Mathematics 9709/32 October/November 2024: question paper and mark scheme

Mark scheme

1 Obtain a correct unsimplified version of the x or x 2 term of the expansion of M1 1 1 1 ×1 1 1 E.g. − × x or − 2 2 × x 2 or ( 9 − 3 x ) 2 or 1 − x 2 1 1 2 3 2 9 −1 −1 −3 3 1 2 1 × 9 ( −3x ) or 2 2 9 2 ( −3 x ) . 1 2 2 2 n Not for symbolic coefficients in the form C r . State correct first term 3 B1 1 1 A1 A1 A1 for each term correct. Obtain the next two terms − x − x 2 Do not ISW. 2 24 1 1 SC M1A1 for 1 − x − x 2 seen on its own or 6 72 as a factor. 4

Mark scheme

2(a) Sketch a relevant graph, e.g. y = cot 2x B1 Alt: use tan 2x and cos x . And only one root in π range. (also cross at 2 ) Sketch a second relevant graph on the same axes, e.g. y = sec x and justify the given B1 Need to mark intersection with a dot, a cross, or statement say roots at points of intersection, OE. 2 2(b) 1 B1 Should see tan 2x = cos x before the given State x = tan −1 ( cos x ) 2 conclusion. 1 and rearrange to the given equation cot 2x = sec x Or rearrange cot 2x = sec x to x = tan −1 ( cos x ) 2 and state iterative formula 1 Note: If using the alternative approach in (a), can stop at tan 2x = cos x xn+1 = tan −1 ( cos xn ) . 2 1

Mark scheme

3 Square x + iy and equate real and imaginary parts to 6 and −8 respectively *M1 Condone +8 in place of -8 and/or i 2 = 1. Obtain equations x2 – y2 = 6 and 2xy = –8 A1 OE Eliminate one variable and find an equation in the other (from 2 equations each in 2 DM1 Condone a slip but not seriously incorrect unknowns) algebra, e.g. use of x = −4 y is M0. Obtain x4 – 6x2 – 16 = 0 or y4 + 6y2 – 16 = 0 A1 Accept 3-term equivalents e.g. x 4 = 6 x 2 + 16. Condone missing ‘= 0’ if implied by subsequent work. ( ) Obtain answers ± 2 2 − 2i or exact equivalents A1 Allow if values of x and y stated separately but the pairing is clear. Ignore additional correct solutions for x and y not real, but A0 if any additional incorrect answers. 5

Mark scheme

4 Use laws of indices correctly and solve for 5 x *M1 5 E.g. obtain 5 x = OE. 12 Allow for y = … if they have previously stated y = 5 x. Could be implied if they have a correct simplified equation in 5x, e.g. 12 × 5x = 5. Use a correct method for solving an equation of the form 5 x = a, where a > 0 DM1 Allow x ln5 = ln 10 . 24 Obtain answer – 0.544 A1 CWO. If no working shown, 0/3. Note: 3 dp required. 3

Mark scheme

5(a) 4π π M1 SOI and/or − 4 1 4 1 7 7 Allow π − − π or π − π 7 7 7 7 Note: Many multiply top and bottom by the conjugate, which is fine, but to score the M1 they need to state or imply the argument of a complex number. 5 A1 Do not accept degrees. Obtain arg u = π 7 2 5(b) Reflection (in the) real axis B1 Correct non-contradictory statement. Condone x–axis or horizontal axis. Need ‘reflection’. Not ‘mirror’, ‘flip’. 5 B1FT FT their exact (a). Accept 2π − their exact (a). arg u* = − π 7 Accept an ‘exact’ expression in place of an exact value. Need to see a value or an expression. Do not accept arg u* = − arg u without a value seen. 2

Mark scheme

6 Form a pair of equations in a and b *M1 Condone sign slips but must be using the given coordinates correctly. ln a + 8.27 = 3.4ln b e.g. ln a + 2.24 = 0.5ln b ae2.24 = b0.5 or ae 8.27 = b3.4 Carry out a correct method for finding lna or lnb or a or b DM1 Condone sign slip. Obtain value a = 0.3 A1 (0.30109…) Obtain value b = 8 A1 (7.99895…) Allow A0A1 if both values ‘correct’ but not rounded to 1 sf. Allow 4/4 for 0.3 y = 8 x with correct working shown. 6 Alternative Method for Question 6: Carry out a correct method for finding lnb or b *M1 Condone sign slips but must be using the given (Need to link the gradient to lnb at some point) coordinates correctly. 8.27 − 2.24 ln b = ( = 2.079....) 3.4 − 0.5 Obtain value b = 8 A1 Correct method to find ± ln a or a DM1 Condone sign slip ( ln a = −1.200...) . Obtain value a = 0.3 A1 Allow A0A1 if both values ‘correct’ but not rounded to 1 sf. Allow 4/4 for 0.3 y = 8 x with correct working shown. 4

