Must Know iii – x Chapter 1 Functions 1 – 11 NOTES 1 Paper 1 2 Paper 2 9 Chapter 2 Quadratic Functions 12 – 21 NOTES 12 Paper 1 14 Paper 2 19 Chapter 3 Systems of Equations 22 – 26 NOTES 22 Paper 2 23 Chapter 4 Indices, Surds and Logarithms 27 – 38 NOTES 27 Paper 1 28 Paper 2 37 Chapter 5 Progressions 39 – 48 NOTES 39 Paper 1 40 Paper 2 47 Chapter 6 Linear Law 49 – 55 NOTES 49 Paper 1 49 Paper 2 54 Chapter 7 Coordinate Geometry 56 – 64 NOTES 56 Paper 1 57 Paper 2 61 Chapter 8 Vectors 65 – 76 NOTES 65 Paper 1 66 Paper 2 74 Chapter 9 Solution of Triangles 77 – 82 NOTES 77 Paper 2 78 Chapter 10 Index Number 83 – 88 NOTES 83 Paper 2 83 Form 4 Assessment 89 – 97 Answers 98 – 142 ii Contents 00A_1202 QB AMath F4.indd 2 09/05/2022 11:29 AM
Important Facts (Chapter 1) 1 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 7 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 1) 3 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 9 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 1) 5 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 11 @ Pan Asia Publications Sdn. Bhd. Composite Function and Inverse Function Functions Identify a Function MUST KNOW Important Facts 1. Function: 2. 4 types of relations: Domain = {a, b, c} Codomain = {1, 2, 3} Objects = a, b, c Images = 1, 2, 3 Range = {1, 3} A function Not a function The vertical line cuts the graph at two points Not a function A function One-to-one One-to-many Many-to-one Many-to-many Forms of Quadratic Functions 1. General form f(x) = ax2 + bx + c, a ≠ 0, b and c are constant 2. Vertex form f(x) = a(x + h)2 + k, a ≠ 0, h and k are constant 3. Intercept form f(x) = a(x – p)(x – q), a ≠ 0, p and q are constant f(x) = a(x – h)2 + k f(x) = ax2 + bx + c f(x) = a(x – p)(x – q) Expansion Factorisation or formula Completing the square Expansion 1. If f : x → y, then f(x) = y. 2. By using vertical line test: If any vertical line intersects f(x) graph at not more than one point, then it is a function. A function Not a function 1. Composite function: f f –1 g–1 f –1g –1 = (gf ) –1 g gf ● f(x) ● x ● g[f(x)] 2. Characteristics of inverse function: (a) Only one-to-one function has an inverse function. (b) If (a, b) is a point on the graph f(x), then (b, a) is its corresponding point on f –1(x). gf(x) ≠ fg(x) f 2 (x) = ff(x), f 3 (x) = fff(x) = f 2 f(x) = ff 2 (x) Type of Roots for Quadratic Equations The type of roots obtained depends on the discriminant, D = b2 – 4ac Discriminant D = b2 – 4ac a . 0 a , 0 D . 0 Two real and distinct roots y x y x D = 0 Two real and equal roots y x y x D , 0 No real roots y x y x a ● ● 1 b ● ● 2 c ● ● 3 x y x y x y y = x f –1(x) f(x) x y Solving Quadratic Equation 1. Three ways to solve the quadratic equations: (a) Factorisation (Use the principle “If pq = 0, then p = 0 or q = 0”) (b) Completing the squares (c) Formula x = –b + ABBBBBB b2 – 4ac 2a 2. If the roots are given, then the equation can be obtained by: x2 – (Sum of roots)x + (Product of roots) = 0 Solving Quadratic Inequalities 1. Three ways to solve the quadratic inequalities: (a) Graph sketching (b) Number line (c) Table y y = f(x) = (x + 2)(x – 4) y > 0 y > 0 x < –2 x > 4 –2< x < 4 y < 0 –2 0 4 x 00B_1202 QB AMath F4.indd 3 09/05/2022 11:30 AM
Common Mistakes (Chapter 2) 8 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 1) 2 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 2) 10 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 1) 4 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 2) 12 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 1) 6 @ Pan Asia Publications Sdn. Bhd. Vertex Form of Quadratic Equations Roots for Quadratic Equations Solving Quadratic Equations Use the method of completing the squares to solve 2x2 + 5x + 1 = 0 . Correct Wrong 2x2 + 5x + 1 = 0 21x2 + 5 2 x2 + 1 = 0 231x + 5 4 2 2 – 25 16 4 + 1 = 0 21x + 5 4 2 2 – 25 8 + 1 = 0 21x + 5 4 2 2 = 17 8 1x + 5 4 2 2 = 17 16 x = – 5 4 ± ABBB17 16 = –0.219, –2.281 2x2 + 5x + 1 = 0 12x + 5 2 2 2 + 1 = 0 Correct Wrong f(x) = 2x2 + 4x – 3 = 21x2 + 2x – 3 2 2 = 21(x + 1)2 – 1 – 3 2 2 = 2(x + 1)2 – 5 f(x) = 2x2 + 4x – 3 = (2x + 2)2 – 4 – 3 = (2x + 2)2 – 7 The coordinates of the minimum point is (–1, –5). The coordinates of the minimum point is (1, –5). Correct Wrong Sum of roots = – 1 2 Product of roots = – 6 2 = –3 Sum of roots = 1 or –1 Product of roots = –6 Correct Wrong fg(x) = 2x – 9 Let y = 2x – 9 then x = y + 9 2 (fg)–1(y) = y + 9 2 (fg)–1(x) = x + 9 2 (fg)–1(x) = f –1 g–1(x) = f –1(x + 5) = (x + 5) – 1 2 = x + 4 2 Correct Wrong fg(x) = f(x – 5) = 2(x – 5) + 1 = 2x – 10 + 1 = 2x – 9 fg(x) = g(2x + 1) = 2x + 1 – 5 = 2x – 4 Complete the square for f(x) = 2x2 + 4x – 3. Hence, find the