Cambridge A Level Mathematics 9709 — 2017 Oct/Nov Paper 1 · Variant 3

1 An arithmetic progression has first term −12 and common difference 6. The sum of the first n terms exceeds 3000. Calculate the least possible value of n. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 ½ 24 1 6 ~ 3000 − + −    n n Note: ~ denotes any inequality or equality ( )( ) ( ) 2 3 5 1000 ~ 0 − − n n A1 Rearrange into a 3-term quadratic. ( ) ~ 34.2 & 29.2 − n A1 35. Allow 35 n. A1 4 *M1 Rearrange into a 3-term quadratic.

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2 Find the set of values of a for which the curve y = −2 and the straight line y = ax + 3a meet at two x distinct points. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 ( ) 2 2 3 3 2 0 ax a ax ax x + = − → + + = Apply 2 4 0 − > b ac SOI DM1 Allow ⩾. If no inequalities seen, M1 is implied by 2 correct final answers in a or x. a < 0, a > 8 9 (or 0.889) OE A1 A1 For final answers accept 0 > a > 8 9 but not ⩽, ⩾. 4

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cos 1 + 4 5 (i) Show that the equation + 5 sin 1 −5 = 0 may be expressed as 5 cos21 −cos 1 −4 = 0. sin 1 + 1 [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ cos 1 + 4 (ii) Hence solve the equation + 5 sin 1 −5 = 0 for 0Å ≤1 ≤360Å. [4] sin 1 + 1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) 2 cos 4 5sin 5sin 5sin 5 0 θ θ θ θ + + + − − = is not seen ( ) ( ) 2 5 1 cos cos 1 0 θ θ − + − = M1 Use 2 2 1 = − s c 2 5cos cos 4 0 θ θ − − = AG A1 Rearrange to AG 3 5(ii) cos 1 and 0.8 θ = − B1 Both required [ ] [ ] [ ] 0 , 360 , 143.1 , 216.9 θ = ° ° ° ° B1 B1 B1 FT Both solutions required for 1st mark. For 3rd mark FT for ( ) 360 1 43.1 °− ° their Extra solution(s) in range (e.g. 180º) among 4 correct solutions scores 3 4 4

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7 A 5 B 3 D P Q C The diagram shows a rectangle ABCD in which AB = 5 units and BC = 3 units. Point P lies on DC and AP is an arc of a circle with centre B. Point Q lies on DC and AQ is an arc of a circle with centre D. (i) Show that angle ABP = 0.6435 radians, correct to 4 decimal places. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Calculate the areas of the sectors BAP and DAQ. 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(iii) Calculate the area of the shaded region. 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Mark scheme: 7(i) 1 3 sin 0.6435 5 − =     AG M1 OR ( ) ( ) 1 3 cos 0.9273 0.9273 0.6435 5 2 PBC ABP π −  = = ⇒ = − =     Or other valid method. Check working and diagram for evidence of incorrect method 7(ii) Use (once) of sector area 2 ½ θ = r M1 Area sector 2 ½ 5 0.6435 = × × BAP = 8.04 A1 Area sector 2 ½ ½ 3 π = × × DAQ = 7.07 , Allow 9 4 π A1 3 Question Answer Marks Guidance 7(iii) EITHER: Region = sect + sect ‒ (rect ‒ ∆) or sect ‒ [rect ‒ (sect + ∆)] (M1 Use of correct strategy (Area ∆ BPC =) ½ × 3 × 4 = 6 Seen A1 8.04 + 7.07 ‒ (15 ‒ 6) = 6.11 A1) OR1: Region = sector ADQ ‒ (trap ABPD ‒ sector ABP). (M1 Use of correct strategy (Area trap ABPD = ) ½ (5 + 1) ×3 = 9 Seen A1 7.07 ‒ (9 ‒ 8.04) = 7.07 ‒ 0.96 = 6.11 A1) OR2: Area segment AP= 2.5686 Area segment AQ = 0.5438 Region = segment AP + segment AQ + ∆APQ. (M1 Use of correct strategy (Area ∆APQ =) ½ × 2 × 3 = 3 Seen A1 2.57 + 0.54 + 3 = 6.11 A1) 3

