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Checklist and Resources for H1&H2 Maths/Physics/IB HL Maths

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    wee_ws's Avatar
    1,524 posts since Mar '08
    • Hi,

        JC 2 students may consider practising on the following papers when preparing for the coming Prelims:

        - 2008 IJC

        - 2008 HCI

        - 2008 JJC

        - 2008 PJC

        - 2007 VJC

        - 2007 YJC

        - 2007 PJC

        Thanks and jiayou!

      P.S. This recommendation is made based on personal preference and is not meant to endorse certain schools. Thank you.

      Wen Shih

      Edited by wee_ws 19 Jul `09, 9:02AM
    • Hi,

        I'd like to share some practical strategies when preparing for H2 maths exam (school or national):

        0. Ideally you should start early, at least 2 months ahead of the exam.

        1. Look at a couple of past year papers and analyse the questions in detail. For school exams, refer to your school's past papers. For national exams, refer to Cambridge past papers. Your analysis should focus on

      - weightage of marks by topics (see an example on my website at:


      - specific skills/concepts assessed (e.g. place ticks on points within the checklists I have developed previously).

        You should then have a really big picture of question trends and level of difficulty. You may even predict questions that may arise in your coming exam.

        2. In your practices, attempt the questions by topics or by level of personal ease (i.e. starting from the most manageable problems in your view). Complete each question under time limit (1.8 minutes per mark) as far as possible.

        3. Before you do each question:

        - scan the problem statement quickly and note its requirements;

        - pay attention to key skills/concepts the problem requires;

        - make some quick notes about skills/concepts so that you can refer to them, if you have yet to internalise them.

        4. When the time is up and you are not done with the question, analyse what held you back (e.g. lack of understanding because even cannot start, lack of speed in manipulation/evaluation, etc.). Pen your analyses down.

        If you managed to complete the question, review your solution and analyse what went right and what is needed for you to improve upon (e.g. clearer and bigger diagrams, clearer working without skipping steps, carelessness in steps to be avoided, etc.). Pen your analyses down.

        I must highlight that self-reflection is a necessary part for any real progress to be made. Patience and perseverence are also key ingredients of personal success. 

        5. Keep all these hard work you have done, because they help to serve as great reminders (of your good habits and undesirable practices that should be avoided) when you revise some time later.


      Wen Shih

      Edited by wee_ws 19 Jul `09, 9:02AM
    • Hi,

        I'd like to elaborate on point 3 that was mentioned in the last post.

        Suppose one is going to attempt this question on statistics:

        The random variable X has the binomial distribution B(n, p) where 0 < p < 1.

        (a) Given that Var(X) = 4/5 E(X), find the least value of n such that P(X >= 1) > 0.92.

        (b) Given that n = 8 and p = 1/3, the random variable S is the sum of 60 independent observations of X. Find the approximate value of P(S > 162).

        A quick scan will reveal that the problem focuses on the Binomial distribution and the Central Limit Theorem.

        Quick notes may include the following points:

        - E(X) = np, Var(X) = npq for any Binomial random variable X.

        - P(X = x) = nCx p^x q^(n - x) for any Binomial random variable X.

        - X_1 + X_2 + ... + X_m ~ N( m E(X), m Var(X) ) for large n (i.e. >= 50).

        - TI-GC's command normalcdf (a, b, mu, sigma) to compute P(a < X < b).


      Wen Shih

      Edited by wee_ws 19 Jul `09, 9:22AM
    • Hi,

        My latest version (3rd revision) of H2 maths checklists are available at:


        The content has been reorganised in line with the topics in the syllabus to allow for easy reference by students. In addition, some new points have been added to enhance the lists further :)

        Thanks, and best of luck to all in your A-level exams!

      Wen Shih

    • Hi,

        I'd like to propose this sequence of topics for teaching and learning H2 Maths. It may be useful for incoming JC 1 students as well as JC 2 students in 2010.

        0. Review of O-level maths concepts (e.g. partial fractions, completion of square, nature of roots, elementary trigonometric knowledge, basic graphs and linear transformations, differentiation and integration, number patterns, basic statistics and probability concepts, equations of straight lines, etc.).

        1. Binomial expansion.
        Use of prior knowledge: Binomial expansion (at O-level), where n is a positive integer.

        2. AP/GP.
        Use of prior knowledge: Number patterns at O-level.

        3. Summation of series.
        Motivating factor: One can write sums more efficiently.

        4. Mathematical induction.
        Motivating factor: One can prove results involving sums in a powerful way.

        5. Graphing of basic functions.
        Use of prior knowledge: Graphs at O-level.

        6. Linear transformations.
        Use of prior knowledge: Linear transformations (at O-level), involving trigo graphs.

        7. Functions.
        Use of prior knowledge: Functions at O-level.

        8. Further transformations.
        Motivating factor: Graphs...transform, more can meet the eye!

        9. Inequalities.
        Use of prior knowledge: Inequalities at O-level.

        10. System of linear equations.
        Use of prior knowledge: Matrices at O-level.

        11. Differentiation and its applications.
        Use of prior knowledge: Differentiation and its applications at O-level.

