PELANGI ONLINE TEST PELANGI ONLINE TEST 4 5 KSSM FORM Bonus SPM QUICK REVISION Chew Su Lian Ong Yunn Tyug Tee Hock Tian Moh Sin Yee Dr. Pauline Wong Mee Kiong (Textbook Writer) ADDITIONAL MATHEMATICS SPM Practice & SPM Model Paper FULL SOLUTIONS PELANGI
Formulae vi–viii Form 4 1 CHAPTER Functions 1 Concept Map 1 1.1 Functions 2 1.2 Composite Functions 5 1.3 Inverse Functions 7 SPM Mastery 8 HOTS Example 11 SPM Practice 1 12 2 CHAPTER Quadratic Functions 14 Concept Map 14 2.1 Quadratic Equations and Inequalities 15 2.2 Types of Roots of Quadratic Equations 18 2.3 Quadratic Functions 19 SPM Mastery 27 HOTS Example 28 SPM Practice 2 29 3 CHAPTER Systems of Equations 33 i-THINK Gallery 33 3.1 Systems of Linear Equations in Three Variables 34 3.2 Simultaneous Equations involving One Linear Equation and One Non-Linear Equation 37 SPM Mastery 39 HOTS Example 41 SPM Practice 3 43 4 CHAPTER Indices, Surds and Logarithms 45 Concept Map 45 4.1 Laws of Indices 46 4.2 Laws of Surds 47 4.3 Laws of Logarithms 51 4.4 Applications of Indices, Surds and Logarithms 55 SPM Mastery 55 HOTS Example 57 SPM Practice 4 59 5 CHAPTER Progressions 61 Concept Map 61 5.1 Arithmetic Progressions 62 5.2 Geometric Progressions 66 SPM Mastery 70 HOTS Example 71 SPM Practice 5 73 6 CHAPTER Linear Law 77 Concept Map 77 6.1 Linear and Non-Linear Relations 78 6.2 Linear Law and Non-Linear Relations 81 6.3 Applications of Linear Law 82 SPM Mastery 84 HOTS Example 86 SPM Practice 6 87 iii
7 CHAPTER Coordinate Geometry 92 i-THINK Gallery 92 7.1 Divisor of a Line Segment 93 7.2 Parallel Lines and Perpendicular Lines 94 7.3 Areas of Polygons 96 7.4 Equations of Loci 97 SPM Mastery 98 HOTS Example 101 SPM Practice 7 102 8 CHAPTER Vectors 105 i-THINK Gallery 105 8.1 Vectors 106 8.2 Addition and Subtraction of Vectors 108 8.3 Vectors in a Cartesian Plane 111 SPM Mastery 112 HOTS Example 118 SPM Practice 8 119 9 CHAPTER Solution of Triangles 124 Concept Map 124 9.1 Sine Rule 125 9.2 Cosine Rule 128 9.3 Area of a Triangle 130 9.4 Application of Sine Rule, Cosine Rule and Area of a Triangle 132 SPM Mastery 133 HOTS Example 135 SPM Practice 9 137 10 CHAPTER Index Numbers 142 Concept Map 142 10.1 Index Numbers 143 10.2 Composite Index 144 SPM Mastery 147 HOTS Example 150 SPM Practice 10 152 Form 5 1 CHAPTER Circular Measure 162 Concept Map 162 1.1 Radian 163 1.2 Arc Length of a Circle 163 1.3 Area of Sector of a Circle 165 1.4 Application of Circular Measures 166 SPM Mastery 168 HOTS Example 171 SPM Practice 1 173 2 CHAPTER Differentiation 179 Concept Map 179 2.1 Limit and its Relation to Differentiation 180 2.2 The First Derivative 181 2.3 The Second Derivative 184 2.4 Application of Differentiation 185 SPM Mastery 188 HOTS Example 191 SPM Practice 2 192 3 CHAPTER Integration 196 Concept Map 196 3.1 Integration as the Inverse of Differentiation 197 3.2 Indefinite Integral 198 3.3 Definite Integral 200 3.4 Application of Integration 207 SPM Mastery 208 iv
HOTS Example 211 SPM Practice 3 212 4 CHAPTER Permutation and Combination 216 Concept Map 216 4.1 Permutation 217 4.2 Combination 220 SPM Mastery 223 HOTS Example 224 SPM Practice 4 225 5 CHAPTER Probability Distribution 228 Concept Map 228 5.1 Random Variable 229 5.2 Binomial Distribution 231 5.3 Normal Distribution 234 SPM Mastery 238 HOTS Example 241 SPM Practice 5 242 6 CHAPTER Trigonometric Functions 247 Concept Map 247 6.1 Positive Angles and Negative Angles 248 6.2 Trigonometric Ratios of any Angle 249 6.3 Graphs of Sine, Cosine and Tangent Functions 251 6.4 Basic Identities 254 6.5 Addition Formulae and Double Angle Formulae 255 6.6 Application of Trigonometric Functions 256 SPM Mastery 258 HOTS Example 261 SPM Practice 6 262 7 CHAPTER Linear Programming 265 Concept Map 265 7.1 Linear Programming Model 266 7.2 Application of Linear Programming 269 SPM Mastery 270 HOTS Example 272 SPM Practice 7 274 8 CHAPTER Kinematics of Linear Motion 278 Concept Map 278 8.1 Displacement, Velocity and Acceleration as a Function of Time 279 8.2 Differentiation in Kinematics of Linear Motion 279 8.3 Integration in Kinematics of Linear Motion 282 8.4 Application of Kinematics of Linear Motion 282 SPM Mastery 284 HOTS Example 287 SPM Practice 8 287 SPM Model Paper 290 Answers 306 Full Solutions for SPM Practice and SPM Model Paper v
2 CHAPTER Quadratic Functions x = –b ± √b2 – 4ac 2a 4 CHAPTER Indices, Surds and Logarithms am × an = am + n am ÷ an = am – n (am) n = amn √a × √b = √ab √a ÷ √b = a b loga mn = loga m + loga n loga m n = loga m – loga n loga mn = n loga m loga b = logc b logc a 5 CHAPTER Progressions Arithmetic progressions, Tn = a + (n – 1)d Sn = n 2 [2a + (n – 1)d] Sn = n 2 [a + l] Geometric progressions, Tn = arn – 1 Sn = a(r n – 1) r – 1 for |r| . 1 or Sn = a(1 – r n ) 1 – r for |r| , 1 S∞ = a 1 – r for |r| , 1 7 CHAPTER Coordinate Geometry Divisor of a line segment, (x, y) = 1 nx1 + mx2 m + n , ny1 + my2 m + n 2 Area of a triangle = 1 2 u(x1 y2 + x2 y3 + x3 y1 ) – (x2 y1 + x3 y2 + x1 y3 )u Area of a quadrilateral = 1 2 u(x1 y2 + x2 y3 + x3 y4 + x4 y1 ) – (x2 y1 + x3 y2 + x4 y3 + x1 y4 )u Area of a polygon = 1 2 x1 x2 … xn x1 y1 y2 … yn y1 Form 4 vi
Additional Mathematics SPM Formulae 8 CHAPTER Vectors u r ~u = √x2 + y2 ^ r ~ = r ~ u r ~u 9 CHAPTER Solution of Triangles a sin A = b sin B = c sin C a2 = b2 + c2 – 2bc cos A b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos C Area of a triangle = 1 2 ab sin C = 1 2 bc sin A = 1 2 ac sin B Heron’s formula = √s(s – a)(s – b)(s – c), s = a + b + c 2 10 CHAPTER Index Numbers I = Q1 Q0 × 100 – I = ∑I i wi ∑wi Form 5 1 CHAPTER Circular Measure Arc length, s = rq Area of sector, L = 1 2 r 2 q Area of triangle = 1 2 r 2 sin q 2 CHAPTER Differentiation y = uv, dy dx = u dv dx + v du dx y = u v , dy dx = v du dx – u dv dx v2 dy dx = dy du × du dx 3 CHAPTER Integration Area under a curve = b a y dx or = b a x dy Generated volume = b a πy2 dx or = b a πx2 dy vii
Additional Mathematics SPM Formulae 4 CHAPTER Permutation and Combination n Pr = n! (n – r)! n Cr = n! (n – r)!r! Formula involving identical objects = n! p!q!r!… 5 CHAPTER Probability Distribution P(X = r) = n Cr pr qn – r, p + q = 1 Mean, µ = np s = √npq Z = X – µ s 6 CHAPTER Trigonometric Functions sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A sin 2A = 2 sin A cos A cos 2A = cos2 A – sin2 A = 2 cos2 A – 1 = 1 – 2 sin2 A tan 2A = 2 tan A 1 – tan2 A sin (A ± B) = sin A cos B ± cos A sin B cos (A ± B) = cos A cos B sin A sin B tan (A ± B) = tan A ± tan B 1 tan A tan B viii
