Jφss Sticks » Tuition Notes http://www.exampaper.com.sg Sassy O Level Maths Tuition, Questions & Tips from Singapore's Favourite Private Tutor Thu, 09 Feb 2012 05:31:17 +0000 en hourly 1 The Road To Mathematical Reincarnation – Modulus Functions http://www.exampaper.com.sg/tuition-notes/a-maths-tips/the-road-to-mathematical-reincarnation-modulus-functions http://www.exampaper.com.sg/tuition-notes/a-maths-tips/the-road-to-mathematical-reincarnation-modulus-functions#comments Mon, 18 Oct 2010 15:22:59 +0000 Sadako Loi http://www.exampaper.com.sg/?p=1941 *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! SIC sauce leh ~~~ " *tap tap tap* "this zehzeh i got from exampap ..." WAKE UP!!!!!!!!!!

© 2007-2011 Jφss Sticks. Copycats will be mercilessly hunted down.
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]]> *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 …

© 2007-2011 Jφss Sticks. Copycats will be mercilessly hunted down.
Please 'like' the Joss Sticks Society on Facebook! Miss Loi's Twitter

]]> http://www.exampaper.com.sg/tuition-notes/a-maths-tips/the-road-to-mathematical-reincarnation-modulus-functions/feed 1 The Road To Mathematical Reincarnation – Binomial Theorem http://www.exampaper.com.sg/tuition-notes/a-maths-tips/the-road-to-mathematical-reincarnation-binomial-theorem http://www.exampaper.com.sg/tuition-notes/a-maths-tips/the-road-to-mathematical-reincarnation-binomial-theorem#comments Fri, 08 Oct 2010 16:51:17 +0000 Sadako Loi http://www.exampaper.com.sg/?p=1755 Everything you need to know about scary-looking Binomial Theorem questions for this 'O' Levels! I hope :P

© 2007-2011 Jφss Sticks. Copycats will be mercilessly hunted down.
Please 'like' the Joss Sticks Society on Facebook! Miss Loi's Twitter

]]> “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 …

© 2007-2011 Jφss Sticks. Copycats will be mercilessly hunted down.
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]]> http://www.exampaper.com.sg/tuition-notes/a-maths-tips/the-road-to-mathematical-reincarnation-binomial-theorem/feed 5 Of Approximation And That Granny On A Trishaw http://www.exampaper.com.sg/tuition-notes/e-maths-tips/of-approximation-and-that-granny-on-a-trishaw http://www.exampaper.com.sg/tuition-notes/e-maths-tips/of-approximation-and-that-granny-on-a-trishaw#comments Thu, 23 Jul 2009 13:25:37 +0000 Miss Loi http://www.exampaper.com.sg/tuition-notes/e-maths-tips/of-approximation-and-that-granny-on-a-trishaw 三轮车，跑得快，上面坐個老太太。要五毛，給一块，你说奇怪不奇怪！

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]]> Pi turned rational
You DO get this, right?

As if the world hasn’t had enough of the geek calendar, today (being 22 July) (Edit: Actually it was yesterday – couldn’t post this in time :( ) happens to be Pi Approximation Day – a day to honour that often-used 22/7 fraction to approximate the value of π, though in reality 333/106, 355/113, 52163/16604 etc. are actually better approximations.

Speaking of approximation, Miss Loi would like to pay a little homage on this day to this most primal and basic of mathematical topics. A topic that many of us first may have first grasped from the Granny of that classic folk song:

『三轮车，跑得快，上面坐個老太太。
要五毛，給一块，你说奇怪不奇怪！』

Translation: Granny was so impressed with her fast & furious trishaw that she rounded off her $0.50 fare to the nearest dollar.

A topic that resurfaced later in life when a mean taxi uncle (who knew you haven’t been taught approximation in school yet) asked for $10 when the meter showed $9.70.

Or when you somehow instinctively brought along $50 to a sale to grab five items at $9.90 each, for nothing would be left if you had to return home to get more money.

Or that Miss Loi’s weight will always be 50 kg (correct to the nearest 10 kg) everytime you asked her …

In any case, do take a moment today to reflect on the following rules which are, well, supposedly so simple it’s laughable for anyone old enough not to be cheated by mean taxi uncles.

Rounding Off A Number

  1. Take the digit to the right of the specified (decimal) place/significant figure.

  2. If digit < 5, drop this digit/replace with zeros to keep place value. If digit ≥ 5, add 1 to digit on the left before dropping/replace with zeros to keep place value.

    要五毛，給一塊, and the 三輪車 Granny is always right!

  3. ALWAYS use/show at least 1 more decimal place/sig. fig. in your intermediate workings and only round off to the required decimal places/sig. fig. in your final answer.

    CASE-IN-POINT: See Miss Loi’s O-Level ‘significant’ careless mistake

Rules of Significant Figures

  1. All non-zero digits (i.e. 1-9) and zeros in between them are significant.
    e.g. 12 (2 sig. fig.), 12.5 (3 sig. fig.), 1.025 (4 sig. fig.)

  2. Zeros ARE significant UNLESS

    1. they are at the beginning of a decimal less than 1
      e.g. 0.007 (1 sig. fig.), 0.071 (2 sig. fig.), 1.007 (4 sig. fig.)
    2. *they are at the end of a whole number
      e.g. *87 000 (2, 3, 4 or 5 sig. fig.), 8.7000 (5 sig. fig.)

    *Depends on how estimation is made e.g.

    86 999.5 ≈ 87 000 (correct to 5 sig. fig.)
    86 995 ≈ 87 000 (correct to 4 sig. fig.)
    86 990 ≈ 87 000 (correct to 3 sig. fig.)
    86 900 ≈ 87 000 (correct to 2 sig. fig.)
    86 000 ≈ 90 000 (correct to 1 sig. fig.)

    Note how we round off the digit to the right of the required no. of significant figures.

Ultimately, however, do be aware that this seemingly-innocuous topic actually harvests itself in more than half of the questions in any O-Level exam that require numerical answers, as pre-warned on the cover page of every paper (which some of you never ever read :? ):

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. Or unless you want to self-PWN.
  • In addition, what has not (but should have) been stated in the instruction is that when it comes to money, the final answer should be in 2 decimal places (i.e. to the nearest cent) – a leading cause of grief in many exams :(

  • Also do note the following additional note in your syllabus document:

    “Unless stated otherwise within a question, three-figure accuracy will be required for answers. This means that four-figure accuracy should be shown throughout the working, including cases where answers are used in subsequent parts of the question. Premature approximation will be penalised, where appropriate. Angles in degrees should be given to one decimal place.”

And in this era of foreign cyborgs and “90% to get A1!” moderation, failure to adhere to the above Commandment has often led to minor loss of marks here and there that could sadly mean the difference between grades, euphoria and despair.

Significant Figure Careless Mistake
Be careful when using the value of an answer from an earlier part. Even Miss Loi fell prey to this ‘significant’ careless mistake in the 2008 O-Level EMaths Paper 2 *sigh*

But having said this, Miss Loi and some of her students have always wondered what exactly is a “non-exact numerical” answer?

What if an answer showed exactly 1.2345678 in your calculator – should you leave as such or round it off to 1.23 (3 sig. fig.)?

Hmmm … 你说奇怪不奇怪?

