JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Afternoon)

  • question_answer
    If \[\cos x\frac{dy}{dx}-y\sin x=6x,(0<x<\frac{\pi }{2})\] and\[y\left( \frac{\pi }{3} \right)=0,\]then\[y\left( \frac{\pi }{6} \right)\]is equal to :-             [JEE Main 9-4-2019 Afternoon]

    A) \[-\frac{{{\pi }^{2}}}{4\sqrt{3}}\]             

    B) \[-\frac{{{\pi }^{2}}}{2}\]

    C) \[-\frac{{{\pi }^{2}}}{2\sqrt{3}}\]

    D) \[\frac{{{\pi }^{2}}}{2\sqrt{3}}\]

    Correct Answer: C

    Solution :

    \[\frac{dy}{dx}-y\tan x=6x\sec x\]           \[y\left( \frac{\pi }{3} \right)=0;y\left( \frac{\pi }{6} \right)=7\]           \[{{e}^{\int_{{}}^{{}}{pdx}}}={{e}^{-\int_{{}}^{{}}{\tan \,xdx}=}}{{e}^{\ell n\,\cos x}}=\cos x\]           \[y.\cos x=\int_{{}}^{{}}{6x\sec x\cos xdx}\]           \[y.\cos x=\frac{6{{x}^{2}}}{2}+C\]           \[y=3{{x}^{2}}\sec x+C\sec x\]           0\[=3.\frac{{{\pi }^{2}}}{9}.(2)+C(2)\]           \[2C=\frac{-2{{\pi }^{2}}}{3}\Rightarrow \]           \[y(\pi /6)=3.\frac{{{\pi }^{2}}}{36}.\left( \frac{2}{\sqrt{3}} \right)+\left( \frac{2}{\sqrt{3}} \right).\left( -\frac{{{\pi }^{2}}}{3} \right)\]           \[\Rightarrow y=-\frac{{{\pi }^{2}}}{2\sqrt{3}}\]                 


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