Use the second derivative to justify that B is a maximum.

Given that \(f'(x) = \frac{3}{2}{x^2} - x + p\) , show that \(p = - 4\) .

Find \(f(x)\) .

substituting into the second derivative     M1

e.g. \(3 \times \left( { - \frac{4}{3}} \right) - 1\)

\(f''\left( { - \frac{4}{3}} \right) = - 5\)     A1

since the second derivative is negative, B is a maximum     R1     N0

[3 marks]

setting \(f'(x)\) equal to zero     (M1)

evidence of substituting \(x = 2\) (or \(x = - \frac{4}{3}\) )     (M1)

e.g. \(f'(2)\)

correct substitution     A1

e.g. \(\frac{3}{2}{(2)^2} - 2 + p\) , \(\frac{3}{2}{\left( { - \frac{4}{3}} \right)^2} - \left( { - \frac{4}{3}} \right) + p\)

correct simplification

e.g. \(6 - 2 + p = 0\) , \(\frac{8}{3} + \frac{4}{3} + p = 0\) , \(4 + p = 0\)     A1

\(p = - 4\)     AG     N0

[4 marks]

evidence of integration     (M1)

\(f(x) = \frac{1}{2}{x^3} - \frac{1}{2}{x^2} - 4x + c\)     A1A1A1

substituting (2, 4) or \(\left( { - \frac{4}{3},\frac{{358}}{{27}}} \right)\) into their expression     (M1)

correct equation     A1

e.g. \(\frac{1}{2} \times {2^3} - \frac{1}{2} \times {2^2} - 4 \times 2 + c = 4\) , \(\frac{1}{2} \times 8 - \frac{1}{2} \times 4 - 4 \times 2 + c = 4\) , \(4 - 2 - 8 + c = 4\)

\(f(x) = \frac{1}{2}{x^3} - \frac{1}{2}{x^2} - 4x + 10\)     A1     N4

[7 marks]

Many candidates were successful with this question. In part (a), some candidates found \(f''\left( { - \frac{4}{3}} \right)\) and were unclear how to conclude, but most demonstrated a good understanding of the second derivative test.

A large percentage of candidates were successful in showing that \(p = - 4\) but there were still some who worked backwards from the answer. Others did not use the given information and worked from the second derivative, integrated, and then realized that p was the constant of integration. Candidates who evaluated the derivative at \(x = 2\) but set the result equal to 4 clearly did not understand the concept being assessed. Few candidates used the point B with fractional coordinates.

Candidates often did well on the first part of (c), knowing to integrate and successfully finding some or all terms. Some had trouble with the fractions or made careless errors with the signs; others did not use the value of \(p = - 4\) and so could not find the third term when integrating. It was very common for candidates to either forget the constant of integration or to leave it in without finding its value.