Cambridge A Level Mathematics 9709 — 2022 Feb/March Paper 3 · Variant 2

1 Solve the inequality 2x 3 3 x 2 . [4] + > + ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 State or imply non-modular inequality ( ) ( ) 2 2 2 2 3 3 2 + > + x x , or corresponding quadratic equation, or pair of linear equations B1 Make a reasonable attempt at solving a 3-term quadratic, or solve two linear equations for x M1 Quadratic formula or (5x + 9)(x +3) Obtain critical values x = – 3 and x = 9 5 − A1 OE State final answer 9 3 5 −< < − x or 3 > − x and 9 5 < − x A1 [Do not condone ⩽ for < in the final answer.] No ISW Alternative method for question 1 Obtain critical value x = – 3 from a graphical method, or by solving a linear equation or linear inequality B1 2x + 3 = 3(x + 2)  x = − 3 Obtain critical value x = 9 5 − similarly B2 State final answer 9 3 5 −< < − x or 3 > − x and 9 5 < − x B1 [Do not condone ⩽ for < in the final answer.] No ISW 4

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2 On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying the inequalities z 2 and arg z [4] + −3i ≤2 ≤34π.

Mark scheme: 2 Show a circle with centre – 2 + 3i B1 Must see (− 2, 3) or appropriate marks on axes Show a circle of radius 2 and centre not at the origin. B1 Show correct half line from the origin B1 3π 4 or π 4 seen, or half line that approximately bisects angle π 2 . Shade the correct region. B1 4 N.B. Maximum 3 out of 4 if any errors seen.

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4 The parametric equations of a curve are x 1 y cos cos 4 = −cos 1, = 1 −1 21. dy 1 Show that sin2 . [5] dx = −2 21 ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 State d dθ dθ y = 1 sin sin 2 2 θ θ − + Use d d y x = d dθ y ÷ d dθ x M1 Obtain correct answer in any form A1 e.g. 1 sin sin 2 2 sin θ θ θ − + Use double angle correctly to obtain d d y x in terms of θ M1 sin 2θ = 2sin θ cos θ Obtain the given answer with no errors seen − 2sin2 1 2θ       A1 AG. Requires correct cancellation of ALL sin θ terms and cos θ = 1 − 2sin2 1 2θ       seen SC For incorrect signs, consistent throughout max. B0, M1, A0, M1, A1 5

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6 Find the complex numbers w which satisfy the equation w2 2iw* 1 and are such that Re w Give your answers in the form x iy, where x and y are real. + = ≤0.[6] + ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 6 Substitute and obtain a correct equation in x and y Use 2i 1 = − at least once and equate real and imaginary parts M1 Obtain two correct equations, e.g. 2 2 2 1 − + = x y y and 2 2 0 + = xy x A1 Solve for x or for y M1 Using y = – 1, obtain answer w = – 2 – i only A1 A0 if w = 2 – i as well Using x = 0, obtain answer w = i A1 6

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7 (a) By sketching a suitable pair of graphs, show that the equation 4 sec 2x1 has exactly one root in the interval 0 −x2 = [2] ≤x < π. (b) Verify by calculation that this root lies between 1 and 2. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 2xn (c) Use the iterative formula xn+1 = ?4 −sec 1 to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) Sketch a relevant graph, e.g. 2 4 = − y x Sketch a second relevant graph, e.g. 1 sec 2 = y x , and justify the given statement B1 Needs (0, 1) or mark on axis and (π, 0) Asymptote NOT required, but must NOT reach x = π. Sec graph must exist over at least interval 3π 0, 4 é ù ê ú ê ú ë û and quadratic graph over [0, 2.5]. 2 7(b) Calculate the value of a relevant expression or values of a pair of relevant expressions at x = 1 and x = 2. M1 Need all 4 values or the 2 values correct for M1. Angles in degrees score M0. Complete the argument with correct calculated values A1 2 Question Answer Marks Guidance 7(c) Use the iterative process correctly at least twice M1 Obtain final answer 1.60 A1 Must be 2 d.p. Show sufficient iterations to 4 d.p.to justify 1.60 to 2 d.p. or show there is a sign change in the interval (1.595, 1.605) A1 3

