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Edexcel Advanced Extension Award Mathematics ^b^ péÉÅáãÉå=m~éÉê=~åÇ=j~êâ=pÅÜÉãÉ bÇÉñÅÉä=^Çî~åÅÉÇ=bñíÉåëáçå=^ï~êÇ j~íÜÉã~íáÅë=EVUMNF cçê=cáêëí=bñ~ãáå~íáçå pìããÉê=OMMO bÇÉñÅÉä=áë=çåÉ=çÑ=íÜÉ=äÉ~ÇáåÖ=Éñ~ãáåáåÖ=~åÇ=~ï~êÇáåÖ=ÄçÇáÉë=áå=íÜÉ=rh=~åÇ=íÜêçìÖÜçìí íÜÉ=ïçêäÇK==tÉ=éêçîáÇÉ=~=ïáÇÉ=ê~åÖÉ=çÑ=èì~äáÑáÅ~íáçåë=áåÅäìÇáåÖ=~Å~ÇÉãáÅI=îçÅ~íáçå~äI çÅÅìé~íáçå~ä=~åÇ=ëéÉÅáÑáÅ=éêçÖê~ããÉë=Ñçê=ÉãéäçóÉêëK qÜêçìÖÜ=~=åÉíïçêâ=çÑ=rh=~åÇ=çîÉêëÉ~ë=çÑÑáÅÉëI=bÇÉñÅÉäDë=ÅÉåíêÉë=êÉÅÉáîÉ=íÜÉ=ëìééçêí=íÜÉó åÉÉÇ=íç=ÜÉäé=íÜÉã=ÇÉäáîÉê=íÜÉáê=ÉÇìÅ~íáçå=~åÇ=íê~áåáåÖ=éêçÖê~ããÉë=íç=äÉ~êåÉêëK cçê=ÑìêíÜÉê=áåÑçêã~íáçå=éäÉ~ëÉ=Å~ää=çìê=`ìëíçãÉê=oÉëéçåëÉ=`ÉåíêÉ=çå=MUTM=OQM=VUMMI=çê îáëáí=çìê=ïÉÄëáíÉ=~í=ïïïKÉÇÉñÅÉäKçêÖKìâ g~åì~êó=OMMO mìÄäáÅ~íáçåë=`çÇÉ=r^ ^ää=íÜÉ=ã~íÉêá~ä=áå=íÜáë=éìÄäáÅ~íáçå=áë=ÅçéóêáÖÜí «=g~åì~êó=OMMO=bÇÉñÅÉä bÇÉñÅÉä=cçìåÇ~íáçå=áë=~=êÉÖáëíÉêÉÇ=ÅÜ~êáíó=~åÇ=~=Åçãé~åó=äáãáíÉÇ=Äó=Öì~ê~åíÉÉK oÉÖáëíÉêÉÇ=áå=båÖä~åÇ=kçK=NSUSNSQ CONTENTS Page 9801 AEA Mathematics Specimen Paper 5 Specimen Markscheme 10 Guidance for centres 16 Guidance for students 19 Paper Reference(s) 9801 Mathematics Advanced Extension Award Materials required for examination Items included with question papers Answer Book (AB16) Nil Graph Paper (ASG2) Mathematical Formulae (Lilac) Candidates may NOT use calculators in answering this paper. Instructions to Candidates Full marks may be obtained for answers to ALL questions. In the boxes on the answer book provided, write the name of the examining body (Edexcel), your centre number, candidate number, the paper title, the paper reference (9801), your surname, other names and signature. Information for Candidates A booklet ‘Mathematical Formulae including Statistical Formulae and Tables’ is provided. Full marks may be obtained for answers to ALL questions. This paper has seven questions. Advice to Candidates You must ensure that your answers to parts of questions are clearly labelled. You must show sufficient working to make your methods clear to the examiner. Answers without working may gain no credit. Printer’s Log No. -----/----- This publication may only be reproduced in accordance with QCA copyright policy. Turn over © 2001 QCA. 1. (a) By considering the series 1+ t + t 2 + t 3 +K+ t n , or otherwise, sum the series 1 + 2t + 3t 2 + 4t 3 + K + nt n −1 for t ≠ 1. (5) (b) Hence find and simplify an expression for 1+ 2× 3 + 3× 32 + 4 × 33 +K+ 2001× 32000 . (1) (c) Write down an expression for both the sums of the series in part (a) for the case where t = 1. (1) π π 2 