2
18.(14分)已知各项均为正数的数列{an}满足a26,1log2anlog2an1,
nN.
(1)求数列{a}的通项公式及前n项和Sn;
n
2
a
n
(2)若n(),求数列{bn}的前n项和Tn.
blog2nN
9
完美格式可编辑
WORD资料整理
19.(12分)某校从初三年级体育加试百米测试成绩中抽取100个样本,所有样
本成绩全部在11秒到19秒之间.现将样本成绩按如下方式分为四组:
第一组[11,
13),第二组[13,15),第三组[15,17),第四组[17,19],题19图是根据上述分组
得到的频率分布直方图.
(1)若成绩小于13秒被认定为优秀,求该样本
在这次百米测试中成绩优秀的人数;
(2)是估算本次测试的平均成绩;
(3)若第四组恰有3名男生,现从该组随机抽
取3名学生,求所抽取的学生中至多有1名女
生的概率.
20.(12分)已知正弦型函数f(x)Hsin(x),其中常数H0,0,
0,若函数的一个最高点与其相邻的最低点的坐标分别是,3,
212
7
12
,3.
(1)求f(x)的解析式;
(2)求f(x)的单调递增区间;
(3)在△ABC中A为锐角,且f(A)0.若AB3,BC33,求△ABC的面积
S.
完美格式可编辑
WORD资料整理
21.(10分)某学校计划购买x咯篮球和y个足球.
2xy5
(1)若x,y满足约束条件
xy2,问该校计划购买这两种球的总数最多是
x7
多少个?
2xy5
(2)若x,y满足约束条件,已知每个篮球100元,每个足球70元,
xy2
x7
求该校最少要投入多少元?
22.(10分)某辆汽车以x千米/小时x60,120的速度在高速公路上匀速行驶,
每小时的耗油量为
1
5
x
k
3600
x
升,其中k为常数.若该汽车以120千米/小
时的速度匀速行驶时,每小时的耗油量是12升.
(1)求常数k值;
(2)欲使每小时的耗油量不超过8升,求x的取值范围;
(3)求该汽车匀速行驶100千米的耗油量y(升)的最小值和此时的速度.
2y
2
x
23.(14分)已知椭圆C:
1和直线l:
yxm,直线l与椭圆C交于A,B
23
两点.
(1)求椭圆C的准线方程;
(2)求△ABO(O为坐标原点)面积S的最大值;
(3)如果椭圆C上存在两个不同的点关于直线l对称,求m的取值范围.
完美格式可编辑
WORD资料整理
江苏省2018年普通高校对口单招文化统考
数学试题答案及评分参考
一、单项选择题(本大题共10小题,每小题4分,共40分)
题号12345678910
答案BCDCBCCADA
二、填空题(本大题共5小题,每小题4分,共20分)
11.612.
2
5
5
13.4814.515.a4
三、解答题(本大题共8小题,共90分)
16.(8分)
解:
(1)由题意知:
2a32,·····························2分
即1a5.··········································2分
(2)因为1a5,所以
2x273,······················2分
13
3
于是2x13,故x1.·······························2分
17.(10分)
解:
(1)因为当x20,即x0时,····························1分
g(x)12,···········································1分
所以定点A的坐标为(2,12).·························1分
(2)因为f(x)是奇函数,
所以f
(2)f
(2),·································2分
于是(42m)12,即m4.·······················2分
(3)由题意知:
773311
f()f
(2)f()f
(2)f()f()
222222
1
(32.···························3分
2)
2
18.(14分)
a
n1
解:
(1)由题意知log2an1log2an1,得2
a
n
,
a
2
所以数列{an}是公比q=2,a3的等比数列,·······2分
1
2
于是
an132n,·····························3分
aq
n1
1
完美格式可编辑
WORD资料整理
S
n
n
(312)n
3(2
12
1).
