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1 Q1. A trapezium has parallel sides of length (x + 1) cm and (x + 2) cm. The perpendicular distance between the parallel sides is x cm. The area of the trapezium is 10 cm 2. Not drawn accurately Find the value of x. Answer x =... cm (Total 5 marks) Page 1 of 64

2 Q2. Here are the equations of four straight lines. Line 1: y = x + 4 Line 2: y = 3x Line 3: y = 3x + 5 Line 4: y = x + 5 (a) Which two lines are parallel? Answer... and... (1) (b) Which two lines intersect the y axis at the same point? Answer... and... (1) (Total 2 marks) Q3. (a) Simplify fully (2) (b) Given that work out the value of Write your answer in its simplest form. (2) (Total 4 marks) Page 2 of 64

3 Q4. Solve the equation Answer x =... (Total 4 marks) Q5. (a) Simplify fully You must show your working. (2) (b) Rationalise the denominator and simplify (2) (Total 4 marks) Page 3 of 64

4 Q6. On Friday the ratio of the time Priya is sleeping to the time she is awake is 3 : 5. She is sleeping for less time than she is awake. (a) Work out the number of hours that she is sleeping on Friday. hours (2) (b) On Saturday she sleeps for one hour more than she did on Friday. Show that the ratio of the time she is sleeping to the time she is awake on Saturday is 5 : 7 (3) (Total 5 marks) Q7. Show that is an integer. (Total 2 marks) Page 4 of 64

5 Q8. Multiply out and simplify (2p 5q)(3p + q) (Total 3 marks) Q9. The line PQ is shown on the grid. (a) Find the gradient of a line which is perpendicular to PQ.... (3) Page 5 of 64

6 (b) Hence find the equation of the perpendicular bisector of the line PQ.... (2) (Total 5 marks) Q10. (a) Find the values of a and b such that x 2 + 6x 3 = (x + a) 2 + b Answer a =..., b =... (2) Page 6 of 64

7 (b) Hence, or otherwise, solve the equation x 2 + 6x 3 = 0 giving your answers in surd form (3) (Total 5 marks) Q11. Each term of a Fibonacci sequence is formed by adding the previous two terms. A Fibonacci sequence starts a, b, a + b, 1, 1, 2, 3, 5, 8, 13, 21, (a) Use algebra to show that the 6th term of this Fibonacci sequence is 3a + 5b (2) Page 7 of 64

8 (b) Use algebra to prove that the difference between the 9th term and 3rd term of this sequence is four times the 6th term. (3) (Total 5 marks) Q12. The table gives the diameter, in metres, of planets in the solar system. The diameters are given to an accuracy of 3 significant figures. Planet Diameter (metres) Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto Page 8 of 64

9 (a) Which planet has the largest diameter? (1) (b) Which planet has the smallest diameter? (1) (c) Which planet has a diameter approximately 10 times that of Venus? (1) (d) Write as an ordinary number. (1) (e) What is the diameter of Pluto in kilometres? Give your answer in standard form. km (2) (Total 6 marks) Q13. (a) Simplify.... (2) Page 9 of 64

10 (b) Simplify (3) (Total 5 marks) Q14. (a) Write down whether each of the following is an expression (X), an identity (I), an equation (E) or a formula (F). X, I, E or F v = u + at 3n + 2n 5n 3x + 2 = 7 + 2x 3 (3) (b) Show clearly that (2) (Total 5 marks) Page 10 of 64

11 Q15. Rearrange to make x the subject. Simplify your answer as much as possible. (Total 4 marks) Q16. Write each of these in the form p, where p is an integer. (a) (2) (b) (2) Page 11 of 64

12 (c) (2) (Total 6 marks) Q17. (a) Show clearly that (p + q) 2 p 2 + 2pq + q 2 (1) (b) Hence, or otherwise, write the expression below in the form ax 2 + bx + c (2x + 3) 2 + 2(2x + 3)(x 1) + (x 1) 2 (3) (Total 4 marks) Page 12 of 64

