Mathematics 9709/11 October/November 2021: question paper and mark scheme

Mark scheme

1(a) 1 1 B1 OE. Multiply or use binomial expansion. Allow 1− + 2 unsimplified. x 4x 1 1(b) 1 + 12 x + 60 x 2 + 160 x 3 B2, 1, 0 Withhold 1 mark for each error; B2, 1, 0. ISW if more than 4 terms in the expansion. 2 1(c) their (1× 12 ) + their ( −1× 60 ) + their ( 14 × 160 ) M1 Attempts at least 2 products where each product contains one term from each expansion. [12 − 60 + 40 =] −8 A1 Allow ‒8x. 2

Mark scheme

2 kx 2 + 2 x − k = kx − 2 leading to kx 2 + ( − k + 2 ) x − k + 2 [ = 0] *M1 Eliminate y and form 3-term quadratic. Allow 1 error. ( − k + 2 ) 2 − 4k ( − k + 2 ) DM1 Apply b 2 − 4 ac ; allow 1 error but a, b and c must be correct for their quadratic. 5k 2 − 12k + 4 or ( −k + 2 )( −k + 2 − 4k ) A1 May be shown in quadratic formula. ( −k + 2 )( −5k + 2 ) DM1 Solving a 3-term quadratic in k (all terms on one side) by factorising, use of formula or completing the square. Factors must expand to give their coefficient of k2. 2 A1 WWW, accept two separate correct inequalities. <k<2 If M0 for solving quadratic, SC B1 can be awarded for 5 correct final answer. 5

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3 3cosθ (2 tan θ − 1) + 2(2 tan θ − 1) [ = 0] M1 Or similar partial factorisation; condone sign errors. (2 tan θ − 1) (3cosθ + 2) [ = 0] M1 OE. At least 2 out of 4 products correct. 1 2 [leading to tanθ = , cosθ = − ] 2 3 26.6º, 131.8º A1 A1 WWW. Must be 1 d.p. or better. Final A0 if extra solution within the interval. SC B1 No factorisation: Division by 2tan θ ‒1 leading to 131.8° or division by 3cosθ + 2 or similar leading to 26.6°. Alternative method for question 3 sinθ sin θ M1 sinθ 6cos θ − 3cosθ + 4 − 2 [ = 0] Using tanθ = cosθ and reaching a partial factorisation; cosθ cos θ 6cos θ sin θ − 3cos 2θ + 4sin θ − 2cosθ [ = 0] condone sign errors. 2sinθ ( 3cosθ + 2 ) − cosθ ( 3cosθ + 2 ) [ = 0] (2 sinθ − cosθ ) ( 3cosθ + 2 ) [ = 0] M1 At least 2 out of 4 products correct. 1 2 [leading to tanθ = , cosθ = − ] 2 3 26.6º, 131.8º A1 A1 WWW. Must be 1 d.p. or better. Final A0 if extra solution within the interval. SC B1 No factorisation: Division by 2tan θ ‒1 leading to 131.8° or division by 3cosθ + 2 or similar leading to 26.6°. 4

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4(a) 5a B1 Use of correct formula for sum to infinity. 1 − ( ± 14 ) 8 *M1 Use of correct formula for sum of 8 terms and form 2a + 7 ( −4 ) 2 equation; allow 1 error. 4a = 8a − 112 leading to a = [ 28] DM1 Solve equation to reach a value of a. a = 28 A1 Correct value. 4 4(b) their 28 + ( k − 1)( −4 ) = 0 M1 Use of correct method with their a. [ k =] 8 A1 2

