Condensed. Mathematics. General Certificate of Education Advanced Level Examination January Unit Further Pure 4.

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1 Geeral Certificate of Educatio Advaced Level Examiatio Jauary 0 Mathematics MFP4 Uit Further Pure 4 Friday Jauary am to 0.30 am For this paper you must have: the blue AQA booklet of formulae ad statistical tables. You may use a graphics calculator. Time allowed hour 30 miutes Istructios Use black ik or black ball-poit pe. Pecil should oly be used for drawig. Fill i the boxes at the top of this page. Aswer all questios. Write the questio part referece (eg (a), (b)(i) etc) i the left-had margi. You must aswer the questios i the spaces provided. Do ot write outside the box aroud each page. Show all ecessary workig; otherwise marks for method may be lost. Do all rough work i this book. Cross through ay work that you do ot wat to be marked. Codesed Iformatio The marks for questios are show i brackets. The maximum mark for this paper is 5. Advice Uless stated otherwise, you may quote formulae, without proof, from the booklet. You do ot ecessarily eed to use all the space provided. P4589/Ja/MFP4 6/6/6/ MFP4

2 p The vectors a ad b are such that a.b¼, jaj ¼5 ffiffiffi ad jbj ¼3. Determie the exact value of ja bj. (5 marks) Describe the sigle trasformatio represeted by each of the matrices: (a) (b) ; 0 0 ( marks) 0: : (3 marks) 0:8 0 0:6 3 (a) Fid the eigevalues ad correspodig eigevectors of the matrix M ¼ (6 marks) (b) The plae trasformatio T is give by the matrix M. Write dow the coordiates of the ivariat poit of T. ( mark) 4 Let X ¼ 3 x. (a) Determie XX T. ( marks) (b) Show that DetðXX T X T XÞ 4 0 for all real values of x. (4 marks) (c) Fid the value of x for which the matrix ðxx T X T XÞ is sigular. ( mark) (0) P4589/Ja/MFP4

3 3 5 (a) Determie the two values of the iteger for which the system of equatios x þ y þ z ¼ 5 3x y þ z ¼ x þ y þ z ¼ does ot have a uique solutio. (4 marks) (b) For the positive value of foud i part (a), determie whether the system is cosistet or icosistet, ad iterpret this result geometrically. (6 marks) 6 The plaes P ad P have equatios respectively. 3 r. 4 5 ¼ 0 ad r ¼ (a) Determie, to the earest degree, the acute agle betwee P ad P. (4 marks) 4 (b) By settig z ¼ t, fid cartesia equatios for the lie of itersectio of P ad P i the form x a l ¼ y b m ¼ z ¼ t (6 marks) (c) The lie L, with equatio r ¼ 4 5 þ l4 9 5, itersects P at the poit P ad P 4 at the poit Q. p Show that PQ ¼ k ffiffiffi, where k is a iteger. (6 marks) Tur over s (03) P4589/Ja/MFP4

4 4 The plae trasformatio T is a rotatio through y radias aticlockwise about O, ad maps poits ðx, yþ oto image poits ðx, YÞ such that X Y x y ¼ c s s c where c ¼ cos y ad s ¼ si y. (a) Write dow the iverse of the matrix c s s ad hece show that c (b) The curve C has equatio x 6xy y ¼ 8. x ¼ cx þ sy ad y ¼ sx þ cy (3 marks) The image of C uder T is the curve C with equatio px þ qxy þ ry ¼ 8. (i) Use the results of part (a) to show that q ¼ 6s þ 6sc 6c ad express p ad r similarly i terms of c ad s. (4 marks) (ii) Give that y is a acute agle, fid the values of c ad s for which q ¼ 0 ad hece i this case express the equatio of C i the form (iii) Hece explai why C is a hyperbola. X a Y b ¼ (8 marks) ( mark) 8 For 6¼, the vectors a, b ad c are such that a ¼ 4 5, 6 b ¼ 4 þ 5 ad 6 c ¼ 4 5 Determie the value of for which a, b ad c are liearly depedet. (9 marks) Copyright ª 0 AQA ad its licesors. All rights reserved. (04) P4589/Ja/MFP4

