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澳大利亚2007年数学高考卷


EXAMINERS' REPORT ON 2007 TERTIARY ENTRANCE EXAMINATION SUBJECT: CALCULUS Statistics
Year 2007 2006 2005 Number who sat 1502 1434 1601 Non-examination candidates 6 7 5 Absent 83 68 69

The examiners' report is written by the chief examiner to comment on matters relating to the Tertiary Entrance Examination in this subject. The opinions and recommendations expressed of those of the chief examiner alone and are not necessarily representative of or endorsed by the Curriculum Council. The marking guideline 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 Anecdotal evidence suggests that the 2007 Calculus paper was generally well received by candidates and the subsequent final exam average of 55.4% was therefore somewhat disappointing (and a little down on last year’s result). The examiners had hoped for an average nearer 60% (compared with the Curriculum Council’s recommended mean of 58%) but there was one very poorly answered question that depressed the results. The majority of teachers and markers seemed to think the paper a fair one and most of the criticisms of the examination were relatively minor. The paper distinguished the weaker from the stronger candidates well and, just as last year, many candidates lost many marks due to insufficient reasoning – a large proportion seemed quite incapable of presenting a logical argument consisting of more than a couple of steps. The length of the paper was felt to be about right and there was scant evidence that timing was a problem as the least popular question was still attempted by over 90% of candidates. The wide range of raw scores (0%–99%) and the high standard deviation (19.9%) compared with 2006 raw scores (0%–98%) and standard deviation of 19.0%, show that the paper was an effective discriminator. The paper had an excellent internal reliability coefficient (0.94). The general standard of performance on the paper was similar to last year. Standard type questions and those requiring graphics calculators were done well. Algebraic manipulation showed some improvement over recent years, but the quality of sketch diagrams and of reasoning ability was less than should be expected at this level.

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus
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General comments The first half of the paper was intended to be quite standard and straightforward to give the candidates the opportunity to develop some confidence and demonstrate their knowledge via fairly familiar tasks. This goal seems to have been achieved as the average candidate scored about 70% of the maximum marks on Q1–10. Questions 11–20 were meant to be more challenging with the latter questions designed to sort out the candidates at the upper end of the scale. Had the examination average been slightly higher then we would have been very satisfied with the overall outcome but Q12 was particularly badly answered. After marking, it was pointed out to the examiners that Q13 and Q20 were essentially identical to questions from the 2000 paper. Although the examiners would prefer not to repeat questions, they do not think it is a disaster so long as there is a considerable time-lag between the occurrences. However, the examiners wish to emphasise that this repetition of questions was completely unintentional and, had it been spotted earlier, would not have occurred. Teachers and future calculus candidates should be aware that this recycling of questions was accidental and are advised not to spend time trying to guess repeat questions that might appear in later years. The examiners’ intention is that this will not recur. Most markers felt that the paper was not difficult to assess although, as always, the reasoning type of task like Q20 required some care to ensure that arguments presented were watertight. The paper tested most parts of the syllabus providing a balanced coverage of the broad content sections specified in the assessment table as shown below. Syllabus content Calculus of trig. functions Functions & limits Theory/techniques calc. Applications of calculus Vector calculus Complex numbers Total Recommended weighting (%) 5–10 15–20 15–20 25–30 5–10 20–25 Relevant question numbers 1,4,15 1,3,7,10,11 4,8,11,18 1,2,5,12,13,14,17 19 6,9,16,20 Marks available 13 35 27 50 16 39 180 % weighting 7 19 15 28 9 22 100

As has been mentioned in previous reports, there was some evidence that an appreciable number of candidates are too reliant on their calculator and prefer to obtain approximate numerical answers when proper reasoning would give accurate and exact results. The authors of last year’s report went to great pains to amplify the meanings of common words used in questions such as: show, hence, deduce, use the results, etc. It is essential that candidates understand these terms and how they differ; disappointingly, many candidates still appear to ignore the precise wording used. Candidates must appreciate that the language used in questions is formulated very carefully and that answers will only be given full credit if it is clear that the requirements of the question have been completed as asked (rather than as the candidate may have hoped it had been asked). As in previous examinations, many candidates did not explain themselves properly and wrote rather vague answers. It is not clear whether such responses are deliberate, or whether this is just a consequence of poor or sloppy linguistic style. Whichever, this type of answer leaves the marker in a quandary as to whether the candidate really understands the crux of the problem but cannot express themselves well, or whether this is a conscious attempt to ‘muddy the waters’ in the hope that some lack of understanding can be disguised. As this has been raised
Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus 2

in consecutive years, the examiners wish to emphasise that it is the responsibility of the candidate to demonstrate fully that the requirements of the question have been fulfilled completely. Markers will be instructed that in future exams no benefit of the doubt should be exercised. Comments on specific questions Question 1 Mean 6.12/7 (87.38%) Standard deviation 19.72%

A simple first question. In a) some sort of explanation was required for full marks; merely stating the solutions was not sufficient. In b) the most common errors were not to put x=3 in the derivative or recognise that cos(4 π)=1. Question 2 Mean 3.33/4 (83.30%) Standard deviation 26.26%

Another easy question. For full marks, candidates had to demonstrate the formula used; merely quoting two iterates lost marks. As might be expected, some rounding errors occurred; others omitted the 7 in the definition of the function and some were penalised for quoting something other than 6 dp. Question 3 Mean 5.74/9 (63.81%) Standard deviation 26.78%

Most of the marks lost resulted from not reading the question properly. In a) many candidates failed to sketch the functions within the domain given. Many sketches were of unacceptably low quality; straight lines were not shown as such and the fact that g(x)=1 for all x≥0 was not obvious. In b) and c) some candidates discussed the continuity of the functions rather than their differentiability and several seemed to think that continuous functions must be differentiable. In order to obtain full marks it was necessary to calculate values of the left and right derivatives of the functions at x=0; for instance, merely stating that g'(0+)=g'(0-) was not sufficient. Question 4 Mean 4.78/6 (79.59%) Standard deviation 27.53%

