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Semester One Examination, 2017

Question/Answer booklet

SOLUTIONSSOLUTIONSMATHEMATICS

SPECIALIST

UNIT 1

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% (52 Marks)

This section has six (6) questions. Answer all questions. Write your answers in the spaces provided.

Working time: 50 minutes.

Question 1 (7 marks)

It can be shown that for all n≥0,

n+1Pr=n+1n-r+1× nPr

(a) Show that the identity is true when n=4 and r=2. (2 marks)

SolutionLHS= 5P2=5×4×3×2×13×2×1=20RHS=53× 4P2=53×4×3×2×12×1=20Specific behaviours evaluates LHS evaluates RHSSolutionLHS= 5P2=5×4×3×2×13×2×1=20RHS=53× 4P2=53×4×3×2×12×1=20Specific behaviours evaluates LHS evaluates RHS

Given that 8P4=1 680, 12P5=95 040 and 12P6=665 280, evaluate

(b) 11P6. (2 marks)

Solution 12P6=126× 11P6
11P6=6652802=332 640Specific behaviours relates 12P6 and 11P6 evaluatesSolution 12P6=126× 11P6
11P6=6652802=332 640Specific behaviours relates 12P6 and 11P6 evaluates

(c) 10P4. (3 marks)

Solution 10P4=106× 9P4
=106×95× 8P4
=3×1680=5040Specific behaviours expresses in terms of 9P4 expresses in terms of 8P4 evaluatesSolution 10P4=106× 9P4
=106×95× 8P4
=3×1680=5040Specific behaviours expresses in terms of 9P4 expresses in terms of 8P4 evaluates

Question 2 (11 marks)

Three vectors are given by a=3i-5j, b=-2i+7j and c=6i+j.

(a) Determine

Solutiona+b+c=7i+3jSpecific behaviours states vectorSolutiona+b+c=7i+3jSpecific behaviours states vector(i) a+b+c. (1 mark)

Solutionc=36+1=37Specific behaviours states exact valueSolutionc=36+1=37Specific behaviours states exact value(ii) |c|. (1 mark)

Solution2a=6i-10j3b=-6i+21j2a+3b=11jSpecific behaviours determines scalar multiples determines sumSolution2a=6i-10j3b=-6i+21j2a+3b=11jSpecific behaviours determines scalar multiples determines sum(iii) 2a+3b. (2 marks)

(b) Determine the unit vector d that is parallel and in the same direction as b-a. (3 marks)

Solutiond=b-a=-5i+12jb-a=13d=-513i+1213jSpecific behaviours determines b-a determines magnitude states unit vectorSolutiond=b-a=-5i+12jb-a=13d=-513i+1213jSpecific behaviours determines b-a determines magnitude states unit vector

(c) Express c in terms of a and b. (4 marks)

SolutionLet c=xa+ybi-coeffs:3x-2y=6j-coeffs: -5x+7y=115x-10y=30-15x+21y=311y=33⇒y=3x=4c=4a+3bSpecific behaviours equates i-coeffs equates j-coeffs solves equations for first variable solves equations for second variable and states cSolutionLet c=xa+ybi-coeffs:3x-2y=6j-coeffs: -5x+7y=115x-10y=30-15x+21y=311y=33⇒y=3x=4c=4a+3bSpecific behaviours equates i-coeffs equates j-coeffs solves equations for first variable solves equations for second variable and states c

Question 3 (8 marks)

(a) Write the inverse of the following true statement and comment on the truth of the inverse statement. (2 marks)

"If the discriminant of the quadratic formula is zero, then the quadratic has just one real root."

SolutionIf the discriminant of the quadratic formula is not zero, then the quadratic does not have just one real root.Statement is true.Specific behaviours changes 'if P then Q' to 'if not P then not Q' indicates statement is trueSolutionIf the discriminant of the quadratic formula is not zero, then the quadratic does not have just one real root.Statement is true.Specific behaviours changes 'if P then Q' to 'if not P then not Q' indicates statement is true

(b) Write the converse of the following true statement and comment on the truth of the converse statement. (2 marks)

"If x>3 then x>2."

SolutionIf x>2 then x>3.Statement is false.Specific behaviours changes 'if P then Q' to 'if Q then P' indicates statement is falseSolutionIf x>2 then x>3.Statement is false.Specific behaviours changes 'if P then Q' to 'if Q then P' indicates statement is false

(c) Determine the truth of the following statements, using an example or counter-example to support each answer.

(i) If z∈R and z3 is an even number then z is an even number. (2 marks)

SolutionStatement is false.If z3=6 (even) then z=36 (irrational, not even).Specific behaviours states false supplies counter-exampleSolutionStatement is false.If z3=6 (even) then z=36 (irrational, not even).Specific behaviours states false supplies counter-example

(ii) If x,y∈Z and x>y then x2>y2. (2 marks)

SolutionStatement is false.If x=2, y=-3 then 2>-3 but 22≯(-3)2.Specific behaviours states false supplies counter-example using integersSolutionStatement is false.If x=2, y=-3 then 2>-3 but 22≯(-3)2.Specific behaviours states false supplies counter-example using integers

Question 4 (7 marks)

(a) A body moves from P(2, -3) to Q(-2, 1).

