Mathematics 9709/35 May/June 2025: question paper and mark scheme

Mark scheme

1 Use law of logarithm of a product or quotient M1 ln 5 + ln 6x−1 or ln 34−2x – ln 5 or ln 34−2x − ln 6x−1 Allow logs to any base but must be consistent throughout the equation. Use law of logarithm of a power twice M1 (4 – 2x)ln 3 and (x – 1)ln 6. Omission of bracket(s) is an accuracy error if not corrected later. Can have M0M1, i.e. go wrong with product but powers dealt with correctly. E.g. (4 – 2x)ln 3 = ln 5×(x – 1)ln 6, or (4 – 2x)ln 3 = (x – 1)ln 30 or 33x×2x written as 3xln 3×xln 2. Obtain a correct equation in any form, A1 May see (4 – 2x)ln 3 written as (2 – x)ln 9. e.g. (4 – 2x)ln 3 = ln 5 + (x – 1)ln 6 or 1.10(4 – 2x) = 1.61 + 1.79(x – 1) Obtain x = 1.15 A1 Must be 3 s.f. If no working seen, no marks available. 4

Mark scheme

2 Reduce to an equation in a single trig function by use of correct trig formula(s) M1 E.g. cosec2 θ = cot2 θ +1. Obtain correct simplified solvable equation in one trig function with all terms on A1 Other equations may be possible. one side, any order, including θ, e.g. one of 4cot2 θ – 3cot θ – 1 = 0 tan2 θ + 3tan θ − 4 = 0 34sin4 θ − 49sin 2 θ + 16 = 0 34cos 4θ −19cos2 θ + 1 = 0 *For this equation, it is not necessary to have all terms on 34 9 cos 2 2θ + 103 cos 2θ = 0 ( ) * 34 sin 2θ − tan −1 ( 53 ) = 3 one side. Solve their equation correctly, by formula, factorisation or calculator, to obtain M1 M0M1 is available if only sign slip(s) in trig formula(s). two values for a trig function, e.g. one of Incorrect solvable equation from use of correct trig 1 formula(s) can score M1M1. cot θ = 1 and – 0 tan θ = 1 and – 4 4 4 1 1 1 sin θ = ± and ± cos θ = ± and ± 17 2 17 2 15 5 3 cos 2θ = 0 and − *sin 2θ − tan −1 = 17 3 34 *For this equation, only one value is needed. Obtain two of answers, 14 π, − 34 π, –1.33, 1.82 A1 ISW Accept 0.785 for 14 π and –2.36 for − 34 π. Second M1 can be implied by 14 π or − 34 π, and –1.33 or 1.82 AWRT –1.33, 1.82, 0.785, –2.36. Obtain all answers 14 π, − 34 π, –1.33, 1.82 and no others in the interval A1 ISW Accept 0.785 for 14 π and –2.36 for − 34 π. AWRT –1.33, 1.82, 0.785, –2.36. Ignore answers outside the given interval. 5

Mark scheme

3(a) Correct modulus = 56 or 0.833 or 56 e iθ or r = 56 only B1 Note 0.83 scores B0. Do not ISW other than converting fractions to decimals. Correct arg = –2.75 or − 114 or re− i2.75 or θ = –2.75 only B1 0.25 – 3 must be evaluated. B0 for θ = –2.75i. Do not ISW other than converting fractions to decimals. 2 3(b) Enlargement [scale] factor 16 B1 Need both parts. Allow expansion or stretch for enlargement Not required to state centre (0, 0), but if incorrect then B0. Do not allow e.g. ‘reduction’ or ‘shrink’ or ‘compression’ for enlargement Clockwise rotation of 3 [radians] B1 Need both parts. or rotation of –3 [radians], or [anticlockwise] rotation of (2π – 3) [radians] Not required to state centre (0, 0), but if incorrect then B0. If B0B0 scored, allow SC B1 for enlargement AND rotation stated. 2

