已知函数y=2sin(π3-2x),①求其对称轴方程;②求其单调增区间.-数学

题目简介

已知函数y=2sin(π3-2x),①求其对称轴方程;②求其单调增区间.-数学

题目详情

已知函数y=2sin(
π
3
-2x),
①求其对称轴方程;
②求其单调增区间.
题型:解答题难度:中档来源:不详

答案

①∵y=sin(class="stub"π
3
-2x)
=-2sin(2x-class="stub"π
3
),
令2x-class="stub"π
3
=kπ+class="stub"π
2
可得对称轴方程为:x=class="stub"kπ
2
+class="stub"5π
12
,k∈Z
②解法一:∵正弦函数y=sinx单调减区间是[2kπ+class="stub"π
2
2kπ+class="stub"3π
2
],k∈Z
∴令 2kπ+class="stub"π
2
≤2x-class="stub"π
3
2kπ+class="stub"3π
2

则有2kπ+class="stub"5π
6
≤2x≤2kπ+class="stub"11π
6

kπ+class="stub"5π
12
≤x≤kπ+class="stub"11π
12

∴函数的单调递减区间是[kπ+class="stub"5π
12
kπ+class="stub"11π
12
],k∈Z
解法二:∵函数y=-2sin(2x-class="stub"π
3
)的最大点(取最大值时的x的值)为2x-class="stub"π
3
=2kπ+class="stub"3π
2

取k=0可得x=class="stub"11π
12
,(增区间的右端点的特解)
∵函数的周期为T=π
∴左端点的特解为x=class="stub"11π
12
-class="stub"T
2
=class="stub"11π
12
-class="stub"π
2
=class="stub"5π
12

则函数y=2sin(class="stub"π
3
-2x)的单调增区间是[kπ+class="stub"5π
12
kπ+class="stub"11π
12
],k∈Z

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