$(a)\;\frac{-f V_0}{u- f} \\ (b)\;\bigg[\frac{f}{u+ f}\bigg]^2 V_0 \\ (c)\;\bigg[\frac{f}{u- f}\bigg]^2. v_0 \\ (d)\;- \bigg[\frac{f}{u- f}\bigg]^2 V_0 $

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As we know,

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

differentiating with respect to time

we obtain , $-\large\frac{-1}{V^2} \frac{dv}{dt} -\frac{1}{u^2} \frac{du}{dt}$$=0$

or $\large\frac{-1}{V^2} V_{image}- \large\frac{1}{u^2} V_{object}=0$

But from mirror equation $v= \large\frac{uf}{u-f}$

So $V_{image}= - \bigg(\large\frac{V}{u} \bigg)^2 V_{object}$

$\qquad= - \bigg[ \large\frac{f}{u-f}\bigg]^2 V_0$

Hence d is the correct answer.

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