© 2007-2011 Jφss Sticks. Copycats will be mercilessly hunted down.
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]]> http://www.exampaper.com.sg/tuition-notes/e-maths-tips/of-approximation-and-that-granny-on-a-trishaw/feed 8 The 阴阳眼 Of Plane Geometry http://www.exampaper.com.sg/tuition-notes/a-maths-tips/the-yin-yang-eye-of-plane-geometry http://www.exampaper.com.sg/tuition-notes/a-maths-tips/the-yin-yang-eye-of-plane-geometry#comments Sun, 19 Oct 2008 12:34:10 +0000 Miss Loi http://www.exampaper.com.sg/tuition-notes/a-maths-tips/the-yin-yang-eye-of-plane-geometry A long time ago (okay not that long) in an old A-Maths Syllabus far, far away, there was an evil topic called Relative Velocity – where a mere mention of its name would strike fear into the hearts of students, caused teachers to quit, and made babies cry at night. So one can imagine the [...]

© 2007-2011 Jφss Sticks. Copycats will be mercilessly hunted down.
Please 'like' the Joss Sticks Society on Facebook! Miss Loi's Twitter

]]> A long time ago (okay not that long) in an old A-Maths Syllabus far, far away, there was an evil topic called Relative Velocity – where a mere mention of its name would strike fear into the hearts of students, caused teachers to quit, and made babies cry at night.

So one can imagine the national euphoria that reached epic proportions when the said topic was removed from the New Syllabus …

… only to be replaced by another fiendish nightmare called Plane Geometry (or Geometric Proofs to some) that’s causing widespread pandemonium for students across the nation, making teachers stutter in class, and … umm … the babies haven’t stopped crying.

While it’s easy to say that everything you need to know is summarized in the following (along with this Similarity & Congruency chart):

Geometric Formulae For Plane Geometry
Go on. Click to download and print this out.

The reality, however, is quite often that mortals like you and me will probably stare and stare at the Plane Geometry question till we end up like this:

Stressed StudentStressed Student II
Crazyhamster & DK, don’t mind Miss Loi borrow your handsome faces for illustrative purposes ok?

But we suppose that a ‘killer’ topic like this is meant to separate the haves from the have-nots in that cumulative frquency curve, for only those on the brink of their Mathematical Nirvana will get to experience that life-changing moment of sudden realization, when you reach your Mathematical high, and your knees begin to tremble as you feel the power of the Midpoint/Intercept Theorem flowing through your body as you look up to see a glowing image of Miss Loi amongst the clouds in the sky.

Mathematical Nirvana
They have just found the proof in their Plane Geometry question.

If you desire to join Stephen Chow, Cristiano Ronaldo and whoever’s that Miss Universe in these emotive Plane Geometry moments, you must develop the 阴阳眼 of Plane Geometry, for the signs within those complex geometrical diagrams will only reveal themselves magically to those who possess it, and only then will you be able to ‘see’ eerie things that others don’t.

Contrary to the myth that one can only be born with it (as suspected in most foreign cyborgs cases), the 阴阳眼 (lit. Yin Yang Eye) can actually be acquired through long hours of old-fashioned practice.

To that end, Miss Loi has toiled through the week to string together the following series of Plane Geometry questions that she sincerely hope can help improve your ‘eyesight’ in time for your O-Level A-Maths papers this week.

THE PLANE GEOMETRY EYESIGHT TEST

Warning PLEASE DO NOT CLICK ON THE ANSWERS BEFORE YOU’VE ATTEMPTED THE QUESTION! You will NEVER be able to develop the 阴阳眼 by merely reading through the answers!

For a start, don’t always insist on jumping in with the latest theorems. Sometimes all you need are just the basic geometrical properties you’ve learnt in your primary school and Sec One Maths (see part 1 of this question).

Also when you see an obvious lack of circles in the diagram, be prepared to seek out and use the Intercept Theorem or the Midpoint Theorem at some point.

Triangle Diagram

In the given figure, RT = RQ. If SR bisects ∠PRQ,

  1. Prove that ∠PRS = ∠RTQ.
  2. Show that RP/RQ = SP/SQ.



[+] Warning ANSWER - you shall suffer a terrible fate if you click and peek at the answer before you’ve tried the question!

Answer Diagram

  1. Let ∠RTQ = ∠RQT = θ (∠s of isosceles Δ)
    ⇒ ∠PRQ = 2θ (ext. ∠ = sum of interior opposite ∠s of Δ)

    Since SQ bisects ∠PRQ,
    ⇒ ∠PRS = ∠QRS = θ
    ⇒ ∠PRS = ∠RTQ

  2. And since ∠PRS = ∠RTQ
    ⇒ RS // TQ (corresponding ∠s)

    Using Intercept Theorem,
    RP/RT = SP/SQ doubleright RP/RQ = SP/SQ (∵ RQ = RT)

Instead of folding your arms and stare and stare at the diagram till you turn into stone, things can magically appear when you finally decide to lift up that 100kg pencil of yours to draw a couple of lines to join up all points in the diagram, for fortune tends to favour those who can lift and use their pencils.

Question 1 Diagram

Given a quadrilateral HIJK, let L, M, N and O denote the respective midpoints. Show that LMNO is a parallelogram.

[+] Warning ANSWER - you shall suffer a terrible fate if you click and peek at the answer before you’ve tried the question!

Answer 1 Diagram

MN // MN // LO (Midpoint Theorem)
NO // IK // ML (Midpoint Theorem)
⇒ LMNO is a parallelogram (∵ MN // LO & NO // ML

Ditto for this next question. You’ll be amazed how much 天机 (Heaven’s Will) is revealed by a single (correct) stroke of fate. Moreover, you’ll also discover that most Plane Geometry solutions are pretty short (typically 1-2 steps) since each part usually carries only 3-5 marks.

It’s the 阴阳眼 that you need!

Circle 1 Diagram

The diagram shows a circle with centre O and a chord AB. P is on AB and C is on the circle such that ∠OPC = 90°.

Prove PC2 = PA∙PB

[+] Warning ANSWER - you shall suffer a terrible fate if you click and peek at the answer before you’ve tried the question!

Answer Diagram

Extending the line PC to C’,
C’P = PC (Perdendicular bisector of chord)
⇒ C’P∙PC = PA∙PB (Intersecting Chords Theorem)
⇒ PC2 = PA∙PB (∵ C’P = PC)

Once again, join up the points dear – join up those points!

Circle 3 Diagram

ABCD is a cyclic quadrilateral. CE is a tangent to the circle at C. Given that AD // BC, prove that

  1. ∠BCA = ∠DCE
  2. ∠ABC = ∠BCD

[+] Warning ANSWER - you shall suffer a terrible fate if you click and peek at the answer before you’ve tried the question!

Answer Diagram

  1. ∠DCE = ∠DBC (Alternate Segment Theorem)
    ∠DBC = ∠DAC (∠s in same segment)
    ∠DAC = ∠BCA (alternate ∠s)
    ⇒ ∠BCA = ∠DCE

  2. Let ∠DCE = x
    ⇒ ∠DBC = x (Alternate Segment Theorem)
    ⇒ ∠ACB = x (alternate ∠s)
    Let ∠ABD = y
    ⇒ ∠ACD = y (∠s in same segment)

    ∴ ∠ABC = x + y = ∠BCD

Have your eyes warmed up sufficiently by now? Can you now see ‘them’ now in the more complicated diagram below? Yes, they’re everywhere, crying out to you.

Question Diagram ABT is a tangent to circle PBCQ. The circles ABCD and PBCQ intersect at B and C. APC and BQD are straight lines intersecting at X.

  1. Prove that PQ is parallel to AD.
  2. Show that XP = {AB^2~XQ}/{AC~DQ}.

[+] Warning ANSWER - you shall suffer a terrible fate if you click and peek at the answer before you’ve tried the question!

