Cambridge IGCSE Maths Core Extended Study Revision Guide 3rd edition (John Jeskins, Powell) (Z-Library)

44 It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Simultaneous equations Simultaneous equations PAGES 130–137 To solve simultaneous equations: l Step 1: Make the coefficients of x or y equal by multiplying one or both of the equations. Do not forget to multiply every term. l Step 2: Eliminate the term with equal coefficients: – if the signs are the same, subtract – if the signs are different, add. l Step 3: Substitute the value back into one of the original equations to find the value of the other letter. l Step 4: Write down both values as your answer. l Step 5: Check your answer by substituting both values back into the equation not used in step 3. Sample question and answer 1 Solve the simultaneous equations: 3x − 5y = 1 (1) 2x + 3y = 7 (2) 1 Make y the same by multiplying (1) × 3 and (2) × 5. 9x – 15y = 3 10x + 15y = 35 19x = 38 Add x = 2 2 × 2 + 3y = 7, y = 1 Substitute in (2) 3 × 2 – 5 × 1 = 1 Check in (1) Test yourself 26 Solve these simultaneous equations: a 3x + 2y = 13 4x + y = 14 d 3x - 2y = 7 y = 2 - 4x b 2x + y = 7 3x - 2y = 28 e x + y = -1 x - y = 6 c 3x - 2y = 8 4x + 3y = 5 f 5x - 3y = 29 3x + 2y = 6 Answers on page 173 Sample exam question and answer 1 2x + 5y = 90 (1) 3x + 6y = 123 (2) 6x + 15y = 270 (1) × 3 6x + 12y = 246 (2) × 2 3y = 24 Subtract y = 8 2x + 5 × 8 = 90 Substitute y = 8 into (1) 2x = 50 x = 25 x = 25, y = 8 3 × 25 + 6 × 8 = 123 Check in (2) 75 + 48 = 123 1 A gardener finds that two apple trees and five pear trees cost $90 and that three apple trees and six pear trees cost $123. If the cost of one apple tree is $x and the cost of one pear tree is $y, write down two equations in x and y. Then solve them to find x and y. [5] 9781510421714.indb 44 11/07/19 2:30 PM

45 It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 13 Equations and inequalities Solving quadratic equations by factorising If the product of two numbers is 0, then one of the numbers is 0. If a quadratic expression equals 0, then one or other of its factors equals 0. Sample questions and answers 1 Solve (x - 2)(x + 3) = 0. 2 Solve x 2 - 3x = 0. 1 Either x – 2 = 0 or x + 3 = 0, so x = 2 or x = –3. 2 x(x – 3) = 0 Then either x = 0 or x – 3 = 0, so x = 0 or x = 3. Test yourself Solve these quadratic equations by factorising: 27 (x - 3)(x - 6) = 0 32 2x 2 - 13x - 7 = 0 28 (x + 4)(x - 1) = 0 33 3x 2 - x - 10 = 0 29 x 2 + 5x + 4 = 0 34 6x 2 - 17x + 5 = 0 30 x 2 - 7x + 12 = 0 35 x 2 + 4x = 0 31 x 2 + 3x - 10 = 0 36 2x 2 - 50 = 0 Answers on page 173 Completing the square l If the coefficient of x 2 is not 1, divide through the equation by the coefficient of x 2 . l Make the x 2 + bx terms into a complete square and correct the term added by subtracting 2 2 b ( ) . l Rearrange the equation. l Take the square root, remembering ±. Sample question and answer 1 x 2 – 6x + 0.5 = 0 Divide by 2 (x – 3)2 – 9 + 0.5 = 0 (x – 3)2 = x 2 – 6x + 9 (x – 3)2 – 8.5 = 0 (x – 3)2 = 8.5 x – 3 = ± 8.5 x = 3 ± 8.5 = 5.92 or 0.08 to 2 d.p. 1 Solve 2x 2 - 12x + 1 = 0. Sample exam question and answer 1 a x 2 + 6x + 2 = (x + 3)2 – 9 + 2 = (x + 3)2 – 7 b The least value of (x + 3)2 is 0, so the least value of y is –7. c x 2 + 6x + 2 = 0 (x + 3)2 – 7 = 0 (x + 3)2 = 7 x + 3 = ± 7 x = – 3 ± 7 x = –0.35 or –5.65 to 2d.p. 1 a Write x 2 + 6x + 2 in the form (x + a) 2 + b. [2] b Hence state the minimum value of y on the curve y = x 2 + 6x + 2.[1] c Solve the equation x 2 + 6x + 2 = 0. [4] PAGES 137–139 9781510421714.indb 45 11/07/19 2:30 PM

46 It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide The quadratic formula The quadratic formula PAGE 140 l When ax 2 + bx + c = 0, x = – ( – 4 ) 2 . 2 b b ac a ± l When using this formula, take care with the signs and brackets. Make sure you know how to use your calculator to work it out correctly. Sample question and answer 1 a = 2, b = –7, c = –3 ( ) = ± × × × x 7 –7 – 4 2 – 3 2 2 2 = 7 4 ± + 9 24 4 = 7 7 ± 3 4 = 3.89 or –0.39 to 2 d.p. 1 Solve 2x 2 − 7x − 3 = 0. Test yourself Solve these by completing the square: 37 x 2 − 4x − 6 = 0 41 4x 2 − 8x + 1 = 0 38 x 2 + 8x + 2 = 0 42 2x 2 − 4x − 7 = 0 39 x 2 − 10x + 7 = 0 43 3x 2 + 12x − 2 = 0 40 x 2 − 3x + 1 = 0 44 5x 2 + 30x + 4 = 0 Solve these using the quadratic formula: 45 x 2 − 5x + 3 = 0 49 4x 2 − 7x + 2 = 0 46 2x 2 + 8x + 1 = 0 50 2x 2 − 9x − 7 = 0 47 x 2 − 6x + 4 = 0 51 3x 2 + 12x + 2 = 0 48 3x 2 − 5x − 1 = 0 52 5x 2 + 20x + 4 = 0 53 Try to solve x 2 − 4x + 7 = 0 using the quadratic formula or completing the square. What happens? Answers on page 173 Simultaneous equations involving one linear and one non-linear equation PAGES 141–142 Sample question and answer 1 Solve the following simultaneous equations: y = x - 5 y = x 2 + 6x + 1 1 Since both expressions in x are equal to y, they must be equal to each other. x 2 + 6x + 1 = x – 5 x 2 + 5x + 6 = 0 (x + 3)(x + 2) = 0 x = −2 or x = −3 When x = −2, y = −7 When x = −3, y = −8 9781510421714.indb 46 11/07/19 2:30 PM

47 It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 Test yourself Solve these simultaneous equations: 54 y = 2x + 3 and y = x 2 - x - 1 55 y = -3x + 4 and y = x 2 + 2x - 3 Answers on page 173 Linear inequalities PAGES 142–144 The symbols used are: l >, which means ‘greater than’ l <,which means ‘less than’ l >,which means ‘greater than or equal to’ l <,which means ‘less than or equal to’. Inequalities are solved using the same rules as when solving equations except when dividing or multiplying by a negative number. Then the inequality sign is reversed. Sample questions and answers 1 Find the largest integer that satisfies 4x + 10 < 2x + 14. 2 Solve the inequality 4x < 6x + 2. 3 Solve x 2  16. 4 Solve -10 < 2x + 1  15. 1 4x – 2x < 14 – 10 2x < 4 x < 2 The largest integer value less than 2 is 1. 2 –2x < 2 x >–1 Divided by –2, so reverse the inequality sign. 3 x < 4 and x  –4 Turn the inequality round for the negative solution. The answer is –4 < x < 4. 4 –11 < 2x < 14 Subtract 1 from –10 and 15. –5.5 < x < 7 Test yourself 56 Solve these inequalities: a 3x < x + 4 b 3 - 4x > 6 + x 57 Find the smallest integer value that satisfies 3(9 - 4x) < 9 + 6x. 58 Solve x 2  25. 59 Solve 5 < 3 - x < 9. Answers on page 173 c 2(3x - 4)  3(4x - 3) Sample exam questions and answers 1 2x – 5x < 12 – 3 –3x < 9 x > –3 1 Solve 2x + 3 < 5x + 12. [2] 13 Equations and inequalities 9781510421714.indb 47 11/07/19 2:30 PM

48 It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Linear inequalities Exam-style questions 1 Solve 3x = x + 1. [2] 2 Solve 3p - 4 = p + 8. [2] 3 Solve 3m 4 = 9. [2] 4 Solve 2(y + 3) = 5y. [3] 5 Solve 4(x - 1) = 2x + 3. [3] 6 Solve 4(x + 2) + 2(3x - 2) = 14. [3] 7 A rectangle has its longer side 2cm greater than its shorter side. Its perimeter is 36cm. Let x cm be the length of the shorter side. a Write down an equation in x. [3] b Solve your equation to find x. [2] c Find the area of the rectangle. [1] 8 Large radiators cost $45 more than small radiators. Glen buys three small and two large radiators for $415. Write down an equation and solve it to find the cost of a small radiator. [3] 9 Solve these simultaneous equations: 5x + 4y = 13 3x + 8y = 5 [3] 10 Solve these simultaneous equations: 4x + 3y = 5 2x + y = 1 [3] 11 The equation of the straight line passing through the points (3, 2) and (9, 11) is given by px + qy = 5. a Explain why 3p + 2q = 5. [1] b Write down another equation in terms of p and q. [1] c Solve the two equations to find the value of p and the value of q. [3] 12 Solve these equations by factorising: a x 2 - 6x + 8 = 0 [2] b 2x 2 + 3x - 9 = 0 [3] 13 a Factorise completely 5x 2 - 20. [2] b (i) Factorise x 2 - 9x + 8. [2] (ii) Hence solve x 2 - 9x + 8 = 0. [1] 14 a Multiply out and simplify the expression (2x + 7) (3x - 6). [3] b (i) Factorise x 2 + 6x. [1] (ii) Solve the equation x 2 + 6x = 0. [1] 15 Solve the equation 2x 2 - 38x + 45 = 0. [2] 16 a Write 3x 2 - 12x + 2 in the form 3(x - a) 2 - b. [2] b Hence solve the equation 3x 2 - 12x + 2 = 0. [3] 17 The length of a rectangle is y cm, the perimeter is 30cm and the area is 55cm2 . a Form an equation in y and show that it can be simplified to y 2 - 15y + 55 = 0. [3] b Solve the equation y 2 - 15y + 55 = 0 to find the length and width of the rectangle. Give your answers correct to 2 d.p. Do not use a trial and improvement method. [3] 18 Bill is making a rectangular chicken pen against a wall. The other three sides will be made from wire netting. Here is the plan. Wire netting Wall x The total length of wire netting is 22m. The area inside the pen must be 60m2 . a Show that x 2 - 11x + 30 = 0. [3] b Solve the equation. [3] c Describe the size of the pen. [1] 19 Solve the simultaneous equations. [6] y = 2x - 6 and y = x 2 − 4x + 2. 20 Solve 8x + 5 > 25. [2] 21 Solve 2x + 17 > 4x + 6. [3] 22 Solve x 2 < 36. [2] Answers on page 174 9781510421714.indb 48 11/07/19 2:30 PM

