《2025年对点对题高考状元训练手册高三数学人教版》

根据下面的题设,完成34～36小题.
已知函数$f(x)=2\sqrt{3}\sin(\frac{\pi}{4}+\frac{x}{2})\sin(\frac{\pi}{4}-\frac{x}{2})-\sin(\pi + x)$,且函数$y = g(x)$的图象与函数$y = f(x)$的图象关于直线$x=\frac{\pi}{4}$对称.
34.函数$g(x)$的解析式为( )
A. $g(x)=2\sin(x-\frac{\pi}{3})$
B. $g(x)=2\sin(x+\frac{\pi}{3})$
C. $g(x)=2\sin(x-\frac{\pi}{6})$
D. $g(x)=2\sin(x+\frac{\pi}{6})$
35.若存在$x\in[0,\frac{\pi}{2}]$,使等式$[g(x)]^{2}-mg(x)+2 = 0$成立,则实数$m$的最小值是( )
A. $\sqrt{2}$
B. $2\sqrt{2}$
C. 3
D. $3\sqrt{2}$
36.若$x\in[-\frac{\pi}{3},\frac{2\pi}{3}]$,不等式$\frac{1}{2}f(x)-ag(-x)\gt a - 2$恒成立,则实数$a$的取值范围是( )
A. $(-2,-\frac{2}{3})$
B. $(-\frac{2}{3},2)$
C. $(\frac{2}{3},2)$
D. $(-2,\frac{2}{3})$

答案： 函数$f(x)=2\sqrt{3}\sin(\frac{\pi}{4}+\frac{x}{2})\cdot\sin(\frac{\pi}{4}-\frac{x}{2})-\sin(\pi + x)$，化简得$f(x)=2\sqrt{3}\sin(\frac{\pi}{4}+\frac{x}{2})\cdot\cos(\frac{\pi}{4}+\frac{x}{2})+\sin x=\sqrt{3}\sin(\frac{\pi}{2}+x)+\sin x=\sqrt{3}\cos x+\sin x=2\sin(x+\frac{\pi}{3})$.函数$y = g(x)$的图象与函数$y = f(x)$的图象关于直线$x=\frac{\pi}{4}$对称，即$g(x)=f(\frac{\pi}{2}-x)$，$\therefore g(x)=2\sin(\frac{\pi}{2}-x+\frac{\pi}{3})=2\sin(\frac{5\pi}{6}-x)=2\sin(x+\frac{\pi}{6})$，故选D.
@@$\because x\in[0,\frac{\pi}{2}]$，$g(x)=2\sin(x+\frac{\pi}{6})$，$\therefore x+\frac{\pi}{6}\in[\frac{\pi}{6},\frac{2\pi}{3})$，$\therefore1\leqslant g(x)\leqslant2$，令$g(x)=t$，则$1\leqslant t\leqslant2$，那么$[g(x)]^{2}-mg(x)+2 = 0$，得$t^{2}-mt + 2 = 0$成立，即$m=t+\frac{2}{t}\geqslant2\sqrt{2}$，当且仅当$t=\sqrt{2}$时取等号，$\therefore m$的最小值为$2\sqrt{2}$，故选B.
@@不等式$\frac{1}{2}f(x)-ag(-x)＞a - 2$恒成立，即$\sin(x+\frac{\pi}{3})+2a\sin(x-\frac{\pi}{6})＞a - 2$恒成立.$\because x\in[-\frac{\pi}{3},\frac{2\pi}{3}]$，$\therefore0\leqslant x+\frac{\pi}{3}\leqslant\pi$，$-\frac{\pi}{2}\leqslant x-\frac{\pi}{6}\leqslant\frac{\pi}{2}$.当$a = 0$时，显然$\sin(x+\frac{\pi}{3})＞-2$恒成立；当$a＞0$时，当$x=-\frac{\pi}{3}$时，$\sin(x+\frac{\pi}{3})+2a\sin(x-\frac{\pi}{6})$取得最小值，即$\sin(-\frac{\pi}{3}+\frac{\pi}{3})+2a\sin(-\frac{\pi}{3}-\frac{\pi}{6})＞a - 2$成立，得$-2a＞a - 2$，解得$0＜a＜\frac{2}{3}$；当$a＜0$时，当$x=\frac{2\pi}{3}$时，$\sin(x+\frac{\pi}{3})+2a\sin(x-\frac{\pi}{6})$取得最小值，即$\sin(\frac{2\pi}{3}+\frac{\pi}{3})+2a\sin(\frac{2\pi}{3}-\frac{\pi}{6})＞a - 2$成立，得$a＞-2$，$\therefore-2＜a＜0$，综上$a$的范围是$(-2,\frac{2}{3})$，故选D.