Mathematics 9709/32 October/November 2025: question paper and mark scheme

Mark scheme

1(a) y B1 Roughly symmetrical. Condone some inaccuracy, but needs to look as if they intended symmetry and straight lines. If not, then B0. Needs to be in the correct position. 3a Needs to have two solid line segments. Needs to exist in both quadrants above the axis. Ignore dotted lines below the axis. Solid line below the axis is B0. O -3a x Condone if no scale shown, but need to see 3a and -3a marked. Ignore y = a − 2 x if seen. 1 1(b) Obtain critical value − 23a from x + 3a = a − 2x B1 Ignore x = 4a if seen. State final answer x > − 23a B1 Need a clear conclusion – must imply rejection of x = 4a. B0 if using ⩾. Alternative Method for Question 1(b) Obtain critical value − 23a from ( x + 3a ) = ( a − 2 x ) 2 2 B1 Ignore x = 4a if seen. State final answer x > − 23a B1 Need a clear conclusion – must imply rejection of x = 4a. B0 if using ⩾. 2

Mark scheme

2 Use a correct law for the logarithm of a product or the logarithm of a power *M1 Not available after incorrect manipulation, e.g. 6 x +1 = 122 x −3 Obtain ln 3 + ( x + 1) ln 2 = ln 4 + ( 2 x − 3) ln 3 A1 Or equivalent with no powers Solve for x DM1 2 ln 81 E.g. x = ln 92 Allow with logarithms evaluated. Obtain 2.46 only A1 Alternative Method for Question 2 Use a correct law of indices for a product or a power *M1 E.g. 2x+1 = 2 × 2x Not available after incorrect manipulation, e.g. 6 x +1 = 122 x −3. 4 = 22 on its own is not enough. ( 92 ) = 812 A1 OE with powers of x only, e.g. 6 × 2x = 4 × 9x x Obtain 27 ( ) is not far enough. 2 Note: 3x Solve for x DM1 2 ln 81 E.g. x = or x = ln 2 812 . ln 92 9 Allow with logarithms evaluated. Obtain 2.46 only A1 4

Mark scheme

3(a) Complete method to obtain equation in y only M1 E.g. ( 2 − 3)2 + ( y − 1)2 = 4 or Obtain the y coordinate of P Or consider triangle and state ± ( 3 − 1) . 1 3 2 ( Obtain 2 + i 1 − 3 ) A1 Or exact equivalent in the form x + iy. ( Allow 2 − i 3 − 1 . ) Must not be coordinates, and not x, y stated separately. 2 3(b) 1 2 B1 State one relevant angle, e.g. tan −1 or sin −1 Im(z) 3 10 2 i 10 1 O 3 Re(z) 1 1 Note: tan −1 = sin −1 3 10 Complete method to obtain the required angle M1 1 2 tan −1 + sin −1 3 10 Obtain 1.01 radians or 57.7° A1 Accept AWRT Alternative Method for Question 3(b) If y = mx is a tangent to the circle ( x − 3) + ( mx − 1) = 4 2 2 B1 Must be choosing the positive root. ((1 + m ) x + ( −6 − 2m) x + 6 = 0) , 2 2 then ( −6 − 2m ) − 24 (1 + m2 ) = 0 and m = 2 3+ 2 6 5 Substitute into the quadratic and solve for x, or use gradient to obtain tangent M1 9− 6 9− 6 Note x = 5 Required angle is − sin −1 5 OE π 2 6 Obtain 1.01 radians or 57.7° A1 Accept AWRT 3

Mark scheme

4 2 M1* Condone sign error in formula for integration by parts. Begin integration by parts and obtain px 2 tan −1 x + q x dx 1 + x2 x2 1 x2 A1 OE Obtain tan −1 x − dx Allow with arctan x or tan-1 x. 2 2 1 + x2 2 DM1 Split the fraction and integrate. Use x 1 dx = −1 1− dx and integrate to obtain β x + γ tan x 1 + x2 1 + x2 A1 OE ( x − tan x ) 2 x 1 −1 Obtain tan x − −1 2 2 Substitute correct limits correctly in an expression of the form DM1 π 1 π α x 2 tan −1 x + β x + γ tan −1 x 8 −0− 2 + 8 ( ) + 0 −0 Dependent on both previous M marks. Need to see evidence of the use of the lower limit at least once. Must evaluate the trigonometry using radians. π 1 A1 ISW Obtain − Or exact two term equivalent 4 2 6

