Making Use of this September Holidays

Making Use of this September Holidays

JC Mathematics, Mathematics

This is a little reminder and advice to students that are cheong-ing for their Prelims or A’levels…

For students who have not taken any H2 Math Paper 1 or 2, I strongly advise you start waking at up 730am and try some papers at 8am. I gave my own students similar advices and even hand them 4 sets of 3 hours practice papers. Students need to grind themselves to be able to handle the paper at 8am. It is really different. Not to mention, this September Holidays is probably your last chance to be able to give yourself timed practices.

For students who took H2 Math Paper 1, you might be stunned with the application questions that came out. For NJC, its Economics. For YJC, its LASER. For CJC, a wild dolphin appeared. And more. These application questions are possible, due to the inclusion of the problems in real world context in your syllabus. You can see the syllabus for yourself. I’ve attached the picture below. So for Paper 2, expect these application questions to be from statistics mainly, as suggested in your scheme of work below.

Scheme of Examination Source: SEAB

For students that have took H2 Math paper 1 & 2, and this is probably ACJC. The paper was slightly stressful, given the mark distributions, but most of the things tested are still technically “within syllabus”. For one, the directional cosine question, is a good reminder to students that they should not leave any pages un-highlighted. AC students should be able to properly identify their weaknesses and strengths this time round. If its time management, then start honing that skill this holidays – by having timed practice. A quick reminder that the TYS papers are not 3 hours, since some of the questions are out of H2 Mathematics 9758 syllabus. Students can consider the ratio of 1 mark to 1.5 min to gauge how much time they have for each paper.

R-Formulae seems to be popular about the prelims exams this time round, making waves in various schools. Perhaps it was because it appeared in the specimen paper, and if you’re keen on how it can be integrated or need a refresher. I did it recently here.

Lastly, for the students that are very concerned on application questions. Check the picture below. It contains some examples that SEAB has given. Students should also be clear about the difference between a contextual question and an application question.

Integration & Applications Source: SEAB

With that, all the best to your revision! 🙂

Trigonometry Formulae & Applications (Part 2)

Trigonometry Formulae & Applications (Part 2)

JC Mathematics, Secondary Math

I meant to share more on factor Formulae today. However, a few students are not so sure how to get the R-formulae correctly during their preliminary exams recently. So I thought that I’ll share how they can derive the R-Formulae from the MF26.

The following is the R-Formulae which students should have memorised. It is under assumed knowledge, just saying…

a \text{cos} \theta \pm b \text{sin} \theta = R \text{cos} (\theta \mp \alpha)

a \text{sin} \theta \pm b \text{cos} \theta = R \text{sin} (\theta \pm \alpha)

where R = \sqrt{a^2 + b^2} and \text{tan} \alpha = \frac{b}{a} for a > 0, b > 0 and \alpha is acute.

So here, I’ll write the addition formulae that’s found in MF26.

\text{sin}(A \pm B) \equiv \text{sin}A \text{cos} B \pm \text{cos} A \text{sin} B

\text{cos}(A \pm B) \equiv \text{cos}A \text{cos} B \mp \text{sin} A \text{sin} B

I’ll use an example I discussed previously.

f(x) = 3 \text{cos}t - 2 \text{sin}t

Write f(x) as a single trigonometric function exactly.

Lets consider the formulae from MF26.

\text{cos}(A \pm B) \equiv \text{cos}A \text{cos} B \mp \text{sin} A \text{sin} B

R\text{cos}(A \pm B) \equiv R \text{cos}A \text{cos} B \mp R \text{sin} A \text{sin} B

We can let

3 = R \text{cos} B ---(1)

2 = R \text{sin} B ---(2)

\Rightarrow \sqrt{ 3^2 + 2^2 } = \sqrt{ R^2 \text{cos}^2 B + R^2 \text{sin}^2 B}

\Rightarrow \sqrt{13} = R

\Rightarrow \frac{R \text{sin} B}{R \text{cos} B} = \frac{2}{3}

\Rightarrow \text{tan} B = \frac{2}{3}

Putting things together, we have that

 f(x) = \sqrt{13} \text{cos} ( t + \text{tan}^{\text{-1}} (\frac{2}{3}))

Students, test your Vectors!

Students, test your Vectors!

JC Mathematics

As the prelims examinations draw really close, many students were asking me to give questions to test their concepts for several topics. In class, I had the opportunity to explore several applications questions too. We saw several physics concepts mixed. We also have some conceptual questions that need students to be able to use the entire topic to solve it.

