Mathematics 9709/12 May/June 2023: question paper and mark scheme

Mark scheme

1 4 4 B1 4 2[ [ y] = ( x − 3)−3+1 or +c ] OE Allow and −3 + 1 for the power. −2 −2 ( x − 3) −3 + 1 4 4 M1 Correct use of (4, 5) to find c in an integrated expression 5= ( 4 − 3)−2 + c or 5 = + c leading to c = (defined by the correct power and no extra x’s or terms). −2 ( 4 − 3) 2 −2 −2 A1 4 + 7 or y = −2 ( x − 3) + 7 −2 y= OE − must be simplified to –2. Condone c = 7 as their ( x − 3) 2 2 final line as long as either y or f(x) = is seen elsewhere. Do not ISW if the result is of the form y = mx+c. 3

Mark scheme

2 [Coefficient of x4 = p =] 15a 2 B1 May be seen in an expansion or with x4. [Coefficient of x2 = q =] 54 a 2 B1 May be seen in an expansion or with x2. Equating their p + their q to 276 leading to an equation in a 2 only M1 No x terms and no extra terms. If p and q are not identified then it needs to be clear from the expansion that the appropriate coefficients are being used. 69 a 2 = 276 implies the first 3 marks. a = ±2 A1 CAO 4

Mark scheme

3(a) 4 ( x − 3 ) seen or a = 4 and b = −3 B1 OE Award marks for the correct expression or their values 2 p a, b and c. Condone 4 ( x − 3) + p − 36 = 0 and 4 − 9 . –36 + p or p – 36 seen or c = p – 36 B1 4 2 3(b) p − 36 > 0 leading to p > 36 or 242 − 4 × 4 p 0 p 36 or 36 < p B1 Allow (36, ∞) or 36 < p < ∞ . Consider final answer only. 1

Mark scheme

4 ( )( ) [8 x6 + 215 x3 − 27 = 0] leading to 8 x3 − 1 x 3 + 27 [ = 0] M1 OE If a substitution is used then the correct coefficients must OR be retained. Condone substitution of x = x 3 . −215 ± 2152 − 4.8. − 27 −215 ± 47089 or 2.8 2.8 1 A1 Both correct values seen. , − 27 8 1 SC: if M0 scored SC B1 is available for sight of and 8 −27 OE 1 A1 SC: if M0SCB1 scored then SCB1 is available for the or 0.5 , −3 correct answers and no others. Do not ISW if answers 2 given as a range. 3

Mark scheme

5 B1 B1 B1 for contents of each { } then ISW. 1 5 3 10 3 5 5 20 3 5 [ 10 x 2 − x 2 =] x 2 − 2 3 2 × 5 x2 = 3 x 2 − x2 2 2 20 M1 Using limit(s) correctly in an integrated expression = their × 8 − 32 [ −0] (defined by one correct power). Minimum acceptable 3 160 working is their ( − 32 ). 3 64 1 A1 Condone the presence of π for the first 3 marks. [Area of shaded region =] , 21 or 21.3[333…] Condone using the limits the wrong way around for the M 3 3 mark and if −21.3 is corrected to 21.3 allow the A mark. SC: if M0 scored SCB1 is available for correct final answer 1 5 3 If 10 x 2 − x 2 = 21.3 and no integration seen B1 only. 2 4

Mark scheme

6(a) 1 OA x *B1 1 OA = x cosθ or = or Correct expression containing OA , OA or OA 2 (allow p, 2 sin ( π − 2θ ) sin θ 2 a or r for OA) containing only terms with x and θ but not OA2 = x 2 + x 2 − 2 x 2 cos ( π − 2θ ) or just OA = 2 x cos θ . x 2 = r 2 + x 2 − 2 rx cos θ or other valid method. Do not condone sin π − 2θ until missing brackets recovered or cos(180 −2θ ) until it becomes −cos 2θ etc. OA = 2 x cos θ leading to Arc length = 2 xθ cos θ DB1 AG Complete correct method showing all necessary working. Condone 2 x cos θ ×θ . 2 If B0 but www then SCB1 for OA = 2 x cos θ leading to Arc length = 2 xθ cos θ . 6(b) 1 M1 OE Using sector formula with a correct OA. Condone Sector area = ( 2 x cosθ )2 × θ cosθ 2 for cos 2θ and missing brackets. 2 1 1 M1 Using a correct triangle formula for the correct triangle. Triangle area = × 2 x cosθ x sin θ OR x 2 sin ( π − 2θ ) Condone missing brackets and 180 for π. 2 2 [Area APB =] Their sector area – their triangle area M1 Both expressions must be areas involving terms with x2 and θ only. Condone missing brackets and 180 for π for the triangle. Condone calling the sector a segment. 1 1 A1 OE [Area APB =] ( 2 x cosθ )2 × θ − x 2 sin ( π − 2θ ) A correct expression. Mark the first unsimplified result of 2 2 1 subtraction and ISW any incorrect ‘simplifications’. [ = x 2 (2θ cos 2 θ − sin 2θ ) or x 2 cos θ (2θ cos θ − sin θ ) ] 2 4

