Mathematics 9709/12 May/June 2024: question paper and mark scheme

Mark scheme

1 240 x 2 or 80a 2 [x2] B1 May be seen in an expansion. 240 = 12 × 80a 2 M1 Their 240 equated to 12 × their 80a 2 which must contain a2. 0.5 A1 OE Condone ± 0.5 3

Mark scheme

2 Stretch factor 4 in y-direction/parallel to the y axis/vertically. B1 Allow use of SF in place of factor. Allow in/on/along the y axis or ‘the x axis is invariant.’ Translation B2 Condone ‘Shift’. 3 3 or 3 parallel to the x axis or in the x direction, allow horizontally. These translations can be combined as , this 0 −8 0 counts as 2 elements. or −8 parallel to the y axis or in the y direction, allow vertically. Give priority to a correct vector over any incorrect −8 wording. B2 for all 3 B1 for 2 out of 3 Two translations, one in each direction, and a stretch only. M1 Condone inaccurate terminology, such as up, down, left and right, if the intention is clear. Correct order of operations. The stretch which must be in the in the y direction A1 Condone inaccurate terminology if the intention is must come before the translation in the y direction. clear but numerical values must be correct. 2 Alternative Method for Question 2 Translation (B2) Condone ‘Shift’. 3 3 or 3 parallel to the x axis or in the x direction, allow horizontally. These translations can be combined as , this 0 −2 0 counts as 2 elements. or −2 parallel to the y axis or in the y direction, allow vertically. Give priority to a correct vector over any incorrect −2 wording. B2 for all 3. B1 for 2 out of 3. Stretch factor 4 in y-direction/parallel to the y axis/vertically. (B1) Allow use of SF in place of factor. Allow in/on/along the y axis or “the x axis is invariant.” Two translations, one in each direction, and a stretch only. (M1) Condone inaccurate terminology, such as transform, move, up, down, left and right, if the intention is clear. Correct order of operations. The stretch which must be in the in the y direction (A1) Condone inaccurate terminology if the intention is must come after the translation in the y direction. clear but numerical values must be correct. 5

Mark scheme

3(a) sin θ sin θ M1* OE 7 ÷ cos θ + 12 [ = 0] leading to 7 + 12cosθ = 0 sin θ cos θ cos θ Use of tan θ = . cos θ ( ) 7 sin θ + 12 1 − sin 2 θ [ = 0] DM1 Use of s 2 + c 2 = 1 . ⇒ 12sin 2 θ − 7sin θ − 12 = 0 A1 AG, WWW Condone use of s, c and t and/or omission of θ throughout working but the A1 is for cao. 3 3(b) 12sin 2 θ − 7sin θ − 12 = 0 leading to ( 4sin θ + 3)( 3sin θ − 4 ) M1 3 4 B1 OE, WWW sin θ = − or 3 4 3 Can be implied by a correct value for sin −1 − 4 e.g. − 48.6°. [θ = ] 228.6°,311.4° B1 AWRT, WWW No others in the range 0° ≤ θ ≤ 360° . Ignore any answers outside this range. Condone 229°, 311°. 3

Mark scheme

4(a) f −1 ( x ) = ( x + 1)2 B1 ISW Condone ‘y =’. 1 4(b) 1 1 1 B1 1 0 < g ( x) or g ( x ) > 0 and g ( x ) or 0, Do not allow g ( x ) > 0, g ( x ) . 2 2 2 2 1 Do not allow g ( x ) > 0 or g ( x ) . 2 Condone g or y in place of g ( x ) . g–1 does not exist because it is one to many or B1 g–1 can be replaced by ‘It’ throughout. g–1 does not exist because it is not one to one. Or A correct statement followed by any further incorrect g–1 does not exist because g is not one to one or explanation can be awarded B1. g–1 does not exist because g is many to one or g–1 does not exist because g fails the horizontal line test. 2 4(c) 25 1 B1 SOI f = 16 4 1 1 M1 1 = Equating , or their ‘simplified’ ( ) ( ) 2 2 x −1 + 2 4 x −1 + 2 25 version, to their f . 16 ( ) x − 1 2 + 2 = 4 leading to ( ) A1 Simplification as far as x =… 2 x − 1 = 2 leading to x = 1 ± 2 Or Allow just + in the results because − can be disregarded at this stage. ( ) 2 x − 2 x + 1 + 2 = 4 leading to x − 2 x − 1 = 0 leading to x = 1 ± √ 2 Can be implied by the final answer. Or 6 ± 36 − 4 Note: x = 1 ± 2 scores A0. x − 1 = 2 x leading to x 2 − 6 x + 1 = 0 leading to x = 2 3+ 2 2 A1 Must discount the solution 3 −2 2 . 4

