Mathematics 9709/12 February/March 2024: question paper and mark scheme

Mark scheme

1 Integrate to obtain −2x −1 B1 OE Substitute limits correctly with clear indication seen that upper limit gives 0 M1 For integral of form − k x − n , where k > 0 , n > 0 . Obtain 23 A1 WWW Accept 0.667. 3

Mark scheme

2(a) State (3π, − k ) B1 1 2(b) 1 M1 Any non-zero c. Obtain equation of form [ y =] c ± k sin x 2 1 A1 OE Obtain correct equation [ y =] 2 − k sin x 2 State (3π, 2 + k ) B1 FT Following part (a), i.e. (their x, 2 – their y). 3 3

Mark scheme

3 Integrate to obtain form k (4x + 5) 2 *M1 Obtain correct 12 (4x + 5) 2 3 A1 Or (unsimplified) equivalent. Condone missing ... + c so far. Substitute x = 1, y = 9 to form an equation in c DM1 3 A1 3 Obtain or imply [ y =] 1 2 ( 4 x + 5) 2 − 9 2 May be implied by [ a =] 1 2 ( 4 (1) + 5 ) 2 − 9 2 . Substitute x = 5 to obtain a = 58 A1 5

Mark scheme

4(a) Expand bracket to obtain 3 terms and use correct identity M1 θ may be missing or another symbol used. sin θ M1 Does not require any further explanation. Use identity = tan θ θ may be missing or another symbol used. cosθ Conclude with 2tanθ A1 WWW AG 3 4(b) Attempt solution of 5tan 3 θ = 2 tan θ to obtain at least one value of tan θ M1 SOI Can be awarded if tan θ is cancelled and ignored. Obtain at least two of 0, ± 32.3 A1 Or greater accuracy. SC B1 if no method shown. Obtain all three values A1 Or greater accuracy; and no others in −90° < θ < 90° range. Other units SC B1 only for all 3 angles. SC B1 if no method shown. 3

Mark scheme

5 Differentiate to obtain form kx(2 x 2 − 5) −2 M1 Obtain correct −12 x(2 x 2 − 5) −2 A1 OE Substitute (2, 1) to obtain gradient − 249 A1 8 OE e.g. − . Allow − 2.67. 3 Apply negative reciprocal to their numerical gradient to obtain gradient of *M1 3 normal Must have been some attempt at differentiation. Expect 8 Attempt equation of normal using their gradient of the normal and (2, 1) DM1 3 Expect y − 1 = ( x − 2) . 8 Obtain 3x − 8 y + 2 = 0 (allow multiples) A1 Or equivalent of requested form e.g. 8 y − 3x − 2 = 0 . 6

Mark scheme

6 4 2 2 4 [1] B1 B1 OE 2 ( ax ) , 2 ( ax ) 3 Expect 24a 2 x 2 ,8a3 x3 (may be seen in an expansion). 2 3 Multiply terms involving x 2 and x 3 by 5 − ax to obtain x 3 term *M1 Must find two products only (may be seen in an expansion). Equate coefficient of x 3 to 432 and solve for a DM1 Ignore inclusion of x 3 at this stage. Obtain a = 3 only A1 5

Mark scheme

7(a) Attempt substitution for y in quadratic equation *M1 Or substitution for x … Obtain 5x2 + 30x + 75 − k [= 0] or 5 y2 − 20 y + 50 − k [= 0] A1 OE e.g. x 2 + 6 x + 15 − k (all terms gathered together). 5 Use b 2 − 4ac = 0 with their a, b and c DM1 ‘ = 0’ may be implied in subsequent working or the answer. Obtain 900 − 20(75 − k ) = 0 or equivalent and hence k = 30 A1 … obtaining 400 − 20(50 − k ) = 0 and k = 30 . 4 7(b) Substitute their value of k in equation from part (a) and attempt solution M1 Expect 5x2 + 30x + 45[= 0] or 5 y2 − 20 y + 20[= 0] . Obtain coordinates (−3, 2) A1 SC B1 only (−3, 2) without attempt at quadratic solution. 2

Mark scheme

8(a) Substitute n = 10 and a = 6 into un = a + ( n − 1) d *M1 Expect 6 + 9d = 19.5 or equivalent. [ d =]1.5 A1 Substitute a = 6 and their d into correct formula for the sum of 100 terms DM1 Obtain 8025 A1 4 8(b) Obtain S = 48 B1 Identify for S E first term 12 and common ratio 14 B1 Attempt sum to infinity, S E , with at least one of first term and common ratio M1 Only awarded if |r| < 1. correct Obtain S E = 16 A1 4

