Earnest Wong Derek Yim

MATHEMATICSIBDP

STUDY • PRACTICE • SUCCEED

Analysis and Approaches

2Practice tests

included

IB 360

Chapter 1 Reasoning and proof1.1 Introduction 2

1.2 Proving identities 4

Chapter 2 Exponents and logarithms2.1 Surds 14

2.2 Integer exponents 18

2.3 Standard form (scientific notation) 20

2.4 Rational exponents 22

2.5 Logarithms with base 10 and base e 27

2.6 Logarithms with other bases 30

2.7 Logarithm laws 32

2.8 Solving exponential equations without using logarithms 37

2.9 Solving exponential equations using logarithms 38

Chapter 3 Sequences and series3.1 Arithmetic sequences and series 46

3.2 Geometric sequences and series 50

3.3 Sigma notation 56

3.4 Applications 59

Chapter 4 The binomial theorem4.1 Pascal’s triangle 72

4.2 Factorials and nCr 73

4.3 The binomial theorem 76

4.4 The general term in a binomial expansion 79

Chapter 5 Gradients and straight lines5.1 Distance, mid-point and gradient 86

5.2 Equation of a straight line 91

Chapter 6 Functions6.1 Key function concepts 106

6.2 Composite functions 114

6.3 Inverse functions 116

Chapter 7 Quadratic functions7.1 Quadratic functions and their graphs 136

7.2 The quadratic formula 142

Chapter 8 Reciprocal functions and rational functions8.1 The reciprocal function 152

8.2 Rational functions 153

Topic 1 Number and algebra

Topic 2 Functions

C o n te n ts

Chapter 9 Exponential and logarithmic functions9.1 Exponential functions 160

9.2 Logarithmic functions 161

9.3 Relationship between exponential and logarithmic functions 162

9.4 Applications 166

Chapter 10 Solving equations by technology10.1 Solving polynomial equations, with and without technology 172

10.2 Use of technology to solve equations 174

10.3 Applications of graphing skills and solving equations related to real-life situations 176

Chapter 11 Transformations of graphs11.1 Translations 180

11.2 Reflections 182

11.3 Stretches 183

11.4 Composite transformations 185

Chapter 12 Right-angled and non-right-angled trigonometry12.1 Right-angled trigonometry 194

12.2 Non-right-angled trigonometry 198

12.3 Applications of trigonometry 210

Chapter 13 Geometry13.1 Distance and mid-point between two points 220

13.2 Surface areas and volumes of three-dimensional solids 222

13.3 Angles between lines and planes 229

Chapter 14 Radian measure, arc length and sector area14.1 Radian angle measure 238

14.2 Arc length and sector area 240

Chapter 15 Trigonometric functions, identities and equations15.1 Sine, cosine and tangent functions 250

15.2 Trigonometric identities 259

15.3 Trigonometric function graphs 263

15.4 Trigonometric equations 273

Chapter 16 Sampling and statistics16.1 Basic concepts 284

16.2 Reliability of data sources, sampling techniques and sampling bias 285

16.3 Measures of central tendency 288

16.4 Measures of location and dispersion 293

16.5 Transformations of data sets 297

16.6 Graphical presentations of data 300

Chapter 17 Bivariate statistics17.1 Bivariate data and linear correlation 318

17.2 Meaning of the parameters and prediction from the equation of the regression line 323

Topic 3 Geometry and trigonometry

Topic 4 Statistics and probability

Chapter 18 Probability18.1 Introduction to probability 338

18.2 Venn diagrams, combined events and mutually exclusive events 341

18.3 Tables and tree diagrams, conditional probabilities and independent events 346

Chapter 19 Discrete random variables and binomial distribution19.1 Discrete random variables 366

19.2 Binomial probability distributions 372

Chapter 20 Normal distribution20.1 Properties of normal probability distributions 382

20.2 Calculator functions and normal curves 385

20.3 The standard normal distribution 390

Chapter 21 Limits and rules of differentiation21.1 Limits of a function 398

21.2 Introduction to differentiation 399

21.3 Further differentiation rules 409

21.4 Higher order derivatives 415

Chapter 22 Applications of differentiation22.1 Tangents and normals 420

22.2 Increasing and decreasing functions 424

22.3 Sign diagrams 425

22.4 Local maximum and minimum points 427

22.5 Concavity of functions 430

22.6 Graphical behaviour of functions 437

22.7 Optimization 443

Chapter 23 Rules of integration and indefinite integrals23.1 Anti-differentiation and indefinite integrals 452

23.2 Integration by inspection and substitution 462

Chapter 24 Definite integrals and applications of integration24.1 Definite integrals 472

24.2 Applications of integration 477

Chapter 25 Kinematics25.1 Terminology 494

25.2 Properties 495

Practice Test 1 Paper 1 & 2 509

Practice Test 2 Paper 1 & 2 529

Answers and markschemes can be downloaded for free at www.ntk.edu.hk.

