∴a n =5n -2 ············································································· 6分
（2）b =?1
=1? 1 - 1 ? n
(5n -2)(5n +3) 5?5n -2 5n +3?
T =1?1
- 1 ?=1- 1 ··································10分 n 5 3 5n +3?
15 5(5n +3) 1 T n <15 ··················································································
12分
18.（1） 平面EDAF ⊥平面ABCD ，DE ?平面EDAF 平面EDAF 平面ABCD =AD ，DE ⊥AD
∴DE ⊥平面ABCD ·
······························································································· 2分 AC ?平面ABCD ∴DE ⊥AC
四边形ABCD 是正方形∴AC ⊥BD ······································································ 3分
DE 、BD ?平面BDE ，DE BD =D ，∴AC ⊥平面BDE ··································· 4分
AC ?平面ACE ∴平面AEC ⊥平面BDE ··································································· 5分
（2）建系如图
平面BEF 的法向量n =(1,1,1) ····················································································· 7分
E (0,0,2)设H (a ,a ,0)，EH =(a ,a ,-2)
1
7
cos <
=
，解得a =
或a =
（舍） ············ 10分
9 2
4
∴H ?11 ?
，∴DH =
2 ··
·········································12分 , ,0?
2
?22 ?
19.（1）y ?=0.5x 2
+0.8 ·············································································· 6分
（2）模型二的残差平方和为0.42<3.7，模型二的拟合效果更好 ········· 6分 x 2 2
20.（1）M :
+y 4
=1 ············································································ 3分
（2）F 2
(
3,0)
，设直线l :x =my +
3，A (x 1,y 1)，B (x 2,y 2)，
??x 2+4y 2-4=0 ? ??x =my +
，代入得：(
m 2
+4)
y 2
+23my -1=0
由于?=16(
m 2
+1)
>0恒成立
1 3
'e -e
则有y1+y2=
m2+4 ，y1y2=
m2+4 ·················································
5分
4(m2+1)
AB= y-y= =?6分
1 2
点O到直线l的距离d=
m2+4 ······································7分
则S =1
?AOB 2AB?d= m2+4
23
≤1·············10分3
当且仅当：=3
，即m=
+
m2+1
时取等号，
又由于OC=-OA，知S 2 ?ABC =
5
S
2
?AOB ≤
5
，此时l:x=±
2
2y+.···12分
21.（1）a=1,f(x)的最大值为1 ····································································3分
（2）方法一：
由题意知，eλx2≤2
e x+e-x
恒成立
只需λx2≤ln 2
e x+e-x
恒成立
设g(x)=λx2-ln2+ln(e x+e-x)
由于g(x)为偶函数，只需x≥0时g(x)≤0，而g(0)=0 x -x
g(x)=+2λ
e x+e-x
x,g'(0)=0
g'(x)=
4
(e x+e-x)2+2λ
=
