Must Know iii – viii Chapter 1 Circular Measure 1 – 11 NOTES 1 Paper 1 1 Paper 2 8 Chapter 2 Differentiation 12 – 21 NOTES 12 Paper 1 13 Paper 2 18 Chapter 3 Integration 22 – 33 NOTES 22 Paper 1 24 Paper 2 30 Chapter 4 Permutation and Combination 34 – 42 NOTES 34 Paper 1 35 Paper 2 41 Chapter 5 Probability Distribution 43 – 54 NOTES 43 Paper 1 44 Paper 2 51 Chapter 6 Trigonometric Functions 55 – 65 NOTES 55 Paper 1 58 Paper 2 63 Chapter 7 Linear Programming 66 – 72 NOTES 66 Paper 1 67 Paper 2 69 Chapter 8 Kinematics of Linear Motion 73 – 81 NOTES 73 Paper 1 74 Paper 2 78 SPM Assessment 82 – 97 Answers 98 – 136 ii Contents 00A_1202 QB AMath F5.indd 2 24/11/2021 4:28 PM
Important Facts (Chapter 1) 1 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 7 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 1) 3 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 9 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 5 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 11 @ Pan Asia Publications Sdn. Bhd. Relationships between Angle in Degrees, Angle in Radians, Arc Length and Area of Sector Arc Length AB, Length of Chord AB, Area of Sector AOB and Area of the Shaded Segment Stationary Point Techniques of Differentiation Small Changes and Approximation of Certain Quantities Limits and Its Relation to Differentiation ——θ° 360° = θ rad –——2π = Arc length, s –——————–––— 2πr (Circumference) = Area of sector –————–—–—– πr2 (Area of circle) If y = f(x), then lim δx → 0 f(x + δx) – f(x) ——————— δx = lim δx → 0 δy —– δx = dy —– dx = f ʹ(x) where δx is a small change in x. • Arc length AB, s = rθ • Length of chord AB: ✤ AB2 = r2 + r2 – 2r2 cos θ° (Cosine rule) ✤ ———AB sin θ° = ———–———— r sin 1 ————– 180° – θ° 2 2 (Sine rule) • Area of sector AOB = —1 2 r2 θ • Area of the shaded segment = Area of sector AOB – Area of triangle AOB = —1 2 r2 θ – —1 2 r2 sin θ A point P(x, y) is a stationary point if dy —– dx = 0. The stationary point P(x, y) is • a maximum point if d2 y —–– dx2 , 0. • a minimum point if d 2 y —–– dx2 . 0. • a point of inflection if d 2 y —–– dx2 = 0. • If y = axn , then dy —– dx = anxn – 1 • If y = a, where a is a constant, then dy —– dx = 0 • If y = f(x) + g(x), then dy —– dx = f ʹ(x) + gʹ(x) • If y = g(u), where u = h(x), then dy —– dx = dy —– du × du —– dx • If y = uv, where u = f(x) and v = g(x), then dy —– dx = u dv —– dx + v du —– dx • If y = —u v , where u = f(x) and v = g(x), then dy —– dx = v du —– dx – u dv —– dx —————– v2 • If y = f(x) and δx is a small change in x, then δy = dy —– dx × δx • If y = f(x) and x = g(t), thus the rate of change of y is: dy —– dt = dy —– dx × dx —– dt • If x changes from x to x + δx, then: ✤ The percentage change in x = δx —–x × 100% ✤ The percentage change in y = δy —–y × 100% θ r s O A B θ r s O A B MUST KNOW Important Facts 00B_1202 QB AMath F5.indd 3 02/12/2021 8:24 PM
Common Mistakes (Chapter 2) 8 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 1) 2 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 2) 10 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 1) 4 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 3) 12 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 2) 6 @ Pan Asia Publications Sdn. Bhd. Quotient Rule Converting Radians into Degrees and Vice Versa Second Derivative Method Indefinite Integral Chain Rule Area of ∆AOB Given y = x + 1 —–—– 2x – 1 , find —–d dx 1 x + 1 —–—– 2x – 1 2. Correct Wrong Use dy —– dx = v du —– dx – u dv —– dx —————– v2 where u = x + 1 and v = 2x – 1. • Use dy —– dx = u dv —– dx – v du —– dx —————– v2 where u = x + 1 and v = 2x – 1. • Use dy —– dx = v du —– dx + u dv —– dx —————– v2 where u = x + 1 and v = 2x – 1. Correct Wrong Given a stationary point (a, b), then it is a maximum point if d 2 y —–– dx2 , 0. Given a stationary point (a, b), then it is a maximum point if d2 y —–– dx2 . 0. Given a stationary point (a, b), then it is a minimum point if d 2 y —–– dx2 . 