IGCSE Additional Mathematics (0606) Revision Notes

IGCSE Additional Mathematics (0606)

Revision Notes CAIE 2020 syllabus

By: Steven Zhou, find more at ste-z.com

Table of Contents

Legend ........................................................................................................................................................................ 2

1. Functions........................................................................................................................................................... 3

2. Quadratic Function ......................................................................................................................................... 5

3. Equations, Inequalities And Graphs............................................................................................................. 8

4. Indices And Surds ......................................................................................................................................... 11

5. Factors Of Polynomials................................................................................................................................ 12

6. Simultaneous Equations ............................................................................................................................... 15

7. Logarithmic & Exponential Functions ......................................................................................................... 16

8. Straight Line Graph ..................................................................................................................................... 18

9. Circular Measures ........................................................................................................................................ 20

10. Trigonometry ........................................................................................................................................... 21

11. Permutation & Combination .................................................................................................................. 26

12. Series ........................................................................................................................................................ 32

13. Vectors In 2 Dimensions .......................................................................................................................... 34

14. Differentiation & Integration ................................................................................................................. 36

Appendix A: Formula Sheet .................................................................................................................................. 44

Appendix B: Operations On Graphs ................................................................................................................... 44

Acknowledgement................................................................................................................................................... 45

http://ste-z.com/

CAIE IGCSE Additional Mathematics (0606) Revision Notes

By Steven Zhou 2

Legend

Syllabus objective

Notation

Note and tip

Formula and Laws

Key concepts

Information and explanation

Information and explanation

1. Step 1

a) Step 1.1

b) Step 1.2

2. Step 2


By Steven Zhou 3

1. Functions

understand the terms: function, domain, range (image set), one-one function, inverse

function and composition of functions

use the notation 𝑓(𝑥) = 𝑠𝑖𝑛 𝑥, 𝑓: 𝑥 ↦ 𝑙𝑔 𝑥, (𝑥 > 0), 𝑓−1(𝑥) and 𝑓2(𝑥) [=

𝑓(𝑓(𝑥))]

Function Notation

e.g. 𝑓(𝑥) = 𝑠𝑖𝑛 𝑥

e.g. 𝑓: 𝑥 ↦ 𝑙𝑔 𝑥

One-one function

function with no 2 points have the same y-coordinate

Can be tested with the horizontal line test

draw a horizontal line and move it across

a function is one-one is the line does not touch two point on the curve at

once

explain in words why a given function does not have an inverse

find the inverse of a one-one function

use sketch graphs to show the relationship between a function and its inverse

Inverse function

𝑥 and 𝑦 value exchanges when 𝑓(𝑥) become 𝑓−1(𝑥)

The domain and range exchanges

Notation - 𝑓−1(𝑥)

A function has an inverse when it is one-one

Finding the inverse function

Graph between function and its inverse (Figure 1)

The inverse function is the reflection of the function in 𝑦 = 𝑥

form composite functions

Figure 1


By Steven Zhou 4

Composition of Functions

Notation

When plugging 𝑔(𝑥) into 𝑓(𝑥) – 𝑓𝑔(𝑥) or 𝑓(𝑔(𝑥))

When plugging 𝑓(𝑥) into 𝑓(𝑥) - 𝑓2(𝑥) or 𝑓(𝑓(𝑥))

For fg(k) and k is a given value, first calculate g(k), and then plug the result of g(k)

into f(x)

why a given function is a function – one x value only corresponds to one y

Domain and Range

Domain

set of values of x

Notation e.g. (𝑥 > 0)

Range

set of values of y

Notation e.g. (𝑦 > 0)

understand the relationship between 𝑦 = 𝑓(𝑥) and 𝑦 = |𝑓(𝑥)|, where 𝑓(𝑥) may

be linear, quadratic or trigonometric

Modulo functions

Notation: 𝑦 = |𝑓(𝑥)|

any part of 𝑦 = 𝑓(𝑥) below the x axis is reflected

upwards


By Steven Zhou 5

2. Quadratic Function

solve quadratic equations for real roots

Solving quadratic equations for real roots

By factorizing

1. Rearrange the equation so one side is zero

2. Factorize into the form (𝑎𝑥 − 𝑏)(𝑐𝑥 − 𝑑) (a,b,c and d is constants)

3. Solve the 2 (or 1) linear equation 𝑎𝑥 − 𝑏 = 0 and 𝑐𝑥 − 𝑑 = 0

By completing the square

1. Put the constant at the right of the equal sign

2. If the coefficient of 𝑥2 is not 1, divide through to make it 1

3. Find and add the square number (= coefficient of 𝑥2

2) to both sides

4. Complete the square

5. Square root both sides and solve

NOTE: when square rooting both sides, make sure

RHS have ±

By the Quadratic formula

Arrange the equation into the form 𝑎𝑥2 +

𝑏𝑥 + 𝑐 = 0

𝑥 =−𝑏±√∆

2𝑎, where ∆ (discriminant) = 𝑏2 − 4𝑎𝑐 (This will be provided

on formula sheet)

NOTE: do not divide each side by x

find the maximum or minimum value of the quadratic function 𝑓: 𝑥 ↦ 𝑎𝑥2 + 𝑏𝑥 + 𝑐

by any method

Maximum and minimum values

1. Find the vertex of the parabola

a) Method 1

i. Arrange the function into the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0

ii. The axis of symmetry is 𝑥 = −𝑏

2𝑎 : Therefore, the vertex is


By Steven Zhou 6

(−𝑏

2𝑎, 𝑓(−

𝑏

2𝑎))

b) Method 2

i. Find the two roots – 𝑝 and 𝑞

ii. The axis of symmetry is 𝑥 =𝑝+𝑞

2: Therefore, the vertex is

(𝑝 + 𝑞

2, 𝑓(𝑝 + 𝑞

2))

c) Method 3

i. Arrange into the form 𝑦 = 𝑎(𝑥 − ℎ)2 + 𝑘 by by completing the

square

ii. The vertex is (ℎ, 𝑘)

2. Determine if the vertex is the maximum or minimum

a) Arrange the function into the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0

b) If 𝑎 > 0 then it is a minimum, if 𝑎 < 0 then it is a maximum

use the maximum or minimum value of 𝑓(𝑥) to sketch the graph or determine the

range for a given domain

Sketching the graph of quadratic function

1. For 𝑦 = 𝑎(𝑥 − 𝑏)2 + 𝑐, 𝑦 = 𝑎(𝑥 + 𝑏)(𝑥 + 𝑐), 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐:

2. Find the x-intercept (the roots) by finding 𝑥 for 𝑓(𝑥) = 0

3. Find the y-intercept by finding the value of 𝑓(0)

4. Find the vertex

5. Pinpoint the important points – intercepts and vertex

6. Draw using the fact that the graph of quadratic function is always symmetric

7. Label the axis

MARKING: 1 point for correct shape + max/min in correct quadrant; 1 mark

for labeling all the intercept

Determining the range

If the vertex is a maximum: 𝑦 < (𝑦 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑒𝑟𝑡𝑒𝑥)


By Steven Zhou 7

If the vertex is a minimum: 𝑦 > (𝑦 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑒𝑟𝑡𝑒𝑥)

