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KVPY Sample Paper KVPY Stream-SX Model Paper-18

  • question_answer
    If \[\alpha =\int\limits_{0}^{1}{({{e}^{9x+3{{\tan }^{-1}}x}})\left( \frac{12+9{{x}^{2}}}{1+{{x}^{2}}} \right)}\,dx,\] where \[{{\tan }^{-1}}x\] takes only principal values, then the value of \[\left( {{\log }_{e}}\left| 1+\alpha  \right|-\frac{3\pi }{4} \right)\]

    A) 6

    B) 7       

    C) 8

    D) 9

    Correct Answer: D

    Solution :

    We have,
    \[\alpha =\int\limits_{0}^{1}{({{e}^{9x+3{{\tan }^{-1}}x}})\left( \frac{12+9{{x}^{2}}}{1+{{x}^{2}}} \right)dx}\]
    \[\alpha =\int\limits_{0}^{1}{({{e}^{9x+3{{\tan }^{-1}}x}})\left( 9+\frac{3}{1+{{x}^{2}}} \right)dx}\]
    Put \[9x+3{{\tan }^{-\,1}}x=t\]
    \[\Rightarrow \]   \[\left( 9+\frac{3}{1+{{x}^{2}}} \right)dx=dt\]
    \[\therefore \]\[\alpha =\int\limits_{0}^{9+\frac{3\pi }{4}}{{{e}^{t}}dt=[{{e}^{t}}]_{0}^{9+\frac{3\pi }{4}}}\]
    \[\Rightarrow \]\[\alpha ={{e}^{9+\frac{3\pi }{4}}}-1\]\[\Rightarrow \]\[(\alpha +1)={{e}^{9+\frac{3\pi }{4}}}\]\[\Rightarrow \]\[{{\log }_{e}}(\alpha +1)-\frac{3\pi }{4}=9\]


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