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7. She always thinks of____ more than herself.

A. other 　B. others　 C. the other 　D. the others

6. - ___is it from here?　 -Only half an hour’s ride.

A. How far 　B. How long　 C. How soon 　D. Flow much

5. - Shall we go on a picnic?　　 -That’s going to be ______

A.　　　 fun 　B. funny 　C. fun 　D. very fun

4. Jim and Bill don’t live _____ the middle school.

A. away from 　B. far from　 C. far away 　D. far

3. They have decided to go to work _____every day.

A. by bikes 　B. on feet　 C. by bus 　D. in car

2. Jack began to do his homework as soon as he____ home.

A. came to 　B. reached　 C. arrived at 　D. got to

1. The bus station is about five hundred meters____ here，it’s within walking distance.

A. away from 　B. away　 C. far away from 　D. far from

12.設(shè)函數(shù)f(x)在(-∞,+∞)上滿(mǎn)足f(2-x)=f(2+x)，f(7-x)=f(7+x)，且在閉區(qū)間［0，7］上，只有f(1)=f(3)=0.

(1)試判斷函數(shù)y=f(x)的奇偶性；　

(2)試求方程f(x)=0在閉區(qū)間［-2 005，2 005］上的根的個(gè)數(shù)，并證明你的結(jié)論.

解 (1)由

　從而知函數(shù)y=f(x)的周期為T(mén)=10.又f(3)=f(1)=0,而f(7)≠0,故f(-3)≠0.　

故函數(shù)y=f(x)是非奇非偶函數(shù).　

(2)由(1)知y=f(x)的周期為10.　

又f(3)=f(1)=0,f(11)=f(13)=f(-7)=f(-9)=0,　

故f(x)在［0，10］和［-10，0］上均有兩個(gè)解，從而可知函數(shù)y=f(x)在［0，2 005］上有402個(gè)解，在［-2 005，0］上有400個(gè)解，所以函數(shù)y=f(x)在［-2 005，2 005］上有802個(gè)解.

11.已知函數(shù)f(x)=x²+|x-a|+1,a∈R.　

(1)試判斷f(x)的奇偶性；　

(2)若-≤a≤，求f(x)的最小值.

解 　(1)當(dāng)a=0時(shí)，函數(shù)f(-x)=(-x)²+|-x|+1=f(x),　

此時(shí),f(x)為偶函數(shù).當(dāng)a≠0時(shí)，f(a)=a²+1,f(-a)=a²+2|a|+1,　

f(a)≠f(-a),f(a)≠-f(-a),此時(shí)，f(x) 為非奇非偶函數(shù).　

(2)當(dāng)x≤a時(shí),f(x)=x²-x+a+1=(x-)²+a+,　

∵a≤,故函數(shù)f(x)在(-∞，a］上單調(diào)遞減，　

從而函數(shù)f(x)在(-∞，a］上的最小值為f(a)=a²+1.　

當(dāng)x≥a時(shí)，函數(shù)f(x)=x²+x-a+1=(x+)²-a+,　

∵a≥-，故函數(shù)f(x)在［a，+∞)上單調(diào)遞增，從而函數(shù)f(x)在［a，+∞)上的最小值為f(a)=a²+1.　

綜上得，當(dāng)-≤a≤時(shí)，函數(shù)f(x)的最小值為a²+1.　

10.已知f(x)是R上的奇函數(shù)，且當(dāng)x∈(-∞,0)時(shí)，f(x)=-xlg(2-x),求f(x)的解析式.　

解　 ∵f(x)是奇函數(shù)，可得f(0)=-f(0),∴f(0)=0.　

當(dāng)x＞0時(shí)，-x＜0,由已知f(-x)=xlg(2+x),∴-f(x)=xlg(2+x)，　

即f(x)=-xlg(2+x) (x＞0).∴f(x)=

即f(x)=-xlg(2+|x|) (x∈R).