jlgdr wrote:
Could someone please elaborate more on Statement 2 about the range?
Thanks
Cheers
J
Hi J
Set X consists of 9 positive elements. Set Y consists of the squares of the elements of X. Is the average of the elements in Y greater than that of set X?
Statement 2: The range of X is 2.
If the range is 2 and all elements are positive in X,
range = Maximum element value - Minimum element value = 2---> Maximum element value = 2+ Minimum element value = 2 + k , as k >0 [given all elements are positive]
The corresponding element in Y is \((2+k)^2= 4+4*k+k^2\) --------------(i)
Also,
The difference f(x)=
x^2 - x = x (x-1) is a parabola which opens upward,
so the minimum value of f(x) when x=1/2, the midpoint of the two x-intercepts.
f(1/2)=1/4 which can also be derived using differentiation eq : 1-2*x = 0 ----------> x= 1/2
Now,taking two cases:
a)when all elements in X > 1
In that case: each elements in Y > each elements in X so, the avg(Y)> avg(X)
b)When at least on element in X < 1
Looking at the boundary case:
If each of the 9 elements in X is 1/2, then this case should give the minimum value of \(x^2 - x\) as depicted by parabola
then, Sum of Y - Sum of X
= \([9*(4+4*k+k^2) ]- [9*1/4]\) > 0 as k>0
and thus the (avg)Y>avg(X). Sufficient in all cases. So,B.
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