Analysis SPM_Add Mathematics Form 4&5

PENERBIT ILMU BAKTI SDN. BHD.

iii Contents Standard Normal Distribution Table v Additional Mathematics SPM Format vi Formulae vii – viii FORM 4 Chapter 1 Functions 1 HOTS Zone 6 SPM Practice 1 6 Answers 10 Chapter 2 Quadratic Functions 11 HOTS Zone 20 SPM Practice 2 20 Answers 24 Chapter 3 Systems of Equations 25 HOTS Zone 29 SPM Practice 3 29 Answers 32 Chapter 4 Indices, Surds and Logarithms 33 HOTS Zone 38 SPM Practice 4 38 Answers 41 Chapter 5 Progressions 42 HOTS Zone 51 SPM Practice 5 52 Answers 60 Chapter 6 Linear Law 61 HOTS Zone 69 SPM Practice 6 71 Answers 78 Chapter 7 Coordinate Geometry 83 HOTS Zone 91 SPM Practice 7 92 Answers 100 Chapter 8 Vectors 101 HOTS Zone 110 SPM Practice 8 111 Answers 120 Analisis&Tip SPM Add Maths-Prelim 3rd.indd 3 16-Feb-23 7:32:16 PM PENERBIT ILMU BAKTI SDN. BHD.

iv Chapter 9 Solution of Triangles 122 HOTS Zone 128 SPM Practice 9 129 Answers 133 Chapter 10 Index Numbers 134 HOTS Zone 137 SPM Practice 10 138 Answers 144 FORM 5 Chapter 1 Circular Measure 145 HOTS Zone 151 SPM Practice 1 152 Answers 160 Chapter 2 Differentiation 161 HOTS Zone 170 SPM Practice 2 171 Answers 178 Chapter 3 Integration 179 HOTS Zone 189 SPM Practice 3 190 Answers 197 Chapter 4 Permutation and Combination 198 HOTS Zone 201 SPM Practice 4 201 Answers 205 Chapter 5 Probability Distribution 206 HOTS Zone 213 SPM Practice 5 214 Answers 219 Chapter 6 Trigonometric Functions 220 HOTS Zone 229 SPM Practice 6 230 Answers 233 Chapter 7 Linear Programming 235 HOTS Zone 241 SPM Practice 7 243 Answers 247 Chapter 8 Kinematics of Linear Motion 250 HOTS Zone 256 SPM Practice 8 257 Answers 259 SPM Model Test 260 Answers 269 Analisis&Tip SPM Add Maths-Prelim 3rd.indd 4 16-Feb-23 7:32:16 PM PENERBIT ILMU BAKTI SDN. BHD.

1 1.1 Functions 1 Function is a special relation such that each element in the input set is related to one and only one element in the output set. 2 In a function between an input set (A) and an output set (B), • the input set (A) is known as the domain, • the output set (B) is known as the codomain, • the elements in the domain are known as objects, • the elements in the codomain which are related to objects are known as images, • the set of images is known as range. 3 If a function maps x onto x 2 – 2, it is denoted by a function notation e.g. f : x → x2 – 2 or f(x) = x 2 – 2. 4 A function is undefined if its denominator is zero. 5 The absolute value of a number x is the numerical value of x and is represented by |x|. |x| is known as the modulus of x. For example, |–3| = 3 and |5| = 5. 6 An absolute value function is defined by f(x) if f(x)  0 |f(x)| =    –f(x) if f(x)  0 7 The graph of a linear absolute value function has the shape of V. 1.2 Composite Functions 1 If f is a function which maps set A onto set B and g is a function which maps set B onto set C, then gf is a composite function of f followed by g which maps set A straight away onto set C. g[f(x)] = gf(x) gf x A B C f ➤ ➤ f(x) ➤ g 1.3 Inverse Functions 1 If f : x → y is a function which maps x onto y, then its inverse function is denoted by f –1. The inverse function is a function which maps y back to x. f x y f –1 ➤ ➤ 2 ff –1(x) = f –1f(x) = x 3 (a) The domain of f –1 is the range of f. (b) The range of f –1 is the domain of f. 4 The graph of f –1 is the reflection of the graph f in the straight line y = x. Object Image Range Nonimage Domain Codomain P Q multiplied by 2 –2 3 4 –4 6 8 9 Functions Learning Area: Algebra EXPRESS NOTES Chapter 1 Analisis&Tip SPM Add Maths-F4-C1 2nd.indd 1 16-Feb-23 7:19:46 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 1 2 1.1 Functions Example 1 The arrow diagram below represents the function f(x) = a x + b. x 1 2 5 a x + b –2 f Find (a) the value of a and of b, (b) the value of x when the function f is undefined, (c) the image of 6, (d) the object that has the image –5, (e) the values of k such that f(k) = 2k. Solution (a) Given the function f(x) = a x + b, f(1) = 2 f(–2) = 5 a 1 + b = 2 a –2 + b = 5 Substitute x = 1 Substitute x = –2 a = 2 + 2b a = –10 + 5b a – 2b = 2 ... ➀ a – 5b = –10 ... ➁ ➀ – ➁ : 3b = 12 ⇒ b = 4 From ➀ : a – 2(4) = 2 ⇒ a = 10 (b) f(x) = a x + b = 10 x + 4 The function f is undefined when its denominator of 10 x + 4 is 0. When denominator = 0, x + 4 = 0 x = –4 Hence, the function f in undefined when x = –4. (c) f(6) = 10 6 + 4 = 1 Hence, the image of 6 is 1. (d) f(x) = –5 Object Image 10 x + 4 = –5 10 = –5x – 20 5x = –20 – 10 5x = –30 x = –6 Hence, the object that has the image –5 is –6. (e) Given that f(k) = 2k, 10 k + 4 = 2k 10 = 2k(k + 4) 10 = 2k2 + 8k 2k2 + 8k – 10 = 0 Quadratic equation in the form ax2 + bx + c = 0. k2 + 4k – 5 = 0 Simplity the equation by dividing throughout by 2. (k + 5)(k – 1) = 0 k 5 × × k –1 5k + –k k2 k = –5 or 1 –5 4k Example 2 Given the absolute value function f(x) = |2x – 1|, (a) sketch the graph of the function f(x) for the domain –2  x  2. State the corresponding range of values of f(x), (b) find the value of x that has the image 7. Solution (a) Prepare a table as follows. x –2 –1 0 1 2 f(x) 5 3 1 1 3 When x = –2, f(x) =|2(–2) – 1| =|–5| = 5 Analisis&Tip SPM Add Maths-F4-C1 2nd.indd 2 16-Feb-23 7:19:46 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 1 3 The value of x where the graph touches the x-axis has to be determined as follows. On the x-axis, y = f(x) = 0 |2x – 1| = 0 2x – 1 = 0 x = 1 2 The sketch of the graph of the function f(x) is as follows. x O 1 1 2 –1 1 2 3 4 5 –2 2 y ! Attention! Note that the graph of a linear absolute value function always has a V shape. The corresponding range of values of f(x) is 0  f(x)  5. The range of values of f(x) is the range from the smallest value of y to the largest value of y covered by the graph. (b) Given that the image of the function f(x) is 7, thus f(x) = 7. f(x) = 7 |2x – 1| = 7 There are two equations that can be formed, i.e.: 2x – 1 = 7 or 2x – 1 = –7 x = 4 x = –3 Therefore, the values of x are x = 4 or x = –3. 1.2 Composite Functions Example 3 The functions f and g are defined by f : x → 2x + 4 and g : x → 1 7 x respectively. (a) For the function f, find the value of x that is mapped onto itself. (b) Calculate gf (5). (c) Find fg. (d) Find f 2 . Solution (a) f(x) = x Self-mapping is written as f(x) = x. 2x + 4 = x x = –4 Hence, the value of x that is mapped onto itself under the function f is –4. (b) gf(5) = g[2(5) + 4] = g(14) = 1 7 × 14 = 2 (c) fg(x) = f 1 1 7 x2 = 21 1 7 x2 + 4 Substitute x with 1 1 7 x2 = 2x + 28 7 ∴fg : x → 2x + 28 7 Caution! fg(x)= f(x) × g(x) = (2x + 4)1 1 7 x2 = 2x2 + 4x 7 Incorrect (d) f 2 (x) = ff(x) = f(2x + 4) = 2(2x + 4) + 4 Substitute x with (2x + 4) = 4x + 12 Example 4 Given the function f : x → x – 5, find the function g if (a) fg : x → x2 + 1, (b) gf : x → 3 x – 2 , x ≠ 2 Solution (a) fg : x → x2 + 1 fg(x) = x2 + 1 g(x) – 5 = x2 + 1 Substitute x with g(x) g(x) = x2 + 1 + 5 g(x) = x2 + 6 ∴g : x → x2 + 6 Analisis&Tip SPM Add Maths-F4-C1 2nd.indd 3 16-Feb-23 7:19:47 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 1 4 (b) gf : x → 3 x – 2 gf(x) = 3 x – 2 g(x – 5) = 3 x – 2 Let x – 5 = u, x = u + 5 g(u) = 3 (u + 5) –2 Each x is substituted by u + 5 g(u) = 3 u + 3 g(x) = 3 x + 3 ∴ g : x → 3 x + 3 , x ≠ –3 1.3 Inverse Functions Example 5 Given the functions f(x) = x – 2 x + 1 , x = –1 and g(x) = 2x – 6, find (a) g–1(4), (b) f –1, (c) f –1g. Solution (a) Let g–1(4) = u g(u) = 4 2u – 6 = 4 2u = 10 u = 5 ∴g–1(4) = 5 (b) Method 1 Let f –1(x) = y f(y) = x y – 2 y + 1 = x Rearrange the equation in order that y becomes the subject of the formula. y – 2 = x(y + 1) Cross multiplication. y – 2 = xy + x Expand. y – xy = x + 2 Collect all the terms in y on the left-hand side of the equation. y(1 – x) = x + 2 Factorise. y = x + 2 1 – x ∴ f –1(x) = x + 2 1 – x , x ≠ 1 Method 2 Let y = x – 2 x + 1 Rearrange the equation in order that x becomes the subject of the formula. y(x + 1) = x – 2 xy + y = x – 2 y + 2 = x – xy x(1 – y) = y + 2 x = y + 2 1 – y f –1(y) = y + 2 1 – y ∴ f –1(x) = x + 2 1 – x , x ≠ 1 (c) f –1g(x) = f –1(2x – 6) Composite function operation of g followed by f –1. = (2x – 6) + 2 1 – (2x – 6) = 2x – 4 7 – 2x , x ≠ 7 2 Example 6 Given that f –1(x) = 1 k – x , x ≠ k and g(x) = x + 2, find the value of k if (a) f(x) = 3x – 1 x , x ≠ 0, (b) ff –1 (k2 + 2) = g[(k + 2)2 ]. Solution (a) f(x) = 3x – 1 x , x = 0 Let f –1(x) = y f(y) = x 3y – 1 y = x Rearrange the equation in order that y becomes the subject of the formula. 3y – 1 = xy 3y – xy = 1 y(3 – x) = 1 y = 1 3 – x ∴f –1(x) = 1 3 – x , x ≠ 3 But it is given that f –1(x) = 1 k – x , x ≠ k. Hence, by comparison, k = 3. Analisis&Tip SPM Add Maths-F4-C1 2nd.indd 4 16-Feb-23 7:19:47 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 1 5 (b) ff –1 (k2 + 2) = g[(k + 2)2 ]. k2 + 2 = (k + 2)2 + 2 For any function f(x), ff –1(x) = x is always true. Hence, ff –1(k2 + 2) = k2 + 2. k2 + 2 = k2 + 4k + 4 + 2 4k + 4 = 0 k = –1 Example 7 A function f is defined by f(x) = –2x + 5 for the domain 2  x  5. (a) Find the values of p if ff –1(p2 ) = f(p – 5). (b) Find f –1(x). Hence, sketch the straight line y = x, y = f(x) and y = f –1(x) on the same Cartesian plane. What is the relation between the graph of f(x) and f –1(x)? (c) Determine the corresponding range of f(x) for the domain 2  x  5. Hence, determine the domain and range of f –1(x). What is the relation between the domain and the range of f(x) and f –1(x)? HOTS Applying HOTS Analysing HOTS Evaluating Solution ff –1(x) = x (a) ff –1(p2 ) = f(p – 5) p2 = –2(p – 5) + 5 p2 = –2p + 10 + 5 p2 + 2p – 15 = 0 (p – 3)(p + 5) = 0 p = 3 or –5 (b) Let f –1(x) = y f(y) = x –2y + 5 = x 2y = 5 – x y = 5 – x 2 y = – 1 2 x + 5 2 Hence, f –1(x) = – 1 2 x + 5 2 y y = x y = f(x) = –2x + 5 y = f –1(x) = – 1 2 x + 5 2 x –2 O 2 –2 2 4 –4 –4 4 The graph of f –1 is the reflection of the graph of f in the straight line y = x. (c) The range of f(x) is –5  x  1. The domain of f –1(x) is –5  f –1(x)  1. The range of f –1(x) is 2  f –1(x)  5. The domain of f –1(x) is the range of f(x). The range of f –1(x) is the domain of f(x). Analisis&Tip SPM Add Maths-F4-C1 2nd.indd 5 16-Feb-23 7:19:48 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 1 6 1.1 Functions 1 Given the function f(x) = 2x2 + 3x, find the objects that have the image 2. 2 Given the function g(x) = 4x + 7 x – 2 , x ≠ 2, find the objects that are mapped onto themselves. 