Cambridge A Level Mathematics 9709 — 2017 Oct/Nov Paper 3 · Variant 2

2 x 14 The curve with equation y has one stationary point in the interval x −sin cosx = −120 < < 20. (i) Find the exact coordinates of this point. 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(ii) Determine whether this point is a maximum or a minimum point. 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Mark scheme: 4(i) Use correct product or quotient rule or rewrite as 2sec x − tan x and differentiate M1 Obtain correct derivative in any form A1 Equate the derivative to zero and solve for x M1 Obtain x = 16 π A1 Obtain y = 3 A1 5 4(ii) Carry out an appropriate method for determining the nature of a stationary point M1 Show the point is a minimum point with no errors seen A1 2

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6 The equation of a curve is x3y 2a4, where a is a non-zero constant. −3xy3 = dy 3x2y (i) Show that . [4] −3y3 dx = 9xy2 −x3 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Hence show that there are only two points on the curve at which the tangent is parallel to the x-axis and find the coordinates of these points. 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Mark scheme: 6(i) 2 3 dy B1 State or imply 3 x y + x as derivative of 3x y dx 2 dy 3 3 B1 State or imply 9 xy + 3 y as derivative of 3xy dx d y M1 Equate derivative of the LHS to zero and solve for d x Obtain the given answer AG A1 4 6(ii) Equate numerator to zero and use x = – y to obtain an equation in x or in y M1 Obtain answer x = a and y = – a A1 Obtain answer x = – a and y = a A1 Consider and reject y = 0 and x = y as possibilities B1 4

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7 Throughout this question the use of a calculator is not permitted. The complex number 1 i is denoted by u. − ï3 (i) Find the modulus and argument of u. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Show that u3 8 0. [2] + = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying both the inequalities z and Re z where Re z denotes the real part of z. −u ≤2 ≥2, [4]

Mark scheme: 7(i) State modulus 2 B1 State argument − 13π or −60° ( 53π or 300°) B1 2 7(ii) EITHER: Expand (1 − ( 3)i) 3 completely and process i2 and i3 (M1 Verify that the given relation is satisfied A1) OR: u 3 = 23 ( cos ( −π) + i sin ( −π) ) or equivalent: follow their answers to (i) (M1 Verify that the given relation is satisfied A1) 2 7(iii) Show a circle with centre 1 − ( 3)i in a relatively correct position B1 Show a circle with radius 2 passing through the origin B1 Show the line Re z = 2 B1 Shade the correct region B1 4

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a 1 9 It is given that x 2 ln x dx 2, where a 1. Ó 1 = > 3 3 7 2a 2 (i) Show that a 2 . [5] + 3 ln a = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Show by calculation that a lies between 2 and 4. 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(iii) Use the iterative formula ` 3 a23 7 2a2n + 3 ln an an+1 = to determine a correct to 3 decimal places. Give the result of each iteration to 5 decimal places. 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Mark scheme: 3 9(i) 3 2 2 1 *M1 x . dx Integrate by parts and reach ax ln x + b ∫ x 2 32 2 12 A1 Obtain 3 x ln x 3 x d x −∫ 2 32 4 32 A1 Obtain integral 3 x ln x − 9 x , or equivalent Substitute limits correctly and equate to 2 DM1 Obtain the given answer correctly AG A1 5 9(ii) Evaluate a relevant expression or pair of expressions at x = 2 and x = 4 M1 Complete the argument correctly with correct calculated values A1 2 9(iii) Use the iterative formula correctly at least once M1 Obtain final answer 3.031 A1 Show sufficient iterations to 5 d.p. to justify 3.031 to 3 d.p., or show there is a sign A1 change in the interval (3.0305, 3.0315) 3

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10 Two planes p and q have equations x y 3z 8 and 2x z 3 respectively. + + = −2y + = (i) Calculate the acute angle between the planes p and q. 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(ii) The point A on the line of intersection of p and q has y-coordinate equal to 2. Find the equation of the plane which contains the point A and is perpendicular to both the planes p and q. Give your answer in the form ax by cz d. [7] + + = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 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Mark scheme: 10(i) State or imply a correct normal vector to either plane, e.g. i + j + 3k or 2i − 2 j + k B1 Carry out correct process for evaluating the scalar product of two normal vectors M1 Using the correct process for the moduli, divide the scalar product of the two M1 normals by the product of their moduli and evaluate the inverse cosine of the result Obtain final answer 72.5° or 1.26 radians A1 4 10(ii) EITHER: Substitute y = 2 in both plane equations and solve for x or for z (M1 Obtain x = 3 and z = 1 A1) OR: Find the equation of the line of intersection of the planes Substitute y = 2 in line equation and solve for x or for z (M1 Obtain x = 3 and z = 1 A1) EITHER: Use scalar product to obtain an equation in a, b and c, e.g. a + b + 3c = 0 (B1 Form a second relevant equation, e.g. 2 a − 2b + c = 0 , and solve for one *M1 ratio, e.g. a : b Obtain final answer a : b : c = 7 : 5 : – 4 A1 Use coordinates of A and values of a, b and c in general equation and find DM1 jjjjjjjjjjjjjjjd Obtain answer 7 x + 5 y − 4 z = 27 , or equivalent A1 FT) OR1: Calculate the vector product of relevant vectors, e.g. (*M1 ( i + j + 3k ) × (2 i − 2 j + k ) Obtain two correct components A1 Obtain correct answer, e.g. 7 i + 5 j − 4k A1 Substitute coordinates of A in plane equation with their normal and find d DM1 Obtain answer 7 x + 5 y − 4 z = 27 , or equivalent A1 FT) OR2: Using relevant vectors, form a two-parameter equation for the plane (*M1 State a correct equation, e.g. r = 3i + 2 j + k + λ( i + j + 3k ) + µ(2i − 2 j + k ) A1 FT State 3 correct equations in x, y, z, λand µ A1 FT Eliminate λand µ DM1 Obtain answer 7 x + 5 y − 4 z = 27 , or equivalent A1 FT) OR3: Use the direction vector of the line of intersection of the two planes as (*M1 normal vector to the plane Two correct components A1 Three correct components A1 Substitute coordinates of A in plane equation with their normal and find d DM1 Obtain answer 7 x + 5 y − 4 z = 27 , or equivalent A1 FT) 7

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