2019 Maths Specialist Unit 2 Sem 2 Exam CA Marking Key.docx
Semester Two Examination, 2019
Question/Answer booklet
MATHEMATICS SPECIALIST
UNIT 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 |
| 10 | 4 | 17 | 8 | ||
| 11 | 7 | 18 | 5 | ||
| 12 | 6 | 19 | 10 | ||
| 13 | 10 | 20 | 10 | ||
| 14 | 7 | 21 | 12 | ||
| 15 | 5 | 22 | 6 | ||
| 16 | 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 | 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 Two: Calculator-assumed (94 Marks)
This section has thirteen (13) 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 10 (4 marks)
Given the following graph for y=asinsin bx-c+d , state the values of a,b, c and d
(all unknowns are positive numbers).
| Solution |
| a=5 b=3c= π4 d=2 |
| Specific behaviours |
| ✔ correct value for a✔ correct value for b✔ correct value for c✔ correct value for d |
Question 11 (7 marks)
The water depth, h metres, measured from the bottom of a lake t hours after 6 am is modelled by the function
h=8-3sinsin π3t-12
Determine the water depth at 12 noon. (2 marks)
| Solution |
| When t=6,h=8-3sinsin π36-12 =9.5 m |
| Specific behaviours |
| ✔ uses correct t value✔ states correct answer |
Determine the first time (after 6 am) when the water depth is at a maximum. (3 marks)
| Solution |
| h is at a maximum when sinsin π3t-12 =-1This first occurs when π3t-12=3π2, i.e. when t=5.So the first time is 11 am. |
| Specific behaviours |
| ✔ states the condition on the sine function for h to be a maximum✔ solves for t✔ states the correct time |
Determine the length of time before noon (in hours) when the water depth is less than 8 m. (2 marks)
| Solution |
| 8=8-sinsin π3t-12 π3t-12=kπ, k∈ZSince t<6, t=12 or t=72.∴ duration = 3 hours |
| Specific behaviours |
| ✔ solves for t when h=8✔ states the duration in hours |
Question 12 (6 marks)
Determine the number of four letter “words” (i.e. distinct arrangements of 4 letters) that can be made using the letters of the word FOURTH if:
there are no restrictions. (1 mark)
| Solution |
| Number of words =6!2!=360 |
| Specific behaviours |
| ✔ states 360 |
the T and H must be together, in that order, at the start of the word. (2 marks)
| Solution |
| Number of words =1×4×3=12 |
| Specific behaviours |
| ✔ calculates by grouping T and H together✔ states correct value |
the letters T and H must be together, in any order, anywhere in the word, and there must be at least one vowel in the word. (3 marks)
| Solution |
| Number of words =2!×3×4×3-2!×3×2×1=60 |
| Specific behaviours |
| ✔ writes expression for total number of words with T and H together✔ subtracts number of words with T and H together but containing no vowels✔ states correct value |
Question 13 (10 marks)
Airports A and B are such that the vector AB=(-350i+650j) km. A helicopter is to be flown directly from A to B, and in still air the helicopter can maintain a steady speed of 180km/h. There is a wind blowing with a velocity of -12i-3j km/h.
Draw a diagram to represent this information, using ai+bj to represent the velocity vector the helicopter should set to follow the shortest path from A to B. (1 mark)
Determine the velocity vector ai+bj defined in part (a) (assuming the wind maintains the same strength and direction for the duration of the journey). Give answers to 2 decimal places. (4 marks)
| Solution |
| Let the required velocity vector be ai+bj.Thenai+bj-12i-3j=λ(-350i+650j)Equating coefficients givesa-12=-350λb-3=650λand soa-12b-3=-350650In additiona2+b2=1802Solving simultaneously,a=-74.59 and b=163.82 or a=95.70 and b=-152.45From diagramai+bj=-74.59i+163.82j |
| Specific behaviours |
| ✔ equates resultant velocity vector to a scalar multiple of displacement vector (or equates a scalar multiple of velocity vector with displacement vector)✔ eliminates scalar✔ solves simultaneously with equation for square of magnitude of velocity vector✔ states correct velocity vector |
Calculate the resultant speed of the helicopter, giving your answer to 2 decimal places. (2 marks)
| Solution |
| Resultant velocity =-74.59-12i+163.82-3j=-86.59i+160.82jSpeed=-86.592+160.822=182.65 km/h |
| Specific behaviours |
| ✔ calculates resultant velocity vector✔ states correct speed |
Determine to the nearest minute the time it will take for the helicopter to travel from A to B.
(2 marks)
| Solution |
| Distance=-350i+650j=738.24 kmTime=738.24182.65=4.04 h=4 h 3 min |
| Specific behaviours |
| ✔ calculates distance✔ states time to the nearest minute |
Assuming the wind maintains the same strength and direction, find, in the form ci+dj, the velocity vector the pilot should take for the return journey from B to A, giving answers to 2 decimal places.
