问题 解答题

已知函数f(x)的定义域为R,对任意s,t∈R都有f(s+t)=f(s)+f(t),且对任意x>0,都有f(x)<0,且已知f(3)=-3.

(1)求证:f(x)是R上的单调递减函数;

(2)求证:f(x)是奇函数;

(3)求f(x)在[m,n](m,n∈Z且m>0)上的值域.

答案

(1)在R任取x1,x2,且x1<x2,则f(x2)=f[(x2-x1)+x1]=f(x2-x1)+f(x1),…(1分)

∴f(x2)-f(x1)=f(x2-x1).…(2分)

∵x2-x1>0,∴f(x2-x1)<0,…(3分)∴f(x2)-f(x1)<0,即f(x2)<f(x1),∴f(x)是R上的单调递减函数. …(4分)

(2)令s=t=0,则f(0)=f(0)+f(0),∴f(0)=0.…(5分)

又令s=x,t=-x,则f(0)=f(x)+f(-x),∴f(x)+f(-x)=0,…(6分)

∴f(-x)=-f(x),∴f(x)是奇函数.…(7分)

(3)∵f(x)是R上的单调递减函数,∴f(x)在[m,n]上也为减函数,…(8分)

∴f(x)在[m,n]上的最大值为f(m),最小值为f(n).…(9分)

又m,n∈Z,∴f(m)=f[1+(m-1)]=f(1)+f(m-1)=2f(1)+f(m-2)=…=mf(1).

同理f(n)=nf(1),…(11分)

已知f(3)=-3得f(3)=3f(1)=-3,∴f(1)=-1,…(12分)∴f(n)=-n,f(m)=-m,…(13分)

所以,函数的值域为[-n,-m].…(14分)

单项选择题 A1/A2型题
完形填空
“Sixteen,” I said. I have forgotten the math question my second-grade teacher, Joyce Cooper, asked that day, but I will never forget my   36  . As soon as the number left my mouth, the whole class started  37  . I felt like the   38  person in the world..
Mrs. Cooper fixed them with a firm look. Then she said: “We’re all here to  39 .”
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Not long after I started  50 , something happened that brought the  51  rushing back to me. I had written a sentence with grammatical errors on the blackboard. Then I asked students to come and  52 the mistakes.
One girl got halfway through, became  53  and stopped. As the other children laughed, 54  rolled down her cheeks. I gave her a hug and told her to have a drink of water. Then, remembering Mrs. Cooper, I fixed  55  of the class with a firm look. “We are all here to learn,” I said.
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