Mathematics 9709/21 May/June 2024: question paper and mark scheme

Mark scheme

1 Differentiate to obtain 2sec 2 x − 5cos x B1 Equate first derivative to zero and solve correctly at least as far as cos x = ... M1 B1 M1 for x = 42.5o . Obtain cos x = 3 0.4 and hence x = 0.742 A1 Or greater accuracy (0.74261…). 3

Mark scheme

2 Attempt use of product rule for differentiating x 2 ln y M1 dy Allow if missing. dx x 2 dy A1 Obtain 2 x ln y + y dx dy B1 Obtain + 2 y dx dy M1 Correct equation: Substitute x = 2, y = 1 in equation equal to 0 involving at least one and solve for dx x 2 dy dy 2 x ln y + + 2y +4=0 dy y dx dx dx Obtain − 23 A1 Or exact equivalent. Not from wrong working. 5

Mark scheme

3(a) Draw V-shaped graph with vertex on positive x-axis in the first quadrant. B1 Draw correct graph of y = 5 − x correctly positioned with respect to modulus graph. B1 Two points of intersection. 2 3(b) Solve 3x − 8 = 5 − x to obtain 134 B1 Or inequality. Solve linear equation or inequality with signs of 3x and x the same M1 Obtain 32 A1 Conclude 32 < x < 134 or x > 32 and x < 134 A1 Allow alternative notation e.g. ( 32 , 134 ) . Alternative Method for Question 3(b) State or imply non-modulus equation (or inequality) (3 x − 8) 2 = (5 − x ) 2 (B1) Attempt solution of three-term equation (or inequality) (M1) Obtain 32 and 134 (A1) Conclude 32 < x < 134 or x > 32 and x < 134 (A1) Allow alternative notation e.g. ( 32 , 134 ) . 4 3(c) Attempt value of N (maybe non-integer at this stage) for e 0.1N < their 134 M1 Allow e0.1N = their 13 (or inequality). 4 Conclude with single integer 11 A1 2

Mark scheme

4(a) 6 tan θ B1 2 tan θ State 3tan 2θ = Allow for 3tan 2θ = 3 . 1 − tan 2 θ 1 − tan 2 θ tan θ + 1 B1 State tan(θ + 45°) = 1 − tan θ Attempt to express left-hand side in terms of tan θ as a single fraction M1 Condone sign errors in identities. Allow if tan 45o is not evaluated. Allow if 2 separate terms with a common denominator. tan 2 θ + 8 tan θ + 1 A1 Answer given – necessary detail needed. Confirm 1 − tan 2 θ 4 4(b) Solve quadratic equation 5 tan 2 θ + 8 tan θ − 3 = 0 to obtain at least one value of θ M1 Must be using a correct method. Obtain 17.4 A1 Or greater accuracy. Obtain 117.6 A1 Or greater accuracy; and no others between 0° and 180° . 3

Mark scheme

5(a) Differentiate using quotient rule *M1 OE (1 + 3 x ) 2e 2 x − (1 + e 2 x ) 3 A1 OE Obtain (1 + 3 x ) 2 Equate first derivative to zero and arrange as far as x = ... DM1 −2 x A1 Answer given – necessary detail needed. Confirm x = 16 + 12 e Must have exact terms. 4 5(b) −2 x M1 OE Consider sign of x − 16 − 12 e Obtain −0.06... (– 0.064959...) and 0.08... (0.0800…) or equivalents and justify A1 Answer given – necessary detail needed. conclusion Alternative Method 1 for Question 5(b) Consider f ( x ) = 16 − 12 e and obtain f ( 0.35 ) = 0.42 (0.4149…) and f ( 0.45 ) = 0.37 −2 x (M1) (0.36995….) Conclude f ( 0.35 ) < 0.45 and f ( 0.45 ) > 0.35 so root lies in given interval. (A1) Alternative Method 2 for Question 5(b) dy (M1) Consider the sign of their from part (a) dx Obtain −0.19... (– 0.187...) and 0.21... (0.2139…) or equivalents and justify (A1) conclusion 2 5(c) Use iterative process correctly at least once M1 Obtain final answer 0.394 A1 Answer required to exactly 3sf. Show sufficient iterations to 5 sf to justify answer or show sign change in interval A1 [0.3935, 0.3945] 3

Mark scheme

6(a) Use y-values (0), sin 16 π + sin 2 16 π, sin 13 π + sin 2 13 π or decimal equivalents B1 ( 0) , 0.75 or 0.866, 1.616 or 1.271 . Use correct formula, or equivalent, with h = 121 π M1 Must be using ‘y’ values. May do as 2 separate trapezia (0.113359 + 0.27976). Obtain 0.39 A1 Allow 0.393 but not greater accuracy. 3 6(b) Use ( π ) (sin 2 x + sin 2 2 x) dx M1 OE Express integrand in the form k1 sin 2 x + k2 + k3 cos 4 x *M1 Where k1k 2 k3 ≠ 0 . Obtain correct sin 2 x + 12 − 12 cos 4 x A1 Or π times this. Integrate to obtain − 12 cos 2 x + 12 x − 18 sin 4 x A1 FT Following their integrand only if of correct form. Apply limits correctly to obtain exact value DM1 Obtain volume = 14 π + 12 1 2 π − 161 π 3 A1 Or exact equivalent. 6

Mark scheme

7(a) Carry out division at least as far as 3x 2 + n1 M1 Or equivalent (inspection, …). Obtain quotient 3 x 2 + 4 A1 Confirm remainder is k − 8 A1 Answer given – necessary detail needed. SC B1 for correct use of factor theorem to show remainder is k − 8 . Alternative Method for Question 7(a) Synthetic division (M1) Allow one sign error. –2/3 9 6 12 k –6 0 –8 9 0 12 k −8 Obtain quotient 3 x 2 + 4 (A1) Confirm remainder is k − 8 (A1) 3 7(b) Integrate to obtain at least a term in x 3 and term of form n2 ln(3 x + 2) *M1 Need to be using their answer to part (a). 3 Obtain x + 4 x + 13 (k − 8)ln(3x + 2) A1 FT on a quotient of 9 x 2 + 12 . Apply limits correctly to expression with three terms DM1 Obtain a = 235 A1 FT on a quotient of 9 x 2 + 12 . Equate logarithm term to ln 64 and apply appropriate logarithm properties DM1 Obtain k = 17 A1 6 SC 2 marks for use of quotient 3 x 2 + 4 or 9 x 2 + 12 to obtain either 235 or 705 if no other marks are available.