Cambridge A Level Mathematics 9709 — 2021 May/June Paper 1 · Variant 1

dy 3 1 The equation of a curve is such that = + 32x3. It is given that the curve passes through the point dx x4 12, 4 . Find the equation of the curve. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 [ ] 4 3 1 8 = − + y x x [+ c] B1 B1 OE. Accept unsimplified. 4 = –8 + 1 2 + c M1 Substituting 1 , 4 2       into an integrated expression 4 3 1 23 8 2 = − + + y x x A1 OE. Accept 3 − −x ; must be 8; y = must be seen in working. 4

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2 The sum of the first 20 terms of an arithmetic progression is 405 and the sum of the first 40 terms is 1410. Find the 60th term of the progression. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 10(2a + 19d) = 405 B1 20(2a + 39d) = 1410 B1 Solving simultaneously two equations obtained from using the correct sum formulae [a = 6, d = 1.5] M1 Reach = a or = d Using the correct formula for 60th term with their a and d M1 60th term = 94.5 A1 OE, e.g. 189 2 5

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5 The fifth, sixth and seventh terms of a geometric progression are 8k, −12 and 2k respectively. Given that k is negative, find the sum to infinity of the progression. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 5 M1 Forming an equation in k k = −3 A1 Using correct formula for S∞ [r = 0.5, a = −384] M1 With 1 1 −< < r S∞ = −768 A1 Alternative method for Question 5 2 2 8 = k r k M1 [ ]0.5 = ± r A1 Using correct formula for S∞ [r = 0.5, a = −384] M1 1 1 −< < r S∞ = −768 A1 4

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1 −2 sin21 7 (a) Prove the identity  1 −tan21. [2] 1 −sin21 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 1 −2 sin21 (b) Hence solve the equation = 2 tan41 for 0Å ≤1 ≤180Å. [3] 1 −sin21 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) Reach 2 2 2 cos θ sin θ cos θ − or 2 2 2 2 1 sin θ sin θ 1 sin θ cos θ − − − or 2 2 2 2 sin θ cos θ 2tan cos θ θ + − or 2 2 2 2sin θ sec cos θ θ − or 2 2 sec θ − or 2 cos2 cos θ θ M1 May start with 2 1 tan θ − 2 1 tan θ − A1 AG, must show sufficient stages 2 7(b) 2 4 1 tan 2tan θ θ − = ⇒ 4 2tan θ + 2 tan θ −1 [= 0] M1 Forming a 3-term quadratic in 2 tan θ or e.g. u 2 tan θ = 0.5 or −1 leading to [ ] tan 0.5 θ = ± M1 θ = 35.3° and 144.7° (AWRT) A1 Both correct. Radians 0.615, 2.53 scores A0. 3

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8 P Q C S R The diagram shows a symmetrical metal plate. The plate is made by removing two identical pieces from a circular disc with centre C. The boundary of the plate consists of two arcs PS and QR of the original circle and two semicircles with PQ and RS as diameters. The radius of the circle with centre C is 4 cm, and PQ = RS = 4 cm also. (a) Show that angle PCS = 23π radians. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the exact perimeter of the plate. 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(c) Show that the area of the plate is 20 π + 8 3 cm2. 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Mark scheme: 8(a) Either Let midpoint of PQ be H: sin HCP = 2 4 ⇒ Angle HCP = π 6 Or sin PSQ = 4 8 ⇒ Angle PSQ = π 6 Or using cosine rule: angle PCQ = π 3 Or by inspection: triangle PCQ or PCT is equilateral so angle PCQ = π 3 M1 Angle PCS = π π 2 π π 6 6 3 − − = A1 AG 2 8(b) Perimeter = 2π 2 4 3 × × or 8π 8π 3 − M1 Length of two arcs PS and QR 2π 2 + × M1 Adding circumference of two semicircles 28π 3 A1 Must be a single term 3 Question Answer Marks Guidance 8(c) Area sector CPQ = 2 1 π 8π 4 2 3 3 × × = M1 Uses correct formula for sector Area of segment of large circle beyond CPQ = 2 8π 1 π 8π 4 sin 4 3 3 2 3 3   − × × = −     M1 Attempts to find area of segment Area of small semicircle = π 2 × or area of small circle = 2 π 2 × M1 Area of plate = Large circle – [2 ×] small semicircle – [2 ×] segment area M1 2 2 8π 20π π 4 π 2 2 4 3 8 3 3 3   × − × −× − = +     A1 AG Alternative method for Question 8(c) Area of sector PCS 2 1 2π 16π 4 2 3 3 = × × = M1 Uses correct formula for sector Area of triangle PCQ = 2 1 π 4 sin 4 3 2 3 × × = M1 Uses correct formula for triangle Area of small semicircle = π 2 × or area of circle = 2 π 2 × M1 Area of plate = [2 ×] large sector + [2 ×] triangle – [2 ×] small semicircle M1 ( ) 2 16π 20π 2 2 4 3 π 2 8 3 3 3  + − × = +     A1 AG 5

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10 The equation of a circle is x2 + y2 −4x + 6y −77 = 0. (a) Find the x-coordinates of the points A and B where the circle intersects the x-axis. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the point of intersection of the tangents to the circle at A and B. 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Mark scheme: 10(a) When y = 0 2 4 77 0 − − = x x [⇒ ( )( ) 7 11 0 + − = x x or ( ) 2 2 81 − = x ] M1 Substituting = So x-coordinates are −7 and 11 A1 2 Question Answer Marks Guidance 10(b) Centre of circle C is (2, −3) B1 Gradient of AC is 1 3 − or Gradient of BC is 1 3 M1 For either gradient (M1 sign error, M0 if x-coordinate(s) in numerator) Gradient of tangent at A is 3 or Gradient of tangent at B is −3 M1 For either perpendicular gradient Equations of tangents are y = 3x + 21, y = −3x + 33 A1 For either equation Meet when 3x + 21 = −3x + 33 M1 OR: (centre of circle has x coordinate 2) so x coordinate of point of intersection is 2 Coordinates of point of intersection (2, 27) A1 Alternative method for Question 10(b) Implicit differentiation: d 2 d y y x seen B1 d d 2 4 2 6 0 d d − + + = y y x y x x M1 Fully differentiated 0 = with at least one term involving y differentiated correctly Gradient of tangent at A is 3 or Gradient of tangent at B is −3 M1 For either gradient Equations of tangents are y = 3x + 21, y = −3x + 33 A1 For either equation Meet when 3x + 21 = −3x + 33 M1 OR: (centre of circle has x coordinate 2) so x coordinate of point of intersection is 2 Coordinates of point of intersection (2, 27) A1 6

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