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GCE O Level Oct/Nov 2010 E-Maths 4016 Paper 1 Suggested Answers & Solutions???

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Posted at 11:40 pm by Miss Loi in Miss Loi the Tutor

The first skirmish in this Week of Mathematical Armageddon occurred this afternoon, and since then the Battle Control Centre that is The Temple has gone into overdrive with reports from the battlefields across Singapore.

Thank you Ms Loi. Today’s paper was relatively easy.

Paper one went relatively well =)

2day was easy!

I found the paper very manageable. I had time to re-check.

Ms Loi!! The paper hard siak!!!!!!! (Miss Loi: Aiyooo … *shakes head*)

Interspersed among these frantic messages were more frantic messages and calls from unknown students who allegedly “couldn’t stand the uncertainty anymore” and kept pestering asking Miss Loi when would the EMaths Paper 1 solutions be posted up.

Unfortunately, since the answers were written on and handed in together with the question paper, the ‘O’ Level EMaths Paper 1 question paper tends to be a bit of a holy grail of exam papers these days.

And since The Temple does not have the evil policy of asking its students to steal, smuggle or even copy out the questions during the exam itself (obviously, ANY spare time is better spent on checking the answers more than anything else), Miss Loi unfortunately couldn’t produce any solution without the questions.

But as fate would have it, just as Miss Loi was letting out a sigh of resignation, a student strode into The Temple, held a couple of pieces of paper high in the air and exclaimed “MY SOLUTIONS!!!!!!!!!”

And so in order the pacify the anxious callers, this is what Miss Loi can offer here for now as far as 2010 EMaths Paper 1 goes (in the absence of the question paper):

Please understand that these solutions have NOT BEEN CHECKED BY ANYONE AT ALL, apart from the student who claimed to have 45 mins of spare time to do this during the exam (aiyohh …), and there are bound to be some mistakes (in fact the student-in-question has already admitted some have been spotted). But given the situation this would have to do for all of you who are anxious to see if anyone gets the same answers as you. Alright the verified ones are up above (albeit late – sorry Miss Loi’s students don’t possess superhuman mega memory abilities to recall the entire set of EMaths Paper 1 questions :P ) So do leave a comment should you spot something that isn’t right, and do pardon the messy chunk of words towards the end.

But MOST IMPORTANTLY, it’s time to turn your attention to tomorrow’s EMaths Paper 2, and here are some of the remaining topics that you may want to focus on:

One down, one or three (depending on whether you take AMaths) more maths papers to go! It’s coming thick and fast now!

ALL THE BEST & GANBATTE!
がんばって!!!
Miss Loi
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The Week Of Mathematical Armageddon Reincarnation

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Posted at 11:27 pm by Sadako Loi in Miss Loi the Tutor

A brilliant haze-enhanced sunset greeted them as they stood on the top floor of his HDB flat.

She turned to look into his panda eyes (formed from weeks of burning midnight joss sticks).

Boy Boy, this is it.

Have you finished your TYS, gone through my Sensei’s solutions, and cleared EVERY SINGLE doubt that you have?

He nodded his head.

Have you read my Sensei’s last words (ignore topics which are already out of syllabus), and uttered her E-Maths & AMaths prayers?

*nods again*

Have you made sure you know your significant figures?

*nods and nods*

Have you put in fresh batteries for your calculator (plus backups just in case)?

*nods and nods and nods*

Have you put your construction set (i.e. compass, protractor etc.), curve rule, and all your favourite stationery in your bag?

*nodding like a Heavy Metal headbanger*

Most importantly, have you brought along your SELF BELIEF and CONFIDENCE that you will be able do the questions, since they will NEVER be as ridiculously difficult as your school’s sadistic prelim papers and mock exams, and that you stand an equal chance of scoring as those foreign ghosts cyborgs in you class?

“GIRL GIRL! You very lor sor leh!!!“

The sun had set by then, making way for an apocalyptic darkness that lingered on the distant horizon.

It’s dark there and it looks like it’s gonna rain, and the wind is blowing like it’s the end of the world.

She held his hand and looked once more into his eyes.

Boy Boy, are you ready?

He gripped her tender hand tightly (not realizing it was stone cold) and said

Yes.

And so, like those who had bade their final farewell to Miss Loi and The Temple this weekend, they made their way towards the darkness, towards their respective schools, and towards their destinies …

Mathematical Armageddon 2010 Mathematical Armageddon 2010 Mathematical Armageddon 2010 Mathematical Black Friday 2010

All The Best!
GANBATTE!!! がんばって!!!
Ema 2010

Will hang them up when Miss Loi finds the time ...

