Question Working Answer Mark AO Notes Grade 1 a b c d Example 2 × 3 = 6 2 + 3 = 5 3(c + 5) = 3c + 15 Example 32 = 3 × 3 = 9 2 × 3 = 6 Example 2n2 = 2 × (32) = 2 × 9 = 18 (2 × 3)2 = 6 × 6 = 36 2n means 2 × n which is different from n + 2. n2 means n× n which is different from 2n. BIDMAS for 2n2 tells you to calculate the power first. BIDMAS for (2n)2 tells you that you do the calculation inside the bracket first. B1 B1 M1 B1 B1 B1 2 B1 for explanation; an example could be given to support the argument An additional mark can be given for identifying the exception, which is when n = 2 M1 for multiplying out the brackets to show that the two expressions are not equivalent B1 for explanation; an example could be given to support the argument An additional mark can be given for identifying the exception which is when n = 2 B1 for an explanation; an example could be given to support the argument B 6 2 A letter, say f, stands for an unknown if it is in an equation such as 3f + 2 = 14. Then f = 4 is the only number that satisfies this equation. A letter stands for a variable if it is part of an equation that has more than two letters, e.g. A = πr2 , where both A and r are variables that will be different for different values of A or r. B1 B1 B1 B1 2 B1 for a clear explanation B1 for an example alongside the explanation B1 for a clear explanation B1 for an example alongside the explanation B 4
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 A1 M1 A1 M1 A1 M1 A1 2 M1 for correctly expanding the brackets A1 for suitable feedback M1 for correctly expanding the brackets A1 for suitable feedback M1 for correctly expanding the brackets A1 for suitable feedback M1 for correctly expanding the brackets A1 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 A1 B1 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 A1 for an example that works B1 for a clear, complete solution showing two different methods and two examples B 5 5 = = n + 3 M1 A1 2 M1 for factorising A1 for any correct expression B 2
9 a b c Need to find both times when h = 0. Substitute u = 16 m/s into the equation. 16t – 5t2 = 0 t(16 – 5t) = 0 so t = 0 or (16 – 5t) = 0 t = 0 or 5t = 16 t = 3.2 Maximum height = 16 × 1.6 – 5 × 1.62 = 25.6 – 12.8 = 12.8 m h = ut + 5t2 + 1 3.2 s B1 M1 A1 A1 B1 B1 B1 3 B1 for clear explanation M1 for setting h = 0 A1 for 0 and 3.2 A1 for 3.2 seconds M1 for substituting t = 1.6 B1 for 12.8 m B1 cao 7 10 a b c Use s = ut – gt2. Assuming g = 10, given u = 8 sin θ and assuming suitable value for θ, for example, 30° sin 30° = 0.5 So s = –5t2 + 4t Complete the square to give: = 0 Comparing this to the equation of y = x2. Then the horse will reach its maximum at t = 0.4. Substitute this into s = –5t2 + 4t to give s = 0.8 m. Parabola/quadratic equation s = 0.8 m Suitable justification, e.g. Yes it does, the horse is in the air for 0.8 s and jumps 0.8 m into the air. B1 B1 M1 M1 A1 A1 3 B1 for either of these EC for stating suitable assumptions for the starting point M1 for a suitable method for finding greatest height, could also be sketch graph M1 for suitable comparison A1 for ft from the initial assumption A1 cao B 6
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. B1 M1 A1 B1 B2 M1 M1 A1 B2 B1 3 B1 for a clearly graph drawn M1 for method of solving equation when demand = 1 A1 for answer £257 B1 for a clear explanation B1 for explaining the relationship B1 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 B1 for a drawn graph of the quadratic equation found B
13 c and d can be difficult because they contain minus signs; errors are often made when combining minus signs. In substituting x = –3 into t = –2(3 – x), a common error is to assume 3 – –3 is 0. In substituting x = –3 into z = a typical error is to assume a negative divided by a negative gives a negative answer. A suggestion to avoid these errors is to remember that when multiplying or dividing with positive and negative numbers, same signs means positive, different signs means negative. B1 B2 B1 2 B1 for identifying some examples with a valid reason B1 for clear identification of one typical error with one equation. B1 for another typical error B1 for a satisfactory suggestion M 4 14 The similarities are that both include an equals sign and both require the manipulation of terms. The difference is that in solving an equation you reach a numerical answer, but in rearranging you still have a formula. B1 B1 2 B1 for clear explanation of similarities B1 for clear explanation of differences M 2 15 In line 2 Phillip has initially rearranged x2 + 2x – 3 to x(x + 2) – 3 when he should have factorised it as (x + 3)(x – 1). He has incorrectly simplified in line 3. He should have factorised (x2 – 9) to (x + 3)(x – 3). Philip has cancelled incorrectly just by looking at the different numbers and not realising that you can only cancel a number on both numerator and denominator if it is a factor of the complete expression. B1 B1 B1 2 B1 for identifying the first error B1 for identifying the second error B1 for a clear explanation of the errors made M 3
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
