Mathematics 9709/33 October/November 2024: question paper and mark scheme

Mark scheme

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 1 4 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 1 2 π with any radius B1 Max B1 if sector shaded 2

Mark scheme

2(a) 4 *B1 Could work with xn +1 throughout or with xn State or imply the equation x = and square the equation 5 − 2x throughout instead of x. Rearrange this with at least one intermediate step in the form 2 x3 − 5 x 2 + 4 = 0 DB1 Alternative Method 1 for Question 2(a) Rearrange 2 x3 − 5 x 2 + 4 = 0 to x2(5 – 2x) = 4 (or a different intermediate step) B2 4 4 and to either x 2 = or x = 5 − 2x 5 − 2x 4 and then obtain the iterative formula xn +1 = 5 − 2 xn Alternative Method 2 for Question 2(a) Rearrange 2 x3 − 5 x 2 + 4 = 0 to x2(5 – 2x) = 4 (or a different intermediate step) and *B1 Must have introduced xn +1 and xn . 4 4 to x 2 = and to xn +12 = 5 − 2x 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

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 1 P = e3e 20 gets B1B1. ln P = t + 3 B0 until associated with a and /or k 20 1 dP B1 OE. Can be embedded in P = aekt . 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 = aekt . 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. 2a = aekt , 2 = ekt , kt = ln 2. are both numerical) Obtain t = 14 [ hours ] A1 Allow 13.75 [hrs] (13 hrs 45 min) to 14 [hrs]. ISW 2

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 i2 = −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. 3x − 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 i2 = −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

Mark scheme

5(a) 1 + cos 2θ 1 − cos 2θ B1 Starting on left. 2 2 Rewrite cos θ as 4 or sin θ as 4 or Double angle for one term. 2 2 4sin θ cos θ as sin 2θ 2 2 2 1 + cos 2θ 1 − cos 2θ 2 2 B1 OE, e.g. 1× cos 2θ − sin 2 2θ . − − sin 2θ 2 Obtain 2 2 Expand to B1 AG 1 1 1 1 1 1 + cos 2θ + cos 2 2θ − − cos 2θ + cos 2 2θ − (1 − cos 2 2θ ) 4 2 4 4 2 4 and simplify to obtain cos2 2θ + cos 2θ − 1 Alternative Method 1 for Question 5(a): ( )( Express cos 4 θ − sin 4 θ as cos2 θ + sin 2 θ cos2 θ − sin 2 θ ) B1 Starting on left. or rewrite 4sin 2 θ cos2 θ as sin 2 2θ Simplify to cos 2θ − sin 2 2θ B1 If cos4 θ − sin 4 θ = cos 2θ instead of ( cos θ + sin θ )( cos θ − sin θ ) = cos 2θ , B0. 2 2 2 2 Use sin 2 2θ = 1 − cos2 2θ to obtain cos2 2θ + cos 2θ − 1 B1 AG 5(a) Alternative Method 2 for Question 5(a): Use correct double angle formulae once e.g. replace cos2θ with cos 2 θ − sin 2 θ B1 Starting on right. ( cos θ − sin θ ) + ( cos θ − sin θ ) − 1 2 2 2 2 2 Double angle for one term. Expand to obtain B1 cos4 θ − 2sin 2 θ cos2 θ − sin4θ + 2sin4θ + cos2θ − sin 2 θ − 1 * Write sin 4 θ as − sin 4 θ + 2sin 4 θ . or cos4 θ − 2sin 2 θ cos 2 θ + sin 4 θ + cos 2θ − sin 2 θ − 1 leading to cos4 θ − 2sin 2 θ cos2 θ + sin 4 θ − 2sin 2 θ leading to Write 2sin 2 θ as 2sin 2 θ (cos2 θ + sin2 θ ). ( cos4 θ − 2sin 2 θ cos2 θ + sin 4 θ −2sin2θ cos2θ + sin2θ ** ) Rewrite as B1 * cos4 θ − 2sin 2 θ cos2 θ − sin 4 θ + 2sin 4 θ − 2sin 2 θ leading to ( ) cos4 θ − 2sin 2 θ cos2 θ − sin 4 θ + 2sin 2 θ sin 2 θ −1 leading to cos4 θ − 4sin 2 θ cos2 θ − sin 4 θ ** cos4 θ − 2sin 2 θ cos2 θ + sin 4 θ − 2sin 2 θ cos2 θ − 2sin4 θ leading to cos4 θ − 4sin 2 θ cos2 θ − sin 4 θ 3 5(b) State a quadratic equation in cos2α and solve for α M1 Alternative: form a quadratic in tan 2 α and solve ( cos 2α + cos 2α − 1 = 0) 2 ( for α tan 4 α + 4 tan 2 α − 1 = 0 . ) Obtain α = 25.9° or α = 154.1° A1 May be more accurate. Allow 154 for 154.1. Obtain α = 25.9° and α = 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

