Cambridge A Level Mathematics 9709 — 2025 Oct/Nov Paper 2 · Variant 1

1 Find 6 sin 2 x dx . [3] y .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Use identity to express integrand in the form k1 + k 2 cos2 x M1 Obtain correct 3 − 3cos2 x A1 Integrate to obtain 3 x − 32 sin 2 x A1 Condone omission of + .c 3

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2 Solve the equation e 2 x `e 2 x - 8j = 48 . [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 Attempt solution of quadratic equation in e 2 x to obtain at least one value of e 2 x M1 Obtain (e 2 x − 12)(e 2 x + 4) = 0 OE, and hence at least e 2 x = 12 A1 Obtain x = 12 ln12 or 1.24 and no other solution A1 3

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3 (a) Solve the equation 2x - 3 = 5 x + 2 . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence solve the equation 2 sec i - 3 = 5 sec i + 2 for r 1 i 1 2 r . Give your answer correct to 3 significant figures. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) Solve 2 x −=3 5 x + 2 to obtain − 53 B1 Or exact equivalent. Attempt solution of linear equation where 2x and 5x have different signs M1 Obtain 1 A1 OE 7 Alternative Method for Question 3(a) State or imply non-modulus equation (2 x − 3) 2 = (5 x + 2) 2 B1 Attempt complete solution of three-term quadratic equation M1 Obtain − 53 and 17 or equivalents A1 OE 3 3(b) State or imply cos= − 53 B1 FT Following an appropriate answer from part (a). Attempt correct process for finding third quadrant angle M1 Condone working in degrees for this mark. Obtain 4.07 A1 Or greater accuracy 4.0688… 3

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4 Solve the equation cot i tan ( i+ 45 °) = 7 for 0° 1 i 1 90° . [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4 Attempt to express equation in terms of tan only M1* 1 tan+ tan45 A1 OE Obtain  = 7 tan 1 − tantan45 Simplify to obtain equation 7tan 2 − 6tan+ 1 = 0 A1 Attempt solution of three-term quadratic equation in tan to obtain at least one DM1 value of  Obtain tan= 17 (3  2) or equivalent and hence 12.8 and 32.2, and no others in the A1 Or greater accuracy 12.764…, 32.235… given range 5

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5 y A B x O - 21 x The diagram shows the curve with equation y = 8e - 1. The curve meets the axes at the points A and B. The shaded region is bounded by the curve and the line segment AB. (a) Show that the x-coordinate of B is 6 ln 2. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the area of the shaded region. Give your answer in the form p ln 2- q , where p and q are positive integers. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ .......................... ............................................................................................................................

Mark scheme: 5(a) − 12 x M1 1 Attempt solution of 8e −=1 0 with use of a relevant logarithm property E.g. ln x = ln8 2 Confirm x = 6ln2 A1 AG – necessary detail needed. Need to see ln8 = 3ln2 or 64 = 2 6 OE used. Alternative Method for Question 5(a) − 12 x M1 Substitute x = 6ln2 in 8e − 1 and show use of a relevant logarithm property −3ln2 ln 18 A1 AG – necessary detail needed. Obtain 8e − 1 and hence 8e − 1 or equivalent and verify answer 0 2 5(b) − 12 x 12 x M1 Integrate 8e − 1 to obtain the form k e− − x 2 x − x A1 Obtain correct −16e − 1 2 6ln2 − 6ln2 + 16 A1 OE Apply limits 0 and 6ln2 to obtain −16e −1 14 − 6ln2 Attempt area of triangle minus area under curve M1 Allow decimals for this mark (14.56). Obtain 12  6ln 2  7 − (14 − 6ln 2) or equivalent, and hence 27ln2 − 14 A1 5

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7 The polynomial p ( )x is defined by p ( )x = 2 x 4 + kx 3 + kx 2 + 17x + 18 , where k is a constant. It is given that ( x + 2 ) is a factor of p ( )x . (a) Find the value of k. 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It is given that the equation p ( )x = 0 has exactly two real roots, denoted by a and b, where a is an integer and b is not an integer. (b) State the value of a and show that b satisfies the equation x = 3 - 2x - 4 .5 . 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(c) Show by calculation that -1.4 1 b 1 -1 .0 . 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(d) Use an iterative formula, based on the equation in part (b), to find the value of b correct to 3 significant figures. Give the result of each iteration to 5 significant figures. 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Mark scheme: 7(a) Substitute x = −2 , equate to zero and attempt solution M1 Obtain 32 − 8k + 4k − 34 + 18 = 0 or equivalent and hence k = 4 A1 2 7(b) State = −2 B1 Divide by x + 2 at least as far as 2 x 3 + mx M1 Or equivalent method, e.g. inspection. Obtain 2 x 3 + 4 x + 9 A1 Rearrange 2 x 3 + 4 x + 9 = 0 to confirm x = 3 −2 x − 4.5 A1 AG – necessary detail needed. 4 7(c) Consider sign of x − 3 −2 x − 4.5 or equivalent for −1.4 and −1.0 M1 Or Example 1: f ( x ) = 2 x3 + 4 x + 9, finding f ( − 1.4 ) and f ( − 1) . Example 2: f ( x ) = 3 −2 x − 4.5, finding f ( − 1.4 ) and f ( − 1) . Obtain −0.2... and 0.3… or equivalents and justify conclusion A1 Example 1: f ( −1) = 3, f ( −1.4 ) = −1.2... with a correct conclusion. Example 2: f ( −1) = − 1.357.... so −−1 1.357... f ( −1.4 ) = −1.193.... so −1.4 −1.19... with a correct conclusion. AG – necessary detail needed. 2 7(d) Use iterative process correctly at least once M1 Obtain final answer −1.26 A1 Answer required to exactly 3 significant figures. Show sufficient iterations to 5 sf to justify answer or show sign change in interval A1 [ −1.265, − 1.255] 3

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8 The equation of a curve is y = 4e 1 - 2x 3x - 1 . Find the exact coordinates of the stationary point of the curve. 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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: 28 2 *M1 Where k1k 2  0. Differentiate to obtain the form k1e1− 2 x (3x − 1) 1 + k2 e1− 2 x (3x − 1)− 1 1− 2 x 12 A1 OE. Allow unsimplified Obtain correct first term −8e (3x − 1) 1− 2 x 12 A1 OE. Allow unsimplified Obtain correct second term + 6e (3x − 1)− Equate first derivative to zero and attempt solution as far as 3x −=1 k DM1 OE. Must be in the form 2 2 k1e1− 2 x (3x − 1) 1 + k2 e1− 2 x (3x − 1)− 1 Obtain 3 x −=1 34 OE, and hence x = 127 A1 Or exact equivalent. − 16 A1 Or exact equivalent. Obtain y = 2 3e 6

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