ZNOTES.ORG
UPDATED TO 2023-25 SYLLABUS
CAIE AS LEVEL
MATHS
SUMMARIZED NOTES ON THE PURE 1 SYLLABUS
CAIE AS LEVEL MATHS
1. Quadratics
1.1. Completing the square
The equation y = ax 2 + bx + c written in the form of
y = p(x + q)2 + r is called the complete square form
The vertex is (−q, r) .
Example
Example
Express 3x 2 + 9x + 5 in the form of p(x + q)2 + r , where
p, q and r are constants and find its vertex
Solution:
If kx 2 + 4kx + 3k = 0 , find the inequality in terms of k for
which the equation has distinct real roots.
Solution:
Apply b2 − 4ac > 0 for the equation to have two distinct real
roots:
3x 2 + 9x + 5
(4k)2 − 4(k)(3k) > 0
= 3(x 2 + 3x) + 5
⟹ 16k 2 − 12k 2 > 0
3 2
3 2
= 3 ((x + ) − ( ) ) + 5
2
2
​
= 3 (x +
⟹ k < 0 and k > 0
2
3
7
) −
2
4
​
​
Hence its vertex is (− 32 , − 74 ) .
​
⟹ 4k 2 > 0
​
1.4. Quadratic Inequalities
​
1.2. Sketching the Graph
y-intercept
x -intercept
Case 1: Assuming d < β,
(x − d) (x − β ) < 0 ⟹ d < x < β
(x − d ) (x − β ) > 0 ⟹ x < d or x > β
Vertex (turning point)
Case 2: When no x term,
x2 − c > 0
1.3. Discriminant of a Quadratic
Expression
⟹ x < − c or x >
⟹− c≤x≤
The discriminant of ax 2 + bx + c is
If b2 − 4ac = 0 , real and equal (repeated) roots
If b2 − 4ac < 0 , no real roots
If b2 − 4ac > 0 , real and distinct roots
​
x2 − c ≤ 0
The discriminant describes the roots for a quadratic
equation ax 2 + bc + x = 0
b2 − 4ac
c
​
​
c
​
Example
x 2 + 6x + 8 < 0
factorize: x 2 + 6x + 8 = (x + 4)(x + 2)
find roots:
x = -4
x = -2
sketch parabola which touches the x-axis at -4 and -2
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CAIE AS LEVEL MATHS
as
Reject
∴ x = 19
x = u → x = u2
​
x = − 12 as it has no solutions,
​
​
​
2. Functions
2.1. Terms
the blue region represents the section of the parabola where
the value of the quadratic is < 0
therefore, the answer of x 2 + 6x + 8 < 0 is: −4 < x <
−2
1.5. Solving Equations in Quadratic
Form
Function: is a relation that uniquely associates one set of
values to another set
Domain: is the set of values that are the inputs of the
function
Range: is the set of values that are the outputs of the
function
Inverse Function: The function which maps the Range back
into its Domain.
Mapping:
Takes a value from the domain and linking it to the
range
they can be:
one-to-many
many-to-one
one-to-one
or many-to-many
in cases like x 4 − 5x 2 + 4 = 0 and 6x + x − 1 = 0 ,
they are not quadratic equations but can be converted
into quadratic equations in some function of x
​
Example
Solve the equation x 4 − 5x 2 + 4 = 0
Solution:
We let u = x 2 , this gives us: u 2 − 5u + 4 = 0 , which is a
quadratic equation and can be solved
(u − 4)(u − 1) = 0
u = 4, 1
as x 2 = u → x = ± u
​
∴ x = ±2 or ±1
Example
Solve the equation 6x + x − 1 = 0
Solution:
We let u = x , this gives us 6u 2 + u − 1 = 0 , which is a
quadratic equation and can be solved
​
Notations
​
Functios can be either written as f(x), g(x), etc. (e.g.
(3u − 1)(2u + 1) = 0
f(x) = 2x + 5 )
1
1
u = ,−
3
2
2.2. Find Range
​
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​
Or they can be written as f : x ↦ 2x + 5
CAIE AS LEVEL MATHS
Find the highest possible y -value and lowest possible y value based on the domain
f (x ) = 3x + 4
y = 3x + 4
For Quadratic functions, such as f (x ) = 3x 2 + 5x − 6 ,
complete square first to find vertex and use it to find its
range.
If coefficient of x 2 is positive, vertex is minimum
If coefficient of x 2 is negative, vertex is maximum
2.3. Composition of 2 Functions
y − 4 = 3x
2.4. One-One Functions
Definition: One x value substitutes to give one y value
y−4
3
y=
x−4
3
​
Swap all the x with y ,
Definition: a function with another function as an input
fg (x ) ⇒ f (g (x ))
E.g. f (x ) = 4x + 5
g (x ) = x 2 − 5
2
Then fg (x ) = 4 (x − 5 ) + 5
A composite function like fg (x ) can only be formed when
the range of g (x ) is within the domain of f (x )
x=
​
Replace y with f −1 (x ) ,
f −1 (x ) =
x−4
3
​
Example:
Make f (x ) = x 2 + 1 a one-to-one function.
Solution:
x 2 + 1, x ∈ R
One value of x that doesn’t have alternate value of x which
maps same value of y is 0
∴ We separate the function into two functions
No indices
If function is not one-to-one, restrict the function in a
domain such that the function is one-to-one under that
domain.
