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Semester Two Examination, 2021

Question/Answer booklet

SOLUTIONSSOLUTIONSMATHEMATICS
SPECIALIST
UNITS 1&2

Section One:
Calculator-free

In words

Your name

Time allowed for this section

Reading time before commencing work: five minutes

Working time: fifty minutes

Materials required/recommended for this section

To be provided by the supervisor

This Question/Answer booklet

Formula sheet

To be provided by the candidate

Standard items: pens (blue/black preferred), pencils (including coloured), sharpener,
correction fluid/tape, eraser, ruler, highlighters

Special items: nil

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 One: Calculator-free 35% (50 Marks)

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

Working time: 50 minutes.

Question 1 (5 marks)

Let matrix A=-2031 and matrix B=2k-112k+1, where k is a constant.

(a) When k=-1, determine

SolutionAB=-2031×-3120
=6-2-73Specific behavioursü correct productSolutionAB=-2031×-3120
=6-2-73Specific behavioursü correct product(i) AB. (1 mark)

Solution3A-2B=-6093--6240
=0-253Specific behaviours correct scalar multiplesü correct differenceSolution3A-2B=-6093--6240
=0-253Specific behaviours correct scalar multiplesü correct difference(ii) 3A-2B. (2 marks)

(b) Determine the value(s) of k if matrix B is singular. (2 marks)

SolutiondetB=0
2k-1k+1-2=0
2k2+k-3=0
2k+3k-1=0
k=1, k=-32Specific behaviours equates expression for determinant to zeroü correct values of kSolutiondetB=0
2k-1k+1-2=0
2k2+k-3=0
2k+3k-1=0
k=1, k=-32Specific behaviours equates expression for determinant to zeroü correct values of k

Question 2 (6 marks)

(a) Sketch the graph of y=secx2 on the axes below for 0≤x≤4π. (3 marks)

SolutionSee graphSpecific behaviours asymptotesü y-intercept, maxü correct curvatureSolutionSee graphSpecific behaviours asymptotesü y-intercept, maxü correct curvature

(b) Prove the identity cosec2A-cot2A=tanA. (3 marks)

SolutionLHS=cosec2A-cot2A
=1sin2A-cos2Asin2A
=1-cos2Asin2A
=2sin2A2sinAcosA
=sinAcosA
=tanA
=RHSSpecific behaviours converts between reciprocal and basic trig functionsü writes as single fractionü uses double angle identities and simplifiesSolutionLHS=cosec2A-cot2A
=1sin2A-cos2Asin2A
=1-cos2Asin2A
=2sin2A2sinAcosA
=sinAcosA
=tanA
=RHSSpecific behaviours converts between reciprocal and basic trig functionsü writes as single fractionü uses double angle identities and simplifies

Question 3 (6 marks)

Let z1=5+3i and z2=5-i. Determine each of the following in the form a+bi.

Solution25+6i-5-i=5+7iSpecific behaviours correct resultSolution25+6i-5-i=5+7iSpecific behaviours correct result(a) 2z1-z2. (1 mark)

Solutioni5-3i=3+5iSpecific behavioursü correct resultSolutioni5-3i=3+5iSpecific behavioursü correct result(b) iz1. (1 mark)

Solution5+3i5-i=5+35i-5i+3
=8+25iSpecific behaviours correctly expandsü correct resultSolution5+3i5-i=5+35i-5i+3
=8+25iSpecific behaviours correctly expandsü correct result(c) z1×z2. (2 marks)

Solution5+3i5-i×5+i5+i=2+45i6
=13+253iSpecific behaviours uses conjugate correctlyü correct resultSolution5+3i5-i×5+i5+i=2+45i6
=13+253iSpecific behaviours uses conjugate correctlyü correct result(d) z1÷z2. (2 marks)

Question 4 (6 marks)

(a) Determine the value(s) of the constant t given that -24t8tt=2t33. (2 marks)

Solutiont2+8t=33
t2+8t-33=0
t+11t-3=0
t=-11, t=3Specific behaviours forms quadraticü correct valuesSolutiont2+8t=33
t2+8t-33=0
t+11t-3=0
t=-11, t=3Specific behaviours forms quadraticü correct values

(b) Determine A-1 when A=73-22. (2 marks)

Solution73-22-1=1202-327Specific behaviours indicates determinantü correct inverseSolution73-22-1=1202-327Specific behaviours indicates determinantü correct inverse

(c) Show use of matrix methods to solve the following system of linear equations:

7x+3y-25=0
2y-2x+10=0

Solution73-22xy=25-10xy=1202-32725-10
=4-1Specific behaviours writes as matrix equationü shows use of inverse and obtains correct solutionSolution73-22xy=25-10xy=1202-32725-10
=4-1Specific behaviours writes as matrix equationü shows use of inverse and obtains correct solution (2 marks)

Question 5 (6 marks)

(a) Using a product identity, or otherwise, evaluate sin5π12+sinπ12. (3 marks)

Solutionsin5π12+sinπ12=sin3π+2π12+sin3π-2π12
=2sin3π12cos2π12
=2×22×32
=62Specific behaviours indicates appropriate sum and differenceü uses identityü evaluatesSolutionsin5π12+sinπ12=sin3π+2π12+sin3π-2π12
=2sin3π12cos2π12
=2×22×32
=62Specific behaviours indicates appropriate sum and differenceü uses identityü evaluates

(b) Solve the equation 2sin22x=3cos2x, 0≤x≤2π. (3 marks)

