Mathematics 9709/12 October/November 2025: question paper and mark scheme

Mark scheme

1(a) 9 {(x − 2)2}−28 B2,1,0 B2 for all 3 correct values. B1 for any two correct. Allow p = 9, q = −2, r = −28 . If contradictory, give preference to the expression. Condone correct values in an expression of the correct form, even if missing the square on the bracket. Allow B1 for 9{(x − 2) 2 + k}. 2 1(b) k < −28 B1FT k < their r. 1 1(c) 13 36 ± 362 − 4 × 9 × 23 M1 Solving as far as ' ( x + q )2 = '. 9 ( x − their 2 ) − their 28 = −15 ( x − 2 ) = 2 2 or x = 9 2×9 Allow sign errors only in the rearrangement, or rearranging and using the quadratic formula (see quadratics guidance). 13 6 ± 13 13 36 ± 468 A1 ISW if decimals given after the exact answers. x =2± or or 2 ± or oe 3 3 9 18 2

Mark scheme

2 6 6− 2 B1 6 6 2 2 3 × (2 x ) × − may be replaced by or 15. 2 x 2 4 6 6− 2 May be seen in a list. or × 26 × ( x2 )2 × − 3 Condone missing brackets only if recovered later. 2 2x Identifying the relevant term of the form: M1 May still be part of a list if, e.g., underlining or 3 6− 2 arrow. k × (2 x2 )2 × − p x k can be any constant or any . 3 6− 2 q or k ×26 × ( x2 )2 × − May be implied by correct final answer. 2x If constants evaluated, then condone errors for this mark. = 4860 A1 Accept 4860x 0 SC B2 for – 4860 3

Mark scheme

3(a) 0 B1 Condone ‘shift’ and vector written as co-ordinates. Translation , or Allow any mention of the y-axis but not just ‘up’. 2 Translation 2 ‘parallel to the y-axis’ or ‘in the y-direction’ or ‘vertically’. 1 B2,1,0 B2 for 3 correct components. B1 for 2. {Stretch} {factor } {parallel to x-axis or in x-direction or horizontally} 3 Allow ‘the y-axis is invariant’. 3 3(b) [ y =] 3g ( − x ) B2,1,0 B1 for 3 outside the function and nowhere else, or – inside the function and nowhere else. So –3g(x), x 3(–gx), g(–3x), g , 3g(x) +2, –3g(–x) and −3 3g ( −3x ) all score B1. But 3g ( 3x ) , −g ( −x ) score B0. SC B1 for 3f(–x). 2

Mark scheme

4(a) [k =] 4 B1 1 4(b) k 4 2x −2 +1 B2,1,0FT FT their k from 4(a) or the letter k. x + [ +c ] B2 for both correct components and no other x-terms. 4 −2 + 1 B1 for one correct. their k 4 2 M1 Substituting x = 2 and y = 20 into their integrated 20 = ×2 − +c 4 2 expression (defined by having at least one correct power and including a '+ c ' ). Allow one slip. c =5 A1 [t = ] 4 A1 5

Mark scheme

5(a) dy − 12 B1 dx = 2 x − 1 −1 1 2x 2 −1 = 0 x 2 = 2 M1 dy of the form k x − 2 − 1 equal to 0 and 1 Setting their dx 1 solving as far as x 2 = d , condoning sign errors only. Can be implied by correct final answer but not if clearly following wrong working. [ a =] 4 A1 Alternative Method for Question 5(a): y = 4 x − ( x )2 B1 Recognising the quadratic in x. −b −4 M1 Condone sign errors only. Max at x= = =2 2a −2 Allow z = x , −( z − 2) 2 + 4 max at 2. 1 or −( x − 2) 2 + 4 x 2 = 2 [ a =] 4 A1 3 5(b) [ x = or b =] 16 B1 SOI It may be found in 5(a), but must be seen in 5(b). Condone extra ‘solution’ x = 0. B2, 1, 0 B2 for both correct components and no other x terms. ( ) 4 x 12 − x dx = 4 32 3 2 1 x + − x 2 2 B1 for one correct term. Allow any correct unsimplified form. 8 3 1 8 3 1 M1 Substituting their a (from part (a)) and their b (from [Area =] × 16 2 − × 162 − × 4 2 − × 42 1 3 2 3 2 an attempt to solve 4 x 2 − x = 0 ) into an integrated expression (defined by having at least one correct power) and subtracting. If correct limits and integration, then minimum 128 40 acceptable working is − . 3 3 If incorrect limits or integration, then full substitution of every term must be seen. Note: needs 0 < a < b, otherwise M0, but allow limits applied either way round. Allow missing brackets if recovered. 128 40 88 1 DB1 88 88 = − = or 29 Must be exact. Allow − if it becomes . 3 3 3 3 3 3 Do not ISW if a further area is added or subtracted. Dependent upon B1B2 scored earlier. 5

