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Applecross Senior High School

Semester One Examination, 2020

Question/Answer booklet

SOLUTIONSSOLUTIONSMATHEMATICS

SPECIALIST

UNIT 1

Section One:

Calculator-free

In words

Your 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.

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 preferably using a blue/black pen.
Do not use erasable or gel pens.

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

4. 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.

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

6. Supplementary pages for planning/continuing your answers to questions are 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.

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

Section One: Calculator-free 35% (52 Marks)

This section has eight questions. Answer all questions. Write your answers in the spaces provided.

Working time: 50 minutes.

Question 1 (3 marks)

Two vectors a and b have magnitudes 3 and 3 respectively. The angle between the two vectors is measured at 30°. Find the magnitude of the resultant of the two vectors.

Question 2 (7 marks)

Two forces are given by F1=-3i+5j N and F2=2i-j N.

(a) Determine

Solution-35-2-1=-56 NSpecific behaviours correct vectorSolution-35-2-1=-56 NSpecific behaviours correct vector(i) F1-F2. (1 mark)

(ii) 5F1+10F2. (2 marks)

Solution5-35+102-1=-1525+20-10=515 NSpecific behaviours one correct multiple correct vectorSolution5-35+102-1=-1525+20-10=515 NSpecific behaviours one correct multiple correct vector

Solution-35=-32+52=34 NSpecific behaviours correct valueSolution-35=-32+52=34 NSpecific behaviours correct value(iii) |F1|. (1 mark)

(b) The resultant of 3F1, 6F2 and a third force is 5i+4j N. Determine the magnitude of the third force. (3 marks)

SolutionF3=54--915-12-6
=2-5|F3|=4+25=29Specific behaviours correct vector equation correct third force correct magnitudeSolutionF3=54--915-12-6
=2-5|F3|=4+25=29Specific behaviours correct vector equation correct third force correct magnitude

Question 3 (8 marks)

(a) Consider the statement: n=2⇒n2=4.

(i) Write the inverse statement. (1 mark)

Solutionn≠2⇒n2≠4Specific behaviours correct statementSolutionn≠2⇒n2≠4Specific behaviours correct statement

(ii) Write the converse statement. (1 mark)

Solutionn2=4⇒n=2Specific behaviours correct statementSolutionn2=4⇒n=2Specific behaviours correct statement

(b) State whether each of the following statements are true or false, supporting each answer with an example or counterexample.

(i) ∀ positive integer x, x≤x. (2 marks)

SolutionTrue.If x=1, x=1 and 1≤1.Specific behaviours states true, with counterexample counterexample using positive integerSolutionTrue.If x=1, x=1 and 1≤1.Specific behaviours states true, with counterexample counterexample using positive integer

(ii) ∀a∈R, ∃b∈R such that ab=24. (2 marks)

SolutionFalse.If a=0 then no value for b exists so that ab=24.Specific behaviours states false, with counterexample counterexample with a=0SolutionFalse.If a=0 then no value for b exists so that ab=24.Specific behaviours states false, with counterexample counterexample with a=0

(c) If a true statement is negated, explain whether the contrapositive of the negated statement will also be true.

(2 marks)

Question 4 (6 marks)

The position vectors of points A and B are rA=-83 and rB=7-2.

(a) Determine the position vector of point P that divides AB internally in the ratio 2:3.

SolutionAB=7-2--83=15-5P=A+25AB
=-83+2515-5
=-21Specific behaviours vector AB indicates appropriate method correct position vectorSolutionAB=7-2--83=15-5P=A+25AB
=-83+2515-5
=-21Specific behaviours vector AB indicates appropriate method correct position vector (3 marks)

SolutionAC=69--83=146But 146≠k21 and so body will not pass through C as AC is not parallel to 21.Specific behaviours vector AC indicates AC not parallel to velocity states will not pass through CSolutionAC=69--83=146But 146≠k21 and so body will not pass through C as AC is not parallel to 21.Specific behaviours vector AC indicates AC not parallel to velocity states will not pass through C(b) A small body leaves A and moves with a constant velocity in a direction parallel to 21. Determine, with reasons, whether the body will pass through point C with position vector rC=69. (3 marks)

Question 5 (7 marks)

(a) 4 different letters are chosen from the 7 in the word PAYMENT and then arranged to form a password. Determine how many different passwords are possible that

Solutionn---T=6×5×4×1=120Specific behaviours correct numberSolutionn---T=6×5×4×1=120Specific behaviours correct number(i) end in T. (1 mark)

(ii) end in T or start with P. (3 marks)

SolutionnP---=120nP--T=1×5×4×1=20n=120+120-20=220Specific behaviours number for P and T uses inclusion-exclusion principal correct numberSolutionnP---=120nP--T=1×5×4×1=20n=120+120-20=220Specific behaviours number for P and T uses inclusion-exclusion principal correct number

(b) Determine the number of two letter permutations that can be made using letters from the word REPAYMENT. (3 marks)

Solution8 letters - 7 singles and 1 double (E).Both different: 8×7=56Both the same: 1×1=1Total permutations: 56+1=57.Specific behaviours breaks into exclusive cases correct calculation for each case correct numberSolution8 letters - 7 singles and 1 double (E).Both different: 8×7=56Both the same: 1×1=1Total permutations: 56+1=57.Specific behaviours breaks into exclusive cases correct calculation for each case correct number

Question 6 (7 marks)

Trapezium OABC is such that AB=3OC.