Mark scheme

4 Use the correct product rule *M1 (Note: may start with 12 x3 ln x) Attempt ln x 4 d dx ( ) 3x3 + 3x3 d dx ( ) ln x 4 with their derivatives. Obtain the correct derivative in any form e.g. 9x2 ln x4 + 12x2 A1 4 x3 E.g. 9 x 2 ln x 4 + 3x3 × . (If starting with 12 x3 ln x, should get e.g. 36x2 ln x +12x2 ) x4 May see d dx 4 ln x 4 = . x ( ) 4 − 4 DM1 E.g. ax4 = eb, or cx3 = ed or x = e f. Equate to zero and eliminate ln, e.g. x = e 3 Allow this mark even if in decimals. Allow SCB1 if incorrect sign in product rule resulting in 4 x4 = e 3 , OE. − 1 A1 ISW Obtain x = e 3 only or exact simplified equivalent 1 − 43 4 x = e scores A0. 1 − x = ±e 3 scores A0. Answers with no working score no marks. 4 A1 ISW Obtain y = − only or exact simplified equivalent 4 e − y = 3e−1 ln e 3 scores A0. Answers with no working score no marks. 5

Mark scheme

5(a) Attempt at method for calculating the distance from the origin to the centre of M1 E.g. 4 + 5 or 2 2 41 may be seen on diagram. the circle Obtain distance 41 – 3 A1 ISW AWRT 3.4(0). Need not identify as maximum and minimum, but if swapped, M1 SCB1 max 2/3. Obtain distance 41 + 3 A1 ISW AWRT 9.4(0). 2 2 3 3 Note: 5 + +4+ = 9.391 scores M0A0A0. 2 2 3 5(b) Correct tangent and sight of correct trigonometry B1FT Identify or imply correct point on the circle and sight of 4 tan −1 or 0.674 ( 7 ) or 38.7 5 5 or tan −1 or 0.896 or 51.3 4 −1 3 or sin OE, or 0.487(6) or 27.9 41 −1 5 or 4 5 or 4 Allow sin or cos −1 instead of inverse tan 41 41 If no correct tangent on diagram: 4 5 4 tan −1 or tan −1 or tan −1 (or above alternative trigonometry) 5 4 −5 3 FT their 41 found from a correct method for B1 and M1 AND sin −1 OE B1FT marks, e.g. by drawing correct tangent and sight of 41 4 5 3 tan −1 or tan −1 or sin −1 OE. 5 4 41 4 3 tan −1 − sin −1 scores B1FT M0 A0 3 4 2 3 5 41 Note sin −1 may be cos −1 or tan −1 or 41 41 4 2 4 3 3 2 tan −1 − sin −1 scores B1FT M1 1.98 A1 tan −1 . −5 41 8 −1 3 −1 1 −2 sin − tan scores B0 M0 A0 SC Thinks min arg z is actually max arg z: 41 5 π 5 3 + tan −1 + sin −1 B1FT 2.95447c or 169.279° B1. 2 4 41 5(b) Attempt at correct method to find the required argument, M1 π 5 3 4 3 Or + tan −1 − sin −1 . e.g. π − tan −1 − sin −1 2 4 41 5 41 FT their 41 found from a correct method for B1 and M1 marks. Previous B1 may be implied by a correct method here. Obtain 1.98 or 113.4 A1 AWRT E.g. 1.97923… or 113.401… 3

Mark scheme

6(a) dy B1 OE Obtain = 3sec 2 3t dt dx M1 dx d Attempt chain rule on 2 ( cos 3t ) for −1 = −2 ( cos 3t ) ( cos3t ) with their derivative. −2 or attempt to differentiate 2sec 3t Attempt dt dt dt dx A1 OE, e.g. 6 cos 3t −2 sin 3t. Obtain = 6sec 3t tan 3t ( ) dt Allow unsimplified, e.g. (−1) × 2 × (−1) × 3 for 6. dy dy dt dy dt M1 1 Use = × with their and their Expect, for example, 3sec 2 3t × dx dt dx dt dx 6sec 3t tan 3t 1 or 3sec 3t × 2 if correct. 6 ( cos 3t ) sin 3t −2 dy 1 A1 WWW Obtain = cosec3t Must be in this form of answer given in the question. dx 2 Not required to state A = 12 . dy dx dy Allow slips in notation for , and . dt dt dx 6(a) Alternative Method for Question 6(a) x2 B1 Convert to Cartesian form e.g. 1 + y = 2 4 d dy B1 Correct use of implicit differentiation e.g. y2 = 2 y dy dx dy x B1 OE Obtain 2 y = dx 2 dy 2sec 3t M1 Convert to parametric form e.g. 2 tan 3t = dx 2 dy 1 A1 WWW Obtain = cosec3t Must be in this form of answer given in the question. dx 2 5 6(b) Obtain x = 2 2 and y = 1 when t = 121 π B1 Must be exact. dy M1 Expect gradient of normal = − 2 if correct. Substitute t = 121 π into −1 ÷ their dx Allow if in decimals. Allow a small slip, but not with their coefficient A If their coefficient A is dealt with incorrectly M0, but allow second M1. Form equation of the normal with their (x, y), found using x = 2 and ( ) M1 E.g. y – 1 = − 2 x − 2 2 if correct or find c in equation cos 3t of line. dy Allow M1 even if decimals. y = tan 3t , and −1 ÷ their dx M0 if using gradient of tangent. Obtain equation of normal y = − 2 x + 5 A1 CAO Require y = mx + c and exact m and c 2 Accept y = − x + 5. 2 4