3 Given the function h : x → x – 6 x2 – 4, state the values of x such that the function h is undefined. 4 The following arrow diagram represents the function f : x → m 2x – k , x ≠ k 2 , where m and k are constants. m 2x – k – 4 11 –1 1 3 f x 2 –2 Find the value of m and of k. The diagram below represents the mapping of y onto x by the function f : y → 2y + h and the mapping of y onto z by the function g : y → (y + 1)2 – k. 2 5 7 ➤ ➤ x y z Find (a) the value of h and of k, (b) the function that maps x onto y, (c) the function that maps x onto z. HOTS Applying HOTS Analysing HOTS Evaluating Solution (a) f : y → 2y + h g : y → (y + 1)2 – k f(y) = 2y + h g(y) = (y + 1)2 – k f(2) = 7 g(2) = 5 2(2) + h = 7 (2 + 1)2 – k = 5 h = 3 9 – k = 5 k = 4 (b) The function that maps x onto y is f –1(x). It is found that f(y) = 2y + 3. Let w = 2y + 3, y = w – 3 2 ∴ f –1 : x → x – 3 2 (c) x f –1 gf –1 g y ➤ ➤ ➤ z Hence, the function that maps x onto z is gf –1(x). Composite function of f –1 followed by g. gf –1(x) = g1 x – 3 2 2 = 1 x – 3 2 + 12 2 – 4 = 1 x – 3 2 2 2 + 2(1)1 x – 3 2 2 + 1 – 4 = x2 – 6x + 9 4 + (x – 3) – 3 = x2 – 6x + 9 4 + x – 6 = x2 – 6x + 9 + 4x – 24 4 = x2 – 2x – 15 4 ∴gf –1 : x → x2 – 2x – 15 4 Caution! The function that maps x onto z is gf –1(x) and not f –1g(x). HOTS Zone SPM Practice 1 Analisis&Tip SPM Add Maths-F4-C1 2nd.indd 6 16-Feb-23 7:19:48 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 1 7 5 Given the function g(x) = 3x2 – 5, find the values of t such that g(x) = – 7 2 t. 6 On the same axes in the answer space, sketch the graph of the function f(x) =|x + 2| for the domain –3  x  1. y x O 7 On the same axes in the answer space, sketch the graph of the function f(x) =|4x – 3| for the domain –1  x  3. y x O 8 Given the function f(x) =|3x – 7|, find the values of x that satisfy f(x) = 4. 1.2 Composite Functions 9 Given the functions f(x) = x + 3 x – 1 , x ≠ 1 and g(x) = 2x + 5, calculate the value of gf(3). 10 Given that f(x) =|x| and g(x) = ! x – 1 , x  1, calculate the value of gf(–26). 11 Given that h(x) = (x + 1)2 , calculate the value of h2 (–3). 12 Given the functions f(x) = p – 2x and g(x) = 3x2 – 7, find the value of p if fg(2) = 4. 13 The functions f and g are defined by f : x → 2x x – 2 , x ≠ 2 and g : x → x + 4 respectively, find gf. 14 Given that g : x → x2 – 3, find g2 . 15 The function f is defined by f : x → x + 2. Find the function g such that fg : x → 1 x – 1, x ≠ 1. 16 The function f is defined by f : x → x + 2. Find the function g such that gf : x → x2 – 3. 17 The function f is defined by f(x) = x – 9. Another function g is such that fg(x) = 2 3 x. Find the value of g(–3). 18 Given that f : x → 4x + k and g : x → x – 2, find the value of m and of k if fg : x → mx + 8. 19 In the following arrow diagram, the function f : x → x – 1 2 maps x onto y and the function g : y → 3y + 4 maps y onto z. x y z f g Find the function that maps x straight away to z. 20 It is given that f(x) = 9 – 2x and g(x) = ax + b, where a and b are constants. If fg(x) = 1 – 6x, find the value of a and of b. 21 The function g is defined by g : x → x + 3 and the function f is such that fg : x → x2 + 6x + 7. Find (a) the function f, (b) the value of k such that f(2k) = 8k + 30. 22 Given that m(t) = 4t2 + 4t and n(t) = 2 + 3t, find (a) the value of t such that m(t) = 15, (b) the function m(n) in terms of t, (c) the value of m(t) if n(t) = 17. 23 Given the functions f : x → hx + k and f 2 : x → 25x + 48, find (a) the values of h and the corresponding values of k, (b) the value of x if f(x) = f(3x + 1), such that h . 0. Analisis&Tip SPM Add Maths-F4-C1 2nd.indd 7 16-Feb-23 7:19:49 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 1 8 24 Given the function f : x → 2x + 5 and the composite function gf : x → 2x + 9, (a) express fg in the same form, (b) find the values of k such that fg(k2 + 1) = 5k + 27. HOTS Applying HOTS Analysing 25 The functions f and g are defined by f : x → 4 – x and g : x → hx2 + k. If the composite function gf is given by gf : x → 2x2 – 16x + 26, find (a) the value of h and of k, (b) the value of g2 (–1). 26 The function m is defined by m(x) = x – 6. If p is another function such that mp(x) = x2 + 8, find the function p. Hence, find the values of x that satisfy pm(x) = 23. 1.3 Inverse Functions 27 Given the function f : x → 2x – 5, find f –1(3). 28 Given the function f : x → x x – 2, x ≠ 2, find f –1. 29 Given the function g : x → 4x + 1, find (g–1)2 . 30 Given the functions f(x) = x + 1 x – 1 , x ≠ 1 and g(x) = 4x, find fg–1(x). 31 Given the functions m : x → 1 4 x and n : x → 2x + 5, find m–1n–1. 32 The functions f and g are defined by f(x) = x2 + x + 2 and g(x) = x – 1, find g–1f(–1). 33 Given that g–1 : x → x + 3 2x , x ≠ 0, find g(x). 34 Given the function f(x) = 2x + p 5 and its inverse function f –1(x) = 5x + 3 q , find the value of p and of q. 35 Given the function g : x → x k2 – x , x ≠ k 2 and its inverse function g–1 : x → 9x x + 1, x ≠ –1, find the values of k. 36 Given that f(x) = x + 2 x , x ≠ 0, find the values of k such that f –1(3k) = k. HOTS Applying 37 Given that f –1(x) = 1 m – x, x ≠ m and g(x) = 1 + x, find (a) f(x) in terms of m, (b) the value of m if ff –1(m2 – 2) = g[(1 – m) 2 ]. HOTS Applying HOTS Analysing 38 Given the functions f(x) = mx + n, m . 0 and f 2 (x) = 4x – 9, find (a) the value of m and of n, (b) (f –1)2 (x). HOTS Applying HOTS Analysing 39 Given that f –1(x) = 4 – px 2 and g(x) = 2x2 – 4, find (a) f(x) in terms of p, (b) the value of p such that f(x2 ) = 2g(–x). 40 Given the function f : x → h – kx, find (a) f –1(x) in terms of h and k, (b) the value of h and of k if f –1(8) = –1 and f(4) = –2. 41 Given that f(x) = 9x + 4 and g(x) = 3x + 8, find (a) fg–1(x), (b) the value of x such that gf(–x) = 23. 42 The diagram below shows the mapping of x onto y by the function f : x → a b – x , x ≠ b and the mapping of z onto y by the function g : z → 5 + bz 3z , z ≠ 0. 2 1 x y z –5 Analisis&Tip SPM Add Maths-F4-C1 2nd.indd 8 16-Feb-23 7:19:49 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 1 9 Find (a) the value of a and of b, (b) the function that maps y onto z, (c) the function that maps x onto z. HOTS Applying HOTS Analysing 43 Given the function f : x → 3x + 5 x – h , x ≠ h and its inverse function f –1 : x → 2x + 5 x – k , x ≠ k find (a) the value of h and of k, (b) the values of x such that f –1(x) = 3x + 1. 44 Given the functions f : x → px + q, g : x → (x – 1)2 – 3 and fg : x → 3(x – 1)2 – 11, find (a) the value of g2 (–1), (b) the value of p and of q, (c) gf –1. 45 Given the function f : x → 2x – m and its inverse function f –1 : x → nx + 7 2 , find (a) the value of m and of n, (b) (i) the value of f(2), (ii) the value of f –1f(2). 46 A function f is defined by f : x → k – x 5 + 2x for all values of x except x = h, where k is a constant. (a) State the value of h. (b) Given that 1 is mapped onto itself under the function f, find (i) the value of k, (ii) another value of x that is mapped onto itself, other than 1, (iii) the value of f –1(2). HOTS Applying HOTS Analysing 47 A function f is defined by f(x) = 2x x – 4, x ≠ 4. (a) Find f 2 and f –1. State the value of x such that the function is undefined. (b) Hence, find the positive value of x such that f 2 = f –1. HOTS Applying 48 A function f is defined by f (x) = 2x – 4 for the domain 0  x  4. (a) Find f –1(x) (b) Sketch the straight lines y = x, y = f(x) and y = f –1(x) on the same Cartesian plane. What is the relation between the graphs of f(x) and f –1(x)? (c) Determine (i) the domain and the range of f(x), (ii) the domain and the range of f –1(x). Make a conclusion regarding the relation between your answers in (c)(i) and (c)(ii). HOTS Applying HOTS Analysing HOTS Evaluating 49 A function f is defined by f : x → 3x + 4. Another function g is such that gf : x → 9x2 + 24x + 22. Find (a) the function g, (b) the values of p such that g–1(5p) = p, (c) the values of x if fg(x) = gf(x). HOTS Applying HOTS Analysing 50 (a) A function h is defined by h : x → 12 x – 4 , x ≠ 4. Find the values of x which are mapped onto themselves. (b) A function f is defined by f : x → 2 – 2x. Find (i) f –1, (ii) the function g such that gf –1 : x → x2 – 4x + 5. HOTS Applying HOTS Analysing Analisis&Tip SPM Add Maths-F4-C1 2nd.indd 9 16-Feb-23 7:19:49 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 1 10 1 1 2 or –2 2 –1 or 7 3 x = ±2 4 m = 4, k = 7 5 t = 5 6 or –2 6 (–3, 1) y x (1, 3) 2 –2 O 7 y (3, 9) 3 3 4 (–1, 7) x O 8 x = 3 2 3 or 1 9 11 10 5 11 25 12 p = 14 13 gf : x → 6x – 8 x – 2 , x ≠ 2 14 g2 : x → x4 – 6x2 + 6 15 g : x → 3 – 2x x – 1 , x ≠ 1 16 g : x → x2 – 4x + 1 17 7 18 m = 4, k = 16 19 gf : x → 3x + 5 2 20 a = 3, b = 4 21 (a) f : x → x2 – 2 (b) k = –2 or 4 22 (a) t = 1 1 2 or –2 1 2 (b) 36t2 + 60t + 24 (c) 120 23 (a) h = 5, k = 8 or h = –5, k = –12 (b) x = – 1 2 24 (a) fg : x → 2x + 13 (b) k = –1 1 2 or 4 25 (a) h = 2, k = –6 (b) 26 26 p(x) = x2 + 14 x = 3 or 9 27 4 28 f –1 : x → 2x x – 1, x ≠ 1 29 (g–1)2 (x) = x – 5 16 30 fg–1(x) = x + 4 x – 4 , x ≠ 4 31 m–1n–1 : x → 2x – 10 32 3 33 g(x) = 3 2x – 1, x ≠ 1 2 34 p = –3, q = 2 35 k = ±3 36 k = – 2 3 or 1 37 (a) f(x) = mx – 1 x (b) m = 2 38 (a) m = 2, n = –3 (b) (f –1)2 (x) = x + 9 4 39 (a) f(x) = 4 – 2x p (b) p = – 1 2 40 (a) f –1(x) = h – x k (b) h = 6, k = 2 41 (a) fg–1(x) = 3x – 20 (b) x = – 1 9 42 (a) a = 2, b = 4 (b) g–1(y) = 5 3y – 4, y ≠ 4 3 (c) g–1 f(x) = 20 – 5x 4x – 10, x ≠ 5 2 43 (a) h = 2, k = 3 (b) x = – 2 3 or 4 44 (a) –3 (b) p = 3, q = –2 (c) gf –1(x) = x2 – 2x – 26 9 45 (a) m = 7, n = 1 2 (b) (i) –3 (ii) 2 46 (a) h = –2 1 2 (b) (i) k = 8 (ii) x = –4 (iii) – 2 5 47 (a) f 2 (x) = 2x 8 – x , x ≠ 8 f –1(x) = 4x x – 2, x ≠ 2 (b) x = 6 48 (a) f –1(x) = x + 4 2 (b) –2 2 –2 –4 O 4 2 4 x –4 y y = x y = f(x) = 2x – 4 y = f –1(x) = x + 4 2 The graph of f –1 is the reflection of the graph of f in the straight line y = x. (c) (i) The domain of f(x) is 0  x  4. The range of f(x) is –4  f(x)  4. (ii) The domain of f –1(x) is –4  x  4. The range of f –1(x) is 0  f –1(x)  4. Conclusion: The domain of f –1(x) is the range of f(x). The range of f –1(x) is the domain of f(x). 49 (a) g(x) = x2 + 6 (b) p = 2 or 3 (c) x = 0 or –4 50 (a) x = 6 or –2 (b) (i) f –1(x) = 2 – x 2 (ii) g(x) = 4x2 + 1 Answers Analisis&Tip SPM Add Maths-F4-C1 2nd.indd 10 16-Feb-23 7:19:50 PM PENERBIT ILMU BAKTI SDN. BHD.

