高一理科下学期期末考试数学试题

资阳市高一理科第二学期期末测试数学试题本试卷分选择题和非选择题两部分，第Ⅰ卷（选择题）1至2页，第Ⅱ卷（非选择题）3至8页，共8页。满分150分，考试时间120分钟。注意事项：1．答第Ⅰ卷前，考生务必将自己的姓名、考号、考试科目用铅笔涂写在答题卡上。2．每小题选出答案后，用铅笔把答题卡上对应题目的答案标号涂黑.如需改动，用橡皮擦干净后，再选涂其它答案，不能答在试题卷上。3．考试结束时，将本试卷和答题卡一并收回。第Ⅰ卷（选择题，共50分）一、选择题：本大题共10小题，每小题5分，共50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．sin15cos15A．12B．14C．32D．342．直线134xy与两坐标轴围成的三角形的周长为A．6B．7C．12D．143．下列向量中，与向量c(23)，不．共线的一个向量pA．(32)，B．3(1)2，C．2(1)3，D．11()32，4．若0ab，则下列不等式成立的是A．baabB．abbaC．baabD．22baab5．已知等差数列{}na的首项11a，公差15d，则{}na的第一个正数项是A．4aB．5aC．6aD．7a6．若直线1:1lykx与2:10lxy的交点在第一象限内，则k的取值范围是A．1kB．11kC．11kk或D．1k7．如图，ABC的AB边长为2，PQ，分别是ACBC，中点，记ABAPBABQm，ABAQBABPn，则A．24mn，B．31mn，C．26mn，D．3mn，但mn，的值不确定8．如图，在5个并排的正方形图案中作出一个135nAOB（1,2,3,4,5,6n），则nA．1，6B．2，5C．3，4D．2，3，4，59．设01x，函数411yxx的最小值为A．10B．9C．8D．27210．已知{}na，{}nb都是等比数列，它们的前n项和分别为nnST，，且314nnnST对N*n恒成立，则11nnabA．3nB．4nC．3n或4nD．4()3n资阳市2013—2014学年度高中一年级第二学期期末质量检测数学第Ⅱ卷（非选择题，共100分）题号[来源:学,科,网Z,X,X,K]二[来源:学科网ZXXK]三总分[来源:学*科*网]总分人[来源:学科网][来源:学|科|网Z|X|X|K]161718192021得分注意事项：1．第Ⅱ卷共6页，用钢笔或圆珠笔直接答在试题卷上。2．答卷前将密封线内的项目填写清楚。二．填空题：本大题共5小题，每小题5分，共25分。11．不等式234xx的解集为__________．12．等比数列{}na中，223aa，48a，则nS___________．13．已知向量ab，满足||||||1abab，则||ab___________．14．若实数x，y满足线性约束条件3122xyxyx，则z2xy的最大值为________．15．给出以下结论：①直线12ll，的倾斜角分别为12，，若12ll，则12||90；②对任意角，向量1e(cossin)，与2(cos33cossin)sin，e的夹角都为3；③若ABC满足coscosabBA，则ABC一定是等腰三角形；④对任意的正数ab，，都有12abab．其中所有正确结论的编号是_____________．三．解答题：本大题共6个小题，共75分。解答应写出文字说明，证明过程或演算步骤。16．（本小题满分12分）如图，矩形OABC的顶点O为原点，AB边所在直线的方程为34250xy，顶点B的纵坐标为10．（Ⅰ）求OAOC，边所在直线的方程；（Ⅱ）求矩形OABC的面积．17．（本小题满分12分）已知向量a(1，2)，b(3，4)．（Ⅰ）求ab与ab的夹角；（Ⅱ）若a(ab)，求实数的值．18．（本小题满分12分）已知函数2()2sincos()42fxxx．（Ⅰ）求()fx的最小正周期；（Ⅱ）设(0)2，，且3()285f，求tan()4．19．（本小题满分12分）已知ABC的三个内角ABC，，成等差数列，它们的对边分别为abc，，，且满足:2:3ab，2c．（Ⅰ）求ABC，，；（Ⅱ）求ABC的面积S．20．（本小题满分13分）等差数列{}na中，11a，221nnaa（*nN），nS是数列{}na的前n项和.（Ⅰ）求nnaS，；（Ⅱ）设数列{}nb满足1212112nnnbbbaaa（*nN），求{}nb的前n项和nT．21．（本小题满分14分）已知2()fxxpxq，其中00pq，．（Ⅰ）当pq时，证明()()fqfppq；（Ⅱ）若()0fx在区间(01]，，(12]，内各有一个根，求pq的取值范围；（Ⅲ）设数列{}na的首项11a，前n项和()nSfn，*nN，求na，并判断{}na是否为等差数列？资阳市2013—2014学年度高中一年级第二学期期末质量检测数学参考答案及评分意见三．解答题：本大题共6个小题，共75分。解答应写出文字说明，证明过程或演算步骤。16．（Ⅰ）∵OABC是矩形，∴//OAABOCAB，．····································1分由直线AB的方程34250xy可知，34ABk，∴4334OAOCkk，，······························································4分∴OA边所在直线的方程为43yx，即430xy；·······································5分OC边所在直线的方程为34yx，即340xy．·······································6分（Ⅱ）∵点B在直线AB上，且纵坐标为10，∴点B的横坐标由3410250x解得为5，即(510)B，．·························7分22|030425|||534OA，22|4(5)310|||104(3)AB，···························11分∴||||50OABCSOAAB．····································································12分17．（Ⅰ）∵(1a，2)，(3b，4)，∴(2ab，6)，(4ab，2)，·························································2分∴(26)(42)202cos240204020abab，，，．······························5分∴34abab，．···········································································6分【另】22()()5252cos||||||||24020ababababababababab，，····5分∴34abab，．···········································································6分（Ⅱ）当()aab时，()0aab，····················································8分∴(12)(1324)0，，，则13480，∴1．··························12分【另】当()aab时，()0aab，····················································8分∴20aab，则5[1(3)24]0，∴1．······························12分18．（Ⅰ）2()2sin(coscossinsin)442fxxxx······································2分2211cos222(sincossin)2(sin2)2222xxxxx······························4分222(sin2cos21)(sin2cos2)222xxxxsin(2)4x．·······················································································6分∴()fx的最小正周期为．······································································7分（Ⅱ）3()sin[2()]sin282845f，············································8分由(0)2，可知，4cos5，3tan4．··················································10分∴3tantan144tan()7341tantan144．···················································12分19．（Ⅰ）∵ABC，，成等差数列，∴2ACB，又180ABC，∴60120BAC，，···············································2分由正弦定理sinsinsinabcABC可知，sinsinaAbB，∴2sinsin2sinsin602332AAA．··························································4分∵0120A，∴45A，12075CA．综上，456075ABC，，．······························································6分【另】∵ABC，，成等差数列，∴2ACB，又180ABC，∴60120BAC，，···············································2分设23akbk，，其中0k．由余弦定理可知，2222431cos2240622222kkBkkkk，∴2(31)6(31)ab，，∴2(31)6(31)2sinsin232AA，·······················································4分∵0120A，∴45A，12075CA，综上，456075ABC，，．······························································6分（Ⅱ）62sinCsin75sin(3045)4，·············································8分由22sin45sin60sin752362224abab，得2(31)6(31)ab，，······