WACEhub

Semester Two Examination, 2016

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 for section: 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 the WACE examinations

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 notes or other items of a non-personal nature in the examination room. 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. 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.

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/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 for this section is 100 minutes.

Question 8 (5 marks)

Points B, C and E lie on the circle with diameter AOD as shown below. ∠ABC=115°, ∠BAC=20° and ∠ACE=60°.

Determine the size of the following angles.

Solution60° Specific behaviours states angle Solution60° Specific behaviours states angle (a) ∠ADE. (1 mark)

Solution30° Specific behaviours states angle Solution30° Specific behaviours states angle (b) ∠EAD. (1 mark)

Solution65° Specific behaviours states angle Solution65° Specific behaviours states angle (c) ∠AEC. (1 mark)

Solution25° Specific behaviours states angle Solution25° Specific behaviours states angle (d) ∠CAD. (1 mark)

Solution25° Specific behaviours states angle Solution25° Specific behaviours states angle (e) ∠CED. (1 mark)

Question 9 (7 marks)

(a) ABCDEF is a regular hexagon. The midpoint of side DE is M.

Let a=AB and b=AF. Express each of the following in terms of a and b.

(i) BC. (1 mark)

SolutionBC=a+b Specific behaviours states correct expressionSolutionBC=a+b Specific behaviours states correct expression

SolutionAE=AF+FE=b+(a+b)=a+2b Specific behaviours states correct expressionSolutionAE=AF+FE=b+(a+b)=a+2b Specific behaviours states correct expression(ii) AE. (1 mark)

SolutionMB=ME+BE=-0.5a-2b Specific behaviours states correct expressionSolutionMB=ME+BE=-0.5a-2b Specific behaviours states correct expression(iii) MB. (1 mark)

(b) Three forces, F1, F2 and F3 act on a body that remains in equilibrium.

F1 has a magnitude of 400 N. The angle between the directions of F1 and F2 is 150°, between F1 and F3 is 135° and between F2 and F3 is 75°.

Determine the magnitudes of F2 and F3, rounding your answers to the nearest whole number. (4 marks)

SolutionFor equilibrium, F1+F2+F3=0 - nose to tail vectors in triangle.If magnitudes of F2=x and F3=y then xsin45=ysin30=400sin105Solving gives x=F2=293 N and y=F3=207 N.Specific behaviours sketch diagram triangle or equation indicates force vectors must sum to zero solves for x solves for ySolutionFor equilibrium, F1+F2+F3=0 - nose to tail vectors in triangle.If magnitudes of F2=x and F3=y then xsin45=ysin30=400sin105Solving gives x=F2=293 N and y=F3=207 N.Specific behaviours sketch diagram triangle or equation indicates force vectors must sum to zero solves for x solves for y

Question 10 (7 marks)

(a) A number is to be formed by randomly selecting three different digits from those in the number 93265. Determine how many different numbers

(i) start with an odd digit. (1 mark)

Solution3×4×3=36 numbersSpecific behaviours calculates correct numberSolution3×4×3=36 numbersSpecific behaviours calculates correct number

(ii) end with an even digit. (1 mark)

Solution2×4×3=24 numbersSpecific behaviours calculates correct numberSolution2×4×3=24 numbersSpecific behaviours calculates correct number

(iii) start with an odd digit or end in an even digit. (2 marks)

Solution3×2×3=18 numbers start with an odd digit and end in an even digit36+24-18=42 numbersSpecific behaviours calculates set intersection calculates correct number for set unionSolution3×2×3=18 numbers start with an odd digit and end in an even digit36+24-18=42 numbersSpecific behaviours calculates set intersection calculates correct number for set union

(b) A computer user has forgotten their six character, case-sensitive password, but know that they always use a permutation of F, F, 1, 9, 9, and 9 - their initials and the year they were born. Determine how many passwords are possible if

(i) the F's must both be uppercase. (2 marks)

Solution6!2!×3!=6×5×42=60 passwordsSpecific behaviours shows correct method calculates correct numberSolution6!2!×3!=6×5×42=60 passwordsSpecific behaviours shows correct method calculates correct number

(ii) either F can be lowercase or uppercase. (1 mark)

SolutionFF can be replaced with Ff, fF or ff, so 4×60=240 passwordsSpecific behaviours calculates correct numberSolutionFF can be replaced with Ff, fF or ff, so 4×60=240 passwordsSpecific behaviours calculates correct number

Question 11 (8 marks)

(a) Triangle BCE is such that B, C and E lie on a circle with centre O and radius 29 cm. Diameter AD and chord CE intersect at F, so that DF=8.5 cm and EF=25.5 cm.

