General Certificate of Education Advanced Subsidiary Eamination January 0 Mathematics MPC Unit Pure Core Monday 0 January 0 9.00 am to 0.0 am For this paper you must have: the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed hour 0 minutes Instructions Use black ink or black ball-point pen. Pencil should only be used for drawing. Fill in the boes at the top of this page. Answer all questions. Write the question part reference (eg (a), (b)(i) etc) in the left-hand margin. You must answer the questions in the spaces provided. Do not write outside the bo 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 maimum mark for this paper is 7. Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. P6487/Jan/MPC 6/6/ MPC
The diagram shows a sector OAB of a circle with centre O and radius cm. cm A O y 4cm cm B The angle between the radii OA and OB is y radians. The length of the arc AB is 4 cm. (a) Find the value of y. ( marks) (b) Find the area of the sector OAB. ( marks) (a) Write down the values of p, q and r given that: (i) 8 ¼ p ; ( mark) (ii) (iii) 8 ¼ q ; pffiffi ¼ r. ( mark) ( mark) (b) Find the value of for which pffiffi ¼ 8. ( marks) P6487/Jan/MPC
The triangle ABC, shown in the diagram, is such that AB ¼ cm, AC ¼ 8 cm, BC ¼ 0 cm and angle BAC ¼ y. A cm y 8cm B 0 cm C (a) Show that y ¼ 97:9, correct to the nearest 0.. ( marks) (b) (i) (ii) Calculate the area of triangle ABC, giving your answer, in cm, to three significant figures. ( marks) The line through A, perpendicular to BC, meets BC at the point D. Calculate the length of AD, giving your answer, in cm, to three significant figures. ( marks) 4 (a) Use the trapezium rule with four ordinates (three strips) to find an approimate value ð : pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi for 7 þ 4 d, giving your answer to three significant figures. (4 marks) 0 (b) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi The curve with equation y ¼ 7 þ 4 is stretched parallel to the -ais with scale factor to give the curve with equation y ¼ gðþ. Write down an epression for gðþ. ( marks) (a) Using the binomial epansion, or otherwise, epress ð Þ in ascending powers of. ( marks) (b) Show that the epansion of ð þ yþ 4 ð yþ is 7y þ py þ qy þ y 4 (c) where p and q are constants to be found. (4 marks) ðh p Hence find þ ffiffi 4 pffiffi i d, epressing each coefficient in its simplest form. (4 marks) Turn over s P6487/Jan/MPC
4 6 A geometric series has third term 6 and sith term 97. (a) (i) Show that the common ratio of the series is. ( marks) (ii) Find the first term of the series. ( marks) (b) The nth term of the series is u n. (i) Show that X0 n¼ u n ¼ ð 0 Þ. ( marks) (ii) Find the least value of n such that u n > 4 0. ( marks) 7 A curve C is defined for > 0 by the equation y ¼ þ þ 8 and is sketched 4 below. y O (a) Given that y ¼ þ þ 8 dy, find 4 d. ( marks) (b) Find an equation of the tangent at the point on the curve C where ¼. ( marks) (c) The curve C has a minimum point M. Find the coordinates of M. (4 marks) (d) (i) ð Find þ þ 8 4 d. ( marks) (e) (ii) Hence find the area of the region bounded by the curve C, the -ais and the lines ¼ and ¼. ( marks) 0 The curve C is translated by to give the curve y ¼ fðþ. Given that the -ais k is a tangent to the curve y ¼ fðþ, state the value of the constant k. ( mark) P6487/Jan/MPC
8 (a) Given that log k log k ¼, epress k in terms of. Give your answer in a form not involving logarithms. (4 marks) (b) Given that log a y ¼ and that log 4 a ¼ b þ, show that y ¼ p, where p is an epression in terms of b. ( marks) 9 (a) Solve the equation tan ¼ in the interval 0 4 4 60, giving your answers to the nearest degree. ( marks) (b) (i) Given that 7 sin y þ sin y cos y ¼ 6 show that tan y þ tan y 6 ¼ 0 ( marks) (ii) Hence solve the equation 7 sin y þ sin y cos y ¼ 6 in the interval 0 4y460, giving your answers to the nearest degree. (4 marks) Copyright ª 0 AQA and its licensors. All rights reserved. P6487/Jan/MPC
Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 Key to mark scheme abbreviations M mark is for method m or dm mark is dependent on one or more M marks and is for method A mark is dependent on M or m marks and is for accuracy B mark is independent of M or m marks and is for method and accuracy E mark is for eplanation or ft or F follow through from previous incorrect result CAO correct answer only CSO correct solution only AWFW anything which falls within AWRT anything which rounds to ACF any correct form AG answer given SC special case OE or equivalent A, or (or 0) accuracy marks EE deduct marks for each error NMS no method shown PI possibly implied SCA substantially correct approach c candidate sf significant figure(s) dp decimal place(s) No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded.
Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 MPC Q Solution Marks Total Comments (a) Arc = rθ arc = rθ seen or used. PI by correct θ 4 4= θ θ = = 0.8 A (θ =) 4 OE (b) Area of sector = r θ Area = = 0.8 0 = (cm ) AF Total 4 (a)(i) (p =) B r θ seen or used within (b). PI Ft on. c s eact value for θ in part (a) provided c s area 0 (ii) (q =) BF If not correct, ft on p (iii) ( r = ) B OE + Using a law of indices or logs correctly to combine at least two of the powers of PI If not correct, ft on = q r provided method shown (b) = = = AF Total (a) 0 = 8 + 8 cosθ Use of the cosine rule PI by net line 8 + 0 cos θ = ( = = 0.7) m Rearrangement 8 80 θ = 97.90(...) = 97.9º (to nearest 0.º) A CSO (Must see either eact value for cosθ or at least 4sf value for either cosθ or θ before the printed answer 97.9º) AG (b)(i) Area = 8 sinθ OE = 9.80 = 9.8 (cm ) to sf A Condone > sf (ii) Area of triangle = 0. BC AD Or valid method to find sin B or sin C or B or C AD = [Ans.(b)(i)] [0. BC] m Or AD = sin B; or AD = 8 sin C OE 9.80.. AD = =.96.. =.96 (cm) to sf A Condone > sf Total 8 4
Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 MPC (cont) Q Solution Marks Total Comments 4(a) h = 0. B PI f() = 7 + 4 I h/{ } { }= f(0)+ f(.)+[f(0.)+ f()] OE summing of areas of the trapezia.. { }= 4 + 9. + ( 7.7 + ) = +9.7... +(.76 +.677 ) A OE Accept dp rounded or truncated as evidence for surds (I ) 0. 8.0 = 7.08 (to sf) A 4 Must be 7.08 (b) g() = 7 4 + = + 4 k + 4, Any form which simplifies to k 7, k 0 or which simplifies to + 4 A ACF Total 6 (a) ( ) = + terms correct or (±) (±) (±) seen A All correct (b) 4 4 ( + y) = + 4 y+ 6 y + 4 y + y 4 terms correct, accept unsimplified A All terms correct and simplified at some stage 4 ( + y) ( y) = 4 (4y+ y) + (6y y ) + (4 y + y ) + y 4 = 7y+ y + y + y A, 4 A Be convinced as part answer is given (as required with p= and q=) (A for three terms found correctly or if found correct values for p and q but did not show 7y+y 4.) (c) ( + ) ( ) 4 d= ( 7 + + + ) d ( 7 0. + +. + ) d.. 7 = + + + (+c) m.. Use of part (b).y OE before any integration Correct integration of an k term where k is non-integer = 4.. + + + (+c) A,F 4 Coeffs simplified; condone absent (+c) Ft on c s p and q ie nd p term + and rd q. term is +. (AF for three of these four ft terms or for four correct ft terms unsimplified) Total 0
Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 MPC (cont) Q Solution Marks Total Comments 6(a)(i) ar = 6; ar = 97; For ar = 6 or ar = 97 or for seeing 6r = 97 r = 97 6 (= 7) r = A CSO AG Full valid completion. (ii) a = 6 OE. PI a = 4 A Correct answer without working scores the two marks (b)(i) 0 un = n= S 0 a r = r 0 ( ) OE = 0 4( ) = ( 0 0 ) = ( ) A CSO AG Be convinced (ii) un a n = B Seen or used 4 > 4 0 > 0 n n ( n )log ( > ) log0 n > ; n >.4... log 0 ( n >.4... and n is an integer so least value of n is) n = A CSO Total 9 Or finds values of un for appropriate adjacent integer values of n so that u n s are either side of 4 0 6
Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 MPC (cont) Q Solution Marks Total Comments 7(a) 8 4 8 y = + + = + + 8 4 B For 4 4 = 8 PI by correct differentiation of rd term dy = or d A k OE For either (b) When =, y = B When =, d y d y when = d Tangent: y = ( ) AF Only ft on c s answer to (a). Any correct (ft on c s (a) ) form. (c) = 0 = or c s d y 0 d = 0 = m Attempt to form net line = A CSO (Coordinates of M) (,.) A 4 CSO n = const ( 0). PI by 8 (d)(i) + + d 4 8 = + + c A B Power correctly obtained 8 + + c (ii) Area = = 8 + 8 + 6 + = 9 + 7 = 4 A 6 Attempting to calculate F() F() where F() is c s answer to part (d)(i) provided F is not just the c s integrand (++8/ 4 ) OE Accept 6.8 or better provided d(i) used (e) k =. BF Ft on y M from part (c). Total 6 7
Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 MPC (cont) Q Solution Marks Total Comments 8(a) log log = A valid law of logs used correctly k k = logk Another valid law of logs used correctly or correct method to reach log f() = log k log k = log k [or log = logk ] A PI by net line k = k ie k = A 4 Accept either of these two forms. (b) log a log 4 a = b+ y = a a 4 b + = For either equation Elimination of a from two correct b+ y = (4 ) m equations not involving logarithms y ( ) b + = ; y 6 b + = A CSO Either form acceptable Total 7 8
Mark Scheme General Certificate of Education (A-level) Mathematics Pure Core January 0 MPC (cont) Q Solution Marks Total Comments 9(a) tan = = tan ( ) ( = 7.6...) PI eg by 7(.6..) or 7(.6..) seen = 08º, 88º A,A Condone more accurate answers. (08.449, 88.449 ). [Ignore answers outside interval; If more than answers inside interval from A marks for each etra to a min of 0] (b)(i) 7sin θ + sinθcosθ = 6(cos θ + sin θ) θ + θ = used; OE cos sin 7sin θ 6sin θ + sinθcosθ 6cos θ = 0 sin θ + sinθcosθ 6cos θ = 0 sin θ sinθ sinθ + 6= 0 tanθ cos θ cosθ cosθ = used tan θ + tanθ 6 = 0 A CSO AG (ii) ( θ )( θ ) tan + tan = 0 tanθ = or tanθ = A Need both Factorise or other valid method to solve quadratic θ = 08º, 88º; θ = 6º, 4º; BF,F 4 Only ft on (a) for the c s two + ve tan ( ) vals. [B if correct (ft)] Condone more accurate answers. (08.449, 88.449 ; 6.449, 4.449 ) [Ignore answers outside interval; If more than answers for each inside interval, for each etra from Bs to a min of 0] Total 0 TOTAL 7 9
Scaled mark unit grade boundaries - January 0 eams A-level Ma. Scaled Mark Grade Boundaries and A Conversion Points Code Title Scaled Mark A A B C D E MD0 GCE MATHEMATICS UNIT D0 7-6 49 4 7 MFP GCE MATHEMATICS UNIT FP 7-6 6 49 4 7 MA GCE MATHEMATICS UNIT A 00 no candidates were entered for this unit MB GCE MATHEMATICS UNIT B 7-6 46 9 MPC GCE MATHEMATICS UNIT PC 7-6 49 4 6 0 MSA GCE MATHEMATICS UNIT SA 00-84 74 64 4 44 MS/SSA/W GCE MATHEMATICS UNIT SA - WRITTEN 7 64 4 MS/SSA/C GCE MATHEMATICS UNIT SA - COURSEWORK 0 0 MSB GCE MATHEMATICS UNIT SB 7-6 46 9 MD0 GCE MATHEMATICS UNIT D0 7 69 6 6 0 44 8 MFP GCE MATHEMATICS UNIT FP 7 67 60 4 4 6 MMB GCE MATHEMATICS UNIT MB 7 6 47 40 6 MPC GCE MATHEMATICS UNIT PC 7-6 4 47 40 MSB GCE MATHEMATICS UNIT SB 7 66 9 4 8 XMCA GCE MATHEMATICS UNIT XMCA 0 9 8 70 9 48 MFP GCE MATHEMATICS UNIT FP 7 66 9 4 8 MPC GCE MATHEMATICS UNIT PC 7 66 9 4 8 MFP4 GCE MATHEMATICS UNIT FP4 7 6 47 40 6 MPC4 GCE MATHEMATICS UNIT PC4 7 68 6 4 47 4 MEST GCE MEDIA STUDIES UNIT 80-6 49 4 8 MEST GCE MEDIA STUDIES UNIT 80-6 4 4 6 8 MEST GCE MEDIA STUDIES UNIT 80 69 8 47 7 7 7 MEST4 GCE MEDIA STUDIES UNIT 4 80 7 6 4 0 PHIL GCE PHILOSOPHY UNIT 90-4 48 4 7 PHIL GCE PHILOSOPHY UNIT 90-6 6 0 44 8