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JEE Main & Advanced Mathematics Differential Equations Question Bank Critical Thinking

  • question_answer
    The slope of the tangent at \[(x,y)\]to a curve passing through \[\left( 1,\frac{\pi }{4} \right)\]is given by \[\frac{y}{x}-{{\cos }^{2}}\left( \frac{y}{x} \right)\], then the equation of the curve is [Kurukshetra CEE 2002]

    A) \[y={{\tan }^{-1}}\left[ \log \left( \frac{e}{x} \right) \right]\]

    B) \[y=x{{\tan }^{-1}}\left[ \log \left( \frac{x}{e} \right) \right]\]

    C) \[y=x{{\tan }^{-1}}\left[ \log \left( \frac{e}{x} \right) \right]\]            

    D) None of these

    Correct Answer: C

    Solution :

    • We have \[\frac{dy}{dx}=\frac{y}{x}-{{\cos }^{2}}\left( \frac{y}{x} \right)\]        
    • Putting \[y=vx\]so that \[\frac{dy}{dx}=v+x\frac{dv}{dx}\], we get        
    • \[v+x\frac{dv}{dx}=v-{{\cos }^{2}}v\] or \[\frac{dv}{{{\cos }^{2}}v}=-\frac{dx}{x}\]        
    • On integrating, we get \[\tan v=-\log x+\log c\]         Þ \[\tan \left( \frac{y}{x} \right)=-\log x+\log C\]        
    • This passes through \[\left( 1,\frac{\pi }{4} \right)\], therefore\[1=\log c\]                   
    • Þ\[\tan \left( \frac{y}{x} \right)=-\log x+\log e\]Þ\[y=x{{\tan }^{-1}}\left[ \log \left( \frac{e}{x} \right) \right]\].


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