Cambridge A Level Mathematics 9709 — 2025 May/June Paper 3 · Variant 1

1 (a) Sketch the graph of y = 2 x - 3 . [1] (b) Solve the inequality 3x - 1 1 2 x - 3 . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 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 = 3 x − 1 if seen. x O 3 2 1 1(b) Obtain critical value 54 from 3x −=1 3 − 2 x B1 State final answer x  54 B1 Alternative Method for Question 1(b) 4 2 2 B1 Ignore x = − 2 if seen. Obtain critical value 5 from ( 3 x − 1) = ( 3 − 2 x ) State final answer x  54 B1 2

More questions on Quadratics

2 It is given that 2 ln p + ln ( p - 1 ) - 1 ln ( q + 1) = 3 . 2 Find q in terms of p. [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

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 e 3 . Obtain p 2 ( p − 1) = e 3 q + 1 A1 OE without logs. 2 2 A1 OE, ISW  p ( p − 1)  Obtain q =  3  − 1  e  3

More questions on Logarithmic and exponential functions

z + 5i 3 Find the complex numbers z for which is real and z = 17 . Give your answers in the form z - 5 z = x + yi , where x and y are real. 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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 2 2 2 x + ( x − 5 ) = 17  x − 5 x + 4 = 0 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

More questions on Complex numbers

4 The parametric equations of a curve are x = etant , y = 3 tan 2 t . Find the equation of the tangent to the curve at the point (e, 3). Give your answer in the form y = mx + c , where m and c are exact. 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Mark scheme: π B1 SOI4 At ( e, 3 ) , tan t = 1 or t = 4 d x 2 tan t B1 = sec t  e d t dy 2 B1 = 6tan t  sec t dt dy 6tan t  sec 2 t − tan t *M1 Correct use of their derivatives. = = 6tan t  e 2 tan t ( ) dx sec t  e y − 3 = 6 x − e ) DM1 Substitute for t, and use correct method for the e ( equation of the line. y = 6e x − 3 A1 Or exact equivalent. Alternative Method for Question 4 y y 2 B1 SOI 3 Correct cartesian form, e.g. x = e or ln x = 3 or y = 3 ( ln x ) Differentiate function of a function *M1 Complete method. y A1 dx e 3 d y 1 Obtain = k y e or = k  ln x dy d x x y A1 dx 1 3 3 d y 1 Obtain = 6 y e or = 6  ln x dy d x x y − 3 = 6 x − e ) DM1 Use correct method for the equation of the line. e ( 4 y = 6e x − 3 A1 Or exact equivalent. 6

More questions on Differentiation

5 The polynomial 3 x 3 + pax 2 + 7a 2 x + qa 3 is denoted by f ( )x , where p, q and a are constants and a ! 0 . When f ( )x is divided by ( x + 2a ) the remainder is -22a 3 . When f ( )x is divided by ( 3x - a ) the remainder is -a3 . Find the values of p and q. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 5 Use f ( − 2 a ) = − 22 a 3 M1 Or use long division and equate a constant remainder to −22a3 . Obtain −24 a 3 + 4 pa 3 − 14 a 3 + qa 3 = −22 a 3 A1 Must evaluate the terms. OE, e.g. 4 p + q = 16.  a  3 M1 Or use long division and equate a constant Use f   = − a  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 + 9 q = −31 Obtain p = 5, q = −4 A1 5

More questions on Quadratics

4 1 ri 3 61 ri 21 ri6 It is given that z = 3e , z = e and ~ = 2e . 1 2 2 (a) State the values of ~z1 and ~z2 . Give your answers in the form reii, where r 2 0 and - r 1 i G r . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) On a sketch of an Argand diagram with origin O, show the points A, B, C and D representing the complex numbers z1, z2, ~z1 and ~z2 respectively. [2] (c) State the geometric effects of multiplying z1 and z2 by ~. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) Obtain z1 = 6e 4i3 π B1 Obtain z 2 = 3e 3i2 π 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 3 A 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. 3π Not required to state the centre of the rotation, but or rotation 2 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

More questions on Complex numbers

7 (a) Express 5 sin b x + 1 rl - 4 cos x in the form R sin ( x - a) , where R 2 0 and 0 1 a 1 1 r . State the 6 2 exact value of R and give the value of a correct to 3 decimal places. 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(b) Hence solve the equation 5 sin b2i + 1 rl - 4 cos 2i = 7 for 0 G i G r . Give your answers correct 6 to 2 decimal places. 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Mark scheme: 7(a)   5 3 B1 Or exact 2 term equivalent. Expand 5sin  x +  − 4cos x to obtain 2 3sin x − 2 cos x  6  State R = 21 B1ft Follow their 52 3 and 32 . Use correct trig formulae to obtain tan M1 OE, WWW. 3  3  E.g. tan =  =  ,   5 3  5  3  3  sin =  =  ,   2 21  2 7  5 3  5  cos= = .   2 21  2 7  Obtain = 0.333 A1 4 7(b) −1 7  B1FT SOI sin   0.615... can be implied by one correct answer.  21  Follow their 21. Use a correct method to obtain a value of  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

