[AM_20160125DAHT] Horizontal Tangents via Quadratic Discriminants

Problem

Introduction

     This is a Additional Mathematics textbook problem.  This question is of an intermediate level of difficulty.  The general method is by differentiation.  The equation of the curve happens to be capable of being put into a quadratic equation in  x.  Hence we can also use the theory of quadratic discriminants.  I present both methods of solution.

Method 1 (Using differential calculus)

Method 2 (Using quadratic discriminants)

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

H12* Think of a related problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* other syllabuses that involve differentiation or quadratic discriminants

* any independent learner who is interested

[AM_20151226EIQR] Looking for a Pea among Quadratic Roots?

Question

Introduction

     This question is about finding the parameter  p, and not about solving for the “unknown”  x.  It is heavy on algebra, one has to be patient, careful and meticulous.  Please refer to this article for a recapitulation of (Vieta’s) theory of Quadratic Roots.

Solution

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* other syllabuses that involve quadratic roots
* any learner who is interested

[AM_20151224QERI] Quadratic Roots and Use of Identities

Problem

Introduction

     Here is a fairly standard question on roots of quadratic equations, except that part (iii) is slightly more challenging.  To solve this question, one must know the square of sum identity well.

Recapitulation

     Please refer to this previous article  and  this article  for the theory on quadratic roots.

Solution

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘A’ Levels H2 Mathematics

* other syllabuses that involve roots of quadratic equations

* any learner who is willing to learn

[AM_20151130IAXS] A Motif for the Absolutely Absolute

Problem

Introduction

     This question would pose a challenge for many students, although theoretically it is within reach of a good Additional Mathematics student (~ grade 10).  Graphs of both  sin x  and  cos x  are waves that oscillate up and down.  There are many pairs of vertical bars, indicating the absolute values or modulus, and these seem confusing.

Strategy

     Let us graph the functions  y = |cos x|   and  y = |sin x|.   Note that  ||sin x| – |cos x|| = ||cos x| – |sin x||.   The absolute difference of  |cos x|   and  |sin x|  is the difference between them ignoring the negative sign (if any) of the result.  And this is just the difference between the higher value and the lower value. 

Can you see any repeating patterns?  [H04]  Can you visualise the required area?  How many times is that of the basic pattern (known as “motif” in art)?  [H09, H10, H11]

Solution

Remarks

     Our total area is made up of four congruent pieces.  When  0 < x < ^p/₄,  cos x  is higher than  sin x.  That allows us to strip away all the absolute signs and do the calculation.

Heuristics Used

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* challenge for GCE ‘O’ Additional Mathematics  * IB Mathematics SL HL

* GCE ‘A’ Level H2 Mathematics  * IB Mathematics HL

* AP Calculus AB & BC

* University / College calculus

* other syllabuses that involve integration and area

* whoever is game for a challenge in integration

[AM_20151127DTCR] The “Onion” Method for Differentiation

Question

Introduction

     The differentiation of the secant function is not taught directly as part of the Additional Mathematics syllabus.  It can be derived from known facts.  I first show the standard application of the (extended) Chain Rule for novices, and then show a more effective way of applying the chain rule, which I called the “Onion Method”.  This looks like peeling onions or unpacking Matryoshka dolls (“Russian dolls”). 

Reminders

Solution 1  (for beginners)

Solution 2  (a more expedient way)

Final Remarks

     When we peel onions, we peel from the outer layer inwards.  Likewise, when we have a composite function, we differentiate from the outer layer first, and then work to the inner layers.  Every time we differentiate a layer, we write down the changed layer and then copy and paste everything within that layer.  With regular practice, this should become second nature.
     For another example of the “onion”, take a look at the derivative of the arcsecant function.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* GCE ‘A’ Level H1 Mathematics

* GCE ‘A’ Level H2 Mathematics (revision)

* International Baccalaureate SL & HL Mathematics

* AP Calculus AB & BC

* other syllabuses that calculus

* anyone who loves to learn!

[AM_20151117] An “Unorthodox” Technique for Trigonometric Proof

Question

Introduction

     When proving trigonometrical identities, one usually starts from the “more difficult” side and try to work towards the other side.  The above identity looks like a tough nut to crack.  Both sides look equally complicated.  Where do we even begin?