49 It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 Key objectives ● To represent inequalities graphically and use this representation to solve simple linear programming problems. Linear programming deals with problems in which a number of simple conditions have to be satisfied at the same time. There are three steps in solving a problem by linear programming. l Step 1: Set up the inequalities from the information. l Step 2: Draw the graph. l Step 3: Find the solution that best satisfies the requirements. Graphical solution of a set of inequalities The solution will be represented by a region of the graph. l Step 1: Draw the boundary lines. Change the inequality sign to an equals sign. Use a continuous line if there is an equals sign in the inequality and a dotted line if there is no equals sign. l Step 2: Substitute the coordinates of the origin, (0, 0), to test each inequality. If the result is true, shade the side of the line opposite the origin. If the result is false, shade the side of the line the same as the origin. l Step 3: The remaining, unshaded area is the region of points satisfying the set of inequalities. Sample question and answer 1 Find the region satisfied by x  1, x + y < 5 and y  2x - 6. 1 x = 1 is the line through (1, 0) parallel to the y-axis. The line x + y = 5 passes through (0, 5) and (5, 0), and y = 2x – 6 passes through (0, –6), (1, –4) and (2, –2). x  1 ⇒ 0  1 False, shade origin side. x + y < 5 ⇒ 0 < 5 True, shade opposite side. y  2x – 6 ⇒ 0  –6 True, shade opposite side. Test yourself 1 Find the region satisfied by each set of inequalities. a y  4 x + y  4 y  3x - 1 b x  2 2x + 3y  12 5x + 2y  10 Answers on page 174 PAGES 147–150 Tip Shade the unwanted region unless the question states otherwise. 0 4 6 y x 2 – 2 4 6 – 4 – 6 2 14 Linear programming 9781510421714.indb 49 11/07/19 2:30 PM

50 Graphing the inequalities It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Linear programming – setting up inequalities PAGES 150–151 Use a letter for each of the two variables and write an inequality for each restriction. Often two trivial inequalities, x  0 and y  0, are given. Sample question and answer 1 Let there be x doughnuts and y muffins. There must be at least 12 items: x + y  12. Ahmed can spend up to $18 but no more. So 1x + 1.50y < 18, or 2x + 3y < 36. There must be more muffins than doughnuts: y > x. 1 Ahmed buys some doughnuts and some muffins for a party. Doughnuts cost $1 each, and muffins cost $1.50 each. He wants at least 12 items. He has $18 to spend. He wants more muffins than doughnuts. Write down an inequality to represent each of these three conditions. Test yourself 2 A committee must have fewer than 15 members. There must be at least five men and at least five women. There are x men and y women on the committee. Write down three inequalities that x and y must satisfy. 3 A theatre holds 600 people. Adult tickets cost $8 and child tickets cost $5. At a performance at least $3200 must be taken to make a profit. There must be more adults than children. If x children and y adults are at a performance that makes a profit, write down three inequalities that x and y must satisfy. Answers on page 174 Graphing the inequalities PAGES 147–150 For each inequality draw the line of the associated equation. Use a continuous line if the line is included in the required region. Use a dotted line if the line is not included in the required region. Shade the unwanted region. Sample question and answer 1 Draw axes taking values of both x and y from 0 to 20 and show by shading the region satisfying the inequalities from the previous sample question: x + y  12, 2x + 3y  36 and y > x. 1 Use a continuous line for x + y = 12 and 2x + 3y = 36. Use a dotted line for y = x. The required region is the unshaded polygon. x y 0 15 x = y 2x + 3y = 36 x + y = 12 5 10 5 10 2015 9781510421714.indb 50 11/07/19 2:30 PM

51 14 Linear programming It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 Test yourself For each question, draw and label axes taking values of both x and y from 0 to 12 and show by shading the region satisfied by these inequalities. 4 y  3, 4x + 3y  36, 2x > y 5 y  0, x < 8, y  x, 2x + 3y  18 6 4x + 5y  40, y  3x, y  3 Answers on page 174 Finding the solution l Step 1: Identify the possible solutions in the required region. l Step 2: Choose the one that best satisfies the requirements. Sample question and answer 1 The greatest number of items is six doughnuts and eight muffins. These cost $18. x y 0 15 x = y 2x + 3 = 36 x + y = 12 10 5 5 10 15 20 1 In the previous example, find the greatest number of items that can be bought and the cost. Mark the integer points inside the required region. Test yourself 7 a On the graph for question 4, mark all the integer values of (x, y) in the region. b (i) Find which of these points gives the greatest value of x + 2y. (ii) State the value. 8 a Using the graph for question 5, find the integer value of (x, y) in the region that gives the least value of 2x + 3y. b Write down this value. 9 Using the graph for question 6, list the integer values of (x, y) in the region where x = y. Answers on page 174 Tip You do not include (6, 6) or (7, 7) as they are on the dotted boundary. 9781510421714.indb 51 11/07/19 2:30 PM

52 It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Finding the solution Sample exam questions and answers 1 Write down the three inequalities that satisfy the shaded region. [3] 0 4 6 y x 2 – 2 1 2 8 10 3 4 x = 3 y = 2x + 1 x + y = 4 2 There are x girls and y boys in a school choir. a (i) The number of girls in the choir is more than 1.5 times the number of boys. Show that y < 2 3 x . [1] (ii) There are more than 12 girls in the choir. There are more than five boys in the choir. The maximum number of children in the choir is 35. Write down three more inequalities. [3] b (i) Draw axes for 0  x  40 and for 0  y  40. [1] (ii) Draw four lines on your graph to represent the inequalities in part a. Shade the unwanted parts of the grid. [4] c The school buys a uniform for each choir member. A girl’s uniform costs $25. A boy’s uniform costs $20. Find the maximum possible cost for the choir uniforms. Mark clearly the point, P, on your graph that you use to calculate the cost. [2] 1 Choose a point in the region e.g. (2, 3) so x = 2, y = 3 x = 3 y = 2x + 1 x + y = 4 2 < 3 3 < 4 + 1 5 > 4 so x < 3 y < 2x + 1 x + y > 4 2 a (i) x > 1.5y = 3 2 y, so y < 2x 3 (ii) x > 12, y > 5, x + y < 35 b y 0 x 10 20 30 40 10 20 30 40 P y =2x 3 x = 12 x + y = 35 y = 5 c (29, 6) marked on graph. Cost = 29 × 25 + 6 × 20 = 725 + 120 = $845 9781510421714.indb 52 11/07/19 2:30 PM

53 14 Linear programming It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 Exam-style questions 1 Find the region satisfied by the inequalities: x + y  5 y  2x - 1 y  0 [3] 2 Look at the graph and write down the three inequalities that satisfy the shaded region. [3] 2 3 y x 1 21 4 3 y = x x + y = 3 –2 –3 –2 x = –2 –1 0 –1 3 A factory produces two types of garden ornaments, type A and type B. Type A takes 1 hour of machine time and 3 hours of workers’ time. Type B takes 2 hours of machine time and 1 hour of workers’ time. In a day there are 28 hours of machine time and 24 hours of workers’ time available. a If x of type A and y of type B are produced in a day, write down two inequalities satisfied by x and y. [2] b Draw these inequalities on a graph, shading the unwanted regions. [3] c The profit made on each type A is $20 and on each type B is $10. (i) Find the greatest profitthat can be made. [2] (ii) How many of each type are produced to make this profit? [1] 4 A school library is given $1000 for new books. They buy x paperback and y hardback books. There must be at least 50 books and there must be more hardback than paperback books. Paperback books cost $10 and hardback books cost $25. a (i) Explain why 2x + 5y  200. [2] (ii) Write down two more inequalities involving x and y, other than x  0, y > 0. [2] b Represent these inequalities on a graph, shading the unwanted regions. [4] c The school orders the maximum number of hardback books, subject to these conditions. Find this number from your graph. [1] 5 A photographer charges $10 for an individual photo and $30 for a set of six. She always sells more individual photos than sets in a week. She sells x individual photos and y sets of photos in a week. She needs to earn at least $300 a week. a Write down two inequalities involving x and y, other than x  0, y  0. [2] b Represent these inequalities on a graph, shading the unwanted regions. [3] c To produce an individual photo takes 1 hour and to produce a set takes 2 hours. (i) What is the least time she can work and earn at least $300? [2] (ii) How many individual photos and how many sets does she produce in this time? [2] Answers on pages 174–175 9781510421714.indb 53 11/07/19 2:30 PM

54 It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide 15 Sequences Key objectives ● To continue a given number sequence. ● To recognise patterns in sequences, including the term-to-term rule, and relationships between different sequences. ● To find and use the nth term of sequences. Term-to-term rules PAGES 153–156 Every number in a sequence is called a term of the sequence. There is often a simple rule for going from one term to the next. Use the term-to-term rule to find the formula for a sequence. Sample questions and answers 1 Here are the first four terms of a sequence: 3, 7, 11, 15. a Find the term-to-term rule. b Find the 20th term. c Find a formula for the nth term of the sequence. 1 a Add 4 b To get the 20th term, add four 19 times: 3 + (19 × 4) = 3 + 76 = 79 c The formula is 3 + (n – 1) × 4 = 3 + 4n – 4 = 4n – 1 Test yourself 1 Find the first four terms for the sequence given by 3n + 3. 2 Find the term-to-term rule for the sequence 2, 7, 12, 17, … . 3 Find the 15th term and the formula for the nth term for the sequence 5, 11, 17, 23, … . 4 Find the formula for the nth term for the sequence 40, 36, 32, … . Answers on page 175 Common sequences PAGES 153–156 You should know and recognise common sequences. Examples The powers of 2: 1, 2, 4, 8, 16, 32, …, 2n The powers of 10: 1, 10, 100, 1000, …, 10n The square numbers: 1, 4, 9, 16, …, n 2 The cube numbers: 1, 8, 27, 64, …, n 3 The triangle numbers: 1, 3, 6, 10, …, n n 1 2 ( ) + 9781510421714.indb 54 11/07/19 2:30 PM