Mark scheme

6(a) B2, 1, 0 B2 Fully correct curve, including 2, 5 and − 1 clearly labelled or implied on the y-axis and the curve horizontal only at ( 12 π, 5) and ( 32 π, − 1) (approximately). Condone the x-axis intercepts and the stationary points not being in exactly the correct positions and y-axis not to scale. B1 for one full cycle from 0 to 2π starting on the positive y-axis and initially increasing. Note: Starting at the origin or finishing at (2π, 0) scores B0. 2 6(b)(i) 1 B1 Ignore any working out seen or graphs drawn. 1 6(b)(ii) 3 B1 1 6(c) ( 3sin x + 2 = 5 1 − sin 2 x − 1 ) M1* For replacing cos 2 x with 1 − sin 2 x. 5sin 2 x + 3sin x − 2 [= 0] DM1 For simplifying to a 3-term quadratic. The terms don’t all need to be on the same side. Condone sign errors only. Candidates who miss out the ‘ −1 ’ can score the M1DM1 only. sin x = 0.4 and −1 A1 x = 0.41, 2.73, 32 π AWRT A2, 1, 0 A2 for three correct values and no others in the range 0 x 2π. or 0.13π, 0.87π, 1.5π 59 391 3 A1 for 2 correct values. or π, π, π Ignore ° symbol if seen with answers. 450 450 2 Ignore answers outside the domain. Allow 4.71 or equivalent fractions instead of 32 π SC A1 for 23.6 ° , 156.4 ° AWRT and 270 ° or 32 π only. SC A1 for 0.4, 2.7 and 4.7 or 32 π only. 5

Mark scheme

7(a) [Centre is] ( 4, 6 ) B1 SOI 32 + 52 B1 SOI Finding PQ or PQ 2 or 12 PQ or ( 12 PQ ) . 2 or 32 + 52 or 62 + 102 Allow 5.83 or 5.832 or 11.7 or 11.7 2 AWRT. or 62 + 102 oe Allow unsimplified, e.g. ( 4 − 1)2 + ( 6 − 1)2 . ( x − their 4)2 + ( y − their 6)2 = (their radius)2 M1 Using what they think that the centre and radius are, correctly, in the equation of a circle. Do not allow P or Q as the centre or the diameter being used as the radius. ( x − 4)2 + ( y − 6)2 = 34 or x 2 + y 2 − 8 x − 12 y + 18 = 0 A1 May be done by other equivalent method. ( 34 ) or 5.83 for the final answer. 2 2 Do not allow 4 7(b) 11 − 1 5 B1 Allow unsimplified. [Gradient of PQ =] or oe 7 −1 3 3 M1 Use of m1m2 = −1 with their gradient PQ. [Gradient of tangent =] − 5 y − 11 3 3 76 A1 ISW = − oe e.g. y = − x + x−7 5 5 5 76 3 Condone stopping at c = if y = − x + c stated 5 5 earlier. Alternative Method for Question 7(b): dy B1 Or rearranging to form ' y = ' and differentiating 2 ( x − 4) + 2 ( y − 6) =0 dx using the chain rule. dy dy 3 M1 Substituting x = 7 and y = 11 into a correct Substitute x = 7, y = 11 6 + 10 =0 = − dx dx 5 dy differential equation and rearranging to give dx allowing sign errors only. y − 11 3 3 76 A1 = − oe e.g. y = − x + x−7 5 5 5 3 7(c) y-coordinate of R is 1 B1 y −1 3 3 16 B1 [Equation of tangent at R is] = oe eg y = x − x−7 5 5 5 3 3 M1 Using their stated line and the equation from part (b) ( x − 7 ) + 1 = − ( x − 7 ) + 11 x = … or y = … to reach a value of x or y. 5 5 Do not allow if the gradients are the same. 46 A1 x= oe and y = 6 3 Alternate Method for Question 7(c): y-coordinate of R is 1 B1 Can be implied by correct y-coordinate of point of intersection. y-coordinate of point of intersection is 6 B1 Then either 3 M1 Using y = 6 in their equation of tangent at R. 6 − 11 = − ( x − 7) 5 Or ( x − 4) 2 = 34 + ( x − 7) 2 + 25 (M1) Using the right-angled triangle formed by the centre, Q and the point of intersection of the tangents. 46 46 A1 Allow 15.3 AWRT. x= ,6 oe 3 3 4

