6.设变量x y ,满足约束条件:222y x x y x ??
+??-?
,
,.≥≤≥,则y x z 3-=的最小值为( )
A .2-
B .4-
C .6-
D .8-
7.设曲线2ax y =在点(1,a )处的切线与直线062=--y x 平行,则=a ( )
A .1
B .
1
2
C .12
-
D .1-
8.正四棱锥的侧棱长为32,侧棱与底面所成的角为?60,则该棱锥的体积为( )
A .3
B .6
C .9
D .18
9.44)1()1(x x +
-的展开式中x 的系数是( )
A .4-
B .3-
C .3
D .4
10.函数x x x f cos sin )(-=的最大值为( )
A .1
B .
2 C .3
D .2
11.设ABC △是等腰三角形,120ABC ∠=
,则以A B ,为焦点且过点C 的双曲线的离心率为( )
A .
2
2
1+ B .
2
3
1+ C . 21+ D .31+
12.已知球的半径为2,相互垂直的两个平面分别截球面得两个圆.若两圆的公共弦长为2,则两圆的圆心距等于( )
A .1
B .2
C .3
D .2
第Ⅱ卷
二、填空题:本大题共4小题,每小题5分,共20分.把答案填在题中横线上.
13.设向量(1
2)(23)==,,,a b ,若向量λ+a b 与向量(47)=--,c 共线,则=λ . 14.从10名男同学,6名女同学中选3名参加体能测试,则选到的3名同学中既有男同学
又有女同学的不同选法共有 种(用数字作答)
15.已知F 是抛物线24C y x =:的焦点,A B ,是C 上的两个点,线段AB 的中点为
(22)M ,,则ABF △的面积等于 .
16.平面内的一个四边形为平行四边形的充要条件有多个,如两组对边分别平行,类似地,写出空间中的一个四棱柱为平行六面体的两个充要条件:
充要条件① ; 充要条件② . (写出你认为正确的两个充要条件)
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.(本小题满分10分) 在ABC △中,5cos 13A =-
,3
cos 5
B =. (Ⅰ)求sin
C 的值;
(Ⅱ)设5BC =,求ABC △的面积. 18.(本小题满分12分)
等差数列{}n a 中,410a =且3610a a a ,,成等比数列,求数列{}n a 前20项的和20S . 19.(本小题满分12分)
甲、乙两人进行射击比赛,在一轮比赛中,甲、乙各射击一发子弹.根据以往资料知,甲击中8环,9环,10环的概率分别为0.6,0.3,0.1,乙击中8环,9环,10环的概率分别为0.4,0.4,0.2.
设甲、乙的射击相互独立.
(Ⅰ)求在一轮比赛中甲击中的环数多于乙击中环数的概率;
(Ⅱ)求在独立的三轮比赛中,至少有两轮甲击中的环数多于乙击中环数的概率. 20.(本小题满分12分)
如图,正四棱柱1111ABCD A BC D -中,124AA AB ==,点
E 在1CC 上且EC E C 31=. (Ⅰ)证明:1AC ⊥平面BED ; (Ⅱ)求二面角1A DE B --的大小. 21.(本小题满分12分)
设a ∈R ,函数2
3
3)(x ax x f -=.
(Ⅰ)若2=x 是函数)(x f y =的极值点,求a 的值;
(Ⅱ)若函数()()()[02]g x f x f x x '=+∈,,
,在0=x 处取得最大值,求a 的取值范围. 22.(本小题满分12分)
A
B C
D E
A 1
B 1
C 1
D 1
设椭圆中心在坐标原点,(20)(01)A B ,,,是它的两个顶点,直线)0(>=k kx y 与AB 相交于点D ,与椭圆相交于E 、F 两点.
(Ⅰ)若6ED DF =
,求k 的值;
(Ⅱ)求四边形AEBF 面积的最大值.
2008年普通高等学校招生全国统一考试 文科数学试题(必修+选修Ⅰ)参考答案和评分参考
评分说明:
1.本解答给出了一种或几种解法供参考,如果考生的解法与本解答不同,可根据试题的主要 考查内容比照评分参考制订相应的评分细则.
2.对计算题,当考生的解答在某一步出现错误时,如果后继部分的解答未改变该题的内容和 难度,可视影响的程度决定后继部分的给分,但不得超过该部分正确解答应得分数的一半;如果后继部分的解答有较严重的错误,就不再给分.
3.解答右端所注分数,表示考生正确做到这一步应得的累加分数. 4.只给整数分数.选择题不给中间分.
一、选择题
1.C 2.B 3.D 4.C 5.C 6.D 7.A 8.B 9.A 10.B 11.B 12.C 二、填空题
13.2 14.420 15.2
16.两组相对侧面分别平行;一组相对侧面平行且全等;对角线交于一点;底面是平行四边形.
注:上面给出了四个充要条件.如果考生写出其他正确答案,同样给分. 三、解答题 17.解:
(Ⅰ)由5cos 13A =-
,得12sin 13
A =, 由3cos 5
B =,得4
sin 5
B =. ······························································································· 2分
所以16
sin sin()sin cos cos sin 65
C A B A B A B =+=+=. ··············································· 5分
(Ⅱ)由正弦定理得45sin 13512sin 313
BC B AC A ?