     Here is one way to “cheat”.  Starting from, say, the LHS, we multiply the RHS expression and also multiply by its reciprocal i.e. dividing by the same amount.  It is like using a magic wand to create something out of nothing (无中生有 in Chinese), but doing so does not change the value of the LHS expression.  Now we do not touch the part that is equal to the RHS (highlighted in yellow below), but we try to find a way to cancel away the other stuff, as shown here.

Solution

Remarks

     In the third step, I had replaced  cos²A  with  1 – sin²A  and  sin²B  with  1 – cos²B.  This leads to the required cancellation and we are left with the yellow patch, which was never touched since the first step and it is the RHS.  This completes the proof.

     Just as in martial arts where deadly opponents require deadly strokes to counter them, evil questions require “unorthodox” techniques.  Even though most people would not have thought of it, all the steps presented above are actually legitimate.  This is because at every step, the equality is preserved.  All the “=” are really equal, and it’s legit (either work hard or you might as well quit), although my “can’t touch this” tactic seems a bit clairvoyant.

. 

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H08. Make suppositions

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* other syllabuses that involve proof of trigonometric identities

* anyone who loves mathematical challenges

[AM_20151105QFER] New Quadratic Equation satisfied from New Roots

Question

The roots of the quadratic equation  2x2 – 3x + 6 = 0  are  a  and  b. (i)   Without finding the value of  a,  show that  8a4 = 18 – 45a. (ii)  Find the quadratic equation whose roots are  (a2 + 1)  and  (b 2 + 1).

Introduction

     Do you know what a “root” is?  Is it like radish or ginseng?  Do you know what “satisfied” means?  Is it that nice feeling you get when you eat carrots?  Read on!

     The featured problem above is modified from an original question that contained an error.  The modified part is shown in red.  I present two solutions.  The first solution uses pretty much standard theory, and I use notations  a’  and  b’  to denote the new roots  (a ² + 1)  and  (b ² + 1)  respectively.  For the second solution, I present an alternative working part (i), and one using the method of substitution for obtaining new equations (not usually taught in schools at the secondary level) for part (ii).  But before that let me first explain what “root” and “satisfied” means.

Recapitulation of Standard Theory

Solution 1 – Using Standard Theory

Solution 2 – Using the Method of Substitution for part (ii)

 Remarks

     Once again we can see that there are many ways to skin the cat, as it were.  Mathematics is not about following a fixed procedure.  There are various truths, notions and rules that are inviolable.  But other than that, you can have as much creativity as you want!

HeuristicsUsed

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* other syllabuses that involve quadratic roots

* whoever loves roots and enjoy being satisfied by conquering mathematical challenges J

[AM_20151027DAX] Quadratic quagmire?

Question

Introduction

     This question was most probably taken from an Integrated Programme (IP) school in Singapore.  For your information, students in Integrated Programme schools do not take the GCE ‘O’ Levels, and each IP school is free to design its own individual curriculum.  In practice, they incorporate the mainstream GCE ‘O’ Level topics, as well as additional topics, and/or teach topics in advance, and may call their syllabuses by different names.  They tend to set more challenging questions than the mainstream schools, which are already targetting their internal examination standards above the ‘O’ Levels.  In other words, they tend to cram in more, but never less.  Part (d) tests approximate change, which had been taken out of the mainstream syllabus at this time of writing.
     Note that the IP schools tend to be schools that traditionally attract the academically best students from each cohort.  Even before the IP programme was introduced, these schools were already setting harder questions.  Anyway, this blog welcomes everybody from all around the world who is willing to learn, regardless of the type of school they are from, even home-schoolers and independent learners!  Let us see how we can employ re-usable tactics to tackle this question.

Get to the root of the matter, fast!

     The question is on the applications of differential calculus on a quadratic curve (a parabola).  Observe that the equation of the curve is given in completed square form.  From the equation, can you spot the line of symmetry (centre line) and the  y-intercept immediately (like within 5 seconds)?  [You need to know all the basic facts at your finger tips and make observations.]

The line of symmetry always passes through the maximum or (in this case) minimum point.  We know that the minimum is when the squared term  (x – 3)²  is zero.  So

                              the line of symmetry is  x = 3.

