范文一:2015 朝阳高三一模 数学 文 试题
北京市朝阳区高三年级第一次综合练习
数学试卷(文史类) 2015. 4 (考试时间 120分钟 满分 150分)
一、选择题:本大题共 8小题,每小题 5分,共 40分.在每小题给出的四个选项中,选出 符合题目要求的一项.
(1)已知全集 {, , , }U a b c d =,集合 {, },{, }A a b B b c ==,则 () U
A B e等于
A . {}b
B . {}d
C . {, , }a c d D . {, , }a b c
(2)已知命题 :p x ?∈R , sin 1x ≤,则
A . :p ?x ?∈R , sin 1x ≥ B . :p ?x ?∈R , sin 1x > C . :
p ?0x ?∈R , 0sin 1x ≥ D . :p ? 0x ?∈R , 0sin 1x >
(3)若抛物线 22(0) y px p =>的焦点与双曲线
222x y -=的右焦点重合,则 的值为 A
B . 2 C . 4 D
.
(4)如图所示的程序框图表示的算法功能是
A .计算 123456S =?????的值 B .计算 12345S =????的值 C .计算 1234S =???的值 D .计算 1357S =???的值
p
第(4)题图
(5)已知
113
log 2
x =, 1
2
22x -=, 3x 满足 333
1() log 3x x =,则
A . 123x x x < b="" .="" 132x="" x="" x="">< c="" .="" 213x="" x="" x="">< d="" .="" 312x="" x="" x=""><>
ππ
() 2sin()cos()
66f x x x =--图象的一条对称轴方程是 A .
π6x = B . π3x = C . 5π12x = D . 2π3x =
(7)已知实数 x , y 满足 20, 20, 0, x y x y y t +≥??
-≤??≤≤?
其中 0t >.若 3z x y =+的最大值为 5,则 z 的最
小值为
A . 5
2 B . 1 C . 0 D . 1-
(8) 已知边长为 3的正方形 ABCD 与正方形 CDEF 所在的平面互相垂直, M 为线段 CD 上的动点(不含端点) ,过 M 作 //MH DE 交 CE 于 H ,作 //MG AD 交 BD 于 G ,连结
GH . 设 CM x =(03) x <, 则下面四个图象中大致描绘了三棱锥="" c="" ghm="" -的体积="" y="">,>
变量 x 变化关系的是
A . B .
C . D .
二、填空题:本大题共 6小题,每小题 5分,共 30分.把答案填在答题卡上.
(9) i 为虚数单位,计算 1i
1i +-=________
(10) 已知平面向量 a , b 满足
1==a b , a 与 b
的夹角为 60?,则 () ?+=a a b ________
(11)圆
22
:(2) (2) 8C x y -+-=与 y 轴相交于 , A B 两点,则弦 AB 所对的圆心角的大小 为 __________
(12) 一个四棱锥的三视图如图所示, 其中侧视图为正三角形, 则该四棱锥的体积是 ______, 四棱锥侧面中最大侧面的面积是 _________
(13) 稿酬所得以个人每次取得的收入, 定额或定率减除规定费用后的余额为应纳税所得额, 每次收入不超过 4000元, 定额减除费用 800元; 每次收入在 4000元以上的, 定率减除 20%的费用 . 适用 20%的比例税率,并按规定对应纳税额减征 30%,计算公式为: (1)每次收入不超过 4000元的:应纳税额 =(每次收入额-800) ×20%×(1-30%) (2)每次收入在 4000元以上的:应纳税额 =每次收入额 ×(1-20%) ×20%×(1-30%) . 已知某人出版一份书稿,共纳税 280元,这个人应得稿费 (扣税前 ) 为 __________元
(14)记 12x x -为区间 12[, ]x x 的长度.已知函数 2x
y =, x ∈
[]2, a -(0a ≥) ,其值域为 [], m n ,则区间 [], m n 的长度的最小值是 3.
三、解答题:本大题共 6小题,共 80分.解答应写出文字说明,演算步骤或证明过程. (15) (本小题满分 13分)
在 ABC ?中,
π3A =
, cos B =, 6BC =.
(Ⅰ)求 AC 的长; (Ⅱ)求 ABC ?的面积.
第(12)题图
正视图 侧视图
俯视图
某次考试结束后,为了解甲、乙两所学校学生的数学考试情况,随机抽取甲、乙两校各 10名 学 生 的 考 试 成 绩 , 得 茎 叶 图 如 图 所 示 (部 分 数 据 不 清 晰 ):
(Ⅰ)请根据茎叶图判断哪个学校的数学成绩平均水平较高(直接写出结果) ;
(Ⅱ) 若在抽到的这 20名学生中, 分别从甲、 乙两校随机各抽取 1名成绩不低于 90分的学 生,求抽到的学生中, 甲校学生成绩高于乙校学生成绩的概率.
如图, 在三棱柱 111C B A ABC -中, 各个侧面均是边长为 2的正方形,
D 为线段 AC 的中点.
(Ⅰ)求证:BD ⊥平面 11A ACC ; (Ⅱ)求证:直线 1AB ∥平面 D BC 1; (Ⅲ) 设 M 为线段
1BC 上任意一点,
在 D D BC 1内的平面区域 (包括边界) 是否存在点 E ,
使 CE ⊥DM ,并说明理由.
A
B
C
A 1
B 1
C 1
设数列
{}n a 的前 n 项和为 n S ,且 14a =, 1n n a S +=, n *∈N .
(Ⅰ)写出
2a , 3a , 4a 的值;
(Ⅱ)求数列
{}n a 的通项公式;
(Ⅲ)已知等差数列 {}n b 中,有 22b a =, 33b a =,求数列 {}n n a b ?的前 n 项和 n T .
已 知 椭圆 22
22:1(0) x y C a b a b +=>>的 两 个焦 点分别 为 12(2,0), (2,0)F F -, 离 心率
为 . 过焦点 2F 的直线 l (斜率不为 0) 与椭圆 C 交于 , A B 两点, 线段 AB 的中点为 D , O
为坐标原点,直线 OD 交椭圆于 , M N 两点.
(Ⅰ)求椭圆 C 的方程; (Ⅱ)当四边形 12MF NF 为矩形时,求直线 l 的方程.
已知函数
() ()e x
a
f x x x =+, a ∈R . (Ⅰ)当 0a =时,求曲线 () y f x =在点 (1,(1))f 处的切线方程; (Ⅱ)当 1a =-时,求证:() f x 在 (0,) +∞上为增函数;
(Ⅲ)若 () f x 在区间 (0,1)上有且只有一个极值点,求 a 的取值范围.
范文二:2015朝阳高三一模数学(理科)
北京市朝阳区高三年级第一次综合练习
数学学科测试(理工类)
2015. 4
(考试时间 120分钟 满分 150分)
本试卷分为选择题(共 40分)和非选择题(共 110分)两部分
第一部分(选择题 共 40分)
一、选择题 :本大题共 8小题,每小题 5分,共 40分 . 在每小题给出的四个选项中,选出
符合题目要求的一项 . 1. 已知集合
{}
21,2, A m =,
{}
1, B m =. 若 B A ?,则 m =
A. 0 B. 2 C. 0 或 2 D. 1或 2
2. 已知点 0(1,) A y 0(0) y >为抛物线 2
2y px
=()0p >上一点 . 若点 A 到该抛物线焦点的距离为
3,则 0y =
A.
B. 2
C. D. 4
3. 在 ABC ?中,若 π3
A =
,
cos B =6BC =, 则 AC =
A. B.4
C.
4. “ x ?∈R , 210x ax ++≥成立”是“ 2
a ≤”的 A .充分必要条件 B .必要而不充分条件 C . 充分而不必要条件 D. 既不充分也不必要条件
5. 某商场每天上午 10点开门, 晚上 19点停止进入. 在
如图所示的框图中, t 表示整点时刻, () a t 表示时间 段 [1, ) t t -内进入商场人次, S 表示某天某整点时刻 前进入商场人次总和,为了统计某天进入商场的总 人次数,则判断框内可以填
A. 17? t ≤ B . 19? t ≥ C . 18? t ≥ D . 18? t ≤
6. 设 123, , x x x 均为实数,且 1211log (1) 3x x ??=+ ???, 2321log 3x x ??= ???, 3
23
1log 3x x ??
= ???则
A. 132x x x < b.="" 321x="" x="" x="">< c.="" 312x="" x="" x="">< d.="" 213x="" x="" x="">
7. 在平面直角坐标系中, O 为坐标原点,已知两点 (1,0)A , (1,1)B ,且 90BOP ∠=. 设 OP OA kOB =+() k ∈R ,则
OP =
A . 1
2
B. C.
D. 2
8. 设集合 M =
{}2
20
000(, ) 20, , x y x
y x y +≤∈∈Z Z
, 则 M 中元素的个数为
A. 61 B. 65 C. 69 D.84
第二部分(非选择题 共 110分)
二、填空题:本大题共 6小题,每小题 5分,共 30分 . 把答案填在答题卡上 .
