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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% (98 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(4, -7) is translated by the column vectors xy and -811 to P'(17, -13). Determine the values of the constants x and y. (2 marks)

Solution4-7+xy+-811=17-13xy=17-13-4-7--811xy=21-17Specific behaviours value of x value of ySolution4-7+xy+-811=17-13xy=17-13-4-7--811xy=21-17Specific 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=3x followed by a rotation of 180° about the origin. Express matrix coefficients in exact form. (4 marks)

Solutiontanθ=3⇒θ=60°-100-1×cos120sin120sin120-cos120=12-32-32-12Specific behaviours matrix for reflection matrix for rotation multiplies in correct order correct matrixSolutiontanθ=3⇒θ=60°-100-1×cos120sin120sin120-cos120=12-32-32-12Specific behaviours matrix for reflection matrix for rotation 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 inverse of this statement. (1 mark)

SolutionIf chords of a circle are of not of equal length then they do not subtend equal angles at the centre.Specific behaviours writes converseSolutionIf chords of a circle are of not of equal length then they do not subtend equal angles at the centre.Specific behaviours writes converse

(ii) Draw a diagram to illustrate the inverse statement and state whether it is true.

SolutionInverse is true.Specific behaviours diagram states inverse trueSolutionInverse is true.Specific behaviours diagram states inverse 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 C, ∠PCD=26°, ∠QCB=37° and ∠ADB=65°.

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

Solutionx=180-180-26-37=63°y=180-63-65=52°z=∠PCD=26°Specific behaviours each angle Solutionx=180-180-26-37=63°y=180-63-65=52°z=∠PCD=26°Specific behaviours each angle

Question 12 (9 marks)

(a) If p=13i-11j and q=15i+4j determine

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

SolutionUsing CAS angle is 55.168≈55.2° (1dp)Specific behaviours states angle rounds correctlySolutionUsing CAS angle is 55.168≈55.2° (1dp)Specific behaviours states angle rounds correctly (2 marks)

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

Solution132+(-11)2×cos55.168≈9.73Exact: 151241241Specific behaviours expression for projection states valueSolution132+(-11)2×cos55.168≈9.73Exact: 151241241Specific behaviours expression for projection states value

(b) The vector 45i-4aj has a magnitude of 53 and is perpendicular to the vector 3i-5bj. Determine the values of the constants a and b, where a>b. (5 marks)

Solution452+(-4a)2=532a=±7453+(-4a)(-5b)=0135+20(±7)b=0b=∓2728a=7, b=-2728Specific behaviours uses magnitude to form equation calculate values of a uses dot product to form equation calculate values of b chooses correct pairingSolution452+(-4a)2=532a=±7453+(-4a)(-5b)=0135+20(±7)b=0b=∓2728a=7, b=-2728Specific 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 (7 marks)

(a) Point R lies on the circumference of a circle with diameter PQ=51 cm, so that PR=4RQ. Determine the exact length RQ. (3 marks)

SolutionLet RQ=x, then x2+(4x)2=512x=317 cmSpecific behaviours diagram uses Pythagoras' Theorem states lengthSolutionLet RQ=x, then x2+(4x)2=512x=317 cmSpecific behaviours diagram uses Pythagoras' Theorem states length

(b) Use a vector method to prove that the angle in a semi-circle is a right-angle. (4 marks)

Let OC=c and OB=b.

SolutionAC=b+c
BC=c-b
AC∙BC=b+c∙c-b
=c2-b2
=0, as c=b=radiusHence ∠ACB=90°Specific behaviours vectors for AC, BC forms scalar product simplifies scalar product, with reasons concludes angle is rightSolutionAC=b+c
BC=c-b
AC∙BC=b+c∙c-b
=c2-b2
=0, as c=b=radiusHence ∠ACB=90°Specific behaviours vectors for AC, BC forms scalar product simplifies scalar product, with reasons concludes angle is right

Question 14 (6 marks)

(a) Prove that sin3A=3sinA-4sin3A. (4 marks)

