Cambridge A Level Mathematics 9709 — 2022 Oct/Nov Paper 3 · Variant 1
9709/31/O/N/22 · 8 questions · 75 marks · ≈84 min
The question paper and its mark scheme, free to read here and free to download. This is Cambridge’s own paper, exactly as it was sat.
Question paper20 pages




















Mark scheme17 pages
Answers below. Sit the paper first if you are practising.

















Questions as text
Q1 · Sketch the graph of y = 2x + 1
1 (a) Sketch the graph of y = 2x + 1 . [1] (b) Solve the inequality 3x + 5 < 2x + 1 . [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Mark scheme: Question Answer Marks Guidance 1(a) Show a recognisable sketch graph of y = 2 x + 1 B1 y 1 x 1 - 2 Ignore y = 3 x + 5 if also drawn on the sketch. 1 1(b) Find x-coordinate of intersection with y = 3x + 5 M1 6 A1 Obtain x = − 5 6 A1 Do not condone for < in the final answer. State final answer x − only 5 Alternative method 1 for question 1(b) Solve the linear inequality 3 x + 5 − ( 2 x + 1) , or corresponding equation M1 Must solve the relevant equation. 6 A1 Ignore – 4 if seen. Obtain critical value x = − 5 6 A1 State final answer x − only 5 Alternative method 2 for question 1(b) 2 2 M1 2 Solve the quadratic inequality ( 3x + 5 ) ( 2 x + 1) , or corresponding 5 x + 26 x + 24 0 equation 6 A1 Ignore – 4 if seen. Obtain critical value x = − 5 6 A1 State final answer x − only 5 3
Q2 · On a sketch of an Argand diagram shade the region whose points represent complex numbers…
2 On a sketch of an Argand diagram shade the region whose points represent complex numbers z satisfying the inequalities z ≤3, Rez ≥−2 and 14π ≤arg z ≤π. [4]
Mark scheme: 2 Show a circle with radius 3 and centre the origin B1 Im(z) Show the line x = – 2 B1 π B1 Show the correct half line for 4 Shade the correct region B1 -3 -2 0 Re(z) For the vertical line and the circle, allow the B1 marks if all you see is the relevant part. 4
Q5 · Πi 13πi.5 The complex numbers u and w are defined by u = 2e and w = 3e u2 (a) Find…
14πi 13πi.5 The complex numbers u and w are defined by u = 2e and w = 3e u2 (a) Find , giving your answer in the form rei1, where r > 0 and −π < 1 ≤π. Give the exact values w of r and 1. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) State the least positive integer n such that both Imwn = 0 and Re wn > 0. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Mark scheme: 5(a) 1πi B1 State or imply u 2 = 4e 2 πi B1 + B1 For the modulus and the argument. 4 16 Obtain answer v = e 3 3 5(b) State n = 6 B1 1
Q6 · Prove the identity cos 41 + 4 cos 21 + 3 8 cos41
6 (a) Prove the identity cos 41 + 4 cos 21 + 3 8 cos41. 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(b) Hence solve the equation cos 41 + 4 cos 21 = 4 for 0Å ≤1 ≤180Å. 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Mark scheme: 6(a) Express cos4 in terms of cos2 and/or sin2 B1 Express cos2 in terms of cos and/or sin B1 Anywhere 2 Expand to obtain a correct expression in terms of cos B1 2 2cos − 1 −+1 4 2cos − 1 + 3 e.g. 2 ( 2 ) ( ) 4 B1 AG Reduce correctly to cos4+ 4cos2+ 3 8cos 4 6(b) Use the identity and carry out method to calculate a root M1 4 8cos − 3 = 4 Obtain answer, e.g. 14.7 A1 Obtain second answer, e.g.165.3, and no other in the given interval A1 FT Ignore answers outside the given interval. Treat answers in radians as a misread. 3
Q7 · X 1 7 The equation of a curve is y = cos2x, for 0 ≤x < 2π
x 1 7 The equation of a curve is y = cos2x, for 0 ≤x < 2π. At the point where x = a, the tangent to the curve has gradient equal to 12. ` _ a 3 cos a + 2a sin a (a) Show that a = cos−1 . 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(b) Verify by calculation that a lies between 0.9 and 1. 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(c) Use an iterative formula based on the equation in part (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. 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Mark scheme: 7(a) Use correct product or quotient rule M1 Obtain correct derivative in any form A1 2 dy cos x + 2 x sin x cos x e.g. = 4 or dx cos x dy 2 2 = sec x + 2 x sec x tan x dx A1 AG cos a + 2 a sin a Equate derivative at x = a to 12 and obtain a = cos −1 3 12 3 7(b) Evaluate a relevant expression or pair of expressions at a = 0.9 and a = 1 M1 Must be calculated in radians. Complete the argument correctly with correct calculated values A1 cos 0.9 = 0.622 0.553 0.9 0.985 0.0846 0 e.g. or or cos1 = 0.540 0.570 1 0.964 −0.0358 0 or could be looking at values of the gradient 8.46 & 14.1 2 7(c) M1 Must be working in radians. cosan + 2 an sinan 3 correctly at least once Use the process a n +1 = cos−1 12 Obtain final answer 0.97 A1 Show sufficient iterations to 4 d.p. to justify 0.97 to 2 d.p., or show there A1 e.g. 0.95, 0.9743, 0.9694, 0.9704 is a sign change in the interval (0.965, 0.975) 3