Only one-to-one functions are invertible
2.5. Finding Inverse
Definition: An inverse function shows what the input is
based from the output e.g. if f (3 ) = 5 then f −1 (5 ) = 3
. In other words, it reverses the process. The graph of
y = f (x ) and y = f −1 (x ) is symmetrical by the line
y = x.
An inverse function has a property such that:
ff −1 (x ) = f −1 f (x ) = x
Make sure that it is a one-to-one function if it is then,
Write f (x ) as y
Make x the subject
Swap every single x with y . By now you should have y as
the subject
Replace y withf −1 (x ) . Read as “The f inverse of x ”
Example:
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x2 + 1, x ≤ 0 and x2 + 1, 0 ≤ x
CAIE AS LEVEL MATHS
4 [(x − 3 ) − 3 2 ] + 11
2
4 (x − 3 )2 − 25
Part (ii)
Observe given domain, x ≤ 1 .
Substitute highest value of x
g (x ) = 4 (1 − 3 )2 − 25 = −9
Substitute next 3 whole numbers in domain:
x = 0, −1, −2
g (x ) = 11, 39, 75
Thus, they are increasing
∴ g (x ) ≥ −9
Part (iii)
Let y = g(x), make x the subject
y = 4 (x − 3 )2 − 25
y + 25
2
= (x − 3 )
4
​
y + 25
4
x=3+
​
​
Can be simplified more
x=3±
2.6. Relationship of Function & its
Inverse
y + 25
​
∴x=3−
1
2
​
∴ g −1 (x ) = 3 −
graph of the function in y = x
​
Positive variant is not possible because x ≤ 1 and using
positive variant would give values above 3
The graph of the inverse of a function is the reflection of a
y + 25
1
2
​
​
x + 25
​
Domain of g −1 (x ) = Range of g (x ) ∴ x ≥ −9
{W12-P11} Question 10:
f (x ) = 4x 2 − 24x + 11, for x∈ R
g (x ) = 4x 2 − 24x + 11, for x ≤ 1
2.7. Translation
1. Express f(x) in the form a (x − b)
2
+ c, hence state
coordinates of the vertex of the graph y = f(x)
2. State the range of g
3. Find an expression for g −1 (x) and state its domain
Solution:
First pull out constant, 4 , from x related terms:
4 (x 2 − 6x ) + 11
Use following formula to simplify the bracket only:
n 2
n 2
) −( )
2
2
​
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Let y = f (x )
Shift along x-axis by a units to the right: f (x − a)
Part (i)
(x −
1
2
​
CAIE AS LEVEL MATHS
Shift along y-axis by b units upwards: f (x ) + b
Example
The graph of y = x 2 + 3x + 2 has been translated 1 unit
left and 3 units upward, find the equation of the resulting
graph.
Answer:
The translation vector would be [1, 3], so where f(x) =
If a > 1 it will shrink the graph sideways
If 0 < a < 1 it will expand the graph sideways
Stretches upwards and downwards: af (x )
x 2 + 3x + 2
y = f(x − (−1)) + 3
y = f(x + 1) + 3
y = (x + 1)2 + 3(x + 1) + 2 + 3
y = x 2 + 2x + 1 + 3x + 3 + 2 + 3
∴ y = x 2 + 5x + 9
2.8. Stretch
Stretches the graph sideways:f(ax)
If a > 1 it will expand the graph up & downwards
If 0 < a < 1 it will shrink the graph up & downwards
Example
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CAIE AS LEVEL MATHS
the graph of y = 5x 2 + 2 is stretched by a factor of 2 along
the y-axis, find the resulting equation of the graph.
PQ =
(x 2 − x 1 )2 + (y2 − y1 )2
​
​
​
(a − 0)2 + (1 − 4)2
5=
for this stretch factor, y = 2f(x)
​
​
​
25 = a2 + 9
y = 2(5x 2 + 2)
a2 = 16
∴ y = 10x 2 + 4
∴a= 4
Example
the graph of y = x 2 + 3x + 2 is stretched by a factor of 13
3.2. Midpoint of a Line Segment
​
along the x-axis, find the resulting equation of the graph.
for this stretch factor,
The midpoint between two points (x 1 , y1 ) and (x 2 , y2 ) is
​
(
y = f(3x)
y = (3x)2 + 3(3x) + 2
​
​
​
x 1 + x 2 y1 + y2
,
)
2
2
​
​
​
​
​
​
Example
∴ y = 9x 2 + 9x + 2
A line segment joining points P (2, −3) and Q(4, a) has a
midpoint M (b, −2) . Find the values of a and b.
2.9. Reflections
For reflection of the equation y = f(x) in the x-axis, the
resulting equation is y = −f(x)
For reflection of the equation y = f(x) in the y-axis, the
resulting equation is y = f(−x)
Answer:
M =(
x 1 + x 2 y1 + y2
,
)
2
2
​
​
​
y1 + y2
2
−3 + a
−2 =
2
−2 =
y = 2(−x)2 + 3(−x) + 2
​
thus
Example
Find the reflection of the graph equation y = 2x 2 + 3x + 2
in the y-axis and x-axis and write their equations.