  1. Answer 1 Diagram

    ∠ACB = ∠ADB (∠s in same segment)
    ∠PQB = ∠ACB (∠s in same segment)
    ⇒ ∠ADB = ∠PQB (corresponding ∠s)

    ∴ PQ // AD





  2. Answer 2 Diagram

    Since PQ // AD,
    XP/PA = XQ/DQ (Intercept Theorem)
    doubleright XP = PA {XQ/DQ}

    Also,
    AC∙PA = AB2 (Tangent Secant Theorem)
    doubleright PA = AB^2/AC
    ∴ XP = {AB^2~XQ}/{AC~DQ}

And now one of the so-called ‘killer’ question to wrap things up.

Question 2 Diagram ABCD is a rectangle whose vertices lie on the circumference of a circle, centre O. X is the midpoint of AB, and AD = AX. Y lies on BD and XYC is a straight line. Z lies on the circumference of the circle and DXZ is a straight line. AC and BD intersect at O. The line CZ meets AB at P.

  1. Prove that ΔDXY is similar to ΔDZB.
  2. Explain why a circle that passes through A, P and Z can be drawn.

[+] Warning ANSWER - you shall suffer a terrible fate if you click and peek at the answer before you’ve tried the question!

  1. Answer 2a Diagram

    ΔDXY and ΔDZB share ∠ZDB.
    ∠DZB = 90° (∠ in semi-circle)
    ∠AXD = ∠BXC = (180°-90°)/2 = 45° (interior ∠s of isosceles Δ)
    ⇒ ∠DXC = 180°-45°-45° = 90° (adjacent ∠s)
    ⇒ ∠DXC = DZB
    ⇒ ∠DYX = DBZ (sum of interior ∠s of Δ)
    ⇒ ΔDXY similar to ΔDZB (AAA)

  2. Answer 2b Diagram

    ∠AZP = 180°-∠ADC (=90°) = 90° (Opposite ∠s of cyclic quadrilateral)
    ⇒ a circle that passes through A, P and Z (∵∠AZP is ∠ in semi-circle)

Most of the theorems have been covered till this point. And if you’ve successfully proven all the questions ON YOUR OWN, your knees will tremble as you receive that almighty gift of the 阴阳眼 Of Plane Geometry, knowing that no Plane Geometry question can stop you from now on.

If you don’t, you’ll just have to try the above questions all over again till you do ;)

Most importantly to all O-Level A-Maths students, may you find your 阴阳眼 in your A-Maths Papers this week!

GOOD LUCK!!!

P.S. Little Miss Loi wishes all Science students good luck for your O-Levels too!

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]]> http://www.exampaper.com.sg/tuition-notes/a-maths-tips/the-yin-yang-eye-of-plane-geometry/feed 24 Integration Amid Global Disintegration http://www.exampaper.com.sg/tuition-notes/a-maths-tips/integration-amid-global-disintegration http://www.exampaper.com.sg/tuition-notes/a-maths-tips/integration-amid-global-disintegration#comments Sun, 12 Oct 2008 11:31:14 +0000 Miss Loi http://www.exampaper.com.sg/tuition-notes/a-maths-tips/integration-amid-global-disintegration While the world deals with the small matter of the possible disintegration of its global financial system, students at Miss Loi's Temple have far more important things to worry about, like that big A-Maths Integration chapter. "Eh Miss Loi, how come these two Integration questions I never see before in my TYS one?"

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]]> Death By Integration
May the The Temple never see this
(click to enlarge)

With less than ten days (at the time of writing) to go, the skies darkened as the wind picked up, tossing helpless NTUC Fairprice plastic bags up and down like a stock market, as old symptoms (and a couple of new ones) began to surface once more.

While the world deals with the small matter of the possible disintegration of its global financial system, students at Miss Loi’s Temple have far more important things to worry about, like that big A-Maths Integration chapter.

Eh Miss Loi, how come these two Integration questions I never see before in my TYS one?

The One That Got Away
Latest Edition of Shing Lee Textbook
that missed its intended target

*Narrowly dodging the big Shing Lee A-Maths textbook that came hurling straight at his face He sat upright and was all ears …

Ooi! Dreaming of your crush again?!

You should know by now that two elements from the New A-Maths Syllabus have impishly made their way into the Integration chapter!

GEN-Y INTEGRATON

A. Integration of 1/(ax + b)

Starting this year, we have a new entry to the Integration scene:

For x > 0,
int{}{}{1/{ax + b}} dx = {1/a} ln (ax + b) + c —– (A1)

Having signed a formidable pact with Partial Fractions, the integrals will generally appear in the form of
int{}{}{1/{(ax+b)(cx+d)}} dx or int{}{}{{kx+h}/{(ax+b)(cx+d)}} dx
i.e. only linear factors/repeated linear factors in the denominator – very unlikely you’re see quadratic factors like (x2+b2) unless it’s a Hence question with a big clue in the first part.

So whenever you encounter something like the above, regardless of whether you’ve been explicitly told by the question to do so, you MUST perform your Partial Fractions 三大招式 (Three Devastating Moves) to express it in partial fractions first before proceeding to integrate using (A1) above.

SAMPLE QUESTION

  1. Express {8x+13}/{(1+2x)(2+x)^2} in partial fractions. Hence evaluate int{1}{2}{{8x+13}/{(1+2x)(2+x)^2}} dx.

    Ans:

    Expressing in partial fractions (reference here for the steps if you’re unsure):

    1. Check: degree of (8x+13) = 1 < degree of (1+2x)(2+x)2 = 3
      ⇒ PROPER (YAY!)

    2. {8x+13}/{(1+2x)(2+x)^2} = A/{1+2x} + B/{2+x} + C/(2+x)^2
      → 1 x linear factor (1+2x) and
      → 1 x repeated linear factor (2+x)2

    3. Multiply both sides by (1+2x)(2+x)2,
      8x+13 = A(2+x)2 + B(1+2x)(2+x) + C(1+2x)

      Sub x = -2, -½ … blah blah blah … we get

      {8x+13}/{(1+2x)(2+x)^2} = 4/{1+2x} - 2/{2+x} + 1/(2+x)^2

      Attention For all rogue Cover-Up Rule users, note that you can only use it to find the corresponding partial fractions for the terms with linear factors and the higher order of repeated linear factors

      e.g. For this question you can use the Cover-Up rule to find A/{1+2x} and C/(2+x)^2 but NOT B/{2+x}.

    So,
    int{1}{2}{{8x+13}/{(1+2x)(2+x)^2}} dx = int{1}{2}{4/{1+2x} - 2/{2+x} + 1/(2+x)^2} dx
    {} = 4 int{1}{2}{1/{1+2x}} dx - 2 int{1}{2}{1/{2+x}} dx + int{1}{2}{1/(2+x)^2} dx

    And now you can unleash the expression (in (A1) above) upon these integrals!

    Attention At this point, many students will ‘ln ln ln’ till they’ve forgotten how to integrate the 3rd term i.e. the 1/(2+x)2.

