Mathematics 9709/12 October/November 2023: question paper and mark scheme

Mark scheme

1 [Coefficient of x3 from ( 3 + 2ax ) =] 10 × 9 × 8a3 = 720a3 B1 May be seen in an expansion or with x3. 5 [Coefficient of x2 from ( 2 + ax ) =] 15 × 16 × a 2 = 240a 2 B1 May be seen in an expansion or with x2. 6 ( ) ( their 10 × 9 × 8a3 = 6 × their 15 ×16 × a 2 ) M1 OE Equating their coefficient of x3 and 6 × their coefficient of x2. 720a3 = 1440a 2 a=2 A1 Condone extra solution a = 0. 4

Mark scheme

2 π π −1 π M1 OE [tan −1 4 x =] their − ± tan 4 x = ± 3 , ±1.047 or 0 6 6 3 π Evaluating − cos −1 in rad and adding or subtracting . 2 6 Allow working with both angles in degrees. A1 4 x = − 3, x = − 3 Note: answer of −0.43 or 3 implies M1 4 4 2

Mark scheme

3(a) M1 Tangent gradient must come from x = 2 substituted into the −1 −1 2 given expression. =− 11 11 [Gradient of normal =] Their 11 2 2 y −8 2 2 x 92 A1 OE =− or 11y + 2 x = 92 or y = − + x−2 11 11 11 2 3(b) 1 2 72 x 2 24 B1, B1 One mark for each correct unsimplified { }. [ ] y = x ÷ 2 + 3 ÷ −3 [ ] − 3 + c + c 2 x 4 x 1 24 M1 Substitution of x = 2, y = 8 into their integrated expression, 8= ×4− +c defined by at least one correct power. Two terms and + c 4 8 needed. 1 24 A1 Both coefficients must be simplified but allow x −3 . Condone y = or 0.25 x 2 − 3 + 10 4 x c = 10 as line as long as either y or f(x) = is seen elsewhere. 4

Mark scheme

4(a) π 60 B1 Finding one correct arc length – may be implied by correct [ Arc length = ] 2 × or × 2π × 2 final answer. 3 360 [ Perimeter =] 2π or 6.28 B1 AWRT 2 4(b) 1 2 π 60 2π B1 SOI AWRT [Area of one sector =] × 2 × or × π × 22 = or 2.09 2 3 360 3 1 π B1 AWRT [Area of triangle = ] × 22 × sin or other valid method Allow use of 60 ° 2 3 = 3 or1.73 [ Area of coin = 3 segments + triangle ] 3 2π M1 OE − 3 + 3 [= 2.82] 3 2π Or 3 sectors – 2 triangles 3 × − 2 × 3 or 3 2π 2π Sector + 2 segments + 2 − 3 3 3 2π − 2 3 or 2( π - 3 ) A1 Must be one of these simplified versions but equivalent decimal answers can score B1B1M1 4

Mark scheme

5(a) cosθ 2 − sin θ *M1 OE. Forming a correct equation in θ only using the terms of = leading to cos2 θ [sinθ ] = sinθ ( 2 − sinθ )[sinθ ] sin θ cosθ the GP and an attempt to clear fractions. 1 DM1 Correct use of cos2 θ + sin 2 θ = 1 and attempt to solve for cos2 θ + sin 2 θ = 2sin θ leading to sin θ = 2 sinθ . π A1 AWRT [θ =] or 0.524 5π 6 A0 for θ = 30° . Condone inclusion of and/or 6 3 5(b) 1 B1 OE SOI a= r= 3 Trigonometric values need to have been evaluated but allow 2 decimal equivalents (0.5 and 1.73 AWRT) M1 Use of a correct formula for S10, with their value of θ . ( ) 10 π 1 − their 3 S10 = sin their Their 3 needs to come from ( 6 1 − their 3 ) cos ( theirθ ) 2 − sin ( theirθ ) or OE sin ( theirθ ) cos ( theirθ ) 121 A1 −121 [ S10 =] or 165 AWRT scores B1M1A0. 3 −1 1− 3 3 ( )