9. i 为虚数单位,计算 12i
1i -=
+ ______.
10. 设 n S 为等差数列 {}n a 的前 n 项和 . 若 383a a +=, 31S =, 则通项公式 n a =______.
11. 在极坐标中,设 002πρθ>≤<>
,曲线 2ρ=与曲线 sin 2ρθ=交点的极坐标为 ______. 12. 已知有身穿两种不同队服的球迷各有三人,现将这六人排成一排照相,要求身穿同一种
队服的球迷均不能相邻,则不同的排法种数为 . (用数字作答)
13. 设 3z x y =+,实数 x , y 满足
20, 20, 0, x y x y y t +≥??
-≤??≤≤?
其中 0t >.若 z 的最大值为 5,则实数 t 的
值为 ______,此时 z 的最小值为 ______.
14. 将体积为 1的四面体第一次挖去以各棱中点为顶点的构成的多面体, 第二次再将剩余的
每个四面体均挖去以各棱中点为顶点的构成的多面体,如此下去,共进行了 n (n *
∈N ) 次 . 则第一次挖去的几何体的体积是 ______;这 n 次共挖去的所有几何体的体积和是
______.
三、解答题:本大题共 6小题,共 80分 . 解答应写出文字说明,演算步骤或证明过程 . 15. (本小题满分 13分)
已知函数
2
() cos cos f x x x x =, x ∈R . (Ⅰ)求 () f x 的最小正周期和单调递减区间;
(Ⅱ)设 x m =() m ∈R 是函数 () y f x =图象的对称轴,求 sin 4m 的值.
16. (本小题满分 13分)
如图所示, 某班一次数学测试成绩的茎叶图和频率分布直方图都受到不同程度的污损, 其中,频率分布直方图的分组区间分别为 [)
50,60, [
)
60,70, [
)
70,80, [)80,90, [90,100].
据此解答如下问题.
(Ⅰ)求全班人数及分数在 [80,100]之间的频率;
(Ⅱ)现从分数在 [80,100]之间的试卷中任取 3份分析学生失分情况,设抽取的试卷分
数在 [90,100]的份数为 X ,求 X 的分布列和数学期望.
17. (本小题满分 14分)
如图,正方形 ADEF 与梯形 ABCD 所在平面互相垂直, 已知 //, AB CD AD CD ⊥, 12AB AD CD
==.
(Ⅰ)求证 :BF //平面 CDE ;
(Ⅱ)求平面 BDF 与平面 CDE 所成锐二面角的余弦值 ; (Ⅲ ) 线 段 EC 上 是 否 存 在 点 M , 使 得 平 面 BDM ⊥平 面
0.0375 0.0125
0.025 E
BDF ? 若存在,求出 EM
EC 的值;若不存在,说明理由 .
18. (本小题满分 13分)
已知函数 2
() ln (1) 2x f x a x a x
=+-+, a ∈R .
(Ⅰ) 当 1a =-时,求函数 () f x 的最小值; (Ⅱ) 当 1a ≤时,讨论函数 () f x 的零点个数 .
19. (本小题满分 14分)
已知椭圆 2222:1(0) x y C a b a b +=>>的一个焦点为 (2,0)F
,离心率为 3
F
的直线 l 与椭圆 C 交于 , A B 两点,线段 AB 中点为 D , O 为坐标原点,过 O , D 的直线 交椭圆于 , M N 两点.
(Ⅰ)求椭圆 C 的方程;
(Ⅱ)求四边形 AMBN 面积的最大值.
20. (本小题满分 13分)
若数列 {}n a 中不超过 () f m 的项数恰为 m b () m ∈*N , 则称数列 {}m b 是数列 {}n a 的生成数 列,称相应的函数 () f m 是 {}n a 生成 {}m b 的控制函数.设 2() f m m =. (Ⅰ)若数列 {}n a 单调递增,且所有项都是自然数, 11=b ,求 1a ; (Ⅱ)若数列 {}n a 单调递增,且所有项都是自然数, , 11b a =求 1a ;
(Ⅲ ) 若 2(1,2,3) n a n n ==,是否存在 {}m b 生成 {}n a 的控制函数 2() g n pn qn r =++(其中
常数 , , p q r ∈Z )?使得数列 {}n a 也是数列 {}m b 的生成数列?若存在,求出 ) (n g ;若 不存在,说明理由 .
北京市朝阳区高三年级第一次综合练习
数学答案(理工类) 2015. 4
二、填空题(满分 30分)
三、解答题 (满分 80分) 15. (本小题满分 13分)
解:(Ⅰ)由已知,函数 2() cos cos f x x x x = 1
(1cos2) 2
x =+2x =π1
sin(2) 62
x ++.
函数 () f x 的最小正周期为 πT =.
当 ππ3π2π22π262k x k +
≤+≤+
时(k ∈Z ) ,即 π2π
π+π+63
k x k ≤≤时,函数 () f x 为减函数 . 即 函数 () f x 的单调减区间为 π2ππ+, π+63k k ?
?????, k ∈Z . ………………… .9分
(Ⅱ) 由 x m =是函数 () y f x =图象的对称轴, 则 ππ2=π62m k +
+(k ∈Z )
, 即 126
m k π
=π+, k ∈Z . 则 423
m k 2π
=π+
. 则 sin 4m = ………………… .13分
16. (本小题满分 13分)
解 :(Ⅰ ) 由 茎 叶 图 可 知 , 分 布 在 [50,60)之 间 的 频 数 为 4, 由 直 方 图 , 频 率 为
0.0125100.125?=,
所以全班人数为
4
320.125
=人. 所以分数在 [80,100]之间的人数为 32(4810) 10-++=人 . 分数在 [80,100]之间的频率为
10
0.312532
= ………………… .4分 (Ⅱ)由(Ⅰ)知,分数在 [80,100]之间的有 10份,分数在 [90,100]之间的人数有
0.01251032=4创 份,由题意, X 的取值可为 0,1, 2,3.
363101(0) 6C P X C ===, 12
463101
(1) 2C C P X C ===,
21463103(2) 10C C P X C ===, 343101
(3) 30
C P X C ===.
所以随机变量 X 的分布列为
随机变量 X 的数学期望为 01236210305
EX =?+?+?
+?=. ………………… .13分 17. (本小题满分 14分)
解:(Ⅰ)因为 //, AB CD AB ?平面 , CDE CD ?平面 CDE ,
所以 //AB 平面 CDE ,同理, //AF 平面 CDE , 又 , AB AF A =所以平面 //ABF 平面 CDE ,
因为 BF ?平面 , ABF 所以 //BF 平面 CDE . ……………… .4分
(Ⅱ)因为平面 ADEF ^平面 ABCD ,平面 ADEF I 平面 ABCD =AD , C D A D ^
, CD ì平面 ABCD ,
所以 CD ^平面 ADEF . 又 DE ì平面 ADEF ,故 CD ED ^. 而四边形 ADEF 为正方形,所以 AD DE ^又 AD CD ^,
以 D 为原点, DA , DC , DE 所在直线分别为 x 轴, y 轴, z 轴,建立空间直角坐标
系 D xyz -. 设 1
AD =,则 (0,0,0),(1,1,0),(1,0,1),(0,2,0),(0,0,1)D B F C E , 取平面 CDE 的一个法向量 (1,0,0) DA =, 设平面 BDF 的一个法向量 (, , ) x y z =n ,
则 00DB DF ??=???=??n n ,即 00x y x z +=??+=?,令 1x =,则 1y z ==-, 所以 (1,1, 1) =--n .
设平面 BDF 与平面 CDE 所成锐二面角的大小为 θ,
则 cos |cos , |DA θ=<>=
=
n . ……………… .9分 所以平面 BDF 与平面 CDE
. (Ⅲ )
若 M 与 C 重合,则平面 () BDM C 的一个法向量 0(0,0,1)=m ,由(Ⅱ)知平面 BDF 的一个法向量 (1,1, 1) =--n ,则 010? m n =,则此时平面 BDF 与平面 BDM 不垂直 .
若 M 与 C 不重合,如图设 EM
EC
λ=() 01λ? ,则 (0,2,1) M λλ-,设平面 BDM 的一个法
向量 000(, , ) x y z =m ,
则 00DB DM ??=???=??m m ,即 000002(1) 0x y y z λλ+=??+-=?,令 01x =,则 0021, 1y z λλ=-=-,
所以 2(1, 1,
) 1λ
λ
=--m , 若平面 BDF ⊥平面 BDM 等价于 0?=m n ,即 2110, 1λλ+-
=-所以 []1
0,12λ=∈. 所以, EC 上存在点 M 使平面 BDF ⊥平面 BDM ,且 1
2
EM EC =. ……………… .14分 18. (本小题满分 13分)
解:(Ⅰ)函数 () f x 的定义域为 {}0x x >.
