Hi,
If you are asked to find the foot of perpendicular from a point to the line or plane (topic: vectors), do you have the intuitive feeling that you may be asked next about
1. shortest distance,
2. image of a point?
If you are asked to show a certain probability (section: statistics), do you think it will be applied in another part of the same question?
If a question suggests that you consider a certain approach, say, x_{n + 1} - x_n (topic: recurrences), do you have any idea what it should lead you to, say, increasing or decreasing sequence?
It is, therefore, important that one has an inkling of what concepts and ideas exam questions could possibly assess. This knowledge can only come about after one's reflection upon tons of past questions that he/she has solved.
Thanks.
Cheers,
Wen Shih
Hi,
I am leaving this forum for good. It would be useful to have the notes on Maths, that I have written in the past, kept for future students' reference.
I also welcome other Maths educators to keep the Maths content alive with new articles.
Thanks and take good care.
Cheers,
Wen Shih
Hi,
I have returned :)
During the break from the forum (May - Jul), I have conceptualised a package of 50 essential questions (complete with authentically written solutions) for JC 2 students who are preparing for their H2 Maths exam come Nov 2011.
For details about this package, please visit my website:
wenshih.wordpress.com
Thanks.
Cheers,
Wen Shih
Hi,
Students may be interested to download a phone App for A-level Maths by UK university:
http://www.mathscard.co.uk/mobile/?pageurl=mobile
Thanks.
Cheers,
Wen Shih
Hi,
A common application of a binomial expansion is approximation.
Consider the expansion of sqrt(1 + 3x) in ascending powers of x up to x^3, i.e.,
1 + (3/2)x - (9/8)x^2 + (27/16)x^3.
By a suitable value of x (where |x| < 1/3), one can find an estimate of the square root of some number:
x = 1/6, 3/sqrt(6) = 157/128 => sqrt(6) = 384/157 approximately.
It is often difficult for us to arrive at a suitable value of x to obtain the square root of some value one is looking for. The use of an abitrary value for x may be a practical approach to solve this difficulty.
Since |x| < 1/3, we consider a substitution for x having the form 1/a, where a is some positive integer and a > 3.
When x = 1/a, sqrt(1 + 3x) becomes sqrt(a + 3) / sqrt(a).
Consider two cases.
For case 1, we have the value a taking some perfect square while the value of sqrt(a + 3) is to be estimated:
a = 2^2, 3^2, 4^2, 5^2, 6^2, ...
and correspondingly,
sqrt(a + 3) / sqrt(a) = sqrt(7)/2, sqrt(12)/3, sqrt(19)/4, sqrt(28)/5, sqrt(39)/6, ...,
from which the estimates sqrt(7), sqrt(19), sqrt(28), sqrt(39), ... can be obtained using the expansion.
For case 2, we have the value (a + 3) taking some perfect square while the value of sqrt(a) is to be estimated:
a + 3 = 3^2, 4^2, 5^2, 6^2, 7^2, ...
and correspondingly,
sqrt(a + 3) / sqrt(a) = 3/sqrt(6), 4/sqrt(13), 5/sqrt(22), 6/sqrt(33), 7/sqrt(46), ...,
from which the estimates sqrt(6), sqrt(13), sqrt(22), sqrt(33), sqrt(46), ... can be obatined using the expansion.
With this simple investigation, it becomes easy for us to come up systematically with suitable substitution values in future when we wish to estimate some given values.
Is there another substitution for x that will allow one to find sqrt(6)? Yes, let x = -1/9. I would encourage the interested student to continue with further research into x taking a substitution of the form -1/a, using the same approach we have discussed above.
Thanks for reading!
Cheers,
Wen Shih
Hi,
How could one find sqrt(15) using a suitable value of x? One may explore the possibility of letting x = a/b, where a and b are positive integers and a/b < 1/3. It is not difficult, using the same type of investigative work mentioned previously, to arrive at x = 2/9 to obtain an estimate of sqrt(15). Thanks.
Cheers,
Wen Shih
Hi,
Students are often confused with various quantities for sampling and hypothesis testing. They include:
1. μ: refers to the population mean;
2. σ^2: refers to the population variance;
3. x-bar: refers to the sample mean, it is also the unbiased estimate of population mean;
4. s^2: refers to the unbiased estimate of population variance, it is sometimes found via the relationship s^2 = { n / (n - 1) } x sample variance;
5. sample variance: self-explanatory, it is sometimes used to find s^2.
Thanks for reading!