13 Q18. A is the point (2, 9) B is the point (8, 7) M is the midpoint of AB C is the point (8, 18) Not drawn accurately Is MC perpendicular to AB? You must justify your answer. Do not use graph paper to answer this question. (Total 4 marks) Q19. (a) Factorise 5x x (1) (b) Factorise x 2 49 (1) Page 13 of 64

14 (c) Factorise fully (3x + 4) 2 (2x + 1) 2 (3) (Total 5 marks) Q20. Evaluate (a) (3) (b) (2) (Total 5 marks) Page 14 of 64

15 Q21. A is the point (1, 2). B is the point (5, 4). Find the equation of the line perpendicular to AB, passing through the mid-point of AB. (Total 4 marks) Page 15 of 64

16 Q22. Find the values of a and b such that x 2 10x + 18 = (x a) 2 + b Answer a =..., b =... (Total 2 marks) Q23. Find the equation of the line through (0, 2) and (4, 18). (Total 3 marks) Page 16 of 64

17 Q24. Solve the simultaneous equations y = x + 2 y = 3x 2 You must show your working. Do not use trial and improvement.... (Total 5 marks) Q25. A shape is made from two trapezia. Not drawn accurately The area of this shape is given by A = (a + b) + (a + h) Page 17 of 64

18 Rearrange the formula to make a the subject. Answer a =... (Total 4 marks) Page 18 of 64

19 Q26. Match each of the shaded regions to one of these inequalities. A y + 2 D y 2x 4 B y + 2 E y 2x 4 C y 2x + 4 Region 1... Region 2... Region 3... Region 4... (Total 4 marks) Page 19 of 64

20 Q27. Julie has a bag containing x blue marbles and y red marbles. The ratio of blue marbles to red marbles is 2:3 She adds z blue marbles. The ratio of blue marbles to red marbles is now 2:1 What is the ratio between x and z? (Total 3 marks) Q28. Make x the subject of the formula a(x b) = a 2 + bx... (Total 4 marks) Page 20 of 64

21 Q29. The triangle number sequence is 1, 3, 6, 10, 15, 21,... The nth term of this sequence is given by n(n + 1) (a) Write down an algebraic expression for the (n 1)th term of the sequence. (1) (b) Prove that the sum of any two consecutive triangle numbers is a square number. (3) (Total 4 marks) Page 21 of 64

22 Q30. (a) This is a page from Zoe s exercise book. Give a counter example to show that Zoe is wrong. Justify your answer. (2) (b) Prove that (n + 5) 2 (n + 3) 2 = 4(n + 4) (3) (Total 5 marks) Page 22 of 64

23 Q31. Solve these simultaneous equations x + 3.6y = 2 x 2.4y = 5 You must show all your working. Do not use trial and improvement. Answer x =... y =... (Total 3 marks) Q32. (a) Find the value of (1) (b) Find the value of 8x 0 (1) (Total 2 marks) Page 23 of 64

24 Q33. The diagram shows the graph of an equation of the form y = x 2 + bx + c Find the values of b and c. You must show your method. Answer b =..., c =... (Total 3 marks) Q34. Some large numbers are written below. 1 million = billion = trillion = (a) How many millions are there in one trillion?... (1) Page 24 of 64

25 (b) Write 8 billion in standard form.... (1) (c) Work out 8 billion multiplied by 3 trillion. Give your answer in standard form.... (2) (Total 4 marks) Q35. Annie, Bert and Charu are investigating the number sequence 21, 40, 65, 96, 133,... (a) Annie has found the following pattern. 1st term = 21 2nd term = 40 3rd term = 65 4th term = 96 5th term = 133 Complete the nth term for Annie s pattern. nth term n (n + 1) (2) Page 25 of 64