        12. Maclaurin's expansion.
        Use of prior knowledge: Topics 1 and 11.

        13. Integration and its applications.
        Use of prior knowledge: Integration and its applications at O-level.

        14. Differential equations.
        Use of prior knowledge: Topic 13.

        15. Vectors.
        Use of prior knowledge: Vectors at O-level.

        16. Complex numbers.
        Use of prior knowledge: Topic 15.

        17. Permutations and combinations.
        Motivating factor: Can we count possibilities?

        18. Probability.
        Motivating factor: Chances are ...?

        19. Binomial and Poisson distributions.
        Motivating factor: Can certain types of occurrence be modelled mathematically?

        20. Normal distribution.
        Motivating factor: How are grades awarded?

        21. Sampling and sampling distribution.
        Motivating factor: What the small can say about the large?

        22. Hypothesis testing.
        Motivating factor: How can claims be made to convince?

        23. Correlation and regression.
        Use of prior knowledge: Equations of straight lines.


      Wen Shih

      P.S. This page will be updated several times. Do be patient, thanks!

      Edited by wee_ws 24 Nov `09, 9:22AM
    • Hi,

        Students should realise that there is a rising trend of questions involving arbitrary variables. Thanks!

      Wen Shih

    • Hi,

        Other than the use of arbitrary variables, questions can also be set in such a way that they look unfamiliar and atypical to students. Despite that, students should still apply the normal approaches that they have been taught to tackle such problems.

        Take, for example, a question about the method of differences. Most students are comfortable when it's an algebraic expression involving the use of partial fractions, e.g. 1/(4r^2 - 1) = 1/2 [ 1/(2r - 1) - 1/(2r + 1) ] .

        What if one is asked to find the sum of { sin (k + 1)x - sin (k - 1)x }? It may get many lost souls stuck at wondering what to do?

        Now if a learner realises the working principle of the method of differences, then he/she will ALWAYS know what to do when expressions change.

        The method of differences relies on the difference of at least two SIMILAR expressions, e.g. f(r) - f(r + 1)  or f(r + 1) - f(r). Notice that nothing is said about the expression for f , so it can be ANYTHING.

        We can use the familiar LIATE rule for f, so f may be

        a LOGARITHMIC expression,

        an INVERSE TRIGONOMETRIC expression,

        an ALGEBRAIC expression,

        a TRIGONOMETRIC expression,

        or an EXPONENTIAL expression.

        Let's look at some concrete examples.

        Example 1: f(r) = e^r.

        Example 2: f(r) = ln r.

        Example 3: f(r) = sqrt(2r - 1).

        Example 4: f(r) = 1/(r!)(r + 1).

        Example 5: f(r) = cos rx.

        Example 6: f(r) = arctan (2r).

        Regardless of what forms f may take, we just proceed with the usual cancellation of terms, leaving just the top-left/bottom-right or top-right/bottom-left portions.


      Wen Shih   

      Edited by wee_ws 03 Dec `09, 6:30AM
    • Hi,

        Continuing our discussion about the method of differences, it can be assessed along with the topics of recurrences and mathematical induction. Do check out N07/P2/Q2 for details. Thanks!

      Wen Shih

    • Hi,

        Here is a quick summary on Transformations of graphs:

      Types include:
      1. Translations.
      (i) Parallel to the x-axis, i.e. y = f(x - a).
          a > 0: a units right.
          a < 0: a units left.

      (ii) Parallel to the y-axis, i.e. y = f(x) + a.
           a > 0: a units up.
           a < 0: a units down.

      2. Scalings.
      (i) Of factor 1/a, parallel to the x-axis, i.e. y = f(ax).
          0 < a < 1: graph appears 'fatter'.
          a > 1: graph appears 'thinner'.
          a < 0: perform a reflection about the y-axis first.

      (ii) Of scale factor a, parallel to the y-axis, i.e. y = af(x).
           0 < a < 1: graph appears 'shorter'.
           a > 1: graph appears 'taller'.
           a < 0: perform a reflection about the x-axis first.

      3. Reflections.
      (i) About the x-axis, i.e. y = -f(x).

      (ii) About the y-axis, i.e. y = f(-x).

      4. Modulus.
      (i) Applied to x, i.e. y = f(|x|).
          - Drop the portion of the graph where x < 0.
          - Keep the portion of the graph where x >= 0.
          - Reflect the kept portion about the y-axis.

      (ii) Applied to y, i.e. y = |f(x)|.
           - Reflect, about the x-axis, the portion of the graph below the y-axis.

