Mark scheme - NLCS Maths Department



Mark schemeQuestionAnswer/Indicative contentMarksPart marks and guidance1y = x2 tan 2xM1product ruleu × their v′ + v × their u′ attemptedM1d / du(tan u) = sec2u soiM0 if d / dx (tan 2x) = (2) sec2x? dy / dx = 2x2sec22x + 2xtan 2xA1caoor 2x2 / cos22x + 2xtan 2xiswM1product rulesee additional notes for complete solutionu × their v′ + v × their u′ attemptedA1correct expression= … = 2x2sec22x + 2xtan 2xA1caoor 2x2 / cos22x + 2xtan 2x (isw)or (2x2 + 2xsin2xcos2x) / cos22xor 2x2 / cos22x + 2xsin2x / cos2xM1quotient rule(v × their (u′ ? u × their v′) / v2 attemptedA1correct expression= … = 2x2sec22x + 2xtan 2xA1caoor 2x2 / cos22x + 2xtan 2x (isw)Examiner's CommentsThe derivative of tan x was usually familiar, but those candidates who started with?sin 2x/cos 2x usually got lost in algebraic complexity. A surprising number lost marks through giving the derivative of tan 2x as sec2x, or omitting the ‘2’ in 2 sec22x.However, better candidates just wrote the result down.or (2x2 + 2xsin2xcos2x) / cos22xor 2x2 / cos22x + 2xsin2x / cos2xAdditional notes and solutions???? ? ? ? ? ? ????Total32iat A y = 3B1iB1itheir M1*must follow from attempt at differentiationigrad of normal = ?1/their4M1dep*iy ? 3 = (??) × (x ? 4) oe iswA1isubstitution of y = 0 and completion to given result with at least 1 correct intierim step wwwA1or substitution of x = 16 to obtain y = 0Examiner's CommentsThis was done extremely well, with the majority of even the weakest candidates scoring full marks. A few wrote 2x ? 4 = 0 to incorrectly obtain m = 2 and made no further progress, and a very small minority tried to answer the question without using calculus and working backwards.correct interim step may occur before substitutioniiat B, x = 3B1may be embeddediiM1*condone one error, must be three terms, ignore + ciiF[4] ? F[their 3]M1*depdependent on integration attemptediiarea of triangle = 18 soiB1may be embedded in final answeriiArea of region = oe iswA119.3 or betterExaminer's CommentsNearly all candidates identified the coordinates of B correctly. However, most — as if by rote — subtracted the equation of the line from the equation of the curve and then integrated. Some candidates integrated the equation of the curve correctly, but used the wrong limits (usually 3 to 16) and made no further progress, and of those that did adopt the correct approach, a large number were unable to find the area of the triangle correctly (? × 12 × 4 was common).Total113iM1Rearranging for y and differentiating explicitly is M0iA1correct equationIgnore superfluous dy/dx = … unless used subsequentlyiA1o.e., but mark final answer Examiner's CommentsThis relatively simple implicit differentiation was very well done by almost all candidates.iiB1depdep correct derivativeii? 4 + 2y2 = 8 ? y2 = 2, y = √2 or ?√2B1B1√2, ?√2 Examiner's CommentsMost candidates scored two out of three for the point (2, √2), but missed the y = .√2 solution. In a few cases, the denominator was set to zero, giving y = 0.can isw, penalise inexact answers of ±1.41 or better once only?1 for extra solutions found from using y = 0Total64iAt P(a, a) g(a) = a so ?(ea ?1) = ai? ea = 1 + 2a*B1NB AG Examiner's CommentsThis mark was usually earned.iiM1correct integral and limitslimits can be implied from subsequent workiiB1integral of ex ? 1 is ex ? xii= ? (ea ?a e0)A1ii= ? (1 + 2a ?a ?1) = ? a*A1NB AGiiarea of triangle = ? a2B1iiarea between curve and line = ? a2?? aB1caomark final answer Examiner's CommentsVirtually everyone scored M1 for writing down the correct integral and limits, but many candidates made a meal of trying to integrate ? (ex – 1) , with ? (ex – 1)2 not an uncommon wrong answer. Having successfully negotiated this hurdle, using part (i) to derive ? a was spotted by about 50% of the candidates. Quite a few candidates managed to recover to earn the final 2 marks for ? (a2 – a) (without incorrectly simplifying this to? a!).iiiy = ?