Mathematics (JAN13MPC301) General Certificate of Education Advanced Level Examination January Unit Pure Core TOTAL

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1 Centre Number Candidate Number For Eaminer s Use Surname Other Names Candidate Signature Eaminer s Initials Mathematics Unit Pure Core Wednesday January General Certificate of Education Advanced Level Eamination January 9. 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 hour minutes MPC 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), (i) etc) in the left-hand margin. You must answer each question in the space provided for that question. If you require etra space, use an AQA supplementary answer book; do not use the space provided for a different question. 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. 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. You do not necessarily need to use all the space provided. Question TOTAL Mark (JANMPC) P5899/Jan/MPC 6/6/6/ MPC

2 Do not write outside the bo Answer all questions. Answer each question in the space provided for that question. (a) Show that the equation 6 þ ¼ has a root a, where < a <. ( marks) Show that the equation 6 þ ¼ can be rearranged into the form ¼ 6 ( mark) (c) Use the recurrence relation nþ ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffi 6, with ¼ :5, to find the value of, n giving your answer to four significant figures. ( marks) QUESTION PART REFERENCE Answer space for question () P5899/Jan/MPC

3 Do not write outside the bo (a) Use Simpson s rule, with five ordinates (four strips), to calculate an estimate for ð þ d Give your answer to four significant figures. ( marks) Show that the eact value of ð þ d is ln k, where k is an integer. (5 marks) QUESTION PART REFERENCE Answer space for question () P5899/Jan/MPC

4 6 Do not write outside the bo (a) Find dy d when y ¼ e þ ln ( marks) QUESTION PART REFERENCE (i) Given that u ¼ sin du, show that þ cos d ¼ þ cos (ii) Hence show that if y ¼ ln Answer space for question sin þ cos, then dy d ¼ cosec.. ( marks) ( marks) (6) P5899/Jan/MPC

5 8 Do not write outside the bo The diagram shows a sketch of the curve with equation y ¼ fðþ. y O (a) On the aes below, sketch the curve with equation y ¼jfðÞj. ( marks) Describe a sequence of two geometrical transformations that maps the graph of y ¼ fðþ onto the graph of y ¼ fð Þ. ( marks) QUESTION PART REFERENCE (a) Answer space for question y O (8) P5899/Jan/MPC

6 Do not write outside the bo 5 The function f is defined by fðþ ¼, for real values of, where (a) State the range of f. ( marks) The inverse of f is f. (i) Write down the domain of f. ( mark) (ii) Find an epression for f ðþ. ( marks) (c) The function g is defined by gðþ ¼ln j j, for real values of, where 6¼ The curve with equation y ¼ gðþ is sketched below. y O P (i) The curve y ¼ gðþ intersects the -ais at the origin and at the point P. Find the -coordinate of P. ( marks) (ii) State whether the function g has an inverse. Give a reason for your answer. ( mark) (iii) Show that gfðþ ¼ln j kj, stating the value of the constant k. (iv) Solve the equation gf ðþ ¼. ( marks) ( marks) () P5899/Jan/MPC

7 Do not write outside the bo 6 (a) Show that sec ðsec þ Þðsec Þ can be written as cosec. ( marks) Hence solve the equation sec ðsec þ Þðsec Þ ¼ cosec þ giving the values of to the nearest degree in the interval 8 < < 8. (6 marks) (c) Hence solve the equation sec ðy 6 Þ ¼ cosecðy 6 Þþ ðsecðy 6 ÞþÞðsecðy 6 ÞÞ giving the values of y to the nearest degree in the interval < y < 9. ( marks) QUESTION PART REFERENCE Answer space for question 6 () P5899/Jan/MPC

8 8 Do not write outside the bo 7 A curve has equation y ¼ cos. (a) Find an eact equation of the tangent to the curve at the point on the curve where ¼ p. (5 marks) The region shaded on the diagram below is bounded by the curve y ¼ cos and the -ais from ¼ to¼ p. y O p By using integration by parts, find the eact value of the area of the shaded region. (5 marks) QUESTION PART REFERENCE Answer space for question 7 (8) P5899/Jan/MPC

9 Do not write outside the bo 8 (a) Show that ð ln e d ¼ 8 e ( marks) Use the substitution u ¼ tan to find the eact value of ð p sec p ffiffiffiffiffiffiffiffiffi tan d (8 marks) QUESTION PART REFERENCE Answer space for question 8 () P5899/Jan/MPC

10 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 ) 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.

11 MPC - AQA GCE Mark Scheme January series MPC Q Solution Marks Total Comments (a) f f 6 must have both values f M correct allow f and f only if f is defined and no errors seen change of sign A must have both statement and interval which may be written in words/symbols 6 or or B AG (c) 6.66 must see one of these lines and no errors at least sf needed B.5 PI by correct.6 B SC if BB scored and =.67 Total 5

12 MPC - AQA GCE Mark Scheme January series Q Solution Marks Total Comments (a) y y. y. B all 5 -values PI by 5 correct y-values y.7 y. at least y-values eact or rounded or B 8 truncated to at least sf...7. M correct use of Simpson s rule using and and correctly with candidate s 5 y-values =. A CAO (must be eactly this value) d ln M A for k ln all correct; limits not needed ln8 ln AF For k (ln8 ln) ln 9 AF combining candidate s logarithms correctly (must be seen) ln A 5 CAO (must be eactly this) NMS scores /5 Total 9