Mark scheme

5(a) Use correct product rule M1 f ′ ( x ) = ( x − a ) g′ ( x ) + 2 ( x − a ) g ( x ) 2 Allow incorrect chain rule. Obtain correct derivative and state a clear conclusion A1 AG E.g. take out a factor of (x – a) and show the factorised form as far as ( x − a ) h( x) correctly (no further comment needed), or show that f ′ ( a ) = 0 and state that ‘(x – a) is a factor’. 2 5(b) Use f ( 3) = 0 M1 2 × 33 − 4 × 32 + 3 p + q = 0 Obtain 3 p + q = −18 A1 OE Powers should be evaluated but do not need to be simplified. Use f ′ ( 3 ) = 0 M1 54 − 24 + p = 0 Obtain p = −30 A1 OE Obtain p = −30, q = 72 A1 Correct only. Alternative Method for Question 5(b) Attempt division of f(x) by ( x − 3) as far as ( 2 x + Q ) 2 M1 Q≠0 Obtain quotient ( 2 x + 8 ) A1 Equate linear remainder to zero and compare coefficients M1 Or expand ( 2 x + 8)( x − 3)2 and compare coefficients. Method to obtain an equation in p or q. obtain one of p = −30, q = 72 A1 Obtain p = −30, q = 72 A1 Correct answers only. 5(b) Alternative Method 2 for Question 5(b) Use f ( 3) = 0 M1 54 − 36 + 3 p + q = 0 Obtain 3 p + q = −18 A1 OE Powers evaluated. Divide f(x) by ( x − 3) as far as ( 2 x + Q ) and form an equation in p only or q 2 M1 Using this as part of a hybrid method they need to be working towards a second equation by considering only or in p and q coefficients in the remainder or use f ( −4 ) = 0 or equivalent for their ( 2 x + 8 ) . Obtain correct equation in p or q A1 Obtain p = −30, q = 72 A1 Correct answer only. Alternative Method 3 for Question 5(b) f ′ ( x ) = 6 x 2 − 8x + p B1 = ( x − 3)( 6 x + 10 ) M1 Use the factor x − 3. Obtain p = −30 A1 f ( x ) = ( x − 3) ( 2 x + Q ) 2 ( = 2x − 4x − 30x + q ) 3 2 M1 Use the factor ( x − 3)2 . Q = 8, q = 72 A1 5(b) Alternative Method 4 for Question 5(b) Expand f ( x ) = ( x − 3) ( Ax + B ) and compare at least one coefficient other 2 M1 A = 2 can be found by inspection. than for x3 Obtain q = 9B and p = 9 A − 6 B A1 or obtain B = 8 Use their A and B to solve for p or q M1 A=2 B =8 Obtain one of p = −30, q = 72 A1 Obtain p = −30, q = 72 A1 Correct answer only. 5 f(x) = 2∙sin(2∙x) 1

Mark scheme

6(a) Sketch a relevant graph, e.g. y = cot 2 x 1 B1 2 cot(2x) g( x ) = Scales must be correct if seen. Do not need a full scale, but tan ( 2∙x) do need to be able to verify the key points mentioned below: 1 For y = cot 2 x, correct intercept on x-axis, asymptotes at x = 0, x = 12 π implied 2sin(2x)-1 (condone large gap between the curve and the asymptote on the RHS). Must be continuous. π 3π π π O π π 3π π For 2sin 2 x − 1, correct shape and position. No incorrect curvature. 4 Correct 2 4 4 2 4 maximum implied. Intercept at (0, -1). 1 Sketch a second relevant graph, e.g. y = 2sin 2 x − 1 and justify the given B1 Need 2 to see enough of both graphs to confirm there is statement not a second intersection in the given interval. Need to mark the intersection with a dot or a cross, or say that the root lies at the point of intersection (or equivalent). 2 6(b) Calculate the values of a relevant expression or pair of expressions at M1 E.g. 0.9712 > 0.435 and 0.389 < 0.864 x = 0.4and x = 0.6 0.537 > 0 and − 0.475 < 0 Or using the iterative formula, 0.18 > 0, –0.17 < 0 Allow working on a smaller interval contained within the given interval Or using tan 2 x ( 2sin 2 x − 1) − 1 = 0 – 0.55 < 0, and 1.22 > 0. M0 if working in degrees. Justify the given statement with correct calculated values A1 Values to 2 sf or better. Accept values with statement f ( 0.4 ) × f ( 0.6 ) < 0. 2 6(c) 1 1 M1 M0 if working in degrees. Use the iterative process xn+1 = tan −1 correctly at least once Need to get as far as a second iteration completed. 2 2sin 2 xn − 1 Obtain final answer 0.49 A1 Show sufficient iterations to 4 d.p. to justify 0.49 to 2 d.p. or show that there A1 E.g. 0.5,0.4858,0.4967,0.4883,0.4947,.... is a sign change in ( 0.485, 0.495 ) or 0.4, 0.5804, 0.4378, 0.5395, 0.4595, 0.5188, 0.4726, –0.0157 < 0, 0.355 > 0 0.5075, 0.4804, 0.5009, 0.4851, 0.4972, 0.4879, 0.4950, 0.4895, 0.4937 or 0.6, 0.4291, 0.5483, 0.4544, 0.5235, 0.4695, 0.5101, 0.4786, 0.5025, 0.4840, 0.4981, 0.4872, 0.4955, 0.4891, 0.4940 The last 2 options are a lot of iterations. Allow the marks if they start correctly and finish correctly. Condone if there are some missing in the middle. 3

