General Certificate of Education Advanced Level Examination January 2010

Transcription

1 General Certificate of Education Advanced Level Eamination January 00 Mathematics MPC3 Unit Pure Core 3 Friday 5 January pm to 3.00 pm For this paper you must have: an 8-page answer book the blue AQA booklet of formulae and statistical tables an insert for use in Question (enclosed). You may use a graphics calculator. Time allowed hour 30 minutes Instructions Use black ink or black ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Eamining Body for this paper is AQA. The Paper Reference is MPC3. Answer all questions. Show all necessary working; otherwise marks for method may be lost. Fill in the boes at the top of the insert. Information The marks for questions are shown in brackets. The maimum mark for this paper is 75. Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. P090/Jan0/MPC3 6/6/6/ MPC3

2 Answer all questions. A curve has equation y ¼ e ð þ Þ. (a) (b) Show that dy d ¼ e ð5 3 Þ. Find the eact values of the coordinates of the stationary points of the curve. (3 marks) (5 marks) [Figure, printed on the insert, is provided for use in this question.] (a) (i) Sketch the graph of y ¼ sin, where y is in radians. State the coordinates of the end points of the graph. (3 marks) (ii) By drawing a suitable straight line on your sketch, show that the equation sin ¼ þ has only one solution. ( marks) (b) The root of the equation sin ¼ þ isa. Show that 0:5 < a <. ( marks) (c) The equation sin ¼ þ can be rewritten as ¼ sin þ. (i) (ii) Use the iteration nþ ¼ sin n þ with ¼ 0:5 to find the values of and 3, giving your answers to three decimal places. ( marks) The sketch on Figure shows parts of the graphs of y ¼ sin þ and y ¼, and the position of. On Figure, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of and 3 on the -ais. ( marks) P090/Jan0/MPC3

3 3 3 (a) Solve the equation cosec ¼ 3 giving all values of in radians to two decimal places, in the interval 0 p. ( marks) (b) By using a suitable trigonometric identity, solve the equation cot ¼ cosec giving all values of in radians to two decimal places, in the interval 0 p. (6 marks) (a) Sketch the graph of y ¼j8 j. ( marks) (b) Solve the equation j 8 j¼. ( marks) (c) Solve the inequality j 8 j >. ( marks) 5 (a) Use the mid-ordinate rule with four strips to find an estimate for giving your answer to three significant figures. ð 0 lnð þ 5Þ d, ( marks) (b) A curve has equation y ¼ lnð þ 5Þ. (i) Show that this equation can be rewritten as ¼ e y 5. ( mark) (ii) The region bounded by the curve, the lines y ¼ 5 and y ¼ 0 and the y-ais is rotated through 360 about the y-ais. Find the eact value of the volume of the solid generated. ( marks) (c) The graph with equation y ¼ lnð þ 5Þ is stretched with scale factor parallel to the 0 -ais, and then translated through to give the graph with equation y ¼ fðþ. 3 Write down an epression for fðþ. (3 marks) Turn over for the net question P090/Jan0/MPC3 Turn over s

4 6 The functions f and g are defined with their respective domains by fðþ ¼e 3, for all real values of gðþ ¼ 3 þ, for real values of, 6¼ 3 (a) Find the range of f. ( marks) (b) The inverse of f is f. (i) Find f ðþ. (3 marks) (ii) Solve the equation f ðþ ¼0. ( marks) (c) (i) Find an epression for gf ðþ. ( mark) (ii) Solve the equation gfðþ ¼, giving your answer in an eact form. (3 marks) 7 It is given that y ¼ tan. (a) sin dy By writing tan as, use the quotient rule to show that cos d ¼ pð þ tan Þ, where p is a number to be determined. (3 marks) (b) Show that d y d ¼ qyð þ y Þ, where q is a number to be determined. (5 marks) 8 (a) Using integration by parts, find ð sinð Þ d. (5 marks) (b) Use the substitution u ¼ to find ð d, giving your answer in terms of. (6 marks) END OF QUESTIONS Copyright Ó 00 AQA and its licensors. All rights reserved. P090/Jan0/MPC3

5 MPC3 - AQA GCE Mark Scheme 00 January series Key to mark scheme and abbreviations used in marking M m or dm A B E mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for eplanation or ft or F follow through from previous incorrect result MC mis-copy CAO correct answer only MR mis-read CSO correct solution only RA required accuracy AWFW anything which falls within FW further work AWRT anything which rounds to ISW ignore subsequent work ACF any correct form FIW from incorrect work AG answer given BOD given benefit of doubt SC special case WR work replaced by candidate OE or equivalent FB formulae book A, or (or 0) accuracy marks NOS not on scheme EE deduct marks for each error G graph NMS no method shown c candidate PI possibly implied sf significant figure(s) SCA substantially correct approach 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. However, there are situations in some units where part marks would be appropriate, particularly when similar techniques are involved. Your Principal Eaminer will alert you to these and details will be provided on the mark scheme. 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. 3

