riz_gmat wrote:What is the value of x?
1) |x² + 1| < 1
2) |x² - 1| < 1

Q: x = ?

St1: |x² + 1| < 1
x = 2; |x² + 1| < 1 becomes 5 < 1 INVALID, We cant chose a large positive number lets decrease our test value
x = 1; |x² + 1| < 1 becomes 2 < 1 INVALID
x = 0; |x² + 1| < 1 becomes 1 < 1 OK; lets keep this for now and test simple negative values knowing that a square of a number can't be negative.
x = -0.5; |x² + 1| < 1 becomes 1.25 < 1 INVALID, so x cant be negative.
The only possible option is x = 0
SUFFICIENT

St2: |x² - 1| < 1
We can easily choose two test values to prove that this statement is not sufficient.
x = -1; |0| = 0; 0 < 1 VALID
x = 1; |0| = 0; 0 < 1 VALID
INSUFFICIENT

Answer A

You can also do straight algebra:

S1::

|x² + 1| ≤ 1

x² is always nonnegative, so x² + 1 is also nonnegative, so |x² + 1| = x² + 1.

So we really have x² + 1 ≤ 1, or x² ≤ 0, or x² = 0, or x = 0. SUFFICIENT!

S2::

|x² - 1| ≤ 1

is really

x² - 1 ≤ 1 (if x² - 1 is nonnegative)
or 0 ≤ x² ≤ 2,
or -√2 ≤ x ≤ √2

We don't even need to consider the other case (x² - 1 is negative) as we already have more than one possible value of x. INSUFFICIENT

Since x² is always greater than or equal to 0, we can see that x must equal 0

Hi Brent ,

Can you please explain this above part ; I didn't understand.

As we have -2<=x²<=0.

Please explain .

Thanks in advance.

SJ

jain2016 wrote:

Since x² is always greater than or equal to 0, we can see that x must equal 0

Hi Brent ,

Can you please explain this above part ; I didn't understand.

As we have -2<=x²<=0.

Please explain .

Thanks in advance.

SJ

Algebraically, if you have the two equations

x² ≥ 0
and
x² ≤ 0

The only overlap is x² = 0. So x² = 0 is our only solution, and x = 0.