IMO MOCK-5(ii)
- Sazid Akhter Turzo
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Determine with proof all functions $f:\mathbb{R}_{0} \rightarrow \mathbb{R}_{0}$ such that
$4f(x) \geq 3x$ and $f(4f(x)-3x)=x$ $\forall x \in \mathbb{R}_{0}$
$4f(x) \geq 3x$ and $f(4f(x)-3x)=x$ $\forall x \in \mathbb{R}_{0}$
- Sazid Akhter Turzo
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Re: IMO MOCK-5(ii)
here's my solution!
- Phlembac Adib Hasan
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Re: IMO MOCK-5(ii)
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 $f(x)=x\forall x\in \mathbb {R}_0$.
- Tahmid Hasan
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Re: IMO MOCK-5(ii)
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.
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.
বড় ভালবাসি তোমায়,মা
Re: IMO MOCK-5(ii)
Please explain more.Tahmid Hasan wrote: Let $g(x')>x'$ for some $x' \in \mathbb{R}_{0}$
But then $g(g(x'))+3g(x')>4x'$
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- Tahmid Hasan
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Re: IMO MOCK-5(ii)
Sorry, my bad. I assumed if $g(x')>x'$for some $x'$, then $g(g(x))>x'$, but that is not necessarily true.*Mahi* wrote:Please explain more.Tahmid Hasan wrote: Let $g(x')>x'$ for some $x' \in \mathbb{R}_{0}$
But then $g(g(x'))+3g(x')>4x'$
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.
বড় ভালবাসি তোমায়,মা
Re: IMO MOCK-5(ii)
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 $f(x)=x\forall x\in \mathbb R$.
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