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Câu hỏi: Cho hàm số $y=f\left( x \right)$ xác định $R\backslash \left\{ 0 \right\}$ thoả mãn ${f}'\left( x \right)=\dfrac{x+1}{{{x}^{2}}},f\left( -2 \right)=\dfrac{3}{2}$ và $f\left( 2 \right)=2\ln 2-\dfrac{3}{2}$.Tính giá trị biểu thức $f\left( -1 \right)+f\left( 4 \right)$ bằng.
A. $\dfrac{6\ln 2-3}{4}$.
B. $\dfrac{6\ln 2+3}{4}$.
C. $\dfrac{8\ln 2+3}{4}$.
D. $\dfrac{8\ln 2-3}{4}$.
A. $\dfrac{6\ln 2-3}{4}$.
B. $\dfrac{6\ln 2+3}{4}$.
C. $\dfrac{8\ln 2+3}{4}$.
D. $\dfrac{8\ln 2-3}{4}$.
$\begin{aligned}
& f\left( x \right)=\int{{{f}^{'}}\left( x \right)dx=\int{\dfrac{x+1}{{{x}^{2}}}dx=\int{\left( \dfrac{1}{x}+\dfrac{1}{{{x}^{2}}} \right)}}}dx=\ln \left| x \right|-\dfrac{1}{x}+C \\
& \Rightarrow f\left( x \right)=\left\{ \begin{aligned}
& \ln \left( x \right)-\dfrac{1}{x}+{{C}_{1}}khix>0 \\
& \ln \left( -x \right)-\dfrac{1}{x}+{{C}_{2}}khix<0 \\
\end{aligned} \right. \\
\end{aligned}$
Do $f\left( -2 \right)=\dfrac{3}{2}\Rightarrow \ln \left( -\left( -2 \right) \right)-\dfrac{1}{-2}+{{C}_{2}}=\dfrac{3}{2}\Rightarrow \ln 2+\dfrac{1}{2}+{{C}_{2}}=\dfrac{3}{2}\Rightarrow {{C}_{2}}=1-\ln 2$
Do $f\left( 2 \right)=2\ln 2-\dfrac{3}{2}\Rightarrow \ln \left( 2 \right)-\dfrac{1}{2}+{{C}_{1}}=2\ln 2-\dfrac{3}{2}\Rightarrow \ln 2-\dfrac{1}{2}+{{C}_{1}}=2\ln 2-\dfrac{3}{2}\Rightarrow {{C}_{1}}=\ln 2-1$
Như vậy $f\left( x \right)=\left\{ \begin{aligned}
& \ln \left( x \right)-\dfrac{1}{x}+\ln 2-1khix>0 \\
& \ln \left( -x \right)-\dfrac{1}{x}+1-\ln 2khix<0 \\
\end{aligned} \right.$
Vậy ta có
$\begin{aligned}
& f\left( -1 \right)+f\left( 4 \right)=\left[ \ln \left( -\left( -1 \right) \right)-\dfrac{1}{-1}+1-\ln 2 \right]+\left[ \ln \left( 4 \right)-\dfrac{1}{4}+\ln 2-1 \right] \\
& =0+1+1-\ln 2+2\ln 2-\dfrac{1}{4}+\ln 2-1=2\ln 2+\dfrac{3}{4}=\dfrac{8\ln 2+3}{4} \\
\end{aligned}$
& f\left( x \right)=\int{{{f}^{'}}\left( x \right)dx=\int{\dfrac{x+1}{{{x}^{2}}}dx=\int{\left( \dfrac{1}{x}+\dfrac{1}{{{x}^{2}}} \right)}}}dx=\ln \left| x \right|-\dfrac{1}{x}+C \\
& \Rightarrow f\left( x \right)=\left\{ \begin{aligned}
& \ln \left( x \right)-\dfrac{1}{x}+{{C}_{1}}khix>0 \\
& \ln \left( -x \right)-\dfrac{1}{x}+{{C}_{2}}khix<0 \\
\end{aligned} \right. \\
\end{aligned}$
Do $f\left( -2 \right)=\dfrac{3}{2}\Rightarrow \ln \left( -\left( -2 \right) \right)-\dfrac{1}{-2}+{{C}_{2}}=\dfrac{3}{2}\Rightarrow \ln 2+\dfrac{1}{2}+{{C}_{2}}=\dfrac{3}{2}\Rightarrow {{C}_{2}}=1-\ln 2$
Do $f\left( 2 \right)=2\ln 2-\dfrac{3}{2}\Rightarrow \ln \left( 2 \right)-\dfrac{1}{2}+{{C}_{1}}=2\ln 2-\dfrac{3}{2}\Rightarrow \ln 2-\dfrac{1}{2}+{{C}_{1}}=2\ln 2-\dfrac{3}{2}\Rightarrow {{C}_{1}}=\ln 2-1$
Như vậy $f\left( x \right)=\left\{ \begin{aligned}
& \ln \left( x \right)-\dfrac{1}{x}+\ln 2-1khix>0 \\
& \ln \left( -x \right)-\dfrac{1}{x}+1-\ln 2khix<0 \\
\end{aligned} \right.$
Vậy ta có
$\begin{aligned}
& f\left( -1 \right)+f\left( 4 \right)=\left[ \ln \left( -\left( -1 \right) \right)-\dfrac{1}{-1}+1-\ln 2 \right]+\left[ \ln \left( 4 \right)-\dfrac{1}{4}+\ln 2-1 \right] \\
& =0+1+1-\ln 2+2\ln 2-\dfrac{1}{4}+\ln 2-1=2\ln 2+\dfrac{3}{4}=\dfrac{8\ln 2+3}{4} \\
\end{aligned}$
Đáp án C.