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Maths Help Guide

Edexcel Core 1 Help Guide

1

Content
Indices
Surds
Quadratic Equations and Expressions
Simultaneous Equations
Inequalities
Graph Sketching
Coordinate Geometry
Sequences and Series
Differentiation
Integration

Formulae Given in Exam

Mensuration
Surface area of sphere = 4 r 2
Area of curved surface of cone =  rl (where l is the slant height)

Arithmetic series
an  a  (n  1)d
n
Sn  (2a  (n  1)d )
2
n
Sn  (a  l )
2

2

Indices
The rules of indices
The index is simply the power a base is raised to
...

Most questions will ask you to simplify expression or find numerical values and use one of the rules below
...
If they are worth 2 marks show workings e
...
Find  
 25 


3

3



3
2

3

 36  2  25  2  5  125
 25    36    6   216
 
 
 
Another typical question might ask us to Find the value of x given 23 x  4 x1
...

3x  2( x  1)
3x  2 x  2
x2
To check my answer I can substitute x  2 back in to the equation see if it makes the equation true
...
Do check your answer!
We may have to expand brackets and simplify
...
This is a common error
...
Remember a  m  m
...
2 3  3 
a
2
8
1

One important rule to remember is a  a 2
...
This will
be very important when it comes to differentiation and integration in the section on calculus
...
Don’t be tempted to try and write a truncated or
rounded decimal answer
...
Calculations in this form will be more accurate & easy
to perform
...

Some Basic Surd Laws
a a a

a c  b d  ab cd
a a 2 a
Just try these with numeric values
...
g
...
g
...

Sometimes you will have to simplify first
e
...
50  8  ( 5  5  2 )  ( 2  2  2 )  5 2  2 2  7 2
Expanding Brackets
You may be expected to expand single or double brackets
...
g
...
Do check first!
Rationalising the Denominator
Mathematicians hate having an irrational denominator
...
You will come across
two types of fractions where you will have to rationalise the denominator:
1
4
2 3
1 When the denominator is a single surd value such as
or
or

...
For the first one
multiply numerator and denominator by 3 , the second 7 and the third 5
...

4
4 7
4 7 4 7
An example could be



35
5 7 5 7  7 5(7)
4
2 When the denominator has two values and an additional or subtraction sign e
...

or
2 3
9
2 3
or

...
This will create a ‘difference of squares’ and allow you to simplify to give a
rational, denominator
...

3
3( 3  2)
3( 3  2)
3( 3  2)



 3( 3  2)  3 3  6
1
3  2 ( 3  2)( 3  2) 3  2 3  2 3  4
There may be some practical
length ×width = area  rearranging, rationalising & simplifying
applications of surds such as areas or
lengths
...
g
...
An expression is a collection of
terms, an equation will have an equals (=) sign in and we could look to use a range of techniques to solve
...
g
...
A common error is to simply give the answer of 5
...
An example would be a ‘negative length’ in an area
of a rectangle question
...

When trying to solve a quadratic equation go through a checklist of methods you can use to solve an equation
...

Type 2 ax 2  bx  0
Type 3 ax 2  bx  c  0 , a  1 that
Simply square root both sides
...
Set the RHS side to 0,
2
2
solve
...
DON’T factor
...
Completing the
the formula or you can complete
square is an option
...
When a quadratic
of the term in x 2 must be 1
...
Simply rearrange the
2
2
2 x  5x  9  0
5
97

equation into the form
x  
2
4  16
5  ( 5)  4(2)( 9)

ax 2  bx  c  0 and look to solve
...
The
x
or x 
4
2
question might, for example, state
x  3 which rules one solution out
...
In the exam you might be asked to
write a quadratic expression the form a ( x  b)2  c
...

Completing the square when the quadratic expression is in the form ax 2  bx  c , a  1
If you want an algorithm: “take half the coefficient of the term in x into the bracket with an x , square the
bracket, subtract the squared value away”
x2  4 x  5

( x  2)2  4  5
( x  2)2  9
This is a positive quadratic expression
...

6

Completing the square when the quadratic expression is in the form ax 2  bx  c , a  1
In order the complete the square the coefficient of the term in x 2 must be 1
...
If you are simply solving an equation divide through the equation by a
...

3x 2  5 x  4

5 

3 x 2  x   4
3 

2

5  25 
3  x      4
6  36 



2

5  35

3 x     4
6  12

2

5  19

3 x   
6  12

If you graphed this the parabola would open upwards (as it’s a positive value of a ) and the minimum point will
5
4
 5 19 
have coordinates  ,  
...