Discriminant

know the conditions for 𝑓(𝑥) = 0 to have (i) two real roots, (ii) two equal roots, (iii) no

real roots

conditions of root for 𝑓(𝑥) = 0

1. Rearrange into the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0

2. Find ∆

∆ CONDITIONS OF ROOT FOR 𝒇(𝒙) = 𝟎

> 𝟎 two real roots

= 𝟎 two equal roots

< 𝟎 no real roots

related conditions for a given line to

(i) intersect a given curve, (ii) be a tangent to a given curve, (iii) not intersect a given

curve

related conditions for a given line to (i) intersect a given curve, (ii) be a

tangent to a given curve, (iii) not intersect a given curve

1. Substitute the function of the line into the quadratic function

2. Rearrange into the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0

3. Find ∆

∆ CONDITIONS

> 𝟎 The line intersects a given curve

= 𝟎 The line is a tangent to the given curve

< 𝟎 The line did not intersect the curve

find the solution set for quadratic inequalities

Solving quadratic inequalities

1. Solve for x

2. Sketch a graph

3. Determine the values according to the graph


By Steven Zhou 8

3. Equations, inequalities and graphs

solve graphically or algebraically equations of the type |𝑎𝑥 + 𝑏| = 𝑐(𝑐 ⩾ 0)

and |𝑎𝑥 + 𝑏| = |𝑐𝑥 + 𝑑|

Solving modulo equations

Graphically

1. Graph both side of the equal sign (e.g. graph 𝑦 = |𝑎𝑥 + 𝑏| and 𝑦 = 𝑐

for the equation |𝑎𝑥 + 𝑏| = 𝑐 )

2. the point(s) of intersection is the solution

Algebraically

When on one side the modulo sign is removed, a ± has to be added to

the other side

𝐼𝑓 |𝑎𝑥 + 𝑏| = 𝑐

𝑇ℎ𝑒𝑛 𝑎𝑥 + 𝑏 = ±𝑐

When removing modulo on both sides, only one ± has to be added

𝐼𝑓 |𝑎𝑥 + 𝑏| = |𝑐𝑥 + 𝑑|

𝑇ℎ𝑒𝑛 𝑎𝑥 + 𝑏 = ±(𝑐𝑥 + 𝑑) 𝑜𝑟 ± (𝑎𝑥 + 𝑏) = 𝑐𝑥 + 𝑑

solve graphically or algebraically inequalities of the type |𝑎𝑥 + 𝑏| > 𝑐(𝑐 ⩾ 0),

|𝑎𝑥 + 𝑏| ⩽ 𝑐(𝑐 > 0) and |𝑎𝑥 + 𝑏| ⩽ |𝑐𝑥 + 𝑑|

Solving modulo inequalities

Graphically

Draw each side of the equation

(e.g. for |𝑎𝑥 + 𝑏| ⩽ |𝑐𝑥 + 𝑑|, draw 𝑦 = |𝑎𝑥 + 𝑏|, and 𝑦 =

|𝑐𝑥 + 𝑑| on the same axis, and parts of the 𝑦 = |𝑎𝑥 + 𝑏| graph

under 𝑦 = |𝑐𝑥 + 𝑑| is the solution)

Algebraically - Follow the same when solving modulo equations – expect the

sign need to change direction (e.g. < 𝑡𝑜 >,≥ 𝑡𝑜 ≤) when both sides is

divided by a negative number

sketch the graphs of cubic polynomials and their moduli, when given in factorised form

𝑦 = 𝑘(𝑥 – 𝑎)(𝑥 – 𝑏)(𝑥 – 𝑐)

Graphing Cubic polynomials in form 𝑦 = 𝑘(𝑥 – 𝑎)(𝑥 – 𝑏)(𝑥 – 𝑐)

1. Find the x-intercept (root): x=a, x=b, x=c


By Steven Zhou 9

2. Find the y intercept by substituting 0 for x

3. 𝑘

k>0 k<0

k affects the expansion vertically – do not affect x-intercept

4. For 𝑦 = 𝑘(𝑥 − 𝑎)2(𝑥 − 𝑏)

Curve will touch the x-axis at 𝑥 = 𝑎, and cut the

curve at 𝑥 = 𝑏

5. For 𝑦 = 𝑘(𝑥 − 𝑎)3

Curve will meet the curve at x=a:

solve cubic inequalities in the form 𝑘(𝑥 – 𝑎)(𝑥 – 𝑏)(𝑥 – 𝑐) ⩽ 𝑑 graphically

Solving cubic inequalities in the form 𝑘(𝑥 – 𝑎)(𝑥 – 𝑏)(𝑥 – 𝑐) ⩽ 𝑑 by graphing

1. Graph the cubic curve

2. Graph the line y=d

3. Any part of the curve that is under the line x=d is the solution of the

inequality

use substitution to form and solve a quadratic equation in order to solve a related

equation

Problem solving with Quadratics

1. Translate the words to algebraic equation, define what x is

2. Solve the equation


By Steven Zhou 10

3. Check if the results are practical (e.g. numbers of objects present cant be

negative)

4. Give the answers in a sentence


By Steven Zhou 11

4. Indices and surds

perform simple operations with indices and with surds, including rationalising the

denominator

Operations with indices

A negative base raised to odd index is always negative; a negative base

raised to even index is always positive

𝑎𝑚 × 𝑎𝑛 = 𝑎𝑚+𝑛

𝑎𝑚

𝑎𝑛= 𝑎𝑚−𝑛 (𝑎 ≠ 0)

(𝑎𝑚)𝑛 = 𝑎𝑚×𝑛

(𝑎𝑏)𝑛 = 𝑎𝑛𝑏𝑛

(𝑎

𝑏)𝑛 =

𝑎𝑛

𝑏𝑛 (𝑏 ≠ 0)

𝑎0 = 1 (𝑎 ≠ 0)

𝑎−𝑛 =1

𝑎𝑛

𝑎𝑚

𝑛 = √𝑎𝑚𝑛

(𝑎 > 0, 𝑛 ∈ ℤ+,

𝑚 ∈ ℤ)

Operation with surds

√𝑎 ≥ 0, 𝑎 𝑚𝑢𝑠𝑡 ≥ 0

√𝑎𝑏 = √𝑎 × √𝑏 (for 𝑎, 𝑏 ≥ 0) √

𝑎

𝑏=√𝑎

√𝑏

𝑎√𝑘 + 𝑏√𝑘 = (𝑎 + 𝑏)√𝑘

Rationalizing the denominator

for 𝑏

√𝑎 : multiply by

√𝑎

√𝑎 (since it equals to 1)

for 𝑐

𝑎+√𝑏 : multiply by

𝑎−√𝑏

𝑎−√𝑏


By Steven Zhou 12

5. Factors of polynomials

BASE KNOWLEDGE

Zero for polynomial and root for equations

Adding and subtraction of polynomials can be done by collecting like

terms

Scalar can be multiplied to a polynomial by multiplying each term

Multiplying polynomials - each term of the first polynomial is multiplied

with each term of the second polynomial

Dividing polynomials

By long division


By Steven Zhou 13

TIP: 𝟎𝒙 can be inserted when there are no that kind of term

find factors of polynomials

Factors of a polynomial

If 𝑥 − 𝑎 is a factor of 𝑃(𝑥), then there exists a polynomial 𝑄(𝑥) such that

𝑃(𝑥) = (𝑥 − 𝑎)𝑄(𝑥)

