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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Critical Thinking

  • question_answer
    If \[\sqrt{(1-{{x}^{6}})}+\sqrt{(1-{{y}^{6}})}={{a}^{3}}({{x}^{3}}-{{y}^{3}})\], then \[\frac{dy}{dx}=\] [Roorkee 1994]

    A) \[\frac{{{x}^{2}}}{{{y}^{2}}}\sqrt{\frac{1-{{x}^{6}}}{1-{{y}^{6}}}}\]

    B) \[\frac{{{y}^{2}}}{{{x}^{2}}}\sqrt{\frac{1-{{y}^{6}}}{1-{{x}^{6}}}}\]

    C) \[\frac{{{x}^{2}}}{{{y}^{2}}}\sqrt{\frac{1-{{y}^{6}}}{1-{{x}^{6}}}}\]

    D) None of these

    Correct Answer: C

    Solution :

    • Put \[{{x}^{3}}=\sin \theta ,\,\,{{y}^{3}}=\sin \varphi \]                   
    • \[\therefore \sqrt{1-{{x}^{6}}}+\sqrt{1-{{y}^{6}}}={{a}^{3}}({{x}^{3}}-{{y}^{3}})\]                    Þ
    • \[\cos \theta +\cos \varphi ={{a}^{3}}(\sin \theta -\sin \varphi )\]                   
    • or \[2\cos \frac{\theta +\varphi }{2}\cos \frac{\theta -\varphi }{2}=2{{a}^{3}}\sin \frac{\theta -\varphi }{2}\cos \frac{\theta +\varphi }{2}\]                   
    • or \[\cos \frac{\theta +\varphi }{2}\left[ \cos \frac{\theta -\varphi }{2}-{{a}^{3}}\sin \frac{\theta -\varphi }{2} \right]=0\]                   
    • If \[\cos \frac{\theta +\varphi }{2}=0,\,\,\]then \[\frac{\theta +\varphi }{2}=\frac{\pi }{2}\]                   
    • \[\therefore \theta =\pi -\varphi \] or \[\sin \theta =\sin \varphi \] or \[x=y\]                   
    • But if we put \[x=y\]in the given equation it is not satisfied and hence we must have                   
    • \[\cos \frac{\theta -\varphi }{2}-{{a}^{3}}\sin \frac{\theta -\varphi }{2}=0\]or \[\cot \frac{\theta -\varphi }{2}={{a}^{3}}\]                   
    • \[\therefore \theta -\varphi =2{{\cot }^{-1}}{{a}^{3}}\]or \[{{\sin }^{-1}}{{x}^{3}}-{{\sin }^{-1}}{{y}^{3}}=2{{\cot }^{-1}}{{a}^{3}}\]                   
    • Differentiating w.r.t. x, we get \[\frac{3{{x}^{2}}}{\sqrt{1-{{x}^{6}}}}-\frac{3{{y}^{2}}}{\sqrt{1-{{y}^{6}}}}\frac{dy}{dx}=0\]                   
    • \[\Rightarrow \,\,\frac{dy}{dx}=\frac{{{x}^{2}}}{{{y}^{2}}}\sqrt{\frac{1-{{y}^{6}}}{1-{{x}^{6}}}}\].


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