Analysis SPM Mathematics Form 4&5

PENERBIT ILMU BAKTI SDN. BHD.

ii Contents Formulae iv Formative Practice • Forms 1, 2 & 3 R1 Answers R29 FORM 4 Chapter 1 Quadratic Functions and Equations in One Variable 1 HOTS Zone 7 SPM Practice 1 8 Answers 13 Chapter 2 Number Bases 14 HOTS Zone 18 SPM Practice 2 18 Answers 22 Chapter 3 Logical Reasoning 23 HOTS Zone 30 SPM Practice 3 31 Answers 37 Chapter 4 Operations on Sets 40 HOTS Zone 53 SPM Practice 4 53 Answers 65 Chapter 5 Network in Graph Theory 68 HOTS Zone 73 SPM Practice 5 74 Answers 80 Chapter 6 Linear Inequalities in Two Variables 81 HOTS Zone 85 SPM Practice 6 86 Answers 91 Chapter 7 Graphs of Motion 95 HOTS Zone 101 SPM Practice 7 101 Answers 106 Chapter 8 Measures of Dispersion for Ungrouped Data 107 HOTS Zone 114 SPM Practice 8 114 Answers 117 Chapter 9 Probability of Combined Events 119 HOTS Zone 125 SPM Practice 9 126 Answers 129 Chapter 10 Consumer Mathematics: Financial Management 130 HOTS Zone 132 SPM Practice 10 132 Answers 135 FORM 5 Chapter 1 Variation 136 HOTS Zone 139 SPM Practice 1 139 Answers 143 Analysis SPM Maths 2023 Eng Con 3rd.indd 2 17/2/2023 9:10:03 PM PENERBIT ILMU BAKTI SDN. BHD.

iii Chapter 2 Matrices 144 HOTS Zone 148 SPM Practice 2 148 Answers 153 Chapter 3 Consumer Mathematics: Insurance 155 HOTS Zone 161 SPM Practice 3 161 Answers 165 Chapter 4 Consumer Mathematics: Taxation 166 HOTS Zone 176 SPM Practice 4 178 Answers 182 Chapter 5 Congruency, Enlargement and Combined Transformations 183 HOTS Zone 190 SPM Practice 5 191 Answers 203 Chapter 6 Ratios and Graphs of Trigonometric Functions 206 HOTS Zone 218 SPM Practice 6 218 Answers 223 Chapter 7 Measures of Dispersion for Grouped Data 225 HOTS Zone 239 SPM Practice 7 240 Answers 244 Chapter 8 Mathematical Modeling 248 SPM Practice 8 251 Answers 254 SPM Model Test 255 Answers 267 Analysis SPM Maths 2023 Eng Con 3rd.indd 3 17/2/2023 9:10:04 PM PENERBIT ILMU BAKTI SDN. BHD.

1 (b) If a vertical line which is parallel to the y-axis intersects the curve only once, the curve is a graph of a function. (c) If the vertical line intersects the curve more than once, then the curve is not a graph of a function. SMART TIP (a) y y = x² – 6x + 5 x O 1 5 (3, –4) 5 Vertical line Since the vertical line intersects the graph of the curve at only one point, then y = x2 – 6x + 5 is a function. The function is a quadratic function. (b) O y y = 2 x x Vertical line Since the vertical line intersects the graph of the curve at more than one point, then y = ±2 x is not a function. The effects of the change of the values of a, b and c to a graph of a quadratic function f(x) = ax2 + bx + c 5 The values of a, b and c in a quadratic function f(x) = ax2 + bx + c: 1.1 Quadratic Functions and Equations Quadratic expressions 1 A quadratic expression is an expression in the form ax2 + bx + c, a ≠ 0, where a, b and c are constants and x is a variable. SMART TIP • A quadratic expression must have only one variable (can be represented by any letter). • The highest degree of the variable is 2. Quadratic functions SMART TIP A function is a type of special relation such that each object in the domain has only one image in the codomain. 2 A quadratic function is f(x) = ax2 + bx + c, a ≠ 0, where a, b and c are constants and the highest power of the variable x is 2. 3 Graphs of quadratic functions (a) The graph of the quadratic function f(x) = ax2 + bx + c, a ≠ 0 is a symmetrical smooth curve known as a parabola. (b) If the coefficient of x2 is positive, that is a > 0, the shape of the graph of the quadratic function is a parabola with the shape of ∪. (c) If the coefficient of x2 is negative, that is a < 0, the shape of the graph of the quadratic function is a parabola with the shape of ∩. 4 Vertical line test (a) Vertical line test is a graphical method that can be used to determine whether a curve is a graph of a function or not. Quadratic Functions and Equations in One Variable Chapter 1 Learning Area: Relationship and Algebra EXPRESS NOTES 1 Analysis SPM Maths 2023 Eng F4 C1 3rd.indd 1 17/2/2023 9:11:23 PM PENERBIT ILMU BAKTI SDN. BHD.

2 Form 4 CHAPTER 1 (a) If the value of a is positive, the graph of the quadratic function is concave upward. The more positive the value of a, the shape of the graph becomes steeper such that its parabolic shape becomes narrower and vice versa. (b) If the value of a is negative, the graph is concave downward. The more negative the value of a, the shape of the graph becomes steeper such that its parabolic shape becomes narrower and vice versa. (c) The value of b determines the axis of symmetry of the graph of the quadratic function such that its equation is given by x = – b 2a . (d) The value of c determines the y-intercept of the graph of the quadratic function. SMART TIP Minimum point Axis of symmetry a > 0 a < 0 Minimum point Axis of symmetry Relating quadratic functions to quadratic equations 6 Quadratic equation is an algebraic equation that has only one variable such that its highest power is 2, that is ax2 + bx + c = 0, a ≠ 0, where a, b and c are constants and x is a variable. Roots of a quadratic equation 7 The values of the variables that satisfy a quadratic equation are known as the roots of the quadratic equation. Solving quadratic equations by factorisation 8 There are four methods to determine the roots of a quadratic equation ax2 + bx + c = 0, a ≠ 0, which are (a) trial and improvement method, (b) completing the square, (c) quadratic equation and (d) factorisation. Main method used. SMART TIP The roots of a quadratic equation ax2 + bx + c = 0, a ≠ 0, is the x-intercept of the graph f(x) = ax2 + bx + c. Sketching graphs of quadratic function 9 For the graph of y = ax2 + c, the curve is ∪ if a is positive and ∩ if a is negative. c is the y-intercept of the curve. Solving quadratic equation problems 10 After you have learnt how to form a quadratic function related to certain situation and hence relate it to a quadratic equation, you will now learn the method to solve various problems in our daily real-life situations which are related to the quadratic equations. Analysis SPM Maths 2023 Eng F4 C1 3rd.indd 2 17/2/2023 9:11:23 PM PENERBIT ILMU BAKTI SDN. BHD.

3 Form 4 CHAPTER 1 1.1 Quadratic Functions and Equations Example 1 State whether each of the following expressions is a quadratic expression in one variable or not. (a) 12w2 + 4w – 5 (b) 5h3 + 3h – 12 (c) p2 – 7pq + 10 Solution (a) 12w2 + 4w – 5 is a quadratic expression. This is because the highest power of the variable w is 2. (b) 5h3 + 3h – 12 is not a quadratic expression. This is because the highest power of the variable h is 3 and not 2. (c) p2 – 7pq + 10 is not a quadratic expression. This is because the expression consists of two variables, which are p and q and not one variable only. Example 2 Determine whether each of the following functions is a quadratic function or not. (a) f(x) = (3x – 2)(x + 1) (b) g(x) = x3 – x2 – 6x (c) h(x) = 6 x2 Solution (a) f(x) = (3x – 2)(x + 1) = 3x2 + x – 2 Since the highest power of x is 2, then f(x) = (3x – 2)(x + 1) is a quadratic function. (b) g(x) = x3 – x2 – 6x Since the highest power of x is 3, then g(x) = x3 – x2 – 6x is not a quadratic function. (c) h(x) = 6 x2 = 6x–2 Since the highest power of x is –2, then h(x) = 6 x2 is not a quadratic function. Example 3 Determine whether the general shape of the graph of each of the following quadratic functions is ∪ or ∩? (a) f(x) = 2x2 + 7x + 5 (b) g(x) = –x2 + 4x + 5 Solution (a) Since the coefficient of x2 for f(x) = 2x2 + 7x + 5 is positive, then the general shape of the graph of f(x) is ∪. (b) Since the coefficient of x2 for g(x) = –x2 + 4x + 5 is negative, then the general shape of the graph of g(x) is ∩. Example 4 The following diagram shows the sketch of the graph of the quadratic function f(x) = x2 + 6x + 5. y x –5 –1 O f(x) = x² + 6x + 5 Determine (a) the equation of the axis of symmetry of the graph, (b) the coordinates of the minimum point of the graph, (c) the value of a, of b and of c if the graph of f(x) = ax2 + bx + c is reflected in the x-axis, (d) the value of b if the graph of f(x) = ax2 + bx + c is reflected in the y-axis. Solution y –5 –1 O Axis of symmetry f(x) = x² + 6x + 5 x Analysis SPM Maths 2023 Eng F4 C1 3rd.indd 3 17/2/2023 9:11:24 PM PENERBIT ILMU BAKTI SDN. BHD.

4 Form 4 CHAPTER 1 (a) The average of –5 and –1 is –5 + (–1) 2 = –3 Hence, the equation of the axis of symmetry is x = –3. (b) When x = –3, y = f(–3) = (–3)2 + 6(–3) + 5 = –4 Hence, the coordinates of the minimum point are (–3, –4). (c) If the graph of f(x) is reflected in the x-axis, the sign of each term is changed, i.e. a = –1, b = –6 and c = –5. (d) If the graph of f(x) is reflected in the y-axis, the sign of b is changed, i.e. b = –6. Example 5 The following diagram shows a drawing which is fixed into a photograph frame. (x – 3) cm (4x – 5) cm 4x cm The measurements of the drawing are 4x cm × (4x – 5) cm. The frame around the rectangular drawing has a uniform width of (x – 3) cm. If the area, L(x) cm2 , of the frame around the drawing is 156 cm2 , form a (a) quadratic function, (b) quadratic equation in the general form. Solution (a) L(x) = Area of the larger rectangle – Area of the smaller rectangle = [4x + 2(x – 3)][(4x – 5) + 2(x – 3)] – 4x(4x – 5) = (4x + 2x – 6)(4x – 5 + 2x – 6) – 16x2 + 20x = (6x – 6)(6x – 11) – 16x2 + 20x = 36x2 – 66x – 36x + 66 – 16x2 + 20x = 36x2 – 16x2 – 66x – 36x + 20x + 66 = 20x2 – 82x + 66 (b) Given L(x) = 156, 20x2 – 82x + 66 = 156 20x2 – 82x + 66 – 156 = 0 20x2 – 82x – 90 = 0 10x2 – 41x – 45 = 0 Divide each term by 2. Quadratic equation in the general form. SMART TIP The general form of a quadratic equation is ax2 + bx + c = 0, a ≠ 0, where a, b and c are constants and x is a variable. Example 6 Determine whether each of the following values is a root of the quadratic equation x2 + 4x – 12 = 0 or not. (a) x = 2 (b) x = 3 Solution (a) When x = 2, LHS = 22 + 4(2) – 12 = 0 = RHS LHS represents ‘left-hand side’ while RHS represents ‘right-hand side’. Hence, 2 is a root of the quadratic equation x2 + 4x – 12 = 0. (b) When x = 3, LHS = 32 + 4(3) – 12 = 9 ≠ RHS Hence, 3 is not a root of the quadratic equation x2 + 4x – 12 = 0. Example 7 Solve each of the following quadratic equations by factorisation. (a) 3x2 – 7x = 0 (d) 8q2 + 2q – 15 = 0 (b) 1 9 p2 – 49 = 0 (e) 2m – 1 = 11 m + 4 (c) 6h2 – 19h + 15 = 0 SMART TIP Solving a quadratic equation means to determine the roots of the quadratic equation. Solution (a) 3x2 – 7x = 0 x(3x – 7) = 0 x is the common factor of the quadratic expression of 3x2 – 7x. x = 0 or 3x – 7 = 0 3x = 7 x = 7 3 Hence, x = 0 or 7 3 . Analysis SPM Maths 2023 Eng F4 C1 3rd.indd 4 17/2/2023 9:11:25 PM PENERBIT ILMU BAKTI SDN. BHD.

5 Form 4 CHAPTER 1 SMART TIP If two numbers c and d are such that cd = 0, then c = 0 or d = 0. This property is known as the zero product property. (b) 1 9 p2 – 49 = 0 1 1 3 p2 2 – 72 = 0 1 1 3 p + 721 1 3 p – 72 = 0 a2 – b2 = (a + b)(a – b) 1 3 p + 7 = 0 or 1 3 p – 7 = 0 1 3 p = –7 or 1 3 p = 7 p = –21 or p = 21 Hence, p = –21 or 21. (c) 6h2 – 19h + 15 = 0 Firstly, we have to factorise the quadratic expression 6h2 – 19h + 15 [a > 1]. List the possible combinations of the factors of 6h2 dan +15. 6h2 = h × 6h = 2h × 3h Since the value of c is positive, we should consider the factors of c that have the same sign with b (= –19), that is negative. 15 = (–1) × (–15) = (–3) × (–5) 2h –3 ⊗ ⊗ 3h –5 –9h ⊕ –10h 6h2 15 –19h Hence, 6h2 – 19h + 15 = (2h – 3)(3h – 5). Thus, 6h2 – 19h + 15 = 0 (2h – 3)(3h – 5) = 0 2h – 3 = 0 or 3h – 5 = 0 2h = 3 or 3h = 5 h = 3 2 or h = 5 3 Hence, h = 3 2 or 5 3 . (d) 8q2 + 2q – 15 = 0 First of all, we have to factorise the quadratic expression 8q2 + 2q – 15 = 0 [a > 1]. List the possible combinations of the factors of 8q2 and –15. 8q2 = q × 8q = 2q × 4q Since the value of c (= –15) is negative, the sign in the first brackets and the sign in the second brackets must be opposite. –15 = (–1) × 15 = 1 × (–15) = (–3) × 5 = 3 × (–5) 2q 3 ⊗ ⊗ 4q –5 12q ⊕ –10q 8q2 –15 2q Hence, 8q2 + 2q – 15 = (2q + 3)(4q – 5). SMART TIP Not all quadratic equations ax2 + bx + c = 0 can be factorised. For quadratic equations that cannot be factorised, the roots can be determined using the quadratic formula x = –b ±  b2 – 4ac 2a . Thus, 8q2 + 2q – 15 = 0 (2q + 3)(4q – 5) = 0 2q + 3 = 0 or 4q – 5 = 0 2q = –3 or 4q = 5 q = – 3 2 or q = 5 4 Hence, q = – 3 2 or 5 4 . (e) 2m – 1 = 11 m + 4 (2m – 1)(m + 4) = 11 Cross multiplication 2m2 + 8m – m – 4 = 11 2m2 + 7m – 4 – 11= 0 2m2 + 7m – 15 = 0 Express in the general form first. Analysis SPM Maths 2023 Eng F4 C1 3rd.indd 5 17/2/2023 9:11:25 PM PENERBIT ILMU BAKTI SDN. BHD.

