Hi,
Here is a quick analysis of paper 1:
Q1 Inequalities.
Q2 System of linear equations.
Q3 Tangent to a curve defined parametrically.
Q4 Maclaurin's expansion and approximations.
Q5 Graphs of y = f(|x|) and y = |f(x)| and definite integrals involving them.
Q6 Method of differences & MI involving trigonometric functions.
Q7 Vector product related to area, unit vector, length of projection.
Q8 Solve given DE involving a real-life problem.
Q9 AP/GP applied to real-life.
Q10 Solve given equation involving complex roots, then deduce roots to another. Sketch perpendicular bisectors which do not intersect.
Q11 Planes and lines.
Possibilities for paper 2 are:
1. Graph of a rational function.
2. Area and volume.
3. Connected rates of change.
4. Modulus and argument of complex number.
5. Solve DE by substitution.
6. Inverse and composite functions.
Jiayou to all!
Cheers,
Wen Shih
the chem paper and math paper are so depressing.....
Hi qdtimes,
Stay hopeful and focus on doing well for the other papers!
Jiayou!
Cheers,
Wen Shih
I find Paper 1 relatively standard.
Remember you don't need 100 marks to score an A. A number of my students claimed they lost between 20 to 40 marks for Paper 1. That is between A and B, borderlining on a C I think.
Good luck to those taking paper 2 right now!
thanks mr.ws...
wah really very sian...did so many stupid mistakes in chem and the 2 math papers...
most of my peers got about 70+ for math paper 1, easily 80 and above for paper 2...
doesn't this makes the requirement for A to be easily above 70-80..? i will probably get 72.5% overall which is so unsafe.
worst thing for math paper 2 is that my compass spoilt halfway when i was drawing the locus -_- and many mistakes followed after. >=( !!
Originally posted by eagle:I find Paper 1 relatively standard.
I disagree. The papers are far from "standard". They are too deviant from any past year papers (clearly) and any school prelim paper archetypes.
People I know from many JCs lost marks ranging from only 5 to 40. I wonder what the bell curve's going to be like this year, and how much will be required for a distinction.
some ppl thinks that the cut-off point would be 80+ this year due to math paper 2 -_-. not surprising to see many people getting above 80 and 90 for paper 2, though i screwed up so badly, losing around 25 marks....
Nothing left to do but wait and see.
And hope. A lot.
The ever-so-brilliant education system in Singapore means anything more than 2 "B"s in your A' Levels kills off eligibility for 90% of the NUS courses immediately.
Sh!t...I counted about 12 marks gone due to careless mistakes. FML.
Most people I know lost at least 25 marks for paper 1, maybe 15 for paper 2? Praying damn hard..
ok la personally i think cutoff point is around 75-79, considering there will be also some people who screwed up as badly as me. it's oxymoronic that i shamelessly hope the bell-curve do wonders for me while i dislike competing with each other in things like that.
Hi Dejomel,
The examiners are mathematicians and researchers themselves (I believe), so their questions are likely to be refreshing and yet made accessible to students (who has a good conceptual understanding and who inquire further in a mathematical sense) within the boundaries of the syllabus.
Our schools tend to prepare students for the 'standard' fare :P
Thanks.
Cheers,
Wen Shih
Hi,
In paper 1 Q4(ii)(b), we are asked for the reason why the approximation is not very good. I believe the 'standard' response by some (or most) students would be that the value of the substitution (for x) is not sufficiently small (a well-learnt answer) or that the percentage difference (i.e., 13.8%) is large (a desperate attempt to give a vague reply but not incorrect in a broad sense).
If one inquires further, one notes that x can take any value, since the power series for cos x is valid for all real values of x. One should then look at another line of reasoning - whether there is a sufficient number of terms in the original expansion to allow the value of x to converge (i.e., pi/4 raised to some odd power greater than 5) and approach the exact value of the integral.
Out of curiosity, I went to investigate the value of n that would give a very small value of (pi/4)^n and noted that when n = 19, we obtain a value of 0.01016. Indeed, through this crude (but quick) check, we will need more terms for the approximation to be improved within reasonable accuracy.
Thanks.
Cheers,
Wen Shih
Originally posted by wee_ws:Hi,
In paper 1 Q4(ii)(b), we are asked for the reason why the approximation is not very good. I believe the 'standard' response by some (or most) students would be that the value of the substitution (for x) is not sufficiently small (a well-learnt answer) or that the percentage difference (i.e., 13.8%) is large (a desperate attempt to give a vague reply but not incorrect in a broad sense).
If one inquires further, one notes that x can take any value, since the power series for cos x is valid for all real values of x. One should then look at another line of reasoning - whether there is a sufficient number of terms in the original expansion to allow the value of x to converge (i.e., pi/4 raised to some odd power greater than 5) and approach the exact value of the integral.