      5. Reciprocal, i.e. y = 1/f(x).
         - Turning points (not on x-axis): min -> max or max -> min.
         - Asymptotes:
           vertical asymptote x = a -> x-intercept x = a.
           horizontal asymptote y = b (non-zero) -> horizontal asymptote y = 1/b.
         - Intercepts:
           x-intercept x = a (non-zero) -> vertical asymptote x = a.
           y-intercept y = b (non-zero) -> y-intercept y = 1/b.
         - Shape:
           Increasing -> decreasing.
           Decreasing -> increasing.

      6. y^2 = f(x).
         - Drop the portion of the graph below the y-axis.
         - Vertical asymptote stays as it is.
         - Horizontal asymptote y = a -> horizontal asymptotes y = -sqrt(a) and y = sqrt(a).
         - y = f(x) touches the x-axis at x = a -> y^2 = f(x) is 'sharp' at x = a.
         - y = f(x) crosses the x-axis at x = a -> y^2 = f(x) is 'smooth' at x = a.

      7. Combinations of the above.
         - Use algebraic substitutions to help determine the order of transformations.
         - Example 1: y = f(1 - |x|).
           x -> x + 1,
           so y = f(x) -> y = f(x + 1),
           so we have a translation to the left by 1 unit.

           x -> -x,
           so y = f(x + 1) -> y = f(-x + 1),
           so we have a reflection about the y-axis.

           x -> |x|,
           so y = f(-x + 1) -> y = f(-|x| + 1),
           so we carry out the actions of modulus applied to x.

           In this example, correct order matters!

         - Example 2: y = f(2x) - 1.
           x -> 2x,
           so y = f(x) -> y = f(2x),
           so we have a scaling of factor 1/2, parallel to the x-axis.

           y -> y - 1,
           so y = f(2x) -> y = f(2x) - 1,
           so we have a translation of 1 unit down.

           In this example, order does not matter.

      Common exam questions include:
      1. Given y = f(x), sketch the transformed graphs.

      2. Given transformed graphs (e.g. y^2 = f(x) and y = |f(x)|), sketch y = f(x).

      3. Given the equation of a curve, find the resulting equation after performing some transformations, which may have to be described in words. [ Refer to N07/P1/Q5. ]

      4. Given the equation of a curve that has gone through some transformations, find the original equation by undoing the transformations in reverse order, which may have to be described in words.

         Example 3: Suppose y = f(x) goes through

         (A) a scaling of factor 1/2, parallel to the x-axis;
         (B) a translation of 1 unit down.

         The resulting equation becomes y = x^2. Find f(x).

         After undoing (B),
         y -> y + 1, so y = x^2 -> y = x^2 + 1.

         After undoing (A),
         x -> x/2, so y = x^2 -> y = (x/2)^2 + 1.

         So f(x) = (x^2)/4 + 1.

      5. Given the graph of a rational function, sketch its transformations. [ Refer to N08/P1/Q9(iv)(b). ]


      Wen Shih

      Edited by wee_ws 02 Dec `09, 2:07PM
    • Hi,

        Here is a quick summary on Maclaurin's series.

      1. Maclaurin's series is essentially a polynomial given by

         f(x) = f(0) + f'(0) x + f''(0)/2! x^2 + f'''(0)/3! x^3 + ... + f^(n)(0)/n! x^n + ...

      2. From the expression, we see the need for us to find the values of

         f(0), f''(0), f'''(0), ..., f^(n)(0)

         by (i) repeated differentiation of f, and (ii) letting x = 0.

      3. Often, implicit differentiation is carried out. The use of suitable substitutions can make implicit differentiation less tedious.

      4. We can also make use of the standard Maclaurin's series in MF15.

      5. When x is sufficiently small so that x^3 and higher powers of x can be ignored, the series for sin x and cos x can be simplified to

         sin x ~= x,

         cos x ~= 1 - x^2/2.

         In addition, we note the series for tan x, i.e.

         tan x ~= x.

      6. The use of trigonometric identities (see MF15) and geometrical results (e.g. Pythagoras' theorem, sine and cosine rules) may be necessary when we solve a question involving a trigonometric expression.

      7. Maclaurin's expansion of (1 + x)^n is given by

         (1 + x)^n = 1 + nx + n(n-1)/2 x^2 + ... + n(n-1)...(n-r+1)/n! x^n + ...,

        where |x| < 1.

         We have used it in a restricted way (where n is a positive integer) at O-level Additional Mathematics. When n is a positive integer, the series has a fixed number of terms (i.e. n + 1) and the series is always true for any value of x.

      8. Often, we see a more general expression, i.e.

         (a + bx)^n,

         which may be rewritten as

         a^n (1 + b/a x)^n -- (1)

         or as

         (bx)^n (1 + a/bx)^n -- (2).

         Result (1) gives us a series in increasing powers of x where x is small; result (2) gives us a series in decreasing powers of x where x is large.

      9. We may be able to obtain an expression for the general term of a binomial expansion by observation or using the result

         n(n - 1)(n - 2)...(n - r + 1)/r! x^r.

