Mathematics 9709/12 October/November 2022: question paper and mark scheme

Mark scheme

1(a) 10 + 5 2 − 1 15 1 B1 Accept unsimplified. Mid-point AB is = , 2 2 2 , 2 −1 − 2 −3 5 M1 Change in y Gradient of AB = = Gradient perpendicular = For use of , condone inconsistent order of x and y, 10 − 5 5 3 Change in x and m1m2 = –1. 1 A1 OE ISW y− 2 = 5 y − 1 = 5 x − 15 5 Any correct version e.g. y = x − 12 or 5 x − 3 y = 36 . 15 3 x− 2 3 2 3 2 3 1(b) [Radius =] 34 or 5.8 AWRT or [(radius)2 =] 34 B1 Sight of 34 or 34. Condone confusion of r and r 2 . ( x − 5 )2 + ( y − 2 )2 B1 Sight of ( x − 5)2 + ( y − 2)2 ( x − 5)2 + ( y − 2)2 = 34 B1 CAO ISW Alternative method for Question 1(b) x 2 + y 2 − 10 x − 4 y B1 [c =] 5 or [c =] − 5 B1 Substitution of (10, –1) into x 2 + y 2 − 10 x − 4 y + c = 0. x 2 + y 2 − 10 x − 4 y − 5 = 0 B1 3

Mark scheme

2 2a − a = a 2 − 2a B1 OE An unsimplified correct equation in a or d only, e.g. a 2 + a = 4a . Can be implied by correct values for a or d. a = 3 or d = 3 B1 Condone ‘extra’ solution of a = 0 or d = 0. a = 3 and d = 3 B1 SOI 50 M1 May be done using 50th term (=150). Their a and d must be S50 = ( 2 × their a + 49 × their d ) numerical. 2 3825 A1 ISW SC B2 for 1275 a or 1275d 5

Mark scheme

3(a) k 2 − 4 × 8 × 2 [< 0] M1 Use of b2 – 4 a c but not just in the quadratic formula. −8 < k < 8 or − 8 < k , k < 8 or k < 8 or (–8, 8) A1 Condone ‘ − 8 < k or k < 8’, ‘ − 8 < k and k < 8’ but not 64 . 2 3(b) 2 ( 4cos θ − 1)( cos θ − 1) or ( 4cos θ − 1)( cos θ − 1) M1 OE Or use of formula or completing the square. Allow use of replacement variable. 2 A1 OE For both answers. cosθ = , cosθ = 1 2 8 SC: If M0, SC B1 available for sight of cos θ = and 1 8 [θ =] 0°, 75.5° A1 AWRT ISW rejection of 0°. For both answers and no others in the range 0° θ 180° , must be in degrees. SC: If M0 B1 scored, SC B1 available for correct answers. 2 SC: If M1 A0 scored, SC B1 available for cos θ = and θ = 8 75.5° only, WWW. 3

Mark scheme

4 a r2 = 1764 and a r + a r2 = 3444 or a r = 1680 or B1 Two correct algebraic statements. ( a 1− r 3 ) − a = 3444 1− r Attempt to solve as far as r = or a = M1 Any valid method, e.g. 1764 ÷1680 or from 20 r 2 − 41r + 21 OE (condone solving using a calculator). 1764 21 A1 1764 r= = or 1.05 [ a = 1600] Note: r = www implies B1 and M1. 1680 20 3444 − 1764 17 500 A1 AWRT e.g. 17 474.1….. 4

Mark scheme

5(a) Three points at the bottom of their transformed graph plotted at y B1 All 5 points of the graph must be connected. =2 Bottom three points of /\/\ at x = 0, x = 1 & x = 2 B1 Must be this shape. All correct B1 Condone extra cycles outside 0 x 2. 3 SC: If B0 B0 scored, B1 available for /\ in one of correct positions or all 5 points correctly plotted and not connected or correctly sized shape in the wrong position. 5(b) [g( x ) =] f(2x) + 1 B1 B1 Award marks for their final answer as follows: f(2x) B1, + 1 B1. Condone y = or f ( x ) = . 2

