湖北省沙市中学2019届高三高考冲刺卷(五)数学(理)试卷 下载本文

2016级高三五月理数冲刺卷(五)答案

1 D 2 B 3 C 4 A 5 B 6 D 7 C 8 B 9 A 10 A 11 C 12 13.

D 3

214.-122 15.(??,]U[2,??) 16.6?2

3

?a1?a1?4d?2217.(1)由已知,得:?..........................2分解得:

a?3d?15?1?a1?3.........................4分所以:an?4n?1.................5分,??d?4bn?2n?1...............6分

(2)令数列?bn?的前n项和为Sn.所以:

Tn?nb1?(n?1)b2??S1?S2??(2?22?分

?2bn?1?bn?b1?(b1?b2)??(2n?1)?(b1?b2??bn)........12

?Sn?(2?1)?(22?1)??2n)?n?2n?1?n?218.(1)证明:连接A1C.四边形ABCD是平行四边形?AB∥CD

∠CDA?60,AC⊥AB,?AC⊥CD,AD?2CD.AA1⊥平面ABC,?AA1⊥AB,?AA1⊥CD. ACAA1=A,?CD⊥平面ACC1A1,?CD⊥AC1........................3分

AC1CD=C,?AC1⊥平面A1B1CD.

四边形ACC1A1是菱形,?AC1⊥AC1.?平面ACC1⊥平面A1B1CD......................................................

..................4分 (2)

AA1⊥平面ABC,四边形ABCD为平行四边形,AD?2CD,AC⊥CD.

?建立以C为坐标原点,CD,CA,CC1分别为x,y,z轴的空间直角坐标系如图所示:.

因AB?CD?1,AD?2,AC?3,AA1?CC1?3.

则C(0,0,0),D(10,,0),A(0,3,0),C1(0,0,3),A1(0,3,3),A1D?(?1,?3,?3), AC?(0,?3,?3),AC,?3,0)...........................................111?(0......6分

设平面A1DC的法向量n?(x,y,z).

??n?A1D?x?3y?3z?0则?.取n?(0,1,?1)....................................

??3y?3z?0??n?AC1..8分

设平面A1DC1的法向量m?(a,b,c).

??m?A1D?a?3b?3c?0则?.取m?(3,0,1)................................10

m?AC??11??3b?0分

设二面角C?A1D?C1的平面角为?.则cos??m?nmn?2. 4?二面角C?A1D?C1的余弦值为2................................12分 4x19.解:(1)根据散点图判断,y?c?d适宜作为扫码支付的人数y关于活动推出天数x的

型. ........ .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... ............2分

(2)因为y?c?d,两边同时取常用对数,得:lgy?lgc?xlgd

x设lgy?v,?v?lgc?xlgd.....................................................

.......................3分

x?4,v?1.54,?xi2?140.

i?17?lgd??xv?7xviii?177?xi2?7xi?12?50.12?7?4?1.547??0.25.............................2140?7?428

.............4分

将样本中心点(4,1.54)代入v?lgc?xlgd,得lgc?0.54.?v?0.54?0.25x.

?lgy?0.54?0.25x............................................................

.......................................5分

?y关于x的回归方程为y?100.54?0.25x?100.54?(100.25)x?3.47?100.25. 把x?8代入上式,y?3.47?102?347.

8

使

3470. .....................................7分

(3)记一名乘客乘车支付的费用为?,则?的所有可能取值为2,1.8,1.6,1.4.

1P(??2)?0.1,P(??1.8)?0.3??0.15,

211P(??1.6)?0.6?0.3??0.7,P(??1.4)?0.3??0.05,........................

36..10分 ?的分布列为:? P 2 0.1 1.8 0.15 1.6 0.7 1.4 0.05 ?E(?)?2?0.1?1.8?0.15?1.6?0.7?1.4?0.05?1.66(元)....................12分

20.(1)依题意,F(1,,抛物线0)C的准线方程为x??1...........................1分

记点A,B到抛物线C的准线的距离分别为d1,d2.故AB+AF=AB?d1?d2?2,

故AB+AF的最小值为2......................................................

..................3分

(2)设直线MN的方程为x?my?1(m?0),

将x?my?1(m?0)代入y2?4x,整理得:y2?4my?4?0,................4分

?16m2?16>0,解得m<?1或m?1,y1?y2?4m,y1y2?4..............5分

y12y2242x1?x2?(my1?1)?(my2?1)?4m?2,x1x2????1........6分

44162FM?(x1?1,y1),FN?(x2?1,y2),

?FM?FN?(x1?1)(x2?1)?y1y2?x1x2?(x1?x2)?1?4?8?m2,

84故8?4m2?,解得m??.

93故直线MN的方程为3x?4y?3?0或3x?4y?3?0....................8分

又y2?y1?(4m)2?4?4?..9分

47...............................................3?直线NP的斜率为kNP?y1?y243???.

x2?x1y2?y17令g(x)?2lnx?x,g'(x)?2?x xy224x?y1y244(x?1)?y?(x?)?y2,?y?,即y?.

y2?y14y2?y1y2?y1令y?0,解得x?1,故直线NP过点(1,0)........................................11

故直线NP的方程为3x?7y?3?0或3x?7y?3?0............................

...........12分

21.(1)当k?0时,f(x)?(x?2x)lnx?213x?2x 3f'(x)?2(x?1)lnx?x2?x?(x?1)(2lnx?x)....................................

................1分

令g(x)?2lnx?x,则g'(x)?2?xx

当x?(0,2)时,g'(x)?0,当x?(2,??)时,g'(x)?0, ?g(x)在(0,2)单调递增,在(2,??)单调递减. ?当x?0时,g(x)?g(2)?2(ln2?1)?0.?f(x)在(0,1)单调递增,在(1,??)单调递减.

5?f(x)max?f(1)?............................................................

3...........5分

11(2)f(x)?(x2?2x)lnx?kx4?(5k?1)x3?2kx2?2x,0?x?4.

43f'(x)?2(x?1)lnx?kx3?(5k?1)x2?(4k?1)x?(x?1)[2lnx?kx2?(4k?1)x]. 令h(x)?2lnx?kx2?(4k?1)x.h'(x)?......6分

(2kx?1)(x?2)............................

x当k?0时,2kx?1?0,x?(0,2)时h'(x)?0,h(x)在(0,2)上递增, x?(2,4)时h'(x)?0,h(x)在(2,4)上递减.

又2kh(4)?2(ln2?1)?0,0?e2k?1,4k2k2kh(e)?4k?ke?4ke?e?4k?4ke2k?4k(1?e)?0.?h(e)?0.2k2k

(A).当h(x)max?h(2)?0,即ln2?1?k?0时,h(x)在(0,1)递增,(1,4)递减. 2此时x?1是h(x)在(0,4)内唯一极值点,且为极大值点............................

.8分

(B).当h(x)max?h(2)?0,即k?ln2?1时,h(x)在(0,2),(2,4)上分别存在唯一零点x1,x2.21若x1?1,即k??时,在(0,1)上,x?1?0,h(x)?0,f'(x)?0,3在(1,x2)上,x?1?0,h(x)?0,f'(x)?0, 在(x2,4)上,x?1?0,h(x)?0,f'(x)?0,