(ex ? 1) swap x and yx = ? (ey ? 1)iii? 2x = ey ? 1M1Attempt to invert — one valid stepmerely swapping x and y is not ‘one step’iii? 2x + 1 = eyA1iii? ln(2x + 1) = y*? g(x) = ln(2x + 1)A1y = ln(2x + 1) org(x) = ln(2x + 1) AGapply a similar scheme if they start with g(x) and invert to get f(x).or g f(x) = g((ex ? 1)/2) M1iiiSketch: recognisable attempt to reflect iniiiy = xM1through O and (a, a)= ln(1 + ex ? 1) = ln(ex) A1 = x A1iiiGood shapeA1no obvious inflexion or TP, extends to third quadrant, without gradient becoming too negative Examiner's CommentsFinding the inverse function proved to be an easy 3 marks for most candidates – candidates are clearly well practiced in this. The graphs were usually recognisable reflections in y = x, but only well drawn examples – without unnecessary maxima or inflections – were awarded the ‘A’ mark.similar scheme for fgSee appendix for examplesivf ′(x) = ? exB1ivg ′(x) = 2/(2x + 1)M11/(2x + 1) (or 1/u withu = 2x + 1) …ivA1… × 2 to get 2/(2x + 1)ivg ′(a) = 2/(2a + 1) , f ′(a) = ? eaB1either g′(a) or f ′(a) correct soiivso g ′(a) = 2/ea????or f ′(a) = ? (2a + 1)M1substituting ea = 1 + 2aiv= 1/(?ea)??????= (2a + 1)/2[= 1/f ′(a)]??????[= 1/g ′(a])A1establishing f ′(a) = 1/g ′(a)either way roundivtangents are reflections in y = xB1must mention tangents Examiner's CommentsThis proved to be more difficult, as intended for the final question in the paper. As with the integral, many candidates struggled to differentiate ? (ex – 1) correctly, and equally many omitted the ‘2’ in the numerator of the derivative of ln(1 + 2x). Once these were established correctly the substitution of x = a and establishing of f'(a) = 1/g'(a) was generally done well, though sometimes the arguments using the result in part (i) were either inconclusive or done ‘backwards’. The final mark proved to be elusive for most, as we needed the word ‘tangent’ used here to provide a geometric interpretation of the reciprocal gradients.Total1956x2 + 18x – 24B1their 6x2 + 18x ? 24 = 0 or >0 or ? 0M1or sketch of y = 6x2 + 18x ? 24 with attempt to find x-intercepts? 4 and + 1 identified oeA1x < ? 4 and x > 1 caoA1or x ? ? 4 and x ≤ 1Examiner's CommentsMost candidates differentiated correctly and identified the correct values of x. The final mark was often lost, either due to a misunderstanding of what had been found — answer given as ?4 < x < 1 or poor notation — answer given as ?4 > x > 1. Those who used a graphical approach with the derivative generally scored full marks. A few candidates missed the last term out, converted the first plus sign to a minus sign or failed to multiply 2 by 3 correctly, and lost the first mark.if B0M0 then SC2 for fully correct answerTotal46u = x, du / dx = 1, dv / dx = cos ? x, v = 2sin ? xM1correct u, u′, v, v′but allow v to be any multiple of sin ? xM0 if u = cos ? x, v′ = xA1ftconsistent with their u, vA12x sin ? x + 4 cos ? x oe (no ft)M1substituting correct limits into correct expressioncan be implied by one correct intermediate stepA1caoNB AGExaminer's CommentsThere was a mixed response to the question, with plenty of faultless answers, but others with errors in v = 2sin ? x, e.g. v = sin ? x or –2 sin ? x or ? sin ? x.?Occasionally there was insufficient working to show that the given result had been established: candidates are well advised to include ample working.Total57ih = 20, stops growingB1AG need interpretationExaminer's CommentsMost candidates correctly wrote down the value of h but quite a number failed to give the interpretation that the tree stopped growing when its height was 20m.iih = 20 ? 