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Semester One Examination, 2017

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.

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

4. Additional working space pages at the end of this Question/Answer booklet are for planning or continuing an answer. If you use these pages, indicate at the original answer, the page number it is planned/continued on and write the question number being planned/continued on the additional working space page.

5. 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.

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

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

Section Two: Calculator-assumed 65% (96 Marks)

This section has twelve (12) questions. Answer all questions. Write your answers in the spaces provided.

Working time: 100 minutes.

Question 7 (6 marks)

A music playlist contains nine different tracks, including one called First Night and another called Last Night. Each track is three minutes long.

(a) A shuffle feature randomly arranges the nine tracks. Determine the number of all possible arrangements that

Solution1×8!=40320 waysSpecific behaviours states numberSolution1×8!=40320 waysSpecific behaviours states number(i) start with First Night. (1 mark)

(ii) start with First Night and end with Last Night. (1 mark)

Solution1×7!×1=5040 waysSpecific behaviours states numberSolution1×7!×1=5040 waysSpecific behaviours states number

(iii) start with First Night or end with Last Night. (2 marks)

Solution40320+40320-5040=75600Specific behaviours uses inclusion-exclusion principle states correct numberSolution40320+40320-5040=75600Specific behaviours uses inclusion-exclusion principle states correct number

(b) Determine the number of selections of different tracks from the playlist that do not include First Night and Last Night and have a total playtime of 15 minutes. (2 marks)

Solution15÷3=5 tracks 7C5=21 selectionsSpecific behaviours calculates number of tracks states number of selectionsSolution15÷3=5 tracks 7C5=21 selectionsSpecific behaviours calculates number of tracks states number of selections

Question 8 (5 marks)

In the diagram below, AD and BE are diameters of the circle with centre O, C lies on the circumference and ∠COD=28°.

Determine the sizes of the following angles.

Solution∠AOB=∠BOC
∠AOB=180-282=76°Specific behaviours indicates congruent angles calculates angleSolution∠AOB=∠BOC
∠AOB=180-282=76°Specific behaviours indicates congruent angles calculates angle(a) ∠AOB. (2 marks)

Solution∠AEB=∠AOB2=762=38°Specific behaviours states angleSolution∠AEB=∠AOB2=762=38°Specific behaviours states angle(b) ∠AEB. (1 mark)

Solution∠EAO=∠AEO=38
∠ADE=90-38=52°Specific behaviours indicates size of ∠AEO calculates angleSolution∠EAO=∠AEO=38
∠ADE=90-38=52°Specific behaviours indicates size of ∠AEO calculates angle(c) ∠ADE. (2 marks)

Question 9 (8 marks)

Two tugs pull an offshore drilling rig. The first tug applies a force of 5 500 N in direction 122° and the second tug applies a force of 6 000 N in direction 088°.

(a) Show that the resultant force applied by the two tugs has magnitude close to 11 000 N, and determine the angle that the resultant force makes with the direction of the force applied by the first tug boat. (5 marks)

SolutionR2=55002+60002-255006000cos146
R=10998.48≈11 000 N6000sinθ=11000sin146
θ=17.76≈18°Specific behaviours sketch uses cosine rule shows magnitude before rounding uses sine rule determines angleSolutionR2=55002+60002-255006000cos146
R=10998.48≈11 000 N6000sinθ=11000sin146
θ=17.76≈18°Specific behaviours sketch uses cosine rule shows magnitude before rounding uses sine rule determines angle

(b) The second tug boat is asked to decrease the magnitude of the force it applies to reduce the resultant force to 9 000 N. Determine the percentage decrease required. (3 marks)

Solution90002=55002+x2-25500xcos146
x=38986000-38986000×100=35% reductionSpecific behaviours uses cosine rule solves for magnitude determines % reductionSolution90002=55002+x2-25500xcos146
x=38986000-38986000×100=35% reductionSpecific behaviours uses cosine rule solves for magnitude determines % reduction

Question 10 (8 marks)

Three vectors a, b and c are non-zero and non-parallel.