Determine the lengths OF, CF and BC. (5 marks)

SolutionOF=29-8.5=20.5 cmCF×EF=DF×AF⇒CF=8.5×29+20.525.5=16.5 cmΔBCE is right-angled ⇒BC=29+292-16.5+25.52=40 cmSpecific behaviours calculates OF uses intersecting chord theorem calculates CF uses angle in semicircle calculates BCSolutionOF=29-8.5=20.5 cmCF×EF=DF×AF⇒CF=8.5×29+20.525.5=16.5 cmΔBCE is right-angled ⇒BC=29+292-16.5+25.52=40 cmSpecific behaviours calculates OF uses intersecting chord theorem calculates CF uses angle in semicircle calculates BC

(b) In the diagram below, points B, C and D lie on a circle with centre O. The tangents to the circle at B and D intersect at point A. If ∠BAD=x, prove that ∠BCD=90°-x2. (3 marks)

SolutionIn quadrilateral ABOD, ∠ABO=∠ADO=90° (tangent-radius angle)∠BOD=360-90-90-x=180-x (angle sum in quadrilateral)∠BCD=12×∠BOD=12×180-x=90°-x2 (angle at centre twice that on circ.)Specific behaviours adds radii to diagram noting right-angles determines ∠BOD with reason determines ∠BCD with reasonSolutionIn quadrilateral ABOD, ∠ABO=∠ADO=90° (tangent-radius angle)∠BOD=360-90-90-x=180-x (angle sum in quadrilateral)∠BCD=12×∠BOD=12×180-x=90°-x2 (angle at centre twice that on circ.)Specific behaviours adds radii to diagram noting right-angles determines ∠BOD with reason determines ∠BCD with reason

Question 12 (9 marks)

Transformation A is an anti-clockwise rotation about the origin of 90° and matrix B=2003.

(a) Represent transformation A as a 2×2 matrix. (2 marks)

SolutionA=0-110Specific behaviours writes matrix for 90° rotation matrix is for anticlockwise rotationSolutionA=0-110Specific behaviours writes matrix for 90° rotation matrix is for anticlockwise rotation

(b) Describe the transformation represented by matrix B. (2 marks)

SolutionA dilation parallel to x-axis of scale factor 2 and dilation parallel to y-axis of scale factor 3.Specific behaviours two dilations fully qualifies both dilations with directions and scale factorsSolutionA dilation parallel to x-axis of scale factor 2 and dilation parallel to y-axis of scale factor 3.Specific behaviours two dilations fully qualifies both dilations with directions and scale factors

(c) Determine the coordinates of the point P(-15, -11) following transformation A and then transformation B. (2 marks)

SolutionP'=2003×0-110×-15-11=2003×11-15=22-45Specific behaviours calculates first image using A calculates P'SolutionP'=2003×0-110×-15-11=2003×11-15=22-45Specific behaviours calculates first image using A calculates P'

(d) Following transformation B and then transformation A, point Q is transformed to point Q'(12, 7).

Determine the single matrix that will transform Q' back to Q and hence determine the coordinates of point Q. (3 marks)

SolutionQ'=ABQ⇒Q=B-1A-1Q'
B-1A-1=120013×01-10=012-130
Q=012-130×127=3.5-4Specific behaviours determines inverse of B and inverse of A determines single matrix determines coordinates of QSolutionQ'=ABQ⇒Q=B-1A-1Q'
B-1A-1=120013×01-10=012-130
Q=012-130×127=3.5-4Specific behaviours determines inverse of B and inverse of A determines single matrix determines coordinates of Q

Question 13 (8 marks)

(a) On the axes below sketch the graph of y=12secx-π2. (3 marks)

SolutionSee graphSpecific behaviours max/min correct locations vertical asymptotes smooth curvesSolutionSee graphSpecific behaviours max/min correct locations vertical asymptotes smooth curves

(b) Consider the function ft=2sint-5cost, t≥0.

