题目内容
已知函数

(1)求函数f(x)的单调递减区间;
(2)在¬p的条件下,求f(x)的值域;
(3)若条件q:-2<f(x)-m<2,且¬p是q的充分条件,求实数m的取值范围.
【答案】分析:(1)把给出的函数先降幂再化积,然后运用复合函数的单调性求减区间;
(2)根据给出的条件p得到¬p:
,代入函数解析式后求值域;
(3)把条件q整理后得到f(x)的范围,由¬p是q的充分条件,说明(2)中求出的函数值域是条件q得到的f(x)的范围的子集,比较区间端点值可得m的范围.
解答:解:(1)f(x)=5-
=5-2
+
=-2sin2x+
cos2x+3=-2
=-4
由
,得:
.
所以原函数的单调减区间为{x|kπ-
≤x≤kπ+
,k∈Z};
(2)由于给定条件p:
.
则¬p:
,所以
,所以-1≤
.
所以函数f(x)的值域为[-1,2];
(3)由-2<f(x)-m<2,即m-2<f(x)<m+2,
又¬p是q的充分条件,即当-1≤f(x)≤2时,必有m-2<f(x)<m+2,
所以
,解得:0<m<1.
所以实数m的取值范围是(0,1).
点评:本题考查了复合命题的真假,考查了三角函数中的恒等变换的应用,本题考查了数学转化思想,解答(3)的关键是把充分条件问题转化为集合间的关系求解,此题为中档题.
(2)根据给出的条件p得到¬p:

(3)把条件q整理后得到f(x)的范围,由¬p是q的充分条件,说明(2)中求出的函数值域是条件q得到的f(x)的范围的子集,比较区间端点值可得m的范围.
解答:解:(1)f(x)=5-



=-2sin2x+



由


所以原函数的单调减区间为{x|kπ-


(2)由于给定条件p:

则¬p:



所以函数f(x)的值域为[-1,2];
(3)由-2<f(x)-m<2,即m-2<f(x)<m+2,
又¬p是q的充分条件,即当-1≤f(x)≤2时,必有m-2<f(x)<m+2,
所以

所以实数m的取值范围是(0,1).
点评:本题考查了复合命题的真假,考查了三角函数中的恒等变换的应用,本题考查了数学转化思想,解答(3)的关键是把充分条件问题转化为集合间的关系求解,此题为中档题.

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