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Transcription

1 QUESTION 8 a. (i) Vertical translation of a so a. Or as t, f( t) then e 0a a (ii) f t bt e.4 ( ) ( 0.8) t Using ( 3, 086) when t 3, f( t) 086 b e.4( 3) 086 ( 3 0.8) Gives b 5.70 b. f t t e.4 ( ) ( ) t Gives /.4.4 ( ) t t f t e (5.70 t).4(5.70t 0.8) e /.4t f t e t t ( ) (5.70.4( )) /.4t f t e t ( ) (.8.) f / () The School For Excellence 05 Unit 3 & 4 Mathematical Methods Maximising Exam Marks Page

2 c. Find the lowest point on the graph of f '( t ). This point occurs at the stationary point on the graph of f '( t ). i.e. at t.88. or So on the graph of f () t, f (.88).34 Coordinates are (.88,.34) d. (i) y t - Area (8.87) (.8) (.609) 3.73 units The School For Excellence 05 Unit 3 & 4 Mathematical Methods Maximising Exam Marks Page 3

3 (ii) Find f() t 0. or The School For Excellence 05 Unit 3 & 4 Mathematical Methods Maximising Exam Marks Page

4 t-intercept at (-0.69, 0) Integral = 0.69 f () tdt f() tdt 0.69 Area = 0.8 square units (iii) Average value f () tdt ( ) Answer = The School For Excellence 05 Unit 3 & 4 Mathematical Methods Maximising Exam Marks Page

5 e. (i) gt () is reflection over y-axis of f () t. This is f ( t)..4t gt ( ) ( 5.70t0.8) e (ii) k (t) is reflection over y-axis of positive t values of f () t. This is f ( t ). k t t e.4 ( ) t The School For Excellence 05 Unit 3 & 4 Mathematical Methods Maximising Exam Marks Page

6 (iii) h (t) is f ( t ) then translated vertically downwards by unit. This is f( t)..4 t h(t) (5.70 t 0.8) e (iv) k(t).4 t (5.70 t 0.8) e and f t t e.4 ( ) ( ) t When t t then k( t) f ( t) when t 0.4 t k(t) (5.70 t 0.8) e and gt t e.4 ( ) ( ) t When t t then k( t) g( t) when t 0 The School For Excellence 05 Unit 3 & 4 Mathematical Methods Maximising Exam Marks Page 3

7 QUESTION 9 The graph of de y is shown below: dt a. (i) de dt 0.4t y tae b y(0) 5, 5 0 bb 5 de (ii) At t = 0, 05 5 people/hr dt b. (i) dy dt e tae ae t 0.4t 0.4t 0.4t a 0.4 ( 0.4 ) Solve 0.4 t ae (0.4 t) 0 0.4t ae 0 no solution (0.4 t) 0 0.4t t.5 Giving :30 pm The School For Excellence 05 Unit 3 & 4 Mathematical Methods Maximising Exam Marks Page 4

8 (ii) Let y = t tae When t =.5 Giving a 80.93e c. (i) Solve ete 0.4t Gives t ,7.374 The School For Excellence 05 Unit 3 & 4 Mathematical Methods Maximising Exam Marks Page 5

9 Or graphically: Need times after midday Times (to the nearest minute) at :8pm and 7:pm (ii) As the rate is positive for all values of t (the graph is always located above the t axis), this means that more people are entering the venue as opposed to leaving, across all values over hours. e te dt 0.4t d. (i) people The School For Excellence 05 Unit 3 & 4 Mathematical Methods Maximising Exam Marks Page

10 (ii) At t.5 hrs dt people 0.4t te 0 A graph The School For Excellence 05 Unit 3 & 4 Mathematical Methods Maximising Exam Marks Page 3

11 de e. E is found by integrating y dt Screen shows a function with +c found by substituting the point (.5, 40) The School For Excellence 05 Unit 3 & 4 Mathematical Methods Maximising Exam Marks Page 4

12 Graph matches part d. The average value of a function f ( x ) over the interval [ ab, ] is given by b ( ) b a f x dx a Define this complicated function E as g(x) for ClassPad or e(t) for TI-inspire and find average value. Average value of E is people The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page

