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M19#14

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M19#14 [#permalink] New post 07 Feb 2009, 15:57
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A
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Question Stats:

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f(x) = \frac{x}{x + 1} . What is f(\frac{1}{x}) in terms of f(x) ?

(A) f(x)
(B) -f(x)
(C) \frac{1}{f(x)}
(D) 1 - f(x)
(E) none of the above

[Reveal] Spoiler: OA
D

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[Reveal] Spoiler: OA
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Re: M19#14 [#permalink] New post 08 Feb 2009, 10:50
topmbaseeker wrote:
f(x) = \frac{x}{x + 1} . What is f(\frac{1}{x}) in terms of f(x) ?

(C) 2008 GMAT Club - m19#14

* f(x)
* -f(x)
* \frac{1}{f(x)}
* 1 - f(x)
* none of the above


f(\frac{1}{x}) = \frac{x}{x + 1}
f(\frac{1}{x}) = f(x)

A.
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Re: M19#14 [#permalink] New post 01 Feb 2010, 03:48
the answer is surely D.
without a doubt.

(1/x) / (1+1/x) = 1/x+1 :twisted: :twisted: :twisted: :twisted:
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Re: M19#14 [#permalink] New post 01 Feb 2010, 06:32
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f(1/x) = (1/x)/[(1+x)/x] = 1/(1+x)

D. 1 - f(x) = (x+1-x)/(x+1) = 1/(1+x)

So, Answer is D.
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Re: M19#14 [#permalink] New post 01 Feb 2010, 07:04
GMAT TIGER, your mistake is that f(1/x) = 1/(1+x) not x/(1+x)

Answer is D as shown by above explanations.
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Re: M19#14 [#permalink] New post 01 Feb 2010, 10:34
f(1/x)=1/(x+1)=(x+1-x)/(x+1)=1 - x/(x+1) = 1- f(x).
D.
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Re: M19#14 [#permalink] New post 08 Feb 2011, 08:05
f(1/x) = (1/x)/(1/x+1)
f(1/x) = (1/x)/((x+1)/x)
f(1/x) = (1/x)(x/(x+1))
f(1/x) = 1/(x+1)

Just by looking at the answer choices, we can immediately eliminate all but D. Solve for D to eliminate E.

1-f(x) = 1 -x/(x+1)
1-f(x) = 1 -x/(x+1)
1-f(x) = x+1-x/(x+1)
1-f(x) = 1/(x+1)
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Re: M19#14 [#permalink] New post 17 Feb 2011, 02:35
I reached till \frac{1}{(1+x)} thats alright.. But how did you guys know you had to subtract this value from 1 to get the answer in terms of f(x)?
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Re: M19#14 [#permalink] New post 09 Feb 2012, 15:46
@aks1985

so f(x)= x/x+1. it is asking what f(1/x) is in terms of f(x).

So to find that you plug 1/x in for x in f(x). the result is,

(1/x)/[(1/x)+1]

since you have two fractions you will want to multiply by a reciprocal, but before that lets simply the denominator.

denominator= [(1/x)+1] you can simplify that by adding them together to create one term but to do that the denominators must be the same, so you end up with [(1/x)+(x/x)].
*note that (x/x)=1

now your equation for f(1/x) looks like (1/x)/[(1+x)/x]

multiply by the reciprocal so (1/x)*(x/1+x), the x in the denominator of 1/x cancels out with the x in the numerator of x/(1+x) and you end up with 1/(1+x)

now you have to find out what that equals in terms of f(x).

A) cant be because it is not the same equation of f(x)
B) This is not the cause because it is not -(x/(x+1))
C) Hold off on this one because a little algebra needs to be down
D)1-f(x)= 1- (x/x+1) which equals ((x+1)/(x+1))-(1/(x+1))---->x/(x+1) which is f(x)


So the answer is D
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Re: M19#14 [#permalink] New post 08 Feb 2013, 01:35
Hi guys.

For Algebra questions, I think the easiest way is plug-in. Clearly see f(1/x) = (1/x) / ((1/x) +1) = 1/(x+1). Don't take time to convert 1/(x+1) to a formula in terms of f(x), just examine each answer.

Ans A: f(x) = x/(x+1) wrong
Ans B: -f(x) = -x/(x+1) Wrong
Ans C: 1/f(x) = (x+1)/x Wrong
Ans D: 1 - f(x) = 1 - x/(x+1) = 1/(x+1) CORRECT
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Re: M19#14 [#permalink] New post 08 Feb 2013, 02:29
f(x) = x / (x+1)

=> f(1/x) = (1/x) / ((1/x) + 1) = 1/(x+1)

now 1-f(1/x) = 1- [1 / (x+1)] = x / (x+1) = f(x)

So, we have f(x) = 1 - f(1/x)

or f(1/x) = 1 - f(x)

Hence D is the answer.
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Re: M19#14 [#permalink] New post 08 Feb 2013, 20:41
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topmbaseeker wrote:
f(x) = \frac{x}{x + 1} . What is f(\frac{1}{x}) in terms of f(x) ?

(A) f(x)
(B) -f(x)
(C) \frac{1}{f(x)}
(D) 1 - f(x)
(E) none of the above

[Reveal] Spoiler: OA
D

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The best way to solve these kind of questions is to assume values...
For example take x=2...with this value of x you will get f(x) as 2/3...
Now find f(1/x) i.e. f(1/2), which will come out to be 1/3....
Now check which option satisfies the above relation...the only option that satisfies is D
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Re: M19#14 [#permalink] New post 27 Jan 2014, 23:33
f(x) = \frac{x}{x + 1} . What is f(\frac{1}{x}) in terms off(x) ?

(A) f(x)
(B) -f(x)
(C) \frac{1}{f(x)}
(D)1 - f(x)
(E) none of the above


It may be noted that the function f(x) = \frac{x}{x + 1} is not defined for x = -1 and x= 0

f(x) + f(\frac{1}{x}) = \frac{x}{(1+x)} +\frac{1}{(1+x)} = \frac{(1+x)}{(1+x)}=1

Answer: (D)
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Re: M19#14 [#permalink] New post 21 Apr 2014, 03:59
f(1/x)=(1/x)/(1+1/x)=1/(x+1)=1-x/x+1=1-f(x)
Re: M19#14   [#permalink] 21 Apr 2014, 03:59
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