2019秋高三文数参考答案及评分细则
一、选择题：本大题共12小题，每小题5分，共60分.
1.答案：C，注意是求z的共轭复数.
2.答案：D，集合，故
3.答案：B，列举后容易知道，基本事件总数有10种，恰有2件检测过该成分含量的事件共有3种，所以所求概率为
4.答案：B，
5.答案：D，如下图，所围成的图形的面积，
6.答案：B，易知即
7.答案：C，①l不变，有l∥α且l⊥βα⊥β；②m不变，有m∥α且α⊥βm⊥β；③n不变，有α∥β且n⊥αn⊥β；分析知①，③正确.
8.答案：A，
9.答案：C，由对数运算的性质知，，所以l=n，又为增函数，时，，所以m>l,所以有
10答案：C，（1）错，（2）（3）（4）对
11.答案：A，假设C为钝角，则，，显然充分性不成立，又由可知，即，此时有，即A为钝角或B为钝角，从而△ABC为钝角三角形，必要性成立
12.答案：D，由知，令，则所以有
，即的图像关于直线对称.当时，；当时，。作出的图像可知，当时，有两个零点.
二、填空题（本大题共4小题，每小题5分，共20分）
13.答案：
14答案：
15. 答案:,
解析：由题意有:
故最小正周期为,最小值为.
16. 答案：
解析：设每天生产A药品x吨，B药品y吨，利润，则有作出可行域知，z在点处取得最大值.
三、解答题：本大题共6小题，共70
分。解答应写出必要的文字说明。证明过程或者演算步骤。第17-21题为必考题，每个试题考生必须作答，第22,23题为选考题，考生根据要求作答。
（一）必考题：共60分
17.解：（I）由直方图可知，用户所用流量在区间内的频率依次是0.1,0.15,0.2,0.25,0.15，········································································································3分
所以该月所用流量不超过3GB的用户占85%，所用流量不超过2GB的用户占45%，故k至少定为3；
·····································································································································6分
（II）由所用流量的频率分布图及题意，用户该月的人均流量费用估计为：
2×1×0.1+2×1.5×0.15+2×2×0.2+2×2.5×0.25+3×2×0.15+(3×2+0.5×4)×0.05+(3×2+1×4)×0.05+(3×2+1.5×4)×0.05=5.1元······················································································12分
18.解：（I）设等差数列的公差为d，因为，所以
又，所以d=2，即，································ ··································3分
设正项等比数列的公比为q，因为即，由，知，所以·················································································································6分
（II）······················································································8分
设，则
····································································································································12分
19. 解：（I）证明:如图,由直三棱柱知，··············································2分
又M为BC的中点知AM⊥BC,又,所以·······································4分
又AMÌ平面AMN,所以平面AMN⊥平面B1BCC1·······································································6分
(II)如图：设AB的中点为D,连接A1D,CD.因为△ABC是正三角形,所以CD⊥AB.由直三棱柱知CD⊥AA1.所以CD⊥平面A1ABB1,所以∠CA1D为直线A1C与平面A1ABB1所成的角.即∠CA1D=30°,···8分
所以A1C=2CD=2×=,所以A1D=6,在Rt△AA1D中，AA1=，NC=·······························································································10分
三棱锥的体积即为三棱锥的体积,所以V=···································································12分
20.解：（I）由，，知,··································2分
由kOM=知······································································································4分
所以,,所以e=.·································································6分
(II)证明:由N是AC的中点知,点N,所以，···············································8分
又，所以·····················································10分
由（I）知，即，所以=0，
即MN⊥AB. ····················································································································12分
21.解:(I)易知切线的斜率为2,即,又,所以a=1; ················· ·············4分
(II)设,
当时,.又所以存在,使得.······················································································································6分
又
··························································································8分
所以当时,
,
当[2,)时,即时,为增函数,所以时,方程在内存在唯一的根. ···············································································································12分
（二）选考题：共10分.请考生在第22,23题中任选一题作答，如果多做，则按所做的第一题计分.
22.解: （I）由知所以所以⊙C的直角坐标方程为··········································································································5分
(II)由(I)知⊙C的标准方程为,即圆心,设P点坐标为，则，所以当t=0时，|PC|有最小值，此时P点坐标为(6,0).·····························································································································10分
23.解：（I）由知，所以即；······························5分
（II）依题意知:
····························································8分
当且仅当即时等号成立，
所以所求式子的最大值为.··························································································10分
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