Topic 5 Calculus

2.1 Introduction

How would you prove that the following equation is true for all values of the unknown x?

x x x1 2 12 2( )+ = + +

If we substitute x = 1 we find that the value on the left hand side is equal to the value on the right hand side. We get a similar outcome if we successively substitute x = 2, x = 3, x = 4, x = 0, x = −4 and so on. This gives us good evidence that the left side is identical to the right side for all values of x. However, no mathematician would agree that the two sides are actually the same based on these substitutions.

What makes mathematics different from other subjects is that before a conjecture is accepted, there has to be a logical proof of it.

Of course, in the situation above, we could prove that the left side is always equal to the right side by expanding x 12( )+

and then simplifying using the accepted methods of algebra.

In this chapter, we will consider the difference between equations and identities, followed by techniques for proving identities.

Equations and identities

Some algebraic formulas work for a limited number of values only. For example, consider this equation:

x3 2 5+ =

This equation is true for x = 1 but not for x = 2 or x = 3. In fact, it is true only for x = 1.

Now consider this equation:

x x x2 42( )+ = +

You will find that the equation is true for x = 2, but false for all other values of x.

Some equations may have no solution at all. Consider the equation:

x1 0=

There is no numerical value that can be substituted for the unknown x to make this true.

1.1

Chapter 1Reasoning and proof

Number and algebra2

Finally, consider this equation:

x x x x2 22( )+ = +

This time you will find that any value of x you pick will work. This is because x x x x2 22( )+ = + is not an equation, but an identity. In an algebraic identity, the left hand side (LHS) and the right hand side (RHS) are identical, no matter what value is substituted for the variable. For an equation, we use the symbol ‘=’ to separate the LHS and the RHS; for an identity, we use the symbol ‘≡’ instead.

In summary:

An equation (using =) is true for only a limited number of values of the unknown(s) (or possibly not true for any value at all).

An identity (using ≡) is true for any value of the unknown(s).

Example 1-1

Decide whether x x x6 2 3 2 2 3( ) ( )− − = + might be an identity by substituting for the unknown. Comment on your results.

SolutionWe will replace x with some arbitrarily chosen numerical values.

When x = 0,

LHS 6 0 2 0 3 6( ) ( )= − − =

RHS = 2 2 0( ) + 3 = 6

The statement is true when x = 0.

When x = 1,

LHS 6 1 2 1 3 10( ) ( )= − − =

RHS = 2 2 1( ) + 3 = 10

The statement is true when x = 1.

When x = 31,

LHS 6 31 2 31 3 130( ) ( )= − − =

RHS = 2 2 31( ) + 3 = 130

The statement is true when x = 31.

Chapter 1 Reasoning and proof 3

When x = −4,

LHS 6 4 2 4 3 10( ) ( )= − − − − = −

RHS = 2 2 4( ) + 3 = 10

The statement is true when x = −4.

From the results of our substitutions we have a reasonable amount of evidence to claim that the statement is an identity. However, we cannot state definitely that it is an identity as we do not yet have a logical proof.

1.2 Proving identities

Reducing the LHS to the RHS or the RHS to the LHS

A common method of proving a proposed identity is to start with the expression on one side of the identity and, using standard rules of algebra, eventually arriving at the expression which is on the other side.

Example 1-2

Prove the identity x x x6 2 3 2 2 3( ) ( )− − ≡ + .

Solution

x x

x xx

x

LHS 6 2 3

6 2 64 62 2 3

RHS

( )

( )

≡ − −

≡ − +≡ +≡ +

≡

Therefore, x x x6 2 3 2 2 3( ) ( )− − ≡ + .

Although we should indicate identities and steps in an identity proof with the ‘≡’ symbol, teachers, books, online sites, and even IBSL mathematics papers will frequently use the ‘=’ sign instead. We will continue to use ‘≡’ when stating identities, but will follow the common practice of using ‘=’ for the steps in the proof.

1.2

Number and algebra4

Example 1-3

Prove the identity x x x x

11 2

11

12( )( )+ +

≡+

−+

.

SolutionIn this case, it is easier to prove the identity if we start from the RHS and finish the proof with the expression on the LHS.

x xx xx x

x xx x

x x

RHS 11

12

2 11 2

2 11 2

11 2

LHS

( ) ( )( )( )

( )( )

( )( )

=+

−+

=+ − ++ +

=+ − −+ +

=+ +

=

Therefore, x x x x

11 2

11

12( )( )+ +

≡+

−+

.

Example 1-4

Prove the identity x x x x2 8 4 22 ( )( )− − ≡ − + .

