A-LEVEL Mathematics. Paper 1 Mark scheme. Specimen. Version 1.2

Transcription

1 A-LEVEL Mathematics Paper Mark scheme Specimen Version.

2 Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students scripts. Alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students reactions to a particular paper. Assumptions about future mark schemes on the basis of one year s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. Further copies of this mark scheme are available from aqa.org.uk Copyright 07 AQA and its licensors. All rights reserved. AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre.

3 Mark scheme instructions to examiners General The mark scheme for each question shows: the marks available for each part of the question the total marks available for the question marking instructions that indicate when marks should be awarded or withheld including the principle on which each mark is awarded. Information is included to help the examiner make his or her judgement and to delineate what is creditworthy from that not worthy of credit a typical solution. This response is one we expect to see frequently. However credit must be given on the basis of the marking instructions. If a student uses a method which is not explicitly covered by the marking instructions the same principles of marking should be applied. Credit should be given to any valid methods. Examiners should seek advice from their senior examiner if in any doubt. Key to mark types M dm R A B E F mark is for method mark is dependent on one or more M marks and is for method mark is for reasoning 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 explanation follow through from previous incorrect result Key to mark scheme abbreviations CAO CSO ft their AWFW AWRT ACF AG SC OE NMS PI SCA sf dp correct answer only correct solution only follow through from previous incorrect result Indicates that credit can be given from previous incorrect result anything which falls within anything which rounds to any correct form answer given special case or equivalent no method shown possibly implied substantially correct approach significant figure(s) decimal place(s)

4 Examiners should consistently apply the following general marking principles 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 students showing no working is that incorrect answers, however close, earn no marks. Where a question asks the student 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. Diagrams Diagrams that have working on them should be treated like normal responses. If a diagram has been written on but the correct response is within the answer space, the work within the answer space should be marked. Working on diagrams that contradicts work within the answer space is not to be considered as choice but as working, and is not, therefore, penalised. Work erased or crossed out Erased or crossed out work that is still legible and has not been replaced should be marked. Erased or crossed out work that has been replaced can be ignored. Choice When a choice of answers and/or methods is given and the student has not clearly indicated which answer they want to be marked, only the last complete attempt should be awarded marks.

5 Circles correct answer B 5 Total Circles correct answer B x x Total Uses either cosx x or sin x x (PI) Substitutes and into their expression AO. B AO.a cos sin 5 Obtains correct answer A Total 5 of 9

6 4(a) Demonstrates p( ) 0 B p( ) ( ) 7( ) ( ) Constructs rigorous mathematical proof (to achieve this mark, the student must clearly calculate and state that p( ) 0and clearly state that this implies that x + is a factor) AO. R p( ) = 0 x is a factor (b) Factorises the numerator and denominator (this mark is achieved for any reasonable attempt at factorisation through the selection of an appropriate method, eg long division) Finds second factor in numerator or fully factorises denominator (PI by complete factorisation) AO.a ( x)( x x) ( x)( x) ( x)( x)( x+ ) ( x)( x) ( x )( x ), x A ( x ) Finds fully correct factorised expression (PI by complete factorisation) A Obtains a completely correct solution with restriction on domain stated A Total 6 6 of 9

7 5(a) Recalls A r or l r PI by use in equation AO. B Area of sector gives 8 r 9, r Constructs two equations at least one correct AO.a Perimeter of sector gives rr 5 Eliminates FT incorrect equations AO.a 8 r 5 r Constructs a rigorous mathematical argument to show required result, clearly constructing two correct simultaneous equations and eliminating AG AO. R r 85r r r (AG) (b) Solves a quadratic equation and finds two values of Finds two correct values of r B Finds both values of A Gives a valid reason for rejecting one of their values AO.a r 6, r AO.4 R Total 8 r 6 r so only one possible value of 7 of 9

8 6 (a) Translates rate of change into d m dt AO. dm k dt m Translates inverse proportionality by using in an equation m (no need to see minus sign or k to earn this mark) AO. Forms correct equation with correct notation d m k or equivalent dt m d eg m k dt m or d m d km t A (b) Gives a relevant criticism of the assumption AO.5b E Sam s mass is unlikely to follow this model all the time, when he eats his mass will go up. OR Sam s assumption predicts that his mass will decrease indefinitely. Total 4 7 Clearly states that equal roots b 4ac0 Forms quadratic expression in k (allow one error) AO.4 B b 4ac0for equal roots k 4 ( k)( k) 0 7k 0k 0 AO.a 6 k, k 7 Obtains correct quadratic equation in k (PI by correct values for k) A Obtains correct values for k for their quadratic equation AF Total 4 8 of 9

