Google
Câu hỏi: Cho hàm số $f\left( x \right)$ xác định trên $\mathbb{R}\backslash \left\{ 0 \right\}$ thỏa mãn ${f}'\left( x \right)=\dfrac{x+1}{{{x}^{2}}}$, $f\left( -2 \right)=\dfrac{3}{2}$ và $f\left( 2 \right)=2\ln 2-\dfrac{3}{2}$. Giá trị của 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. $\dfrac{8\ln 2-3}{4}$.
Có $f\left( x \right)=\int{{f}'\left( x \right)\text{d}x=\int{\dfrac{x+1}{{{x}^{2}}}\text{d}x=\ln \left| x \right|-\dfrac{1}{x}+C}}$
$\Rightarrow f\left( x \right)=\left\{ \begin{aligned}
& \ln \left( -x \right)-\dfrac{1}{x}+{{C}_{1}}\ \ \ \ \ \ \text{khi}\ x<0 \\
& \ \ln x-\dfrac{1}{x}+{{C}_{2}}\ \ \ \ \ \text{khi}\ x>0\ \\
\end{aligned} \right.$
Do $f\left( -2 \right)=\dfrac{3}{2}$ $\Rightarrow \ln 2+\dfrac{1}{2}+{{C}_{1}}=\dfrac{3}{2}\Rightarrow {{C}_{1}}=1-\ln 2$.
Do $f\left( 2 \right)=2\ln 2-\dfrac{3}{2}$ $\Rightarrow \ln 2-\dfrac{1}{2}+{{C}_{2}}=2\ln 2-\dfrac{3}{2}\Rightarrow {{C}_{2}}=\ln 2-1$
Như vậy, $f\left( x \right)=\left\{ \begin{aligned}
& \ln \left( -x \right)-\dfrac{1}{x}+1-\ln 2\ \ \ \ \ \ \ \ \text{khi}\ x<0 \\
& \ \ln x-\dfrac{1}{x}+\ln 2-1\ \ \ \ \ \ \ \text{khi}\ x>0\ \\
\end{aligned} \right.$
Vậy $f\left( -1 \right)+f\left( 4 \right)=\left( 2-\ln 2 \right)+\left( \ln 4-\dfrac{1}{4}+\ln 2-1 \right)=\dfrac{8\ln 2+3}{4}$.
$\Rightarrow f\left( x \right)=\left\{ \begin{aligned}
& \ln \left( -x \right)-\dfrac{1}{x}+{{C}_{1}}\ \ \ \ \ \ \text{khi}\ x<0 \\
& \ \ln x-\dfrac{1}{x}+{{C}_{2}}\ \ \ \ \ \text{khi}\ x>0\ \\
\end{aligned} \right.$
Do $f\left( -2 \right)=\dfrac{3}{2}$ $\Rightarrow \ln 2+\dfrac{1}{2}+{{C}_{1}}=\dfrac{3}{2}\Rightarrow {{C}_{1}}=1-\ln 2$.
Do $f\left( 2 \right)=2\ln 2-\dfrac{3}{2}$ $\Rightarrow \ln 2-\dfrac{1}{2}+{{C}_{2}}=2\ln 2-\dfrac{3}{2}\Rightarrow {{C}_{2}}=\ln 2-1$
Như vậy, $f\left( x \right)=\left\{ \begin{aligned}
& \ln \left( -x \right)-\dfrac{1}{x}+1-\ln 2\ \ \ \ \ \ \ \ \text{khi}\ x<0 \\
& \ \ln x-\dfrac{1}{x}+\ln 2-1\ \ \ \ \ \ \ \text{khi}\ x>0\ \\
\end{aligned} \right.$
Vậy $f\left( -1 \right)+f\left( 4 \right)=\left( 2-\ln 2 \right)+\left( \ln 4-\dfrac{1}{4}+\ln 2-1 \right)=\dfrac{8\ln 2+3}{4}$.
Đáp án C.