( The square of a number is always a positive, and the modulus of a number is also positive number)

Statement 1) |x - |x^2|| = 2.

so, |X^2| = x^2

We can square on both sides.

=> ( x - x^2)^2 = 4

=> ( x - x^2) = ±2

when

=> ( x - x^2 ) = +2 => has no integer solutions

when

=> ( x - x^2 ) = -2

=> x^2 -x -2 => x = -1 or 2

This is not sufficient, we have two values

Now let us pick up statement 2) |x^2 - |x|| = 2

the |x| is confusing, but if we assume |x| as y, then x^2 = y^2

=> | y^2 - y | = 2

there can be two cases

Case 1)

=> y^2 - y = +2

=> y^2 - y -2 = 0 => y = (y - 2) ( y + 1) = 0 = y = 2 or -1

But since y = |x| so y cannot be -1 , hence x can be +2, -2

Case 2)

=> y^2 - y = -2

=> No integer solutions

This is not sufficient, we have two values

But if we combine both these statements, we figure out a common value +2.

hence C is answer.

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