若角α,β满足-π2<α<β<π2,则2α-β的取值范围是()A.(-π,0)B.(-π,π)C.(-3π2,π2)D.(-32π,3π2)-数学

题目简介

若角α,β满足-π2<α<β<π2,则2α-β的取值范围是()A.(-π,0)B.(-π,π)C.(-3π2,π2)D.(-32π,3π2)-数学

题目详情

若角α,β满足-
π
2
<α<β<
π
2
,则2α-β的取值范围是(  )
A.(-π,0)B.(-π,π)C.(-
2
π
2
D.(-
3
2
π
2
题型:单选题难度:中档来源:不详

答案

由题意可得-class="stub"π
2
<α<class="stub"π
2
,-class="stub"π
2
<β<class="stub"π
2

故-π<2α<π,-class="stub"π
2
<-β<class="stub"π
2

由不等式的性质可得-class="stub"3π
2
<2α-β<class="stub"3π
2

又可得-π<α-β<0,和-class="stub"π
2
<α<class="stub"π
2
可得-class="stub"3π
2
<2α-β<class="stub"π
2

综合可得-class="stub"3π
2
<2α-β<class="stub"π
2

故选C

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