Inequalities

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Inequalities

by riz_gmat » Mon Oct 28, 2013 5:10 am
What is the value of x?

1) |x^2 + 1| <= 1
2) |x^2 - 1| <= 1


[spoiler]OA: A[/spoiler]
Did not understand how they got this OA.
Source: — Data Sufficiency |

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by Brent@GMATPrepNow » Mon Oct 28, 2013 5:55 am
riz_gmat wrote:What is the value of x?

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

Statement 1: |x² + 1| < 1
Rule: If |something| < k (where k > 0), then -k < something < k
So, we can take |x² + 1| < 1 and rewrite it as -1 < x² + 1 < 1
Subtract 1 from both sides to get -2 < x² < 0
Since x² is always greater than or equal to 0, we can see that x must equal 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: |x² - 1| < 1
Rewrite as -1 < x² - 1 < 1
Add 1 to both sides to get: 0 < x² < 2
There are many values of x that satisfy this inequality.
For example, x = 0 and x = 1 both work.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer = A

Cheers,
Brent
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by mevicks » Mon Oct 28, 2013 6:18 am
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

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by Matt@VeritasPrep » Mon Oct 28, 2013 10:52 am
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

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by jain2016 » Sat Dec 05, 2015 9:03 pm
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

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by Brent@GMATPrepNow » Sat Dec 05, 2015 9:19 pm
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
We concluded that x² < 0 (less than or equal to zero)
We know that x² can never be less than zero, so x² must equal zero.
If x² equals zero, then x = 0

Does that help?
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by Matt@VeritasPrep » Fri Dec 11, 2015 1:58 pm
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.