Reasonably well answered. Most realised that the fundamental theorem of calculus was required here but some did not incorporate the chain rule correctly. Part b) was an easy example of implicit differentiation and, apart from a few careless errors, seemed to cause no real problems. Question 5 Mean 4.49/6 (74.91%) Standard deviation 29.42%

A simple volume question that was poorly answered by some. Most candidates found the coordinates of P but then too many made mistakes in the formulation of the volume integrals. A common error was to try to evaluate the integral of the square of the difference of the functions between Q and P rather than the difference of the squares. This resulted in a horrible integral that was often integrated incorrectly. Disappointingly, many candidates resorted to the calculator to evaluate the integrals which could be simply done by standard techniques. Question 6 Mean 5.97/10 (59.70%) Standard deviation 23.81%

Another easy question that was not well handled. Most candidates sketched the points of the parallelogram but some failed to join them up. A common error in b) was not to compute |z?+z?| squared but rather (z?+z?) squared etc. so that complex rather than real numbers were quoted as answers. (This type of mistake shows a real lack of understanding!) Part c) was badly answered; the majority wrote very poor descriptions of the connection between the answers in b) and the geometry of the parallelogram. There were 10 gift marks on offer here; it is noted that the average performance here was barely above the average score on the paper as a whole.
Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus 3

Question 7

Mean 5.02/7 (71.70%)

Standard deviation 29.82%

A standard type of question that should have been an easy source of marks. In a) answers were not well reasoned. Many candidates resorted to L'Hopital's rule without commenting that this method was being used. In b) a significant minority of candidates did not make the connection with the definition of a derivative – of those that did, more could not apply the definition correctly. Question 8 Mean 3.60/6 (59.96%) Standard deviation 37.72%

Satisfactorily handled, although several tried to integrate a) by parts and made problems for themselves. The multiple angle in b) confused a number of candidates. Again, given the standard nature of these two integrals, one could have hoped for a better outcome. Question 9 Mean 7.29/10 (72.91%) Standard deviation 26.95%

Three fairly standard locus questions. As always, candidates ought to be more precise in sketching and make sure they show all the relevant details. For instance, in a i) it is vital that the points -3i and 5-4i are shown and that the required locus is the perpendicular bisector of the segment joining these points. In ii) a good answer must identify the location of the vertex of the wedge, the size of the included angle, and the region satisfying the inequality shaded. Many answers for part b) were given in terms of r and θ. As the question asked for inequalities on the modulus and argument of z, answers should be given in terms of these quantities even though it can be inferred what might be intended by the symbols r and θ. Question 10 Mean 7.71/12 (64.28%) Standard deviation 24.59%

Part a) was generally well done. although several candidates lost marks unnecessarily through sloppy sketching. In b i), many found the domain even though answers were not written in a mathematically careful manner. The range proved to be very much more troublesome and not many candidates acquired marks here. Very few scripts gave any working at all for the range and the vast majority of answers appeared from thin air – unless some explanation is given, no marks can be awarded for an incorrect answer and even a correct response is unlikely to be given full marks if no supporting reasoning is included. The algebraic manipulation in ii) bamboozled some candidates and many made silly errors as a result of being careless with minus signs. In iii), the majority realised that the domain of the inverse function follows from the range of f(x) and picked up the mark on offer. Question 11 Mean 4.76/8 (59.44%) Standard deviation 26.99%

The start of the second half of the paper and the first non-standard question. This task was actually very easy but a lack of algebraic skills and an inability to follow the argument through proved the undoing of many. Common errors included difficulties in differentiating c/(1-dx) and, worryingly, misunderstanding of what was happening near x=0. There was the perception that since f(0)=7, f'(0) must be zero (presumably differentiating a constant). Of those that did find the four constants a-d a large proportion could not find x? properly and said that x?<0 even though the cubic used to find the zero of f(x) was defined only for x>0. Question 12 Mean 2.47/9 (27.41%) Standard deviation 36.45%

In the main, the answers to this question were extremely disappointing – this should have been a routine optimisation task. It is a major worry that the overwhelming majority of candidates were frankly clueless when it came to this problem. They were unable to formulate the problem in a way appropriate for subsequent application of calculus techniques. In this subject, which should represent the 'gold standard' of the three mathematical subjects examined at TEE, it is expected that candidates will be able to model and then solve simple optimisation problems. Future candidates should ensure that questions like this will not trip them up, and it is a safe assumption that questions of this type will continue to appear until standards improve significantly.
Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus 4

Question 13

Mean 5.79/10 (57.91%)

Standard deviation 28.76%

This question was satisfactorily answered although there were still many errors that crept in due to lack of attention to detail. Many statements made in a), which should have been an easy couple of marks, were imprecise – it seemed that many candidates think that a reduction of (1/16)th volume and a reduction to (1/16)th volume are the same. There was considerable fuzziness in answers to b); while most candidates knew the increment formula, the examiners and markers remained unconvinced that they really knew what to do with it. Many neglected to include negative signs to indicate a reduction in volume. Very few indicated why the reduction predicted in c) was physically absurd although most did appreciate that the increment formula is only likely to be useful when dealing with small changes in quantities. Question 14 Mean 5.38/10 (53.83%) Standard deviation 29.67%

Again not as well answered as it should have been. The formulation of the problem was generally poor and insufficient attention to detail meant that the phase angle in a) was often wrongly calculated. Many candidates did not appreciate that both the initial displacement and the direction of movement is required to tie down the phase uniquely and the information provided in the question concerning the initial velocity was often ignored. Many candidates attempted to answer b) using the fact that the distance travelled may be found by integrating the magnitude of the velocity between t=0 and t=4. The examiners had intended this part of the problem to be answered by finding the displacement at the beginning and the end of the period and working out how many times the particle had reversed direction. The integral approach is of course acceptable if it is followed through correctly. Too many candidates integrated |x(t)| rather than |v(t)|; others forgot the modulus signs. Such mistakes were heavily penalised; if candidates opt to rely on memorised formulae they must be correct – ‘near misses’ are not good enough. In my opinion, good mathematics requires a minimum of rote learning of formulae; much better to solve problems from ‘first principles’ each time. Question 15 Mean 4.57/9 (50.75%) Standard deviation 27.44%

