Cambridge A Level Mathematics 9709 — 2025 Oct/Nov Paper 3 · Variant 1

1 Find the exact value of ln 3x dx . Give your answer in the form a + ln b , where a and b are integers. 2y 1 [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Commence integration by parts and reach Ax ln3 x +  Bdx *M1 1 Accept Ax ln3 x − B x  dx. Allow unsimplified.  x Obtain x ln3x – x A1 Substitute limits correctly in an expression of the form Ax ln3x + Bx DM1 2ln6 −−2 ln3 + 1 Obtain –1 + ln12 A1 If no working seen, no marks are available. 4

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2 (a) Show that the equation log ( 2x + 1 ) = 2 log ( 3 x - 1 ) - 2 can be written as a quadratic equation 4 4 in x. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence solve the equation log ( 2x + 1 ) = 2 log ( 3 x - 1 ) - 2 . [2] 4 4 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) State or imply 2 = log4 42 B1 Use log power, product or quotient law e.g. 2log4 (3x – 1) = log4 (3x – 1)2 M1 ( 3 x − 1) 2 E.g. log4 (3x – 1)2 – log4 42 = log4 4 2 Obtain a correct equation in any form, free of logs A1 E.g. 16(2x + 1) = (3x – 1)2 or 9x2 – 38x – 15 = 0 3 2(b) Solve 3-term quadratic M1 Obtain answer 4.59 only A1 19 + 4 31 OE. 9 Accept AWRT 4.59. 2

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3 (a) Express 3 2 sin ( x + 45°) + cos x in the form R cos ( x - a) , where R 2 0 and 0° 1 a 1 90° . 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(b) Hence solve the equation 3 2 sin ( 3i + 45°) + cos 3i =-4 for 0° 1 i 1 180° . 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Mark scheme: 3(a) Obtain 3sinx + 4cosx B1 State R = 5 B1 FT FT their 3 and 4. −1 3  M1 Using their 3 and 4. Use trig formula to find α e.g.  = tan   If using cos= 4 and sin= 3 seen, this scores  4  M0A0. Obtain = 36.9 (or awrt 36.9) A1 4 3(b) − 1  − 4  B1 FT FT their R from (a). Evaluate cos    − 4   R  Allow cos − 1   for B1 143.13…..  5  Carry out a correct method to find a value of θ in the given interval M1 Obtain answer, e.g. 60(.0)° A1 Answers may be more accurate for A marks. Accept AWRT 60.0. Obtain a second answer, e.g. 84.6° and no other in the interval A1 Ignore answers outside the given interval. Accept AWRT 84.6. 4

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4 y M O 1 x r 4 The diagram shows the graph of y = e sin 2 x cos 4x for 0 G x G 1 r , and its maximum point M. 4 Find the x-coordinate of M. 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Mark scheme: 4 Use correct product rule: p sin 4 xe sin2 x + q cos2 x cos4 xe sin2 x M1 Condone sign error. Condone if no attempt at chain rule seen. Obtain derivative −4sin4 xesin2 x + 2cos2 x cos4 xe sin2 x A1 OE Equate derivative to zero and obtain an equation in one trig function M1 Allow arithmetical errors. Obtain 2sin 2 2 x + 4sin2 x −=1 0 A1 OE Obtain x = 0.113 and no other root A1 May be more accurate. 5

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5 Im 0 −1 1 2 3 Re − i − 2i − 3i − 4i The shaded region on the Argand diagram shows points representing complex numbers z defined by two inequalities. The shaded region is bounded by a circle and a line parallel to the imaginary axis. The boundaries of the region are included in the shaded region. (a) Find two inequalities in terms of z that define the shaded region. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Calculate the least value of arg z for points in this region. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) Obtain Re z  2 B1 Allow 2 Re( z ) 3 Condone strict inequalities throughout (a). The shading is to the RHS of the diagram. Obtain answer of the form z − a  b M1 Accept equation or any inequality sign. Must have a complex and b real. Condone brackets in place of modulus for this mark. Obtain answer z −+1 2i  2 A1 OE z − (1 − 2i )  2 3 5(b) Identify the correct point B1 May be clearly indicated as a point on the diagram. May be 2, − 3 − 2 . ( ) May be by attempt to solve (x – 1)² + (y + 2)² = 2² with x = 2. Carry out a correct method for finding the least value of arg z M1 Obtain answer –61.8° or 298.2° or –1.08c or 5.2(0)c A1 3

