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标题: [求助]PP3的4道题(2天就考了,拜托了!谢谢!) [打印本页]

作者: 深谷幽兰 时间: 2006-4-23 10:51
标题: [求助]PP3的4道题(2天就考了,拜托了!谢谢!)
1. Which of the following inequlities has a solusion set that when graphed on the number line, is a single segment of finite length?
a. x^4>=1
b. x^3<=27
c. x^2>=16
d. 2<=I1I<=5
e. 2<=3x+4<=6
ANS: E
我觉得BCDE都可以呀, 都是有限的.

2.IF r and s are intergers and rs+s is odd, which of the following must be even?
a.r
b.s
c, r+s
d. rs-r
e. r^2+s
ANA: C
RS+S 是偶数, 则RS和S必须都是偶数=> S是偶数或R是偶数或都是ODD. => R+S不一定是偶数呀?

3. IF the operation is defined for all integers a and b by a@b=a+b-ab, which of the following statement must be true for all a, b,c,
ANS: (A@B)@C=A@(B@C)
不明白这是如何得来的?

4.a certain library accesses fines for overdue books as follows. on the first day that book is overdued, the total fine is $0.1, for each additional day that the book is overdue, the total fine is either increased by $0.30 or doubled,whichever results in the lesser amount. What is the total fine for a book on the fourth day it is overdue?
ANS: $0.7
我算了半天都是0.6.

还有两天考了, 那位帮忙回答一下.谢谢!

作者: scujean 时间: 2006-4-23 13:02

1. Which of the following inequlities has a solusion set that when graphed on the number line, is a single segment of finite length?
a. x^4>=1
b. x^3<=27
c. x^2>=16
d. 2<=I1I<=5
e. 2<=3x+4<=6
ANS: E
我觉得BCDE都可以呀, 都是有限的.

a. x^4>=1 →(x^4-1)>=0→(x^2-1)(x^2+1)>=0→(x+1)(x-1) (x^2+1)>=0

→因为(x^2+1)>0，所以(x+1)(x-1)>=0→x<-1,x>1

b. x^3<=27→(x^3-3^3)<=0→(x-3)(X^2+3x+3^2)<=0→用判别式d=b^2-4ac→得知(X^2+3x+3^2)没有实根，但(X^2+3x+3^2)=[(x+3/2)^2+27/4]>0→所以(x-3)<=0→x<=3

c. x^2>=16→[x^2-4^2]>=0→(x-4)(x+4)>=0→x>4,x<-4

d. 2<=I1I<=5--- 2<=I1I，2小于觉对值1，是否抄错？

作者: 深谷幽兰 时间: 2006-4-23 13:07

d. 2<=I1I<=5--- 2<=I1I，2小于觉对值1，是否抄错？

谢谢. 是2<=IXI<=5

作者: 深谷幽兰 时间: 2006-4-23 13:11

b.
x^3<=27→(x^3-3^3)<=0→(x-3)(X^2+3x+3^2)<=0→用判别式d=b^2-4ac→得知(X^2+
3x+3^2)没有实根，但(X^2+3x+3^2)=[(x+3/2)^2+27/4]>0→所以(x-3)<=0→x<=3

问题是X应该在-3<=X<=3呀?考试时什么时候需要象你这样分解什么时候就直接看呢?谢谢.

作者: scujean 时间: 2006-4-23 13:13

4.a certain library accesses fines for overdue books as follows. on the first day that book is overdued, the total fine is $0.1, for each additional day that the book is overdue, the total fine is either increased by $0.30 or doubled,whichever results in the lesser amount. What is the total fine for a book on the fourth day it is overdue?
ANS: $0.7
我算了半天都是0.6.

那么第一天是0.1，第二、第三天显然是double比较合算，到第四天是+$0.3合算，也就是这样的一个罚款变化，0.1->0.2->0.4->0.7

作者: scujean 时间: 2006-4-23 13:17

以下是引用深谷幽兰在2006-4-23 13:11:00的发言：

b.
x^3<=27→(x^3-3^3)<=0→(x-3)(X^2+3x+3^2)<=0→用判别式d=b^2-4ac→得知(X^2+
3x+3^2)没有实根，但(X^2+3x+3^2)=[(x+3/2)^2+27/4]>0→所以(x-3)<=0→x<=3

问题是X应该在-3<=X<=3呀?考试时什么时候需要象你这样分解什么时候就直接看呢?谢谢.

你的想法应是→觉对值X<a→-a<x<a

这种的，我只会这样算，无法贪快..看看其他人有没其他办法..

2<=IXI<=5

用数字想，要分成x>0或X<0讨论

当x>0，把X想成3→2<=X(=3) <=5

X<0→X想成-3→X(=-3) <-2

3. IF the operation is defined for all integers a and b by a@b=a+b-ab, which of the following statement must be true for all a, b,c,
ANS: (A@B)@C=A@(B@C)
不明白这是如何得来的?

a@b=a+b-ab

其运算为前+后-前*后

(A@B)@C=A@(B@C)

(a@b)=a+b-ab

(a@b)@c=(a+b-ab)+c-(a+b-ab)*c---1式

(b@c)=b+c-bc

a@(b@c)=a+(b+c-bc)-a*(b+c-bc)----2式

1式=2式

[此贴子已经被作者于2006-4-23 13:48:36编辑过]

作者: 深谷幽兰 时间: 2006-4-23 13:43
非常感谢!