20e?t/10dh/dt = 2e?t/10M1A1differentiation (for M1 need ke?t/10, k const)ii20e?t/10 = 20 ? 20(1 ? e?t/10) = 20 ? h= 10dh/dtM1iiA1oe eg 20 ? h = 20 ? 20(1 ? e?t/10) = 20e?t/10= 10dh/dt (showing sides equivalent)iiwhen t = 0, h = 20(1 ? 1) = 0B1initial conditionsii...........................................................OR verifying by integration...........................................................iiM1sep correctly and intending to integrateii? ?ln (20 ? h) = 0.1t + cA1correct result (condone omission of c, although no further marks are possible)condone ln (h ? 20) as part of the solution at this stageiih = 0,t = 0, ? c = ?ln 20? ln(20 ? h) = ?0.1t + ln 20B1constant found from expression of correct form (at any stage) but B0 if say c = ln (?20) (found using ln (h ? 20))ii? 20 ? h = 20e?0.1tM1combining logs and anti-logging (correct rules)ii? h = 20(1 ? e?0.1t)A1correct form (do not award if B0 above)Examiner's CommentsThose who approached the verification by integration were quite successful. The common errors were:-omitting the negative sign when integrating 1/(20?h)omitting the constant of integrationgiving ln(h?20) in their answers (without modulus signs) despite having usually given h=20 as a maximum value in (i)incorrect anti-logging.Those who approached from differentiation usually obtained some marks, particularly the mark for checking the initial conditions but many gave insufficient detail when verifying the given result.iii? 200 = A(20 ? h) + B(20 + h)M1cover up, substitution or equating coeffsiiih = 20 ? 200 = 40 B, B = 5A1iiih = ?20 ? 200 = 40A, A = 5200 dh/dt = 400 ? h2A1iiiM1separating variables and intending to integrate (condone sign error)iii?substituting partial fractionsiii? 5ln(20 + h) ? 5ln(20 ? h) = t + cA1ft their A, B, condone absence of c, Do not allow ln (h-20) for A1.iiiWhen t = 0, h = 0 ? 0 = 0 + c ? c = 0B1cao need to show this. c can be found at any stage. NB c = ln (?1) (from ln (h ? 20)) or similar scores B0.iii?iiiM1anti-logging an equation of the correct form . Allow if c = 0 clearly stated (provided that c = 0) even if B mark is not awarded, but do not allow if c omitted. Can ft their c.iii? 20 + h = (20 ? h)et/5 = 20 et/5 ? h et/5? h + h et/5 = 20 et/5 ? 20? h(et/5 + 1) = 20(et/5 ? 1)DM1making h the subject, dependent on previous markiii?NB method marks can be in either order, in which case the dependence is the other way around.(In which case, 20 + h is divided by 20 ? h first to isolate h).iiiA1AG must have obtained B1 (for c) in order to obtain final A1.Examiner's CommentsThere were a few completely correct solutions to this part. However, many different errors were seen from the majority of candidates. There was also a lot of confused work.Those who started with the correct partial fractions, from 200/(20?h)(20+h) or 1/(20?h)(20+h), usually obtained the first three marks and then integrated having scored M1A1A1M1 thus far. Common errors then included omitting the negative sign when integrating 5/(20?h) (ie giving 5ln(20?h) and hence A0) or failing to state and then evaluate a constant. Those who had no constant were unable to score further marks. Those who did score the first 5 or 6 marks (dependent upon when the constant was evaluated) often used the laws of logarithms correctly and anti-logged although some fiddled the signs when subsequently making h the subject.Some candidates thought that 1/(400?h2)=1/(h?20)(h+20). Marks were scored for using partial fractions on 1/(h?20)(h+20) but logarithms such as ln (h?20) for h<20 and constants such as ln(?1) could not obtain accuracy marks although the marks for anti-logging and making h the subject were still available.There were also a number who felt that 1/(400?h2)=1/(200?h)(200+h).The use of modulus signs was rarely seen.ivAs t → ∞, h → 20. So long-term height is 20m.B1wwwExaminer's CommentsUsually correct.v1st model h = 20(1 ? e?0.1) = 1.90..B1Or 1st model h = 2 gives t = 1.05..v2nd model h = 20(e1/5 ? 