Mathematics 9709/33 October/November 2023: question paper and mark scheme

Mark scheme

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, ( ) 2 or 2 x − 1 < 0.252 or corresponding pair of linear ( ) < 0.5 2 or 2 x+1 − 2 2 equations 0.25 = 2x – 1 and −0.25 = 2x – 1 or quadratic – 2 and − 0.5 = 2 –2 ( ) x+1 x+1 or corresponding pair of linear equations 0.5 = 2 2 equation 2 x − 1 = 0.252 . ( ) = 0.5 2 or quadratic equation 2 x+1 − 2 2 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 2 x +1 = a or 2 x = b where a, b > 0 x=… Obtain critical values x = 0.322 and –0.415 A1 ln 2.5 ln1.5 or awrt x = 0.32 and –0.42 e.g. − 1 and −1 . ln 2 ln 2 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 . 2 x +1 = a or 2 x = 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 or awrt x = 0.32 e.g. −1 . ln 2 or exact equivalent Obtain the other critical value e.g. –0.415 or awrt x = –0.42 or exact A1 ln1.5 equivalent e.g. −1 . ln 2 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

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

Mark scheme

3 2(−2)3 + a(−2)2 +b(−2) + 6 = −38 M1 Substitute x = –2 and equate the result to –38 Allow errors or divide by x + 2 to obtain quadratic quotient, and equate 2 x 2 + ( a − 4 ) x + b − 2a + 8 constant remainder to –38. x+2 3 2 x + ax 2 + bx +6 3 2 2x + 4x (a – 4)x2 + bx (a – 4)x2 + (2a – 8)x (b – 2a + 8)x + 6 (b – 2a+8)x + 2b – 4a + 16 4a – 2b –10 3 Obtain a correct evaluated equation, e.g. –16 + 4a – 2b + 6 = –38 or 4a – A1 2b = −28 1 3 1 2 1 M1 1 19 2 + a +b + 6 = 19 Substitute x = and equate the result to 2 2 2 2 2 2 Allow errors or divide by 2x – 1 to obtain quadratic quotient, and a +1 b a 1 x2 + x+ + + 19 2 2 4 4 equate constant remainder to . 2x −1 3 2 2 x + ax 2 + bx + 6 2x – x 3 2 (a + 1)x2 + bx ( a + 1) x 2 − + x a 1 2 2 a 1 b + + x + 6 2 2 a 1 b a 1 b + + x − + + 2 2 2 4 4 b a 1 6+ + + 2 4 4 1 a b 19 a b 13 A1 Obtain a correct evaluated equation, e.g. + + +6= or + = 4 4 2 2 4 2 4 Obtain a = –3 and b = 8 A1 ISW 5 4 2± ( −2 )2 − 4 ( 3 + i )( 3 − i ) M1 Use quadratic formula to solve for w 2 (3 + i ) Use i2 = –1 in (3 + i)(3 − i) M1 2 + 6i 2 − 6i A1 Must be simplified to this form. Obtain one of the answers w = or w = 6 + 2i 6 + 2i Show intention to multiply numerator and denominator by conjugate of M1 Independent of previous M marks but must be of the same their denominator. a form, e.g. . b + ci 3 4 A1 2 + 6i Obtain final answers + i and – i SC Both correct final answers from w = and 5 5 6 + 2i Accept 0.6 + 0.8i and 0 – i 2 − 6i w= seen, no evidence of conjugate, then SC B1 6 + 2i for both. 3 4 Allow x = , y = or x = 0, y = –1. 5 5 3 + 4i A0 for . 5 Alternative method for Question 4 Multiply the equation by 3 – i M1 Use i2 = –1 in (3 + i)(3 − i) M1 Obtain 10w2 – 2(3 – i)w + (3 – i)2 = 0 or equivalent A1 Use quadratic formula or factorise to solve for w M1

