Step 1: Recall what a convex mirror always does.
A convex mirror always forms a virtual, erect image no matter where the object is kept, because it is a diverging mirror, so the image type will not flip to real at any stage.
Step 2: Test the trend with real numbers.
Take a convex mirror of focal length $f = 15\text{ cm}$, so $f = +15\text{ cm}$. First place the object at $u = -15\text{ cm}$: \[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} + \frac{1}{15} = \frac{2}{15} \implies v = 7.5\text{ cm}, \quad m = -\frac{v}{u} = 0.5 \]
Step 3: Move the object closer and compare.
Now bring the object to $u = -5\text{ cm}$: \[ \frac{1}{v} = \frac{1}{15} + \frac{1}{5} = \frac{4}{15} \implies v = 3.75\text{ cm}, \quad m = -\frac{v}{u} = 0.75 \] Both values of $v$ are positive, meaning the image stays behind the mirror, that is virtual, and the magnification has grown from $0.5$ to $0.75$ as the object came closer.
Step 4: State the trend.
As the object keeps moving towards the pole, the image keeps growing in size while remaining virtual and erect throughout.
\[ \boxed{\text{enlarged and virtual}} \]