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

2 A function f is defined by f : x →x3 −x2 −8x + 5 for x < a. It is given that f is an increasing function. Find the largest possible value of the constant a. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 f′(x) 2 3 2 8 = − − x x M1 4 3 − , 2 SOI A1 f′(x) > 0 4 3 x ⇒ < − SOI M1 Accept x > 2 in addition. FT their solutions Largest value of a is 4 3 − A1 Statement in terms of a. Accept a ⩽ 4 3 − or 4 3 a < − . Penalise extra solutions 4

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3 (a) A geometric progression has first term 3a and common ratio r. A second geometric progression has first term a and common ratio −2r. The two progressions have the same sum to infinity. Find the value of r. 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(b) The first two terms of an arithmetic progression are 15 and 19 respectively. The first two terms of a second arithmetic progression are 420 and 415 respectively. The two progressions have the same sum of the first n terms. Find the value of n. 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Mark scheme: 3(i) 3 1 1 2 = − + a a r r M1 Attempt to equate 2 sums to infinity. At least one correct 3 6 1 + = − r r DM1 Elimination of 1 variable (a) at any stage and multiplication 2 7 r = − A1 3 3(ii) ( ) ] [ ( )( ) ½ 2 15 1 4 ½ 2 420 1 5   × + − = × + − −   n n n n M1A1 Attempt to equate 2 sum to n terms, at least one correct (M1). Both correct (A1) 91 = n A1 3

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4 Machines in a factory make cardboard cones of base radius r cm and vertical height h cm. The volume, V cm3, of such a cone is given by V = 130r2h. The machines produce cones for which h + r = 18. (i) Show that V = 60r2 −130r3. 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(ii) Given that r can vary, find the non-zero value of r for which V has a stationary value and show that the stationary value is a maximum. 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(iii) Find the maximum volume of a cone that can be made by these machines. 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Mark scheme: 4(i) ( ) 2 2 3 1 1 18 6 3 3 V r r r r π π π = − = − B1 1 4(ii) 2 d 12 0 d V r r r π π = − = M1 Differentiate and set = 0 ( ) 12 0 12 π − = → = r r r A1 2 2 d 12 2 d V r r π π = − M1 Sub r = 12 → 12 24 12 MAX π π π − = − → A1 AG 4 4(iii) Sub 12, 6 Max 288 or 905 π = = → = r h V B1 1

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5 B 10 cm 10 cm A D E C 16 cm The diagram shows an isosceles triangle ABC in which AC = 16 cm and AB = BC = 10 cm. The circular arcs BE and BD have centres at A and C respectively, where D and E lie on AC. (i) Show that angle BAC = 0.6435 radians, correct to 4 decimal places. 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(ii) Find the area of the shaded region. 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Mark scheme: 5(i) B1 AG Allow other valid methods e.g. sin 6 /10 A 1 5(ii) EITHER: Area ½ 16 6 or ½ 10 16sin0.6435 48 ∆ = × × × × = ABC (M1A1 Area 1 sector 2 ½ 10 0.6435 × × M1 Shaded area 2 sector = × − ∆ their their ABC M1) OR: 12, 30 ∆ = ∆ = BDE BDC (B1 B1 Sector = 32.18 M1 2×segment + ∆BDE M1) =16.4 A1 5

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6 The points A 1, 1 and B 5, 9 lie on the curve 6y = 5x2 −18x + 19. (i) Show that the equation of the perpendicular bisector of AB is 2y = 13 −x. 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The perpendicular bisector of AB meets the curve at C and D. O@p A (ii) Find, by calculation, the distance CD, giving your answer in the form , where p and q are q integers. 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Mark scheme: 6(i) Mid-point of AB = (3, 5) B1 Answers may be derived from simultaneous equations Gradient of AB = 2 B1 Eqn of perp. bisector is ( ) 5 ½ 3 − = − − y x → 2 13 = − y x M1A1 AG For M1 FT from mid-point and gradient of AB 4 6(ii) ( )( )( ) 2 2 3 39 5 18 19 5 3 4 0 − + = − + → − − = x x x x x M1 Equate equations and form 3-term quadratic 4 or 1 = − x A1 4½ or 7 = y A1 2 2 2 5 2½ 12 4 5 CD CD = + → = M1A1 Or equivalent integer fractions ISW 5

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8 (a) Relative to an origin O, the position vectors of two points P and Q are p and q respectively. The point R is such that PQR is a straight line with Q the mid-point of PR. Find the position vector of R in terms of p and q, simplifying your answer. 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(b) The vector 6i + aj + bk has magnitude 21 and is perpendicular to 3i + 2j + 2k. Find the possible values of a and b, showing all necessary working. 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Mark scheme: 8(a) EITHER: ( ) 2 2 = = − JJJG JJJG PR PQ q p (B1 JJJG OR =p+2q‒2p = 2q‒p M1A1) OR: = JJJG JJJG QR PQ = q ‒ p (B1 JJJG OR = + JJJG JJJG OQ QR =q+q‒p=2q‒p M1A1) Or other valid method 3 8(b) 2 2 2 2 6 21 + + = a b SOI B1 18 2 2 0 + + = a b B1 ( ) 2 2 9 405 + −− = a a M1 Correct method for elimination of a variable. (Or same equation in b) ( )( )( ) 2 2 9 162 0 + − = a a A1 Or same equation in b 9 or 18 = − a A1 18 or 9 = − b A1 6

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10 y y = 12 x4 −1 x O 1 −1 2 The diagram shows part of the curve y = 1 x4 −1 , defined for x ≥0. 2 (i) Find, showing all necessary working, the area of the shaded region. 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(ii) Find, showing all necessary working, the volume obtained when the shaded region is rotated through 360Å about the x-axis. 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(iii) Find, showing all necessary working, the volume obtained when the shaded region is rotated through 360Å about the y-axis. 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Mark scheme: 10(i) Area ( ) 5 4 ½ 1 d ½ 5   = ∫ − = −     x x x x *B1 ( ) 1 2 ½ 1 0 5 5   − − = −     DM1A1 Apply limits 0→1 3 10(ii) Vol ( ) ( ) 2 8 4 d ¼ 2 1 d π π = ∫ = ∫ − + y x x x x M1 (If middle term missed out can only gain the M marks) ( ) 9 5 2 ¼ 9 5 π   − +     x x x *A1 ( ) 1 2 ¼ [ 1 0 9 5 π   − + −     DM1 8 45 π or 0.559 A1 4 Question Answer Marks Guidance 10(iii) Vol ( ) ( ) 1/2 2d 2 1 d π π = ∫ = ∫ + x y y y M1 Condone use of x if integral is correct ( ) ( ) [ ] 3/2 2 1 2 3 / 2 π   +  ÷     y *A1A1 Expect ( ) ( ) 3/2 2 1 3 π   +       y ( ) 1 0 3 π   −     DM1 3 π or 1.05 A1 Apply 1 0 2 − → 5

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