3A MAS Calc-assumed marking key.docx

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3A MAS Calc-assumed marking key.docx

Revision Examination Assessment Papers (REAP)

Semester 1 Examination 2012

Question/Answer Booklet

(This paper is not to be released to take home before 25/6/2012)

MATHEMATICS:

SPECIALIST 3A

Section Two:

Calculator-assumed

Name of Student: ______________Marking Key________________________________

Time allowed for this section

Reading time before commencing work: 10 minutes

Working time for this section: 100 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 student

Standard items: pens, pencils, pencil sharpener, eraser, correction fluid/tape, ruler,

highlighters

Special items: drawing instruments, templates, notes on two unfolded sheets of A4 paper,

and up to three calculators satisfying the conditions set by the Curriculum

Council for this examination

Important note to students

No other items may be used in this section of the examination. It is your responsibility to ensure

that you do not have any unauthorised notes or other items 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

Section	Number of questions available	Number of questions to be answered	Working time(minutes)	Marks available	Percentage of exam
Section OneCalculator-free	6	6	50	50
Section TwoCalculator-assumed	12	12	100	100

Total

150

100

Instructions to students

1 Write your answers in the spaces provided in this Question/Answer Booklet. 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. 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(s) that you are continuing to answer at the top of the page.

2 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 an answer to any question, ensure that you cancel the answer you do not wish to have marked.

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

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

Section Two: Calculator-assumed (100 marks)

This section has twelve (12) questions. Answer all questions. Write your answers in the spaces provided.

Working time: 100 minutes

_________________________________________________________________________________

Question 7 (10 marks)

Determine Leave your answer as a surd (2)

Solution

Specific behaviours

simplify to determines magnitude of

Determine (2)

Solution

Specific behaviours

correct expression for simplifies to correct answer

Determine a vector, (in exact simplest form) in the direction of

such that . (3)

Solution

Unit vector = Or

Specific behaviours

magnitude of unit vector of times 5 simplified answer

Find the value of k if and are parallel vectors.

, (3)

Solution

Specific behaviours

Question 8 (8 marks)

(a) Beijing in China is located at 40oN, 116oE. Perth in Western Australia is located at

32oS, 116oE. Karen’s direct flight from Beijing to Perth departed at 11.00pm on Sunday. If the plane had an average flying speed of 900km/h, how long does the flight take?

Express your answer in hours and minutes.

Take the radius of the earth to be 6360km. (4)

Solution

Arc BP = = 7992.21 kmTime taken to reach Perth = = 8.88 hr = 8 hr 53 min

Specific behaviours

angle of 720 arc length BP time in hour time in hour and mins

(b) Determine the equation for each of the following graphs by selecting from the list below.

(4)

(i) (ii)

(iii) (iv)

Solution

Graph (i ) - Graph (ii) – Graph (iii) – Graph (iv) -

Specific behaviours

for each correct answer

Question 9 (10 marks)

(a) Point A has position vector . Point B has position vector .

If , find the value(s) of k. (4)

Solution

Specific behaviours

express in i – j component correctly for solves for the two “k” values

(b) Let A=(5,1), B=(0,4), C=(-1,0)

(Hint: Use the grid below to help you find the points)

					X FX F
		X BX B
D XD X

				X AX A
	C -1C -1
			X GX G

Find

D such that (2)

Solution

D = ( -6, 3)

Specific behaviours

for x-coordinate for y- coordinate

F such that (2)

Solution

F = (6, 5)

Specific behaviours

for x-coordinate for y-coordinate

G such that (2)

Solution

G = (1.5,-1.5)

Specific behaviours

for x-coordinate for y-coordinate

Question 10 (10 marks)

The area of PQR is 16cm2. PQ=8cm, RPQ=30o.

Find the length of PR. (2)

Solution

PR = 8 cm

Specific behaviours

uses the Area of non right triangle formula correct answer for PR

Hence, find the size of PQR. (2)

Solution

As is an isosceles triangle

Specific behaviours

identifying triangle is isosceles size of

Using the SINE rule and , show that Show all working steps. (6)

Solution

Specific behaviours

expression for the Sine Rule substitute expression for sin 750 simplify expression multiply by conjugate of denominator simplify and gets to

Question 11 (9 marks)

The diagram shows the graph of y=f(x), xR.

bb(1,a)(1,a)

If ,

(i) Find the value of a and the value of b. (2)

Solution

f(0) = b ---- b = -1f(1) = a ---- a = -2

Specific behaviours

for each “a” and “b” value

(ii) Solve algebraically the value of x for which f(x) = 5x (3)

Solution

Specific behaviours

correct piece of f(x) used i.e. –(x-1) - 2 expand and rearrange solve correctly for value of x

(iii) Determine the solution set for (2)

Solution

Specific behaviours

for for value of 1

(iv) Find the value of k for which has no solution. (2)

Solution

Specific behaviours

for < for ( -3)

Question 12 (8 marks)

(a) The population, P, of cane toads in Australia has been growing at a rate proportional to P,

such that where k is a positive constant. There were 102 cane toads brought to Australia in 1935. Seventy six years later, in 2011, it is estimated that there are 243 million cane toads in Australia.

