3A MAS Calc-free 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 One:
Calculator-free
Name of Student: ________________________Marking key___________________
Time allowed for this section
Reading time before commencing work: 5 minutes
Working time for this section: 50 minutes
Materials required/recommended for this section
To be provided by the supervisor
This Question/Answer Booklet
Formula Sheet
To be provided by the student
Standard items: pens, pencils, pencil sharpener, eraser, correction fluid/tape, ruler,
highlighters
Special items: nil
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.
Section One: Calculator-free (50 marks)
This section has six (6) questions. Answer all questions. Write your answers in the spaces provided.
Working time: 50 minutes
_________________________________________________________________________________
Question 1 (10 marks)
Let P be point with polar coordinates and O with co-ordinates (0,0 )
(i) State the Cartesian co-ordinates of P. (3)
| Solution |
| x: , x = 2y: , Cartesian coordinates of P = |
| Specific behaviours |
| x-coordinate of 2 negative sign for x-coordinate y-coordinate of |
(ii) Q is a point which lies in the first quadrant such that it has Cartesian
co-ordinates . State the polar co-ordinates of Q. (2)
| Solution |
| , or , Polar coordinates of Q = |
| Specific behaviours |
| “r” value of 8 argument of or |
(iii) Hence, calculate the exact distance between P and Q. (3)
| Solution |
| PQ2 = PQ2 = PQ2 = 48PQ = or |
| Specific behaviours |
| use of Cosine Rule simplify the equation correct answer of or |
(iv) Hence, or otherwise state what type of triangle is ? Justify your answer. (2)
| Solution |
| PQ2 + PO2 = 48 +16 = 64 = OQ2By Pythagoras’ theorem, triangle is a right-angled triangle with angle P as 900 |
| Specific behaviours |
| state triangle is a right-angled triangle evidence of Pythagoras Theorem |
Question 2 (10 marks)
X QX Q(a)
X CX CAA
X PX P
BBOO
It is given that 4a and 4b.
If , mark the point P on the grid. (2)
| Solution |
| = = |
| Specific behaviours |
| calculate correctly point P correctly marked on grid |
Q is the point of intersection of OA and BP produced. Mark the point Q clearly on the grid. Hence state the value of n given that . (2)
| Solution |
| n = 2 |
| Specific behaviours |
| point Q correctly marked on grid correct value of n |
If , mark the point C on the grid. (1)
| Solution |
| = |
| Specific behaviours |
| point C correctly marked on grid |
If , express in component. (1)
| Solution |
| From diagram |
| Specific behaviours |
| correct expression for |
(b) Solve the equation. (4)
| Solution |
| Let or i.e. or (not possible) |
| Specific behaviours |
| multiply both sides by expresses equation as a quadratic equal to zero factorises correctly finds a solution which is |
Question 3 (9 marks)
The graphs of f(x), g(x) and h(x) are drawn below such that
f(x) is a reciprocal function, x>0.
g(x) is a quadratic function such that 6y = -x2+28
h(x) is the absolute value function , where m is a constant
The point P(4,2) is the point of intersection of the three graphs.
h(x)h(x)
f(x)f(x)QQP(4,2)P(4,2)
g(x)g(x)
(a) Determine the equation of f(x) (2)
| Solution |
| Specific behaviours |
| general equation of a reciprocal function substitutes (4,2) to get k = 8 , hence |
(b) Show that the value of m is 2. (3)
| Solution |
| Solve and simultaneouslyi.e. But (4,2) lies on f(x) and h(x)i.e. 16 – 8 = 4m m= 2OR i.e. or From diagram because the corner point is to the left of P |
| Specific behaviours |
| use and sub (4,2) into equation solve for m correctly |
(c) Calculate the co-ordinates of Q. (4)
| Solution |
| For Q, , y = 4Coordinates of Q = ( -2, 4) |
| Specific behaviours |
| equate h(x) and g(x) solve for x ignore x= 8 as Q lies in 2nd quadrant coordinates of Q as (-2, 4) |
Question 4 (6 marks)
If and
(i) Find f(-2). (1)
| Solution |
| Specific behaviours |
| correct answer of 5 |
(ii) Determine and simplify . (3)
| Solution |
| ===== for |
| Specific behaviours |
| determines the correct function composition simplifies correctly correct answer of for |
(iii) State the natural domain and corresponding range of . (2)
| Solution |
| Specific behaviours |
| correct domain correct range |
Question 5 (6 marks)
In the diagram below, LMNP is a square whose diagonals are each 2cm long. MP and LP are diameters of the bigger and smaller circles respectively. Find the perimeter of the shaded region, expressing your answer in surd form.
XX1111
| Solution |
| Note: The diagonals of a square bisect each other at right anglesLet X be the centre of smaller circleLP =PX=LX = Semi- circle of smaller circle: LP = = Arc LP of bigger circle = Perimeter of shaded region = |
| Specific behaviours |
| length of LP semi –circle LP arc LP perimeter of shade region |
Question 6 (9 marks)
(a) Show that. (3)
| Solution |
| LHS = = = 1 – 1 = 0 |
| Specific behaviours |
| express as , as , as simplify to zero |
(b) Solve for x. (3)
| Solution |
| OR |
| Specific behaviours |
| express group LIKE terms simplify to |
(c) If 25.322=40 determine log280. (3)
Show all calculations. (Hint: Let x=log280)
| Solution |
| i.e. |
| Specific behaviours |
| express “log” equation as an exponential equation simplify to and equate correct answer of |