Mathematics 9709/11 May/June 2023: question paper and mark scheme

Mark scheme

1 sinθ M1 sin θ 4sinθ + tanθ = 0 ⇒ 4sinθ + [ = 0] For use of tan θ = . BOD if θ missing. cosθ cos θ 1 M1 WWW Factorise, not divide by sinθ or tanθ . May ⇒ sinθ ( 4cosθ + 1) [= 0 ⇒ sinθ = 0 or ] cosθ = − 4 see tanθ ( 4cosθ + 1) [ = 0] or sinθ ( 4 + secθ ) [ = 0] . θ = 104.5° A1 AWRT 1.82 rads A0. Ignore answers outside (0, 180°). If M1 M0, SC B1 for θ = 104.5° max 2/3. 3

Mark scheme

2(a) 16 + 96 x + 216 x 2 B2, 1, 0 ISW (higher powers of x). Terms may be in any order or presented as a list. 2 2(b) 1 − 10 x + 40 x 2 B2, 1, 0 ISW (higher powers of x). Terms may be in any order or presented as a list. 2 2(c) (16 × 40 ) − (10 × 96 ) + (1 × 216 ) M1 Their 3 products which would give the term in x 2 (FT their values). Look for 640 − 960 + 216 . – 104 A1 Condone −104 x 2 . 2

Mark scheme

3 {Stretch} {factor 2} {in y-direction} B2, 1, 0 B2 for fully correct, B1 with two elements correct. {} indicates different elements. −6 B2, 1, 0 B2 for fully correct, B1 with two elements correct. {Translation} {} indicates different elements. 0 4 Transformations may be in either order.

Mark scheme

4 1 2 16π π B1 2π × 8 ×θ = ⇒θ= SOI OE e.g. , 0.524 ( 3 s.f .) 2 3 6 12 Use of degrees acceptable throughout provided conversion used in formulae for sector area and arc length. π M1 4π Arc length = 8 × their [= 4.1887…] OE FT their θ. Look for . 6 3 1 π M1 Attempt to find BC or BC 2 (see alt. methods below) [BC =] 2 × 8 sin × their [= 4.1411…] 2 6 π FT their θ. Look for 16sin or 4 6 − 4 2 . 12 Perimeter = 8.33 A1 AWRT Must be combined into one term. 4 Alternative methods for Question 4: 2nd M1 mark (use normal scheme for the other marks) π ALT 1 Substitute into correct cosine rule. FT their θ ALT 1 BC 2 = 82 + 82 − 2 × 8 × 8cos their [ BC = 4.14] 6 Look for 128 − 64 3 ( ) + 4 [ BC = 4.14] 2 ALT 2 BC 2 = 8 − 4 3 2 ALT 2 Find lengths 4 and 4 3 then use Pythagoras in the left hand triangle. 8 4 4√3 π / 6 BC 8 ALT 3 = [ BC = 4.14] π 5π ALT 3 Substitute into correct sine rule. sin sin 6 12 4

Mark scheme

5 1 *M1 1 1 kx − k = − ⇒ 2kx 2 − 2kx + 1[ = 0] OE e.g. kx 2 − kx + [ = 0 ] , x 2 − x + [ = 0] 2x 2 2k y+k 1 Equate line and curve to form 3-term quadratic (all OR quadratic in y : x = y=− 2 y 2 + 2ky + k = 0 terms on one side). k y+k 2 k b 2 − 4ac [ = 0] ⇒ ([ − ] 2 k ) − 4 ( 2 k )(1) [ = 0 ] DM1 Use discriminant correctly with their a, b, c not in 2 quadratic formula. DM0 if x still present. or 4k 2 − 8k [ = 0 ⇒ 4k ( k − 2 ) = 0 ] 1 1 May see k 2 − 4 ( k ) = 0 or 1 − 4 = 0 . OR using equation in y : ( 2k )2 − 4 ( 2 )( k ) = 0 2 2k k = 2 only A1 If DM0 then k = 2 , award A0 XP then B0 B0 Allow A1 even if divides by k to solve. If k = 0 also present but uses k = 2 , award A1. 1 B1 4 x 2 − 4 x + 1 = 0 ( 2 x − 1) = 0 ⇒ x = 2 2 1 B1 y = 2× − 2 = −1 2 5 Alternative method for Q5 dy 1 1 *M1 1 = 2 or x −2 Differentiate − M0 for 2 x −2 . No errors. dx 2 x 2 2x 1 1 1 1 1 DM1 dy [ y =] 2 x− 2 =− or = 2 2 x 2 − x = 0 Sub their into equation of line or set gradient = k 2x 2x 2x x 2x dx to form equation in x . 1 A1 1 x= only If DM0 then x = , award A0XP then B0 B0. 2 2 1 B1 y = 2 × − 2 = −1 2 k =2 B1 5