Additional Mathematics SPM Chapter 1 Functions 4 Form Example: f(x) = 2x + 1 x 1• •3 •5 •6 •7 •9 2• 3• 4• X Y 2x + 1 f The function from set X to set Y is a special relation that maps every element x X to only one element y Y. Composite function X x y = g(x) z = f fig(x)ff f g Y Z g f = f(y) • The result of the composition of two functions above is written as fg(x) • Normally, fg(x) gf(x) Vertical line test Horizontal line test • Domain • Codomain • Range • Object • Image Inverse function x f f –1 y Inverse function, f–1 exists if f is a one-to-one function. Functions Concept Map Functions 1 CHAPTER Learning Area Algebra Form 4 1
Additional Mathematics SPM Chapter 1 Functions 4 Form 1. The function from set X to set Y is a special relation that maps every element x X to only one element y Y. For example, Graphically: 8 6 4 2 0 (1, 3) (2, 5) (3, 7) 2 4 6 y x y = 2x +1 Arrow diagram: X Y 1• •3 •5 •7 2• 3• The value of 1 is mapped to the value of 3, the value of 2 is mapped to the value of 5 and so on. 2. The function in note 1 can be written as f : x → 2x + 1 or f(x) = 2x + 1 such that x is the object and 2x + 1 is the image. A relation is a function if every object has only one image. SPM TipS 3. Vertical line test (a) If the vertical line cuts the graph at one point only, then the relation is a function. y x 0 y = x + 5 Function (b) If the vertical line cuts the graph more than one point, then the relation is not a function. y x y2 = x – 1 Not a 0 function 4. The existence of a function Example: Given the function f(x) = (2x – 3). Notice that 2x – 3 0, so that the square root of it can be obtained. Hence, f(x) exists if and only if x 3 2 . 5. Undefined Example: Given the function f(x) = 1 x – 2 . Notice that the graph of the function only approaching but will never touch the line x = 2. Hence, the function is not defined at x = 2. 6. Absolute function For absolute function, we accept the magnitude only, by ignoring the related signs. For example, x if x 0 f(x) =│x│= –x if x 0 1.1 Functions 2
Additional Mathematics SPM Chapter 1 Functions 4 Form Example: f(–4) = │–4│= 4 f(3) = │3│= 3 7. Domain, codomain and range of a function (a) Domain is the values of x in set X, and a function is defined. (b) Codomain is the possible values of y in set Y of the function. (c) Range is the values of y in set Y, which are obtained by substituting all the possible values of x. 8. If the object of a function is given, then the corresponding image can be obtained using substitution method. Example 1 The following diagram shows two graphs. Which graph is a function? Give your reason. y 0 x y = –x2 + 4x – 2 y 0 x y2 = –x2 + 1 Graph (a) Graph (b) Solution Graph (a) is a function. When it is tested with vertical line test, it cuts only one point on graph (a) whereas it cuts two points on graph (b). This means for every object of graph (a) has only one image but for graph (b), there are objects which have two images. Quiz Based on the arrow diagram of f(x) = 2x + 1 on page 1, state the (a) domain; (b) codomain; (c) range; (d) object of 3; (e) image of 3 Example 2 Determine the domain, codomain and range of each of the following function f. (a) f(x) x 2 0 2 4 –2 –2 (b) x y y = f(x) 16 –2 0 2 8 Solution (a) Domain = {–3, –2, 0, 1, 3} Codomain = {–2, –1, 0, 1, 2} Range = {–2, –1, 0, 1, 2} (b) Domain of f is –2 x 2 Codomain of f is 0 f(x) 16 Range of f is 0 f(x) 16 Function of Example 2(a) is a discrete function. Function of Example 2(b) is a continuous function. SPM TipS 3
Additional Mathematics SPM Chapter 1 Functions 4 Form Example 3 The diagram below shows function f(x) = 1. x y 1 2 3 0 1 2 3 f(x) = 1 Give a reason why f is a function. Hence, determine (a) the domain, (b) the range. Solution f is a function because it passes the vertical line test where it cuts only one point on the graph. Every object has only one image. (a) Domain of f is 0 x 3. (b) Range of f is f(x) = 1. 3 is not the image of f. Thus, the range of f does not include 3. However, 3 is the element in the codomain. SPM TipS Example 4 Sketch the graph of function f(x) = │x + 1│ for the domain –3 x 1. Hence, state the corresponding range of the given domain. Solution f(x) f(x) = fix + 1fi –3 –1 0 1 1 2 x The corresponding range is 0 f(x) 2. Example 5 Given f(x) = 3x – 1, determine the value of (a) the image of object 2, (b) the object when the image is –13. Solution (a) f(2) = 3(2) – 1 = 5 (b) f(x) = –13 3x – 1 = –13 3x = –13 + 1 3x = –12 x = –12 3 x = – 4 Object means the value of x given. Image means f(x) or the algebraic expression. SPM TipS Example 6 Function f is defined by f : x → |2 − 3x |. Find (a) f(2), (b) the domain for f(x) 2. Solution (a) f(2) = | 2 − 3(2)| = | –4 | = 4 (b) f(x) 2 |2 − 3x | 2 −2 2 − 3x 2 −4 −3x 0 0 x 4 3 4
Additional Mathematics SPM Chapter 1 Functions 4 Form 4. For example, given f(x) = x2 + 3 and g(x) = 5x fg(x) gf(x) machine g machine f x 5x 25x2 + 3 5 times input Square the input then add 3 Thus, fg(x) = 25x2 + 3 machine f machine g x x2 + 3 5x2 + 15 Square the input then add 3 5 times input Thus, gf(x) = 5x2 + 15 5. If an object is given, then the image of the composite function can be obtained by substituting a function by another function. 6. If the composite function and one of the functions are given, then the other function can be determined. 1. Composite function is obtained when two functions are carried out one after another with the same function or different one. 2. Given f(x) and g(x), function fg is known as the composite function such that x will be replaced by g(x), that is f[g(x)]. This means function g is carried out first and then followed by function f on the image obtained from function g. fg(x) means x is the object of function g and its image, g(x) is the object of function f. SPM TipS 3. Commonly, fg(x) ≠ gf(x). fg(x) represents the composite function where function g is carried out on x and then function f is carried out on the image obtained in g(x). gf(x) is the reverse of fg(x). 1.2 Composite Functions 5
Additional Mathematics SPM Chapter 1 Functions 4 Form Example 7 Given f(x) = 2x – 1 and g(x) = x2 . Find (a) fg(x) (b) gf(x) Solution (a) fg(x) = f [g(x)] = f(x2 ) = 2x2 – 1 Thus, fg(x) = 2x2 – 1 (b) gf(x) = g[f(x)] = g(2x – 1) = (2x – 1)2 Thus, gf(x) = (2x – 1)2 Example 8 Given f(x) = 5x – 6 and g(x) = x2 – x. Determine (a) f 2 (1) (b) fg(–2) (c) gf(0) (d) the values of x when fg(x) = –6 (e) the values of x when gf(x) = 0 Solution (a) f 2 (1) = ff(1) = f [5(1) – 6] = f(–1) = 5(–1) – 6 = –11 (b) fg(–2) = f[(–2)2 – (–2)] = f(6) = 5(6) – 6 = 24 (c) gf(0) = g[5(0) – 6] = g(–6) = (–6)2 – (–6) = 42 (d) fg(x) = f(x2 – x) = 5(x2 – x) – 6 fg(x) = –6 Then, 5(x2 – x) – 6 = –6 5(x2 – x) = 0 (x2 – x) = 0 x(x – 1) = 0 x = 0 or x – 1 = 0 x = 1 Thus, the values of x are 0 and 1. (e) gf(x) = 0 g(5x – 6) = 0 (5x – 6)2 – (5x – 6) = 0 (5x – 6)(5x – 6 – 1) = 0 5x – 6 = 0 or 5x – 7 = 0 x = 6 5 x = 7 5 Thus, the values of x are 6 5 and 7 5 . (5x – 6) is the factor of these two algebraic terms Example 9 Given fg(x) = 2 – x2 and f(x) = 3x, determine the function g. Solution f[g(x)] = 2 – x2 3[g(x)] = 2 – x2 g(x) = 2 – x2 3 Thus, g : x → 2 – x2 3 Example 10 Given gf(x) = 2 – x2 and f(x) = 3x, determine the function g. Solution Let y = f(x) y = 3x x = y 3 gf(x) = 2 – x2 g(y) = 2 – y 3 2 y is the input of the function g g(x) = 2 – x 3 2 Thus, g : x → 2 – x 3 2 6