Out of curiosity, I went to investigate the value of n that would give a very small value of (pi/4)^n and noted that when n = 19, we obtain a value of 0.01016. Indeed, through this crude (but quick) check, we will need more terms for the approximation to be improved within reasonable accuracy.
Thanks.
Cheers,
Wen Shih
Hi Wen Shih,
The larger the value of x (while being less than one), the more the number of terms required for a more accurate estimation. These two go hand in hand. I would think it is the same line of reasoning, but it requires good conceptual understanding by students to link them together.
Hence, I believe that the answer that x is not sufficiently small should suffice as a correct answer.
Hi,
For a power series to be used to achieve a good approximation, we can now fully characterise two situations:
1. If we have a few terms in the series, we will need |x| to be as small as possible (i.e., close to zero).
2. If |x| is rather close to 1 (as in the question we discussed), we will need to have more terms in the series.
P.S. Eagle: I believe that Cambridge markers are compassionate and will award marks for any decent attempt to answer the questions.
Thanks.
Cheers,
Wen Shih
hello, i realised that for the complex number question, if we use the quadratic formula for part i and ii), we will still get the same answer. Will it still be considered correct? i always thought that the coefficients need to be real.. :/
Originally posted by qdtimes2:ok la personally i think cutoff point is around 75-79, considering there will be also some people who screwed up as badly as me. it's oxymoronic that i shamelessly hope the bell-curve do wonders for me while i dislike competing with each other in things like that.
I think that will be the range for getting a distinction this year, too.
Which means that, sadly, I'll probably be getting a "B".
If only there was an "A2" safety net to catch us just like in the O' Levels :(
Hi qdtimes2,
The quadratic formula can be applied, the coefficients can be imaginary.
You probably got mixed up with the fact that a polynomial with real coefficients has pairs of complex roots.
Thanks.
Cheers,
Wen Shih
Originally posted by Plaxo:Thanks wee_ws.
Is it possible for u to post the answers for the paper 1 /
Hi,
Q1 -2 < x < 1
Q2(i) a = 0.215, b = -0.490, c = 3.281
(ii) {x is real : x > 1.14}
Q3(i) y = -(1/p^3)x + (3/p)
(ii) Q (3p^2, 0), R (0, 3/p)
(iii) 8xy^2 = 27
Q4(i) 1 - 3x^2 + 4x^4
(ii)(a) 0.540
(b) 0.475, Series does not have sufficient number of terms.
Q5(ii) {x is real : 0 <= x <= 2}
(iii) 2 + sqrt(5)
Q6(ii) {sin (n + 0.5)theta} / {2 sin 0.5theta} - (1/2)
Q7(i) (1/6)a + (3/10)b, k = 1/20
(ii)(a) 1/7
(b) Length of projection of b onto a.
(c) (1/7)(9i + 7j + 8k)
Q8(i) (1/20) ln |(10 + v)/(10 - v)| + c
(ii)(a) t = (1/2) ln |(10 + v)/(10 - v)|, (1/2)ln3 s
(b) 7.62 m/s
(c) Approaches/increases to a limit of 10 m/s.
Q9(i) 193m, 4810m
(ii) 40 days
Q10(i) -2 + 2i, 2 - 2i
(ii) -3 + i, -1 - i
(iii)(a) Perpendicular bisector passing through the midpoint of (-2, 2) & (2, -2).
(b) Perpendicular bisector passing through the midpoint of (-3, 1) & (-1, -1).
(iv) Both loci are parallel with gradient 1.
Q11(i) x + y + 2z = -3
(ii) k = -7
(iii) (-1, 6, -4)
(iv) 22.2 degrees
Cheers,
Wen Shih
Hi,
It is an important skill to look back at one's solution after solving a problem.
Two methods for checking are:
1. Backward substitution;
2. Solving the problem in another way.
I will illustrate their application with selected questions from paper 1.
For Q1, we could sketch the graph of y = 1 / (x^2 + x - 2) and observe that it is indeed negative for the range -2 < x < 1.
For Q2(i), we could use the values found for a, b and c to verify that the three given points can be obtained.
For Q3(i), the graphs can be checked by means of GC. In (iii), the value of a found may be used to check whether the given equation is satisfied.
For Q10(i) & (ii), the values found for z and w can be used to verify that the equations involving z and w are satisfied.
For Q11(i), we could subtitute the three points into the equation of p to see if it is satisfied. We could use (iii) to ensure that (ii) is correctly obtained.
Thanks.
