(a) Calculation of distance of an object and location of image
(b) Reason for virtual image, through convex mirror
(a) Given $\mathit{R}=-20\mathrm{cm}$ , and magnification $\mathit{m}=-2$
Focal length of the mirror $\mathit{f}=\frac{\mathit{R}}{2}=-10\mathrm{cm}$
$\Rightarrow -\frac{1}{10}=\frac{1}{2\mathit{u}}+\frac{1}{\mathit{u}}$
$\Rightarrow \mathit{u}=-15\mathrm{cm}$
$\therefore \mathit{v}=2\times -15\mathrm{cm}=-30\mathrm{cm}$

$-2=-\frac{\mathit{v}}{\mathit{u}}$
$\Rightarrow \mathit{v}=2\mathit{u}$
Using mirror formula
$\frac{1}{\mathit{f}}=\frac{1}{\mathit{v}}+\frac{1}{\mathit{u}}$
(b) $\frac{1}{\mathit{f}}=\frac{1}{\mathit{v}}+\frac{1}{\mathit{u}}$
Using sign convention, for convex mirror, we have

From the formula
$\frac{1}{\mathit{v}}=\frac{1}{\mathit{f}}-\frac{1}{\mathit{u}}$
$\because \mathit{f}$ is positive and $\mathit{u}$ is negative,
$\Rightarrow \mathit{v}$ is always positive, hence image is always virtual.