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=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).
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