Google
Câu hỏi: Cho hàm số f(x) có đạo hàm ${f}'\left( x \right)={{\left( x-1 \right)}^{3}}{{\left( x+3 \right)}^{5}}\left( x+1 \right)g\left( x \right)-\dfrac{2}{\sqrt{{{x}^{2}}-2x+5}},\forall x\in \mathbb{R}.$ Trong đó $g\left( x \right)>0$, $\forall x\in \mathbb{R}.$ Hàm số $y=f\left( 2x+1 \right)+\ln \left( x+\sqrt{{{x}^{2}}+1} \right)$ nghịch biến trên khoảng nào dưới đây?
A. $\left( -\infty ;-\dfrac{3}{2} \right).$
B. $\left( -\dfrac{3}{2};-1 \right).$
C. $\left( 0;+\infty \right).$
D. $\left( -1;0 \right).$
A. $\left( -\infty ;-\dfrac{3}{2} \right).$
B. $\left( -\dfrac{3}{2};-1 \right).$
C. $\left( 0;+\infty \right).$
D. $\left( -1;0 \right).$
Ta có ${y}'=2{f}'\left( 2x+1 \right)+\dfrac{1}{x+\sqrt{{{x}^{2}}+1}}.\left( 1+\dfrac{x}{\sqrt{{{x}^{2}}+1}} \right)$
$=2{{\left( 2x \right)}^{3}}{{\left( 2x+4 \right)}^{5}}\left( 2x+2 \right)g\left( 2x+1 \right)-\dfrac{2}{\sqrt{{{\left( 2x+1 \right)}^{2}}-2\left( 2x+1 \right)+5}}+\dfrac{1}{\sqrt{{{x}^{2}}+1}}$
$=2{{\left( 2x \right)}^{3}}{{\left( 2x+4 \right)}^{5}}\left( 2x+2 \right)g\left( 2x+1 \right)-\dfrac{2}{\sqrt{4{{x}^{2}}+4}}+\dfrac{1}{\sqrt{{{x}^{2}}+1}}<0$
$=2{{\left( 2x \right)}^{3}}{{\left( 2x+4 \right)}^{5}}\left( 2x+2 \right)<0\Leftrightarrow \left[ \begin{array}{*{35}{l}}
x<-2 \\
-1<x<0 \\
\end{array} \right..$
$=2{{\left( 2x \right)}^{3}}{{\left( 2x+4 \right)}^{5}}\left( 2x+2 \right)g\left( 2x+1 \right)-\dfrac{2}{\sqrt{{{\left( 2x+1 \right)}^{2}}-2\left( 2x+1 \right)+5}}+\dfrac{1}{\sqrt{{{x}^{2}}+1}}$
$=2{{\left( 2x \right)}^{3}}{{\left( 2x+4 \right)}^{5}}\left( 2x+2 \right)g\left( 2x+1 \right)-\dfrac{2}{\sqrt{4{{x}^{2}}+4}}+\dfrac{1}{\sqrt{{{x}^{2}}+1}}<0$
$=2{{\left( 2x \right)}^{3}}{{\left( 2x+4 \right)}^{5}}\left( 2x+2 \right)<0\Leftrightarrow \left[ \begin{array}{*{35}{l}}
x<-2 \\
-1<x<0 \\
\end{array} \right..$
Đáp án D.