1)/(e1/5 + 1) = 1.99..B1????2nd model h = 2 gives t = 1.003..vso 2nd model fits data betterB1 depdep previous B1s correctExaminer's CommentsMost candidates scored all three marks.Total198iB1iM1iA1oeiA1agExaminer's CommentsThe differentiation was usually correct and the use of the chain rule usually lead to full marks for those that started correctly.ii∫πxdx = ∫kdtM1separate variables and attempt integration of both sidesii? ? πx2 = kt + cA1condone absence of ciiWhen t = 0, x = 0 ? c = 0? ? πx2 = ktB1c = 0 wwwiiFull when x = 10, t = T? 50π = kTM1substitute t or T = 50 π/k or x = 10 and rearranging for the other (dependent on first M1) oeii? T = 50π/k *A1.ag, need to have c = 0Examiner's CommentsThe integration here was not difficult. Most candidates scored either three marks or five marks depending upon whether they included a constant of integration. It was very disappointing to see how common this error was.iiidV/dt = ?kxB1correctiiiM1dV/dx.dx/dt = ±kx ftiiiA1agExaminer's CommentsThose who started correctly with dV/dt = ?kx usually obtained full marks. Some candidates had given up by this point.iv∫ π(20 ? x) d x = ∫ ?k dtM1separate variables and intend to integrate both sidesivπ(20x ? ?x2) = ?kt + cB1LHS (not dependent on M1)ivA1RHS i.e. ?kt + c (condone absence of c)ivWhen t = 0, x = 10? π(200 ? 50) = c? c = 150πA1evaluation of c cao oe (x = 10, t = 0)iv? π(20x ? ?x2) = 150π ? ktx = 0 when 150π ? kt = 0M1substitute x = 0 and rearrange for t -dependent on first M1 and non-zero c, oeiv? t = 150π/k = 3T*A1agExaminer's CommentsThe separation of variables and integration were again generally well answered by those who attempted them. As before, the constant was rarely included or found and as it was non-zero in this case, some confused attempts at the final part were seen. As a result, three marks were usually lost here. Those who did include the constant were usually successful in scoring all six marks.Total189e2y = 5 ? e?xB1or y = ln√(5 ? e?x) o.e (e.g.? ln(5 ? e?x))B1? dy/dx = e?x/[2(5 ? e?x)] o.e. B1B1= e?x(but must be correct)M1depsubstituting x = 0, y = ln 2 into their dy/dx dep 1st B1 ? allow one slipor substituting x = 0 into their correct dy/dxA1caoExaminer's CommentsThis implicit differentiation was generally well done. The most common error was d/dx(e-x) = ex instead of e-x. Some candidates re-arranged the original equation correctly to give y = ? ln(5 – e?x), though log errors were quite common here; however, many went on from here by differentiating this incorrectly.Total410iy = 2 arc sin ? = 2 x π/6M1y = 2 arcsin ?i= π/3A1must be in terms of π? can isw approximate answersExaminer's CommentsThis was generally answered successfully, with only a few failing to give the exact value π/3.1.047... implies M1iiy = 2 arcsin x????????????? x ? yii? ??x = 2 arcsin yii? ??x/2 = arcsin yM1or y/2 = arcsin xii? ??y = sin (x/2) [so g(x) = sin (x/2)]A1but must interchange x and y at some stageii? ??dy/dx = ? cos(? x)A1caoiiAt Q, x = π/3M1substituting their π/3 into their derivativeii? ??dy/dx = ? cos π/6 = ? √3/2 = √3/4A1must be exact, with their cos(π/6) evaluatedii? ??gradient at P = 4/√3B1 fto.e. e.g. 4√3/3 but must be exact ft their √3/4 unless 1or f′(x) = 2/√(1?x2)f′(?) = 2/√? = 4/√3 caoiiExaminer's CommentsMost candidates successfully found the inverse function, but ? sin x was occasionally seen. Once that hurdle was crossed, most differentiated sin ? x correctly, though cos(?x) and 2 cos(?x) were seen. The substitution of x = π/3 was usually correct, though a small number used x = 1. The