(a) Sketch a diagram using the parallelogram rule to show that vector addition is commutative, that is a+b=b+a. (2 marks)

SolutionSpecific behaviours shows a and b nose to tail with resultant PQ completes parallelogram with b and a nose to tailSolutionSpecific behaviours shows a and b nose to tail with resultant PQ completes parallelogram with b and a nose to tail

(b) Sketch a diagram to clearly illustrate each of the following vector equations.

(i) a+b=c. (2 marks)

SolutionSpecific behaviours closed triangle vectors labelled and directedSolutionSpecific behaviours closed triangle vectors labelled and directed

Solutionc=2b+aSpecific behaviours closed triangle vectors labelled and directedSolutionc=2b+aSpecific behaviours closed triangle vectors labelled and directed(ii) c-a=2b. (2 marks)

(c) If a+b+c=0, then is it also true that a+b+c=0? Explain your answer. (2 marks)

SolutionFor most cases, it is not true.However, it is true when a=b=c.Specific behaviours states for most case (but not all) states case when is trueSolutionFor most cases, it is not true.However, it is true when a=b=c.Specific behaviours states for most case (but not all) states case when is true

Question 11 (8 marks)

(a) In the diagram, A , B, C and D lie on the circumference of circle with centre O. Given that ∠ADC=67°, ∠BCO=63° and ∠DCO=23° determine the values of a, b and c. (3 marks)

Solutiona=2×67=134°
b=180-67=113°
c=360-63-113-134=50°Specific behaviours a b cSolutiona=2×67=134°
b=180-67=113°
c=360-63-113-134=50°Specific behaviours a b c

(b) In the diagram below, points B, C and D lie on the circumference of circle centre O and AB and AD are tangents to the circle.

(i) Prove that ABOD is a cyclic quadrilateral. (3 marks)

Solution∠ABO=∠ADO=90° (tangent-radii angle).Hence ∠ABO+∠ADO=180°.Hence ABOD is a cyclic quadrilateral, as opposite angles ∠ABO and ∠ABO are supplementary.Specific behaviours indicates tangent-radii are at 90° indicates opposite pair of angles are supplementary writes conclusionSolution∠ABO=∠ADO=90° (tangent-radii angle).Hence ∠ABO+∠ADO=180°.Hence ABOD is a cyclic quadrilateral, as opposite angles ∠ABO and ∠ABO are supplementary.Specific behaviours indicates tangent-radii are at 90° indicates opposite pair of angles are supplementary writes conclusion

(ii) Determine the size of ∠BAD if the size of ∠BCD=78°. (2 marks)

Solution∠BOD=2×78=156°∠BAD=180-156=24°Specific behaviours determines ∠BOD determines ∠BADSolution∠BOD=2×78=156°∠BAD=180-156=24°Specific behaviours determines ∠BOD determines ∠BAD

Question 12 (8 marks)

A seaplane with a cruising speed of 250 kmh-1 is required to fly to a location 355 km away on a bearing of 305°. A wind of 36 kmh-1 is blowing from bearing 020°.

(a) Sketch a diagram to show this information. (2 marks)

SolutionSpecific behaviours includes required path includes bearings or 105° angle in triangleSolutionSpecific behaviours includes required path includes bearings or 105° angle in triangle

(b) Determine the bearing that the seaplane should steer. (3 marks)

Solution250tsin105=36tsinθθ=8.00°Bearing is 305+8=313°Specific behaviours uses sine rule solves triangle states bearingSolution250tsin105=36tsinθθ=8.00°Bearing is 305+8=313°Specific behaviours uses sine rule solves triangle states bearing

(c) Determine the flight time, in hours and minutes. (3 marks)

Solution250tsin105=355sin⁡(180-105-8)=355sin67t=1.49 h0.49×60=29.4t=1 h 29 mSpecific behaviours uses sin or cosine rule determines t states t in h:mSolution250tsin105=355sin⁡(180-105-8)=355sin67t=1.49 h0.49×60=29.4t=1 h 29 mSpecific behaviours uses sin or cosine rule determines t states t in h:m

Question 13 (6 marks)

Seven teams from WA, six teams from SA and five teams from NT apply for eight available places in a league competition. The league is run so that every team plays every other team exactly once and no game ends in a tie.

