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Semester Two Examination, 2021

Question/Answer booklet

SOLUTIONSSOLUTIONSMATHEMATICS
SPECIALIST
UNITS 1&2

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, which can include scientific, graphic and Computer Algebra System (CAS) calculators, are permitted in this ATAR course 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% (92 Marks)

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

Working time: 100 minutes.

Question 9 (6 marks)

(a) Determine the vector projection of 5-5 on -12. (2 marks)

Solution5-5∙-12-12∙-12-12=-155-12
=-3-12
=3-6Specific behaviours indicates appropriate methodü correct vectorSolution5-5∙-12-12∙-12-12=-155-12
=-3-12
=3-6Specific behaviours indicates appropriate methodü correct vector

(b) Determine the value(s) of t so that the vectors t-3 and 3t+81 are

(i) parallel. (2 marks)

Solutiont-3=3t+81⇒t=-125=-2.4Specific behaviours indicates equation using of ratio of coefficientsü correct valueSolutiont-3=3t+81⇒t=-125=-2.4Specific behaviours indicates equation using of ratio of coefficientsü correct value

(ii) perpendicular. (2 marks)

Solutiont-3∙3t+81=0
3t2+8t-3=0
t=-3, t=13Specific behaviours indicates equation using scalar productü correct valuesSolutiont-3∙3t+81=0
3t2+8t-3=0
t=-3, t=13Specific behaviours indicates equation using scalar productü correct values

Question 10 (7 marks)

(a) Five-digit odd numbers are to be made using the digits 1, 2, 3, 4, 5, 6 and 7. Determine how many such numbers exist if the number must exceed 50 000 and no digit may be used more than once in a number. (3 marks)

SolutionEnd with 1 or 3: 2×3×5×4×3=360End with 5 or 7: 2×2×5×4×3=240Total possible numbers: 360+240=600Specific behaviours splits into mutually exclusive casesü correctly counts at least one caseü calculates totalSolutionEnd with 1 or 3: 2×3×5×4×3=360End with 5 or 7: 2×2×5×4×3=240Total possible numbers: 360+240=600Specific behaviours splits into mutually exclusive casesü correctly counts at least one caseü calculates total

(b) The library in a small guesthouse has 32 different books, of which 21 are non-fiction and the remainder fiction. Determine the number of different ways that a guest can select four books if they want

(i) the same number of fiction and non-fiction books. (2 marks)

Solutionn=212112
=210×55
=11 550 waysSpecific behaviours indicates correct methodü correct number of waysSolutionn=212112
=210×55
=11 550 waysSpecific behaviours indicates correct methodü correct number of ways

(ii) more fiction than non-fiction books. (2 marks)

Solutionn=211113+210114
=21×165+1×330
=3465+330
=3795 waysSpecific behaviours indicates correct methodü correct number of waysSolutionn=211113+210114
=21×165+1×330
=3465+330
=3795 waysSpecific behaviours indicates correct methodü correct number of ways

Question 11 (7 marks)

Two transformation matrices are M=0110 and N=22-44.

Triangle PQR has an area of 39 cm2, with vertices at P5, 3, Q-2, 8 and R-5, -1.

(a) Determine the coordinates of PQR after the triangle has been transformed by matrix M.

Solution01105-2-538-1=38-15-2-5P'3, 5, Q'8, -2, R'-1, -5.Specific behaviours indicates pre-multiplication by Mü correct matrix productü correctly lists set of coordinatesSolution01105-2-538-1=38-15-2-5P'3, 5, Q'8, -2, R'-1, -5.Specific behaviours indicates pre-multiplication by Mü correct matrix productü correctly lists set of coordinates (3 marks)

(b) Use the geometric transformation to explain why the determinant of M is 1. (1 mark)

SolutionM represents a reflection, the area of triangle does not change and so determinant is 1.Specific behaviours reflection will not change areaSolutionM represents a reflection, the area of triangle does not change and so determinant is 1.Specific behaviours reflection will not change area

(c) Use the geometric transformation to explain why M2=I, where I is the 2×2 identity matrix. (1 mark)

SolutionM2 represents two reflections in the same line, and so the triangle will be back where it started, with the same coordinates.Specific behaviours two reflections in same lineSolutionM2 represents two reflections in the same line, and so the triangle will be back where it started, with the same coordinates.Specific behaviours two reflections in same line

(d) Determine the area of PQR after the triangle has been transformed by matrix N. (2 marks)

SolutiondetN=8+8=16New area=16×39=624cm2Specific behaviours calculates determinantü calculates new areaSolutiondetN=8+8=16New area=16×39=624cm2Specific behaviours calculates determinantü calculates new area

Question 12 (8 marks)

(a) Write the converse of the true statement 'if a figure is a square then it has four congruent sides' and use an example or counter-example to briefly discuss the truth of the converse.

