Cambridge A Level Mathematics 9709 — 2023 May/June Paper 1 · Variant 3

4 (a) Show that the equation 3 tan2x −3 sin2x −4 = 0 may be expressed in the form a cos4x + b cos2x + c = 0, where a, b and c are constants to be found. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence solve the equation 3 tan2x −3 sin2x −4 = 0 for 0Å ≤x ≤180Å. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) 2 2 2 2 3sin 3sin cos 4cos 0    x x x x M1 Replace 2 tan x with 2 2 sin cos x x and multiply by 2 cos x .       2 2 2 2 3 1 cos 3 1 cos cos 4cos 0      x x x x M1 Replace 2 2 sin by 1 cos  x x twice. 4 2 3cos 10cos 3 0    x x or 4 2 3cos 10cos 3 0     x x A1 Or multiple of these equations. 3 Question Answer Marks Guidance 4(b)     2 2 3cos 1 cos 3 0    x x M1 OE, using their equation in the given form. Allow unusual notation if meaning is clear. 1 cos 3  x A1 SOI Answer only SC B1. 54.7º, A1 125.3º A1 FT Only other answer and must be from correct factorisation for A1. FT for 180 first answer their . Answers only SC B1, SC B1 FT. 4

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5 A circle has equation x −1 2 + y + 4 2 = 40. A line with equation y = x −9 intersects the circle at points A and B. (a) Find the coordinates of the two points of intersection. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find an equation of the circle with diameter AB. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) 2 2 1 9 4 40      x x   2 6 7 0 x x    leading to     1 7 0 x x    M1 Simplify to 3-term quadratic and factorise OE. (‒1, ‒10), (7, ‒2) or x = ‒1 and 7, y = ‒10 and ‒2 A1 A1 Answers only SC B1, SC B1 but must see a correct quadratic equation. 4 Question Answer Marks Guidance 5(b) [C is mid-point =] ( 1 2 1 2 , 2 2   their x their x their y their y ) M1 Expect (3, ‒6). Radius =       2 2 3 6     their x their their y their OR           2 2 7 1 2 10 / 2    their M1 Expect 32 .     2 2 3 6 32     x y A1 OE 3

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6 A r cm O 1 rad r cm B The diagram shows a sector OAB of a circle with centre O and radius r cm. Angle AOB = 1 radians. It is given that the length of the arc AB is 9.6cm and that the area of the sector OAB is 76.8cm2. (a) Find the area of the shaded region. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the perimeter of the shaded region. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) 2 ½ 76.8 9.6    r r or 2 2 1 9.6 76.8 2          16  r A1 0.6  A1 Accept 34.4o OAB = ½  their 162  sin their 0.6 M1 Allow Segment = 76.8 –½  their 162  sin their 0.6. Expect 72.27 . [Area = 76.8 ‒ 72.27 =] 4.53 A1 AWRT 5 6(b) 2 16 sin 0.3    AB OR 2 2 2 2 16 16 2 16 cos0.6     AB M1 Any valid method with their r, θ. Expect AB = 9.46. Perimeter = 9.6 + 9.46 = 19.1 A1 AWRT 2

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a2 8 A progression has first term a and second term where a is a positive constant. a + 2, (a) For the case where the progression is geometric and the sum to infinity is 264, find the value of a. 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(b) For the case where the progression is arithmetic and a = 6, determine the least value of n required for the sum of the first n terms to be less than −480. 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Mark scheme: 8(a) 2   a r a 264 1 2    a a a M1 Use of S∞ formula.   2 264 2 a a a a     leading to   2 264 2 a a   leading to   2 2 528 0 a a    M1* Process to a 3 term quadratic or a 3 term cubic. May contain terms on LHS and RHS.     22 24 0    a a DM1 Attempt to solve. 22  a (only) A1 22 without working SC DB1 (dep on 2nd M1). 5 8(b) 2 6 3 6 6 2 2     d B1   3 12 1 [ ] 480 2 2                n n M1* Forming an inequation with their numerical d. May use an equality.   2 3 9 640 [ 0]    n n A1 OE May contain terms on LHS and RHS.  9 81 2560 2    n DM1 OE. Expect 30.19 . Working for solution must be shown. 31 only A1 Must come from a correct first inequality (or an equality). 31 no working SC DB1 (dep on correct quadratic and correct inequality/equality). 5

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10 y 3 y = 9x − 2x + 1 2 A 112, 512 B 712, 312 x O The diagram shows the points A 112, 512 and B 712, 312 lying on the curve with equation 3 y = 9x − 2x + 1 2. (a) Find the coordinates of the maximum point of the curve. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Verify that the line AB is the normal to the curve at A. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the area of the shaded region. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 10(a)    1/2 d 3 9 2 1 2 d 2                 y x x B1, B1 Including ‘+c’ makes the second term B0.   1/2 9 3 2 1 0 x    leading to 2 1 9 x  M1 Set differential to zero and solve by squaring SOI. Beware   2 2 9 3 2 1 0    x M0A0. 2 1 3 2 1 9   or x x get M0. Max point = (4, 9) A1 WWW y = 9 must come from original equation. 4 10(b) When x = 1½, shows substitution or d 3 d  y x M1 Substituting x = 1½ into their d d y x . Gradient of AB is 5½ 3½ 1 1½ 7½ 3           M1 Substituting into a correct expression for mAB. 1 x3 1 3  . [Hence AB is the normal] A1 Alternative method for Question 10(b) When x = 1½ d 3 d  y x ,[ perpendicular gradient is -1/3] M1 Perpendicular through A has equation 3   x y + 6 which contains B(7.5,3.5) leading to AB is a normal to the curve at A M1 A1 3 Question Answer Marks Guidance 10(c)   5 2 2 2 1 9 5 2 2 2                     x x B1 B1 Integrating y with respect to x.     2.5 2.5 2 2 9 1 9 1 7.5 2 7.5 1 1.5 2 1.5 1 2 5 2 5                    or 9 225 1024 81 32 2 4 5 8 5                 or 1933 149 40 40  or 48.325 – 3.725 M1 OE Apply limits 1½ to 7½ to an integral. Working must be seen. Expect 44.6 . 1 1 1 5 3 6 2 2 2         or 15 2 3 2 1 ( 6)d 3    x x = 2 2 1 15 15 1 3 3 6 6 6 2 2 6 2 2                                      or 285 69 [ 8 8  = 27] B1 SOI Area of trapezium. May be seen combined with the area under the curve integral. [Shaded area = 44.6 – 27 =] 17.6 A1 SC B1 if no substitution of the limits seen. 5 Question Answer Marks Guidance 10(c) Alternative method for Question 10(c) A =   15 2 3 2 3 2 1 ((9 2 1 ) 6 )d 3              x x x x   15 2 3 2 3 2 28 (( 2 1 6)d 3     x x x M1 Finding the equation of AB and subtracting from the equation of the curve.   5 2 2 2 1 28 6 5 3 2 2 2                        x x x A1 A1 127 49 10 10   M1 Apply limits 1½ to 7½ to an integral. Working must be seen. 17.6 A1 SC B1 if no substitution of limits seen. 5

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