Cho hàm số

D

- Hướng dẫn giải

Điều kiện $\frac{1}{{\left(3-x\right)}^3}>0\Rightarrow x<3$

$$\begin{gathered} f\left(x\right)=\ln \left(\frac{1}{{\left(3-x\right)}^2}\right) \\ =\ln \left(1\right)-3\ln \left(3-x\right) \\ =-3\ln \left(3-x\right) \\ f\text{'}\left(x\right)=-3\frac{1}{\left(3-x\right)}\left(3-x\right)\text{'} \\ =\frac{3}{3-x} \end{gathered}$$

Ta có

$$\begin{gathered} \frac{6}{\mathrm{\pi }}{\int }_0^{\mathrm{\pi }}{\sin }^2\left(\frac{t}{2}\right)dt \\ =\frac{6}{\mathrm{\pi }}{\int }_0^{\mathrm{\pi }}\frac{1-\cos \left(t\right)}{2}dt \\ =\frac{3}{\mathrm{\pi }}{\left(t-\sin \left(t\right)\right)}_0^{\mathrm{\pi }} \\ =\frac{3}{\mathrm{\pi }}\left[\left(\mathrm{\pi }-\sin \left(\mathrm{\pi }\right)\right)\right]=3 \end{gathered}$$

Khi đó

$$\begin{gathered} f\text{'}\left(x\right)>\frac{{\displaystyle \frac{6}{\mathrm{\pi }}}{\int }_0^{\mathrm{\pi }}{\sin }^2\left({\displaystyle \frac{t}{2}}\right)dt}{x+2} \\ \iff \left\{\begin{array}{l}\frac{3}{3-x}>\frac{3}{x+2}\\ x<3;x\ne -2\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}\frac{2x-1}{\left(x-3\right)\left(x+2\right)}<0\\ x<3;x\ne -2\end{array}\right. \\ \iff \left\{\begin{array}{l}x<-2\\ \frac{1}{2}<x<3\end{array}\right. \end{gathered}$$

Vậy tập nghiệm của bất phương trình là

$S=\left(-\infty ;-2\right)\cup \left(\frac{1}{2};3\right)$

Đáp án D