(i) ft can be expressed in the form rsin(t-α), where r>0 and 0≤α≤π2. Determine the values of r and α, rounded to 2 decimal places. (3 marks)

Solution2sint-5cost= rsin(t-α)
=rsintcosα-rcostsinαrcosα=2 and rsinα=5 ⇒r=29≈5.39, α=tan-152≈1.19Specific behaviours uses difference identity determines r determines αSolution2sint-5cost= rsin(t-α)
=rsintcosα-rcostsinαrcosα=2 and rsinα=5 ⇒r=29≈5.39, α=tan-152≈1.19Specific behaviours uses difference identity determines r determines α

(ii) Hence or otherwise determine the minimum value of f(t) and the smallest value of t for this minimum to occur. (2 marks)

SolutionMinimum value is -29 Occurs when t-tan-152=3π2⇒t≈5.90Specific behaviours states minimum value determines first time, t≥0SolutionMinimum value is -29 Occurs when t-tan-152=3π2⇒t≈5.90Specific behaviours states minimum value determines first time, t≥0

Question 14 (8 marks)

(a) Consider the vectors p=(24, -143) and q=(20, -21). Determine

(i) the angle between the directions of vectors p and q. (1 mark)

SolutionUsing CAS, angle is 34.1°Specific behaviours states correct angle.SolutionUsing CAS, angle is 34.1°Specific behaviours states correct angle.

(ii) two vectors that are perpendicular to q and have the same magnitude as p.

(3 marks)

SolutionMagnitude of required vectors are p=145Unit vectors ⊥ to q are ±12921, 20Required vectors are ±14529×21, 20=105, 100 and (-105, -100)Specific behaviours calculates magnitudes of p and q determines at least one perpendicular vector states both required vectorsSolutionMagnitude of required vectors are p=145Unit vectors ⊥ to q are ±12921, 20Required vectors are ±14529×21, 20=105, 100 and (-105, -100)Specific behaviours calculates magnitudes of p and q determines at least one perpendicular vector states both required vectors

(b) If AB=(3, 4) and AC=(-2, 1), determine

(i) the component of AB parallel to AC. (2 marks)

SolutionVector projection of AB on AC:3, 4⋅-2, 1-2, 1⋅-2, 1×-2, 1=(0.8,-0.4)Specific behaviours substitutes correctly into vector projection formula evaluates componentSolutionVector projection of AB on AC:3, 4⋅-2, 1-2, 1⋅-2, 1×-2, 1=(0.8,-0.4)Specific behaviours substitutes correctly into vector projection formula evaluates component

(ii) the component of AB perpendicular to AC. (2 marks)

SolutionLet component be r.r+0.8, -0.4=AB
r=2.2, 4.4Specific behaviours shows use of vector addition evaluates componentSolutionLet component be r.r+0.8, -0.4=AB
r=2.2, 4.4Specific behaviours shows use of vector addition evaluates component

Question 15 (8 marks)

(a) Express the recurring decimal 1.158 as a rational number. (2 marks)

SolutionLet x=1.158585858585… then 100x=115.858585858…So 100x-x=99x=114.7⇒x=1147990Specific behaviours multiplies by 100 expresses as rational number in lowest termsSolutionLet x=1.158585858585… then 100x=115.858585858…So 100x-x=99x=114.7⇒x=1147990Specific behaviours multiplies by 100 expresses as rational number in lowest terms

(b) Use a counterexample to explain why the statement (∀x∈Z)(∃y∈Z)(2xy=24) is false.

SolutionIf x=24, then y=12 and so statement is false, as an integer value for y clearly does not exist for all values of x.Specific behaviours counterexample explanationSolutionIf x=24, then y=12 and so statement is false, as an integer value for y clearly does not exist for all values of x.Specific behaviours counterexample explanation (2 marks)

(c) Prove, by contradiction, that 6 is irrational. (4 marks)

SolutionAssume that 6 is rational and can be expressed in the form ab where a and b are both integers with no common factors.Then 6=a2b2⇒a2=6b2=2(3b2), so that a2 and hence a must be even.If a=2k (k an integer) then 2k2=6b2⇒3b2=2k2, so that 3b2 and hence b2 and b must also be even.But if both a and b are multiples of 2, this contradicts the original assumption, which means it is false and so 6 is not rational, and so must be irrational.Specific behaviours clearly makes rational assumption deduces that a must be even deduces that b must be even explains contradictionSolutionAssume that 6 is rational and can be expressed in the form ab where a and b are both integers with no common factors.Then 6=a2b2⇒a2=6b2=2(3b2), so that a2 and hence a must be even.If a=2k (k an integer) then 2k2=6b2⇒3b2=2k2, so that 3b2 and hence b2 and b must also be even.But if both a and b are multiples of 2, this contradicts the original assumption, which means it is false and so 6 is not rational, and so must be irrational.Specific behaviours clearly makes rational assumption deduces that a must be even deduces that b must be even explains contradiction

Question 16 (7 marks)

(a) Let the angle θ=π3-π4=π12.