11 2.1 Quadratic Equations and Inequalities 1 The roots of a quadratic equation can be determined using the following methods: (a) completing the square, (b) quadratic formula. 2 The formula to find the roots of a quadratic equation ax 2 + bx + c = 0 is x = –b ± ! b2 – 4ac 2a 3 If x = α and x = β are the roots of a quadratic equation, then the quadratic equation can be formed as follows: (x – α)(x – β) = 0 x 2 – (α + β)x + αβ = 0 OR x2 – (sum of roots)x + (product of roots) = 0 ... ➀ 4 For the quadratic equation ax 2 + bx + c = 0 ⇒ x 2 + b a x + c a = 0 ... ➁, Comparing ➀ and ➁: Sum of roots = – b a Product of roots = c a 5 A quadratic inequality can be solved using the graph sketching method as follows. (a) h k x (x – h)(x – k)  0 (x – h)(x – k)  0 h  x  k h  x  k (b) h k x (x – h)(x – k)  0 (x – h)(x – k)  0 x < h or x  k x  h or x  k 2.2 Types of Roots of Quadratic Equations 1 The types of roots of a quadratic equation can be determined based on the value of b2 – 4ac. b2 – 4ac  0 b2 – 4ac = 0 b2 – 4ac  0 Real and distinct roots Real and equal roots No real roots b2 – 4ac  0 Real roots Quadratic Functions Learning Area: Algebra EXPRESS NOTES Chapter 2 Analisis&Tip SPM Add Maths-F4-C2 2nd.indd 11 16-Feb-23 7:20:18 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 2 12 2.3 Quadratic Functions 1 The position of the graph of the quadratic function f(x) = ax 2 + bx + c with respect to the x-axis depends on the types of roots of the quadratic equation ax 2 + bx + c = 0. Types of roots a is positive (a  0) a is negative (a  0) b2 – 4ac  0 (Two real and distinct roots) The graph intersects the x-axis at two different points. x x b2 – 4ac = 0 (Two real and equal roots) The graph touches the x-axis at only one point. x x b2 – 4ac  0 (No real roots) The graph does not meet the x-axis. x x 5 When a  0 (i.e. the sign in front of the brackets is negative), the function a(x – h)2 + k has a maximum value. The maximum value is k and this occurs when x – h = 0, i.e. x = h. For the case where a is negative (a  0) y x O (h, k) Maximum value = k x = h 6 The steps to sketch a quadratic function graph are: (a) Find the coordinates of the maximum or minimum point by completing the square. (b) For a minimum point, make an early sketch of ∪. For a maximum point, make an early sketch of ∩. (c) If the curve is expected to intersect the y-axis and the x-axis based on 2 The quadratic function f(x) = ax2 + bx + c has a maximum value if the coefficient of x 2 is negative, i.e. a  0 and a minimum value if the coefficient of x 2 is positive, i.e. a  0. 3 The maximum value or the minimum value of a quadratic function f(x) = ax 2 + bx + c can be determined by completing the square, i.e. by expressing ax 2 + bx + c in the vertex form a(x – h)2 + k. 4 When a  0 (i.e. the sign in front of the brackets is positive), the function a(x – h)2 + k has a minimum value. The minimum value is k and this occurs when x – h = 0, i.e. x = h. For the case where a is positive (a  0) y x O (h, k) Minimum value = k x = h Analisis&Tip SPM Add Maths-F4-C2 2nd.indd 12 16-Feb-23 7:20:18 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 2 13 the early sketch, find the points where it intersects the y-axis and the x-axis. (d) If the curve is expected to intersect only the y-axis and will not intersect the x-axis, then the working to find the points where it intersects the x-axis is not required. (e) Sketch the curve completely and label the coordinates of the maximum or minimum point clearly and the points where it intersects the y-axis and / or the x-axis. (f) Sketch and label the equation of the axis of symmetry if required by the question. The axis of symmetry is a straight line that is parallel to the y-axis and passes through the maximum or minimum point. For the quadratic function f(x) = ax 2 + bx + c = a(x – p)2 + q, the equation of the axis of symmetry is x = p = – b 2a . 13 (a) In cases where a straight line intersects a curve at two different points, then b2 – 4ac  0 is applied. (b) In cases where a straight line touches a curve at only one point, then b2 – 4ac = 0 is applied. In these cases, the straight line is a tangent to the curve. Tangent (c) In cases where a straight line does not meet a curve, then b2 – 4ac , 0 is applied. 2.1 Quadratic Equations and Inequalities Example 1 Solve each of the following quadratic equations using the method stated. (a) 3x2 + 5x – 2 = 0 (Use completing the square method) (b) 2x2 – 4x + 1 = 0 (Use quadratic formula) Solution (a) 3x2 + 5x – 2 = 0 3x2 + 5x = 2 Rewrite in the form ax2 + bx = c. x2 + 5 3 x = 2 3 Divide both sides of the equation by 3 so that the coefficient x2 becomes 1. x2 + 5 3 x + 25 36 = 2 3 + 25 36 Add the term 1 coefficient of x 2 2 2 to both sides of the equation. 1x + 5 6 2 2 = 49 36 Express all terms in the left-hand side as a perfect square. x + 5 6 = ±! 49 36 x + 5 6 = ± 7 6 Analisis&Tip SPM Add Maths-F4-C2 2nd.indd 13 16-Feb-23 7:20:18 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 2 14 x + 5 6 = 7 6 or x + 5 6 = – 7 6 x = 7 6 – 5 6 x = – 7 6 – 5 6 x = 1 3 x = –2 (b) 2x2 – 4x + 1 = 0 x = –b ± ! b2 – 4ac 2a x = –(–4) ± ! (–4)2 – 4(2)(1) 2(2) Use the quadratic formula. x = 4 ± ! 8 4 x = 4 ± 2.8284 4 x = 4 + 2.8284 4 or x = 4 – 2.8284 4 x = 1.71 or x = 0.29 SMART TIP The quadratic formula is suitable to be used when (a) the quadratic equation ax2 + bx + c = 0 is difficult to be factorised because the values of a, b and c are very large. (b) the quadratic equation cannot be factorised. Example 2 Form a quadratic equation that has the roots – 1 4 and 3 5 . Solution Sum of roots = – 1 4 + 3 5 = 7 20 Product of roots = – 1 4 × 3 5 = – 3 20 The quadratic equation that is formed is x2 – 7 20 x – 3 20 = 0 20x2 – 7x – 3 = 0 x2 – (sum of roots)x + (product of roots) = 0 Example 3 If α and β are the roots of the quadratic equation 2x2 + 5x – 12 = 0, form a quadratic equation that has the roots 1 α and 1 β . Solution 2x2 + 5x – 12 = 0 where a = 2, b = 5 and c = –12. Sum of roots: Product of roots: α + β = – b a αβ = c a = – 5 2 = – 12 2 = –6 The new roots are 1 α and 1 β . Sum of new roots: Product of new roots: 1 α + 1 β = β + α αβ 1 1 α2 1 1 β2 = 1 αβ = – 5 2 –6 = 1 –6 = 5 12 = – 1 6 Thus, the new quadratic equation with roots 1 α and 1 β is x2 – 5 12 x – 1 6 = 0 x2 – (sum of roofs)x + (product of roots) = 0 ⇒ 12x2 – 5x – 2 = 0 Example 4 One of the roots of the quadratic equation x2 + 8 = (p + 1)x is two times of the other root. Find the possible values of p. Solution x2 + 8 = (p + 1)x x2 – (p + 1)x + 8 = 0 Arrange the equation to be in the general form. Hence, a = 1, b = –(p + 1) and c = 8. Let α and 2α as the roots. Analisis&Tip SPM Add Maths-F4-C2 2nd.indd 14 16-Feb-23 7:20:19 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 2 15 Sum of roots, α + 2α = – b a 3α = – 1 –(p + 1) 1 2 3α = p + 1 α = p + 1 3 ...... ➀ Product of roots, (α)(2α) = c a 2α2 = 8 1 α2 = 4 ...... ➁ Subtitute ➀ into ➁: 1 p + 1 3 2 2 = 4 1 p2 + 2p + 1 9 2 = 4 p2 + 2p + 1 = 36 p2 + 2p – 35 = 0 (p – 5)(p + 7) = 0 Using factorisation ∴ p = 5 or –7 Example 5 Find the range of values of x that satisfies the quadratic inequality 2x2 + 5x – 12  0. Solution 2x2 + 5x – 12  0 (x + 4)(2x – 3)  0 Use factorisation Let (x + 4)(2x – 3) = 0 ⇒ x = –4 or 1 1 2 Hence, the graph intersects the x-axis at the points (–4, 0) and 11 1 2 , 02. The region below the x-axis that has to be shaded because the inequality is . –4 x 11 2 Hence, the range of values of x that satisfies the quadratic inequality 2x2 + 5x – 12  0 is –4  x  1 1 2 . Example 6 Find the range of values of x that satisfies the quadratic inequality –6x2 – 5x + 6  0. Solution SMART TIP Change the sign of each term in order that the coefficient of x2 is changed from negative to positive to simplify the factorisation. But you must remember to reverse the inequality sign. –6x2 – 5x + 6  0 6x2 + 5x – 6 > 0 (2x + 3)(3x – 2)  0 ... * Using factorisation Let (2x + 3)(3x – 2) = 0 ⇒ x = – 3 2 or 2 3 Hence, the graph will intersect the x-axis at the points 1– 3 2 , 02 and 1 2 3 , 02. 2 3 – 3 2 x The region above the x-axis that has to be shaded because the inequality obtained in the step * is . Hence, the range of values of x that satisfies the quadratic inequality –6x2 – 5x + 6  0 is x  – 3 2 or x  2 3 . If the question uses the inequality  (i.e. including the ‘equal to’ sign), then the answer has to have the ‘equal to’ sign too. Analisis&Tip SPM Add Maths-F4-C2 2nd.indd 15 16-Feb-23 7:20:19 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 2 16 Caution! 6x2 + 5x – 6  0 (2x + 3)(3x – 2)  0 x  – 3 2 or x  2 3 Incorrect because the step of graph sketching is not carried out. 2.2 Types of Roots of Quadratic Equations Example 8 Given that the quadratic equation 2x2 – (2 + h)x + 2 = 0 has real and equal roots, find the values of h. Solution Given that 2x2 – (2 + h)x + 2 = 0, where a = 2, b = –(2 + h) and c = 2. For the case of ‘real and equal roots’ b2 – 4ac = 0, is used. b2 – 4ac = 0 [–(2 + h)]2 – 4(2)(2) = 0 (2 + h)2 – 16 = 0 4 + 4h + h2 – 16 = 0 h2 + 4h – 12 = 0 (h – 2)(h + 6) = 0 Use factorisation ∴h = 2 or –6 Example 9 Find the range of values of h if the quadratic equation x(x + h + 3) = –1 has two real and distinct roots. Solution x(x + h + 3) = –1 x2 + hx + 3x + 1 = 0 x2 + (h + 3)x + 1 = 0 Arrange the equation to be in the general form. Hence, a = 1, b = h + 3 and c = 1. If the quadratic equation has two real and distinct roots, then b2 – 4ac  0 (h + 3)2 – 4(1)(1)  0 h2 + 6h + 9 – 4  0 h2 + 6h + 5  0 (h + 5)(h + 1)  0 –5 –1 h Hence, the range of values of h is h  –5 or h  –1. Example 10 Find the range of values of k if the quadratic equation x2 + kx + 16 = 2(x + k) does not have real roots. Solution x2 + kx + 16 = 2(x + k) x2 + kx + 16 = 2x + 2k x2 + kx – 2x + 16 – 2k = 0 x2 + (k – 2)x + 16 – 2k = 0 Arrange the equation to be in the general form. Hence, a = 1, b = k – 2 and c = 16 – 2k If the quadratic equation does not have real roots, then b2 – 4ac  0 (k – 2)2 – 4(1)(16 – 2k)  0 k2 – 4k + 4 – 64 + 8k  0 k2 + 4k – 60  0 (k + 10)(k – 6)  0 –10 6 k Hence, the range of values of k is –10  k  6. 2.3 Quadratic Functions Example 11 Express the quadratic function f(x) = 2x2 – 3x – 5 in the form a(x – h) 2 + k, where a, h and k are constants. Hence, state the maximum or minimum value of the function and the corresponding value of x. Analisis&Tip SPM Add Maths-F4-C2 2nd.indd 16 16-Feb-23 7:20:20 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 2 17 Solution SMART TIP In the process of 'completing the square', the coefficient of x2 must be 1. Hence, if the coefficient of x2 is not 1, then it must be reduced to 1 first, by factorisation. f(x) = 2x2 – 3x – 5 = 21 x2 – 3 2 x – 5 2 2 The number 2 is factorised out to reduce the coefficient of x2 to 1. = 21 x2 – 3 2 x + 9 16 – 9 16 – 5 2 2 Add and subtract 1 coefficient of x 2 2 2 i.e. 3 1 2 × 1– 3 2 24 2 = 9 16 = 231x – 3 4 2 2 – 9 16 – 5 2 4 The first three terms, i.e. 1 x2 – 3 2 x + 9 162 are factorised to form a perfect square, i.e. 1 x – 3 4 2 2 = 2 . 31x – 3 4 2 2 – 49 164 Do not forget to multiply 2 with each term in the brackets at the last step. = 21 x – 3 4 2 2 – 49 8 The form a(x – h) 2 + k, where a = 2, h = 3 4 and k = – 49 8 SMART TIP Since the number in front of the brackets (i.e. a = 2) is positive, then the value of f (x) is minimum. ∴ Minimum value of f(x) = – 49 8 = –6 1 8 when x – 3 4 = 0 ⇒ x = 3 4 . Example 12 (a) Express the quadratic function f(x) = –2x2 + 5x + 3 in the form a(x – p) 2 + q, where a, p and q are constants. Hence, state the maximum or minimum value of the function and the corresponding value of x. (b) Sketch the graph of the function f(x) = –2x2 + 5x + 3. (c) State the equation of the axis of symmetry of the graph of f(x). (d) Hence, sketch the graph of the absolute value function g(x) = |–2x2 + 5x + 3|. Solution (a) f(x) = –2x2 + 5x + 3 = –21x2 – 5 2 x – 3 2 2 = –21x2 – 5 2 x + 25 16 – 25 16 – 3 2 2 Add and subtract 1 coefficient of x 2 2 2 = 3 1 2 × 1– 5 2 24 2 = 25 16 = –231x – 5 4 2 2 – 25 16 – 24 164 Factorised the first three terms to form a perfect square. = –231x – 5 4 2 2 – 49 164 = –21x – 5 4 2 2 + 49 8 The form a(x – p) 2 + q, where a = –2, p = 5 4 and q = 49 8 . SMART TIP Since the number in front of the brackets (i.e., a = –2) is negative, then the value of the function f (x) is a maximum. ∴ Maximum value of the function f(x) = 49 8 = 6 1 8 when x – 5 4 = 0 ⇒ x = 5 4 = 1 1 4 . Analisis&Tip SPM Add Maths-F4-C2 2nd.indd 17 16-Feb-23 7:20:20 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 2 18 (b) Step 1 From (a), it is determined that the maximum value of f(x) is 6 1 8 when x = 1 1 4 . Hence, the maximum point of the graph is 1 1 1 4 , 6 1 8 2 . The early sketch of the graph is as follows. y x O 111 4 , 61 82 Step 2 Next, the points where the graph intersects the x-axis and the y-axis have to be determined. On the x-axis, y = 0. When y = 0, –2x2 + 5x + 3 = 0 2x2 – 5x – 3 = 0 (2x + 1)(x – 3) = 0 x = – 1 2 or 3 Hence, the graph intersects the x-axis at the points 1– 1 2 , 02 and (3, 0). On the y-axis, x = 0. When x = 0, y = –2(0)2 + 5(0) + 3 = 3 Hence, the graph intersects the y-axis at the point (0, 3). Step 3 The complete sketch of the graph is as follows. 