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quantumliner
If Y = |X + 1| - |X-2|, then

(a) -3 <= Y <=0
(b) -3 <= Y <=3
(c) Y <=-3
(d) Y >=-3
(e) No Solution


3 ranges to define

1) x< -1 then y= -3
2) -1<=x<2 then y = 2x-1
3) x>= 2 then y=3

as second condition dependent on value of x
thus from range 1 & 2
-3 <= Y <=3

Ans B
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rohit8865

3 ranges to define

1) x< -1 then y= -3
2) -1<=x<2 then y = 2x-1
3) x>= 2 then y=3

as second condition dependent on value of x
thus from range 1 & 2
-3 <= Y <=3

Ans B

How did you came up with the 3 ranges and how to define them? I dont understand what you are doing.
Thanks
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quantumliner
If Y = |X + 1| - |X-2|, then

(a) -3 <= Y <=0
(b) -3 <= Y <=3
(c) Y <=-3
(d) Y >=-3
(e) No Solution

Since the expressions in the absolute value will equal zero when X = -1 and X = 2, we need to investigate the following three cases:

i) X < -1, ii) -1 < X < 2, and iii) X > 2

Now let’s analyze each case.

i) If X < -1, we see that both of the expressions inside the absolute values are negative, and thus:

Y = -(X - 1) - [-(X - 2)]

Y = - X -1 + X - 2

Y = -3

ii) If -1 < X < 2, then the first expression is positive but the second expression is negative; thus:

Y = X + 1 - [-(X - 2)]

Y = X + 1 + X - 2

Y = 2X -1

Now, recall that -1 < X < 2, so

-2 < 2X < 4

-3 < 2X - 1 < 3

Since Y = 2X - 1, -3 < Y < 3.

iii) If X > 2, then both expressions inside the absolute value are positive; thus:

Y = X + 1 - (X - 2)

Y = X + 1 - X + 2

Y = 3

Combining the results of the three cases, we see that -3 ≤ Y ≤ 3.

Answer: B


I have a query regarding the range. Shouldn't it be -1 <= X < 2 and X >= 2
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warriorguy

I have a query regarding the range. Shouldn't it be -1 <= X < 2 and X >= 2

Dear warriorguy, Graphical illustration will clear your doubt.

When \(x = 2\)

\(Y = |x+1| - |x-2| = |2+1| - |2-2| = |3| + |0| = 3\)

Coordinate 1 \((2, 3)\)

When \(x = -1\)

\(Y = |x+1| - |x-2| = |-1+1| - |-1-2| = |0| - |3| = -3\)

Coordinate 2 \((-1, -3)\)

Therefore, the solution is \(-3 ≤ y ≤ 3\)
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Hi All,

This question can be solved by TESTing VALUES.

We're told that Y = |X + 1| - |X-2|. We're asked to find the range of value for Y.

Let's start with something easy....

IF....
X = 0, then Y = |1| - |-2| = -1, so Y can equal -1. Eliminate Answers C and E. Considering the three remaining answers, we know that Y either has an "upper limit" or it does not, so let's see what happens if we make X really big or really small....

IF...
X = 100, then Y = |101| - |98| = 3. This is interesting, since we've appeared to now randomly hit the 'upper limit' in Answer B. Eliminate Answer A. What if we try something even bigger....
X = 1000, then Y = |1001| - |998| = 3. This is the exact SAME value... It certainly looks like we've found the upper limit, but just to be sure, I'll do one more Test...

IF....
X = -50, then Y = |-49| - |-52| = -3. We clearly have the range now.

Final Answer:
GMAT assassins aren't born, they're made,
Rich
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EMPOWERgmatRichC
...

IF....
X = 0, then Y = |1| - |-2| = 1, so Y can equal 1.

...
Rich

How so? How does one approach this via a number line?
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mikemcgarry - Could you help me with plotting a solution on the number line, please?
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Hi Blackbox,

Thanks for catching the error; I've updated my explanation.

GMAT assassins aren't born, they're made,
Rich
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The maximum/Minimum DIFFERENCE between two ABSOLUTE values is THE ABSOLUTE VALUE of their individual DIFFERENCE .
For example, if two values are a&b.
we know, their individual absolute values are /a/ and /b/ and
DIFFERENCE between a & b= a-b.
The maximum/Minimum DIFFERENCE between two ABSOLUTE values= /(a-b)/
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Back to the question:
Max and min value of y=
y =|X+1|−|X−2|
[The maximum/Minimum DIFFERENCE between two ABSOLUTE values is THE ABSOLUTE VALUE of their individual DIFFERENCE].
=|(x+1)-(x-2)|
=|3|
=+3 or -3
So maximum value of y =3 and minimum value of y= -3.
In other words, Y must be in the range of -3 and +3.
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quantumliner
If \(Y = |X + 1| - |X-2|\), then

(a) \(-3 \leq Y \leq 0\)
(b) \(-3 \leq Y \leq 3\)
(c) \(Y \leq -3\)
(d) \(Y \geq -3\)
(e) No Solution

Asked: If \(Y = |X + 1| - |X-2|\), then

Y = |X+1| - |X-2|

Case 1: X<=-1
Y = (-1-X) - (2-X) = -3

Case 2: -1<X<=2
Y = (X+1) - (2-X) = 2X -1
-3<Y<=3

Case 3: X>2
Y = (X+1) - (X-2) = 3

-3<=Y<=3

IMO B
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