Mark scheme

6(a) 5 2 There is no requirement for the candidate to list a , b and c. y = 4 x + − 19 Look at values in their final expression, condone omission of 2 , 2 and award marks as follows: B1 a=4 B1 5 b= OE 2 B1 c = −19 3 6(b) 5 2 5 2 *M1 Equate their quadratic completed square form from 6(a) to 45 or Their 4 x + − 19 = 45 x + = 16 re-start and use completing the square. 2 2 Solve as far as x = DM1 Any valid method leading to two answers. 3 13 A1 SC: If M0 or M1 DM0 awarded, B1 available for correct final [ x =] , − answers. 2 2 3 6(c) Quadratic curve that is the right way up (must be seen either side B1 No axes required, ignore any axes even if incorrect. of stationary point) Stationary point stated using any valid method or correctly B1 FT FT their values from 6(a) as long as their expression is of the labelled on their diagram. 5 B1 FT form p ( qx + r ) + s . Expect − , −19 . 2 2 Condone if stated correctly but plotted incorrectly. 3

Mark scheme

7(a) sin θ ( sin θ − cosθ ) + cosθ ( sin θ + cosθ ) sin 2 θ + cos2 θ *M1 Sight of a correct common denominator, either in one or two = 2 2 fractions, condone missing brackets if recovered. In the numerator ( sin θ + cosθ )( sin θ − cosθ ) sin θ − cos θ condone ± sign errors only. sin 2 θ cos 2 θ DM1 Divide throughout by cos 2 θ . + cos 2 θ cos 2 θ sin 2 θ cos 2 θ − cos 2 θ cos 2 θ tan 2 θ + 1 A1 AG tan 2 θ − 1 Alternative method for Question 7(a) sin 2 θ *M1 sin 2 θ +1 sin 2 θ + cos 2 θ Replace tan2 θ with and multiply top and bottom by cos 2 θ cos 2 θ cos 2 θ × or the equivalent step = 2 sin θ 2 cos θ 2 sin θ − cos θ 2 cos2 θ . Condone ± sign errors. − 1 cos 2 θ Sight of convincing use of partial fractions DM1 sinθ cosθ A1 + AG sinθ + cosθ sinθ − cosθ 3 Note: M1 DM1 A1 for working on both sides at the same time and finishing at the same correct expression. M1 DM1 for starting separately and finishing at the same correct expression and A1 if there is a final conclusion e.g. QED. Do not allow cross multiplication. Condone use of s, c and t and omission of θ . 7(b) tan 2 θ + 1 *M1 Equate expression from (a) to 2 and clear fraction. tan θ − 1 2 ( = 2 tan 2 θ + 1 = 2 tan 2 θ − 1 ) tanθ = [±] 3 DM1 Simplify as far as tanθ = . May be implied by a correct final answer in degrees or radians. Alternative method for first two marks of Question 7(b) sin 2 θ + cos 2 θ *M1 Equate expression to 2, clear fraction and use trig identities to sin 2 θ − cos 2 θ ( = 2 1 = 2 sin 2 θ − 2 1 − sin 2 θ ) form an equation in sinθ or cosθ only. 3 1 DM1 Simplify as far as sinθ = , or cosθ = . sinθ = [ ± ] or cosθ = [ ± ] 4 4 1 2 A1 A1 for either correct answer then A1FT For their second value θ = π, π being π − (their first) and no others in range 0 θ π , both 3 3 A1 FT values must be exact and in radians. SC: B1 for θ = 60°,120° or 0.333π , 0.667π AWRT. or 1.05, 2.09 AWRT. 4

Mark scheme

8(a) 3 1 B1 B1 Marks can be awarded for correct unsimplified expressions, 1 3x 2 3x 2 3 1 mark each for contents of { } ISW. [ ] 3 1 [ ] y = + − + c = 2 x 2 − 6x 2 2 2 3 1 M1 Correct use of (3,5) in an integrated expression (defined by at least 5 = 2 × 32 − 6 × 32 + c one correct power) including + c. 3 1 A1 Condone c = 5 as their final line if either y = or f(x) = seen y = 2x 2 − 6x 2 + 5 elsewhere in the solution, but coefficients must not contain unresolved double fractions. 4 8(b) 1 − 1 M1 Setting given differential to 0. 3x 2 − 3x 2 = 0 [x=] 1 A1 CAO WWW Condone extra solution of —1 only if it is rejected. 2 8(c) x>1 or x> “their 8(b)” B1FT Allow ⩾ 1

