Cambridge A Level Mathematics 9709 — 2019 Oct/Nov Paper 3 · Variant 2

1 Solve the equation 5 ln 4 6. Show all necessary working and give the answer correct to −3x = 3 decimal places. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 Remove logarithms and state 1.2 4 3 e x − = , or equivalent B1 Accept 4 3 3.32 01169..... x − = 3 s.f. or better Use correct method to solve an equation of the form 3x a = , where a > 0. M1 ( ) 3 0.67988.. x = Complete method to x =… If using log3 the subscript can be implied Obtain answer x = – 0.351 only A1 CAO must be to 3 d.p. 3

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e−2x dy 2 The curve with equation y has a stationary point in the interval x 1. Find and = 1 −1 < < dx −x2 hence find the x-coordinate of this stationary point, giving the answer correct to 3 decimal places. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Use correct quotient rule or correct product rule M1 Obtain correct derivative in any form A1 ( ) ( ) 2 2 2 2 2 2e 1 2 e d d 1 x x x x y x x − − − − + = − Equate derivative to zero and obtain a 3 term quadratic in x M1 Obtain a correct 3-term equation e.g. 2 2 2 2 0 x x + − = or 2 1 x x + = A1 From correct work only Solve and obtain x = 0.618 only A1 From correct work only 5

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5 The equation of a curve is 2x2y a3, where a is a positive constant. Show that there is only one −xy2 = point on the curve at which the tangent is parallel to the x-axis and find the y-coordinate of this point. 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Mark scheme: 5 State 2 4 2 xy x + d d y x , or equivalent, as derivative of 2 2x y B1 State 2 2 y xy + d d y x , or equivalent, as derivative of 2 xy B1 Equate attempted derivative of LHS to zero and set d d y x equal to zero (or set numerator equal to zero) *M1 2 2 d 4 d 2 2 y y xy x x xy − = − Reject y = 0 B1 Allow from 2 0 y kxy − = Obtain y = 4x A1 OE from correct numerator. ISW Obtain an equation in y (or in x) and solve for y (or for x) in terms of a DM1 3 3 3 3 3 3 8 16 or 8 4 y y x x a a − = − = Obtain y = – 2a A1 With no errors seen 7 Question Answer Marks Guidance 5 Alternative method for question 5 Rewrite as 3 2 2 a y x xy = − and differentiate M1 Correct use of function of a function and implicit differentiation Obtain correct derivative (in any form) A1 ( ) 3 2 2 d 4 d d d 2 y a x y x y x x x xy   − − −     = − set d d y x equal to zero (or set numerator equal to zero) *M1 Obtain 4x – y = 0 A1 Confirm 2 2 0 x xy − ≠ B1 0 x = and 2x y = both give 0 a = Obtain an equation in y (or in x) and solve for y (or for x) DM1 3 3 3 3 3 3 8 16 or 8 4 y y x x a a − = − = Obtain y = – 2a A1 With no errors seen 7

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7 (a) Find the complex number z satisfying the equation iz z 0, + z* −2 = where z* denotes the complex conjugate of z. Give your answer in the form x iy, where x and + y are real. 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(b) (i) On a single Argand diagram sketch the loci given by the equations z 2 and Im z 3, −2i = = where Imz denotes the imaginary part of z. [2] (ii) In the first quadrant the two loci intersect at the point P. Find the exact argument of the complex number represented by P. 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Mark scheme: 7(a) Substitute and obtain a correct horizontal equation in x and y in any form B1 2 2 i 2 2i 0 x y x y x y + + − − + = Allow if still includes brackets and/or 2i Use 2i 1 =− and equate real and imaginary parts to zero OE *M1 For their horizontal equation Obtain two correct equations e.g. 2 2 2 0 x y y x + − − = and x + 2y = 0 A1 Allow i 2i 0 x y + = Solve for x or for y DM1 Obtain answer 6 3i 5 5 − and no other A1 OE, condone ( ) 1 6 3i 5 − 5 Question Answer Marks Guidance 7(b)(i) Show a circle with centre 2i and radius 2 B1 Show horizontal line y = 3 – in first and second quadrant B1 SC: For clearly labelled axes not in the conventional directions, allow B1 for a fully ‘correct’ diagram. 2 7(b)(ii) Carry out a complete method for finding the argument. (Not by measuring the sketch) M1 ( ) 3 3i z = + Must show working if using 1.7 in place of 3 . Obtain answer 1 π 3 (or 60 )° A1 SC: Allow B2 for 60° with no working 2 Im(z) Re(z) P 3i 2i

