贵阳市2013年高三适应性考试(一)参考答案及评分标准文科数学

文科数学试题参考答案第1页共5页[试卷免费提供]贵阳市2013年高三适应性监测考试（一）文科数学参考答案与评分建议2013年2月一、选择题题号123456789101112答案ADABCCBBDDBA二、填空题（13）102（14）52（15）1（16）33三、解答题（17）解：（Ⅰ）设数列}{na的首项为1a，公差为d，由题意知).6)(()2(,106411211dadadada······································3分解得123ab·····························································5分所以35nan.……………………………………………………………………6分（Ⅱ）35112284nannnb∴数列{bBnB}是首项为41，公比为8的等比数列，…………………………………9分所以;281881)81(41nnnS………………………………………12分文科数学试题参考答案第2页共5页（18）解：（Ⅰ）四天的发芽总数为33392646144，这四天的平均发芽率为144100%36%400············································6分（Ⅱ）任选两天种子的发芽数为,mn，因为mn，用(,)mn的形式列出所有的基本事件有：(26,33),(26,39),(26,46),(33,39),(33,46),(39,46)，所有基本事件总数为6。则事件A包含的基本事件为(33,46),(39,46)所以21()63PA··········································································12分（19）解：（Ⅰ）在Rt△ABC中，AB＝1，解：（Ⅰ）在RtABC中，1AB，60BAC∴3,2BCAC在RtACD中，2,60ACCAD∴23,4CDAD∴1111513223322222ABCDSABBCACCD·······3分则155332323V·························································6分（Ⅱ）∵,PACAF为PC的中点，∴AFPC··········································7分∵PA平面ABCD，∴PACD∵,ACCDPAACA，∴CD平面PAC，∴CDPC∵E为PD中点，F为PC中点，∴EF//CD，则EFPC·····························································11分∵AFEFF∴PC平面AEF·········································································12分文科数学试题参考答案第3页共5页（20）解：（Ⅰ）设焦距为2c，由已知可得直线l的方程为)(3cxy···························2分点1F到直线l：)(3cxy的距离d323,2.cc故··················4分所以椭圆C的焦距为4.········································································6分（Ⅱ）设112212(,),(,),0,0,AxyBxyyy由题意知直线l的方程为3(2).yx联立2222422223(2),(3)4330.1yxabybybxyab得………………………7分解得221222223(22)3(22),.33babayyabab………………………8分因为22122,2.AFFByy所以即2222223(22)3(22)2.33babaabab··················································10分得223.4,5.aabb而所以故椭圆C的方程为221.95xy·························12分（21）解：（Ⅰ）函数()fx的定义域为(0,).··········································1分求导数，得11'()axfxaxx.··················2分①若0a≤，则'()0,()fxfx是(0,)上的增函数，无极值；················3分②若0a，令'()0fx，得1xa.当1(0,)xa时，'()0fx，()fx是增函数；当1(,)xa时，'()0fx，()fx是减函数.···················5分所以当1xa时，()fx有极大值，极大值为11()ln1ln1faaa.文科数学试题参考答案第4页共5页综上所述，当0a≤时，()fx的递增区间为(0,)，无极值；当0a时，()fx的递增区间为1(0,)a，递减区间为1(,)a，极大值为ln1a······································6分（Ⅱ）因为1xe是函数()fx的零点，所以()0fe，即102ae，解得1e=2e2ea.·····················8分所以1()ln2fxxxe.因为2352235()0,()02222eefefe，所以3522()()0fefe·············10分由（Ⅰ）知，函数()fx在(2,)e上单调递减，所以函数()fx在区间3522(,)ee上有唯一零点，因此322xe.························12分（22）解：（Ⅰ）连结AC，因为OAOC，所以OACOCA，······························2分为CD为半圆的切线，所以OCCD，又因为ADCD，所以OC∥AD，所以OCACAD，OACCAD，所以AC平分BAD．·········4分（Ⅱ）由OACCAD知BCCE，······················································6分连结CE，因为ABCE四点共圆，BCED，所以coscosBCED，······································································································8分所以DECBCEAB，所以2BC．·······················································10分（23）解：（Ⅰ）2cos,2sin2.xy且参数0,2，所以点P的轨迹方程为22(2)4xy．··········································3分（Ⅱ）因为)4sin(210，所以2sin()104，所以sincos10，所以直线l的直角坐标方程为100xy．·6分文科数学试题参考答案第5页共5页法一：由（Ⅰ）点P的轨迹方程为22(2)4xy，圆心为(0,2)，半径为2.221012104211d，所以点P到直线l距离的最大值422.····10分法二：222cos2sin21022cos()4411d，当74，max422d，即点P到直线l距离的最大值422.···············10分（24）解：（Ⅰ）由26xaa≤得26xaa≤，∴626axaa≤≤，即33ax≤≤，∴32a，∴1a．·········································5分（Ⅱ）由（Ⅰ）知211fxx,令nfnfn，则124,211212124,22124,n2nnnnnnn≤≤∴n的最小值为4，故实数m的取值范围是4,．··························10分