Mathematics 9709/11 May/June 2022: question paper and mark scheme

Mark scheme

1(a) x2 – 8x + 11 = (x – 4)2 … or p = − 4 B1 If p and q-values given after their completed square expression, mark the expression and ISW. … −5 or q = − 5 B1 2 1(b) (x – 4)2 − 5 = 1 so (x – 4)2 = 6 so x − 4 = ± [] 6 M1 Using their p and q values or by quadratic formula 8 ± 24 A1 Or exact equivalent. x = 4 ± 6 or No FT; must have ± for this mark. 2 ISW decimals 1.55, 6.45 if exact answers seen. If M0, SC B1 possible for correct answers. 2

Mark scheme

2 a + 12 d = 12 B1 For correct equation. 30 B1 n 2 ( 2a + ( 30 − 1) d ) = −15 For correct equation in a and d. If using 2 ( a + l ) , must replace l with an expression involving a and d. a = 72, d = −5 B1 Both values correct SOI. 50 M1 Using sum formula with their a and d values obtained S50 = 2 ( 2 ( their a ) + 49 ( their d ) ) via a valid method. S50 = −2525 A1 5

Mark scheme

3(a) 2 M1 For selecting the term in x 4 . 2 3k 2 x 4 term is [10 ×] ( 2 x ) x 80k 4 x 4 a = 80k 4 A1 For correct value of a. Allow 8 0 k 4 x 4 . [ x 2 term is [6 ×](2kx)2 × 1 = 24k2x2 ] b = 2 4 k 2 B1 For correct value of b. Allow 2 4 k 2 x 2 . 3 3(b) 80k 4 + 24k 2 − 216[ = 0] 10k 4 + 3k 2 − 27 = 0 M1 Forming a 3-term equation in k (all terms on one side) with their a and b and no x’s. ( 2k − 3)( 5k + 9) [= 0] [⇒ k = 32 or − 95 ] 2 2 2 M1 Attempt to solve 3-term quartic (or quadratic in another variable) by factorisation, formula or completing the square – see guidance. 3 A1 6 [k ] = ± OE e.g. ± , ± 1 .5 , AWRT ± 1 .22 2 2 Omission of ± A0. Additional answers A0. If M1 M0, SC B1 can be awarded for correct final answer, max 2/3. 3

Mark scheme

4(a) sin 3θ sin 2θ sin 3θ (1 + sin θ ) sin 2θ ( sin θ − 1) *M1 Using a common denominator. − = − sin θ − 1 1 + sin θ ( sin θ − 1)(1 + sin θ ) ( sin θ − 1)(1 + sin θ ) sin 3θ (1 + sin θ ) − sin 2θ ( sin θ − 1) = ( sin θ − 1)(1 + sin θ ) − sin 2θ + sin 4θ 1 − sin 2θ DM1 ( ) Reaching ± 1 − sin θ in denominator. 2 SOI by ± c o s θ . 2 − ( sin 2θ 1 + sin 2θ ) DM1 Using s i n 2 θ + c o s 2 θ = 1 in denominator and isolating s i n 2 θ in numerator. cos 2θ ( − tan 2 θ 1 + sin 2 θ ) A1 AG - Using/stating tanθ = sinθ is sufficient for A1. cosθ May be working from both sides provided the argument is complete. A0 if θ or brackets missing throughout, or sign errors. Allow recovery if AG follows from their working. Alternative method for Q4(a) − tan θ (1 + sin θ ) = − 2 2 ( sin 2θ 1 + sin 2θ ) *M1 Using tanθ = sinθ and s i n 2 θ + c o s 2 θ = 1 . 1 − sin 2θ cosθ −sin 2θ − sin 4θ DM1 Factorising denominator. (1 − sin θ )(1 + sin θ ) sin 2θ + sin 3θ − sin 3θ + sin 4θ sin θ (1 + sin θ ) − sin θ ( sin θ − 1) DM1 Factorising numerator. 3 2 = ( sin θ − 1)(1 + sin θ ) ( sin θ − 1)(1 + sin θ ) 4(a) sin 3θ sin 2θ A1 AG − A0 if θ or brackets missing throughout, or sign errors. sin θ − 1 1 + sin θ Allow recovery if AG follows from their working. 4 4(b) ( ) ( − tan 2 θ 1 + sin 2 θ = tan 2 θ 1 − sin 2 θ ) leading to [2] tan θ = 0 2 M1 Obtaining a (trig function)2 = 0 WWW. tan θ = 0 leading to [θ =] π A1 Ignore extra solutions outside the interval ( 0, 2π ) . Alternative method for Q4(b) sin 2θ sin 2θ M1 Obtaining a (trig function)2 = 0 WWW. − (1 + sin 2 θ ) = (1 − sin 2θ ) leading to cos θ 2 cos θ 2 −sin 2θ − sin 4θ = sin 2θ − sin 4θ leading to [ 2] sin 2 θ = 0 sin θ = 0 leading to [θ =] π A1 Ignore extra solutions outside the interval ( 0, 2π ) . 2

