Last Hustle for A’levels 2017

Last Hustle for A’levels 2017

JC General Paper, JC Mathematics, Mathematics

As we are all busy counting down to A’levels, The Culture SG Team will like to share the preparatory course that we have for students.
The lessons will all be $70 for each session and the max class size will be 15 students.

Lessons will be held at:
Newton Apple Learning Hub
Blk 131, Jurong Gateway Road #03-263/265/267 Singapore 600131
Tel: +65 6567 3606

You may contact Tutor KS or Tutor Christine for further questions.

Details are as follow:

Last Hustle for A’levels

“Lets’ Hustle!”

Some TYS Questions worth looking at

Some TYS Questions worth looking at

JC Mathematics

Prelims Exams was scary. H2 Mathematics isn’t that easy.

Students that had difficulties finishing their prelims exams, should consider working on their time management. The best way to do it, practice 3 hour paper… in a single sitting. And students should note to modify their TYS slightly as several questions in each paper are out of syllabus. In general, we give ourselves 1.5min for every 1 mark.

So here, I’ll share a list of questions that Mr. Wee has compiled. Mr. Wee also wrote e-books recently on solving non-routine problems. They are very interesting and provides the learners a new perspective to solving problems.

Non-routine Problems (Click to link to the solutions)

Application Questions
Specimen P1/Q9
Specimen P1/Q11
Specimen P2/Q9
Specimen P2/Q10

All the best for your revision!

APGP Interest Rate Question

APGP Interest Rate Question

JC Mathematics, Mathematics

This is a question from TJC Promotion 2017 Question 10. Thank you Mr. Wee for sharing.

Mr. Scrimp started a savings account which pays compound interest at a rate of r% per year on the last day of each year.

He made an initial deposit of $ x on 1 January 2000. From 1 January 2001 onwards, he makes a deposit of $ x at the start of each year.

(i) Show that the total amount in the savings account at the end of the n ^{th} year is $ \frac{k}{k-1} (k^n - 1)x , where k = \frac{100 + r}{100}.

(ii) At the end of the year 2019, Mr. Scrimp has a total amount of $ 22x in the savings account. Find the value of r, giving your answer correct to one decimal place.

Assume that the last deposit is made on 1 January 2019 and that the total amount in the savings account is $50000 on 1 January 2020.

For a period of N years, where 1 \le N \le 20, Mr. Scrimp can either continue to keep this amount in the savings account to earn interest or invest this amount in a financial product. The financial product pays an upfront sign-up bonus of $2000 and a year-end profit of $200 in the first year. At the end of each subsequent year, the financial product pays $20 more profit that in the previous year.

(iii) Find the total amount Mr. Scrimp will have at the end of N years if he invests in the financial product.

(iv) Using the value of r found in (ii), find the maximum number of years Mr. Scrimp should invest in the financial product for it to be more profitable than keeping the money in the savings account.

(1 + \frac{r}{100})x  + (1 + \frac{r}{100})^2 x + (1 + \frac{r}{100})^3 x + \ldots + (1 + \frac{r}{100})^n x

= \frac{(1 + \frac{r}{100})x[(1 + \frac{r}{100})^n - 1]}{(1 + \frac{r}{100}) - 1}

= \frac{(\frac{100 + r}{100})x[(\frac{100 + r}{100})^n - 1]}{(\frac{100 + r}{100}) - 1}

Let k = \frac{100 + r}{100}

\Rightarrow \frac{k x[k^n - 1]}{k - 1}

Total Amount = \$ [\frac{k}{k-1}(k^n - 1 ) x ]


\frac{k}{k-1}(k^{20} - 1 ) x  = 22 x

\frac{k}{k-1}(k^{20} - 1 )  = 22

Using GC, k = 1.0089905

\Rightarrow \frac{100 + r}{100} = 1.0089905

\therefore r = 0.89905 \approx 0.9 (1 decimal place)


Total amount = 50000 + 2000 + \frac{N}{2}[2(200) + (N-1)(2)]

= 52000 + N(200 + N-1)

= 52000 + 199N + N^2


With the savings account, he has $ [50000(1.009)^N] at the end of N years.

With the financial product, he has $ [52000 + 199N + N^2] at the end of N years.

For it to be more profitable,

[52000 + 199N + N^2] > [50000(1.009)^N]

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}))

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.

Trigonometry Formulae & Applications (Part 1)

Trigonometry Formulae & Applications (Part 1)

JC Mathematics, Secondary Math

Upon request by some students, I’ll discuss a few trigonometry formulae here and also some of their uses in A’levels. I’ve previously discussed the use of factor formulae here under integration.

I’ll start with the R-Formulae. It should require no introduction as it is from secondary Add Math. This formulae is not given in MF26, although students can derive it out using existing formulae in MF26.

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.

Here is a quick example,

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

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

Here, we observe, we have to use the R-Formulae where

R = \sqrt{3^2 + 2^2} = \sqrt{13}

\alpha = \text{tan}^{\text{-1}} (\frac{2}{3})

We have that

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

I’ll end with a question from HCI Midyear 2017 that uses R-Formulae in one part of the question.

A curve D has parametric equations

x = e^{t} \text{sin}t, y = e^{t} \text{cos}t, \text{~for~} 0 \le t \le \frac{\pi}{2}

(i) Prove that \frac{dy}{dx}  = \text{tan} (\frac{\pi}{4} - t).

I’ll discuss about Factor Formulae soon.  And then the difference and application between this two formulae.

Random Questions from 2017 Prelims #2

Random Questions from 2017 Prelims #2

JC Mathematics, Mathematics, Secondary Math

Today I’ll share a question that came out of CJC Prelim 2017 Paper 1 for H2 Mathematics 9758. I think some of my student would have seen this question before and we discussed it in class before. Very technical question. This is question 11, I’ll share only the first part which is on the application of ratio theorem or mid point theorem. The second part is on application: Ray Tracing which I’ll discuss in class.

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. For the triangle show below, O, A and B are vertices, where O is the origin. \vec{OA} = a and \vec{OB} = b. The midpoints of OB, OA and AB are M, N and T respectively.

It is given that X is the point of intersection between the medians of triangle OAB from vertices A and B.

(i) Show that \vec{OX} = \frac{1}{3} (a +b)

(ii) Prove that X also lies on OT, the median of triangle OAB from vertex O.

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!