2016上海旅游高等专科学校自主招生数学模拟试题及答案

考单招——上高职单招网 1113P(??4)?C3?()3??.

2432因此其概率分布为：

P ? 0 3 321 10 322 12 323 6 324 1 32 ····························································································· 11分

所以在5月13日抵达灾区的队伍数?的数学期望为：

E?=0×

31012617+ 1× + 2× + 3×+ 4×=. 323232323247答：在5月13日抵达灾区的队伍数?的数学期望E?=. ························ 12

4分

解法二：设5月13日抵达灾区的队伍数为?，则?=0、1、2、3、4. ········· 7分 1113由已知有：P(??0)?(1?)2?(1?)?(1?)?；

22432P(??1)?10； 3211111111111122110P(??2)?C2?()2?(1?)?(1?)?C2??(1?)?[(1?)???(1?)]?C2?(1?)2???；

22422242422432111111111621P(??3)?C2?()2?[?(1?)?(1?)?]?C2?()?(1?)???；

2242422243211112P(??4)?C2?()2???.

24232因此其概率分布为：

P ? 0 3 321 10 322 12 323 6 324 1 32 ····························································································· 11分

所以在5月13日抵达灾区的队伍数?的数学期望为：

考单招——上高职单招网 E?=0×

31012617+ 1× + 2× + 3×+ 4×=. 323232323247答：在5月13日抵达灾区的队伍数?的数学期望E?=. ························ 12

4分

20．（Ⅰ）证法一：由an?1?2an?n?1可得an?1?(n?1)?2(an?n)，又a1?2，则a1?1?1，

∴数列{an?n}是以a1?1?1为首项，且公比为2的等比数列， ··················· 4分 则an?n?1?2n?1，∴an?2n?1?n.······················································ 6分 证法二：

an?1?(n?1)2an?n?1?(n?1)2an?2n???2，又a1?2，则a1?1?1，

an?nan?nan?n∴数列{an?n}是以a1?1?1为首项，且公比为2的等比数列， ··················· 4分 则an?n?1?2n?1，∴an?2n?1?n.······················································ 6分 （Ⅱ）解：∵bn?nnn?n ，∴bn?··································· 7分

2an?2n2an?2n2?Sn?b1?b2???bn?111?2?()2???n?()n?????① 22211111∴Sn?()2?2?()3???(n?1)()n?n?()n?1????② 2222211[1?()n]1111112?n?(1)n?1?1?(n?2)(1)n?1， 由①?②，得Sn??()2?()3???()n?n?()n?1?21222222221?21?Sn?2?(n?2)()n． ·································································· 9分

23n1n3nn?21n(n?2)[2n?(2n?1)]?Sn??2?(n?2)()???(n?2)()?，

2n?122n?12n?12(2n?1)?2nn?1时，Sn?3n3n；n?2时，Sn?； 2n?12n?13n3n． ?0.∴Sn?2n?12n?111n?3时，2n?Cn0?Cn???Cnn?1?Cnn?Cn0?Cn?Cnn?1?2n?1，则?Sn?

考单招——上高职单招网 综上：n?1或2时，Sn?分

21．（Ⅰ）∵f(x)?lnx?x2?x?2，其定义域为(0,??).

1分

3n；n?3时Sn?3n. ································ 122n?12n?11?2x2?x?1?(2x?1)(x?1)∴f?(x)??2x?1?. ··································· 2分 ?xxx∵x?0，∴当0?x?1时，f?(x)?0；当x?1时，f?(x)?0.

故函数f(x)的单调递增区间是(0,1)；单调递减区间是(1,??). ···················· 4分 （Ⅱ）由（Ⅰ）知，函数f(x)的单调递增区间是(0,1)；单调递减区间是(1,??). 当0?a?1时，f(x)在区间(0,a]上单调递增，f(x)的最大值

f(x)max?f(a)?lna?a2?a?2；

当a?1时，f(x)在区间(0,1)上单调递增，在(1,a)上单调递减，则f(x)在x?1处

取得极大值，也即该函数在(0,a]上的最大值，此时f(x)的最大值f(x)max?f(1)?2；

?lna?a2?a?2,0?a?1,∴f(x)在区间(0,a]上的最大值f(x)max?? ······················· 8分

2,a?1.?（Ⅲ）讨论函数f(x)与g(x)图象交点的个数，即讨论方程f(x)?g(x)在(0,??)上

根的个数.

该方程为lnx?x2?x?2?x3?(1?2e)x2?(m?1)x?2，即lnx?x3?2ex2?mx. 只需讨论方程

lnx··························· 9分 ?x2?2ex?m在(0,??)上根的个数，

x令u(x)?lnx(x?0)，v(x)?x2?2ex?m. x1?x?lnx1?lnxlnxx?因u(x)?，令u?(x)?0，得x?e， (x?0)，u?(x)?2xx2x1当x?e时,u?(x)?0；当0?x?e时,u?(x)?0. ∴u(x)极大?u(e)?，

e当x?0?时,u(x)?且以x轴为渐近线.

lnxlnx?0， 但此时u(x)?0，???； 当x???时,limu(x)?limx???x???xx

考单招——上高职单招网 如图构造u(x)?lnx的图象，并作出函数v(x)?x2?2ex?m的图象. x22①当m?e?即m?e?时，方程无根，没有公共点；

1e1e22②当m?e?即m?e?时，方程只有一个根，有一个公共点；

1e1e22③当m?e?即m?e?时，方程有两个根，有两个公共点． ·················· 12

1e1e分

22．（Ⅰ）令y?yx，解得x?，由0?x?1，解得y?0，

1?y1?x?1∴函数f(x)的反函数f(x)?x(x?0)． ········································· 2分 1?x则an?1?f?1(an)?an11??1. ，得·············································· 3分

1?anan?1an11． ······················ 4分 ?{}是以2为首项，l为公差的等差数列，故an?n?1an（Ⅱ）∵f?1(x)?x1(x?0)，∴[f?1(x)]??， ····························· 5分

(1?x)21?xn1?(x?n)， 2n?1(1?n)∴y?f?1(x)在点(n,f?1(n))处的切线方程为y?n2令x?0， 得bn?. ····························································· 6分

(1?n)2bn??2?22?n??(n?1)?(n?)???∴2?，

anan24∵仅当n?5时取得最小值，∴4.5??2?5.5，解之9???11.

∴?的取值范围为(9,11)． ····························································· 8分

xx1?x22x1?x2?[?]??g(x)?[f(x)?f(x)]?（Ⅲ）···· 9分 22，x?(0,1). ·21?x1?x1?x1?x1?x?1

考单招——上高职单招网 则xn?1?xn?xn(1?xn)?1?xn1，因0

1?xn11112?1 ······· 10?????22xn?14x?1?8?2422?2nxn?1(xn?1?xn)2xn?1?xn112?111∴?(xn?1?xn)?(xn?1?xn)(?)?(?)

xnxn?1xnxn?1xnxn?18xnxn?12?1111111(xn?1?xn)2(x1?x2)2(x2?x3)2?[(?)?(?)???(?)] ∴????8x1x2x2x3xnxn?1x1x2x2x3xnxn?1?2?1112?11················································· 12(?)?(2?)，

8x1xn?18xn?1分

1111?2，∴0?2??1 ∵x1?,xn?1?xn，∴?xn?1?1，∴1?xn?1xn?1223?1(xn?1?xn)(x2?x1)(x3?x2)2?112?125∴14?????(2?)???. ·········

x1x2x2x3xnxn?18xn?18816222分