Question

已知向量a=(sinx,1)b=(sinx,cosx)f(x)=ab

f(x)>=1求x范围
求f(x)最大值及此时x的值

Answer 1

f(x)=ab
=sin^2 x+cosx>=1
1-cos^2x+cosx>=1
cos^2x-cosx<=0
cosx(cosx-1)<=0
0<=cosx<=1
2kπ-π/2<=x<=2kπ+π/2 其中k为一切实数

f(x)=sin^2 x+cosx
=1-cos^2 x+cosx
=-(cos^2 x-cosx+1/4)+5/4
=-(cosx-1/2)^2+5/4
当cosx=1/2时f(x)得最大值5/4
x=2kπ±π/3 其中k为一切实数

Answer 2

思路：
(1)f(x)=-(cosx-1/2)^2+5/4
f(x)>=1=>0≤cox≤1
=>x范围[2kπ-π/2,2kπ+π/2)
(2)f(x)最大值=5/4
此时x=2kπ+π/3

Answer 3

f(x)=sin²x+cosx
=1-cos²x+cosx
令cosx=t(-1≤t≤1)
则f(t)=1-t²+t
f(t)≥1
-1≤t≤1
0≤t≤1