Mathematics 9709/12 October/November 2020: question paper and mark scheme

Mark scheme

1 Coefficient of x³ in (1 − 2 x ) is −80 5 B1 Can be seen in an expansion but must be simplified correctly. Coefficient of x² in (1 − 2 x ) is 40 5 B1 Coefficient of x³ in (1 + kx )(1 − 2 x ) is 40k −80 = 20 5 M1 Uses the relevant two terms to form an equation = 20 and solves to find k. Condone x³ appearing in some terms if recovered. 5 A1 (k =) 2 4

Mark scheme

2 −2 p p+2 M1 OE. Using “a, b, c then b² = ac” or a =2p+6 , ar = −2p and (−2p)² = (2p + 6) × (p + 2) or = ar² = p + 2 to form a correct relationship in terms of p only 2 p + 6 −2 p ( 2 p ² − 10 p − 12 = 0 ) p = 6 A1 a = 18 and r = −⅔ A1 (s∞ ) = their a ÷ (1 − their r) M1 Correct formula used with their values for a and r , r < 1 5 Both a & r from the same value of p. = 18 ÷ 3 (s∞ = )10.8 A1 OE. A0 if an extra solution given 2p + 6 2p + 6 SC B2 for s∞ = or ignore any −2 p p+2 1− 1− 2p + 6 −2 p subsequent algebraic simplification. 5

Mark scheme

3 2 x 2 + m ( 2 x + 1) − 6 x − 4 ( = 0 ) *M1 y eliminated and all terms on one side with correct algebraic steps. Condone ± errors Using b ² − 4ac on 2 x 2 + x ( 2m − 6 ) + m − 4 ( = 0 ) DM1 Any use of discriminant with their a, b and c identified correctly. 4m2 − 32m + 68 or 2m2 − 16m + 34 or m2 − 8m + 17 A1 ( 2m − 8 )2 + k or ( m − 4 )2 + k or minimum point ( 4, k ) DM1 OE. Any valid method attempted on their 3-term quadratic or finds b ² − 4ac ( = −4, −16 , −64 ) ( m − 4 )2 + 1 oe + always > 0 → 2 solutions for all values of m A1 Clear and correct reasoning and conclusion without wrong working. or Minimum point (4,1) + (fn) always > 0 → 2 solutions for all values of m or b ² − 4ac < 0 + no solutions → 2 solutions for the original equation for all values of m 5

Mark scheme

4 S x and S x +1 M1 Using two values of n in the given formula a = 5, d = 2 A1 A1 a + (n – 1) d > 200 → 5 + 2(k – 1) > 200 M1 Correct formula used with their a and d to form an equation or inequality with 200, condone use of n (k =) 99 A1 Condone ⩾ 99 Alternative method for question 4 n d 1 M1 Equating two correct expressions of S n and equating ( 2a + (n – 1) d ) ≡ n2 + 4n → = 1 , a − d = 4 2 2 2 coefficients of n and n2 d = 2, a = 5 A1 A1 a + (n – 1) d > 200 → 5 + 2(k – 1) > 200 M1 Correct formula used with their a and d to form an equation or inequality with 200, condone use of n (k =) 99 A1 Condone ⩾ 99 Alternative method for question 4 sumk − sumk −1 → k 2 + 4 k − ( k − 1) 2 − 4 ( k − 1) M1 A1 Using given formula with consecutive expressions subtracted. Allow k+1 and k. 2k + 3 > 200 or = 200 M1 A1 Simplifying to a linear equation or inequality (k =) 99 A1 Condone ⩾ 99 5

Mark scheme

5(a) 0 B1 1 5(b) x+2 4− x 4 B1 B1 OE. Sight of correct inverses. (f−1(x)) = , (g−1(x)) = or −1 4 x x x² + 6x – 16 (= 0) B1 Equating inverses and simplifying. ( x + 8 ) and ( x − 2 ) M1 Correct attempt at solution of their 3-term quadratic- factorising, completing the square or use of formula. (x =) 2 or −8 A1 Do not accept answers obtained with no method shown. 5

