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

  • question_answer
    If \[y={{\sec }^{-1}}\frac{2x}{1+{{x}^{2}}}+{{\sin }^{-1}}\frac{x-1}{x+1}\],then \[\frac{dy}{dx}\]is equal to [Pb. CET 2000]

    A) 1

    B) \[\frac{x-1}{x+1}\]

    C) Does not exist

    D) None of these

    Correct Answer: C

    Solution :

    • Let \[y={{\sec }^{-1}}\frac{2x}{1+{{x}^{2}}}+{{\sin }^{-1}}\left( \frac{x-1}{x+1} \right)\]                   
    • \[\because \] \[-1\le \frac{2x}{1+{{x}^{2}}}\le 1\]; So, \[{{\sec }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\] is defined only at \[x=-1\] and 1. So, \[y\] is not differentiable.                   
    • \[\therefore \] \[\frac{dy}{dx}\] does not exist.


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