Cambridge A Level Mathematics 9709 — 2014 Oct/Nov Paper 1 · Variant 3

1 In the expansion of 2 + ax 6, the coefficient of x2 is equal to the coefficient of x3. Find the value of the non-zero constant a. [4]

Mark scheme: 3 × (ax )3 B1B11 (15 or 4 × (ax )2 , (20 or 16 C 2 ) × 2 6 C 3 ) × 2 15 × 2 4 3 a = = M1A1 240 a = 160 a is M0 3 20 × 2 2 [4] 3 π

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2 C 1 rad B 3 D O 3 cm A In the diagram, OADC is a sector of a circle with centre O and radius 3 cm. AB and CB are tangents to the circle and angle ABC = 1 radians. Find, giving your answer in terms of 3 and , 3 (i) the perimeter of the shaded region, [3] (ii) the area of the shaded region. [3]

Mark scheme: 3 π 2 (i) CB or AB = or 3 tan B1 Allow throughout for e.g. 3 3 , π 3 tan 3 9 3 6 3 , B1 27, 3 , ( ) 3  2π π  Arc or AC = 3 × or (= 2π or π ) B1 After B0B0 SCB1 for 16.7  3 3  [3] Perimeter = 6 3 + 2π oe B1 Their AB in form k√3 1 (ii) Area OABC (2 ) × × 3 × their AB 2 9 3 (=9√3 or ) B1 2 1 2  2π π  3π  Area OADC × 3 ×  or  = 3π or  B1 After B0B0 SCB1 for 6.16 or 6.17. 2  2 3   2  5 Allow ( 3 ) − 3π Shaded area 9 3 − 3π oe [3] ( )2 B1B1B1 F ith f 1st 2 k b k t

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3 (i) Express 9x2 −12x + 5 in the form ax + b 2 + c. [3] (ii) Determine whether 3x3 −6x2 + 5x −12 is an increasing function, a decreasing function or neither. [3]

Mark scheme: 3 (i) (3 x − 2 ) 2 + 1 B1B1B1 For either of 1st 2 marks bracket must be in the form (ax + b )2 except for 2  2  SCB2 for 9 x −  + 1  3  [3] (ii) f ′( x ) = 9 x 2 − 12 x + 5 B1 = their (3 x − 2 ) 2 + 1 M1 Ft from (i). Some > 0 (or > 1) hence an increasing function A1 reference/recognition [3] Allow > 1. Allow their 1 provided positive. Allow a complete alt method (2/2 or 0/2) 2 3

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4 Three geometric progressions, P, Q and R, are such that their sums to infinity are the first three terms respectively of an arithmetic progression. Progression P is 2, 1, 12, 14, . Progression Q is 3, 1, 13, 19, . (i) Find the sum to infinity of progression R. [3] (ii) Given that the first term of R is 4, find the sum of the first three terms of R. [3]

Mark scheme: 2 3 4 (i) S P = , S P = M1 At least one correct 1 1 1 − 1 − 2 3 9 S P = ,4 S Q = A1 At least one correct 2 S R = 5 cao A1 [3] 4 (ii) = their S R M1 1 − r 1 r = A1 5 4 4 24 R = 4 + + = 4 or .496 cao A1 5 25 25 [3] ( 2 2 )( 2 2 ) 2 ( 2 ) 2 ( 2 ) 4 2 2

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5 (i) Show that sin4 −cos4 2 sin2 −1. [3] (ii) Hence solve the equation sin4 −cos4 = 1 for 0 ≤ ≤360 . [4] 2

Mark scheme: M1 OR sin4θ – (1 – sin2θ)25 (i) ( s 2 − c 2 )( s 2 + c 2 ) OR s 2 (1 − c 2 ) − c 2 (1 − s 2 ) sin2θ – cos2θ A1 sin4θ – (1 – 2sin2θ + sin4θ) 2sin2θ – 1 www AG A1 = 2sin2θ – 1 AG [3] 1 3 1 (ii) 2sin2θ – 1 = ⇒ sinθ = (± ) or (± ).0 866 B1 OR cos 2θ = − → 2θ = 120, 240 2 2 2 etc. B1 Ft for 180 – their 60 θ = 60 ° B1 Ft for 180 + their 60, 360 – their θ =120 ° 60 π 2π B1 Allow , etc. Extra sols in θ = 240°, 300° 3 3 [4] range −1 3a + 9 − (2 a − 1) a + 10 −a − 10 ll i i f b k

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6 A is the point a, 2a −1 and B is the point 2a + 4, 3a + 9 , where a is a constant. (i) Find, in terms of a, the gradient of a line perpendicular to AB. [3] (ii) Given that the distance AB is 260 , find the possible values of a. [4]

Mark scheme: 3a + 9 (2 a 1) a + 10 a 10 6 (i) m = = oe e.g. M1A1 cao Allow omission of brackets 2 a + 4 − a a + 4 −a − 4 for M1 − (a + 4 ) Gradient of perpendicular = oe but A1 Do not ISW. Max penalty for a + 10 erroneous cancellation 1 mark − 1 not  a + 10     a + 4  [3] (ii) (√)[(a + 4)2 + (a + 10)2] = (√)260 M1 Allow their (a + 4), (a +10) from (i). Allow (–a – 4)2 etc. Allow omission of brackets (√)[(a + 4)2 + (a + 10)2] cao A1 (2 )(a 2 + 14 a − 72 ) (= 0 ) A1 A1 a = 4 or − 18 cao [4] 2

