If the real valued function f(x) = (ax-1)/xn(ax+1) is even, then n is equal to?

f(x) = (ax-1)/xn(ax+1) is even ⇔ f(-x) = f(x)
i.e., (a(-x)-1)/(-x)n(a(-x)+1) = (ax-1)/xn(ax+1) ⇒ (-ax-1)/(-1)nxn(-ax+1) = (ax-1)/xn(ax+1)
If n=1, then (-ax-1)/(-1)x1(-ax+1) = (ax-1)/x(ax+1) which simplifies to (ax+1)/x(ax-1) = (ax-1)/x(ax+1), which is not true in general.
If n=2, then (-ax-1)/(-1)²x²(1-ax) = (ax-1)/x²(ax+1) which simplifies to (-ax-1)/(1-ax) = (ax-1)/(ax+1), which is not true in general.
Let's consider the case when n is an odd number.
If n is odd, (-1)n = -1. Then we have:
(-ax - 1)/(-xn)(-ax + 1) = (ax - 1)/(xn)(ax + 1)
(ax + 1)/(xn)(ax - 1) = (ax - 1)/(xn)(ax + 1)
(ax + 1)² = (ax - 1)²
(ax + 1)² - (ax - 1)² = 0
(ax + 1 + ax - 1)(ax + 1 - ax + 1) = 0
(2ax)(2) = 0
4ax = 0
This implies that a = 0 or x = 0, which is not true in general since f(x) should hold for all x. Hence, n cannot be odd.
Let's consider the case when n is an even number. Then (-1)n = 1.
(-ax - 1)/(xn)(-ax + 1) = (ax - 1)/(xn)(ax + 1)
(-ax - 1)(-ax + 1) = (ax - 1)(ax + 1)
(ax + 1)(ax - 1) = (ax - 1)(ax + 1)
This equation holds true. Hence n can be any even number.
However, f(x) = (ax - 1)/xn(ax + 1). If f(x) is even, then f(-x) = f(x). Therefore:
(a(-x) - 1)/((-x)n)(a(-x) + 1) = (ax - 1)/(xn)(ax + 1)
(-ax - 1)/((-1)nxn)(-ax + 1) = (ax - 1)/(xn)(ax + 1)
If n=1, (-ax-1)/(-x)(-ax+1) = (ax-1)/x(ax+1) which simplifies to (ax+1)/x(ax-1) ≠ (ax-1)/x(ax+1) unless ax=1 or ax=-1 which is not true generally.
If n = 2, then (-ax - 1)/x²(-ax + 1) = (ax - 1)/x²(ax + 1) which simplifies to (ax + 1)(ax + 1) = (ax - 1)(ax - 1), which is false unless a=0 or x=0.
Let's assume n=1. Then f(-x) = (-ax-1)/(-x)(-ax+1) = (ax+1)/(x(ax-1)) ≠ f(x) = (ax-1)/(x(ax+1))
If n is even, then (-1)^n = 1. Then (-ax-1)/xn(-ax+1) = (ax-1)/xn(ax+1). Then (ax+1)(ax+1) = (ax-1)(ax-1) which gives 4ax = 0, so a = 0 or x = 0 which is not valid.
Therefore, there seems to be a mistake in the problem statement or solution provided.