This document contains a summary of all the knowledge needed for an A-Level Mathematics examination under the exam board of Pearson Edexcel. It contains the key facts in a way that's easily readable, and compressed to include only the main information needed for exams. It contains information for ...
2a
62 -uac > O - 2 Distinct roots
↑
used to find the
turning point
* substitution * Elimination
To solve Simultaneous Equations , make lory) the subject of an equation and substitute into another a make the <
Cory) coefficient have the same coefficient from the other , and subtract(only for linear) .
↳Solutions show where the INTERSECT
graphs .
factor Theorem f(x)) is f(p) D2-p) is factor off
It polynomial then 0 &
: =
a ,
a
Sketching Polynomials
&
C intercepts by 0
"I
- =
in even , and o , anco
M
a + a x a ,x + x y intercept y do/constant term
↓ ↓
+
y
y
-
= a anx
:
+... +
, ,
Turning Point to Differentiation
If n= odd , and ot" , anso
h
= (x -
P,)(x -
Pc) ... (x -
Pr)
Reciprocal functions proportionality ~ /INVERSEL How to solve modulus
↓ Square both sides then solve
·
↓ neveodd
.
yxx
,
yax
f(x) =
a ·
Graph & find intersecting points &
y = kx
y
=
x identify a values corresponding to
7gative
(a , n - 0)
= y,x
=
y232
+ ve values of the functions .
Modulus Graphs Defining functions Inverse functions
Domain for "DC) their fix)'s
w
:
Inputs (x) values values
y
f(x)s
=
,
Range :
outputs (y) forf"(2) ,
their values =
f(x)'s y values
·
nu reflected
One-to-one ↳ Reflected IN y =
C
If()l < in the
: each
axis
xdy is unique
Partial fractions
f((x1) reflected Many-to-one : In a
- in they domain () of f , there's f(xC) A B C
axis = + +
also another possible domain (ax + b) ((x + d)2 ax + b a +d (x + d)2
(x2) f(x) =
f(x2) * When should
Improper algebraic division
·
,
be used
First .
Graph Transformations
y
=
f(x) + 6 -
up
by b //
y
=
f(x) -> scale
vertically by 6 //y =
- fD) >
-
reflect in axis
y f
(s) + 6)
= -
LEfTbya//y =
flas) - Scale
horizontally by "a /y =
f(x) >
- reflect in yaxis
to make 3 sing to make 4 cos8
anAItan B
tan(AIB) =
1 tan Atan B simplify to RCOS(OIC
-
>
R =
b2 )
.
3 + 4 = RCOSC) + Rsince =
R(cos + since)
3 + 42 = R2(cos" < + Sin = c) =
R (1)
=
↳ c =
tan" (f)
~
R2 =
32 + 42 = 25
R =
c =
5---- -
tanceSin
a = tan" (b) =
53 1
.
3 sinO 4 cos E
formula
Angle
+
Double Trigequations 1)
=
Ssin 10 53
solve + .
-
Sin (2A) = 2SinACOSA EG : find solutions to sin (2x +
*
2) =
2 1 -
>)
2 = x -
cos(2A) = cos A-sin
? ?
A .
1 Adjust the domain - 2x
+
2a + 7/2
2COSPA-1
=
?
- +
= 1 -
2 sin A
. +an A
2
solvable state sin" (2)
tan (2A) =
.
2
Rearrange to a 2x + = =
=
1 - tan ? A
I
3 and
.
Use
symmetry (x + =
-
periodicity to find
solutions.
x=
52
-
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