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# M19#14

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07 Feb 2009, 16:57
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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

Source: GMAT Club Tests - hardest GMAT questions
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08 Feb 2009, 11:50
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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01 Feb 2010, 04:48
the answer is surely D.
without a doubt.

(1/x) / (1+1/x) = 1/x+1
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01 Feb 2010, 07:32
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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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01 Feb 2010, 08:04
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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01 Feb 2010, 11:34
f(1/x)=1/(x+1)=(x+1-x)/(x+1)=1 - x/(x+1) = 1- f(x).
D.
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08 Feb 2011, 09:05
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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17 Feb 2011, 03:35
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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09 Feb 2012, 16:46
@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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08 Feb 2013, 02:35
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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08 Feb 2013, 03:29
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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08 Feb 2013, 21:41
2
KUDOS
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

Source: GMAT Club Tests - hardest GMAT questions

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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28 Jan 2014, 00:33
$$f(x) = \frac{x}{x + 1}$$ . 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

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

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

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21 Apr 2014, 04:59
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, 04:59
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# M19#14

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