(i) Determine the displacement vector PQ in component form. (1 mark)

SolutionPQ=-2-21--3=-44Specific behaviours expresses in component formSolutionPQ=-2-21--3=-44Specific behaviours expresses in component form

(ii) Determine the magnitude of the vector PQ. (1 mark)

SolutionPQ=(-4)2+42=32=42Specific behaviours states magnitudeSolutionPQ=(-4)2+42=32=42Specific behaviours states magnitude

(b) A force of 6i-63j N acts on a body. Determine the magnitude of the force and the angle its direction makes with the positive x-axis. (2 marks)

SolutionF=6(i-3)j
F=12 N
θ=-60°Specific behaviours states magnitude states angleSolutionF=6(i-3)j
F=12 N
θ=-60°Specific behaviours states magnitude states angle

(c) A body moves with a velocity of 20 ms-1 at an angle of 135° with the positive x-axis. Express the velocity of the body in the form ai+bj, where a and b are constants.

Solutiona=20cos135=-2022=-102
b=20sin135=2022=102
v=-102i+102j m/sSpecific behaviours determines expressions for a and b simplifies a and b states in required formSolutiona=20cos135=-2022=-102
b=20sin135=2022=102
v=-102i+102j m/sSpecific behaviours determines expressions for a and b simplifies a and b states in required form (3 marks)

Question 5 (10 marks)

(a) In the diagram below, not drawn to scale, PQRS is a cyclic quadrilateral such that PS=QS, ∠RPQ=34° and ∠PQR is a right-angle.

Determine the sizes of

Solution∠PRQ=90-34=56°∠PSQ=∠PRQ=56°Specific behaviours determines ∠PRQ states ∠PSQSolution∠PRQ=90-34=56°∠PSQ=∠PRQ=56°Specific behaviours determines ∠PRQ states ∠PSQ

(i) ∠PSQ. (2 marks)

Solution∠SPQ=180-562=62°∠RPS=62-34=28°Specific behaviours determines ∠SPQ determines ∠RPSSolution∠SPQ=180-562=62°∠RPS=62-34=28°Specific behaviours determines ∠SPQ determines ∠RPS(ii) ∠RPS. (2 marks)

(b) In the circle with centre O drawn below, chord AC intersects chord BD at E. Explain, with reasoning, why triangles AED and BEC are similar. (3 marks)

Solution∠AED=∠BEC (vertically opposite angles)∠CBD=∠CAD (angles stand on same arc)Hence triangles are similar as three pairs of equal angles.Specific behaviours one pair of angles, with reason second pair of angles, with reason summary, using AAA reasoningSolution∠AED=∠BEC (vertically opposite angles)∠CBD=∠CAD (angles stand on same arc)Hence triangles are similar as three pairs of equal angles.Specific behaviours one pair of angles, with reason second pair of angles, with reason summary, using AAA reasoning

(c) Prove that when two chords of a circle intersect, the product of the lengths of the intervals on one chord equals the product of the lengths of the intervals on the other chord.

(3 marks)

SolutionUsing diagram from (b), where ∆AED~ΔBEC.Then from ratio of corresponding sides BEAE=CEDEHence, BE×DE=AE×CE and proof is complete.Specific behaviours uses diagram/similar triangles uses ratio of sides shows products are equalSolutionUsing diagram from (b), where ∆AED~ΔBEC.Then from ratio of corresponding sides BEAE=CEDEHence, BE×DE=AE×CE and proof is complete.Specific behaviours uses diagram/similar triangles uses ratio of sides shows products are equal

Question 6 (9 marks)

(a) Determine the number of different four-letter passwords that can be made by arranging a selection of four letters chosen from the list P, Q, R, R, R, R and S. (4 marks)

SolutionIn each case, ways to choose R's×ways to choose others×arrangements:1 R: 1×1×4!=242 R's: 1×3×4!2!=363 R's: 1×3×4!3!=124 R's: 1×1×1=1Number of different passwords is 73.Specific behaviours breaks in to cases calculates one case correctly calculates at least three cases correctly correct totalSolutionIn each case, ways to choose R's×ways to choose others×arrangements:1 R: 1×1×4!=242 R's: 1×3×4!2!=363 R's: 1×3×4!3!=124 R's: 1×1×1=1Number of different passwords is 73.Specific behaviours breaks in to cases calculates one case correctly calculates at least three cases correctly correct total

(b) Determine the number of positive integers between 1 and 240 inclusive that are not divisible by at least one of the integers 4, 5 or 6. (5 marks)

SolutionMultiples of 4, 5 or 6: 2404+2405+2406=60+48+40=148Multiples of 12, 20 or 30: 24012+24020+24030=20+12+10+8=40Multiples of 60: 24060=4Total divisible: 148-40+4=112Total not divisible: 240-112=128Specific behaviours uses multiples of one, two and three numbers uses LCM for two and three numbers calculates at least one set of multiples correctly uses inclusion-exclusion principle correct answerSolutionMultiples of 4, 5 or 6: 2404+2405+2406=60+48+40=148Multiples of 12, 20 or 30: 24012+24020+24030=20+12+10+8=40Multiples of 60: 24060=4Total divisible: 148-40+4=112Total not divisible: 240-112=128Specific behaviours uses multiples of one, two and three numbers uses LCM for two and three numbers calculates at least one set of multiples correctly uses inclusion-exclusion principle correct answer

Additional working space

Question number: _________

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