Mathematics 9709/31 May/June 2025: question paper and mark scheme

Mark scheme

1(a) y B1 Symmetrical. In correct position. Condone if no complete scale shown, but must see 3 and 32 marked. 3 Needs to exist for negative x. Must be intending straight lines. Ignore y = 3x − 1 if seen. x O 3 2 1 1(b) Obtain critical value 54 from 3x − 1 = 3 − 2x B1 State final answer x < 54 B1 Alternative Method for Question 1(b) Obtain critical value 54 from ( 3x − 1) = ( 3 − 2 x ) 2 2 B1 Ignore x = −2 if seen. State final answer x < 54 B1 2

Mark scheme

2 Use logarithm of a root or a power M1 E.g. 2ln p = ln p 2 , 12 ln ( q + 1) = ln q + 1 or 3 = ln e3 . Obtain p 2 ( p − 1) = e3 q + 1 A1 OE without logs. p 2 ( p − 1) 2 A1 OE, ISW Obtain q = −1 e3 3

Mark scheme

3 x + i ( y + 5) ( x − 5) − iy *M1 Multiply numerator (and denominator) by the × conjugate of the denominator. ( x − 5) + iy ( x − 5) − iy Equate imaginary part of numerator to zero DM1 Numerator = x ( x − 5) + y ( y + 5) + i ( ( y + 5)( x − 5) − xy ). Obtain x − y = 5 A1 OE Correct use of modulus and solve for x and y M1 x2 + ( x − 5) = 17 x2 − 5x + 4 = 0 2 Or x − y = 5 and xy = −4. z = 4−i A1 z = 1 − 4i A1 Alternative Method for the first two marks of Question 3 x y+5 M1 Quotient real = x+5 y Rearrange to linear form and simplify M1 6

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4 At ( e, 3) , tan t = 1 or t = π4 B1 SOI dx B1 = sec 2 t × e tan t dt dy B1 = 6 tan t × sec 2 t dt dy 6 tan t × sec 2 t *M1 Correct use of their derivatives. = dx sec2 t × e tan t ( = 6 tan t × e ) − tan t y − 3 = 6e ( x − e ) DM1 Substitute for t, and use correct method for the equation of the line. y = 6e x − 3 A1 Or exact equivalent. Alternative Method for Question 4 y B1 SOI or y = 3( ln x ) y 2 Correct cartesian form, e.g. x = e 3 or ln x = 3 Differentiate function of a function *M1 Complete method. y dy 1 A1 dx Obtain =k e y e 3 or = k × ln x dy dx x y dy 1 A1 dx 1 Obtain = 3 e 3 or = 6 × ln x dy 6 y dx x y − 3 = 6e ( x − e ) DM1 Use correct method for the equation of the line. 4 y = 6e x − 3 A1 Or exact equivalent. 6

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5 Use f ( −2a ) = −22a 3 M1 Or use long division and equate a constant remainder to −22a3. Obtain −24a3 + 4 pa3 − 14a3 + qa 3 = −22a 3 A1 Must evaluate the terms. OE, e.g. 4 p + q = 16. a M1 Or use long division and equate a constant Use f = −a3 3 remainder to −a3. Obtain 19 a 3 + 91 pa 3 + 73 a 3 + qa 3 = − a 3 A1 Must evaluate the terms. OE, e.g. p + 9q = −31 Obtain p = 5, q = −4 A1 5

Mark scheme

6(a) Obtain ω z1 = 6e 4 i3π B1 Obtain ω z2 = 3e 3 i2π B1 SC B1 for both moduli correct or both arguments correct. 2 6(b) A and B plotted correctly B1 Im(z) C D 6 A 3 3 B 1.5 O Re(z) Follow their answers to part (a) C and D plotted with angles relatively correct and the same stretch implied. B1FT C and D correct, or FT their A and B. 2 6(c) Rotation π2 radians (anticlockwise), B1 Accept 90°. Not required to state the centre of the rotation, but or rotation 32π radians clockwise B0 if an incorrect statement seen. Enlargement (scale) factor 2 B1 Not required to state the centre of the enlargement, but B0 if an incorrect statement seen. Allow ‘expansion’ or ‘stretch’. 2

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7(a) π B1 Or exact 2 term equivalent. Expand 5sin x + − 4cos x to obtain 2 3 sin x − 2 cos x 5 3 6 State R = 21 B1ft Follow their 5 3 and 3 . 2 2 Use correct trig formulae to obtain tan α M1 OE, WWW. 3 3 E.g. tan α = = , 5 3 5 3 3 sin α = = , 2 21 2 7 5 3 5 cos α = = . 2 21 2 7 Obtain α = 0.333 A1 4 7(b) 7 B1FT SOI sin −1 0.615... can be implied by one correct answer. 21 Follow their 21. Use a correct method to obtain a value of θ in the interval M1 Obtain one correct answer, e.g. 0.47 A1 Obtain second correct answer, e.g. 1.43, and no others in the interval A1 4