Mark scheme

7(a) k1 k2 M1 k1 , k2 ≠ 1 and B ≠ 1 or 4. Differentiate to obtain or 1 + Bx 2 1 x2 + B 4 A1 1 1 Obtain OE, e.g. × . 1 + 16x 2 4 1 x2 + 16 dy 1 15 15 A1 Or exact equivalent. Set = and obtain x = ± or ± dx 4 4 16 Alternative Method for Question 7(a) dx M1 Rewrite as tan y = 4x and obtain sec2 y = k dy dx A1 OE Replace sec2 y with 1 + tan2 y and tan y with 4x to obtain 1 + 16x2 = 4 dy dx 15 15 A1 Or exact equivalent. Set = 4 and obtain x = ± or ± dy 4 16 3 7(b) B *M1 OE Commence integration by parts and reach Ax tan −1 ( 4 x ) ± x dx 1 + Cx 2 A, B, C ≠ 0 4x A1 OE Obtain x tan −1 ( 4 x ) − dx 1 + 16 x 2 B A1FT OE SOI Complete integration and obtain x tan −1 ( 4 x ) ± 2C ( ln 1 + 16 x 2 ) FT their B and their C. 1 ( ) Expect x tan −1 ( 4 x ) − ln 1 + 16 x 2 if correct. 8 Substitute limits correctly in an expression of the form DM1 D ≠ 0 ( Ax tan −1 ( 4 x ) − D ln 1 + Cx 2 ) Allow omission of substitution of x = 0. 1 1 A1 ISW Obtain answer π − ln 2 or exact equivalent 16 8 1 1 1 Allow, for example, π+ ln or 16 4 2 1 1 √2 1 1 1 π + ln or π + ln . 16 4 2 16 8 2 Alternative Method for Question 7(b) *M1 OE Ay sec y dy and 2 Use substitution 4x = tan y to find A, B ≠ 0 commence integration by parts to reach Ay tan y ± B tan ydy 1 1 A1 OE Obtain y tan y − tan ydy 4 4 7(b) Complete integration and obtain Ay tan y B ln cos y A1FT OE FT their B. 1 1 Expect y tan y + ln cos y if correct. 4 4 Allow A1FT even if errors in substitution of limits into 1 y tan y if correct earlier. 4 Substitute limits correctly in an expression of the form Ay tan y ± B ln cos y DM1 Allow omission of substitution of x = 0. 1 1 A1 1 1 √2 1 1 2 Obtain answer π − ln 2 or exact equivalent Allow e.g. π+ ln or π − ln 16 8 or 16 4 2 16 4 √2 1 1 1 π + ln . 16 8 2 5

