【文档说明】江苏省连云港市2021高二下学期数学期末试题（及答案）.pdf，共(8)页，1.877 MB，由baby熊上传

转载请保留链接：https://www.ichengzhen.cn/view-83912.html

以下为本文档部分文字说明：

高二数学参考答案及评分建议0627一、单项选择题（本大题共8个小题，每小题5分，共40分）1．B2．D3．C4．B5．C6．A7．D8．A二、多项选择题（本大题共4个小题，每小题5分，共20分，在每小题给出的选项中，有多项是符合题目要求．全选对的得5分，

部分选对的得2分，有选错的得0分）9．AC10．ABD11．BCD12．ACD三、填空题（本大题共4个小题，每小题5分，共20分）13．1014．8856315．1016．243（第一空2分，第二空3分）四、解答题17．解：若选

择①，设i(,,0)zababRb,则2222(i)()2i=724izababab由22=7224abab解得3344aabb或，„„„„„„„„„„„„„„„„„„„„5分所以

34i34izz或，则5.z„„„„„„„„„„„„„„„„„„„„„10分若选择②，设i(,,0)zababRb则22i(1)5i=15izabzab由22=15aabb解得125ab，„„„„

„„„„„„„„„„„„„„„„„„„„5分所以125iz，则13.z„„„„„„„„„„„„„„„„„„„„„„„„„„10分若选择③，设i(,,0)zababRb，则2211iiabzabab2222221ii+()()iababzababzababab

是实数，则220bbab，„„„„5分又0b，所以221ab，则1.z„„„„„„„„„„„„„„„„„„„„„10分18．解：6126012666(1)0.996=10.0040.0040.0

04CCC10.0240.000240.976„„„„„„„„„„„„„„„„„„„„„„„„6分2021202102021120202201920201202120212021202120212021(2)5152152525252aaCCCCCa

其中02021120202201920201202120212021202152525252CCCC能被13整除，„„„„„„„„10分只需20212021Ca能被13整除，由013a，得1a=0，故1a.„„„„„„„„„12分19．解：（1）连接AC，过点

B作直线MN，分别交直线DC，DA的延长线于N，M两点，连接EM，FN分别交1AA，1CC与P，Q两点，连接PB，BQ，则五边形EPBQF为所求截面„„„„„„„„3分在正方形1111ABCD中，111222EFA

C，在Rt△AMB中，∠AMB=∠DAC=45°，∠ABM=45°,故AM=AB=4，由△AMP∽△1AEP，故1112APAEPAAM，故12,4APAP，故221122PEAEAP,2242PBABAP„„„„„„„„„5分同理，可求

得22FQ，42BQ,故五边形EPBQF周长为：142EFEPPBBQQF，则截面周长为142„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„6分（2）分别取AD，DC的中点R，T，连接ER，FT，在Rt△

ABR中，2225BRABAR在Rt△ERB，22214BEERBR，同理214BF求得等腰△EBF的面积为63EBFS△，求得△E1BF的面积为16EBFS△„„„„„„„„„9分设B1到平面BEF的距离为h，由11BEBFBEBFVV

，得111133EBFEBFShSBB△△故11662363EBFEBFSBBhS△△，故B1到平面BEF的距离为23„„„„„„„„„„„„12分（本题第（2）问，也可以利用“综合法”或者“向量法”求出结果

）20．解：（1）提出假设0H：给药方式和药的效果无关，由表格数据得：22100(40203010)1003.8417030505021K，„„„„„„„„„„„„4分因为当0H成立时，23.841K的概率约为0.05，所以，我们有95％的把握认为给药方式和药的效果有

关.„„„„„„„„„„„„„„„„6分TRQPA1FEAD1C1B1NDCBM（2）依题意，从样本的注射病人（50人）中按分层抽样的方法取出的5人中，有效的305350人，无效的有2人，记抽取的3人中有i人有效的为事件2,3iAi，则2132235320.610CCPAC

；„„„„„„„„„„„„„„„„„„„„„„„„„„„8分3333510.110CPAC„„„„„„„„„„„„„„„„„„„„„„„„„„„„„10分因为1A和2A互斥,所以抽取的这3个病人中至少有2人有效的概率为22230.60.10.7PAAPA

PA.答：其中至少2个病人有效的概率为0.7.„„„„„„„„„„„„„„„„„„„„„„„12分21．解（1）∵四棱锥SABCD的底面是矩形，∴ADAB，又∵平面SAB平面ABCD，平面ABCD平面

SAB=AB，AD平面ABCD，∴AD平面SAB，又BS平面SAB，∴ADBS，„„„„„„„„„„„„„„„„„„2分∵ASBABS，∴AS=AB，又E为BS的中点，∴AEBS，又ADAE=A，∴BS平面DAE，„„„„„„„„„„„„„„

„„„„„„„„„„„4分∵BS平面SBC，∴平面DAE平面SBC.„„„„„„„„„„„„„„„„„„„„„„5分（2）如图，连接CA，CE，在平面ABS内作AB的垂线，建立空间直角坐标系Axyz-，„„„„6分设2

4ABADa，14SESB，∴(000)A，，，(040)Ba，，，(042)Caa，，，(002)Da，，，(2320)Saa，，,(23,6,0)SBaa33(0)22aEa，，，则=(

042)ACaa，，，33=(0)22aAEa，，，=(002)ADa，，设平面CAE的法向量为=()xyz，，n，∴00ACAE，，nn即42033022ay+az=aaxy，，令1x，则33,63yz，∴=

(13363)，，n是平面CAE的一个法向量，„„„„„„„„„„„„„„„„„„9分设平面DAE的法向量为=()xyz，，n，∴00ADAE，，nn即2033022az=aaxy

，，得：=(1330)，，n„„„„„„„„„„„„„„„10分28238cos,||||3428136nnnnnn，∴锐二面角CAED的余弦值为23834„„„„„„„„„„„„„„„„„„„„„12分

22．解：（1）由题意知1~3,2XB,则30311(0)C28PX，213113(1)C228PX，223113(2)C228PX，33

311(3)C28PX，„„„„„„„„„„„„4分所以X的分布列为13()322EX．„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„6分（不列表不扣分，分布列每对1个，得1分）（2）由（1）可知在一局游戏中，甲得3分的概率为31

1882，得1分的概率为131882，若选择nk，此时要能获得大奖，则需2k次游戏的总得分大于4k，设2k局游戏中，得3分的局数为m，则3(2)4mkmk，即mk．易知1~2,2mBk，故此时获大奖的概率1122

2122212211111()CCC22222kkkkkkkkkkkPPmk…+212222220122222222)1()2

1=(CCC21CC1(2C)2C21C2kkkkkkkkkkkkkkkkkk…+…+22C1(1)22kkk„„„„„„„„„„„„„„„„„„„„„„„„„9分同理可以求出当

1nk，获大奖的概率为21(12P122222Ckkk）„„„„„„„„„„„„10分因为222211222222(2)!44(1)2(1)(!)(!)21(22)!(22)(21)21[(1)!][(1)!]2C4CCCkkkkkkkkkkkkkkkkkkkkk

，所以12222222C2Ckkkkkk，则21PP答：甲选择1nk时，获奖的概率更大．„„„„„„„„„„„„„„„„„„„„12分X0123P18383818