IMO MOCK-5(ii)

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Sazid Akhter Turzo
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IMO MOCK-5(ii)

Unread post by Sazid Akhter Turzo » Wed Oct 17, 2012 7:55 am

Determine with proof all functions $f:\mathbb{R}_{0} \rightarrow \mathbb{R}_{0}$ such that
$4f(x) \geq 3x$ and $f(4f(x)-3x)=x$ $\forall x \in \mathbb{R}_{0}$

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Sazid Akhter Turzo
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Re: IMO MOCK-5(ii)

Unread post by Sazid Akhter Turzo » Wed Oct 17, 2012 10:58 am

here's my solution!
The solution is $f(x)=x$ ; $\forall x \in \mathbb{R}_{0}$
$P(4f(x)-3x) \Rightarrow f(4f(4f(x)-3x)-3(4f(x)-3x))=4f(x)-3x and f(4f(4f(x)-3x)-3(4f(x)-3x)) \geq 0$
$\Rightarrow f(13x-12f(x))=4f(x)-3x and 13x-12f(x) \geq 0 \cdot \cdot \cdot (i)$
now define $g:\mathbb{R}_{0} \rightarrow \mathbb{R}_{0}$ such that $g(x)=13x-12f(x)$
now $(i)$ becomes,
$g(g(x))=17g(x)-16x \cdot \cdot \cdot (ii)$ implying that $g$ is injective....
now it suffices to show that $g(x)=x$ ; $\forall x \in \mathbb{R}_{0}$
clearly $g(x)=x$ ; $\forall x \in \mathbb{R}_{0}$ is a solution to $(ii)$
if it's not the only solution, let $g(x) \neq x$ for some $x \in \mathbb{R}_{0}$.
$Case-1:$
let $\exists r \in \mathbb{R}_{+}$ such that $g(x)=x-r < x$
$(i)$ becomes $g(x-r)=x-17x$ similarly $g(x-17r)=x-(17^{2}-16)r$ and by induction, $g(x-(17^{n}-16(n-1))r)=x-(17^{n+1}-16n)r$
let ${\alpha _{n}}_{n=0}^{\infty}$ be defined by $\alpha _{n}=17^{n+1}-16n$
so $f(x- \alpha _{n}r)=x- \alpha _{n+1}r$
clearly this sequence is strictly increasing
now $\exists n \in \mathbb{N}$ such that $x- \alpha _{n}r \geq 0 > x- \alpha _{n+1}r$
so $13(x- \alpha _{n}r)- 12f(x- \alpha _{n}r)=g(x- \alpha _{n}r)=x- \alpha _{n+1}r < 0$
which contradicts the second part of $(i)$

case-2:
let $\exists r \in \mathbb{R}_{+}$ such that $g(x)=x+r > x$
similar approach implies $f(x+ \alpha _{n}r)=x+ \alpha _{n+1}r$
let $y$ be of the form $x+ \alpha _{n}r$
then $g(y)=y+(\alpha _{n+1}-\alpha _{n})r$
using $g(x)=13x-12f(x)$ ,
$4f(y)-3y= \frac{4\alpha _{n}-\alpha _{n+1}}{3}r+x \cdot \cdot \cdot (iii)$ where $y=x+\alpha _{n}r$
now the LHS of $(iii)$ is nonnegative(by the given condition of the problem) but the RHS is strictly decreasing, and for sufficiently large $n$ , it becomes negative... a contradiction!!!
sp our assumption was wrong!

so $g(x)=x$ ; $\forall x \in \mathbb{R}_{0}$ and we are done!! :P :P

NB: যথেষ্ট পরিমাণ typo থাকতে পারে :(

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Phlembac Adib Hasan
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Re: IMO MOCK-5(ii)

Unread post by Phlembac Adib Hasan » Wed Oct 17, 2012 9:37 pm

We can extend the given equation in the following way: if $f(4kf(x)-(4k-1)x)=kx-(k-1)f(x)$ for some $k>0$, then $f((64k-12)f(x)-x(64k-13))=something\; \; huge>0$.So $(64k-12)f(x)-x(64k-13)\ge 0$.Now after taking limit, we find that $f(x)\ge x$.Now show $f(13x-12f(x))=4f(x)-3x$ and extend it in a similar way.Then taking limit will give $x\ge f(x)$.So we must have $f(x)=x\forall x\in \mathbb {R}_0$.