If one factor is found, other factor could be obtained by performing division –

the quotient is the other factor

know and use the remainder and factor theorems

Remainder theorem

When polynomial 𝑃(𝑥) is divided by 𝑥 − 𝑘 until a constant remainder 𝑅 is

obtained ⟺ 𝑅 = 𝑃(𝑘)

We can use the theorem to determine the value of the remainder

Factor theorem

𝑘 is a zero of the polynomials 𝑃(𝑥) ⟺ 𝑥 − 𝑘 is a factor of 𝑃(𝑥)

TIP: if 𝑎𝑥 − 𝑏 is a factor of 𝑃(𝑥) ⟺ one of the zero is 𝑏

𝑎

We can use the theorem to determine whether x-k is a factor of P(x)

solve cubic equations

Solving cubic equations

1. Identify the constant term

2. Factorize the constant term

3. Substitute the possible factor of constant term into the cubic equation until

a) the factor that allow the equation to =0 is one of the root of the cubic

equation

4. One factor is obtained by utilizing the factor theorem


By Steven Zhou 14

5. The other quadratic factor is obtained by performing long division of the

equation with the factor

6. Solve the quadratic equation


By Steven Zhou 15

6. Simultaneous equations

solve simple simultaneous equations in two unknowns by elimination or substitution

solve simple simultaneous equations in two unknowns

by elimination

𝑒𝑥𝑚𝑎𝑝𝑙𝑒 {4𝑥 − 𝑦 = 0 𝑦 − 𝑥 = 3

𝐼𝑓 𝑤𝑒 𝑎𝑑𝑑 4𝑥 − 𝑦 = 0 𝑡𝑜 𝑦 − 𝑥 = 3, 𝑡ℎ𝑒𝑛:

3𝑥 = 3

𝑥 = 1

𝐴𝑛𝑑 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑔 𝑥 = 1 𝑖𝑛𝑡𝑜 𝑦 − 𝑥 = 3

𝑦 − 1 = 3

𝑦 = 4

by substitution

𝑒𝑥𝑎𝑚𝑝𝑙𝑒 {2𝑥 − 𝑦 = 0𝑦 = 5

𝑆𝑢𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑦 = 5 𝑖𝑛𝑡𝑜 2𝑥 − 𝑦 = 0:

2𝑥 − 5 = 0

2𝑥 = 5

𝑥 =5

2, 𝑦 = 5


By Steven Zhou 16

7. Logarithmic & exponential functions

know simple properties and graphs of the logarithmic and exponential functions

including ln 𝑥 and 𝑒𝑥 (series expansions are not required) and graphs of 𝑘𝑒𝑛𝑥 +

𝑎 and 𝑘 ln(𝑎𝑥 + 𝑏) where 𝑛, 𝑘, 𝑎 and 𝑏 are integers

graphs of the exponential functions

for 𝑦 = 𝑎𝑏𝑥−𝑐 + 𝑑

𝑏 controls the steepness

𝑐 controls horizontal translation

𝑑 controls vertical translation, so that the horizontal asymptote of the

graph is 𝑦 = 𝑑

MARKING: 1 mark for correct shape; 1 mark for intercept labeled; 1 mark for the

asymptote

Logarithms

If 𝑓(𝑥) = 𝑎𝑥 𝑡ℎ𝑒𝑛 𝑓−1(𝑥) = log𝑎 𝑥

Logarithm in base 10 is written as lg 𝑎

Logarithm in base 𝑒 is written as ln 𝑎

Simple properties

log𝑎(𝑔(𝑥)) is defined only when 𝑎 and 𝑔(𝑥) > 0

TIP: This can be used to determine the domain/range of a logarithmic

function

log𝑎 𝑎𝑥 = 𝑥

𝑎log𝑎 𝑥 = 𝑥 (𝑓𝑜𝑟 𝑥 > 0)

𝐼𝑓 𝑏 = 𝑎𝑥, 𝑡ℎ𝑒𝑛 𝑥 = log𝑎 𝑏 (𝑓𝑜𝑟 𝑎, 𝑏 > 0)

ln 𝑥 = log𝑒 𝑥

lg 𝑥 = log10 𝑥


By Steven Zhou 17

graphs of the logarithmic functions

find the inverse function (the exponential function) of the given logarithmic

function

𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑦 = 𝑘 ln(𝑎𝑥 + 𝑏)

𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑠: 𝑥 = 𝑘 ln(𝑎𝑦 + 𝑏)

ln(𝑎𝑦 + 𝑏) =𝑥

𝑘

𝑎𝑦 + 𝑏 = 𝑒𝑥𝑘

𝑎𝑦 = 𝑒𝑥𝑘 − 𝑏

𝑦 =1

𝑎𝑒𝑥𝑘 −

𝑏

𝑎

graph the inverse function

reflect the graph of the inverse function in the line y=x

know and use the laws of logarithms (including change of base of logarithms) solve

equations of the form 𝑎𝑥 = 𝑏

laws of logarithms

log𝑐 𝑎 + log𝑐 𝑏 = log𝑐 𝑎𝑏

log𝑐 𝑎 − log𝑐 𝑏 = log𝑐𝑎

𝑏

nlog𝑐 𝑎 = log𝑐( 𝑎)𝑛

log𝑏 𝑎 =log𝑐 𝑎

log𝑐 𝑏 (𝑓𝑜𝑟 𝑎, 𝑏, 𝑐 > 0, 𝑎𝑛𝑑 𝑏, 𝑐 ≠ 1) (Change of base rule)

Nature logarithms follows the same rule

Solving equations

By equating indices

𝐼𝑓 𝑎𝑥 = 𝑎𝑘 , 𝑡ℎ𝑒𝑛 𝑥 = 𝑘

By using logarithms

Logarithms can be added to both sides of the equal sign

Then solve the logarithmic equation formed - Utilize the laws of

logarithms to solve


By Steven Zhou 18

8. Straight line graph

interpret the equation of a straight-line graph in the form 𝑦 = 𝑚𝑥 + 𝑐

Equation of a straight-line graph is 𝑦 = 𝑚𝑥 + 𝑐

Slope = 𝑚 =∆𝑦

∆𝑥=

𝑦2−𝑦1

𝑥2−𝑥1

y-intercept = 𝑐

transform given relationships, including 𝑦 = 𝑎𝑥𝑛 and 𝑦 = 𝐴𝑏𝑥 , to straight line form

and hence determine unknown constants by calculating the gradient or intercept of the

transformed graph

Transforming 𝑦 = 𝑎𝑥𝑛 and 𝑦 = 𝐴𝑏𝑥 to to straight line form

Add logarithms to both side of the equarion (𝑙𝑔, 𝑙𝑛 etc)

E.g. for 𝑦 = 𝑎𝑥𝑛

ln 𝑦 = ln 𝑎𝑥𝑛

ln 𝑦 = ln 𝑎 + ln 𝑥𝑛

ln 𝑦 = ln 𝑎 + 𝑛 ln 𝑥

Gradient: 𝑛;

y-intercept: ln 𝑎

Axis: ln 𝑦 on vertical, ln 𝑥

on horizontal

E.g. for 𝑦 = 𝐴𝑏𝑥

ln 𝑦 = ln 𝐴𝑏𝑥

ln 𝑦 = ln 𝐴 + ln 𝑏𝑥

ln 𝑦 = ln 𝐴 + 𝑥 ln 𝑏

Gradient: ln 𝑏;

y-intercept: ln 𝐴

Axis: ln 𝑦 on vertical, 𝑥 on

horizontal

determine unknown constants by calculating the gradient or intercept

by substituting x and y value of two point (may be given by the question

or may need to be obtained from a graph)