Additional Mathematics SPM Chapter 1 Functions 4 Form Example 11 Given f : x → 3x2 – 2. Determine the values of a if f : a → a. Solution f(a) = a 3a2 – 2 = a 3a2 – a – 2 = 0 (3a + 2)(a – 1) = 0 3a + 2 = 0 or a – 1 = 0 a = – 2 3 a = 1 f : a → a also means mapping to itself. SPM TipS 1. Assume x is the input of a function which will give an output y, then the inverse function upon y will result in x. x f f –1 y 2. If f has an inverse function, then (a) the inverse function is represented by f –1 (b) the graph of f –1 is the reflection of the graph of f in the line y = x (c) domain of f –1 = range of f and range of f –1 = domain of f (d) ff –1(x) = x and f –1f(x) = x 3. f –1 exists if and only if f is a one-toone function. 4. Horizontal line test (a) If the horizontal line cuts the graph at one point only, then it is a one-to-one function and hence, the function has an inverse function. y x y = f(x) f has an inverse function 0 (b) If the horizontal line cuts the graph at more than one point, then it is not a one-to-one function and hence, the function does not have an inverse function. y x y = f(x) f does not have an inverse function. 0 1.3 Inverse Functions 7
Additional Mathematics SPM Chapter 1 Functions 4 Form Example 12 Verify that the inverse function of f(x) = 3x – 1 is g(x) = x + 1 3 . Solution fg(x) = f x + 1 3 = 3 x + 1 3 – 1 = (x + 1) – 1 = x gf(x) = g(3x – 1) = (3x – 1) + 1 3 = 3x 3 = x f –1f(x) = x Since fg(x) = gf(x) = x, thus the inverse function of f(x) = 3x – 1 is g(x) = x + 1 3 . f f –1(x) = x Example 13 Given that f : x → 3 – 4x, determine f –1 in the same form. Solution Let y = f(x) Then, x = f –1(y) y = 3 – 4x 4x = 3 – y x = 3 – y 4 Express x in terms of y f –1(y) = 3 – y 4 f –1(x) = 3 – x 4 Substitute y with x Thus, f –1 : x → 3 – x 4 f x ⎯→ y means f(x) = y x ←⎯ y means f –1(y) = x f –1 SPM TipS Quiz A function is defined by f(x) = 2x + 1 for the domain –2 x 3, state (a) the domain for f –1(x). (b) the range for f –1(x). 1. Given function f(x) = a – 6x, such that a is a constant, find the value of a such that f(a) = 15. Solution f(a) = 15 a – 6a = 15 –5a = 15 a = –3 2. The diagram below shows the function m : x → 3x + n, such that n is a constant. –1 2 x m 3x + n Find the value of n. SPM Mastery 8
Additional Mathematics SPM Chapter 1 Functions 4 Form Solution m(–1) = 2 3(–1) + n = 2 –3 + n = 2 n = 5 3. Given the function f : x → 2x – 3. Find the value of x when f(x) is mapped onto itself. Solution f(x) = x 2x – 3 = x x = 3 “mapped onto itself” means the object is the image. SPM TipS 4. The diagram below shows a composite function, hk that maps p to r. p q hk r State (a) the function that maps p to q, (b) h–1(r). Solution (a) Function k (b) h–1(r) = q The function k maps p to q whereas the function h maps q to r. Hence, the function hk maps p to r. SPM TipS 5. Given f(x) = 4x and g(x) = a + bx, such that a and b are constants. Express a in terms of b when fg(–1) = 5. Solution f[g(–1)] = 5 f[a + b(–1)] = 5 f(a – b) = 5 4(a – b) = 5 a – b = 5 4 a = 5 4 + b 6. Given f : x → x – 2, find (a) f –1(x), (b) the value of t such that f 2 3t 5 = 12. Solution (a) Let y = f(x) Then, x = f –1(y) y = x – 2 y + 2 = x f –1(y) = y + 2 Thus, f –1(x) = x + 2 (b) f f 3t 5 = 12 f 3t 5 – 2 = 12 3t 5 – 2 – 2 = 12 3t 5 = 16 t = 26 2 3 f 2 (x) means f[f(x)]. SPM TipS 9
Additional Mathematics SPM Chapter 1 Functions 4 Form 7. In the diagram below, function g maps set P to set Q and function h maps set Q to set R. x 4 – 3x 9 – 6x P Q g h R Find (a) in terms of x, the function (i) that maps set Q to set P, (ii) h(x), (b) the value of x when gh(x) = 2x + 3. Solution (a) (i) g(x) = 4 – 3x Let y = g(x) Then, g–1(y) = x y = 4 – 3x 3x = 4 – y x = 4 – y 3 g–1(y) = 4 – y 3 Thus, g–1(x) = 4 – x 3 (ii) h(4 – 3x) = 9 – 6x Let y = 4 – 3x 3x = 4 – y x = 4 – y 3 h(y) = 9 – 6 4 – y 3 = 9 – (8 – 2y) = 1 + 2y Thus, h(x) = 1 + 2x (b) g[h(x)] = 2x + 3 4 – 3h(x) = 2x + 3 4 – 3(1 + 2x) = 2x + 3 4 – 3 – 6x = 2x + 3 –8x = 2 x = – 1 4 The function that maps set Q to set P is g–1. 8. Given that f : x → 3x – 2 and g : x → 4 + 3x. (a) Find (i) g(7), (ii) the value of p if f( p + 3) = 1 5 g(7), (iii) gf(x). (b) Sketch the graph y = | gf(x)| for –1 x 1. Hence, state the range of y. Solution (a) (i) g(x) = 4 + 3x g(7) = 4 + 3(7) = 25 (ii) f(p + 3) = 1 5 g(7) 3(p + 3) – 2 = 1 5 (25) 3p + 9 – 2 = 5 3p + 7 = 5 3p = –2 p = – 2 3 (iii) gf(x) = g(3x – 2) = 4 + 3(3x – 2) = 4 + 9x – 6 = 9x – 2 (b) x –1 0 2 9 1 y 11 2 0 7 11 7 2 0 –1 x y 2 1 9 y = 9x – 2 Thus, the range of y is 0 y 11. Alternative Method The graph y = |9x – 2| can be obtained by drawing the graph of y = 9x – 2 and the part of the graph y = 9x – 2 which is below the x-axis is reflected in the x-axis. 10
Additional Mathematics SPM Chapter 1 Functions 4 Form Solution (a) f(2) = p 2 + 3(2) = p p = 8 The value is incorrect because f(2) = p = 8. (b) g –1(2 + 3x) = 2 x + 2 Let y = 2 + 3x x = y – 2 3 g–1(y) = 2 y – 2 3 + 2 = 2 y – 2 + 6 3 = 6 y + 4 Thus, g–1(x) = 6 x + 4 HOTS Example 3 Given the inverse function of f, f –1 : x → 1 2 – 3x , x ≠ m. State the value of m. Hence, find f. Solution 2 – 3x ≠ 0 → x ≠ 2 3 → Thus, m = 2 3 Let y = f –1(x). Then, f(y) = x y = 1 2 – 3x 2 – 3x = 1 y 3x = 2 – 1 y x = 2 3 – 1 3y f( y) = 2y – 1 3y Thus, f : x → 2x – 1 3x HOTS Example 1 The diagram below shows three balls which are labelled with numbers that are picked from a box. 1 2 5 The colours of the balls are yellow, blue and orange respectively. (a) Draw an arrow diagram to represent the relation ‘the colour balls picked to the numbers on the balls’. (b) Hence, state the type of relation. Is the relation a function? Solution (a) Yellow Blue Orange 1 2 5 (b) One-to-one relation. It is a function. HOTS Example 2 The diagram below shows the relation of set X, set Y and set Z. 2 Set X Set Y Set Z 1 2 p f g–1 Given that f(x) = 2 + 3x and g –1f(x) = 2 x + 2 , x ≠ –2. (a) If Aini writes p = 7, determine whether the value is correct. Give your reason. (b) Find g–1(x). 11