5 Key to mark scheme abbreviatios M mark is for method m or dm mark is depedet o oe or more M marks ad is for method A mark is depedet o M or m marks ad is for accuracy B mark is idepedet of M or m marks ad is for method ad accuracy E mark is for explaatio or ft or F follow through from previous icorrect result CAO correct aswer oly CSO correct solutio oly AWFW aythig which falls withi AWRT aythig which rouds to ACF ay correct form AG aswer give SC special case OE or equivalet A, or (or 0) accuracy marks x EE deduct x marks for each error NMS o method show PI possibly implied SCA substatially correct approach c cadidate sf sigificat figure(s) dp decimal place(s) No Method Show Where the questio specifically requires a particular method to be used, we must usually see evidece of use of this method for ay marks to be awarded. Where the aswer ca be reasoably obtaied without showig workig ad it is very ulikely that the correct aswer ca be obtaied by usig a icorrect method, we must award full marks. However, the obvious pealty to cadidates showig o workig is that icorrect aswers, however close, ear o marks. Where a questio asks the cadidate to state or write dow a result, o method eed be show for full marks. Where the permitted calculator has fuctios which reasoably allow the solutio of the questio directly, the correct aswer without workig ears full marks, uless it is give to less tha the degree of accuracy accepted i the mark scheme, whe it gais o marks. Otherwise we require evidece of a correct method for ay marks to be awarded.

6 MFP4 Q Solutio Marks Total Commets Use of ab cosθ = a. b = cos θ = 5 A si θ = 5 B ft FT exact oly Use of a b = ab siθ = 3 A 5 CSO Total 5 (a) Reflectio i x = z A (b) Rotatio about the y-axis A Through cos 0.6 ( 53.3 o ) A 3 Igore directio Total 5 3(a) Char. Eq. is λ 8λ 9 = 0 Attempted Quadratic solved to get two roots d λ = 9, A Subst g. back λ (at least oce) : λ = 9 x + y = 0 A ay α 0 λ = 9 has evecs. α λ = x + y = 0 A 6 ay β 0 λ = has evecs.β (b) (0, 0) B Total

7 MFP4 (cot) Q Solutio Marks Total Commets 4(a) X X T = 3 x 3 x Attempted mult. with X T correct x + 9 x 3 = x 3 50 A (b) X T X = 3 x 3 x 0 3x = 3x x + 49 Good attempt X X T X T X = x 4x + 4 4x + 4 x Good attempt Det(X X T X T X) = x 4x + 4 x 4 = (x + ) 4 x 4x + 4 x Good attempt to factorise/expad the determiat {( ) + 6} = ( x + ) x 0 for all real x E 4 Explaied/demostrated fully (c) x = B CSO Total

8 MFP4 (cot) Q Solutio Marks Total Commets 5(a) 3 Expadig the det. of the coefft. mtx. Settig it = 0 Obtaiig & solvig a quadratic eq. i 0 = + 8 = ( + 8)( ) =, 8 A 4 CSO (b) = gives x + y + z = 5 3x y + z = x + y + z = Elimiatig oe variable from a pair of equatios, twice e.g. x y = 4 ad 4x 8y = 0 Icosistecy clearly demostrated from fully correct workig B A ft A ft E ft their chose iteger 3 plaes have o commo itersectio (or form a r prism) B ft 6 Also ft 3 plaes meet i a commo lie or 3 plaes form a sheaf if cosistecy coclusio made Total 0

