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EXAMINERS’ REPORT ON 2006 TERTIARY ENTRANCE EXAMINATION SUBJECT: CALCULUS STATISTICS Year 2006 2005 2004 Number Who Sat 1434 1601 1574 Non-Examination Candidates 7 5 0 Did Not Sit 68 69 94

The Examiners’ Report is written jointly by the Chief Examiner and his Deputy to comment on matters relating to the Tertiary Entrance Examination in this subject. The opinions and recommendations expressed of those of the Chief Examiner and Deputy and are not necessarily representative of or endorsed by the Curriculum Council. The Marking Guide provided at the end of this report was prepared by the examiners and may have been substantially amplified by discussions held in the pre-marking meeting. It is not intended as a set of model answers and is not exhaustive as regards alternative answers. Some of the answers are less than perfect, but represent a standard of response that the examiners deemed sufficient, on this occasion, to earn full marks. Teachers who use this guide should do so with its original purpose in mind. SUMMARY/ ABSTRACT The 2006 Calculus paper met with satisfactory acceptance by candidates. The exam average of 57.3% is up on last years result and close to the ideal mark of 58%. Overall teachers and markers seemed to think the paper fair but there was some feeling that there were too many questions asking candidates to show given results rather than find solutions. The paper sorted the weaker from the stronger students well. Many candidates lost a considerable number of marks due to insufficient explanations and reasoning. Often the justifications given by candidates were at best sketchy and, at worst, non-existent. Teachers should emphasise that in Calculus the ability to explain what you are doing is vital (in many cases more important than the final answer itself!) and candidates must get more practised at this skill so that, come the examination, it is almost second nature. The length of the paper was felt to be about right – there is little evidence that timing was a problem as the least favoured (hardest and last) question was still attempted by 85% of candidates. The wide range of raw scores (0%- 98%) and the high standard deviation (19.0%) show that the paper was an effective discriminator. The paper had an excellent internal reliability coefficient of 0.94 (an improvement on last year’s also excellent result). The general standard of performance on the paper was better than last year. Some routine questions and those requiring graphics calculators were generally done well (but with some notable exceptions). The general level of competence in algebra was noticeably better but examination technique is still lacking. GENERAL COMMENTS Given the difficulty of last year’s paper, the examiners made a conscious effort to make a somewhat easier and more straightforward paper. That said, there was the desire to include a few questions to sort out the candidates at the upper end of the scale and Questions 15c, 17 and 18 appeared to do this. Had the examination average been slightly higher then we would have been very satisfied with the overall outcome but again, just as last year, marks were lost through carelessness and sloppy work. We have detected that both teachers and candidates might appreciate some guidance as to the meanings of some of the ‘keywords’ used in the examination paper. Later in this report we will comment further on

Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus 1

these aspects, but it must be realised that some of the definitions are problem-dependent and cannot be given a dry interpretation that holds in every single circumstance. One aspect that examiners continued to note is the over-reliance of candidates on the use of the graphical calculator. Some of the questions were best tackled using the calculator, but others were specifically designed to test analytical and reasoning skills. Many candidates appeared to use their calculators at every possible opportunity, irrespective of whether it was the best strategy. These candidates were penalised in a number of questions where exact values were asked for. Both candidates and their teachers are reminded that one of the cognitive objectives for the course is to “select and use different technologies appropriately”. An appreciable minority of students appear to approach questions thinking “how do I use my calculator to solve this” rather than deciding which strategy is best in each individual case. Overall it was felt that the paper was much easier to mark than last year, despite the presence of several question asking candidates to show, prove or deduce a given result rather than merely finding some quantity. As expected, this led to solutions containing a good deal of dubious reasoning. Teachers would be well advised to make candidates more aware that attempting to fool markers and examiners by blatantly asserting something has been shown, when a cursory read shows it has not, is unlikely to succeed. There is some feeling that the style of examination should move away from the “Show” approach back to the “Find” style of question but this is not likely to happen in the near future for reasons we discussed last year. The paper tested most of the parts of the syllabus providing a good balanced coverage of the broad content sections specified in the assessment table as shown below. Syllabus content Recommended weighting (%) 5--10 Relevant Question No’s 1,14* 2,3,9,18 4,6,10,17* 11,12,13,16,17* 14* 5,7,8,15 Marks allocated 9 33 33 47 14 44 180 % weighting 5 18 18 26 8 25 100

Calculus of trigonometric functions Functions and limits 15--20 Theory and techniques 15--20 of calculus Applications of calculus 25--30 Vector calculus 5--10 Complex numbers 20--25 Total * marks apportioned to both content sections

The candidates did reasonably well on the routine questions but, as in previous years, the paper showed up a number of weaknesses. We have already mentioned that the general competence in algebra is improved (though there is still plenty to be worked on). The most noticeable defect this year was the unsatisfactory level of explanation and reasoning given throughout many papers. It is sometimes thought that this complaint discriminates against good students as they can take bigger steps in their working and can “see” their way to the answer without needing to detail what they are doing. The evidence, however, is that it is precisely the very good students who write the most by way of explanation and rely the least on memorised formulae etc – they derive things they need from first principles each time. Once again, we ask that students be made more aware of the basics of good examination technique. Above all, read the question carefully. The wording of the questions is normally very carefully considered and has very specific meaning. Candidates should read the key words in any question to make quite sure they know what is required before starting out.

Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

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Multi-part questions with a common thread running through are normally presented sequentially with each part following from the previous. Often the earlier parts of such questions will be of “Show that” type – not to be mean or awkward but precisely the opposite. If a candidate cannot derive the answer to an early part of the question, they are expected to use the given result to help them answer later parts. It is not advised to “fudge” show that questions – this is likely to be detected and any goodwill that a sympathetic marker might give later in the paper will be surely lost. Answers were often presented sloppily and in a way which made them very difficult to read. Either way, working needs to be legible for it to be considered. COMMENTS ON SPECIFIC SECTIONS/QUESTIONS Question 1 Mean 4.37/5 (87.38%) Standard deviation 23.32%

An easy starter. Some students were unable to differentiate x cos x and some lost marks through sloppy algebra. Question 2 Mean 5.02/7 (71.71%) Standard deviation 21.50%

Several candidates intepreted the composite function as a product. The domain generally caused little difficulty but finding the range was less well done. Here was an example where the limits of the range could be found easily by knowing the range of the sine function but too many resorted to the calculator to deduce approximate answers. The examiners intended that exact values be required for full marks, but given this was so early in the paper approximate answers were not penalized. Question 3 Mean 4.71/7 (67.34%) Standard deviation 27.38%

The main problem with this question was the general lack of explanation or reasoning. The key words here are that exact values are sought and some justification needs to be given. Many ignored the latter request and were penalised accordingly. Some candidates clearly have been taught L’Hopital’s rule (off syllabus), but were unable to sort out when it can be used (e.g. part c) and when not (e.g. part b). Pleasingly, many students did identify that part c) was related to the definition of an appropriate derivative although many also could not finish the calculation properly, while others took the derivative with respect to a, which was a given constant. Question 4 Mean 5.12/9 (56.85%) Standard deviation 34.47%

A large number of candidates were unable to differentiate the function as a product. This made the remainder of the calculation either trivial, impossible or non-sensical depending on the type of error made. Given this, part b) was poorly answered. Question 5 Mean 8.06/12 (67.17%) Standard deviation 23.08%

The first few parts of this question were designed to be straightforward and, although the overall mean score was high, it should have been higher. Many candidates lost marks because they failed to answer the question as posed. In a), the question asks for the real and imaginary parts of the expression; the answer therefore requires the candidate to finish with a statement along the lines “Real part =...”, “Imag. part =...”. Too many worked out a complex number and left it to the marker to decide whether they knew what the real and imaginary parts were. Equally, in part c), the question explicitly asks for the modulus and argument for the four sub-cases but many candidates left the results in “cis” notation. Question 6 Mean 6.07/9 (67.39%) Standard deviation 27.16%

Parts (a) and (b) were generally well done. Many made careless mistakes in determining the second derivative although many of these errors resolved themselves when the specific value dy/dx = 0 was substituted! Too many candidates tried to do more than was required and embarked on relatively lengthy and unnecessary algebraic manipulations. All that was asked for was the value of the second derivative at a specific point so there was no need to try to derive a general formula. Question 7 Mean 3.33/7 (47.59%) Standard deviation 38.41%

Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus 3

This question was badly done by many. This is nothing more than a textbook example that should have been an easy source of marks. Many students expanded the fifth power from first principles which proved lengthy, algebraically messy, and usually ended in error. Those candidates that got further often did not provide adequate explanation of what they were doing at each stage. All in all, disappointing. Question 8 Mean 6.73/9 (74.74%) Standard deviation 25.01%

Generally well handled by most students although some of the sketching left a lot to be desired. With unlabelled axes, candidates need to provide a suitable scale themselves. The sketching itself should be done in pencil. A significant minority of sketches were at best only vague approximations to the actual regions! a) b) Most knew this was the interior of a circle, though the location of its centre proved troublesome. Some didn’t identify the key points as - 4 and 3i. Many did not adequately define the perpendicular bisector – there were many lines drawn that were roughly perpendicular to the base line and roughly bisecting it. It was easy for candidates to show conclusively that they are sketching the perpendicular bisector and they should have ensured that the marker is in no doubt that is what they are describing. Lack of knowledge as to what to do with the translation of 3 in the definition of the argument was the most common mistake here.

Tip: Candidates should be advised that with this type of question there is no objection at all to adding a written description of the region alongside the sketch. This makes it clear that they know what the answer is; if their sketch is then not particularly accurate they will still get full marks as what is being tested is their understanding of regions of the complex plane and not their artistic ability. However, with no accompanying working or description, many of the sketches drawn were penalised as it was far from obvious whether the appropriate locus had been correctly identified. Question 9 Mean 9.24/12 (76.99%) Standard deviation 22.72%

Part b) was rather well done overall; the sketch of the functions was almost universally correct and many could use the diagram to deduce the solution of the inequality in part ii). The examiners had intended candidates to solve the inequality by looking at the appropriate trigonometric equations but many circumvented this by clever use of the calculator. Some were caught out by the fact that the exact solution of the inequality was requested and lost marks accordingly. On the other hand, part a) was rather less well done. Many candidates could find the four important values but didn’t really know how to combine all the information. Tip: The relevant part of the syllabus decrees that students should be able to solve, algebraically and geometrically, simple equations and inequalities involving absolute values of linear functions. Attention should be drawn to the fact that students are expected to be proficient at both methods of solution. In other questions candidates may be given a choice of solution methods, but in this instance the examiners wished to force the algebraic approach, given that the solution to part b) was intended to be graphically based. Candidates who attempted part a) graphically were very heavily penalised. Once again it is a case of reading the question carefully and answering the question as asked, not as one may have hoped the question might have been phrased. Question 10 Mean 7.36/11 (66.86%) Standard deviation 29.01%

Parts a) and b) were generally well done. Part c) was less well done – although many could show the result required in i) rather few could use it to deduce the result ii). Some “solved” ii) by quoting results from the tables, but this scored zero as this was not the solution method asked for. Occasionally candidates can be expected to be asked to derive some standard results from first principles and quoting results from other sources is not a valid way of tackling this.

Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

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Question 11

Mean 6.80/11 (56.69%) Standard deviation 23.90%

A classic type of optimisation question that was not as well answered as was hoped. Part a) was generally well done though many did not properly deduce the restriction on y. There was some fudging of the algebra to express the area in terms of y in part b): although most could write down the correct form of the area in terms of x and y, too many could not then proceed to eliminate x properly. A high proportion of candidates could not make much of part c); a good number did not recognise that any differentiation was required, and several who did differentiate were subsequently bamboozled with the algebra. A significant minority of candidates appeared to wrongly believe that the maximum area must be associated with the greatest possible value of y as derived in part a). Question 12 Mean 7.23/9 (80.33%) Standard deviation 28.81%

Apart from the first question, this was the best answered question on the paper. It should have been routine and well-practiced -- there were some problems in b) with solving the pair of equations for the two constants k and C; but, pleasingly, many of those who got confused in b) ploughed on into the later parts with assumed values of the constants so that follow through marks were scored. Question 13 Mean 5.06/13 (38.90%) Standard deviation 26.26%

This question was rather poorly done. It might have been expected that the algebraic manipulation required in part a) might catch some; but in the main the candidates made valiant attempts to simplify as required and many did it successfully. In contrast, part b) was beyond the vast majority of the candidates. Despite it being a standard type of related rates question, many candidates were all at sea with this problem. They did not see any connection at all with a), did not appreciate that calculus techniques might be helpful and could not relate rate of change of volume to rate of change of depth. This question highlighted a real problem area in understanding. Question 14 Mean 10.57/18 (58.74%) Standard deviation 23.89%

This question proved to be the one that correlated most closely with the overall total for the paper and was a bit of a mixed bag. Generally parts a) and b) were well done save a few slips with signs and carelessness. Too many candidates appeared to not know how to prove two vectors are parallel; there seems to be a view that the dot product should be used! The sketches in d) were of variable quality. Some were disgracefully drawn, indicating that candidates ought to take more pride in their work. Others had no scale; many drew a half-ellipse (probably indicating poorly thought out calculator work just using default parameter ranges). The calculation in e) was reasonably well done, although the interpretation gave the impression that the candidates didn’t really know what the result meant in reality. The answers in f) were of very variable quality; some candidates used degrees rather than radians, others imagined that the integrand was a vector(?) and there were many incorrect answers. Again the interpretation was not good. Tip: If asked to integrate numerically, indicate on the script what you are integrating and what is being put into the calculator. In part F here the overwhelming majority of candidates gave an answer only; if it was wrong they had to be awarded zero. Marks were available for getting the correct expression for the integrand, working in radians etc. - if this information isn’t evident then it cannot be rewarded. Question 15 Mean 8.00/16 (50.00%) Standard deviation 20.64%

Part a (i) was very easy, and answered well. Many candidates fudged around in part a (ii) and not many gave answers that were convincing. Part b was a standard calculation that was generally well done. Some did not express answers in the range requested; most of the sketches did at least have the fifth roots of unity spaced evenly even if not quite orientated correctly. But, once again, even allowing for the pressure of an examination, the quality of many of the diagrams was poor. Part c was intended to be a discriminator; and it proved to serve its purpose. It was beyond most candidates; but the small number who did attempt it realised that effectively all the information required had already been done in the earlier parts. Question 16 Mean 4.73/10 (47.25%) Standard deviation 32.86%

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Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

Generally badly done. This, together with Q7, were the two most disappointingly answered questions on the paper. It is fairly routine; the only slight twist in the formulation was that the phase angle was not in the first quadrant. Candidates should not assume that phase angles are always in that quadrant and far too many attempts seemed to not even consider the possibility that anything else might occur. In some ways the style and content of this question paralleled that in Q12 but with one important difference; that question was scaffolded and the candidates only needed to proceed in small steps. Here a longer, coherent approach was required and the difference was noticeable. In the main, the explanations and commentary can only be described as poor; the majority of the answers consisted of little more than a morass of equations (most of them irrelevant or wrong) with no indication what they relate to. This question was nowhere near as difficult as candidates appeared to find it, and good Calculus students should have been able to score highly. Question 17 Mean 2.17/7 (31.05%) Standard deviation 29.60%

The second “discriminator” question. A few candidates decided to find the volume of revolution rather than the area – whether this was a genuine misunderstanding or a subtle way of simplifying the subsequent integration is hard to tell. Of those who found the correct integrand, many forgot the factor 2 arising from the symmetry of the loop. Some identified the correct substitution, but then didn’t really know how to proceed. On a more positive note, some students did not find the substitution intended but nevertheless made some other appropriate substitution and pushed through to the correct conclusion. Question 18 Mean 1.32/7 (18.83%) Standard deviation 25.80%