See you guys again in Halloween!

がんばって!!!
Sadako Loi
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The Road To Mathematical Reincarnation – Modulus Functions

(1)
Posted at 11:22 pm by Sadako Loi in A-Maths Notes & Tips

*tap tap tap* “moi ish sad … feel sho moddy … ”

*tap tap tap* “GPGT! here’s a chio bu peekture to cheer u up”

*tap tap tap* “wah! LoveLoveLove SIC sauce leh ~~~ Boing boing!”

*tap tap tap* “this zehzeh i got from exampap …”

WAKE UP!!!

Once again, he nearly fell off his chair.

Your O Levels are barely a week away and you still have time to chat in that Hardwarezone forum?!

Sorry lah Miss Loi. I’m still trying to get over my breakup okay? My grades used to be good when I studied together with my Girl Girl. But now I’m just feeling lonely :(

Suddenly he felt his neck being shaken violently by a pair of invisible hands.

For the sake of your future, you must not be distracted and keep your FOCUS for the next few weeks! After that, you can surf all the 变态 Hardwarezone forums to your heart’s content!

Upon seeing his soulless lovesick eyes, however, she knew that her words (and violent shaking) had no effect on him and decided that some drastic action needed to be taken. So she took a glance at his Neoprint, removed her glasses and chanted something unintelligible to herself …

CHIJ Sadako Loi

Boy Boy …

When he turned around and saw the girl standing before him, he was stunned.

Girl Girl! It’s you!

Boy Boy Darling, now that I’m here to 陪 you to study, are you able to focus now? Good. Since you’re feeling … umm … MODDY let’s start talking about Modulus Functions today shall we? It’s getting late.

Modulus Functions

Re-introduced into the New AMaths 4038 Syllabus since 2008 following the removal of a once-mighty chapter called Functions *distant thunder could be heard at the mention of this word*, modulus functions these days mostly involve freeing variables trapped within |modulus signs|, and the sketching of V-, W-shaped modulus graphs.

[+] The Basics

  1. |a| ≥ 0
    ⇒ |a| = a if a ≥ 0 → useful in simplifications to immediately remove mod brackets when a is confirm chop stamp positive

    e.g. |√6 − √3| + |√3 − √6| = √6 − √3 + |√6 − √3| = 2√6 − 2√3

  2. |a| = |−a| (NOTE: |a| ≠ −|a|!)
    → this gives us |a−b| = |b−a| which is useful for bringing out common factors

  3. |ab| = |a||b|
    → always remember to keep the mod sign when bringing out constants & variables i.e. DON’T do this:

    • |−2(x−1)| = −2|(x−1)| → WRONG!
      |−2(x−1)| = |−2||(x−1)| = 2|(x−1)| ✔

    • |x(x−1)| = x|(x−1)| → WRONG!
      |x(x−1)| = |x||(x−1)| ✔

  4. delim{|}{a/b}{|}=delim{|}{a}{|}/delim{|}{b}{|}
    → same warning applies as per Property 2 above.

  5. |an| = |a|n
    → this gives us |a|2 = a2 which is useful for removing the mod sign via ‘squaring both sides’. However the mod sign stays for odd powers i.e. |a|3 ≠ a3!

[+] Removing Modulus Signs and Solving Modulus Equations

  1. Rearrange/simplify (e.g. via bringing out common factors) so that at most one pair of mod signs appears on either side i.e.

    |f(x)| = g(x), f(x) = |g(x)|, |f(x)| = |g(x)|

  2. Slowly … but surely … slide those modulus signs off to form two equations i.e.

    |f(x)| = |g(x)| ⇒ f(x) = g(x) or f(x) = −g(x)

    Solve the two equations.

    Alternatively, you may also remove the modulus signs via squaring both sides i.e.

    |f(a)| = |g(x)| ⇒ [f(x)]2 = [g(x)]2

  3. *Check and reject any value of x that is not valid for the original equation.
    → many missed out this step which is especially IMPORTANT for equations with x on both sides, where one side must be positive but not the other.

