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Date	November 2020	Marks available	3	Reference code	20N.1.AHL.TZ0.H_10
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	State	Question number	H_10	Adapted from	N/A

Question

Consider the function $f (x) = a x^{3} + b x^{2} + c x + d$ , where $x \in ℝ$ and $a, b, c, d \in ℝ$ .

Consider the function $g (x) = \frac{1}{2} x^{3} - 3 x^{2} + 6 x - 8$ , where $x \in ℝ$ .

The graph of $y = g (x)$ may be obtained by transforming the graph of $y = x^{3}$ using a sequence of three transformations.

Write down an expression for $f' (x)$ .

[1]

a.i.

Hence, given that $f^{- 1}$ does not exist, show that $b^{2} - 3 a c > 0$ .

[3]

a.ii.

Show that $g^{- 1}$ exists.

[2]

b.i.

$g (x)$ can be written in the form $p (x - 2)^{3} + q$ , where $p, q \in ℝ$ .

Find the value of $p$ and the value of $q$ .

[3]

b.ii.

Hence find $g^{- 1} (x)$ .

[3]

b.iii.

State each of the transformations in the order in which they are applied.

[3]

Sketch the graphs of $y = g (x)$ and $y = g^{- 1} (x)$ on the same set of axes, indicating the points where each graph crosses the coordinate axes.

[5]

Markscheme

$f' (x) = 3 a x^{2} + 2 b x + c$ A1

[1 mark]

a.i.

since $f^{- 1}$ does not exist, there must be two turning points R1

( $\Rightarrow f' (x) = 0$ has more than one solution)

using the discriminant $Δ > 0$ M1

$4 b^{2} - 12 a c > 0$ A1

$b^{2} - 3 a c > 0$ AG

[4 marks]

a.ii.

METHOD 1

$b^{2} - 3 a c = {(- 3)}^{2} - 3 \times \frac{1}{2} \times 6$ M1

$= 9 - 9$

$= 0$ A1

hence $g^{- 1}$ exists AG

METHOD 2

$g' (x) = \frac{3}{2} x^{2} - 6 x + 6$ M1

$Δ = {(- 6)}^{2} - 4 \times \frac{3}{2} \times 6$

$Δ = 36 - 36 = 0 \Rightarrow$ there is (only) one point with gradient of $0$ and this must be a point of inflexion (since $g (x)$ is a cubic.) R1

hence $g^{- 1}$ exists AG

[2 marks]

b.i.

$p = \frac{1}{2}$ A1

${(x - 2)}^{3} = x^{3} - 6 x^{2} + 12 x - 8$ (M1)

$\frac{1}{2} (x^{3} - 6 x^{2} + 12 x - 8) = \frac{1}{2} x^{3} - 3 x^{2} + 6 x - 4$

$g (x) = \frac{1}{2} {(x - 2)}^{3} - 4 \Rightarrow q = - 4$ A1

[3 marks]

b.ii.

$x = \frac{1}{2} {(y - 2)}^{3} - 4$ (M1)

Note: Interchanging $x$ and $y$ can be done at any stage.

$2 (x + 4) = {(y - 2)}^{3}$ (M1)

$\sqrt[3]{2 (x + 4)} = y - 2$

$y = \sqrt[3]{2 (x + 4)} + 2$

$g^{- 1} (x) = \sqrt[3]{2 (x + 4)} + 2$ A1

Note: $g^{- 1} (x) = \dots$ must be seen for the final A mark.

[3 marks]

b.iii.

translation through $(\begin{matrix} 2 \\ 0 \end{matrix})$ , A1

Note: This can be seen anywhere.

EITHER
a stretch scale factor $\frac{1}{2}$ parallel to the $y$ -axis then a translation through $(\begin{matrix} 0 \\ - 4 \end{matrix})$ A2
OR
a translation through $(\begin{matrix} 0 \\ - 8 \end{matrix})$ then a stretch scale factor $\frac{1}{2}$ parallel to the $y$ -axis A2

Note: Accept ‘shift’ for translation, but do not accept ‘move’. Accept ‘scaling’ for ‘stretch’.

[3 marks]

A1A1A1M1A1

Note: Award A1 for correct ‘shape’ of $g$ (allow non-stationary point of inflexion)
Award A1 for each correct intercept of $g$
Award M1 for attempt to reflect their graph in $y = x$ , A1 for completely correct $g^{- 1}$ including intercepts