SolutionConverse: If a figure has four congruent sides, then it is a square.The converse is false, as the figure could be a rhombus.Specific behaviours correct converseü states false with counter-exampleSolutionConverse: If a figure has four congruent sides, then it is a square.The converse is false, as the figure could be a rhombus.Specific behaviours correct converseü states false with counter-example (2 marks)

(b) Points A, B, C and D lie as shown on a circle
with centre O so that AD is a diameter,
AB=CD and ∠COD=42°.

Determine the size of

Solution∠CAD=42°÷2=21°∠BAD=∠CDA=90°-21°=69°∠CAB=69°-21°=48°Specific behaviours ∠CAD ∠BAD ∠CABSolution∠CAD=42°÷2=21°∠BAD=∠CDA=90°-21°=69°∠CAB=69°-21°=48°Specific behaviours ∠CAD ∠BAD ∠CAB(i) ∠CAD. (1 mark)

(ii) ∠BAD. (1 mark)

(iii) ∠CAB. (1 mark)

(c) Two chords of a circle, LM and PQ, intersect at N so that LM=41 cm, NM=20 cm and PQ=44 cm. Determine all possible lengths of QN. (3 marks)

SolutionUsing intersecting chord theorem, LN×NM=PN×NQ.LN=LM-NM=41-20=21Let NQ=x, so that PN=44-x21×20=x(44-x)
x=14, 30Hence QN=14 cm or QN=30 cm.Specific behaviours identifies required lengthsü forms quadratic equationü states both valuesSolutionUsing intersecting chord theorem, LN×NM=PN×NQ.LN=LM-NM=41-20=21Let NQ=x, so that PN=44-x21×20=x(44-x)
x=14, 30Hence QN=14 cm or QN=30 cm.Specific behaviours identifies required lengthsü forms quadratic equationü states both values

Question 13 (8 marks)

In triangle OAB, P is the midpoint of OA and M is the midpoint of PB. Let OA=a and OB=b.

(a) Show that OM=14a+12b. (2 marks)

SolutionOM=OP+PM, OP=12 OA, PM=12 PB=12 OB-OP
=12 a+12b-12 a
=14 a+12 bSpecific behaviours indicates logical stepsü uses correct vector notation throughoutSolutionOM=OP+PM, OP=12 OA, PM=12 PB=12 OB-OP
=12 a+12b-12 a
=14 a+12 bSpecific behaviours indicates logical stepsü uses correct vector notation throughout

The position vector of A is 64, position vector of B is 7-4 and O is the origin.

(b) Determine a unit vector u in the same direction as OM. (2 marks)

SolutionOM=1464+127-4=5-1
u=26265-1*NB might use CAS for last stepSpecific behaviours calculates OMü states unit vectorSolutionOM=1464+127-4=5-1
u=26265-1*NB might use CAS for last stepSpecific behaviours calculates OMü states unit vector

(c) Show that OA is perpendicular to PM. (2 marks)

SolutionOA=64, PM=127-4-1464=2-3OA∙PM=64∙2-3=12-12=0Hence vectors are perpendicular.Specific behaviours calculates PMü shows scalar product is zeroSolutionOA=64, PM=127-4-1464=2-3OA∙PM=64∙2-3=12-12=0Hence vectors are perpendicular.Specific behaviours calculates PMü shows scalar product is zero

(d) Determine the size of ∠AOB. (2 marks)

Solution∠AOB=cos-1a∙bab
=cos-155
=63.4°*NB might use CASSpecific behaviours indicates suitable methodü correct angleSolution∠AOB=cos-1a∙bab
=cos-155
=63.4°*NB might use CASSpecific behaviours indicates suitable methodü correct angle

Question 14 (7 marks)

(a) In trapezium ABCD, AC and BD are diagonals, and AB is parallel to CD. Use a vector method to prove that AC+DB=AB+DC. (3 marks)

SolutionNote that AC=AB+BC and DB=DC-BC.LHS=AC+DB
=AB+BC+DC-BC
=AB+DC
=RHSSpecific behaviours expression for ACü expression for DBü adds expressions and simplifiesSolutionNote that AC=AB+BC and DB=DC-BC.LHS=AC+DB
=AB+BC+DC-BC
=AB+DC
=RHSSpecific behaviours expression for ACü expression for DBü adds expressions and simplifies

