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

Looking individually at the two statements, both are clearly not sufficient as each is a quadratic equation and will have two roots.....

Combined..
\(|x-|x^2||=|x^2-|x||\)...
this shows that x is +ive, otherwise \(|x-|x^2||>|x^2-|x||\)...
Now lets solve any equation
\(|x^2 -|x|| =2... ..............\\
x^2-x=2..\)
or \(x^2-x-2 =0...................x^2-2x+x-2=0...................(x-2)(x+1) = 0...................... x = 2 ...or... -1\), BUT x is +ive so x=2
Suff
C

a.
squaring both sides we get
x^2 + x^4 - 2x* x^2 = 4
meaning, x^2 (x-1)^2 = 4
thus x = 2 | -1 not sufficient.

b same process and we get x = 2| -2
not sufficient.

a+b gives x = 2.

thus C it is.

If x is an integer, what is the value of x?
1) |x - |x2|| = 2
2) |x2 - |x|| = 2

Hi Rohit,these type of questions are extremely easy.they just seem to be intimidating but they are not .
You just need to know one concept
[x]= x if x is positive
[x]=-x if x is negative.

Now take 1) |x - |x2|| = 2

|x2| is always positive. |x - |x2|| is negative since x^2>x

x^2-x=2.The value of X can be obtained as 2,-1.
Statement alone is not sufficient

From 2) Similarly we get 2 equations x^2-x=2 and x^2+x=2 depeding upon whether X is positive or negative respectively which we dont know .
Statement 2 alone is not sufficient .

Thanks for the reply Bunuel. I didnt understand the factorization part in your post. It would be great if you can simplify the parts.
But is there a mistake in the below post?

First question: factoring out.
\((x-x^2)^2=4\) --> \((x*(1-x))^2=4\) --> \(x^2*(1-x)^2=4\);
\((x^2-|x|)^2=4\) --> now, we want to factor out \(|x|\) (notice x^2=|x|*|x| and we are factoring out one |x|) --> \((|x|*(|x|-1))^2=4\) --> \(x^2*(|x|-1)^2=4\).

Second question: other solutions.
amit2k9 corrected his solution after my post so the answer there is correct.
You also quote there saikarthikreddy's solution which I don't really understand as there are some parts in reasoning missing. Also it's not clear what is saikarthikreddy's answer. E? C?

Please ask if anything remains unclear in my post.

Bunnel if you dont mind can you explain the factorizing part bit elaborately...Am totally not able to understand the second statement factorization

(2) |x^2 - |x|| = 2 --> square both sides: \((x^2-|x|)^2=4\). Since \(x^2=|x|^2\), then we have that \((|x|^2-|x|)^2=4\). Factor out \(|x|\): \(|x|^2*(|x|-1)^2=4\) --> \(x^2*(|x|-1)^2=4\).

Hi Bunuel,

I understood what you did, what I didnt understand is why you squared the equation before simplifying it. What I knew about Mods, my line of reasoning is similar to what Apex231 did. I was just wondering about the approach that you took, squaring and then moving forward, clearly I am missing something here.. can you please explain.

1) \(|x - |x^2|| = 2\)
Putting x as 2 -> \(|2 - 4| = 2\)
Putting x as -1 -> \(|-1 - 1| = 2\)
x has two values (2, -1)
Hence NOT SUFFICIENT.

2) \(|x^2 - |x|| = 2\)
Putting x as 2 -> \(|4 - 2| = 2\)
Putting x as -2 -> \(|4 - 3| = 2\)
x has two values (2, -2)
Hence NOT SUFFICIENT.

Combining 1 & 2 gives \(x =2\).
Hence (C) is the answer.

PS: how did we arrive into 2, -2, -1 roots?
Either pick numbers that are + or - integers or use algebraic approach below:

For choice 1 consider positive absolute value:
\(x - x^2 = 2\)
\(x^2 - x + 2 = 0\)
- no possible roots for this equation.

For choice 1 consider negative value:
\(-(x - x^2) = 2\)
\(x^2 - x -2 = 0\)
\(x = 2\) and \(x = -1\)

A tough one indeed.
This question is not at all a sub-700 level question. IMO it is atleast 730 level question.

Statement 1 yields 2 cases, among which one provides non-real numbers. The two real number values of x are 2,-1.
Insufficient.

Statement 2 yields 4 cases as well, among which 2 provide non real numbers. The 2 real number solutions of x are (-2 and 1) and (2 and -1) respectively. One doesn't satisfies the statement 2 and thus is not considered.
3 values, hence insufficient.

