Explanation

(b)
Given, $f(x) = x^{3} - 3x^{2} + 2x$
$\Rightarrow f^{'}(x) = 3x^{2} - 6x + 2$
Now, $f$(a)=f(0)=0
And $f(b) = f\left( \frac{1}{2} \right)$
$= \frac{1}{2}\left( \frac{1}{2} - 1 \right)\left( \frac{1}{2} - 2 \right) = \frac{3}{8}$
By Lagrange’s Mean Value Theorem
$\frac{f(b) - f(a)}{b - a} = f^{'}(c)$
$\Rightarrow \frac{\frac{3}{8} - 0}{\frac{1}{2} - 0} = 3c^{2} - 6c + 2$
$\Rightarrow 12c^{2} - 24x + 5 = 0$
This is a quadratic equation in c.
$c = \frac{24 \pm \sqrt{576 - 240}}{24}$
$= 1 \pm \frac{\sqrt{21}}{6}$
But c lies between 0 to$\frac{1}{2}$
$\therefore we\ take,\ c = 1 - \frac{\sqrt{21}}{6}$