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Câu hỏi: Cho hàm số $f\left( x \right)$ xác định trên $\mathbb{R}\backslash \left\{ -1;1 \right\}$ thỏa mãn ${f}'\left( x \right)=\dfrac{2}{{{x}^{2}}-1}$, $f\left( -2 \right)+f\left( 2 \right)=0$ và $f\left( -\dfrac{1}{2} \right)+f\left( \dfrac{1}{2} \right)=2$. Tính $f\left( -3 \right)+f\left( 0 \right)+f\left( 4 \right)$ được kết quả
A. $\ln \dfrac{6}{5}+1$.
B. $\ln \dfrac{6}{5}-1$.
C. $\ln \dfrac{4}{5}+1$.
D. $\ln \dfrac{4}{5}-1$.
A. $\ln \dfrac{6}{5}+1$.
B. $\ln \dfrac{6}{5}-1$.
C. $\ln \dfrac{4}{5}+1$.
D. $\ln \dfrac{4}{5}-1$.
Ta có $f\left( x \right)=\int{{f}'\left( x \right)dx}=\int{\dfrac{2}{{{x}^{2}}-1}}dx=\int{\left( \dfrac{1}{x-1}-\dfrac{1}{x+1} \right)dx}$
$=\left\{ \begin{aligned}
& \ln \left( \dfrac{x-1}{x+1} \right)+{{C}_{1}} khi x<-1 \\
& \ln \left( \dfrac{x+1}{x+1} \right)+{{C}_{2}} khi -1<x<1. \\
& \ln \left( \dfrac{x-1}{x+1} \right)+{{C}_{3}} khi x>1 \\
\end{aligned} \right.$
Khi đó $\left\{ \begin{aligned}
& f\left( -2 \right)+f\left( 2 \right)=0 \\
& f\left( -\dfrac{1}{2} \right)+f\left( \dfrac{1}{2} \right)=2 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \ln 3+{{C}_{1}}+\ln \dfrac{1}{3}+{{C}_{3}}=0 \\
& \ln 3+{{C}_{2}}+\ln \dfrac{1}{3}+{{C}_{2}}=2 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{C}_{1}}+{{C}_{3}}=0 \\
& {{C}_{2}}=1 \\
\end{aligned} \right.$
Do đó $f\left( -3 \right)+f\left( 0 \right)+f\left( 4 \right)=\ln 2+{{C}_{1}}+{{C}_{2}}+\ln \dfrac{3}{5}+{{C}_{3}}=\ln \dfrac{6}{5}+1.$
$=\left\{ \begin{aligned}
& \ln \left( \dfrac{x-1}{x+1} \right)+{{C}_{1}} khi x<-1 \\
& \ln \left( \dfrac{x+1}{x+1} \right)+{{C}_{2}} khi -1<x<1. \\
& \ln \left( \dfrac{x-1}{x+1} \right)+{{C}_{3}} khi x>1 \\
\end{aligned} \right.$
Khi đó $\left\{ \begin{aligned}
& f\left( -2 \right)+f\left( 2 \right)=0 \\
& f\left( -\dfrac{1}{2} \right)+f\left( \dfrac{1}{2} \right)=2 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \ln 3+{{C}_{1}}+\ln \dfrac{1}{3}+{{C}_{3}}=0 \\
& \ln 3+{{C}_{2}}+\ln \dfrac{1}{3}+{{C}_{2}}=2 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& {{C}_{1}}+{{C}_{3}}=0 \\
& {{C}_{2}}=1 \\
\end{aligned} \right.$
Do đó $f\left( -3 \right)+f\left( 0 \right)+f\left( 4 \right)=\ln 2+{{C}_{1}}+{{C}_{2}}+\ln \dfrac{3}{5}+{{C}_{3}}=\ln \dfrac{6}{5}+1.$
Đáp án A.