Mastering Mathematics Smartly by Wee Wen Shih

Part 1

Mathematics is a discipline of patterns and due to this nature, the process of generalisation gets its way frequently into the subject.

In 2008 H2 maths exam paper 1, a question (Q9) was asked about generalising the behaviour of the graph of y = (ax + b)/(cx + e). The general features include:

1. If ad - bc is not equal to zero, then the graph has no turning points.

2. If ad - bc is equal to zero, then the graph is a horizontal line parallel to the x-axis.

Now, let's consider the graph of ax^2 + by^2 = c where a, b and c are non-zero constants. What general features (related to the syllabus content on graphing) can we observe?

1. If a, b > 0 and c = 1, then we obtain the graph of an ellipse.

For example, x^2 + 4y^2 = 9 is an ellipse with centre (0, 0), horizontal length of 6 units and vertical length of 3 units.

Note: Substituting convenient values for variables is a very practical strategy that I'd like to encourage students to carry out when investigating mathematical properties.

2. If a = b = 1 and c > 0, then we obtain the graph of a circle (whose centre is at the origin and radius of sqrt(c) units) which is a special case of an ellipse.

3. If a > 0, b < 0 and c = 1, then we obtain the graph of a hyperbola. The graph of a hyperbola can also be obtained by setting a < 0, b > 0 and c = 1.

For example, x^2 - 4y^2 = 1 is a hyperbola with centre (0, 0), oblique asymptotes y = x/2 and y = -x/2 and x-intercepts x = 1 and x = -1.

Next, we consider the graph of y = (ax^2 + bx + c)/(dx + e) where b, c, d, e are non-zero constants.

1. If a = 0, then we have the graph of a rectangular hyperbola.

2. If a is non-zero, then we have a graph with

- an oblique asymptote,

- a vertical asymptote,

- two turning points or no turning points.

There have been questions set in the past that required students to arrive at the condition involving the constants for which the graph has stationary points or none. I will say more in part 9.

Cultivate the habit of generalising results to help you master mathematics smartly!

Part 2

In another question (Q11) of 2008 exam paper 1, we saw yet another form of generalisation involving a system of linear equations.

Consider the equations:

2x - 5y + 3z = 3,

3x + 2y - 5z = -5,

5x + (lambda)y + 17z = mu.

Students were led to investigate different values of lambda and mu:

1. When lambda = -20.9 and mu = 16.6, the planes intersect at a unique point.

2. When lambda = -22 and mu = 17, the planes meet in a line.

3. When lambda = -22 and mu is not 17, the planes have no point in common.

In conclusion, we will always encounter three cases, i.e. unique solution, infinitely many solutions and no solution, for a system of three linear equations.

Part 3

Let me illustrate another application of generalisation involving loci with this question:

In an Argand diagram, the point A is represented by the fixed complex number a, where 0 < arg(a) < pi/2. Sketch on a single Argand diagram, the loci representing the complex numbers w and z for which |w - a| = |ia| and |z| = |z + 2a|.

If we analyse the question carefully, we see that

1. A lies in the first quadrant because we are given that 0 < arg(a) < pi/2.

2. |w - a| = |ia| = |i||a| = |a| represents a circle whose centre is at point A and having a radius of |a|, the distance OA. Take note of the steps we took to make the locus equation look like the standard form of a circle.

3. |z| = |z + 2a| = |z - (-2a)| represents the perpendicular bisector of OA', where A' is a point in the third quadrant and OA' = 2OA. Note also that A', O and A are collinear. Observe the steps we took to make the locus equation look like the standard form of the perpendicular bisector. It is a common error of students to ignore the sign and they end up drawing a perpendicular bisector passing through the wrong midpoint.

If it is not easy to observe these, then a practical way is to set A to be the point (1, 1) and things will be much clearer.

Part 4

A fourth application of generalisation is the transformed graph of y = a f(bx + c) + d based on the graph of y = f(x).

This has been asked in 2007 exam paper 1 Q5: State a sequence of transformations which map the graph of y = 1/x to the graph of y = (2x + 7)/(x + 2).

Firstly, we express y = (2x + 7)/(x + 2) in terms of y = 1/x, since y = 1/x is the starting point. So y = 2 + 3/(x + 2).

Now we want to obtain the transformations involved to obtain the graph of y = 2 + 3 f(x) from the graph of y = f(x):

1. Scaling of factor 3, parallel to the y-axis.

2. Translation of 2 units, parallel to the y-axis.

Can the ordering be 2 followed by 1? Try to think of examples.