This scaffolded question contained a number of key words and phrases to link the various parts together. Inexperience with this type of more formal question was evident by the way many candidates were unable to connect the sub-tasks and avail themselves of the help provided. Many attempted to justify their answers by recourse to sketch graphs and, while this was acceptable in a), it is insufficient for the remainder of the question. Candidates should note that each part of the question from b) onwards demands reference to results from preceding parts; since the words ‘or otherwise’ were not used, marks could only be awarded if the appropriate linking was made. Question 16 Mean 4.25/8 (53.12%) Standard deviation 37.80%

A fairly standard exercise involving use of DeMoivre's theorem which was not done as well as should have been the case. Many tried to solve z? = +1 rather than z? = -1 and few candidates gave proper statements of DeMoivre's theorem itself. Many good attempts were made at b) although, once again, marks were lost due to scrappy diagrams. Question 17 Mean 4.74/8 (59.21%) Standard deviation 38.72%

Very standard question that should have been better answered. Too many candidates attempted to write down (ie. without any working at all) the general solution of the differential equation in a) and virtually all of these were deficient in some way. There was much confusion caused in integrating dC/(1-C); the minus sign was often omitted. Many also neglected to add in an arbitrary constant so that the most frequently quoted general ‘solution’ to a) had the incorrect exponential and no constant. Fortunately for these candidates these errors were not catastrophic for the remainder of the question and b) and c) were well done.

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

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Question 18

Mean 7.18/14 (51.29%)

Standard deviation 25.87%

This question proved to be the most correlated with overall performance on the paper. The sketches in a) were generally adequate though, again, some lost marks through untidiness and lack of attention to detail. The explanations in b) were of very variable quality; surprisingly many candidates did not make it clear that the value of the integral was determined by the area under the curve. The easy part c) was well done but the wording in d) tripped many. There are two tasks to be tackled here and it was surprising just how many ignored the first and concentrated only on the second. Most candidates did tackle e) and an appreciable minority made a mess of the arithmetic. Others ignored completely the instruction to use parts c) and d) and evaluated the integral directly using their calculator; such an approach was worth no marks at all. Question 19 Mean 5.16 /16 (32.26%) Standard deviation 25.00%

Part a) was not done well considering there were 6 straightforward marks on offer. Some candidates tried to quote the results directly and made an error – others tried to integrate the given equations but ignored constants along the way. The instructions for b) were quite explicit but many couldn't perform the manipulations; it was also interesting how many with the wrong answers for a) could use them to derive the correct (and given) answer to b). Parts c) and d) were beyond most candidates but that was as expected; these parts were the hardest ones on the entire paper. Question 20 Mean 4.18/11 (38.01%) Standard deviation 28.71%

Not a difficult question for the end of the paper which required a (short) formal argument to deduce the given results. Overall, the standard of deductive reasoning was very poor; many candidates gave arguments with great holes in the logic. Not many recognised the connection between a) and b) and tackled the latter almost ‘from first principles’. In this case that was acceptable, and the careful candidates did obtain the correct solution to b). Nevertheless, in the main the scripts left me with the impression that scant attention had been paid to accuracy and detail. Issues for consideration Nil. Acknowledgement and 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 December 2007 2007 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 2007 Tertiary Entrance Examination – Calculus

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Tertiary Entrance Examination, 2007 CALCULUS

Marking Guidelines

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

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Calculus TEE Marking Guidelines 2007 As a general marking rule – any question or part question worth 1 or 2 marks was given full marks even if there was no working. Any question or part question worth more than 2 marks required some valid working/justification to receive full marks. 1. (7 marks) (a) Solve |2x+1| = x+3. (3)

Sketches of the functions | 2 x + 1 | and x + 3 shown above. From graph, or algebraically, 2 x + 1 = x + 3 ? x = 2 or ? (2 x + 1) = x + 3 ? 3x = ?4 ? x = ?4 / 3 . (2 marks) Hence solution of equality is x = ?4 / 3 or x = 2 . Find the equation of the tangent to the curve y = 3 sin ? (1 mark)

(b)

? 4πx ? ? at x = 3. ? 3 ?

(4)

For y = 3 sin ?

dy ? 4πx ? ? 4πx ? = 4π cos? ? we have ? dx ? 3 ? ? 3 ?

(1 mark) (1 mark) (2 marks)

which equals 4π cos 4π = 4π at x = 3 . Hence equation of required tangent is y ? 0 = 4π ( x ? 3) or y = 4πx ? 12 Ξ. No penalty for giving answer as a decimal (e.g. y = 12.566 x ? 37.699 ).

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

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Calculus TEE Marking Guidelines 2007 2. (4 marks) A first approximation x1 to the root of x 4 + x 3 + x 2 = 7 is x1 = 1.2 Use the Newton-Raphson algorithm twice to find the next two approximations x 2 and x 3 to the root. Give your answers correct to 6 decimal places.
4 3 2 ? xn + xn + xn ? 7 ? f ( xn ) ? = xn ? = xn ? ? 3 ? 4 x + 3x 2 + 2 x ? f ' ( xn ) n n ? ? n

Newton-Raphson formula is x n +1

(1+1 mark)

If we start with the iterate x1 = 1.2 then, correct to six decimal places, the next iterates are

x 2 = 1.328991
and x3 = 1.315557

(1 mark) (1 mark)

Candidates had to show that Newton-Raphson was used to gain full marks – by stating the derivative and the function or the formulae as above. If x2 and x3 only shown then 2 marks were awarded.

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

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Calculus TEE Marking Guidelines 2007 3. (9 marks) (a) Sketch the graphs of the functions f ( x) = | x | + 2 and g ( x) = 1 +

| x| ≤ 2.

