Cambridge A Level Mathematics 9709 — 2022 Oct/Nov Paper 3 · Variant 1

9709/31/O/N/22 · 8 questions · 75 marks · ≈84 min

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Mark scheme17 pages

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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

More questions on Quadratics

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

More questions on Complex numbers

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

More questions on Complex numbers

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

More questions on Trigonometry

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

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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

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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

More questions on Differentiation

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

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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.

A58/75
B51/75
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D33/75
E23/75