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Semester Two Examination, 2018

Question/Answer booklet

SOLUTIONSSOLUTIONSMATHEMATICS

SPECIALIST

UNITS 1 AND 2

Section Two:

Calculator-assumed

In words

Your 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.

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.

3. You must be careful to confine your response to the specific question asked and to follow any instructions that are specified to a particular question.

4. Supplementary pages for the use of planning/continuing your answer to a question
have been 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.

5. 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.

6. It is recommended that you do not use pencil, except in diagrams.

7. The Formula sheet is not to be handed in with your Question/Answer booklet.

Section Two: Calculator-assumed 65% (98 Marks)

This section has thirteen (13) questions. Answer all questions. Write your answers in the spaces provided.

Working time: 100 minutes.

Question 9 (7 marks)

(a) Given that 20×19×1819×18×17×16= aPb cP4, determine the values of a, b and c. (3 marks)

Solution20×19×1819×18×17×16=20!19!×15!17!=20!17!×15!19!20!17!= 20P3, 19!15!= 19P4a=20, b=3, c=19Specific behaviours expresses fraction with factorials expresses as permutations lists all valuesSolution20×19×1819×18×17×16=20!19!×15!17!=20!17!×15!19!20!17!= 20P3, 19!15!= 19P4a=20, b=3, c=19Specific behaviours expresses fraction with factorials expresses as permutations lists all values

(b) Determine how many integers between 1 and 100 inclusive are divisible by 2, 3 or 13.

Solution100÷2+100÷3+100÷13=50+33+7=90100÷6+100÷26+100÷39=16+3+2=21100÷78=1n=90-21+1=70 integersSpecific behaviours correct number divisible singly correct number divisible by pairs correct number divisible by all three correct totalSolution100÷2+100÷3+100÷13=50+33+7=90100÷6+100÷26+100÷39=16+3+2=21100÷78=1n=90-21+1=70 integersSpecific behaviours correct number divisible singly correct number divisible by pairs correct number divisible by all three correct total (4 marks)

Question 10 (6 marks)

(a) In the circle shown below, minor arc PR subtends an angle of 120° at O, the centre of the circle, and the size of angle RPQ is 55°. Determine the size of angle POQ. (2 marks)

Solution∠ROQ=2×55=110°
∠POQ=360-110-120=130°Specific behaviours indicates size of ∠ROQ correct ∠PORSolution∠ROQ=2×55=110°
∠POQ=360-110-120=130°Specific behaviours indicates size of ∠ROQ correct ∠POR

(b) In the diagram below, AB is tangent to the circle with centre O at A, secant BD intersects the circle at C and D, and the sizes of angles AOC and COD are 72° and 104° respectively. Determine the size of angle ABC. (4 marks)

Solution∠ODC=180-1042=38°
∠DOA=72+104=176°∠OAB=90°Using OABD:∠ABC=360-38-176-90
=56°Specific behaviours correct ∠ODC correct ∠DOA indicates ∠OABA is right-angle correct ∠ABCSolution∠ODC=180-1042=38°
∠DOA=72+104=176°∠OAB=90°Using OABD:∠ABC=360-38-176-90
=56°Specific behaviours correct ∠ODC correct ∠DOA indicates ∠OABA is right-angle correct ∠ABC

Question 11 (8 marks)

(a) Show how to express 0.23 as a rational number. (2 marks)

SolutionIf x=0.232323… then 100x=23.232323…Hence by subtraction 99x=23⇒x=2399, which is rational.Specific behaviours expresses as x and 100x uses subtraction to express as rationalSolutionIf x=0.232323… then 100x=23.232323…Hence by subtraction 99x=23⇒x=2399, which is rational.Specific behaviours expresses as x and 100x uses subtraction to express as rational

(b) Prove that the sum of any three consecutive integers is always a multiple of three.

