Find $f'(x)$ .

There is a minimum value of $f(x)$ when $x = - 2$ . Find the value of $p$ .

$f'(x) = 2x - \frac{p}{{{x^2}}}$     A1A1     N2

Note: Award A1 for $2x$ , A1 for $- \frac{p}{{{x^2}}}$ .

[2 marks]

evidence of equating derivative to 0 (seen anywhere)     (M1)

evidence of finding $f'( - 2)$ (seen anywhere)     (M1)

correct equation     A1

e.g. $- 4 - \frac{p}{4} = 0$ , $- 16 - p = 0$

$p = - 16$     A1     N3

[4 marks]

Candidates did well on (a).

For (b), a great number of candidates substituted into the function instead of into the derivative.

The derivate of ${x^2}$ was calculated without difficulties, but there were numerous problems regarding the derivative of $\frac{p}{x}$ . There were several candidates who considered both p and x as variables; some tried to use the quotient rule and had difficulties, others used negative exponents and were not successful.