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Given

The function f(x) = |x + 3| + |x + 2| ...... Equation A

To Find

For which value of x, does f(x) = f(x - 1) ?

Let us replace "x" by "x-1" in the function f(x) to get the function f(x-1):

f(x-1) = | (x-1) + 3 | + | (x-1) + 2 |

f(x-1) = |x+2| + |x+1| ...... Equation B

To find the value of x:

Equation A = Equation B

|x + 3| + |x + 2| = |x+2| + |x+1|

|x+3| = |x+1| ...... Equation C

Now, let us start substituting the values from the options to check if it fits the Equation c

1) When x = -3

|-3+3| = |-3+1|

|0| = |-2|

0 = 2 (Absurd)

So x cannot be -3

2) When x = -2

|-2+3| = |-2+1|

|1| = |-1|

1 = 1 ( Makes sense )

Since LHS = RHS , x=-2 is a sloution of the Equation C.

We do not need to check for the rest of the options as we have already got our answer, x=-2

So the correct answer is B.
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Bunuel

FRESH GMAT CLUB TESTS QUESTION



The function f is defined for all numbers x by f(x) = |x + 3| + |x + 2|. For which value of x, does f(x) = f(x - 1) ?

A. -3
B. -2
C. -1
D. 1
E. 2

equate for values f(x) = |x + 3| + |x + 2|

f(x) = f(x - 1)

test with answer values we see at -2 we get f(-2)=f(-3)
IMO B
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Bunuel

FRESH GMAT CLUB TESTS QUESTION



The function f is defined for all numbers x by f(x) = |x + 3| + |x + 2|. For which value of x, does f(x) = f(x - 1) ?

A. -3
B. -2
C. -1
D. 1
E. 2

I started with 2 in a jiffy, putting values I realized that if we put x -ve then we can cancel out terms to get f(x) = f(x - 1)
Substituting x= -2 we get it.

It all boils down to how fat we can get to the answer. I'm sharing a snapshot of my notepad.
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|something|=|something|
√(Something)^2= √(something)^2

After putting the values
|X+3|=|X+1|
(X+3)^2=(X+1)^2


So , X =-2

Posted from my mobile device
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Bunuel

FRESH GMAT CLUB TESTS QUESTION



The function f is defined for all numbers x by f(x) = |x + 3| + |x + 2|. For which value of x, does f(x) = f(x - 1) ?

A. -3
B. -2
C. -1
D. 1
E. 2

f(x)=f(x-1):
f(x)=|x+3|+|x+2|
f(x-1)=|x-1+3|+|x-1+2|=|x+2|+|x+1|
|x+3|+|x+2|=|x+2|+|x+1|…|x+3|-|x+1|=|x+2|-|x+2|…|x+3|-|x+1|=|x+2|(1-1)…|x+3|=|x+1|
|x+3|≥0 (positive) x≥-3 and |x+3|<0 (negative) x<-3
|x+1|≥0 (positive) x≥-1 and |x+1|<0 (negative) x<-1
range x≥-1: |x+3|=|x+1|…x-x=1-3…0=-2 invalid;
range -3≤x<-1: |x+3|=|x+1|…x+3=-x-1…2x=-1-3…x=-2 valid (inside range);
range x<-3: |x+3|=|x+1|…-x-3=-x-1…0=2 invalid;

Ans (B)
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Bunuel

FRESH GMAT CLUB TESTS QUESTION



The function f is defined for all numbers x by f(x) = |x + 3| + |x + 2|. For which value of x, does f(x) = f(x - 1) ?

A. -3
B. -2
C. -1
D. 1
E. 2

equate for values f(x) = |x + 3| + |x + 2|

f(x) = f(x - 1)

test with answer values we see at -2 we get f(-2)=f(-3)
IMO B

Can you do this quicker with any other method other than plugging in?
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Bunuel

FRESH GMAT CLUB TESTS QUESTION



The function f is defined for all numbers x by f(x) = |x + 3| + |x + 2|. For which value of x, does f(x) = f(x - 1) ?

A. -3
B. -2
C. -1
D. 1
E. 2

equate for values f(x) = |x + 3| + |x + 2|

f(x) = f(x - 1)

test with answer values we see at -2 we get f(-2)=f(-3)
IMO B

Can you do this quicker with any other method other than plugging in?



Hi,

Yes you can do it faster, in few seconds, and without plugging in numbers :
Ix+3I=Ix+1I

now drop the brackets, you can see that it can't be x+3=x+1, therefore, one side of the equation has to be negative

-x-3=x+1
-2x=4
x=-2

Answer B
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MUCH quicker way:

|x+3| + |x+2| find the zeros of each module: x= -3 and -2

substitute x-1 into x:

|x-1+3| + |x-1+2| = |x+2| + |x+1|

|x+2| + |x+1| find the zeros of each module: x = -2 or 1


x = -2 is common, plug it in and it works
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chetan2u Bunuel

Please help solving this graphically?
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Bunuel

FRESH GMAT CLUB TESTS QUESTION



The function f is defined for all numbers x by f(x) = |x + 3| + |x + 2|. For which value of x, does f(x) = f(x - 1) ?