gradient at P was usually the reciprocal of that at Q, with –1/m (instead of 1/m) being the most common error. A few candidates differentiated f(x) directly, often with success.Total811iM1d/dx(sin 2x) = 2cos 2x soican be inferred from dy/dx = 2x cos 2xiA1cao, mark final answere.g. dy/dx = tan 2x + 2x is A0idy/dx = 0 when sin 2x + 2x cos 2xM1equating their derivative to zero,iprovided it has two termsi? ???? tan 2x + 2x = 0*A1must show evidence of division by cos 2xExaminer's CommentsThe vast majority differentiated correctly - though 2xcos2x was seen occasionally - and equated their derivative to zero. Most then succeeded in dividing by cos 2x to arrive at the required result. Some candidates, however, divided before equating the derivative to zero, and gave the derivative as 2x + tan 2x.iiAt P, x sin 2x = 0M1Finding x = π/2 using the given lineii? ??sin 2x = 0, 2x = (0), π ? x = π/2A1x = π/2equation is M0iiAt P, dy/dx = sin π + 2(π/2) cos π = ?πB1 ftft their π/2 and their derivativeiiEqn of tangent: y ? 0 = ?π (x ? π/2)???? y = ?πx + π2/2M1substituting 0, their π/2 and their ?π into y ? y1 = m(x ? x1)or their ?π into y = mx + c, and then evaluating c: y = (?π)x + c,ii? ??2πx + 2y = π2 *A1NB AG0 = (?π) (π/2) + c M1? ??c = π2/2iiWhen x = 0, y = π2/2, so Q is (0, π2/2)M1A1can isw inexact answers from π2/2Examiner's CommentsMost candidates solved x sin 2x = 0 to obtain x = π/2 at P. The derivative was then required to obtain the gradient of the tangent and hence its equation, but some used the given tangent equation itself to find the gradient. The last part was successfully completed by nearly all candidates, with the given tangent equation being used to obtain the correct y-coordinate at Q of π2/2.? ??y = ?πx + π2/2 ? 2πx + 2y = π2 *A1iiiArea = triangle OPQ ? area under curveM1soi (or area under PQ ? area under curve)area under line may be expressed in integral formiiiTriangle OPQ = ? × π/2 × π2/2 [π3/8]B1caoallow art 3.9or using integral:iiiParts: u = x , dv/dx = sin 2x?????????du/dx = 1, v = ?? cos2xM1condone v = k cos2x soiv can be inferred from their ‘uv’iiiA1ftft their v = ?? cos2x, ignore limitsiiiA1correct at this stage, ignore limitsiiiA1cao(so dep previous A1)iiiSo shaded area = π3/8 ? π/4 = π(π2 ? 2)/8*A1NB AG must be from fully correct workExaminer's CommentsMost candidates attempted find the area of the triangle and the area under the curve, though a clear statement of method was not always given. Quite a few candidates attempted to find the triangle area by integration, and came unstuck in the process. The area under the curve was generally recognised as integration by parts, but marks were lost through incorrect v’, or mistakes with signs. Some tried to combine both integrals (for line and curve), and got into a muddle by stock-piling negative signs, rather than simplifying these on a step-by-step basis. Nevertheless, good candidates had little trouble in supplying a fluent solution.Total1812iWhen t = 2, r = 20(1 ? e?0.4) = 6.59 mM1A16.6 or art 6.59idr / dt = ?20 × (?0.2e?0.2t)?????= 4e?0.2tM1?0.2e?0.2t?soiiWhen t = 2, dr / dt = 2.68A12.7 or art 2.68 or 4e?0.4Examiner's CommentsIn part (i), the first two marks for finding the radius when t = 2 were readily achieved. Not so the next two, with some generally rather poor attempts to differentiate?20(1 ? e?0.2t). Quite a few candidates substituted t = 2 into e?0.2t to get e?0.4, then differentiated this as ?0.4e?0.4. Some simply divided their value of r by 2.mark final answeriiA = πr2M1attempt to differentiate πr2or differentiating 400π(1 ? e?0.2t)2 M1ii????dA / dr = 2πr (= 41.428…)A1dA / dr = 2πr (not dA / dt, dr / dA etc.)dA / dt = 400π.2(1 ? e?0.2t).