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高二数学参考答案（第1页共9页）质量抽测数学参考答案及评分细则评分说明：1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制定相应的评分细则。2．对计算题

，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分。3．解答右端所注分数，表示考生正确做到这一步应得

的累加分数。4．只给整数分数。一、选择题：本题共8小题，每小题5分，共40分．1．B2．C3．A4．A5．D6．C7．D8．C二、选择题：本题共4小题，每小题5分，共20分．9．AD10．BC11．ABD12．BC三、填空题：

本大题共4小题，每小题5分，共20分．13．0.114．2x，3460xy（写其中一条直线方程即可）15．910π16．2,3四、解答题：本大题共6小题，共70分．17．【命题意图】本小

题主要考查三角恒等变换、三角函数及其性质等基础知识；考查运算求解能力、数学建模能力；考查化归与转化思想、函数与方程思想；导向对发展直观想象、逻辑推理、数学运算、数学建模等核心素养的关注；体现基础性和应用性，满

分10分．【解答】（1）由ABDx，则π0,3x，····················································1分因为π6CAEx，········

··········································································2分所以1πcos6ACx，····················································

··························3分2sinABx，·····························································································4分

高二数学参考答案（第2页共9页）所以1π3()sinπ232cossin6SxACABxx，π0,3x.······························5分（2）由

（1），3312sincossin22Sxxxx················································6分331cos2sin222xx················

····································7分23π2sin216x．···········································

············8分因为π0,3x，ππ5π2666x，，所以π2sin216x的取值范围是0,1，······9分所以Sx的最小值为23，此时π6x．···························

··························10分18．【命题意图】本小题主要考查随机变量的分布列与期望、条件概率与全概率公式等知识；考查数学建模能力、运算求解能力、逻辑思维能力；考查统计与概率思想、分类与整合思想；导向对发展逻辑

推理、数学运算、数学建模、数学抽象、数据分析等核心素养的关注；体现综合性和应用性，满分12分．【解答】设Ai=“第i天去A餐厅用餐”（i=1,2），Bj=“第j天去B餐厅用餐”（j=1,2），·····································

············································································1分则A1与B1对立，A2与B2对立．（1）依题意得，X=0,1,

2．··········································································2分121211310|1(1)346PXPBBPBPBB

，······························3分121212121211211)(||PXPABBAPABPBAPAPBAPBPAB，所以1313191(1)1353430PX，·

··················································4分121211312|355PXPAAPAPAA，·······························

············5分则X的分布列为：X012P16193015所以119131012630530EX．··························································6分（2）由全概率公式，得2

121121||PBPAPBAPBPBB13233(1)(1)353410，······························8分高二数学参考答案（第3页共9页）所以121211222131|435|3910PABPAPBAPABPBPB

，·····························9分所以121245|1|199PBBPAB，·······················································10分

所以1212||PABPBB，·········································································11分所以如果周同学第2天去B餐厅，那么第1天去B餐厅的可能性更大．···········

··12分解法二：（1）同解法一.···············································································6分（2）12121212

()()(|)5315PABPAPBA，···················································8分12211121()(|)()436PBBPBBPB

，···························································9分所以1212()()PABPBB，······················································

·····················10分因为12122()(|)()PABPABPB，12122()(|)()PBBPBBPB，所以1212(|)(|)PABPBB，·················································

·······················11分所以如果周同学第2天去B餐厅，那么第1天去B餐厅的可能性更大．·············12分19．【命题意图】本小题主要考查直线与直线、直线与平面、平面与平面的位置关系等基础知识；考查空间想象能力、逻辑思维能力、运算求解能力

；考查化归与转化思想、函数与方程思想；涉及的核心素养有直观想象、逻辑推理、数学运算等，体现基础性、综合性．满分12分．【解答】（1）设圆柱的高BB1=h，连接A1B交AB1于点E，连接DE，····························1分因为A1C//平面AB1D，平

面A1CB∩平面AB1D=DE，A1C平面A1CB，所以A1C//DE，·························································································3分又

因为E是A1B的中点，所以D是BC中点．················································4分所以12ABDABCSS△△，所以21112233ABCABDVShShV△△．····

···················································5分（2）如图，分别以CB，CA，1CC为x，y，z轴的正方向，建立空间直角坐标系Cxyz，则(0,2,0)A，(4,0,0)B，(2

,0,0)D，1(4,0,)Bh，高二数学参考答案（第4页共9页）所以1(4,2,)ABh，(2,2,0)AD，(2,0,0)CD．·······································

