Equation of a Circle

•  Edexcel, Module - C2, Chapter - Coordinate Geometry
• AQA, Module - C1, Chapter - Coordinate Geometry
• OCR, Module - C1, Chapter - Coordinate Geometry

In C1, part of the Coordinate Geometry Chapter, we learnt about linear equations and graphing them. In C2, we learn about the equations of circles, from which we can graph them, aswell as solving problems again involving tangents. Remember in GCSE, we learn about Circle Theorems, here are a few we need to know, which will come in use in this chapter :

Circle Terminology !

Circumference - The distance around a circle, given by the formula 2pi * r.

Diameter - The distance from one point on the circumference, to another, passing through the centre of the circle. It is twice the radius (2r)

Radius - A line which is from the centre of the circle, to any point of on the circumference of the circle. It is half the diameter.

Area of a circle is given by the formula - pi * (r^2)

Circle Theorems Recap !

1) The angle in a semicircle is a right angle (An angle formed by drawing lines from the ends of a diameter to its circumference, will form a right angle)

2) The perpendicular from the centre of the chord, bisects the chord.

(A chord is a line that has both endpoints on the circumference on the circle.)

3)The perpendicularity of the radius and tangent.
(A tangent to the circle, forms a right angle, with the circle's radius, at the point of contact with a tagnent).

Showing the 3rd theorem, which shows a line from the centre of the circle till the circumference (Radius), when it meets with a tangent to the circle at a point, it will form a right angle !

Equation of a Circle

The equation of a circle is given in the form : (x-a)² + (y-b)² = r² 

^{a = x -coordinate of centre}

^{b= y-coordinate of centre}

^{r= radius}

<sup>e.g.1 ) What is the equation of a circle when the centre of the circle is (-4,5) and the diameter is 10.

centre of circle (-4,5), so a = -4  and b = 5
radius = half the diameter so 10/2 = 5
so</sup>   (x + 4)² + (y - 5)² = 5<sup>2 


Just leave it in the form 5 squared, rather than writing 25.

e.g.2) What is the centre of the circle, and radius, given this equation of the circle :</sup>

(x)² + (y+8)² = 49


<sup>Centre of Circle -a= 0, and b = 8. So centre of circle is (0, -8 )
Radius is the square root of 49, some people still write + / - 7, even knowing a length can't be negative. so the radius is 7.

We've met our first point, knowing how to derive the equation of a circle, given the centre points and the radius, and vice versa.

Finding the centre of a circle, through Completing the Square


If we are given the equation of the circle to be :</sup>

x² + 6x + y² - 8y - 11 = 0 , and we have to determine the centre of the circle and radius. Here, we can see it's not in the usual form, but still it is an equation of a circle. We just have to complete the square (look under the c1 completing the square post if you've forgotten) :

<sup>complete the square for x terms and y terms seperately:

so</sup>  x² + 6x  becomes = (x+3)² - 9
so y² - 8y becomes = (y - 4)² - 16

we move the -11 to the other side, so we get :

(x+3)² - 9 + (y - 4)² - 16 = 11

(x+3)² + (y - 4)² - 25 = 11

(x+3)² - 9 + (y - 4)² = 36

(x+3)² + (y - 4)² = 6<sup>2

Now, we have to the equation of the cirlce in the desired form :</sup>
(x- a)² + (y - b)² = r<sup>2

The centre of the circle is (-3,4) and the radius is 6.

This is all to the coordinate geometry chapter, in terms of what we need to know. In the next post, I will post some questions and their model solutions that are quite common, which require some knowledge of GCSE theorems!</sup>

 

Algebraic Division,Factor Theorem and Remainder Theorem

•  Edexcel, Module - C2, Chapter - Algebra and Functions
• AQA, Module - C1, Chapter - Algebra
• OCR, Module - C2, Chapter - Algebra

First chapter of C2 is Algebra and Functions again. We firstly start off with Algebraic Division, this is basically the long division we've done with numbers, except now we have polynomials. Before proceeding, it's good to know what a polynomial is.. and how to factorise certain polynomials (cubics, quadratics).. if you've forgotten. look into my c1 posts.

We are required to divide polynomials by (x-a ) or (x+ a), where a is some integer. This division process can either result in (a being a factor, which will mean after the division is complete, the remainder will be 0, and we will obtain a quotient (which is usually a degree less of the polynomial being divided). OR the division can result in a remainder, in which a is not a factor.