(?0.2e?0.2t) A1iidA / dt = (dA / dr) × (dr / dt)= 41.428… × 2.68M1(o.e.) chain rule expressed in terms of their A, r or impliedsubstitute t = 2 into correct dA / dt M1(Could use another letter for A)ii= 111 m2 / hrA1110 or art 111Examiner's CommentsPart (ii) offered some accessible marks for stating the chain rule, and for dA/dr = 2πr. The final mark depended on getting dr/dt = 2.68 from part (i).Total813ix3 + y3 = 3xy????3x2 + 3y2(dy / dx) = 3x(dy / dx) + 3yB1B1LHS, RHSCondone 3x dy / dx + y (i.e. with missing bracket) if recovered thereafteror equivalent if re-arrangedi????(3y2 ? 3x)(dy / dx) = 3y ? 3x2????dy / dx = (3y ? 3x2) / (3y2 ? 3x)M1collecting terms in dy / dx and factorisingft correct algebra on incorrect expressions with two dy / dx termsi= (y ? x2)/(y2 ? x)*A1caoNB AGExaminer's CommentsPart (i) was very well done – it is pleasing to see how well implicit differentiation is understood, and the algebra to derive the given result was generally done well.Ignore starting with ‘dy / dx = …’ unless pursuediiTP when y ? x2 = 0ii????y = x2M1or x = √yii????x3 + x6 = 3x.x2????x6 = 2x3M1substituting for y in implicit eqn (allow one slip, e.g. x5)or x for y (i.e. y3/2 + y3 = 3y1/2y o.e.)ii????x3 = 2 (or x = 0)A1o.e. (soi)or y3/2 = 2ii????A1caomust be exactExaminer's CommentsIn part (ii), many fully correct answers notwithstanding, some failed to get beyond the first M1 for y = x2; others who substituted for y in the implicit function sometimes erred with (x2)3 = x5.x = 1.2599… is A0 (but can isw )Total814iWhen x = 3, y = 3/√(3 ? 2) = 3M1substituting x = 3 (both x's)or x = x/√(x ? 2) M1iSo P is (3, 3) which lies on y = xA1y = 3 and completion (‘3 = 3’ is enough)Examiner's CommentsPart (i) was an easy two marks for nearly all candidates. However, sometimes it was difficult to tell whether it was made clear that the point (3, 3) lies on the line y = x.? x = 3 A1(by solving or verifying)iiM1Quotient or product rulePR: ??x(x ? 2)?3/2 + (x ? 2)?1/2If correct formula stated, allow one error; otherwise QR must be on correct u and v,iiA1correct expressionwith numerator consistent with their derivatives and denominator correct initiallyiiM1× top and bottom by √(x ? 2) o.e. e.g. taking out factor of (x ? 2)?3/2allow ft on correct equivalent algebra from their incorrect expressioniiA1NB AGiiWhen x = 3, dy / dx = ?? × 13/2M1substituting x = 3ii= ??A1iiThis gradient would be ?1 if curve were symmetrical about y = xA1caoor an equivalent valid argumentExaminer's CommentsIn part (ii), both the product and quotient rules were seen – perhaps the product rule is slightly easier to sort out in this case. Although most gained the initial M1A1 for this, the algebra required to derive the given answer, either by using a common denominator or factoring out (x – 2)? ? , was poorly done. Most candidates should have been able to recover to get the derivative at x = 3, and 4/7 was a common mark for the part. The final mark, using this result to examine the symmetry of the function, was the preserve of more able candidates. Many thought that the P had to be a turning point for the graph to be symmetrical about y = x.iiiu = x ? 2 ? du / dx = 1 ? du = dxWhen x = 3, u = 1 when x = 11, u = 9B1or dx / du = 1No credit for integrating initial integral by parts. Condone du = 1. Condone missing du's in subsequent working.iiiB1iiiM1splitting their fraction (correctly) and u/u1/2 = u1/2 (or √u)or integration by parts: ?(must be fully correct – condone missing bracketiiiA1by parts: [2u1/2 (u + 2) ? 4u3/2/3]iii= (18 + 12) – (2/3 + 4)M1substituting correct limitsiiiA1caoNB AGdep substitution and integration attemptediiiArea under y = x is ? (3 + 11) × 8 = 56B1o.e. (e.g. 60.5 ? 