Cambridge A Level Mathematics 9709 — 2023 May/June Paper 3 · Variant 2

1 Solve the inequality 5x 2 3x . [4] −3 < −7 ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 State or imply non-modular inequality 2 2 2 (5 3) 2 (3 7) x x    , or corresponding quadratic equation, or pair of linear equations (5x – 3) = ± 2(3x – 7) B1 2 11 138 187 0 x x    . Solve a 3-term quadratic, or solve two linear equations for x M1 If no working is shown, the M1 is implied by the correct roots for an incorrect quadratic. Obtain critical values 17 and 11 11 x x   A1 Accept 1.55 or better. State final answer 17 , 11 11 x x   A1 Strict inequality required. In set notation, allow notation for open sets but not for closed sets e.g. accept     17 , 11, 11   or 17 ( , [ ]11, ) 11   but not 17 ( , ] [11, ) 11   . Allow ‘or’ but not ‘and’. Accept .  Final A0 for 17 11 11 x   . Exact values expected but ISW if exact inequalities seen followed by decimal approx. Alternative Method for Question 1 Obtain critical value x = 11 from a graphical method, or by inspection, or by solving a linear equation or an inequality B1 Obtain critical value 17 11 x  similarly B2 Accept decimal value. State final answer 17 , 11 11 x x   B1 Strict inequality required. See notes above. 4

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3 (a) On an Argand diagram, sketch the locus of points representing complex numbers z satisfying z 3 2. [2] + −2i = (b) Find the least value of z for points on this locus, giving your answer in an exact form. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) B1 Show a circle with centre –3 + 2i. Allow for a curved figure with ‘centre’ in roughly the correct position. Accept marks or numbers on axes, coordinates of centre shown. B0B1 available for axes the wrong way round (and M1 A1 in part (b)). Show a circle with radius 2 B1 FT FT centre not at the origin. Allow ‘near miss’ on x axis. Different scales on axes require an ellipse for B1 B1. Scales on the axes and any label of the radius must be consistent for B1 B1. Correct circle shaded scores B1 B0. 2 Re(z) bold distance is the length required for part (b) 2i -3 Question Answer Marks Guidance 3(b) Carry out a correct method for finding the least value of |z| M1 e.g. distance of centre from origin – radius or find point of intersection of circle and 3y = -2x and use Pythagoras. If they subtract the wrong way round M0. If their diagram is a reflection or a rotation of the correct diagram, M1 A1 is available (requires equivalent work). Any other circle M0. Obtain answer 13 2  or 17 4 13  A1 Or exact equivalent e.g. 26 36 17 3 13  . Correct solution only. Allow A1 if exact answer seen and then decimal given. 2

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1 1 for 0 [5]4 Solve the equation 2 cosx −cos 2x = ≤x ≤2π. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 Use correct double angle formula to obtain an equation in cos 2 x       only e.g. 2 2 2cos 1 cos 1 2 2 x x                      . Obtain a 3 term quadratic in cos 2 x       , A1 e.g. 2 4cos cos 3 0 2 2 x x                . Allow 2 4cos cos 3 0 u u    . Condone 2 x x  . Obtain 3 cos andcos 1 2 4 2 x x               A1 Allow answer in u e.g.    4cos 3 cos 1 u u   and condone 2 x x  . Solve for the original x DM1 Must see evidence of doubling, not halving. Obtain x = 0 and 4.84 and no others in the interval A1 Ignore any answers outside interval. Accept AWRT 4.84. Accept 1.54. Must be in radians. 277.2 indicates M1 but is A0. Alternative Method for Question 4 Use correct double angle formula to obtain an equation in cosx only *M1 e.g. cos 1 2cos 1 2 x x   . Obtain a 3 term quadratic in cosx , A1 e.g. 2 8cos 9cos 1 0 x x   . Obtain 1 cos 8 x  and cos x = 1 A1 Solve for x DM1 Obtain answers x = 0 and 4.84 and no others in the interval A1 Ignore any answers outside interval. Accept AWRT 4.84. Must be in radians. 277.2 is A0. 5