Mark scheme

8(a) −3c − b = b − a B1 For one correct equation connecting a, b and c only. or 2b = a − 3c Allow a + 2 ( b − a ) = −3c. b c or = a b or b 2 = ac oe 4ac = ( a − 3c ) or 2 ac = a − 3c oe 2 M1 Forming an equation in a and c only using both the AP and GP. Condone sign errors only. ⇒ a 2 − 6ac + 9c 2 = 4ac ⇒ a 2 − 10ac + 9c 2 = 0 A1 AG Convincing proof needed, e.g. ( a − 3c ) expanded. 2 3 8(b)(i) 90 ± 5184 90 ± 72 M1 Factorising or other accepted method for solving. 81 − 90c + 9c 2 = 0 ⇒ [9] ( c − 9 )( c − 1) = 0 or = 18 See quadratics guidance but condone missing the 18 factor 9. Allow presence of a. c =1 B1 Condone extra ‘solution’ c = 9. B0 if any error seen leading to c = 1 but all further marks are still available in this part and (b)(ii). their 1 1 M1* Correctly finding a value for r from their c. r2 = ⇒ r = , 1 1 1 9 3 May be implied by r = or r = ± or r = − . 1 3 3 3 or b2 = their 1× 9 ⇒ b = 3 ⇒ r = 3 9 DM1 Use of correct sum to infinity formula with 9 and S∞ = 1 − 13 their r < 1. Ignore any extra working with a negative value of r. 27 A1 Do not condone extra ‘answer’. S∞ = oe This mark is not dependent upon the first M1. 2 5 8(b)(ii) d = −6 B1 SOI 20 M1 Use of a correct sum formula with n = 20 , a = 9 and S20 = 2 ( 2 × 9 + ( 20 − 1) × ( −6)) their non-zero d stated in this part (can be implied by quoting the correct formula then substituting). But allow use of ±6 without stating either the formula or ‘d =’. S20 = −960 A1 3

Mark scheme

9(a) 4 × 3 × ( −2 ) 1× 3 × ( −3) B1B1B1 B1 for the correct powers, B1 for ×3 in at least one + term, B1 for all correct which can be unsimplified. ( 3 x − 6 )3 ( 3 x − 6 ) 4 or 4 × 3 × ( −2)(3x − 6) + 1× 3 × ( −3)(3x − 6) −3 −4 Decreasing. B1* This mark is only available if f ' ( x ) is of the form ( − p )(3x − 6)−3 + ( −q )(3x − 6)−4 , where p and q are positive coefficients. 4 9(b) f −1 exists because f is a decreasing function DB1 Or f −1 exists because it is one-to-one, or passes the horizontal line test. 1 9(c) g ( x ) > 4a − 3 B1 Allow ‘ y > ’ or ‘ g > ’ only. Accept 4a − 3 < y < ∞ or ( 4a − 3 , ∞ ) . Condone ( 4 a − 3 , ∞ ]. Accept g ( x ) > g ( a ) , but not if they make an error ‘simplifying’ it. 1 9(d) Either 4a − 3 > 2 allow with x or a or ≥ M1 Do not allow = or ≠, unless they reach a correct inequality later. 5 5 A1 5 5 5 5 a> or a oe Accept < a < ∞, a < ∞ , , ∞ or , ∞ . 4 4 4 4 4 4 Or 3 × ( 4a − 3) − 6 > 0 allow with x or a or M1 Do not allow = or ≠, unless they reach a correct inequality later. 5 5 A1 5 5 5 5 a> or a oe Accept < a < ∞, a < ∞ , , ∞ or , ∞ . 4 4 4 4 4 4 2

Mark scheme

10(a) 2π M1 Forming a correct equation. s 2 = r 2 + r 2 − 2r 2 cos Allow use of degrees. 3 r s or = π 2π sin sin 6 3 π or s = 2r × cos 6 2 2 r s or + = r 2 2 2 2 2 3r s or + = s 2 oe 2 2 3 A1 AG ⇒ E.g. s = 2r × s = 3r Convincing proof required. E.g., for those using trig, 2 the exact value of the trig ratio substituted before the final answer is needed. 2 10(b) 2π B1 SOI, unless clearly associated with the wrong angle. [Angle BAD, BAC or CAD =] or 120 ° 3 Condone 2.09 for this mark only. 1 2 2π B1 [Sector BAD =] r × oe 2 3 1 π B1* [Sector BCD =] s 2 × oe 2 3 3 2 1 DB1 Sector BCD simplified in terms of r. = πr or πr 2 6 2 1 2π r s B1* May be multiplied by 2 or by 3. Either [ Area triangle ABD,ABC or ACD =] r 2 × sin or × oe 2 3 2 2 1 π 3 2 Note: r 2 × sin = r is B0DB0. 2 3 4 3 2 DB1 May be multiplied by 2 or by 3. = r 4 1 2 π 3r s (B1* 1 2 2π 3 3 2 Or [Area triangle CBD =] s × sin or × oe Note: s × sin = r is B0DB0. 2 3 2 2 2 3 4 3 3 2 DB1) = r 4 3 π 2 12 3 − 4π 2 B1 Fully correct solution. − r or e.g. r oe in the required exact form 3 π 2 6 24 Accept a = , b=− . 2 6 7