SolutionLet the integers be n, n+1, n+2 and their sum be S.S=n+n+1+n+2
=3n+3
=3(n+1)Hence S is always a multiple of 3.Specific behaviours clearly indicates three consecutive integers creates sum factors out 3 and makes conclusionSolutionLet the integers be n, n+1, n+2 and their sum be S.S=n+n+1+n+2
=3n+3
=3(n+1)Hence S is always a multiple of 3.Specific behaviours clearly indicates three consecutive integers creates sum factors out 3 and makes conclusion (3 marks)

(c) Prove by contradiction that 7 is irrational. (3 marks)

SolutionAssume that 7 is rational and can be expressed in the form ab, where a and b are integers with no common factor greater than 1.7=ab⇒a2=7b2, so that a2 and hence a must be a multiple of 7.Since a=7k (k an integer) then 7k2=7b2⇒7k2=b2, so that b2 and hence b must be a multiple of 7.Since a and b are both multiples of 7, the assumption they have no common factor is contradicted and so 7 must be irrational.Specific behaviours makes rational assumption including bolded condition deduces that a and b must both be multiples of 7 explains contradictionSolutionAssume that 7 is rational and can be expressed in the form ab, where a and b are integers with no common factor greater than 1.7=ab⇒a2=7b2, so that a2 and hence a must be a multiple of 7.Since a=7k (k an integer) then 7k2=7b2⇒7k2=b2, so that b2 and hence b must be a multiple of 7.Since a and b are both multiples of 7, the assumption they have no common factor is contradicted and so 7 must be irrational.Specific behaviours makes rational assumption including bolded condition deduces that a and b must both be multiples of 7 explains contradiction

Question 12 (8 marks)

Let vector a=4i-6j.

(a) Determine the angle between a and -7i-10j. (1 mark)

SolutionUsing CASθ=68.7°Specific behaviours correct angleSolutionUsing CASθ=68.7°Specific behaviours correct angle

(b) Let vector b=14i+tj. Determine the value of t so that a is

(i) parallel to b. (2 marks)

Solution414=-6t⇒t=-6×144=-21Specific behaviours indicates method correct valueSolution414=-6t⇒t=-6×144=-21Specific behaviours indicates method correct value

(ii) perpendicular to b. (2 marks)

Solutiona∙b=0⇒414+-6t=0t=283=9.3Specific behaviours indicates method correct valueSolutiona∙b=0⇒414+-6t=0t=283=9.3Specific behaviours indicates method correct value

(c) Determine the vector projection of a on -6i+8j. (3 marks)

SolutionLet c=-6i+8j. Then c=-0.6i+0.8j.Using CAS, a∙cc=10825i-14425j=4.32i-5.76jSpecific behaviours indicates unit vector c indicates method correct projectionSolutionLet c=-6i+8j. Then c=-0.6i+0.8j.Using CAS, a∙cc=10825i-14425j=4.32i-5.76jSpecific behaviours indicates unit vector c indicates method correct projection

Question 13 (8 marks)

Two matrices are given by P=47-83 and Q=3-784.

SolutionPQ=680068Specific behaviours correct productSolutionPQ=680068Specific behaviours correct product(a) Determine PQ. (1 mark)

(b) Given that Q-1=kP, determine the exact value of the constant k. (2 marks)

SolutionQ-1Q=kPQ⇒I=kPQk=168Specific behaviours uses matrix algebra or states Q-1 correct valueSolutionQ-1Q=kPQ⇒I=kPQk=168Specific behaviours uses matrix algebra or states Q-1 correct value

The system of equations 3a=7b+102 and 8a+4b+34=0 can be expressed as a matrix equation in the form QX=R.

(c) Determine matrices X and R. (2 marks)

Solution3-784×ab=102-34X=ab, R=102-34Specific behaviours correct matrix X correct matrix RSolution3-784×ab=102-34X=ab, R=102-34Specific behaviours correct matrix X correct matrix R

(d) Express matrix X in terms of matrices P and R. (2 marks)

SolutionQX=RQ-1QX=Q-1RX=168PRSpecific behaviours pre-multiplies by Q-1 correct expressionSolutionQX=RQ-1QX=Q-1RX=168PRSpecific behaviours pre-multiplies by Q-1 correct expression

(e) Solve the system of equations. (1 mark)

Solutiona=2.5, b=-13.5Specific behaviours correct solutionSolutiona=2.5, b=-13.5Specific behaviours correct solution

Question 14 (6 marks)

A segment of a circle has a perpendicular height of h and width w.