6 Form 4 CHAPTER 1 (2m – 1)(m + 4) = 11 2m – 1 = 11 or m + 4 = 11 Caution Incorrect because it does not satisfy the zero product property. 2m –3 ⊗ ⊗ m +5 –3m ⊕ 10m 2m2 –15 7m (2m – 3)(m + 5) = 0 2m – 3 = 0 or m + 5 = 0 2m = 3 or m = –5 m = 3 2 or m = –5 Hence, m = 3 2 or –5. Example 8 Sketch the graph of each of the following quadratic functions. (a) f(x) = x2 + 1 (b) g(x) = x2 – 6x + 8 (c) h(x) = –x2 – 2x + 8 Solution (a) f(x) = x2 + 1. Since a > 0, the shape of the graph is ∪. On the y-axis, x = 0. Thus, y = f(0) = (0)2 + 1 = 1 Hence, the graph will intersect the y-axis at (0, 1). When x = ±1, y = (±1)2 + 1 = 2 When x = ±2, y = (±2)2 + 1 = 5 Hence, the graph will pass through the points (–2, 5), (–1, 2), (1, 2) and (2, 5). The graph of y = f(x) = x2 + 1 is as follows. y (–2, 5) (2, 5) f(x) = x² + 1 x O 1 (b) g(x) = x2 – 6x + 8. Since a > 0, the graph has the shape ∪. At the y-axis, x = 0. Thus, y = g(0) = 02 – 6(0) + 8 = 8 Hence, the graph will intersect the y-axis at (0, 8). At the x-axis, y = 0. Thus, x2 – 6x + 8 = 0 (x – 2)(x – 4) = 0 x = 2 or x = 4 Hence, the graph will intersect the x-axis at (2, 0) and (4, 0). The graph of y = g(x) = x2 – 6x + 8 is as follows: y O 4 g(x) = x2 – 6x + 8 2 8 x (c) h(x) = –x2 – 2x + 8. Since a < 0, the graph has the shape ∩. At the y-axis, x = 0. Thus, y = h(0) = –02 + 2(0) + 8 = 8 Hence, the graph will intersect the y-axis at (0, 8). At the x-axis, y = 0. Thus, –x2 – 2x + 8 = 0 x2 – 2x – 8 = 0 (x + 2)(x – 4) = 0 x = –2 or x = 4 Hence, the graph will intersect the x-axis at (–2, 0) and (4, 0). The graph of y = h(x) = –x2 – 2x + 8 is as follows: y O 4 h(x) = –x2 – 2x + 8 8 –2 x Analysis SPM Maths 2023 Eng F4 C1 3rd.indd 6 17/2/2023 9:11:26 PM PENERBIT ILMU BAKTI SDN. BHD.

7 Form 4 CHAPTER 1 Example 9 In one of the stage of the le Tour de Langkawi road cycling championship with a distance of 150 km, the average speed of the cyclist is (3x + 10) km hour–1 . If the average time taken by them to complete the race is 1 1 5 x + 7 4 2 hours, find the average speed of the cyclist. Solution Understanding the problem Distance = 150 km Average speed = (3x + 10) km h–1 Time = 1 1 5 x + 7 4 2 hours Planning a strategy (a) Form a quadratic equation. (b) Solve the quadratic equation that is formed. (c) Substitute the value of x into the given formula to find the average speed. Carry out the strategy Time × Speed = Distance 1 1 5 x + 7 4 2 × (3x + 10) = 150 km (4x + 35)(3x + 10) = 3000 The whole equation is 12x multiplied by 20. 2 + 145x + 350 = 3000 12x2 + 145x – 2650 = 0 (x – 10)(12x + 265) = 0 x = 10 or x = – 265 12 x = – 265 12 is not accepted. This is because the time and speed cannot take a negative Thus, x value. = 10 Making conclusion Hence, the average speed = 3(10) + 10 = 40 km h–1. The following diagram shows a piece of cloth. A B C D E 25x cm (2x + 20) cm 7x cm Given that the area of the cloth is 1.61 m2 , find the value of x. HOTS Applying HOTS Analysing HOTS Evaluating Solution Area of cloth = 1.61 m2 Area of rectangle ABCE – Area of triangle CDE = (1.61 × 10 000) cm2 (25x × 7x) – 3 1 2 × 7x × (2x + 20)4 = 16 100 175x2 – x + 10 1 3 1 2 × 7x × (2x + 20)4 = 16 100 175x2 – [7x × (x + 10)] = 16 100 175x2 – 7x2 – 70x = 16 100 168x2 – 70x – 16 100 = 0 Express in the general form. 84x2 – 35x – 8050 = 0 12x2 – 5x – 1150 = 0 (x – 10)(12x + 115) = 0 x –10 ⊗ ⊗ 12x 115 –120x ⊕ 115x 12x2 –1150 –5x x – 10 = 0 or 12x + 115 = 0 x = 10 12x = –115 x = – 115 12 Hence, x = 10. Not accepted because length cannot take a negative value. HOTS Zone Analysis SPM Maths 2023 Eng F4 C1 3rd.indd 7 17/2/2023 9:11:27 PM PENERBIT ILMU BAKTI SDN. BHD.

8 Form 4 CHAPTER 1 Objective Questions 1 Given that –4 is one of the roots of the quadratic equation 2x2 + hx – 12 = 0, where h is a constant, find the value of h. A 2 C 4 B 3 D 5 2 In a football free kick, the height of the ball from the ground, h(x) m, is given by h(x) = – 1 5 x2 + 47 20 x, where x is the horizontal distance of the ball from the player who takes the free kick. Find the possible horizontal distance, in m, when the height of the ball is 6 m from the ground. A 7 C 9 B 8 D 10 3 Which of the following graph represents f(x) = –x2 – 4? A y 4 –2 O 2 x B y –2 O 2 x –4 C y 4 O x D y x O –4 4 The following diagram shows a Jalur Gemilang. (x – 1) m (x + 1) m 2x m x m2 If the area of the cloth used to sew the Jalur Gemilang is 35 m2 , find the value of x. 3Use p = 22 7 4 A 1 C 3 B 2 D 4 5 If the radius and the surface area of a handball are (x + 3) cm and 616 cm2 respectively, what is the value of x? A 4 C 6 B 5 D 7 SPM Practice 1 Analysis SPM Maths 2023 Eng F4 C1 3rd.indd 8 17/2/2023 9:11:28 PM PENERBIT ILMU BAKTI SDN. BHD.

9 Form 4 CHAPTER 1 1.1 Quadratic Functions and Equations 1 State whether each of the following expressions is a quadratic expression (in one variable) or not. (a) x + 8 (b) b2 – 9 (c) c2 + 10c (d) k2 + 4k + 5 (e) 4x3 + 3x2 – 6x + 8 (f) 0h2 + 18h + 3 (g) 12 – 5pq + 2p2 (h) x2 + 4x – 5 x 2 Determine whether each of the following functions is a quadratic function or not. (a) f(x) = (3 – x)(2 + 5x) (b) g(x) = 4 x2 – 5 (c) h(x) = (2 – 3x) 2 (d) m(x) = x – 3 x (e) n(x) = x(3x + 1)2 (f) p(x) = x2 + 5 x2 3 Is the general shape of the graph of each of the following quadratic functions has a shape of ∪ or ∩? (a) f(x) = 6x2 + 7x + 2 (b) g(x) = –3x2 + 5x – 2 (c) m(x) = 15x2 – 4x – 6 (d) n(x) = –10x2 – 13x – 3 4 (a) The following diagram shows the sketch of the graph of the quadratic function f(x) = –x2 + 2x + 3. y f(x) = –x² + 2x + 3 3 x Axis of symmetry –1 O Determine (i) the equation of the axis of symmetry of the graph, (ii) the coordinates of the maximum point of the graph, (iii) the value of a, of b and of c if the graph of f(x) = ax2 + bx + c is reflected in the x-axis. (b) The following diagram shows the sketch of the graph of the quadratic function g(x) = x2 – 8x + 7. y Axis of symmetry O 1 7 x g(x) = x² – 8x + 7 Determine (i) the equation of the axis of symmetry of the graph, (ii) the coordinates of the minimum point of the graph, (iii) the value of b if the graph of f(x) = ax2 + bx + c is reflected in the y-axis. 5 It is given that h(x) represents the product of two consecutive numbers. The product of two positive consecutive numbers is 132. Find a quadratic expression of h(x). Hence, form a quadratic equation from the given information. 6 Zaki is 2 years older than his sister, Lila. It is given that h(x) represents the product of the age of Zaki and Lila. Find a quadratic expression of h(x). The product of their age is 120. Hence, form a quadratic equation from the given information. 7 The following diagram shows a triangle PQR. It is given that L(x) represents the area, in cm2 , of the triangle. Subjective Questions Analysis SPM Maths 2023 Eng F4 C1 3rd.indd 9 17/2/2023 9:11:28 PM PENERBIT ILMU BAKTI SDN. BHD.

10 Form 4 CHAPTER 1 P Q R 4y cm (y + 8) cm Find a quadratic expression of L(x). Hence, form a quadratic equation if the area of the triangle PQR is 40 cm2 . 8 The following diagram shows a trapezium PQRS. P Q R S (5t + 2) cm 4t cm 3t cm It is given that L(x) represents the area, in cm2 . Find a quadratic expression of L(x). Hence, if the area of the trapezium is 80 cm2 , form a quadratic equation. 9 The following diagram shows a cuboid. (x + 4) cm 5 cm 2x cm It is given that V(x) represents the volume, in cm3 . Find a quadratic expression of V(x). Given that the volume of the cuboid is 600 cm3 , form a quadratic equation. 10 Aina is 2 years younger than her sister, Latifah. The age of their mother, Puan Noriah, is the square of the age of Latifah. It is given that h(x) represents the sum of the age of Aina, Latifah and Puan Noriah. If the age of Latifah is x, find a quadratic expression of h(x). Hence, form a quadratic equation if the sum of the age of all three of them is 33. 11 In a road cycling event, Ismail cycles with a speed of 3x km h–1 for (x – 8) hours. He then cycles with a speed of (3x + 5) km h–1 for (x – 9) hours. It is given that J(x) represents the distance, in km. Find a quadratic expression of J(x). If the distance from the starting line until the ending line is 95 km, form a quadratic equation. 12 The following table shows the price per kg and the number of kg bought by a restaurant owner for three types of cooking items. Cooking items Price (RM) per kg Number of kg bought Ginger x + 5 x – 4 Garlic x + 10 x – 5 Onion x – 4 x – 6 It is given that W(x) represents the total amount of money, in RM. Find a quadratic expression for W(x). If the total amount of money spent by the restaurant owner is RM214, form a quadratic equation. 13 Determine whether each of the following values is a root for the quadratic equation 3x2 – 5x – 2 = 0 or not. (a) x = 2 (b) x = –1 (c) x = – 1 3 14 Determine whether each of the following values is a root for the quadratic equation –2x2 + 3x – 1 = 0 or not. (a) x = 2 (b) x = 1 (c) x = 1 2 15 Solve each of the following quadratic equations by factorisation. (a) 3x2 – 5x = 0 (b) 5k2 + 45k = 0 (c) 16t2 – 25 = 0 (d) 36y2 – 16 = 0 (e) 12t2 – 28t + 15 = 0 (f) 8m2 – 51m + 18 = 0 (g) 6n2 + 5n – 6 = 0 (h) 10s2 – 7s – 12 = 0 (i) –12j 2 – 11j + 36 = 0 Analysis SPM Maths 2023 Eng F4 C1 3rd.indd 10 17/2/2023 9:11:29 PM PENERBIT ILMU BAKTI SDN. BHD.

11 Form 4 CHAPTER 1 16 Solve each of the following quadratic equations by factorisation. (a) –3x2 = 4 – 13x (b) (2y + 1)2 = 16 (c) 3r + 1 = 7 r – 1 (d) k – 1 = k + 20 6k (e) (p – 3)(p + 2) = 1 2 p(p – 3) (f) h – 1 6 – 2h – 1 5h = 0 17 Sketch the graph of each of the following quadratic functions. (a) f(x) = 2x2 + 2 (b) g(x) = x2 – 6x + 5 (c) h(x) = –x2 – 8x – 12 18 A rectangle has a length of p cm and a width of (2p – 3) cm. It is given that the area of the rectangle is 77 cm2 . (a) Form a quadratic equation, (b) Find the value of p, (c) Calculate the width of the rectangle. HOTS Applying 19 The following diagram shows a rectangle ABCD. A 6 cm B C D (x + 2) cm x cm R It is given that the area of the shaded region in the diagram is 20 cm2 , (a) Form a quadratic equation, (b) Find the value of x. (c) Calculate the width of the rectangle. HOTS Applying 20 The following diagram shows a rightangled triangle ABC. A B C x cm (2x – 4) cm 10 cm (a) Form a quadratic equation. (b) Find the value of x. (c) Calculate the length of BC. HOTS Applying 21 The following diagram shows a trapezium PQRS. S P Q R 3x cm 9x cm (10 – x) cm It is given that the area of the trapezium PQRS is 96 cm2 , (a) Form a quadratic equation. (b) Find the possible values of x. (c) Calculate the lengths of PQ, QR and RS using the smaller value of x. HOTS Applying 22 The following diagram shows a water tank in the shape of a cuboid with a volume of of 240 m3 . 4x m 3 m (x + 8) m (a) Form a quadratic equation. (b) Find the value of x. (c) Calculate the height of the water tank. HOTS Applying Analysis SPM Maths 2023 Eng F4 C1 3rd.indd 11 17/2/2023 9:11:30 PM PENERBIT ILMU BAKTI SDN. BHD.

