Mathematics 9709/12 May/June 2022: question paper and mark scheme

Mark scheme

1 Coefficient of x 4 = 15 B1 Condone inclusion of x4 . Can be seen as part of an expansion. Coefficient of x2 = 240a 2 B1 Condone inclusion of x 2 . Can be seen as part of an expansion. ‘Their 240’a2 – ‘their 15’ M1 Forming an equation of the form pa2 = q, where p and q are constants. Condone inclusion of powers of x as long as they then disappear. 1 A1 OE a= or 0.25 4 1 1 Do not condone extra ‘answer’ of − ,or allow or similar. 4 16 4

Mark scheme

2 r = 0.8 B1 OE a = 12.5 B1 OE S∞ = 12.5 ÷ (1 − 0.8 ) M1 a Using with ‘their a’ and ‘their r’ but r must be < 1 . 1− r 125 1 A1 1 S∞ = , 62 or 62.5 12 2 2 2 or similar does not get A1. 1 5 4

Mark scheme

3 3 B1 B1 Marks can be awarded for correct unsimplified expressions ISW. 3(4 x − 7) 2 4 12 1 3 1 [ ] 3 y = + − 1 x (4 x − 7) 2 − 8x 2 [ +c ] ×4 2 2 2 5 1 3 1 M1 5 = ( 9 ) 2 − 8 × 4 2 + c [⇒c = 5] Using (4, ) in an integrated expression (defined by at least one 2 2 2 correct power) including + c. 3 3 1 A1 Condone c = 5 as their final line if either y = or f(x) = seen y= ( 4 x − 7 ) 2 − 8x 2 + 5 . elsewhere in the solution. Coefficients must not contain 6 unresolved double fractions. 4

Mark scheme

4(a) 2 × 6k = k + k + 6 or 6k − k = k + 6 − 6k B1 OE or 2d = 6 leading to d = 3, ⸫ 6k – 3 = k A correct equation in k only. Can be implied by correct final answer. 6 B1 OE k= or 0.6 10 2 4(b) d=3 B1 Correct value of d can be implied by a correct final answer. Working may be seen in part (a) but must be used in (b). 30 M1 It needs to be clear that the candidate is using a correct sum S30 = ( 2 × ‘their k’ + 29 × ‘their d’) formula. There is no requirement to check the candidates working 2 for d but it must be clearly identified. S30 = 1323 A1 ISW if corrected to 1320. 3

