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INSIGHT YEAR Trial Exam Paper 03 MATHEMATICAL METHODS (CAS) Written examination s This book presents: correct solutions with full working mark allocations tips This trial examination produced by

1 INSIGHT YEAR Trial Exam Paper 03 MATHEMATICAL METHODS (CAS) Written examination s This book presents: correct solutions with full working mark allocations tips This trial examination produced by Insight Publications is NOT an official VCAA paper for the 03 Mathematical Methods CAS written examination. This examination paper is licensed to be printed, photocopied or placed on the school intranet and used only within the confines of the purchasing school for examining their students. No trial examination or part thereof may be issued or passed on to any other party including other schools, practising or non-practising teachers, tutors, parents, websites or publishing agencies without the written consent of Insight Publications.

2 SECTION Question Answer is C Period is and the amplitude is the coefficient of the sine term, which is. n Tip Amplitude and period are never negative. Question Answer is D Choose some sample values for a and b. Using CAS, the graph of f ( x) loge ( x) 5 looks like This shows the lowest point that the graph reaches is 5, so the range is [ 5, ) or in the general sense [ b, ). SECTION

3 3 Question 3 Answer is D The amplitude is ( b), the period is (so, so n ), the graph has been reflected n across the x-axis, and shifted down units. These values give an equation of y ( b)sin( x), which simplifies to give y ( b)sin( x). Question 4 Answer is E The graph of y tan(ax) has a period of and asymptotes at x a a. So, in this case Question 5 Answer is B z 3 xz 5 x y0 can be written as 0x0yz 3 x0yz 5 xy0z 0 a a a 6 Moving line to the bottom gives x0yz 5 xy0z 0 0x0yz 3 In matrix form, this is x 5 0 y 0. z 3 SECTION TURN OVER

4 4 Question 6 Answer is A y e x 4 3 can be written as So y y 3 and equations is A. x 4 y3 e and therefore y y 3 and xx 4. x ( x4) x. The matrix equation that corresponds to these Question 7 Answer is E Overall, the graph represents a negative polynomial of odd degree. It has x-intercepts at a stationary point of inflection at x a, x 0, x b. These give factors of ( x a), x, and ( x b). There is x a, so the factor ( x a) becomes 3 ( x a). 3 This means the equation of the graph could be y x( x a) ( x b) and a version of this is E. Question 8 Answer is D Reflect the graph in the line y-axis to get the graph of y f ( x). y x to get the graph of the inverse and then reflect in the Question 9 Answer is D The period has been halved, therefore there is a dilation of factor from the y-axis. The graph has been reflected in the x-axis. SECTION

5 5 Question 0 Answer is E Let b be any value, say b. Using CAS, the graph of y x e looks like The graph is one to one for the domain [ b, ). Question Answer is D g( f ( x)) and with x 0, g ( f ( x)) x e e 0 SECTION TURN OVER

6 6 Question Answer is D Using the chain rule dy dx f ( x) f ( x) f ( x) f ( x) f ( x) Question 3 Answer is D Using CAS x So the equation of the normal is y. Rearranging, this gives ( y ) x. SECTION

7 7 Question 4 Answer is D The average value of a function is calculated as Using CAS, this gives b x f ( x) dx e( e ) dx a 0 ba 0. Question 5 Answer is C This question represents the fundamental theorem of integral calculus with n 8 i x0 i 0 lim (4 x x) 4x dx Using CAS, this is SECTION TURN OVER

8 8 Question 6 Answer is D 5 5 For f ( x), f( x) dx cos ( kx) cos ( kx) Using CAS, this gives Question 7 Answer is C As q ( x) p( x), this means that q ( x) p( x) dx. b. So p( xdx ) q( x) q( b) q( a) a b a SECTION

9 9 Question 8 Answer is D The graph of y f ( x) will give the gradient function of g. A quick sketch of y f ( x) gives Over interval [ a, b] the y-values are positive, therefore the gradient values are positive. Question 9 Answer is B Let the number of defectives in the box be X. X ~ Bi( n, p ) We are given E( X) np and Var( X) 0 npq. 5 So, q 0, q and 6 p. 6 The chance of not being defective is given by q and Question 0 Answer is D 5 q. 6 Only in a symmetrical distribution is the probability of being less than the mean equal to 0.5. SECTION TURN OVER

10 0 Question Answer is B px ( ) ab0. and EX ( ) a4b0.6.6 Solving the simultaneous equations gives Question Answer is C X ~ N( 76, unknown) Pr( X 75) 0.0 This can be solved using CAS. SECTION

