Giải bất phương trình \({\varphi '\left( x \right) < 0}\) với \(\varphi \left( x \right) = \frac{{2x - 1}}{{{x^2} + 1}}\)
A. \(\left( { - \infty ;\frac{{1 - \sqrt 5 }}{2}} \right) \cup \left( {\frac{{1 + \sqrt 5 }}{2} + \infty } \right)\)
B. \(\left( { - \infty ; - \sqrt 3 } \right)\mathop \cup \nolimits^ \left( {\sqrt 5 ; + \infty } \right)\)
C. \(\left( { - \infty ; - 1} \right)\mathop \cup \nolimits^ \left( {1; + \infty } \right)\)
D.
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
- {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
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