3 a b c d 5(c + 4) = 5c + 20 Feedback: Don’t forget to multiply out both terms in the brackets. 6(t – 2) = 6t – 12 Feedback: Don’t forget 6(…..) means multiply both terms by 6. –3(4 – s) = –12 + 3s Feedback: Don’t forget –3(……) means multiply both terms by 6 and minus × minus = … 15 – (n – 4) = 15 – n + 4 = 15 + 4 – n = 19 – n Feedback: Don’t forget –(n – 4) means multiply each term inside the brackets by –1 and that the – inside the brackets belongs to the 4 to make it – 4. M1 C1 M1 C1 M1 C1 M1 C1 2 M1 for correctly expanding the brackets C1 for suitable feedback M1 for correctly expanding the brackets C1 for suitable feedback M1 for correctly expanding the brackets C1 for suitable feedback M1 for correctly expanding the brackets C1 for suitable feedback B 8 4 Start with numbers that work. = 2.5 So z = will satisfy conditions. Start with a formula. e.g. z = Substitute z = 2.5, s = 6, t = 2 to find x. 5 = 18 – 8 + x x = –5 so z = satisfies conditions. M1 A1 M1 B1 C1 2 3 M1 for first method, e.g. starting with numbers A1 for an example that works M1 for second method, e.g. starting with a formula B1 for an example that works C1 for a clear, complete solution showing two different methods and two examples B 5 5 = = n + 3 M1 B1 2 M1 for factorising B1 for any correct expression B 2
11 a b c d This graph shows expected sales for the different prices charged. If he prices his snowboard at more than £257 demand will be 0. The graph also shows that the cheaper the snowboards, the more he will sell. But he needs to consider his charges to make sense of it all. Number sold = 450 00 – 175P Sales = number sold × P = (45 000 – 175P)P Costs = set-up fees + manufacturing costs per board Minimum demand = 1 So 1 = 45 000 – 175P 175P = 44 999 P = = £257 Demand = 45 000 + 95P Profit = sales – costs Profit = (45 000 – 175P)P – (45 000 + 95P) = 45 000P – 175P2 – 45 000 – 95P = 44 905P – 175P2 – 45 000 From a graph of this quadratic function, his maximum profit would be approximately £2 850 000 if he sold his boards at £135. C1 M1 A1 C1 C2 M1 M1 A1 B2 C1 3 C1 for a clearly graph drawn M1 for method of solving equation when demand = 1 A1 for answer £257 C1 for a clear explanation C1 for explaining the relationship C1 for commenting that there will be other costs to take into account M1 for setting up the sales equation M1 for setting up the profit equation A1 for cao B1 for profit between £2 800 000 and £2 900 000 B1 for cost of boards between £120 and £140 C1 for a drawn graph of the quadratic equation found B
10 12 a b ii, v and vi might be difficult as they all involve squaring a term. The typical error made in ii will be to calculate half of at and then to square that. The same error can be found in vi where 2πr can be calculated first and then squared. ii and vi are also difficult to rearrange as they involve a quadratic element and it’s not easy to make each variable the subject of the formula. Typical errors in rearranging the equation s = ut + at2 to make a the subject include: incorrect sign when changing sides, e.g. s + ut = at2 incorrect removal of fraction, e.g. to leave (s + t) = at2. B2 B2 2 B1 for identifying some examples with a valid reason B1 for clear identification and explanation of classic errors B1 for identifying some examples with a valid reason B1 for clear identification and explanation of classic errors M 4
16 a b i ii iii ‘I think of a number and double it’ just means an expression of 2x, where x is the number I thought of – still unknown at the moment. ‘I think of a number and double it – the answer is 12’ has a solution that I know is 6. One e.g. 10 = p + 3 Because each solution is p = 7. C1 B1 B1 C1 2 C1 for clear explanation of the difference B1 cao B1 a correct example C1 a clear explanation M 4