3 (x? | x |)2 for 16
(3)

!2 Correct form of | x | +2 .

0

2

x (1 mark) (1 mark) (1 mark)

Identification that the squared function is constant for x > 0 . Quadratic form in x<0. -1 if the given domain was ignored or if no vertical scale was given or at least implied.

(b)

Is f (x) differentiable at x = 0 ? Justify your answer.

(3)

lim
x →0 +

f ( x ) ? f ( 0) f ( x) ? 2 x = lim = lim = 1 = f ' (0 + ) x?0 x x →0 + x →0 + x f ( x ) ? f ( 0) f ( x) ? 2 ?x = lim = lim = ?1 = f ' (0 ? ) x?0 x x x →0? x →0 ?

(1 mark)

lim
x →0 ?

(1 mark) (1 mark) (3)

As these limits are different, f(x) is not differentiable at x=0. (c) Is g (x) differentiable at x = 0 ? Justify your answer.

lim
x →0+

g ( x ) ? g ( 0) 1?1 0 = lim = lim = 0 x?0 x x →0+ x →0 + x

= g ' (0 + )

(1 mark)

[(3 / 16) * (2 x) 2 + 1] ? 1 3x 2 g ( x) ? g (0) = lim = lim =0 lim x ? 0 x x →0 ? x →0 ? x →0? 4 x
Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus 10

Calculus TEE Marking Guidelines 2007

= g ' (0 ? )
As these limits are the same g(x) is differentiable at x=0.

(1 mark) (1 mark)

If candidates answered (b) NO and (c) YES – without any justification they were awarded 1 mark out of the 6 available. If candidates answered (b) NO and (c) NO or (b) YES and (c) YES - they scored 0 marks. If candidates answered (b) as NO because the limit from the left ≠ limit from the right and (c) as YES because the limit from the left = limit from the right, without actually evaluating the limits in each case, they were awarded 3 out of the 6 marks available for both parts of the question. 4. (6 marks) Find (a)

dy given that: dx
x3

y=


1

1 + t 2 dt

(2)

By the chain rule and fundamental rule of calculus

d ? ? dx ? ?

F ( x)

∫ g (t )dt ? = g ( F ( x)) dx . ?
a

? ?

dF

x3

Hence if y =


1

1 + t 2 dt then

dy d = 1 + ( x 3 ) 2 × ( x 3 ) = 3x 2 1 + x 6 . dx dx

(2 marks)

(b) If y 2 = x sin y

y 2 = x sin y
then implicit differentiation gives

(4)

2 yy ' = xy ' cos y + sin y
whence

(2 marks) (2 marks)

dy sin y = . dx 2 y ? x cos y

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

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Calculus TEE Marking Guidelines 2007 5. (6 marks) The region OPQ sketched below is bounded by the x-axis, the parabola y = x and the line y = x ? 6 . Find exactly the volume generated when OPQ is rotated completely about the x-axis. y

P

O

Q

x

The line PQ meets the parabola y =

x where ( x ? 6) 2 = x ? x 2 ? 13 x + 36 = 0 .

(1 mark) (1 mark)

Hence ( x ? 9)( x ? 4) = 0 . Clearly the root of interest is x = 9 .

Both of the above marks awarded if limits of 9 and 6 found, irrespective of how done (ie by use of calculators) Now as the volume of revolution about the x-axis is π y 2 dx
9 9



Volume generated when OPQ rotated is

V = π ∫ ( x ) 2 dx ? π ∫ ( x ? 6) 2 dx
0 6

(2 mark)

?1 ? ?1 ? = π ? x 2 ? ? π ? ( x ? 6) 3 ? ?2 ?0 ?3 ?6

9

9

(1 mark)

=
-1 if π is left out.

81 27 63 π? π = π. 2 3 2

(1 mark)

An answer of 98.96 without working was awarded 1 mark only as an exact answer was asked for and working should have been shown.

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

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Calculus TEE Marking Guidelines 2007 6. (10 marks) The vertices of a parallelogram P in the complex plane are the numbers 0, z1, z2 and z1 + z2 , where z1 = 2 + i and z2 = 1 + 4i . (a) Sketch the location of P in the complex plane. (2)

R(z)

1 mark for positioning of z1 and z2 and 1 mark for location of z1 + z2 (b) Evaluate (i) If z1 = 2 + i

(1+1 mark)

z1 + z2 + z1 ? z2

2

2

(2)

and z2 = 1 + 4i then (2 marks) (2) (2 marks)

| z1 + z2 |2 + | z1 ? z2 |2 =| 3 + 5i |2 + | 1 ? 3i |2 = 32 + 52 + 12 + ( ?3)2 = 9 + 25 + 1 + 9 = 44.

(ii)

2 | z1 |2 + | z2 |2

(

)

2{| z1 |2 + | z2 |2 } = 2{| 2 + i |2 + | 1 + 4i |2 } = 2{22 + 12 + 12 + 42 } = 2{22} = 44 .

If substitution only is correct, I mark allocated in total for both parts. (c)
| z1 + z2 |2 | z1 ? z2 |2

Describe the results of (b) as a relationship between the lengths of the sides and the diagonals of the parallelogram P. (4) is the square of the length of the diagonal of P joining O and the point z1 + z2 . is the square of the length of the other diagonal of P. (1 mark) (1 mark)

Thus | z1 + z2 |2 + | z1 ? z2 |2 is the sum of the squares of the diagonal. Also | z1 |2 + | z2 |2 is the sum of squares of adjacent sides of P.

Hence the result of (b) shows that the sum of squares of the diagonals equals twice the sum of squares of adjacent sides of the parallelogram. (2 marks) Full marks awarded if a correct statement given.

