Bài tập 5.37 trang 205 SBT Toán 11

Ta có:

\(\begin{array}{l}
\varphi \prime (x) = \frac{{(2x - 1)\prime ({x^2} + 1) - (2x - 1)({x^2} + 1)\prime }}{{{{({x^2} + 1)}^2}}}\\
 = \frac{{2({x^2} + 1) - 2x(2x - 1)}}{{{{({x^2} + 1)}^2}}}\\
 = \frac{{ - 2{x^2} + 2x + 2}}{{{{({x^2} + 1)}^2}}}
\end{array}\)

\( \Rightarrow \varphi \prime (x) < 0 \Leftrightarrow \frac{{ - 2{x^2} + 2x + 2}}{{{{({x^2} + 1)}^2}}} < 0 \Rightarrow  - {x^2} + x + 1 < 0\)

Vì 

\(\begin{array}{l}
 - {x^2} + x + 1 =  - \left( {x - \frac{{1 + \sqrt 5 }}{2}} \right)\left( {x - \frac{{1 - \sqrt 5 }}{2}} \right) \Rightarrow  - {x^2} + x + 1 < 0\\
 \Leftrightarrow x \in \left( { - \infty ;\frac{{1 - \sqrt 5 }}{2}} \right)\mathop  \cup \nolimits^ \left( {\frac{{1 + \sqrt 5 }}{2}; + \infty } \right)
\end{array}\)
Chọn A.

-- Mod Toán 11 HỌC247