山东省枣庄市薛城区2021-2022学年高一上学期期中考试数学试卷含答案

2021～2022学年度模块检测试题

高一数学试题参考答案及评分标准    2021.11

一、单项选择题(本题共8小题，每小题5分，共40分)

1~4  CBCA   5~8  BCDC

二.多项选择题（本题共4小题，每小题5分，共20分）

9. CD       10. ABD       11. AD       12. BC

三、填空题（本题共4小题，每小题5分，共20分）

13.   14.    15.     16．① , ②或者 

四、解答题（本题共6小题，共70分）

17. 解（1）时，集合，·················································1分

，·······························································3分

A∪B=. ····························································4分

（2）若选择①A∩B=A，则，

当，即时，，满足题意；··············································6分

当时，应满足，解得:；···············································9分

综上知，实数a的取值范围是(-∞,-4]∪.  ··································10分

若选择②A∩(CRB)=A，则A是CRB的子集，CRB=(-∞,-2)∪(4,+ ∞)

当，即时，，满足题意；··············································6分

当时，或解得：-4＜a≤或a≥4. 9分

综合得：a的取值范围是：(-∞,]∪[5,+ ∞). ································10分

若选择③A∩B=∅，则当，即时，，满足题意；·····························6分

当时，应满足或者解得：-4＜a≤或a≥5.··································9分

综上知，实数a的取值范围是：(-∞,] ∪[5,+ ∞). ····························10分

18. 解: （1）函数f（x）是定义在上的奇函数，所以当f（0）=0.

由时，f（x）＝x2+2x—1得时f（x）＝﹣x2+2x+1    

所以 .····························································4分

（2）如图单调递增区间是，单调减区间（-3，-1），（1,3）

值域是 ，(图像略2分，单调区间和值域3分 ) ···································9分

（3）  由   得  

由对称性得

由得或.

由图得到不等式的解集是·············································12分

19 . 解：（1）当时，不等式为，整理得             

等价于，    ······················································  2分                                   

所以解集  .·························································4分                                          

（2）当时，整理得        

 等价于.  ··························································· 6分                                       

 当时，，解集为  ····················································8要                              

 当时，解集为.  ····················································· 10分                                       

  当时，解集为.   ···················································12分             

20.解：(1)当x∈[30，50]时，设该工厂获利S，

则 ·······························································  2分

所以当x∈[30，50]时，Smax=-700<0 .  ·····································4分

因此该工厂不会获利，国家至少需要补贴700万元，该工厂才不会亏损. 5分

(2)由题易知，二氧化碳的平均处理成本

(x∈[30，50) .  ······················································7分

当x∈[30，50]时，

.······························································· 10分

当且仅当，即x=40时等号成立，

故P(x)取得最小值为P(40)=40.

所以当处理量为40吨时，每吨的平均处理成本最少.  ·······················12分

21. 解：（1）若，，                                       

因为的定义域为，所以时，，                 

 所以是偶函数．·····················································4分

（2）.  ···························································· 5分                        

 当时，在上，，的最小值在处取得，

 令，解得，符合条件;   ···············································7分                      

当时，在上，，的最小值在处取得，

令，解得，符合条件; ················································9分                       

当时，在上单调递减，在上单调递增，所以的最小值在处取得，，所以此时最小值不可能是 . ······ 11分              综上，存在，使得在区间上的最小值为.······12分