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

e x + 2e -x 1 Solve the equation x = 4 . Give your answer correct to 3 decimal places. [5] e - 3 .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................. ............................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Obtain a 3-term quadratic in ex *M1 Obtain e.g. 3e2x – 12ex – 2 = 0 or 3-term equivalent A1 2 E.g. 3m − 12m − 2 = 0. ‘= 0’ could be implied by subsequent working. Solve a 3-term quadratic to obtain a value for x or ex DM1 Need to get as far as a value for x or ex. 6 + 42 A1 OE Obtain root or 4.16… Ignore second root if seen. 3 Obtain answer 1.426 only A1 CAO, must be 3 d.p. 0/5 for answer with no working. Second root must be rejected if seen. 5

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- 232 (a) Expand ( 6 - x)( 1 - 2x) in ascending powers of x, up to and including the term in x2, simplifying the coefficients. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) State the set of values of x for which the expansion is valid. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ .......................... ............................................................................................................................

Mark scheme: 2(a) − 3 B1 2 Find the first two terms of the expansion of (1 − 2x ) B1 15 3 3 3   3   2 x 2 − − 1 − 1  −  −  −  2  2 2 2  2  2 Ignore extra terms. Obtain correct third term −( 2 x ) or ( 2 x ) 2! 2! 2 M1 2 2 Multiply their 3 term expansion a + bx + cx by (6 – x) obtaining all necessary 6 + 18 x + 45 x −−x 3x .... terms Ignore extra terms. 6 + 17x + 42x2 A1 Ignore extra terms. Allow with the terms in any order. 4 2(b) 1 1 1 B1 OE |x| < or −  x  or (-0.5, 0.5) or ]-0.5, 0.5[ B0 for an ambiguous statement. 2 2 2 Must be strict inequality. 1

More questions on Series

4 Solve the equation 3 cot x - 4 cot 2 x = 3 for 0° G x G 180° . [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4 Use correct trigonometric formulae to form an equation in tan x only, or an *M1 Condone one slip in manipulating the original equation in terms of sin x and cos x only equation provided correct trig formulae used. 3  1 − tan 2 x  e.g. − 4   = 3 tan x  2tan x  3 4 − = 3 tan x 2tan2x 1− tan x cos x cos 2 x − sin 2 x 3 − 4 = 3 sin x 2sin x cos x  cot 2 x − 1  3cot x − 4   = 3  2cot x  Obtain a correct horizontal equation, in tan x or in cos x and sin x, in any form A1 E.g. 3 – (2 – 2 tan2 x) = 3 tan x or 2 sin² x – 3 sin x cos x + cos² x = 0. Reduce equation to a 3-term quadratic A1 E.g. 2 tan2 x – 3 tan x + 1 = 0, 2 sin² x – 3 sin x cos x + cos² x = 0. Allow if they square both sides and obtain a quartic in sin x or cos x Allow if they leave the equation as a cubic with a factor of tan x and go on to give the correct solutions to the quadratic. Solve a 3-term quadratic to obtain a value for x DM1 If the initial equation is correct, then M1 is implied The quadratic could be arrived at from a correct cubic by a correct solution. Not available if they have an incorrect cubic with an incorrect quadratic factor. Obtain one answer, e.g. (x =) 45° A1 Obtain a second answer, e.g. AWRT (x =) 26.6° and no other in the given A1 Ignore answers outside the given interval. interval, e.g. x = 0 6

More questions on Trigonometry

5 The square roots of -1 - 4 5i can be expressed in the Cartesian form x + yi , where x and y are real and exact. By first forming a quartic equation in x or y, find the square roots of -1 - 4 5i in exact Cartesian form. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 5 Square x + iy and equate real and imaginary parts to –1 and −4 5 respectively *M1 Obtain equations x2 – y2 = –1 and 2xy = −4 5 from their expansion A1 Working from two equations in two unknowns, eliminate one variable and find DM1 Do not condone incorrect algebra, e.g. x = −2 5 y. an equation in the other Obtain x4 + x2 – 20 = 0 or y4 – y2 – 20 = 0 A1 Accept 3-term equivalents. Condone missing “= 0” if implied by subsequent working. A1 Must state the square roots, not x and y separately, Obtain answers  2 − 5i and no others ( ) and not coordinates. Do not allow 4 in place of 2. A0 if there are additional incorrect solutions. No working seen scores 0/5. 5

