## Secondary Maths Tuition: Kinematics Question

Solution:

acceleration of car $=\frac{12}{6}=2m/s^2$

$\frac{v}{15-5}=2$

$v=2\times 10=20$

Let $T$ be the time (in seconds) when the car overtakes the truck.

Total distance travelled by car at T seconds = area under graph = $\frac{1}{2}(T-5)(2(T-5))$

($2(T-5)$ is the velocity of car at T seconds, it is obtained in the same way as we calculated v.)

Total distance travelled by truck at T seconds = $\frac{1}{2}(6)(12)+\frac{1}{2}(15-6)(12+16)+16(T-15)$

Equating the two distances will lead to a quadratic equation $T^2-26T+103=0$

Solving that gives $T=21.12s$ or $T=4.876s$ (rejected as car only starts at t=5)

$21.12-5=16.1s$ (3 s.f.)

## Make (y) the subject of the formulae (O Level Maths Tuition)

Question:

Make (y) the subject of the formulae

(1) $\displaystyle x =\frac{y}{y+1}$

Solution:

$\displaystyle \frac{x}{1}=\frac{y}{y+1}$

Cross multiply,

$xy+x=y$

$xy-y=-x$

$y(x-1)=-x$

$\displaystyle y=\frac{-x}{x-1}=\frac{x}{1-x}$

## O Level 2007 E Maths Paper 2 Q3 Solution (Geometry Question)

(a)
$\angle DAB=\angle DCB$
(opposite angles of parallelogram)

$\displaystyle\angle PAD=\frac{1}{2}\angle DAB=\frac{1}{2}\angle DCB=\angle RCB$
(shown)

(b)
$AD=BC$

$\angle PAD=\angle RCB$
(from part a)

$\angle ADC=\angle ABC$
(opposite angles of parallelogram)

$\displaystyle\angle ADP=\frac{1}{2}\angle ADC=\frac{1}{2} ABC=\angle RBC$

Thus, triangles ADP and CBR are congruent (ASA).

(c)(i)
$\angle ADC+\angle DAB=180^\circ$
$\angle DAP+\angle ADP=90^\circ$

$\angle DPA=180^\circ-90^\circ=90^\circ$
(shown)

(ii)
$\angle DAB+\angle ABC=180^\circ$

$\angle BAQ+\angle ABQ=90^\circ$

Considering the triangle ABQ,
$\angle PQR=180^\circ-90^\circ=90^\circ$
(shown)

## 积少成多: How can doing at least one Maths question per day help you improve! (Maths Tuition Revision Strategy)

We all know the saying “an apple a day keeps the doctor away“. Many essential activities, like eating, exercising, sleeping, needs to be done on a daily basis.

Mathematics is no different!

Here is a surprising fact of how much students can achieve if they do at least one Maths question per day. (the question must be substantial and worth at least 5 marks)

This study plan is based on the concept of 积少成多, or “Many little things add up“. Also, this method prevents students from getting rusty in older topics, or totally forgetting the earlier topics. Also, this method makes use of the fact that the human brain learns during sleep, so if you do mathematics everyday, you are letting your brain learn during sleep everyday.

Let’s take the example of Additional Mathematics.

Exam is on 24/25 October 2013.

Let’s say the student starts the “One Question per day” Strategy on 20 May 2013

Days till exam: 157 days  (22 weeks or 5 months, 4 days)

So, 157 days = 157 questions (or more!)

Each paper in Ten Year Series has around 25 questions (Paper 1 & Paper 2), so 157 questions translates to more than 6 years worth of practice papers! And all that is achieved by just doing at least one Maths question per day!

A sample daily revision plan can look like this. (I create a customized revision plan for each of my students, based on their weaknesses).

