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Câu hỏi: Cho hàm số $f\left( x \right)$ xác định trên $\mathbb{R}\backslash \left\{ 1 \right\}$ thỏa 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. $S=\ln 4035$.
B. $S=\ln 2$.
C. $S=4$.
D. $S=1$.
A. $S=\ln 4035$.
B. $S=\ln 2$.
C. $S=4$.
D. $S=1$.
Ta có: $f\left( x \right)=\int{\dfrac{1}{x-1}\text{d}x=}\ln \left| x-1 \right|+C=\left\{ \begin{aligned}
& \ln \left( x-1 \right)+{{C}_{1}} \text{khi} x>1 \\
& \ln \left( 1-x \right)+{{C}_{2}} \text{khi} x<1 \\
\end{aligned} \right.$.
Do $f\left( 2 \right)=2023\Rightarrow \ln \left( 2-1 \right)+{{C}_{1}}=2023\Rightarrow {{C}_{1}}=2023\Rightarrow f\left( 3 \right)=\ln 2+2023$.
Mặt khác $f\left( 0 \right)=2022\Rightarrow \ln \left( 1-0 \right)+{{C}_{2}}=2022\Rightarrow {{C}_{2}}=2022\Rightarrow f\left( -1 \right)=\ln 2+2022$.
Từ đó ta có: $S=f\left( 3 \right)-f\left( -1 \right)=1$.
& \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.$.
Do $f\left( 2 \right)=2023\Rightarrow \ln \left( 2-1 \right)+{{C}_{1}}=2023\Rightarrow {{C}_{1}}=2023\Rightarrow f\left( 3 \right)=\ln 2+2023$.
Mặt khác $f\left( 0 \right)=2022\Rightarrow \ln \left( 1-0 \right)+{{C}_{2}}=2022\Rightarrow {{C}_{2}}=2022\Rightarrow f\left( -1 \right)=\ln 2+2022$.
Từ đó ta có: $S=f\left( 3 \right)-f\left( -1 \right)=1$.
Đáp án A.