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1.下列关系式中正确的是
A. B. C. D. 2.函数 的定义域为
A. B. C. D. 3.在一次游戏中,获奖者按系统抽样的方法从编号为1~56的56种不同奖品中抽取4件,已知编号为6、20、48的奖品已被抽出,则被抽出的4件奖品中还有一件奖品的编号是
A.32 B.33 C.34 D.35
4.若函数 唯一的零点同时在(1,1.5),(1.25,1.5),(1.375,1.5),(1.4375,1.5)内,则该零点(精确度为0.1)的一个近似值约为
A.1.02 B.1.27 C.1.39 D.1.45
5.函数 的图像大致是
A. B. C. D.
6.如图,函数 的图像过矩形OABC的顶点B,且 .
若在矩形OABC内随机地撒100粒豆子,落在图中阴影部分
的豆子有67粒,则据此可以估算出图中阴影部分的面积约为
a=3
b=1
IF a<b THEN
c=a
ELSE
c=b
END IF
PRINT c
END
A.2.64 B.2.68 C.5.36 D.6.64
7.运行右图所示的程序,最后输出的结果是
A.3 B.1
C. D. 8.某校为了解高三学生英语听力情况,抽查了甲、乙两班各
十名学生的一次英语听力成绩,并将所得数据用茎叶图表示
(如图所示),则以下判断正确的是
A.甲组数据的众数为28
B.甲组数据的中位数是22
甲
乙
8 0
9 7 7 7 1 6 7 8 9 9
8 8 6 5 2 1 1 2 3 4
0 3
C.乙组数据的最大值为30
D.乙组数据的极差为16
9.某市刑警队对警员进行技能测试,测试成绩分为优秀、良好、
合格三个等级,测试结果如下表:(单位:人)
优秀良好合格
男40105
开始
S=0,i=2
是
否
输入m
S=S+i
i=i+2
i >m ?
输出S
结束
25
女a155
若按优秀、良好、合格三个等级分层,从中抽取40人,成绩
为良好的有24人,则 等于
A.10 B.15 C.20 D.30
10.如图是运算 的程序框图,则其中实数m的取值范围是
A. B.
C. D.
11.某商场为了解商品销售情况,对某种电器今年一至六月份的
月销售量 (台) 进行统计,得数据如下:
(月份)123456
(台)6910862
根据上表中的数据,你认为能较好描述月销售量 (台)与时间 (月份)变化关系的模拟函数是
A. B. C. D. 12.已知函数 满足 ,且 时, ,函数 为偶函数,且 时, ,则函数 的图像与函数 的图像的所有交点的横坐标之和等于
A.0 B. 2 C. 4 D.6
第Ⅱ卷(非选择题 90分)
二、填空题:本大题共4小题,每小题5分,共20分.请把答案填在答题卷的相应位置.
13. 口袋内装有形状、大小完全相同的红球、白球和黑球,它们的个数分别为3、2、1,从中随机摸出1个球,则摸出的球不是白球的概率为 .
14. 已知定义在 上的奇函数 满足 ,且 ,则 = .
15.在某次飞镖集训中,甲、乙、丙三人10次飞镖成绩的条形图如下所示,则他们三人中成绩最稳定的是 .
乙
甲
丙
16. 已知方程 的两根为 ,且 ,则 的大小关系为
.(用“<”号连接)
宁德市2015-2016学年度第一学期高一质量检测(二)
(本题满分10分)
(Ⅰ)已知全集 , , .
求集合 和集合 ;
(Ⅱ)计算:
(本题满分12分)
已知 为 上的偶函数.
(Ⅰ)求实数 的值;
(Ⅱ)判断函数 在 上的单调性,并利用定义证明.
(本题满分12分)
某公司为确定下一年度投入某种产品的宣传费,需了解年宣传费 (单位:万元)对年销售量 (单位:吨)的影响,为此对近6年的年宣传费 (单位:万元)和年销售量 (单位:吨)的数据进行整理,得如下统计表:
x(万元)234.557.58
y(吨)33.53.5467
(Ⅰ)由表中数据求得线性回归方程 中的 ,试求出 的值;
(Ⅱ)已知这种产品的年利润 (单位:万元)与 、 之间的关系为 ,根据(Ⅰ)中所求的回归方程,求年宣传费x为何值时,年利润 的预估值最大?
(本题满分12分)
开始
输入x
x<1?
输出y
结束
是
否
阅读如图所示程序框图,根据框图的算法功能回答下列问题:
(Ⅰ)当输入的 时,求输出 的值组成的集合;
(Ⅱ)已知输入的 时,输出 的最大值为8,最小值
为3,求实数 的值.
.(本题满分12分)
为了调查某校2000名高中生的体能情况,从中随机选取m名学生进行体能测试,将得到的成绩分成 , ,…, 六个组,并作出如下频率分布直方图,已知第四组的频数为12,图中从左到右的第一、二个矩形的面积比为4:5.规定:成绩在 、 、 、 的分别记为“不合格”、“合格”、“良好”,“优秀”,根据图中的信息,回答下列问题.
(Ⅰ)求x和m的值,并补全这个频率分布直方图;
(Ⅱ)利用样本估计总体的思想,估计该校学生体能情况为“优秀或良好”的人数;
(Ⅲ)根据频率分布直方图,从“不合格”和“优秀”的两组学生中随机抽取2人,求所抽取的2人恰好形成“一帮一” (一个优秀、一个不合格)互助小组的概率.
