广东省佛山市南海区桂城中学等七校联合体2019届高三冲刺模拟数学(文)试题含答案

（一）必考题：共60分。

17.【命题意图】本题考查解三角形的基础知识和基本运算，中等题． 【解】

2?，根据余弦定理：a2?b2?c2?2bccosA, tanA??3,A?(0,?),?A?（1）3得：19?9?c2?2?3?c?(?),根据正弦定理：

12c2?3c?10?0,c?2,c??5（舍去）， （3分）

ac19257?,?,?sinC?sinAsinC19 3sinC257； （6分） 19综上，c?2,sinC?（2）由sinC?357AD33,得出tanC?tanC?,?AD?△ADC，在直角中，， （9分） 419AC41?13313333（12分） ?S?ABD??AB?AD?sin??2???△ABD的面积为，即2624288 .

18．【命题意图】本题考查立体几何的线面位置关系，几何作图，直线与平面所成角的计算，中等题． 【解】

（1）过点M在平面A1B1C1D1内作一条直线B1D1即为所求． （2分） 理由如下：

连接C1M，在直角△CC1M中，可计算C1M?CM2?CC12?2. 又A1M?2，A1C1?22，所以点M是A1C1的中点，

所以B1D1?C1M，B1D1?CC1，C1MICC1?C1，所以B1D1?平面CC1M，进而可得B1D1?CM. （6分） （2）连接AC与BD交于点O，易证AC?平面BDD1B1， 所以直线CM在平面BDD1B1内的射影是OM，

所以?CMO就是直线CM与平面BDD1B1所成角. （9分） 在VCMO中，cos?CMO?OM26. ??CM36D1MA1B1C1DCA第18题图

B故直线CM与平面BDD1B1所成角的余弦值为19． 解:令t?lnx则y?bt?a.

6. （12分） 3- 9 -

b???ti?110i?t(yi?y)i???ti?110?t?2?24.2?5, 4.84i155.515.1i?1y???15.55,t???1.51

10101010i?1?y10i?t10a?y?bt?15.55?5?1.51?8，

所以y关于t的回归方程为y?5t?8,

因为t?lnx,从而y关于x的回归方程为y?5lnx?8.

(2)由(1)得该IT从业者在36岁时月平均收入为: y?5ln11?8?5?2.4?8?20 (千元) 旧个税政策下缴交的个人所得税为：

1500?3%?3000?10%?4500?20%?(20000?3500?9000)?25%?3120(元)

新个税政策下缴交的个人所得税为：

1500?3%?3000?10%?(20000?5000?3000?3000)?10%?990(元)

故根据新旧个税政策，该IT从业者在36岁时每个月少缴交的个人所得税为

3120?990?2130(元).

20.（1）【解析】：依题意设直线l1的方程为y?x?p， ··············································· 1分 22（x+1）+y2?2的圆心C2(?1，0)，半径r?2 ···························· 2分 由已知得：圆C2：

因为直线l1与圆C2相切，

p|p2所以圆心到直线l1:y?x?的距离d?··································· 3分 ?2． ·

2221?(?1)|?1?|?1?p|2?2，解得p?6或p??2（舍去）．： ··············································· 4分

即2所以p?6． ·································································································· 5分

xx2（m,?3)，（2）解法一：依题意设M由（1）知抛物线C1方程为x?12y，所以y?，所以y??，设A(x1,y1)，

126x则以A为切点的切线l2的斜率为k?1， ·································································· 6分

61所以切线l2的方程为y?x1(x?x1)?y1． ································································ 7分

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11令x?0，y??x12?y1=??12y1?y1=?y1，即l2交y轴于B点坐标为(0,?y1)， ·········· 8分

66所以MA?(x1?m,y1?3)， ····················································································· 9分 MB?(?m,?y1?3)，···························································································· 10分

∴MN?MA?MB=(x1?2m,6)， ·············································································· 11分 ∴ON?OM?MN?(x1?m,3). 设N点坐标为(x,y)，则y?3，

