2020 Maths Specialist Unit 2 Sem 2 Exam CA DRAFT.docx
Semester Two Examination, 2020
Question/Answer booklet
MATHEMATICS SPECIALIST
UNIT 1&2
Section Two:
Calculator-assumed
Your Name
Your Teacher’s Name
Time allowed for this section
Reading time before commencing work: ten minutes
Working time: one hundred minutes
Materials required/recommended for this section
To be provided by the supervisor
This Question/Answer booklet
Formula sheet (retained from Section One)
To be provided by the candidate
Standard items: pens (blue/black preferred), pencils (including coloured), sharpener, correction fluid/tape, eraser, ruler, highlighters
Special items: drawing instruments, templates, notes on two unfolded sheets of A4 paper, and up to three calculators approved for use in this examination
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.
| Question | Marks | Max | Question | Marks | Max |
| 9 | 10 | 16 | 6 | ||
| 10 | 6 | 17 | 6 | ||
| 11 | 5 | 18 | 8 | ||
| 12 | 9 | 19 | 8 | ||
| 13 | 7 | 20 | 8 | ||
| 14 | 10 | 21 | 7 | ||
| 15 | 10 |
Structure of this paper
| Section | Number of questions available | Number of questions to be answered | Working time (minutes) | Marks available | Percentage of examination |
| Section One:Calculator-free | 8 | 8 | 50 | 54 | 36 |
| Section Two:Calculator-assumed | 13 | 13 | 100 | 100 | 64 |
| Total | 100 |
Instructions to candidates
The rules for the conduct of the Western Australian Certificate of Education ATAR course examinations are detailed in the Year 12 Information Handbook 2019. Sitting this examination implies that you agree to abide by these rules.
Write your answers in this Question/Answer booklet.
You must be careful to confine your answers to the specific questions asked and to follow any instructions that are specific to a particular question.
Additional pages for the use of planning your answer to a question or continuing your answer to a question have been provided at the end of this Question/Answer booklet. If you use the space to continue an answer, indicate in the original answer space where the answer is continued, i.e. give the page number.
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.
It is recommended that you do not use pencil, except in diagrams.
The Formula sheet is not to be handed in with your Question/Answer booklet.
See Next Page
Section Two: Calculator-assumed (100 Marks)
This section has 13 (thirteen) questions. Answer all questions. Write your answers in the spaces provided.
Spare pages are included at the end of this booklet. They can be used for planning your responses and/or as additional space if required to continue an answer.
● Planning: If you use the spare pages for planning, indicate this clearly at the top of the page.
● Continuing an answer: If you need to use the space to continue an answer, indicate in the original answer space where the answer is continued, i.e. give the page number. Fill in the number of the question that you are continuing to answer at the top of the page.
Working time: 100 minutes.
Question 9 (2.2.1, 2.2.5-2.2.7, 2.2.9, 2.2.10) (10 marks)
Consider the triangle with vertices A(0,0), B(1,0) and C(1,4), plotted below.
The triangle is transformed by a matrix M to give an image with vertices A'(0,0), B'(0,1) and C'(-4,1). Write down the matrix M. (2 marks)
| Solution |
| M is a rotation by π2, and so M=0-110 |
| Specific behaviours |
| states transformation or sketches diagram states correct matrix |
Question 9 continued
Triangle A'B'C' (the image from part (a) ) is transformed by a matrix N to give an image with vertices A''(0,0), B''(0,-1) and C''(-4,-1). Write down the matrix N. (2 marks)
| Solution |
| N is a reflection through y=0, and so N=100-1 |
| Specific behaviours |
| states transformation or sketches diagram states correct matrix |
Hence write down the matrix P which would transform triangle ABC to triangle A''B''C'', showing your working. (3 marks)
| Solution |
| P=NM=100-10-110=0-1-10 |
| Specific behaviours |
| multiplies N and M (in either order) multiplies in correct order states correct matrix |
The triangle ABC is transformed by a matrix Q to a triangle with coordinates A'''(0,0), B'''(2,1) and C'''(2,7). State the value of detQ, given that detQ>0, justifying your answer.
