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高一数学答案第1页（共6页）PQCDAB2020～2021学年度第二学期期末考试高一数学参考答案及评分标准2021．7一、选择题：本题共8小题，每小题5分，共40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．1～4：DBCC

5～8：AABC二、选择题：本题共4小题，每小题5分，共20分．在每小题给出的选项中，有多项符合题目要求．全部选对的得5分，部分选对的得2分，有选错的得0分．9．BCD10．AD11．ABD12．BC三、填空题：本题共4小题，每小题5分，共20分．13．1i14．18π15．1(2，

1)216．262四、解答题：本题共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．（本题满分10分）解：（1）因为P、Q分别为BC、CD的中点，所以12APABBPABAD；12BQCQCBABAD

．······················································2分于是ACAPBQ11()()22ABADABAD11()()22ABAD

．············································3分又ACABAD，由平面向量基本定理，得11,211.2··················

···4分解得65，25．·····································································5分（2）由（1）可知12APABAD，12BQABAD，高一数学答案第2页（共6页）PQCDAB所以

22211121||()22422APABADABADABAD；········6分222111||()21242BQABADABABADAD；················7分11()()22APBQABADABAD2

2131242ABABADAD221313221cos6012424．·····························8分所以3214cos,14||||2112APBQAPBQAPBQ．··

························10分18．（本题满分12分）解：（1）因为PC平面ABC，AC平面ABC，所以ACPC．·······························2分因为C是以AB为直

径的圆O上的点，所以ACBC．·······························4分又PCBCC，所以AC平面PBC．.......................................

...................5分因为E，F分别是PA，PC的中点，所以EFAC．...................................6分所以EF平面PBC．.................................................

.........................................7分又EF平面BEF，故平面BEF平面PBC．..............................................

..8分（2）EFl．...................................................................................................................9分证明如下：由（

1），EFAC．又AC平面ABC，EF平面ABC，所以EF平面ABC．.............................................................................

............10分又EF平面BEF，平面BEF平面ABCl，所以EFl．.....................................................................................................

...12分FECAPB高一数学答案第3页（共6页）19．（本题满分12分）解：（1）由题意，得1(1)(1),61(1)(1).2pqpqpq·····················

·····························2分解得13pq，76pq．所以,pq是方程271063xx的两个实根．又pq，解得23p，12q．···························

···························4分（2）解法一：设1A，2A分别表示甲两轮猜对1个，2个谜语的事件，1B，2B分别表示乙两轮猜对1个，2个谜语的事件，则121124()33339PA，2224()3

39PA，111111()22222PB，2111()224PB．·································8分设M表示“星队”在前两轮活动中猜对3个谜语的事件，由题意，1221()()PMPABA

B····················································9分1221()()PABPAB···············································10分1221()()()()PAPBPAPB····

··································11分4141194923．·············································12分解法二：设1A，2A分别表示第一轮两人猜对1个，2个谜语的事件，1B，2B分别表示第二轮两人

猜对1个，2个谜语的事件，则121111()32322PA，2211()323PA，121111()32322PB，2211()323PB．································8分设M表示“星队”在前两轮活

动中猜对3个谜语的事件，由题意，1221()()PMPABAB····················································9分1221()()PABPAB···································

············10分1221()()()()PAPBPAPB······································11分1111123323．·············································12

分高一数学答案第4页（共6页）20．（本题满分12分）解：（1）由cos3sinbAbAca及正弦定理，得sincos3sinsinsinsinBABACA①·········································2分因为π()CAB，所以

sinsin()sincoscossinCABABAB②·······························4分将②代入①得sincos3sinsinsincoscossinsinBABAABAB

A，即3sinsinsincossinBAABA．··············································5分又sin0A，所以有3sincos1BB，即223sincos2cos222

BBB．因为π(0,)22B，所以cos02B，于是有3sincos22BB，即3tan23B．········································7分又π(0,)22B，所以π26B，即π3B．···············

·····························8分（2）由△ABC的面积为3，得1πsin323ac，即4ac．·····················9分由余弦定理，得222π2cos3bacac，即22π()2(1cos)3acacb．···········

··········································10分将4ac，2b代入上式，得21()24(1)42ac．解得4ac．································

··········································11分所以a，c是方程2440xx的两个实根，显然2ac．·············12分高一数学答案第5页（共6页）EDACBB'C'A'EDACBB'C'A'21．（本题满

分12分）解：（1）取线段AC的中点E，连接BE，CB，CE，则平面BEC平面ABD．BE，CB，CE即为应画的线．·····················································2分证明

如下：因为D为AC的中点，E为AC的中点，所以AEDC，且AEDC，所以四边形AECD为平行四边形，所以DACE．·································3分又DA平面ABD，CE平面AB

D，所以CE平面ABD．··································································4分连接DE，则DEAA，DEAA．又BBAA

，BBAA，所以DEBB，DEBB，所以四边形DEBB是平行四边形，所以BEBD．···········································································5分又BD平面ABD，BE平面ABD，

所以BE平面ABD．··································································6分又CEBEE,CE平面BEC,BE平面BE

C，故平面BEC平面ABD．·····························································8分（2）设棱柱ABCABC的底面积为S，高为h，则30SVh三棱柱．111()5326BCAABECDVVShSh

三棱锥三棱锥．······························10分所以三棱柱夹在平面BEC与平面ABD之间部分的体积305520ABCABCAABBCDCEVVVV

三棱柱三棱锥三棱锥．···············12分高一数学答案第6页（共6页）22．（本题满分12分）解：（1）因为频率分布直方图中，所有小矩形的面积之和为1，所以40.0640.0840.0440.044

0.02441aa，即0.9681a．·········································································1分故0.005a

．·············································································3分该市居民月均用水量的平均值3.2(40.06)7.2(40.08)11.2(40.04)15.2(40.04

)x19.2(40.02)23.2(40.005)27.2(40.005)···················4分9.84．························································

···························6分（2）由频率分布直方图知月均用水量在13.2t以下的居民用户所占的比例为40.0640.0840.040.72．·····································7分月均用水量在17

.2t以下的居民用户所占的比例为0.7240.040.88．·····················································8分因此，第80百分位数一定位于[13.2,17.2)内，0.80.7213.24

15.20.880.72．············································10分因此，要使80%的居民用户生活用水费用支出不受影响，一户居民月均用水量

的标准A为15.2t．··········································································12分