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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