Explanation

(d)
$\because\ \ \ y = x^{3} - 6x^{2} + 9x + 4$
Now, $\frac{dy}{dx} = 3x^{2} - 12x + 9$
Let $u = \frac{dy}{dx} = 3x^{2} - 12x + 9$
Now, $\frac{du}{dx} = 6x - 12$
Put $\frac{du}{dx} = 0$ for maximum or minimum
$\therefore\ \ 6x - 12 = 0$
$\Rightarrow \ \ x = 2$
Now, at $x = 0,\ \ \ u = 9$
at $x = 2,\ \ u = - 3$
and at $x = 5,\ \ u = 24$
Thus, the maximum of $u(x),\ 0 \leq x \leq 5$ is $u(5)$
Hence, $x = 5$