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第1页共6页数学参考答案题号123456789101112答案DCCBABACABACDADACD1.【解析】由2{|280}{|24},{3,2,0,2,3}MxxxxxN,得MN{0,2,3}.2.【解析】由复数(1)(1)i(
R)zmmm为纯虚数,可得复数1m,所以复数(3i)i(3i)13iz,所以在复平面内对应的点为(1,3),在第三象限,故选C.3.【解析】由题意可得(2,2)ab,因为(
)//abb,所以4(2)4,解得1,故选C.4.【解析】由题意可得tan2,所以原式=333333sincostan1813sin2costan2822,故选B.5.A【解析】由题意可得正方体卡片纸盒如图所示,则易知CMEF,//CMAB,CM与
CD,MN的夹角均为60,故选A.6.【解析】由题意估计高一高二日阅读时间的平均数为4060504044100100x,方差为2224060[4(5044)][6(4044)]29.2100100s,故选B.7.【解析】由5ln41logln5c,得55lo
g(5)log4c,即54c,即45c.555455444444log3log3log243log256log45a,555455777774log5log5log3125log2401log75b,综上可知acb,故选A.8.【解析】因
为,||||ABACABAC分别是向量,ABAC的单位向量,所以由()(0)||||ABACAQABAC,易知AQ是ABC中角A平分线,于是结合ABCABQBCQSSS,43A
Q可得,111sin6043sin3043sin30222bccb,化简得4()bcbc,故1114bc,所以1116164(16)()4(17)100bcbcbcbccb
,当且仅当20,5bc时等号成立,故选C.9.【解析】A选项中的极差为2.8-0.1=2.7(%),故A正确;B选项中的数据由小到大排列为0.1%,0.2%,0.7%,1.0%,2.1%,由i=5×75%=3.75,不是整数,所以75%分位数为第4个数1.0%,故
B正确;C选项中2月至4月均为负数,说明下降,故C错误;D选项等价于从5个数字中随机选取2个,样本空间包含的样本点总数为10,其中随机事件“选到4月和5月”包含的样本点数为1,古典概型概率计算公式可得所求{#{QQABLYYUggggABAAABgCQQWCCgOQ
kAECAIgORBAIIAABSQNABAA=}#}第2页共6页概率为110P,故D错误.综上,选AB.10.【解析】由题意知//BCx,所以()fx的图像的一条对称轴方程为2723212x,71212344,所以2.由于函数fx图
像过π(,0)3,由223k,kZ,且02,得3,所以()sin(2)3fxx,故A正确,B错误;()fx的图象向右平移6个单位长度得到()sin(2)sin233gxxx,是奇函数,故C正确;()fx的图象向左平移12个单
位长度得到()sin(2)cos263hxxx,是偶函数,故D正确.故选ACD.11.【解析】连接PQ,4,//ADBCADBC,所以4DPBP,又因为4DQQS,所以//PQSB,故A正确;过Q作//QMAD交SA于点M,由4DQQS,可得1455QMADBC
,所以四边形BCQM是梯形,CQ与BM的延长线必定相交,故CQ与平面SAB必定相交,故B错误;设直四棱锥SABCD的高为H,底面梯形ABCD的高为h,则易得三棱锥SABC和三棱锥QACD的高分别为4,5HH,所以12111532,11441643255BChHVVADhH
所以C错误,D正确.综上,选AD.12.【解析】因为2sin63sin642sin(4518)2sin(4519)cos18cos19cos18cos19xsin18cos18sin18cos18(1tan18)(1
tan19)cos18cos19,所以A正确;由正切函数在(0,)2上恒为正且单调递增得(1tan18)(1tan19)(1tan26)(1tan27)xy,所
以B错误;注意到tan18tan271tan(1827)1tan18tan27,所以tan18tan27tan18tan271,同理tan19tan26tan19tan261,于是(1tan1
8)(1tan27)(1tan19)(1tan26)xy(1tan18tan27tan18tan27)(1tan19tan26tan19tan26)224,故C正确;由基本不等式可得24xyxy,故D正确.综上,选ACD.13.【答
案】15【解析】3ii(43i)34i43i(43i)(43i)25z,所以1||5z.{#{QQABLYYUggggABAAABgCQQWCCgOQkAECAIgORBAIIAABSQNABAA=}#}第3页共6页14.【答案】112【解析】由题意可知正四棱台的高为2253
4,所以该四棱台的体积为22221(2828)41123.15.【答案】40【解析】由题意知,60,30,120ADBBCADAC,设DAx,则3,3ABxACx,在ABC,由余弦定理可得22240391()(3)23()32xxxx
,解得4033x,所以340ABx(米).16.【答案】12,1128【解析】在1()()32xffx中,分别令0x,得(0)0f,在(1)1()fxfx中,令11,3xx,得12(1)1,()()133fff.又令13
x,得111()(1)322ff,所以111()()322ff,结合对于1201xx,都有12()()fxfx,可得当11[,]32x时,康托尔函数1()2fx,反复利用1()()32xffx,
可得223311311320232023()()()()()20233220232342023fffff34345633113113131320232023()()()()()()438202382162023322023642023ffffff,注意到6
666666133332323333729320233337292,所以1111()2023642128f,故第二空填1128.(注:第一空2分,第二空3分)17.【解析】(1)因为(2)()3abab,所
以2223aabb,又||1,||2ab,所以1ab,························2分因为22222212123|ab|aab+b,所以3|ab|=.·····················
