Cambridge A Level Mathematics 9709 — 2025 May/June Paper 2 · Variant 3

11 8 1 Show that dx = ln a , where a is an integer to be found. [3] y 2 4x + 1 .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Integrate to obtain 2ln(4 x + 1) B1 Apply limits correctly to k ln(4 x + 1) and use at least one relevant logarithm property M1 Obtain 2ln45 − 2ln9 = 2ln5 or equivalent, and conclude ln25 A1 3

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2 (a) Sketch on the same diagram the graphs of y = 2x - 9 and y = 4x - 5 . [2] (b) Solve the inequality 2x - 9 1 4 x - 5 . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw correct graph of y = 4 x − 5 B1 Correctly placed with reference to modulus graph and with steeper gradient. 2 2(b) Solve linear equation or inequality with signs of 2x and 4x different M1 Obtain −2 x + 9 = 4 x − 5 and hence x = 73 A1 OE Conclude x  7 A1  7   7  3 OE, e.g. ,  , or ,  .      3   3  Alternative Method for Question 2(b) State or imply non-modulus equation (or inequality) (2 x − 9) 2 = (4 x − 5) 2 B1* Attempt solution of three-term quadratic equation (or inequality) DM1 12 x 2 − 4 x − 56 = 0 leading to 73 and −2. Obtain at least 7 and conclude x  7 A1 Must be from correct work. 3 3  7   7  OE, e.g. ,  or ,       3   3  3

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8x 3 Find the coordinates of the stationary points of the curve with equation y = - 6 x + 5 . [5] 2x + 3 .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3 8 x −12 x 2 + 15 *M1 Attempt use of quotient rule (or equivalent) to differentiate or 2 x + 3 ( 2 x + 3 ) d y 24 A1 OE Obtain = − 6 d x (2 x + 3) 2 Allow unsimplified. Equate first derivative to zero, and attempt solution of a quadratic equation to find DM1 4 x 2 + 12 x + 5 = 0 two values of x Obtain at least one of the stationary points ( − 5 , 30) and ( − 1 , 6), A1 2 2 or both x-values, − 5 and − 1 2 2 Obtain both stationary points A1 5

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4 y P O x The diagram shows parts of the curves with equations y = 4e -2 x and y = 1 + 0.5 sin 3x . Point P is a point of intersection of the curves, and the shaded region is bounded by the two curves and the y-axis. (a) Show that the x-coordinate of P satisfies the equation x =-0. 5 ln ( 0. 25 + 0. 125 sin 3 x) . 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(b) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of P correct to 4 significant figures. Use an initial value of 0.5 and give the result of each iteration to 6 significant figures. 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(c) Hence find the area of the shaded region. Give your answer correct to 2 significant figures. 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Mark scheme: 4(a) State 4e−2 x = 1 + 0.5sin3x and confirm x = −0.5ln(0.25 + 0.125sin3 x ) B1 AG Necessary detail needed. 1 4(b) Use iterative process correctly at least once M1 Obtain final answer 0.4912 A1 Answer required to exactly 4 significant figures. Show sufficient iterations to 6 significant figures to justify answer A1 or show a sign change in the interval [0.49115, 0.49125] 3 4(c) Integrate to obtain the form k1e −2 x + k 2 x + k3 cos3 x *M1 k1 , k2 , k3  0 OE, with separate integrals or ‘reversed’. Obtain correct −2e −2 x − x + 1 cos3 x A1 OE 6 −2 x 1 Allow for 2e + x − cos3 x 6 Apply limits 0 and their part (b) answer correctly and attempt evaluation DM1 Obtain 0.61 A1 4

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5 The polynomial p ( )x is defined by p ( )x = ax 4 + bx 3 + 13 x 2 - 35 x + 15 , where a and b are constants. It is given that ( 2x - 1 ) and ( x - 3 ) are factors of p ( )x . (a) Find the values of a and b. 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(b) Hence factorise p ( )x . 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(c) Find the least positive value of i in radians such that p ( cot2 i) = 0 . 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Mark scheme: 5(a) Substitute x = 1 and x = 3, and equate each to zero to produce two equations M1 SC B1 for a correct equation, if M0 otherwise. 2 Obtain 1 a + 1 b = − 3 A1 OE 16 8 4 a + 2b + 12 = 0 Obtain 81a + 27b = −27 A1 OE 3a + b + 1 = 0 Solve simultaneous equations to obtain a = 2 and b = −7 A1 4 5(b) Divide by 2 x 2 − 7 x + 3 or successively by 2 x − 1 and x − 3 M1 OE method, such as inspection. from synthetic division. Obtain quotient x 2 + 5 A1 Condone 2 ( x 2 + 5 ) State fully factorised form (2 x − 1)( x − 3)( x 2 + 5) A1 3 5(c) Attempt solution of at least cot2= 3 M1 Obtain tan 2= 1 and hence = 0.161 A1 Or greater accuracy 0.16087… 3 2

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6 A curve has equation ( x 2 - 3 )ln y + 6 x = 14 . (a) Show that there is no point on the curve at which the y-coordinate is e -1 . 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(b) Find the equation of the tangent to the curve at the point ( 2, e2 ) . Give your answer in the form y = mx + c , where m and c are exact constants. 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Mark scheme: 6(a) Substitute y = e − 1 in equation and simplify to quadratic equation in x *M1 Evaluate discriminant DM1 OE, such as completion of square or use of formula. Obtain − x 2 + 6 x − 11 = 0, giving discriminant −8 and confirm result A1 OE, such as ( x − 3) 2 + 2 = 0, etc. AG – necessary detail needed. 3 6(b) Attempt use of product rule for differentiation of ( x 2 − 3)ln y *M1 May see in part (a) x 2 − 3 dy A1 OE Obtain 2 x ln y + y dx x 2 − 3 dy A1 FT OE Obtain complete 2 x ln y + + 6 = 0 2 Following their derivative of ( x − 3)ln y. y dx Substitute x = 2 and y = e 2 to find value of gradient of tangent DM1 Need to see attempt at substitution unless correct. −103 implies M1. Obtain gradient −14e 2 A1 Obtain equation y = −14e 2 x + 29e 2 or equivalent of required form A1 6

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7 (a) Express 4 cos i sin ( i+ 30°) in the form R cos ( 2i - a) + k , where R 2 0 , 0° 1 a 1 90° and k is a constant. 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(b) Hence solve the equation 12 cos 2 z sin ( 2 z+ 30° ) = 5 for 0° 1 z 1 90 ° . 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Mark scheme: 7(a) Expand to obtain the form k1 cossin+ k 2 cos 2  M1 2 A1 Obtain 2 3cossin+ 2cos  with exact coefficients Obtain 3sin2+ cos2+ 1 A1 FT Following their expression in sin and cos. State R = 2 B1 FT Following their expression in sin2 and cos2. Use appropriate trigonometry to find value of  M1 Must be of the form a sin2+ b cos2. Conclude 2cos(2− 60) + 1 A1 6 7(b) Attempt to express equation in form cos(4− ) = k *M1 Use correct process to find one value of  DM1 Obtain cos(4− 60) = 1 and hence 32.6 A1 Or greater accuracy 32.6321… 3 Use correct process to find second value of  DM1 Obtain 87.4 and no others between 0 and 90 A1 Or greater accuracy 87.3678… 5

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