Find the value of ‘k’ to 4 decimal places. (2)

Solution

In 2011, t = 76Using CAS to 4 decimal places

Specific behaviours

correct k value to 4 decimal places

If the population continues to grow at this rate, how many cane toads will there be in Australia in 2035? (2)

Solution

In 2035, t = 100P= 2.5 x 1010 or P = 25071 million

Specific behaviours

t = 100 correct value for P

(b) In 2005 there were 13.8million registered motor vehicles in Australia. The number of registered motor vehicles is increasing at a rate of 2.3o/o per year.

Write an expression to represent the number (in millions) of registered motor vehicles, N, if t represents the number of years after 2005. (2)

Solution

Specific behaviours

exponential equation, y = k. a x k= 13.8, a = 1.023

Hence determine the number of registered vehicles in Australia in 2011. (2)

Solution

In 2011, t = 6N = 13.8(1.023)6N= 15.8 million

Specific behaviours

for t = 6 for correct value of N

Question 13 (6 marks)

In the diagram

(a) Express as simply as possible, in terms of (2)

Solution

Specific behaviours

correct answer

Solution

Specific behaviours

correct answer

(b) If , express in terms of and (1)

Solution

Specific behaviours

correct answer

Solution

Equating coefficients: Or

Specific behaviours

replace equation by simplify solve correctly for value of h

Question 14 (11 marks)

The diagram shows the graph of f(x)=ax and g(x)=bx2. The graph f(x), intersects g(x) at

(a) Calculate the value of a and b. (2)

Solution

Specific behaviours

correct vale for “a” and “b”

(b) Write down the equation of the inverse function, f -1(x). (3)

Solution

OR OR using CAS Solve(x=0.1y,y) results in

Specific behaviours

take the “log” of both sides of equation re arrange and correct answer for the inverse function

Solution

g(x) is a many-to-one function OR For a function to have an inverse function, it must be a one-to-one functionOR g(1)=0.1 and g(-1) = 0.1, hence g(x) cannot have an inverse function

Specific behaviours

for anyone of the above reasons

(d) Write down two ways in which the domain of g(x) could be restricted in order that g(x) can have an inverse function. (2)

Solution

(i) or (ii) or

Specific behaviours

for each of the two ways

(e) Determine the x-values for which

f -1(x) >0 (2)

Solution

0 < x < 1

Specific behaviours

correct interval

f(x )- 1=g(x) (1)

Solution

x = 0

Specific behaviours

or X

Question 15 (6 marks)

The diagram shows the graph of y=f(x)

(i) Sketch the graph of

showing clearly the coordinates of the turning points and the intersections with the axes. (3)

Solution

As shown in diagram above

Specific behaviours

shape points (-3,0),(-2,1),(-1,0),(0,1),(1,0)

(ii) f(x) has been transformed and the transformed graph is shown below.

State the equation of the transformed graph in terms of f(x). (3)

Show how the point (-3, -1) is transformed to (-2, -2)

Solution

Specific behaviours

for equation for

Question 16 (6 marks)

(a) Find a simple expression for e2lnx (1)

Solution

Let OR Using CAS simplify results in i.e.

Specific behaviours

or X

(b) Solve the equation graphically 2ex=4x+3.

Express your answers to 2 decimal places. (2)

Solution

and

Specific behaviours

correct answers for x

Solution

Specific behaviours

expresses as powers of 3 equate solves correctly for x

Question 17 (7 marks)

M is the mid-point of line segment AB.

(a) Show that (4)

Solution

Specific behaviours

correctly shows each line with no shortcut as it is a “show”

(b) Hence, or otherwise state the coordinates of S if S divides the line segment joining F(1,4) to G(6,9) in the ratio 1:1. (3)

Solution

S is the mid point of FG Or = = Coordinates of S = (3.5, 6.5)

Specific behaviours

identifies that S is the mid-point of FG hence use result from (a) simplifies and correct coordinates for S

Question 18 (9 marks)

The diagram shows a circle of radius 10cm, centre O and a tangent BC of length 16cm, AB=DC, calculate

(i) AOD in radians, to 2 decimal places. (4)

Solution

AD = BCLet M be mid -point of AD =

Specific behaviours

AM = 8 angle AOM angle AOD in degrees angle AOD in radians to two decimal places

(ii) the area of the shaded region. (5)

Solution

Area of OABCD = 64 + (0.5 x 16 x 6) = 112Area of sector AOD = = 92.73 or 92.5 if used 1.85 radians Shaded area = = 9.64 cm2If used 1.85 radians shaded area = 9.75 cm2

Specific behaviours

area of sector AOD area of OABCD shaded area