Mark scheme

7(a) Use correct double angle formula to obtain an equation in tan x M1 2 × 2 tan x e.g. tan 3 x + − tan x ( = 0 ) . 1 − tan 2 x Allow if the correct formula is quoted but then they lose the 2 from the numerator when they use the formula. Obtain a correct equation in tan x in any form without fractions A1 E.g. tan3 x − tan5 x + 4tan x − tan x + tan3 x ( = 0). Condone if ‘= 0’ is missing here. Reduce to the given answer of tan 4 x − 2tan 2 x − 3 = 0 correctly A1 Obtain given answer from correct working but condone if never mention tan x ≠ 0. Condone the right terms in a different order ‘Show that’ so each line must be correct. 3 7(b) A complete correct method to solve the equation to obtain a value for θ M1 ( tan 2θ = ± 3 ) Allow if they make a slip in copying the equation but do have a complete method to obtain a value of θ . M0 if they get a value for 2θ but never halve it. 1 1 2 5 A1 Obtain two of ( θ =) π, π, π and π 6 3 3 6 1 1 2 5 A1 Exact, ignore any answers outside interval Obtain the other two of (θ =) π, π, π and π and no others in the interval 6 3 3 6 2 1 4 2 Accept π for π and π for π 6 3 6 3 π Do not need to see θ = (from tan 2θ = 0). 2 3

Mark scheme

8(a) Use correct product rule or chain rule to find derivative of x with respect to t M1 Obtain k tan 2t sec2 2t. dx A1 Obtain = 4 tan 2t sec2 2t oe dt dy B1 = −2sin 2t dt dy dy dt dy 1 B1 dy Use = × to obtain given answer = − cos3 2t Condone if missing. dx dt dx dx 2 dx 4 8(b) 2 B1 Accept y = 0.707… Obtain x = 1 and y = 2 − 2 B1 5 State or imply gradient of tangent is or gradient of normal is 4 2 Any equivalent form, e.g. 2 2 . 8 Accept – 0.177 or 5.66. Use correct method to find equation of normal using their values M1 Need a fully substituted equation for the normal (in any form) or to get at least as far as finding value for m and expression for c. 7 2 A1 E.g., y = 5.66x – 4.95. Obtain equation of normal is y = 4 2 x − or equivalent 3 term equation Must be y = … 2 4

Mark scheme

9(a) Use a correct method to find OD ( M1 E.g. OC + 3 OA − OB = ) (–3i – 2j + 2k) + 3((2i + j – 3k) – (4j + k)) ( AB = −2i + 3j + 4k) Accept column vectors throughout. Obtain position vector of D is 3i – 11j – 10k A1 Accept coordinates. 2 9(b) Carry out correct method for finding a vector equation for AC or BD *M1 E.g. 2i + j – 3k + λ (5i + 3j – 5k) or 4j + k + µ (3i – 15j – 11k). Condone missing r = … Both diagonal equations correct. A1ft Seen or implied. Follow their D. Condone missing r = … Equate at least two pairs of corresponding components and solve for λ or for µ DM1 Dependent on using relevant lines and two different parameters. 1 1 A1 The values will depend on the directions of their Obtain λ = – or µ = lines 4 4 3 1 7 A1 OE Obtain position vector of P is i+ j– k Accept coordinates. 4 4 4 Do not ISW. 9(b) Alternative Method for Question 9(b): State or imply AC = 5i − 3j + 5k B1 FT Or BD = 3i − 15 j − 11k Follow their D if used. Identify similar triangles with ratio 1 : 3 M1 1 M1 Must be correct fraction. Use similar triangles to obtain OP , e.g. OP = OA + AC 4 3 1 7 A2 OE Obtain position vector of P is i+ j– k Allow A1A0 if any two values are correct. 4 4 4 5 9(c) Find direction vector BA = 2i – 3j – 4k and BC = –3i – 6j + k or equivalent B1FT Or AB and CB. FT if using an incorrect AB from earlier work. Carry out correct process for evaluating the scalar product of two relevant vectors M1 Allow if one is going in the negative direction, e.g. AB and BC . Using the correct process for the moduli, divide their scalar product by the product M1 Independent of the first M1. of their moduli and evaluate the inverse cosine of the result to obtain an angle For their two vectors 8 θ = cos −1 = ... 29 46 Obtain answer 77.3° (or 1.35 radians) A1 77.347… Correctly rounded to more than 3 sf or AWRT 77.3. 4

