Cambridge A Level Mathematics 9709 — 2019 May/June Paper 3 · Variant 2

2 Showing all necessary working, solve the equation 9x 3x 12. Give your answer correct to 2 decimal = + places. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 State or imply 2 12 0 − − = u u B1 Need to be convinced they know 2 23 3 = x x Solve for u, or for 3x , and obtain root 4 B1 Use a correct method to solve an equation of the form 3 = x a where a >0 M1 Need to see evidence of method. Do not penalise an attempt to use the negative root as well. e.g. 3 ln3 ln , log = = x a x a If seen, accept solution of straight forward cases such as 3x = 3 , x = 1 without working Obtain final answer x = 1.26 only A1 The Q asks for 2 dp 4

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3 Showing all necessary working, solve the equation cot 2 tan for [5] 21 = 1 0Å < 1 < 180Å. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Use correct trig formulae to obtain an equation in tan θ or equivalent (e.g all in sin θ or all in cos θ) *M1 2 1 tan 2tan 2tan θ θ θ − = . Allow 2 cot 1 2 2cot cot θ θ θ −= Obtain a correct simplified equation A1 2 5tan 1 θ = or 2 2 1 5 sin or cos 6 6 θ θ = = Solve for θ DM1 Dependent on the first M1 Obtain answer 24.1° (or 155.9°) A1 One correct in range to at least 3 sf Obtain second answer A1 FT 180 24.1 their ° − ° and no others in range. Correct to at least 3 sf. Accept 156° but not 156.0 Ignore values outside range If working in tanθ or cosθ need to be considering both square roots to score the second A1 Mark 0.421, 2.72 as a MR, so A0A1 5

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x 4 Find the exact coordinates of the point on the curve y at which the gradient of the tangent = 1 ln x 1 + is equal to 4. [7] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 Use correct quotient rule M1 Allow use of correct product rule on 1 1 ln − × + x x Obtain correct derivative in any form A1 ( ) ( ) ( ) 2 2 1 1 ln d 1 1 d 1 ln 1 ln 1 ln + −×     = = −   + + +   x x y x x x x x Equate derivative to 1 4 and obtain a quadratic in ln x or (1+ ln x) M1 Horizontal form. Accept ( ) 2 1 ln 1 ln 4 = + x x Reduce to 2 (ln ) 2ln 1 0 − + = x x A1 or 3-term equivalent. Condone 2 lnx if later used correctly Solve a 3-term quadratic in ln x for x M1 Must see working if solving incorrect quadratic Obtain answer x = e A1 Accept 1e Obtain answer y = 1 2 e A1 Exact only with no decimals seen before the exact value. Accept 1e 2 but not e 1 ln e + 7

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5 Throughout this question the use of a calculator is not permitted. It is given that the complex number i is a root of the equation −1 + ï3 kx3 5x2 10x 4 0, + + + = where k is a real constant. (i) Write down another root of the equation. 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Mark scheme: 5(i) State answer 1 3i −− B1 If 1 2 − given as well at this point, still just B1 1 Question Answer Marks Guidance 5(ii) Substitute x = 1 3i −+ in the equation and attempt expansions of 2x and 3x M1 Need to see sufficient working to be convinced that a calculator has not been used. Use 2i 1 = − correctly at least once M1 Allow for relevant use at any point in the solution Obtain k = 2 A1 Carry out a complete method for finding a quadratic factor with zeros 1 3i −+ and 1 3i −− M1 Could use factor theorem from this point. Need to see working. M1 for correct testing of correct root or allow M1 for three unsuccessful valid attempts. Obtain 2 2 4 + + x x A1 Using factor theorem, obtain 1 f 0 2   − =     Obtain root x = 1 2 − , or equivalent, via division or inspection A1 Final answer Question Answer Marks Guidance 5(ii) Alternative method 1 Carry out a complete method for finding a quadratic factor with zeros 1 3i −+ and 1 3i −− (multiplying two linear factors or using sum and product of roots) M1 Need to see sufficient working to be convinced that a calculator has not been used. Use 2i 1 = − correctly at least once M1 Allow for relevant use at any point in the solution Obtain 2 2 4 + + x x A1 Allow M1A0 for 2 2 3 + + x x Obtain linear factor 1 kx + and compare coefficients of x or x2 and solve for k M1 Can find the factor by inspection or by long division Must get to zero remainder Obtain k = 2 A1 Obtain root x = 1 2 − A1 Final answer Note: Verification that x = 1 2 − is a root is worth no marks without a clear demonstration of how the root was obtained

