Congruent Triangles Geometry Maths Tuition: Solution

Solution:

(a) $\angle EBD=\angle BEC$ (given)

BE is a common side for both triangles $\triangle BCE$ and $\triangle EDB$

BD=EC (given)

Therefore, $\triangle BCE \equiv \triangle EDB$ (SAS)

(proved)

(b)

Since $\triangle BCE\equiv\triangle EDB$ we have $\angle CBE=\triangle DEB$

Thus, $\begin{array}{rcl}\angle ABE&=&180^\circ - \angle CBE\\ &=&180^\circ -\angle DEB\\ &=& \angle AEB \end{array}$

Therefore, $\triangle ABE$ is an isosceles triangle.

Thus, $AB=AE$

(proved)

Congruent Triangles Maths Tuition: Solution

Solution:

BS=BE+ES=ST+ES=ET

$\angle DES=\angle ESA=90^\circ$

BA=DT (given)

Thus, $\triangle ASB$ is congruent to $\triangle DET$ (RHS)

Hence $\angle DTE=\angle SBA$

Thus DT//BA (alt. angles)

(Proved)

By Pythagoras’ Theorem, we have

$\begin{array}{rcl}DB&=&\sqrt{DE^2+BE^2}\\ &=&\sqrt{SA^2+ST^2}\\ &=&TA \end{array}$

Hence $\triangle DEB$ and $\triangle AST$ are congruent (SSS).

Hence $\angle DBE=\angle STA$

Thus DB//TA (alt. angles)

Therefore, ABDT is a parallelogram since it has two pairs of parallel sides.

(shown)

French Curve

Math Online Tom Circle

The French method of drawing curves is very systematic:

“Pratique de l’etude d’une fonction”

Let f be the function represented by the curve C

Steps:

1. Simplify f(x). Determine the Domain of definition (D) of f;
2. Determine the sub-domain E of D, taking into account of the periodicity (eg. cos, sin, etc) and symmetry of f;
3. Study the Continuity of f;
4. Study the derivative of fand determine f'(x);
5. Find the limits of fwithin the boundary of the intervals in E;
6. Construct the Table of Variation;
7. Study the infinite branches;
8. Study the remarkable points: point of inflection, intersection points with the X and Y axes;
9. Draw the representative curve C.

Example:

$latex \displaystyle\text{f: } x \mapsto \frac{2x^{3}+27}{2x^2}$
Step 1: Determine the Domain of Definition D
D = R* = R –…

View original post 454 more words

Cut a cake 1/5

Math Online Tom Circle

Visually cut a cake 1/5 portions of equal size:

1) divide into half:

2) divide 1/5 of the right half:

3) divide half, obtain 1/5 = right of (3)

$latex \frac{1}{5}= \frac{1}{2} (\frac{1}{2}(1- \frac{1}{5}))= \frac{1}{2} (\frac{1}{2} (\frac{4}{5}))=\frac{1}{2}(\frac{2}{5})$

4) By symmetry another 1/5 at (2)=(4)

5) divide left into 3 portions, each 1/5

$latex \frac{1}{5}= \frac{1}{3}(\frac{1}{2}+ \frac{1}{2}.\frac{1}{5}) = \frac{1}{3}.\frac{6}{10}$

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Tuition That We May Have To Believe In

This insightful article makes a really good read.

Quotes from the article:

To be honest, the amount to be learnt at each level of education is constantly increasing, and tuition could just help you get that edge over others. After all, it was meant to be supplementary in nature.

The toughest part at the end of the day however, is probably this: getting the right tutor.

guanyinmiao's musings (Archived: July 2009 to July 2019)

This commentary, “Tuition That We May Have To Believe In”, is a reply to a previous article on tuition by Howard Chiu (Mr.), “Tuition We Don’t Have To Believe In” (Read).

I must say Howard’s article had me on his side for a moment. He appealed to me emotively. Nothing like a mental picture of some kid attending hours and hours of tuition immediately after school when he could well be enjoying himself thoroughly with… an iPhone or iPad (I highly doubt kids these days still indulge their time at playgrounds). But the second time I read his article, I silenced the part of my brain which still prays the best for children, so do pardon me if I sound a tad too pragmatic at times.