20 a i ii b c The highest power will be 2 with no negative powers, e.g. y = x2 + 3x – 1, where 2 is the highest power. The highest power will be 3 with no negative powers, e.g. y = x3 + 5x2 – 6, the 3 being the highest power. Find points that cross the axes where possible and then create a table of values including for the turning point and the axis intercepts so that you have sufficient points to plot the curve. 1: x2 is always positive for positive or negative values of x, hence 2x2 will also be positive as positive multiplied by positive is positive. The +5 moves the graph up 5, so y = 2x2 + 5 will also always be positive for all values of x. 2: Draw the graph and illustrate all the points on the graph are above the x axis. C1 C1 C1 C1 C1 C1 C1 2 C1 for clear explanation C1 for also using an example C1 for clear explanation C1 for also using an example C1 for clear explanation C1 for first explanation C1 for second explanation M 7 21 a b c A quadratic function always has line symmetry because x2 and (–x)2 have the same y value. A cubic equation does not have a line of symmetry as A3 will have different values depending on whether A is negative or positive. Rotational symmetry of order 2 about the point of inflection. C1 C1 C1 2 C1 for clear explanation C1 for clear explanation C1 for clear description M 3 22 i ii iii iv D, y = 5x2 B, y = A, y = 0.5x + 2 C, d = √A B1 B1 B1 B1 3 B1 correct letter with correct example B1 correct letter with correct example B1 correct letter with correct example B1 correct letter with correct example M 4
27 Drawing a velocity/time graph allows us to illustrate the journey. The area under the graph is the distance travelled. Distance travelled = (15 × (u + 3u)) + (10 × 3u) + (20 × 3u) = 30u + 30u +30u = 90u The assumption is that acceleration is at a steady rate when the motorbike speeds up and slows down. C1 M1 M1 A1 A1 3 C1 for a good diagram illustrating the journey M1 for method of using diagram M1 for correct area equation A1 cao A1 for clear explanation of assumption. M 5 28 a b c x = 0 The circle has a radius of 5 and its centre is at (0, 0), halfway between D and E. Find the distance of F(–3, 4) from the origin. If it is 5 units from the origin, it is on the circumference of the circle. Using Pythagoras’ theorem: distance = = 5 So F(–3, 4) is on the circumference of the circle. The tangent at F will be at right angles to the line joining the point to the origin. Gradient of the line is –. The product of the gradient of this line and the tangent is –1. Use this to work out the gradient of the tangent. Substitute the gradient and the coordinates of the point into the general equation of a straight line (y = mx + c) to find the y-intercept c. B1 C2 C3 2 B1 cao C1 for explanation referring to a right angle triangle C1 for explaining how using Pythagoras’ theorem helps in finding the length 5 C1 for explaining the gradient is tangent of the angle C1 for showing y = mx + c C1 for complete explanation showing how to work out the equation of the line M 6
30 3x – 4y = 13 (1) 2x + 3y = 20 (2) Using elimination: Multiply (1) by 2 and (2) by 3: 6x – 8y = 26 6x + 9y = 60 Subtract the first equation from the second equation: 9y – –8y = 60 – 26 17y = 34 y = 2 Substitute y = 2 into (2): 2x + 3 × 2 = 20 2x = 14 x = 7 Check by substituting both values into (1) 3 × 7 – 4 × 2 = 21 – 8 = 13 Correct. Using substitution: Rearrange one of the equations: 2x + 3y = 20 y = Substitute into the other equation and rearrange. 3x – 4() = 13 3x – + = 13 3x + = 13 + = = 119 x = 7 Substitute this into other equation as M1 A1 M1 A1 M1 A1 M1 A1 M1 A1 3 M1 for method of changing equations in order to be able to eliminate A1 for correct equations M1 for subtracting equations A1 cao M1 for substitution A1 cao M1 for rearrangement to get one variable as a subject A1 cao M1 for substitution of the one variable into the other equation A1 cao M