NOTE: the two point you choose should be as far from each other as

possible (to be precise)

solve questions involving mid-point and length of a line

Midpoint of points (𝑥1, 𝑦1) and (𝑥2, 𝑦2) is

𝑀(𝑥1+𝑥2

2,𝑦1+𝑦2

2)

Length of a line between points (𝑥1, 𝑦1) and (𝑥2, 𝑦2) is

𝑑 = √(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2

know and use the condition for two lines to be parallel or perpendicular, including


By Steven Zhou 19

finding the equation of perpendicular bisectors

condition for two lines to be parallel

the slope of the two-line equals

condition for two lines to be perpendicular

if one line has slope 𝑚, then the second line have the slope −1

𝑚 (Opposite

reciprocal)

when multiplying the slope of two perpendicular lines together, the result is -1

Finding the equation of perpendicular bisector

1. Find the gradient of the given line by using the 2 points

2. Find the gradient of the perpendicular bisector, 𝑚

3. Find the midpoint, (𝑥1, 𝑦1), of the given line (since it the perpendicular

bisector bisects the line)

4. Plug the midpoint, gradient into the equation 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1) or 𝑦−𝑦1

𝑥−𝑥1= 𝑚


By Steven Zhou 20

9. Circular measures

solve problems involving the arc length and sector area of a circle, including

knowledge and use of radian measure

Radian and Degree measure of angles

Degree - 1° is 1

360𝑡ℎ of one revolution of circle

Radian - 1𝑐 is an angle which results the arc length to equal the radius, if the

angle is at the center of a circle

𝜋 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 = 180°

Arc length of a circle

𝜃 in radians: 𝑙 = 𝜃𝑟

𝜃 in degrees: 𝑙 =𝜃

3602𝜋𝑟

Sector area of a circle

𝜃 in radians: 𝐴 =1

2𝜃𝑟2

𝜃 in degrees: 𝐴 =𝜃

360𝜋𝑟2


By Steven Zhou 21

10. Trigonometry

know the six trigonometric functions of angles of any magnitude (sine, cosine, tangent,

secant, cosecant, cotangent)

BASE KNOWLEDGE

Unit circle – circle with center (0,0) and radius 1

Equation of a circle with center (0,0) and radius r - 𝒙𝟐 + 𝒚𝟐 = 𝒓𝟐

Therefore, the equation of unit circle is 𝒙𝟐 +

𝒚𝟐 = 𝟏

Angle measure

Positive for clockwise, negative for

anticlockwise

Six Trigonometric functions

Sine (sin)

Right triangle trigonometry: sin 𝜃 =𝑂𝑃𝑃

𝐻𝑌𝑃

In unit circle: sin 𝜃 is the y-coordinate

of P

−1 ≤ sin 𝜃 ≤ 1 in unit circle

Cosine (cos)

Right triangle trigonometry: cos 𝜃 =𝐴𝐷𝐽

𝐻𝑌𝑃

In unit circle: cos 𝜃 is the x-coordinate of P

−1 ≤ cos 𝜃 ≤ 1 in unit circle

Tangent (tan)

Right triangle trigonometry: 𝑡𝑎𝑛 =𝑂𝑃𝑃

𝐴𝐷𝐽

tan 𝜃 =sin𝜃

cos 𝜃

Secant (sec)

sec 𝜃 =1

sin𝜃

Cosecant (cosec)

cosec 𝜃 =1

cos𝜃

Cotangent (cot)

cot 𝜃 =1

tan𝜃=cos 𝜃

sin𝜃


By Steven Zhou 22

Special angles (use when the question said “find exact value of sin 𝑥 , cos 𝑥 etc.”)

Angle is multiple of 𝜋

2 (90 degrees)

Coordinates in unit circle consists of 0 and ±1


4 (45 degrees) except the

𝜋

2s

Coordinates in unit circle consists of ±1

√2


6 (30 degrees) except the

𝜋

2s

Coordinates in unit circle consists of ±1

2 and ±

√3

2

understand amplitude and periodicity and the relationship between graphs of related

trigonometric functions, e.g. 𝑠𝑖𝑛𝑥 and 𝑠𝑖𝑛2𝑥

Periodicity

A function is periodic when 𝑓(𝑥 + 𝑝) = 𝑓(𝑥)

Since 2𝜋 is one revolution, there is no change when adding 2𝜋 to 𝜃

sin 𝜃 = sin(𝜃 + 𝑘2𝜋) and cos 𝜃 = cos(𝜃 + 𝑘2𝜋)

(𝑓𝑜𝑟 𝜃 𝑖𝑛 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 𝑎𝑛𝑑 𝑘 ∈ ℤ)

tan 𝜃 = tan(𝜃 + 𝑘𝜋) (𝜃 𝑖𝑛 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 𝑎𝑛𝑑 𝑘 ∈ ℤ)

NOTE: When performing inverse trigonometric functions, there could be

multiple answers – need to use periodicity and check the domain

Amplitude for a graph - 𝑀𝑎𝑥−𝑀𝑖𝑛

2

draw and use the graphs of

𝑦 = 𝑎 𝑠𝑖𝑛 𝑏𝑥 + 𝑐

𝑦 = 𝑎 𝑐𝑜𝑠 𝑏𝑥 + 𝑐

𝑦 = 𝑎 𝑡𝑎𝑛 𝑏𝑥 + 𝑐

where 𝑎 is a positive integer, 𝑏 is a simple fraction or integer (fractions will have

a denominator of 2, 3, 4, 6 or 8 only), and 𝑐 is an integer


By Steven Zhou 23

Graph of sine - 𝑦 = 𝑎 sin 𝑏𝑥 + 𝑐

Amplitude = 𝑎

Period = 2𝜋

𝑏

Principle axis: 𝑦 = 𝑐

Graph of cosine – 𝑦 = 𝑎 cos 𝑏𝑥 + 𝑐

Amplitude = 𝑎

Period = 2𝜋

𝑏


Graph of tangent - 𝑦 = 𝑎 tan 𝑏𝑥 + 𝑐

Amplitude undefined – do not affect the graph

Period = 𝜋

𝑏


use the relationships

sin2 𝐴 + cos2 𝐴 = 1

sec2 𝐴 = 1 + tan2 𝐴, cosec2 𝐴 = 1 + cot2 𝐴


By Steven Zhou 24

tan 𝐴 =sin𝐴

cos 𝐴, cot 𝐴 =

cos 𝐴

sin𝐴

solve simple trigonometric equations involving the six trigonometric functions and the

above relationships (not including general solution of trigonometric equations)

prove simple trigonometric identities

Trigonometric relations and identities – used to prove other identities, solve

equations

sin2 𝐴 + cos2 𝐴 = 1 (given in formula sheet)

sec2 𝐴 = 1 + tan2 𝐴 (given in formula sheet)

cosec2 𝐴 = 1 + cot2 𝐴 (given in formula sheet)

tan 𝜃 =sin𝜃

cos 𝜃

cot 𝜃 =cos𝜃

sin𝜃

Solving Trigonometric equations

1. When there is multiple trigonometric functions (e.g. having both sin 𝜃 and

cos 𝜃), simplify until only have one trigonometric functions (by e.g. dividing

each side of the equal sign by cos 𝜃 so the sin 𝜃 is turned into tan 𝜃, and

cos 𝜃 will become 1)

2. If there is reciprocal trigonometric functions (cot 𝜃 , sec 𝜃 , cosec 𝜃) turn them

into tan 𝜃 , cos 𝜃 , or sin 𝜃 (so that calculator can calculate)

3. Move the trigonometric function to LHS and all other numbers (including the

coefficient of the trigonometric function) to RHS by moving terms, dividing

both side by a number etc.