Answer:
for the reflection in y-axis, y = f(−x)
​
​
​
​
​
​
a − 3 = −4
∴ a = −1
x1 + x2
2
2+4
b=
2
b=
∴ y = 2x 2 − 3x + 2
for the reflection in x-axis, y = −f(x)
​
​
​
​
2
y = −1(2x + 3 − x + 2)
∴b= 3
∴ y = −2x 2 − 3x − 2
3.3. Equation of a Straight Line
3. Coordinate Geometry
Straight line equations are linear equations in the form of
3.1. Length of a line segment
The distance between two points (x 1 , y1 ) and (x 2 , y2 ) is
​
​
​
y = mx + c
m is the gradient
and c is the y -intercept
If a single point A(x 1 , y1 ) on the line with a gradient m is
given and the general point is P (x, y) , then the equation
of the line can be written in the form of y − y1 =
m(x − x 1 )
​
(x 2 − x 1 )2 + (y2 − y1 )2
​
​
​
​
​
​
​
Example
the distance between points P (0, 4) and Q(a, 1) is 5 units.
Find the positive value of a.
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​
Example
CAIE AS LEVEL MATHS
Find the equation of the straight line with gradient 3, which
passes through the point (1,6).
Solution:
m × m1 = −1 and so m1 = 32
​
​
Perpendicular general equation:
y=
A(1, 6)m = 3
3
x+c
2
​
Substitute known values
3 = 32 (−1 ) + c and so c = 92
Final perpendicular equation:
y − 6 = 3(x − 1)
​
y = 3x − 3 + 6
​
2y = 3x + 9
∴ y = 3x + 3
Find the point of intersection by equating two equations
Example
2
3x + 9
x=
3
2
13
13 =
x
3
11 −
Find the equation of the straight line that passes through the
points (-5, 3) and (-4, 1).
Solution:
m=
y2 − y1
1−3
=
x2 − x1
−4 − (−5)
​
​
​
​
x = 3, y = 9
​
​
​
​
​
m=
−2
1
Vector change from (−1, 3 ) to (3, 9 ) is the vector change
from (3, 9 ) to R
Finding the vector change:
​
m = −2
C hange in x = 3 − −1 = 4
y = −2x + c
Substitute any point in this equation to find the value of c
1 = −2(−4) + c
C hange in y = 9 − 3 = 6
Thus R
x = 3 + 4 = 7 and y = 9 + 6 = 15
1=8+c
R = (7, 15 )
c = −7
3.5. Equation of a circle
∴ y = −2x − 7
Standard Form: (x − a)
Parallel lines: m1 = m2
Perpendicular lines: m1 × m2 = −1
The gradient at any point on a curve is the gradient of the
tangent to the curve at that point
The gradient of a tangent at the vertex of a curve is equal
to zero – stationary point
​
​
{S13-P12} Question 7:
Point R is a reflection of the point (−1, 3 ) in the line
3y + 2x = 33 .
Find by calculation the coordinates of R
Solution:
Find the equation of line perpendicular to 3y + 2x = 33
intersecting point (−1, 3 )
General form: x 2 + y 2 + ax + by + c = 0
Centre = (− a2 , − 2b )
​
​
2
2
Radius =( a2 ) + ( 2b ) − c 2
​
​
Note: if eqn. of circle is in general form, it’s highly
recommended to convert it into its standard form by
completing square to easily find center and radius
Tangents on a circle are always perpendicular to its
radius
If a right-angled triangle is inscribed in a circle, its
hypotenuse is the diameter of the circle
Example
The equation of a circle: x 2 + y 2 + 4x + 2y − 20 = 0 The
3y + 2x = 33 ⇔ y = 11 −
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+ (y − b)2 = r 2
Centre = (a, b)
Radius = r
3.4. Special Gradients
m=−
2
2
3
​
2
x
3
​
line L has the equation 7x + y = 10 intersects the circle at
point A and B . The x -coordinate of A is less than the x coordinate of B .
1. Find the center and the length of diameter of the circle
2. Find the coordinates of A and B
CAIE AS LEVEL MATHS
Solution:
i. Rearrange the equation to standard form by using
completing square:
Example
convert 3π
4 radians into degrees and 45° into radians
Answer:
​
2
2
x + 4x + y + 2y = 20
(x + 2 )2 − 4 + (y + 1 )2 − 1 = 20
⇒ (x + 2 )2 + (y + 1 )2 = 25
∴ its center: (−2, −1 ). Its diameter: 2 × 5 = 10
(x + 2 )2 + (y + 1 )2 = 25 & y = −7x + 10
2
2
3π
180
×
4
π
=
3 × 180
4
​
​
​
∴ = 135°
ii.Do simultaneous equation
Use substitution y = −7x + 10 onto (x + 2 )
=
+
= 45 ×
∴ = π4
(y + 1 ) = 25 .
