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Date	May 2022	Marks available	6	Reference code	22M.1.AHL.TZ2.11
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 2
Command term	State and Sketch	Question number	11	Adapted from	N/A

Question

A function $f$ is defined by $f (x) = \frac{1}{x^{2} - 2 x - 3}$ , where $x \in ℝ, x \neq - 1, x \neq 3$ .

A function $g$ is defined by $g (x) = \frac{1}{x^{2} - 2 x - 3}$ , where $x \in ℝ, x > 3$ .

The inverse of $g$ is $g^{- 1}$ .

A function $h$ is defined by $h (x) = arctan \frac{x}{2}$ , where $x \in ℝ$ .

Sketch the curve $y = f (x)$ , clearly indicating any asymptotes with their equations. State the coordinates of any local maximum or minimum points and any points of intersection with the coordinate axes.

[6]

Show that $g^{- 1} (x) = 1 + \frac{\sqrt{4 x^{2} + x}}{x}$ .

[6]

b.i.

State the domain of $g^{- 1}$ .

[1]

b.ii.

Given that $(h \circ g) (a) = \frac{π}{4}$ , find the value of $a$ .

Give your answer in the form $p + \frac{q}{2} \sqrt{r}$ , where $p, q, r \in ℤ^{+}$ .

[7]

Markscheme

$y$ -intercept $(0, - \frac{1}{3})$ A1

Note: Accept an indication of $- \frac{1}{3}$ on the $y$ -axis.

vertical asymptotes $x = - 1$ and $x = 3$ A1

horizontal asymptote $y = 0$ A1

uses a valid method to find the $x$ -coordinate of the local maximum point (M1)

Note: For example, uses the axis of symmetry or attempts to solve $f' (x) = 0$ .

local maximum point $(1, - \frac{1}{4})$ A1

Note: Award (M1)A0 for a local maximum point at $x = 1$ and coordinates not given.

three correct branches with correct asymptotic behaviour and the key features in approximately correct relative positions to each other A1

[6 marks]

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

Note: Award M1 for interchanging $x$ and $y$ (this can be done at a later stage).

EITHER

attempts to complete the square M1

$y^{2} - 2 y - 3 = {(y - 1)}^{2} - 4$ A1

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

${(y - 1)}^{2} - 4 = \frac{1}{x} ({(y - 1)}^{2} = 4 + \frac{1}{x})$ A1

$y - 1 = \pm \sqrt{4 + \frac{1}{x}} (= \pm \sqrt{\frac{4 x + 1}{x}})$

attempts to solve $x y^{2} - 2 x y - 3 x - 1 = 0$ for $y$ M1

$y = \frac{- (- 2 x) \pm \sqrt{{(- 2 x)}^{2} + 4 x (3 x + 1)}}{2 x}$ A1

Note: Award A1 even if $-$ (in $\pm$ ) is missing

$= \frac{2 x \pm \sqrt{16 x^{2} + 4 x}}{2 x}$ A1

THEN

$= 1 \pm \frac{\sqrt{4 x^{2} + x}}{x}$ A1

$y > 3$ and hence $y = 1 - \frac{\sqrt{4 x^{2} + x}}{x}$ is rejected R1

Note: Award R1 for concluding that the expression for $y$ must have the ‘ $+$ ’ sign.
The R1 may be awarded earlier for using the condition $x > 3$ .

$y = 1 + \frac{\sqrt{4 x^{2} + x}}{x}$

$g^{- 1} (x) = 1 + \frac{\sqrt{4 x^{2} + x}}{x}$ AG

[6 marks]

b.i.

domain of $g^{- 1}$ is $x > 0$ A1

[1 mark]

b.ii.

attempts to find $(h \circ g) (a)$ (M1)

$(h \circ g) (a) = arctan (\frac{g (a)}{2}) ((h \circ g) (a) = arctan (\frac{1}{2 (a^{2} - 2 a - 3)}))$ (A1)

$arctan (\frac{g (a)}{2}) = \frac{π}{4} (arctan (\frac{1}{2 (a^{2} - 2 a - 3)}) = \frac{π}{4})$

attempts to solve for $g (a)$ M1

$\Rightarrow g (a) = 2 (\frac{1}{(a^{2} - 2 a - 3)} = 2)$

EITHER

$\Rightarrow a = g^{- 1} (2)$ A1

attempts to find their $g^{- 1} (2)$ M1

$a = 1 + \frac{\sqrt{4 {(2)}^{2} + 2}}{2}$ A1

Note: Award all available marks to this stage if $x$ is used instead of $a$ .

$\Rightarrow 2 a^{2} - 4 a - 7 = 0$ A1

attempts to solve their quadratic equation M1

$a = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} + 4 (2) (7)}}{4} (= \frac{4 \pm \sqrt{72}}{4})$ A1

Note: Award all available marks to this stage if $x$ is used instead of $a$ .

THEN

$a = 1 + \frac{3}{2} \sqrt{2}$ (as $a > 3$ ) A1

$(p = 1, q = 3, r = 2)$

Note: Award A1 for $a = 1 + \frac{1}{2} \sqrt{18}$ $(p = 1, q = 1, r = 18)$

[7 marks]

Examiners report

Part (a) was generally well done. It was pleasing to see how often candidates presented complete sketches here. Several decided to sketch using the reciprocal function. Occasionally, candidates omitted the upper branches or forgot to calculate the y-coordinate of the maximum.

Part (b): The majority of candidates knew how to start finding the inverse, and those who attempted completing the square or using the quadratic formula to solve for y made good progress (both methods equally seen). Otherwise, they got lost in the algebra. Very few explicitly justified the rejection of the negative root.

Part (c) was well done in general, with some algebraic errors seen in occasions.

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b.i.

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b.ii.

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