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f(x,y)= {x+y, if x+y < 1 0, if x+y = 1 xy ,if x+y > 1 [#permalink]
23 Jan 2011, 02:04

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f(x,y)={x+y, if x+y<1; 0, if x+y=1; xy, if x+y>1. Where x and y are real numbers. If f(x,1/2)= 3/4 which of the following can be the values of x? I. 1/2 II. 3/4 III. 3/2 IV. 1/4

Answer:- III & IV (Sorry I don't have the answer choices for the question above.)

Please Explain

Last edited by MichelleSavina on 25 Jan 2011, 20:46, edited 3 times in total.

Re: f(x,y)= {x+y, if x+y < 1 0, if x+y = 1 xy ,if x+y > 1 [#permalink]
23 Jan 2011, 02:54

MichelleSavina wrote:

f(x,y)= {x+y, if x+y < 1 0, if x+y = 1 xy ,if x+y > 1. Where x and y are real numbers. If f(x,1/2)= 3/4 which of the following can be the values of x ? I. 1/2 II. 3/4 III. 3/2 IV. ¼

Note that the function f(x, y) has three different expressions for three different ranges of values of of (x + y).

Now, f(x, 1/2) = 3/4 As f(x, 1/2) ≠ 0, (x + 1/2) ≠ 1 => x ≠ 1/2

Let's analyze it for other two regions.

1. (x + 1/2) < 1 .... f(x, 1/2) = (x + 1/2) = 3/4 => x = (3/4 - 1/2) = 1/4 Also (1/4 + 1/2) < 1 as we assumed. Hence 1/4 is a possible value of x

2. (x + 1/2) > 1 .... f(x, 1/2) = x/2 = 3/4 => x = 3/2 Also (3/2 + 1/2) = 2 > 1 as we assumed. Hence 3/2 is a possible value of x

Thus, the correct answer is III and IV only. _________________

Anurag Mairal, Ph.D., MBA GMAT Expert, Admissions and Career Guidance Gurome, Inc. 1-800-566-4043 (USA) +91-99201 32411 (India) http://www.facebook.com/Gurome

Re: f(x,y)= {x+y, if x+y < 1 0, if x+y = 1 xy ,if x+y > 1 [#permalink]
24 Jan 2011, 09:15

Expert's post

diebeatsthegmat wrote:

to be honest, i dont understand this question. can you tell me which source is it from?

f(x,y)={x+y, if x+y<1; 0, if x+y=1; xy, if x+y>1. Where x and y are real numbers. If f(x,1/2)= 3/4 which of the following can be the values of x? I. 1/2 II. 3/4 III. 3/2 IV. ¼

We are given some function f(x,y) and are told that it can have the following 3 values:

1. \(f(x,y)=x+y\) when \(x+y<1\). For example if \(x=y=-1\) then \(x+y=-2<1\) and thus \(f(-1, -1)=-1+(-1)=-2\);

2. \(f(x,y)=0\) when \(x+y=1\). For example if \(x=2\) and \(y=-1\) then \(x+y=1\) and thus \(f(2, -1)=0\);

3. \(f(x,y)=xy\) when \(x+y>1\). For example if \(x=y=1\) then \(x+y=2>1\) and thus \(f(1, 1)=1*1=1\).

Now we are told that \(f(x,\frac{1}{2})=\frac{3}{4}\) and are asked to determine possible values of \(x\) (\(y\) is given as \(\frac{1}{2}\)). We can notice that the value of the function does not equal to zero so we don't have the second case thus \(x+y\neq{1}\) --> \(x+\frac{1}{2}\neq{1}\) --> \(x\neq{\frac{1}{2}}\);

If we have the first case then \(f(x,\frac{1}{2})=x+\frac{1}{2}=\frac{3}{4}\) --> \(x=\frac{1}{4}\), note that \(x+y=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}<1\) so \(\frac{1}{4}\) is a possible value of \(x\);

If we have the third case then \(f(x,\frac{1}{2})=x*\frac{1}{2}=\frac{3}{4}\) --> \(x=\frac{3}{2}\). note that \(x+y=\frac{3}{2}+\frac{1}{2}=2>1\), so \(\frac{3}{2}\) is a possible value of \(x\).

Re: f(x,y)= {x+y, if x+y < 1 0, if x+y = 1 xy ,if x+y > 1 [#permalink]
25 Jan 2011, 07:43

Bunuel wrote:

diebeatsthegmat wrote:

to be honest, i dont understand this question. can you tell me which source is it from?

f(x,y)={x+y, if x+y<1; 0, if x+y=1; xy, if x+y>1. Where x and y are real numbers. If f(x,1/2)= 3/4 which of the following can be the values of x? I. 1/2 II. 3/4 III. 3/2 IV. ¼

We are given some function f(x,y) and are told that it can have the following 3 values:

1. \(f(x,y)=x+y\) when \(x+y<1\). For example if \(x=y=-1\) then \(x+y=-2<1\) and thus \(f(-1, -1)=-1+(-1)=-2\);

2. \(f(x,y)=0\) when \(x+y=1\). For example if \(x=2\) and \(y=-1\) then \(x+y=1\) and thus \(f(2, -1)=0\);

3. \(f(x,y)=xy\) when \(x+y>1\). For example if \(x=y=1\) then \(x+y=2>1\) and thus \(f(1, 1)=1*1=1\).

Now we are told that \(f(x,\frac{1}{2})=\frac{3}{4}\) and are asked to determine possible values of \(x\) (\(y\) is given as \(\frac{1}{2}\)). We can notice that the value of the function does not equal to zero so we don't have the second case thus \(x+y\neq{1}\) --> \(x+\frac{1}{2}\neq{1}\) --> \(x\neq{\frac{1}{2}}\);

If we have the first case then \(f(x,\frac{1}{2})=x+\frac{1}{2}=\frac{3}{4}\) --> \(x=\frac{1}{4}\), note that \(x+y=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}<1\) so \(\frac{1}{4}\) is a possible value of \(x\);

If we have the third case then \(f(x,\frac{1}{2})=x*\frac{1}{2}=\frac{3}{4}\) --> \(x=\frac{3}{2}\). note that \(x+y=\frac{3}{2}+\frac{1}{2}=2>1\), so \(\frac{3}{2}\) is a possible value of \(x\).

Answer: III and IV only.

Hope it's clear.

thank you. it is much clear with exact problem posted correctly.

gmatclubot

Re: f(x,y)= {x+y, if x+y < 1 0, if x+y = 1 xy ,if x+y > 1
[#permalink]
25 Jan 2011, 07:43

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