I will show you the actual process and examples in my image notes, it is hard to write it on here. The process is always the same. A couple of common errors :

* Say we have x^3 + 3x + 2. and we have to divide it by some factor (x-a).. some students may not be able to continue after some point, it is simply because we have to include the x^2 term in the division, so we rewrite the polynomial as :  x^3 + 0x^2 + 3x + 2 divided by (x-a).  [Always include terms that are not present by writing 0x^a (a being whatever power).. this will line up the columns]

General Teminology

Factor - Something that is completely divisible by another thing, (no remainder) x+4 is a factor of x^2 + 8x + 16, 2 is a factor of 10, 10 is a factor of 150.. etc.

Remainder - The value obtained / left over when one polynomial is divided by another, where it is not a factor. The Remainder Theorem can be used to find the remainder, instead of long division.

Divisor - The polynomial you are dividing by. e.g x^2 - x + 1 divided by (x - 2 < divisor)

Quotient - The expression obtained as a result of dividing polynomials. e.g. ( in numbers 8 /4 = 2 - quotient)

Quotient * Divisor = Dividend (original expression)
or
(Quotient*Divisor) + Remainder = Dividend (original expression)


Factor Theorem

The Factor Theorem states, that if f(a)=0, then (x-a) is a factor. This is another way of checking if some expression is completely divisible by another.

Remainder Theorem
Same as the factor theorem, when you divide a polynomial by (x-a), the remainder will be f(a).

Notes of Algebraic Division 2

Notes for Algebraic Divsion 1

C2 - Introduction

C2 is the second module, of the Alevel Maths which is core. It is in my opinion the hardest module of AS, for Edexcel, it is one of the longer modules of the whole course.The topics are :

Edexcel

1. Algebra and Functions - Algebraic Division, Factor Theorem, Remainder Theorem.
2. Coordinate Geometry - Equation of Circle, Use of certain circle properties
3. Sequences and Series - Geometric Series {Sum of a g. series, Sum to infinity, Proof of sum formula}, Binomial Expansion for (1+x)^n and (a+b) ^n.
4. Trigonometry - Sine and Cosine Rule, Area of Triangle, Radians, Arc length, area of a sector and segment, Sin,Cos and Tan Graphs, Use of Trig. Identities, Solving simple equations.
5. Exponentials and Logarithms - y=a* and the graph, Laws of Logarithms and Change of base formula.
6. Differentiation - Maxima and Minima problems, Stationery points, Increasing and Decreasing Functions.
7. Integration - Definite Integrals, Area under a curve is a definite integral, Trapezium Rule, Area bounded between a curve and line.

AQA

1. Algebra and Functions - Laws of Indices, Simple transformations on graphs
2. Sequences and Series - Arithmetic series {nth term, Sum of a. series}, Geometric series {Sum of finite g. series, sum to infinity}, Binomial Expansion (1+x)^n.
3. Trigonometry - Sine and Cosine Rule, Area of Triangle, Radians, Arc length, area of a sector and segment, Sin,Cos and Tan Graphs, Use of Trig. Identities, Solving simple equations.
4.  Exponentials and Logarithms - y=a* and the graph, Laws of Logarithms and Change of base formula.
5. Differentiation - Differentiate functions of form x^n. where n is a rational number.
6.  Integration - Integration of functions of form x^n, Trapezium Rule

OCR

1. Algebra - Factor theorem, Remainder theorem, algebraic division, a* graph, laws of logarithms, change of base of logs, and solving log equations.
2. Sequences and Series - Sigma Notation, Arithmetic & Geometric Series, Sum of arithmetic and geometric series, sum to infinity, Binomial Expansion.
3. Trigonometry - Sine and Cosine Rule, Area of Triangle, Radians, Arc length, area of a sector and segment, Sin,Cos and Tan Graphs, Use of Trig. Identities, Solving simple equations.
4.  Integration - Indefinite Integration, Evaluation of definite integrals, Finding area of a region bounded by a curve and line,. Trapezium Rule.

There is alot of new content, and alot which is expanded from the topics learnt in C1. To Achieve a top grade, you should ensure you can apply everything used in C1, these modules are synoptic, and they can ask you a c1 question in c2, or ask a question which relates to c1 content.

Indefinite Integration

•  Edexcel - C1, Integration
• AQA - C1, Integration
• OCR - C2, Integration

We already discussed one of the branches of calculus. The other is Integration, it's general purpose is to find the area between some given intervals. It can be used to find the area bounded by a curve, you will learn this in C2. Right now, we need to know what Indefinite Integration is. It is simply the reverse of Differentiation, if dy/dx is the derivative of some function, we can use integration to obtain the function y. It's also called Anti differentiation. When will integrate some function, we will add a constant c. (The constant of Integration).

