Thoughts on H2 Mathematics (9758) 2017 Paper 1

Thoughts on H2 Mathematics (9758) 2017 Paper 1

JC Mathematics, Mathematics

This is a new syllabus and this is the first time it will be tested. Personally, I don’t think it will be easy and students should not underestimate this upcoming A’levels. And I’m referring to the A’levels, on the whole. We saw how the Science Paper 4 were… unexpected.

The new H2 Mathematics (9758) syllabus has several topics removed, and these were mostly topics that were “drill-able”, aside from complex numbers. The new syllabus added in mainly, new integration forms, focus on parametric Equations with Cartesian equations, and of course, Discrete R.V. But let us leave the statistics out.

Students should familiarise themselves with the trigonometry Formulae in MF26. There are several topics that can be linked up with trigonometry, makes me wonder why it isn’t a chapter by itself. Complex numbers has a trigonometry form too, so make sure students know how to manipulate it, given the trigonometry Formulae.

Next, students should understand the use of Maclaurin’s. What does it mean for x to be small, and the implications when they say h is small compared to R… And also finding the general term of a Maclaurin’s Expansion.

It won’t hurt to review how to find the Area using Shoe-lace method. And not forgetting our Sine Rule and Cosine Rule.

Do know how to prove a one-one function… Non-graphically. (i.e. not using the Horizontal Line Test)

Do know that the oblique asymptote of f(x) becomes y=0 when we do the y = \frac{1}{f(x)} transformation too.

Lastly, students must READ really carefully and discern every information. Having marked many scripts, many students do not read carefully and lose marks here and there. And they do add up… Be alert and read, take note of the forms that they want. Here are 10 little things to take note when you read the question.

  1. Cartesian/ Polar/ Exponential for complex
  2. Scalar/ Parametric/ Cartesian for vectors
  3. Set/ range/ interval of values
  4. Algebraically => show all the workings without a GC.. usually discriminant, completing the square or maybe some differentiation will be involved.
  5. Without using a calculator => show your workings and check with a GC (secretly)
  6. Decimal places, etc…
  7. Rounding off when you’re dealing with an inequality
  8. Units used in the questions, (ten thousands, etc)
  9. Rate of change; leaking means the rate is negative…
  10. All answers should be in 3 SF UNLESS OTHERWISE STATED. Degrees to 1 DP. RADIANS to 3 SF.

Have fun and all the best!

Random Questions from 2017 Prelims #3

Random Questions from 2017 Prelims #3

JC Mathematics

Here is another question that is from CJC H2 Mathematics 9758 Prelim Paper 1. Its a question on differentiation. I think it is simple enough and tests student on their thinking comprehension skills. This is question 6.

A straight line passes through the point with coordinates (4, 3) cuts the positive x-axis at point P and positive y-axis at point Q. It is given that \angle PQO = \theta, where 0 < \theta < \frac{\pi}{2} and O is the origin.

(i) Show that equation of line PQ is given by y = (4-x) \text{cot} \theta +3.

(ii) By finding an expression for OP + OQ, show that as \theta varies, the stationary value of OP + OQ is a + b \sqrt{3}, where a and b are constants to be determined.

Integration By Parts

Integration By Parts

JC Mathematics

Today, I’ll share a little something about Integration by Parts. I want to share this because I observe that several students are over-reliant on the LIATE to perform Integration by Parts. This caused them to overlook/ appreciate its use.

Lets start by reviewing the formula for Integration by Parts

\int v \frac{du}{dx} ~dx = uv - \int u \frac{dv}{dx} ~dx

I like to share with students that Integration by Parts have two interesting facts.

  1. It allows us to integrate expression that we cannot integrate. Eg. \text{ln}x or inverse trigonometric functions. This is also closely related to how LIATE is established.

  2. This is closely related to point 1. That is, instead of trying to integrate the expression, we are differentiating it instead. And that itself, is a very important aspect.

Next, I like to point out to students that LIATE is a general rule of thumb.
GENERAL – because it does not work 100% of the times. And today, I’ll use an example to illustrate how LIATE actually fails.