The overarching assertion that Howard projects his points from is that there is “huge over consumption of this good”. Firstly, private tutoring…

View original post 1,472 more words

Generalized Analytic Geometry

Math Online Tom Circle

Generalized Analytic Geometry

Find the equation of the circle which cuts the tangent 2x-y=0 at M(1,4), passing thru point A(4,-1).

Solution:

1st generalization:
Let the point circle be:
(x-1)² + (y-4)² =0

2nd generalization:
It cuts the tangent 2x-y=0
(x-1)² + (y-4)² +k(2x-y) =0 …(C)

Pass thru A(4,-1)
x=4, y= -1
=> k= -2
(C): (x-3)² + (y-1)² =…
[QED]

View original post 1 more word

Math Chants

Math Online Tom Circle

Math Chants make learning Math formulas or Math properties fun and easy for memory . Some of them we learned in secondary school stay in the brain for whole life, even after leaving schools for decades.

Math chant is particularly easy in Chinese language because of its single syllable sound with 4 musical tones (like do-rei-mi-fa) – which may explain why Chinese students are good in Math, as shown in the International Math Olympiad championships frequently won by China and Singapore school students.

1. A crude example is the quadratic formula which people may remember as a little chant:
“ex equals minus bee plus or minus the square root of bee squared minus four ay see all over two ay.”

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

2. $latex \mathbb{NZQRC}$
Nine Zulu Queens Rule China

3. $latex \boxed {\cos 3A = 4\cos^{3}…

View original post 199 more words

What is “sin A”

Math Online Tom Circle

What is “sin A” concretely ?

1. Draw a circle (diameter 1)
2. Connect any 3 points on the circle to form a triangle of angles A, B, C.
3. The length of sides opposite A, B, C are sin A, sin B, sin C, respectively.

Proof:
By Sine Rule:

$latex \frac{a}{sin A} = \frac{b}{sin B} =\frac{c}{sin C} = 2R = 1$
where sides a,b,c opposite angles A, B, C respectively.
a = sin A
b = sin B
c = sin C

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成语数学

An alternative answer to Q1) 20 除 3 is “陆续不断”.

20除以3，因为它的答案接近于6.6666，所以这道题的答案是陆续不断，或者是六六大顺都行，百分之一就是百里挑一，9寸加1寸等于一尺即是得寸进尺，12345609，七零八落，1、3、5、7、9无双数所以叫做举世无双，或者你把它答出天下无双都行，如此小升初的难题您答对了吗？

Source: http://politics.people.com.cn/n/2013/0528/c70731-21638931.html

Math Online Tom Circle

中国的小学离校考试 (PSLE) 「神题」: 猜成语
1) 20 除 3
2）1 除100
3）9寸+1寸=1尺
4）12345609
5）1,3,5,7,9

答案：:
1) 20/3= 6.666 六六大顺
2）百中挑一
3）得寸進尺
4）七零八落
5）举世无双

View original post

O Level E Maths Tuition: Statistics Question

Solution:

From the graph,

Median = 50th percentile = $22,000 (approximately)

The mean is lower than $22000 because from the graph, there is a large number of people with income less than $22000, and fewer with income more than $22000. (From the wording of the question, calculation does not seem necessary)

Hence, the median is higher.

The mean is a better measure of central tendency, as it is a better representative of the gross annual income of the people. This is because more people have an income closer to the mean, rather than the median.

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.)

Maths Resources available for sale!

Do check out our Maths Resources available for sale! All Maths notes and worksheets are priced affordably, starting from just $0.99!
All the resources are personally written by our principal tutor Mr Wu.

https://mathtuition88.com/maths-notes-worksheets-sale/

Tip: Using Google “filetype” filter to search for Free Exam Papers

This is a handy tip to search for Free Exam Papers on Google.

(Note: Google is very powerful, but it can only search for exam papers that are already online in the first place)

Since most exam papers are in PDF format, we can restrict our search to PDF files by adding the phrase “filetype:pdf” to the Google search.

For example, searching “hwa chong maths sec 2” in Google does not yield many exam papers.

Searching “hwa chong maths sec 2 filetype:pdf” returns a much better result, including some worksheets and test papers.