Mark scheme

6(a) d 2 M1 dy x − 8x + 5 = 0 [ 2 x − 8 = 0 ] Correct differentiation of x2 and equating their to 0. dx dx Alternative method 1 for first mark of Question 6(a) y = ( x − 4) − 11 M1 Attempt to complete the square as far as y = ( x − 4)2 ± k . 2 Alternative method 2 for first mark of Question 6(a) −b ±8 M1 x= = 2a 2 x = 4, y = -11 A1 8 ± 64 − 20 Answers from x = leading to x = 4 ± 11 2 scores M0A0 2 6(b) x = ( their x value from a ) + 4 [=8] B1 FT Can be from finding the equation of the transformed curve, dy differentiating and putting = 0. dx y = (their y value from a ) × 2 +1 [ − 21] B1 FT Can be from putting x = 8 in the equation of the transformed curve. 2 If B0B0 scored, SC B1 for sight of ( 4, −22 ) . 6(c) ( ) 2 x 2 − 8 x + 5 or 2{ ( x − 4) − 11} 2 B1 Can be implied if both transformations done together: ( ) 2 ( x − 4 ) − 8 ( x − 4 ) + 5 + 1 OE. 2 (( x − 4) − 8 ( x − 4) + 5) + 1 or ( x − 4 − 4) − their11 + 1 2 2 M1 For the x translation, each x becomes ( x − 4 ) . M1 For the y translation of +1. y = 2 x 2 − 32 x + 107 or a = 2,b = −32, c = 107 A1 Evidence to support their answer may be in (b) but answer must be seen in (c). 4

Mark scheme

7(a) ( 2 x − 1) ( 4 x2 + 2 x − 1) = 8x3 + 4 x2 − 2 x − 4 x2 − 2 x + 1 = 8x3 − 4 x + 1 B1 AG Six correct terms leading to the correct answer. 1 7(b) sin 2 θ *M1 sin θ +1 For use of tan θ = in the numerator and denominator. cos 2 θ sin θ + cos θ 2 2 cosθ Starting with the LHS = sin 2 θ sin θ − cos θ 2 2 −1 cos 2 θ 1 DM1 For use of sin 2 θ + cos 2 θ = 1 twice, in a correct expression, = need to see clear evidence of this step 1 − cos 2 θ − cos 2 θ resulting in an expression in cos2 θ . 1 A1 AG = 1 − 2cos 2 θ Alternative method 1 for Question 7(b) sin 2 θ + cos 2θ sin 2 θ + cos 2θ *M1 For use of sin 2 θ + cos 2 θ = 1 twice. Starting with the RHS = sin 2 θ + cos 2 θ − 2cos 2 θ sin 2 θ − cos 2 θ sin 2 θ DM1 Dividing throughout by cos 2 θ . +1 = cos2 θ 2 need to see clear evidence of this step sin θ −1 cos 2 θ tan 2 θ + 1 A1 AG = tan 2 θ − 1 7(b) Alternative method 2 for Question 7(b) sec 2 θ *M1 For use of 1+ tan 2 θ = sec2 θ twice. Starting with the LHS sec 2 θ − 2 1 DM1 AG For multiplying throughout by cos 2 θ to give the RHS. Clear statement 1 − 2cos 2 θ A1 3 ( ) 7(c) 1 B1 Replace LHS with RHS from (b) and clear fractions. = 4cosθ leading to 1 = 4cosθ 1 − 2cos 2 θ 1 − 2cos 2 θ 8cos3 θ − 4cosθ + 1= 0 ( 2cosθ − 1) ( 4cos2θ + 2cosθ − 1)[= 0] *B1 Use of the expression from (a) with x = cosθ . 1 −2 ± 4 + 16 DB1 OE [ x or cosθ =] and OR 0.31, −0.81 AWRT For all three values. 2 8 [θ =] 60°,72°,144° B2,1,0 B2 for three correct answers only, B1 for two correct answers and no others (but allow 36 ° instead of 144 ° ) in the given range or 3 correct answers plus other values in the given range. Ignore answers outside of the given range. Accept AWRT 72.0, 144.0 . SC B1 for all 3 correct answers in radians and no others: π 2π 4π , and . 3 5 5 5

