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Quadratic

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Quadratic

by tanyajoseph » Thu Aug 09, 2007 11:17 am
Can someone explain the attached problem?
I have attached an image so that there is no confusion with the expressions.
Thanx
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by mindruna » Thu Aug 09, 2007 11:42 am
Is the answer C?
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by tanyajoseph » Fri Aug 10, 2007 8:39 am
The answer is II only . Can u please explain in detail? Thanx
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by givemeanid » Fri Aug 10, 2007 10:01 am
x +ve, y +ve

I. Let's assume sqrt(x+y)/2x > 1/sqrt(x+y)
sqrt(x+y) > 0 since x and y are +ve
sqrt(x+y)*sqrt(x+y) > 2x
x+y > 2x
y > x
But x could be > y and our assumption falls flat.
NOT TRUE.


II. Lets assume (sqrt(x) + sqrt(y))/(x+y) > 1/sqrt(x+y)
sqrt(xx + xy) + sqrt(xy + yy) > x+y

Since x and y are +ve, sqrt(xx + xy) > x and sqrt(xy + yy) > y
TRUE.


III. Lets assume (sqrt(x) - sqrt(y))/(x+y) > 1/sqrt(x+y)
Right side is always +ve
However, if y > x, left side is -ve. So, we cannot assume this to be true.
NOT TRUE.


II only. Answer is C.
So It Goes
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