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JEE Main & Advanced Sample Paper JEE Main Sample Paper-2

  • question_answer
    \[\underset{n\to \infty }{\mathop{\lim }}\,\,{{\left\{ \left( 1+\frac{1}{{{n}^{2}}} \right)\left( 1+\frac{{{2}^{2}}}{{{n}^{2}}} \right)\left( 1+\frac{{{3}^{2}}}{{{n}^{2}}} \right).....\left( 1+\frac{{{n}^{2}}}{{{n}^{2}}} \right) \right\}}^{1/n}}=\]

    A)  \[{{e}^{\frac{\pi -4}{2}}}\]                          

    B)  \[2{{e}^{\frac{\pi -4}{2}}}\]

    C)  \[-{{e}^{\frac{\pi -4}{2}}}\]                        

    D)  \[\frac{1}{2}{{e}^{\frac{\pi -4}{2}}}\]

    Correct Answer: B

    Solution :

    \[\ell ny=\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{1}{n}\sum\limits_{r=1}^{n}{\ell n\left( 1+{{\left( \frac{r}{n} \right)}^{2}} \right)}\] \[=\int\limits_{0}^{1}{\ell n(1+{{x}^{2}})dx=|\ell n(1+{{x}^{2}})\cdot \,|_{0}^{1}-2\int\limits_{0}^{1}{\frac{{{x}^{2}}}{1+{{x}^{2}}}}dx}\] \[=\ell n2-2|x-{{\tan }^{-1}}x|_{0}^{1}\,=\ell n2-2\left( 1-\frac{\pi }{4} \right)\] \[\Rightarrow \,\,y=2\cdot \,{{e}^{\frac{\pi -4}{2}}}\]


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