A-LEVEL Further Mathematics

A-LEVEL Further Mathematics F1 Mark scheme Specimen Version 1.1

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 017 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.

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)

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 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. 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.

1 Circles correct answer AO.a B1 5 1 3 Total 1 Recalls correct definitions of cosh x and sinh x Demonstrates clearly that cosh x sinh x 1 AG Award only for completely correct argument including expansion and simplification AO1. B1 x x x x e e e e cosh xsinh x AO.1 e e e e 4 4 4 4 1 x x x x Total 5 of 3

3(a) Forms identity using the numerators from each side AO1.1a A B ( r1)( r)( r3) ( r1)( r) ( r)( r3) Obtains the correct values of A and B A1 Ar ( 3) Br ( 1) A1, B1 (b) Uses their result from part (a) to write fraction as sum of differences. Ignore 1 at this stage. AO1.1a Clearly shows step of cancelling of terms Obtains correct two term difference Ft their values for A and B provided that their A = their B States that method of differences gives 97 1 r ( r1)( r)( r3) 9 so divides their answer by to obtain correct rational solution from fully correct working AG (If student merely divides by without justification withhold this mark) AO.4 AO.1 A1 97 97 1 1 1 1 ( r1)( r)( r3) ( r1)( r) ( r)( r3) r9 r9 97 1 1 r9 ( r1)( r) ( r)( r3) 1 10 11 1 1 111 111 1 1 1 13 1 13... 1 1 98 99 98 99 1 99 100 1 1 10 11 99 100 89 9900 97 1 r9 ( r1)( r)( r3) 1 89 9900 89 19800 Total 6 6 of 3

4 Correctly states that the student s argument is invalid because the integrand is undefined at a value in the interval (0, ) AO.3 E1 π When x, 4 cos x sin x 0 Correctly justifies the reason why it is undefined and states a correct conclusion AO.4 E1 the integrand is undefined at this point and the integral is improper Total 7 of 3

5(a) Makes a correct deduction about another root (PI) AO.a B1 ( z( 3i))( z( 3i)) z 4z13 Finds quadratic factor by expanding brackets or using sum and product of roots AO1.1a p( z) ( z 4z13)( z czd) 4 ( z 4z13)( z czd) z 3z azb Finds a correct quadratic factor A1 Compares coefficients with quartic z 4 3 z azb AO1.1a c 40 13 4cd 3 c 4, d 6 States the correct product of quadratic factors A1 p () z z 4z13 z 4z6 ALT Makes a correct deduction about another root AO.a B1 ( z( 3i))( z( 3i)) z 4z13 p( z) ( z 4z13)( z czd) Finds quadratic factor by expanding brackets or using sum and product of roots AO1.1a 4 ( z 4z 13)( z cz d) z 3z az b Obtains a correct quadratic factor A1 0 4 c 4 Uses coefficients/roots to set up equations and find required coefficients States the correct product of quadratic factors AO1.1a A1 0= + 3 0= 3i 3i 6 γ δ 4 16 4 d 6 z z z pz 4 13 4z6 8 of 3

(b) States all four correct solutions B1F z 3i, i FT their two quadratic factors from part (a) provided both marks have been awarded Total 6 9 of 3

6(a) Selects appropriate method for example by changing to reduced equation by dividing by tan x AO3.1a d y ( x) y x dx cot sin Finds correct integrating factor B1 Integrating factor ln( sinx) e sin x e ( cot x)dx ALT Alt. finds an integrating factor by inspection, using original equation. (PI) AO3.1a dy cos xtan x y x x x x dx cos sin tan cos dy x xy x dx sin cos sin finds integrating factor = cos x B1 Multiples reduced or original equation by their integrating factor and identifies LHS as differential of ysin x PI AO1.1a d y x ( x) y x dx sin cos sin d [ ] ysin x sin x dx Uses appropriate integration 1 method for RHS of their equation AO1.1a y sin x ( 1cosx) dx Integrates correctly to obtain correct solution A1 1 1 y sin x x sin x C 4 (b) Uses boundary condition after integration completed in either y sin x... or y... form OE AO1.1a π 1 1 π 1 π sin.. sin C 4 4 4 1 π C 8 States fully correct particular solution A1 1 1 1 π ysin x x sinx 4 8 Total 7 10 of 3

