 Maths Core One Core One is the non-calculator exam. But don't panic! Keep breathing! Everything will be fine if you just revise! Indices ... Well, there's no worse place to start, I suppose ...

Indices are powers. For example: 4 is 4 to the power of 4. Look at this pretty pattern: 4 = 64 4 = 16 4 = 4 4 = 1 4 = 4 =  From this, we can see why anything to the power of 0 is always 1.

General Rules

1. Multiplying means adding the powers. E.g. 33 x 33 = 36 (notice the 3 remains the same.)
a x a = a(m+n)
2. And so dividing means taking away the powers. E.g. 33 32 = 31. = a(m-n)
3. Raising by a power means multiplying. E.g. (33)3 = 39.
(am )n = amn
4. Anything to the power of 1 is just itself.
5. Anything to the power of 0 is 1.

Fractional powers

These are pretty obviously where you have a number to the power of a fraction. It just means the same as taking a root. The denominator tells us the root and the numerator then tells us to what power it needs to b. For example:

a = a

a = a

a = a

a = ( a)² or a²

64 = ( 64)2 = 42 =16

The proof of this is simple too: If we have a x a = a( + ) = a1 = a.

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Simplifying Surds

What are quadratics, you might ask? If you recall GCSE, they're a form of polynomial, the one with the x2 term in them. For example: x2 - 4x + 4 is a quadratic equation.

How do we solve them?

There are three main methods and all three are cool. Depending on the question, it might be easier to use one over the other.

• Factorising
• General Formula
• Completing the Square

Factorising is the simplest. It just means sticking some brackets in.

Sketching Curves

One of the worst parts of Maths... graphs. *shudders* Well, let's get it over with.

The general quadratic curve is:
y = ax² + bx + c

• If a is positive, you have a happy graph shape: • If a is negative, you have a sad graph shape: • As usual, c is where the graph crosses the y-axis.
• The discriminant tells us how many roots the graph has (how many times it crosses the x-axis.)
• If b²-4ac = 0 there is 1 root.
• If b²-4ac > 0 there are 2 roots.
• If b²-4ac < 0 there are no real roots.
• The x-value for line of symmetry is: .
• If you have one coordinate, it's pretty obvious you need to sub in for the other.
• To find where it crosses x, use general formula, completing the square or factorising.
• To find a maximum/minimum point, put it into the completing square form:
(x + a)² + b.

Example - Sketch the graph y = 15x - 2x² and write down the coordinates of where it crosses x and y, the line of symmetry and the minimum point.

First, what do we know? Well, it's a sad graph, because it has a = - 2.

We also know it crosses the y-axis at 0 because there's no c. Therefore, we have one point it crosses the x-axis: (0,0).

To find the second point, we put y to 0 and solve the quadratic:

-2x² + 15x = 0
-x(2x - 15) = 0
=> x = 0 or x = 7 The second point it crosses x = (7 , 0).

The line of symmetry is: x = - 15 -4 = 3 To work out minimum point of y = 15x - 2x², put in completing the square form:

(x - 3 )² - (3 )² = (x - 3 )² - 14.0625

=> Minimum value is - 14.0625 when x = 3 . Sub 3 into y = 15x - 2x² to get y-value of 28 . Coordinate (3 , 28 ).

Finally, we do a sketch, remembering to include all these: Curves you need to know y = x2 or its reflection y = -x2  y = x3 or its reflection y = -x3  y = or it's reflection y = -  Transformations

The quick hand way of writing y=x-whatever, is to use call it f(x). f(x) just means 'function of x'. This is how we display transformations, using the general graph y=f(x).

There are 5 transformations and you must learn them! The red line shows the new position.

1. f(x) + a
This moves the graph up if a>0 and down if a<0.
For example: f(x) + 3, where y=x3. 2. f(x+a)
This moves the graph left if a>1 and right if a>1.
For example: f(x+3), where y=x2. 3. af(x)
This stretches the graph by scale factor 3 in the y direction. (This means that every y value is x3).
For example: 3f(x), where y=x2-1. 4. f(ax)
This is a stretch in the x-direction. If a>1, the curve appears to shrink, whereas if 0<a<1, the curve appears to expand.
For example: f( x), where y=x2. (in other words, a stretch of scale factor 2). 5. Reflections:
-f(x) is a reflection in the y-axis.
f(-x) is a reflection in the x-axix.
(Pretty obvious why, if you consider the previous two rules.)

Simultaneous Equations
Series and Sequence

A series is something like this:

5, 8, 11, 14, 17... etc.

You can obviously see that we're adding +3 each time. But how do you write that in algebra?

