Mathematics 9709/32 May/June 2025: question paper and mark scheme

Mark scheme

1 Obtain a 3-term quadratic in ex *M1 Obtain e.g. 3e2x – 12ex – 2 = 0 or 3-term equivalent A1 E.g. 3m2 − 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

Mark scheme

2(a) 3 B1 Find the first two terms of the expansion of (1 − 2x ) 2 − 3 3 3 3 B1 15 2 x − − − 1 − − − 1 2 2 2 Obtain correct third term ( −2 x )2 or 2 2 ( 2 x )2 Ignore extra terms. 2! 2! Multiply their 3 term expansion a + bx + cx2 by (6 – x) obtaining all necessary M1 6 + 18x + 45x2 − x − 3x2 .... 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

Mark scheme

3 Show a circle centre (0, 3) B1 Allow for a circle in the correct place even if their scales require an ellipse. ‘3’ marked on the axis, centre implied by vertical intercepts or dashes to indicate a scale. Show a circle with radius 2 B1FT FT centre not at the origin. Consistent with their scale if they have unequal scales on the axes. Show the point representing (1, 2) B1 Position can be implied provided there is some indication of scale, e.g. numbers or marks on the axes. Show correct half-lines from (1, 2), one at an angle of 14 π and the other at an B1FT FT from their (1, 2) not at the origin. Correct lines can imply previous B1. angle of 34 π Half-lines must be intended to be ‘symmetrical’. Angles consistent with their scale: correct lines need to pass very close to (0, 3) and (2, 3). Shade the correct region B1 Dependent on all previous marks. 5

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 2 tan x 3 4 − 2tan x = 3 tan x 2 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

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. ( ) Obtain answers ± 2 − 5i and no others A1 Must state the square roots, not x and y separately, 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

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 3π 2π π π 2π 3π 4 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 2 Sketch second relevant graph and confirm root. B1 Needs to mark intersection with a dot, a cross, or say 4 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 π they do not need to make a comment. 2 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 1 x. 2 x = 1.5 Using 1: x – 2 = 1, 2sin 12 x = 0.958 e.g. f(x) = |x – 2| – 2sin 12 x 1 > 0.958… f(1) = 0.0411… > 0 f(1.5) = –0.863… < 0 Using 1.5: x – 2 = 0.5 2sin 12 x = 1.36 0.5 < 1.36 e.g. f ( x ) = ( x − 2 ) − 4sin 2 ( 12 x ) f (1) = 0.0806..., f (1.5) = −1.60... Need all values but condone one error. 2 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

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. 7 M1 If cos α = 24 and sin α = 7 seen, then M0 A0. Use correct trig formula to find α, e.g. α = tan −1 24 Obtain α = 0.2838 A1 CAO 3 7(b) 24.5 B1 FT Can be implied by ± 0.2(0033…), or by cos −1 1 x = 0.08... or 0.48... or a correct value of x 25 3 following M1. FT their R from (a). 24.5 Allow B1 if cos −1 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 Sight of a correct equation using (a) is needed, e.g. cos ( 13 x − 0.2838) = 0.98 OE answers and insufficient evidence scores M0. 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

Mark scheme

8 Separate variables correctly B1 1 cos 2θ dx = dθ 4x + 3 sin 2θ Can be implied by obtaining both correct integrals. Obtain term 14 ln ( 4 x + 3) B1 OE or 14 ln ( x + 43 ). Obtain term of the form A ln ( sin 2θ ) M1 Or A ln ( k sin 2θ ) OE, e.g. P ln a sin θ + Q ln b cosθ from using the tan 2θ formula. Or expanding cos 2θ as cos 2 θ − sin 2 θ . Obtain term 12 ln ( sin 2θ ) A1 OE Correct in any form, e.g. 12 ln a sin θ + 12 ln b cosθ . Use x = 0 when θ = 121 π to evaluate a constant or as limits in a solution M1 E.g. c = 1 ln3 − 1 lnsin ( 1 π ) 4 2 6 containing terms of the form ln ( sin 2θ ) and ln ( 4 x + 3) 1 ln ( 4 x + 3) − 14 ln 3 = 12 ln ( sin 2θ ) − 12 ln 12 . 4 c = … seen or implied Note that the constant may be expressed as a logarithm Obtain correct answer in any form with the trigonometry evaluated A1 E.g. 14 ln ( 4 x + 3) = 12 ln ( sin 2θ ) + 14 ln12 1 4 ln( 4 x + 3) = 12 ln ( sin 2θ ) + 0.621 2 ( ) 1 ln 4 x +3 = 1 ln 2sin 2θ 4 3 12sin 2 2θ − 3 A1 OE Obtain final answer x = 4 e2.48.... sin 2 2θ − 3 Allow . 4 Allow 2.48. ( Allow x = 14 e2ln(sin 2θ )+ 2.48... − 3 . ) 7

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 Obtain r = −4 + λ 5 Must have r = … not l = …. 2 1 Accept rAB = ... Accept R = …. 2 9(b) 1 B1 OE Obtain a direction vector for AC = 7 Or CA. 3 Carry out correct process for evaluating the scalar product of AB and AC or M1 −3 × 1 + 5 × 7 + 1 × 3 = 35 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 θ = cos −1 35 59 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. Use sin x = 1 − cos 2 x or an equivalent exact method with their cos x (< 1) to M1 NB: These two M marks are independent. 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 = 1 352 1 840 59 2 35 59 1 − = 35 × 59 2 35 59 35 59 ( sin cos−1 35 59 ) is allowed for the first M1, but not 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 sin cos −1 35 with no evidence of the method of 59 evaluation. The answer needs to come from using angle BAC. 3

Mark scheme

10(a) 1 B1 Could be found by using long division or by writing Obtain quotient 4 ( ) x 2 = q 1 + 4 x 2 + 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) B *M1 OE Commence integration by parts and reach Ax 2 tan −1 2 x ± x 2 C + Dx2 dx x2 A1 OE Obtain 12 x 2 tan −1 2 x − 1 + 4 x2 dx x2 DM1 Integrate 1 + 4 x2 dx to obtain an expression of the form p tan −1 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 and k8 tan −1 2 x − k4 x from their kx 2 . 1 + 4x Substitute limits correctly in an expression of the form DM1 Need some evidence that they have considered the Fx2 tan−1 2x + G tan−1 2x + 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

Mark scheme

11(a) Differentiate cos2 x to obtain −2sin x cos x B1 OE Could be stated as − sin 2x. Use correct product rule *M1 With a ‘+’ in the middle. Obtain derivative −10sin 2x sin x cos x + 10cos2x cos2 x A1 OE 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 cos3 x B1 Use correct product rule *M1 Obtain derivative 10cos4 x − 30sin 2 x cos2 x A1 OE 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 sin 2 x ( cos 2 x + 1) B1 Use double angle formula to obtain y = 54 sin 4 x + 52 sin 2 x *M1 Obtain derivative 5cos 4 x + 5cos 2 x A1 Equate derivative to zero and obtain an equation in cos2x DM1 2cos2 2x + cos2x −1 = 0 Obtain cos 2x = 12 only A1 Obtain x = 16 π only A1 6 11(b) du B1 SOI = − sin x dx *M1 OE Reach an integral of the form Au du 3 The question requires use of the substitution method. A1 OE Obtain −10 u3du 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 cos4 x DM1 2 u = 1 and u = 2 1 x = 0 and x = π 4 1 1 −10 u 3du or 10 1 u 3du 2 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