Mark scheme

7(a) State or imply 6 y 2 ddyx as the derivative of 2 y 3 B1 State or imply 3x 2 ddyx + 6 xy as the derivative of 3x 2 y B1 Complete the differentiation and equate the derivative of the LHS to zero and M1 Needs to be clear how the value of dy was obtained, dy dx solve for dx (the = 0 can be implied) e.g. collecting like terms and using brackets. dy x 2 + 2 xy A1 AG Obtain = from correct working dx 2 y2 − x 2 dy 2 xy + x 2 Accept = . dx 2 y 2 − x2 No incorrect statements, e.g. ‘cancel by 3’ or ‘divide top and bottom by 3’ are acceptable, but to have dy 3x2 + 6 xy = and say “divide by 3” at the final step dx 6 y 2 − 3x2 scores A0. 4 7(b) Equate derivative to 0 and solve for x or for x in terms of y. *M1 E.g., x 2 + 2 xy = 0 x = 0 or x = −2 y. Must be using the numerator. Obtain ( 0, 2 ) B1 Allow if the values are stated separately. Do not ISW. Use their x = −2 y to form an equation in one unknown dM1 E.g., 2 y 3 − 12 y 3 + 8 y 3 = 16 or 2 Or equivalent e.g. substitute y = − 2x x 1 1 −2 x3 + 3 x3 − x3 = 16. 8 2 Obtain ( 4, − 2 ) A1 Allow if the values are stated separately. Do not ISW 4

Mark scheme

8(a) Use correct double angle formula to expand sin 4x *M1 2sin 2x cos2x OE Use correct double angle formulae to obtain an expression in sin x and cos x dM1 E.g., 4sin x cos x ( 2cos 2 x − 1) OE. ( ) Obtain sin 4 x ≡ 4sin x 2cos3 x − cos x from fully correct working A1 AG Alternative Method for Question 8(a) Use correct double angle formula to rearrange the RHS ( *M1 E.g. 2sin 2 x 2cos2 x − 1 ) Use correct double angle formula to obtain an expression in sin 2 x and cos 2 x dM1 E.g. 2sin 2x cos2x ( ) Obtain sin 4 x ≡ 4sin x 2cos3 x − cos x from fully correct working A1 AG Alternative Method 2 for Question 8(a) Use correct angle sum formulae to expand sin 4x as far as an expression in *M1 E.g. sin x ( cos x cos2x − sin x sin2x ) sin x, cos x, sin 2 x and cos 2 x + cos x ( sin x cos 2 x + cos x sin 2 x ) Use correct double angle formulae to obtain an expression in sin x and cos x ( ) dM1 E.g., sin x cos x 2cos2 x − 1 − 2sin 3 x cos x ( ) + sin x cos x 2cos2 x − 1 + 2sin x cos3 x ( ) Obtain sin 4 x ≡ 4sin x 2cos3 x − cos x from fully correct working A1 3 8(b) B1 Use the identity to obtain p sin x cos6 x dx + q sin x cos 4 x dx 8 sin x cos6 x dx − 4 sin x cos 4 x dx Accept terms of the correct form but without the integral signs or the dx. Obtain r cos7 x + s cos5 x B1 Obtain − 87 cos7 x + 54 cos5 x B1 If using the substitution u = cos x, accept the form 8 4 − u7 + u5 . 7 5 Use the correct limits correctly in an expression of the form r cos7 x + s cos5 x M1 OE 8 1 4 1 8 4 Must be evaluated, e.g. − × + × + − . 7 8 2 5 4 2 7 5 1 Obtain 35 ( 2 + 12) A1 Or exact simplified equivalent (e.g. as 2 terms). 5