Mark scheme

8(a) Use a correct method to form an equation for l1 M1 Accept column vectors. Obtain r = ( 2i + 4k ) + λ ( 3i + j + 2k ) A1 OE, e.g. r = ( 5i + j + 6k ) + λ ( 3i + j + 2k ) . Must have r = …, or in x, y, z or R = …. 2 8(b) Express general point of a line in component form B1ft 2+μ 2 + 3λ 5 + 3λ E.g. 1 + 2 μ or λ or 1 + λ . 5 + 3μ 4 + 2λ 6 + 2λ Equate two pairs of components of l2 and their l1 , and solve for λ or μ M1 2 + μ 2 + 3λ 1 + 2μ = λ 5 + 3μ 4 + 2λ Obtain e.g. λ = − 15 , μ = − 53 A1 Or λ = − 17 , μ = − 73 or λ = −1, μ = −1. Show that this does not fit the third component and hence the lines do not intersect. A1 16 5 ≠ 185 or 17 ≠ − 17 or 1 ≠ −1. 4 8(c) Carry out the correct process for evaluating the scalar product of the direction vector M1 E.g. ( i + 2 j + 3k ) · ( 3i + j + 2k ) = 3 + 2 + 6. of l1 and l2 Using the correct process for the moduli, divide their scalar product by the product M1 3+ 2+6 of the moduli of their vectors and evaluate the inverse cosine of the result θ = cos −1 14 × 14 Obtain AWRT 38.2° or 0.667 radians A1 3

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9(a) M1* Commence integration and reach px2 ln x + q x dx Allow M1 with q x2 x dx. Obtain 3x2 ln x − 3 x dx A1 x 2 must be simplified. x Complete integration and obtain 3 x 2 ln x − 32 x 2 A1 Use limits correctly in an expression of the form ax2 ln x − bx2 and equate to 4, having DM1 3a 2 ln a − 32 a 2 − 0 + 32 = 4 integrated twice 1 5 Obtain a = exp 2 + 3 correctly A1 AG 6 a 5 9(b) Calculate the values of a relevant expression or pair of expressions at M1 Not from using calculator to evaluate the original a = 2 and a = 2.1 integral. Justify the given statement with correct calculated values A1 E.g., using f ( x) = 6 x 2 ln x − 3x 2 f (2) = 4.636 < 5 and f (2.1) = 6.402 > 5, or using the exponential, 2 < 2.0306 and 2.1 > 1.9917 or 0.0306 > 0 and −0.1083 < 0. . 2 9(c) 1 5 M1 Use the iterative process an +1 = exp 2 + 3 correctly at least once. 6 an Obtain final answer 2.02 A1 Show sufficient iterations to at least 4 d.p. to justify 2.02 to 2 d.p. A1 E.g. 2 → 2.0306 → 2.0180 → 2.0231.... or show that there is a sign change in ( 2.015, 2.025) 2.05 → 2.0103 → 2.0263 → 2.0197 → 2.0224.. 2.1 →1.9917 → 2.0342 → 2.0166 → 2.0237.. 3

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10(a) Divide by ( x 2 − 3) to obtain x + k ( k ≠ 0 ) M1 x +5 x2 − 3 x3 +5x 2 −2 x −15 x3 −3x +5x 2 +x +5x 2 −15 x Obtain quotient x + 5 A1 Obtain remainder x A1 ISW x Allow 2 . x −3 3 10(b) B1 Separate variables correctly and obtain 6 y dy = 3 y 2 OE from 6 y dy = x 3 + 5 x 2 − 2 x − 15 dx. x2 − 3 Obtain 12 x 2 + 5 x B1ft Follow their linear quotient. ( Obtain 12 ln x 2 − 3 ) B1ft From the x term in their remainder ax + b. Use y = 2, x = 2 to evaluate the constant of integration in an integral containing M1 12 = 2 + 10 + 12 ln1 + C C = 0 ( k ln x 2 − 3 ) ( Obtain y 2 = 16 x 2 + 53 x + 16 ln x 2 − 3 ) A1 OE 5

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11(a) d 2cos 2 x cos 2 x B1 SOI sin 2 x = Accept k dx 2 sin 2 x sin 2 x Use correct product rule M1* Obtain correct derivative in any form A1 dy 2cos 2 x E.g. = − sin x sin 2 x + cos x × . dx 2 sin 2 x Equate the derivative to zero and obtain a horizontal equation DM1 E.g. − sin a sin 2a + cos a cos2a = 0 Use correct trig formulae to obtain an equation in one trig function DM1 E.g. cos3a = 0 or sin 2 a = 1 . 4 Obtain a = 16 π A1 Exact answer in radians only. 6 11(b) Use [ π] y 2 dx M1* Use double angle formula to obtain a form that can be integrated directly, DM1 Or sin 4x + sin 2x dx, 1 4 1 2 2 3 e.g. cos x sin 2 x dx = 2cos x sin x dx or 2 ( u − u ) du by substituting u = sin x. 3 Obtain − ( π ) × 12 cos 4 x A1 Or − [ π] ( 1 cos 4 x + 1 cos 2 x ) , OE. 16 4 Use correct limits correctly DM1 ( ) − [ π ] × 12 cos 4 x 2 = − [ π ] 12 cos ( 12 π ) − 12 , π 0 ( or − [ π] 16 1 cos 2π + 1 cos π − 1 − 1 OE. 4 16 4 ) Obtain final answer 12 π A1 If the factor of π is missing throughout, allow the first 4 marks and A0 here. 5