Cambridge A Level Mathematics 9709 — 2019 Oct/Nov Paper 1 · Variant 3

3 The equation of a curve is y = x3 + x2 −8x + 7. The curve has no stationary points in the interval a < x < b. Find the least possible value of a and the greatest possible value of b. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 d d y x = 3x2 + 2x ‒ 8 B1 Set to zero (SOI) and solve M1 (Min) a = ‒2, (Max) b = 4/3. – in terms of a and b. A1 A1 Accept 4 3 2, a b − . - SC: A1 for 4 3 2, a b > − < or for 4 3 2 x −< < 4

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4 C A r O r B The diagram shows a semicircle ACB with centre O and radius r. Arc OC is part of a circle with centre A. (i) Express angle CAO in radians in terms of 0. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the area of the shaded region in terms of r, 0 and ï3, simplifying your answer. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) Angle CAO = π 3 B1 1 4(ii) (Sector AOC) = 2 3 1 2 π r their × M1 SOI (∆ ABC) = ( )( ) π 1 2 sin 2 3 r r their       or ( )( ) 1 3 2 2 2 r r or ( )( ) 1 3 2 r r M1 For M1M1, π 3 their must be of the form kπ where 0 < k < ½ (∆ ABC) = ( )( ) π 1 3 2 sin 2 r r       or ( )( ) 1 3 2 2 2 r r or ( )( ) 1 3 2 r r A1 All correct 2 2 2 π 3 3 1 2 r r     −           A1 4

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5 x cm 4x cm 2x cm The dimensions of a cuboid are x cm, 2x cm and 4x cm, as shown in the diagram. (i) Show that the surface area S cm2 and the volume V cm3 are connected by the relation 2 S = 7V 3. 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(ii) When the volume of the cuboid is 1000 cm3 the surface area is increasing at 2 cm2 s−1. Find the rate of increase of the volume at this instant. 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Mark scheme: 5(i) B1B1 SOI 2 2 3 7 7 4 V x S = × = B1 AG, WWW 3 5(ii) 1 3 d 14 14 d 3 30 S V V −  = =     SOI when V = 1000 *M1 A1 Attempt to differentiate For M mark d d S V       to be of form 1 3 kV − d d d d d d V S V t t S   = ×     OE used with d 2 d S t = and 14 30 1 their DM1 30 7 or 4.29 A1 OE Alternative method for question 5(ii) 3 2 1 2 d 3 1 30 d 2 14 7 7 7 7 S V V S S   = → = × × =     SOI when S = 700 *M1 A1 Attempt to differentiate For M mark 1 2 d to be of form d V kS S       d d d d d d V S V t t S   = ×     OE used with d 2 d S t = and 14 30 1 their DM1 30 7 or 4.29 A1 OE Question Answer Marks Guidance 5(ii) Alternative method for question 5(ii) Attempt to find either d d V x or d d and d d S V x S       together with either d d x t or x *M1 d d V x = 24x2 or d d 3 56 and d d 7 S V x x x S   = =     , d d x t = 1 140 or x = 5 (A1) A1 Correct method for d d V t DM1 30 7 or 4.29 A1 OE 4

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6 A line has equation y = 3kx −2k and a curve has equation y = x2 −kx + 2, where k is a constant. (i) Find the set of values of k for which the line and curve meet at two distinct points. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) For each of two particular values of k, the line is a tangent to the curve. Show that these two tangents meet on the x-axis. 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Mark scheme: 6(i) 2 2 3 2 2 4 2 2 0 kx k x kx x kx k − = − + → − + + = B1 kx terms combined correctly-implied by correct 2 4 b ac − Attempt to find 2 4 b ac − M1 Form a quadratic equation in k 1 and 1 2 − A1 SOI 1 2 1, k k > < − A1 Allow 1, 1/ 2 x x > < − 4 6(ii) 3 3 2, 1 2 y x y x = − = − + M1 Use of their k values (twice) in 3 2 y kx k = − 3 3 2 1 2 x x −= − + OR 2 2 2 y y + = − M1 Equate their tangent equations OR substitute y = 0 into both lines x = 2 3 , → y = 0 in one or both lines A1 Substitute x = 2 3 in one or both lines 3

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7 (i) Show that the equation 3 cos41 + 4 sin21 −3 = 0 can be expressed as 3x2 −4x + 1 = 0, where x = cos21. 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(ii) Hence solve the equation 3 cos41 + 4 sin21 −3 = 0 for 0Å ≤1 ≤180Å. 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Mark scheme: 7(i) 4 2 3cos 4 1 cos 3 0 θ θ + − − = M1 Use 2 2 1 s c = − ( ) ( ) ( ) 2 2 3 4 1 3 0 3 4 1 0 x x x x + − − = → − + = A1 AG 2 7(ii) Attempt to solve for x M1 Expect x = 1, 1/3 ( ) ( ) cos 1, 0.5774 θ = ± ± A1 Accept ( ) 1 3   ±     SOI (θ = ) 0º, 180º, 54.7º, 125.3º A3,2,1,0 A2,1,0 if more than 4 solutions in range 5

