General Certificate of Education Advanced Level Examination January 0 Mathematics MFP Unit Further Pure Monday 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 boxes 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 box 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 maximum mark for this paper is 75. Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. P8505/Jan/MFP 6/6/ MFP
The function yðxþ satisfies the differential equation ¼ fðx, yþ p where fðx, yþ ¼x þ ffiffi y and yðþ ¼ Use the improved Euler formula y rþ ¼ y r þ ðk þ k Þ where k ¼ hfðx r, y r Þ and k ¼ hfðx r þ h, y r þ k Þ and h ¼ 0:, to obtain an approximation to yð:þ, giving your answer to three decimal places. (5 marks) (a) Find the values of the constants p and q for which p sin x þ q cos x is a particular integral of the differential equation þ 5y ¼ cos x ( marks) (b) Hence find the general solution of this differential equation. ( marks) A curve C has polar equation rð þ cos yþ ¼. (a) Find the cartesian equation of C, giving your answer in the form y ¼ fðxþ. (5 marks) (b) The straight line with polar equation r ¼ sec y intersects the curve C at the points P and Q. Find the length of PQ. ( marks) By using an integrating factor, find the solution of the differential equation x y ¼ x e x given that y ¼ e when x ¼. Give your answer in the form y ¼ fðxþ. (9 marks) P8505/Jan/MFP
5 (a) Write x þ x þ in the form C ðx þ Þðx þ Þ, where C is a constant. ( mark) (b) Evaluate the improper integral ð 0 ðx þ Þðx þ Þ showing the limiting process used and giving your answer in the form ln k, where k is a constant. (6 marks) 6 The diagram shows a sketch of a curve C. O Initial line The polar equation of the curve is p r ¼ sin y ffiffiffiffiffiffiffiffiffiffi cos y, 0y p Show that the area of the region bounded by C is 6. (7 marks) 5 7 (a) Write down the expansions in ascending powers of x up to and including the term in x of: (i) cos x þ sin x ; (ii) lnð þ xþ. ( mark) ( mark) (b) It is given that y ¼ e tan x. (i) (ii) Find and show that d y ¼ð þ tan xþ. Find the value of when x ¼ 0. (5 marks) ( marks) Turn over s P8505/Jan/MFP
(iii) Hence, by using Maclaurin s theorem, show that the first four terms in the expansion, in ascending powers of x, of e tan x are þ x þ x þ x ( marks) (c) Find lim x! 0 e tan x ðcos x þ sin xþ x lnð þ xþ ( marks) 8 (a) Given that x ¼ e t and that y is a function of x, show that x ¼ ( marks) (b) Hence show that the substitution x ¼ e t transforms the differential equation x x þ y ¼ lnx into d y þ y ¼ t (5 marks) (c) Find the general solution of the differential equation (d) d y þ y ¼ t Hence solve the differential equation x x þ y ¼ lnx, given that y ¼ and ¼ when x ¼. (6 marks) (5 marks) Copyright ª 0 AQA and its licensors. All rights reserved. P8505/Jan/MFP
Mark Scheme General Certificate of Education (A-level) Mathematics Further Pure 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 explanation 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 x EE deduct x 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 Further Pure January 0 MFP Q Solution Marks Total Comments k ( ) = 0. + (=0.5) M k = 0. f (.,.5) k = 0. (.+.5) = 0.5... A PI accept dp or better y(.) = y() + [ + ] M k k = + 0.5.0. m Dep on previous two Ms and numerical values for k s y(.) =.5 A 5 Must be.5 Total 5 (a) pcos x qsin x+ 5 psin x+ 5qcos x= cos x M Differentiation and subst. into DE p+ 5q= ; 5 p q= 0 m Equating coeffs. 5 p = ; q = A OE Need both (b) Aux. eqn. m + 5 = 0 M PI. Or solving y (x)+5y=0 as far as y= ( y 5 = ) Ae x A OE CF 5x 5 c s CF + c s PI with exactly one ( ygs = ) Ae + sinx+ cosx BF arbitrary constant OE Total 6 (a) r + rcosθ = M r + x = B rcosθ = x stated or used r = x A x + y = ( x) M r = x + y used y = x A 5 Must be in the form y = f(x) but accept ACF for f(x). (b) Equation of line: rcosθ = x= M Use of rcosθ = x A x= OE y = = y =± ; [Pts, ± ] M Distance between pts (0.75, ) and (0.75, ) is A Altn: 5 (M elimination of either r or θ) At pts of intersection, r = and cosθ = OE (MA) 5 (For A condone slight prem approx.) Distance PQ = r sinθ (M) Or use of cosine rule or Pythag. 5 = = 5 (A) Must be from exact values. Total 9
