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Semester Two Examination, 2017

Question/Answer booklet

SOLUTIONSSOLUTIONSMATHEMATICS

SPECIALIST

UNITS 1 AND 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 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% (95 Marks)

This section has thirteen (13) questions. Answer all questions. Write your answers in the spaces provided.

Working time: 100 minutes.

Question 9 (6 marks)

An exam has two parts, I and II, containing 15 and 8 questions respectively.

Determine the number of different combinations of questions a candidate could choose if they must answer

(a) 5 questions from part I and 4 questions from part II. (2 marks)

Solution15584=3003×70
=210 210 waysSpecific behaviours uses multiplication of combinations correct numberSolution15584=3003×70
=210 210 waysSpecific behaviours uses multiplication of combinations correct number

(b) 3 questions, all chosen from the same part. (2 marks)

Solution15083+15380=56+455
=511 waysSpecific behaviours uses addition of combinations correct numberSolution15083+15380=56+455
=511 waysSpecific behaviours uses addition of combinations correct number

(c) 3 questions, with at least one question from each part. (2 marks)

Solution233=17711771-511=1 260 waysSpecific behaviours calculates total ways, no restriction subtracts answer from (b) Solution233=17711771-511=1 260 waysSpecific behaviours calculates total ways, no restriction subtracts answer from (b)

Question 10 (6 marks)

(a) The point P(21, -11) is translated by the column vectors xy and -713 to P'(-6, 5). Determine the values of the constants x and y. (2 marks)

Solution21-11+xy+-713=-65xy=-65--713-21-11xy=-203Specific behaviours value of x value of ySolution21-11+xy+-713=-65xy=-65--713-21-11xy=-203Specific behaviours value of x value of y

(b) Determine the single matrix that represents, in order, the composition of a reflection in the line y=-x followed by an anti-clockwise rotation of 135° about the origin. Express matrix coefficients in exact form. (4 marks)

Solutioncos135-sin135sin135cos135×0-1-10=222222-22Specific behaviours matrix for rotation matrix for reflection multiplies in correct order correct matrixSolutioncos135-sin135sin135cos135×0-1-10=222222-22Specific behaviours matrix for rotation matrix for reflection multiplies in correct order correct matrix

Question 11 (6 marks)

(a) A circle property says that if chords of a circle are of equal length then they subtend equal angles at the centre.

(i) Write the converse of this statement. (1 mark)

SolutionIf chords of a circle subtend equal angles at the centre then they are of equal length.Specific behaviours writes converseSolutionIf chords of a circle subtend equal angles at the centre then they are of equal length.Specific behaviours writes converse

SolutionConverse is true.Specific behaviours diagram states converse trueSolutionConverse is true.Specific behaviours diagram states converse true(ii) Draw a diagram to illustrate the converse statement and state whether the converse is also true. (2 marks)

(b) The diagram below shows four points A, B, C and D lying on the circumference of a circle. The line PQ is a tangent to the circle at A, ∠BDC=21°, ∠PAD=35° and ∠QAB=62°.

Determine the size of angles x, y and z. (3 marks)

Solutionx=180-180-35-62=97°y=180-97-21=62°z=∠PAD=35°Specific behaviours each angle Solutionx=180-180-35-62=97°y=180-97-21=62°z=∠PAD=35°Specific behaviours each angle

Question 12 (9 marks)

(a) If p=4i-2j and q=3i+2j determine

(i) the angle between the directions of p and q, to the nearest tenth of a degree.

SolutionUsing CAS angle is 60.255≈60.3° (1dp)Specific behaviours states angle rounds correctlySolutionUsing CAS angle is 60.255≈60.3° (1dp)Specific behaviours states angle rounds correctly (2 marks)

(ii) the scalar projection of q on p. (2 marks)

Solution32+22×cos60.255≈1.79Exact: 455Specific behaviours expression for projection states valueSolution32+22×cos60.255≈1.79Exact: 455Specific behaviours expression for projection states value

(b) The vector 21i+5aj has a magnitude of 29 and is perpendicular to the vector 4i-2bj. Determine the values of the constants a and b, where a<b. (5 marks)

Solution212+(5a)2=292a=±4214+(5a)(-2b)=084-10(±4)b=0b=±2110a=-4, b=-2110Specific behaviours uses magnitude to form equation calculate values of a uses dot product to form equation calculate values of b chooses correct pairingSolution212+(5a)2=292a=±4214+(5a)(-2b)=084-10(±4)b=0b=±2110a=-4, b=-2110Specific behaviours uses magnitude to form equation calculate values of a uses dot product to form equation calculate values of b chooses correct pairing

Question 13 (8 marks)

In the diagram below, forces F1 and F2 act on a body at the origin.

