Mathematics 9709/32 February/March 2024: question paper and mark scheme

Mark scheme

1 Commence division and reach partial quotient of the form x 2 ± 3 x M1 ( )( ) or x 4 − 3 x3 + 9 x 2 − 12 x + 27 = x 2 + 5 Ax 2 + Bx + C + Dx + E or Ax4 + Bx3 + (5A + C)x2 + 5Bx + 5C and reach A = 1 and B = ±3 Obtain quotient x 2 − 3 x + 4 A1 A = 1, B = −3 [5A + C = 9 so C = 4; 5B + D = − 12 so D = 3; 5C + E = 27 so E = 7]. A pair of incorrect statements ‘remainder x 2 − 3 x + 4 ’ and ‘quotient 3 x + 7 ’ score M1 A1 A0. Obtain remainder 3 x + 7 A1 x2 − 3x + 4 3 x2 + 5 x4 – 3x3 + 9x2 – 12x + 27 x4 + 5x2 − 3x + 4x2 3 − 3x3 − 15x + 4x2 + 3x + 4x2 + 20 + 3x +7

Mark scheme

2(a) State unsimplified term in x, or its coefficient, in the expansion of ( 4 − x ) 2 1 B1 1 1  −x  −x 42 × ×   = . 2  4  4 State unsimplifed term in x 2 , or its coefficient, in the expansion of ( 4 − x ) 2 1 B1 2 2 1 1 × −21  − x  − x2 x . Allow   . 42 × 2 ×  = 2  4  64 4 1 M1 Allow unsimplified 2x. Multiply by ( 2 x − 5 ) and obtain 2 terms in x 2 , allow even if errors in 4 2 , signs, etc. 1 1  −x  4 2 × ×  − 5. 2  4  1 −1 1 × 2 2 2 2  −x   x 4 × ×  . Allow   . 2 2  4  4 − x2 −1 −1  −x  2x ×   ( −5 ) × or 2 ×   ( −5 ) ×   .  4  64  4   64  27 54 A1 Allow in a full expansion up to x2 , ignore extra Obtain − or –0.421875 or − terms even if they contain errors. 64 128 4 2(b) x <4 B1 or −4 < x < 4 . 1 (( − 3 ) + 1 ) so r =z =

Mark scheme

3(a) Obtain r = 4 B1 (− 3 ) +1 . 2 2 2 2 2 z = Correct method for the argument M1 − 3 5π θ = 2 tan −1   or 2 × .  1  6 π A1 Arg with no working B1 instead of M1 A1. Obtain θ = − A0 if decimals. 3 Allow separate mod and arg to gain full marks Alternative solution for Question 3(a) B1 ( ) =4 2 z2 = 2 – 2 3 i so r = 22 + – 2 3 Correct method for the argument M1 –2 3 arg z2 = tan −1 2 π A1 Arg with no working B1 instead of M1 A1. Obtain θ = − A0 if decimals. 3 Allow separate mod and arg to gain full marks 3 3(b) Use of α + their θ = 0 or α + their θ = − π or α + their θ = π M1 Seen or implied. Using their θ or new value calculated in (b). their r M1 Seen or implied. Use of R = 12 2π Obtain 13 e −i 3 and 13 ei 3 π A1 3

Mark scheme

4 Obtain ln p − ln q = a B1 p a =e . q Obtain ln p + 2ln q = b B1 pq2 = eb . Completed method to obtain ln p 7 q( ) M1 E.g. ln q = b−a , ln p = 2a + b 3 3 and attempt 7ln p + ln q. All exponentials must be removed to obtain M1. 13a + 8b A1 Obtain 3 Alternative solution for Question 4  p x B1 p ( ) Or ln= p 7 q x ln + y ln q 2 p . y State p q =   q 2 p 7 q q Equate indices to form simultaneous equations in x and y, can have errors M1 x + y = 7 and − x + 2y = 1. Obtain 7= x + y and = 1 2y − x A1 Leading to = x 13 = , y 8. 3 3 13a + 8b A1 Evaluate x × a + y × b to obtain 3 4

Mark scheme

5(a) Show a circle with centre 4 + 2i B1 Im(z) Show a circle with radius 3 and centre not at the origin B1 Show the straight line Re ( z ) = 5 B1 2i Shade the correct region B1 O 4 5 Allow even if radius 3 mark not gained or shown incorrectly Re(z) 4 If 4 and 6 seen on diagram and line is at mid point, but 5 not marked, allow final two B1 marks. 5(b) Carry out a complete method for finding the greatest value of arg z M1 2+2 2 e.g. tan −1 . 5 Allow 2√2 as √(32 − 12 ). Obtain answer 0.768 radians or 44.0° A1 2 SC B1 tan−1 (2/4) + sin−1 (3/√(42 + 22 )) = 26.565° + 42.130° = 68.695° 68.7° or [1.19896] 1.20 radians.