(1 mark)
| Solution |
| Velocity vector = 95.70i-152.45j |
| Specific behaviours |
| ✔ states correct velocity vector |
Question 14 (7 marks)
Of the ten players in the squad for a basketball team, three are primarily guards, two are primarily centres and five are primarily forwards. How many different teams of five could be chosen if:
There are no restrictions. (1 mark)
| Solution |
| 105=252 |
| Specific behaviours |
| ✔ states correct value |
There must be one centre, two forwards and two guards. (2 marks)
| Solution |
| 21×52×32=60 |
| Specific behaviours |
| ✔ multiplies the numbers of ways of choosing players of each type✔ states correct value |
There must be at least one guard. (2 marks)
| Solution |
| 105-75=231 |
| Specific behaviours |
| ✔ subtracts number of teams with no guard from total number of teams✔ states correct value |
There must be at least one guard and no centres. (2 marks)
| Solution |
| 85-55=55 |
| Specific behaviours |
| ✔ subtracts number of teams with no guards and no centres from number with no centres✔ states correct value |
Question 15 (5 marks)
Prove the following identity:
csc(x) +cotcot (x) =2tantan (x) coscos (x) +sinsin (2x) 1-coscos 2x
| Solution |
| RHS= 2tantan x coscos x +sinsin 2x 1-coscos 2x =2sinsin x coscos x coscos x +2sinsin x coscos x 1-(1-2sin2x ) |
| =2sinsin x +2sinsin x coscos x sin2x =1sinsin x +coscos x sinsin x =csccsc x +cotcot x =LHS |
| Specific behaviours |
| ✔ rearranges RHS (or LHS) to obtain LHS (or RHS)✔ writes tantan x as sinsin x coscos x ✔ uses double angle formula for sinsin 2x or coscos 2x ✔ divides all terms by 2sinsin x ✔ simplifies to LHS (or RHS) |
Question 16 (4 marks)
How many positive integers strictly less than 1000 are divisible by 5, 7, or 9?
| Solution |
| Number of integers divisible by 5, 7 or 9= # divisible by 5 + # divisible by 7 + # divisible by 9– # divisible by 5×7 – # divisible by 5×9 – # divisible by 7×9+ # divisible by 5×7×9= 199 + 142 +111 – 28 – 22 – 15 +3= 390 |
| Specific behaviours |
| ✔ uses inclusion/exclusion principle✔ demonstrates understanding that (in this case where numbers are pairwise coprime) n is divisible by a and b iff n is divisible by ab.✔ correctly counts number of integers in each set✔ states correct value |
Question 17 (8 marks)
Identify the matrix T that will rotate points 90° anticlockwise about the origin. (1 mark)
| Solution |
| T= 0 -1 1 0 |
| Specific behaviours |
| ✔ states correct matrix |
Find the image of (-4,7) following a transformation by T. (2 marks)
| Solution |
| T-4 7 = 0 -1 1 0 -4 7 =-7 -4 Therefore the image of -4,7 is (-7,-4). |
| Specific behaviours |
| ✔ writes point as a column vector and pre-multiplies by T✔ states correct point |
Identify the matrix Q that represents the single transformation with the same effect as first reflecting in the line y=3x and then dilating vertically by a factor of 3. (5 marks)
| Solution |
| Reflection in line y=3x:Angle of line is 3 =60°.Hence matrix is R=coscos 120° sinsin 120° sinsin 120° -coscos 120° =-12 32 32 12 Dilation vertically by factor 3 is D=1 0 0 3 Therefore Q=DR=-12 32 332 32 |
| Specific behaviours |
| ✔ identifies angle of line of reflection✔ determines reflection matrix✔ determines dilation matrix✔ multiplies in correct order to obtain Q✔ states correct matrix Q |
Question 18 (5 marks)
Simplify the following expressions into their simplest factorial form. (2 marks)
100!-99!99
| Solution |
| 100!-99!99=99!100-199=99! |
| Specific behaviours |
| ✔ factorises numerator✔ states correct answer |
Prove that
n+3!+(n+2)!+(n+1)!n+1! =n+32
(3 marks)
| Solution |
| LHS= n+3!+n+2!+n+1!n+1!=n+1!n+3n+2+n+2+1n+1!=n+3n+2+n+2+1=n2+5n+6+n+3=n+32=RHS |
| Specific behaviours |
| ✔ correctly factorises numerator of LHS✔ cancels n+1!✔ simplifies to obtain RHS |
Question 19 (10 marks)
Two perpendicular lines are each tangent to a circle of radius 1 at points A and B. The centre of the circle is C.
a) Prove that ∠ACB=90°. (3 marks)
| Solution |
| ∠AOB=90° (Given)∠OAC=∠OBC=90° (Tangent and radius to point of intersection are perpendicular)∠ACB=90° (sum of angles in a quadrilateral is 360°). |
| Specific behaviours |
| ✔ states that ∠AOB=90°✔ states with reason that ∠OAC=∠OBC=90°✔ concludes with reason that ∠ACB=90° |
b) Prove that ∠OBA=45°. (2 marks)
| Solution |
| OA=OB (Tangents to common point have equal length)ΔOBA is isosceles (OA=OB)∠OBA=180-902=45° (angles in a triangle add to 180°) |
| Specific behaviours |
| ✔ states with reason that OA=OB✔ states that Δ is isosceles or uses calculation to justify that ∠OBA=45° |
c) Point D lies on the circle such that ∠OBD=105°. Find the distance AD as an exact value.
