Explanation

(d) The average speed
$v_{av} = \frac{length\ of\ path\ ACB}{time\ interval\ \left( t_{2} - t_{1} \right)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
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And average velocity,
$\ \ \ \ \ \ \ \ \ \ \mathbf{v}_{av} = \frac{displacement\ }{time\ interval} = \frac{\mathbf{r}_{2} - \mathbf{r}_{1}}{t_{2} - t_{1}}\ \ \ \ldots(ii)$
But we know that distance is always be greater than or equal to magnitude of displacement. So the average speed will always be greater than or equal to the magnitude of average velocity.
From Eqs. (i) and (ii)
$\ \ \ \ \ \ \ \ \ \ \frac{\mathbf{v}_{av}}{v_{av}} = \frac{displacement\ }{length\ of\ path\ (distance)}\ \leq 1$