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GCE AS Mathematics (8MA0) – Paper 1 Pure Mathematics Summer 2019 student-friendly mark scheme Please note that this mark scheme is not the one used by examiners for making scripts. It is intended more as a guide to good practice, indicating where marks are given for model solutions. As such, it doesn’t show follow-through marks (marks that are awarded despite errors being made) or special cases. It should also be noted that for many questions, there may be alternative methods of finding correct solutions that are not shown here – they will be covered in the formal mark scheme. Guidance on the use of codes within this document M1 – method mark. This mark is generally given for an appropriate method in the context of the question. This mark is given for showing your working and may be awarded even if working is incorrect. A1 – accuracy mark. This mark is generally given for a correct answer following correct working. B1 – working mark. This mark is usually given when working and the answer cannot easily be separated. Some questions require all working to be shown; in such questions, no marks will be given for an answer with no working (even if it is a correct answer).
GCE AS Mathematics (8MA0) – Paper 1 Pure Mathematics model solutions (Version 1.0) 2 Part Working or answer an examiner might expect to see Mark Notes (a) y^ =^ –^ 2
x + 4 3 M1 This mark is given for a method to rearrange to find an equation for l 1 in terms of y = m = 2 A1 This mark is given for deducing the gradient of the perpendicular line l 2 (b) Substituting y = 2 x + 7 into 2 x + 4 y – 3 = 0 gives 2 x + 4(2 x + 7) – 3 = 0 M1 This mark is given for a method to substitute to form and solve an equation in a single variable. 10 x + 25 x = – 2. A1 This mark is given for solving to find the value of the x - coordinate of the point P. Question 2 (Total 8 marks) Part Working or answer an examiner might expect to see Mark Notes (i) 16 a^2 = 2√ a so 2 1 2 2
a a = 1 2 3 8 a = 1^ so^2 3 a =^ 8
M1 This mark is given for a method to find an equation to solve with the terms in a on one side a = 2 3 8
M1 This mark is given for finding a way to deal with the indices when solving the equation a = 4 1 A1^ This mark is given for finding one correct solution to the equation a = 0 is also a solution B1 This mark is given for deducing that a = 0 is also a solution (ii) b 4
- 18 = 0 factorises to ( b^2 + 9)( b^2 – 2) = 0 M1 This mark is given for factorising the equation given b 2 = – 9, 2 A1 This mark is given for finding two correct solutions for b 2 For real solutions, b 2 = 2 only M1 This mark is given for recognising that b = √–9 is not a real solution b = √2, – √ 2 A1^ This mark is given for finding the two real solutions to the equation
GCE AS Mathematics (8MA0) – Paper 1 Pure Mathematics model solutions (Version 1.0) 4 Part Working or answer an examiner might expect to see Mark Notes (a) M1 This mark is given for recognising that x n becomes xn^ –^1 when differentiating x y d d = 6 x – (^2)
x A1 This mark is given for one of the two terms 6 x or – (^2)
x given correctly A This mark is given for x y d d given fully correct (b) 6 x – (^2)
x
0 (^) M1 This mark is given for setting x y d d greater than 0 (allow ≥) 6 x 3
x > 3
A1 This mark is given for the exact range of values of x for which the curve is increasing (allow ≥) Question 6 (Total 6 marks) Part Working or answer an examiner might expect to see Mark Notes (a) (^18) √3 = 2
× 2 x × 3 x × sin 60° M1 This mark is given for use of the formula A = 2
ab sin C for the area of the triangle 18 √3 = 3 x 2 × 2
x 2 = 12 M1 This mark is given for using a value of sin 60° to find a value for x 2 x = √ 12 = √(4 × 3) = 2√ 3 A1 This mark is given for a full solution to show that x = 2√ 3 (b) BC 2 = ( 6 √3)^2 + (4√3)^2 – 2 × 6 √ 3 × 4 √ 3 × cos 60° M1 This mark is given for using the cosine rule to start to find the length BC BC 2 = 84 A1 This mark is given for finding a value for BC^2 BC = 2√ 21 A1^ This mark is given for a correct answer presented as a simplified surd
GCE AS Mathematics (8MA0) – Paper 1 Pure Mathematics model solutions (Version 1.0) 5 Part Working or answer an examiner might expect to see Mark Notes (a) M1 This mark is given for a graph with shape x
in the first quadrant A1 This mark is given for a fully correct sketch B1 This mark is given for the asymptote y = 1 correctly shown on the sletch (b) x k^2
- 1 = – 2 x + 5 M1 This mark is given for deducing the point of intersection k^2 + x = – 2 x^2 + 5 x
- 2 x 2 + 5 x – x – k 2 = 0 2 x 2 - 4 x + k 2 = 0 A1 This mark is given for correct working to show the result required (c) (^) 16 = 4 × 2 × k^2 16 = 8 k 2 M1 This mark is given for deducing that the equation has a single root and setting b 2
- 4 ac = 0 A1 This mark is given for correctly finding b 2 - 4 ac = 0 k^2 = ±√ 2 A1^ This mark is given for finding the two exact values of k
GCE AS Mathematics (8MA0) – Paper 1 Pure Mathematics model solutions (Version 1.0) 7 Part Working or answer an examiner might expect to see Mark Notes (a) 1200 – 3(1 – 20) 2 = 1200 – 3(–19)^2 = 1200 – 1083 = 117 tonnes B1 This mark is given for the correct answer (b) 1200 tonnes B1 This mark is given for deducing that ( n – 20) 2 is always positive, and so deducing the maximum value for T Units (tonnes) must be stated (c) [1200 – 3(5 – 20) 2 ] – [1200 – 3(4 – 20) 2 ] = 525 – 432 M1 This mark is given for a method to find the mass of tin that will be mined in 2023 = 93 tonnes A1 This mark is given for the correct answer (units need not be given) (d) (^) n ≤ 20 B1 This mark is given for an appreciation that the model only predicts the mass of tin mined for the next 20 years This model predicts that the mass of tin mined will increase each year B1 This mark is given for an appreciation that the total mass of tin mined cannot decrease but that for n > 20 the value of T decreases as n increases.