4.5)iiiArea = (area under y = x) ? (area under curve)M1soi from workingmust be trapezium area: iiiA1cao30.7 or betterExaminer's CommentsPart (iii) achieved mixed success. It was pleasing to see that most gained the B1 for du = dx; most got the second B1 for (u + 2)√u; thereafter, the ‘M’ for splitting the fraction was often lost – some used integration by parts here with some success (a sledgehammer to crack a nut?). Those who got beyond this hurdle often gained all 6 marks. The final 3 marks were often omitted, but the best candidates got all 9 marks; the most common error here was to use the triangle with vertices (0, 0), (11, 0) and (11, 11) rather than the trapezium formed by removing the triangle with vertices (0, 0), (3, 0) and (3, 3).Total1815M1[k (3x ? 2)1/2]A2k = 2/3M1depsubstituting limits dep 1st M1= 2/3*A1NB AGORM1A1× 1/3 (du)× 2/3 w (dw)A1M1depsubstituting correct limits dep 1st M1upper ? lower, 1 to 4 for u or 1 to 2 for w or= 2/3*A1NB AGExaminer's CommentsThis was a straightforward starter question, for which many candidates scored full marks. The most popular strategy was to use the substitution u = 3x – 2, and candidates were generally adept at replacing dx with 1/3 du, integrating correctly and substituting correct limits. In a few cases ln u? was obtained after integration. The substitution u = (3x – 2)? was less common and caused greater difficulty. Relatively few students attempted to integrate directly without substitution, but those that did often succeeded, and gained the 5 marks with ease.substituting back (correctly) for x and using 1 to 2Total516iWhen x = 1, f(1) = ln(2/2) = ln 1 = 0 so P is (1, 0)B1or ln(2x/1 + x) = 0 ? 2x/(1 + x) = 1? 2x = 1 + x ? x = 1if(2) = ln(4/3)B1Examiner's CommentsPart (i) offered two straightforward marks. Many approximated for ln (4/3), but we ignored this in subsequent working.if approximated, can isw after ln(4/3)iiy = ln (2x) ln(1 + x)M1condone lack of bracketsiiM1one term correct2/2x or ?1/(1 + x)iiA1caomark final ansiiB1correct quotient or product ruleneed not be simplifiedii?M1chain rule attemptediiA1o.e., but mark final ansneed not be simplifiediiAt P, dy/dx = 1 ??? = ?A1caoExaminer's CommentsIn part (ii), the hint proved valuable and was taken by nearly all candidates. However, many found the derivative of ln(2x) as 1/(2x) and lost two marks. Those who avoided this error usually scored all 4 marks.iiix = ln[2y/(1 + y)] or(x?y here or at end to complete)x = ey/(2 ? ey)iii? ex = 2y / (1 + y)B1x(2 ? ey) = ey????B1iii? ex(1 + y) = 2yB12x = ey+ xey = ey(1 + x)??B1iii? ex = 2y ? exy = y(2 ? ex)B12x/(1 + x) = ey???B1iii? y = ex/(2 ? ex) [= g(x)]B1completionln[2x/(1 + x)] = y [= f(x)]??B1iiiOR gf(x) = g(2x/(1 + x)) =eln[2x/(1 + x)]/{2 ? eln[2x/(1 + x)]}M1forming gf or fgfg(x) = ln{2ex/(2 ? ex)/[1 + ex/(2 ? ex)]} M1iiiA1= ln[2ex/(2 ? ex + ex)]???A1iiiM1A1= ln(ex) = x?????M1A1iiigradient at R = 1/ ? = 2B1 ft1 / their ans in (ii) unless ±1 or 0Examiner's CommentsInverting the function in part (iii) was less successful than usual. This might have been caused by candidates using the ‘hint’ from the previous part to write x = ln 2y – ln (1 + y), and then getting stuck. The gradient in the last part as the reciprocal of that in part (ii) was better answered than in previous papers.2 must follow ? for (ii) unless g′(x) usedAdditional notes and solutionslast part: g(x) = ex/(2 ? ex) ? g′(x) = [(2 ? ex)ex ? ex(?ex)]/(2 ? ex)2 = 2ex /(2 ? ex)2or g′(x) = ex(?1) (?ex)](2 ? ex)?2 + ex(2 ? ex)?1g′(0) = 2.1/12 = 2 B1ivlet u = 2 ? ex ? du / dx = ?exx = 0, u = 1, x = ln(4/3), u = 2 ? 