设函数f(x)=(sinωx+cosωx)2+2cos2ωx(ω>0)的最小正周期为2π3.(Ⅰ)求ω的值;(Ⅱ)若函数y=g(x)的图象是由y=f(x)的图象向右平移π2个单位长度得到,求y=g(x

题目简介

设函数f(x)=(sinωx+cosωx)2+2cos2ωx(ω>0)的最小正周期为2π3.(Ⅰ)求ω的值;(Ⅱ)若函数y=g(x)的图象是由y=f(x)的图象向右平移π2个单位长度得到,求y=g(x

题目详情

设函数f(x)=(sinωx+cosωx)2+2cos2ωx(ω>0)的最小正周期为
3

(Ⅰ)求ω的值;
(Ⅱ)若函数y=g(x)的图象是由y=f(x)的图象向右平移
π
2
个单位长度得到,求y=g(x)的单调增区间.
题型:解答题难度:中档来源:重庆

答案

(Ⅰ)f(x)=(sinωx+cosωx)2+2cos2ωx=sin2ωx+cos2ωx+sin2ωx+1+2cos2ωx
=sin2ωx+cos2ωx+2=
2
sin(2ωx+class="stub"π
4
)+2

依题意得class="stub"2π
=class="stub"2π
3
,故ω的值为class="stub"3
2

(Ⅱ)依题意得:g(x)=
2
sin[3(x-class="stub"π
2
)+class="stub"π
4
]+2=
2
sin(3x-class="stub"5π
4
)+2

2kπ-class="stub"π
2
≤3x-class="stub"5π
4
≤2kπ+class="stub"π
2
(k∈Z)

解得class="stub"2
3
kπ+class="stub"π
4
≤x≤class="stub"2
3
kπ+class="stub"7π
12
(k∈Z)

故y=g(x)的单调增区间为:[class="stub"2
3
kπ+class="stub"π
4
,class="stub"2
3
kπ+class="stub"7π
12
](k∈Z)

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