Mark scheme

5(a) [ d = ] sin θ − tan θ M1 For subtraction of their first term from their second term. Condone incorrect evaluation before subtraction. 2 A1 OE [ d =] −1 Sight of −0.29 AWRT can be awarded M1A1. 2 40 M1 Use of a correct formula for S40. Condone use of [ S40 = ] ( 2 × tan θ + 39 ( sin θ − tan θ ) ) their, clearly identified, incorrect values for a and d 2 for this mark. 780 A1 ISW 390 2 − 740 or − 740 If A0 then sight of − 188 AWRT, − 188.5 or −189 2 should be awarded M1A1M1A0. 4 5(b)(i) sin θ B1 Condone omission of θ . r= [ = cosθ ] tan θ tan θ sin θ tan 2 θ B1 ISW S∞ = or or Do not allow fractions within fractions nor omission 1 − cos θ cos θ − cos 2θ tan θ − sin θ of θ . 2 5(b)(ii) 1 B1 OE, SOI. a = 3 (1.73..) and r = 2 1 10 M1 This mark can be awarded for a correct formula with 1− their values for a and r or 2 [ S10 =] 3 sin θ 1 10 1 a = tan θ and r = or cos θ . Condone . 1 − 2 tanθ 2 = 3.46 A1 AWRT 1023 3 Condone . 512 3 Note: S9 gives the same answer but scores B1M0A0.

Mark scheme

6(a) dy 1 − 1 B1 = 2 − ×8 x 2 dx 2 − 1 M1 dy 2 − 4x 2 =0 Equating their two term , with at least one term dx correct, to 0. [A is] ( 4, −8 ) or x = 4, y = −8 A1 [B is] (16, 0 ) or x = 16, y = 0 B1 4 dy Note: Correct answers without use of can be dx awarded 4/4. 6(b) 2x2 8 2 3 B1 Seen correct in unsimplified form or better. [±] − x 3 [ +C ] 2 2 x 2 − 32 x ( 2 x − 32 ) 2 B1 Seen correct in unsimplified form or better. [±] or [ +C ] 3 12 Attempt to integrate, defined by at least one correct power in each expression, M1 Multiplying by 3 before integration scores M0. and then subtract. M1 Use of their x values, > 0, from (a) as limits in their 2 8 3 8 2 3 162 − 32 × 16 42 − 32 × 4 integrated expressions. Allow, for correct limits, 16 − 3 .16 2 − 42 − 3 .4 [ ] − − sight of 3 3 256 80 256 112 − − [ −] − 2 2 ± − −− . 3 3 3 3 If incorrect limits are used, then clear substitution must be seen. 6(b) Alternative Method 1 for first 4 marks of Question 6(b) 1 2 8 3 (B1) Seen correct in unsimplified form or better. [ ± ] 2 x − 8 x 2 dx = x − x 2 [ +C ] 3 2 [Area of triangle =] 48 (B1) Attempt to integrate, defined by at least one correct power, and then subtract (M1) their triangle area. (M1) Use of their x values, > 0, from (a) as limits in their 2 8 3 2 8 3 integrated expression. Allow sight of 16 − 3 .16 2 − 4 − 3 .4 2 256 80 ± − − − . 2 2 3 3 If incorrect limits are used, then clear substitution must be seen. 6(b) Alternative Method 2 for first 4 marks of Question 6(b) Subtract and then integrate, defined by at least two correct powers. (M1) If terms in x have not been combined use the first Condone functions being the wrong way round. scheme. (B2,1,0) B2 for 3 correct terms, B1 for any 2 correct terms. 4 2 8 23 32 x [±] x − x + 3 3× 2 3 2 (M1) Use of their x values, >0, from (a) as limits in their 4 8 3 32 × 16 4 8 3 32 × 4 32 [±] × 162 − × 16 2 + 3 − × 42 − × 4 2 + 3 integrated expression. Allow sight of ± 0 − . 3 × 2 3 3× 2 3 3 2 2 If incorrect limits are used, then clear substitution must be seen. 32 2 (B1) AWRT , 10 or 10.7 3 3 32 32 32 Allow − or − changed to + for this mark. 3 3 3 (5) Condone the inclusion of π for the first 4 marks but use of y 2 scores a maximum of B1 for the triangle.

Mark scheme

7(a) ( x − 6 ) + ( 2a − x + a ) = 18 M1* Replacing y with 2a – x in the circle equation, 2 2 condone incorrect expansion before substitution. 2 x 2 − 12 x − 6ax + 9a 2 + 36 − 18[ = 0] A1 All terms collected on one side of the equation. May be implied by the discriminant. (12 + 6a )2 − 4 × 2 × ( 9a 2 + 18 ) [ = 0] DM1 Correct use of “b2 — 4ac” from their 3 term quadratic equation in x , with an x term of the form ( m + na ) x with both m and n ≠ 0. −36a 2 + 144a [ +0 = 0] A1 a = 0, a = 4 A1 5 7(b) [Centre is] (6, −4) or [Point of intersection is] (9, −1) B1 [Gradient of diameter] = 1 B1 y + 4 = x − 6 or y + 1 = x − 9 [leading to y = x − 10 ] B1FT FT on their point of intersection or their centre with an x co-ordinate of ±6 and gradient = 1. 3

