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高二数学答案第1页（共5页）2021—2022青岛市期末考试高二数学答案学年度第二学期第二学段模块检测高二数学答案及评分标准一、单项选择题：本大题共8小题．每小题5分，共40分．BABCDABC二、多项选择题：本大题共4小题．每小题5分，共20分．9．AB；10．BC；11．BCD；12．ACD.

三、填空题：本大题共4小题，每小题5分，共20分．13．240；14．1；15．5(,]4；16．（1）14；（2）34.四、解答题：本大题共6小题，共70分，解答应写出文字说明、证明过程或演算步骤．17．（本小题满分10分）解：由题知：每瓶饮料的利润33224π()

0.20.8π0.8π()33rryfrrr，06r·································3分所以，2()0.8π(2)0.8π(2)frrrrr····························

··························4分令()0fr，解得2r················································································5分当(0,2)r时，()0fr，()fr在

(0,2)上单调递减········································6分当(2,6]r时，()0fr，()fr在(2,6]上单调递增··························

···············7分（1）因为(6)0f所以，当6cmr时，每瓶饮料的利润最大···············································8分（2）当2cmr时，每瓶饮料的利润最小······

···················································9分（3）由32()0.8π()0(06)3rfrrr，解得36r故，所求瓶子的半径取值范围是：3cm6cmr························

·············10分18.（本小题满分12分）解：（1）由题意，原始平均分450.1550.15650.20750.3850.20950.0570x···············3分（2）优秀等级最低分约为样本数据的80%分位数·······

·······································4分80分以下的学生所占的比例为10%15%20%30%75%90分以下的学生所占的比例为100%5%95%高二数

学答案第2页（共5页）所以，80%分位数一定位于[80,90)内·····························································5分由0.800.75801082.50.9

50.75可以估计优秀等级最低分约为82.5分·······························································6分（3）用分层抽样的方法在分数段为[60,80)的学生中抽取一个容量

为5的样本则分数段[60,70)中抽取的学生数为：0.020520.0200.030人·····························7分分数段[70,80)中抽取的学生数为：0.030530.0200.030人······························

··8分则从5人中任意抽取2人的样本空间的样本点个数为2510C································9分记事件“这2人中至多有1人在分数段[60,70)内”为C，记事件“这2人中有1人在分数段[60,70

)内”为1C，记事件“这2人中没有人在分数段[60,70)内”为2C，则12CCC，且1C与2C互斥····································································10分所以11022

323121222559()()()()10CCCCPCPCCPCPCCC··························12分19．（本小题满分12分）解：（1）根据22列联表：所以222()200(924968)1.4182.072()()

()()10010018812nadbcKabacbdcd····3分依据0.15的独立性检验，不能认为产品的包装合格与装流水线的选择有关联··········································

····4分（2）由题知：282220082212122200()14(|)33()CCCPABPABCCPBC·········································6分（3）由已知可得：14781065x·················

····································7分214243540235y······································································

·8分51124147248351040906iiixy········································9分52222221147810230ii

x·····························································10分高二数学答案第3页（共5页）由回归直线的系数公式，51522222222159065623216ˆ4.32(147810)56505iiiiix

yxybxx····················11分ˆ234.3262.92aybx所以ˆˆ4.322.92ybxax当20x（百件）

时，4.32202.9283.4883y件所以估计一小时生产2000件时的不合格品数约为83件······································12分20．（本小题满分12分）解：（1）因为()xfxea····

·······································································1分若0a，则()0fx，()fx在(,)上单调递增······················

·················2分若0a，令()0fx，解得lnxa······························································3分当(,ln)xa时，()0fx，()fx在(,

ln)a上单调递减························4分当(ln,)xa时，()0fx，()fx在(ln,)a上单调递增························5分（2）由（1）知：（法一）若0a，则()fx在(,)上单调递增，且(0)1f

，······················6分所以，当0x时，()1fx，不合题意························································7分若0a，因为()(0)1fxf，所以0(

0)0fea，解得1a·································································8分此时()xfxex，则()1xfxe·················

··········································9分当(,0)x时，()0fx，()fx在(,0)上单调递减；当(0,)x时，()0fx，()

fx在(0,)上单调递增；所以()(0)1fxf，符合题意·································································11分综上，1a·················