(3 marks)
| (Solution |
| ΔABC has area 12×1×4=2 and ΔA'''B'''C''' has area 12×6×2=6. Hence detQ=62=3 and since detQ>0, it follows that detQ=3. |
| Specific behaviours |
| determines areas of ΔABC and ΔA'''B'''C''' divides 6 by 2 states detQ=3 |
Question 10 (1.1.1, 1.1.2, 1.1.3, 1.1.4) (6 marks)
The genetic code is a set of rules defined by the four nucleotides of DNA, represented by the letters A, T, C and G. Three-letter nucleotide sequences are made from the four nucleotides.
With no restrictions, how many 3-letter nucleotide sequences are possible in DNA?
(1 mark)
| Solution |
| 4×4×4=64 |
| Specific Behaviours |
| correct number |
How many 3-letter nucleotide sequences start with A and end with C? (2 marks)
| Solution |
| 1×4×1=4 |
| Specific Behaviours |
| uses multiplicative reasoning correct number |
How many 3-letter nucleotide sequences have a G at least twice? (3 marks)
| Solution |
| 1+31×3=10 |
| Specific Behaviours |
| identify two cases uses addition principle correct number |
Question 11 (2.3.7-2.311, 2.3.13-2.3.16) (5 marks)
Consider the following quadratic equation where c is a real number.
x2+12x+c=0
One of the solutions to this equation is z=-6+3i.
Write down the other solution w of the equation, and plot (and label) both solutions in the complex plane below. (3 marks)
| Solution |
| w=z=-6-3iRe(z)Im(z)Re(z)Im(z) |
| Specific behaviours |
| writes correct value for w plots and labels z in correct position plots and labels w in correct position |
Hence (or otherwise) determine the value of c. (2 marks)
| Solution |
| c=z×w=-6+3i-6-3i=36+9=45 |
| Specific behaviours |
| multiplies z and w states c=45 |
Question 12 (2.1.1-2.1.2, 2.1.9) (9 marks)
A roofing panel with the dimensions shown below has been left on the ground. An ant is walking across the top of the panel from the left end to the right end.
Write a function in the form
h=acos(b(x-c))+d
modelling the height h mm that the ant is above the ground in terms of the horizontal distance x mm that the ant is from the left end of the panel. Specify the domain of the function. (Assume the panel has negligible thickness.) (3 marks)
Graph the function on the axes below. (3 marks)
The ant gets tired and stops to rest the third time he is climbing at a height of 12 mm. How far (horizontally) does he have left to walk? (3 marks)
Question 13 (2.1.4) (7 marks)
Describe the transformation of y=cosecx to y=2cosecx-4. (2 marks)
Sketch y=2cosecx-4 on the graph shown, labelling all key features.
(5 marks)
Question 14 (1.2.6-1.2.13) (10 marks)
Three vectors are given by a=5i-12j, b=-15i+10j and c=-7i+yj where y is a constant.
Determine the vector projection of b on a (give components as exact values).
(3 marks)
| Solution |
| a=1135ⅈ-12jb⋅a=-15b⋅aa= -7513ⅈ+18013j |
| Specific behaviours |
| States unit vector for aStates b⋅aStates projection as a vector |
Find y if the angle between b and c is 45°. (3 marks)
| Solutioncos(45)=-15i+10j.(-7i+yj)|-15i+10j|.|-7i+yj|y=35 or y= -75Specific behavioursUses scalar productStates one solutionStates second solution |
Vectors ai+a-3j and (a-7)i+5j are perpendicular. Find the value(s) of a and the corresponding pairs of vectors. (4 marks)
| Solution |
| aa-3⋅a-75=a2-2a-15=0a+3a-5=0a=-3 or a=5if a=-3, the vectors are -3i-6j and -10i+5jif a=5, the vectors are 5+2j and -2i+5j |
| Specific behaviours |
| Uses (dot product = 0) to form quadratic equationSolves for two values of aStates one pair of vectorsStates two pairs of vectors |
Question 15 (1.3.6-1.3.15) (10 marks)
Consider the diagram below. AD and BE are diameters of the circle with centre O, ∠COD= 32° and C lies on the circumference of the circle such that AB=BC.
