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Manager  P
Joined: 28 May 2018
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Location: India
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GMAT 1: 640 Q45 V35 GMAT 2: 670 Q45 V37 GMAT 3: 730 Q50 V40 The function f is defined as f(x,y) = xy for positive numbers x and y.  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 39% (02:35) correct 61% (01:49) wrong based on 28 sessions

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The function f is defined as $$f(x,y) = xy$$ for positive numbers $$x$$ and $$y$$. If $$f(x,\frac{1}{x}) = f(x,y)$$, which of the following must be true?

A) $$f(x,x) = f(y,y)$$

B) $$f(x,\frac{1}{x}) = f(x,\frac{1}{y})$$

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

D) $$f(x^2,y^2) = f(y(x+1),\frac{1}{(y+1)})$$

E) $$f(x,\frac{1}{y}) = 1$$

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Senior Manager  V
Joined: 25 Dec 2018
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Concentration: General Management, Finance
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Re: The function f is defined as f(x,y) = xy for positive numbers x and y.  [#permalink]

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PriyankaPalit7 wrote:
The function f is defined as $$f(x,y) = xy$$ for positive numbers $$x$$ and $$y$$. If $$f(x,\frac{1}{x}) = f(x,y)$$, which of the following must be true?

A) $$f(x,x) = f(y,y)$$

B) $$f(x,\frac{1}{x}) = f(x,\frac{1}{y})$$

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

D) $$f(x^2,y^2) = f(y(x+1),\frac{1}{(y+1)})$$

E) $$f(x,\frac{1}{y}) = 1$$

Experts please explain this question.
Intern  B
Joined: 12 Feb 2018
Posts: 12
Re: The function f is defined as f(x,y) = xy for positive numbers x and y.  [#permalink]

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As per option D

x^2*y^2 = (y*(x+1))/(y+1)

-------> Bring the y on RHS to denominator

x^2*y^2 = (x+1)/(1+1/y)

-------> Subtract 1 from both num and den on RHS

x^2*y^2 = x/(1/y)
x^2*y^2 = xy
xy = 1

---> as is the case in the function equivalence mentioned in the question
Intern  B
Joined: 04 Aug 2016
Posts: 8
The function f is defined as f(x,y) = xy for positive numbers x and y.  [#permalink]

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2
As the statement mentions that function f(x,y)=xy
And f(x,1/x)=f(x,y)
Going by the formula, we can write the above eqn as

x*1/x=xy
1=xy
So we just need to find from the below options which must be true.

A. f(X,X)=f(y,y)
Simplifying we get x^2=y^2
As X and y are positive, we get X=y which is not the answer as we xy=1 from the question.

B. f(x,1/x)=f(x,1/y)
X*1/X = x*1/y
1=X/y
Wrong

C. f(x+y,1/x)=f(x+y,1/y)
(X+Y)1/X = (X+y)1/y
1+y/X=X/y +1
Y/X = X/y
X^2 = Y^2
X=y
Wrong

D. f(x2,y2)=f(y(x+1),1/(y+1))
X^2*y^2= y(X+1)/(y+1)
(Xy)^2= y(X+1)/(y+1)
As we know xy=1 and substitute in above eqn
We get y+1=yx+y
Xy=1, substituting xy again, we get 1=1

E. f(x,1/y)=1
X/y=1
Wrong.

So D is the answer

Posted from my mobile device The function f is defined as f(x,y) = xy for positive numbers x and y.   [#permalink] 26 Mar 2019, 20:56
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