WACEhub

Applecross Senior High School

Semester One Examination, 2019

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 preferably using a blue/black pen.
Do not use erasable or gel pens.

3. You must be careful to confine your answer to the specific question asked and to follow any instructions that are specified 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% (52 Marks)

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

Working time: 50 minutes.

Question 1 (4 marks)

In the diagram below (not drawn to scale) A, B and C lie on the circle with centre O and OA is parallel to CB.

Determine, with reasons, the size of ∠OBA and the size of ∠ABC when ∠OAC=23°.

Solution∠ACB=∠OAC=23 (Alternate angles)∠AOB=2×∠ACB=46 (Angle at centre)∠OBA=180-46÷2=67° (Isosceles)∠OBC=∠BOA=46 (Alternate angles)∠ABC=46+67=113°Specific behaviours ∠ACB with reason ∠AOB with reason ∠OBA with reason ∠ABCSolution∠ACB=∠OAC=23 (Alternate angles)∠AOB=2×∠ACB=46 (Angle at centre)∠OBA=180-46÷2=67° (Isosceles)∠OBC=∠BOA=46 (Alternate angles)∠ABC=46+67=113°Specific behaviours ∠ACB with reason ∠AOB with reason ∠OBA with reason ∠ABC

Question 2 (8 marks)

Let a=4i-8j, b=-3i+6j and c=2i+3j.

(a) Determine

Solution-36-23=-53Specific behaviours correct vectorSolution-36-23=-53Specific behaviours correct vector(i) b-c. (1 mark)

Solution3-36+44-8=-918+16-32=7-14Specific behaviours determines scalar multiples correct vectorSolution3-36+44-8=-918+16-32=7-14Specific behaviours determines scalar multiples correct vector(ii) 3b+4a. (2 marks)

Solution4-8+23=6-562+-52=61Specific behaviours determines sum correct valueSolution4-8+23=6-562+-52=61Specific behaviours determines sum correct value(iii) |a+c|. (2 marks)

(b) Determine a unit vector that is parallel to a+b but in the opposite direction. (3 marks)

Solution-4-8+-36=-12-12=5Soln:15-12Specific behaviours determines -(a+b) determines magnitude correct unit vectorSolution-4-8+-36=-12-12=5Soln:15-12Specific behaviours determines -(a+b) determines magnitude correct unit vector

Question 3 (6 marks)

(a) Body A moves 40 m on a bearing of 315°. Express this displacement in component form using unit vectors i and j. (3 marks)

Solution∠x-axis=135°r=40cos135°i+40sin135°j
=-202i+202jSpecific behaviours correct angle from x-axis correct i-coefficient correct j-coefficientSolution∠x-axis=135°r=40cos135°i+40sin135°j
=-202i+202jSpecific behaviours correct angle from x-axis correct i-coefficient correct j-coefficient

(b) Body B moves with a velocity of 43i-4j ms-1. Determine the speed of this body and the bearing it is travelling in. (3 marks)

Solutions2=432+42
s=8 m/s∠x-axis=-30°Bearing=90+30=120°Specific behaviours correct speed angle with x-axis correct bearingSolutions2=432+42
s=8 m/s∠x-axis=-30°Bearing=90+30=120°Specific behaviours correct speed angle with x-axis correct bearing

Question 4 (7 marks)

Quadrilateral ABCD is shown below. The midpoints of sides AB, BC, CD and DA are P, Q, R and S respectively. Let AB=2b, AC=2c and AD=2d.

Solution (a)See diagramSpecific behaviours correct quadrilateralSolution (a)See diagramSpecific behaviours correct quadrilateral

(a) Sketch quadrilateral PQRS on the diagram above. (1 mark)

(b) Determine expressions for AQ, AR and QR in terms of b, c and d. (3 marks)

SolutionAQ=2b+122c-2b=c+bAR=2d+122c-2d=c+dQR=QA+AR
=-c-b+c+d=d-bSpecific behaviours derives expression for AQ derives expression for AR derives expression for QRSolutionAQ=2b+122c-2b=c+bAR=2d+122c-2d=c+dQR=QA+AR
=-c-b+c+d=d-bSpecific behaviours derives expression for AQ derives expression for AR derives expression for QR

(c) Prove that PQ=SR and PS=QR. (3 marks)

SolutionPS=AS-AP
=d-b
=QRPQ=AQ-AP
= c+b-b
=cSR=AR-AS
=c+d-d
=c
=PQSpecific behaviours derives expression for PS and equates to QR derives expression for PQ derives expression for SR and equates to PQSolutionPS=AS-AP
=d-b
=QRPQ=AQ-AP
= c+b-b
=cSR=AR-AS
=c+d-d
=c
=PQSpecific behaviours derives expression for PS and equates to QR derives expression for PQ derives expression for SR and equates to PQ

Question 5 (6 marks)

Consider the following statement that refers to two isosceles triangles.

If the triangles have the same area, then the triangles are congruent.