当 1a =-时, 2() ln 2
x f x x =-+.
211(1)(1)
() x x x f x x x x x -+-'=-+==
. 由 (1)(1) 0x x x +->() 0x >解得 1x >;由 (1)(1) 0x x x
+-<() 0x="">解得 01x <>
所以 () f x 在区间 (0,1)单调递减 , 在区间 (1,) +∞单调递增 .
所以 1x =时 , 函数 () f x 取得最小值 1
(1)2
f =. ……………… .5分 (Ⅱ) (1)()
() x x a f x x
--'=
, 0x >.
(1)当 0a ≤时,
(0,1)x ∈时 , () 0f x '<, ()="" f="" x="" 为减函数="" ;="" (1,)="" x="" ∈+∞时,="" ()="" 0f="" x="" '="">, () f x 为增函数 . 所以 () f x 在 1x =时取得最小值 1(1)2
f a =--
. (ⅰ)当 0a =时, 2
() 2
x f x x =-,由于 0x >,令 () 0f x =, 2x =,则 () f x 在 (0,) +∞上
有一个零点;
(ⅱ)当 1
2
a =-时,即 (1)0f =时, () f x 有一个零点;
(ⅲ)当 1
2
a <-时,即 (1)0f="">时, () f x 无零点 .
(ⅳ)当 1
02
a -<时,即 (1)0f="">时,即><>
由于 0x →(从右侧趋近 0)时, () f x →+∞; x →+∞时, () f x →+∞, 所以 () f x 有两个零点 .
(2)当 01a <>
(0,) x a ∈时 , () 0f x '>, () f x 为增函数 ; (,1) x a ∈时, () 0f x '<, ()="" f="" x="" 为减函数;="" (1,)="" x="" ∈+∞时,="" ()="" 0f="" x="" '="">, () f x 为增函数 .
所以 () f x 在 x a =处取极大值, () f x 在 1x =处取极小值 .
21() ln (1) 2f a a a a a a =+-+21
ln 2
a a a a =--.
当 01a <时, ()="" 0f="" a="">时,><, 即在="" (0,1)x="" ∈时,="" ()="" 0f="" x="">,><>
而 () f x 在 (1,) x ∈+∞时为增函数,且 x →+∞时, () f x →+∞, 所以此时 () f x 有一个零点 .
(3)当 1a =时, 2
(1) () 0x f x x
-'=≥在 ()0, +∞上恒成立,所以 () f x 为增函数 .
且 0x →(从右侧趋近 0)时, () f x →-∞; x →+∞时, () f x →+∞. 所以 () f x 有一个零点 .
综上所述, 01a ≤≤或 12a =-时 () f x 有一个零点; 12a <-时, ()="" f="" x="" 无零点;="">-时,>
02
a -
() f x 有两个零点 .
……………… .13分
19. (本小题满分 14分) 解:(Ⅰ)由题意可得
2
222,
, c c a a b c =??
?=??=+??
解得 a =
b 故椭圆的方程为 22
162
x y +=. …… .4分
(Ⅱ)当直线 l 斜率不存在时 , A B
的坐标分别为
, (2,
, ||MN = 四边形 AMBN 面积为 1
||||42
AMBN S MN AB =
?=. 当直线 l 斜率存在时, 设其方程为 (2) y k x =-, 点 11(, ) A x y , 22(, ) B x y , 33(, ) M x y , 33(, ) N x y --,点 , M N 到直线 l 的距离分别为 12, d d ,则四边形 AMBN 面积为
121
||() 2
AMBN S AB d d =
+. 由 22
1, 62
(2), x y y k x ?+
=???=-?
得 2222(13) 121260k x k x k +-+-=, 则 21221213k x x k +=+, 2122
126
13k x x k -=+,
所以 ||AB
=
=
因为 12122
4(4) 13k
y y k x x k -+=+-=
+, 所以 AB 中点 222
62(, ) 1313k k
D k k
-++. 当 0k 1时,直线 OD 方程为 30x ky +=, 由 2230,
1, 6
2x ky x y +=???+=??解得 333, x ky =-2
32
213y k =+.
所以 121
||() 2
AMBN S AB d d =
+
12=
=
=
==.
当 0k =时,四边形 AMBN
面积的最大值 AMBN S =综上四边形 AMBN
面积的最大值为 . ?????????? 14分
20. (本小题满分 13分) 解:
(Ⅰ)若 11b =,因为数列 {}n a 单调递增,所以 211a ≤,又 1a 是自然数,所以 10a =
或 1. ??? 2分 (Ⅱ)因为数列 {}n a 的每项都是自然数,
若 2101a =≤,则 11b ≥,与 11a b =矛盾;
若 12a ≥,则因 {}n a 单调递增,故不存在 21n a ≤,即 10b =,也与 11a b =矛盾 . 当 11=a 时,因 {}n a 单调递增,故 2≥n 时, 1>n a ,所以 11b =,符合条件, 所以, 11a =. ??? 6分 (Ⅲ)若 2(1,2, ) n a n n ==,则数列 {}n a 单调递增,显然数列 {}m b 也单调递增,
由 2n a m ≤,即 22n m ≤,得 2
12n m ≤,
所以, m b 为不超过 2
12
m 的最大整数,
当 21m k =-() k *?N 时,因为 2222
11222222122k k m k k k k -<><>
所以 222m b k k =-;
当 2m k =() k *?N 时, 22
122
m k =,所以, 22m b k =.
综上, 2222, 21(2,
2(m k k m k k b k m k k **ì?-=- ?=í?= ??N ) N ) , 即当 0m >且 m 为奇数时, 212m m b -=;当 0m >且 m 为偶数时, 22
m m b =. 若数列 {}n a 是数列 {}m b 的生成数列,且 {}m b 生成 {}n a 的控制函数为 () g n , 则 m b 中不超过 () g n 的项数恰为 n a ,即 m b 中不超过 () g n 的项数恰为 2n , 所以 221() n n b g n b +≤<,即 222222n="" pn="" qn="" r="" n="" n="">,即><+对一切正整数 n="">+对一切正整数>
即 22(2) 0(2) (2) 0
p n qn r p n q n r ?-++≥??-+-->??对一切正整数 n 都成立, 故得 2p =,且 0(2) 0qn r q n r +≥??-->?
对一切正整数 n 都成立,故 02q ≤≤, q Z ∈. 又常数 r Z ∈,
当 0q =时, 02(1) r n n ≤<≥,所以 0r=",或" 1r=";" 当="" 1q="时," (1)="" n="" r="" n="" n="">≥,所以><≥,所以 0r=",或" 1r="-;" 当="" 2q="时," 20(1)="" n="" r="" n="">≥,所以><≥,所以 2r="-,或" 1r="">≥,所以>
所 以 2() 2g n n =, 或 221n +, 或 221n n +-, 或 22n n +, 或 2222n n +-, 或 2221
n n +-(n *
?N ) . ??? 13分
范文三:2015 朝阳高三一模 数学 文 答案
答案:B
ABabcCABd::,,,,,,,,,,,U解析:
答案:D
解析:命题的否定要注意“全称量词更改为存在量词,否定原命题的结论” 答案:C
22p,,xy,0,,1,,22,0,,ypxp,,2(0)2,,22解析:双曲线 的右焦点是 ,抛物线的焦点 故
p,4
答案:B
St步骤
12开始
n=123
n=264
n=3245
n=41206解析:此程序计算过程如下表所示 答案:A
1,12x,,,log2log10,,,x211112233解析:,,
x1,,yyx,,,log3,,3,,作出函数 的图像:
x,13可知 答案:C
ππ,,,5,,,,,,,,fxxxx()2sin()cos()sin22xk,,,,,xk,,,,663,,3212解析:,当时,即
为此函数对称轴
答案:D
1
t,2解析:作出可行域:当 时恰好满足zxy,,3 的最大值为5,此时z 的最小值是-1
答案:A
解析:如图所示
HMCMx,,MGDMx,,,3由题可知
1112112VHMCMGMxxxxx,,,,,,,,,,,33,,,,CGMH,326322
112xxx,,,3x,2322当且仅当 即时体积取得最大值 由此可以判断A为正确选项 答案:i
1i1i,,,,,,,1i,,,i1i1i1i,,,,,,,,解析:
2
3
2答案:
,,,,,,213aabaab,,,,,,,,1,,22解析:
90:答案:
解析:根据题意作图:易知所求角为直角
37
64答案: ,
解析:由题作出直观图:
133V,,,,1326其体积为
7S, AOB4最大侧面积为
答案:
解析:稿酬为4000元时,应纳税额为(4000-800)×20%×(1,30%)=448>280
3
故纳税额280元的稿酬低于4000元,根据公式可知,此人稿酬2800元
答案:3
2,a解析:如图所示,当 时,
a,,1,21,4,,,,a,2函数值域为 其区间长度大于3;当 时,函数值域为 其区间长度等于3,
mn,,,故区间的长度的最小值是3
23+62答案:(?)4(?)
