CONTENTS CHAPTER 1 Functions..............................................1 CHAPTER 2 Quadratic Functions ...........................17 CHAPTER 3 Systems of Equations.........................31 CHAPTER 4 Indices, Surds and Logarithms...........37 CHAPTER 5 Progressions ......................................52 CHAPTER 6 Linear Law..........................................64 CHAPTER 7 Coordinate Geometry .........................72 CHAPTER 8 Vectors ...............................................86 CHAPTER 9 Solution of Triangles...........................99 CHAPTER 10 Index Number................................... 113 Form 4 Form 5 CHAPTER 1 Circular Measure .............................. 119 CHAPTER 2 Differentiation ...................................129 CHAPTER 3 Integration ........................................147 CHAPTER 4 Permutation and Combination..........162 CHAPTER 5 Probability Distribution......................170 CHAPTER 6 Trigonometric Functions ...................187 CHAPTER 7 Linear Programming.........................205 CHAPTER 8 Kinematics of Linear Motion .............212 ii Content GrabME SPM AddMaths.indd 2 01/04/2022 12:07 PM
CHAPTER 1 FUNCTIONS Algebra • Function is a one-to-one relation or a many-to-one relation such that one object has only one image. • Functions may be discrete functions or continuous functions. • Vertical line test is used to determine whether a relation is a function or not. • Function notation: Alphabets such as f, g or h can be used to represent a function that maps set A to set B. For example, f : A ˜ B f : x ˜ f(x) Let f(x) = 3x + 1 Therefore, f(4) = 3(4) + 1 = 13 • Absolute value function can be written as f : x ˜ |ax + b|. The shape of the graph is . 1.1 Functions • Set A is a domain whereas set B is a codomain. • x is an object whereas f(x) is an image. • Set of all images is known as range. REMEMBER Functions representation Arrow diagram Graph Ordered pair 1 Ch 1_Grab Me SPM AddMaths F4.indd 1 13/05/2022 8:56 AM
Example 1 Example 2 Determine whether the following relations is a function or not. Give your justification. (a) a • b • c • • 1 • 2 • 3 ➤ ➤ ➤ (b) {(1, 2), (2, 3), (2, 4), (3, 5)} (c) y 1 1 0 2 3 2 3 x Solution (a) A function. Every object has only one image, that is, {(a, 2), (b, 2), (c, 1)}. (b) Not a function. Object 2 has two images. (c) A function. Every object has only one image. Determine whether the following graphs represents a function or not. Give your justification. (a) y 0 x (b) y 0 x Solution (a) No. The vertical line cuts the graph at more than one point. (b) Yes. The vertical line cuts the graph at one point only. y 0 x y 0 x 2 Ch 1_Grab Me SPM AddMaths F4.indd 2 13/05/2022 8:56 AM
TIPS Given the function f(x) = |4x – 3|. (a) Find (i) the image of –2, (ii) the objects of 5, (iii) the value of x when f(x) = 0. (b) (i) Sketch the graph of f(x) for the domain –2 x 2. (ii) Hence, state the corresponding range. f(x) 0 Domain Range x Solution (a) (i) f(–2) = |4(–2) – 3| = |–11| = 11 (ii) If f(x) = 5, then |4x – 3| = 5 4x – 3 = 5 or 4x – 3 = –5 4x = 8 4x = –2 x = 2 x = – 1 2 (iii) f(x) = 0 4x – 3 = 0 x = 3 4 (b) (i) y 0 2 5 11 –2 x 3 – 4 3 f(x) = |4x – 3| (ii) The range of f is 0 f(x) 11. Example 4 4 Ch 1_Grab Me SPM AddMaths F4.indd 4 13/05/2022 8:56 AM
5.1 Arithmetic Progressions First term, T1 = a The nth term, Tn = a + (n – 1)d Example: 2, 4, 6, 8, … Common difference, d = Tn – Tn - 1 CHAPTER 5 PROGRESSIONS Algebra Arithmetic progressions • The sum of the first n terms, Sn is given by the formula: Sn = n 2[2a + (n – 1)d] = n 2[a + l], l = Tn • Given Sn , (a) a = T1 = S1 (b) Tn = Sn – Sn – 1 (c) The sum from the nth term to the (n + k)th term, where k is a positive integer = Sn + k – Sn – 1 52 Ch 5_Grab Me SPM AddMaths F4.indd 52 13/05/2022 9:01 AM