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

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Calculus TEE Marking Guidelines 2007 7. (7 marks) Find the exact values of the following limits. Show your reasoning in each case. (a)

? x2 ? 6x ? ? lim ? 2 x→ 6 ? x ? 5 x ? 6 ? ? ? ? x( x ? 6) ? x 6 lim? ? ( x ? 6)( x + 1) ? = lim x + 1 = 7 . ? x →6 x →6 ? ?

(3)

? x 2 ? 6x ? ?= lim? 2 x →6 ? x ? 5 x ? 6 ? ? ?

(1+1+1 mark)

Alternatively, can evaluate ratio using at least 2 values near x = 6 and inferring that limit is approximately 0.8571. But lose one mark if result not stated as 6/7 as exact values asked for. Correct answer with no working, 1 mark only awarded. Working must be clear - if use L’Hopital’s rule – must clearly state it or lose 1 mark.

(b)

h→ 0

lim

a+h ? a h

where a is a positive constant.

(4)

By definition of the derivative, f ' (a ) = lim? So lim?

? f ( a + h) ? f ( a ) ? ? h →0 h ? ?
(2 marks)

? a+h ? a? ? = g ' (a) where g ( x) = x . ? h →0 ? h ? ?
1 1 ?1 / 2 x | x =a = . 2 2 a

Hence limit is

(2 marks)

Answer only, awarded 2 marks. 8. (6 marks) Determine the indefinite integrals: (a)



(ln x )2
x

dx

(3)

Putting u = ln x ?

du 1 = so dx x

(ln x) 2 1 3 1 2 3 ∫ x dx = ∫ u du = 3u + C = 3 (ln x) + C .

(3 marks)

(1 for identifying substitution, 1 for implementing substitution and 1 for the integration.) A correct answer with no working was awarded full marks in this case.

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

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Calculus TEE Marking Guidelines 2007 (b)

∫ cos

2

3x dx

(3) (1 marks) (2 marks)

∫ cos

2

3 xdx =

1 (1 + cos 6 x)dx 2∫

=

1? sin 6 x ? x sin 6 x ? x + 6 ? + C = 2 + 12 + C . 2? ?

Penalise 1 mark for omitting constant – but only once in entire paper.

9.

(10 marks) (a) Sketch the set of points in the complex plane that satisfy: (i)

| z + 3i | = | z ? 5 + 4i |
I ()

(3)

Re(z)

Straight line, which is the perpendicular bisector of the line joining the points ? 3i and 5-4i For identifying the points ? 3i and 5-4i. For recognising locus as a straight line For indicating the line is a perpendicular bisector (1 mark) (1 mark) (1 mark)

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

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Calculus TEE Marking Guidelines 2007

(ii)

0 ≤ arg( z + 1) ≤

3π 4

(3)

Re(z

Wedge-shaped region. Vertex located at ? 1 and edges through this point angled as shown (of course domain extends outwards in the obvious way) For identifying the importance of ? 1 For stating region of wedge shape For indicating orientation of edges (b)

(1 mark) (1 mark) (1 mark)

Write down inequalities on the modulus and argument of the complex number z that together describe the set of points shaded below. (4)

Im(z)

!2

!1

1

2
Re(z)

Half annulus described by the pair of inequalities

1 ≤| z |≤ 2

(2 marks) (2 marks)

and

(or anything equivalent: eg. limits ? π and 0.)
Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

π ≤ arg( z ) ≤ 2π

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Calculus TEE Marking Guidelines 2007 10. (12 marks) (a) The graph of a function y = g(x) is sketched below. On the second set of axes (4) sketch a graph of the inverse function y = g ?1 ( x) .

y . . . . . . . . . . . . . . . . . . . . . .

4

1 0
!2

1

x

. . . . . . . . . . . . . . . . . . . . . .

Inverse function y = g ?1 ( x)

y

0

x

Graph is reflection of the original in the line y = x . Main features: i) passes through (1,1); ii) vertical asymptote at x = ?2 and vertical asymptote at x = 4 ; iii) approximately correct shape to curve, and iv) correct orientation (1,1,1,1 mark)

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

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Calculus TEE Marking Guidelines 2007 (b) A function f(x) is defined by the formula f ( x) = (i) Find the domain and range of f(x).

1 . 3 ? e?x

(4)

If f ( x) =

1 then f(x) well-defined everywhere except at e ? x = 3 ? x = ? ln 3 . 3 ? e ?x Hence domain is x ∈ R \ {? ln 3} (or {x : x ≠ ? ln 3, x ∈ R}). ? ? 1? 3?

(1 mark) (1 mark)

As e ? x >0 for all x ∈ R and e ? x → ∞ as x → ?∞ , denominator of f(x) takes all values <3. (1 mark) Thus range of f(x) is { f ( x) < 0} ∪ ? f ( x) > ? . (A graphical solution perfectly acceptable)(1 mark)

(ii)

Determine the inverse function f

?1

( x) .

(3)

1 (1 mark) ? 3 y ? ye ? x = 1 ? ye ? x = 3 y ? 1 . 3 ? e ?x ? 1? 1? 1 ? Thus e ? x = 3 ? so x = ? ln? 3 ? ? . Hence inverse function is f ?1 ( x) = ? ln? 3 ? ? .(1+1 mark) ? ? y? y x? ? ?
If y =

(iii)
?1

What is the domain of f

?1

( x) ?

(1)

Domain of f

( x) is the range of f(x). ? ? 1? 3?

(1 mark)

Hence from (i) domain is {x < 0} ∪ ? x > ? . A correct follow through answer from part (i) was accepted.

