Want to ace your A Maths at the O levels....???

...then you would have to practise.  Sorry, there are no short cuts here. Practise like it's a competition. It is a tug of war between the math question and you. Do you want to let the question prevail, or will you better it?

I read an article about a Chinese lady who shares her parenting role. She says the Chinese parents will always push their children to greater heights. They will do everything in their power to ensure that their children put in 100% effort to get things right. If they are not right, they will look at what happened and will set the motion to correct it. Then and there! Not tomorrow, not next week. Now! They will master it before they retire.

While I don't fully subscribe to the lady's child-rearing methods, some things she said did make sense. If you want to enjoy doing something, you must first be good at it. Hence, you would have to break into the system and get the body and mind moving together as one. It is the same for music, for sport and for A Maths.  You must first put in the hours. You have to master the subject; not let the subject master you.

No one who ever amounted to anything took the short cut. There is no greatness in the easy way. As the Chinese say, "You have to eat hardship before you can enjoy the spoils."  There is certainly truth in that.  So, this is the first step kids. Learn, practise, preactise and practise some more. Soon you will see the patterns and the answer will jump at you.

Don't believe me? Why don't you try it out?

You will be amazed at what you can achieve by it.

Take care!

For those of you starting your A Maths journey this year...

I thought I'd start the 2011 series of A Maths blogs on an advice: "Don't give up". I know this is a cliche and we have heard it too many times before, but it is so true, especially for A Maths. And for those of you who are just starting on it, you might feel overwhelmed by it in the beginning.

Don't give up.

It is important to understand that A Maths is training you for higher order thinking. You would not have had any such subjects before and it may be daunting.

Don't give up.

You must always look for connections; understand the basics and.... you got it -

don't give up.

Lastly, you must bone up on your algebraic manipulations, solving of algebraic equations and the like. A Maths takes your understanding of these things and projects them to a higher plane. So if you think that your algebra is shaking, it would be best to start practising on it. It will save your A Maths life!

Have a great 2011!

Quadratic functions - nature of roots

Find the range of values of k for which the quadratic equation

2x² – 2x + k + 2 = kx has real roots.

This is a classic question that makes use of the determinant of the quadratic equation. The determinant of a quadratic equation is D = b² – 4ac.

When D = 0, then it means that the equation has equal roots; i.e. the two answers of x for the quadratic equation are the same.

When D > 0, the roots will be real and distinct; i.e. there will be two different answers for x.

When D < 0, then the roots will be imaginary. This means that there are no answers to the quadratic equation and graphically, it means that the curve does not intersect the x axis at all.

Now, what about when the roots are simply real?  Then it means that D >= 0.  So, apply this to our question, we will first rearrange the equation so that it will equal to zero. Hence,

2x² – 2x  - kx + k + 2 = 0

or

2x² – (2 + k)x + (k + 2) = 0

For real roots,

Since this is a positive k² function, then

Logarithms - change of base

Solve  log₅x – 2log_x5 = 1

In order for us to solve this question, we need to recall the change of base formula, which is:

We also know that we cannot have an unknown base. Hence, we need to change the log_x5 into log₅x.  Therefore, the equation can be expressed as:

Solving this gives:

And this is how we solve for logarithm questions with different base.

Hope this helps!

How to solve indices equations...

Today I want to talk about solving indices questions. There are basically two types of questions. The first is when 

                                                a ^{f (x)} = a ^g(x)

the base of each side of the equation is the same  (a). Hence,  f (x) = g(x). Of course, in this case, you must be able to simplify each side of the equation to the same base.  One hint: all the terms on each side are multiplied or divided with each other.

Let’s look at an example:

Solve:

 

In this case, you will see that the terms on LHS of the equation are multiplied with each other. Hence, we can use this form of solution; and first transform them into the same base viz,

The next type of question is the substitution question. Basically, this is when the terms in the equation are added (or subtracted) together, not multiplied. In this case, we are unable to simplify them into the same base for both sides (even if they are already of the same base!) because the terms are separated by a “+” or “-“ sign.  In this case, we will use the substitution method; as such:

 So now, we have seen both types of questions that usually come with indices questions. Master them and you will be on your way to A Maths mastery….!

Have a great weekend!

Solving absolute value equations...

Today I want to discuss a topic which I am sure not many students are clear about. How to solve absolute value equations. Although it may be simple, there are some equations that are a little tricky. I shall talk about all of them here today.

First, what is an absolute value function?  It is a function that will always yield positive results. It is written as such:  | f(x)|.  So |f(x)| = f(x) if  the value of f(x) is positive, but is -f(x) if the value of f(x) is negative.

Let me illustrate with an example.

Say I have an absolute value function  |x - 1|.  Now, x - 1 will be positive (or zero) if x = 1 or more. But if x is less than 1, then it will return a negative value. Correct?  So, then

|x - 1| = x - 1 when x ≥1; but is  -(x – 1) when x < 1
Hence, |x - 1| will return two functions, depending on what the value of x is.

Understand so far?

If you do, then let us look at the first equation to solve:    |x + 3| = 5

Then, it means that  x + 3 = 5   or    - (x + 3) = 5
Since the negative sign can be brought over to the other side of the equation, then the solution for

|x + 3| = 5 is   x + 3 = ± 5
Which means that x + 3 = 5   or    x + 3 = -5
Hence, x = 2 or x = -8.

So far so good, right?

