Calculate the value of \(q\).

The vertex of the function is \((3,{\text{ }}27)\).

Find the value of \(p\).

The vertex of the function is \((3,{\text{ }}27)\).

Write down the range of \(f\).

\(0 = p(6)(q - 6)\)      (M1)

\(q = 6\)     (A1)

OR

\(f(x) =  - p{x^2} + pqx\)

\(3 = \frac{{ - pq}}{{ - 2p}}\)     (M1)

\(q = 6\)     (A1)

OR

\(f(x) =  - p{x^2} + pqx\)

\(f'(x) = pq - 2px\)     (M1)

\(pq - 2p(3) = 0\)

\(q = 6\)     (A1)     (C2)

\(27 = p(3)(6 - 3)\)     (M1)

Note: Award (M1) for correct substitution of the vertex \((3,{\text{ }}27)\) and their \(q\) into or equivalent \(f(x) = px(q - x)\) or equivalent.

\(p = 3\)     (A1)(ft)     (C2)

Note: Follow through from part (a).

\(y \leqslant 27\;\;\;\left( {f(x) \leqslant 27} \right)\)     (A1)(A1)     (C2)

Notes: Award (A1) for \(y \leqslant {\text{ }}\left( {{\text{or }}f(x) \leqslant } \right)\), (A1) for \(27\) as part of an inequality.

Accept alternative notation: \(( - \infty ,{\text{ }}27],{\text{ }}] - \infty ,{\text{ }}27]\).

Award (A0)(A1) for \([27,{\text{ }}- \infty )\).

Award (A0)(A0) for \(( - \infty ,{\text{ }}\infty )\).

This question was left unanswered by many candidates. For candidates who attempted the question, one method mark was often awarded for a correct equation resulting from substitution of the point (6, 0). Many were unable to find the value of q, and therefore did not continue to find the value of p.

Many were unable to find the value of q, and therefore did not continue to find the value of p.

Part (c) was poorly attempted, although the range was independent of the values of p and q. The most common error was confusion between domain and range, resulting in an answer of (-∞, ∞).