Cambridge A Level Mathematics 9709 — 2021 Oct/Nov Paper 3 · Variant 2

2 Solve the inequality 3x 2 x 2a , where a is a positive constant. [4] −a > + ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 State or imply non-modular inequality ( ) ( ) 2 2 2 3 2 2 x a x a − > + , or corresponding quadratic equation, or pair of linear equations or linear inequalities B1 Need 2 2 seen or implied. Make reasonable attempt to solve a 3-term quadratic, or solve two linear equations for x in terms of a M1 ( ) 2 2 5 22 15 0 x ax a − − = Obtain critical values x = 5a and x = 3 5 a − and no others A1 OE Accept incorrect inequalities with correct critical values. Must state 2 values i.e. a b c ± is not sufficient. State final answer x > 5a, x < 3 5 a − A1 Do not condone ⩾ for > or ⩽ for < in the final answer. 3 5 5a x a < < − is A0, ‘and’ is A0. Alternative method for Question 2 Obtain critical value x = 5a from a graphical method, or by solving a linear equation or linear inequality B1 Obtain critical value x = 3 5 a − similarly B2 Maximum 2 marks if more than 2 critical values. State final answer x > 5a , x < 3 5 a − B1 Do not condone ≥ for > or ≤ for < in the final answer. 3 5 5a x a < < − is B0, ‘and’ is B0. 4

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5 (a) On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying the inequalities z and Imz [4] −3 −2i ≤1 ≥2. (b) Find the greatest value of arg z for points in the shaded region, giving your answer in degrees. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Show circle with centre 3 + 2i B1 Show circle with radius 1. Must match their scales: if scales not identical should have an ellipse. B1 Show line y = 2 in at least the diameter of a circle in the first quadrant B1 Shade the correct region in a correct diagram B1 4 5(b) Identify the correct point B1 Carry out a correct method for finding the argument M1 e.g. 1 1 2 1 arg tan sin 3 13 x − − = + Exact working required. Obtain answer 49.8° A1 Or better. 0.869 radians scores B1M1A0. 3 Special Case 1: B1M0 for 45° if they have shaded the wrong half of the circle. Special Case 2: 3 out of 3 available if they identify the correct point on the correct circle and it is consistent with their shading. O 2i 3

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7 The variables x and y satisfy the differential equation dy e2x 4xy2, dx = and it is given that y 1 when x 0. = = Solve the differential equation, obtaining an expression for y in terms of x. 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Mark scheme: 7 Separate variables correctly B1 2 2 1 d 4 e d x y x x y − =   2 1 dy y =  1 y − B1 OE Commence the other integration and reach 2 2 e e d x x ax b x − − +  M1 Obtain 2 2 2 e 2 e d x x x x − − − +  or 2 2 1 1 e e d 2 2 x x x x − − − +  A1 SOI (might have taken out factor of 4) Complete integration and obtain 2 2 2 e e x x x − − − − A1 Evaluate a constant or use x = 0 and y = 1 as limits in a solution containing terms of the form 2 2 , e , e x x p qx r y − − , or equivalent. M1 Obtain y = 2e 2 1 x x + , or equivalent expression for y A1 ISW 7

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8 (a) By first expanding show that cos21 + sin21 2, 1 sin2 [3] cos41 + sin41  −12 21. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence solve the equation 59, cos41 + sin41 = for [4] 0Å < 1 < 180Å. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(a) Expand the square and equate to 1 B1 Use correct double angle formula M1 Need to see 4 2 or sin2 2sin cos θ θ θ = stated. Obtain 4 4 2 1 2 cos sin 1 sin 2 θ θ θ + = − A1 Obtain the given result correctly. 3 8(b) Use the identity and carry out a method for finding a root M1 ( ) 2 5 1 1 sin 2 2 9 θ − = Obtain answer 35.3° A1 Must be correct if overspecified: 35.264... Obtain a second answer, e.g. 54.7° A1 FT [ ] e.g 90 35.3 their ° − ° Do not FT if mixing degrees and radians. Obtain the remaining answers, e.g. 144.7° and 125.3° and no others in the given interval A1 FT [ ] e.g.180 .. and180 .. ° − ° − Ignore answers outside the given interval. Treat answers in radians as a misread. ( ) 0.615, 0.955, 2.19, 2.53 Do not FT if mixing degrees and radians. 4