The Discriminant
The Discriminant determines the nature of the roots of a quadratic equation
...
The 3 scenarios are listed below:
2 Distinct Real Roots
No Real Roots
Repeated or Equal Roots
2
2
ax  bx  c  0
ax  bx  c  0
ax 2  bx  c  0
b2  4ac  0

b2  4ac  0

b2  4ac  0

An example could be find the set of values of k for which the quadratic equation x 2  kx  2k  0 has two
distinct real roots:
x 2  kx  2k  0
a  1, b  k , c  2k
b2  4ac  0 for real roots
k 2  4(1)( 2k )  0
k ( k  8)  0
k  0, k  8
You can use a sketch to help on the last part (covered in the next section on quadratic inequalities)
A quick tip! – A tangent touches a curve
...
Sketching a quick diagram should make sense
of this!
7

Simultaneous Equations
Linear Simultaneous Equations
Simultaneous equations are simply 2 or more equations that share common solutions
...
You will need to be able deal with 3 different scenarios in the module
...

So what does linear mean? The powers of x and y will both be one and there will be no terms in x and y that are
multiplied or divided by one another
...

(1) 2 x  y  4
Labelling the equations
(1) 2 x  y  4
(2) 3x  2 y  13
I am going to write equation (1) to make y the subject such
(2) 3x  2 y  13
I am going to multiply equation (1) by 2
that (1) y  2 x  4
(1) 4 x  2 y  8
I can now substitute this into equation (2) to eliminate
y from the equation such that
(2) 3x  2 y  13
(2) 3x  2(2 x  4)  13
Adding the two equations will
eliminate y leading to:
If I simplify this I have a linear equation in x such
that (2) 7 x  21
...

7 x  21 , which gives x  3
...

for y
...

2y  4
Check your answers for both x and y satisfy both equations
(1) and (2)
...

This wasn’t the only way I could have done the as y could be written in terms of x with no fractions on the
RHS
...
I could have multiplied equation (1)
by 3 and equation (2) by 2 to make the terms in
x the same and then subtracted equation (2)
from (1)
...
These equations would, in the main, be solve algebraically in C1

One Linear Equation, One Non Linear
Typical examples will include a circle and a line, a parabola and a line or a line and another type of equation
where y is implicitly defined as a function of x (usually where there is a product of terms in x and y ) such as
xy  y 2  3 y  4 x and 3x  y  6
...

Find the coordinates of points of intersection of the equations 3 y  4 x and x 2  y 2  25
...
The first equation is the equation
a straight line and the second equation is a circle, centred at the origin and has a radius of 5 units
...

3
2
4 
(1) x 2   x   25
3 
16
(1) x 2  x 2  25
9
25 2
(1)
x  25
9
(1) x 2  9
Solving for x we can see x  3
...
(2) y  x
3
When x  3, y  4 and when x  3, y  4
...

You can of course make x the subject of equation (2) at the start and solve the quadratic equation in y THEN
find the values of x from the linear equation
...
These are relatively
straight forward and we will work though a typical question: Solve the simultaneous equations:
y  x 2  5x  3
y  2x2  4x  3
We have expressions for y in terms of x for both equations so we can simply either set the equations equal or
subtract one equation from the other to eliminate y
...

At this stage some students finish the question without solving for y
...


9

Inequalities
Linear Inequalities
Inequalities tell us about the relative size of two values
...
Mathematically we could write x  4 and we would
read this as “ x is greater than 4”
...
In C1 you will be
expected to solve linear and quadratic inequalities and sometimes state the set of values that satisfies both a
linear and quadratic inequality
...

An example of a linear inequality may be “Find the set of values of x that for which 2 x  1  4 x  5 ”
2 x  1  5x  4
1  3 x  4
3  3x
1 x
Quadratic Inequalities
Quadratic inequalities are dealt with in a similar manner to quadratic equations (and will usually factor when
written on the form y  ax 2  bx  c ) and solved often with aid of a sketch
...
Remember! The x axis is the line y  0
If we were asked to find the set of
x2  x  6  0
x2  x  6  0
values that satisfy BOTH
( x  2)( x  3)  0
( x  2)( x  3)  0
inequalities below we could add
The critical values are 2 and -3
The critical values are 2 and -3
the linear inequality to our sketch
...
In this case it would be
x  3

I have simply used the information
from previous questions to graph
the inequalities above
Be careful with strict and inclusive inequalities and their respective notation,  / ,  / 
...

x  a can be read “ x is greater than a ” An open dot would be used on a number line
...
The values that satisfy this inequality are all those
strictly less than 3
...

p  2 can be read “ p is equal to or greater than 2”
...
A closed dot would be used on a number line
...
A closed dot would be
used on a number line
...
Quadratic graphs (parabolas), cubic graphs
and reciprocal graphs
...
The examiner is looking for a basic
understanding of the shape, key features, any asymptotes and points of intersection
...