Sample Questions
Solve 2|x − 2| + |14 − 7x| = 9

[+] ANSWER

2|x − 2| + |7||2 − x| = 9 → bring out the common factor x−2 via |ab|=|a||b|
2|x − 2| + 7|x − 2| = 9
→ |x − 2|=|2 − x| (now you know why they’re common factors ;) )
9|x − 2| = 9
|x − 2| = 1

x − 2 = 1 or x − 2 = −1 → Modulus sign removed!
∴ x = 3 or 1

Solve |x + √6||x − √6| = −5x

[+] ANSWER

|(x + √6)(x − √6)| = −5x → |a||b| = |ab|
|x2 − 6| = −5x

x2 − 6 = −5x or x2 − 6 = 5x → Modulus sign removed!
x2 + 5x − 6 = 0 or x2 − 5x − 6 = 0
(x − 6)(x + 1) = 0 or (x + 6)(x − 1) = 0
x = 6, −1, −6, 1

Attention Before you merrily hop away, STOP & CHECK!
The LHS of the equation is always positive but the RHS can be negative if x is 1 or 6. Hence we reject 1 and 6.

∴ x = −1, −6

[+] Removing Modulus Signs in Inequalities

In general, modulus inequalities in ‘O’ Level AMaths are usually solved via the graphical approach. But just in case some question-setter wakes up on the wrong side of the bed, these two expressions are almost always used to remove the modulus signs should you ever encounter the need to solve a modulus inequality via a non-graphical approach:

  1. |f(x)| < k ⇔ −k < f(x) < k [explanation diagram]
  2. |f(x)| > k ⇔ f(x) > k or f(x) < −k [explanation diagram]

Check out this past question to see how this is applied. Else skip to the graphical approach that is more likely to be used in your exam (but don’t blame me if it turns out otherwise!).

Understand so far?

My Girl Girl … when can we go Bugis Junction to take Neoprints again?

Not till your O Levels are over Dear … now let’s move on to Modulus Graphs, shall we?

Graphs of Modulus Functions

[+] Sketch → Reflect → Flip → Shift!

AMaths modulus graphs are usually of the form

y = ±|f(x)| + C, where f(x) can be linear or quadratic.

To sketch them, we do NOT waste time drawing up a table of values but instead perform the following steps (which must be done in order):

  1. SKETCH: sketch only the part of the expression within the modulus signs i.e. f(x)
  2. REFLECT: reflect in the x-axis only the portion below it − you should get a V-shape graph (if f(x) is linear) or a W-shape graph (if f(x) is quadratic)
  3. FLIP: flip the entire graph in the x-axis to get a Λ-shape or M-shape graph. This step is only required if there’s a negative sign (−) before the modulus sign!
  4. SHIFT: shift the entire graph upwards(+C units) or downwards (−C units) along the y-axis. This step is only required if C is not zero!
Sample Sketches

Pay attention to how the equation changes in the diagrams.

Sketch y = |2x − 6| − 4

[+] ANSWER

  1. SKETCH!

    Sketch the graph for the part of the expression that’s within the modulus sign.

    Modulus Graph Sketching (Step 1 - Sketch)

  2. REFLECT!

    Using the x-axis like a mirror, reflect in the x-axis the portion of the line that’s below it. Notice the sign of the y-intercept changing from −6 to 6.

    Modulus Graph Sketching (Step 2 - Reflect)

  3. FLIP!

    There’s no negative sign before the modulus sign, so we skip the flipping step.

  4. SHIFT!

    There’s a constant (−4) outside the modulus sign, so we shift the graph downwards by 4 units along the y-axis. Notice all the y-coordinates have changed by −4 units.

    Modulus Graph Sketching (Step 3 - Shift)

    N.B. Remember to calculate (by setting x/y = 0 etc., as learnt in your EMaths) and label all your intercept and turning points and axes in your final sketch as a good graph sketcher ought to do!

Sketch y = −|x2 − 1| + 2

[+] ANSWER

  1. SKETCH!

    Sketch the graph for the part of the expression that’s within the modulus sign.

    Modulus Graph Sketching (Step 1 - Sketch)

  2. REFLECT!

    Mirror, mirror on the x-axis, reflect all the negativity that’s below it!

    Modulus Graph Sketching (Step 2 - Reflect)

  3. The Olympic-Style FLIP!

    There’s a negative sign before the modulus sign, so we must flip the entire graph in the x-axis. This is somewhat similar to the Reflect step before this, only that the entire graph is reflected here (vs only the negative portion). Notice that the only point(s) that remain(s) unchanged after the Olympic-style flip is/are the x-intercept(s).

    Modulus Graph Sketching (Step 3 - Flip)

  4. SHIFT!

    There’s a constant (+2) outside the modulus sign, so we shift the graph upwards by 2 units along the y-axis.

    Modulus Graph Sketching (Step 4 - Shift)

    Once again, remember to label all your intercept and turning points and axes in your final sketch as a 乖乖 graph sketcher ought to do!