(b) In rectangle OPQR, let OP=p and OR=r. Use a vector method to prove that if the diagonals OQ and PR are perpendicular, then OPQR is a square. (4 marks)

SolutionThe diagonals are the vectors OQ=p+r and PR=r-p.If diagonals are perpendicular, then OQ∙PR=0 and so:p+r∙r-p=0
p∙r-p∙p+r∙r-r∙p=0
|r​2-|p​2=0
|r​2=|p​2⇒r=|p|Hence OPQR must be a square since it is a rectangle with equal length sides.Specific behaviours determines vectors for diagonals uses scalar product expands and simplifies scalar product explains that sides are equal lengthSolutionThe diagonals are the vectors OQ=p+r and PR=r-p.If diagonals are perpendicular, then OQ∙PR=0 and so:p+r∙r-p=0
p∙r-p∙p+r∙r-r∙p=0
|r​2-|p​2=0
|r​2=|p​2⇒r=|p|Hence OPQR must be a square since it is a rectangle with equal length sides.Specific behaviours determines vectors for diagonals uses scalar product expands and simplifies scalar product explains that sides are equal length

Question 15 (6 marks)

Consider the following statement:

For two integers a,b if 3a2-2b2 is a multiple of 4 then at least one of a,b is even.

(a) Write the contrapositive of the statement. (1 mark)

SolutionFor two integers a, b if both are odd then 3a2-2b2 is not a multiple of 4.Specific behaviours correct contrapositiveSolutionFor two integers a, b if both are odd then 3a2-2b2 is not a multiple of 4.Specific behaviours correct contrapositive

(b) Prove that the statement is true. (5 marks)

SolutionProof of contrapositive:If a, b both odd, then a=2m+1 and b=2n+1, where n,m both integers.Hence3a2-2b2=32m+12-22n+12
=12m2+12m-8n2-8n+1
=43m2+3m-2n2-2n+1Hence 3a2-2b2 will never be a multiple of 4 as it is always one more than a multiple of 4.Since the contrapositive statement has been proved to be true then it follows that the original statement must also be true.Specific behaviours attempts to prove contrapositive and states truth of contrapositive implies truth of original statementü uses form 2k+1 form for odd numbers a, bü substitutes for a, b and expandsü factors out 4ü explains why contrapositive trueSolutionProof of contrapositive:If a, b both odd, then a=2m+1 and b=2n+1, where n,m both integers.Hence3a2-2b2=32m+12-22n+12
=12m2+12m-8n2-8n+1
=43m2+3m-2n2-2n+1Hence 3a2-2b2 will never be a multiple of 4 as it is always one more than a multiple of 4.Since the contrapositive statement has been proved to be true then it follows that the original statement must also be true.Specific behaviours attempts to prove contrapositive and states truth of contrapositive implies truth of original statementü uses form 2k+1 form for odd numbers a, bü substitutes for a, b and expandsü factors out 4ü explains why contrapositive true

Question 16 (8 marks)

The height of the tide, h cm, of the sea above the mean level at time t hours after midnight one day is given by

ht=28cosπt6+45sinπt6 .

(a) Express h in the form acos(bt-θ), where a, b>0 and 0≤θ≤2π. (3 marks)

Solutiona=282+452=53h=53cosπt6-θ
=53cosπt6cosθ+sinπt6sinθcosθ=2853, sinθ=4553⇒θ=1.0142h=53cosπt6-1.0142Specific behaviours value of aü value of θü correct expression for hSolutiona=282+452=53h=53cosπt6-θ
=53cosπt6cosθ+sinπt6sinθcosθ=2853, sinθ=4553⇒θ=1.0142h=53cosπt6-1.0142Specific behaviours value of aü value of θü correct expression for h

(b) Determine, to the nearest minute, the time of the first high tide. (2 marks)

SolutionRequire πt6-1.0142=0⇒t=1.937 h. Hence at 0156 or 1:56 am.Specific behaviours time in hoursü time of day, to nearest minuteSolutionRequire πt6-1.0142=0⇒t=1.937 h. Hence at 0156 or 1:56 am.Specific behaviours time in hoursü time of day, to nearest minute

(c) Sketch the graph of the height of the tide on the axes below. (3 marks)

SolutionSee graphSpecific behaviours vertical scale and interceptü rootsü smooth sinusoidal curveSolutionSee graphSpecific behaviours vertical scale and interceptü rootsü smooth sinusoidal curve

Question 17 (7 marks)

Three forces F1, F2 and F3 act on a small body, where F1=4i-10j N, F2=-8i+16j N and
F3=9i-15j N.