More questions on Complex numbers

6 (a) By sketching a suitable pair of graphs, show that the equation x - 2 = 2 sin 1 x 2 has only one root in the interval 0 1 x 1 r. [2] (b) Show by calculation that this root lies between 1 and 1.5. 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(c) Use the iterative formula x = 2 - 2 sin 1 x with an initial value of 1.03 to calculate the root n + 1 2 n correct to 2 decimal places. Give the result of each iteration to 4 decimal places. 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Mark scheme: 6(a) Sketch a relevant graph for 0  x  π B1 2 For y = |x – 2| graph should be symmetrical and have correct intercepts on the axes π For y = 2 sin 12 x , graph should pass through the origin, have correct curvature Ignore anything outside 0  x  π. Ignore what happens in y < 0. and max y = 2 when x = π Sketch second relevant graph and confirm root. B1 Needs to mark intersection with a dot, a cross, or say roots at points of intersection, OE. The vertex of the modulus graph in roughly correct position relative to π and/or If the intersection is highlighted in some way, then 1 2 π they do not need to make a comment. If no mark on the graph, check to see if they have written something below the graph. SC A sketch y = x − 2 and y = 2sin 12 x (above and below the x-axis) scores B1. A clear indication of the root scores second B1. 2 6(b) Calculate the values of a relevant expression or pair of expressions at x = 1 and M1 Or comparing x – 2 and 2sin 12 x. x = 1.5 1 Using 1: x – 2 = 1, 2sin 2 x = 0.958 1  1 > 0.958… e.g. f(x) = |x – 2| – 2sin 2 x Using 1.5: x – 2 = 0.5 2sin 12 x = 1.36  f(1) = 0.0411… > 0 f(1.5) = –0.863… < 0  0.5 < 1.36 2 2 1 Need all values but condone one error. e.g. f ( x ) = ( x − 2 ) − 4sin ( 2 x )  f (1) = 0.0806..., f (1.5 ) = −1.60... If the solution involves 4 values, the pairing must be clear. Embedded values are not sufficient, e.g. f1(1) = … and f2(1) =… etc. M0 if working in degrees (gives 0.98… and 0.47… if using f(x) = 0). Allow if working on a smaller interval. Complete the argument correctly with correct calculated values A1 Values correct. They must have a conclusion in words or symbols, but they do not need to say that the function is continuous. A correct statement with correct inequalities is sufficient. 2 6(c) Use the iterative process correctly at least once starting at 1.03 (get as far as M1 M0 if working in degrees. 1.0281) Obtain final answer 1.02 A1 No working seen at all scores 0/3. Show sufficient iterations to 4 d.p. to justify 1.02 to 2 d.p. A1 1.03, 1.0149, 1.0281, 1.0166, 1.0266, 1.0179, 1.0255, 1.0189, 1.0246 or show there is a sign change in the interval (1.015, 1.025) (or a smaller interval Incorrect starting point is M0. containing the root) Once the convergence is established ISW. 3

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7 (a) Express 7 sin i + 24 cos i in the form R cos ( i - a) , where R 2 0 and 0 1 a 1 1 r . Give the value 2 of a correct to 4 decimal places. 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(b) Hence solve the equation 7 sin 1 x + 24 cos 1 x = 24. 5 for 0 1 x 1 r. 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Mark scheme: 7(a) State R = 25 B1 From correct work. BOD if correct value follows use of their α, but not if a decimal approximation to R is seen first. −1 7  M1 If cos = 24 and sin = 7 seen, then M0 A0. Use correct trig formula to find α, e.g. = tan    24  Obtain α = 0.2838 A1 CAO 3 7(b) −1 24.5  B1 FT Can be implied by 0.2(0033…), or by 1 cos   x = 0.08...or 0.48... or a correct value of x 3  25  following M1. FT their R from (a). −1 24.5  Allow B1 if cos   is not evaluated, provided  R  R  24.5. Carry out a complete correct method to find a value of x M1 Need some method shown, but might not show interim values if working on a calculator. Incorrect 1 answers and insufficient evidence scores M0. Sight of a correct equation using (a) is needed, e.g. cos ( 3 x − 0.2838 ) = 0.98 OE Obtain answer (x = ) 1.45 A1 AWRT Need 3 sf or better. Obtain (x = ) 0.250 or (x = ) 0.251, and no other in the interval A1 AWRT Ignore answers outside the given interval. 4

