Cambridge A Level Mathematics 9709 — 2021 May/June Paper 1 · Variant 2

1 (a) Express 16x2 −24x + 10 in the form 4x + a 2 + b. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) It is given that the equation 16x2 −24x + 10 = k, where k is a constant, has exactly one root. Find the value of this root. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) recovery. + 1 or b = 1 B1 2 1(b) [For one root] k = 1 or ‘their b’ B1 FT Either by inspection or solving or from 242 – 4 × 16 × (10 – k) = 0 WWW [ ] 3 Root or or 0.75 4 = x B1 SC B2 for correct final answer WWW. 2

More questions on Quadratics

3 The equation of a curve is y = x −3 x + 1 + 3. The following points lie on the curve. Non-exact values are rounded to 4 decimal places. A 2, k B 2.9, 2.8025 C 2.99, 2.9800 D 2.999, 2.9980 E 3, 3 (a) Find k, giving your answer correct to 4 decimal places. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the gradient of AE, giving your answer correct to 4 decimal places. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The gradients of BE, CE and DE, rounded to 4 decimal places, are 1.9748, 1.9975 and 1.9997 respectively. (c) State, giving a reason for your answer, what the values of the four gradients suggest about the gradient of the curve at the point E. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) 1.2679 B1 AWRT. ISW if correct answer seen. 3 – 3 scores B0 1 3(b) 1.7321 B1 AWRT. ISW if correct answer seen. 1 3(c) Sight of 2 or 2.0000 or two in reference to the gradient *B1 This is because the gradient at E is the limit of the gradients of the chords as the x-value tends to 3 or ꝺx tends to 0. DB1 Allow it gets nearer/approaches/tends/almost/approximately 2 2

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10 3 1 4 The coefficient of x in the expansion of 4x + is p. The coefficient of in the expansion of x x @ A5 k 2x + is q. x2 Given that p = 6q, find the possible values of k. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 [Coefficient of x or p =] 480 B1 SOI. Allow 480x even in an expansion. [ ]( ) 2 3 2 1 Term in 10 2     = ×         k or q x x x M1 Appropriate term identified and selected. [10 × 23k2 =] 80k2 A1 Allow 2 80k x p = 6q used (480 = 6 × 80k2 or 80 = 80k2) M1 Correct link used for their coefficient of x and 1 x (p and q) with no x’s. [k2 = 1 ⇒] k = ±1 A1 A0 if a range of values given. Do not allow 1 ± . 5

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7 The point A has coordinates 1, 5 and the line l has gradient −2 and passes through A. A circle has 3 centre 5, 11 and radius 52. (a) Show that l is the tangent to the circle at A. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the equation of the other circle of radius 52 for which l is also the tangent at A. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) (5 – 1)2 + (11 – 5)2 = 52 or 11 5 5 1 − − M1 For substituting (1,5) into circle equation or showing gradient 3 2 = . For both circle equation and gradient, and proving line is perpendicular and stating that A lies on the circle A1 Clear reasoning. Alternative method for Question 7(a) ( ) ( ) 2 2 5 11 52 − + − = x y and ( ) 2 5 1 3 − = − − y x M1 Both equations seen and attempt to solve. May see 2 17 3 3 = − + y x Solving simultaneously to obtain (y – 5)2 = 0 or (x – 1)2 = 0 ⇒ 1 root or tangent or discriminant = 0 ⇒ 1 root or tangent A1 Clear reasoning. Alternative method for Question 7(a) d 10 2 10 2 d 2 22 10 22 − − = = − − y x x y M1 Attempting implicit differentiation of circle equation and substitute x = 1 and y = 5. Showing gradient of circle at A is 2 3 − A1 Clear reasoning. 2 7(b) Centre is (−3, −1) B1 B1 B1 for each correct co-ordinate. Equation is (x + 3)2 + (y + 1)2 = 52 B1 FT FT their centre, but not if either (1, 5) or (5, 11). Do not accept 2 52 . 3