    Remember this?

    int{}{}{(ax+b)^n} dx = 1/a {(ax+b)^{n+1}}/{n+1} + c

    {} = 4 delim{[}{{1/2}ln(1+2x)}{]}^2_1 - 2delim{[}{ln(2+x)}{]}^2_1 - delim{[}{1/{2+x}}{]}^2_1
    {} = 2 delim{[}{ln 5 - ln 3}{]} - 2 delim{[}{ln 4 - ln 3}{]} - delim{[}{1/4 - 1/3}{]}
    {} = 2 ln (5/4) + 1/12

  2. Now can you try evaluating int{2}{3}{{9-4x}/{(2x+3)(x-1)^2}} dx? ;)

B. Integrating Trigonometric Functions Using FURTHER TRIGONOMETRIC IDENTITIES

Basic Rules of Integrating Trigonometric Functions

Firstly, you should all know these by now:

  1. int{}{}{cos (ax + b)} dx = 1/a sin (ax + b) + c
  2. int{}{}{sin (ax + b)} dx = -1/a cos (ax + b) + c - note the minus sign!
  3. int{}{}{sec^2 (ax + b)} dx = 1/a tan (ax + b) + c
  4. int{}{}{tan^2 (ax + b)} dx
    {}= int{}{}{delim{[}{sec^2 (ax + b) - 1}{]}} dx (∵ sec2x = 1 + tan2x)
    {} = 1/a tan (ax + b) - x + c

Starting this year, however, strange forms of trigonometric functions may be pleading for your integrating wizardry in the O-Levels e.g. int{}{}{sin^2 x} dx, int{}{}{sin x cos x} dx, int{}{}{sin 2x cos 3x} dx etc.

These are brought to you courtesy of the new Trigonometric Identities of the new syllabus, namely the Double Angle Formulae, Addition Formulae and the Factor Formulae.

While these formulae will be provided in your exam, you’re likely to find yourself frequently using the following derived Double Angle expressions to solve these new wave integration problems:

  1. sin x cos x = {1/2} sin 2x —– (B1)
  2. cos^2 x = {1 + cos 2x}/2 —– (B2)
  3. sin^2 x = {1 - cos 2x}/2 —–(B3)

Armed with these, we can now join the Gen-Y Integration Movement!

SAMPLE QUESTIONS

  1. int{}{}{sin^2 x} dx = int{}{}{{1 - cos 2x}/2} dx - Using (B3) above
    {} = {1/2} int{}{}{delim{[}{1 - cos 2x}{]}} dx
    {} = {1/2} (x - {1/2} sin 2x) + c

  2. int{}{}{sin x cos x} dx = int{}{}{{1/2} sin 2x} dx - Using (B1) above
    {} = {1/2}int{}{}{sin 2x} dx
    {} = {1/2}({-1/2}cos 2x) +c = {-1/4} cos 2x +c

  3. int{}{}{sin 2x cos 3x} dx
    {} = {1/2}int{}{}{2 cos 3x sin 2x} dx
    {} = {1/2}int{}{}{delim{[}{sin 5x - sin x}{]}} dx
    (Using Factor Formula sin A sin B = 2 cos ½(A+B) sin ½(A-B) )
    {} = {1/2}({-1/5}cos 5x + cos x) + c = {cos x}/2 - {cos 5x}/10 + c

Now can you evaluate these? ;)

  1. int{pi/4}{pi/2}{delim{[}{sin^2x - 4 sin x cos x}{]}} dx
  2. int{}{}{(sin x - cos x)^2} dx

P.S. Check out this online integrator for quick answers to your integration problems.

Attention*LAST BUT NOT LEAST Miss Loi is too gentle and demure to even harm an ant, much less hurl big Shing Lee textbooks at daydreaming students.

© 2007-2011 Jφss Sticks. Copycats will be mercilessly hunted down.
Please 'like' the Joss Sticks Society on Facebook! Miss Loi's Twitter

]]> http://www.exampaper.com.sg/tuition-notes/a-maths-tips/integration-amid-global-disintegration/feed 5 Trigonometry Type R http://www.exampaper.com.sg/tuition-notes/a-maths-tips/trigonometry-type-r http://www.exampaper.com.sg/tuition-notes/a-maths-tips/trigonometry-type-r#comments Tue, 07 Oct 2008 06:05:22 +0000 Miss Loi http://www.exampaper.com.sg/tuition-notes/a-maths-tips/trigonometry-type-r The pride of racers and the probable source ofnoise pollution in Miss Loi’s estate at 2am Like Honda’s Type R variant of cars that appeal to Ah Beng racer boys who like to bully weave in front of Miss Loi’s demure little car on the expressways, Trigonometry has its own “Type R” model in the [...]

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]]> Type R Logo
The pride of racers and the probable source of
noise pollution in Miss Loi’s estate at 2am

Like Honda’s Type R variant of cars that appeal to Ah Beng racer boys who like to bully weave in front of Miss Loi’s demure little car on the expressways, Trigonometry has its own “Type R” model in the form of the R-Formula that was unveiled in this year’s New AMaths Syllabus.

As such students are warned to listen out for the roar of the R-Formula whenever you encounter a trigonometric question:

  1. with an expression that contains a constant, like this:

    Solve for 0° < x < 360°,
    2 cot x = 3 + 2 cosec x

  2. or where you’re asked to find an expression’s max/min value, like in this rather cheong hei (long-winded) question:

    Trignometry R-Formula Question Diagram

    The diagram shows two right-angled triangles with their right angles at B and N. The sides OB and AB are of length 6 cm and 2 cm respectively. The lines OB is inclined at angle θ to ON. The line AM is perpendicular to ON.

    1. Express OM in the form OM = R cos (θ + α) and calculate the values of R and α.
    2. Find the value of θ for which OM = 5 cm.
    3. In the diagram, state the line with a length of R cm and the angle with a value α.
    4. Express AM in terms of R and (θ + α) and show that the area of triangle OAM is 10 sin 2(θ + α).
    5. Given that θ can vary, find the maximum value of the area of triangle OAM and the corresponding value of θ when this occurs.

To handle the new R-Formula questions, students need to recall the following expressions, as they will not be appearing on your formula sheet during your actual exam:

The R-Formula

R-Formula Triangle

  • a cos θ ± b sin θ = R cos (θ ∓ α) - look out for that inverted plus/minus!
  • a sin θ ± b cos θ = R sin (θ ± α)

where R = sqrt{a^2 + b^2} and tan alpha = b/a
and a > 0, b > 0, α is acute

Maximum/Minimum Values of R-Formula Expressions

  • The maximum value is R which occurs when cos/sin (θ ± α) = 1
  • the minimum value -R which occurs when cos/sin (θ ± α) = -1

Deriving The R-Formula

Do note that the R-Formula highlighted above is NOT a given, especially when the relevant part of the question carries a high weightage of marks e.g.

Express 4 cos θ – 3 sin θ in the form of R cos (θ+α) [6 Marks]

There’s a bit of debate on this but don’t think you can get away with all 6 marks here by simply substituting the values into the R-Formula! So to be a little kiasu, it’s advisable to derive the whole thing as per the following steps:

4 cos θ – 3 sin θ
= R cos (θ+α)
= R [cosθcosα-sinθsinα] (Addition Formula)
= (R cosα)cosθ-(R sinα)sinθ

Comparing coefficients between LHS & RHS,
R cosα = 4 —– (1)
R sinα = 3 —– (2)

(2)/(1): tan α = 3/4 ⇒ α = 36.9°

(1)2+(2)2:
R2cos2α+R2sin2α = 42+32
R2[cos2α+sin2α] = 25 ⇒ R = 5

∴ 4 cos θ – 3 sin θ = 5 cos(θ+36.9°)

Important Notes

Warning! To reiterate what was stated above, a and b MUST be positive. DON’T do this:

(-3) cos θ + (-2) sin θ = 3 → WRONG!

Instead, move the terms around to ensure that the first term is always positive:
3 cos θ + 2 sin θ = -3 → CORRECT! YAY!

Warning! Please, please, please, please, please for the addition cases (i.e. a cos θ + b sin θ or a sin θ + b cos θ) you can simply re-arrange the sine and cosine terms depending on how you wish to express the R function in your question.