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8 The variables x and i satisfy the differential equation d x sin 2i = ( 4x + 3) cos 2i , d i and x = 0 when i = 1 r . 12 Solve the differential equation and obtain an expression for x in terms of i. 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Mark scheme: 8 Separate variables correctly B1  1  cos2  dx =  d  4 x + 3  sin2 Can be implied by obtaining both correct integrals. Obtain term 14 ln ( 4 x + 3 ) B1 OE 3 or 1 ) . 4 ln ( x + 4 Obtain term of the form A ln ( sin2) M1 Or A ln ( k sin2OE,) e.g. P ln a sin+ Q ln b cos from using the tan2 formula. Or expanding cos2as cos 2 − sin 2 . Obtain term 12 ln ( sin 2) A1 OE Correct in any form, e.g. 12 ln a sin+ 12 ln b cos. 1 Use x = 0 when θ = 121 π to evaluate a constant or as limits in a solution M1 E.g. c = 14 ln3 − 12 lnsin ( 6 π ) 1 1 containing terms of the form ln ( sin 2) and ln ( 4 x + 3 ) . ln 1 4 ln ( 4 x + 3 ) − 4 ln3 = 12 ln ( sin 2) − 12 2 c = … seen or implied Note that the constant may be expressed as a logarithm 1 1 Obtain correct answer in any form with the trigonometry evaluated A1 E.g. 14 ln ( 4 x + 3 ) = 2 ln ( sin2) + 4 ln12 1 1 4 ln ( 4 x + 3 ) = 2 ln ( sin2) + 0.621 1 ln 4 x + 3 = 1 ln ( 2sin2) 4 3 2 12sin 2 2− 3 A1 OE Obtain final answer x = 2.48.... 2 e sin 2− 3 4 Allow . 4 Allow 2.48. 1 2ln ( sin2+) 2.48... e − 3 . Allow x = 4 ( ) 7

More questions on Differential equations

9 With respect to the origin O, the points A, B and C have position vectors given by 1 - 2 2 OA = f- 4p, OB = f 1p and OC = f 3 p. 2 3 5 (a) Find a vector equation for the line through A and B. 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(b) Using a scalar product, find the exact value of cos BAC. 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(c) Hence find the exact area of triangle ABC. 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Mark scheme: 9(a) Carry out a correct method for finding a vector equation for the line through A M1 A complete method. and B Can be working from any point on AB.  1   −3  A1 OE     Must have r = … not l = …. Obtain r = −4 +  5         Accept rAB = ...  2   1  Accept R = …. 2 9(b) 1 B1 OE  Obtain a direction vector for AC = 7 Or CA.   3 M1 −+3 1 5  7 + 1 =3 35 Carry out correct process for evaluating the scalar product of AB and AC or Allow for correct answer and no working seen. BA and CA Using the correct process for the moduli, divide their scalar product by the M1 Independent M0M1 is possible. product of the moduli ISW finding the angle. 35 A1 From correct working. The answer needs to come Obtain (cos BAC =) or exact simplified equivalent from using a scalar product. 59 35 35 2065 Accept or or 2065 35 59 59 = cos −1 5935 without a statement of cos BAC scores A0. ISW finding the angle. 4 9(c) 1 M1 For “hence”, must be using the angle at A. Use area = AB AC sin BAC with their AB AC Need not substitute for the trigonometry. 2 Accept any equivalent form for their sin BAC. 2 M1 NB: These two M marks are independent. Use sin x = 1 − cos x or an equivalent exact method with their cos x (< 1) to Might not quote the formula. Could draw a triangle obtain an exact value for sin x and use Pythagoras, which is equivalent. 24 sin x = 2 59 1 35 1 840 35 59 1 − = 35 59 1 35 is allowed for the first M1, but not 2 35  59 2 35 59 sin ( cos− 59 ) for the second. Obtain answer 210 from correct working A1 Accept simplified equivalent exact forms, e.g. 1 2 840. Watch out for fortuitous answers from negative value of the cosine. Do not accept an answer coming from  −1 35  sin  cos  with no evidence of the method of  59  evaluation. The answer needs to come from using angle BAC. 3

