WACEhub

Applecross Senior High School

Semester One Examination, 2020

Question/Answer booklet

SOLUTIONSSOLUTIONSMATHEMATICS

SPECIALIST

UNIT 1

Section Two:

Calculator-assumed

In words

Your name

Time allowed for this section

Reading time before commencing work: ten minutes

Working time: one hundred minutes

Materials required/recommended for this section

To be provided by the supervisor

This Question/Answer booklet

Formula sheet (retained from Section One)

To be provided by the candidate

Standard items: pens (blue/black preferred), pencils (including coloured), sharpener,
correction fluid/tape, eraser, ruler, highlighters

Special items: drawing instruments, templates, notes on two unfolded sheets of A4 paper, and up to three calculators approved for use in this examination

Important note to candidates

No other items may be taken into the examination room. It is your responsibility to ensure that you do not have any unauthorised material. If you have any unauthorised material with you, hand it to the supervisor before reading any further.

Structure of this paper

Instructions to candidates

1. The rules for the conduct of examinations are detailed in the school handbook. Sitting this examination implies that you agree to abide by these rules.

2. Write your answers in this Question/Answer booklet preferably using a blue/black pen.
Do not use erasable or gel pens.

3. You must be careful to confine your answers to the specific question asked and to follow any instructions that are specific to a particular question.

4. Show all your working clearly. Your working should be in sufficient detail to allow your answers to be checked readily and for marks to be awarded for reasoning. Incorrect answers given without supporting reasoning cannot be allocated any marks. For any question or part question worth more than two marks, valid working or justification is required to receive full marks. If you repeat any question, ensure that you cancel the answer you do not wish to have marked.

5. It is recommended that you do not use pencil, except in diagrams.

6. Supplementary pages for planning/continuing your answers to questions are provided at the end of this Question/Answer booklet. If you use these pages to continue an answer, indicate at the original answer where the answer is continued, i.e. give the page number.

7. The Formula sheet is not to be handed in with your Question/Answer booklet.

Section Two: Calculator-assumed 65% (98 Marks)

This section has thirteen questions. Answer all questions. Write your answers in the spaces provided.

Working time: 100 minutes.

Question 9 (6 marks)

Determine the size of the angles marked a, b, c, d, e and f shown in the circles below. Where marked, O is the centre of the circle.

Solutiona=90°-52°=38°b=2×52°=104°c=54°d=86°-54°=32°e=32°f=12360°-192°=84°Specific behaviours each correct angleSolutiona=90°-52°=38°b=2×52°=104°c=54°d=86°-54°=32°e=32°f=12360°-192°=84°Specific behaviours each correct angle

Question 10 (5 marks)

Three forces act on an object so that it remains in equilibrium. Two of the forces have magnitudes of 80 N and 110 N and the angle between their directions is 105°. Determine the magnitude of the third force and the angle its direction makes with the smaller force.

SolutionR2=802+1102-280110cos75°
R=118.1 Nsinθ110=sin75°118.1
θ=64.1°Hence angle between directions is 180-64=116°.Specific behaviours diagram showing vector sum is zero uses cosine rule to solve triangle magnitude of resultant uses sine rule direction with smaller forceSolutionR2=802+1102-280110cos75°
R=118.1 Nsinθ110=sin75°118.1
θ=64.1°Hence angle between directions is 180-64=116°.Specific behaviours diagram showing vector sum is zero uses cosine rule to solve triangle magnitude of resultant uses sine rule direction with smaller force

Question 11 (8 marks)

(a) An art gallery plans to display a single painting on each of the three walls in a room. Determine how many arrangements of paintings are possible in the room if they have a selection of 24 different paintings to choose from. (2 marks)

Solution 24P3=24×23×22=12 144Specific behaviours indicates method correct number of arrangementsSolution 24P3=24×23×22=12 144Specific behaviours indicates method correct number of arrangements

(b) In another room, the gallery plan to hang 8 different paintings in a row. If 2 of the paintings are by the artist Marr, determine the number of different arrangements of paintings that are possible when

(i) the paintings by Marr must be at the ends. (2 marks)

Solution2×6!=1 440Specific behaviours uses 6! correct number of arrangementsSolution2×6!=1 440Specific behaviours uses 6! correct number of arrangements

(ii) the paintings by Marr must be next to each other. (2 marks)

Solution2!×7!=10 080Specific behaviours groups Marr together correct number of arrangementsSolution2!×7!=10 080Specific behaviours groups Marr together correct number of arrangements

(iii) the paintings by Marr must be apart and not at the ends. (2 marks)

Solution6 non-Marr leave 5 spaces to hang Marr in between (N_N_N_N_N_N):n=6!×5×4=14 400Specific behaviours indicates method correct number of arrangementsSolution6 non-Marr leave 5 spaces to hang Marr in between (N_N_N_N_N_N):n=6!×5×4=14 400Specific behaviours indicates method correct number of arrangements

Question 12 (8 marks)

(a) Prove that chords of equal length subtend equal angles at the centre of a circle.