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denoted by 3x has one root in the interval 0 x6 The equation cot 2x1 = < < π, !. (a) Show by calculation that lies between 0.5 and 1. 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(b) Show that, if a sequence of positive values given by the iterative formula P @ AQ 1 1 4 tan−1 xn + xn+1 = 3 3xn converges, then it converges to [2] !. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Use this iterative formula to calculate correct to 2 decimal places. Give the result of each ! iteration to 4 decimal places. 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Mark scheme: 6(a) Calculate the values of a relevant expression or pair of expressions at x = 0.5 and x = 1 M1 Need to evaluate at both points, but M1 still available if one value incorrect. Use of degrees is M0. Correct use of a smaller interval is M1. If using g(x) – f(x), there needs to be a clear indication of the comparison being made e.g. by listing values in a table. Embedded values 0.5 and 1 are not sufficient. 3.92 and 1.83 alone are not sufficient. Complete the argument correctly with conclusion about change of sign or change of inequalities and with correct calculated values. Can all be in symbols – an explanation in words is not required. A1 e.g. 3.92 > 1.5, 1.83 < 3 or 2.42 > 0, –1.17 < 0. 2 6(b) State 1 1 1 4tan 3 3 x x x          M1 Or rearrange cot 3 2 x x       as far as 1 1 2 4tan 3 x x        Rearrange to the given equation cot 3 2 x x       Need intermediate step between 1 1 tan 2 3 x x   and cot 3 2 x x       A1 Or continue rearrangement to 1 1 1 4tan 3 3 x x x          and state iterative formula of 1 1 1 1 4tan 3 3 n n n x x x           AG 2 Question Answer Marks Guidance 6(c) Use the iterative process correctly at least once M1 Obtain one value and substitute that back in to obtain a second value. Working in degrees is M0. Obtain final answer 0.79 A1 Must be to 2 d.p. Show sufficient iterations to at least 4 d.p. to justify 0.79 to 2 d.p. or show there is a sign change in the interval (0.785, 0.795) A1 e.g. 1, 0.7623, 0.8037, 0.7921, 0.7951, 0.7943, 0.7945 or 0.5, 0.9506, 0.7665, 0.8024, 0.7924, 0.7950, 0.7944, 0.7945 or 0.75, 0.8076, 0.7911, 0.7954, 0.7943, 0.7946, 0.7945 . Condone truncation. Allow recovery. Condone minor differences in the final d.p. 3 If they do the iteration in (b) but restate the conclusion here, no marks in (b) but could score 3/3 for (c).

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7 The equation of a curve is 3x2 4xy 3y2 5. + + = dy 2y (a) Show that + [4] dx = −3x2x 3y. + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence find the exact coordinates of the two points on the curve at which the tangent is parallel to y 2x 0. 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Mark scheme: 7(a) State or imply d 6 d y y x as the derivative of 3y2 B1 Allow y for d d y x throughout. Accept 6 4 f x y x   . State or imply d 4 4 d y x y x  as the derivative of 4xy B1 Accept 4 6 f x y y    . Equate derivative of LHS to zero and solve for d d y x M1 Allow an extra d d y x in front of their differentiated equation. Allow if ‘= 0’ is implied but not seen. Allow d d f y x f x y      Obtain d 3 2 d 2 3 y x y x x y    A1 AG – must come from correct working. The position of the negative must be clear. 4 Question Answer Marks Guidance 7(b) Equate d d y x to –2 and solve for x in terms of y or for y in terms of x *M1 Must be using the given derivative. Obtain 4 x y  or 4 x y  A1 Seen or implied by correct later work. Substitute 4 their x y  or 4 x their y  in curve equation DM1 Allow unsimplified. Obtain 1 7 y  or 4 7 x  A1 Or exact equivalent. Or 4 1 and 7 7 x y   or exact equivalent. Obtain both pairs of values A1 Or 4 1 and 7 7 x y   or exact equivalent. A1 A0 for incorrect final pairing. 5

More questions on Differentiation

8 (a) The variables x and y satisfy the differential equation dy 4 9y2 + . dx = e2x+1 It is given that y 0 when x 1. = = Solve the differential equation, obtaining an expression for y in terms of x. 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(b) State what happens to the value of y as x tends to infinity. Give your answer in an exact form. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(a) Separate variables correctly B1   2 1 2 1 d e d 4 9 x y x y       . Condone missing integral signs or dx and dy missing. Obtain term 2 1 1 e 2 x    B1 OE e.g. 2 1 e 2e x   . Obtain term of the form 1 3 tan 2 y a       M1 Obtain term 1 1 3 tan 6 2 y       A1 OE e.g. 1 1 3 3 tan 9 2 2 y   . Use x = 1, y = 0 to evaluate a constant or as limits in a solution containing or derived from terms of the form   1 tan a by  and   2 1 e x c   M1 If they rearrange before evaluating the constant, the constant must be of the correct form. Obtain correct answer in any form A1 e.g.   2 1 1 3 1 3 1 1 tan e e 6 2 2 2 x y       . Obtain final answer   3 2 1 2 tan 3e 3e 3 x y      A1 OE Allow with 3 3e 0.149....  7 8(b) State that y approaches   3 2 tan 3e 3  B1 FT Or exact equivalent. The FT is on correct work on a solution containing e–2x–1. Condone y = … Accept correct answer stated with minimal wording. 0.10032… is not exact so B0. 1

