Cambridge A Level Mathematics 9709 — 2023 Oct/Nov Paper 3 · Variant 3

1 Find the set of values of x satisfying the inequality 2x+1 −2 < 0.5, giving your answer to 3 significant figures. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 State or imply non-modular inequality −0.5  2 x+1 − 2  0.5 , can be in two B1 −0.25  2 x −1 0.25 , can be in two separate statements, separate statements, x 2 2 or 2 − 1  0.25 or corresponding pair of linear 2 ( ) or 2 x+1 − 2  0.5 2 ( ) equations 0.25 = 2x – 1 and −0.25 = 2x – 1 or quadratic or corresponding pair of linear equations 0.5 = 2x+1 – 2 and − 0.5 = 2x+1 – 2 x 2 2 equation 2 − 1 = 0.25 . 2 ( ) or quadratic equation 2 x+1 − 2 = 0.52 ( ) Incorrect inequality mark recoverable by correct final answer or x < 0.32 and x > –0.42 . Use correct method for solving an equation or inequality of the form M1 Reach (x + 1)ln2 = lna or equivalent, do not need to reach 2x +1 = a or 2x = b where a, b > 0 x = … Obtain critical values x = 0.322 and –0.415 A1 ln 2.5 ln1.5 e.g. − 1 and −.1 or awrt x = 0.32 and –0.42 ln 2 ln2 or exact equivalents State final answer –0.415 < x < 0.322 or (–0.415, 0.322) A1 Need 3 significant figures. Need combined result, not x < 0.32 and x >–0.42 . Must be strict inequalities. No working, 0/4. Alternative method for Question 1 Use correct method for solving an equation or inequality of the form M1 May see 2x+1 = 1.5 and 2x+1 = 2.5 . 2x +1 = a or 2x = b where a, b > 0 Reach (x + 1)ln2 = lna or equivalent, don’t need to reach x = … Obtain one critical value, e.g. 0.322 A1 ln 2.5 e.g. −.1 or awrt x = 0.32 ln 2 or exact equivalent Obtain the other critical value e.g. –0.415 or awrt x = –0.42 or exact A1 ln1.5 e.g. −.1 equivalent ln2 1 State final answer –0.415 < x < 0.322 or (–0.415, 0.322) A1 Need 3 significant figures. Need combined result, not x < 0.32 and x > – 0.42 . Must be strict inequalities. No working, 0/4. 4

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2 On an Argand diagram, shade the region whose points represent complex numbers z satisfying the inequalities z −1 + 2i ≤ z and z −2 ≤1. [5]

Mark scheme: 2 Show a circle centre (2, 0) B1 Show the relevant part of a circle with radius 1 B1 FT FT centre not at the origin even if centre at 1 – 2i. Must clearly go through (1, 0) or (3, 0) (oe for FT mark). Show the point representing 1 – 2i B1 Can be implied by correct perpendicular bisector Show the perpendicular bisector of the line joining 1 – 2i and the origin. B1 FT FT on the position of 1 – 2i. Perpendicular to OP by eye and at midpoint of OP by eye sufficient. Must reach midpoint of OP and if extended will cut BE. 2 Shade the correct region. Dependent on all previous marks, except in case B1 3 below, and the perpendicular must cut axes between CF and BE, but not actually through C or F and not through B or E Scale can be implied by dashes 1 Scale only on y-axis and 2OA = OC B1, B1FT, B1, B1FT, B1 2 Scale only on x-axis and 2OB = OE B1, B1FT, B1, B1FT, B1 3 No scale on either axis, but 2OA = OC B0, B1FT, B0, B1FT, B1 then 2OB = OE 5

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e3x2−1 5 Find the exact coordinates of the stationary points of the curve y = . [6] 1 −x2 ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 5 Use correct product or quotient rule M1 Need attempt at both derivatives condone errors in chain rule. In quotient rule allow BOD in formula if ± 2x seen unless clear that incorrect formula has been used. If omit denominator or forget to square or complete reversal of signs then M0 A0 M1 A1 A1 A1. 2 3 x 2 −1 3 x 2 −1 A1 2 3 x 2 −1 3 x 2 −1 6 x (1 − x )e + 2 xe If 6 x 1 − x e + 2 xe = 0 from the start, with no ( ) Obtain correct derivative in any form, e.g. 2 1 − x 2 wrong formula seen, award M1A1. ( ) Equate derivative (or its numerator) to zero and solve for x M1 6x – 6x3 + 2x = 0 and solve. Allow for just one x value. Allow if from solution of 3 term quadratic equation, but if they get x = 0 the x must factorise out Obtain the point (0,e −1 ) or exact equivalent A1 2 3 Or for all three x coordinates found 0,  oe and no 3 extras but if this is the case then one pair of correct coordinates A1 and both other pairs of correct coordinates A1. Accept, e.g. x = 0, y = e–1 ISW for last 3 marks.  2 3 3  A1 Allow √(4/3). Obtain the point  , −3e  or exact equivalent    3   2 3 3  A1 Obtain the point  − , −3e  or exact equivalent    3  6

