Mathematics 9709/12 October/November 2024: question paper and mark scheme

Mark scheme

1(a) a=4 B1 Allow 4sin ( 2 x ) + 3 if values of a, b and c are not stated. b=2 B1 c =3 B1 3 1(b)(i) 5 B1 Ignore attempts at finding solutions. 1 1(b)(ii) 1 B1 Ignore attempts at finding solutions. 1

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2(a) M1 Correct use of either S20 formula with a = — 20 and d = 5. 20 2 ( 2 × −20 + ( 20 − 1) × 5) or 20 2 ( −20 + 75) 550 A1 2 2(b) M1* Correct use of Sn formula with a = −20 , d = 5 and either k or 2k 2 ( −40 + ( 2k − 1) × 5) or ( −40 + ( k − 1) × 5) k 2 2k. n This mark can be awarded for clear use of ( a + l ) when 2 correct values of a and d are used. −40k + 10k 2 − 5k = −200k + 25k 2 − 25k 15k 2 − 180k = 0 DM1 Equating their S2k to 10 × their Sk and reaching a 2-term quadratic or 2 term linear equation if k has been cancelled. Condone errors in simplification. k = 12 A1 Condone extra solution k = 0. 3

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3(a) f ( 2 + h ) = 2 ( 2 + h ) − 3 B1 SOI 2 ( 2 ( 2 + h) − 3) − 5 = 2h + 8h 2 2 M1 their ( 2( 2 + h) − 3) − their5 can be implied by the 2 ( 2 + h) − 2 h ( 2 + h) − 2 simplified expression or the correct answer. Their 5 must come from 2 ( 2) − 3. 2 2h + 8 or 2 ( h + 4 ) A1 3 3(b) h → 0 , or chord [AB] → tangent [at A] B1 Either of these statements or any sight of h = 0.

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4(a) 6 B1 OE 2 2 3 6×5 3 15 or × x 4 2 or x6 × May be in a list. 2 x 2 x3 6 Allow . 4 135 B1 Correct term must be identified if in a list. Allow 135x0. 2 4(b) 6 B1 OE 3 3 3 6.5.4 3 20 or × x3 2 or x6 × May be in a list. 3 x 3! x3 540 B1 1 = Identifying term. x3 x3 This can be implied by sight of 2160 as part of the constant term. 4 × 540 − 5 × 135 M1 4 × their 540 − 5 × their135 1485 A1 Allow 1485x0. 4 4 4 OR Then using b 2 − 4ac = 0 on their quadratic to form an equation 5 y2 − y (38 + 2 p ) + 100 + 10 p + q = 0 (38 + 2 p)2 − 4 × 5 × (100 + 10 p + q ) = 0 in p and q. Each of these p − 12 p − 139 − 5q = 0 2 Solving the equations simultaneously to find p or q DM1 p = −4 A1 q = −15 A1 5

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5(a)(i) 1 B1 Condone 0.333. f ( −1) = 3 1 5(a)(ii) 4 y B1 For showing the correct mirror line. 2 B1 For correct shape: the curves should intersect in the first square in the third quadrant. To the left of the point of intersection, the x −10 −8 −6 −4 −2 2 4 6 8 10 reflection is below the original and crosses the x-axis. To the right of the point of intersection, the reflection is to the right the −2 original. −4 2 5(a)(iii) 2x + 1 M1* Equating y to the given function and clearing of fractions. = y 2 x + 1 = y ( 2 x − 1) x and y may be interchanged at this stage. 2x − 1 2 xy − 2 x = y + 1 DM1 Condone ± errors during simplification. x +1 −x −1 A1 Allow ‘ f −1 ’or ‘y =’ but NOT ‘x =’, nor fractions within , 2 ( x − 1) 2 − 2 x fractions. [Domain of f −1 is] x < 1 B1 Accept — ∞ < x <1 or (— ∞, 1), condone [— ∞, 1). Alternative Method for Question 5(a)(iii) 2 2 M1* Equating y to the given function after division by 2x −1. y =1+ y −1 = 2x − 1 2x −1 Isolating the term in x. x and y may be interchanged at this stage. 2 DM1 Condone ± errors during simplification. 2x = +1 y −1 1 1 A1 OE + x −1 2 Allow ‘ f −1 ’or ‘y =’ but NOT ‘x =’, nor fractions within fractions. [Domain of f −1 is] x < 1 B1 Accept — ∞ < x <1 or (— ∞, 1), condone [— ∞, 1). 4 5(b) 1 B1 gf = −7 4 2x + 1 M1 2x + 1 1 = −7 Equating to their gf . 2x −1 2x −1 4 3 A1 OE [ x =] 8 Alternative solution for Question 5(b) 1 B1 gf = −7 4 x = f −1 ( −7 ) M1 1 x = f −1 their gf 4 3 A1 OE [ x =] 8 3