(a) If F1=50 N, F2=65 N, α=22° and β=32°, determine the magnitude of the resultant force and the angle it makes with the positive x axis. (5 marks)

SolutionFR=502+652-25065cos126FR=102.7 Nsinθ65=sin126102.7θ=30.8°ϕ=22°-30.8°=-8.8°Specific behaviours sketch with forces nose to tail uses cosine rule for magnitude states magnitude uses sine rule for angle states angle with x axisSolutionFR=502+652-25065cos126FR=102.7 Nsinθ65=sin126102.7θ=30.8°ϕ=22°-30.8°=-8.8°Specific behaviours sketch with forces nose to tail uses cosine rule for magnitude states magnitude uses sine rule for angle states angle with x axis

(b) If F1=75 N and F2=95 N, determine the angles α and β so that the resultant force is directed along the positive x axis and has a magnitude of 155 N. (3 marks)

Solutionα=cos-1752+1552-952275155α=27.5°β=cos-1952+1552-752295155β=21.4°Specific behaviours sketch with resultant on x axis uses cosine rule for angle uses sine or cosine rule for second angleSolutionα=cos-1752+1552-952275155α=27.5°β=cos-1952+1552-752295155β=21.4°Specific behaviours sketch with resultant on x axis uses cosine rule for angle uses sine or cosine rule for second angle

Question 14 (6 marks)

(a) Prove that tan3A=3tanA-tan3A 1-3tan2A . (4 marks)

SolutionLHS=tan3A
=tanA+2A
=tanA+tan2A1-tanAtan2A
=tanA+2tanA1-tan2A÷1-2tan2A1-tan2A
=tanA-tan3A+2tanA1-tan2A×1-tan2A1-tan2A-2tan2A
=3tanA-tan3A1-3tan2A
=RHSSpecific behaviours expands sum of A and 2A uses double angle identity for tan2A eliminates 1-tan2A simplifiesSolutionLHS=tan3A
=tanA+2A
=tanA+tan2A1-tanAtan2A
=tanA+2tanA1-tan2A÷1-2tan2A1-tan2A
=tanA-tan3A+2tanA1-tan2A×1-tan2A1-tan2A-2tan2A
=3tanA-tan3A1-3tan2A
=RHSSpecific behaviours expands sum of A and 2A uses double angle identity for tan2A eliminates 1-tan2A simplifies

(b) Hence, or otherwise, solve 3tanA-tan3A=1-3tan2A, 0≤A≤π6. (2 marks)

Solutiontan3A=1
3A=π4
A=π12Specific behaviours writes as tan3A=1 solutionSolutiontan3A=1
3A=π4
A=π12Specific behaviours writes as tan3A=1 solution

Question 15 (7 marks)

(a) The diagram below shows vertices A, B and C of rhombus OABC lying on the circumference of circle centre O and point M lying on the major arc AC. Determine the size of angle AMC. (3 marks)

SolutionOA=OB=AB radii and sides of rhombus∠AOB=60⇒∠AOC=120∠AMC=12×120=60°Specific behaviours indicates triangle OAB equilateral angle AOB angle BMCSolutionOA=OB=AB radii and sides of rhombus∠AOB=60⇒∠AOC=120∠AMC=12×120=60°Specific behaviours indicates triangle OAB equilateral angle AOB angle BMC

(b) Use a vector method to prove that the diagonals of a rhombus are perpendicular.

(4 marks)

Let OC=c and OA=a.

SolutionAC=c-a
OB=c+a
AC∙OB=c-a∙c+a
=c2-a2
=0, as c=a (Side length)Hence AC⊥OBSpecific behaviours vectors for AC, OB forms scalar product simplifies scalar product, with reasons concludes perpendicularSolutionAC=c-a
OB=c+a
AC∙OB=c-a∙c+a
=c2-a2
=0, as c=a (Side length)Hence AC⊥OBSpecific behaviours vectors for AC, OB forms scalar product simplifies scalar product, with reasons concludes perpendicular

Question 16 (10 marks)

(a) The graph of y=cosax+b+c is shown below for 0≤x≤2π.

