Mathematics 9709/31 October/November 2025: question paper and mark scheme

Mark scheme

1 Commence integration by parts and reach Ax ln 3x + Bdx *M1 1 Accept Ax ln 3x − B x × dx. Allow unsimplified. x Obtain x ln3x – x A1 Substitute limits correctly in an expression of the form Ax ln3x + Bx DM1 2ln 6 − 2 − ln3 + 1 Obtain –1 + ln12 A1 If no working seen, no marks are available. 4

Mark scheme

2(a) State or imply 2 = log4 42 B1 Use log power, product or quotient law e.g. 2log 4 (3x – 1) = log4 (3x – 1)2 M1 E.g. log4 (3x – 1)2 – log 42 = log ( 3x − 1)2 4 4 42 Obtain a correct equation in any form, free of logs A1 E.g. 16(2x + 1) = (3x – 1)2 or 9x2 – 38x – 15 = 0 3 2(b) Solve 3-term quadratic M1 Obtain answer 4.59 only A1 19 + 4 31 OE. 9 Accept AWRT 4.59. 2

Mark scheme

3(a) Obtain 3sinx + 4cosx B1 State R = 5 B1 FT FT their 3 and 4. 3 M1 Using their 3 and 4. Use trig formula to find α e.g. α = tan −1 If using cosα = 4 and sinα = 3 seen, this scores 4 M0A0. Obtain α = 36.9° (or awrt 36.9° ) A1 4 3(b) −4 B1 FT FT their R from (a). Evaluate cos −1 R −4 Allow cos −1 for B1 143.13….. 5 Carry out a correct method to find a value of θ in the given interval M1 Obtain answer, e.g. 60(.0)° A1 Answers may be more accurate for A marks. Accept AWRT 60.0. Obtain a second answer, e.g. 84.6° and no other in the interval A1 Ignore answers outside the given interval. Accept AWRT 84.6. 4

Mark scheme

4 Use correct product rule: p sin 4 xesin 2 x + q cos 2 x cos 4 xesin 2 x M1 Condone sign error. Condone if no attempt at chain rule seen. Obtain derivative −4sin 4 xesin 2 x + 2cos 2 x cos 4 xesin 2 x A1 OE Equate derivative to zero and obtain an equation in one trig function M1 Allow arithmetical errors. Obtain 2sin 2 2 x + 4sin 2 x − 1 = 0 A1 OE Obtain x = 0.113 and no other root A1 May be more accurate. 5

Mark scheme

5(a) Obtain Re z 2 B1 Allow 2 Re( z ) 3 Condone strict inequalities throughout (a). The shading is to the RHS of the diagram. Obtain answer of the form z − a b M1 Accept equation or any inequality sign. Must have a complex and b real. Condone brackets in place of modulus for this mark. Obtain answer z − 1 + 2i 2 A1 OE z − (1 − 2i ) 2 3 5(b) Identify the correct point B1 May be clearly indicated as a point on the diagram. ( May be 2, − 3 − 2 . ) May be by attempt to solve (x – 1)² + (y + 2)² = 2² with x = 2. Carry out a correct method for finding the least value of arg z M1 Obtain answer –61.8° or 298.2° or –1.08c or 5.2(0) c A1 3

Mark scheme

6 Use quadratic formula to solve for w M1 −4 ± 16 − 4 ( 2 − i )( 2 + i ) w= 4 + 2i Use i2 = –1 M1 −4 − 2i −4 + 2i A1 OE Obtain one of the answers w = or w = 4 + 2i 4 + 2i Multiply numerator and denominator by conjugate of their denominator M1 Need some evidence of multiplying. 3 4 A1 Obtain final answers − + i and –1 [+ 0i] 5 5 Alternative Method for Question 6 Multiply the equation by 2 – i M1 Use i2 = –1 M1 Obtain 5w2 + 4(2 – i)w + (2 – i)2 A1 OE Use quadratic formula or factorise to solve for w M1 3 4 A1 Obtain final answers − + i and –1 [+ 0i] 5 5 Alternative Method 2 for Question 6 Substitute w = x + iy and form equations for real and imaginary parts M1 Two separate equations. Use i2 = –1 M1 Obtain 2(x2 – y2 ) – 2xy + 4x + 2 = 0 and x2 – y2 + 4xy + 4y – 1 = 0 A1 OE Form equation in x only or y only and solve M1 3 4 A1 Obtain final answers − + i and –1 [+ 0i] 5 5 5