π
180
​
​
4.2. Arc length
(x + 2 )2 + (−7x + 11 )2 = 25
Arc length = rθ ( θ is in radians)
Find x
x 2 + 4x + 4 + 49x 2 − 154x + 121 = 25
Example
50x 2 − 150x + 100 = 0
find the arc length subtended at an angle of π4 radians of a
circle with radius 4
Answer:
Arc length = 4 × π4
∴ Arc Length = π
​
2
x − 3x + 2 = 0
∴x=1 x=2
​
Put x values back into y = −7x + 10 to find y value:
4.3. Area of a Sector
∴ A (1, 3 ) B (2, −4 )
A=
4. Circular Measure
1 2
r θ
2
​
In Radians
{S11-P11} Question 9:
4.1. Radians
Radians is an angle measurement (just like degrees)
defined in which the arc length is the same as the radius
is an angle of 1 radian at the center
Triangle OAB is isosceles, OA = OB and ASB is a
tangent to PST
1. Find the total area of the shaded region in terms of r
and π
2. When θ = π3 and r = 6 , find the total perimeter of the
shaded region in terms of 3 and π
​
​
Solution:
Part (i)
Use trigonometric ratios to form the following:
AS = r tan θ
π radians is equal to 180° and 2π radians is equal to 360°
to change from radians to degrees, multiply by 180
π
π
to change from degrees to radians, multiply by 180
Find the area of triangle OAS:
​
​
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θ
CAIE AS LEVEL MATHS
r tan θ × r
1
= r 2 tan θ
2
2
OAS =
​
​
5.1.
Use the formula of the sector to find the area of OPS:
1 2
r θ
2
OPS =
​
Area of ASP is OAS − OPS :
Example
1
1
1
∴ ASP = r 2 tan θ − r 2 θ = r 2 (tan θ − θ )
2
2
2
​
​
Multiply final by 2 because BST is the same and shaded is
ASP and BST
Area = 2 ×
Draw the graph of y = 3cos(x − 12 π) − 1
Solution:
​
Draw the graph of y = cos(x)
1 2
r (tan θ − θ ) = r 2 (tan θ − θ )
2
​
Part (ii)
Use trigonometric ratios to get the following:
π
6
cos ( ) =
3
AO
​
​
∴ AO = 12
Finding AP:
AP = AO − r = 12 − 6 = 6
Finding AS:
π
AS = 6 tan ( ) = 6 3
3
​
​
Finding arc PS:
Arc PS = rθ
PS = 6 ×
π
= 2π
3
​
The perimeter of 1 side of the shaded region:
Pe 1 = 6 + 6 3 + 2π
​
​
Perimeter of the entire shaded region is double:
2 × Pe 1 = 12 + 12 3 + 4π
​
​
5. Trigonometry
opposite
sin(θ) = hypotenuse
adjacent
cos(θ) = hypotenuse
opposite
tan(θ) = adjacent
​
​
​
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Apply the transformation inside the brackets ...(x −
90)..., which translates the graph by
(90, 0)
CAIE AS LEVEL MATHS
Apply the transformation which is being multiplied/divided
to the trig function, in this case 3, which should stretch the
graph by a factor of 3 along the y-axis
5.2. Sine Curve
5.3. Cosine Curve
Finally, apply the transformation that is being
added/subtracted to the expression, in this case it is −1 ,
which would translate the graph by (0, −1) .
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CAIE AS LEVEL MATHS
To find the other values, draw a graph and draw a
horizontal line for the value given to find the angles at
which the trigonometric function give the value stated. in
the previous example, the graph would look like this
5.4. Tangent Curve
5.5. Exact values of Trigonometric
Functions
As you can see there are also other points on the graph
which intersect with the horizontal line, these are also
additional solutions
In the question, a range might be given and you would
have to find all the solutions in that given range.
In this case, to find the other point A we can do 2π − π6 ,
which equals to 11π
6
Here we are using the property of trigonometry functions
that they are symmetrical, and thus such calculations can
be used to find the angles
​
​
5.8. Identities
5.6. When sin, cos and tan are positive
tan θ ≡
sin θ
cos θ
​
sin2 θ + cos2 θ ≡ 1
Example
Prove the identity
5.7. Finding other Angles from the
Principal Angle
In case of cos(x) = 23 , we know using a calculator or by
memory that x = 30° or x = π6 , but this is only the
principle angle and there are many more solutions to this.
​
​
​
​
​
Solution:
sin2 (x) can be written as 1 − cos2 (x) so we obtain
2
2
(x))
1
= cos (x)−(1−cos
+ cos(x)
cos(x)
​
2
2
(x))
1
= cos (x)−1+cos
+ cos(x)
cos(x)
2
(x)−1
1
= 2 cos
+ cos(x)
cos(x)
2
(x)−1+1
= 2 coscos(x)
cos2 (x)
= 2 cos(x)
​
​
​
​
​
= 2 cos(x)
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cos2 (x)−sin 2 (x)
1
+ cos(x)
≡ 2 cos(x)
cos(x)
​
​
CAIE AS LEVEL MATHS
∴
cos2 (x)−sin2 (x)
1
+ cos(x)
≡ 2 cos(x)
cos(x)
​
​
5.9. Trigonometry with Non RightAngled Triangles
In these cases, you can either use the sine rule or the
cosine rule
Example
Find the value of x
Solution:
In this question, we are given two side lengths and the angle
between them
In this case, we shall use the cosine rule a2 = b2 + c 2 −
(2bc ∗ cos(A))
we can write this as x 2 = 4 2 + 4 2 − (2(4)(4)cos(60))
x 2 = 16 + 16 − (16)
x 2 = 16
∴x = 4
5.10. Inverse Functions
If trig (θ ) = a, then θ = trig −1 (a)
Where “trig” represents any Trigonometric Function
Inverse trigonometric functions are used to find angle
Find the value of x
Solution:
in this question, we are given two angles and one of the sides,
where we have to find the length of the other side.
a
b
So we know we should use the sine rule sin(A) = sin(B) =
​
c
sin(C)
​
8
x
we can write this as sin(30)
= sin(60)
​
cross multiplying it would give us
8 sin(60) = x sin(30)
8 × 23 = 12 x
8 3=x
∴x = 8 3
6. Series
6.1. Binomial Expansion
A neat way of expanding terms with high powers.
