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

  • question_answer
    \[f(x)=({{x}^{2}}-4)|{{x}^{3}}-6{{x}^{2}}+11x-6|+\frac{x}{1+|x|}\] then the set of the points at which fix) is not differentiable is-

    A) {-2, 2, 1, 3}    

    B)        {-2, 0, 3}

    C) {-2, 2, 0}       

    D)        {1, 3}

    Correct Answer: D

    Solution :

    [d] \[f\left( x \right)=\left( x+2 \right)\left( x-2 \right)|\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)|+\frac{x}{1+|x|}\]\[\left( y-1 \right)(x-2)\left( x-3 \right)\] is not differentiable at x= 1,3 only. \[\frac{x}{1+|x|}\] is always differentiable.


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