Mark scheme

6(a) State or imply 4 y ddyx as the derivative of 2 y 2 B1 SC If dy introduced instead of d then allow B1 dx dx for both, followed by correct method M1 Max 2. State or imply 3 y + 3 x ddyx as the derivative of 3xy B1 Allow extra dy = correct expression to collect all dx marks if correct. Complete the differentiation, all 4 terms, isolate 2 ddyx terms on LHS or bracket ddyx M1 terms and solve for ddyx dy 2 x − 3 y − 1 A1 Answer Given – need to have seen 4y dy + 3x dy Obtain = dx dx dx 4 y + 3x = 2x −3y −1 or (4y + 3x) ddyx − 2x +3y = −1. Need to see = 2x or = 0 consistently throughout otherwise M1 A0. No recovery allowed. When all terms are included then must be an equation. 4 Allow all marks if using dx and dy. 6(b) Equate numerator to zero, obtaining 2x = 3y + 1 or 3y = 2x −1 and form equation in M1* e.g. 2 ( 2 x − 1)2 + x ( 2 x − 1) + x =x2 x only or y only from 2y2 +3xy + x = x2 9 or 2 y 2 + 32 (1 + 3 y ) y + 12 (1 + 3 y ) = 14 (1 + 3 y ) . 2 Allow errors. Obtain 92 ( 2 x − 1) = 2 − x 2 or a 3 term quadratic in one unknown and try to solve. DM1 e.g. 17 x 2 − 8 x + 2 = ( 0 b 2 − 4ac = −72 ) If errors in quadratic formulation allow solution, applying usual rules for solution of quadratic equation, and allow M1 or 17 y 2 + 6 y + 1 = ( −32 . 0 b 2 − 4ac = ) x = 4/17 ± (3√2/17)i, y = − 3//17 ± (2√2/17)i . Conclude that the equation has no [real] roots A1 Given Answer. CWO 3

Mark scheme

7(a) Use correct product rule M1 Obtain correct derivative in any form A1 e.g. dy = e 2 x + 2 xe 2 x − 5 dx 1  5  A1 Given answer – need to see e2 x= 5/(1 + 2x) Equate derivative to zero and obtain α = ln   or ln e2 x = ln (5/(1 + 2x)) in working. 2  1 + 2α  Must be in terms of α not x. Allow α to be used before equating to 0. 3 7(b) Calculate the value of a relevant expression or values of a pair of expressions at M1 Need to attempt BOTH values and have one x = 0.4 and x = 0.5 correct. Complete the argument correctly with correct calculated values A1 e.g. 0.4 < 0.51[ 08] and 0.5 > 0.458 or 0.46 or 0.45 or – 0.11[08] < 0 and 0.042 > 0 If use original derivative −0.994 (0.4) and 0.437 (0.5). 2 7(c) 1  5  M1 Obtain one value and then substitute it into the Use the iterative process α n +1 = ln   correctly at least twice anywhere in formula to obtain a second value. 2  1 + 2α n  iteration process Obtain final answer 0.47 A1 Show sufficient iterations to 4 d.p. to justify 0.47 to 2 d.p. or show there is a sign A1 0.4,0.5108,0.4528,0.4823,0.4670,0.4749 change in the interval ( 0.465, 0.475 ) 0.45,0.4838,0.4663,0.4753,0.4707,0.4730 0.5,0.4581,0.4795,0.4685,0.4742 Allow self correction. 3 SC B1 No working 0.47

Mark scheme

8(a) Use the correct expansion of cos ( x + 14 π ) to obtain sin x + 2cos x B1  1 1  3sin x + 2 2  cos x − sin x  .  2 2  State R = 5 B1 FT ISW FT their a sin x + b cos x provided this expression obtained by correct method. Use correct trig formulae to find α M1 α = tan−1 (b/a) from their a sin x + b cos x or sin−1 or cos−1 provided this expression obtained by correct method. NB If cos α = 1 and sin α = 2 then M0 A0. Obtain α = 1.107 A1 3 d.p. CAO Treat answer in degrees as a misread ( 63.435° ) . 4 8(b)  1.5  B1 FT Follow their R. sin −1    R  Use a correct method to obtain an un-simplified value of θ with their α M1  1.5     −1  1.5   2  sin −1   − α  or 2  π − sin   −α  .   R     R   Obtain one correct answer e.g. −0.74 in the interval A1 Obtain second correct answer e.g. 2.60 (2.5986) or 4π – 0.74 = 11.8 A1 If uses 1.11° withhold first accuracy mark gained, or 2.60 − 4π = −9.97 in the interval but allow rest of accuracy marks. Allow 2.6(0). Obtain two more correct answers e.g. −9.97 and 11.8 and no others in the interval A1 Ignore answers outside the interval. Treat answers in degrees as a misread. ( −571.1°, − 42.6°,148.9°, 677.2° ) . 5 ρ ρ

