Mathematics 9709/33 October/November 2025: question paper and mark scheme

Mark scheme

1 State or imply non-modular inequality (3 x + 2) 2 < 32 (2 x − 1) 2 , B1 Allow ‘=’, or any inequality sign. or pair of linear equations (3x + 2) = ± 3(2x – 1) Make reasonable attempt at solving a 3-term quadratic, or solve two linear M1 E.g. 27x2 – 48x + 5 = 0, (9x – 1)(3x – 5) = 0, x = … equations for x Allow even if quadratic not given in a 3-term form. See guidelines document for solving a quadratic. 1 5 A1 Allow ‘=’, or any inequality sign. Obtain critical values x = and x = 9 3 1 5 A1 Allow ‘OR’ but not ‘AND’. State final answer x < , x > only Allow ‘ ∪ ’ but not ‘ ∩ ’. 9 3 No marks can be scored if no working is seen. Alternative Method for Question 1 5 B1 Allow ‘=’, or any inequality sign. Obtain critical value x = from a graphical method, or by inspection, or by 3 solving a linear equation or an inequality 1 B2 Allow ‘=’, or any inequality sign. Obtain critical value x = similarly 9 1 5 B1 Allow ‘OR’ but not ‘AND’. State final answer x < , x > only Allow ‘ ∪ ’ but not ‘ ∩ ’. 9 3 No marks can be scored if no working is seen. 4

Mark scheme

2 Commence division and reach partial quotient of the form 3x3 ± 3x2 M1 May be seen in synthetic division. or 3x4 – 2x2 ≡ (x + 1)(Ax3 + Bx2 + Cx + D) + Ex + F, and reach A = 3 and B = ± 3 Obtain quotient 3x3 – 3x2 + x – 1 Do not ISW A1 Don’t need to state which is the quotient and which is remainder. However, if clearly muddled, then M1A1A0 for both expressions correct. Obtain remainder of 1 A1 Do not ISW. 1 Allow e.g. 3x3 – 3x2 + x – 1 + but NOT x +1 1 remainder = . x +1 Alternative Method for Question 2 f (–1) = 3 – 2 = 1 = remainder B1 Do not ISW. Use division or inspection or compare coefficients M1 3x4 − 2x2 – 1 ≡ (x + 1)(3x3 – 3x2 + x – 1) Obtain quotient 3x3 – 3x2 + x – 1 A1 Do not ISW. 1 Allow e.g. 3x3 – 3x2 + x – 1 + but NOT x +1 1 remainder = . x +1 3

Mark scheme

3 3 M1 May work in log to any base for first 3 marks. Use log quotient law, e.g. ln x = ln 3 – ln 5 x 5 or log product law, e.g. ln5x + ln(23 x – 4 ) Use log power law, e.g. ln 5x = x ln5 or ln (23 x – 4 ) = (3x – 4) ln 2 M1 Condone missing brackets if recovered at some point. ln 3 + 4 ln 2 A1 Obtain correct expression for x in any exact form, e.g. 3ln 2 + ln 5 ln 48 A1 Do not ISW. Obtain final answer Final answer of log40 48 scores 3 marks only. ln 40 No working award 0 marks. Alternative Method for Question 3 Rearrange to obtain (23 × 5)x = 3 × 24 B1 OE Use log power law on an equation of the form abx = c M1 May work in log to any base for first 3 marks. e.g. x ln (23 × 5) = ln (3 × 24 ) Obtain correct expression for x in any exact form, e.g. x = ( ) ln 3 × 24 A1 ln ( 2 × 5 ) 3 ln 48 A1 Do not ISW. Obtain final answer Final answer of log40 48 scores 3 marks only. ln 40 No working award 0 marks. 4