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9 The first, second and third terms of a geometric progression are 3k, 5k −6 and 6k −4, respectively. (i) Show that k satisfies the equation 7k2 −48k + 36 = 0. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find, showing all necessary working, the exact values of the common ratio corresponding to each of the possible values of k. 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(iii) One of these ratios gives a progression which is convergent. Find the sum to infinity. 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Mark scheme: 9(i) ( ) ( ) 2 5 6 6 4 5 6 3 6 4 3 5 6 k k k k k k k = → − = − − M1 OR any valid relationship 2 2 2 25 60 36 18 12 7 48 36 k k k k k k − + = − → − + A1 AG 2 9(ii) k = 6 7 , 6 B1B1 Allow 0.857(1) for 6 7 When k = 6 7 , r = 2 3 − B1 Must be exact When k = 6, r = 4 3 B1 4 9(iii) Use of 1 a S r ∞= − with 2 3 r their = − and 6 3 7 a their = × M1 Provided 0 < |their ‒2/3| < 1 18 2 54 1 7 3 35   ÷ + =     or 1.54 A1 FT if 0.857(1) has been used in part (ii). 2

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10 Relative to an origin O, the position vectors of the points A, B and X are given by ` a ` a ` a −8 10 −2 −−→ −−→ −−→ OA = −4 , OB = 2 and OX = −2 . 2 11 5 −−→ (i) Find AX and show that AXB is a straight line. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ` a 1 −−→ The position vector of a point C is given by OC = −8 . 3 (ii) Show that CX is perpendicular to AX. 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(iii) Find the area of triangle ABC. 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Mark scheme: 10(i) 6 18 12 12 2 , and one of 6 , 4 , 4 3 9 6 6 −               = = = = −               −        AX AB XB BX B1B1 State 3 = AB AX ( 3 or 2 or 2 = = XB AX AB XB etc) hence straight line OR . AX AB AX AB = 1 ( θ → = 0) or . AX BX AX BX = –1 ( θ → = 180) hence straight line B1 WWW A conclusion (i.e. a straight line) is required. 3 10(ii) 3 6 2 −     =       CX B1 . 18 12 6 = − + + CX AX M1 = 0 (hence CX is perpendicular to AX) A1 3 10(iii) 2 2 2 2 2 2 3 6 2 , 18 6 9 = + + = + + CX AB Both attempted M1 Area ∆ ABC = 1 2 × their 21 × their 7 = 73 1 2 M1A1 Accept answers which round to 73.5 3

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11 y A 2, 3 B y = x −1 −2 + 2 x O 1 3 The diagram shows part of the curve y = x −1 −2 + 2, and the lines x = 1 and x = 3. The point A on the curve has coordinates 2, 3 . The normal to the curve at A crosses the line x = 1 at B. (i) Show that the normal AB has equation y = 12x + 2. 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(ii) Find, showing all necessary working, the volume of revolution obtained when the shaded region is rotated through 360Å about the x-axis. 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Mark scheme: 11(i) ( ) 3 d 2 1 d y x x − = − − B1 When x = 2, m = ‒2 → gradient of normal = 1 m − M1 m must come from differentiation Equation of normal is ( ) 3 ½ 2 ½ 2 y x y x − = − → = + A1 AG Through (2, 3) with gradient 1 m − . Simplify to AG 3 Question Answer Marks Guidance 11(ii) ( ( ) ( ) ( ) 2 2 1 2 π) d , π d y x y x ∫ ∫ *M1 Attempt to integrate 2 y for at least one of the functions ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 1 2 4 4 2 π 2 or 2 4 π 1 4 1 4 x x x x x − − + + + − + − + ∫ ∫ A1A1 A1 for ( ) 2 1 2 2 x + depends on an attempt to integrate this form later ( ) ( ) ( ) ( ) ( ) 3 3 2 2 1 1 3 2 12 3 1 π 2 or 4 1 4 1 π 4 3 1 x x x x x x x − − + + + − − + + − −             A1A1 Must have at least 2 terms correct for each integral (π) 125 2 1 18 4 8 1 4 12 3 12 or − + + − + +             1 1 2 12 4 8 24 3 − − − + − − +             DM1 Apply limits to at least 1 integrated expansion Attempt to add 2 volume integrals (or 1 volume integral + frustum) π{ } 7 7 7 6 12 24 + DM1 13 7 8 π or 111π 8 or 13.9π or 43.6 A1 2 1 4 8 1 4 3 12 + + − + +       1 1 2 12 4 8 24 3 − − − + − − +       8

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