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

这是一份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，文件包含202111楂樹竴鏁板绛旀doc、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题doc等2份试卷配套教学资源，其中试卷共11页， 欢迎下载使用。
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