6cosB,22B,,(0,)sincos1BB,,3解析:(?)因为,,又,
3sinB,3所以.
ACBC,sinsinBA由正弦定理得,.
AC6,
33
32所以.
AC,4所以. ??? 6分
,sinsin(60)CB,,,ABC(?)在中,
,,,,sincos60cossin60BB
4
13,,sincosBB22
1336,,+2323 =
3+32
6 =.
113+32SACBCC,,,,,46sin,ABC23+62226所以==. ??13分
2
5答案:(?)乙校的数学成绩整体水平较高(?)
解析:(?)从茎叶图可以看出,乙校10名学生的考试成绩的平均分高于甲校10名学生的考试成绩平均分,故乙校的数学成绩整体水平较高( ??? 4分
M(?)设事件:分别从甲、乙两校随机各抽取1名成绩不低于90分的同学,抽到的学生中,甲校学生成绩高于乙校学生成绩(
AA,12由茎叶图可知,甲校成绩不低于90分的同学有2人,从小到大依次记为;乙校成绩
BBBBB,,,,12345不低于90分的同学有5人,从小到大依次记为(
AABBBBB=======92,93,90,91,95,96,98.1212345其中
分别从甲、乙两校各随机抽取1名成绩不低于90分的同学共有ABABABABABABABABABAB,,,,,,,,,11121314152122232425这10种可能(
ABABABAB,,,11122122其中满足“抽到的学生中,甲校学生成绩高于乙校学生成绩”共有这4种可能(
42PM(),,105所以( 即分别从甲、乙两校随机各抽取1名成绩不低于90分的同学,抽到的学生中,甲校学生成
2
5绩高于乙校学生成绩的概率为. ??? 13分
5
答案:(?)略(?)略(?)存在
解析:(?)证明:因为三棱柱的侧面是正方形,
CCBCCCAC^^,BCACCI=11所以,.
CC^1ABC所以底面(
CCBD^BDìABC1因为底面,所以(
ABC由已知可得,底面为正三角形(
ACBDAC^D因为是中点,所以(
ACCCC?ACCA111BD^因为,所以平面( ??? 5分
C1
BA1 1
O
C
D A B BCBCOOD11(?)证明:如图,连接交于点,连接(
BCO1显然点为的中点(
ABOD//1ACD因为是中点, 所以(
BCDBCDAB?111ODì又因为平面,平面,
AB//BCD11所以直线平面( ??? 10分
BCDCE,DEDME1(?)在内的平面区域(包括边界)存在一点,使(此时点是在
C1
BA1 1
M
E C
D A B CD1线段上. 证明如下:
6
CECD,CDC11E过作交线段于,
ACCAACCACE1111BD^,由(?)可知平面,而平面, BDCE^所以(
CECD,BDCDDI=BCDCE111^又,,所以平面(
BCDCE,DM,DM1又平面,所以( ??? 14分
4,1,n,,a,,nn,3na,4a,8a,16Tn,,,,16(2)22,2.n,243,n答案:(?),,((?)(?)
a,4aS,nn,11解析:(?)解:因为,,
aSa,,,4aSaa,,,,,,4482113212 所以,,
aSaaa,,,,,,,,4481643123( ??? 3分
nnn,1aSS,,,,,222n,2nnn,1(?)当时,(
aS,,4n,111又当时,(
4,1,n,,a,,nn2,2.n,,所以 ??? 6分
ba,,4ba,,82233(?)依题意,,.
bd,,4,1,bd,,28b,0bn,,4(1)1,d,41n则由得,,,则.
0,1,n,,ab,,,nnn,2(1)2,2.nn,,,所以
n,2abnn,,,,(1)2(*)Nnn所以.
Tabababababab,,,,,,...1122334411nnnn,,n因为=
45612nn,,,,,,,,,,,,,,,,0122232...(2)2(1)2nn,
56723nn,,2122232...(2)2(1)2Tnn,,,,,,,,,,,,,n所以.
456723nn,,,,,,,,,,,,Tn2222...2(1)2n所以
7
41n,2(12),nn,,33nn,,,,,,,,,(1)216(2)212, .
n,3Tn,,,,16(2)2n所以. ??? 13分
22xy3,,1yx,,,(2)623答案:(?)(?)
解析:(?)由题意可得
c,2,,
,c6,,,,a3,222abc,,,,a,6,b,2解得,.
22xy,,162故椭圆的方程为( ??? 5分
Axy(,)Bxy(,)ykx,,(2)l1122(?)由题意可知直线斜率存在,设其方程为,点,,
Mxy(,)Nxy(,),,3333,,
22,xy,,1,,62,
2222,ykx,,(2),(13)121260,,,,,kxkxk,由得,
212kxx,,12213,k所以(
,4kyykxx,,,,,(4)1212213,k因为,
262kk,D(,)221313,,kkAB所以中点(
k?0()xky,,30OD因此直线方程为(
xky,,30,,,22,xy22,,1,y,,32xky,,362,13,k33由解得,(
,,,,,,,,,,
MFNFFMFN,,01222因为四边形为矩形,所以,
8
(2,)(2,)0xyxy,,,,,,3333即(
2240,,,xy33所以(
22(91)k,40,,213,k所以(
33yx,,,(2)k,,l33 解得(故直线的方程为( ??? 14分
2ee=0xy,,a,0答案:(?)(?)略(?))
32xxaxa,,,x,()efx,2{0}xx,fx()x解析:函数定义域为,.
xx,fx(),fxx()e,,(1)ex,a,0(?)当时,,.
,ff(1)e,(1)2e,,所以.
yfx,()(1,(1))fyx,,,e2e(1)所以曲线在点处的切线方程是, 2ee=0xy,,即. ??? 3分
32xxx,,,1xe2,fx(),a,,1x(?) 当时,.
232,gx(),gxxxxx()321(31)(1),,,,,,xxx,,,1设,则.
11x,x,,gxxx()(31)(1)0,,,,x,,1x,033令得,或,注意到,所以.
10,,x,gxxx()(31)(1)0,,,,x,03令得,注意到,得.
11(0,)(,),,gx()33所以函数在上是减函数,在上是增函数.
1122x,g()0,,gx()3327所以函数在时取得最小值,且. gx()(0,),,所以在上恒大于零.
9
32xxx,,,1xe0,2,x,,,(0,)fx(),x于是,当,恒成立.
0,,,,,fx()a,,1所以当时,函数在上为增函数. ??? 7分
32xxx,,,1xe2,fx(),a,,1x(?)问另一方法提示:当时,.
320,,,0,,,,,,,fx()xxx,,,,10由于在上成立,即可证明函数在上为增函数.
32xxaxa,,,x,()e()fx,2x(?)(?).
232,hx(),hxxxa()32,,,xxaxa,,,设,.
,(0,),,hx()0,a,0当时,在上恒成立,
(0,),,hx()即函数在上为增函数.
