I’ve been looking at exam questions a lot recently in this diary – well it is coming up to exam time!

In June, the focus is going to be on famous Mathematicians, Over the summer I’m looking to write some entries on mathematical puzzles – If you know some good ones please send them in!

Today though, I’m looking at curves – the lines we can draw to represent certain mathematical equations.

In particular, today and tomorrow I am writing about curves that are ‘continuous’ and curves which are not, which we call ‘discontinuous’. Also, what that means for how we can use these curves to answer questions – That bit is for tomorrow!

What do I mean by continuous? Basically this means that an ant can follow the curve and get to all parts of the curve, without flying off the page or leaving the line. There is a Mathematically precise way of defining ‘continuous’ but I think my ant gives you the idea.

Lets look at some examples. (For this I am going to borrow screenprints from my favourite websites, the link on the Links page)

y = x²

You can imagine the any being able to walk around this line. In fact any line of the form
Ax^a + Bx^b + Cx^c+ ….   will be continuous so long as a, b, c etc are positive integers.

 

 

 

 

y = sin (x)

 

And again the ant can walk up and down these curves.

 

Since the line for y = cos x looks similar, we can say that is continuous too.

Lets have a look at some Non-Continuous curves now, The easiest to show is  y = 1/x

This curve is in two parts. Our ant isn’t going to get from one part of the curve to the other without a jump.

 

 

 

 

 

Going back to our trigonometry, Tan (x)  – unlike Sine and cosine – is discontinuous. There are many and regular breaks in the line.

 

 

In tomorrow’s post I will look at some more eccentric examples and show why its important to understand if a curve is continuous or not.

 

 

 

Yesterday’s post built up some of the things we need to answer a question where the power is negative and a fraction.

One key point is that for all numbers n

n⁰ = 1  and n¹ = n.

With negative powers we still need to maintain the rule of adding powers.

2^x x 2^-x  = 2^x-x  (and since x – x x = 0)  = 2⁰ = 1

Re-arrange this and we get  2^-x = 1/2^x

And there the first new rule – a negative power is 1/the positive power

3² = 9 so 3^-2 = 1/9

To get the second rule we need to consider how powers can be combined.

(n²)³ = n⁶  – When you raise a power to a power – multiply the powers

[Consider  n x  n     x     n x n     x    n x n]

 

Now look at

(n²)^1/2  =  n¹  = n

So what does raise to power of half mean if this involves getting from n² to n?   Taking the square root!  We could replace 2 and 1/2 with 3 and 1/3 – so see n^1/3 means take the cube root – and so on.

n^1/x means take the xth root of n.

Just before we get back to the question given, lets just complete the patterns by considering what x^3/2 means. I have seen some exam questions that do pose questions like this.

I suggest you split the 3/2  into  1/2 x 3  or,  n^3/2 = (n^1/2)³

so 4^3/2  =  (sqrt(4)³)  = 2³ = 8. It is generally easier to take the ‘root part first. In a non calculator paper you’ll only be asked about roots you know.

Let’s get back, at last, to the original question.

64^-1/2–   Take the 1/2 part first, that means square root, and the square root of 64 is 8.  The – part means take the reciprocal  1/8

What Carol did was take 1/2 of 64, not the square root of 64, so your answer should include a sentence.  ‘Carol didn’t know that a power of 1/2 means square root, not multiply by 1/2’ – then give the correct answer of 1/8.

 

I’ve made this into 2 blogs posts with lots of background but if you can remember the rules given in these posts, this question need not take you long in an exam,

 

Not because I’ve been watching election results today, but because I was discussing ‘powers’ with a student this week – Today’s subject is ‘powers of numbers’…  Its the first of a two-parter.

 

Raising a number by a fraction

 

 

 

 

Before starting on the matters of powers, let’s consider the form of this question.

It’s another of those ones that suggests someone has done the question, and you have to give comment on their answer.  In this case the question does say that Carol has made an error, but this won’t always be the case. I have done a question recently where the ‘answer’ given  was correct, and the marks were to be gained by saying so.

With these question types, I think you just have to do the question yourself. You may be able to spot an error in something you haven’t tried yourself, but I’ll be honest, as a Maths tutor I can’t always do so myself.

Let’s get back to the question in hand and consider what a power means if is a)  negative and b) a fraction – since in this case the power is negative and a fraction.  I could tell you both answers, but I like to show how these answers fit into the wider picture.

You will recall that if were want to multiply two ‘powers’ together you can do this by adding the powers  – so long as the ‘base’ is the same. The base is the number ‘raised to’ the power. In this case the base is 2.

2² x  2³  = 2⁵.

This makes sense when you think of 2² as 2 x 2 and 2³ as 2 x 2 x 2.  Put the big multiply together as

2 x 2       x      2 x 2 x 2 and you can see this can be written as 2⁵

That rule of powers works well when the powers are positive whole numbers. What we want is for it to work with other numbers.

Powers of 0 and 1

First we need to consider two special cases.