[+] Finding Coordinates in Modulus Graphs

Finding Coordinates of 'Sharp' Turning Point of a Modulus Graph

From the modulus graph sketches above, it should be fairly obvious in the final graphs that, for the x-intercepts, the x-coordinates remain unchanged while the y-coordinates are equal to the constant outside the modulus sign.

This enables us to determine quickly the coordinates of the sharp turning point(s) of modulus graphs.

Sample Question

Modulus Graph Coordinates Q1 Diagram

The diagram shows part of the graph of y = delim{|}{{x/2}+3}{|} - 2. Find the coordinates of points A, B and C.

[+] ANSWER

From the equation, you should be able straightaway obtain the coordinates C(−6, −2) in a record-breaking 1.0 × 10−100 seconds!

Too fast for you? OK …

x-intercept of y = {x/2} + 3 is (−6, 0) → obtained via x/2 + 3 = 0
y = delim{|}{x/2 + 3}{|} - 2 basically shifts y = {x/2} + 3 downwards by 2 units
⇒ its y-coordinate ↓ 2 units, its x-coordinate is unchanged → it has become the ‘sharp’ turning point of the modulus graph
⇒ C(−6, −2)

Finding Modulus Coordinates Answer Diagram

Coordinates for the intercepts A and B are found via simply setting x=0 and y=0 respectively, as learnt in your EMaths Coordinate Geometry.

Sub x = 0: y = |0/2 + 3| − 2 = 1 ⇒ A(0, 1)
Sub y = 0: |x/2 + 3| − 2 = 0
|x/2 + 3| = 2
x/2 + 3 = 2 or x/2 + 3 = −2 → you still remember how to remove the modulus sign right?
x = −2 or −10

From the graph, B(−10, 0)

Revision Question

Modulus Graph Coordinates Revision Question Diagram

The diagram shows part of the graph of y = |px − 2| + q where A(−1, −1) is the minimum point of the graph.

  1. State the value of q.
  2. Find the value of p.
  3. Find the coordinates of points B, C and D.

[Answer: 1. −1 2. −2 3. B(−3/2, 0), C(−1/2, 0), D(0, 1)]

[+] Modulus Graph Ranges & Inequalities

As mentioned previously, you’re more likely to solve a modulus inequality using a graphical approach, which is arguably a ‘safer’ method since you can really ‘see’ and convince yourself of the validity of the ranges in question.

This usually entails

  1. Sketching a modulus graph for a range of values of x – you’ll still be performing the same Sketch → Reflect → Flip → Shift acrobatic sequence but now there’s an extra step of including only the part of the graph bounded by the given range of x – which means you’ll also have to calculate and include in the final sketch the corresponding range of values of y.
  2. STARING at the graph(s) with your eyes open till you attain enlightenment as to the range you’re looking for.
Sample Question

Sketch the graph of y = −|2x − 5| for −1 < x ≤ 4. State the range of values of x for which y > −3.

[+] ANSWER

We’ll fast-forward and condense the Sketch → Reflect → Flip → Shift sequence and present the sketch for y = −|2x − 5| in a single diagram. TADA!

Modulus Inequality Graph Sketch

Now we calculate the values of y when x = −1 and 4.

Sub x = −1: y = −|2(−1) − 5| = −7
Sub x = 4: y = −|2(4) − 5| = −3

So we take only the portion of the graph that falls within −1 < x < 4 as our final sketch, with the corresponding values of y as indicated.

Modulus Inequality Graph Sketch With Boundaries

NOTE:

  1. Though many TYS solutions don’t do this, it’s always a good practice to insert the relevant circles at the extreme ends of the graph to indicate whether it’s ● (inclusive) or ○ (not inclusive).
  2. Some may prefer to calculate the boundaries and sketch y = 2x − 5 for the range first, before proceeding to Reflect → Flip → Shift within the boundaries to obtain the final sketch. There’s no right or wrong sequence to this as long as you don’t end up plotting a table of values!

Potential Goondu Moment: If you’re asked to state the corresponding range of y in this case, it’s −7 < y ≤ 0 NOT −7 < y ≤ −3!!! That’s why you must always STARE at the graph with your eyes open!

To find the range of x where y > −3, instead of trying to solve the inequality −|2x − 5| > −3 (and wondering at the end if your > < signs are pointing in the right directions), we simply draw the horizontal line y = −3 across the sketched graph and then proceed to STARE at it with your eyes big big:

Solving Modulus Inequality via Graphical Approach

If you’ve stared at it long enough, it’ll be as clear as the diamond on my Sensei‘s earring that you’re looking at the portion of y = −|2x − 5| that’s above y = −3 (shown in red).