Quadratic Graphs
Quadratic equations can be written in the form y  ax 2  bx  c and their graphs are symmetric parabolas
...
Negative (when a  0 ) will
open downwards and have a maximum
...

These solutions, or roots, can be found using the techniques above & max/min from completing the square
...

The completed square form y  a ( x  b) 2  c can help sketch the main features of a quadratic graph e
...

y  ( x  3)2  2 will have a minimum point at (3, 2), open upwards (as it’s positive) and the axis of symmetry
will be the line x  3
...
The sketch should be smooth, not a
collection of straight lines
...

Cubic Graphs
Cubic equations can be written in the form y  ax 3  bx 2  cx  d
...
g
...
g
...
If the equation was such that y  0 it will have the solutions
x  0, x  2 or x  3 which will assist in a sketch
...
Negative cubic graphs
(when a  0 ) will start in the 2nd quadrant and leave in the 4th
...
We can see if we expanded the
brackets the term in x 3 would be negative
...
These points would be plotted on the x axis
...

Positive Cubic Graph
Negative Cubic Graph

11

Some cubic graphs have repeated roots
...
If the equation had been such that y  ( x  2)2 ( x  3) then it would touch at 2 and pass through
at 3
...
The graph will have
x
two asymptotes, the x axis (or y  0 ) and the y axis (or x  0 )
...
These will be
transformations such that a reciprocal function would be written in the form y 
( x  b)
1
covered in the section on transformations below
...
As x gets
x
large y gets small both in the positive and negative direction
...
Translations, Reflections and
Stretches
...
Using numeric values
are often a good way of confirming this
...
The sketches don’t have to be
perfect and you may even find describing what you have done may help out if your sketch is as bad as mine!
For example “Scale factor stretch 2 in the x direction” or “Scale factor stretch of ½ in the negative y direction”
12

Translations in the x direction
...


Stretches in the x direction
...


1
scale factor stretch in
a
the x direction
...
If you are
given f(2 x ) for example,
the x coordinates are divided by 2
...

1 
f  x  would see the x divided
2 
by ½, or multiplied by 2, such
that the graph looks more
stretched out
...


This is a

If you try this with f( x )  x 2 you
may be a little disappointed!
The graph has moved 3 units to the
right
...

Translations in the y direction
...


Stretches in the y direction
...

a

The sign of the y coordinate
changes and the result is a
reflection in the x axis
...
You can
simply multiply the y coordinates
by a
...

13

Coordinate Geometry
Before we start…One tip! “If in doubt, sketch it out”
...
The whole topic is simply about straight lines in
the x , y plane
...

The Gradients of a Line or Line Segment
y  y2
The gradient is the change in y over the change in x such that the gradient m is given as m  1

...

42
2
1  3 4
Often errors are made with signs
...

2  4 2
The Equation of a Straight Line
For the equation of a straight line you need 2 things
...
A
typical question might be: Find an equation of the straight line passing through the points A(4, 3) and B(2, 1)
...
We can now choose either A(4, 3) or B(2, 1) as a
point the line passes through
...
If I had chosen B(2, 1) my final answer would be
the same
...

y  mx  c
y  y1  m ( x  x1 )
Using A(4, 3) and m  2 ,
Using A(4, 3) and m  2 ,
3  2(4)  c
y  3  2( x  4)
38c
y  3  2x  8
5  c
y  2x  5
 y  2x  5
I have written the line in the form y  mx  c
...
The example above would be 2 x  y  5  0
...

We need a gradient and a point the line passes through
...