Mark scheme

6(a) p2 p2 *M1 Correct method leading to formation of a 3-term 2 ( 2 p − 6) = p + − 3 p + 12 [ = 0] quadratic in p (all terms on one side) or 2-term 6 6 quadratic in d . p2 p2 OR ( 2 p − 6 ) − = p − ( 2 p − 6) − 3 p + 12 [ = 0] 1 6 6 OE e.g. p 2 − 18 p + 72 [ = 0 ] , p 2 − 9 p + 36 [ = 0] . 1 2 OR d 2 + d [ = 0] 6 DM1 Solve a 3-term quadratic in p by factorisation, 18 ± ( −18)2 − 4 (1)( 72) p − 18 p + 72 [ = 0 ] ⇒ ( p − 6 )( p − 12 ) [ = 0 ] or 2 formula or completing the square or solve a 2-term 2 quadratic in d by factorisation. 1 OR d d + 1 [ = 0] d = −6 6 p = 12 only A1 Since p = 6 gives d = 0. If *M1 DM0 then p = 12 only, award SC B1, max 2/3 marks. A0 XP if error in either factor and p = 12 only. p = 12 only by trial and improvement 3/3. 3 6(b) B1 OE SOI. 2 p − 6 18 3 For GP r = 2 = = p 24 4 6 24 B1 FT FT their value of p if used correctly to find r (B0 if Sum to infinity = = 96 3 ‘ p ’ used) provided r < 1 . 1− 4 54 e.g. p = 18 ⇒ [ S∞ = ] = 121.5 . 5 1− 9 2

Mark scheme

7(a) [Greatest =] 5 B1 No inequality required. [Least =] ‒1 B1 No inequality required. Condone ( −1, 5 ) or equivalent. 2 7(b) y B1 One complete cycle starting and finishing at y = 2. 4 Maximum and minimum in correct quadrants. 2 Shape and curvature approximately correct. B1 FT Maximum and minimum (indicated on y -axis with x π 2π 3π 4π −2 numbers or lines, or labelled on graph). FT their greatest and least values. Award B1 for 5 and −1 even if their values were incorrect in (a). 2 7(c) 1 B1 WWW 1

Mark scheme

8(a) 2a 5 2a 3 M1 OE Substitute x = 7 then solve for a via legitimate 1+ = = 7 a = 21 a = 7 − a 2 7−a 2 mathematical steps. Condone sign errors only. 2a 5 5 OR 1 + = ( 7 − a ) + 2a = ( 7 − a ) [ 7 a = 21 a = 7 − a 2 2 a=3 A1 If M0, SC B1 for a = 3 with no working. 2 ( their 3) M1 Evaluate f(5), either separately or within gf then solve f (5) = 1 + = 4 [ 4b − 2 = 4] b = for b via legitimate mathematical steps. Condone sign 5 − their 3 errors only. FT their a value. 2 ( their 3) OR gf ( 5 ) = b 1 + − 2 [ 4b − 2 = 4] b = 5 − their 3 3 A1 6 b= OE e.g. , 1.5 . 2 4 4 8(b) x>1 B1 Accept (1, ∞ ) or {*: * > 1} where * is any variable. B0 for f −1 ( x ) > 1 or f ( x ) > 1 or y > 1 . 1 8(c) 6 *M1 6 EITHER x − 1 = ( y − 3 )( x − 1) = 6 OE y − 1 = ⇒ ( x − 3 )( y − 1) = 6 . y −3 x−3 6 6 OR x = 1 + x ( y − 3) = ( y − 3) + 6 OE y = 1 + y ( x − 3) = ( x − 3) + 6 . y −3 x−3 Allow *M1 for use of their 3 from (a). 6 DM1 6 y −3= or y ( x − 1) = 3 x + 3 OE x − 3 = or x ( y − 1) = 3 y + 3 . x −1 y −1 Allow DM1 for use of their 3 from (a). 6 A1 3x + 3 f −1 ( x ) = 3 + OE Correct answer e.g. ISW. x −1 x −1 Must be in terms of x . *M1 DM1 possible for ‘ a ’ used, but A0 so max 2/3. 3