Cheers,
Wen Shih
Dear students,
School exam questions are typically set based on past Cambridge question trends. I will quote specific examples from the most recent YJC Prelim 2011 paper 1:
Q1 Behaviour of a recurrence relation
This question has been set in the same flavour as N07/P1/Q9.
Q4 Method of differences
The use of partial fractions has appeared in N09/P1/Q3 and N10/P2/Q2(ii).
Q6(a) Sequence of transformations
This question is similar to N07/P1/Q5.
Q8(b) Tangent to a curve defined implicitly
This question assesses the same concepts and skills as N98/P1/Q16.
Q10 Vector involving planes
Part (ii) resembles N08/P1/Q11(ii). Finding the image of a point in the plane in (iii) has been asked in N10/P1/Q10(iii).
It is important that you know the Cambridge type of questions as you prepare to sit for your school exams. Thanks.
Cheers,
Wen Shih
Hi,
Here is a detailed analysis of another prelim paper 1 from ACJC:
- The last part of Q5 looks like N08/P2/Q1(iv).
- Q6 (nth roots and factorisation) is similar to N07/P1/Q7.
- Q9 about a self-inverse function f bears some resemblance to N09/P2/Q3.
- Q11 (formulation of DE via Newton's cooling model) has been set,
since N10/P1/Q7 appeared last year.
- Q12(ii) about the graph of y = f ' (x) has been set, since N10/P2/Q3(iii) appeared last year.
- Q13(b) about compound interest looks like N08/P1/Q10.
Thanks.
Cheers,
Wen Shih
Hi,
It is not easy, in my view, to use the Casio GC to investigate the behaviour of a sequence as there is a need to go to different screens to enter the expression and to set the table parameters. I have written a set of clear, simple steps to help students overcome this difficulty:
wenshih.files.wordpress.com/2010/02/using-the-casio-gc-to-investigate-the-behaviour-of-a-sequence.pdf
Thanks!
Cheers,
Wen Shih
Hi,
Consider the problem of finding the line of intersection between two planes.
Two common methods are:
1. Use a GC.
2. Find a point by substituting a convenient value for one variable. Then consider the vector product of the normals of the two planes.
It is possible for one to apply another approach, without the need to use a GC and to carry out a vector product. We will look at this example:
Find the line of intersection between the planes whose equations are
x - 5y + 3z = 1 and -x + y + 2z = 11.
Let z = 0, we get the equations
x - 5y = 1 and -x + y = 11.
Adding the equations, we obtain y = -3 and x = -14, giving us a point (-14, -3, 0).
Let y = 0, we have the equations
x + 3z = 1 and -x + 2z = 11.
Adding the equations, we obtain z = 12/5 and x = -31/5, giving us another point (39/5, 3, 12/5).
With these two points, we are able to obtain the equation of the line containing them.
Thanks.
Cheers,
Wen Shih
Hi,
Usually, when we are given the graph of y = f(x) and asked to
sketch the graph of y = f(-|x|), we would have the following
transformations:
1. Reflection about the y-axis (or we could see it as x = 0).
2. Keep the graph for x >= 0 and then carry out reflection about the y-axis.
Suppose we are given the graph of y = f(2x - 1) and asked to
sketch the graph of y = f(-|2x -1|), which NYJC set in its prelim exam this year. The sequence of transformations
would then be:
1. Reflection about
the line 2x - 1 = 0 or x = 1/2.
2. Keep the graph for 2x - 1 >= 0 or x >= 1/2 and then carry out reflection about the line x = 1/2.
Now it is not difficult to sketch the graph of y = f(-|ax + b|) from the graph of y = f(ax + b).
Thanks.
Cheers,
Wen Shih
Is this under JC syllabus?
1) To prove whether a function is a concave or convex function.
2) Hyperbola with asymptotes
Hi HyuugaNeji,
1) Yes, we consider the value of the second derivative to determine the nature.
Second derivative < 0 => we have a concave down (or concave function).
Second derivative > 0 => we have a concave up (or convex function).
2) Yes, a question on that was asked in the 2010 exam.
Thanks.
Cheers,
Wen Shih
Dear students,
Do pay attention to these comments from the examiners, so that you do not risk losing precious points unwittingly:
1. Candidates lacked precision in their solutions.
Personal advice: Read the question carefully to understand its requirement on the answer.
2. Some of the algebraic manipulation was poor.
Personal advice: Strengthen this skill, perhaps by recalling knowledge from secondary algebra, say, factorisation, results like x^2 - y^2 = (x + y)(x - y), etc.
