山东省济南市2018届高考考前针对性练习(二模)数学(理)试卷及答案 下载本文

????4a2?8a?0?只需?,即a?0,

????1??0????2??a1??1???1?Q???1????0??1,?????1??0,?x1???1,??,?x2???,0?

22??2???2??g?x?在??1,x1?上单调递增,在?x1,x2?上单调递减,在?x2,???上单调递增; ?g?x?在x?x1处取得极大值,在x?x2处取得极小值,符合题意;

Q??x2??2ax2?2ax2?1?0

22?1n?1?x2???g?x2?=1n?1?x2??ax2112? ?x?1n1?x??2222x2?2x22x2?2设m?t?=1n?1?t??1112t?1?1?,t???,0?,m'?t?=???0, 2222t?21?t??2?t?1?2?t?1?111?1??1??m?t?在??,0?上是增函数, ?m?t??m???=1n+=?1n2

222?2??2??g?x2??1?1n2. 2【解法二】

2ax2?2ax?11, ?2ax?g?x?=1n?1?x??ax,?g'?x?=x?11?x2设??x??2ax?2ax?1,

2若a?0,??x??1?0,?g'?x??0,

?g?x?在??1,???上单调递增,不合题意;

当a?0时, Q???1????0??1,

???x??0在??1,???上只有一个根,不合题意;

当a?0时, Q???1????0??1,要使方程??x??2ax?2ax+1=0有两个实根x1,x2,

2???=4a2?8a?0?只需?,即a?2

????1??0?????2?a1??1???1?Q???1????0??1,?????1??0,?x1???1,??,?x2???,0?

22??2???2??g(x)在??1,x1?上单调递增,在?x1,x2?单调递减,在?x2,???上单调递增; ?g(x)在x?x1处取最大值,在x?x2处取最小值,符合题意;

2Q??x2??2ax2?2ax2?1?0

设2ax2?t,则tx2?t?1?0,?t??1???2,?1?, x2?12?g?x2??1n?1?x2? ?ax2??1n??t??

1t?,t???2,?1? 221t11t?2设m?t???1n??t???,t???2,?1?,m'?t???????0,

22t22t1?m?t?在??2,?1?单调递增,?m?t??m??2???1n2

21?g?x2???1n2.

222.[选修4-4:坐标系与参数方程] 解:(1) l的普通方程为: x?y?1?0; 又Q???sin??2, ?x?y?y?2

222222x2?y2?1 即曲线C的直角坐标方程为: 2?1x???2??11?(2)解法一: P?,?在直线l上,直线l的参数方程为??22?1?y???2?2't22't2(t为参数),代入曲线C的直

'?1?13'22'52'?2'?t?t??0, t?2?t?2?0角坐标方程得??,即??22???22??224????''?t1't2?PA?PB? t1'?t2225. 6?4?y?1?x解法二: ?? 3x2?4x?0 ?x1?0,x2?

223??x?2y?2?41??A?0,1?,B?,?? ,

?33?1??1?252??41??11??PA??0????1???,?PB??????????,

22232326????????PA?PB?2525?? 266222223.[选修4-5:不等式选]

解:(1) f(x)?f(2x?5)?x?1?2x?4?x?9

当x??2时,不等式为4x??12?x??3,?x????,?3?; 当?2?x?1时,不等式为5?9,不成立;

当x?1时,不等式为2x?6?x?3,?x????,?3?, 综上所述,不等式的解集为???,?3?U?3,???;

(2)解法一: f(x?a)?f(x?b)?x?a?1?x?b?1?x?a?1??x?b?1??a?b,

?12?5b2a? a?b??a?b???a?b???????2ab?22ab?5b2a9?2?? 22ab2当且仅当

b2a?,即b?2a时“?”成立; 2ab??3?b?2a由?可得:a?,b?3.

2?1?2?1??2ab解法二:f(x?a)?f(x?b)?x?a?1?x?b?1,

当x?1?a时,f(x?a)?f(x?b)??x?a?1?x?b?1??2x?2?a?b?a?b; 当1?a?x?1?b时,f(x?a)?f(x?b)?x?a?1?x?b?1?a?b; 当x?1?b时,f(x?a)?f(x?b)?x?a?1?x?b?1?2x?2?a?b?a?b

?f(x?a)?f(x?b)的最小值为a?b,

?a?b???a?b???当且仅当

b2a912?5b2a5??, ???????22ab22ab??22ab2b2?,即b?2a时“?”成立; 2ab??3?b?2a由?可得:a?,b?3.

2?1?2?1??2ab