That image is the general notation used for Integration. When we integrate, we first draw the S kind of line, then write the function, and then dx. (for now anyway). This shows we are integrating. We need to know how to integrate functions in the form : x^n.

As I said Integration is the reverse of Differentiation. So when we differentiate we multiply the power by the coefficient of the function, and subtract the power by 1. When integrating, we firstly raise the power by 1, and divide by this new power.


a)

      *We usually write c (instead of constant) [  + c ]


I will also post some further examples in my image notes. In my image notes, I will also show you how to derive a the equation of the curve, when given dy/dx and a point on the curve, these question are usually 5/6 easy marks.

With this we end C1 here. I shall start C2 next week... I have done all model solutions for the Edexcel C1 Papers (From Jan 2005 - June 2011), these are available on request. I will start working on the model solutions for Solomon Papers aswell.

Differentiating Functions with Examples and Applications of Differentiation

I will use two examples from the specification itself, which you should be able to differentiate :

1) y= (2x+5)(x-1)
First we expand the brackets, and simplify :

2x^2 - 2x + 5x -5
y = 2x^2 + 3x -5, now we can differentiate so :
dy/dx = 4x + 3

I will show the 2nd example, in my set of image notes.

Application of Differentiation to Equations of Tangents

If we have a curve, and a tangent to some point on the curve. We can find the equation of the tangent. How ?

If we know the coordinates of that point. Firstly we differentiate the curve, and substitute the x-coordinate to get the gradient of the curve at the point. Say, if the tangent is perpendicular to the curve. We know that the gradient of this tangent, will be -1 divided by the gradient of the curve, we just got. Using the gradient, and the coordinates, we can use the y -y1 = m(x-x1) formula to find out the equation.

Example

This question is taken from the Edexcel C1 Jan 2006 Paper. (Question 9)

Sorry for the bad quality

Model Solution 

I have ignored part a).. the rest of the question is more relevant to the post.

Notes on Differentiation with Four Examples

Application of Differentiation for Tangents,Gradients and Normals

Differentiation

• Edexcel- C1 Differentiation
• AQA- C1 Differentiation
• OCR- C1 Differentiation

One of the new topics, we learn in A-Level Maths is Differentiation. It is one of the branches of Calculus, which is a major field in Mathematics, and almost is a useful application in loads of other fields.. Engineering, Economics, Physics and Chemistry. It is concerned with how one thing changes, as a result of another quantity changes. E.g. How displacement changes, as time changes (dd/ dt) would be the velocity.... We shall look for now at how y changes with respect to x (we call this dy/dx) = d (delta which means change)

We will look at curves, where the gradient is changing at each point on the curve. So dy/dx on each point is different, and not constant. First, i'll introduce you to a tangent. A tangent, is a straight line which touches a point on the curve, it only touches that point though.

The green line is the tangent to the curve( in black), this tangent only touches the yellow point on the curve. Note the gradient at the yellow point, is different to the rest of the points on the curve, because the gradient changes as x changes.

The derivative of a curve is the same as the dy/dx of a function e.g.

The derivative of x^3 = 3x^2
dy/dx (x^3) = 3x^2
This is the first derivative, if differentiate again, we would get the second derivative, again.. the third.. and so on..

dy / dx means differentiating y with respect to x. (what is happening to y, as x changes)

How to Differentiate

To differentiate a function, you reduce the power by 1, and multiply by the new power :

Function          Derivative
axⁿ                 anx<sup>n-1

e.g. x^2 
dy/dx = 2x


x^3 
dy/dx = 3x^2


* If we had to differentiate anything to the first power e.g. x , 3x, 5x... it would be 1,3 and 5 respectively. Why ? 
Because it is to the power 1, reducing the power to 0.. anything to the power of 0 equals 1 .. so we just multiply the coefficient of x by 1.. which is the same as taking away the x.

*Differentiating a number .. gives 0. Think about it, if draw a graph of say y = 5, the gradient is 0.

Other Notation 

A function can be written as y =... or f(x) = ..., if we have a function defined as f(x) =...., then the derivative of that is f'(x)=...

f(x) = 5x ... f '(x) = 5 (this is called f prime)

*For the first derivative we use one dash.. second derivative two..etc 
We can only differentiate functions in the form of</sup> axⁿ    , so if it looks any different, we have to rearrange to get in that form, using rules of indices.

