鼠穴太太乐446

astronautking

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？

sdcar2010

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.

astronautking

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.

-- by 会员 sdcar2010 (2011/5/2 12:00:50)

Got it! Thanks a lot