(a) Determine the radius of the arc of the segment when h=3 cm and w=8 cm. (3 marks)

Solutionr2=r-32+426r=9+16r=256=4.16 cmSpecific behaviours relevant sketch uses Pythagoras' Theorem correct radiusSolutionr2=r-32+426r=9+16r=256=4.16 cmSpecific behaviours relevant sketch uses Pythagoras' Theorem correct radius

(b) Use the intersecting chord theorem to derive a formula for the radius of the arc of a segment of width w and height h, where the chords are the straight edge of the segment and the diameter of the circle. (3 marks)

Solutionh2r-h=w22
2rh=h2+w24
r=h2+w28hSpecific behaviours labelled sketch of intersecting chords uses theorem to form equation correct formulaSolutionh2r-h=w22
2rh=h2+w24
r=h2+w28hSpecific behaviours labelled sketch of intersecting chords uses theorem to form equation correct formula

Question 15 (8 marks)

Circle C has equation x-22+y+62=16.

(a) Circle C is transformed by the matrix M=0110 to circle C'. Describe transformation M and state the equation of circle C'. (3 marks)

SolutionM is a reflection in the line y=x.Centre: 01102-6=-62Equation: x+62+y-22=42=16Specific behaviours states reflection with equation of line identifies new centre correct equationSolutionM is a reflection in the line y=x.Centre: 01102-6=-62Equation: x+62+y-22=42=16Specific behaviours states reflection with equation of line identifies new centre correct equation

(b) Circle C' is then transformed by the matrix N=3003 to circle C''. Describe transformation N and state the equation of circle C''. (3 marks)

SolutionN is a dilation about (0, 0) of scale factor 3.Centre: 3003-62=-186Equation: x+182+y-62=4×32=122=144Specific behaviours states dilation with scale factor (dilation centre not required) identifies new centre correct equationSolutionN is a dilation about (0, 0) of scale factor 3.Centre: 3003-62=-186Equation: x+182+y-62=4×32=122=144Specific behaviours states dilation with scale factor (dilation centre not required) identifies new centre correct equation

(c) Determine the single matrix P that will transform circle C'' back to circle C. (2 marks)

SolutionNM-1=013130Specific behaviours indicates correct method correct matrix PSolutionNM-1=013130Specific behaviours indicates correct method correct matrix P

Question 16 (11 marks)

Two forces act on a body. F1 has a magnitude of 335 N and acts on a bearing of 145. F2 has a magnitude of 295 N and acts on a bearing of 080.

(a) Determine

(i) the magnitude and direction of the sum of the two forces. (6 marks)

SolutionR2=3452+2952-2345295cos115R=531.8 N531.8sin115=295sinαα=30.2°Bearing:180-35-30.2≈115Specific behaviours triangle showing sum of two forces use of cosine rule with correct angle correct magnitude indicates use of sine rule correct value of α correct bearingSolutionR2=3452+2952-2345295cos115R=531.8 N531.8sin115=295sinαα=30.2°Bearing:180-35-30.2≈115Specific behaviours triangle showing sum of two forces use of cosine rule with correct angle correct magnitude indicates use of sine rule correct value of α correct bearingCAS SolutionSpecific behaviours correct polar angles converts F1 to Cartesian converts F2 to Cartesian adds forces converts back to polar states magnitude and bearingCAS SolutionSpecific behaviours correct polar angles converts F1 to Cartesian converts F2 to Cartesian adds forces converts back to polar states magnitude and bearing

Solution115+180=295F3=531.8 N on bearing 295Specific behaviours correct magnitude and bearingSolution115+180=295F3=531.8 N on bearing 295Specific behaviours correct magnitude and bearing(ii) the magnitude and direction of a third force that would keep the body in equilibrium. (1 mark)