3. Evidence of an over-reliance on the use of graphic calculators.
Personal advice: GC is a checking tool. One must still know, say, properties of basic curves.
4. Full and precise working is needed in questions in which an answer to be proven is given.
Personal advice: Show all details of working in any case.
5. Some time was wasted by not using the clue in the question.
Personal advice: Always identify, at the start, what information is given in the question and how it could be used.
6. Majority of candidates did not understand the notation.
Personal advice: Always fall back on what you already know or have seen before.
7. A lack of appreciation as to the relevance of earlier parts of the question.
Personal advice: Remember that a question with parts (i), (ii), (iii), etc. may be related.
8. Candidates lost track of what they were doing.
Personal advice: See advice 1 and stay mentally alert during the examination.
9. Candidates quoted and used wrong formulas.
Personal advice: Check your working and ask whether your answers make sense.
10. Difficulty arriving at the correct variance.
Personal advice: Recall that Var(aX) = a^2 Var(X) and Var(X - Y) = Var(X) + Var(Y).
11. Main error being to ignore the continuity correction, or to use the wrong value.
Personal advice: Recall that continuity correction is necessary when B(n, p) and Po(lambda) becomes normal in the approximation process. As for the value, consider the method of rounding up or rounding down, e.g., P(X < 7) = P(X <= 6) which becomes P(X < 6.5) via correction as a result of rounding any value less than 6.5 down to 6.
Jiayou and good luck!
Cheers,
Wen Shih
Hi,
Questions with no guidance may be asked in which students are expected to be familiar with the problem-solving approach.
Example 1: N2010/P1/Q9(i)
In this minimisation problem, a student has to form a condition connecting relevant variables, set up the expression of the external surface area and apply differentiation to determine the stationary value and its nature.
Example 2: N2010/P2/Q2(ii)(a)
In this method of differences problem, a student has to consider the use of partial fractions to obtain a difference of two similar expressions before systematic cancellation of terms can be made possible.
Thanks.
Cheers,
Wen Shih
Hi,
Both teachers and students of H2 Maths should take note of what a lecturer from NUS has to say:
http://math.nie.edu.sg/ame/mtc11/pdf/Lecture/MTC%202011_mattan.pdf
Thanks.
Cheers,
Wen Shih
Dear students,
It may be a good idea to review the following problems with your remaining time:
1. Induction based on a
conjecture.
2. Solving a DE by the method of substitution.
3. Formulating a DE, perhaps, based on the difference of rate in (or birth rate) and rate out (or death rate).
4. Hypothesis testing involving an unknown alpha (level of significance), x-bar, mu-zero, n.
5. Area of a region involving a curve defined parametrically.
6. Area approximated by means of rectangles under or above the curve.
7. Probability requiring the need to use P&C.
8. Connected rates of change.
9. Maclaurin's expansion involving the need to use the result to find a definite integral or find an approximation.
10. Graph of a rational function with arbitrary constant(s).
11. Application of the De Moivre's theorem where we solve an equation of the form
(z + c)^n = z^n, where c is a complex number.
12. Vectors involving a given 3-dimensional figure.
Good luck and jiayou!
Cheers,
Wen Shih
Hi JC 1 students,
Here is an interesting question about graphical properties from a foreign exam board:
A certain function, f(x), has the following characteristics:
Determine a possible expression for f(x).
Enjoy, while preparing for OP!
Cheers,
Wen Shih
Hi,
Let's extend the previous question further. Suppose we add this condition:
Exactly one turning point at one of the roots.
Determine a possible expression for f(x).
Thanks!
Cheers,
Wen Shih
Hi,
Comments to specific questions in paper 1 assessed today.
Q1 Students should notice that the numerator is always positive. This could be checked by completing the square.
Q6(ii) Looking at (i), students should realise that it is a question on method of differences.
Q7(i) Students will handle vector algebra for this question, this is a type of question from the distant past.
Q9(ii) Students should realise that the theoretical maximum refers to the sum to infinity in the context of the question.
It seems that the examiners are really out to test students' level of mathematical maturity and sensitivity to given information. The flavour of the paper seems to be the same as that of N2010. Here are two examples.
In N2010, the question on the method of differences did not prompt students to find partial fractions. In 2011, the question did not mention about the method of differences.
In 2010, the question on vectors involving p referred to length and there was a part on the dot product result. In 2011, the question involving p referred to a unit vector and there was a part on the vector product to find the area.
Thanks.