Solution21-cos22x-3cos2x=0
2cos22x+3cos2x-2=0
2cos2x-1cos2x+2=0
cos2x=12
2x=π3,5π3,7π3,11π3
x=π6, 5π6, 7π6, 11π6Specific behaviours uses Pythagorean identity to form quadraticü factors quadratic and indicates one solutionü all correct solutionsSolution21-cos22x-3cos2x=0
2cos22x+3cos2x-2=0
2cos2x-1cos2x+2=0
cos2x=12
2x=π3,5π3,7π3,11π3
x=π6, 5π6, 7π6, 11π6Specific behaviours uses Pythagorean identity to form quadraticü factors quadratic and indicates one solutionü all correct solutions

Question 6 (7 marks)

(a) Determine all complex solutions to the equation z2-10z+27=0. (2 marks)

Solutionz-52=25-27
z-52=2i2
z=5±2iSpecific behaviours completes squareü both correct solutionsSolutionz-52=25-27
z-52=2i2
z=5±2iSpecific behaviours completes squareü both correct solutions

(b) z1=-4-i is a solution to fz=0, where f(z) is a real quadratic polynomial.

(i) State z2, another solution to fz=0. (1 mark)

Solutionz=-4+iSpecific behaviours correct complex numberSolutionz=-4+iSpecific behaviours correct complex number

(ii) Let z3=z2-z1. Plot and label z1, z2 and z3 in the complex plane below. (2 marks)

SolutionSee graphSpecific behaviours sketches roots as conjugate pairü correctly locates z3SolutionSee graphSpecific behaviours sketches roots as conjugate pairü correctly locates z3

(iii) Determine f(z), given that the coefficient of its z2 term is 1. (2 marks)

SolutionLet fz=z2+bz+c.Then b=-2-4=8 and c=42+12=17.fz=z2+8z+17Specific behaviours shows sum and product of roots (or product of factors)ü correct equationSolutionLet fz=z2+bz+c.Then b=-2-4=8 and c=42+12=17.fz=z2+8z+17Specific behaviours shows sum and product of roots (or product of factors)ü correct equation

Question 7 (6 marks)

Use mathematical induction to prove that 25n-5n is divisible by 9 for all integers n≥1.

SolutionLet fn=25n-5n.When n=1 thenf(1)=32-5
=27
=3×9 which is divisible by 9Assume true for n=k so that 25k-5k=9I for some integer I.When n=k+1 thenfk+1=25k+5-5k+1
=25∙25k-5∙5k
=329I+5k-5∙5k (using assumption)
=329I+275k
=932I+35k which is divisible by 9Hence f(n) is divisible by 9 for n=k+1 and as demonstrated divisible for n=1 then will be divisible for all n≥1.Specific behaviours demonstrates true for n=1ü makes assumption for n=kü expression for f(k+1)ü uses assumption to replace 25kü factors out 9 from f(k+1)ü concluding statementSolutionLet fn=25n-5n.When n=1 thenf(1)=32-5
=27
=3×9 which is divisible by 9Assume true for n=k so that 25k-5k=9I for some integer I.When n=k+1 thenfk+1=25k+5-5k+1
=25∙25k-5∙5k
=329I+5k-5∙5k (using assumption)
=329I+275k
=932I+35k which is divisible by 9Hence f(n) is divisible by 9 for n=k+1 and as demonstrated divisible for n=1 then will be divisible for all n≥1.Specific behaviours demonstrates true for n=1ü makes assumption for n=kü expression for f(k+1)ü uses assumption to replace 25kü factors out 9 from f(k+1)ü concluding statement

Question 8 (8 marks)

(a) Points A, B and C lie on a circle.

The tangent to the circle at A
intersects secant BC at point D.

Prove that AD2=BD×CD. (4 marks)

SolutionFirst prove that ΔADC~ΔBDA:∠ADC=∠BDA (common)
∠DAC=∠DBA (angles in opposite segments)Hence ΔADC~ΔBDA as two pairs of congruent angles.Using ratios of corresponding sides, ADBD=CDAD⇒AD2=BD×CD.Specific behaviours shows congruency of one pair of angles, with reasoning shows congruency of second pair of angles, with reasoningü establishes similarity, with reasoningü completes proof using ratio of sidesSolutionFirst prove that ΔADC~ΔBDA:∠ADC=∠BDA (common)
∠DAC=∠DBA (angles in opposite segments)Hence ΔADC~ΔBDA as two pairs of congruent angles.Using ratios of corresponding sides, ADBD=CDAD⇒AD2=BD×CD.Specific behaviours shows congruency of one pair of angles, with reasoning shows congruency of second pair of angles, with reasoningü establishes similarity, with reasoningü completes proof using ratio of sides

(b) Two unequal circles intersect at P and Q. A common tangent touches one circle at R and the other circle at S. PQ produced intersects RS at X. Prove that X bisects RS. (4 marks)

SolutionLet PQ and RS intersect at X.In smaller circle, RX2=XQ×XP (tangent secant theorem)In larger circle, SX2=XQ×XP (tangent secant theorem)Hence RX2=SX2⇒RX=SX and so PQ produced bisects RS.Specific behaviours labelled diagramü uses tangent-secant theorem in one circleü uses tangent-secant theorem in the other circleü concludes proofSolutionLet PQ and RS intersect at X.In smaller circle, RX2=XQ×XP (tangent secant theorem)In larger circle, SX2=XQ×XP (tangent secant theorem)Hence RX2=SX2⇒RX=SX and so PQ produced bisects RS.Specific behaviours labelled diagramü uses tangent-secant theorem in one circleü uses tangent-secant theorem in the other circleü concludes proof

Supplementary page

Question number: _________

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