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2022年专项练习题集-不等式恒成立问题 .pdf

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1、20XX 年专项练习题集-不等式恒成立问题三级知识点：不等式恒成立问题介绍：不等式恒成立问题以含参不等式“恒成立”为载体，镶嵌函数、方程、不等式等内容，综合性强，能力要求高，为历年高考试题的热点。选择题1不等式2230mxmx对一切 xR 恒成立，则实数m的取值范围是（）A30mB30mC30mD30m【分值】5【答案】D【易错点】容易忽略0m的情形。【考查方向】本题主要考查了含参数的二次不等式的恒成立问题。【解题思路】对m的分类讨论，（1）0m，（2）当0m时，结合二次函数图象，二次函数应该开口向下，判别式小于等于零，列出满足的条件求解【解析】当0m时不等式化为30恒成立；当0m时需满足00

2、m，所以30m，综上可知实数a的取值范围是30m.2已知2()3f xaxbx，不等式0)(xf的解集是(1,3)，若对于任意 1,2x，不等式()10f xm恒成立，m的取值范围是（）A 14,10B(,10C(,14D 14,14【分值】5【答案】C【易错点】不会求出a，b 的值，不会转化恒成立问题。【考查方向】本题主要考查了函数的解析式，考查恒成立问题，解题的关键是利用好不等式的解集与方程解之间的关系，将恒成立问题转化为函数的最值加以解决【解题思路】（1）根据不等式的解集与方程解之间的关系可知230axbx的两根为1，3，从而可求,a b 的值，进而可求fx的解析式；（2）要使对于任意

3、1,2x，不等式()10f xm恒成立，只需min()10f xm即可，从而可求m的范围【解析】不等式()0f x的解集是(1,3)，所以1和3是方程230axbx的两个根，由韦达定理得1,2ab 所以2()23f xxx，所以()10fxm恒成立等价于2213xxm恒成立，由22213(1)1414xxx，所以14m选 C3对任意的实数x，不等式30 xxa恒成立，则实数a的取值范围是（）A0aB03aC3aD3a文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9

4、Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O

5、6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K

6、6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7

7、J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P

8、5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J1

9、0K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6

10、O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8【分值】5【答案】D【易错点】不会去掉绝对值，函数的最值。【考查方向】本题主要考查了含参数的绝对值不等式的恒成立问题。【解题思路】令()3f xxx，依题意，只需求得min()f x即可求得a的取值范围【解析】令3,3()323,3xfxxxxx，则min()3f x，所以min()3af x，即3a，故选 C.4若不等式290 xtx对于任意(0,)x都成立，则t的最大值是（）A 0 B-6 C6 D 9【分值】5【答

11、案】C【易错点】不会将变量t 分离出来。【考查方向】本题主要考查了含参数的二次不等式的恒成立问题以及分类变量法。【解题思路】首先根据不等式将t 分离出来，即9txx对任意(0,)x都成立，即min9txx【解析】不等式290 xtx对于任意(0,)x都成立等价于9txx对任意文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F

12、7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9

13、Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O

14、6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K

15、6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7

16、J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P

17、5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J1

18、0K6F7V4 ZA8Y6O7J9Y8(0,)x都成立因为9926xxxx，所以只需6t即可故C 正确5若关于x的不等式2(2)120 xaxa对任意的 2,2a均成立，则x的取值范围是（）A(,1)(3,)B(,5)(5,)C(,5)(3,)D(5,3)【分值】5【答案】C【易错点】不知道讲原不等式转化为关于a 的一次函数。【考查方向】本题主要考查了一元二次不等式恒成立问题，将恒成立问题转化为函数的最值加以解决【解题思路】可将a 视作自变量，则上述问题即可转化为在-2，2内关于 a 的一次函数大于 0 恒成立的问题.解：原不等式转化为2(2)210a xxx在 2,2a时恒成立,设2()(2

