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41. （本小题满分12分）
已知f?x??2xlnx?x2?ax?3．
（1）当a?1时，求曲线y?f?x?在x?1处的切线方程； ?1?（2）若存在x0??,e?，使得f?x0?≥0成立，求a的取值范围．
?e?【命题意图】本题主要考查函数和导数及其应用等基础知识，意在考查直观想象、逻辑推理与数学运算等数学核心素养．满分12分.
【解答】f??x??2?lnx?1??2x?a． ······················································· 1分 （1）当a?1时，f?x??2xlnx?x2?x?3,f??x??2?lnx?1??2x?1，
所以f?1??5,f??1??5， ········································································ 3分 所以曲线y?f?x?在x?1处的切线方程为y?5?5?x?1?，即y?5x． ············ 5分 ?1?（2）存在x0??,e?，使得f?x0?≥0成立，
?e?2xlnx?x2?3?1?等价于不等式a≥?在?,e?有解． ······································ 6分
x?e??x?3??x?1?， ·x2?2x?32xlnx?x2?3设h?x???，则h??x???············· 7分 ??22xxx1当?x?1时，h??x??0，h?x?为增函数；当1?x?e时，h??x??0，h?x?为减函数．e ·············································································································· 8分
3e2?2e?1e2?2e?3?1??1?又h????，h?e???，故h???h?e?＜0 ················· 10分
eeee????3e2?2e?1?1??1?所以当x??,e?时，h?x?＞h????，··································· 11分
e?e??e??3e2?2e?1?3e2?2e?1,???． 所以a＞?，即a的取值范围为??··················· 12分 ee??42. （本小题满分12分）
6x2y2已知椭圆C:2?2?1(a＞b＞0)的离心率为，以C的短轴为直径的圆与直线
3abl:3x?4y?5?0相切．
（1）求C的方程；
（2）直线y?x?m交椭圆C于M?x1,y1?，N?x2,y2?两点，且x1＞x2．已知l上存在点P，使得△PMN是以?PMN为顶角的等腰直角三角形．若P在直线MN右下方，求m的值．
【命题意图】本题主要考查椭圆的标准方程、直线与椭圆的位置关系、直线和圆的位置关系等基础知识，意在考查直观想象、逻辑推理与数学运算等数学核心素养．满分12分.
【解答】（1）依题意，b?0?0?53?422?1， ·················································· 2分
ca2?b26因为离心率e??， ?aa3a2?16?所以，解得a?3， ··························································· 4分 a3x2所以椭圆C的标准方程为?y2?1． ····················································· 5分
3（2）因为直线y?x?m的倾斜角为45?，且△PMN是以?PMN为顶角的等腰直角三角形，P在直线MN右下方，所以NP∥x轴． ············································· 6分
过M作NP的垂线，垂足为Q，则Q为线段NP的中点，所以Q?x1,y2?，故
P?2x1?x2,y2?， ························································································ 7分
所以3?2x1?x2??4y2?5?0， 即3?2x1?x2??4?x2?m??5?0，
整理得6x1?x2?4m?5?0．① ·································· 8分
ONQyMxPl?x?3y?3,由?得4x2?6mx?3m2?3?0. ?y?x?m22所以??36m2?48m2?48＞0，解得?2＜m＜2， ····································· 9分 3所以x1?x2??m，②
2x1x2?32·········································································· 10分 ?m?1?，③ ·
4m，④ 2由①-②得，x1?1?将④代入②得x2??1?m，⑤ ······························································· 11分
3?m?将④⑤代入③得??1??m?1???m?1??m?1?，解得m??1．
4?2?综上，m的值为?1． ········································································· 12分 （二）选考题：共10分．请考生在第22，23两题中任选一题作答．如果多做，则按所做第一个题目计分，作答时请用2B铅笔在答题卡上将所选题号后的方框涂黑． 22.（本小题满分10分）选修4?4：坐标系与参数方程
?x??3?t,已知直角坐标系xOy中，曲线C1的参数方程为?（t为参数）．以O为极点，
y?t?x轴的正半轴为极轴，建立极坐标系，曲线C2的极坐标方程为?2?1?2?cos?．
（1）写出C1的普通方程和C2的直角坐标方程；
（2）设点P为C2上的任意一点，求P到C1距离的取值范围．
【命题意图】本题主要考查直线的参数方程、曲线直角坐标方程、极坐标方程的互化，圆的极坐标方程等基础知识，意在考查直观想象、逻辑推理与数学运算的数学核心素养．满分10分.
························· 2分 【解答】（1）C1的普通方程为x?y??3，即x?y?3?0． ·
2···················· 5分 曲线C2的直角坐标方程为x2?y2?1?2x，即?x?1??y?2．·
2·························· 6分 （2）由（1）知，C2是以?1,0?为圆心，半径r?2的圆， ·圆心C2?1,0?到C1的距离d?1?0?32?22＞2， ···································· 7分
所以直线C1与圆C2相离，P到曲线C1距离的最小值为d?r?22?2?2；最大值d?r?22?··········································································· 9分 2?32， ·
?······································ 10分 所以P到曲线C1距离的取值范围为??2,32?． ·
23.（本小题满分10分）选修4?5：不等式选讲
已知a＞0,b＞0,c＞0，且a?b?c?2． （1）求a2?b?c的取值范围； （2）求证：
149??≥18． abc【命题意图】本题主要考查配方法、基本不等式和不等式证明等基础知识，意在考查直观想象、逻辑推理与数学运算的数学核心素养．满分10分.
····························· 1分 【解答】（1）依题意，2?a?b?c＞0，故0＜a＜2． ·1?7?············································· 3分 所以a?b?c?a??2?a???a???， ·
2?4?2227?7?22········ 5分 所以≤a?b?c＜2??2?2??4，即a2?b?c的取值范围为?,4?． ·
4?4?（2）因为a＞0,b＞0,c＞0，
b4ac9a4c9b?149????? ···························· 7分 所以?a?b?c??????14??abcabacbc??≥14?2b4ac9a4c9b??2??2? ···················· 8分 abacbc···························· 9分 ?14?24?29?236?36． ·
又因为a?b?c?2， 所以
149??≥18． ·········································································· 10分 abc