13 QUESTION 0 a. (i) Initial number = 400 Solve t e t = to the nearest week, 3 weeks. (ii) dp 0.t 80e dt (iii) Rate of decrease after 0 weeks is rabbits per week. The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page

14 b. Use average value function = 5 0.t 400e dt To nearest rabbit = 443 c. (i) To find P o find P (5) where 0.t P 400e. P(5) So Po 70 to the nearest rabbit and P 70 (ii) Using P P 8t 5log t o So P 70 8t 5log t for t 5. Find P (30) : e e o The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 3

15 P(30) 47 rabbits d. (i) P 70 8t 5log t (ii) find dp 8 t 5 8loge t dt t e dp for t 5. dt A Stationary point found at (3.85, 64.34) Not in domain for t 5 e. Average rate of change = P(30) P(0) 30 0 where P(30) is from nd function and P(0) is from st function M 30 0 Average rate of change = 3.77 The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 4

16 f. (i) Use your CAS to define the piece-wise function 0.t P400e Pt () Po 8 t5 loge t30 0 t 5 t 5 Showing meeting point at (5, 70). (ii) Differentiable for t 0,55,. 4A The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 5

17 QUESTION a. b. c. (i) (ii) The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 6

18 d. e. (i) (ii) Use technology to find the points of intersection. x x x e 3 ( 3 ) x x 0, Answer: (0.87,.6) The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 7

19 QUESTION a. (i) When t 4, A 9 : a(4) 9 3(4) 4 4a 9 6 a 9 4 a 36 M (ii) 36t A 3 t 4 Using polynomial long division: 3t 4 36t _ 36t A M 3t 4 4 3t 4 b. D A B 4 3 D 3t 4 t 3 D 3 4 t 3 3t 4 M 3 4 c. D t 3 3t 4 D (3( t 3) 4(3t 4) ) dd dt 3 ( t 3) (3t 4) A The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 8

20 dd d. Maximum value of D occurs when 0 : dt 3 0, 0 t 4. M ( t 3) (3t 4) 3 0 ( t 3) 0 ( t 3) (3t 4) 4 (3t 4) ( t 3) 4 (3t 4) t ( 3 4) 4( 3t 4 ( t 3), t 3) 3t 4 t 6 t M 3 4 When t : D = Hence the maximum difference in new influenza cases between those previously unexposed and those exposed to the virus is A Note: D is measured in hundreds. t k e. C( t) 4( t t) e C( t) 4(t ) e t k e k t k ( t t) 4 t e k t ( t k t) For maximum concentration C ( t) 0 : M t t ( t t) 0 or e k 0 M k tk k t t 0 No solution When t = 4, C ( t) 0 : 8k k k 0 0 k A 9 The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 9

21 f. Let A C(t) Substitute 0 k : 9 A 4e 4e e 5 9t 0 9t 0 9t 0 t 9 0 ( t 3t 9t 0 0 3t 0 9t t) da 9 e dt 5 0 da For maximum absorption 0 : dt 9t 3t 0 9t e 38t 9t 0 0 M 9t 9 0 e t 0 9t t 3t 0 9t e 38t 0 0 t t or 7. 3 M From the graph of C (t) below it can be seen that the maximum rate of absorption occurs after 0.76 hours. A The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 0

22 QUESTION 3 0 a. min C 0 max C 0 b. (i) Let T 4 C t 0 4cos 4 t 4cos 4 t cos t t t 3 pm (ii) 0 at t 3 T 4 C Period t 3 t 3 4 hrs no. cycles 0 T C at t n Z U 0 i.e. At pm every day. n, where n represents the number of cycles and The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page

