Mathematics B. Statistics. General comments. Characteristics of good responses Senior External Examination Assessment report

Transcription

1 Mathematics B 0 Senior Eternal Eamination Assessment report Statistics Year Number of candidates Level of achievement VHA HA SA LA VLA General comments Both eamination papers sampled subject matter and learning eperiences across the full range of syllabus contets. Not all candidates attempted every question. Candidates responses generally indicated that the objectives of each of the three assessment criteria were addressed. In criteria Knowledge and procedures () and Modelling and problem solving (MP), candidates provided better quality responses in Paper One than in Paper Two. In the criterion Communication and justification (CJ), the standard achieved was generally consistent across both papers. Candidates who performed poorly in also performed poorly in MP as the knowledge to solve problems is implicit in MP. Candidates are encouraged to attempt every question in both papers. Characteristics of good responses In the criterion, candidates with good responses characteristically provided: clear justification and working an effective use of diagrams the significant intermediate calculation steps. In the MP criterion, candidates were required to demonstrate mathematical thinking across the contets and topics of the syllabus. The basic requirements of this criterion were satisfactorily met by candidates who demonstrated: interpreting and clarifying selecting and using strategies.

2 Eamples of these include correctly identifying or recalling a rule, formula or procedure, providing an effective diagram or creating a list/table of information that may be used to solve problems. Candidates who performed better in this criterion also demonstrated mathematical insight in eploring problems and were able to identify, in contet, mathematical strengths and limitations of simple models. In the CJ criterion, many candidates appreciated the need to justify and validate their solutions. This was clearly evident in those responses which ehibited a developed argument and an eamination of the effects of assumptions. Common weaknesses In the criterion, candidates were required to demonstrate accurate recall, selection and use of basic procedures. This was not evident in poorer responses. Candidates are required to know basic formulas and how they are used. These issues were evident in the use and manipulation of inde and logarithmic laws, rules of calculus involving differentiation and integration and in the solving of equations where simple substitution was required. In the MP criterion, candidates must be able to apply subject matter in a variety of contets. All contets require some measure of careful reading, interpretation and planning. This was not evident in some responses. Candidates must remember to use clearly labelled and neatly drawn diagrams to support arguments and conclusions reached. These diagrams are frequently a significant part of the justification of a response and, in that respect, were deficient in some responses. These issues were very evident in applications of trigonometry and applied statistics. In the CJ criterion, poorer-performing candidates did not demonstrate organisational and presentation skills; correct knowledge and use of mathematical symbols; the correct use of units of measure; and the correct spelling of words even when they appeared in the eamination paper. These issues were evident across all topics and contets. Sample solutions The following solutions are not necessarily prescriptive model responses and are not necessarily the only way of solving a problem. Other approaches and problem-solving strategies may be just as acceptable. Mathematics B 0 Senior Eternal Eamination Assessment report

3 Paper One Question a. i. Car speeds Number of cars Speed ii. = 49.8 km/h σ n = 5.5 or σ n = 5.6 b. i. Sprinter A Sprinter B Min..0 Q.. Median.45.5 Q.7.4 Ma. 6.8 Queensland Studies Authority February 04

4 ii. The coach is correct because within each of the quartiles sprinter B has a smaller range of values. The interquartile range is lower as a result. The outlier appears to be an anomalous result and should be ignored. MP Question a. i. The functions are B, D, and E. [No credit if more than three relations are identified as functions.] ii. B is continuous, D is discontinuous, and E is discrete. iii., 0 and y = if < 0 and y = if > 0. b. i. Independent variable is the number of leaflets, say n. The dependent variable is the income earned in dollars, say I. ii. iii. The independent variable is discrete since only countable numbers of leaflets are delivered. If n is the number of leaflets delivered and I is the income earned in dollars, then I = 0.n+. c. Vertical shift +, horizontal shift +, vertical dilation of +. 4 Mathematics B 0 Senior Eternal Eamination Assessment report

5 d. Let the daughter s age be years. Let Jenny s age be y years. y = 5 ( y 4) ( 4) + = + ( 5+ 4) = ( + 4) 5+ 4 = + = 4 and y = 0 If n is the number of additional years, then: ( n) 0 + n= n= 8 + n n = In twelve years time, Jenny will be and her daughter 6 years old as required. MP Question a. b. i. 5π 5 = or π radians cos θ + sinθ = + sinθ sin θ Consider the LHS. = θ + LHS cos sin Since θ cos θ sin θ + = by the Pythagorean identity, then ( θ) LHS = sin + sinθ = + sinθ sin θ LHS = RHS cos θ + sinθ = + sinθ sin θ as required. Queensland Studies Authority February 04 5

6 ii. Given that sinθ sin θ + =, then sin θ sinθ = 0 sin + sin = 0 and hence, So ( θ )( θ ) sinθ + = 0 or sinθ = 0 sinθ = or sinθ = 7π π π θ =, or θ = The solution set is π, π, π θ = 6 6 c. For y 6cos( π ) = +, π i. Amplitude is 6, period is π, horizontal shift is to the left, and the vertical shift is +. ii. Using features of the graphics calculator, A(.047, 9 ) and B(.0944, 9) 6 Mathematics B 0 Senior Eternal Eamination Assessment report

