Mathematics 9709/22 February/March 2025: question paper and mark scheme

Mark scheme

1 Apply appropriate logarithm property M1 3x + 1 A1 OE Obtain =7 x −5 Solve to obtain x = 9 A1 3

Mark scheme

2 Integrate to obtain the form y = k tan 12 x *M1 No need for c yet. Substitute x = 12 π and y = 5 to determine value of c DM1 Obtain y = 8 tan 12 x − 3 A1 3

Mark scheme

3(a) State e2 x = 8e− x and obtain 3x = ln8 or e x = 2 or equivalent B1 Confirm x = ln 2 B1 AG Necessary detail needed. Alternative Method for Question 3(a) Substitute x = ln 2 in either e2 x or 8e − x and obtain value 4 B1 Substitute in other expression, obtain value 4 and conclude B1 AG appropriately Necessary detail needed. 2 3(b) Integrate 8e− x − e2 x to obtain −8e− x − 12 e2 x B1 OE, involving separate integrations. Apply limits correctly and simplify to eliminate e and ln M1 Obtain −8 × 12 − 12 × 4 + 8 + 12 or equivalent, and hence 52 or A1 equivalent 3

Mark scheme

4(a) Attempt use of quotient rule or equivalent *M1 4cos x(3 + cos 2 x) + 8sin x sin 2 x A1 Obtain (3 + cos 2 x) 2 2 4(b) Equate first derivative to zero and attempt to express numerator in DM1 terms of cos x Obtain 24cos x − 8cos3 x = 0 A1 OE Attempt solution to find coordinates of at least one stationary point M1 From equation involving two powers of cos x. Obtain ( 12 π, 2) and ( 32 π, − 2) and no others A1 4

Mark scheme

5(a) Show an increasing curve through the origin for y = ln( x + 1) B1 Appearing in first and third quadrants. Show V-shaped graph with vertex on positive x-axis and showing B1 two intersections 2 5(b) Equate − (2 x − 3) and ln( x + 1) or equivalent and confirm result B1 AG (using x or α ) Necessary detail needed. 1 5(c) Use iterative process correctly at least once M1 Obtain final answer 1.12 A1 Answer required to 3 significant figures only. Show sufficient iterations to justify answer or show a sign change A1 in the interval [1.115, 1.125] 3 5(d) Consider sign of 2 x − 3 − ln( x + 1) or equivalent for 2.055 and M1 But not for − (2 x − 3) − ln( x + 1), nor for calculations based on 2.065 equation in part (b). Obtain − 0.006... and 0.009... or equivalents and justify A1 conclusion 2

Mark scheme

6(a) Carry out division at least as far as 6 x 2 + k1 M1 Obtain quotient 6 x 2 − 10 A1 Obtain remainder − 6 A1 3 6(b) 6 B1 FT Following their quotient and remainder. Identify integrand as 6 x 2 − 10 − 3x − 1 Integrate to obtain at least 2x3 and term of the form k2 ln(3x − 1) *M1 Obtain 2 x − 10 x − 2ln(3 x − 1) A1 FT Following their quotient and remainder. 3 Apply limits and appropriate logarithm properties DM1 Obtain 208 − ln 49 A1 5

Mark scheme

7(a) State R = 52 or 2 13 B1 Use appropriate trigonometry to find α M1 Obtain α = 33.69 A1 Or greater accuracy. 3 7(b) Use correct method to find one value of θ M1 Obtain 257.6 (or 349.8) A1 Or greater accuracy 257.587… (or 349.792…). Use correct method to find second value of θ between 0 and 360 M1 Obtain 349.8 (or 257.6) and no others between 0 and 360 A1 Or greater accuracy. 4 7(c) Obtain greatest value 28 B1 Attempt complete method to find positive value of β using M1 sin(4β − α ) = 1 Obtain 14 (90 + 33.69) and hence 30.9 A1 Or greater accuracy (30.922…). 3

Mark scheme

8(a) Attempt use of product rule for differentiating 3e 2 x y M1 dy A1 Obtain 6e2 x y + 3e2 x dx dy B1 Obtain remaining terms 12e3 x + 3 y 2 dx Equate sum of correct terms to zero and rearrange to confirm A1 AG given result Necessary detail needed. 4 8(b) Equate derivative to zero and obtain y = −2e x or equivalent B1 Substitute for y in original equation and attempt appropriate M1 simplification Obtain −10e3 x = 18 or equivalent and conclude no possible value A1 AG of x Necessary detail needed. 3