Mark scheme

9(a) B C B1 State or imply the form A + + x + 2a x + 3a Use a correct method for finding a coefficient M1 Obtain one of A = 1, B = 2a and C = −3a A1 SC: If B0 is scored because of an omission of ‘A’, then maximum M1A1 is available for one constant correct. SC: If they substitute a value for a, or if their working implies the use of a = 1, they can score maximum B1M1. Obtain a second value A1 Obtain the third value A1 Alternative Method for Question 9(a) linear expression B1 − ax State or imply 1 + 1+ ( x + 2a )( x + 3a ) ( x + 2a )( x + 3a ) B C B1 State or imply the form 1 + + x + 2a x + 3a Use a correct method for finding B or C M1 Obtain one of B = 2a and C = −3a A1 Obtain the second value A1 5 9(b) Integrate and obtain terms Ax + B ln ( x + 2a ) + C ln ( x + 3a ) B2FT B1 for any two terms correct, B2 for all three terms correct. The FT is on A, B and C: x + 2a ln ( x + 2a ) − 3a ln ( x + 3a ) . Allow for FT on a split completed in (b). Substitute limits correctly in an integral containing at least 2 terms from the form rx + s ln ( x + 2a ) + t ln ( x + 3a ) () M1 E.g. 2a + 2a ln 3 − 3a ln 4 1 2 ( ) ( Obtain a 2 + ln 89 ( )) from correct working A1 Accept equivalent fractions with integers. ( ( )) is following an error in (a). A0 XP if a 2 + ln 89 4

Mark scheme

10(a) dV B1 SOI State = 5000 − kh2 dt dV *M1 OE their dh 5000 − kh 2 dt Use V = 2500h and chain rule to obtain = 2500 dt 2500 dh dM1 5000 − 400k Use h = 20, = 0.4 to obtain k 0.4 = k = 10 dt 2500 dh 500 − h 2 A1 AG Obtain = from correct working There needs to be a complete statement for A1. dt 250 Condone ‘divide by 10’ in place of showing the cancelling. 4 10(b) Separate variables correctly B1 1 1 E.g. dh = 1dt 500 − h 2 250 Condone missing integral signs or missing dh, dt, but not both. Obtain 1 t B1 OE (depending on how they use the 250). 250 1 500 + h B1 1 10 5 + h Obtain ln OE, e.g. ln . 2 500 500 − h 20 5 10 5 − h Coefficients will depend on where the 250 is. Condone missing modulus signs. Use t = 0, h = 0 in an expression of the form pt + q ln ( a + h) + r ln (b − h) to M1 Requires a value for the constant. M0 if their equation has no solution. obtain the constant of integration or as limits on a definite integral M0 if they never had a constant. Need some indication that the conclusion is based on using t = 0, h = 0, e.g. "t = 0, h = 0 C = 0". Obtain t = 16.1(s) A1 Accept AWRT. 5

Mark scheme

11(a) MB = −i + 5 j + 2k B1 Do not ISW. Accept as a column vector. MC = −5i − 3j + 4k B1 Do not ISW. Accept as a column vector. SC: B0B1 for column vectors with i, j, k (or coordinates) (allow one slip). 2 11(b) Carry out the correct process for evaluating the scalar product of M1 Or BM and CM . MB and MC Need some evidence of an attempt to evaluate the scalar product. Using the correct process for the moduli, divide the scalar product by the M1 5 − 15 + 8 product of the moduli for their two vectors. ± 30. 50 The scalar product could be incorrect, but it must be a scalar. Condone a sign change. 15 A1 1 −2 Obtain − Or exact equivalent, e.g. − , . 75 5 15 30 50 A0 if they never have a statement "cosθ = ....". ISW if they have correct "cosθ = ...." and then state the angle. Alternative Method for Question 11(b) Carry out the correct process for finding the lengths of the sides of triangle M1 MB = 30, MC = 50, BC = 84 BMC Correct use of the cosine rule as far as cosθ = .... M1 2 their MB + their MC − 84 2 cos M = 2their MB × their MC 50 + 30 − 84 cos M = 2 50 30 −1 A1 Or exact equivalent. Obtain A0 if they never have a statement "cosθ = ....". 375 ISW if they have correct "cosθ = ...." and then state the angle. 3 11(c) Working with triangle MBC: Correct method to calculate the exact value of M1 sin ( CMB ) 374 A1 5400 sin ( CMB ) = Or equivalent, e.g. sin ( CMB ) = . 375 5625 A0 XP if they had changed the sign in their working for the cosine in (b) and for the final answer. Correct method to calculate the area of triangle ABC using MB and MC M1 2 × area of CMB. 374 A1 Or exact equivalent, e.g. 2 2 11 17 2 × 12 30 50 × =2 374 375 3366000000 or . 1500 Alternative Method for Question 11(c) Working with triangle ABC: correct method to calculate the exact sine of an M1 56 64 20 angle in the triangle ABC cos A = , cos B = , cos C = 120 × 76 120 × 84 76 × 84 4 374 4 374 4 374 A1 Or exact equivalent. Obtain sin A = or sin B = or sin C = 120 76 120 84 76 84 Correct method to calculate the area of triangle ABC M1 Must have a matching set of sides for their angle. 1 4 374 A1 Or exact equivalent. Area = 120 76 or 2 120 76 1 4 374 area = 120 84 or 2 120 84 1 4 374 area = 76 84 = 2 374 2 76 84 4