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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    The equation of the tangent to the curve \[y=4{{e}^{\frac{x}{4}}}\]at the point where the curve crosses \[y-\]axis is equal to

    A)  \[3x+4y=16\]                   

    B)  \[4x+y=4\]

    C)  \[x+y=4\]        

    D)         \[4x-3y=-12\]

    E)  \[x-y=-4\]

    Correct Answer: C

    Solution :

    As the curve crosses\[y-\]axis ie,\[x=0\] \[\therefore \]  \[y=4{{e}^{-0}}\Rightarrow y=4\] Given,         \[y=4{{e}^{-\frac{x}{4}}}\] \[\Rightarrow \]               \[\frac{dy}{dx}=4{{e}^{-\frac{x}{4}}}\left( -\frac{1}{4} \right)=-{{e}^{-\frac{x}{4}}}\] \[\Rightarrow \]               \[{{\left( \frac{dy}{dx} \right)}_{(0,4)}}=-{{e}^{-0}}=-1\] \[\therefore \]Equation of tangent at (0, 4) is \[y-4=-1(x-0)\] \[\Rightarrow \]               \[x+y=4\]


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