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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
      If\[x{{e}^{xy}}=y+si{{n}^{2}}x,\]then\[\frac{dy}{dx}\]at\[x=0\]is:

    A)  \[-1\]                   

    B)         0

    C)  1                            

    D)         2

    E)  \[-2\]

    Correct Answer: C

    Solution :

    Give that,\[x{{e}^{xy}}=y+si{{n}^{2}}x\] at\[x=0,\text{ }y=0\] On differentiating w.r.t.\[x,\]we get \[x{{e}^{xy}}\left( x\frac{dy}{dx}+y \right)+{{e}^{xy}}=\frac{dy}{dx}+2\sin x\,\cos x\] \[\Rightarrow \]\[{{x}^{2}}{{e}^{xy}}\frac{dy}{dx}-\frac{dy}{dx}=2\sin x\cos x\]                                                 \[-{{e}^{xy}}-x{{e}^{xy}}y\] \[\Rightarrow \]\[\frac{dy}{dx}=\frac{\sin 2x-{{e}^{xy}}(1+xy)}{({{x}^{2}}{{e}^{xy}}-1)}\] \[\therefore \]\[{{\left. \frac{dy}{dx} \right|}_{x=0}}=\frac{0-1(1)}{(-1)}=1\]


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