The hardest question on the paper, which required little mathematical working but did need careful argument and deductive reasoning. Not unexpectedly, this proved beyond most candidates who could not relate how the graph of f’(x) to f(x). Some realised the importance of the points x=0, 6 and 10 but didn’t proceed any further. ISSUES FOR THE SYLLABUS COMMITTEE TO CONDISER Nil. ACKNOWLEDGEMENTS & THANKS The members of the Examining Panel would like to express our sincere appreciation to all the other markers for the thoroughly professional manner in which they carried out their task. Andrew Bassom Rom Cirillo December 2006 2006 Examining Panel Chief Examiner: Professor Andrew Bassom Deputy: Mr Rom Cirillo Third Member: A/Professor Ken Harrison Chief Marker: Mr Rom Cirillo

Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

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AN INFORMAL VOCABULARY OF SOME MATHEMATICAL TERMS In response to several requests, here we outline our interpretations of several of the keywords that have been used in questions on the Calculus paper. It is important to emphasise that these notes are our thoughts on the topic; they do not necessarily carry over to other mathematical TEE papers nor need be common to other examiners (or future examiners) of the Calculus paper. 1. SKETCH If asked to sketch a function or graph, what is being asked for is a diagram that shows the main features of the function under consideration. If the axes supplied do not have a scale then one needs to be added, to give an idea of the dimensions involved. For a function some of the key features that should be included are: intercepts on the axes, asymptotes (if any), location of turning points, whether they are maxima or minima. Unless the required differentiation is particularly easy, or unless you’ve been asked to find them explicitly, points of inflexion need not be found. For a sketch accuracy is not vital in the sense that individual points do not need to be precisely located to the nearest millimetre, but the overall shape and geometry of the graph needs to be identifiable. Often when asked for a sketch some of the preliminary work will have been asked for in earlier parts of the question and does not need to be repeated.

To give some ideas of the important points of a good sketch in terms of questions on the 2006 paper: Q4. Function goes through origin, vanishes for large x, and has maximum at the given point. Asked to show the point where f’’(x)=0 (which is at x = 1)- the ‘action’ is at values of x up to 1, but to show the correct behaviour at infinity the x-axis ought to be taken to about x = 2 or 3. Q8. (a) Interior of a circle centre (3,-2) of radius 5. Shade the interior, choose labelling on axes so that the circle isn’t tiny, but equally fits comfortably in the space provided. Use compasses or template to make the shape look circular rather than an indefinable blob. Use pencil – its unlikely you’ll get a good sketch first time and will need to correct it. (b) Indicate the points -4 and 3i on the axes – making sure you get these points roughly in the right positions. Dot a line joining these points and then draw a solid line at right angles. Add appropriate symbols to indicate that you know that the line drawn is perpendicular and is a bisector. (c) Identify the significance of the point 3 on the real axis. If you know you are looking for the interior of a ‘wedge’ make sure that the inclined side of the wedge is roughly in the right direction. Mark the size of the included angle so there is no doubt as to what it is intended to be Q9. (b) Axes labeled this time. This is an easy sketch – but make sure that the two curves are distinguished and it is clear which is which. Q14. (d) Here the dimensions are what is crucial. Show that the curve cuts the x-axis at +/- 10 and the yaxis at +/-8. Ensure that the figure drawn looks elliptical and has roughly the correct aspect ratio. Q15. (b) Make sure that the five points are all the same distance from the origin (again, use a compass to construct a circle so that the points are in the right place). Make sure that the points shown are equally spaced (as you know they should be) and possibly mark in a few angles to emphasise this. 2. FIND EXACT VALUES Generally, this expression means that the examiner expects the problem to be answered without recourse to the calculator. In some questions your calculator is the only way that it is possible to derive an approximation to the answer of a problem but you must remember that in almost every instance the calculator will give you precisely that – an approximation to an answer (albeit a very good one) and not the exact solution. Sometimes it is not sensible to tackle a problem without the calculator as the algebra and working required will be excessive but it is safe to assume that on TEE papers, if asked to find exact values it ought to be relatively straightforward to do this by reasoning and algebraic manipulation alone. If you think you have deduced the exact value requested, it can often be checked on your calculator that it is right, but the calculator alone is unlikely to be able to find the exact value unaided.

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Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

In terms of the 2006 questions, all the limits in Q3 can be deduced from first principles. The division of the complex numbers in Q4 a) ought to be done by hand. The limits on the solutions of the inequality in Q9 b)ii ought to be deduced in terms of π rather than approximate decimals. 3. SHOW THAT In many ways this term is the hardest to characterise. Depending on the context it can assume several shades of meaning, but the most common requirement of a ‘Show that’ question is that you use the information already provided to proceed through a series of logical steps to deduce the result that is to be shown. Remember that often the purpose of a ‘Show that’ statement, especially in a multi-part question, is that is for any reason you are unable to get the required result, you do know what it should be and can use it for answering later parts of the question. In terms of the 2006 paper, just one example of this is Q11. The examiners are asking for the maximum area of the window and thus it is likely that an expression for this area will need to be differentiated and the result made zero. If you do not have the necessary formula then part c) of the question is impossible as you’ve nothing to work from; so the Show that of part b) at least provides you with the starting point of c) even if you can’t do b) itself. A good test for most ‘Show that’ questions would be to mentally replace those words with ‘Find’. For instance, in Q13 a) it would be perfectly possible to ask the question ‘Find the volume of the solid’. All the ‘Show that’ is doing is giving the correct answer; in this case so that progress could be made in part b). Care needs to be exercised if tempted to use a calculator to show the truth of a statement. In the main (but not absolutely always) calculators need to be used with caution. For instance, for any of the integrals in Q10 on the 2006 paper, integration using the calculator is not appropriate and is unlikely to be worth many marks. If one used a calculator to approximate the integral in part a) it would be possible to derive a decimal approximation to its value, which might be identical to the decimal expansion of the given result to many significant figures, but this would not be sufficient evidence that the two values are the same (e.g. they could differ in the 100th decimal place). A calculator could show that the integral is very close to 16√2 -12; but it couldn’t prove that it is exactly this value. If one were asked to find the value of the integral correct to a few decimal places then the calculator route is viable; if the question is of the ‘Show that’ variety (or said ‘Evaluate exactly’) then the expected answer involves integration from first principles. If a question expects an integral to be evaluated numerically it will normally make this clear (see e.g. Q15 f). One last point to be made in connection with the ‘Show that’ instruction is that important not to make work and do more than is strictly required. On the 2006 paper, the best example of this is Q 12a); the question asks the candidate to show that the given function satisfies the given differential equation. It is sufficient to show that if one substitutes the given function in both sides of the differential equation then they are the same; so the equation is satisfied. In particular, it is not necessary to solve the differential equation from first principles. 4. USE THIS RESULT This expression asks the candidate to use some information just derived to answer the next part of the question. There may be other ways to tackle what is wanted; but doing this will be rewarded with zero marks. Generally, ‘Use the result’ indicates that the examiner wants the candidate to proceed with the next step in a line of reasoning that has already started. The one example of this instruction in the 2006 paper appeared in Q 10 c ii); here the requirement was to show how knowledge of the derivative of 2^x in Q 10 c i) enables one to deduce its integral as well. The integral of this function could be read straight from the tables; but as this approach does not use what has been already shown it is liable to be awarded few marks.

Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

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5.

VERIFY In most cases the word ‘verify’ is synonymous with ‘check’. The request to verify the truth of a statement is often the weakest form of proof. In Q 6a), to check P(2,1) lies on the curve merely requires a demonstration that the equation given is satisfied by x=2, y=1. In Q 15 a) all that needs to be done is to check that multiplying out the left hand sides of the expressions does give the right hand sides. DEDUCE In most cases the work required to deduce a result from previous workings should be a quick and relatively painless exercise. The mathematical word, deduce, invites the candidate to use previous information/answers in a question to find a new result with almost no, or at worst minimal, additional workings and effort. See, for example, Q 6c); once the second derivative is known at the point P, it is then known immediately what the nature of the stationary point must be. HENCE In many ways the word hence has a similar meaning to deduce – the only difference being that there is less expectation that the new result ought to be obtained quickly. While answers to a deduce question should be very swift, responses to a ‘Hence’ may well be more involved. The word ‘Hence’ implies that previous findings should lead directly to the new desired result, but there could be some effort required to get there. DETERMINE Synonymous with ‘find’. STATE If asked to state an answer this implies that the examiner is expecting minimal (and quite possibly, no) working. Depending on the context in the question, the candidate ought to be able to find the desired answer almost by inspection.

6.

7.

8. 9.

Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

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Calculus TEE 2006 – Outline solutions 1. [5 marks] For x < π / 2 have f(x)=ax+b and f ' ( x) = a .

[1]

[1]

For x ≥ π / 2 have f ( x) = x cos x and f ' ( x) = cos x ? x sin x .

Continuity of function and derivative at x = π / 2 requires

aπ / 2 + b = 0

and

a=?

π

2

.

[1+1]

Hence b =

π2

4

.

[1]

_____________________________________________________________________________ 2. [7 marks]

a) [2 marks]

( f o g )( x) = exp[2 sin x]

so

[1] [1]

( f o g )(π / 6) = exp(1) = e .

Domain is ? π < x < π and range is

b) [3 marks]

1 ≤ y ≤ e2 . e2

[1+2]

c) [2 marks]

d d d [( f o g )( x)] = [exp(2 sin x)] = exp(2 sin x) (2 sin x) dx dx dx = 2 cos x exp(2 sin x) .

[1]

[1]

_____________________________________________________________________________

Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

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3. [7 marks]

a)

? ex ? 5 ? 1 1 ? = lim 1 ? 5e ? x = lim? x →∞ 3 x →∞ ? 3e x ? 3 ? ?

(

)

since e ? x → 0 as x → ∞ .

[2]

(Alternatively, if evaluated at at least 2 large values of x and limit deduced then full marks.) b) [2 marks]

? x3 ? 3 ? ?8?3 (?2) 3 ? 3 11 ?= lim ? 3 ? x ? x 2 ? 5 ? (?2) 3 ? (?2) 2 ? 5 = ? 8 ? 4 ? 5 = 17 . x → ?2 ? ?

[2]

c) [3 marks]

? f ( a + h) ? f ( a ) ? By definition we have f ' (a ) = lim? ? h →0 h ? ?

[1]

? cos 2(a + h) ? cos 2a ? so with f ( x) = cos 2 x it follows that lim? ? = f ' (a ) = ?2 sin 2a . h →0 h ? ?

[1+1]

_____________________________________________________________________________ 4. [9 marks]

a) [5 marks] If f ( x) = ax exp(bx) then f ' ( x) = a (1 + bx) exp(bx) [2] [2] [1]

and if f(x) has a turning point at x = 1 / 2 so f ' (1 / 2) = 0 and 1 + b / 2 = 0 ? b = ?2 . As f (1 / 2) = 1 then (a / 2) exp(?1) = 1 and a = 2e . b) [4 marks] Curve is f ( x) = 2 x exp(1 ? 2 x) , f ' ( x) = 2(1 ? 2 x) exp(1 ? 2 x) and so

f ' ' ( x) = 2[?2 ? 2(1 ? 2 x)] exp(1 ? 2 x) = 8( x ? 1) exp(1 ? 2 x) ? f' ' (x) = 0 at x = 1 .