Mark scheme

8(a)(i) ˆ =π C XCE 6 ˆ = π CEX 3 X E XE π XE π M1 A correct trig expression involving XE. = sin or = cos [XE = 0.2] Do not condone a mixture of degrees and radians. 0.4 6 0.4 3 Length EF = 2 + 2 × 0.2 = 2.4 A1 AG 2 8(a)(ii) π π B1 OE, SOI [CX = ] 0.4cos or 0.4sin or 0.42 − 0.22 3 6 3 Expect or 0.3464. 5 1 π B1 SOI [Sector = ] × ( 0.4 )2 × 2π 60 2 3 Expect 0.0838 or . Allow use of π ( 0.4 ) . 2 75 360 Either Area of their (rectangle + two triangles + two sectors) M1 Either implied by a correct answer or areas clearly labelled. Or Area of their (trapezium + two sectors) Expect 0.6928 + 0.06928 + 0.1676 or 2 3 3 4π + + . 5 25 75 11 3 4π Or 0.7621 + 0.1676 or + . 25 75 0.930 A1 AWRT 11 3 4π Condone + . 25 75 4 8(b) [Length AD =] 2 + 2r B1 Must be seen alone or part of a list and not part of a product. π B1 π [Arc length =] r × May be implied by r × × 2 . 3 3 Must be seen alone or part of a list. π π B1 Must be seen alone or part of a list and not part of a [ EF =] 2 + 2r sin or 2 + 2r cos or 2 + r product. 6 3 2πr B1 AWRT [4+ 3r + = 6 leading to ] 0.393 3 6 Condone . 2π + 9 NB: Using EF = 2.4 gives 0.391. 4

Mark scheme

9(a) 6 ( 2 x − 3) − 6 x < 0 or = 0 B1* Condone ⩽ 0. If 6 ( 2 x − 3)2 only or 6 ( 2 x − 3)2 − 6 is 2 used, do not treat as a MR. 24 x 2 − 78 x + 54 or 4 x 2 − 13x + 9 or ( x − 1)( 4 x − 9 ) M1 Expanding brackets and collecting terms to arrive at a three term quadratic, only condone sign errors. OR 6 ( 2 x − 3) < 6 x leading to ( 2 x − 3) < x leading to 2 x − x − 3 2 9 B1 [ x =]1 , 4 9 9 9 DB1FT OE 1< x < or x > 1 and x < or 1, Condone consistent use of ⩽ and ⩾ or [ ]. 4 4 4 9 9 Do not allow x > 1or x < nor x > 1, x < . 4 4 FT on their values coming from a correct initial statement. 4 9(b) 6 B1 B1 B1 for each Correct integral . ( 2 x − 3)3 6 2 f ( x ) = − x [ +C ] 3 × 2 2 −1 = ( −1) − 3 × 12 + C 3 M1 f ( x ) = −1 equated to their integrated expression, defined by two terms with at least one correct power + C, with x = 1. f ( x ) = ( 2 x − 3) − 3x 2 + 3 A1 CAO 3 Only condone C = 3 as final answer if coefficients have been simplified earlier. Do not ISW if the result is of the form y = mx + c . Alternative method for Question 9(b) f ′( x ) = 24 x − 78 x + 54 leading to f ( x ) = 8 x − 39 x + 54 x [ +C ] 2 3 2 (B2,1,0) B2 completely correct, B1 any two correct terms. −1 = 8 − 39 + 54 + C (M1) f ( x ) = −1 equated to their integrated expression, defined by three terms with at least one correct power + C, with x = 1. f ( x ) = 8 x3 − 39 x 2 + 54 x − 24 (A1) Only condone C = − 24 as final answer if coefficients have been simplified earlier. Do not ISW if the result is of the form y = mx + c. 4

Mark scheme

10(a) [ x =] − 2 B1 dy 1 3 1 M1* OE = k (5 − 2x ) 2 = −2 × 2 ( 5 − 2 x ) 2 1 dx Differentiating to get k ( 5 − 2 x ) 2 only. dy dy dt dt DM1 Correct statement linking their numerical expression dx = dt × dx leading to −9 = ± 5 × dx dy dt for with and ± 5. dx dx 5 A1 AWRT or 0.556 = 9 4 10(b) 1 M1 dy 1 k ( 5 − 2 x ) 2 = −3 Equating their of the form k ( 5 − 2 x ) 2 to −3 . dx [B is] ( 2, 6 ) A1 32 − 6 1 4 M1* For A, y must be 32. Gradient AB = m1 = , gradient of perpendicular = − = difference in y co-ordinates −2 − 2 m1 26 Clear use of for points difference in x co-ordinates A and B, condone inconsistent order, and using m1m2 = −1. If incorrect values or another complete method used, then working must be clear. 2 − 2 6 + 32 M1* Finding the midpoint of AB using A and B. If Mid point is , = ( 0,19 ) incorrect values used then all working must be clear. 2 2 For A, y must be 32. 2 DM1 Finding the equation of the perpendicular bisector y − 19 = ( x − 0) using their midpoint and their perpendicular 13 gradient. 2 x − 13 y + 247 = 0 or ± integer multiples of this. A1 6