9 8(a) States the correct derivative B x ln (b) Selects an appropriate method for integrating, which could lead to a correct exact solution (this could be indicated by an attempt at a substitution or attempting to write the integrand in the form f'( )f( ) n x x ) Correctly writes integrand in a form which can be integrated (condone missing or incorrect limits) AO.a x Let u Then d u x dx I = A du And ln dx du ( u) dx ln dx = ( u) du ln x Integrates their expression (allow one error) AO.a = ( u) ln c Substitutes correct limits corresponding to their method AO.a Sub limits: ( u) ln Obtains correct value in an exact form A ln ALT (direct inspection) Mark awarded if they have a completely correct solution, which is clear, easy to follow and contains no slips AO. R x x dx x x ln dx ln Substitution should be clearly stated in exact form and change of variable or solution by direct inspection should be achieved correctly with correct use of symbols and connecting language ln x x ln ( ) dx x ( ) ln x ( ) ln 0 ( ) ln Total 7 9 of 9

10 9(a)(i) Selects an appropriate routine procedure; evidence of quotient rule or product rule AO.a dy 4 ( x 7) 8x( x) dx ( 4x 7) Obtains correct derivative (no need for simplification) A (a)(ii) States clearly that AO.4 R dy dx Forms inequality from their AO.a B dy dx dy y is increasing 0 dx 4 ( x 7) 8x( x) 0 ( 4x 7) ( 4x 7) 0for all x 4 ( x 7) 8x( x) 0 Deduces numerator must be positive AO.a R 8x 46x 4x0 x x (AG) Considers denominator alone and sets out clear argument to justify given inequality AG AO. R Only award this mark if they have a completely correct solution, which is clear, easy to follow and contains no slips (b) Solves the correct quadratic inequality (accept evidence of factorising, completing the square, use of formula, or correct critical values stated) AO.a (x + 7)(x ) x 7, Obtains fully correct answer, given as an inequality or using set notation A 7 < x < 7 Or x, Or x x: 7 x Total 8 0 of 9

11 0(a) Makes a deduction about the lower bound of the function (4) AO.a B Correctly states the range of f using set notation AO.5 B The range of f is the set x: x4, x (b)(i) States correctly the set they gave in part (a) AO. BF x: x4, x (b)(ii) Interchanges x and y at any stage AO.a y = 4 + x x = 4 + y Rearranges and takes logs AO.a y = x 4 y = log (x 4) Obtains correct expression from completely correct working for f ( x), notation correct throughout A f (x) = log (x 4) (c)(i) Obtains gf(x) B gf(x) = g(4 + x ) = 5 (4 + x ) 0.5 (c)(ii) Forms equation and rearranges using their gf(x)= AO.a 5 (4 + x ) 0.5 = (4 + x ) = 9 Correctly rearranges to get a single exponential term where logs can be taken. (Follow through provided their equation requires the use of logs.) AF x = 5 x = log 5 Obtains correct solution A Total 0 of 9

12 (a) Completes the square twice or applies standard formula AO.a ( x4) ( y6) 6 6 ( x4) ( y6 ) = 64 Obtains correct equation A Centre ( 4, 6) Radius = 8 Obtains correct radius and correct coordinates of C Follow through their equation AF (b) Demonstrates a method to find the length OP or OQ (or their squares), or the coordinates P or Q using their values from part (a) AO.a OC 4 6 = 5 OP r OC = 64 5 = Uses a circle property that may lead to a solution, eg radius and chord meet at right-angles (evidence for this could be the use of Pythagoras or perpendicular gradients) AO.a PQ = OP 4 Finds OP or OQ or coordinates of P or Q CAO A Obtains length of PQ Follow through from their coordinate of P and Q (Does not need to be in the required form) AF Sets out a well-constructed mathematical argument, using precise statements and correct use of symbols throughout to show the correct required result in required form AO. R Only award if they have a completely correct solution, which is clear, easy to follow and contains no slips Total 8 of 9

13 Finds the difference between the AO.b x xyy 0 maximum and minimum values of y dy dy x y x y dx 4 dx 0 Uses implicit differentiation AO.a Highest and lowest points occur when d y Differentiates correctly A dx 0 dy x y dx 0 States stationary points occur when dy y y y 0 0 AO.4 R dx y 0 Height 0 0 Uses d y 0 to find x in terms of y (or dx vice versa) AO.a = 0 = 6. m Finds x = y A Deduces maximum and minimum values of y FT their expression provided all marks have been awarded AO.a AF States the height of the sculpture above the platform FT their max and min values for y provided all marks have been awarded AO.a AF Total 8 of 9

