Explanation

(c)
From mean value theorem $f^{'}(c) = \frac{f(b) - f(a)}{b - a}$
Given, $a = 0\ \ \Rightarrow \ \ f(a) = 0$
and $b = \frac{1}{2}\ \ \ \ \Rightarrow \ \ f(b) = \frac{3}{8}$
Now, $f^{'}(x) = (x - 1)(x - 2) + x(x - 2) + x(x - 1)$
$\therefore\ \ f^{'}(c) = (c - 1)(c - 2) + c(c - 2) + c(c - 1)$
$= c^{2} - 3c + 2 + c^{2} - 2c + c^{2} - c$
$\Rightarrow \ \ f^{'}(c) = 3c^{2} - 6c + 2$
By definition of mean value theorem
$f^{'}(c) = \frac{f(b) - f(a)}{b - a}$
$\Rightarrow \ \ \ 3c^{2} - 6c + 2 = \frac{\left( \frac{3}{8} \right) - 0}{\left( \frac{1}{2} \right) - 0} = \frac{3}{4}$
$\Rightarrow \ \ 3c^{2} - 6c + \frac{5}{4} = 0$
This is a quadratic equation in $c$
$\therefore\ \ \ c = \frac{6 \pm \sqrt{36 - 15}}{2 \times 3} = \frac{6 \pm \sqrt{21}}{6} = 1 \pm \frac{\sqrt{21}}{6}$
Since, $'c'$ lies between $\left\lbrack 0,\frac{1}{2} \right\rbrack$
$\therefore\ \ \ c = 1 - \frac{\sqrt{21}}{6}\ \ \ \left( neglecting\ \ c = 1 + \frac{\sqrt{21}}{6} \right)$