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−−→ −−→ 7 Three points, O, A and B, are such that OA = i + 3j + pk and OB = −7i + 1 −p j + pk, where p is a constant. −−→ −−→ (i) Find the values of p for which OA is perpendicular to OB. [3] −−→ −−→ (ii) The magnitudes of OA and OB are a and b respectively. Find the value of p for which b2 = 2a2. [2] −−→ (iii) Find the unit vector in the direction of AB when p = −8. [3]

Mark scheme: 7 (i) OA.OB = − 7 + 3 − 3 p + p 2 M1 Correct method for scalar product ( p + 1)( p − 4 ) = 0 DM1 Equate to zero & attempt to factorise/solve p = −1 or 4 A1 ‘= 0’ implied by answers [3] 2 2 (ii) 49 + (1 − p ) + p 2 = 2 (1 + 9 + p ) M1 Scalar result required p = 15 A1 [2] (iii) AB = −8i + 6j B1 p = 15 used – treat as MR Divide AB by│AB│ = (− 8 ) 2 + 6 2 = 10 soi  − 8  M1 1   →  − 17  1 Unit vector = (− 8i + 6 )j oe cao A1 353   0 10   [3]

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8 A curve y = f x has a stationary point at 3, 7 and is such that f ′′ x = 36x−3. (i) State, with a reason, whether this stationary point is a maximum or a minimum. [1] (ii) Find f ′ x and f x . [7]

Mark scheme: 8 (i) Minimum since f ″(3) (= 4/3) > 0 www B1 [1] (ii) f ′( x ) = −18 x − 2 (+ c ) B1 0 = − 2 + c M1 Sub f 3 0. (dep c present) c = 2 (→ f ′( x ) = − 18 x − 2 + 2 ) A1 c = 2 sufficient at this stage f ( x ) = 18 x −1 + 2 x (+ k ) B1 B1 Allow cx at this stage 7 = 6 + 6 + k M1 Sub f(3) = 3 (k present & numeric −1 (or no) c) k = −→5 ( f ( x ) = 18 x + 2 x − 5 ) cao A1 [7] 2 2 ( ) ( )2

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9 y B y = 3 x y = x + 2 A x O The diagram shows parts of the graphs of y = x + 2 and y = 3 x intersecting at points A and B. (i) Write down an equation satisfied by the x-coordinates of A and B. Solve this equation and hence find the coordinates of A and B. [4] (ii) Find by integration the area of the shaded region. [6] [Question 10 is printed on the next page.]

Mark scheme: 2 9 (i) x − 3 x + 2 or k 2 − 3k + 2 or (3 x ) = ( x + 2 )2 M1 OR attempt to eliminate x eg sub y 2 x = 9 x = 1 or 2 or k = 1 or 2 or x 2 −x5 + 4 (= 0 ) A1 y 2 − 9 y + 18 = 0 y = 3 or 6 x = 1 or 4 A1 x = 1 or 4 y = 3 or 6 A1 [4] 1 dx –  M1DM1 Attempt to integrate. Subtract at (ii) ∫ 3 x  ∫ ( x + 2 ) dx or attempt at trapezium  2 some stage 3  1 2  1  2x −  x + 2 x  or ( y 2 + y1 )( x 2 − x1 A1A1 Where ( x1 , y 1 ), ( x 2 , y 2 ) is their 2  2  2 ) (1, 3), (4, 6)    1   1  (16 − 2) –  (8 + 8 ) −  + 2  or their × 9 × 3 DM1 Apply their 1→4 limits correctly    2   2  to curve 1 A1 For A mark allow reverse subtn→ 2 −1 →1 but not reversed limits [6] 2 2 OR y 2  M1DM1  ∫ ( y − 2 ) d y or attempt at trap  − ∫ 9 d y  1 2 1  y 3 y − 2 y or ( x1 + x 2 )( y 2 − y1 ) − A1A1  2 2  27   1  1  (18 − 12 ) −  4 − 6  or × 5 × 3 − [8 − ]1 DM1 Apply their 3→6 limits correctly   2  2  to curve 1 A1 2 1 2

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10 (a) The functions f and g are defined for x ≥0 by 1 f : x → ax + b 3, where a and b are positive constants, g : x →x2. Given that fg 1 = 2 and gf 9 = 16, (i) calculate the values of a and b, [4] (ii) obtain an expression for f −1 x and state the domain of f −1. [4] 1 (b) A point P travels along the curve y = 7x2 + 1 3 in such a way that the x-coordinate of P at time t minutes is increasing at a constant rate of 8 units per minute. Find the rate of increase of the y-coordinate of P at the instant when P is at the point 3, 4 . [5]

Mark scheme: 1 2 3 = 16 B1B1 Ignore 2nd soln (–9, 17) throughout10 (a) (i) ( a + b ) 3 = 2, ( 9 a + b ) a + b = 8, 9a + b = 64 M1 Cube etc. & attempt to solve A1 Correct answers without any a = 7, b = 1 working 0/4 [4] 1 (ii) x = (7 y + 1)3 (x/y interchange as first or last B1 ft on from their a, b or in terms of a, b step) x 3 = 7 y + 1 or y 3 = 7 x + 1 B1 ft on from their a, b or in terms of a, b −1 1 3 f ( x ) = (x − )1 cao B1 A function of x required 7 Domain of f − 1 is x > 1 cao B1 Accept >. Must be x [4] 2 1   dy 3 (b) = × [14 x ] B1B1 ( 7 x 2 + 1) −   d x  3  dy 1 − 2 7  When x = 3 , = × (64 ) 3 × 42 =  M1 dx 3  8  dy dy dx 7 = × = × 8 DM1 Use chain rule dt dx dt 8 A1 7 [5]

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