11 SECTION Instructions for Section Answer all questions in the spaces provided. A decimal answer will not be accepted if an exact answer is required to a question. For questions where more than one mark is available, appropriate working must be shown. Unless otherwise stated, diagrams are not drawn to scale. Question a. The information given, together with the transition matrix, suggests Pr( Li Li) Pr( Li Ri) Pr( Ri Li) Pr( Ri Ri), 5 3 where L i is the event of buying lilies one week and R i is the event of buying roses one week. So the chance that Rebecca buys lilies one week, given that she bought roses the previous week is 3. Mark allocation mark for the correct answer. SECTION TURN OVER

12 Question b. i. The fifth Saturday means that there will be four transitions. The probabilities will be Using CAS, this gives So, in 4 weeks time, the probability that Rebecca buys lilies is Mark allocation: + = marks mark for setting the transition matrix to the power of 4. mark for the correct answer. Tip This needs to be the exact answer. A decimal approximation, regardless of the number of decimal places, is not acceptable. Question b. ii. This means that Rebecca buys lilies the first week, then roses for the next 3 weeks, and, finally, lilies the last week. i.e. Mark allocation L R R R L mark for the correct answer. (Note: Answer must be exact.) SECTION

13 Question c. This can be calculated using the probabilities shown on the diagonals So the long term probability of buying roses is Mark allocation: + = marks mark for method. mark for the correct answer. Tip An alternative method is to raise the power of the transition matrix to a sufficiently large power and then evaluate. Question d. As the function is symmetrical around x 6, the mean, median (and mode) are concurrent and occur at x 6. Mark allocation: + = marks mark for the mean. mark for the median. SECTION TURN OVER

14 4 Question e. This is a conditional probability question; i.e. Pr( T 9 T 6). Pr( TR 9) Pr( TR 9 TR 6) Pr( T 6) The probability of lasting longer than 9 days = R R R 5 t( t) dt Pr( TR 9 TR 6) Mark allocation: + = marks 5 mark for finding Pr( TR 9). 3 mark for the correct answer. SECTION

15 5 Question f. 7 For roses, Pr( TR 9) , correct to 4 decimal places. 3 For lilies, Pr ( TL 9) 0.964, correct to 4 decimal places. Mark allocation: + = marks mark for each correct answer. Tip Answers must be in decimal form for this question; 7 3 is not acceptable. SECTION TURN OVER

16 6 Question g. For the probabilities to be equal, t( t) dt normcdf ( k, 00000,, 7.). k 88 Check whether to let k = 7.77 or If 7.77 k then correct to decimal places the integral probability is 0.9, whereas the normal distribution probability is 0.8 (so the two are not equal at the decimal places level). SECTION

17 7 However if k 7.76, the integral probability gives 0.9 and the normal distribution probability gives 0.9 (correct to decimal places) and so they are equal. So, k = Mark allocation: + = marks mark for method. mark for the correct answer. SECTION TURN OVER

18 8 Question a. f is strictly increasing for x : f ( x) 0. The minimum turning point occurs at ( 4, ). So f is strictly increasing for [ 4, ). Mark allocation: + = marks mark for finding the turning point. mark for the correct answer. Note: Must have square bracket at 4. Question b. Mark allocation: + = marks mark for the correct shape. mark for co-ordinates labelled correctly. SECTION

19 9 Question c. Let the width of the rectangle be w, so 9 w ( 4 3) 8 x x dx 3 9 x 4 x 3 dx w(9). So w, therefore D has the co-ordinates of 3 Mark allocation: + = marks mark for method. mark for the correct answer. (, ). 3 SECTION TURN OVER

20 Question d The shaded area is, therefore half the area is 3 3. The graph of y f (x) intersects the line y p at x p 4 p 5 and x p 4 p 5. p4 p5 8 So the integral is ( p( x4 x 3)) dx and we need to find p. p4 p5 3 SECTION

21 Using the graph screen of CAS, this is found to be p Mark allocation: + + = 3 marks mark for stating the area as mark for determining the points of intersection of the line y p with the graph y f (x). mark for finding the value of p. SECTION TURN OVER

22 Question e. i. f g x x x ( ( )) 4 3 x x 4 3 Mark allocation: + = marks mark for method. mark for the correct answer. Note: Must show a working step and not just the answer. Question e. ii. f g x x x So ( ( )) 4 3 d dx x x 4 3 x 4x3 x 0 x 4x3, x0 x 4, ( f ( g( x)) x 4, x 0 x 0 Mark allocation: + = marks mark for setting up the hybrid function of f ( g( x)). mark for the correct derivative with domains specified. SECTION

