Cambridge A Level Mathematics 9709 — 2023 Feb/March Paper 1 · Variant 2

1 A line has equation y = 3x −2k and a curve has equation y = x2 −kx + 2, where k is a constant. Show that the line and the curve meet for all values of k. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 x 2 − kx + 2 = 3 x − 2k leading to x 2 − x ( k + 3) + ( 2 + 2k )  = 0 M1 3-term quadratic, may be implied in the discriminant. 2 2 DM1 Cannot just be seen in the quadratic formula. b − 4ac = ( k + 3) − 8 (1 + k ) (ignore ‘= 0’ at this stage) = ( k − 1) 2 accept ( k − 1)( k − 1) A1 Or use of calculus to show minimum2 of zero at k = 1 or sketch of f ( k ) = k − 2k + 1. 0 Hence will meet for all values of k A1 Clear conclusion. 4

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4 The circumference round the trunk of a large tree is measured and found to be 5.00m. After one year the circumference is measured again and found to be 5.02m. (a) Given that the circumferences at yearly intervals form an arithmetic progression, find the circumference 20 years after the first measurement. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Given instead that the circumferences at yearly intervals form a geometric progression, find the circumference 20 years after the first measurement. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) 5.00 + 20  0.02 or 5.02 + 19  0.02 M1 Allow for a = 5, n = 20 with d = 0.02 only. a = 5, n = 21(OE) with d = 0.2 gets M1 only. 5.40 A1 2 4(b) 5.02 251 B1 r = = 1.004 or 5 250 20 19 M1 Allow a = 5, n = 20. 5.00  ( their1.004 ) or 5.02  ( their1.004 ) 5.42 A1 Any correct rounding of 5.41557108 . 3

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@ A7 x a 6 In the expansion of + , it is given that a x2 the coefficient of x4 = 3. the coefficient of x Find the possible values of the constant a. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 6 6 6 5 2 5 2 B1 B1 4  x  a   x  a   x  a   x  a  Coefficients x & x . Can be seen in an 7C1   2  or 7C6   2  7C2   2  or 7C5   2  expansion.  a  x   a  x   a  x   a  x   7  M1 OE. Allow extraneous 4x and x at this stage;  5   a  numerator and denominator must be functions of = 3 .a  21   3  Allow errors in evaluation of the combinations.  a  A1 Completely correct. 2 1 A1 1 a = SOI (implied by a = ). 9 3 1 A1 Allow ± 0.333 . a =  3 6

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7 (a) By first obtaining a quadratic equation in cos 1, solve the equation tan 1 sin 1 = 1 for 0Å < 1 < 360Å. 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(b) Show that tan 1 −sin 1  tan 1 sin 1. [3] sin 1 tan 1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) tansin= 1 leading to sin 2 = cos M1 sin  Use of tan  = and multiplication by cos cos. 1 − cos 2= cos or cos 2+ cos− 1  = 0 M1 Use of trig identity to form a 3-term quadratic. −1 5 M1 Use of formula or completion of the square must [cos=] be seen on a 3-term quadratic. Expect 0.6180 . 2 51.8º, A1 Both A marks dependent on the 2nd M1. 308.2º A1 FT FT for (360º ‒ 1st soln), A0 if extra solutions in range. Radians 0.905 and 5.38, A1 only for both. 5 7(b) tan sin sin sincos 1 M1 sin  − = − = − cos Use tan  = twice with correct use of sin tan sincos sin cos cos fractions. 1 − cos 2 sin 2 M1 Use 1 − cos 2= sin 2 with correct use of = = cos cos fractions. = tansin A1 WWW 3