 Topic Monday Algebra Tuesday Geometry and Trigonometry Wednesday Calculus Thursday Algebra Friday Geometry and Trigonometry Saturday Calculus Sunday Geometry and Trigonometry

(Calculus means anything that involves differentiation, integration)

(Geometry and Trigonometry means anything that involves diagrams, sin, cos, tan, etc. )

(Algebra is everything else, eg. Polynomials, Indices, Partial Fractions)

By following this method, using a TYS, the student can cover all topics, up to 6 years worth of papers!

Usually, students may accumulate a lot of questions if they are stuck. This is where a tutor comes in. The tutor can go through all the questions during the tuition time. This method makes full use of the tuition time, and is highly efficient.

Personally, I used this method of studying and found it very effective. This method is suitable for disciplined students who are aiming to improve, whether from fail to pass or from B/C to A. The earlier you start the better, for this strategy. For students really aiming for A, you can modify this strategy to do at least 2 to 3 Maths questions per day. From experience, my best students practice Maths everyday. Practicing Ten Year Series (TYS) is the best, as everyone knows that school prelims/exams often copy TYS questions exactly, or just modify them a bit.

The role of the parent is to remind the child to practice maths everyday. From experience, my best students usually have proactive parents who pay close attention to their child’s revision, and play an active role in their child’s education.

This study strategy is very flexible, you can modify it based on your own situation. But the most important thing is, practice Maths everyday! (For Maths, practicing is twice as important as studying notes.) And fully understand each question you practice, not just memorizing the answer. Also, doing a TYS question twice (or more) is perfectly acceptable, it helps to reinforce your technique for answering that question.

If you truly follow this strategy, and practice Maths everyday, you will definitely improve!

Hardwork $\times$ 100% = Success! (^_^)

There is no substitute for hard work.” – Thomas Edison

## A Maths Tuition: Trigonometry Formulas

Many students find Trigonometry in A Maths challenging.

This is a list of Trigonometry Formulas that I compiled for A Maths. Students in my A Maths tuition class will get a copy of this, neatly formatted into one A4 size page for easy viewing.

A Maths: Trigonometry Formulas

$\mathit{cosec}x=\frac{1}{\sin x}$$\mathit{sec}x=\frac{1}{\cos x}$

$\cot x=\frac{1}{\tan x}$$\tan x=\frac{\sin x}{\cos x}$

(All Science Teachers Crazy)

$y=\sin x$

$y=\cos x$

$y=\tan x$

$\frac{d}{\mathit{dx}}(\sin x)=\cos x$

$\frac{d}{\mathit{dx}}(\cos x)=-\sin x$

$\frac{d}{\mathit{dx}}(\tan x)=\mathit{sec}^{2}x$

$\int {\sin x\mathit{dx}}=-\cos x+c$

$\int \cos x\mathit{dx}=\sin x+c$

$\int \mathit{sec}^{2}x\mathit{dx}=\tan x+c$

Special Angles:

$\cos 45^\circ=\frac{1}{\sqrt{2}}$

$\cos 60^\circ=\frac{1}{2}$

$\cos 30^\circ=\frac{\sqrt{3}}{2}$

$\sin 45^\circ=\frac{1}{\sqrt{2}}$

$\sin 60^\circ=\frac{\sqrt{3}}{2}$

$\sin 30^\circ=\frac{1}{2}$

$\tan 45^\circ=1$

$\tan 60^\circ=\sqrt{3}$

$\tan 30^\circ=\frac{1}{\sqrt{3}}$

$y=a\sin (\mathit{bx})+c$ Amplitude: $a$; Period: $\frac{2\pi }{b}$

$y=a\cos (\mathit{bx})+c$ Amplitude: $a$; Period: $\frac{2\pi }{b}$

$y=a\tan (\mathit{bx})+c$ Period: $\frac{\pi }{b}$

$\pi \mathit{rad}=180^\circ$

Area of  $\triangle \mathit{ABC}=\frac{1}{2}\mathit{ab}\sin C$

Sine Rule:  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Cosine Rule:  $c^{2}=a^{2}+b^{2}-2\mathit{ab}\cos C$