已知函数f(x)=2sinxcosx+sin(2x+π2).(1)若x∈R,求f(x)的最小正周期和单调递增区间;(2)设x∈[0,π3],求f(x)的值域.-数学

题目简介

已知函数f(x)=2sinxcosx+sin(2x+π2).(1)若x∈R,求f(x)的最小正周期和单调递增区间;(2)设x∈[0,π3],求f(x)的值域.-数学

题目详情

已知函数f(x)=2sinxcosx+sin(2x+
π
2
)

(1)若x∈R,求f(x)的最小正周期和单调递增区间;
(2)设x∈[0, 
π
3
]
,求f(x)的值域.
题型:解答题难度:中档来源:深圳模拟

答案

(1)f(x)=sin2x+cos2x=
2
sin(2x+class="stub"π
4
)

周期T=class="stub"2π
2

2kπ-class="stub"π
2
≤2x+class="stub"π
4
≤class="stub"π
2
+2kπ
,得kπ-class="stub"3π
8
≤x≤kπ+class="stub"π
8

所以,单调递增区间为[kπ-class="stub"3π
8
,kπ+class="stub"π
8
],k∈Z

(2)若0≤x≤class="stub"π
3
,则class="stub"π
4
≤2x+class="stub"π
4
≤class="stub"11π
12
sinclass="stub"11π
12
=sinclass="stub"π
12
=sin(class="stub"π
4
-class="stub"π
6
)=
6
-
2
4
<sinclass="stub"π
4
6
-
2
4
≤sin(2x+class="stub"π
4
)≤1
3
-1
2
2
sin(2x+class="stub"π
4
)≤
2

即f(x)的值域为[
3
-1
2
, 
2
]

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