衡中同卷2024高三一轮复习周测卷(小题量) 全国版三数学试题

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2022年春季鄂东南省级示范高中教育教学改革联盟学校五月模拟考8,【答案】A【详解】因为公比q±0,所以a1>a=⊙am9>aa9g>g台g0-9)>0高三数学参考答案一、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9g1-gl+g+g)>0g0-g)>005g<1:题目要求的L.【答案】C4g2x0xg3>4c9'台q2>g2g20-q)>0eg<1q*0:【详解】因为M=(0.2),N=(1,+x),所以MAN=(1,2),故症:C所以“a>ax”是“am>ae"的充分不必要条件故选:A2.【答案】D二、多选题:本题共4小题,每小题5分,共20分.在每小题给出的选项中,有多项符合题日要求.全【f解】设:=a+bi,其中a,bcR,则2Ξ-:=2a-bi)-(a+i)=a-3i=1+3i,则a=1,b=-1,郁选对的得5分,有选错的得·分,部分选对的得2分.此时:在复平面内对应的点为ZL,-),位于第同象限,故达:D.9.【答案】AC3.【答案】A【详解】由题意得:2弘=2互,所以b=√2,因为2-1-号故。=3,四为焦点,5在y输621[x+y=2=上所以桥国C的方程为号号-1,由庭径长可有,阳2沙4,△PF0的周长为4a=45,a 3故近:AC时。等号成立,故远:A10.【容案】BC4.【答案】B【译解】设圆锥的厚线为!,即侧面展开图的半径为1,义园维的收而半径为1,则侧面展开图的瓜【解行】随机抽州的10名学生,回答第一个同题的概水是宁其解号是奇数的藏水电是子,所长为2x,又侧展开图是半個,测1=2x,测1=2,所以侧1面积为=2x,成选:B以国答何遇1月区答的是的学4人煮为100×20,同溶利题2同斧的是的人藏为5.【答案】B265-250=15,从面告计该地区中学生吸相人数的面分比为5-3%,估计被调查着中吸智的人数500【详解】山题意知,一所学校有2名志愿素,一所学校有1名志愿紧,总的况有C:,=6种,中被为1000×3%=30,故选:BC派到A学校的情况有C+1=3种,故甲被派到A学校的板率为2-}故远:B11.【答案】AB62【解断】出i正实数a.b,c,以及e<1,b<1可得c,be(0,1),又log,a>1=log.c,所以a<c<1,所6.【答案】B【详解】由抛物线定义可知PF=Pg,∠POF=∠PFQ-区,△PQF为正以d<c,又<,所以d<b,即bna<ahh,等价于血”e血b,由于函数f-hr在0.na b三角形设准线!与x轴交于点A,由抛物线可:AF=2,PQAF上递增,从而a<b,又取b=e时,原式为b°<b“<I<g,a同样成立,故CD不正确,从而本题选∠FQ=∠Por=号lor=2A=4,PF=Or=4,放流:BAB.12.【答案】ACD7.【答案】B【答案】ACD【详解】因为a公+c2-a)-bc,所以2a×+C-d=b,即6=2acos4,所以【解析】由题可知,AB⊥C,DC⊥C,则可将四面体A6CD的四个项点放入如下图所示的直三simB-2sin4cos4=in24,所以B-24或B+24=及.若B+24-元,侧C-A,这与趣设不合:若棱柱中,老虑到直线4B与CD所成角为,故有如下两种情况:B-2A,又B=C,所以A+B+C=SA=:,即A=石放选:B.对于左图。∠8E=号则E=2,D=+孕=25:此时D与5C所成角余弦值为25

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