Solution23-14=6-24 Specific behaviours writes an exact valueSolution23-14=6-24 Specific behaviours writes an exact value(i) Use your calculator to write down an exact value for sinπ12. (1 mark)

(ii) Use an angle sum or difference identity to show how to obtain the above exact value for sinπ12. (3 marks)

Solutionsinπ3-π4=sinπ3cosπ4-cosπ3sinπ4
=32×22-12×22
=6-24Specific behaviours uses difference identity shows substitution of four correct values shows simplificationSolutionsinπ3-π4=sinπ3cosπ4-cosπ3sinπ4
=32×22-12×22
=6-24Specific behaviours uses difference identity shows substitution of four correct values shows simplification

(b) Prove the identity sinx+sin2x+sin3x=1+2cosxsin2x. (3 marks)

SolutionRHS=1+2cosxsin2x
=sin2x+2sin2xcosx
=sin2x +sin2x+x+sin2x-x
=sinx+sin2x+sin3x
=LHSSpecific behaviours expands RHS uses product as sum and difference completes proof in logical stepsSolutionRHS=1+2cosxsin2x
=sin2x+2sin2xcosx
=sin2x +sin2x+x+sin2x-x
=sinx+sin2x+sin3x
=LHSSpecific behaviours expands RHS uses product as sum and difference completes proof in logical steps

Question 17 (9 marks)

Trapezium OPQR has parallel sides PQ and OR such that OR=k|PQ|. Let OP=a and PQ=b.

(a) Sketch the trapezium. (1 mark)

SolutionSpecific behaviours draws labelled trapeziumSolutionSpecific behaviours draws labelled trapezium

(b) Determine vectors for OQ and PR in terms of k, a and b. (2 marks)

SolutionOQ=a+b and PR=kb-a.Specific behaviours states first vector states second vectorSolutionOQ=a+b and PR=kb-a.Specific behaviours states first vector states second vector

(c) Show that the scalar product of OQ and PR is kb2-a2+k-1a⋅b. (2 marks)

Solutiona+b⋅kb-a=ka⋅b-a.a+kb.b-a.b
=kb2-a2+k-1a⋅bSpecific behaviours expands scalar product simplifies scalar productSolutiona+b⋅kb-a=ka⋅b-a.a+kb.b-a.b
=kb2-a2+k-1a⋅bSpecific behaviours expands scalar product simplifies scalar product

(d) Simplify your result from (c) if k=1, a=i+4j and b=3i-22j. (2 marks)

Solutionkb2-a2+k-1a⋅b=b2-a2+1-1a⋅b
=9+8-1+16+0
=0Specific behaviours substitutes k=1 to eliminate a⋅b determines magnitudes and simplifies expression to zeroSolutionkb2-a2+k-1a⋅b=b2-a2+1-1a⋅b
=9+8-1+16+0
=0Specific behaviours substitutes k=1 to eliminate a⋅b determines magnitudes and simplifies expression to zero

(e) Explain the geometric significance of your result from (d). (2 marks)

SolutionThe values of k, a and b have turned the trapezium into a rhombus and as the scalar product is zero, the diagonals must intersect at right angles.Specific behaviours identifies significance of values uses scalar product to conclude that diagonals intersect at right anglesSolutionThe values of k, a and b have turned the trapezium into a rhombus and as the scalar product is zero, the diagonals must intersect at right angles.Specific behaviours identifies significance of values uses scalar product to conclude that diagonals intersect at right angles

Question 18 (7 marks)

(a) The work done, in joules, by a force F Newtons in changing the displacement of an object s metres is given by the scalar product of F and s. Calculate the work done when a force of 750 N moves an object a distance of 85 cm at an angle of 5° to the force.

(2 marks)

Solution750×0.85×cos5°=635.1 NSpecific behaviours substitutes correct values into scalar productevaluates work doneSolution750×0.85×cos5°=635.1 NSpecific behaviours substitutes correct values into scalar productevaluates work done

(b) A drone flies with a constant velocity and height above level ground, over which a wind blows from the north west at 3.5 metres per second. After 15 seconds, the drone reaches a point 85 metres on a bearing of 020° from where it was launched. Determine the velocity of the drone, giving its magnitude to two decimal places and bearing to the nearest degree. (5 marks)

