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Semester Two Examination, 2018

Question/Answer booklet

SOLUTIONSSOLUTIONSMATHEMATICS

SPECIALIST

UNITS 1 AND 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.

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. Supplementary pages for the use of planning/continuing your answer to a question
have been 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.

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 One: Calculator-free 35% (52 Marks)

This section has eight (8) questions. Answer all questions. Write your answers in the spaces provided.

Working time: 50 minutes.

Question 1 (7 marks)

Let A=2-1-30 and B=5476.

Determine

SolutionAA-1=I=1001Specific behaviours identity matrixSolutionAA-1=I=1001Specific behaviours identity matrix(a) AA-1. (1 mark)

Solution2×2-1-30+5476=4-2-60+5476=9216Specific behaviours multiple of A correct sumSolution2×2-1-30+5476=4-2-60+5476=9216Specific behaviours multiple of A correct sum(b) 2A+B. (2 marks)

(c) AB. (2 marks)

Solution2-1-305476=32-15-12Specific behaviours at least two correct elements correct productSolution2-1-305476=32-15-12Specific behaviours at least two correct elements correct product

Solution5476-1=130-28×6-4-75=3-2-3.52.5Specific behaviours indicates determinant correct inverseSolution5476-1=130-28×6-4-75=3-2-3.52.5Specific behaviours indicates determinant correct inverse(d) B-1. (2 marks)

Question 2 (4 marks)

z1 and z2 are the complex solutions to the equation z2-2z+10=0.

The location of z1 is shown in the complex plane below.

Add, with a label, the following complex numbers to the complex plane above.

Solutionz2=z1=1-3i
u=2
v=i1+3i=-3+i
w=-1+i=-1-iSpecific behaviours plots z2 plots u plots v plots wSolutionz2=z1=1-3i
u=2
v=i1+3i=-3+i
w=-1+i=-1-iSpecific behaviours plots z2 plots u plots v plots w

(a) z2. (1 mark)

(b) u=z1+z2. (1 mark)

(c) v=iz1. (1 mark)

(d) w=u+v. (1 mark)

Question 3 (6 marks)

(a) Draw a neat diagram to illustrate the angle in the alternate segment theorem. (2 marks)

SolutionSpecific behaviours diagram correctly shows circle, tangent and triangle  clearly marks angles that are equalSolutionSpecific behaviours diagram correctly shows circle, tangent and triangle  clearly marks angles that are equal

(b) Consider the statement 'if a pentagon is regular, then it has sides of equal length'.

(i) Write the inverse of the statement. (1 mark)

SolutionIf a pentagon is not regular, then it does not have sides of equal length.Specific behaviours changes P⇒Q to P⇒QSolutionIf a pentagon is not regular, then it does not have sides of equal length.Specific behaviours changes P⇒Q to P⇒Q

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

SolutionIf a pentagon does not have sides of equal length, then it is not regular.Specific behaviours changes P⇒Q to Q⇒PSolutionIf a pentagon does not have sides of equal length, then it is not regular.Specific behaviours changes P⇒Q to Q⇒P

(iii) Write the converse of the statement and neatly sketch a counter-example to show that the converse is not true. (2 marks)

SolutionIf a pentagon has sides of equal length, then it is regular.Specific behaviours reverses P⇒Q to Q⇒P diagram shows irregular pentagon with equal sidesSolutionIf a pentagon has sides of equal length, then it is regular.Specific behaviours reverses P⇒Q to Q⇒P diagram shows irregular pentagon with equal sides

Question 4 (8 marks)

(a) On the axes below, sketch the graphs of y=tanx2 and y=cotx2, where 0≤x≤360°, labelling all asymptotes. Note that tan76°≈4. (5 marks)

SolutionSee graphSpecific behaviours tan graph through A, B, D, E asymptote for tan graph  cot graph through B, C, D two asymptotes for cot graph accurately uses tan76°≈4 for bothNB solution assumes cot=cos/sin, not 1/tanSolutionSee graphSpecific behaviours tan graph through A, B, D, E asymptote for tan graph  cot graph through B, C, D two asymptotes for cot graph accurately uses tan76°≈4 for bothNB solution assumes cot=cos/sin, not 1/tan

(b) Solve the equation sec2(x)=3tan2(x)-1 over the interval 0≤x≤360°. (3 marks)

Solution1+tan2(x)=3tan2(x)-1tan2x=1tanx=±1x=45°, 135°, 225°, 315°Specific behaviours uses Pythagorean identity correct values for tanx correct solutionsSolution1+tan2(x)=3tan2(x)-1tan2x=1tanx=±1x=45°, 135°, 225°, 315°Specific behaviours uses Pythagorean identity correct values for tanx correct solutions

Question 5 (6 marks)

The position vectors of points P and Q are p=31 and q=1-1 respectively.