Mark scheme

9(a) Attempt to form expression for gf ( x ) *M1 ( ) Expect 5 ( 3 x − 2 ) + k − 1; fg(x) is M0. 2 Do not allow algebraic errors. Obtain 5( 3x − 2) + 5k − 1 A1 OE e.g. 45x2 − 60 x + 5k + 19 . 2 Their 5k − 1 = 39 or 5k − 1 ⩾ 39 DM1 Or use b 2 − 4ac = 0 (must be ‘= 0’, could be implied later) on 45 x 2 − 60 x + 5k + 19 − 39 0 OE. Obtain k = 8 A1 Do not accept k 8. 4 9(b) Obtaining ( 3 ( 5 x − 1) − 2 ) + their k M1 May simplify and/or use k at this stage; k may have come 2 from an inequality in (a). Conclude fg ( x ) 8 allow [ y ] 8 A1 FT OE Following their value of k; must be ⩾, not >. Allow an accurate written description. 2 9(c) State g−1 ( x) = 15 ( x + 1) B1 OE 1 ( x + 1) must be indicated as the inverse. 5 h ( x ) = 7 x + 4 B1B1 If 7 x + 4 only, it must be clear that this is h ( x ) . 3

Mark scheme

10(a) Obtain gradient of relevant radius is –2 B1 Using m1m2 = −1 obtain the gradient of the tangent and use it to form a M1 m1 must be from an attempt to find the gradient of the straight line equation for a line containing (–6, 9) radius using the centre and the given point. Obtain y = 12 x + 12 A1 1 OE e.g. y − 9 = ( x + 6) . 2 3 10(b) State or imply ( x + 4) 2 + ( y − 5) 2 = 20 B1 If x 2 + y 2 − 2 gx − 2 fy + c = 0 is used correctly with ( − g , − f ) = ( −4, 5) and c = g 2 + f 2 − r 2 then M1. Obtain x 2 + y 2 + 8 x − 10 y + 21 = 0 B1 A1 if above method used. 2 10(c) Substitute x = 0 in equation of circle to find y-values 3 and 7 B1 May be implied by AB = 4 or use of |x-coordinate of C|. or state C to AB = 4 Attempt value of θ either using cosine rule or via 12 θ using right-angled M1 Using their AB. If θ /2 used, must be multiplied by 2. triangle Obtain θ = 0.9273 A1 Or greater accuracy. A correct answer implies the M1. 3 10(d) Attempt arc length using rθ formula with their θ (not their θ /2) and M1 Expect 4.15. r = √ 20 Obtain perimeter = 8.15 or greater accuracy A1 Condone missing units or incorrect units. Attempt area using 12 r 2 (θ − sinθ ) formula or equivalent with their θ and M1 If sector – triangle used, both formulae must be correct. If triangle ACM used, area must be multiplied by 2. r = √ 20 Obtain area = 1.27 or greater accuracy A1 Condone missing units or incorrect units. 4

Mark scheme

11(a) Differentiate to obtain − 43 x − 3 + x − 3 B1 5 4 2 −1 − 1 Expect quadratic 2 x 3 − 3x 3 + 1 OE − 1 1 or rewrite as a quadratic equation in x 3 or x3 Allow 2 x − 3x + 1 . 2 − 1 1 M1 Substitution SOI if dealt with correctly later Equate first derivative to zero and reach a solution for x 3 or x3 with no error in use of indices 2 1 3 1 − or complete square to find minimum point 2 a − − where a = x 3 4 8 Obtain x = 64 27 A1 Or exact equivalent. SC B1 if no working shown. Ignore extra solution x = 0 . y = − 18 seen B1 Or exact equivalent. Allow −0.125 . 4 11(b) −1 M1 1 Recognise equation as quadratic in x 3 or equivalent and attempt solution − 2a2 − 3a + 1[= 0] where a = x 3 . −1 −1 A1 OE Obtain x 3 = 1 and x 3 = 12 SC B1 if no M mark awarded. Obtain 1 and 8 A1 SC B1 if no M mark awarded. 1 2 *M1 1 9 23 Integrate to obtain form k1 x 3 + k2 x 3 + x or 2 out of 3 correct terms Expect 6 x 3 − x + x. 2 1 2 A1 No other terms from a second integral. Obtain correct 6x 3 − 92 x 3 + x Apply their limits correctly DM1 Their limits must be from their working. [Obtain –0.5 and conclude area is] 0.5 A1 7