Sirajgonj Secondary 2013/8
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
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Fatin Farhan
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For an injective function $$f:R \rightarrow R$$, $$f(x + f(y)) = 2012 + f (x + y) $$ then $$f (2013)=?$$
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Re: Sirajgonj Secondary 2013/8
Let, $\mathcal P(x,y)\Rightarrow f(x+f(y))=2012+f(x+y)$.
Since, the R.H.S. is symmetric, substitute $\mathcal P(y,x)$, and, show that,$f(x+f(y))=f(y+f(x))$.
Now, the function is injective. So, $x+f(y)=y+f(x) \Rightarrow f(x)=x+f(0)$.
Using $f(x)=x+f(0)$, from, $\mathcal P(x,y)$, we get,
$x+f(y)+f(0)=2012+x+y+f(0) \Rightarrow f(y)=2012+y$.
By putting, $y=2013$, $f(2013)=2012+2013=\boxed{4025}. $
Since, the R.H.S. is symmetric, substitute $\mathcal P(y,x)$, and, show that,$f(x+f(y))=f(y+f(x))$.
Now, the function is injective. So, $x+f(y)=y+f(x) \Rightarrow f(x)=x+f(0)$.
Using $f(x)=x+f(0)$, from, $\mathcal P(x,y)$, we get,
$x+f(y)+f(0)=2012+x+y+f(0) \Rightarrow f(y)=2012+y$.
By putting, $y=2013$, $f(2013)=2012+2013=\boxed{4025}. $