Additional Mathematics SPM Chapter 1 Functions 4 Form Paper 1 1. Function f is defined by f(x) = 3x – 5 for the domain –2 x 4. Determine the range of the function f. 2. Given f(x) = px + q, such that p and q are constants. If f(1) = 7 and f –1(–5) = 3, find the value of p and of q. 3. Determine whether the following relation is a function. If the relation is a function, determine the type of relation. y x (2, 4) (–2, –3) 0 4. The diagram below shows the function f(x). If y = f(x) has an inverse function, sketch the graph of y = f –1(x) on the same diagram. y x 0 2 y = f(x) 5. Given that f : x → 3 + 2x 5 , find (a) f(–5), (b) the value of x such that f(x) = 3. 6. Given that f(x) = 3x x – 2, x ≠ p. Find (a) the value of p, (b) the image of –1. 7. The functions f and g are defined by f(x) = x + 2 and g(x) = x 3 – 5. Find the composite functions (a) f 2(x), (b) gf(x). 8. Given that f(x) = x 3 and composite function fg(x) = 2x2 . Find the function g(x). 9. Function f is defined by f(x) = │2x│– 5 for the domain –3 x 2. Find the range of f(x). 10. Given f(x) = 3x – 2 and g(x) = 4x – 1 2 . Find the value of x, such that x = fg(x). 11. Given f –1(x) = 2 3 – px , x ≠ 3 p , such that p is a constant. Find the value p if f(2) = 3. 12. Given h(x) = x 2 – 3x , x ≠ 2 3 . Find h(–1). 13. Given k(x) = 4 x – 2, x ≠ 2. Find k –1(5). 14. Given f(x) = 4x – 1 x + 2 , x ≠ –2 and g(x) = 3x. Find the value of m if gf(m) = 9. Paper 2 1. The diagram below shows the relation between set X and set Y. Set X Set Y Square of –1 1 2 3 1 4 9 16 (a) The relation is a function. Given a reason why it is a function. (b) Hence, write the relation using function notation. SPM Practice 1 12
Additional Mathematics SPM Chapter 1 Functions 4 Form (c) State (i) the domain of the relation, (ii) the range of the relation, (iii) the object of 1. 2. (a) Given the function f : x → 3x + 5. Find (i) the value of x when f(x) maps onto itself, (ii) the value of s, such that f(2s – 1) = 4s. (b) Given f : x → 5x – 2 and g : x → 3x 4 , show that f –1g–1(x) = (gf ) –1(x). 3. (a) Given h(x) = 9 – 3x, find h(–3). (b) Given f(x) = x2 and g(x) = 1 – 6x. (i) Show that f(–2) = g– 1 2 . (ii) Find fg(–3). (iii) Find the value of x, such that g(x) = f(4). 4. (a) Given f(x) = px2 + qx + r, such that f(0) = 5, f(–2) = 21 and f(3) = –4. Find the values of p, q and r. (b) Given h(x) = 8 – 3x, find (i) h(2x – 1), (ii) h–1(4). 5. (a) Given h(x) = │5x – 1│, find the values of x if h(x) = x. (b) Given f(x) = 9x2 – 1, such that x 0 and g f(x) = 2 x – 3, such that x ≠ 0. Find HOTS (i) g(x), such that x p, (ii) the value of p. 6. (a) The function f and the function g are defined by f(x) = 4 – 3x and g(x) = 5x – 2. Find fg –1 (x). (b) The function m is defined by m : x → │9x – 13│, determine HOTS (i) m(1), (ii) the domain for m(x) 13 and sketch the graph of the function for m(x) 13. 7. (a) Given that f(x) = 3x – 5 and g(x) = 2x + m. Find the value of m if fg(x) = g f(x). (b) Given g –1(x) = 3x – 2, such that x p. HOTS (i) State the value of p. (ii) Determine the inverse function of g –1(x). 8. (a) The diagram below shows the function f(x). x y 7 –1 0 1 (i) Determine the domain of f. (ii) Determine the range of f. (iii) Explain why f –1(x) does not exist. (b) Given f(x) = 4 x , x ≠ 0. (i) Determine f –1(x). (ii) Show that ff –1(x) = f –1f(x) = x. 9. Given f(x) = x 3 – 2 and g(x) = 6x + 4. Find (a) f –1g–1(x), (b) ( fg)–1(x), (c) the value of r, such that f –1g–1(r) = ( fg) –1(2r). 13
Additional Mathematics SPM Chapter 1 Circular Measure 5 Form Circular Measure 1 CHAPTER Learning Area Geometry Form 5 Concept Map Circular Measure radian L degree Arc Length s = rθ Perimeter of Segment = Length of arc PQR + Length of chord PR Area of Sector A = 1 2 r2θ Area of Segment = Area of sector OPQR – Area of ∆OPR r s θ O Q P R θ r 162
Additional Mathematics SPM Chapter 1 Circular Measure 5 Form 1.2 Arc Length of a Circle 1.1 Radian 1. 360° = 2π radian 180° = π radian 2. 1° = π 180° and 1 rad = 180° π 3. θ° = θ × π 180° rad 4. θ rad = θ × 180° π Example 1 (a) Convert π 5 rad into degrees. (b) Convert 0.8 rad into degrees. Give answer correct to one decimal place. [Use π = 3.142] (c) Convert 40° into radians, in terms of π. (d) Convert 80°50' into radians. Give answer correct to three significant figures. [Use π = 3.142] Solution (a) π 5 rad = π 5 × 180° π = 36° (b) 0.8 rad = 0.8 × 180° 3.142 = 45.8° (c) 40° = 40° × π 180° rad = 2 9 π rad (d) 80°50ʹ = 80°50ʹ × 3.142 180° rad = 1.41 rad 2. Perimeter of segment = Length of arc PQR + Length of chord PR 3. Formula in finding the length of chord PR: (a) PR = 2r sin θ 2 (b) PR2 = 2r 2 – 2r2 cos θ (c) PR sin θ = r sin P = r sin R r O s Q P R θ O θ s Q P r R 1. Arc length PQR, s = rθ where θ in radians. INFO Derive the formula PR = 2r sin θ 2 163
Additional Mathematics SPM Chapter 1 Circular Measure 5 Form Example 2 The diagram below shows sector POQ and sector ROS for two concentric circles with centre O. The ratio of the radius of sector POQ to the radius of sector ROS is 1 : 2. Find the angle θ, in radians, where the length of arc RS is equal to the perimeter of sector POQ. O Q S R P θ Solution Let the radius of sector POQ is r. Arc length RS = Arc length PQ + OP + OQ 2rθ = rθ + 2r 2rθ – rθ = 2r rθ = 2r θ = 2 rad Example 3 The diagram below shows a semicircle with centre O. Q O R P 0.3 rad It is given that ∠RPQ = 0.3 rad and the length of POQ is 20 cm. Calculate (a) ∠POR, in radians, (b) the perimeter of the shaded region, in cm. [Use π = 3.142] Solution (a) ∠RPQ = ∠PRO = 0.3 rad ∠POR = π – ∠RPQ – ∠PRO = 3.142 – 0.3 – 0.3 = 2.542 rad (b) The length of chord PR = 102 + 102 – 2(10)(10)(cos 2.542) = 19.11 cm The perimeter of the shaded region = 20 2 (2.542) + 19.11 = 25.42 + 19.11 = 44.53 cm Example 4 The city developer wants to design a garden surrounded by a jogging track. The shaded region in the diagram below shows the plan of the garden where two arcs KL and MN of two concentric circles with centre O. L N K M O 0.5 rad It is given that ∠KOL = 0.5 rad, arc length KL = 9 m and OM : OK = 5 : 3, find (a) the radius, OK, in m, (b) the length of the jogging track, in m. Solution (a) Let the radius OK = r r × 0.5 = 9 r = 18 m (b) OM OK = 5 3 OM = 5 3 × 18 = 30 m The length of the jogging track = Arc length KL + Arc length MN + 2(30 – 18) = 9 + 30(0.5) + 24 = 48 m 164
Additional Mathematics SPM Chapter 1 Circular Measure 5 Form 1.3 Area of Sector of a Circle Example 5 The diagram below shows sector AOB and quadrant of a circle COD with centre O. It is given that ∠AOB = π 4 rad, OC = r cm and OA : OC = 2 : 3. Find the area of the shaded region in terms of π and r. O r cm C D B A Solution OA OC = 2 3 OA = 2 3 r The area of the shaded region = Area of quadrant COD – Area of sector AOB = 1 2 r2 π 2 – 1 2 2 3 r 2 π 4 = πr2 4 – πr2 18 = πr2 1 4 – 1 18 = 7 36 πr2 cm2 Example 6 The diagram below shows a sector OPQR with a radius of 16 cm. O P R Q 20 cm 16 cm Given the length of chord PR = 20 cm, find the area of segment PQR. Solution 202 = 162 + 162 – 2(16)(16)(cos ∠POR) cos ∠POR = 162 + 162 – 202 2(16)(16) ∠POR = 1.350 rad Area of sector OPQR = 1 2 (16)2 (1.350) = 172.8 cm 2. Area of segment = Area of sector OPQR – Area of triangle POR 3. Formula in finding the area of triangle POR: (a) Area = 1 2 r2 sin θ (b) Heron’s formula Area = s(s – p)(s – r)2 , where s = 1 2 (2r + p) r O θ Q P R r p O θ Q P R 1. Area of sector OPQR, A = 1 2 r2 θ where θ in radians. 165