3 y 3 x O 111 4 , 61 82 – 1 2 (c) The equation of the axis of symmetry is x = 1 1 4 . The axis of symmetry is a vertical line that passes through the maximum point. The equation of the axis of symmetry is x = – b 2a = –1 5 2(–2) 2 = 1 1 4 Alternative Method (d) The graph of the absolute value function g(x) = |–2x2 + 5x + 3| is as follows. The part of the graph of the function f(x) = –2x2 + 5x + 3 which is below the x-axis is reflected in the x-axis. 3 y 3 x O 111 4 , 61 82 – 1 2 Example 13 Find the range of values of t if the graph of the quadratic function f(x) = tx2 + 9x + t + 12 intersects the x-axis at two different points. Solution f(x) = tx2 + 9x + t + 12 where a = t, b = 9 and c = t + 12. In cases where the graph of a quadratic function intersects the x-axis at two different points, b2 – 4ac  0 is applied. b2 – 4ac  0 92 – 4t(t + 12)  0 81 – 4t2 – 48t  0 –4t 2 – 48t + 81  0 4t 2 + 48t – 81 , 0 (2t + 27)(2t – 3) , 0 Arrange the equation to be in the general form. If the sign of each term is changed, then the inequality sign must be reversed too. Use factorisation Analisis&Tip SPM Add Maths-F4-C2 2nd.indd 18 16-Feb-23 7:20:21 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 2 19 3 2 – 27 2 t Hence, the required range of value of t is – 27 2  t  3 2 ⇒ –13 1 2  t  1 1 2 . Example 14 Find the range of values of m if the straight line y = m(4x + 5) intersects the curve y = 4x2 + 12x + 15 at two different points. Solution y = 4x2 + 12x + 15 y = m(4x + 5) y = m(4x + 5) … ➀ y = 4x2 + 12x + 15 … ➁ Substitute ➁ into ➀: 4x2 + 12x + 15 = m(4x + 5) 4x2 + 12x + 15 = 4mx + 5m 4x2 + 12x – 4mx + 15 – 5m = 0 4x2 + (12 – 4m)x + 15 – 5m = 0 4x2 + 4(3 – m)x + 15 – 5m = 0 a = 4, b = 4(3 – m), c = 15 – 5m Arrange the equation to be in the general form. In cases where a straight line intersects a curve at two different points, b2 – 4ac  0 is applied. b2 – 4ac  0 [4(3 – m)]2 – 4(4)(15 – 5m)  0 16(3 – m) 2 – 16(15 – 5m)  0 (3 – m) 2 – (15 – 5m)  0 The whole inequality is divided by 16. 9 – 6m + m2 –15 + 5m  0 m2 – m – 6  0 (m + 2)(m – 3)  0 –2 3 m Hence, the required range of values of m is m  –2 or m  3. Example 15 Given that the quadratic function f(x) = x2 – tx + t + 3 is always positive, find the range of values of t. Solution SMART TIP If a quadratic function is always positive, its graph is always above the x-axis, i.e. it does not meet the x-axis. Hence, b2 – 4ac  0 is applied. x f(x) = x2 – tx + t + 3 a = 1, b = –t, c = t + 3 b2 – 4ac  0 (–t)2 – 4(1)(t + 3)  0 t 2 – 4t – 12  0 (t + 2)(t – 6)  0 t –2 6 Hence, the range of values of t such that f(x) = x2 – tx + t + 3 is always positive is –2  t  6. Caution! If a quadratic function is always positive or always negative, b2 – 4ac  0 is applied and not b2 – 4ac  0. Analisis&Tip SPM Add Maths-F4-C2 2nd.indd 19 16-Feb-23 7:20:21 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 2 20 Find the range of values of k if the quadratic function f(x) = 5x2 + 4kx + k2 – 20 is always positive for all values of x. HOTS Applying HOTS Analysing HOTS Evaluating Solution SMART TIP If a quadratic function is always positive, its graph is always above the x-axis, i.e. it does not intersect the x-axis. Hence, b2 – 4ac  0 is used. x y = f(x) f(x) = 5x2 + 4kx + k2 – 20 where a = 5, b = 4k and c = k2 – 20 b2 – 4ac  0 (4k) 2 – 4(5)(k2 – 20)  0 16k2 – 20(k2 – 20)  0 16k2 – 20k2 + 400  0 –4k2 + 400  0 –k2 + 100  0 k2 – 100 . 0 (k + 10)(k – 10) . 0 Divide throughout by 4. When the sign of each term is changed, then the inequality sign has to be reversed. k –10 10 Hence, the required range of values of k is k  –10 or k  10. 2.1 Quadratic Equations and Inequalities 1 Solve the quadratic equation k2 – k – 12 = 0 by completing the square. 2 Find the roots of the quadratic equation 2x2 – 9x + 6 = 0. State your answer correct to two decimal places. 3 The following diagram shows the front view of four pieces of woods with the same width. The total area of all the front surface of the woods is 80 cm2 . 26 cm The four pieces of woods are used to make a rectangular photo frame as shown in the following diagram. 26 cm 16 cm Find the width, in cm, of each piece of wood. HOTS Applying HOTS Analysing HOTS Evaluating 4 Form the quadratic equation that has the roots – 2 3 and 3 2 . 5 Form the quadratic equation that has the repeated roots 4. 6 Given that one of the roots of the quadratic equation px2 – px + 2 = 0 is two times of the other root, find the value of p. HOTS Zone SPM Practice 2 Analisis&Tip SPM Add Maths-F4-C2 2nd.indd 20 16-Feb-23 7:20:22 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 2 21 7 If one of the roots of the quadratic equation ax2 + bx + c = 0 is two times of the other root, find an expression in the terms of a, b and c. HOTS Applying 8 If m and n are the roots of the quadratic equation x2 + 2x – 8 = 0, form a quadratic equation that has the roots 2m and 2n. 9 One of the roots of the quadratic equation x2 – px + 8 = 0 is the square of the other root. Find the value of p. HOTS Applying 10 The quadratic equation x2 + kx + 3 = 0 has the roots α and β. The quadratic equation 2x2 + 2 = 8x + k also has the roots α and β. Find the value of k. HOTS Applying 11 It is given that 2 and m + 1 are the roots of the quadratic equation x2 + (n – 2)x + 6 = 0, where m and n are constants. Find the value of m and of n. HOTS Applying 12 It is given that –3 and k are the roots of the quadratic equation (2x + 1)(x – 1) = p(1 – 3x), where p is a constant. Find the value of p and of k. 13 Given that p + 1 and q – 1 are the roots of the quadratic equation x2 + 3x = 10, find the values of p and the corresponding values of q. HOTS Applying HOTS Analysing 14 Given that p and q are the roots of the quadratic equation 2x2 – x – 6 = 0, form the quadratic equation that has the roots 2p + 1 and 2q + 1. 15 Given that α and β are the roots of the quadratic equation 2x2 + 5x – 2 = 0, find the quadratic equation that has the roots 1 α + 1 and 1 β + 1 . 16 One of the roots of the quadratic equation 3x2 + (p + 1)x + p + 6 = 0 is 1 3 times of the other root. Find the values of p. 17 One of the roots of the quadratic equation 2kx2 + 1 = kx is two times of the other root. (a) Find the value of k. (b) Hence, find the roots of the quadratic equation. 18 Given that the roots of the quadratic equation kx2 – 5x + k + 1 = 0, k ≠ 0 are in the ratio 2 : 3, find the values of k. HOTS Applying HOTS Analysing 19 Given that k 3 and k 5 are the roots of the quadratic equation 15x2 + 16x + m = 0, find the value of k and of m. HOTS Applying HOTS Analysing 20 The quadratic equation px2 – (p + 4)x + (2q + 1) = 0, p ≠ 0 has the roots 1 p and q. (a) Find the value of p and of q. (b) Using the value of p and of q in (a), form a quadratic equation that has the roots –p and 2 3 q. HOTS Applying HOTS Analysing 21 α 2 and β 2 are the roots of the quadratic equation kx2 + 2m = (k + 5)x. If α + β = 7 and αβ = 12, find the value of k and of m. HOTS Applying HOTS Analysing 22 Find the range of values of x that satisfies the inequality 3x2 – 5x – 2  0. 23 Find the range of values of x that satisfies the inequality (x – 1)(x – 2)  6. 24 Find the range of values of x that satisfies the inequality –x2 + 2x + 35  0. 25 Find the range of values of x that satisfies the inequality 10  3x2 – x. 26 Find the range of values of x that satisfies the inequality (x – 1)2  (x – 1)(2x + 3). 27 Find the range of values of x that satisfies the inequality (x + 3)(1 – 2x)  0. 28 Find the range of values of x that satisfies the inequalities xy  2 and y – 2x + 3 = 0. Analisis&Tip SPM Add Maths-F4-C2 2nd.indd 21 16-Feb-23 7:20:22 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 2 22 2.2 Types of Roots of Quadratic Equations 29 Determine the type of roots of the quadratic equation 3x2 + 4 = 7x. 30 Determine the type of roots of the quadratic equation 2x2 + 3 = 4x. 31 Determine the type of roots of the quadratic equation 4x2 + 9 = 12x. 32 Find the value of p if the quadratic equation 4x2 – (4 + p)x + p = 0 has real and equal roots. 33 Find the values of h if the quadratic equation (h + 3)x2 – 3hx + 9 = 0 has real and equal roots. 34 If the quadratic equation p2 x2 + 4qx + 25 = 0 has real and equal roots where p  0 and q  0, find p : q. HOTS Applying 35 Find the range of values of t if the quadratic equation x2 + tx + 2t – 3 = 0 has real and distinct roots. 36 Find the range of values of q if the quadratic equation x2 – qx + q + 3 = 0 does not have real roots. 37 Find the range of values of k if the quadratic equation 2x2 + 20 = k(2x + 3) has two real and distinct roots. 38 Find the range of value of k if the quadratic equation 3x2 – 3kx + (k2 – k – 3) = 0 does not have real roots. 39 Find the range of value of t if the quadratic equation (2t + 1)x2 + 2t = 4 – 3tx has two real roots. 2.3 Quadratic Functions 40 If the curve y = –x2 + nx – 9 touches the x-axis at only one point, find the values of n. HOTS Applying 41 Find the range of values of k if the graph of the quadratic function f(x) = 2x2 + kx + 2 intersects the x-axis at two different points. 42 Find the range of values of m if the graph of the quadratic function g(x) = x2 + mx + 16 does not meet the x-axis. 43 State the maximum value of the quadratic function f(x) = –3(x – 4)2 – 5 and the equation of its axis of symmetry. 44 Express the quadratic function f(x) = 2x2 + 3x – 4 in the form a(x – h) 2 + k, where a, h and k are constants. 45 Express the quadratic function g(x) = 4 – x – 3x2 in the form a(x – h) 2 + k, where a, h and k are constants. 46 By expressing f(x) = 2x2 – 3x + 1 in the form a(x – h) 2 + k, where a, h and k are constants, find the minimum value of f(x). 47 By expressing g(x) = 3 + 2x – 2x2 in the a(x – h) 2 + k, where a, h and k are constants., find the maximum value of g(x). 48 By expressing f(x) = 3x2 – 6x + 8 in the form a(x – h) 2 + k, where a, h and k are constants, find the value of x when f(x) has a minimum value. 49 By expressing g(x) = –x2 + 3x – 7 in the form a(x – h)2 + k, where a, h and k are constants. find the value of x when g(x) has a maximum value. 50 Find the values of k if the quadratic function f(x) = x2 + 2kx + k has a minimum value of –2. HOTS Applying 51 Find the equation of the axis of symmetry of the graph of the quadratic function f(x) = 2x2 – 5x + 6. 52 Find the equation of the axis of symmetry of the graph of the quadratic function g(x) = 7 – 2x – 3x2 . Analisis&Tip SPM Add Maths-F4-C2 2nd.indd 22 16-Feb-23 7:20:23 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 2 23 53 Find the range of values of k if the quadratic function f(x) = 2x2 – x + k is always positive. HOTS Applying 54 Find the range of values of h if the quadratic function g(x) = h + 3x – 5x2 is always negative. HOTS Applying 55 The function f(x) = –x2 + 4kx – 5k2 – 1 has the maximum value –r 2 – 2k, where r and k are constants. (a) By completing the square, show that r = k – 1. (b) Hence, find the value of k and of r if the graph of the function is symmetrical about the axis x = r 2 – 1, where k ≠ 0. HOTS Applying HOTS Analysing HOTS Evaluating 56 Given the quadratic function f(x) = 5x – 3 – 2x2 , (a) express the quadratic function in the form a(x – p) 2 + q, where a, p and q are constants, (b) determine the maximum or the minimum value of f(x) and state the corresponding value of x, (c) sketch the graph of the function f(x). 57 Find the maximum or the minimum value of the quadratic function f(x) = (2 – 3x)(x + 1) + 2(5x – 1) – 6. Hence, sketch the graph of the function f(x). 58 (a) Sketch the graph of the function f(x) = x2 + 2x – 3 for the domain –4  x  3. (b) Hence, sketch the graph of g(x) = |x2 + 2x – 3| for the domain –4  x  3 and state the corresponding range of values of g(x) for the given domain. HOTS Applying HOTS Analysing 59 It is given that y = 3 + hx – x2 = –(x – k) 2 + 4, where h  0 and k  0. (a) Find the value of h and of k. (b) Find the equation of the axis of symmetry. (c) Sketch the graph of the function. (d) If the straight line y = mx + 4 is a tangent to the curve, find the possible values of m. HOTS Applying HOTS Analysing 60 y = x2 + 2px + 4p has a minimum value of 4. (a) Find the possible value of p. (b) By using the value of p in (a), sketch the graph of y = x2 + 2px + 4p. HOTS Applying HOTS Analysing 61 The diagram below shows the curve of the quadratic function f(x) = –x2 + mx – 6 that has a maximum point Q(2, k). The curve intersects the f(x)-axis at point P. O P Q(2, k) x f(x) (a) State the coordinates of point P. (b) By completing the square, find the value of m and of k. (c) Find the range of values of x such that f (x)  –6. HOTS Applying HOTS Analysing 62 Express y = 2x2 + 7x + 3 in the vertex form. Hence, sketch the graph of y = 2x2 + 7x + 3 for the domain –4  x  2 and state the corresponding range of y. 63 The quadratic function f (x) = x2 + (k – 2)x + 16 – 2k, where k is a constant, is always positive when p  k  q. Find the value of p and of q. HOTS Applying HOTS Analysing 64 Find the range of values of p if the graph of y = x2 – (2p + 4)x does not meet the straight line y = 1 – 10p. 65 Find the range of values of m such that the straight line y = 2x + m does not intersect the circle x2 + y2 = 4. Analisis&Tip SPM Add Maths-F4-C2 2nd.indd 23 16-Feb-23 7:20:23 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 2 24 1 k = –3 or 4 2 x = 3.69 or 0.81 3 1 cm 4 6x2 – 5x – 6 = 0 5 x2 – 8x + 16 = 0 6 p = 9 7 2b2 = 9ac 8 x2 + 4x – 32 = 0 9 p = 6 10 k = –4 11 m = 2, n = –3 12 p = 2, k = 1 2 13 p = 1, q = –4 or p = –6, q = 3 14 x2 – 3x – 10 = 0 15 2x2 – 9x + 5 = 0 16 p = 19 or –5 17 (a) k = 9 (b) x = 1 3 or 1 6 18 k = 2 or –3 19 k = –2, m = 4 20 (a) q = –1, p = –1 1 2 (b) 6x2 – 5x – 6 = 0 21 k = 2, m = 3 22 2 3  x  1 23 x  –1 or x  4 24 –5  x  7 25 x  –1 2 3 or x  2 26 –4  x  1 27 x  –3 or x  1 2 28 x  – 1 2 or x  2 29 Real and distinct roots 30 No real roots 31 Real and equal roots 32 p = 4 33 h = –2 or 6 34 2 : 5 35 t  2 or t  6 36 –2  q  6 37 k  –10 or k  4 38 k  –2 or k  6 39 – 4 7  t  4 40 n = ±6 41 k  –4 or k  4 42 –8  m  8 43 –5, x = 4 44 f(x) = 21x + 3 4 2 2 – 41 8 45 g(x) = –31x + 1 6 2 2 + 49 12 46 21x – 3 4 2 2 – 1 8 ; – 1 8 47 –21x – 1 2 2 2 + 7 2 ; 3 1 2 48 3(x – 1)2 + 5; x = 1 49 –1x – 3 2 2 2 – 19 4 ; x = 1 1 2 50 k = 2 or –1 51 x = 1 1 4 52 x = – 1 3 53 k  1 8 54 h  – 9 20 55 (b) k = 4, r = 3 56 (a) f(x) = –21x – 5 4 2 2 + 1 8 (b) Maximum value = 1 8 ; x = 1 1 4 (c) y x O –3 1 1 11 4, 1 82 11 2 57 Maximum value = 3 4 y x O –6 1 2 1 11 2, 3 42 58 (a) y (3, 12) (–4, 5) (–1, –4) –3 –3 1 x O (b) y x O 1 3 –3 (3, 12) (–1, 4) (–4, 5) 0  g(x)  12 59 (a) h = 2, k = 1 (b) x = 1 (c) y (1, 4) 3 –1 3 x O (d) m = 0 or 4 60 (a) p = 2 (b) y x O (–2, 4) 8 61 (a) (0, –6) (b) m = 4, k = –2 (c) 0 < x < 4 62 y = 21x + 7 4 2 – 25 8 f(x) (2, 25) (–4, 7) (0, 3) –3 x 0 1 – 7 4, – 25 8 2 – 1 2 –3 1 8  y  25 63 p = –10, q = 6 64 1  p  5 65 m  –2!5 or m  2!5 Answers Analisis&Tip SPM Add Maths-F4-C2 2nd.indd 24 16-Feb-23 7:20:24 PM PENERBIT ILMU BAKTI SDN. BHD.