6分设平面AB1D的一个法向量为111(,,)xyzm，则11111420,220,xyhzxy取12z，得(,,2)hhm，·························

···············8分因为A1C到平面AB1D的距离即点C到平面AB1D的距离，所以||4||3CDmm，即222434hhh，解得4h，·······································9分所以

(4,4,2)m，因为1(0,0,2)CC，所以111||221|cos,|3||||62CCCCCCmmm，···················

·····························11分所以直线CC1与平面AB1D所成角的正弦值为13．··········································12分解法二：（1）同解法一；·················

···························································5分（2）如图，分别以CB，CA，1CC为x，y，z轴的正方向，建立空间直角坐标系Cxyz，则(0,2,0)A，(4,0,0)

B，(2,0,0)D，1(4,0,)Bh，所以1(4,2,)ABh，(2,2,0)AD，(2,0,0)CD．·······································6分设平面AB

1D的一个法向量为111(,,)xyzm，则11111420,220,xyhzxy取12z，得(,,2)hhm，········································8分因为A1C到平面AB1D的距离即点C到平面AB1D的距离，高二数学参

考答案（第5页共9页）所以||1||CDmm，即222434hhh，解得4h，········································9分因为CC1//BB1，所

以直线CC1与平面AB1D所成角与直线BB1与平面所成角相等，设为θ．··············10分因为D是BC的中点，所以点B到平面AB1D的距离d与C到平面AB1D的距离相等，即43d．所以1413sin4

3dBB，所以直线CC1与平面AB1D所成角的正弦值为13．··········································12分20．【命题意图】本小题主要考查数列的通项、数列的单调性和最值、数列求和等基础知识；考查逻辑

思维能力、运算求解能力；考查化归与转化思想、函数与方程思想等；导向对发展数学抽象、逻辑推理、数学运算等核心素养的关注；体现基础性、综合性和创新性，满分12分．【解答】（1）依题意，10b，·········································

····························1分1210nnbbn，··································································

··················2分于是当2n≥时，111111210nniiinibbbbi2422101nn21110nn．········································

·······································5分即21110nbnn，又10b也符合上式，所以21110nbnn．·············································

···6分（2）由（1）可知1110nnnbaann，·············································7分当29n≤≤时，0nb，即1nnaa，当11

n≥时，0nb，即1nnaa，······························································10分当1n或10时，0nb，即1nnaa

，·························································11分所以na取得最小值时10n或11．······························································12分21

．【命题意图】本小题主要考查双曲线的标准方程和简单几何性质，直线与双曲线的位置关系等基础知识；考查运算求解能力，逻辑思维能力，空间想象能力和创新能力等；考查数形结合思想，函数与方程思想，化归与转化思想；导向对发展直观想象，逻辑推理，高二数学参考答案（第6页共9页）数学运

算等核心素养的关注；体现综合性与创新性，满分12分．【解答】（1）因为(1,1)P在C的渐近线byxa上，所以ab，···························1分因为(,0)Aa，所以PAO△的面积为122a，········

··········································2分解得1a，所以1b，······························································

················3分所以C的方程为221xy.··········································································4分（2）当直线l的斜率不存在时，不符合题

意，舍去；当直线l的斜率存在时，设直线l的方程为1(1)ykx，11(,)Mxy，22(,)Nxy，由221(1),1,ykxxy得222(1)2(1)220kxkkxkk，·······························5分222

24(1)4(1)(22)88kkkkkk，由210,0,k得1k且1k，则1221kxxk，2122221kkxxk．·······························

·····························6分直线AM的方程为11(1)1yyxx，令2xx，得1221(1)(,)1yxGxx，·····················

················································7分因为H为NG的中点，所以12212(1)1(,)2yxyxHx，所以122112212(1)1121211AH

yxyxyykxxx，···············································8分因为1212121212(1)1(1)1112111

111yykxkxkxxxxxx，·····················9分又1212121221111()1xxxxxxxx··········

···················································10分22221222111kkkkkkk22k，·······················

····················································11分所以1AHk，所以直线AH的斜率为定值．···································

··································12分22.【命题意图】本小题主要考查导数及其应用、函数的单调性、零点等基础知识；考查运算求解能力、逻辑思维能力、空间想象能力和创新能力等；考查数形结合思想、函数与方程思想、化归与转化思想、