Now all that’s left to do is to find the value(s) of x where the lines meet i.e.
−|2x − 5| = −3
|2x − 5| = 3
2x − 5 = 3 or 2x − 5 = −3
x = 4, 1.

∴ From the sketch, when y > −3, 1 < x < 4

Wow this turned out to be quite a long session today. I certainly hope my 口水 has not been in vain.

My Girl Girl … when can we go Bugis Junction to take Neoprints again?

*facepalm* Is that all you can say?!

Now do this question as part of your homework tonight!

Summary Question

  1. Solve |(x − 1)(x − 4)| = 2
  2. Sketch the graph of y = |(x − 1)(x − 4)|. Find the range of values of x for which y = |(x − 1)(x − 4)| < 2.

[Answer: i. 0.439, 2, 3, 4.56 ii. 0.439 < x < 2 or 3 < x < 4.56]

Oh by the way, where’s your completed homework for last week’s Binomial Theorem?

…

Where is it huh? Huh? HUH???!!! You want to feel moddy again?

… errr … my Girl Girl … when can we go Bugis Junction to take Neoprints again?

An unearthly scream was heard in the neighbourhood that night …

がんばって!!!
Sadako Loi
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The Road To Mathematical Reincarnation – Binomial Theorem

(5)
Posted at 12:51 am by Sadako Loi in A-Maths Notes & Tips

“Baby, baby, baby ohh … I thought you’d always be mine”, those emo words from Justin Bear Bear Bebe Barber Whoever’s Youtube video rang agonizingly off the study room walls as he stared in vain at his A-Maths notes.

Fresh from a breakup with a girl he’d known since he was 13, he found his eyes constantly flitting from his depressingly boring binomial theorem notes towards a depressingly heartbreaking Neoprint of the two of them taken at Bugis Junction in happier times.

With less than a month to the A-Maths Paper, he’s deep into Last-Minute Buddha Foot Hugging territory. But tried as he might, the scholarly, geeky student just couldn’t focus and that endlessly-looping chorus certainly didn’t help his droopy eyes …

The geeky tutor

The Geeky Tutor

Suddenly the song stopped and a breeze chilled the air in the room, following which he nearly jumped off his chair when he opened his eyes to the sight of a young bespectacled girl seated next to him.

*GASP* Who on EARTH are you?!!!

I’m Miss Loi. Your new maths tutor. Sorry to have startled you.

What??? Tuition now? It’s almost midnight!

With the exams so near and seeing you struggling to get over your breakup, your Mom called my tuition centre for help. Unfortunately we are full but my Sensei decided to send me, an intern tutor, to help you as part of my training. However I can only make it on very late nights so here I am.

Anyway since you seem to be diligently onto your Binomial Theorem notes right now (an oft-misunderstood topic that scared off lots of students due to its complicated and tedious-looking workings at first glance), let’s start on that shall we?

DEMYSTIFYING BINOMIAL THEOREM

The Binomial Expansion

(a+b)^n = a^n + (matrix{2}{1}{n 1})a^{n-1}b + (matrix{2}{1}{n 2})a^{n-2}b^2 + cdots + (matrix{2}{1}{n r})a^{n-r}b^r + cdots + b^n
This is given in the formula sheet – don’t waste your brain cells memorising it!

Expression for the r+1th term

T_{r+1} = (matrix{2}{1}{n r})a^{n-r}b^r

This is NOT given in the formula sheet – but if you can’t remember it you can easily ‘extract’ it from the main binomial expansion expression: Term formula hidden within Binomial Expansion Formula

Attention! Don’t simply assume the r+1th term Tr+1 is “the term in xr “!

Attention! The enigmatic “term independent of x“ and the “constant term” both refer to the term that contains x0.

Attention! Don’t always rush in headlong like a mad dog unleashed to expand the entire expression. Instead, keep the Tr+1 expression in mind when asked to find or compare coefficients of specific terms.

The nCr Formula

That (matrix{2}{1}{n r}) in the binomial expansion is not a matrix. Instead it’s given in the formula sheet as (matrix{2}{1}{n r})={n!}/{r!(n-r)!}={n(n-1) cdots (n-r+1)}/{r!}

Many students, however, ignore the above KPKB exclamation-laden expression and simply press the nCr button on their calculators – only to meet a tragic end when n is unknown in the question.

