JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    Let y = y(x) be a function of x satisfying \[y\sqrt{1-{{x}^{2}}}=k-x\sqrt{1-{{y}^{2}}}\] where k is a constant and \[y\left( \frac{1}{2} \right)=-\frac{1}{4}.\]Then \[\frac{dy}{dx}\]at \[x=\frac{1}{2}\], is equal to [JEE MAIN Held on 07-01-2020 Evening]

    A) \[-\frac{\sqrt{5}}{2}\]

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

    C) \[\frac{\sqrt{5}}{2}\] 

    D) \[-\frac{\sqrt{5}}{4}\]

    Correct Answer: A

    Solution :

    [a] \[\because \,\,\,\,y\sqrt{1-{{x}^{2}}}=k-x\sqrt{1-{{y}^{2}}}\]                ...(i) On differentiating both side of eq. (i) w.r.t. x we get, \[\frac{dy}{dx}\sqrt{1-{{x}^{2}}}-y\frac{2x}{2\sqrt{1-{{x}^{2}}}}=0-\sqrt{1-{{y}^{2}}}+\frac{x\cdot y}{\sqrt{1-{{y}^{2}}}}\frac{dy}{dx}\] Put \[x=\frac{1}{2}\]and \[y=-\frac{1}{4}\] we get \[\frac{dy}{dx}\cdot \frac{\sqrt{3}}{2}-\left( -\frac{1}{4} \right)\cdot \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=-\frac{\sqrt{15}}{4}+\frac{-\frac{1}{8}}{\frac{\sqrt{15}}{4}}\cdot \frac{dy}{dx}\] \[\therefore \,\,\,\frac{dy}{dx}=-\frac{\sqrt{5}}{2}\]

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