​
n
n
n
(x + y)n = ( )x n + ( )x n−1 y + ( )x n−2 y2 + … +
0
1
2
​
​
​
​
n
n!
( ) = n Cr =
r
r! (n − r )!
​
​
​
​
​
​
​
Example
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n
n
In summation : (a + b)n = ∑ ( )an−k bk
k
​
​
k=0
(The summation form is just another way to express
(a + b)n , it’s not important but some students may like to
see it that way)
CAIE AS LEVEL MATHS
This only works when n is a positive integer
Patterns:
make it to the power of 3 and thus, r should be 3
thus, the term would be 8 C 3 ⋅ (1)8−3 ⋅ (2x)3
​
n
C0 = n Cn = 1
nC = nC
1
n−1 = n
nC = nC
r
n−r
​
= 56 ⋅ 1 ⋅ 8x 3
​
​
​
​
​
∴ coefficient of x 3 = 448
To understand these patterns, you may use the
Pascal’s triangle
6.2. Arithmetic Progression
Definition: Sequence where successive terms are gained
from adding same value E.g. 1,3,5,7,9,11…
u n = a + (𝑛 − 1)d
​
Sn = 12 n[2a + (n − 1)d] or Sn = n2 (a + l)
u n = the n-th term of the sequence
​
​
​
​
​
In the Pascals triangle for n C r
𝑎 = First term of the sequence
n = The 𝑛-th term
𝑑 = Main difference
Sn = Sum from 1st term to 𝑛-th term
​
n is the row (starting from 0)
r is the position of the value from the left (starting
​
Example
from 0)
Example
Find the 120th term of the arithmetic sequence which follows
the following pattern: 2, 4, 6, 8, 10 …
Answer:
Expand the expression (a + b)5
the first term gives us the value for a = 2
the common difference is d = 2
Solution:
Using the binomial theorem, we get:
thus u
= 2 + (120 − 1)2
5 5−1
5 5−2 2
5 5−3 3
5 5−41204
5
2 + 238
(a + b) = a + ( )a b + ( )a b + ( )a b + ( =)a
b + ( )a5−5 b5
1
2
3
4∴ the 120th term
5 of the arithmetic sequence is = 240
​
5
5
​
​
​
​
(a + b)5 = a5 + 5a4 b + 10a3 b2 + 10a2 b3 + 5a1 b4 + 1a0 b5
∴ (a + b)5 = a5 + 5a4 b + 10a3 b2 + 10a2 b3 + 5ab4 + b5
​
Example
Find the sum of the first 20th term of the arithmetic
progression with its first term being 7 and its 8th term being
28.
Answer:
the first term gives us the value for a = 7
Example
Expand the expression (3x + 2)4
Solution:
Using the binomial theorem, we get:
u 8 = 28 = 7 + (8 − 1)d
4
4
4
4
28 4−4
− 7(2)4
(3x + 2)4 = (3x)4 + ( )(3x)4−1 (2) + ( )(3x)4−2 (2)2 + ( )(3x)4−3 (2)3 + ( )(3x)
1
2
3
4d =
7
(3x + 2)4 = 81x 4 + (4)(3x)3 (2) + (6)(3x)2 (4) + (4)(3x)(8) + 16
d=3
​
​
​
​
​
​
∴ (3x + 2)4 = 81x 4 + 216x 3 + 216x 2 + 96x + 16
20
thus S20 = 2 (2(7) + (20 − 1)3)
​
​
S20 = 10(14 + 57)
Example
​
Find the coefficient of x 3 in the expansion of (1 + 2x)8
Solution:
To find the coefficient of x 3 , we have to see at which term
does the expansion have x 3
∴ S20 = 710
​
6.3. Geometric Progression
Definition: Sequence where successive terms are gained
n
n
n
from multiplying the same value E.g. 2,4,8,16,32…
(a + b)n = an + ( )an−1 b + ( )an−2 b2 + ( )an−3 b3 ... + bn
1
2
3
u n = ar n−1
n
The pattern we can see here is that in ( r ), r is always the
value of the power of b. As b is 2x in the question, we have to
​
​
​
​
WWW.ZNOTES.ORG
CAIE AS LEVEL MATHS
a (1 − r n )
(1 − r )
Sn =
​
1
​
u n = the n-th term of the sequence
a = First term of the sequence
n = The n-th term
r = Common Ratio
7.2. Chain Rule
​
When ∣r∣ < 1 , Sum to infinity:
​
a
1−r
​
f(x) = 5, then f ′ (x) = 0
Sn = Sum from 1st term to n-th term
S∞ =
1
Remember that x can be written as x −1 and xn can be
written as x −n and vice versa
It is also important to remember that the differentiation of
constants gives us 0
e.g.