I have written more about transformations that one may be interested to study more deeply about:

http://www.freewebs.com/weews/graphtransformations.htm

Part 5

Solution curves of differential equations is another type of generalisation.

In 2008 exam paper 1 Q4, we saw that y = 3/2 ln (x^2 + 1) + c is a general solution to the differential equation. With different values of c (say c = 0, c > 0 and c < 0), we can sketch members of the family of solution curves. In this case, we observe that all members will look alike in terms of shape. However, this may not so in general.

Consider the DE (x^2)(dy/dx) -2xy + 6 = 0 whose general solution is y = cx^2 + 2/x. If c = 0, we have the familiar rectangular hyperbola y = 2/x; otherwise we obtain very different graphs for c > 0 and c < 0. To see that, set c = 1, 2, 3 and c = -1, -2, -3 on your GC and compare the graphs.

Part 6

In the Nov '05 exam paper, we see a generalisation involving indefinite integrals. I have modified the question a little for this discussion.

Let $I = \int \frac{\text{P}(x)}{x^3 - 1} \text{d}x$ . We consider three simple cases and then make use of the results to find $\int \frac{1}{x^3 - 1} \text{d}x$ . Traditionally, we would have employed the technique of integration by partial fractions.

Case 1: $\text{P}(x) = x^2$ .

It is easy to find I via

$I = \frac{1}{3}\int \frac{3x^2}{x^3 - 1} \text{d}x = \frac{1}{3}ln |x^3 - 1| + c$ .

Case 2: $\text{P}(x) = x^2 + x + 1$ .

First, we note that $x^3 - 1 = (x^2 + x + 1)(x - 1)$ .

It is again straightforward to integrate the expression via $I = \int \frac{1}{x - 1} \text{d}x = ln |x - 1| + c$ .

Case 3: $\text{P}(x) = x - 1$ .

We obtain $I = \int \frac{1}{x^2 + x + 1} \text{d}x$ . In this case, we need to complete the square of the denominator first, i.e.

$I = \int \frac{1}{\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}} \text{d}x$

$= \frac{2}{\sqrt{3}} tan^{-1} \left(\frac{2x + 1}{\sqrt{3}}\right) + c$ .

Finally, we observe that $\int \frac{1}{x^3 - 1} \text{d}x = \frac{1}{2}\int \left[\frac{x^2 + x + 1}{x^3 - 1} - \frac{x - 1}{x^3 - 1} - \frac{x^2}{x^3 - 1} \right] \text{d}x$ , noting that the first, second and third terms correspond to cases 2, 3 and 1 respectively. We are done.

Part 7

This article discusses about generalisations involving inequalities.

It is easy to solve the inequality $\frac{5x}{x - 2} \ge 3$ , where a unique solution set involving concrete values can be obtained.

Sometimes, school examiners like to reframe the problem to something like this: solve the inequality $\frac{ax}{x - 2} \ge 3$ for which $a > 3$ .

Let's investigate it further. Firstly, we note that $x \ne 2$ . It is always a good practice to exclude unacceptable solutions at the start.

We then multiply both sides by $(x - 2)^2$ to obtain $ax(x - 2) \ge 3(x - 2)^2$ .

Factorising and simplifying, we arrive at $(x - 2)\{(3 - a)x - 6\} \leq 0$ .

Since $a > 3$ , we have an inverted 'U' shaped parabola (why?) whose roots are $\frac{6}{3 - a}$ and $2$ (draw it!). Thus, we obtain $x \leq \frac{6}{3 - a}$ or $x > 2$ as our solution set.

It is interesting to ask what happens when $0 < a < 3$ . In this case, we will have a 'U' shaped parabola (why?) whose roots are $2$ and $\frac{6}{3 - a}$ . We will then arrive at the solution set $2 < x \leq \frac{6}{3 - a}$ .

How about $a = 0$ ? We do not have any real solutions (why?).

When $a < 0$ , the solution set is $\frac{6}{3 - a} \le x < 2$ (why?).

What if we consider the case $a = 3$ ? It is easy for one to sketch (go ahead and do it!) the graph of $y = \frac{3x}{x - 2}$ (a rectangular hyperbola) and see that the solution set is simply $x > 2$ .

Thanks to Tan Junwei for pointing out my slips.