Mark scheme

4 3 4 A1 SC Both correct final answers from Obtain final answers + i and – i 10w2 – 2(3 – i)w + (3 – i)2 = 0 with no working then SC 5 5 Accept 0.6 + 0.8i and 0 – i B1 for both. 3 4 Allow x = , y = or x = 0, y = –1. 5 5 3 + 4i A0 for . 5 Alternative method for Question 4 Substitute w = x + iy and form equations for real and imaginary parts M1 Use i2 = –1 in (x + iy)2 M1 Obtain 3(x2 – y2) – 2xy – 2x + 3 = 0 and x2 – y2 + 6xy – 2y – 1 = 0 A1 OE Form quartic equation in x only or y only using the correct substitution M1 (3 − x ) and solve for x or y Use correct y = to attempt to form and solve 10 x − 3 50x4 – 60x3 + 63x2 – 27x = 0 x (5x – 3)(10x2 − 6x + 9) = 0. 3(1 + y ) Use correct x = to attempt to form and solve 10 y + 1 100y4 + 40y3 – 66y2 – 14y – 8 = 0 (y + 1)(5y – 4)(20y2 + 4y + 2) = 0. 3 4 A1 SC Both correct final answers from Obtain final answers + i and – i 3(x2 – y2) – 2xy – 2x + 3 = 0 and x2 – y2 + 6xy – 2y – 1 = 0 5 5 Accept 0.6 + 0.8i and 0 – i with no working then SC B1 for both. 3 4 Allow x = , y = or x = 0, y = –1. 5 5 3 + 4i A0 for . 5 5

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. 6 x(1 − x 2 )e3 x −1 + 2 xe3 x −1 ( ) If 6 x 1 − x 2 e3 x −1 + 2 xe3 x −1 = 0 from the start, with no 2 2 2 2 A1 Obtain correct derivative in any form, e.g. (1 − x ) 2 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 A1 Allow √(4/3). Obtain the point , −3e3 or exact equivalent 3 2 3 A1 Obtain the point − , −3e3 or exact equivalent 3 6

Mark scheme

6(a) Use correct Pythagoras cot2θ = cosec2θ – 1 or cot2θ = 1/sin2θ – 1 M1 If consistent omission of brackets, e.g. (sinθ)2 written as or cot²θ = cos²θ /sin²θ and then cos²θ = 1 – sin²θ , sinθ2 then SC B1 in place of M1A1. together with double angle formula cos2θ = 1 – 2sin2θ , to obtain an equation in sin 𝜃 or sin 𝜃 and cosec2θ Obtain a correct equation in sin 𝜃 in any form A1 e.g. 1/sin2θ − 1 + 2(1 – 2sin2θ ) = 4 or 1 – sin² sin² ( ) + 2 1 – 2sin 2 = 4 . If cos² sin² ( ) + 2 1 – 2sin 2 = 4 then ( ) e.g. 1 − sin² + 2 1 – 2sin 2 sin² = 4 . 2 (missing sin on right) allow M1A1A0. Reduce to the given answer of 4sin 4 θ + 3sin 2 θ − 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 (4sin2θ − 1)(sin2θ + 1) = 0 and solve for 𝜃. Incorrect sign in solution of quadratic seen, e.g. (4sin2θ − 1)(sin2θ – 1) = 0 then M0 A0 A0 but if only see (4sin2θ − 1) = 0 and nothing incorrect seen allow 3/3. Obtain answer, e.g. 𝜃 = 30° A1 π/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 5π/6 7π/6 11π/6 award A1. 3

Mark scheme

7(a) State or imply 2 y dy as the derivative of y2 B1 Allow for 3x2dx + 2ydy or Fx = 3x 2 + 6 x and Fy = 2 y + 3 . dx dy M1 dy dy Equate derivative of LHS to zero and solve for 3x2 + 2 y + 6x + 3 =0 dx dx dx dy F or 3x2dx + 2 ydy + 6xdx + 3dy = 0 or = − x need dx Fy evidence from B1 mark or formula must be seen. Allow errors. Obtain the given answer A1 dy 3x 2 + 6 x −3x 2 − 6 x AG =− not . dx 2y + 3 2y + 3 dy dy Must factorise with e.g. 3x2 + 6x + (2 y + 3) = 0 dx dx or 3x2dx + 6xdx + dy ( 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

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 B1 Can recover the previous B1 if de–4x seen here. Obtain term − e −4 x 4 Obtain only a term of the form a tan 3 y M1 Can recover the first B1 if a tan 3 y seen here. 1 A1 Obtain term tan 3 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 −4x or tan3y = − e + 8.39 × 10–5 3 4 1 3 3 A1 ISW Obtain final answer y = tan −1 e −8 − e −4 x 1 3 3 4 4 OE e.g. y = tan −1 2.52 ×10−4 − e−4 x 3 4 7

