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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    The point of the curve\[2y=3-{{x}^{2}}\]at which the tangent is parallel to the line\[x+y=0,\]is:

    A)  (1, 1)                    

    B)         \[(1,-1)\]

    C)  \[(-1,1)\]            

    D)         (0, 1)

    E)  (0, 1)

    Correct Answer: A

    Solution :

    Slope of line\[x+y=0\]is\[-1,\]therefore slope of required tangent\[=-1\] Given curve is\[2y=3-{{x}^{2}}\Rightarrow 2\frac{dy}{dx}=-2x\] \[\Rightarrow \]               \[\frac{dy}{dx}=-x=-1\] \[\Rightarrow \]               \[x=1\] and        \[2y=3-1\Rightarrow y=1\] \[\therefore \] Required point is (1, 1).


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