ó 2x ó 2 2. Given that S = ô e sin x dx and C = ô e 2x cos x dx, õ0 õ0 (a) show that S = 1 + 2C, (3) (b) find the exact value of S. (6) 3. Solve for values of θ, in degrees, in the range 0 ≤ θ ≤ 360, √2 × (sin 2θ + cos θ ) + cos 3θ = sin 2θ + cosθ. (12) 6 4. A curve C has equation y = f(x) with f ′(x) > 0. The x-coordinate of the point P on the curve is a. The tangent and the normal to C are drawn at P. The tangent cuts the x-axis at the point A and the normal cuts the x-axis at the point B. (a) Show that the area of ∆APB is æ []′ 2 + ö 1 2 ç f (a) 1÷ []f (a) ç ÷ . 2 è f ′(a) ø (8) (b) Given that f(x) = e5x and the area of ∆APB is e5a , find and simplify the exact value of a. (4) 5. The function f is defined on the domain [–2, 2] by: ì− kx(2 + x) if − 2 ≤ x < 0, f (x) = í î kx(2 − x) if 0 ≤ x ≤ 2, where k is a positive constant. The function g is defined on the domain [–2, 2] by g(x) = (2.5) 2 − x 2 . (a) Prove that there is a value of k such that the graph of f touches the graph of g. (8) (b) For this value of k sketch the graphs of the functions f and g on the same axes, stating clearly where the graphs touch. (4) (c) Find the exact area of the region bounded by the two graphs. (5) 7 6. Given that the coefficients of x, x² and x 4 in the expansion of ()1+ kx n , where n ≥ 4 and k is a positive constant, are the consecutive terms of a geometric series, 6(n −1) (a) show that k = . (n − 2)(n − 3) (6) (b) Given further that both n and k are positive integers, find all possible pairs of values for n and k. You should show clearly how you know that you have found all possible pairs of values. (5) (c) For the case where k = 1.4, find the value of the positive integer n. (2) (d) Given that k = 1.4, n is a positive integer and that the first term of the geometric series is the coefficient of x, estimate how many terms are required for the sum of the × 12 ≈ geometric series to exceed 1.12 10 . [You may assume that log10 4 0.6 and ≈ log10 5 0.7. ] (5) 8 7. The variable y is defined by = 2 + 2 < < π y ln (sec x cosec x) for 0 x 2 . A student was asked to prove that dy = −4 cot 2x. dx The attempted proof was as follows: y = ln (sec 2 x + cosec 2 x) = ln (sec 2 x) + ln (cosec 2 x) = 2ln sec x + 2ln cosec x dy = 2 tan x − 2cot x dx 2 − 2 = 2(sin x cos x) sin x cos x − = 2cos 2x 1 2 sin 2x = −4cot 2x (a) Identify the error in this attempt at a proof. (1) (b) Give a correct version of the proof. (5) (c) Find and simplify a general relationship between p and q, where p and q are variables that depend on x, such