Mark scheme

8(a) y = ( x + a ) − a leading to ( x + a ) = y ± a *M1 x and y may be interchanged initially. 2 2 Allow ± errors for these method marks. x = [ ±] y ± a ± a DM1 Alternative method for first 2 marks of Question 8(a) x = ( y + a ) − a leading to y2 + 2ay + a2 − a − x [= 0] 2 *M1 Allow ± errors for this method mark. y= ( −2a ± 4a 2 − 4 a 2 − a − x ) DM1 2 [y or f −1 (x) =] − x + a − a A1 OE Must choose negative root. 3 8(b)(i) x −a B1 Ignore infinity limit if included. 1 8(b)(ii) y or f −1 ( x ) −a B1 Ignore negative infinity limit if included. 1 8(c) 7 7 7 B1 OE 2 2 7 7 7 = 2 2 gf x + − − 1 or 2 x 2 + 4 x + 2 − 2 − 1[ = 0] 1 2 2 2 2 2 Alternatively, gf ( x ) = 0 f ( x ) = . 2 33 M1 OE −14 ± 142 − 4 × 2 × Solving their three term quadratic equation as far as two 7 [ x =] − ± 2 or 2 solutions or correctly selecting the negative root only. 2 4 1 7 7 −14 ± 64 Alternatively, ± + − . or factorising 2 2 2 4 11 A1 If B1M0 scored then award SCB1 for the correct final [ x =] − answer. 2 3

Mark scheme

9(a) 1 − 1 − 1 1 1 *M1 OE 2 x 2 + 13x 2 = 3x 2 + 12 all × x 2 x − 6 x 2 + 5 = 0 Equating the two expressions in x and then multiplying each 1 1 term by x 2 or by their substitution for x 2 . Coefficients need to be retained but condone +/– sign errors. 1 Allow x 2 replaced by x. 1 1 6 ± 36 − 4 × 1 × 5 DM1 OE x 2 − 1 x 2 − 5 [= 0] or [x=] Solving their three-term quadratic. 2 Alternative method for first 2 marks of Question 9(a) 1 − 1 − 1 1 1 *M1 Equating the two expressions in x and isolating their term in 2 x 2 + 13x 2 = 3x 2 + 12 all × x 2 leading to 2 x + 10 = 12 x 2 1 x2 . ( ) (2 x + 10) 2 = 144 x leading to [4] x2 − 26 x + 25 [ = 0] DM1 OE Squaring both sides, rearranging and solving a three-term 26 ± 676 − 4 × 1 × 25 leading to [4] ( x − 25 )( x − 1) [= 0] or [x=] quadratic. 2 3 A1, A1 A1 for both x-values and A1 for both y values. x = 1 and 25 , y = 15 and 12 If M1DM0 scored then SCB1B1 is available for final 5 answers. 4 Answers without working score 0/4 9(b) −1 1 − 1 1 − 1 M1 Attempt to integrate, defined by at least one correct fractional Area = 3x 2 + 12 − 2 x 2 + 13x 2 [ dx ] = −2 x 2 + 12 − 10 x 2 power, and subtract – condone the wrong way round. 3 1 B1 B1 B1 for either { }. 2x 2 10 x 2 B1 for completely correct integration of their expression = − + 12 x − 1 following through +/– sign errors from the subtraction. 3 2 2 4 3 1 M1 OE − ( their 25 ) 2 + 12 ( their 25 ) − 20 ( their 25 ) 2 − Substitution of their positive limits from part (a) in their 3 integrated expression, defined by at least one correct 4 3 1 − ( their 1) 2 + 12 ( their 1) − 20 ( their 1) 2 fractional power, and subtraction. 3 9(b) Alternative method for first 4 marks of Question 9(b) −1 1 − 1 M1 Attempt to integrate, defined by at least one correct fractional Area = 3x 2 + 12 [dx ] − 2 x 2 + 13x 2 [dx ] power, and subtract – condone the wrong way round. 1 3 1 B1 B1 OE 3x 2 2 x 2 13x 2 One mark for each correct expression. = + 12 x [ −] + 1 3 1 2 2 2 1 1 M1 OE 6 ( their 25) 2 + 12 ( their 25) − 6 ( their 1) 2 + 12 ( their 1) [ − ] Substitution of their positive limits from part (a) in both of their integrated expressions, defined by at least one correct 4 3 1 4 3 1 fractional power, and subtraction. ( their 25) 2 + 26 ( their 25) 2 − ( their 1) 2 + 26 ( their 1) 2 3 3 128 2 A1 AWRT [Area =] , 42 , 42.7 If M1B1B1M0 then SC B1 available for correct final answer. 3 3 Condone negative answer if corrected. 5 Condone the presence of π for the first 4 marks but use of y 2 scores 0/5