26 (b) Bert has found this formula for the nth term (3n + 1)(n + 3) + 5 Charu has found this formula for the nth term (2n + 3) 2 (n + 1) 2 Prove that these two formulae are equivalent. (3) (Total 5 marks) Page 26 of 64

27 Q36. (a) Find the equation of the line AB. (3) (b) Give the y-coordinate of the point on the line with an x-coordinate of 6. (2) (c) Write down the gradient of a line perpendicular to AB. (1) (Total 6 marks) Page 27 of 64

28 Q37. (a) Factorise 2n 2 + 5n (2) (b) Hence, or otherwise, write 253 as the product of two prime factors.... (1) (Total 3 marks) Q38. (a) n is a positive integer. (i) Explain why n(n + 1) must be an even number (1) (ii) Explain why 2n + 1 must be an odd number (1) (b) Expand and simplify (2n + 1) (2) Page 28 of 64

29 (c) Prove that the square of any odd number is always 1 more than a multiple of (3) (Total 7 marks) Q39. (a) (i) Factorise x 2 10x (2) (ii) Hence, or otherwise, solve the equation (y 3) 2 10(y 3) + 25 = Answer y =... (2) Page 29 of 64

30 (b) Simplify (3) (Total 7 marks) Q40. Make x the subject of the formula Answer x =... (Total 4 marks) Page 30 of 64

31 Q41. Find the equation of the straight line passing through the point (0, 5) which is perpendicular to the line y = x + 3 (Total 2 marks) Q42. Make x the subject of the formula w = x 2 + y.... Answer x =... (Total 2 marks) Page 31 of 64

32 Q43. Solve the simultaneous equations 4x + 3y = 14 2x + y = 5 You must show your working. Do not use trial and improvement. Answer x =..., y =... (Total 3 marks) Q44. A special packet of breakfast cereal contains 20% more than a normal packet. The special packet contains 600 g of cereal. How much cereal does the normal packet contain? g (Total 3 marks) Q45. Two gas supply companies have different ways of charging for the gas they supply. Alpha gasco Fixed Charge 9.60 Price per kilowatt hour of gas First 5 kilowatt hours free then 1.30 for every kilowatt hour over 5. Beta gasco Fixed Charge No fixed charge Price per kilowatt hour of gas 1.50 for every kilowatt hour. Page 32 of 64

33 Find the number of kilowatt hours after which Alpha gasco becomes cheaper than Beta gasco. You might want to use some graph paper. You must show your method clearly. Answer... kilowatt hours (Total 4 marks) Q46. The region R is shown shaded below. Page 33 of 64

34 Write down three inequalities which together describe the shaded region (Total 3 marks) Q47. Solve the equation (Total 5 marks) Page 34 of 64

35 Q48. Simplify fully Answer... (Total 4 marks) Q49. (a) (i) Evaluate 13z 0 Answer... (1) (ii) Evaluate (13z) 0 Answer... (1) (b) If 3 x =, find the value of x. Answer x =... (2) (c) If 4 y =, find the value of y. Answer y =... (2) (Total 6 marks) Page 35 of 64

36 Q50. On the grid below, indicate clearly the region defined by the three inequalities Mark the region with an R. x 1 y x 1 x + y 7 (Total 3 marks) Page 36 of 64

37 Q51. Solve the equation (Total 5 marks) Q52. Simplify (Total 4 marks) Page 37 of 64

38 . (Area =) x (x x + 2) 2x 2 + 3x 20 = 0 oe (x + 1) + x (1) oe eg x x 10 = 0 (2x 5)(x + 4) = 0 for an attempt at using an algebraic method such as factorising, formula (allow one error) or completing the square (allow one error) to solve the quadratic eg for (2x + a)(x + b) where ab = ± 20 for a completely correct method dep x = 2.5 Do not award last if a negative value given as possible answer eg if 4 given 2.5 seen with no or incomplete work SC2 2.5 after first, give 5/5 [5] M2. (a) 2 and 3 oe (b) 3 and 4 oe [2] Page 38 of 64