Additional Mathematics SPM Chapter 1 Circular Measure 5 Form s = 1 2 [2(16) + 20] = 26 cm Area of triangle OPR = (26)(26 – 20)(26 – 16)2 = 124.9 cm2 Area of segment PQR = 172.8 – 124.9 = 47.9 cm2 Example 7 The diagram below shows a horse is grazing on the field with the shape of sector of a circle. The horse is tied to a pole at point O with a rope of 5 m length. O P T S R Q OPQT is a semicircle with centre O and radius 5 m. RPS is a sector with centre P and radius 11 m. If the chord PQ = 5 m, calculate the area, in m2 , of the field where the horse is not able to graze. [Use π = 3.142] Solution Since OP = OQ = PQ = 5 m, then ∆POQ is an equilateral triangle. ∠RPS = ∠QPO = π 3 rad The area of the field where the horse is not able to graze = Area of region QRST = Area of sector RPS – Area of sector QOT – Area of triangle POQ = 1 2 (11)2 π 3 – 1 2 (5)2 π – π 3 – 1 2 (5)2 sin π 3 = 63.36 – 26.18 – 10.83 = 26.35 m2 Example 8 The diagram below shows a lawn sprinkler spraying water on the lawn while rotating at an angle of 3 2 rad. If the perimeter of the lawn where the sprinkler sprays the water is 28 m, calculate its area. 3 2 rad Solution Let the radius of the lawn = r r + r + 3 2 r= 28 7 2 r = 28 r = 8 Area of the lawn where the sprinkler sprays the water = 1 2 (8)2 3 2 = 48 m2 1.4 Application of Circular Measures 166
Additional Mathematics SPM Chapter 1 Circular Measure 5 Form Example 10 Wankel engine is a type of internal combustion engine used on various vehicles due to its rotary design. The design of Wankel engine is based on an equilateral triangle as shown in the diagram below. A C 20 cm B In the Wankel engine, each of its sides is an arc of a circle with the respective centre at the opposite vertex. It is given that the length of each chord forming the triangle ABC is 20 cm. By using π = 3.142, calculate the area, in cm2 , of the Wankel engine. Solution ∆ABC is an equilateral triangle, so ∠ABC = ∠BAC = ∠ACB = π 3 rad. The area of the Wankel engine = Area of sector ABC with centre A + 2(Area of segment) = 1 2 (20)2 π 3 + 2 1 2 (20)2 π 3 – sin π 3 = 209.47 + 72.50 = 281.97 cm2 Example 9 The diagram below shows a solar station that collects solar energy to generate electricity. The shaded region in the circle with centre O and a radius of 3 km represents the region of solar panels. O A B C 1 rad 3 km Region of solar panels Given ∠BOC = 1 rad, find the perimeter, in km, of the region of solar panels. [Use π = 3.142] Solution O A B r C 0.5 rad 3 km Let the radius of sector BAC = r r2 = 32 + 32 – 2(3)(3)(cos 0.5) r = 2.204 km The perimeter of the region of solar panels = Major arc length with centre O – Arc length with centre A = 3(2π – 1) – (2.204) 2π – 1 2 = 15.852 – 5.823 = 10.03 km CAUTION ! Make sure your calculator is in the correct mode to get the value of sine or cosine. Use Rad mode when θ is in radians and Deg mode when θ is in degrees. 167
Additional Mathematics SPM Chapter 1 Circular Measure 5 Form 1. The diagram below shows sector MOQ with centre O. N P 30° 7 cm Q M O Given that the arc length PQ is 2.8 cm. Calculate (a) ∠POQ, in radians, (b) the area of the shaded region. [Use π = 3.142] Solution (a) 7(∠POQ) = 2.8 ∠POQ = 0.4 rad (b) tan 30° = MN 7 MN = 4.041 cm The area of region MPN = Area of ∆MON – Area of sector MOP = 1 2 (7)(4.041) – 1 2 (7)2 30° × 3.142 180° = 1.314 cm2 Area of sector POQ = 1 2 (7)2 (0.4) = 9.8 cm2 Area of the shaded region = 1.314 + 9.8 = 11.11 cm2 2. The diagram below shows two sectors JOM and KOL with a common centre O. O J M K L The angle subtended at the centre O by the major arc JM is 8v rad and the perimeter of the whole diagram is 100 cm. Given that OK = u cm, 2OJ = 3OK and ∠KOL = 2v, express u in terms of v. Solution OJ OK = 3 2 OJ = 3 2 u KJ OK = 1 2 KJ = 1 2 u Major arc length JM + Arc length KL + 2 1 2 u = 100 3 2 u(8v) + u(2v) + u = 100 14uv + u = 100 u(14v + 1) = 100 u = 100 14v + 1 3. The diagram below shows a circle with centre O. O A B α rad C AB and BC are the tangents to the circle at the point A and point C respectively. Given that the length of minor arc AC is 12 cm and OB = 13 α cm. Express in terms of α, (a) the radius, r, of the circle, (b) the area, A, of the shaded region. SPM Mastery 168
Additional Mathematics SPM Chapter 1 Circular Measure 5 Form Solution (a) Arc length AC = 12 rα = 12 r = 12 α cm (b) ∆OAB and ∆OCB are right-angled triangles. AB = OB2 – OA2 = 13 α 2 – 12 α 2 = 5 α The area of the shaded region = 2(Area of triangle OAB) – Area of sector AOC = 2 1 2 12 α 5 α – 1 2 12 α 2 (α) = 60 – 72α α2 cm2 4. The diagram below shows a rhombus OKLM that is inscribed in the sector KOM with centre O and radius r cm. M K O L θ rad Given the area of sector KOM is 24 cm2 , express (a) θ in terms of r, (b) the perimeter of the shaded region in terms of r. Solution (a) Area of sector KOM = 24 1 2 r2 θ = 24 θ = 48 r2 rad (b) OK = KL = LM = OM = r The perimeter of the shaded region = Arc length KLM + 2r = rθ + 2r = r 48 r2 + 2r = 48 + 2r2 r cm 5. The diagram below shows a circle and a sector with a common centre O. The radius of the circle is u cm. X Y V O α rad W It is given that the arc lengths VW and XY are 6 cm and 16 cm respectively. VX = 5 cm. Find (a) the value of u and of α, (b) the area of the shaded region, in cm2 . [Use π = 3.142] Solution (a) Arc length VW = 6 uα = 6 .........a Arc length XY = 16 (u + 5)α = 16 .......b b a : (u + 5)α uα = 16 6 u + 5 u = 8 3 8u = 3(u + 5) u = 3 cm Substitute u = 3 into a 3α = 6 α = 2 (b) Area of the shaded region = Area of ∆OXY – Area of sector OVW = 1 2 (3 + 5)2 (sin 2) – 1 2 (3)2 (2) = 20.10 cm2 169
Additional Mathematics SPM Chapter 1 Circular Measure 5 Form 6. The diagram below shows a circular logo used by Gladiator E-Sports Club in a school. All the three red coloured regions are congruent. GLADIATOR E–SUKAN KELAB ESPORTS CLUB Given the perimeter of the red region is 16π cm, find (a) the radius, in cm, of the logo to the nearest whole number, (b) the area, in cm2 , of the yellow region. [Use π = 3.142] Solution (a) 60° Let the radius of the logo = r cm The circumference of the circle = The perimeter of all three red coloured regions 2πr = 16π r = 8 cm (b) Area of segment = 1 2 (8)2 60° × 3.142 180° – 1 2 (8)2 sin 60° = 33.5147 – 27.7128 = 5.8019 cm2 Area of the yellow region = Area of circle – 12 (Area of segment) = 3.142(8)2 – 12(5.8019) = 131.47 cm2 Alternative Method Area of the yellow region = 6[Area of triangle – (Area of sector – Area of triangle)] = 6[2(Area of triangle) – Area of sector] = 62 1 2 (8)2 sin 60° – 1 2 (8)2 60° × 3.142 180° = 131.47 cm2 7. The diagram below shows sector POQ with centre O and sector RPQ with centre P. P O Q R 8 cm 45° It is given that OP = 10.4 cm and PQ = 8 cm. Find (a) ∠OPQ, in radians, (b) the perimeter of sector RPQ, in cm, (c) the area of the shaded region, in cm2 . [Use π = 3.142] Solution (a) ∠OPQ = 180° – 45° 2 × 3.142 180° = 1.178 rad (b) The perimeter of sector RPQ = 8(1.178) + 2(8) = 25.42 cm (c) The area of segment PQ = 1 2 (10.4)2 45° × 3.142 180° – 1 2 (10.4)2 sin 45° = 4.240 cm2 The area of the shaded region = 1 2 (8)2 (1.178) + 4.240 = 41.94 cm2 170
