why is (3mu)mu > (3mu)(3mu), given x(mu) =x(x+1)?

Question

this is question from a GRE prep book that i'm stuck with. according to my deduction, (3mu)(3mu) could be bigger or smaller (indeterminate that is). could they have the wrong answer?

Answer 1

x(mu)=x(x+1)
so that means mu=x+1
so mu >x
3(mu)(mu)>(3mu)(3mu) as mu>3mu
by the same logic as above

Answer 2

Looks weird to me. But it looks like an unusual way of defining a function where the subscript mu means to do what you've indicated. So now we just assume that you can compose mu with mu for the lhs and regular multiplication on the rhs.

Like so:

3mu = 3(3+1)

and 3mu(mu) is [3(3+1)[3(3+1)+1]

and (3mu)(3mu) = [3(3+1)]^2

so the inequality would be true.

Answer 3

The inequality is impossible since if mu is not zero, then
3m^2 > 9m^2 will imply 1 > 3, which is absurd (here I am using
m for mu). If m = 0, then the inequality becomes 0 > 0,
which is also absurd. I don't see why the condition
mx = x(x + 1) is given. Is it possible that you didn't write the
question correctly?