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6 [Perimeter =] r + rθ + r + 2r × 2θ + r + rθ + r [= 4r + 6rθ ] B1 1 2 1 1 B1 r θ + ( 2r ) × 2θ + r 2θ = 5r 2θ 2 [Area =] 2 2 2 4r + 6rθ = 14 and 5r 2θ = 10 M1* ar + brθ = 14 and cr 2θ = 10 where a, b and c are constants ≠ 0. Terms may be uncollected. EITHER 14 − 4r 10 DM1 Eliminate θ to get an equation in r. 5r 2 = 10 or 4r + 6r 2 = 14 6r 5r 2r 2 − 7r + 6 = 0 ( r − 2 )( 2r − 3) = 0 DM1 Factorise or other accepted method for solving their 3-term quadratic. OR 14 2 10 10 DM1 Eliminate r to get an equation in θ . 5 θ = 10 or 4 5θ + 6 5θ θ = 14 4 + 6θ [ 18θ 2 − 25θ + 8 = 0 ] (9θ − 8)( 2θ − 1) = 0 DM1 Factorise or other accepted method for solving their 3-term quadratic. Then r = 2 and θ = 0.5 B1 3 8 Condone extra answers r = and θ = . 2 9 6

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7(a) ( ) −2 ( x ± p ) ± q or −2 ( x ± p ) ± q 2 2 M1* p ≠ 0. ( ) −2 ( x − 2 ) ± q or −2 ( x − 2) ± q 2 2 DM1 −2 ( x − 2) + 19 and (2, 19) 2 A1 Accept x = 2, y = 19 or 2, 19. 3 7(b) Method 1 [ x =] ± 1 B1* Both x co-ordinates for the points of intersection. Subtract and attempt to integrate M1* ( ) B1* Both terms correct. −2 x 2 + 2 dx − 2 x3 + 2 x 3 2 2 M1 Apply their limits, one positive and one negative, obtained − + 2 − − 2 from equating the line and the curve to their integrated 3 3 expression. 8 2 DB1 AWRT 2.67 WWW. = , 2 −8 8 3 3 Condone → . 3 3 11 SC B1 for mistaking triangle for trapezium leading to , i.e. 3 a total of 2/5. Method 2 [ x =] ± 1 B1* Both x co-ordinates for the points of intersection. Attempt to integrate and subtract M1* The second integral can be replaced with what is clearly their area of a trapezium. −2 x3 8 2 8 B1* OE + x + 11x [ −] x 2 + 9 x All terms correct. 3 2 2 1 The second integral can be replaced by (1 + 17 ) × 2 OE. 2 7(b) −2 2 M1 Apply their limits, one positive and one negative, obtained + 4 + 11 − + 4 − 11 [ −] ( 4 + 9 ) − ( 4 − 9 ) from equating the line and the curve, to their integrated 3 3 expressions. If the trapezium has been used, the second integral can be replaced by their 18. 8 2 DB1 AWRT 2.67 WWW. = , 2 −8 8 3 3 Condone → . 3 3 11 SC B1 for mistaking triangle for trapezium leading to , i.e. 3 a total of 2/5. Method 3 [ x =] ± 1 B1* Both x co-ordinates for the points of intersection. Subtract and attempt to integrate M1* 2 8 B1* All terms correct. − ( x − 2)3 − x2 + 10 x 3 2 2 M1 Apply their limits, one positive and one negative, obtained − 4 + 10 − (18 − 4 − 10 ) from equating the line and the curve, to their integrated 3 expression. 8 2 DB1 AWRT 2.67 WWW. = , 2 3 3 7(b) Method 4 [ x =] ± 1 B1* Both x co-ordinates for the points of intersection. Attempt to integrate and subtract M1* The second integral can be replaced with what is clearly their area of a trapezium. 2 8 2 B1* All terms correct. − ( x − 2 ) + 19 x [ − ] x + 9 x 3 1 3 2 The second integral can be replaced with (1 + 17 ) × 2 OE. 2 2 M1 Apply their limits, one positive and one negative, obtained + 19 − (18 − 19 ) [ −] ( 4 + 9 ) − ( 4 − 9 ) from equating the line and the curve, to their integrated 3 expression. If the trapezium has been used the second integral can be replaced with their 18 OE. 8 2 DB1 AWRT 2.67 WWW. = , 2 −8 8 3 3 Condone → . 3 3 11 SC B1 for mistaking triangle for trapezium leading to , i.e. 3 a total of 2/5. 5