Q8 · In a certain chemical reaction the amount, x grams, of a substance is increasing
8 In a certain chemical reaction the amount, x grams, of a substance is increasing. The differential equation satisfied by x and t, the time in seconds since the reaction began, is dx = kxe−0.1t, dt where k is a positive constant. It is given that x = 20 at the start of the reaction. (a) Solve the differential equation, obtaining a relation between x, t and k. 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(b) Given that x = 40 when t = 10, find the value of k and find the value approached by x as t becomes large. 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Mark scheme: 8(a) Separate variables correctly B1 1 −0.1t dx = ke dt x Obtain term ln x B1 0.1t B1 −0.1t xe dt Obtain term −10 ke− Not from Use x = 20, t = 0 to evaluate a constant or as limits in a solution containing M1 terms alnx , be −0.1t where ab 0 −0.1t A1 or equivalent ISW Obtain ln x = 10k 1 − e + ln20 ( ) 5 8(b) Use x = 40, t = 10 to find k or 10k M1 Available for their function of the correct structure even if they found no constant in (a). Obtain 10k = 1.09654 A1 ln 2 or equivalent e.g. 10 k = −1 1 − e State that x tends to 59.9 A1 Need a number, not an expression for that value 3 sf or better 59.87595….. 3
Q9 · Y x O M The diagram shows part of the curve y = 3 −x e−13x for x ≥0, and its minimum…
9 y x O M The diagram shows part of the curve y = 3 −x e−13x for x ≥0, and its minimum point M. (a) Find the exact coordinates of M. 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(b) Find the area of the shaded region bounded by the curve and the axes, giving your answer in terms of e. 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Mark scheme: 9(a) Use correct product or quotient rule *M1 Obtain correct derivative in any form A1 dy − 3x 1 − 3x = −e − 3 − x ) e e.g. dx 3 ( Equate their derivative to zero and solve for x DM1 Obtain x = 6 A1 Obtain y = − 3e−2 A1 Or exact equivalent. 5 9(b) − 1 x − 1 x *M1 3 + b e 3 dx, where ab 0 Commence integration and reach a ( 3 − x ) e 1 1 − x − x A1 3 −3 e 3 dx, or equivalent Obtain −3 ( 3 − x ) e 1 A1 − 3x − 3x −x 3 −3e ( 3 − x ) + 9e Complete integration and obtain 3 xe , or equivalent Substitute limits x = 0 and x = 3, having integrated twice DM1 9 A1 Obtain answer , or exact equivalent e 5
Q11 · D C k N O i j B M A In the diagram, OABCD is a solid figure in which OA = OB = 4 units and…
11 D C k N O i j B M A In the diagram, OABCD is a solid figure in which OA = OB = 4 units and OD = 3 units. The edge OD is vertical, DC is parallel to OB and DC = 1 unit. The base, OAB, is horizontal and angle AOB = 90Å. Unit vectors i, j and k are parallel to OA, OB and OD respectively. The midpoint of AB is M and the point N on BC is such that CN = 2NB. −−−¿ −−¿ (a) Express vectors MD and ON in terms of i, j and k. 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(c) Show that the length of the perpendicular from M to ON is 22 . 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Mark scheme: 11(a) State OM = 2i + 2 j or equivalent B1 Can be implied by MB = − 2 i + 2 j or MA = 2 i − 2 j. Obtain MD = – 2i – 2j + 3k B1 Use a correct method to find ON M1 e.g. OC + 23 CB Obtain answer 3j + k A1 4 11(b) Use the correct process for evaluating the scalar product of MD and ON M1 Using the correct process for the moduli, divide the scalar product by the M1 1 −+6 3 product of the moduli and reach the inverse cosine of the result cos− 10 17 Obtain final answer 103.3 A1 3 11(c) Taking a general point P of ON to have position vector λ ( 3 j + k ) , form M1 −2 0 an equation in λ by either equating the scalar product of ON and MP to e.g. −+2 3 . 3 = 0 zero, or applying Pythagoras to triangle OMP, or equating the derivative 1 of MP to zero 3 A1 Solve and obtain λ = 5 Substitute for λ and calculate MP M1 1 3 MP = −2i − j + k 5 5 22 A1 AG Obtain 5 Alternative method for question 11(c) Use a scalar product to find the projection OQ of OM on OM M1 6 A1 Obtain OQ = 10 Use Pythagoras in triangle OMQ to find MQ M1 22 A1 AG Obtain 5 4
What was in this paper
The subtopics covered by these 8 questions, and how many questions each got. Open one in a new tab to see every Cambridge question on it.
What you needed in this session
Cambridge’s own grade thresholds for 2022 Oct/Nov, Paper 3 · Variant 1. A higher threshold means an easier paper — the bar moves with how the cohort did.