所以点N在定直线y?3上． ·················································································· 12分 （2）解法二：设M（m,?3)，由（1）知抛物线C1方程为x2?12y，① 设A(x1,y1)，以A为切点的切线l2的方程为y?k(x?x1)?y1②， 联立①②得：x2?12[k(x?x1)?12··································································· 6分 x1]， ·

12x1， 6因为?=144k2?48kx1?4x12?0，所以k=所以切线l2的方程为y?1······························································· 7分 x1(x?x1)?y1． ·

6令x?0，得切线l2交y轴的B点坐标为(0,?y1)， ····················································· 8分 所以MA?(x1?m,y1?3)， MB?(?m,?y1?3)， ······················································· 10分 ∴MN?MA?MB=(x1?2m,6) ················································································· 11分 ∴ON?OM?MN?(x1?m,3)， 设N点坐标为(x,y)，则y?3，

所以点N在定直线y?3上． ·················································································· 12分

21.【命题意图】本题考查导数的综合应用，函数单调性及函数零点问题，考查学生运算能力及综合运用数学知识分析问题、解决问题的能力，较难题． x2【解】（1）当k?1时，f(x)?xe??x，

2x?f?(x)?(x?1)ex?(x?1)?(x?1)(ex?1)，

故x?(??,?1),f/(x)?0,f(x)为增，

x?(?1,0),f/(x)?0,f(x)为减，

x?(0,??),f?(x)?0,f(x)为增，

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故函数f(x)的单调增区间为(??,?1)和(0,??)；单调减区间为(?1,0)． （4分）

f(x)极大=f(?1)?11?，f(x)极小?f(0)?0. （5分） 2exx2（2）解法一：由已知，f(x)?kxe??x，g(x)?kex?x

2x2x2xx?F(x)?kxe??x?ke?x?k(x?1)e?

22x?F?(x)?kxex?x?x(kex?1)

①当k?0时，F(x)在(??,0)为增，在(0,??)为减，

且注意到F(0)??k?0，函数F(x)的图像两边向下无限伸展， 故此时F(x)存在两个零点，适合题意。 ②当k?0时，F(x)在(??,0)为增，在(0,??)为减，

且F(0)?0，故此时F(x)只有一个零点。

③当k?1时，?F/(x)?xex?x?x(ex?1)，故函数(??,??)为增，易知函数F(x)只有一个零点 ④当k?(0,1)时，ln?0，F(x)在(??,0)为增，(0,ln)为减，(ln,??)为增，

且F(0)??k?0易知F(x)只有一个零点。

⑤当k?(1,??)时，ln?0，F(x)在(??,ln)为增，(ln,0)为减，(0,??)为增，

1(ln?1)2?11且F(ln)??k?0，F(0)??k?0易知F(x)只有一个零点。

k21k1k1k1k1k1k综上，k?(??,0)时，函数F(x)?f(x)?g(x)存在两个零点。 （12分） x2x2x2xx解法二：依题F(x)?kxe??x?ke?x?k(x?1)e??k(x?1)?2222ex来源:Zxxk.Com]

x2依题函数F(x)?f(x)?g(x)存在两个零点，即方程k(x?1)?2有两个根

2ex2也即直线y?k(x?1)与函数y?x的图像有两个交点

2ex2x(2?x)记h(x)?x?h?(x)?，

2e2ex由h?(x)?0?x(2?x)?0?0?x?2，由h?(x)?0?x(2?x)?0?x?0,x?2 故h(x)在(??,0)上单调递减，在(0,2)上单调递增，在(2,??)上单调递减 且h(0)?0，x?0时h(x)?0 又直线y?k(x?1)过(1,0)，斜率为k

x2由图像观察知：当k?0时直线y?k(x?1)与h(x)?x的图像必有两个交点，

2ex2当k?0时直线y?k(x?1)与h(x)?x的图像只有一个交点

2e- 12 -