MATHEMATICS
GCE Advanced Subsidiary Level Paper 8709/01Paper 1
General comments
The respon to this paper was pleasing. There were many excellent scripts and the prentation was generally good. Candidates emed to have sufficient time to answer all the questions and there was little evidence of later questions being rushed. Much time was lost, however, on Question 3, by candidates who interpreted the word ‘sketch’ as ‘accurate graph’. Many candidates also need to read the rubric which requests that answers to angles should be given correct to 1 decimal place, unless otherwi requested.Comments on specific questions
Question 1
The majority of candidates elected to form a quadratic equation in x by eliminating y from the given equations. Only about half of the recognid the need to t b ² − 4ac to 0, and many others failed to t the quadratic to 0 before applying b ² − 4ac = 0. Many candidates preferred to equate the gradient of the line with the differential of x ² − 6x + 14 and then to obtain x , y and finally k . This method was usually successful,but tting the gradient to either 0 or to 2 were common errors.
Answer : k = 10.
Question 2
(i)
Most candidates realid the need to take 2 out of the expression, but errors such as ‘2x 2 − 12x + 11 = 2(x 2 − 12x + 5.5) leading to 2(x − 6)2 + …’ were common.(ii)This was badly answered with many candidates either substituting x = 0 or obtaining a table of
values. The answer −7 ≤ f (x ) ≤ 11 was common, as was f (x ) ≥ 11. Many candidates failed to reali that there was a link with part (i), (i.e. f(x ) ≥ c ) and preferred to u calculus to find the minimum point (3, −7). Even then it was not automatic to state f (x ) ≥ −7.
Answers : (i) 2(x − 3)2 − 7; (ii) f(x ) ≥ −7.
Question 3
(i) Most candidates correctly drew the graph of y = cos x , although ‘V’ shapes were common. Very few
drew the graph of y = cos3x correctly, most thinking either that the graph lay between −3 and +3 or that the graph of y = cos x had the same shape as y = cos3x between 0 and 2p . Many candidates wasted considerable time by ignoring the instruction ‘sketch’ and instead drawing accurate graphs.(ii)
Only a handful of candidates realid that for f to have an inver it needed to be 1 : 1 and that this only occurred for 0 ≤ x ≤ p , leading to k = p .Answers : (i) Sketch; (ii) k = p .
Question 4
Candidates should reali that a request for an answer in terms of p and 3means that u of a calculator will lead to loss of marks. Many candidates failed to reali that angle POQ = 31p or that cos30º = 213.
U of s = r θ with θ in degrees occurred occasionally and a common error was to assume that the radius of the arc PXQ was PS rather than PO . The most common error however was to express cos30° = OS
6 as OS = 6cos30°.
Answer : 4p + 83.
Question 5
This was well answered and a half of all attempts were completely correct. Occasionally the surface area was given as 3x 3 and there were solutions in which areas of only 3 or 5 faces were considered.Differentiation was usually correct. In part (ii), although the chain rule was usually correctly quoted for the variables concerned, there were misunderstandings over t
A d d = 0.14 and often the rate of decrea of x was given as x t d d rather than t x d d .Answers : (i) A = 14x ² and
x A d d = 28x ; (ii) 0.0025.
Question 6At least a half of all attempts were completely correct. There were occasional errors in the calculation of the gradient of AB , but most candidates realid that the gradient of the line L 2 was −12
1¸. There were a few misunderstandings of the relationship between L 1 and L 2, particularly from weaker candidates who automatically involved the mid-point of the line AB . The solution of the simu
ltaneous equations was very well done.
Answer : (4, 6).
Question 7
(i)Although a few candidates took ‘(2s + c )² = 4s ² + c ² ’ and ‘(2c − s )² = 4c ² ± s ² ’, most correctly
cancelled the 4cos θsin θ and reduced the expression to 5s ² + 5c ² and from there to 5. Others however divided throughout by 5 and stated the answer as 1.
style是什么意思中文(ii)
Most realid the need to collect terms and to express 7sin θ = 4cos θ as tan θ =74. Despite having the formula on the formula sheet, many expresd tan θ as cos θ ÷ sin θ, or were unable to manipulate the expressions correctly. Candidates need to read questions cloly, for many omitted to find the corresponding values of θ. It was very rare for candidates to go wrong over which quadrants to u.
Answers : (i) a ² + b ² = 5; (ii) tan θ =
7
4, θ = 29.7º or 209.7º. Question 8At least a quarter of all candidates failed to read the question carefully and took the progression as being arithmetic. Of the other attempts, many were completely correct and it was surprising to e some weaker candidates scoring highly on this question. Some wrote down all the amounts for the first 10 years of operation, and others incorrectly took the n th term as ar n . In part (ii) common errors were to take the 20th term rather than the sum of 20 terms and the formula for sum of 20 terms was often taken as S n = 21
n (1 - r n - 1)/(1 - r ). It was pleasing that in part (iii) the majority of candidates realid the need to find
the sum to infinity.
Answers : (i) 775kg; (ii) 17 600kg; (iii) 20 000kg.
Question 9
(i)About a half of all candidates took the equation of the curve to be the equation of the tangent and a
straight line with gradient 21 was depressingly common. Surprisingly enough, many of the candidates then realid the need to integrate the given expression for x
y d d and produced this in part (ii). The integration was generally good, though common errors were to fail to cope with the negative power, to ignore the integral of −3 or to fail to reali the need to include and evaluate the constant of integration.
(ii)Most realid the need to t
x y d d to 0 to find the stationary point and despite a few algebraic slips,most obtained x = 2. Answers : (i) y = 212x -
− 3x + 31; (ii) (2, 22).