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8 C 4 cm A 5 cm D 3 cm B The diagram shows triangle ABC in which angle B is a right angle. The length of AB is 8cm and the length of BC is 4cm. The point D on AB is such that AD = 5cm. The sector DAC is part of a circle with centre D. (a) Find the perimeter of the shaded region. 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(b) Find the area of the shaded region. 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Mark scheme: 8(a) 4 4 3 M1 −1 4 tan BDC = or sin BDC = or cos BDC = used to find ADC May use cosine rule or CAD = tan . 3 5 5 8 BDC = 0.927 3 → ADC = π − 0.927 3 [= 2.214 to 2.215 ] A1 Allow degrees, 126.87, and 0.7048 π or 0.705 π . Arc AC = 5  their 2.214 M1  Use of r or .2πr Expect 11.07 . 360 2 2 M1 Expect 8.94 . AC = 8 + 4 or 2 5 sin1.107  Perimeter =11.07 + 8.94 = 20.0 A1 Accept AWRT [20.01, 20.02]. 5 8(b) Sector ACD = ½  52  their 2.214 M1 1 2  2 See use of r  or .π r . Expect 27.7 . 2 360 1 2 M1 Subtracting the area of ADC, expect −10. Subtracting the area of ADC = ½ 5 4 or 5 sin their 2.214 or 2 1 1 −8 4 3 4 2 2 Shaded area = 27.7 − 10 =17.7 A1 Accept AWRT [17.67, 17.68]. Correct answer cannot come from an angle of 2.215 . 3

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dy10 At the point 4, −1 on a curve, the gradient of the curve is −3 It is given that = x−1 2. 2 + k, where k dx is a constant. (a) Show that k = −2. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the equation of the curve. 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(c) Find the coordinates of the stationary point. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (d) Determine the nature of the stationary point. 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Mark scheme: 110(a) B1 − 3 1 1 − = + k leading to k = −2 2 evaluated as or better. AG Need to see 4 1 2 2 4 2 1 10(b) 1 M1 A1 1 2 y = 2 x 2 − 2 x  + c  x Allow − 2 x . 12 −=1 4 −+c8 M1 Substitute x = 4, y = −1 (c present) Expect c = 3. 1 A1 Allow if f(x) = or y = anywhere in the solution. y = 2 x 2 − 2 x + 3 or y = 2 x − 2 x + 3 4 10(c) x −1/2 − 2 = 0 M1 d y Set their to zero. d x 1 A1 2 x =  1 1  1 1  =  max of M1A1 if If    ,3  seen. 4  2  4  4 2  (¼, 3½) A1 3 10(d) d 2 y 1 − 32 B1 2 = − x d x 2 < 0 (or −)4 hence Maximum DB1 1 WWW Ignore extra solutions from x = − . 4 2

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11 y A 5, 2 x O B 2, −1 x = y2 + 1 The diagram shows the curve with equation x = y2 + 1. The points A 5, 2 and B 2, −1 lie on the curve. (a) Find an equation of the line AB. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the volume of revolution when the region between the curve and the line AB is rotated through 360Å about the y-axis. 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Mark scheme: M1 Expect 1, must be from y / x .11(a) 2 −−( 1) Gradient of AB = 5 − 2 OE. Expect y = x − 3 . Equation of AB is y − 2 = 1( x − 5 ) or y + 1 = 1( x − 2 ) A1 2 π  y + 1 dy = π  y + 2 y + 1 dy11(b) π  x 2 dy = ( 2 2 4 2 M1 For curve: Attempt to square y 2 + 1 and attempt ) ( ) integration. Subtracting curve equation from line equation before squaring is M0. Integration before squaring M0.  y 5 2 y 3  A2, 1, 0 π  + + y   5 3  π  y + 6 y + 9 dy (π  y + 3 ) 2 dy = ( 2 M1 For line: Attempt to square their y + 3 and attempt ) integration.  y 3 2   ( y + 3 ) 3  A2, 1, 0 Not available for incorrect line equations. π  + 3 y + 9 y  or [ π]    3   3      1    16  1 2   DM1 Apply limits −→1 2 to either integral providing + 3 − 9 8π  + 12 + 18 −−   or 32 π  + + 2 −− − − 1   3  3  3    5 3  5 3   they have been awarded M1. Expect 15 [ π] 5 and/or 39[ π]. Some evidence of substitution of both −1 and 2 must be seen. Dependent on at least one of the first 2 M1 marks. 3 DM1 Appropriate subtraction. Dependent on at least one Volume = π (39 ‒ 15 ) of the first 2 M1 marks. 5 2 117 A1 = 23 π or π or awrt 73.5[1327] 5 5 9

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