An object is placed between infinity and the pole (P) of a convex...

Using the mirror formula:

$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

$\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$

For a convex mirror, the focal length $f$ is positive ($f > 0$). The object is placed in front of the mirror, so the object distance $u$ is negative ($u < 0$).

Since $f > 0$ and $u < 0$, the term $-\dfrac{1}{u}$ is positive. Therefore, $\dfrac{1}{v}$ is the sum of two positive quantities, which means $\dfrac{1}{v} > \dfrac{1}{f}$.

This implies that $v < f$. Also, since $\dfrac{1}{v} > 0$, we have $v > 0$.

Thus, $0 < v < f$. The image is formed between the pole (P) and the focus (F), behind the mirror.

Answer: between pole (P) and the focus (F), behind the mirror