高考数学（重庆专用 理科）一轮复习题库 选考部分选修4－5不等式选讲 Word版含解析

1、课时作业71不等式选讲填空题1不等式|2x1|x的解集是_2已知实数集r，集合mx|x2|2，集合n，则m(rn)_.3对于实数x，y，若|x1|1，|y2|1，则|x2y1|的最大值为_4不等式|xx22|x23x4的解集是_5如果关于x的不等式|xa|x4|1的解集是全体实数，则实数a的取值范围是_6已知函数，则不等式f(x)f的解集为_7不等式|x3|x2|3的解集为_8若不等式|x1|x3|a对任意的实数x恒成立，则实数a的取值范围是_9如果存在实数x使不等式|x1|x2|k成立，则实数k的取值范围是_10已知函数f(x)|x2a|，不等式f(x)4的解集为x|2x6，则a_；若存在x

2、r，使不等式f(x)f(x2)m成立，则实数m的取值范围是_11设函数f(x)，当m4时，函数f(x)的定义域为_；若不等式f (x)0的解集为r，则m的取值范围是_12不等式|x29|3x的解集是_13对于任意实数a(a0)和b，不等式|ab|ab|a|(|x1|x2|)恒成立，则实数x的取值范围是_14关于x的不等式|3x2|2m1的解集是_15如果关于x的不等式|x3|x4|a的解集不是空集，则参数a的取值范围是_参考答案填空题1解析：不等式|2x1|x等价于解得由此可得不等式|2x1|x的解集为2x|0x1解析：由已知得mx|0x4，nx|x1，m(rn)x|0x4x|x1x|0x13

3、5解析：|x2y1|x12(y2)2|x1|2|y2|21225当且仅当x0，y3时取等号4x|x3解析：|xx22|x2x2|，而x2x20恒成立，原不等式等价于x2x2x23x4，即2x6，x3故原不等式的解集为x|x35(，53，)解析：在数轴上，结合绝对值的几何意义，可知a5或a36解析：f，当x1时，f(x)xx，由可解得1x2；当0x1时，f(x)xx，由x可解得x1，综上可得不等式f(x)f的解集为7x|x1解析：原不等式可化为或或解得x或1x2或x2所以原不等式的解集为x|x18(，0)2解析：当a0时，显然成立；当a0时，|x1|x3|的最小值为4，a4a2综上可知a(，0)

4、29(3，)解析：令f(x)|x1|x2|，则f(x)作出其图象，可知f(x)min3，即k3101m2解析：由f(x)4得|x2a|4，解得2a4x2a4，又已知不等式f(x)4的解集为x|2x6，所以解得a1所以f(x)|x2|设g(x)f(x)f(x2)，即g(x)|x2|x|当x0时，g(x)2；当0x2时，g(x)2；当x2时，g(x)2综上，g(x)2，故m211x|x1或x3(，2解析：当m4时，|x|x2|40故有x|x1或或x|x3，函数f(x)的定义域为x|x1或x3由题知0的解集为r，即m|x|x2|的解集为r设g(x)|x|x2|xx2|2，则mgmin(x)2，即m(

5、，212x|x3，或2x4131415(1，)高考资源网版权所有！投稿可联系qq：1084591801c29c91afe4ced1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afedd1b6c87d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91af95ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9d1b6c8795ad9c91afe4ced1b6c8

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