iv

Non-repeated quadratic factor:

px qx rax b x c

Aax b

Bx Cx c

2

2 2 2 2+ +

+( ) +( ) =+( ) + +

+( )

Trigonometry

sin sin cos cos sinA B A B A B±( ) = ±

cos( ) cos cos sin sinA B A B A B± =

tan( ) tan tantan tanA B A B

A B± = ±1

sin sin cos2 2A A A=

cos cos sin sin cos2 1 2 2 12 2 2 2A A A A A= - = - = -

tan tantan

2 21 2A A

A=

-

sin sin sin cosP Q P Q P Q+ = +( ) -( )2 12

12

sin sin cos sinP Q P Q P Q- = +( ) -( )2 12

12

cos cos cos cosP Q P Q P Q+ = +( ) -( )2 12

12

cos cos sin sinP Q P Q P Q- = - +( ) -( )2 12

12

Principal values:

- ≤ ≤-12

12

1p psin x x ≤( )1

0 1≤ ≤-cos x p x ≤( )1

- < <-12

12

1p ptan x

Derivatives

f( )x f ( )' x

sin-1 x 11 2- x

cos-1 x --1

1 2x

389_AL025_ALevelMathsComp_Guide-FinalPass 17/06/19

v

tan-1 x 11 2+ x

cosec x - cosec cotx xsec x sec tanx x

Integrals (Arbitrary constants are omitted; a denotes a positive constant.)

f( )x f( )x x d∫1

2 2x a+ 1 1

axatan- ( )

12 2a x-

sin- ( )1 xa x a<( )

12 2x a-

12a

x ax aln -

+( ) x a>( )

12 2a x-

12a

a xa xln +

-( ) x a<( )tan x ln sec x( ) x <( )1

2 p

cot x ln sin x( ) 0 < <( )x p

cosec x - +( )ln cos cotecx x 0 < <( )x p

sec x ln sec tanx x+( ) x <( )12 p

Vectors

The point dividing AB in the ratio λ m: has position vector m λλ ma b+

+

Vector product:

a b× =

×

=

---

aaa

bbb

a b a ba b a ba b a

1

2

3

1

2

3

2 3 3 2

3 1 1 3

1 2 22 1b

389_AL025_ALevelMathsComp_Guide-FinalPass 17/06/19

CONTENTS1 Function 12 Inequalities 83 System of Linear Equations 154 Sequences 175 Series 186 Arithmetic and Geometric Progressions 247 Curve Sketching 288 Transformations 339 Parametric Curves 39

10 Conics 4111 *Binomial Expansion 4512 Differentiation 5013 Application of Differentiation 5314 Maclaurin’s Series 6015 Small Angles Approximations 6316 Integration 6417 Application of Integration 7418 Differential Equations 8019 Vectors 8520 Complex Numbers 10321 Permutations & Combinations 11222 Probability 11623 Discrete Random Variable and Binomial Distribution 12024 Normal Distribution 13025 Sampling 13526 Hypothesis Testing 13827 Correlation and Regression 14228 Examination Tips 149

Pg. No.

389_AL025_ALevelMathsComp_Guide-FinalPass 17/06/19

1A Level Maths Comprehensive Guide

What is a function?

Every input (x) has only one output (y).

Use Vertical line test to check for functions.Two functions with the same equation are different as long as their domains are different.

Finding range of function

Plot graph on GC and using domain of graph, find its corresponding set of y values.

Inverse Functions

Inverse functions are 1–1 functions.

Use Horizontal line test to check for 1–1 functions.‘Since every horizontal line y k= , k R∈ f cuts the graph of y x= f( ) at exactly one point, f is 1–1.’

Function 1389_AL025_ALevelMathsComp_Guide-FinalPass 17/06/19

© Fairfield Book Publishers2 Function

Finding Inverse Function

Set y x= f( ) and make x the subject.

Example

The function f is defined by f : ,x x x -( ) + ≤3 1 32 .

Show that the inverse function of f exists and find f -1 in similar form.