To get the  y-intercept, put  x = 0.  This gives  y = (-3)² = 9.  So 

                              C = (0, 9)  and equation of  CD is  y = 9

Since  PQ = 2k,  distance from  P  to the centre line = k.  All the above are basic observations that should be carried out mentally within one minute and you should be able to mark the diagram with pencil notes (shown above in blue).  Once this is done, let us get on to the real business.

Solution

Final Remarks

     For part (c), they have already told you it’s maximum, so you do not need to prove that it is maximum.  However, intuitively it is obvious there is a maximum: imagine if  k = 0  or  k = 3, then we get very thin rectangles with area zero.  As  k  increases from  0, the area gets bigger and after that shrinks towards zero again. 

     Actually, was this problem really so difficult?  What did we do to solve it?  Let’s review

· Know all basic facts and skills thoroughly at immediate recall (e.g. extremum of parabola lies on line of symmetry, how to spot that from completed square form, how to find y-intercept)

· Use heuristics: e.g. make observations.  Use simple facts you already know (e.g. subtract lengths, substitute values of  x  to find  y  which is “height” have x-axis etc,).  Practice positive psychology: instead of worrying, write down everything you can deduce.  Then try to find connections.

· Apply the formulas

· Use your intuition to see whether your answer makes sense.

· For part (d), if it is not in your syllabus, do not worry about it.  But if you are curious or feel the itch to learn more, it is also not too difficult.  See the boxed formulas in the solution above.  Just remember that the ratio of small changes  ^DA/_Dk  is approximately equal to the derivative  ^dA/_dk.  You can detach the  Dk,  bring it to the other side of the equation, and that allows you to approximate  DA.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

· GCE ‘O’ Level Additional Mathematics

· other syllabuses that involve applications of differentiation

· anyone who is interested in calculus!

[AM_20150510PLRF27] Looking for a Polynomial's Missing Link

Question

Introduction

     The first part of this question is rather standard.  The second part is quite challenging, especially if you are trying to connect with the earlier part.

Reminder

     To solve polynomial equations of degree  3  or higher (that are tested in school tests and exams), we often need to guess a rational (fraction or integer) root.  By the way, whole numbers are rational numbers because we can always put them into fractions upon  1  as denominator.  So how do we guess the roots?  The following is a very important theorem that guides us as to what numbers to try.

So we consider all the possible factors of the constant term  a₀  for the numerator and

all the possible factors of the coefficient  a_n  of the highest power for the denominator and consider the + and the – of all the possible fractions formed.  Usually, we try those with denominator 1 i.e. the integers first.

Solution

Remarks

     Note that the solution consists of only the part in blue.  Black is used for explanations, which are lengthy because of the dense interplay of ideas and subtleties involved.

     For the second part, if you are not able to see the connection, then use the standard method to solve the equation.  Here we realise that when the  x  is replaced by  ^v/₂,  the coefficients are reduced to the original coefficients.  However, these are in reverse order.  This indicates that one needs to use the reciprocal, so you divide throughout by  v³,  so that the highest power becomes just a constant.  I know you would not have thought of this if you have not seen this kind of question before, but this is the trick to use.

     Do not be discouraged by difficult question.  Have a growth mindset.  Every time you encounter a difficult question, learn how the trick ticks.  Your brain muscles will get stronger.  Try to apply the same trick when you see a similar question next time.

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* GCE ‘A’ Level H2 Mathematics (revision)

* IB Mathematics (revision)

* other syllabuses that involve polynomials, Remainder and Factor Theorem

[IBHL_SOTA201304_1B10c] Quadratic Discriminants

Question

Important Reminders

Solution

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

*  IB Mathematics HL / SL

* other syllabuses that involve quadratics

[IBHL_SOTA201304_1B07] Quadratic Equations and Roots

Question

Important Reminders

Solution

     Actually we could have multiplied by  -4  or any multiple of  4  for that matter, but this is the set of integer solutions for which  a  is the least positive.

     For part (b), if we can solve the first equation easily, then the roots of the second equation can be obtained by just squaring your answers.  However, the LHS of the first equation cannot be factorised nicely, so we might as well use the quadratic formula to solve the second equation directly.

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

*  IB Mathematics HL / SL

* other syllabuses that involve quadratics