0,1,,xha(0)0,,,h(1)20,,hx()0而,,则函数在区间上有且只有一个零点,使
x,1,()??fx()0,xfx()0 20,10,1,,,,,hx()hxxx()320,,,a,0x?(2)当时,当时,成立,函数在区间上为 0,1?,,fx()0>h(0)0,hx()0,增函数,又此时,所以函数在区间恒成立,即, 0,10,1,,,,fx()fx()故函数在区间为单调递增函数,所以在区间上无极值; 3232hx(),xxaxaxxax,,,,,,,(1)a,0(3)当时,. x,0,10,1,,,,,hx()0,fx()0,fx()当时,总有成立,即成立,故函数在区间上为单调 0,1,,fx()递增函数,所以在区间上无极值. a,0综上所述. ??? 13分 10 北京市朝阳区高三年级第一次综合练习 数学试卷(文史类) 2015. 4 (考试时间 120分钟 满分 150分) 本试卷分为选择题(共 40分)和非选择题(共 110分)两部分 第一部分(选择题 共 40分) 一、选择题 :本大题共 8小题,每小题 5分,共 40分 . 在每小题给出的四个选项中,选出 符合题目要求的一项 . 1.已知全集 {, , , }U a b c d =,集合 {, },{, }A a b B b c ==,则 () U A B e等于 A . {}b B . {}d C . {, , }a c d D . {, , }a b c 2.已知命题 :p x ?∈R , sin 1x ≤,则 A . :p ?x ?∈R , sin 1x ≥ B . :p ?x ?∈R , sin 1x > C . : p ?0x ?∈R , 0sin 1x ≥ D . :p ? 0x ?∈R , 0sin 1x > 3.若抛物线 2 2(0) y px p =>的焦点与双曲线 222 x y -=的右焦点重合,则 p 的值为 A B . 2 C . 4 D . 4.如图所示的程序框图表示的算法功能是 A .计算 123456S =?????的值 B .计算 12345S =????的值 C .计算 1234S =???的值 D .计算 1357S =???的值 5.已知 113 log 2x =, 1 2 22 x -=, 3x 满足 3 331 () log 3 x x =,则 A . 123x x x < b="" .="" 132x="" x="" x=""> 第(4)题图 C . 213x x x < d="" .="" 312x="" x="" x=""> 6.函数 ππ () 2sin(66f x x x =--图象的一条对称轴方程是 A . π6x = B. π3x = C. 5π12x = D. 2π3x = 7.已知实数 x , y 满足 20, 20, 0, x y x y y t +≥?? -≤??≤≤? 其中 0t >.若 3z x y =+的最大值为 5,则 z 的最小 值为 A . 5 2 B . 1 C . 0 D . 1- 8. 已知边长为 3的正方形 ABCD 与正方形 CDEF 所在的平面互相垂直, M 为线段 CD 上 的动点(不含端点) ,过 M 作 //MH DE 交 CE 于 H ,作 //MG AD 交 BD 于 G ,连结 GH .设 CM x =(03) x <,则下面四个图象中大致描绘了三棱锥 c="" ghm="">,则下面四个图象中大致描绘了三棱锥> 积 y 与变量 x 变化关系的是 A B C D 第二部分(非选择题 共 110分) 二、填空题:本大题共 6小题,每小题 5分,共 30分 . 把答案填在答题卡上 . 9. i 为虚数单位,计算 1i 1i +- 10.已知平面向量 a , b 满足 1==a b , a 与 b 的夹角为 60?, 则 () ?+=a a b . 11. 圆 22 :(2) (2) 8C x y -+-=与 y 轴相交于 , A B 两点, 则 弦 AB 所对的圆心角的大小为 . 12.一个四棱锥的三视图如图所示,其中侧视图为正三角形,则 该四棱锥的体积是 ,四棱锥侧面中最大侧面的面积 是 . 13.稿酬所得以个人每次取得的收入,定额或定率减除规定费用后的余额为应纳税所得额,每 次收入不超过 4000元,定额减除费用 800元;每次收入在 4000元以上的,定率减除 20%的费用 . 适用 20%的比例税率,并按规定对应纳税额减征 30%,计算公式为: (1)每次收入不超过 4000元的:应纳税额 =(每次收入额-800) ×20%×(1-30%) (2) 每次收入在 4000元以上的:应纳税额 =每次收入额 ×(1-20%) ×20%×(1-30%) . 已知某人出版一份书稿,共纳税 280元,这个人应得稿费 (扣税前 ... ) 为 14. 记 12x x -为区间 12[, ]x x 的长度. 已知函数 2x y =, x ∈[]2, a -(0a ≥) , 其值域为 [], m n , 则区间 [], m n 的长度的最小值是 . 三、解答题:本大题共 6小题,共 80分 . 解答应写出文字说明,演算步骤或证明过程 . 15. (本小题满分 13分) 在 ABC ?中, π3A = , cos B =, 6BC =. (Ⅰ)求 AC 的长; 第(12)题图 正视图 侧视图 俯视图 (Ⅱ)求 ABC ?的面积. 16. (本小题满分 13分) 某次考试结束后,为了解甲、乙两所学校学生的数学考试情况,随机抽取甲、乙两校 各 10名学生的考试成绩, 得茎叶图如图所示 (部分数据 不清晰) : (Ⅰ )请根据茎叶图判断哪个学校的数学成绩平均水 平较高(直接写出结果) ; (Ⅱ )若在抽到的这 20名学生中,分别从甲、乙两校 随机各抽取 1名成绩不低于 90分的学生, 求抽到的学生中, 甲校学生成绩高于乙校 学生成绩的概率. 17. (本小题满分 14分) 如图,在三棱柱 111C B A ABC - 中,各个侧面均是边长为 2的 正方形, D 为线段 AC 的中点. (Ⅰ)求证:BD ⊥平面 11A ACC ; (Ⅱ)求证:直线 1AB ∥平面 D BC 1; (Ⅲ) 设 M 为线段 1BC 上任意一点, 在 D D BC 1内的平面区域 (包括边界) 是否存在点 E , 使 CE ⊥DM ,并说明理由. 18. (本小题满分 13分) 设数列 {}n a 的前 n 项和为 n S ,且 14a =, 1n n a S +=, n *∈N . (Ⅰ )写出 2a , 3a , 4a 的值; (Ⅱ )求数列 {}n a 的通项公式; (Ⅲ )已知等差数列 {}n b 中,有 22b a =, 33b a =,求数列 {}n n a b ?的前 n 项和 n T . A B C A 1 B 1 C 1 19. (本小题满分 14分) 已知椭圆 22 22:1(0) x y C a b a b +=>>的两个焦点分别为 12(2,0), (2,0)F F -,离心率 为 3 2F 的直线 l (斜率不为 0)与椭圆 C 交于 , A B 两点,线段 AB 的中点为 D , O 为坐标原点,直线 OD 交椭圆于 , M N 两点. (Ⅰ )求椭圆 C 的方程; (Ⅱ )当四边形 12MF NF 为矩形时,求直线 l 的方程. 20. (本小题满分 13分) 已知函数 () ()e x a f x x x =+, a ∈R . (Ⅰ )当 0a =时,求曲线 () y f x =在点 (1,(1))f 处的切线方程; (Ⅱ )当 1a =-时,求证:() f x 在 (0,) +∞上为增函数; (Ⅲ )若 () f x 在区间 (0,1)上有且只有一个极值点,求 a 的取值范围. 北京市朝阳区高三年级第一次综合练习 数学 参考答案(文史类) 2015.4 一、选择题 :本大题共 8小题,每小题 5分,共 40分 . 在每小题给出的四个选项中,选出 二、填空题:本大题共 6小题,每小题 5分,共 30分 . 把答案填在答题卡上 . 三、解答题:本大题共 6小题,共 80分 . 解答应写出文字说明,演算步骤或证明过程 . 15. (本小题满分 13分) (Ⅰ)因为 cos 3 B =, (0,) B ∈π,又 22 sin cos 1 B B +=,所以 sin 3 B =. 由正弦定理得, sin sin AC BC B A = . 所以 = . 所以 4 AC =. ??? 6分 (Ⅱ)在 ABC ?中, sin sin(60 ) C B =+ sin cos60cos sin 60 B B =+ 1 sin 2 B B = = 1 + 2323 ? =. 所以 1 sin 2 ABC S AC BC C ? =? = 1 46 2 ??? 6=?? 13分 16. (本小题满分 13分) 解:(Ⅰ)从茎叶图可以看出,乙校 10名学生的考试成绩的平均分高于甲校 10名学生的考 试成绩平均分,故乙校的数学成绩整体水平较高. ??? 4分 (Ⅱ) 设事件 M :分别从甲、 乙两校随机各抽取 1名成绩不低于 90分的同学,抽到的学 生中,甲校学生成绩高于乙校学生成绩. 由茎叶图可知,甲校成绩不低于 90分的同学有 2人,从小到大依次记为 12, A A ;乙 校成绩不低于 90分的同学有 5人,从小到大依次记为 12345, , , , B B B B B . 其中 121234592, 93, 90, 91, 95, 96, 98. A A B B B B B ======= 分 别 从 甲 、 乙 两 校 各 随 机 抽 取 1名 成 绩 不 低 于 90分 的 同 学 共 有 11121314152122232425, , , , , , , , , A B A B A B A B A B A B A B A B A B A B 这 10种可能. 其 中 满 足 “ 抽 到 的 学 生 中 , 甲 校 学 生 成 绩 高 于 乙 校 学 生 成 绩 ” 共 有 11122122, , , A B A B A B A B 这 4种可能. 所以 42() 105P M = = . 即分别从甲、 乙两校随机各抽取 1名成绩不低于 90分的同学, 抽到的学生中, 甲校 学生成绩高于乙校学生成绩的概率为 2 5. ??? 13分 17. (本小题满分 14分) 解:(Ⅰ)证明:因为三棱柱的侧面是正方形, 所以 11, CC BC CC AC ^^, BC AC C =I . 所以 1CC ^底面 ABC . 因为 BD ì底面 ABC ,所以 1CC BD ^. 由已知可得,底面 ABC 为正三角形. 因为 D 是 AC 中点,所以 BD AC ^. 因为 1 AC CC C ? ,所以 BD ^平面 11ACC A . ??? 5分 (Ⅱ)证明:如图,连接 1B C 交 1BC 于点 O ,连接 OD . 显然点 O 为 1B C 的中点. 因为 D 是 AC 中点, 所以 1//AB OD . 又因为 OD ì平面 1BC D , 1AB ?平面 1BC D , 所以直线 1//AB 平面 1BC D . ??? 10分 A B A 1 B 1 C 1 (Ⅲ)在 D D BC 1内的平面区域(包括边界)存在一点 E ,使 CE ⊥DM . 此时点 E 是在线段 1C D 上 . 证明如下: 过 C 作 1CE C D ⊥交线段 1C D 于 E , 由 (Ⅰ) 可知 BD ^平面 11ACC A , 而 CE ?平面 11ACC A , 所以 BD CE ^. 又 1CE C D ⊥, 1BD C D D =I , 所以 CE ^平面 D BC 1. 又 DM ?平面 D BC 1,所以 CE ⊥DM . ??? 14分 18. (本小题满分 13分) (Ⅰ)解:因为 14a =, 1n n a S +=, 所以 2114a S a ===, 3212448a S a a ==+=+=, 4312344816a S a a a ==++=++=. ??? 3分 (Ⅱ )当 2n ≥时, 11222n n n n n n a S S +-=-=-=. 又当 1n =时, 114a S ==. 所以 4, 1, 2, 2. n n n a n =?=?≥? ??? 6分 (Ⅲ )依题意, 224b a ==, 338b a ==. 则由 114 28 b d b d +=?? +=?得, 10b =, 4d =, 则 4(1) n b n =-. 所以 2 0, 1, (1)2, 2. n n n n a b n n +=? ?=?-≥? 所以 2 (1)2(*)n n n a b n n +?=-∈N . 因为 n T =1122334411... n n n n a b a b a b a b a b a b --++++++ 456120122232... (2) 2(1) 2n n n n ++=+?+?+?++-?+-?, 所以 567232122232... (2) 2(1) 2n n n T n n ++=?+?+?++-?+-?. C 1 A B A 1 B 1 M 所以 456723 2222... 2(1) 2n n n T n ++-=+++++--? 4133 2(12) (1) 216(2) 212n n n n n -++-=--?=---?- . 所以 3 16(2) 2n n T n +=+-?. ??? 13分 19. (本小题满分 14分) 解:(Ⅰ)由题意可得 2 222, , c c a a b c =?? ?=??=+?? 解得 a = b = 故椭圆的方程为 22 162 x y +=. ??? 5分 (Ⅱ) 由题意可知直线 l 斜率存在,设其方程为 (2) y k x =-,点 11(, ) A x y , 22(, ) B x y , 33(, ) M x y , 33(, ) N x y --, 由 22 1, 62(2), x y y k x ?+=???=-? 得 2222 (13) 121260k x k x k +-+-=, 所以 2 122 1213k x x k +=+. 因为 12122 4(4) 13k y y k x x k -+=+-= +, 所以 AB 中点 222 62(, ) 1313k k D k k -++. 因此直线 OD 方程为 30x ky +=() 0k 1. 由 2230, 1, 62x ky x y +=???+ =?? 解得 2 32 213y k =+, 333x ky =-. 因为四边形 12MF NF 为矩形,所以 220F M F N ?=, 即 3333(2, ) (2, ) 0x y x y -?---=. 所以 22 3340x y --=. 所以 22 2(91) 4013k k +- =+. 解得 3k =± .故直线 l 的方程为 (2) 3 y x =±-. ??? 14分 20. (本小题满分 13分) 解:函数 () f x 定义域为 {0}x x ≠, 322 () e x x x ax a f x x ++-'=. (Ⅰ)当 0a =时, () e x f x x =?, () f x '=(1)e x x +. 所以 (1)e, (1)2e f f '==. 所以曲线 () y f x =在点 (1,(1))f 处的切线方程是 e 2e(1) y x -=-, 即 2e e =0x y --. ??? 3分 (Ⅱ) 当 1a =-时, () f x '=3221e x x x x x +-+. 设 () g x =321x x x +-+,则 2() 321(31)(1) g x x x x x '=+-=-+. 令 () (31)(1) 0g x x x '=-+>得, 13x > 或 1x <-,注意到 0x="">,所以 1 3x > . 令 () (31)(1) 0g x x x '=-+<得,注意到 0x="">,得 1 03x . 所以函数 () g x 在 1(0,) 3上是减函数,在 1 (, ) 3+∞上是增函数 . 所以函数 () g x 在 13x = 时取得最小值,且 122 () 0 327g =>. 所以 () g x 在 (0,) +∞上恒大于零 . 于是,当 (0,) x ∈+∞, () f x '=3221e 0x x x x x +-+>恒成立 . 所以当 1a =-时,函数 () f x 在 ()0, +∞上为增函数 . ??? 7分 (Ⅱ)问另一方法提示:当 1a =-时, () f x '=3221e x x x x x +-+. 由于 3210x x x +-+>在 ()0, +∞上成立,即可证明函数 () f x 在 ()0, +∞上为增函数 . (Ⅲ) (Ⅱ) 322() e () x x x ax a f x x ++-'=. 设 () h x =32x x ax a ++-, 2() 32h x x x a '=++. (1) 当 0a >时, () 0h x '>在 (0,) +∞上恒成立, 即函数 () h x 在 (0,) +∞上为增函数 . 而 (0)0h a =-<, (1)20h="">,则函数 () h x 在区间 ()0,1上有且只有一个零点 0x ,使 0() 0f x '=, 且在 0(0,) x 上, () 0f x ¢<,在 ()="" 0,1x="" 上,="" ()="" 0f="" x="" ¢="">,故 0x 为函数 () f x 在 区间 ()0,1上唯一的极小值点 ; (2)当 0a =时,当 x ?