Cambridge A Level Mathematics 9709 — 2019 May/June Paper 3 · Variant 1

1 Use the trapezium rule with 3 intervals to estimate the value of 3 2x dx. [3] Ó0 −4 ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 State or imply ordinates 3, 2, 0, 4 B1 These and no more Accept in unsimplified form 0 2 4 − etc. Use correct formula, or equivalent, with h = 1 and four ordinates M1 Obtain answer 5.5 A1 3

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1 5 (i) Differentiate with respect to [2] 1. sin21 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) The variables x and satisfy the differential equation 1 dx x tan 0, 1 + cosec21 = d1 for 0 1 and x 0. It is given that x 4 when 1 Solve the differential equation, < 1 < 20 > = 1 = 60. obtaining an expression for x in terms of [6] 1. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 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Mark scheme: 5(i) Use chain rule M1 3 2 cos sin cosec cot θ θ θ θ − = − k k Allow M1 for 1 2cos sin θ θ − − Obtain correct answer in any form A1 e.g. 2 2cosec cot θ θ − , 3 2cos sin θ θ − Accept 4 2sin cos sin θ θ θ − 2 5(ii) Separate variables correctly and integrate at least one side B1 2 d cosec cot d θ θ θ = − ∫ ∫ x x Obtain term 2 1 2 x B1 Obtain term of the form 2 sin k θ M1* or equivalent Obtain term 2 1 2sin θ A1 or equivalent Use x = 4, 1 6 θ π = to evaluate a constant, or as limits, in a solution with terms 2 ax and 2 sin b θ , where ab ≠ 0 DM1 Dependent on the preceding M1 Obtain solution ( ) 2 cosec 12 x θ = + A1 or equivalent 6

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7 y x O a 4 1 1 for 0 The diagram shows the curves y 4. When x = ≤x < = 4 cos 2x and y = 4 −x, a, the tangents to the curves are perpendicular. 4 (i) Show that a / 2 sin 12a . 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(ii) Verify by calculation that a lies between 2 and 3. 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(iii) Use an iterative formula based on the equation in part (i) to determine a correct to 3 decimal places. Give the result of each iteration to 5 decimal places. 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Mark scheme: 7(i) State at least one correct derivative B1 ( ) 2 1 1 2sin , 2 4 − − x x Equate product of derivatives to – 1 M1 or equivalent Obtain a correct equation, e.g. ( ) 2 1 2sin 4 2 = − x x A1 Rearrange correctly to obtain 4 2sin 2 = − a a AG A1 4 7(ii) Calculate values of a relevant expression or pair of expressions at a = 2 and a = 3 M1 e.g. 2 2 2.7027.. 3 3 2.587.. = < = > a a 0.703 2.317 0.412 0.995         − −     Values correct to at least 2 dp Complete the argument correctly with correct calculated values A1 2 7(iii) Use the iterative formula 1 ) 1 4 (2sin 2 n n a a + = − correctly at least once M1 Obtain final answer 2.611 A1 Show sufficient iterations to 5 d.p. to justify 2.611 to 3 d.p., or show there is a sign change in the interval (2.6105, 2.6115) A1 2, 2.70272, 2.60285, 2.61152, 2.61070, 2.61077 2.5, 2.62233, 2.60969, 2.61087, 2.61076 3, 2.58756, 2.61301, 2.61056, 2.61079 Condone truncation. Accept more than 5 dp 3