23 c. (i) T ABsinCt D If T T 0 then T T t 0 4cos ABsin CtD Need to identify what transformations would be required to convert the cosine curve to a sine curve. For curves to be equal, the amplitude, period and vertical translation must be the same. Sine curve must have: Amp 4 Vert Trans 0 Reflection in x axis Equation becomes 0 4sin tc A 0 B 4 C (ii) If c was equal to t i.e. 0 4sin 0 4sin 4 t i.e. sin t 3 t t = 9 hours A Bsin Ct D, the first maximum would occur at t 9 hours. Use this to identify the translation (horizontal) required to move the first sin max (at t 9 ) to the maxima on the cos curve (at t 3 ). i.e. Need to move sin 6 units to left. i.e. t6t 5 The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page

24 t Sine curve equation is: 0 4sin 5 Expand: Equate: 5 0 4sin t 5 D d. Let T3 T or T3 T Safer as both equations have been given and are error free. Alternatively: Move sin curve 4-6 hours = 8 hours to right t 8 t 9 t t t 8 cos sin 0 0 4cos 3 8 cos ( t ) sin ( t ) 4cos ( t ) 3 8 cos ( t ) sin ( t ) 4cos ( t ) cos ( t) sin ( t) 40 3 Case : cos ( t ) 0 3 ( t ), t 6,8 t 7,9 8 Case : sin ( t ) sin ( t ) ( t ), 3 3 t 6, 0 t 7, Therefore, the first two times are t 7, 7. The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 3

25 e. (i) fog is defined if r g d f dg S d (0,] r r (,0] max, g f f y x r g d f -4 (ii) (0,] (0,] This is the max possible range of g. Find the corresponding domain - S. Let t cos 0 t t 6 t 7 t S x:0t 7 t f g t loge cos d d [0, 7) fog g y x The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 4

26 QUESTION 4 a. (i) PM Using right-angled triangle PMT, tan( ) x Giving PM = x tan( ) (ii) Using Pythagoras in right-angled triangle, PT = x +PM Giving PT = So PT = x x tan( ) and PT = x tan ( ) x + taking positive root. tan ( ) b. (i) L( ) = with x = 0 0 tan ( ) and simplify dl 0sin( ) Giving d cos ( ) The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 5

27 dl 0sin( ) (ii) For minimum let 0 d cos ( ) Solution is 0,,,... this gives no actual triangle. Minimum is 0 so PT = TM = 0 metres. c. (i) Using chain rule dx dx d dt d dt dx d 0.5 where cos d dt Giving x x PT and cos dx d cos and sin 0.5 sin d dt d 3M dt 4sin x (ii) d dt 4sin d When then 4 dt 4sin 4 4 d radians/sec. dt The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 6

28 dt d d. (i) 4sin (ii) dt 4sin d d t 4sin and t 4cos c 0 4 cos cc 3 at t = 0 3 Giving t 4cos M Rearrange to give in terms of t: t cos 4 t 4 cos The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 7

29 e. or Graph to find absolute maximum at point 36,. Maximum time Tom can walk is 36 secs. The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page

30 QUESTION 5 a. (i) metres. M a amplitude (ii) Minimum height = 8 + = 0 metres = (-Amplitude + Vertical Translation) 0 b a b 70 M b. Period = 80 seconds = 3 4 minutes 4 c 3 Therefore c. h70 50sin td When 3 c M 3 t h If constn()=0 or If constant n= d is a possible value. M M 4 d. Period = minutes. 3 4 The point P reaches the maximum height once every minutes: Therefore the point P reaches its maximum height 7 times during a 36 minute ride. A The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 3

31 sin 95 e. Solve h t, letting d = for 0 t 36. CAS gives 4 n 8 4 n t 4 where n Z or n n t where n Z 9 9 M When n, 4 t The smallest value of t satisfying 0 t 36 is t minutes. A 9 Therefore the point P first reaches a height of at least 95 metres above ground level after 9 4 minutes. f. During the first rotation P is at a height of 95 metres above the ground a second time 8 when t minutes (see part e) seconds 3 h = 95 M The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 4

32 From the graph it is clear that the number of minutes during one rotation that the point P is at least 95 metres above ground level is minutes: minutes seconds. 9 3 To nearest second, answer = 7 seconds 3 g. h70 50sin t A dh 3 75 cos t dt or dh 3 t 75 sin dt M Solve 00 dh dt graphically: The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 5

