An object is placed in front of a concave mirror of focal length ‘f’ as shown if figure. Choose the correct shape of the image :  

With concave mirror and u > 2f , V decreases with increase in u.

$\text{ As }{\mathrm{u}}_{\mathrm{ab}}>{\mathrm{u}}_{\mathrm{cd}},{\mathrm{V}}_{\mathrm{ab}}<{\mathrm{V}}_{\mathrm{cd}}$. Hence image (a'b') of ab will be closer than image (c'd') of cd.

magnification $\mathrm{m}=\frac{\mathrm{f}}{\mathrm{f}-\mathrm{u}}=\frac{-\mathrm{f}}{-\mathrm{f}-(-\mathrm{u})}=\frac{\mathrm{f}}{\mathrm{f}-\mathrm{u}}.$
The value of ‘m’ decreases as u increases. Hence size of a'b'  will be less than that of c'd'.
$\text{ As }\mathrm{u}>\mathrm{f},\mathrm{m}\text{ will be -ve. Hence }{\mathrm{a}}^1{\mathrm{b}}^1\text{ and }{\mathrm{c}}^1{\mathrm{d}}^1$ are inverted when compared with ab and cd