Mark scheme

5(a) 1 2 π π 2 B1 1 2 Sector area = r = r Using r θ with θ in radians SOI. 2 6 12 2 B0 if using a value for r. π 1 π 3 B1 SOI Finding triangle area. BD = sin r = r and AD = cos r = r Decimals B0 unless exact values seen in working. 6 2 6 2 1 π π 1 1 3 so triangle area = sin r cos r = × r × r 2 6 6 2 2 2 1 π π 1 3 1 or r cos r sin = r × r× 2 6 6 2 2 2 1 2 3 2 B1 r2 π 3 π π Area of BCD = πr − r OE e.g. − with cos and sin evaluated. 12 8 4 3 2 6 6 Must be exact, in terms of r 2 . ISW 3 5(b) 3 B1 SOI by length of AD, CD or arc, or by perimeter. r π −1 2 = Angle BAC = sin r 3 1 M1 SOI Finding length by Pythagoras, or by trigonometry Length AD = cos r = r π 1 [so length CD = r] π 3 2 2 with their angle BAC, provided BAC ≠ . 6 π M1 π Length of arc BC = r × SOI Using rθ with θ in radians. Condone θ = . 3 6 3 1 π A1 3 +1 π π Perimeter of BCD = r+ r+ r with e.g. cos evaluated. 2 + 3 OE e.g. r 2 2 3 3 Must be exact, in terms of r. ISW 4

Mark scheme

6(a) x2 − 4 *M1 For clearing denominator and expanding brackets. y= x2 + 4 leading to ( x + 4 ) y = ( x − 4 ) leading to x 2 y + 4 y = x2 − 4 2 2 If swap variables first, look for y 2 x + 4 x = y 2 − 4 . x 2 y − x 2 = −4 y − 4 leading to x 2 (1 − y ) = 4 y + 4 leading to x 2 = ... DM1 For making x 2 the subject. If swap variables first, look for y2 (1− x) = 4x + 4 y2 = 4y + 4 4y + 4 4x + 4 A1 −4 x − 4 x2 = leading to x = leading to [f −1 ( x )] = OE e.g. x −1 without ± in final answer. 1− y 1− y 1− x Alternative method for Q6(a) y2 − 4 8 −8 *M1 For division and reaching x − 1 = ... ( or y − 1 = ...) x= 2 leading to x = 1 − 2 leading to x − 1 = 2 y +4 y +4 y +4 y2 + 4 = −8 leading to y 2 = −8 −4 ( ) DM1 For making y 2 or x 2 the subject. x −1 x −1 −8 A1 OE without ± in final answer. [ y = ] f −1 ( x ) = −4 x −1 3 6(b) 8 x2 + 4 8 x2 + 4 − 8 x2 − 4 M1 A1 Using common denominator or division to reach 1. 1− = − = 2 = Remainder −8 . x2 + 4 x2 + 4 x2 + 4 x + 4 x2 + 4 WWW 0 < f ( x) < 1 B1 B1 B1 for each correct inequality. B0 if contradictory statement seen. Accept f ( x ) > 0, f ( x ) < 1; 1 > f ( x ) > 0; ( 0,1) SC B1 for 0 ⩽ f(x) ⩽ 1. 4 6(c) Because the range of f does not include the whole of the domain of f (or any B1 Accept an answer that includes an example outside the of it) 12 domain of f, e.g. f(4) = . Must refer to the domain or > 20 2. Need not explicitly use the term ‘domain’ but must not refer just to the range. 1

Mark scheme

7(a) 2 M1 1 2 1 1 1 1 x + 1 M0 ( 3x − 2) 2 = x + 1 3x − 2 = x + 1 = x2 + x + 1 Equating curve and line, attempt to square; 4 2 2 4 1 M1 Forming and solving a 3TQ by factorisation, formula or x 2 − 2 x + 3[ = 0] x 2 − 8 x + 12 = 0 ( x − 6 )( x − 2 ) [ = 0] completing the square – see guidance. 4 (2, 2) and (6, 4) A1 A1 A1 for each point, or A1 A0 for two correct x-values. If M0 for solving, SC B2 possible: B1 for each point or B1 B0 for two correct x-values. 4 7(b) [ 6] *M1 For intention to integrate and subtract (M0 if squared). 1 1 Area = ± ( 3x − 2 ) 2 − x + 1 [dx] [ 2] 2 2 3 1 6 B1 B1 B1 for each bracket integrated correctly (in any form). ± ( 3x − 2 ) 2 − x 2 + x 9 4 2 2 3 1 2 3 1 DM1 ± ( F(their 6) – F(their 2)) with their integral. ± (16 ) 2 − × 36 + 6 − ( 4 ) 2 − × 4 + 2 Allow 1 sign error. 9 4 9 4 4 A1 AWRT 0.444. 9 4 SC1 B1 for if *M1 B1 B1 DM0 . 9 4 SC2 B1 for if *M1 B0 B0 DM0, provided limits 9 stated. Alternative method for question 7(b) [6] 1 *M1 For intention to integrate and subtract (M0 if squared). Area = ± ( 3 x − 2 ) 2 [dx] − area of trapezium (or triangle + rectangle) [2] 2 3 6 2+4 2 3 6 2+4 B1 B1 for bracket integrated correctly (in any form). ± (3x − 2 )2 − 4 or ± ( 3x − 2 ) 2 − + ( 2 × 4) B1 FT B1 FT for using correct formula with their values. 9 2 2 9 2 2 2 3 2 3 DM1 ±(F(their 6) – F(their 2)) using their integral. ± (16 ) 2 − ( 4 ) 2 − 12 Allow 1 sign error. 9 9 7(b) 4 A1 AWRT 0.444. 9 4 SC1 B1 for if *M1 B1 B1 DM0 . 9 4 SC2 B1 for if *M1 B0 B0 DM0, provided limits 9 stated. 5