Mark scheme

5(a) 1 M1 Equating the equations of curve and line and substituting x = 0 . 4 × 02 − 0 + k 2 = 0 − a Condone slight errors e.g. ± sign errors. 2 3 3 2 1 3 M1 3 4× − k + k2 = − a Equating the equations of curve and line and substituting x = . 4 4 2 4 4 Condone slight errors e.g. ± sign errors. k = 2, a = − 2 A1 A1 WWW Alternative method for question 5(a) *M1 3 ( x − 0 ) x − 3 = 0 or x ( 4 x − 3) = 0 [⟹ 4 x − 3 x = 0] 2 Use 0, to form a quadratic equation. Do not allow 4 4 ( x + 0) x + = 0 . 3 4 1 1 DM1 Equating the equations of curve and line and rearranging so that 4 x 2 − kx + k 2 = x − a leading to 4 x 2 − ( k + 1) x + k 2 + a [ = 0] terms are all on same side. Condone slight errors e.g. ± sign errors. 2 2 k = 2, a = − 2 A1 A1 WWW Alternative method for question 5(a) b 3 c 3 *M1 Using sum and product of roots. Condone ± sign errors. − = + 0 and = 0 × a 4 a 4 1 2 DM1 3 k +1 3 k +a Equating the equations of curve and line and equating to and 0. = and 2 =0 4 4 4 4 k = 2, a = − 2 A1 A1 WWW 4 5(b) 1 7 1 7 *M1 7 4 x 2 − kx + k 2 = x + 4 x 2 − kx − x + k 2 − [ = 0] OE Substitute a = − and rearrange so that terms are all on same 2 2 2 2 2 side, condone ± sign errors. Watch for multiples. *DM1 Use of b2 – 4ac with the coefficients from their 3-term quadratic. ( k + 1) − 4 × 4 k 2 − 2 1 7 Both coefficients ‘b’ and ‘c’ must consist of two components. 2 2 7 k 2 − 2k − 57 A1 OE ( k − 3)( 7 k + 19 ) or other valid method DM1 Factorising or use of the formula or completing the square. Must be evidence of an attempt to solve for this mark. Dependent upon both previous method marks. 19 A1 OE e.g. AWRT − 2.71. No ISW if inequalities used. k = 3, k = − SC: If second DM1 not scored, SC B1 available for correct final 7 answers. Alternative method for question 5(b) 1 7 *M1 Equating gradients and equating line and curve. 8 x − k = 1 and 4 x 2 − kx + k 2 = x + 2 2 1 7 *DM1 Forming an equation in x or k only. 4 x 2 − ( 8 x − 1) x + (8 x − 1)2 = x + or 2 2 k +1 2 k +1 1 2 k +1 7 4( ) −k + k = + 8 8 2 8 2 28 x 2 − 8 x − 3 or 7 k 2 − 2k − 57 A1 OE A correct 3 term quadratic in x or k only. (14 x + 3)( 2 x − 1) or ( k − 3)( 7k + 19 ) or other valid method DM1 OE Factorising or use of the formula or completing the square. Must be evidence of an attempt to solve for this mark. Dependent upon both previous method marks. 5(b) 19 A1 OE e.g. AWRT − 2.71. No ISW if inequalities used. k = 3, k = − SC: If second DM1 not scored, SC B1 available for correct final 7 answers. 5

Mark scheme

6 Line meets curve when: M1 Equating line and curve and rearranging so that terms are all on 1 1 same side, condone sign errors, and making a valid attempt to 2 x + 2 = 5x 2 leading to [ ] 2 x — 5 x 2 +2 = 0 solve by factorising, using the formula or completing the square. 1 1 or 4 x + 8 x + 4 = 25 x leading to 4 x 2 − 17 x + 4 [ = 0 ] 2 Factors are:(2 x 2 -1)( x 2 -2), (4x-1)(x-4) and (2y-5)(y-10). y2 or x = leading to 2 y 2 − 25 y + 50 [ = 0 ] 25 1 A1 SC: If M1 not scored, SC B1 available for correct answers, could x= ,x = 4 just be seen as limits. 4 1 1 *M1 Intention to integrate and subtract areas. Condone missing brackets Area = 5 x 2 − (2 x + 2) dx = 5x 2 − 2 x − 2 dx and/or subtraction wrong way around. 10 3 4 DM1 3 10 10 1 1 1 = x 2 − x 2 − 2 x = × 8 − 16 − 8 − × − − Integrating( kx 2 seen) and substituting ‘their points of 3 1 3 3 8 16 2 intersection’ (but limits need to be found, not assumed to be 0 and 4 something else). 45 13 A1 OE exact answer. or 2 or 2.8125 45 45 16 16 Condone − if corrected to . A0 for inclusion of π. 16 16 SC: If *M1 DM0 scored, SC B1 available for correct answer. 6 Alternative method for question 6 1 1 M1 Equating line and curve and rearranging so that terms are all on Line meets curve when: 2 x + 2 = 5 x 2 ⇒ 2 x — 5 x 2 +2 [ = 0 ] same side, condone sign errors, and making a valid attempt to or 4 x 2 + 8 x + 4 = 25 x ⇒ 4 x 2 − 17 x + 4 [ = 0 ] solve by factorising, using the formula or completing the square. 1 1 2 y Factors are:(2 x -1)( x 2 -2), (4x-1)(x-4) and (2y-5)(y-10). 2 or x = ⇒ 2 y 2 − 25 y + 50 [ = 0 ] 25 1 A1 SC: If M1 not scored, SC B1 available for correct answers, could x= ,x = 4 just be seen as limits. 4 1 *M1 Intention to integrate and subtract areas. Or integrate curve and Area = 5 x 2 dx − { (2 x + 2)dx or area of trapezium} subtract area of trapezium. 4 10 3 2 DM1 3 4 1 15 x 2 − x + 2 x 1 or ( sum of ‘their y values’) ‘their ’ Integrating ( kx 2 seen) and substituting ‘their points of 3 1 4 2 4 intersection’ (but limits need to be found, not assumed to be 0 and 4 15 10 10 1 1 1 1 5 15 something) or a trapezium using the correct formula (‘their ’ = × 8 − × − (16 + 8 ) − + or + 10 4 3 3 8 16 2 2 2 4 1 must be ‘their 4’ – ‘their ’ but not 0). 4 45 13 A1 OE exact answer. or 2 or 2.8125 16 16 45 45 Condone − if corrected to . A0 for inclusion of π. 16 16 SC: If *M1 DM0 scored, SC B1 available for correct answer. 5