12 Form 4 CHAPTER 1 23 The following diagram shows rectangles ABCD and EFGH. A B D C x cm (2x + 3) cm E F H G 3x cm (x – 2) cm It is given that the two rectangles have a same area. (a) Form a quadratic equation. (b) Find the value of x. (c) Calculate the length and width of (i) the rectangle ABCD, (ii) the rectangle EFGH. HOTS Applying HOTS Analysing 24 A swimming pool has a length of (6x – 10) m and a width of (2x + 5) m. A pathway with a uniform width of (x – 5) m surrounds the swimming pool. The pathway is represented by the shaded region in the following diagram. (x – 5) m (2x + 5) m (6x – 10) m Swimming pool Pathway It is given that the area of the shaded region is 850 m2 . (a) Form a quadratic equation. (b) Find the value of x. (c) Calculate (i) the length of the swimming pool, (ii) the width of the swimming pool, (iii) the uniform width of the pathway. HOTS Applying HOTS Analysing 25 The following photograph shows a free throw in basketball. It is given that the height, h(x) feet, of the ball from the ground level is given by h(x) = – 16 225 x2 + 4 3 x + 6, such that x is the horizontal distance, in feet, from the player who takes the free throw. If the height of the basket from the ground level is 10 feet, find the horizontal distance, in feet, of the basket from the player. HOTS Applying HOTS Analysing HOTS Evaluating Analysis SPM Maths 2023 Eng F4 C1 3rd.indd 12 17/2/2023 9:11:30 PM PENERBIT ILMU BAKTI SDN. BHD.

13 Form 4 CHAPTER 1 Objective Questions 1 D 2 B 3 D 4 C 5 A Subjective Questions 1 (a) No (b) Yes (c) Yes (d) Yes (e) No (f) No (g) No (h) No 2 (a) Yes (b) No (c) Yes (d) No (e) No (f) No 3 (a) ∪ (b) ∩ (c) ∪ (d) ∩ 4 (a) (i) x = 1 (ii) (1, 4) (iii) a = 1, b = –2, c = –3 (b) (i) x = 4 (ii) (4, –9) (iii) b = 8 5 h(x) = x2 + x x2 + x – 132 = 0 6 h(x) = x(x + 2) = x2 + 2x x2 + 2x – 120 = 0 7 L(y) = 2y2 + 16y y2 + 8y – 20 = 0 8 L(t) = 16t2 + 4t 4t2 + t – 20 = 0 9 V(x) = 10x2 + 40x x2 + 4x – 60 = 0 10 h(x) = x2 + 2x – 2 x2 + 2x – 35 = 0 11 J(x) = 6x2 – 46x – 45 3x2 – 23x – 70 = 0 12 W(x) = 3x2 – 4x – 46 3x2 – 4x – 260 = 0 13 (a) Yes (b) No (c) Yes 14 (a) No (b) Yes (c) Yes 15 (a) x = 0 or 5 3 (b) k = 0 or –9 (c) t = 5 4 or – 5 4 (c) y x O –12 –6 –2 18 (a) 2p2 – 3p – 77 = 0 (b) p = 7 (c) 11 cm 19 (a) x2 – 10x + 16 = 0 (b) x = 2 (x = 8 is not accepted because x < 6) (c) 4 cm 20 (a) 5x2 – 16x – 84 = 0 (b) x = 6 (c) 8 cm 21 (a) x2 – 10x + 16 = 0 (b) x = 2 or 8 (c) PQ = 6 cm, QR = 8 cm, RS = 18 cm 22 (a) x2 + 8x – 20 = 0 (b) x = 2 (c) 10 m 23 (a) x2 – 9x = 0 (b) x = 9 (c) (i) 21 cm, 9 cm (ii) 27 cm, 7 cm 24 (a) 2x2 – 13x – 70 = 0 (b) x = 10 (c) (i) 50 m (ii) 25 m (iii) 5 m 25 15 feet (x = 3.75 is not accepted.) (d) y = 2 3 or – 2 3 (e) t = 3 2 or 5 6 (f) m = 6 or 3 8 (g) n = 2 3 or – 3 2 (h) s = 3 2 or – 4 5 (i) j = – 9 4 or 4 3 16 (a) x = 4 or 1 3 (b) y = 3 2 or – 5 2 (c) r = 2 or – 4 3 (d) k = 5 2 or – 4 3 (e) p = 3 or –4 (f) h = 3 or 2 5 17 (a) y O (–2, 10) (2, 10) 2 f(x) = 2x² + 2 x (b) y x O 5 1 5 g(x) = x² – 6x + 5 Answers Analysis SPM Maths 2023 Eng F4 C1 3rd.indd 13 17/2/2023 9:11:31 PM PENERBIT ILMU BAKTI SDN. BHD.

14 2.1 Number Bases 1 The digit used in the base one number system is 0. 2 The digits used in the base two number system are 0 and 1. 3 The base two number system is also known as the binary number system. 4 The digits used in the base three number system are 0, 1 and 2. 5 The digits used in the base four number system are 0, 1, 2 and 3. 6 The digits used in the base six number system are 0, 1, 2, 3, 4 and 5. 7 The digits used in the base seven number system are 0, 1, 2, 3, 4, 5 and 6. 8 The digits used in the base eight number system are 0, 1, 2, 3, 4, 5, 6 and 7. 9 The base eight number system is also known as the octal number system. 10 The digits used in the base nine number system are 0, 1, 2, 3, 4, 5, 6, 7 and 8. 11 Each digit of the numbers in base two has place values as shown in the following table. Place value 24 23 22 21 20 16 8 4 2 1 12 Each digit of the numbers in base five has place values as shown in the following table. Place value 54 53 52 51 50 625 125 25 5 1 13 Each digit of the numbers in base eight has place values as shown in the following table. Place value 84 83 82 81 80 4 096 512 64 8 1 14 The value of a digit of a number in various bases is determined by multiplying the digit with its place value. 15 A number can be expanded based on the place value of its digits as the sum of the values of digits which form the number. Changing a number from one base to another base 16 The steps to change a number in various bases to a number in base 10 are as follows: Expand the given number based on the place value of its digits. Calculate the value in Step to produce a single number. 17 A number in base ten can be changed to a number in any base x by dividing the number with x repeatedly until the quotient is zero. The number in base x is then determined by writing its remainder from bottom to top. 18 To change a number from base two to a number in base eight, arrange the digits in groups of three digits, from right to left. Then, use the combination of the numbers 4, 2 and 1. 19 To change a number in base eight to a number in base two, change each digit in base eight to the combination of three digits in base two using the combination of the numbers 4, 2 and 1. EXPRESS NOTES Number Bases Chapter 2 Learning Area: Number and Operations 14 Analysis SPM Maths 2023 Eng F4 C2 2nd.indd 14 17/2/2023 9:15:13 PM PENERBIT ILMU BAKTI SDN. BHD.

15 Form 4 CHAPTER 2 20 A number in base x can be changed to a number in base y (x and y are not ten) by changing the given number to base ten (by expanding the number based on the place value of its digits) and next change the number in base ten to a number in base y using repeated division until its quotient is zero. Add/Subtract the given digits from right to left. Change the two given numbers (in various bases) which are to be added or subtracted to base 10. Change the obtained sum or difference back to the original base. Calculate the sum or difference of the two numbers in base 10. The process is repeated until every digit have been added or subtracted. Addition The sum of the digits in base 10 is converted to the given base. Subtraction The answer of the subtraction is written in the answer. The answer shall not exceed the given base and its value is equal to the number in base 10. (a) Common form Solving problems involving number bases 22 The concept of number bases can be used to solve various problems in the daily real-life situations. Addition and subtraction of numbers in various bases 21 The addition and subtraction operations for numbers in the base range can be completed using basic conversion methods: (b) Change of base 2.1 Number Bases Example 1 Give an example of a three-digit number in (a) base 2 (b) base 5 (c) base 9 Solution (a) 1012 (b) 1435 (c) 3589 Example 2 Calculate the value of the underlined digit, in base ten, for the number 51326 . Solution Place value 63 62 61 60 Digit 5 1 3 2 The value of the underlined digit = 5 × 63 = 5 × 216 = 1 080 The place value of the underlined digit 5 is 63 = 216. Analysis SPM Maths 2023 Eng F4 C2 2nd.indd 15 17/2/2023 9:15:13 PM PENERBIT ILMU BAKTI SDN. BHD.

16 Form 4 CHAPTER 2 Example 7 Change the number 718 to a number in base two. Solution 7 4 + 2 + 1 22 21 20 1 1 1 1 0 + 0 + 1 22 21 20 0 0 1 Base 2 Base 8 Two zeros is added to form a group of three digits. ∴ 718 = 1110012 Example 8 Change 32014 to a number in base 6. Solution (a) Step 1 • Change the number in base four to a number in base 10. Place value 43 42 41 40 Digit 3 2 0 1 32014 = (3 × 43 ) + (2 × 42 ) + (0 × 41 ) + (1 × 40 ) = 22510 (b) Step 2 • Change the number in base ten to a number in base six. 6 225 Remainder 6 37 3 6 6 1 6 1 0 0 1 ∴ 32014 = 10136 Example 9 Calculate each of the following (a) 1345 + 2015 (b) 7538 – 2368 Solution (a) Change of base method Step 1 Place value 52 51 50 Digit 1 3 4 1345 = (1 × 52 ) + (3 × 51 ) + (4 × 50 ) = 4410 Example 3 Expand 312045 based on the place value of its digits. Solution Place value 54 53 52 51 50 Digit 3 1 2 0 4 312045 = (3 × 54 ) + (1 × 53 ) + (2 × 52 ) + (0 × 51 ) + (4 × 50 ) Example 4 Change the number 216358 to a number in base ten. Solution Place value 84 83 82 81 80 Digit 2 1 6 3 5 216358 = (2 × 84 ) + (1 × 83 ) + (6 × 82 ) + (3 × 81 ) + (5 × 80 ) = 8192 + 512 + 384 + 24 + 5 = 911710 Example 5 Change 88810 to a number in base 9. Solution 9 888 Remainder 9 98 6 9 10 8 9 1 1 0 1 ∴ 88810 = 11869 Example 6 Change the number 10001102 to a number in base eight. Solution 1 0 0 0 1 1 0 1 1 1 4 2 1 0 + 0 + 0 0 4 2 1 4 + 2 + 0 6 Base 2 Combination of the numbers 4, 2 and 1. Base 8 ∴ 10001102 = 1068 Analysis SPM Maths 2023 Eng F4 C2 2nd.indd 16 17/2/2023 9:15:14 PM PENERBIT ILMU BAKTI SDN. BHD.

17 Form 4 CHAPTER 2 Place value 52 51 50 Digit 2 0 1 2015 = (2 × 52 ) + (0 × 51 ) + (1 × 50 ) = 5110 Step 2 4410 + 5110 = 9510 Step 3 5 95 Remainder 5 19 0 5 3 4 0 3 Hence, 1345 + 2015 = 3405 Common form method 1 1345 +2015 3405 1 + 2 = 310 = 35 1 + 3 + 0 = 410 = 45 4 + 1 = 510 = 105 5 5 Remainder 5 1 0 0 1 5 4 Remainder 0 4 5 3 Remainder 0 3 Write 0 in the answer section, bring 1 to the next place value. Hence, 1345 + 2015 = 3405 Alternative Method (b) Change of base method Step 1 Place value 82 81 80 Digit 7 5 3 7538 = (7 × 82 ) + (5 × 81 ) + (3 × 80 ) = 49110 Place value 82 81 80 Digit 2 3 6 2368 = (2 × 82 ) + (3 × 81 ) + (6 × 80 ) = 15810 Step 2 49110 – 15810 = 33310 Step 3 8 333 Remainder 8 41 5 8 5 1 0 5 Hence, 7538 – 2368 = 5158 Common form method 4 8 7538 – 2368 5158 7 – 2 = 58 4 – 3 = 18 8 + 3 – 6 = 58 Hence, 7538 – 2368 = 5158 Alternative Method Example 10 Johari bought 1112 gel pens and 112 long rulers and paid RM518 . David bought 1102 gel pens and 112 long rulers and paid RM1215 . Find the price, in RM, of a gel pen and a long ruler, in base 10. HOTS Applying HOTS Analysing HOTS Evaluating Solution Understanding the problem The total price of 1112 pens and 112 long ruler is RM518 . The total price of 1102 pens and 112 long ruler is RM1215 . Planning a strategy • Change all the numbers in bases two, five and eight to base ten. • Form simultaneous linear equations. • Solve the simultaneous linear equations. Implementing the strategy The numbers in base 2 that need to be changed to base 10 are: 1112 = (1 × 22 ) + (1 × 21 ) + (1 × 20 ) = 710 112 = (1 × 21 ) + (1 × 20 ) = 310 1102 = (1 × 22 ) + (1 × 21 ) + (0 × 20 ) = 610 The number in base 8 that need to be changed to base 10 is: 518 = (5 × 81 ) + (1 × 80 ) = 4110 The number in base 5 that need to be changed to base 10 is: 1215 = (1 × 52 ) + (2 × 51 ) + (1 × 50 ) = 3610 Let x represent the price of a gel pen and y represent the price of a long ruler. Analysis SPM Maths 2023 Eng F4 C2 2nd.indd 17 17/2/2023 9:15:14 PM PENERBIT ILMU BAKTI SDN. BHD.