··············································································12分（法二）若0a，则()fx在(,)

上单调递增，且(0)1f，······················6分所以，当0x时，()1fx，不合题意························································7分若0a，则()(ln)ln

fxfaaaa所以ln1aaa，即11ln0aa···························································8分令1()1ln,0gaaaa，则22111()agaaaa

·································9分当(0,1)a时，()0ga，()ga在(0,1)上单调递增；当(1,)a时，()0ga，()ga在(1,)上单调递减；所以()

(1)0gag·················································································11分综上，1a·············

··················································································12分高二数学答案第4页（共5页）21．（本小题满分12分）解：（1）由题知：0123(

)01231.1EXpppp，·························3分（2）因为230123pppppppp························································

·····5分所以，p是方程230123ppxpxpxx的正实根··········································6分令230123()(0)fxppxpxpxxx，则2123()2

31fxppxpx令2123()231gxppxpx，所以23()260gxppx·····························7分所以()fx在区间[0,1]上单调递增又因为1(0)10fp，123(1)231()1fpp

pEX··························8分当()1EX时，(1)()10fEX存在(0,1)，使得()0f··············································

·····················9分当(0,)x时，()0fx，所以()fx在0,（）上单调递减；当(,1)x时，()0fx，所以()fx在,1（）上单调递增；·······················10分又因为00123(

0)0,()(1)10fpffpppp··························11分所以()fx在(0,)x上存在唯一零点xp，综上，所以p是方程230123ppxpxpxx的最小正实根··························

··12分22.（本小题满分12分）解：（1）由题知：21ln()xfxx···································································1分当xe时，()0fx··············

····································································2分当0xe时，()0fx；当xe时，()0fx·············

······························3分所以，()fx的单调递增区间为(0,]e，单调递减区间为[,)e·····························4分（2）22()|ln||()|Fxxxaxxfx·······················

·····································5分由（1）知：()fx在[1,]e上单调递增，所以1(1)()()affxfeae·········6分（ⅰ）当10ae，即1ae时，1()0fxe，所

以2()lnxFxxxe，2()ln1xFxxe令()()GxFx，所以21()Gxex令()0Gx，得2ex；令()0Gx，得2ex；所以()Fx在区间[1,]2e单调递减，在区

间[,]2ee单调递增；又因为2(1)10,()0FFee，所以()0Fx；高二数学答案第5页（共5页）所以()Fx在[1,]e上单调递减，()Fx无极值·······································

···············8分（ⅱ）当10ae，即1ae时，()0fx，所以2()lnFxaxxx，则()2ln1Fxaxx，令()()GxFx，所以1()2Gxax，因为[1,]xe，所以11[,1]xe①当

21a，即12a时，则()0Gx，所以()Fx在区间[1,]e单调递增，所以()(1)210FxFa所以()Fx在[1,]e上单调递增，()Fx无极值················

······································9分②当112ae，即221ae时，令()0Gx，得12xa所以，当1[1,]2xa时，()0Gx，()Fx在区间1[1,]2a单调递减；当1[,]2xe

a时，()0Gx，()Fx在区间1[,]2ea单调递增；又因为(1)210,()220FaFeae，所以存在0(1,)xe使得0()0Fx所以，()Fx在0[1,]x上单调递减，在0[,]xe上单调递增所以()

Fx在[1,]e上有极小值········································································10分（ⅲ）当10ae时，因为()fx在[1,]e上单调递增，且1(1)0,()0fafea

e，所以，存在0(1,)xe使得0()0fx，所以，当0[1,)xx时，()0fx；当0[,]xxe时，()0fx所以函数2020ln,1,()ln,,axxxxxFxxxaxxxe

························································11分所以002ln1,1,()ln12,,axxxxFxxaxxxe设ln1()xxx（[1,

]xe），则2ln()0xxx，所以()x在[1,]e上单调递减所以，2()()2xeae，即当[1,]xe时，2ln10axx，ln120xax所以，()Fx在0[1,]x上单调递减，在0[,]xe上单调递增，所

以()Fx在[1,]e上有极小值综上，a的取值范围是111(0,)(,)2ee·····························································12分

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