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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：DBCC5～8：AABC二、选择题：本题共4小题，每小题5分，共2

0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；12BQCQCBA

BAD．······················································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

22131242ABABADAD221313221cos6012424．·····························8分所以3214cos,14||||2112

APBQAPBQAPBQ．··························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()339PA，111111()22222

PB，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()()PMPABAB················

····································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cos3sinbAbA

ca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sincos3sinsinsincoscossinsinBABAABABA，即3sinsinsincossinBAABA

．··············································5分又sin0A，所以有3sincos1BB，即223sincos2cos222BBB．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ABD，所以CE平面ABD．······

····························································4分连接DE，则DEAA，DEAA．又BBAA，BBAA，所以DEBB，D

EBB，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BEC，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40.0640.0840.0440.0440.02441aa，即0.968

1a．·········································································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，0.80.7213.2415.20.880.72．············································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