      10. We may use the binomial expansion to approximate values by means of suitable values of x.

      11. Other typical applications involving Maclaurin's series include:

          (a) Finding the equation of tangent at the origin. It is given by y = f(0) + f'(0)/2 x.

          (b) Finding the Maclaurin's series by integration and differentiation.   

          (c) Evaluating definite integrals and limits.

          (d) Comparing the graphs of f and its Maclaurin's series. 

          (e) Solving equations. 

          (f) Solving inequalities.


      Wen Shih

      Edited by wee_ws 20 Dec `09, 8:18PM
    • Hi,

        Here is a quick summary on Functions.

      I Functions

      1. A function is defined by a domain and a rule. The domain is a set of input values; the rule is an expression involving any input value. Two ways of defining a function are:

        f : x -> RULE, DOMAIN

        f(x) = RULE, DOMAIN

      2. When we graph a function y = f(x), we can obtain the range of f (i.e. the set of y-values) corresponding to the set of input x-values. The range of f may be described by using the interval notation, e.g.

        (a, b] = {y \in R : a < y <= b}.

      3. Since one needs to sketch graphs, one has to be familiar with

        - parabolas,
        - rectangular hyperbolas,
        - logarithmic/exponential graphs.

      4. The vertical line test helps one to determine whether f is a function or a relation. The line x = k must intersect the graph of y = f(x) once, so that f is a function.

      II Inverse functions

      5. The horizontal line test allows one to decide whether f is a one-one function. The line y = k must intersect the graph of y = f(x) once, so that f is one-one.

      6. When f is one-one, its inverse f^(-1) exists. The idea of the inverse is this: find x, given y. Thus, one makes x the subject when it is necessary to find the rule of f^(-1).

      7. We may encounter the situation where x takes two expressions. The suitable expression is one which satisfies the domain of f.

      8. To define the inverse function f^(-1), one needs to state its domain and rule. The domain of f^(-1) corresponds to the range of f.

      9. To find the range of f^(-1), one determines the domain of f.

      10. Graphically speaking, the relationship between y = f(x) and y = f^(-1) is a reflection in the line y = x. So any point (a, b) on y = f(x) becomes the point (b, a) on y = f^(-1)(x). Since the graphs of y = f(x) and y = f^(-1)(x) intersect along the line of reflection y = x, the solution to f(x) = f^(-1)(x) may be found by f(x) = x or f^(-1)(x) = x.

      11. If the horizontal line fails, one will need to change the domain of f to make f one-one.

      III Composite functions

      12. A composite function fg goes through two functions in this order: g first, followed by f.

      13. Since the range of g becomes input to f, it is necessary for us to have

        range of g \subseteq domain of f.

      14. Since g is the first function, it is obvious that

        domain of fg = domain of g.

      15. To find the range of fg, we take these steps:

        i. Graph y = g(x). Find the range of g.

        ii. Graph y = f(x). Find the range of f, corresponding to the range of g.

      16. If range of g \subseteq domain of f fails, then one has to change the domain of g to make the range of g fit the condition.

      IV Other applications

      17. A piecewise function contains several rules and domains. A familiar example is y = |x| where x is real, since

        y = x if x >= 0,
        y = -x if x < 0.

      18. The topic of functions can be assessed along with the topic of areas [ refer to N09/P1/Q4 ].


      Wen Shih

    • Hi,

        In the topic of vectors, many students find it difficult to determine whether two points are on the same side of a given plane or they are on opposite sides. We shall consider some simple numerical examples to overcome this difficulty.

        Consider the x-y plane, whose equation is given by r.(0, 0, 1) = 0. Also, consider two points P(1, 2, 3) and Q(1, 2, -3). Substituting each point into the equation of the plane, we obtain

        (1, 2, 3).(0, 0, 1) = 3 (i.e. > 0) and (1, 2, -3).(0, 0, 1) = -3 (i.e. < 0),

        meaning that P lies 3 units above the x-y plane and Q is 3 units below the x-y plane. Thus P and Q are on opposite sides of the x-y plane.

        Next, we look at P(1, 2, 3) and R(3, 4, 5). When we substitute R into the equation of the plane, we have

        (3, 4, 5).(0, 0, 1) = 5 (i.e. > 0),

        so both P and R lie above the x-y plane. Thus P and R are located at the same side of the x-y plane.

        These numerical examples help us to understand:

        1. The need for us to substitute each point into the equation of a given plane r.n = k.

        2. The need for us to interpret several cases geometrically, i.e.

            (a) both substitutions give RHS values > k,
                  will mean that both points lie above the given plane,
                  so both points lie on the same side of the plane.

            (b) both substitutions give RHS values < k,
                  will mean that both points lie below the given plane,
                  so both points lie on the same side of the plane.

