函数y=2+sinx-cosx的最大值是______,最小值是______,最小正周期为______,单调增区间为______,减区间为______.-数学

题目简介

函数y=2+sinx-cosx的最大值是______,最小值是______,最小正周期为______,单调增区间为______,减区间为______.-数学

题目详情

函数y=2+sinx-cosx的最大值是______,最小值是______,最小正周期为______,单调增区间为______,减区间为______.
题型:填空题难度:中档来源:不详

答案

∵y=2+
2
sin(x-class="stub"π
4
)
,∴①当sin(x-class="stub"π
4
)
=1时,ymax=2+
2
;②当sin(x-class="stub"π
4
)
=-1时,ymin=2-
2
;③函数的最小正周期为2π;
④由-class="stub"π
2
+2kπ≤x-class="stub"π
4
≤2kπ+class="stub"π
2
,解得-class="stub"π
4
+2kπ≤x≤2kπ+class="stub"3π
4
(k∈Z).∴函数f(x)的单调递增区间[-class="stub"π
4
+2kπ,2kπ+class="stub"3π
4
]
(k∈Z);
⑤由class="stub"π
2
+2kπ≤x-class="stub"π
4
≤2kπ+class="stub"3π
2
,解得class="stub"3π
4
+2kπ≤x≤2kπ+class="stub"7π
4
(k∈Z).∴函数f(x)的单调递减区间为[2kπ+class="stub"3π
4
,2kπ+class="stub"7π
4
]
 (k∈Z).
故答案分别为2+
2
2-
2
,2π,[-class="stub"π
4
+2kπ,2kπ+class="stub"3π
4
]
(k∈Z),[2kπ+class="stub"3π
4
,2kπ+class="stub"7π
4
]
 (k∈Z).

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