C1 - Algebra and Functions - Polynomials

Sorry to update after almost a week, was really busy with other work. In this post, I shall talk about Polynomials. A polynomial is basically, a sequence of monomials, which is a term of product of powers of variables. Confused much ?

In simple terms, we just need to know about two polynomials, one we've already encountered, is a Quadratic, this is a polynomial of degree 2. The other we learn about in this post is a Cubic, which is a polynomial of degree 3.

We are expected to factorise cubics, aswell sketch them, and manipulate them algebraically, i.e. Factorise, Collect like terms and expand brackets. A cubic takes the form :

where a is not 0, and b,c,d are constants.

Cubics, can be factorised into 3 brackets, as Quadratics can be factorised using two. Where a cubic function crosses the x-axis is known as the roots. In terms of an equation, (the solutions), when the cubic is set equal to 0.

An sketch of some cubic function looks like :

A cubic function, this one crosse the x-axis 3 times, meaning it has 3 roots/ 3 solutions.Try looking a the graph with your head tilted, and notice the kind of S shape, it is a feature of a cubic. Not all cubic functions will look like the S shape though !

Multiplying out Polynomials (Expanding Brackets)

If we have two polynomials (one a quadratic, the other linear) e.g.

(3x^2 + 2x + 8)  (x + 5), if we had to expand them, its basically saying multiply each term in one bracket with the other term in the other bracket :

Multiply 3x^2 by the second bracket we get :

3x^3 + 15x^2

Multiply 2x by the second bracket we get :

2x^2 + 10x

Multiply 8 by the second bracket we get:

8x + 40

Collect like terms and simplify :

3x^3 + 15x^2 + 2x^2 + 10x + 8x + 40

= 3x^3 + 17x^2 + 18x + 40

Notice, its in the form of a cubic, where a = 3, b = 17, c = 18 and d = 40.
Though I don't think we need to be able to solve cubics, we can if the original quadratic we had can be factorised. If it can be factorised, we would have three roots, in this case the solutions we be :

x = -5, and x = one of the solutions of quadratic, x = other solution of quadratic

C1-Algebra and Functions - Solving Simultaneous Equations

We now,move on to Solving Simultaneous Equations. We will usually encounter questions where 1 equation is a quadratic and the other being a linear. From Gcse, you should know how to solve simultaneous equations by either Substituition or Elimination. Let's Recap :

1) y = 4x + 2
2) 3x + 6y =39

Let's Solve this firstly by Substituition:

* In General, we rearrange one of the equations to make x or y the subject, we then substitute this expression in the 2nd equation, to find one of the variables.

*Simply then plug into any equation, to find the value of the other variable.

We see y is already the subject of the first equation, let's substitute this equation as y in the 2nd equation so :
3x + 6(4x+2) = 39
Now expand and simplify:
3x + 24x + 12 = 39
27x =27
x= 1
Now Substitute this value of x=1, into any equation to find the value for y:
y=4(1) + 2
y=6

Solutions {x=1,y=6}

By Elimination
*In General, We usually eliminate one of the variables x or y, we do this by making the coefficient of the x's or y's same and the subtracting or adding them to make it 0.
*Now we only have one variable, so we solve for this variable. And substitute the value in one of the equations to get the value for the other variable.

1) y = 4x + 2
2) 3x + 6y =39

We can use either make the coefficient of x or y the same, we'll stick for y, it's easier in this case. Multiply the first equation by 6 :
6y = 24x + 12

Now rearrange the second equation to make 6y the subject :
6y = 39 - 3x

Subtract these two new equations :
       6y = 12 + 24x

   -   6y = 39 - 3x 
 which gives you :
0 = -27 + 27x
rearrange to solve for x:
27= 27x
x= 27/27 = 1


Now substitute this value in any of our two original equations for y :
y= 4(1)+2
y=6


Solutions = {1,6}


That shows how to solve two linear equations / simultaneous equations. To Solve a Simultaneous equation system with a quadratic, simply use Substituition :

1) Rearrange the the linear equation to make x or y the subject.
2) Substitute this expression for x/y, in the quadratic equation. Solve for this variable
3)Substitute the value you get, into either of the two original equations {linear or quadratic} to find the other variable's value.

I don't want to make this post too long, so i will add an example in my set of image notes.

