2019 Maths Specialist Unit 2 Sem 2 Exam CF Marking Key.docx
Semester Two Examination, 2019
Question/Answer booklet
MATHEMATICS SPECIALIST
UNIT 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 | 3 | 6 | 8 | ||
| 2 | 6 | 7 | 6 | ||
| 3 | 7 | 8 | 4 | ||
| 4 | 8 | 9 | 3 | ||
| 5 | 7 |
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 | 9 | 9 | 50 | 52 | 36 |
| Section Two:Calculator-assumed | 13 | 13 | 100 | 94 | 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 One: Calculator-free (52 Marks)
This section has nine (9) 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 (3 marks)
Prove the following statement:
If a and b are each 1 less than a multiple of 3, then ab is 1 more than a multiple of 3.
| Solution |
| Assume that a and b are each 1 less than a multiple of 3.Then a=3k-1 and b=3l-1 for some k,l∈Z.Hence ab=3k-13l-1=9kl-3k-3l+1=33kl-k-l+1which is 1 more than a multiple of 3 since 3kl-k-l is an integer.QED |
| Specific behaviours |
| writes a and b as 3k-1 and 3l-1 multiplies and simplifies to 33kl-k-l+1 concludes that ab is 1 more than a multiple of 3 |
Question 2 (6 marks)
Consider the system of simultaneous linear equations:
3x-ay=6
-6x+4y=b
Write down the matrix A such that the equation
Axy=6b
is equivalent to the system of equations above. (1 mark)
| Solution |
| 3-a-64 |
| Specific behaviours |
| writes correct matrix |
Suppose that A is singular (non-invertible).
Determine the value of a (show working). (2 marks)
| Solution |
| A singular⇔detA=0⇔3×4--a-6=0⇔a=2 |
| Specific behaviours |
| attempts to solve detA=0 for a states correct value |
State the possible number(s) of solutions that the system of equations could have with the value of a you just found. (2 marks)
| Solution |
| Either no solutions or infinitely many solutions. |
| Specific behaviours |
| states at least one of ‘no solutions’ and ‘infinitely many solutions’ states both possibilitiesNo marks if implies that there can be exactly one solution |
State the number of solutions the system has if a has the value found above and b=11. (1 mark)
| Solution |
| No solutions. |
| Specific behaviours |
| States no solutions. |
Question 3 (7 marks)
Write 3cos5x+33sin5x in the form asin(bx+α). (3 marks)
| Solution |
| a=32+332=36=6b=5α=sin-136=π6So 3cos5x+33sin5x=6sin5x+π6 |
| Specific behaviours |
| writes correct value for a writes correct value for b writes correct value for α |
Hence, solve the equation 3cos5x+33sin5x=33 for -π2≤x≤π2. (4 marks)
| Solution |
| 3cos5x+33sin5x=336sin5x+π6=33sin5x+π6=325x+π6=π3+k2π or 5x+π6=2π3+k2π, k∈Zx=π30,13π30,-11π30 or x=π10,5π10,-3π10x=-11π30,-3π10,π30,π10,13π30,π2 |
| Specific behaviours |
| equates expression from part (a) to 33 obtains at least one general solution (using π3 or 2π3) for 5x+π6 obtains at least 3 correct solutions for x (in correct domain) obtains 6 correct solutions for x (in correct domain) |
Question 4 (8 marks)
A 2×2 real matrix A can ‘transform’ a complex number if we view the complex number as a column vector. That is, for any complex number z=a+bi, the matrix A transforms z to c+di where cd=Aab.