23 3 Question e. iii. Mark allocation: + + = 3 marks mark for y x 4, x 0 section. mark for y x 4, x 0 section. mark for open circles and correctly labelled intercepts. SECTION TURN OVER

24 4 Question 3a. f Using CAS, f 0.5 and Mark allocation: + = marks mark for each correct answer. Question 3b. i. Using CAS gives d x x x (cos ) cos( )sin( ). dx Mark allocation mark for the correct answer. SECTION

25 Question 3b. ii. 5 Gradient of normal is, therefore gradient of tangent is x x cos sin 4 4 Using CAS to solve over the domain [, 6 ] gives. So there are two points on the curve where the gradient of the normal is ; i.e. at (, ) and (5, ). Mark allocation: + = marks mark for gradient of the tangent equals. mark for giving answer as co-ordinates. SECTION TURN OVER

26 6 Question 3c. Using CAS gives x 3 x 5 y and y Mark allocation: + = marks mark for each correct answer. SECTION

27 7 Question 3d. The tangent lines intersect at x 4, y. So if the graph of y f (x) is shifted down by the x-axis, therefore b. Mark allocation: + = marks mark for finding the point of intersection. mark for the correct answer. units, then the tangents will intersect at SECTION TURN OVER

28 8 CONTINUES OVER PAGE

29 9 Question 3e. Using CAS to simplify gives x x cos cos. 4 So a, b. Period 4 n Mark allocation: + = marks mark for finding a and b values. mark for showing period. SECTION TURN OVER

30 Question 3f. i. Need the lowest point in the cycle to occur at x 6, so half the period needs to be Therefore, the period of the transformed graph is. So Using kx kx k f( kx) cos cos, so 4 Using a graph to verify this shows, 6, n n k Mark allocation mark for the correct answer. SECTION

31 3 Question 3f. ii. The equation has one solution at the end of the domain (at x 6 ) for x x i.e. f( kx) cos cos 6 k ; 3 The equation has two solutions in the domain, with one at x 6, when the period of the graph is 4. Period 4, n, so k. n k 3 Mark allocation: + = marks mark for identifying k. mark for the correct answer. SECTION TURN OVER

32 3 Question 3f. iii. 5x The graph of y cos 6 5 when y, so k. 3 shows that there would be three points in the domain The graph of y cos( x) has four points in the domain where y, so k. 5 k 3 Mark allocation: + = marks 5 mark for obtaining either or. 3 mark for the correct answer. SECTION

33 33 Question 4a. au au i. f ( u) e and f( u) e, so au au auau f( u) f( u) e e e e 0 Mark allocation mark for obtaining e 0. Question 4a. ii. au ( v) auav au av f ( uv) e e e. e f( u) f( v) Mark allocation: + = marks mark for finding f ( u v). mark for forming f ( u). f ( v). Question 4b. Using CAS gives 0.x So the character intersects the obstacle when 9 e 3; i.e. at x 5log 6, y 3. (5log 6, 3) e e Mark allocation: + = marks 0.x mark for setting 9 e 3. mark for the correct answer. SECTION TURN OVER

34 34 Question 4c. To land on the horizontal top section, he needs to land between (5, 3) and (0, 3). At (5, 3), y 9 e ax At (0, 3), y 9 e ax 3 9 e 5a 6 e 5a a log e e 0a 6 e 0a a log e 6 0 log 6 e a loge6 0 5 Mark allocation: + = marks mark for finding either endpoint. mark for the correct interval values. Question 4d. To clear the obstacle, he must jump beyond (3, 0). 3a 3a 9 e 0, e 9 a loge 9 3 So, 0 a log e 9 to clear the obstacle. 3 Mark allocation mark for correct interval values. SECTION

35 35 Question 4e. x 0x y 0 3 y x x x y x, y y 3y 3 So the equation y 0.x 9 e becomes y 3 0.x 0.x 9e y 7 3e. Mark allocation: + = marks x y mark for getting x, y. 3 mark for finding new equation. Question 4f. 0.x The graph of y 9 e shows that Super Marius will clear the base of the obstacle, so just the height of the obstacle will pose a problem. The rising section extends from x 5 to x 0. At x 0 the graph has a y value of 9 e, so to clear the obstacle the horizontal top section has to be less than 9 e units high. 0 p 9 e Mark allocation: + = marks mark for finding 9 e. mark for correct interval values. END OF SOLUTIONS BOOK

INSIGHT YEAR 12 Trial Exam Paper

INSIGHT YEAR 12 Trial Exam Paper 2013 MATHEMATICAL METHODS (CAS) STUDENT NAME: Written examination 1 QUESTION AND ANSWER BOOK Reading time: 15 minutes Writing time: 1 hour Structure of book Number of questions