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9 z D y C B O A x The diagram shows a set of rectangular axes Ox, Oy and Oz, and four points A, B, C and D with position vectors −−→OA 3i, −−→OB 3i 4j, −−→OC i 3j and −−→OD 2i 3j 5k. = = + = + = + + (i) Find the equation of the plane BCD, giving your answer in the form ax by cz d. 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(ii) Calculate the acute angle between the planes BCD and OABC. 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Mark scheme: 9(i) Obtain a vector parallel to the plane, e.g. 2 = + CB i j B1 Use scalar product to obtain an equation in a, b, c, M1 e.g. 2a + b = 0, a + 5c = 0, a + b – 5c = 0 Obtain two correct equations in a, b, c A1 Solve to obtain a : b : c, M1 or equivalent Obtain a : b : c = 5 : – 10 : – 1, A1 or equivalent Obtain equation 5 10 25 − − = − x y z , A1 or equivalent Alternative method 1 Obtain a vector parallel to the plane, e.g. 5 = + JJJG CD i k B1 5 = −−+ JJJG BD i j k Obtain a second such vector and calculate their vector product, e.g. ( ) ( ) 2 5 + × + i j i k M1 Obtain two correct components A1 Obtain correct answer, e.g. 5i – 10j – k A1 Substitute to find d M1 Obtain equation 5 10 25 − − = − x y z , A1 or equivalent Question Answer Marks Guidance 9(i) Alternative method 2 Obtain a vector parallel to the plane, e.g. 5 = + − JJJG DB i j k B1 Obtain a second such vector and form correctly a 2-parameter equation for the plane M1 State a correct equation, e.g. ( ) ( ) 3 4 5 5 µ = + + + + + − r i j i k i j k λ A1 State three equations in x, y, z, λ and µ A1 Eliminate λ and µ M1 Obtain equation 5 10 25 − − = − x y z A1 or equivalent Alternative method 3 Substitute for B and C and obtain 3a + 4b = d and a + 3b = d B1 Substitute for D to obtain a third equation and eliminate one unknown (a, b, or d) entirely M1 Obtain two correct equations in two unknowns, e.g. a, b, c A1 Solve to obtain their ratio, e.g. a : b : c M1 Obtain a : b : c = 5 : – 10 : – 1, a : c : d = 5 : – 1: – 25, or b : c : d = 10 : 1 : 25 A1 or equivalent Obtain equation 5 10 25 − − = − x y z A1 or equivalent Question Answer Marks Guidance 9(i) Alternative method 4 Substitute for B and C and obtain 3a + 4b = d and a +3b = d B1 Solve to obtain a : b : d M2 or equivalent Obtain a : b : d = 1 : – 2: – 5 A1 or equivalent Substitute for C to obtain c M1 Obtain equation 5 10 25 − − = − x y z A1 or equivalent 6 9(ii) State or imply a normal vector for the plane OABC is k B1 Carry out correct process for evaluating a scalar product of two relevant vectors, e.g. (5i – 10j – k).(k) M1 i.e. correct process using k and their normal Using the correct process for calculating the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result M1 Allow M1M1 for clear use of an incorrect vector that has been stated to be the normal to OABC Obtain answer 84.9° or 1.48 radians A1 4

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10 Throughout this question the use of a calculator is not permitted. The complex number i is denoted by u. ï3 + 0 and Hence giving the exact values of r and (i) Express u in the form rei1, where r > −0 < 1 ≤0, 1. or otherwise state the exact values of the modulus and argument of u4. 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(ii) Verify that u is a root of the equation z3 0 and state the other complex root of this −8z + 8ï3 = equation. 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(iii) On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying the inequalities z and Imz where Im z denotes the imaginary part of z. −u ≤2 ≥2, [5]

Mark scheme: 10(i) State or imply r = 2 B1 Accept 4 State or imply θ = 1 6π B1 Use a correct method for finding the modulus or the argument of 4 u M1 Allow correct answers from correct u with minimal working shown Obtain modulus 16 A1 Obtain argument 2 3π A1 Accept 2 3 16e π i 5 10(ii) Substitute u and carry out a correct method for finding 3 u M1 ( ) 3 8 = u i Follow their 3 u if found in part (i) Verify u is a root of the given equation A1 State that the other root is 3 i − B1 Alternative method State that the other root is 3 i − B1 Form quadratic factor and divide cubic by quadratic M1 ( )( )( ) 2 3 3 2 3 4 − − − + = − + z i z i z z Verify that remainder is zero and hence that u is a root of the given equation A1 3 Question Answer Marks Guidance 10(iii) Show the point representing u in a relatively correct position B1 Show a circle with centre u and radius 2 B1 FT on the point representing u. Condone near miss of origin Show the line y = 2 B1 Shade the correct region B1 Show that the line and circle intersect on x = 0 B1 Condone near miss 5 Im y = 2 Re shaded u

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