0. Given a stationary point (a, b), then it is a minimum point if d2 y —–– dx2 , 0. Correct Wrong ∫ ———— 4 (3x + 1)2 dx = ∫ 4(3x + 1)–2 dx = 4(3x + 1)–2 + 1 —————– (–2 + 1)(3) + c = 4(3x + 1)–1 ————– –3 + c = – 4 ——–—– 3(3x + 1) + c ∫ ———— 4 (3x + 1)2 dx = ∫ 4(3x + 1)–2 dx = 4(3x + 1)–2 + 1 —————– –2 + 1 + c = 4(3x + 1)–1 ————– –1 + c = – 4 —––— 3x + 1 + c Correct Wrong d —– dx [3(2x2 – x)4 ] = 3(4)(2x2 – x) 4 – 1 d —– dx (2x2 – x) = 12(2x2 – x)3 (4x – 1) = 12(4x – 1)(2x2 – x)3 d —– dx [3(2x2 – x)4 ] = 3(4)(2x2 – x)4 – 1 = 12(2x2 – x)3 or d —– dx [3(2x2 – x)4 ] = d —– dx [(6x2 – 3x)4 ] Correct Wrong Must convert an angle in radians into degrees first. 1.45 rad = 1.45 × ——– 180° π = 83.08° Area of ΔAOB = —1 2 r2 sin θ = —1 2 r2 sin 83.08° = 0.4964r2 Angle in radians is not converted into degrees. Area of ΔAOB = —1 2 r2 sin θ = —1 2 r2 sin (1.45) = —1 2 r2 (0.0253) = 0.0127r2 • Converting radians into degrees For example: 1.35 rad Correct Wrong 1.35 × ——– 180° π 1.35 × ——– π 180° • Converting degrees into radians For example: 46° Correct Wrong 46° × ——– π 180° 46° × ——– 180° π One of the test to determine whether a stationary point is a maximum point or a minimum point. ∫ ———— 4 (3x + 1)2 dx = —–d dx (3x + 1) = 3 —–d dx (3x + 1) = 3 is missing d —– dx [3(2x2 – x)4 ] = 1.45 rad B r A O MUST KNOW Common Mistakes 00B_1202 QB AMath F5.indd 4 02/12/2021 8:24 PM
NOTES 1 1SOS TIP 1.1 Radian 1. One radian is the measure of an angle subtended at the centre O of a circle where the arc length, s is the same as the radius of the circle, r, that is, s = r = 1 rad. A B s 1 rad r O 2. 360° = 2π rad 1.2 Arc Length of a Circle 1. Given a circle with centre O and a radius of r units where ∠AOB = θ rad (or θ°) and the arc length AB is s units, then A θ B s r O (a) —–—θ° 360° = —–— θ rad 2π = —–————–——– Arc length, s 2πr (Circumference) (b) Arc length AB, s = rθ (c) Length of chord AB: (i) AB2 = r2 + r2 – 2r2 cos θ° (Cosine rule) (ii) ——– AB sin θ° = r ————––— sin ———–– 180° – θ° 2 2 (Sine rule) 1.3 Area of Sector of a Circle 1. Given a circle with centre O and a radius of r units where ∠AOB = θ rad (or θ°), then A θ B r O (a) —–—θ° 360° = —–— θ rad 2π = —–————–—–—–– Area of the sector πr2 (Area of the circle) (b) Area of sector AOB = —1 2 r2 θ (c) Area of the shaded segment = Area of the sector AOB – Area of triangle AOB = —1 2 r 2 θ – —1 2 r 2 sin θ° Section A Question 1: (a) Perimeter of a sector = r + r + Arc length (b) Use s = rθ 1. The perimeter of a sector AOB of a circle with centre O and an arc length of 5.4 cm is 23.4 cm. Find (a) the radius of the circle, [3 marks] (b) the angle AOB subtended at the centre of the circle. [2 marks] CLONE SPM Answer: (a) (b) Chapter 1 Circular Measure PAPER 1 01_1202 QB AMath F5.indd 1 02/12/2021 8:44 PM
22 2 SOS TIP 2. The area of a sector KOL with centre O and a radius of 11 cm is 160 cm2 . (a) Find ∠KOL, in radians. [2 marks] (b) If the sector KOL is folded to form a cone, find the base radius of the cone. [3 marks] Answer: (a) (b) 3. Given AOB is a sector of a circle with centre O and a radius of r cm where ∠AOB = θ rad. (a) Show that the length of the chord AB is 2r sin —θ 2 . [2 marks] (b) Given r = 5 cm and ∠AOB = 1.2 rad, find the difference in length between the arc length AB and the chord AB. [3 marks] Answer: (a) (b) 4. The diagram shows a paper fan consisting of two sectors, POQ and AOB. The shaded region is covered by paper. P A O B Q Question 2: (b) Find the arc length of the sector. When the sector is folded, the arc length of the sector is the circumference of the base of the cone. Question 5: Use the cosine rule, a2 = b2 + c2 – 2bc cos θ Given that OA : OP = 1 : 3, ∠POQ = —2 3 π rad and OA = 15 cm, find (a) the perimeter of the area covered by the paper, [3 marks] (b) the area of the paper used. [2 marks] Answer: (a) (b) 5. The diagram shows a sector AOB with centre O and a radius of 15 cm. A B C 15 cm O θ Given that C divides the line OB in the ratio 3 : 2 and the length of chord AB is 10 cm. Find (a) the angle θ, in radians, [2 marks] (b) the perimeter of the shaded region. [3 marks] Answer: (a) (b) CLONE SPM 01_1202 QB AMath F5.indd 2 02/12/2021 8:44 PM
