2020 Maths Specialist Unit 2 Sem 2 Exam CF DRAFT (1).docx
Semester Two Examination, 2020
Question/Answer booklet
MATHEMATICS SPECIALIST
UNITS 1&2
Section One:
Calculator-free
Your Name_______________________________
Your Teacher’s 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.
| Question | Mark | Max | Question | Mark | Max |
| 1 | 4 | 6 | 5 | ||
| 2 | 9 | 7 | 7 | ||
| 3 | 7 | 8 | 5 | ||
| 4 | 9 | ||||
| 5 | 4 |
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 | 50 | 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 2020. 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 One: Calculator-free (50 marks)
This section has 8 (eight) 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: 50 minutes.
Question 1 (2.3.1) (4 marks)
Prove the following statement:
If m and n are odd integers, then m2-n2 is divisible by 4.
| Solution |
| Assume that m and n are both odd integers.Then m=2k+1 and n=2l+1 for some k,l∈Z, and som2-n2=2k+12-2l+12=4k2+4k+1-4l2+4l+1=4k2+4k-4l2-4l=4k2+k-l2-lwhich is divisible by 4.Hence m2-n2 is divisible by 4.QED |
| Specific behaviours |
| assumes m and n are odd expresses m as 2k+1 and n as 2l+1expands and factorises 2k+12-2l+12concludes that m2-n2 is divisible by 4 |
Question 2 (2.2.1 - 2.2.3, 2.2.11) (9 marks)
Let M= 341-2.
Determine M-1. (2 marks)
| Solution |
| detM=3×-2-4×1=-10Hence M-1=-110-2-4-13 |
| Specific behaviours |
| calculates determinant writes correct expression for M-1 |
Showing use of an appropriate matrix equation together with your answer to part (i), determine the coordinates of the point of intersection of the lines
3x+4y=-1 and x-2y=8.
(3 marks)
| Solution |
| Lines intersect at (x,y) where xy satisfies:341-2xy=-18NowM-1341-2xy=M-1-18xy=-110-2-4-13-18=-110-3025Hence the lines intersect at (3,-52) |
| Specific behaviours |
| writes correct matrix equation multiplies both sides by M-1 states coordinates of point of intersection |
Consider the equation
64k-2xy=-84
where k∈R.
Determine the value of k such that the equation does not have a unique solution for xy. (2 marks)
| Solution |
| No unique solution if det64k-2=0, which is true if -12-4k=0.Hence if k=-3, the equation does not have a unique solution. |
| Specific behaviours |
| equates determinant to 0 states k=-3 |
With the value of k obtained in part (i), what is the geometrical relationship between the lines 6x+4y=-8 and kx-2y=4? (2 marks)
| Solution |
| With k=-3, the first equation is -2 times the second. Hence the lines are coincident. |
| Specific behaviours |
| states that equations are scalar multiples of each other states that lines are coincident/same |
Question 3 (2.3.4, 2.3.5) (7 marks)
Use the principle of mathematical induction to prove that
4+32+108+…+4n3=n2n+12
for all integers n≥1.
| Solution |
| Let P(n) stand for the statement ‘4+32+108+…+4n3=n2n+12’ for all n∈Z+.In P(1):LHS=4×13=4RHS=121+12=4Hence LHS = RHS and so P(1) is true.Now assume that P(k) is true for some integer k≥1.Then 4+32+108+…+4k3=k2k+12.Now LHS of Pk+1=4+32+108+…+4k3+4k+13=k2k+12+4k+13=k+12(k2+4k+1)=k+12(k2+4k+4)=k+12k+22=k+12((k+1)+1)2=RHS of Pk+1This shows that P(k+1) is also true.Hence, by PMI, P(n) is true for all integers n≥1. |
| Specific behaviours |
| defines P(n) shows that P(1) is true by evaluating LHS and RHS separately assumes P(k) is true writes LHS of P(k+1) in terms of RHS of P(k) simplifies expression to obtaink+12k+22 concludes that P(k+1) is also true concludes proof by referring to PMI |
Question 4 (2.3.7-2.3.10) (9 marks)
Let z= 3-5i and w=-2+i. Write each of the following in the form a+bi where a,b∈R.