33 or M t 0. 5 minutes and t minutes minutes 4. 8 seconds. Therefore dh dt 00 m/s for 4.8 seconds, which is less than 0 seconds. The average person will not feel sick on the Southern Star Observation Wheel. A The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 6

34 h. h70 50sin ct dh 50ccosct dt dh The smallest positive value of c such that 00 for no more than dt 0 seconds at a time and the wheel turns as quickly as possible is required. 0 seconds = minutes. 6 Therefore the positive value of c is required such that if t and t t are two consecutive solutions to 50c cosc( t ) 00, then t t. 6 c( t ) 00 4 c( t ) 50c cos cos c 4 c( t ) cos c 4 t cos c c 4 t cos c c and 4 t cos c c 4 t t cos c c cos 6 c c cos 4 c 4 c M M c 4.67, correct to three decimal places. A The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 7

35 Further clarification about why the expression for t is made equal to. 4 i.e. c( t ) cos c As shown in screen below we have a ve solution and a +ve solution. These are: 4 4 cos cos c c and. c c 4 Rearranging the equation gave us cos( ct ( )). If we were to solve this equation or c the equation cos x 0.5, we would be concerned with finding solutions in the quadrants where cosine is positive i.e. Quadrants and 4. The rules describing the angles in Quadrants and 4 are and which is equal to. Therefore, when we rearrange the equation to create an expression for inverse cosine, these two angle solutions must be taken into consideration. Therefore, 4 t cos where the positive represents the st Quadrant angle and the c c negative sign represents the angle in the 4th Quadrant.) The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 8

36 QUESTION 6 a. Amplitude =. Without a vertical translation, minimum and maximum values would be. To reach values of 0 and, the curve must be raised by, hence a. b. (i) x E sin n Let n0, x 5 5 E sin 0 4 (ii) x sin n0, E 0.75 x sin x 3.33 Therefore, a student must answer at least 4 questions. c. There are a variety of ways to score 50 marks. The student that answers the minimum number of questions would need to correctly answer the six questions that are worth 7 marks each (total = 4 marks) + one 5 mark question + one 3 mark question. Minimum Number Correct Questions = 8 Score (6 7) (5) (3) 50 When x 8, E 0.35 Maximum Number Correct Questions = i.e. The six questions that are worth 3 marks each (total = 8 marks) + five 5 mark questions + one 7 mark question. Score (63) (55) (7) 50 When x, E 0.65 Answer is 0.35 E 0.65 The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 9

37 d. (i) (ii) x' 0.5 0x y' 0 3 y x' 0.5x 0.5x y' 3y 3y x' 0.5x y' 3y x4 x' 8 y ' y 3 Substitute x and y into E( x ): y' ( 4 x' 8) sin 3 n 3 4 ( x ' ) 3 y ' sin n 3 4 ( x ) 5 y sin n x e. (i) Dx ( ) e (ii) ( ) E D x exists when range D( x) domain E( x) domain E( x ): As 0n 00 then 0 x 00 range D( x) [0,00] Maximum range Dx ( ) (0,00] x (Note that e 0 ). The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 0

38 Corresponding domain: x Minimum occurs when e 0 i.e.. x Maximum occurs when e 00 i.e. x loge 00 Gives domain (,loge 00] But answer is [0, loge 00] as x represents the number of correct questions, which cannot be a negative number. de( D( x)) f. Rate of change in cheating dx Using chain rule: de( D( x)) dx x x e e cos n n Allow n 0 and find derivative when x 5, de( D( x)) dx To nearest whole number = 0985 The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page

39 QUESTION 7 t b x Ae cos( ct). The graph of x (t) is illustrated below. a. (i) When t 0, x Ae 0 cos(0) A Also, when t 0, x 5 so A 5. (ii) Using c and f ( 4). 39 t b xt 5e cos( t) 4 b.39 5e cos(8) 4.39 b e 5cos(8) 4.39 loge b 5cos(8) b 4.39 loge 5cos(8) Gives 4 b.39 loge 5cos(8) The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page

