Mathematics 9709/13 October/November 2024: question paper and mark scheme

Mark scheme

1 a + 3d = 15 and a + 7d = 25 M1 Or forming any valid equations which can be used to find d or a. 5 15 DM1 Or any valid method to find a using their d or d using their a Finding a and d d = , a = 2 2 or finding u30 directly from either u 4 or u8 and d. 15 5 A1 u30 = + 29 × = 80 2 2 3

Mark scheme

2 π 3 M1 Making tan 2x the subject. tan 2x = 0 is M0. cos + tan 2 x + = 0 tan 2 x = − 3 Accept decimals and one sign error. 6 2 π π A1 May come from non-exact working. 2x = − x=− Ignore answers outside the given range. 3 6 2

Mark scheme

3(a) 5 5 M1 Allow for either term, allow sign error and combination x3 : 32 ( −ax ) −10 × 32 × a3 or x4 : 3( −ax ) 5 × 3 × a 4 3 4 notation. 3 4 x3 : −90a3 A1 Allow in the full expansion. x 4 : 15a 4 A1 Allow in the full expansion. 3 3(b) Coefficient of x 4 is a × their − 90a3 + 7 × their 15a 4 = 15a 4 M1 Must select two appropriate terms only. 15 a 4 = 240 DM1 Reducing to a simple quartic equation in a. a 4 = 16 a = 2 A1 A0 if a = −2 is given as a solution. 3

Mark scheme

4 Let x = sin 2 θ M1 Or equivalent method. ( 2 x + 7 )( 2 x − 1) = 0 or ( 2sin 2 θ + 7 )( 2sin 2 θ − 1) 1 1 M1 Finding sin 2 θ and then sinθ (may be implied). ⇒ sin 2 θ = ⇒ sin θ = [ ± ] 2 2 θ = 45°, 135°, 225°, 315° A1 A1 A1 for any two correct values. A1 for all correct and no others within the range. For answers in radians, A1 only for all 4 angles. If no (correct) working, then SC B1 for all 4 solutions. 4

Mark scheme

5(a) Reflection [in] y-axis B1 B1 B1 for reflection B1 mention of y-axis, OE. SC B2 for stretch, SF –1, parallel to x-axis. −1 B1* B1 for ‘translation’ and a correct vector/description. Translation or shift Do not accept ‘left’/’right’. 0 If two translations then B0 and B0 for the order. Stretch, factor 2, parallel to y-axis B2,1,0 B2 all correct OE. B1 any 2 parts correct. This can be at any point in the sequence. Correct order and three correctly named transformations only DB1 If a fourth transformation is given this mark is not awarded and no marks are given for the two transformations of the same type, except where the reflection is described as a stretch. If any transformation is incorrectly named this cannot be given. −1 1 If translation is not or then DB0 is given. 0 0 Alternative Solution for first 3 marks 1 B1* B1 for ‘translation’ and correct vector/description. Translation or shift 0 Reflection [in] y-axis B1 B1 B1 for ‘reflection’, B1 for ‘in y-axis’. Alternative solutions There are alternative solutions which can be marked in the same way −4 e.g. the given stretch, translation , reflect in x = −2.5 0 6 5(b) g ( x ) = 2f ( − x − 1) or a = 2, b = −1, c = −1 B1 First B1 for a = 2 and no additional terms added to the function. a = −2 is B0. B1 Second B1 for b = −1 and c = −1. 2

Mark scheme

6 ( 10 1 − r 8 ) M1* 17 ( ) ( ) OE, i.e. substituting p and q expressions into ratio . 1− r 1− r 8 4 16 1− r = 17 17 = ×a ( 10 1 − r 4 ) 16 [a (1 − r ) 16 (1 − r ) ] 16 = a (1 − r ) , 17 = a (1 − r ) gets M0 unless recovered later. 4 8 1− r (1 − r ) (1 − r ) Simplifying to 16 r8 − 17r 4 + 1 [= 0 ] (or equivalent form) DM1 (1 − r ) = 1 + r = 17 . 8 Or ( ) 16 4 (1 − r )4 ( )( ) A1 16r 4 − 1 r 4 − 1 = 0 r = ± 1 Or r 4 = 1 1 r = ± (condone extra r = ±1 solution). 2 16 2 10 DM1 Use of correct sum to infinity formula with either of their r S∞ = 1 values providing r < 1 . 1 − [±] 2 20 A1 Allow 6.67 or better. S∞ = 20 and 3 A0 if there is only one or more than two S∞ values. 5