​
dy
dy
du
=
×
dx
du
dx
​
​
{W05-P01} Question 6:
A small trading company made a profit of 250000 dollars in
the year 2000 . The company considered two different plans,
plan A and plan B , for increasing its profits. Under plan A,
the annual profit would increase each year by 5% of its value
in the preceding year. Under plan B , the annual profit would
increase each year by a constant amount of D
​
(f (g (x ))) ′ = f′ (g (x )) × g′ (x )
Example
5
Differentiate y = (x + x 3 )
Solution:
Let u = x + x 3 , then find du
dx
​
1. Find for plan A, the profit for the year 2008
u = x + x3
2. Find for plan A, the total profit for the 10 years 2000
to 2009 inclusive
3. Find for plan B the value of D for which the total profit
du
= 1 + 3x 2
dx
for the 10 years 2000 to 2009 inclusive would be the
same for plan A
Solution:
Part (i)
Increases are exponential ∴ it is a geometric sequence:
2008 is the 9 th term:
∴ u 9 = 250000 × 1.05 9−1 = 369000 (3s.f.)
Part (ii)
Use sum of geometric sequence formula:
250000 (1 − 1.05 10 )
S10 =
= 3140000
1 − 1.05
​
​
Now y = u 5
dy
= 5u 4
du
​
Multiply them together
dy
dy
du
4
=
×
= (1 + 3x 2 ) × 5 (x + x 3 )
dx
du
dx
​
​
​
Part (iii)
Plan B arithmetic; equate 3140000 with sum formula
1
3140000 = (10 ) (2 (250000 ) + (10 − 1 ) D)
2
​
​
​
Another quick way is to:
1. Take the derivative of the “inside”
2. Then take the derivative of the “outside”
3. Multiply them together
In our case:
The inside: x + x 3
The outside: u 5
So, differentiating will give us (1 + 3x 2 ) × 5 (x + x 3 )
4
​
D = 14300
7. Differentiation
n−1
When y = x n , dy
dx = nx
​
1st Derivative = dx = f′ (x )
dy
​
2
d y
2nd Derivative = dx
2 = f′′ ( x )
​
Sometimes negative powers may also be presented in a
question, or there might even be x in the denominator.
For these questions, you have to take the x to the
numerator and change the power sign
1
WWW.ZNOTES.ORG
1
7.3. Finding gradient using
differentiation
For y = f(x), f ′ (x) would give you the gradient of the
curve/line at point x on the graph
To find the normal of a curve at a point, its gradient is the
reciprocal of the gradient of the tangent and multiplied by
-1
When you know the gradient of the tangent or normal to a
point on a curve, it is easy to find its formula as you can
use y = mx + c and by using the point on the curve, you
can find the value of c and complete the equation
CAIE AS LEVEL MATHS
Example
Find the gradient of the tangent of the curve y = 3x 2 at point
x = 2 and the gradient of the normal to that point
Answer:
dy
dx = 6x
​
gradient = 6(2) = 12
∴ gradient of tangent = 12
1
∴ gradient of normal = − 12
​
7.4. Stationary points, Increasing and
Decreasing Functions
When f ′ (x) > 0 , it is increasing
When f ′ (x) < 0 , it is decreasing
When f′(x) = 0 , it is a stationary point
Example
A graph is plotted for the equation y = x 3 + 3x 2 − 9x + 4 ,
find the values of x for which the graph is an increasing
function and values of x for which the graph is a decreasing
function
Answer:
find the derivative first
∴ Gradient is increasing for x < −3 and x > 1 , and
gradient is decreasing for −3 < x < 1
dy
= 3x 2 + 6x − 9
dx
7.5. Nature of Stationary Point
The graph is a parabola in a U shape and we need to find the
values of x for which it gives us a positive value. To do this we
find the roots of the derivative
Find second derivative dx2
Substitute x -value of stationary point
​
d2 y
​
d2 y
If value +ve → min. point, dx2 > 0
​
2
3x + 6x − 9 = 0
d2 y
If value –ve → max. point dx2 < 0
​
x = 1 and x = −3
dy
Now for the values where dx is positive, knowing that it is a U
​
shaped parabola we know that it is positive for x < −3 and
x>1
We also know that the graph is negative for any x values
between -3 and 1 ( −3 < x < 1 )
Example
WWW.ZNOTES.ORG
CAIE AS LEVEL MATHS
dy
= 12 (5 − 2 (1 ))−2
dx
Find the stationary points on the equation y = x 3 + 3x 2 and
identify the maximum and minimum stationary point
Solution:
We first find its first derivative
dy
= 3x 2 + 6x
dx
​
dy
4
=
dx
3
​
Thus, the gradient is equal to 43 at this point
Part (ii)
Rate of increase in time can be written as:
​
​
Make it equal to 0 to find the stationary points
dx
dt
3x 2 + 6x = 0
x = −2 and x = 0
Corresponding values of y for the values of x are:
(−2)3 + 3(−2)2 = 4
dy
4
=
dx
3
​
Thus, we can formulate an equation:
dy
dy
dx
=
÷
dx
dt
dt
​
2
d y
dx2 = 6x + 62
d y
for x = −2 , dx2 = 6(−2) + 6 = −6
​
​
Rearranging the formula, we get:
​
dx
dy
dy
=
÷
dt
dt
dx
​
∴ The maximum stationary point is (−2, 4) , as the value is
negative
d2 y
​
​
​
Substitute values into the formula
For x = 0 , dx2 = 6(0) + 6 = 6
​
∴ The minimum stationary point is (0, 0) , as the value is
positive
dx
4
= 0.02 ÷
dt
3
​
dy
dy dx
dy
dy
dt
=
/
or
=
×
dx
dt dt
dx
dt
dx
​
​
​
​
​
8. Integration
{W05-P01} Question 6:
The equation of a curve is given by the formula:
∫ ax n dx =
​
6
y=
5 − 2x
​
∫ (ax + b) dx =
n
​
1. Calculate the gradient of the curve at the point where
x=1
2. A point with coordinates (x, y) moves along a curve in
such a way that the rate of increase of y has a
constant value of 0.02 units per second. Find the rate
of increase of x when x = 1
​
dx
3
= 0.02 × = 0.015
dt
4
7.6. Connected Rates of Change
​
dy
= 0.02
dt
and
​
The stationary points are (−2, 4) and (0, 0) , find the second
derivative now
​
​
We know the following:
​
(0)3 + 3(0)2 = 0
​
​
ax n+1
+c
n+1
​
n+1
(ax + b)
+c
a (n + 1 )
Integration is the reverse process of differentiation
The "S" shaped symbol is used to mean the integral of,
and dx is written at the end of the terms to be integrated,
meaning "with respect to x ". This is the same "dx" that
appears in dy
dx .
