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  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-32

  • question_answer
    Let \[(x)={{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)+{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right),x\in [-1,0]\] Then the number of points where g(x) is non-differentiable in [-1,0] is

    A) 0     

    B)    1

    C) 2                             

    D)    3

    Correct Answer: A

    Solution :

    \[g(x)=-2\,\tan {{\,}^{-1}}x+2\,\tan {{\,}^{-1}}x=0\,\,\forall \,\,x\,\,\in \,\,[-1,0]\]


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