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by Suyog » Fri Oct 05, 2007 1:29 pm
lets take 1st case:

x > y > z
this is simple
Assume z = 2, y = 5, and x =26
26 > 25 > 4
This cond is TRUE

now for 2 and 3, we know that the square of positive or negative even number is always positive and cube of negative number is negative, which is not the case here.

So for the 3rd Case.
assume x = 5, y = -2 and z = -1
5 > 4 > 1
This cond is also TRUE

Case 2 dont satisfy the equation for diff values.

Ans B
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by nidhanoor » Fri Oct 05, 2007 2:06 pm
case 2 hold true in this case

x = 0.5 y =0.6 z=0.7
y^2 = 0.36 z^4 = 0.2401

here x > y^2 > z^4

but z>y>x

So E
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