Part 8

The topic of geometric progression related to compound interest problems is another type of generalisation, which may be helpful for students to learn about. We shall discuss three cases.

Case 1 (one deposit): Suppose we deposit $10 on the first day of the month and the bank pays compound interest of 2% per month on the last day of each month. What is the amount at the end of n months?

It is always helpful to tabulate with "Month", "Start" and "End" columns to aid in arriving at a general formula. For the convenience of typing, I will use a linear (or sequential) representation instead.

Month: 1
Start: 10
End: 1.02(10)

Month: 2
Start: 1.02(10)
End: (1.02^2)(10)

Month: 3
Start: (1.02^2)(10)
End: (1.02^3)(10)

:

Month: n
End: (1.02^n)(10) <-- our desired answer!

Case 2 (multiple deposits): Suppose we deposit $10 on the first day of the month and the bank pays compound interest of 2% per month on the last day of each month. We put a further $10 into the account on the first day of each subsequent month. What is the amount at the end of n months?

Month: 1
Start: 10
End: 1.02(10)

Month: 2
Start: 1.02(10) + 10 <-- do not simplify, or the pattern may be lost.
End: (1.02^2)(10) + 1.02(10)

Month: 3
Start: (1.02^2)(10) + 1.02(10) + 10
End: (1.02^3)(10) + (1.02^2)(10) + 1.02(10)

:

Month: n
End: (1.02^n)(10) + {1.02^(n - 1)}(10) + ... + 1.02(10) <-- our required answer, which is a GP sum.

Case 3 (multiple withdrawals): Suppose we deposit $100,000 at the beginning of the year and the bank pays compound interest of 10% per annum at the end of each year. After the interest is credited, we withdraw $12,000 immediately. We repeat the same action of withdrawal for subsequent years. What is the amount at the end of n withdrawals?

The column headings are now revised to "Year", "Start" and "End".

Year: 1
Start: 100,000
End: 1.1(100,000) - 12,000

Year: 2
Start: 1.1(100,000) - 12,000
End: (1.1^2)(100,000) - 1.1(12,000) - 12,000

Year: 3
Start: (1.1^2)(100,000) - 1.1(12,000) - 12,000
End: (1.1^3)(100,000) - (1.1^2)(12,000) - 1.1(12,000) - 12,000

:

Year: n
End: (1.1^n)(100,000) - {(1.1^(n - 1)}(12,000) - {1.1^(n - 2)}(12,000) - ... - 12,000, our required amount. Notice that

- {(1.1^(n - 1)}(12,000) - {1.1^(n - 2)}(12,000) - ... - 12,000 is a GP sum if we see it as

-12,000 {1.1^(n - 1) + 1.1^(n - 2) + ... + 1}.

Additional remarks:

1. We can link our discussion to the topic of recurrence relations as well.

2. For case 1: $U_1 = 1.02(10), U_{n + 1} = 1.02U_n$ for $n \ge 1$ .

3. For case 2: $U_1 = 1.02(10), U_{n + 1} = 1.02(U_n + 10)$ for $n \ge 1$ .

4. For case 3: $U_1 = 1.1(100000) - 12000, U_{n + 1} = 1.1U_n - 12000$ for $n \ge 1$ .

Part 9

We shall investigate the graph of $y = \frac{x^2 + ax + 1}{x - 2}$ based on different values of a. In particular, we are interested to know when the curve has stationary points or none at all.

First we find the expression of the derivative using the quotient rule of differentiation, i.e. $\frac{\text{d}y}{\text{d}x} = \frac{x^2 - 4x - 2a - 1}{(x - 2)^2}$ after simplification.

If the curve has stationary points, we will need to have a numerator that takes a value of zero. This means that the discriminant of $x^2 - 4x - 2a - 1$ is greater than zero (why?). So $(-4)^2 - 4(1)(-2a - 1) > 0$ , leading us to conclude that $a > -\frac{5}{2}$ .

It is now easy for us to conclude that when $a < -\frac{5}{2}$ the curve has no stationary points. But what about the case where $a = -\frac{5}{2}$ ? By long division, the equation of the curve becomes $y = x - \frac{1}{2}$ , a straight line!

In summary, we have three families of graphs of $y = \frac{x^2 + ax + 1}{x - 2}$ . One family consists of members whose graphs have stationary points; a second family consists of members whose graphs have no stationary points; a third family has a single member whose graph is simply a straight line.