14 Recalls a correct trig identity, which could lead to a correct answer Performs some correct algebraic manipulation and uses second identity to commence proof (at least two lines of argument) Concludes a rigorous mathematical argument to prove given identity AG Must start with one side and through clear logical steps arrive at the other side. In order to be sufficiently clear, each line should be a single step, unless clear further explanation is given. AO. B ( LHS ) AO. AO. R R cot cos cos cos sin cos sin cos cosec cos cot ( RHS ) (AG) Total 4 of 9

15 4(a) Identifies and clearly defines variables. Models the cost with an expression of the form ax bxh AO.b AO. B Let C = total cost x = length of base edges h = length of height C 5x xh Eliminates either variable, using volume equation, to form a model for the cost in one variable. AO. xh h x Obtains a correct equation to model cost in one variable Uses their model to find minimum. (at least one term correctly differentiated and equated to zero) AO.b AO.4 A 90 C 5x (*) x Differentiating 90 0x 0 x x = 64 x = 4 Obtains correct equation A Obtains correct value for h with correct units in context Award FT from correct substitution into incorrect equation for h but only if all three marks have been awarded, must have correct units. Performs a correct test of their solution: uses the second derivative of their expression for C in terms of x or h to justify that a minimum value for h has been found OE (Second derivative > 0 or test gradient/values either side) AO.a AO.4 AF R 60 h Height of tank is.75 m d C dx x d C x 4 > 0 therefore minimum dx ALT from (*) onwards C 900h 60h h 0 900h 6 60 Height of tank is.75 m d C 800h 8 60 h dh d C h 75.. dh 56 0 minimum 5 of 9

16 4(b)(i) Explains that two of the sides of the tank will need to have length x 005. in order to join them AO.5c E The sides will need to overlap to be joined, so two of the side lengths will need to be x 005. (b)(ii) Explains that the refinement is relatively small and unlikely to have a significant effect on the result. AO.5a R The minimum cost is likely to increase slightly, but relative to the size of the tank, an extra 5cm is unlikely to make a significant difference. Total 0 6 of 9

17 5(a) Separates variables, at least one side correct. AO.a dx x t dt 8sin Obtains correct separation PI A x dx 8sin t dt integrates their expressions, at least one of their sides correct AO.a x dx 8sin t dt Obtains correct integral (condone missing + c) CAO A x 4 cost( c) Substitutes initial conditions, to find + c. AO.b ( 0) 4cos( 0 ) c c = 4 Obtains a correct solution ACF A x = cos t Obtains correct solution of the form x f( t) AO.5 A x cos t (b) Obtains correct max height, in cm Award FT from correct substitution into incorrect equation x f( t) but only if all three marks have been awarded, must have correct units. AO.4 AF Max height 4 5 cm Total 8 7 of 9

18 6(a) Identifies zero as number for which student s argument is not true AO. B 0 (b) Uses proof by contradiction Must see commencement of argument including stated assumption and at least two lines of argument Represents product of rational and irrational numbers in symbolic form Correctly deduces that the product must be irrational Completes a rigorous mathematical argument, proving that a non-zero rational multiplied by an irrational is irrational AO. AO.5 AO.a AO. A R Let a be irrational, and b be a non-zero rational, so b c, where c, d ; c, d 0 d Assume ab is rational, so p ab, where p, q ; q 0 q ac p d q pd a qc so a is rational, which is a contradiction Hence ab must be irrational Must start with initial assumptions and prove the result convincingly Must define p q c d as integers Total 5 8 of 9

19 7 Translates f' 6 into sin hsin 6 6 lim h0 h Uses sin (A + B) identity to replace sin h, to 6 commence argument (at least two lines of argument seen) Obtains correct two term expression involving cos h and sin h Deduce what happens as h 0, for one part of their expression sinh using the limit of h OR by using small angles approximations AO.a AO. AO.a A R sin hsin f' 6 6 limh0 6 h sin coshcos sinhsin lim h0 h cos h sinh lim h0 h cosh sinh lim h0 h h h sin h sin lim h0 h h h h sin sin sinh lim lim lim h0 h h h 0 0 h Deduce what happens as h 0, for the second part of their expression using the limit of (cos h )h OR by using small angle approximations AO.a R 0 Completes a rigorous argument leading to the correct exact value, with all the steps in the method clearly shown. AO. R Total 6 TOTAL 00 9 of 9