Mark scheme

7(a) 1 M1 Or combination of large semi-circle and small sector: Area of sector BOF = × 202 × ( 2π − 2.4 ) [ = 776.63] 1 1 2 × 202 × π + × 202 × ( π − 2.4 ) . 2 2 Length BD = DF = 2 × 20sin0.6 or 202 + 202 − 2 × 20 × 20cos1.2 M1* Length of radius of small circles is acceptable for M1. [ = 22.58] Area of two semicircles = π × ( 20sin0.6) 2 [= 400.64] DM1 1 M1 Area of triangles = 2 × × 20 × 20sin1.2 [ = 372.81] 2 Total area = 1550 [cm2 ] A1 Expect 1550.09 but accept AWRT to 3sf. 5 7(b) 1 2 B1 θ π πr = 50π r = 10 May be seen as 20sin , where θ = . 2 2 3 π M1* OE ⇒θ= Finding θ using their r. Allow working in degrees. 3 2π DM1 Arc length of sector BOF = 20 × 2π − their 3 2π DM1 Dependent on the first dM1. Total perimeter = 20 × 2π − their + 2π × their 10 3 140π 2 A1 Must be a single exact term. or 46 π 3 3 5

Mark scheme

8(a) 3( x − 2) + 2 or a = −2, b = 2 B1 B1 2 2 8(b) 2 or k = 2 or k ≥ 2 B1FT FT on their a. Do not accept x = 2 or x ⩾ 2. 1 8(c) y−2 M1 Using their completed square form. 3( x − 2 ) + 14 − 12 = y ( x − 2 ) = 2 2 3 y−2 DM1 x = [±] +2 3 x−2 A1 3x − 6 f −1 ( x ) = +2 OE, e.g. y = + 2. 3 3 3 8(d) Finding f −1 ( 29) [= 5] M1 Or solving f(x) = 29 [using their completed square form, OE]. Finding f −1 ( their 5) M1 Or solving f(x) = their 5. x =3 A1 If using f(x) method, x = 1 must be discarded. Alternative solution for Question 8(d) ( ) 3 3 ( x − 2 ) + 2 − 2) 2 + 2 = 29 using their completed square form 2 M1 ( Or 3 3x 2 − 12 x + 14 ) − 12 (3x − 12 x + 14) + 14 = 29. 2 2 Allow if the ' = 29' appears later in the working. Solving as far as 9 ( x − 2) = 9 or x 2 − 4 x + 3 = 0 DM1 OE 4 ( ) Or [ 27] x 4 − 8x3 + 24 x 2 − 32 x + 15 = 0. x = 3 only A1 WWW Only dependent on the first M1. 3

Mark scheme

9(a) y = x 3 − 3 x + 3 and y = 2x3 − 4x2 + 3 x3 − 4x2 + 3x [= 0] M1 Reducing to 3-term cubic or quadratic if x cancelled. [ x ] ( x − 1)( x − 3) [ = 0] DM1 Factorising the cubic or quadratic. x = 0, 1 and 3 {x = 0 may be seen in the working} A1 SC B1 for x = 1, 3 only, with no M marks awarded. 3 ( ( x − 3x + 3 ) − ( 2 x − 4 x + 3)) dx or 9(b) Attempt at integration of both functions. Can be before or after subtraction M1 Expect integration of 3 3 2 of the functions or integrals ( − x + 4x − 3x ) dx. 3 2 At this stage, subtraction can be done either way. x 4 4 x3 3x 2 x 4 3 2 2 4 4 3 A1 OE = ± − + − or ± − x + 3x − x − x + 3x ± covers A1 being awarded to those who subtract the ‘other’ 4 3 2 4 2 4 3 way. 81 108 27 1 4 3 DM1 OE = − + − − − + − , 63 7 27 13 4 3 2 4 3 2 Minimum required is − − − , i.e. four fractions. or 4 4 2 6 Correctly apply limits their 1 and 3. 81 27 1 3 81 108 1 4 − + 9 − − +3 − − + 9 − − + 3 Do not allow if x = 0 used. 4 2 4 2 2 3 2 3 Need at least one correct substitution in every bracket. If two integrals, need to see substitution into both. Allow one sign error only in each expression, if brackets are not shown. 8 A1 Accept if this comes from use of limits f (1) − f ( 3) or = 3 −8 ( x − 4x + 3x ) dx, if 3 used. 3 2 Only dependent on the first method mark. Accept AWRT 2.67. 4