(a) The organisers decide that there must be at least four teams from WA and an equal number of teams from SA and NT. Determine the total number of ways in which the organisers can select the eight teams for the league. (3 marks)

Solutionn=74×62×52+76×61×51n=5250+210=5460 waysSpecific behaviours identifies ways to select teams using combinations shows use of multiplication and addition calculates correct numberSolutionn=74×62×52+76×61×51n=5250+210=5460 waysSpecific behaviours identifies ways to select teams using combinations shows use of multiplication and addition calculates correct number

Assume the eight teams have already been chosen.

(b) Determine the number of games that will be played in the league and hence the number of schedules possible for the first three games. (3 marks)

Solution 8C2=28 games required 28P3=19656 waysSpecific behaviours calculates games required uses permutation to arrange evaluates number of waysSolution 8C2=28 games required 28P3=19656 waysSpecific behaviours calculates games required uses permutation to arrange evaluates number of ways

Question 14 (8 marks)

Three vectors are given by a=3i-4j, b=-3i+1.5j and c=-2i+yj, where y is a constant.

(a) Determine the vector projection of b on a. (3 marks)

Solutiona=35i-45jb∙a=-3b∙a×a=-95i+125jSpecific behaviours states unit vector for a states b∙a states projection as vectorSolutiona=35i-45jb∙a=-3b∙a×a=-95i+125jSpecific behaviours states unit vector for a states b∙a states projection as vector

(b) Determine the value(s) of y if

(i) a and c are perpendicular. (2 marks)

Solutiona∙c=0⇒-6-4y=0y=-32Specific behaviours uses scalar product states value of ySolutiona∙c=0⇒-6-4y=0y=-32Specific behaviours uses scalar product states value of y

(ii) the angle between the directions of b and c is 45°. (3 marks)

Solutioncos45=-3i+1.5j∙(-2i+yj)|-3i+1.5j|×|-2i+yj|y=-23, y=6 (using CAS)Specific behaviours uses scalar product states one solution states second solution(CAS is quickest if use numerical solve as shown)Solutioncos45=-3i+1.5j∙(-2i+yj)|-3i+1.5j|×|-2i+yj|y=-23, y=6 (using CAS)Specific behaviours uses scalar product states one solution states second solution(CAS is quickest if use numerical solve as shown)

Question 15 (9 marks)

(a) The work done, in joules, by a force of F Newtons in changing the displacement of an object by s metres, is given by the scalar product of F and s.

Solutionwd=250×4.3×cos45
=760 NSpecific behaviours substitutes correctly evaluates work doneSolutionwd=250×4.3×cos45
=760 NSpecific behaviours substitutes correctly evaluates work done(i) A force of 250 N acting due south moves an object 4.3 m in a south-westerly direction. Determine the work done. (2 marks)

(ii) Another force of 155 N does 269 joules of work in moving an object 190 cm. Determine the angle between the force and the direction of movement. (2 marks)

Solution269=155×1.9×cosθ
θ=24°Specific behaviours substitutes correctly evaluates angleSolution269=155×1.9×cosθ
θ=24°Specific behaviours substitutes correctly evaluates angle

SolutionSpecific behaviours sketchSolutionSpecific behaviours sketch(b) A triangle is formed by three non-zero vectors a, b and c, so that c=a-b, and θ is the angle between a and b.