Mark scheme

8(a) Sketch y = sec 2x for 0 ⩽ x ⩽ 12 π M1 Need 1 or –1 and 1 π or 1 π. 4 2 Ignore regions outside 0 ⩽ x ⩽ 12 π. Sketch y = –2x – 12 for 0 ⩽ x ⩽ 12 π and justify the given statement A1 Need a dot at the intersection of graphs, or dotted line parallel to the y-axis from where graphs cross to the x-axis, or state only one point of intersection OE. Do not allow, e.g. ‘only one root’. Ignore regions outside 0 ⩽ x ⩽ 12 π. Diagram for reference 2 8(b) Calculate the values of a relevant expression or pair of expressions at x = 0.8 and M1 M1 two values attempted and at least one correct in x = 1.2 f(x) = sec 2x +2 x + 12 : f(0.8) = –32.1 < 0, f(1.2) = 1.54 > 0. Can use smaller interval provided it contains root M1 four values attempted and at least three correct M0 if working in degrees when comparing sec 2 x and –2 x – 12 : 1 Using 0.8: sec 2 x = –34.2 –2 x – = –2.1 , 2 so –34.2 < –2.1. 1 Using 1.2: sec 2 x = –1.36 –2 x – = –2.9, 2 so –1.36 > –2.9. M1 two values attempted and at least one correct in 1 −2 f(x) = cos −1 −x: 2 4x + 1 f(0.8) = 0.234 > 0, f(1.2) = – 0.239 < 0. M1 four values attempted (must see 0.8 and 1.2 explicitly, not just embedded) and at least three correct when −2 comparing x and 12 cos −1 : 4x + 1 1 −2 Using 0.8: cos −1 = 1.03, so 0.8 < 1.03. 2 4x + 1 1 −2 Using x = 1.2 cos −1 = 0.961 , so 1.2 > 0.961 2 4x + 1 Complete the argument correctly with correct calculated values A1 If accurate to only 1sf, M1A0. < 0 and > 0 or change of sign is sufficient For A1, answers must be correct to at least 2 sf 2 8(c) 1 −2 1 −2 M1 Need consistent variable, could be xn or xn+1. Express xn +1 = cos −1 as x = cos −1 2 4 xn + 1 2 4x + 1 1 −2 1 A1 AG Rearrange x = cos −1 to sec 2 x = −2 x − with full and correct working, 2 4x + 1 2 −2 Full working should include cos2x = or no slips allowed 4x + 1 1 1 1 −2 x − = or cos2 x = . 2 cos 2x 1 −2 x − 2 Alternative Method for Question 8(c) 1 1 −2 M1 −2 1 Rearrange sec 2 x = −2 x − to x = cos −1 after full and correct working, Full working should include = or 2 2 4x + 1 4x + 1 sec 2x no slips allowed −2 1 = cos 2x or cos2 x = . 4x + 1 1 −2 x − 2 1 −2 1 −2 A1 AG Express x = cos −1 as xn +1 = cos −1 2 4x + 1 2 4 xn + 1 2 8(d) Use the iterative formula correctly at least twice (consecutive) even if only 3 M1 M0 if first value is not between 0.8 and 1.2 inclusive. decimal places M0 if working in degrees. Obtain final answer [x = or α = ] 0.992 A1 A0 if state xn = 0.992. Show sufficient iterations to at least 5 d.p. to justify 0.992 to 3 d.p. or show there A1 Iterations for each starting value: is a sign change in the interval (0.9915, 0.9925) 0.8, 1.03356, 0.98547, 0.99373, 0.99226, 0.99252, 0.99247, 0.99248 0.9, 1.1030, 0.98938, 0.99303, 0.99238, 0.99250, 0.99248, 0.99248 1, 0.99116, 0.99271, 0.99244, 0.99249 1.1, 0.97510, 0.99560, 0.99193, 0.999258, 0.99246, 0.99248 1.2, 0.96143, 0.99813, 0.99148, 0.99265, 0.99245, 0.99248 3

Mark scheme

9(a) B C B1 Dx + E F P Qx + R State or imply the form A + + + and + B0 3x − 2 x + 6 3x − 2 x + 6 3x − 2 x + 6 Values for A, B and C do not need to be substituted into this form to gain full However, can recover all marks from these marks S T + B0 and can only gain maximum M1A1 3x − 2 x + 6 Use a correct method for finding a constant M1 Obtain one of A = 4, B = 6 and C = –5 A1 Allow maximum M1A1 for one or more ‘correct’ values or F = − 5 or P = 6 S T or S = 6 or T = –5 after B0, even if from + . 3x − 2 x + 6 D = 12 E = − 2 F = − 5 P = 6 Q = 4 R = 19 S = 6 T = –5 Dx + E B Qx + R C A1 Obtain a second value from = A+ or = A+ 3x − 2 3x − 2 x+6 x+6 Obtain a third value A1 Alternative Method for Question 9(a) Divide numerator by denominator and reach quotient of 4 and remainder of M1 Or by inspection Px + Q P or Q ≠ 0 −9 x + 46 −9 x + 46 A1 Obtain 4 + or 4 + ( 3x − 2 )( x + 6 ) 3x + 16 x – 12 2 D E B1 State or imply their remainder is of form + 3x − 2 x + 6 Values for D and E do not need to be substituted into this form to gain full marks 9(a) Obtain one of D = 6, E = –5 A1 Obtain a second value A1 5 9(b) Use the correct method to find the first two unsimplified terms of the expansion M1 E.g. –2−1 – 2−2 (3x), or 6−1 − 6−2(x), or 1 + 32 x or 1 − 16 x. of ( 3x − 2 ) or ( x + 6 ) or (1 − 32 x ) or (1 + 16 x ) −1 −1 −1 −1 Obtain correct unsimplified expansions up to the term in x2 of each partial A1FT The FT is on B and C, fraction B 3 C 1 1 2 2 3 B=6C=−5A=4 A1FT e.g. 1 + x + x + 1 − x + x OE. −2 2 2 6 6 6 1 157 1463 2 A1 OE − x− x Do not ISW, e.g. multiplication through by 216. 6 36 216 Allow terms in any order. 4