Question 10There were many excellent respons to this question and the candidates’ ability to calculate ‘scalar product’was impressive. Unfortunately, too many marks were lost in the first part through failure to obtain correct expressions for the vectors MN and MD . A surprising number of candidates totally ignored the dimensions of the problem (i.e. OA = 6cm, OC = 8cm and OB = 16cm) and gave their answers with coefficients of i , j and k as ±1 or ±2
racist
1. Others were able to cope with but struggled with the fact that N was the mid-point of AC . Even then, all the method marks available for part (ii) were usually obtained.
Answers : (i) MN = −3i − 8j + 4k , MD = −6i + 8j + 8k ; (ii) −14, 97º.
Question 11
Many candidates scored highly on the question, but for weaker candidates, failure to recogni the need for ‘function of a function’ led to loss of marks. Surprisingly, veral candidates took the gradient of the tangent
to be the gradient of the normal and calculated −1 ÷ x
y d d , but otherwi part (i) was usually correct. The follow through mark for part (ii) was nearly always obtained, though a few candidates put y = 0 rather than x = 0.
In part (iii), the integration was surprisingly better than the differentiation in part (i) and more candidates included the ‘8
1’ than the ‘8’ in part (i). Most candidates worked about the x -axis and realid the need to subtract
areas. U of limits was generally correct, though 0 to 5 was often ud instead of 0 to 3. The most common error was to automatically assume that the value of any expression at 0 is 0. Candidates choosing to work about the y -axis fared badly becau it was very rarely realid that the limits for the line were different from the limits for the curve.
Answers : (i) 5y = 4x + 13; (ii) (0, 2.6); (iii) 15
16 or 1.07 unit 2 .
Paper 8709/02
Paper 2
General comments
Few high marks were scored by candidates, largely due to candidates’ weakness in certain areas of the syllabus, such as integration and iteration techniques. There was a general tendency to u degrees, rather than radians, for angles.
Converly, there were certain key techniques, such as differentiation of functions, that were commonly the source of many marks for candidates.
Work was clear and neat, with no evidence of candidates running out of time at the end of the paper. Comments on specific questions
Question 1
Many candidates scored full marks and recognid that each value of tan x gave ri to two values of x. Weaker solutions, that scored no marks, showed no u of the condition c2x = 1 + tan2x and intractable equations involving both cos x and sin x, with neither of the having been eliminated, were common.
Answer: x = 135º or 315º, 56.3º or 236.3º.
Question 2
(i)As the answer was given, weaker candidates struggled to produce it via errors such as u2 = 2x+1
and 4x = 2.2x = 2u. However, most candidates obtained the given result correctly.
(ii)Many candidates obtained the correct result u = 1 + 13» 4.6055, but then failed to solve for x = ln u¸ ln2.
Answer: (ii)x = 2.20.
Question 3
(i)Most candidates correctly sketched the line 2y = x + 1, but very few obtained a plot of 2y = ½x− 4½,
and most graphs showed portions of that line below the x-axis.
(ii)There were a pleasing number of fully correct solutions, sometimes bad on ‘trial and error’, or simply by quoting the answer. Weaker solutions involved squaring one or both of the equations of the two lines, but with only one side being squared, e.g. 2y = (x− 4)2.
Answer: (ii)x = 1.5, y = 1.25.
Question 4
Although many candidates noted that ln y = lnA + n ln x, wrong answers were often bad on fal variants, such as ln y = (A n)ln x. There was a marked tendency to substitute the values (1, 2.4) and (4, 0.6) into the original equation y = A x n rather than into the ln y versus ln x relation.
Answer: n = −0.6, A = e3 = 20.1
Question 5
(a)This part was very well done, with a sound grasp of all the esntial ideas.
(b)There were many excellent solutions, and this question was admirably handled. Only occasionally
was the derivative of xy given as x
x y d d only.Answers : (a)英语副词
honeycomb
94, (b) y + 4x = 14.Question 6
(i)
Although this part prented little or no problems, candidates struggled thereafter.(ii)
Almost all candidates expresd u 2 = tan −1
1speaking2 in degrees , without noting the vast disparity between this value and that of u 1 = 1. (iii)Almost no one could cope with this part, failing to spot that, as n ® ¥, u n and u n + 1 tend to the same
limiting value, this latter being the required value x m .Answer : (ii) 1.08.
Question 7
点钟(i)There were many sign errors, for example, cos 2x ≡
2
1(1 − cos2x ), and often a factor ‘2’ was omitted. (ii)Candidates failed to remove brackets and u the twin results of (i). Many candidates believed
that ()()[]f f f f ¢
==+òò3,3d d cos 3sin 23李阳疯狂英语口语速成
322x x x x x or f ¢2.Answer : (ii) ()8
3413+p » 9.36.Paper 8709/04Paper 4
General comments
都柏林大学The paper was generally well attempted. However candidates often failed to obtain answers correct to three significant figures, even when correct methods were ud. This is a problem which Centres are urged to address. Most inaccuracy aris from premature approximation.
There is clearly a problem too, with terminology. Familiarity with and understanding of terms germane to the syllabus are expected. In very many cas candidates attempted to find forces when work done by forces was required, and coefficient of friction when frictional force was required.
Inappropriate u was made of µR F = in both Question 5 and Question 7.
Comments on specific questions
Question 1
crossingMost candidates answered this question correctly.
A few candidates ud sine instead of cosine; some did not u the given angle, obtaining the incorrect answer of 2400 J.
Mis-reading 30 N and 10o as 10 N and 30o
to be or not to bewas fairly common.Answer : 2360 J.