So I’ll share one here. This involves several concepts put together. I’ll put the solution up once I find the time. Concepts that will be involved, will be

  1. Vector Product
  2. Equations of Plane
  3. Finding foot of perpendicular of point

The question in one a reflection of a plane in another plane. I think such questions will come out in a few guided steps in exams. But should a student be able to solve it independently, it shows that he has good understanding.

The plane p has equation x + y + z = 9 and the plane p_1 contains the lines passing through (0, 2, 3) and are parallel to (1, -1, 0) and (0, 1, 1) respectively. Find, in scalar product form, the equation of the plane which is the reflection of p_1 in p.

Quick Summary (Probability)

Quick Summary (Probability)

JC Mathematics, Mathematics, University Mathematics

University is starting for some students who took A’levels in 2016. And, one of my ex-students told me to share/ summarise the things to know for probability at University level. Hopefully this helps. H2 Further Mathematics Students will find some of these helpful.


Random Variables

Suppose X is a random variable which can takes values x \in \chi.

X is a discrete r.v. is \chi is countable.
\Rightarrow p(x) is the probability of a value of x and is called the probability mass function.

X is a continuous r.v. is \chi is uncountable.
\Rightarrow f(x) is the probability density function and can be thought of as the probability of a value x.

Probability Mass Function

For a discrete r.v. the probability mass function (PMF) is

p(a) = P(X=a), where a \in \mathbb{R}.

Probability Density Function

If B = (a, b)

P(X \in B) = P(a \le X \le b) = \int_a^b f(x) ~dx.

And strictly speaking,

P(X = a) = \int_a^a f(x) ~dx = 0.

Intuitively,

f(a) = P(X = a).

Properties of Distributions

For discrete r.v.
p(x) \ge 0 ~ \forall x \in \chi.
\sum_{x \in \chi} p(x) = 1.

For continuous r.v.
f(x) \ge 0 ~ \forall x \in \chi.
\int_{x \in \chi} f(x) ~dx = 1.

Cumulative Distribution Function

For discrete r.v., the Cumulative Distribution Function (CDF) is
F(a) = P(X \le a) = \sum_{x \le a} p(x).

For continuous r.v., the CDF is
F(a) = P(X \le a ) = \int_{- \infty}^a f(x) ~dx.

Expected Value

For a discrete r.v. X, the expected value is
\mathbb{E} (X) = \sum_{x \in \chi} x p(x).

For a continuous r.v. X, the expected value is
\mathbb{E} (X) = \int_{x \in \chi} x f(x) ~dx.

If Y = g(X), then

For a discrete r.v. X,
\mathbb{E} (Y) = \mathbb{E} [g(X)] = \sum_{x \in \chi} g(x) p(x).

For a continuous r.v. X,
\mathbb{E} (Y) = \mathbb{E} [g(X)] = \int_{x \in \chi} g(x) f(x) ~dx.

Properties of Expectation

For random variables X and Y and constants a, b, \in \mathbb{R}, the expected value has the following properties (applicable to both discrete and continuous r.v.s)

\mathbb{E}(aX + b) = a \mathbb{E}(X) + b

\mathbb{E}(X + Y) = \mathbb{E}(X) + \mathbb{E}(Y)

Realisations of X, denoted by x, may be larger or smaller than \mathbb{E}(X),

If you observed many realisations of X, \mathbb{E}(X) is roughly an average of the values you would observe.

\mathbb{E} (aX + b)
= \int_{- \infty}^{\infty} (ax+b)f(x) ~dx
= \int_{- \infty}^{\infty} axf(x) ~dx + \int_{- \infty}^{\infty} bf(x) ~dx
= a \int_{- \infty}^{\infty} xf(x) ~dx + b \int_{- \infty}^{\infty} f(x) ~dx
= a \mathbb{E} (X) + b

Variance

Generally speaking, variance is defined as

Var(X) = \mathbb{E}[(X- \mathbb{E}(X)^2] = \mathbb{E}[X^2] - \mathbb{E}[X]^2

If X is discrete:

Var(X) = \sum_{x \in \chi} ( x - \mathbb{E}[X])^2 p(x)

If X is continuous:

Var(X) = \int_{x \in \chi} ( x - \mathbb{E}[X])^2 f(x) ~dx

Using the properties of expectations, we can show Var(X) = \mathbb{E}(X^2) - \mathbb{E}(X)^2.

