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JEE Main & Advanced Sample Paper JEE Main Sample Paper-4

  • question_answer
    If\[\int_{{}}^{{}}{\frac{{{e}^{x}}(1+100{{x}^{99}}-{{x}^{200}})}{\left( 1-{{x}^{100}} \right)\sqrt{1-{{x}^{200}}}}}dx={{e}^{x}}{{\left( \frac{1-{{x}^{k}}}{1+{{x}^{k}}} \right)}^{m}}+c\]then

    A) \[k=100,m=1\] 

    B) \[k=99,m=\frac{1}{2}\]

    C) \[k=99,m=1\]                    

    D) \[k=100,m=-\frac{1}{2}\]

    Correct Answer: D

    Solution :

    \[I=\int_{{}}^{{}}{{{e}^{x}}}\left( \sqrt{\frac{1+{{x}^{100}}}{1-{{x}^{100}}}}+\frac{100{{x}^{99}}}{(1-{{x}^{100}})\sqrt{1-{{x}^{200}}}} \right)dx\] Now\[\frac{d}{dx}\left( \sqrt{\frac{1+{{x}^{100}}}{1-{{x}^{100}}}} \right)=\frac{100{{x}^{99}}}{\left( 1-{{x}^{100}} \right)\sqrt{1-{{x}^{200}}}}\] Use\[I=\int_{{}}^{{}}{{{e}^{x}}(f(x)+f(x))={{e}^{x}}f(x)+C}\]


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