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a 39 It is given that 3, where the constant a is such that 0 a x cos 3x1 dx = < < 20. Ó 0 (i) Show that a satisfies the equation 4 cos 13a −3 . 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(ii) Verify by calculation that a lies between 2.5 and 3. 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(iii) Use an iterative formula based on the equation in part (i) to calculate a correct to 3 decimal places. Give the result of each iteration to 5 decimal places. 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Mark scheme: 9(i) Commence integration by parts, reaching 1 1 sin sin d 3 3 ax x b x x −∫ *M1 Obtain 1 1 3 sin 3 sin d 3 3 x x x x −∫ A1 Complete integration and obtain 1 1 3 sin 9cos 3 3 x x x + A1 Substitute limits correctly and equate result to 3 in an integral of the form 1 1 sin cos 3 3 px x q x + DM1 ( ) 3 3 sin 9cos 0 9 3 3 a a a = + − − Obtain 4 3cos 3 sin 3 a a a − = correctly A1 With sufficient evidence to show how they reach the given equation 5 9(ii) Calculate values at a = 2.5 and a = 3 of a relevant expression or pair of expressions. M1 2.5 2.679 and 3 2.827 < > If using 2.679 and 2.827 must be linked explicitly to 2.5 and 3. Solving f(a) = 0, f(2.5) = 0.179. and f(3) = –0.173 or if 1 1 f ( ) sin 3cos 4 f (2.5) 0.13..,f (3) 0.145... 3 3 a a a a = + − ⇒ = − = Complete the argument correctly with correct calculated values A1 Accept values to 1 sf. or better 2 Question Answer Marks Guidance 9(iii) Use the iterative process 1 na + = 1 3 1 1 3 4 3cos sin n n n a a a + − correctly at least once M1 Show sufficient iterations to at least 5 d.p. to justify 2.736 to 3d.p., or show a sign change in the interval (2.7355, 2.7365) A1 Obtain final answer 2.736 A1 3

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10 The line l has equation r i 3j i 3k . The plane p has equation 2x y 5. = + −2k + , −2j + + −3z = (i) Find the position vector of the point of intersection of l and p. 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(ii) Calculate the acute angle between l and p. 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(iii) A second plane q is perpendicular to the plane p and contains the line l. Find the equation of q, giving your answer in the form ax by cz d. 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Mark scheme: 10(i) Express general point of l in component form e.g. (1 + λ, 3 – 2λ, – 2 + 3λ) B1 Substitute in equation of p and solve for λ M1 Obtain final answer 5 5 3 3 + i j from λ = 2 3 A1 OE Accept 1.67 1.67 + i j or better 3 Question Answer Marks Guidance 10(ii) Use correct method to evaluate a scalar product of relevant vectors e.g. (i – 2j + 3k).(2i + j – 3k) M1 Using the correct process for calculating the moduli, divide the scalar product by the product of the moduli and evaluate the inverse sine or cosine of the result M1 9 sin 14 θ = Obtain answer 40.0° or 0.698 radians A1 AWRT 3 Alternative method for question 10(ii) Use correct method to evaluate a vector product of relevant vectors e.g. (i – 2j + 3k)x(2i + j – 3k) M1 Using the correct process for calculating the moduli, divide the modulus of the vector product by the product of the moduli of the two vectors and evaluate the inverse sine or cosine of the result M1 115 cos 14 θ = Obtain answer 40.0° or 0.698 radians A1 AWRT 3 Question Answer Marks Guidance 10(iii) State a – 2b + 3c = 0 or 2a + b – 3c = 0 B1 Obtain two relevant equations and solve for one ratio, e.g. a : b M1 Could use 2a + b – 3c = 0 and 3 2 5 5 3 3 a b c d a b d + − =  + =  i.e. use two points on the line rather than the direction of the line. The second M1 is not scored until they solve for d. Obtain a : b : c = 3 : 9 : 5 A1 OE Substitute a, b, c and a relevant point in the plane equation and evaluate d M1 Using their calculated normal and a relevant point Obtain answer 3x + 9y + 5z = 20 A1 OE Alternative method for question 10(iii) Attempt to calculate vector product of relevant vectors, e.g. (i – 2j + 3k) × (2i + j – 3k) M1 Obtain two correct components A1 Obtain correct answer, e.g. 3i + 9j + 5k A1 Use the product and a relevant point to find d M1 Using their calculated normal and a relevant point Obtain answer 3x + 9y + 5z = 20, or equivalent A1 OE Question Answer Marks Guidance 10(iii) Alternative method for question 10(iii) Attempt to form a 2-parameter equation with relevant vectors M1 State a correct equation e.g. r = i + 3j – 2k + λ(i – 2j + 3k) + µ(2i + j – 3k) A1 State 3 equations in x, y, z, λ and µ A1 Eliminate λ and µ M1 Obtain answer 3x + 9y + 5z = 2 A1 OE 5

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