Mark scheme

4 Show a circle centre (0, –2) B1 For all marks: Accept a scale or dashes representing a scale or points labelled. If scale only on one axis, allow this to imply the same scale on the other axis. Condone dashed circle and dashed perpendicular bisector for all marks Show a circle with radius 3 B1FT FT centre not at the origin. Allow circle with 3 radii correct out of 4 ‘compass directions’. If no indication of scale, allow SC B1FT only for circle. Show the point representing (2, – 4) or the midpoint (1, –3) B1 May be implied by a correct perpendicular bisector. Condone if (2, –4) on the circle. Show the perpendicular bisector of the line joining (2, – 4) and (0, –2) B1FT FT is on the positions of (2, – 4) and (0, –2) or on the or the perpendicular bisector of the line joining (2, – 4) and centre of their circle position of (2, – 4) and centre of their circle or on the or the perpendicular bisector going through (1, –3) position of (1, –3). Cuts (or would cut) x-axis between 3 and 5 and y-axis between –5 and –3 if everything else correct. If no indication of scale, allow SC B1FT only for perpendicular bisector. Shade the correct region B1 Dependent on all previous marks. Allow SC B1 for correct shading if the perpendicular bisector looks correct and the only error is that it is slightly out when crossing the axes. If no indication of scale, allow SC B1 for shading if everything is relatively correct. 5

Mark scheme

5(a) Use correct double angle formula to express cos 4x in terms of sin2 2x B1 Need to see 1 – 2 sin2 2x or (1 – sin2 2x) – sin2 2x. May be implied by further work. Use correct double angle formula to express cos 4x in terms of single angles M1 Allow with cos x but not cos 2x. E.g. 1 – 2(2sin x cos x)2 or 2(1 – 2sin2 x)2 – 1. Obtain a correct expression in sin x in any form A1 E.g. 1 – 2[4 sin² x(1 – sin² x)] + 2 sin² x – 1, or 2(1 – 2 sin2 x)2 – 1 + 2 sin² x – 1. Obtain the given answer 8 sin4 x – 6 sin² x A1 AG Must show at least one intermediate line of working including sin4 x between first A1 and AG. Allow, e.g., A for x except in the final answer. Allow recovery on the next line after a slip. Allow recovery on the next line after missing x. Alternative Method for Question 5(a) Use correct double angle formula to express whole expression in terms of co s2x M1 E.g. (1 – cos 2x)(2 – 2 cos 2x – 3). Use correct double angle formula to express 2 cos2 2x – 1 as cos 4x B1 Use correct double angle formula to express whole expression as cos 4x – cos2x A1 Obtain the given answer cos 4x + 2sin2 x – 1 A1 AG Allow, e.g., A for x except in the final answer. Allow recovery on the next line after a slip. Allow recovery on the next line after missing x. 4 5(b) Obtain answers 0°, 180° and –180° B1 Carry out a correct method to find a value of x in the given interval for M1 3 8 sin2 x – 6 = 0 Condone a wrong value of x if sin x = OE 4 seen. Allow M1A1A1 if dividing by sin2 x, but B1 is not scored. Obtain answer, e.g. 60° A1 In radians, would be 13 π. Obtain remaining answers, e.g. –60°, 120° and –120° and no other in the interval A1 Ignore answers outside the given interval. 4

Mark scheme

6 Commence integration by parts and reach Ax 2 cos 2 x − ∫ Bx cos 2 xdx *M1 OE Condone sign error in formula. 1 2 A1 OE Obtain − x cos 2 x + ∫ x cos 2 xdx Unsimplified. 2 Complete integration by parts and reach Ax 2 cos 2 x + Bx sin 2 x + C cos 2 x *M1 OE ∫ Bx cos 2 xdx = Bx sin 2 x + C cos 2 x may be written separately to Ax 2 cos 2 x for M1, but not for A1. 1 2 1 1 A1 OE Obtain − x cos 2 x + x sin 2 x + cos 2 x Unsimplified. 2 2 4 Substitute limits correctly in an expression of the form DM1 2 π 1  π  3  1 Ax 2 cos 2 x + Bx sin 2 x + C cos 2 x and evaluate the trig expressions A     + B     + C   − C 6 2  6  2  2 Allow one slip, including omitting the ‘– C’. Do not allow only decimals 1 2 3 1 A1 ISW Obtain answer − π + π − or exact three-term equivalent Allow equivalent fractions. 144 24 8 1 Allow 0.125 for . 8 6

