### Random Questions from 2017 Prelims #1

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.

7
(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

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

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

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.

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.

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

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

$(2)^{-4-r}[(-1)^{r}(r+1)(r+2)]x^{r}$

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

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

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

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

Now we all know that $|x| gives us $-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. Similarly, $|x|>a$ is true only when $x>a$ or $x<-a$. I do hope that the graphical representation helps here.

### An easier approach to remembering discriminant

I notice many students forget how discriminant works! I think I should inform A-level students that they need to know how it works and that is one thing they should not return to their O-level teachers. So I thought I share a bit on how to effectively, get it correct and also use it. At the same time, I hope to better the students’ understanding too.

Let us first look at the quadratic formula that is well engrained in our heads.

$x= \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Now do we see a very familiar formula in there? Yes, its our discriminant formula, $b^{2}-4ac$!!! So how does it work, alongside with roots. Let’s look at what a root is next, a root is loosely put, a solution to the equation. I hope it is slowly making sense.

When the discriminant is negative, we have a $\sqrt{-\mathrm{number}}$ which is imaginary, that means we have NO real roots!!

And when the discriminant is zero, we have $\sqrt{0}=0$ so we only have one single root, that is $-\frac{b}{2a}$.

BUT if we have a positive discriminant, than due to the $\pm{ }$ there, we end up with two roots.

So I do hope this clears up the air as to how we can relate some algebra with roots. 🙂

This is a question worth looking at from A-levels which test us on discriminant.

### Display of arithmetic skills

This is a video of an arithmetic challenge that took place in China in 2015. It is China against Japan, and the title is called the best brain of China. I shared a video on arithmetic previously, and a student mentioned that this only possible after years of training. So here is some kids showing us how math is done!

As you observe from the video, the participants are really young, as young as 9 years old! And the mental arithmetic skills exhibited here is truly insane. I shared this video to show students how “cool” is it to be able to do mental calculations really fast. At the same time, I hope to motivate students to brush up on their arithmetic skills.

From my experience, students with strong arithmetic backgrounds, are able to be more meticulous with their work. They can check their own work efficiently. Furthermore, it removes over-reliance on calculator. It pains me to see students do $2-6$ with calculator. Just a few days back, one students got a question wrong and realised she calculated it correctly but keyed in wrongly in calculator. She trusted the calculator and went ahead. This is really a waste, and I notice this in many students.

If I do have kids, I will hope to train their arithmetic skills up. And that will begin by putting the calculator away.

### The Modulus Sign #4

I received this from a student of mine who was really confused with the transition from line 2 to line 3. Students should give it a thought before reading my explanation below.

Firstly, my student felt that $|(2x-1)+2yi| \le |(x+1)+yi|$ was being squared on both sides to remove the modulus and the next step should be $[(2x-1)+2yi]^{2} \le [(x+1)+yi]^{2}$. She is very mistaken, as $(2x-1)+2yi$ is a complex number, which is a vector, and not a mere scalar. So the transition from step 2 to step 3 was really an evaluation of the modulus. If we do not skip any steps, the following should clarify how we got from step 2 to step 3.

$|(2x-1)+2yi| \le |(x+1)+yi|$

$\sqrt{(2x-1)^{2}+(2yi)^{2}} \le \sqrt{(x+1)^{2}+(yi)^{2}}$

$\sqrt{(2x-1)^{2}+(2yi)^{2}} \le \sqrt{(x+1)^{2}+(yi)^{2}}$

$(2x-1)^{2}+(2yi)^{2} \le (x+1)^{2}+(yi)^{2}$

I do hope this raise awareness for students to treat modulus carefully and question themselves, when they are dealing with a scalar or vector.