(a) Determine the magnitude of the displacement vector PQ. (2 marks)

SolutionPQ=1-1-31=-2-2PQ=22Specific behaviours correct PQ correct magnitudeSolutionPQ=1-1-31=-2-2PQ=22Specific behaviours correct PQ correct magnitude

(b) Determine the values of λ so that p-λq=4. (4 marks)

Solutionp-λq=3-λ1+λ∴3- λ2+1+λ2=16λ2-6λ+9+λ2+2λ+1=162λ2-4λ-6=0λ2-2λ-3=0λ+1λ-3=0λ=-1, λ=3Specific behaviours expression for vector in terms of λ equation for magnitude of vector simplifies and factorises equation states both values for λSolutionp-λq=3-λ1+λ∴3- λ2+1+λ2=16λ2-6λ+9+λ2+2λ+1=162λ2-4λ-6=0λ2-2λ-3=0λ+1λ-3=0λ=-1, λ=3Specific behaviours expression for vector in terms of λ equation for magnitude of vector simplifies and factorises equation states both values for λ

Question 6 (8 marks)

Let fx=2sinx+2cosx.

(a) Express f(x) in the form rsinx+α where 0<α<π2. (4 marks)

Solutionr=22+22=22sinx+α=222sinx+222cosx
=sinxcosα+cosxsinα∴cosα=222=12, sinα=222=12⇒α=π4fx=22sinx+π4Specific behaviours value of r values of sinα and cosα value of α expression for f(x)Solutionr=22+22=22sinx+α=222sinx+222cosx
=sinxcosα+cosxsinα∴cosα=222=12, sinα=222=12⇒α=π4fx=22sinx+π4Specific behaviours value of r values of sinα and cosα value of α expression for f(x)

(b) Sketch the graph of y=f(x). (4 marks)

SolutionSee graphSpecific behaviours y-intercept roots maximum and minimum smooth sinusoidal curveSolutionSee graphSpecific behaviours y-intercept roots maximum and minimum smooth sinusoidal curve

Question 7 (5 marks)

Let OAPB be a parallelogram where OA=a and OB=b.

Use a vector method to prove that the sum of the squares of the lengths of the diagonals of a parallelogram is equal to the sum of the squares of the lengths of the sides.

SolutionRTP:OP2+BA2=OA2+OB2+BP2+AP2Note:r2=r∙rLHS=OP2+BA2
=a+b2+a-b2
=a+b∙a+b+a-b∙a-b
= a∙a+a∙b+b∙a+b∙b+a∙a-a∙b-b∙a+b∙b
=a2+b2+a2+b2
=OA2+OB2+BP2+AP2
=RHSSpecific behaviours expresses OP and BA in terms of a and b uses scalar product to expand sums and differences simplifies scalar products as magnitudes expresses in terms of sides logical presentation of proofSolutionRTP:OP2+BA2=OA2+OB2+BP2+AP2Note:r2=r∙rLHS=OP2+BA2
=a+b2+a-b2
=a+b∙a+b+a-b∙a-b
= a∙a+a∙b+b∙a+b∙b+a∙a-a∙b-b∙a+b∙b
=a2+b2+a2+b2
=OA2+OB2+BP2+AP2
=RHSSpecific behaviours expresses OP and BA in terms of a and b uses scalar product to expand sums and differences simplifies scalar products as magnitudes expresses in terms of sides logical presentation of proof

Question 8 (8 marks)

(a) Determine all complex solutions to the equation z2-2z+5=0. (3 marks)

Solutionz-12-1=-5z-12=4i2
z-1=±2iz=1+2i, z=1-2iSpecific behaviours completes square simplifies square roots of both sides states both solutionsSolutionz-12-1=-5z-12=4i2
z-1=±2iz=1+2i, z=1-2iSpecific behaviours completes square simplifies square roots of both sides states both solutions

(b) Determine the values of the positive real constants p and q so that p-2i is a solution to the equation 2z2-qz+26=0. (5 marks)

Solution2p-2i2-qp-2i+26=0
2p2-4pi-4-pq+2qi+26=02p2-8pi-8-pq+2qi+26=0
2p2-pq+18+2q-8pi=0Im: 2q-8p=0⇒q=4pRe: 2p2-4p2+18=0⇒p2=9p>0⇒p=3, q=12Specific behaviours substitutes, expanding square fully expands and simplifies uses imaginary terms to express q in terms of p uses real terms to determine p2 states valuesSolution2p-2i2-qp-2i+26=0
2p2-4pi-4-pq+2qi+26=02p2-8pi-8-pq+2qi+26=0
2p2-pq+18+2q-8pi=0Im: 2q-8p=0⇒q=4pRe: 2p2-4p2+18=0⇒p2=9p>0⇒p=3, q=12Specific behaviours substitutes, expanding square fully expands and simplifies uses imaginary terms to express q in terms of p uses real terms to determine p2 states values

Supplementary page

Question number: _________

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