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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

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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 [#permalink]

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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

[Reveal] Spoiler:
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, 21:46, edited 3 times in total.

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Re: f(x,y) = x+y, if x+y<1; 0, if x+y=1; xy, if x+y>1. Where x and y are [#permalink]

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New post 23 Jan 2011, 03:54
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.
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Re: f(x,y) = x+y, if x+y<1; 0, if x+y=1; xy, if x+y>1. Where x and y are [#permalink]

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New post 24 Jan 2011, 09:15
to be honest, i dont understand this question. can you tell me which source is it from?

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Re: f(x,y) = x+y, if x+y<1; 0, if x+y=1; xy, if x+y>1. Where x and y are [#permalink]

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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.
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Re: f(x,y) = x+y, if x+y<1; 0, if x+y=1; xy, if x+y>1. Where x and y are [#permalink]

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New post 25 Jan 2011, 08:43
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.

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Re: f(x,y) = x+y, if x+y<1; 0, if x+y=1; xy, if x+y>1. Where x and y are [#permalink]

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Re: f(x,y) = x+y, if x+y<1; 0, if x+y=1; xy, if x+y>1. Where x and y are   [#permalink] 03 Oct 2017, 23:09
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