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f(2x-1) = ( x + 2 ) / ( x – 2 ). What is f(x)?

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f(2x-1) = ( x + 2 ) / ( x – 2 ). What is f(x)?  [#permalink]

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New post 31 Jan 2019, 01:16
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A
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C
D
E

Difficulty:

  45% (medium)

Question Stats:

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[GMAT math practice question]

\(f(2x-1) = \frac{( x + 2 )}{( x – 2 )}\). What is \(f(x)\)?

\(A. f(x) = \frac{( x + 5 )}{( x – 3 )}\)
\(B. f(x) = \frac{( x - 5 )}{( x + 3 )}\)
\(C. f(x) = \frac{( x + 5 )}{( x + 3 )}\)
\(D. f(x) = \frac{( x – 5 )}{( x – 3 )}\)
\(E. f(x) = \frac{( x + 3 )}{( x – 3 )}\)

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Re: f(2x-1) = ( x + 2 ) / ( x – 2 ). What is f(x)?  [#permalink]

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New post 31 Jan 2019, 02:04
f(2x-1) = f(x)
Now we have to replace 2x-1 by x, i.e, 2x-1 = x
then x = 1
f(2x-1) = -3

Now put x = 1 in all the options, Only Option A yields -3.

A is the answer.
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Re: f(2x-1) = ( x + 2 ) / ( x – 2 ). What is f(x)?  [#permalink]

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New post 31 Jan 2019, 02:06
MathRevolution wrote:
[GMAT math practice question]

\(f(2x-1) = \frac{( x + 2 )}{( x – 2 )}\). What is \(f(x)\)?

\(A. f(x) = \frac{( x + 5 )}{( x – 3 )}\)
\(B. f(x) = \frac{( x - 5 )}{( x + 3 )}\)
\(C. f(x) = \frac{( x + 5 )}{( x + 3 )}\)
\(D. f(x) = \frac{( x – 5 )}{( x – 3 )}\)
\(E. f(x) = \frac{( x + 3 )}{( x – 3 )}\)


from given answer options
do plugin of f(x) as f(2x-1)
in option A
we get
\(f(2x-1) = \frac{( x + 2 )}{( x – 2 )}\)

IMO A
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Re: f(2x-1) = ( x + 2 ) / ( x – 2 ). What is f(x)?  [#permalink]

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New post 31 Jan 2019, 03:21
MathRevolution wrote:
[GMAT math practice question]

\(f(2x-1) = \frac{( x + 2 )}{( x – 2 )}\). What is \(f(x)\)?

\(A. f(x) = \frac{( x + 5 )}{( x – 3 )}\)
\(B. f(x) = \frac{( x - 5 )}{( x + 3 )}\)
\(C. f(x) = \frac{( x + 5 )}{( x + 3 )}\)
\(D. f(x) = \frac{( x – 5 )}{( x – 3 )}\)
\(E. f(x) = \frac{( x + 3 )}{( x – 3 )}\)


Let x=1.
Plugging x=1 into \(f(2x-1) = \frac{x + 2}{x – 2}\), we get:
\(f(2*1-1) = \frac{1 + 2}{1 – 2}\)
\(f(1) = -3\)

When x=1, the question stem becomes:
What is \(f(1)\)?
Since f(1) = -3, the correct answer must yield -3 when x=1.
Only A works:
\(f(1) = \frac{1 + 5}{1-3}\) = -3


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Re: f(2x-1) = ( x + 2 ) / ( x – 2 ). What is f(x)?  [#permalink]

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New post 31 Jan 2019, 04:39
MathRevolution wrote:
[GMAT math practice question]

\(f(2x-1) = \frac{( x + 2 )}{( x – 2 )}\). What is \(f(x)\)?