SolutionLHS=sin3A
=sinA+2A
=sinAcos2A+cosAsin2A
=sinA1-2sin2A+cosA2sinAcosA
=sinA-2sin3A+2sinA1-sin2A
=3sinA-4sin3A
=RHSSpecific behaviours expands sum of A and 2A uses double angle identity for cos2A uses double angle identity for sin2A expands and simplifiesSolutionLHS=sin3A
=sinA+2A
=sinAcos2A+cosAsin2A
=sinA1-2sin2A+cosA2sinAcosA
=sinA-2sin3A+2sinA1-sin2A
=3sinA-4sin3A
=RHSSpecific behaviours expands sum of A and 2A uses double angle identity for cos2A uses double angle identity for sin2A expands and simplifies

(b) Hence, or otherwise, solve 3sinA-4sin3A=12, 0≤A≤π3. (2 marks)

Solutionsin3A=12
3A=π6,5π6
A=π18,5π18Specific behaviours one correct solution all solutions within required domainSolutionsin3A=12
3A=π6,5π6
A=π18,5π18Specific behaviours one correct solution all solutions within required domain

Question 15 (8 marks)

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

(a) If F1=85 N, F2=105 N, α=21° and β=35°, determine the magnitude of the resultant force and the angle it makes with the positive x axis. (5 marks)

SolutionFR=852+1052-285105cos124
FR=168 Nsinθ105=sin124168
θ=31.2°ϕ=21°-31.2°=-10.2°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=852+1052-285105cos124
FR=168 Nsinθ105=sin124168
θ=31.2°ϕ=21°-31.2°=-10.2°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=145 N and F2=180 N, determine the angles α and β so that the resultant force is directed along the positive x axis and has a magnitude of 310 N. (3 marks)

Solutionα=cos-11452+3102-18022145310
α=19.5°β=cos-11802+3102-14522180310
β=15.6°Specific behaviours sketch with resultant on x axis uses cosine rule for angle uses sine or cosine rule for second angleSolutionα=cos-11452+3102-18022145310
α=19.5°β=cos-11802+3102-14522180310
β=15.6°Specific behaviours sketch with resultant on x axis uses cosine rule for angle uses sine or cosine rule for second angle

Question 16 (10 marks)

(a) The graph of y=sinax-b+c is shown below for -π≤x≤π.

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

Solutiona=2, b=π3, c=14Specific behaviours value of a, value of b, value of cSolutiona=2, b=π3, c=14Specific behaviours value of a, value of b, value of c

(b) On the axes below, sketch the graph of y=3secx-π2, 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=2sin(4πt)+3cos4πt, t≥0. Determine

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

Solutionx=3Specific behaviours correct valueSolutionx=3Specific behaviours correct value

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

SolutionA=22+32=13 cmSpecific behaviours correct valueSolutionA=22+32=13 cmSpecific behaviours correct value

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

SolutionP=2π4π=12 sSpecific behaviours correct valueSolutionP=2π4π=12 sSpecific behaviours correct value

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

Solutiont=0.17 sSpecific behaviours correct valueSolutiont=0.17 sSpecific behaviours correct value (1 mark)

Question 17 (9 marks)

Triangle ABC has vertices A-1, 5, B2, 7 and C4, 4.

(a) The vertices ABC are transformed to A'B'C' using matrix 0110. Write down the new coordinates of the vertices and describe the transformation. (4 marks)

Solution0110-124574=574-124A'5, -1, B'7, 2, C'4, 4Transformation is a reflection in the line y=x.Specific behaviours matrix product writes as coordinates states reflection states equation of line of reflectionSolution0110-124574=574-124A'5, -1, B'7, 2, C'4, 4Transformation is a reflection in the line y=x.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'' using matrix M so that the new coordinates of the vertices are A''25, 2, B''35, -4 and C''20, -8.