coordinates of the minimum point. MUST KNOW Common Mistakes Note: When 12x + 5 2 2 2 is expended, it becomes 4x2 + 10x + 25 4 . Functions The diagram shows the function f maps set A to set B and g maps set B to set C. Find g(x). Correct Wrong g(x + 2) = 5x + 4 Let y = x + 2 Then, x = y − 2 g(y) = 5(y – 2) + 4 = 5y – 10 + 4 = 5y – 6 ∴ g(x) = 5x – 6 g(x + 2) = 5x + 4 g(y) = 5y + 4 Composite Function If f(x) = 2x + 1 and g(x) = x – 5, find fg(x). Determine the sum of roots and product of roots for the quadratic equation 2x2 + x – 6 = 0. Note: For ax2 + bx + c = 0. Then, Sum of roots = – b a Product of roots = c a Note: (fg)–1(x) ≠ f –1 g–1(x) (fg) –1(x) = g–1 f –1(x) fg(x) ≠ gf(x) Inverse Function If f –1(x) = x – 1 2 , g–1(x) = x + 5 and fg(x) = 2x – 9, find (fg)–1 (x). x f g x + 2 5x + 4 Substitute g(x) first followed by f(x). Note: Do not substitute f(x) first. (2x + 2)2 ≠ 2x2 + 4x + 4 Note: The vertex for f(x) = a(x – h) 2 + k is (h, k). A B C 00B_1202 QB AMath F4.indd 4 09/05/2022 11:30 AM
NOTES 1 Chapter 1 Functions 1.1 Functions 1. Function is a relation between two sets, domain and codomain. 2. Elements in the domain are called objects while the elements in the codomain are called images. 3. The diagram shows a function that connecting set X to set Y which represents in an arrow diagram. Set X Set Y Function a ● ● 1 b ● ● 2 c ● ● 3 The function is denoted by f : x → y or f(x) = y. Then object = {a, b, c} image = {1, 2, 3} 4. There are four types of relations: (a) One-to-one (b) Many-to-one a ● ● 1 b ● ● 2 c ● ● 3 a ● ● 1 b ● ● 2 c ● ● 3 (c) One-to-many (d) Many-to-many a ● ● 1 b ● ● 2 c ● ● 3 a ● ● 1 b ● ● 2 c ● ● 3 5. Function is a special relation such that: (a) Every element in the domain must map to one element in the codomain. (b) More than one element in the domain map to one element in the codomain. 6. A function maps onto itself if f : x → x or f(x) = x 7. Relation representations: (a) Arrow diagram (b) Graph p ● ● 4 q ● ● 6 r ● ● 8 p q r x y 2 4 6 8 (c) Ordered pairs (p, 6), (q, 4) , (r, 8) 8. A function is not defined if there exists a function in fraction form and its denominator is zero. For example, f : x → x 2x – 1 if x = 1 2 , then f(x) = 1 2 21 1 2 2 – 1 = 1 2 0 , no solution 9. Vertical line test: If a vertical line parallel to the y-axis intersects a graph more than once, then the algebraic expression is not a function. If it intersects once, then it is a function. x f y 0 Vertical line test 10. An absolute function is a function whose value is positive only. f(x) = |x| = x, if x > 0 –x, if x , 0 x f(x) f(x) = |x| f(x) = x Graph f(x) is always positive 11. Discrete function is a function where the points on the graph is real, separated and not connected by a straight line or curve. x f(x) 0 1 1 2 3 –1 2 3 Domain = {−1, 1, 2, 3} Codomain = {1, 2, 3} Range = {1, 2, 3} 01_1202 QB AMath F4.indd 1 09/05/2022 11:30 AM
22 2 SOS TIP Question 1(a): Write f(x) = ax + b, then find a and b. 12. Continuous function is a function where the points are connected by a straight line or a curve within the given interval. x f(x) 0 3 4 2 4 Domain of f is 0 < x < 4 Codomain of f is 0 < f(x) < 4 Range of f is 0 < f(x) < 4 1.2 Composite Functions 1. Composite function: y = f(x) z = g(f(x)) gf A x B f g C 2. If a function f maps x to y, written as f(x) = y and another function g maps y to z, written as g(y) = z, then gf(x) maps x to z and is written as gf(x) = z. 3. gf(x) ≠ fg(x) 4. f 3 (x) = fff(x) = ff 2 (x) = f 2 f(x) 1.3 Inverse Functions 1. An inverse function maps an image to its object. x ● f f –1 ● y If f(x) = y, then f –1(y) = x 2. Properties of inverse function: (a) A function f that maps set A to set B has an inverse function f –1 if f is a one-to-one function. (b) fg(x) = x where x in the domain of g and gf(x) = x where x in the domain of f. (c) If two functions, f and g are inverse to each other, then (i) domain of f = range of g, (ii) domain of g = range of f, (iii) graph g is a reflection of graph f. (d) For any real numbers, a and b, if the point (a, b) is on the graph f, then the point (b, a) is on the graph g. 3. Horizontal line test: If the horizontal line intersects the graph of a function at only one point, then this type of function is one-to-one function and it has an inverse function. x f(x) f 0 Horizontal line test 1. The diagram shows a relation between set A and set B. p ● f ● q 6 ● ● 13 8 ● ● 17 A B State (a) the function of f(p), [2 marks] (b) q in terms of p. [3 marks] CLONE SPM Section A PAPER 1 Answer: (a) (b) 01_1202 QB AMath F4.indd 2 09/05/2022 11:30 AM