More questions on Differentiation

6 (a) Show that the equation cot21 + 2 cos 21 = 4 can be written in the form 4 sin41 + 3 sin21 −1 = 0. 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(b) Hence solve the equation cot21 + 2 cos 21 = 4, for 0Å < 1 < 360Å. 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Mark scheme: 6(a) Use correct Pythagoras cot2 = cosec2– 1 or cot2 = 1/sin2– 1 M1 If consistent omission of brackets, e.g. (sinθ)2 written as or cot²= cos²/sin² and then cos²= 1 – sin², sinθ2 then SC B1 in place of M1A1. together with double angle formula cos2 = 1 – 2sin2, to obtain an equation in sin 𝜃 or sin 𝜃 and cosec2 Obtain a correct equation in sin 𝜃 in any form A1 e.g. 1/sin2 − 1 + 2(1 – 2sin2) = 4 1– sin² 2 or + 2 1– 2sin = 4 . ( ) sin² cos² 2 If + 2 1– 2sin = 4 then ( ) sin² e.g. 1 − sin² + 2 1– 2sin 2 sin² = 4 . ( ) (missing sin 2 on right) allow M1A1A0. Reduce to the given answer of 4sin 4 + 3sin 2 −=1 0 correctly A1 AG Must follow from a horizontal equation (no denominators). If s = sin 𝜃 used and defined, allow all marks. If not defined, award M1A1A0. 3 6(b) Solve the given quadratic to obtain a value for 𝜃 M1 (4sin2 − 1)(sin2 + 1) = 0 and solve for 𝜃. Incorrect sign in solution of quadratic seen, e.g. (4sin2 − 1)(sin2 – 1) = 0 then M0 A0 A0 but if only see (4sin2− 1) = 0 and nothing incorrect seen allow 3/3. Obtain answer, e.g. 𝜃 = 30° A1 /6 award A0 Obtain three further answers, e.g. 𝜃 = 150°, 210° and 330° and no others A1 Ignore any answers outside interval. in the interval 5/6 7/6 11/6 award A1. 3

More questions on Trigonometry

7 The equation of a curve is x3 + y2 + 3x2 + 3y = 4. dy + 6x (a) Show that = −3x2 . 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(b) Hence find the coordinates of the points on the curve at which the tangent is parallel to the x-axis. 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Mark scheme: 7(a) d y B1 Allow for 3x2dx + 2ydy or Fx = 3 x 2 + 6 x and Fy = 2 y + 3 . as the derivative of y2 State or imply 2 y dx d y M1 dy d y Equate derivative of LHS to zero and solve for 3x2 + 2 y + 6x + 3 = 0 d x dx d x dy Fx or 3x2dx + 2 ydy + 6xdx + 3dy = 0 or = − need dx Fy evidence from B1 mark or formula must be seen. Allow errors. Obtain the given answer A1 dy 3 x 2 + 6 x −3 x 2 − 6 x AG = − not . dx 2 y + 3 2 y + 3 d y dy Must factorise with e.g. 3x2 + 6x + (2y + 3) = 0 d x dx or 3x2dx + 6xdx + d y ( 2 y + 3 ) = 0. 3 7(b) Equate numerator to zero and solve for x *M1 Allow for just one x value. Obtain x = 0 and x = –2 only A1 Substitute their x, [x = 0 or x = –2] in curve equation to obtain quadratic DM1 y2 + 3y – 4 = 0 or y2 + 3y = 0. equation in y equal to 0 Obtain y = 1 and y = –4 [when x = 0] A1 Obtain y = 0 and y = –3 [when x = –2] A1 ISW If forget x = 0 then max 3/5. 5

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8 The variables x and y satisfy the differential equation dy e4x = cos2 3y. dx It is given that y = 0 when x = 2. Solve the differential equation, obtaining an expression for y in terms of x. 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Mark scheme: 8 Separate variables correctly and reach asec2 3y or be−4x B1 Condone missing integral signs or dy and dx, but allow if recognisable integrals follow. Not for 1/cos2 3y and 1/e4x. 1 −4 x B1 Can recover the previous B1 if de–4x seen here. Obtain term − e 4 Obtain only a term of the form a tan3 y M1 Can recover the first B1 if a tan3 y seen here. 1 A1 Obtain term tan3 y 3 Use x = 2, y = 0 to evaluate a constant or as limits in a solution containing M1 May see tan by and e  4 x here. terms of the form a tan by and ce 4 x Obtain correct answer in any form A1 1 1 1 e.g. tan3y = − e−4x + e−8 3 4 4 1 1 or tan3y = − e−4x + 8.39  10–5 3 4 1 −1  3 −8 3 −4 x  A1 ISW e − Obtain final answer y = tan  e  3  −1  3 4  4 e −4 x  OE e.g. y = 1 tan  2.52 10 −4 −  3  4  7

