Condensed. Mathematics. General Certificate of Education Advanced Level Examination January Unit Pure Core 3. Time allowed * 1 hour 30 minutes

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Barrie James
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1 Gnral Crtificat of Education Advancd Lvl Eamination January 0 Mathmatics MPC Unit Pur Cor Friday 0 January 0.0 pm to.00 pm For this papr you must hav: th blu AQA booklt of formula and statistical tabls. You may us a graphics calculator. Tim allowd hour 0 minuts Instructions Us black ink or black ball-point pn. Pncil should only b usd for drawing. Fill in th bos at th top of this pag. Answr all qustions. Writ th qustion part rfrnc (g (a), (i) tc) in th lft-hand margin. You must answr th qustions in th spacs providd. Do not writ outsid th bo around ach pag. Show all ncssary working; othrwis marks for mthod may b lost. Do all rough work in this book. Cross through any work that you do not want to b markd. Condnsd Information Th marks for qustions ar shown in brackts. Th maimum mark for this papr is 75. Advic Unlss statd othrwis, you may quot formula, without proof, from th booklt. You do not ncssarily nd to us all th spac providd. P506/Jan/MPC 6/6/6/ MPC

2 (a) Us Simpson s rul with 7 ordinats (6 strips) to find an stimat for ð 0 d. ( marks) A curv is dfind by th quation y ¼. Th curv intrscts th lin y ¼ 8 at a singl point whr ¼ a. (i) Show that a lis btwn. and.. ( marks) (ii) Th quation ¼ 8 can b rarrangd into th form ¼ lnð8 Þ. ln Us th itrativ formula nþ ¼ lnð8 nþ with ln ¼ : to find th valus of and, giving your answrs to thr dcimal placs. ( marks) Th curv with quation y ¼ 6 y is sktchd blow for 6. O 6 Th function f is dfind by fðþ ¼ 6 for 6. (a) Find th rang of f. ( marks) Th invrs of f is f. (i) Find f ðþ. ( marks) (ii) Solv th quation f ðþ ¼. ( marks) (c) Th function g is dfind by gðþ ¼ for. (i) Writ down an prssion for fgðþ. ( mark) (ii) Solv th quation fgðþ ¼. ( marks) (0) P506/Jan/MPC

3 (a) Givn that y ¼ 6 þ, find dy d. ( mark) ð Hnc find d, giving your answr in th form p ln q, whr 6 þ p and q ar rational numbrs. (5 marks) (a) By using a suitabl trigonomtrical idntity, solv th quation tan y ¼ ð sc yþ giving all solutions to th narst 0. in th intrval 0 < y < 60. (6 marks) Hnc solv th quation tan ð 0 Þ ¼½ scð 0 ÞŠ giving all solutions to th narst 0. in th intrval 0 < < 90. ( marks) 5 (a) Dscrib a squnc of two gomtrical transformations that maps th graph of y ¼ ln onto th graph of y ¼ lnð Þ. ( marks) Sktch, on th as givn blow, th graph of y ¼jlnð Þj, indicating th act valu of th -coordinat whr th curv mts th -ais. ( marks) (c) (i) Solv th quation jlnð Þj ¼. ( marks) (ii) Hnc, or othrwis, solv th inquality jlnð Þj 5. ( marks) y O Turn ovr s (0) P506/Jan/MPC

4 6 (a) Givn that ¼ d, us th quotint rul to show that ¼ cosc y cot y. sin y dy ( marks) Us th substitution ¼ cosc y to find thr significant figurs. ð pffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi d, giving your answr to (9 marks) 7 (a) A curv has quation y ¼. Show that th curv has actly two stationary points and find th act valus of thir coordinats. (7 marks) (i) Us intgration by parts twic to find th act valu of ð 0 d. (7 marks) (ii) Th rgion boundd by th curv y ¼ 8, th -ais from 0 to and th lin ¼ is rotatd through 60 about th -ais to form a solid. Us your answr to part (i) to find th act valu of th volum of th solid gnratd. ( marks) Copyright ª 0 AQA and its licnsors. All rights rsrvd. (0) P506/Jan/MPC

