Cambridge A Level Mathematics 9709 — 2023 Feb/March Paper 3 · Variant 2

4 Solve the equation 5z 30 10i 0, 1 2i −zz* + + = + giving your answers in the form x iy, where x and y are real. [5] + ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 Substitute z = x + iy and z* = x − iy to obtain a correct equation, horizontal B1 5(x + iy) – (x + iy)( x – iy )(1 + 2i) + (30 +10i)(1 + 2i) = 0 or with (1– 2i)/( 1 – 2i) seen, in x and y 5(x + iy)(1 – 2i)/[(1 + 2i)(1 – 2i)]– (x + iy)( x – iy )+ (30 +10i) = 0 x − 2i x + i y + 2 y − x 2 − y 2 + 30 + 10i = 0 . Use i2 = –1 at least once and equate real and imaginary parts to zero *M1 OE For their horizontal equation. Obtain two correct equations A1 5x – (x2 + y2) + 10 = 0 e.g. x + 2y – x2 – y2 + 30 = 0 and –2x + y + 10 = 0 5y – 2(x2 + y2) +70 = 0 5y – 10x + 50 = 0 x + 2y – (x2 + y2) + 30 = 0 Allow –2ix + iy + 10i = 0. Solve quadratic equation for x or for y DM1 x2 – 9x + 18 = (x – 3)(x – 6) = 0 y2 + 2y – 8 = (y + 4)(y – 2) = 0 DM0 If x or y imaginary. Obtain answers 3 – 4i and 6 + 2i A1 5

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5 The parametric equations of a curve are x te2t, y t2 t 3. = = + + dy (a) Show that [3] dx = e−2t. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 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........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ @ A (b) Hence show that the normal to the curve, where t passes through the point 0, 3 . = −1, −1e4 [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 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Mark scheme: 5(a) dx 2 t 2 t B1 OE Obtain = e + 2te dt dy dy dt M1 d y 2t + 1 Use =  = 2 t . dx dt dx d x e (1 + 2t ) d y −2t A1 AG Need to see e2t (1 + 2t) in denominator. Obtain the given answer = e d x 3 5(b) 2 1 B1 Obtain x = − e− or − and y = 3 at t = − 1 e 2 2 1 B1 Obtain gradient of normal = − e− or − e 2 x = 0 substituted into equation of normal or use of gradients to give B1 −2 −2 Equation of normal y −=3 −e x −−e . ( ) 1 y = 3 − with no errors AG 4 e SC Decimals B0 B1 B0 − 0.135 . 3

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6 (a) Express 5 sin 12 cos in the form R cos , where R 0 and 0 1 [3] 1 + 1 1 −! > < ! < 2π. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence solve the equation 5 sin 2x 12 cos 2x 6 for 0 [4] + = ≤x ≤π. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) State R = 13 B1 Allow if (122 + (− 5)2) seen. Use correct trig formulae to find M1 cos(α) = 12 and sin(α) = 5 M0 α = tan-1(±5/12) = cos-1(±12/13) = sin-1(±5/13) However, sin(α)/cos(α) = 5/12 or − 5/12 with no error seen, or tan(α) = 5/12 or − 5/12 quoted then allow. Obtain α = 0.395 A1 CWO If negative sign seen when finding R then A0 here. If degrees 22.6 A0 MR. Only penalise degrees once in (a) and (b). Note α = 0.39479… 3 B1FT SOI 6  16(b) cos−   1.0910… FT their incorrect R.  R  Use correct method to find a value of 2x in the interval M1 6  6  1 1 2x = cos−   + α or 2 − cos−   + α.  R   R  Allow if cos(2x + 0.395) seen Obtain answer, e.g. x = 0.743 or 0.742 A1 42.5 or 42.6 degrees. Obtain second answer, e.g. x = 2.79 and no others in the interval A1 159.8, 159.9 or 160.0 degrees all possible depending whether using 3 dp or 4 dp. 4

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7 B r O x rad A The diagram shows a circle with centre O and radius r. The angle of the minor sector AOB of the circle is x radians. The area of the major sector of the circle is 3 times the area of the shaded region. (a) Show that x 3 sin x 1 [4] = 4 + 2π. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Show by calculation that the root of the equation in (a) lies between 2 and 2.5. 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(c) Use an iterative formula based on the equation in (a) to calculate this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. 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Mark scheme: 7(a) 1 2 B1 OE State or imply area of major sector = r (2− x ) 2 1 2 1 2 B1 OE State or imply area of shaded segment = r x − r sin x r2 sin(x/2) cos(x/2) B0 until changed to (1/2)r2 sin x. 2 2 1 2  1 2 1 2  M1 OE State r (2− x ) = 3  r x − r sin x  Area of major sector = 3 times (area of minor sector – area of 2  2 2  triangle). Allow r2 sin(x/2) cos(x/2). 3 1 A1 AG Allow rectified slip if before penultimate line. Obtain the given answer x = sin x + after full and correct working 4 2 4 7(b) Calculate the values of a relevant expression or pair of expressions at x = M1 x= 2 x = 2.5 2 and x = 2.5 (3/4) sin x + (1/2)π 2.2(5277) 2.0(197) 2 < 2.2 or 2.3 2.5 > 2.0 x – (3/4) sin x – (1/2)π − 0.2(5277) < 0 + 0.4(803) > 0 or change of sign Attempt both values and one correct for M1. Complete the argument correctly with correct calculated values A1 Degrees award 0/2 2 7(c) Use the iterative formula correctly at least twice M1 Obtain final answer 2.18 A1 Show sufficient iterations to 4 d.p. to justify 2.18 to 2 d.p. or show there is A1 a sign change in the interval (2.175, 2.185) 2 2.25 2.5 2.2528 2.1543(5) 2.0196(5) 2.1530 2.1967 2.2465 2.1972 2.1786 2.1560 2.1784 2.1865 2.1960 2.1866 2.1831 2.1789 2.1830 2.1845 2.1863 2.1846 2.1831 2.1845 Degrees award 0/3 3