(5 marks)
| Solution |
| ∠ABD=105-45=60° ∠ADB=45° (angle at centre is twice angle at circumference) OR (alternate segment theorem)AB=2 (Pythagoras’ Theorem)ADsinsin 60° =2sinsin 45° (sine rule)AD=3 |
| Specific behaviours |
| ✔ calculates that ∠ABD=60°✔ states with reason that ∠ADB=45°✔ determines length AB✔ uses sine rule✔ obtains correct value for AD |
Question 20 (10 marks)
If u=8i-4j and v=6i+4j determine
u∙v (1 mark)
| Solution |
| 8×6+-4×4=32 |
| Specific behaviours |
| ✔ correct value for scalar product |
the angle between the vectors u and v, to the nearest tenth of a degree.
(2 marks)
| Solution |
| u=82+(-4)2=45v=62+42=213θ=u∙vuv=3245×213=60.3° |
| Specific behaviours |
| ✔ correct expression for scalar product with cos✔ states angle |
the scalar projection of v on u. (2 marks)
| Solution |
| v∙uu=3245=855 |
| Specific behaviours |
| ✔ correct expression for scalar projection✔ states value |
The vector 21i+7mj has a magnitude of 35 and is perpendicular to the vector
5i-4nj. Determine the values of the constants m and n, where m>n. (5 marks)
| Solution |
| 212+(7m)2=352m=±421×5+7m×-4n=0105-28×±4n=0n=±1516∵m>n, m=4, n=1516 |
| Specific behaviours |
| ✔ uses magnitude to form an equation for m✔ calculates values of m✔ uses scalar product equal to zero✔ calculates values of n✔ chooses correct pairing |
Question 21 (12 marks)
Solve the matrix equation BA+A=B-I, where I=1 0 0 1 and B=1 2 -3 1 , to determine the matrix A. (4 marks)
| Solution |
| B+IA=B-I2 2 -3 2 A=0 2 -3 0 A=2 2 -3 2 -10 2 -3 0 =35 25 -35 35 |
| Specific behaviours |
| ✔ factorises✔ determines (B+I) and (B-I)✔ pre-multiplies by (B+I)-1✔ determines P |
Let P=k-1 1 6 4 and Q=4 -1 -6 k-1 . Determine the value of k if P×Q=P+Q. (3 marks)
| Solution |
| 4k-10 0 0 4k-10 =k+3 0 0 k+3 4k-10=k+3k=133 |
| Specific behaviours |
| ✔ determines P×Q and P+Q✔ writes equation for k✔ correct value for k |
Preimage ABCDEFG is transformed by the single matrix T=a b c d to Image A’B’C’D’E’F’G’.
Given A'-32,-72 and G'2,-72, determine the matrix T. (2 marks)
| Solution |
| T5 3 2 4 =-32 2 -72 -72 T=-32 2 -72 -72 5 3 2 4 -1=-2 2 -2 -2 |
| Specific behaviours |
| ✔ writes matrix equation✔ determines T |
Find the area of the Image A’B’C’D’E’F’G’. (3 marks)
| Solution |
| Area of Preimage = 6×2+12×4×2=16 units2det(T)=4Area of Image =4×16=64 units2 |
| Specific behaviours |
| ✔ correct area of preimage✔ correct det(T)✔ correct area of image |
Question 22 (6 marks)
Three forces act on an object at point P such that the system is in equilibrium. These forces are represented with vectors as shown in the diagram, where θ is acute. Determine the values of θ and k, giving answers to 2 decimal places.
| Solution |
| 10coscos θ i+10sinsin θ j-35coscos 2θ i+35sinsin 2θ j+6i+kj=0i+0j10coscos θ i-35coscos 2θ i+6i=0i10coscos θ -35coscos 2θ +6=010coscos θ -352θ -1+6=0-70θ +10coscos θ +41=0coscos θ =-0.6972 or coscos θ =0.8401Since θ is acute, coscos θ =0.8401 and θ=32.8510sinsin 32.85 j+35sinsin 65.70 j+kj=0jk=-37.32 |
| Specific behaviours |
| ✔ writes equation for equilibrium state✔ equates i components✔ uses double angle formula and obtains quadratic equation in coscos θ ✔ solves for coscos θ and obtains correct value for θ✔ equates j components✔ solves for k |
END OF QUESTIONS
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