55 It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 15 Sequences Position-to-term rules PAGES 153–156 Every number in a sequence can be expressed in terms of its position. Use the position-to-term rule to find the formula for the nth term of a sequence: l look at the difference between successive terms l multiply this difference by n l add or subtract a constant to give the required results. Sample question and answer 1 Find the rule for this sequence: 7, 11, 15, 19, … . 1 It is sometimes useful to write the sequence out in columns. Position Term Difference between terms 1 7 2 11 4 3 15 4 This shows that you multiply by 4. The term 4n will be in the formula. Look at position 1. The term is 7. 4 × n = 4 × 1 = 4. You need to add 3 to give the required result of 7. Check for position 2: 4 × 2 + 3 = 11 So the nth term is 4n + 3. Test yourself Find a formula for the nth term of each of these sequences: 5 3, 5, 7, 9, … . 6 10, 20, 30, 40, … . 7 –4, 1, 6, 11, … . 8 32, 29, 26, 23, … . Answers on page 175 Quadratic sequences PAGES 156–162 When the differences between the terms of the sequence are not the same, the sequence may be a quadratic sequence. For this, check second differences. 9781510421714.indb 55 11/07/19 2:30 PM

56 It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Sample question and answer 1 Find the rule for this sequence: 3, 6, 11, 18, 27, … . 1 Again, it is sometimes useful to write the sequence out in columns. When the second differences are the same then the sequence is a quadratic one. Let the nth term of the sequence be an2 + bn + c. Substitute n = 1, 2, 3. Position Term 1st difference 2nd difference 1 3 = a + b + c 2 6 = 4a + 2b + c 3 = 3a + b 3 11 = 9a + 3b + c 5 = 5a + b 2 = 2a 4 18 7 2 5 27 9 2 Using the formulae in the table, a = 1, b = 0 and c = 2. So, the sequence is n 2 + 2. Other sequences PAGES 163–164 Some sequences do not have a common difference between terms. For example, if each term is the previous term multiplied by a constant, the sequence is an exponential sequence, sometimes called a geometric sequence. The rules for other sequences can often be found on inspection. Sample question and answer 1 Each term is 5 × the previous one. 2nd term = 4 × 5 3rd term = 4 × 5 × 5 = 4 × 52 4th term = 4 × 5 × 5 × 5 = 4 × 53 Un = 4 × 5n – 1 The nth term is also written Un . 1 Find a formula for the nth term of the sequence 4, 20, 100, 500, … . Sample question and answer 1 U1 = 2 × 3 = (1 + 1) (1 + 2) U2 = 3 × 4 = (2 + 1) (2 + 2) Un = (n + 1) (n + 2) 1 Find a formula for the nth term of the sequence 2 × 3, 3 × 4, 4 × 5, … . Test yourself 9 Find the formula for the nth term of the following sequences: a 3, 9, 17, 27, 39, … c 1, 5, 25, 125, … b 2, 8, 32, 128, … d 3 × 5, 4 × 6, 5 × 7, … Answers on page 175 Other sequences 9781510421714.indb 56 11/07/19 2:30 PM

57 It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 Sample exam questions and answers 1 a Write down the 10th term for the sequence 3, 7, 11, 15, … .[1] b Write down an expression for the nth term. [2] c Show that 137 cannot be a term in this sequence. [2] 1 a 39 10th term = 3 + 9 × 4 b 4n – 1 Either: the difference between terms is 4, so the expression will start 4n. If n = 1, then 4n = 4, therefore subtract 1 to get 3. Therefore the expression is 4n – 1. Or: 1st term is 3, add 4 (n – 1) times, therefore the nth term is 3 + 4(n – 1) = 3 + 4n – 4 = 4n – 1. c If 137 is in the sequence, then 4n – 1 = 137. Therefore 4n = 138 and n = 138 ÷ 4 = 34.5 This is not a whole number. Therefore 137 cannot be in the sequence. Exam-style questions 1 The first four terms of a sequence are 3, 8, 13, 18. Find the 50th term and the nth term of this sequence. [3] 2 The first five terms of a sequence are 1, 6, 11, 16, 21. Find the formula for the nth term. [2] 3 The first four terms of a sequence are 2, 9, 16, 23. a Find the nth term of this sequence. [2] b Show that 300 is not in this sequence. [2] 4 a Write down the formula for the nth term of the sequence 1, 4, 9, 16, 25, … . [1] b Hence, or otherwise, find the formula for the nth term of the sequence 4, 13, 28, 49, 76, … . [3] 5 a Write down the formula for the nth term of the sequence 3, 5, 7, 9, … . [2] b Hence, or otherwise, find the formula for the nth term of the sequence 1 × 3 + 1, 2 × 5 + 1, 3 × 7 + 1, … . [3] 6 The three sequences below are linked. Write down the formula for the nth term of the sequence a. Use the answer to write down the formula for sequence b and hence find the formula for sequence c. a 1, 8, 27, 64, … [1] b 2, 16, 54, 128, … [1] c –1, 10, 45, … [2] 7 Find a formula for the nth term of this quadratic sequence: 2, 14, 34, 62, 98, … . [3] Answers on page 175 15 Sequences 9781510421714.indb 57 11/07/19 2:30 PM

58 It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide 16 Proportion Key objectives ● To express direct and inverse proportion in algebraic terms and to use this form of expression to find unknown quantities. Direct proportion PAGES 166–168 l In direct proportion both variables change in the same way – either both getting larger or both getting smaller. l Using symbols, direct proportion is written as y ∝ x or y ∝ x 2 . l The formulae for these are y = kx or y = kx2 , respectively, where k is a constant. Sample questions and answers 1 a V ∝ W2 or V = kW2 490 = k × 142 , k = 2.5 So V = 2.5 × 402 , i.e. V = $4000 b 6000 = 2.5 × W2 W2 = 2400, so W = 48.99 1 The value, $V, of a diamond is proportional to the square of its mass, Wg. A diamond weighing 14g is worth $490. a Find the value of a diamond weighing 40 g. b Find the mass of a diamond worth $6000. Test yourself 1 If y ∝ x 3 and y = 24 when x = 2, find y when x = 4. 2 If y ∝ x and y = 8 when x = 20, find x when y = 36. 3 A car uses 14 litres of fuel to travel 80 kilometres. How much fuel will it use to travel 250 kilometres? 4 When an object is dropped, the distance, dmetres, which it falls in t  seconds, is proportional to t 2 . If d = 122.5 metres when t = 5 seconds, calculate d when t = 7 seconds. Answers on page 175 Inverse proportion PAGES 168–170 l In inverse proportion when one variable increases the other variable decreases. l Using symbols, inverse proportion is written as y ∝ x 1  or y ∝ x 1 2 . l The formulae for these are y = k x or y = k x 2  respectively, where k is a constant. Tip These are the most common direct proportions but you could also meet y ∝ x 3 and y ∝ x . Tip These are the most common inverse proportions, but you could also meet y ∝ x 1 .   9781510421714.indb 58 11/07/19 2:30 PM

59 16 Proportion It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 Sample questions and answers 1 a V ∝ P 1 or V = k P 2 = k 500, k = 1000 So V = 1000 400 = 2.5 m3 b P = 200 N/m2 1 The volume, V, of a given mass of gas varies inversely as the pressure, P. When V = 2m3 , P = 500 N/m2 . a Find the volume when the pressure is 400 N/m2 . b Find the pressure when the volume is 5m3 . Test yourself 5 y is inversely proportional to x 2 , and y = 4 when x = 1. Find y when x = 2. 6 y is inversely proportional to x . y = 1.2 when x = 100. a Calculate y when x = 4. b Calculate x when y = 3. 7 The wavelength of a sound-wave, Wm, is inversely proportional to the frequency, FHz. A sound-wave with a frequency of 300Hz has a wavelength of 1.1m. Calculate the wavelength of a sound-wave with frequency of 660Hz. 8 The brightness of a light varies inversely as the square of the distance from the source. A lamp has a brightness of 20 lumens at a distance of 1 metre. How bright is it at 10 metres? Answers on page 175 Sample exam question and answer 1 d ∝ h or d = k h 35 = k 100, k = 3.5 So d = 3.5 × 25 17.5kilometres 1 From a point h metres above sea level the distance, d kilometres, to the horizon is given by d ∝ h. When h = 100 metres, d = 35kilometres. Find d when h = 25 metres. [3] Exam-style questions 1 The variable y is directly proportional to x 2 . Given that y = 75 when x = 5, find the value of y when x = 10. [3] 2 The number of coins, N, that can be made from a given volume of metal is given by N ∝ d 1 2 , where d cm is the diameter. Given that 8000 coins with a diameter of 2 cm can be made from the volume of metal, how many coins with a diameter of 4 cm can made from the same volume of metal? [3] 3 The distance travelled by a car after the brakes are applied is proportional to the square of the initial speed. If it takes 12.5 metres to stop when travelling at 50 km/h, how far will a car travel if its initial speed is 120 km/h? [3] Answers on page 175 9781510421714.indb 59 11/07/19 2:30 PM

60 It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide 17 Graphs in practical situations Key objectives l To interpret and use graphs in practical situations, including travel graphs and conversion graphs. l To draw graphs from given data. l To apply the idea of rate of change to simple kinematics involving distance–time and speed– time graphs, acceleration and deceleration. l To calculate distance travelled as area under a linear speed–time graph. Graphs in real situations PAGES 172–188 To interpret real-life graphs: l look at the labels on the axes – they tell you what the graph is about l notice whether the graph is a straight line or a curve l the slope of the graph gives you the rate of change l for distance–time graphs, the rate of change is the velocity l for speed–time graphs, the rate of change is the acceleration l the area under a speed–time graph = distance travelled. Constant rate of change x y Rate of change increasing x y Rate of change decreasing x y No change for the variable on the y axis x y Sample questions and answers 1 Asif walked to the bus stop, waited for the bus, and then travelled on the bus to school. Opposite is a distance–time graph for Asif’s journey. a How long did Asif wait at the bus stop? b How far did Asif travel: (i) on foot (ii) by bus? c How fast did Asif walk? d What was the average speed of the bus? e Why are the sections of the graph unlikely, in reality, to be totally straight? 1 a 5 minutes b (i) 1km (ii) 6km c 1 15 km/min or 4 km/h d 6 10 km/min or 36 km/h e Stops and starts are likely to be gradual, not sudden. Tip Remember to find the rate of change for each part of the graph. If it is zero, say how long this lasts. 8 6 2 10 25 Time in minutes Distance in km 0 5 20 4 15 30 35 9781510421714.indb 60 11/07/19 2:30 PM

61 17 Graphs in practical situations It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 Test yourself 1 Describe what is really happening in these graphs. a 20 15 10 5 Time (seconds) Speed (m/s) 0 2010 30 40 50 Answers on page 175 Drawing graphs When drawing a graph, always label your axes with the quantity and the unit. Sometimes only a sketch is asked for, not an accurate plot. You should always check whether the rate of change is constant, increasing or decreasing. Sample question and answer 1 As the radius is decreasing at first, the rate of change will increase and then become constant. t (s) d (cm) 1 Water is poured into this vessel at a constant rate. Sketch a graph of depth of water (dcm) against time (t seconds). Test yourself 2 A car is travelling at 70 km/h, when it approaches some roadworks where the speed limit is 50 km/h. It slows down from 70 km/h to 50 km/h in 2 minutes. It goes through the roadworks at 50 km/h in 5 minutes and then accelerates back to 70 km/h in 30 seconds. Draw a velocity–time graph to illustrate this. 3 Find the distance travelled in the first 40 seconds in ‘Test yourself’ question 1a. 4 Water is poured into this vessel at a constant rate. Sketch a graph of depth of water (d cm) against time (t seconds). Answers on pages 175–176 b 40 30 20 10 5 10 15 20 25 Time (minutes) Volume of bath water (litres) 0 9781510421714.indb 61 11/07/19 2:30 PM

