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Inequality

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by garima99 » Tue Jun 21, 2011 7:41 pm
MBA.Aspirant wrote:if
-2<= n < 3
&
m <= n <=5
what is the greatest possible value of (n+m)(n-m)?

a.9
b.5
c.16
d.25
e.none
IMO Ebecause max values n,m can take is 1 and min n can take is -2 so 2(1+2)=6
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by Frankenstein » Tue Jun 21, 2011 7:58 pm
MBA.Aspirant wrote:if
-2<= n < 3
&
m <= n <=5
what is the greatest possible value of (n+m)(n-m)?

a.9
b.5
c.16
d.25
e.none
Hi,
-2<= n < 3
& n <=5
So, -2<= n < 3
(n+m)(n-m) = n^2 - m^2
For this to be maximum n^2 should be maximum and m^2 should be minimum
n^2<9(but equality doesn't hold) and min. m^2 = 0
So, n^2 - m^2 <9
As equality doesn't hold, we cannot certainly say what the maximum value is.
Cheers!

Things are not what they appear to be... nor are they otherwise
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by MBA.Aspirant » Tue Jun 21, 2011 8:11 pm
my question is how n<= 5 and at the same time <3?
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by Frankenstein » Tue Jun 21, 2011 8:24 pm
MBA.Aspirant wrote:my question is how n<= 5 and at the same time <3?
Hi,
The question is not framed in a great way.
Even if we consider 2<= n < 3 & n <=5, because of 'and' we have to find the common part of the inequalities: 2<= n < 3, n <=5 which is 2<= n < 3.
Cheers!

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by shoot4greatness » Tue Jul 12, 2011 4:24 pm
Hi, just found this post while searching for inequality problems. for n^2-m^2 to have maximum value, we need two things. The maximum value of n and the minimum value of m. From the given inequalities, the maximum value of n is 5, and the maximum value of m is 2 (we also know that m <= n, so the minimum value of n will be the maximum value of m). However, we are only given the maximum value of m, not the minimum value of m. So right away, I concluded e.
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by shoot4greatness » Tue Jul 12, 2011 4:27 pm
Another thing, since we're squaring, the sign doesn't matter. The values of n are 2,1,0,-1,and -2. Either way, we can't come up with the minimum value of m.
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