Mark Scheme General Certificate of Education (A-level) Mathematics Further Pure January 0 MFP(cont) Q Solution Marks Total Comments d x x IF is e M Award even if negative sign missing ln( x) ( + c) ln( ) ( + ) = e = A OE Condone missing c e x c = (k)x AF Ft earlier sign error x x y = xe d ( x y) = x e x x M LHS as d/(y IF) PI x y = x e x = x x d(e ) = x e x e x M A Integration by parts in correct dirn x x x y= xe e (+c) A ACF When x =, y = e so m Boundary condition used e = e e + c to find c after integration. 5 e c = x x 5 y = x e x e x e A 9 Must be in the form y = f(x) Total 9 5
Mark Scheme General Certificate of Education (A-level) Mathematics Further Pure January 0 MFP(cont) Q Solution Marks Total Comments 5(a) x+ 8 x 5 = (x+ )(x+ ) (x+ )(x+ ) B Accept C = 5 (b) 0 = (x+ )(x+ ) x+ x+ M = [ ln(x ) ln(x ) ] + + (+c) A OE I = lim a a 0 (x+ )(x+ ) M replaced by a and lim a (OE) [ a a ] a = lim = lim a = ln( + ) ln( + ) (ln 5 ln 5) ln a + a + 6 ln = ln 9 = lim a + ln a + a m,m A 6 CSO Total 7 Limiting process shown. Dependent on the previous MM 6
Mark Scheme General Certificate of Education (A-level) Mathematics Further Pure January 0 MFP(cont) Q Solution Marks Total Comments 6 Area = ( sin cos ) θ θ dθ M Use of r dθ = ( cos sin ) d θ θ θ 0 B r = cosθsin θ or better = ( ) B Correct limits 6sin θ cos θ dθ 0 M sin k sin cos = ( ) = θ = θ θ (k>0) 8sin θ (-sin θ) d sinθ 0 m Substitution or another valid method to 8sin θ 8sin θ 5 0 5 8 8 6 = 0 = 5 5 AF integrate sin A 7 CSO AG θcos θ Correct integration of p sin θ cos θ Alternatives for the last four marks cosθ 0 cos θ cos θ dθ (M) Area = ( ) ( cos θ cos θ) dθ = (cos θsinθ sin θcos θ) 5 Area = ( 0) + ( ) [ 0 (0)] = 6 5 5 (m) (AF) (A) cosθsin θ = λcosθ + μ cos θcosθ ( λ, μ 0) Integration by parts twice or use of cos θ cosθ = ( cos5θ + cosθ) Correct integration of p cos θcosθ [eg p sin 5θ + sin θ 0 6 ] CSO AG 6 { + = } 0 6 5 Total 7 7
Mark Scheme General Certificate of Education (A-level) Mathematics Further Pure January 0 MFP(cont) Q Solution Marks Total Comments 7(a)(i) B Accept coeffs unsimplified, even! cos x + sin x= + x x x 6 for 6. (ii) 9 ln(+ ) x = x () x + () x = x x + 9x B Accept coeffs unsimplified (b)(i) tan y = e x, d d y tanx sec x e x = M A Chain rule ACF eg ysec x d d y tanx tanx sec xtan x e sec x e x = + m A Product rule OE ACF tanx = sec x e ( tan x+ sec x) = (tan x + + tan x ) = ( + tan x) A 5 AG Completion; CSO any valid d x method. (ii) ( + tan x) sec x + ( + tan x) M When x = 0, A CSO (iii) y(0) = ; y (0) = ; y (0) = ; y (0) = ; y(x) y(0) + x y (0) + x y (0) +! x y (0) M (c) tan e x + x + x + x tan x lim e (cosx + sin x) x 0 xln( + x) x x x x lim + x+ + x+ + = 6 x 0 9 x x x +... = lim x + x +.. x 0 9 = lim + x +.. x x... x 0 9 x... = A CSO AG M m A Using series expns. Dividing numerator and denominator by x to get constant terms. OE following a slip. Total 8
Mark Scheme General Certificate of Education (A-level) Mathematics Further Pure January 0 MFP(cont) Q Solution Marks Total Comments 8(a) y M Chain rule e t y d y x d t A CSO AG (b) d x ; d d d x y x d M OE x = + x = m Product rule (dep on previous M) x + x = d x A OE x x + y = lnx becomes x x + y = lnx d x t + y = lne (using (a) + y = t m A 5 CSO AG (c) Auxl eqn m m + = 0 M PI (m ) = 0, m = A PI CF: ( y ) ( )e t C = At+ B M Ft wrong value of m provided equal roots and arb. constants in CF. Condone x for t here PI Try ( y = ) at+ b M If extras, coeffs. must be shown to be 0. P a+ at+ b= t a= b= A Correct PI. Condone x for t here GS {y} = (At +B)e t +0.5(t + ) BF 6 Ft on c s CF + PI, provided PI is non-zero and CF has two arbitrary constants and RHS is fn of t only (d) y = (Alnx + B)x +0.5(lnx + ) M y =.5 when x = B = AF Ft one earlier slip y (x) = (A lnx + B ) x + Ax + 0.5 x m Product rule y () = 0.5 A = AF Ft one earlier slip y = ( ln x) x + (ln x+ ) A 5 ACF Total 8 TOTAL 75 9
