Manipal Engineering Manipal Engineering Solved Paper-2015

  • question_answer
    The value of \[4R\Omega .\]is

    A) \[{{w}_{1}}\]     

    B) \[{{w}_{2}}\]

    C) \[\frac{-({{w}_{2}}-{{w}_{1}})}{5Rt}\]     

    D) \[\frac{-n({{w}_{2}}-{{w}_{1}})}{5Rt}\]

    Correct Answer: D

    Solution :

    Let\[x=\frac{1}{y}.\]Then, \[\underset{x\to \infty }{\mathop{\lim }}\,{{\left\{ \frac{a_{1}^{1/x}+a_{2}^{1/x}+...+a_{n}^{1/x}}{n} \right\}}^{nx}}\] \[=\underset{y\to 0}{\mathop{\lim }}\,{{\left\{ \frac{a_{1}^{y}+a_{2}^{y}+...+a_{n}^{y}}{n} \right\}}^{n/y}}\] \[=\underset{y\to 0}{\mathop{\lim }}\,{{\left\{ 1+\frac{a_{1}^{y}+a_{2}^{y}+...+a_{n}^{y}-n}{n} \right\}}^{n/y}}\] \[={{e}^{\underset{y\to 0}{\mathop{\lim }}\,}}\left\{ \frac{a_{1}^{y}-1}{y}+\frac{a_{1}^{y}-1}{y}+...+\frac{a_{n}^{y}-1}{y} \right\}\] \[={{e}^{\log {{a}_{1}}+\log {{a}_{2}}+...+\log {{a}_{n}}}}\] \[={{e}^{\log ({{a}_{1}}{{a}_{2}}...{{a}_{n}})}}\] \[={{a}_{1}}{{a}_{2}}{{a}_{3}}...{{a}_{n}}\]


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