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Applecross Senior High School

Semester One Examination, 2019

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 answer to the specific question asked and to follow any instructions that are specified 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 (13) questions. Answer all questions. Write your answers in the spaces provided.

Working time: 100 minutes.

Question 9 (5 marks)

In the diagram below, M is the midpoint of QR.

If OP=p, OQ=q and OR=r, express the following in terms of p, q and r.

SolutionPR=r-pSpecific behaviours correct expressionSolutionPR=r-pSpecific behaviours correct expression(a) PR. (1 mark)

SolutionOM=OQ+12 QR
=q+12 (r-q)
=12 q+12 rSpecific behaviours indicates correct method correct expressionSolutionOM=OQ+12 QR
=q+12 (r-q)
=12 q+12 rSpecific behaviours indicates correct method correct expression(b) OM. (2 marks)

SolutionMP=MO+OP=p-12 q-12 rMP=6p-3q-3rSpecific behaviours indicates MP correct expressionSolutionMP=MO+OP=p-12 q-12 rMP=6p-3q-3rSpecific behaviours indicates MP correct expression

(c) 6MP. (2 marks)

Question 10 (8 marks)

Points P, Q and R have coordinates -2, 11, (8, 15) and (17, 3) respectively. Determine

SolutionPQ=8, 15--2, 11
=10, 4Specific behaviours correct vectorSolutionPQ=8, 15--2, 11
=10, 4Specific behaviours correct vector(a) PQ. (1 mark)

SolutionQR=17, 3-8,15=9, -12QR=15Specific behaviours correct vector magnitudeSolutionQR=17, 3-8,15=9, -12QR=15Specific behaviours correct vector magnitude(b) |QR|. (2 marks)

(c) 2PQ-60u, where u is a unit vector in the direction QR. (3 marks)

Solutionu=115(9, -12)2PQ-60u=210, 4-60159, -12
=-16, 56Specific behaviours indicates unit vector expression for result correct vectorSolutionu=115(9, -12)2PQ-60u=210, 4-60159, -12
=-16, 56Specific behaviours indicates unit vector expression for result correct vector

(d) The coordinates of point S, given that RS=QP. (2 marks)

SolutionOS=OR+RS
=OR-PQ
=21, 3-(10, 4)
=(7, -1)Specific behaviours expression for result correct coordinatesSolutionOS=OR+RS
=OR-PQ
=21, 3-(10, 4)
=(7, -1)Specific behaviours expression for result correct coordinates

Question 11 (7 marks)

(a) In the diagram below (not drawn to scale) A,B, C and D lie on a circle and EB and ED are tangents to the circle. If ∠BED=54° and ∠CDB=20°, determine the size of ∠CBD.

(3 marks)

Solution∠BDE=(180-54)÷2=63∠CDE=63-20=43∠CBD=∠CDE=43° (AltSegment)Specific behaviours ∠BDE ∠CDE ∠CBDSolution∠BDE=(180-54)÷2=63∠CDE=63-20=43∠CBD=∠CDE=43° (AltSegment)Specific behaviours ∠BDE ∠CDE ∠CBD

(b) Quadrilateral ABCD is such that CB=CD, ∠BAD=96° and ∠BDC=48°.

(i) Sketch a diagram to show this information. (1 mark)

SolutionSpecific behaviours correct diagramSolutionSpecific behaviours correct diagram

(ii) Show that ABCD is cyclic and hence determine the size of ∠CAD. (3 marks)

Solution∠CBD=∠CDB=48∠BCD=180-2×48=84∠BAD+∠BCD=96+84=180Hence cyclic as opposite angles supplementary.∠CAD=∠CBD=48° (Same arc)Specific behaviours use isosceles triangle for ∠BCD uses supplementary angles for cyclic correct size of ∠CADSolution∠CBD=∠CDB=48∠BCD=180-2×48=84∠BAD+∠BCD=96+84=180Hence cyclic as opposite angles supplementary.∠CAD=∠CBD=48° (Same arc)Specific behaviours use isosceles triangle for ∠BCD uses supplementary angles for cyclic correct size of ∠CAD

