Function $F(x)=i \cos ^{2} x+2 \sin x \cos x-\sin ^{2} x+A x+B$, on the interval $\left[0, \frac{3}{2} \pi\right]$, the maximum value $M(A, B)$, when the real numbers $A, B$ vary, find the minimum value of $M(A, B)$.
13. $\mathcal{U} f(x)=\cos ^{2} x+2 \sin x \cos x-\sin ^{2} x+A x+B=\cos 2 x+\sin 2 x+A x+B=$ $\sqrt{2} \sin \left(2 x+\frac{\pi}{4}\right)-A x+B$. From the definition of $M(A, B)$, we know that $M(A, B) \geqslant \frac{1}{4}\left(\left|f\left(\frac{\pi}{8}\right)\right|+2\left|f\left(\frac{5 \pi}{8}\right)\right|+\right.$ $\left.\left|f\left(\frac{9 \pi}{8}\right)\right|\right\rangle$, so $M(A, B) \geqslant \frac{1}{4}\left|f\left(\frac{\pi}{8}\right)+f\left(\frac{9 \pi}{8}\right)-2 f\left(\frac{5 \pi}{8}\right)\right|=\sqrt{2}$. On the other hand, when $A=B=0$, $F(x)=\sqrt{2} \left\lvert\, \sin \left(2 x+\frac{\pi}{4}\right)_{\mathrm{i}}^{\mid} \leqslant \sqrt{2}\right.$. Therefore, the minimum value of $M(A, B)$ is $\sqrt{2}$.