Step 1: Note the object position.
Object distance satisfies \(2f < |u| < \infty\), meaning the object is farther than C but nearer than infinity.
Step 2: Apply the mirror equation for a check.
Using \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\), when \(|u|\) lies between \(2f\) and infinity, the image distance \(|v|\) comes out between \(f\) and \(2f\). This confirms the image sits between F and C.
Step 3: Construct the rays.
Take one incident ray parallel to the axis; after reflection it goes through F. Take a second incident ray directed toward F; after reflection it emerges parallel to the axis. The point where the reflected rays cross fixes the image tip.
Step 4: Read off the nature.
Magnification \(m=-v/u\) has magnitude less than one, so the image is diminished; the negative sign shows it is inverted; the rays actually cross, so it is real.
\[\boxed{\text{Real, inverted, diminished image formed between F and C}}\]