好题集

1.已知函数\(\large\)\(\large f(x)=2\cos(\omega x-\frac{\pi}{6})-1(\omega>0)\)在\([0,\pi]\)上仅有\(2\)个不同零点，则

A.\(\omega\)的最小值为\(\frac{11}{6}\)

B.函数\(f(x)\)在\([0,\pi/5]\)上单调递增

C.函数\(f(x)\)在\([0,\pi]\)上有且仅有一个最大值点

D.\(\exists m,n\in(0,\pi),\)使得\(f(m)=f(n)+4\)

\(\Large \text{answer}:\)

\[f(x)=2\sin(\omega x+\frac{\pi}{3})-1\\ \]

令\(x=0,f(0)=\frac{1}{2}\)

因为\(x\in [0,\pi]\),所以

\[\omega x+\frac{\pi}{3}\in[\frac{\pi}{3},\omega\pi+\frac{1}{3}] \]

令

\[z=\omega x+ \frac{\pi}{3}\in[\frac{\pi}{3},\omega\pi+\frac{1}{3}] \]

由题意,\(\sin z = \frac{1}{2}\)在\(\large [\frac{\pi}{3},\omega\pi+\frac{1}{3}]\)上只能有两解\(x=\frac{5\pi}{6},x=\frac{13\pi}{6}\)

则

\[\frac{13\pi}{6}\le \omega x+\frac{\pi}{3}<\frac{17\pi}{6} \]

解得

\[\frac{11}{6}\le \omega<\frac{5}{2} \]

所以\(A\)正确

当\(\large 0<x< \frac{1}{5}\)时,易得\(z\)的范围的右端点大于\(\large \frac{\pi}{2}\),所以\(B\)错误

在\(z\)的区间上有

\[\sin\frac{\pi}{2}-\sin\frac{3\pi}{2} = 2 \]

故\(\exists m,n\in(0,\pi),\)使得\(f(m)=f(n)+4\)

\(D\)正确

\(\large z=\frac{\pi}{2}\)与\(\large z=\frac{5\pi}{2}\)都可能在区间上,所以不一定只有一个最大值点,C错误

总结:这是一道难度较大,讨论较多的试题,非常耗费考场时间
2.

解析: