 Solution

A suitable continuous function isFor $t_k=\pi /2+2\pi k$ and $u_k=3\pi /2+2\pi k$,All these arguments lie in $(0,1]$. Its image is a real interval by the continuous image of a real interval result, and it is unbounded both above and below. Given any real $y$, choose two of the displayed points with function values on opposite sides of $y$, then apply the intermediate value theorem on the segment joining them. Hence the image is all of $\mathbb{R}$. A monotone example would not suffice here; the oscillation gives both directions of unboundedness near the same missing endpoint.