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8 B1FT Could come from anywhere except wrong working. Either correct or FT their linear expression from (a). 2 8(a) 1 2 B1* 1 1 x − − p + ( y − ( −1) ) OE 2 Allow a = − p and b = −1, or centre is − p, −1 . 2 2 2 2 2 DB1 1 1 x − − p + ( y − ( −1) ) = −q + 1 + − P OE 2 2 2 2 8(b)(i) 1 B1 OE [Gradient of tangent =] − SOI 2 [Gradient of normal =] 2 M1 Use of m1m2= − 1with their numeric tangent gradient. y −3 A1 OE = 2 [ y = 2 x − 5] ISW x−4 Allow y = 2 x + c, 3 = 2 × 4 + c c = −5. 3 8(b)(ii) Method 1 for the first two marks: 1 M1* ±p −1 − 3 = 2 − p − 4 or −1 = − p − 5 Using their stated centre or , ± 1 in their equation of the 2 2 normal. p = −4 A1 Method 2 for the first two marks: 1 M1* Using their normal equation and their stated centre or −1 = 2x − 5 x = 2 − p=2 2 ±p , ± 1 . 2 p = −4 A1 Method 3 for the first two marks: dy dy dy M1* 2x + 2 y + p + 2 = 0 p = −8 − 8 dx dx dx dy 1 A1 dx = − 2 p = −4 8(b)(ii) Method 1 for the last 3 marks: r 2 = ( 4 − 2) + ( 3 − ( −1) ) [= 20] M1* ±their p 2 2 Using (4, 3) and their centre or , ± 1 to find r2 or r. 2 1 2 DM1 OE −q + 1 + p = 20 Using their expression for r2 (from (a)) equated to their 20. 4 q = −15 A1 Method 2 for the last 3 marks: 2 − 2 − 10 10 M1* Using ( 2, −1) and x + 2 y − 10 = 0 (distance from a point to a r= = 5 5 line). 1 10 2 DM1 OE −q + 1 + p 2 = 2 4 5 2 10 Using their expression for r equated to their . 5 q = −15 A1 Method 3 for the last 3 marks: 42 + 32 + 4 p + 6 + q = 0 [ 4 p + q + 31 = 0] M1* Substituting ( 4,3) into their circle equation. OR 2 2 1 1 4 − − p + ( 3 − ( −1) ) = −q + 1 + − p 2 2 2 4 ( −4 ) + q + 31 = 0 DM1 Substituting their p = −4. q = −15 A1 8(b)(ii) Alternative Method for Question 8(b)(ii) 42 + 32 + 4 p + 6 + q = 0 M1* Substituting ( 4, 3 ) into their circle equation, or x2 + ( 2x − 5) + px + 2 ( 2x − 5) + q = 0 with x = 4 2 replacing y with 2x − 5 from the normal equation, or 2 10 − x 10 − x 10 − x replacing y with from the tangent equation, or x + + px + 2 + q = 0 with x = 4 2 2 2 2 y+5 2 replacing x with from the normal equation, or y +5 y +5 2 + y + p + 2 y + q = 0 with y = 3 2 replacing x with 10 − 2 y from the tangent equation, and using 2 2 (10 − 2 y )2 + y2 + p (10 − 2 y ) + 2 y + q = 0 with y = 3 either x = 4 or y = 3 to form an equation in p and q. Each of these 4 p + q + 31 = 0 5 2 5 M1* Solving the tangent and circle equations simultaneously to form x + ( p − 6 ) x + 35 + q = 0 ( p − 6)2 − 4 × × ( 35 + q ) = 0 a quadratic equation in either x or y.

