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Semester Two Examination, 2016

Question/Answer Booklet

SOLUTIONSSOLUTIONSMATHEMATICS

SPECIALIST

UNITS 1 AND 2

Section One:

Calculator-free

In words

Your name

Time allowed for this section

Reading time before commencing work: five minutes

Working time for section: 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 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 One: Calculator-free 35% (51 Marks)

This section has seven (7) questions. Answer all questions. Write your answers in the spaces provided.

Working time for this section is 50 minutes.

Question 1 (6 marks)

(a) Determine the number of real solutions to the equation x2+x+1=0. (1 mark)

Solutionb2-4ac=-3⇒ no real solutionsSpecific behaviours uses discriminant, or otherwise, to determine no solutionsSolutionb2-4ac=-3⇒ no real solutionsSpecific behaviours uses discriminant, or otherwise, to determine no solutions

(b) Determine all complex solutions to the equation x2+2x+10=0. (2 marks)

Solutionx+12=-9=9i2
x=-1±3iSpecific behaviours completes square or uses quadratic formula states both complex solutionsSolutionx+12=-9=9i2
x=-1±3iSpecific behaviours completes square or uses quadratic formula states both complex solutions

(c) x1 and x2 are the complex solutions to the equation 4x2=20x-41. If x1=2.5+2i, plot x1, x2 and x1+x2 in the complex plane below. (3 marks)

Solutionx2 must be conjugate, so x2=2.5-2i and x1+x2=5Specific behaviours determines x2 plots x1 and x2 determines sum and plotsSolutionx2 must be conjugate, so x2=2.5-2i and x1+x2=5Specific behaviours determines x2 plots x1 and x2 determines sum and plots

Question 2 (7 marks)

Three vectors are given by a=2i-2j, b=i-3j and c=3i+j.

Determine

(a) a unit vector d, parallel to a+2b. (3 marks)

Solutiond=a+2b=4-8 and d=80=45 d=1454-8=15-25 Specific behaviours calculates a+2b calculates magnitude states unit vector in simplified formSolutiond=a+2b=4-8 and d=80=45 d=1454-8=15-25 Specific behaviours calculates a+2b calculates magnitude states unit vector in simplified form

(b) the value(s) of k so that the magnitude of the vector a+kb is 4. (4 marks)

Solutiona+kb=2+k-2-3k Require 2+k2+-2-3k2=424+4k+k2+4+12k+9k2-16=0
10k2+16k-8=0
5k2+8k-4=0
5k-2k+2=0k=25 or k=-2Specific behaviours writes magnitude equation expands and simplifies equation factorises equation states both solutionsSolutiona+kb=2+k-2-3k Require 2+k2+-2-3k2=424+4k+k2+4+12k+9k2-16=0
10k2+16k-8=0
5k2+8k-4=0
5k-2k+2=0k=25 or k=-2Specific behaviours writes magnitude equation expands and simplifies equation factorises equation states both solutions

Question 3 (9 marks)

Consider the matrices A=2-3-24, B=-32, C=10-102-2 and D=4-5.

(a) It is possible to form the product of all four matrices. State the dimensions of the resulting product. (2 marks)

SolutionABDC or BDAC are possible, both resulting in a 2×3 matrix.Specific behaviours lists possible product states dimensions of productSolutionABDC or BDAC are possible, both resulting in a 2×3 matrix.Specific behaviours lists possible product states dimensions of product

(b) Determine the matrix 12DC. (2 marks)

Solution12×4-5×10-102-2=12×4-106
=[2-53]Specific behaviours calculates DC calculates required resultSolution12×4-5×10-102-2=12×4-106
=[2-53]Specific behaviours calculates DC calculates required result

(c) Determine the inverse of matrix A. (2 marks)

SolutionA-1=18--64322
=21.511Specific behaviours uses determinant determines inverseSolutionA-1=18--64322
=21.511Specific behaviours uses determinant determines inverse

(d) Clearly show use of matrix algebra to solve the system of equations 2x-3y+3=0 and 4y=2x+2. (3 marks)

Solution2x-3y=-3-2x+4y=2⇒AX=B, where X=xyX=A-1B=21.511-32=-3-1 x=-3, y=-1 Specific behaviours shows system can be written as matrix equation shows pre-multiplication of equation by inverse from (c) states solution of systemSolution2x-3y=-3-2x+4y=2⇒AX=B, where X=xyX=A-1B=21.511-32=-3-1 x=-3, y=-1 Specific behaviours shows system can be written as matrix equation shows pre-multiplication of equation by inverse from (c) states solution of system

Question 4 (7 marks)

Let z1=2-2i and z2=3+i.

