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A function f(x) is defined as f(x+y)=f(x)+f(y) for all real value of x

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nick1816
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A function f(x) is defined as f(x+y)=f(x)+f(y) for all real value of x [#permalink] New post   01 Dec 2019, 20:55
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A function f(x) is defined as \(f(x+y)=f(x)+f(y)\) for all real value of x and y, and \(f(1)=3\). What is the value of\( f(\frac{2}{3})\)?

A. 0
B. \(\frac{2}{3}\)
C. 1
D. \(\frac{4}{3}\)
E. 2
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E
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nick1816
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Re: A function f(x) is defined as f(x+y)=f(x)+f(y) for all real value of x [#permalink] New post   02 Dec 2019, 10:40
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rajatchopra1994 Though you got correct answer, your approach is not correct.

\(f(\frac{1}{3}+\frac{1}{3})= f(\frac{1}{3})+f(\frac{1}{3})\)

\(f(\frac{2}{3})= 2f(\frac{1}{3})\)......(1)

also, \(f(\frac{2}{3}+\frac{1}{3})= f(\frac{2}{3})+f(\frac{1}{3})\)

\(f(1)= 2f(\frac{1}{3})+f(\frac{1}{3})\)

\(f(1)= 3f(\frac{1}{3})\)

\(f(\frac{1}{3})= \frac{3}{3}=1\)

and \( f(\frac{2}{3})= 2*1=2\)




rajatchopra1994 wrote:
nick1816 wrote:
A function f(x) is defined as \(f(x+y)=f(x)+f(y)\) for all real value of x and y, and \(f(1)=3\). What is the value of\( f(\frac{2}{3})\)?

A. 0
B. \(\frac{2}{3}\)
C. 1
D. \(\frac{4}{3}\)
E. 2


IMO-E

I am not sure. As Given F(1) = 3
We need for F(2/3), By Ratio Proportionate, its is 66.6%
So , 66.6% of 3 is 2.

:please Please give Kudos, If you find my explanation Good Enough :please
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Re: A function f(x) is defined as f(x+y)=f(x)+f(y) for all real value of x [#permalink] New post   01 Dec 2019, 23:34
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nick1816 wrote:
A function f(x) is defined as \(f(x+y)=f(x)+f(y)\) for all real value of x and y, and \(f(1)=3\). What is the value of\( f(\frac{2}{3})\)?

A. 0
B. \(\frac{2}{3}\)
C. 1
D. \(\frac{4}{3}\)
E. 2


IMO-E

I am not sure. As Given F(1) = 3
We need for F(2/3), By Ratio Proportionate, its is 66.6%
So , 66.6% of 3 is 2.

:please Please give Kudos, If you find my explanation Good Enough :please
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allkagupta
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Re: A function f(x) is defined as f(x+y)=f(x)+f(y) for all real value of x [#permalink] New post   21 Dec 2019, 08:03
Can somebody explain this
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A function f(x) is defined as f(x+y)=f(x)+f(y) for all real value of x [#permalink] New post   21 Dec 2019, 08:55
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nick1816 wrote:
A function f(x) is defined as \(f(x+y)=f(x)+f(y)\) for all real value of x and y, and \(f(1)=3\). What is the value of\( f(\frac{2}{3})\)?

A. 0
B. \(\frac{2}{3}\)
C. 1
D. \(\frac{4}{3}\)
E. 2


We don't know any of the function values besides f(1) = 3 so we have to start from there. The target is to get the input value of \(\frac{2}{3}\) instead of 1. We can observe \(f(x+x) = f(x) + f(x)\), or \(f(2x) = 2f(x)\). This can be expanded to \(f(3x) = 3f(x)\) (from \(f(x+y)=f(x)+f(y)\) let y = 2x and plug in \(f(2x) = 2f(x)\)).

Therefore we can start from x = 1/3 and plug in \(f(3x) = 3f(x)\) to get \(f(3 * \frac{1}{3}) = 3f(\frac{1}{3}) = 3\).

\(f(\frac{1}{3}) = 1\)

\(f(\frac{2}{3}) = 2f(\frac{1}{3}) = 2\)

Ans: E
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A function f(x) is defined as f(x+y)=f(x)+f(y) for all real value of x [#permalink] New post   11 Dec 2021, 11:53
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Given that \(f(x+y)=f(x)+f(y)\) and \(f(1)=3\) and we need to find the value of\( f(\frac{2}{3})\)

\(\frac{2}{3}\) = \(\frac{1}{3}\) + \(\frac{1}{3}\)
So, f(\(\frac{1}{3}\) + \(\frac{1}{3}\)) = f(\(\frac{1}{3}\)) + f(\(\frac{1}{3}\))
=> f(\(\frac{2}{3}\)) = 2*f(\(\frac{1}{3}\))
=> f(\(\frac{1}{3}\)) = \(\frac{1}{2}\) * f(\(\frac{2}{3}\))

Now, 1 = \(\frac{1}{3}\) + \(\frac{2}{3}\)
=> f(1) = f(\(\frac{1}{3}\) + \(\frac{2}{3}\)) = f(\(\frac{1}{3}\)) + f(\(\frac{2}{3}\)) = \(\frac{1}{2}\) * f(\(\frac{2}{3}\)) + f(\(\frac{2}{3}\))
= \(\frac{3}{2}\) * f(\(\frac{2}{3}\))

And f(1) = 3 (given)
=> \(\frac{3}{2}\) * f(\(\frac{2}{3}\)) = 3
=> f(\(\frac{2}{3}\)) = 2

So, Answer will be E
Hope it helps!

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Re: A function f(x) is defined as f(x+y)=f(x)+f(y) for all real value of x [#permalink] New post   20 Dec 2021, 03:59
nick1816, is it allowed to use ratio proportion method for functions ?
2. How did you get f(1/3)=3/3=1 (2nd last step)
Is it because,
f(1)=3f(1/3)
3 = 3f(1/3) (as qs states f(1) = 3)
3/3 = f(1/3)
1= f(1/3)
If there's any other reason please correct me, thanks
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Re: A function f(x) is defined as f(x+y)=f(x)+f(y) for all real value of x [#permalink] New post   09 Apr 2024, 07:23
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One example of this functional equation is y=ax; a=1/3, f(2/3)=2
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Re: A function f(x) is defined as f(x+y)=f(x)+f(y) for all real value of x [#permalink]
09 Apr 2024, 07:23
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