# calculas

let f : $\mathbb{R}\implies\mathbb{R}$ be a monotonic function for which there exist a,b , c, d $\in$ with a $\neq$ 0 , c $\neq$ 0 such that for all x the following equalities hold :
$\int_{x}^{x+\sqrt{3}}$f(t)dt = ax+b and $\int_{x}^{x+\sqrt{2}}$f(t)dt =cx+d.
Prove that f is a linear function , i.e, f(x)=mx+n for some m,n$\in$$\mathbb{R}$

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1. Solution :(AOPS)
#6
VPM Thursday at 6:27 PM
neelanjan wrote:
let f : $\mathbb{R}\implies\mathbb{R}$ be a monotonic function for which there exist a,b , c, d $\in$ with a $\neq$ 0 , c $\neq$ 0 such that for all x the following equalities hold :
$\int_{x}^{x+\sqrt{3}}$f(t)dt = ax+b and $\int_{x}^{x+\sqrt{2}}$f(t)dt =cx+d.
Prove that f is a linear function , i.e, f(x)=mx+n for some m,n$\in$$\mathbb{R}$

$\int_{x}^{x+\sqrt{3}}f(t)dt=ax+b;\int_{x}^{x+\sqrt{2}}f(t)dt =cx+d$

$\\ \Rightarrow \left ( \int_{x}^{x+\sqrt{3}}f(t)dt \right )’=(ax+b)’ \Leftrightarrow f(x+\sqrt{3})-f(x)=a;f(x+\sqrt{2})-f(x)=c$

$f(x+\sqrt{3})-f(x)=a$ g(x)=a $\Rightarrow g(x)$ is diffrenable$\Rightarrow \exists$diffrenable $(f(x+\sqrt{3})-f(x))$

Diffrenable is lim ;$\lim {(f(x+\sqrt{3})-f(x))} =\lim{f(x+\sqrt{3})}+\lim{f(x)} \Rightarrow \exists f'(x)$

$\\ \Leftrightarrow f'(x+\sqrt{3})=f'(x);f'(x+\sqrt{2})=f'(x)$

$\\ g(x)=f'(x) \Rightarrow g(x+\sqrt{2})=g(x);g(x+\sqrt{3})=g(x) \forall x \in R$

$\\ \sqrt{3};\sqrt{2}$ are period of $g(x) \Rightarrow \sqrt{3}=n\sqrt{2} (n\in Z)$

It is nonsense $\Rightarrow g(x)=const \Rightarrow f'(x)=c \Rightarrow f(x)=mx+n$

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