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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