So for the sake of the questions where n is unknown, it’s worthwhile to be familiar with the following expressions for up till *nC4:

(matrix{2}{1}{n 1})=n/{1!}=n
(matrix{2}{1}{n 2})={n(n-1)}/{2!}
(matrix{2}{1}{n 3})={n(n-1)(n-2)}/{3!}
(matrix{2}{1}{n 4})={n(n-1)(n-2)(n-3)}/{4!}

*Most O Level AMaths binomial questions only require you to deal with up to the first 4 terms, but don’t take my word for it :) In any case, you should be able to ‘see’ the pattern for the expressions and use the x! button on your calculator to find that factorial (!) in the denominator if you need to obtain nCr expressions beyond the first 4 terms.

OK this, plus a firm grasp of your Rules of Indices, is ALL you need to know for Binomial Theorem!

Huh? That’s all? Like this I can also be a tutor myself! I wonder why Mom decided to engage you … hey where have you gone???

The student looked up to find that she had suddenly vanished.

I’m not done yet.

Turning around, he nearly fell off from his chair again at the sight of her seated behind him.

The Four Basic Binomial Theorem Questions

‘O’ Level AMaths binomial questions generally fall into four types, though a combination of them may be asked within the same question:

[+] A. Expand & Multiply

This is the classic binomial question where you expand an expression up to the first couple of terms, and then use your partial expansion to obtain the expansion of more a complicated expression. You may or may not be told how many terms to expand to, but you almost always should NEVER attempt to expand the entire expansion, especially if the powers are high.

Find, in ascending powers of x, the first 3 terms in the expansion of
(a) (1 + 4x)6 (b) (1 − 2x)14

Hence find the expansion of (1 + 4x)6(1 − 2x)14 up to the terms in x2.

ANS: Using the Binomial Expansion formula given in your formula sheet,

(a) (1 + 4x)6
= 16 + 6C1(15)(4x) + 6C2(14)(4x)2 + …
= 1 + 6(1)(4x) + 15(1)(16x2) + …
= 1 + 24x + 240x2 + …

(b) (1 − 2x)14
= 114 + 14C1(113)(−2x) + 14C2(112)(−2x)2 + …
= 1 − 14(1)(2x) + 91(1)(4x2) + …
= 1 − 28x + 364x2 + …

To obtain the first 3 terms of (1 + 4x)6(1 − 2x)14, we multiply and expand (1 + 24x + 240x2 + …)(1 − 28x + 364x2 + …) but we must know when to stop multiplying!

Know when enough is enough!

Since we’re only interested in terms up to x2, we ignore all combos that result in powers of x > 2 and kick them into oblivion for trying to waste your precious exam time!

(1 + 24x + 240x2 + …)(1 − 28x + 364x2 + …)
= 1(1 − 28x + 364x2) + 24x(1 − 28x + 364x2) + 240x2(1 − 28x + 364x2)
= 1 − 28x + 364x2 + 24x − 672x2 + 240x2
= 1 − 4x − 68x2

Attention! This is when your intimate knowledge of indices is called upon, especially with trickier expressions containing more than one x and inverse powers of x e.g. (x^3 - 2/{x^2})^10

Attention! Sometimes you’re NOT told how many terms to expand, but simply told to e.g. “expand up to and including the terms in x3” or “given that (1 + 4x)6(1 − 2x)14 = a + bx + cx2 + …, find a, b and c.” In such cases, you’re expected to know from the start how many terms to expand to, but for the sake of your future NEVER, EVER kill yourself by trying to expand the entire expression, especially when the powers are high e.g. (1 − 2x)14

Tekan Revision Exercise

Find the term independent of x in the expansion (2+1/{2x})^2(1-2x)^7.

[Answer: −3]

[+] B. Expand & Substitute

Instead of ‘Expand & Multiply’ describe above, this time you expand and substitute a suitable value/expression into your expansion in order to expand a more complicated expression, estimate a value, and achieve world peace.

Expand (1 − p)5. Use this result to find the expansion of (1 − x + x2)5 as far as the term in x3. Use a suitable value of x to estimate (1.11)5 from this expansion.

ANS: Using the Binomial Expansion formula given in your formula sheet,

(1 − p)5
= 15 + 5C1(14)(−p) + 5C2(13)(−p)2 + 5C3(12)(−p)3 + 5C4(12)(−p)4 + (−p)5
= 1 − 5p + 10p2 − 10p3 + 5p4 − p5

Do the Substitution!