Screenshot:

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

Considering the triangle ADP,
$\angle DPA=180^\circ-90^\circ=90^\circ$
(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)

Logarithm and Exponential Question: A Maths Question

Question:

Solve $(4x)^{\lg 5} = (5x)^{\lg 7}$

Solution:

$4^{\lg 5}\cdot x^{\lg 5}=5^{\lg 7}\cdot x^{\lg 7}$

$\displaystyle\frac{4^{\lg 5}}{5^{\lg 7}}=\frac{x^{\lg 7}}{x^{\lg 5}}=x^{\lg 7-\lg 5}$

Using calculator, and leaving answers to at least 4 s.f.,

$0.6763=x^{0.1461}$

Lg both sides,

$\lg 0.6763=0.1461\lg x$

$\lg x=\frac{\lg 0.6763}{0.1461}=-1.1626$

$x=10^{-1.1626}=0.0688$ (3 s.f.)

Check answer (to prevent careless mistakes):

$LHS=(4\times 0.0688)^{\lg 5}=0.406$

$RHS=(5\times 0.0688)^{\lg 7}=0.406$

Since LHS=RHS, we have checked that our answer is valid.

Maths Tutor Singapore, H2 Maths, A Maths, E Maths

If you or a friend are looking for maths tuition: o level, a level, IB, IP, olympiad, GEP and any other form of mathematics you can think of

Experienced, qualified (Raffles GEP, Deans List, NUS Deans List, Olympiads etc) and most importantly patient even with the most mathematically challenged.

so if you are in need of the solution to your mathematical woes, drop me a message!

Tutor: Mr Wu

Email: mathtuition88@gmail.com

Website: https://mathtuition88.com/

Sec 3 Hwa Chong Institution Maths Test Papers and Resources

Site: http://anglc.wiki.hci.edu.sg/space/content

Description: Includes Sec 3 Hwa Chong Institution Maths Test Papers and Resources

Please visit our site https://mathtuition88.com/free-exam-papers for updates on more Free Exam Papers and Maths Tips.

Top Maths Tuition

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Thank you for your support.

Please continue to visit our site for Free Exam Papers, and tips regarding Maths.

Free Exam Paper: CHIJ Katong Convent Sec 3 Express Additional Mathematics

Katong Convent Exam Paper

(Sec 3 A Maths, with answers)

H2 Maths Tuition: Complex Numbers Notes

H2 Maths: Complex Numbers 1 Page Notes

	Modulus	Argument
Cartesian Form		Draw diagram first, then find the appropriate quadrant and use (can use GC to double check)
Polar Form
Exponential Form

☺ When question involvespowers, multiplication or division, it may be helpful toconvert to exponential form.

☺ Please write Ƶ and 2 differently.

☺ De Moivre’s Theorem

Equivalent to

☺

Memory tip: Notice that arg behaves similarly to log.

☺ Locusof z is aset of pointssatisfying certain given conditions.

☺ in English means:The distance between (the point representing)and (the point representing)

☺ Means the distance offromis a constant,.

So this is acircular loci.

Centre:, radius =

☺ means that the distance offromis equal to its distance from

In other words, the locus is theperpendicular bisectorof the line segment joiningand.

☺ represents ahalf-linestarting frommaking an anglewith the positive Re-axis.

(Exclude the point (a,b) )

Common Errors

– Some candidates thought thatis the same asand thatis the same as.

– The “formula”for argumentsdoes not workfor points in the 2^ndand 3^rdquadrant.

– Very many candidates seem unaware that their calculators will work in radians mode and there were many unnecessary “manual” conversions from degrees to radians.

Maths Tuition Free Exam Papers (Primary, Secondary, O Levels, A Levels) Links

The Free Exam Papers Database has moved to:

https://mathtuition88.com/free-exam-papers

(Click here to go to Free Exam Papers Database)

Ten Year Series: How many questions or papers to practice for Maths O Levels / A Levels?

This is a question to ponder about, how many questions or papers to practice for Maths O Levels / A Levels for the Ten Year Series?

If you searched Google, you will find that there is no definitive answer of how many questions to practice for Maths O Levels/ A Levels anywhere on the web.

For O Level / A Level, practicing the Ten Year Series is really helpful, as it helps students to gain confidence in solving exam-type questions.

Here are some tips about how to practice the Ten Year Series (TYS):

1) Do a variety of questions from each topic. This will help you to gain familiarity with all the topics tested, and also revise the older topics.

2) Fully understand each question. If necessary, practice the same question again until you get it right. There is a sense of satisfaction when you finally master a tough question.