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 ± cosθ using M1 their 8 consistent vectors throughout their 5 their 30 8 A1 8 4 6 Obtain cos θ = 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. cosθ = 2 5 30 8 A1 8 4 6 8 Obtain cos θ = 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 λ = 2 and μ = −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

Mark scheme

7(a) dx B1 Allow 3.2 cos 2t. Obtain = 6cos 2t dx = 6cos2t is B0. dt dy B1 Any equivalent form. Obtain = sec2 t − cosec2t dy = sec2t – cosec2t is B0, dt dy sec2 t − cosec2t but = can go on to gain dx 6cos 2t M1M1A1, so 3/5 possible. dy dy dx = ÷ *M1 dy sec2 t − cosec2t Use = dx dt dt dx 6cos 2t Express as a single fraction DM1 sin 2 t − cos 2 t 1 dy Allow 2 2 × With correctly simplified as a single fraction in terms of sin t and cos t cos t sin t 6cos 2t dt sin 2 t − cos 2 t 1 sin t − cos t 2 2 2 2 Allow with 6 cos 2t expressed as outside bracket or ÷ 6cos 2t or cos t sin t 6cos 2t 2 cos t sin t 2 6cos 2t −4cos 2t −2 A1 AG Obtain = from full and correct working Allow slips in θ and x for t to recover earlier 6cos 2t × sin 2t 3sin 2t 2 2 marks, provided these are corrected before the Numerator and denominator must have identical terms before cancelling, both cos2t final line. or both + and – (sin2t – cos2t) Do not allow serious errors in working for the final mark. 7(a) Alternative Method for Question 7(a): y= 6 B2 tan 2 t + 1 sec2 t 1 Using y = = = x tan t sin t sin t cos t cos t dy B1 = −6 x −2 dx dy 6 M1 =− dx 9sin 2 2t dy −2 A1 Obtain = from full and correct working dx 3sin 2 2t 5 7(b) 3 B1 Gradient of normal = 2 Use correct method to find the equation of the normal M1 3 3 E.g. ( y − 2 ) =( x − 3) or find c in y = x + c. 2 2 Allow a wrong value for x or y but not both, with their normal gradient. Obtain 2 y − 3x + 5 = 0 A1 Or k ( 2 y − 3x + 5 ) = 0, where k is an integer. 3

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 Use a correct method to obtain the first two terms of the expansion of ( a − 2 x ) 8(b) −1 M1 or −1 −1 2x x or ( 3a + x ) −1 1 − or 1 + a 3a 2 x 4x2 A1ft OE. May be unsimplified. Obtain +2 1 + + 2 + ... Follow their A, B for an expansion involving a. a a 1 x x2 A1ft OE. May be unsimplified. Obtain + 1 − + 2 + ... Follow their A, B for an expansion involving a. 3 3a 9a 7 35 x 217 x 2 A1 Or simplified equivalent. Final answer. Obtain + + + Ignore any terms in higher powers of x. 3 9a 27a 2 Do not ISW, e.g. multiplying by 27a2. Condone different order of terms. 8(b) Alternative Method for Question 8(b) Expanding 7a2 ( a − 2x ) (3a + x )−1 from the original question. −1 M1 Use a correct method to obtain the first two terms of the expansion of ( a − 2 x ) or −1 −1 −1 2x x 1 − or ( 3a + x ) or 1 + −1 a 3a 2 x 4 x2 7a x x2 A1 OE. May be unsimplified. Obtain +7a 1 + + 2 + ... or + 1 − + 2 + ... May be implied by the expression shown for the a a 3 3a 9a next A1. 7 2 x 4 x2 x x2 A1 OE. May be unsimplified. Obtain + 1 + + + ... 1 − + + ... 3 a a2 3a 9a 2 7 35 x 217 x 2 A1 Or simplified equivalent. Final answer. Obtain + + + Ignore any terms in higher powers of x. 3 9a 27a 2 Do not ISW, e.g. multiplying by 27a2. Condone different order of terms. 4 8(c) a B1 a a x< Or − < x < . 2 2 2 Mark final answer. Must make a clear statement. 1

