IMO longlist 1989 [reformed]

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sourav das
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IMO longlist 1989 [reformed]

Unread post by sourav das » Thu Aug 04, 2011 11:35 am

$0<a<1\ constant,\ f:R\mapsto R, \ f(0)=0,\ f(1)=1,\ f(\frac{x+y}{2})=(1-a)f(x) +$ $af(y) $ for $x\leq y$. if $f(x)$ is not equal$\ 0$ for all $x$, prove that $a = \frac{1}{2}$
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Re: IMO longlist 1989 [reformed]

Unread post by *Mahi* » Thu Aug 25, 2011 10:05 pm

Let $m=2n$ be positive real numbers.
Now,replacing $y$ with $m$ and $x$ with $0$ in the main equation,
And replacing $x$ with $-m$ and $y$ with $0$,
Adding $i$ and $ii$
But again,replacing $x$ with $-m$ and $y$ with $m$,
So,$-f(x)=f(-x)$ for all $x \in \mathbb{R}$.
And now we just put this in $(iii)$,
Or, $af(x)+af(x)=f(x)$
So, $a= \frac 1 2$
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