4. Apply the inverse trigonometric functions (e.g. cos−1 𝜃)

NOTE: be aware of the mode (radian/degrees)

Do NOT move the term that is previously inside the trig function at LHS to RHS

at this moment

5. Find the other possible angle in the unit circle by:

6. Apply periodicity to find all possible angles within the domain

(adding/subtracting answer by 2𝜋’s for sin and cos, and 𝜋’s for tan)

7. Move the terms (except the unknown) at LHR to RHS

8. Check to see if the answer is in the domain again

Solving quadratic trigonometric equations


By Steven Zhou 25

See the trig function as a whole (e.g. letting cos 𝜃 to be 𝑏) and solve

Proving simple identities (e.g. prove that…./ show that ), often needs:

The trigonometric identities and relations above

Turning sec 𝜃 into 1

cos 𝜃, cosec 𝜃 into

1

sin𝜃

Multiplying sin 𝜃 and cos 𝜃 to both the denominator and numerator (to turn

cot 𝜃 into cos 𝜃 and tan 𝜃 into sin 𝜃)

Combining fractions by making a common denominator

Splitting fraction into 2


By Steven Zhou 26

11. Permutation & combination

BASE KNOWLEDGE

If there are 𝒎 ways of performing an operation, and for each of the way there

are also 𝒏 ways of performing a second independent operation, then there are

𝒎𝒏 number of ways of performing the 2 operation in succession

“and”/ “or”

“and” means multiplying the possibility

“or” means adding the possibility

recognise and distinguish between a permutation case and a combination case

distinguishing between permutation and combination case

permutation – arrangement with a definite order

combination – selection of objects without regard to order

know and use the notation 𝑛! (with 0! = 1), and the expressions for permutations

and combinations of 𝑛 items taken 𝑟 at a time

Factorial notation

Notation 𝑛!

𝑛! Is the product of the first n positive integers

𝑛! = 𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)…× 3 × 2 × 1 (𝑓𝑜𝑟 𝑛 ≥ 1)

0! = 1

Expression for permutations and combinations of 𝑛 items taken 𝑟 at a time

Permutations

𝑃𝑟 𝑛 or 𝑃𝑟

𝑛 = 𝑛!

(𝑛−𝑟)!

Combinations

𝐶𝑟 𝑛 or 𝐶𝑟

𝑛 or (𝑛𝑟) =

𝑛!

𝑟!(𝑛−𝑟)!(given in formula sheet)

answer simple problems on arrangement and selection (cases with repetition of objects,

or with objects arranged in a circle, or involving both permutations and combinations,

are excluded)


By Steven Zhou 27

Permutation selection problems

Choosing passwords from characters (digit/numbers, letters and symbols)

A 𝐤-character password is to be chosen from the following 𝐦 characters

Letter: ……….

Digit: ……….

Symbol: ……….

(when each character may only be used once)

No restriction: 𝑷 𝒎

𝒌

Contain 𝒙 letter, 𝒚 digit, 𝒛 symbol, in this order

𝑷 𝒏𝒖𝒎𝒃𝒆𝒓𝒔 𝒐𝒇 𝒍𝒆𝒕𝒕𝒆𝒓 𝒂𝒗𝒂𝒊𝒍𝒂𝒃𝒍𝒆

𝒙 × 𝑷 𝒏𝒖𝒎𝒃𝒆𝒓𝒔 𝒐𝒇 𝒅𝒊𝒈𝒊𝒕 𝒂𝒗𝒂𝒊𝒍𝒂𝒃𝒍𝒆

𝒚 ×

𝑷 𝒏𝒖𝒎𝒃𝒆𝒓𝒔 𝒐𝒇 𝒔𝒚𝒎𝒃𝒐𝒍 𝒂𝒗𝒂𝒊𝒍𝒂𝒃𝒍𝒆

𝒛

Start with an [letter/number/symbol] and end with an [letter/number/symbol]

𝒏𝒖𝒎𝒃𝒆𝒓𝒔 𝒐𝒇 [𝐥𝐞𝐭𝐭𝐞𝐫/𝐧𝐮𝐦𝐛𝐞𝐫/𝐬𝐲𝐦𝐛𝐨𝐥] 𝐚𝐯𝐚𝐢𝐥𝐢𝐛𝐥𝐞 × 𝑷 𝒎−𝟐

𝒌−𝟐 ×

𝒏𝒖𝒎𝒃𝒆𝒓𝒔 𝒐𝒇 [𝐥𝐞𝐭𝐭𝐞𝐫/𝐧𝐮𝐦𝐛𝐞𝐫/𝐬𝐲𝐦𝐛𝐨𝐥] 𝐚𝐯𝐚𝐢𝐥𝐢𝐛𝐥𝐞

Contain at least 𝒙 [letter/number/symbol]

METHOD ONE: total number of cases minus the number of cases where

there are no [letter/number/symbol] included

𝑷 𝒎

𝒌 − 𝑷 (𝒎−𝒏𝒖𝒎𝒃𝒆𝒓𝒔 𝒐𝒇 𝐥𝐞𝐭𝐭𝐞𝐫/𝐧𝐮𝐦𝐛𝐞𝐫/𝐬𝐲𝐦𝐛𝐨𝐥 𝒂𝒗𝒂𝒊𝒍𝒂𝒃𝒍𝒆)

𝒌

METHOD TWO: Adding the cases of having (1, 2, …, numbers of

[letter/number/symbol] available)character together


By Steven Zhou 28

Combination selection problems

Choosing men and women to form a group (e.g. a committee, team)

A team/committee of 𝒌 is to be chosen from 𝒎 people (with 𝒃 boys/men and 𝒈

girls/women).

No restriction: 𝑪 𝒎

𝒌

Have 𝒙 boys/mens and 𝒚 girls/womens: 𝑪 𝒃𝒙 × 𝑪

𝒈𝒚

Have at least 𝒙 [boys/girls]

METHOD ONE: total number of cases minus the number of cases where


By Steven Zhou 29

there are no 𝒙 [boys/girls] included

METHOD TWO: Adding the cases of having different combination of

numbers of girls and boys that satisfies the condition (e.g. 1B2G, 2B1G, 3B

if must have at least 1 boy)

A (specific person) or B, but not both, must be included

total number of cases minus the number of cases where both of the person

is selected/not selected

When two people cannot be separated – see them as one person


By Steven Zhou 30

Choosing questions in exams

An exam consists of section A (containing 𝒂 question), and section B (containing 𝒃

questions)

Find the numbers of possible questions selection that can be made

Have to choose 𝒌 question in section A and 𝒎 question in section B

𝑪 𝒂𝒌 × 𝑪

𝒃𝒎

Must answer 𝒙 specific questions in Section A and 𝒚 specific question in section

B AND Have to choose 𝒌 question in section A and 𝒎 question in section B

𝑪 𝒂−𝒙

𝒌−𝒙 × 𝑪 𝒃−𝒚

𝒎−𝒚

Arrangements problems

Forming numbers form digits

𝒌 digit numbers can be formed using 𝒎 digit listed (no zero)

How many numbers can be formed?