Notes for Just "Simultaneous Equations"

Here is an example with Simultaneous with a quadratic involved:

Simultaneous With a Quadratic Equation

*Note - The example I have used kind of uses everything we learnt from the previous topics - Using how to solve a quadratic to manipulating surds. The answers are left in "Exact Values", i.e. not in decimals. Be sure that edexcel can test you in anyway way.

C1- Algebra and Functions - Quadratic Graphs and The Discriminant

After knowing how to solve the Quadratic Equations by all methods. I shall now introduce some key points of a Quadratic Function, and how it looks on graphs.. while introducing the Discriminant.

The Quadratic Equation has a degree to the power 2 (when we say x squared). We say quadratics are polynomials of the degree 2, in the form of ax^2 + bx + c. Where a,b,c are constants, and a is not 0. The graph of a quadratic is  a parabola, almost a U shape, like this :

The shape of a parabola 

When a (coefficient of  is x²  )is positive, a U shaped parabola will be the graph. When a is negative, it will be an upside down U. This is one of the transformation of functions, I will be showing you in a later post ( which results in a reflection in the x-axis)


















<sup>Where this curve, crosses the x-axis, are known as the roots. These roots are the "solutions" when we solve these functions as equations by equalling them to 0, and using one of the 3 methods.

The Discriminant 
The Discriminant, is an expression, which allows us to find out the nature of the roots of a quadratic equation. It is given by :</sup> 

 <sup>There are 3 possibilities of the discriminant :
* Discriminant > 0 (This means the quadratic has two distinct real roots)
*Discriminant < 0 (This means the quadratic has two distinct complex / no real roots)
*Discriminant = 0 (This means the quadratic has 1 real root / repeared root)

I shall provide all three posibilities with an example, an my image notes, along with some graphs.</sup>

C1 - Quadratic Equations - Completing the Square

Final method of solving quadratic equations is Completing the Square. By completing the square for :
 ax² + bx + c, you will generate the Quadratic Formula by solving for x.

The general method consists of changing the quadratic in the form ax² + bx + c into (x+d)²+ e =0

where d = b/2 and e = b² / 4 
* This method requires a = 1, so if the coefficient of x is not 1 . Divide by a to get it to be 1.
* Then rewrite the Formula in the form shown above. Solve for x by :
1) Taking e to the otherside and squarerooting it, and then subtracting b.

My image notes show two examples. 1) How to derive the quadratic formula 2) A numerical example :

Completing the Square Notes

     

C1 - Algebra and Functions - Quadratic Equations (Solving using Formula)

One of the methods, apart from Factorisation is using the Quadratic Formula. Usually, when a quadratic equations cannot be factorised, it is a good idea to use the Formula. Though the formula works for all quadratic equations. We have to use the form ax^2 + bx + c. Identity our coefficients and simply substitute into the formula which is :

There's nothing else to know I reckon for the Quadratic Equation. Though I will show you how to derive it by Completing the Square (which i will cover in the next post). Another fundamental point that comes out of the Formula, is the b^2 - 4ac, this is called the Discriminant. We shall look it later, it is important because we can determine how a quadratic function may look, and the number and types of roots assosciated with the equation.

Quadratic Formula Notes

TO SEE ANY OF MY IMAGE NOTES IN FULL, right click and click on view image.




 

C1- Algebra and Functions - Quadratic Functions

Now we learn about Quadratic Functions. They're arguably one of the most important types of functions, and you will need to be able to solve them and master the concept, as they will pop up later in other modules. A Quadratic equation, have a degree of 2, takes the form :

ax^2 + bx + c = 0 , where a,b,c are real numbers.

We can solve, quadratic equations using three methods : a) Factorisation, b) Completing the Square and c)The Quadratic Formula.

Quadratic Functions when sketched are of a parabola shape ( a U shape), the coefficient of x^2 will decide the shape of the parabola. If a is negative, the shape will be an upside down U, if positive it will be a U.

The solutions of a quadratic equations are called ROOTS. If you're dealing with a quadratic inequality we call the roots (CRITICAL VALUES).

1) Factorisation... Let ax^2 + bx + c = 0

find two numbers that multiply to give c , and add to give b. (These numbers must be the same for the product and sum)
*Make sure you take care, when you have negative numbers. (Remember a negative * negative = positive)

If you find two numbers that satisfy the product for c, and sum of b: simply put it in the form :

(x + ...) (x + ...) = 0  (... are the two numbers, it does not matter which order they are put)

To find the solution simply put (x +...) = 0, for both of them and solve for x. I shall provide an examples on my
handwritten image.