18 Form 4 CHAPTER 2 The simultaneous linear equations that need to be solved are: 1112 x + 112 y = 518 , i.e. 7x + 3y = 41 …… 1102 x + 112 y = 1215 , i.e. 6x + 3y = 36 …… – : x = 5 Substitute x = 5 into . 7(5) + 3y = 41 3y = 41 – 35 3y = 6 y = 2 Making a conclusion Hence, the prices of a gel pen and a long ruler are RM5 and RM2 respectively. 1 Calculate 7128 – 4325 giving your answers in base 7. Solution 82 81 80 7 1 2 7128 = (7 × 82 ) + (1 × 81 ) + (2 × 80 ) = 45810 52 51 50 4 3 2 4325 = (4 × 52 ) + (3 × 51 ) + (2 × 50 ) = 11710 7128 – 4325 = 45810 – 11710 = 34110 7 341 Remainder 7 48 5 7 6 6 0 6 Hence, 7128 – 4325 = 6657 2 A secret code is given by the following: 2314 5426 7268 Convert the numbers given into base 5 to break the secret code. What is the code from left to right? Solution 42 41 40 2 3 1 2314 = (2 × 42 ) + (3 × 41 ) + (1 × 40 ) = 4510 62 61 60 5 4 2 5426 = (5 × 62 ) + (4 × 61 ) + (2 × 60 ) = 20610 82 81 80 7 2 6 7268 = (7 × 82 ) + (2 × 81 ) + (6 × 80 ) = 47010 5 206 Remainder 5 41 1 5 8 1 5 1 3 0 1 5 470 Remainder 5 94 0 5 18 4 5 3 3 0 3 5 45 Remainder 5 9 0 5 1 4 0 1 1405 13115 33405 Hence, the secret code is 14013113340 1 Given that 2536 = x7 , find the value of x. A 200 C 210 B 205 D 215 HOTS Applying 2 Given that 3268 – 4325 = k9 , find the value of k. A 117 C 135 B 126 D 144 HOTS Applying HOTS Analysing 3 Puan Ng bought 102 kg of bitter gourd and 112 kg of cucumber with a sum payable of RM2235 . Puan Zakiah bought 1002 kg of bitter gourd and 1012 kg of cucumber with a sum payable of RM1618 . How much is the prices, in RM, for a kilogram of bitter gourd and a kilogram of cucumber? A 10 and 11 C 14 and 15 B 12 and 13 D 16 and 17 HOTS Applying HOTS Analysing HOTS Evaluating Objective Questions HOTS Zone SPM Practice 2 Analysis SPM Maths 2023 Eng F4 C2 2nd.indd 18 17/2/2023 9:15:15 PM PENERBIT ILMU BAKTI SDN. BHD.

19 Form 4 CHAPTER 2 2.1 Number Bases 1 Calculate the value of the underlined digit, in base ten, for each of the following numbers. (a) 10110112 (c) 10011002 (b) 11100102 (d) 110101012 2 Calculate the value of the underlined digit, in base ten, for each of the following numbers. (a) 5348 (c) 635428 (b) 32508 (d) 7350618 3 Calculate the value of the underlined digit, in base ten, for each of the following numbers. (a) 4135 (c) 321345 (b) 51235 (d) 241305 4 Calculate the value of the underlined digit, in base ten, for each of the following numbers. (a) 2013 (d) 623507 (b) 32104 (e) 241359 (c) 23546 5 What is the value of the underlined 1, in base 10, of the number 1001010102 ? 6 What is the value of the digit 7, in base 10, of the number 73518 ? 7 What is the value of the digit 2, in base 10, of the number 24315 ? 8 Expand each of the following numbers based on the place value of its digits. (a) 1023 (d) 64307 (b) 41234 (e) 87659 (c) 54316 9 Change each of the following numbers to a number in base ten. (a) 1101012 (c) 110110112 (b) 10111102 10 Change each of the following numbers to a number in base ten. (a) 21378 (c) 765438 (b) 111018 11 Change each of the following numbers to a number in base ten. (a) 3045 (c) 121045 (b) 10235 12 Change each of the following numbers to a number in base ten. (a) 2123 (d) 26407 (b) 12034 (e) 24689 (c) 53216 13 Change 22113 to a number in base ten. 14 Change 23014 to a number in base ten. 15 Change 53216 to a number in base ten. 16 Change 23567 to a number in base ten. 17 Change 84269 to a number in base ten. 18 Change each of the following numbers in base ten to a number in base two. (a) 4710 (c) 15710 (b) 9910 19 Change each of the following numbers in base ten to a number in base eight. (a) 12810 (c) 1010010 (b) 205010 4 What is the value of the digit 6, in base ten, for 632307 ? A 1 296 C 7 776 B 2 058 D 14 406 5 If 53126 = h4 , what is the value of h? A 1223 C 102230 B 3221 D 103320 6 Given that 1111000101102 = p8 , what is the value of p? A 7246 C 6247 B 7426 D 6427 7 Calculate 7539 + 4269 . A 12809 C 30039 B 20029 D 40049 8 Calculate 5416 – 4325 stating your answer in base 7. A 1347 C 3147 B 1547 D 4137 Subjective Questions Analysis SPM Maths 2023 Eng F4 C2 2nd.indd 19 17/2/2023 9:15:15 PM PENERBIT ILMU BAKTI SDN. BHD.

20 Form 4 CHAPTER 2 20 Change each of the following numbers in base ten to a number in base five. (a) 4310 (b) 40210 (c) 123410 21 (a) Given that 2610 = h3 , find the value of h. (b) Given that 5710 = k4 , find the value of k. (c) Given that 42210 = m6 , find the value of m. (d) Given that 209610 = p7 , find the value of p. (e) Given that 632310 = q9 , find the value of q. 22 Complete the following number sequence in base two. 1012 , , 1112 , , , 23 Determine 8(82 + 8 + 3) as a number in base eight. 24 Complete the following number sequence in base eight. 2478 , , , 2528 , 25 Determine 55 + 53 + 4 as a number in base five. 26 Complete the following number sequence in base five. 1315 , , , , 27 Given that 7410 = x5 , find the value of x. 28 Given that 97110 = h7 , find the value of h. 29 Given that 186310 = k9 , find the value of k. 30 Determine 8[7(84 ) + 6(82 ) + 5] as a number in base eight. 31 (a) Given that 21510 = h3 , where h is a constant, find the value of h. (b) Determine 3(94 ) + 5(93 ) + 2 as a number in base nine. 32 Determine 7[4(73 ) + 3(72 ) + 2(7)] as a number in base seven. 33 (a) Given that 19510 = k4 , where k is a constant, find the value of k. (b) Determine 6[1(64 ) + 2(63 ) + 3] as a number in base six. 34 Change each of the following numbers to a number in base eight. (a) 10110102 (b) 1100000012 (c) 1001111012 35 Change each of the following numbers to a number in base two. (a) 318 (b) 7048 (c) 6528 36 Change (a) 11112 to a number in base five, (b) 435 to a number in base two, (c) 758 to a number in base five, (d) 445 to a number in base eight. 37 Change (a) 2314 to a number in base six, (b) 32145 to a number in base three, (c) 34569 to a number in base seven. 38 Given that 7358 = w2 where w is an integer, find the value of w. 39 Given that 1111101010112 = p8 where p is an integer, find the value of p. 40 Given that 11012 = m5 , where m is an integer, find the value of m. 41 Given that 4445 = q2 , where q is an integer, find the value of q. 42 Determine 3045 as a number in base eight. 43 Change 528 to a number in base five. 44 Given that 11025 = k10 = x3 = h9 , find the value of k, of x and of h. HOTS Applying HOTS Analysing HOTS Evaluating 45 Calculate the sum of the following: (a) 101012 + 1112 (b) 1110012 + 101102 46 Calculate (a) 10112 – 1102 (b) 1011102 – 100102 Analysis SPM Maths 2023 Eng F4 C2 2nd.indd 20 17/2/2023 9:15:16 PM PENERBIT ILMU BAKTI SDN. BHD.

21 Form 4 CHAPTER 2 47 Calculate (a) 1023 + 2223 (b) 5236 + 4016 (c) 3578 + 4168 48 Calculate (a) 3334 – 2014 (b) 6327 – 1457 (c) 8539 – 4169 49 Calculate 4216 + 7658 , giving your answer in base five. 50 Calculate 5126 – 3334 , giving your answer in base three. 51 Calculate 11023 + 1113 . 52 Calculate 33214 – 12344 . 53 Calculate 45216 + 33336 . 54 Calculate 61237 – 55447 . 55 Calculate 54126 + 32124 , giving your answer in base five. 56 Calculate 24619 – 23518 , giving your answer in base seven. 57 Puan Faziatul bought 1002 red tilipia fish and 102 black tilipia fish and paid RM1124 . Puan Rosnani bought 112 red tilipia fish and 102 black tilipia fish and paid RM2003 . Find the price, in RM, of a red tilipia fish and a black tilipia fish, in base 10. HOTS Applying HOTS Analysing HOTS Evaluating 58 The total price of 1102 chairs and 112 tables is RM21206 . The total price of 10002 chairs dan 12 table is RM11117 . Find the price, in RM, of a chair and a table, in base ten. HOTS Applying HOTS Analysing HOTS Evaluating 59 Puan Harizah bought 1102 pineapples and 112 honey dews and paid RM3145 . Puan Chang bought 1112 pineapples and 1002 honey dews and paid RM3006 . Find the prices, in RM, of a pineapple and a honey dew, in base 10. HOTS Applying HOTS Analysing HOTS Evaluating 60 Letchumy bought 10012 bottles of 500 ml mineral water and 1102 bottles of 1 500 ml of mineral water and paid RM1114 . Catherine bought 10002 bottles of 500 ml mineral water and 1102 bottles 1 500 ml mineral water and paid RM2023 . Find the prices, in RM, of a bottle of 500 ml mineral water and a bottle of 1 500 ml mineral water in base 10. HOTS Applying HOTS Analysing HOTS Evaluating 61 Restaurants A and B bought garlic and onions as cooking materials. Restaurant A bought 1102 kg of garlic and 125 kg of onions with a total payment of RM3506 . Restaurant B bought 223 kg of garlic and 214 kg of onion with a total payment of RM3507 . Find the price, in RM, of a kilogram of garlic and a kilogram of onions, in base 10. HOTS Applying HOTS Analysing HOTS Evaluating 62 The total price of 1012 2B pencils and 1102 short rulers is RM1004 while the total price of 102 2B pencils and 12 short ruler is RM123 . Find the price of a 2B pencil and a short ruler, in base 10. HOTS Applying HOTS Analysing HOTS Evaluating Analysis SPM Maths 2023 Eng F4 C2 2nd.indd 21 17/2/2023 9:15:16 PM PENERBIT ILMU BAKTI SDN. BHD.

22 Form 4 CHAPTER 2 12 (a) 2310 (b) 9910 (c) 1 20110 (d) 1 00810 (e) 1 84410 13 7610 14 17710 15 1 20110 16 87410 17 6 18010 18 (a) 1011112 (b) 11000112 (c) 100111012 19 (a) 2008 (b) 40028 (c) 235648 20 (a) 1335 (b) 31025 (c) 144145 21 (a) h = 222 (b) k = 321 (c) m = 1542 (d) p = 6053 (e) q = 8605 22 1102 , 10002 , 10012 , 10102 23 11308 24 2508 , 2518 , 2538 25 1010045 26 1325 , 1335 , 1345 , 1405 27 x = 244 28 h = 2555 29 k = 2500 30 7060508 31 (a) h = 21222 (b) 350029 32 432007 33 (a) k = 3003 (b) 1200306 34 (a) 1328 (b) 6018 (c) 4758 Objective Questions 1 C 2 A 3 B 4 D 5 C 6 B 7 A 8 B Subjective Questions 1 (a) 8 (b) 16 (c) 0 (d) 128 2 (a) 320 (b) 1 536 (c) 24 576 (d) 229 376 3 (a) 100 (b) 625 (c) 1 875 (d) 1 250 4 (a) 18 (b) 192 (c) 108 (d) 14 406 (e) 2 916 5 256 6 3 584 7 250 8 (a) (1 × 32 ) + (0 × 31 ) + (2 × 30 ) (b) (4 × 43 ) + (1 × 42 ) + (2 × 41 ) + (3 × 40 ) (c) (5 × 63 ) + (4 × 62 ) + (3 × 61 ) + (1 × 60 ) (d) (6 × 73 ) + (4 × 72 ) + (3 × 71 ) + (0 × 70 ) (e) (8 × 93 ) + (7 × 92 ) + (6 × 91 ) + (5 × 90 ) 9 (a) 5310 (b) 9410 (c) 21910 10 (a) 1 11910 (b) 4 67310 (c) 32 09910 11 (a) 7910 (b) 13810 (c) 90410 35 (a) 110012 (b) 1110001002 (c) 1101010102 36 (a) 305 (b) 101112 (c) 2215 (d) 308 37 (a) 1136 (b) 1210023 (c) 103207 38 w = 111011101 39 p = 7653 40 m = 23 41 q = 1111100 42 1178 43 1325 44 k = 152, x = 12122, h = 178 45 (a) 111002 (b) 10011112 46 (a) 1012 (b) 111002 47 (a) 11013 (b) 13246 (c) 7758 48 (a) 1324 (b) 4547 (c) 4369 49 101135 50 111223 51 12203 52 20214 53 122546 54 2467 55 213225 56 14567 57 RM4, RM3 58 RM40, RM80 59 RM4, RM20 60 RM1, RM2 61 RM16, RM6 62 RM2, RM1 Answers Analysis SPM Maths 2023 Eng F4 C2 2nd.indd 22 17/2/2023 9:15:16 PM PENERBIT ILMU BAKTI SDN. BHD.

