Mode critical points

[vipulgoyal]

If x is an intiger, what is the value of X
1. |x - |x^2| | = 2
2. |x^2 -|x| | =2

[mkdureja]

Difference between x and x² can be 2 only when x = -1 or x = 2

Statement 1:
If x=-1, |x-x²| = |-1-1| = 2.
If x=2, |x-x²| = |2-4| = 2. NOT SUFFICIENT (x can be -1 or 2)
Statement 2:
If x=-1, |x²-|x|| = |1-1| = 0.
If x=2, |x²-|x|| = |4-2| =2. SUFFICIENT (x can be 2 only)

B

[Atekihcan]

If x is an intiger, what is the value of X
1. |x - |x^2| | = 2
2. |x^2 -|x| | =2

As x² ≥ 0, |x²| = x²

Statement 1: |x - |x²|| = |x - x²| = 2
This means the distance between x and x² is 2.
This is possible only if x = -1 and x = 2
So, statement is not sufficient

Statement 2: |x² - |x|| = 2
This means the distance between |x| and x² is 2.
This is possible only if x = -2 and x = 2
So, statement is not sufficient

Both statements together: x = 2
So, both statements together is sufficient

Answer : C

[vipulgoyal]

Hi Atekihcan, got the point but please explain how you are opening modes to get these values
This is possible only if x = -2 and x = 2.

[Atekihcan]

Hi Atekihcan, got the point but please explain how you are opening modes to get these values
This is possible only if x = -2 and x = 2.

I did not solve this problem by opening absolute value problem. My logic is as follows,
|x² - |x|| = 2 means the distance between |x| and x² is 2 and 0it is given that x is an integer.
As the expression involves x² and |x|, it does not matter whether x is negative or positive.
If some x = k satisfy the equation, x = -k will also satisfy.

Now, if x ≥ 3, the distance between x² and |x| will be always greater than 2.
And, if x = 0 or x = 1, x² = |x|

Only, possible solution is x = 2
So, x = -2 is another solution.


If you want to solve this problem by opening modulus brackets, here you go...
As x is an integer and x = 0 or x = 1 cannot be the solution of the equation, |x| must be greater than 1.
So, x² is always greater than |x|
So, |x² - |x|| = x² - |x| = 2

Now, if x > 1, x² - |x| = x² - x = 2 ---> x² - x - 2 = 0 ---> (x + 1)(x - 2) = 0 ---> x = 2
And, if x < -1, x² - |x| = x² - (-x) = 2 ---> x² + x - 2 = 0 ---> (x - 1)(x + 2) = 0 ---> x = -2


Aside 'mode' is very different thing (a statistical measure).
This problems are on absolute values or modulus. Some people refer it as 'mod' but that has another different interpretation in number theory.

[GMATGuruNY]

If x is an integer, what is the value of x?

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

In statement 1, |x²| is redundant: since x² cannot be negative, |x²| = x².

Statement 1: |x-x²|=2
x - x² = ±2
x(1-x) = ±2.

Since x must be an integer, x=±1 or x=±2.
Check which of these values are valid solutions for |x-x²| = 2.

If x=1, then |x-x²| = |1 - 1²| = 0.
If x=-1, then |x-x²| = |-1 - (-1)²| = 2.
If x=2, then |x-x²| = |2 - 2²| = 2.
If x=-2, then |x-x²| = |-2 - (-2)²| = 6.

Since it's possible that x=-1 or that x=2, INSUFFICIENT.

Statement 2: |x² -|x||=2
x²-|x| = ±2
Since x² = |x|*|x|, we can factor out |x|:
|x| (|x|-1) = ±2.

Since x must be an integer, |x|=1 or |x|=2, implying that x=±1 or x=±2.
Check which of these values are valid solutions for |x² -|x||=2.

If x=-1, then |x² -|x|| = |(-1)² - |-1|| = 0.
If x=1, then |x² -|x|| = |1² - |1|| = 0.
If x=2, then |x² -|x|| = |2² - |2|| = 2.
If x=-2, then |x² -|x|| = |(-2)² - |-2|| = 2.

Since it's possible that x=2 or that x=-2, INSUFFICIENT.

Statements 1 and 2 combined:
Both statements are satisfied only by x=2.
SUFFICIENT.

The correct answer is C.

[vipulgoyal]

Thanks Mitch for fresh approach