Cambridge A Level Mathematics 9709 — 2025 Oct/Nov Paper 2 · Variant 2

1 Solve the equation ln ( 3x + 5) - ln ( x - 2) = 4 . Give your answer in an exact form. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Apply relevant logarithm property *M1 3 x + 5 4 A1 Allow recovery from an incorrect log statement. Obtain correct equation without logarithms = e or equivalent x − 2 Solve equation to find exact value of x DM1 Must be in correct form, condone one sign slip. 2e 4 + 5 A1 Or exact equivalent. Obtain e 4 − 3 4

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2 Solve the equation 2 tan 2i + 3 sec i = 18 for –180° 1 i 1 180° . [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 Attempt identity involving tan 2  and sec 2  to obtain quadratic equation in sec or *M1 cos Obtain 2sec 2 + 3sec− 20 = 0 or 20cos 2 − 3cos− 2 = 0 A1 Solve three-term quadratic equation to obtain at least one value of  DM1 Obtain at least two values of 66.4 and 104.5 A1 Or greater accuracy 66.421…, 104.477… Obtain all four values and no others between –180 and 180 A1 Or greater accuracy. 5

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4 The polynomial p ( )x is defined by p ( )x = x 4 - 10x 3 + 20 x 2 - 30 x + 40 . (a) Find the quotient when p ( )x is divided by (x 2 + 3 ) and show that the remainder is –11. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence find the real roots of the equation p ( )x + 11 = 0 . Give your answers in exact form. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) Carry out division as far as x 2  10 x M1 Or equivalent method, e.g. inspection. Obtain quotient x 2 − 10 x + 17 A1 WWW Confirm remainder is –11 A1 WWW AG – necessary detail needed. 3 4(b) State or imply that equation is ( x 2 + 3)( x 2 − 10 x + 17) = 0 B1 FT Following their quotient from part (a). Attempt solution of their three-term quadratic quotient from (a) to give two exact M1 roots Obtain 5  8 or 5  2 2 and no other solutions A1 Do not ISW. 16  2 2 gets A0. 3

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5 y A B O x The diagram shows the curve with equation y = 4 cos 2 x + 8 sin x for 0 G x G r . The maximum points on the curve are denoted by A and B, and the shaded region is bounded by the line segment AB and the curve. (a) Find the coordinates of A and B. 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(b) Find the exact area of the shaded region. 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Mark scheme: 5(a) Differentiate to obtain −8sin2 x + 8cos x B1 Attempt to solve equation of form k1 sin2 x + k 2 cos x = 0 using correct identity *M1 Must have attempted differentiation. Obtain at least sin x = 12 A1 Attempt to find both coordinates of at least one maximum point DM1 5 1 and no others between 0 and π A1 Angles must be in radians. Obtain ( and ( 6 π, 6 ) 6 π, 6 ) 5 5(b) Integrate to obtain expression of form k1 sin2 x + k 2 cos x *M1 Obtain correct 2sin2 x − 8cos x A1 Condone poor notation for this mark. Apply their x-limits correctly to evaluate area under curve between A and B *DM1 Condone use of degrees for this mark. Obtain 6 3 or exact unsimplified equivalent A1 Carry out correct process to find area of shaded region DDM1 The y-coordinates of both points must be the same. Rectangle – their 6 3. 5 and hence 4π − 6 3 A1 Or exact equivalent. Obtain 6 ( 6 π − 16 π ) − 6 3, 6

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a 6 (a) Given that y b 12 e 2 x + 14 e – xl dx = 5 , where a is a positive constant, show that – 2 a a = 1 ln b10 + 1 e – a + 1 e – 4 al. 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(b) Hence show by calculation that the value of a lies between 1.0 and 1.2. 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(c) Use the iterative formula a = 1 ln b10 + 1 e – a n + 1 e – 4 a nl n + 1 2 2 2 to find the value of a correct to 4 significant figures. Give the result of each iteration to 6 significant figures. 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Mark scheme: 6(a) Integrate to obtain expression of form k1e 2 x + k 2 e− x *M1 Obtain correct 1 4 e2 x − 14 e− x A1 Apply limits and arrange at least as far as 1 2 e 2 a = ... or 2e a = ... DM1 Confirm a = 12 ln(10 + 12 e− a + 12 e−4 a ) A1 AG – necessary detail needed. 4 16(b) Consider sign of a − 2 ln(10 + 12 e − a + 12 e −4 a ) or equivalent for values 1.0 and 1.2 M1 May use an intermediate equation from part (a). Example 1: consider 1 2 ln(10 + 12 e− a + 12 e −4 a ) for values 1.0 and 1.2. 1 2 a 1 − a 1 −4 a Example 2: consider e − e − e − 5 2 4 4 for values 1.0 and 1.2. Obtain –0.16… and 0.041… or equivalents and justify conclusion A1 Example 1: 1.0  1.1608... and 1.2  1.1589... (allow to 2dp and truncation), so 1.0   1.2 OE. Example 2: f (1.0 ) = −1.40... f (1.2 ) = 0.43... and change of sign so 1.0   1.2 OE. 2 6(c) Use iteration process correctly at least once M1 Obtain final answer 1.159 A1 Required to precisely 4 significant figures. Show sufficient iterations to justify answer or show a sign change in the interval A1 [1.1585, 1.1595] 3

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7 A curve has equation 5x 2 y + 4e 2 y - 7x + 10 = 0 . dy (a) Find an expression in terms of x and y for and hence find the gradient of the curve at the point dx for which y = 0 . 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(b) Show that there is no point on the curve at which the tangent is parallel to the y-axis. 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Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 7(a) Attempt use of product rule for differentiating 5x 2 y *M1 Must have evidence of implicit differentiation. 2 dy A1 Obtain 10 xy + 5 x dx 2 y 2 y dy B1 Differentiate 4e and obtain 8e dx 2 dy 2 y dy dy 7 − 10 xy A1 OE Obtain 10 xy + 5 x + 8e − 7 = 0 and hence = dx dx dx 5 x 2 + 8e 2 y Attempt to find relevant value of x and substitute that and y = 0 to find gradient DM1 x = 2 Obtain x = 2 and gradient 14 A1 OE 6 7(b) Attempt to show that denominator of derivative cannot be zero M1 d y Must not be using when y = 0. d x State that both terms are non-negative or equivalent and conclude appropriately A1 FT Following their denominator of form k1 x 2 + k 2e 2 y , provided that k1k 2  0. 2

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