3) Quality is more important than quantity. It is better to do and understand 1 question completely than do many questions but not understanding them.

Back to the original query of how many questions or papers to practice for Maths O Levels / A Levels for the Ten Year Series, I will attempt to give a rough estimate here, based on personal experience.

5 Questions done (full questions worth more than 5 marks) will result in an improvement of roughly 1 mark in the final exam.

(The 5 Questions must be fully understood. )

So, if a student wants to improve from 40 marks to 70 marks, he/she should try to do 30×5=150 questions (around 7 years worth of past year papers). Repeated questions are counted too, so doing 75 questions (around 3 years worth of past year papers) twice will also count as doing 150 questions. In fact, that is better for students with weak foundation, as the repetition reinforces their understanding of the techniques used to solve the question.

If the student starts revision early, this may work out to just 1 question per day for 5 months. Of course, the 150 questions must be varied, and from different subject topics.

Marks improved by	Long Questions to be done	Approx. Number of years of TYS	OR (even better)
10	50	2	1 year TYS practice twice
20	100	4	2 year TYS practice twice
30	150	6	3 year TYS practice twice
40	200	8	4 year TYS practice twice
50	250	10	5 year TYS practice twice

This estimate only works up to a certain limit (obviously we can’t exceed 100 marks). To get the highest grade (A1 or A), mastery of the subject is needed, and the ability to solve creative questions and think out of the box.

When a student practices TYS questions, it is essential that he/she fully understands the question. This is where a tutor is helpful, to go through the doubts that the student has. Doing a question without understanding it is essentially of little use, as it does not help the student to solve similar questions should they come out in the exam.

Hope this information will help your revision.

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).

Suggested daily revision (Additional Mathematics):

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

Exam Time Management and Speed in Maths (Primary, O Level, A Level)

Time management is a common problem for Maths, along with careless mistakes.

For Exam Time management, here are some useful tips:
1) If stuck at a question for some time, it is better to skip it and go back to it later, rather than spend too much time on it. I recall for PSLE one year, there was a question about adding 1+2+…+100 early in the paper, and some children unfortunately spent a lot of time adding it manually.
2) Use a exam half-time strategy. At the half-time mark of the exam, one should finish at least half of the paper. If no, then need to speed up and skip hard questions if necessary.

To improve speed:
1) Practice. It is really important to practice as practice increases speed and accuracy.
2) Learn the faster methods for each type of question. For example, guess and check is considered a slower method, as most questions are designed to make guess and check difficult.

Sincerely hope it helps.
For dealing with careless mistakes (more for O and A levels), you may read my post on How to avoid Careless Mistakes for O-Level / A-Level Maths?

English: an animated clock (Photo credit: Wikipedia)

Challenging Geometry E Maths Question — St Andrew’s Sec 3 Maths Tuition Question

Question:

ABCD is a rectangle. M and N are points on AB and DC respectively. MC and BN meet at X. M is the midpoint of AB.

(a) Prove that $\Delta CXN$ and $\Delta MXB$ are similar.

(b) Given that area of $\triangle CXN$ : area of $\triangle MXB$ =9:4, find the ratio of,

(i) DN: NC

(ii) area of rectangle ABCD: area of $\triangle XBC$ . (Challenging)

[Answer Key] (b) (i) 1:3

(ii) 20:3

Suggested Solutions:

(a)
$\angle MXB=\angle NXC$ (vert. opp. angles)

$\angle MBX = \angle XNC$ (alt. angles)

$\angle BMX = \angle XCN$ (alt. angles)

Therefore, $\Delta CXN$ and $\Delta MXB$ are similar (AAA).