E.g.

3 cos θ + 2 sin θ → if you require R cos (θ – α)
2 sin θ + 3 cos θ → if you require R sin (θ + α)

Hailed as the pinnacle of O-Level Trigonometry, a typical R-Formula question may also require your skills on lesser trigonometry variants such as the Addition Formulae, or the Double Angle Formulae, or the Factor Formulae, together with your knowledge of basic angles (e.g. in question 1 above when you’re asked to find all angles within the range of blah blah blah).

But once you tame it, you’ll be rewarded with a straightforward ride and zoom away with plenty of marks.

So are you ready to test-drive your R-Formula with the two questions above? Miss Loi doesn’t expect you to crash this in your O-Level in two weeks’ time!

*Zooms away for Jφss Sticks Sessions (Type-R variant)*

N.B. Once again, Miss Loi does NOT advocate nor encourage speeding or any form of street racing in Singapore.

© 2007-2011 Jφss Sticks. Copycats will be mercilessly hunted down.
Please 'like' the Joss Sticks Society on Facebook! Miss Loi's Twitter

]]> http://www.exampaper.com.sg/tuition-notes/a-maths-tips/trigonometry-type-r/feed 25 Sergeant Loi’s Mid-Year Boot Camp 2008 – Finding Your Roots With Remainder & Factor Theorems http://www.exampaper.com.sg/tuition-notes/a-maths-tips/sergeant-lois-mid-year-boot-camp-2008-finding-your-roots-with-remainder-factor-theorems http://www.exampaper.com.sg/tuition-notes/a-maths-tips/sergeant-lois-mid-year-boot-camp-2008-finding-your-roots-with-remainder-factor-theorems#comments Tue, 06 May 2008 09:30:09 +0000 Sergeant Loi http://www.exampaper.com.sg/tuition-notes/a-maths-tips/sergeant-lois-mid-year-boot-camp-2008-finding-your-roots-with-remainder-factor-theorems This heavy thing is causing permanent damageto her previously-rebonded hair One more time … *Puts on helmet* *SHOWS STERN & MEAN & TIRED FACE* *Sounds bugle* EVERYBODY FALL IN! Most of you would’ve started your Mid-Year Exams by now – a series of no-holds-barred trials to determine once and for all if indeed a wind [...]

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Please 'like' the Joss Sticks Society on Facebook! Miss Loi's Twitter

]]> Sergeant Loi's Helmet
This heavy thing is causing permanent damage
to her previously-rebonded hair

One more time …

*Puts on helmet*

*SHOWS STERN & MEAN & TIRED FACE*

*Sounds bugle* EVERYBODY FALL IN!

Most of you would’ve started your Mid-Year Exams by now – a series of no-holds-barred trials to determine once and for all if indeed a wind tunnel exists between your ears to test your understanding of topics taught in Semester 1.

For some, this could also be a time for your ‘chers to avenge all the tortures you’ve subjected them to throughout the term. As such, the Mids always tend to be a little on the sadistic side, and strewn with devious tricks around every turn and corner.

That’s why Miss Sergeant Loi (whose Teresa Teng voice is now hoarse from all the shouting) is here – to hopefully help save you a mark or two, to give you that little bit of edge from being pwned by your ‘chers.

So let’s do this one more time (to complete the chapter on Polynomials) … for now …

Remainder & Factor Theorems

A. REMAINDER THEOREM

If f(x) is divided by (x - a) ⇒ the remainder is f(a)

e.g. Find the remainder when 4x3 – 5x + 1 is divided by:
i. x-2, ii. x+3, iii. 2x-1

Ans: Let f(x) = 4x3 – 5x + 1. Remainder, R =

  1. f(2) = 4(2)3-5(2)+1 = 23
  2. f(-3) = 4(-3)3-5(-3)+1 = -92
    → note it’s divided by x (+) 3 so you’re have to sub in (-)3 instead
  3. f(½) = 4(½)3-5(½)+1 = -1
    → note when divided by (2x-1) → you’ll have to convert it to the form (x-½) first and then sub in the ½

Warning! DON’T waste time doing long division in remainder theorem questions!!!

B. FACTOR THEOREM

If f(x) is divided by (x – a) and the remainder is 0 ⇔ f(a)=0
⇒ (x – a) is a factor of f(x)
⇒ f(x) is exactly divisible by (x – a)

From your Sec Two Expansion & Factorisation chapter:

  1. Expansion → remove brackets
  2. Factorisation → put back brackets ⇒ final answer must always be in brackets!

e.g. Factorise x2 – 5x + 6

Ans via Trial & Error (try getting this under 10 sec :D ):

Choose 2 factors of the constant 6
try: 1 x 6 → 1x + 6x ≠ -5x (reject)
try: 2 x 3 → 2x + 3x ≠ -5x (reject)
try: (-2) x (-3) → -2x + (-3)x = -5x (YAY!)
⇒ cross-check: f(3) = f(2) = 0 (YAY!)
⇒ x2 – 5x + 6 = (x – 3)(x – 2)

Warning! When you see the keyword Factorise, final answer must be in (brackets) i.e. don’t try to be funny and write x = 3, 2 → minus marks!

C. SOLUTION OF EQUATIONS

When you spot the keywords Solve and/or = 0 in your exam question, it means you’ll normally have to:

  1. Find the factors of an equation f(x) (usually cubic)
  2. Find the roots of f(x)=0 (i.e. final answer must be in the form: x = a, b … where a, b, … are the roots)

e.g. Solve 3x3 – 10x2 + 9x – 2 = 0

Ans: Let f(x) = 3x3 – 10x2 + 9x – 2.

  1. Find the first factor via trial and error

    Try x=1: f(1) = 3(1)3 – 10(1)2 + 9(1) – 2 = 0 (YAY!)
    ⇒ (x-1) is a factor

  2. Find the remainder expression by either COMPARING COEFFICIENTS:

    Factorization by Comparing Coefficients

    OR LONG DIVISION (if you’re a long division aficionado)

    Factorization by Long Division

    You should get the SAME expression either way – use which ever method you’re more comfortable with (use one method to cross check the other if you’re one of those with lotsa free time left in your exam).

  3. Factorize the remaining quadratic expression 3x2-7x+2 (via quick Trial and Error method described in B above):

    Choose 2 factors of the constant 2
    try: (-2) x (-1) → (3)(-2)x + (-1)x = -7x (YAY!)
    → Note the coefficient of 3 of the x2 term
    ⇒ 3x2 – 7x + 2 = (3x – 1)(x – 2)
    → Note it’s NOT (3x – 2)(x – 1) coz you need to corss-multiply
    ⇒ cross-check: f(⅓) = f(2) = 0 (YAY!)

    ⇒ 3x3 – 10x2 + 9x – 2 = (x-1)(3x-1)(x-2) = 0
    ⇒ x = 1, 2, ⅓

Warning! When you see the keywords Solve and/or = 0, final answer must be in the form x = a, b … i.e. don’t stop at factorising → minus marks!
Warning! Sometimes the quadratic equation in Step 3 cannot be easily factorised → you’ll have to use the Quadratic Formula to find the two solutions. You’ll normally get the hint when you see terms like ±√ within the question.

SAMPLE PRACTICE QUESTION

The cubic polynomial f(x) is such that the coefficient of x3 is -1 and the roots of the equation f(x) = 0 are 1, 2 and k. Given that f(x) has a remainder of 8 when divided by x-3, find

  1. the value of k,
  2. the remainder when f(x) is divided by x+3

Ans:

Since 1, 2 and k are roots, a(x-1)(x-2)(x-k) = 0
→ straightaway write down in factorized form once roots are known
→ always remember to include the coefficient a for x3 for it may not always be 1!