5 5SOS TIP 13. The diagram shows a rhombus inscribed in a sector AOC with centre O and a radius of r. O β C B A r rad Given that ∠AOC = β rad and the area of the sector is 20 cm2 , express each of the following in terms of r. (a) The angle β. [3 marks] (b) The perimeter, in cm, of the shaded region. [2 marks] Answer: (a) (b) CLONE SPM 14. The diagram shows a circle with centre O. PT and QT are tangents to the circle at points P and Q respectively. P Q O r cm T θ Given that the minor arc length PQ is 5 cm and OT = — 6 θ cm, (a) express the radius, r of the circle in terms of θ, [1 mark] (b) find the area of the shaded region. [4 marks] Answer: (a) (b) CLONE SPM Question 14: Tangent PT = Tangent QT and ∠TPO = 90° Question 15: (b) Area of the segment = Area of the sector AOB – Area of the triangle AOB Section B 15. The diagram shows a circular cross-section of a container with centre O resting on two supports, each of height 25 cm. HOTS Applying O A B 25 cm 10 cm 32 cm The shortest distance of the container to the horizontal surface is 10 cm. Given that the radius of the container is 32 cm. Find (a) the angle AOB, in radians, [4 marks] (b) the area between the supports and the container. [4 marks] Answer: (a) (b) 01_1202 QB AMath F5.indd 5 02/12/2021 8:44 PM
66 6 SOS TIP 16. Two connecting gears are rotating simultaneoulsy. The smaller gear has a radius of 5 cm whereas the larger gear’s radius is 9 cm. (a) What is the angle subtended by the larger gear if the smaller gear has made one complete rotation? [4 marks] (b) How many rotations will the smaller gear make if the larger gear makes one complete rotation? [4 marks] Answer: (a) (b) 17. A sector of a circle with angle θ and a radius of r cm has an area of 5 cm2 and its perimeter is 9 cm. Find the possible values of r and θ. [8 marks] Answer: 18. The diagram shows the cross-section of a tunnel with its width AC is the diameter of a semicircle ABC. The shaded region is the wall of the tunnel and is fill up with cement. The cross-section of the tunnel is in the form of a sector with centre P and a radius of r m . Question 17: Relate the arc length to the area of the sector. C B A P 2θ rad r m Given that ∠APC = 2θ rad, express the width AC of the tunnel and the area of the wall filled up by the cement, in terms of r and θ respectively. HOTS Analysing [8 marks] Answer: 19. The length of a chord PQ of a circle with centre O and a radius of 3 cm is 4.4 cm. Calculate (a) the angle subtended at the centre of the circle, in radians, [4 marks] (b) the area of the segment enclosed by the chord PQ and the arc PQ. [4 marks] Answer: (a) (b) 01_1202 QB AMath F5.indd 6 02/12/2021 8:44 PM
88 8 SOS TIP Question 3: The diameter of the small circle is the same as the radius of the semicircle and AB is the tangent to the small circle. Question 6: (b) Cross-sectional area in the water = Cross-sectional area of a circle – Area of segment above the surface of water 1. The diagram shows a circle with centre O and a radius of 25 cm. A B 25 cm O θ α A sector with angle θ = 1.2 rad at the centre is removed. Then, the end A is joined to B to form a cone. Calculate (a) the base radius, in cm, of the cone, [3 marks] (b) the angle α, in degrees. [4 marks] 2. Point A is a fixed point on the circumference of a circle with centre O and a radius of 10 cm. Point P moves along the circumference of the circle at a speed of 3 cm per second. Given the angle AOP is θ rad, find (a) the rate of change of θ, in radians per second, [3 marks] (b) the rate of change of the area of sector AOP. [3 marks] 3. The diagram shows one of the pattern of a mural drawn by a pupil on the wall of a school canteen. C D O A B 10 cm Given AB = 10 cm is a chord of the major sector ACB with centre O and a radius of 15 cm. AB is also the diameter of a semicircle ADB. A small circle with centre O inscribed in the semicircle. Calculate (a) the arc length ACB, [4 marks] (b) the area of the shaded region. [4 marks] CLONE SPM CLONE SPM 4. The diagram shows a trapezium ABCD where AB is parallel to DC. HOTS Applying A D B C E 5 cm 12 cm 6 cm rad —π 6 rad —π 3 Given AB = 5 cm, DC = 12 cm, ∠ADC = —π 6 rad, ∠BCD = — π 3 rad and E is the midpoint of BC. Calculate (a) the perimeter of the shaded region, [4 marks] (b) the area of the shaded region. [4 marks] 5. The diagram shows a circle with centre O and a radius of 6 cm and a rectangle ABCO with an area of 48 cm2 . A 6 cm O D C E B Calculate (a) ∠AOB, in radians, [3 marks] (b) the area of the sector EOD. [4 marks] 6. A ball is floating on the surface of water such that the highest point of the ball from the surface of the water is half of the radius, r cm, of the ball. HOTS Applying P Q r cm O (a) Find the length of the chord PQ, in terms of r. [3 marks] (b) Find the area, in cm2 , of the cross-section of the ball below the water if r = 25 cm. [4 marks] Section A PAPER 2 01_1202 QB AMath F5.indd 8 02/12/2021 8:44 PM
