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12021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0.152

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2

0.（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3（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4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5

综上，···································································12分