(i) Sketch the triangle. (1 mark)

Solutionusing definition of scalar product, since θ=0 and cos0=1, then c∙c=c2.Specific behaviours explanationSolutionusing definition of scalar product, since θ=0 and cos0=1, then c∙c=c2.Specific behaviours explanation(ii) Explain why c∙c=c2. (1 mark)

(iii) Use c∙c=a-b∙a-b to deduce the cosine rule. (3 marks)

Solutionc∙c=a-b∙a-b
c∙c=a∙a-a∙b-b∙a+b∙b
c2=a2+b2-2abcosθSpecific behaviours expands scalar product uses result from (ii) uses scalar product definitionSolutionc∙c=a-b∙a-b
c∙c=a∙a-a∙b-b∙a+b∙b
c2=a2+b2-2abcosθSpecific behaviours expands scalar product uses result from (ii) uses scalar product definition

Question 16 (10 marks)

In the diagram below, POT is a diameter of circle with centre O, QP is a tangent to the circle at P, QR is a tangent to the circle at R and PT is extended to meet QR extended at S. You may want to let ∠OTR=θ.

(a) Prove that ΔOPQ is congruent to ΔORQ. (3 marks)

Solution(i) OP=OR (radii)(ii) QP=QR (tangent length from external point)(iii) OQ (common to both triangles)(iv) ∠OPQ=∠ORQ (tangent-radius angle 90°)Using various combinations, reason one of SSS, SAS or RHS.Specific behaviours first statement with reason second and third statements with reasons relevant conclusionSolution(i) OP=OR (radii)(ii) QP=QR (tangent length from external point)(iii) OQ (common to both triangles)(iv) ∠OPQ=∠ORQ (tangent-radius angle 90°)Using various combinations, reason one of SSS, SAS or RHS.Specific behaviours first statement with reason second and third statements with reasons relevant conclusion

(b) Prove that OQ is parallel to TR. (4 marks)

Solution (one of many possibilities)Given ∠OTR=θ, then ∠ORT=θ (as ΔOTR isosceles - OT and OR radii)Hence ∠POR=2θ (sum of interior angles = opposite exterior angle)Hence ∠POQ=θ (congruent triangles, ∠POQ=∠ROQ)Hence ∠OTR=∠POQ (corresponding angles equal)Hence OQ is parallel to TR Specific behaviours uses isosceles triangle determines ∠POR determines ∠POQ uses corresponding angles to make conclusionSolution (one of many possibilities)Given ∠OTR=θ, then ∠ORT=θ (as ΔOTR isosceles - OT and OR radii)Hence ∠POR=2θ (sum of interior angles = opposite exterior angle)Hence ∠POQ=θ (congruent triangles, ∠POQ=∠ROQ)Hence ∠OTR=∠POQ (corresponding angles equal)Hence OQ is parallel to TR Specific behaviours uses isosceles triangle determines ∠POR determines ∠POQ uses corresponding angles to make conclusion

(c) If TR=TS, deduce that ΔOTR is equilateral. (3 marks)

SolutionΔTSR is isosceles.Hence ∠TSR=∠TRS=θ2 (sum of interior angles = opposite exterior angle)But ∠TRS=90-∠ORT=90-θ (tangent-radius angle 90°)Hence 90-θ=θ2⇒θ=60°.Hence ∠OTR=∠ORT=∠TOR=60° - ΔOTR is equilateral.Specific behaviours expresses ∠TRS using isosceles triangle expresses ∠TRS using radii-tangent angle solves for θ and deduces triangle is equilateralSolutionΔTSR is isosceles.Hence ∠TSR=∠TRS=θ2 (sum of interior angles = opposite exterior angle)But ∠TRS=90-∠ORT=90-θ (tangent-radius angle 90°)Hence 90-θ=θ2⇒θ=60°.Hence ∠OTR=∠ORT=∠TOR=60° - ΔOTR is equilateral.Specific behaviours expresses ∠TRS using isosceles triangle expresses ∠TRS using radii-tangent angle solves for θ and deduces triangle is equilateral

Question 17 (11 marks)

A small boat that can maintain a steady speed of 5 ms-1 is to cross a river from A to B, where AB=35i-105j m.