No restriction: 𝑷 𝒎

𝒌

The number formed must be [even/odd]

𝑵𝒖𝒎𝒃𝒆𝒓𝒔 𝒐𝒇 𝒅𝒊𝒈𝒊𝒕 𝒍𝒊𝒔𝒕𝒆𝒅 𝒕𝒉𝒂𝒕 𝒂𝒓𝒆 𝒆𝒗𝒆𝒏 × 𝑷 𝒎−𝟏

𝒌−𝟏


By Steven Zhou 31

Arranging books on shelfs OR Arranging people in seats

𝒌 different books are to be arranged on a shelf. There are 𝒂 math books and 𝒃

history books. Find the number of arrangements of books

[Math/History] books need to be together

[Math/History] books cannot be together


By Steven Zhou 32

12. Series

use the Binomial Theorem for expansion of (𝑎 + 𝑏)𝑛 for positive integer 𝑛

Binomial theorem

(𝑎 + 𝑏)𝑛 = 𝑎𝑛 + (𝑛1)𝑎𝑛−1𝑏 + ⋯+ (𝑛

𝑟)𝑎𝑛−𝑟𝑏𝑟 +⋯+ 𝑏𝑛 for 𝑟 =

0,1,2,… , 𝑛 (given in formula sheet)

Sigma form:

∑(𝑛

𝑟)𝑎𝑛−𝑟𝑏𝑟

𝑛

𝑟=0

use the general term (𝑛𝑟)𝑎𝑛−𝑟𝑏𝑟 , 0 ≤ 𝑟 ≤ 𝑛 (knowledge of the greatest term and

properties of the coefficients is not required)

General term, (𝑟 + 1)𝑡ℎ term, in binomial expansion: (𝑛𝑟)𝑎𝑛−𝑟𝑏𝑟

NOTE: when finding a specific term, minus the term number by one and then use

the formula

recognise arithmetic and geometric progressions

use the formulae for the 𝑛th term and for the sum of the first 𝑛 terms to solve

problems involving arithmetic or geometric progressions

use the condition for the convergence of a geometric progression, and the formula for

the sum to infinity of a convergent geometric progression

Arithmetic progressions

Have a constant difference (common difference, 𝒅) between two consecutive

number in a sequence

𝑛th term, 𝑢𝑛

𝑢𝑛 = 𝑎 + (𝑛 − 1)𝑑

𝑎 is the first term, 𝑑 is the common difference

Sum of the first 𝑛 terms, 𝑆𝑛

𝑆𝑛 =1

2𝑛(𝑎 + 𝑙)

𝑆𝑛 =1

2𝑛[2𝑎 + (𝑛 − 1)𝑑]

𝑎 is the first term, 𝑙 is the last term, 𝑑 is the common difference

Geometric progressions

Have a constant ratio (common ratio, 𝒓) between two consecutive number in a

sequence


By Steven Zhou 33

𝑛th term, 𝑢𝑛

𝑢𝑛 = 𝑎𝑟𝑛−1

𝑎 is the first term, 𝑟 is the common ratio

Sum of the first 𝑛 terms, 𝑆𝑛

𝑆𝑛 =𝑎(1−𝑟𝑛)

1−𝑟 (𝑓𝑜𝑟 𝑟 ≠ 1)


Sum to infinity, 𝑆∞ , of a convergent geometric progression

convergence of a geometric progression

when the sum trends to a finite value

𝑆∞ =𝑎

1−𝑟 (𝑓𝑜𝑟 |𝑟| < 1)



By Steven Zhou 34

13. Vectors in 2 dimensions

use vectors in any form, e.g. (𝑎𝑏), 𝐴𝐵⃗⃗⃗⃗ ⃗, 𝒑, 𝑎𝒊 − 𝑏𝒋

Forms of vector

𝐴𝐵⃗⃗⃗⃗ ⃗

𝒑 or 𝑝

Component form - (𝑎𝑏)

Unit vector form – 𝑎𝒊 − 𝑏𝒋

𝑖 = (10) and 𝑗 = (

01)

know and use position vectors and unit vectors

Position vector

Position vector of B relative to A is 𝐴𝐵⃗⃗⃗⃗ ⃗

𝐴𝐵⃗⃗⃗⃗ ⃗ = 𝑂𝐵⃗⃗ ⃗⃗ ⃗ − 𝑂𝐴⃗⃗ ⃗⃗ ⃗

Point 𝑃(𝑥, 𝑦) has position vector 𝑂𝑃⃗⃗ ⃗⃗ ⃗ 𝑜𝑟 (𝑥𝑦) 𝑜𝑟 𝑥𝒊 + 𝑦𝒋

Unit vector

Any vector which has a length of one unit

Base unit vectors in the x or y positive direction

𝑖 = (10)

𝑗 = (01)

Unit vector in the direction 𝑎 is 𝒂

|𝒂|

A vector of length 𝑘 in direction 𝑎 is 𝑘𝒂

|𝒂|

A vector of length 𝑘 parallel to 𝑎 is ±𝑘𝒂

|𝒂|

find the magnitude of a vector;


By Steven Zhou 35

Magnitude of a vector

Notation: |𝐴𝐵⃗⃗⃗⃗ ⃗| or |𝒂|

The length of a vector

If 𝒂 = (𝑎𝑥𝑎𝑦) = 𝑥𝑖 + 𝑦𝑗, then its magnitude is √𝑎𝑥2 + 𝑎𝑦2

add and subtract vectors and multiply vectors by scalars

Adding vectors

If 𝒂 = (𝑎𝑥𝑎𝑦) and 𝒃 = (

𝑏𝑥𝑏𝑦), then 𝒂 + 𝒃 = (

𝑎𝑥 + 𝑏𝑥𝑎𝑦 + 𝑏𝑦

),

Subtracting vectors

If 𝒂 = (𝑎𝑥𝑎𝑦) and 𝒃 = (

𝑏𝑥𝑏𝑦), then 𝒂 − 𝒃 = (

𝑎𝑥 − 𝑏𝑥𝑎𝑦 − 𝑏𝑦

),

Multiplying vectors by scalars

Scalar – a non-vector quantity – only has magnitude but do not have direction

If 𝑘 Is and scalar and 𝒗 = (𝑣𝑥𝑣𝑦), then 𝑘𝒗 = (

𝑘𝑣𝑥𝑘𝑣𝑦)

Parallelism – two non-zero vector are parallel to each other if one is a scalar

multiple of another

If 𝒂 = 𝑘𝒃, then 𝒂//𝒃 and |𝒂| = |𝑘||𝒃| (for 𝑘 is an scalar, and 𝒂, 𝒃

is non-zero vector)

