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JEE Main & Advanced Sample Paper JEE Main Sample Paper-16

  • question_answer
    The coordinate of one of the points on the curve \[f(x)=\frac{x}{1-{{x}^{2}}}\] tor which normal is parallel to bisector of IInd and IVth quadrant, is

    A) \[\left( 2,\frac{-2}{3} \right)\]                    

    B) \[\left( 3,\frac{-3}{8} \right)\]

    C) \[\left( -2,\frac{2}{3} \right)\]                    

    D) \[\left( -\sqrt{3},\frac{\sqrt{3}}{2} \right)\]

    Correct Answer: D

    Solution :

    Slope of normal is \[=-1\Rightarrow \]slope of tangent =1 \[f'(x)=\frac{1+{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}}=1\]i.e.\[x=0,\pm \sqrt{3}\] \[\therefore \]\[\left( -\sqrt{3},\frac{\sqrt{3}}{2} \right)\]is a point


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