·····5分(2)设,ab的夹角,由(1)中可得1cos||||2abab,故3.因为(3,1)b,所以31(,)||22bb,·······················7分{#{QQABLYYUggggABAAABgCQQWCCgOQkAECAIgORB
AIIAABSQNABAA=}#}第4页共6页所以向量a在向量b上的投影向量为13131cos(,)(,)||22244b|a|b········10分18.【解析】(1)由22abcbaab
,得2222abcab,所以222222abcab,结合余弦定理可知2cos2C,而0C,所以4C.·············································5分(2)由正弦定理可得5225252a,解得45a,···
······················································7分又由余弦定理可得2250802452bb,即2410300bb,解得310b,或10b,而52c
b,所以310b,········································································10分所以ABC的面积112sin4531030222SabC···········
······························12分19.【解析】设事件iA=“甲答对了i道题”,事件iB=“乙答对了i道题”,0,1,2i,由题意0111()224PA,111111()22222PA,2111()224PA,0111()339PB
,112214()33339PB,2224()339PB,(4分)(1)由题意得,甲,乙都通过考试的概率;222211221()()()22339PABPAPB.······8分(2)由题意得,112002EABABAB,所
以112002()()()()()()()PEPAPBPAPBPAPB1411141329494936.··········12分20.【解析】(1)连接OB,由题意知DO平面ABC,故DOBO,则直线PB与底面的夹角为45PBO
.····················································2分因为底面圆的半径为2,所以2POBO,底面圆的周长为4,而P为DO的中点,所以4DO,所以圆锥的母线长为224225,所以圆锥侧面展开图的面积
为1425452.·······················································6分(2)证明:连接,OAOC,∵P在DO上,OA=OB=OC,∴PA=PB=
PC,{#{QQABLYYUggggABAAABgCQQWCCgOQkAECAIgORBAIIAABSQNABAA=}#}第5页共6页∵△ABC是圆内接正三角形,∴AB=BC=CA,所以,PABPBCPABPAC,∴∠BPC=∠APC=9
0°,即PB⊥PC,PA⊥PC,PA∩PB=P,∴PC⊥平面PAB,PC⊂平面PAC,∴平面PBC平面PAB.···························································12分21.【解析】(1)当13t时,13APAQ
,因为AQ为边BC的中线,所以1111()2222AQABBQABBCABACABABAC,所以1166APABAC.···
···········································································5分(2)证明:由(1)可知1122AQABAC,所以()2tAPtAQABAC
.···········································································6分而MPxMN,,AMABANAC
,所以APAMxANxAM,即()2tABACABxACxAB,整理可得()()22ttxABxAC
,·····································································9分而,ABAC是不共线向量,所以022ttxx,消去x,可得2tt(
定值).··············································································12分22.【解析】(1)2422()23cossincossincossin
fxxxxxxx22223sin2cossin(sincos)xxxxx,3sin2cos2xx2sin(2)6x,因为1,所以2sin(2)6fxx.······················2
分{#{QQABLYYUggggABAAABgCQQWCCgOQkAECAIgORBAIIAABSQNABAA=}#}第6页共6页由5()6212f得3sin5,而(,)2,所以4cos5,所以24
sin22sincos25,27cos22cos125,············3分故cos(2)cos2cossin2sin666732417324()25225250.···················
····5分(2)由(1)知()sin026()()xfx,因为函数333(,),(,),(,)222262626xxx,()sin026()()xfx在区间3(,)22上没有零点,所以2
63126kkkZ,即1722,393kkkZ,·············8分因为7212933kk,所以13k,又因为0,所以72093k,所以76k,所以7163k,······
······10分因为kZ,所以1k或0k,当1k时,5139;当0k时,1739,又因为0,所以的取值范围是117(0,][,]939.·················12分{#{QQABLYYUggggABAAABgCQ
QWCCgOQkAECAIgORBAIIAABSQNABAA=}#}获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com