19、)21f aa xxx,则()f a在-2,2 上恒大于 0，故有：(2)0(2)0ff即2243050 xxx解得：3155xxxx或或所以35xx或，故选 C.文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z

20、6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F

21、7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9

22、Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O

23、6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K

24、6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7

25、J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8填空6若函数()sincos3f xxax的图象始终在直线1y的上方，则a 的取值范围是_【分值】5

26、【答案】(3,3)【易错点】不会利用辅助角公式对()sincos3f xxax进行变形，不会将()f x在1y的上方转化成()1f x恒成立。【考查方向】三角恒等变换和不等式恒成立问题。【解题思路】问题转化为()1f x恒成立，利用三角恒等变形以及三角函数的最值建立不等式，求出a 的范围。【解析】由()f x的图象始终在1y的上方，即()1f x恒成立，2()sincos31sin()31f xxaxax，即21sin()2ax恒成立，即22sin()1xa恒成立，所以2211a，解得33a7若关于x的不等式21xmxm恒成立，则实数m【分值】5【答案】2【易错点】判别式容易容易出现0。【考查

27、方向】二次不等式恒成立问题。【解题思路】将不等式右边项移到左边，利用判别式0，求出m的值.文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P

28、5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J1

29、0K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6

30、O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO

31、3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2

32、J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8

33、Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8【解析】原不等式可变为210 xmxm，0，2410mm，220m，2m.8已知1a，若关于x不等式1logaxx在区间0,2上恒成立，则实数a的取值范围是【分

34、值】5【答案】4,【易错点】不会转化原不等式，不会利用数形结合处理本题。【考查方向】本题主要考查了反比例函数及其单调性、不等式恒成立问题，同时考查了数形结合的思想。【解题思路】由题意可转化为不等式1logaxx在区间0,2上恒成立，由图象可知，在区间0,2上，函数1yx的图象在函数logayx的图象的上方，从而可得解【解析】依题意1logaxx，则必有则必有11log 22aa，解得4a，所以实数a的取值范围4,综合题文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO

35、3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2

36、J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8

37、Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:

38、CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 H

39、E2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 Z

40、A8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编

41、码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y89已知函数321()13f xaxx()xR其中0a，若在区间1,22上，()0f x恒成立，求a的取值范围【分值】6【答案】2 3183a【易错点】导数的计算与分类讨论。【考查方向】导数与不等式恒成立问题。【解题思路】对321()13fxaxx进行求导，判断利用导数求出321()13f xaxx的极值点，利用极值点与端点值的函数值大于0，解不等式，得到a 的取值范围。【解析】22()2()fxaxxax xa，由于0a，所以20a，对

42、a 进行讨论：（1）若01a，22a，于是当10 x时，()0fx；当02x时，()0fx。由1()022()0ffa，即1898aa，由01a，故无解。（2）若1a，22a，于是当10 x时，()0fx；当20 xa时，()0fx，当22xa时，()0fx。由1()02(2)0ff，即21843aa，解得2 3183a。综合（1）（2）得2 3183a。10 已知不等式2310axxa对于所有的实数x不等式恒成立，求a的取值范围.【分值】6 文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6

43、O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO

44、3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2

45、J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8

46、Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:

47、CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 H

48、E2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 Z

49、A8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8【答案】152a【易错点】讨论时容易忽略0a的情形。【考查方向】本题主要考查了一元二次不等式恒成立问题。【解题思路】当0a时，经检验不满足条件；解得0a时，设2()31f xaxxa，则由题意可得094(1)0aa a，解出a 的范围即可【解析】当0a时,310 x，即当13x时不等式不恒成立，不满足条件；当0a时，设2()31f xaxxa，由于()0f x恒成立，则有094(1)0aa a，解得152a。文档

50、编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R8 HE2J10K6F7V4 ZA8Y6O7J9Y8文档编码:CO3P5O6Z6R

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