underset{x\to 0}{\mathop{\lim }}\,\frac{2f(x)-3f(2x)+f(4x)}{{{x}^{2}}}\left[ \frac{0}{0}\text{form} \right]\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2f'(x)-6f'(2x)+4f'(4x)}{2x}\]                                 [using L-Hospital's rule] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2f''(x)-12f''(2x)+16f''(4x)}{2}\]                             [using L?Hospital?s rule] \[=\frac{2f''(0)-12f''(0)+16f''(0)}{2}\] \[=3f''(0)=3\times 2=6\] 3


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