32 a b i ii iii c i ii iii d If one of the equations is a multiple of the other then there will be an infinite number of solutions, e.g. x + y = 5 2x + 2y = 10 Every point on the line x + y = 5 is a solution, giving an infinite number of solutions. If the equations have graphs that are parallel to each other then there will be no intersection and so no solution, e.g. x + y = 5 x + y = 6 One solution, as one equation is not a multiple of the other and they are not parallel. None –the gradient is the same but the intercepts are different so they are parallel. Infinite number of solutions as the first equation is a multiple of the second, so drawing graphs gives the same line. y – 2x = –5 y = 0.5x + 1 Substitute for y in the first equation: 0.5x + 1 – 2x = –5 –1.5x = –6 x = 4 Substitute into second equation: y = 0.5 × 4 + 1 = 3 Check by substituting both values in the first equation 3 – 2 × 4 = 3 – 8 = –5 Correct. No solution. Infinite number of solutions. You can see how many times the graphs cross each other. B1 B1 B1 B1 B1 B1 B1 M2 A1 B1 2 B1 for clear explanation B1 for use of a good example accompanying the explanation B1 for clear explanation B1 for use of a good example accompanying the explanation B1 for clear explanation B1 for clear explanation B1 for clear explanation M1 for arranging equations in a suitable format M1 equating both equations A1 cao B1 clear explanation M 11
40 a b c d If the sum of the whole numbers from 1 to 50 is 1275 the sum from 2 to 51 will be 1275 + 51 – 1 = 1325. Check method, using simple examples. Using Sn = S52 == 1326 1326 – 1 + 1325 as above. If 1 is the first term then the nth term will be n, so their sum is 1 + n. Because they come in pairs, there will be of these pairs adding to the total. So the total = (n + 1) × =, the formula given. B1 B1 B2 B2 2 B1 for clear explanation. B1 for clear explanation B1 for clear use of the formula B1 for showing same answer as above B1 for clear explanation. B1 for use of the generalization and good mathematical language. M 6
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 B1 2 B1 cao B1 cao B1 cao B1 for no B1 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 B1 B1 B1 B1 B1 B1 B1 2 B1 for neither B1 for clear explanation B1 for geometric B1 for clear explanation B1 for arithmetic B1 for clear explanation B1 for arithmetic B1 for arithmetic M 8
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 = B1 B1 B2 B1 B1 B1 2 3 B1 for equation showing at least up to ar3 and the generalisation B1 for equation showing at least up to ar4 and the two generalisations B1 for top two rows shown correctly B1 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 B1 B2 B1 2 B1 for explaining the sides of each square are (x + 1) B1 for creating the square divided into the rectangles, using the x and the 1 B1 for areas of each rectangle indicated in the rectangles B1 for clear explanation of required result M 4
48 a b c For example, 2x > 10 Divide both sides by 2 to get x > 5. We show the solution with a line and a circle at each end point. A solid circle means that the solution includes the end point; an open circle means that the solution does not include the end point. For example: The top diagram shows x ≤ 2. It has a solid circle at the end point x = 2 because that is part of the solution. The bottom diagram shows x > 1, it has an open circle at the end point x = 1 because x = 1 is not part of the solution. Starting with an equation 10 – x > 4 and solving by adding x to both sides gives the solution 6 > x. This can also be given as x < 6 ….. (1) Consider again 10 – x > 4. This time multiply throughout by –1. Keeping the inequality sign the same gives: –10 + x > -4 Add 10 to each side to give x > 10 – 4. This gives the solution as x > 6 ….. (2) But comparing this with equation (1) we see that the signs are the other way round, this illustrates that when we multiplied through by a negative number, we should have changed the sign from > to <. B1 B1 B1 B1 B1 B1 B2 2 B1 for a correct inequality B1 for clear explanation of how to solve the inequality chosen B1 for explanation about circles at the end of each line B1 for clear explanation differentiating between solid and open circles B1 for use of a clear diagram to support the explanation B1 for a full, clear explanation showing both aspects of the circles B1 for clear explanation B1 for using an example in a way that illustrates the principle M 8
51 a b c For example, n2 can generate a quadratic sequence: 1, 4, 9, 16, 25 ……. n2 Similarities: you find the nth term for both types of sequence by looking at the differences between terms. Differences: in a linear sequence you find the first term by subtracting the difference from the second term. In a quadratic sequence you also have to look at the second differences. This allows you to extend the differences backwards to find the values of a, b and c in the nth term of an2 + bn + c. For a linear sequence, just keep on adding 6 each time to give: 2, 8, 14, 20 ………..(6n – 4) The nth term includes 6n because we add 6 each time, 6n – 4 because (2 – 6) = –4. For a quadratic equation, we build up the series by again having the first differences as 6, then choosing a second difference, say 2. This will give a table such as: Start with the two terms in positions n = 1 (2) and n = 2 (8) in the sequence. Put each second difference as 2. Then complete the first differences as shown. B1 B1 B1 B1 B1 B3 B2 2 B1 for a valid quadratic sequence B1 for a clear explanation B1 for a clear explanation B1 for explaining how you woud find the sequence B1 for cao B1 for explaining second differences B1 for use of a table or equivalent B1 for correctly finding a quadratic equation with 2 and 8 as starting terms B1 for explaining how you would find the nth term M