And since coefficient of x3 = -1
⇒ a = -1
⇒ (-1)(x-1)(x-2)(x-k) = 0

Let f(x) = (-1)(x-1)(x-2)(x-k)

  1. Since remainder is 8 when divided by (x-3),
    f(3) = (-1)(3-1)(3-2)(3-k) = 8
    (using Remainder Theorem from A above)
    ⇒ k = 7
  2. Now using k = 7 above, f(x) = (-1)(x-1)(x-2)(x-7)

    Remainder when divided by x+3:
    → f(-3) = (-1)((-3)-1)((-3)-2)((-3)-7) = 200

*For some reason, students have a habit of expanding the entire expression after they’ve written down everything in factorized form = what a waste of time. Tsk.

As always, get these rules drilled into your head! Spot the pointers and common mistakes in red! Understand the representative sample question! Check out further questions on factor theorem!

Print this out if necessary and remember the above procedures by heart … and do let Miss Loi know which topics and stuffs you would like to see in her next set of Maths Notes ;)

Till then, understand that the ultimate root of your own equation is to prepare youself in mind and in soul for the Great War at year’s end. So don’t be afraid to make all the mistakes you need to make now (as long as you know what mistakes you’re making!).

Good Luck For Your Mids!
P.S. To the reader who longs and yearns to see Miss Loi’s divine face again, a very heartbroken and upset Sergeant Loi was last seen charging out of camp with her Katana sword, vowing to hunt this reader down, and slice him into many pieces and use the Remainder Theorem to turn whatever that remains of him into 人肉叉烧包 (human buns)!

© 2007-2011 Jφss Sticks. Copycats will be mercilessly hunted down.
Please 'like' the Joss Sticks Society on Facebook! Miss Loi's Twitter

]]> http://www.exampaper.com.sg/tuition-notes/a-maths-tips/sergeant-lois-mid-year-boot-camp-2008-finding-your-roots-with-remainder-factor-theorems/feed 11 Sergeant Loi’s Mid-Year Boot Camp 2008 – Rationalising The Rationale Of Surds http://www.exampaper.com.sg/tuition-notes/a-maths-tips/sergeant-lois-mid-year-boot-camp-2008-rationalising-the-rationale-of-surds http://www.exampaper.com.sg/tuition-notes/a-maths-tips/sergeant-lois-mid-year-boot-camp-2008-rationalising-the-rationale-of-surds#comments Sat, 03 May 2008 16:38:09 +0000 Sergeant Loi http://www.exampaper.com.sg/tuition-notes/a-maths-tips/sergeant-lois-mid-year-boot-camp-2008-rationalising-the-rationale-of-surds Don’t worry.These aliens are often friendlier than they look. *Puts on helmet* *SHOWS STERN & MEAN FACE* *Sounds bugle* EVERYBODY WAKE UP! Sergeant Loi is back from her May Day leave – only to find The Temple in a state of 兵慌马乱 as the mid-year exams loom. To whip everyone back to shape, Sergeant Loi [...]

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Please 'like' the Joss Sticks Society on Facebook! Miss Loi's Twitter

]]> Surds
Don’t worry.
These aliens are often friendlier than they look.

*Puts on helmet*

*SHOWS STERN & MEAN FACE*

*Sounds bugle* EVERYBODY WAKE UP!

Sergeant Loi is back from her May Day leave – only to find The Temple in a state of 兵慌马乱 as the mid-year exams loom.

To whip everyone back to shape, Sergeant Loi shall complete the trinity (which began with indices, and followed by logarithms) with the concluding drill on surds today!

Despite the scary-looking longish workings, surds are the most straightforward part of the trinity. For most parts, the TWO main things you’ll ever need to know are how to simplify (with a dose of trial-and-error common sense) and rationalise surd expressions.

That’s it! Now let’s go eat some surds for breakfast!

*Cracks whip*

The Rationale of Surds

A. WHAT IS A SURD?

High-class definition:
An irrational root of a real number. *sweats*

Simple O Level definition:
A *square-root of a number that results in endless decimal places when you press your calculator e.g.

  1. √2 is a surd (=1.4142135623 … when you press your calculator)
  2. √9 is NOT a surd (= 3 when you press your calculator – WOW NO DECIMALS!)

*Actually can also be cube-root, nth-root etc. but let’s not go there …

B. THE OPERATIONS OF SURDS

  1. sqrt{a}*sqrt{a} = a
  2. sqrt{a}*sqrt{b} = sqrt{ab}
  3. sqrt{a}/sqrt{b} = sqrt{a/b}
  4. m sqrt{a}+n sqrt{a} = (m+n)sqrt{a}
  5. m sqrt{a}-n sqrt{a} = (m-n)sqrt{a}

*Compare operations B(2) & B(3) to the Rules of Indices (Part C) and you’ll realize they’re like long-lost twins!

e.g. Simplifying surd expressions:

  1. √18 = √(9 × 2) = √9 × √2 = 3√2
    → Using B(2) and a tiny bit of trial and error to get the (9 × 2)
  2. 3√2 + 5√2 = (3 + 5)√2 = 8√2 → Using B(4)
  3. (√5 – √2)2 = (√5 – √2)(√5 – √2)
    = √5√5 – √5√2 – √2√5 + √2√2
    = 5 – √10 – √10 + 2 → Using B(1) + B(2)
    = 7 – 2√10 → Using B(4)
Warning! DON’T ever do this:

√(a+b) ≠ √a + √b → WRONG!
√(a-b) ≠ √a – √b → WRONG!

e.g.
√64 ≠ √32 + √32!
√25 ≠ √12 + √13!

C. CONJUGATE SURDS

(h√a + k√b)(h√a – k√b) = h2a - k2b
(Note the MINUS sign on the RHS)

e.g. (√3 + √2)(√3 – √2) = 3 – 2 = 1

Warning! DON’T ever do these:

(√a – √b)2 ≠ (√a2 – √b2) → WRONG!
(√a + √b)(√a – √b) ≠ (√a)2 + (√b)2 → WRONG!

D. RATIONALISING THE DENOMINATORS OF SURDS

Having a surd in the denominator is IMPROPER, UNGLAM, SINFUL & IMMORAL!!!

So we need to rationalise the denominators by multiplying the top and bottom by the same number with respect to the denominator i.e.