62 Non-linear graphs It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Sample exam questions and answers 1 Tom leaves home at 8.20 a.m. and goes to school on a moped. The graph shows his distance from the school in kilometres. a How far does Tom live from school? [1] b Write down the time that Tom arrives at the school. [1] c Tom stopped three times on the journey. For how many minutes was he at the last stop? [1] d Calculate his speed in kilometres per hour between 8.20 a.m. and 8.30 a.m. [2] 1 a 7.6km b 8.47 a.m. c 2.5 minutes d Distance = 7.6 – 4.6 = 3km; time = 10 min; speed = 3 10 × 60 = 18 km/h Area under a graph PAGES 180–183 To find the area under straight-line graphs, use the formulae for areas of a triangle, rectangle and trapezium. For the area under a curve, you need to estimate by counting squares or by dividing up the area into strips and approximating with trapezia. The area under a speed–time graph represents distance. Non-linear graphs PAGES 183–188 l When the graph is a practical graph the gradient is called a rate of change and has units. l The gradient of a distance–time graph represents speed (usual units m/s). l The gradient of a speed–time graph represents acceleration (usual units m/s2 ). Sample question and answer 1 The graph shows the area (Acm2 ) of an ink blot against time (t seconds). Find the rate of change when t = 4. 1 Draw a tangent to the curve at t = 4. Rate of change = − − 18.5 3.5 6 2 = 15 4 = 3.75cm2 /s. 0 5 10 15 20 25 2 4 6 t A 1 3 5 8.20 a.m. 8.50 a.m. Time 0 2 4 6 8 Distance from school (km) 8.30 a.m. 8.40 a.m. Tip Remember, when reading off y increases and x increases, use the scales; do not count squares. 9781510421714.indb 62 11/07/19 2:30 PM

63 17 Graphs in practical situations It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 Test yourself 5 The graph shows distance (d m) against time (t seconds). Estimate the speed at: a t = 2 b t = 4 0 50 100 150 2 4 6 t d 1 3 5 6 The graph shows speed (vm/s) against time (t seconds). Estimate the acceleration at: a t = 2 b t = 8 0 10 20 30 t v 2 4 6 8 10 12 14 16 Answers on page 176 Exam-style questions 1 Steve travelled from home to school by walking to a bus stop and then catching a school bus. a Use the following information to construct a distance–time graph for Steve’s journey. Steve left home at 8.00a.m. He walked at 6 km/h for 10 minutes. He then waited for 5 minutes before catching the bus. The bus took him a further 8km to school at a steady speed of 32 km/h. [4] b How far was Steve from home at 8.20a.m? [3] 2 The graph on the right describes a real-life situation. Describe what is possibly happening. [3] 3 The diagrams below show the cross-sections of three swimming pools. Water is pumped into all three at a constant rate. Sketch graphs of depth against time for each. a [2] b [2] c [2] 4 Ameni is cycling at 4 metres per second. After 3.5 seconds she starts to decelerate and after a further 2.5 seconds she stops. The diagram on the right shows the speed–time graph of her journey. a Calculate the distance travelled during the 6 seconds. [2] b Calculate the acceleration between t = 3.5 and t = 6. [2] 5 The number of bacteria in a colony multiplies by a factor of 3 every hour. Initially there are 20 bacteria. a Copy and complete the table: [1] Time (t, hours) 0 1 2 3 4 5 Number of bacteria (n) 20 60 b Write down the formula for the number of bacteria, n, after t hours. [2] c Draw a graph showing how the number of bacteria changes in the 5 hours. [2] d (i) By drawing a tangent to the curve in part c, calculate the gradient of the curve at the point where t = 3.5. [2] (ii) What information does this gradient represent? [1] Answers on page 176 t v 0 1 1 4 4 5 6 3 3 2 2 Time (seconds) Speed (m/s) Speed Time 9781510421714.indb 63 11/07/19 2:30 PM

64 It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Key objectives l To construct tables of values for functions of the form ax + b, + x 2 ± ax + b, a x (x ≠ 0), where a and b are integral constants. l To draw and interpret such graphs. l To solve linear and quadratic equations approximately, including finding and interpreting roots by graphical methods. l To construct tables of values and draw graphs for functions of the form axn , where a is a rational constant and n = -2, -1, 0, 1, 2, 3, and simple sums of not more than three of these, and for functions of the form ax , where a is a positive integer. l To solve associated equations approximately by graphical methods. l To draw and interpret graphs representing exponential growth and decay problems. l To recognise, sketch and interpret graphs of functions. l To estimate gradients of curves by drawing tangents. Quadratic graphs PAGES 193–195 Graphs of equations 2 y=ax +bx+c are parabolas. Their shape is: a 0For a 0For All parabolas are symmetrical. This symmetry can be seen on the graph and can often be seen in the table as well. Sample question and answer 1 Complete a table of values for y = x 2 - 4x + 3 and draw the graph for -1  x  5. 1 –2 –1–2 2 4 6 8 10 x y 0 1 2 3 4 5 6 x –1 0 1 2 3 4 5 y 8 3 0 –1 0 3 8 Tip When drawing an accurate graph, use a table to work out the points, even if you are not asked to do so. As well as doing it by hand, you can also use a spreadsheet to work out the values and draw these graphs. Tip Notice the symmetry. 18 Graphs of functions 9781510421714.indb 64 11/07/19 2:30 PM

65 18 Graphs of functions It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 Test yourself 1 a Complete this table for y = x 2 - 3x + 4. x -1 0 1 2 3 4 y b Explain why it is a good idea to add an extra column to the table and work out the value of y when x = 1.5. c Draw the graph for -1 < x < 4. 2 Complete this table and draw the graph for y = 2x 2 - 5x, for -2  x  5. x -2 -1 0 1 2 3 4 5 y Answers on page 176 Using graphs to solve quadratic equations PAGES 195–196 The solution of ax 2 + bx + c = k is given by the values of x where the graph of y = ax 2 + bx + c crosses the line y = k. The solution of the simultaneous equations y = ax 2 + bx + c and y = mx + k is given by the values of x where the two graphs cross. Sample questions and answers 1 Use the graph of y = x 2 - 4x + 3 (drawn above) to solve the following equations. a x 2 - 4x + 3 = 0 b x 2 - 4x + 3 = 6 c x 2 - 4x + 3 = x d x 2 - 2x - 2 = 0 1 a Look at where the graph crosses y = 0, the x-axis; the solution is x = 1 or 3. b Look at where the graph crosses the line y = 6; the solution is x = –0.6 or 4.6, to 1 d.p. c This is where the graph crosses the line y = x. When this line is added to the graph, it can be seen that the solution is x = 0.7 or 4.3, to 1 d.p. d Manipulating x 2 – 2x – 2 = 0 to give x 2 – 4x + 3 on the left-hand side gives x 2 – 4x + 3 = 5 – 2x. So the line to draw is y = 5 – 2x. When this line is added to the graph, it can be seen that the solution is x = 2.7 or –0.7, to 1 d.p. Test yourself 3 Use the graph you drew in ‘Test yourself’ question 1 to solve the following equations. a x 2 - 3x + 4 = 4 b x 2 - 3x + 4 = 7 c x 2 - 3x + 4 = x + 2 4 Use the graph you drew in ‘Test yourself’ question 2 to solve the following equations. a 2x 2 - 5x = 0 b 2x 2 - 5x = -1 c 2x 2 - 7x = 1 5 Solve graphically, drawing the graphs for values of x from -1 to 3, these simultaneous equations: y = x 2 - 2x + 1, y = 2 - x. Answers on page 176 9781510421714.indb 65 11/07/19 2:30 PM

66 Cubic, reciprocal and exponential graphs It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Completed square form and the graph of a quadratic equation PAGES 197–198 For the shaped quadratic graph, the turning point is known as the minimum point, while for the shaped graph, the turning point is known as the maximum point. In general, for a quadratic curve with equation y = a(x − b) 2 + c, the coordinate of the turning point will be (b, c). If a is positive, the turning point is a minimum. If a is negative, the turning point is a maximum. Sample questions and answers 1 Without drawing the graph, find the coordinates and type of turning point shown by these graphs. a y = (x + 4)2 + 7 b y = x 2 - 2x + 5 c y = -x 2 - 2x + 2 1 a Minimum point at (–4, 7) b y = (x – 1)2 – 1 + 5 y = (x – 1)2 + 4 Minimum point at (1, 4) c y = –(x 2 + 2x) + 2 y = –[(x + 1)2 – 1] + 2 y = –(x – 1)2 + 3 Maximum point at (1, 3) Cubic, reciprocal and exponential graphs PAGES 198–209 You also need to be able to recognise the shape of cubic, reciprocal and exponential graphs and to draw these graphs. The cubic graph y = ax 3 has this shape when a > 0, and is reflected in the x-axis when a < 0. The reciprocal graph y = a x  has this shape when a > 0, and is reflected in the x-axis when a < 0. The exponential graph y = k x , for k > 1, has this shape. You will be asked to work out only the values for integer values of x. All exponential graphs go through the point (0, 1), as k0 = 1 for all positive values of k. When k > 1, they increase steeply for x > 0, and are small when x is negative. This is reversed for 0 < k < 1. Sample question and answer 1 Draw the graph of y = x 3 - 3x 2 for values of x from -2 to 4. Use your graph to solve the equation x 3 - 3x 2 = -1. Make a table for the values. x y x y 0 1 y x 9781510421714.indb 66 11/07/19 2:30 PM

67 18 Graphs of functions It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 1 x –2 –1 0 1 2 3 4 y –20 –4 0 –2 –4 0 16 –2 –20 –15 –10 –5 0 4 5 x y 10 15 20 –1 1 2 3 The solution of the equation is where the curve crosses the line y = –1. The solution is x = –0.5, 0.7 or 2.9. Sample question and answer 1 Draw y = 3x and y = x 1 ( ) 2 on the same grid. 1 –2 1 5 x 10 15 20 –3 25 y = 3x y = ( ) 1 x 2 30 y –1 0 2 3 Test yourself 6 Draw the graph of y = x 12 for values of x from -12 to 12. Use the same scale on both axes. 7 Draw the graph of y = x 3 - 2x for values of x from -2 to 2. Use your graph to find the roots of the equation x 3 - 2x = 0 to 1 d.p. Then use trial and improvement to find the positive root correct to 2 d.p. 8 Draw the graph of y = 2x for values of x from -1 to 4. Use your graph to find, to 1 d.p., the value of x for which 2x = 10. Answers on page 177 Tip In this example the shape of the ‘double bend’ is more pronounced than in the graph of y = ax 3 . This is because of the extra term −3x 2 in the equation. 9781510421714.indb 67 11/07/19 2:30 PM

