75th Romanian Mathematical Olympiad

a) The function $f$ being continuous on $[0,1]$, it follows that the function $F:[0,1]\to \mathbb{R}$ defined by $F(x)={\int }_0^{x}f(t)dt$ is differentiable on $[0,1]$, with $F^{\prime }=f$. It follows that the function $\tilde{f}$ is differentiable on $(0,1]$, as a product of differentiable functions. It remains to show that $\tilde{f}$ is continuous at $0$. Since $f$ is continuous, ${\lim }_{x\to 0}f(x)=f(0)$, so that, applying l'H\^opital's rule, we have: $$\underset{x\to 0}{\lim }\tilde{f}(x)=\underset{x\to 0}{\lim }\frac{F(x)}{x}=\underset{x\to 0}{\lim }\frac{F^{\prime }(x)}{x^{\prime }}=\underset{x\to 0}{\lim }f(x)=f(0)=\tilde{f}(0),$$ and it follows that $\tilde{f}$ is continuous at $0$.