CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    Let a be a positive number such that the arithmetic mean of a and 2 exceeds their geometric mean by 1. Then, the value of a is

    A)  3                                            

    B)  5

    C)  9                                            

    D)  8

    E)  10

    Correct Answer: D

    Solution :

    \[\frac{a+2}{2}=\sqrt{2a}+1\] \[\Rightarrow \]               \[\frac{a}{2}+1=\sqrt{2a}+1\] \[\Rightarrow \]               \[\frac{a}{2}=\sqrt{2a}\] \[\Rightarrow \]               \[\frac{{{a}^{2}}}{4}=2a\] \[\Rightarrow \]               \[a\left( \frac{a}{4}-2 \right)=0\] \[\Rightarrow \]               \[a=0,a=8\]


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