Mark scheme

8(a) 30° 60° B2,1,0 B2 for fully correct, B1 with two elements correct. EITHER (1){Translation} OR (2){Translation} { } indicates different elements. 0 0 Accept angle in radians. (3){Stretch} {factor 2} {in x-direction} B2,1,0 B2 for fully correct, B1 with two elements correct. { } indicates different elements. (4) Stretch factor 4 in y-direction and correct order B1 Stretch, y-direction and factor and correct order. Correct order is either (1) then (3) or (3) then (2). (4) can be anywhere in the sequence. 5 8(b) 1 2 M1 SOI 4sin x − 30° = 2 2 sin −1 [ = 45] 2 2 1 M1 SOI. The M marks are independent. x − 30 = 45 or 135 x = 2 ( 45 + 30 ) or x = 2 (135 + 30 ) 2 x = 150°, x = 330° A1 5π 11π Both exact values, condone , . 6 6 A0 if extra solutions in the interval. Ignore other solutions outside [0°,360°] . 3

Mark scheme

9(a) Express as ( x + 3) + ( y −1) = 26 + 9 +1 = 36 2 2 [ ] M1 Completing the square on x and y or using the form x2 + y2 +2gx +2fy +c =0, centre ( −g,−f ) and radius g2 + f 2 − c . SOI by correct answer. Centre (−3, 1) B1 Radius 6 B1 So lowest point is (−3, −5) A1 FT FT on their centre and their radius. 4 9(b) Intersects when x 2 + ( kx − 5) + 6 x − 2 ( kx − 5) − 26 = 0 2 *M1 Substituting y = kx − 5 into their circle equation or rearranging and equating y. or ( x + 3) + ( kx − 5 − 1) = 36 2 2 x 2 + k 2 x 2 − 10kx + 25 + 6 x − 2kx + 10 − 26 = 0 DM1 Rearranging to 3-term quadratic (terms grouped, all on one side). Allow 1 error. or x 2 + 6 x + 9 + k 2 x 2 − 12kx + 36 = 36 A1 Correct quadratic (need to see 9 as constant term). ( ) leading to k 2 x 2 + x 2 + 6 x − 12kx + 9 [ = 0] or k 2 + 1 x 2 + ( 6 − 12k ) x + 9 [ = 0] ( 6 −12k ) − 4( k 2 + 1) × 9 [> 0] DM1 Using discriminant b 2 − 4ac [ > 0] with their values. 2 Allow if in square root. leading to 144k 2 − 144k + 36 − 36k 2 − 36 > 0 4 A1 Need not see method for solving. [108k2 – 144k = 0 leading to] k = 0 or k = 3 4 A1 4 k < 0, k > Do not accept < k < 0. 3 3 6

Mark scheme

10(a) d2 y 4 d2 y B1 d2 y = 6 ( −1) − 2 > 0 minimum or = 10 minimum Sub x = −1 into , correct conclusion. WWW dx 2 ( −1) 3 dx 2 dx 2 1 10(b) dy 2 *M1 d2 y = 2 x3 + 2 [ +c ] Integrating (at least one term correct). dx x dx 2 0 = −2 + 2 + c leading to c = [0] DM1 dy Substituting x = −1, = 0 (need to see) to evaluate c. dx DM0 if simply state c = 0 or omit +c. 1 4 2 A1 FT Integrated. FT their non-zero value of c if DM1 y= x − + ( their c ) x + k awarded. 2 x 9 1 DM1 9 = + 2 + k leading to k = [2] Substituting x = –1, y = to evaluate k (dep on *M1). 2 2 2 1 2 A1 4 y = x4 − + 2 OE e.g. 2 x − 1 or . 2 x 2 A0 (wrong process) if c not evaluated but correct answer obtained. 5 10(c) dy 2 M1 dy = 2 x3 + 2 = 0 Their = 0. dx x dx Leading to x 5 = −1 M1 Reaching equation of the form x5 = a . So only stationary point is when x = −1 A1 x = −1 and stating e.g. ‘only’ or ‘no other solutions. 3 10(d) dy *M1 dy At x = 1, = [ 4] Substituting x = 1 into their . dx dx dx dx dy 1 DM1 OE Using chain rule correctly SOI. = × = ×5 dt dy dt 4 5 A1 OE e.g. 1.25. 4 3