z+w (2 marks)
| Solution |
| z+w=3-2-5i+i=1-4i |
| Specific behaviours |
| adds real and imaginary parts separately states 1-4i |
3zw (2 marks)
| Solution |
| 3zw=33-5i-2+i=3(-6+3i+10i-5i2)=-3+39i |
| Specific behaviours |
| expands correctly states -3+39i |
z+z (2 marks)
| Solution |
| z+z=3-5i+3+5i=6 |
| Specific behaviours |
| determines conjugate of z or uses z+z=2Re(z) states 6 |
zw (3 marks)
| Solution |
| zw=3-5i-2+i×-2-i-2-i=3-5i-2-i5=-11+7i5=-115+75i |
| Specific behaviours |
| multiplies numerator and denominator by w expands states -115+75i (must be in form a+bi i.e. don’t accept -11+7i5 as final answer) |
Question 5 (1.2.11, 1.2.12) (4 marks)
Let ΔOAB be an isosceles triangle with OA=OB, and let X, Y and Z be the midpoints of OA, OB and AB respectively. Let a=OA and b=OB.
Use a vector method to prove that ΔXYZ is isosceles.
| Solution |
| AB=b-aXZ=12OA+12AB=12a+12b-a=12bYZ=12OB-12AB=12b-12b-a=12aThus XZ=12b and YZ=12a.But since ΔOAB is isosceles, a=|b|, and hence XZ=YZ, meaning that ΔXYZ is isosceles. |
| Specific behaviours |
| writes AB as b-a shows that XZ=12b shows that YZ=12a deduces from a=|b| that XZ=YZ |
Question 6 (2.1.3 - 2.1.6, 2.1.8) (5 marks)
Prove that the following is true for all θ.
cotθ-2cot2θ=tanθ
| Solution |
| LHS=cotθ-2cot2θ=cosθsinθ-2cos2θsin2θ=cosθsinθ-2cos2θ2sinθcosθ=2cos2θ-22cos2θ-12sinθcosθ=cos2θ-2cos2θ+1sinθcosθ=1-cos2θsinθcosθ=sin2θsinθcosθ=sinθcosθ=tanθ=RHSHence cotθ-2cot2θ=tanθ for all θ. |
| Specific behaviours |
| writes cot expressions using cossin combines to give a single fraction using double angle formula for cos simplifies to sinθcosθ works from LHS to RHS (or vice versa) |
Question 7 (1.3.2, 1.3.5, 2.3.1) (7 marks)
Let p be an irrational number and q a rational number. Use the method of proof by contradiction to prove that pq is irrational. (4 marks)
| Solution |
| Assume that p is irrational and q is rational, but that pq is rational.Then q=ab and pq=cd for some integers a, b, c and d.Now p=pq÷q=cd÷ab=cd×ba=cbdawhich is rational since cb and da are integers. This contradicts the assumption that p is irrational; hence pq must be irrational. |
| Specific behaviours |
| assumes that pq is rational writes q and pq as a ratio of integers shows that pqq is rational notes contradiction and concludes that pq is irrational |
State whether the following is true or false and prove or disprove it accordingly:
‘If p is irrational and q is rational, then p+q2 is irrational.’
(3 marks)
| Solution |
| The statement is false.E.g. let p=2-1 and q=1. Then p is irrational and q is rational, but p+q2=2, which is rational. |
| Specific behaviours |
| states false gives counterexample values for p and q shows that p+q2 is rational for those values |
Question 8 (2.1.7) (5 marks)
The diagram below shows the graphs of two functions fx=3cos(x) and gx=asin(x) where a is a real constant.
Below is the graph of y=fx+g(x), which can also be expressed as y=bsin(x+α) where b and α are positive real constants.
Determine an exact value for α. (1 mark)
| Solution |
| α=5π6 |
| Specific behaviours |
| states correct value |
Determine the values of a and b. (4 marks)
| Solution |
| Equation of graph isy=bsinx+5π6=bsinxcos5π6+bcosxsin5π6=b-32sinx+b12cosxSince the equation of the graph is also y=asinx+3cosx, we have b2=3, so b=6.Hence a=6-32=-33. |
| Specific behaviours |
| notes that y=bsinx+5π6 expands using compound angle formula solves for b and states correct value solves for a and states correct value |
END OF SECTION ONE
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