Mark scheme

7(a)(i) cos 2θ + 2sinθ cosθ + sin 2θ = 1 leading to 2sinθ cosθ = 0 or sin 2θ = 0 *B1 Or arriving at cosθ = 0 or sin θ = 0 or tan θ = 0 after first expanding and www. π DB B2 for three correct answers only. [θ =] 0 , ,π 2,1,0 B1 for two correct answers and one incorrect or 3 correct 2 answers plus other values in the range. SC DB1 for correct 3 answers in degrees and no others. Ignore extras outside of the range and allow decimal equivalents. 3 Verifying 3 answers rather than expanding and solving 0/3. 7(a)(ii) π π B1 Checking both correct values. Do not allow solving an cos 0 + sin 0 = [1 + 0 =]1 and cos + sin [ = 0 + 1] = 1 equation. 2 2 Condone use of 90 degrees. cosπ + sin π [ = −1 + 0 ] = −1 or ≠ 1 B1 www 2 7(b) ( cosθ − sinθ ) sinθ + ( cosθ + sinθ )(1 − cosθ ) M1 Correct common denominator and correct products in the numerator and no missing terms. Correct factors in the ( cosθ + sinθ )( cosθ − sinθ ) denominator can be implied by cos 2θ − sin 2θ . Condone brackets missing if recovered. cosθ sinθ − sin 2θ + cosθ − cos 2θ + sinθ −sinθ cosθ A1 = cos 2θ − sin 2θ sinθ + cosθ − cos 2θ − sin 2θ cosθ + sinθ − 1 A1 AG = = Clear evidence of using sin 2θ + cos 2θ = 1 in either the cos 2θ − sin 2θ 1 − 2sin 2θ numerator or denominator. Condone c, s and/or omission of θ . Working from both sides of the identity and correctly arriving at the same expression can score M1A1. A final statement is then required for the A1. 3 7(c) cosθ + sinθ − 1 *M1 Replacing LHS with the expression from (b) and = 2 ( cosθ + sinθ − 1) attempting to simplify i.e. condone omission of 1 − 2sin 2θ leading to 1 = 2( 1 − 2sin 2θ ) ( cosθ + sinθ − 1) = 0 at this stage. M0 for 0 = 2( 1 − 2sin 2θ ) 1 or 3 DM1 Dividing by k and taking the square root of a positive value ksin 2θ =1 or 3 leading to sinθ = [ ± ] < 1. k 1 5 1 This mark can be implied by the solutions π, π . 4 sin θ = 1 leading to sin θ = ± 2 2 6 6 1 1 5 A1 Allow 0, 0.524, 1.57, 2.62 AWRT. Solutions 0, π, π , π 6 2 6 1 If M0 SCB1 for ( cosθ + sinθ − 1) = 0 ⇒ 0, π. 2 If M0 SCB1 for all four correct answers and no others. Ignore answers outside of the range. Answers in degrees A0. 3