To make (1 − x + x2)5 be like (1 − p)5, we equate 1 − p with 1 − x + x2:
⇒ 1 − p = 1 − x + x2
⇒ p = x − x2 → our substitute expression!

So sub p = x − x2 into the expansion:
= 1 − 5(x − x2) + 10(x − x2)2 − 10(x − x2)3 + 5(x − x2)4 − (x − x2)5
→ we’re only interested in terms up to x3, so we ignore all combos that result in powers of x > 3 to save time
= 1 − 5x + 5x2 + 10(x2 − 2x3 + x4) − 10(x3 + …)
→ Yay! Don’t have to expand that cubic expression as there’s only one x3 term in (x2 − 2x3 + x4)(x − x2)
= 1 − 5x + 5x2 + 10x2 − 20x3 − 10x3 + …
= 1 − 5x + 15x2 − 30x3 + …

Substitute again!

To estimate (1.11)5 using x, we equate 1.11 with 1 − x + x2
⇒ 1.11 = 1 − x + x2
⇒ x2 − x − 0.11 = 0
Solving the quadratic equation, x = −0.1 or 1.1

In order to use (1 − x + x2)5 = 1 − 5x + 15x2 − 30x3 + … to estimate (1.11)5, it’s better to use x = −0.1 since its higher powers are comparatively smaller enough to be ignored i.e. −0.14 = −0.0001 vs 1.14 = 1.4641

∴ (1.11)5 ≈ 1 − 5(−0.1) + 15(−0.1)2 − 30(−0.1)3 ≈ 1.68

[+] C. Finding Specific Terms

This is the kind of question where you’re asked to find “the term in x10“, “the term independent of x“, “the constant term”, “the coefficient of x4” etc. from a straightforward (a + b)n expression (vs the multiplied (a + b)n(c + d)m in preceding examples).

This is when the r+1th Term formula must instantly possess your mind, so that you curb your expansionary instincts and control yourself – don’t rush in blindly like a mad dog unleashed to expand the entire expression to the detriment of your precious exam time!

In the expansion of (x^3 - 2/{x^2})^10, find
(a) the term in x10
(b) the coefficient of 1/{x^5}
(c) the constant term

ANS: This is a straightforward (a + b)n expression ⇒ NO expansion is necessary (Control yourself! You don’t want to die young expanding this to the power of 10!)

Derive the General Tr+1 ‘Formula’ for the Expression

The key is to first derive a general Tr+1 ‘formula’ for this from the r+1th Term formula and simplify it using, once again, some indices magic:
T_{r+1} = (matrix{2}{1}{10 r})(x^3)^{10-r}(-2/{x^2})^r
{} = (matrix{2}{1}{10 r})x^{30-3r} (-2)^r x^{-2r}
{} = (matrix{2}{1}{10 r}) (-2)^r x^{30-5r}

⇒ power of x = 30 − 5r

And now, armed with this almighty expression for the power of x, you find yourself in ‘God Mode’ where, by a simple substitution of r with a suitable value, nothing, absolute nothing, can stand between you and the solutions to parts (a), (b) and (c) :twisted: Muahahaha!

(a) For x10, 30 − 5r = 10 ⇒ r = 4
∴ term in x10 = T_5 = (matrix{2}{1}{10 4})(-2)^4 x^10 = 3360x^10 (Muahahaha!)
→ x10 is included in the final answer since we are finding the term

(b) For x−5, 30 − 5r = −5 ⇒ r = 7
∴ coefficient of x−5 = (matrix{2}{1}{10 7})(-2)^7 = -15360 (Muahahahaha!!!)
→ x−5 is NOT included in the final answer since we are finding the coefficient

(c) For the constant term x0 (which is also the oft-used “term independent of x“), 30 − 5r = 0 ⇒ r = 6
∴ constant term = T_7 = (matrix{2}{1}{10 6})(-2)^6 x^0 = 13 440 (MUAHAHAHAHA!!! :twisted: )
→ we don’t include x0 in the final answer since it is a constant

Feels great to be powerful isn’t it? But please don’t write “Muahahaha” on your actual answer script :P

Tekan Revision Exercise

(i) In the binomial expansion of (x + k/x)^7, where k is a positive constant, the coefficients of x3 and x are the same. Find the value of k.

(ii) Using the value of k found in part (i), find the coefficient of x7 in the expansion of (1 - 5x^2)(x + k/x)^7

[Answer: k = 3/5, Coefficient of x7 = −20]

[+] D. When n is Unknown …

Whenever n is unknown, it’s almost certain that you’ll have to expand the nCr Formula into the {n(n-1) cdots (n-r+1)}/{r!} form within an expansion or Tr+1 formula (that nCr calculator button will suddenly become useless for those who like to keep pressing it), to obtain the value of n and some other variables via the comparison of the coefficients of suitable terms.