23 3.1 Statements 1 A statement is a sentence such that we can determine whether it is true or false. 2 A sentence which is a question, an instruction and an exclamation is not a statement because we cannot determine whether it is true or false. 3 The quantifier “all” means that every object or case satisfies a certain condition. 4 The quantifier “some” means a few and not every object or case satisfies a certain condition. Negating a statement 5 Negation is a process which denies a statement using the word “not” or “no”. A true statement can be changed to a false statement and vice versa using the word “not” or “no”. 6 Negation of a statement can change the truth value of the statement. Determining the truth value of a compound statement 7 A compound statement can be formed by combining two statements using the word “and”. 8 A compound statement can also be formed by combining two statements using the word “or”. 9 The truth value of a compound statement “p and q” is as shown in the following truth table: p q p and q True True True True False False False True False False False False 10 The truth value of a compound statement “p or q” is as shows in the following truth table: p q p or q True True True True False True False True True False False False Implications Antecedent and Consequent of an Implication 11 For two statements, p and q, the statement “if p, then q” is an implication such that p is the antecedent and q is the consequent. Implication “if and only if” 12 For two statements p and q, the statement “p if and only if q” is the short form of “if p, then q” and “if q, then p”. 13 For a “p if and only if q” statement, two implications can be written as follows: Implication 1: If p, then q. Implication 2: If q, then p. 23 EXPRESS NOTES Logical Reasoning Chapter 3 Learning Area: Discrete Mathematics Analysis SPM Maths 2023 Eng F4 C3 3rd.indd 23 17/2/2023 9:20:28 PM PENERBIT ILMU BAKTI SDN. BHD.

24 Form 4 CHAPTER 3 Converse, inverse and contrapositive of an implication Converse of an implication 14 The converse of an implication “if p, then q” is “if q, then p”. 15 The converse of an implication may not be true. Inverse of an implication 16 The inverse of an implication “if p, then q” is “if not p, then not q”. 17 The inverse of an implication may not be true. Contrapositive of an implication 18 The contrapositive of an implication “if p, then q” is “if not q, then not p”. 19 The contrapositive of an implication is always true if the implication is true. 3.2 Arguments 20 A premise is a few (usually two) statements that can justify a conclusion. 21 An argument is a process to make a conclusion based on a few given statements. The given statements are known as premises. Deductive argument and inductive argument 22 Deductive argument is an argument such that its premises are surely true to guarantee the truth of a conclusion. SMART TIP Deductive argument is an argument such that its premises are general which produce a specific conclusion. 23 In an inductive argument, its premises are only meant to be made as convincing as possible such that, if it is cogent, then its conclusion can be strong. SMART TIP Inductive argument is an argument such that its premises are specific which produce a general conclusion. Determine wether a valid deductive argument is reasonable 24 The three forms of a deductive argument are: Deductive Argument Form I Premise 1: All A is B. Premise 2: C is A. Conclusion: C is B. Deductive Argument Form II Premise 1: If p, then q. Premise 2: p is true. Conclusion: q is true. Deductive Argument Form III Premise 1: If p, then q. Premise 2: Not q is true. Conclusion: Not p is true. 25 A deductive argument is valid (correct in terms of Form I, II or III) will become unsound if premise 1 or premise 2 is not true. Valid deductive arguments 26 For a valid deductive argument, based on the two true premises, a true conclusion can be made. Strength of an inductive argument 27 The strength of an inductive argument is determined by the reliability level of the premises and how logic of the conclusion made. (a) A strong inductive argument has cogent premises and high reliability level such that a logic and convincing conclusion can be made. (b) A weak inductive argument has doubtful premises causing the conclusion made to be not logic and not cogent. Solving problems involving logical reasoning 28 Deductive reasoning is a process of making a specific conclusion based on general statements. 29 Inductive reasoning is a process of making a general conclusion based on specific cases. Analysis SPM Maths 2023 Eng F4 C3 3rd.indd 24 17/2/2023 9:20:28 PM PENERBIT ILMU BAKTI SDN. BHD.

25 Form 4 CHAPTER 3 3.1 Statements Example 1 Determine whether each of the following sentences is a statement or not. Give your reason. (a) The angle at the circumference of a semicircle is 90°. (b) What is the root of the quadratic equation (x – 3)2 = 0? (c) 1 kg is equal to 100 g. (d) Please fill in this form. (e) Wow! Your dress is really beautiful! (f) 5k + 2k – 3k Solution (a) Statement because we can determine that the sentence is true. (b) Not a statement because the sentence is a question. (c) Statement because we can determine that the sentence is false. (d) Not a statement because the sentence is an instruction. (e) Not a statement because the sentence is an exclamation. (f) Not a statement because we cannot determine whether the sentence is true or false. Example 2 Determine whether each of the following statements is true or false. (a) The number 0.005620 has three significant figures. (b) 23 + 27 3 > 10 (c) All quadratic expressions can be factorised. (d) Some prime numbers are odd. Solution (a) False statement 0.005620 has four significant figures because the digit 0 at the end (after the decimal point) is a significant figure. (b) True statement 23 + 27 3 = 8 + 3 = 11, and 11 is greater than 10. (c) False statement A quadratic expression such as x2 + x – 3 cannot be factorised. (d) True statement 2 is a prime number which is even. Example 3 Determine whether each of the following statements is true or false. Write a statement which changes the truth value of each given statement using the word “no” or “not”. (a) The maximum number of roots of a quadratic equation is 2. (b) If x < –6, then x can take the value –4. Solution (a) The maximum number of roots of a quadratic equation is 2. [True] The maximum number of roots of a quadratic equation is not 2. [False] (b) If x < –6, then x can take the value –4. [False] If x < –6, then x cannot take the value –4. [True] Example 4 Determine whether each of the following compound statements is true or false. (a) 0.00204 has 3 significant figures and 6450 = 6.45 × 103 . (b) 8 is a root of the equation 3x – 5 = x + 9 and the solution of x = 3 ialah x = 9. (c) 5 7 > 2 7 and 11 8 > 11 5 . (d) A straight line with the equation y = –2x + 5 as a positive gradient and the sum of the lengths of two sides of a triangle is less than the length of the third side. Solution True (a) 0.00204 has 3 significant figures and 6450 = 6.45 × 103 is true. True ‘True’ and ‘True’ is ‘True’. False (b) 8 is a root of the equation 3x – 5 = x + 9 and and the solution of x = 3 ialah x = 9 is false. True ‘False’ and ‘True’ is ‘False’. (c) 5 7 > 2 7 and 11 8 > 11 5 is false. ‘True’ and ‘False’ True False is ‘False’. Analysis SPM Maths 2023 Eng F4 C3 3rd.indd 25 17/2/2023 9:20:28 PM PENERBIT ILMU BAKTI SDN. BHD.

26 Form 4 CHAPTER 3 False (d) A straight line with the equation y = –2x + 5 has a positive gradient and the sum of lengths of two sides of a triangle is less than the length of the third side is false. False ‘False’ and ‘False’is ‘False’. Example 5 Determine whether each of the following compound statements is true or false. (a) An octagon has eight sides or 121 is a perfect square. (b) The volume of a cuboid is the product of length, width and height or all multiples of 3 are multiples of 6. (c) (125 3 )–2 = 25 or 0.000128 = 1.28 × 10–4. (d) The straight line with the equation y = 2x + 5 intersects the y-axis at the point (0, 2) or sin 30° = 3 2 . Solution True (a) An octagon has eight sides or 121 is a perfect square is true. True ‘True’ or ‘True’ is ‘True’. True (b) The volume of a cuboid is the product of length, width and height or all multiples of 3 are multiples of 6 is true. False ‘True’ or ‘False’ is ‘True’. False True (c) (125 3 )–2 = 25 or 0.000128 = 1.28 × 10–4 is true. ‘False’ or ‘True’ is ‘True’. False (d) The straight line with the equation y = 2x + 5 intersects the y-axis at the point (0, 2) or sin 30° = 3 2 is false. False ‘False’ or ‘False’ is ‘False’. Example 6 State the antecedent and consequent of each of the following implications. (a) If k is a factor of 5, then k is also a factor of 10. (b) If θ = 30°, then cos θ = 3 2 . Solution (a) The antecedent is k is a factor of 5. The consequent is k is a factor of 10. (b) The antecedent is θ = 30°. The consequent is cos θ = 3 2 . Example 7 Construct an implication based on the given antecedent and consequent. Antecedent: h is a multiple of 6. Consequent: h is also a multiple of 3. Solution If h is a multiple of 6, then h is also a multiple of 3. Example 8 Write two implications based on each of the following statements: (a) x < 0 if and only if x is a negative number. (b) (x – 2)(x + 4) = 0 if and only if x = 2 or x = –4. Solution (a) Implication 1: If x < 0, then x is a negative number. Implication 2: If x is a negative number, then x < 0. (b) Implication 1: If (x – 2)(x + 4) = 0, then x = 2 or x = –4. Implication 2: If x = 2 or x = –4, then (x – 2)(x + 4) = 0. Example 9 Construct a statement in the form “if and only if” from each of the following pairs of implications. (a) If the radius of a sphere is 7 cm, then the volume of the sphere is 4 3 π(7)3 cm3 . (b) If the area of a circle is 25π cm2 , then the radius of the circle is 5 cm. Analysis SPM Maths 2023 Eng F4 C3 3rd.indd 26 17/2/2023 9:20:29 PM PENERBIT ILMU BAKTI SDN. BHD.

27 Form 4 CHAPTER 3 Solution (a) The radius of a sphere is 7 if and only if the volume of the sphere is 4 3 π(7)3 cm3 . (b) The area of a circle is 25π cm2 if and only if its area is 5 cm. Example 10 State the converse of each of the following implications and hence, determine whether the converse is true or false. (a) If 7x is an even number, then x is an even number. (b) If y > 8, then y > 5. Solution (a) Converse: If x is an even number, then 7x is an even number. The converse is true. (b) Converse: If y > 5, then y > 8. The converse is false. y > 5: y = 6, 7, 8, … y > 8: y = 9, 10, 11, … If y = 6, then y is not greater than 8. Hence, the converse ‘If y > 5, then y > 8” is false. SMART TIP One case is sufficient to deny the truth value of a statement. Example 11 State the inverse of the following implication and hence, determine whether the inverse is true or false. “If y > x, then –y < –x.” Solution Not p (antecedent) Not q (consequence) Inverse: “If y  x, then –y  –x.” The inverse is true. Example 12 State the contrapositive of the following implication and hence, determine whether the contrapositive is true or false. “If Pandelela Rinong originates from Bau, then she originates from Sarawak.” Solution Not p (antecedent) Contrapositive: If Pandelela Rinong does not originate from Sarawak, then she does not originate from Bau. The contrapositive is true. Not q (consequence) 3.2 Arguments Example 13 Determine whether each of the following arguments is a deductive argument or an inductive argument. (a) The sum of all interior angles in a triangle is 180°. ABC is a triangle. Hence, the sum of all interior angles in triangle ABC is 180°. (b) 20° and 160° are complimentary angles. 110° and 70° are complimentary angles. Hence, the sum of two complimentary angles is 180°. (c) Salmon fish has high content of omega-3. The high level of omega-3 helps to maintain our cardiovascular health. Hence, eating lots of salmon fish could help to reduce the risk of heart attack. Solution (a) Deductive argument The premises are general and the conclusion is specific. (b) Inductive argument The premises are specific and the conclusion is general. (c) Inductive argument A strong conclusion based on convincing premises. Analysis SPM Maths 2023 Eng F4 C3 3rd.indd 27 17/2/2023 9:20:29 PM PENERBIT ILMU BAKTI SDN. BHD.

28 Form 4 CHAPTER 3 Example 14 Determine whether each of the following deductive arguments is sound or unsound. Give your reason. (a) Premise 1: All heptagons have nine sides. Premise 2: PQRSTUV is a heptagon. Conclusion: PQRSTUV has nine sides. (b) Premise 1: If the last two digits of a number are divisible by 4, then the number is divisible by 4. Premise 2: The two last digits of 924 is divisible by 4. Conclusion: 924 is divisible by 4. (c) Premise 1: All similar triangles are congruent triangles. Premise 2: ABC and XYZ are similar triangles. Conclusion: ABC and XYZ are congruent triangles. (d) Premise 1: If the gradients of two straight lines are equal, they are parallel. Premise 2: The straight line y = 2x + 4 and y = –2x + 4 has the same gradient. Conclusion: The straight line y = 2x + 4 and y = –2x + 4 is parallel. Solution (a) Unsound because the first premise is not true. This is because a heptagon has seven sides and not nine sides. (b) Sound because the deductive argument is valid and both premises and conclusion are true. (c) Unsound because the first premise is not true. Not reasonable because premise 1 is false. It should be ‘all congruent triangles are similar triangles’. (d) Unsound because premise two is not true. The gradient of the straight line y = 2x + 4 and y = –2x + 4 are 2 and –2 respectively, i.e. not the same. Example 15 Write a conclusion for each of the following deductive arguments. (a) Premise 1: All birds lay eggs. Premise 2: Penguin is a bird. Premise 3: Parrot is a bird. Conclusion: (b) Premise 1: If each interior angle of a regular polygon is 108°, then the polygon is a regular pentagon. Premise 2: Each interior angle of a regular polygon PQRST is 108°. Conclusion: (c) Premise 1: If x and y are opposite angles of a cyclic quadrilateral, then x + y is equal to 180°. Premise 2: x + y is not equal to 180°. Conclusion: Solution (a) Premise 1: All birds lay eggs. A B Premise 2: Penguin is a bird. C1 A Premise 3: Parrot is a bird. C2 A Conclusion: Penguin and parrots lay eggs. C1 C2 B p (b) Premise 1: If each interior angle of a regular polygon is 108°, then the polygon is a regular pentagon. q Premise 2: Each interior angle of a regular polygon PQRST is 108°. p is true. Conclusion: PQRST is a regular pentagon. q is true. Analysis SPM Maths 2023 Eng F4 C3 3rd.indd 28 17/2/2023 9:20:29 PM PENERBIT ILMU BAKTI SDN. BHD.

