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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    If a tangent of the curve\[y=2+\sqrt{4x+1}\]has slope\[\frac{2}{5}\]at a point, then the point is

    A)  (0, 2)                                    

    B)  \[\left( \frac{3}{4},4 \right)\]

    C)  (2, 5)                    

    D)         (7, 6)

    E)  (6, 7)

    Correct Answer: E

    Solution :

    Curve,\[y=2+\sqrt{4x+1}\]has slope 2/5. Differentiating w.r.t.\[x,\] \[\frac{dy}{dx}=\frac{2}{\sqrt{4x+1}}=\frac{2}{5}\]          \[\left( \because \frac{dy}{dx}=\frac{2}{5} \right)\] \[\Rightarrow \]               \[5=\sqrt{4x+1}\] \[\Rightarrow \]               \[25=4x+1\] \[\Rightarrow \]               \[x=6\] and        \[y=2+\sqrt{25}=2+5\] \[\Rightarrow \]               \[y=7\] Thus, the required point is (6, 7).


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