Given that (1 + ax)n = 1 − 12x + 63x2 + …, find a and n.

ANS: (1 + ax)n = 1n + nC11n−1(ax) + nC21n−2(ax)2 + …
→ since there’s only one x term with non-negative power in the expression, we only need to expand to the first 3 terms
= 1 + nax + {n(n-1)}/{2!}a2x2 + …
→ instead of trying to derive from the definition of nCr given in the formula sheet, being familiar with the nC1 to nC4 expressions will be handy here.

Compare Coefficients

Comparing our own expansion with the given expansion,

na = −12 —– (1) → compare coefficients of x
{n(n-1)}/{2!}a^2 = 63 —– (2) → compare coefficients of x2
Sub a = −12/n into (2):
{n(n-1)}/{2}(-12/n)^2 = 63
9n = 72
∴ n = 8
Sub n = 72 into (1):
∴ a = −12/8 = 3/2

Tekan Revision Exercise

In the expansion of (2 + 3x)n, the coefficients of x3 and x4 are in the ratio 8 : 15. Find the value of n.

[Answer: n = 8]

[+] Summary Exercise

To round things off, now that you have understood the 4 kinds of binomial questions I’ve just taught you, you should be able to solve (with one of your eyes closed) this question from the 2009 GCE ‘O’ Level AMaths Paper 2 that had somehow resulted in massive outpouring of grief among those who took the paper that year.

(i) Write down the first three terms in the expansion, in ascending powers of x, of (2-x/4)^n, where n is a positive integer greater than 2.

The first two terms in the expansion, in ascending powers of x, of (1+x)(2-x/4)^n are a + bx2, where a and b are constants.

(ii) Find the value of n.
(iii) Hence find the value of a and b.

[Answer: n = 8, a = 256, b = −144]

Now make sure you hand in your completed homework the next time I see you or you’ll be surprised at the nasty stuff I can do to you! I’ll be watching you ….

The Justin Bleah Bleah song started playing again, waking him from his slumber.

Realizing that he must have been talking in his sleep again, he wiped away the disgusting saliva that had stained his sleeve, before proceeding to attempt a couple of those depressingly-humorless questions in his binomial notes.

And then he suddenly realized that those questions actually fall into four main types …

がんばって!!!
Sadako Loi
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Miss Loi Is NOT A Million-Dollar Tutor!

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Posted at 11:16 pm by Miss Loi in Miss Loi the Tutor

Amid the ubiquitous images of F1 cars zipping through the pages of today’s Sunday Times, you’ll find a million-dollar link on the front page

Sunday Times, 26 September 2010
It’ F1 weekend … VROOM VROOOM!

to a million-dollar article on page 12,

Sunday Times Million-Dollar Tutors Article

about million-dollar tutors and their supposed million-dollar salaries,

The Bionic Million-Dollar Tutors
The Knights of the Million Dollar Round Table

where, according to the solitary source, there are whisperings of a secretive Million Dollar Round Table of Tutors, like the legendary King Arthur’s Knights of the Round Table.

The Knights of the Million Dollar Round Table
The Bionic Million-Dollar Tutors – they tend to last longer with Duracells.
Mic sold separately for sessions of >70 students.

Reading through the article, however, the math doesn’t seem to add up, as none of the figures put forth (from those of that champion economics tutor mass lecturer to the relatively tiny classes of yours truly) are anywhere near that magic number.

'Million-dollar Tutor' Celine Loi
Finally it’s the turn of the humble thousannaire tutor to speak

In any case, with the ‘O’ Levels looming in a month’s time, Miss Loi had better focus on her teaching her flock of 160 students who sometimes like to ‘fly Miss Loi’s aeroplane’ rather than ponder about monetary fantasies, and be contented with her life as a thousannaire just like all everyone else.

*wipes sweat from forehead*

But in any case, a warm welcome from Jφss Sticks to all Sunday Times readers , but too bad there’s no sexy maths tutor photo of her this time!

.
.
.

*clears throat*

*** BEGIN CODED MESSAGE

Now that I’ve seen them,
SOMEONE STOLE MY STICKERS!!!

:evil: :evil: :evil: :evil: :evil: :evil: :evil: :evil: :evil: :evil: :evil: :evil:

*** END OF CODED MESSAGE

*tidies hair*

がんばって!!!
Miss Loi