Function F(x) is defined as follows: if x is positive or 0 then F(x) =

Official Solution:


Function \(F(x)\) is defined as follows:

if \(x\) is positive or 0 then \(F(x) = 1 - x\);

if \(x\) is negative then \(F(x) = x - 1\).

Which of the following is true about \(F(x)\)?

I. \(F(|x|) = |F(x)|\)

II. \(F(x) = F(-x)\)

III. \(|F(x + 1)| = |x|\)


A. I only
B. II only
C. III only
D. I and III only
E. not I, II, or III


I is not necessarily true. Consider \(x = -1\). \(F(|x|) = 0\) does not equal \(|F(x)| = 2\).

II is not necessarily true. Consider \(x = 1\). \(F(x) = 0\) does not equal \(F(-x) = -2\).

III is always true:

if \(x + 1\) is non-negative then \(|F(x + 1)| = |1 - (x + 1)| = |-x| = |x|\);

if \(x + 1\) is negative then \(|F(x + 1)| = |(x + 1) - 1| = |x|\)


Answer: C
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If x is +/0 = F(x) = 1-x
If x is - = F(x) = x-1

1) F(|x|) = |F(x)|

Case 1: x=1
F(|x|) = |F(x)|
F(|1|) = |F(1)|
1-|x| = |1-x|
1-|1| = |1-1|
0 = 0 (sufficient)

Case 2: x=-1
F(|x|) = |F(x)|
F(|-1|) = |F(-1)|
|x|-1 = |x-1|
|-1|-1 = |-1-1|
0 = 2 (Insufficient)

2) F(x) = F(-x)
Case 1: x=1
F(x) = F(-x)
F(1) = F(-1)
1-x = 1-(-1)
1-1 = 1+1
0 = 2 (Insufficient)

Case 2: x=-1
F(x) = F(-x)
F(-1) = F[-(-1)]
x-1 = x-1
-1-1 = +1-1
-2 = 0 (Insufficient)

3) |F(x+1)| = |x|
Case 1: x=1
|F(x+1)| = |x|
|F(1+1)| = |1|
|1-x| = 1
|1-2| = 1
1 = 1 (Sufficient)

Case 2: x=-1
|F(x+1)| = |x|
|F(-1+1)| = |-1|
|0-1| = |-1|
+1 = +1 (Sufficient)

So only option 3 is sufficient.

Answer is C

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Hi Bunuel,

In the OE, you have taken x+1 non negative, then converted |-x| = |x|. I am confused here.
Usually, |A| = A, only if A>0. Here x≥ -1, still |-x| = |x|.

Please explain.

|-x| = |x| for any value of x. |x| is the distance from x to 0, and |-x| is the distance from -x and 0. Obviously these distances are equal.
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IanStewart can you please explain how to solve this because i am not able to understand the explanations provided?

I wouldn't worry about this question much, because it's not like any real GMAT question. Really the things you'd want to know that we use in this question are:

- how to plug algebraic things into functions, so how to work out what f(x+1) is when a question tells you what f(x) is
- that |x| = |-x|, and that |x| = -x when x is negative

If you think about what the function looks like in coordinate geometry, you can evaluate items I and II quickly, so I'll describe that, but you can also just plug in numbers to see what happens. The function is in two parts:

• when x is positive (or zero, though zero doesn't really matter here), the function is the line y = 1 - x, so it has a y-intercept of 1, and a slope of -1.
• when x is negative, the function is the line y = x - 1, so it has a y-intercept of -1 and a slope of 1

We then need to see which of the three roman numeral items must be true. Absolute values are always positive (or zero), so looking at I:

F(|x|) = | F(x) |

here we're plugging a positive thing into the function on the left, and item I says we're always getting a (specific) positive result. But we know when we plug in positive things the function is a line with slope -1, so the function will take negative values for sure if we plug in something large enough, and this can't be right. You could also see that by plugging in x = 2, say; then F(|x|) = F(|2|) = F(2) = -1, which is not equal to |F(2)| = |1 - 2| = |-1| = 1.

Looking at II. if F(X) = F(-X), then we get the same result when we plug in a number or its negative. That would mean the graph is symmetric if we reflect the function through the y-axis, but that's not the case here, if you visualize the lines that make up the function. Or you can again plug in x = 2, and then F(X) = F(2) = -1, while F(-2) = x - 1 = -3, which aren't equal.

For III, when x is positive, |x| = x, and x+1 is positive too. So when x > 0, then |F(x+1)| = |x| just says, plugging "x+1" into the definition of the function for positive values, |1 - (x+1)| = x, or |-x| = x, and since |-x| = |x| = x when x is positive, this is true.

When x is negative, then |x| = -x. We have two cases now, when we plug into the function. Either x + 1 is also negative (x < -1) so we'll use the negative definition of the function, or x + 1 > 0, or -1 < x < 0, and we'll use the positive definition of the function. When x and x + 1 are both negative, item III becomes

|F(x + 1)| = |x|
|(x + 1) - 1| = -x
|x| = -x

and since x < 0, the left and right sides are equal. When x is negative and x+1 is positive, then using the positive definition of the function, we have

|F(x + 1)| = |x|
|1 - (x + 1)| = -x
|-x| = -x
|x| = -x

and again since x < 0, these are equal. So III is true in every case. I've never seen an official question like this one though.
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Just plug in the value
1) x= 5 , -4 and the other function is a positive value not possible
2) x= 2 , -1 and -3 not equal
Then only one option left 3 yeah Jackpot
Therefore IMO C

S1) defined F(|x|) and |F(x)| and equated both. they are not equal

S2) defined F(-x) and F(x) and equated both. they are not equal

S3) defined F(x+1) then |F(x+1)|. its equal to |x|

Solution:

I. \(F(|x|) = |F(x)|\)

• If x is negative,
  • LHS: \(F(|x|)=1-(-x)=1+x\)
  • RHS: \(|F(x)|=|x-1|\)
• We can see that LHS ≠ RHS and there is no need to check for when x is positive

II. \(F(x) = F(-x)\)

• If x is negative,
  • LHS: \(F(x)=x-1\)
  • RHS: \(|F(-x)|=1-x\)
• We can see that LHS ≠ RHS and there is no need to check for when x is positive

III. \(|F(x + 1)| = |x|\)

• LHS: \(|F(x+1)|=\) either \(|1-(x+1)|=|-x|=x\) or \(|x+1-1|=|x|=x\)
• RHS: \(|x|=x\)
• We can see that LHS = RHS always

Hence the right answer is Option C
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