There are two ways of doing it, and the first is by the nth term. You should be familiar with it from GCSE.

nth term = dn + a

d is the common difference. In the example above, we had +3.

To find out a, look at the 1st term. Times 1 by d. a is the number you need to add to make the first term. In the example above 1 x 3 = 3 + 2 = 5 and therefore a = 2.

For the series above, we have the nth term = 3d + 2.

Instead of writing "nth term", we use the denotation: un. The same applies for the first term: u1, second term: u2, etc...

The difference between terms is called "first difference". This is important with numbers with powers, because you then need the "second difference".

To be a series, there has to be an nth term.

By knowing the first term, we can work out how to get to the next. Looking at the example above, we can say:

un+1 = un + 3

(in other words, the next number is just the number you have add 3).

If u1 = 5, we can say:

u2 = u1 + 3 => u2 = 5 + 3 = 8

u3 = u2 + 3 => u3 = 8 + 3 = 11

etc...

This is called a recurrence relation.

An Arithmetic Series produced from a recurrence relation like the one above. It is produced by adding the same number (which could be negative) to each term, to get the next term.

un+1 = un + d ('d' stands for difference).

In general

1st term: u1 = a
2nd term: u2 = a + d
3rd term: u3 = a + 2d
4th term: u3 = a + 3d
nth term: un = a + (n - 1)d

a is just the number you start with. n is obviously the number of terms. If you're ever unsure of how to find the common difference, d, just take two consecutive terms and take the first from the second.

How is this useful? Well, a question might ask:

The 4th term in an Arithmetic Series is 8.1 and the 8th term is 17.3. Find the 100th term.

We can look above and see that the 4th term would be:

u4: a + 3d = 8.1

While the 8th term would be:

u8: a + 7d = 17.3

From here, we can use simultaneous equations to work out a and d.

4d = 9.2
d = 2.3

a + 6.9 = 8.1 (subbing into a + 3d = 8.1)
a = 1.3

Now we have a and d it's a simple matter of putting them into the formula to work out the 100th term:

u100 = 1.3 + (99 x 2.3)

u100 = 228.9

Note: n must be an integer and can't be negative. Pretty obvious why if you think about it!

Arithmetic Progression

This is where you add up all the numbers in an arithmetic series. While there's a formula, you need to learn the proof (they once asked poor AS students to write it in the exam!)

The notation for sum of n terms is Sn. So we can say:

Sn = u1 + u2 + u3 + .... un-1 + un

A quick way of adding them all up is to put them all in order, and then below, put them in the reverse order:

1. Sn = u1 + u2 + u3 + .... un-1 + un
2. Sn = un + un-1 + .... u3 + u2 + u1

Each little group is equal to the first term add the last term. (a + l). There are n of these. Add them all together and you 2Sn.

2Sn = n(a + l)

Therefore:

Sn = (a + l)

Now that's fine, but what if you don't know what the last term is? Well, we know from earlier what un is:

un = a + (n - 1)d

If we sub this into the Sn equation for l we get:

Sn = (2a + [n - 1]d)

This is the formula usually used to work out the sum of an arithmetic series.

For example:

30
Σ 3r + 2
r=1

The Σ symbol is just another way of writing 'sum of'. r tells us where in the series we're adding numbers from, in this case from 1, the first term. The number above the Stigma (Σ) tells us that there are 30 terms.

From this, we can write out a few terms to make things easier:

r=1, r=2, r=3
5 + 8 + 11 + .....

Now we can see that a=5, d=3, n=30. Put these into the formula:

Sn = (2a + [n - 1]d)

S30 = 15(10 + 29x3)

S30 = 15 x 97

S30 = 1455

A trickier example might involve inequalities:

What n first exceeds 1000 in:

n
Σ 5r + 3
r=1

The series is:

r=1, r=2, r=3
8 + 13 + 18 + .....

We know: a=8 and d=5.

Translated into maths, the question is looking for Sn > 1000.

Therefore:

Sn = (2a + [n - 1]d)

1000 > 2(16 + [n - 1]5)
2000 > n(16 + 5n - 5)
2000 > 16n + 5n2 - 5n
2000 > 5n2 + 11n

5n2 + 11n - 2000 > 0

Solve as you know from inequalities using general formula:

5n2 + 11n - 2000 = 0

-11  40,121
---------------------
_______10

n = -21.13 or 18.93022716

Draw a sketch of the graph and interpret when n>0: From this, we can see that n < -21.13, n > 18.93.

n can't be negative, and so n > 18.93. It must be a whole number, so n = 19.

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