33 a b They are the same equation. Multiply the first equation by 3 and it is the same as the second equation, so they have an infinite number of solutions. Treble the first equation to get 15x – 3y = 27 They have the same coefficients of x and y but a different constant so they are parallel lines with no intersections and so no solutions. C1 C1 3 C1 for clear explanation C1 for clear explanation M 2 34 a b x2 + 2x – 5 = 6x – 9 x2 – 4x + 4 = 0 (x – 2)(x – 2) = 0 x = 2 y = 6 × 2 – 9 = 3 y = 3 There is just one intersection of the two graphs, so it has to be sketch iii, as the straight line touches the curve once. x = 2 y = 3 M1 M1 M1 A1 A1 C1 C1 2 3 M1 for equating both equations M1 for arranging to equal 0 M1 for factorising A1 for x = 2 cao A1 for y = 3 cao C1 for sketch iii C1 for clear explanation M 7
36 a b S = B =C S + C = 75 C = B B + = 75 =75 B = 75 × = 20 S = 1.25B S = 1.25 × 20 = £25 C = 75 – 25 = £50 The method used was to find equations linking each two persons at a time, and then use those to create one equation that could be solved. Once you had one solution you could find the rest. Answer checked by going back to beginning statements and ensuring each one works. They do. B = £20 S = £25 C = £50 C1 C1 C1 M1 A1 A1 A1 C2 C1 3 C1 for first equation C1 for second equation C1 for third equation M1 for creating single equation with one unknown A1 for £20 cao A1 for £25 cao A1 for £50 cao C1 for clear explanation. C1 for matching explanation with the work done C1 for clear explanation M 10
38 Area of large square = 72 = 49 To find the length (z) of the side of the small square: z2 = z2 = z2 = 5 So the area of the small square is 25. Shaded area is equal to area large square – area small square. 49 – 25 = 24 So less than half is shaded, as required. B1 M1 A1 A1 A1 C2 2 3 B1 for area of large square M1 for use of Pythagoras’ theorem to help find the side length of inner square A1 cao A1 cao A1 for area of shaded part C1 for clear explanation C1 for complete explanation with correct mathematical notation throughout M 7 39 Set up two simultaneous equations, using the information given. 5 is the first term; from the rule for the sequence, the next term is: 5 × a – b, which equals 23 Hence 5a – b = 23 …… (1) Doing the same to the next term gives: 23a – b = 113 ……. (2) Subtract (1) from (2) to give: 18a = 90 a = = 5 Substitute in (1): 25 – b = 23 b = 2 So a = 5, b = 2 Hence, the next 2 terms are 563 and 2813. P1 B1 B1 M1 A1 M1 A1 C1 2 P1 for clear initial explanation B1 for first equation B1 for second equation M1 for clear method of elimination A1 cao M1 for substitution A1 cao C1 for next two terms correctly found M 8
42 a b c d 2n + 1 3n + 4 No, because will always equal no matter what n is, and the denominator increase of 4 will always give a larger increase than the numerator increase of 1, hence the fraction can never be larger than . B1 B1 B1 B1 C1 2 B1 cao B1 cao B1 cao B1 for no C1 for a clear concise explanation M 5 43 a i ii iii b c Neither – it’s the Fibonacci series, where each term is found by adding the previous two. Geometric – because each term is multiplied by 2 to find the next term. Arithmetic – because to find the next term you add 4 to the previous term. Arithmetic. Arithmetic. B1 C1 B1 C1 B1 C1 B1 B1 2 B1 for neither C1 for clear explanation B1 for geometric C1 for clear explanation B1 for arithmetic C1 for clear explanation B1 for arithmetic B1 for arithmetic M 8
44 a b c d a is the first term. d is the amount added each time. For example: use a = 2 and d = 3 to generate the sequence in part a as: 2, 5, 8, 11, 14, … The 5th term is 14. Using the given Xn = a + (n - 1)d: the 5th term will be 2 + 4 × 3 = 14, the same value. Xn = arn-1 It is a quadratic sequence because it contains a term in n2. B1 B1 C1 C1 P1 B1 C1 2 C1 for explaining a C1 for explaining d C1 for creating a suitable example C1 for generating a value higher than the third term P1 for showing the Xn formula gives the same value B1 cao C1 cao M 7 45 a b Evidence of reproducing proof as given in question. Sn = (2a + (n – 1)d) for the set of integers, a = 1 and d = 1. Hence Sn = (2 + (n – 1)) = (n + 1) C1 C1 C1 P1 2 C1 for correct proof clearly explained C1 for using a and d equal to 1 C1 for correct substitution into the formula P1 for showing how this simplifies to the desired formula M 4