Now, let’s go one step further. What happens when |f(x)| = g(x)?
Well, the same argument still applies.
The solution will be  f (x) = g(x)  or  f(x) = -g(x) and we solve for x for each of these cases.

For example,  solve for   |3x – 2| = 1 – x
In this case,
            3x – 2 = 1 – x              or                     3x – 2 = - (1 – x)
            3x + x = 3                                            3x – 2 = - 1 + x
            4x = 3                                                  2x = 1
            x = ¾                                                   x = ½
Hence,
            x = ½  or  ¾.

So far still here?
I hope so, because this is not rocket science.

Now, what happens if our question is  |f(x)| = |g(x)|?
Well in this case, the solution still remains the same way as above; i.e.
             f (x) = g(x)  or  f(x) = -g(x)
So there won’t be any examples for this.

But watch what happens with we have a constant term added to the above equation with two absolute value function, like this |f(x)| = a + |g(x)|? This will make things a little more troublesome.

In this case, we will need to identify which range of values of x there are and apply the right combination of positive or negative functions.  Let’s see an example of this to understand what I am trying to say…

Example we have
|3x – 2| = |1 – x | - 2

In this case, we will first need to see what combinations of positive and negative functions to use. We will take each of the functions in turn and determine the range of values of x for which they are positive.

So, if  3x – 2 > 0, then  x > ⅔
And if  1 – x > 0, then x < 1
So, in this case we have three regions:
Less than ⅔,   between ⅔ and 1 and greater than 1.

We have seen that for 3x – 2, it will yield a positive function only when x > ⅔. It means that for the region where x < ⅔, it would be negative.
For the case of 1 – x, we have seen that it will yield a positive value only when x < 1. Hence, for the region for x > 1, it will yield a negative value.
This now gives us the possibilities of values of x.

So, for the region x < ⅔, we will have the case of –(3x – 2) but +(1 – x). Hence, we will solve the equation:

-(3x – 2) = (1 – x) – 2

-3x + 2 = 1 – x – 2

-2x = - 3

x = 3/2

For the region of x in between ⅔ and 1,
            (3x – 2) = (1 – x) – 2
            3x – 2 = - x – 1
            4x = 1
            x = ¼

And finally, for the region where x > 1, we have
            3x – 2 = - (1 – x) – 2
            3x – 2 = -1 + x – 2
            2x = -1
            x = -½
So now, we have three solutions:  x = 3/2,  ¼  or  -½

This is not the end.  We now need to confirm which of these is valid. We will need to put it back into the equation  |3x – 2| = |1 – x | - 2

So, when x = 3/2, we have  5/2 = ½ - 2  and this is NOT valid.
When x = ¼ , we have   1¼ = - 1 ¼   so this is also not valid.
Lastly, when x = - ½   we have  5/2 = ½  and this is ALSO not valid.

In this case, there are no valid solutions to the equation, even if we thought we could find it. To confirm this, check out wolfram alpha at http://www.wolframalpha.com/input/?i=solve+abs%283x-1%29%3D%281-x%29-2
 
So here you have it. How to solve absolute value functions. The mathematics is easy but the understanding needs time to sink it. So let it sink in…

If you have any queries, please ask away…

2010 Paper 2 Q11 - trigonometry

The diagram shows the curves y = 4 cosx  and y = 2 + 3sin x  for 0 ≤ x ≤ 2π radians. The points A and B are turning points on the curve y = 2 + 3sin x  and the point C is a turning point on the curve y = 4 cosx.  The curves intersect at the points D and E.

(i) Write down the coordinates of A, B and C.

(ii) Express the equation 4 cos x = 2 + 3 sin x in the form cos (x + α) = k, where α and k are constants to be found.

(iii) Hence, find in radian, the x-coordinate of D and of E.

(i)  Some of you might be tempted to use differentiation to solve for the coordinates of A, B and C. While that is not wrong, it is the worst possible way to do it. It will eat up all your time for something as fundamental as this.

To find the coordinates of A and B, we must remember that the sine curve has turning points at π/2 and 3π/2. Also, the max value of 3 sinx is 3 and the min value is -3. However, since the curve has moved up the y-axis by 2 units, then the max value of 2 + 3sin x is 5 and the min value of 2 + 3sin x is -1.

Therefore, the coordinates of A and B are (π/2 , 5) and (3π/2 , -1) respectively.

For C, the min point for cos x occurs at x = π.  The min value of 4 cos x = -4.  Hence, the coordinates of C are (π , -4)

(ii)  Given 4 cos x = 2 + 3 sin x

Therefore,  4 cos x – 3 sin x = 2

Let 4 cos x – 3 sin x ≡ R cos (x + α), where R² = 4² + 3² = 25.  Hence, R = 5

                                                                                and tan α = ¾; i.e.  α = 0.644 radians

Hence  4 cos x – 3 sin x ≡ 5 cos (x + 0.644).

(iii)  To find the x-coordinate of D and E, therefore 5 cos (x + 0.644) = 2

                                cos (x + 0.644) = 0.4

                                x + 0.644 = 1.159 ,   2π – 1.159

                                                  = 1.159 ,  5.124

Therefore,

                                x = 1.159 - 0.644    or   x = 5.124 – 0.644

                                   = 0.515 radians               = 4.45 radians

The x-coordinates of D is 0.515 rad, and the x-coordinates of E is 4.45 rad.