25 3.1 Systems of Linear Equations in Three Variables 1 A system of linear equations in three variables consists of three linear simultaneous equations to be solved, as follows: 3.2 Simultaneous Equations Involving One Linear Equation and One Non-Linear Equation 1 The steps to solve simultaneous equations involving one linear equation and one non-linear equation are as follows: Step 1: From the linear equation, a variable (let it be y) is expressed in terms of the other variable (let it be x). Step 2: The variable (y) is substituted into the non-linear equation and a quadratic equation in terms of the other variable (x) is formed. ➊ Choose two equations and use them to eliminate one variable. ➋ Choose one more pair of equations and use them to eliminate the same variable. ➍ Solve the last equation formed to determine the value of the first variable. ➎ Find the value of the second variable. Do this using either one of the equations formed at steps ➊ and ➋ and the value of the variable determined at step ➍. ➏ Find the value of the third variable using either one of the original equations and the values of the two variables determined at steps ➍ and ➎. ➌ Use the two pairs of equations formed at the two initial steps to eliminate either one of the two variables that are left. Step 3: Simplify and solve the quadratic equation using factorisation or the quadratic formula to obtain the values of the first variable (x). Step 4: Determine the values of the second variable (y) by substituting the values of the first variable (x), one by one, into the linear equation. 2 Simultaneous equations can be used to solve problems in daily real-life situations. Systems of Equations Learning Area: Algebra EXPRESS NOTES Chapter 3 Analisis&Tip SPM Add Maths-F4-C3 2nd.indd 25 16-Feb-23 7:20:53 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 3 26 3.1 Systems of Linear Equations in Three Variables Example 1 Solve the following simultaneous equations: 2x + y + z = 0 3x + 2y – z = 7 4x + y – 5z = 22 Solution 2x + y + z = 0 ... ➀ 3x + 2y – z = 7 ... ➁ 4x + y – 5z = 22 ... ➂ Eliminate the variable z by adding ➀ and ➁. ➀ + ➁: 5x + 3y = 7 ... ➃ Multiply equation ➁ by 5 so that the coefficients of variable z are equal. ➁ × 5 : 15x + 10y – 5z = 35 ... ➄ Eliminate the variable z by substracting ➂ from ➄. ➄ – ➂ : 11x + 9y = 13 ... ➅ Multiply equation ➃ by 3 so that the coefficients of variable y are equal. ➃ × 3: 15x + 9y = 21 ... ➆ Solve the simultaneous equations ➅ and ➆. ➆ – ➅ : 4x = 8 x = 2 Substitute x = 2 into equation ➃. 5(2) + 3y = 7 3y = –3 y = –1 Substitute x = 2 and y = –1 into either one of the three original equations, let’s say ➀. 2(2) + (–1) + z = 0 z = –3 Example 2 Encik Zainal will be retiring. He is the form teacher of the class 4S4 that consists of 38 students. He wants to buy postcards to be given to his students as memory. From a stationery shop, Encik Zainal chooses three types of postcards, i.e. A, B and C, with different prices. The price of each postcard B is the average of the prices of each the postcards A and C. If Encik Zainal bought 20 pieces of postcard A, 12 pieces of postcard B and 6 pieces of postcard C, then he needs to pay RM18.10. If Encik Zainal bought 19 pieces of postcard A and 19 pieces of postcard B, then he needs to pay RM17.10. If x, y and z represent the price (in sen) of each post card A, B and C, form three linear equations. Hence, solve the simultaneous linear equation equations and state the price of each type of postcard. HOTS Applying HOTS Analysing HOTS Evaluating Solution The price of each postcard B is the average of the prices of each the postcards A and C. The equation is y = x + z 2 2y = x + z x – 2y + z = 0 … ➀ The total price of 20 pieces of postcards A, 12 pieces of postcards B and 6 pieces of postcards C is RM18.10. The equation (in sen) is 20x + 12y + 6z = 1 810 10x + 6y + 3z = 905 … ➁ The total price of 19 pieces of postcard A and 19 pieces of postcards B is RM17.10. The equation (in sen) is 19x + 19y = 1 710 x + y = 90 … ➂ ➀ × 3 : 3x – 6y + 3z = 0 … ➃ ➃ – ➁ : –7x – 12y = –905 … ➄ ➂ × 7 : 7x + 7y = 630 … ➅ ➄ + ➅ : –5y = –275 y = – 275 –5 y = 55 Multiply equation ➀ by 3 to eliminate the variable z. Multiply equation ➂ by 7 to eliminate the variable x. Substitute y = 55 into equation ➂: x + 55 = 90 x = 35 Analisis&Tip SPM Add Maths-F4-C3 2nd.indd 26 16-Feb-23 7:20:53 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 3 27 Substitute x = 35 and y = 55 into equation ➀: 35 – 2(55) + z = 0 –75 + z = 0 z = 75 Hence, the price of each of the postcards A, B and C are 35 sen, 55 sen and 75 sen respectively. 3.2 Simultaneous Equations Involving One Linear Equation and One Non-Linear Equation Example 3 Solve the following simultaneous equations: 3x + 2y = 1 3x2 – 5x – y2 – 3y = 0 Solution This is a linear equation because the variable x and the variable y have a degree of one. 3x + 2y = 1 … ➀ 3x2 – 5x – y2 – 3y = 0 … ➁ This is a non-linear equation because there are variables that have a degree of more than one, i.e. x2 and y2 with a degree of two. From ➀: 2y = 1 – 3x y = 1 – 3x 2 … ➂ Make y the subject of the formula. Substitute ➂ into ➁: 3x2 – 5x – 1 1 – 3x 2 2 2 – 31 1 – 3x 2 2 = 0 3x2 – 5x – 1 (1 – 3x)2 4 2 – 31 1 – 3x 2 2 = 0 ** 12x2 – 20x – (1 – 3x)2 – 6(1 – 3x) = 0 Multiply the equation throughout by 4 to eliminate fractions. Solve the quadratic equation by factorisation. 12x2 – 20x – (1 – 6x + 9x2 ) – 6 + 18x = 0 12x2 – 20x – 1 + 6x – 9x2 – 6 + 18x = 0 3x2 + 4x – 7 = 0 (x – 1)(3x + 7) = 0 x = 1 or – 7 3 Subtitute x = 1 and x = – 7 3 into ➂: When x = 1, y = 1 – 3(1) 2 = –1 When x = – 7 3 , y = 1 – 31– 7 3 2 2 = 4 Hence, the solutions are x = 1, y = –1 or x = –2 1 3 , y = 4. Caution! From the step ** 3x2 – 5x – 1 (1 – 3x)2 4 2 – 31 1 – 3x 2 2 = 0 3x2 – 5x – (1 – 3x)2 – 3(1 – 3x) = 0 Incorrect because the terms 3x2 – 5x are not multiplied by 4. 3x2 – 5x – (1 – 6x + 9x2 ) – 3 + 9x = 0 3x2 – 5x – 1 + 6x – 9x2 – 3 + 9x = 0 –6x2 + 10x – 4 = 0 6x2 – 10x + 4 = 0 Stuck because the quadratic equation cannot be factorised. SMART TIP It is a good practice to check the answers obtained by substituting the values of x and y into both original equations to make sure that the values satisfy the equations. Example 4 Solve the following simultaneous equations: x + 4y = 6 5 x – 2 y + 1 = 0 Solution x + 4y = 6 … ➀ 5 x – 2 y + 1 = 0 … ➁ Analisis&Tip SPM Add Maths-F4-C3 2nd.indd 27 16-Feb-23 7:20:54 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 3 28 Eliminate fractions in ➁: 5 x – 2 y + 1 = 0 5y – 2x xy + 1 = 0 5y – 2x + xy xy = 0 5y – 2x + xy = 0 … ➂ A non-linear equation with degree of one is formed. From ➀: x = 6 – 4y … ➃ Make x the subject of the formula. Substitute ➃ into ➂: 5y – 2(6 – 4y) + y(6 – 4y) = 0 5y – 12 + 8y + 6y – 4y2 = 0 19y – 12 – 4y2 = 0 4y2 – 19y + 12 = 0 Solve the quadratic equation by (4y – 3)(y – 4) = 0 factorisation y = 3 4 or 4 Subtitute y = 3 4 and y = 4 into ➃: When y = 3 4 , x = 6 – 41 3 4 2 = 3 When y = 4, x = 6 – 4(4) = –10 Hence, the solutions are x = 3, y = 3 4 or x = –10, y = 4. Example 5 The diagram below shows a rectangular piece of aluminium sheet ABCD with an area of 70 cm2 . A quadrant PBC is cut from the aluminium sheet and the perimeter of the part of the aluminium sheet left is 31 cm. x cm A D C B P y cm Using π = 22 7 , find the integer value of x and the integer value of y. Solution x cm A D C B P y cm y cm y cm (x + y) cm Area of ABCD = 70 cm2 y(x + y) = 70 xy + y2 = 70 … ➀ The perimeter of the part of the aluminium sheet that is left after the quadrant PBC is removed is 31 cm AP + AD + DC + Length of arc PC = 31 Multiply both sides by 7 to eliminate fractions. x + y + (x + y) + 90 360 × 2 × 22 7 × y = 31 2x + 2y + 11 7 y = 31 14x + 14y + 11y = 217 14x + 25y = 217 … ➁ x = 217 – 25y 14 …➂ Substitute ➂ into ➀: Multiply both sides by 14 to eliminate fractions. y1 217 – 25y 14 2 + y2 = 70 y(217 – 25y) + 14y2 = 980 217y – 25y2 + 14y2 = 980 –11y2 + 217y – 980 = 0 11y2 – 217y + 980 = 0 SMART TIP Since the values of a, b and c in the equation are large, it is very difficult for it to be factorised. Hence, it is simpler to solve it using the quadratic formula. y = 217 ± ! (–217)2 – 4(11)(980) 2(11) Use the quadratic formula: y = –b ± ! b2 – 4ac 2a = 217 ± 63 22 = 12 8 11 or 7 = 12 8 11 is not accepted because the question requires the integer value of y. ∴y = 7 Subtitute y = 7 into ➂, x = 217 – 25(7) 14 = 3 Analisis&Tip SPM Add Maths-F4-C3 2nd.indd 28 16-Feb-23 7:20:54 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 3 29 The following diagram shows two circular mini swimming pools with centres P and Q respectively. The circles touch each other on a rectangular plot of land ABCD with a length of 4π m and a width of y m. The radius of the circle with centre P is x m. P A D B C Q x m y m 4π m If the distance between P and Q is 7 m and the area of the shaded region is 7π m2 , find the value of x and of y. HOTS Applying HOTS Analysing HOTS Evaluating Solution P A D S B R C Q x m x m (7 – x) m y m 4π m Area of the shaded region = 7π m2 4πy – πx2 – π(7 – x) 2 = 7π 4y – x2 – (7 – x)2 = 7 Divide throughout by π. 4y – x2 – (49 – 14x + x2 ) = 7 4y – x2 - 49 + 14x – x2 – 7 = 0 –2x2 + 14x + 4y – 56 = 0 x2 – 7x – 2y + 28 = 0 … ➀ Divide throughout by –2 to make the coefficient of x2 be 1. SQR = AD 2(7 – x) = y y = 14 – 2x … ➁ Make y the subject of the formula. Substitute ➁ into ➀: x2 – 7x – 2(14 – 2x) + 28 = 0 x2 – 7x – 28 + 4x + 28 = 0 x2 – 3x = 0 x(x – 3) = 0 x = 0 or 3 x = 0 is not accepted. ∴x = 3 From ➁, y = 14 – 2(3) = 8 (a) 2x + y = 2 – z x + z = –2y 3y + 3z = 1 – x (b) x + 2y = z + 2 2x – z = 1 – y 2y + 3z = 14 – x 3 Solve the following simultaneous linear equations: x 3 + y 6 + z = 1 x + y + 2z = 11 3 3x + 2y = 7 3.1 Systems of Linear Equations in Three Variables 1 Solve each of the following simultaneous linear equations: (a) x + y + z = 3 2x + y – z = 3 x + 2y + 3z = 6 (b) x + y + 2z = 3 2x + y – z = 5 3x + 2y + 5z = 8 2 Solve each of the following simultaneous linear equations: HOTS Zone SPM Practice 3 Analisis&Tip SPM Add Maths-F4-C3 2nd.indd 29 16-Feb-23 7:20:55 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 3 30 4 At a shopping centre, there are three packages of promotion, i.e. package A, package B and package C, which offers shirts, pants and neckties. The number of each item and the promotion price of each package are shown in the following table. Package Number of shirts Number of pants Number of neckties Price (RM) A 3 2 2 256 B 4 1 2 218 C 2 1 3 173 If the prices of the shirts, pants and neckties are RMx, RMy and RMz respectively, form three linear equations. Hence, solve the simultaneous linear equations. HOTS Applying HOTS Analysing HOTS Evaluating 5 Azman, Gunasegaran and Chee Meng bought three subjects of academic books at a bookshop. The following table shows the number of books bought by them for each subject. Mathematics Bahasa Melayu English Azman 3 2 5 Gunasegaran 2 3 1 Chee Meng 1 5 4 The total payment made by Azman, Gunasegaran and Chee Meng are RM267, RM145 and RM230 respectively. The prices of a Mathematics, Bahasa Melayu and English books are RMx, RMy and RMz respectively. Form three linear equations and hence, solve the simultaneous linear equation. HOTS Applying HOTS Analysing HOTS Evaluating 6 Three transport companies, P, Q and R, bought RON95 petrol, RON97 petrol and diesel from a fuel company. The following table shows the volumes of the RON95 petrol, RON97 petrol and diesel bought by each company. Company Volume (in litres) RON95 petrol RON97 petrol Diesel P 500 100 200 Q 100 200 500 R 100 600 400 The bills of the companies P, Q and R are RM1 786, RM1 810 and RM2 592 respectively. The price of each litre of the RON95 petrol, RON97 petrol and diesel are RMx, RMy and RMz respectively. Form three linear equations and hence, solve the simultaneous linear equation. 