Magnitude of focal length of a spherical  mirror is $f$ and magnitude of linear magnification is $\frac{1}{2}$

Inverted and real image is formed by concave mirror.

Let $u=-x$, then $v=+\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$$\begin{gathered} \therefore \frac{1}{{\displaystyle \raisebox{1ex}{$+x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+\frac{1}{-x}=\frac{1}{+f} \\ \Rightarrow x=3f \end{gathered}$$

Erect and virtual image is formed by convex mirror.

Let $u=-x$, then $v=-\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$$\begin{gathered} \therefore \frac{1}{{\displaystyle \raisebox{1ex}{$-x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+\frac{1}{-x}=\frac{1}{+f} \\ \Rightarrow x=f \end{gathered}$$