Solutionx x

x x xx x

RHS 4 2

2 4 82 8

LHS

2

2

( )( )= − +

= + − −= − −=

Therefore, x x x x2 8 4 22 ( )( )− − ≡ − + .

Reducing both the LHS and the RHS to a common expression

Sometimes it can be difficult to get from the RHS to the LHS or from the LHS to the RHS. In these situations we try to simplify each side down to a common expression.

Chapter 1 Reasoning and proof 5

Example 1-5

Prove the identity x y x y x y x y5 7 2 2 5 4 2 3( ) ( ) ( ) ( )− + + ≡ − + + .

Solution

x y x y

x y x yx y

LHS 5 7

5 5 7 712 2

( ) ( )= − + +

= − + += +

x y x y

x y x yx y

RHS 2 2 5 4 2 3

4 10 8 1212 2

( ) ( )= − + +

= − + += +

Since the result from simplifying the LHS is equal to that from simplifying the RHS, we have shown that both sides are equivalent to each other, so the identity is proved.

Interpretations and applications of proofs

Example 1-6

Prove that n n n2 1 2 1 8 22 2 2( ) ( )− + + ≡ + , where n ∈ .

Hence, or otherwise, prove that the sum of the squares of any two consecutive odd numbers is even.

Solution

n n

n n n nn

LHS 2 1 2 1

4 4 1 4 4 18 2RHS

2 2

2 2

2

( ) ( )= − + +

= − + + + += +=

n n n2 1 2 1 8 22 2 2( ) ( )∴ − + + ≡ +

When n is an integer, 2n − 1 and 2n + 1 represent two consecutive odd numbers.

Since n n n n2 1 2 1 8 2 2 4 12 2 2 2( )( ) ( )− + + ≡ + ≡ + , we see that 2 is a factor of n8 22 + , as

well as a factor of n n2 1 2 12 2( ) ( )− + + . Therefore, we can conclude that the sum of the

squares of any two consecutive odd numbers is even.

Number and algebra6

Example 1-7

Prove that n n n n1 2 2 12 2 2( )+ + ≡ + + , where n ∈ .

Hence, or otherwise, prove that the sum of the squares of any two consecutive integers is odd.

Solution

n n

n n nn n

LHS 1

2 12 2 1RHS

2 2

2 2

2

( )= + +

= + + += + +=

n n n n1 2 2 12 2 2( )∴ + + ≡ + +

When n is an integer, n and n + 1 represent two consecutive numbers.

Since n n n n n n1 2 2 1 2 12 2 2 2( )( )+ + ≡ + + ≡ + + , n n 12 2( )+ + can be expressed as the sum of an even number and 1. As the sum of any even number and 1 must be an odd number, we can conclude that the sum of the squares of any two consecutive integers is odd.

Example 1-8

Prove that a b a b a b2 2( )( )+ − ≡ − .

Hence, evaluate 55 452 2− .

Solution

a b a b

a ab ab ba b

LHS

RHS

2 2

2 2

( )( )= + −

= − + −= −=

a b a b a b2 2( )( )∴ + − ≡ −

( )( ) ( )( )−

= + − − = + −

= ×=

a b a b a b55 4555 45 55 45 (As we proved that )

100 101000

2 2

2 2

Chapter 1 Reasoning and proof 7

Exercise 1A 1. Determine which of the following are equations and which are identities:

a x3 2 6− = b x x4 6 3 7( )+ = −

c x x3 3+ = − d x x x x4 42 ( )+ = +

e 3x2 + bx + c = 3 x2 + b3

x + c3

f x x5 10 5 2 1( )− = −

g a b a ab b22 2 2( )+ = + + h x x2 3 4 9

2 2( )+ = +

i x x x2 3 4 12 92 2( )+ = − + j x x x2 3 4 12 9

2 2( )+ = + +

k a b a b a b2 2( )( )− + = − l x x x x3 3 6 92( )( )− + = − +

2. a Show that the values y = 1, y = 4 and y = 6 all satisfy the equation:

y y y8 4 5 12 3 28( ) ( )− − − = +

b Prove the identity:

y y y8 4 5 12 3 28( ) ( )− − − ≡ +

3. a Show that the values x = 1, x = 2 and x = 4 all satisfy the equation:

x x x7 14 83 2− + =

b Show, by substituting the value x = 6, that x x x7 14 83 2− + = is not an identity.

