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JEE Main & Advanced JEE Main Solved Paper-2017

  • question_answer
    Let\[{{I}_{n}}=\int_{{}}^{{}}{{{\tan }^{n}}xdx,(n>1).{{I}_{4}}+{{I}_{6}}}\]\[=a{{\tan }^{5}}x+b{{x}^{5}}+C,\]where C is a constant of integration, then ordered pair (a, b) is equal to:-                                          JEE Main Solved Paper-2017 ]

    A)  \[\left( -\frac{1}{5},0 \right)\]                   

    B)  \[\left( -\frac{1}{5},1 \right)\]

    C)  \[\left( \frac{1}{5},0 \right)\]                    

    D)  \[\left( \frac{1}{5},-1 \right)\]

    Correct Answer: C

    Solution :

     \[{{I}_{4}}+{{I}_{6}}=\int_{{}}^{{}}{({{\tan }^{4}}x+{{\tan }^{6}}x)}\,dx=\int_{{}}^{{}}{{{\tan }^{4}}x{{\sec }^{2}}x\,dx}\]                 \[=\frac{1}{5}{{\tan }^{5}}x+c\Rightarrow \,\,a=\frac{1}{5},b=0\]


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