不等式中的取值范围求法
不等式是高中数学的重要内容，与各部分联系紧密，是历年高考的命题重点，在考查不等式的命题中以求取值范围问题居多，解决此类问题的方法体现了等价转换、函数与方程、分类讨论、数形结合等数学思想。
1、 不等式的性质法
利用不等式的基本性质，注意性质运用的前提条件。
例1：已知![](https://wkrtcs.bdimg.com/rtcs/image?w=344&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"344","h":"24","dataType":"wmf","c":"word/media/image1.png","imgOriW":688,"imgOriH":48})
，试求的取值范围。
解：由![](https://wkrtcs.bdimg.com/rtcs/image?w=97&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"97","h":"48","dataType":"gif","c":"word/media/image3.png"})
解得![](https://wkrtcs.bdimg.com/rtcs/image?w=140&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"140","h":"88","dataType":"gif","c":"word/media/image4.png"})
![](https://wkrtcs.bdimg.com/rtcs/image?w=275&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"275","h":"244","dataType":"wmf","c":"word/media/image5.png","imgOriW":550,"imgOriH":488})
评：解此类题常见的错误是：依题意得
![](https://wkrtcs.bdimg.com/rtcs/image?w=180&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"180","h":"48","dataType":"wmf","c":"word/media/image6.png","imgOriW":360,"imgOriH":96})
用（1）（2）进行加减消元，得
![](https://wkrtcs.bdimg.com/rtcs/image?w=199&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"199","h":"23","dataType":"wmf","c":"word/media/image7.png","imgOriW":398,"imgOriH":46})
由![](https://wkrtcs.bdimg.com/rtcs/image?w=212&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"212","h":"23","dataType":"wmf","c":"word/media/image8.png","imgOriW":424,"imgOriH":46})
其错误原因在于由（1）（2）得（3）时，不是等价变形，使范围越加越大。
2、 转换主元法
确定题目中的主元，化归成初等函数求解。此方法通常化为一次函数。
例2：若不等式 2x－1>m(x2-1)对满足－2![](https://wkrtcs.bdimg.com/rtcs/image?w=13&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"13","h":"20","dataType":"gif","c":"word/media/image9.png"})
m2的所有m都成立，求x的取值范围。
解：原不等式化为 (x2－1)m－(2x－1)<0 记f(m)= (x2－1)m－(2x－1) (－2m2)
根据题意有：![](https://wkrtcs.bdimg.com/rtcs/image?w=193&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"193","h":"52","dataType":"gif","c":"word/media/image10.png"})
即：
解得![](https://wkrtcs.bdimg.com/rtcs/image?w=142&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"142","h":"57","dataType":"gif","c":"word/media/image12.png"})
所以x的取值范围为![](https://wkrtcs.bdimg.com/rtcs/image?w=116&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"116","h":"45","dataType":"gif","c":"word/media/image13.png"})
3、化归二次函数法
根据题目要求，构造二次函数，结合二次函数实根分布等相关知识，求出参数取值范围。
例3：在R上定义运算![](https://wkrtcs.bdimg.com/rtcs/image?w=18&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"18","h":"19","dataType":"gif","c":"word/media/image14.png"})
：xy＝(1－y) 若不等式(x－a) (x＋a)<1对任意实数x成立，则 ( )
(A)－1![](https://wkrtcs.bdimg.com/rtcs/image?w=80&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"80","h":"41","dataType":"gif","c":"word/media/image15.png"})
(D)
解：由题意可知 (x-a)[1-(x+a)] <1对任意x成立
即![](https://wkrtcs.bdimg.com/rtcs/image?w=137&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"137","h":"21","dataType":"gif","c":"word/media/image17.png"})
对xR恒成立
记![](https://wkrtcs.bdimg.com/rtcs/image?w=160&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"160","h":"24","dataType":"gif","c":"word/media/image19.png"})
则应满足![](https://wkrtcs.bdimg.com/rtcs/image?w=39&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"39","h":"19","dataType":"gif","c":"word/media/image20.png"})
即：
解得 ,故选择C。
例4：若不等式![](https://wkrtcs.bdimg.com/rtcs/image?w=109&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"109","h":"44","dataType":"gif","c":"word/media/image22.png"})
对一切恒成立，求实数的取值范围。
解：由![](https://wkrtcs.bdimg.com/rtcs/image?w=190&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"190","h":"24","dataType":"gif","c":"word/media/image25.png"})
，知原不等式恒成立等价于恒成立，那么