5 5SOS TIP Question 11(a)(iii): 0 < f(x) < 1 is the same as 0 < |3 – 1 2 x| < 1. Question 12: (a) Sketch the graph y = 2x – 1. Then, reflect the part of the graph that is below the x-axis on the x-axis. (b) After finding the expression for ff(x) and fff(x), observe for the pattern formed. Section B 11. (a) The diagram shows a graph for the function f : x → |3 – 1 2 x| for the domain –2 < x < 9. x 0 (–2, 4) 1 4 9 f(x) State (i) the object for 4, [1 mark] (ii) the image for 1, [1 mark] (iii) the domain for 0 < f(x) < 1. [2 marks] (b) The diagram shows an inverse function where x ≠ –q. x g–1 1 – 2 3 p –––– q + x –4 –2 Find (i) the value of p and q, [1 mark] (ii) g(x) with the value of p and q from (i), [2 marks] (iii) the value of x if g(x) = 8. [1 mark] Answer: (a) (i) (ii) (iii) (b) (i) CLONE SPM (ii) (iii) 12. (a) Sketch the graph f(x) = |2x – 1| for –1 < x < 2. [3 marks] (b) Given f : x → 2x. (i) Find (a) f 2 , (b) f 3 . [2 marks] (ii) Hence, find the expression for f n , in terms of n where n = 1, 2, 3,…. [1 mark] (iii) Find the value of x if f 5 (x) = 16. [2 marks] Answer: (a) (b) (i) (a) (b) (ii) (iiii) 01_1202 QB AMath F4.indd 5 09/05/2022 11:30 AM
9 9SOS TIP Question 2(a)(ii) : To find g–1(x), let y = x + 1 x – 2 Question 7(b) : Find h–1(x) and then solve h–1(x) = h(x) to find the value of x which is given to be 3. Section A PAPER 2 1. Given two functions defined by f : x → 3 4 x + 1 2 and g : x → 5 4 – 2 3 x. (a) Is f(2) + f(3) = f(2 + 3)? Explain your answer. [2 marks] (b) Is g(4) − g(2) = g(4 − 2)? Show your working. [2 marks] (c) Find the value of k if f(2k) = 6g(k). [2 marks] (d) Find the value of k if f(k) + g(k) = 5. [2 marks] 2. (a) The function g is defined by g : x → x + 1 x – 2 , x ≠ 2, find (i) g2 , [2 marks] (ii) g−1. [2 marks] (b) The function h as defined by h : x → ax + 1 x , x ≠ 0 is given by hg–1(4) = 6, find the value of a. [2 marks] 3. Given g : x → x2 + 5, find (a) an expression for each of the following. (i) g(a + 1), [1 mark] (ii) g(a2 ), [2 marks] (iii) g(2b – 1) – g(b). [2 marks] (b) the possible values of x if g(x) = 5x – 1. [2 marks] 4. The diagram shows a part of the mapping for the function f : x → ax2 + b where a and b are constants. x f ax 2 + b 3 –2 –10 10 (a) Find the value of a and of b. [2 marks] (b) Given the mapping starts with x = 1 2 , where will be the end of the arrow point? [2 marks] (c) Find another value of x so that the function f will map to −10. [2 marks] 5. The diagram shows the graph f : x → 2x + 3 for –1 < x < 3. x 0 2 –2 2 4 6 8 10 –2 4 6 8 10 f(x) f(x) = 2x + 3 (a) Find f –1(x). [1 mark] (b) Based on (a), find the corresponding coordinates for the coordinates (1, 5). [1 mark] (c) On the same axes, sketch the graph f –1(x) and state its domain. [3 marks] (d) Hence, draw a line of symmetry for f and f –1. [2 marks] 6. (a) The functions f and g are defined by f : x → 3x – a and g : x → b x , x ≠ 0 where a and b are constants. Given that f 2 (2) = 0 and fg(2) = 16, find the values of a and b. [4 marks] (b) Hence, find the value of g2 f(x). [3 marks] 7. (a) The function g is defined by g : x → 8 – 3x. Find (i) the expression for g–1 and g2 , (ii) the value of x if g–1(x) = g2 (x). [4 marks] (b) The function h is defined by h : x → ax + b, a ≠ –1 for the domain 0 < x < 5 . Given that the graph y = h(x) passes through the point (8, 5) and the graph y = h(x) and y = h−1(x) intersects at the point whose x-coordinate is 3. Find the value of a and of b. [3 marks] HOTS Analysing 01_1202 QB AMath F4.indd 9 09/05/2022 11:30 AM
1010 10SOS TIP Question 12: (a) V(t) = t 2 + 1, t(r) = 1 2 r + 4. Thus Vt(r) will express V in terms of r. (b) Find V when t = 6.5. Find r when t = 6.5. Section B 8. The function f is defined by f(x) = 1 2 x2 + x – 12. HOTS Analysing (a) Find, in a similar form, (i) f(x + a), (ii) f(x + a) – f(x) a . [4 marks] (b) Hence, find the value for f(x + a) – f(x) a if x = 0.1 when a = 2. [3 marks] 9. A train moves on a straight line. Its acceleration, a m s−2, is depends on time t, in seconds, and is given by a(t) = pt + q where p and q are constants. (a) Given that when the time t = 0 s, the acceleration is −5 m s−2 and when the time t = 4 s, the acceleration is 15 m s−2. Find the values of p and q. [4 marks] (b) Find the time when the acceleration is 25 m s−2. [3 marks] 10. The expenditure, RMC, for an annual dinner of a company depends on the number of employees in the factory. In a certain year, the number of employees is x and the expenditure of employees per head for the annual dinner is RM(x + p). (a) Express the total expenditure in that year, in terms of p. [3 marks] (b) Find the value of p if the expenditure is RM2 650 and the number of employees is 50. [4 marks] 11. The number of story books read by Amin depends on his spare time and his spare time depends on the amount