Solution:

Since every horizontal line y k k= ≥, 1 cuts the graph y x= f( ) exactly once, f is 1-1 and f -1 exists.Let y x= - +( )3 12

( )x yx y- = -

⇒ = ± -

3 13 1

2

Since x x y≤ = - -3 3 1,

f : , .- - - ≥1 3 1 1x x x

Notes when finding inverse

׀׀ If a function is a quadratic equation in terms of x, either use completing the square or apply the formula for finding roots.׀׀ When you apply square root in this case, remember to add ± . Choose the sign by looking at the domain of f׀׀ D R R Df f f f,- -= =1 1

Notes

׀׀ y x= -f ( )1 , y x= f( ) intersects y x= at the same point.׀׀ f f( )- =1 x x and f f ( )- =1 x x by definition. There is no need to prove this. However, the two functions may be different if their domains are different.

389_AL025_ALevelMathsComp_Guide-FinalPass 17/06/19

3A Level Maths Comprehensive Guide

Composite Functions

g f( ) g(f( ))x x=

g acts on the output of fFacts about composite functionsŵ׀ gf exists when R Df g⊆

ŵ׀ D Dgf f=

ŵ׀ R Rggf ⊆

Maximal domain of f for gf to exist

Find range of f such that R Df g⊆ . Use the graph of f to find the corresponding domain.

Example

The functions f and g are defined byf e: x x

3 - , x ∈

g : ( )x x - +1 22 , x ∈

(i) Sketch the graph of y = f(x), and show that f does not have an inverse.

(ii) The function f has an inverse if its domain is restricted to x b≥ . State the smallest possible value of b and find the domain of g such that the composite function fg exists.

Solution:

(i)

D D

g

g

f

fR , R ,= =

= ∞[ ) = ∞[ )

1 2Min point of f(x) is (3, 1)

389_AL025_ALevelMathsComp_Guide-FinalPass 17/06/19

© Fairfield Book Publishers4 Function

Any horizontal line y k= , where k > 1, cuts the graph more than once. This implies that f is not a 1–1 function and thus f -1 does not exist.

(ii) From graph, we can deduce that the smallest possible value of b is 3.For fg to exist, Rg ⊆ Df.Now, R Dg f= ∞ ⊆ = ∞[ , ) [ , ).2 3Hence, restricted Rg = ∞[ , ).3

3 1 2

1 10

2

2

= -( ) +

-( ) ==

x

xx or 2.From the graph of g, maximal domain of g is ( , ] [ , ).-∞ ∪ ∞0 2

Example

The functions f and g are defined as follows:f : x x x x� �+( ) -( ) ∈1 12 , ;

g : x x x x x� �2 2 1+ ∈ < -, , .Show that the composite function fg exists, and define fg in a similar form.

Solution:

g( ) , ,( )

x x x x xx

= + ∈ < -= + -

2

2

2 11 1

∴ = - ∞Rg ( , )1

R Dfg exists

g f= - ∞ ⊆ =⇒

( , )1

fg: f(x x x xx x x x x

2

2 2

2 12 1 2 12 1+ < -

= + + + - < -),

( )( ),

Find range of gf

ŵ׀ Find range of f.ŵ׀ On the graph of g, put range of f as the domain and find the corresponding range of g.

ORŵ׀ Sketch the graph of gf.ŵ׀ Find the range of gf based on the domain of gf (i.e. domain of f).

389_AL025_ALevelMathsComp_Guide-FinalPass 17/06/19

5A Level Maths Comprehensive Guide

Example

The functions f and g are defined by

f : ,x xx x→ +

-- ≤ ≤2

2 2 1 ,

g : ,x x x x→ + + ∈2 2 2 .Show that the composite function gf exists and find the range of gf.

Solution:

Using GC, Rf = [ ]0 3, and Rg = ∞[ )1,

gf exists if R Df g⊆

Since 0 3,[ ] = ⊆ =R Dgf

∴gf exists-[ ] → [ ] → [ ]2 1 0 3 2 17, , ,f g

Rgf = [ ]2 17,

Example

The functions f and g are defined asf : x x 1 - for x ≤ 1,

g e: x x

- -1 for x > 0.

(i) Define f -1 in a similar form, including its domain. (ii) State the relationship between f and f -1 , and sketch the graphs of f

and f -1 on the same diagram. (iii) Find the exact solutions of the equation f f( ) ( )x x= -1 .

