Đáp án đúng: A
Giải chi tiết:\(1)\,\,f\left( x \right) = \dfrac{1}{{3 + {2^x}}} + \dfrac{1}{{3 + {2^{ - x}}}} = \dfrac{1}{{3 + {2^x}}} + \dfrac{{{2^x}}}{{{{3.2}^x} + 1}} = \dfrac{{{4^x} + {{6.2}^x} + 1}}{{{{3.4}^x} + {{10.2}^x} + 3}}\)\(\begin{array}{l} \Rightarrow f'\left( x \right) = \dfrac{{\left( {{{2.4}^x}.\ln 2 + {{6.2}^x}.\ln 2} \right)\left( {{{3.4}^x} + {{10.2}^x} + 3} \right) - \left( {{{6.4}^x}.\ln 2 + {{10.2}^x}.\ln 2} \right)\left( {{4^x} + {{6.2}^x} + 1} \right)}}{{{{\left( {{{3.4}^x} + {{10.2}^x} + 3} \right)}^2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{\left( {{{2.2}^x} + 6} \right)\left( {{{3.4}^x} + {{10.2}^x} + 3} \right) - \left( {{{6.2}^x} + 10} \right)\left( {{4^x} + {{6.2}^x} + 1} \right)}}{{{{\left( {{{3.4}^x} + {{10.2}^x} + 3} \right)}^2}}}{.2^x}.\ln 2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{ - {{8.4}^x} + 8}}{{{{\left( {{{3.4}^x} + {{10.2}^x} + 3} \right)}^2}}}{.2^x}.\ln 2\\f'\left( x \right) = 0 \Leftrightarrow - {8.4^x} + 8 = 0 \Leftrightarrow {4^x} = 1 \Leftrightarrow x = 0\end{array}\)
\(2)\,\,f\left( x \right) = \dfrac{{{4^x} + {{6.2}^x} + 1}}{{{{3.4}^x} + {{10.2}^x} + 3}}\).
Ta có: \(f\left( x \right) - \dfrac{1}{3} = \dfrac{{{4^x} + {{6.2}^x} + 1}}{{{{3.4}^x} + {{10.2}^x} + 3}} - 1 = \dfrac{{ - {{2.4}^x} - {{4.2}^x} - 2}}{{{{3.4}^x} + {{10.2}^x} + 3}} < 0,\,\,\forall x \Rightarrow f\left( x \right) < 1,\,\forall x\)
\(\begin{array}{l} \Rightarrow f\left( 1 \right) + f\left( 2 \right) + ... + f\left( {2017} \right) < 1 + 1 + ... + 1 = 2017\\ \Rightarrow f\left( 1 \right) + f\left( 2 \right) + ... + f\left( {2017} \right) = 2017\end{array}\)
\( \Rightarrow 2)\) sai.
\(3)\,\,f\left( {{x^2}} \right) = \dfrac{1}{{3 + {2^{{x^2}}}}} + \dfrac{1}{{3 + {2^{ - {x^2}}}}}\)\( \Rightarrow f\left( {{x^2}} \right) = \dfrac{1}{{3 + {4^x}}} + \dfrac{1}{{3 + {4^{ - x}}}}\) là sai.
Chọn: A
