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# ISI B.STAT PAPPER 2018 |SUBJECTIVE ## Problem

Let $f$:$\mathbb{R} \rightarrow \mathbb{R}$ be a continous function such that for all$x \in \mathbb{R}$ and all $t\geq 0$

f(x)=f(ktx)
where $k>1$ is a fixed constant

## Case-1

choose any 2 arbitary nos $x,y$ using the functional relationship prove that $f(x)=f(y)$

## Case-2

when $x,y$ are of opposite signs then show that $$f(x)=f(\frac{x}{2})=f(\frac{x}{4})\dots$$
use continuity to show that $f(x)=f(0)$

## Solution

Let us take any $2$ real nos $x$ and $y$.

## Case-1

$x$ and $y$ are of same sign . WLG $0<x<y$

Then$\frac{y}{x}>1$
so there is a no $t\geq 0$ such that
$\frac{y}{x}=k^t$
$f(y)=f(k^tx)=f(x)$ [using$f(x)=f(k^tx)$]

## case-2

$x,y$ are of opposite sign. WLG $x<0<y$
Then $f(x)=f(k^tx)$

$\Rightarrow f(k^tx)=f(k^t2\frac{1}{2}x)$

$\Rightarrow f(k^t2\frac{1}{2}x)=f(k^tk^{log_k2}\frac{x}{2})$

$\Rightarrow f(k^tk^{log_k2}\frac{x}{2})=f(k^{t+log_k2}\frac{x}{2})$

$\Rightarrow f(k^{t+log_k2}\frac{x}{2})=f(\frac{x}{2})$

Using this logic repeatedly we get

$f(x)=f(\frac{x}{2})=f(\frac{x}{4})\dots =f(\frac{x}{2^n})$

Now $\frac{x}{2^n}\rightarrow0$ and $f$ is a continous function hence $\lim_{n\to\infty}f(\frac{x}{2^n})=f(0)$.

[Because we know if $f$ is a continous function and $x_n$ is a sequence that converges to $x$ then $\lim_{n\to\infty}f(x_n)=f(x)$]

using similar logic we can show that $f(y)=f(0)$ so $f(x)=f(y)$ for any $x,y\in \mathbb{R}$

## Problem

Let $f$:$\mathbb{R} \rightarrow \mathbb{R}$ be a continous function such that for all$x \in \mathbb{R}$ and all $t\geq 0$

f(x)=f(ktx)
where $k>1$ is a fixed constant

## Case-1

choose any 2 arbitary nos $x,y$ using the functional relationship prove that $f(x)=f(y)$

## Case-2

when $x,y$ are of opposite signs then show that $$f(x)=f(\frac{x}{2})=f(\frac{x}{4})\dots$$
use continuity to show that $f(x)=f(0)$

## Solution

Let us take any $2$ real nos $x$ and $y$.

## Case-1

$x$ and $y$ are of same sign . WLG $0<x<y$

Then$\frac{y}{x}>1$
so there is a no $t\geq 0$ such that
$\frac{y}{x}=k^t$
$f(y)=f(k^tx)=f(x)$ [using$f(x)=f(k^tx)$]

## case-2

$x,y$ are of opposite sign. WLG $x<0<y$
Then $f(x)=f(k^tx)$

$\Rightarrow f(k^tx)=f(k^t2\frac{1}{2}x)$

$\Rightarrow f(k^t2\frac{1}{2}x)=f(k^tk^{log_k2}\frac{x}{2})$

$\Rightarrow f(k^tk^{log_k2}\frac{x}{2})=f(k^{t+log_k2}\frac{x}{2})$

$\Rightarrow f(k^{t+log_k2}\frac{x}{2})=f(\frac{x}{2})$

Using this logic repeatedly we get

$f(x)=f(\frac{x}{2})=f(\frac{x}{4})\dots =f(\frac{x}{2^n})$

Now $\frac{x}{2^n}\rightarrow0$ and $f$ is a continous function hence $\lim_{n\to\infty}f(\frac{x}{2^n})=f(0)$.

[Because we know if $f$ is a continous function and $x_n$ is a sequence that converges to $x$ then $\lim_{n\to\infty}f(x_n)=f(x)$]

using similar logic we can show that $f(y)=f(0)$ so $f(x)=f(y)$ for any $x,y\in \mathbb{R}$

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