B1. First of all, let's consider two functions: z=x and z=x^2 at positive x. The function coincides at x=0 and x=1. In the range of (0,1) x > x^2 but in the range of (1,+inf) x<x^2. What relation does it have to our problem? if all numbers lies within 0..1, the average of set X will be greater than that of set Y. And vice-versa, if all numbers lies within 1..+inf, the average of set Y will be greater than that of set X.
2. Let's consider first condition. We can construct two examples:
a) X = {5,5,5,5,5,5,5,5,5} - av(Y)=25>av(X)=5
b) X = {0.5, 0.5, 0.5, 0.5, 1.00001, 1.00001, 1.00001, 1.00001, 1.00001} - av(Y)=6/9<av(X)=7/9
insufficient.
3. Let's consider second condition and try to construct two examples:
a) X = {0.000001,2,2,2,2,2,2,2,2} - av(Y)=32/9>av(X)=16/9
b) Here we have a problem. In order to fit condition we should have 0 and 2 as members of X. (other numbers cannot help us to build example for which av(Y) < av(X)). So, we have X = {0,?????,2}. Now, we need to compensate difference between 2^2 and 2 by adding 0.5 --->
---> x e {0.000001, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 2}. But even such set cannot make av(X) greater than av(Y): av(Y)=5.75/9>av(X)=5.5/9
sufficient.
P.S. You may ask me why I use 0.5 but not 0.6 or 0.4 in my way to construct the example?
Let's consider a few numbers:
0: x=0; x^2=0; delta=x^2-x=0
0: x=0.4; x^2=0.16; delta=x^2-x=-0.24
0: x=0.5; x^2=0.25; delta=x^2-x=-0.25
0: x=0.6; x^2=0.36; delta=x^2-x=-0.24
0: x=1; x^2=1; delta=x^2-x=0
So, x=0.5 is the number for which difference between x^2 and x is the least (-0.25).
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