= Solution
{c}
The unit circle is <path-connected>, since it is the continuous image of the interval $[0,2\pi]$ under $\theta\mapsto(\cos\theta,\sin\theta)$. It is therefore connected, and part (ii) makes $f(C)$ connected. A connected subset of $\mathbb R$ is an interval: omission of a point lying between two of its points would separate it by the two open rays about that point.
Moreover, the circle is compact, so the <continuous image of a compact space> is compact. Hence here the image is \b[a closed bounded interval, possibly a singleton].
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