Data Sufficiency

If n is a prime number greater than 2, is 1/x > 1?

(1) x^n < x < x^{\frac{1}{n}}

(2)x^{n-1} > x^{2n-2}

OA later

If n is a prime number greater than 2, is 1/x > 1?

(1) x^n < x < x^{\frac{1}{n}}

(2)x^{n-1} > x^{2n-2}

This is a nasty looking question, so let's rephrase it. Let's say n = 3. (The important thing to note is that n will have to be odd.) And what we're really asking is: Is 0<x<1? (because that's the only time that 1/x would be greater than 1.)

1) Let's call it x^3 < x < x^(1/3)
In this case x could be -2 or 1/2, so not sufficient.

2) Let's call it x^2 > x^4.
In this case x could be 1/2 or -1/2, so not sufficient.

T)Well, it looks like x will have to be between 0 and 1 to satisfy both inequalities, so Sufficient. The answer is C

[/spoiler]

I'm assuming he meant to write x^(1/n) and the notation came out funky.

yes it is x^1/n

Thanks OA is C