2019届广东省佛山市南海区桂城中学等七校联合体高三冲刺模拟数学（理）试题Word版含答案

（2）设g(x)?ln(1?x)?x，则g'(x)?1x， ?1??x?1x?1当x???1,0?时，g'(x)?0，所以g(x)在??1,0?单调递增； 当x??0,???时，g'(x)?0，所以g(x)在?0,???单调递减；

所以g(x)?g?0??0········································································································ 6分 所以a??2时，f(x)?xln(1?x)?a?ln(1?x)?x??xln(1?x)?2?ln(1?x)?x?， 即f?x???x?2?ln(1?x)?2x，要证明f(x)?2x1?e?x， 只需证明?x?2?ln(1?x)??2xe?x??， ··················································································· 7分

由（1）知，f(x)??x?2?ln(1?x)?2x在??1,???单调递增， 所以，当x???1,0?时，?x?2?ln(1?x)?2x?0，即ln(1?x)?所以当x???1,0?时，?x?2?ln(1?x)?所以只需证明

2x···························· 9分 ?0， ·

x?22x?x?2?，

x?2x?2x?2x····························································· 10分 ??e?x，即证明e??1， ·

x?2x?2x2exx?2x?0， ·设h?x??···································································· 11分 e，则h'?x??2x?2?x?2?所以h?x?在??1,0?单调递增，所以h?x??h?0???1，所以原不等式成立，

综上，当a??2，x???1,0?时，f(x)?2x1?e?x． ························································ 12分 解法二：（1）同解法一． ········································································································· 5分 （2）同解法一得只需证明?x?2?ln(1?x)??2xe?x?x??， ·································································· 7分

设??x???x?2?ln(1?x)?2xe，则?'?x??ln(1?x)?x?2?2?1?x?e?x， x?1?''?x??13?x， ············································································· 8分 ??2x?2e??2x?1?x?1?由g(x)?0得，ex?x?1?0，即ex?x?1， 因为x???1,0?，所以e?x?1， ···················································································· 9分 x?12?x?2?2x2?x13???又因为x?2?0，所以?''?x??， 2x?1?x?1?2x?1?x?1?因为x???1,0?，所以2x?x?x?2x?1??0， ·································································· 10分

2所以?''?x??0，?'?x?在??1,0?单调递增，所以?'?x???'?0??0， 所以??x?在??1,0?单调递减，所以??x????0??0，即f(x)?2x1?e?x．

综上，当a??2，x???1,0?时，f(x)?2x1?e?x． ···················································· 12分

????[来源学科网]

解法三：（1）同解法一． ········································································································· 5分 （2）同解法一得要证明f(x)?2x1?e?x，只需证明

即证明?x?2?e?x??x?2·········································· 10分 ??e?x， ·

x?2?x?2?0，设h?x???x?2?e?x?x?2，

?xex?x?1则h'?x????x?1?e?1?， xe由g(x)?0得，ex?x?1?0，即ex?x?1，所以h'?x??0， ·············································· 11分 所以h?x?在??1,0?单调递增，所以h?x??h?0??0， 即?x?2?e?x?x?2?0，所以f(x)?2x?1?e?x?．

综上，当a??2，x???1,0?时，f(x)?2x1?e?x． ························································ 12分 解法四：（1）同解法一． ········································································································· 5分 （2）同解法一得要证明f(x)?2x1?e?x，只需证明

xx????x?2··········································· 10分 ??e?x， ·

x?2即证明?x?2?e?x?2?0，设h?x???x?2?e?x?2， 则h'?x???x?1?e?1，h''?x??xe，

xx因为x???1,0?，所以h''?x??0，所以h'?x?在??1,0?单调递减，

所以h'?x??h'?0??0， ································································································ 11分 所以h?x?在??1,0?单调递增，所以h?x??h?0??0， 即?x?2?e?x?2?0，所以f(x)?2x1?e?x．

x??综上，当a??2，x???1,0?时，f(x)?2x1?e?x． ························································ 12分

??22．本题考查极坐标方程、直线的参数方程、椭圆的参数方程等基础知识；考查运算求解能力、推理论证能力；考查数

形结合思想、化归与转化思想、分类与整合思想等．满分10分． 解：（1）当cos??0，即??kπ?π··········································· 1分 ?k?Z?时，l的普通方程为x?0； ·

2当cos??0，即??kπ?π············································ 2分 ?k?Z?时，l的普通方程为y?xtan?. ·

2[来源:学科网]?x??cos?,422x+4y?4，由?及?2?得，2y??sin?1?3sin??

x22即C的直角坐标方程为+y?1. ······················································································· 5分

4（2）依题意，设l：y?kx，

[来源:Zxxk.Com]

所以C上恰有2个点到l的离等于2等价于C上的点到l的距离的最大值为2. ························· 6分

sin??,则P到l的距离 设C上任一点P?2cos?，d?sin??2kcos?1?k2?1?4k2sin(???)1?k2（其中sin???2k4k2?1,cos??14k2?1················· 8分 ）， ·

当sin(???)=?1时，dmax?1?4k21?k2?2， ········································································ 9分

解得，k??22，所以l的斜率为?． ··········································································· 10分 2223．本题考查绝对值不等式的解法、绝对值不等式的性质等基础知识；考查推理论证能力、运算求解能力；考查化归

与转化，分类与整合思想等．满分10分．

解：（1）当x?4时，原不等式等价于x?2?x?4?3x，解得x??2，所以x?4；

当x??2时，原不等式等价于?x?2?x?4?3x，解得x?2，所以此时不等式无解； 5当?2?x?4时，原不等式等价于x?2?x?4?3x，解得x?2，所以2?x?4； ··············· 4分 综上所述，不等式解集为2,???． ····················································································· 5分 （2）由

?f?x??kx?1，得x?2?x?4?kx?1，

当x?1时，6?0恒成立，所以k?R； ·············································································· 6分 当x?1时，k?x?2?x?4x?1?3?x?1?333． ··························· 7分 ??1??1?x?1x?1x?1x?1因为1?333??3??······················································ 8分 ?1???1??1?????2， ·

x?1x?1?x?1??x?1???3??3?1?········································ 9分 ????0即x?4或x??2时，等号成立， ·

x?1??x?1?当且仅当?1?所以，k?2；

综上，k的取值范围是???,2． ······················································································· 10分

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