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

18

Calculus TEE Marking Guidelines 2007 11. (8 marks) A function f(x) is defined by

?1 3 5 2 ? 4 x ? 2 x + ax + b ? ? c f (x) = ? ?1 ? dx ? ?7 ?

x>0 x<0 x=0

where a, b, c and d are constants. The value x1 is the least number such that f ( x1 ) = 0 . Given that f(x) is differentiable everywhere and that f ′(0) =

17 , find a, b, c, d and x1 . 4

If f ( x) =

3 1 3 5 2 x ? x + ax + b for x > 0 then f ' ( x) = x 2 ? 5 x + a . 4 4 2 17 17 it follows that b = 7 and a = . 4 4

(1 mark)

As f (0) = 7 and f ' (0) =

(1+1 mark)

Since f ( x) =

cd c for x < 0 then f ' ( x) = . 1 ? dx (1 ? dx) 2 17 17 ? c = 7 and cd = 4 4 ?d = 17 ≈ 0.6071 . 28

(1 mark)

Then f (0) = 7 and f ' (0) =

(1 mark)

(1 mark)

From its definition it is clear that f ( x) ≠ 0 when x < 0 so x1 > 0 . When x>0 then

(1 mark)

f ( x) =

1 3 5 2 17 1 1 x ? x + x + 7 = x 3 ? 10 x 2 + 17 x + 28 = ( x + 1)( x ? 4)( x ? 7) ? x1 = 4 . 4 2 4 4 4
(1 mark)

(

)

Correct values without working – penalty (-3)

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

19

Calculus TEE Marking Guidelines 2007 12. (9 marks) A cylinder sits inside a right cone of base radius 1 metre and height 2 metres. The axes of the cylinder and cone coincide. Use calculus techniques to find the greatest possible volume of the cylinder.

Let the inscribed cylinder have radius r and height h. By similar triangles it is clear that

(1 mark) (2 marks)

h = 2 ? h = 2(1 ? r ) . 1? r

Volume of cylinder V = πr 2 h = 2πr 2 (1 ? r ) . Then

(1 mark)

dV 2 = 2π (2r ? 3r 2 ) which is zero if r = 0 or r = . dr 3

(1+1 mark)

Can dismiss solution r = 0 as then V = 0 ; then

EITHER show that

2 d 2V = 2π (2 ? 6r ) < 0 for r = 2 3 dr

OR state that r =

2 must correspond to maximum volume as V = 0 when r = 0 or 1 3 dV . dr
?2? ? ?3? ?
2

OR use a sign test for

(2 marks)

Hence greatest possible volume is V = 2π ? ? ?1 ? ? =

1? 3?

8π cubic metres. 27

(1 mark)

(≈ 0.931 cubic metres)

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

20

Calculus TEE Marking Guidelines 2007 13. (10 marks) The volume V of blood flowing through an artery in unit time can be modelled by the formula V = kr 4 , where r is the radius of the artery and k is a constant. (a) What is the effect on the volume of blood flow if the radius of the artery is halved?
4

(2)

If V = kr 4 then when r → r / 2 we have V = k ? ? =

?r? ?2?

1 4 kr . 16 1 th of its original value. 16

(1 mark) (1 mark)

Hence halving r results in blood flow being decreased to

(b)

Use the incremental formula to estimate the percentage decrease in the radius of a partially clogged artery that will produce a 10% decrease in the flow of blood.

(5)

By the increments formula the change δV resulting from a change δr in r is δV ≈

dV δr . dr
(1 mark)

Now

dV = 4kr 3 dr 4V δV δr δr ? ≈4 . r V r 100δV 100δr = ?10 corresponds to = ?2.5 ; V r

(1 mark) (1 mark) (1 mark) (1 mark)

so δV ≈ 4kr 3δr =

Hence a 10% decrease in V, that is

that is a 2.5% reduction in the radius of the artery. Leaving out a negative sign to indicate decrease – penalty of 1.

(c)

Show that the incremental formula gives a physically absurd estimate for the change in V resulting from a halving of the radius of the artery. Explain why this estimate is so poor compared to the true answer found in (a). (3)

The increments formula suggests that if

100δV = ?4(50) = ?200 . A 200% decrease in blood flow is impossible!! V

100δr = ?50 , (ie. a halving of the radius) then r
(2 marks)

From part a) we actually know that a halving of the radius reduces the flow volume by a factor of 16. The increments formula is so poor because it is only valid for small increments δr and δV . If these increments are not small then the increments formula is not applicable and can give unreliable predictions. (1 mark)

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

21

Calculus TEE Marking Guidelines 2007 14. (10 marks) A particle moves in simple harmonic motion in a straight line with a period of 5 seconds and amplitude of 4 metres. Initially the particle is 1 metre from its equilibrium point and is moving towards it. Find: (a) its distance from its equilibrium point after 3 seconds. (6)

If particle moves in SHM with period 5 seconds and amplitude 4 metres then displacement from equilibrium x(t ) = 4 sin(ωt + α ) where ω = 2π / 5 and α is a phase angle. (2 marks) Now dx / dt = 4ω cos(ωt + α ) so if initially x = ?1 and dx / dt > 0 (could choose other signs) we have sin α = ?1 / 4 and cos α > 0 . (2 marks) Hence α lies in the 4th quadrant and α = ? sin ?1 (1 / 4) = ?0.2527 . Then when t = 3 have x = 4 sin(1.2π + α ) = ?1.4674 . Distance is 1.4674 m. (1 mark) (1 mark)

If

x = 4 cos( 0.4πt + α ) , and initially x = 1 and dx / dt < 0

have cos α = 1 / 4 and α = 1.318 If
x = 4 cos( 0.4πt + α ) , and initially x = ?1 and dx / dt > 0

have cos α = ?1/ 4 , sin α < 0 and α = 4.4596 If
x = 4 sin( 0.4πt + α ) , and initially x = ?1 and dx / dt > 0

have sin α = ?1/ 4 , cos α > 0 and α = ?0.2527 If
x = 4 sin( 0.4πt + α ) , and initially x = 1 and dx / dt < 0

have sin α = 1/ 4 , cos α < 0 and α = 2.889 (b) the total distance traveled in the first 4 seconds. (4) (1 mark)

When t = 4 have x = 4 sin(1.6π + α ) = ?3.9924 m.