Cheers,
Wen Shih
All answers are fine except 8(ii)(c)
The concept of asymptote/tendency is tested, so do use the word 'approaches' or 'tends to' instead of 'reaches'
Originally posted by wee_ws:Hi,
Q1 -2 < x < 1
Q2(i) a = 0.215, b = -0.490, c = 3.281
(ii) {x is real : x > 1.14}
Q3(i) y = -(1/p^3)x + (3/p)
(ii) Q (3p^2, 0), R (0, 3/p)
(iii) 8xy^2 = 27
Q4(i) 1 - 3x^2 + 4x^4
(ii)(a) 0.540
(b) 0.475, Series does not have sufficient number of terms.
Q5(ii) {x is real : 0 <= x <= 2}
(iii) 2 + sqrt(5)
Q6(ii) {sin (n + 0.5)theta} / {2 sin 0.5theta} - (1/2)
Q7(i) (1/6)a + (3/10)b, k = 1/20
(ii)(a) 1/7
(b) Length of projection of b onto a.
(c) (1/7)(9i + 7j + 8k)
Q8(i) (1/20) ln |(10 + v)/(10 - v)| + c
(ii)(a) t = (1/2) ln |(10 + v)/(10 - v)|, (1/2)ln3 s
(b) 7.62 m/s
(c) Reaches a limit of 10 m/s
Q9(i) 193m, 4810m
(ii) 40 days
Q10(i) -2 + 2i, 2 - 2i
(ii) -3 + i, -1 - i
(iii)(a) Perpendicular bisector passing through the midpoint of (-2, 2) & (2, -2).
(b) Perpendicular bisector passing through the midpoint of (-3, 1) & (-1, -1).
(iv) Both loci are parallel with gradient 1.
Q11(i) x + y + 2z = -3
(ii) k = -7
(iii) (-1, 6, -4)
(iv) 22.2 degrees
Cheers,
Wen Shih
Dear frekiwang,
Thanks :)
Cheers,
Wen Shih
Hi,
Paper 2 answers:
Q1(i) z lies inside or on the circumference of the circle with centre (2, 5) and radius 3 units.
(ii) Maximum |z| = sqrt(29) - 3 units, minimum |z| = sqrt(29) + 3 units
(iii) sqrt(17) units, P refers to the points (2, 2) and (5, 5).
Q2(ii) Because x < n/2, x = { (1/2) - sqrt(3)/6 }n
Q3(i) (Inverse of f)(x) = (1/2){e^(x - 3) - 1}, domain = real set, range = (-1/2, inf)
(ii) Asymptotes: x = -1/2, y = -1/2, intercepts on y = f(x): (0, 3), ((e^(-3) - 1)/2, 0), intercepts on y = (inverse of f)(x): (3, 0), (0, (e^(-3) - 1)/2)
(iii) Curves intersect along y = x. x = -0.4847 or x = 5.482
Q4(a)(i) -(1/2)(n^2)e^(-2n) - (1/2)(n)e^(-2n) - (1/4)e^(-2n) + (1/4)
(ii) 1/4
(b) 2pi(pi - 2) cubic units
Q5 mu = 53.7, sigma = 8.32
Q6(i) Choose 5 males and 5 females for the age categories <20, 20-29, 30-39, 40-49, 50-59, >=60.
(ii) Sample does not reflect the actual composition of residents.
(iii) Stratefied sampling. Unrealistic as it is time-consuming getting a representative sample if the suburb covers a large area.
Q7(i) The attempts to contact friends are independent. The attempts to contact friends are identical.
(ii) If a particular friend is uncontactable on an attempt, a different friend is called the next attempt, so the attempts are not identical at all.
(iii) 0.552
(iv) 0.114
Q8(ii) r = -0.992. The value indicates that as x increases y decreases and other models could also give this behaviour.
(iii) For the quadratic model, r = -1.00. Since its |r| value is closer to 1 than the |r| value of the linear model, we decide to take the quadratic model as the better one.
(iv) y = -0.856x^2 + 22.2, y = 13.5
Q9(i)(a) 0.058 (exactly)
(b) 15/29 (or 0.517)
(ii)(a) 0.109
(b) 0.313
Q10(i) Let mu be the expected value of T. Ho : mu = 38.0, H1 : mu < 38.0.
(ii) {t-bar is a positive real number : t-bar <= 36.8}
(iii) {n is a positive integer : n <= 83}
Q11(i) 816/8671 (or 0.0941)
(ii) r = 6
Q12(i) 0.113
(ii) t = 55s
(iii) 0.617
(iv) Over several hours, there are lull and peak periods so the mean number of people would not be constant from one minute to the other.
Thanks.
Cheers,
Wen Shih
Hi,
It is interesting to note that H2 Maths P2/Q2 and H1 Maths P1/Q4 are almost identical. It leads me to think that the same examiner set the papers.
Cheers,
Wen Shih