2018年6月广西高中学业水平考试-数学(附答案)

俯视图正视图侧视图（第3题图）2018年6月广西壮族自治区普通高中学业水平考试数学（全卷满分100分，考试时间120分钟）注意事项：1．答题前，考生务必将自己的姓名、座位号、考籍号填写在答题卡上．2．考生作答时，请在答题卡上作答（答题注意事项见答题卡），在本试题上作答无效．一、选择题：本大题共20小题，每小题3分，共60分．在每小题给出的四个选项中，只有一项是符合题目要求的．1．已知集合A={0，1，2}，则A．0AB．1AC．2AD．3A2．2的角度数是A．30B．60C．90D．1003．某几何体的三视图如图所示，则该几何体是A．圆锥B．圆柱C．棱柱D．棱锥4．已知i是虚数单位，那么312ii=A．23iB.4iC.42iD.43i5．在平面直角坐标系中，指数函数2xy的大致图象是A．B.C.D.6．圆22121xy的半径长等于A．2B．3C．2D．17．已知向量21=，a，02=，b，则+ab＝A．23，B．02，C．03，D．26，8.在程序框图中，下列图形符号表示流程线的是9．在平面直角坐标系中，不等式yx≥表示的平面区域是A．B.C.D.10．下列函数中，是对数函数的是A．2logyxB．1yxC．sinyxD．2yx11．一商店为了研究气温对某冷饮销售的影响，对出售的冷饮杯数y（杯）和当天最高气温x（C）的数据进行了统计，得到了回归直线方程ˆ1.0412yx．据此预测：最高气温为30C时，当天出售的冷饮杯数大约是A．33B．43C．53D．6312．直线30xy与直线10xy的交点坐标是A．35，B．12，C．53，D．45，13．直线21yx的斜率等于A．4B．2C．3D．414．“同位角相等”是“两直线平行”的A．充分而不必要条件B．必要而不充分条件C．充要条件D．既不充分也不必要条件15．已知函数32fxxx，那么2f=A．20B．12C．3D．1A．B．C．D．16．已知函数sin203yAxA的部分图象如图所示，那么AA．6B．3C．1D．217．在△ABC中，已知角A，B，C的对边分别为a，b，c.若1a，2c，30A，则角C=A．15ﾟB．45ﾟC．75ﾟD．90ﾟ18．已知函数yfx的图象如图所示，那么方程0fx在区间ab，内的根的个数为A．2B．3C．4D．519．椭圆221259xy的两个焦点的坐标分别为A．5335，，，B．5353，，，C．4040，，，D．3535，，，20．已知3cos2，且0，那么sin2A．12B．22C．32D．1（第18题图）（第16题图）（第16题图）（第24题图）二、填空题：本大题共6小题，每小题2分，共12分．21．如图，①②③④都是由小正方形组成的图案，照此规律，图案⑤中的小正方形个数为.22．在△ABC中，AB=a，AC=b，若0g=ab，则△ABC是三角形（填“钝角”、“直角”或“锐角”）.23．等比数列1，2，4，8，…的公比q=.24．如图是正方形及其内切圆，向正方形内随机撒一粒“豆子”，它落到阴影部分的概率是.25．函数221fxxx在区间03，上的最大值是.26．设双曲线C：2213yx的左、右焦点分别为1F、2F，P是双曲线C右支上一点，若25PF，则△PF1F2的面积为.三、解答题：本大题共4小题，共28分．解答应写出文字说明、证明过程或演算步骤．27．（本小题满分6分）在我国，9为数字之极，寓意尊贵吉祥、长久恒远，所以在许多建筑中包含了与9相关的设计.某小区拟修建一个地面由扇环形的石板铺成的休闲广场(如图)，广场中心是一圆形喷泉，围绕它的第一圈需要9块石板，从第二圈开始，每一圈比前一圈多9块，共有9圈.问：修建这个广场共需要多少块扇环形石板？①②③④⑤（第21题图）→→（第27题图）28．（本小题满分6分）某商场在“五一”促销活动中，为了了解消费额在5千元以下（含5千元）的顾客的消费分布情况，从这些顾客中随机抽取了100位顾客的消费数据（单位：千元），按01，，12，，23，，34，，45，分成5组，制成了如图所示的频率分布直方图.现采用分层抽样的方法从01，和23，两组顾客中抽取4人进行满意度调查，再从这4人中随机抽取2人作为幸运顾客，求所抽取的2位幸运顾客都来自23，组的概率.29．（本小题满分8分）在三棱柱111ABCABC中，已知底面ABC是等边三角形，1AA底面ABC，D是BC的中点.（1）求证：1ADBC；（2）设12AAAB，求三棱锥11BADC的体积.（参考公式：锥体体积公式13VSh，其中S为底面面积，h为高.）（第28题图）（第29题图）30．（本小题满分8分）已知函数()xfxxe，其中2.71828e…为自然对数的底数.（1）求曲线()yfx在点()()11f，处的切线方程；（2）证明：()ln1fxx＞.2018年6月广西壮族自治区普通高中学业水平考试数学参考答案及评分标准一、选择题（共20小题，每小题3分，共60分）题号12345678910答案ACBDADACAA题号11121314151617181920答案BBBCBDDBCC二、填空题（共6小题，每小题2分，共12分）21．25；22．直角；23．2；24．4；25.2；26．46.三、解答题（共4小题，共28分）27．（本题满分6分）解法一：设从第1圈到第9圈石板数所构成的数列为na，由题意知，na是等差数列，·······································································1分其中19a，公差9d.