Example 17 The sum of the first three terms of a geometric progression with a common ratio – 2 3 is 63. Find the first term. Solution S3 = 63 a(1 – (– 2 3 ) 3 ) 1 – (– 2 3 ) = 63 35 27 a = 63( 5 3 ) a = 63( 5 3 )( 27 35 ) a = 81 Another Method T1 + T2 + T3 = 63 a + (– 2a 3 ) + (– 2 3 )(– 2a 3 ) = 63 9a – 6a + 4a 9 = 63 a = 81 Example 18 Given a geometric progression with T1 = 1 and r = 2. Calculate the minimum value of n if the sum of the first n terms is more than 2 000. Solution Sn 2 000 1(2n – 1) 2 – 1 2 000 2n 2 001 n log10 2 log10 2 001 n log10 2 001 log10 2 n 10.97 Therefore, n = 11. Example 19 The nth term of a geometric progression is Tn = 4(41 – n ). Calculate (a) the common ratio, (b) the sum to infinity. 60 Ch 5_Grab Me SPM AddMaths F4.indd 60 13/05/2022 9:01 AM
Example 8 The diagram below shows part of the graph log10 (1 – y) against log10 x. Given the variables x and y are related by the equation y = 1 – axb . (1, 0) (3, 2) 0 log10 (1 – y) log10 x Find the constants a and b. Solution From the graph, m = 2 – 0 3 – 1 = 1. Thus, Y = X + c. At point (1, 0), 0 = 1 + c Ú c = –1 Hence, log10 (1 – y) = log10 x – 1 … Given y = 1 – axb , thus 1 – y = axb log10 (1 – y) = log10 a + b log10 x … Compare with : log10 a = –1 a = 1 10 , b = 1 6.3 Application of Linear Law Example 9 The table below shows the values of two variables u and v obtained from an experiment. Given that the variables u and v are related by the equation uv = pu + qv, where p and q are constants. u 20 25 30 35 40 45 v 20.0 16.7 15.0 14.0 13.3 12.8 (a) Based on the above table, construct a table for the values of v u . (b) Plot v against v u using a scale of 2 cm to 0.2 unit on the v u -axis and 2 cm to 2 units on the v-axis. Hence, draw the line of best fit. (c) Using the graph in (b), find the values of p and q. 69 Ch 6_Grab Me SPM AddMaths F4.indd 69 13/05/2022 9:02 AM
Solution (a) v u 1.00 0.67 0.50 v 20.0 16.7 15.0 v u 0.40 0.33 0.28 v 14.0 13.3 12.8 (b) 0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 0.2 0.4 0.6 0.8 1.0 1.2 v Graph of v against v – u v – u (c) uv = pu + qv v = p + q v u p = v-intercept and q = gradient Therefore, p = c = 10 and q = 20 – 10 1.0 – 0 = 10 Example 10 The table below shows the values of two variables x and y obtained from an experiment. x 1 2 3 4 5 6 y 36.0 16.7 7.29 3.28 1.48 0.67 Given that x and y are related by the equation y = abx – 1 , where p and q are constants. (a) Convert y = abx – 1 to the linear form. (b) Using a scale of 2 cm to 1 unit on the X-axis and 2 cm to 0.2 unit on the Y-axis, plot Y against X. Hence, draw the line of best fit. (c) From the graph in (b), find (i) the values of a and b, (ii) the value of y when x = 5.5. (d) Find the value of y when x = 7 and the value of x when y = 0.7. 70 Ch 6_Grab Me SPM AddMaths F4.indd 70 13/05/2022 9:02 AM
CHAPTER 8 VECTORS Geometry 8.1 Vectors • The vector notation is →AB or a ~ with magnitude | →AB| or |a ~ |. A →AB or a ~ B ➤ • Vector → BA is a negative vector of vector → AB, that is, – →AB = →BA = –a~ and |– →AB| = |→AB| but both are moving in the opposite direction. • Zero vectors, 0 ~ has zero magnitude and no direction. Scalar quantity A quantity that has magnitude. Example: Distance, speed, mass and height Vector quantity A quantity that has magnitude and direction. Example: Velocity and weight • If a ~ = b ~ , then |a ~ | = |b ~ | and the directions for both vectors are the same. • →AB = l →PQ Í →AB is parallel to →PQ where l is a constant. • Points A, B and C are collinear if →AB = l →BC, →AB = b →AC, →AC = m →BC and there is a common point. • Given a ~ and b ~ are two non-zero vectors and are not parallel. If ha ~ = kb ~ , then h = k = 0. 86 Ch 8_Grab Me SPM AddMaths F4.indd 86 26/04/2022 9:16 AM