[2]

[2]

Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus 11

5. [12 marks]

a) [2 marks]

[1+1] b) [2 marks] Modulus of 3 ? 4i = 3 2 + 4 2 = 5 . Lies in fourth quadrant so argument = ? tan ?1 (4 / 3) = ?0.9273 rads (= ?53.1o ) . c) [8 marks] i) If z = r exp(iθ ) then z1 = z 2 = r 2 exp(2iθ ) . Modulus r 2 , argument 2θ . [1+1] ii)

z 2 = 2 z = 2r exp(iθ ) . Modulus 2r, argument θ .

4 + 5i (4 + 5i )(2 + 3i ) ? 7 + 22i 7 22 = = . Real = ? , Imag = . 2 ? 3i (2 ? 3i )(2 + 3i ) 13 13 13

[1] [1]

[1+1] [1+1]

iii) z 3 = 1 / z = 1 /[ r exp(?iθ )] = r ?1 exp(iθ ) . Modulus r ?1 , argument θ . iv) z 4 = ?iz = r exp(?iπ / 2) exp(iθ ) = r exp(i[θ ? π / 2]) . Modulus r, argument θ ?

π

2

.

[1+1] _____________________________________________________________________________

6. [9 marks]

a) [1 mark] If x = 2 and y = 1 then x 2 y + y 3 = 4 + 1 = 5 and 4 x ? 3 = 5 so P(2,1) is on curve. b) [3 marks] Differentiating equation of curve implicitly gives 2 xy + x 2 y '+3 y 2 y ' = 4 so at P have 2(2)(1) + 4 y '+3 y ' = 4 whence y ' = 0 and hence P is a turning point. c) [5 marks] Differentiating equation implicitly once more gives 2 y + 2 xy '+ x 2 y ' '+2 xy '+6 y ( y ' ) 2 + 3 y 2 y ' ' = 0 At P have x = 2 , y = 1 and y ' = 0 so [2] [2] [1]

[1]

2 + 4 y ' '+3 y ' ' = 0 ? y ' ' = ?2 / 7

and since this is <0, P is a maximum.

Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

[2] [1]

12

7. [7 marks]

With n=5 in De Moivre’s theorem then cos 5θ + i sin 5θ = (cos θ + i sin θ ) 5 = cos 5 θ + 5 cos 4 θ (i sin θ ) + 10 cos 3 θ (i sin θ ) 2 + 10 cos 2 θ (i sin θ ) 3 + 5 cos θ (i sin θ ) 4 + (i sin θ ) 5 [3] Taking imaginary parts of both sides gives that sin 5θ = 5 cos 4 θ sin θ ? 10 cos 2 θ sin 3 θ + sin 5 θ

= 5 sin θ 1 ? sin 2 θ

[1] [1]

(

)

2

? 10 sin 3 θ (1 ? sin 2 θ ) + sin 5 θ

= 5 sin θ (1 ? 2 sin 2 θ + sin 4 θ ) ? 10 sin 3 θ + 11sin 5 θ = 5 sin θ ? 20 sin 3 θ + 16 sin 5 θ [2]

_____________________________________________________________________________

8. [9 marks]

a) [3 marks] | z ? 3 + 2i |< 5 Interior of circle centre 3 ? 2i , radius 5 1 mark for identifying circle 1 mark for finding correct centre and radius 1 mark for indicating interior of circle [3]

Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

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b) [3 marks]

| z ? 3i |=| z + 4 |

Straight line, which is the perpendicular bisector of the line joining the points ? 4 and 3i 1 mark for identifying the points ? 4 and 3i 1 mark for recognising locus as a straight line 1 mark for indicating the line is a perpendicular bisector. [3]

c) [3 marks]

0 ≤ arg( z ? 3) ≤ 3π / 4 Wedge-shaped region. Vertex located at (0,3) and edges through this point angled as shown (of course domain extends outwards in the obvious way) 1 mark for identifying the importance of 3 1 mark for stating region of wedge shape 1 mark for finding the configuration of the edges [3]

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Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

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9. [12 marks]

a) [5 marks] If 1 < | 5 ? 3x | <9 then i) 3 x ? 5 > 1 or 3x ? 5 < ?1 ? x > 2 or x <

4 3

[2]

and ii) ? 9 < 3 x ? 5 < 9 ? ?4 < 3 x < 14 ? ?

4 14 <x< . 3 3 4 4 14 < x < or 2 < x < . 3 3 3

[2]

We need both i) and ii) to hold implying that ?

[1]

b)

[7 marks] Alongside are plots of sin x and | sin 2 x | . (Note horizontal axis is x / π in this diagram.) 1 mark for the correct form of the former and 2 marks for the latter. [3]

If | sin 2 x |≤ sin x then 2 sin x | cos x |≤ sin x as sin x > 0 across the whole of the interval. [1] Thus need | cos x |≤ 1 / 2 implying that [1] [2]

π

3

≤x≤

2π . 3

(NOTE: if solve the inequality part by reading 1.05<x<2.09 off the graph then 1 mark only.) ______________________________________________________________________

Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

15

10. [11 marks]

a) [4 marks]

∫

1

2

6x + 4

dx = ∫ ? 6 x ? ? x 1

2

1

2

+ 4x

?1

2

?dx = ?4 x 3 2 + 8 x 1 2 ? ? ? ?1 ? ? ?