Mark scheme

9(a) Find the scalar product of a pair of adjacent sides M1 OA = (5, − 2, 1), OB = (8, 2, − 6), ρ ρ OC = (3, 4, −7), CB = (5, −2, 1), ρ AB = (3, 4, −7). ρ ρ Show that the sides are perpendicular A1 e.g. OA.OC = 15 − 8 − 7 = 0 . Need to see working of numerator, ignore denominator. ρ ρ ρ ρ Compare a pair of opposite sides M1 OA and CB or OC and AB . ρ ρ ρ ρ Show that they are parallel and equal in length and hence OABC is a rectangle A1 e.g. AB = AO + OB = 3i + 4 j − 7k = OC . ρ ρ If show AB =3i + 4 j − 7 k =OC , then M1 A1 since this implies parallel and of equal length. If only show lengths equal M1. If repeat for other pair of opposite sides then A1. Alternative solution for Question 9(a) ρ ρ ρ Show the diagonals OB and AC are equal in length ( 104 ) AC = ( −2, 6, −8). ρ Show the diagonals bisect each other at ( 4,1, −3) OB ρ 1 ρ ρ = OC + ( OA − OC ) = (4, 1, −3). 2 2 ρ ρ ρ Show the quadrilateral is a parallelogram e.g. OB =OA + OC . Show both pairs of opposite sides are equal in length and a pair of adjacent sides are perpendicular 4 Without calculation of scalar product max is M1 A1. ρ 9(b) ρ AC B1 Seen or implied using diagonals. AC =± ( −2i + 6 j − 8k ) or =± ( −1i + 3 j − 4k ) 2 ρ ρ Scalar product of a pair of relevant vectors M1 e.g. AC.OB = −16 + 12 + 48 . Using the correct process for the moduli, divide the scalar product by the product of M1  44  the moduli and obtain the inverse cosine of the result. ± cos −1  .  104  For any two vectors. Obtain answer 65. ( 0 ) ° A1 Accept 1.13 radians. Alternative solution for Question 9(b) ρ ρ Scalar product of a pair of relevant vectors M1 e.g. OA.OB = 40 − 4 − 6 using one side and a diagonal. ρ ρ or OC.OB = 24 + 8 + 42. Must use scalar product. Using the correct process for the moduli, divide the scalar product by the product of M1  30   74  the moduli and obtain the inverse cosine of the result. Any two vectors. ± cos −1  −1  or cos   .  104   104  Required angle = 180° − 2 × 57.5° or 180° − 2 × 32.5° = 115° and 180° − 115° or B1 OE SOI 2× 32.5° Complete method to find the acute angle. Obtain answer 65.0° A1 Accept 1.13 radians. 4

Mark scheme

10(a) A B C B1 Allow if seen prior to assigning a value for a. State or imply the form + + 2a + x 2 a − x 5a − 2 x Use a correct method for finding a coefficient M1 Obtain one of A = 1, B = 9, C = −16 A1 Obtain a second value A1 Obtain the third value A1 5 Dx + E C SC + B0 M1 and C = −16 4 a ^ 2 − x ^ 2 5a − 2 x A1 Max 2/5. SC Allow M1 only for other incorrect partial fraction. 10(b) Integrate and obtain one of the terms ln 2a + x − 9ln 2a − x + 8ln 5a − 2 x B1 FT Condone missing modulus signs. Use their A, B and C. Obtain a second correct term B1 FT Obtain the third correct term B1 FT Max 3/5 if value is assigned for a (award M0 A0). Substitute limits correctly in an integral of the form M1 Either (i) collect terms with same coeeficient and p ln 2a + x + q ln 2a − x + + r ln 5a − 2 x and remove all a’s remove all a’s e.g. pln 3a −pln a + qln a − qln 3a + rln 3a – rln7a hence pln 3 − qln 3 + rln 3 − rln7 or (ii) collect same ln terms and remove all a’s e.g. (p – q + r) ln 3a – ( p – q) ln a – rln7a and − (p − q) ln a = (−p + q − r ) ln a + r ln a hence p ln 3 – q ln3 + rln 3 – r ln 7. Obtain 18ln 3 − 8ln 7 from correct working A1 A0 if the solution involves logarithms of negative numbers. 5

Mark scheme

11 Separate variables correctly B1 −3 y 1 ∫ (1 + y ) e dy = ∫ 1 + cos 2θ dθ . Allow 1/e3 y and missing integral signs. Integrate to obtain p (1 + y ) e −3 y + qe −3 y dy M1 Allow unless clear evidence that formula used has ∫ a + sign. −1 1 A1 Allow unsimplified. Obtain (1 + y ) e−3 y + e−3 y dy ∫ 3 3 −1 1 A1 Condone no constant of integration. Obtain (1 + y ) e−3 y − e−3 y ( + A) 3 9 1 B1 Use correct double angle formula to obtain ∫ 2cos 2 θ dθ Obtain ktanθ [ + B ] B1 Condone no constant of integration. π M1* 1 1 1  17  Use y = 0, θ = to evaluate a constant of integration in an expression of the form =− − + C C =  4 2 3 9  18  y y α ye −3 y , β e −3 y and γ tan θ only. Allow αye and βe . Must have integrated LHS 3 3 twice. Use y = 1 DM1 − (1 + 1) 1 17 3 − 1( 9e3 ) = tan θ − . 3e 2 18 Must have integrated LHS. 17 14 −3 A1 Or exact equivalent . Exact ISW. Obtain tan= θ − e 9 9  17 14  Allow θ = tan −1  − e −3  .  9 9  If x instead of θ then withhold final A1. 9