More questions on Coordinate geometry

8 The first, second and third terms of an arithmetic progression are a, 32a and b respectively, where a and b are positive constants. The first, second and third terms of a geometric progression are a, 18 and b + 3 respectively. (a) Find the values of a and b. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the sum of the first 20 terms of the arithmetic progression. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(a) 3 2 2 2   + = ×  =     a b a b a 182 = a(b + 3) OE or 2 correct statements about r from the GP, e.g. 18 = r a and b + 3 = 18r or 2 3 + = b r a B1 SOI 324 = a(2a + 3) ⇒ 2a2 + 3a – 324[= 0] or b2 + 3b – 648[= 0] or 6r2 – r – 12[= 0] or 4d2 + 3d – 162[= 0] M1 Using the correct connection between AP and GP to form a 3-term quadratic with all terms on one side. (a – 12)(2a + 27)[= 0] or ( )( )[ ] 24 27 0 − + = b b or ( )( )[ ] 2 3 3 4 0 − + = r r or ( )( )[ ] 6 4 27 0 − + = d d M1 Solving their 3-term quadratic by factorisation, formula or completing the square to obtain answers for a, b, r or d. a = 12, b = 24 A1 WWW. Condone extra ‘solution’ 13.5, 27 =− = − a b only. 5 Question Answer Marks Guidance 8(b) Common difference d = 6 B1 FT SOI. FT their 2 a ( ) 20 20 S 2 12 19 6 2 = × + × M1 Using correct sum formula with their a, their calculated d and 20. 1380 A1 3

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9 y y2 = x −2 1 x 0 5 The diagram shows part of the curve with equation y2 = x −2 and the lines x = 5 and y = 1. The shaded region enclosed by the curve and the lines is rotated through 360Å about the x-axis. Find the volume obtained. 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Mark scheme: 9 Curve intersects y = 1 at (3, 1) B1 Throughout Question 9: 1 < their 3 < 5 Sight of x = 3 Volume = [ ] ( )[ ] π 2 d  − x x M1 M1 for showing the intention to integrate ( ) 2 − x . Condone missing π or using 2 π. [ ] 2 1 π 2 2   −     x x or [ ] ( ) 2 1 π 2 2   −     x A1 Correct integral. Condone missing π or using 2 π. [ ] 2 2 5 3 π 2 5 2 3 2 2       = − × − − ×               their their [ ] 5 3 π 2 2   = +     as a minimum requirement for their values M1 Correct use of ‘their 3’ and 5 in an integrated expression. Condone missing π or using 2 π. Condone +c. Can be obtained by integrating and substituting between 5 and 2 and then 3 and 2 then subtracting. Volume of cylinder = ( )[ ] 2 π 1 5 3 2π × × − = their B1 FT Or by integrating 1 to obtain x (condone y if 5 and their 3 used). [Volume of solid = 4 π 2π ] 2π − = or 6.28 A1 AWRT Question Answer Marks Guidance 9 Alternative method for Question 9 Curve intersects y = 1 at (3, 1) B1 Sight of x = 3 Volume of solid = ( ) [ ] π 2 1 d  − − x x M1 B1 M1 for showing the intention to integrate ( ) 2 − x B1 for correct integration of –1. Condone missing π or 2 π for M1 but not for B1. [ ] 2 1 π 3 2   −     x x or [ ] ( ) 2 1 π 3 2   −     x A1 Correct integral, allow as two integrals. Condone missing π or using 2 π. [ ] 2 2 5 3 π 3 5 3 3 2 2       = −× − −×               their their M1 Correct use of ‘their 3’ and 5 in an integrated expression. Condone missing π or using 2 π. Condone +c. Can be obtained by integrating and substituting between 5 and 2 and then 3 and 2 then subtracting. [Volume of solid = 4 π 2π ] 2π − = or 6.28 A1 AWRT 6

More questions on Integration

10 (a) Prove the identity 1 + sin x −1 −sin x  4 tan x . [4] 1 −sin x 1 + sin x cosx ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence solve the equation 1 + sin x −1 −sin x = 8 tan x for 0 ≤x ≤1 [3] 2π. 1 −sin x 1 + sin x ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 10(a) ( ) ( ) ( )( ) 2 2 1 sin 1 sin 1 sin 1 sin 1 sin 1 sin 1 sin 1 sin + − − + − − ≡ − + − + x x x x x x x x )( ) 1 sin 1 sin − + x x and reasonable attempt at the numerator(s). ( ) ( )( ) 2 2 1 2sin sin 1 2sin sin 1 sin 1 sin + + − − + ≡ − + x x x x x x DM1 For multiplying out the numerators correctly. Condone sign errors for this mark. 2 2 4sin 4sin 1 sin cos ≡ ≡ − x x x x DM1 For simplifying denominator to 2 cos x . 4sin 4tan cos cos cos ≡ ≡ x x x x x A1 AG. Do not award A1 if undefined notation such as s, c, t or missing x’s used throughout or brackets are missing. Alternative method for Question 10(a) 4tan cos ≡ x x 2 2 4sin 4sin cos 1 sin ≡ − x x x x *M1 Using 2 sin tan andcos cos = = x x x x 2 1 sin − x 2 2 1 in 1 sin − ≡ + + − s x x DM1 Separating into partial fractions. 2 2 1 1 1 in 1 sin − ≡+ + − + − s x x DM1 Use of 1-1 or similar 1 sin 1 sin − ≡−+ x x + 1 sin 1 sin + − x x A1 4 Question Answer Marks Guidance 10(b) 1 cos 2 = x *B1 OE. WWW. π 3 x = DB1 Or AWRT 1.05 0from tan 0 or sin 0 = = = x x x B1 WWW. Condone extra solutions outside the domain 0 to π 2 but B0 if any inside. 3 Question Answer Marks Guidance

