Find \(f'(x)\).

Given that \(f'(x) \geqslant 0\), show that \({p^2} \leqslant 3pq\).

\(f'(x) = 3p{x^2} + 2px + q\)     A2     N2

Note:     Award A1 if only 1 error.

[2 marks]

evidence of discriminant (must be seen explicitly, not in quadratic formula)     (M1)

eg     \({b^2} - 4ac\)

correct substitution into discriminant (may be seen in inequality)     A1

eg     \({(2p)^2} - 4 \times 3p \times q,{\text{ }}4{p^2} - 12pq\)

\(f'(x) \geqslant 0\) then \(f'\) has two equal roots or no roots     (R1)

recognizing discriminant less or equal than zero     R1

eg     \(\Delta  \leqslant 0,{\text{ }}4{p^2} - 12pq \leqslant 0\)

correct working that clearly leads to the required answer     A1

eg     \({p^2} - 3pq \leqslant 0,{\text{ }}4{p^2} \leqslant 12pq\)

\({p^2} \leqslant 3pq\)     AG     N0

[5 marks]