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江西省九校2022届高三数学(文)上学期期中联考试题(带答案)

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江西省九校2022届高三上学期期中联考文科数学试卷总分:150分考试时间:120分钟注意事项:1.答题前填写好自己的姓名、班级、考号等信息2.请将答案正确填写在答题卡上一、选择题(本题共12小题,每小题5分,共60分。在每小题给出的四个选项中,只有一项是符号题目要求的)1.已知全集U={-1,0,1,2,3},集合A={0,1,2},B={-1,0,1},则()A.{-1}B.{0,1}C.{-1,2,3}D.{-1,0,1,3}2.命题“”的否定是()A.B.C.D.3.已知复数满足,则复数对应的点位于()A.第一象限B.第二象限C.第三象限D.第四象限4.设,则f(f(2))的值为()A.0B.1C.2D.35.已知,,,则()A.B.C.D.6.如图所示,在中,,,若,,则()A.B.C.D.7.数列是公差不为零的等差数列,且,数列是等比数列,且,则()A.B.C.D.,8.函数的图像大致为()A.B.C.D.9.的内角,,的对边分别为,,.若,,则为()A.等边三角形B.等腰三角形C.直角三角形D.等腰直角三角形10.已知函数的最小正周期为,其图象关于直线对称.给出下面四个结论:①将的图象向左平移个单位长度后得到的函数图象关于y轴对称;②点为图象的一个对称中心;③;④在区间上单调递增.其中正确的结论为()A.①②B.②③C.②④D.①④11.定义为个正数的“均倒数”.若已知数列的前项的“均倒数”为,又,则().A.B.C.D.12.已知函数(为自然对数的底数),若在上恒成立,则实数的取值范围是()A.B.C.D.二、填空题(本题共4小题,每小题5分,共20分)13.已知向量,向量,与共线,则___________.,14.已知,则___________.15.已知中,,,点是线段的中点,则______.16.已知数列满足:,(,),则___________.三、解答题(共70分,解答应写出文字说明,证明过程或演算步骤)17(10分).已知(1)若p为真命题,求x的取值范围;(2)若p是q的必要不充分条件,求实数a的取值范围.18(12分).已知数列{}是首项=,公差为的等差数列,数列{}是首项=,公比为的正项等比数列,且公比等于公差,+=.(1)求数列{},{}的通项公式;(2)若数列{}满足=·(),求数列{}的前项和.19(12分).已知函数.(1)求f(x)的最小正周期;(2)若任意,恒成立,求范围.20(12分).在中,所对的边分别为,向量,且.(1)求角A的大小;(2)若外接圆的半径为2,求面积的最大值.,21(12分).已知函数,曲线在点处切线方程为.(1)求的值;(2)讨论的单调性,并求的极大值.22(12分).设函数.(Ⅰ)讨论的导函数的零点的个数;(Ⅱ)证明:当时.,文科数学答案一.选择题123456789101112CCDBBDADAABD12.【详解】在上恒成立,等价于在上恒成立,构造,则当时,;当时,故在单调递减,在单调递增的最小值为实数的取值范围是.所以选D.二、填空题13.-214.15.16.三、解答题17.(1){x|1≤x≤4};(2).【详解】(1)若p为真命题,则x2≤5x﹣4,即x2﹣5x+4≤0,即(x﹣1)(x﹣4)≤0,即1≤x≤4,······································3分所以x的取值范围{x|1≤x≤4}.··········································4分(2)记A={x|1≤x≤4}.q:x2﹣(a+2)x+2a0(a>2)故当a>2时,B={x|2<x<a}.········································7分,因为p是q的必要不充分条件,所以BÜA,所以,所以2<a≤4,·············································9分故实数a的取值范围为.···································10分18.【详解】解:(1)由题意,可得,因为,则,解得或,·····················2分因为等比数列各项为正项,所以,则,;··········································5分(2)因为,,故,··················6分,①,②··········8分将①-②得:即有··············11分所以.········································12分,19.【详解】解(1)=sin2x+cos2x-=2············································3分f(x)的最小正周期为π;·········································4分(2),······························6分当,即时,············9分,使恒成立················11分.··························································12分20.【详解】(1)依题意得:,则,····································2分∴,又,∴,,故.·········································5分(2)法一:由正弦定理得,,∴面积·······8,分由得:,则,·······························10分∴,故,即时,.··············12分法二:由正弦定理得:,由余弦定理,∴,当且仅当时取等号,····························8分∴,.······································12分21.【详解】(1).································1分由已知得,.·············································2分故,.从而,.·······································4分(2)由(1)知,,.····························6分令得,或,.····································7分从而当时,;当时,.········································10分故在,上单调递增,在上单调递减.·····11分当时,函数取得极大值,极大值为.···········12分22.【详解】(Ⅰ)的定义域为,.············1分当时,,没有零点;····································2分当时,因为单调递增,单调递增,所以在单调递增.··3分又,当b满足且时,,·······················4分故当时,存在唯一零点.······································5分(Ⅱ)由(Ⅰ),可设在的唯一零点为,当时,;当时,.故在单调递减,在单调递增,··7分所以当时,取得最小值,最小值为.························8分由于,所以,.···············11分故当时,.·········································12分

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所属: 高中 - 数学
发布时间:2021-12-22 11:43:57 页数:10
价格:¥3 大小:517.45 KB
文章作者:随遇而安

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