0.020
频率/组距
100
0.005
0.010
x
60
分数
70
80
90
0.040
0.030
110
0
120
0.0225
(本题满分12分)
已知函数 的图象过点 ,且满足 .
(Ⅰ)求函数 的解析式;
(Ⅱ)若函数 在 上的最大与最小值之和为 ,求实数 的值;
(Ⅲ)若实数 为函数 且 的一个零点,求证:函数 的图象恒在函数 图象的上方.
宁德市2015-2016学年度第一学期高一质量检测(三)
一、选择题:本大题共12小题,每小题5分,共60分
1.D 2.C 3.C 4.D 5.A 6.C 7.B 8.B 9.A 10.B 11.C 12.C
二、填空题:本大题共4小题,每小题5分,共20分.
13. 14.3 15.丙 16. 三、解答题:本大题共6小题,共70分.
17.(本题满分10分)
解:(Ⅰ)由已知得 ,······································· 2分
,······································································ 4分
∴ .······································································ 5分
(Ⅱ) = ······································································· 8分
=4- ················································································ 9分
=5- ··················································································· 10分
(注:原式4个加数每个化简正确得1分,即 =4- …1分; = …1分; …1分; =1…1分.)
18.(本题满分12分)
解:(Ⅰ)法一:∵ 为 上的偶函数
∴ ······································································· 2分
∴ ···································································· 3分
∴ , ∴ ················································································ 4分
∴ ·················································································· 5分
法二:∵ 为偶函数
∴ ········································································ 2分
∴ ························································ 3分
∴ ·················································································· 4分
经检验得: 时, 为偶函数
∴ ················································································ 5分
(Ⅱ)函数 在 上单调递减··········································· 6分
证明:设 ,则
······················································· 7分
························································ 8分
···················································· 9分
∵ ∴ , , , ······························· 10分
∴ ,得 ,
∴ ······································································· 11分
∴函数 在 上是单调递减函数········································· 12分
19.(本题满分12分)
解:(Ⅰ)由已知得
,······················································ 2分
,······················································ 4分
因为线性回归直线过点 ,且 ,
所以 ,解得 ,················································· 6分
(Ⅱ)由(Ⅰ)得 ,························································· 7分
∴ ,······················································· 9分
···································································· 10分
当 时, 取得最大值,························································· 11分
所以年宣传费为9万元时,年利润的预估值最大. ······························ 12分
(注:用公式法求“ 时, 取得最大值”同样给2分)
20.(本题满分12分)
解:(Ⅰ)由程序框图可知, ············································· 1分
当 时, ,函数在[-1,1)上是减函数,······················ 2分
∴ ,即 ······························································· 3分
当 时, ,函数在[1,3]上是增函数································ 4分
∴ ,即 ······························································· 5分
综上得,输入x∈[-1,3],输出 的值组成的集合为 ························ 6分
(Ⅱ)当 时,输入 ,输出 ,不合题意,∴ ······ 7分
当 时, ,函数在 上是减函数,由已知得 ·· 8分
解之得 ·········································································· 9分
当 时, ,函数在 上是增函数,由已知得 ·· 10分
解之得 ··········································································· 11分
0.020
频率/组距
100
0.005
0.010
x
60
分数
70
80
90
0.040
0.030
110
0
120
0.0225
综上得,所求实数 的值为 或 ···································· 12分
21. (本题满分12分)
解:(Ⅰ)依题意得: ,……1分
解得 ……………………………2分
∴第四组的频率为
……………………………………………3分
∴ ∴ ………………………………… 4分
补全频率分布直方图如图…………………5分
(Ⅱ)由图估计“优秀或良好”的人数为
························································· 6分
···················································································· 7分
(Ⅲ)“不合格”的人数为 ,
“优秀”的人数为 ,·················································· 8分
设“不合格”的4人分别为 ,“优秀”的2人分别为 ,从中任取2人的所有基本事件为: , , , , , , , , , , , , , , ,共15种···················································································· 10分
设所抽取的2人恰好形成“一帮一”互助小组为事件A,其中包含的基本事件为: , , , , , , , ,共有8种:··································· 11分
故所抽取的2人恰好形成“一帮一”互助小组的概率 ················· 12分
(注:15种基本事件,全对得2分,列错1~7种扣1分,错8种及以上不给分)
22. (本题满分12分)
解:(Ⅰ)依题意得: ··························································· 2分
解得 ∴ . ··························································· 3分
(Ⅱ)∵ 又 在 单调递增,
∴ , ································································ 4分
∴ ∴ ···································································· 5分
∴ ∴ ····················································································· 6分
(Ⅲ)∵ , (i)当 时, 在 上单调递增
由 在 上单调递增,得 在 上单调递增
∴ 在 上单调递增
又 (或 )
∴ 在 上存在唯一零点,
∴当 时, ······························································· 8分
(ⅱ) 当 时, 在 上单调递减
由 在 上单调递增,得 在 上单调递减
∴ 在 上单调递减
又∵ , ∴ 在 上存在唯一零点,
∴当 时, ·························································· 10分
(注:因 时,由对数函数性质可得函数 的图像以 为渐进线,故此题用“ 趋于-1时函数 趋于正无穷大”来求解,同样可以得分.)
综上得, 且 ,(此步没写不扣分)
∴ ··························································· 11分
即对于任意 ,都有 ∴函数 的图象恒在 的图象上方.···························· 12分
(注意:答案只要出现 ,均可得1分)
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