Determine the sizes of the following angles
∠AOB (2 marks)
| Solution |
| ∠AOB=∠BOC ∠AOB+∠BOC+32 =180∠AOB=180-322∠AOB=74° |
| Specific behaviours |
| Uses congruent anglesCalculates angle |
∠CAE (3 marks)
| Solution |
| ∠DOE=∠AOB=74°∠COE=74+32=106°∠CAE=106°2=53° |
| Specific behaviours |
| Indicates size of ∠DOEUses relationship between angle at centre and at circumferenceCalculates angle |
In the diagram below, W is the single point of intersection of the two circles. The segment XY is tangent to both circles, intersecting with the circles at X and Y. Segment WZ is also tangent to both circles, intersecting with XY at Z. Prove that ΔXWY is a right triangle. (5 marks)
| Solution |
| ZX=ZW and ZW=ZY (tangents from a common point)Hence X, W and Y lie on a circle with centre Z.XY is a diameter of this circle and so ∠XWY is an angle in a semicircle.Hence ∠XWY=90°.It follows that ΔXWY is a right triangle. |
| Specific behaviours |
| Notes that XY=ZW=ZY Gives reason for aboveStates that X, W and Y lie on a circle with centre ZStates that ∠XWY is an angle in a semicircleConcludes that ∠XWY=90° |
Question 16 (1.2.2, 1.2.8, 1.2.14) (6 marks)
Three forces act on the point A as shown. What is the magnitude of the resultant force acting on A, and in what direction would A move under these three forces?
Give your answers to 2 decimal places, with the direction as an angle measured anticlockwise from the right (like the 30° angle for Force 1).
| Solution |
| Force 1 =3cos30i+3sin30jForce 2 =4cos60i-4sin60jForce 3 =-5cos30i+5sin30jResultant force:F=332+42-532i+32-432+52j=2-3i+4-23jMagnitude of force:F=2-32+4-232≈0.60 NDirection of force:tan-14-232-3≈63.43°Therefore the direction is 63.43° |
| Specific behaviours |
| writes correct vector expression for at least one force (accept expressions using polar angles e.g. Force 2 =4cos300i+4sin300j) writes correct vector expressions for at least two forces writes correct vector expressions for all three forces determines vector expression for resultant force states correct magnitude of resultant force state correct direction of resultant force |
Question 17 (2.2.1, 2.2.2) (6 marks)
Given invertible n×n matrices A, B, C and X with AX-B=CBX, write X in terms of A, B and C.
(3 marks)
| Solution |
| AX-B=CBXAX-CBX=BA-CBX=BX=A-CB-1B |
| Specific behaviours |
| collects terms with X on LHS factorises X out on the right multiplies both sides by A-CB-1 |
Solve the following matrix equation for Y
3Y-Y1-316=5I
(3 marks)
| Solution |
| 3Y-Y1-316=5IY3I-1-316=5IY23-1-3=5IY=5I23-1-3-1=5333-1-2 |
| Specific behaviours |
| factorises Y out on the left evaluates 3I-1-316 multiplies both sides by 23-1-3-1 and obtains correct answer |
Question 18 (1.1.7, 1.1.8) (8 marks)
Four Year 10 students and eleven Year 11 students from Western Australia are nominated as candidates for a Mathematics Summer Camp. How many ways can a group of four participants be selected:
without restriction? (2 marks)
| Solution |
| 154=1365 |
| Specific Behaviours |
| correct expression correct number |
if the only student from Bunbury must be included? (2 marks)
| Solution |
| 11143=364 |
| Specific Behaviours |
| correct expression correct number |
if there must be exactly two Year 11 students? (2 marks)
| Solution |
| 11242=330 |
| Specific Behaviours |
| correct expression correct number |
if there must be at least one Year 10 student? (2 marks)
| Solution |
| 154-114=1035 |
| Specific Behaviours |
| correct expression correct number |
Question 19 (1.1.5, 1.1.9) (8 marks)
How many integers between 1 and 101 are multiples of 5, 6 or 7? (4 marks)
| Solution |
| Multiples of 5: 100÷5=20 Multiples of 6: 100÷6=16 (rounded down)Multiples of 7: 100÷7=14Multiples of 30 (5 and 6): 100÷30=3Multiples of 35 (5 and 7): 100÷35=2Multiples of 42 (6 and 7): 100÷42=2Multiples of 210 (5, 6 and 7): 0Multiples of 5, 6, or 7: 20+16+14-3-2-2+0=43 |
| Specific Behaviours |
| finds multiples of 5, 6, 7 respectively finds multiples of 30, 35, 42 respectively uses inclusion-exclusion principle correct number |
Use the fact that nCr=n!n-r!r! to show that n-1Cr-1×n=nCr×r. (4 marks)
| Solution |
| LHS=(n-1)![n-1-(r-1)]!(r-1)!×n =(n-1)!×n(n-r)!(r-1)! =n!(n-r)!(r-1)! =n!r(n-r)!r(r-1)! =n!(n-r)!r!×r =nCr×r =RHS |
| Specific Behaviours |
| uses nCr=n!n-r!r! writes n(n-1)! as n! multiplies by rr writes r(r-1)! as r! |
Question 20 (2.2.1-2.2.10) (8 marks)
Let A=cosα-sinαsinαcosα.