(a) Determine the magnitude of the resultant force and the angle between the resultant force and the vector i. (3 marks)

SolutionR=F1+F2+F3
=4-10+-816+9-15=5-9R=52+92=106≈10.3 N∠=tan-1-95≈-60.9°Hence resultant has a magnitude of 10.3 N and makes an angle of 60.9° with i.Specific behaviours correct sum in component formü calculates magnitudeü calculates angleSolutionR=F1+F2+F3
=4-10+-816+9-15=5-9R=52+92=106≈10.3 N∠=tan-1-95≈-60.9°Hence resultant has a magnitude of 10.3 N and makes an angle of 60.9° with i.Specific behaviours correct sum in component formü calculates magnitudeü calculates angle

(b) Two of the forces, F2 and F3, can be multiplied by scalars λ and μ respectively so that the three forces are in equilibrium. Determine the value of λ and the value of μ. (4 marks)

SolutionF1+λF2+μF3=0
4-10+λ-816+μ9-15=00Resolving in i and j directions:4-8λ+9μ=0
-10+16λ-15μ=0Solving simultaneously givesλ=54=1.25, μ=23Specific behaviours writes vector equation equal to 0ü forms two equationsü value of λü value of μSolutionF1+λF2+μF3=0
4-10+λ-816+μ9-15=00Resolving in i and j directions:4-8λ+9μ=0
-10+16λ-15μ=0Solving simultaneously givesλ=54=1.25, μ=23Specific behaviours writes vector equation equal to 0ü forms two equationsü value of λü value of μ

Question 18 (7 marks)

(a) 90 people are asked to choose two different letters from those in the word GAMBLER and write them down in order. Use the pigeonhole principle to prove that at least three people will write the same pair of letters in the same order. (3 marks)

SolutionThere are 7P2=42 different ordered pairs of letters, each of which is a pigeonhole.Using the pigeonhole principle, if 90 pigeons (the number of pairs of letters written by people) are placed into 42 pigeonholes, then at least one pigeonhole will contain 90÷42=3 or more pigeons.Hence at least 3 people will write the same pair of letters in the same order.Specific behaviours obtains number of permutationsü indicates pigeons and pigeonholesü uses pigeonhole principle to complete proofSolutionThere are 7P2=42 different ordered pairs of letters, each of which is a pigeonhole.Using the pigeonhole principle, if 90 pigeons (the number of pairs of letters written by people) are placed into 42 pigeonholes, then at least one pigeonhole will contain 90÷42=3 or more pigeons.Hence at least 3 people will write the same pair of letters in the same order.Specific behaviours obtains number of permutationsü indicates pigeons and pigeonholesü uses pigeonhole principle to complete proof

(b) Three character codes, such as TCU, are made using three different letters chosen from the word DISCOUNT. Determine the proportion of all possible codes that start with a D or end with a T. (4 marks)

SolutionStart with D: nD=1× 7P2=42 codes.End with T: nT=1× 7P2=42 codes.Start with D and end with T: nD∩T=1×1× 6P1=6 codes.Start with D or end with T: nD∪T=42+42-6=78 codes.There are 8P3=336 different codes.Hence required proportion is 78336=1356≈0.232.Specific behavioursü nD and nTü nD∩Tü nD∪T number of possible codes and writes proportionSolutionStart with D: nD=1× 7P2=42 codes.End with T: nT=1× 7P2=42 codes.Start with D and end with T: nD∩T=1×1× 6P1=6 codes.Start with D or end with T: nD∪T=42+42-6=78 codes.There are 8P3=336 different codes.Hence required proportion is 78336=1356≈0.232.Specific behavioursü nD and nTü nD∩Tü nD∪T number of possible codes and writes proportion

Question 19 (7 marks)

Airport B lies 170 km due east of airport A, and in the region of the airports a wind of 25 km/h is blowing from the northeast.

A small plane, with a cruising speed of 120 km/h, leaves airport A. The pilot, not aware of the wind and intending to fly to airport B, steered the plane on a bearing of 090°.

Assuming that the pilot does not realise their mistake, determine how close the plane will come to airport B if it continues to fly for several hours on the same bearing.