2

[3]

= 8 2 + 8 2 ? 4 ? 8 = 16 2 ? 12 .

[1]

b) [3 marks] 1 1 x ?1 2 ? ∫ 1 + x 2 dx = ? 2 ln(1 + x )? 0 = 2 [ln 2 ? ln 1] = 2 ln(2) = ln 2 . ? ? 0

1 1

[2+1]

c) [4 marks] i) If y = 2 x then ln y = x ln 2 . Differentiating gives

dy 1 dy = ln 2 and thus = y ln 2 . [2] dx y dx

ii) With y as above then

∫ ydx =

0

1

1 dy 1 1 1 1 x =1 0 ∫ dx dx = ln 2 [y ]x=0 = ln 2 [2 ? 2 ] = ln 2 . ln 2 0

1

[2]

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Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

16

11. [12 marks]

a) [3 marks] Length of rectangular part of the perimeter is 2(x+y). Edge of the semicircle has length πy so that

L = 2 x + 2 y + πy = πy + 2( x + y ).

Hence x = 1 ( L ? πy ? 2 y ) and for x ≥ 0 need y ≤ L /(π + 2) . 2

[2] [1]

b) [4 marks] 1 Area of window A = x(2 y ) + πy 2 , 2 (area of rectangle + area of semicircle) [2]

1 1 ? ?π so A = y ( L ? πy ? 2 y ) + πy 2 = Ly ? πy 2 ? 2 y 2 = Ly ? ? + 2 ? y 2 . 2 2 2 ? ? c) [5 marks] At extreme values of area have

[2]

dA dA = 0 with = L ? (π + 4) y dy dy L . π +2

[1]

whence y = L /( 4 + π ) which is in the permitted range 0 ≤ y ≤

2

[1]

Then A =

L2 L2 L2 L2 1 ? L ? ? (4 + π )? ? = . ? = (4 + π ) 2 (4 + π ) 2(4 + π ) 2(4 + π ) ? 4+π ?

[2]

As d 2 A / dy 2 = ?(π + 4) < 0 , turning point is a maximum.

[1]

_____________________________________________________________________________

Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

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12. [9 marks]

a) [2 marks] If M (t ) = C exp(? kt ) then and so satisfies the DE.

dM = ?kC exp(? kt ) = ?kM dt

[2]

b) [3 marks] If M = 150 when t = 2 then 150 = C exp(?2k ) and if M = 100 when t = 3.5 [1] then 100 = C exp(?3.5k ) . Dividing the equations gives 3 / 2 = exp(1.5k ) so k = (2 / 3) ln(3 / 2) ≈ 0.27031 . In turn it follows that C ≈ 257.56 . [1] [1]

c) [2 marks] When M = 40 have 40 = C exp(?kt ) ? ?kt = ln(40 / C ) so that t ≈ 6.890 . Hence substance reduces to 40 grams after approximately 6.89 years. [2] d) [2 marks] Initial mass of substance is C; half-life is defined to be time taken for mass to [1] drop to C / 2 . Then

C / 2 = C exp(? kt ) ? ? kt = ln(1 / 2) ? t = k ?1 ln 2 ≈ 2.564 years.

[1]

____________________________________________________________________________

Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

18

13. [13 marks]

a) [5 marks] 1 ? ? Volume of solid = π ∫ y dx = π ∫ (1 ? x )dx = π ? x ? x 3 ? 3 ? 1? h ? 1? h

1 2 2 1

[1+1]

1 1 ? 1 ? ? 1 ? = π ?1 ? ? (1 ? h) + (1 ? h) 3 ? = π ?? + h + (1 ? 3h + 3h 2 ? h 3 )? 3 3 ? 3 ? ? 3 ? 1 ? 1 ? = π ? h 2 ? h 3 ? = πh 2 (3 ? h) . 3 ? 3 ? b) [8 marks] By the related rates formula we know that [3]

dV dV dh dh = and we need to find . dt dh dt dt

[2]

From formula in previous part

dV = π (2h ? h 2 ). dh

[1]

Volume of hemisphere is

2 π cubic metres (either known or put h = 1 in formula of a). 3 dV 2 ? 1 ? π As bowl is filled in 10 minutes, have cubic metres per minute. = π? ? = dt 3 ? 10 ? 15

dh dV ? dV ? = ? ? dt dt ? dh ?

?1

[2]

Thus

=

π

15π (2h ? h )

2

=

1 . 15h(2 ? h)

[1]

When h = 1 / 2 , have

1 4 dh = = dt 15(1 / 2)(3 / 2) 45 =

metres/minute

400 80 = cm/minute. 45 9

[2]

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Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

19

14. [18 marks]

a) [2 marks] acceleration If v (t ) = ?5 sin (t / 2 ) i + 4 cos(t / 2) j then differentiation gives 5 a (t ) = ? cos(t / 2) i ? 2 sin(t / 2) j. 2

[2]

b) [3 marks] Integrating the velocity gives the position vector

r (t ) = 10 cos(t / 2) i + 8 sin(t / 2) j + C

where C is a constant. As r= 10i initially so C=0 and r (t ) = 10 cos(t / 2) i + 8 sin(t / 2) j. c) [2 marks] From the formulae above it is clear that a(t)= ? always parallel. d) [3 marks]

[2] [1]

1 r(t) so these vectors are 4 [2]

Path of object is an ellipse whose shape is sketched here. [2]

Object moves around the ellipse anticlockwise. [1]

(Object starts at point P and returns in time 4π .) e) [3 marks]

4π

∫

0

v(t) dt = [10 cos(t / 2)]0 i + [8 sin(t / 2)]0 j = 10( cos 2π ? 1 )i +8( sin 2π ? 0) j=0

4π 4π

[2]

This means that the displacement after time 4π is zero so particle is at its starting point. [1] f) [5 marks] ? 2π ? 25 sin 2 (t / 2) + 16 cos 2 (t / 2) dt ? = 2 ∫ 25 sin 2 ? + 16 cos 2 ? d? ? ∫ ∫ ? ? 0 0 0 ? ? Using the calculator we find that this integral is 56.723 approximately.