Question 12 (8 marks)

(a) Show that the vectors 8, -5 and 2.5, 4 are perpendicular. (2 marks)

Solution8-5∙2.54=20-20=0Hence perpendicular as scalar (dot) product is 0.Specific behaviours uses dot product explains resultSolution8-5∙2.54=20-20=0Hence perpendicular as scalar (dot) product is 0.Specific behaviours uses dot product explains result

(b) Determine, to the nearest degree, the angle between the vectors 3, -2 and -2, -4.

SolutionUsing CAS: θ=82.87≈83°Or: θ=cos-1213×25Specific behaviours indicates method correct angleSolutionUsing CAS: θ=82.87≈83°Or: θ=cos-1213×25Specific behaviours indicates method correct angle (2 marks)

(c) The vectors a, 2a+3 and a+3, -2 are perpendicular, where a is a constant. Determine the value(s) of a and the corresponding pair(s) of vectors. (4 marks)

Solutiona2a+3∙a+3-2=a2+3a-4a-6=0a+2a-3=0⇒a=-2, a=3a=-2⇒-2-1 and1-2a=3⇒39 and6-2Specific behaviours uses dot product to form equation solves equation states one pair of vectors states both pairs of vectorsSolutiona2a+3∙a+3-2=a2+3a-4a-6=0a+2a-3=0⇒a=-2, a=3a=-2⇒-2-1 and1-2a=3⇒39 and6-2Specific behaviours uses dot product to form equation solves equation states one pair of vectors states both pairs of vectors

Question 13 (8 marks)

(a) A, B, C and D lie on a circle with diameter AC (diagram not to scale). Determine the size of ∠BDC when ∠BCA=25°. (2 marks)

Solution∠BAC=90-25=65∠BAD=∠BAC=65°Specific behaviours uses angle in semi-circle for ∠BAC correct valueSolution∠BAC=90-25=65∠BAD=∠BAC=65°Specific behaviours uses angle in semi-circle for ∠BAC correct value

(b) K, L and M lie on a circle (diagram not to scale). Secant KN cuts the circle at L and JN is a tangent to the circle at M. Given that ∠LNM=37° and ∠LMN=48°, determine the size of ∠MKL and the size of ∠KMJ. (3 marks)

Solution∠MKL=∠LMN=48° (Alternate segments)∠KLM=37+48=85 (Exterior angle)∠KMJ=∠KLM=85° (Alternate segments)Specific behaviours ∠MKL ∠KLM ∠KMJSolution∠MKL=∠LMN=48° (Alternate segments)∠KLM=37+48=85 (Exterior angle)∠KMJ=∠KLM=85° (Alternate segments)Specific behaviours ∠MKL ∠KLM ∠KMJ

(c) P, Q and R lie on a circle of radius 85 mm (diagram not to scale) and PQ=PR=116 mm. Determine the size of angle ∠QPR, to the nearest degree. (3 marks)

Solution116÷2=58θ=cos-15885=47.0°∠QPR=2θ=94°Specific behaviours completes diagram uses trig ratio for half-angle correct angleSolution116÷2=58θ=cos-15885=47.0°∠QPR=2θ=94°Specific behaviours completes diagram uses trig ratio for half-angle correct angle

Question 14 (9 marks)

The parts of this question refer to the word AERIFICATION. It has 5 different consonants and 7 vowels, some of which are repeated.

(a) Determine the number of ways that 3 different consonants chosen from the letters of the word can be arranged in a row. (1 mark)

Solution 5P3=60Specific behaviours correct numberSolution 5P3=60Specific behaviours correct number

(b) Determine the number of ways that all the letters of the word can be arranged in a row.

