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Date	November 2018	Marks available	4	Reference code	18N.2.AHL.TZ0.H_11
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Determine and Sketch	Question number	H_11	Adapted from	N/A

Question

Consider the rectangle OABC such that AB = OC = 10 and BC = OA = 1 , with the points P , Q and R placed on the line OC such that OP = $p$ , OQ = $q$ and OR = $r$ , such that 0 < $p$ < $q$ < $r$ < 10.

Let ${\theta _p}$ be the angle APO, ${\theta _q}$ be the angle AQO and ${\theta _r}$ be the angle ARO.

Consider the case when ${\theta _p} = {\theta _q} + {\theta _r}$ and QR = 1.

Find an expression for ${\theta _p}$ in terms of $p$ .

[3]

a.

Show that $p = \frac{{{q^2} + q - 1}}{{2q + 1}}$ .

[6]

b.

By sketching the graph of $p$ as a function of $q$ , determine the range of values of $p$ for which there are possible values of $q$ .

[4]

c.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

use of tan (M1)

${\text{tan}}\,{\theta _p} = \frac{1}{p}$ (A1)

${\theta _p} = {\text{arctan}}\left( {\frac{1}{p}} \right)$ A1

METHOD 2

AP $= \sqrt {{p^2} + 1}$ (A1)

use of sin, cos, sine rule or cosine rule using the correct length of AP (M1)

${\theta _p} = {\text{arcsin}}\left( {\frac{1}{{\sqrt {{p^2} + 1} }}} \right)$ or ${\theta _p} = {\text{arccos}}\left( {\frac{p}{{\sqrt {{p^2} + 1} }}} \right)$ A1

[3 marks]

a.

QR = 1 ⇒ $r = q + 1$ (A1)

Note: This may be seen anywhere.

${\text{tan}}\,{\theta _p} = {\text{tan}}\left( {{\theta _q} + {\theta _r}} \right)$

attempt to use compound angle formula for tan M1

${\text{tan}}\,{\theta _p} = \frac{{{\text{tan}}\,{\theta _q} + {\text{tan}}\,{\theta _r}}}{{1 - {\text{tan}}\,{\theta _q}\,{\text{tan}}\,{\theta _r}}}$ (A1)

$\frac{1}{p} = \frac{{\frac{1}{q} + \frac{1}{r}}}{{1 - \left( {\frac{1}{q}} \right)\left( {\frac{1}{r}} \right)}}$ (M1)

$\frac{1}{p} = \frac{{\frac{1}{q} + \frac{1}{{q + 1}}}}{{1 - \left( {\frac{1}{q}} \right)\left( {\frac{1}{{q + 1}}} \right)}}$ or $p = \frac{{1 - \left( {\frac{1}{q}} \right)\left( {\frac{1}{{q + 1}}} \right)}}{{\left( {\frac{1}{q}} \right) + \left( {\frac{1}{{q + 1}}} \right)}}$ A1

$\frac{1}{p} = \frac{{q + q + 1}}{{q\left( {q + 1} \right) - 1}}$ M1

Note: Award M1 for multiplying top and bottom by ${q\left( {q + 1} \right)}$ .

$p = \frac{{{q^2} + q - 1}}{{2q + 1}}$ AG

[6 marks]

b.

increasing function with positive $q$ -intercept A1

Note: Accept curves which extend beyond the domain shown above.

(0.618 <) $q$ < 9 (A1)

⇒ range is (0 <) $p$ < 4.68 (A1)

0 < $p$ < 4.68 A1

[4 marks]

c.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.2—2d and 3d trig, sine rule, cosine rule, area

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