\int \frac{te^t}{(t+1)^2} ~ dt

Here we observe two terms \frac{t}{(t+1)^2} and e^t. Going by LIATE, we should be differentiating \frac{t}{(t+1)^2} and integrating e^t.

\int t^2 e^{t} ~ dt
= \frac{t}{(t+1)^2} e^{t} - \int \frac{(1)(t+1)^2 - t(2)(t+1)}{(t+1)^4} e^t ~dt

Now observe what happened here, after applying integration by parts, it got “worst”. We are stuck to integrating \frac{(1)(t+1)^2 - t(2)(t+1)}{(t+1)^4} with an exponential… This should raise some question marks. But we did follow LIATE.

I’ll leave students to try out this on their own. And feel free to ask questions here. Have fun!

Random Questions from 2017 Prelims #1

Random Questions from 2017 Prelims #1

JC Mathematics

Last year, I shared a handful of random interesting questions from the 2016 Prelims. Students feedback that they were quite helpful and gave them good exposure. I thought I share some that I’ve seen this year. I know, its a bit early for Prelims. But ACJC just had their paper 1. 🙂

This is from ACJC 2017 Prelims Paper 1 Question 7. And it is on complex numbers.

(a) Given that 2z + 1 = |w| and 2w-z = 4+8i, solve for w and z.

(b) Find the exact values of x and y, where x, y \in \mathbb{R} such that 2e^{-(\frac{3+x+iy}{i})} = 1 -i

I’ll put the solutions up if I’m free.

But for students stuck, consider checking this link here for (a) and this link here for (b). These links hopefully enlightens students.

Just FYI, you cannot \text{ln} complex numbers as they are not real…

Probability Question #4

Probability Question #4

JC Mathematics

A gambler bets on one of the integers from 1 to 6. Three fair dice are then rolled. If the gambler’s number appears k times (k = 1, 2, 3), he wins $ k. If his number fails to appear, he loses $1. Calculate the gambler’s expected winnings

Problem Solving Tips #1

Problem Solving Tips #1

JC Mathematics, Studying Tips

So these past months, I have been focusing on harnessing students’ abilities to interpret questions properly. Every line in a question is there for a reason and hold little pieces of information. As a student, you need to piece these information together.

Take the following question as an example:

We have a, b and c=a+2b.
Given further that M is on OC, and point A, B, and M are collinear. Find the ratio of OM:OC.

Now this question looks rather short. Many students will first start by drawing to help them see. To be honest, I was tell students that drawing out vectors is not necessary since it doesn’t yield any marks and we spend 10 minutes trying to figure out how it is supposed to look. We are better learning how to read questions.

Let’s start dissecting

M is on OC tells us that \vec{OM} = \lambda \vec{OC}.

A, B, and M are collinear tells us that the points are parallel with a common point. NOTE: Collinear is different from parallel. The former is a proper subset of the latter actually.
This tells us that \vec{AB} = \mu \vec{AM}

Students should have no trouble continuing to solve for \lambda and \mu.

So this little exercise is to simply illustrate the importance of learning how to read questions and of course, writing it out.

Finding the coefficient of terms

JC Mathematics

This problem seem to bug many students, especially J1s who are doing Maclaurin’s series now. Many of the students are not sure how to find the coefficient of x^r. They wonder how to “see” it. So here, I’ll attempt to show a direct method that doesn’t require us to stare and think.

Consider (2+x)^{-3} and we are interested to find the coefficient of x^r.

First of all, to apply (1+x)^n formula, we rewrite to 2^{-3}(1+\frac{x}{2})^{-3}. This step is based on indices and students should always check that the copied the correct power.

Credits: MF15
Credits: MF15

Now consider the formula above that is found in the MF15. We will take the coefficient of x^r that’s found there, taking note that n = -3 here.

We will end of with the following

(2)^{-3}[\frac{(-3)(-4) \ldots (-3-r+1)}{r!}](\frac{x}{2})^{r}

Take note we must copy the (2)^{-3} alongside too and we substitute the n in and preserve the r.