(b) (i) $\displaystyle\frac{NC}{BM}=\sqrt{\frac{9}{4}}=\frac{3}{2}$

Let $BM=2u$ and $NC=3u$

Then $DC=2\times 2u=4u$

So $DN=4u-3u=u$

Thus, $DN:NC=1u:3u=1:3$

(ii)

We now have a shorter solution, thanks to a visitor to our site! (see comments below)

From part (a), since $\Delta CXN$ and $\Delta MXB$ are similar, we have $MX:XC=2:3$

This means that $MC:XC=5:3$

Thus $\triangle MBC:\triangle XBC=5:3$ (the two triangles share a common height)

Now, note that $\displaystyle\frac{\text{area of }ABCD}{\triangle MBC}=\frac{BC\times AB}{0.5 \times BC \times MB}=\frac{AB}{0.5MB}=\frac{2MB}{0.5MB}=4$

Hence area of $ABCD=4\times\triangle MBC$

We conclude that area of rectangle ABCD: area of $\triangle XBC=4(5):3=20:3$

Here is a longer solution, for those who are interested:

Let area of $\triangle XBC =S$

Let area of $\triangle MXB=4u$

Let area of $\triangle CXN=9u$

We have $\displaystyle\frac{S+9u}{S+4u}=\frac{3}{2}$ since $\triangle NCB$ and $\triangle CMB$ have the same base BC and their heights have ratio 3:2.

Cross-multiplying, we get $2S+18u=3S+12u$

So $\boxed{S=6u}$

$\displaystyle\frac{\triangle BCN}{\triangle BDC}=\frac{3}{4}$ since $\triangle BCN$ and $\triangle BDC$ have the same base BC and their heights have ratio 3:4.

Hence,

$\begin{array}{rcl} \triangle BDC &=& \frac{4}{3} \triangle BCN\\ &=& \frac{4}{3} (9u+6u)\\ &=& 20u \end{array}$

Thus, area of $ABCD=2 \triangle BDC=40u$

area of rectangle ABCD: area of $\triangle XBC$ =40:6=20:3

O Level Logarithm Question (Challenging)

Question:

Given $\displaystyle\log_9{a} = \log_{12}{b} =\log_{16}{(a+b)}$ , find the value of $\displaystyle\frac{a}{b}$ .

Solution:

Working with logarithm is tricky, we try to transform the question to an exponential question.

Let $\displaystyle y=\log_9{a} = \log_{12}{b} =\log_{16}{(a+b)}$

Then, we have $a=9^y=3^{2y}$

$b=12^y=3^y\cdot 2^{2y}$

$a+b=16^y=2^{4y}$ .

Here comes the critical observation:

Observe that $\boxed{a(a+b)=b^2}$ .

Divide throughout by $b^2$ , we get $\displaystyle (\frac{a}{b})^2+\frac{a}{b}=1$ .

Hence, $\displaystyle (\frac{a}{b})^2+\frac{a}{b}-1=0$ .

Solving using quadratic formula (and reject the negative value since $a$ and $b$ has to be positive for their logarithm to exist),

We get $\displaystyle\frac{a}{b}=\frac{-1+\sqrt{5}}{2}$ .

If you have any questions, please feel free to ask me by posting a comment, or emailing me.

(I will usually explain in much more detail if I teach in person, than when I type the solution)

Hwa Chong IP Sec 2 Maths Question – Equation of Parabola

Question:

Given that a parabola intersects the x-axis at x=-4 and x=2, and intersects the y-axis at y=-16, find the equation of the parabola.

Solution:

Sketch of graph:

Now, there is a fast and slow method to this question. The slower method is to let $y=ax^2+bx+c$ , and solve 3 simultaneous equations.

The faster method is to let $y=k(x+4)(x-2)$ .

Why? We know that x=4 is a root of the polynomial, so it has a factor of (x-4). Similarly, the polynomial has a factor of (x-2). The constant k (to be determined) is added to scale the graph, so that the graph will satisfy y=-16 when x=0.

So, we just substitute in y=-16, x=0 into our new equation.

$-16=k(4)(-2)$ .

$-16=-8k$ .

So $k=2$ .

In conclusion, the equation of the parabola is $y=2(x+4)(x-2)$ .

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H2 Maths Tuition: Foot of Perpendicular (from point to plane) (Part II)

This is a continuation from H2 Maths Tuition: Foot of Perpendicular (from point to line) (Part I).

Foot of Perpendicular (from point to plane)

From point (B) to Plane ( $p$ )

Equation (I):

Where does F lie?

F lies on the plane $p$ .

$\overrightarrow{\mathit{OF}}\cdot \mathbf{n}=d$

Equation (II):

Perpendicular

$\overrightarrow{\mathit{BF}}=k\mathbf{n}$

$\overrightarrow{\mathit{OF}}-\overrightarrow{\mathit{OB}}=k\mathbf{n}$

$\overrightarrow{\mathit{OF}}=k\mathbf{n}+\overrightarrow{OB}$

Final Step

Substitute Equation (II) into Equation (I) and solve for k.

