【文档说明】广西钦州市2019-2020学年高二下学期期末教学质量监测数学（理）试题扫描版含答案.doc，共(7)页，2.573 MB，由小赞的店铺上传

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钦州市2020年春季学期教学质量监测参考答案高二数学（理科）一、选择题答案：（每小题5分，共60分）题号123456789101112答案BBCCDCBACBAD二、填空题答案：（每小题5分，共20分）13.(1,2)；14．5；15．2或3；

16．1283三、解答题：本大题共6题，共70分．解答应写出文字说明、证明过程或演算步骤．17．证明：要证511313只须证22511313······················································

················2分只须证525511323913···································································

4分只须证5539···························································································6分只须证5539···························

··································································8分因为5539成立所以511313····················

············································10分18．解：（1）所有感染病毒的小白鼠共有50只，其中注射疫苗的共有y只，∴1505yP···············································

······························································2分∴10,501040,yx402060,301040,mn············

·················4分∴22列联表如下：未感染病毒感染病毒总计未注射疫苗204060注射疫苗301040总计5050100··············································································

·······················6分（2）∵22100(20103040)1005016.6675050604063K····································10分∵16.6676.635∴有99%把握认为注射此种疫苗

对预防新型冠状病毒有效.························12分19．解：（1）由题意，样本产品为“一等品”的频率为0.350.250.6，所以样本产品为“一等品”的数量为200.612（件）.········································

·························5分（2）由题意，流水线上任取1件产品为“非一等品”的概率为82,205P···············7分设取到“非一等品”的件数为X由已知，25,5XB～，·····················

······································································9分故32352144(3)556253PXC，∴恰有3件产品为“非一等品

”的概率144625·······················································12分20.解：（1）（1）∵cosx，siny，··············································

···········2分∴1C的极坐标方程为sin2．·······································································4分（2）∵直线3C的极坐标方程为6πR∴124si

n6π，······································································································7分24sin()23,66···

·····················································································10分∴21423MN···························

························································12分21.解：（1）当1a时，()|2|1fxxx.①当1x时，原不等式可化为217,x解得[

3,1]x；②当12x时，原不等式可化为37,解得(1,2)x；···································4分③当2x时，不等式可化为217,x解得[2,4]x；综上，原不等式的解集为[3,4]························

·····················································6分（2）当[2,4]x时，()|2|||2||,fxxxaxxa·····························8分()fx

x恒成立，即||2xa恒成立，化得22,xaxmaxmin[2][2],xax解得42a，a的取值范围为[4,2].·································

·······12分22.解：(1)3a时,2()6ln,(1)1,fxxxf6()2,fxxx···············································

···························································3分切线斜率(1)264kf曲线yfx在点(1,(1))Af处的切线方程为：14(1)yx,曲线在点

A处的切线方程为450xy·····························5分（2）222()2(0)xaafxxxxx··········································

······························7分①当0a时,()0fx恒成立fx在(0,)单调递增,fx无最小值···································

·························9分②当0a时,由()0fx得xa或xa（舍）0,xa时,()0fx,fx在0,a单调递减,xa时,()0fx,fx在,a单调递增所以fx存在最小值,()lnha

faaaa，(0)a()1(ln1)lnahaa由()0ha得1a,易知()ha在(0,1)单调递增,在(1,)单调递减所以()ha的最大值为(1)1.h又()haM恒成立，M取值范围为[1,).·············

····································12分