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8 (a) Find the quotient and remainder when 8x3 4x2 2x 7 is divided by 4x2 1. [3] + + + + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 1 2 8x3 4x2 2x 7 (b) Hence find the exact value of dx. [5] + + + 4x2 1 Ô0 + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(a) Commence division and reach quotient of the form 2x ± 1 + r Obtain (quotient) 2x + 1 A1 Obtain (remainder) 6 A1 3 Question Answer Marks Guidance 8(b) Obtain terms 2 + x x B1 OE Obtain term of the form 1 tan 2 − a x M1 Obtain term 1 3tan 2 − x A1 OE Use x = 0 and 1 2 = x as limits in a solution containing a term of the form 1 tan 2 − a x M1 2 1 1 π 2 2 4 a æ ö÷ ç + + ÷ ç ÷ çè ø , need π 4 seen or implied Obtain final answer 3 (1 π 4 + ), or exact equivalent A1 ISW, Answers in degrees score A0. 5

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10 The points A and B have position vectors 2i j k and i 2k respectively. The line l has vector equation r i 2j i . + + −2j + = + −3k + - −3j −2k (a) Find a vector equation for the line through A and B. 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(b) Find the acute angle between the directions of AB and l, giving your answer in degrees. 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(c) Show that the line through A and B does not intersect the line l. 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Mark scheme: 10(a) Obtain direction vector −− + B1 OE Use a correct method to form a vector equation M1 Obtain answer ( ) 2 3 λ = + + + −− + r i j k i j k or ( ) 2 2 3 λ = + + −− + r i j k i j k A1 Need r or r on LHS 3 10(b) Carry out the correct process for evaluating the scalar product of the direction vectors. M1 (−1, − 3, 1) . (1, −3, −2) = − 1 + 9 − 2 Using the correct process for the moduli, divide the scalar product by the product of the moduli and find the inverse cosine of the result for any 2 vectors M1 ( )( ) 1 1 9 2 cos 1 9 1 1 9 4 ) −  +     + + + +   Obtain answer 61.1° A1 61.086° 3 Question Answer Marks Guidance 10(c) Express general point of AB or l in component form, e.g. (2 – λ, 1 – 3λ, 1 + λ) or (1 + µ, 2 – 3µ, – 3 – 2µ) B1 Equate at least two pairs of components and solve for λ or for µ M1 Obtain a correct answer for λ or µ, e.g. λ = 6, 1 3 , or 14 9 − ; µ = – 5, 2 3 or 11 9 − A1 Verify that all three equations are not satisfied, and the lines do not intersect A1 Express general point of AB or l in component form, e.g. (1 – λ*, − 2 – 3λ*, 2 + λ*) or (1 + µ*, 2 – 3µ*, – 3 – 2µ*) 4

More questions on Vectors

11 y M x O 1 2π The diagram shows the curve y sin x cos 2x for 0 and its maximum point M. = ≤x ≤12π, (a) Find the x-coordinate of M, giving your answer correct to 3 significant figures. 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(b) Using the substitution u cosx, find the area of the shaded region enclosed by the curve and the x-axis in the first quadrant,= giving your answer in a simplified exact form. 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Mark scheme: 11(a) Use correct product rule or chain rule M1 Obtain correct derivative in any form A1 cos x.cos 2x − sin x.2sin 2x Equate derivative to zero and use a correct double angle formula *M1 If chain rule used then derivative set to 0 gains M1 since correct double angle formula has already been used. Obtain an equation in one trigonometric variable DM1 Allow following from coefficient errors in differentiation only Obtain 2 6sin 1 = x , 2 6cos 5 = x or 2 5tan 1 = x A1 One of these 3 expressions Obtain final answer x = 0.421 A1 Must be 3s.f. 6 Question Answer Marks Guidance 11(b) State or imply du = sin − x dx B1 Using double angle formula, express integral in terms of u and du M1 Use cos2x = 2cos2x − 1 Integrate and obtain ± 3 2 3   −     u u A1 Use limits u = 1, u = 1 2 in an integral of the form 3 + au bu , where ab ≠ 0 M1 Require both limits substituted twice in 3 + au bu for M1. Do not condone decimals. Obtain ( ) 1 2 1 3 − or 1 1 2 1 1 2 or or 3 3 3 3 2       simplified equivalent A1 ISW 5

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