39 M3. (a) 16 4 (= 4 2) or or 2(2 2 2) = 2( 2) both steps needed or Both steps needed 2 (b) or or or Do not allow for oe [4] M4. 3(3x + 1) 2 (2x + 5) 9x + 3 4x 10 (their 5x 7) = 6 x = 2.6 Could have 6 as denominator here Condone lack of brackets or dep [4] Page 39 of 64

40 M5. (a) 5 3 or (b) 3 7 [4] M6. (a) Condone 3 unsupported is M0 9 Do not allow (of a day) SC1 Answer 15 or 9 and 15 (b) (their 9 + 1) : 24 (their 9 + 1) 10 and 14 seen 10:14 Must be integers ft 5:7 Must have seen previous ratio [5] Page 40 of 64

41 M ( 2 = 4 2) If attempts to square the bracket 2500 ± 50 2 ± 50 2 ± [2] M8. 6p 2 + 2pq 15pq 5q 2 For 3 correct terms 6p 2 + 2pq 15pq 5q 2 Fully correct 6p 2 13pq 5q 2 From 4 terms Do not ignore fw ft [3] M9. (a) Gradient of PQ = Perp. grad. = (= oe) Drawing method: Perpendicular line drawn and attempt at finding its gradient M2 dep oe Page 41 of 64

42 (b) y = (their ) x + c y = x + oe Accept 1.4 to 1.6 for from graph [5] 0. (a) (a =) 3 (b =) 12 Allow 12 if 12 given in working (b) (x + 3) 2 = 12 or (x =) Using their values from (a) Substitution into formula (allow 1 error) x + 3 = 12 or (x =) Using their values from (a) dep (x =) ± 12 3 or [5] Page 42 of 64

43 1. (a) 4 th term = a + 2b or (a = 1 and b = 1 and) 3(1) + 5(1) oe Accept 5 th term = 2a + 3b (oe) for if 4 th term not seen. 6 th term Must see 4 th and 5 th terms (b) Continuing sequence to 9 th term = 3a + 5b, 5a + 8b, 8a + 13b, 13a + 21b Must come from continuing sequence and not from 4 6 th 3 rd Allow subtraction to be assumed. Condone missing bracket if answer correct Either way round, expansion or factorisation [5] 2. (a) Jupiter (b) (c) Pluto Saturn (d) (e) or oe [6] Page 43 of 64

44 3. (a) 3(x + 5) or 3x + 15 for 3 for x + 5 for B2 (b) (x 3)(x + 3) x(x + 3) Do not ignore further working [5] 4. Allow embedded solutions, but if contradicted M marks only (a) F, I, E, X 1eeoo B3 (b) Must have 4 terms Condone 1 sign error Must show cancellation, either by crossing out or stating ab ab = 0 [5] Page 44 of 64

45 5. y(3x 4) = xy + 2 3xy 4y = xy + 2 y 3x 4 = xy + 2 is M0 unless recovered 2xy = 4y + 2 dep 3xy xy = 4y + 2 Allow one sign error oe Do not award if x = not written SC x = B2 Alt. y(3x 4) = xy + 2 y 3x 4 = xy + 2 is M0 unless recovered 3x 4 = x + 3x 4 = 2x = 4 + 3x x = 4 + Allow one sign error dep x = 2 + oe Deduct mark if x = not written SC x = B2 [4] 6. (a) 300 oe eg, (2 3) (2 25) or ( ) (correct product of factors which includes 3 ) 10 3 SC1 for 5 12 or 2 75 (b) 4 3 or 5 3 seen 9 3 Page 45 of 64

46 (c) Attempt to rationalise ie, Multiply num. and denom. By 3 oe eg, scores 6 3 [6] 7. (a) Convincing algebra Must see box method and or (b) Allow one sign or coefficient error For middle term accept or ft if awarded and no further errors ft [4] Page 46 of 64