4. a Show that the values p = 8 and q = 4 satisfy the equation:

p q p q p q p q6 4 3 3 7( ) ( ) ( ) ( )+ + − = + + −

b Show that the values p = 10 and q = −5 also satisfy the equation in a.

c Prove the identity:

p q p q p q p q6 4 3 3 7( ) ( ) ( ) ( )+ + − ≡ + + −

5. Prove the following identities:

a a b a ab b22 2 2( )+ ≡ + +

b a b a a b ab b3 33 3 2 2 3( )+ ≡ + + +

c a b a a b a b ab b4 6 44 4 3 2 2 3 4( )+ ≡ + + + +

d a b a b a ab b3 3 2 2( )( )+ ≡ + − +

Number and algebra8

6. Prove the identity:

b x y a x y y a b x a b( ) ( ) ( ) ( )− + + ≡ − + +

7. Prove the identity:

x x x x2 5 3 2 1 32 ( )( )− − ≡ + −

8. a Expand and simplify x x3 8( )( )− + .

b Expand and simplify x x6 4( )( )− − .

c Hence, prove the identity:

x x x x x3 8 6 4 15 48( )( ) ( )( )− + − − − ≡ −

9. a Expand and simplify x x2 3 1( )( )− + .

b Expand and simplify x x2 6 3 4( )( )− − .

c Hence, expand and simplify x x x x2 3 1 2 6 3 4( )( )( )( )− + − − .

d Hence, prove the identity:

x x x xx x

x x12 58 56 54 722 6 3 4

2 34 3 2

2

( )( )− + + −

− −≡ − − provided x 4

3 or 3≠

10. a Prove the identity:

x y x y x y2 2 ( )( )− ≡ − +

b Hence, calculate the following without the use of a calculator:

i 71 702 2− ii 201 1912 2−

iii 646 3542 2− iv 999.99 0.012 2−

Chapter 1 Reasoning and proof 9

Equations

A given equation will be true for only a limited number of values of the unknown(s), or for no values at all. We use the symbol ‘=’ for equations.

Identity

A given identity is true for all substituted values of the unknown(s). We frequently use the symbol ‘≡’ for identities.

Proving identities

■ From LHS to RHS (or vice versa)

LHS to RHS (or RHS to LHS) proofs require us to begin with the left hand side expression and transform it to the right hand side expression (or vice versa) using correct algebraic steps.

■ From LHS and RHS at the same time

If the result derived from LHS is equal to that from RHS, we can say that the original expression on the LHS is equal to the original expression on the RHS, so proving the identity.

Summary

Number and algebra10

Review Set 1

Short Questions

1. a Show that n n n n2 2 2 8 8 42 2 2( ) ( )+ + ≡ + + , where n ∈ . [2 marks]

b Hence, or otherwise, prove that the sum of the squares of any two consecutive even numbers is even. [3 marks]

2. a Show that n n n n n2 2 2 16 24 24 83 3 3 2( ) ( )+ + ≡ + + + , where n ∈ . [3 marks]

b Hence, or otherwise, prove that the sum of the cubes of any two consecutive even numbers is divisible by 8. [3 marks]

3. a Show that n n n n n2 1 2 1 2 3 12 12 112 2 2 2( ) ( ) ( )− + + + + ≡ + + , where n ∈ .

[3 marks]

b Hence, or otherwise, prove that the sum of the squares of any three consecutive odd numbers is odd. [3 marks]

4. a Show that n n n n3 3 3 18 18 92 2 2( ) ( )+ + ≡ + + , where n ∈ . [2 marks]

b Hence, or otherwise, prove that the sum of the squares of any two consecutive multiples of 3 is divisible by 9. [3 marks]

5. a Show that 3x2 4x + 9 3 x 23

2

+ 233

. [3 marks]

b Hence, prove that ( ) ( )+ − − +x x x3 7 1 22 is always positive. [4 marks]

6. a Show that ( )( )− ≡ + −x y x y x y2 4 2 2 . [2 marks]

b Hence, evaluate −181 92 4. [3 marks]

7. a Show that ( )( )− ≡ − + +a b a b a ab b3 3 2 2 . [3 marks]

b Hence, evaluate −54 443 3. [4 marks]

Chapter 1 Reasoning and proof 11

8. a Show that ( ) ( )+ − ≡ +n n n2 2 2 8 42 2

, where ∈n . [2 marks]

b Hence, or otherwise, prove that the difference of the squares of any two consecutive even numbers is divisible by 4. [3 marks]

c Expand and simplify ( ) ( )+ − −n n2 1 2 12 2

. [2 marks]

d Hence, determine if the following statement is true:

‘ The difference of the squares of any two consecutive odd numbers is divisible by 4.’ [3 marks]

9. a Show that ( ) ( )+ − − ≡k k k2 1 2 1 82 2

. [3 marks]

b Hence, or otherwise, find two square numbers that have a difference of:

i 48

ii 160 [4 marks]

10. a Show that ( ) ( )+ − − ≡ +k k k1 1 6 23 3 2 . [3 marks]

b Hence, or otherwise, find two positive cube numbers that have a difference of:

i 56

ii 152 [4 marks]

Number and algebra12