5 Ky to mark schm abbrviations M mark is for mthod m or dm mark is dpndnt on on or mor M marks and is for mthod A mark is dpndnt on M or m marks and is for accuracy B mark is indpndnt of M or m marks and is for mthod and accuracy E mark is for planation or ft or F follow through from prvious incorrct rsult CAO corrct answr only CSO corrct solution only AWFW anything which falls within AWRT anything which rounds to ACF any corrct form AG answr givn SC spcial cas OE or quivalnt A, or (or 0) accuracy marks EE dduct marks for ach rror NMS no mthod shown PI possibly implid SCA substantially corrct approach c candidat sf significant figur(s) dp dcimal plac(s) No Mthod Shown Whr th qustion spcifically rquirs a particular mthod to b usd, w must usually s vidnc of us of this mthod for any marks to b awardd. Whr th answr can b rasonably obtaind without showing working and it is vry unlikly that th corrct answr can b obtaind by using an incorrct mthod, w must award full marks. Howvr, th obvious pnalty to candidats showing no working is that incorrct answrs, howvr clos, arn no marks. Whr a qustion asks th candidat to stat or writ down a rsult, no mthod nd b shown for full marks. Whr th prmittd calculator has functions which rasonably allow th solution of th qustion dirctly, th corrct answr without working arns full marks, unlss it is givn to lss than th dgr of accuracy accptd in th mark schm, whn it gains no marks. Othrwis w rquir vidnc of a corrct mthod for any marks to b awardd.

6 MPC Q Solution Marks Total Commnts (a) y B B all 7 valus corrct (and no tra) (PI by 7 corrct y valus) 5 or mor corrct y valus, act,... or valuatd (in tabl or in formula) A = [ corrct substitution of thir 7 y-valus into ] M Simpson s rul = 9 or 5.5 or 7 A CAO 6 (i) f( ) = + 8 or g( ) = 8 f (. ) = 0. or g(. ) = 0. f (. ) = 0.7 or g(. ) = 0.7 M attmpt at valuating f(.) and f(.) AWRT ± 0. and ± 0.7 altrnativ mthod condon f (.) < 0, f (.) > 0 if f is. = 5., 8. = 5.6 dfind. = 6., 8. = 5. M chang of sign. < α <. A at. LHS < RHS (f() must b dfind and all working corrct) at. LHS > RHS. < α <. A (ii) ( = ). B ( = ). B ths valus only Total 8

7 MPC (cont) Q Solution Marks Total Commnts (a) f () = f (6) = M sight of and f ( ) A allow f() rplacd by f, y (i) y = 6 = 6 y ( y ) = 6 or bttr M M rvrs, y on corrct stp Eithr ordr f 6 ( ) = + OE A condon y = (ii) 6 + = 6, or bttr + = on corrct stp from thir (i) =, M or = f() ( = ) A not: scors / fg( ) = 6 B (c)(i) ( ) (ii) 6 = = 6 or bttr M on corrct stp from thir (c)(i) = = 6 OE A g ( + 8) ( 8) = 0, or = ± = ONLY A Total

8 MPC (cont) Q Solution Marks Total Commnts (a) dy = 6 d B do not ISW 6+ d ( ) () ln 6 M = + 6 () A = ln ( 6 + ) 6 ln ( 6 + ) 6 m = ln 9 ln 6 6 AF 9 = ln or = ln 6 6 A 5 Total 6 ( + ) k ln 6, k is a constant k = 6 corrct substitution in F() F(). condon poor us or lack of brackts. kln 9 kln only follow through on thir k or if using th substitution u = 6+ du = k M u = ln u A 6 thn, ithr chang limits to and 9 m thn AF Aas schm or changing back to, thn m AF A as schm (a) sc θ =... B corrct us of sc θ = + tan θ quadratic prssion in scθ with all sc θ + sc θ 0 ( = 0) M trms on on sid ( θ )( θ ) sc + 5 sc = 0 m attmpt at factors of thir quadratic, ( scθ ± 5)( scθ ± ), or corrct us of quadratic formula sc θ = 5, A cos θ =, 5 B 60, 00,0.5, 58.5 (AWRT) B 6 corrct, ignor answrs outsid intrval all corrct, no tras in intrval 0 = 60, 0 5,58 5,00 M 0 = any of thir (60), all thir answrs from (a), BUT must hav = 70, 5,68 5,0 AF scord B = 7 5, 7 9,67.,77 5 (AWRT) A CAO, ignor answrs outsid intrval Total 9