  1. If denominator consists of a single term → multiply top & bottom by denominator term e.g.
    1/sqrt{3} = {1/sqrt{3}} * {sqrt{3}/sqrt{3}} = sqrt{3}/3

  2. If denominator consists of 2 terms → multiply top & bottom by conjugate of denominator e.g.
    1/{sqrt{2}-1} = {1/{sqrt{2}-1}} * {{sqrt{2}+1}/{sqrt{2}+1}} = {sqrt{2}+1}/{(sqrt{2}-1)(sqrt{2}+1)}
    {}= {sqrt{2}+1}/{sqrt{2}^2-1^2} = {sqrt{2}+1}/{2-1} = sqrt{2}+1

SAMPLE PRACTICE QUESTIONS

  1. Simplify {4-sqrt{5}}/{2sqrt{5}+3}.

    Ans:

    {{4-sqrt{5}}/{2sqrt{5}+3}} * {{2sqrt{5}-3}/{2sqrt{5}-3}}
    → multiply top & bottom with conjugate surds (C)
    {} = {(4-sqrt{5})(2sqrt{5}-3)}/{(2sqrt{5})^2 - (3)^2}
    {} = {8 sqrt{5} - 12 - 10 + 3 sqrt{5}}/{4(5)-9} = {11 sqrt{5} - 22}/{11} = sqrt{5} - 2

  2. Given that {2/sqrt{2}}({{3sqrt{54}}/4}-{15/sqrt{150}}-{14/sqrt{294}})=k sqrt{3}, find the value of k.

    Ans:

    → Rationalize all the sinful terms with surds in denominator!
    {2/sqrt{2}}*{sqrt{2}/sqrt{2}}({{3sqrt{54}}/4}-{15/sqrt{150}}*{sqrt{150}/sqrt{150}}-{14/sqrt{294}}*{sqrt{294}/sqrt{294}})
    {}=k sqrt{3}
    doubleright {{2 sqrt{2}}/2}({{3sqrt{54}}/4}-{{15 sqrt{150}}/150}-{{14 sqrt{294}}/294})=k sqrt{3}
    → Now simplify the surds using B(2) above and a tiny bit of trial and error
    doubleright sqrt{2}({{3sqrt{9*6}}/4}-{{15 sqrt{25*6}}/150}-{{14 sqrt{49*6}}/294})=k sqrt{3}
    doubleright sqrt{2}({{9sqrt{6}}/4}-{{sqrt{6}}/2}-{{sqrt{6}}/3})=k sqrt{3}
    doubleright sqrt{2}(17/12)sqrt{6}=k sqrt{3}
    → split √6 into √2√3 since we’re going to compare with the √3 term on RHS
    doubleright sqrt{2}(17/12)sqrt{2}sqrt{3}=k sqrt{3}
    doubleright (34/12)sqrt{3} = (17/6)sqrt{3} =k sqrt{3}
    doubleright k = 17/6 via comparing coefficient of √3

As always, get these rules drilled into your head! Spot the pointers and common mistakes in red! Understand the representative sample questions! Check out more surds in action!

Print this out if necessary and remember the above procedures by heart, for if you fail in rationalizing surds in your exams, you’ll need to write a 1000000-word essay to Sergeant Loi explaining the rationale for not punishing you!

*Cracks whip*

© 2007-2011 Jφss Sticks. Copycats will be mercilessly hunted down.
Please 'like' the Joss Sticks Society on Facebook! Miss Loi's Twitter

]]> http://www.exampaper.com.sg/tuition-notes/a-maths-tips/sergeant-lois-mid-year-boot-camp-2008-rationalising-the-rationale-of-surds/feed 7 Of Matrices & Miss Loi’s Day Of Labour http://www.exampaper.com.sg/tuition-notes/a-maths-tips/of-matrices-miss-lois-day-of-labour http://www.exampaper.com.sg/tuition-notes/a-maths-tips/of-matrices-miss-lois-day-of-labour#comments Thu, 01 May 2008 13:19:35 +0000 Miss Loi http://www.exampaper.com.sg/tuition-notes/a-maths-tips/of-matrices-miss-lois-day-of-labour May 1st (by virtue of its proximity to the Mids) has traditionally been Miss Loi's Day of Labour. With the advent of The Temple, this year's Great Labour Day Racing Circuit certainly looks tame compared to last year's.

© 2007-2011 Jφss Sticks. Copycats will be mercilessly hunted down.
Please 'like' the Joss Sticks Society on Facebook! Miss Loi's Twitter

]]> Miss Loi's 2008 Day of Labour Route
This year’s Great Labour Day Racing Circuit
certainly looks tame compared to last year’s

*Takes off helmet*

*Shows kindly & benign face*

On a day when Singapore’s working class was left to reflect on their social standing in the current status quo, May 1st (by virtue of its proximity to the Mids) has traditionally been Miss Loi’s Day of Labour.

But with the advent of The Temple, the distance travelled is much less this year, with a corresponding drop in the number of stunts overtaking manoevuers needed in order to get to her students’ place on time.

Which also means that Miss Loi actually has time to help that butchy Sergeant Loi (who is on leave today and thereby sparing the serene Temple Grounds from her KPKB for the first time this week) post her Matrices notes:

Ever since the bulk of the Matrices chapter has been migrated over to the New EMaths Syllabus, according to Section 1.4 of the AMaths Syllabus, about the only time you’ll ever see matrices appearing in your AMaths papers would be when you’re asked to solve a pair of simultaneous equations using the inverse matrix method.

The following, then, is really all you need to know :)

Solving Simultaneous Equations Via Inverse Matrix

A. HOW TO MULTIPLY MATRICES

Multiplication of Matrices

B. DETERMINANT OF A 2X2 MATRIX

Determinant of matrix M = ({matrix{2}{2}{a b c d}}) is given by:
|M| = delim{|}{matrix{2}{2}{a b c d}}{|} = ad – bc

If |M| = 0 ⇒ M is singular.
If |M| ≠ 0 ⇒ M is non-singular.

C. INVERSE OF A 2X2 MATRIX

For a non-singular matrix M = ({matrix{2}{2}{a b c d}}),
Inverse Matrix

D. THE INVERSE MATRIX METHOD

AX = B ⇒ X = A-1B

*Note: AX = B does NOT ⇒ X = BA-1!

SAMPLE PRACTICE QUESTION

Use the inverse matrix method to solve the simultaneous equations:

5x + y = -8
-x + 2y = 17

Ans:

*See the words inverse matrix method in question – SIANZ!*

  1. Can you see that the pair of simultaneous equations are actually part of matrix multiplications shown in A above? So you can convert to matrix form:
    (matrix{2}{2}{5 1 {-1} 2})(matrix{2}{1}{x y}) = (matrix{2}{1}{{-8} 17})

    You can then move the 2 x 2 matrix to the RHS of the equation as highlighted in D above:
    doubleright (matrix{2}{1}{x y}) = (matrix{2}{2}{5 1 {-1} 2})^-1(matrix{2}{1}{{-8} 17})

  2. Now to find the inverse matrix (matrix{2}{2}{5 1 {-1} 2})^-1:

    As described in B above:
    Determinant → delim{|}{matrix{2}{2}{5 1 {-1} 2}}{|} = (5)(2)-(1)(-1) = 11 ≠ 0
    ⇒ non-singular
    ⇒ inverse exists!

    As described in C above:
    So (matrix{2}{2}{5 1 {-1} 2})^-1 = {1/11}(matrix{2}{2}{2 {-1} 1 5})

  3. Sub the inverse into the equation in step i:
    (matrix{2}{1}{x y}) = {1/11}(matrix{2}{2}{2 {-1} 1 5})(matrix{2}{1}{{-8} 17})

    Do your multiplication! As described in A above:
    (matrix{2}{1}{x y}) = {1/11}(matrix{2}{1}{{(2)(-8)+(-1)(17)} {(1)(-8)+(5)(17)}})
    doubleright (matrix{2}{1}{x y}) = {1/11}(matrix{2}{1}{{-33} 77}) = (matrix{2}{1}{{-3} 7})
    ⇒ x = -3, y = -7

    Note: most of the time, you should end up with a 2 x 1 matrix after multiplying.

Know with true faith the methods above. Be cautious with the pointers and common mistakes in red. Understand with thy heart the representative sample question at hand.

Print this out if necessary and commit to memory the above procedures, and be comforted that no matter what happens, Miss Loi shall always be here to support you, in spirit and in soul.

HAPPY LABOUR DAY!

