Use the mirror equation to deduce that:
$(a)$ an object placed between $f$ and $2f$ of a concave mirror produces a real image beyond $2f$.
$(b)$ a convex mirror always produces a virtual image independent of the location of the object.
$(c)$ the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
$(d)$ an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

(N/A) For a concave mirror,the focal length $(f)$ is negative,$f < 0$. When the object is placed on the left side of the mirror,the object distance $(u)$ is negative,$u < 0$. Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we get $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$. Since the object lies between $f$ and $2f$,we have $2f < u < f$ (considering magnitudes). This leads to $\frac{1}{2f} > \frac{1}{u} > \frac{1}{f}$,which implies $\frac{1}{v} < 0$. Thus,$v$ is negative,and the image is real and formed beyond $2f$.
$(b)$ For a convex mirror,$f > 0$ and $u < 0$. From $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$,since $f > 0$ and $u < 0$,$\frac{1}{v}$ is always positive. Thus,$v > 0$,meaning the image is always virtual and formed behind the mirror,regardless of $u$.
$(c)$ For a convex mirror,$f > 0$ and $u < 0$. From $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$,we see $\frac{1}{v} > \frac{1}{f}$,which implies $v < f$. Thus,the image is always located between the pole and the focus. Since magnification $m = -\frac{v}{u}$ and $|v| < |u|$,the image is diminished.
$(d)$ For a concave mirror,$f < 0$. When $0 < |u| < |f|$,then $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ results in $\frac{1}{v} > 0$,so $v > 0$. The image is virtual. Since $|v| > |u|$,the magnification $m = |\frac{v}{u}| > 1$,so the image is enlarged.