Additional Mathematics SPM Chapter 1 Circular Measure 5 Form 8. Lai wants to make a cone as shown in the diagram below to conduct an experiment about waves. The height of the cone is 13.5 cm. The following diagram shows the net of the cone drawn on a piece of rectangular paper. 3.44 cm O Length Width (a) Calculate the minimum value of the length and width of the paper to the nearest integer. (b) Hence, find the area, in cm2 , of the unused paper. [Use π = 3.142] Solution (a) 1.72 cm 13.5 cm s The base radius of the cone = 3.44 2 = 1.72 cm Let the length of the paper = p p = The slant height of the cone, s = 13.52 + 1.722 = 13.61 So, the minimum length of the paper = 14 cm O 13.61 cm w θ Arc length = Circumference of the circular base of the cone 13.61θ = π(3.44) θ = 0.7942 rad Let the width of the paper = w w 13.61 = sin 0.7942 w = 9.708 cm So, the minimum width of the paper = 10 cm (b) The area of the unused paper = Area of the paper – Area of sector = (14)(10) – 1 2 (13.61)2 (0.7942) = 66.44 cm2 HOTS Example 1 Wahid cuts part of a cake bought as shown in the diagram below. The piece of cake has a surface area of 286.26 cm2 . 8 cm θ 12 cm By using π = 3.142, calculate the value of angle θ, in degrees. Hence, determine the percentage of the piece of cake from the cake that was originally. 171
Additional Mathematics SPM Chapter 1 Circular Measure 5 Form (a) Calculate the perimeter, in cm, of every step. (b) Determine whether a roll of carpet with the area of 3 m2 is enough to line all the steps or not. [Use π = 3.142] Solution (a) The perimeter of every step = Arc length + 2(Radius) = 10025° × 3.142 180° + 2(100) = 243.6 cm (b) The angle subtended by 12 steps = 1225° × 3.142 180° = 5.237 rad The area of all the steps = 1 2 (1)2 (5.237) = 2.6185 m2 The area of all the steps is 2.6185 m2 which is less than the area of the roll of carpet of 3 m2 . Therefore, the roll of carpet is enough to line all the steps. Solution D F E 12θ cm B A C 2(Area of sector BAC) + 2(Luas ABED) + Area BCFE = 286.26 2 1 2 (12)2 (θ) + 2[12(8)] + 12(8)(θ) = 286.26 144θ + 192 + 96θ = 286.26 240θ = 94.26 θ = 0.39275 rad = 0.39275 × 180° 3.142 = 22.5° The percentage of the piece of cake = 22.5° 360° × 100% = 6.25% HOTS Example 2 The diagram below shows a spiral staircase that has 12 steps and is lined with carpet such that every step has the shape of a sector of circle with radius 100 cm and angle of 25°. 25° 100 cm Quiz It is given that the arc length and perimeter of a sector of circle are 12 cm and 28 cm respectively. Calculate the value of angle subtended at centre of the sector, in radians. 172
Additional Mathematics SPM Chapter 1 Circular Measure 5 Form Paper 1 1. The diagram below shows sector POQ and sector SOR with centre O. O Q R S P 5 cm θ It is given that the lengths of arcs PQ and RS are 2.5 cm and 8 cm respectively. Calculate (a) the value of θ, in radians, (b) the perimeter of the whole diagram, in cm. 2. The diagram below shows a rhombus OCDE inscribed in a sector EOC with centre O. C E O θ D If the perimeter of the rhombus OCDE is 36 cm, express, in terms of π, (a) the value of θ, in radians, (b) the length of arc EDC, in cm. 3. The diagram below shows a semicircle with centre A and a radius of r cm inscribed in a sector POQ with centre O. B A P C Q O Straight line OQ is a tangent to the semicircle at point C. It is given that ∠POQ = π 6 rad, express the perimeter of the shaded region in terms of π and r. 4. The diagram below shows two sectors consisting of sector MON with centre O and sector PMN with centre M. M N P O Given that OM = 10 cm and ∠MON = 120°. By using π = 3.142, calculate the perimeter, in cm, of the whole diagram. 5. The diagram below shows sector OPRT with centre O and a radius of 5 cm inscribed in an isosceles triangle QOS. P T Q R S O It is given that R is the midpoint of QS and ∠POT = 40°. By using π = 3.142, calculate the perimeter, in cm, of the shaded region. 6. The following diagram shows sectors of two concentric circles with centre O such that the line OP is perpendicular to the line OS. SPM Practice 1 173
Additional Mathematics SPM Chapter 1 Circular Measure 5 Form P 0.65 rad S Q R O It is given that the area of sector ROS is 11.7 cm2 and the radius of sector POQ is 2.5 cm. Calculate (a) the radius of sector ROS, (b) the area of sector POQ. [Use π = 3.142] 7. The diagram below shows sector AOB with centre O and sector PAQ with centre A. B A Q P O Given that OA = 10 cm, AP = 5.6 cm and ∠PAQ = 70°. If the length of arc AB = 14.5 cm, calculate the area of the shaded region. [Use π = 3.142] 8. The diagram below shows a semicircle ABC. A B 1.025 rad C It is given that AB = 10 cm. By using π = 3.142, calculate the area of the shaded region. 9. The following diagram shows sector ABC with centre B inscribed in a circle with centre O. Points A, B and C are located on the circumference of the circle. A O 1.4 rad B C It is given that the length of arc AC of sector ABC is 7 cm. Calculate (a) the radius of the circle, (b) the area of the shaded region. [Use π = 3.142] 10. The diagram below shows a 90° longitude on the Earth that produces an arc of 9 990 km along the Equator. Radius of the Earth 90° longitude Equator By using π = 3.142, calculate (a) the radius of the Earth, (b) the percentage of difference between the radius of the Earth obtained and the actual radius of the Earth which is 6 371 km. 11. The diagram below shows a closed circuit television (CCTV) camera installed to monitor an area by visual surveillance. 80° If the area of the field of vision that can be covered by the camera is 120 m2 , calculate the perimeter of the field of vision. [Use π = 3.142] 174