7(a) Uses an appropriate method for finding the values of k ( for example expanding appropriate determinant) AO1.1a 1 1 k k 3 5 0 1 3 Obtains a quadratic equation in k AO1.1a 3 5 k k 5 k 3 0 3 1 3 1 Obtains two correct values for k A1 1 +3k 5 + k ( k + 3)=0 k 6k4 0 k 3k0 ( k)( k1)=0 k = or 1 (b) Selects an appropriate method to determine the appropriate geometrical configuration and substitutes their first value of k Eliminates one variable or uses row reduction AO3.1a when k =1 x y z 3 x3y5z 1 xy3z 4 AO1.1a y4z 4 yz 7 Obtains a contradiction and makes correct deduction about the geometric configuration (must have correct value for k) Substitutes their nd value of k into selected method to determine the appropriate geometrical configuration AO.a yz; yz7 Hence equations are inconsistent and the three planes form a prism when k = AO1.1a x yz3 x3y5z1 xy3z4 Obtains a consistent set of equations and makes correct deduction about geometric configuration (must have correct value for k) AO.a R : yz 7 R3 : yz 7 Hence equations are consistent and the three planes form a sheaf they meet in line 11 of 3

(c) Deduces that the planes must meet in a line and hence that k = AO.a x yz3 x3y5z1 xy3z4 yz7 Let z Then y 7 and x 3 yz 3 7 10 Selects method to find solution: For example, sets one variable =, substitutes and attempts to find other variables in terms of Fully states correct solution CAO AO1.1a A1 ALT 1 1 1 11 3 1 x 10 1 y 7 1 z 0 1 Total 11 1 of 3

8(a) Indicates correct coordinate or intercept on x and y axes B1 x 0 y 5 (0, 5) y 0 5 4x 0 5 x 4 5 (,0) 4 Indicates correct vertical or horizontal asymptote B1 x 1 As x, y 4 y 4 Sketches correct shape of curve AO1. B1 Draws fully correct sketch including intercepts and both asymptotes marked B1 (b) Draws sketch fully correct including shape at x-intercept both asymptotes marked B1 Total 5 13 of 3

Let 9(a) Uses an appropriate method for y ensuring the line lies in the plane x p 3 z AO3.1a q, then x p, yq, z3 sub into equation of plane Obtains equation(s) in p and q AO1.1a p q 3 3 Deduces the values of p and q AO.a A1 ( 3q) ( p1) 0 this is true for all λ therefore p = 1 and q = 3 ALT vector equation of line is p 1 r q 3 1 p therefore lies on the plane 3 p 1. 1 + 3=0 3 1 1 And q is perpendicular to 1 1 1 1 q. therefore 1 0 1 q 3 and p 1 (b) States that to have a solution the coefficient of in equation from (a) cannot be 0 OR dot product must 0 AO.4 Deduces the range of values for q AO.a 1 1 q. 1 0 q 3 1 Deduces correct range of values for p AO.a p can take any value 14 of 3

(c)(i) Finds the correct scalar product of the normal to the plane and the direction vector Correctly deduces the value of cos Forms an equation connecting all relevant parts using n.d n d cos AO.a AO3.1a B1 Obtains correct value for q A1 1 1 n = = q 1 d 1 nd. 3q Let be angle between the line and the normal to the plane 1 1 sin cos 6 6 q3 6 q 6 3q q 1 6q 7 giving q 7 6 (c)(ii) Uses their expressions for x and y and their value for q and the equation of the plane to form an equation to find p AO3.1a y x p 3 z 7 6 z 0x p3, y1.5 Uses z = 0 to deduce expressions for x and y in terms of p and q AO.a Obtains the correct value of p A1 CAO p 3 1 1.5. 1 3 0 p 31.53 p 4.5 Total 13 15 of 3