(3 marks)

SolutionAB=DC (given)OA=OB=OC=OD=r (all radii)Hence ΔOAB≡ΔOCD (SSS)Hence ∠AOB=∠COD - chords of equal length subtend equal angles at the centre.Specific behaviours establishes congruency of sides establishes congruency of triangles concludes equal anglesSolutionAB=DC (given)OA=OB=OC=OD=r (all radii)Hence ΔOAB≡ΔOCD (SSS)Hence ∠AOB=∠COD - chords of equal length subtend equal angles at the centre.Specific behaviours establishes congruency of sides establishes congruency of triangles concludes equal angles

(b) Points P and Q lie on a circle of radius 23.3 cm so that PQ=21 cm. Determine

(i) the distance of chord PQ from the centre of the circle. (3 marks)

SolutionLet midpoint of chord be M. ThenOM2=r2-PM2
OM=23.32-10.52
=20.8 cmSpecific behaviours uses/defines midpoint or sketch diagram indicates correct method correct distanceSolutionLet midpoint of chord be M. ThenOM2=r2-PM2
OM=23.32-10.52
=20.8 cmSpecific behaviours uses/defines midpoint or sketch diagram indicates correct method correct distance

(ii) the angle subtended by chord PQ at the centre of the circle. (2 marks)

SolutionLet θ=∠POM (half angle required). Thenθ=sin-110.523.3
=26.78°
∠POQ=2θ
≈53.6°Specific behaviours indicates correct method correct angleSolutionLet θ=∠POM (half angle required). Thenθ=sin-110.523.3
=26.78°
∠POQ=2θ
≈53.6°Specific behaviours indicates correct method correct angle

Question 13 (7 marks)

(a) The diagram shows points P, Q, R and S that
lie on the circumference of a circle centre O.
PR is a diameter and the size of ∠QPR=27°.

Determine, with reasons, the size of ∠PSQ. (3 marks)

Solution∠PQR=90° (angle in semicircle)∠PRQ=180°-90°-27°=63° (angle sum in triangle)∠PSQ=∠PRQ=63° (angles on same arc)Specific behaviours uses angle in semicircle uses angle sum in triangle correct size of angle, with reasonSolution∠PQR=90° (angle in semicircle)∠PRQ=180°-90°-27°=63° (angle sum in triangle)∠PSQ=∠PRQ=63° (angles on same arc)Specific behaviours uses angle in semicircle uses angle sum in triangle correct size of angle, with reason

(b) In the diagram shown, A, B, C and D are points
on the circumference of a circle with centre O.

Tangents to the circle at B and D intersect at E.

Determine, with justification, the size of ∠BCD
when ∠BED=72°. (4 marks)

Solution∠OBE=∠ODE=90° (radius-tangent angle)∠BOD=360°-180°-72°=108° (angle sum of quadrilateral BODE)∠BAD=12108°=54° (centre-circumference angles)∠BCD=180°-54°=126° (opposite angles in cyclic quadrilateral)Specific behaviours uses radius-tangent angle correct ∠BOD uses angle at centre-circumference correct angleSolution∠OBE=∠ODE=90° (radius-tangent angle)∠BOD=360°-180°-72°=108° (angle sum of quadrilateral BODE)∠BAD=12108°=54° (centre-circumference angles)∠BCD=180°-54°=126° (opposite angles in cyclic quadrilateral)Specific behaviours uses radius-tangent angle correct ∠BOD uses angle at centre-circumference correct angle

Question 14 (8 marks)

In quadrilateral OPQR shown below, M lies on QR so that QM=3|MR|.

(a) If OP=p, OQ=q and OR=r, express the following in terms of p, q and/or r.