Example

[VJC 2010 P1Q8i]

The planes $\Pi _{1}$ and $\Pi _{2}$ have equations $\mathbf{r\cdot(i+j-k)}=6$ and $\mathbf{r\cdot(2i-4j+k)}=-12$ respectively. The point $A$ has position vector $\mathbf{{9i-7j+5k}}$ .

(i) Find the position vector of the foot of perpendicular from $A$ to $\Pi _{2}$ .

Solution

Let the foot of perpendicular be F.

Equation (I)

$\overrightarrow{\mathit{OF}}\cdot \left(\begin{matrix}2\\-4\\1\end{matrix}\right)=-12$

Equation (II)

$\overrightarrow{\mathit{OF}}=k\left(\begin{matrix}2\\-4\\1\end{matrix}\right)+\left(\begin{matrix}9\\-7\\5\end{matrix}\right)=\left(\begin{matrix}2k+9\\-4k-7\\k+5\end{matrix}\right)$

Subst. (II) into (I)

$2(2k+9)-4(-4k-7)+(k+5)=-12$

Solve for k, $k=-3$ .

$\overrightarrow{\mathit{OF}}=\left(\begin{matrix}3\\5\\2\end{matrix}\right)$

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H2 Maths Tuition: Foot of Perpendicular (from point to line) (Part I)

Foot of Perpendicular is a hot topic for H2 Prelims and A Levels. It comes out almost every year.

There are two versions of Foot of Perpendicular, from point to line, and from point to plane. However, the two are highly similar, and the following article will teach how to understand and remember them.

H2: Vectors (Foot of perpendicular)

From point (B) to Line ( $l$ )

(Picture)

Equation (I):

Where does F lie? F lies on the line $l$ .

$\overrightarrow{\mathit{OF}}=\mathbf{a}+\lambda \mathbf{m}$

Equation (II):

Perpendicular:

$\overrightarrow{\mathit{BF}}\cdot \mathbf{m}=0$

$(\overrightarrow{\mathit{OF}}-\overrightarrow{\mathit{OB}})\cdot \mathbf{m}=0$

Final Step

Substitute Equation (I) into Equation (II) and solve for $\lambda$ .

Example:

[CJC 2010 P1Q7iii]

Relative to the origin $O$ , the points $A$ , $B$ and $C$ have position vectors $\left(\begin{matrix}1\\2\\1\end{matrix}\right)$ , $\left(\begin{matrix}2\\1\\3\end{matrix}\right)$ and $\left(\begin{matrix}-1\\2\\3\end{matrix}\right)$ Find the shortest distance from $C$ to $\mathit{AB}$ . Hence or otherwise, find the area of triangle $\mathit{ABC}$ .

[Note: There is a 2nd method to this question. (cross product method)]

Solution:

Let the foot of perpendicular from C to AB be F.

Equation (I):

$\overrightarrow{\mathit{OF}}=\overrightarrow{\mathit{OA}}+\lambda \overrightarrow{\mathit{AB}}=\left(\begin{matrix}1+\lambda \\2-\lambda \\1+2\lambda \end{matrix}\right)$

Equation (II):

$(\overrightarrow{\mathit{OF}}-\overrightarrow{\mathit{OC}})\cdot \overrightarrow{\mathit{AB}}=0$

$\left(\begin{matrix}2+\lambda \\-\lambda \\-2+2\lambda \end{matrix}\right)\cdot \left(\begin{matrix}1\\-1\\2\end{matrix}\right)=0$

$\lambda =\frac{1}{3}$

$\overrightarrow{\mathit{CF}}=\overrightarrow{\mathit{OF}}-\overrightarrow{\mathit{OC}}=\left(\begin{matrix}2\frac{1}{3}\\-{\frac{1}{3}}\\-1\frac{1}{3}\end{matrix}\right)$

$\left|{\overrightarrow{{\mathit{CF}}}}\right|=\sqrt{\frac{22}{3}}$

Area of $\Delta \mathit{ABC}=\frac{1}{2}\left|{\overrightarrow{\mathit{AB}}}\right|\left|{\overrightarrow{\mathit{CF}}}\right|=\sqrt{11}$

For the next part, please read our article on Foot of Perpendicular (from point to plane).