At this time dx / dt = 4ω cos(1.6π + α ) > 0 . Hence the particle is moving back towards the equilibrium point having nearly completed one whole period. (1 mark) Distance to be covered to return to starting point is 3.9924-1 m ≈ 2.992 m so distance travelled in first 4 seconds is 4(4)-2.992 = 13.008 m. (2 marks) Correct answer only without working – 2 marks Alternative method:
4 0

∫ 1.6π cos(0.4πt ? 0.2527 ) dt = 13.008 m

(by calculator)

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

22

Calculus TEE Marking Guidelines 2007

15.

(9 marks) Two functions are defined by f ( x) = sin x ? x cos x and g ( x) = x ? sin x . (a) Show that f ′(x) > 0 and g' (x) > 0, for 0 < x <

π
2

.

(4)

If f ( x) = sin x ? x cos x then f ' ( x) = cos x ? cos x + x sin x

(1 marks) (1 mark) (1 +1 mark)

= x sin x > 0 for 0 < x < π / 2 .
If g ( x) = x ? sin x then g ' ( x) = 1 ? cos x > 0 as cos x < 1 for 0 < x < π / 2

1 mark for each derivative and 1 mark each for showing the derivative is positive over the given domain. A graphical illustration of this was accepted in this part but not in subsequent parts such as (c).

(b)

Use the results from (a) to show that f ( x) > 0 and g ( x) > 0 when 0 < x <

π
2

.
(1)

As f (0) = g (0) = 0 and f ' ( x), g ' ( x) > 0 for 0 < x < π / 2 then it follows that f ( x), g ( x) > 0 . (1 mark)

(c)

Deduce that

sin x π sin x > cos x and < 1 for 0 < x < . x x 2 sin x > cos x x

(2)

Since f ( x) > 0 ? sin x > x cos x ? and g ( x) > 0 ? sin x < x ?

(1 mark) (1 mark)

sin x <1. x

(d)

Hence show that lim

x→ 0

sin x = 1. x

(2)

Results of c) imply that cos x <

sin x < 1 for 0 < x < π / 2 . x

(1 mark)

As x → 0 so cos x → 1 and by the sandwich/squeeze principle

sin x → 1 in this limit. (1 mark) x

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

23

Calculus TEE Marking Guidelines 2007

16.

(8 marks) (a) Solve the equation z 4 + 1 = 0, where z is a complex number, giving answers as exact values. (4) (1 mark)

If z 4 + 1 = 0 then z 4 = ?1 = cis (π + 2kπ ) where k is an integer. Then, by de Moivre’s theorem, z = cis ? choices k = 0,1,2 and 3.

? (2k + 1)π ? ? where distinct values are given by the 4 ? ?
(1 mark)

Hence the four roots are z= cis( π / 4 ), cis( 3π / 4 ), cis( 5π / 4 ) and cis( 7π / 4 ) and so

z=±

1+ i 2

or z = ±

1? i 2

.

(2 marks)

(b)

Hence, or otherwise, solve ( z ? 2) 4 + 1 = 0 , and indicate the approximate locations of these solutions in the complex plane.

(4)

Putting Z = z ? 2 shows that Z 4 + 1 = 0 which is the equation solved in part a). Thus the roots Z are as stated in a) so that z = 2 ±

(1 mark) (1 mark)

1+ i 2

or z = 2 ±

1? i 2

.

(Important features: the roots are “centres” at 2, the roots are roughly the correct distance from this point and the orientation of the roots are correct.)

Re(

(2 marks)

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

24

Calculus TEE Marking Guidelines 2007

17.

(8 marks) After t hours the concentration C(t) of a certain drug in a bacteria cell satisfies

dC = a(1 ? C ) dt
for some constant a. (a) Find the general solution of the differential equation for C(t). (3)

If

dC = a(1 ? C ) then dt

∫ 1 ? C = a ∫ dt
? ? ln | 1 ? C |= at + D .

dC

(1 mark) (1 mark)

Then 1 ? C = exp(?at ? D) = B exp(?at ) so that C = 1 ? B exp(? at ) for an arbitrary constant B (1 mark)

(b)

Initially C = 0. After 2 hours, C = 0.3. Find the value of a.

(3) (1 mark) (1 mark) (1 mark)

As C = 0 when t = 0 so B = 1 and C = 1 ? exp(? at ) . If C = 0.3 when t = 2 then exp(?2a) = 0.7

1 ? a = ? ln 0.7 = 0.1783 . 2

(c)

After how long will C = 0.6?

(2) (1 mark) (1 mark)

When C = 0.6 we have exp(?at ) = 0.4

1 ? t = ? ln 0.4 = 5.139 hours. a

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

25

Calculus TEE Marking Guidelines 2007

18.

(14 marks) (a) Sketch the graphs of f n ( x) = tan n x for 0 ≤ x ≤

π
4

and n = 0, 1, 2 and 3.

(3)

y 1

0

π 4

x

The graph of tan n x for n ≥ 1 joins (0,0) to (π / 4,1) ; for n=0 the function is constant. The graphs correspond to increasing n down the figure. (3 marks) -1 mark per omission
π 4 0

(b)

Suppose that I n = tan n x dx for integer n. Explain, with reference to the sketch in (a), why I n > I n +1 for all n. (3)



For all x ∈ (0, π / 4) we have tan x < 1 . Thus for all such values

tan n +1 x = tan x(tan n x) < tan n x .
π /4

(1 mark)

Since tan n +1 x = tan n x at the ends of the integration range and tan n +1 x < tan n x across the interior it follows that I n +1 ≡

∫ tan
0

n +1

xdx < I n .