分类与整合思想；导向对发展数学抽象、直观想象、逻辑推理、数学运算等核心素养的关注；体现综合性与创新性，满分12分．高二数学参考答案（第7页共9页）【解答】解法一：（1）由1()()e,xfxxaxb，···············

······································1分当ab≥时，()0fx，故()fx在(,)b上单调递减；·····································2分当

ab<时，令()0fx，得xa=，故当(,)xaÎ-¥时，()0fx，则()fx在(,)a-¥上单调递减；当(,)xabÎ时，()0fx，则()fx在(,)ab上单调递增．···································

4分综上所述，当ab<时，()fx在(,)a-¥上单调递减，在(,)ab上单调递增；当ab≥时，()fx在(,)b上单调递减．（2）设切点为()11,Qxy，因为1exy，所以切线的斜率为11ex-；则切线方程为()1111exyyxx--=-，················

··············································5分因为切线经过(,)Pmm，所以()11111(e1)exxmmx---+=-，即()1111(1)e1013xxmmx---+-=-<<，若过点P可以作两条直线与曲线()1e

113xyx-=+-<<相切，则上述关于1x的方程至少有两个不等的实根．令()1(1)e1xGxxmm-=--+-，13x，·················································6分由（1）取3b可得，当3m≥时，()Gx在(1,3)

-上单调递减，故()Gx在(1,3)-至多1个零点，不合题意，舍去；··········································7分当13m-<<时，()Gx在(1,)m-上单调递减；()Gx在(,3)m上单调递增，故()Gx有最小值()Gm，()1e1mGmm-=-+-，

以下证明()0Gm.事实上，设1()e1xhxx，13x，则1()e1xhx，由()hx单调递减，且10h，所以当11x时，()0hx，()hx单调递增，当13x时，()0hx，()hx单调递减，所以()(1)0hxh≤，即()0Gm．

因此()Gx存在两个零点，当且仅当(1)0,(3)0,GG由()22(2)e10,2e10,13,mmmmm-ìï--+->ïïï-+->íïïï-<<ïî解得2222e22e1e1e1m+-<<--，此时

过点P可以作两条直线与曲线()1e113xyx-=+-<<相切；·······················9分当1m≤-时，()Gx在(1,3)-上单调递增，故()Gx在(1,3)-至多1个零点，

不合题意，舍去．高二数学参考答案（第8页共9页）综上所述，m的取值范围是2222e22e1,e1e1+-÷ç÷ç÷ç÷ç--．················································10分（ii）m的取值范围是2

222e22e1,e1e1+-÷ç÷ç÷ç÷ç--．·······················································12分解法二：（1）同解法一；···················································

·························4分（2）设切点为()11,Qxy，因为1exy，所以切线的斜率为11ex-；则切线方程为()1111exyyxx--=-，···············

···············································5分因为切线经过(,)Pmm，所以()11111(e1)exxmmx---+=-，即()1111(1)e1013xxmmx---+-=-<

<，若过点P可以作两条直线与曲线()1e113xyx-=+-<<相切，则上述关于1x的方程至少有两个不等的实根．显然11x=不是该方程的实根，所以关于x的方程()11e111exxxm---+=-在(1,1)(1,3)-U上至少有两个不等的实数根．··················

·····························································6分令()()11e11,[1,1)(1,3]1exxxGxx---+=Î--U，则1121ee1

(),(1,1)(1,3)1exxxxGxx，···················································7分令()1e1,(1,1)(1,3)xgxxx-=+-Î-U，而1()e1,(1,1)(1,3)

xgxx，所以当(1,1)xÎ-时，()0gx，故()gx在(1,1)-上单调递减，当(1,3)xÎ时，()0gx，故()gx在(1,3)上单调递增，所以()()10gxg>=，

所以当(1,1)(1,3)xÎ-U时，()0Gx，所以()Gx在(1,1)-单调递增，在(1,3)单调递增，············································8分()Gx图象如下

图所示：因为e1(0)e1G，222e1(3)e1G，(0)(3)GG，高二数学参考答案（第9页共9页）22e2(1)e1G，(2)1G，(1)(2)GG，所以当2222e2

2e1e1e1m+-<<--时，关于x的方程()11e111exxxm---+=-在(1,1)(1,3)-U上有两个不等的实数根，此时过点P可以作两条直线与曲线()1e113xyx-=+-<<相切，所以m的取值范围是2222e22e1,e1e1+-÷ç÷ç÷ç÷ç--

．························································10分（ii）m的取值范围是2222e22e1,e1e1+-÷ç÷ç÷ç÷ç--．··················

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