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8 y 1 2 x O M The diagram shows the curve y x3 ln x, for x 0, and its minimum point M. = > (a) Find the exact coordinates of M. 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(b) Find the exact area of the shaded region bounded by the curve, the x-axis and the line x 12. 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Mark scheme: 8(a) Use the product rule correctly *M1 x3 d/dx(lnx) + d/dx(x3) lnx. Obtain the correct derivative in any form A1 x 3 2 e.g. + 3 x ln x . x Equate derivative to zero and solve exactly for x DM1 Reaching x = ea.  1 1  A1 ISW Obtain answer  3 , −  or exact equivalent  e 3e  4 8(b) Integrate by parts and reach ax 4 ln x + b  ( x 4 / x )dx *M1 x 4 1 4 A1 OE Obtain ln x −  ( x / x )dx 4 4 x 4 x 4 A1 OE Complete integration and obtain ln x − 4 16 1 DM1 Correct substitution [(1/4)ln1 or 0 − 1/16] – [(1/64)ln(1/2) – Use limits of x = and x = 1 in the correct order, having integrated twice (1/16)2] or minus this value CWO. 2 Allow omission of (1/4)ln1 or 0. 15 1 A1 Obtain answer − ln2 or exact equivalent final answer 256 64 5

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9 The variables x and y satisfy the differential equation dy e3y sin2 2x. dx = It is given that y 0 when x 0. = = Solve the differential equation and find the value of y when x 12. 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Mark scheme: 9 Separate variables correctly and obtain e−y3 and sin 2 2x on the opposite B1 sides 1 −3 y B1 Obtain term − e 3 Use correct double angle formula for sin 2 2x = (1/2)[1 – cos 4x] M1 1  1  A1 Obtain terms x − sin4 x oe   2  4  Use x = 0, y = 0 to evaluate a constant or as limits in a solution containing M1 terms of the form ax and b sin4 x and cey3 Obtain correct answer in any form A1 1 −3 y 1  1  1 e.g. − e = x − sin4 x −   3 2  4  3 1 Substitute x = 12 and obtain y = 0.175 or − 13 ln ( 4 + 83 sin2 ) A1 OE ISW 7

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10 With respect to the origin O, the points A, B, C and D have position vectors given by ` a ` a ` a ` a −−¿OA 3 , −−¿OB 12 , −−¿OC 1 and −−¿OD 5 . = −1 = = −2 = −6 2 5 11 −3 (a) Find the obtuse angle between the vectors −−¿OA and −−¿OB. 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The line l passes through the points A and B. (b) Find a vector equation for the line l. 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(c) Find the position vector of the point of intersection of the line l and the line passing through C and D. 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Mark scheme: 10(a) Carry out correct process for evaluating the scalar product of OA and OB M1 ± (3, − 1, 2).(1, 2, − 3) = ±(3 – 2 – 6) = [− 5]. Using the correct process for the moduli, divide the scalar product by the A1 product of the moduli and obtain cos-1{±(3 – 2 – 6)/[(32 + (–1)2 +22) (12 + 22 + (–3)2)]} Obtain answer 110.9° or 1.94c A1 3 10(b) Use a correct method to form an equation for line through AB M1 Obtain r = 3i – j + 2k + μ1 (2i – 3j + 5k) A1 OE e.g. r = i + 2j – 3k + μ2 (–2i + 3j – 5k). Need r or (x, y, z). 2 10(c) Obtain a correct equation for line through CD B1 OE e.g. [r = ] 5i – 6j + 11k + λ2(–4i + 4j – 6k). e.g. [r = ] i – 2j + 5k + λ1(–4i + 4j – 6k) r can be omitted or another symbol used. Equate two pairs of components of general points on their l and their CD M1 and solve for λ or for μ Obtain e.g. λ1 = –2 or μ1 = 3 or λ2 = –1 or μ2 = − 4 A1 Obtain position vector 9i – 10j + 17k A1 Condone (9, –10, 17) but not (9i, – 10j, 17k). 4

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