Calculate A2 (that is, A×A). Show working and simplify your answer. (3 marks)
| Solution |
| A2=cosα-sinαsinαcosαcosα-sinαsinαcosα=cos2α-sin2α-cosαsinα-cosαsinαcosαsinα+cosαsinα-sin2α+cos2α=cos2α-sin2αsin2αcos2α |
| Specific behaviours |
| writes unsimplified product (2nd line) with at least 2 entries correct writes unsimplified product (2nd line) with all entries correct simplifies using double angle formulas |
Calculate the product A3 by multiplying your answer to part (a) by A (you do not need to simplify your answer). (2 marks)
| Solution |
| A3=cos2α-sin2αsin2αcos2αcosα-sinαsinαcosα=cos2αcosα-sin2αsinα-cos2αsinα-sin2αcosαsin2αcosα+cos2αsinα-sin2αsinα+cos2αcosα |
| Specific behaviours |
| at least 2 entries correct in product all entries correct in product |
Determine a value of α (with 0<α<2π) such that A3=I. Justify your answer by referring to the linear transformation corresponding to the matrix A. (3 marks)
| Solution |
| If α=2π3 then A3=I.Since A represents a rotation by α, A3 represents 3 rotations by α applied in sequence; that is, A3 is a rotation by 3α. Thus if α=2π3, A3 is a rotation by 2π, which is equivalent to a rotation by 0, and therefore A3=I. |
| Specific behaviours |
| states α=2π3 notes that A represents a rotation by α notes that A3 is 3 rotations by α, or a single rotation by 3α |
Question 21 (1.2.2, 1.2.3, 1.2.7, 1.2.9, 1.2.14) (7 marks)
An octopus, which can swim with a steady speed of 3.5 m/s through still water, leaves its home at O to visit a sea anemone at S.
The position vector of S relative to O is 165i-212j m, and a current with velocity -2.3i+1.1j m/s is flowing.
Find the velocity vector v, in the form ai+bj, that the octopus should aim to swim with in order to reach the sea anemone in the shortest possible time. (Give a and b to 2 decimal places.) (5 marks)
| Solution |
| We requireai+bj+-2.3i+1.1j=λ165i-212jsoa-2.3=165λb+1.1= -212λWe also know thata2+b2=3.52Solving these three equations simultaneously and taking the solution with λ>0 givesv=2.93i-1.91j |
| Specific behaviours |
| equates sum of v and current velocity to a scalar multiple of displacement vector equates components to get 2 linear equations states equation a2+b2=3.52 states at least one of v=2.93i-1.91j and v=-1.13i+3.31j states v=2.93i-1.91j (i.e. solution corresponding to positive value of λ) |
Determine the time taken (to the nearest second) for the octopus to reach the sea anemone if it aims to swim with the velocity found in part (a). (2 marks)
| Solution |
| EITHERλ=3.8×10-3 (from part (a))Hence total time will be 1λ=261 sORHorizontal component of resultant velocity = 2.93-2.3=0.631650.63= 262 sORv+-2.3i+1.1j=1.0262165i-212j=268.6429Total time =268.64291.0262=262 s |
| Specific behaviours |
| shows appropriate calculation states 261 s or 262 s |
END OF QUESTIONS
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