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6 A x rad B C r O In the diagram, A is the mid-point of the semicircle with centre O and radius r. A circular arc with centre A meets the semicircle at B and C. The angle OAB is equal to x radians. The area of the shaded region bounded by AB, AC and the arc with centre A is equal to half the area of the semicircle. (i) Use triangle OAB to show that AB 2r cosx. [1] = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ O@ 0 A. [2] (ii) Hence show that x cos−1 = 16x ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Verify by calculation that x lies between 1 and 1.5. 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(iv) Use an iterative formula based on the equation in part (ii) to determine x correct to 3 decimal places. Give the result of each iteration to 5 decimal places. 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Mark scheme: 6(i) Correct use of trigonometry to obtain B1 AG 1 Question Answer Marks Guidance 6(ii) Use correct method for finding the area of the sector and the semicircle and form an equation in x M1 ( ) 2 2 1 1 1 2 cos 2 2 2 2 π × = r r x x Obtain 1 cos 16 π − = x x correctly AG A1 Via correct simplification e.g. from 2 cos 16 π = x x 2 6(iii) Calculate values of a relevant expression or pair of expressions at x = 1 and x = 1.5 Must be working in radians M1 e.g. 1 1 1.11 1.5 1.5 1.20 = → = → x x Accept ( ) ( ) f 1 1.11 f 1.5 1.20 = = ( ) 1 f cos 16 π − = − x x x : ( ) ( ) f 1 0.111., f 1.5 0.3.. = − = f( ) cos 16 π = − x x x :f (1) 0.097.,f (1.5) 0.291. = = − For 2 16 cos π − x x f (1) 1.529..,f (1.5) 3.02.. = = − Must find values. M1 if at least one value correct Correct values and complete the argument correctly A1 2 Question Answer Marks Guidance 6(iv) Use 1 1 π cos 16 n n x x − +   =     correctly at least twice Must be working in radians M1 1,1.11173,1.13707,1.14225,1.14329,1.14349, 1.14354,1.14354 1.25,1.16328,1.14742,1.14432,1.14370 1.5,1.20060,1.15447,1.14570,1.14397,1.14363 Obtain final answer 1.144 A1 Show sufficient iterations to at least 5 d.p. to justify 1.144 to 3 d.p. or show there is a sign change in the interval (1.1435, 1.1445) A1 3

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dy 0.7 The variables x and y satisfy the differential equation It is given that y 0 when x = = dx = xex+y. (i) Solve the differential equation, obtaining y in terms of x. 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(ii) Explain why x can only take values that are less than 1. 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Mark scheme: 7(i) Separate variables correctly and attempt integration of at least one side B1 e d e d − = ∫ ∫ y x y x x Obtain term e− − y B1 B0B1 is possible Commence integration by parts and reach e e d ± ∫ x x x x M1 B0B0M1A1 is possible Obtain e e − x x x A1 or equivalent B1B1M1A1 is available if there is no constant of integration Use x = 0, y = 0 to evaluate a constant, or as limits in a definite integral, in a solution with terms e−y a , ex bx and ex c , where abc ≠ 0 M1 Must see this step Obtain correct solution in any form A1 e.g. e e e −= − y x x x Rearrange as ( ) ln 1 = − − − y x x A1 or equivalent e.g. ( ) 1 ln e 1 = − x y x ISW 7 7(ii) Justify the given statement B1 e.g. require 1 0 − > x for the ln term to exist, hence 1 < x Must be considering the range of values of x, and must be relevant to their y involving ( ) ln 1−x 1

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9 The points A and B have position vectors i 2j and 3i j k respectively. The line l has equation + −k + + r 2i j k i j 2k . = + + + - + + (i) Show that l does not intersect the line passing through A and B. 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(ii) The plane m is perpendicular to AB and passes through the mid-point of AB. The plane m intersects the line l at the point P. Find the equation of m and the position vector of P. 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Mark scheme: 9(i) Carry out correct method for finding a vector equation for AB M1 Obtain ( ) ( ) 2 2 2 λ = + − + −+ r i j k i j k , or equivalent A1 Equate two pairs of components of general points on their AB and l and solve for λ or forµ M1 1 2 2 2 1 1 2 1 2 λ µ λ µ λ µ + +         − = +         −+ +     Obtain correct answer for λ or µ, e.g. λ = 0, µ = – 1 A1 Verify that all three equations are not satisfied and the lines fail to intersect ( ≠ is sufficient justification e.g.2 0 ≠ ) Conclusion needs to follow correct values A1 Alternatives A λ µ B λ µ ij 23 13 1 5 3 3 ≠ 13 − 13 1 5 3 3 ≠ ik 0 –1 2 0 ≠ –1 –1 2 0 ≠ jk 1 0 3 2 ≠ 0 0 3 2 ≠ 5 Question Answer Marks Guidance 9(ii) State or imply midpoint has position vector 3 2 2 + i j B1 Substitute in 2x – y + 2z = d and find d M1 Correct use of their direction for AB and their midpoint Obtain plane equation 4x – 2y + 4z = 5 A1 or equivalent e.g. 2 5 . 1 2 2     − =       r Substitute components of l in plane equation and solve for µ M1 Correct use of their plane equation. Obtain µ = 1 2 − and position vector 3 1 2 2 + i j for the point P A1 Final answer Accept coordinates in place of position vector 5

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