What does 2¹ mean?  And what does 2⁰ mean?

2¹ means 2 multiplied together one time – and that is just 2!  Remember this holds for all base numbers and you can’t go wrong.

4¹  = 4    36¹  = 36   4000¹ = 4000

With 20 we want to keep to the ‘adding indexes’ rule so that

2³ x 2⁰ = 2³⁺⁰ = 2³

8 x 2⁰ = 8  – We can soon see that for this to work, 2⁰ must have the value 1

Again, that works for all numbers

4⁰  = 1    36⁰  = 1   4000⁰ = 1

Now we know what powers of 0 and 1 are, can you see what consequence this has for negative and fraction powers?  This is what I’ll be looking at in the next post.

 

More details on this subject can be found here

 

 

Today’s post is about an exam question about a speed time graph I  was prepping for my students yesterday. I was surprised to see this question because it covers two points I associate more with A-Level work.

 

What can we find from the area under a curve?

 

a)  Area under the curve as the distance
b) Using the ‘ trapezium rule’

Before I say more, here is the question

 

 

 

 

 

 

 

The question asks for the distance the car travels, but the graph is of speed over time. This is where the first of my ‘surprise facts’ comes in – You can find the distance travelled by measuring the ‘area under the curve’. I won’t here show why that is the case – that is for A-Level students (OK I might post on that later). For now though, I will just apply the rule.

The area ‘under the curve’  is the area of the shape made by the curve at the top and the line ‘speed = 0’ at the bottom, bound to the left and right by the times we are interested in : In this case t=0 and t=20.

The Trapezium Rule

We can only estimate the area of this shape, given how irregular it is – the top side anyway – and this is where the ‘Trapezium rule’ comes in. To apply this we have to divide the area into strips, each one a trapezium (OK, the first one is a triangle, but we can apply a similar rule. To be fair, the question does give a clue to what to do by saying we should use ‘4 strips of equal width’

We divide up like this.

The first area is a triange with area

1/2 x 5 x 22*

*The value of the speed at t=5

 

The other areas are trapeziums with area 5 x 1/2(Height at start + height at end).

The ‘height’ in each case is the speed values at t= 10, 15 and 20.

If we look at the calculation closely we can save time by noting the middle times are included twice each, 1/2 each time.

so the area of the trapeziums are 5 x (22 + 28 + 32 + 35 x 1/2)

This is a quicker way that fiddling about with all the 1/2s. The 35 – value at t=20 – still needs the 1/2 because it is a measurement only on one trapezium.

We get the value 5 x 99.5.   This is where I took into account that we are only estimating, so using a value of 100 instead of 99.5 is justifiable.    This gives us an estimate of 500m

There is a part(b) for this question on the next page.

Over and under estimates

 

“Is this an over-estimate or an under-estimate for the distance. Give a reason for you answer.’

 

The answer in this case is that it is an underestimate, because all those trapeziums still left a small space under the curve. I don’t think our rounding up to 500m stops this being the answer.

 

More about the trapezium rule can be found here

As we get close to the exams, my students mostly want to do previous exam questions with me, so today I am sharing one which I found rather interesting – Algebraic proof. Perhaps because it foreshadows the kind of Maths one does at A-Level and degree level but its aimed at GCSE students.

The skills required are not that different to solving equations, but the level of thinking required is perhaps one level further on.

The question is…

Prove that the square of any odd number is one more than a multiple of 4.

The idea of an algebraic proof

The first thing to notice are those words ‘prove that’..  for students who are used to Maths being about doing a sum and finding an answer, this could seem quite new.  We are told what to expect. We just need to show the statement is true.

One of my students said ‘does this mean I have to look at every odd number’.  That would be quite difficult because there is a never -ending list of them. But its not a bad way to start, to convince yourself it might be true.

5² = 25  – and that is 1 more than 24 which is 4 time 6.

11² = 121 and that is 1 more than 120 which is 4 times 30

I could keep going like that – and you should do maybe a couple of your own – but I will never prove its true for ALL numbers like that.

 

What is easier : prove its true or prove it isn’t?

 

Its easier in Maths – and in life – to prove something isn’t true. We only need to find ONE odd number where the square is not 1 more than a multiple of 4 to prove that statement is not true*. A person only needs to show he was in Birmingham on Tuesday evening to show he didn’t commit the crime in Paris.

(*You could try but don’t waste your time looking,  it is true!)

Proving things are true is harder.  But we can still do it. We are going to do this with algebra, looking at the ‘general case’. That is why we call this algebraic proof.

We need to think of numbers more generally.  What is the ‘general’ odd number.  Take a number, any number. Will it be odd? Well there is a 50-50 chance.  But if we double the number we know it will be even (131 doubled is 262), then add 1. We then know the answer we get will be odd.

And in fact, all odd numbers can be found using  2n + 1 – That is a general odd number!!

So take our general odd number, and square it.  (2n+1)(2n + 1).  I could have written this with a ² but I find its easier to multiply brackets if you can see both.