(b) The bearing F2 acts on is changed so that the direction of F1+F2 is due east. Determine the new bearing of F2. (4 marks)

Solution335sinλ=295sin55λ=68.5° or 111.5°Bearings:90-68.5=021.5
or:360-111.5-90=338.5Specific behaviours diagram indicates use of sine rule correct values of λ both possible bearingsSolution335sinλ=295sin55λ=68.5° or 111.5°Bearings:90-68.5=021.5
or:360-111.5-90=338.5Specific behaviours diagram indicates use of sine rule correct values of λ both possible bearings

Question 17 (8 marks)

A small body P moves in a straight line. The displacement of the body from a fixed point O is given by x=asinbt+c+d, where x is in centimetres, t is the time in seconds. The graph of x against t is shown below.

(a) Determine the values of the positive constants a, b, c and d. (4 marks)

Solutiona=24-6÷2=9
b=2π6=π3
c=12 (or 6.5, 12.5, …)
d=24-9=15Specific behaviours each correct valueSolutiona=24-6÷2=9
b=2π6=π3
c=12 (or 6.5, 12.5, …)
d=24-9=15Specific behaviours each correct value

(b) Express the relationship between x and t as a cosine function. (2 marks)

Solutionc=12-146=-1 (or-7, -1, 5, …)
x=9cosπ3t-1+15Specific behaviours only changes value of c correct functionSolutionc=12-146=-1 (or-7, -1, 5, …)
x=9cosπ3t-1+15Specific behaviours only changes value of c correct function

Solution9sinπ3t-12+15=18t=152.18 sSpecific behaviours method correct timeSolution9sinπ3t-12+15=18t=152.18 sSpecific behaviours method correct time(c) Determine the first time that P is 18 cm from O after 150 seconds, giving your answer to two decimal places. (2 marks)

Question 18 (7 marks)

Let N={1, 2, 3, 4, 5, 6, 7, 8}.

(a) Three or four-digit codes are to be formed using integers selected from N, such as 287 or 1381.

Determine the number of codes that can be formed if

(i) there are no restrictions. (2 marks)

Solution83+84=512+4096
=4608 codesSpecific behaviours indicates number of 3- and 4-digit codes correct totalSolution83+84=512+4096
=4608 codesSpecific behaviours indicates number of 3- and 4-digit codes correct total

(ii) no integer may be used more than once in a code. (2 marks)

Solution 8P3+ 8P4=336+1680
=2016 codesSpecific behaviours uses permutations for 3- and 4-digit codes correct totalSolution 8P3+ 8P4=336+1680
=2016 codesSpecific behaviours uses permutations for 3- and 4-digit codes correct total

(b) Using the pigeon-hole principle or otherwise, prove that when five integers are selected from N, at least one pair of the integers will have a sum of 9. (3 marks)

SolutionPartition N into 4 pigeon-holes with sums of 9: 1, 8, 2, 7, 3, 6, {4, 5}If 5 integers (pigeons) are selected from N then by the pigeon-hole principle, at least 2 must be in the same pigeon-hole.Hence at least one pair of the integers will have a sum of 9.Specific behaviours lists pigeonholes uses pigeonhole principle makes conclusionSolutionPartition N into 4 pigeon-holes with sums of 9: 1, 8, 2, 7, 3, 6, {4, 5}If 5 integers (pigeons) are selected from N then by the pigeon-hole principle, at least 2 must be in the same pigeon-hole.Hence at least one pair of the integers will have a sum of 9.Specific behaviours lists pigeonholes uses pigeonhole principle makes conclusion

Question 19 (8 marks)

(a) The four points K, L, M and N lie in that order on the circumference of a circle. Chords KM and LN intersect at X. Prove that ∆KXL~∆NXM. (4 marks)

