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山东省2016年冬季普通高中学业水平考试
数学试题参考答案及评分标准
第Ⅰ卷(共 60 分)
一、选择题(本大题共20个小题,每小题3分,共60分)
第Ⅱ卷(共 40 分)
二、填空题(本大题共5个小题,每小题3分,共15分)
21.
1
2
22.3 23.2 24 25.10
三、解答题(本大题共3个小题,共25分)
本大题的每小题给出一种(两种)解法,考生的其他解法,只要步骤合理,解答正
确,均应参照下述评分说明,给予相应的分数.
26.证明:在△ABC 中,因为E ,F 分别是棱AB ,AC 的中点,
所以EF ∥BC . ··············································································· 4分
又因为EF ?平面BCD , ···································································· 5分
BC ?平面BCD , ···································································· 6分
所以EF ∥平面BCD . ······································································· 8分
27.解:22()cos sin cos2f x x x x =-=. ························································· 2分
(1)()cos(2)cos 12126f πππ=?== ························································ 5分
(2)由2k π-π≤2x ≤2k k π∈Z ,
, 得2
k π
π-
≤x ≤k π,k ∈Z . ····························································· 7分 所以()f x 的单调递增区间为[2
k k ππ-π],,k ∈Z . ································· 8分 28.解:(1)因为函数()f x 有零点, 所以方程21
04
x ax ++
=有实数根. 所以21a ?-=≥0,解得a ≤1-,或a ≥1.
因此,所求a 的取值范围是a ≤1-,或a ≥1. ······································· 2分 (2)注意到1(0)04f =>,函数()f x 图象的对称轴为直线2
a
x =-. 解法一:
①若02
a
-…,即0a …时,
()f x 在区间(01),内没有零点. ·
························································· 3分
②若12
a -…,即2a -…时,
当5(1)04f a =+
<时,即5
4a <-,所以2a -…时,()f x 在区间(01),
内有1个零点; 当5(1)04f a =+…时,即5
4a -…,与2a -…矛盾,无解. ······················· 4分
③若012a
<-<时,即20a -<<时,
当221
()02424
a a a f -=
-+>,解得11a -<<. 所以10a -<<时,()f x 在区间(01),内没有零点. ·
······························· 5分 当221
()02424
a a a f -=
-+=,解得1a =±. 所以1a =-时,()f x 在区间(01),内有1个零点. ·
································· 6分 当221
()02424
a a a f -=
-+<,解得1a <-,或1a >. 若5
(1)04
f a =+
>, 得5
14
a -<<-,此时()f x 在区间(01),内有2个零点.
··························· 7分 若5(1)04f a =+…,
得5
24
a -<-…,此时()f x 在区间(01),内有1个零点. ·
························· 8分 综上,当1a >-时,()f x 在区间(01),内没有零点; 当1a =-,或5
4
a -…时,()f x 在区间(01),内有1个零点;
当514
a -<<-时,()f x 在区间(01),内有2个零点. ······························ 9分 解法二: ①当5(1)04f a =+
<时,即5
4
a <-时, ()f x 在区间(01),内有1个零点. ······················································· 3分
②当5(1)04f a =+
>时,即5
4a >-时, (ⅰ)若012
a
<-<时,即20a -<<时,
若221
()02424
a a a f -=
-+>,解得11a -<<. 所以当10a -<<时,()f x 在区间(01),内没有零点.
若221
()02424
a a a f -=
-+=,解得1a =±. 所以当1a =-时,()f x 在区间(01),内有1个零点.
若221
()02424
a a a f -=
-+<,解得1a <-,或1a >. 所以当514
a -<<-时,()f x 在区间(01),内有2个零点. ·
······················· 5分 (ⅱ)若12a -…,即2a -…时,与54a >-矛盾,无解. ·························· 6分
(ⅲ)若02a
-…,即0a …时,()f x 在区间(01),内没有零点. ·
··············· 7分 ③当5(1)04f a =+=时,即5
4
a =-时,
()f x 在区间(01),内有1个零点. ·
······················································ 8分 综上,当1a >-时,()f x 在区间(01),内没有零点; 当1a =-,或5
4
a -…时,()f x 在(01),内有1个零点;
当514
a -<<-时,()f x 在(01),内有2个零点. ·
··································· 9分