Cambridge A Level Mathematics 9709 — 2024 Oct/Nov Paper 3 · Variant 3

1 The complex number z satisfies z = 2 and 0 G argz G 1 r . 4 (a) On the Argand diagram below, sketch the locus of the points representing z. [2] (b) On the same diagram, sketch the locus of the points representing z2. [2] lm(z) O Re(z)

Mark scheme: Question Answer Marks Guidance 1(a) For all 4 marks, scales must be approximately equal, dashes can replace numbers. Im(z) Arcs don’t have to be perfectly circular, mark intention. 4i Show an arc of a circle, centre the origin and radius 2. B1 Only need 2 on Re(z) or 2i on Im(z) or r = 2 to show correct radius Show an arc centre the origin for 0 arg z 14 π with any radius B1 Max B1 if sector shaded 1 π 4 Re(z) 2 O 2 4 1(b) Show an arc of a circle, centre the origin and radius 4. B1 Only need 4 on Re(z) or 4i on Im(z) or r = 4 to show correct radius Show an arc centre the origin for 0 arg z 12 π with any radius B1 Max B1 if sector shaded 2

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2 Let f ( )x = 2x 3 - 5 x 2 + 4 . (a) Show that if a sequence of values given by the iterative formula 4 x = n + 1 5 - 2x n converges, then it converges to a root of the equation f ( )x = 0 . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) The equation has a root close to 1.2 . Use the iterative formula from part (a) and an initial value of 1.2 to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ .......................... ............................................................................................................................

Mark scheme: 2(a) 4 *B1 Could work with nx +1 throughout or with nx State or imply the equation x = and square the equation 5 − 2 x throughout instead of x. Rearrange this with at least one intermediate step in the form 2 x 3 − 5 x 2 + 4 = 0 DB1 Alternative Method 1 for Question 2(a) Rearrange 2 x 3 − 5 x 2 + 4 = 0 to x2(5 – 2x) = 4 (or a different intermediate step) B2 2 4 4 and to either x = or x = 5 − 2 x 5 − 2 x 4 and then obtain the iterative formula xn +1 = 5 − 2 xn Alternative Method 2 for Question 2(a) Rearrange 2 x 3 − 5 x 2 + 4 = 0 to x2(5 – 2x) = 4 (or a different intermediate step) and *B1 Must have introduced nx +1 and nx . 2 4 2 4 to x = and to xn +1 = 5 − 2 x 5 − 2 xn 4 DB1 Obtain the iterative formula xn +1 = 5 − 2 xn 2 2(b) Use the iterative process correctly at least once M1 The question specifies initial value 1.2, so must use the formula to obtain a value and then use this value in the formula. Obtain final answer 1.28 A1 Can gain this mark even if less than 4 dp shown in iteration. Show sufficient iterations to at least 4 dp to justify 1.28 to 2 dp or show that there is A1 1.2, 1.2403, 1.2601, 1.2700,1.2752,1.2778,... a sign change in the interval (1.275,1.285 ) Allow small errors, truncation and recovery. 3

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3 lnP 3 O t The number of bacteria in a population, P, at time t hours is modelled by the equation P = aekt , where a and k are constants. The graph of lnP against t, shown in the diagram, has gradient 1 and intersects 20 the vertical axis at ( 0, 3) . (a) State the value of k and find the value of a correct to 2 significant figures. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the time taken for P to double. Give your answer correct to the nearest hour. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) State or imply that ln P = ln a + kt or ln P = ln a + k(ln e)t B1 Can be implied by both a and k correct. 1 t P = e 3 e 20 gets B1B1. 1 ln P = t + 3 B0 until associated with a and /or k 20 1 dP B1 OE. Can be embedded in P = ae kt . State k = , not from = k 20 dt ln a = 3  a = 20 to 2 sf B1 Must be 2 sf, can be embedded in P = ae kt . 3 3(b) Form a correct equation in t using a and k, or their a and k where a will cancel (or M1 E.g. 2 a = ae kt , 2 = e kt , kt = ln 2. are both numerical) Obtain t = 14  hours  A1 Allow 13.75 [hrs] (13 hrs 45 min) to 14 [hrs]. ISW 2

