辽宁省大连市2013届高三双基测试数学文试题

23．（本小题满分10分）选修4－4：坐标系与参数方程

在直角坐标系xoy中，以原点o为极点，x轴的正半轴为极轴建立极坐标系. 已知射线

?x?t?1,（t为参数）,相交于A,B两点. l:??与曲线C:?2y?(t?1),4?（Ⅰ）写出射线l的参数方程和曲线C的直角坐标系方程； （Ⅱ）求线段AB的中点极坐标.

? 24．（本小题满分10分）选修4－5：不等式选讲

已知实数t，若存在t?[,3]使得不等式t?1?2t?5?x?1?x?2 成立，求实数x的取值范围． ．

122013年大连市高三双基测试

数学（文科）参考答案与评分标准

说明：

一、本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制订相应的评分细则．

二、对解答题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应得分数的一半；如果后继部分的解答有较严重的错误，就不再给分．

三、解答右端所注分数，表示考生正确做到这一步应得的累加分数． 四、只给整数分数，选择题和填空题不给中间分． 一．选择题

1.A；2.B；3.C；4.B；5.A；6.B；7.D；8.C；9.B；10.C；11.A；12.D. 二、填空题

x2 13．24?；14．242.8；15．?y2?1；16．3n.

4三.解答题

17.解：（Ⅰ）依题意得sin2A?sin2B=2sinAsinC?sin2C， ··············································· 2分

由正弦定理得：a?b?∴a?c?b?22222······································································· 4分 2ac?c2． ·

2ac．

a2?c2?b22?由余弦定理知：cosB?，∴B?. ·············································· 6分 ?2ac24（Ⅱ）∵sinA?又B?23,∴sinA?,∴A?B. ······························································ 8分

25?4,∴A??4,∴cosA?4， ·············································································· 10分 5∴cosC?cos(3?3?3?2. ······························· 12分 ?A)?coscosA?sinsinA??4441018．解：（Ⅰ）由频率分布表可知，样本容量为n，由

2＝0.04，得n＝50． ···· 2分 n25y14∴x?··················· 4分 ?0.5，y?50?3?6?25?2?14，z???0.28． ·

50n50（Ⅱ）记样本中视力在（3.9，4.2]的3人为a,b,c，在（5.1，5.4]的2人为d,e． 由题意，从5人中随机抽取两人，所有可能的结果有：?a,b?，?a,c?，?a,d?，?a,e?，?b,c?，

?b,d?，?b,e?，?c,d?，?c,e?，?d,e?，共10种． ·················································· 7分

设事件A表示“两人的视力差的绝对值低于0.5”，则事件A包含的可能的结果有：?a,b?，

?a,c?，?b,c?，?d,e?，共4种． ·················································································· 9分

∴P(A)?422····················· 12分 ?．故两人的视力差的绝对值低于0.5的概率为． ·

105519．解:(Ⅰ)证明：∵四边形ABCD是平行四边形， ?ACB?900，∴?DAC?900.

∵PA?平面ABCD,DA?平面ABCD,∴PA?DA，

又AC?DA，ACIPA?A，

∴DA?平面PAC. ······································································································· 6分 (Ⅱ)设PD的中点为G，在平面PAD内作GH?PA于H，

则GH平行且等于

1······························································································· 8分 AD. ·

2连接FH，则四边形FCGH为平行四边形，

∴GC∥FH，∵FH?平面PAE，CG?平面PAE，

∴CG∥平面PAE，∴G为PD中点时，CG∥平面PAE. ··································· 10分 设S为AD的中点，连结GS，则GS平行且等于∵PA?平面ABCD，∴GS?平面ABCD，

11PA?， 2211. ········································································ 12分 ?VA?CDG?VG?ACD?SVACDGS?31220.解：（Ⅰ）函数

f(x)?lnx?ax2的定义域为(0,??)，

1?2ax2?1， ··················································································· 1分 ?f?(x)??2ax?xx2∴①当a?0时，f?(x)?0，所以函数f(x)?lnx?ax的增区间为(0,??)， ········ 3分

②当a?0时，若f?(x)?0有0?x?2a2a,若f?(x)?0有x?, 2a2a2a2a2所以函数f(x)?lnx?ax的减区间为(,??)，增区间为(0,)，

2a2a由①②得当a?0时，函数f(x)的增区间为(0,??)，当a?0时，函数f(x)的减区间为

2a2a·················································································· 6分 ,??)，增区间为(0,)． ·

2a2a?x2?41证明（Ⅱ）当a?时，f?(x)?，

4x8∴x?(0,2)时函数f(x)是增函数，x?(2,??)时函数f(x)是减函数， ······················ 8分

1∴函数f(x)的最大值为f(2)?ln2?，

21?f(1)??，

8在(2,??)取x?e4， (e828?4???28?f(1)， ·计算得f(e)?4?·························································· 10分 884（也可以选取其它有效值）．

∴f(e)?f(1)?f(2)，

?x?(0,2)时函数f(x)是增函数，x?(2,??)时函数f(x)是减函数， ∴存在x0?(2,e4)，使f(x0)?f(1)，

∴存在x0?(2,??)，使f(x0)?f(1)． ········································································ 12分

21.解（Ⅰ）设A(x1,4y1)，由对称性可得B(?x1,?y1)

x12y12将A(x1,y1)带入椭圆可得2?2?1，

abx12b(1?2)y1?y1y12b2a直线PA和PB斜率乘积············· 2分 ???2??2． ·x1?a?x1?ax12?a2x1?a2a2b21c211由直线PA和PB斜率乘积为?，所以2?，所以2?，

222aa所以椭圆M离心率为

22． ······························································································ 5分 222（Ⅱ）椭圆方程可化为x?2y?a，

?x2?2y2?a2a2k2a222联立?，可得x?，y?， ····································· 7分 221?2k1?2ky?kx?设O为坐标原点，则|OA|?2a(1?k)2|OC|?，同理可得21?2k22a2(1?1)k2． 21?2k1)2a(1?k)2k所以|AC|? ?221?2k1?2k22a2(1?3k4?6k2?3?a?4?a2?22k?5k?22?2311k2?2?2k?42·················································· 10分 a． ·

34a28当且仅当k??1时取等号，所以?，

33x2即a?2，所以椭圆M的方程为··························································· 12分 ?y2?1． ·

221)2a(1?k)2k (另解：所以|AC|??221?2k1?2k22a2(1?3(k2?1)23(k2?1)2422?a??a??a） 22222k?1?k?223(2k?1)(k?2)()2222．解: (Ⅰ) 连结OC，因为OA?OB,CA?CB,则OC?AB． ································· 2分