GCE AS Mathematics (8MA0) – Paper 1 Pure Mathematics model solutions (Version 1.0) 8 Part Working or answer an examiner might expect to see Mark Notes (a) ( x – 2) 2
- 4 – 16 – 8 = 0 M1 This mark is given for a method to complete the square Centre at x = 2 and y = – 4, (2, – 4) A1 This mark is given for finding the correct coordinates of the centre of the circle ( x – 2) 2
- 28 = 0 Radius = √28 = 2√ 7 A1 This mark is given for finding the exact radius of the circle (b) Tangent of x = k touches circle at 2 + √ 28 and 2 – √ 28 M1 This mark is given for adding or subtracting the length of the radius of the circle from 2 A1 This mark is given for deducing both values of k
GCE AS Mathematics (8MA0) – Paper 1 Pure Mathematics model solutions (Version 1.0) 10 Part Working or answer an examiner might expect to see Mark Notes (a) θ θ θ 3 2 cos 10 sin^27 cos 2
θ θ θ 3 2 cos 10 ( 1 cos ) 7 cos 2 2
M1 This mark is given for using the identity sin θ = 1 – cos^2 θ in the fraction ≡ θ θ θ 3 2 cos 12 7 cos 10 cos 2
− − A1^ This mark is given for finding a simplified expression in terms of cos θ only ≡ θ θ θ 3 2 cos ( 3 2 cos )( 4 5 cos )
- − M1 This mark is given for factorising the numerator of the expression ≡ 4 – 5 cos θ A1^ This^ mark is given for a fully correct proof with correct notation and no errors. (b) 4 – 5 cos x = 4 + 3 sin x tan x = – 3
M1 This mark is given for substituting for the fraction and rearranging the equation, using x x cos sin = tan x x = 121 ° A1^ This mark is given for one correct value of x x = 301° A1^ This mark is given for the other correct value of x
GCE AS Mathematics (8MA0) – Paper 1 Pure Mathematics model solutions (Version 1.0) 11 Part Working or answer an examiner might expect to see Mark Notes x y d d = 6 x^2 – 34 x + 40 B1 This mark is given for the equation correctly differentiated x y d d = 0 when 6 x^2 – 34 x + 40 = 0 M1 This mark is given for setting the equation equal to zero 2(3 x – 5)( x – 4) = 0 M1 This mark is given for factorising the expression ( x = 3
), x = 4, A1 This mark is given for finding two solutions and choosing x = 4 as the upper limit of the integral R = (^2) x 17 x 40 x d x 4 0 3 2 ⎮ ⌡
4 0 4 3 2 20 3
x − x + x B1 This mark is given for integrating the expression from 0 to 4 R = ( 2
× 4
4 ) – ( 3
× 4
3 ) + (20 × 4 2 ) M1 This mark is given for a calculation for find the area R = 127 – 3
256 A1^ This mark is^ given for a full proof with correct notation and no errors
GCE AS Mathematics (8MA0) – Paper 1 Pure Mathematics model solutions (Version 1.0) 13 Part Working or answer an examiner might expect to see Mark Notes If n is even, n = 2 k and n^3 + 2 = (2 k )^3 + 2 = 8 k^3 + 2 M1 This mark is given for finding expressions for n and n^3 + 2 when n is even 8 k^3 + 2 is two more than a multiple of 8 and so not divisible by 8 A1 This mark is given for a correct conclusion following correct working If n is odd, n = 2 k + 1 and n^3 + 2 = (2 k + 1)^3 + 2 = 8 k 3
- 12 k 2
- 6 k + 3 M1 This mark is given for finding expressions for n and n^3 + 2 when n is odd 8 k^3 + 12 k^2 + 6 k + 3 is an odd number and so not divisible by 8 A1 This mark is given for a correct conclusion following correct working Question 16 (Total 5 marks) Part Working or answer an examiner might expect to see Mark Notes (i) a and b lie in the same direction B1 This mark is given for a valid explanation (ii) M1 This mark is given for showing the vector problem graphically (may be implied) 6 sin 30 ° = 3 sin θ sin θ = 6
M1 This mark is given for using the sine rule as a method to find the angle between – n and n – m θ = 14.5° A1^ This mark is given for finding the the angle between – n and n – m Angle between m and m – n = (180 – 30 – 14.5) = 135.5° A1 This mark is given for the angle between vector m and vector m – n