Mark scheme

9(a) dV 4 B1 dV = × 3 ( 25 + h ) [= 4900 when h = 10] 2 Correct expression for . dh 3 dh dV dh dV dh dh 500 M1 Use chain rule correctly to find a numerical expression their "4 ( 25 + 10 ) "× 2 × = = 500 = dh 500 dh dt dt dt dt 4900 for . Accept e.g. . dt 2500 + 2000 + 400 dh A1 5 = 0.102cms −1 AWRT OE e.g. ISW. dt 49 3 9(b) dV dV dh *M1 SOI Use chain rule correctly to form equation in h . × 500 = their "4 ( 25 + h ) "× 0.075 2 = dt dh dt 5000 DM1 Solve quadratic to find h . ( 25 + h ) = 3 h = [15.8 248…] 2 5000 50 6 Exact value of h is − 25 or − 25 3 3 h + 25 = 40.82 V = 69900 cm3 A1 AWRT ISW Look for 698(88.5) . 3

Mark scheme

10(a) *M1 Integrate y 2 (power incr. by 1 or div by their new 16 16 [ π] 4[ dx ] = [ π ] 16 ( 2 x − 1) −4 [dx ] = [ π ] − 3 16 ( 2 x − 1) 3 × 2 × ( 2 x − 1) power). M0 if more than 1 error or − x ( 2 x − 1) . −3 6 A1 8 −3 [ π] − 16 OE e.g. − ( 2 x − 1) . 3 3 3 × 2 × ( 2 x − 1) DM1 [ π ] − 16 16 112 7 3 + = [ π] = [ π] Sub correct limits into their integral: F − F(1). 6 × 8 6 ×1 48 3 2 1 8 Must see at least − + . Allow 1 sign error. 3 3 Decimal: 2.33 π or 7.33 . 1 1 1.5 1 B1 1 3 Volume of cylinder = π × 12 × = π OR [ π ] 1 [ dx ] = π π or ± π − 1 seen. 2 2 1 2 2 2 7 1 11 A1 A0 for 5.76 (not exact). Volume of revolution = π − π = π If DM0 for insufficient substitution, or B0, SC B1 for 3 2 6 11 π. 6 5 10(b) dy B2, 1, 0 OE B1 for each correct element in {}. dx = −8 ( 2 x − 1) ×2 −3 At B gradient = −2 B1 3 M1 SOI Following differentiation OE e.g. y = −2 x + 4 or Eqn of tangent y − 1 = their "− 2" x − 2 1 1 1 y = x + . (Must have mN = − for M1). 1 3 2 4 mT OR Eqn of normal y − 1 = their " " x − 2 2 1 A1 SOI For at least one intercept correct or correct Tangent crosses x-axis at 2 or normal crosses x-axis at − integration. 2 5 A1 1 5 5 1 5 Area = From intercepts:× × 1 = or 1 + = , 4 2 2 4 4 4 1 5 5 from lengths: × 5 × = or by integration. 2 2 4 6