            (c) one substitution gives RHS value < k and
                  the other substitution gives RHS value > k,
                  will correspondingly mean that one point is below the plane and
                  the other is above the plane,
                  so the points are located on opposite sides of the plane.

        Thanks, and have a happy lunar new year!

      P.S. A handy tip for students is to consider concrete examples (using simple numerical values) to help in the understanding of abstract mathematical results which may not be obvious.

      Wen Shih

      Edited by wee_ws 13 Feb `10, 2:27PM
    • Dear JC 1 students and H2 Maths tutors,

        Topics (not in order) that may be covered in JC 1 (across most schools) may include:

        1. Graphing techniques
        2. Transformations of graphs
        3. Binomial expansion
        4. Inequalities
        5. System of linear equations
        6. AP/GP
        7. Sigma notation
        8. Method of differences
        9. Recurrences
        10. Mathematical induction
        11. Functions
        12. Differentiation and its applications
        13. Maclaurin's expansion
        14. Integration techniques
        15. Areas and volumes
        16. Differential equations
        17. Vectors - points and lines
        18. Vectors - lines and planes
        19. Permutations and combinations
        20. Statistics

        Hope this list will help you prepare in advance. Feel free to point out if I have missed any topics.


      Wen Shih

    • Hi,

        Many interesting questions on integration techniques can be formed by two simple expressions, e.g. x and ln x. See if you could identify the integration technique to apply in each case:

        Find the integral of

        (a) ln x,

        (b) x - ln x,

        (c) x ln x,

        (d) (ln x) / x,

        (e) 1 / (x ln x),

        (f) 1 / [ x sqrt(ln x) ],

        (g) x^2 ln x,

        (h) x (ln x)^2,

        (i) sqrt(x) ln x.

        Try it, thanks!

      Wen Shih

    • Hi,

        Why do we learn how to sketch basic graphs? I'd like to raise a couple of applications so that students can better appreciate the content they're learning.

        These include:

        1. determining the number of solutions of an equation,

        2. determining the solutions of an inequality,

        3. determining domain and range of a function,

        4. determining some region bounded by graphs to find area and volume of revolution,

        5. determining the graph after undergoing a transformation.

        Do learn graphs well, for they contribute to the learning and mastery of subsequent topics.

      Wen Shih

    • Hi,

        A useful way to learn all graphs and not miss any out, is to consider the LIATE categories, i.e.

        L, for logarithmic graphs;

        I, for inverse trigonometric graphs;

        A, for algebraic graphs;

        T, for trigonometric graphs;

        E, for exponential graphs.

        Next, we can establish useful relationships between graphs of two categories. One notes that the graph of y = ln x (from L) is a reflection of the graph of y = e^x (from E) in the line y = x. Similarly, one can make a similar observation between the graphs of y = sin x (from T) and y = arcsin x (from I).

        As for A, we may subdivide into the following items:

        linear expressions,
        quadratic expressions,
        cubic expressions,
        quartic expressions,
        circles and ellipses,
        rectangular hyperbolas and hyperbolas,
        rational expressions.

        Finally, I'd add two new categories to complete the 'graphical' picture:

        C, for combinations of LIATE;
        P, for parametric equations.

        An example of C may be the graph of y = x + ln x, y = x ln x, or y = x / (ln x).


      Wen Shih

    • Hi,

        I'd like to discuss about hyperbolas this time.

        A standard representation is given by 

        x^2 / a^2 - y^2 / b^2 = 1.

        The basic properties of this graph include:

        1. Two oblique asymptotes are y = b/a x and y = -b/a x.

        2. Asymptotes intersect at the origin.

        3. Two x-intercepts (-a, 0) and (a, 0), which are aligned horizontally with the intersection point of the asymptotes. No y-intercepts at all.

        What happens when the standard representation changes to the equation

        (x - h)^2 / a^2 - (y - k)^2 / b^2 = 1?

        Based on the original basic properties, we just need to make some minor adjustments, i.e.

        1. Two oblique asymptotes are y - k = b/a (x - h) and y - k = -b/a (x -h), since the original variables x, y have been changed to x - h, y - k correspondingly.

        2. Asymptotes intersect at the point (h, k), since it is clear that (h, k) satisfies the equations of the two new oblique asymptotes.

        3. Since the intersection point of the asymptotes has shifted to (h, k), the original points (-a, 0) and (a, 0) have to be shifted likewise to (-a + h, k) and (a + h, k) respectively so that the horizontal alignment can be maintained.

        What if our hyperbola is expressed as

        y^2 / a^2 - x^2 / b^2 = 1?

        Again, we go back to the basic properties and tweak them a little, i.e.

        1. Two oblique asymptotes are x = b/a y and x = -b/a y, because x is changed to y and y is changed to x. Making y the subject, the equations become y = a/b x and y = -a/b x.

