Cambridge A Level Mathematics 9709 — 2013 Oct/Nov Paper 1 · Variant 1

3 D 3 C B k E j 4 O i 6 A The diagram shows a pyramid OABCD in which the vertical edge OD is 3 units in length. The point E is the centre of the horizontal rectangular base OABC. The sides OA and AB have lengths of 6 units −−→ −−→ −−→ and 4 units respectively. The unit vectors i, j and k are parallel to OA, OC and OD respectively. −−→ −−→ (i) Express each of the vectors DB and DE in terms of i, j and k. [2] (ii) Use a scalar product to find angle BDE. [4]

Mark scheme: 3 (i) DB = 6i + 4j – 3k cao B1 DE = 3i +2j – 3k cao B1 [2] (ii) DB.DE = 18 + 8 + 9 = 35 M1 Use of x1 x 2 + y1 y 2 + z1 z 2 │DB│= √61 or │DE│= √22 M1 Correct method for moduli 35 = 61 × 22 × cos θ oe M1 All connected correctly θ = 17 2. ° (0.300 rad) cao A1 Use of e.g. BD. DE can score M [4] marks (leads to obtuse angle) 2 2 2 ( )

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4 (i) Solve the equation 4 sin2x + 8 cos x −7 = 0 for 0Å ≤x ≤360Å. [4] (ii) Hence find the solution of the equation 4 sin2 121 + 8 cos 121 −7 = 0 for 0Å ≤1 ≤360Å. [2]

Mark scheme: 2 + s 2 = 14 (i) 4 (1 − cos 2 x ) + 8 cos x − 7 = 0 M1 Use c 4c 2 − 8c + 3 = 0 → (2 cos x − 1)(2 cos x − 3 ) = 0 M1 Attempt to solve x = 60 ° or 300 ° A1A1 [4] (ii) 1 θ = 60 ° (or 300 °) M1 Allow 300° in addition 2 θ = 120 ° only A1 [2] ( ) 2 t

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6 B r A r O a rad E D C The diagram shows a metal plate made by fixing together two pieces, OABCD (shaded) and OAED (unshaded). The piece OABCD is a minor sector of a circle with centre O and radius 2r. The piece OAED is a major sector of a circle with centre O and radius r. Angle AOD is ! radians. Simplifying your answers where possible, find, in terms of !, 0 and r, (i) the perimeter of the metal plate, [3] (ii) the area of the metal plate. [3] It is now given that the shaded and unshaded pieces are equal in area. (iii) Find ! in terms of 0. [2]

Mark scheme: 6 (i) r (2π − α ) + 2 r α + 2 r B1B1 2πr + r α + 2 r B1 ft for rα instead of 2rα or omission 2r SC1 for 2 r α + 4 r . (Plate = shaded [3] part) (ii) 1 (2 r )2 α + πr 2 − 1 r 2α B1B1 Either B1 can be scored in (iii) 2 2 3r 2α 2 + πr B1 2 [3] (iii) πr 2 − 1 r 2α = 2 r 2α M1 For equating their 2 parts from (ii) 2 2 α = π A1 5 [2]

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8 x metres r metres The inside lane of a school running track consists of two straight sections each of length x metres, and two semicircular sections each of radius r metres, as shown in the diagram. The straight sections are perpendicular to the diameters of the semicircular sections. The perimeter of the inside lane is 400 metres. (i) Show that the area, A m2, of the region enclosed by the inside lane is given by A = 400r −0r2. [4] (ii) Given that x and r can vary, show that, when A has a stationary value, there are no straight sections in the track. Determine whether the stationary value is a maximum or a minimum. [5]

Mark scheme: 8 (i) A = 2 xr + πr 2 B1 2 x + 2πr = 400 (⇒ x = 200 − πr ) B1 A = 400 r − πr 2 M1A1 Subst & simplify to AG (www) [4] dA (ii) = 400 − 2πr B1 Differentiate dr = 0 M1 Set to zero and attempt to find r 200 r = oe A1 π x = 0 ⇒ no straight sections AG A1 d 2 A = −2π ( < 0 ) Max B1 Dep on − 2π , or use of other valid 2 dr [5] reason GCE AS/A LEVEL – October/November 2013 9709 11 10 ( )

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9 (a) In an arithmetic progression the sum of the first ten terms is 400 and the sum of the next ten terms is 1000. Find the common difference and the first term. [5] (b) A geometric progression has first term a, common ratio r and sum to infinity 6. A second geometric progression has first term 2a, common ratio r2 and sum to infinity 7. Find the values of a and r. [5]

Mark scheme: 10 9 (a) (2 a + 9 d ) = 400 oe B1 → 2 a + 9 d = 80 2 20 (2 a + 19 d ) = 1400 OR 2 10 [2(a + 10 d ) + 9 d ] = 1000 B1 → 2 a + 19 d = 140 or 2 a + 29 d = 200 2 d = 6 a = 13 M1A1A1 Solve sim. eqns both from S n [5] formulae a 2 a (b) = 6 = 7 B1B1 1 −r 1 −r 2 12(1 − r ) 1 − r 2 12 = 7 or = M1 Substitute or divide 1 − r 2 1 − r 7 5 r = or 0.714 A1 7 12 a = or 1.71(4) A1 Ignore any other solns for r and a 7 [5] dy [ ( )2 ] [ ] 2

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10 y y = (3 – 2 x )3 2(1 , 8) x O The diagram shows the curve y = 3 −2x 3 and the tangent to the curve at the point 12, 8 . (i) Find the equation of this tangent, giving your answer in the form y = mx + c. [5] (ii) Find the area of the shaded region. [6]

Mark scheme: dy 2 2 2 ] B1B1 OR − 54 + 72 x − 24 x B2,1,010 (i) = [ 3(3 − 2 x ) ]× [− dx 1 dy At x = , = −24 M1 2 dx  1  y − 8 = −24 x −  DM1  2  y = −24 x + 20 A1 [5]  (3 − 2 x )4  1  2 3 4 (ii) Area under curve =   × − B1B1 OR 27 x − 27 x + 12 x − 2 x B2,1,0 2   4   81  − 2 − −  M1 Limits 0→ ½ applied to integral with  8  intention of subtraction shown − 24x + 20 ) M1 or area trap =½(20 + 8) × ½ Area under tangent = ∫ ( = − 12 x 2 + 20 x or 7 (from trap) A1 Could be implied 9 or 1.125 A1 Dep on both M marks 8 [6]

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