Mark scheme

7 Attempt to remove fraction by multiplying throughout by 2 – i *M1 Allow if still includes brackets and/or i2 . 2+i E.g. 5z – zz* (2 – i) + (20 + 8i)(2 – i) = 0 or better. or multiply fraction by 2+i 5z 2 + i E.g. × – zz* + 20 + 8i = 0 or better. 2−i 2+i Substitute z = x + iy and z* = x – iy throughout the equation B1 May see 5 x + 5iy − 2( x 2 + y2 ) + i(x2 + y2 ) + 48 – 4i = 0, or 2 x + ix + 2iy − y − x 2 − y 2 + 20 + 8i =0. Use i2 = –1 correctly at least once and equate real and imaginary parts to zero *DM1 OE, e.g. 5 x − 2 x 2 – 2y2 + 48 = 0 and 5 y + x2 + y2 – 4 = 0, or 2 x − y − x 2 − y 2 + 20 =0 and x + 2y +8 =0. For their horizontal equation. Obtain two correct equations e.g. 5x – 2(x2 + y2 ) + 48 = 0 A1 E.g. 2x – y – x2 – y2 + 20 = 0 and x + 2y + 8 = 0. and 5y + (x2 + y2 ) – 4 = 0 Allow 5iy + i(x2 + y2 ) − 4i = 0 or ix + 2iy + 8i = 0. Solve a quadratic and a linear equation for x or for y DM1 16 12 A1 16 12 Obtain answers 2 – 5i and − − i only Accept x = 2 y = –5 and x = − y= − only 5 5 5 5  16 12  OE, or (2, –5) and  − , −  only OE.  5 5 Allow decimals. Do not ISW. 6

Mark scheme

8(a) d d −5 x M1 M0 if y = e−5 p ln 5p seen prior to differentiation. Use the correct product or quotient rule, e.g. e−5 x (ln 5x) + ln 5x (e ) Accept if only seen in actual derivative = 0. dx dx 1 −5 x A1 Obtain the correct derivative in any form e.g. e − 5e −5 x ln 5 x x 1 A1 AG Obtain the given answer ln 5 p = after full and correct working 5p 1 −5 x May go from e − 5e −5 x ln 5 x = 0 , or x 1 −5 x e = 5e −5 x ln 5 x to the given answer without x intermediate working. 3 8(b) 1 M1 For both marks: Sketch an acceptable graph, e.g. y = ln 5x or y = 5x Note: Allow without scale on either axis, but if y = ln5x = 0 identified to be not x = 0.2, then 0 marks for y = ln 5x. Allow graphs not labelled, or labelled with p instead of x. Allow ln 5x starting at the x-axis. If either graph shown in other quadrants, must be correct. 1 For y = , asymptotic behaviour needed for at 5x least one axis. Must not touch axes. 1 A1 Sketch a second acceptable graph, e.g. y = or ln 5x, and justify the given 5x statement by dot, cross or statement only one intersection. 2 8(c) Calculate the values of a relevant expression or pair of expressions at p = 0.2 M1 1 and p = 0.6 f(p) = ln 5 p − 5p Note can use, e.g., p = 0.3 and p = 0.5, or any smaller interval which works f(0.2) = –1 < 0, f(0.6) = 0.765 > 0 At least one correct value to at least 2sf. 1 Or comparing ln 5 p and . 5p At least 3 correct values to at least 2sf. Complete the argument correctly with correct calculated values A1 2 8(d) Use the iterative formula correctly at least twice M1 M0 for 0.3526, 0.3526, 0.3526… Obtain final answer p = 0.35, Answer = 0.35, or just 0.35 stated A1 Allow, e.g., a1 , a2, a3 … or x1 , x2 , x3 … or answer1 , answer2 , answer3 … for M1 and second A1. For first A1, must be p = 0.35 or answer = 0.35 unless just 0.35 is stated, e.g. not x = 0.35, p7 =…, p∞ = … etc. Show sufficient iterations to 4 dp to justify 0.35 to 2 dp or show there is a sign A1 E.g. 0.4, 0.3297, 0.3668, 0.3450, 0.3571, 0.3502, change in the interval (0.345, 0.355) 0.3541. Allow M1(A1 or A0) A1 if more values are to at 0.2, 0.5437, 0.2889, 0.3996, 0.3299, 0.3667, 0.3451, 0.3571, 0.3502, 0.3541 least 4dp than to 3dp. 0.25, 0.4451, 0.3135, 0.3786, 0.3392, 0.3607, 0.3482, 0.3552, 0.3512, 0.3535 SC B1 for starting from either 0.3526 or 0.3527 and 0.3, 0.3895, 0.3342,0.3639, 0.3465, 0.3562, 0.3507, 0.3538 0.3520, 0.3530 using iterative formula correctly at least twice if the 0.45, 0.3119, 0.3797, 0.3387, 0.3610, 0.3480, 0.3553, 0.3512, 0.3535 sequence shows a correct change in the 4th decimal 0.5, 0.2984,0.3910, 0.3336, 0.3643, 0.3463, 0.3563, 0.3506, 0.3538 place (and SC DB1 for getting p = 0.35), but 0 0.55, 0.2877, 0.4008, 0.3294, 0.3670, 0.3449, 0.3572, 0.3501, 0.3541 marks otherwise. 0.6, 0.2791, 0.4095, 0.3260, 0.3694, 0.3437, 0.3579, 0.3497, 0.3543 3

