已知:f(x)=23cos2x+sin2x-3+1(x∈R).求:(Ⅰ)f(x)的最小正周期;(Ⅱ)f(x)的单调增区间;(Ⅲ)若x∈[-π4,π4]时,求f(x)的值域.-数学

题目简介

已知:f(x)=23cos2x+sin2x-3+1(x∈R).求:(Ⅰ)f(x)的最小正周期;(Ⅱ)f(x)的单调增区间;(Ⅲ)若x∈[-π4,π4]时,求f(x)的值域.-数学

题目详情

已知:f(x)=2
3
cos2x+sin2x-
3
+1(x∈R).求:
(Ⅰ)f(x)的最小正周期;
(Ⅱ)f(x)的单调增区间;
(Ⅲ)若x∈[-
π
4
π
4
]时,求f(x)的值域.
题型:解答题难度:中档来源:不详

答案

f(x)=sin2x+
3
(2cos2x-1)+1
=sin2x+
3
cos2x+1
=2sin(2x+class="stub"π
3
)+1---------------------------------------(4分)
(Ⅰ)函数f(x)的最小正周期为T=class="stub"2π
2
=π------------------(5分)
(Ⅱ)由2kπ-class="stub"π
2
≤2x+class="stub"π
3
≤2kπ+class="stub"π
2

得2kπ-class="stub"5π
6
≤2x≤2kπ+class="stub"π
6

∴kπ-class="stub"5π
12
≤x≤kπ+class="stub"π
12
,k∈Z
函数f(x)的单调增区间为[kπ-class="stub"5π
12
,kπ+class="stub"π
12
],k∈Z-----------------(9分)
(Ⅲ)因为x∈[-class="stub"π
4
class="stub"π
4
],∴2x+class="stub"π
3
∈[-class="stub"π
6
class="stub"5π
6
],
∴sin(2x+class="stub"π
3
)∈[-class="stub"1
2
,1],∴f(x)∈[0,3].-----------------------------------(13分)

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