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标题: 鼠穴太太乐446 [打印本页]

作者: imbalancey 时间: 2011-5-13 07:09 标题: 鼠穴太太乐446

401、DS:函数h(x)=x-x^2 问能否判断h(a)<h(b)
(1) a<b
(2) a^2<b^2
我琢磨了好久，看起来不像是难题啊，可是就感觉想不通，最后选了C吧，不是很确定

答案应该选E吧。因为是双曲线，不确定a,b的正负。
这题怎么回事啊？

作者: momohe 时间: 2011-5-13 21:02

h(a) = a - a^2
h(b) = b- b^2

We just need to evaluate h(a) - h(b)

h(a) - h(b) = (a-b) - (a^2 - b^2) = (a-b)*(1-a-b)

1) a<b, so (a-b)<0. But the sign of (1-a-b) could not be determined

2) a^2<b^2, then (a-b)*(a+b)<0. So (a-b) and (a+b) are having opposite signs. If (a-b) <0, then (a+b) >0. But we do not know if (1-a-b) is bigger than, equal to, or smaller than zero. If (a-b) >0, then (a+b) <0, then (1-a-b) >0. Only under this condition, you can know h(a)>h(b)

When both 1) and 2) combined, 1) (a-b) <0. 2) We still do not know the sign of (1-a-b). Insufficient.

EEEEEEEEEEEEe

P.S. If condition 1) changes to 1) a>b, then the answer would be C as I explained above.

作者: imbalancey 时间: 2011-5-14 07:19

Got it! Thanks a lot ！

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