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by ssiva » Wed Mar 07, 2007 4:14 pm
It has to be (m+1)^2

as m+1 > 1.

For other anwer choices they are less than 1.
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by jayhawk2001 » Wed Mar 07, 2007 7:21 pm
m^3 should have the least value

For 0 < m < 1,
1 > m > m^2 > m^3

The other 2 are greater than 1. So, m^3 is the least of the lot
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by yvonne12 » Wed Mar 07, 2007 9:12 pm
yes! of course. i understand.

thanks ....
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by Cybermusings » Tue Mar 27, 2007 1:53 am
The answer is m^3

Try and substitute values for m and it'll be easy. Let's say m=.5

Therefore .5^2= .25
.5^3=.125
2-.5=1.5
.5+1= 1.5
Hence m^3 is the smallest
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