abhasjha wrote:

Given : x^n = x^(n+2)
i.e. x^(n+2) - x^n =0
i.e. x^n (x^2 - 1) = 0
i.e. either x^n = 0 or (x^2 - 1)=0
i.e. either x = 0 or (x^2 = 1)
i.e. either x = 0 or x = 1 or x = -1

Question : Is x>0 ?

Statement 1) x = (x^2)-2
i.e. (x^2)-x = 2
i.e. x(x-1) = 2 [Product of 2 consecutive numbers]
i.e. x = -1 or x = 2
Using the Given Information x=-1
SUFFICIENT

Statement 2) 2x < (x^5)

i.e. x is neither 0, nor 1 therefore x is -1 [Using Given information]

SUFFICIENT

Answer: Option D

ccassel wrote:x^n = x^(n+2) for any integer n. Is it true that x>0?

(1) x = x^2 -2
(2) 2x < x^5

Only three values of x will satisfy the condition that x^n = x^(n+2) for any integer n:
x=-1, x=0, x=1.

Let's plug x=-1, x=0, and x=1 into x^n = x^(n+2).

If n=2 and n+2=4:
(-1)² = (-1)�. Yes.
0² = 0�. Yes.
1² = 1�. Yes.

If n=3 and n+2=5:
(-1)³ = (-1)�. Yes.
0³ = 0�. Yes.
1³ = 1�. Yes.

Thus, our only options are x=-1, x=0, or x=1.
Since we want to know whether x>0, the question can be rephrased: Does x=1?

Statement 1: x = x² - 2.
x² - x - 2 = 0.
(x-2)(x+1) = 0.
x=2 or x=-1.
Thus, it is not true that x=1.
Sufficient.

Statement 2: 2x < x�.
Of the 3 possible values x=-1, x=0, and x=1, only x=-1 satisfies statement 2:
2(-1) < (-1)�
-2 < -1.
Thus, it is not true that x=1.
Sufficient.

The correct answer is D.