A current of -i-2j ms-1 flows in the river.

The velocity vector that the pilot of the small boat must set to travel from A to B is ai+bj, where a and b are constants.

(a) Explain why ta-1=35 and tb-2=-105, where t is a constant. (3 marks)

SolutionThe sum of the velocities of the boat and the river must be parallel to AB:tai+bj+-i-2j=35i-105j The two given equations arise by equating i and then j coefficients from this equation.Specific behaviours uses sum of velocities uses equation for parallel condition equates individual coefficientsSolutionThe sum of the velocities of the boat and the river must be parallel to AB:tai+bj+-i-2j=35i-105j The two given equations arise by equating i and then j coefficients from this equation.Specific behaviours uses sum of velocities uses equation for parallel condition equates individual coefficients

(b) Eliminate t from the equations in (a) and hence express b in terms of a, simplifying your expression. (3 marks)

Solutiont=35a-1=-105b-2b-2=-10535a-1b=5-3aSpecific behaviours equates both to t cross-multiplies simplifiesSolutiont=35a-1=-105b-2b-2=-10535a-1b=5-3aSpecific behaviours equates both to t cross-multiplies simplifies

(c) Explain why a2+b2=25. (1 mark)

SolutionThe magnitude of ai+bj is the speed of the boat.Specific behaviours uses magnitude and speedSolutionThe magnitude of ai+bj is the speed of the boat.Specific behaviours uses magnitude and speed

(d) Use your equations from (b) and (c) to determine the values of a and b. (3 marks)

Solutiona2+5-3a2=25a=0, 3b=5, -4v=3i-4jSpecific behaviours writes equation solves for a and b eliminates alternative solution Solutiona2+5-3a2=25a=0, 3b=5, -4v=3i-4jSpecific behaviours writes equation solves for a and b eliminates alternative solution

(e) Determine the time that the small boat will take to travel from A to B. (1 mark)

Solutiont=35a-1=352=17.5 sSpecific behaviours states timeSolutiont=35a-1=352=17.5 sSpecific behaviours states time

Question 18 (6 marks)

Let gx=x2-8x+19, x∈Z.

(a) Use an example to show that when x is odd, gx is even. (1 mark)

SolutionIf x=1 (odd) then g1=1-8+19=12 (even)Specific behaviours suitable exampleSolutionIf x=1 (odd) then g1=1-8+19=12 (even)Specific behaviours suitable example

(b) Write the contrapositive of "if g(x) is an even integer, then x is an odd integer". (1 mark)

SolutionIf x is not odd, then g(x) is not even.Specific behaviours writes contrapositiveSolutionIf x is not odd, then g(x) is not even.Specific behaviours writes contrapositive

Any even integer m can be expressed in the form m=2a, where a∈Z. Similarly, any odd integer n can be expressed in the form n=2a+1.

Solutiong2a=2a2-82a+19=4a2-16a+19Specific behaviours substitutes and simplifiesSolutiong2a=2a2-82a+19=4a2-16a+19Specific behaviours substitutes and simplifies(c) Simplify g(2a). (1 mark)

(d) Express g(2a) in a form that clearly shows it is an odd integer. (1 mark)

Solutiong2a=22a2-8a+9+1Specific behaviours expresses in form 2n+1Solutiong2a=22a2-8a+9+1Specific behaviours expresses in form 2n+1

(e) Use your answers above to prove that if g(x) is even, then x is odd. (2 marks)

Solution(c) and (d) prove that when x is not odd (ie even) then g(x) is not even (ie odd).Hence the contrapositive statement from (b) is true and so the original statement that is to be proved must be true.Specific behaviours uses results from (c) and (d) explains contrapositive is true and hence statement must be trueSolution(c) and (d) prove that when x is not odd (ie even) then g(x) is not even (ie odd).Hence the contrapositive statement from (b) is true and so the original statement that is to be proved must be true.Specific behaviours uses results from (c) and (d) explains contrapositive is true and hence statement must be true

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