f(x,y)= {x+y, if x+y < 1 0, if x+y = 1 xy ,if x+y > 1 : GMAT Problem Solving (PS)
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# f(x,y)= {x+y, if x+y < 1 0, if x+y = 1 xy ,if x+y > 1

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f(x,y)= {x+y, if x+y < 1 0, if x+y = 1 xy ,if x+y > 1 [#permalink]

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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

(Sorry I don't have the answer choices for the question above.)

Last edited by MichelleSavina on 25 Jan 2011, 20:46, 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 [#permalink]

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

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24 Jan 2011, 08:15
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 [#permalink]

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24 Jan 2011, 09:15
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 $$x$$.

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 [#permalink]

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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 $$x$$.

Hope it's clear.

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