设函数f(x)=23sinxcosx+2cos2x-1(x∈R)的最大值为M,最小正周期为T(1)求M,T及函数的单调增区间;(2)10个互不相等的正数xi满足f(xi)=M,且xi<10π(i=1,

题目简介

设函数f(x)=23sinxcosx+2cos2x-1(x∈R)的最大值为M,最小正周期为T(1)求M,T及函数的单调增区间;(2)10个互不相等的正数xi满足f(xi)=M,且xi<10π(i=1,

题目详情

设函数f(x)=2
3
sinxcosx+2cos2x-1(x∈
R)的最大值为M,最小正周期为T
(1)求M,T及函数的单调增区间;
(2)10个互不相等的正数xi满足f(xi)=M,且xi<10π(i=1,2,…,10)求x1+x2+…+x10的值.
题型:解答题难度:中档来源:不详

答案

f(x)=2
3
sinxcosx+2cos2x-1

=
3
sin2x+cos2x=2(cosclass="stub"π
6
sin2x+sinclass="stub"π
6
cos2x)
=2sin(2x+class="stub"π
6

(1)∴函数f(x)的最大值M=2,最小正周期T=class="stub"2π
2

由-class="stub"π
2
+2kπ≤2x+class="stub"π
6
class="stub"π
2
+2kπ,得kπ-class="stub"π
3
≤x≤kπ+class="stub"π
6
,k是整数
∴函数f(x)的单调增区间为[kπ-class="stub"π
3
,kπ+class="stub"π
6
]k∈z
(2)∵f(xi)=M=2
∴2xi+class="stub"π
6
=2kπ+class="stub"π
2
,xi=kπ+class="stub"π
6

∵0<xi<10π,∴0≤k≤9  k∈Z
∴x1+x2+…+x10=(1+2+3+…+9)π+10×class="stub"π
6
=class="stub"140π
3

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