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 32 then written as x 2 − 4 + 2 . x +4 Alternative Method for Question 9(a) Expands brackets to get B = 0 M1 ( x + 4 )( x + Bx + C ) + D = 2 2 x 4 + Bx3 + ( C + 4 ) x 2 + 4 Bx + 4C + D 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 − 4x 3 1 M1 Obtain p tan −1 qx where q = 2 or q = 2 1 A1 FT Follow their constant remainder, Obtain 16 tan −1 x 2 their constant remainder −1 1 i.e. tan x. 2 2 1 M1 Terms need not be evaluated, e.g. Use limits correctly in an expression containing p tan −1 qx where q = 2 or q = 2 8 3 − 8 3 + 16 tan −1 3 − 8 − 8 + 16 tan −1 1 and rx3 + sx 3 8 16 16π or − 8 can be − , 16 tan −1 3 can be , 3 3 3 16 tan −1 1 can be 4π. 4 A1 AG Obtain ( π + 4 ) from full and correct working 3 5

Mark scheme

10(a) dV B1 dV dV State that = 50000 − 600h May be seen as = 50000 and = [ −] 600h. dt dt dt When put together (may be in the chain rule) B1 can be awarded. dV dV M1 dh dV dV [Use V = 40000h to] obtain = 40000 and use this and their in the correct = ÷ dh dt dt dt dh dh 50000 − 600h dh chain rule to obtain E.g. = dt 40000 dt or dV dh dV dh [Use V = 40000h to] obtain = 40000 and equate to their E.g. 40000 = 50000 − 600h dt dt dt dt 10(a) dh A1 AG Obtain 200 = 250 − 3h from full and correct working dt dh 50000 − 600h = OE, leading to given answer dt 40000 with no other working, or no incorrect working seen SC B1 (1/3). dV = 50000 − 600h B1 followed by dt dh 50000 − 600h = OE, leading to given answer dt 40000 with no other working, or no incorrect working seen, SC B1 2/3. dV dV dh = × followed by dt dh dt dh 50000 − 600h = 40000 OE, leading to given dt answer with no other working, or no incorrect dV working seen, B1 for implied and SC B1 2/3. dt 3 10(b) Separate variables correctly and integrate one side correctly M1 1 1 E.g. 250 − 3h dh = 200 dt. Integral signs may be omitted, 200 may be on opposite side. 1 t A1 OE Obtain − ln 250 − 3h = ( +C ) Condone missing “+C” and lack of modulus signs. 3 200 Use t = 0, h = 50 in an expression containing ln ( 250 − 3h ) or ln 250 − 3h to find M1 Or equivalent use of limits 50 and 80. the constant of integration. 1 A1 1 100 t 200 10 Obtain C = − ln100 OE, e.g. ln = , or − ln . 3 3 250 − 3h 200 3 100 With or without modulus signs on the log terms. t = 150 A1 5

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 dy 2sin 2 x A1 OE Obtain = 2cos x 2 + cos x − dx 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 cos a 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 The only error in the form of the differential allowed is for ( 2 + cos x ) 2 to be − 1 3 ( 2 + cos x )+ 2 or ( 2 + cos x )− 2 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 then must have correct limits 2 u d u. 1 (See final M1) Integrate to obtain ku 2 ( +C ) 3 M1 Constant of integration not required 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 OE. Allow, e.g., 3 3 for 27. Obtain 3 3 − 1 or 4 3 − or 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