Mark scheme

5(a) a=5 B1 b=2 B1 c=3 B1 3 5(b)(i) 3 B1 1 5(b)(ii) 2 B1 1

Mark scheme

6(a) π B1 SOI; allow 60°. Recognise that at least one of angles A, B, C is 3 π π M1 SOI e.g. may see 2π or 4π. One arc 6 × their leading to two arcs 2 × 6 × their Use of correct formula for length of arc and multiply by 2. 3 3 Perimeter = 6 + 4π A1 Must be exact value. Alternative method for question 6(a) Calculate circumference of whole circle = 12π B1 1 1 M1 SOI e.g. may see 2π or 4π. One arc × 12π leading to two arcs 2 × × 12π 6 6 Perimeter = 6 + 4π A1 Must be exact value. 3 6(b) 1 2 π M1 Use of correct formula for area of sector. Sector = × 6 × their SOI e.g. may see 6π or 12π. 2 3 1 π 1 π M1 A1 M1 for attempt at strategy with values substituted: 2 ( ) ( ) × 62 × their − × 62 × sin their + 6π = 6π − 9 3 +6π 3 2 3 area of segment + area of sector A1 if correct (unsimplified). Area = 12π − 9 3 A1 Must be simplified exact value. Alternative method for question 6(b) 1 2 π M1 Use of correct formula for area of sector. Sector = × 6 × their SOI e.g. may see 6π or 12π. 2 3 1 π 1 π M1 A1 M1 for attempt at strategy with values substituted: ( ) 2 × × 62 × their − × 62 × sin their 2 3 2 3 2 × sector – triangle A1 if correct (unsimplified). Area = 12π − 9 3 A1 Must be simplified exact value. Alternative method for question 6(b) 1 2 π M1 Use of correct formula for area of sector. Sector = × 6 × their SOI e.g. may see 6π or 12π. 2 3 1 π 1 π M1 A1 M1 for attempt at strategy with values substituted: ( ) ( ) 2 × × 62 × their − × 62 × sin their + 2 × segment + triangle 2 3 2 3 A1 if correct (unsimplified). 1 π 2 ( ) × 62 × sin their = 12π − 18 3 + 9 3 3 Area = 6π − 9 3 + 6π = 12π − 9 3 A1 Must be simplified exact value. 4

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7(a) r 2 = ( 5 − 2 ) + ( 7 − 5) = 13 2 2 B1 r 2 = 13 or r = 13 Equation of circle is ( x − 5) + ( y − 2 ) = 13 2 2 B1 FT OE. FT on their 13 but LHS must be correct. 2 7(b) ( x − 5)2 + ( 5x − 10 − 2 )2 = 13 M1 Substitute y = 5 x − 10 into their equation. 26 x 2 − 130 x + 156 [ = 0] A1 FT OE 3-term quadratic with all terms on one side. FT on their circle equation. [ 26] ( x − 2)( x − 3) [ = 0] M1 Solve 3-term quadratic in x by factorising, using formula or completing the square. Factors must expand to give their coefficient of x2. (2, 0), (3, 5) A1 A1 Coordinates must be clearly paired; A1 for each correct point. A1 A0 available if two x or y values only. If M0 for solving quadratic, SC B2 can be awarded for correct coordinates, SC B1 if two x or y values only. ( AB )2 = ( 3 − 2)2 + ( 5 − 0 )2 M1 SOI. Using their points to find length of AB. AB = 26 A1 ISW. Dependent on final M1 only. 7(b) Alternative method for question 7(b) y + 10 2 M1 y + 10 − 5 + ( y − 2 ) = 13 2 Substitute x = into their equation. 5 5 26 y 2 26 y A1 FT OE 2-term quadratic with all terms on one side. FT on their − [ = 0] circle equation. 25 5 [ 26] y ( y − 5) [ = 0] M1 Solve 2-term quadratic in y by factorising, using formula or completing the square. Factors must expand to give their coefficient of y2. (2, 0), (3, 5) A1 A1 Coordinates must be clearly paired; A1 for each correct point. A1 A0 available if two x or y values only. If M0 for solving quadratic, SC B2 can be awarded for correct coordinates, SC B1 if two x or y values only. ( AB )2 = ( 3 − 2)2 + ( 5 − 0 )2 M1 SOI. Using their points to find length of AB. AB = 26 A1 ISW. Dependent on final M1 only. 7

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8(a) {−3 ( x − 2 ) } 2 {+14} B1 B1 B1 for each correct term; condone a = 2, b = 14. 2 8(b) [k =] 2 B1 Allow [x] ⩽ 2. 1 8(c) [Range is] [y] ⩽ –13 B1 Allow [f(x)] ⩽ –13, [f ] ⩽ –13 but NOT x ⩽ –13. 1 8(d) 14 − y M1 y −14 y = −3 ( x − 2 ) + 14 leading to ( x − 2 )2 = 2 Allow . Allow 1 error in rearrangement if x, y on 3 −3 opposite sides. 14 − y A1 y −14 x = 2(±) Allow . 3 −3 14 − x A1 x −14 [f −1 ( x )] = 2 − OE. Allow . Must be x on RHS; must be negative 3 −3 square root only. Alternative method for question 8(d) 14 − x M1 x −14 x = −3 ( y − 2 ) + 14 leading to ( y − 2 )2 = 2 Allow . Allow 1 error in rearrangement if x, y on 3 −3 opposite sides. 14 − x A1 x −14 = 2(±) Allow . 3 −3 14 − x A1 x −14 [f −1 ( x )] = 2 − OE. Allow . Must be x on RHS; must be negative 3 −3 square root only. 3 8(e) { [g(x) =] −3 ( x + 3 − 2 ) 2 } + {14 + 1} B2, 1, 0 { OR −3 ( x + 3) 2 } + {12 ( x + 3)} + {3} g ( x ) = −3x 2 − 6 x + 12 B1 3