        2. Asymptotes intersect at the origin, no change.

        3. Since x becomes y and y becomes x, we will have two y-intercepts (0, -a) and (0, a), which are aligned vertically with the intersection point of the asymptotes. No x-intercepts at all.

        It is now easy to establish the properties of the hyperbola given by

        (y - k)^2 / a^2 - (x - h)^2 / b^2 = 1,

        using the same approach as with the graph of

        (x - h)^2 / a^2 - (y - k)^2 / b^2 = 1. Try it, as a good form of practice!

        Thus, we see that a good understanding of the basic properties of one graph will help us to appreciate/derive variations of it, without taxing our limited memory.


      Wen Shih

      Edited by wee_ws 02 Mar `10, 10:59PM
    • Hi,

        The graphic calculator may be a powerful aid to mathematics problem solving, but it still possesses limitations that only clever human ingenuity may overcome. We shall discuss some examples.

        Example 1: Sketch the graph of C, defined by the parametric equations

        x = 1/2 ln (t^2 - 1) - ln t, y = t(5t^2 - 8), where t > 1.

        If one were to sketch C with TI graphic calculator using ZOOM-6, one will not be able to see certain details about the graph, e.g. the presence of a horizontal asymptote y = -3. Note that we can only arrive at the equation of the horizontal asymptote by manual means.

        Example 2: Sketch the graph of y = ax / (bx - a), where x is real, except that it cannot take the value of a/b.

        The calculator does not allow us to sketch graphs where arbitrary constants are given. We can only rely on our knowledge of basic properties of this rectangular hyperbola:

        - Vertical asymptote x = a/b;
        - Horizontal asymptote y = a/b;
        - Graph passes through the origin.

        to sketch it by hand.

        We may even have to consider several cases, like:

        - a, b having both positive values;
        - a, b having both negative values;
        - a, b having different signs.

        Regardless of the case, we will have a rectangular hyperbola that appears in the first and third quadrants.

        Thus, we see that the calculator cannot arrive at such analytical reasonings.

        Example 3: It is given that d^2 y/dx^2 = 10 - 6x. Sketch three members of the family of solution curves, given that y = 100 when x = 0.

        The general solution of this second-order differential equation is

        y = 5x^2 - x^3 + cx + 100, where c is an arbitrary contant.

        Three members may be:

        y = 5x^2 - x^3 + 100 (when c = 0),
        y = 5x^2 - x^3 + x + 100 (when c = 1),
        y = 5x^2 - x^3 - 2x + 100 (when c = -2).

        If we sketch these graphs, the calculator's screen size may not allow us to see them clearly in terms of their relationships, even with the use of ZOOM-0, because of the large constant value at the end of each equation.

        The approach then is to use our knowledge of linear transformations of graphs. In this problem, we consider a vertical translation. Now, we can easily compare the graphs of

        y = 5x^2 - x^3,
        y = 5x^2 - x^3 + x, and
        y = 5x^2 - x^3 - 2x,

        without worrying too much about the screen size.

        I hope these examples will hit home the important message that mathematics problem solving is more than just using the graphic calculator. Ultimately, it is the component of critical thinking that gets featured in H2 mathematics exam!

        Thanks for reading!

      Wen Shih

      Edited by wee_ws 19 Mar `10, 10:02AM
    • Dear students and teachers,

        This is an interesting article:


        It is extremely helpful to realise that

        1. students can be encouraged to "use their own mathematical graphics to explore, make and communicate mathematical meanings;"

        2. students "focus on processes of learning, rather than notation as a product;"

        3. mathematics teaching can be made effective when it "focuses on the communicative aspects of mathematics by developing oral and written mathematical language."

        Thank you.

      Wen Shih

    • Hi,

        In this posting, I will clarify some aspects of statistics in point form and in plain simple English, to help clueless students make sense of what they are learning in schools.

        1. Statistics is primarily concerned with us observing some phenomenon, which happens by chance.

        2. One may
            (a) record the number of occurrences of a phenomenon, or
            (b) take measurements of it.

        3. An example of (a) is the number of defective items in an inspection scheme. An example of (b) is the diameter of bolts in a manufacturing process. Both (a) and (b) are specific instances of the random variable. Usually we write the random variable as X.

        4. The random variable X can take values. In the case of (a), the values are just counting numbers 0, 1, 2, 3, ... . For example, if X is the number of defective items, then X can take the values 0, 1, 2, 3, etc. We call X a discrete random variable.

        5. In the case of (b), X could take values that are spread out within some range. For example, if X represents diameters of bolts, then X can take any value, say between 0.78 and 0.82 cm, if a bolt of an acceptable quality is to be manufactured. We call X a continuous random variable.

        6. The random variable X, taking discrete or continuous values, exhibit a certain behaviour. To understand the behaviour, we make use of a suitable model that can explain or reflect it adequately. Therefore, we study the Binomial, Poisson and Normal distributions.

