JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    Let \[M=\int\limits_{0}^{\frac{\pi }{2}}{\frac{\cos x}{x+2}dx}\] and \[N=\int\limits_{0}^{\frac{\pi }{4}}{\frac{\sin x.\cos x}{{{(x+1)}^{2}}}dx}\] then the value of (M - N) equals

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

    B) \[\frac{2}{\pi +2}\]

    C) \[\frac{2}{\pi +4}\]

    D)  0

    Correct Answer: C

    Solution :

                            \[N=\frac{1}{2}\,\int\limits_{0}^{\frac{\pi }{4}}{\frac{\sin \,2x}{{{(x+1)}^{2}}}dx}\] On putting \[2x=1,\] we get \[N=\frac{1}{4}.\,4\int\limits_{0}^{\frac{\pi }{2}}{\,\frac{\sin t}{{{(t+2)}^{2}}}dt}\] \[\left. N=(\sin t)\,\left( \frac{-1}{t+2} \right) \right]_{0}^{\frac{\pi }{2}}+\int\limits_{0}^{\frac{\pi }{2}}{\frac{\cos t}{t+2}dt}\] \[\left. \therefore \,\,M-N\,=\frac{\sin t}{\operatorname{t}+2} \right]_{0}^{\frac{\pi }{2}}\,=\frac{2}{\pi +4}\]

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