Solutionx2=1202+252-212025cos45°
x=103.84sinθ25=sin45°103.84
θ=9.80°d=170sin9.8°
=28.9 kmSpecific behaviours appropriate sketch/explanation for x and θü equation for xü solves for xü equation for θü solves for θ appropriate sketch/explanation for closest distanceü calculates closest distanceSolutionx2=1202+252-212025cos45°
x=103.84sinθ25=sin45°103.84
θ=9.80°d=170sin9.8°
=28.9 kmSpecific behaviours appropriate sketch/explanation for x and θü equation for xü solves for xü equation for θü solves for θ appropriate sketch/explanation for closest distanceü calculates closest distance

Question 20 (7 marks)

Triangle ABC has vertices A2, 2, B(-4, 6) and C(4, 8).

ABC is rotated clockwise 60° about the origin to form triangle A'B'C'.

(a) Determine the coordinates of C'. (2 marks)

Solution1213-3148=2+434-23≈8.9280.536C'2+43, 4-23Specific behaviours rotation matrixü correct coordinatesSolution1213-3148=2+434-23≈8.9280.536C'2+43, 4-23Specific behaviours rotation matrixü correct coordinates

ABC is reflected in the line y=2x to form triangle A''B''C''.

Solution15-3443-46=36525=7.20.4B''7.2, 0.4Specific behaviours reflection matrixü correct coordinatesSolution15-3443-46=36525=7.20.4B''7.2, 0.4Specific behaviours reflection matrixü correct coordinates(b) Determine the coordinates of B''. (2 marks)

(c) Determine matrix T that will transform A'B'C' to A''B''C''. (3 marks)

SolutionMatrix P for A'B'C'→ABC is inverse of that used in (a).Matrix Q for ABC→A''B''C'' is same as used in (b).HenceT=QP-1
=15-3443×1213-31-1
=11043-333+433+4-43+3≈0.3930.9200.920-0.393Specific behaviours matrix for A'B'C'→ABCü indicates correct order of multiplicationü correct matrix TSolutionMatrix P for A'B'C'→ABC is inverse of that used in (a).Matrix Q for ABC→A''B''C'' is same as used in (b).HenceT=QP-1
=15-3443×1213-31-1
=11043-333+433+4-43+3≈0.3930.9200.920-0.393Specific behaviours matrix for A'B'C'→ABCü indicates correct order of multiplicationü correct matrix T

Question 21 (7 marks)

The diagram, not drawn to scale, shows
vertices A, B and D of an isosceles triangle
lying on a circle so that AD=AB.

Lines CE and CF are tangential to the
circle at D and B respectively.

ABCE is a cyclic quadrilateral.

Let ∠FCE=x.

(a) Determine, with reasons, the size of ∠AEC in terms of x. (5 marks)

Solution∠DBC=12180°-x=90°-12x (tangents from C⇒CD=CB)
∠BAD=∠DBC=90°-12x (alternate segment)
∠ABD=12180°-90°-12x=45°+14x (isosceles triangle)
∠ABC=∠ABD+∠DBC (adjacent angles)
=45°+14x+90°-12x=135°-14x
∠AEC=180°-∠ABC (cyclic quadrilateral)
=180°-135°-14x=45°+14xSpecific behaviours expression for ∠DBC with reason expression for ∠BAD with reason expression for ∠ABD with reason expression for ∠ABC with reason expression for ∠AEC with reasonSolution∠DBC=12180°-x=90°-12x (tangents from C⇒CD=CB)
∠BAD=∠DBC=90°-12x (alternate segment)
∠ABD=12180°-90°-12x=45°+14x (isosceles triangle)
∠ABC=∠ABD+∠DBC (adjacent angles)
=45°+14x+90°-12x=135°-14x
∠AEC=180°-∠ABC (cyclic quadrilateral)
=180°-135°-14x=45°+14xSpecific behaviours expression for ∠DBC with reason expression for ∠BAD with reason expression for ∠ABD with reason expression for ∠ABC with reason expression for ∠AEC with reason

(b) Hence determine the range of values for the size of ∠AEC in degrees. (2 marks)

SolutionFor figure to exist, 0°<x<180°.Hence ∠AEC>45°+140° and ∠AEC<45°+14180°Range is 45°<∠AEC<90°.Specific behaviours indicates correct domain for xü correct range, including inequalitiesSolutionFor figure to exist, 0°<x<180°.Hence ∠AEC>45°+140° and ∠AEC<45°+14180°Range is 45°<∠AEC<90°.Specific behaviours indicates correct domain for xü correct range, including inequalities

Supplementary page

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Supplementary page

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