X C. Playground. Y x m. A = x 2 30x [2]

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2 1 In the epansion of ( a )7, the efficient of 5 is 80. Find the value of the nstant a. A function f is such that f() = ( + ) + 1, for. Find (i) f 1 () in the form a + b + c, where a, b and c are nstants, (ii) the domain of f 1. [1] B X m C 40 m Playground A 60 m Y m D The diagram shows a plan for a rectangular park ABCD, in which AB = 40 m and AD = 60 m. Points X and Y lie on BC and CD respectively and AX, XY and YA are paths that surround a triangular playground. The length of DY is m and the length of XC is m. (i) Show that the area, A m, of the playground is given by A = [] (ii) Given that can vary, find the minimum area of the playground. 4 The line y = k + k, where k is a nstant, is a tangent to the curve 4y = at the point P. Find (i) the value of k, (ii) the ordinates of P. Permission to reproduce items where third-party owned material protected by pyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace pyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Eaminations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Eaminations Syndicate (UCLES), which is itself a department of the University of Cambridge. UCLES 01 70/1/O/N/1

3 5 y B (5, 11) C A (1, ) X (4, 4) O The diagram shows a triangle ABC in which A has ordinates (1, ), B has ordinates (5, 11) and angle ABC is 0. The point X (4, 4) lies on AC. Find (i) the equation of BC, (ii) the ordinates of C. 6 (i) Show that the equation s = tan can be written as a quadratic equation in sin. (ii) Solve the equation s y = tan y, for 0 y The position vectors of the points A and B, relative to an origin O, are given by where k is a nstant. 1 OA = ( 0) and k OB = ( k ), k (i) In the case where k =, calculate angle AOB. (ii) Find the values of k for which AB is a unit vector. 8 (a) In a geometric progression, all the terms are positive, the send term is 4 and the fourth term is 1 1. Find (i) the first term, (ii) the sum to infinity of the progression. [] (b) A circle is divided into n sectors in such a way that the angles of the sectors are in arithmetic progression. The smallest two angles are and 5. Find the value of n. [Questions, 10 and 11 are printed on the net page.] UCLES 01 70/1/O/N/1 [Turn over

4 4 y B (0, ) C y = + A (, 1) O The diagram shows part of the curve y =, crossing the y-ais at the point B (0, ). The point + A on the curve has ordinates (, 1) and the tangent to the curve at A crosses the y-ais at C. (i) Find the equation of the tangent to the curve at A. (ii) Determine, showing all necessary working, whether C is nearer to B or to O. [1] (iii) Find, showing all necessary working, the eact volume obtained when the shaded region is rotated through 60 about the -ais. 10 A curve is defined for > 0 and is such that dy d = + 4. The point P (4, 8) lies on the curve. (i) Find the equation of the curve. (ii) Show that the gradient of the curve has a minimum value when = and state this minimum value. 11 Q cm C P S 1. rad R 0 cm O The diagram shows a sector of a circle with centre O and radius 0 cm. A circle with centre C and radius cm lies within the sector and touches it at P, Q and R. Angle POR = 1. radians. (i) Show that = 7.18, rrect to decimal places. (ii) Find the total area of the three parts of the sector lying outside the circle with centre C. [] (iii) Find the perimeter of the region OPSR bounded by the arc PSR and the lines OP and OR. UCLES 01 70/1/O/N/1

5 Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL October/November a Term in 5 is 7 C (²) 4 ( a/)³ This term isolated Equated to 80 a =. Allow on own or in an epansion. Correct term in 5 selected. Equated to 80 + (i) f() = + 1, for Make the subject or interchanges,y ( 1) 4 1 Attempt at as subject and removes +1 Squares both sides and deals with "+" and " ". (ii) domain of f 1 is 1. [1]. ndone >1 (i) A = 400 0(60 ) (40 ) 0 A = ² (uld be trapezium triangle) (ii) da = 0 or ( 15)² + 75 d = 0 when = 15 or Min at = 15 A = 75. [] Needs attempts at all areas answer given - either method okay Sets differential to 0 + solution.. 4 y = + k k (i) 4 4y = = + k k 4 k Uses b² 4ac k = 1 4k = 0 1 (calculus k = 4 1 =, y = k = 1 ) k k (ii) y = 1, 4y = ² = 0 P(, 1) Eliminates or y mpletely. Uses b² 4ac for a quadratic = 0 nb a,b,c must not be f() Elimination of or y Soln of eqn.. Cambridge International Eaminations 01

