Google
Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm ${f}'\left( x \right)=\dfrac{1}{x}$, $\forall x\in \mathbb{R}\backslash \left\{ 0 \right\}$ và $f\left( 1 \right)=2$, $f\left( -e \right)=4$. Giá trị của $f\left( -2 \right)-2f\left( {{e}^{2}} \right)$ bằng
A. $-8+\ln 2$.
B. $-5+\ln 2$.
C. $-2+\ln 2$.
D. $-1+\ln 2$.
A. $-8+\ln 2$.
B. $-5+\ln 2$.
C. $-2+\ln 2$.
D. $-1+\ln 2$.
$f\left( x \right)=\int{{f}'\left( x \right)\text{d}x}=\int{\dfrac{1}{x}\text{d}x}=\left\{ \begin{aligned}
& \ln x+{{C}_{1}}\begin{matrix}
, & \text{ }x>0 \\
\end{matrix} \\
& \ln \left( -x \right)+{{C}_{2}}\begin{matrix}
, & x<0 \\
\end{matrix} \\
\end{aligned} \right.$
$f\left( 1 \right)=2\Rightarrow \ln 1+{{C}_{1}}=2\Leftrightarrow {{C}_{1}}=2$
$f\left( -e \right)=4\Rightarrow \ln \text{e}+{{C}_{2}}=4\Leftrightarrow {{C}_{2}}=3$
Khi đó $f\left( x \right)=\left\{ \begin{aligned}
& \ln x+2\begin{matrix}
, & \text{ }x>0 \\
\end{matrix} \\
& \ln \left( -x \right)+3\begin{matrix}
, & x<0 \\
\end{matrix} \\
\end{aligned} \right.$
$f\left( -2 \right)-2f\left( {{e}^{2}} \right)=\ln 2+3-2\left( 2+2 \right)=-5+\ln 2$.
& \ln x+{{C}_{1}}\begin{matrix}
, & \text{ }x>0 \\
\end{matrix} \\
& \ln \left( -x \right)+{{C}_{2}}\begin{matrix}
, & x<0 \\
\end{matrix} \\
\end{aligned} \right.$
$f\left( 1 \right)=2\Rightarrow \ln 1+{{C}_{1}}=2\Leftrightarrow {{C}_{1}}=2$
$f\left( -e \right)=4\Rightarrow \ln \text{e}+{{C}_{2}}=4\Leftrightarrow {{C}_{2}}=3$
Khi đó $f\left( x \right)=\left\{ \begin{aligned}
& \ln x+2\begin{matrix}
, & \text{ }x>0 \\
\end{matrix} \\
& \ln \left( -x \right)+3\begin{matrix}
, & x<0 \\
\end{matrix} \\
\end{aligned} \right.$
$f\left( -2 \right)-2f\left( {{e}^{2}} \right)=\ln 2+3-2\left( 2+2 \right)=-5+\ln 2$.
Đáp án B.