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9(a) 1 2 25 5 1 M1* 5 1 k × − 2k × + 2 = Using , in the curve equation or equating the line and 2 4 2 2 2 2 OR 5 1 5 the curve and then using x = and p = − k. 1 k 2 × 25 − 2k × 5 + 2 = k × 5 + 1 − 5 k 2 2 2 2 4 2 2 2 2 Simplify to get a three-term quadratic in k. Condone errors in simplification. 25k 2 − 40k + 12 [= 0 ] 2 A1 OE k= 6 5 Condone inclusion of k = . 5 1 2 5 DM1* 5 1 = their + p p = Using , and their k in an equation in p. 2 5 2 2 2 Either the line (as shown) or 4 p 2 + 12 p + 5 = 0 are the most likely and solving for p. 1 A1 OE p=− 5 2 Condone inclusion of p = − . 2 2 2 6 5 DM1 Equating the line and curve using their k and p and simplify to x − x + [ = 0] 4 x2 − 60 x + 125 [ = 0] get a three-term quadratic [= 0]. 9(a) Alternative Method for Question 9(a) 1 2 25 5 5 M1* OE 2 k × 4 − 2k × 2 + 2 = k × 2 + p 5 1 Using , in the curve equation or equating the line and 4 p + 12 p + 5 [= 0] 2 2 2 5 1 2 the curve and then using x = and k = − p. 2 5 5 Simplify to get a three-term quadratic in p [= 0]. 1 A1 5 p=− OE Condone inclusion of p = − . 2 2 1 5 1 DM1* 5 1 = k + their − k = Using , and their p in the line equation and solving for 2 2 2 2 2 k. 2 A1 OE k= 6 5 Condone inclusion of k = . 5 2 2 6 5 DM1 Equating the line and curve using their k and p and simplify to x − x + [ = 0] 4 x 2 − 60 x + 125[ = 0] get a three-term quadratic [= 0]. 9(b) 1 2 2 1 2 2 M1* Equate the original equations of the curve and the line and 2 k x − 2kx + 2 = kx + p 2 k x − 3kx + 2 − p collect like terms; k and p must still be present. 1 DM1 Use of b 2 − 4ac for their quadratic in x to give an expression in 9k 2 − 4 × k 2 ( 2 − p ) 2 k and p. This expression can come from their equation in (a). 5 A1 p<− 2 3

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10(a) −18 B1 SOI 1 M1 Use of m1m2 = −1from f ′ ( x ) with x = 1. 18 y [ − 0] 1 A1 OE = ISW x −1 18 3 10(b) B1B1 B1 for each unsimplified {}. 4 1 1 5 1 Can be implied by equivalent simplified or partly simplified f ( x ) = 8 ( 2 x − 3) 3 . . −10 x 3 . [ +c ] versions. 2 4 5 3 3 4 5 3 ( 2 x − 3 ) 3 − 6 x 3 + c 4 5 M1 Use of x = 1 and y = 0 in their integrated f ′ ( x ) , defined as an 0 = 3 ( 2 (1) − 3) 3 − 6 (1) 3 + c [0 = 3 − 6 + c ] expression with at least one correct power, which must contain + c. 4 5 A1 Only condone c = 3 as their final answer if all coefficients have f ( x ) or y = 3( 2 x − 3) 3 − 6 x 3 + 3 previously been simplified in a correct statement. 4 10(c) b2 − 4ac = 1282 − 4 ×125 ×192 and stating “< 0” M1* b2 − 4ac = −79616 can be accepted in place of working. OR use of the quadratic formula and stating “No solutions” OR completing the square for the given quadratic and stating positive or > 0. OR sketch of the given quadratic and stating positive. No turning points [in the original function.] DM1 Decreasing because f ′ ( any positive x value ) < 0 A1 WWW e.g. f ' (1) = −18. 3

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25 5 2 25 9 A1 A1 OE , 25 9 2 2 Accept x = , y= . 2 2 25 5 2 25 9 A1 A1 OE , 25 9 2 2 Accept x = , y= . 2 2 7