6 MPC3 - AQA GCE Mark Scheme 00 January series MPC3 y e e y = Ae a+ b ± Be + = + ( ) (a) ( ) ( ) + M ( ) ( ) = e or ( ) where A and B are non-zero constants A All correct e - 6 e +0e = e 5 3 A 3 AG; all correct with no errors, or y= e + e e nd line (OE) must be seen Condone incorrect order on final line y = e + e +. e + e + 8e = e 6e + 0e ( ) (M) (A) = e 5 3 (A) A e + Be + Ce + De + Ee All correct AG; all correct with no errors, 3 rd line (OE) must be seen (b) ( )( ) + 5 ( = 0) M OE Attempt at factorisation ( ± ± 5)( ± ± ) or formula with at most one error 5 =, A Both correct and no errors =, y= e m For e b =, y= e AF SC = only scores MA0 y= a attempted Either correct, follow through only from incorrect sign for A 5 CSO solutions only Total 8 Note: withhold final mark for etra solutions Note: approimate values only for y can score m only

7 MPC3 - AQA GCE Mark Scheme 00 January series MPC3 (cont) (a)(i) A B correct shape passing through origin and stopping at A and B B A π, π B, B B 3 SC A(, 90) and B (, 90) scores B (ii) line intersecting their curve (positive gradient, positive y intercept) M Correct statement A one solution only, stated or indicated on sketch - must be in the first quadrant (ie curve intersects line once) (b) ( ) ( ) () () LHS 0.5 = 0.5 RHS 0.5 =. LHS =.6 RHS =.3 At 0.5 LHS < RHS, At LHS > RHS 0.5 < α < or ( ) ( ) f = sin f( 0.5) = 0.6 AWRT f() = 0.3 M Change of sign 0.5 < α < (A) or f ( ) = sin + f ( 0.5) = 0. Attempt (M) f () = 0. Change of sign 0.5 < α < (A) or ( ) A CSO Must have scored B for graph in (a)(i) f( ) must be defined (M) Allow f( 0.5) < 0 f( ) > 0 f( ) must be defined f = sin f( ) must be defined ( ) () f 0.5 =. attempt f =.3 (M) Change of sign 0.5 < α < (A) 5

8 MPC3 - AQA GCE Mark Scheme 00 January series MPC3 (cont) (c)(i) = 0.90 M Sight of AWRT 0.90 or AWRT = 0.9 A These values only (ii) M Staircase, (vertical line) from to curve, horizontal to line, vertical to curve O 3 A, 3 appro correct position on -ais Total 6

9 MPC3 - AQA GCE Mark Scheme 00 January series MPC3 (cont) 3(a) sin =, 3 or sight of ± 0.3, ± 0.π or ± 9.7 (or better) M = 0.3,.8( 0) AWRT A Penalise if incorrect answers in range; ignore answers outside range (b) cosec = cosec M Correct use of cot = cos ec ( ) cosec cosec 0 + = A ( cosec )( cosec 3)( 0) + = m Attempt at Factors Gives cosec or when epanded Formula one error condoned cosec =, 3 A Either Line sin =, 3 sin = = 3.39, 6.03 AWRT 3 correct or their two answers from (a) BF and 3.39, ,.8( 0 ) AWRT B 6 correct and no etras in range ignore answers outside range SC 9.7, 60.53, 9.8, 35.5 B Alternative cos sin = sin cos = sin sin sin = sin sin 0 sin sin (M) = (A) ( )( ) Correct use of trig ratios and multiplying by sin 0 = sin + 3sin (m) Attempt at factors as above sin =, 3 (A) (BF) (B) Total 8 As above 7

10 MPC3 - AQA GCE Mark Scheme 00 January series MPC3 (cont) (a) y Modulus graph V shape in st quad going M into nd quad, touching -ais. Must cross y-ais Condone not ruled A and 8 labelled (b) = B One correct answer = 6 B Second correct answer and no etras Condone answers shown on the graph, if clearly indicated (c) > 6 < 5(a) y B B Total 6 B M A One correct answer Second correct answer and no etras and no further incorrect statement eg 6 < < or < > 6 SC 6, scores B values correct PI 3+ y values correct to sf or better or eact values.98, 3.8 / 9,. / 5,.77 for y ( or better) = 3 y =. A (Note:. with evidence of mid-ordinate rule with four strips scores /) (b)(i) y = ln ( + 5) y e 5 y = + OE = e 5 B AG Must see middle line, and no errors (ii) ( π) ( e y 5) ( dy) (c) 8 M ( ) 0 ( π) e y 5y ( 5) = A 0 5 ( π) ( e 50) ( e 5) = m F ( 0 ) F( 5 ) 0 5 V = π e e 5 A ( y = )ln M Condone omission of brackets around f (y) throughout CSO including correct notation must see dy ISW if evaluated seen, condone ln +... B + 3 A 3 CSO mark final answer (no ISW) Total 8