Mark scheme

6(a) 1 sinx 1 B1 Uses “tanx = sinx ÷ cosx” throughout − + 1 cosx cosx sinx 1 − sinx 1 + sinx 1 − sin² x M1 Correct algebra leading to two or four terms or cosx sinx cosxsinx cos² x A1 OE. A correct expression which can be cancelled directly to cosxsinx cosx e.g. ( ) cos x 1 − sin 2 x sinx ( ) sin x 1 − sin 2 x cos² x cosx 1 A1 AG. Must show cancelling. If x is missing throughout their = = working withhold this mark. cosxsinx sinx tanx 4 6(b) 1 1 M1 Reducing to tan³x = k. Uses (a) → = 2tan² x tan³x = tanx 2 (x =) 38.4° A1 AWRT. Ignore extra answers outside the range 0 to 180° but A0 if within. 2

Mark scheme

7(a) 5 *M1 Substituting 4 into f ' ( x ) f '( 4) = 2 dy dy dx dy 5 DM1 Multiplies their f ' ( 4 ) by 0.12 = × → = × 0.12 dt dx dt dt 2 dy A1 OE = 0.3 dt 3 7(b) 1 − 1 B1 B1 B1 for each unsimplified integral. 6x 2 4x 2 1 − 1 ( +c ) − 2 2 Uses (4, 7) leading to c = (-21) M1 Uses (4, 7) to find a c value 1 − 1 8 A1 Need to see y or f(x) = somewhere in their solution and 12 y or f ( x ) = 12 x 2 + 8 x 2 − 21 or 12 x + − 21 and 8 x 4

Mark scheme

8(a) Use of correct formula for the area of triangle ABC M1 Use of 180–2θ scores M0. Condone 2π–2θ 1 2 ( θ) r sin π − 2 or 12 r 2 sin 2θ or 2 × 12 r × r cos θ × sin θ or A1 OE 2 θ 2 × 12 r cos × r sin θ [Shaded area = triangle – sector] = their triangle area − 12 r 2θ B1 FT FT for their triangle area − 1 r 2θ 2 (Condone use of 180 degrees for triangle area for B1) 3 8(b) Arc BD = rθ = 6 cm B1 SOI AC = 2rcosθ = ( 2×10cos0.6 = 20cos0.6 = 16.506) *M1 1 Finding AC or AC (= 8.25) r × sin ( π – 2θ ) ( or 2r 2 − 2r 2 cos ( π − 2θ ) or ) sin θ 2 DC = 2rcosθ – r or ( 2r − 2r cos ( π − 2θ ) ) – r ( = 6.506) 2 2 DM1 Subtracting r from their AC or r-rcosθ from their half AC (8.25-1.75) (Perimeter = 10 + 6 + 6.506 =) 22.5 A1 AWRT 4