7 Two groups of workers went to take a drink at a shop. The first group of workers which consists of ten persons ordered five cups of coffee, two cups of Milo and three cups of orange juice with a total payment of RM23.60. The second group of workers which consists of six persons ordered three cups of coffee, a cup of Milo and two glasses of orange juice with a total payment RM14.20. The sum of prices of a cup of coffee and three glasses of orange juice is the same as the price of four cups of Milo. If the prices of a cup of coffee, a cup of Milo and a glass of orange juice are RMx, RMy and RMz respectively, form three simultaneous linear equations to represent the given situation and determine the price of each cup or glass of drinks. HOTS Applying HOTS Analysing HOTS Evaluating 3.2 Simultaneous Equations Involving One Linear Equation and One Non-Linear Equation 8 Solve the simultaneous equations: x2 + 2y = 2x + 2y = 6 9 Solve the simultaneous equations: 6 x + 4 y = 3x + 2y = 10 10 Solve the simultaneous equations: x + y + 2 = 0 2 x – 5 y = 6 Analisis&Tip SPM Add Maths-F4-C3 2nd.indd 30 16-Feb-23 7:20:55 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 3 31 11 Solve the simultaneous equations: 3x + 2y = 10 3 x + 2 y = 5 12 Solve simultaneous equations: 2x + 3y = 5 2x + y 3 + 1 = 3 x 13 The following equations are given: P = 2x + 1 Q = 4x + y R = x2 – 2xy + 3y – x Find the values of x and the corresponding values of y if P = Q = R. 14 Solve the simultaneous equations: m – 2n = –1 mn + n – 3m = 0 State your answers correct to three decimal places. 15 Solve the simultaneous equations: 3x + y + 4 = 0 xy + 40 = y2 16 Solve the simultaneous equations: 3x + y = 2 x2 + 2y2 + xy = 4 State your answer correct to three decimal places. 17 The diagram below shows a polygon ABCDEF formed from a piece of wire with a length of 48 cm. ABC and FED are two isosceles triangles while ACDF is a rectangle. y cm y cm A B E C D F 12x cm 10x cm 10x cm 10x cm 10x cm If the area of the polygon is 144 cm2 , find the value of x and of y. HOTS Applying HOTS Analysing 18 The diagram below shows a right prism with a right-angled triangular base PQR, where PQ = 4x cm, QR = 3x cm and the height of the prism is y cm. P S U R T 4x cm Q 3x cm y cm If the sum of the lengths of all the sides of the prism is 66 cm and the sum of all its surfaces is 192 cm2 , find the possible values of x and the corresponding values of y. HOTS Applying HOTS Analysing 19 In the diagram below, KLMN is a rectangular piece of card with an area of 420 cm2 . A semicircle KPN is cut out from the card. The perimeter of the card that is left is 96 cm. 10y cm 7x cm K N P L M Using π = 22 7 , find the integer value of x and the integer value of y. HOTS Applying HOTS Analysing 20 Encik Chuan has a rectangular plot of land in Cameron Highlands. He plants cactuses and roses in the regions as shown in the diagram below. 60 m Roses Cactuses x m y m 20 m Given that the area of the region planted with roses is 1 840 m2 and the perimeter of the rectangular region planted with cactuses is 96 m, find the value of x and of y. HOTS Applying HOTS Analysing Analisis&Tip SPM Add Maths-F4-C3 2nd.indd 31 16-Feb-23 7:20:55 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 3 32 21 The following diagram shows a closed cylinder with a base of radius x cm and a height of y cm. Given that the total surface area of the whole cylinder is 658π cm2 and the circumference of its circular base exceeds its height of 4 cm. x cm y cm Using π = 22 7 , find the value of x and of y. Hence, calculate the volume of the cylinder. HOTS Applying HOTS Analysing HOTS Evaluating 22 Sulaiman plants vegetables on a rightangled triangle piece of land. The longest side of the piece of land is (2x + 3) m. Another two sides of the piece of land are x m and (x + y) m respectively. He uses barbed wire of length 30 m to fence the land. Find the length, in m, of each side of the land. HOTS Applying HOTS Analysing 23 The following diagram shows the plan of a rectangular garden PQRS. The garden consists of a semicircular pond PTS and a grassy area PQRST. P Q T S R It is given that SR = 12y and QR =14x. The area of the rectangular garden PQRS is 672 m2 and the perimeter of the grassy region is 120 m. The pond with uniform depth contains 123.2 m3 of water. Find the depth, in m, of the water in the pond such that PQ  PS. 3Use π = 22 7 4 HOTS Applying HOTS Analysing HOTS Evaluating Answers 1 (a) x = 0, y = 3, z = 0 (b) x = 2, y = 1, z = 0 2 (a) x = 1, y = –1, z = 1 (b) x = 1, y = 2, z = 3 3 x = 1, y = 2, z = 1 3 4 3x + 2y + 2z = 256 4x + y + 2z = 218 2x + y + 3z = 173 x = 30, y = 68, z = 15 5 3x + 2y + 5z = 267 2x + 3y + z = 145 x + 5y + 4z = 230 x = 32, y = 18, z = 27 6 500x + 100y + 200z = 1 786 100x + 200y + 500z = 1 810 100x + 600y + 400z = 2 592 x = 2.20, y = 2.50, z = 2.18 7 5x + 2y + 3z = 23.6 3x + y + 2z = 14.2 x + 3z = 4y ⇒ x – 4y + 3z = 0 RM2.00, RM2.60, RM2.80 8 x = 0, y = 3 or x = 2, y = 1 9 x = 2 3, y = 4 or x = 3, y = 1 2 10 x = 1 2, y = –21 2 or x = –11 3, y = – 2 3 11 x = 2 3, y = 4 or x = 3, y = 1 2 12 x = 9 10, y = 16 15 or x = –2, y = 3 13 x = 1 5, y = 3 5 or x = 2, y = –3 14 m = 3.732, n = 2.366 or m = 0.268, n = 0.634 15 x = 2 3, y = –6 or x = –3, y = 5 16 x = 1.159, y = –1.477 or x = 0.216, y = 1.352 17 x = 1, y = 4 18 x = 1 1 7, y = 126 7 or x = 2, y = 6 19 x = 2, y = 3 20 x = 40, y = 32 21 x = 7, y = 40, Volume = 6 160 cm3 22 5 cm, 12 cm, 13 cm 23 0.9 m Analisis&Tip SPM Add Maths-F4-C3 2nd.indd 32 16-Feb-23 7:20:56 PM PENERBIT ILMU BAKTI SDN. BHD.

33 4.1 Laws of Indices 1 The laws of indices are: • a m × an = a m + n • a m ÷ an = a m – n • (a m) n = a mn • (ab) n = an bn • 1 a b 2 n = an bn 2 A zero index means a° = 1, a ≠ 0. 3 A negative index means a–n = 1 an. 4 A fractional index means a 1 n = ! n a and a m n = 1 ! n a 2 m . 5 An equation that involves indices is known as an index equation. • If an = bn , then a = b such that a  0, b  0 or a  0, b , 0 • If ax = ay , then x = y such that a ≠ 0, a ≠ 1. • If a m n = b, then a = b n m and if a – m n = b, then a = b – n m. 4.2 Laws of Surds 1 For any positive values a and b, (a) !a × !b = !ab (b) !a !b = a ! b (c) !a × !a = 1!a 2 2 = a (d) !a + !a = 2!a 2 To rationalise the denominator of a surd means to eliminate the square root of the denominator of the surd. For example, a !b = a !b × !b !b = a !b b 3 The conjugate surd of (!a + !b ) is 1!a – !b 2 and vice versa. 4 A surd equation can be solved by squaring both sides of the equation twice. 4.3 Laws of Logarithms 1 If ax = y such that a  0, a ≠ 1, then x is the logarithm of y to the base a, i.e. loga y = x. ax = y ⇔ loga y = x 2 loga a = 1 3 loga 1 = 0 4 The laws of logarithms: • loga xy = loga x + loga y • loga 1 x y 2 = loga x – loga y • loga x n = nloga x 5 aloga x = x 6 The logarithm of a number in a certain base can be changed to other bases using the following formulae: • loga b = logc b logc a • loga b = 1 logb a 7 An equation that involves logarithms is known as a logarithmic equation. • If loga x = y, then x = ay . • If loga x = loga y, then x = y. Indices, Surds and Logarithms Learning Area: Algebra EXPRESS NOTES Chapter 4 Analisis&Tip SPM Add Maths-F4-C4 2nd.indd 33 16-Feb-23 7:21:20 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 4 34 4.1 Laws of Indices Example 1 Simplify 63n + 2 × 21 – 3n 32 + 3n . Solution 63n + 2 × 21 – 3n 32 + 3n = (2 × 3)3n + 2 × 21 – 3n 32 + 3n (ab) m = ambm = 23n + 2 × 33n + 2 × 21 – 3n 32 + 3n ap ÷ aq = a p – q = 2(3n + 2) + (1 – 3n) • 3(3n + 2) – (2 + 3n) ap × aq = ap + q = (23 )(30 ) = 8(1) = 8 Example 2 Given that 3 n + 2 – 3 n + 15(3n – 1) = k(3 n ), where k is a constant, find the value of k. Solution 3n+ 2 – 3n + 15(3n – 1) = 3n • 32 – 3n + 15(3n • 3–1) A 'dot' can be used to represent multiplication in Mathematics. a p + q = a p × a q = 9(3n ) – 3n + 15(3n ) 1 1 3 2 = 9(3n ) – 3n + 5(3n ) = (9 – 1 + 5)(3n ) = (13)(3n ) ∴k = 13 Example 3 Solve the equation 81 273x = 1. Solution 81 273x = 1 34 (33 )3x = 30 Express both sides in the same base. 34 39x = 30 (am)n = amn 34 – 9x = 30 1 am an 2 = am – n Equating the indices, 4 – 9x = 0 x = 4 9 4.2 Laws of Surds Example 4 Rationalise the denominator of ! 2 – ! 3 ! 2 + ! 3 . Solution ! 2 – ! 3 ! 2 + ! 3 = 1 ! 2 – ! 3 ! 2 + ! 3 21 ! 2 – ! 3 ! 2 – ! 3 2 Multiply the denominator with the conjugate of 1! 2 + ! 3 2 which is 1! 2 – ! 3 2. = 1! 2 – ! 3 2 2 1! 2 2 2 – 1! 3 2 2 (a – b)(a + b) = a2 + ab – ba – b2 = a2 – b2 = 1! 2 2 2 – 2! 2 ! 3 + 1! 3 2 2 2 – 3 (a + b) 2 = a2 + 2ab + b2 = 2 – 2! 6 + 3 –1 = –5 + 2! 6 Example 5 Solve the equation ! 2x – 1 – ! x + 3 = 1. Solution ! 2x – 1 – ! x + 3 = 1 1! 2x – 1 – ! x + 3 2 2 = 12 Square both sides of the equation. 2x – 1 – 2! 2x – 1 ! x + 3 + x + 3 = 1 2! x + 3 ! 2x – 1 = 3x + 1 Square both sides of the equation again. 12! x + 3 ! 2x – 1 2 2 = (3x + 1)2 4(x + 3)(2x – 1) = (3x + 1)2 4(2x2 + 5x – 3) = 9x2 + 6x + 1 8x2 + 20x – 12 = 9x2 + 6x + 1 Solve by factorisation x2 – 14x + 13 = 0 (x – 13)(x – 1) = 0 x = 13 or 1 x = 1 is not accepted. ∴x = 13 The square root of a negative number is undefined. Analisis&Tip SPM Add Maths-F4-C4 2nd.indd 34 16-Feb-23 7:21:21 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 4 35 4.3 Laws of Logarithms Example 6 Solve the equation 3x + 2 – 3x = 16. Solution 3x + 2 – 3x = 16 3x • 32 – 3x = 16 Express both sides in a same base. a p+q = a p × a q 9(3x ) – 3x = 16 8(3x ) = 16 3x = 16 8 Since we could not make the bases of both sides of the equation to be the same, take lg of both sides of the equation. 3x = 2 lg 3x = lg 2 x lg 3 = lg 2 x = lg 2 lg 3 x = 0.6309 SMART TIP lg x is the short form of log10 x. How to press calculator for lg: Press Display log 2 ÷ log 3 = 0.630929753 Example 7 Given that loga x = r and loga y = s, express each of the following in terms of r and s. (a) loga x3 y4 a 2 , (b) loga x a2 y ! 