of home work given by the school. Given that x(t) = 3t – 5 where x is the number of story books, t is the spare time in hours and t(k) = 4 + 2k where k is the number of homework given. (a) Find the number of story books he can read if he has 2 homeworks. [4 marks] (b) If he can read 7 story books, find the amount of spare time he has and the number of homework given. [4 marks] 12. The diagram shows a cylinder whose volume depends on its radius of the base, r m and its height, t m. V(t) = t 2 + 1 Given that the volume V(t) = (t 2 + 1) m3 and the height t(r) = 1 1 2 r + 42 m. (a) Express the volume, V, in terms of r. [4 marks] (b) Find the volume and radius of the base if the height of the cylinder is 6.5 m. [4 marks] HOTS Analysing 13. A function is defined by HOTS Analysing f(x) = |1 – x| for x < 2 x – 4 for x . 2 (a) Sketch the graph of f(x) for the domain 0 < x < 4. [4 marks] (b) Hence, find the corresponding range for the given domain of f(x). [2 marks] (c) Find the values of x if f(x) = 1 for the domain 0 < x < 4. [4 marks] 14. A function f is defined by f : x → x 2x + 1 , x ≠ k. (a) State the value of k. [1 mark] (b) Find f –11 2 5 2 . [2 marks] (c) Show that f 2 (x) = x 4x + 1 , where x ≠ – 1 4 . [3 marks] (d) Hence, show that f n (x) = x 2nx + 1, where n = 1, 2, 3….. and x ≠ – 1 2n . [4 marks] 01_1202 QB AMath F4.indd 10 09/05/2022 11:30 AM
NOTES 49 49SOS TIP Chapter 6 Linear Law 6.1 Linear and Non-Linear Relations 1. A line of best fit is a line that has the following characteristics: (a) The line passes through as many given points as possible. (b) The points which do not lie on the line are fairly distributed on two sides of the line. 6.2 Linear Law and Non-Linear Relations 1. A non-linear relation of two variables can be reduced to a linear form by substitution method. For example, y = ax2 + b To form a linear equation, let Y = y and X = x2 . We have Y = aX + b where a is the gradient and b is the Y-intercept. y b x 0 y = ax 2 + b Y b X 0 Y = aX + b A non-linear graph A linear graph 1. Reduce the following non-linear relations to the linear form Y = mX + c. Identify and state the quantities represented by Y, X, m and c. (a) yx = 3x3 + x [2 marks] (b) x – y xy = 1 3 [2 marks] (c) y = 100a2x [2 marks] Answer: (a) (b) (c) Section A PAPER 1 2. The diagram shows a line of best fit obtained by plotting a graph of y against x. y x 0 (–1, 3) (1, 1) (3, k) Find (a) the equation of the line of best fit, [2 marks] (b) the value of k. [2 marks] Answer: (a) (b) Question 2: (a) Find the gradient of the line, m. Then, use (y – y1 ) = m(x – x1 ) to find the equation of the line (b) Insert x = 3, y = k in the equation of the line to get the value of k 06_1202 QB AMath F4.indd 49 09/05/2022 11:33 AM
5050 50SOS TIP Question 4: From the graph, find the gradient of the line, m. Then, log10 y = mx – 1 which can be written as 10(mx – 1) = y or 10mx =10y.....… Compare with the given equation ay = bx to determine the values of a and b. 3. The diagram shows a straight line obtained by plotting a graph of y x against x. x 0 1 y —x 1 The variables x and y are related by the equation y = ax + bx2 where a and b are constants. Find the values of (a) a, [3 marks] (b) b. [2 marks] Answer: (a) (b) 4. The diagram shows a straight line obtained by plotting a graph of log10 y against x. log10 y x 0 –1 (1, 3) The variables x and y are related by the equation ay = bx where a and b are constants. Find the values of (a) a, [3 marks] (b) b. [2 marks] Answer: (a) (b) 5. The diagram shows two straight lines with the same equation. x 0 3 –3 y —x —1 x y –—x2 0 (2, h) 1 I II (a) Find the equation. [3 marks] (b) Calculate the value of h. [2 marks] Answer: (a) (b) 6. The diagram shows a straight line obtained by plotting a graph of log10 y against (x – 2). log10 y x – 2 0 (6, 5) (3, 2) Given the variables x and y are related by the equation y = anx – 2 where a and n are constants. Find the value of (a) n, [3 marks] (b) a. [2 marks] Answer: (a) (b) 06_1202 QB AMath F4.indd 50 09/05/2022 11:33 AM