Mark scheme

10(a) dy 5 2 20 B2,1,0 B2 Three correct unsimplified { } and no others. = ( 4 x − 3) 3 ×4 − B1 Two correct { }or three correct { } and an additional term dx 3 3 e.g. + c. B0 More than one error. 20 2 20 M1 dy ( ) = 0 leading to ( 4 x − 3) = k , k > 0 leading to 2 3 4 x − 3 3 − Equating their to 0 and using a valid method to arrive at 3 dx 4 x − 3 = ±m 2 answers. 1 A1 [4 x − 3 = ±1] [ x =] ,1 2 d 2 y 40 1 B1 OE = ( 4 x − 3 ) − 3 ×4 dx 2 9 1 d 2 y 160 1 160 B1 d2 y x = = ( 4 x − 3 ) − 3 <0 or − or − 17.8 so max If evaluated the answers for both must be correct OR 2 dx 9 2 9 dx 2 [ x = 1] 2 = ( 4 x − 3)− 3 > 0 or d2 y 160 1 160 dy or 17.8 so min Clear use of change in sign of correctly for both B1’s. dx 9 9 dx If B1M1A0B0B0 scored then SCB1 can be awarded for: dy 5 2 20 = ( 4 x − 3) 3 − leading to ( 4 x − 3) = 64 leading 2 dx 3 3 5 11 to x = − , . 4 4 d 2 y 10 1 5 d2 y = ( 4 x − 3 ) − 3 , x=− , < 0 so max, dx 2 9 4 dx 2 11 d 2 y x = , 2 > 0 so min. 4 dx 6 10(b) x < 12 , x > 1 B1 Allow ⩽ and/or ⩾. FT only from special case x < − 54 , x > 114 Condone: 1 < x < 12 . 1

Mark scheme

11(a) 7−4 18 − 7 *M1 Difference in the ys their × their = −1 Their gradients must both come from . p−6 14 − p Difference in the xs OR Scalar product leading to (14 − p )( 6 − p ) − 33 = 0 p 2 − 20 p + 84 = 33 leading to p2 − 20 p + 51[= 0]or p2 − 20 p = −51 A1 Clearing of fractions and collecting terms to arrive at the three-term quadratic. Allow integer multiples. Alternative method for first 2 marks of Question 11(a) ( p − 6)2 + ( 7 − 4)2 + (14 − p )2 + (18 − 7)2 = (14 − 6)2 + (18 − 4)2 *M1 For correct use of Pythagoras with A,B and C. OR OR For correct use of Pythagoras with the centre, B and one of E.g. (10 − p ) + 42 = 42+72 2 the other two points. 2 p2 − 40 p + 102[= 0] A1 OE Collecting terms to arrive at the three-term quadratic. 20 ± 202 − 4 × 51 DM1 OE [2] ( p − 3)( p − 17 ) or Solving their three-term quadratic. 2 p=3 A1 If M1A1DM0 scored then SC B1 is available for final answer. 4 11(b) [Midpoint or Centre is] (10, 11) B1 SOI by final answer. 1 M1 Finding half of the length of AC or using their centre, which (14 − 6)2 + (18 − 4)2 or (18 − their11)2 + (14 − their10)2 or cannot be A, B or C, to find r 2 or r. Note: r = 65 is M0. 2 ( their11 − 4)2 + ( their10 − 6)2 r 2 = 65 or r = 65 ( x − 10)2 + ( y − 11)2 = 65 or x 2 + y 2 − 20 x − 22 y + 156 = 0 A1 ( x − 6 )( x − 14 ) + ( y − 4 )( y − 18) = 0 scores 3/3. 3 11(c) 18 − their11 their11 − 4 18 − 4 7 *M1 Gradient of their centre, which cannot be A, B or C, from 14 − their10 or their10 − 6 or 14 − 6 = 4 part (b), to A or C or the gradient of AC but working needed if incorrect centre. OR by clearly differentiating and substitution of (14,18). 1 DM1 OE y − 18 = − 7 ( x − 14) 1 their Using (14,18) and − to form the equation of a 4 7 their 4 straight line. 4 x + 7 y − 182 = 0 A1 All terms on one side in any order. Allow multiples of this format by an integer only. 3