内容正文:
高三数学(理科)答案 第 1 页 共 13 页
2018届高三模拟考试
数学(理科)参考答案及评分标准 2018.3
一、选择题:本大题共 12小题,每小题 5分,共 60分.
ACBD DCBA BBAD
二、填空题:本大题共 4小题,每小题 5分,共 20分.
13. 4 14. 11 15. 2 2( 2) 2x y+ − = 16. 7[
8
,
11]
12
三、解答题:本大题共 6小题,共 70分.解答应写出文字说明、证明过程或演算步骤.
17.(I)当 1n = 时,有 21 1 13 6 4a a a+ = + ,即 1 1( 4)( 1) 0.a a− + =
因为 1 0a > ,所以 1 1 0.a + > 从而 1 4 0a − = ,即 1 4.a = ········································· 2分
由 2 3 6 4n n na a S+ = + ,知
2
1 1 13 6 4.n n na a S+ + ++ = +
两式相减,得 2 21 1 13 3 6 4 6 4n n n n n na a a a S S+ + ++ − − = + − − ····································· 4分
即 2 21 1 13 3 6n n n n na a a a a+ + ++ − − = ,即
2 2
1 13 3 0n n n na a a a+ +− − − = ,
即 1 1( )( 3) 0.n n n na a a a+ ++ − − =
因为 0na > ,所以 1 3 0n na a+ − − = ,即 1 3.n na a+ − = ············································ 6分
所以,数列{ }na 是首项为 4,公差为3的等差数列.
所以 na = 4 3( 1) 3 1.n n+ − = + ················································································· 8分
(II)由(I)知 nb =
3 1 1 .
(3 1)(3 4) 3 1 3 4n n n n
= −
+ + + +
················································ 10分
数列{ }nb 的前 n项和为
1 1 1 1 1 1 1 1( ) ( ) ( ) ( )
4 7 7 10 3 2 3 1 3 1 3 4n n n n n
T = − + − + + − + −
− + + +
L
1 1
4 3 4n
= −
+
. ·································································································· 12分
高三数学(理科)答案 第 2 页 共 13 页
18.(I)证法 1:在平面 ABCD内过点C作两条直线 1 2,l l ,
使得 1l AB⊥ , 2 .l AD⊥ ························································································ 2分
因为 AB AD A=∩ ,所以 1 2,l l 为两条相交直线.
因为平面 SAB ⊥平面 ABCD,平面 SAB ∩平面 ABCD AB= ,
1l ⊂平面 ABCD, 1l AB⊥ ,所以 1l ⊥平面 .SAB
所以 1l ⊥ .SA ……………………………………4分
同理可证 2l ⊥ .SA ………………………………5分
又因为 1l ⊂平面 ABCD, 2l ⊂平面 ABCD, 1 2l l C=∩ ,
所以 SA ⊥平面 .ABCD ·························································································· 6分
证法 2:在平面 SAB内过点 S作 1l AB⊥ ,在平面 SAD内过点 S作 2 .l AD⊥ ············· 2分
因为平面 SAB ⊥平面 ABCD ,平面 SAB ∩平面 ABCD AB= , 1l