AQA Further pure Jan 0 Answers Question : = f ( x, y) = x + y an() = k = hf ( x, y ) = 0. ( + ) = 0.5 0 0 y + k =.5 0 ( y k ) ( ) k = hf x + + = 0..+.5 = 0.5 0 y = y(.) = + ( 0.5 + 0.5 ) =. 5 to d. p. Question : a) y = psinx + qcosx = pcosx qsinx + 5y = pcosx qsinx + 5pSinx + 5qCosx = Cosx ( p + 5 q) Cosx + (5 p q) Sinx = Cosx p = p+ 5q= so 5p q = 0 5 q = 5 A particular integral is y = Sinx + Cosx The auxiliary equation is λ+5=0 λ = 5 The complementary function is y = Ae 5x 5 The general solution is y = Sinx + Cosx + Ae Question : r( + Cosθ ) = a) r + rcosθ = r = rcosθ ( squaring both sides) = + ( θ) r rcos rcos + = + x y x x y = x b)r = Secθ rcosθ = x = Solving simultaneously gives = = = = y x so y or y Thecartesian coordinates of P andq: P(,) and Q(, ) The distance PQ = θ 5x Numerical solutions of first order differential equations continue to be a good source of marks for all candidates, and it was the best answered topic on the paper. However, a few candidates who had an incorrect value for k and just gave a table of values without showing any methods gained no credit. Almost all candidates gave their final answer to the required degree of accuracy. Most candidates differentiated the given expression, substituted into the first order differential equation and equated coefficients to score the method marks. The most common error in solving the resulting equations was to write the solution of 6p = as p =. In part (b), those candidates who wrote down the correct auxiliary equation, m + 5 = 0, generally had no problems scoring all three marks. Those candidates who solved + 5y = 0 sometimes forgot to include the constant of integration and so ended up with a general solution which had no arbitrary constant. There was a minority of candidates who tried to use an integrating factor and appeared not to know the method of solving a first order differential equation by using complementary function and particular integral. This question on polar coordinates was generally answered well. In part (a), most candidates reached the stage r + x =, but some less able candidates squared incorrectly to reach y = x. Those candidates who rearranged the equation to r = x before squaring usually went on to gain all five marks. In part (b), those candidates who wrote the given equation as rcosθ =, converted it to the cartesian equation x = and solved with their answer to part (a) usually had no difficulty in showing that the length of PQ is. The other common approach was to solve the two polar equations simultaneously, but a significant proportion of candidates who used this approach stopped after reaching Cosθ = not even finding the value for r. More able 5 candidates, having found the values for cosθ and r, went on to use basic trigonometry to find the correct length for PQ.
Question : y= xe x x x ln x = = = An Integrating factor is I e e The equation becomes: x y = xe x x d y x = xe x y x x x = xe xe e x = y x x = xe e + c x x x x y = x e e + cx When x =, y = e so e = 8e e c= 5e 5 c = e x 5 y= xe e x x x + c x Most candidates were able to show that they knew how to find and use an integrating factor to solve a first order differential equation, and many gained either full marks for the question or just lost the final accuracy mark. The only other error of note was losing the negative sign in setting up the integrating factor. This led to a more complicated integral which few solved correctly. Question 5: (x+ ) (x+ ) 5 a) = = x+ x+ ( x+ )( x+ ) (x + ) ( x + ) N 0 N 5 b) = (x+ )(x+ ) (x+ )(x+ ) N N = = ln x + ln x + x+ x+ N x+ N + N + = ln = ln ln = ln x + N + N + + N + lim lim N N + = = so lim ln ln = N N + N N + N + N 0 0 6 exists and = ln = ln (x+ )(x+ ) (x+ )( x+ ) 9 Most candidates found the correct value for C in part (a). A large majority of candidates realised that part (a) had some relevance to part (b) and duly wrote the integrand in terms of partial fractions and generally integrated correctly. Those who missed this step gained little or no credit for their later work. Although showing the limiting process is an area of the specification that candidates still find difficult, overall improvement continues to be noted.
