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Câu hỏi: Cho hàm số $y=f\left( x \right)$ xác định trên $\mathbb{R}\backslash \left\{ 1 \right\}$ thoả mãn ${f}'\left( x \right)=\dfrac{1}{x-1} , f\left( 0 \right)=2022 , f\left( 2 \right)=2023$. Tính $S=f\left( 3 \right)-f\left( -1 \right)$.
A. $S=0$.
B. $S=\ln 4045$.
C. $S=1$.
D. $S=\ln 2$.
A. $S=0$.
B. $S=\ln 4045$.
C. $S=1$.
D. $S=\ln 2$.
Ta có ${f}'\left( x \right)=\dfrac{1}{x-1}\Rightarrow f\left( x \right)=\int{\dfrac{1}{x-1}\text{d}x}=\ln \left| x-1 \right|+C=\left\{ \begin{aligned}
& \ln \left( x-1 \right)+{{C}_{1}}\ \ \text{khi}\ \ x>1 \\
& \ln \left( 1-x \right)+{{C}_{2}}\ \ \text{khi}\ \ x<1 \\
\end{aligned} \right.$.
Mặt khác $\left\{ \begin{aligned}
& f\left( 0 \right)=2022 \\
& f\left( 2 \right)=2023 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{C}_{2}}=2022 \\
& {{C}_{1}}=2023 \\
\end{aligned} \right.$.
Vậy $f\left( x \right)=\left\{ \begin{aligned}
& \ln \left( x-1 \right)+2023\ \ \text{khi}\ \ x>1 \\
& \ln \left( 1-x \right)+2022\ \ \text{khi}\ \ x<1 \\
\end{aligned} \right.$.
Do đó $S=f\left( 3 \right)-f\left( -1 \right)=\ln 2+2023-\ln 2-2022=1$.
& \ln \left( x-1 \right)+{{C}_{1}}\ \ \text{khi}\ \ x>1 \\
& \ln \left( 1-x \right)+{{C}_{2}}\ \ \text{khi}\ \ x<1 \\
\end{aligned} \right.$.
Mặt khác $\left\{ \begin{aligned}
& f\left( 0 \right)=2022 \\
& f\left( 2 \right)=2023 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{C}_{2}}=2022 \\
& {{C}_{1}}=2023 \\
\end{aligned} \right.$.
Vậy $f\left( x \right)=\left\{ \begin{aligned}
& \ln \left( x-1 \right)+2023\ \ \text{khi}\ \ x>1 \\
& \ln \left( 1-x \right)+2022\ \ \text{khi}\ \ x<1 \\
\end{aligned} \right.$.
Do đó $S=f\left( 3 \right)-f\left( -1 \right)=\ln 2+2023-\ln 2-2022=1$.
Đáp án C.