46 a B c d e Sn = a + ar + ar2 + ar3 … + arn rSn = ar + ar2 + ar3 + ar4 … + arn + arn + 1 Therefore: Sn - rSn = a – arn + 1 Sn(1 – r) = a(1 – rn + 1) Sn = P1 P1 C2 B1 B1 B1 2 3 P1 for equation showing at least up to ar3 and the generalisation P1 for equation showing at least up to ar4 and the two generalisations C1 for top two rows shown correctly C1 for the bottom row shown correctly B1 cao B1 cao B1 cao M 7 47 Considering the area of an (x + 1) by (x + 1) square: The area of each rectangle created above is shown inside that rectangle, so it can be seen that: (x + 1)2 = x2 + x + x + 1 = x2 + 2x + 1 as required C1 C2 C1 2 C1 for explaining the sides of each square are (x + 1) C1 for creating the square divided into the rectangles, using the x and the 1 C1 for areas of each rectangle indicated in the rectangles C1 for clear explanation of required result M 4
49 a b i ii iii In this example, x and y are values satisfying the conditions: x + y ≤ 5 x > 1 y > 2 These are drawn on the diagram. Any region needs a minimum of three straight lines to enclose it. The region R above is where the solutions satisfying all three inequalities lie. Point (x, y) is inside the region if the point satisfies all three inequalities. For example, (1.5 , 3) is inside the region since 1.5 + 3 ≤ 5, 1.5 >1 and 3 > 2. Point (x, y) is outside the region if it does not satisfy at least one of the inequalities. For example, (2, 4) satisfies two of the conditions (x > 1 and y > 2) but does not satisfy x + y ≤ 5. Point (x,y) is on the boundary of the region if the point satisfies one of the inequalities but only as a equality. For example, (2, 3) is on the boundary of x + y ≤ 5 as 2 + 3 = 5. C1 C1 C1 C1 C1 C1 C1 C1 C1 3 2 C1 for choosing three inequalities that will define a region C1 for a clear diagram illustrating the chosen inequalities C1 for clear explanation linking the chosen inequalities with the diagram C1 for clear correct explanation C1 for use of an example to illustrate this C1 for clear correct explanation C1 for use of an example to illustrate this C1 for clear correct explanation C1 for use of an example to illustrate this M 9
Finally the nth terms can be completed as shown. Extending the table backwards will allow us to find the values of a, b and c in the nth term an2 + bn + c. 2a = 2 → a = 1 a + b = 4 → b = 3 c = –2 Hence nth term is n2 + 3n – 2. C1 for correctly finding the nth term 10 52 (m2 – n2)2 + (2mn)2 = m4 – 2m2n2+ n4 + 4m2n2 = m4 + 2m2n2 + n4 = (m2 + n2)2 M1 A1 A1 C1 2 3 M1 choosing two smallest terms, squaring and adding A1 for cao A1 for cao C1 for showing the factorisation leads to the given result M 4 53 Because the differences between consecutive terms are 2, 3, 4, 5, 6, etc. (even and odd alternately); when generating the triangular number sequence starting with the odd 1, add even to odd to generate odd …3; add odd to odd to generate even …6; add even to even to generate even …10. Add odd to even to generate odd …15. We are now back again where we add even to odd to generate odd, and the whole sequence continues in the same way, continually giving two odd, two even, etc. C3 2 C1 for explaining the about the differences of the terms being the set of integers C1 for explaining the pattern is odd, even, odd, even and so on C1 for explaining the complete sequence of combining odd and even to generate the final sequence M 3