More questions on Trigonometry

12 Q P A B F C E D The diagram shows a cross-section of seven cylindrical pipes, each of radius 20 cm, held together by a thin rope which is wrapped tightly around the pipes. The centres of the six outer pipes are A, B, C, D, E and F. Points P and Q are situated where straight sections of the rope meet the pipe with centre A. (a) Show that angle PAQ = 13π radians. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the length of the rope. 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(c) Find the area of the hexagon ABCDEF, giving your answer in terms of 3. 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(d) Find the area of the complete region enclosed by the rope. 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Mark scheme: 12(a) [By symmetry] [6 × PAQ = 2π], [ PAQ =] 2π÷6, M1 Explaining that there are six sectors around the diagram that make up a complete circle. A1 AG Alternative method for Question 12(a) Using area or circumference of circle centre A ÷ 6 M1 400π 6 or 40π 6 Justification for dividing by 6 followed by comparison with the sector area or arc length. A1 AG Alternative method for Question 12(a) Explain why ∆PAQ is an equilateral triangle M1 Assumption of this scores M0 Using ∆PAQ is an equilateral triangle ⸫ ˆ PAQ = π 3 A1 AG Alternative method for Question 12(a) Using the internal angle of a regular hexagon = 2π 3 Or 3 ˆ ˆ 2π FAO OAB + = , equilateral triangles M1 ˆ PAQ = π 2π π 2π — 2 3 2   + +     = π 3 A1 AG Question Answer Marks Guidance 12(a) Alternative method for Question 12(a) 20, with 40 θ θ = Sin clearly identified M1 π π , 2 6 3 θ θ = = = ˆ FAO and by similar triangles = ˆ PAQ A1 AG 2 12(b) Each straight section of rope has length 40 cm B1 SOI Each curved section round each pipe has length π 20 3 rθ = × *M1 Use of θ r with r = 20 and θ in radians Total length = ( ) ( ) 6 40 π their k × + DM1 6×(their straight section + their curved section). Their curved section must be from acceptable use of θ r – this could now be numeric. 240 40π + or 366 (AWRT) (cm) A1 Or directly: (6 diameter) × + circumference 4 Question Answer Marks Guidance 12(c) [Triangle area =] 1 π 40 40 sin 2 3   × × ×     or 1 40 20 3 2 × × or 400 3 or 693(AWRT) B1 [Total area of hexagon = 6 × 400 3 =] 2400 3 B1 Condone 4800 3 2 Alternative method for Question 12(c) [Trapezium area =] ( ) 1 π 40 80 40sin 2 3   × + ×     or 1200 3 or 2080 (AWRT) B1 [Total area of hexagon = 2 × 1200 3 =] 2400 3 B1 Condone 3 4800 2 √ Alternative method for Question 12(c) Area of triangle ABC = 400 3 or 693 (AWRT) or 4 × Area of half of triangle ABC = 4 200 3 × or 1390 (AWRT) or Area of rectangle ABDE = 1600 3 or 2770 (AWRT) B1 [Total area of hexagon = 2 400 3 × +1600 3 =] 2400 3 Or [= 4 200 3 × +1600 =] 2400 3 B1 Condone 3 4800 2 √ If B0B0, SC B1 can be scored for sight of 4160 (AWRT) as final answer. 2 Question Answer Marks Guidance 12(d) Each rectangle area = 40 × 20 (= 800) B1 SOI, e.g. by sight of 4800 Each sector area = 2 2 1 1 π 200π 20 2 2 3 3 r θ   = × × =     B1 SOI. Total area = 2400 3 4800 400π + + or 10 200 (cm2) (AWRT) B1 Or directly: part (c) + 6800 + area circle radius 20. 3

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