下列函数中,周期为π,且在(π4,π2)上为增函数的是()A.y=sin(2x+π2)B.y=cos(2x+π2)C.y=sin(x+π2)D.y=cos(x+π2)-数学

题目简介

下列函数中,周期为π,且在(π4,π2)上为增函数的是()A.y=sin(2x+π2)B.y=cos(2x+π2)C.y=sin(x+π2)D.y=cos(x+π2)-数学

题目详情

下列函数中,周期为π,且在(
π
4
π
2
)上为增函数的是(  )
A.y=sin(2x+
π
2
)
B.y=cos(2x+
π
2
)
C.y=sin(x+
π
2
)
D.y=cos(x+
π
2
)
题型:单选题难度:中档来源:不详

答案

∵y=sin(x+class="stub"π
2
)与y=cos(x+class="stub"π
2
)的周期均为2π,故可排除C,D;
对于A,∵y=sin(2x+class="stub"π
2
)=cos2x在(class="stub"π
4
class="stub"π
2
)上为减函数,故排除A;
对于B,y=cos(2x+class="stub"π
2
)=-sin2x,T=π,
由2kπ+class="stub"π
2
≤2x≤2kπ+class="stub"3π
2
(k∈Z)得kπ+class="stub"π
4
≤x≤kπ+class="stub"3π
4
,k∈Z
∴y=cos(2x+class="stub"π
2
)的递增区间为[kπ+class="stub"π
4
,kπ+class="stub"3π
4
],k∈Z
∵(class="stub"π
4
class="stub"π
2
)⊂[kπ+class="stub"π
4
,kπ+class="stub"3π
4
],k∈Z
故y=cos(2x+class="stub"π
2
)在(class="stub"π
4
class="stub"π
2
)上为增函数,故B符合题意.
故选B.

更多内容推荐