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

SolutionM-1257=25352-4M=25352-4-1257-1M=05-20Specific behaviours writes matrix equation uses inverse determines MSolutionM-1257=25352-4M=25352-4-1257-1M=05-20Specific behaviours writes matrix equation uses inverse determines M

SolutionM=10New area=10kSpecific behaviours determinant of M expresses areaSolutionM=10New area=10kSpecific 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 18 (7 marks)

(a) How many numbers must be chosen from the set of integers between 1 and 2017 inclusive to be certain that one of the numbers chosen is a multiple of 10. (3 marks)

Solution201 multiples of 10 between 1 to 2017.Require 2017-201=1816 pigeonholes.Hence must choose 1 817 integers.Specific behaviours states # of multiples in set one pigeonhole for every non-multiple uses pigeonhole principle to add oneSolution201 multiples of 10 between 1 to 2017.Require 2017-201=1816 pigeonholes.Hence must choose 1 817 integers.Specific behaviours states # of multiples in set one pigeonhole for every non-multiple uses pigeonhole principle to add one

(b) A number is formed using four different digits chosen from those in the number 23 814. Determine how many different numbers can be formed that are

SolutionnA=3×4×3×2=72Specific behaviours states numberSolutionnA=3×4×3×2=72Specific behaviours states number(i) even. (1 mark)

(ii) greater than 8 000. (1 mark)

SolutionnB=1×4×3×2=24Specific behaviours states numberSolutionnB=1×4×3×2=24Specific behaviours states number

(iii) even or greater than 8 000. (2 marks)

SolutionnA∩B=1×2×3×2=12nA∪B=72+24-12=8484 numbersSpecific behaviours calculates number even and greater than 8 000 states numberSolutionnA∩B=1×2×3×2=12nA∪B=72+24-12=8484 numbersSpecific behaviours calculates number even and greater than 8 000 states number

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=4:1.

Given that OP=p, OR=r and PQ=3r, determine the following in terms of p and r.

(i) OM. (2 marks)

SolutionOM=12OP+PQ
=12p+3r
=12p+32rSpecific behaviours indicates half of OP+PQ correct vectorSolutionOM=12OP+PQ
=12p+3r
=12p+32rSpecific behaviours indicates half of OP+PQ correct vector

(ii) ON. (2 marks)

SolutionON=OR+45RQ
=OR+45RO+OP+PQ
=r+45-r+p+3r
=135r+45pSpecific behaviours indicates OR and four-fifths of RQ correct vectorSolutionON=OR+45RQ
=OR+45RO+OP+PQ
=r+45-r+p+3r
=135r+45pSpecific behaviours indicates OR and four-fifths of RQ correct vector

SolutionNM=OM-ON
=12p+32r-135r+45p
=-310p-1110rSpecific behaviours indicates difference of (i) and (ii) correct vectorSolutionNM=OM-ON
=12p+32r-135r+45p
=-310p-1110rSpecific 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 PQ=SR. (4 marks)

SolutionPQ=a+bOR=2a+2b+12BC
=2a+2b+12-2a-2b+2c
=a+b+cSR=OR-OS
=a+b+c-c
=a+b
=PQSpecific behaviours vector PQ vector 12BC vector OR vector SRSolutionPQ=a+bOR=2a+2b+12BC
=2a+2b+12-2a-2b+2c
=a+b+cSR=OR-OS
=a+b+c-c
=a+b
=PQSpecific behaviours vector PQ vector 12BC vector OR vector SR

Question 20 (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

Question 21 (8 marks)

The sum of the first n terms of the sequence 1+11+21+…+(10n-9) is n(5n-4).

(a) Show that this statement is true when n=4. (2 marks)

SolutionLHS=1+11+21+31=64RHS=454-4=4×16=64Hence statement trueSpecific behaviours shows sum of terms for LHS shows substitution in RHS and states trueSolutionLHS=1+11+21+31=64RHS=454-4=4×16=64Hence 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≥4. (6 marks)

SolutionAssume statement true when n=k:1+11+21+…+10k-9=k(5k-4)When n=k+1:1+11+21+…+10k-9+10k-9+10=k5k-4+10k-9+10
=5k2+6k+1
=k+15k+1
=k+15k+1-4
=n5n-4 when n=k+1The statement is true for n=4 and by induction,the truth when n=k implies the truth when n=k+1and hence the statement is true for n≥4.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=4SolutionAssume statement true when n=k:1+11+21+…+10k-9=k(5k-4)When n=k+1:1+11+21+…+10k-9+10k-9+10=k5k-4+10k-9+10
=5k2+6k+1
=k+15k+1
=k+15k+1-4
=n5n-4 when n=k+1The statement is true for n=4 and by induction,the truth when n=k implies the truth when n=k+1and hence the statement is true for n≥4.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=4

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