47 8. Mid point (5, 8) Gradient AB Accept any indication eg, 6 across, 2 down Attempt to find gradient MC or stepping from M to C for using Their gradient Valid conclusion with justification. eg, No because gradient MC not 3 Accept any indication eg, (5, 8) plus (3, 9) = (8, 17), mm 1 Alt Mid point (5, 8) Use of Pythagoras Three correct lengths Correct conclusion at least 2 correct values [4] 9. (a) 5x (x + 4) (b) (x + 7)(x 7) Page 47 of 64

48 (c) for expanding and collecting to general quad form, allow one error but expansions must have x 2 term, x term and constant term. Allow misuse of minus. eg 9x x x 2 + 4x + 1 Difference of two squares ((3x + 4) (2x + 1)) ((3x + 4) + (2x + 1)) 5x x + 15 for either (x + 3) or (5x + 5) if difference of 2 squares used. 5 (x + 3)(x + 1) Accept (x + 3)(5x + 5) or (5x + 15)(x + 1) [5] M20. (a) (±)6 (±)1.5 oe (b) oe eg, 0.01 for 100 or or B2 [5] Page 48 of 64

49 M21. Attempt to find gradient of perpendicular line Must be negative reciprocal of their gradient for AB (Gradient =) oe eg 0.66, 0.67 Use of midpoint (3, 1) Must be used either on the diagram with an attempt at a perpendicular or in y = mx + c to find c. y = x +3 ft their gradient if first awarded Accept equivalents eg 3y + 2x = 9 ft [4] M22. a = 5 b = 7 from expansion x 2 2ax + a 2 and comparing coeffs. or simply spotting that a = 5 ft. from their a using a 2 + b = 18 ie. b = 18 a 2 or by inspection ft [2] M23. Identifying 2 as constant term in equation y = mx + c Gradient = y = 5x 2 Attempt to find gradient oe [3] Page 49 of 64

50 M24. 3x 2 = x + 2 3x 2 x 2 = 0 y = 3(y 2) 2 3y 2 13y + 12 = 0 (3x + 2)(x 1) = 0 or (x ) 2 = ± ( ) or ± x = (3y 4)(y 3) = 0 (Reverse s below) Must be for factorising a quadratic. x (or y) terms must have product equal to square term and number terms must have a product equal to ± constant term. If completing the square or formula used must be to at least the stage shown for Method mark. or (y ) 2 = ± ( ) or ± y = x = 1 and Need both y = 3 and Must match appropriate values of y with x Must use y = x + 2, or x = y 2. Answers without any working is, otherwise answers must be supported by an algebraic method. Graphical method is M0. Special case: x = 1, y = 3 without working. (Can be guessed). NB only award this if no other marks awarded. ft [5] Page 50 of 64

51 M25. 2A = ah + bh + ab + bh Accept A= ah/2 + bh/2 + ab/2 + bh/2 Allow one error NB 4A = ah + bh + ab + bh is one error. 2A 2bh = ah + ab A bh = ah/2 + ab/2 2A 2bh = a(h + b) For factorising D or equivalent ft if both Ms awarded. oe e.g. ft [4] M26. 1 D 2 C 3 E 4 A 1 y 2x 4 2 y 2x y 2x 4 4 y x + 2 [4] M27. 6:3 or numerical values in the ratios 2:3 and 6:3 (x + z) : y = 2: 1 3x = 2y Page 51 of 64

52 Finding z e.g. 4 or appropriate numerical value x + z = 2y If both correct. Accept x + z = 2y 1: 2 oe Accept words e.g. z is twice x. [3] M28. ax ab = a 2 + bx Allow ax + ab = ax bx = a 2 + ab x(a b) = a 2 + ab For factorising NB sc x(a b) = a 2 + b Allow Ml and Al if D oe, e.g. Follow through on factorisation if D awarded. Do not award if x = not shown, fw such as cancelling a s do not award last Al. ft [4] M29. (a) oe e.g. Page 52 of 64