10(a) Uses quotient or product rule to obtain correct derivative B1 x d y cosh x xsinh x y cosh x dx cosh x Clearly sets their d y dx numerator equal to 0 Rearranges to complete a rigorous argument to show the required result. AG AO.4 AO.1 d y Stationary point 0 dx cosh x xsinh x 0 cosh x cosh xxsinh x 0 sinh x 1 cosh x x 1 tanh x x (b)(i) Sketches tanh x correctly including asymptotes AO1. B1 Sketches correctly 1 AO1. B1 x (ii) Deduces correct number of stationary points AO.a B1F stationary points FT their sketch in (b)(i) 16 of 3

10(c) Finds the second derivative Obtains a correct expression for the second derivative AO1.1a A1 d y cosh x( sinh xxcosh xsinh x) 4 dx cosh x cosh xsinh x( cosh x xsinh x) 4 cosh x Deduces that the second term is zero by using results from part (a) AO.a second term is zero at stationary points d y x y dx cosh x Completes a rigorous argument to show the required result. AG AO.1 d y y dx 0 Mark awarded if they have a completely correct solution, which is clear, easy to follow and contains no slips Total 10 17 of 3

11(a) Commences proof by considering one side of the identity only: if considering LHS combines terms as a single fraction with a common denominator. AO.1 sinh 1 cosh 1 cosh sinh sinh 1cosh cosh ( 1 cosh) sinh If considering RHS writes coth θ as a fraction and introduces factor of (1 + cosh θ to both numerator and denominator) Note alternative valid approaches include commencing proof by considering LHS minus RHS or LHS divided by RHS cosh cosh cosh, ( 1 cosh) sinh 1sinh cosh cosh( 1 cosh) ( 1 cosh) sinh cosh sinh coth Explicitly states identity cosh θ sinh θ 1 and uses it to eliminate (or introduce) sinh θ AO.4 Factorises numerator and cancels correctly for their fraction (if considering RHS rearranges their numerator correctly into two factorised expressions) B1F Completes rigorous proof to obtain result AG AO.1 Only award if they have a completely correct argument, which is clear and contains no slips. (b) Uses result from part (a) to deduce that tanh θ = 1 AO.a coth θ = 4 tanh θ = 1 θ = tanh -1 1 = 1 ln 3 Uses natural log form and substitutes correct value to obtain correct exact form A1 Total 6 18 of 3

1 Correctly manipulates y=f( x) n into a form that can be sketched. Sketches an appropriate section of the graph. Sketch does not need to be accurate but should look symmetrical about y-axis and the area above and below the x-axis should look similar AO3.1a e cosh(i x) ix e ix xi x xi x cos sin cos sin cos sin cos sin xi x xi x cosx n n f( x) cos x Deduces, with fully correct reasoning, the correct mean value AO.a Since the graph has the same shape/area above and below the x-axis over the given domain, the mean value of y= f( x ) n must be 0. Total 3 19 of 3