11 MPC3 - AQA GCE Mark Scheme 00 January series MPC3 (cont) f > 3 M > 3, 3 f 3 6(a) ( ) > or ( ) (b)(i) y = e 3 y + 3= e ( y ) A Allow y > 3 ln + 3 = M swap and y ( f ( ) ) ln( 3) M = + A 3 Alternative e -3 ln + 3 (M) (M) ln( + 3) y = (A) attempt to isolate: ln ( y± A) = B or reverse OE with no further incorrect working Condone y =.. (ii) + 3= M (c)(i) ( ) for putting their p( ) = from kln ( p ( ) ) in their part (b)(i) = A CSO SC: B = - with no working, if full marks gained in part (b)(i) (gf = ) 3e ( 3 ) + substituting f into g either OE B (=) ISW 3e 5 (ii) 3e 5 = = OE M Correct removal of their fraction 3e 5 e = = ln m Correct use of logs leading to k = ln a b = ln OE A 3 CSO No ISW ecept for numerical evaluation Total 9

12 MPC3 - AQA GCE Mark Scheme 00 January series MPC3 (cont) 7(a) dy cos. cos sin. sin ± Acos ± Bsin = M d cos cos cos + sin = or better cos A Both terms correct = + tan CSO A 3 All correct ( ) or dy cos. cos sin. sin = d cos (M) ± Acos ± Bsin cos (b) cos cos sin sin = + cos cos cos cos (A) or better = ( + tan ) CSO (A) All correct d y tan d M A tan f() sec m f() = B sec = 3tan sec AF ft 8 their p ( ) ( y ) = 3 tan + tan m = 3y + A 5 CSO Alternative Solutions sin tan cos from part (a) Previous two method marks must have been earned y = + = + y = 3 3 (M) Acos ± Bsin cos sin cos + s in cos sin where A and B are cos cos constants or trig functions. (m) Where A is msin and B is ncos 8 sin cos cos + sin = cos (AF) ft 8 their p from part (a) = 3 tan sec (m) ktan sec ( y ) = 3y + (A) CSO or dy sec d = d y sec.sectan d = (M) A sec f() (m) f() = B sec tan = 3 sec tan (AF) ft 8 their p from part (a) = 3 ( + tan ) tan Previous two method marks must have (m) been earned = 3 y + y (A) CSO ( ) 0

13 MPC3 (cont) MPC3 - AQA GCE Mark Scheme 00 January series 7(b) or dy = ( + tan ) d u = tan dy = + u d d y du = (8) u d d (M) du tan u d (m) d y 8 u( u ) d (A) = 3 u( + u ) (m) 8(a) sin ( ) = 3 y( + y ) (A) Total 8 d u= dv d = sin ( ) M sin f ( ), ( ) attempted d d du = v= cos( ) d A All correct condone omission of brackets ( = ) cos( ) m correct substitution of their terms into parts cos ( ) ) = cos( ) + cos( ) (d ) A All correct condone omission of brackets CSO condone missing + c and d = cos( ) + sin ( ) + c A 5 Condone missing brackets around if recovered in final line ISW (b) u= 'du= d ' M OE ( u + ) m A d d u = u u + u+ d = u 8 u = u+ + du 8 u u = + u+ lnu 8 ( ) = + ( ) + ln ( ) + c 8 A All in terms of u All correct PI from later working ( ) B u + or + ln u 8 A 6 ( + ) or = + ln ( ) + c 8 CSO condone missing + c only Total TOTAL 75 ISW

14 Scaled mark component grade boundaries - January 00 eams klm A units (legacy) Component Maimum Scaled Mark Grade Boundaries Code Component Title Scaled Mark A B C D E MPC3 GCE MATHEMATICS UNIT PC MPC GCE MATHEMATICS UNIT PC MS/SSA/W GCE MATHEMATICS UNIT SA WRITTEN MS/SSA/C GCE MATHEMATICS UNIT SA CWK MSB GCE MATHEMATICS UNIT SB MSB GCE MATHEMATICS UNIT SB XMCAS GCE MATHEMATICS UNIT XMCAS XMCA GCE MATHEMATICS UNIT XMCA MED GCE MEDIA STUDIES UNIT MED5 GCE MEDIA STUDIES UNIT MED6 GCE MEDIA STUDIES UNIT HEB GCE MODERN HEBREW UNIT MUS GCE MUSIC UNIT MUS6 GCE MUSIC UNIT PAN GCE PANJABI UNIT PLY GCE PHILOSOPHY UNIT PLY5 GCE PHILOSOPHY UNIT PLY6 GCE PHILOSOPHY UNIT PA0/ GCE PHYSICS A UNIT OTQ PA0/ GCE PHYSICS A UNIT WRITTEN PHAP GCE PHYSICS A UNITS 5-9 PRACTICAL PHAC GCE PHYSICS A UNITS 5-9 CWK PHA5/W GCE PHYSICS A UNIT 5 WRITTEN