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Câu hỏi: Cho hàm số $y=g\left( x \right)$ thỏa mãn $2{{g}^{3}}\left( x \right)-6{{g}^{2}}\left( x \right)+7g\left( x \right)=3-\left( 2x-3 \right)\sqrt{1-x}$. Tìm giá trị lớn nhất của biểu thức $P=2g\left( x \right)+x$
A. $0$
B. $1$
C. $4$
D. $6$
A. $0$
B. $1$
C. $4$
D. $6$
$2{{g}^{3}}\left( x \right)-6{{g}^{2}}\left( x \right)+7g\left( x \right)=3-\left( 2x-3 \right)\sqrt{1-x}$
$\Leftrightarrow 2{{g}^{3}}\left( x \right)-6{{g}^{2}}\left( x \right)+6g\left( x \right)-2+g\left( x \right)+2=3-\left( 2x-2 \right)\sqrt{1-x}+\sqrt{1-x}$
$\Leftrightarrow 2{{\left[ g\left( x \right)-1 \right]}^{3}}+g\left( x \right)-1=2\left( 1-x \right)\sqrt{1-x}+\sqrt{1-x}\left( 1 \right)$
Xét hàm số $f\left( t \right)=2{{t}^{3}}+t$
Ta có ${f}'\left( t \right)=6{{t}^{2}}+1>{{0}^{{}}}\forall t\in \mathbb{R}$ $\Rightarrow $ $f\left( t \right)$ đồng biến trên $\left( -\infty ;+\infty \right)$
$\left( 1 \right)\Leftrightarrow g\left( x \right)-1=\sqrt{1-x}\Leftrightarrow g\left( x \right)=\sqrt{1-x}+1\Rightarrow P=2\sqrt{1-x}+2+x$
Ta có $P=2\sqrt{1-x}+2+x=-{{\sqrt{1-x}}^{2}}+2\sqrt{1-x}-1+4=-{{\left( \sqrt{1-x}-1 \right)}^{2}}+4\le 4$
Đẳng thức xảy ra khi $\sqrt{1-x}-1=0\Leftrightarrow x=0$.
$\Leftrightarrow 2{{g}^{3}}\left( x \right)-6{{g}^{2}}\left( x \right)+6g\left( x \right)-2+g\left( x \right)+2=3-\left( 2x-2 \right)\sqrt{1-x}+\sqrt{1-x}$
$\Leftrightarrow 2{{\left[ g\left( x \right)-1 \right]}^{3}}+g\left( x \right)-1=2\left( 1-x \right)\sqrt{1-x}+\sqrt{1-x}\left( 1 \right)$
Xét hàm số $f\left( t \right)=2{{t}^{3}}+t$
Ta có ${f}'\left( t \right)=6{{t}^{2}}+1>{{0}^{{}}}\forall t\in \mathbb{R}$ $\Rightarrow $ $f\left( t \right)$ đồng biến trên $\left( -\infty ;+\infty \right)$
$\left( 1 \right)\Leftrightarrow g\left( x \right)-1=\sqrt{1-x}\Leftrightarrow g\left( x \right)=\sqrt{1-x}+1\Rightarrow P=2\sqrt{1-x}+2+x$
Ta có $P=2\sqrt{1-x}+2+x=-{{\sqrt{1-x}}^{2}}+2\sqrt{1-x}-1+4=-{{\left( \sqrt{1-x}-1 \right)}^{2}}+4\le 4$
Đẳng thức xảy ra khi $\sqrt{1-x}-1=0\Leftrightarrow x=0$.
Đáp án C.