53 (b) oe e.g ft their a n 2 + 2n + 1 n 2 (n + 1) 2 [4] M30. (a) Continuation at least once more e.g = 61, = 91 (allow this to be prime if stated) Correctly evaluated. Justification that the answer is not prime. e.g. 91 = = 169 = Must show the factors. NB = 1 (1 not prime) Ml, (b) n 2 + 5n + 5n + 25 (n 2 + 3n + 3n + 9) Ml for expanding and subtracting (allow 1 arithmetical error). Condone invisible bracket n n + 25 n 2 6n 9 Al for all terms collected and correct signs or clear evidence of subtraction. 4n + 16 = 4(n + 4) Factorisation must be shown. Expanding is AO. [5] Page 53 of 64

54 M31. trial and improvement is 0 1st-2 nd 6y = 3 allow 1 error eg, 12y = 3 6y = y = y allow 1 error or 2.4equation(l) + 3.6equation(2) y = 0.5 or x = 3.8 y = 0.5 and x = 3.8 Must have both. Allow reversed if both seen correct in working ft if Ml awarded ft [3] M32. (a) 4 (b) 8 [2] Page 54 of 64

55 M33. Either, (x + 3)(x 5) = x 2 2x 15 Expansion not necessary for b = 2 c = 15 Note starting with (x 3)(x + 5) could give c = 15 and will score, A0, OR, substituting coordinates ( 3,0) and (5,0) into equation to get: 0 = 9 3b + c and 0 = 5 + 5b + c correct substitution which might be unsimplified eg. 0 = ( 3) 2 3b + c and 0 = b + c Solving to give b = 2 c = 15 [3] M34. (a) 10 6 oe (b) (c) or 22 as power or oe e.g B2 [4] Page 55 of 64

56 M35. (a) (n + 2) 2, (n + 1)(n + 4) 1 eeoo B2 (b) Expand Bert 3n n + 8 Allow one error but not 3n n + 3 Expand Charu 4n n + 9 (n 2 + 2n + 1) 4n 2 + 6n + 6n + 9 n 2 + n + n + 1 Convincing algebra that these are equivalent. Allow dealing correctly with (n 2 + 2n + 1) as minimum. e.g. 4n n + 9 n 2 2n 1 is [5] M36. (a) Intercept = 9 i.e. identifying that 9 is the constant term in the equation. Gradient = = 3 Any attempt at gradient for. i.e ±9/ ±3 y = 3x + 9 Accept equivalent forms. NB y = 3x + 9 is, Ml, A0 (b) Substitute x = 6 into their equation Or recognise that y-step from 0 to 3 is the same as 3 to 6. eg sight of 9. can be implied by answer only. 9 Page 56 of 64

57 (c) ft on their gradient in (a), Allow an 'embedded' answer in an equation, e.g. y = x + 9 ft [6] M37. (a) (2n ± 3)(n ± 1) or (2n ± 1)(n ± 3) (2n + 3)(n + 1) (b) Must see both factors [3] M38. (a) (i) Even odd, so even product or equivalent (ii) 2 n always even, so 2n + 1 is odd or equivalent (b) 4n 2 + 2n + 2n or 4 terms correct 4n 2 + 4n + 1 Must simplify Page 57 of 64

58 (c) Odd 2 1 = (2n + 1) 2 1 = 4n 2 + 4n = 4n(n + 1) Must factorise = 4 even Deduce even connection = multiple of 8 Concluding statement [7] M39. (a) (i) (x 5)(x 5) or (x 5) 2 for incorrect signs (ii) Indicating replacement of x by y 3 y = 8 Might just see (y 3 5) 2 or (y 8) 2 Re-starting?... must get as far as y 2 16y + 64 or (y 8) 2 to score B2 (b) (x 3)(x + 3) x(x + 3) [7] Page 58 of 64

