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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm $f'\left( x \right)=\left( 3-x \right){{\left( 10-3x \right)}^{2}}{{\left( x-2 \right)}^{2}}$ với mọi $x\in \mathbb{R}.$ Hàm số $g\left( x \right)=f\left( 3-x \right)+\dfrac{1}{6}{{\left( {{x}^{2}}-1 \right)}^{3}}$ đồng biến trên khoảng nào trong các khoảng sau?
A. $\left( 1;+\infty \right).$
B. $\left( 0;1 \right).$
C. $\left( -\infty ;0 \right).$
D. $\left( -\infty ;-\dfrac{1}{2} \right).$
A. $\left( 1;+\infty \right).$
B. $\left( 0;1 \right).$
C. $\left( -\infty ;0 \right).$
D. $\left( -\infty ;-\dfrac{1}{2} \right).$
$g'\left( x \right)=-f'\left( 3-x \right)+\dfrac{3}{6}2x{{\left( {{x}^{2}}-1 \right)}^{2}}$
$=-f'\left( 3-x \right)+x{{\left( {{x}^{2}}-1 \right)}^{2}}$
$=-\left[ 3-\left( 3-x \right) \right]{{\left[ 10-3\left( 3-x \right) \right]}^{2}}{{\left( 3-x-2 \right)}^{2}}+x{{\left( {{x}^{2}}-1 \right)}^{2}}$
$=-x{{\left( 1+3x \right)}^{2}}{{\left( 1-x \right)}^{2}}+x{{\left( x-1 \right)}^{2}}{{\left( x+1 \right)}^{2}}$
$={{\left( x-1 \right)}^{2}}\left[ {{x}^{3}}+2{{x}^{2}}+x-x\left( 9{{x}^{2}}+6x+1 \right) \right]$
$={{\left( x-1 \right)}^{2}}\left( -8{{x}^{3}}-4{{x}^{2}} \right)$
$=-4{{x}^{2}}{{\left( x-1 \right)}^{2}}\left( 2x+1 \right)$
$g'\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=0 \\
& x=1 \\
& x=\dfrac{-1}{2} \\
\end{aligned} \right.$
$\Rightarrow g'\left( x \right)>0\Leftrightarrow x\in \left( -\infty ;\dfrac{-1}{2} \right).$
Vậy hàm số đồng biến trên khoảng $\left( -\infty ;-\dfrac{1}{2} \right).$
$=-f'\left( 3-x \right)+x{{\left( {{x}^{2}}-1 \right)}^{2}}$
$=-\left[ 3-\left( 3-x \right) \right]{{\left[ 10-3\left( 3-x \right) \right]}^{2}}{{\left( 3-x-2 \right)}^{2}}+x{{\left( {{x}^{2}}-1 \right)}^{2}}$
$=-x{{\left( 1+3x \right)}^{2}}{{\left( 1-x \right)}^{2}}+x{{\left( x-1 \right)}^{2}}{{\left( x+1 \right)}^{2}}$
$={{\left( x-1 \right)}^{2}}\left[ {{x}^{3}}+2{{x}^{2}}+x-x\left( 9{{x}^{2}}+6x+1 \right) \right]$
$={{\left( x-1 \right)}^{2}}\left( -8{{x}^{3}}-4{{x}^{2}} \right)$
$=-4{{x}^{2}}{{\left( x-1 \right)}^{2}}\left( 2x+1 \right)$
$g'\left( x \right)=0\Leftrightarrow \left[ \begin{aligned}
& x=0 \\
& x=1 \\
& x=\dfrac{-1}{2} \\
\end{aligned} \right.$
Vậy hàm số đồng biến trên khoảng $\left( -\infty ;-\dfrac{1}{2} \right).$
Đáp án D.