6 Page 5 Mark Scheme Syllabus Paper GCE AS/A LEVEL October/November A (1, ), B (5, 11), X (4, 4) (i) Gradient of AB = Gradient of BC = ½ Eqn of BC is y 11 = 1 ( 5) (ii) gradient of AC (or AX) is ⅓ eqn of AC is y = 1 ( 1) 4 = 1 ( 4 or y ) Sim equations C (1,7) For use of m 1 m = 1 unsimplified is fine Correct form of line equation + sim eqns answer only -0/- assumed AB = BC. Uses graph or table and gets eactly (1,7) allow the marks for (ii). 6 s = tan (i) Replaces tan by sin s c² =s s² +s = 0 Uses t = s c Uses s² + c² = 1. Correct eqn. (ii) Soln of quadratic y = 15 º y can also be y = 75 º. D Method for quadratic = 0 and Works with y first before for 0º 1 st answer. (loses mark if etra soln in range) 7 1 k OA = 0 OB = k k 1 (i) 0. = 10 4 = 5 4 s θ θ = 4.1º Use of 1 + y 1 y + z 1 z Product of moduli All nnected rrectly. (ii) k 1 AB = k allow each cpt ± k ( k 1) + k + (k ) 6k 10k + 4 = 0 k = 1 or ⅔ Correct for either AB or BA. Sum of squares (doesn't need =1) Correct quadratic Cambridge International Eaminations 01

7 Page 6 Mark Scheme Syllabus Paper GCE AS/A LEVEL October/November (a) (i) ar = 4, ar³ = 1½ Eliminates a (or r) r = ¾ a = (ii) sum to infinity = ¼ = 18 [] Both needed Method of Solution. Correct formula used. on value of r (b) a =, d = n (6 + ( n 1) ) ( = 60) n + 4n 70=0 n = 18 Correct value for d Correct S n used. no need for 60 here. Correct quadratic y = A (, 1) B (0, ) (i) d y = d ( + ) m = y = ( ) 1 Correct without the. For, independent of first part. Correct form of tan - numerical dy/d For his m following use of dy/d. (normal ma /4, no calculus 0/4) (ii) Meets the y-ais when = 0, y = 1⅔ This is nearer to B than to O. [1] Sets to 0 in his tangent. The 1⅔ and part (i) must be rrect (iii) Integral of = ( + ) + 81 Uses limits 0 to = π 6 Correct without the. For, Use of limits with integral of y² only no π ma ¾. Use of area - 0/4, Cambridge International Eaminations 01

8 Page 7 Mark Scheme Syllabus Paper GCE AS/A LEVEL October/November dy 4 = + and P (4, 8) d 4 (i) y = + (c) Uses (4, 8) c = 1 d y (ii) d = 8 1 = 0 when =. (ignore +c at this stage) Uses the point after integration for c Co Sets to 0 + solution or verifies and states a nclusion (stationary or min) gradient of 8 4 d/d( 1 ) = 4 +ve Min. Allow for = into dy/d. Any valid method - rd differential +ve nd diff goes 0+, or 1st goes >,,> 11 (i) OQ = + OC = 0 sin 0.6 = OC + sin 0.6 OC = sin 0.6 = 0 = 7.18 Used somewhere needs 0. Use of trig in 0º triangle Soln of linear equation. (answer given, ensure there is a rrect method) (ii) Area = ½. 0² 1. π 7.18² = 76. (iii) Angle PCR = π 1. Arc PR = 7.18 (π 1.) = (14.01) OP = OR = tan 0.6 Perimeter of 5.1 cm [] Use of ½r²θ - needs r=0 and θ = 1. Use of s=rθ with r = any θ -even π/ Correct use of trig or Pythagoras Cambridge International Eaminations 01

MARK SCHEME for the October/November 2012 series 9709 MATHEMATICS. 9709/12 Paper 1, maximum raw mark 75

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 01 series 9709 MATHEMATICS 9709/1 Paper 1, maimum raw mar 75 This mar scheme