40 t 0 b. (i) x 5e cos( t) / t t 0 0 So x ( t) 5 e cos( t) e sin( t) 0 / t 0 x ( t) 5e cos( t) sin( t) 0 / t 0 x ( t) 5e cos( t) sin( t) 0 / t 0 x ( t) e cos( t) 40sin( t) t Of form ke ( ksint cos t) Gives k 0.5 and k 40. 3M ( ) cos( ) 40sin( ) No solution for e t 0. 4 / t 0 (ii) Solve x t e t t Solve cos( t) 40sin( t) 0 Simplifies to tan( t). 40 Therefore, and (iii) If tan x a then A 40 x n tan ( a), n Z tan t n n n t 0.05 Therefore, 0.05 The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 3

41 c. (i) Period of x (t) = Period of y (t). (ii) The maximum value on the graph of cos( ct ) is always equal to. t b b Substituting cos( ct) into x Ae cos(ct) gives g( t) Ae. (iii) The minimum stationary points occur at x, which is the reflection of the maximum values, in the Xaxis. Therefore, the equation of the curve on which the minimum stationary points lie will be on the equation which is a reflection of the curve on which the maximum stationary points lie. t t b 0 i.e. gt () Ae 5e t d. / 0 f ( th) f() t hf () t where x f() t 5e cos() t t Using x( 4 ). 39 / f(4 0.) f(4) 0. f (4) Answer = -.53 The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 4

42 60 t t 0 0 e. (i) Area = 5e 5e cos( t) dt 0 (ii) Area = 60 t t cos( ) e e t dt = Area = 60 0 t t 0 0 5e 5e dt = Unshaded region = = Shaded area : unshaded area = : Hence ratio is approximately :. The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 5

43 QUESTION 8 a. i. cos x y e du Let u cos( x) and so sin( x) M dx u dy u y e and so e du dy dx dy du du dx e cos( x) sin( x) e sin( x) cos( x) A sin( x) e dx e 0 A o cos( x) cos( x) ii. ( e e ) A b. i. cos( x) f ( x) sin( x) e uv cos x Let u sin x and v e then dv cos(x) cos( x sin( x) e ) dx dx M du dv cos( x) cos( x) f ( x) v u = cos( x ) e sin ( x ) e dx dx A ii. If f( x) 0 then cos( cos( x) x ) e cos( x) sin ( xe ) 0 cos( x) Therefore e (cos( x) sin ( x )) = 0 Now cos(x) e can never be zero so cos( x ) sin ( x) 0 Hence cos( x ) ( cos ( x)) 0 and so cos ( x ) cos( x) 0 M Using the quadratic formula, 4 cos( x) A One of these values corresponds with what needed to be found. c. f ( g( x)) = sin cos ( x ) x ). e A ( The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 6

44 d. i. ` Intercepts (0,.44), (.98, 0), (6.0, 0). Coordinate format not necessary here. A Shape with two points of intersection at approximately (0.7,.3) and (5.3, -.4) H ii. (0.65,.34) and (5.33,.46) A e. [ f ( x) g( f ( x))] dx f ( x) dx [ f ( x) g( f ( x))] dx f ( x) dx The two difference integrals with correct lower terminals. All four integrals correct. M A Total = 4 marks The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 7

45 QUESTION 9 The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 8

46 QUESTION 0 a. Dt () asin tband D ( ) 35 and D ( ) 5 6 Two equations are formed. 35 asin b and 5 asin( ) b M 6 Solving simultaneously gives a 0 and b 5 (CAS used to solve) M b. Solving D ( t) 0 gives t 7. 6 and t M Tom s parents can leave their books on the beach from 7 : 00am to : 37 pm and from 5 : 3pm to 7 : 00 pm A c. Average value is ( ) 5 0 D t dt M (using av value formula) Average value is at mean point and is therefore the only place where there are 3 distinct solutions, giving m 5 A d. A D 65 S(t) t The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 9