59 M40. y(x 3) = 3x + 4 for cross-multiplying and expanding bracket yx 3y = 3x + 4 yx 3x = 3y + 4 correct expansion x(y 3) = 3y + 4 for clollecting terms and factorising x = (3y + 4)/(y 3) correct factorisation and division [4] M41. Sight of 1 or 1.5 or 3/2 accept 1 / ( ) or 1 / for only y = x + 5 oe eg. 2y = 3x + 10 [2] M42. x 2 = w y. x = (w y) Or equivalent -x 2 = y w Accept ± (w y) and (w y) [2] Page 59 of 64

60 M43. 4x + 3y = 14 4x + 3y = 14 4x + 2y = 10 6x + 3y = 15 allow error in one term y = 4 2x = 1 correct elimination from their equations x = and y = 4 oe SC correct answers with no working or using T & I [3] M % [3] M (x 5) 1.30 Alt: for graph of Alpha parcels = 1.50x 3.10 = 0.20x x = 15.5 for graph of Beta accuracy answer. Accept 16 but not 15. T&I gets iff taken as far as 15. for both schemes at 15 for both schemes at 16 conclusion [4] Page 60 of 64

61 M46. y 0 x 6 Accept y > 0, or 0 y 3, Accept x < 6 or 0 x 6, y Accept y < y = < x x or x 2y or equivalent. Any order. Special case: All three equations given (no inequalities) Special case: All three inequalities the wrong way around B2. [3] M47. (x 2) + 5x(x + 1) = 3(x + 1)(x 2) Allow 1 error 5x 2 + 6x 2 = 3x 2 3x 6 2x 2 + 9x + 4 = 0 (2x + 1)(x + 4) = 0 x = 1/2, 4 [5] Page 61 of 64

62 M48. Numerator = (x + 4)(x 4) Denominator = (3x 2)(x 4) = (3x 2)(x + 4) Answer = (x 4)/(3x 2) or 3x x 2x 8 or 3x 2 2x + 12x 8 [4] M49. (a) (i) 13 (ii) 1 (b) 3 x = 3 3 x = 3 for writing 1/27 as a power of 3, correctly allow embedded answer (c) 4y = 4 1½ y = 1½ for 4 y = 8 allow embedded answer [6] M50. Correct region indicated Award marks dependent upon number of lines drawn correctly and extent of shading B3 [3] Page 62 of 64

63 M51. LHS x(x 1) 2(x + 1) Give for x 2 3x + 2 if first line seen Allow invisible bracket if recovered. LHS = x 2 3x 2 Terms need not be collected. e.g.x 2 x 2x 2 (x 1)(x + 1)(= x 2 1) On RHS or as denominator. x 2 1 can be written as x 2 x + x 1 Their (x 2 3x 2) = their (x 2 1) Dependent on first 2 s D (= 0.33(3...)) Do not follow through. NB cancelling x 2 on top and bottom of Gives correct answer. Give,,. M0, A0. [5] M52. (5x ± a)(x ± b) for attempt to factorise. Must have (5x ± a)(x ± b) where ab = ± 3, a, b must be integers. (5x 1)(x + 3) (x 3)(x + 3) (5x 1)(x 3) Answer seen and further work then deduct last. [4] Page 63 of 64

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Quadratic Equations. All types, factorising, equation, completing the square. 165 minutes. 151 marks. Page 1 of 53

Quadratic Equations. All types, factorising, equation, completing the square. 165 minutes. 151 marks. Page 1 of 53 Quadratic Equations All types, factorising, equation, completing the square 165 minutes 151 marks Page 1 of 53 Q1. (a) Factorise x 2 + 5x 24 Answer... (2) (b) Solve x 2 + 5x 24 = 0 Answer... (1) (Total