68 Recognise, sketch and interpret graphs of functions It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Sample question and answer 1 Plot the graph of y = x 1 2 for -3  x  3. By drawing a suitable line, use your graph to solve the equation 2x 3 - 5x 2 - 1 = 0. 1 To find the equation of the required line, take 2x 3 – 5x 2 – 1 and manipulate it to get x 1 2 . 2x 3 – 5x 2 – 1 = 0 Add 1 to both sides. 2x 3 – 5x 2 =1 Factorise left-hand side. x 2 (2x – 5) = 1 Divide both sides by x 2 . 2x – 5 = x 1 2 The equation of the required line is y = 2x – 5. Plot the curve y = x 1 2 and the line y = 2x – 5. Read off the x values where the curve and the line intersect. –3 –2 –1 –2 2 0 4 6 8 x y y = 1 x 2 y = 2x – 5 1 2 3 The solution of the equation 2x 3 – 5x 2 – 1 = 0 from the graph is x = 2.6. Test yourself 9 Draw the graph of y = x 2 - 3x + 5 for -4  x  5. Using the graph, and by drawing suitable lines, solve the following equations: a x 2 - 3x + 5 = 4 b x 2 - 4x + 2 = 0 c x 2 - 5x = 0 10 Draw the graph of y = x 1 for -5  x  5. Use your graph to solve the following equations: a x 1 = 3x + 1 b x 2 - 2x - 1 = 0 11 Draw the graph of y = x 1 2 for -3  x  3. Use your graph to solve the equation x 3 + 2x 2 - 1 = 0. Answers on page 177 Recognise, sketch and interpret graphs of functions PAGES 206–214 Tips When a sketch graph is required: l the shape of the graph and where it crosses the x-axis and y-axis are the important features. l you should not plot a series of points and join them (though you might work out one or two coordinates as a check). Where a curve approaches a line but never reaches it, the line is called an asymptote. So, in the Sample question above, x = 0 and y = 0 are asymptotes to the curve. 9781510421714.indb 68 11/07/19 2:30 PM

69 18 Graphs of functions It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 Sample questions and answers 1 Draw a sketch graph for each of the following equations: a y = 4 - 2x b y = x 2 + 4x - 5 c y = (x - 2)(x - 4)(x + 1) Test yourself 12 Draw a sketch graph for each of the following equations. a y = 2x + 1 State the equation of any asymptotes. b y = −x 2 + 5x Answers on page 177 Sample exam questions and answers 1 a Complete the table below for y = x 3 - 2x 2 + 1.[1] x -1 -0.5 0 y 0.375 c Use the graph to solve the equation x 3 - 2x 2 + 1 = 0. [3] b Part of the graph is drawn on the grid. Add the three points from the table and complete the curve. [2] d By drawing a suitable straight line on the graph, solve the equation x x –2 – 1 x 0 2 2 + = . [4] 1 a This is a straight-line graph. It crosses the y-axis when x = 0. So y = 4. It crosses the x-axis when y = 0. So 0 = 4 – 2x and x = 2. b This is a quadratic graph with positive x 2 , so it has a ∪ shape. It crosses the y-axis when x = 0. So y = −5. It crosses the x-axis when y = 0. So 0 = x 2 + 4x – 5, 0 = (x – 1)(x + 5) and x = −5 or 1. c This is a cubic graph with positive x 3 This is a cubic graph with positive , so it has a x3. So has a shape. shape. It crosses the y-axis when x = 0. So y = (0 – 2)(0 – 4)(0 + 1) = 8. It crosses the x-axis when y = 0. So 0 = (x – 2)(x – 4)(x + 1) and x = −1, 2 or 4. −5 1 −5 0 x y 0 8 −1 2 4 y x 0 2 4 y x 9781510421714.indb 69 11/07/19 2:30 PM

70 Gradient It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide 1 a x –1 –0.5 0 y –2 0.375 1 b −2 −1.5 −2.5 −1 −0.5 −1.5 −1 0.5 1 1.5 2 2.5 3 3.5 −0.5 0 0.5 x y 1 1.5 2 2.5 y = x3 – 2x2 + 1 y = x c x = –0.6, 1, 1.6 d Manipulating the equation gives x 3 – 2x 2 + 1 = x. So the line to draw is y = x. The solution is x = –0.8, 0.6 or 2.2. Tips l In this question, part of the graph was drawn. This helped not only with drawing the graph but also with some of the answers for parts c and d. l The dashed line shows the part of the graph that was given already. l The solution to part c is where the graph crosses the line y = 0. Gradient PAGES 201–203 Gradient = change in  change in  y x The gradient of a curve varies at different points on the curve. To find the gradient of a curve at a given point, first draw a tangent to the curve at that point. Then find the gradient of the tangent. The gradient of the curve is the same as the gradient of the tangent, but because of inaccuracies in drawing can be regarded only as an estimate. Sample question and answer 1 Find the gradient of the curve y = x 2 at the point (2, 4). 1 Plot the graph and draw a tangent to the curve at (2, 4). Gradient = − − = = 7.8 0 3 1 7.8 2 3.9 0 2 4 6 8 10 1 2 3 x y y = x2 Tip Remember: a line that goes ‘downhill’ from left to right has a negative gradient. Tip When choosing two points on a tangent to find the gradient, make the change in x an easy number. 9781510421714.indb 70 11/07/19 2:30 PM

71 18 Graphs of functions It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 Test yourself 13 Draw a graph of y = x 3 for 0  x  3. Estimate the gradient at: a (1, 1) b (2, 8) 14 Draw a graph of y = x 12 for -12  x  12. Estimate the gradient at: a x = 2 b x = 4 c x = -4 15 Draw a graph of y = x 2 - 3x for -2  x  5. Estimate the gradient at: a x = -1 b x = 2 16 Draw a graph of y = x 3 - 2x for -3  x  3. Estimate the gradient at: a x = -2 b x = 0 c x = 2 Answers on page 178 Exam-style questions 1 a Complete the table for y = 4x - x 2 and draw the graph. [5] x -1 0 1 2 3 4 5 y 3 0 b Use your graph to find: (i) the value of x when 4x - x 2 is as large as possible [1] (ii) between which values of x the value of 4x - x 2 - 2 is larger than 0. [2] 2 The diagram shows the graphs P, Q, R, S, T and U. State which of these graphs could correspond to each of the following equations. a y = x 3 - 1 [1] b y = x 2 - 1 [1] c y = x - 1 [1] y P x y x S y x Q y x T y x R y x U 3 a Complete the table of values and draw the graph of y = x 3 - 4x - 1 for values of x from -3 to 3. [4] x -3 -2 -1 0 1 2 3 y -16 2 -1 -4 -1 14 b Use your graph to solve the equation x 3 - 4x - 1 = 0. [3] c By drawing a suitable straight line on your graph, solve the equation x 3 - 6x - 3 = 0. [4] 4 Draw the graph of y = x 3 for -2  x  2. Use your graph to solve the equation x 3 - 2x -1 = 0. [8] 5 Draw the graph of y = x 2 2 - x for -4  x  3. Use your graph to solve the equation 2x 3 + 7x 2 - 6 = 0. [10] Answers on page 178 9781510421714.indb 71 11/07/19 2:31 PM

72 It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Key objectives l To understand the idea of a derived function. l To be able to use the derivatives of functions of the form axn , and simple sums of not more than three of these. l To be able to apply differentiation to gradients and turning points (stationary points). l To be able to discriminate between maxima and minima by any method. The gradient of a curve PAGES 219–222 The gradient of a curve varies from point to point. The gradient of a curve at a point P on the curve is the same as the gradient of the tangent to the curve at point P. x P y The following sample questions show how the equation of the curve can be used to find the gradient of the tangent at a given point. Sample questions and answers 1 P is the point on the curve y = x 2 + 3 where x = 2. Calculate the gradient of the chord PQ when Q is the point on the curve where: a x = 2.5 b x = 2.1 c x = 2 + h P Q x y 19 Differentiation and the gradient function 9781510421714.indb 72 11/07/19 2:31 PM

73 19 Differentiation and the gradient function It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 1 At P, y = 22 + 3 = 7 a At Q, y = 2.52 + 3 = 9.25 Gradient of chord = − − 2 1 2 1 y y x x = 9.25−7 0.5 = 4.5 b At Q, y = 2.12 + 3 = 7.41 Gradient of chord = − − y y x x 2 1 2 1 = 7.41−7 0.1 = 4.1 c At Q, y = (2 + h) 2 + 3 = 4 + 4h + h 2 + 3 = 7 + 4h + h 2 Gradient of chord = − − y y x x 2 1 2 1 = h h h 4 + 2 = 4 + h Test yourself 1 P is the point on the curve y = 2x 2 + 1 where x = 3. Calculate the gradient of the chord PQ when Q is the point on the curve where: a x = 3.4 b x = 3.1 Answers on page 178 The gradient function PAGES 222–224 Sample question 1 found the gradient of the chord PQ when the x-value of Q has different distances from P. In particular, when the x-difference is h, the gradient of the chord is 4 + h. As h gets nearer to 0, so Q gets nearer to P and the chord gets nearer to the tangent at P. So the gradient of the tangent at P is 4. Using similar methods for different values of x at P, it can be shown that for the curve y = x 2 + 3, the gradient of the curve is 2x. The equation for the gradient of a curve is called the gradient function of the curve. It is also called the derived function of the curve or derivative. Differentiation PAGES 224–228 Finding the gradient function is called differentiation. In general, for a function f(x) = axn , the gradient function is naxn − 1. When you differentiate a multiple of x (e.g. 3x) you get the multiple (i.e. 3) as the answer. When you differentiate any number term you get zero. Sample questions and answers 1 a 12x 1 = 12x c 2x 3 e 2.4x 2 b 50x 4 d 0 f 9 1 Differentiate each of the following expressions. a 6x 2 c 1 2 x 4 e 0.8x 3 b 10x 5 d 7 f 9x Tip Remember that 2x 2 means 2 × x × x or 2 × x 2, not (2 × x) 2 . Tips l Write expressions as sums of terms in powers of x. l The derivative of an expression with several terms is found by differentiating each term separately. 9781510421714.indb 73 11/07/19 2:31 PM