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4 Find the complex number z satisfying the equation z - 3i 2 - 9i = . z + 3i 5 Give your answer in the form x + yi , where x and y are real. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4 Substitute z = x + iy and obtain a horizontal equation *M1 E.g. 5 ( x + ( y − 3) i ) = ( 2 − 9i ) ( x + ( y + 3) i ) Do not allow if this would lead to an equation containing xy terms which do not cancel Use 2i = −1 anywhere M1 Obtain e.g. 5 x + 5 ( y − 3 ) i = ( 2 x + 9 y + 27 ) + i ( 2 y + 6 − 9 x ) A1 Or equivalent expression free of products of complex numbers. or e.g. 3x2 + 3y2 – 12y – 63 + (9x2 + 9y2 – 30x + 54y + 81)i = 0 Terms can be in any order. Obtain simultaneous equations by equating real and imaginary parts DM1 E.g. 3 x − 9 y = 27 and 3 y + 9 x = 21 3x2 + 3y2 – 12y – 63 = 0 and 9x2 + 9y2 – 30x + 54y + 81 = 0 Obtain  z =  3 − 2i only A1 4 Alternative Method for Question 4: Obtain a horizontal equation in z *M1 E.g. 5z – 15i = 2z + 6i − 27i2 – 9iz. Do not allow if it would lead to an equation containing z2 where the xy terms do not Allow errors, but no brackets. cancel Use 2i = −1 anywhere M1 9 + 7i A1 OE Obtain z = (might have an uncancelled factor of 3) 1 + 3i Multiply top and bottom by 1 − 3i or equivalent for their z DM1 Must see working for numerator or denominator, e.g. 9 − 27i + 21 + 7i or 1 + 9 or 10. 9 + 7i If = 3 − 2i M0A0. 1 + 3i 9 + 7i 1 − 3i If  = 3 − 2i M0A0 SC B1. 1 + 3i 1 − 3i 9 + 7i 1 − 3i If  and working in numerator or 1 + 3i 1 − 3i denominator and 3 − 2i M1A1. Obtain  z =  3 − 2i only A1 5

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5 (a) Show that cos 4 i - sin 4 i - 4 sin 2 i cos 2 i / cos 2 2i + cos 2i - 1. 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(b) Solve the equation cos 4 a - sin 4 a = 4 sin 2 a cos 2 a for 0° G a G 180 ° . 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Mark scheme: 5(a) 2 2 B1 Starting on left. 4  1 + cos2 4  1 − cos2 Rewrite cos as   or sin  as   or Double angle for one term.  2   2  4sin 2 cos 2 as sin 2 2  1 + cos2 2  1 − cos2 2 2 B1 OE, e.g. 1  cos2− sin 2 2. Obtain   −   − sin 2  2   2  Expand to B1 AG 1 1 1 2  1 1 1 2  2 + cos2+ cos 2−  − cos2+ cos 2 − (1 − cos 2) 4 2 4  4 2 4  and simplify to obtain cos 2 2+ cos2− 1 Alternative Method 1 for Question 5(a): 4 4 2 2 2 2 B1 Starting on left. Express cos − sin  as cos + sin  cos − sin  ( )( ) or rewrite 4sin 2 cos 2 as sin 2 2 Simplify to cos2− sin 2 2 B1 If cos 4 − sin 4 = cos2 instead of cos 2 + sin 2  cos 2 − sin 2  = cos2, B0. ( )( ) Use sin 2 2= 1 − cos 2 2 to obtain cos 2 2+ cos2− 1 B1 AG 5(a) Alternative Method 2 for Question 5(a): Use correct double angle formulae once e.g. replace cos2 with cos 2 − sin 2  B1 Starting on right. 2 2 2 2 2 Double angle for one term. cos − sin  + cos − sin  − 1 ( ) ( ) Expand to obtain B1 cos 4 − 2sin 2 cos 2 − sin 4 θ + 2sin 4 θ + cos 2− sin 2 − 1 * Write sin 4  as − sin 4 + 2sin 4 . or cos 4 − 2sin 2 cos 2 + sin 4 + cos 2− sin 2 − 1 leading to cos 4 − 2sin 2 cos 2 + sin 4 − 2sin 2  leading to 4 2 2 4 2 2 2 Write 2sin 2  as 2sin 2 (cos2 + sin2 ). cos − 2sin cos + sin  −2sin θ cos θ +sin θ ** ( ) Rewrite as B1 * cos 4 − 2sin 2 cos 2 − sin 4 + 2sin 4 − 2sin 2  leading to cos 4 − 2sin 2 cos 2 − sin 4 + 2sin 2  sin 2 −1 leading to ( ) cos 4 − 4sin 2 cos 2 − sin 4  ** cos 4 − 2sin 2 cos 2 + sin 4 − 2sin 2  cos2 − 2sin4  leading to cos 4 − 4sin 2 cos 2 − sin 4  3 5(b) State a quadratic equation in cos2 and solve for  M1 Alternative: form a quadratic in tan 2  and solve cos 2 2+ cos2−=1 0 tan 4 + 4tan 2 −=1 0 . ( ) for  ( ) Obtain = 25.9 or = 154.1 A1 May be more accurate. Allow 154 for 154.1. Obtain = 25.9 and = 154.1 and no others in range A1 May be more accurate. Allow 154 for 154.1. Mark answers in radians as a misread (0.452, 2.69). 3