Using the first method y  y1  m ( x  x1 ) , where P( 1, 2) and m  3
y  (2)  3( x  ( 2))
y  2  3( x  2)
y  2  3x  6
3x  y  4  0
We could of course write the equation in the form y  mx  c such that y  3x  4
...
These are important
basic facts often overlooked or forgotten by students and may be important parts of exam questions
...
Not
all solutions will have integer values so being confident with fractions is very important
...
Perpendicular lines are at right angles and the product of the
gradients of two perpendicular lines  1 such that m1  m2  1 where m1 and m2 are the gradients of the 2 lines
...
You will generally have
3
2
to state the fact m1  m2  1 in an exam when working with perpendicular lines
...
An example could be m1 

be: The line l1 passes through the points A(4, 3) and B(2, 1)
...

For the equation of a straight line we need a gradient and a point the line passes through
...
We found the gradient of the line passing through A(4, 3) and B(2, 1) in the previous section
...
The gradient of the perpendicular will therefore be the negative reciprocal which
1

...
I can use
gives m2 
2
either method outlined previously
...
In Layman’s terms, add the x coordinates together and
divide by 2, add the y coordinates together and divide by 2
...
The midpoint of the line AB where A(4, 3) and B(2, 1) is
,
 ,  3,1
2
2
2
2
2 2
Expect some non integer answers in exam and be prepared for exam questions that give you the midpoint and
one point where you are expected to find the other
...

The Distance Between Two Points or the Length of a Line Segment
Despite its quite bewildering formula this is simply Pythagoras Theorem
...
The distance formula is such that distance d is d   x1  x2    y1  y2 
Using A(4, 3) and B(2, 1) again we can find the length of the line segment AB by substituting the values in
...
Most questions will ask for the length in the form p q or similar
...

Find the area of the triangle AOB where O is the origin
...
Using the area of a triangle:
2
5
1 5 4
A   1
2 2 5
Here is a quick sketch
15

Sequences and Series
A sequence is an ordered list that follows a given rule
...

The most challenging aspect of this topic for many students is the notation
...
If you struggle with this topic, especially recurrence relations, I
suggest using a table to write down your values
...
Here is a typical question
...
Find the first 4 terms
...

I am going to put these in the boxes below to keep on top of my work
...
You can argue the table is overkill but it can lead to fewer
mistakes
...
an is often used in exams too
...
An
3
example could be an 1   2 , n  0 a1  3 where we might be asked to find the first 4 terms of the sequence
...
We know the first term a1  3 and that n  0 , so
we can start with n  1
...
This is a fairly simple case but gives you an idea on the
5
structure of the questions
...
You will have to solve accordingly
...
(b) Given

u

i

 8 find the value of the constant k
...

We are now going to part b and simply sum the terms we have and set them = 8 as we are summing from
i  1 to i  3
...

Arithmetic sequences and series have a common difference, usually denoted as d
...
The sequence or series will increase or decrease by a fixed amount
...
The first term is given as a or a1
...
You may have to find a term in the sequence, the
number of terms in a sequence or the sum of a series for example
...

a2  a1  a3  a2  a4  a3 
...

Let’s look at a basic example
...

With any question like this we can simply collect the information required and substitute into the formula
...
This will really
help with word based questions too
...

We know an arithmetic sequence has a common difference therefore a2  a1  a3  a2
...
Applying this to the
question:
a=5
n = 12
d=4
Substituting in:
a12  5  (12  1)4
a12  5  44
a12  49
Finding the Sum of a Series
We can use one of 2 formulae (which you may be asked to prove in an exam) to find the sum of a series or
values given a sum
...

2
2
A straight forward example may ask us to find the sum of the first 25 terms of the series 5 + 2 + -1 + -4…
...
Simply collect the information and substitute it into the formula
...

12

Sigma Notation may also be used
...
This is an arithmetic series with
r 1

common difference of 3
...

We can find the first term by substituting in r  1 which gives 2 and find the last term by substituting in
r  12 which gives 35
...
Be careful with the number of terms as r may not start at 1
...
Simply take the values and substitute into the correct
formula and solve checking your answer is logical
...
We might ask ourselves how does one
quantity change in response to another quantity changing? A nice example to look at is displacement ( s ) ,
velocity (v ) and acceleration (a )
...
The rate of change of displacement ‘with
ds
respect to time’ is velocity
...
We know this from
dt
basic work in maths and physics
...

dv
The rate of change of velocity with respect to time is acceleration
...
We are ‘differentiating
dt
velocity with respect to time’
These are basic examples although applications generally will not be tested in C1
...

y  y2
When you find the gradient of a straight line you will use m  1
as we saw in the section on coordinate
x1  x2
geometry
...
These are the same thing using different notation
...
In C1 you will only be expected to differentiate functions of the
form y  x n and will not be expected to prove this from ‘first principles’ (despite it being very interesting!)
Let’s look at a standard result for differentiation:
dy
If y  x n then
 nx n 1
...