\(A. f(x) = \frac{( x + 5 )}{( x – 3 )}\)
\(B. f(x) = \frac{( x - 5 )}{( x + 3 )}\)
\(C. f(x) = \frac{( x + 5 )}{( x + 3 )}\)
\(D. f(x) = \frac{( x – 5 )}{( x – 3 )}\)
\(E. f(x) = \frac{( x + 3 )}{( x – 3 )}\)

\(? = f\left( x \right)\)

\(f\left( {2x - 1} \right) = {{x + 2} \over {x - 2}}\,\,\,\,\,\left( * \right)\)


\(2x - 1 = y\,\,\,\,\, \Rightarrow \,\,\,\,x = {{y + 1} \over 2}\)

\(\left( * \right)\,\,\,\, \Rightarrow \,\,\,\,f\left( y \right)\,\, = \,\,{{{{y + 1} \over 2} + 2} \over {{{y + 1} \over 2} - 2}}\,\, = \,\,{{y + 1 + 4} \over {y + 1 - 4}} = {{y + 5} \over {y - 3}}\)


\(?\,\,\,:\,\,\,f\left( y \right) = {{y + 5} \over {y - 3}}\,\,\,\,\,\,\,\left[ {\,f\left( x \right) = {{x + 5} \over {x - 3}}\,\,\,{\rm{if}}\,\,{\rm{you}}\,\,{\rm{prefer}}!\,} \right]\)


The correct answer is therefore (A).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: f(2x-1) = ( x + 2 ) / ( x – 2 ). What is f(x)?  [#permalink]

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New post 04 Feb 2019, 04:40
=>

We solve this problem using a change of variable. Let \(t = 2x -1.\)
Then \(2x = t + 1\) or \(x = \frac{( t + 1 )}{2.}\)
\(f(2x-1) = f(t) = ( (t+1)/2 + 2 ) / ( (t+1)/2 – 2 ) = ( t + 1 + 4 ) / ( t + 1 – 4 )\)(after multiplying both top and bottom by 2)
\(= \frac{( t + 5 )}{( t – 3 ).}\)
Making the substitution \(t = x\) gives
\(f(x) = \frac{( x + 5 )}{( x – 3 )}.\)

Therefore, the answer is A.
Answer: A
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Re: f(2x-1) = ( x + 2 ) / ( x – 2 ). What is f(x)?  [#permalink]

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New post 06 Feb 2019, 18:34
MathRevolution wrote:
[GMAT math practice question]

\(f(2x-1) = \frac{( x + 2 )}{( x – 2 )}\). What is \(f(x)\)?

\(A. f(x) = \frac{( x + 5 )}{( x – 3 )}\)
\(B. f(x) = \frac{( x - 5 )}{( x + 3 )}\)
\(C. f(x) = \frac{( x + 5 )}{( x + 3 )}\)
\(D. f(x) = \frac{( x – 5 )}{( x – 3 )}\)
\(E. f(x) = \frac{( x + 3 )}{( x – 3 )}\)


Let y = 2x - 1, so x = (y + 1)/2. In other words,

f(y) = [(y + 1)/2 + 2] / [(y + 1)/2 - 2]

f(y) = (y + 1 + 4) / (y + 1 - 4)

f(y) = (y + 5) / (y - 3)

Now we can replace y with x and obtain:

f(x) = (x + 5) / (x - 3)

Alternate Solution:

First, let’s take x = -2 in f(2x - 1) = (x + 2) / (x - 2):

f( 2(-2) - 1) = (-2 + 2) / (-2 - 2)

f( -5) = 0

Looking at the answer choices, we observe that only A and C equal zero when x = -5; therefore we eliminate B, D and E.

Next, let’s take x = 3 in f(2x - 1) = (x + 2) / (x - 2):

f( 2(3) - 1) = (3 + 2) / (3 - 2)

f(5) = 5

In answer choice C, when we take x = 5, we obtain f(5) = 10/8, which is not equal to 5. Therefore, we eliminate C as well. The only remaining answer choice is A. Indeed, for the function in A, if we take x = 5, we obtain f(5) = (5 + 5) / (5 - 3) = 10/2 = 5.

Answer: A
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Re: f(2x-1) = ( x + 2 ) / ( x – 2 ). What is f(x)?   [#permalink] 06 Feb 2019, 18:34
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