Mark scheme

7(a) ˆ = 2 B1 OE AOB 10 Sight of 0.2 from s = rθ but 10 θ = 2 is not enough. 2 π ISW if = . 10 5 5π + 6 1 B1 OE e.g. 0.724c AWRT or 41.5 degrees AWRT. [B Ô C =] or π + 0.2 30 6 5π 2+ But not 3 – fraction within a fraction. 10 ISW incorrect simplifications. Alternative method for question 7(a) 10π 10π B1 10π OR [Arc AC =] or [ Arc BC =] + 2 or 7.2 AWRT. Sight of or 5.2 or 7.2. 6 6 6 5π + 6 1 B1 OE e.g. 0.724c AWRT or 41.5 degrees AWRT. [B Ô C =] or π + 0.2 30 6 5π 2+ But not 3 – fraction within a fraction. 10 ISW incorrect simplifications. 2 7(b) 5π + 6 5π + 6 M1 OE [BP] = 10sin and [OP ] = 10cos Any correct method for both lengths, for their angle BOC (which 30 30 may have been incorrectly ‘simplified’ but not 0.2) or length BP [= 6.6208…] and [= 7.494…] 1 and O B̂ P. May be seen as part of ab sin C . OR 2 Sight of correct method enough. Can be implied by the next A1. 5π + 6 5π − 3 [BP] = 10sin and [O B̂ P] = 30 15 [= 6.6208…] and [= 0.84719…] 1 5π + 6 5π + 6 A1 OE Can be implied by any answer in range (24.7, 24.9) or a final Area of ∆OBP = × 10sin ×10cos or answer in the range (11.3, 11.5) WWW. 2 30 30 1 5π + 6 5π − 3 × 10 × 10sin × sin 2 30 15 [=24.809] 1 5π + 6 M1 1 2 [Sector BOC] = × 102 × their Use of r θ with their angle BOC (may have been incorrectly 2 30 2 ‘simplified’ but not 0.2). 5π + 6 = 50 30 = 36.1799... Area of region BPC = 11.4 A1 CAO 4