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Maths Tuition Flyer

This is a Maths Tuition Flyer I created using the Maths software $\LaTeX$ , for distribution in the following areas:

Lorong Chuan
Bishan
Ang Mo Kio
Toa Payoh
Serangoon Gardens

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Sec 2 IP (HCI) Revision 1: Expansion and Factorisation

This is a nice worksheet on Expansion and Factorisation by Hwa Chong Institution (HCI).

There are no solutions, but if you have any questions you are welcome to ask me, by leaving a comment, or by email.

Hope you enjoy practising Expansion and Factorisation.

The worksheet may be downloaded here:

Expansion and Factorisation by Hwa Chong Institution (HCI)

If your child needs help in Maths, please feel free to contact us for Maths Tuition. 🙂

Is d/dx (a^x)=x a^{x-1}? (a conceptual error in O/A Level Math)

In O Level, students are taught that $\boxed{\frac{d}{dx}(x^{n})=nx^{n-1}}$

So naturally, students may think that $\displaystyle\frac{d}{dx}(a^{x})=xa^{x-1}??$ (a is a constant)

Well, actually that is good pattern spotting, but unfortunately it is incorrect. Do not be too disheartened if you make this mistake, it is a common mistake.

The above is a conceptual error as $\boxed{\frac{d}{dx}(x^{n})=nx^{n-1}}$ only holds when n is a constant.

Fortunately, this question is rarely tested, though it is quite possible that it can come up in A Levels.

To fully understand the following steps, it would help read my other post (Why is e^(ln x)=x?) first.

First, we write $\displaystyle a^x=e^{\ln a^x}=e^{x\ln a}$ .

Hence
$\displaystyle \begin{array}{rcl} \frac{d}{dx} (a^x)&=&\frac{d}{dx} (e^{x\ln a})\\ &=&e^{x\ln a}(\ln a)\\ &=&e^{\ln a^x}(\ln a)\\ &=&a^x(\ln a) \end{array}$

After fully understanding the above steps, you may memorize the formula if you wish:

$\boxed{\frac{d}{dx} (a^x)=a^x(\ln a)}$

Memory Tip: If you let a=e, you should get $\boxed{\frac{d}{dx} (e^x)=e^x(\ln e)=e^x}$

The above steps involve the chain rule, which I will cover in a subsequent post.

How to avoid Careless Mistakes for Maths?

Many parents have feedback to me that their child often makes careless mistakes in Maths, at all levels, from Primary, Secondary, to JC Level. I truly empathize with them, as it often leads to marks being lost unnecessarily. Not to mention, it is discouraging for the child.

Also, making careless mistakes is most common in the subject of mathematics, it is rare to hear of students making careless mistakes in say, History or English.
Fortunately, it is possible to prevent careless mistakes for mathematics, or at least reduce the rates of careless mistakes.

From experience, the ways to prevent careless mistakes for mathematics can be classified into 3 categories, Common Sense, Psychological, and Math Tips.

Common Sense

Firstly, write as neatly as possible. Often, students write their “5” like “6”. Mathematics in Singapore is highly computational in nature, such errors may lead to loss of marks. Also, for Algebra, it is recommended that students write l (for length) in a cursive manner, like $\ell$ to prevent confusion with 1. Also, in Complex Numbers in H2 Math, write z with a line in the middle, like Ƶ, to avoid confusion with 2.
Leave some time for checking. This is easier said than done, as speed requires practice. But leaving some time, at least 5-10 minutes to check the entire paper is a good strategy. It can spot obvious errors, like leaving out an entire question.

Psychological

Look at the number of marks. If the question is 5 marks, and your solution is very short, something may be wrong. Also if the question is just 1 mark, and it took a long time to solve it, that may ring a bell.
See if the final answer is a “nice number“. For questions that are about whole numbers, like number of apples, the answer should clearly be a whole number. At higher levels, especially with questions that require answers in 3 significant figures, the number may not be so nice though. However, from experience, some questions even in A Levels, like vectors where one is suppose to solve for a constant $\lambda$ , it turns out that the constant is a “nice number”.

Mathematical Tips

Mathematical Tips are harder to apply, unlike the above which are straightforward. Usually students will have to be taught and guided by a teacher or tutor.