​
Solution:
Part (i)
Differentiate given equation
6 (5 − 2x )−1
dy
= 6 (5 − 2x )−2 × −2 × −1
dx
​
= 12 (5 − 2x )−2
Now we substitute the given x value:
Indefinite Integrals: Integrals without limits of integration
(the numbers by the integral sign), don’t forget to include
+c
Definite Integrals: Integrals with limits of integration, no
need of putting +c
Use coordinates of a point on the curve to find c when
integrating a derivative to find equation of the curve.
8.1. Example
Find the equation of y in terms of x which passes through the
dy
point (1, 3) , given that dx = 6x
​
d
WWW.ZNOTES.ORG
2
CAIE AS LEVEL MATHS
Solution:
we need to integrate the derivative to find y in terms of x
y = ∫ 6x 2 dx
2+1
y=
6x
+c
2+1
​
Find the integral of each term to give us:
(Note: it is important to remember that constants like 5 in this
case can also be written as 5x 0 )
4
∫ (3x 3 − 4x 2 + 2x + 5), dx = [
​
2
y = 2x 3 + c
substitute the coordinates the graph passes through to find
the value of c
3 = 2(1)3 + c
c=3−2
c=1
∴ y = 2x
3
=[
​
​
4
3x 4
4x 3
2x 2
−
+
+ 5x ]
4
3
2
2
​
​
​
Now substitute the limits and subtract those values
=[
3(4)4
4(4)3
2(4)2
3(2)4
4(2)3
−
+
+ 5(4)] − [
−
4
3
2
4
3
​
​
​
​
256
32
+ 16 + 20 ] − [12 −
+ 4 + 10 ]
3
3
= [192 −
​
​
+1
=[
8.2. Area Under a Curve
428
46
]−[ ]
3
3
​
=
Area bounded by the curve to the x -axis
This is the most common integrals being used
Use dx
Make y the subject in the equation then input it into
your integral
3x 3+1
4x 2+1
2x 1+
−
+
3+1
2+1
1+1
​
382
3
​
4
382
∴ ∫ (3x 3 − 4x 2 + 2x + 5) dx = 3
​
​
2
8.3. Area Between Two Curves
b
∫ y dx
​
a
Area between two curves with respect to x
Just like finding the area under a curve, this time you
subtract the first curve by the second curve
Use dx
Make sure both equations have y as the subject
b
∫ y1 − y2 dx
​
​
b
or
a
b
∫ y1 dx − ∫ y2 dx
​
​
​
a
​
a
Area bounded by the curve to the y -axis
Use dy
Make x the subject of the equation and then input it
into the integral
b
∫ x dy
​
a
Area between two curves with respect to y
Make x the subject in both equations then integrate
its difference
Use dy
b
∫ x 1 − x 2 dy quador
​
a
Example
4
Find the value of ∫ (3x 3 − 4x 2 + 2x + 5) dx
Answer:
2
WWW.ZNOTES.ORG
​
​
b
b
∫ x 1 dy − ∫ x 2 dy
​
a
​
​
a
​
​
CAIE AS LEVEL MATHS
Combine the fractions:
8x − 16
3
(4x + 1 ) 2
​
=0
​
⇒ 8x − 16 = 0
⇒x=2
{S19-P01} Question 11:
Putting the x -value back to the equation of the curve will give
us:
9
4 (2 ) + 1 +
4 (2 ) + 1
​
​
=6
​
∴M (2, 6 )
The diagram shows part of the curve:
y=
9
and the minimum point M .
4x+1
4x + 1 +
​
​
​
iii. The line passing through M is parallel to the x -axis which
means its equation is simply:
dy
1. Find expressions for dx and ∫ y dx
2. Find the coordinates of M
3. The shaded region is bounded by the curve, the y -axis
and the line through M parallel to the x -axis. Find,
y=6
​
showing all necessary working, the area of the shaded
region.