Mark scheme

8(a) 1 + 1 + a + b − 12 = 0 [ a + b = 10] B1 B1 B1 for each equation. Allow unsimplified. Can be implied by correct values for a and b. 4 + 36 + 2a − 6b − 12 = 0 [ 2a − 6b = −28] a = 4, b = 6 B1 their a their b B1 FT Or x = −2, y = −3 Centre is − ,− [ −2, −3] 2 2 4 8(b) 1 − their y 1 − −3 1 + 3 4 *M1 Using their centre correctly. Gradient of AC is [= = = ] 1 − their x 1 − −2 1 + 2 3 −1 3 A1 FT Use of m1m2 = −1 to obtain the gradient of the tangent. Gradient of tangent is = =− 4 4 their 3 3 3 7 DM1 Using (1,1) with their gradient of the tangent at A. Equation: y − 1 = ‘their − ’ ( x − 1) or y = − x + 4 4 4 3 x + 4 y = 7 or 4 y + 3 x = 7 . or integer multiples of these A1 Alternative method for question 8(b) dy dy *M1 Implicit differentiation with at least one y term differentiated 2x + 2 y +4+6 =0 correctly. dx dx dy dy 6 A1 8 = −6 = − dx dx 8 3 3 7 DM1 Using (1,1) with their gradient of the tangent at A. Equation: y − 1 = ‘their − ’ ( x − 1) or y = − x + 4 4 4 3 x + 4 y = 7 or 4 y + 3 x = 7 . or integer multiples of these A1 Alternative method for question 8(b) dy 1 − 1 *M1 Rearranging to form y = and differentiating using the chain rule. = {25 − ( x + 2)2 } 2 ( −2 x − 4 ) dx 2 dy 1 − 1 6 A1 = (25 − 9) 2 ( −6 ) = − dx 2 8 8(b) 3 3 7 DM1 Using (1,1) with their gradient of the tangent at A. Equation: y − 1 = ‘their − ’ ( x − 1) or y = − x + 4 4 4 3 x + 4 y = 7 or 4 y + 3 x = 7 . or integer multiples of these A1 4

Mark scheme

9(a) dy 1 − 1 − 1 B1 B1 Correct differentiation of 3 x + 1 and no other terms and correct = 3 + −4 × (3x + 1) × 3 = 3 − 6(3x + 1) 2 2 1 dx 2 differentiation of −4 ( 3x + 1) 2 . Accept unsimplified. d2 y 1 3 3 B1 dy 2 = − × −6 ( 3 x + 1) − 2 × 3 [ = 9 ( 3 x + 1) − 2] WWW. Accept unsimplified. Do not award if dx is incorrect. dx 2 3 9(b) dy 1 M1 dy = 0 leading to 3 − 6 ( 3x + 1) 2 = 0 − Setting their = 0. dx dx 1 A1 CAO – do not ISW for a second answer. ( 3x + 1) 2 = 2 3x + 1 = 4 leading to x = 1 y = −4 [coordinates (1, −4)] A1 Condone inclusion of second value from a second answer. d2 y 3 9 A1 d2 y ( ) − 2 = 9 3 × 1 + 1 2 = or > 0 so minimum Some evidence of substitution needed but . Do not award if dx 8 dx 2 d2 y is incorrect or wrongly evaluated. Accept correct dx 2 consideration of gradients either side of x = 1 . 4