5252 52SOS TIP Question 13: The quadratic function in vertex form is y = –1x – 3 2 2 2 + 9 4 Then y = –x2 + 3x........... Since the Y-intercept is 3, then needs to be divided by x. Section B 11. The variables x and y are related by the equation y = px qx + 2 where p and q are constants. If the graph of y against x is plotted, a curve which passes through point (2, 2) is obtained. If the graph of 1 y against 1 x is drawn, a straight line with a gradient of 1 3 is obtained. Find the values of (a) p, [4 marks] (b) q. [4 marks] Answer: (a) (b) 12. The diagram shows a straight line graph of 1 xy against 1 x3 . 0 (k, 1) (2, h) 1 —–xy 1 –—x3 Given that the variables x and y are related by y = 4x2 5 – x3 , find the value of (a) k, [4 marks] (b) h. [4 marks] Answer: (a) (b) CLONE SPM 13. The diagram shows the graph of a quadratic function. 0 (3, 0) x y —9 4 (a) Find the equation of the quadratic function. [4 marks] (b) If a straight line is obtained by plotting a graph of Y against X such that the Y-intercept is equal to 3, state the quantities represented by X and Y. [4 marks] Answer: (a) (b) 14. The variables x and y are related by the equation y = hx3 where h is a constant. (a) Convert the equation y = hx3 into the linear form Y = mX + c. [3 marks] (b) The diagram shows a straight line obtained by plotting a graph of log10 y against log10 x. log10 y log10 x 0 –1 (5, k) Find the value of (i) log10 h, [3 marks] (ii) k. [2 marks] Answer: (a) (b) (i) (ii) 06_1202 QB AMath F4.indd 52 09/05/2022 11:33 AM
5454 54SOS TIP Question 4: Given y = qpx + 1 . Take logarithm on both sides, to get log10 y = log10 q + (x + 1) log10 p The suitable graph will be log10 y against (x + 1). 19. The variables x and y are related by the equation y = x2 + 5x – 1. (a) Convert the equation y = x2 + 5x – 1 into the linear form Y = mX + c such that the straight line has a gradient of 5. [4 marks] (b) Hence, state the quantities represented by X, Y and c. [2 marks] (c) Find the value of X in (b) if Y = 6 cm. [2 marks] Answer: (a) (b) (c) 20. Diagram I shows the curve y = 2x2 – 4. Diagram II shows a straight line obtained when y = 2x2 – 4 is expressed in the linear form Y = –4X + c. 0 y = 2x 2 – 4 x y 0 2 X Y I II In terms of x and/or y, express (a) X, [3 marks] (b) Y. [2 marks] Answer: (a) (b) CLONE SPM Section B PAPER 2 1. The table shows the values of two variables, x and y obtained from an experiment. A straight line is obtained when graph y2 x against 1 x is plotted. x 1 2 3 4 5 y 1.87 2.24 2.55 2.83 3.08 (a) Based on the table given, construct another table for the values 1 x and y2 x . [2 marks] (b) Plot y2 x against 1 x and draw the line of best fit. [3 marks] (c) By using the graph on (b), (i) find the value of y when x = 3.5. [2 marks] (ii) express y in terms of x. [3 marks] 2. The table shows the values of two variables, x and y, obtained from an experiment. It is known that the variables x and y are related by the equation y = p + qx x where p and q are constants. x 1 2 3 4 5 6 y 6 4 3.3 3 2.8 2.7 CLONE SPM (a) Plot xy against x. Hence, draw the line of best fit. [5 marks] (b) From the graph in (a), find the value of (i) p, (ii) q. [5 marks] 3. The table shows the values of two variables, V and P, obtained from an experiment. It is predicted that the variables P and V are related by the equation P = a(V + 1)b where a and b are constants. V 1 2 3 4 5 6 P 4 13.5 32 62.5 108 171.5 (a) Convert the equation P = a(V + 1)b into the linear form Y = mX + c. [2 marks] (b) Plot Y against X. Hence, draw the line of best fit. [3 marks] (c) From the graph in (b), determine the value of (i) a, (ii) b. [5 marks] 4. The table shows the values of two variables, x and y, obtained from an experiment. It is predicted that the variables x and y are related by the equation y = qpx + 1 where p and q are constants. x 1 2 3 4 5 y 2.25 6.75 20.25 60.75 182.25 06_1202 QB AMath F4.indd 54 09/05/2022 11:33 AM
55 55SOS TIP Question 10: Given y a + x = 3bx Then y = 3bx(a + x) y = 3abx +3bx2 Divide both sides by x and continue from there. (a) Plot a suitable straight line graph that relates x and y. [6 marks] (b) From the graph in (a), determine the values of p and q. [4 marks] 5. The table shows the values of two variables, x and y, obtained from an experiment. It is known that the variables x and y are related by the equation y = (q + 1)px2 where p and q are constants. x 1 1.5 2 2.5 3 3.5 y 2.25 3.74 7.6 18.9 57.7 215.4 (a) Plot log10 y against x2 . Hence, draw the line of best fit. [5 marks] (b) Use the graph in (a) to estimate the value of (i) p, (ii) q. [5 marks] 6. The table shows the values of two variables, x and y, obtained from an experiment. The variables x and y are related by the equation y = axb – 1 where a and b are constants. x 1 2 3 4 5 6 y 100 25 11.1 6.3 4 2.8 (a) Plot log10 y against log10 x. Hence, draw the line of best fit. [6 marks] (b) Use your graph in (a) to find the value of (i) a, (ii) b. [4 marks] 7. The table shows the values of two variables, x and y, obtained from an experiment. The variables x and y are related by the equation y = hk2x – 1 where h and k are constants. x 1 2 3 4 5 6 y 0.24 0.35 0.5 0.72 1.03 1.49 (a) Plot log10 y against (2x – 1). Hence, draw the line of best fit. [6 marks] (b) Use your graph in (a) to find the value of (i) h, (ii) k. [4 marks] 8. The table shows the values