···············································································2分99(91)9a81，·················································································3分数列na的前9项和199()92aaS··························································································4分(981)92························································································5分=405.答：修建这个广场共需要用405块扇环形石板.····················································6分解法二：依题意，广场从第1圈到第9圈所需的石板数依次为9，18，27，…，81.······················································································3分第1圈到第9圈的石板数之和99182781S··················································································4分99812····························································································5分405.所以，修建这个广场共需要扇环形石板405块.····················································6分28．（本题满分6分）解：根据频率分布直方图，01，组的顾客有1000.1010人，··············1分23，组的顾客有1000.3030人.···············2分用分层抽样的方法从两组顾客中抽取4人，则从01，组抽取1人，记为A；从23，组抽取3人，分别记为123BBB,,.·················································3分于是，从这4人中随机抽取2人的所有可能结果为1AB，2AB，3AB，12BB，13BB，23BB共6种.························································4分设所抽取的2人都来自23，组为事件C，所包含的结果为12BB，13BB，23BB共3种.···························································································5分因此，所抽取的2位幸运顾客都来自23，组的概率31()62PC.···········6分29．（本题满分8分）（1）证明：在三棱柱111ABCABC中，由1AA平面ABC，知1CC平面ABC.∵AD平面ABC，∴1ADCC.·························································1分△ABC是等边三角形，D是BC的中点，∴ADBC.·························································2分又1CC∩BCC，∴AD平面11BCCB.··············································3分又1BC平面11BCCB，∴1ADBC.···························································4分（2）解法一：在三棱柱111ABCABC中，由1AA平面ABC，知1BB平面ABC.∵111111122222BCDSBCBB，，···········································6分∴11111113BADCABCDBCDVVSAD································································7分23.33AD（第28题图）（第29题图）∴三棱锥11BADC的体积为233.································································8分解法二：在三棱柱111ABCABC中，由1AA平面ABC，知1CC平面ABC.∵1111ABCABCABCVSAA，111113AABCABCVSAA，·······································5分1111111326BABDCADCABCABCVVSCCSAA，···········································6分∴1111111111BADCABCABCAABCBABDCADCVVVVV··············································7分112333ABCSAA.∴三棱锥11BADC的体积为233.·························································