Cambridge A Level Mathematics 9709 — 2022 May/June Paper 3 · Variant 2

4 The equation of a curve is y cos3x sin x. It is given that the curve has one stationary point in the 1 = interval 0 x < < 2π. Find the x-coordinate of this stationary point, giving your answer correct to 3 significant figures. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 Use the correct product rule and then the chain rule to differentiate either 3 cos or sin x x M1 e.g. two terms with one part of 3 2 d cos cos cos sin sin d sin y x x p x x x q x x   . Obtain correct derivative in any form e.g. 3 2 d cos cos 3cos sin sin d 2 sin y x x x x x x x   A1 A1 A1 for each correct term substituted in the complete derivative. Equate their derivative to zero and obtain a horizontal equation with positive integer powers of sin x and/or cos x from an equation including sin x or sin 1 x using sensible algebra. M1 e.g. 2 2 4 1 2 3cos sin cos 0 x x x    Use correct formula(s) to express their equation/derivative in terms of one trigonometric function M1 Can be awarded before the previous M1. May involve more than one trigonometric term. Obtain 2 7cos 6 x  , 2 7sin 1 x , or 2 6tan 1 x , or equivalent, and obtain answer x = 0.388 A1 CAO. The question asks for 3 sf. Ignore additional answers outside   π 2 0, . 22.2 is A0. 6

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5 (a) By sketching a suitable pair of graphs, show that the equation ln x 3x has one real root. = −x2 [2] (b) Verify by calculation that the root lies between 2 and 2.8. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Use the iterative formula /3xn xn xn+1 = −ln to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Sketch a relevant graph, e.g. y = ln x B1 ln( ) x : sketch should imply y-axis is an asymptote. Through (1, 0) if marked. Correct shape. 2 3x x  : Symmetrical. Through (0, 0) and (3, 0) if marked. If ln(x) correct accept parabola for +ve y only. If ln(x) incorrect then need parabola in 3 quadrants. Sketch a second relevant graph, e.g. 2 3 y x x   , and justify the given statement by marking the root on the sketch or by use of a suitable comment B1 2 5(b) Calculate the values of a relevant expression or pair of expressions at x = 2 and x = 2.8 M1 Allow for a smaller interval. At least one value correct if comparing with 0. If using pairs then the pairing must be clear. Complete the argument correctly with correct calculated values A1 e.g. 0.693 2 and1.03 0.56   or 1.307 0, 0.47 0    using 3 ln x x  0.304 0, 0.085 0    . Need to have calculated values to at least 2 sf. 2 2 2 Question Answer Marks Guidance 5(c) Use the iterative process correctly at least once M1 Obtain final answer 2.63 A1 Show sufficient iterations to at least 4 dp to justify 2.63 to 2 dp or show there is a sign change in the interval (2.625, 2.635) A1 SC Allow M1 A1 A0 to a candidate who starts at a point in the interval and reaches a premature conclusion 3

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6 The variables x and y satisfy the differential equation dy = xey−x, dx and y 0 when x 0. = = (a) Solve the differential equation, obtaining an expression for y in terms of x. 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(b) Find the value of y when x 1, giving your answer in the form a b, where a and b are = −ln integers. 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Mark scheme: 6(a) Correct separation of variables B1 e d e d y x y x x      Condone missing integral signs. Obtain term e y   B1 Commence integration by parts and reach e e d x x x x     *M1 M0 if clearly using differentiation of a product. Complete integration and obtain e e x x x     A1 Use x = 0 and y = 0 to evaluate a constant or as limits in a solution containing or derived from terms e , e and e y x x a bx c   , where 0 abc  DM1 Must see working for this. In a correct solution they should have e e e y x x C x        or equivalent. If they take logarithms before finding the constant, the constant must be of the right form. Correct solution in any form Must follow from correct working A1 e.g. e e e y x x x       A0 if constant of integration ignored or assumed to be zero. Obtain final answer     ln 1 e x y x    from correct working A1 OE e.g.   ln 1 y x x    , e ln 1 x y x         . A0 if constant of integration ignored or assumed to be zero. 7 6(b) Obtain answer  1 ln2 y   B1 Must follow from at least 6 or7 obtained in part 6(a). 1