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6 The lines l and m have vector equations l: r = 2 i + j - 3k + m ( - i + 2k ) and m: r = 2 i + j - 3k + n (2i - j + 5k ) . Lines l and m intersect at the point P. (a) State the coordinates of P. 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(b) Find the exact value of the cosine of the acute angle between l and m. 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(c) The point A on line l has coordinates ( 0, 1, 1) . The point B on line m has coordinates ( 0, 2, - 8) . Find the exact area of triangle APB. 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Mark scheme: 6(a) P ( 2, 1, − 3 ) B1 Accept x = 2, y = 1, z = –3.  2    Do not accept 2i + j – 3k or 1 .      − 3  1 6(b) Use the correct method to find the scalar product of the direction vectors M1 −( 1 2 + 2  5 ) = 8 Allow error of 0 × –1 = –1. Divide the scalar product by the product of the moduli to obtain  cos using M1 their 8 consistent vectors throughout their 5 their 30 8 A1 8 4 6 Obtain cos= OE, e.g. or . 5 6 150 15 8 If no seen, just 49.2, then A0. 5 6 Decimal only seen, A0. ISW 6(b) Alternative Method for Question 6(b): Use of cosine rule: e.g. sides of 5, 30 and 19 found B1 Could use other points. 5 + 30 − 19 M1 e.g. cos= 2 5 30 8 A1 8 4 6 8 Obtain cos= OE, e.g. or or . 5 6 150 15 5 30 8 If no seen, just 49.2, then A0. 5 6 Decimal only seen, A0. ISW 3 6(c) Any two of PA = 2 5 PB = 30 or AB = 82 seen B1 May be seen by stating or implying that = 2 and = −1. 1 64 M1 Correct method for the exact area of the triangle. Area =  2 5  30  1 − Note that: 2 150 129 sin APB = 15 86 sin ABP = 615 46 cos ABP = 2460 2580 Perp A to BP = 15 430 Perp B to AP = 5 = 86 A1 Or simplified exact equivalent. ISW Alternative Method for Question 6(c) PA  PB = − 4i − 18j − 2k B1 PA = −2i + 4k, PB = −2i + j − 5k. 1 1 M1 Correct method for the exact area of the triangle. Area = PA  PB = 16 + 324 + 4 2 2 = 86 A1 Or simplified exact equivalent. ISW 3

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7a 2 8 Let f ( x) = , where a is a positive constant. ( a - 2x)( 3a + x) (a) Express f ( )x in partial fractions. 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(b) Hence obtain the expansion of f ( )x in ascending powers of x, up to and including the term in x2. 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(c) State the set of values of x for which the expansion in part (b) is valid. 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Mark scheme: 8(a) A B M1 State or imply the form + and use a correct method to find a constant a − 2 x 3a + x Obtain A = 2a or B = a A1 Obtain A = 2a and B = a A1 3 8(b) −1 M1 Use a correct method to obtain the first two terms of the expansion of ( a − 2 x ) or −1 −1  2 x  −1  x   1 −  or ( 3a + x ) or  1 +   a   3a   2 x 4 x 2  A1ft OE. May be unsimplified. Obtain +2  1 + + 2 + ...  Follow their A, B for an expansion involving a.  a a  1  x x 2  A1ft OE. May be unsimplified. Obtain +  1 − + 2 + ...  Follow their A, B for an expansion involving a. 3  3a 9 a  7 35 x 217 x 2 A1 Or simplified equivalent. Final answer. Obtain + + + 2 Ignore any terms in higher powers of x. 3 9 a 27 a Do not ISW, e.g. multiplying by 27a2. Condone different order of terms. 8(b) Alternative Method for Question 8(b) Expanding 7a 2 ( a − 2 x )−1 ( 3a + x )−1 from the original question. M1 Use a correct method to obtain the first two terms of the expansion of ( a − 2 x ) −1 or −1 −1  2 x  −1  x   1 −  or ( 3a + x ) or  1 +   a   3a   2 x 4 x 2  7 a  x x 2  A1 OE. May be unsimplified. Obtain +7 a  1 + + 2 + ...  or +  1 − + 2 + ...  May be implied by the expression shown for the  a a  3  3a 9 a  next A1. 7  2 x 4 x 2  x x 2  A1 OE. May be unsimplified. Obtain +  1 + + 2 + ...  1 − + 2 + ...  3  a a  3a 9 a  7 35 x 217 x 2 A1 Or simplified equivalent. Final answer. Obtain + + + 2 Ignore any terms in higher powers of x. 3 9 a 27 a Do not ISW, e.g. multiplying by 27a2. Condone different order of terms. 4 8(c) B1 a a x a Or −  x  . 2 2 2 Mark final answer. Must make a clear statement. 1

