2021年学业水平考试数学答案一、选择题ACCDDCADBADA二、填空题13.14.15.16.三、解答题17.(1)过线段中点.................................................................................................1分由坐标可得...........................................................................................2分......................................................................................................................3分则整理成一般式为:.........................................4分(2)设①......................................................................................................5分②.........................................................................................7分由①②消可得......................................................................................9分则点坐标为或.......................................................................................10分18(1)第一组的频率为0.15,第二组的频率为0.25所以第30百分位数·······················································6分(2)日销售量都不低于100个的概率为0.6,日销售量低于50的概率为0.151,设A=“在未来连续3天里,有连续2天的日销售量都不低于100个且另1天的日销售量低于50”···············································································································8分P(A)=2×0.6×0.6×0.15=0.10812分19.(1)设圆心,则半径为................................................................................1分圆心到直线的距离为.......................................................2分.......................................................................................3分则.....................................................................4分由,解得............................................................................................5分圆方程:或.........................................6分(2)当圆心在第一象限时,圆心,.........................................7分①斜率不存在时,直线方程:,圆心到直线的距离是2,是圆的切线..........9分②斜率存在时,设过点的圆的切线方程为,即..10分圆心到直线距离,解得,切线方程为.综上,过点的圆的切线方程为和......................................12分20.(1)设A=“从袋中一次摸出2个小球恰为异色球”.............................................................1分..........................................................................................4分(2)设B=“3个球中,黑球与白球的个数都没有超过红球的个数”...........................................5分.......................................................8分2,(3)设C=“得分为8分”..............................................................................................................9分..............................................................12分21.(1)·····················································································4分(2)设直线,,,,代入得:,·····································································6分····························································································8分点到直线的距离·································································10分·······································································12分22.(1)·····················································································2分(2)由(1)得,,又直线的斜率不为零,故可设的方程为,由,代入得,由于恒成立,故可设交点,,又直线,所以,则有,,···············································4分3,相除可得,又直线的方程为,又由,所以,···································6分所以直线的方程为,故直线恒过定点.···································7分(3)设交点,,设①直线斜率存在时,代入得,,·······················································9分,得:所以,··················································11分②直线斜率不存在时,或4,综上,···································································12分5