![](https://wkrtcs.bdimg.com/rtcs/image?w=15&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"15","h":"20","dataType":"gif","c":"word/media/image27.png"})
当时，，不等式成立；
![](https://wkrtcs.bdimg.com/rtcs/image?w=18&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"18","h":"20","dataType":"gif","c":"word/media/image30.png"})
当时，要使不等式恒成立，
应有![](https://wkrtcs.bdimg.com/rtcs/image?w=116&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"116","h":"48","dataType":"gif","c":"word/media/image32.png"})
解得
综上所述：![](https://wkrtcs.bdimg.com/rtcs/image?w=18&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"18","h":"15","dataType":"gif","c":"word/media/image24.png"})
的取值范围为
评：二次项系数含有参数时，要对参数进行讨论等于零是否成立。
4、反解参数法
在题目中反解出参数，化成a>f(x) （a
例5：若不等式![](https://wkrtcs.bdimg.com/rtcs/image?w=104&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"104","h":"21","dataType":"gif","c":"word/media/image35.png"})
对一切恒成立，求的取值范围。
解：因为![](https://wkrtcs.bdimg.com/rtcs/image?w=57&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"57","h":"19","dataType":"gif","c":"word/media/image36.png"})
，所以可转化成
所以要使原不等式恒成立，则需![](https://wkrtcs.bdimg.com/rtcs/image?w=25&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"25","h":"19","dataType":"gif","c":"word/media/image38.png"})
小于的最小值，
令![](https://wkrtcs.bdimg.com/rtcs/image?w=63&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"63","h":"41","dataType":"gif","c":"word/media/image40.png"})
，则此函数在时为增函数，
所以![](https://wkrtcs.bdimg.com/rtcs/image?w=126&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"126","h":"41","dataType":"gif","c":"word/media/image41.png"})
所以![](https://wkrtcs.bdimg.com/rtcs/image?w=48&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"48","h":"19","dataType":"gif","c":"word/media/image42.png"})
，即，故的取值范围为
评：本题也可利用方法3和方法5求解。
例6：已知函数![](https://wkrtcs.bdimg.com/rtcs/image?w=144&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"144","h":"41","dataType":"gif","c":"word/media/image45.png"})
，若在上恒成立，求的取值范围。
解：若![](https://wkrtcs.bdimg.com/rtcs/image?w=89&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"89","h":"21","dataType":"gif","c":"word/media/image46.png"})
在上恒成立，
即![](https://wkrtcs.bdimg.com/rtcs/image?w=107&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"107","h":"41","dataType":"gif","c":"word/media/image49.png"})
，
![](https://wkrtcs.bdimg.com/rtcs/image?w=114&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"114","h":"41","dataType":"gif","c":"word/media/image51.png"})
的最小值为4，
![](https://wkrtcs.bdimg.com/rtcs/image?w=52&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"52","h":"41","dataType":"gif","c":"word/media/image52.png"})
，解得或
所以![](https://wkrtcs.bdimg.com/rtcs/image?w=13&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"13","h":"15","dataType":"gif","c":"word/media/image48.png"})
的取值范围为。
5、 数形结合法
运用数形结合，不仅直观，易发现解题途径，而且能避免复杂的计算与推理，简化了解题过程，在选择和填空中更显其优越。
例7：如果对任意实数x，不等式![](https://wkrtcs.bdimg.com/rtcs/image?w=71&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"71","h":"27","dataType":"gif","c":"word/media/image56.png"})
恒成立，则实数k的取值范围是
解析：画出y1=![](https://wkrtcs.bdimg.com/rtcs/image?w=37&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"37","h":"27","dataType":"gif","c":"word/media/image58.png"})
,y2=kx的图像，由图可看出 0k1
![](https://wkrtcs.bdimg.com/rtcs/image?w=300&md5sum=30d2306efa991c5ba4f82d78ce23f55d&sign=fc111a6d07&rtcs_flag=1&rtcs_ver=3.2&l=webapp&bucketNum=10&ipr={"t":"img","w":"300","h":"221.33333333333","dataType":"gif","c":"word/media/image60.png"})
由于不等式的综合性和灵活性，一道题往往有多种解法，所以要根据题目的情况，选择恰当的方法，不要拘泥一种形式，要灵活多变。
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