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JCECE Engineering JCECE Engineering Solved Paper-2006

  • question_answer
    \[{{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}x}\,\,dx\], then\[\underset{n\to \infty }{\mathop{\lim }}\,n[{{I}_{n}}+{{I}_{n+2}}]\] equal to:

    A) \[\frac{1}{2}\]                                   

    B) \[1\]

    C) \[\infty \]                                           

    D) \[zero\]

    Correct Answer: B

    Solution :

    \[{{I}_{n}}+{{I}_{n+2}}=\int_{0}^{\pi /4}{{{\tan }^{n}}}x(1+{{\tan }^{2}}x)dx\]                 \[=\int_{0}^{\pi /4}{{{\tan }^{n}}x}{{\sec }^{2}}x\,\,dx\]                 \[=\int_{0}^{1}{{{t}^{n}}dt}\]where\[t=\tan x\] \[\Rightarrow \]               \[{{I}_{n}}+{{I}_{n+2}}=\frac{1}{n+1}\] \[\therefore \]  \[\underset{n\to \infty }{\mathop{\lim }}\,n[{{I}_{n}}+{{I}_{n+2}}]\]                 \[=\underset{n\to \infty }{\mathop{\lim }}\,n\cdot \frac{1}{n+1}\]                 \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{n\left( 1+\frac{1}{n} \right)}=1\]


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