54 The assumption is that p and q are integers. 10p + q = 7n where n is also an integer. 7p + 3p + q = 7n 3p + q = 7n – 7p = 7(n – p) As n and p are integers then n – p is also an integer hence 7(n – p) is a multiple of 7 and so 3p + q must be as well. P1 C1 M1 A1 C1 2 P1 for giving the assumption about p and q C1 for expressing 10p + q as a multiple of 7 M1 for expressing 3p + q in terms of 10p + q A1 for similar expression C1 for clear full explanation M 5 55 2(5(x – 2) + y) = 2(5x – 10 + y) = 10x – 20 + 2y …….(1) 10(x – 1) + 2y – 10 = 10x – 10 + 2y – 10 = 10x – 20 + 2y ….(2) Equation (1) = equation (2) Hence the two expressions are equal. M1 A1 M1 A1 P1 2 M1 for expanding A1 cao M1 for expanding A1 cao P1 for explaining the two expressions are the same M 5
56 a b Take two numbers x and y where x > y. First step: 5(x – 2) = 5x – 10 Second step: 2(5x – 10 + y) = 10x – 20 + 2y Third step: 10x – 20 + 2y + 9 – y = 10x – 11 + y Fourth step: 10x – 11 + y + 11 = 10x + y Hence where the first two numbers might have been 7 and 3, the final outcome would be 70 + 3 = 73. Let the single-digit number be x and the two-digit number be 10a + b. First step: 10(10a + b) – 9x = 100a + 10b – 9x = 100a + 10b + x – 10x = 100a + 10b – 10x + x = 100a + 10(b – x) + x The hundreds unit is a. The tens unit is (b – x). The units term is x, the same as the single digit we started with. The split then becomes 10a + (b – x) and x. Adding these two gives 10a + b – x + x which is 10a + b, the two-digit term. C1 B1 B1 B1 B1 P1 C1 P1 C1 C1 C1 C1 P1 2 C1 for initial explanation B1 cao B1 cao B1 cao B1 cao P1 for explanation of how this shows the final result C1 for defining each digit P1 for expressing the first manipulation algebraically C1 for showing it in terms of hundreds, tens and units C1 for correct hundreds C1 for correct tens C1 for correct unit explanation P1 for clear explanation as to how the second manipulation gives the two-digit term M 13 57 Expand each square and add: n2 – 2n + 1 + n2 + n2 + 2n + 1 = 3n2 + 2n – 2n + 2 = 3n2 + 2 as given. M1 A1 C1 C1 2 M1 for expanding brackets A1 for correct expansion of brackets C1 for showing how the n terms cancel C1 for complete solution with no incorrect notation or terminology M 4
58 The difference is 5 so nth term is 5n + c where c = first term – 5 = 4 – 5 = –1 So nth term is 5n – 1. Check the 4th term gives 19. When n = 4, 4n – 1 = 20 – 1 = 19, correct. C1 M1 A1 A1 C1 2 C1 for obtaining the difference of 5 M1 for finding c A1 cao A1 cao C1 for showing a check works M 5 59 Triangular numbers are 1, 3, 6, 10, 15 ….. This will give a table as: Extending the table backwards will allow us to find the values of a, b and c in the nth term an2 + bn + c. . 2a = 1 → a = a + b = 1 → b = c = 0 Hence nth term is n2 + n = (n2 + n) = n(n + 1) B1 M1 A1 C1 M1 A1 B1 B1 B1 P1 2 3 B1 for showing triangular numbers M1 for method of finding differences A1 for correct table C1 for explanation of extending table backwards M1 for method of extending table A1 for correct table B1 correct a B1 correct b B1 correct c P1 for correct evaluation of generalisation to show given result M 10
60 Tn = n(n + 1) T2n + 1 = (2n + 1)(2n + 1 + 1) = (2n + 1)(2n + 2) = (2n + 1)(n + 1) = 2n2 + 3n + 1 …….. (1) Tn + 1 = (n + 1)(n + 1 + 1) = (n + 1)(n + 2) = (n2 + 3n + 2) …….(2) So T2n + 1 – Tn + 1 = 2n2 + 3n + 1 – n2 – n – 1 = n2 + n = (n2 + n) = (n + 1) but Tn = n(n + 1) = 3Tn as given B1 M1 A1 M1 A1 B1 P1 2 B1 for correct Tn formula M1 for substituting 2n + 1 A1 cao M1 for substituting n + 1 A1 cao B1 for subtracting each equation P1 for clear full explanation of proving the final connection M 7 61 Tn = n(n + 1) = == But n2 + n – 2 factorises to (n – 1)(n + 2) So final expression is B1 C3 B1 P1 2 B1 for Tn formula C1 for numerator expansion C1 for denominator expansion C1 for cancelling B1 for correct factorisation P1 for fully clear correct proof with no mathematical notational errors M 6