1010 10SOS TIP Section B 13. Ani wants to make a cone-shaped cap for her children’s party. The height of the cone is 25 cm and the base radius of the cone is 8.5 cm. She takes a piece of cardboard measuring 27 cm × 35 cm to cut out the net of the cone which is a sector of a circle. HOTS Analysing (a) What is the angle, in radians, subtended at the centre of the circle by the arc length of the sector of the net? [4 marks] (b) Based on the calculation in (a), determine whether the cardboard is big enough to make a cone-shaped cap. [6 marks] 14. The diagram shows a semicircle PORQS with centre O and a radius of 9 cm. HOTS Applying θ P 9 cm O M R Q S RPS is an inscribed sector in the semicircle with centre P. A perpendicular line from S to PQ divides the radius of the semicircle into half at M. Find (a) the angle θ, in radians, [3 marks] (b) the arc length RS, [3 marks] (c) the area of the shaded region. [4 marks] 15. The diagram shows the sector AOB with centre O, a radius OB = 8 cm and ∠AOB = — π 3 rad. C A B D E 8 cm P O rad —π 3 OC is the bisector of ∠AOB and P is the midpoint of OC. An arc DCE of a circle is drawn with centre P to meet OA and OB at D and E respectively. Find (a) the angle OPD, in radians, [3 marks] (b) the area of the shaded region. [7 marks] CLONE SPM Question 13: The radius of the sector of the net is the inclined side of the cone. Sketch the net of the cone as well as the cone before calculating the length and width of the cardboard. Question 16: Area of ABCD = Length of AD × Height of B to AD 16. The diagram shows a rhombus ABCD with sides of x cm and ∠A = θ rad. HOTS Analysing B x cm C D A θ rad Four arcs each with radius —x 3 cm are drawn with centres A, B, C and D respectively. Given the shaded area is half of the area of the rhombus, (a) show that sin θ = — 2 9 π, [5 marks] (b) find two possible values of θ. [5 marks] 17. The diagram shows two sectors with centre O. L M 1.5 rad 2 cm N P O Given that ∠LOP = 1.5 rad, LM = NP = 2 cm and the area of the sector LOP is 20.75 cm2 , find (a) the radius OM, [4 marks] (b) the arc length LP, [2 marks] (c) the area of the shaded region. [4 marks] 18. The diagram shows a sector of a circle with centre A. HOTS Analysing y B O x D C A(4, 0) 3y + x = 9 The equation of BD is 3y + x = 9. Find (a) the radius of the sector ABCD, [4 marks] (b) the angle BAD, in radians, [2 marks] (c) the area of the shaded region. [4 marks] 01_1202 QB AMath F5.indd 10 02/12/2021 8:44 PM
82 PAPER 1 Time: 2 hours Section A [64 marks] Answer all questions. 1. (a) Diagram 1 on the answer space shows a part of the graph of a function y = f(x) for the domain 0 < x < 3. On the same axes, sketch the corresponding part of the graph y1 = f –1(x) and state its domain. [2 marks] (b) The function F maps (x, y) onto (x – y, x + 2y) and A is the point (2, 3). F maps A onto B and B onto C. Find the coordinates of points B and C. [2 marks] (c) If f : x → 3 – 4x, find f –1 (–3). [1 mark] Answer: (a) y x 0 1 3 y = f(x) 2 1 2 3 4 Diagram 1 (b) (c) 2. (a) Given that the range of y = f(x) + 1 is –2 < y < 3, find the range of f(x). [1 mark] (b) Given that f : x → x + 2 and gf : x → x2 + 4x + 2, find (i) g(2), (ii) the values of x if fg(x) = 9. [4 marks] Answer: (a) (b) (i) (ii) SPM Assessment 09_1202 QB AMath F5.indd 82 04/12/2021 9:25 AM
87 Section B [16 marks] Answer any two questions from this section. 13. (a) A rectangular container without a lid is made up of thin aluminium sheet. The sides of the base are 2x cm and 3x cm and the height is h cm. If the total surface area is 200 cm2 , (i) show that h = —– 20 x – —– 3x 5 , [2 marks] (ii) find the dimensions of the container such that the volume is maximum, [2 marks] (iii) hence, find the maximum volume of the container. [1 mark] (b) If water drips into the container in (a) at a constant rate of 21 cm3 s–1, find the rate of change of the height of water in the container when h = 1 cm. [3 marks] Answer: (a) (i) (ii) (iii) (b) 09_1202 QB AMath F5.indd 87 04/12/2021 9:25 AM