(2 marks)

Sentence explanation with reference to areas was accepted. Show, by direct evaluation, that I 0 =

(c)

π . 4

(1)

π /4

I0 =

0 ∫ tan xdx = 0

π /4

∫ 1.dx = 4 .
0

π

(1 mark)

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

26

Calculus TEE Marking Guidelines 2007
π

(d)

Show that I n + I n + 2 that I n + 2 =

tan n x =∫ dx and use the substitution u = tan x to deduce 2 0 cos x
4

1 ? In . n +1

(4)

π /4

I n + I n+2 =

∫ (tan
0

n

x + tan

n+2

π /4

x)dx =

∫ tan
0

n

x(1 + tan 2 x)dx

(1 mark)

π /4

=


0

tan n x dx . cos 2 x

(1mark)

If u = tan x then

du 1 = and when x=0, u =0; x = π / 4 , u = 1 . dx cos 2 x
1 1 n

(1 mark)

Then I n + I n + 2

? u n +1 ? 1 1 = ∫ u du = ? ? I n+2 = ? I n as required. ? = n +1 ? n + 1? 0 n + 1 0

(1 mark)

(e)

Use the results in (c) and (d) to calculate I 6 .

(3)

Using result of d) we have I 2 = 1 ? I 0 = 1 ? Repeating gives I 4 =

π
4

(putting n=0).

(1 mark) (2 marks)

1 1 1 2 π 13 π π 2 ? I 2 = ? and I 6 = ? I 4 = + ? = ? . 3 4 3 5 5 3 4 15 4

Must show use of iterations as asked to get full marks. Calculator answer of 0.08127 gained ‘0’ marks and an exact answer without working was awarded 1 mark. 19. (16 marks) A ball is thrown from an origin O, on level ground, with an initial velocity v(0) = V cosΠ i + V sin Π j where the unit vectors i and j are directed horizontally and vertically upwards respectively. At a later time t the ball is at position r(t) = x(t) i + y(t) j. (a)

d2y d 2x = 0 and 2 = ? g, where g is the dt dt 2 (6) acceleration due to gravity, find x(t ) and y (t ) .
Given that the equations of motion are

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

27

Calculus TEE Marking Guidelines 2007

If

dx d 2x = 0 then = cons. = V cos θ by initial conditions. 2 dt dt
(since x = 0 when t = 0 ).

(1 +1 mark) (1 mark) (1+1 mark) (1 mark)

Integrating again gives x = tV cos θ Similarly

dy d2y = ?g ? = ? gt + cons. = ? gt + V sin θ by initial conditions. 2 dt dt

Then y = Vt sin θ ?

1 2 gt . 2

Answer only – 2 marks. (b) Eliminate t from the expressions for x(t) and y(t) to show that the equation for the path of the ball is

y = x tan θ ?

gx 2 2V 2 cos 2 θ .

(2) (1 mark)

From the x-equation we see that t = x / V cos θ . Substituting in the y-equation we have

gx 2 ? x ? 1 ? x ? y = V sin θ ? . ? ? g? ? = x tan θ ? 2V 2 cos 2 θ ? V cos θ ? 2 ? V cos θ ?
(c)

2

(1 mark)

A wall of height h is built at distance a from O. The ball is projected at an angle Ι such that it just clears the wall. Use the result of (b), together with the identity

to derive a quadratic equation for tan Ι.

1 = 1 + tan 2 θ , cos 2 θ
(3) (1 mark) (1 mark) (1 mark)

If ball just clears the wall then y = h when x = a if θ = α .

ga 2 ga 2 Hence we have h = a tan α ? = a tan α ? (1 + tan 2 α ). 2 2 2 2V cos α 2V
Thus our quadratic is

? ga 2 ga 2 ? ? = 0. tan 2 α ? a tan α + ? h + ? 2V 2 2V 2 ? ? ?

Penalty of 1 mark if answer left in original variables of y, x and ?. (d) Deduce that it is impossible to throw the ball over the wall, whatever the angle of

V 2 ga 2 projection, if h > . ? 2 g 2V 2
To throw the ball over the wall - the above quadratic must admit real roots for tan α .
2

(5) (2 marks) (1

Thus it is impossible to launch the ball if there are only complex roots s.t. " B < 4 AC" . mark)
?

a2 <

2 ga 2 V2

? V 2 ga 2 ga 2 V 2 ga 2 ? ?h + ?? h+ > ?h> ? ? 2 g 2V 2 2V 2 ? 2V 2 2 g ? ?

as required. (2 marks)

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

28

Calculus TEE Marking Guidelines 2007

20.

(11 marks) Suppose that z and w are complex numbers such that z + w = 1 and | z | = | w | . (a) Show that Re( z ) = Re( w) =

1 and Im( z ) = ? Im(w) . 2

(6)

Suppose that the complex numbers z and w have real and imaginary parts z = α + iβ , w = γ + i δ . If z + w = 1 then (α + γ ) + i ( β + δ ) = 1 ? α + γ = 1 and δ = ? β .

(1 mark) (1 mark)

If | z |=| w |? α 2 + β 2 = γ 2 + δ 2 ? α 2 + β 2 = γ 2 + δ 2 . As δ = ? β so β 2 = δ 2 and hence

α 2 = γ 2 . We conclude that α = ±γ .
Since we already have the result α + γ = 1 it is not possible for α = ?γ and hence

(2 marks) (1 mark) (1 mark)

α = γ ? 2α = 1 ? α =
Thus α = γ =

1 . 2

1 1 and β = ?δ so Re(z)=Re(w)= and Im(z) = ? Im(w) as required. 2 2

Logic was looked at carefully and penalties applied for making assumptions that needed to be proved.

(b)

Suppose that e iθ + e iφ = 1 for ? π < φ ≤ θ ≤ π . Find θ and φ .

(5)

If z = e iθ and w = e iφ with z + w = 1 and | z |=| w | (=1) then these satisfy the requirements of a). (1 mark) Then Re(z)=Re(w)= and If cos ξ =

1 1 implies that cos θ = cos φ = 2 2

(1 mark) (1 mark)

Im(z) = ? Im(w) ? sin θ = ? sin φ .

1 π π π then ξ = ± ; as ? π < φ ≤ θ ≤ π we conclude that θ = and φ = ? . 2 3 3 3
(2 marks)

Correct answer without supporting argument awarded 2 marks.

Examiners’ Report on 2007 Tertiary Entrance Examination – Calculus

29


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