Mathematics 9709/21 October/November 2023: question paper and mark scheme

Mark scheme

1 1 B1 or exact equivalent. State or imply that cosθ = 5 3 Substitute appropriate values into sinθ cos60° + cosθ sin60° M1 1 1 A1 or exact equivalent. Obtain + 15 3 6 3

Mark scheme

2 1 1 1 B1 State or imply derivative of tan x is sec2 x or derivative of 2 2 2 cos2x is −2sin 2x Attempt use of product rule to find first derivative *M1 3 21 1 A1 or (unsimplified) equivalent. Obtain correct sec x cos 2 x − 6 tan x sin 2 x 2 2 2 1 DM1 Substitute π into attempt at first derivative and attempt evaluation to 3 find the gradient Obtain −4 A1 5

Mark scheme

3(a) Obtain 2ln(2 x − 5) B1 Apply limits correctly M1 For integral of form k ln(2 x − 5) . Use one relevant logarithm property correctly M1 For integral of form k ln(2 x − 5) . Apply second logarithm property correctly and obtain ln 25 A1 4 3(b) 1 2 x−5 B1 Integrate to obtain e 2 Obtain final answer 1 15 1 3 e − e B1FT or exact equivalent, FT on their ke 2 x −5 . 2 2 2

Mark scheme

4(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw (more or less) correct graph of y = 2 x + 7 with smaller gradient B1 And crossing y-axis above y-intercept of modulus graph. 2 4(b) Solve 3x − 5 = 2x + 7 to obtain x = 12 B1 Attempt solution of linear equation where signs of 3x and 2x are M1 3x − 5 = −2x − 7 OE. different 2 A1 Obtain x = − 5 Alternative solution for question 4(b) State or imply non-modulus equation (3x − 5) 2 = (2 x + 7) 2 B1 Must be working with (3x − 5) 2 = (2 x + 7) 2 Attempt solution of 3-term quadratic equation M1 2 A1 Obtain − and 12 5 3 4(c) Apply logarithms and use power law for 3 y = k where k > 0 or M1 Using their positive answer from part (b) correct equivalent or greater accuracy; and no other values. Obtain 2.26 A1 2

Mark scheme

5(a) Substitute x = −2 and equate to zero *M1 Substitute x = −1 and equate to −11 *M1 Obtain 4a − 2b − 68 = 0 and a − b − 26 = −11 or equivalents A1 Solve a pair of relevant simultaneous linear equations to find a or b DM1 Dependent at least one M1 mark. Obtain a = 19 and b = 4 A1 5 5(b) Divide by x + 2 at least as far as the x term M1 or equivalent (inspection, …). Obtain ( x + 2) 2 (6 x − 5) A1 OE Replace (or imply replacement of) x by 3x in factorised form M1 2 5 A1 and no others. Obtain − and 3 18 4

Mark scheme

6(a) 1 B1 Use cosecθ = sin θ Express in terms of sinθ and cosθ only M1 Dependent on B1. Obtain given result 4 + 6cosθ − 4cos2 θ with sufficient detail A1 AG 3 6(b) Attempt use of formula to solve 3-term quadratic equation as far as M1 where −1 < k1 < 1 . cosθ = k1 Solve 4cos2 θ − 6cosθ − 7 = 0 to obtain at least cosθ = − 0.770... A1 6 − 148 or exact equivalent cosθ = . 8 Obtain −2.45 A1 or greater accuracy; and no others between − π and 0. 3 6(c) Express cos 2 θ term in the form k2 + k3 cos 2θ M1 where k2 k3 ≠ 0 . Obtain integrand 6cosθ + 2 − 2cos2θ A1 Following the 3-term expression in cosθ from part (a). Integrate to obtain correct 6sinθ + 2θ − sin 2θ A1 Condone absence of + c , but all in terms of θ . 3

Mark scheme

7(a) dy B1 Differentiate y 3 to obtain 3 y 2 dx Differentiate complete equation to produce at least one term involving M1 dy using implicit differentiation. dx dy dy A1 Obtain 2e2 x − 18 + 3 y 2 + =0 dx dx dy A1 Substitute = 0 to obtain either p = 12 ln 9 or p = ln 3 dx 4 7(b) Substitute value of p in original equation and rearrange as far as y 3 = ... M1 Allow in terms of ln9 . or q3 = … Obtain given result q = 3 2 + 18ln3 − q or y = 3 2 + 18ln3 − y with A1 AG sufficient detail 2 7(c) Consider sign of q − 3 2 + 18ln3 − q or equivalent for 2.5 and 3.0 M1 Obtain −0.18... and 0.34... with sufficient detail and justify A1 OE conclusion 2 7(d) Use iteration process correctly at least once M1 Obtain final answer q = 2.673 A1 Answer required to exactly 4 s.f. Show sufficient iterations to 6 sf to justify answer or show sign change A1 in the interval [2.6725, 2.6735] 3