(2 marks)

Solution12!3!×2!=39 916 800Specific behaviours attempts to account for repeated letters correct numberSolution12!3!×2!=39 916 800Specific behaviours attempts to account for repeated letters correct number

(c) Determine the number of ways that all the letters of the word can be arranged in a row if the vowels must all be adjacent. (3 marks)

Solution(5+1)!×7!3!×2!=302 400Specific behaviours counts vowels as single group counts ways to arrange vowels correct numberSolution(5+1)!×7!3!×2!=302 400Specific behaviours counts vowels as single group counts ways to arrange vowels correct number

(d) Determine how many 3 letter permutations (e.g. TFI, IRI, etc) can be made using the letters of the word. (3 marks)

SolutionAll different: 9×8×7=504Two A's and one other: 3×8=24Two I's and one other: 3×8=24Three I's: 1Total: n=504+24+24+1=553Specific behaviours attempts to consider separate cases correct number containing 2 A's and 2 I's correct totalSolutionAll different: 9×8×7=504Two A's and one other: 3×8=24Two I's and one other: 3×8=24Three I's: 1Total: n=504+24+24+1=553Specific behaviours attempts to consider separate cases correct number containing 2 A's and 2 I's correct total

Question 15 (8 marks)

(a) In the diagram below (not drawn to scale) P, Q, R and S lie on the circle with centre O. Determine the size of angles α, β and γ given that ∠PQR=105° and 2β=3γ. (4 marks)

Solutionα=180-105=75°β+γ=105
2β+2γ=210⇒5γ=210⇒γ=42°β=63°Specific behaviours correct α equation for β+γ correct γ correct βSolutionα=180-105=75°β+γ=105
2β+2γ=210⇒5γ=210⇒γ=42°β=63°Specific behaviours correct α equation for β+γ correct γ correct β

SolutionWhen opposite angles in a quadrilateral are supplementary, the quadrilateral is cyclic.Specific behaviours correct statementSolutionWhen opposite angles in a quadrilateral are supplementary, the quadrilateral is cyclic.Specific behaviours correct statement(b) Write the converse of the theorem that states the opposite angles of a cyclic quadrilateral are supplementary. (1 mark)

(c) Prove by contradiction that the converse you wrote in (b) is true. Start by assuming that there is a quadrilateral that does have supplementary opposite angles but is not cyclic, such as STUW shown below. (3 marks)

SolutionFrom assumption, ∠W=180°-∠T.But from regular theorem, ∠V=180°-∠T.Hence ∠W=∠V, but this is impossible (as SW and SV would then be parallel and triangle SVW would not exist). Thus, our original assumption must be wrong, and the converse must be true.Specific behaviours uses assumption develops contradiction explains contradiction and makes deductionSolutionFrom assumption, ∠W=180°-∠T.But from regular theorem, ∠V=180°-∠T.Hence ∠W=∠V, but this is impossible (as SW and SV would then be parallel and triangle SVW would not exist). Thus, our original assumption must be wrong, and the converse must be true.Specific behaviours uses assumption develops contradiction explains contradiction and makes deduction

Question 16 (7 marks)

Three forces a, b and c act on a point in a plane.

The forces are a=-44i+66j N, b=-12i-75j N and c=180i+102j N.

(a) Determine the magnitude of the resultant force and the direction, to the nearest degree, that the resultant makes with the vector i. (3 marks)

Solutionr=-4466+-12-75+180102=12493|r|=155 N∠=36.9≈37°Specific behaviours resultant correct magnitude correct angleSolutionr=-4466+-12-75+180102=12493|r|=155 N∠=36.9≈37°Specific behaviours resultant correct magnitude correct angle

When λa+μb+c=0, the forces are in equilibrium.