SolutionPR=r-pSpecific behaviours correct expressionSolutionPR=r-pSpecific behaviours correct expression

(i) PR. (1 mark)

SolutionRM=14RQ=14(q-r)Specific behaviours uses correct vector notation correct expressionSolutionRM=14RQ=14(q-r)Specific behaviours uses correct vector notation correct expression

(ii) RM. (2 marks)

SolutionPM=PR+RM
=r-p+14(q-r)
=34r+14q-pSpecific behaviours indicates suitable vector sum correct expressionSolutionPM=PR+RM
=r-p+14(q-r)
=34r+14q-pSpecific behaviours indicates suitable vector sum correct expression(iii) PM. (2 marks)

(b) If O is the origin and points P, Q and R have coordinates -2, 39, (28, -14) and (32, -18) respectively, determine the distance PM. (3 marks)

SolutionPM=3432-18+1428-14--239
=33-56
33-56=65Specific behaviours substitutes into expression for PM PM correct magnitudeSolutionPM=3432-18+1428-14--239
=33-56
33-56=65Specific behaviours substitutes into expression for PM PM correct magnitude

Question 15 (8 marks)

(a) The vertices of quadrilateral ABCD lie on the circumference of a circle centre O shown below. Given that ∠ADC=95° and ∠AOB=84°, determine with reasoning the size of angle BCO. (4 marks)

Solution∠AOC=2×∠ADC=190° (angles at centre-circumference)∠BOC=∠AOC-∠AOB=190°-84°=106° (adjacent angles)∠BCO=12180°-∠BOC (isosceles triangle)∠BCO=12180°-106°=37°Specific behaviours uses angles at centre-circumference uses adjacent angles uses isosceles triangles correct angleSolution∠AOC=2×∠ADC=190° (angles at centre-circumference)∠BOC=∠AOC-∠AOB=190°-84°=106° (adjacent angles)∠BCO=12180°-∠BOC (isosceles triangle)∠BCO=12180°-106°=37°Specific behaviours uses angles at centre-circumference uses adjacent angles uses isosceles triangles correct angle

(b) The vertices of triangle ABC lie on the circumference of a circle. Given that AB=10 cm, AC=7 cm and BC=6 cm, prove by contradiction that AB is not a diameter of the circle.

(4 marks)

SolutionAssume that AB is a diameter of the circle, so that the angle in a semicircle theorem implies that ΔABC must be right angled at C.If ΔABC is right angled, then Pythagoras' theorem implies that AC2+BC2=AB2.But AC2+BC2=72+62=49+36=85 and AB2=102=100.This result contradicts our assumption that AB is a diameter and so AB cannot be a diameter of the circle.Specific behaviours states assumption and uses angle in semicircle theorem uses Pythagoras' theorem to state relationship between side lengths shows relationship is false explains contradictionSolutionAssume that AB is a diameter of the circle, so that the angle in a semicircle theorem implies that ΔABC must be right angled at C.If ΔABC is right angled, then Pythagoras' theorem implies that AC2+BC2=AB2.But AC2+BC2=72+62=49+36=85 and AB2=102=100.This result contradicts our assumption that AB is a diameter and so AB cannot be a diameter of the circle.Specific behaviours states assumption and uses angle in semicircle theorem uses Pythagoras' theorem to state relationship between side lengths shows relationship is false explains contradiction

Question 16 (7 marks)

(a) A calculator can generate random integers between 10 and 25. Use the pigeonhole principle to explain why 49 random integers should be generated to be certain that at least 4 of them are the same. (3 marks)

SolutionThere are 16 pigeonholes (integers from 10 to 25) and each random integer produced is a pigeon.By the pigeonhole principle:If only 48 integers are produced, there will be at least 48÷16=3 pigeons in at least one pigeonhole, but if 49 integers are produced then there will be at least 49÷16=4 pigeons in at least one pigeonhole.Hence 49 integers should be produced to be certain that at least 4 of them are the same.Specific behaviours defines pigeonholes shows 48 insufficient shows 49 sufficientSolutionThere are 16 pigeonholes (integers from 10 to 25) and each random integer produced is a pigeon.By the pigeonhole principle:If only 48 integers are produced, there will be at least 48÷16=3 pigeons in at least one pigeonhole, but if 49 integers are produced then there will be at least 49÷16=4 pigeons in at least one pigeonhole.Hence 49 integers should be produced to be certain that at least 4 of them are the same.Specific behaviours defines pigeonholes shows 48 insufficient shows 49 sufficient

(b) 16 customers bought a total of 130 items from a supermarket. Given that each customer bought at least one item, show that at least two of the customers bought the same number of items. (4 marks)

