题目
题型:不详难度:来源:
3 |
π |
4 |
π |
4 |
(1)求f(x)的单调递增区间;
(2)试画出函数f(x)在区间[0,π]上的图象.
答案
3 |
π |
4 |
π |
4 |
=
3 |
π |
2 |
=sin2x-
3 |
=2sin(2x-
π |
3 |
由-
π |
2 |
π |
3 |
π |
2 |
得-
π |
12 |
5π |
12 |
故函数的单调增区间是[-
π |
12 |
5π |
12 |
(2)函数f(x)=2sin(2x-
π |
3 |
3 |
π |
4 |
π |
4 |
3 |
π |
4 |
π |
4 |
3 |
π |
2 |
3 |
π |
3 |
π |
2 |
π |
3 |
π |
2 |
π |
12 |
5π |
12 |
π |
12 |
5π |
12 |
π |
3 |