江西省重点九江六校2021-2022学年高二文科数学下学期期末联考试题（PDF版附答案）

九江“六校”2021-2022学年度第二学期高二期末联考数学(文科)本试卷共4页,23题(含选考题)。全卷满分150分,考试时间120分钟。考生注意事项:1.答题前,先将自己的姓名、准考证号填写在试卷和答题卡上,并将准考证号条形码粘贴在答题卡上的指定位置。2.选择题的作答:每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑。写在试卷、草稿纸和答题卡上的非答题区域均无效。3.非选择题的作答:用黑色签字笔直接答在答题卡上对应的答题区域内。写在试卷、草稿纸和答题卡上的非答题区域均无效。4.选考题的作答:先把所选题目的题号在答题卡上指定的位置用2B铅笔涂黑。答案写在答题卡上对应的答题区域内,写在试卷、草稿纸和答题卡上的非答题区域均无效。5.考试结束后,请将本试卷和答题卡一并上交。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,只有一项是符合题目要求的。1.已知全集U={0,1,2,3,4},集合M={1,2,3},N={2,3,4},则{0,1,4}=A.CMB.CNC.C(M∪N)D.C(M∩N)UUUU52.已知复数z=2i-1,则z+=zA.4iB.-4iC.2D.-223.已知p:x>1,q:x+x-2>0,则p是q的A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件4.在一组样本数据(x,y),(x,y),…,(x,y)(n≥2,x,x,…,x不全相等)的散点图中,若所有样本1122nn12n点(x,y)(i=1,2,…,n)都在直线y=2x+1上,则这组样本数据的样本相关系数为iiA.-1B.0C.2D.1π5.已知原命题为:在△ABC中,B=,则三个内角A,B,C成等差数列,则下列说法错误的是3A.原命题与逆命题同为真命题B.原命题与否命题同为真命题C.逆命题与否命题同为假命题D.逆命题与否命题同为真命题-0.26.已知a=0.3,b=log6,c=log0.6,则0.20.3A.a>b>cB.a>c>bC.b>c>aD.c>b>aππ7.正弦函数是奇函数,f(x)=sin(x+)是正弦函数,因此f(x)=sin(x+)是奇函数,以上推理66A.结论正确B.大前提错误C.小前提错误D.推理形式错误2233445510108.观察下列各式:a+b=1,a+b=3,a+b=4,a+b=7,a+b=11,…,则a+b=A.123B.121C.231D.211【九江“六校”2021-2022学年度第二学期高二期末联考·文科数学试卷第1页(共4页)】\nx9.函数f(x)=ecosx的部分图象大致为yy11ππ22πOππxππOπx442244A.B.yyππ42πOπxππOππx242442-1-1C.D.10.设复数z满足|z|=|z+1+i|,z在复平面内对应的点为(x,y),则A.y=x-1B.y=-x-1C.y=x+1D.y=-x+1x11.已知函数y=(x+a)e的最小值为-1,则实数a=A.-1B.0C.1D.212.已知函数f(x)是R上可导函数,满足f(x)-xf′(x)>0,则A.3f(1)<f(3)B.3f(1)>f(3)C.3f(1)=f(3)D.f(1)=f(3)二、填空题:本题共4小题,每小题5分,共20分。13.2022年5月10日,某市物价部门对本市的5家商场的某商品的一天销售量及其价格进行调查,5家商场的售价x元和销售量y件之间的一组数据如下表所示:价格x99.51010.511销售量y11n865由散点图可知,销售量y与价格x之间有较强的线性相关关系,其线性回归方程是y^=-3.2x+40,则n=.x-x14.已知f(x)为奇函数,当x>0时,f(x)=e+e,则当x<0时,f(x)=.15.执行如图所示的程序框图,输出的s值为.)!"k=0,s=1k＜3?&'s#$(1s=1+k=k+1s16.已知直线y=ax(a∈R)与曲线y=lnx相交于两点,则a的取值范围是.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。第17-21题为必考题,每个试题考生都必须作答。第22,23题为选考题,考生根据要求作答。(一)必考题:共60分。17.(12分)22已知m∈R,命题p:∃x∈R,x-3x+m≤0;命题q:∀x∈R,x-2mx+9≥0.000(1)若命题p为假命题,求实数m的取值范围;(2)若命题p∧q为真命题,求实数m的取值范围.【九江“六校”2021-2022学年度第二学期高二期末联考·文科数学试卷第2页(共4页)】\n18.(12分)2已知复数z=m(1+i)-m(5+3i)+6,m∈R.(1)若复数z为纯虚数,求m的值;(2)若复数z在复平面内所对应的点Z位于第三象限,求m的取值范围.19.(12分)3已知函数f(x)=x-3ax+2(a∈R)的极小值为0.(1)求实数a的值;(2)若过点P(2,-4)的直线l与曲线y=f(x)相切,求直线l的方程.20.(12分)2021年9月教育部发布了第八次全国学生体质与健康调研结果,根据调研结果数据显示,我国大中小学生的健康情况有了明显改善,学生总体身高水平有所增加.