Condensed. Mathematics. General Certificate of Education Advanced Level Examination June Unit Further Pure 2. Time allowed * 1 hour 30 minutes

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1 General Certificate of Education Advanced Level Examination June 011 Mathematics MFP Unit Further Pure Monday 13 June am to am For this paper you must have: the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed 1 hour 30 minutes Instructions Use black ink or black ball-point pen. Pencil should only be used for drawing. Fill in the boxes at the top of this page. Answer all questions. Write the question part reference (eg (a), (i) etc) in the left-hand margin. You must answer the questions in the spaces provided. Do not write outside the box around each page. Show all necessary working; otherwise marks for method may be lost. Do all rough work in this book. Cross through any work that you do not want to be marked. Condensed Information The marks for questions are shown in brackets. The maximum mark for this paper is 7. Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. P3831/Jun11/MFP 6/6/ MFP

2 1 (a) Draw on the same Argand diagram: (i) the locus of points for which j z i j¼ (3 marks) the locus of points for which argðz þ iþ ¼ p 4 (3 marks) Indicate on your diagram the set of points satisfying both and j z i j 4 argðz þ iþ ¼ p 4 ( marks) (a) Use the definitions of cosh y and sinh y in terms of e y to show that cosh x cosh y sinh x sinh y ¼ coshðx yþ (4 marks) It is given that x satisfies the equation coshðx ln Þ ¼sinh x (i) Show that tanh x ¼. (4 marks) 7 Express x in the form 1 ln a. ( marks) 3 (a) Show that ðr þ 1Þ! ðr 1Þ! ¼ðr þ r 1Þðr 1Þ! ( marks) Hence show that Xn r¼1 ðr þ r 1Þðr 1Þ! ¼ðn þ Þn! (4 marks) (0) P3831/Jun11/MFP

3 3 4 The cubic equation z 3 z þ k ¼ 0 ðk 6¼ 0Þ has roots a, b and g. (a) (i) Write down the values of a þ b þ g and ab þ bg þ ga. ( marks) Show that a þ b þ g ¼ 4. ( marks) (iii) Explain why a 3 a þ k ¼ 0. (iv) Show that a 3 þ b 3 þ g 3 ¼ 8 3k. (1 mark) ( marks) Given that a 4 þ b 4 þ g 4 ¼ 0: (i) show that k ¼ ; (4 marks) find the value of a þ b þ g. (3 marks) (a) The arc of the curve y ¼ x þ 8 between the points where x ¼ 0 and x ¼ 6is rotated through p radians about the x-axis. Show that the area S of the curved surface formed is given by p S ¼ ffiffiffi ð 6 pffiffiffiffiffiffiffiffiffiffiffiffiffi p x þ 4 dx By means of the substitution x ¼ sinh y, show that p S ¼ pð4 ffiffiffi pffiffiffi þ 4 sinh 1 3Þ 0 ( marks) (8 marks) 6 (a) Show that ðk þ 1Þð4ðk þ 1Þ 1Þ ¼4k 3 þ 1k þ 11k þ 3 ( marks) Prove by induction that, for all integers n 1, 1 þ 3 þ þ ::: þðn 1Þ ¼ 1 3 nð4n 1Þ (6 marks) Turn over s (03) P3831/Jun11/MFP

4 4 7 (a) (i) Use de Moivre s Theorem to show that cos y ¼ cos y 10 cos 3 y sin y þ cos y sin 4 y and find a similar expression for sin y. ( marks) Deduce that tan y ¼ tan yð 10 tan y þ tan 4 yþ 1 10 tan y þ tan 4 y (3 marks) Explain why t ¼ tan p is a root of the equation t 4 10t þ ¼ 0 and write down the three other roots of this equation in trigonometrical form. (3 marks) (c) Deduce that tan p tan p ¼ p ffiffi ( marks) END OF QUESTIONS Copyright Ó 011 AQA and its licensors. All rights reserved. (04) P3831/Jun11/MFP

5 MFP Q Solution Marks Total Comments 1(a) Im Use average of whole question if diagrams used Re (i) Circle Circle in any position correct centre Must be shown touching x-axis F 3 ft incorrect centre half-line through (0, ) Can be inferred through point of contact of circle with x-axis 3 Inside circle On line F ft errors in position of line and circle Total 8 x x y y x x y y e + e e + e e e e e M0 if sinh and cosh confused A1 for formula quoted correctly Correct expansions A1 Use of e xy A0 1 ( ) ( e x y x y = + e ) = cosh( x y ) A1 4 AG (a) ( ) ( ) ( ) ( ) (i) ( x ) = x ( ) cosh ln cosh cosh ln sinhx sinh ln ( ) Alternative: x ln x+ ln x x e + e e e cosh ( ln ) = x 4 x ln e Both any method e = or e 3 sinh ( ln ) = correct 4 7 cosh x = sinh x A1F e x = 6 A1 4 4 tanh x = A1 4 AG tanh x = A1 7 7 = = e x+ ln 1 used x ln 7 = 1 7 or e x e x = x x Could be embedded in (i) e+e 7 1 = ln 6 A1 Total 10