·······························3分
2n12
a(32)
2n2
(2)因为nlog222,·······2分
n
blog2logn
22
99
所以数列{
b}是首项为0,公差为2的等差数列,·········2分
n
2n22
于是Tnn.·····························2分
nn
2
19.(12分)
解:
(1)由频率分布直方图可得成绩优秀的人数为
0.1×2×100=20.······································4分
(2)因为12×0.1+14×0.15+16×0.2+18×0.05=7.4,·············2分
所以本次测试的平均成绩为7.4×2=14.8秒.··············2分
(3)由频率分布直方图得第四组有100×0.05×2=10人,其中由7名女
生,3名男生.·········································1分
设“所抽取的3名学生中至多有1名女生”记作事件A
321
CCC11
337
所求事件的概率为P(A).·················3分
3
C60
10
20.(12分)
解:
(1)由题意知H3,········································1分
因为
T
2
7
12
122
2
,所以T,即2
T
,··········1分
于是f(x)3sin(2x),把点(,3)
代入可得
12
3
,
即f).·································2分
(x)3sin(2x
3
(2)由kx2k
22
232
,························2分
5
解得kxk
1212
,kZ,
5
f(x)的单调递增区间为k,k,kZ.······2分
1212
(3)由f)0,A为锐角,得
(A)3sin(2A
3
A,··········1分
3
2
9AC271
在△ABC中,,解得AC6.·······1分
cos
6AC2
完美格式可编辑
WORD资料整理
193
故.
S36sin····························2分
232
21.(10分)
解:
(1)设该校一共购买z个球,则目标函数是zxy,··········1分
作出约束条件所表示的平面区域(答21图),
解方程组
2x
x
y
7
5
得
x
y
7
9
,···········2分
图中阴影部分是问题的可行域,根据题意
xN,yN,
从图中看出目标函数在点A(7,9)处取得最大值,
即maxz=7+9=16个,
所以该校最多一共可购买16个球.········3分
(2)设该校需要投入w元,则目标函数是
w100x70y,·························1分
约束条件的可行域是答21图中不包含边界的部分,根据xN,yN,
容易得到满足条件的整数点只有三个,分别是(5,4),(6,5),(6,6),
·························································2分
显然点(5,4)是最优解,此时minw=100×5+70×4=780元,
所以该校最少投资780元.··································1分
22.(10分)
13600
解:
(1)由题意知:
12(120k),解得k90.···········3分
5120
13600
(2)由题意知(x90)8,··························2分
5x
2x
化简得13036000
x,
解得40x90,·····································1分
因为x[60,120],
故x的范围是60x90.······························1分
(3)由题意知
y
100
x
1
5
(
x
90
3600
x
)
·····························1分
完美格式可编辑
WORD资料整理
201
(
90
x
3600
)
2
x
111
令t,t(,)
x12060
,
2t
则72000180020
yt
当
135
t时,即x80千米/小时,最低耗油量y8.75升.
804
···················································2分
23.(14分)
22
解:
(1)易知3b,得c1,·······················2分
a,2
2
a
所以准线方程为y3.·····················2分
c
yxm
(2)联立方程组
2y2
x
23
2mxm
,化简得54260
x2,
1
2
由24m1200得5m5
设(,)(,)
Ax1y,Bxy,
122
则
4m
xx,
12
5
2
2m6
xx,
12
5
于是|AB|=
11|
x
1
x|
2
2
2
16m
2
20(2m
5
6)
43
5
5
2
m,·························2分
又原点O到直线yxm的距离
|m|
d,············1分
2
所以
S
1
2
43
5
5
|m|6
2||5
mm
5
2
2
m
6
5
(5
22
265mm6
2265mm6
m)m,
522
完美格式可编辑
WORD资料整理
当
10
m时,等号成立,
2
即△ABO面积的最大值为
6
2
.·····················3分
(3)M(x3,y3),N(x4,y4)是椭圆上不同的两点,它们关于直线l
对称,所以直线MN的方程可设为yxn,
yxn
2nxn2
2y,化简得5x4260,
2
x
联立方程组1
23
2n2
n,解得5n5,·····1分
于是16401200
4n
x,
x
又5
34
6n
y,
y-x3nx4n
34
5
2n3n
因此MN的中点坐标)
P(,,点P必在直线l上,
55
n
m,····························1分
代入直线方程得5
又5n5,
55
m.·······························2分
所以5
5
完美格式可编辑