SolutionWind component is 15×3.5=52.5 ma2=52.52+852-252.585cos115⇒a=117.2737 Speed of drone is 117.2737÷15=7.82 m/s (2dp).52.5sinx=106.4211sin115⇒x=24° and so bearing is 020-24=356°.Specific behaviours sketch displacement vector diagram uses cosine rule to determine drone displacement calculates drone speed uses sine rule to determine angle determines bearingSolutionWind component is 15×3.5=52.5 ma2=52.52+852-252.585cos115⇒a=117.2737 Speed of drone is 117.2737÷15=7.82 m/s (2dp).52.5sinx=106.4211sin115⇒x=24° and so bearing is 020-24=356°.Specific behaviours sketch displacement vector diagram uses cosine rule to determine drone displacement calculates drone speed uses sine rule to determine angle determines bearing

Question 19 (8 marks)

(a) A high school has 5 male and 9 female volunteers from which to choose a debating team of 5 students. Determine the number of different teams that can be formed if

(i) there are no special requirements. (1 mark)

Solution145=2002 teamsSpecific behaviours evaluates correct numberSolution145=2002 teamsSpecific behaviours evaluates correct number

(ii) there must be a captain and a vice-captain. (2 marks)

Solution141×131×123=40040 teamsSpecific behaviours shows a suitable method evaluates correct numberSolution141×131×123=40040 teamsSpecific behaviours shows a suitable method evaluates correct number

(iii) there must be more females than males, but at least one male. (2 marks)

Solution51×94+52×93=1470 teamsSpecific behaviours shows a suitable method evaluates correct numberSolution51×94+52×93=1470 teamsSpecific behaviours shows a suitable method evaluates correct number

(b) Determine how many different numbers must be selected from the first 25 positive integers to be certain that at least one of them will be twice the other. (3 marks)

SolutionPartition integers into pigeonholes, where double included if possible:{1, 2}, {3, 6}, {4, 8}, {5, 10}, {7, 14}, {9, 18}, {11, 22}, {12, 24},{13}, {15}, {16}, {17}, {19}, {20}, {21}, {23}, {25}.NB Other partitions possible.There are 17 partitions (pigeonholes) and so 18 numbers (pigeons) are required to ensure that at least one will be twice the other. Specific behaviours partitions most integers systematically partitions all integers applies pigeonhole principle to get correct numberSolutionPartition integers into pigeonholes, where double included if possible:{1, 2}, {3, 6}, {4, 8}, {5, 10}, {7, 14}, {9, 18}, {11, 22}, {12, 24},{13}, {15}, {16}, {17}, {19}, {20}, {21}, {23}, {25}.NB Other partitions possible.There are 17 partitions (pigeonholes) and so 18 numbers (pigeons) are required to ensure that at least one will be twice the other. Specific behaviours partitions most integers systematically partitions all integers applies pigeonhole principle to get correct number

Question 20 (7 marks)

In the diagram below, the tangents from point A touch the circle at B and F. Point E lies on the major arc BF and D lies on BF so that DE⊥BF. Points C and G lie on AB and AF extended respectively such that EC⊥AC and EG⊥AG.

(a) Show that ΔBCE and ΔFDE are similar. (3 marks)

Solution∠CBE=∠DFE (alternate segment theorem)∠BCE=∠FDE (both right)Hence ΔBCE∼ΔFDE (AA)Specific behaviours shows one pair of angles equal with reason shows second pair of angles equal with reason makes conclusion that similarSolution∠CBE=∠DFE (alternate segment theorem)∠BCE=∠FDE (both right)Hence ΔBCE∼ΔFDE (AA)Specific behaviours shows one pair of angles equal with reason shows second pair of angles equal with reason makes conclusion that similar

(b) Show that DE2=CE×GE. (4 marks)

Solution∠GFE=∠DBE (alternate segment theorem)∠BDE=∠FGE (both right)Hence ΔBDE∼ΔFGE (AA)So ratio of sides is DEGE=BEFEBut ratio of sides from (a) is CEDE=BEFE⇒CEDE=DEGE⇒DE2=CE×GE.Specific behaviours realises second pair of similar triangles required uses same reasoning from (a) to show ΔBDE∼ΔFGE states ratio of corresponding sides for both pairs of triangles uses common ratio to obtain resultSolution∠GFE=∠DBE (alternate segment theorem)∠BDE=∠FGE (both right)Hence ΔBDE∼ΔFGE (AA)So ratio of sides is DEGE=BEFEBut ratio of sides from (a) is CEDE=BEFE⇒CEDE=DEGE⇒DE2=CE×GE.Specific behaviours realises second pair of similar triangles required uses same reasoning from (a) to show ΔBDE∼ΔFGE states ratio of corresponding sides for both pairs of triangles uses common ratio to obtain result

Additional working space

Question number: _________

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