More questions on Differential equations

10 y M x O The diagram shows the curve y x 5 3 and its maximum point M. = + −2x (a) Find the exact coordinates of M. 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(b) Using the substitution u 3 find by integration the area of the shaded region bounded by = −2x, the curve and the x-axis. Give your answer in the form a 13, where a is a rational number. 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Mark scheme: 10(a) Use the product rule correctly to obtain 1 2 ( 5)(3 2 ) (3 2 ) n p x x q x     BOD over sign errors unless an incorrect rule is quoted. Obtain correct derivative in any form A1 e.g. 1 1 2 2 ( 5)(3 2 ) (3 2 ) x x x       . Equate derivative to zero and obtain a linear equation DM1 Allow with surd factor e.g.        1 2 3 2 5 3 2 0 x x x        . Obtain a correct linear equation. A1 e.g. –(x + 5) + 3 – 2x = 0. Obtain answer 2 13 39 , 3 9          . A1 Or exact equivalent e.g. 2 13 13 , 3 3 3          or 2 2197 , 3 27          . Accept with x, y stated separately. ISW Alternative Method for Question 10(a) Obtain y2 and differentiate *M1 Ignore their left hand side i.e. their 2 d d y x . Obtain correct derivative in any form A1 e.g. 2 6 34 20 x x    . Equate derivative to zero and solve for x DM1 Obtain 2 3  A1 Ignore –5 if seen. Obtain answer 2 13 39 , 3 9          only A1 Or exact equivalent e.g. 2 13 13 , 3 3 3          or 2 2197 , 3 27          . ISW 5 Question Answer Marks Guidance 10(b) Use the given substitution and reach 1 2 13 d 2 2 u a u u         *M1 OE Need to see -2 or -½ used. Condone if du missing or the integral sign is missing. Allow M1A0 for complete substitution into 3 2 d x x x   to obtain first term of the line below. Obtain correct integral 1 2 1 13 d 2 2 2 u u u          A1 OE e.g. 1 3 d 5 d 2 2 u u u u u            . Ignore limits at this stage. Condone if du missing. x = –5 and 3 2 B1 SOI e.g. by u = 13 and 0. In any order and at any stage. Use correct limits the right way round in an integral of the form 3 5 2 2 26 2 3 5 a u u          DM1 Obtain answer 169 13 15 or a = 169 15 A1 or exact equivalents. 5

More questions on Differentiation

11 The points A and B have position vectors i 2j and 2i k respectively. The line l has equation + −2k −j + r i 3k 2i 4k . = −j + + - −3j + (a) Show that l does not intersect the line passing through A and B. 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(b) Find the position vector of the foot of the perpendicular from A to l. 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Mark scheme: 11(a) Carry out correct method for finding a vector equation for AB M1 Obtain [r =] i + 2j – 2k + λ (i – 3j + 3k) A1 OE e.g.   2 3 3       r i j k i j k . Equate two pairs of components of general points on their AB and l and evaluate λ or μ M1 1 1 2 2 3 1 3 2 3 3 4                                 . Obtain correct answer for λ or μ, e.g. λ = –1, μ = –2 A1 Correct value from two correct component equations. Verify that all three equations are not satisfied and the lines fail to intersect (≠ is sufficient justification e.g. 0 ≠ –3). A1 Conclusion needs to follow correct values. Hybrid versions are possible e.g. using j and k to get one parameter and then i to obtain the other. or e.g. solving two pairs of simultaneous equations and showing that the results are not the same. Alternatives: A λ μ B λ μ ij 2 1 4 ≠ 7 ij 1 1 4 ≠ 7 ik 5 5/2 –13≠- 17/2 ik 4 5/2 –13≠- 17/2 jk –1 –2 0 ≠ –3 jk –2 –2 0 ≠ –3 5 Question Answer Marks Guidance 11(b) Find AP  for a general point P on l, e.g. –3j + 5k + μ(2i – 3j + 4k) B1 Or equivalent e.g.     2 3 3 4 5 PA         i j k  . Calculate scalar product of their AP  and a direction vector for l and equate the result to zero M1 e.g.    4 9 9 20 16 0         . M0 if using OP  . M0 if using parallel line through A. Obtain = –1 A1 Obtain answer –i + 2j – k A1 Accept coordinates in place of position vector. Alternative Method for Question 11(b) Find AP  for a general point P on l, e.g. –3j + 5k + μ(2i – 3j + 4k) B1 Or equivalent e.g.     2 3 3 4 5 PA         i j k  . Use Pythagoras and differentiate with respect to to obtain value of  corresponding to minimum distance. (No need to prove it is a minimum) M1       2 2 2 d 4 9 1 4 5 0 d          . Obtain μ= –1 A1 Obtain answer –i + 2j – k A1 Accept coordinates in place of position vector. 4

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