7 d. h tan4 = h= tan4 Now, in ABC, 58.4 Hence, 00 00sin 5 = = sin 5 sin80 sin80 h = 58.4 tan4 h 9.8 metres. The height of the tower is approimately 9. metres. MP Question 4 a. i. f ( ) = 4+ and g( ) = 5 ( ) f g = g 4g+ ( ( ) ) ( 5) ( 5) f g = 4 + ( ( )) f g = ii. g( f ) = f 5 ( ( )) = ( ) 5 g f f ( ( )) g f = 4 ( f ( )) g = 7 Queensland Studies Authority February 04 7

8 iii. g( ) = 5 = y = 5 Let y g( ) ( ) = g y = y+ 5 g ( y) Hence, g ( ) = y + 5. = + 5 b. Given that 4= 7 ( ) = = 0, = ± c. The curves = + 4 and y 6 y + 4 = = = 0 ( )( ) 5 = 0 = and 5 = (, 5 ) and ( ) = meet when 5, 9 8 Mathematics B 0 Senior Eternal Eamination Assessment report

9 dy y = d = d. ( ) dy d = = The gradient of the tangent is and the gradient of the normal is When, y. = = The equation of the tangent is y = ( ) Hence, y =. The y-intercept of the tangent is B( 0, ) The equation of the normal at the same point is:. y = ( ) or y = +. The y-intercept of the normal is A ( 0, ). AP = 4 + AP = 5 and BP = AP = 0 The area of the 0 5 ABP = or 5 square units. MP Question 5 f = + 6 a. ( ) f ( + h) = ( + h) + 6 f = + h + h + 6 ( ) = lim h 0 ( + ) ( ) f h f h = = = lim h 0 lim h 0 lim h 0 h + h h ( + ) h h h + h ( ) f = Queensland Studies Authority February 04 9

10 b. i. 4 y = 5 + dy 0 d = + dy 4 7 d = + ii. y = ( 7+ ) 6 ( ) 5 iii. y = e cos( ) Let u = e u = e Let v = cos( ) v = sin ( ) By the Product rule, dy = uv + u v d dy e sin e cos d = + ( ) ( ) dy = e d ( cos( ) sin ( )) iv. y loge ( 5) = + dy d = + 5 v. y = e 4 Let Let u = e u = e v = 4 v = dy vu uv = d v By the Quotient rule, dy d = dy = d ( ) ( 4) e 4 e ( 8) ( 4) e 0 Mathematics B 0 Senior Eternal Eamination Assessment report

11 D = t 4 +, c. Given ( ) i. Average velocity, v = D ( 8) D( 4) 4 50 v = or m/s 4 ii. velocity, dd v = dt t= 6 ( t ) = 6 = 6 4 t = m/s d. For the given configuration, A = y and + y = 700. Therefore, 700 A= or A ( 700 ) =. da da = and a stationary point eists when 0 d d =. Since A < 0 for all, A ma eists when 50 = metres. For this value of, A ma 408. square metres. MP Question P blue red = a. (, ) 5 P( blue, red ) = or 0.65 b. P( X > 8) = P( X 8) ( ) = P X > 8 = 0.05 Queensland Studies Authority February 04

12 c. i. P(.5 z.56) = P(.5 z 0) + P( 0 z.56) ( z ) P.5.56 = 0.90 = ii. P k 00 X > k = P z> 5 ( ) k 00 P z > = k 00 P 0 < z < = k 00 = k = k = 4.5 Mathematics B 0 Senior Eternal Eamination Assessment report

13 d. µ = kwh and σ = Cost is.5 cents/kwh up to kwh and then 5 cents/kwh thereafter. If kwh are used then the cost will be = $ Since $65 < $ = 49000kWh and $7450 > = 58000kWh 5 ( $65 $ $7 450) ( ) P X = P kwh X kwh kwh Converting to the standard normal variable, z: z = and z = 5000 =. = 0.6 ( ) (. 0.6) P kwh X kwh kwh = P < z < (. < < 0.6) = (. < < 0) + ( 0 < < 0.6) P z P z P z = = The electricity bill will fall between $6 5 and $7 450 with about a 6% chance. Assuming that the electricity consumption is normally distributed and that the pricing tariffs remain unchanged, An increase would create problems for BBC manufacturing unless the costs could be offset somehow. MP Queensland Studies Authority February 04

14 Paper Two Question Given y = ( ) ( + ) a. i. y = ( 4+ 4)( + ) = + 8 dy 8 d = dy Stationary points eist when 0 d = 8= 0 ( )( ) + 4 = 0 = 4 or = The stationary points are A(,8 4 ) and ( ) 7 B,0. Classifying the stationary points using the second derivative test. d y d = 6 At A(,8 4 ), 7 d y d = 4 < 0 A(,8 4 ) is a maimum. 7 At B(,0 ), d y d = > 0 B(,0 ) is a minimum. 4 Mathematics B 0 Senior Eternal Eamination Assessment report