Mark scheme

10(a) For reference A = (2, −1, − 6), B = (b, − 2, 3), C = (−4, 5, − 2) Allow column vectors throughout. Allow coordinates throughout except in 10(b). Allow their notation for vectors throughout Carry out a correct method for finding AB or BC M1 E.g. AB = ( b − 2 ) i + ( −2 + 1) j + ( 3 − −6 ) k Allow if use BA for AB or CB for BC or BC = ( −4 − b ) i + ( 5 − −2 ) j + ( −2 − 3) k. Correct method to form equation with their AB = their BC M1 ( b − 2 ) + 12 + 92 = ( −4 − b ) + 72 + 52 allow 2 2 May see one further slip. Note that M0M1 is possible. Obtain [b =] − 13 A1 3 10(b) Find OA + their BC or OC − their AB M1 E.g. ( 2 − 11 ) i + ( −1 + 7 ) j + ( −6 − 5) k 3 OD = OC + BA M1 or ( −4i + 5 j − 2k ) − ( ( b − 2 ) i + ( −2 + 1) j + ( 3 − −6 ) k ) OD = OC − BA M0 = ( −4 + 73 ) i + ( 5 + 1) j + ( −2 − 9 ) k. OD = OA + BC M1 May equate the midpoint of AC and BD to find OD: OD = OA − BC M0 1 2 (OA + OC ) = (OB + OD ) . 1 2 Incorrect order of vertices scores M0. Obtain OD = − 53 i + 6 j − 11k A1 – 53 – 53 i Allow 6 but not 6 j . −11 −11k OD = + 53 i − 6 j +11k scores M1 A0. OD = ( – 53 , 6, −11) scores M1 A0. 2 10(c) Carry out correct process for evaluating the scalar product of their ± BA and M1 E.g. ( 7 × − 11 ) + (1× 7 ) + ( −9 × −5) , 3 3 their ± BC 77 391 or − + 7 + 45 or . 9 9 Using the correct process for the moduli, divide the scalar product by the M1 391 391 product of the moduli for their pair of vectors and obtain cosine of angle (allow E.g. cosine of angle = 9 or 9 unsimplified form as in above scalar product) AB BC AB AB 7 × − 11 + 1× 7 + −9 × −5 ( ) ( ) = 3 3 49 121 + 1 + 81 + 49 + 25 9 9 391 9 391 787 = = 0.4968 787 9 AB = BC , so may be expressed differently. Obtain answer 60.2 or 1.05 A1 3

Mark scheme

11 Separate variables correctly B1 1 Sight of sufficient for f(y) in f(y)dy = g(x)dx to obtain e3 y B1. 1 B1 Obtain term − e −3 y 3 1 B1 Separate fractions and obtain term 2 ( ln x 2 + 3 ) Separate fractions and obtain term of the form a tan −1 bx M1 2 x A1 OE Obtain term − tan −1 3 3 Use x = 0, y = 0 to evaluate a constant or as limits in a solution containing terms M1 ( ) of the form a e ±3 y , b ln x 2 + 3 and c tan −1 mx Obtain correct solution in any form relating x and y A1 1 −3 y 1 2 x 1 1 E.g. − e = ln x + 3 − 3 2 2 ( 3 ) tan −1 − − ln 3. 3 3 2 1 1 Constant − − ln 3 may be – 0.883. 3 2 Obtain – 0.331 A1 OE AWRT. E.g. –0.33084… Note A0A1 is possible. 8