Mark scheme

9(a) 1 B1 ISW a x + +1 x 1 9(b) 1 M1 Substitute x = 2 into their expression from (a) and equate to 11. a 2 + + 1 = 11 2 This may be done in 2 stages: f(2)= 2.5, g ( 2.5 ) = 11 . [a =] 4 A1 2 9(c) No,[because it is] not one-one B1 Or other suitable explanation that may include one to many or many to one. 1 9(d) x −1 B1 Condone use of a instead of 5. [g −1 ( x )] = WWW 5 1 M1 Correct combination of their g−1 ( x ) with given f ( x ) x+ −1 [g −1f ( x ) =] x OE Condone use of a instead of 5. 5 x2 − x + 1 1 1 x + x −1 − 1 ( ) 1 A1 or x + − 1 or x + x −1 − 1 OE ISW Must not contain unresolved fractions e.g. . 5x 5 x 5 5 3 9(e) The domain of f does not include the whole of the range of g. B1 Accept an answer that includes an example outside the domain of Or f, e.g. g(—1) = — 4 but for f, x > 0 . The range of g does not lie in the domain of f. 1

Mark scheme

10(a) 4π 5π 10π 28π B1 For either arc correct. Arc ARB could be AR+RB. 2.5 × + 2.24 × [= 10.47[2] + 5.86[4] or + ] 3 6 3 15 M1 For adding two (or three) arc lengths using different radii and angles and nothing else. SOI 26π A1 AWRT 16.34 or Condone 16.33 only. 5 3 10(b) 1 2π M1 For either ΔAOB or ΔAPB (AB = 4.33, h= 1.25, 0.58) or any other Area AOB = × 2.52 sin [=2.706] valid method. 2 3 1 5π Area APB = × 2.242 sin [=1.254] 2 6 [Difference =] 1.45 A1 AWRT Condone 1.46 only. 2 10(c) 1 4π B1 For either sector area correct Area AOB = × 2.52 × [=13.09] 2 3 1 5π Area APB = × 2.242 × [=6.57] 2 6 [Area of cross section =] M1 Adding two sector areas from different sectors and ‘ their 10(b) ’ 1 4π 1 5π × 2.52 × + × 2.242 × + “their 10 ( b )” and nothing else. SOI 2 3 2 6 13.09 + 6.57 + “their 10 ( b )” = 21.1 A1 CAO Condone slight inaccuracies in intermediate working if the correct answer is arrived at. 3

Mark scheme

11(a) 9 8 8 9 B1 dy 2 2 dy 9 8 = dx 2 x − 12 [ = 0 ] or [y =] x − + or x − +2 OE Either dx or a correct expression in completed square form. 4 3 9 4 3 Allow unsimplified. 24 B1 OE Condone 2.67 AWRT. x= 9 y=2 B1 CAO −b 8 8 Note: x = = B1; substitute for x in y = B1; y =2 B1. 2a 3 3 3 For 11(b) look for working to be marked on page 19 or annotate it as BP or SEEN 11(b) 3 5 9 M1 Intention to integrate and subtract areas (either way around). Can [Area =] 18 − x 2 − x 2 − 12 x + 18 dx be two separate functions or combined. 8 4 Using y 2 scores 0/5 but condone inclusion of π except for the final mark. Note: Subtraction not required for these marks. B1,B1 One mark for correct integration of each curve, allow 7 unsimplified. 3x 2 9 x3 12 x 2 3 2 3 3 7 Either separately [18 x ] − , − [ +18 x] [ ] 18 x − x x − 6 x [ +18 x ] 2 7 8× 4×3 2 28 4 2 7 3 2 3 3 [18 x] − x − x + 6 x 2 [ −18 x ] BUT condone sign errors 7 3 2 or 3x 2 9x 12 x Or combined [18x] − − + [ −18x] 28 4 8× 7 4×3 2 that are only due to missing brackets. 2 3 7 3 M1 Clear substitution of 4 into at least one integrated expression = − × 4 2 − × 43 + 6 × 42 28 4 ( 0 ) − (defined by at least one correct power) which can be unsimplified. 240 A1 SC: If all marks awarded except the final M1, SCB1 is available = or 34.3 AWRT for the correct final answer. 7 5 dy −5 × 3 2 15 2 11(c) 3 3 B1 Allow unsimplified. = dx 2 × 8 x = − x 16 dy dy dx dy 15 M1 dy = × ⇒ = − ×8× 2 Substitute x = 4 into their and multiply by 2. dt dx dt dt 16 dx −15 A1 Accept decreasing [at/by] 15 3 Note: If incorrect curve used, this is not a MR and only M1 mark 9 ( 4) is available. Expect ( − 12) × 2[=12] 2