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已知函数 ${\displaystyle f(x)={\cos }^{4}x+{\sin }^{4}x+a\sin 4x-b}$, 且 ${\displaystyle f(x+\frac{\pi }{6})}$ 为奇函数. 若方程 ${\displaystyle f(x)+m=0}$ 在 ${\displaystyle [0,\pi ]}$ 上有四个不同的实数解 ${\displaystyle x_1,x_2,x_3,x_4}$, 则 ${\displaystyle f\left(\frac{x_1+x_2+x_3+x_4}{4}\right)}$ 的平方值为_____.

化简 $$\begin{array}{rl}f(x)& =({\cos }^{2}x+{\sin }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x+a\sin 4x-b\\ & =1-\frac{1}{2}{\sin }^{2}2x+a\sin 4x-b=1-\frac{1}{2}\frac{1-\cos 2x}{2}+a\sin 4x-b\\ & =\frac{1}{4}\cos 4x+a\sin 4x+\frac{3}{4}-b.\end{array}$$ 因为 ${\displaystyle f(x+\frac{\pi }{6})}$ 是奇函数, 所以赋值 $$\{\begin{array}{rl}& f(0+\frac{\pi }{6})=0,\\ & f(-\frac{\pi }{6}+\frac{\pi }{6})=-f(\frac{\pi }{6}+\frac{\pi }{6})=0\end{array}\Rightarrow \{\begin{array}{rl}& -\frac{1}{8}+\frac{\sqrt{3}}{2}a+\frac{3}{4}-b=0,\\ & \frac{1}{4}=\frac{1}{8}+\frac{\sqrt{3}}{2}a,\end{array}$$ 解得 ${\displaystyle a=\frac{\sqrt{3}}{12}}$, ${\displaystyle b=\frac{3}{4}}$.
下面我们根据对称中心 ${\displaystyle (\frac{\pi }{6},0)}$ 和周期 ${\displaystyle \frac{2\pi }{4}=\frac{\pi }{2}}$ 画图:

可以看出要有四个解, ${\displaystyle m\in (\frac{1}{4},1)\cup (-1,\frac{1}{4})}$, 以及 ${\displaystyle \frac{x_1+x_2+x_3+x_4}{4}=\frac{2\pi }{3}\pm \frac{\pi }{8}}$. 所以 $$\begin{array}{rl}{f}^2(\frac{2\pi }{3}\pm \frac{\pi }{8})& ={(\frac{1}{4}\cos (\frac{8\pi }{3}\pm \frac{\pi }{2})+\frac{\sqrt{3}}{12}\sin (\frac{8\pi }{3}\pm \frac{\pi }{2}))}^2\\ & ={(\mp \frac{1}{4}\cos \frac{8\pi }{3}\pm \frac{\sqrt{3}}{12}\sin \frac{8\pi }{3})}^2=\frac{1}{12}.\end{array}$$