(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.