This gives us the ‘gradient function’ and we can find the gradient of the tangent to the curve at any point by
dy
simply substituting the x coordinate of that point into either or f '( x )
...
We say we are ‘differentiating both sides of the equation
with respect to x ’ in each given case
...
We might have to find
ds
given s  4t 5  2t  1 where we would be differentiating s with respect (WRT) to t
...
Many marks are lost in exam questions through sloppy fraction
work
...

Find when y  5  3 x
Find f '( x ) given f ( x )  4 x 3  6 x 2
dx
dx
x
1

2
2
Using the rules of indices to first
y  3x
f '( x )  12 x  3x 2
simplify:
dy
1
2
 2(3 x )
I have not shown full workings
y  5 3 x
dx
x
here
...

1
dy
3 2
6
 10 x  x
dx
2
At this stage I feel it’s important to discuss 2 results that may be intuitive but often cause some confusion
...

dy
For example, if y  3x then
 3
...
The gradient of the line y  3x is 3
...
Alternatively you can say initially the power of x is 1 such that y  3x1
...

19

dy
 0
...
Alternatively you could see this as y  5 x 0 and when you multiply down by the power
it will = 0
...
If we wanted to find the gradient of the tangent at a given
point we could simply substitute in the given x coordinate
...

dy
We need to find
dx
2
y  3x

Here is an example of differentiating a constant: If y  5 then

dy
 2(3 x1 )
dx
dy
 6x
dx
dy
 6(1)
dx
This gives a gradient of 6
...
The sketch below
shows a graphical representation
...
As we have seen before, we need two things for the equation of a straight
line
...
We can find the gradient of a tangent using the
dy
gradient function ( or f '( x ) ) and then simply substitute the values into the equation of a straight line
...

We need to find the gradient and the y coordinate of the point P
...
We need the gradient function so need to differentiate the function
with respect to x
...
We say “f dashed of x ”
...

f '(1)  8(1) This gives a gradient of 8
...

P(1,1) and m  8
P(1,1) and m  8
y  mx  c
y  y1  m ( x  x1 )
1  8(1)  c
y  1  8( x  1)
7  c
y  8x  7
y  8x  7
The Equation of a Normal
The normal is a straight line perpendicular to the tangent
...
Once we have this gradient we can simply substitute the values we are given (or have
to find) into the equation of a straight line to find an equation for the normal
...

We have a point the curve passes through so all we need is the gradient
...
We have a gradient of
...
It’s the negative reciprocal
...
The question will guide you in
terms of the form required
...

Some questions will extend beyond these basics concepts and ask, for example, you may be asked “What are
coordinates of the other point on the curve where the gradient is also 2?” or Where does the normal intersect
the curve again? A quick sketch and basic applications of either algebra (mainly simultaneous equations) or the
dy
use of
will allow you to find the given coordinates
...
Applications of
integration are not considered until later units
...
The first part should make sense as it’s the reverse of differentiation
x n 1
n
 c , n  1
...
The formal result is  x dx 
n 1
n  1 (as division by 0 is undefined)
...
Often questions will ask you to find an equation for y given
...

dy
We can simply write y   dx
...
An example might be y  2 x 2  3x  5 which differentiates to give
 4x  3
...
This would also have the derivative
dx
is a constant
...
Integrating gives us a general solution and a
k
family of curves as we don’t yet know the value of the constant we ‘lost’ when differentiating the original
function
...
e
...

Let’s consider basic some examples below:
Given the curve C ,
 2

Find  (3x 5  4x 2  2)dx
y    4  5 x dx
where y  f( x ) passes through the
x

3 6 4 3 2 1
5
2
point P(1, 2) and is such
1
 (3x  4x  2)dx  6 x  3 x  1 x  c
 4

2
y    2 x  5 x dx
dy
that
 4 x  3
...

y
x  x c
2
3
3
3
y    4 x  3 dx
2
4
3
Which simplifies to give:
y  x 2  x1  c
3
2
1
2 3 10 2
2
y  x  x c
y  2 x  3x  c
3
3
Substituting in the values for
x and y :
2  2(1)  3(1)  c
c3
This gives us an equation for
C which can be
written y  2 x 2  3x  3
As with questions on differentiation, you may have to use the rules of indices to simplify your equation or
expression first
...


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