Solving equations involving component form

(𝑥𝑦) = (

𝑎1 − 𝑏1𝑎2 − 𝑏2

) can form the simultaneous equation {𝑥 = 𝑎1 − 𝑏1𝑦 = 𝑎2 − 𝑏2

compose and resolve velocities

Constant velocity questions

If an object initially has position vector 𝒂, and move with a constant velocity

𝒃 (speed is |𝒃|), its position vectorx after time t is 𝒓 = 𝒂 + 𝑡𝒃


By Steven Zhou 36

14. Differentiation & Integration

Differentiation

understand the idea of a derived function, use the notations 𝑓′(𝑥), 𝑓′′(𝑥), 𝑑𝑦

𝑑𝑥, 𝑑2𝑦

𝑑𝑥2,

(= 𝑑

𝑑𝑥(𝑑𝑦

𝑑𝑥))

derived function – gradient function

Notation: 𝑓′(𝑥) or 𝑑𝑦

𝑑𝑥

use the derivatives of the standard functions 𝑥𝑛(for any rational 𝑛), sin 𝑥, tan 𝑥,

cos 𝑥, 𝑒𝑥, ln 𝑥, together with constant multiples, sums and composite functions of these

differentiate products and quotients of functions

Simple Rules of differentiation

FUNCTION DERIVITIVE

𝒌 (a constant) 0

𝒙𝒏 (for any rational 𝑛) 𝑛𝑥𝑛−1

𝐬𝐢𝐧𝒙 cos 𝑥

𝐭𝐚𝐧 𝒙 sec2 𝑥

𝐜𝐨𝐬 𝒙 −sin𝒙

𝒆𝒙 𝑒𝑥

𝐥𝐧 𝒙 1

𝑥

Deriving sums of functions: 𝑢(𝑥) + 𝑣(𝑥) 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑡𝑒⇒ 𝑢′(𝑥) + 𝑣′(𝑥)

Deriving functions with constant multiples: 𝑐𝑢(𝑥) 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑡𝑒⇒ 𝑐𝑢′(𝑥)

Chain rule (use when differentiating composite functions) :

𝑦 = 𝑓(𝑢(𝑥)) 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑡𝑒⇒ 𝑓′(𝑢(𝑥)) × 𝑢′(𝑥)

Product rule: 𝑢(𝑥)𝑣(𝑥) 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑡𝑒⇒ 𝑢′(𝑥)𝑣(𝑥) + 𝑣′(𝑥)𝑢(𝑥)

Quotient rule: 𝑢(𝑥)

𝑣(𝑥) 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑡𝑒⇒

𝑢′(𝑥)𝑣(𝑥)−𝑣′(𝑥)𝑢(𝑥)

[𝑣(𝑥)]2

apply differentiation to gradients, tangents and normals, stationary points, connected

rates of change, small increments and approximations and practical maxima and

minima problems


By Steven Zhou 37

use the first and second derivative tests to discriminate between maxima and minima

Second derivative

Notation: 𝑓′′(𝑥) or 𝑑2𝑦

𝑑𝑥2

Derivative of 𝑓′(𝑥) is 𝑓′′(𝑥) (= 𝑑

𝑑𝑥(𝑑𝑦

𝑑𝑥))

Application of differentiation

Gradients - The value of 𝑓′(𝑎) is the gradient of the tangent to 𝑦 = 𝑓(𝑥)

at the point 𝑎

tangents and normal

Equation of tangent to 𝑓(𝑥) at 𝑥 =

𝑎 is: 𝑦 − 𝑓(𝑎) = 𝑓′(𝑎)(𝑥 − 𝑎)

Equation of normal to 𝑓(𝑥) at 𝑥 = 𝑎 is:

𝑦 − 𝑓(𝑎) = −1

𝑓′(𝑎)(𝑥 − 𝑎)

stationary points (where gradient is 0) - 𝑓(𝑥) has stationary point at x-

coordinate of 𝑓′(𝑥) = 0

Determining whether a stationary point is a maxima or minima

By drawing sign diagrams


By Steven Zhou 38

By using second derivative (For a function with stationary point at 𝑥 = 𝑎)

𝑓′′(𝑥) > 0: minima

𝑓′′(𝑥) < 0: maxima

NOTE: This method does NOT work when 𝑓′′(𝑥) = 0

connected rates of change

TIP: 𝑑𝑦

𝑑𝑥 gives the rate of change of 𝑦 with respect to 𝑥

Draw a diagram that clearly shows the situation (if not given)

Label the numbers (+ distinguish between the constant and the changing

quantity)

Write an equation that connects the variables (e.g. using Pythagoreans,

trig, similar triangles)

Differentiate the equation with respect to 𝑡

NOTE: the changing quantity is a function – need to apply chain rule

Plug into the number of the particular case given

NOTE: do NOT plug the numbers of the particular case before you have

get a generalized differential equation

(Example on next page)

Give the answer in a sentence


By Steven Zhou 39

small increments and approximations

The small change in y: δy small change in x: δx

𝛿𝑦

𝛿𝑥≈𝑑𝑦

𝑑𝑥 (if the increment is small)

Finding the small change of y when x increases by small amount

𝛿𝑦

𝛿𝑥≈𝑑𝑦

𝑑𝑥

𝛿𝑦 = 𝛿𝑥 ×𝑑𝑦

𝑑𝑥

Finding the small change of x when y increases by small amount

𝛿𝑦

𝛿𝑥≈𝑑𝑦

𝑑𝑥

𝛿𝑥 = 𝛿𝑦 ×𝑑𝑥

𝑑𝑦

𝛿𝑥 = 𝛿𝑦 ÷𝑑𝑦

𝑑𝑥

practical maxima and minima problems

1. Draw a diagram (if not given) + Label

2. Write an equation that connects the variables

3. Find the stationary points

4. Determine the nature of the stationary points and identify the asked point

according to the question (maxima/minima)

5. Give the answer in a sentence


By Steven Zhou 40

Integration

BASE KNOWLEDGE

F’(x)=f(x)

understand integration as the reverse process of differentiation

Integration - the reverse process of differentiation

If F’(x)=f(x), then ∫𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥)

integrate sums of terms in powers of 𝑥 including 1

𝑥,

1

𝑎𝑥+𝑏

Simple rules of integration

FUNCTION INTEGRAL

𝒌 (a constant) 𝑘𝑥 + 𝑐

𝒙𝒏 (for 𝒏 ≠ −𝟏) 𝑥𝑛+1

𝑛 + 1+ 𝑐

𝐬𝐢𝐧𝒙 −cos 𝑥 + 𝑐

𝐜𝐨𝐬 𝒙 sin 𝑥 + 𝑐

𝒆𝒙 𝑒𝑥 + 𝑐

Integrating sums of functions: ∫[𝑓(𝑥) + 𝑔(𝑥)]𝑑𝑥 = ∫𝑓(𝑥)𝑑𝑥 + ∫𝑔(𝑥)𝑑𝑥

Integrating functions with constant multiples: ∫𝑘𝑓(𝑥)𝑑𝑥 = 𝑘 ∫𝑓(𝑥)𝑑𝑥

integrate functions of the form (𝑎𝑥 + 𝑏)𝑛(for any rational 𝑛), sin(𝑎𝑥 + 𝑏),

tan(𝑎𝑥 + 𝑏), cos(𝑎𝑥 + 𝑏), 𝑒𝑎𝑥+𝑏

Integrating 𝑓(𝑎𝑥 + 𝑏): ∫[𝑓(𝑎𝑥 + 𝑏)]𝑑𝑥 =1

𝑎𝐹(𝑎𝑥 + 𝑏) + 𝑐

evaluate definite integrals and apply integration to the evaluation of plane areas

