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高一数学参考答案第1页(共5页)ABCDEFGH2020-2021学年度第二学期期末学业水平诊断高一数学参考答案一、选择题CACDBCDA二、选择题9.AC10.BCD11.ABD12.ACD三、填空题13.0.941

4.366615.2222ABCACDADBBCDSSSS∆∆∆∆++=16.283π四、解答题17.证明：（1）连接BD，因为,EF分别为棱,ABAD的中点，所以11//,22EFBDEFBD=.·································2分同理11//

,22HGBDHGBD=.································4分所以//EFGH且EFGH=.··································5分所以四边形EFGH是平行四边

形.····························6分（2）当ACBD⊥且ACBD=时，四边形EFGH为正方形.·····················10分18.解：（1）设事件,,,ABCD分别表示“被评为等级,,,ABCD”.由

题意，事件,,,ABCD两两互斥，所以3131()1483232PD=−−−=.·························································2分又CD=“延迟送达且

被罚款”,所以1()()()8PCDPCPD=+=.······················································4分因此“延迟送达且被罚款”的概率为18.······

···········································5分高一数学参考答案第2页(共5页)（2）设事件,,,iiiiABCD表示“第i单被评为等级,,,ABCD”，1,2i=.则“两单共获得的奖励为0元”即事件221221()()()AB

ACAC，·············7分且事件221221,,ABACAC互斥，又22111()8864PAB=×=.····································································8分又

1221339()()432128PACPAC==×=.··················································9分所以221221[()()()]PPABA

CAC=221221()()()PABPACPAC=++··············································10分1133528843232=×+××=.··················

·······································12分19.解：（1）由题意，435761179869440y×+×+×+×+×+×=··································

·········2分6.5=.··························································································3分22221[3(

46.5)7(56.5)11(66.5)40ys=×−+×−+×−2229(76.5)6(86.5)4(96.5)]+×−+×−+×−·······························5分1.95=.····································

····················································6分（2）由（1）可得，1(5040)90zxy=+·······························

·······················8分1(507.4406.5)790=×+×=.·····································9分2225040[2.6(7.47)][1.95(6.5

7)]9090zs=+−+×+−·································11分11345=.·············································

·····································12分高一数学参考答案第3页(共5页)20.解（1）证明：因为1ABAA⊥，ABAC⊥，1ACAAA=所以AB⊥平面11ACCA，·············

····················································2分1AC⊂平面11ACCA，所以1ABAC⊥.···············································3分又因为直三

棱柱111ABCABC−中，1AAAC=，所以四边形11ACCA为正方形，所以11ACAC⊥.···································4分因为1ACABA=,所以1AC⊥平面1ABC，···············

························5分1BC⊂平面1ABC,所以11ACBC⊥.···················································6分（2）过1A作111ADBC⊥，垂足为D，连CD，则1AD⊥平面11BCCB，1ACD∠为1AC与

平面11BCCB所成的角.···················8分因为11AAAC==，则12AC=，所以11112sin42ADADACDACa∠===，所以112AD=.··················

········9分在11RtACD∆中，111111sin2ADACDAC∠==，所以1130ACD∠=.在111RtABC∆中，11113tan303ABAC==.······································10分所以1111331132318AA

BCBAACVV−−==××××=.·································12分21.解：（1）由已知()0.0040.0060.020.030.024101m+++++×=，···

··········2分解得0.016m=.·············································································3分（2）

测试成绩的平均数450.04550.06650.2750.3850.24950.16x=×+×+×+×+×+×·········4分76.2=.··································

···················································5分测试成绩落在区间[40,70)的频率为()0.0040.0060.02100.3++×=，落在在区间[4

0,80)的频率为()0.0040.0060.020.03100.6+++×=，所以设第57百分位数为a，有()0.3700.030.57a+−×=，··················6分解得79a=.·

·················································································7分DCBAC1B1A1高一数学参考答案第4页(共5页)FGMEDCB

ANNABCDEM（3）由题知，测试分数位于区间[80,90)、[90,100)的人数之比为0.2430.162=，所以采用分层随机抽样确定的5人，在区间[80,90)中3人，用123,,AAA表示，在区间[90,100)中2人，用12,BB表示.········

············································8分从这5人中抽取2人的所有可能情况有：12(,)AA，13(,)AA，11(,)AB，12(,)AB，23(,)AA，21(,)AB，22(,)AB，31(,)AB，32(

,)AB，12(,)BB，共10种.······10分其中“落在区间[80,90)和[90,100)”有6种.···········································11分所以3()5PA=.··

·················································································12分22.解：（1）证明：取AE中点N，连MN，则//MNDE，12

MNDE=.······1分又因为//BCDE，12BCDE=，所以//BCMN,BCMN=.···································2分所以四边形BCMN为平行四边形，所

以//BNCM.····3分又因为BN⊂平面ABE,CM⊄平面ABE，所以//CM平面ABE.···········································5分（2）延长,EBDC交于点F，则AF为平面ABE与平面ACD的交线.·············6分因为/

/BCDE，12BCDE=,所以2BFBE==.三角形ADE中，因为2ADDEAE===,N为AE的中点，所以DNAE⊥,又因为DEBE⊥，AEBE⊥，DEAEE=所以BE⊥平面ADE，DN⊂平面ADE，所以DNBE⊥.又因为BEAEE=，所以D

N⊥平面ABE.AF⊂平面ABE，所以DNAF⊥.························································8分在三角形AEF中，过N作NGAF⊥，垂足为G，连接DG，因为DNN

GN=.所以AF⊥平面DNG，DG⊂平面DNG，所以AFDG⊥.所以DGN∠为二面角DAFE−−的平面角.············································9分高一数学参考答案第5

页(共5页)在RTAEF∆中，2,4AEEF==,41625AF=+=,由AEFAGH∆∆,所以AFANAEGN=,255NG=.···································10分在RTDNG∆中，253,5

DNNG==,22955DGDNNG=+=，所以219cos19NGDGNDG∠==.···························································11分即

面ABE与面ACD所成二面角（锐角）的余弦值为21919.······················12分