(b) Determine the values of the scalar constants λ and μ for equilibrium to occur. (4 marks)

Solutionλ-4466+μ-12-75+180102=0-44λ-12μ+180=066λ-75μ+102=0λ=3, μ=4Specific behaviours equation using i-coefficients equation using j-coefficients solves for λ solves for μSolutionλ-4466+μ-12-75+180102=0-44λ-12μ+180=066λ-75μ+102=0λ=3, μ=4Specific behaviours equation using i-coefficients equation using j-coefficients solves for λ solves for μ

Question 17 (8 marks)

(a) A set of cards is numbered from 100 to 999. Determine the minimum number of cards that must be selected to ensure that at least 3 cards in the selection have the same last digit. Justify your answer using the pigeonhole principle. (3 marks)

SolutionLet pigeonholes be digits 0, 1, 2, …, 9 and pigeons be the last digit of number on card.Then fill all pigeonholes with 2 pigeons, a total of 20 pigeons.The next pigeon will fill one of the pigeonholes with 3 pigeons, and so the minimum number is 21.Specific behaviours defines pigeons and pigeonholes clear explanation correct numberSolutionLet pigeonholes be digits 0, 1, 2, …, 9 and pigeons be the last digit of number on card.Then fill all pigeonholes with 2 pigeons, a total of 20 pigeons.The next pigeon will fill one of the pigeonholes with 3 pigeons, and so the minimum number is 21.Specific behaviours defines pigeons and pigeonholes clear explanation correct number

(b) Eight different books sit on a shelf, one of which has a hardcover and the rest softcovers. A student is told they can take away as many of them as they like but must not leave empty handed. Determine how many different selections can be made

Solution83=56Specific behaviours correct numberSolution83=56Specific behaviours correct number(i) of exactly 3 books. (1 mark)

SolutionChoose either 1, 2, … up to all 8 books:n=188n=28-1=255Specific behaviours uses property of Pascals triangle correct numberSolutionChoose either 1, 2, … up to all 8 books:n=188n=28-1=255Specific behaviours uses property of Pascals triangle correct number(ii) altogether. (2 marks)

(iii) that include the hardcover. (2 marks)

SolutionChoose hardcover and then 0, 1, … up to 7 others:11×n=077n=27=128Specific behaviours indicates method correct numberSolutionChoose hardcover and then 0, 1, … up to 7 others:11×n=077n=27=128Specific behaviours indicates method correct number

Question 18 (8 marks)

Relative to the origin, A and B have position vectors 18i+18j and 21i-15j respectively.

Particle P is initially at A and moves with a constant velocity of 8i-15j ms-1.

(a) Calculate

Solutions=82+(-15)2=17 m/sSpecific behaviours correct speedSolutions=82+(-15)2=17 m/sSpecific behaviours correct speed

(i) the speed of P. (1 mark)

(ii) the position vector of P after 4 seconds. (1 mark)

Solution1818+48-15=50-42Specific behaviours correct positionSolution1818+48-15=50-42Specific behaviours correct position

(iii) the distance of P from B after 4 seconds. (2 marks)

SolutionPB=21-15-50-42=-2927PB=-292+272=1570≈39.6 mSpecific behaviours vector PB correct distanceSolutionPB=21-15-50-42=-2927PB=-292+272=1570≈39.6 mSpecific behaviours vector PB correct distance

(b) Determine how long after leaving A that P is 157 m from B. (4 marks)

SolutionOP=1818+t8-15PB=21-15-18+8t18-15t|PB​2=3-8t2+-33+15t2=1572t=11Specific behaviours expression for OP expression for PB equation using distance correct timeSolutionOP=1818+t8-15PB=21-15-18+8t18-15t|PB​2=3-8t2+-33+15t2=1572t=11Specific behaviours expression for OP expression for PB equation using distance correct time

Question 19 (7 marks)

ABCD is a trapezium with AB parallel and in the same direction to DC.