Solution:
i. Differentiate the equation:
∫
2
​
0
9
− 6 dx
4x + 1
4x + 1 +
​
​
​
Which simplifies to:
dy
d
=
( 4x + 1 +
dx
dx
​
We know that: 1. This is an area between two curves 2. It
ranges from x = 0 to x = 2 which means our integral will
be:
​
9
)
4x + 1
​
​
​
3
(4x + 1 ) 2
9
[
+
6
2
​
​
Use the Chain Rule:
4x + 1 − 6x ]
​
​
0
Compute its value
1
1
d
((4x + 1 ) 2 + 9 (4x + 1 )− 2 )
dx
​
​
2
​
​
2
3
(4x + 1 ) 2
9
[
+
6
2
4x + 1 − 6x ] =
​
dy
=
dx
​
2
18
−
3
4x + 1 (4x + 1 ) 2
​
​
​
​
​
​
0
​
∴ The area is
Integrate the equation:
∫ y dx = ∫
9
dx
4x + 1
4x + 1 +
​
​
​
Apply the reverse chain rule:
= ∫ (4x + 1 ) + 9 (4x + 1 )
1
2
​
4
3
4
3
​
​
Note: You can integrate the two equations separately and
then subtract the area, you will still get the same answer
8.4. Volume of Revolution
− 12
​
dx
With respect to x
Use dx
Make y the subject of the equation of the curve then
Don’t forget to include +c
3
(4x + 1 ) 2
9
∫ y dx =
+
4x + 1 +c
6
2
​
​
​
​
2
input πy in the integral
​
b
∫ πy2 dx
​
ii. Since M is minimum point, find its coordinates by using
dy
dx = 0
​
2
18
−
3 = 0
4x + 1
(4x + 1 ) 2
​
​
​
​
WWW.ZNOTES.ORG
a
CAIE AS LEVEL MATHS
b
π ∫ x 21 − x 22 dy
​
​
​
b
b
∫ πx 21 dy − ∫ πx 22 dy
or
​
a
​
​
a
​
a
{W18-P01} Question 10:
‎
With respect to y
1
Use dy
Make x the subject of the equation of the curve and
input πx 2 in the integral
The diagram shows part of the curve y = 2 (3x − 1 ) 3
​
and the lines x = 23 and x = 3 . The curve and the line x =
2
3 intersect at the Point A.
​
​
Find, showing all necessary working, the volume obtained
when the shaded region is rotated 360 ∘ about the x -axis
b
∫ πx 2 dy
Solution:
Using the formula for Volume of Revolution:
​
a
b
∫ πy2 dx
​
a
We will get:
∫
3
2
3
​
=∫
2
π (2 (3x − 1 ) 3 ) dx
−1
​
3
​
π (4 (3x − 1 ) 3 ) dx
−2
​
2
3
​
​
Integrate it:
8.5. Volume of Revolution Between 2
Curves
With respect to x
Just like a normal Volume of Revolution, this time we
subtract two volumes off each other
Use dx
Make sure that y is the subject of the equations of the
two curves
b
π ∫ y12 − y22 dx
​
​
​
a
b
or
∫ πy12 dx − ∫ πy22 dx
a
​
1
​
​
3
8.6. Improper Integrals
It is when the one or both the limits of a definite integral
are:
Positive or negative infinity
Undefined on the graph
e.g.
​
a
∫ x1 dx
​
​
0
∞
∫ x1 dx
​
​
4
With respect to y
Use dy
Make x the subject of the equations of the two curves
b
WWW.ZNOTES.ORG
b
b
​
2
b
​
3
[4π (3x − 1 ) 3 ] 2 = 4
To calculate these kind of questions, you would have to
consider what would happen when the limits reach near
the actual value (like how it would be if it reaches close to
infinity, meaning a very high number, or how low the value
would be if the limit is 0 and how you should think of
putting a very small value near zero for it)
e.g. if you think about it, for x1 if the limit was 0, then we
know that the value of it would be ∞ as the denominator is
so small, in the same case if the limit would be close to ∞,
​
CAIE AS LEVEL MATHS
we know that the denominator is way too high and thus he
actual value would be close to 0
Example
2
∴∫
1
x dx = 2
​
​
0
2
Example
2
Find the value of ∫
0
∞
1
x dx
Find the value of ∫ 4 1 x dx
​
​
​
Solution:
We know this is an improper integral as the expression is
undefined at the limit 0, so instead of 0 lets take another
variable a, which is a value very close to 0
2
Solution:
​
​
∞
∫
2
2
= ∫ x − 2 dx
1
x− 2
dx
4
​
​
​
1
1
x − 2 +1
=[
]
4(− 12 + 1)
​
​
a
​
1
x − 2 +1
=[ 1
]
−2 + 1
2
​
​
​
1
a
​
0
1
The value of 2x 2 when limit approaches 2 would be
2(2) = 2 2
1
The value of 2x 2 when limit approaches a or zero would be
1
would be 2(a) 2 = 0 , as a is a very small number close to 0
Subtracting these gives us the value of 2 2 − 0
​
​
WWW.ZNOTES.ORG
​
​
​
x
2 would be close to infinity too, so at the limit ∞, the value
​
​
should also be ∞ as the limit reaches 2 , here we can just
x
​
​
x
]
2 2
As the value of x reaches infinity, the value of
2
​
​
​
2
∞
=[
​
= [2x 2 ]
∞
​
​
1
2
​
​
2
substitute the value 2 into 2 to give us 2 now when we
2
subtract both values we get ∞ − 2 ,
​
​
​
​
​
∞
∴ ∫ 4 1 x dx = ∞
​
2
​
​
​
CAIE AS LEVEL
Maths
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