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KVPY Sample Paper KVPY Stream-SX Model Paper-25

  • question_answer
    Let \[f:\text{ }R\to R\] be such that \[f\left( 1 \right)=3\] and \[f'\left( 1 \right)=6.\] Then \[\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{f\left( 1+x \right)}{f\left( 1 \right)} \right)}^{1/x}}\] equals

    A) 1

    B) \[{{e}^{1/2}}\]

    C) \[{{e}^{2}}\]  

    D) \[{{e}^{3}}\]

    Correct Answer: C

    Solution :

    Given that \[f:R\to R\] such that \[f(1)=3\]and\[f'(1)=6\]
    Then \[\underset{x\to 0}{\mathop{\lim }}\,{{\left[ \frac{f\left( 1+x \right)}{f\left( 1 \right)} \right]}^{1/x}}\]\[={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{1}{f(1+x)}f'(1+x)}{1}}}={{e}^{\frac{f'(1)}{f(1)}}}={{e}^{6/3}}={{e}^{2}}\]
    [Using L Hospital rule]


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