Manipal Engineering Manipal Engineering Solved Paper-2012

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

    A) \[1\]                                     

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

    C) \[{{e}^{2}}\]                      

    D)        \[{{e}^{3}}\]

    Correct Answer: C

    Solution :

    Let\[y={{\left[ \frac{f(1+x)}{f(1)} \right]}^{1/x}}\] \[\Rightarrow \]               \[\log y=\frac{1}{x}[\log f(1+x)-\log f(1)]\] \[\Rightarrow \]\[\underset{x\to 0}{\mathop{\lim }}\,\log y=\underset{x\to 0}{\mathop{\lim }}\,\frac{[\log f(1+x)-\log f(1)]}{x}\] \[\Rightarrow \]\[\underset{x\to 0}{\mathop{\lim }}\,\log y=\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{1}{f(1+x)}f(1+x) \right]\]                                 (using L Hospitals rule) \[\Rightarrow \]               \[\underset{x\to 0}{\mathop{\lim }}\,y={{e}^{2}}\]


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