The common value in statement 1 solution and statement 2 solution is -2.
Hence x=2 .
+1C

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

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

|x - |x^2|| = 2
(x^2 is ALWAYS greater than or equal to zero so we can drop the absolute value sign)
|x-x^2| = 2
Two cases, positive and negative
Positive: x>1
x-x^2 = 2
-x^2 + x - 2 = 0
x^2 - x + 2 = 0
(can this be factored out?)

Negative: x<1
-x + x^2 = 2
x^2 - x -2 = 0
(x-2) * (x+1) = 0
x=2, x=-1

Here, we have two possible values for x.
INSUFFICIENT

(2) |x^2 - |x|| = 2

|x^2 - |x|| = 2
Two cases for x, positive and negative.
x>= 0
x^2 - x = 2
x^2 - x - 2 = 0
(x-2) * (x+1) = 0
x = 2, x = -1
two values, one less than zero one greater than zero.
INSUFFICIENT

x<0
x^2 - -x =2
x^2 + x - 2 = 0
(x+2) * (x-1) = 0
x=-2, x=1
two values for x but one is greater than zero and one is less than zero.
INSUFFICIENT

See, I would say E in this case. Where did I go wrong?

Solving it another way...

(1) |x - |x^2|| = 2
A stated above x^2 is greater than or equal to 0 so we can drop the absolute value signs.

(x-x^2)^2 = 4
x=2, x=-1

We get two values.
INSUFFICIENT

(2) |x^2 - |x|| = 2
As with #1 we can get rid of the outer absolute value signs by squaring.
|x^2 - |x|| = 2
(x^2 - |x|) = 2
(x^2 - |x|)^2 = 4
(remember, x^2 = |x|^2)
(|x|^2 - |x|)^2 = 4
|x|(|x| - 1)^2 = 4
x= 2, x=-2

we get two values
INSUFFICIENT

1+2 we get an intersection of x=-2
SUFFICIENT

Here is my question. There are many times where I have correctly used the first method (taking the positive and negative cases to solve) to solve problems and this seems like it could be one of those problems. Why is it that with this problem, that method appears to be incorrect?

Thanks!

This is not the best way to solve this question.

Also, notice that:
x-x^2<0 for x<0 and x>1 and
x-x^2>0 for 0<x<1.

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

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

|x - |x^2|| = 2
(we can drop the inner absolute value signs because x^2 is always >= 0)
|x - x^2| = 2

Normally, there would be a positive and a negative case
Positive: 0<x<1
Negative: x>1, x<0

However, because x must be an integer, there is no positive case to test because only a fraction between 0 and 1 will provide a positive case.

Negative: x>1, x<0
|x - x^2| = 2
-(x-x^2) = 2
-x + x^2 = 2
x^2 - x - 2 = 0
(x - 2)(x + 1) = 0
x=2, x=-1
Both values of x satisfy their given ranges (2>1 and -1<-0) So we are left with two possible correct answers
INSUFFICIENT

(2) |x^2 - |x|| = 2

|x^2 - |x|| = 2
Two cases:

X is an integer so it must be greater than or equal to 1 or less than or equal to -1. This means that |x^2 - |x|| = 2 will always be positive but |x| could be positive or negative.

|x^2 - |x|| = 2
x>0
(x^2 - x) = 2
x^2 - x - 2 = 0
(x - 2)(x + 1) = 0
x=2, x=-1
x=2 is Valid
OR
x<0
(x^2 - (-x)) = 2
x^2 + x - 2 = 0
(x + 2) (x - 1) = 0
x=-2, x=1
x=-2 is Valid

So, as with #1, we have two valid solutions for x, 2, -2

1+2) The valid solutions for #1 are 2 and -1, the valid solutions for #2 are 2 and -2. The only common number between them is 2.
SUFFICIENT


Could someone show me how to factor out both sides like Bunuel did in his explanation?

Thanks!

Hey Bunuel,

I drew up the same solution but I chose E because after combining (1) and (2) I saw it as x=2 or x=-2 or x=1. Can you explain why this is wrong?

x = -1 does not satisfy either of the statements.
x = -2 does not satisfy the first statement.

Hey bunuel,
I got x=-1,2 from the first statement.
But in the second statement, i took positive and negative possibilities and came up with 4 equations.
the 4 equations are
1.(x^2)-x=2
2.(x^2)+x=2
3. -(x^2)+x=2
4. -(x^2)-x=2
Out of these four equations, according to me, only the 1st and the second have real solutions.
So from the 1st equation above i got x=-1,2(Same as the result from our First Statement )
and from the second equation i got x=1,-2.
But everywhere people have written that they got x=-2,2 from the second statement. Can you please explain it to me as to what the issue is.