()0,1时, 2() 320h x x x '=+>成立,函数 () h x 在区间 ()0,1上为 增函数,又此时 (0)0h =,所以函数 () 0h x >在区间 ()0,1恒成立,即 () 0f x ¢>, 故函数 () f x 在区间 ()0,1为单调递增函数,所以 () f x 在区间 ()0,1上无极值; (3)当 0a <> () h x =3232(1) x x ax a x x a x ++-=++-. 当 ()0,1x ∈时,总有 () 0h x >成立,即 () 0f x '>成立,故函数 () f x 在区间 ()0,1上 为单调递增函数,所以 () f x 在区间 ()0,1上无极值 . 综上所述 0a >. ??? 13分 2015北京朝阳高三一模 单项选择 1. —Are you doing your homework? —No, I'm writing a short play. It __________ at the Christmas party. A. will be put on B. will put on C. puts on D. is put on 2. In order to keep fit, the old man makes it a rule __________ for a walk after supper every day. A. going B. to go C. go D. gone 3. —Steve, the vacation is coming soon. Have you found a summer job yet? —I suppose I can work at the boy's camp __________ I worked last summer. A. that B. where C. which D. what 4. Sally was excited to meet Susan at the party last night. They __________ each other since they graduated from Oxford University in 2010. A. haven't seen B. hadn't seen C. didn't see D. don't see 5. —We've only got this small bookcase—will that do? —No. __________ I was looking for was something much bigger and stronger. A. What B. Where C. That D. Which 6. I will wait __________ 6:30, but then I'm going home. A. from B. at C. after D. until 7. Studies show that people who have a glass of wine per day do better than __________ who don't. A. that B. the one C. ones D. those 8. Oh! I can feel something __________ up my leg! It must be an insect. A. to climb B. climbing C. climb D. climbed 9. She doesn't speak our language, __________ she seems to understand what we say. A. for B. and C. yet D. or 10. If it __________ earlier, the printing machine would not have broken down. A. has been repaired B. is repaired C. had been repaired D. was repaired 11. —Congratulations! I hear you've won the first prize in the singing competition. —You __________ be mistaken. I'm in the dance class. A. must B. may C. should D. could 12. A public health campaign __________ the number of heart disease deaths by 80 percent over the past three decades. A. had reduced B. had been reduced C. has reduced D. has been reduced 13. —How are you getting along with your German, Kate? —Oh,Mr. Black,I'm so tired of it. Maybe I should drop out __________ it kills me. A. when B. after C. while D. before 14. —What do you do, Rita? —I'm a clerk in a foreign company now. But I __________ English in a high school for 10 years. A. teach B. have taught C. taught D. am teaching 15. Comparison may make something appear more beautiful than it is when __________ alone. A. seen B. seeing C. see D. to see 完型填空 1. Appeal to Emotion (情感) My childhood left three months ago on a plane to Austria. It was a 1 day when my baby cousins moved away. My cousins were a fundamental part of my life; 2 they were not with me, they were on my mind. They brought back the untroubled days of my childhood, 3 games, adventures and silliness, but they also helped 4 me from an at-times selfish teenager into a responsible adult. My aunt and uncle moved here from Boston with their 1-year-old daughter Yasmeen. They lived in an apartment on the side of our house and I was 5 to have our family, especially a baby girl, so close to us. Yasmeen had close friendships with each of my sisters, but I knew the one that 6 between us was the strongest. As she began to walk on her own and talk in full sentences, I realized my great 7 upon her. I noticed that she'd 8 some of my unpleasant actions, 9 with her mother after I had done the same. Yasmeen made me realize what being a good 10 really was. When Maya was born, Yasmeen had a hard time 11 . She was jealous(妒忌)of the attention we all paid to her new younger sister, so I did my best to keep her company when she might have not been 12 . 0nce Yasmeen overcame her jealousy, she was able to 13 Maya's presence, like we all did. I tried to take advantage of our times together, doing my best to 14 my schedule for my two favorite people. My 15 with Yasmeen and Maya has taught me the importance of influenUal people. I know that I have helped Yasmeen and Maya 16 , but "the babies" have had an even greater influence on my own life. They have shown me how to be a parent, a sister, a child, and most importantly, a friend. Our 17 has made me a better person—a more patient person, with the ability to 18 endless questions and spilled (洒)juice; an 19 person, able to have fun and be happy with others; a role model, 20 the babies ideas about right and wrong, and a compassionate person, able to forgive and love. 1. A. sadB. busyC. fineD. relaxing 2. A. untilB. beforeC. whenD. unless 3. A. againstB. throughC. besidesD. for 4. A. leadB. challengeC. matureD. encourage 5. A. overjoyedB. embarrassedC. surprisedD. afraid 6. A. endedB. startedC. survivedD. developed 7. A. pressureB. dependenceC. mercyD. influence 8. A. doubtB. regretC. copyD. guide 9. A. debatingB. arguingC. negotiatingD. checking 10. A. leaderB. partnerC. friendD. example 11. A. sharingB. adjustingC. recognizingD. struggling 12. A. noticedB. admiredC. comfortedD. protected 13. A. ignoreB. mindC. senseD. enjoy 14. A. affectB. acceptC. freeD. rain 15. A. instructionB. experimentC. experienceD. interview 16. A. growB. succeedC. thinkD. behave 17. A. meetingB. relationshipC. responsibilityD. respect 18. A. reviewB. tolerateC. repeatD. avoid 19. A. honestB. independentC. aggressiveD. enthusiastic 20. A. showingB. writingC. awardingD. permitting 阅读理解 1. A Let's Go Fly a Kite ... —at Piedmont Middle School's celebration of kites! Come and learn how to build all sorts of kites, from the simplest diamond-shaped kites to the most complex box kites. Stay as long as you like and build as many kites as you want. Once you have finished a kite, get advice on flying techniques from kite expert Lorena Hallsberg. The celebration will be at Piedmont Middle School, 151 Piedmont School Drive. The Piedmont Middle School Parent Teacher Organization ( PTO) has organized a refreshment (茶点)tent. All profits will benefit future PTO activities. Take a break from kite flying and drink some lemonade~ While you are doing so, why not join the PTO? Membership is free; you just donate your time. Show your support for Piedmont Middle School by joining the PTO this Saturday ! When:Saturday, April 11,from 9:00 am to 5:00 pm Where:Piedmont Middle School Why :For fun ~ Cost:Free, thanks to a generous gift from Bizarco Kite Company~ Schedule 9:00 am — Kite-building booths open. All materials are supplied for kites. 10:00 am — Kite-building shows by Lorena Hallsberg in the courtyard. Come by and learn how to build box kites and kites that look and fly like butterflies. 11:00 am — Kite-flying shows on the school track. Learn all the most important skills. 12:00 pm — Kite-flying competitions on the school track. 1:00 pm — Presentation by Dr. Brian Lehrman in the show tent: "The History of Kites". 2:00 pm — Best Kite competitions and judging in the show tent. Come see the most artistic kites and the most interesting theme kites. 3:00 pm — Presentation by Dr. Lehnnan in the show tent: "Kites and Science". 