(i)     Show that the gradient of [PQ] is \(\frac{{{a^3}}}{{a - \frac{2}{3}}}\) .

(ii)    Find \(f'(a)\) .

(iii)   Hence show that \(a = 1\) .

Given that the area of T is \(2k + 4\) , show that k satisfies the equation \({k^4} - 6{k^2} + 8 = 0\) .

(i) substitute into gradient \( = \frac{{{y_1} - {y_2}}}{{{x_1} - {x_2}}}\)     (M1)

e.g. \(\frac{{f(a) - 0}}{{a - \frac{2}{3}}}\)

substituting \(f(a) = {a^3}\)

e.g. \(\frac{{{a^3} - 0}}{{a - \frac{2}{3}}}\)     A1

gradient \(\frac{{{a^3}}}{{a - \frac{2}{3}}}\)     AG     N0

(ii) correct answer     A1     N1

e.g. \(3{a^2}\) , \(f'(a) = 3\) , \(f'(a) = \frac{{{a^3}}}{{a - \frac{2}{3}}}\)

(iii) METHOD 1

evidence of approach     (M1)

e.g. \(f'(a) = {\rm{gradient}}\) , \(3{a^2} = \frac{{{a^3}}}{{a - \frac{2}{3}}}\)

simplify     A1

e.g. \(3{a^2}\left( {a - \frac{2}{3}} \right) = {a^3}\)

rearrange     A1

e.g. \(3{a^3} - 2{a^2} = {a^3}\)

evidence of solving     A1

e.g. \(2{a^3} - 2{a^2} = 2{a^2}(a - 1) = 0\)

\(a = 1\)     AG     N0

METHOD 2

gradient RQ \( = \frac{{ - 8}}{{ - 2 - \frac{2}{3}}}\)     A1

simplify     A1

e.g. \(\frac{{ - 8}}{{ - \frac{8}{3}}},3\)

evidence of approach     (M1)

e.g. \(f'(a) = {\rm{gradient}}\) , \(3{a^2} = \frac{{ - 8}}{{ - 2 - \frac{2}{3}}}\) , \(\frac{{{a^3}}}{{a - \frac{2}{3}}} = 3\)

simplify     A1

e.g. \(3{a^2} = 3\) , \({a^2} = 1\)

\(a = 1\)     AG     N0

[7 marks]

approach to find area of T involving subtraction and integrals    (M1)

e.g. \(\int {f - (3x - 2){\rm{d}}x} \) , \(\int_{ - 2}^k {(3x - 2) - \int_{ - 2}^k {{x^3}} } \) , \(\int {({x^3} - 3x + 2)} \)

correct integration with correct signs     A1A1A1

e.g. \(\frac{1}{4}{x^4} - \frac{3}{2}{x^2} + 2x\) , \(\frac{3}{2}{x^2} - 2x - \frac{1}{4}{x^4}\)

correct limits \( - 2\) and k (seen anywhere)     A1

e.g. \(\int_{ - 2}^k {({x^3} - 3x + 2){\rm{d}}x} \) , \(\left[ {\frac{1}{4}{x^4} - \frac{3}{2}{x^2} + 2x} \right]_{ - 2}^k\)

attempt to substitute k and \( - 2\)     (M1)

correct substitution into their integral if 2 or more terms     A1

e.g. \(\left( {\frac{1}{4}{k^4} - \frac{3}{2}{k^2} + 2k} \right) - (4 - 6 - 4)\)

setting their integral expression equal to \(2k + 4\) (seen anywhere)     (M1)

simplifying     A1

e.g. \(\frac{1}{4}{k^4} - \frac{3}{2}{k^2} + 2 = 0\)

\({k^4} - 6{k^2} + 8 = 0\)     AG     N0

[9 marks]

Part (a) seemed to be well-understood by many candidates, and most were able to earn at least partial marks here. Part (ai) was a "show that" question, and unfortunately there were some candidates who did not show how they arrived at the given expression.

In part (b), the concept seemed to be well-understood. Most candidates saw the necessity of using definite integrals and subtracting the two functions, and the integration was generally done correctly. However, there were a number of algebraic and arithmetic errors which prevented candidates from correctly showing the desired final result.