A. -3
B. -2
C. -1
D. 1
E. 2


If \(f(x) = |x + 3| + |x + 2|\), then \(f(x-1)= |x-1 + 3| + |x-1 + 2|=|x+2|+|x+1| \)

Given that \( f(x)=f(x-1) \) => \(|x + 3| + |x + 2|=|x + 1| + |x + 2|\)
\(|x + 3|=|x + 1|\)

We can square both sides as both sides are non-negative
\(|x + 3|^2=|x + 1|^2\)
\(x^2+6x+9=x^2+2x+1\)
\(4x=-8\)
\(x=-2\)

B

wishmasterdj
It would be better to follow approach as above rather than graphical.

If you wish to do it graphically, plot the graphs as following
1) Take y=|x+3|+|x+2|....brown line
2) Take y=|x-1+3|+|x-1+2|=|x+2|+|x+1|....blue line
They intersect at x=-2, where y=1 in each case.
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we can always solve such question using conventional methods but by looking at the question and going through the options i could easily figure out that we can use option to arrive at the solution in less time.
To begin with x must be a negative number so option D and E are ruled out now trying with option A,B & C i easily found that -2 is the correct answer.

Hope this helps for those looking for tricks to solve questions in less time.
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Given that f(x) = |x + 3| + |x + 2| and we need to find the value of x for which f(x) = f(x - 1)

To find f(x - 1) we need to compare what is inside the bracket in f(x - 1) and f(x)

=> We need to substitute x with x-1 in f(x) = |x + 3| + |x + 2| to get the value of f(x - 1)

=> f(x-1) = |x-1 + 3| + |x-1 + 2| = |x + 2| + |x + 1| = f(x) (given)

=> |x + 2| + |x + 1| = |x + 3| + |x + 2|

Cancelling |x +2| from both the sides we get

|x +1| = |x + 3|

Squaring both the sides we get (Watch this video to learn about the Basics of Absolute Values)

\((x +1)^2\) = \((x +3)^2\)
=> \(x^2 + 2*x*1 + 1^2\) = \(x^2 + 2*x*3 + 3^2\)
=> 2x + 1 = 6x + 9
=> 6x - 2x = 1-9 = -8
=> 4x = -8
=> x = \(\frac{-8}{4}\) = -2

So, Answer will be B
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

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BrushMyQuant
Given that f(x) = |x + 3| + |x + 2| and we need to find the value of x for which f(x) = f(x - 1)

To find f(x - 1) we need to compare what is inside the bracket in f(x - 1) and f(x)

=> We need to substitute x with x-1 in f(x) = |x + 3| + |x + 2| to get the value of f(x - 1)

=> f(x-1) = |x-1 + 3| + |x-1 + 2| = |x + 2| + |x + 1| = f(x) (given)

=> |x + 2| + |x + 1| = |x + 3| + |x + 2|

Cancelling |x +2| from both the sides we get

|x +1| = |x + 3|

Squaring both the sides we get (learn about the Basics of Absolute Values)

\((x +1)^2\) = \((x +3)^2\)
=> \(x^2 + 2*x*1 + 1^2\) = \(x^2 + 2*x*3 + 3^2\)
=> 2x + 1 = 6x + 9
=> 6x - 2x = 1-9 = -8
=> 4x = -8
=> x = \(\frac{-8}{4}\) = -2

So, Answer will be B
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

hey everything is understanble, but why did we square both sides? :)
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BrushMyQuant
Given that f(x) = |x + 3| + |x + 2| and we need to find the value of x for which f(x) = f(x - 1)

To find f(x - 1) we need to compare what is inside the bracket in f(x - 1) and f(x)

=> We need to substitute x with x-1 in f(x) = |x + 3| + |x + 2| to get the value of f(x - 1)

=> f(x-1) = |x-1 + 3| + |x-1 + 2| = |x + 2| + |x + 1| = f(x) (given)

=> |x + 2| + |x + 1| = |x + 3| + |x + 2|

Cancelling |x +2| from both the sides we get

|x +1| = |x + 3|

Squaring both the sides we get (learn about the Basics of Absolute Values)

\((x +1)^2\) = \((x +3)^2\)
=> \(x^2 + 2*x*1 + 1^2\) = \(x^2 + 2*x*3 + 3^2\)
=> 2x + 1 = 6x + 9
=> 6x - 2x = 1-9 = -8
=> 4x = -8
=> x = \(\frac{-8}{4}\) = -2

So, Answer will be B
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

hey everything is understanble, but why did we square both sides? :)
Squaring is a common technique to eliminate absolute value signs because (|x|)^2 = x^2. For example, if we have |x| = |y|, squaring both sides gives x^2 = y^2, removing the modulus signs to simplify.

10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.­
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