Find the matrix A which (according to this rule) will transform any complex number z to:
3z (2 marks)
| Solution |
| Multiplication by 3 corresponds to dilation by factor 3.3003 |
| Specific behaviours |
| identifies corresponding transformation writes correct matrix |
z (2 marks)
| Solution |
| Conjugation corresponds to reflection in x-axis.100-1 |
| Specific behaviours |
| identifies corresponding transformation writes correct matrix |
iz (2 marks)
| Solution |
| Multiplying by i corresponds to rotating anti-clockwise by 90°.0-110 |
| Specific behaviours |
| identifies corresponding transformation writes correct matrix |
iz (2 marks)
| Solution |
| 0-110×100-1=0110 |
| Specific behaviours |
| writes product of matrices OR identifies corresponding transformation sequence writes correct matrix |
Question 5 (7 marks)
Evaluate the following for complex numbers z=2+5i and w=1-4i
z-w (2 marks)
| Solution |
| z-w=2+5i-1-4i=1+9i |
| Specific behaviours |
| writes correct calculation writes correct answer |
z(w+w) (2 marks)
| Solution |
| z(w+w)=(2+5i)(1-4i+1+4i)=4+10i |
| Specific behaviours |
| writes correct expression for conjugate of w writes correct answer |
wz (3 marks)
| Solution |
| wz=1-4i2+5i=1-4i2-5i2+5i2-5i=2-8i-5i+20i222+52=-1829-1329i |
| Specific behaviours |
| multiplies numerator and denominator by w expands numerator and denominator correctly writes correct answer |
Question 6 (8 marks)
In this question, a proper factor is a factor greater than 1.
Assume that a and b are both integers, and consider the following statement:
If ab has no proper square factors, then neither a nor b has a proper square factor.
Prove the statement using the method of proof by contradiction. (3 marks)
| Solution |
| Assume that ab has no proper square factors, but that either a or b has a proper square factor.Then either a=nk2 or b=ml2 for some k,l,m,n∈Z.Hence either ab=nk2b or ab=aml2, and in each case ab has a proper square factor.This is a contradiction, and so neither a nor b has a proper square factor. |
| Specific behaviours |
| assumes negation of statement writes a and/or b as a product of an integer and a square shows that ab must therefore have a square factor and notes contradictionAccept argument for a alone having a square factor together with recognition that a similar argument will apply if b has a square factor. |
Write the converse of the statement. (2 marks)
| Solution |
| If neither a nor b has a proper square factor, then ab has no square factors. |
| Specific behaviours |
| writes correct converse statement |
Write whether the converse is true or false and prove or disprove it accordingly.
(3 marks)
| Solution |
| False. E.g. if a=6 and b=10, then neither a nor b has a proper square factor, but ab=60=22×15. |
| Specific behaviours |
| states false gives correct counterexample |
Question 7 (6 marks)
Let O be the origin, let A and B be points such that OA=OB, and let C be a point on AB such that OC bisects ∠AOB.
Let a=OA, b=OB and c=OC.
a) Show that a⋅c=b⋅c. (3 marks)
| Solution |
| Let θ=∠COB. Then ∠COA=θ.Now a⋅c=accosθ and b⋅c=bccosθ.Since a=|b| it follows that a⋅c=b⋅c. |
| Specific behaviours |
| writes expressions for a⋅c and b⋅c uses the fact that ∠COB=∠COA uses the fact that a=|b| |
b) Hence, prove that OC is perpendicular to AB. (3 marks)
| Solution |
| AB=b-ab-a⋅c=b⋅c-a⋅c =0 (by part a)Hence AB is perpendicular to OC. |
| Specific behaviours |
| writes AB as b-a attempts to determine dot product b-a⋅c shows that b-a⋅c=0 |
Question 8 (4 marks)
Prove the following identity.
sin7θ-sin2θcos2θ+cos7θ=tan5θ2
| Solution |
| LHS=sin7θ-sin2θcos2θ+cos7θ=2sin5θ2cos9θ22cos9θ2cos5θ2=sin5θ2cos5θ2=tan5θ2=RHSHence LHS = RHS QED |
| Specific behaviours |
| rearranges RHS (or LHS) to obtain LHS (or RHS) correctly uses sum-to-product identities cancels factor of 2cos9θ2 simplifies to RHS (or LHS) |
Question 9 (3 marks)
Let l be a line containing a point P, and let Q be a point not on l. Suppose that n is a unit vector perpendicular to the line l. Prove that the perpendicular distance from Q to l is PQ⋅n.
| Solution |
| Distance from Q to l is d=PQcosθPQ⋅n=PQncosθ=PQcosθHence d=PQ⋅n (or -PQ⋅n if n has opposite direction). |
| Specific behaviours |
| writes PQ⋅n as PQncosθ uses the fact that n=1 uses the fact that the shortest distance from Q to l is PQcosθ (Note that only d=PQ⋅n is required for marks.) |
END OF SECTION ONE
Additional working space
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Additional working space
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