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10 (a) Find the quotient and remainder when x2 is divided by 1 + 4x 2 . 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Mark scheme: 10(a) 1 B1 Could be found by using long division or by writing Obtain quotient 2 2 4 x = q 1 + 4 x + r and comparing coefficients: ( ) 1 = 4q, 0 = q + r. 1 B1 Allow B1B1 if implied by correct division and no Obtain remainder − further working, but do not ISW. 4 Allow for a correct statement of the identity, but not for an incorrect statement of the remainder. 2 10(b) 2 −1 2 B *M1 OE 2 Commence integration by parts and reach Ax tan 2 x   x C + Dx dx 1 2 −1 x 2 A1 OE dx 2 Obtain 2 x tan 2 x −  1 + 4 x x 2 −1 DM1 2 Integrate  1 + 4 x dx to obtain an expression of the form p tan 2 x + qx Complete integration and obtain 12 x 2 tan −1 2 x + 18 tan −1 2 x − 14 x A1 FT OE FT their constant quotient and remainder from (a), 2 k − 1 k kx and 8 tan 2 x − 4 x from their 2 . 1 + 4 x Substitute limits correctly in an expression of the form DM1 Need some evidence that they have considered the Fx 2 tan −1 2 x + G tan −1 2 x + Hx lower limit, e.g. sight of 0 in the working. No need to evaluate trigonometry. If in stages, then the 0 needs to be seen for each part. Obtain answer 161 π − 81 or exact one- or two-term equivalent with trigonometry A1 evaluated 6

More questions on Algebra

11 y M O 1 r x 2 The diagram shows the graph of y = 5 sin 2x cos 2 x for 0 G x G 1 r and its maximum point M. 2 (a) Find the exact x-coordinate of M. 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(b) By using the substitution u = cos x , find the area of the region bounded by the curve, the x-axis between x = 0 and x = 1 r , and the line x = 1 r . 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Mark scheme: 11(a) 2 B1 OE Differentiate cos x to obtain −2sin x cos x Could be stated as − sin2 .x Use correct product rule *M1 With a ‘+’ in the middle. 2 A1 OE Obtain derivative −10sin2 x sin x cos x + 10cos2 x cos x Equate derivative to zero and obtain an equation in one trig function DM1 Trigonometry formulas used need to be correct. Obtain 3 tan2 x = 1, 4 sin2 x = 1 or 4 cos2 x = 3 or cos3x = 0 A1 OE Obtain x = 16 π only A1 Alternative Method for the Question 11(a) Use double angle formula to obtain y = 10sin x cos 3 x B1 Use correct product rule *M1 4 2 2 A1 OE Obtain derivative 10cos x − 30sin x cos x Equate derivative to zero and obtain an equation in one trig function DM1 Trigonometry formulas used need to be correct. Obtain 3 tan2 x = 1, 4 sin2 x = 1 or 4 cos2 x = 3 or cos3x = 0 A1 OE Obtain x = 16 π only A1 11(a) Alternative Method 2 for the Question 11(a) Use double angle formula to obtain y = 52 sin2 x ( cos2 x + 1) B1 Use double angle formula to obtain y = 54 sin4 x + 52 sin2 x *M1 Obtain derivative 5cos4 x + 5cos2 x A1 Equate derivative to zero and obtain an equation in cos2x DM1 2cos 2 2 x + cos2 x −=1 0 Obtain cos2x = 12 only A1 Obtain x = 16 π only A1 6 11(b) d u B1 SOI = − sin x d x 3 *M1 OE Reach an integral of the form Au du  The question requires use of the substitution method. 3 A1 OE Obtain 10 u du − Ignore limits, but check order of limits if no minus sign. Substitute correct limits correctly in an expression of the form Cu 4 or C cos 4 x DM1 2 u = 1 and u = 2 1 x = 0 and x = π 4 1 1 10 2 u 3 du or 10 u 3 du 1 −1  2 Allow the correct answer from the correct integration and relevant limits to imply M1. 15 A1 WWW Obtain answer or 1.875 ISW 8 5

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