74 The second derivative It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Sample questions and answers 1 Find the first derivative of each of the following: a y = 7x 4 + 2x 3 − 3x + 1 b s = 5 + 6t − 4t 2 [Notice that the variables can be any letter.] c g(x) = (x − 1)(2x + 1) d y = 5 3 3 2 x x x x + − 1 a 28x 3 + 6x 2 – 3 b –8t + 6 c g(x) = 2x2 – x – 1 so derivative is 4x – 1 d y = 5x 2 + 3x – 1 so derivative is 10x + 3 Test yourself 2 Differentiate each of the following expressions: a y = 12x 3 c s = 1 2 t 4 - 6t 2 + 1 4 e f(x) = x x x x 4 3 2 2 3 2 − + b f(x) = 3x 5 + 7x 3 - 2x 2 + 4x - 5 d y = (2x + 5)(3x - 1) f y = (x + 8) 2 Answers on page 178 The second derivative PAGES 228–229 The second derivative is found by differentiating the first derivative. This notation is sometimes used. Function First derivative Second derivative y dy dx 2 2 d y dx f(x) f′(x) f′′(x) Sample questions and answers 1 For each of the following, find 2 2 d y dx : a y = x 5 + 6x 4 + 3x 3 b y = 7x 2 − 6x + 5 c dy dx = 3x 2 + 9x + 2 2 Find the second derivative of each of the following: a y = x 3 (2x 2 + 5x) b f(x) = x x x 3 2 − c s = t 3 − 5t 2 + 4t + 3 1 a dy dx =5x 4 + 24x 3 +9x 2 d y dx 2 2 = 20x 3 + 72x 2 +18x b dy dx = 14x – 6 d y dx 2 2 = 14 c d y dx 2 2 = 6x + 9 2 a y = 2x 5 + 5x 4 dy dx = 10x 4 + 20x 3 d y dx 2 2 = 40x 3 + 60x 2 b f(x) = x 2 – x f´(x) = 2x – 1 f˝(x) = 2 c ds dt = 3t 2 − 10t + 4 d s dt 2 2 = 6t – 10 Tip When using function notation, one dash means differentiate once and two dashes means differentiate twice. 9781510421714.indb 74 11/07/19 2:31 PM

75 19 Differentiation and the gradient function It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 Test yourself 3 For each of the following, find d y dx 2 2 : a y = 4x 6 c y = x − 2 3 x3 b y = 4x 3 − 15x 2 – 18x + 12 d y = (1 – x 2 ) 2 4 Find the second derivative of each of the following: a y = 2x 2 (x 2 + x − 3) c f(x) = 3x 2 + 5x b s = t − 3t 3 d y = 2x + x 2 − 8x 3 Answers on page 178 Sample exam questions and answers 1 a dy dx = 9x 2 + 4x b d y dx 2 2 = 18x + 4 2 a f(x) = x 2 – 8x + 21, so f´(x) = 2x – 8 b f´´(x) = 2 1 For y = 3x 3 + 2x 2 − 1 find: a dy dx [2] b 2 2 d y dx [1] 2 For f(x) = (x − 4)2 + 5 find: a f′(x) [2] b f′′ (x) [1] Gradient of a curve PAGES 229–231 To find the gradient of a curve at a given point: l Find the gradient function. l Substitute the x-coordinate of the given point into the gradient function to find the gradient at that point. Sample question and answer 1 Find the gradient of the curve y = 2x 3 - 5x 2  + 7x at the point (2, 10). 1 Gradient function = 6x 2 – 10x + 7 Gradient at the point (2, 10) = 6x 2 – 10x + 7 = (6 × 22 ) – (10 × 2) + 7 = 24 – 20 + 7 = 11 Test yourself 5 Find the gradient of the curve y = 3x 2 − 4x − 5 at the point (1, −6). 6 Find the gradient of the curve y = x 3 + 4x 2 − 10 at the point (−3, −1). Answers on page 178 Equation of a tangent to a curve at a given point PAGES 233–234 Find the gradient function. Substitute the x-coordinate of the given point into the gradient function to find the gradient of the curve at that point. The gradient of the tangent is the same as the gradient of the curve at that point. Replace the m in y = mx + c by the gradient (see Chapter 21). Substitute the coordinates of the given point to find the value of c. 9781510421714.indb 75 11/07/19 2:31 PM

76 Maxima, minima and points of inflexion It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Sample question and answer 1 Find the equation of the tangent of the curve y = 5x 2 − 7x + 3 at the point (2, 9). 1 Gradient function = 10x – 7 Gradient at (2, 9) = (10 × 2) – 7 = 20 – 7 = 13 Equation of tangent is y = 13x + c Substituting x = 2 and y = 9 gives 9 = (13 × 2) + c, so c = 9 – 26 = –17 Equation of tangent is y = 13x – 17 Test yourself 7 Find the equation of the tangent to the curve y = 2x 3 − 3x 2 + 5x at the point (1, 4). 8 Find the equation of the tangent to the curve y = 3 + 7x − x 3 at the point (−2, −3). Answers on page 178 Maxima, minima and points of inflexion PAGES 234–237 You need to know that any point on a curve where the gradient is zero is known as a stationary point or turning point. A curve can have a number of stationary points. Finding if stationary points are local maxima or local minima can be done by looking at the second derivative of the function. A point of inflexion is a stationary point that is not a local maximum or local minimum. Sample questions and answers 1 A curve has the equation y = x 2 − 2x + 3. a Calculate the gradient function. b Find the coordinates of the stationary point of the curve. 1 a If y = x 2 − 2x + 3, then − dy dx =2x 2 b When − dy dx =2x 2 =0 2x = 2 x = 1 Substitute x = 1 into the equation of the curve to find the y-coordinate. When x = 1 y = 12 − 2(1) + 3 = 2 The coordinate of the stationary point is (1, 2). 9781510421714.indb 76 11/07/19 2:31 PM

77 19 Differentiation and the gradient function It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 2 A curve has the equation y x = − 9 2 x + 7 1 3 3 . a Find the coordinates of the stationary points of the curve. b For each stationary point, state whether it is a maxima or a minima. 2 a If y x = 9 1 − x +27 3 3 , then = − dy dx x 9 2 When x 2 − 9 = 0 x 2 = 9 x = ±3 Substitute x = 3 and x = −3 into the equation of the curve to find the y-coordinates. When x = 3, y = ( 1 3) −9 3( )+27 =9 3 3 When x = −3, y = ( 1 − − 3) 9 3 ( ) − +27 =45 3 3 The coordinates of the stationary points are (3, 9) and (−3, 45). b When − dy dx = 9 x 2 , d y dx = 2x 2 2 For the stationary point (3, 9), d y dx =2( ) 3 =6 2 2 , which is >0; therefore this point is a minima. For the stationary point (−3, 9), = − = − d y dx 2( 3) 6 2 2 , which is <0; therefore this point is a maxima. Test yourself 9 For the function f 3 ( ) 4 1 2 x x = − x − , find the coordinates of the stationary point. 10 For the curve y x2 3x x 36 7 3 2 = + − + , find the coordinates of any turning points indicating if each is a minima or maxima. Answers on page 178 Tip Although there are other methods, the most efficient way to check maxima and minima is by using the second derivative. Sample exam questions and answers 1 Find the coordinates of the turning point of the curve f ( ) x x = − x x + 1 3 3 2 . [5] 2 Find the coordinates of the stationary points of the curves: a y = 3x 2 − 4x + 5 [4] b y = x 3 − 2x 2 + 2 [5] 9781510421714.indb 77 11/07/19 2:31 PM

78 Maxima, minima and points of inflexion It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide 1 f(x x )= −x x + 1 3 3 2 , f'( )x x = −2 1 x + 2 When x x − + 2 1 = 0 2 (x – 1)(x – 1) = 0 x = 1 When x = 1, y = − (1) (1) +1= 1 3 3 2 1 3 Coordinates of turning point are ( ) 1 1 3 , ; this is a point of inflexion. 2 a y = − 3 4 x x + 5 2 , = − dy dx 6 4 x When 6 4 x − = 0 = 2 3 x When = 2 3 x , = − 3 4 () ()+ = 5 3 2 3 2 2 3 2 y 3 Coordinates of stationary point are ( ) 3 2 3 2 3 , . b y x = −2 + x 2 3 2 , = − dy dx 3 4 x x 2 so 3 4 x x − =0 2 x x ( ) 3 4− =0 x = 0 or (3x – 4) = 0 x = 0 or x = 4 3 When x =0, y = − 0 2( ) 0 2+ =2 3 2 When x = 4 3, y = 2 () () − +2= 4 3 3 4 3 2 22 27 Coordinates of stationary points are (0, 2) and ( ) 4 3 22 27 , Exam-style questions 1 Find: a the first derivative of y = x 2 − 4x + 5 [1] b the second derivative of y = (1 - x 2 )(2x + 1). [4] 2 For y = 4x 3 − 15x 2 - 18x + 12 find: a the gradient function [2] b the values of x for which the gradient function is 0. [3] 3 a Show that the curve y = 5 + 3x - 2x 3 passes through the point A with coordinates (2, -5). [1] b Find the equation of the tangent to the curve at the point A. [4] 4 For the curve y = x 3 - x 2 - x- 1, find the coordinates of the stationary points. [5] 5 For the function f 9 ( ) 2: 1 3 3 x x = − x + a calculate f'(x) [1] b find the coordinates of the stationary points [4] c sketch the graph of f(x). [2] 6 For the curve y = 2x 2 - x 4 , find the coordinates of the stationary points that are local maxima. [6] Answers on pages 178–179 9781510421714.indb 78 11/07/19 2:31 PM

79 It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 20 Functions Key objectives l To use function notation, e.g. f(x) = 3x - 5, f: x → 3x - 5. l To find inverse functions f –1(x). l To form composite functions as defined by gf(x) = g(f(x)). Notation PAGE 239 f: x → 3x + 2 means: f is the function that maps x on to 3x + 2. It is more commonly written as: f(x) = 3x + 2. This is read: f of x equals 3x + 2. Using functions PAGES 239–241 f(7) means: the value of the function when x = 7. To find the value of a function, substitute the number or expression in the brackets for x in the formula. When you are given the value of the function and want to find the value of x, put the formula for the function equal to the number or expression. Then solve the equation to find the value of x. Sample questions and answers 1 For the function f(x) = 3x + 2, find: a f(7) b f( ) 1 2 c f(2x) 2 For the function g(x) = 2x - 1, find the value of x for which: a g(x) = 11 b g(x) = x + 7 1 a f(7) = 3 × 7 + 2 = 23 b f( )1 2 = 3 × 1 2 + 2 = 31 2 c f(2x) = 3 × (2x) + 2 = 6x + 2 2 a g(x) = 11 2x – 1 = 11 2x = 12 x = 6 b g(x) = x + 7 2x – 1 = x + 7 2x – x = 7 + 1 x = 8 9781510421714.indb 79 11/07/19 2:31 PM

