f(x) |
cosx |
则g′(x)=
f′(x)cosx−f(x)(cosx)′ |
cos2x |
1 |
cos2x |
∵对任意的x∈(-
π |
2 |
π |
2 |
∴g′(x)>0,即函数g(x)在x∈(-
π |
2 |
π |
2 |
则g(-
π |
3 |
π |
6 |
f(−
| ||
cos(−
|
f(−
| ||
cos(−
|
3 |
π |
3 |
π |
6 |
故A正确,故选:A.
π |
2 |
π |
2 |
3 |
π |
3 |
π |
6 |
π |
6 |
| ||
2 |
π |
4 |
2 |
π |
3 |
2 |
π |
4 |
f(x) |
cosx |
f′(x)cosx−f(x)(cosx)′ |
cos2x |
1 |
cos2x |
π |
2 |
π |
2 |
π |
2 |
π |
2 |
π |
3 |
π |
6 |
f(−
| ||
cos(−
|
f(−
| ||
cos(−
|
3 |
π |
3 |
π |
6 |