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9 (a) Find the quotient and remainder when x 4 + 16 is divided by x 2 + 4 . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ c 2 3 4 x + 16 4 (b) Hence show that dd 2 dx = ( r + 4 ) . 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Mark scheme: 9(a) Divide to obtain quotient x 2 + k M1 k is a constant. Obtain quotient x 2 − 4 A1 If quotient stated separately, mark at this stage. Obtain remainder 32 A1 If remainder stated separately, mark at this stage. Need not state which is quotient and remainder, but if stated wrongly, max 2/3. After a correct division, still allow the marks if 2 32 then written as x − 4 + . x 2 + 4 Alternative Method for Question 9(a) Expands brackets to get B = 0 M1 ( x 2 + 4 )( x 2 + Bx + C ) + D = 2 + 4 Bx + 4C + D x 4 + Bx 3 + ( C + 4 ) x C = – 4 A1 D = 32 A1 Need not state which is quotient and remainder, but if stated wrongly, max 2/3. 3 9(b) 1 3 B1 FT Follow their quotient of form Ax2 + B. x − 4 x 3 1 1 M1 Obtain p tan− qx where q = 2 or q = 2 −1 1 A1 FT Follow their constant remainder, Obtain 16tan x 2  their constant remainder  −1 1 i.e.   tan x.  2  2 1 1 M1 Terms need not be evaluated, e.g. Use limits correctly in an expression containing p tan− qx where q = 2 or q = 8   2  8 3 − 8 3  + 16tan −1 3 − − 8 + 16tan −1 1   3   and rx + sx  3  8 16 −1 16π or − 8 can be − , 16tan 3 can be , 3 3 3 16tan −1 1 can be 4π. 4 A1 AG Obtain ( π + 4 ) from full and correct working 3 5

More questions on Integration

11 y R a O r 2r x M The diagram shows the curve y = 2 sin x 2 + cos x , for 0 G x G 2 r , and its minimum point M, where x = a . (a) Find the value of a correct to 2 decimal places. 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(b) Use the substitution u = 2 + cos x to find the exact area of the shaded region R. 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Mark scheme: 11(a) Use of correct product rule and correct chain rule M1 dy Bsin xsin x = A cos x 2 + cos x + dx 2 + cos x d y 2sin 2 x A1 OE Obtain = 2cos x 2 + cos x − d x 2 2 + cos x Equate the derivative to zero and obtain a horizontal 3 term quadratic equation or 4 *M1 Accept in cos .x term quartic equation in cosa E.g. 3cos2x + 4cosx – 1 = 0. If M0 earlier then needs that expression to be such that arrive at 3 term quadratic or E.g. 3cos4x + 16cos3x + 18cos2x – 1 = 0. 4 term quartic equation in cos x without further trig errors. 1 − 2 to be The only error in the form of the differential allowed is for ( 2 + cos x ) 1 3 2 ( 2 + cos x ) + − 2 or ( 2 + cos x ) Solve for cos a DM1  −+2 7   cos a = or 0.215     3  Allow presence of other solution(s). Obtain a = 4.93 A1 Allow more accurate, e.g. 4.929… even though question states 2 dp. If x = 1.35 leads to x = 4.93 award A1 BOD. If x = 1.35 and x = 4.93 award A0. 5 11(b) State or imply du = − sin x dx B1 OE If B0, max M1M1M1. Substitute throughout for u and du M1 Obtain − 2 udu A1 OE. Ignore limits if − 2 udu , but if + 2 udu , 3 u d u . then must have correct limits  12 (See final M1) 3 2 M1 Constant of integration not required Integrate to obtain ku ( +C ) 3 3 M1 1 and 3 for u, or 0 and π for x. Use correct limits correctly in an expression of the form ku 2 or k (2 + cos x ) 2 4 4 4 4 A1 3 Obtain 3 3 − 1 or 4 3 − or 27 − OE. Allow, e.g., 3 for 27. ( ) 3 3 3 3 ISW but don’t ignore e.g. multiplying throughout by 3. If the answer is changed from negative to positive value at end, then A0. Last M1A1 can use modulus, providing no errors seen. 6

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