Simplifying, we have

(2)^{-3}[\frac{(-3)(-4) \ldots (-2-r)}{r!}](\frac{x}{2})^{r}

(2)^{-3}[\frac{(-1)^{r}(3)(4) \ldots (r+2)}{r!}](\frac{x}{2})^{r}

Here we attempt to factorise (-1)^r out, we need to figure how many (-1)’s we have here to factorise. To find out, we can take (r+2)-3+1 = r to find out how many terms there are. Next, we try to simplify the factorials.

(2)^{-3}[\frac{(-1)^{r}(3)(4) \ldots (r+2)}{r!}](\frac{1}{2})^{r}x^{r}



Therefore, the coefficient is (2)^{-4-r}[(-1)^{r}(r+1)(r+2)].

Note: (-1)^r cannot be further simplified as we do not know if r is even or odd. In the event that r is even, then (-1)^r = 1 and if r is odd, then (-1)^r = -1. An example is if we consider (2-x)^{-3} instead of (2+x)^{-3}, we observe that we will find a (-1)^{2r} which is equals to 1 since (-1)^2r = ((-1)^2)^r = 1^r = 1.

Topics for A-levels H2 Mathematics

JC Mathematics

The following are a list of A-levels topics in H2 Mathematics. I’ve added what I feel is the hardest part of each topic amongst students.

  1. Functions: Finding the range of a composite
  2. Curve Sketching: Sketching curves with unknowns and solving conics’ roots.
  3. Graphing Transformations: Finding the correct transformation the new graph went through from the original.
  4. Inequalities: When there are modulus
  5. Arithmetic Progression & Geometric Progression: Finding the general formula for those interest rate questions
  6. Summation and Series: MI with Trigo
  7. Recurrence Relation: Conjecture
  8. Differentiation: Parametric equations
  9. Maclaurins’ Series: “Hence” parts which require students to either approximate or relate to standard series.
  10. Integration: With trigonometry
  11. Definite Integral: Identify the region they want and crafting the correct rotation formula
  12. Differential Equations: with given substitution. General vs Particular Solution
  13. Vectors I: When no numbers are given, and its just all a, b, c.
  14. Vectors II: Distance between two planes, geometrical interpretations.
  15. Complex Numbers: Locus of half line. Questions combined with vectors.
  16. Probability: story questions which students are still unable to identify the clear events
  17. Permutation & Combination: Honestly, don’t spend too long on this questions, beyond 15 minutes. There are too many ways you can suffer at the hands of this question. I can’t even permute it.
  18. Normal Distribution: Continuity Correction after approximation
  19. Binomial & Poisson Distribution: When we need to find X - Y for a passion distribution we know that we cannot take the difference of two poisson distributions.
  20. Hypothesis Testing: Wrong alternate hypothesis
  21. Correlation & Regression: Choosing the appropriate model

The Modulus Sign #3

JC Mathematics

Here, we shall discuss something called the Triangle inequality which states that |x+y| \le |x| + |y|. I thought this is rather important as I see too many students writing things like |z_1 + z_2| = |z_1| + |z_2|$ in complex numbers. This is really scary because it shows that they are not thinking when they are doing the question.

Students can argue that adding two number and applying modulus should give positive so we can split the positive numbers up. Now this is sufficiently true. But then what if we have negative numbers coming into play? Consider this

|2+(-2)| = 0

Then we see that |2|+|-2| = 4.

So this is a big problem as 4 \ne 0 and thus we cannot assume that relation above! So students please be really careful!

The Modulus Sign #2

Secondary Math

So now let us look at a simple relationship that many students memorise instead of understand. Below, we see a absolute function graph.

Credits: Wikipedia
Credits: Wikipedia

Now we all know that |x|<a gives us -a<x<a. So how did that happen? Consider drawing a horizontal line, y=a in the graph above. |x| is less than a only when -a<x<a. Similarly, |x|>a is true only when x>a or x<-a. I do hope that the graphical representation helps here.