Cheers,
Wen Shih
Hi,
I mentioned about two important attributes that students will need to develop in order to do well for Maths. Let me elaborate further in this posting.
Specific skills to increase one's level of mathematical maturity:
1. show an appreciation of the motivation behind or (better still) derive basic mathematical results;
2. show an appreciation of the interconnections between mathematical topics;
3. show an appreciation of the interconnectedness between algebraic, graphical and geometrical approaches to mathematical problem-solving;
4. possess the ability to work with both concrete (numerical-based problems) and abstract (problems with arbitrary constants or variables) mathematics;
5. possess a broad knowledge of mathematical problem-solving strategies (acquired from the primary level up through to college level) and have a good intuition of applying them;
6. show an appreciation of the application of mathematical theories in real-life contexts;
7. possess the ability to evaluate correctness of mathematical solutions.
Specific skills to increase one's sensitivity to given information:
1. possess a good understanding of mathematical definitions and terms;
2. possess the ability to translate information into mathematical expressions, conditions, equations which will then facilitate mathematical problem-solving;
3. show an appreciation of the rationale behind these given information.
Thanks.
Cheers,
Wen Shih
Hi,
Here are two other examples to show the close relationship between 2010 and 2011 exams - 2010/P2/Q10 vs 2011/P2/Q8, 2010/P2/Q8 vs 2011/P2/Q11.
In example 1, both questions cover the comparison between a linear and a quadratic model.
In example 2, both questions cover the idea of using P&C method of obtaining the probability.
Thanks.
Cheers,
Wen Shih
Hi,
Questions that require the application of P&C and probability concepts have appeared in the last 2 years.
We will discuss an example here.
I need to arrange three red cubes, four blue cubes and six yellow cubes in a row. Find the probability that the cubes at each end are the same colour.
This probability is
the number of ways of arranging the cubes such that the cubes at the ends have the same colour
divided by
the number of ways of arranging the cubes without any restriction.
It is easy to begin by finding the denominator, that is,
13! / (3! 4! 6!) = 60060,
since we have 13 items of which 3 are identically red, 4 are identically blue and 6 are identically yellow.
To find the numerator, we first list 3 possibilities:
1) Red cubes at the ends
Number of ways = 11! / (1! 4! 6!) = 2310,
since we fix 2 red cubes at the ends and have the remaining 11 items to arrange, with 4 items that are identically blue and 6 items that are identically yellow.
2) Blue cubes at the ends
Number of ways = 11! / (3! 2! 6!) = 4620,
by a similar reasoning as above.
3) Yellow cubes at the ends
Number of ways = 11! / (3! 4! 4!) = 11550.
The numerator is the sum of the ways (i.e., 18480), by the addition principle.
The final answer can then be arrived at (i.e., 4/13).
As practice, you may try this question:
Find the probability that the cubes at each end are of a different colour.
Thanks and Merry Christmas in advance!
Cheers,
Wen Shih
Hi,
This discussion has been motivated by N2011/P1/Q4(ii)(b) about using a Maclaurin series with a given substitution to find an approximate value.
For a Maclaurin series to be used to achieve a good approximation, we can 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, we will need to have more terms in the series.
Let's consider a trivial example where we take the Maclaurin series of 1/(1 - x) which is our familiar infinite geometric series 1 + x + x^2 + x^3 + ... + x^n + ... .
To illustrate situation 1, suppose we substitute x = 1/9 and simply consider 4 terms of the series, we obtain
1 / {1 - (1/9)} is approximately equal to 1 + (1/9) + (1/9)^2 + (1/9)^3
=> 9/8 is approximately equal to 820/729
=> 9 is approximately equal to 6560/729 (or 8.9986).
Before we illustrate situation 2, we recall the sum of n terms of a GP with first term 1 and common ratio x, i.e., (1 - x^n) / (1 - x).
Suppose we substitute x = 8/9 and consider n terms of the series, we obtain
1 / {1 - (8/9)} is approximately equal to {1 - (8/9)^n} / {1 - (8/9)}
=> 9 is approximately equal to 9 {1 - (8/9)^n}.
If we want the approximation in situation 2 to be as good as that of situation 1 (i.e., 8.9986), one could use GC and find that n must at least be 75:
n = 74, 9 {1 - (8/9)^n} = 8.9985
n = 75, 9 {1 - (8/9)^n} = 8.9987.
To conclude, one should pay particular attention to the substitution value as well as the number of terms of the series in order to arrive at a good approximation.
Thanks for reading.
Cheers,
Wen Shih