You are sitting inside a parked car and observing the things behind the car in rear view mirror. For this situation mark the correct statement(s) 

The rear view mirror of the car is of convex type. From mirror formula

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\begin{array}{l}\Rightarrow -\frac{1}{v^2}\frac{dv}{dt}-\frac{1}{u^2}\frac{du}{dt}=0\\ \frac{dv}{dt}=-\frac{v^2}{u^2}\frac{du}{dt}\end{array}$

$v_i=-\frac{v^2}{u^2}\times v_0\text{ where }\frac{dv}{dt}$ is velocity of image with respect to mirror, as mirror is at rest it can be considered as velocity of image with respect to ground and similarly $\frac{du}{dt}$ is velocity of object with respect to ground. From $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{u-f}{uf}=\frac{v}{u}=\frac{f}{u-f}$

As object (here jogger) approaches mirror (car) the value of u decreases. Here u is negative and f is positive, So $u-f$ would be negative.

As jogger approaches mirror with constant velocity the value of u decreases and hence $u-f$ decreases and hence $\frac{v}{u}$ increases. So $V_1$ increases, hence a is correct $V_1$ would be constant if ${\left(\frac{v}{u}\right)}^2\times v_0$ is constant

which may be possible for situation of option (4).