9 MFP4 (cot) Q Solutio Marks Total Commets 6(a) Use of scalar product o 3 ad 4 Sc.Prod. = ± B Moduli 54 ad 6 correct B Accept.348 & AWRT 56 o A 4 From correct workig (b) x + y + t = 0 ad 3x + y 4t = oted or used Elimiatig (say) y to get x as a f. of t x = t 3 A CAO Subst g. back for y y = 6 9t A CAO x + 3 = y 6 = z 9 ( = t) B ft 6 (c) λ + 0 Attempt at either 9λ 4λ + λ λ = 4λ + 4 = 0 or Solvig either 40 + λ + 9λ λ = 0 or λ + 9λ 8 6λ = λ = λ = 6 AA P = (8, 9, ) ad Q = (6, 53, 3) PQ = = 6 = 56 A 6 NB P, Q ot required: d = λ λ i + 9j + 4k = 8 = 56 A Total 6

10 MFP4 (cot) Q Solutio Marks Total Commets (a) c s s c B x c = y s s X c Y cx + sy = sx + cy A 3 (b)(i) ( cx sy ) ( cx sy )( sx cy ) ( sx cy ) = 8 ( c X + csxy + s Y ) 6( c s XY + sc Y X ) ( s X csxy + c Y ) = 8 Subst. for x & y i eq. ad multiplyig out p = c + 6sc s A q = 6cs 6(c s ) A AG r = s 6sc c A 4 (ii) Factorisig: 3s + 8sc 3c = (3s c)(s + 3c) = 0 Deducig a ta value cosθ = taθ = 3 (θ acute) 3, siθ = 0 0 Subst g. sesible values back for p ad r X 8Y = 8 A A A A Or by double agles Both X Y = A 8 CSO (iii) Sice C is a hyperbola, ad it is just C rotated, it follows that C is a hyperbola E Total 6

11 MFP4 (cot) Q Solutio Marks Total Commets 8 For cosiderig + B Or by scalar triple product = ( ) + + A For st factor = + + ( ) ( + ) ( + )( ) 0 Row ops. for d factor R ' = R + R 3 3 = ( )( + ) A = ( )( + ) 3 3 { } OR = ( )( + ) 0 0 R' = R R = ( )( + ){ + } Full method for remaiig factors = ( )( + )( ) A = B 9 CSO Total 9 TOTAL 5 Note: Expadig straightaway scores B ad the A for Thereafter, A for st factor, for d factor attempted ad for full method for remaiig factors plus A ad B cso at the ed, as above.

12 Scaled mark uit grade boudaries - Jauary 0 exams A-level Max. Scaled Mark Grade Boudaries ad A Coversio Poits Code Title Scaled Mark A A B C D E LAW0 LAW UNIT LAW03 LAW UNIT MD0 MATHEMATICS UNIT MD MFP MATHEMATICS UNIT MFP MA MATHEMATICS UNIT MA 00 o cadidates were etered for this uit MB MATHEMATICS UNIT MB MPC MATHEMATICS UNIT MPC MSA MATHEMATICS UNIT MSA MS/SSA/W MATHEMATICS UNIT SA - WRITTEN MS/SSA/C MATHEMATICS UNIT SA - COURSEWORK MSB MATHEMATICS UNIT MSB MD0 MATHEMATICS UNIT MD MFP MATHEMATICS UNIT MFP MMB MATHEMATICS UNIT MMB MPC MATHEMATICS UNIT MPC MSB MATHEMATICS UNIT MSB MFP3 MATHEMATICS UNIT MFP MPC3 MATHEMATICS UNIT MPC MFP4 MATHEMATICS UNIT MFP MPC4 MATHEMATICS UNIT MPC MEST MEDIA STUDIES UNIT MEST MEDIA STUDIES UNIT MEST3 MEDIA STUDIES UNIT MEST4 MEDIA STUDIES UNIT PHIL PHILOSOPHY UNIT

Version 1.0: abc. General Certificate of Education. Mathematics MFP2 Further Pure 2. Mark Scheme examination - January series

Version 1.0: abc. General Certificate of Education. Mathematics MFP2 Further Pure 2. Mark Scheme examination - January series Versio.0: 008 abc Geeral Certificate of Educatio Mathematics 660 MFP Further Pure Mark Scheme 008 examiatio - Jauary series Mark schemes are prepared by the Pricipal Examier ad cosidered, together with