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20．（本小题满分13分）若抛物线C的顶点在坐标原点O，其图象关于x轴对称，且经过点M(1,2)．

（Ⅰ）若一个等边三角形的一个顶点位于坐标原点，另外两个顶点在该抛物线上，求该等边三角形的边长；

（Ⅱ）过点M作抛物线C的两条弦MA,MB，设MA,MB所在直线的斜率分别为k1,k2, 当k1,k2变化且满足k1?k2??1时，证明直线AB恒过定点，并求出该定点坐标．

121．（本小题满分14分）已知函数f1(x)?x2，f2(x)?alnx（其中a?0）．

2（Ⅰ）求函数f(x)?f1(x)?f2(x)的极值；

1（Ⅱ）若函数g(x)?f1(x)?f2(x)?(a?1)x在区间(,e)内有两个零点，求正实数a的取值

e范围；

31（Ⅲ）求证：当x?0时，lnx?2?x?0．（说明：e是自然对数的底数，e=2.71828…）

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资阳市2012—2013学年度高中三年级第二次高考模拟考试

数学（理工农医类）参考答案及评分意见

一、选择题：本大题共10个小题，每小题5分，共50分．

1－5. CABCB；6－10.DADCC.

二、填空题：本大题共5个小题，每小题5分，共25分．

11．3； 12．?316?； 13．12?；14．3?1；15．①②④． 33三、解答题：本大题共6个小题，共75分．解答要写出文字说明，证明过程或演算步骤．

16．解析 （Ⅰ）由3a?2csinA?0及正弦定理， 得3sinA?2sinCsinA?0（sinA?0）， ∴sinC?∴C?3，∵△ABC是锐角三角形， 2?3． ···················································································································· 6分

（Ⅱ）∵c?2，C??3，由余弦定理，a2?b2?2abcos?3?4，即a2?b2?ab?4．

······································································································································ 8分 ∴(a?b)2?4?3ab?4?3?(a?b2)，即(a?b)2?16， 2∴a?b?4，当且仅当a?b?2取“=”，故a?b的最大值是4． ······························· 12分 17．解析 （Ⅰ）记“A、B两小区已经收到邀请的人选择“幸福愿景”座谈会的人数相等”为

3135113事件A，则P(A)?(1?)?C30()3??C3． ······················································ 4分 ()?424216（Ⅱ）随机变量?的可能值为0，1，2，3，4．

311； P(??0)?(1?)?(1?)3?42323136113； P(??1)??(1?)3?(1?)?C3()?42423233112113； P(??2)??C3()?(1?)?C32()3?424232313110； P(??3)??C32()3?(1?)?()3?424232313．（每对一个给1分） ··························································· 9分 P(??4)??()3?4232?的分布列如下：

? 0 1 2 3 4 1612103 P 3232323232 ······································································································································ 10分

16121039∴?的数学期望E??0??1??2??3??4?··························· 12分 ?． ·

32323232324

18．（Ⅰ）证明：取AB的中点M，?AF?AB，

4?F为AM的中点，又?E为AA1的中点，∴EF//A1M， 在三棱柱ABC?A1B1C1中，D,M分别为A1B1,AB的中点， ?A1D//BM，且A1D?BM，

1则四边形A1DBM为平行四边形，?A1M//BD， ?EF//BD，又?BD?平面BC1D，EF?平面BC1D，

··································································································· 6分 ?EF//平面BC1D． ·

??????????????（Ⅱ）连接DM，分别以MB、MC、MD所在直线为x轴、y轴、z轴，建立如图空间直角坐标系，则B(1,0,0)，E(?1,0,1)，D(0,0,2)，C1(0,3,2)， ?????????????∴BD?(?1,0,2)，BE?(?2,0,1)，BC1?(?1,3,2)． 设面BC1D的一个法向量为m?(x1,y1,z1)，面BC1E的一个法向量为n?(x2,y2,z2)， A1????????x1?2z1?0,?m?BD?0,则由?????得?取m?(2,0,1)， ??x?3y?2z?0,m?BC?0,???111?1????E?2x?z?0,??n?BE?0,22??又由?????得?取n?(1,?3,2)， ????x2?3y2?2z2?0,?n?BC1?0,?C1B1DCm?n410??则cos?m,n??，

|m||n|55?8故二面角E－BC1－D的余弦值为AFMB10． ··································································· 12分 519．解析 （Ⅰ）由2an?1?3Sn?3n?4，得2an?3Sn?1?3n?1（n?2）， 两式相减得2an?1?2an?3(Sn?Sn?1)?3，即2an?1?an?3， ····································· 2分

13∴an?1??an?，

221则an?1?1??(an?1)（n?2）， ················································································ 4分