4π 4π

|v(t)| dt =

[1] [2]

This is the circumference of the ellipse – the distance travelled by the object in one circuit. [2]

Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

20

15. [16 marks]

a) [5 marks] i) ( z ? 1)( z 4 + z 3 + z 2 + z + 1) = ( z 5 + z 4 + z 3 + z 2 + z ) ? ( z 4 + z 3 + z 2 + z + 1) = z 5 ? 1 , as required ii) [ z ? cis (θ )][ z ? cis (?θ )] = z 2 ? z[ cis (θ ) + cis (?θ )] + cis (θ ) cis (?θ ). Now cis( θ )+cis( ? θ ) = (cos θ + i sin θ ) + [cos(?θ ) + i sin(?θ )] = cos θ + i sin θ + cos θ ? i sin θ = 2 cos θ and cis( θ )cis( ? θ )= ( cos θ + i sin θ )(cos θ ? i sin θ ) = cos 2 θ + sin 2 θ = 1 so that [ z ? cis (θ )][ z ? cis (?θ )] = z 2 ? 2 z cos θ + 1

[2] [1]

[1] [1]

as required.

b) [5 marks] If z 5 = 1 then the roots are z=cis( 2kπ / 5) for k=0….4. In order to express all angles in the range prescribed we write z = cis( θ ) where θ = ±4π / 5, ± 2π / 5 and zero. [3] In the complex plane these roots are represented by five unit length lines emanating from the origin; one line lies along the positive real axis and the five lines are equally spaced with angles 2π / 5 between them. [2] c) [6 marks] Given the factorisation in a i) and the fact that z = 1 is one root of z 5 = 1 , the remaining four roots must be the solutions of z 4 + z 3 + z 2 + z + 1 = 0 . [2] Hence ? ? 2π z 4 + z 3 + z 2 + z + 1 = ? z ? cis? ? ? 5 ? ? 4π ? ?? ? ?? z ? cis? ?? ? 5 ? ?? ? 2π ? ?? ? ?? z ? cis? ? ?? ? 5 ? ?? ? 4π ? ?? ? ?? z ? cis? ? ?? ? 5 ? ?? ?? ?? . ? ?? [2]

Using result a ii) above and pairing up the factors in the obvious way, we deduce that

? ? 2π ? ?? 2 ? 4π ? ? z 4 + z 3 + z 2 + z + 1 = ? z 2 ? 2 z cos? ? + 1?? z ? 2 z cos? ? + 1? . ? ?? ? ? 5 ? ?? ? 5 ? ? ? [2]

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Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

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16. [10 marks]

If x(t ) = A sin(ωt + α ) then max. value taken by x is A so with given information A = 5 . d 2x π2 π = ?ω 2 x and so comparing with given DE shows that ω 2 = ?ω = . 2 36 6 dt dx = +2 so A sin α < 0 and Aω cos α = +2 . dt

[1]

Also

[1]

When t = 0 are given that x < 0 and

[2] [1] [1]

Now signs of sin α and cos α mean that α lies in the fourth quadrant with cos α = 2 / Aω = 0.76394 .

Then α = ?0.7014 radians and x(t ) = A sin(ωt ? 0.7014) . At high tide have x = 5 and this next occurs when

[1]

ωt ? 0.7014 =

π

2

? ωt = 2.2722 ? t = 4.340 hours.

[2] [1]

As 0.340 hours is equivalent to 20.4 minutes, next high tide at 4.20 pm approx.

17. [7 marks]

Area of loop = 2 ∫ y dx = 2 ∫ x a ? x dx

0 0

a

a

[1+1]

Substitute u = a ? x

A = 2∫ (a ? u ) u (? du ) = 2∫ (au

a

[2]

a

0

whence

1/ 2

?u

3/ 2

0

2 ?2 ? )du = 2? au 3 / 2 ? u 5 / 2 ? 5 ?3 ?0 sq. units.

a

8 ?2 2? ?4? = 2? ? ?a 5 / 2 = 2? ?a 5 / 2 = a 5 / 2 15 ?3 5? ? 15 ?

[3]

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Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

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18. [7 marks]

From the graph of f ' ( x) it is clear that f(x) is increasing on 0 ≤ x ≤ 6 and decreasing on 6 ≤ x ≤ 10 . Thus maximum value is thus taken at x = 6 where

9 1 f ( x) = f (0) + ∫ f ' ( x)dx = 1 + π (3) 2 = 1 + π . 2 2 0 Using the observation above, f(x) must be least at one end of the interval. Clearly the magnitude of

6

[2]

[2]

[1]

∫

0

6

10

f ' ( x)dx is greater than that of

∫ f ' ( x)dx so f(10) must exceed f(0).

6

[1] Hence minimum value of f ( x) for 0 ≤ x ≤ 10 is f (0) = 1 . [1] ____________________________________________________________________________

175786-1

Examiners’ Report on 2006 Tertiary Entrance Examination – Calculus

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