Mark scheme

8(a) *B1 The line y = x correctly drawn. Can be implied by reasonably correct graph of f −1 ( x ) . DB1 Fully correct (needs to reach y = 2 π and x-axis and cross the line y = x in the correct squares). 2 8(b) 1 1 y±3 M1 1 y = 3 + 2sin x leading to sin x = Attempting to arrive at an expression for sin x; condone 4 4 2 4 ± sign errors. Variables may be interchanged initially. y±3 M1 not implied by x = . 1 2sin 4 y −3 −1 x − 3 A1 x −3 sin −1 ( x − 3) x = 4sin −1 leading to [f ( x ) or y =] 4sin −1 ISW Must clearly be sin −1 NOT . 2 2 2 2 3− x 1 Allow but not ÷ . −2 4 2 8(c) B1 Continuing given graph from y intercept to –2π. The correct shape needed between 0 and -2π, including starting to level off (gradient in the final two squares needs to be reducing) as -2π is approached. The y co-ordinate at-2π must be in the correct square. Yes it does have an inverse, because the graph is always increasing B1 FT If there is no graph to the left of the y axis, no mark is OR because it is one-one OR because it passes the horizontal line test OR available. it is not a many to one [function]. FT an incorrect graph and if the answer is now ‘No’ provide an appropriate reason. 2 8(d) { } indicates different elements throughout. {Stretch} {factor 4} {in x-direction} B2, 1, B2 for fully correct, B1 with two elements correct. 0 Condone use of ‘sf’ instead of factor and ‘co-ordinates’ stretched instead of graph stretched. Allow any mention of x-axis, horizontally or y-axis invariant. Wavelength or period increased by a factor of 4 for B2 or by 4 for B1. {Stretch} {factor 2} {in y-direction} B2, 1, B2 for fully correct, B1 with two elements correct. 0 Condone use of ‘sf’ instead of factor and ‘co-ordinates’ stretched instead of graph stretched. Allow any mention of y-axis, vertically or x-axis invariant. Allow y ‘co-ordinates’ doubled or amplitude doubled for B2. 0 B2, 1, B2 for fully correct, B1 with two elements correct. {Translation} 0 Allow shift. Any mention of y axis, y-direction or 3 vertically implies 0 , so shift by 3 vertically is B2, but shift by a factor of 3 vertically or a translation of 3 ‘up’ is B1. 6 After scoring B2, B2 the final transformation can only be awarded B2 if the order is fully correct i.e. the translation must not be applied before the y stretch. If all correct except the order award B2B2B1.

Mark scheme

9(a) a 16 B1 Rearranging two algebraic statements to give a = . ar = 16 , 1 − r = 100 leading to a = r and a = 100 (1 − r ) These can be implied by a correct equation in one variable. 100 (1 − r ) r = 16 leading to 100r 2 − 100r + 16 [ = 0] *M1 Using their two expressions and rearranging to get a 3-term quadratic expression with all of the terms on one side. Condone sign errors only. ( 5r − 4 )( 5r − 1) = 0 DM1 Condone ( 5r − 4 )( 5r − 1) following 100 r 2 − 100 r + 16 . OR 4 1 leading to r = or 2 25 ± 25 − 4.25.4 5 5 2.25 a = 20, a = 80 A1 SC: if DM0 scored SCB1 is available for sight of 20 and 80. Alternative Method for Question 9(a) a 16 100 − a B1 Rearranging two algebraic statements to give r = . ar = 16, 1 − r = 100 leading to r = a and r = 100 These can be implied by a correct equation in one variable. 1600 = 100 a − a 2 leading to a 2 − 100a + 1600 [ = 0] *M1 Using their two expressions and rearranging to get a 3-term quadratic expression with all of the terms on one side. Condone sign errors and 160 instead of 1600 only. 100 ± 1002 − 4.1600 DM1 ( a − 20 )( a − 80 ) = 0 OR 2 a = 20, a = 80 A1 SC: if DM0 scored SCB1 is available for sight of 20 and 80. 4 9(b) 4 1 B1 OE SOI r= , 5 5 4 n −1 n −1 B1FT 2 expressions for the nth term FT their values from part (a) [vn =] their80 × their 1 [un =] their 20 × their if r less than 1. 5 5 Method 1 for final 2 marks 1 n −1 M1 Correctly separating the numerator and denominator of 20 × × 4n −1 n −1 5 4 their or one correct step towards the solution eg 5 4n −2 un = 80 × . 5n −1 1 1 n −1 1 n −1 A1 AG Given result clearly shown un = × 80 × × 4n −1 = 4n − 2 × 80 × = 4n − 2 × vn 4 5 5 Method 2 for final 2 marks 20 × 0.8n −1 1 n −1 M1 Dividing two nth terms of the correct format and = ×4 simplifying their terms in r. 80 × 0.2n −1 4 = 4−1 × 4n −1 = 4n − 2 A1 AG 4