62 Let the first number be x, then the next four are x + 1, x + 2, x + 3 and x + 4. The sum of these is 5x + 1 + 2 + 3 + 4 which is 5x +10 = 5(x + 2), a multiple of 5. P1 M1 P1 2 P1 for stating each term in algebraic form M1 for adding all 5 terms P1 for showing they are multiple of 5 M 3 63 10x = Hence b × 10x = a Substitute this into 10y = to give 10y = Hence 10y = So 10y × 10x = 1 So 10(x + y) = 1 But 100 = 1 And so x + y = 0. P1 M1 C1 C1 C1 B1 P1 2 P1 for expressing a as subject M1 for substituting into other expression C1 for showing the correct substitution C1 for showing the product of the two terms equal to 1 C1 for showing combination of indices B1 for 100 = 1 P1 for complete, clear proof with clear mathematical statements M 7
64 Where two terms are x and x + 1 the expression required is: (x + x + 1)2 – (x2 + (x + 1)2) = (2x + 1)2 – (x2 + x2 + 2x + 1) = 4x2 + 4x + 1 – 2x2 – 2x – 1 = 2x2 + 2x = 2x(x + 1) …..(1) Where Tx = x(x + 1) 4Tx = 2x(x + 1), which is same as the result in equation (1). P1 C1 C1 C1 C1 P1 2 P1 for identifying the two terms C1 for correct expression as asked C1 for correct expansion of all brackets C1 for correct simplification C1 for correct factorisation P1 for clear complete proof with correct mathematical notation M 6 65 Let p = x Then q = x + 1 and r = x + 2 so pr = x(x + 2) = x2 + 2x q2 – 1 = (x + 1)2 – 1 = x2 + 2x + 1 – 1 = x2 + 2x = pr B1 B1 B1 B1 B1 P1 2 B1 for q expressed algebraically B1 for r expressed algebraically B1 for product pr expressed algebraically B1 for q2 – 1 expressed algebraically B1 for simplification P1 for complete clear proof M 6 66 a b i ii There are many equivalent expressions. For example, expand the bracketed term: – q2 – q – 4 For example: For example: 4x2 + 10x B1 B1 B1 2 3 B1 for a correct example B1 for a correct example B1 for a correct example H 3 67 a b c d To make it a product of two linear expressions. The quadratic expression. That the signs of the numbers in the brackets are different. One factor of the constant term is zero. There is only one set of brackets. C1 C1 C1 C1 2 3 C1 for clear explanation C1 for clear explanation C1 for clear explanation C1 for clear explanation H 4
71 (2a + b)(2a + b) = 4a2 + 4ab + b2 (2a + b)(2a – b) = 4a2 – b2 (2a – b)(2a – b) = 4a2 –4ab + b2 (a + b)(a + b) = a2 + 2ab + b2 (a + b)(a – b) = a2 – b2 (a – b)(a – b) = a2 – 2ab + b2 The difference in the two is that in the (2a ± b) product, both the a2 and ab terms have a coefficient of 4 (when the ab term is not zero), but in the (a ± b) product, the a2 term has a coefficient of 1 and the ab term has a coefficient of 2 (when the ab term is not zero). C1 C1 C1 2 C1 for showing all the possibilities C1 for showing all the possibilities C1 for clear explanation H 3
73 f(x) = 3 – 7x Find f–1(x) from y = 3 – 7x: 7x = 3 – y x = So f–1(x) = Find g–1(x) from y = 7x + 3: 7x = y – 3 x = So g-1(x) = So f-1(x) + g-1(x) = + = = = 0 as required. M1 A1 M1 A1 P1 C1 2 M1 for method of finding inverse A1 cao M1 for method of finding inverse A1 cao P1 for showing how the two functions can be added together C1 for complete explanation of how they sum to 0 H 6
74 a i ii iii b i ii c It means that the constant terms in both expressions are positive, or if one is positive and one is negative, their sum is positive. It means that the constant terms in both expressions are negative, or if one is positive and one is negative their sum is negative. The expression is the difference of two squares. For example, (4x + 2)(x – 1) = 4x2 – 2x – 2 as required. For example, (3x + 1)(x + 1) = 3x2 + 4x + 1 as required. If it is positive then both expressions have the same sign. If it is negative then the expressions have different signs. P1 P1 P1 P1 C1 C1 C1 C1 2 P1 for first condition P1 for second condition P1 for first condition P1 for second condition C1 for clear explanation C1 for correct example explained C1 for correct example explained C1 for complete clear explanation H 8