        7. Models need to meet certain assumptions, or they will not be good enough to be used to explain behaviours. For example, the Poisson distribution may not be a good model for the number of grand piano sold by a shop in a year, because the probability of a piano being sold may not be constant throughout the year since there are variations in the demand for pianos.
        8. If X follows a Binomial distribution, we write
            X ~ B(n, p),
      where n (i.e., the number of experiments or trials) and p (i.e., the probability of a successful outcome) are parameters that the Binomial distribution depends on.

        9. If X follows a Poisson distribution, we write
            X ~ Po(lambda),
      where lambda (i.e., mean) is a parameter that the Poisson distribution relies on.

        10. Given a problem description, we may need to look for specific keywords that will help us decide what model is to be used to explain the situation. We look at two specific examples.

        Example 1: Plane crashes involving the loss of life may be assumed to occur at random at an average rate, for all major airlines taken together, of 1.5 per year.

        We will use Poisson distribution as the model, because of the presence of the keywords "average rate". In particular, lambda = 1.5.

        Example 2: Samples, each of 8 articles, are taken at random from a large consignment in which 20% of articles are defective.

        We will apply Binomial distribution in this situation, because of the keywords "8 articles" and "20%...defective". In particular, n = 8 and p = 0.2.

        More points will come in my next posting. Thanks.

      Wen Shih

    • Hi,

        This is part 2 of our earlier discussion.

        1. Let's look at another example, where we will determine the distribution that the random variable should likely follow:

        An airline always sells 108 tickets for its daily 6 a.m. flight from Singapore to Los Angeles and, on average, 4% of customers who have purchased tickets do not turn up. Let X be the number of customers who do not turn up for this flight. State the distribution of X and give a necessary assumption for the model to be valid.

        The problem description gives us two important pieces of information: "108 tickets" and "4% of customers...do not turn up". So we can write down the distribution of X, i.e.,

        X ~ B(108, 0.04).

        For the Binomial model to hold, it is necessary for us to assume that customers are not turning up independently of one another.

        2. The purpose of using models is to compute probabilities and to better understand likelihoods of events.

        3. We use a probability distribution function (PDF in short) to find the probability of a single event, i.e., P(X = x). Both Binomial and Poisson models have their formulae listed in the MF15 booklet for reference.

        4. To calculate the probability of a number of possible events up to a particular point, i.e.,

        P(X <= x) = P(X = 0) + P(X = 1) + ... + P(X = x),

      we use a cumulative distribution function (CDF in short). Note that the word "cumulative" suggests the need for us to add items.

        5. We will use the following TI graphic calculator commands to find probabilities:

        (a) binompdf(n, p, x) is equivalent to P(X = x),
            where X ~ B(n, p).

        (b) binomcdf(n, p, x) is equivalent to P(X <= x),
            where X ~ B(n, p).

        (c) poissonpdf(lambda, x) is equivalent to P(X = x),
            where X ~ Po(lambda).

        (d) poissoncdf(lambda, x) is equivalent to P(X <= x),
            where X ~ Po(lambda).

        6. In O-level E. Maths., we have come already across the notions of central tendency (i.e., mean, mode and median) and spread (i.e., variance or standard deviation). In H2 Maths., we study the mean and variance of the random variable X, i.e., E(X) and Var(X) respectively:

        (a) If X ~ B(n, p), then
            E(X) = np and Var(X) = np(1 - p).

        (b) If X ~ Po(lambda), then
            E(X) = Var(X) = lambda.

        These results can be found in MF15.

        7. We now look at some useful results of mean and variance of any distribution:

        (a) E(a) = a, where a is some constant.

        (b) E(aX) = a E(X), where a is some constant.

        (c) E(aX +/- bY) = a E(X) +/- b E(Y), where a and b are some constants.

        (d) Var(a) = 0, where a is some constant.

        (e) Var(aX) = a^2 Var(X), where a is some constant.

        (f) Var(aX +/- bY) = a^2 Var(X) + b^2 Var(Y),
            where a and b are some constants
            and X, Y are independent random variables.

        8. It is interesting to ask this question: If X and Y follow Poisson distributions, is X + Y still a Poisson-related random variable? It turns out that the answer is positive. So we have the following result:

        If X ~ Po(lambda_1) and Y ~ Po(lambda_2), then
        X + Y ~ Po(lambda_1 + lambda_2).

        Note: The answer to the question is negative if X and Y follow Binomial distributions.

        9. We end this discussion with 2 final examples, which deal with the need to interpret English descriptions into their equivalent mathematical forms.

        Example: A baker, on average, sells 4 out of 5 muffins a day. He prepares enough ingredients to sell exactly 200 muffins a day. The cost price of the ingredients for 200 muffins is $270. In a randomly chosen day, find the probability that he can make a profit if he were to sell the muffins at $1.80 each.

        Discussion: Let X be the number of muffins sold in a day, out of 200. Then
        X ~ B(200, 4/5).