90 PAPER 2 Time: 2 hours 30 minutes Section A [50 marks] Answer all questions. 1. The sum of the digits of three-digit number is 16. The unit digit is 2 more than the sum of the other two digits. The tens digit is 5 more than the hundreds digit. What is the number? [7 marks] 2. (a) Diagram 1 shows the net of an open box. x cm x cm 5 cm (15 – 2x) cm Diagram 1 (i) Show that the volume of the box, in cm3 , is given by V = 75x – 10x2 . [2 marks] (ii) Hence, find the value of x, in cm, such that the volume of the box is maximum. State the maximum volume of the box. [3 marks] (b) Find the possible range of values of x if the volume of the box is between 90 cm3 and 125 cm3 . [2 marks] 3. In Diagram 2, A, B and C are the vertices of a right-angled triangle. C x cm α G E B F D H A Diagram 2 Given that ∠ACB = α and BC = x cm. (a) Express BD and DE in terms of x and α. [2 marks] (b) Show that BD, DE and EF form the first three terms of a geometric progression and state the common ratio. [3 marks] (c) Find the length of HG in terms of x and α. [1 mark] (d) Given that x = 8 cm and α = 60°, find the sum to infinity of the geometric progression. [2 marks] 09_1202 QB AMath F5.indd 90 04/12/2021 9:25 AM
93 Section B [30 marks] Answer any three questions from this section. 8. (a) (i) Find lim x → 3 3 + 2x – x2 ——–—— x – 3 . [1 mark] (ii) Given that x = t + t 2 and y = 2t + 1, where t . 0, find dy —– dx in terms of y and hence, find the approximate change in y if x decreases from 2 to 1.98 when t = 1. [3 marks] (b) Diagram 4 shows a curve y = x(x – 3)2 intersects with the straight line y = 4x at O, A and B. y A O B y = x(x – 3)2 y = 4x x Diagram 4 (i) Find the coordinates of A and B. [2 marks] (ii) Calculate the area of the shaded region. [4 marks] 9. In Diagram 5, O →P = 2x ~, O →Q = 3y ~ and Q →R = x ~ – y ~ . The lines PQ and OR intersect at X. Q O R P X Diagram 5 Given P →X = hP →Q and O →X = kO →R, (a) Express P →X in terms of h, x ~ and y ~ . [2 marks] (b) Show that O →X = 2(1 – h)x ~ + 3hy ~ . [2 marks] (c) Find the values of h and k. [3 marks] (d) If the area of the triangle QOX is 24 units2 , find the area of the triangle XOP. [3 marks] 09_1202 QB AMath F5.indd 93 04/12/2021 9:25 AM
95 Section C [20 marks] Answer any two questions from this section. 12. (a) In Diagram 7, ABC is a triangle where AB = 13 cm, AC = 8.5 cm and ∠ABC = 38°. B 13 cm 8.5 cm A C Diagram 7 Calculate (i) the angle BAC where ACB is an obtuse angle, (ii) the length of BC. [4 marks] (b) A new triangle is formed such that AB, AC and the size of the angle ABC remain unchanged. (i) Sketch the new triangle. [1 mark] (ii) Calculate the area of the new triangle. [3 marks] (iii) Hence, find the shortest distance from A to BC. [2 marks] 13. Table 2 shows the price indices, the changes in price index and the quantities required for the four ingredients, A, B, C and D used to make a product. Ingredient Price index for the year 2019 based on the year 2015 Changes in price index from the year 2019 to 2020 Quantity (g) A 125 Increased by 20% 500 B 120 Decreased by 15% 200 C 90 No change 200 D 150 Increased by 10% 100 Table 2 (a) Calculate (i) the price of 1 kg of ingredient A in the year 2015 if the price in the year 2019 is RM15, (ii) the price index of ingredient B in the year 2020 based on the year 2015. [4 marks] (b) Calculate the composite index for the cost of the product in the year 2020 based on the year 2019. [2 marks] (c) If the cost of the product in the year 2019 is RM50, find the corresponding cost in the year 2020. [2 marks] (d) The composite index for the cost of making the product from the year 2020 to the year 2021 is expected to increase at the same rate as the rate of increase from the year 2015 to the year 2019. Find the expected composite index for the cost of making the product for the year 2021 based on the year 2015. [2 marks] 09_1202 QB AMath F5.indd 95 04/12/2021 9:25 AM
98 Answers 10_1202 QB AMath F5.indd 98 10/01/2022 4:49 PM