3:30 pm — Awards ceremony conducted by Headmaster Seward on the football field. The results of the day's judging will be announced, with awards such as Best of Show, Most Artistic, Highest Flyer, aird others. Winners will receive gifts from the Bizarco Kite Company ~ 4:00 - 5:00 pm — Let's all go fly a kite! Everyone flies kites at the same time, creating a wonderful sight for all to enjoy. Come to the kite celebration, enjoy yourself and learn more. (1) The main purpose of Paragraph 2 is to ask people to __________. A. build a kite B. support the PTO C. take a break D. join the fun (2) Which times are most important for people who want to join in kite competitions? A. 10:00 am and 11:00 am. B. 12:00 pm and 2:00 pm. C. 1:00 pm and 3:00 pm. D. 2:00 pm and 4:00 pm. (3) From the passage, we know that the kite celebration __________ A. is enjoyable and educational B. is strict about the shapes of kites C. gets money from PTO of Piedmont Middle School D. gives people a chance to see kites from around the world (4) The passage is intended for __________. A. school staff B. kite experts C. students and parents D. kite companies 2. B "The 13th of June, 1325 , I left Tangier, my birthplace, with the intention of making the pilgrimage (朝圣)to Mecca... to leave all my friends,to abandon my home as birds abandon their nests." So begins an old manuscript in a library in Paris—the travel diary of Ibn Battuta. Almost two centuries before Columbus, this young Moroccan set off for Mecca, returning home three decades later as one of history's great travelers. Driven by curiosity, he journeyed to remote corners of the Islamic world,traveling through 44 modern countries, three times as far as Marco Polo. Little celebrated in the West, his name is well known among Arabs. In his hometown of Tangier, a square, a hotel, a cafe, a ferry boat, and even a hamburger are named after him. Ibn Battuta stayed in Mecca as a student for several years, but the urge to travel soon took over. In one adventure, he traveled to India seeking profitable employment with the sultan—the Muslim ruler of Delhi. On the way,he described his group being attacked in the open country by 80 men on foot, and two horsemen: "...I was hit by an arrow and my horse by another, but God in his grace preserved me..." In Delhi, the sultan gave him the position of judge, based on his previous study at Mecca. But the sultan had an unpredictable character, and Ibn Battuta looked for an opportunity to leave. When the sultan offered to finance a trip to China, he agreed. Ibn Battuta set off in three ships, but misfortune struck while he was still on the shore. A sudden storm grounded and broke up two ships, scattering (散播)treasure and drowning many people and horses. As he watched, the third ship, with all his belongings and slaves—one carrying his child—was carried out to sea and never heard from again. After a lifetime of amazing adventures, Ibn Battuta was finally ordered by the Sultan of Morocco to return home to share his wisdom with the world. Fortunately, he agreed and wrote a book that has been translated into numerous languages, allowing people everywhere to read about his unparalleled journeys. (1) What can we learn about Ibn Battuta from the passage? A. He had great interest in the Islamic world. B. He returned to his homeland to write a book. C. His journeys were less important than Marco Polo's. D. His journeys were very common for people of that time. (2) The Sultan of Delhi gave Ibn Battuta a position of judge because __________. A. Ibn Battuta had studied in Mecca B. Ibn Battula had been a judge before C. Ibn Battuta had worked as a translator D. Ibn Battuta had traveled to many countries (3) Which is the best title for the passage? A. The Learned Ibn Battuta B. A Visitor to Mecca C. The Travels of Ibn Battuta D. Desire for Adventures 3. C Culture and Cuisine The United States is known for jazz and blue jeans. But travel to Paris and ask your average French citizen about American cuisine (烹饪)and he's likely to answer, “ McDonalds. ” Ask the same thing of any American citizen on any American street and I'm afraid you'd get the same answer, or something close to it. Hamburgers and hotdogs and fries are all fine, but with American malls and other outlets standardizing everything from clothing to food, the sad truth is that American cuisine is becoming more homogeneous—all the same—no matter where you live. True, many Americans are eating more varied foods these days, but these are largely the cuisines of immigrant groups, and they are quite likely to be affected by homogenization of American cuisine. So what exactly is American cuisine? Well, to some extent it is a reflection of our melting pot culture meaning that Europeans made huge contributions in the form of wheat, dairy products, pork beef and poultry. But American cuisine also includes products that once were known only to the New World, including potatoes, corn, pumpkin, sweet potatoes, and peanuts. The one region of the country where you still find all these things in daily use is the Deep South. The South lost the Civil War, but children of the southerners are winning the battle to preserve and advance their cooking traditions—and in this case one of the few cuisines can truly be called American, which is why we're pleased to have Low Country cuisine in this issue of food creation. That's right, grits and gravy are back in a big way in cities like Charleston and Savannah. Tnith is, they never really left, but up until a decade ago Low Country cuisine was more common at home than in restaurants. In fact, a large number of tourists now go to the lower Atlantic region in order to experience this extraordinary cuisine for themselves. Time will tell whether Low Country cuisine becomes popular in other regions of the country in the way that, say, Italian cuisine has, but it's amazing and heartening to see one of our true cultural treasures enjoying renewed popularity in these increasingly homogeneous times. (1) According to the passage, American cuisine impresses people as being _________. A. dull and changeless B. rich and various C. popular and delicious D. disagreeable and unpleasant (2) It can be seen that the writer feels regretful that __________. A. cuisines of other countries play a more important role in America B. American cuisine has become increasingly lacking in variety C. American cuisine tends to vary because of immigration D. American cuisine is being changed by foreign cuisines (3) From the passage, we know that grits and gravy __________. A. were cooked with new materials after the Civil War B. are gaining popularity in the south of America C. were more popular over ten years ago D. are seldom served in restaurants (4) What's the writer's attitude towards the renewal of Low Country cuisine in America? A. Concerned and cautious. B. Hopeless and doubtful. C. Positive and supportive. D. Critical and disapproving. 