Cho hàm số

D

- Hướng dẫn giải

Đáp án D.

Ta có: 

$$\begin{gathered} \underset{4}{\overset{8}{\int }}\frac{f\text{'}\left(x\right)}{{\left[f\left(x\right)\right]}^2}dx \\ =\underset{4}{\overset{8}{\int }}{\left[f\left(x\right)\right]}^{-2}d\left[f\left(x\right)\right]={\left.\frac{{\left[f\left(x\right)\right]}^{-1}}{-1}\right|}_4^8 \\ =-\frac{1}{f\left(8\right)}+\frac{1}{f\left(4\right)}=-2+4=2. \end{gathered}$$

Gọi k là 1 hằng số thực. Xét

$$\begin{gathered} \underset{4}{\overset{8}{\int }}{\left(\frac{f\text{'}\left(x\right)}{f^2\left(x\right)}+k\right)}^{2}dx \\ =\underset{4}{\overset{8}{\int }}\frac{{\left[f\text{'}\left(x\right)\right]}^2}{{\left[f\left(x\right)\right]}^4}dx+2k\underset{4}{\overset{8}{\int }}\frac{f\text{'}\left(x\right)}{f^2\left(x\right)}dx+k^2\underset{4}{\overset{8}{\int }}dx \\ =1+2k.k+4k^2={\left(2k+1\right)}^2. \end{gathered}$$

Chọn $k=\frac{-1}{2},$ ta có $\underset{4}{\overset{8}{\int }}{\left(\frac{f\text{'}\left(x\right)}{f^2\left(x\right)}-\frac{1}{2}\right)}^{2}dx=0,$ mà ${\left(\frac{f\text{'}\left(x\right)}{f^2\left(x\right)}-\frac{1}{2}\right)}^2\ge 0$ nên $$\begin{gathered} {\left(\frac{f\text{'}\left(x\right)}{f^2\left(x\right)}-\frac{1}{2}\right)}^2=0 \\ \iff \frac{f\text{'}\left(x\right)}{f^2\left(x\right)}=\frac{1}{2} \end{gathered}$$

$\Rightarrow \int \frac{f\text{'}\left(x\right)}{f^2\left(x\right)}dx=\frac{x}{2}+C\Rightarrow -\frac{1}{f\left(x\right)}=\frac{x}{2}+C.$

 Với $x=4$, ta có

$$\begin{gathered} -\frac{1}{f\left(4\right)}=2+C \\ \iff -4=2+C\iff C=-6. \end{gathered}$$

Do đó: $f\left(x\right)=\frac{-1}{\frac{x}{2}-6}=\frac{2}{12-x}.$ Do đó $f\left(6\right)=\frac{2}{12-6}=\frac{2}{6}=\frac{1}{3}.$