47 e. (i) Pr( X 8) normalcdf (8,, 5, 3) A (ii) Pr( X k X 8) 0.3 M Pr( X kx 8) 0.3 Pr( X 8) Pr( X k) 0.3 Pr( X 8) Note: If the numerator was equal to Pr( X 8), the resultant fraction would be equal to, which is not the case. Hence the numerator must equal Pr( X k). Pr( X k) Pr( X k) M invnorm( , 5, 3) k 3.55 pounds A The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 0

48 QUESTION a. If Wt () is increasing then W '( t) 0 for t 0. ct ct W '( t) 3 ( c be ) a( be ) ct ct W '( t) 3 abce ( be ) i.e. ct ct 3 abce ( be ) 0 M ct As abc,, 0 then 3abce 0. As the square of a number is greater than or equal to zero, then ( be ct ) 0. Therefore, W '( t) 0 for all values of t. A b. lim Wt ( ) lim a be t t As t then e 0. ct 3 a be 3 Therefore, 3 3 a be a( 0) a A As an animal ages, its weight aproaches the value a. c. (i) y A A (ii) Substitute t 0 and f( t) 0.49 : 0 be 0.49 b 0.49 b 0.5 M (iii) f () t 0.5e ct Substitute t and f( t) 0.45 : e c c e c log e(.07843) M d. Find the Interquartile range (IQR). The interquartile range describes the values of t for which the middle 50% of the distribution lies. i.e. The difference between the 75 th percentile and the 5 th percentile. i.e. The 75 th percentile (a t value) The 5 th percentile (a t value) The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page

49 Let the 75 th percentile equal q. Let the 5 th percentile equal p. q Wt ( ) dt 0.75 M 0 q , 5.85 As t 0, q 5.85 p Wt ( ) dt 0.5 M 0 p 6.478,.36 As t 0, p.36 IQR A e. Binomial with n 6 and p. binompdf 6,, Pr( X 3) 0.9 A f. (i) Binomial with n n and p 0. Pr( X 3) 0. n 3 3 n3 (0.) (0.8) 0. M n (use technolgy) n 7 rats A (ii) Let X Number of overweight rats n n p 0. np( p).5 M n(0.)(0.8) (.5) n n 39 A The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page

50 QUESTION a. (i) Pr( W 00) (ii) Pr( W a) 0.75 giving Pr( W a) 0.5 Using Inverse Normal: a b. Binomial distribution. Let X = Number of underweight packs Let p = probability of pack being underweight = n = 5 Box rejected is X 3 Pr( X 3) c. (i) Binomial application but as order is important, do not apply usual rules. Use multiplication principle. Pr (Box accepted) = = Pr (Box rejected) = = Pr (First boxes only accepted) = Pr(Accepted) Pr(Accepted) Pr (Rejected) (ii) Binomial Let 0.5 Pr (Rejected) Pr (Rejected) 3 ( ) (0.066) np n p 0.5 n n Answer: n = 5 boxes. The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 3

51 d. (i) Type Type Type 3 Profit ($) x 0 0.8x 0 0 Probability (Pr) Pr (Box accepted) = = Pr (Box rejected) = = Type = 85% of = Type = 5% of = Type 3 = Probability of box being rejected = A (ii) ( x0) 0.340(0.8x0) 0(0.06) x 0 (iii) Profit occurs when x x 0 x.5340 Profit occurs when x $.54 The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 4

52 QUESTION a. (i) Pr( tuna from Boat ) (ii) Let X Number of tuna. E ( X ) b. (i) Pr( Boat / tuna) Pr( Boat tuna) Pr( tuna) Alternatively the probability may be obtained directly from the given table. Number tuna caught by Boat Total number tuna caught (ii) Pr( one is a tuna) Pr( tuna from Boat AND not from Boat ) OR Pr( tuna from Boat AND not from Boat ) c. (i) Tuna Salmon Other Number of fish Price per kilo (x$) 0 5 Probability (Pr(X=x) Pr(X=x) Price (x$) M (ii) E($) (0 0.6) ( 0.3) (5 0.) $0. 0 / kg The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 5