Mark scheme

7 dx 2 B1 State = 2t − dt 2t + 1 Use correct quotient rule or correct product rule M1 Might split first as 12 − 4t1+ 2 . dy ( 2t + 1) .1 − t ( 2 ) A1 Simplifies to 1 Obtain = oe dt ( 2t + 1)2 ( 2t + 1)2 dy dy dx M1 dx Use = ÷ Allow for . dx dt dt dy 1 A1 OE. Must have 1 in numerator. Obtain answer 2 ( 2t + 1)( 2t − 1)( t + 1) 1 May have, e.g., . ISW. ( 2t + 1) 4t 2 + 2t − 2( ) y 2y 2 B1 y 2 1 Obtain x = − ln + 1 OE, e.g. x = − ln . 1− 2y 1− 2y 1− 2y 1− 2y Use correct quotient rule to differentiate M1 dx 2 y (1 − 2 y ) + 2 y 2 A1 dy (1 − 2 y )3 (1 − 2 y )3 OE, = = Obtain = − 2 y − 2 (1 − 2 y ) −2 (1 − 4 y )(1 − y ) dy 1 − 2 y (1 − 2 y ) 2 2 1− 2y dx dy (1 − 2t ) 3 1 M1 Express in terms of t. 2 t +1 = = dx −2 (1 − 24t +t 1 )(1 − 2 tt+1 ) −2 ( 2t + 1)(1 − 2t )( t + 1) 1 A1 OE Obtain answer 2 ( 2t + 1)( 2t − 1)( t + 1) 5

Mark scheme

8 Separate variables correctly B1 kx 2 dx = e−2 y dy x +1 Condone missing integral signs or missing dx, dy, but not both. Obtain term Ae −2 y B1 OE − 12 e−2 y ( Obtain term B ln x2 + 1 ) B1 OE k 2 ( ln x 2 + 1 ) ( ) Obtain − 12 e−2 y = k2 ln x2 + 1 ( +C ) B1 OE Use y = 0 when x = 0 to evaluate c or as limits in a solution containing terms of M1 ( the form e −2 y and ln x 2 + 1 ) Use y = − 12 when x = 1 to evaluate k or as limits in a solution containing terms of M1 ( the form e −2 y and ln x 2 + 1 ) 1− e A1 1− e Obtain c = – 1 2 and k = ln 2 1 OE, e.g. − e−2 y = 2 2ln 2 ( ) 1 ln x2 + 1 − . 2 1 A1 Or exact equivalent. Obtain final answer y = − ln ( 2e − 1) 2 8

Mark scheme

9(c) ( ) and rearrange to obtain sec 2x = –e State the equation x = 12 cos−1 −e− x x B1 Or rearrange sec 2x = –e as x = cos ( −e ) and state x = cos ( −e ) x 1 2 −1 −x n +1 1 2 −1 − xn 1 9(d) Use the iterative process correctly at least once M1 Obtain final answer 0.978 A1 Show sufficient iterations to at least 5 d.p. to justify 0.978 to 3 d.p. or show there A1 E.g. is a sign change in the interval (0.9775, 0.9785) 1, 0.97376, 0.97903,0.97796, 0.97813 0.95, 0.98395, 0.97697, 0.97838, 0.97809 0.9, 0.99475, 0.97480, 0.97882, 0.97800, 0.97817 3

Mark scheme

10(a) B Cx + D B1 State or imply the form A + + 3 + x 2 + x2 Use a correct method for finding a constant M1 −3x2 − 17 Might be working from 1 + . (3 + x)(2 + x2 ) Obtain one of A = 1, B = –4, C = 1 and D = –3 A1 SC: If B0 scored due to missing term(s), then a maximum of M1A1 is available for a correct method leading to a correct value. Obtain a second value A1 SC if obtaining A = 1 and then scoring B0, they can score maximum M1A1 + A1 for a correct value for one other constant. Obtain a third value A1 Obtain all four correct values A1 6 10(b) Use a correct method to find the first two terms in the expansion of (3 + x) –1 , M1 Symbolic binomial coefficients not sufficient for the −1 −1 M1. x x2 1 + , (2 + x ) or 1 + 2 –1 3 2 Obtain correct unsimplified expansions up to the term in x2 of each partial A1 FT FT B, C and D. fraction A1 FT B x x2 Cx + D x 2 E.g. 1 − + and 1 − . 3 3 9 2 2 x2 −1 M1 x2 −1 Multiply (Cx + D) by the expansion of 1 + up to the term in x2 where Expansion of 1 + . 2 2 CD ≠ 0 Must be of the form 1 + Qx². Multiplication must include 3 relevant terms. 11 17 65 2 A1 Or exact equivalent. Obtain final answer − + x+ x Do not ISW attempts to multiply expression by, e.g., 6 18 108 108. 5

Mark scheme

11(a) Carry out a correct method for finding a vector equation for m M1 1 1 A1 0 1 Obtain r = 5 + λ 1 Or r = 4 + λ 1 3 2 1 2 OE. Must have r = … 2 11(b) Express general point of the second line in component form, B1 Or ( 3 + 2μ , − 5 − 2μ , 4 + 3μ ) . i.e. (1 + 2µ, –3 – 2µ, 1 + 3µ) Equate at least two pairs of corresponding components and solve for λ or for µ M1 Obtain λ = –4 or µ = –2 A1 −1 Note if m has direction vector −1 , then λ = 4. −2 −2 If line through CD has direction vector 2 , then −3 µ = 2. −3 A1 Accept coordinates. Obtain position vector of point of intersection is 1 −5 4 11(c) 0 1 B1 −1 1 Find CP for a general point P on m, e.g. 8 + λ 1 Or 7 + λ 1 . 2 2 0 2 Calculate scalar product of CP and a direction vector for m and equate the result M1 to zero Obtain λ = –2 A1 OE −1 A1 OE Obtain answer 3 −1 4