Mark scheme

10(a) dV B1 Need a complete correct statement seen or Obtain = 40π − 0.8πr or equivalent implied. dt dV dV dr B1 Need a complete correct statement seen or Obtain = 4πr 2 or equivalent e.g. = 4πr 2 implied. dr dt dt Use the chain rule to obtain given answer (including the derivative) B1 dr 50 − r dr 40 − 0.8r Allow if = 2 follows = dt 5r dt 4r 2 without further explanation (π already cancelled) and no incorrect statements seen. 3 10(b) Commence division and reach quotient of the form M1 Allow M1 if divide by r − 50 to obtain –5r ± 250 5r ± 250 . or 5r2 = (50 – r)(Ar + B) + C and reach A = –5 and B = ± 250 Obtain quotient –5r – 250 A1 Do not need to state which is quotient and which is remainder. However, if clearly muddled, then M1A1A0 for both expressions correct. Obtain remainder 12 500 A1 Note: 12 500 following division by r – 50 is correct and scores this A1 ISW. SC B1 only for correct use of remainder theorem to obtain correct remainder. 3 10(c) Prepare to integrate e.g. separate variables correctly B1FT 5r dr = 1dt 2 50 − r dt 5r 2 12500 Or express in the form = = − ( 5r + 250 ) + Condone missing dr, dt or missing integral dr 50 − r 50 − r signs, but not both. Follow their division in (b) if substitute before separating. Obtain term t DB1 A 2 M1 C Obtain terms r + Br − Cln ( 50 − r ) From their Ar + B + in (b) where 2 50 − r ABC ≠ 0. Allow a single slip in the coefficients. 5 A1FT FT their (b), provided of the correct form. Obtain terms − r 2 − 250r − 12500ln(50 − r ) 2 Use t = 0, r = 0 to evaluate a constant or as limits in a solution containing terms of M1 the form r2, r, ln(50 – r) and t 5 A1 OE Obtain final answer t = − r 2 − 250r − 12500ln(50 − r ) + 12500ln 50 Must be t = ….. 2 Allow with 12500ln50 = 48900 or better. 6 10(d) Obtain t = 70.5 B1 May be more accurate (70.4605…). 1

Mark scheme

11(a) Use correct quotient rule NB the question asks for f ′ ( x ) so need complete form M1 Or correct product rule. 4e 2 x (e 2 x − 3e x + 2) − 2e 2 x (2e 2 x − 3e x ) A1 Obtain correct derivative in any form, e.g. ( e − 3e + 2 ) 2x x 2 Equate their derivative to zero *M1 Can be implied by numerator equated to zero for ( quotient rule. 8 = 6e x ) Solve for x to obtain x = lna DM1 a positive. 4 A1 No errors seen. Obtain x = ln and y = –16 3 8 Accept equivalent exact forms, e.g. x = ln . 6 11(a) Alternative Method for Question 11(a) Complete method to express f ( x ) in partial fractions M1 As far as p + q + x r with values for p, e − 2 e −1 x q and r 2 + 8 − x 2 . e − 2 e −1 x ( Allow in u u = e x . ) se x te x *M1 Note: the question requires f’(x) so if they have Differentiate to obtain f ′ ( x ) = + substituted for ex, they will also need chain rule. ( e − 2 ) ( e − 1) x 2 x 2 −8e x 2e x A1 From correct work. Obtain f ′ ( x ) = + ( e − 2 ) ( e − 1) x 2 x 2 Equate derivative to zero and solve for x to obtain x = lna DM1 Must follow correctly to give a positive value of a. 4 A1 No errors seen. Obtain x = ln and y = –16 3 8 Accept x = ln , or equivalent. 6 5 11(b) du B1 State or imply = ex dx 2u B1 Correct expression in u. Obtain du or equivalent u − 3u + 2 2 Condone missing du or missing integral but not both. 2 + 8 − 2 du Or1 Allow FT if using their partial fractions from u u ( u − 2 ) u ( u − 1) (a). A B B1 FT Complete reduction to partial fractions. + State or imply partial fractions of the form u −1 u − 2 Correct form for their integrand. C D E 2u − 3 3 2u − 3 F G Or1 + + Or2 2 + 2 = 2 + + u u − 2 u −1 u − 3u + 2 u − 3u + 2 u − 3u + 2 u − 2 u −1 Use a correct method for finding a constant M1 Available if they have incorrect form. −2 4 A1 Obtain correct + u −1 u − 2 2u − 3 3 3 Or2 2 + − u − 3u + 2 u − 2 u − 1 Integrate to obtain a ln (u – 1) + b ln (u – 2) or equivalent *M1 M0 if they have additional terms that do not cancel out. Obtain correct –2 ln (u – 1) + 4 ln (u – 2) or equivalent A1FT FT values of their partial fraction coefficients. Correctly use limits u = 5 and 3 in an expression of the form a ln (u – 1) + b ln (u – 2) DM1 or x = ln 5 and ln 3 in an expression of the form a ln (ex – 1) + b ln (ex – 2) 81 A1 Accept ln 20.25. Obtain ln 4 9