Additional Mathematics SPM Chapter 1 Circular Measure 5 Form 12. The diagram below shows that the rear windshield wiper of a car rotates at angle of 125°. 125° 63.5 cm 35 cm It is given that the length of the wiper blade is 35 cm and the length of the whole wiper is 63.5 cm. Find the area of the rear windshield that can be cleaned by the wiper. [Use π = 3.142] 13. A piece of paper was originally a semicircular with O modified as shown in the diagram below. HOTS O P T R Q S It is given that the radius of the semicircle is 10 cm. By using π = 3.142, calculate the area of the whole diagram. 14. The diagram below shows a redesigned softball field with the shape of a quadrant of circle with centre O and a radius of 72 m. The softball field consists of the outfield that covered with grass and the infield that covered with sand. 26 m Outfield fence Outfield Infield Grass line O By using π = 3.142, calculate (a) the length of the outfield fence, (b) the area of the outfield that covered with grass. Paper 2 1. The diagram below shows a sector BOD with centre O, a sector DAC with centre A and an equilateral triangle OAB with the length of each side is 5 cm. D O A B C It is given that O and B are the midpoints of the straight lines DA and AC respectively. Calculate (a) ∠DOB, in terms of π radians, (b) the perimeter of the shaded region, (c) the area of the shaded region. [Use π = 3.142] 2. The following diagram shows a sector MOQ of circle with centre O and a radius of r cm. NP is perpendicular to MO. 45° O Q P N M It is given that N is the midpoint of MO. (a) Show that the perimeter of the shaded region is π + 8 – 22 4 r cm. (b) Express the area of the shaded region in terms of π and r. 175
Additional Mathematics SPM Chapter 1 Circular Measure 5 Form 3. The diagram below shows a semicircle AFEDC with the length of diameter AC is 20 cm. ABC is an arc with centre O. 1.71 rad F E B D A C O It is given that O is the midpoint of straight line FD. The arcs AF, FED and DC are of the same length. Calculate (a) the length of arc ABC, (b) the perimeter of segment FED, (c) the area of the whole diagram. [Use π = 3.142] 4. The diagram below shows a semicircle PQRO with centre O and radius 9 cm, and sector PRA of circle with centre R. Q R P A O It is given that the length of arc PQ is 15.75 cm. Calculate (a) ∠PRA, in radians, (b) the perimeter of the whole diagram, (c) the area of the shaded region. [Use π = 3.142] 5. The diagram below shows a circle with centre O and a radius of 8 cm. DA and DC are tangents to the circle at point A and point C respectively. A C D B O α It is given that the area of major sector ABC is 128 cm2 . Calculate (a) the value of α, in radians, (b) the perimeter of the whole diagram, (c) the area of the shaded region. [Use π = 3.142] 6. The diagram below shows a sector AOC with centre O. MC is a perpendicular bisector of straight line OB at point M. 0.2π rad A C M B O Given that the length of arc AB is 2π. (a) Find the radius of sector AOC. (b) Express the length of arc BC in terms of π. (c) Show that the area of the shaded region is 1 6 100π – 753 cm2 . [Use π = 3.14] 7. The diagram below shows a sector POQ of circle with centre O and a rhombus OABC with the length of side of 4 cm. 110° A P Q C B O It is given that the ratio of the length of OA to the length of AP is 2 : 3. Calculate (a) ∠POQ, in radians, (b) the perimeter of the shaded region, (c) the area of the shaded region. [Use π = 3.14] 176
Additional Mathematics SPM Chapter 1 Circular Measure 5 Form 8. The diagram below shows a sector ROP of circle with centre O and a sector PQR of circle with centre Q. The lines RQ and PQ are tangents to the sector ROP at point R and point P respectively. HOTS 45° P Q R O It is given that the radius of sector ROP is 10 cm. (a) Find ∠ROP, in π radians. (b) Show that the radius of sector PQR 24.14 cm. (c) Calculate the perimeter of the shaded region. (d) Calculate the area of the shaded region. [Use π = 3.142] 9. The diagram below shows four sectors AOB, COD, EOF and GOH with a common centre, O. The angle subtended at centre O by each sector is 30°. A B C D E F G H O It is given that the perimeter of the whole diagram is 14π + 60 3 cm where BC, DE and FG are of the same length of 2 cm. Find (a) the radius of sector AOB, (b) the area of the whole diagram. [Use π = 3.142] 10. The diagram below shows a semicircle OPQR with centre O and a sector QAOB of circle with centre Q. OQ is a perpendicular bisector of RP. A B R P Q O It is given that semicircle OPQR and sector QAOB has radius of 5 cm respectively. Calculate (a) ∠AQB, in radians, (b) the perimeter of the shaded region, (c) the area of the shaded region. [Use π = 3.142] 11. The following diagram shows a region of solar panels PQRS surrounded by two arcs PQ and RS of circle with a common centre O and a radius of 2.5 km, and two straight lines QR and PS that are of the same length. HOTS R P S Q O 60° 100° By using π = 3.142, calculate (a) the perimeter of the region of solar panels PQRS, (b) the area of the region of solar panels PQRS, (c) the amount of electricity that can be generated if it is estimated that the region of solar panels with an area of 0.01 km2 can generate 1 megawatt of electricity. 177
Additional Mathematics SPM Chapter 1 Circular Measure 5 Form 12. The diagram below shows a circular cross section with centre O and radius 20 cm of a timber floating in water. Both point P and point R are on the surface of water and the highest point Q is 12 cm above the surface of water. P 12 cm 20 cm R Q O (a) Find ∠POR, in radians. (b) Calculate the length of arc PQR. (c) Determine the cross-sectional area of the timber floating below the surface of water. (d) How many percentage of the timber is below the surface of water? [Use π = 3.142] 13. The diagram below shows two types of pizza slices of different sizes and prices. HOTS RM7.50 per slice RM8.50 per slice 15 cm 18 cm 60° 45° Pizza slice A Pizza slice B Determine which pizza slice gives the best offer. Give your justification. 14. The following diagram shows the shape of logo of a computer company. MPR is a sector of circle with centre P. Both LKP and PTS are identical segments of circle with centre O. Whereas, PNO and PQO are two identical right-angled triangles. O L S P N K T Q M R Given that OP = 6 cm, PM = PR = 10 cm and ∠LPS = 60°. Calculate (a) ∠POS, in radians, (b) the perimeter of the shaded region, (c) the area of the whole diagram. [Use π = 3.142] 15. The diagram below shows a floodlight installed at point O to provide illumination in the shape of a sector POQ with centre O in the rectangular dark region ABCD. HOTS A P B 32 m 22 m Floodlight D Q C O It is given that O is the midpoint of AD. AP = DQ and PB = QC such that AP : PB = 6 : 5. By using π = 3.142, calculate (a) ∠POQ, in radians, (b) the perimeter of the region that acquires the illumination of floodlight, (c) the area of the region that does not acquire the illumination of floodlight. 178
Paper 1 2 hours Section A [64 marks] Answer all questions. 1. It is given that p : x → 1 (x – 1)2 , x ≠ k and q : x → 2x + 1. (a) State the value of k. [1 mark] (b) Find the function r(x) if rq −1(x) = p(x). [2 marks] (c) The following dialogue occurs. Amin : I think that p−1(x) is a function. Razif : I don’t agree. The function p(x) has no inverse. Determine who is correct and give your reason. [3 marks] Answers: (a) (b) (c) 2. Diagram 1 shows the relation between set X, set Y and set Z. It is given that set X is mapped to set Z by the function 2x – 1 and is mapped to set Y by gf(x) = 4 – (x + 1)2 . (a) Write the function which maps set X to set Z by using function notation. (b) Find the function that maps set Y to set X. (c) Find the function that maps set Z to set Y. [6 marks] X Y Z gf(x) = 4 – (x + 1)2 Diagram 1 SPM MODEL Paper 290