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Tahmid Hasan
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Re: IMO MOCK-5(ii)

Unread post by Tahmid Hasan » Thu Oct 18, 2012 1:08 pm

I am a little confused about my solution. Please check it.
Let $g(x)=4f(x)-3x[g: \mathbb{R}_{0} \rightarrow \mathbb{R}_{0}]$
$g(g(x))=13x-12f(x)$
So $g(g(x))+3g(x)=4x \forall x \in \mathbb{R}_{0}$.....$(1)$
Let $g(x')>x'$ for some $x' \in \mathbb{R}_{0}$
But then $g(g(x'))+3g(x')>4x'$ which is a contradiction according to $(1)$.
So there is no such $x'$.
Let $g(x")<x"$ for some $x" \in \mathbb{R}_{0}$
But then $g(g(x"))+3g(x")<4x"$ which is a contradiction according to $(1)$.
So there is no such $x"$.
Hence $g(x)=x \forall x \in \mathbb{R}_{0}$
So $4f(x)-3x=x \Rightarrow f(x)=x$ which is indeed a solution.
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Re: IMO MOCK-5(ii)

Unread post by *Mahi* » Thu Oct 18, 2012 2:28 pm

Tahmid Hasan wrote: Let $g(x')>x'$ for some $x' \in \mathbb{R}_{0}$
But then $g(g(x'))+3g(x')>4x'$
Please explain more.
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Tahmid Hasan
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Re: IMO MOCK-5(ii)

Unread post by Tahmid Hasan » Thu Oct 18, 2012 5:29 pm

*Mahi* wrote:
Tahmid Hasan wrote: Let $g(x')>x'$ for some $x' \in \mathbb{R}_{0}$
But then $g(g(x'))+3g(x')>4x'$
Please explain more.
Sorry, my bad. I assumed if $g(x')>x'$for some $x'$, then $g(g(x))>x'$, but that is not necessarily true.
Here's another solution.
$4f(x) \ge 3x$, plugging $x \rightarrow 4f(x)-3x$ twice yields
$f(x) \ge \frac{51}{52}x$, we call this operation double iteration!
Now we prove by induction for all double iteration $f(x) \ge \frac{k}{k+1}x, k \in \mathbb{N}$
The base case is proved above. Assume for the $n$-th double iteration $x,f(x) \ge \frac{k}{k+1}x$....$(1)$
Doing a double iteration on $(1)$, we get $f(x) \ge \frac{16k+3}{16k+4}x$
Taking $K \rightarrow \infty$[sufficiently large] we get $f(x) \ge x$....$(2)$
plugging $x \rightarrow 4f(x)-3x$ in $(2)$ we get $x \le f(x)$.
So $x \le f(x) \le x$, hence $f(x)=x$ which is indeed a solution.
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SANZEED
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Re: IMO MOCK-5(ii)

Unread post by SANZEED » Tue Oct 23, 2012 10:46 pm

I have shown first that if there exists $k\in \mathbb N$ such that $(k+1)x\geq kf(x)$ then $(16k+13)x\geq (16k+12)f(x)$. Again we can show that if there exists $k\in \mathbb N$ such that $(k+1)f(x)\geq kx$ then $(16k+4)f(x)\geq (16k+3)x$. This two helps us to attain the limit $x\leq f(x)\leq x$ so $f(x)=x\forall x\in \mathbb R$.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

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