80 Composite functions It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Test yourself 1 For the function f(x) = 5x + 1, find: a f(7) b f(-3) c f(0) 2 For the function g(x) = 3 - 4x, find: a g(6) b g( ) 11 2 c g(2x) 3 For the function h(x) = 3(2x - 7), find: a h(5) b h(3x) c h(x - 1) 4 For the function p(x) = − x 2 1 , find: a p(0) b p(–4) c p( ) 1 2 5 For the function q(x) = 3x 2 - 2, find: a q(5) d q(x + 3) b q(-1) e q(2x) + 5 c q(2x) 6 For the functions f(x) = 5x - 4 and g(x) = 7x + 2, find x for which: a f(x) = 6 c f(x) = g(x) b g(x) = -12 d 3f(x) = 2g(x) Answers on page 179 Composite functions PAGES 243–244 When two or more functions are combined, the result is called a composite function. The composite of the two functions f(x) and g(x) is written fg(x) or f[g(x)]. In general, the composite fg(x) is not the same as gf(x). To find the value of a composite function fg(x), find the value of g(x) first, then put the answer in place of x in the formula for f(x). Sample questions and answers 1 a fg(3) = f[g(3)] =f[32 – 2 × 3 + 1] Find g(3) first. =f[4] = 3 × 4 + 2. Then put the answer in place of x in the formula for f(x). =14 b gf(0) = g[f(0)] =g[3 × 0 + 2] =g[2] =22 – 2 × 2 + 1 =1 2 a fg(x) = f[g(x)] =f[x 2 – 2x + 1]. Replace x in the formula for f(x) with the formula for g(x). =3(x 2 – 2x + 1) + 2 =3x 2 – 6x + 5 b gf(x) = g[f(x)] =g[3x + 2] =(3x + 2)2 – 2(3x + 2) + 1 =(9x 2 +12x +4) –6x –4+1 =9x 2 + 6x + 1 Replace every x in the formula for g(x) with the formula for f(x). f(x) = 3x + 2 and g(x) = x 2 − 2x + 1 1 Find: a fg(3) b gf(0) 2 Find the formula for: a fg(x) b gf(x) 9781510421714.indb 80 11/07/19 2:31 PM

81 20 Functions It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 Test yourself 7 For f(x) = x – 3 and g(x) = x + 1, find: a fg(3) c a formula for b gf(–3) (i) fg(x) 8 For h(x) = 5x + 7 and k(x) = 2x – 4, find: a kh(4) c a formula for: (i) kh(x) 9 For p(x) = 4x + 7 and q(x) = 3 2 x ( ) + , find: a pq(5) c a formula for: b qp( ) 1 2 (i) pq(x) 10 For f(x) = 1 – 3x and g(x) = 2(x + 4), find: a fg(0) c a formula for: b gf(–2) (i) fg(x) 11 For h(x) = 4x – 3 and j(x) = x 2 , find a formula for: a hj(x) b jh(x) 12 For f(x) = 1 + 2x and g(x) = x 2 – 2x + 1, find a formula for: a fg(x) b gf(x) Answers on page 179 (ii) gf(x) b hk – 1 ( ) 2 (ii) hk(x) (ii) qp(x) (ii) gf(x) Inverse functions PAGES 241–242 An inverse function has exactly the opposite effect to that of a given function. For the function f(x), the inverse is written f−1 (x) so that when f(x) = x + 1, f −1 (x) = x − 1. To find the formula of an inverse function, put y in place of f(x). Rearrange the formula to make x the subject. Replacing y with x gives the formula of the inverse function. Sample questions and answers 1 For t(x) = 5x − 4, find: a a formula for t −1 (x) b t −1 (1) c t −1 (−3) 1 a t(x) = 5x – 4 y = 5x – 4 5x = y + 4 x = y + 4 5 , so t–1(x) = x + 4 5 b t –1(1) = 1 4 + 5 = 1 c t –1(–3) = + = –3 4 5 1 5 Alternative method for b and c without finding t–1(x): b 1 = 5x – 4 x = 1, so t–1(1) = 1 c –3 = 5x – 4 x = 1 5, so t–1 (–3) = 1 5 9781510421714.indb 81 11/07/19 2:31 PM

82 Inverse functions It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Test yourself 13 For f(x) = 7x + 2, find: a f –1(x) b f–1(2) c f–1(0) 14 For g(x) = 4 1 x − , find: a g–1(x) b g–1 11 4 ( ) c g–1(-5) 15 Find the inverse of each of the following functions: a p(x) = 3(x + 2) b h(x) = 2 5 x − 4 c k(x) = 4(3x - 6) d t(x) = 5x 2 + 3, where x  0 Answers on page 179 Sample exam question and answer 1 For the function f(x) = px + q, f(5) = 7 and f(1) = 3. Find the values of p and q. [4] 1 You are given the value of the function for two values of x so you can form a pair of simultaneous equations. f(5) = p × 5 + q so 5p + q = 7 f(1) = p × 1 + q so p + q = 3 Subtract the simultaneous equations: 4p = 4, so p = 1. Substitute the value of p: 1 + q = 3, so q = 2. Exam-style questions 1 f: x → 2x - 1 and g: x → x 2 - 1. Find, in their simplest form: a f-1 (x) [2] b gf(x) [3] 2 f(x) = 1 3 x and g(x) = 2x 2 - 5 a Find: (i) g(4) [1] (ii) fg(4) [1] 3 f: x → 3x - 5 and g: x → x + 1 a Calculate f(-1). [1] b Find f -1 (x). [2] c Find fg(x). Give your answer in its simplest form. [2] d Solve the equation 3f(x) = 5g(x). [3] 4 f(x) = 1 3 (2x + 5) a Find f(-4). [1] b Find f -1 (x). [3] 5 f(x) = 3x + 1 and g(x) = 2x 2 . Find, in their simplest form: a f-1 (x) [2] c gf(x) [3] b fg(2) [2] d ff(x) [2] 6 f(x) = 2x + 1 and g(x) = x 3 - 5 a Find f(-1). [1] c Find g-1 (x). [2] b Find f -1 (x). [2] d Find fg(0). [2] e Find fg(x) in its simplest form. [2] Answers on page 179 b Find expressions for: (i) gf(x) [1] (ii) f–1(x) [1] 9781510421714.indb 82 11/07/19 2:31 PM

83 It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 21 Key objectives l To demonstrate familiarity with Cartesian coordinates in two dimensions. l To find the gradient of a straight line. l To interpret and obtain the equation of a straight-line graph in the form y = mx + c. l To determine the equation of a straight line parallel to a given line. l To calculate the gradient of a straight line from the coordinates of two points on it. l To calculate the length and the coordinates of the midpoint of a straight line from the coordinates of its end points. l To find the gradient of parallel and perpendicular lines. Drawing straight-line graphs PAGES 266–267 There are two main ways of writing equations of straight-line graphs: the first type y = 2x + 1, and the second type 3x + 2y = 6. To draw the first type: l Step 1: Draw up a table of values using three x numbers. l Step 2: Substitute these into the formula to find the corresponding y values. l Step 3: Plot the coordinate pairs. They should lie in a straight line. To draw the second type: l Step 1: Draw up a table of values. Use x = 0, y = 0 and one other x value. l Step 2: Substitute these into the formula to find the other values. l Step 3: Plot the coordinate pairs. Sample question and answer 1 Draw the line y = 2x + 1. 1 x 0 1 2 y 1 3 5 x = 2, y = 2 × 2 + 1 = 5 y 0 1 x 1 2 2 3 3 4 5 Straight-line graphs 9781510421714.indb 83 11/07/19 2:31 PM

84 Graphical solution of simultaneous equations It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Sample question and answer 1 Draw the line 3x + 2y = 6. 1 x 0 2 1 y 3 0 11 2 For x = 0, 2y = 6 so y = 3 For y = 0, 3x = 6 so x = 2 For x = 1, 3 + 2y = 6 so y = 11 2 y x 0 1 1 2 2 3 3 4 Test yourself Draw the following straight-line graphs: 1 y = 3x − 2 2 2x + 5y = 10 3 y = 6 − 2x Answers on pages 179–180 Graphical solution of simultaneous equations PAGES 267–268 Simultaneous equations can also be solved by graphical methods. The solution is found where the graphs intersect. l Step 1: Work out some values for each equation. l Step 2: Draw the two graphs on the same grid. l Step 3: Find where the two lines cross. Write down the x and y values of the point of intersection. Sample question and answer 1 Solve graphically y = 4x − 1 and x + y = 4. 1 The line y = 4x – 1 passes through (0, –1) and (2, 7), and x + y = 4 passes through (0, 4) and (4, 0). Solution is x = 1, y = 3. 0 2 4 6 8 y x 2 –2 4 6 –4 (1, 3) Test yourself 4 Solve these simultaneous equations graphically: a y = 2x − 1, y = x + 2 b y = 2x − 5, 2x + y = 3 Answers on page 180 9781510421714.indb 84 11/07/19 2:31 PM

85 21 Straight-line graphs It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 Sample exam question and answer 1 Solve these simultaneous equations graphically: y = 3x − 1, y = 3 − 2x [6] 1 The line y = 3x – 1 passes through (0, –1), (1, 2) and (2, 5). The line y = 3 – 2x passes through (0, 3), (1,1) and (2, –1). Solution is x = 0.8, y = 1.4. 0 2 3 y x 1 –1 1 2 3 4 (0.8, 1.4) y = 3 – 2x y = 3x – 1 5 Gradient and y-intercept PAGES 252–257 l The gradient of a line is a number indicating how steep it is. The larger the number, the steeper the line. l Lines with positive gradient slope forwards (/); lines with negative gradient slope backwards (\). l Gradient y x change in  change in  = . l For a line through (x 1 , y 1 ) and (x 2 , y 2 ), the gradient is − − y y x x 2 1 2 1 . Sample questions and answers Write down the gradient and y-intercept of these lines. 1 0 1 2 3 4 y x −1 −2 increase in y is 6 2 increase in x is 2 31 2 0 1 2 3 4 y x 1 −1 5 2 3 1 Gradient = 6 2 = 3; y-intercept = –2 OR using (2, 4) and (0, –2), gradient = 6 2 =3 2 Gradient = −4 2 = –2; y-intercept = 5 OR using (0, 5) and (2, 1), gradient = −4 2 = –2 Test yourself 5 Find the gradient and y-intercept of each of these lines. a 6 Find the gradient of the line that passes through each pair of coordinate points. a (−4, 2) and (4, 0) c (−1, −3) and (2, 3) b (0, −2) and (4, 4) d (−6, 4) and (6, −2) 7 For each part of question 6, draw a diagram to find the y-intercept of the line. Answers on page 180 b 0 4 6 8 y x 2 21 3 0 2 3 y x 1 1 2 3 Tip The y-intercept is the value where the line crosses the y-axis. 9781510421714.indb 85 11/07/19 2:31 PM