Definite integrals

Definite integral of 𝑓(𝑥) (where 𝑓(𝑥) is continuous) in the interval 𝑎 ≤ 𝑥 ≤

𝑏 is ∫ 𝑓(𝑥)𝑏

𝑎𝑑𝑥 = [𝐹(𝑥)]𝑎

𝑏 = 𝐹(𝑏) − 𝐹(𝑎)

Rules of definite integrals

∫ 𝑓(𝑥)𝑎

𝑎𝑑𝑥 = 0

∫ 𝑓(𝑥)𝑏

𝑎𝑑𝑥 = −∫ 𝑓(𝑥)

𝑎

𝑏𝑑𝑥


By Steven Zhou 41

∫ 𝑘𝑓(𝑥)𝑏

𝑎𝑑𝑥 = 𝑘 ∫ 𝑓(𝑥)

𝑏

𝑎𝑑𝑥

∫ 𝑓(𝑥)𝑏

𝑎𝑑𝑥 + ∫ 𝑓(𝑥)

𝑐

𝑏𝑑𝑥 = ∫ 𝑓(𝑥)

𝑐

𝑎𝑑𝑥

∫ [𝑓(𝑥) + 𝑔(𝑥)]𝑏

𝑎𝑑𝑥 = ∫ 𝑓(𝑥)

𝑏

𝑎𝑑𝑥 + ∫ 𝑔(𝑥)

𝑏

𝑎𝑑𝑥

Evaluation of plane areas using definite integrals

If 2 functions 𝑓(𝑥) and 𝑔(𝑥), intersecting at 𝑥 = 𝑎 and 𝑥 = 𝑏, and

𝑓(𝑥) > 𝑔(𝑥) for all 𝑎 ≤ 𝑥 ≤ 𝑏, then the area of the region between th

intersection is 𝐴 = ∫ [𝑓(𝑥) − 𝑔(𝑥)]𝑑𝑥𝑏

𝑎

TIP: x-axis’ equation is 𝑦 = 0

apply differentiation and integration to kinematics problems that involve displacement,

velocity and acceleration of a particle moving in a straight line with variable or

constant acceleration, and the use of 𝑥– 𝑡 and 𝑣– 𝑡 graphs

BASE KNOWLEDGE

Displacement & distance

Displacement – the straight-line distance relative to the initial position

Interpreting displacement of a Particle P

moving in a straight line

Velocity and speed

Velocity – vector quantity (both magnitude and direction)

Average 𝑽𝒆𝒍𝒐𝒄𝒊𝒕𝒚 =𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒕𝒓𝒂𝒗𝒍𝒆𝒅

𝒕𝒊𝒎𝒆 𝒕𝒂𝒌𝒆𝒏

Interpreting velocity of a Particle P moving

in a straight line

Speed – scalar quantity (only magnitude)

Has unit 𝒎𝒔−𝟏

Acceleration


By Steven Zhou 42

Average 𝑨𝒄𝒄𝒆𝒍𝒂𝒓𝒂𝒕𝒊𝒐𝒏 =𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚

𝒕𝒊𝒎𝒆 𝒕𝒂𝒌𝒆𝒏

Interpreting acceleration of a Particle P moving in a straight line

Has unit 𝒎𝒔−𝟐

Application of differentiation and integration to kinematics problems (involving

displacement, velocity and acceleration of a particle moving in a straight line with

variable or constant acceleration)

Obtaining displacement function from velocity function and velocity function

from acceleration function

Integrate

𝑐 can be found by plugging the value given

Finding the displacement in a time interval 𝑡1 ≤ 𝑡 ≤ 𝑡2

𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 = 𝑠(𝑡2) − 𝑠(𝑡1) = ∫ 𝑣(𝑡)𝑡2𝑡1

Finding the distance traveled in a time interval 𝑡1 ≤ 𝑡 ≤ 𝑡2

1. Find 𝑣(𝑡)

2. Find the values when 𝑣(𝑡) = 0 and draw the sign diagram to find, if

present, the time when the particle changes direction

3. Find 𝑠(𝑡) (the constant of integration, c, may be included)

4. Find 𝑠(𝑡2), 𝑠(𝑡1) and 𝑠(𝑡) for each time the direction changes

5. Draw a motion diagram

6. Identify + calculate the total distance travled

Finding the specific value of velocity/acceleration

1. Find the function (𝑣(𝑡) or 𝑎(𝑡))

2. Plug in the specific time value


By Steven Zhou 43

3. Solve for 𝑣(𝑡) or 𝑎(𝑡)

Finding the time when displacement relative to initial

position/velocity/acceleration equals to a specific value

1. Find the function (𝑣(𝑡), 𝑎(𝑡) or 𝑠(𝑡))

2. Plug in the specific initial position/velocity/acceleration value

3. Solve for 𝑡

TIP: “Stationary” means 𝑣(𝑡) = 0; “At the origin"/”initial position” means

𝑠(𝑡) = 0

Finding the time when the particle reverses/changes direction

1. Find 𝑣(𝑡)

2. Find value of 𝑡 when 𝑣(𝑡) = 0

3. Draw a sign diagram – maxima/minima point is where the particle

reverses direction

Finding the time when the particle is at its maxima/minima displacement

1. Find 𝑣(𝑡)

2. Find value of 𝑡 when 𝑣(𝑡) = 0 and draw a sign diagram to find the

maxima/minima

Finding the time when the particle is at its maxima/minima velocity

1. Find 𝑎(𝑡)

2. Find value of 𝑡 when 𝑎(𝑡) = 0 and draw a sign diagram to find the

maxima/minima

𝑥– 𝑡 (𝑠-𝑡) and 𝑣– 𝑡 graphs

displacement-time graph (𝑥– 𝑡 or 𝑠-𝑡 graph)

slope=velocity

velocity-time graph (𝑣– 𝑡 graph)

area under the line=distance traveled

slope=acceleration


By Steven Zhou 44

Appendix A: Formula Sheet

Appendix B: Operations on graphs

𝑓(−𝑥): reflection in the 𝑦-axis

−𝑓(𝑥): reflection in the 𝑥-axis

𝑓(𝑥)+𝑎: translation of 𝑎 units

parallel to 𝑦-axis

𝑓(𝑥 + 𝑎): translation of – 𝑎 units

parallel to 𝑥-axis

𝑓(𝑎𝑥): stretch, scale factor 1/𝑎

parallel to 𝑥-axis

𝑎𝑓(𝑥): stretch, scale factor 𝑎 parallel

to 𝑦-axis

1/ 𝑓(𝑥): every point become (1/x,

1/y)


By Steven Zhou 45

Acknowledgement

This work is licensed under a Creative Commons Attribution-

NonCommercial 4.0 International License.

Some Information and diagrams come from:

YK Pao school’s Teacher’s resources (worksheets, PowerPoints etc.)

Cambridge Assessment International Education (CAIE) International General

Certificate of Secondary Education (IGCSE) Additional Mathematics (0606)

Syllabus for examinations from 2020

CAIE IGCSE Additional Mathematics Past papers & corresponding mark schemes

(2013-2016)

Haese Mathematics’ Textbook

http://creativecommons.org/licenses/by-nc/4.0/



IGCSE Additional Mathematics (0606) Revision Notes

Documents

Transcript of IGCSE Additional Mathematics (0606) Revision Notes