Var(X)
= \mathbb{E} [(X - \mathbb{E}[X])^2]
= \mathbb{E} [(X^2 - 2X \mathbb{E}[X]) + \mathbb{E}[X]^2]
= \mathbb{E}[X^2] - 2\mathbb{E}[X]\mathbb{E}[X] + \mathbb{E}[X]^2
= \mathbb{E}[X^2] - \mathbb{E}[X]^2

Standard Deviation

The standard deviation is defined as

std(X) = \sqrt{Var(X)}

Covariance

For two random variables X and Y, the covariance is generally defined as

Cov(X, Y) = \mathbb{E}[(X - \mathbb{E}[X])(Y - \mathbb{E}[Y])]

Note that Cov(X, X) = Var(X)

Cov(X, Y) = \mathbb{E}[XY] - \mathbb{E}[X] \mathbb{E}[Y]

Properties of Variance

Given random variables X and Y, and constants a, b, c \in \mathbb{R},

Var(aX \pm bY \pm b ) = a^2 Var(X) + b^2 Var(Y) + 2ab Cov(X, Y)

This proof for the above can be done using definitions of expectations and variance.

Properties of Covariance

Given random variables W, X, Y and Z and constants a, b, \in \mathbb{R}

Cov(X, a) = 0

Cov(aX, bY) = ab Cov(X, Y)

Cov(W+X, Y+Z) = Cov(W, Y) + Cov(W, Z) + Cov(X, Y) + Cov(X, Z)

Correlation

Correlation is defined as

Corr(X, Y) = \dfrac{Cov(X, Y)}{Std(X) Std(Y)}

It is clear the -1 \le Corr(X, Y) \le 1.

The properties of correlations of sums of random variables follow from those of covariance and standard deviations above.

Thinking Math@TheCulture #2

Thinking [email protected] #2

JC Mathematics, Mathematics

[email protected] is a series of questions that we, as tutors feel that are useful in helping students think and improve their understanding.

Thinking [email protected] is curated by KS. More of him can be found here.


(i) Find the two possible values of z such that z^2 = 1 + \sqrt{3}i, leaving your answer in exact form a + bi, where a and b are real numbers.

(ii) Hence or otherwise, find the exact roots of the equation

2w^2 + 2 \sqrt{6}w + 1 - 2 \sqrt{3} i = 0

Junior College education: A path increasingly less traveled?

Junior College education: A path increasingly less traveled?

JC General Paper

Following up from the previous blog post about JC merger, it seems that there are more interesting and candid responses from the public. Here is one example that could interest you: <a href=”http://mothership.sg/2017/04/ex-nmp-calvin-cheng-makes-the-point-about-the-jc-mergers-that-no-one-wants-to-talk-about/”>http://mothership.sg/2017/04/ex-nmp-calvin-cheng-makes-the-point-about-the-jc-mergers-that-no-one-wants-to-talk-about/</a>

This begs the question whether the merger is an attempt to make JC education more exclusive/competitive, considering that 7 out of the 8 JCs involved in the merger “happened” to have the worst cut off points. Also, this attempt is seen as an elitist move to improve the quality of education for this system.

Considering this trend, poly education would be where the majority is heading towards… what then would be the purpose of JC education? We all know that for the more popular courses in poly such as accountancy, business admin and even engineering, some of the best students from the O levels are vying hard to get in. It is not uncommon to hear of 6 pointers who want to go to poly since they find the education there to be more broad-based,  Is it to train the academically inclined ppl to pursue academia in universities? would this system be slowly phased out in the years to come? (Many have criticized the JC education to be too theory-based; does not teach real life skills and applications. With all the technology-disruptions today and the move towards a sharing economy, where does the JC education fit?)

Thinking [email protected] #1

JC Mathematics, Mathematics

[email protected] is a series of questions that we, as tutors feel that are useful in helping students think and improve their understanding.

Thinking [email protected] is curated by KS. More of him can be found here.


Each card in a deck of cards bear a single number from 1 to 5 such that there are n cards bearing the number n, where n = 1, 2, 3, 4, 5. One card is randomly drawn from the deck. Let X be the number on the card drawn.

(i) Find the probability distribution of X.

(ii) Show that \mathbb{E}(X) = \frac{11}{3} and find \text{Var}(X).

Andrew draws one card from the deck, notes the number and replaces it. The deck is shuffled and Beth also draws on card from the deck and notes the number. Andrew’s score is k times the number on teh card he draws, while Beth’s score is the square of the number on the card she draws. Find the value of k so that the game is a fair one.