99 CHAPTER 1 Paper 1 Section A 1. rθ r r A O B θ P = 2r + rθ = 23.4 (a) 2r = 23.4 – 5.4 r = 9 cm (b) 9θ = 5.4 θ = 0.6 rad ∴ ∠AOB = 0.6 rad 2. K L O 11 cm θ (a) —1 2 (11)2 θ = 160 θ = 2.64 rad ∴ ∠KOL = 2.64 rad (b) r Circumference, KL = 11(2.64) = 29.04 cm 2πr = 29.04 r = 4.62 cm 3. r cm O B θ rad A (a) sin —θ 2 = ——AB 2r 2r sin —θ 2 = AB (b) ∠AOB = 1.2 rad = 68.75° Difference = Arc length AB – Length of the chord AB = 5(1.2) – 2(5) sin 34.38° = 0.353 cm 4. P A B Q O 30 cm 15 cm π rad 2 —3 (a) Perimeter = 15—2 3 π2 + 2(30) + 45—2 3 π2 = 185.66 cm (b) Area of the paper = —1 2 (45)2 1—2 3 π2 – —1 2 (15)2 1—2 3 π2 = 600π cm2 5. 15 cm 9 cm A B C O θ (a) ——OC OB = —3 5 OC = —3 5 × 15 = 9 cm AB2 = 152 + 152 – 2(15)2 cos θ cos θ = 152 + 152 – 102 ——————– 2(15)2 θ = 38° 57ʹ θ = 0.68 rad (b) AC2 = 152 + 92 – 2(15)(9) cos 38° 57ʹ AC = 9.8 cm Perimeter of the shaded region = 6 + 15(0.68) + 9.8 = 26 cm 6. A B C O 12 cm (a) Area of ΔAOB= 72 —1 2 (12)AB= 72 AB= 12 cm ∠AOB = 45° = —π 4 rad (b) Area of minor sector AOC = —1 2 (12)2 1 —π 4 2 = 18π cm2 7. 12 1 2 3 4 5 6 7 8 9 10 11 P R Q O ∠POQ = —1 3 × 360° = 120° ∠ROP = —2 3 × 30° + 30° = 50° Total angle = 170° = 2.97 rad 8. L 41.44° R O 18 cm (a) Arc length = 100 cm, r = 18 cm rθ = 100 θ = ——100 18 θ = 5.56 rad θ = 318.56° (b) OL = 18 cos 41.44° Height above PQ = 18 – 18 cos 41.44° = 4.51 cm 9. 60 seconds = 33—1 3 revolutions 1 second = ——100 3 × —–1 60 = —5 9 revolution (a) —5 9 × 2π = —– 10 9 π radians per second (b) r = 20 cm Speed = —– 10 9 π × 20 = 69.81 cm s–1 = 0.698 m s–1 10. 50 cm 24 cm C A B D O 5 —9 π rad (a) Area traversed by the wiper = —1 2 (74)2 1 —5 9 π2 – —1 2 (24)2 1 —5 9 π2 = 4 276.06 cm2 (b) Perimeter of the area traversed by the wiper = 241 —5 9 π2 + 2(50) + 741 —5 9 π2 = 271.04 cm 11. (a) Perimeter of the shaded region = 101 —2 5 π2 + 10 sin 72° + (10 – 10 cos 72°) = 28.99 cm (b) Area of the shaded region = —1 2 (10)2 1 —2 5 π2 – —1 2 (10 cos 72°)(10 sin 72°) = 48.14 cm2 12. T U r rad π –– 6 O (a) Area = 20 cm2 20 = —1 2 r2 1 —π 6 2 – —1 2 r2 sin 30° 20 = r2 1 —– π 12 – ———– sin 30° 2 2 r2 = ——————– 20 —– π 12 – ———– sin 30° 2 r = 41.17 cm (b) Length of the chord UT = 2(41.17) sin 15° = 21.31 cm Perimeter of the segment = 21.31 + 41.171 —π 6 2 = 42.87 cm 10_1202 QB AMath F5.indd 99 10/01/2022 4:49 PM
100 13. O β β C B A r r (a) 20 = —1 2 r2 β β = —– 40 r2 (b) Perimeter of the shaded region = 2r + rβ = 2r + r1 —– 40 r2 2 = 12r + —– 40 r 2 cm 14. P Q T O r —θ 2 (a) rθ = 5 r = —5 θ cm (b) OT = —6 θ cos —θ 2 = —5 θ ÷ —6 θ = —5 θ × —θ 6 = —5 6 θ = 67.11° θ = 1.17 rad Area of ∆POT = —1 2 r1 —6 θ 2 sin 33.56° = —1 2 1 —5 θ 21 —6 θ 2 sin 33.56° = —1 2 1 ——5 1.17 21 ——6 1.17 2 sin 33.56° = 6.06 cm2 Area of the shaded region = 2(6.06) – —1 2 r2 θ = 2(6.06) – —1 2 1 52 ——1.17 2 = 1.44 cm2 Section B 15. A B O D 25 cm 32 cm 10 cm (a) OD = 32 + 10 – 25 = 17 cm 32 cos ——— ∠AOB 2 = 17 ——— ∠AOB 2 = 57.91° ∠AOB = 115.82° ∠AOB = 2.02 rad (b) Area of the segment = 1 —2 (32)2 (2.02) – 1 —2 (32)2 sin 115.82° = 573.35 cm2 Area of rectangle AOB = 25(32 sin 57.91°) = 677.77 cm2 Area required = 677.77 – 573.35 = 104.42 cm2 16. 5 cm 9 cm O A (a) Circumference of the small gear = 2π(5) = 10π cm 10π = 9θ θ = —– 10 9 π rad (b) Circumference of the big gear = 2π(9) = 18π cm Number of rotations = ——18π 10π = 1.8 rotations 17. θ r cm —1 2 r2 θ = 5 .................1 2r + rθ = 9 .................2 rθ = 9 – 2r..........3 From 1, —1 2 r(9 – 2r) = 5 9r – 2r2 = 10 2r2 – 9r + 10 = 0 (2r – 5)(r – 2) = 0 r = —5 2 or r = 2 —1 2 1 —5 2 2 2 θ = 5 or —1 2 (2)2 θ = 5 θ = —8 5 rad θ = —5 2 rad 18. AC = 2r sin θ Area = —π 2 (r sin θ)2 – 3 —1 2 r2 (2θ) – —1 2 r2 sin 2θ4 = —1 2 r2 (π sin2 θ – 2θ + sin 2θ) 19. θ 3 cm 4.4 cm O P Q (a) sin —θ 2 = ——2.2 3 —θ 2 = 47° 10’ θ = 94.33° θ = 1.65 rad (b) Area of the segment = —1 2 (3)2 (1.65) – —1 2 (3)2 sin 94.33° = 2.94 cm2 20. B A P Q O r 4 cm 5.6 cm 4 cm θ (a) rθ = 4 (r + 4)θ = 5.6 4 + 4θ = 5.6 4θ = 1.6 θ = 0.4 rad r = OP = ——4 0.4 = 10 cm (b) ∠POQ = 0.4 rad (c) Area of ABQP = —1 2 (14)2 (0.4) – —1 2 (10)2 (0.4) = 19.2 cm2 21. P Q S R T O 6 cm 6 cm 6 cm 9 cm (a) sin ——— ∠SOR 2 = —– 4.5 6 ——— ∠SOR 2 = 48.59° ∠SOR = 97.18° ∠SOR = 1.7 rad (b) Area of the segment STR = —1 2 (6)2 (1.7) – —1 2 (6)2 sin 97.18° = 12.74 cm2 22. M P Q N 6 cm O 120° 60° 60° rad —π – 3 (a) Arc length PQ = π(6) – Arc length MP – Arc length QN = 6π – 61—π 3 2 – 61—π 3 2 = 2π cm (b) Perimeter of the shaded region = 12 + 61—π 3 2 + 61—π 3 2 + 2(6 cos 60°) = 12 + 4π + 121 —1 2 2 = 18 + 4π = 30.57 cm Paper 2 Section A 1. (a) Circumference = 2πr (2π – 1.2)(25) = 2πr r = (2π – 1.2)(25) —————— 2π r = 2.02 cm 10_1202 QB AMath F5.indd 100 10/01/2022 4:49 PM
