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Câu hỏi: Cho hàm số $f\left( x \right)$ xác định trên $\mathbb{R}\backslash \left\{ 2 \right\}$ thỏa mãn ${f}'\left( x \right)=\dfrac{1}{x-2} , f\left( 1 \right)=2021 , f\left( 3 \right)=2022$.
Giá trị $P=\dfrac{f\left( 2023 \right)}{f\left( -2019 \right)}$ là
A. $P=\dfrac{\ln 2021}{\ln 2022}$.
B. $P=\dfrac{2022+\ln 2021}{2021+\ln 2021}$.
C. $P=\ln 4042$.
D. $P=\ln \dfrac{2021}{2022}$.
Giá trị $P=\dfrac{f\left( 2023 \right)}{f\left( -2019 \right)}$ là
A. $P=\dfrac{\ln 2021}{\ln 2022}$.
B. $P=\dfrac{2022+\ln 2021}{2021+\ln 2021}$.
C. $P=\ln 4042$.
D. $P=\ln \dfrac{2021}{2022}$.
Ta có ${f}'\left( x \right)=\dfrac{1}{x-2}\Rightarrow f\left( x \right)=\ln \left| x-2 \right|+C=\left\{ \begin{aligned}
& \ln \left( x-2 \right)+{{C}_{1}} \text{khi} x>2 \\
& \ln \left( 2-x \right)+{{C}_{2}} \text{khi} x<2 \\
\end{aligned} \right.$.
$f\left( 1 \right)=2021\Rightarrow {{C}_{2}}=2021 ; f\left( 3 \right)=2022\Rightarrow {{C}_{1}}=2022$.
Từ đó, ta được: $P=\dfrac{f\left( 2023 \right)}{f\left( -2019 \right)}=\dfrac{2022+\ln 2021}{2021+\ln 2021}$.
& \ln \left( x-2 \right)+{{C}_{1}} \text{khi} x>2 \\
& \ln \left( 2-x \right)+{{C}_{2}} \text{khi} x<2 \\
\end{aligned} \right.$.
$f\left( 1 \right)=2021\Rightarrow {{C}_{2}}=2021 ; f\left( 3 \right)=2022\Rightarrow {{C}_{1}}=2022$.
Từ đó, ta được: $P=\dfrac{f\left( 2023 \right)}{f\left( -2019 \right)}=\dfrac{2022+\ln 2021}{2021+\ln 2021}$.
Đáp án B.