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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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VP
Joined: 19 Oct 2018
Posts: 1289
Location: India
A function f(x) is defined as f(x+y)=f(x)+f(y) for all real value of x  [#permalink]

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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
Senior Manager
Joined: 16 Feb 2015
Posts: 252
Location: United States
Concentration: Finance, Operations
Re: A function f(x) is defined as f(x+y)=f(x)+f(y) for all real value of x  [#permalink]

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01 Dec 2019, 23:34
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 give Kudos, If you find my explanation Good Enough
VP
Joined: 19 Oct 2018
Posts: 1289
Location: India
Re: A function f(x) is defined as f(x+y)=f(x)+f(y) for all real value of x  [#permalink]

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02 Dec 2019, 10:40
2
2
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 give Kudos, If you find my explanation Good Enough
Intern
Joined: 16 Dec 2019
Posts: 13
Re: A function f(x) is defined as f(x+y)=f(x)+f(y) for all real value of x  [#permalink]

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21 Dec 2019, 08:03
Can somebody explain this
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Joined: 17 Sep 2014
Posts: 327
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GMAT 1: 780 Q51 V45
GRE 1: Q170 V167
A function f(x) is defined as f(x+y)=f(x)+f(y) for all real value of x  [#permalink]

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21 Dec 2019, 08:55
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] 21 Dec 2019, 08:55
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# A function f(x) is defined as f(x+y)=f(x)+f(y) for all real value of x

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