已经函数f(x)=2sinxcosx+sin2x-cos2x.(1)求f(x)递增区间;(2)求f(x)当x∈[0,π2]时的值域.-数学

题目简介

已经函数f(x)=2sinxcosx+sin2x-cos2x.(1)求f(x)递增区间;(2)求f(x)当x∈[0,π2]时的值域.-数学

题目详情

已经函数f(x)=2sinxcosx+sin2x-cos2x.
(1)求f(x)递增区间;
(2)求f(x)当x∈[0,
π
2
]时的值域.
题型:解答题难度:中档来源:不详

答案

∵f(x)=2sinxcosx+sin2x-cos2x
=sin2x-cos2x
=
2
(cos(-class="stub"π
4
)sin2x+sin(-class="stub"π
4
)cos2x)
=
2
sin(2x-class="stub"π
4

(1)f(x)递增区间为2x-class="stub"π
4
∈[-class="stub"π
2
+2kπ,class="stub"π
2
+2kπ]
  k∈Z
即递增区间为x∈[-class="stub"π
8
+kπ,class="stub"3π
8
+kπ
]k∈Z)
(2)当x∈[0,class="stub"π
2
]
即2x-class="stub"π
4
∈[-class="stub"π
4
class="stub"3π
4
]
∴f(x)min=
2
sin(-class="stub"π
4
)=-1
f(x)max=
2
sin(class="stub"π
2
)=
2

即f(x)当x∈[0,class="stub"π
2
]时的值域为[-1,
2
]

更多内容推荐