(a) Simplify

(i) 2z1-z2. (1 mark)

Solution4-4i-3-i=1-5iSpecific behaviours simplifies resultSolution4-4i-3-i=1-5iSpecific behaviours simplifies result

(ii) z13. (2 marks)

Solution2-2i2-2i2-2i=-8i2-2i
=-16-16iSpecific behaviours simplifies z12 simplifies z13Solution2-2i2-2i2-2i=-8i2-2i
=-16-16iSpecific behaviours simplifies z12 simplifies z13

(iii) z1z2. (2 marks)

Solution2-2i3-i3+i3-i=4-8i9+1
=25-45iSpecific behaviours multiplies by conjugate simplifiesSolution2-2i3-i3+i3-i=4-8i9+1
=25-45iSpecific behaviours multiplies by conjugate simplifies

(b) Show that z1×z2=z1×z2. (2 marks)

SolutionLHS=2+2i3-i=8+4i
RHS=2-2i3+i=8-4i=8+4i Specific behaviours evaluates RHS evaluates LHSSolutionLHS=2+2i3-i=8+4i
RHS=2-2i3+i=8-4i=8+4i Specific behaviours evaluates RHS evaluates LHS

Question 5 (7 marks)

(a) Solve the equation tanx+25°2=3 for 0°≤x≤540°. (3 marks)

Solution0°≤x≤540°⇒0°≤x2≤270°x+25°2=60°, 240°x=95°, x=455°Specific behaviours uses tan60°=3 determines first solution determines second solutionSolution0°≤x≤540°⇒0°≤x2≤270°x+25°2=60°, 240°x=95°, x=455°Specific behaviours uses tan60°=3 determines first solution determines second solution

(b) Prove that 1-cosx1+secx=sinxtanx. (4 marks)

SolutionLHS=1+secx-cosx-cosxsecx
=secx-cosx
=1-cos2xcosx
=sin2xcosx
=sinxtanx
=RHSSpecific behaviours expands and simplifies LHS combines into single fraction uses Pythagorean identity simplifies to RHSSolutionLHS=1+secx-cosx-cosxsecx
=secx-cosx
=1-cos2xcosx
=sin2xcosx
=sinxtanx
=RHSSpecific behaviours expands and simplifies LHS combines into single fraction uses Pythagorean identity simplifies to RHS

Question 6 (7 marks)

(a) Determine the value(s) of a for which the matrix aa32a is singular. (2 marks)

SolutionSingular ⇒ determinant is zero, so require 2a2-3a=0.a2a-3=0⇒a=0 or a=32 Specific behaviours writes determinant in terms of a and equates to 0 solves equation for aSolutionSingular ⇒ determinant is zero, so require 2a2-3a=0.a2a-3=0⇒a=0 or a=32 Specific behaviours writes determinant in terms of a and equates to 0 solves equation for a

(b) The non-singular matrix B is such that [-32]×B=[83] and [26]×B=[104].

(i) Use these results to show that [-18]×B=[187]. (2 marks)

Solution-32×B+[26]×B=[83]+[104]
-32+[26]×B=[187]
[-18]×B=[187]Specific behaviours uses sum of equations illustrates distributive lawSolution-32×B+[26]×B=[83]+[104]
-32+[26]×B=[187]
[-18]×B=[187]Specific behaviours uses sum of equations illustrates distributive law

(ii) Determine [21]×B-1. (3 marks)

Solution[26]-[-32]×B=104-[83]
[54]×B=[21]
[54]×B×B-1=[21]×B-1
[54]=[21]×B-1Specific behaviours uses difference of equations shows post-multiplication by inverse clearly shows resultSolution[26]-[-32]×B=104-[83]
[54]×B=[21]
[54]×B×B-1=[21]×B-1
[54]=[21]×B-1Specific behaviours uses difference of equations shows post-multiplication by inverse clearly shows result

Question 7 (8 marks)

(a) Prove that the sum of any three consecutive terms of an arithmetic sequence with first term a and common difference d is always a multiple of three, for a, d ∈N. (3 marks)

SolutionLet Tn=a+n-1d so that Tn+Tn+1+Tn+2=(a+nd-d)+(a+nd)+(a+nd+d) =3a+3nd=3a+nd⇒ always a multiple of 3Specific behaviours writes expression for three consecutive terms of arithmetic sequence simplifies expression factors 3 out and states conclusionSolutionLet Tn=a+n-1d so that Tn+Tn+1+Tn+2=(a+nd-d)+(a+nd)+(a+nd+d) =3a+3nd=3a+nd⇒ always a multiple of 3Specific behaviours writes expression for three consecutive terms of arithmetic sequence simplifies expression factors 3 out and states conclusion

(b) Use mathematical induction to prove that 72n-1+5 is always divisible by 12, for n∈N.

(5 marks)

SolutionLet fn=72n-1+5, so clearly true when n=1 as f1=12.Assume that f(k) is always true, so that fk=72k-1+5=12I, where I is an integer.fk+1=72k+1-1+5
=72+2k-1+5
=72×72k-1+5
=49×72k-1+5
=48×72k-1+72k-1+5
=48×72k-1+12I
=12(4×72k-1+I)Since f(1) is divisible by 12, and it has been shown that if f(k) is, so is f(k+1), then 72n-1+5 is divisible by 12 for all n≥1.Specific behaviours shows true for initial case assumes true for n=k and equates result to multiple of 12 uses index laws to achieve 49×72k-1+5 factors 12 out of expression makes summary statementSolutionLet fn=72n-1+5, so clearly true when n=1 as f1=12.Assume that f(k) is always true, so that fk=72k-1+5=12I, where I is an integer.fk+1=72k+1-1+5
=72+2k-1+5
=72×72k-1+5
=49×72k-1+5
=48×72k-1+72k-1+5
=48×72k-1+12I
=12(4×72k-1+I)Since f(1) is divisible by 12, and it has been shown that if f(k) is, so is f(k+1), then 72n-1+5 is divisible by 12 for all n≥1.Specific behaviours shows true for initial case assumes true for n=k and equates result to multiple of 12 uses index laws to achieve 49×72k-1+5 factors 12 out of expression makes summary statement

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