      We need to sell more than some number of muffins, say n, in order to make a profit. So
        (1.80)n > 270 (i.e., sales > cost)
        => n > 150.

      Thus, the probability of making a profit gets translated in mathematics as
        P(X > 150).

        Example: A garage has 2 vans and 3 cars, which can be hired out for a day at a time. Requests for the hire of a van follow a Poisson distribution with a mean of 1.5 requests per day and requests for the hire of a car follow an independent Poisson distribution with a mean of 4 requests per day.

        (i) Find the least number of vans that the garage should have so that, on any particular day, the probability that a request for the hire of a van for that day has to be refused is less than 0.1.

        (ii) Find the probability that, on any particular day, there is at least one request for a van and at least two requests for a car, given that there are a total of 4 requests on that day.

        Discussion of (i): Let V be the number of requests for vans per day. Then
        V ~ Po(1.5).

      Let n be the least number. A request is refused when requests (i.e. demand) exceed n (i.e. supply). So
        P(V > n) < 0.1
      is the mathematical expression we need to arrive at. Upon simplification, we have
        1 - P(V <= n) < 0.1
        => P(V <= n) > 0.9,
      from which the calculator can be used to find the desired n, by checking different values of n.

        Discussion of (ii): Let C be the number of requests for cars per day. Then
        C ~ Po(4).

      The keyword "given" indicates a conditional probability, which you would have studied earlier:
        P(V >= 1 and C >= 2 | V + C = 4),
      which can be simplified to
        [P(V >= 1 and C >= 2 and V + C = 4)] / P(V + C = 4).

      The numerator may further be broken down into two possible cases that satisfy the 3 conditions:
        P(V = 1 and C = 3) and P(V = 2 and C = 2).

      Now P(V = 1 and C = 3) = P(V = 1).P(C = 3), since V, C are independent random variables. The same principle is applied to find P(V = 2 and C = 2).

      To find P(V + C = 4), we note that V + C ~ Po(5.5), because the sum of 2 Poisson random variables gives a Poisson random variable, as we have pointed out in point (8).


      Wen Shih

    • Hi,

        I would encourage all students to look closely at the H2 Mathematics syllabus document on SEAB's site:


        On page 4 of the document, the syllabus describes 3 assessment objectives:

        A01 understand and apply mathematical concepts and skills in a variety of contexts, including the manipulation of mathematical expressions and use of graphic calculators;

        A02 reason and communicate mathematically through writing mathematical explanation,arguments and proofs, and inferences;

        A03 solve unfamiliar problems; translate common realistic contexts into mathematics; interpret and evaluate mathematical results, and use the results to make predictions, or comment on the context.

        With these objectives explicitly stated, it makes sense to me why certain types of questions appear in the examination. I will now cite specific examples from the latest exam that support the objectives.

        Examples in support of A01: Paper 2/Q7(ii) in which the concept of differentiation was applied in the context of probability. Paper 1/Q11(iii) in which the concept of limits was necessary to find an improper integral.

        Examples in support of A02: Paper 1/Q9(iii) and Q10(iii) in which students were asked to provide explanations.

        Example in support of A03: Paper 1/Q4(i), (ii) in which students were given an unfamiliar piecewise function that is periodic.

        Thank you.

      Wen Shih

    • Hi,

        I have developed a checklist of concepts and skills for H1 Maths, based on the latest syllabus. May teachers and students find it helpful :)

        Here's the link to the document:



      Wen Shih

    • Hi,

        How can one maximise learning after completing a tutorial set of questions on a particular topic? Here are some general questions to enable one to reflect on:

        1. What types of question are covered for the topic?

        2. What other topics go along with this topic?

        3. What are the important keywords encountered within the questions?

        4. What are my common mistakes and areas of difficulty?

        5. What are the common or unusual working steps observed?

        6. Under what situations was the graphic calculator used in problem solving?

        7. Can I see any trends in the questions set by my school or set by Cambridge/SEAB?

        8. Can I get further insights from the discussions with friends, classmates and teachers?

        9. Can I brainstorm about possible questions that may be asked?

        I believe these questions, with the exception of Q6, can be applied to most subjects :)


      Wen Shih

    • Hi,

        In this post, I wish to clarify a common misconception from the topic of Functions. Many students think (or may have been taught) that the range of the composite function fg is the range of f, which is incorrect in general.

        Consider the following example involving two given functions:

        f(x) = x + 1, x is real,
        g(x) = e^x, x is real.

        fg exists because range of g = (0, \infty) is a subset of the real set = domain of f.

        Now range of fg = (1, \infty) but range of f is the real set, so that both are unequal.

        If the range of g is equal to the domain of f, then the range of fg is exactly the range of f. You may convince yourself with this next example:

        f(x) = x + 1, x > 0,
        g(x) = e^x, x is real.

        Thank you.

      Wen Shih

      Edited by wee_ws 05 May `10, 8:19AM
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