SolutionAssume that each customer bought a different number of items.Then the minimum number of items bought would be:1+2+3+…+15+16=136But the number of items bought (130) was less than this minimum, which contradicts the assumption made.Hence at least two customers bought the same number of items.Specific behaviours states assumption uses assumption to calculate minimum states contradiction summary statementSolutionAssume that each customer bought a different number of items.Then the minimum number of items bought would be:1+2+3+…+15+16=136But the number of items bought (130) was less than this minimum, which contradicts the assumption made.Hence at least two customers bought the same number of items.Specific behaviours states assumption uses assumption to calculate minimum states contradiction summary statement

Question 17 (9 marks)

(a) Determine the scalar product of

(i) 3.5i+6.5j and 8i-2j. (1 mark)

Solution3.5×8+6.5-2=15Specific behaviours correct valueSolution3.5×8+6.5-2=15Specific behaviours correct value

(ii) two vectors with directions 60° apart that have magnitudes of 15 and 18. (1 mark)

Solution15×18×cos60°=135Specific behaviours correct valueSolution15×18×cos60°=135Specific behaviours correct value

(b) Given that a=3 and b=7 simplify a+b⋅a+b+a⋅(a-2b). (3 marks)

Solutiona⋅a+2a⋅b+b⋅b+a⋅a-2a⋅b
=2a2+b2
=2×32+72=67Specific behaviours expands using scalar products simplifies using magnitudes correct valueSolutiona⋅a+2a⋅b+b⋅b+a⋅a-2a⋅b
=2a2+b2
=2×32+72=67Specific behaviours expands using scalar products simplifies using magnitudes correct value

(c) The position vectors of points P, Q and R are 3-2, -2-1 and -53. Show use of a vector method to determine the size of angle PQR. (4 marks)

SolutionQR=-53--2-1=-34QP=3-2--2-1=5-1cos∠PQR=-34⋅5-15×26=-19526∠PQR=138°Specific behaviours vectors QR and QP shows magnitudes shows scalar product correct angleSolutionQR=-53--2-1=-34QP=3-2--2-1=5-1cos∠PQR=-34⋅5-15×26=-19526∠PQR=138°Specific behaviours vectors QR and QP shows magnitudes shows scalar product correct angle

Question 18 (8 marks)

A school yearbook is produced by a committee of 3 teachers and 8 students. 5 teachers and 17 students have nominated for the committee.

(a) Determine how many different committees could be formed from the nominations.

(2 marks)

Solution53178=10×24310=243 100Specific behaviours chooses teachers and students separately correct numberSolution53178=10×24310=243 100Specific behaviours chooses teachers and students separately correct number

(b) The student nominations include two sets of twins. Determine how many different committees could be chosen that include at least one set of twins. (4 marks)

SolutionChoose students with at least one set of twins (Set A, Set B, Others):n=2222134+2221+2122135+2220+2022136
=715+5148+3432
=9295Ways to choose whole committee: 53×9295=92 950Specific behaviours indicates isolation of cases uses systematic approach correct ways to choose students correct number of committeesSolutionChoose students with at least one set of twins (Set A, Set B, Others):n=2222134+2221+2122135+2220+2022136
=715+5148+3432
=9295Ways to choose whole committee: 53×9295=92 950Specific behaviours indicates isolation of cases uses systematic approach correct ways to choose students correct number of committees

(c) Suppose one of the teachers in the committee will be appointed as treasurer and one of the students will be appointed as secretary. Determine how many different committees can be formed with this structure. (2 marks)

SolutionSelect a teacher and others, select a student and others:5142×171167=30×194480
=5 834 400Specific behaviours indicates correct method correct numberSolutionSelect a teacher and others, select a student and others:5142×171167=30×194480
=5 834 400Specific behaviours indicates correct method correct number

Question 19 (8 marks)

Oil platform T lies 66.5 km away from another oil platform F on a bearing of 215°. A steady current of 4.5 km per hour flows between the platforms on a bearing of 100°. A small boat at F, with a cruising speed of 12 km per hour, needs to arrive at T by 4 pm.

Determine the bearing that the boat should steer and the latest time it should depart from F.

