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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Critical Thinking

  • question_answer
    \[\frac{d}{dx}\left[ {{\tan }^{-1}}\frac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right]=\] [Roorkee 1980; Karnataka CET 2005]

    A) \[\frac{-x}{\sqrt{1-{{x}^{4}}}}\]

    B) \[\frac{x}{\sqrt{1-{{x}^{4}}}}\]

    C) \[\frac{-1}{2\sqrt{1-{{x}^{4}}}}\]

    D) \[\frac{1}{2\sqrt{1-{{x}^{4}}}}\]

    Correct Answer: A

    Solution :

    • Let \[\frac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}}=A\]           
    • Putting \[{{x}^{2}}=\cos 2\theta ,\] we get           
    • \[A=\frac{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }}=\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }\]           
    • \[A=\frac{1+\tan \theta }{1-\tan \theta }=\tan \left( \theta +\frac{\pi }{4} \right)\]           
    • Now \[y={{\tan }^{-1}}\tan \left( \theta +\frac{\pi }{4} \right)=\theta +\frac{\pi }{4}=\frac{\pi }{4}+\frac{1}{2}{{\cos }^{-1}}{{x}^{2}}\]                   
    • \[\therefore \frac{dA}{dx}=\frac{1}{2}\frac{-1}{\sqrt{1-{{x}^{4}}}}2x=\frac{-x}{\sqrt{1-{{x}^{4}}}}\]


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