Mark scheme

9(a) r= ( 6 + 3 ) or r² = 45 2 2 B1 Sight of r = 6.7 implies B1 (x − 5)² + (y − 1)² = r² or x 2 − 10 x + y 2 − 2 y = r 2 − 26 M1 Using centre given and their radius or r in correct formula (x − 5)² + (y − 1)² = 45 or x 2 − 10 x + y 2 − 2 y = 19 A1 ( 45 ) for r 2 2 Do not allow 3 9(b) C has coordinates (11, 4) B1 0.5 B1 OE, Gradient of AB, BC or AC. Grad of CD = −2 M1 Calculation of gradient needs to be shown for this M1. 1 A1 Clear reasoning needed. ( × −2 = −1 ) then states + perpendicular → hence shown or tangent 2 Alternative method for question 9(b) C has coordinates (11, 4) B1 0.5 B1 OE, Gradient of AB, BC or AC. Gradient of the perpendicular is −2 M1 Use of m1m2 = −1 with their gradient of AB, BC or AC and → Equation of the perpendicular is y − 4 = −2 ( x − 11) correct method for the equation of the perpendicular. Could use D(5, 16) instead of C(11,4). Checks D(5, 16) or checks gradient of CD and then states A1 Clear check and reasoning needed. Checks that the other D lies on the line or CD has gradient −2→ hence shown or tangent point lies on the line or checks gradient. 9(b) Alternative method for question 9(b) C has coordinates (11, 4) or Gradient of AB, BC or AC = 0.5 B1 Only one of AB, BC or AC needed. Equation of the perpendicular is y − 4 = −2 ( x − 11) B1 Finding equation of CD. ( x − 5) 2 + ( −2 x + 26 − 1) 2 = 45 → ( x 2 − 22 x + 121 = 0) M1 Solving simultaneously with the equation of the circle. ( x − 11 ) 2 = 0 or b ² − 4ac = 0 → repeated root → hence shown or tangent A1 Must state repeated root. Alternative method for question 9(b) C has coordinates (11, 4) B1 Finding CD = 180 and BD = 225 B1 OE. Calculated from the co-ordinates of B, C & D without using r. Checking (their BD)2 – (their CD)2 is the same as (their r)2 M1 ⸫ Pythagoras valid ⸫ perpendicular → hence shown or tangent A1 Triangle ACD could be used instead. Alternative method for question 9(b) C has coordinates (11, 4) B1 ρ ρ ρ ρ Finding vectors AC and CD or BC and CD B1 Must be correct pairing. 6 −6 12 −6 (= and or and ) 3 12 6 12 ρ ρ ρ ρ Applying the scalar product to one of these pairs of vectors M1 Accept their AC and CD or their BC and CD Scalar product = 0 then states ⸫ perpendicular → hence shown or tangent A1 4 9(c) E (−1, 4) B1 B1 WWW B1 for each coordinate Note: Equation of DE which is y = 2 x + 6 may be used to find E 2

Mark scheme

10(a) dy 8 B2, 1, 0 B2 for all three elements correct, B1 for two elements ( ( ) = [8] ×[ 3 − 2 x) −3 + [ −1] = −1 correct, B0 for only one or no elements correct. dx (3 − 2x) ³ B1 FT FT providing their bracket is to a negative power d² y 48 = −3 × 8 × (3 − 2 x) −4 × ( −2 ) = dx² ( 3 − 2x )4 1 1 B1 B1 B1 Simplification not needed, B1 for each correct element ∫ydx = [(3− 2x)−1 ] [2 ÷ (−1 × −2)] [ − ½x²] (+c) = − x2 + c 3 − 2x 2 6 10(b) dy M1 Setting their 2-term differential to 0 and attempts to solve = 0 → (3 – 2x)³ = 8 → 3 – 2x = k → x = as far as x = dx 1 A1 2 Alternative method for question 10(b) 2 M1 Setting y to 0 and attempts to solve a cubic as far as x = y = 0→ 2 − x = 0 → ( x − 2 ) (2 x − 1) 2 = 0 → x = (3 factors needed) (3 − 2 x) 1 A1 2 2 10(c) 1 M1 Using their integral, their positive x limit from part (b) and ² 0 correctly. 1 2 1 Area under curve = their − − − 0 3 − 2× 1 2 3 − 2 × 0 2 1 A1 24 2

Mark scheme

11(a) 5, –1 B1 B1 Sight of each value 2 11(b) *B1 Needs to be a curve, not straight lines. One complete cycle starting and finishing at their largest value. DB1 One complete cycle starting and finishing at y = 5 and going down to y = −1 and starting to level off at least one end. 2 11(c)(i) 0 solution B1 1 11(c)(ii) 2 solutions B1 1 11(c)(iii) 1 solution B1 1 11(d) 1 B1 Stretch by (scale factor) , parallel to x-axis or in x direction (or 2 horizontally) 0 B1 Accept translation/shift Translation of Accept translation 4 units in positive y-direction. 4 2 11(e) π B1 Accept translation/shift − Translation of 2 π Accept translation − units in x-direction. 0 2 Stretch by (scale factor) 2 parallel to y-axis (or vertically). B1 2