东北三省三校高三第二次联合模拟考试理科数学试题及答案.doc
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1、二模理科数学参考答案题号123456789101112答案ADCBCBDAABCC13 14 15 1617()解:当时, 2分又 4分 数列是首项为,公比为的等比数列, 6分(), 8分所以 10分12分18()解:第三组的频率是0.1502=0.3;第四组的频率是0.1002=0.2;第五组的频率是0.0502=0.1 3分()由题意可知,在分层抽样的过程中第三组应抽到60.5=3个,而第三组共有1000.3=30个,所以甲乙两产品同时被选中的概率为 7分第四组共有X个产品被购买,所以X的取值为0,1,2;所以X的分布列为 10分 12分19()证明: 连结 3分()于是7分()如图建立直
2、角坐标系,设平面的法向量为 9分 11分所以,直线与平面所成角的正弦值为 12分20()设,则 4分()设直线:,将直线代入到中得,所以5分又因为所以8分或 10分所以恒过定点 12分21 ()令 2分有两实根不妨记为极小极大所以,有两个极值点 ,一个极大值点一个极小值点 4分(),由韦达定理得 6分,所以 7分()因为,所以 8分又因为当时,不等式恒成立所以,原问题对一切恒成立法一、设()设,当时,所以,当时,所以,所以在上单调递增,又因为所以当时, ,当时, 所以在上递减,递增,所以 10分所以当时, ,当时, 所以在上递减,递增,所以所以 12分法二不妨设,当时,所以在上单调递增,所以在
3、上单调递增, ,所以当时成立10分当时得当时所以在上单调递减,所以在上单调递减,与条件矛盾,同理时亦如此综上 12分22 ()5分()为切线又因为为切线 10分23(), 5分()设,则点到直线的距离 8分当且仅当,即()时取等 10分24解:()由柯西不等式得, 所以的取值范围是 5分()同理, 7分若不等式对一切实数恒成立,则,解集为 10分6a*CZ7H$dq8KqqfHVZFedswSyXTy#&QA9wkxFyeQ!djs#XuyUP2kNXpRWXmA&UE9aQGn8xp$R#͑GxGjqv$UE9wEwZ#QcUE%&qYpEh5pDx2zVkum&gTXRm6X4NG
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