We get

Factorise just part of this you get 4(n² + n) + 1

And that is actually what we are looking for! Let me explain. In 2n + 1, our ‘general’ odd number, n is just any whole number, and so n² + n is also just a  whole number.

So 4(n² + n) must be a multiple of 4 [ 4 times (n² + n)] and 4(n² + n) + 1 is 1 more!

This completes our example of algebraic proof.

For today’s post I’m going to look at an exam question I saw posted on Facebook.

I confess this is the sort of stuff I get nerdy about! I love the way that equations can describe pictures, and vice versa.

 

 

 

Before we start, lets make sure we understand the terms used. What exactly is a ‘turning point’?   Well, it is the point where the line stops going down and starts going up (see diagram below).   That point at the bottom of the smile.

There are three different ways to find that but in all cases, we need to start by finding the equation – finding out the values of ‘a’  and ‘b’.

‘b’ is the easier. take the case where x=0 and the two terms with x can both be disregarded and so b = -5 = the point where the curve crosses the y-axis (also known as the line ‘X = 0’.  That is similar to the reason why the term ‘c’ is often called ‘the y intercept’ for straight lines, equation y = mx+c

The other point we know is (5,0)  so we can create the equation

25 + 5a – 5 = 0 (By substituting the value of 5 in for x)

We can solve this for a giving a=-4

 

The full equation is  y = x² – 4x – 5

I usually check my work at this stage  5² – 4 x 5 – 5 = 0 – as required

Now, I said there were 3 ways to find the turning point. I will give all three here bu, be warned, the third does require some A-Level maths. This diary is more aimed at GCSE students but method 3 is actually the way I would usually find the turning point, so I will give a brief description here.

Method 1 – The ‘parabola’ is symmetrical.

A Parabola is the name of the shape formed by an x² formula

 

On this version of the graph

I’ve marked the turning point with an X and the line of symmetry in green. 

This tells us the value of x on the turning point lies halfway between the two places where y=0 (These are solutions, or roots, of x² – 4x – 5 = 0

We know one of these is is x=5. We can get the other by factorising  to give (x-5)(x+1) = 0. So x = -1 is the other solution. Halfway between x = -1 and x = 5 is x = 2.

when x = 2, y = 2² – 4 x 2 – 5 = -9.  So the turning point is (2,-9).

Method 2  Complete the Square

If we ‘complete the square’ on this equation we get

x² – 4x + 4 – 4 – 5  –  I’ve added in 4 and taken 4 away which looks like an eccentric thing to do, but this means we can factorise the first part into a square

(x – 2)² – 9 (also combined the -4 and – 5 to make -9)

This where we use our knowledge that a ‘square’ is never less than 0, but it can be 0. So its minimum value is when that square = 0 – so x – 2 = , so x = 2. And we can also see the value of the whole thing there is 0 – 9 = -9

Method 3 – By differentiation

This is A-Level stuff really, so I’ll only give an overview. This is the way I would usually do this, but then I have studied Maths through A-level (and beyond!)

Differentiation is one branch of Calculus, the mathematics of measuring change.

By a rule you will learn if and when you first study calculus, the equation of how much x² – 4x – 5 is changing is given by 2x – 4.

The turning point is where the line isn’t changing, so 2x – 4 = 0 (Zero change) so 2x = 4 and x = 2. y = 9 can be as above.

 

Summary

I’ve given three methods here. Fortunately they all give the same answer! (They wouldn’t be good alternatives otherwise!)

Choose which of the first two you feel most comfortable with.  The third method is the easiest to extend to where we have equations of than an a x²

Let’s consider this question from a Higher GCSE paper.

 

 

 

First thing I think is – Wow, that looks complicated.

[Actually, the first thing I should think is, what sign is that. It is a divide sum in the middle, not an add sign as I first thought]

Algebra fractions can stump the brighter students, but its worth remembering to rules are just the same as fractions with numbers.

We could do this divide sum by flipping the second and multiplying the numerators together and then the denominators. But given that we are told the final answer is simple, there is certain to be a lot of simplifying we can we first.

The first thing I noitices was that the bottom of the second fraction was going to factorise. It looks so close to the ‘difference of 2 squares’ rules.

Indeed it does factorise to

x(x+5)(x-5)

And immediately we see there is some cancelling to be done with the top to give 1/x(x-5) for the second fraction

What about the bottom of the first fraction. Looks like there could be some factorising to be done there. And indeed, taking my cue from the (x-5) we have already seen, this factorises to

(x – 5)(x + 2)

Something I missed before because I was so busy factorising the higher powers of x, but the top can be factorised as 3(x+2) and so we can cancel to

3/(x-5)

NOW we can do the divide by flipping the 2nd fraction, and the operation is now much easier than before

3/(x – 5) *  x(x – 5)   [Using an * for multiply here to stop confusion with the x’s]

which is then 3x = so a = 3.

 

 

;