WACEhub

Semester Two Examination, 2017

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. 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 One: Calculator-free 35% (48 Marks)

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

Working time: 50 minutes.

Question 1 (6 marks)

(a) Determine the values of the real constants b and c if z=1+3i is a solution of the equation z2+bz+c=0. (3 marks)

Solutionb=-1+3i+1-3i=-2c=1+3i1-3i=10Specific behaviours indicates other (conjugate) solution states value of b states value of cSolutionb=-1+3i+1-3i=-2c=1+3i1-3i=10Specific behaviours indicates other (conjugate) solution states value of b states value of c

(b) Express the real quadratic polynomial z2-4z+8 as a product of its linear factors.

Solutionz2-4z+8=z-22-4+8
=z-22--4
=z-22-2i2
=z-2-2iz-2+2iSpecific behaviours complete square express as difference of squares write as requiredSolutionz2-4z+8=z-22-4+8
=z-22--4
=z-22-2i2
=z-2-2iz-2+2iSpecific behaviours complete square express as difference of squares write as required (3 marks)

Question 2 (4 marks)

The complex numbers u and v are shown in the complex plane below.

Plot and label the following complex numbers:

(a) z1=u-v. (1 mark)

SolutionSee graphSpecific behaviours z1 z2 z3 z4SolutionSee graphSpecific behaviours z1 z2 z3 z4

(b) z2=2v+u. (1 mark)

(c) z3=u. (1 mark)

(d) z4=u-v-u-v. (1 mark)

Question 3 (6 marks)

(a) A set of real numbers is given by 2, 3.14, π, 314 . Clearly show that one of the numbers in the set is rational. (3 marks)

Solutionlet x=3.14Then 100x-x=314.14-3.1499x=311x=31199 and hence is rationalSpecific behaviours chooses rational number indicates use of 100x-x writes as rationalSolutionlet x=3.14Then 100x-x=314.14-3.1499x=311x=31199 and hence is rationalSpecific behaviours chooses rational number indicates use of 100x-x writes as rational

(b) Show that if n is one more than a multiple of three, then n2 will also be one more than a multiple of three, where n∈Z. (3 marks)

SolutionLet n=3k+1, k∈ZThen n2=9k2+6k+1
=33k2+2k+1Hence trueSpecific behaviours writes n in required form squares n writes n2 in required form SolutionLet n=3k+1, k∈ZThen n2=9k2+6k+1
=33k2+2k+1Hence trueSpecific behaviours writes n in required form squares n writes n2 in required form

Question 4 (9 marks)

Let A=8352 and B=611-37.

(a) Determine

Solution3A-B=249156-611-37=18-218-1Specific behaviours multiple of A differenceSolution3A-B=249156-611-37=18-218-1Specific behaviours multiple of A difference

(i) 3A-B. (2 marks)

SolutionBA=611-378352=391092469Specific behaviours at least two elements correct all elements correctSolutionBA=611-378352=391092469Specific behaviours at least two elements correct all elements correct(ii) BA. (2 marks)

SolutionA=16-15=1A-1=2-3-58Specific behaviours indicates use of determinant correct inverseSolutionA=16-15=1A-1=2-3-58Specific behaviours indicates use of determinant correct inverse(iii) A-1. (2 marks)

(b) Use a matrix method to solve the system of equations 8x+3y=10 and 5x+2y=7.

(3 marks)

Solution8352xy=107xy=2-3-58107xy=-16Specific behaviours writes as matrix equation shows correct use of inverse states solutionSolution8352xy=107xy=2-3-58107xy=-16Specific behaviours writes as matrix equation shows correct use of inverse states solution

Question 5 (4 marks)

Prove that 1+cosxsinx+tanx=cotx. (4 marks)

SolutionLHS=1+cosxsinx+sinxcosx
=1+cosx1÷sinxcosx+sinxcosx
=cosx1+cosxsinx1+cosx
=cotx
Specific behaviours eliminates tanx and combines denominator into single fraction divides and simplifies factorises numerator and denominator simplifies and writes as requiredSolutionLHS=1+cosxsinx+sinxcosx
=1+cosx1÷sinxcosx+sinxcosx
=cosx1+cosxsinx1+cosx
=cotx
Specific behaviours eliminates tanx and combines denominator into single fraction divides and simplifies factorises numerator and denominator simplifies and writes as required

Question 6 (8 marks)

Relative to the origin O, the points A, B and C have position vectors a=5i-6j, b=i-3j and c=-8i+15j respectively.

(a) Determine in Cartesian form

(i) the vector AB. (1 mark)

SolutionAB=b-a
=-4i+3jSpecific behaviours subtracts position vectorsSolutionAB=b-a
=-4i+3jSpecific behaviours subtracts position vectors

(ii) a vector d, parallel to AB and of magnitude 5. (3 marks)

SolutionAB=5d=55-4i+3jSpecific behaviours states magnitude indicates unit vector states required vector (in either direction)SolutionAB=5d=55-4i+3jSpecific behaviours states magnitude indicates unit vector states required vector (in either direction)

(b) If c=λa+μb, determine the values of the constants λ and μ. (4 marks)

Solutioni-coeff: 5λ+μ=-8j-coeff: -6λ-3μ=1515λ+3μ=-24-6λ-3μ=159λ=-9⇒λ=-1μ=-8+5=-3Specific behaviours uses coefficients to form equations uses elimination or substitution states λ states μSolutioni-coeff: 5λ+μ=-8j-coeff: -6λ-3μ=1515λ+3μ=-24-6λ-3μ=159λ=-9⇒λ=-1μ=-8+5=-3Specific behaviours uses coefficients to form equations uses elimination or substitution states λ states μ

Question 7 (6 marks)

Let z1 and z2 be complex numbers such that 3z1-2z2=7 and z1+iz2=3i.

Determine z1 and z2 in the form z=a+bi, where a, b∈Z.

Solution3z1-2z2=7
3z1+3iz2=9iz22+3i=-7+9iz2=-7+9i2+3iz2=-7+9i2+3i×2-3i2-3iz2=13+39i13z2=1+3iz1=3i-i(1+3i)z1=3+2iSpecific behaviours eliminate z1 express z2 as quotient realise denominator state z2 substitute for z1 state z1Solution3z1-2z2=7
3z1+3iz2=9iz22+3i=-7+9iz2=-7+9i2+3iz2=-7+9i2+3i×2-3i2-3iz2=13+39i13z2=1+3iz1=3i-i(1+3i)z1=3+2iSpecific behaviours eliminate z1 express z2 as quotient realise denominator state z2 substitute for z1 state z1

Question 8 (5 marks)

Cyclic quadrilateral ABCD has diagonals AC and BD that intersect at M. Given that BM=12 cm, DM=7 cm and AC=20 cm, determine the largest possible length of AM.

SolutionLet AM=x, then MC=20-xx20-x=12×7=84x2-20x-84=0x-102=16x=10±4⇒AM=14 cmSpecific behaviours annotated diagram expressions for AM and MC uses intersecting chord theorem to form equation factors equation states required solutionSolutionLet AM=x, then MC=20-xx20-x=12×7=84x2-20x-84=0x-102=16x=10±4⇒AM=14 cmSpecific behaviours annotated diagram expressions for AM and MC uses intersecting chord theorem to form equation factors equation states required solution

Additional working space

Question number: _________