The midpoints of sides OA, AB, BC and OC are P,Q, R and S.

Let OA=a and OC=c. Use a vector method to prove that PQRS is a parallelogram.

SolutionThen OP=12a, OQ=a+32c.Hence PQ=OQ-OP=a+32c-12a=12a+32c.Note that CB=-c+a+3c=a+2c.Also OS=12c, OR=c+12a+2c=12a+2c.Hence SR=OR-OS=12a+2c-12c=12a+32c.Hence PQRS is a parallelogram as PQ=SR
(has a pair of equal length, parallel sides).Specific behaviours diagram of trapezium, roughly to scale uses correct vector notation throughout vectors OP, OQ vector PQ vectors OS, OR vector SR conclusionSolutionThen OP=12a, OQ=a+32c.Hence PQ=OQ-OP=a+32c-12a=12a+32c.Note that CB=-c+a+3c=a+2c.Also OS=12c, OR=c+12a+2c=12a+2c.Hence SR=OR-OS=12a+2c-12c=12a+32c.Hence PQRS is a parallelogram as PQ=SR
(has a pair of equal length, parallel sides).Specific behaviours diagram of trapezium, roughly to scale uses correct vector notation throughout vectors OP, OQ vector PQ vectors OS, OR vector SR conclusion

Question 7 (7 marks)

Consider the vectors p=-78, q=3-4 and r=1-2.

(a) Determine the vector projection of r onto q. (3 marks)

Solutionq=153-4153-4⋅1-2=115Hence required vector is115×153-4=3325-4425Specific behaviours unit vector for q scalar product correct vectorSolutionq=153-4153-4⋅1-2=115Hence required vector is115×153-4=3325-4425Specific behaviours unit vector for q scalar product correct vector

(b) Given that p=λq+μr, determine the value of λ and the value of μ. (4 marks)

Solutionλ3-4+μ1-2=-78Equating i and j coefficients:3λ+μ=-7
-4λ-2μ=8Hence6λ+2μ=-14
-4λ-2μ=82λ=-6⇒λ=-3
μ=-7-3-3=2λ=-3, μ=2Specific behaviours equation using i-coefficients equation using j-coefficients value of λ value of μSolutionλ3-4+μ1-2=-78Equating i and j coefficients:3λ+μ=-7
-4λ-2μ=8Hence6λ+2μ=-14
-4λ-2μ=82λ=-6⇒λ=-3
μ=-7-3-3=2λ=-3, μ=2Specific behaviours equation using i-coefficients equation using j-coefficients value of λ value of μ

Question 8 (7 marks)

In the diagram shown, A, B and C lie on a circle.

The tangent at C and secant BA intersect at D.

Point E lies on AB so that CE bisects ∠ACB.

(a) Show that ∠DEC=∠DCE. (3 marks)

SolutionGiven ∠ACE=∠BCE=α.Let ∠ACD=β, so that ∠DCE=α+β.∠CBE=∠ACD=β (alternate segment)∠DEC=α+β (sum of opposite interior angles)Hence ∠DEC=∠DCE as required.Specific behaviours uses alternate segment theorem uses triangle properties logical explanationSolutionGiven ∠ACE=∠BCE=α.Let ∠ACD=β, so that ∠DCE=α+β.∠CBE=∠ACD=β (alternate segment)∠DEC=α+β (sum of opposite interior angles)Hence ∠DEC=∠DCE as required.Specific behaviours uses alternate segment theorem uses triangle properties logical explanation

(b) Given that AE=4 cm and BE=9 cm, determine the length of DC. (4 marks)

SolutionDC2=DA×DB (intersecting secants)Let DC=DE=x (isosceles triangle) so that DA=x-4 and DB=x+9.Thenx2=x-4x+9
x2=x2+5x-36
5x=36
x=365 cm (=7.2)Specific behaviours uses intersecting secants theorem uses isosceles triangles to express required lengths forms equation correct lengthSolutionDC2=DA×DB (intersecting secants)Let DC=DE=x (isosceles triangle) so that DA=x-4 and DB=x+9.Thenx2=x-4x+9
x2=x2+5x-36
5x=36
x=365 cm (=7.2)Specific behaviours uses intersecting secants theorem uses isosceles triangles to express required lengths forms equation correct length

Supplementary page

Question number: _________

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