9 MPC (cont) Q Solution Marks Total Commnts 5(a) strtch I SF II MA I + (II or III) in y-dirction III ithr ordr translat E 0 B accpt in positiv -dirction M A mod graph, in connctd sctions, both in th first quadrant, touching -ais curv touchs -ais at + (or.7 or bttr), and lablld (ignor scal) + A corrct curvatur, including at thir +, appro. asymptot at = (c)(i) ln ( ) = ( ) ( ) ln = ln = or bttr M must s quations, condon omission of brackts ( =) do not ISW A accpt valus of AWRT 5., 5., 5. = + or ( = ) + do not ISW A accpt valus of AWRT.08,.09 if M0 thn = with or without working scors SC ( ) (ii) B accpt valus of AWRT 5., 5., 5. < + B accpt valus of AWRT.7,.08,.09 if B not arnd, thn SCfor any of +, < < +, < + Total

10 MPC (cont) Q Solution Marks Total Commnts 6(a) ( ) d sinθ 0 cosθ = dθ sin θ M ± sinθ k ± cosθ quotint rul sin θ whr k = 0 or must s th 0 ithr in th quotint or in A g d u 0 dθ = cosθ cosθ or = sin θ sinθsinθ or quivalnt = coscθ cotθ A CSO, AG must s on of th prvious prssions = coscθ d coscθ cotθ dθ B OE, g d = coscθ cotθdθ Rplacing cosc θ by cot θ, or bttr B at any stag of solution ( ) = coscθcotθ ( θ ) cosc θ cosc dθ M all in trms ofθ, and including thir attmpt at d, but condon omission of dθ A coscθ cotθ ( dθ ) cosc θ cotθ ( dθ ) coscθ = cosθ A =, θ = 0.5 AWRT B =, θ = AWRT fully corrct and must includ dθ (at som stag in solution) = A OE g sinθ( dθ) corrct chang of limits ± cos = ± OE or ( ) θ ( ) m c's F( 0.5 ) F( 0.79 ) = 0.59 A 9 Total substitution into ± cos θ only or

11 MPC (cont) Q Solution Marks Total Commnts 7(a) M p, q constants dy = p + q d A p = and q = + = 0 0 E or = 0 impossibl OE (may b sn latr) ( a + b ) = 0 m or ( a + b ) = 0 = 0, 8 A = 0, y = 0 A = 8, y = 6 B 7 condon 8 y = 8 tc ignor furthr numrical valuation (i) d d v u = = d du v k d M whr k is a constant k = A (d ), or bttr AF corrct substitution of thir trms u = m dv = n d du m v n d = = m both diffrntiation and intgration must b corrct = d 8 = ( ) ( 0 ) Al = + [ ] [ ] m (dp on M only) corrct substitution and attmpt at subtraction in a + b + c (may b in stags) = 8 0 A 7 or 8 0 ignor furthr numrical valuation (ii) ( ) ( 0 ) ( ) v= π 9 d M condon omission of brackts, limits = 9π 8 0 AF 9π (thir act b(i)) Total 6 TOTAL 75

12 Scald mark unit grad boundaris - January 0 ams A-lvl Ma. Scald Mark Grad Boundaris and A Convrsion Points Cod Titl Scald Mark A A B C D E LAW0 LAW UNIT LAW0 LAW UNIT MD0 MATHEMATICS UNIT MD MFP MATHEMATICS UNIT MFP MMA MATHEMATICS UNIT MMA 00 no candidats wr ntrd for this unit MMB MATHEMATICS UNIT MMB 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 MFP MATHEMATICS UNIT MFP MPC MATHEMATICS UNIT MPC MFP MATHEMATICS UNIT MFP MPC MATHEMATICS UNIT MPC MEST MEDIA STUDIES UNIT MEST MEDIA STUDIES UNIT MEST MEDIA STUDIES UNIT MEST MEDIA STUDIES UNIT PHIL PHILOSOPHY UNIT

Note If the candidate believes that e x = 0 solves to x = 0 or gives an extra solution of x = 0, then withhold the final accuracy mark.

Note If the candidate believes that e x = 0 solves to x = 0 or gives an extra solution of x = 0, then withhold the final accuracy mark. . (a) Eithr y = or ( 0, ) (b) Whn =, y = ( 0 + ) = 0 = 0 ( + ) = 0 ( )( ) = 0 Eithr = (for possibly abov) or = A 3. Not If th candidat blivs that = 0 solvs to = 0 or givs an tra solution of = 0, thn withhold