Mark scheme

9(a) Ax + B C B1 If incorrect partial fractions e.g. A = 0 or Ax2 + B then State or imply the form + M1, A1 A0 for correct C. Only allow single A1 even if 2 + 3x 2 2− x other coefficients correct. B1 recoverable by a correct form end statement. Use a correct method for finding a coefficient M1 e.g. (Ax + B)(2 – x) + C(2 + 3x2) = (3C −A)x2 + (2A −B)x + (2B + 2C) = 17x2 – 7x + 16. Obtain one of A = –2, B = 3 and C = 5 A1 If error present in above still allow A1 for C. Obtain a second value A1 Obtain the third value A1 Extra term in partial fractions, D/(2 + 3x2), that is 4 unknowns A, B, C and D then B0 unless recover at end, e.g. by setting B or D = 0. If B or D set to any value other than 0 and all coefficients correctly found to their new values then allow all A marks, but still B0 for partial fraction expression unless B + D combined. Hence A1 for each coefficient, but nothing for coefficient set to specific value. Another case of extra term in partial fraction expression, namely + F, mark as above but need F = 0 to recover B1. 5 9(b) Use a correct method to find the first two terms of the expansion M1 Symbolic coefficients are not sufficient for the M1. (2 − x)−1 = 2−1 + (−1) 2−2(−x) + [(−1)(−2)2−3(−x)2/2!], −1 −1 3x 2 3x2 x −x 1 + =1– or 1 − = 1 – 2 2 2 2 Ax + B 3x 2 A1 FT Obtain correct un-simplified expansions up to the term in 1 + ( −1) A = –2 B = 3 x 3 of each partial fraction. 2 2 − x ( −1)( −2 ) − x 2 1 + ( −1) + C 2 2 2 C=5 2 + ( −1)( −2 )( −3) − x 3 6 2 3 − 2 x 3x 2 5 x x 2 x 3 A1 FT Unsimplified (2 − x)−1 expanded correctly, error in = 1 − + 1 + + + simplifying before their C is involved in the expression, 2 2 2 2 4 8 allow A1FT when their C is introduced. 3 5 5 9 5 3 5 The FT is on A, B, C. = + + −1 + x + − + x 2 + + x3 2 2 4 4 8 2 16 3x 2 −1 3x 2 M1 Allow either ±2 or ±2−1 outside bracket or missing. Multiply expansion of 1 + must reach 1± by Ax + B, Allow one error in actual multiplication to acquire the 4 2 2 terms [all terms needed]. where AB ≠ 0, up to the term in x 3 . Allow if used Cx + D (Ax + B Ignore errors in higher powers. miscopied). 9(b) 1 13 29 3 A1 Maclaurin’s Series: Obtain final answer 4 + x − x2 + x , or equivalent f(0) = 4 B1 f’(0) = 1/4 B1. 4 8 16 f”(0) = –13/4 and f’”(0) = 87/8 B1. [If final answer has been multiplied throughout, e.g. by 16 then A0 at the 13 2 87 3 x x end] 1 4+ x− 4 + 8 or equivalent M1 A1. 4 2 6 13 2 87 3 x x 1 If 1 + x − 4 + 8 then M0 A0 unless their f(0) 4 2 6 actually is 1. 5 9(c) 2 2 2 B1 Or exact equivalent. State answer |x| < or − <x< clear conclusion required Strict inequality. 3 3 3 1

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 dy π Obtain y = − and = −1 when x = FT their with x = substituted. 2 dx 2 dx 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 sin 2 x + b sin 2 xdx *M1 1 1 A1 OE Obtain x sin 2 x − sin 2 xdx 2 2 1 1 A1 OE Complete integration and obtain x sin 2 x + cos 2 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 − cos 0 4 2 4 4 4 4 4 to obtain ax sin 2x + ccos 2x 1 π 2π 1 or sin − cos 0 . 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 sin 2 x + cos 2 x then final answer. 2 4 5

Mark scheme

11(a) Use correct process for modulus on direction vector of l, e.g. M1 SOI Allow −12. (−1)2 + 12 + 22 (−λ ) 2 + λ 2 + ( 2λ ) . 2 Allow 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