C1- Algebra and Functions - Rationalising the Denominator

Now that we can manipulate surds, by adding/subtracting/expanding and simplifying. Our next concept, is Rationalising the Denominator. This simply means, making the denominator of a fraction "surd free". There are 3 general rules outlined in the image below with the some examples :

Rationalise Denominator Notes

C1 - Algebra and Functions - Surds

Our next topic is in C1 is Surds. Surds are simply numbers left in the the root form, they are irrational numbers. Firstly we need to know a few rules of surds.

I shall just post images of my scanned written notes, as they are clearer and concise. With Examples.

C1 - Algebra and Functions - Laws of Indices

In this First Post for C1. I shall be posting some notes and examples about the 1st topic in Algebra and Functions. This is about the laws of indices and surds.

Laws of Indices.
A variable raised to the power, is called an indice / index. We have certain rules we can apply, to simplify / manipulate, usually if the base of the power is the same.

1) x^a * x ^b = x^(a+b)   (X to the power of a multiplied by X to the power of b) = X to the power of (a+b)

2)x^a /  x^b = x^(a-b)   (X to the power of a divided by X to the power of b) = X to the power of (a-b)

3) (x^a)^b = x^(a*b) (X^a to the power of b) = X to the power of (a*b)

*** For the 3rd rule, the b will be adjacent to the a, I can't write it in the handwriting form. If there is a number in front of the x, this number will be raised to the power of b.

e.g. (3x^4)*2 = This means 3x to the power of 4 , the 2 is outside the bracket.

The answer would be 3 to the power of 2,multiplied by, x to the power of (4*2) =  (9x^8)

4) (x/m)^n = x^n / m^n (x divided by m, all the power of n = x to the power of n divided by m to the power of n).

General Facts

Anything to the power of 0 will be 1. I mean anything from a number to a variable.

Anythign to the power of 1 will be itself. e.g. 2^1 = 2 ,   (x^2) to the power of 1 = x^2

Other Rules
I have uploaded an image file explaining the rules, as it is not appropriate to write them up, as the typos can be hard to understand.

Handwritten Notes on Laws of Indices (Printable)

If any problem understanding the notes, or feedback email : mathematicsthesolver@gmail.com

C1 Introduction

C1. The first module of AS, of any Alevel Maths Course. Suppose to be the easiest, it generally forms on from the A-A* topics of gcse. However, you don't have a calculator... so brush up on general arithmetic skills, and fractions. Topics :

1)  Algebra and Functions
*Laws of Indices, Surds, Rationalising the Denominator, Quadratic Equations (Discriminant, Completing the Square, Quadratic Formula,Factorisation), Sketching Graphs and Transformations of Graphs.

2) Coordinate Geometry
*Equation of a Straight Line, Gradients, Perpendicular Lines

3) Sequences and Series
*Terms, nth Term, Arithmetic Series, Sum of Terms and Stigma Notations.

4)Differentiation
*Differentiating basic functions, and relation to Gradients.

5)Integration
*Indefinite Integration

Introduction to A-Level Maths

Hello and Welcome to my blog for A-Level Maths. This blog shall cover material / notes from C1 - C4, updating weekly. I am focusing for now on Edexcel's new specification (from 2008 onwards), and hopefully later on will go on to OCR and AQA.

Firstly, I'd like to make clear, if hardwork is put it at any stage, the good results will be the output. It does not really matter whether you have an A* or a C at gcse. Alevel does require independant study, and Practice makes Maths, not rewriting notes hundreds of times. I stress, that it is important to understand a key concept, rather than trying to solve and generalise examples from textbooks. Here is the specification link for Edexcel :

http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/UA024850%20GCE%20in%20Mathematics%20issue%202%20180510.pdf

Here is the link to the Formula Booklet, you will have access to in the exams :

http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/N38210A-GCE-Mathematical-Formulae-Statistical-Tables.pdf

For A-Level Maths (Edxcel), you will have to take 4 core modules {C1,C2,C3,C4}. And two applied modules from {D1,D2,M1,M2,S1,S2}. Usually candidates choose D,M,S 1 then D,M,S 2. But you are allowed to take part 1 of two modules, if your school permits.

*C2,C3,C4 are synoptic modules, each requiring candidates to understand content from previous modules.

 AS - (C1,C2 ,1st applied module)
A2 - (C3,C4, 2nd applied module)

I will write up model solutions on request by email. So email me your problems on mathematicsthesolver@gmail.com  {ONLY C1- C4 Problems}.