4. D The Enigma (谜)of Beauty The search for beauty spans centuries and continents. Paintings of Egyptians dating back over 4,000 years show both men and women painting their nails and wearing makeup. In 18th-century France, wealthy noblemen wore large wigs (假发) of long, white hair to make themselves attractive. Today, people continue to devote a lot of time and money to their appearance. There is at least one good reason for the desire to be attractive : beauty is power Studies suggest that good-looking people make more money, get called on more often in class, and are regarded as friendlier. But what exactly is beauty? It's difficult to describe it clearly, and yet we know it when we see it. And our awareness of it may start at a very early age. In one set of studies, six-month-old babies were shown a series of photographs, The faces on the pictures had been rated for attractiveness by a group of coUege students. In the studies, the babies spent more time looking at the attractive faces than the unattractive ones. The idea that even babies can judge appearance makes perfect sense to many researchers. In studies by psychologists, men consistently showed a preference for women with larger eyes, fuller lips, and a smaller nose and chin while women prefer men with large shoulders and a narrow waist. According to scientists, the mind unconsciously tells men and women that these traits—the full lips,clear skin, strong shoulders—equal health and genetic well-being. Not everyone thinks the same way, however. “Our liardwiredness can be changed by all sorts of expectations一mostly cultural,” says C(Loring Brace, an anthropologist at the University of Michigan. What is considered attractive in one culture might not be in another. Look at most Western fashion magazines: the women on the pages are thin. But is this “perfect” body type for women worldwide? Scientists' answer is no; what is considered beautiful is subjective and varies around the world. They found native peoples in southeast Peru preferred shapes regarded overweight in Western cultures. For better or worse, beauty plays a role in our lives. But it is extremely difficult to describe exactly what makes one person attractive to another. Although there do seem to be certain physical traits considered universally appealing, it is also true that beauty does not always keep to a single, uniform standard. Beauty really is, as the saying goes, in the eye of the beholder. (1) People's ideas about beauty __________. A. have existed since ancient times B. can be easily described C. have little influence on a person's success D. are based upon strict criteria (2) In Paragraph 3, the babies in the study __________. A. were rated for their appearance B. were entered in a beauty contest C. were shown photos of a group of college students D. were able to tell attractive faces from unattractive ones (3) The underlined word “traits”in Paragraph 4 probably means __________. A. qualities B. measurements C. judgments D. standards (4) We can learn from the passage that __________. A. the ideas of beauty vary as people grow up B. the search for beauty is rooted in lack of confidence C. the standards for beauty are based on scientific researches D. the underetanding of beauty depends on cultural backgrounds 5. Virtual (虚拟)Teams Virtual teams are a great way to enable teamwork in situations where people are not sitting in the same office at the same time. 1 This is particularly so for businesses that use virtual teams to build global presence, or need less common skills or knowledge from people who are unwilling to travel. Virtual teams are governed by the same basic principles as traditional teams. 2 It is the way the team members communicate. They rely on special communication channels enabled by modern technologies, such as emails, faxes, and teleconferences, and alike. Due to more limited communication channels, the success of virtual teams is much more sensitive to the type of project the group works on, what people are selected,and how the team is managed. 3 One challenging case is the projects that rely heavily on integrated work. That is to say, when each person's work depends significantly on what someone else is doing at the same moment, like in a sports team, there is an ongoing heavy exchange of information in real time, and the tasks have to go through a strict order within a short time. Not everyone can perform well in a virtual team environment. 4 Another important quality is communication skills. The team members must be able to communicate clearly and positively. Managers of virtual teams need to pay much more attention to having clear goals, performance standards, and communication rules. People have various assumptions on what to expect from each other. 5 One of the biggest challenges of virtual teams is building trust between the team members. Trust is important for unblocking communication between members and increasing motivation of each person in the team. The issue of trust needs special attention at any stage of team existence. A. Yet, there is one significant difference. B. Not every type of project is suitable for a virtual team. C. A virtual team can choose whatever project they like to work on. D. The members must be self-motivated and able to work independently. E. Such teams are now widely used by companies and organizations to cut business costs. F. Members of virtual teams communicate quite well although they never meet face-to-face. G. To avoid misunderstanding, clear rules that everyone understands and agrees on are necessary. 书面表达 1. 你的美国朋友Chris目前在北京学习汉语,他对中国传统文化非常感兴趣。本周五下午你校将要举办一个文化讲座,请你根据以下提示给他写一封电子邮件,邀请他来参加。 1.农业大学张教授讲解中国茶文化的历史和传播; 2.讲座后有交流和品茶活动; 3.你将陪同Chris并帮其翻译讲解。 注意:1.词数不少于50。 2.可适当增加细节,以使行文连贯。 3.开头和结尾已给出,不计入总词数。 Hi Chris, _____________________________________________________________________ ___________ Yours, Joe 2. 假设你是红星中学高二(1)班的学生李华,上周日你班组织到首都图书馆参加一天的文化志愿活动,请根据以下四幅图的先后顺序,用英语写一篇日记,记述活动的全过程。 注意:1.词数不少于60。 2.日记的开头已给出,不计入总词数。 March 29th Sunday Fine Today some classmates and I served at Capital Library as volunteers. 2015北京朝阳高三一模 单项选择 1. 【答案】 A 2. 【答案】 B 3. 【答案】 B 4. 【答案】 B 5. 【答案】 A 6. 【答案】 D 7. 【答案】 D 8. 【答案】 B 9. 【答案】 C 10. 【答案】 C 11. 【答案】 A 12. 【答案】 C 13. 【答案】 D 14. 【答案】 C 15. 【答案】 A 完型填空 1. 【答案】 A,C,B,C,A,D,D,C,B,D,B,A,D,C,C,A,B,B,D,A 阅读理解 1. 【答案】 1. B 2. B 3. A 4. C 2. 【答案】 1. A 2. A 3. C 3. 【答案】 1. A 2. B 3. B 4. C 4. 【答案】 1. A 2. D 3. A 4. D 5. 【答案】 E,A,B,D,G 书面表达 1. 【答案】 Hi Chris, Good news! There will be a lecture in our school this Friday afternoon. Professor Zhang from University of Agriculture will tell us about the history and spread of Chinese tea. This will be followed by a tea party and you can taste different kinds of tea while chatting with teachers and students of our school. I wonder if you want to participate in it. Don’t worry about the language. I’ll be with you and explain what you don’t understand. If you do not have any prior appointment then, I am looking forward to your coming. Yours, Joe 2. 【答案】 March 29th, Sunday Fine Today some classmates and I served at Capital Library as volunteers. At 9 o’clock this morning, we gathered together at the gate of the library. After getting into the library, we were first trained by a librarian on how to operate the computers to borrow or return books so that we could help readers later. Then, wearing the volunteer uniform and volunteer identity card, we got down to our work. Some of us helped with shelving books and some offered assistance to the people who had problems with using the computers. All of us took our tasks seriously and tried to do them well. Time passed quickly. Before we knew it, it was time to say goodbye to the staff. On the way back, we shared our feelings happily. When doing voluntary work, we did our part in helping others. Meanwhile, it promoted our social awareness. We all believe volunteering is rewarding and look forward to more of it. 转载请注明出处范文大全网 » 2015朝阳高三一模数学文试范文四:2015朝阳高三一模数学(文科)
范文五:2015朝阳高三英语一模