27．（2023春•蚌埠期末）已知函数f(x)=22sinxcosx+2cos2x−2sin2xf(x)=2\sqrt{2}\sin x\cos x+\sqrt{2}{\cos ^2}x-\sqrt{2}{\sin ^2}xf(x)=22sinxcosx+2cos2x−2sin2x．（1）求f(x)f(x)f(x)的最小正周期及单调递增区间．（2）证明：当x∈[−π4,π4]x\in [{-\frac{\pi }{4},\frac{\pi }{4}}]x∈[−4π,4π]时，f(x)<x2−2x+3f(x)<x^{2}-2x+3f(x)<x2−2x+3．