• Addition and subtraction of non-parallel vectors. Triangle law A B C a ~ a + b ~ ~ b ~ A B C a ~ –a + b ~ ~ b ~ →AC = →AB + →BC →BC = →BA + →AC = →AC – →AB Parallelogram law A B D C a ~ a + b b ~ ~ ~ →AC = →AB + →AD Polygon law →AF = →AB + →BC + →CD + →DE + →EF A B E F C D a ~ b ~ c ~ e ~ f ~ d ~ 91 Ch 8_Grab Me SPM AddMaths F4.indd 91 26/04/2022 9:16 AM
CHAPTER 1 CIRCULAR MEASURE Geometry 1.1 Radian • One radian is a measure of the angle subtended at the centre of a circle by an arc of equal length to the radius of the circle. O B A r r r 1 rad Hence,1 rad = 180° π ≈ 57.29° and 1° = π 180° ≈ 0.01746 rad • Conversion of degrees to radians and vice versa: Degrees × 180° π × π 180° Radians Example: (a) 100° = 100° × π 180° = 1.745 rad (b) 0.7 rad = 0.7 × 180° π = 40.11° Example 1 The diagram on the right shows a circle with centre O which is divided into eight equal parts. y β θ α x O p rad = 180° 119 Ch 1_Grab Me SPM AddMaths F5.indd 119 13/05/2022 9:06 AM
State the magnitude of angles (a) a, (b) b and (c) q (i) in degrees, (ii) in radians, in terms of π, (iii) in radians. Give the correct answer to four significant figures. [Use π = 3.142] Solution (a) (i) a = 45° (ii) a = 45° × π 180° = π 4 rad (iii) a = 3.142 4 = 0.7855 rad (b) (i) b = 90° (ii) b = 90° × π 180° = π 2 rad (iii) b = 3.142 2 = 1.571 rad (c) (i) q = 135° (ii) q = 135° × π 180° = 3 4 π rad (iii) q = 3 4 (3.142) = 2.357 rad 1.2 Arc Length of a Circle • The formula for the arc length of a circle is given by: Arc length, s = rq where q is the angle in radians. • Note the diagram below. O B A C r θ Perimeter of the segment ACB = Arc length ACB + Length of chord AB 120 Ch 1_Grab Me SPM AddMaths F5.indd 120 13/05/2022 9:06 AM
Example 2 The diagram below shows a sector AOB with centre O and a radius of 8 cm. O A B θ 8 cm 6 cm Given the arc length AB is 6 cm. Find (a) the angle q, in radians, (b) the angle q, in degrees. Solution (a) s = rq (b) q = 0.75 × 180° π 8q = 6 q = 42.97° q = 0.75 rad Example 3 The diagram below shows a sector AOB with centre O and a radius of 5 cm. O A C B θ 5 cm Find the perimeter of the shaded segment ACB for each of the following values of q. (a) 2.09 rad (b) 144° REMEMBER O Major sector Minor sector Minor arc Major arc Segment Chord 121 Ch 1_Grab Me SPM AddMaths F5.indd 121 13/05/2022 9:06 AM
2.2 The First Derivative • Differentiation is a process of determining the gradient function of a function y = f(x). • The gradient function is also known as the first derivative of a function or the derivative function or the differential coefficient of y with respect to x. • Techniques of differentiation for an algebraic function: General formula If y = axn where a and n are constants, then dy dx = naxn – 1 Algebraic sum and difference If y = f(x) ± g(x), then dy dx = fʹ(x) ± gʹ(x) Chain rule (a) If y = f(u) where u = g(x), then dy dx = dy du × du dx (b) If y = [f(x)]n where n is a constant, then dy dx = n[f(x)]n – 1 fʹ(x) Product rule If y = uv where u = f(x) and v = g(x), then dy dx = udv dx + v du dx = uvʹ + vuʹ Quotient rule If y = u v where u = f(x) and v = g(x), then dy dx = v du dx – udv dx v2 = vuʹ – uvʹ v2 132 Ch 2_Grab Me SPM AddMaths F5.indd 132 13/05/2022 9:09 AM