that the student would obtain the correct result when differentiating ln (p + q) with respect to x by the above incorrect method. (8) (d) Given that p(x) = k sec rx and q(x) = cosec² x, where k and r are positive integers, find the values of k and r such that p and q satisfy the relationship found in part (c). (4) END Marks for presentation: 7 TOTAL MARKS: 100 9 ADVANCED EXTENSION AWARD in MATHEMATICS SPECIMEN PAPER – MARK SCHEME No Working Marks 1(a) n+1 − M1 = t 1 G.P. a = 1, r = t, S n A1 t −1 − d S =1+ 2t +K+ nt n 1 = (1st series) M1 dt d ()()t −1 n +1 t n − (t n+1 −1).1 Quotient M1 ()S = dt n ()t −1 2 rule A1 n+1 − ()+ n + = nt n 1 t 1 ()t −1 2 (5) (b) t = 3, n = 2001 2001.32002 − 2002.32001 +1 4001.32001 +1 B1 Þ Sum = = (1) 22 4 + − n()n +1 (c) 1+ t +K+ t n 1 → n + 1 ; 1+ 2t +K + nt n 1 → B1 (both) 2 (1) (7) 2(a)π π M1 2 π 2 = 2x ()− = []− 2x 2 + 2x A1 (ignore S ò e d cos x dx cos x e 0 2ò e cos x dx 0 0 limits) S = ()()− 0 − −1 + 2C or 1+ 2C A1 cso (3) π π (b) 2 π 2 M1 = 2x ()= []2x 2 − 2x C ò e d sin x sin xe 0 2ò e sin xdx A1 (ignore 0 0 limits) C = ()eπ − ()0 − 2S i.e. C = eπ − 2S A1 Solving S =1+ 2()eπ − 2S M1 =1+ 2eπ − 4S π 1 π M1, A1 5S =1+ 2e and S = ()1+ 2e 5 (6) (9) 10 ADVANCED EXTENSION AWARD in MATHEMATICS SPECIMEN PAPER – MARK SCHEME No Working Marks 3 ( 2 −1)()sin 2θ + cosθ + cos3θ = 0 M1 ()2 −1 cosθ ()1+ 2sinθ + cos3θ = 0 cos3θ ≡ cos 2θ cosθ − sin 2θ sinθ M1 ≡ cosθ []cos 2θ − 2sin 2 θ M1 ≡ cosθ []1− 4sin 2 θ A1 So: cosθ (( 2 −1)()()()1+ 2sinθ + 1− 2sinθ 1+ 2sinθ )= 0 M1 i.e. cosθ ()1+ 2sinθ ( 2 −1+1− 2sinθ )= 0 M1 i.e. 2 cosθ ()1+ 2sinθ (1− 2 sinθ )= 0 M1 cosθ = 0 Þ θ = 90°, 270° A1 (both) + θ = Þ θ = − 1 Þ θ = ° ° A1, A1 1 2sin 0 sin 2 210 , 330 A1, A1 1− 2 sinθ = 0 Þ sinθ = 1 Þ θ = 45°, 135° 2 (12) 4(a) y Equation of tangent: y – f(a) = f ′(a)(x – a) M1 P æ f ()a ö A1 Þ A is ça − ,0÷ f(x) è f ′()a ø O A B x 1 Gradient of normal is − ; f ′()a M1 1 Equation of normal : y − f ()a = − ()x − a M1 f ′()a Þ B is ()a + f ()()a f ′ a ,0 A1 é f ()a ù M1, A1 Area ∆APB is 1 AB.f ()a = 1 f ()a êf ()a f ′ ()a + ú 2 2 ë f ′()a û æ []f ′()a 2 +1ö = 1 []()2 ç ÷ A1 c.s.o. 2 f a ç ÷ è f ′()a ø (8) æ 5a 2 ö 2 25()e +1 (b) f ′()a = 5e5a ∴e5a = 1 ()e5a ç ÷ M1, A1 2 ç 5a ÷ è 5e ø 2 2 10()e5a = ()(e5a 25e10a +1 ) 10a = 9 Þ = 1 ()9 = 1 3 e 25 a 10 ln 25 5 ln 5 M1 A1 c.s.o. (4) (12) 11 ADVANCED EXTENSION AWARD in MATHEMATICS SPECIMEN PAPER – MARK SCHEME No Working Marks 5(a) 25 − 2 = − 2 M1 By symmetry consider RHS: 4 x 2kx kx ()− 2 − + 25 = M1 i.e.
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