More questions on Trigonometry

8 With respect to the origin O, the points A and B have position vectors 2i + 4k and 5i + j + 6k respectively. The line l1 passes through the points A and B. (a) Find a vector equation for the line l1. 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The line l2 has equation r = 2i + j + 5k + n ( i + 2j + 3k ) . (b) Show that l1 and l2 do not intersect. 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(c) Find the acute angle between the directions of l1 and l2. 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Mark scheme: 8(a) Use a correct method to form an equation for 1l M1 Accept column vectors. Obtain r = ( 2i + 4k ) + ( 3i + j + 2k ) A1 OE, e.g. r = ( 5i + j + 6k ) + ( 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 +    2 + 3  5 + 3       E.g. 1 + 2 or  or 1 +  .              5 + 3  4 + 2  6 + 2 Equate two pairs of components of 2l and their l1, and solve for  or  M1  2 +    2 + 3     1 + 2 =           5 + 3  4 + 2 Obtain e.g. = − 15 , = − 53 A1 Or = − 17 , = − 73 or = −1, = −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 1l and 2l Using the correct process for the moduli, divide their scalar product by the product M1 −1  3 + 2 + 6   = cos of the moduli of their vectors and evaluate the inverse cosine of the result    14  14  Obtain AWRT 38.2 or 0.667 radians A1 3

More questions on Vectors

9 The constant a is such that ; 6x ln x dx = 4 . 1 1 5 (a) Show that a = exp f e 2 + 3op, where exp(x) denotes ex. 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(b) Verify by calculation that a lies between 2 and 2.1. 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(c) Use an iterative formula based on the equation in part (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. 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Mark scheme: 9(a) M1* x 2 x dx  d x. Allow M1 with q Commence integration and reach px 2 ln x + q   x A1 x 2 3 x dx must be simplified. Obtain 3 x 2 ln x − x Complete integration and obtain 3 x 2 ln x − 32 x 2 A1 Use limits correctly in an expression of the form ax 2 ln x − bx 2 and equate to 4, having DM1 3a 2 ln a − 32 a 2 − 0 + 32 = 4 integrated twice  1  5   + 3 Obtain a = exp   correctly A1 AG   2  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 − 3 x 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 a n +1 = exp   2 + 3   correctly at least once.  6  a n   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

More questions on Integration

10 (a) Find the quotient and remainder when x 3 + 5x 2 - 2x - 15 is divided by x 2 - 3 . 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(b) The variables x and y satisfy the differential equation d y x 3 + 5 x 2 - 2 x - 15 = . dx 6y ( x 2 - 3 ) It is given that y = 2 when x = 2 . Solve the differential equation to obtain an expression for y2 in terms of x. 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Mark scheme: M1 x + 510(a) Divide by ( x 2 − 3 ) to obtain x + k ( k  0 ) x 2 − 3 3x +5x 2 − 2x −15 3x −3x +5x 2 + x +5x 2 −15 x Obtain quotient x + 5 A1 Obtain remainder x A1 ISW x Allow . x 2 − 3 3 2 10(b) x 3 + 5 x − 2 x − 15 Separate variables correctly and obtain 6 y dy = 3 y 2 B1  6 y dy = dx. OE from  2  x − 3  Obtain 12 x 2 + 5 x B1ft Follow their linear quotient. 1 2 B1ft From the x term in their remainder ax + b. x − 3 Obtain 2 ln ( ) C = 0 Use y = 2, x = 2 to evaluate the constant of integration in an integral containing M1 12 = 2 + 10 + 12 ln1 + C  k ln x 2 − 3 ( ) 2 1 2 5 1 2 A1 OE x − 3 Obtain y = 6 x + 3 x + 6 ln ( ) 5

More questions on Differential equations

11 y M x O a 1 r 2 The diagram shows the curve y = cos x sin 2x for 0 G x G 1 r. The curve has a maximum point at M, 2 where x = a . (a) Find the exact value of a. 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(b) The region enclosed between the x-axis and the curve is rotated through 2r radians about the x-axis. Find the exact volume of the solid generated. 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Mark scheme: 11(a) d 2cos2 x cos2 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 2cos2 x E.g. = − sin x sin2 x + cos x  . dx 2 sin2 x Equate the derivative to zero and obtain a horizontal equation DM1 E.g. − sin a sin2a + 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 = 14 . Obtain a = 16 π A1 Exact answer in radians only. 6 11(b) 2 M1* y dx Use π  Use double angle formula to obtain a form that can be integrated directly, DM1 1 1 Or 4 sin4 x + 2 sin2 x dx,  2 3 e.g.  cos x sin2 x dx =  2cos x sin x dx 3 or  2 ( u − u ) du by substituting u = sin x. 1 Obtain − ( π )  2 cos 4 x A1 Or − (π 16 1 cos4 x + 14 cos2 x ) , OE. Use correct limits correctly DM1 1 4 π2 1 1 1  π , ( 2 cos ( 2 π ) − 2 )  − π  2 cos x  0 = −  or − π 1 cos2π + 1 cosπ − 1 − 1 OE. ( 16 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

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