Cambridge A Level Mathematics 9709 — 2017 May/June Paper 1 · Variant 1

2 Relative to an origin O, the position vectors of points A and B are given by ` a ` a 3 2 −−→ −−→ OA = −6 and OB = −6 , p −7 and angle AOB = 90Å. (i) Find the value of p. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ −−→ −−→ The point C is such that OC = 2 OA. 3 −−→ (ii) Find the unit vector in the direction of BC. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2 3 6 OA p     = −       uuur and 2 6 7 OB     = −     −   uuur 2(i) Angle AOB = 90° → 6 + 36 −7p = 0 M1 Use of x1x2 + y1y2 + z1z2 = 0 or Pythagoras → p = 6 A1 Total: 2 Question Answer Marks Guidance 2(ii) 3 2 6 3 OC p     = −       uuur = 2 4 4     −       B1 FT CAO FT on their value of p BC uuur = c – b = 0 2 11           ; magnitude = √125 M1 M1 Use of c – b. Allow magnitude of b + c or b – c Allow first M1 in terms of p Unit vector = 0 1 2 125 11           A1 OE Allow ± and decimal equivalent θ θ +

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1 + cos 1 sin 1 2 3 (i) Prove the identity +  [3] sin 1 1 + cos 1 sin 1. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 1 + cos 1 sin 1 3 (ii) Hence solve the equation + = for 0Å ≤1 ≤360Å. [3] sin 1 1 + cos 1 cos 1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) 1 cos sin 2 sin 1 cos sin θ θ θ + ≡ + . ( ) ( ) 2 1 ² 1 c s s c + + + = ( ) 1 2 ² ² 1 c c s s c + + + + M1 Correct use of fractions = ( ) ( ) ( ) 2 1 2 2 1 1 c c s c s c + + = + + → 2 s M1 A1 Use of trig identity, A1 needs evidence of cancelling Total: 3 3(ii) 2 3 = s c → 2 3 t = M1 Use part (i) and t = s ÷ c, may restart from given equation → θ = 33.7° or 213.7° A1 A1FT FT for 180° + 1st answer. 2nd A1 lost for extra solns in range Total: 3

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4 (a) An arithmetic progression has a first term of 32, a 5th term of 22 and a last term of −28. Find the sum of all the terms in the progression. 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(b) Each year a school allocates a sum of money for the library. The amount allocated each year increases by 2.5% of the amount allocated the previous year. In 2005 the school allocated $2000. Find the total amount allocated in the years 2005 to 2014 inclusive. 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Mark scheme: 4(a) a = 32, a + 4d = 22, → d = −2.5 B1 a + (n – 1)d = −28 → n = 25 B1 S25 = ( ) 25 64 2.5 24 2 − × = 50 M1 A1 M1 for correct formula with n = 24 or n = 25 Total: 4 4(b) a = 2000, r = 1.025 B1 1 2.5% r = + ok if used correctly in Sn formula S10 = 2000( 10 1 .025 1 1.025 1 − − ) = 22400 or a value which rounds to this M1 A1 M1 for correct formula with n = 9 or n = 10 and their a and r SR: correct answer only for n = 10 B3, for n = 9, B1 (£19 900) Total: 3

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6 The horizontal base of a solid prism is an equilateral triangle of side x cm. The sides of the prism are vertical. The height of the prism is h cm and the volume of the prism is 2000 cm3. (i) Express h in terms of x and show that the total surface area of the prism, A cm2, is given by ï3 24 000 A = x2 + x−1. [3] 2 ï3 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Given that x can vary, find the value of x for which A has a stationary value. 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(iii) Determine, showing all necessary working, the nature of this stationary value. 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Mark scheme: 6(i) Volume = 1 3 ² 2 2 x h       = 2000 → h = 8000 3 ²x √ M1 Use of (area of triangle, with attempt at ht) × h =2000, f h x = A = ( ) 2 1 3 3 2 2 2 xh x   + × × ×     M1 Uses 3 rectangles and at least one triangle Sub for h → 2 1 3 24 000 2 3 A x x− √ = + A1 AG Total: 3 6(ii) 2 d 3 24000 2 d 2 3 A x x x − = − B1 CAO, allow decimal equivalent = 0 when x³ = 8000 → x = 20 M1 A1 Sets their d d A x to 0 and attempt to solve for x Total: 3 Question Answer Marks Guidance 6(iii) 3 d² 3 48000 2 d ² 2 3 A x x − = + > 0 M1 Any valid method, ignore value of d² d ² A x providing it is positive → Minimum A1 FT FT on their x providing it is positive Total: 2

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8 X M A B 12 cm 10 cm O In the diagram, OAXB is a sector of a circle with centre O and radius 10 cm. The length of the chord AB is 12 cm. The line OX passes through M, the mid-point of AB, and OX is perpendicular to AB. The shaded region is bounded by the chord AB and by the arc of a circle with centre X and radius XA. (i) Show that angle AXB is 2.498 radians, correct to 3 decimal places. 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(ii) Find the perimeter of the shaded region. 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(iii) Find the area of the shaded region. 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Mark scheme: 8(i) Letting M be midpoint of AB OM = 8 (Pythagoras) → XM = 2 B1 (could find √40 and use sin−1or cos−1) tan AXM = 6 2 AXB = 2tan−13 = 2.498 M1 A1 AG Needs × 2 and correct trig for M1 (Alternative 1: 6 sin , 0.6435, 0.6435 10 AOM AOM AXB π = = = − ) (Alternative 1: Use of isosceles triangles, B1 for AOM, M1,A1 for completion) (Alternative 2: Use of circle theorem, B1 for AOB, M1,A1for completion) Total: 3 8(ii) AX = √(6² + 2²) = √40 B1 CAO, could be gained in part (i) or part (iii) Arc AYB = rθ = √40 × 2.498 M1 Allow for incorrect √40 (not 6 1 2 1 0 r or or = ) Perimeter = 12 + arc = 27.8 cm A1 Total: 3 8(iii) area of sector AXBY = ½ × (√40)² × 2.498 M1 Use of ½r²θ with their r , (not 6 10 r or r = = ) Area of triangle AXB = ½ × 12 × 2, Subtract these → 38.0 cm² M1 A1 Use of ½bh and subtraction. Could gain M1 with 10 r = . Total: 3

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10 y 4 y = 5 −3x x O 1 4 The diagram shows part of the curve y = 5 −3x. (i) Find the equation of the normal to the curve at the point where x = 1 in the form y = mx + c, where m and c are constants. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The shaded region is bounded by the curve, the coordinate axes and the line x = 1. (ii) Find, showing all necessary working, the volume obtained when this shaded region is rotated through 360Å about the x-axis. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 10(i) ( ) d 4 d 5 3 ² y x x = − × (−3) B1 B1 B1 without ×(−3) B1 For ×(−3) Gradient of tangent = 3, Gradient of normal – ⅓ *M1 Use of m1m2 = −1 after calculus → eqn: ( ) 1 2 1 3 y x − = − − DM1 Correct form of equation, with (1, their y), not (1,0) → 1 7 3 3 y x = − + A1 This mark needs to have come from y = 2, y must be subject Total: 5 10(ii) Vol = π ( ) 1 0 16 d 5 3 ² x x − ∫ M1 Use of ²d V y x π = ∫ with an attempt at integration π ( ) 16 3 5 3x   − ÷ −   −     A1 A1 A1 without( ÷ −3), A1 for (÷ −3) = ( π 16 16 6 15   −     ) = 8 5 π (if limits switched must show – to +) M1 A1 Use of both correct limits M1 Total: 5

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