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7 The equation of a curve is x3 3x2y 3. + −y3 = dy x2 2xy (a) Show that + . 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(b) Find the coordinates of the points on the curve where the tangent is parallel to the x-axis. 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Mark scheme: 7(a) State or imply 2 6 3 xy x  d d y x as derivative of 2 3x y State or imply 2 3y d d y x as derivative of 3y B1 Allow B1 B1 for     2 2 2 3 d 6 d 3 d 3 d 0 x x xy x x y y y     Equate attempted derivative of left-hand side to zero and solve to obtain an equation with d d y x as subject M1 Allow if zero implied by subsequent working. Allow if recover from an extra d d y x  at the beginning of the left-hand side. Obtain 2 2 2 d 2 d y x xy x y x    correctly A1 AG Accept d d for y x y . 4 7(b) Equate numerator to zero *M1 Must be using the given derivative. Obtain x = – 2y, or equivalent A1 An equation with x or y as the subject SOI. Use 3 2 3 3 3 x x y y    to obtain an equation in x or y DM1 3 3 3 8 12 3 y y y     or 3 3 3 3 1 2 8 3 x x x    or any equivalent form (do not need to evaluate powers). Obtain the point (– 2, 1) and no others from solving their cubic equation A1 Allow if each component stated separately. ISW. State the point (0, 3 3  ), or equivalent from correct work B1 Accept (0, 3 3 ), or (0, – 1.44) (-1.44225). Allow if each component stated separately. ISW. 5

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9 The lines l and m have vector equations r 3j 4k 2i and r 5i 4j 3k ai bj k = −i + + + , −j −k = + + + - + + respectively, where a and b are constants. (a) Given that l and m intersect, show that 2b 4. 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(b) Given also that l and m are perpendicular, find the values of a and b. 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(c) When a and b have these values, find the position vector of the point of intersection of l and m. 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Mark scheme: 9(a) Express general point of l or m in component form, i.e.   1 2 , 3 ,4       or   5 , 4 , 3 a b       B1 Equate components and eliminate either λ or  M1 e.g. 1 6 2 4 1 1 2 2 , , , b a b b a a                Eliminate the other parameter or obtain a second expression in the first M1 λ and  are not required to be the subject of the equations. Show intermediate steps to obtain 2 4 b a   A1 AG Alternative method for question 9(a) Express general point of l or m in component form, i.e.   1 2 , 3 ,4       or   5 , 4 , 3 a b       B1 Express a or b in terms of λ and  M1 2 6 1 , a b         Use 1    M1 Obtain 2 4 b a   A1 AG 4 9(b) Using the correct process equate the scalar product of the direction vectors to zero *M1    2 . 0 a b     i j k i j k SOI. Obtain 2 1 0 a b   A1 OE e.g.   2 2 4 1 0 b b   Solve simultaneous equations for a or for b DM1 Obtain a = 2, b = 3 A1 4 Question Answer Marks Guidance 9(c) Substitute found values in component equations and solve for λ or for µ M1 Obtain answer 3i + j + 2k from either λ = 2 or µ = – 1 A1 Accept as coordinates or equivalent. 2

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10 The complex number 7i is denoted by u. It is given that u is a root of the equation −1 + 2x3 3x2 14x k 0, + + + = where k is a real constant. (a) Find the value of k. 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(b) Find the other two roots of the equation. 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(c) On an Argand diagram, sketch the locus of points representing complex numbers z satisfying the equation z 2. [2] −u = (d) Determine the greatest value of arg z for points on this locus, giving your answer in radians. 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Mark scheme: 10(a) Substitute x = 1 7i  in the equation and attempt expansions of 2 x and 3x *M1 Use 2i 1  correctly at least once and solve for k DM1       2 20 4 7i 3 6 2 7i 14 1 7i 0 k        Obtain answer k = – 8 A1 SC B1 only for those who show no working for the cube and square and obtain answer k = – 8. Alternative method for question 10(a) Attempt division by   1 7i x  as far as 2 1 2 ... x z x   *M1 See division on next page. Use 2i 1  correctly at least once and obtain 2 1 2 2x z x z   + remainder DM1 Obtain answer k = – 8 A1 3 Question Answer Marks Guidance 10(b) State answer 1 7i  B1 Can be seen simply stated on its own, or in a list of roots. Allow if stated clearly in part 10(a). Carry out a method for finding a quadratic factor with zeros 1 7i  and 1 7i  M1 Or state          1 7i 1 7i 2 x x x p    Obtain 2 2 8 x x   A1 Or obtain    1 7i 1 7i 8 p    Or obtain    3 2 2 1 7i 1 7i p     Obtain root x = 1 2 , or equivalent, via division or inspection A1 Needs to follow from the working. 4 Question Answer Marks Guidance 10(c) Show a circle with centre 1 7i  B1 If the scales are very different from each other then B1 for centre in the correct position and B1 for an ellipse. If there is more than one circle the max score is B1. Show circle with radius 2 and centre not at the origin There needs to be some evidence of scale e.g. radius marked or a scale on the axes B1 2 10(d) Carry out a complete method for calculating the maximum value of arg z for correct circle M1 e.g. 1 π π 1 2 4 7 tan   Can be implied by 155.7. Obtain answer 2.72 radians A1 CAO. The question requires radians. 2 Im Re 7 -1

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