Passage: Let f(x)= cases ax(x-1), & x 3 cases Given that f(x) is...

For $f(x)$ to be continuous at $x=3$, the left-hand limit and right-hand limit at $x=3$ must be equal.

$\lim_{x \to 3^-} f(x) = 3 - 1 = 2$

$\lim_{x \to 3^+} f(x) = p(3)^2 + q(3) + 2 = 9p + 3q + 2$

Equating them, we get:

$9p + 3q + 2 = 2 \Rightarrow 9p + 3q = 0 \Rightarrow q = -3p$

The derivative of the function for $x \in (1, 3)$ is $f'(x) = 1$, and for $x > 3$ is $f'(x) = 2px + q$.

Since $f'(x)$ is continuous at $x=3$, the left-hand derivative and right-hand derivative at $x=3$ must be equal.

$\lim_{x \to 3^-} f'(x) = 1$

$\lim_{x \to 3^+} f'(x) = 2p(3) + q = 6p + q$

Equating them, we get:

$6p + q = 1$

Substituting $q = -3p$ into the above equation:

$6p - 3p = 1 \Rightarrow 3p = 1 \Rightarrow p = \dfrac{1}{3}$

Answer: $\dfrac{1}{3}$