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10 y x O The diagram shows the curve y = x cos 2x, for x ≥0. (a) Find the equation of the tangent to the curve at the point where x = 12π. 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(b) Find the exact area of the shaded region shown in the diagram, bounded by the curve and the x-axis. 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Mark scheme: 10(a) Use the product rule correctly on y = x cos 2x M1 dx/dx cos 2x + x d/dx(cos 2x) attempted. Obtain the correct derivative in any form A1 e.g. cos 2x – 2x sin 2x. If cos 2x + x–2sin 2x, not recovered, max M1A0A1FTA0 but can recover for full marks by seeing correct substitution. π dy π A1FT d y π Obtain y = − and = −1 when x = FT their with x = substituted. 2 dx 2 d x 2 Obtain answer x + y = 0 A1 π OE CWO Need to see y and dy/dx at x = . 2 4 10(b) Integrate by parts and reach ax sin2 x + b  sin2 xdx *M1 1 1 A1 OE Obtain x sin2 x −  sin2 xdx 2 2 1 1 A1 OE Complete integration and obtain x sin2 x + cos2 x 2 4 π DM1 1  π  2π 1 2π 1 Use limits of x = 0 and x = in the correct order, having integrated twice If correct,   sin + cos − cos0 4 2  4  4 4 4 4 to obtain ax sin 2x + ccos 2x 1  π  2π 1 or   sin − cos0 . 2  4  4 4 Max one substitution error. π 1 A1 π − 2 Obtain answer − or exact simplified two term equivalent ISW Accept . 8 4 8 1 1 Accept x sin2 x + cos2 x then final answer. 2 4 5

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11 The line l has equation r = i −2j −3k + , −i + j + 2k . The points A and B have position vectors −2i + 2j −k and 3i −j + k respectively. (a) Find a unit vector in the direction of l. 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The line m passes through the points A and B. (b) Find a vector equation for m. 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(c) Determine whether lines l and m are parallel, intersect or are skew. 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Mark scheme: 11(a) Use correct process for modulus on direction vector of l, e.g. M1 SOI Allow −12. ( −1) 2 + 12 + 2 2 Allow ( −) 2 + 2 + ( 2) 2 . 1 A1 OE Allow coordinates as row or column, but not row or  ( −+i j + 2k ) column with i, j and k included. 6 2 11(b) Use a correct method to form an equation for line m M1 Allow even if all signs of point incorrect, namely use +2i − 2j + k or –3i + j – k. Obtain r = –2i + 2j – k + μ1(–5i + 3j – 2k) A1 OE, e.g. r = 3i – j + k + μ2(−5i + 3j − 2k) Must have r = … 2 11(c) Justify lines are not parallel B1 ( −5, 3, −2) ≠ d (− 1, 1, 2) or ( −5, 3, −2)x(− 1, 1, 2) ≠ 0. Can find angle (105°, 74.6°, 1.84c or 1.3(0)c) instead but if incorrect B0 and A0 at end. Accept direction vectors don’t have common factor but not direction vectors are not equal or direction vectors are different or μ ≠ λ or scalar product ≠ 0. Not the line equations are not multiples of each other. Express l or m in component form B1 e.g. (–2 – 5μ1, 2 + 3μ1, –1 – 2μ1) or (3 – 5μ2, − 1 + 3μ2, 1 – 2μ2) or (1 – λ, –2 + λ, –3 + 2λ) Equate two pairs of components of general points on l and their m M1 and solve simultaneously for λ or for μ 11 1 A1 Obtain correct answer for λ or μ, e.g. λ = , μ1 = 2 2 Determine that all three equations are not satisfied and the lines fail to A1 1 λ μ1 2 λ μ2 intersect and conclude the lines are skew. Conclusion needs to follow correct working ij 11/2 1/2 8 ≠ –2 ij 11/2 3/2 8 ≠ –2 ik 4/3 –1/3 –2/3 ≠ 1 ik 4/3 2/3 –2/3 ≠ 1 jk 7/4 –3/4 –3/4≠7/4 jk 7/4 1/4 –3/4≠7/4 Dependent on 4 previous marks gained. 5

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