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Câu hỏi: Cho hàm số $f\left( x \right)$ xác định trên $\left( 0;+\infty \right)\backslash \left\{ e \right\}$, thỏa mãn ${f}'\left( x \right)=\dfrac{1}{x\left( \ln x-1 \right)}$, $f\left( \dfrac{1}{{{e}^{2}}} \right)=\ln 6$ và $f\left( {{e}^{2}} \right)=3$. Giá trị biểu thức $f\left( \dfrac{1}{e} \right)+f\left( {{e}^{3}} \right)$ bằng
A. $3\left( \ln 2+1 \right)$.
B. $2\ln 2$.
C. $3\ln 2+1$.
D. $\ln 2+3$.
A. $3\left( \ln 2+1 \right)$.
B. $2\ln 2$.
C. $3\ln 2+1$.
D. $\ln 2+3$.
Ta có ${f}'\left( x \right)=\dfrac{1}{x\left( \ln x-1 \right)}$
$\Rightarrow f\left( x \right)=\int{\dfrac{1}{x\left( \ln x-1 \right)}dx=\ln \left| \ln x-1 \right|+C=\left\{ \begin{aligned}
& \ln \left( 1-\ln x \right)+{{C}_{1}} khi x\in \left( 0;e \right) \\
& \ln \left( \ln x-1 \right)+{{C}_{2}} khi x\in \left( e;+\infty \right) \\
\end{aligned} \right.}$.
+) $f\left( \dfrac{1}{{{e}^{2}}} \right)=\ln 6\Rightarrow {{C}_{1}}=\ln 2$.
+) $f\left( {{e}^{2}} \right)=3\Rightarrow {{C}_{2}}=3$.
Do đó $f\left( x \right)=\left\{ \begin{aligned}
& \ln \left( 1-\ln x \right)+\ln 2 khi x\in \left( 0;e \right) \\
& \ln \left( \ln x-1 \right)+3 khi x\in \left( e;+\infty \right) \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& f\left( \dfrac{1}{e} \right)=\ln 2+\ln 2 \\
& f\left( {{e}^{3}} \right)=\ln 2+3 \\
\end{aligned} \right.$
$\xrightarrow{{}}f\left( \dfrac{1}{e} \right)+f\left( {{e}^{3}} \right)=3\left( \ln 2+1 \right)$.
$\Rightarrow f\left( x \right)=\int{\dfrac{1}{x\left( \ln x-1 \right)}dx=\ln \left| \ln x-1 \right|+C=\left\{ \begin{aligned}
& \ln \left( 1-\ln x \right)+{{C}_{1}} khi x\in \left( 0;e \right) \\
& \ln \left( \ln x-1 \right)+{{C}_{2}} khi x\in \left( e;+\infty \right) \\
\end{aligned} \right.}$.
+) $f\left( \dfrac{1}{{{e}^{2}}} \right)=\ln 6\Rightarrow {{C}_{1}}=\ln 2$.
+) $f\left( {{e}^{2}} \right)=3\Rightarrow {{C}_{2}}=3$.
Do đó $f\left( x \right)=\left\{ \begin{aligned}
& \ln \left( 1-\ln x \right)+\ln 2 khi x\in \left( 0;e \right) \\
& \ln \left( \ln x-1 \right)+3 khi x\in \left( e;+\infty \right) \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& f\left( \dfrac{1}{e} \right)=\ln 2+\ln 2 \\
& f\left( {{e}^{3}} \right)=\ln 2+3 \\
\end{aligned} \right.$
$\xrightarrow{{}}f\left( \dfrac{1}{e} \right)+f\left( {{e}^{3}} \right)=3\left( \ln 2+1 \right)$.
Đáp án A.