13 MPC - AQA GCE Mark Scheme January series Q Solution Marks Total Comments dy B B for one term correct (a) e d B B all correct (i) cos cos sin sin cos du d cos cos cos sin sin cos cos sin cos cos cos M A Acso cos clear attempt at quotient/product rule condone poor use of brackets any correct form seen AG be convinced correct use of brackets and correct notation used throughout (eg A if cos etc seen) (ii) dy cos d sin cos sin OE M correct use of chain rule cosec A AG, must see Total 7 and no errors seen; sin condone incorrect use of brackets only if penalised in part (i)

14 MPC - AQA GCE Mark Scheme January series Q Solution Marks Total Comments (a) M reflection in the -ais for the negative f and remainder as given on sketch A correct curvatures, correct cusp at = condone straight lines for < and > marked on -ais Either. Stretch M and either or. -ais. by factor.5 A, and (followed by) translation E.5 B or translation (E) (B) (followed by). Stretch (M) and either or. -ais. by factor.5 (A), and Total 6

15 MPC - AQA GCE Mark Scheme January series Q Solution Marks Total Comments 5(a) M f,f,range f A (i) (ii) BF correct or FT from (a) y y M either order M for correctly changing the subject or reversing f M operations; M for replacing y with f A (dependent on both M marks) correct sign (c)(i) M Or e or OE A CAO, NMS OE scores / (ii) g has NO inverse because two values of map to one value (of y) or it is many-one or it is not oneone or it is two-one B must indicate no inverse with valid reason; do not accept contradictory reasons (iii) ln M ln 5 A NMS scores /, condone k = 5 after correct epression seen (iv) ln (or or e or e seen) M k etc, for candidate s positive integer, k 6, or candidate s k or k 6, 6, AF AF eact values PI by correct answers 6, A CAO, rejecting the positive Total 5 5

16 MPC - AQA GCE Mark Scheme January series Q Solution Marks Total Comments 6(a) sec sec sec sec sec sec tan used M M for correct use of sec tan at least once or cosec cot sec tan or tan tan cos tan or cot sin A Shown, with no errors cosec A AG (No errors, omissions or poor notations seen) cosec cosec cosec cosec B must have = correct solution of the quadratic, or by completing the square cosec or (... and...) M cosec PI by values for sin sin BF BF for cosec seen or implied sin. and.768 or.767 A PI 6,5, 5, B B 6 B for any three values correct AWRT B for all four values correct AWRT and no etras in the interval 8 8 (c) 6 M where is a written value from candidate s in degrees PI by their answer, 5 A CSO Ignore solutions outside interval 9 Total 6

17 MPC - AQA GCE Mark Scheme January series Q Solution Marks Total Comments 7(a) y cos dy cos sin d gradient of the tangent π π π Acos B sin π A must have π an equation of the tangent is π yπ M A m anything reducible to Acos Bsin where A and B are non-zero integers OE, all correct substituting π into candidate s derived function using correct d y d A 5 OE, dependent on previous A π cosd u A dv cos d M all terms in this form seen or used du A v Bsin d B or A = and B =, etc π π sin sind m π π sin cos AF π A 5 OE, eact value correct substitution of candidate s terms into integration by parts formula condone missing limits candidate's second integration completed correctly FT on one error including coefficient condone missing limits Total 7

18 MPC - AQA GCE Mark Scheme January series Q Solution Marks Total Comments 8(a) e d ke or e e k M where k is a rational number ln ln e d e or e e ln A correct integration condone missing limits e e eliminating ln e 8 A AG, be convinced ln () e e A correct (no decimals) u tan du sec d M PI below, condone du sec d Replacing d by du in integral A or du sec u sec u B PI below u this could be gained by changing u to tan after the integration and using π B u and π sec tan d du all in terms of u including replacing d u ud u or u u M u all correct, condone omission of du 5 u u du A must be in this form 7 u u 7 A accept correct unsimplified form A 8 CAO Total TOTAL 75 8

19 Scaled mark unit grade boundaries - January eams A-level Maimum Scaled Mark Grade Boundaries and A Conversion Points Code Title Scaled Mark A A B C D E LAW LAW UNIT MD MATHEMATICS UNIT MD MD MATHEMATICS UNIT MD MFP MATHEMATICS UNIT MFP MFP MATHEMATICS UNIT MFP MFP MATHEMATICS UNIT MFP MFP MATHEMATICS UNIT MFP MMB MATHEMATICS UNIT MMB MMB MATHEMATICS UNIT MMB MPC MATHEMATICS UNIT MPC MPC MATHEMATICS UNIT MPC MPC MATHEMATICS UNIT MPC MPC MATHEMATICS UNIT MPC MSA MATHEMATICS UNIT MSA MS/SSA/W MATHEMATICS UNIT SA - WRITTEN MS/SSA/C MATHEMATICS UNIT SA - COURSEWORK 5 MSB MATHEMATICS UNIT MSB MSB MATHEMATICS UNIT MSB MEST MEDIA STUDIES UNIT MEST MEDIA STUDIES UNIT MEST MEDIA STUDIES UNIT MEST MEDIA STUDIES UNIT

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