Mark scheme

10(a) −5 − 3 M1* Gradient of AB = [ = −2] 8−4 8 + 4 −5 + 3 M1 Midpoint AB = , ( 6, −1) 2 2 1 1 DM1 Must be used to find equation of perpendicular through their Gradient of normal = − = and an attempt to find the required −2 2 ( 6, −1) . equation 1 1 A1 WWW Equation of perpendicular bisector is y + 1 = ( x − 6) , so y = x − 4 AG – working involving the perpendicular bisector must be 2 2 seen. Alternative Method for Question 10(a) AC2 = ( a − 4) + ( b − 3) , BC2 = ( a − 8) + ( b + 5) both expanded M1* 2 2 2 2 Solving AC = BC [= 10] DM1 Only allow a single sign error. Eliminating a 2 and b2 DM1 May be awarded before the previous DM1. x A1 WWW a = 2b + 8, concluding y = −4 2 4 10(b) 1 M1 May see centre as (2y + 8, y) OE. Using the centre as a, a − 4 May be seen in an incorrect equation. 2 ( 4 − a )2 + (3 − 0.5a + 4)2 = 100 M1 Sub in (4, 3) or (8, −5). Could use circle with ( 6, −1) and r = 80. ( 1.25a 2 − 15a − 35 [ = 0 ] a 2 − 12a − 28 [ = 0] or b2 + 2b − 15 [= 0] ) DM1 Obtain a 3-term quadratic in their x or y. ( a − 14)( a + 2) = 0 a = 14, a = −2 A1 Or ( b − 3)( b + 5) = [0] b = 3, b = −5. ⇒ ( x − 14) + ( y − 3) = 100 and ( x + 2) + ( y + 5) = 100 A1 2 2 2 2 Alternative Method 1 for the first 3 marks: Make a or b the subject from a circle centre ( a , b ) using A or B M1 E.g. b = 100 − ( y − 3) + 4 from circle through A. 2 These equations may have been found in part (a). Form an equation in a or b only M1 Substitute their a or b into their second circle equation. Simplify to a quadratic in a or b DM1 Expect a 2 − 12a − 28 = 0 or b2 + 2b − 15 = 0, OE. Alternative Method 2 for the first 3 marks: Obtaining CM (C, centre; M, mid-point of AB) M1 Expect 80. Must be clear this is CM, not AB. Using the triangle CMT, where CT is parallel to the x-axis, to find the DM1 Expect MT = 4. vertical distance of C from M, MT Using the triangle CMT, where MT is parallel to the y-axis, to find the DM1 Expect CT = 8. horizontal distance of C from M, CT 5

Mark scheme

11(a) dy 1 − 2 1 B1 = kx − 8 x dx 2 d2 y 1 −2 3 B1 = − kx − 8 dx 2 4 2 11(b) − 1 3 1 3 M1 OE x 2 − 8 x = 0 ⇒ 1 − 8 x 2 = 0 or x −1 = 64 x 2 x3 = or 8 x 2 = 1 1 64 Award if working leads to x = WWW. 4 dy − 1 Setting their to zero and solving, providing their only error(s) are dx Squaring x 2 − 8 x 2 = 0 to x −1 − 64 x 2 = 0 gets M0. incorrect coefficients 1 A1 If x = 0 included, A0 and max of 3/4. x= only 1 4 1 − SC B1 only for x = only from squaring x 2 − 8 x 2 = 0 4 −1 directly to x − 64 x 2 = 0 (SC B1 replacing the M1A1). 11 A1 11 − 1 y= SC B1 for y = from squaring x 2 − 8 x 2 = 0 to 4 4 −1 x − 64 x = 0. 2 d2 y 1 − 3 B1 FT WWW 2 = − x 2 − 8 which is negative, so maximum d2 y dx 2 FT their x-value and their . dx 2 No FT if x = 0 is the only solution. 4 11(c) 1 M1* OE When x = 1, attempting to find y = k − 2 and gradient = k −8 SC B1 if both correct gradients only, or both correct 2 y-coordinates only. 1 A1 k k k k Equation of tangent is y − k + 2 = k − 8 ( x − 1) OE, e.g. y = − 8 x + + 6 or y = x − 8 x + + 6. 2 2 2 2 2 1 1 M1* OE When x = ,attempting to find y = k + 1.75 and gradient = k − 2 4 2 1 A1 k 9 k 9 Equation of tangent is y − k − 1.75 = ( k − 2 )( x − 0.25) OE, e.g. y = ( k − 2 ) x + + or y = kx − 2 x + + . 2 4 4 4 4 1 1 DM1 k k k 9 Meet at k − 8 ( 0.6 − 1) + k − 2 = ( k − 2 )( 0.6 − 0.25 ) + k + 1.75 OE, e.g. − 8 0.6 + + 6 = ( k − 2 ) 0.6 + + . 2 2 2 2 4 4 Equate two tangent equations and substitute x = 0.6 M0 if constants in both equations are the same. ⇒ [ −0.2k + k + 3.2 − 2 = 0.35k − 0.7 + 0.5k + 1.75 ] A1 ⇒ 0.05k = 0.15 k =3 6