53 d. (i) Area = a 35 kg 35 a X z c Pr( X 35 a) 0.85 Pr( z c) 0.85 Using Inverse Normal: c c X a a.0364 (ii) Binomial. X Number acceptable fish p probability acceptable 0.70 n 5 Find Pr( X ) e. (i) Let X number of acceptable fish p p n 5 Pr( X ) p p 0 p ( p ) (ii) P' ( p) 3 0 p 3( p) ( p) 0 p and simplify 0 p( p) ( 5p) Let P '( p) 0 : 0 p ( p) ( 5p) 0 p 0,, 5 As 0 p, p. 5 The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 6

54 QUESTION 4 a. 99 Binomial, p, q 300, n,000, Where np, 000, b. 99 p, n Pr( X 99) Pr( X 98) c. (i) $x Profit $.0 Loss $0.50 Pr(X=x) E(P) = ($.0 x ) - ($0.50 x ) = $. (ii) Binomial distribution: X Bi(5, ) Let X = Number of packets Pr( X 5) The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 7

55 d. (i) k x f( x) 0 0 x elsewhere f ( x) k ( x) k(x) k( x) for x0, x Graph is and f ( x) k ( x ) kx for x 0, 0 x k 4A (ii) For PDF Area = Symmetric graph so area of triangle = 0.5 Area = k So k = e. (i) Need PrX 7 PrX xdx xdx 8 The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 8

56 (ii) Var( X ) 0 x f( x) dx x xdx x ( x) dx 0 where by symmetry mean = Gives Var(X) Gives sd(x) % confidence interval: X X 6 6 f. Pr( X p) 0.6 p The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 9

57 From graph means Pr( X p) 0. Need p ( xdx ) 0. Within domain gives 0 5 p. 5 The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 30

58 QUESTION 5 a. Range is b. Pr(h > 9) = Pr(z > = 0 5. = 9.8 m to 30. m 9 0 ) 5. = Pr(z > 0.96) = Binomial : Pr(h 65) = 00 C 65 (0.5777) 65 (0.43) C 66 (0.5777) 66 (0.43) 34 + = c. (i) Pr(at least yellow) = Pr(y=) + Pr(y=3) = ( 3 C (0.8) (0.7) ) = 0.93 M Expected number = = 93 (ii) Pr(h 9 Y ) = ( ) {3( ) } = ( from part b) d. Conditional probability: Pr( h 6 Y 3) Pr( h 9 Y ) = = Pr( h 6 Y 3) Pr( h 9 Y ) 6 0 Pr( z ) Pr( Y 3) Pr( z ) 0.8 = e. Pr(z > ) = 0.5 z =.0364 f. 50 Pr(z < 3 = ) = 0.38 z = Solving simultaneously, 49., 9.8 k (0.93 from part c) 0.93 k M 0.7 M = 8.45 Million dollars The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 3

59 QUESTION 6 n a. (i) Sn T S0 Sn n car( C) public( P) A (ii) S T S 0 S S car( C) 87.5 public( P) Answer: 87 are catching public transport b. Use trial and error: Car users drop below 450 at S 8, meaning at 9 th festival. The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 3

60 c. Could check with steady state S first, where S T S. 0 S car( C) public( P) P P T Q Q where T = So P Q = P Q Where Q = 500 P Giving P P = P 500 P Equate and solve: 0.75P0.(500 P) P Meaning P = 48 travelling by car. The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 33

61 d. Find, using integration where area = a Solve for a, cos x If constn()=0, a = or If n=, a= d dx e. (i) xsin x6 sin x6 x cos x6 sin x6 xcos x (ii) Mean of PDF is xf( x) dx So mean of f (x) is xf( x) dx x x dx x x dx Giving cos 6 cos 6 Using integration by recognition we know that: sin x6 xcos x6dx xsin x Simplifying and separating terms gives: sin x6dx xcos x6dx xsin x x cos x 6 dx xsin x 6 sin x6 dx The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 34

62 x cos x 6 dx 0 0 x cos x 6 dx xsin x 6 sin x 6 dx x RHS sin sin cos cos cos M Checking with CAS: Mean = 6 The School For Excellence 05 Year Mathematical Methods Exam Analysis Lecture Page 35