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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13.0.9414.36

6615.2222ABCACDADBBCDSSSS∆∆∆∆++=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//,22HGB

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

···································8分又1221339()()432128PACPAC==×=.··················································9分所以221221[()()()]PPABAC

AC=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(6.57)]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1111331132318AABCBAACVV−−==××××=.··

·······························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[40,80)的频率为()0.0040.0060.020.03100.6+++×=，所以设第57百分位数为a，有()0.3700.030.57

a+−×=，··················6分解得79a=.··················································································7分D

CBAC1B1A1高一数学参考答案第4页(共5页)FGMEDCBANNABCDEM（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=.···················································································1

2分22.解：（1）证明：取AE中点N，连MN，则//MNDE，12MNDE=.······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DN⊥平面ABE.AF⊂平面ABE，所以DNAF⊥.························································8分在三角形AEF中，过N作N

GAF⊥，垂足为G，连接DG，因为DNNGN=.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，253,5DN

NG==,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