Mark scheme

9(a) Obtain r = 3i + j – 6k + λ (2i + j + 4k) B1 OE Must have r = …, but penalise missing r = only once in (a). Do not allow column vectors with i, j and k included. Carry out a correct method for finding a direction vector M1 E.g. –2y + z = 0 SOI. for l2, (3i – 2j +k)·(yj + zk) = 0 Obtain r = –i + 3j – 6k + µ(j + 2k) A1 OE Must have r = …, but penalise missing r = only once in (a). Do not allow column vectors with i, j and k included. 3 9(b) Carry out correct process for evaluating the scalar product of the direction *M1 Using their direction vectors from (a). vectors of l1 and l2 Ignore symbol if state e.g. ‘×’ in place of ‘·’. Allow the same parameter for both lines here. Using the correct process for the moduli, divide the scalar product by the product DM1 Using their direction vectors from (a). of the moduli and state cos θ = the result Obtain answer 28.6° or 0.498c A1 3 9(c) Equate components of general points on their l1 and their l2 , provided these are M1 (3 + 2λ, 1 + λ, −6 + 4λ) = (−1, 3 + μ, −6 +2μ) both equations of lines and solve for λ or for μ Do not allow the same parameter for both lines here if solving using two linear equations. Allow M1 for 3 + 2 λ = –1 ⇒ λ = … even if the other equations are incorrect or not stated. Obtain correct answer for λ or μ, e.g. λ = –2, μ = – 4 A1 Allow M1A1 for 3 + 2λ = –1 leading to λ = –2, even if the other equations are incorrect or not stated. Obtain position vector of point of intersection is –i – j – 14k A1 OE Do not accept coordinates. Allow even if j and k equations are incorrect or not stated. 3