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6 Solve the quadratic equation ( 2 + )i w 2 + 4 w + 2 - i = 0 . Give your answers in the form x + yi , where x and y are real. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 6 Use quadratic formula to solve for w M1 −4 16 − 4 ( 2 − i )( 2 + i ) w = 4 + 2i Use i2 = –1 M1 −−4 2i −+4 2i A1 OE Obtain one of the answers w = or w = 4 + 2i 4 + 2i Multiply numerator and denominator by conjugate of their denominator M1 Need some evidence of multiplying. 3 A1 Obtain final answers − + 4i and –1 [+ 0i] 5 5 Alternative Method for Question 6 Multiply the equation by 2 – i M1 Use i2 = –1 M1 Obtain 5w2 + 4(2 – i)w + (2 – i)2 A1 OE Use quadratic formula or factorise to solve for w M1 3 A1 Obtain final answers − + 4i and –1 [+ 0i] 5 5 Alternative Method 2 for Question 6 Substitute w = x + iy and form equations for real and imaginary parts M1 Two separate equations. Use i2 = –1 M1 Obtain 2(x2 – y2) – 2xy + 4x + 2 = 0 and x2 – y2 + 4xy + 4y – 1 = 0 A1 OE Form equation in x only or y only and solve M1 3 A1 Obtain final answers − + 4i and –1 [+ 0i] 5 5 5

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7 The parametric equations of a curve are 2 t x = t - ln ( 2 t + 1), y = . 2t + 1 dy Obtain a simplified expression for in terms of t. [5] dx .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 7 dx 2 B1 State = 2t − dt 2t + 1 Use correct quotient rule or correct product rule M1 Might split first as 1 2 − 4 t+1 2. dy ( 2t + 1) .1 − t ( 2 ) A1 1 Obtain = oe Simplifies to 2 2 dt ( 2t + 1) ( 2t + 1) dy dy dx M1 dx Use =  Allow for . dx dt dt dy 1 A1 OE. Must have 1 in numerator. Obtain answer 2 ( 2t + 1)( 2t − 1)( t + 1) 1 May have, e.g., . ISW. 4t 2 + 2t − 2 ( 2t + 1)( ) 2 B1 2  y   2 y   y   1  Obtain x =   − ln  + 1  OE, e.g. x =   − ln   .  1 − 2 y   1 − 2 y   1 − 2 y   1 − 2 y  Use correct quotient rule to differentiate M1 dy (1 − 2 y ) 3 (1 − 2 y ) 3 d x  2 y  (1 − 2 y ) + 2 y  2 A1 = = OE,   − Obtain =  dx −2 (1 − 4 y )(1 − y )   2 y − 2 (1 − 2 y ) 2 d y 1 − 2 y  1 − 2 y 1 − 2 y ) 2  (  2 t 3 M1 Express in terms of t. dy (1 − 2 t +1 ) 1 = = 4 t t dx −2 (1 − 2 t + 1 )(1 − 2 t +1 ) −2 ( 2t + 1)(1 − 2 t )( t + 1 ) 1 A1 OE Obtain answer 2 ( 2t + 1)( 2t − 1)( t + 1) 5

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8 The variables x and y satisfy the differential equation 2 dy 2 y `x + 1j = kxe , dx where k is a constant. It is given that y = 0 when x = 0 and that y =- 1 when x = 1. 2 Solve the differential equation and find the exact value of y when x = 3 . 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Mark scheme: 8 Separate variables correctly B1 kx −2 y dx = e dy 2  x + 1  Condone missing integral signs or missing dx, dy, but not both. Obtain term Ae−y2 B1 OE − 12 e−2 y 2 B1 OE Obtain term B ln x + 1 ( ) k 2 x + 1 2 ln ( ) 1 −2 y k 2 B1 OE x + 1 +C ) Obtain − 2 e = 2 ln ( )( Use y = 0 when x = 0 to evaluate c or as limits in a solution containing terms of M1 the form e−y2 and ln x 2 + 1 ( ) Use y = − 12 when x = 1 to evaluate k or as limits in a solution containing terms of M1 the form e−y2 and ln x 2 + 1 ( ) 1 1 − e A1 1 −2 y 1 − e 2 1 Obtain c = – and k = OE, e.g. − e = ln x + 1 − . ( ) 2 ln2 2 2ln2 2 1 A1 Or exact equivalent. Obtain final answer y = − ln ( 2e − 1) 2 8

