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JEE Main & Advanced JEE Main Solved Paper-2016

  • question_answer
    If a curve y = f(x) passes through the point (1, -1) and satisfies the differential equation, \[y(1+xy)dx=xdy,\]then \[f\left( -\frac{1}{2} \right)\] is equal to : [JEE Main Solved Paper-2016 ]

    A) \[\frac{4}{5}\]                                   

    B) \[-\frac{2}{5}\]

    C) \[-\frac{4}{5}\]                                 

    D) \[\frac{2}{5}\]

    Correct Answer: A

    Solution :

                    Given differential equation \[ydx+x{{y}^{2}}dx=xdy\] \[\Rightarrow \]\[\frac{xdy-ydx}{{{y}^{2}}}=xdx\]\[\Rightarrow \]\[-d\left( \frac{x}{y} \right)=d\left( \frac{{{x}^{2}}}{2} \right)\] Integrating we get\[-\frac{x}{y}=\frac{{{x}^{2}}}{2}+C\] \[\because \]It passes through (1, ?1) \[\therefore \]\[1=\frac{1}{2}+C\Rightarrow C=\frac{1}{2}\] \[\therefore \]\[{{x}^{2}}+1+\frac{2x}{y}=0\Rightarrow y=\frac{-2x}{{{x}^{2}}+1}\] \[\therefore \]\[f\left( -\frac{1}{2} \right)=\frac{4}{5}\]


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