Substitute back the final answer into the equations. For example, when solving simultaneous equations like x+y=3, x+2y=4, after getting the solution x=2, y=1, you should substitute back into the original two equations to check it.
Substitute in certain values. For example, after finding the partial fraction $\displaystyle\frac{1}{x^2-1} = \frac{1}{2 (x-1)}-\frac{1}{2 (x+1)}$ , you should substitute back a certain value for x, like x=2. Then check if both the left-hand-side and right-hand-side gives the same answer. (LHS=1/3, RHS=1/2-1/6=1/3) This usually gives a very high chance that you are correct.

Thanks for reading this long article! Hope it helps! 🙂

I will add more tips in the future.

Recommended Maths Book:

Math Doesn’t Suck: How to Survive Middle School Math Without Losing Your Mind or Breaking a Nail

This book is a New York Times Bestseller by actress Danica McKellar, who is also an internationally recognized mathematician and advocate for math education. It should be available in the library. Hope it can inspire all to like Maths!

Why is e^(ln x)=x? (O Level Math/ A Level Math Tuition)

Why is $\boxed{e^{\ln x}=x}$ ?

This formula will be useful for some questions in O Level Additional Maths, or A Level H2 Maths.

There are two ways to show or prove this, first we can let

$y=e^{\ln x}$

Taking natural logarithm (ln) on both sides, we get

$\ln y=\ln x\ln e=\ln x$

So $y=x$ . Substitute the very first equation and we get $e^{\ln x}=x$ . 🙂

Alternatively, we can view $e^x$ and $\ln x$ as inverse functions of each other. So, we can let $f(x)=e^x$ and $f^{-1}(x)=\ln x$ . Then, $e^{\ln x}=f(f^{-1})(x)=x$ by definition of inverse functions. This may be a better way to remember the result. 🙂

The above method of inverse functions can be used to remember $\ln (e^x)=x$ too.

Factorize a^3-b^3 (O Level Math Tuition Question)

$(a-b)^3=a^3-3a^2b+3ab^2-b^3$

So,
$\begin{array}{rcl} a^3-b^3&=&(a-b)^3+3a^2b-3ab^2\\ &=&(a-b)(a-b)^2+3a^2b-3ab^2\\ &=&(a-b)(a^2-2ab+b^2)+(a-b)(3ab)\\ &=&(a-b)(a^2+ab+b^2) \end{array}$

The mass of particles of a certain radioactive chemical element (O Level Math Tuition Question)

Question from http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=29&t=8315&start=780

The mass of particles of a certain radioactive chemical element is halved every 10 months. During a chemical experiment, the initial mass of particles of the chemical element is 3mg.

(i) write down an expression, in terms of t, for the mass of particles after t years.
(ii) Hence, find the value of t, if the mass is reduced to 0.046875 mg after t years.

Solution:

(i)

$1\: \text{year} = 12\: \text{months}$

Therefore, $t\: \text{years}=12t\: \text{months}$ .

How many 10 months are there in $t\: \text{years}$ ? (Ans: $\frac{12t}{10}=1.2t$ )

Hence, the mass of particles after $t$ years is $3\times(0.5)^{1.2t}$ mg.

(ii)
We need to solve $3(0.5)^{1.2t}=0.046875$ .

Dividing by 3, we have $(0.5)^{1.2t}=0.015625$ .

Ln both sides, we have $1.2t\ln{0.5}=\ln{0.015625}$ .

Hence, $t=\frac{\ln{0.015625}}{1.2\ln{0.5}}=5$ .

If you liked our solution above, please consider signing up for Maths Tuition with us! 🙂

The Free Exam Papers Database has moved to:

Foot of Perpendicular (from point to plane)

Equation (I):

Equation (II):

Final Step

Example

Solution

Equation (I)

Equation (II)

H2 Maths Tuition

Foot of Perpendicular is a hot topic for H2 Prelims and A Levels. It comes out almost every year.

H2: Vectors (Foot of perpendicular)

Equation (I):

Equation (II):

Final Step

Example:

Solution:

H2 Maths Tuition

Recommended Maths Book: Math Doesn’t Suck: How to Survive Middle School Math Without Losing Your Mind or Breaking a Nail

Recommended Maths Book:

Math Doesn’t Suck: How to Survive Middle School Math Without Losing Your Mind or Breaking a Nail