Mark scheme

10(a) 2 *M1 Obtaining an unsimplified equation in x or y only. ( x − a ) + x + 6 − 3 = 20 or using x = 2 y − 12 2 1 2 5 2 A1 OE e.g. 5x 2 + 4 ( 3 − 2a ) x + 4a 2 − 44 x + ( 3 − 2a ) x + a 2 − 11[ = 0] 4 Rearranging to get a correct 3-term quadratic on one side. Condone terms not grouped together. 5 y 2 − y ( 54 + 4a ) + 133 + a 2 + 24 . DM1 OE Using b 2 − 4ac on their 3 term quadratic [ = 0] . ( 3 − 2a )2 − 4 × ( a 2 − 11) [ = 0] 5 4 Method 1 for final 2 marks Using a = 4: ( 3 − 8 ) − 5 ( 5 ) = 0 2 A1 Clearly substituting a = 4. a = −16 B1 Condone no method shown for this value. Method 2 for final 2 marks − a 2 − 12 a + 64 = 0 ⇒ ( a − 4 )( a + 16 ) = 0 ⇒ a = 4 A1 AG Full method clearly shown. a = −16 B1 Condone no method shown for this value. 5 If M0, SCB1 available for substituting a = 4, finding P(2, 7) and verifying that CP2 = 20. 10(b) Centre (4, 3) identified or used or the point P is (2, 7) B1 ⸫ gradient of normal = −2 B1 SOI Forming normal equation using their gradient (not 0.5) and their centre or M1 Condone use of ( ±4, ±3 ) . P y −3 A1 OE Condone f ( x ) = . = −2 or y − 7 = −2 ( x − 2 ) ( x − 4) 4 10(c) Method 1 for Question 10(c) *M1 1 ( x − 4 ) leading to y = x + 1 1 1 Diameter: y − 3 = Using gradient with their centre. 2 2 2 Or dy 1 2 ( x − 4 ) + 2 ( y − 3) = 0 leading to y = x + 1 By implicit differentiation. dx 2 2 DM1 Obtaining an unsimplified equation in x or y only. ( x − 4 ) + 5 2 2 1 x + 1 − 3 = 20 4 x − 10 x = 0 [ y 2 − 6 y + 5 = 0] . 2 x = 0 or 8, y = 1 or 5 [(0, 1) and (8, 5)] A1 Correct co-ordinates for both points. Condone no method shown for solution. Equations are y − 1 = −2 x and y − 5 = −2 ( x − 8 ) A1 2 x + y = 1 and 2 x + y = 21 . Method 2 for Question 10(c) Coordinates of points at which tangents meet curve are *M1 A1 Vector approach using their centre and gradient = 0.5 . (4+4, 3+2) = (8, 5) and (4 – 4, 3 – 2) = (0, 1) Condone answers only with no working. Equations are y − 5 = −2 ( x − 8 ) and y − 1 = −2 x DM1 A1 Forming equations of tangents using their (0, 1) and (8, 5). Method 3 for Question 10(c) ( x − 4 )2 + ( −2 x + c − 3)2 = 20 *M1 Obtaining an unsimplified equation in x only using equation of circle with y = −2 x + c . 5 x 2 + ( 4 − 4c ) x + ( c − 3)2 − 4 = 0 ( 4 − 4c ) − 20 ( ( c − 3) − 4 ) [ = 0] DM1 Using b 2 − 4ac [ = 0] . 2 2 [leading to −4c 2 − 32c + 120c + 16 − 100 = 0 ] 10(c) 4c 2 − 88c + 84 [ = 0] [leading to c 2 − 22c + 21 = 0 ] A1 c = 21 and c = 1 or y = −2 x + 21 and y = −2 x + 1 A1 Condone no method shown for solution. 4

Mark scheme

11(a) dy 1 − 1 B 2,1,0 1 = k ( 4 x + 1) 2 ×4−1 OE e.g. 2k ( 4 x + 1) 2 − 1 − dx 2 B2 Three correct unsimplified { } and no others. B1 Two correct { } or three correct { } and an additional term e.g. +5. B0 More than one error. 2 11(b) 1 1 2k M1 dy 1 2k ( 4 x + 1) 2 − 1 = 0 leading to ( 4 x + 1) 2 = 2k or of the form ak ( 4 x + 1) 2 − 1 where − − 1 =1 OE Equating their dx ( 4 x + 1) 2 a = 2 or 0.5, to 0 and dealing with the negative power 1 correctly including k not multiplied by ( 4 x + 1) 2 . 4k 2 − 1 A1 CAO x= OE simplified expression ISW. 4 2 11(c) 1 M1 dy 2 × 10.5 ( 4 x + 1) 2 − 1 = 2 − Putting k= 10.5 into their and equating to 2. dx 1 A1 33 1 7 = ( 4 x + 1) 2 leading to 4 x + 1 = 49 leading to x = 12 If M1 earned SCB1 available for x = from a = . 64 2 y = 10.5 4 x + 1 − x + 5 = 66.5 [leading to (12, 66.5)] A1 1 A1 OE y − 66.5 = − ( x − 12 ) 2 4