We have a function
$f(x)=(x-1)(x-2)$
$=x^{2}-3 x+2$ defined on $\left[0, \frac{1}{2}\right]$
According to LMV theorem,
$f'(c) =\frac{f(b)-f(a)}{b-a}=\frac{f\left(\frac{1}{2}\right)-f(0)}{\frac{1}{2}-0}$
$\Rightarrow 2 c-3 =\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)-(-1)(-2)}{\frac{1}{2}}$
$\Rightarrow 2 c-3 =2\left(\frac{3}{4}-2\right)$
$\Rightarrow 2 c-3 =-\frac{5}{2}$
$\Rightarrow 4 c-6=-5$
$\Rightarrow c =\frac{1}{4} \in\left(0, \frac{1}{2}\right)$
Thus, required value of $c$ is $\frac{1}{4}$.