作者: salvador 时间: 2006-4-23 14:17

2.IF r and s are intergers and rs+s is odd, which of the following must be even?
a.r
b.s
c, r+s
d. rs-r
e. r^2+s
ANA: C
RS+S 是偶数, 则RS和S必须都是偶数=> S是偶数或R是偶数或都是ODD. => R+S不一定是偶数呀?

lz，题目说RS+S 是奇数，是敲错了还是看错了？

作者: 深谷幽兰 时间: 2006-4-23 14:24

lz，题目说RS+S 是奇数，是敲错了还是看错了？

没错, 我看了半天

作者: 深谷幽兰 时间: 2006-4-23 14:35

.IF r and s are intergers and rs+s is odd, which of the following must be even?
a.r
b.s
c, r+s
d. rs-r
e. r^2+s
ANA: C
RS+S 是偶数, 则RS和S必须都是偶数=> S是偶数或R是偶数或都是ODD. => R+S不一定是偶数呀?

lz，题目说RS+S 是奇数，是敲错了还是看错了？

明白了. 看的时候把RS+S看成了EVEN.如果是奇数就好做了. 做题真是要仔细呀!

作者: salvador 时间: 2006-4-23 14:40

以下是引用深谷幽兰在2006-4-23 14:35:00的发言：

lz，题目说RS+S 是奇数，是敲错了还是看错了？

明白了. 看的时候把RS+S看成了EVEN.如果是奇数就好做了. 做题真是要仔细呀!

but if rs+s=s(r+1) is odd, s and r+1 should be odd, then r is even and s is odd, r+s is odd.

answer is D? open to discuss

作者: 深谷幽兰 时间: 2006-4-23 14:49

以下是引用深谷幽兰在2006-4-23 14:35:00的发言：

lz，题目说RS+S 是奇数，是敲错了还是看错了？

明白了. 看的时候把RS+S看成了EVEN.如果是奇数就好做了. 做题真是要仔细呀!

but if rs+s=s(r+1) is odd, s and r+1 should be odd, then r is even and s is odd, r+s is odd.

answer is D? open to discuss

SORRY,又看了一下. ANSWER IS "S". 但S可能为ODD,也可能为EVEN

作者: salvador 时间: 2006-4-23 14:51

晕倒了！

这样还是与题目不一致啊！不过这道题我也碰到过，不是很难，到时候仔细一点肯定能作对。

[此贴子已经被作者于2006-4-23 14:52:52编辑过]

作者: scujean 时间: 2006-4-23 19:14

早上想破头就是想不出是C，原来是B

我的做法 rs+s is odd→s(r+1)=积

所以s=积，且(r+1)=积→r=积-1=偶

作者: salvador 时间: 2006-4-23 20:11

以下是引用scujean在2006-4-23 19:14:00的发言：

早上想破头就是想不出是C，原来是B

我的做法 rs+s is odd→s(r+1)=积

所以s=积，且(r+1)=积→r=积-1=偶

your answer is a! but the correct answer is b? why?

[此贴子已经被作者于2006-4-23 20:11:59编辑过]

作者: scujean 时间: 2006-4-23 20:53
可能是题目哪里有问题..我的观念应该正确吧..

作者: hitlzc 时间: 2006-4-24 14:54

不好意思，深谷幽兰，我这两天很忙，才看到你的短信。

支持scujean的看法，选A

3. IF the operation is defined for all integers a and b by a@b=a+b-ab, which of the following statement must be true for all a, b,c,
ANS: (A@B)@C=A@(B@C)
不明白这是如何得来的?
代入去算了，

a@b=a+b-ab, (a@b)@c = (a+b-ab)@c = a@c+b@c-ab@c =a+c-ac+b+c-bc-ab-c+abc

=a+b+c-ac-bc-ab+abc

a@(b @c)= a@(b+c-bc)=a@b+a@c-a@bc =a+b-ab+a+c-ac-a-bc+abc

=a+b+c-ac-bc-ab+abc

另外就是这个算式a@b=a+b-ab和两个集合的并集的规律类似，AUB=A+B-AB

（AUB）UC=AU（BUC）

当然我认为前一种解法更保险一些，就是太慢。

http://forum.chasedream.com/dispbbs.asp?boardid=22&id=152966&star=1#1433693

[此贴子已经被作者于2006-4-24 14:57:26编辑过]

作者: wsdoll 时间: 2007-9-10 08:48

以下是引用hitlzc在2006-4-24 14:54:00的发言：

不好意思，深谷幽兰，我这两天很忙，才看到你的短信。

破解版题意是 rs+r 是 odd., 所以 r is odd, S is even. 选B.

lz 的条件选A 和 D.

作者: phoenixfr 时间: 2008-10-7 19:16

以下是引用wsdoll在2007-9-10 8:48:00的发言：

破解版题意是 rs+r 是 odd., 所以 r is odd, S is even. 选B.

lz 的条件选A 和 D.

我说怎么上面的答案感觉都不只一个……

作者: 5cheng 时间: 2010-11-20 16:25
求助第一个的翻译和步骤！

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