Solution:

(i) Let y x x= = -f ( ) 1 , x ≤ 1⇒ 1 2- =x y

⇒ x y= -1 2

⇒ f 1- ( )x x= -1 2

D Rf f- = =1 0,∞[ )Thus, f - -1 21: x x , x ≥ 0

1 x

y

y = f(x)

−2

3

Note

׀׀ Range of f is now the new domain of g. Hence, the corresponding new range is the range of gf.

389_AL025_ALevelMathsComp_Guide-FinalPass 17/06/19

© Fairfield Book Publishers6 Function

(ii) The graph of f -1 is the reflection of the graph of f in the line y x= .

(iii) f f( ) ( )x x= -1

⇒ f ( )x x=

⇒ 1 - =x x⇒ x x2 1= -

⇒ x x2 1 0+ - =

⇒ x = - ± - -1 1 4 122 ( )

⇒ x = - +1 52 or x = - -1 5

2 (rejected since 0 ≤ x ≤ 1)

From the diagram in part (ii), other solutions are x = 0 and x = 1.

Hence, the solutions are of the equation are 0, - +1 52 1, .

Example

The function h is defined by

h for for 3

x x xx x

( ) - < ≤- < ≤

12 0 33 6 6

2 ..

and h(x) = h( x - 6 ) for all x ∈. (i) Find h(16) + h(25). (ii) Sketch y = h(x) for - ≤ ≤6 12x .

Solution:

(i) h(16) + h(25) = h(4) + h(1) = (12 – 6) + (12 – 1) = 17

y

x

1

1

f

f−1

Note

׀׀ This is a piecewise function.

389_AL025_ALevelMathsComp_Guide-FinalPass 17/06/19

7A Level Maths Comprehensive Guide

(ii)

Example

The function f is defined by f : , .x xx x→ -

->2 3

3 3

Define, in a similar manner, the inverse function f-1 and show that f2 x x( ) = .

Hence determine f13 in a similar manner.

Solution:

f x xx x( ) = -

->2 3

3 3, ,

Let y xx= -

-2 33

3 2 3-( ) = -x y x

3 2 3-( ) = -y x y

x yy= -

-2 33

∴ f - → --

>1 2 33 3: ,x x

x x

f f f2 x x( ) = ( )

= ( ) -f f 1 x ( f = f -1 and f f = f f = - -1 1 x )

= xf f f13 12x x( ) = ( ) = ( )f x

= --

2 33

xx

y

x−6 0

3

12

6 12

Note

׀׀ Recognise the pattern generated.

389_AL025_ALevelMathsComp_Guide-FinalPass 17/06/19

© Fairfield Book Publishers8

2Inequalities

Solving polynomial inequalities

To solve inequalities involving polynomial, bring all the terms to one side and factorise them. After which, sketch the graph of factorised polynomials and solve the inequalities.

Sketching graphs of factorised polynomials

For the factorised polynomial of y x a x b x c= -( ) -( ) -( )2 3 where a, b and c are constants, the following can be observed,

(i) x- intercept at single root i.e. ( )x a-

(ii) turning point at even powered root i.e. ( )x b- 2

(iii) stationary point of inflexion at odd powered root i.e. ( )x c- 3

Example

Solve the inequality ( ) ( )( )x x x+ - + ≤1 2 3 4 02 3

Solution:

∴ ≤ - ≥x x34 2 or

Solving inequalities involving rational functions

Rational functions are functions of the form f( )g( )

xx where both f( )x and g( )x

are polynomials.

−1 2− 34

x

389_AL025_ALevelMathsComp_Guide-FinalPass 17/06/19

9A Level Maths Comprehensive Guide

Polarity of denominator unknown

When the polarity of denominator of a rational function is unknown, multiply throughout the inequality by the lowest even power of the denominator so that the direction of the inequality sign will not be affected.

Example

Solve the inequality x x

x-( ) +( )

+( ) ≤1 2

3 40

Solution:

x xx

x-( ) +( )

+( ) ≤ ≠ -1 2

3 40 4

3,

Multiplying throughout by 3 4 2x +( ) ,

⇒+( ) -( ) +( )

+( ) ≤ +( ) ⋅

⇒ +( ) -( ) +( ) ≤

⇒ ≤ -

3 4 1 23 4

3 4 0

3 4 1 2 0

2

22x x x

xx

x x x

x oor - < ≤43 1x

−2 1− 43

389_AL025_ALevelMathsComp_Guide-FinalPass 17/06/19