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Câu hỏi: Cho hàm số $f\left( x \right)$ xác định trên $\mathbb{R}\backslash \left\{ \dfrac{1}{2} \right\}$ thỏa mãn ${f}'\left( x \right)=\dfrac{2}{2x-1}$ và $f\left( 0 \right)=1,\ f\left( 1 \right)=-2$. Giá trị $f\left( -1 \right)+f\left( 3 \right)$ bằng
A. $2+\ln 15$.
B. $\ln 15-1$.
C. $3-\ln 15$.
D. $\ln 15$.
A. $2+\ln 15$.
B. $\ln 15-1$.
C. $3-\ln 15$.
D. $\ln 15$.
Ta có ${f}'\left( x \right)=\dfrac{2}{2x-1}\Leftrightarrow f\left( x \right)=\int{\dfrac{2}{2x-1}\text{d}x}=\ln \left| 2x-1 \right|+C=\left\{ \begin{aligned}
& \ln \left( 2x-1 \right)+{{C}_{1}},\ x>\dfrac{1}{2} \\
& \ln \left( 1-2x \right)+{{C}_{2}},\ x<\dfrac{1}{2} \\
\end{aligned} \right.$
Do $f\left( 0 \right)=1\Rightarrow {{C}_{2}}=1$ ; $f\left( 1 \right)=-2\Rightarrow {{C}_{1}}=-2$.
Vậy $f\left( x \right)=\left\{ \begin{aligned}
& \ln \left( 2x-1 \right)-2,\ x>\dfrac{1}{2} \\
& \ln \left( 1-2x \right)+1,\ x<\dfrac{1}{2} \\
\end{aligned} \right.$
Do đó $f\left( -1 \right)+f\left( 3 \right)=\ln 3+1+\ln 5-2=\ln 15-1$.
& \ln \left( 2x-1 \right)+{{C}_{1}},\ x>\dfrac{1}{2} \\
& \ln \left( 1-2x \right)+{{C}_{2}},\ x<\dfrac{1}{2} \\
\end{aligned} \right.$
Do $f\left( 0 \right)=1\Rightarrow {{C}_{2}}=1$ ; $f\left( 1 \right)=-2\Rightarrow {{C}_{1}}=-2$.
Vậy $f\left( x \right)=\left\{ \begin{aligned}
& \ln \left( 2x-1 \right)-2,\ x>\dfrac{1}{2} \\
& \ln \left( 1-2x \right)+1,\ x<\dfrac{1}{2} \\
\end{aligned} \right.$
Do đó $f\left( -1 \right)+f\left( 3 \right)=\ln 3+1+\ln 5-2=\ln 15-1$.
Đáp án B.