Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a continuous function such that for any two real numbers $x$ and $y$, $|f(x)-f(y) \leq 10| x-.\left.y\right|^{201}$, then

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Aicharya · Accepted Answer

Given, $f: \mathbf{R} ightarrow \mathbf{R}$ is continuous $ \begin{aligned} & |f(x)-f(y)| \leq 10|x-y|^{201} \ \Rightarrow & \frac{|f(x)-f(y)|}{|x-y|} \leq 10|x-y|^{200} \end{aligned} $ Put, $y=x+h$ and taking limit $h ightarrow 0$ on both sides $ \begin{aligned} & \lim _{h ightarrow 0} \frac{|f(x)-f(x+h)|}{|x-(x+h)|} \leq \lim _{h ightarrow 0}|x-(x+h)|^{200} \ & \lim _{h ightarrow 0}\left|\frac{f(x+h)-f(x)}{h}ight| \leq 0 \ & \Rightarrow \quad\left|f^{\prime}(x)ight| \leq 0 \Rightarrow f^{\prime}(x)=0 \ & 	herefore \quad f(x)=	ext { constant } \ & 	herefore \quad f(2019)=	ext { constant } \ & f(2021)=f(2022)=	ext { constant } \ & \end{aligned} $ So, $f(2019)+f(2022)=2 f(2021)$ $\{\because$ boths are constant $\}$ $	herefore$ Hence, option (2) is correct

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a continuous function such that for any two real numbers $x$…

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