但在超重和肥胖率上,中小学生却有一定程度上升,大学生整体身体素质有所下滑.某市为了调研本市学生体质情况,采用按性别分层抽样的方法进行调查,得到体质检测样本的统计数据(单位:人)如下.优秀良好及格不及格男生100200780120女生120200520120(1)记体质检测结果为优秀、良好或及格的学生为体质达标,否则为体质不达标.根据所给数据,完成下面2×2列联表.达标不达标合计男生女生合计(2)依据(1)的统计结果判断,是否有95%的把握认为该市学生体质检测是否达标与性别有关?请说明理由.2n(ad-bc)2附:K=(a+b)(c+d)(a+c)(b+d)2P(K>k)0.150.100.050.0250.010.0010k02.0722.7063.8415.0246.63510.828【九江“六校”2021-2022学年度第二学期高二期末联考·文科数学试卷第3页(共4页)】\n21.(12分)x已知函数f(x)=x(e-1).(1)求证:f(x)的极小值为0;3(2)讨论方程mf(x)=x-mx(m∈R)实数解的个数.(二)选考题:共10分。请考生在第22、23题中任选一题做答。如果多做,则按所做的第一题计分。22.(10分)选修4-4:坐标系与参数方程在平面直角坐标系xOy中,以原点O为极点,x轴的非负半轴为极轴建立极坐标系,两点P,Q的极3ππ坐标分别为P(4,),Q(2,),以OQ为直径的圆记为☉C.44(1)求☉C的直角坐标方程;2(2)若直线l经过点P与☉C相交于A,B两点,求证:PA·PB=OP.23.(10分)选修4-5:不等式选讲己知函数f(x)=2|x-2|-3|x-1|.(1)求满足f(a)≥-5的最大整数a的值;22mn(2)在(1)的条件下,对于任意正数m,n,若m+n=a,求证:+≥a.nm【九江“六校”2021-2022学年度第二学期高二期末联考·文科数学试卷第4页(共4页)】\n\nxxfxfxee.51＋115.【答案】【解析】运行该程序,k＝0，s＝1,k<3；k＝0＋1＝1,s＝＝2,k<3；k＝1＋1＝2，313＋12＋13255s＝＝，k<3；k＝1＋2＝3，s＝＝，此时不满足循环条件，输出s，故输出的s值为.2233321116.【答案】0,【解析】直线yaxaR与曲线ylnx相切，可得a.由直线yaxee11与曲线ylnx相交于两点,则0a,故a的取值范围是0,.ee217.【解析】（1）由已知，命题xRx,3xm0为真命题，···························3分9故0，即94m0，解得m，49所以实数m的取值范围是，.·····························································5分49（2）由（1）知命题p为真命题，则m；···················································7分42命题q:xRx,2mx90为真命题，2则4m360，解得：3m3······················································10分9由命题pq为真命题，故p真q真，故实数m的取值范围是3,.···········12分42218.【解析】zm5m6m3mi·····················································2分2m5m60（1）因为复数z为纯虚数，则.··············································4分2m3m022由m5m60，解得m2或m3；由m3m0，解得m0或m3.故当复数z为纯虚数时，m2.·······························································6分（2）因为复数z在复平面内所对应的点Z位于第三象限，2m5m60所以.·······························································8分2m3m02由m5m60，解得2m3；2由m3m0，解得0m3.·································································10分故m的取值范围是2,3.··································································12分第2页共5页\n219.