237由a1?t，又2a2?3S1?7，则a2??t?，

2237?t??1a?112?2??，∴t?2． 又∵数列{an?1}是等比数列，∴只需要2a1?1t?121此时，数列{an?1}是以a1?1?1为首项，?为公比的等比数列． ························· 6分

2111（Ⅱ）由（Ⅰ）得，an?1?(a1?1)?(?)n?1?(?)n?1，∴an?(?)n?1?1，·············· 8分

22211bn??[(?)n?1?1]???n2??(?)n?1?n2，

2211由题意得b2n?1?b2n，则有?(?)2n?2?(2n?1)2??(?)2n?1?(2n)2，

2211即?(?)2n?2[1?(?)]?(2n?1)2?(2n)2，

22(4n?1)?4n∴???， ···································································································· 10分

6

(4n?1)?4n(4n?1)?4n(4?1)?4*而?对于n?N时单调递减，则?的最大值为???2，

666故???2． ··················································································································· 12分 20．解析 （Ⅰ）根据题意，设抛物线C的方程为y2?ax(a?0)，点M(1,2)的坐标代入该方程，得a?4，故抛物线C的方程为y2?4x． 2分

设这个等边三角形OEF的顶点E，F在抛物线上，且坐标为(xE,yE)，(xF,yF)．

22则yE?4xE，yF?4xF，又|OE|?|OF|， 222222∴xE，即xE?yE?xF?yF?xF?4xE?4xF?0， ∴(xE?xF)(xE?xF?4)?0，因xE?0，xF?0， ∴xE?xF，即线段EF关于x轴对称．

则?EOx?30?，所以

yE3?tan30??， xE32即xE?3yE，代入yE?4xE得yE?43，

故等边三角形的边长为83． ····················································································· 6分 （Ⅱ）设A（x1,y1)、B（x2,y2)，则直线MA方程y?k1(x?1)?2，MB方程y?k2(x?1)?2，

??y?k1(x?1)?2,联立直线MA方程与抛物线方程，得?2消去x，

y?4x,??得k1y2?4y?8?4k1?0，

4∴y1??2， ①

k14同理y2??2， ②

k2y2?y1y12y2?y1(x?)， 而AB直线方程为y?y1?(x?x1)，消去x1，x2，得y?y1?22y2y14x2?x1?44yy4x?12 ③化简得即y? y1?y2y1?y2k?k2(k1?k2)?446?4，y1y2?4[??1]?4(?1)， 由①、②，得y1+y2＝4?12?4?k1k2k1k2k1k2k1k2k1k2代入③，整理得k1k2(x?y?1)?6?y?0． ?x?y?1?0,?x?5,由?得?故直线AB经过定点(5，－6). ········································ 13分

y?6?0,y??6.??121．解析 （Ⅰ）f(x)?f1(x)?f2(x)?ax2?lnx，

211∴f?(x)?axlnx?ax?ax(2lnx?1)（x?0，a?0），

22由f?(x)?0，得x?e?12?12，由f?(x)?0，得0?x?e，

?12?12故函数f(x)在(0,e)上单调递减，在(e,??)上单调递增， 所以函数f(x)的极小值为f(e)??（Ⅱ）函数g(x)??12a，无极大值． ·············································· 4分 4e12x?alnx?(a?1)x， 2ax2?(a?1)x?a(x?a)(x?1)则g?(x)?x??(a?1)?， ?xxx

令g?(x)?0，∵a?0，解得x?1，或x??a（舍去）， 当0?x?1时，g?(x)?0，g(x)在(0,1)上单调递减； 当x?1时，g?(x)?0，g(x)在(1,??)上单调递增．

1函数g(x)在区间(,e)内有两个零点，

ea?12e?1??1??a?0,a?,?12??2e2g()?0,e2e?2e?e???1?1?只需?g(1)?0,即??a?1?0,∴?a?,

2?g(e)?0,?2?2??e?2e?e2,???(a?1)e?a?0,?a?2e?2?2?2e?11故实数a的取值范围是(2······································································ 9分 ,)． ·

2e?2e2x2312（Ⅲ）问题等价于xlnx?x?．由（Ⅰ）知f(x)?x2lnx的最小值为?．

e42ex23x(x?2)设h(x)?x?，h?(x)??得h(x)在(0,2)上单调递增，在(2,??)上单调递减．

e4ex43∴h(x)max?h(2)?2?，

e43e2?2e?16(3e?8)(e?2)143314??0， ∵??(2?)???2=?224e4e2ee442ee2x331∴f(x)min?h(x)max，∴x2lnx?x?，故当x?0时，lnx?2?x?0． ··········· 14分

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