Mark scheme

10(a) Carry out a relevant method to find A and B such that M1 OE 2 A B A B = 2 + o Allow M1 for finding A and B for + 1− 9 y 1+ 3y 1− 3y 1+ 3 y 3 y −1 and A1 for A = 1, B = –1 if –2 = A(3y – 1) + B(1 + 3y). But, A0 for A = –1, B = 1 if 2 = A(3y – 1) + B(1 + 3y). Obtain A = 1 and B = 1 A1 If work with x and never see y, award M1A0, but allow M1A1 if y is seen anywhere on right hand side. 2 10(b) Separate variables correctly and attempt integration of at least one side M1 Integrate to obtain at least one log term of the form a + by p ln (a + by) OE, q ln on one side, or a tan a − by term on the other, and disregard the 2 if it appears. 2 1 1 A1FT 1  1+ 3y  Integrate to obtain ln (1 + 3 y ) − ln (1 − 3 y ) OE, e.g. ln  . 1− 9 y 2 3 3 3  1− 3y  A B FT ln (1 + 3 y ) − ln (1 − 3 y ) 3 3 A B or ln (1 + 3 y ) + ln ( 3 y − 1) if their partial 3 3 fractions used. The ‘2’ must have been dealt with correctly for this mark (check right hand side for 2 appearing here) Obtain r tan 3x B1 1 B1 1 Obtain term tan 3x Allow tan 3 x if ‘2’ not dealt with correctly 3 6 earlier. 1 M1 0 + 0 +1/3 + C = 0 Use y = 0 when x = π to evaluate a constant or as limits in a solution of the No errors in substitution. 12 form p ln (1 + 3 y ) + q ln (1 − 3 y ) + r tan 3x where p,q,r ≠ 0 1 1 1 1 A1 OE ln (1 + 3 y ) − ln (1 − 3= y) tan 3 x − ISW 3 3 3 3 1 2 e tan 3 x −1 − 1 Obtain answer y= − or y = ( 3 3 1 + e tan 3 x −1) ( ) 3 e tan 3 x −1 + 1 6

Mark scheme

11(a) 1 B1 OE (can be unsimplified). 3 + 2 tan x to obtain sec 2 x ( 3 + 2 tan x ) 2 − Differentiate Not dependent on product rule, but must be convincing or seen in isolation, not as a derivative of the whole expression. B0B0 for e.g. 1 2sec x sec x tan x sec 2 x ( 3 + 2 tan x ) 2 . − Differentiate sec2 x to obtain 2secx secx tanx B1 OE Not dependent on product rule but must be convincing or seen in isolation, not as a derivative of the whole expression. B0B0 for e.g. 1 2sec x sec x tan x sec 2 x ( 3 + 2 tan x ) 2 OE, e.g. − 2sin x . cos3 x Use correct product (or quotient) rule M1 d d sec2 x (√……..) + √……… (sec 2 x) dx dx [Obtain derivative, if correct M1 1 1 Must include (…) 2 and (…) 2 here. − 1 1 2sec 2 x tan x ( 3 + 2 tan x ) + sec 4 x ( 3 + 2 tan x ) ] − 2 2 Arithmetical errors only for this M1. and equate derivative to zero and obtain an equation in one trig function May work in terms of sin x and cos x. Obtain 5tan2 x + 6 tan x + 1 = 0 A1 OE, e.g. 5tan4 x + 6tan3 x + 6tan2 x + 6tan x + 1 = 0 52sin4 x – 28sin2 x + 1 = 0 52cos4 x – 76cos2 x + 25 = 0 cot2 x + 6 cot x + 5 = 0 Obtain AWRT x = – 0.197 only A1 ISW May be more accurate. 6 11(b) du B1 SOI = 2sec 2 x dx *M1 OE Reach an integral of the form ∫ A u du 1 1 A1 OE Obtain ∫ u 2 du 2 1 32 A1FT OE Obtain u FT their coefficient. 3 3 DM1 OE Substitute correct limits correctly in an expression of the form Bu 2 u = 1 and u = 5 3 1 1 or B (3 + 2 tan x) 2 x= − π and x = π 4 4  3 3  Do not allow only decimals. and obtain c  5 2 − 12    5 5 1 A1 125 − 1 Obtain answer − Or exact equivalent, e.g. . 3 3 3 ISW 6