Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5 GMAT data sufficiency

(1) (x+1) (|x| - 1) > 0. Consider two cases:

If x>0 then |x|=x
so (x+1)(|x|−1) > 0 becomes (x+1)(x−1) > 0
--> x^2 − 1 > 0
--> x^2 > 1
--> x < −1 or x > 1.
Since we consider range when x > 0
then we have x > 1
for this case;
If x ≤ 0 then |x| = −x
so (x+1)(|x|−1) > 0
becomes (x+1)(−x−1) > 0
--> −(x+1)(x+1) > 0
--> −(x+1)^2 > 0
--> (x+1)^2 < 0.
Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that (x+1)(|x|−1) > 0
holds true only when x > 1.
Sufficient.

(2) |x| < 5 --> −5 < x < 5.
Not sufficient
The answer is A.

The correct answer is D.

Statement 1:
For (x+1)(lxl-1) > 0, we should have either (x+1) > 0 and (lxl - 1) > 0 or (x+1) < 0 and (lxl -1) < 0
when (x+1) > 0 and (lxl - 1) > 0
(x+1) > 0 => x > -1
(lxl-1) > 0 => x > 1 or x < -1
From above two, possible solution is x > 1
when (x+1) > 0 and (lxl -1) < 0
(x+1)<0 => x < -1
(lxl-1) < 0 => -1 < x < 1
Both of these can not be satisfied by any value of x.
Hence we get only 1 solution, x > 1. which is what we wanted to ascertain. Sufficient.

Statement 2:
|x| < 5
=> -5 < x < 5
Clearly not sufficient to tell whether x > 1 or not.

The correct answer is A.

The expression (x+1)(|x|-1) > 0 can be satisfied when (x+1) > 0 and (|x|-1) > 0, or when (x+1) < 0 and (|x|-1) < 0.

When (x+1) > 0 and (|x|-1) > 0, the solution is x > 1.
When (x+1) > 0 and (|x|-1) < 0, there are no values of x that satisfy both conditions simultaneously.

Therefore, the solution to the inequality (x+1)(|x|-1) > 0 is x > 1.

Regarding Statement 2, the inequality |x| < 5 does not provide enough information to determine whether x > 1 or not.
The inequality -5 < x < 5 represents all values of x that satisfy |x| < 5, but it does not narrow down the range enough to determine the value of x in relation to 1.

In summary:
Statement 1 is sufficient to determine that x > 1.
Statement 2 is not sufficient to determine whether x > 1 or not.

The correct answer is A.