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9 The equation of a curve is ye2x 2. −y2ex = dy 2yex (a) Show that −y2 . [4] dx = 2y −ex ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the exact coordinates of the point on the curve where the tangent is parallel to the y-axis. 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Mark scheme: 9(a) State correct derivative of 2 e x y with respect to x B1 2 2 d 2 e e d x x y y x + State correct derivative of 2ex y with respect to x B1 2 d 2 e e d x x y y y x + Equate attempted derivative of the LHS to zero and solve for d d y x M1 Obtain 2 d 2 e d 2 e x x y y y x y − = − A1 Obtain the given answer correctly. Condone multiplication by 1 1 − − and cancelling of ex without comment. Alternative method for Question 9(a) Rearrange as ( ) 2 2 2 2 d d e e 2e e e d d e e x x x x x x x y y y y x x y =  − = − − − B1 Other rearrangements are possible e.g. ( ) 2 2 2 2 d d 2e e e 2 e e d d x x x x x y y y y y y x x − − − − − = + = − ( ) 2 2 2 d 2 d 2e e e d d e e x x x x x y y y x x y   = − × − −     − B1 2 d d 4 2 e e d d x x x y y e y y x x − − −  = − + − Solve for d d y x M1 Obtain 2 d 2 e d 2 e x x y y y x y − = − A1 Obtain the given answer correctly. 4 Question Answer Marks Guidance 9(b) Equate denominator to zero and substitute for y or for ex in the equation of the curve *M1 Obtain equation of the form 3e x a b = or 3 cy d = DM1 ( ) 3 3 e 8, 1 x y = = SOI Obtain x = ln 2 A1 Accept 1 ln8 3 ISW Obtain y = 1 A1 4

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10 With respect to the origin O, the position vectors of the points A and B are given by −−¿OA 12 and ` a = 0 −1 −−¿OB 3 . = 1 (a) Find a vector equation for the line l through A and B. 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(b) The point C lies on l and is such that −−¿AC 3−−¿AB. = Find the position vector of C. 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(c) Find the possible position vectors of the point P on l such that OP 14. 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Mark scheme: 10(a) Obtain direction vector 2 −+ + i j k , or equivalent B1 Accept answers as column vectors throughout. Use a correct method to form a vector equation M1 State answer r = i + 2j – k + λ (– i + j + 2k), or equivalent correct form A1 e.g. 0 1 3 1 1 2 μ = + − −                r Allow x y z      for r. 3 10(b) Use a correct method to find the position vector of C M1 e.g. 1 3 2 3 1 6 − = + = + −+           OC OA AC Obtain answer – 2i + 5j +5k, or equivalent A1 Accept as coordinates. 2 Question Answer Marks Guidance 10(c) State OP  in component form B1 FT Form an equation in λ by equating the modulus of OP to 14 , or equivalent M1 Simplify and obtain 2 3 4 0 λ λ − − = , or equivalent A1 2 3 4 0 λ λ + − = if using 2 −− i j k in (a). 2 3 5 2 0 μ μ + − = if using 2 −+ + i j k in (a) and OB. Solve a 3-term quadratic and find a position vector M1 ( ) 4 4 1 1 1, or 1, or , 2 or ,2 3 3 3 3 λ λ μ μ = − = − = − = − Obtain answers 2i + j – 3k and 1 10 5 3 3 3 − + + i j k , or equivalent A1 Accept as coordinates. 5

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