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JEE Main & Advanced Sample Paper JEE Main Sample Paper-15

  • question_answer
    A continuous function \[f:R\to R\] satisfy the differential equation \[f(x)=(1+{{x}^{2}})\] \[\left( l+\int\limits_{0}^{x}{\frac{{{f}^{2}}(t)}{1+{{t}^{2}}}dt} \right)\] then the value of f (-2) is

    A)  0                                            

    B) \[\frac{17}{15}\]

    C) \[\frac{-17}{15}\]                                            

    D) \[\frac{15}{17}\]

    Correct Answer: D

    Solution :

    \[\frac{f(x)}{1+{{x}^{2}}}=1+\int\limits_{0}^{x}{\frac{{{f}^{2}}(t)dt}{1+{{t}^{2}}}}\][Note: \[f(0)=1\]] differentiate both sides w.r.t. \[x\] \[\frac{(1+{{x}^{2}})f'(x)-f(x)(2x)}{{{(1+{{x}^{2}})}^{2}}}=\frac{{{f}^{2}}(x)}{(1+{{x}^{2}})}\] \[\frac{dy}{dx}-\left( \frac{2x}{1+{{x}^{2}}} \right)y={{y}^{2}}\] \[\frac{1}{{{y}^{2}}}\frac{dy}{dx}-\left( \frac{2x}{1+{{x}^{2}}} \right)\frac{1}{y}=1\]Let\[\frac{-1}{y}=t\] \[\frac{dt}{dx}+\left( \frac{2x}{1+{{x}^{2}}} \right)t=1\]solving the above LDE, \[f(x)=\frac{-3(1+{{x}^{2}})}{{{x}^{3}}+3x-3}\] \[f(-2)=\frac{-3(5)}{-8-6-3}=\frac{15}{17}\]


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