已知函数f(x)=sin(2x+π6)+2sin2(x+π6)-2cos2x+a-1(a∈R,a为常数)(1)求函数f(x)的最小正周期(2)求函数f(x)的单调递增区间(3)若x∈[0,π2]时,f

题目简介

已知函数f(x)=sin(2x+π6)+2sin2(x+π6)-2cos2x+a-1(a∈R,a为常数)(1)求函数f(x)的最小正周期(2)求函数f(x)的单调递增区间(3)若x∈[0,π2]时,f

题目详情

已知函数f(x)=sin(2x+
π
6
)+2sin2(x+
π
6
)-2cos2x+a-1
(a∈R,a为常数)
(1)求函数f(x)的最小正周期
(2)求函数f(x)的单调递增区间
(3)若x∈[0,
π
2
]时,f(x)的最小值为1,求a的值.
题型:解答题难度:中档来源:不详

答案

(1)f(x)=sin(2x+class="stub"π
6
)+2sin2(x+class="stub"π
6
)-2cos2x+a-1

=sin(2x+class="stub"π
6
)
-cos(2x+class="stub"π
3
)-2cos2x+a
=sin2x•
3
2
+cos2x•class="stub"1
2
-cos2x•class="stub"1
2
+sin2x•
3
2
-2×class="stub"1+cos2x
2
+a 
=
3
sin2x
-cos2x+a-1=2sin(2x-class="stub"π
6
)+a-1.
故函数f(x)的最小正周期等于 class="stub"2π
2
=π.
(2)由2kπ-class="stub"π
2
≤2x-class="stub"π
6
≤2kπ+class="stub"π
2
,k∈z,可得kπ-class="stub"π
6
≤x≤kπ+class="stub"π
3
,k∈z,
故函数f(x)的单调递增区间为[kπ-class="stub"π
6
,kπ+class="stub"π
3
],k∈z.
(3)若x∈[0,class="stub"π
2
]时,有-class="stub"π
6
≤2x-class="stub"π
6
class="stub"5π
6
,故当2x-class="stub"π
6
=-class="stub"π
6
 时,即x=0时,f(x)有最小值为1,
由2×(-class="stub"1
2
)
+a-1=1,∴a=3.

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