Object distance (u) = -4 cm
Radius of curvature (R) = -24 cm
⇒ Focal length (f) = \(\frac{-24}2\) = 12 cm
Let the image distance be 'v' then
∴ The image is formed 6 cm behind the mirror.
Now, magnification (m) = \(-\frac vu\)
Or, m = \(-\frac 6{-4}\) = 1.5
Yes, the image is magnified.