101 (b) 25 cm α r sin —α 2 = ——– 2.02 25 —α 2 = 4° 38ʹ α = 9° 16ʹ 2. A P O 10 cm θ rad (a) 10θ = 3 θ = —–3 10 θ = 0.3 rad s–1 (b) Area of sector AOP = —1 2 r2 θ = 50θ = 501 —–3 10 2 = 15 cm2 s–1 3. sin θ = —–5 15 θ = 19° 28ʹ ∠AOB = 360° – 2(19° 28ʹ) = 321.07° = 5.6 rad C D O A B θ 2.5 cm 15 cm 5 cm 5 cm (a) Arc length ACB = 15(5.6) = 84 cm (b) Area of the segment = ———– 38° 56ʹ 360° × π(15)2 – —1 2 (15)2 sin 38° 56’ = 5.75 cm2 Area of the shaded region = π(15)2 – 5.75 – —π 2 (5)2 + π(2.5)2 = 681.47 cm2 4. (a) h = 6 sin 30° = 3 cm sin 60° = —–3 BC BC = 3.46 cm BE = 1.73 cm A h D B C E 5 cm 12 cm 6 cm 1.73 cm 3 cm rad —π – 6 rad — π – 3 Perimeter of the shaded region = 61—π 6 2 + 5 + 1.73 + 1.731—π 3 2 + (6 – 1.73) = 15.95 cm (b) Area of the shaded region = —1 2 (5 + 12)(3) – —1 2 (6)2 1—π 6 2 – —1 2 (1.73)2 1—π 3 2 = 14.51 cm2 5. 6 cm A O D C E B (a) Area = 48 cm2 6(AB) = 48 AB = 8 cm tan ∠AOB = —8 6 ∠AOB = 53.13° ∠AOB = 0.93 rad (b) ∠EOD = 90° – 53.13° = 36.87° Area of the sector EOD = ——— 36.87° 360° × π(6)2 = 11.58 cm2 6. — 1 2 r cm — 2 3 π rad r cm P Q O (a) Length of the chord PQ = 2ABBBBBB r2 – —1 4 r2 = 2ABBB —3 4 r2 = 2AB3 ——2 r = AB3r cm (b) Area of the circle = πr2 Area under water = πr2 – Area of the segment = πr2 – 3—1 2 r2 θ – —1 2 r2 sin 120°4 = πr2 – 3—1 2 r2 1—2 3 π2 – —1 2 r2 1 AB3 —–2 24 = 2πr2 —–—3 + AB3 —–4 r2 If r = 25 cm, Area under water = 1 579.63 cm2 7. h cm O T Q P rad rad A θ 2θ (a) (i) OP = 2h cos θ (ii) TQ = 2h cos θ – h (b) Perimeter of the shaded region = (2h cos θ – h)(2θ) + (2h cos θ)θ + (h – (2h cosθ – h)) = 4θh cos θ – 2θh + 2θh cos θ + 2h – 2h cos θ = 6θh cos θ – 2θh + 2h – 2h cos θ = h(6θ cos θ – 2θ + 2 – 2 cos θ) If θ = —π 6 , Perimeter of the shaded region = h361—π 6 2 cos 30° – 21—π 6 2 + 2 – 2 cos 30°4 = h(1.94) = 1.94h cm 8. 8 cm —π – 6 rad N r M Q P O 3 cm (a) r1—π 6 2 = 3 r = —– 18 π cm OM1—π 6 2 = 8 OM = —– 48 π OM = 15.28 cm NM = —– 48 π – —– 18 π = 9.55 cm (b) MQ = 2(OM sin 15°) = 2(15.38) sin 15° = 29.52 cm Perimeter of the shaded region = 29.52 + 9.55 × 2 + 3 = 51.62 cm (c) Area of MOQ = —1 2 (15.28)2 sin 30° = 58.37 cm2 Area of the shaded region = 58.37 – —1 2 1—– 18 π 2 2 1—π 6 2 = 49.78 cm2 9. 60° 60° B a a A O P a (a) cos 30° = —1 2 a ——OP OP1 AB3 —–2 2 = —1 2 a OP = —–a AB3 OP = OA Arc length AB with centre O = 1 —–a AB3 21—– 2π 3 2 = 2πa –—–– 3AB3 × AB3 ––– AB3 = —–—– 2AB3πa 9 10_1202 QB AMath F5.indd 101 10/01/2022 4:49 PM
132 = 3 —– 75 2 + 75 – —–– 125 2 4 – 0 = 50 m 25. (a) v = 2t(10 – 3t) v = 20t – 6t 2 a = —– dv dt a = 20 – 12t For maximum velocity, a = 0 20 – 12t = 0 12t = 20 t = —5 3 When t = —5 3 , v = 21 —5 3 2110 – 31 —5 3 22 v = —– 50 3 Therefore, the maximum velocity of the particle is —– 50 3 m s–1. (b) s = ∫ v dt s = ∫ (20t – 6t 2 ) dt s = 10t 2 – 2t 3 + c When t = 0, s = 0 10(0)2 – 2(0)3 + c = 0 c = 0 Hence, s = 10t 2 – 2t 3 . When t = 2, s2 = 10(2)2 – 2(2)3 s2 = 24 When t = 3, s3 = 10(3)2 – 2(3)3 s3 = 36 Total distance travelled = s3 – s2 = 36 – 24 = 12 m (c) When the particle passes through point P, s = 0 10t 2 – 2t 3 = 0 2t 2 (5 – t) = 0 t = 0 or t = 5 Therefore, t = 5 s. (d) When the particle reverses its direction of motion, v = 0 2t(10 – 3t) = 0 t = 0 or t = —– 10 3 Therefore, t = —– 10 3 s. SPM ASSESSMENT Paper 1 Section A 1. (a) 0 < x < 2 y x 0 1 3 y = f(x) y = x y = f –1(x) 2 1 2 3 4 (b) f : (x, y) → (x – y, x + 2y) f : (2, 3) → (–1, 8) f : (–1, 8) → (–9, 15) ∴ B(–1, 8), C(–9, 15) (c) f(x) = 3 – 4x Let y = 3 – 4x = –3 6 = 4x x = —3 2 ∴ f –1(–3) = —3 2 2. (a) y = f(x) + 1 –2 < y < 3 –2 < f(x) + 1 < 3 –3 < f(x) < 2 (b) f(x) = x + 2 gf(x) = x2 + 4x + 2 gf(x) = g(x + 2) Let y = x + 2 x = y – 2 g(y) = (y – 2)2 + 4(y – 2) + 2 = y2 – 4y + 4 + 4y – 8 + 2 = y2 – 2 (i) g(x) = x2 – 2 g(2) = 22 – 2 = 2 (ii) fg(x) = 9 f(x2 – 2) = 9 x2 – 2 + 2 = 9 x2 = 9 x = ±3 3. (a) x2 + 3 = t(x + 1) x2 – tx + 3 – t = 0 b2 – 4ac , 0 (–t)2 – 4(1)(3 – t) , 0 t 2 + 4t – 12 , 0 (t – 2)(t + 6) , 0 –6 , t , 2 (b) y2 = 2x y = mx + c (mx + c)2 = 2x m2 x2 + 2mcx + c2 – 2x = 0 m2 x2 + (2mc – 2)x + c2 = 0 b2 – 4ac = 0 (2mc – 2)2 – 4m2 c2 = 0 4m2 c2 – 8mc + 4 – 4m2 c2 = 0 –8mc = –4 m = —–– –4 –8c m = —–1 2c 4. (a) h(t) = 20t – 5t 2 = 5t(4 – t) h(t) 20 h(t) = 20t – 5t 2 0 t 4 When t = 2, h(2) = 20(2) – 5(2)2 = 20 0 < h(t) < 20 (b) (α + 1) + (β + 1) = —5 2 α + β = —5 2 – 2 = —1 2 (α + 1)(β + 1) = –1 αβ + (α + β) + 1 = –1 αβ = –1 – 1 – —1 2 = – —5 2 If —1 α and —1 β are roots, —1 α + —1 β = α + β ——–– αβ = —1 2 —— – —5 2 = – —1 5 —–1 αβ = ——– 1 – —5 2 = – —2 5 Therefore, the equation is x2 + —1 5 x – —2 5 = 0 5x2 + x – 2 = 0 5. (a) Midpoint AC = (3, 1) A(1, 4) y O C(5, –2) B(–3, 0) D(h, k) x ——– h – 3 2 = 3 h = 9 ——– k + 0 2 = 1 k = 2 ∴ h = 9, k = 2 (b) mPC = 2 – (–2) ———– 9 – 5 = —4 4 = 1 Gradient to DC = –1 Therefore, the equation is y – 4 = –(x – 1) y = –x + 5 6. (a) 22x + 2 + 1 = 5(2x ) 22x ⋅ 22 + 1 – 5(2x ) = 0 22 ⋅ 22x – 5(2x ) + 1 = 0 4(22x ) – 5(2x ) + 1 = 0 (4(2x ) – 1)(2x – 1) = 0 4(2x ) = 1 or 2x = 20 2x = 2–2 x = 0 x = –2 (b) (i) log4 ABBxy = —1 2 1 log2 xy ——–– log2 4 2 = —1 4 (log2 x + log2 y) = —1 4 (p + q) (ii) log8 16y ——x2 = log2 16y ——x2 ———— 3 = —1 3 (log2 16 + log2 y – 2 log2 x) = —1 3 (4 + q – 2p) 10_1202 QB AMath F5.indd 132 10/01/2022 4:49 PM
134 dh = 1– 20 —– x2 – —3 5 2 dx —– dV dh = —– dV dx × —– dx dh = 1120 – 54x2 ——5 2 1 ————–– 1 – 20 —– x2 – —3 5 2 When h = 1, —– 20 x – —– 3x 5 = 1 100 – 3x2 = 5x 3x2 + 5x – 100 = 0 (3x + 20)(x – 5) = 0 x = 5 —– dV dh = 1120 – 54(5)2 —–––– 5 2 1 ————–– 1 – 20 —– 52 – —3 5 2 = –1501– —5 7 2 = 107.14 —– dV dt = 107.141—– dh dt 2 Given —– dV dt = 21 cm3 s–1 21 = 107.141—– dh dt 2 —– dh dt = 0.196 cm s–1 14. (a) BC = 10 sin θ AB = 10 cos θ (i) Area, A = —1 2 (10) sin θ (10 cos θ) = 50 sin θ cos θ = (25 sin 2θ) cm2 (ii) 25 sin 2θ = 25AB3 ——– 2 sin 2θ = AB3 —–2 2 1 2θ fiff3 2θ = 60°, 120° θ = 30°, 60° θ = —π 6 rad, —π 3 rad (b) 2x2 – x – 3 = 0 (2x – 3)(x + 1) = 0 x = —3 2 or x = –1 tan A = —3 2 , tan B = –1 Sum of roots: tan A + tan B = —1 2 Product of roots: tan A tan B = – —3 2 (i) tan (A + B) = tan A + tan B —————— 1 – tan A tan B = —1 2 ————– 1 – 1– —3 2 2 = —1 2 —— —5 2 = —1 5 (ii) tan 2A = ———— 2 tan A 1 – tan2 A = 21 —3 2 2 ———— 1 – 1 —3 2 2 2 = ———3 1 – —9 4 = – —– 12 5 15. P(spoilt) = p —– 10 , n = 5 (a) P(X = 0) = 0.1681 5 C0 1 p —– 10 2 0 1 q —– 10 2 5 = 0.1681 q —– 10 = 0.7 q = 7 Therefore, p = 3 (b) 7 good, 3 spoilt If 3 spoilt bulbs are arranged together, then number of ways = —– 8! 7! = 8 (c) p + q + 0.2 + 2p + q = 1 3p + 2q = 0.8 ...........1 P(X < 2) = p + q = 0.55 – 0.2.....2 2p + 2q = 0.7................3 1 – 3: p = 0.1 q = 0.35 – 0.1 = 0.25 Paper 2 Section A 1. x y z x + y + z = 16 .................................1 z – 2 = x + y x + y – z = –2 .................................2 y – 5 = x y – x = 5 ...................................3 1 + 2: 2x + 2y = 14 x + y = 7 ......................4 –x + y = 5.......................................5 4 + 5: 2y = 12 y = 6 x = 1 1 + 6 + z = 16 z = 9 Therefore, the number is 169. 2. (a) (i) V = 5(15 – 2x)(x) V = 75x – 10x2 (ii) —– dV dx = 75 – 20x = 0 x = —– 75 20 x = —– 15 4 When x = —– 15 4 , V = 751 —– 15 4 2 – 101 —– 15 4 2 2 V = 140—5 8 cm3 (b) 90 , 75x – 10x2 , 140 10x2 – 75x + 90 , 0 2x2 – 15x + 18 , 0 (2x – 3)(x – 6) , 0 —3 2 < x , 6 75x – 10x2 , 125 10x2 – 75x + 125 . 0 2x2 – 15x + 25 . 0 (2x – 5)(x – 5) . 0 —3 2 , x , —5 2 and 5 , x , 6 3. (a) BD = x sin α DE = DB sin α = (x sin α)(sin α) = x sin2 α (b) EF = DE sin α = x sin3 α ∴ x sin α, x sin2 α, x sin3 α r = x sin2 ———–α x sin α = sin α r = x sin3 ———–α x sin2 α = sin α (c) T5 = HG = x sin5 α (d) S∞ = ——– a 1 – r = ———— x sin α 1 – sin α If x = 8, α = 60° S∞ = ————– 8 sin 60° 1 – sin 60° S∞ = ——8AB3 2 ———– 1 – —– AB3 2 S∞ = 8AB3(2 + AB3) ——————— (2 – AB3)(2 + AB3) S∞ = 24 + 16AB3 4. (a) y = abx – 2 log10 y = log10 a + (x – 2) log10 b Y = log10 y, X = x – 2, m = log10 b, c = log10 a (b) x – 2 –2 –1 1 3 4 5 log10 y –0.9 –0.6 0 0.6 0.9 1.2 log10 y –2 –1 –0.2 0.2 0.4 0.6 0.8 1.0 –0.4 –0.6 –0.8 0 1 2 3 4 5 (x – 2) (c) From the graph, m = —– 0.9 3 = 0.3 log10 b = 0.3 b = 2 log10 a = c = –0.3 a = 0.5 ∴ a = 0.5, b = 2 5. (a) —1 2 (3 – cos 2x) = —1 2 (3 – (1 – 2 sin2 x)) = —1 2 (2 + 2 sin2 x) 10_1202 QB AMath F5.indd 134 10/01/2022 4:49 PM