13 Uses proof by induction and investigates formula for n = 1 and n = k (must see evidence of both n = 1 and n = k being considered) Demonstrates that formula is true for n = 1 States assumption that formula true for n = k and uses k 1 k M MM AO3.1a Using induction method, Let P(n) be the statement n1 n1 n1 3 3 3 n n1 n1 n1 M 3 3 3 n1 n1 n1 3 3 3 A1 For n =1 0 0 0 3 3 3 1 1 1 3 3 3 1 1 1 M AO.1 0 0 0 3 3 3 1 1 1 P(1) is true 0 0 0 1 Deduces that formula is also true for n = k + 1 from correct working Completes a rigorous argument and explains how their argument proves the required result. AG AO.a AO.4 Assume P(k) is true k 1 k M MM k1 k1 k1 1 1 1 3 3 3 k k k 1 1 1 1 1 1 3 3 3 k1 k1 k1 1 1 1 3 3 3 (since P(k) is true) k1 k1 k1 k1 k1 ( 3 3 3 ) ( 3..) ( 3.. k1 k1 k1 k1 k1 ( 3 3 3 ) ( 3..) ( 3.. k1 k1 k1 k1 k1 ( ) (..) (.. 3 3 3 3 3 But 3 3 3 33 k1 k1 k1 k1 Hence M k 3 k k k 3 3 3 3 3 3 k k k 3 3 3 k1 k k k 3 3 3 ( k1) 1 ( k1) 1 ( k1) 1 k1 ( k1) 1 ( k1) 1 ( k1) 1 M 3 3 3 ( k1) 1 ( k1) 1 ( k1) 1 3 3 3 P(k +1) is true Since P(1) is true and P(k) P(k + 1), hence, by induction, P(n) is true for all n Total 5 0 of 3

14(a) Models the motion of the particle by forming a second order differential equation. (must have correct terms but allow sign errors AO3.3 d x dx M 4M 8 Mx d x dx 4 8x 0 Obtains correct differential equation A1 480 i Forms and solves auxiliary equation for their D.E. States a correct form of the general solution for their auxiliary solution. (ft only if both marks have been awarded) Uses initial conditions to find arbitrary constants for their solution Obtains correct value for one of their constants (ft only if all marks have been awarded) AO1.1a AO1.1a A1F A1F Complex roots General solution is of the form: t x Ae cos( tb) t t x Ae cos( tb) Ae sin( tb) x ( 0) 0 so, Acos ( B) Asin ( B) 0 tan B 1 π B 4 x( 0) 1 A 1 A Obtains correct value for both of their constants (ft only if all marks have been awarded) A1F x t e cos( t π) 4 the particle oscillates about O, with period π seconds and decreasing amplitude. Uses their model to describe the motion of the particle either as a written description or shown on a clearly labelled graph. AO3.4 A1F 1 of 3

(b) Refines their DE model to account for altered resistive force by introducing a new coefficient for dx Uses or states condition for critical damping Deduces value for coefficient of dx AO3.5c AO1. AO.a B1 B1 d x dx 8x 0 Critical damping roots 3 4 8 0 must have equal Resistive force should have magnitude 4 Mv States resistive force AO3.4 B1 Total 1 of 3

15(a)(i) Forms a differential equation for the foxes. Forms a differential equation for the rabbits. Differentiates their equation that contains y and obtains expression with at least two terms correct. Formulates a second order linear differential equation. Obtains roots of auxiliary equation for their second order differential equation. States correct general solution. FT provided all M marks have been awarded Uses initial population to find equation linking constants for their general solution. AO3.3 B1 dy dy x kx dy AO3.3 B1, x k 0 00 0.1 AO1.1a AO3.1a AO1.1a A1F AO3.4 dy 0.1x dx x y 1. 1.1 1.x 1 dx y 1.1 1.1 dy 1. dx 1 d x 1.1 1.1 1. dx 1 d x 0.1x 1.1 1.1 d x dx 1. 0.11x 0 1.0.110 0.1 or 1.1 0.1t 1.1t x Ae Be t 0, x80 A+B= 80 Obtains initial rate of change for rabbits from their DE and differentiates and obtains a second equations for A and B from their general solution. Finds A and B and states the model for the number of rabbits. AO3.4 AO1.1a dx t 0, 74 74 0.1A1.1B 74 0.1( 80 B) 1.1B B68, A1 x 1e 68e 0.07 0.77 (a)(ii) Substitutes 0.7 and obtains approximately 160. CAO AO3.4 A1 0.07 0.77 1e 68e 159.7 (b) States a suitable refinement about the fact that an increased rabbit population will require more food supply or other valid refinement AO3.5c B1 Take account of the food available for the rabbits as this may limit population growth. Total 11 Total 100 3 of 3