Cambridge A Level Mathematics 9709 — 2025 Feb/March Paper 2 · Variant 2

1 Solve the equation ln ( 3x + 1) - ln ( x - 5) = ln 7 . [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Apply appropriate logarithm property M1 3 x + 1 A1 OE Obtain = 7 x − 5 Solve to obtain x = 9 A1 3

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1 d y 2 12 A curve passes through the point with coordinates b r, 5l and is such that = 4 sec b xl. 2 d x 2 Find the equation of the curve. [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 = 8tan 12 x − 3 A1 3

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3 y x O The diagram shows the curves y = e 2 x and y = 8e -x . The shaded region is bounded by the two curves and the y-axis. (a) Show that the x-coordinate of the point of intersection of the two curves is ln2. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the area of the shaded region. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) State 2e x = 8e− x and obtain 3x = ln8 or e x = 2 or equivalent B1 Confirm x = ln2 B1 AG Necessary detail needed. Alternative Method for Question 3(a) Substitute x = ln2 in either 2e 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) x 2 x − x 1 2 x B1 OE, involving separate integrations. Integrate 8e −− e to obtain −8e − 2 e 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

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4 sin x 4 A curve has equation y = for values of x such that 0 G x G 2 r . 3 + cos 2x dy (a) Find . [2] dx ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence find the coordinates of the stationary points of the curve. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) Attempt use of quotient rule or equivalent *M1 4cos x (3 + cos2 x ) + 8sin x sin2 x A1 Obtain (3 + cos2 x ) 2 2 4(b) Equate first derivative to zero and attempt to express numerator in DM1 terms of cos x Obtain 24cos x − 8cos 3 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 2(1 π, 2) and 2(3 π, − 2) and no others A1 4

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5 (a) Sketch on the same diagram the graphs of y = 2 x - 3 and y = ln ( x + 1 ) . [2] The x-coordinates of the points where the graphs intersect are denoted by a and b, where a 1 b. (b) Show that a = 1.5 - 0 .5 ln ( a + 1 ) . [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Use an iterative formula, based on the equation in part (b), to find the value of a correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (d) Show by calculation that 2.055 1 b 1 2.065 . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

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 ) 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

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6 (a) Find the quotient and remainder when 18x 3 - 6 x 2 - 30x + 4 is divided by ( 3x - 1 ) . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 5 3 2 18x - 6x - 30x + 4 (b) Hence find d x . Give your answer in the form a - ln b , where a and b are y 1 3x - 1 integers. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

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) 2 6 B1 FT Following their quotient and remainder. Identify integrand as 6 x − 10 − 3 x − 1 Integrate to obtain at least 2x 3 and term of the form k 2 ln(3 x − 1) *M1 Obtain 2 x 3 − 10 x − 2ln(3 x − 1) A1 FT Following their quotient and remainder. Apply limits and appropriate logarithm properties DM1 Obtain 208 − ln49 A1 5

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7 (a) Express 6 sin i - 4 cos i in the form R sin ( i - a) , where R 2 0 and 0° 1 a 1 90° . Give the exact value of R and the value of a correct to 2 decimal places. 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(b) Hence solve the equation 6 sin i - 4 cos i + 5 = 0 for 0° 1 i 1 360° . 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(c) As the value of b varies, find the greatest possible value of ( 3 sin 4b - 2 cos 4 b) 2 + 15 and determine the smallest positive value of b, in degrees, for which this greatest value occurs. 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Mark scheme: 7(a) State R = 52 or 2 13 B1 Use appropriate trigonometry to find  M1 Obtain = 33.69 A1 Or greater accuracy. 3 7(b) Use correct method to find one value of  M1 Obtain 257.6 (or 349.8) A1 Or greater accuracy 257.587… (or 349.792…). Use correct method to find second value of  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  using M1 sin(4− ) = 1 Obtain 14 (90 + 33.69) and hence 30.9 A1 Or greater accuracy (30.922…). 3

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8 A curve has equation 3 e 2 x y + 4 e 3 x + y 3 = 18 . dy - 2e 2 x y - 4e 3 x (a) Show that = . [4] dx e 2 x + y 2 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Show that the curve has no stationary points. 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Mark scheme: 8(a) Attempt use of product rule for differentiating 3e 2 x y M1 2 x 2 x dy A1 Obtain 6e y + 3e dx 3 x 2 dy B1 Obtain remaining terms 12e + 3 y 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 Necessary detail needed. of x 3

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