D. Rates of change for related quantities • If y = f(x) and x = g(t), then dy dt = dy dx × dx dt (Chain rule). t TIPS Steps to solve problems involving the rates of change for related quantities: 1. Interpret the information in the form of mathematical symbols. 2. Determine the appropriate chain rule. • Interpretation of rates of change for related quantities: (a) When dy dt 0, y increases when t increases. (b) When dy dt 0, y decreases when t increases. E. Small changes and approximations of certain quantities • For a function y = f(x), dy dx ≈ dy dx dy ≈ dy dx × dx (dx = xnew – xinitial) where dy is a small change in y and dx is a small change in x. (a) When dy 0 or dx 0, there is a small increase in y or x. (b) When dy 0 or dx 0, there is a small decrease in y or x. • The approximation value of y is given by the formula: ynew = yinitial + dy • If y = f(x) and x changes by dx, then (a) the percentage change in x = dx x × 100% (b) the percentage change in y = dy y × 100% 137 Ch 2_Grab Me SPM AddMaths F5.indd 137 13/05/2022 9:09 AM
Example 11 The diagram below shows a part of the graph y = f(x). y y = f(x) x 0 D(6, 18) C(2, –4) B(–2, 4) A(–5, –16 ) 1 – 2 2 –5 –2 6 (a) Find the points such that the gradient of the tangent to the curve is (i) zero, (ii) positive, (iii) negative. (b) Determine the points such that (i) d2 y dx2 0 (ii) d2 y dx2 0 (c) Hence, state the methods for finding the maximum point or the minimum point. TIPS Nature dy dx Sketch of the graph d2 y dx2 Maximum point = 0 0 + – y = f(x) 0 Minimum point 0 – + y = f(x) 0 Point of inflection 0 0 + – + – y = f(x) y = f(x) = 0 Solution (a) (i) B(–2, 4) and C(2, –4) (ii) A(–5, –16 1 2 ) and D(6, 18) (iii) O(0, 0) 138 Ch 2_Grab Me SPM AddMaths F5.indd 138 13/05/2022 9:09 AM
CHAPTER 6 TRIGONOMETRIC FUNCTIONS Trigonometry 6.1 Positive Angles and Negative Angles • In trigonometry, (a) Angles measured in the anticlockwise direction from the positive x-axis is known as positive angles. (b) Angles measured in the clockwise direction from the positive x-axis is known as negative angle. • The position of angles in the Cartesian plane: Quadrant II Quadrant I Quadrant III 90° 270° 180° 0° 360° x y O Quadrant IV Example 1 Convert each of the following angles into radians. Hence, determine the quadrant for each of the angles and represent them in the same Cartesian plane. (a) 123° (b) 251° (b) –35° Solution (a) 123° × p 180° = 2.147 rad 123° lies in quadrant II. (b) 251° × p 180° = 4.381 rad 251° lies in quadrant III. (c) –35° × p 180° = –0.6109 rad –35° lies in quadrant IV. 187 Ch 6_Grab Me SPM AddMaths F5.indd 187 13/05/2022 9:21 AM
188 x y O A B C 1 –1 –1 1 251° 123° –35° 6.2 Trigonometric Ratios of Any Angle • Relation between six trigonometric ratios: sin q = y r cosec q = r y cos q = x r sec q = r x r x y θ tan q = y x cot q = x y • From the relation, it is found that: (a) cosec q = 1 sin q (b) sec q = 1 cos q (c) cot q = 1 tan q • Signs for trigonometric ratios in the quadrants: x y O Quadrant I All (+) Quadrant II sin (+) cosec (+) Quadrant IV cos (+) sec (+) Quadrant III tan (+) cot (+) • Complementary angle formulae: (a) sin q = cos (90° – q) (b) sec q = cosec (90° – q) (c) cos q = sin (90° – q) (d) cosec q = sec (90° – q) (e) tan q = cot (90° – q) (f) cot q = tan (90° – q) 188 Ch 6_Grab Me SPM AddMaths F5.indd 188 13/05/2022 9:21 AM
Example 8 The diagram below shows the graph of y = m cos px – 1 for 0 x p. 1 –� 4 1 –� 2 3 –� 4 � 1 2 –1 –3 –4 –2 y y = m cos px – 1 x 0 (a) State the values of m and p. (b) Find the number of solutions for each of the following. (i) y = –4 (ii) m cos px = –2 Solution (a) When x = 0 and y = 2, m cos 0 – 1 = 2 m = 3 When x = p and y = 2, 3 cos pp – 1 = 2 cos pp = 1 pp = 2p p = 2 (b) (i) m cos px – 1 = –4 Draw a graph of y = –4 on the graph. There is only 1 point of intersection, therefore the number of solution is 1. (ii) m cos px = –2 m cos px – 1 = –2 – 1 m cos px – 1 = –3 Draw a graph of y = –3 on the graph. There are 2 points of intersection, therefore the number of solutions is 2. Example 9 The diagram below shows the graph of y = m sin nx for 0 x 2p. y y = m sin nx x 0 � 2� –3 3 State the values of m dan n. 195 Ch 6_Grab Me SPM AddMaths F5.indd 195 13/05/2022 9:22 AM