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9(a) f ( x ) = 23 x3 − 7 x + 4 x −1 [ +c ] B2, 1, 0 Allow terms on different lines; allow unsimplified. − 13 = 23 − 7 + 4 + c leading to c = [2] M1 Substitute f(1) = − 1 into an integrated expression and 3 evaluate c. f ( x ) = 23 x3 − 7 x + 4 x −1 + 2 A1 OE. 4 9(b) 2 x 4 − 7 x 2 − 4 [ = 0] M1 Forms 3-term quadratic in x2 with all terms on one side. Accept use of substitution e.g. 2 y 2 − 7 y − 4 [ = 0] . ( 2 x + 1)( x − 4 ) [= 0] 2 2 M1 Attempt factors or use formula or complete the square. Allow ± sign errors. Factors must expand to give their coefficient of x2 or e.g. y. Must be quartic equation. Accept use of substitution e.g. ( 2 y + 1)( y − 4 ) . x = [ ±] 2 A1 If M0 for solving quadratic, SC B1 can be awarded for [ ±] 2 . 2 3 4 14 B1 B1 B1 B1 for correct coordinates clearly paired; B1 for each 3 ( 2 ) − 7 ( 2 ) + 2 + 2 leading to 2, − 3 correct point; B1 B0 if additional point. 2 4 26 3 ( −2 ) − 7 ( −2 ) + −2 + 2 leading to −2, 3 3 5 9(c) f ′′ ( x ) = 4 x + 8 x −3 B1 OE 1 9(d) f ′′ ( 2 ) = 9 > 0 MINIMUM at x = their 2 B1 FT FT on their x = [ ± ] 2 provided f ′′ ( x ) is correct. Must have correct value of f ′′ ( x ) if x = 2 . f ′′ ( −2 ) = −9 < 0 MAXIMUM at x = their − 2 B1 FT FT on their x = [ ±] 2 provided f ′′ ( x ) is correct. Must have correct value of f ′′ ( x ) if x = −2 . Special case: If values not shown and B0B0 scored, SC B1 for f ′′ ( 2 ) > 0 MIN and f ′′ ( −2 ) < 0 MAX Alternative method for question 9(d) Evaluate f ' ( x ) for x-values either side of 2 and –2 M1 FT on their x = [ ± ] 2 MINIMUM at x = their 2, MAXIMUM at x = their 2 A1 FT FT on their x = [ ± ] 2 . Must have correct values of f ' ( x ) if shown. Special case: If values not shown and M0A0 scored SC B1 f ′ ( 2 ) − /0/+ MIN and f ' ( −2 ) +/0/ − MAX Alternative method for question 9(d) Justify maximum and minimum using correct sketch graph B1 B1 Need correct coordinates in (b) for this method. 2

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10(a) ( 3x − 2 )− 12 B2, 1, 0 Attempt to integrate ÷ {3} −1/ 2 − 23 [ 0 − 1] M1 M1 for applying limits 1 → ∞ to an integrated expression (either correct power or dividing by their power). 2 A1 3 4 10(b) ( 3x − 2 )−2 *M1 A1 M1 for attempt to integrate y2 (power increases); allow 1 [ ] π y d2 x = [ ]( π 3 x − 2 ) d −3 x = [ ] π error. A1 for correct result in any form. −2 × 3 DM1 Apply limits 1 and 2 to an integrated expression and [ π ] − 1 1 − 1 subtract correctly; allow 1 error. 6 16 5π A1 OE 32 4 10(c) dy 3 − 5 M1 M1 for attempt to differentiate (power decreases); allow 1 = − × 3 ( 3x − 2 ) 2 error. dx 2 dy 9 *M1 Substitute x = 1 into their differentiated expression; allow 1 At x = 1, =− error. dx 2 2 DM1 Forms equation of line or evaluates c using (1, 1) and [Equation of normal is] y − 1 = ( x − 1) OR evaluates c −1 9 gradient . dy their dx 7 A1 OE e.g. AWRT 0.778; must clearly identify y-intercept At A, y = 9 4