2024届福州市5月质检文科数学试题及答案(3)

∴ AC?平面PAB． ····································································· 8分 （Ⅲ）∵ ?PBC为正三角形， ∴ PB?BC?2．

在Rt?PAB中，PA?PB2?AB2?2．

11?12?························· 10分 ∴ VC?PAB?S?PAB?AC????2?2??2?， ·

33?23?11?32?3VA?PBC?S?PBC?h????2?h?h（其中h为三棱锥A?PBC的高）． ??33?43?? ································································································ 11分 ∵ VC?PAB?VA?PBC， ∴ h?6············································································· 12分 ． ·321．（本小题满分12分）

解：（Ⅰ）设圆C的半径为r（r?0），依题意，圆心坐标为(r,2)． ········ 1分 ∵ MN?32

225?3?∴ r????22，解得r2?． ····················································· 3分

4?2?5?252?∴ 圆C的方程为?x????y?2??． ·········································· 4分

24??5?252?（Ⅱ）把y?0代入方程?x????y?2??，解得x?1，或x?4，

2?4?22即点M?1,0?，N?4,0?． ·································································· 5分 ⑴ 当AB?x轴时，由椭圆对称性可知?ANM??BNM． ······················· 6分 ⑵ 当AB与x轴不垂直时，可设直线AB的方程为y?k?x?1?． 联立方程??y?k?x?1?，消去y得，k2?2x2?2k2x?k2?8?0． ··············· 7分 22?2x?y?8??设直线AB交椭圆?于A?x1,y1?、B?x2,y2?两点，则

2k2k2?8，x1?x2?2． ························································ 8分 x1?x2?2k?2k?2∵ y1?k?x1?2?,y2?k?x2?2?， ∴ kAN?kBN??k?x1?1?k?x2?1?y1y2???x1?4x2?4x1?4x2?4

k?x1?1??x2?4??k?x2?1??x1?4??x1?4??x2?4?2?k?8?2．

∵?x1?1??x2?4???x2?1??x1?4??2x1x2?5?x1?x2??8

10k2??2?8?0， ···························································· 10分 k2?2k?2∴ kAN?kBN?0，∴?ANM??BNM． ············································ 11分 综上所述，?ANM??BNM． ························································· 12分 22．（本小题满分14分）

解：（Ⅰ）当a?2时，f(x)?ln(x?1)?∴f?(x)?2x（x??1）， ························ 1分 x?112x?3??， ······················································ 2分 x?1(x?1)2(x?1)2∴ f?(0)?3，所以所求的切线的斜率为3． ········································· 3分 又∵f?0??0，所以切点为?0,0?.

故所求的切线方程为：y?3x． ······················································ 4分 （Ⅱ）∵f(x)?ln(x?1)?∴f?(x)?ax，(x??1) x?11a(x?1)?axx?1?a?? ·················································· 5分 x?1(x?1)2(x?1)2①当a≥0时，∵x??1，∴f?(x)?0； ················································ 6分 ②当a?0时， 由??f?(x)?0?f?(x)?0，得?1?x??1?a；由?，得x??1?a； ··················· 8分

x??1x??1??综上，当a≥0时，函数f(x)在(?1,??)单调递增；

当a?0时，函数f(x)在(?1,?1?a)单调递减，在??1?a,???上单调递增． ··· 9分 （Ⅲ）①当a≥0时，由（Ⅱ）可知，函数f(x)在(?1,??)单调递增．此时，································· 11分 ?a,a?1????1,???，故f?x?在?a,a?1?上为增函数． ·

②当a?0时，由（Ⅱ）可知，函数f(x)在??1?a,???上单调递增． ∵ f?x?在?a,a?1?上为增函数，

∴

1?a,a?1????1?a,???，故a≥?1?a，解得a≥?，

21········································································· 13分 ≤a?0． ·

2?1?2??∴ ?综上所述，a的取值范围为??,???． ············································· 14分

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