Determine the value of the positive constants a, b and c. (3 marks)

Solutiona=3, b=π3-π12=π4, c=12Specific behaviours value of a, value of b, value of cSolutiona=3, b=π3-π12=π4, c=12Specific behaviours value of a, value of b, value of c

(b) On the axes below, sketch the graph of y=2cosec(x-π), 0≤x≤2π. (3 marks)

SolutionSee graphSpecific behaviours asymptotic behaviour locations of max/min smooth curveSolutionSee graphSpecific behaviours asymptotic behaviour locations of max/min smooth curve

(c) The displacement, x cm, of a particle from a fixed point O varies with time, t seconds, according to the model x=4sin(3πt)-7cos3πt, t≥0. Determine

(i) the initial displacement of the particle from O. (1 mark)

Solutionx=-7Specific behaviours correct valueSolutionx=-7Specific behaviours correct value

(ii) the exact amplitude of the motion. (1 mark)

SolutionA=42+72=65 cmSpecific behaviours correct valueSolutionA=42+72=65 cmSpecific behaviours correct value

(iii) the period of motion. (1 mark)

SolutionP=2π3π=23 sSpecific behaviours correct valueSolutionP=2π3π=23 sSpecific behaviours correct value

(iv) the first time that the particle passes through O, rounded to two decimal places.

Solutiont=0.11 sSpecific behaviours correct valueSolutiont=0.11 sSpecific behaviours correct value (1 mark)

Question 17 (4 marks)

A number is formed using five different digits chosen from those in the number 681 429. Determine how many different numbers can be formed that are

SolutionnA=2×5×4×3×2=240Specific behaviours states numberSolutionnA=2×5×4×3×2=240Specific behaviours states number(i) odd. (1 mark)

(ii) greater than 90 000. (1 mark)

SolutionnB=1×5×4×3×2=120Specific behaviours states numberSolutionnB=1×5×4×3×2=120Specific behaviours states number

(iii) odd or greater than 90 000. (2 marks)

SolutionnA∩B=1×1×4×3×2=24nA∪B=240+120-24=336336 numbersSpecific behaviours calculates number odd and greater than 90 000 states numberSolutionnA∩B=1×1×4×3×2=24nA∪B=240+120-24=336336 numbersSpecific behaviours calculates number odd and greater than 90 000 states number

Question 18 (9 marks)

Triangle ABC has vertices A1, -2, B5, 2 and C4, -4.

(a) The vertices ABC are transformed to A'B'C' using matrix 100-1. Write down the new coordinates of the vertices and describe the transformation. (4 marks)

Solution100-1154-22-4=1542-24A'1, 2, B'5, -2, C'4, 4Transformation is a reflection in the line y=0.Specific behaviours matrix product writes as coordinates states reflection states equation of line of reflectionSolution100-1154-22-4=1542-24A'1, 2, B'5, -2, C'4, 4Transformation is a reflection in the line y=0.Specific behaviours matrix product writes as coordinates states reflection states equation of line of reflection

(b) The vertices ABC are transformed to A''B''C'' by matrix M so that the new coordinates of the vertices are A''-6, -4, B''6, -20 and C''-12, -16.

(i) Determine the transformation matrix M. (3 marks)

SolutionM15-22=-66-4-20M=-66-4-2015-22-1M=03-40Specific behaviours writes matrix equation uses inverse determines MSolutionM15-22=-66-4-20M=-66-4-2015-22-1M=03-40Specific behaviours writes matrix equation uses inverse determines M

SolutionM=12New area=12kSpecific behaviours determinant of M expresses areaSolutionM=12New area=12kSpecific behaviours determinant of M expresses area(ii) If the area of triangle ABC is k square units, express the area of triangle A''B''C'' in terms of k. (2 marks)

Question 19 (10 marks)

(a) Trapezium OPQR has parallel sides PQ and OR. M is the midpoint of OQ and N lies on QR so that RN:NQ=1:2.

Given that OP=p, OR=r and PQ=2r, determine the following in terms of p and r.

