202602042

题目

f(x)=cos2x2cosx,xR
(1)f(x)[0,π]
(2)f(x)?,,
,.
(3)f(x)f(x)[0,2026]
644π2023.19,645π2026.33,646π2029.47

解析

(1)f(x)=2cos2x12cosx
t=cosx[1,1]
y=2t22t1=2(t12)232
f(x)max=f(1)=3,f(x)min=f(12)=32

(2)

f(x)=f(x),f(x)y
y=cos2xx=kπ2,y=cosxx=kπ
f(x)x=kπ,kZ

(3)

f(x)2π
[0,2π]
2cos2x12cosx=0cosx=132cosx=1+32
[0,2π],y=cosxcosx=132
y=f(x)[0,2π]
[0,π][π,2π]
132=cosx>cos3π4=22
[0,3π4][5π4,2π]
f(x)[0,644π]644
644π<2026<644π+3π4
[644π,2026]
f(x)[0,2026]645