29 Form 4 CHAPTER 3 p (c) Premise 1: If x and y are opposite angles of a cyclic quadrilateral, then x + y is equal to 180°. q Premise 2: x + y is not equal to 180°. Not q is true. Conclusion: x and y are not opposite angles of a cyclic quadrilateral. Not p is true. Example 16 HOTS Applying Complete each of the following deductive arguments: (a) Premise 1: Premise 2: θ is a reflex angle. Conclusion: θ is greater than 180° and less than 360°. (b)Premise 1: If x 3 – 2 3 = 4, then x = 14. Premise 2: Conclusion: x = 14 (c) Premise 1: If x = 45°, then sin x = 1 2 . Premise 2: Conclusion: x ≠ 45° Solution (a) Deductive argument form I All A is B. Premise 1: Premise 2: θ is a reflex angle. C A C B Conclusion: θ is greater than 180° and smaller than 360°. A ∴ Premise 1: All reflex angles is greater than 180° and less than 360°. All A is B. B (b) Deductive argument form II p q Premise 1: If x 3 – 2 3 = 4, then x = 14. p is true. Premise 2: Conclusion: x = 14 q is true. ∴ Premise 2: x 3 – 2 3 = 4 p is true. (c) Deductive argument form III p q Premise 1: If x = 45º, then sin θ = 1 2 . Not q is true. Premise 2: Conclusion: x ≠ 45° Not p is true. ∴ Premise 2: sin θ ≠ 1 2 Not q is true. Example 17 Determine each of the following inductive arguments is strong or weak. (a) Premise 1: Pak Zainal is a grandfather. Premise 2: Pak Zainal is already half-bald because of hair fall. Conclusion: All grandfathers are half-bald. (b) Premise 1: The Menara Jam Condong is either situated in Teluk Intan or in Ipoh. Premise 2: The Menara Jam Condong is not situated in Ipoh. Conclusion: The Menara Jam Condong is situated in Teluk Intan. Solution (a) Weak There are people who have become a grandfather or grandmother below the age of 40. Hence, the conclusion that all grandfathers are half-bald is not firm. (b) Strong The premise and conclusion are logic and convincing. Analysis SPM Maths 2023 Eng F4 C3 3rd.indd 29 17/2/2023 9:20:30 PM PENERBIT ILMU BAKTI SDN. BHD.

30 Form 4 CHAPTER 3 Example 18 It is given that the volume of a cone is 1 3 πr2 h, such that r is the base-radius and h is the height of the cone. Make a conclusion based on deductive reasoning of the volume of a cone with a base-radius of 7 cm and a height of 12 cm. 3Use π = 22 7 4 Solution Volume of the cone = 1 3 × 22 7 × 72 × 12 = 616 cm3 Example 19 The following table shows two number sequences which follow a certain pattern. n nth term Sum of first n terms 1 10 = 10 + (1 – 1)(5) 10 = 1 2 [2(10) + (1 – 1)(5)] 2 15 = 10 + (2 – 1)(5) 25 = 2 2 [2(10)+(2 – 1)(5)] 3 20 = 10 + (3 – 1)(5) 45 = 3 2 [2(10)+(3 – 1)(5)] 4 25 = 10 + (4 – 1)(5) 70 = 4 2 [2(10)+(4 – 1)(5)]    n Tn Sn A piece of wire with a length of 325 cm is bent to form exactly n parts as shown in the following diagrams such that the lengths of the first three parts are 10 cm, 15 cm and 20 cm respectively. It is given that Tn and Sn represent the length of the nth part and the sum of the first n parts respectively. Find (a) Tn dan Sn in terms of n based on inductive reasoning, (b) the value of n, (c) the length of the last part of the wire. HOTS Applying HOTS Analysing HOTS Evaluating Solution (a) Tn = 10 + (n – 1)(5) = 5 + 5n, n = 1, 2, 3, 4, … Sn = n 2 [ 2(10) + (n – 1)(5)] = n 2 (15 + 5n), n = 1, 2, 3, 4, … (b) Sn = 325 n 2 (15 + 5n) = 325 n(15 + 5n) = 325 × 2 n(5n + 15) = 650 5n2 + 15n = 650 n2 + 3n = 130 n2 + 3n – 130 = 0 (n – 10)(n + 13) = 0 n = 10 or –13 n = –13 is not accepted. n = –13 is not accepted because n refers to the number of parts of the wire. n –10 × × n 13 –10n + 13n n2 –130 3n Divide each term by 5. ∴ n = 10 (c) T10 = 5 + 5(10) = 55 Hence, the length of the last part of the wire is 55 cm. (a) Complete the following deductive argument: Premise 1: Premise 2: θ ≠ 30° Conclusion: tan 30° ≠ 1 3 (b) Based on inductive reasoning, make a conclusion for the number squence 7, 20, 57, 166, …, which follows the following pattern: 7 = 2(3)1 + 1 20 = 2(3)2 + 2 57 = 2(3)3 + 3 166 = 2(3)4 + 4 Hence, determine the eighth term. HOTS Zone Analysis SPM Maths 2023 Eng F4 C3 3rd.indd 30 17/2/2023 9:20:31 PM PENERBIT ILMU BAKTI SDN. BHD.

31 Form 4 CHAPTER 3 1 Which of the following sentences is a statement? A What is the meaning of generalisation? B Go and wash your face. C Perak Wow! D The highest common factor of 18 and 27 is 9. 2 Determine the false statement. A Some quadratic expression cannot be factorised. B The Malaysia flag is also known as Jalur Gemilang and orang utan is the national animal of Malaysia. C The Malayan tiger is the national animal of Malaysia or rafflesia is the national flower of Malaysia. D If x < –5, then x < –2. 3 Given the implication “if m < –7, then m < –5”, determine the false statement. A The converse of the given implication is “if m < –5, then m < –7”. B The inverse of the given implication is “if m  –7, then m  –5”. C The contrapositive of the given statement is “if m  –5, then m  –7”. D The contrapositive of the given statement is false. 4 Which of the following sentences is a statement? A Pentagon is a polygon that has nine sides. B 2x2 + 5x + 2 C Wow! Your batik dress is so beautiful. D Lend teacher your scientific calculator for a while. 5 Premise 1: Patriotism is an important moral value. Premise 2: Love our king and country is a value of patriotism. Conclusion: We have to stand straight and sing together when the Negaraku is played. What is represented by the contents in the above box? A Deductive argument B Inductive argument C Deductive reasoning D Inductive reasoning 6 Which of the following compound statements is false? A –6 > –5 or 3 4 > 2 5 . B 49 is a perfect square or 1 is a prime number. C Kota Kinabalu is the capital of Sabah or Kuching is the capital of Sarawak. D All hexagons have six sides with equal length or some regular octagons have eight axis of symmetry. Solution (a) Argument form III If p, then q. Premise 1: Premise 2: θ ≠ 30° Not q is true. Conclusion: tan θ ≠ 1 3 Not p is true. ∴ Premise 1: If tan θ = 1 3 , then θ = 30°. Premise 1: If θ = 30°, then tan θ = 1  3 . [Incorrect] It must be “if p, then q´and not íf q, then p”. Caution (b) The nth term = 2(3)n + n, n = 1, 2, 3, 4, … The 8th term = 2(3)8 + 8 = 13 130 Objective Questions SPM Practice 3 Analysis SPM Maths 2023 Eng F4 C3 3rd.indd 31 17/2/2023 9:20:31 PM PENERBIT ILMU BAKTI SDN. BHD.

32 Form 4 CHAPTER 3 Subjective Questions 3.1 Statement 1 Determine whether each of the following sentences is a statement or not. Give your reason. (a) A whale is not a mammal. (b) What is the national animal of Malaysia? (c) Draw a stem-and-leaf plot. (d) Malaysia boleh! (e) (x + 2)(x – 2) = 0 is a quadratic equation. (f) 125 3 – 16 (g) The highest common factor of 15, 30 and 45 is 45. 2 Determine whether each of the following statements is true or false. (a) 32 × 16 < 33 + 23 (b) (x – 6)2 = x2 – 12x + 36 (c) The radius of a circle which passes through the contact point of a tangent to the circle is perpendicular to the tangent. (d) 0.0000075 = 7.5 × 10–5 (e) 50 km2 = 50 000 m2 (f) All roots of quadratic equations are real. (g) All mammals live on land. (h) All polygons with n sides have a sum of interior angle which is equal to (n – 2) × 180º. (i) Some decimals are rational numbers. (j) Some quadratic function graphs do not intersect the x-axis. (k) Some tetrahedrons have four faces. 3 Determine whether each of the following statements is true or false. Write a statement which changes the truth value of each given statement using the word “not” or “no”. (a) The Rajang River is situated in the state of Sabah. (b) 2 is the smallest prime number. (c) The sum of all the interior angles of a hexagon is 720°. (d) Sea turtles are reptiles. (e) cos 60° is equal to 3 2 . (f) Tun Dr. Mahathir Mohamad is the fourth and seventh Prime Minister of Malaysia. 7 Given the implication “if h > 8, then h > 5”, which of the following statements is false? A The converse of the implication is “if h > 5, then h > 8”. B The inverse of the implication is “if h  8, then h  5”. C The contrapositive of the implication is “if h < 5, then h < 8”. D The contrapositive of the implication is “if h  5, then h  8”. 8 Which of the following deductive arguments is unsound? A Premise 1: All triangles do not have any diagonal. Premise 2: KMN is a triangle. Conclusion: KMN does not have any diagonal. B Premise 1: If – 3 4 x > 6, then x < –8. Premise 2: – 3 4 x > 6 Conclusion: x < –8 C Premise 1: If x = 45°, then tan x = 1. Premise 2: tan x ≠ 1 Conclusion: x ≠ 45° D Premise 1: If p = 30°, the supplementary angle of p = 60°. Premise 2: p = 30° Conclusion: The supplementary angle of p = 60°. Analysis SPM Maths 2023 Eng F4 C3 3rd.indd 32 17/2/2023 9:20:31 PM PENERBIT ILMU BAKTI SDN. BHD.

33 Form 4 CHAPTER 3 4 Determine whether each of the following compound statements is true or false. (a) Alternate angles are equal and the angle of elevation is the angle between the horizontal line which passes through the eye of an observer and the straight line which connects the observer’s eye with an object situated above the horizontal line. (b) cos 60° = 1 2 and 1 000 3 = 100. (c) – 9 11 > – 6 11 and 9 11 > 6 11 . (d) Some of the straight lines y = mx + c (c ≠ 0) intersect both axes and the sum of all the interior angles of a hexagon is 360°. (e) The highest common factor of 18 and 27 is 9 and – 5 6 > – 1 2 . (f) The largest integer value of k for the inequality – 2 3 k > 4 is –6 and the lowest common multiple of 30 and 36 is 180. 5 Determine whether each of the following compound statements is true or false. (a) The length of each side of all regular polygons are equal or 729 is a perfect cube. (b) –x–1 = x or 1 century = 100 years. (c) A regular hexagon has six axes of symmetry or 2 × 10–3 × 7 × 105 = 1.4 × 102 . (d) The straight line with the equation y = 5 is parallel to the y-axis or sin 45° = 2 . (e) 10m4 n6 5m3 n5 = 2mn or the solution of the inequality –3x < 12 is x < –4. (f) 2 is the smallest prime number or a lizard is an amphibian. 6 State the antecedent and consequent of each of the following implications: (a) If the angle B of a triangle ABC is a right angle, then the triangle ABC is a right-angled triangle. (b) If PQRS is a quadrilateral, then PQRS has four sides. (c) If y is an odd number, then y is not divisible by 2. (d) If m > n, then –n > –m. (e) If two sets are equal, then both sets have the same elements. (f) If set A is a subset of set B, then all the elements of set A are also the elements of set B. 7 Construct an implication for each of the following pairs of antecedent and consequent. (a) Antecedent: x = 125 Consequence: x3 = 5 (b) Antecedent: PQRSTUV is a regular heptagon. Consequence: PQRSTUV has seven axes of symmetry.. (c) Antecedent: –2x > 8 Consequence: x < –4 (d) Antecedent: k is a positive integer. Consequence: 2k + 1 is an odd number. 8 Write two implications based on each of the following statements: (a) x > y if and only if 1 3 x > 1 3 y. (b) x2 = 49 if and only if x = ±7. (c) 3(x + 4) = 18 if and only if jika 3x + 12 = 18. (d) PQR is an equilateral triangle if and only if ∠P = ∠Q = ∠R = 60°. (e) The last two digits of a number n is divisible by 4 if and only if n is divisible by 4. (f) θ is a complementary angle of a if and only if θ + a = 90°. 9 Construct a statement in the form ‘if and only if’ from each of the following pairs of implications. (a) Implication 1: If a b is an improper fraction, then a > b. Implication 2: If a > b, then a b is an improper fraction. (b) Implication 1: If two straight lines are parallel, then their gradients are equal. Implication 2: If the gradients of two straight lines are equal, then the two straight lines are parallel. Analysis SPM Maths 2023 Eng F4 C3 3rd.indd 33 17/2/2023 9:20:31 PM PENERBIT ILMU BAKTI SDN. BHD.

34 Form 4 CHAPTER 3 (c) Implication 1: If a polygon is a quadrilateral, then the sum of its interior angles is 360°. Implication 2: If the sum of the interior angles of a polygon is 360°, then it is a quadrilateral. (d) Implication 1: If tan x = 1, then x = 45°. Implication 2: If x = 45°, then tan x = 1. 10 For each of the following implications, state the converse of the implication and hence, determine whether the converse is true or false. (a) If x > –5, then x > –8. (b) If y = 3x + 2, then x = y – 2 3 . (c) If ABCDEF is a regular hexagon, then each interior angle of ABCDEF is 120°. (d) If last digit of a number is 0, then the number is divisible by 5. (e) If a polygon is a rhombus, square or kite, then the diagonals of the polygon intersect at a right angle. (f) If a number p is a multiple of 6, then p is a multiple of 3. 11 For each of the following implications, state the inverse of the implication and hence, determine whether the inverse is true or false. (a) If h > 7 3 , then h > 4 3 . (b) If k < –15, then k < –10. (c) If p < – 7 9 , then p < – 4 9 . (d) If f(x) is a quadratic function, then its graph is a parabola. (e) If two straight lines do not intersect, then they are parallel. (f) If a number is negative, then the cube of the number is negative. 12 For each of the following implications, state the contrapositive of the implication and hence, determine whether the contrapositive is true or false. (a) If a polygon has five sides of equal length, then it is a regular pentagon. (b) If θ = 45°, then sin θ = 1 2 . (c) If x is a multiple of 12, then it is a multiple of 3. (d) If x < – 8 9 , then x < – 5 9 . (e) If a number is an odd number, then it is not divisible by 2. (f) If k is a factor of 12, then k is a factor of 24. 3.2 Arguments 13 Determine whether each of the following arguments is a deductive argument or inductive argument. (a) All regular polygons with n sides has n axis of symmetry. A regular heptagon has 7 sides. Hence, a regular heptagon has 7 axes of symmetry. (b) In a bowling game, Zahid’s first twenty throws fail to achieve a strike. Zahid’s coach gave him some guidance. Hence, in the following throw, Zahid will sure to achieve a strike. (c) If a road of length 30 km is drawn on a map with a length of 50 cm, then the scale used on the map is 1 : 60 000. The road PQ of length 30 km is drawn on a map with a length of 50 cm. Hence, the scale used on the map is 1 : 60 000. (d) 45° and 135° are two opposite angles in a cyclic quadrilateral. 60° and 120° are two opposite angles in a cyclic quadrilateral. Hence, the sum of two opposite angles in a cyclic quadrilateral is 180°. 14 Determine whether each of the following deductive arguments is sound or unsound. Give your reason. (a) Premise 1: All rational number can be expressed as a fraction a b . Premise 2: 0.8 can be expressed as a fraction 4 5 . Conclusion: 0.8 is a rational number. Analysis SPM Maths 2023 Eng F4 C3 3rd.indd 34 17/2/2023 9:20:32 PM PENERBIT ILMU BAKTI SDN. BHD.

