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

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06 Feb 2017, 13:42
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32% (01:39) wrong

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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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06 Feb 2017, 18:22
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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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09 Feb 2017, 05:30

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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09 Feb 2017, 06:08
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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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13 Feb 2017, 16:05
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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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15 Feb 2017, 03:45

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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15 Feb 2017, 05:04
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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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16 Feb 2017, 13:13
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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Re: If Y = |X + 1| - |X-2|, then [#permalink ]

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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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23 Sep 2017, 12:23

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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24 Sep 2017, 12:29
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
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24 Sep 2017, 12:29