86 It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide The general equation of a straight line y = mx + c Midpoint and length of a line segment A line segment is the part of a line between two points. It has finite length. For the line segment joining (x 1 , y 1 ) and (x 2 , y 2 ): Midpoint = x x     y y 2 ,     2  1 2 + + 1 2      Length = (   x x –  ) ( + − y y  ) 1 2 2 1 2 2 Sample questions and answers 1 A is (1, 5) and B is (7, 2). Find: a the midpoint of AB b the length AB 1 a Midpoint=  + +      1 7 2 , 5 2 2 =(4, 3 1 2 ) b Length= 7 1 2 5 2 2 ( ) − +( ) − = 6 3 + −( ) 2 2 = 45 = 6.7 to 1 d.p. Test yourself 8 Find: (i) the midpoint (ii) the length joining each of these pairs of points: a (4, 5) and (0, 1) c (−6, 1) and (7, 5) e (2, 3) and (−1, −3) b (2, 6) and (−4, 3) d (−3, −2) and (−4, −1) Answers on page 180 The general equation of a straight line y = mx + c PAGES 263–264 In the equation, m stands for the gradient of the line and c is the y-intercept. The equation of a line must be of the form y = mx + c for the two numbers to represent the gradient and y-intercept. Sample questions and answers 1 y = 1x + 4, m = 1, c = 4 2 y = 3x – 1.5,m = 3, c = –1.5 3 –2y = –5x + 12, so y = 2.5x − 6, m = 2.5, c = −6 Find the gradient and y-intercept of these lines. 1 y = x + 4 2 2y = 6x − 3 3 5x − 2y = 12 PAGES 269–270 9781510421714.indb 86 11/07/19 2:31 PM

87 21 Straight-line graphs It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 Sample question and answer 1 Find the equation of the line through the points (10, 20) and (30, 30). 1 m = 30 20 30 10 10 20 − − = = 0.5, so y = 0.5x + c As the line goes through (10, 20), 20 = 0.5 × 10 + c, so 20 = 5 + c, c = 15 So the equation is y = 0.5 x + 15. Test yourself 9 Work out the gradient, m, and y-intercept, c, for each of the following straight lines. a y = 4x − 1 d y + x = 4 b y = 3 + 2x e 2y − 3x = 4 c y = x f x − 4y = 10 10 Find the equation of the line in each part. a Gradient 3 and passing through the point (0, −2) b Gradient 1 2 − and passing through the point (0, 7) c Gradient 2 and passing through the point (−2, 4) d Gradient −4 and passing through (−3, 1) 11 Find the equation of the line passing through the points: a (1, 0) and (4, 9) b (−2, 9) and (7, −9) Answers on page 180 Parallel lines PAGE 265 Lines that are parallel have the same gradient. Sample question and answer 1 Find the equation of the line parallel to y = 2x − 5 and passing through (0, 4). 1 From y = 2x – 5, m = 2 From (0, 4), c = 4 The equation is y = 2x + 4. Test yourself 12 Which of these lines A–F are parallel? A y = 6x − 2 D 6y = 36x − 24 B y = 2x − 6 E y = 6 + x C 6y = 6x − 2 F 6x − y = 2 13 Find the equation of the line passing through the point (0, 7) and parallel to y = 2x. 14 Find the equation of the line passing through the point (5, 6) and parallel to y = 3x − 3. 15 Find the equation of the line passing through the point (−3, 1) and parallel to 4y + 2x = 1. 16 Find the equation of the line passing through the point (5, −1) and parallel to the line that passes through the two points (2, 3) and (5, −6). Answers on page 180 9781510421714.indb 87 11/07/19 2:31 PM

88 Perpendicular lines It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Perpendicular lines PAGES 271–274 Two lines with gradients m1 and m2 are perpendicular if m1 × m 2 = −1. This means that m2 = − m 1 1 . Sample exam question and answer 1 Find the equation of the line perpendicular to y = 4x + 1 that passes through the point (8, 5). [4] 1 m 1 = 4, so m2 = − 1 4 y = –0.25x + c As the line goes through (8, 5), 5 = –0.25 × 8 + c, so 5 = –2 + c, c = 7 So the equation is y = –0.25x + 7. Test yourself 17 Which of these lines A–F are perpendicular? A y = 3x − 5 D 3y = 2x −5 B 2y = 3x − 5 E 2y = 6x − 5 C 3y = 5 − 2x F 6x − y = 2 18 Find the equation of the line passing through the point (0, 3) and perpendicular to y = 4x + 5. 19 Find the equation of the line passing through the point (−5, 2) and perpendicular to 4y + 2x = 1. 20 Find the equation of the line passing through the point (6, −1) and perpendicular to the line that passes through the two points (1, −1) and (3, 5). Answers on page 180 Sample exam question and answer 1 Find the gradient and the equation of the straight line in the diagram. [3] 0 1 2 3 4 y x 1 −1 −2 −1 5 2 3 4 5 (4, 5) 1 Gradient = 3 4 Equation is y = 3 4 x + 2 9781510421714.indb 88 11/07/19 2:31 PM

89 21 Straight-line graphs It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 Exam-style questions 1 a Work out the gradient of this line. [2] b Write down the equation of the line. [2] 0 1 2 3 4 y x 1 −1 5 2 3 2 a Write down the gradient and y-intercept of the line with equation y = 4 − 2x. [2] b Write down the equation of the line parallel to y = 4 − 2x that passes through the point (0, −1). [2] 3 Solve these simultaneous equations graphically: y = 3 − x, y = 3x − 2. [6] 4 Solve these simultaneous equations graphically: y = 3x + 4, x + y = 2. [6] 5 Find: (i) the midpoint of the line (ii) the length of the line joining each of these pairs of points: a (3, 7) and (8, 2) [2] b (4, 1) and (−2, 3) [3] 6 The midpoint of the line segment XY is (−1, 2) X is the point (3, 6). What are the coordinates of Y? [2] 7 Find the equation of the straight line that passes through the two points (−3, 12) and (5, −4). [5] 8 Find the equation of two straight lines that pass through the point (1, 1), one parallel to x + y = 1 and the other perpendicular to x + y = 1. [6] 9 a Find the equation of the line through A (6, −1) and B (3, 2) [5] b Find the equation of the line perpendicular to AB that passes through A. [4] Answers on page 180 9781510421714.indb 89 11/07/19 2:31 PM

90 It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Key objectives ● To use and interpret the geometrical terms: point, line, parallel, bearing, right angle, acute, obtuse and reflex angles, perpendicular, similarity and congruence. ● To use and interpret the vocabulary of triangles, quadrilaterals, circles, polygons and simple solid figures including nets. ● To measure and draw lines and angles. ● To construct a triangle given the three sides using only a ruler and compasses. ● To read and make scale drawings. Angles and lines PAGE 282 l Acute angles are angles between 0° and 90°. l Right angles are 90°. l Obtuse angles are between 90° and 180°. l Reflex angles are between 180° and 360°. l Perpendicular lines meet at a right angle. l Parallel lines never meet and are marked with arrows. Triangles PAGES 283–285 l Isosceles triangles have two equal sides and two equal angles. l Equilateral triangles have three equal sides and three equal angles. l Scalene triangles have all three sides with different lengths. l In a right-angled triangle the longest side is called the hypotenuse. Circles PAGE 285 The names connected with circles are shown in these two diagrams. The segment in the right-hand diagram is called a minor segment. A segment that is more than half the circle is called a major segment. The sector in the right-hand diagram is called a minor sector. A sector that is more than half the circle is called a major sector. Quadrilaterals PAGES 285–286 The table on the next page summarises the special quadrilaterals: Arc Radius Centre Diameter Chord Tangent Segment Sector Centre Circumference Geometrical vocabulary 22 and construction 9781510421714.indb 90 11/07/19 2:31 PM

91 22 Geometrical vocabulary and construction It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 Shape Name Sides Angles Diagonals Square All equal All equal Equal and bisect each other at right angles Rectangle Opposite sides equal All equal Equal and bisect each other Rhombus All equal Opposite angles equal Bisect each other at right angles Parallelogram Opposite sides equal Opposite angles equal Bisect each other Trapezium One pair of opposite sides parallel Kite Two pairs of adjacent sides equal One pair of opposite angles equal Intersect at right angles Polygons PAGES 286–287 Polygons are shapes with straight sides. The table below gives the numbers of sides and names: 3 4 5 6 7 8 10 12 Triangle Quadrilateral Pentagon Hexagon Heptagon Octagon Decagon Dodecagon Nets of solids PAGE 287 The net of a solid is the shape you would cut out and fold to make the solid. This is a possible net for a cube. Sample questions and answers For this accurately drawn diagram write down: 1 an acute angle 2 an obtuse angle 3 two lines perpendicular to each other 4 two parallel lines D A B C 1 angle D 2 angle A 3 AB and BC or DC and BC 4 AB and DC 9781510421714.indb 91 11/07/19 2:31 PM

92 Measuring angles It is illegal to photocopy this page Cambridge IGCSE Mathematics Core and Extended Study and Revision Guide Sample exam question and answer 1 a diameter b arc c chord d radius 1 From this list of words pick the right one for each label on the diagram. Circumference Tangent Chord Sector Radius Segment Arc Diameter [4] d c b a Test yourself 1 State whether these angles are acute, obtuse or reflex: a 63° d 168° b 248° e 339° c 94° 2 Which of these are possible nets for a cube? a b c d 3 Explain why these two rectangles are NOT similar. (The rectangles are not to scale.) 3cm 6cm 6 cm 9cm Answers on page 180 Drawing shapes PAGES 288–289 You should be able to draw and measure lines and angles. You should be able to construct a triangle given three sides using only a ruler and compasses. You should be able to construct other shapes using a ruler, compasses and protractor. Measuring angles PAGES 288–289 Place your protractor so that the zero line is along one arm of the angle and the centre is at the point of the angle. 9781510421714.indb 92 11/07/19 2:31 PM

93 22 Geometrical vocabulary and construction It is illegal to photocopy this page Hodder & Stoughton Limited © Brian Seager et al., 2019 100 80 70 60 50 40 30 20 10 0 110 120 130 140 150 160 170 180 0 10 20 30 40 50 60 70 80 100 110 120 130 140 150 160 170 180 90 R Q P Start at zero. Go round this scale until you reach the other arm of the angle. Then read the size of the angle on the scale. Angle PQR = 117° Drawing angles PAGES 288–289 The following instructions describe how to draw an angle of 45°. Draw a line. Put the centre of the protractor on one end of the line with the zero line over the line you have drawn. 100 80 70 60 50 40 30 20 10 0 110 120 130 140 150 160 170 180 0 10 20 30 40 50 60 70 80 100 110 120 130 140 150 160 170 180 90 Go round the scale until you reach 45° and mark a point. Remove the protractor and draw a straight line from the point to the end of the line. Triangle with three sides given l Step 1: Draw line AB of given length. l Step 2: Use compasses to construct arcs AC and BC with the compasses set to the given lengths. l Step 3: Draw AC and BC. Other shapes Triangle knowing two sides and the included angle l Step 1: Draw line AB of given length. l Step 2: Measure and draw the angle at A. l Step 3: Draw line AC of given length. l Step 4: Join C to B. C A B Arc with centre A Arc with centre B C A B 9781510421714.indb 93 11/07/19 2:31 PM