Cambridge A Level Mathematics 9709 — 2021 Oct/Nov Paper 3 · Variant 3

1 Find the quotient and remainder when 2x4 1 is divided by x2 2. [3] + −x + ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 Commence division and reach partial quotient of the form 2 2x kx + M1 Obtain quotient 2 2 2 2 x x + − A1 Obtain remainder 6 5 x − + A1 3

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2 (a) Sketch the graph of y 2x . [1] = −3 (b) Solve the inequality 2x 3x 2. [3] −3 < + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) Show a recognizable sketch graph of 2 3 y x = − B1 1 Question Answer Marks Guidance 2(b) Find x-coordinate of intersection with y = 3x + 2 M1 Obtain 1 5 x = A1 State final answer 1 5 x > only A1 Alternative method for Question 2(b) Solve the linear inequality 3 2 3 2 x x − < + , or corresponding equation M1 Obtain critical value 1 5 x = A1 State final answer 1 5 x > only A1 Alternative method for Question 2(b) Solve the quadratic inequality ( ) ( ) 2 2 2 3 3 2 x x − < + , or corresponding equation M1 Obtain critical value 1 5 x = A1 State final answer 1 5 x > only A1 3

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4x giving your answer correct to 3 decimal places. [4]3 Solve the equation 4x−2 = −42, ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 M1 Obtain correct solution in any form, e.g. 256 4 15 x = A1 Use a correct method for solving an equation of the form 4x a = , where a > 0 M1 Obtain answer 2.047 A1 4

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π 14 Find the exact value of x sin 2x dx. [5] Ó 1 3π ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 Commence integration and reach 1 1 cos cos d 2 2 ax x b x x +  *M1 Obtain 1 1 2 cos 2 cos d 2 2 x x x x − +  A1 OE Complete integration obtaining 1 1 2 cos 4sin 2 2 x x x − + A1 OE Use limits correctly, having integrated twice DM1 Obtain answer 2 + 3 3 π, or exact equivalent A1 5

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5 Solve the equation sin 3 cos 2, for [5] 1 = 21 + 0Å ≤1 ≤360Å. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 5 Use double angle formula and obtain an equation in sinθ M1 Reduce to 2 6sin sin 5 0 θ θ + − = , or 3-term equivalent A1 Solve a 3-term quadratic in sinθ and calculate θ M1 Obtain answer, e.g. 56.4° A1 Obtain second and third answers, e.g.123.6° and 270° and no others in the given interval A1 Ignore answers outside the interval. Treat answers in radians as a misread. 5

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8 D N k C j B O M i A In the diagram, OABCD is a pyramid with vertex D. The horizontal base OABC is a square of side 4 units. The edge OD is vertical and OD 4 units. The unit vectors i, j and k are parallel to OA, OC = and OD respectively. The midpoint of AB is M and the point N on CD is such that DN 3NC. = (a) Find a vector equation for the line through M and N. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Show that the length of the perpendicular from O to MN is 1 82. 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Mark scheme: 8(a) State 4 2 OM = + i j B1 Use a correct method to find ON  M1 Obtain answer 3j + k A1 Use a correct method to find a line equation for MN M1 Obtain answer r = 3j + k + λ(4i – j – k), or equivalent A1 5 Question Answer Marks Guidance 8(b) Taking a general point P on MN, form an equation in λ by either equating a relevant scalar product to zero or equating the derivative of OP  to zero or using Pythagoras in triangle OPM or OPN M1 Obtain λ = 2 9 A1 OE Use correct method to find OP M1 Obtain the given answer correctly A1 Alternative method to Question 8(b) Use a scalar product to find the projection of OM (or ON) on MN M1 Obtain answer 14 18 4 or 18       A1 Use Pythagoras to obtain the perpendicular M1 Obtain the given answer correctly A1 4

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10 A large plantation of area 20 km2 is becoming infected with a plant disease. At time t years the area infected is x km2 and the rate of increase of x is proportional to the ratio of the area infected to the area not yet infected. dx When t 0, x 1 and 1. = = dt = (a) Show that x and t satisfy the differential equation dx 19x [2] dt = 20 −x. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Solve the differential equation and show that when t 1 the value of x satisfies the equation = [5] x = e0.9+0.05x. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Use an iterative formula based on the equation in part (b), with an initial value of 2, to determine x correct to 2 decimal places. Give the result of each iteration to 4 decimal places. 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(d) Calculate the value of t at which the entire plantation becomes infected. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 10(a) State or imply equation of the form d d x t = k 20 x x − M1 Obtain k = 19 A1 AG 2 10(b) Separate variables and integrate at least one side M1 Obtain terms 20 ln x – x and 19t, or equivalent A1 A1 Evaluate a constant or use t = 0 and x = 1 as limits in a solution containing terms a ln x and bt M1 Substitute t = 1 and rearrange the equation in the given form A1 AG 5 10(c) Use 1 nx + = 0.9 0.05 e n x + correctly at least once M1 Obtain final answer x = 2.83 A1 Show sufficient iterations to 4 decimal places to justify 2.83 to 2 d.p. or show there is a sign change in the interval (2.825, 2.835) A1 3 10(d) Set x = 20 and obtain answer t = 2.15 B1 1

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