35 Form 4 CHAPTER 3 (b) Premise 1: If p is a complementary angle of q, then p is the same as 180° – q. Premise 2: 60° is a complementary angle of q. Conclusion: p is equal to 180° – 60°. (c) Premise 1: Similar triangles have equal corresponding angles and sides. Premise 2: ABC and PQR are similar triangles. Conclusion: ABC and PQR have equal corresponding angles and sides. (d) Premise 1: If the diagonals of a quadrilateral intersect at a right angle, then it is a parallelogram. Premise 2: The diagonals of PQRS intersect at a right angle. Conclusion: The quadrilateral PQRS is a parallelogram. 15 State the conclusion based on the given premises. (a) Premise 1: All trapeziums have two parallel sides. Premise 2: KLMN is a trapezium. Conclusion: (b) Premise 1: If –5x  7, then x  – 7 5 . Premise 2: –5x  7 Conclusion: (c) Premise 1: All kites have only one axis of symmetry. Premise 2: PQRS is a kite. Conclusion: (d) Premise 1: If sin x = cos y = 1 2 , then x + y = 90º. Premise 2: sin x = cos y = 1 2 Conclusion: (e) Premise 1: If set P = φ, then n(P) = 0. Premise 2: n(P) ≠ 0 Conclusion: 16 Complete each of the following arguments: (a) Premise 1: Premise 2: PQRS is a parallelogram. Conclusion: PQRS is a quadrilateral. HOTS Applying HOTS Analysing (b) Premise 1: All pentagons have five sides. Premise 2: Conclusion: RSTUV have five sides. (c) Premise 1: If a m n = b, then a = b n m. Premise 2: Conclusion: a = b n m (d) Premise 1: If tan x = 3 , then x = 60º. Premise 2: Conclusion: tan x ≠ 3 (e) Premise 1: Premise 2: a b not a proper fraction. Conclusion: a > b HOTS Applying HOTS Analysing (f) Premise 1: Premise 2: Crocodiles are reptiles. Conclusion: Crocodiles are coldblooded HOTS Applying HOTS Analysing (g) Premise 1: If x is a supplementary angle of y, then x + y = 90°. Premise 2: Conclusion: x + y = 90° (h) Premise 1: If θ + a = 180°, then θ is a supplementary angle of a. Premise 2: Conclusion: θ + a ≠ 180° (i) Premise 1: Premise 2: h + k = 360° Conclusion: k is a conjugate angle of h. HOTS Applying HOTS Analysing 17 Determine whether each of the following inductive arguments is strong or weak. Give your reason. (a) Premise 1: There are 20 books on the top level of a bookrack and 15 books at the bottom level. Premise 2: There are no other books on the bookrack. Conclusion: There are 35 books on the bookrack. (b) Premise 1: Encik Hashim’s family members consist of his wife, Puan Sofi, his son, Sadini and his daughter, Noriah. Premise 2: Puan Sofi, Sadini and Noriah wear glasses. Conclusion: The whole family of Encik Hashim wear glasses. Analysis SPM Maths 2023 Eng F4 C3 3rd.indd 35 17/2/2023 9:20:32 PM PENERBIT ILMU BAKTI SDN. BHD.

36 Form 4 CHAPTER 3 (c) Premise 1: A scientist claims that chlorine and fluorine in the water supplied in Pulau Pinang can be harmful to the health of the residents in Pulau Pinang. Premise 2: No residents in Pulau Pinang who falls sick because of drinking the water supplied. Conclusion: The claim of the scientist cannot be accepted. (d) Premise 1: All burettes used by Form Six students of a school to conduct Chemistry experiment are in good conditions. Premise 2: All burettes used by Form Four and Form Five students of the school to conduct Chemistry experiment are also in good conditions. Conclusion: All burettes available in the school are in good condition. 18 Write a suitable conclusion so that the following inductive arguments is strong. (a) Premise 1: Drinks with soda have high contents of sugar which can cause diabetes if taken frequently. Premise 2: Encik Steven of age 55 drinks 6 bottles of drinks with soda each day. Conclusion: (b) Premise 1: Exercise is important to maintain a good health. Premise 2: Encik Jamaludin who has just retired at the age of 60 walks 5 km daily. Premis 3: Encik Ong exercise 20 minutes three times a day. Conclusion: (c) Premise 1: 20° and 70° are complimentary angles. Premise 2: 40° and 50° are complimentary angles. Conclusion: 19 The following diagram shows a right pyramid with a horizontal base of a square ABCD. M is the midpoint of the edge AB and V is the peak of the pyramid. V A M B C 10 cm 13 cm D Given that BC = 10 cm, VM = 13 cm and the volume of the pyramid is 1 3 × area of base × height, make a conclusion based on deductive reasoning of the volume of the pyramid. 20 Given that the curved surface area of a hemisphere and a cone are 2πr2 and πrs, respectively, such that r is the radius and s is the slant height of the cone. 50 cm 28 cm Make a conclusion based on deductive reasoning of the total of the curved surface area of the combined solid in the above diagram which consists of the combination of a hemisphere with a diameter of 28 cm and a cone with a slant height of 50 cm. 3Use p = 22 7 4 Analysis SPM Maths 2023 Eng F4 C3 3rd.indd 36 17/2/2023 9:20:32 PM PENERBIT ILMU BAKTI SDN. BHD.

37 Form 4 CHAPTER 3 21 The following table shows two number sequences which follow a certain pattern. n nth term Sum of the first n terms 1 1 = (2)1 – 1 1 = 21 – 1 2 2 = (2)2 – 1 3 = 22 – 1 3 4 = (2)3 – 1 7 = 23 – 1 4 8 = (2)4 – 1 15 = 24 – 1    n Tn Sn The following diagram shows a traditional game set which has holes in it. Suraini puts 1 red seed, 2 red seeds, 4 red seeds, 8 red seeds and so on into the holes. Suraini has 511 red seeds. HOTS Applying HOTS Analysing HOTS Evaluating It is given that Tn represents the number of red seeds put into the nth hole and Sn represents the sum of red seeds put into the first n holes. Find (a) Tn and Sn in terms of n based on deductive reasoning, (b) the value of n, (c) the number of red seeds put into the seventh hole. Objective Questions 1 D 2 B 3 D 4 A 5 B 6 D 7 C 8 D Subjective Questions 1 (a) Statement because we can determine that the sentence is false. (b) Non-statement because the sentence is a question. (c) Non-statement because the sentence is an order. (d) Non-statement because the sentence is an exclamation. (e) Statement because we can determine that the sentence is true. (f) Non-statement because we cannot determine whether the sentence is true or false. (g) Statement because we can determine that the statement is false. 2 (a) False (b) True (c) True (d) False (e) False (f) False (g) False (h) True (i) True (j) True (k) False 3 (a) False. The Rajang River is not situated in Sabah. [True] (b) True. 2 is not the smallest prime number. [False] (c) True. The sum of the interior angles of a hexagon is not 720º. [False] (d) True. Sea turtles are not reptiles. [False] (e) False. cos 60º is not equal to  3 2 . [True] (f) True. Tun Dr Mahathir Mohamad is not the fourth and seventh Prime Minister of Malaysia. [False] 4 (a) True (b) False (c) False (d) False (e) False (f) False 5 (a) True (b) True (c) True (d) False (e) True (f) True 6 (a) Antecedent: The angle B of the triangle ABC is a right angle. Consequent: The triangle ABC is a right angle triangle. (b) Antecedent: PQRS is a quadrilateral. Consequent: PQRS has four sides. (c) Antecedent: y is an odd number. Consequent: y is not divisible by 2. (d) Antecedent: m > n Consequent: –n > –m (e) Antecedent: Two sets are equal. Consequent: Both sets have the same elements. Answers Analysis SPM Maths 2023 Eng F4 C3 3rd.indd 37 17/2/2023 9:20:33 PM PENERBIT ILMU BAKTI SDN. BHD.

38 Form 4 CHAPTER 3 (f) Antecedent: Set A is a subset of set B. Consequent: All the elements of set A are also the elements of set B. 7 (a) If x = 125, then 3x = 5. (b) If PQRSTUV is a regular heptagon, then PQRSTUV has seven axes of symmetry. (c) If –2x > 8, then x < –4. (d) If k is a positive integer, then 2k + 1 is an odd number. 8 (a) Implication 1: If x > y, then 1 3 x > 1 3 y. Implication 2: If 1 3 x > 1 3 y, then x > y. (b) Implication 1: If x2 = 49, then x = ±7. Implication 2: If x = ± 7, then x2 = 49. (c) Implication 1: If 3(x + 4) = 18, then 3x + 12 = 18. Implication 2: If 3x + 12 = 18, then 3(x + 4) = 18. (d) Implication 1: If PQR is an equilateral triangle, then ∠P = ∠Q = ∠R = 60°. Implication 2: If ∠P = ∠Q = ∠R = 60°, then PQR is an equilateral triangle. (e) Implication 1: If the last two digits of a number n is divisible by 4, then n is divisible by 4. Implication 2: If a number n is divisible by 4, then the last two digits of n is divisible by 4. (f) Implication 1: If θ is a complementary angle of α, then θ + α = 90°. Implication 2: If θ + α = 90°, then θ is a complementary angle of α. 9 (a) a b is an improper angle if and only if a > b. (b) Two straight lines are parallel if and only if their gradients are equal. (c) A polygon is a quadrilateral if and only if the sum of its interior angles is 360°. (d) tan x = 1 if and only if x = 45º. 10 (a) If x > –8, then x > –5. [False] (b) If x = y – 2 3 , then y = 3x + 2. [True] (c) If each interior angle of ABCDEF is 120º, then ABCDEF is a regular hexagon. [True] (d) If a number is divisible by 5, then the last digit of the number is 0. [False] (e) If the diagonals of a polygon intersect at a right angle, then the polygon is a rhombus, square or kite. [True] (f) If a number p is a multiple of 3, then p is a multiple of 6. [False] 11 (a) If h  7 3 , then h  4 3 . [False] (b) If k  –15, then k  –10. [False] (c) If p  – 7 9 , then p  – 4 9 . [False] (d) If f(x) is not a quadratic function, then its graph is not a parabola. [True] (e) If two straight lines intersect, then they are not parallel. [True] (f) If a number is not negative, then the cube of the number is not negative. [True] 12 (a) If a polygon is not a regular pentagon, then it cannot have five sides of equal length. [True] (b) If sin θ ≠ 1  2 , then θ ≠ 45°. [True] (c) If x is not a multiple of 3, then it is not a multiple of 12. [True] (d) If x  – 5 9 , then x  – 8 9 . [True] (e) If a number is divisible by 2, then the number is not an odd number. [True] (f) If k is not a factor of 12, then k is not a factor of 24. [True] 13 (a) Deductive argument (b) Inductive argument (c) Deductive argument (d) Inductive argument 14 (a) Sound because the deductive argument is valid and both premises and conclusion are true. (b) Unsound because the first premise is not true. The word “complementary” should be “supplementary”. (c) Unsound because the first premise is not true. The word ‘similar’ should be ‘congruent’. (d) Unsound because Premise 2 may not be true. This is because the diagonals of a parallelogram may not intersect at a right angle unless it is a rhombus or a square. 15 (a) KLMN has two parallel sides. (b) x  – 7 5 (c) PQRS has one axis of symmetry only. (d) x + y = 90° (e) Set P ≠ φ 16 (a) All parallelogram is a quadrilateral. (b) RSTUV is a pentagon. (c) a m n = b (d) x ≠ 60° (e) If a < b, then a b is a proper fraction. (f) All reptiles are coldblooded. (g) x is a complementary angle of y. (h) θ is not a supplementary angle of α. (i) If h + k = 360°, then k is a conjugate angle of h. Analysis SPM Maths 2023 Eng F4 C3 3rd.indd 38 17/2/2023 9:20:33 PM PENERBIT ILMU BAKTI SDN. BHD.

39 Form 4 CHAPTER 3 17 (a) Strong because its premises and conclusion are logic and convincing. (b) Weak because the inductive argument does not state whether Encik Hashim wears glasses or not causing its conclusion to be not cogent. (c) Strong because its premises and conclusion are logic and convincing. (d) Weak because there is a possibility the lower form students do not have a chance to use good condition burettes to conduct their Science experiments. 18 Suggested answers: (a) Encik Steven has high possibility to be diabetic. (b) The possibility of Encik Jamaludin and Encik Ong to maintain a good health is high. (c) The sum of two complimentary angles is 90°. 19 400 cm3 20 3 432 cm2 21 (a) Tn = 2n – 1 , n = 1, 2, 3, 4, … Sn = 2n – 1, n = 1, 2, 3, 4, … (b) n = 9 (c) 64 red seeds Analysis SPM Maths 2023 Eng F4 C3 3rd.indd 39 17/2/2023 9:20:33 PM PENERBIT ILMU BAKTI SDN. BHD.