【解析】（1）fx3xa，当a0时，显然fx0，fx在,上单调递增，无极值；···················2分当a0时，随x的变化fx,fx情况变化如下：x,aaa,aaa,fx00fx极大值极小值所以fx极小值为fa2aa2·························································2分由已知2aa20，解得a1····································································5分32（2）由（1）知：fxx3x2，fx3x132设l切点坐标为Mxfx,，fxx3x2，fx3x10000000切线l方程为：yfxfxxx····················································7分000又因为切线l过点P2,4，所以4fxfx2x00032联立上化简式得：x3x0，解得x0或x3········································10分0000当x0时，l的方程为：y3x2；0当x3时，l的方程为：y24x52.0故直线l的方程为y3x2或y24x52······················································12分20.【解析】（1）22列联表如下：达标不达标合计男生10801201200女生840120960合计19202402160·····················5分22160(1080120840120)k（2）19202409601200····················································8分273.3753.841······································································10分8故没有95%的把握认为该市学生体质检测是否达标与性别有关································12分第3页共5页\nx21.【解析】（1）fxx1e1，····················································2分x所以当x0,时，fxe10，fx在0,单调递增；xx故当x,0时，fxx1e1e10，fx在,0单调递减.所以，fx的极小值为f00.······················································5分32x（2）方程mfxxmx等价于x0或x0时mxe.2x2x令gx，则gx2xxe，由gx0，得x0或x2················7分xe随x的变化可得gxgx,情况变化如下：x,000,222,gx00gx极小值极大值2又gx的极小值g00，极大值g24e，且gx0.当x时，gx；当x时，gx0.···································10分2x所以：当m0时，方程mxe（x0）无实数解；22x当0m4e时，方程mxe（x0）有3个实数解；22x当m4e时，方程mxe（x0）有2个实数解；22x当m4e时，方程mxe（x0）有1个实数解；3综上：当m0时，方程mfxxmx有1个实数解；23当0m4e时，方程mfxxmx有4个实数解；23当m4e时，方程mfxxmx有3个实数解；23当m4e时，方程mfxxmx有2个实数解；··········································12分22.【解析】（1）由已知设C任上一点M,，则OMOQcosMOQ.故2cos，即C极坐标方程为2cos·····························2分442即2cos2sin，所以2cos2sin第4页共5页\n222又xy,cosx,siny，22所以C的直角坐标方程为：xy2x2y，2222即xy1········································································5分223（2）由已知P4,的直角坐标为P22,22，且OP4.4x22tcos设直线l的倾斜角为，则其参数方程为（t为参数），·············7分y22tsin222代入xy2x2y化简得：t32sin52cost160·············8分由已知0，设该方程的两根分别为tt,，则tt16.12122由参数的几何意义知：PAPBtt16OP，122故PAPBOP.·················································································10分注：（1）也可以先求Q的直角坐标，再求圆的直角坐标方程；（2）本质是证明切割线定理，直接用不给分；若是转化为用几何法证明可给满分。x1,x123.【解析】（1）因为fx()2|x2|3|x1|5x7,1x2·······················2分x1,x2x1,1x2x2所以fx()5等价于或或x155x75x15解得6x1或1x2或2x4，···················································4分故fx()5的解集为6,4，故a4··················································5分（2）由（1）可知a4，故mn4.因为mn42mn，（当且仅当mn2时取等号），所以mn4·············7分m2n24n24m216mn64mn16124nmnmmnmn22mn故a.······································································10分nm第5页共5页