4 . Solution (a) loga x3 y4 a 2 = loga x3 + loga y4 + loga a 2 loga a = 1 loga pqr = loga p + loga q + loga r = 3loga x + 4loga y + 2loga a loga pn = nloga p = 3r + 4s + 2 (b) loga ! x a2 y4 = loga 1 x a2 y4 2 1 2 !u = u 1 2 = loga 1 x 1 2 ay2 2 = loga x 1 2 – (loga a + loga y2 ) loga 1 p q 2 = loga p – loga q = 1 2 loga x – loga a – 2loga y = 1 2 r – 1 – 2s Example 8 Given that log2 3 = 1.5850 and log2 5 = 2.3219, without using a calculator, calculate the value of (a) log2 270, (b) log2 3.6. Solution (a) log2 270 = log2 (2 × 33 × 5) = log2 2 + log2 33 + log2 5 = 1 + 3log2 3 + log2 5 = 1 + 3(1.5850) + 2.3219 = 8.0769 2 270 3 135 3 45 3 15 5 5 1 (b) log2 3.6 = log2 3 3 5 = log2 1 18 5 2 Change the decimal to an improper fraction first. = log2 1 2 × 32 5 2 = log2 2 + log2 32 – log2 5 = 1 + 2 log2 3 – log2 5 = 1 + 2(1.5850) – 2.3219 = 1.8481 loga 1 xy z 2 = loga x + loga y – loga z Example 9 Without using a calculator, calculate the value of 3log4 2 + 2log4 3 – 2log4 6. Solution SMART TIP ‘Without using a calculator’ means you cannot use the ‘log’ button on a calculator. Instead, you should use the laws of logarithms. Analisis&Tip SPM Add Maths-F4-C4 2nd.indd 35 16-Feb-23 7:21:21 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 4 36 3 log4 2 + 2 log4 3 – 2 log4 6 = log4 23 + log4 32 – log4 62 = log4 8 + log4 9 – log4 36 = log4 1 8 × 9 36 2 = log4 2 = log4 4 1 2 = 1 2 log4 4 = 1 2 (1) = 1 2 Example 10 Given that 3 = 2p and 5 = 2q , express log4 1.8 in terms of p and q. Solution y = a x ⇒ loga y = x Given that 3 = 2p , then log2 3 = p. Given that 5 = 2q , then log2 5 = q. log4 1.8 = log4 11 4 5 2 = log4 1 9 5 2 Change the decimal to an improper fraction first. = log2 1 9 5 2 log2 4 loga b = logc b logc a = log2 9 – log2 5 log2 22 logc b logc a = logc a – logc b = log2 32 – log2 5 log2 22 = 2log2 3 – log2 5 2log2 2 = 2p – q 2 Example 11 Given that log2 k = r and log3 k = s, express logk 12 in terms of r and s. Solution loga b = 1 logb a Given that log2 k = r, then logk 2 = 1 r . Given that log3 k = s, then logk 3 = 1 s . logk 12 = logk (22 × 3) = logk 22 + logk 3 logk ab = logk a + logk b = 2 logk 2 + logk 3 logk an = nlogk a = 21 1 r 2 + 1 s = 2s + r rs Example 12 Solve the equation log3 (2x + 3) = 2 + log3 (2x – 1). Solution log3 (2x + 3) = 2 + log3 (2x – 1) log3 (2x + 3) – log3 (2x – 1) = 2 log3 1 2x + 3 2x – 1 2 = 2 loga b = c ⇒ b = ac 2x + 3 2x – 1 = 32 2x + 3 = 9(2x – 1) 2x + 3 = 18x – 9 16x = 12 x = 3 4 Group the logarithmic terms on one side of the equation and the constant on the other side. Caution! log3 (2x + 3) = 2 + log3 (2x – 1) log3 2x + log3 3 = 2 + log3 2x – log3 1 log3 2x + 1 = 2 + log3 2x – 0 log3 2x – log3 2x = 2 – 1 0 = 1 Illogical Incorrect Analisis&Tip SPM Add Maths-F4-C4 2nd.indd 36 16-Feb-23 7:21:22 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 4 37 Example 13 Solve the equation log3 x = log9 (x + 6). Solution log3 x = log9 (x + 6) log3 x = log3 (x + 6) log3 9 Express both sides in a same base. log3 x = log3 (x + 6) 2 log3 9 = log3 32 = 2 2log3 x = log3 (x + 6) log3 x2 = log3 (x + 6) x2 = x + 6 x2 – x – 6 = 0 (x + 2)(x – 3) = 0 x = –2 or 3 x = –2 is not accepted. ∴x = 3 Log of a negative number is undefined. Example 14 Solve the simultaneous equations: 16x = 32(2y ) log2 (4 + 4y) = log2 x + 3 Solution 16x = 32(2y) 24x = 25 • 2y Express both sides in a same base. 24x = 25 + y 4x = 5 + y 4x – y = 5 … ➀ log2 (4 + 4y) = log2 x + 3 log2 (4 + 4y) – log2 x = 3 log2 1 4 + 4y x 2 = 3 loga p – loga q = loga 1 p q 2 4 + 4y x = 23 loga x = y ⇒ x = a y 4 + 4y = 8x 8x – 4y = 4 4x – 2y = 2 … ➁ ➀ – ➁ : y = 3 Substitute y = 3 into ➀: 4x – 3 = 5 ∴ x = 2 4.4 Aplication of Indices, Surds and Logarithms Example 15 The amount of money (RM) invested in a finance company after n years is given by 150 000(1 + 0.03)n . Calculate the minimum number of years required for the amount of money to exceed RM240 000. Solution 150 000(1 + 0.03)n  240 000 (1.03)n  240 000 150 000 (1.03)n  1.6 lg (1.03)n  lg 1.6 nlg1.03  lg 1.6 n(0.01284)  0.2041 n  0.2041 0.01284 n  15.90 Smallest integer value of n = 16 Hence, the minimum number of years required for the amount of money to exceed RM240 000 is 16 years. Analisis&Tip SPM Add Maths-F4-C4 2nd.indd 37 16-Feb-23 7:21:22 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 4 38 Given that loga b = logb c = logc a show that a = b = c. HOTS Applying HOTS Analysing HOTS Evaluating Solution Given loga b = logb c, lg b lg a = lg c lg b (lg b)2 = (lg a)(lg c) lg b = !(lg a)(lg c) … ➀ Caution! It is incorrect to write (lg b)2 as lg b2 . Given logb c = logc a, lg c lg b = lg a lg c (lg c)2 = (lg a)(lg b) … ➁ Substitute ➀ into ➁: (lg c)2 = (lg a)1!(lg a)(lg c)2 (lg c)2 lg a = !(lg a)(lg c) (lg c)4 (lg a)2 = (lg a)(lg c) Square both sides of the equation. (lg c)4 lg c = (lg a) 3 (lg c)3 = (lg a) 3 lg c = lg a ∴c = a Substitute c = a into ➀: lg b = !(lg a)(lg c) lg b = !(lg a)(lg a) lg b = (lg a) 2 ! lg b = lg a ∴b = a Hence, a = b = c [Shown] 4.1 Laws of Indices 1 Simplify 54n + 2 × 25n 56n – 1 . 2 If 5n + 2 – 5n + 1 – 5n = p(5n ), where p is a constant, find the value of p. 3 If 56n × 94n × 152n 32n = k 8n , where k is a positive integer, find the value of k. HOTS Applying 4 Simplify 25m × 10m + 1 2m – 1 × 52 + 3m . HOTS Applying 5 Solve the equation 16(83x ) = 1. 6 Find the values of x that satisfy the equation 5x2 – 256 – 2x = 0. 7 Solve the equation 3m – 1 + 3m – 12 = 0. 8 Solve the equation 23x = 12 – 23x – 1. 9 Solve the equation 3x + 2 + 2(3x ) = 11 27 . 10 Solve the equation 3r – 1 + 3r + 2 = 84. 11 Solve the equation 272x – 5 = 1 ! 9x + 1 12 Given that 52x = k, 5y = h and 5y + 2x = 11 + 25x , express k in terms of h. 13 Given that 3p = 7q = 21r , express r in terms of p and q. HOTS Applying HOTS Analysing SPM Practice 4 HOTS Zone Analisis&Tip SPM Add Maths-F4-C4 2nd.indd 38 16-Feb-23 7:21:23 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 4 39 4.2 Laws of Surds 14 Rationalise the denominator of (a) 2 ! 5 – 1, (b) 1 2 – ! 3 . 15 Rationalise the denominator of (a) ! 2 + 1 ! 2 – 1, (b) ! 3 – ! 2 ! 3 + ! 2 . 16 Rationalise the denominator of (a) 3 + ! 2 3 – ! 2 + 3 – ! 2 3 + ! 2 , (b) ! 5 + 1 ! 5 + 2 + ! 5 – 1 ! 5 – 2. 17 Solve the equation of ! x – 2 = ! x – 1. 18 Solve the equation of ! 4x – 9 = 2! x – 1. 19 Solve the equation of ! 3 – 3x – ! 2 – x = 1. 4.3 Laws of Logarithms 20 If loga x = u, find log 1 a x in terms of u. 21 If loga 2 = h and loga 3 = k, express log6 12 in terms of h and k. HOTS Applying 22 Given that a = 3p and b = 3q , find log9 a + log27 b in terms of p and q. HOTS Applying 23 Given that log!x 9 = u, find log9 x3 in terms of u. HOTS Applying 24 Given that logy 2 = k, find log4 y in terms of k. HOTS Applying 25 Solve the equation 4t = 23.4. 26 Solve the equation 2x + 3 – 7 = 0. 27 Solve the equation 3x • 5x = 8x + 1. 28 Given that log27 n = 2 3 , find the value of n. 29 Solve the equation lg 5 + lg(2s – 1) = lg 3 + lg(s + 2). 30 Solve the equation log t (t – 2) + logt (t + 5) = 2. 31 Solve the equation 1 2 logm 9 + 1 4 logm 81 = 2. 32 Solve the equation log2 (5t – 3) + 2 = log2 4t. 33 Solve the equation 3 logx 5 + 2 logx 4 – logx 250 = 3. 34 Solve the equation log4 (1 – x) = 1 2 + log4 3x. 35 If 2 – log10 y = 3 log10 x, express y in terms of x. 36 Solve the equation 9log(2x – 1)7 = 7. 37 Solve the equation 2log3 x = 32. 38 Solve the equation log49 [log2 (5x – 2)] = 1 2 . 39 Find the values of k that satisfy the equation log10 (k2 + 6k + 28) = 2. 40 Solve the equation log3 r = log9 36. 41 Solve the equation log2 x – log4 x = –2. 42 Given that log2 3 = 1.585 and log2 5 = 2.322, without using a calculator, calculate the value of (a) log2 90, (b) log4 113 8 9 2. 43 (a) Given that log5 3 = 0.683 and log5 7 = 1.209, without using a calculator, calculate the value of (i) log5 4.2, (ii) log7 315. (b) Solve the equation 9log3 x = 7. Analisis&Tip SPM Add Maths-F4-C4 2nd.indd 39 16-Feb-23 7:21:23 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 4 40 44 Solve each of the following equations: (a) ! 4x + 1 = 1 8x 2x + 3 (b) log2 x + log4 (4x) = –2 45 Given that loga xy3 = 9 and loga x2 y = 8, find the value of loga ! xy . HOTS Applying HOTS Analysing 46 (a) Solve the equation log!x 54 – log!x 2 = 3. (b) If 2a = 5b = 20c , express c in terms of a and b. HOTS Applying HOTS Analysing 47 Show that log2 xy = 2 log4 x + 2 log4 y. Hence, find the value of x and of y that satisfy the following simultaneous equations: log2 xy = 10 HOTS Applying log4 x log4 y = 3 2 HOTS Analysing HOTS Evaluating 48 (a) Given that 2 log2 (x + y) = 3 + log2 x + log2 y, show that x2 + y2 = 6xy. (b) Without using a calculator, solve the equation log9 [log3 (3x – 6)] = 5log51 1 2 2 HOTS Applying HOTS Analysing 49 Given that log!x 9 = a and logy 3 = b, find log9 xy2 in terms of a and b. HOTS Applying HOTS Analysing HOTS Evaluating 50 Solve the following simultaneous equations: 8x + 1 = 2y log3 y = 2 + log3 (x – 1) HOTS Applying 51 (a) If 23x = a 4 • 32x , show that x = 4 loga 1 8 9 2 . (b) Given that 2 log5 xy2 = 3 + log5 y – log5 x, show that xy = 5. HOTS Applying HOTS Analysing 52 (a) Given that log3 t = m and log5 t = n, express logt 45 in terms of m and n. (b) Solve the equation log9 (3 – x) = log3 (x – 1). HOTS Applying HOTS Analysing 53 (a) Given that 2 log9 y – 3 log27 x = 2, express y in terms of x. (b) Given that logx a = 2.32 and logx b = 2.81, find the value of logab x. HOTS Applying HOTS Analysing 54 (a) Solve the following simultaneous equations: 3x – 1 × 243y + 2 = 81 83 – y 23x = 1 (b) Solve the equation 3 log8 (2x + 14) – 4 log16 (x + 1) = 3. HOTS Applying HOTS Analysing 55 Solve the equation log3 (3x + 1) – log3 x2 + log9 x2 = 2 HOTS Applying HOTS Analysing 56 (a) Solve the equation 7x2 – 496 – 2x = 0. (b) Express 4 + ! 2 2 – ! 2 in the form a + k! b , where a, b and k are constants. (c) Solve the equation log9 x + logx 9 = 5 2 . HOTS Applying HOTS Analysing HOTS Evaluating 4.4 Application of Indices, Surds and Logarithms 57 The value of a house at a strategic location increases by 4% from its original value early each year. If the original value of the house is RM450 000, its value, RMp, after h years is given by p = 450 000(1.04)h . Find the minimum number of years for the value of the house to exceed RM600 000. HOTS Applying HOTS Analysing HOTS Evaluating Analisis&Tip SPM Add Maths-F4-C4 2nd.indd 40 16-Feb-23 7:21:23 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 4 41 1 125 2 p = 19 3 k = 15 4 4 5 5 x = – 4 9 6 x = 2 or –6 7 m = 2 8 x = 1 9 x = –3 10 r = 2 11 x = 2 12 k = 11 h – 11 13 r = pq p + q 14 (a) ! 5 + 1 2 (b) 2 + ! 3 15 (a) 3 + 2! 2 (b) 5 – 2! 6 16 (a) 22 7 (b) 6 17 x = 9 4 18 x = 25 4 19 x = –2 20 –u 21 2h + k h + k 22 3p + 2q 6 23 6 u 24 1 2k 25 t = 2.274 26 x = –0.1926 27 x = 3.308 28 n = 9 29 s = 1 4 7 30 t = 10 3 31 m = 3 32 t = 3 4 33 x = 2 34 x = 1 7 35 y = 100 x3 36 x = 5 37 x = 243 38 x = 26 39 k = –12 or 6 40 r = 6 41 x = 1 16 42 (a) 6.492 (b) 1.898 43 (a) (i) 0.892 (ii) 2.957 (b) 2.646 44 (a) x = – 4 5 (b) x = 1 4 45 2 1 2 46 (a) x = 9 (b) c = ab 2b + a 47 x = 64, y = 16 48 (b) x = 11 49 2b + a ab 50 x = 2, y = 9 52 (a) 2n + m mn (b) x = 2 53 (a) y = 9x (b) 0.1949 54 (a) x = 5, y = –2 (b) x = 1 55 x = 1 6 56 (a) x = 2 or –6 (b) 5 + 3! 2 (c) x = 3 or 81 57 8 years Answers Analisis&Tip SPM Add Maths-F4-C4 2nd.indd 41 16-Feb-23 7:21:24 PM PENERBIT ILMU BAKTI SDN. BHD.