*Drifts away like 倩女幽魂*

© 2007-2011 Jφss Sticks. Copycats will be mercilessly hunted down.
Please 'like' the Joss Sticks Society on Facebook! Miss Loi's Twitter

]]> http://www.exampaper.com.sg/tuition-notes/a-maths-tips/of-matrices-miss-lois-day-of-labour/feed 12 Sergeant Loi’s Mid-Year Boot Camp 2008 – Defeat Partial Fractions In Three Devastating Moves http://www.exampaper.com.sg/tuition-notes/a-maths-tips/sergeant-lois-mid-year-boot-camp-2008-defeat-partial-fractions-in-three-devastating-moves http://www.exampaper.com.sg/tuition-notes/a-maths-tips/sergeant-lois-mid-year-boot-camp-2008-defeat-partial-fractions-in-three-devastating-moves#comments Tue, 29 Apr 2008 17:33:36 +0000 Sergeant Loi http://www.exampaper.com.sg/tuition-notes/a-maths-tips/sergeant-lois-mid-year-boot-camp-2008-defeat-partial-fractions-in-three-devastating-moves Within its pages lies the secretto defeating Partial Fractions *Puts on helmet* *SHOWS STERN & MEAN FACE* *Sounds bugle* EVERYBODY FALL IN! Today you’ll learn to deal with a new enemy originally from the A-Level campaign. Something never been seen by your previous O-Level comrades, nor has it appeared in the pages of the old [...]

© 2007-2011 Jφss Sticks. Copycats will be mercilessly hunted down.
Please 'like' the Joss Sticks Society on Facebook! Miss Loi's Twitter

]]> 武林秘籍
Within its pages lies the secret
to defeating Partial Fractions

*Puts on helmet*

*SHOWS STERN & MEAN FACE*

*Sounds bugle* EVERYBODY FALL IN!

Today you’ll learn to deal with a new enemy originally from the A-Level campaign. Something never been seen by your previous O-Level comrades, nor has it appeared in the pages of the old Ten-Year Series 武林秘籍 (ala Secret Manual).

But fret not! Though it looks formidable at first, its size also makes it unwieldy and predictable, so master the rules of engagement (see A. and B. below) and follow the standard 三大招式 (Three Devastating Moves) detailed in each of the Sample Questions below, and you’ll be on your way to defeating these Partial Fractions!

NOW ALL TURN TO PAGE 1 OF YOUR 武林秘籍 (2008 revised edition)!!!

*Cracks whip*

Frag The Partial Fractions!

A. COMPULSORY PRE-CHECK

  1. The degree of a polynomial is the highest power of x. e.g.

    p(x) = 4x3 – 12x2 – x – 4 ⇒ degree = 3
    q(x) = (x – 1)(x + 2) ⇒ degree = 2 ( ∵ when expanded q(x)=x2+x-2)

  2. For {p(x)}/{q(x)}:

    1. if degree [ p(x) < q(x) ] ⇒ PROPER ⇒ proceed with main Partial Fraction calculations (YAY!)
    2. if degree [ p(x) ≥ q(x) ] ⇒ IMPROPER ⇒ Long Division Time! (see Qn 2 below) (SIANZ!) (note it’s greater OR EQUAL)

B. THE RULES OF PARTIAL FRACTIONS ENGAGEMENT

For {p(x)}/{q(x)}: (*Have you ‘propered’ this first? See above.)

Rule If q(x) contains Partial Fraction must contain
1 linear factor
(ax+b)		 A/{ax+b} for each linear factor
2 repeated linear factor
(ax+b)2		 A/{ax+b} + {B/(ax+b)^2}
for each repeated linear factor
3 quadratic factor
x2+c2		 {Ax+B}/{x^2+c^2}
for each quadratic factor

Attention! (x2 – b2) can be factorized further into (x + b)(x – b)
→ two linear factors! (≠ quadratic/repeated linear factor!)


SAMPLE PRACTICE QUESTIONS

(Note the standard 三大招式 steps).

  1. Express {7x+4}/{(2x+1)(x-2)^2} in partial fractions.

    Ans:

    1. 第一式 (绝招)

      Check: degree of (7x+4) = 1 < degree of (2x+1)(x-2)2 = 3
      ⇒ PROPER (YAY!)

    2. 第二式

      So {7x+4}/{(2x+1)(x-2)^2}
      → 1 x linear factor (2x+1) and
      → 1 x repeated linear factor (x-2)2

      Using Rules 1 & 2 from the table above, we have:
      {7x+4}/{(2x+1)(x-2)^2} = A/{2x+1} + B/{x-2} + C/(x-2)^2

      *Note there’re two components for the repeated linear factor (x-2)2!

    3. 第三式 (绝招)

      Multiply both sides by (2x+1)(x-2)2,
      7x+4 = A(x-2)2 + B(2x+1)(x-2) + C(2x+1)

      To find A, B and C, sub in suitable values for x that makes certain components disappear:

      Sub x = 2,
      doubleright 18 = A(0) + B(5)(0) + C(5)
      doubleright C = 18/5

      Sub x = -1/2,
      doubleright 1/2 = A(25/4) + B(0)(-5/2) + C(0)
      doubleright A = 2/25

      Sub x = 0, A = 2/25, C = 18/5,
      doubleright 4 = (2/25)(4) + B(1)(-2) + (18/5)(1)
      doubleright B = -1/25

      Hence,
      {7x+4}/{(2x+1)(x-2)^2}
      {} = 2/{25(2x+1)} - 1/{25(2x+1)(x-2)} + 18/{5(x-2)^2}

  2. Express {x^4+9}/{x^3+3x} in partial fractions.

    Ans:

    1. 第一式

      Check: degree of (x4+9) = 4 > degree of (x3+3x) = 3
      ⇒ IMPROPER ⇒ LONG DIVISION TIME!!! (SIANZ! :( )

      Partial Fractions Long Division

    2. 第二式

      Now {x^4+9}/{x^3+3x} can be factorized further to {x^4+9}/{x(x^2+3)}
      → 1 x linear factor (x) and
      → 1 x quadratic factor (x2+3)

      Using Rules 1 & 3 from the table above, we have:
      {x^4+9}/{x(x^2+3)} = x + A/x + {Bx+C}/{x^2+3}

      *Note: all we need is the quotient (x) from the long division in i.

    3. 第三式

      Multiply both sides by x(x2+3),
      x4+9 = x2(x2+3) + A(x2+3) + (Bx+C)x

      To find A, B and C, sub in suitable values for x that makes certain components disappear or your life easier:

      Sub x = 0,
      ⇒ 9 = (0)(3) + A(3) + (C)(0)
      ⇒ A = 3

      Oh we can’t reduce other components to 0 with another substitution → need simultaneous equations here:

      Sub x = 1 (to make your life easier), A = 3,
      ⇒ 1+9 = (1)(4) + (3)(4) + B + C
      ⇒ B + C = -6 —– (1)

      Sub x = -1 (to make your life easier), A = 3,
      ⇒ 1+9 = (1)(4) + (3)(4) + (-B + C)(-1)
      ⇒ B – C = -6 —– (2)

      Solving (1) & (2),
      B = -6, C = 0

      Hence,
      {x^4+9}/{x^3+3x} = x + 3/x - {6x}/{x^2+3}

As always, get these rules drilled into your head! Spot the pointers and common mistakes in red! Understand the representative sample questions!

Print this out if necessary and remember the above procedures by heart, for if you don’t Sergeant Loi will unleash her own version of the 三大招式 upon you!

*Cracks whip!*

© 2007-2011 Jφss Sticks. Copycats will be mercilessly hunted down.
Please 'like' the Joss Sticks Society on Facebook! Miss Loi's Twitter

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