Additional Mathematics SPM Model Paper Section B [16 marks] Answer any two questions from this section. 13. (a) Diagram 10 shows a triangle with the area of (1 + 15) cm2 . (2 a + 2 b ) cm (2 a – b ) cm Diagram 10 Find the value of a and of b. [6 marks] (b) Determine whether the set of expressions 245, 60 and 375 are similar surds or not similar surds. Explain your answer. [2 marks] Answers: (a) (b) 14. Diagram 11 shows the straight line x = 6 intersects the curve y = kx2 + 5 and the straight line y = 2kx + 2. y x 0 y = h x = 6 y = 2kx + 2 y = kx2 + 5 Diagram 11 (a) Find the value of k if the area of the shaded region is 27 unit2 . [5 marks] (b) Hence, find the volume of revolution, in terms of π, when the region bounded by the curve y = kx2 + 5 and the straight line y = h is revolved through 360° about y-axis. [3 marks] 298
Additional Mathematics SPM Model Paper Paper 2 2 hours 30 minutes Section A [50 marks] Answer all questions. 1. A study shows that the lifespan of a washing machine, X is normally distributed with X ~ N(µ, σ2 ) as shown in Diagram 1. (a) Find P(X 1515) (b) Given that AB is the axis of symmetry of the graph, find the value of µ and of σ. [6 marks] 2. Given ∫ h 5 f(x) dx = –4, find (a) ∫ 5 h 1 8 f(x) dx, (b) ∫ h 3 f(x) dx if ∫ 3 5 1 3 f(x) dx = 2, (c) the value of h where ∫ h 5 [f(x) – 3] dx = –20. [6 marks] 3. Air escapes from a balloon that has a shape of sphere due to a leak. Find, in terms of π (a) the approximate decrease in the volume of the balloon, in cm3 when its radius decreases from 18 cm to 17.7 cm. (b) the rate of change of radius, in cm s–1 when the volume of the balloon is 4500π cm3 if its volume decreases by 20 cm3 every second. [6 marks] 4. Diagram 2 shows the points A, B and C on a Cartesian plane. (a) Point P(x, y) moves such that it is always equidistant from the straight line OA and straight line OB. Given that triangle OAB is an isosceles triangle and the area of quadrilateral OACB is 10 unit2 , find (i) the gradient of straight line AB, (ii) the y-intercept of straight line AB. [5 marks] (b) Point Q(x, y) moves such that its distance from point C is the equal to its distance from the x-axis. Find the equation of locus of point Q. [3 marks] f(x) B 1474.5 A 1515 x 19.15% 19.77% Diagram 1 y B A P (x, y) C (8, 2) O x Diagram 2 300
Additional Mathematics SPM Model Paper Section B [30 marks] Answer any three questions from this section. 8. Table 1 shows the values of two variables, x and y, obtained from an experiment. The variables x and y are related by the equation y = 10kx + p , where k and p are constants. x 0.1 0.3 0.4 0.6 0.7 0.8 y 4.37 10.96 17.39 43.65 96.18 109.65 Table 1 (a) Plot log10 y against x, using a scale of 2 cm to 0.1 unit on the X-axis and 2 cm to 0.2 unit on the Y-axis. Hence, draw the line of best fit. [4 marks] (b) Using the graph in 8(a), find the value of p and of k. [4 marks] (c) It is found that one of the readings for the value of y has been incorrectly recorded. Identify the incorrect value of y and estimate the nearest correct value of y using the graph in 8(a). [2 marks] 9. Diagram 4 shows a trapezium OABC such that OA = 8~ p, OC = 12~ q and AB = 5 6OC. A B E O D C Diagram 4 It is given that OD : DC = 2 : 1, OE = mEB and AE = nAD. (a) Find OE in terms of (i) m, ~ p and ~ q, (ii) n, ~ p and ~ q. [4 marks] (b) Hence, find the value of m and of n. [4 marks] (c) If BC is the height of the trapezium OABC such that ~ p = 1 unit and ~ q = 2 units, calculate BC . Give your answer in the form of ab. [2 marks] 302
Additional Mathematics SPM Model Paper Section C [20 marks] Asnwer any two questions from this section. 12. Table 2 shows information related to the three types of costs involved in the production of a box of children’s breakfast cereal in terms of price indices and weightages. Type of cost Price index for the year 2015 based on the year 2010 Price index for the year 2020 based on the year 2010 Price index for the year 2020 based on the year 2015 Weightage Raw material 175 182 104 x Manufacturing h 200 125 5 Packaging 145 k 120 2 Table 2 (a) Given the composite index for the production cost of a box of children’s breakfast cereal in the year 2020 based on the year 2015 is 117.7, calculate the value of x. [2 marks] (b) Given the price of a box of children’s breakfast cereal in the year 2015 is RM30, calculate its corresponding price in the year 2020. [2 marks] (c) Calculate the value of h and of k. [6 marks] 13. Diagram 6 shows a transparent prism with a rectangular base ABCD such that AD = 10 cm and BF = 15 cm. Both inclined surfaces ADEF and BCEF are rectangles. ABF is a uniform cross section of the prism. ABE is a shaded plane in the prism. D C B A F E Diagram 6 It is given that ∠FAB = 50° and ∠ABF = 35°. Calculate (a) the length, in cm, of AF, [2 marks] (b) the area, in cm2 , of the shaded region ABE, [6 marks] (c) the shortest length, in cm, from point E to the straight line AB. [2 marks] 304
Form 4 1 CHAPTER Functions Quiz on page 3 (a) {1, 2, 3, 4} (b) {3, 5, 6, 7, 9} (c) {3, 5, 7, 9} (d) 1 (e) 7 Quiz on page 8 (a) -3 x 7 (b) -2 f –1(x) 3 SPM Practice 1 Paper 1 1. –11 f(x) 7 2. p = –6, q = 13 3. function, one-to-one relation 4. y x 0 2 2 y = f –1(x) 5. (a) – 7 5 (b) 6 6. (a) 2 (b) 1 7. (a) x + 4 (b) x – 13 3 8. 6x2 9. –5 f(x) 1 10. 7 10 11. 2 3 12. – 1 5 13. 2 4 5 14. 7 Paper 2 1. (a) The relation is a function because every object has only one image. (b) f : x → x2 or f(x) = x2 (c) (i) {-1, 1, 2, 3} (ii) {1, 4, 9} (iii) -1, 1 2. (a) (i) – 5 2 (ii) -1 3. (a) 18 (b) (ii) 361 (iii) – 5 2 4. (a) p = 1, q = –6, r = 5 (b) (i) 11 – 6x (ii) 4 3 5. (a) 1 4 , 1 6 (b) (i) 6 x + 1 – 3 (ii) -1 6. (a) 14 – 3x 5 (b) (i) 4 (ii) 0 x 2 8 9 y x 0 1 4 9 13 2 8 9 m(x) = |9x –13| 7. (a) – 5 2 (b) (i) 2 3 (ii) x2 + 2 3 306
Bahasa Melayu English Matematik Mathematics Sains Science Sejarah Pendidikan Islam Biologi Biology Fizik Physics Kimia Chemistry Matematik Tambahan Additional Mathematics Ekonomi Perniagaan Prinsip Perakaunan Concise Notes HOTS & i-THINK SPM Practices SPM Mastery SPM Model Paper Answers Info & Quizzes QR code SPM QUICK REVISION PelangiPublishing PelangiBooks PelangiBooks W.M: RM21.95 / E.M: RM22.95 KC118334 ISBN: 978-629-470-485-5 The RANGER SPM series allows students to do express revision before the SPM examination. The content of this book is aligned with the Standard Curriculum for Secondary Schools (DSKP) for Form 4 and Form 5, textbooks, and the latest SPM assessment format. It also includes engaging digital resources to enhance understanding of the subject. Form 4 1SY85g@c Form 5 9sE^3G2m Scan, register and insert Enrolment Key How to Access MCQ Quiz Purchase eBook here!