?==
=. ······················································ 8分
所以ABC △的面积1sin 2S BC AC C =???1131652365=???8
3
=. ··························· 10分
18.解:
设数列{}n a 的公差为d ,则
3410a a d d =-=-, 642102a a d d =+=+,
1046106a a d d =+=+. ·
···································································································· 3分 由3610a a a ,,成等比数列得2
3106a a a =,
即2(10)(106)(102)d d d -+=+, 整理得2
10100d d -=,
解得0d =或1d =. ············································································································· 7分 当0d =时,20420200S a ==. ························································································· 9分 当1d =时,14310317a a d =-=-?=, 于是2012019
202
S a d ?=+207190330=?+=. ·························································· 12分 19.解:
记12A A ,分别表示甲击中9环,10环,
12B B ,分别表示乙击中8环,9环,
A 表示在一轮比赛中甲击中的环数多于乙击中的环数,
B 表示在三轮比赛中至少有两轮甲击中的环数多于乙击中的环数,
12C C ,分别表示三轮中恰有两轮,三轮甲击中环数多于乙击中的环数.
(Ⅰ)112122A A B A B A B =?+?+?,
············································································· 2分 112122()()P A P A B A B A B =++ 112122()()()P A B P A B P A B =++
112122()()()()()()P A P B P A P B P A P B =++
0.30.40.10.40.10.40.2=?+?+?=. ············································································· 6分
(Ⅱ)12B C C =+, ············································································································· 8分
2
2213()[()][1()]30.2(10.2)0.096P C C P A P A =-=??-=,
332()[()]0.20.008P C P A ===,
1212()()()()0.0960.0080.104P B P C C P C P C =+=+=+=.···································· 12分 20.解法一:
依题设,2AB =,1CE =.
(Ⅰ)连结AC 交BD 于点F ,则BD AC ⊥.
由三垂线定理知,1BD AC ⊥.
···························································································· 3分 在平面1ACA 内,连结EF 交1AC 于点
G ,
由于
1AA AC
FC CE
== 故1Rt Rt A AC FCE △∽△,1
AAC CFE ∠=∠, CFE ∠与1FCA ∠互余.
于是1
AC EF ⊥. 1AC 与平面
BED 内两条相交直线BD EF ,都垂直, 所以1
AC ⊥平面BED . ······································································································· 6分 (Ⅱ)作GH DE ⊥,垂足为H ,连结1A H .由三垂线定理知1A H DE ⊥,
故1A HG ∠是二面角1A DE B --的平面角. ·
····································································· 8分
EF =
CE CF CG EF ?=
=
EG ==. 13EG EF =
,13EF FD GH DE ?=?=
又1
AC ==
11AG AC CG =-=.
11tan A G
A HG HG
∠=
= 所以二面角1A DE B --
的大小为 ······························································· 12分
1 A
B C
D
E
A 1
B 1
C 1
D 1 F
H G
解法二:
以D 为坐标原点,射线DA 为x 轴的正半轴, 建立如图所示直角坐标系D xyz -.
依题设,1(220)(020)(021)(204)B C E A ,,,,,,,,,,,.
(021)(220)DE DB ==
,,,,,,11(224)(204)AC DA =--= ,,,,,. ······································ 3分 (Ⅰ)因为10AC DB = ,10AC DE =
, 故1AC BD ⊥,1AC DE ⊥. 又DB DE D = ,
所以1AC ⊥平面DBE . ········································································································ 6分 (Ⅱ)设向量()x y z =,,n 是平面1DA E 的法向量,则
DE ⊥ n ,1DA ⊥ n .
故20y z +=,240x z +=.
令1y =,则2z =-,4x =,(412)=-,,n . ··································································· 9分
1
AC <>
,n 等于二面角1A DE B --的平面角,
1
1
1
cos 42AC AC AC <>==
,n n n 所以二面角1A DE B --
的大小为arccos 42
. ······························································ 12分 21.解:
(Ⅰ)2()363(2)f x ax x x ax '=-=-.
因为2x =是函数()y f x =的极值点,所以(2)0f '=,即6(22)0a -=,因此1a =. 经验证,当1a =时,2x =是函数()y f x =的极值点. ···················································· 4分 (Ⅱ)由题设,3
2
2
2
()336(3)3(2)g x ax x ax x ax x x x =-+-=+-+.
当()g x 在区间[02],
上的最大值为(0)g 时, (0)(2)g g ≥,
即02024a -≥.
故得65
a ≤
. ························································································································· 9分 反之,当6
5a ≤时,对任意[02]x ∈,,
26
()(3)3(2)5g x x x x x +-+≤
23(210)5x
x x =+- 3(25)(2)5
x
x x =+- 0≤,
而(0)0g =,故()g x 在区间[02],上的最大值为(0)g .
综上,a 的取值范围为65?
?-∞ ???
,. ······················································································ 12分
22.(Ⅰ)解:依题设得椭圆的方程为2
214
x y +=, 直线AB EF ,的方程分别为22x y +=,(0)y kx k =>.················································ 2分 如图,设001122()()()D x kx E x kx F x kx ,,,,,,其中12x x <, 且12x x ,满足方程2
2
(14)4k x +=,
故21x x =-=
.①
由6ED DF = 知01206()x x x x -=-
,得021215
(6)77x x x x =+==;
由D 在AB 上知0022x kx +=,得02
12x k
=+. 所以
212k =
+, 化简得2
242560k k -+=,
解得23k =
或38
k =. ············································································································ 6分 (Ⅱ)解法一:根据点到直线的距离公式和①式知,点E F ,到AB 的距离分别为
1
h==
2
h== ·····································································9分
又AB==AEBF的面积为
12
1
()
2
S AB h h
=+
1
2
=
=
=
≤
当21
k=,即当
1
2
k=时,上式取等号.所以S
的最大值为 ·······························12分解法二:由题设,1
BO=,2
AO=.
设
11
y kx
=,
22
y kx
=,由①得
2
x>,
21
y y
=->,
故四边形AEBF的面积为
BEF AEF
S S S
=+
△△
22
2
x y
=+ ······························································································································9分
=
=
=
当
22
2
x y
=时,上式取等号.所以S
的最大值为··················································12分