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]