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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 17

  • question_answer
    Let \[y=f(x)\] be an invertible function such that x-intercept of the tangent at any point \[P(x,y)\] on \[y=f(x)\] is equal to the square of abscissa of the point of tangency. If \[f(2)=1,\] then\[{{f}^{-1}}\left( \frac{5}{8} \right)\]- equals

    A) 3          

    B)                    5             

    C) 6                       

    D)        7

    Correct Answer: B

    Solution :

      [b] Equation of tangent at point \[P(x,y)\] on the curve \[y=f(x)\] is \[Y-y=\frac{dy}{dx}\,(X-x)\] X-intercept of tangent is \[x-\frac{y}{f'(x)}\] According to the question, \[x-\frac{y}{f'(x)}={{x}^{2}}\Rightarrow \frac{xf'(x)-f(x)}{{{x}^{2}}}=f'(x)\] Integrating both sides with respect to x, we get \[\frac{f(x)}{x}=f(x)+C\] Since \[f(2)=1,\,\,C=\frac{-1}{2}.\] Then \[f(x)=\frac{x}{2(x-1)}\] \[\therefore \,\,\,{{f}^{-1}}\left( \frac{5}{8} \right)=5\]      


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