SolutionSpecific behaviours correct diagramSolutionSpecific behaviours correct diagram(a) Sketch a labelled diagram of ABCD. (1 mark)

(b) Show that AC+DB=AB+DC. (2 marks)

SolutionAC+DB=AB+BC+DA+AB
=AB+DA+AB+BC
=AB+DCSpecific behaviours splits AC and DB groups vectors that make DCSolutionAC+DB=AB+BC+DA+AB
=AB+DA+AB+BC
=AB+DCSpecific behaviours splits AC and DB groups vectors that make DC

(c) M lies on AC and N lies on BD so that AM:MC=BN:ND=2:1. Use a vector method to prove that ABNM is a trapezium. (4 marks)

SolutionAM=23AC, AN=AB+23BDNM=AM-AN
=23AC-AB+23BD
=23AC-BD-AB
=23AC+DB-AB
=23AB+DC-AB [from (b)]But DC=kABNM=23AB+kAB-AB
=2k-13AB⇒ABNM is trapeziumSpecific behaviours vectors for M and N obtains NM without M and N obtains NM in terms of AB, DC obtains NM in terms of ABSolutionAM=23AC, AN=AB+23BDNM=AM-AN
=23AC-AB+23BD
=23AC-BD-AB
=23AC+DB-AB
=23AB+DC-AB [from (b)]But DC=kABNM=23AB+kAB-AB
=2k-13AB⇒ABNM is trapeziumSpecific behaviours vectors for M and N obtains NM without M and N obtains NM in terms of AB, DC obtains NM in terms of AB

Question 20 (7 marks)

Farm A lies 95 km away from farm B on a bearing of 062°. A helicopter leaves farm A at 7:30 am to fly to farm B. The helicopter can maintain a speed of 145 kmh-1 and there is a steady wind of 35 kmh-1 blowing from the north.

Determine the bearing that the helicopter should steer and the time of its arrival at farm B, to the nearest minute.

Solutionsin62145t=sinα35tα=12.3°Bearing: 180+62+12.3=254.3°180-62-12.3=105.7sin62145t=sin105.795t=0.601 h
=36 mArrive at 8:06 amSpecific behaviours diagram showing vectors and resultant equation using sin rule for α value of α correct bearing equation using sin rule for t value of t correct arrival timeSolutionsin62145t=sinα35tα=12.3°Bearing: 180+62+12.3=254.3°180-62-12.3=105.7sin62145t=sin105.795t=0.601 h
=36 mArrive at 8:06 amSpecific behaviours diagram showing vectors and resultant equation using sin rule for α value of α correct bearing equation using sin rule for t value of t correct arrival time

Question 21 (8 marks)

Determine how many of the integers between 1 and 340 inclusive are

Solution340÷6=56
n=56Specific behaviours correct numberSolution340÷6=56
n=56Specific behaviours correct number(a) divisible by 6. (1 mark)

SolutionLCM: 6,7=42;340÷7=48
340÷42=8n=56+48-8=96Specific behaviours number divisible by 42 indicates use of inclusion-exclusion correct numberSolutionLCM: 6,7=42;340÷7=48
340÷42=8n=56+48-8=96Specific behaviours number divisible by 42 indicates use of inclusion-exclusion correct number(b) divisible by 6 or 7. (3 marks)

(c) divisible by 6 or 7 but not both. (1 mark)

Solutionn=96-8=88Specific behaviours correct numberSolutionn=96-8=88Specific behaviours correct number

(d) divisible by 6 or 7 but not 4. (3 marks)

SolutionLCM's: 6, 4=12; 7, 4=28; 4, 6, 7=84340÷12=28
340÷28=12
340÷84=4n=96-28-12+4=60Specific behaviours divisible by 12, 28 divisible by 84 correct numberSolutionLCM's: 6, 4=12; 7, 4=28; 4, 6, 7=84340÷12=28
340÷28=12
340÷84=4n=96-28-12+4=60Specific behaviours divisible by 12, 28 divisible by 84 correct number

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