of two variables, x and y, obtained from an experiment. It is known that the variables x and y are related by the equation y = pABx + qx where p and q are constants. x 1 2 3 4 5 6 y –1 –0.24 0.8 2 3.3 4.65 CLONE SPM CLONE SPM CLONE SPM CLONE SPM (a) Plot y ABx against ABx . Hence, draw the line of best fit. [5 marks] (b) Use your graph in (a) to find the value of (i) p, [2 marks] (ii) q, [2 marks] (iii) y if x = 3.61. [1 mark] 9. The table shows the experimental values of two variables, x and y. One of the values of y is wrongly recorded. It is known that the variables x and y are related by the equation 4a2 x = (y + b)2 where a and b are constants. x 1 2 3 4 5 6 y 2 4.1 5.7 6 8.2 9.2 (a) Plot y against ABx. Hence, draw the line of best fit. [6 marks] (b) From the graph in (a), (i) identify the incorrect value of y and determine its correct value, [2 marks] (ii) find the values of a and b. [2 marks] 10. The table shows the values of two variables, x and y, obtained from an experiment. The variables x and y are related by the equation y a + x = 3bx where a and b are constants. HOTS Analysing x 1 2 3 4 5 y 3 0 –9 –24 –45 (a) Convert the equation y a + x = 3bx into the linear form Y = mX + c. Hence, draw the line of best fit. [6 marks] (b) From your graph in (a), find the values of a and b. [4 marks] 11. The table shows the experimental values of two variables, x and y. It is known that the variables x and y are related by the equation px2 y = q where p and q are constants. x 0.2 0.3 0.4 0.5 0.6 y 8.3 3.7 2.1 1.3 0.93 (a) Reduce the equation px2 y = q to the linear form Y = mX + c. Hence, draw the line of best fit. [6 marks] (b) Use your graph in (a) to find (i) the value of q p , [2 marks] (ii) the value of q if p = 6. [2 marks] CLONE SPM 06_1202 QB AMath F4.indd 55 09/05/2022 11:33 AM
89 PAPER 1 Time: 2 hours Section A [64 marks] Answer all questions. Form 4 Assessment 1. (a) Diagram 1 shows a graph f(x) for the domain a < x < b. x f(x) 4 0 –2 Diagram 1 (i) State the value of a and b. (ii) Determine whether f(x) has an inverse function for the domain given. Explain your answer. [3 marks] (b) The function f and g are defined by f : x → 4x + 5 and g : x → x2 + 8. Find the possible values of x so that f(x) = g(x). [2 marks] Answer: (a) (i) (ii) (b) 2. (a) Given the function g : x → 2 – 3x and h–1 : x → x 2 – x , x ≠ k. Find (i) the value of k, (ii) hg(x). [3 marks] (b) Diagram 2 shows a graph of f(x). x f(x) 0 Diagram 2 State whether f(x) is a function. Justify the test used in determining the answer. [2 marks] Answer: (a) (i) (ii) (b) 11_1202 QB AMath F4.indd 89 09/05/2022 11:36 AM
93 13. Diagram 8 shows a graph log10 y against x which relating variables x and y in an experiment. x 1 1 2 1.5 0.5 0 2 3 4 log10 y Diagram 8 Based on the graph above, (a) find the gradient and the y-intercept, (b) form an equation which relating x and y, (c) find the value of y when x = 0.5, (d) find the value of x when y = 40. [8 marks] Answer: (a) (b) (c) (d) 14. Diagram 9 shows a cyclic quadrilateral KLMN with centre of O. NOK is the diameter, ML = 6 cm, LK = MN = 5 cm and ∠MNK = 65°. Diagram 9 N 65° O 6 cm M L K Find (a) the length of KM, [2 marks] (b) ∠KML, [2 marks] (c) the radius of the circle, [2 marks] (d) the area of quadrilateral KLMN. [2 marks] Section B [16 marks] Answer any two questions from this section. Answer: (a) (b) (c) (d) 15. O is the origin. O →A and O →B are 2i ~ + j ~ and 2i ~ + 3j ~ respectively. E is the midpoint of AB. (a) Find the vector O →E. (b) OAPB is a parallelogram, find the vector of O →P. (c) Express B →P – B →E, in terms of i ~ and j ~ . (d) Find ∠AOB. [8 marks] Answer: (a) (b) (c) (d) END OF QUESTION PAPER 11_1202 QB AMath F4.indd 93 09/05/2022 11:36 AM
94 PAPER 2 Time: 2 hours 30 minutes Section A [50 marks] Answer all questions. 1. If the line x + 3y = 1 intercepts the curve y2 – 9 = 1 2 xy at two points P and Q, find (a) the coordinates of P and Q as the x-coordinate of P is positive, (b) the equation of the line that passes through point Q and is perpendicular to x + 3y = 1. [7 marks] 2. The roots of the equation 3x2 + bx + c = 0 are m and 1 m . (a) Find the value of c. [1 mark] (b) Given m2 + 1 m2 = 46 9 , find the possible values of b. [3 marks] (c) Hence, find the possible values of m for b . 0. [3 marks] 3. Diagram 3 shows a piece of land, PQRS drawn on a Cartesian plane. Given the scale 1 unit on the Cartesian plane represents 5 m on the real ground. M P(–5, 0) S Q(5, 10) R(7, y) y x Diagram 3 P lies on the x-axis and M divides the line PQ in the ratio 2 : 3. Find (a) the coordinates of M and R, [2 marks] (b) the equation of the line MR, [3 marks] (c) the actual area of PQRS, in m2 . [3 marks] 4. A coil of wire with length t cm is cut into a few parts. Each part is bent into a square. Given that the length of each side of the square follows an arithmetic progression with the common difference 1.5 cm. The smallest and the largest squares have sides 1.5 cm and 16.5 cm respectively. Find the number of squares can be made and the value of t. [8 marks] 11_1202 QB AMath F4.indd 94 09/05/2022 11:36 AM
