7.若0<x≤2则函数y=-x2+x+2的值域是________.
[解析] y=-x2+x+2=-(x-)2+在(0,2]的图象.知f(x)在(0,]上是增函数,在[,2]上是减函数,所以y∈[0,].
[答案] [0,]
6.(2008·广州二模)设函数f(x)=g(x)=x2f(x-1),则函数g(x)的递减区间是
( )
A.(-∞,0] B.[0,1)
C.[1,+∞) D.[-1,0]
[解析] g(x)=如右图所示,其递减区间是[0,1).故选B.
[答案] B
5.函数y=1- 的单调增区间是
( )
A. B.
C.(-∞,1) D.(2,+∞)
[解析] 由x2-3x+2≥0得x≤1或x≥2,要求原函数的递增区间,只需求t=x2-3x+2的递减区间,即(-∞,1),故选C.
[答案] C
4.(2008·广州一模)函数y=(0<|x|≤1)的值域是
( )
A.(-∞,-8]∪[0,+∞) B.[0,+∞)
C.[0,1] D.(0,1]
[解析] 解法一:y=
=(x+2)+-4,
∵0<|x|≤1,∴x+2∈[1,2)∪(2,3],
又∵函数g(t)=t+-4在[1,2)上是减函数,在(2,3]上是增函数,∴原函数的值域是(0,1],故选D.
解法二:∵y′==,
∴函数y=在[-1,0)上是减函数,在(0,1]上是增函数.
∴原函数的值域为(0,1]∪(0,]=(0,1].
[答案] D
3.(2007·福建)已知f(x)是R上的减函数,则满足
f>f(1)的实数x的取值范围是
( )
A.(-∞,1) B.(1,+∞)
C.(-∞,0)∪(0,1) D.(-∞,0)∪(1,+∞)
[解析] 由f(x)在R上是减函数,所以<1,解得x>1,或x<0,故选D.
[答案] D
2. (2006·陕西)函数f(x)=(x∈R)的值域是
( )
A.(0,1) B.(0,1]
C.[0,1) D.[0,1]
[解析] y=x2+1≥1,
所以∈(0,1].
故选B.
[答案] B
1.(2009·福建)下列函数f(x)中,满足“对任意x1,x2∈(0,+∞),当x1<x2时,都有f(x1)>f(x2)”的是
( )
A.f(x)= B.f(x)=(x-1)2
C.f(x)=ex D.f(x)=ln(x+1)
[答案] A
在世界环保日到来前夕,联合国教科文组织就如何处理城市垃圾向世界中学生征文。 请你以Saving Our City 为题, 写一篇100 字左右的短文。
提示:
城市垃圾的危害
(1)污染环境
(2)有害健康
你所在的城市是如何处理垃圾的
(1)垃圾分类
(2)报纸,玻璃的再利用
(3)有害垃圾填埋
此题要求改正所给短文中的错误。对标有题号的每一行作出判断: 如无错误,在该行右边横线上画一个勾( √ );如有错误(每行只有一个错误),则按下列情况改正:
此行多一个词: 把多余的词用斜线()划掉,在该行右边横线上写出该词,并也用斜线划掉。
此行缺一个词: 在缺词处加一个漏字符号(∧),在该行右边横线上写出该加的词。
此行错一个词: 在错的词下划一横线,在该行右边横线上写出改正后的词。
注意: 原行没有错的不要改。
After I finished the school this year, I began 71.__________
to look for work.Now several month later, I still 72.__________
hadn‘t found the job that I was interested.Last Sunday 73.__________
morning I received a phone call from a man calling him 74.__________
Mr Smith.He said to me on the phone, "I hear that you do 75.__________
very well in your studies.I may have a job for you." 76.__________
I entered his office with a beaten heart.How I hoped 77.__________
that I will go through the job-hunting talk today and he 78.___________
would take me on as a lab assistant.But to my surprised, 79___________
what he said disappointing.He only needed a model. 80.___________
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