42 5.1 Arithmetic Progressions 1 An arithmetic progression is a number sequence such that the difference between each term (after the first term) and its preceding term is a constant. The constant is known as a common difference. 2 The common difference, d, is given by d = Tn + 1 – Tn where Tn + 1 = (n + 1)th term Tn = nth term The common difference cannot take the value 0. 3 The steps to determine whether a number sequence is an arithmetic progression or not are as follows: • Choose any three consecutive terms in a number sequence, e.g. Tn, Tn + 1 and Tn + 2. • Calculate the values of Tn + 1 – Tn and Tn + 2 – Tn + 1. • If Tn + 1 – Tn = Tn + 2 – Tn + 1, then the number sequence is an arithmetic progression. • If Tn + 1 – Tn ≠ Tn + 2 – Tn + 1, then the number sequence is not an arithmetic progression. 4 The nth term, Tn , of an arithmetic progression is given by Tn = a + (n – 1)d where a = first term d = common difference 5 An arithmetic progression can be written as a, a + d, a + 2d, a + 3d, … 6 If a, b and c are three consecutive terms in an arithmetic progression, then b – a = c – b 2b = a + c b = a + c 2 b is the arithmetic mean of a and c. 7 The sum of the first n terms, Sn , of an arithmetic progression is given by Sn = n 2 32a + (n – 1)d4 where a = first term d = common difference or Sn = n 2 (a + l) where a = first term l = last term 8 The nth term of an arithmetic progression can be determined using the formula Tn = Sn – Sn – 1 5.2 Geometric Progressions 1 A geometric progression is a number sequence such that each term (after the first term) is obtained by multiplying the preceding term with a constant. The constant is known as a common ratio. Progressions Learning Area: Algebra EXPRESS NOTES Chapter 5 Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 42 16-Feb-23 7:22:18 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 5 43 5.1 Arithmetic Progressions Example 1 The diagram below shows a series of a few isosceles triangles with a constant height of 8 cm. The base of the first triangle is b cm and the base of each of the following triangles increases by 2 cm successively. 8 cm b cm (b + 2) cm (b + 4) cm Show that the areas (in cm2 ) of the series of triangles form an arithmetic progression. State its common difference. Solution T1 = Area of the first triangle = 1 2 × b × 8 = 4b T2 = Area of the second triangle = 1 2 × (b + 2) × 8 = 4(b + 2) = 4b + 8 2 The common ratio, r, is given by r = Tn + 1 Tn where Tn + 1 = (n + 1)th term Tn = nth term The common ratio cannot take the values 0 or 1. 3 The steps to determine whether a number sequence is a geometric progression or not are as follows: • Choose any three consecutive terms in a number sequence, e.g. Tn , Tn + 1 and Tn + 2. • Calculate the values of Tn + 1 Tn and Tn + 2 Tn + 1 . • If Tn + 1 Tn = Tn + 2 Tn + 1 , then the number sequence is a geometric progression. • If Tn + 1 Tn ≠ Tn + 2 Tn + 1 , then the number sequence is not a geometric progression. 4 The nth term, Tn , of a geometric progression is given by Tn = ar n – 1 where a = first term r = common ratio 5 A geometric progression can be written as a, ar, ar 2 , ar 3 , ... 6 If a, b and c are three consecutive terms in a geometric progression, then b a = c b b2 = ac b = !ac b is the geometric mean of a and c. 7 The sum of the first n terms, Sn, of a geometric progression is given by Sn = a(r n – 1) r – 1 , r  1 or Sn = a(1 – r n ) 1 – r , r  1 where a = first term r = common ratio 8 In the case –1 < r < 1, the sum of the first n terms of a geometric progression by Sn = a(1 – r n ) 1 – r = a 1 – r – ar n 1 – r When n approaches infinity, i.e. as n → ∞, r ∞ → 0 and ar n 1 – r → 0. Hence, S∞ = a 1 – r , –1  r  1 S∞ is read as sum to infinity. Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 43 16-Feb-23 7:22:18 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 5 44 T3 = Area of the third triangle = 1 2 × (b + 4) × 8 = 4(b + 4) = 4b + 16 T2 – T1 = (4b + 8) – 4b = 8 T3 – T2 = (4b + 16) – (4b + 8) = 8 Since T2 – T1 = T3 – T2 = 8 (a constant), the areas (in cm2 ) of the series of triangles form an arithmetic progression. Common difference = 8 Example 2 Calculate the 11th term in the arithmetic progression 5, 8, 11, … Solution Tn = a + (n – 1)d T11 = 5 + (11 – 1)(3) d = T2 – T1 = 8 – 5 = 35 = 3 Example 3 Calculate the number of terms in the arithmetic progression 7, 11, 15, …, 163. Solution SMART TIP To calculate the number of terms is to find the value of n. Tn = 163 a + (n – 1)d = 163 7 + (n – 1)(4) = 163 d = T2 – T1 = 11 – 7 7 + 4n – 4 = 163 = 4 4n = 160 n = 40 Hence, the number of terms = 40 Example 4 Given the arithmetic progression –7, –2, 3, …, calculate the sum of the first 15 terms. Solution Sn = n 2 [2a + (n – 1)d] d = T2 – T1 = –2 – (–7) S = 5 15 = 15 2 [2(–7) + (15 – 1)5] = 15 2 (–14 + 70) = 420 Example 5 Calculate the sum of all the multiples of 13 between 270 and 420. Solution The multiples of 13 between 270 and 420 form an arithmetic progression: 273, 286, 299, …, 416 First of all, find the number of terms. Tn = 416 a + (n – 1)d = 416 273 + (n – 1)(13) = 416 273 + 13n – 13 = 416 13n = 416 – 273 + 13 13n = 156 n = 12 Next, calculate the sum of all the 12 terms. Sn = n 2 (a + l) Since the last term is known, it is faster to use the formula Sn = 1 2 (a + l) compared to the formula Sn = 1 2 [2a + (n – 1)d]. = 12 2 (273 + 416) = 4 134 Hence, the sum of all the multiples of 13 between 270 and 420 is 4 134. Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 44 16-Feb-23 7:22:18 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 5 45 Example 6 If the sum of the first n terms of an arithmetic progression 4, 7, 10, … is 175, find the value of n. Solution Sn = 175 n 2 32a + (n – 1)d4 = 175 n 2 32(4) + (n – 1)(3)4 = 175 n 2 (8 + 3n – 3) = 175 n 2 (5 + 3n) = 175 n(5 + 3n) = 350 5n + 3n2 = 350 3n2 + 5n – 350 = 0 n = –b ± ! b2 – 4ac 2(3) n = –5 ± ! 52 – 4(3)(–350) 2(3) n = –5 ± 65 6 n = 10 or –11 2 3 n = –11 2 3 is not accepted. ∴n = 10 n must be a positive integer. Example 7 Given that the 9th term of an arithmetic progression is 22 and the sum of the first 6 terms is 33, find (a) the first term and the common difference, (b) the sum of all the terms from the 5th term to the 10th term. Solution (a) T9 = 22 a + 8d = 22 … ➀ S6 = 33 6 2 (2a + 5d) = 33 3(2a + 5d) = 33 6a + 15d = 33 2a + 5d = 11 … ➁ ➀ × 2 : 2a + 16d = 44 ... ➂ ➂ – ➁ : 11d = 33 ⇒ d = 3 Subtitute d = 3 into ➀: a + 8(3) = 22 ⇒ a = –2 (b) The sum of all the terms from the 5th term to the 10th term S10 – S4 = 10 2 32(–2) + 9(3)4 – 4 2 32(–2) + 3(3)4 = 115 – 10 = 105 SMART TIP The sum of all the terms from the 5th term to the 10th term = S10 – S4 . S10 S4 S10 – S4 T1 + T2 + T3 + T4 + T5 + T6 + T7 + T8 + T9 + T10                                      Caution! The sum of all the terms from the 5th term to the 10th term = S10 – S5 Incorrect Example 8 The sum of the first n terms of an arithmetic progression is given by Sn = 3n(n + 4). Find (a) the first term, (b) the common difference, (c) the 7th term. Solution (a) T1 = S1 = 3(1)(1 + 4) It is obvious that the sum of one term is the first term itself. Hence, T1 = S1 . = 15 (b) Find T2 first. T2 = S2 – S1 Tn = Sn – Sn – 1 = 3(2)(2 + 4) – 3(1)(1 + 4) = 36 – 15 = 21 Hence, find the common difference. d = T2 – T1 = 21 – 15 = 6 (c) T7 = S7 – S6 Tn = Sn – Sn – 1 = 3(7)(7 + 4) – 3(6)(6 + 4) = 231 – 180 = 51 Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 45 16-Feb-23 7:22:19 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 5 46 Example 9 Encik Radzi rents a house with a monthly rent of RM420. In the agreement, it is stated that the monthly rent will be raised by the same amount of money each year. In the 8th year, the monthly rent paid by Encik Radzi is RM560. Find (a) the total rent that is to be increased each year, (b) the total rent that Encik Radzi has to pay from the 6th year to the 12th year. Solution (a) Number of years 1 … 8 Monthly rent RM420 … RM560 Annual rent RM5 040 … RM6 720 T8 = 6 720 a + (n – 1)d = 6 720 5 040 + (8 – 1)d = 6 720 5 040 + 7d = 6 720 7d = 1 680 d = 240 Hence, the total rent that is to be increased each year is RM240. (b) The total rent that Encik Radzi has to pay from the 6th year to the 12th year a = 5 040, d = 240 S12 – S5 = 12 2 32(5 040) + 11(240)4 – 5 2 32(5 040) + 4(240)4 = 76 320 – 27 600 = RM48 720 Example 10 A piece of wire is bent to form semicircles as shown in the diagram below. The radius of the smallest semicircle is 3 cm and the radii of the following semicircles increase by 2 cm successively. (a) If the radius of the largest semicircle is 41 cm, find the number of semicircles that can be formed. (b) Can the number of semicircles in (a) be formed using a piece of wire with a length of 360π cm? Solution Radius (cm) 3 5 … 41 Perimeter of semicircles (cm) 3π 5π … 41π (a) Tn = 41 a + (n – 1)d = 41 3 + (n – 1)(2) = 41 3 + 2n – 2 = 41 2n = 40 n = 20 Hence, 20 semicircles that can be formed. (b) The length of the wire is 360π cm. Sn = 360π n 2 [2a + (n – 1)d] = 360π a = 3π d = 5π – 3π = 2π n 2 [2(3π) + (n – 1)(2π)] = 360π n 2 [6 + 2(n – 1)] = 360 The equation is divided throughout by π. 3n + n(n – 1) = 360 3n + n2 – n – 360 = 0 n2 + 2n – 360 = 0 n = –b ± ! b2 – 4ac 2a n = –2 ± ! 22 – 4(1)(–360) 2(1) n = –2 ± 38 2 n = 18 or –20 n = –20 is not accepted. ∴n = 18 Only 18 semicircles that can be formed. Hence, the number of semicircles in (a), which consists of 20 semicircles cannot be formed. Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 46 16-Feb-23 7:22:19 PM PENERBIT ILMU BAKTI SDN. BHD.

Form 4 CHAPTER 5 47 5.2 Geometric Progressions Example 11 The following diagram shows a square ABCD of side a cm. The second square, APQR is drawn such that P and R are the midpoints of AB and AD respectively. The third square, AKLM is drawn such that K and M are the midpoints of AP and AR respectively. The other squares are drawn using the same procedure. a cm a cm B A M R K P L Q C D Show that the areas of the squares ABCD, APQR and AKLM (in cm2 ) form a geometric progression. State its common ratio. Solution T1 = Area of ABCD = a2 T2 = Area of APQR = 1 1 2 a2 × 1 1 2 a2 = 1 4 a2 T3 = Area of AKLM = 1 1 4 a2 × 1 1 4 a2 = 1 16 a2 T2 T1 = 1 4 a2 a2 = 1 4 T3 T2 = 1 16 a2 1 4 a2 = 1 4 Since T2 T1 = T3 T2 = 1 4 (a constant), the areas of the squares ABCD, APQR and AKLM (in cm2 ) form a geometric progression. Common ratio = 1 4 Example 12 Calculate the 7th term of the geometric progression 1 3 , 1 6 , 1 12 , … Solution Tn = ar n – 1 T7 = 1 1 3 21 1 2 2 7 – 1 r = T2 T1 = 1 2 = 1 1 3 21 1 2 2 6 = 1 192 Example 13 Calculate the sum of the first 6 terms of the geometric progression 729, 243, 81, … Solution r = T2 T1 = 243 729 = 1 3 Sn = a(1 – r n ) 1 – r Since r  1, the formula Sn = a(1 – rn ) 1 – r is used. S6 = 729 31 – 1 1 3 2 6 4 1 – 1 3 = 1 092 Example 14 Calculate the sum of the geometric progression 2 3 , 1, 1 1 2 , …, 5 1 16. Analisis&Tip SPM Add Maths-F4-C5 3rd.indd 47 16-Feb-23 7:22:20 PM PENERBIT ILMU BAKTI SDN. BHD.