(a) Write the inverse statement and state whether it is true or false. (2 marks)

SolutionIf the triangles do not have the same area, then the triangles are not congruent.This statement is true.Specific behaviours correct inverse statement states trueSolutionIf the triangles do not have the same area, then the triangles are not congruent.This statement is true.Specific behaviours correct inverse statement states true

(b) Write the converse statement and state whether it is true or false. (2 marks)

SolutionIf the triangles are congruent, then the triangles have the same area.This statement is true.Specific behaviours correct inverse statement states trueSolutionIf the triangles are congruent, then the triangles have the same area.This statement is true.Specific behaviours correct inverse statement states true

(c) Write the contrapositive statement and use a counter-example to explain why it is false.

(2 marks)

SolutionIf the triangles are not congruent, then the triangles do not have the same area.The isosceles triangles shown are not congruent but have the same area.Specific behaviours correct contrapositive statement correct example that uses isosceles trianglesOr  shows dimensions that give same areaSolutionIf the triangles are not congruent, then the triangles do not have the same area.The isosceles triangles shown are not congruent but have the same area.Specific behaviours correct contrapositive statement correct example that uses isosceles trianglesOr  shows dimensions that give same area

Question 6 (7 marks)

(a) The work done, in joules, by a force of F Newtons in changing the displacement of an object by s metres, is given by the scalar product of F and s. Determine the work done by

(i) force F=10i+8j N that moves a small body from 2i-8j m to 15i+12j m.

Solution1512-2-8=1320w=108∙1320=130+160=290 JSpecific behaviours displacement vector correct work doneSolution1512-2-8=1320w=108∙1320=130+160=290 JSpecific behaviours displacement vector correct work done (2 marks)

(ii) a horizontal force of 30 N that pushes a small body 1.8 m up a slope inclined at 30° to the horizontal. (2 marks)

Solutionw=30×1.8×cos30
=30×1.8×32
=273 JSpecific behaviours uses correct expression correct work doneSolutionw=30×1.8×cos30
=30×1.8×32
=273 JSpecific behaviours uses correct expression correct work done

(b) Determine the vector projection of 2i+4j on -3i+4j. (3 marks)

Solution24∙-34=10-34∙-34=251025-34=-6585Specific behaviours scalar products substitutes into expression correct vector projectionSolution24∙-34=10-34∙-34=251025-34=-6585Specific behaviours scalar products substitutes into expression correct vector projection

Question 7 (6 marks)

In the diagram below (not drawn to scale), two circles intersect at F and G. AH is a tangent to the circle at H. AE is a straight line that cuts the circles at A, B, D and E and intersects chord GF at C. AB=8, GC=4.5, CF=2, AH=12 and BC<CE.

SolutionAH2=AB×AE
AE=122÷8=18BE=AE-AB
=18-8=10Specific behaviours justifies length of AE justifies length of BESolutionAH2=AB×AE
AE=122÷8=18BE=AE-AB
=18-8=10Specific behaviours justifies length of AE justifies length of BE(a) Deduce that BE=10. (2 marks)

(b) Determine BC and CD, justifying your answers. (4 marks)

SolutionBC×CE=GC×CFx=BCx10-x=4.5×2=9x2-10x+9=0x-1x-9=0x=BC=1AC×CD=GC×CF=9
CD=9÷8+1=1Specific behaviours justifies equation for BC length of BC justifies equation for CD length of CDSolutionBC×CE=GC×CFx=BCx10-x=4.5×2=9x2-10x+9=0x-1x-9=0x=BC=1AC×CD=GC×CF=9
CD=9÷8+1=1Specific behaviours justifies equation for BC length of BC justifies equation for CD length of CD

Question 8 (8 marks)

Solution 2020P2101× 20P1=2020!2018!÷101×20!19!
=2020×2019÷101×20
=2020×2019÷2020
=2019Specific behaviours expresses as factorials eliminates factorials correct valueSolution 2020P2101× 20P1=2020!2018!÷101×20!19!
=2020×2019÷101×20
=2020×2019÷2020
=2019Specific behaviours expresses as factorials eliminates factorials correct value(a) Evaluate 2020P2101× 20P19. (3 marks)

(b) Given that n+1Pr=k× nPr, determine the constant k in terms of n and/or r. (3 marks)

Solution n+1Pr=n+1!n+1-r!=n+1n!n+1-rn-r!∴k=n+1n+1-rSpecific behaviours expresses LHS using factorials factors out term from denominator correct expressionSolution n+1Pr=n+1!n+1-r!=n+1n!n+1-rn-r!∴k=n+1n+1-rSpecific behaviours expresses LHS using factorials factors out term from denominator correct expression

(c) Given that 14P12=43 589 145 600, determine 16P12. (2 marks)

Solution 16P12=164× 15P12=4×153× 14P12=20× 14P1220×43 589 145 600=871 782 912 000Specific behaviours correct multiplier correct valueSolution 16P12=164× 15P12=4×153× 14P12=20× 14P1220×43 589 145 600=871 782 912 000Specific behaviours correct multiplier correct value

Supplementary page

Question number: _________

© 2019 WA Exam Papers. Applecross Senior High School has a non-exclusive licence to copy and communicate this document for non-commercial, educational use within the school. No other copying, communication or use is permitted without the express written permission of WA Exam Papers. SN002-131-2.