Mark scheme

11(a) 6a 2 − 30a + 6a = 0 [ ⇒ 6a ( a − 4 ) = 0] B1 Sub x = a into dy = 0 . May see a 2 − 5a + a = 0 . dx a = 4 only B1 2 11(b) d2 y dy M1 dy = 12 x − 30 or correct values of either side of x = 4 Differentiate (mult. by power or dec. power by 1) dx 2 dx dx dy M0 if no values of , only signs. dx d2 y d2 y A1 WWW A0 XP if a = 4 obtained incorrectly in (a) At x = 4, > 0 minimum or = 18 minimum Must see ‘minimum’. dx 2 dx 2 dy dy If M0, SC B1 for ‘minimum’ from sign diagram. or concludes minimum from values dx dx 2 11(c) 6 3 30 2 B1 FT Expect 2 x 3 − 15 x 2 + 24 x [ + c ] . [ y =] x − x + 6 ( their a ) x [ +c ] 3 2 B1 poss. even if uses ‘ a ’ – no value in (a) – max 1/3. −15 = 2 ( their "4") − 15 ( their "4") + 6 ( their "4") + c 3 2 2 M1 Sub x = their "4" , y = –15 into integral (must incl + c ) Look for –15 = 128 – 240 + 96 + c [⇒ c = 1]. y = 2 x 3 − 15 x 2 + 24 x + 1 A1 Coefficients must be correct and simplified. Need to see ‘ y = ’ or ‘ f ( x ) = ’ in the working. 3 11(d) dy M1 dy = 6 x 2 − 30 x + 6 ( their "4") [ = 0] OE Forming a 3-term quadratic using the given dx dx and solving by factorisation, formula or completing the 30 ± ( −30 )2 − 4 ( 6 )( 24 ) If correct, [ 6] ( x − 1)( x − 4 ) [ = 0] or square. Check for working in (b). 12 Coordinates (1,12 ) A1 Allow x = 1, y = 12 (ignore x = 4 if present). If M0, award SC B1 for (1,12 ) . 2

Mark scheme

12(a) x 2 + ( y − 2 ) = 100 B1 OE e.g. ( x − 0 )2 + ( y − 2 )2 = 10 2 ISW. 2 1 12(b) 10 − 2 4 −3 M1 OE SOI Use coordinates to find gradient of radius or Gradient of radius = = or gradient of tangent = differentiate to find mT 6−0 3 4 dy dy −3 e.g. 2 x + 2 ( y − 2 ) = 0 = at (6, 10) dx dx 4 1 dy 1 ( ) 3 − y = 2 + 100 − x 2 = 100 − x 2 2 ( −2 x ) = − . dx 2 4 3 3 29 A1 58 Equation of tangent is y − 10 = − ( x − 6) y = − 4 x + 2 OE ISW Allow e.g. 4 . 4 2 12(c) 29 M1 SOI From a linear equation in (b). Coordinates of centre of circle Q are 0, their 2 2 2 A1FT OE e.g. ( x − 0 )2 + ( y − 14.5 )2 = 31.25 ISW. 2 29 5 5 125 Equation of circle Q is x + y − their = = 2 2 4 2 29 125 B1 OE e.g. x = [ ± ] 19, x 2 − 19 = x 2 − 19 x 2 + (11 − 2 ) = 100 x 2 = 19 and x 2 + 11 − = 2 x 2 = 19 2 4 Correct argument to verify both y -coords are 11 125 29 2 ISW. − y − + ( y − 2 ) = 100 25 y = 275 y = 11 2 OR e.g. 4 2 3 12(d) 3 2 29 29 125 25 2 125 M1 Substitute equation of their tangent into equation of 2 x +− x + − = x = x 2 = 20 4 2 2 4 16 4 their circle. May see y = 31.25 − x 2 + 14.5 . or y − 29 y + 199 [ = 0] 2 29 μ 3 5 A1 OE e.g. x = ± 20 x = ±2 5 or y = 2 For 2 x -values or 2 y -values or correct ( x, y ) pair. 3 29 29 μ 3 5 A1 58 3 20 58 3 20 y = − × ± 20 + = OE e.g. + , − Correct ( x, y ) pairs. 4 2 2 4 4 4 4 3