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JEE Main & Advanced AIEEE Solved Paper-2010

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    Directions: Questions number 89 are Assertion - Reason type questions. Each of these questions contains two statements: Statement - 1 (Assertion) and Statement - 2 (Reason). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice. Let\[f:R\to R\]be a continuous function defined by \[f(x)=\frac{1}{{{e}^{x}}+2{{e}^{-x}}}\] Statement - 1: \[f(c)=\frac{1}{3},\]for some\[c\in R\]. Statement - 2: \[0<f(x)\le \frac{1}{2\sqrt{2}},\]for all \[x\in R\]. Statement - 1 (Assertion) and Statement - 2 (Reason). Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice.     AIEEE  Solved  Paper-2010

    A) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for Statement -1

    B) Statement -1 is true, Statement -2 is true; Statement -2 is not a correct explanation for Statement -1.

    C) Statement -1 is true, Statement -2 is false.

    D) Statement -1 is false, Statement -2 is true.

    Correct Answer: A

    Solution :

    \[AM\ge GM\] \[\frac{{{e}^{x}}+\frac{2}{{{e}^{x}}}}{2}\ge \sqrt{({{e}^{x}})\left( \frac{2}{{{e}^{x}}} \right)}\] \[{{e}^{x}}+\frac{2}{{{e}^{x}}}\ge 2\sqrt{2}\]                      (1) \[\because \] \[{{e}^{x}}>0\Rightarrow {{e}^{x}}+\frac{2}{{{e}^{x}}}>0\]               (2) \[0<\frac{1}{{{e}^{x}}+\frac{2}{{{e}^{x}}}}\le \frac{1}{2\sqrt{2}}\] also\[f(c)=1/3\]for \[c=0\] so statement 1 : is true statement 2 : is also true with correct explanation


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