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. B1 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 A1 2 3 M1 choosing two smallest terms, squaring and adding A1 for cao A1 for cao A1 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. B3 2 B1 for explaining the about the differences of the terms being the set of integers B1 for explaining the pattern is odd, even, odd, even and so on B1 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. M1 A1 M1 A1 A1 2 M1 for giving the assumption about p and q A1 for expressing 10p + q as a multiple of 7 M1 for expressing 3p + q in terms of 10p + q A1 for similar expression A1 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 A1 2 M1 for expanding A1 cao M1 for expanding A1 cao A1 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. B1 B1 B1 B1 B1 M1 A1 M1 A1 A1 A1 A1 B1 2 B1 for initial explanation B1 cao B1 cao B1 cao B1 cao M1 for explanation of how this shows the final result A1 for defining each digit M1 for expressing the first manipulation algebraically A1 for showing it in terms of hundreds, tens and units A1 for correct hundreds A1 for correct tens A1 for correct unit explanation B1 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 B1 B1 2 M1 for expanding brackets A1 for correct expansion of brackets B1 for showing how the n terms cancel B1 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. B1 M1 A1 A1 B1 2 B1 for obtaining the difference of 5 M1 for finding c A1 cao A1 cao B1 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 B1 M1 A1 B1 B1 B1 B1 2 3 B1 for showing triangular numbers M1 for method of finding differences A1 for correct table B1 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 B1 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 B1 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 B1 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 B3 B1 B1 2 B1 for Tn formula B1 for numerator expansion B1 for denominator expansion B1 for cancelling B1 for correct factorisation B1 for fully clear correct proof with no mathematical notational errors M 6
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). M1 A1 A1 A1 A1 B1 2 M1 for identifying the two terms A1 for correct expression as asked A1 for correct expansion of all brackets A1 for correct simplification A1 for correct factorisation B1 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 B1 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 B1 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. B1 B1 B1 B1 2 3 B1 for clear explanation B1 for clear explanation B1 for clear explanation B1 for clear explanation H 4
68 a b For example: x2 – 1. Because each part is a square, x2 and 12, one is subtracted from the other. Because: 1000 × 998 = (999 + 1) × (999 – 1) = 9992 – 1 B1 B1 3 B1 for clear explanation B1 for clear explanation H 2 69 Two that can be cancelled, for example: and I chose two straightforward ones, one that would cancel by a single letter and one that would cancel by an algebraic term. Two that cannot be cancelled, for example: and I chose two straightforward examples, one being a single term as numerator and denominator, the other one where the denominator was more than a single term. B1 B1 B1 B1 2 3 B1 for two examples that cancel B1 for a clear explanation B1 for two examples that don’t cancel B1 for a clear explanation 4 70 To get such a term on the top this must be the difference of two squares, hence the two expressions both need multiplying by (3x – 4) to give: This expands to: B1 B1 B1 B1 2 B1 for clear explanation B1 for (3x – 4) B1 for setting up the expression B1 for showing how to find the final expression in suitable format 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). B1 B1 B1 2 B1 for showing all the possibilities B1 for showing all the possibilities B1 for clear explanation H 3
72 a b Draw a triangle ABC. Using trigonometric functions: sin C = Therefore: h = b sin C Then using the basic formula for area of a triangle: Area = a × h Substituting for h gives: Area = a × b sin C Area = ab sin C as required. Using the given triangle: C = 45°, a = x + 2 = 6, b = x – 2 = 2 Area = ab sin C =× 6 × 2 × sin 45° = 6 × = Multiply numerator and denominator by . This gives = = 3as required. B1 B1 B1 B1 M1 A1 M1 M1 M1 A1 M1 A1 2 B1 for drawing diagram correctly labelled B1 for trigonometric expression linking C, h and b B1 for h = b sin C B1 formula for area of triangle M1 substitution of h A1 cao M1 for complete correct proof with correct mathematical notation throughout M1 for expressing a, b and C M1 for substituting for a, b and C A1 cao M1 for dealing with the in denominator A1 for full explanation showing given result H 12
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. B1 B1 B1 B1 B1 B1 B1 B1 2 B1 for first condition B1 for second condition B1 for first condition B1 for second condition B1 for clear explanation B1 for correct example explained B1 for correct example explained B1 for complete clear explanation H 8