Things to note for 9758 H2 Mathematics

Things to note for 9758 H2 Mathematics

JC Mathematics, Studying Tips

It has been awhile since the A’levels. We talked and met up with several of our students. Some students are working and some are preparing their Personal Statements for overseas University Applications.

A few of them also remarked that they wish they put more efforts into studying A’levels.  An advice to this year JC2 – Do not wait till it is too late.

Students should also have a clear understanding of their syllabus, especially their scheme of exam. I still have JC2s this year who get stunned by the applications questions I threw at them (P.S. Aside from spending time with family, I wrote many sets of applied questions.). You may read more about your syllabus here. Or see the following images.

Integration & Applications Source: SEAB
Scheme of Examination Source: SEAB

How much practice or emphasis your school put on this, is up to them. But it is clear that application takes up 25% of your marks. The entire syllabus can be found here.

PLEASE DON’T WALK INTO THE EXAM HALL BLUR BLUR…

Checklist for Vectors

Checklist for Vectors

JC Mathematics

Many schools have been doing vectors recently. Thought I’ll share a little summary/ checklist I have done for my students.

Basic Concepts

  • Operations on Vectors
    • Addition & Subtraction
    • Scalar multiplication
    • Dot Product (Scalar)
      1. a • a = |a|2
      2. If a ⊥ b, then a • b = 0
      3. a • b = b • a
    • Cross Product (Vector)
      1. a × a = 0
      2. a × b = − b × a
  • Unit Vectors
  • Parallel Vectors ( a = k)
  • Collinear Vectors ( Parallel with a common point )
  • Ratio Theorem ( Found in MF26)
    • Midpoint Theorem
  • Directional Cosines
  • Angles between two Vectors
  • Length of Projection
  • Perpendicular Distance

Lines

  • Equations
    • Vector Form ( : r = a + λb, λ∈ ℜ )
    • Parametric Form
    • Cartesian Form
  • Line & Line
    • Parallel ( Directions are parallel to each other. )
    • Same ( Same Equations )
    • Intersecting ( There is a unique solution for λ and μ. )
    • Skewed ( Not parallel AND not Intersecting. )
  • Angle between two lines ( Angle between their directions )
  • Point & Line
    • Foot of Perpendicular
    • Perpendicular (Shortest) distance
    • Point on Line

Planes

  • Equations
    • Parametric Form ( π r = a + λb + μc, λ, μ ∈ ℜ )
    • Scalar Product Form ( r • n = a • n  = d )
    • Cartesian Form
  • Point & Plane
    • Foot of Perpendicular
    • Perpendicular (Shortest) distance
    • Distance from O to Plane
    • Point on Plane
    • Reflection of Point
  • Line & Plane
    • Relationships
      1. Parallel
        • Line intersects Plane entirely ( Infinite Solutions )
        • Do not intersect ( No Solution )
      2. Not Parallel
        • Intersects at a point ( One Solution )
    • Intersection Point
    • Angle between Line & Plane
    • Reflection of Line
  • Plane & Plane
    • Relationships
      1. Parallel
        • Same ( Infinite Solutions )
        • Do not intersect ( No Solution )
      2. Not Parallel
        • Intersects at a line ( Infinite Solutions )
    • Intersection Line ( Use of GC )
    • Angle between two Planes ( Angle between their normals )

2016 A Level H2 Physics (9646) Paper 1 Suggested Solutions

JC Physics

All solutions here are SUGGESTED. Casey will hold no liability for any errors. Comments are entirely personal opinions.

  1. B
  2. C
  3. B
  4. D
  5. D
  6. C
  7. D
  8. D
  9. B
  10. C
  11. A
  12. D
  13. D
  14. C
  15. C
  16. A
  17. D
  18. A
  19. B
  20. D
  21. Question 21 is a flawed question. When unpolarised light goes through a polarizer, the I is halved while the A is reduced by a factor of root 2. But based on the information Cambridge provides, the answer is C.
  22. D
  23. D
  24. C
  25. C
  26. B
  27. C
  28. A
  29. B
  30. B
  31. A
  32. C
  33. D
  34. B
  35. B
  36. B
  37. D
  38. C
  39. C
  40. C

Note to all: Casey will not respond to most of the comments as he is busy. You may contact him by SMS at  +65 9474 5005 if you have a burning question.

Feel free to explain the answers, if you are confident. Many thanks.