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

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If Y = |X + 1| - |X-2|, then  [#permalink]

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New post 06 Feb 2017, 14:42
3
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A
B
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D
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  45% (medium)

Question Stats:

68% (01:48) correct 32% (01:39) wrong based on 399 sessions

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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
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Re: If Y = |X + 1| - |X-2|, then  [#permalink]

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New post 13 Feb 2017, 17:05
2
4
quantumliner wrote:
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
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Re: If Y = |X + 1| - |X-2|, then  [#permalink]

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New post 06 Feb 2017, 19:22
1
2
quantumliner wrote:
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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Re: If Y = |X + 1| - |X-2|, then  [#permalink]

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New post 09 Feb 2017, 06:30
rohit8865 wrote:

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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Re: If Y = |X + 1| - |X-2|, then  [#permalink]

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New post 09 Feb 2017, 07:08
3
maxschmid wrote:
rohit8865 wrote:

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


maxschmid

for such questions always do the below suggested

Ix+1I will be 0 when x= -1
similarly Ix-2I will be 0 when x=2

thus we have boundaries defined as -1 and 2
so i have checked on both sides of boundaries defined in three steps for value y

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

hope now u can understand
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Re: If Y = |X + 1| - |X-2|, then  [#permalink]

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New post 15 Feb 2017, 04:45
JeffTargetTestPrep wrote:
quantumliner wrote:
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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Re: If Y = |X + 1| - |X-2|, then  [#permalink]

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New post 15 Feb 2017, 06:04
2
warriorguy wrote:

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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Re: If Y = |X + 1| - |X-2|, then  [#permalink]

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New post Updated on: 24 Sep 2017, 13:27
3
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:

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Originally posted by EMPOWERgmatRichC on 16 Feb 2017, 14:13.
Last edited by EMPOWERgmatRichC on 24 Sep 2017, 13:27, edited 1 time in total.
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Re: If Y = |X + 1| - |X-2|, then  [#permalink]

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New post 23 Sep 2017, 13:15
1
EMPOWERgmatRichC wrote:
...

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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Re: If Y = |X + 1| - |X-2|, then  [#permalink]

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New post 23 Sep 2017, 13:23
mikemcgarry - Could you help me with plotting a solution on the number line, please?
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Re: If Y = |X + 1| - |X-2|, then  [#permalink]

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New post 24 Sep 2017, 13:29
Hi Blackbox,

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

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Rich
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Re: If Y = |X + 1| - |X-2|, then &nbs [#permalink] 24 Sep 2017, 13:29
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