(i) OM. (2 marks)

SolutionOM=12OP+PQ
=12p+2r
=12p+rSpecific behaviours indicates half of OQ correct vectorSolutionOM=12OP+PQ
=12p+2r
=12p+rSpecific behaviours indicates half of OQ correct vector

(ii) ON. (2 marks)

SolutionON=OR+13RQ
=OR+13RO+OP+PQ
=r+13-r+p+2r
=43r+13pSpecific behaviours indicates OR and third of RQ correct vectorSolutionON=OR+13RQ
=OR+13RO+OP+PQ
=r+13-r+p+2r
=43r+13pSpecific behaviours indicates OR and third of RQ correct vector

SolutionNM=OM-ON
=12p+r-43r+13p
=16p-13rSpecific behaviours indicates difference of (i) and (ii) correct vectorSolutionNM=OM-ON
=12p+r-43r+13p
=16p-13rSpecific behaviours indicates difference of (i) and (ii) correct vector(iii) NM. (2 marks)

(b) Quadrilateral OABC is shown below, where P, Q, R and S are the midpoints of the sides OA, AB, BC and OC respectively. Let OP=a, AQ=b and OS=c.

Show that PS=QR. (4 marks)

SolutionPS=c-aOR=2a+2b+12BC
=2a+2b+12-2a-2b+2c
=a+b+cQR=OR-OQ
=a+b+c-(2a+b)
=c-a
=PSSpecific behaviours vector PS vector 12BC vector OR vector QRSolutionPS=c-aOR=2a+2b+12BC
=2a+2b+12-2a-2b+2c
=a+b+cQR=OR-OQ
=a+b+c-(2a+b)
=c-a
=PSSpecific behaviours vector PS vector 12BC vector OR vector QR

Question 20 (8 marks)

The sum of the first n terms of the sequence 1+5+9+13+…+(4n-3) is n(2n-1).

(a) Show that this statement is true when n=6. (2 marks)

SolutionLHS=1+5+9+13+17+21=66RHS=626-1=6×11=66Hence statement trueSpecific behaviours shows sum of terms for LHS shows substitution in RHS and states trueSolutionLHS=1+5+9+13+17+21=66RHS=626-1=6×11=66Hence statement trueSpecific behaviours shows sum of terms for LHS shows substitution in RHS and states true

(b) Use mathematical induction to prove the statement is true for n∈Z ,n≥6. (6 marks)

SolutionAssume statement true when n=k:1+5+9+13+…+4k-3=k(2k-1)When n=k+1:1+5+9+13+…+4k-3+4k-3+4=k2k-1+4k-3+4
=2k2+3k+1
=k+12k+1
=k+12k+1-1
=n2n-1 when n=k+1The statement is true for n=6 and by induction,the truth when n=k implies the truth when n=k+1and hence the statement is true for n≥6.Specific behaviours assumed true for n=k adds next term to both sides simplifies RHS factors k+1 out of RHS indicates true for n=k+1 summary statement including truth of n=6SolutionAssume statement true when n=k:1+5+9+13+…+4k-3=k(2k-1)When n=k+1:1+5+9+13+…+4k-3+4k-3+4=k2k-1+4k-3+4
=2k2+3k+1
=k+12k+1
=k+12k+1-1
=n2n-1 when n=k+1The statement is true for n=6 and by induction,the truth when n=k implies the truth when n=k+1and hence the statement is true for n≥6.Specific behaviours assumed true for n=k adds next term to both sides simplifies RHS factors k+1 out of RHS indicates true for n=k+1 summary statement including truth of n=6

Question 21 (6 marks)

The diagram shows a semi-circle, with diameter SR and centre O, circumscribed by triangle ABC, in which ∠BAC=48° and ∠BCA=36°.

Determine, with reasons, the size of angles ∠PRO and ∠PQR.

Solution∠AOP=90-48=42° (∠APO=90°, tangent-radius)∠PRO=42÷2=21° (centre ∠=2×circumference ∠)∠PSR=90-21=69° (as ∠SPR=90°, angle in semi-circle)∠PQR=180-69=111° (PQRS cyclic quadrilateral)Specific behaviours ∠AOP ∠PRO reasoning ∠PSR ∠PQR reasoningSolution∠AOP=90-48=42° (∠APO=90°, tangent-radius)∠PRO=42÷2=21° (centre ∠=2×circumference ∠)∠PSR=90-21=69° (as ∠SPR=90°, angle in semi-circle)∠PQR=180-69=111° (PQRS cyclic quadrilateral)Specific behaviours ∠AOP ∠PRO reasoning ∠PSR ∠PQR reasoning

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