96 10. Diagram 10 shows a parallelogram ABCD. M N D L A B C Diagram 10 NLM is a straight line and AL is perpendicular to DC. Given D →L = 1 4 D →C, C →B = 4C →M, L →A = 2x ~ and D →A = 8y ~ . (a) Express, in terms of x and y, (i) D →L, (ii) L →M. (b) Given N →D = ky ~ and N →L = hL→M, find the value of h and k. (c) If |x ~| = 3 units and |y ~ | = 2.5 units, find (i) |D →C|, (ii) the area of the parallelogram ABCD. [10 marks] 11. A straight line with gradient of 4 passes through A(–2, 5) and intercepts the x-axis at B. Another straight line passes through A and intersects the x-axis at C(2, 0). (a) Find the equation of the line AB and AC. [2 marks] (b) Calculate the area of the triangle ABC. [2 marks] (c) Find the coordinates of D so that ABCD is a parallelogram. [3 marks] (d) A point P(x, y) moves such that ∠APC is always 90°. Find the equation of locus of point P. [3 marks] Section C [20 marks] Answer any two questions from this section. 12. Diagram 12 shows a pyramid with a equilateral base of side 8 cm. 8 cm M N L P Q V Diagram 12 V is 5 cm vertically above P where P is the midpoint of LM. Given that ∠VNQ = ∠QNM, find (a) the length of VN, [2 marks] (b) ∠QNM, [3 marks] (c) the length of QM, [2 marks] (d) the area of the inclined plane, VNM. [3 marks] 11_1202 QB AMath F4.indd 96 09/05/2022 11:36 AM
97 13. A grocery shop sells three types of can drinks. Table 13 shows the price, the price indices and the percentage of can drinks sold. Type of can drink Price per can (RM) Price index in the year 2018 based on the year 2016 Percentage of can 2016 2018 sold in 2018 P 1.60 1.80 x 25 Q 1.20 1.50 150 p R y 0.96 120 55 Table 13 (a) Find the value of x, y and p. [3 marks] (b) (i) The price of can drink P increased by 5% from 2018 to 2019. Find the price index of can drink P in the year 2019 based on the year 2016. [2 marks] (ii) The price index of can drink R in the year 2016 based on the year 2012 was 130. Calculate the price index of can drink R in the year 2018 based on the year 2012 and its price in the year 2012. [2 marks] (c) Calculate the composite index for the can drinks sold by the grocery shop in the year 2018 based on the year 2016. [3 marks] 14. A carpenter wants to make a sculptor as shown in the Diagram 14. E J H C G A B D F 5 cm 24 cm Diagram 14 Given the height of the sculptor is 5 cm, EH = 24 cm and EJ : JH = 1 : 2. Find (a) the length of CJ, [3 marks] (b) the length of BC if the area of the triangle BCJ is 39 cm2 , [2 marks] (c) the angle between JB and the plane EFGH, [3 marks] (d) ∠ABJ. [2 marks] 15. Diagram 15 shows the front view of a tunnel in the parabolic form with the highest point of 10 m from the ground. O P L1 A B L2 y x 10 m 30 m Diagram 15 Given the width of the tunnel is 30 m. (a) Write three possible equations in the vertex form to represent the shape of the parabola. [6 marks] (b) Two lights is hung from the roof of the tunnel such that AL1 = L1 L2 = L2 B. Find the height of the lights that hung from the ground. [4 marks] END OF QUESTION PAPER 11_1202 QB AMath F4.indd 97 09/05/2022 11:36 AM
Answers 12A_1202 QB AMath F4.indd 98 09/05/2022 11:38 AM
99 CHAPTER 1 Paper 1 Section A 1. (a) 2(6) + 1 = 13 2(8) + 1 = 17 Thus, the function is f(p) = 2p + 1, (b) 2(p) + 1 = q q = 2p + 1 2. (a) 0 2 y x 4 6 2 4 6 8 10 (b) Many-to-many (c) 2, 4 and 8 3. (a) Since –6 < y < 4, thus a = –6 and b = 4. (b) –3 (c) –2 < x < 5 4. (a) f –1(x) (b) g–1f(x) or f –1g(x) 5. (a) f(p) = –6 4p – 5p = –6 p = 6 (b) (i) f(–2) = (–2)2 – 5(–2) = 4 + 10 = 14 (ii) x2 – 5x = 6 x2 – 5x – 6 = 0 (x – 6)(x + 1) = 0 x = 6 or x = –1 6. ts(x) = t[s(x)] = t(6 – 4x) = p(6 – 4x) – 3 = 6p – 4px – 3 Compare with ts(x) = q – 4px: Thus, q = 6p – 3. 7. (a) gf(–2) = 5 g(2(–2) – 6) = 5 g(–10) = 5 h + 10k = 5 h = 5 – 10k (b) (i) k = 1 (ii) 3x x – 1 = x 3x = x2 – x x2 – 4x = 0 x(x – 4) = 0 x = 0, x = 4 (iii) h(x) = 3x x – 1 Let y = 3x x – 1 yx – y = 3x yx – 3x = y x(y – 3) = y x = y y – 3 Hence, h–1(x) = x x – 3 , x ≠ 3. 8. (a) qp1– 1 2 2 = 61– 1 2 2 – 3 = –6 (b) Given qp(x) = 6x – 3 q12 – x 4 2 = 6x – 3 Let y = 2 – x 4 x 4 = 2 – y x = 8 – 4y q(y) = 6(8 – 4y) – 3 = 45 – 24y Thus, q(x) = 45 – 24x. 9. (a) f(x) = 2x – 5 x + 2 Let y = 2x – 5 x + 2 y(x + 2) = 2x – 5 yx + 2y = 2x – 5 yx – 2x = –2y – 5 (y – 2)x = –2y – 5 x = –2y – 5 y – 2 Thus, f –1(y) = –2y – 5 y – 2 f –1(x) = –2x – 5 x – 2 , x ≠ 2 (b) f(x) = 2x – 5 x + 2 f(k) = 3 2k – 5 k + 2 = 3 2k – 5 = 3k + 6 k = –11 Thus, the value of k is –11. 