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9 (a) By sketching a suitable pair of graphs, show that the equation sec 2x =- ex has only one root in the interval 0 1 x 1 1 r . [2] 2 (b) Show by calculation that this root lies between 0.9 and 1. 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(c) Show that if a sequence of values given by the iterative formula nj x = 1 cos -1 `-e -x n + 1 2 converges, then it converges to the root of the equation in part (a). 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(d) Use the iterative formula given in part (c) to calculate x correct to 3 decimal places. Give the result of each iteration to 5 decimal places. 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Mark scheme: 9(a) Sketch a relevant graph, e.g. y = sec 2x B1 2 May be sketching cos2x and –e–x 2 π 2 π 2 2 For y = sec 2x correct shape, asymptote in correct position. Cuts at (0, 1) For −xe correct shape, cuts at (0, 1) 4 For y = cos2 x correct shape, correct max/min, cuts axis at π4 Complete scale not needed, but should have enough x to imply the key points for both graphs. For − e− correct shape, cuts at (0, -1) Sketch a second relevant graph, e.g. y = –ex and justify the given statement B1 Needs to mark intersection with a dot, a cross, or say roots at points of intersection, OE. 2 9(b) Calculate the values of a relevant expression or pair of expressions at M1 −4.401 −2.460 − 1.94  0 E.g. or x = 0.9 and x = 1 −2.403 −2.718 0.315  0 0.179  0 (or for the reciprocal graphs). −0.048  0 Complete the argument correctly with correct calculated values A1 2 9(c) 1 −1 − x B1 and rearrange to obtain sec 2x = –ex −e State the equation x = 2 cos ( ) −e − x −e − nx Or rearrange sec 2x = –ex as x = 12 cos −1 ( ) and state nx +1 = 12 cos −1 ( ) 1 9(d) Use the iterative process correctly at least once M1 Obtain final answer 0.978 A1 Show sufficient iterations to at least 5 d.p. to justify 0.978 to 3 d.p. or show there A1 E.g. is a sign change in the interval (0.9775, 0.9785) 1, 0.97376, 0.97903,0.97796, 0.97813 0.95, 0.98395, 0.97697, 0.97838, 0.97809 0.9, 0.99475, 0.97480, 0.97882, 0.97800, 0.97817 3

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x 3 + 2x - 11 10 Let f ( x) = . ( 3 + x) `2 + x 2j (a) Express f ( x) in partial fractions. 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(b) Hence obtain the expansion of f ( x) in ascending powers of x, up to and including the term in x2. 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Mark scheme: 10(a) B Cx + D B1 State or imply the form A + + 3 + x 2 + x 2 Use a correct method for finding a constant M1 −3 x 2 − 17 Might be working from 1 + . (3 + x )(2 + x 2 ) Obtain one of A = 1, B = –4, C = 1 and D = –3 A1 SC: If B0 scored due to missing term(s), then a maximum of M1A1 is available for a correct method leading to a correct value. Obtain a second value A1 SC if obtaining A = 1 and then scoring B0, they can score maximum M1A1 + A1 for a correct value for one other constant. Obtain a third value A1 Obtain all four correct values A1 6 10(b) Use a correct method to find the first two terms in the expansion of (3 + x) –1, M1 Symbolic binomial coefficients not sufficient for the −1 2 −1 M1.  x  –1  x   1 +  , (2 + x2) or  1 +   3   2  Obtain correct unsimplified expansions up to the term in x2 of each partial A1 FT FT B, C and D. fraction A1 FT B  x x 2  Cx + D  x 2  E.g.  1 − +  and  1 −  . 3  3 9  2  2  −1 −1 2 M1 2  x   x  Multiply (Cx + D) by the expansion of  1 +  up to the term in x2 where Expansion of  1 +  .  2   2  CD ≠ 0 Must be of the form 1 + Qx². Multiplication must include 3 relevant terms. 11 17 65 2 A1 Or exact equivalent. Obtain final answer − + x + x Do not ISW attempts to multiply expression by, e.g., 6 18 108 108. 5

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11 With respect to the origin O, the points A, B, C and D have position vectors given by 1 0 1 3 OA = f 5 p, OB = f 4 p, OC = f- 3p and OD = f- 5p. 3 1 1 4 The line m passes through the points A and B. (a) Find a vector equation for m. 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(b) Find the position vector of the point of intersection of m and the line passing through the points C and D. 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(c) Find the position vector of the foot of the perpendicular from C to m. 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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: 11(a) Carry out a correct method for finding a vector equation for m M1 1 1 A1 0 1     Obtain r = 5 +  1 Or r = 4 +  1         3 2 1 2 OE. Must have r = … 2 11(b) Express general point of the second line in component form, B1 Or ( 3 + 2, − 5 − 2, 4 + 3) . i.e. (1 + 2µ, –3 – 2µ, 1 + 3µ) Equate at least two pairs of corresponding components and solve for λ or for µ M1 Obtain λ = –4 or µ = –2 A1  −1    Note if m has direction vector −1 , then λ = 4.      −2   − 2    If line through CD has direction vector 2 , then      −3  µ = 2.  −3  A1 Accept coordinates.   Obtain position vector of point of intersection is 1      −5  4 11(c) 0 1 B1  −1  1       Find CP for a general point P on m, e.g. 8 +  1 Or 7 +  1 .           2 2  0  2  Calculate scalar product of CP and a direction vector for m and equate the result M1 to zero Obtain λ = –2 A1 OE  −1  A1 OE   Obtain answer 3      −1  4

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