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8 y A y = 3 −2x B y = 4 −3 x x O The diagram shows parts of the graphs of y = 3 −2x and y = 4 −3 x intersecting at points A and B. (i) Find by calculation the x-coordinates of A and B. 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(ii) Find, showing all necessary working, the area of the shaded region. 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Mark scheme: 8(i) EITHER: 4 ‒ 3√x = 3 ‒ 2x → 2x ‒ 3√x + 1 (=0) or e.g. 2k2‒3k + 1 (=0) ½, 1 x = A1 Or ½ or 1 = k (where k = √x). ¼, 1 = x A1) OR1: ( ) 2 2 (3 ) 1 2 = + x x (M1 2 4 5 1 ( − + x x =0) A1 ¼, 1 = x A1) OR2: 2 3 4 2 3 − −   =     y y ( ) ( ) 2 2 7 5 0 → − + = y y (M1 Eliminate x y = 5 2 , 1 A1 ¼, 1 = x A1) 3 Question Answer Marks Guidance 8(ii) EITHER: Area under line = ( ) 2 3 2 d 3 ∫ − = − x x x x (B1 ( ) 3 1 3 1 4 16     = − − −         M1 Apply their limits (e.g. ¼ → 1) after integn. Area under curve ( ) 1/2 3/2 4 3 d 4 2 x x x x = ∫ − = − B1 ( ) ( ) 4 2 1 ¼  − − −    M1 Apply their limits (e.g. ¼ → 1) after integration. Required area = 21 16 ‒ 5 4 = 1 16 (or 0.0625) A1) OR: +/‒ ( ) 1 1 2 2 3 2 4 3 / ( 1 2 3 )   ∫ − − − = + −∫−− +       x x x x (*M1 Subtract functions and then attempt integration +/‒ 3/2 2 3 3 / 2   −− +     x x x A2, 1, 0 FT FT on their subtraction. Deduct 1 mark for each term incorrect +/‒ 1 1 1 1 1 1 2 4 16 8 16     −−+ −− + + =         (or 0.0625) DM1 A1) Apply their limits ¼ → 1 5

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9 Relative to an origin O, the position vectors of the points A, B and C are given by ` a ` a ` a 8 −10 2 −−→ −−→ −−→ OA = −6 , OB = 3 and OC = −3 . 5 −13 −1 −−→ −−→ −−→ A fourth point, D, is such that the magnitudes AB , BC and CD are the first, second and third terms respectively of a geometric progression. −−→ −−→ −−→ (i) Find the magnitudes AB , BC and CD . [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Given that D is a point lying on the line through B and C, find the two possible position vectors of the point D. 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Mark scheme: 9(i) 18 12 / 9 , / 6 , 18 12 AB BC −         = + − = + −−         −     JJJG JJJG 27, 18 AB BC = = JJJG JJJG B1 FT B1 FT FT on their AB JJJG , theirOD JJJG . 18 18 27 CD   = ×     JJJG OR 2 18 27  ×     27 = 12 B1 5 9(ii) ( ) 18 27 CD their their = ± × JJJG BC JJJG SOI M1 Expect ( ) 8 4 8     ± −       . ( ) 2 12 10 6 18 3 6 7 , 1 27 1 12 7 9 OD their −               = − ± − = −               − −        JJJG M1 A1 A1 Other methods possible for OD JJJG , e.g. OB JJJG + 5 2 CD JJJG , OB JJJG + 1 2 CD JJJG (One soln M2A1, 2nd soln A1) OR OB JJJG + 5 3 BC JJJG , OB JJJG + 1 3 BC JJJG (One soln M2A1, 2nd soln A1) 4

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11 y 1 y = x −1 2 B 5, 2 x O A 1, 0 1 The diagram shows the curve y = x −1 2 and points A 1, 0 and B 5, 2 lying on the curve. (i) Find the equation of the line AB, giving your answer in the form y = mx + c. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find, showing all necessary working, the equation of the tangent to the curve which is parallel to AB. 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(iii) Find the perpendicular distance between the line AB and the tangent parallel to AB. Give your answer correct to 2 decimal places. 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Mark scheme: 11(i) Gradient of AB = 1 2 B1 Equation of AB is y = 1 2 x – 1 2 B1 2 11(ii) d d y x ( ) 1 2 ½ 1 x − = − B1 ( ) 1 2 ½ 1 ½ x − − = . Equate their d d y x to their ½ *M1 2, 1 = = x y A1 y ‒ 1 = ½(x ‒ 2) (thro' their(2,1) & their ½) → ½ = y x DM1 A1 5 Question Answer Marks Guidance 11(iii) EITHER: sin sin 1 d d θ θ = → = (M1 Where θ is angle between AB and the x-axis gradient of ( ) ½ tan ½ 26.5 7 AB θ θ = ⇒ = ⇒ = ° B1 ( ) sin26.5 7 0.45 = ° = d (or 1 5 ) A1) OR1: Perpendicular through O has equation 2 = − y x (M1 Intersection with AB: 1 2 2 ½ ½ , 5 5 −   − = − →    x x A1 2 2 1 2 0.45 5 5     = + =         d (or 1 5 ) A1) OR2: Perpendicular through (2, 1) has equation 2 5 = − + y x (M1 Intersection with AB: 11 3 2 5 ½ ½ , 5 5   − + = − →    x x A1 2 2 1 2 5 5     = +         d = 0.45 (or 1/√5) A1) Question Answer Marks Guidance 11(iii) OR3: OAC ∆ has area 1 4 [where C = (0, 1 2 − )] (B1 1 2 × 5 2 × d = 1 4 → d = 1 5 M1 A1) 3

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