6 MFP (cont) Q Solution Marks Total Comments 3(a) ( r+ 1! ) = ( r+ 1 ) r( r 1! ) Result A1 AG Attempt to use method of differences n ( r + r 1)( r 1)! = ( n+ 1 )! + n! 1! 0! A1 r= 1 ( ) ( ) ( n ) n! n+ 1! = n+ 1 n! Must be seen + A1 4 AG Total 6 4(a)(i) α = αβ = 0 (A0) α = ( α ) αβ Used. Watch α = = 4 A1 AG (iii) Clear explanation E1 1 eg α satisfies the cubic equation since it is a root. Accept z = α (iv) (i) 3 α = α 3k Or ( ) = 8 3k A1 AG α = α kα α 3 = α 3 α αβ + 3 αβγ 4 3 α = α k α 4 Or ( ) ( ) ( k) = 8 3 k A1 ft on α = k = A1 4 AG 4 α = α k α Substitution of values A1 = 8 A1 3 Total 14 α = α αβ + 4αβγ α

7 MFP (cont) Q Solution Marks Total Comments (a) dy y x dx dy ( ) 1 x x 8 dx 6 x S = π y 1+ dx for use of formula provided y 0 y dx A1F function of x A1 for substitution for d y (one slip) dx Eliminating all y 6 = π x + 4dx A1 AG 0 dx dx = coshθ dθ or coshθ dθ S = π 4sinh x+ 4. coshθ dθ For eliminating x completely and use of dθ, ie dθ attempted S = ( ) π cosh θ. coshθ dθ Use of cosh θ sinh θ = 1 (ignore limits) = 4 π ( coshθ + 1) dθ Use of formula for cosh θ ; must be correct sinh θ = 4 π + θ F Correct integration of acosh θ + b = 4 π[ sinhθcosh θ + θ] Use of sinh θ = sinhθcoshθ Must be seen x x 1 x = 4 π sinh = π sinh 3 A1 8 AG ( ) 6(a) Expansion of ( k ) ( k ) Total 13 Or change limits Any valid method first step correct 3 = 4k + 1k + 11k + 3 A1 AG Assume true for n= k For n= k + 1: k ( r 1) = k( 4 k 1) + ( k + 1) r= 1 3 A1 No LHS A0 = 1 ( k k + 11 k + 3 ) 3 A1F ft error in (k + 1) 1 = ( 1) ( 4( 1 ) 1) 3 k + k + A1 Using part (a) True for n = 1 shown Proof by induction set out properly E1 6 Dependent on all marks correct (if factorised by 3 linear factors, allow A1 at this particular point) Total 8

8 MFP (cont) Q Solution Marks Total Comments 7(a)(i) ( ) cosθ + isinθ = cosθ + isinθ Attempt to expand 3 correct terms Expansion in any form A1 Correct simplification Equate real parts: 3 4 cosθ = cos θ 10cos θsin θ + cosθsin θ A1 AG Equate imaginary parts: 4 3 sin θ = cos θ sin θ 10cos θ sin θ + sin θ A1 CAO sin θ tan θ = cos θ Used 4 Division by cos θ or by cos θ 4 tanθ( 10tan θ + tan θ) tanθ = tan θ + tan θ A1 3 AG π 4 θ = tanθ = 0 Or for tan θ 10tan θ + = 0 π 4 tan satisfies t 10t + = 0 A1 Or for tanθ = 0 Other roots kπ tan k =, 3, 4 3 OE (c) Product of roots = π 4π tan = tan Or π π tan tan = A1 π π tan tan =+ A1 sign rejected with reason E1 Total 16 TOTAL 7 π 3π tan = tan Alternative (c) Use of quadratic formula t = ± A1 t =± ± Correct selection of +ve values E1 Multiplied together to get A1

9 Scaled mark unit grade boundaries - June 011 exams A-level Max. Scaled Mark Grade Boundaries and A Conversion Points Code Title Scaled Mark A A B C D E MS1B GCE MATHEMATICS UNIT S1B MD0 GCE MATHEMATICS UNIT D MFP GCE MATHEMATICS UNIT FP MMB GCE MATHEMATICS UNIT MB MPC GCE MATHEMATICS UNIT PC MSB GCE MATHEMATICS UNIT SB XMCA GCE MATHEMATICS UNIT XMCA MFP3 GCE MATHEMATICS UNIT FP MM03 GCE MATHEMATICS UNIT M MPC3 GCE MATHEMATICS UNIT PC MS03 GCE MATHEMATICS UNIT S MFP4 GCE MATHEMATICS UNIT FP MM04 GCE MATHEMATICS UNIT M MPC4 GCE MATHEMATICS UNIT PC MS04 GCE MATHEMATICS UNIT S MM0 GCE MATHEMATICS UNIT M MEST1 GCE MEDIA STUDIES UNIT MEST GCE MEDIA STUDIES UNIT MEST3 GCE MEDIA STUDIES UNIT MEST4 GCE MEDIA STUDIES UNIT MHE GCE MODERN HEBREW UNIT MHEB GCE MODERN HEBREW UNIT MUSC1 GCE MUSIC UNIT MUSA GCE MUSIC UNIT A MUSB GCE MUSIC UNIT B MUSC GCE MUSIC UNIT C