Mark scheme

10(a) x ≠ 1 or x < 1, x > 1 or ( −∞,1) , (1, ∞ ) [ x ] B1 Must be x not f −1 ( x ) or y. Do not accept 1 < x < 1 . 1 10(b) 2x + 1 *M1 Setting y = , removing fraction and expanding brackets. y= leading to ( 2 x − 1) y = 2 x + 1 leading to 2 xy − y = 2 x + 1 2x −1 2 xy − 2 x = y + 1 leading to 2 x ( y − 1) = y + 1 DM1 Reorganising to get x =. Condone ± sign errors only. y +1 leading to x = 2 ( y − 1) x +1 x +1 1 1 1 A1 x +1 [f −1 ( x )] = , × or + OE. Must be in terms of x. Do not allow ÷ 2. 2 ( x − 1) x − 1 2 x −1 2 x −1 3 10(c) ( their f ( 3) ) leading to ( their f ( 3) ) + 4 f ( 3) = 1,1 + 4 = M1 Correct order of operations and substitution of x = 3 needed. −1 −1 2 −1 5 A1 2 10(d) Sight of ‘not one to one’ or ‘many to one’ or ‘one to many’ B1 Any reason mentioning 2 values, or + and — , such as: square root gives 2 values or horizontal line test crosses curve twice or 2 values because of turning point or 2 values because it is a quadratic. 1 10(e) 2 2x −1 2 2x + 1 B1 AG f ( x) = 1 + = + = Do not condone equating expressions and verification. 2x −1 2x −1 2x −1 2x − 1 f ′ ( x ) = −4 ( 2 x − 1) *M1 For k ( 2 x − 1)−2 and no other terms or correct use of the product or −2 or ( 2 x − 1(2) 2x−−21)( 2 x + 1) quotient rule then ISW. or 2 ( 2 x − 1) + − ( 2 x + 1) 2 ( 2 x − 1) −1 −2 2 Gradient m = −4 A1 Differentiation must have clearly taken place. Equation of tangent is y − 3 = −4 ( x − 1) [ y = −4 x + 7 ] DM1 Using (1, 3) in the equation of a line with their gradient. 7 A1 FT SOI from their straight line or by integration from 0 to ‘their 7/4’. Crosses axes at ,0 and ( 0, 7 ) 4 49 A1 OE e.g. 6.13 AWRT. [Area =] If M0 A0 DM0, SC B2 available for correct answer. 8 6

Mark scheme

11(a) ( )( ) 4cos 4 x + cos 2 x − 3 = 0 4 cos 2 x − 3 cos 2 x + 1 = 0 M1 Attempt to solve 3 term quartic (or quadratic in another variable). 3 A1 3 cos2 x = cos 2 x = −1 If M0 scored then SC B1 is available for sight of [and −1]. 4 4 3 3 M1 Square rooting ‘their cos 2 x ’. Allow without ±. May be implied cosx = [ ± ] their OE = ± 4 2 by correct final answer(s). Ignore −1 . π 5π 7π 11π A1 Dependent on preceding M1 only. Exact answers needed. [x =] , , , A1 for any 2 correct answers A1 A1 for 4 correct answers and no 6 6 6 6 A1 FT others inside the range 0 ⩽ x ⩽ 2π A0 A1 FT can be awarded for π 5π two exact answers that are 2π − ‘their and ’ within the range 6 6 0 ⩽ x ⩽ 2π. SC: If all 4 answers given in degrees (30, 150, 210, 330) or non- exact (AWRT 0.524, 2.62, 3.67, 5.76 or 0.167π, 0.833π, 1.17π, 1.83) and no others then SC B1. 5 11(b) −1 − 1 + 16k B1 State that this root is less than 0, needs to be linked to cos 2 x . cos2x = < 0 [⸫ no solutions]. 8 Can be achieved by substituting a value for k ⩾ 0. −1 ± 1 + 16k *M1 Must use quadratic formula. Allow any value of k but not ±3. [cos2x] = Condone + rather than ±. 8 Substituting k = 5 and obtain 1 from the formula DM1 Or argue logically if k > 5 ⇒ 1 + 16 k > 81 ⇒ >1. cos2 x = 1 or cos2 x > or ⩾ 1 A1 Needs to be linked to cos2 x. Concluding statement having considered both ± cases. ⸫ no A1 Dependent upon all previous marks having been scored. solutions Alternative method for question 11(b) −1 − 1 + 16k B1 State that this root is less than 0, needs to be linked to cos 2 x . cos2x = < 0 [⸫ no solutions]. 8 Can be achieved by substituting a value for k ⩾ 0. −1 ± 1 + 16k *M1 Must use quadratic formula. Allow any value of k but not ±3. [cos2x] = Condone + rather than ±. 8 −1 + 1 + 16k DM1 ∗ represents any inequality or =. *1 ⇒ −1 + 1 + 16k * 8 ⇒ 1 + 16 k * 81 8 k *5 A1 ∗ represents any inequality or =. Concluding statement having considered both ± cases. ⸫ no A1 Dependent upon all previous marks having been scored. solutions 5