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时间:2019-08-17
《2018-2019学年高一数学上学期期中试题 (IV)》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
1、2018-2019学年高一数学上学期期中试题(IV)一、选择题(本大题共12小题,每小题5分,每小题给出的四个选项中,只有一项符合要求)1、已知则()2、若函数则方程的解是()....3、方程的根所在区间是()....4、下列函数中,与是同一函数的是()5、若,,,则的大小关系是()6、函数恒过定点,则()7.已知函数是幂函数,则()8、函数在区间上是单调函数,则()9、函数使成立的的集合是()10、函数的图像大致是()11、若函数是奇函数,则使成立的的取值范围是()12、某同学在研究函数时,分别给出下面几个结论:①函数是奇函数;②函数的值域为;③函数在上是增函数;其
2、中正确结论的序号是()①②③①③②③①②二、填空题(本大题共4小题,每小题5分)13、设集合___________.14、定义在上的函数是奇函数且每隔2个单位的函数值都相等,则_____________.15、已知集合_____________.16、已知函数是定义在上的偶函数,当时,,如果函数恰有个零点,则实数的取值范围是 .三、解答题(本大题6小题,共70分,解答应写出必要的文字说明,证明过程或演算步骤)17、(本小题10分)(1)已知函数,求的定义域;(2)解不等式18、(本小题12分)已知集合(1)求与.(2)若求实数的取值范围.19、(本小题12分)已
3、知函数(1)求函数的定义域;(2)判断函数在定义域上的单调性,并说明理由;(3)当满足什么关系时,在上恒取正值.20、(本小题12分)已知:函数(且).(1)求函数的定义域;(2)判断函数的奇偶性,并加以证明;(3)设,解不等式.21、(本小题12分)求函数的单调区间(写出解答过程)22、(本小题12分)已知函数对一切实数都满足,且.(1)求的值;(2)求的解析式;(3)当时,恒成立,求的范围.2021届高一上学期期中考试答案一、选择题题号123456789101112答案CADABCDCBDDA二、填空题题号13141516答案三、解答题17、(本小题满分10分)解
4、析:(1)由条件可知,······································2分函数在上单调递增·····················································3分.···························································5分(2)··················································7分又函数在上单调递减,·····················9分·······················
5、·················10分18.(本小题满分12分)解析:(1)由条件可知································3分又.····························································4分·············································6分(2)··············································8分分析可知,解得········································
6、10分·····································································12分19.(本小题满分12分)解析:(1),················································1分又,·····································································2分定义域为.································································
7、··3分(2)函数在定义域上是单调递增函数.证明:,················4分··································································6分·························································7分所以函数在定义域上是单调递增函数.···································8分(3)要使得在上恒为正值,则在上的最小值必须大于0,由(2)知······················
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