10. (a) f(x – 2) = 3 – 2(x – 2) = 3 – 2x + 4 = 7 – 2x (b) Let y = 3 – 2x 2x = 3 – y x = 3 – y 2 So, f –1(x) = 3 – x 2 . 2f –1(k) = f(k – 2) 21 3 – k 2 2 = 7 – 2k 3 – k = 7 – 2k k = 4 Section B 11. (a) (i) –2 (ii) f(x) = |3 – 1 2 x| f(1) = |3 – 1 2 (1)| = 5 2 (iii) |3 – 1 2 x| = 1 3 – 1 2 x = –1 and 3 – 1 2 x = 1 – 1 2 x = –4 – 1 2 x = –2 x = 8 x = 4 Thus, the domain is 4 < x < 8. (b) (i) Given g–1(x) = p q + x , x ≠ –q g–1(3) = –2 p q + 3 = –2 p = –2q – 6............. g–11 1 2 2 = –4 p q + 1 2 = –4 p = –4q – 2 .........2 = 2: –2q – 6 = –4q – 2 2q = 4 q = 2 Substitute q = 2 into 1, p = –2(2) – 6 = –10 (ii) g–1(x) = – 10 2 + x , x ≠ –2 Let y = – 10 2 + x 2y + xy = –10 xy = –10 – 2y x = –10 – 2y y Thus, g(x) = –10 – 2x x , x ≠ 0 (iii) g(x) = 8 –10 – 2x x = 8 –10 – 2x = 8x 10x = –10 x = –1 12. (a) x –1 0 1 2 1 2 f(x) 3 1 0 1 3 0 y x –1 1 2 1 2 3 (b) (i) (a) ff(x) = f(2x) = 4x = 22 x (b) fff(x) = ff(2x) = f(4x) = 8x = 23 x (ii) f n (x) = 2n x where n = 1, 2, 3,…. (iii) f 5 (x) = 16 25 (x) = 16 32x = 16 x = 1 2 13. (a) (i) If f maps set L to M, then the element in set M is represented by f(x). Thus, the function that maps M to N must be g(x) because from L to N is gf(x). Thus, g(x) = x – 1 2 . (ii) gf(x) = x2 – x + 2 f(x) – 1 2 = x2 – x + 2 f(x) – 1 = 2x2 – 2x + 4 f(x) = 2x2 – 2x + 5 (b) (i) Let y = 3x – 4 3x = y + 4 x = y + 4 3 Thus, g–1(x) = x + 4 3 . (ii) g2 1 p 3 2 = g3g1 p 3 24 = g331 p 3 2 – 44 = g(p – 4) = 3(p – 4) – 4 = 3p – 16 12A_1202 QB AMath F4.indd 99 09/05/2022 11:38 AM
100 g2 1 p 3 2 = 20 3p – 16 = 20 3p = 36 p = 12 14. (a) (i) f(–2) = 1 – 2(–2) = 5 (ii) f(3) = 3 – 1 = 2 (iii) x –3 –2 –1 0 f(x) 7 5 3 1 x 1 2 3 4 f(x) –1 1 2 3 0 y x –2 2 4 2 4 6 (b) (i) (kh)–1(–5) = 4 kh(x) = k(y) = z x = (kh) –1(z) Thus, (kh) –1(–5) = 4. (ii) hh–1(2) = 2 (iii) h–1k –1(–5) = h–1(2) = 4 (kh) –1(–5) = 4 Thus, h–1k –1(–5) = (kh)–1(–5). 15. (a) p = 1 (b) Let y = x + 2 1 – x y – xy = x + 2 x + xy = y – 2 (1 + y)x = y – 2 x = y – 2 1 + y h(x) = x – 2 1 + x , Thus, a = –2 , b = 1. (c) q = –1 (d) h–1(x) + h(x) = 0 x + 2 1 – x + x – 2 1 + x = 0 (x – 2)(1 + x) + (1 – x)(x – 2) (1 – x)(1 + x) = 0 6x = 0 x = 0 16. (a) When t = 2, s(2) = 22 – 5(2) + 6 = 0 m (b) t 2 – 5t + 6 = 30 t 2 – 5t – 24 = 0 (t – 8)(t + 3) = 0 t = 8 or t = –3 (Not applicable) Thus, t = 8 seconds. (c) Displacement is negative, s(t) , 0 t 2 – 5t + 6 , 0 (t – 2)(t – 3) , 0 Thus, 2 , t , 3. 17. (a) V(5) = 52 + 4(5) + 3 = 48 m3 (b) t 2 + 4t + 3 = 63 t 2 + 4t – 60 = 0 (t + 10) (t – 6) = 0 t = 6 m or t = –10 (Not applicable) Thus, the height of the tank is 6 m. 18. (a) 12 = 3p + q................ 24 = 5p + q................ 2 2 – 1: 12 = 2p p = 6 Substitute p = 6 into 1: 12 = 3(6) + q q = –6 Thus, p = 6 and q = –6. (b) K(4.5) = 6(4.5) – 6 = 21 Thus, the cost is RM21. 19. (a) 0 y f f –1 y = x x A(2, 0) A'(0, 2) B'(7, 3) B(3, 7) 2 2 4 6 4 6 Thus, Aʹ(0, 2) and Bʹ(7, 3). (b) Domain is 0 < x < 7. 20. (a) By using the vertical line test, the line intersect only once. Hence, f(x) is a function. (b) By using horizontal line test, the line intersects the curve more than once. Hence, f(x) does not have an inverse function. (c) Range of f(x) is f(x) . 0 Paper 2 Section A 1. f(x) = 3 4 x + 1 2 g(x) = 5 4 – 2 3 x (a) f(2) + f(3) = 3 3 4 (2) + 1 2 4 + 3 3 4 (3) + 1 2 4 = 2 + 11 4 = 19 4 f(2 + 3) = f(5) = 3 4 (5) + 1 2 = 15 4 + 1 2 = 17 4 ∴ f(2) + f(3) ≠ f(2 + 3) (b) g(4) – g(2) = 3 5 4 – 2 3 (4)4 – 3 5 4 – 2 3 (2)4 = – 17 12 – 1– 1 12 2 = – 4 3 g(4 – 2) = g(2) = 5 4 – 2 3 (2) = – 1 12 ∴ g(4) – g(2) ≠ g(4 – 2) (c) f(2k) = 6g(k) 3 4 (2k) + 1 2 = 63 5 4 – 2 3 k4 3k 2 + 1 2 = 15 2 – 4k 11k 2 = 14 2 k = 14 11 (d) f(k) + g(k) = 5 3 3 4 k + 1 2 4 + 3 5 4 – 2 3 k4 = 5 7 4 + 1 12 k = 5 1 12 k = 13 4 k = 39 2. (a) g(x) = x + 1 x – 2 x ≠ 2 (i) gg(x) = g3 x + 1 x – 2 4 = x + 1 x – 2 + 1 x + 1 x – 2 – 2 = 2x – 1 5 – x , x ≠ 5 (ii) Let x + 1 x – 2 = y x + 1 = xy – 2y 1 + 2y = x[y – 1] x = 1 + 2y y – 1 ∴ g–1(x) = 1 + 2x x – 1 , x ≠ 1 (b) h(x) = ax + 1 x hg–1(4) = 6 h3 1 + 2(4) 4 – 1 4 = 6 h[3] = 6 3a + 1 3 = 6 3a + 1 = 18 3a = 17 a = 17 3 3. (a) (i) g(a + 1) = (a + 1)2 + 5 = a2 + 2a + 6 (ii) g(a2 ) = a4 + 5 (iii) g(2b – 1) – g(b) = (2b – 1)2 + 5 – [b2 + 5] = 4b2 – 4b + 6 – b2 – 5 = 3b2 – 4b + 1 (b) g(x) = 5x – 1 x2 + 5 = 5x – 1 x2 – 5x + 6 = 0 (x – 3)(x – 2) = 0 x = 3 or x = 2 4. (a) f(x) = ax2 + b f(3) = 9a + b = 10 .......... f(–2) = 4a + b = –10 ........ 2 – 2: 5a = 20 a = 4 Substitute a = 4 into : 9(4) + b = 10 b = 10 – 36 = –26 (b) f(x) = 4x2 – 26 f1 1 2 2 = 41 1 4 2 2 – 26 = –25 12A_1202 QB AMath F4.indd 100 09/05/2022 11:38 AM