济源市、平顶山市、许昌市2021-2022学年高三第二次质量检测理科数学试卷及答案(PDF版带答案)
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平顶山许昌济源2021-2022学年高三第二次质量检测理科数学参考答案一、选择题:本题共12小题,每小题5分,共60分。1.D2.B3.C4.C5.B6.B7.D8.A9.A10.D11.D12.B二、填空题:本题共4小题,每小题5分,共20分。101113.-114.4515.(0,2)16.4045三、解答题:共70分。17.解:2bccosC(1)因为,所以2bccosAacosC,acosA所以2sinBcosAsinAcosCcosAsinCsinACsinB,············(3分)1因为sinB0,所以cosA,2π因为A0,π,所以A;···············································(6分)313(2)因为BDBABC,所以AD3DC,44113所以SSbcsinAbc,·································(8分)△BCD△ABC4816222因为abc2bccosA,22所以16bcbcbc,当且仅当bc时等号成立,3所以Sbc3,················································(11分)△BCD16所以BCD面积的最大值为3············································(12分)18.解:(1)相关系数为1,nnn2xixyiyxixyiyxix22ri1i1i1bˆnsxbˆsxn2n2n2n222nssxxyyxixyyyyiiii1i1i1i12=4.7´=0.94Î[0.75,1]··············································(6分)50故y与x线性相关较强···················································(7分)22100(35301520)(2)K9.0917.879····························(11分)50504555所以有99.5%的把握认为购买电动汽车与性别有关····(12分)1119.解:(1)因为BEEC,DFFC,所以EF//BD,22因为EF平面ABD,所以EF//平面ABD,·····························(2分)1因为G是ABC的重心,BEEC,所以GE∥AB,2因为GE平面ABD,所以GE//平面ABD;因为GEEFE,所以平面GEF//平面ABD·························(5分)(2)因为GE//AB,所以GEBC,因为DC平面ABC,所以GECD,因为BC,CD平面BCD,BCCDC,所以GE平面BCD,因为EF平面BCD,所以GEEF,所以当点P与E重合时,GP最短;·····································(7分)如图,在平面ABC内,作CHBC,以C为坐标原点,CB为x轴,CH为y轴,CD为2z轴,建立空间直角坐标系,则A(1,3,0),C0,0,0,D0,0,2,P(,0,0),312所以AP(,3,0),DP(,0,2),CA(1,3,0),CD(0,0,2)33设m(x,y,z)为平面ADP的一个法向量,1111x3y0mAP0311则,即,mDP02x2z01133令x13,得m(3,,1);3设n(x,y,z)为平面ADC的一个法向量,2222,nCA0x3y022则,即令x23,得n(3,3,0);nCD02z02mn10531cosm,n|m||n|3131·····················(11分)123531所以二面角PADC的余弦值为.·····························(12分)31p20.解:(1)抛物线2F,x2py(p0)的焦点为(0,)2ppp设N(t,),则OF(0,),ON(t,)2222p因为OFON1,所以1,得p242所以抛物线的方程为x4y;····························(4分)(2)依据圆与抛物线的对称性,四边形ABCD是以y轴为对称轴的等腰梯形,不妨设ABCD,A,D在第一象限,A(x1,y1),D(x2,y2),则B(x1,y1),C(x2,y2),y1y2,22x(y3)52联立消去x得y(62p)y40,显然关于y的一元二次方程2x2py有互异二正根,所以262p160y1y262p0,解得p1,而p0,所以0p1..............................(6分)yy412依据对称性可知,点G在y轴上,可设G(0,t),ytyy112由kAGkAC得,,xxx112ytyyyy11212所以2py2p(yy)2p112则ty1y22,所以点G(0,2);························(8分)3,11S2(SS)2[2x(yy)2x(2y)]2x(y2)22py(y2)ABDABG1211112122222p(yyy2y)22p(2y2y)42pyy2yy1221211212p(1p)1=42p(62p)48p(1p)84,等号当且仅当p时成立,221所以p时,S取得最大值4·························(12分)221.(12分)112解:(1)当a时,f(x)xsinxcosxx,441则f'(x)x(cosx),x(,)·····························(2分)2令f'(x)0,得x或0x;33令f'(x)0,得x0或x··························(4分)33所以f(x)在(-π,π)上的增区间为(,),(0,),减区间为(,0),(,)·····(5分)33331(2)由题知:g(x)xsinxtanx,x(,)322①g(0)0,x0是函数g(x)的一个零点;·······················(6分)②当x(,0)时,sinx0,tanx0,g(x)02所以函数g(x)在(,0)上无零点;·······························(7分)2③当x(0,)时,由g(x)0得3xcosx102记h(x)3xcosx1,x(0,),h'(x)3cosx3xsinx,24,记u(x)cosxxsinx,x(0,),则2u'(x)2sinxxcosx0,所以函数u(x)在(0,)单调递减,2(4)2又u()0,u()0,4822所以存在唯一实数x0(,),使得u(x0)0,42且x(0,x0)时,u(x)0,则h'(x)0,所以h(x)在(0,x0)上单调递增,x(x0,)时,u(x)0,则h'(x)0,所以h(x)在(x0,)上单调递减;2232易知:h(x)h()10,又h(0)10,h()100482所以函数h(x)在(0,x0)和(x0,)上各有一个零点,2所以函数g(x)在(0,)上有且仅有两个零点···························(11分)2综上:函数g(x)在(,)上有且仅有三个零点.···················(12分)222xt,222.解:(1)因为直线l的参数方程为(t为参数),2y1t,2则直线l的普通方程为xy10.....................................................................(2分)2222因为曲线C:13sin4,则曲线C的直角坐标方程为x4y4,2x2即y1.......................(5分)4(2)易知,点A(1,0)在直线l上,直线l的斜率为-1,所以可设直线l的参数方程为5,2x1t,2(t为参数),代入曲线C的直角坐标系方程得25t22t60.2yt,2226则t1t2,t1t255tt2所以M对应的参数t12,0252226()4()|AP|+|AQ||t|+|t||tt|55故1212=8............................(10分)|AM||t||t|2005x5,x223.解:(1)f(x)3x3,2x1···············(2分)x5,x1x<-2-2x<1x1由题意得或或-x-5>03x+3>0x+5>0\x<-5或-1
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