15 ii. Points of oblique inflection eist when d y d dy = 0 provided 0 d Since d y = 6, then d Therefore, 7 ( ) 7 d y = 0 = and d C,9 is a point of oblique inflection. dy d = 0 iii. For the given domain 4, y ( 4) = 6 and ( ) y = 6 Considering A(,8 4 ), 7 is an absolute maimum. 8 > 6 A(,8 4 ) 4 7 Considering B(,0 ), 6 < 0 ( 4, 6) is the point where an the absolute minimum occurs. iv. intercepts eist when ( ) ( ) The intercepts are (,0 ) and (,0) The y intercept occurs at 0 Hence, the y intercept is ( 0, ). 7 + = 0 = and =. = ( ) y 0 = Queensland Studies Authority February 04 5

16 b. The container takes 5 seconds to fill assuming a steady rate of fill. The water level is rising so dd dt is always positive. Point A on the graph is the neck of the container and when D = cm, dd dt will be greatest. Point B on the graph represents where the container is widest and so when D = 7 cm, dd dt will be least. MP Question a. i d ii. ( ) = C ( ) sin 4 cos 4. d = + C 8 iii e e. d = + C 5 6 Mathematics B 0 Senior Eternal Eamination Assessment report

17 5 5. d = +. d + b. ( ) ( ) = 4 + = 4( 6 ) = 5 = c. 6. d using five equal strips. + Upper = 6 Lower = n = 5 Width = f() Sum = d = ( ) ( ) 6. d.555 correct to three decimal places. + = Queensland Studies Authority February 04 7

18 d. Area 0 = + a + 5. d a = a = 6 + = = 0 Since the area is given to be 54 then a = 4. The curve has equation: y = The zeros can be found by solving: = 0. ( ) + 9 = 0 = ± 9 b = 9 and c = + 9. MP Question a. Given V t = 0 50 dv dt i. 4 dh = π = 4π 6 t dt 5 50 t = 5 dh = dt.08 4π or m/min t ii. A limitation or restriction in the model is 0 t 50 minutes. MP 50 8 Mathematics B 0 Senior Eternal Eamination Assessment report

19 b. i. Cost of fuel 600 = a or 600 a + Cost of driver = b time 400 b or 400b Commission = c Hence the cost of the trip, C = 600 a + 400b + + c ii. C = 600 a b + c C = 600a + + a 400b + c dc d = 600a + a 400b d C d = 00a + 800b or ( + ) 800 4a b Since each of, ab and d C is greater than zero, then it follows that 0 d > Hence C dc is minimised when 0 d = 600a + a 400b = 0. This implies that ( + ) 400 4a b = a or ( + ) 400 4a b = a =, 6a = 4a+ b b= a When 80 assuming that the speed of 80 km/h is maintained. MP Queensland Studies Authority February 04 9

20 Question 4 a. n n+ n ( n+ ) 4 = n ( n 8 ) = n+ ( n+ ) ( n ) n n+ n+ 4 = or n n b. i. ( ) log + log = ( ) log = ( ) = 4 4= 0 ( )( ) 4 + = 0 = 4 or = With =, log 4 does not eist; hence, the only solution occurs when = 4. ii. ( ) 0 0 dt = ln t + t + ( ) t= t = = ln + ln + = ln When dt =, t ln = + = e = ( e ) 0 Mathematics B 0 Senior Eternal Eamination Assessment report

21 c. i. Given log e V = at + b and using the data provided, log e V = at + b.605 = 0a+ b and.976 = 4a+ b Subtracting the equations, 4a = 0.7 and hence, a = Selecting.605 = 0a+ b, b=.605 0a and substituting for a b = then b =.6745 ii. log = T e V V = e 0.098T T V = e e which approimates to V 0.098T = e litres or T e billion litres In 0, T = e or 7.4 billion litres correct to three significant figures. iii. The reliability of the data is questionable since only two data points were given. With the insertion of etra data points a more reliable model might eist. These models could also include polynomial forms and not just eponential models. Commercial, judicial and legislative factors might be the unknowns that make predicting from the eponential model unreliable. Certainly, broad assumptions regarding the continuity of sales must be assumed. A strength of the model is that it provides a suggestion when sales targets might be reached T 7.07e = T = ln or about 8.7 years and this would mean sometime in 04. MP Queensland Studies Authority February 04

22 Question 5 a. Using the future value of an annuity with (( + i) n ) R S = i Then n n = n 0.06 i =, S = 6000 and R = 450, ( n ) = = n = n (.778) (.005) n.8 or months. b. Tom has taken out a loan of $ at 7% p.a. compounding monthly over 0 years. i. Using the present value of an annuity, R = n ( ( + i) ) i = 0.07 R ii R = R = $79.79 monthly. Having paid the loan for ten years, Tom is still required to keep paying for another 0 years. That debt would stand as: Debt = or $ A single payment of $ will not be enough for Tom to payout his debt. The claim that Tom could pay out his loan is invalid. MP Mathematics B 0 Senior Eternal Eamination Assessment report