40 4.1 Intersection of Sets 1 Intersection of sets P and Q is written as P ∩ Q. It is the set which contains the common elements of both of the sets P and Q. 2 In the following Venn diagram, the shaded region represents P ∩ Q.  P Q The shaded region consists of the common elements of the sets P and Q. 3 The intersection between the sets P, Q and R is written as P ∩ Q ∩ R. It is the set which contains the common elements of all the three sets P, Q and R. 4 In the following Venn diagram, the shaded region represents P ∩ Q ∩ R. P Q R  The shaded region consists of the common elements of the sets P, Q and R. 5 It is given that the universal set, ξ = {x :1  x 10, x is an integer}, set F = {x : x is a prime number} and set G = {x : x is an odd number}. Thus, the universal set, ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} set F = {2, 3, 5, 7} set G = {1, 3, 5, 7, 9} Hence, F ∩ G = {x : x is a prime number and x is an odd number} = {3, 5, 7} Representative using the set builder notation of set F ∩ G. SMART TIP Representative by description of F ∩ G is “a prime number and an odd number”. 6 If P ⊂ Q, then P ∩ Q = P. This relation is shown in the following Venn diagram. P Q  7 If set A and set B do not intersect between each other, then A ∩ B = φ. The relation is shown in the following Venn diagram. A B  40 EXPRESS NOTES Operations on Sets Chapter 4 Learning Area: Discrete Mathematics Analysis SPM Maths 2023 Eng F4 C4 2nd.indd 40 17/2/2023 9:27:23 PM PENERBIT ILMU BAKTI SDN. BHD.

41 Form 4 CHAPTER 4 Complement of the intersection of sets 8 The complement of the intersection of two sets, A and B, is represented by (A ∩ B)'. It is a set that contains all the elements in the universal set, ξ, which are not the elements of A ∩ B. 9 (A ∩ B)' is represented by the shaded region that is shown in the following Venn diagram. A B  11 The following Venn diagram shows three group of students of a school regarding the fruits they like. It is given that the universal set, ξ = {students of a school}, set B = {students who like papayas}, SMART TIP In problems that represent the real-life situations, (A ∩ B)' represents “not the combinations of both sets A and B”. SMART TIP • A ∩ φ = φ • A ∩ ξ = A • A ∩ A' = φ Solving problems involving intersection of sets 10 The following Venn diagram shows two groups of students of a school regarding the types of novels they like. It is given that the universal set, ξ = {students of a school}, set R = {students who like romantic novel} and set P = {student who like investigation novel}. set T = {students who like watermelons} and set J = (students who like guavas}. R P  Students who like investigation novels only. Students who like romantic novel only. Students who like both romantic and investigation novels. Students who do not like romantic nor investigation novels. 41 Analysis SPM Maths 2023 Eng F4 C4 2nd.indd 41 17/2/2023 9:27:23 PM PENERBIT ILMU BAKTI SDN. BHD.

42 Form 4 CHAPTER 4 4.2 Union of Sets 12 The union between two sets, P and Q, is written as P ∪ Q. It is a set which consists of all the elements in set P or set Q or both. 13 In the following Venn diagram, the shaded region represents P ∪ Q. P Q  P only P and Q Q only 14 The union between three sets, P, Q and R, is written as P ∪ Q ∪ R. It is a set which consists of the elements in set P or set Q or set R or all of them. 15 In the following Venn diagram, the shaded region represents P ∪ Q ∪ R. P Q R  16 If P ⊂ Q, then P ∪ Q = Q. The relation is shown in the following Venn diagram. Q P  B T J  Students who like watermelons and guavas only. Students who like watermelons only. Students who like all the three papayas, watermelons and guavas. Students who like papayas and guavas only. Students who like papayas only. Students who like papayas and watermelons only. Students who do not like papayas Students who like guavas only. nor watermelons nor guavas. 42 Analysis SPM Maths 2023 Eng F4 C4 2nd.indd 42 17/2/2023 9:27:24 PM PENERBIT ILMU BAKTI SDN. BHD.

43 Form 4 CHAPTER 4 17 If set A and set B do not intersect between each other, i.e. A ∩ B = φ, then the region which represents A ∪ B is as shown in the following Venn diagram. A B  Complement of the union of sets 18 The complement of the union of two sets, A or B, is represented by (A ∪ B)'. It is a set which contains all the elements in the universal set, ξ, which are not the elements of A ∪ B. 19 (A ∪ B)' is represented by the shaded region shown in the following Venn diagram. A B  SMART TIP In problems which involve real-life situations, (A ∪ B)' represents “not set A nor set B”. SMART TIP • A ∪ φ = A • A ∪ ξ = ξ • A ∪ A' = ξ Solving problems involving union of sets 20 The problems in our daily real-life situations can be solved using a Venn diagram which involves union of sets. 4.3 Combined Operations on Sets 21 Combined operations on sets involve the combination of ∩ and ∪. 22 When implementing combined operations on sets, implement it from left to right. 23 If the combined operations on sets involve braces, implement the operation in the braces first. Complement of the combined operations on sets 24 The combined operations on sets involve the usage of the symbols intersection ‘∩’, union ‘∪’, complement of set e.g. ‘(A')’ and braces ‘( )’. 25 The following Venn diagram shows the set (A ∩ B) ∪ (B ∩ C) ∪ (A ∩ C), such that ξ = A ∪ B ∪ C. A B C Hence, the complement of the set (A ∩ B) ∪ (B ∩ C) ∪ (A ∩ C) is written as [(A ∩ B) ∪ (B ∩ C) ∪ (A ∩ C)]' and the following Venn diagram shows the complement of the combined operations on sets. A B C Solving problems involving combined operations on sets 26 Combined operations on sets can be applied to solve problems in our daily real-life situations. Analysis SPM Maths 2023 Eng F4 C4 2nd.indd 43 17/2/2023 9:27:25 PM PENERBIT ILMU BAKTI SDN. BHD.

44 Form 4 CHAPTER 4 4.1 Intersection of Sets Example 1 It is given that the universal set, ξ ={x : 1  x  8, x is integer}, set P = {x : x is a factor of 8} and set Q = {x : x is an even number}. List the elements of P ∩ Q. Solution ξ = {1, 2, 3, 4, 5, 6, 7, 8} P = {1, 2, 4, 8} Q = {2, 4, 6, 8} P ∩ Q = {2, 4, 8} The common elements of the sets P and set Q. Example 2 It is given that the universal set, ξ = {a, b, c, e, f, o, q, r, u, x, y, z}, set P = {a, e, o, u, q, r}, set Q = {a, b, c, e, o, f} and set R = {a, c, e, u, x, y}. (a) Draw a Venn diagram to represent the elements of the universal set ξ, set P, set Q and set R. (b) Shade the region which represents P ∩ Q ∩ R. Solution (a) P Q R z r q o u e a b f c x y  (b) P ∩ Q = {a, e, o} ∴ P ∩ Q ∩ R = {a, e} P Q R z r q o u e a b f c x y  Example 3 It is given that the universal set, ξ = {x : 2  x  8, x is an integer}, set A = {x : x is a multiple of 2} and set B = {x : x is a multiple of 4}. (a) List the elements of A ∩ B. (b) (i) Draw a Venn diagram to represent the elements of the universal set ξ, set A and set B. (ii) Shade the region which represents A ∩ B. Solution (a) ξ = {2, 3, 4, 5, 6, 7, 8} A = {2, 4, 6, 8} B = {4, 8} A ∩ B = {4, 8} (b) (i) 3 5 7 2 6 4 8 B A  (ii) 3 5 7 2 6 4 8 B  A Example 4 Given that the universal set, ξ = {x : 11  x  18, x is an integer}, set A = {x : x is an odd number} and set B = {x : x is an even number}, (a) Draw a Venn diagram to represent the elements of the universal set ξ, set A and set B. (b) State the set which represents A ∩ B. Solution (a) A = {11, 13, 15, 17} B = {12, 14, 16, 18} A B 11 15 13 17 12 16 14 18  (b) A ∩ B = φ Analysis SPM Maths 2023 Eng F4 C4 2nd.indd 44 17/2/2023 9:27:26 PM PENERBIT ILMU BAKTI SDN. BHD.

45 Form 4 CHAPTER 4 Example 5 The following Venn diagram shows the sets A, B and C. On the Venn diagram, shade the set A ∩ B ∩ C. A B C Solution Step 1: Shade the region which represents A ∩ B using the pattern . A B C Step 2: Shade the region which represents set C using the pattern . A B C Alternative Method Label each region using Roman numerals. A B C I II III IV V List the region of each set. Set A = {II, III, IV, V} Set B = {I, II, III} Set C = (III, IV) A ∩ B ∩ C = {III} Shade region III, as follows: A B C I II III IV V Step 3: Shade the region of intersection between the two different shading patterns at Step 1 and Step 2. A B C Example 6 It is given that the universal set, ξ = {x : 1  x  12, x is an integer}, set A = {3, 5, 6, 7, 9} and set B = {x : x is a multiple of 3}. (a) List the elements of (i) A ∩ B, (ii) (A ∩ B)'. (b) Draw a Venn diagram to represent the universal set ξ, set A and set B. Hence, shade the set (A ∩ B)'. Solution (a) ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} A = {3, 5, 6, 7, 9} B = {3, 6, 9, 12} (i) A ∩ B = {3, 6, 9} (ii) (A ∩ B)' = {1, 2, 4, 5, 7, 8, 10, 11, 12} (b) A B 3 6 9 12 1 2 4 5 7 11 10 8  Example 7 HOTS Applying HOTS Analysing The Venn diagram shows three groups of students of a school based on their cocurriculum activities. It is given that the universal set, ξ = M ∪ E ∪ C, set M = {students who are members of the Mathematics Society}, set E = {students who are members of the English Society} and set C = {students who are members of the Malay Language Society}. Analysis SPM Maths 2023 Eng F4 C4 2nd.indd 45 17/2/2023 9:27:28 PM PENERBIT ILMU BAKTI SDN. BHD.

46 Form 4 CHAPTER 4 50 students are members of the Mathematics Society, 55 students are members of the English Society, 62 students are members of the Malay Language Society, 26 students are members of the Mathematics Society and the English Society, 29 students are members of the Mathematics Society and the Malay Language Society, 31 students are members of the English Society and the Malay Language Society and 15 students are members of all the three societies. M E C Calculate the number of members who are (a) the members of the Mathematics Society and the English Society but not the members of the Malay Language Society, (b) the members of two societies only, (c) the members of one society only. Solution Label each region in the Venn diagram with I, II, III, IV, V, VI and VII respectively, as follows. M E C V II I VI III IV VII • 15 students are members of all the three societies. Hence, n(region I) = 15. • 26 students are members of the Mathematics Society and the English Society. Hence, n(region II) = 26 – 15 = 11. • 29 students are members of the Mathematics Society and the Malay Language Society. Hence, n(region III) = 29 – 15 = 14. • 31 students are members of the English Society and the Malay Language Society. Hence, n(region IV) = 31 – 15 = 16. • 50 students are members of the Mathematics Society. Hence, n(region V) = 50 – n(region I, II and III) = 50 – (15 + 11+ 14) = 10 • 55 students are members of English Society. Hence, n(region VI) = 55 – n(region I, II and IV) = 55 – (15 + 11 + 16) = 13 • 62 students are members of the Malay Language Society. Hence, n(region VII) = 62 – n(region I, III and IV) = 62 – (15 + 14 + 16) = 17 Hence, the complete Venn diagram with the number of the members in each region is as follows. M E C 10 11 15 13 14 16 17 You should pay attention to whether the given information regarding the number of elements contains the word ‘only’ or not. For example, the information ‘26 students are members of the Mathematics Society and the English Society’ involves both of the region I and region II’ but if the given information is ‘26 students are members of the Mathematics Society and the English Society only’, then it involves the region II only. Caution In a Venn diagram, ‘• 5’ (with a dot on the left) represents an element in a set while ‘5’ (without any dot on the left) represents the number of elements in a set. Caution (a) SMART TIP Students who are the members of the Mathematics Society and the English Society but not the members of the Malay Language Society is represented by the set (M ∩ E) ∩ C'. Analysis SPM Maths 2023 Eng F4 C4 2nd.indd 46 17/2/2023 9:27:28 PM PENERBIT ILMU BAKTI SDN. BHD.

47 Form 4 CHAPTER 4 Step 1: Shade the region M ∩ E using the pattern . M E C Step 2: Shade the region C' using the pattern . M E C Step 3: Shade the intersection region between the two different patterns of Step 1 and Step 2. M 10 11 13 14 15 17 16 E C Hence, the number of the Mathematics Society and the English Society members but not the members of the Malay Language Society is 11. (b) The number of members of two societies only = n(region II) + n(region III) + n(region IV) = 11 + 14 + 16 = 41 (c) The number of members of one society only = n(region V) + n(region VI) + n(region VII) = 10 + 13 + 17 = 40 4.2 Union of Sets Example 8 Given the universal set, ξ = {x : 1  x  12, x is an integer}, set P = {x : x is a factor of 12} and set Q = {x : x is a prime number}. List the elements of P ∪ Q. Solution ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} P = {1, 2, 3, 4, 6, 12} Q = {2, 3, 5, 7, 11} P ∪ Q = {1, 2, 3, 4, 5, 6, 7, 11, 12} Example 9 It is given that the universal set, ξ = {3, 4, 5, 8, 9, 10, 11, 12, 13, 15}, set P = {3, 4, 5, 8, 10}, set Q = {3, 5, 9, 12} and set R = {11, 13}. (a) List the elements of (i) P ∩ Q, (ii) P ∪ Q ∪ R. (b) Draw a Venn diagram to represent the elements in the universal set ξ, set P, set Q and set R. (c) Shade the region which represents P ∪ Q ∪ R. Solution (a) (i) P ∩ Q = {3, 5} (ii) P ∪ Q ∪ R = {3, 4, 5, 8, 9, 10, 11, 12, 13} (b) 4 8 10 3 5 9 12 P Q R 15 11 13  (c) 4 8 10 3 5 9 12 P Q R 15 11 13  Example 10 It is given that the universal set, ξ = {x : 2  x  10, x is an integer}, set A = {x : x is an integer between 2 and 11} and set B = {x : x is a perfect square}. Analysis SPM Maths 2023 Eng F4 C4 2nd.indd 47 17/2/2023 9:27:30 PM PENERBIT ILMU BAKTI SDN. BHD.