Explanation

(a)
Given, $2a + 3b + 6c = 0$
Let $f^{'}(x) = ax^{2} + bx + c$
$f(x) = \frac{ax^{3}}{3} + \frac{bx^{2}}{2} + cx$
$\Rightarrow f(x) = \frac{2ax^{3} + 3bx^{2} + 6cx + 6d}{6}$
Now, $f(1) = \frac{2a + 3b + 6c}{6} = 0$
And $f(0) = 0$
$\therefore f(0) = f(1)$
There 0 and 1 are the roots of the polynomial $f(x).$So by Rolle’s theorem, there exists at least one root of the polynomial $f^{'}(x)$ lying between 0 and 1.