Solution∠T=360°-110°-145°=105°If journey takes t hours, then AF=12t and AT=4.5t.∠F using sine rule:sinF4.5t=sin105°12t⇒sinF=4.5sin105°12Hence ∠F=21.24° and ∠A=180°-20.24°-105°=53.76°.Bearing to steer: 215°+21.24°≈236°Distance AF using sine rule:AFsin105°=66.5sin53.76°⇒AF=79.64t=79.64÷12=6.636 h=6 h 38 mHence steer on bearing 236° and leave before 09:22 am.Specific behaviours sketch diagram angle at T equation using sine rule solves angles in triangle solves for second side in triangle journey time in hours correct time to leave correct bearingSolution∠T=360°-110°-145°=105°If journey takes t hours, then AF=12t and AT=4.5t.∠F using sine rule:sinF4.5t=sin105°12t⇒sinF=4.5sin105°12Hence ∠F=21.24° and ∠A=180°-20.24°-105°=53.76°.Bearing to steer: 215°+21.24°≈236°Distance AF using sine rule:AFsin105°=66.5sin53.76°⇒AF=79.64t=79.64÷12=6.636 h=6 h 38 mHence steer on bearing 236° and leave before 09:22 am.Specific behaviours sketch diagram angle at T equation using sine rule solves angles in triangle solves for second side in triangle journey time in hours correct time to leave correct bearing

Question 20 (8 marks)

Circles C1 and C2 intersect at points P and Q. C1 passes through O, the centre of C2. R lies on C2 so that line segment RS is tangential to C1 at Q. Let ∠PRQ=α.

(a) Sketch a diagram to show the above information. (3 marks)

SolutionSpecific behaviours C1 passing through centre of C2 tangent PQS marks angleSolutionSpecific behaviours C1 passing through centre of C2 tangent PQS marks angle

(b) Determine ∠POQ in terms of α. (1 mark)

Solution∠POQ=2α (angle centre-circumference C2)Specific behaviours correct expressionSolution∠POQ=2α (angle centre-circumference C2)Specific behaviours correct expression

(c) Explain why ∠PQS=2α. (1 mark)

Solution∠PQS=∠POQ (alternate angles C1)Specific behaviours states equal to ∠POQ using circle theoremSolution∠PQS=∠POQ (alternate angles C1)Specific behaviours states equal to ∠POQ using circle theorem

(d) Prove that PQ=QR. (3 marks)

Solution∠RPQ=2α-α=α (exterior - interior sum in ΔRPQ)∠RPQ=α=∠PRQ⇒ΔRPQ is isoscelesHence PQ=QR.Specific behaviours deduces ∠RPQ with reason states ΔRPQ isosceles deduces lengths equalSolution∠RPQ=2α-α=α (exterior - interior sum in ΔRPQ)∠RPQ=α=∠PRQ⇒ΔRPQ is isoscelesHence PQ=QR.Specific behaviours deduces ∠RPQ with reason states ΔRPQ isosceles deduces lengths equal

Question 21 (8 marks)

Particle A, initially at the point with position vector 42i-25j cm, moves with a constant velocity of -8i+15j cm/s. Particle B is stationary at the point with position vector -35i+11j.

(a) Determine the initial distance of A from B. (2 marks)

SolutionAB=-3511-42-25=-7736|AB|=85 cmSpecific behaviours vector AB correct distanceSolutionAB=-3511-42-25=-7736|AB|=85 cmSpecific behaviours vector AB correct distance

(b) Determine an expression for the distance d between A and B after t seconds. (3 marks)

SolutionAB=-3511-42-25+t-815
=8t-7736-15td=8t-772+36-15t2
=17t2-8t+25Specific behaviours position vector for A at time t vector AB distance expression (no need to simplify)SolutionAB=-3511-42-25+t-815
=8t-7736-15td=8t-772+36-15t2
=17t2-8t+25Specific behaviours position vector for A at time t vector AB distance expression (no need to simplify)

(c) Sketch a graph of d against t and hence determine the time that minimises d and state what this minimum distance is. (3 marks)

SolutionMinimum when t=4 sMinimum distance is 51 cmSpecific behaviours sketch graph time minimum distanceSolutionMinimum when t=4 sMinimum distance is 51 cmSpecific behaviours sketch graph time minimum distance

Supplementary page

Question number: _________

Supplementary page

Question number: _________

Supplementary page

Question number: _________

Supplementary page

Question number: _________

© 2020 WA Exam Papers. Applecross Senior High School has a non-exclusive licence to copy and communicate this document for non-commercial, educational use within the school. No other copying, communication or use is permitted without the express written permission of WA Exam Papers. SN002-151-4.