Example solution∠KXL=∠NXM - vertically opposite
∠KLN=∠KMN - stand on same arc
∴∆KXL~∆NXM - AAASpecific behaviours labelled diagram one pair of equal angles with reason second pair of equal angles with reason states similarity with reasonExample solution∠KXL=∠NXM - vertically opposite
∠KLN=∠KMN - stand on same arc
∴∆KXL~∆NXM - AAASpecific behaviours labelled diagram one pair of equal angles with reason second pair of equal angles with reason states similarity with reason

(b) In triangle ABC, P, Q and R are the midpoints of AB, AC and BC respectively. If AB=b and AC=c, use a vector method to prove that PBRQ is a parallelogram. (4 marks)

Example solutionPR=PA+AR
=-12b+12cBQ=12BC
=12BA+AC
=12-b+c
=-12b+12c
=PRHence PBRQ is a parallelogram since it has a pair of opposite sides that are parallel and equal in length.Specific behaviours labelled diagram derives vector for one side of parallelogram derives second vector for opposite side shows vectors are equal and makes conclusionExample solutionPR=PA+AR
=-12b+12cBQ=12BC
=12BA+AC
=12-b+c
=-12b+12c
=PRHence PBRQ is a parallelogram since it has a pair of opposite sides that are parallel and equal in length.Specific behaviours labelled diagram derives vector for one side of parallelogram derives second vector for opposite side shows vectors are equal and makes conclusion

Question 20 (6 marks)

Use mathematical induction to prove that for all positive integers n

1×5+2×6+3×7+…+nn+4=n6n+12n+13.

SolutionLet Claim(n) be the statement1×5+2×6+3×7+…+nn+4=n6(n+1)(2n+13)Claim(1) is the statement 1×5=162(15) and so Claim(1) is shown to be true.Assume Claim(k) is true so that1×5+2×6+3×7+…+kk+4=k6(k+1)(2k+13)LHS of Claimk+1=1×5+2×6+…+kk+4+k+1k+1+4
=k6k+12k+13+k+1k+1+4 using Claim(k)
=k+162k2+13k+6k+30
=k+16k+22k+15
=RHS of Claimk+1We have shown that Claim(1) is true and that Claimk⇒Claim(k+1) and so by the principle of mathematical induction it follows that Claim(n) is true.Specific behaviours shows truth of initial case clearly states assumption adds k+1 term to statement, using Claim(k) factors out k+1 completes factorisation closing statementSolutionLet Claim(n) be the statement1×5+2×6+3×7+…+nn+4=n6(n+1)(2n+13)Claim(1) is the statement 1×5=162(15) and so Claim(1) is shown to be true.Assume Claim(k) is true so that1×5+2×6+3×7+…+kk+4=k6(k+1)(2k+13)LHS of Claimk+1=1×5+2×6+…+kk+4+k+1k+1+4
=k6k+12k+13+k+1k+1+4 using Claim(k)
=k+162k2+13k+6k+30
=k+16k+22k+15
=RHS of Claimk+1We have shown that Claim(1) is true and that Claimk⇒Claim(k+1) and so by the principle of mathematical induction it follows that Claim(n) is true.Specific behaviours shows truth of initial case clearly states assumption adds k+1 term to statement, using Claim(k) factors out k+1 completes factorisation closing statement

Question 21 (7 marks)

A small drone is to fly in a straight line and at a constant altitude from P to Q. Q lies 775 m away from P on a bearing of 165° and a steady wind of 6.6 ms-1 is blowing in the area from due east.

If the speed of the drone is set to 14.5 ms-1, determine the bearing it should steer and the time that it will take to reach Q.

Solutionsin10514.5t=sinα6.6tα=26.1°180-105-26.1=48.9°sin10514.5t=sin48.9775t=68.5 secondsBearing: 165-26.1=138.9°Specific behaviours diagram with key elements angle between wind and PQ equation using sin rule for α solves for α equation using sin rule for t correct time correct bearingSolutionsin10514.5t=sinα6.6tα=26.1°180-105-26.1=48.9°sin10514.5t=sin48.9775t=68.5 secondsBearing: 165-26.1=138.9°Specific behaviours diagram with key elements angle between wind and PQ equation using sin rule for α solves for α equation using sin rule for t correct time correct bearing

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