Question 6: r = Sinθ cosθ with 0 θ A = Sin θcosθdθ CosθSin θdθ = 0 0 0 0 A = Cosθ Sin θcos θdθ = 8 CosθSin θ( Sin θ) dθ A = 8 Cos Sin Cos Sin d 0 θ θ θ θ θ n type : f ' f = f n + n+ 5 8 8 6 A = 8 Sin θ Sin θ = 0= 5 5 5 0 Although most candidates scored marks for substituting a correct expression for r into A= rdθ and inserting the correct limits, less able candidates made little further progress. A significant proportion of other candidates went further by either using the identity sin θ = sinθ cosθ or writing sin θ in terms of cos θ. A surprisingly common error amongst those solutions which used the latter approach is illustrated by ( cosθ)cosθ = cosθ cos θ. Some excellent solutions were seen from the more able candidates, which involved a variety of approaches including integration of 8sin θ cos θ by use of the substitution s = sinθ, direct integration by recalling n+ that the integral of sin n Sin θ θ cosθ with respect to θ is or n + use of the identity cosθ cosθ = cos5θ + cosθ.those candidates who used integration by parts were generally less successful. Only a few obtained the correct answer by applying integration by parts twice. Question 7: x x i) Cosx + Sinx = +... + x +... 6 x x Cosx + Sinx = + x +... 6 ( x) ( x) ii)ln( + x) = ( x) + +... 9 = x x + 9 x +... tan x tan x b) y = e i) = ( + Tan x) e tan x tan x = ( + Tan x) Tan( x) e + ( + Tan x) e = Tanx + (+ Tan x) = (+ Tanx + Tan x) = ( + Tanx) ii) = ( + Tan x)( + Tanx) + ( + Tanx) y(0) = y'(0) = y''(0) = y'(0) = y () (0) = + = tan x tan x e = + x+ x + x +... e = + x+ x + x +... 6 tan x c) e ( Cosx + Sinx) = + x + x + x x+ x + x +... 6 = x + x +... 9 xln( + x) = x x +... tan x x + x +... + x+... e ( Cosx + Sinx) so = = xln(+ x) 9 9 x x +... x+... tan x e ( Cosx + Sinx) therefore lim = x 0 xln(+ x) Most candidates were able to write down the required two expansions in parts (a). In part (b)(i), candidates generally used the chain rule to find and then applied the product rule to find. Although a common error was to differentiate sec x as secx tanx, those who did find correctly often produced a convincing solution to reach the printed answer. Some very good solutions were seen for parts (b)(ii) and (b)(iii) but it was disturbing to see some other candidates make no attempt to find the third derivative in part (b)(ii), yet claim that f (0) = and write down the printed result in part (b)(iii). Clearly in such cases marks cannot be awarded. Candidates who attempted part (c) generally showed a thorough understanding of the process required, including the need for explicitly reaching the stage of a constant term in both the numerator and denominator before taking the limit as x 0.
Question 8: t t a) x = e = e = x and = = t e x = = x d d b) = x = = x x x x + = + x = = so x The equation x x + y = ln x becomes : + y = t + y = t c) The auxiliary equation : λ λ+ = 0 ( ) 0 λ = repeated root λ = The complementary function is y = ( At + B) e t The particular integral y = at + b = a, = 0 0 a + at + b = t a = This gives so a = and b = b a = 0 The general solution is y = t + + ( At + B) e d) y= ln x+ + ( Aln x+ Bx ) When x =, y = so = 0 + + B B = y = ln x + + Ax ln x + x = + Ax ln x + Ax + x x x When x =, = so = + A + A = Conclusion : y = ln x+ + ( ln x+ ) x t In part (a), most candidates were able to convincingly show, by use of the chain rule, that x =. Part (b), as expected, was not answered well by the average candidate. Those more able candidates who differentiated the result in part (a) either with respect to x, writing d d as =, or with respect to t, d d writing x = x normally scored all or most of the marks in part (b). In part (c), although most candidates realised what was required, it was not uncommon to see x appear in general solutions. Many candidates correctly tried a particular integral of the form at + b, but a common error was to write y as at + b, which led to an incorrect particular integral. However, this error was classed as a slip and so some follow through was applied in marking part (d). In part (d), those candidates who had a general solution in part (c) that was entirely in terms of t were generally able to pick up at least the method marks in the final part of the question. A common error was to differentiate ln x + as +, which led to x the loss of the last two accuracy marks. Grade boundaries Grade A B C D E Mark Max 75 66 59 5 5 8