4/3 = 2/3B12?e0 = 1, and 2 ? eln(4/3) = 2/3 seenhere or later (i.e. after substituting 0 andln(4/3) into ln(2 ? ex))iv?M1or by inspection [k ln (2 ? ex)]ivA1[?ln(u)] (could be [lnu] if limits swapped)k = ?1ivA1caoNB AGivShaded region = rectangle ? integralM1Allow full marks here for correctly evaluating?iv= 2ln(4/3) ? ln(3/2)B1rectangle area = 2ln(4/3)iv= ln(16/9 × 2/3)iv= ln(32/27)*A1caoNB AG must show at least one step from 2ln(4/3) ? ln(3/2)Examiner's CommentsFinally, part (iv) was the least well answered question. The new ‘u’ limits of 1 and 2/3 were usually present, but many lost the minus sign from du = ?e?xdx , and few gave fully convincing ‘shows’. The last result was rarely done, though it was not possible to gather whether this was due to difficulty or lack of time.Additional notes and solutionslast part?[x ln 2 +?x ln x?? x= 2ln2 + 2ln2 ? 2 ? 3ln 3 + 2 ? (ln2 ? 1 ? 2ln2 + 1) = 5ln2 ? 3ln3 = ln(32/27)Total1817idV/dt = k√VB1cao condone different k (allow MR B1 for = kV2)iM1A1A12(1/2 kt + c) × constant multiple of k (or from multiplying out oe; or implicit differentiation)cao www any equivalent form (including unsimplified)Allow SCB2 if V = (1/2 kt + c)2 fully obtained by integration including convincing change of constant if usedCan score B1 M0 SCB2Examiner's CommentsMost candidates scored the first mark for writing down the differential equation. Those who differentiated often scored full marks. Common errors included, incorrectly differentiating the inside of the bracket- instead of 1/2k, a variety of errors were seen, including functions of t, and, for those who did differentiate correctly, failing to equate this to k√V at the final stage.Quite a number omitted this differentiation. Some others decided to ignore the instruction given and integrate instead in order to derive the given result instead of verifying it. Very few of these attempts gained any further credit as they failed to deal with the change in constant. Those who integrated to reach 2√V=kt+c then, too often, gave √V= [1/2(kt+c) =] 1/2kt+c when trying to establish the given result and obtained no marks unless they explained the change of constant.iiB1substituting any one from t = 1, V = 10,000 or t = 0, V = 0 or t = 2, V = 40,000 into squared form or rooted form of equation(Allow ?/±100 or ?/±200)iiB1M1A1substituting any other from aboveSolving correct equations for both www (possible solutions are (200,0), (?200,0), (600, ?400), (?600,400) (some from –ve root))either form wwwSC B2 for V = (100t)2 oe stated without justificationSCB4 if justification eg showing substitutionSC those working with (k + c)2 = 30,000 can score a maximum of B1B0 M1A0 (leads to k ≈ 146, c ≈ 26.8)Examiner's CommentsThe majority scored two marks for writing down two correct equations. Those who then?square rooted say, (1/2k+c)2?= 10,000 to reach 1/2k+c =100, and the other equation to obtain k+c=200 usually obtained full marks. Those who did not square root the equations were sometimes successful but more often made errors or abandoned their attempts.Some felt that (1/2k+c)2= 1/4k2+c2Total818idF / dv = ?25 v?2M1d / dv(v?1) = ?v?2 soiiA1?25 v?2 o.e mark final ansExaminer's CommentsThis was almost invariably correctly done. No candidates seemed to be put off by the rather excessive speed of the car. Occasionally, the quotient rule was seen, with errors in differentiating the ‘25’.iiWhen v = 50, dF/dv = ?25/502 (= ?0.01)B1?25/502iiM1o.e.ii= ?0.01 × 1.5 = ?0.015A1caoo.e. e.g. ?3/200 iswExaminer's CommentsAgain, this was very well answered, provided part (i) was correct. Almost all candidates scored an M1 for the chain rule.Total519Let u = 1 + x ?M1∫ (u ? 1)u?1/2(du)*condone no du, missing bracket, ignore limitsA1∫ (u1/2 ? u?1/2)(du)A1; ignore limits= (16/3 ? 4) ? (2/3 ? 2)M1depupper–lower dep 1st M1 and integrationwith correct limits e.g. 1, 4 for u or 0, 3 for xA1caoor but must be exactor using w = (1 + x)1/2 ?OR Let u = x, v′ = (1 + x)?1/2M1? u′ = 1, v = 2(1 + x)1/2A1upper–lower with correct limits (w = 1,2) M1? A1ignore limits, condone no dx8/3 A1 caoA1ignore limits*If du done by parts:= (2 × 3 × 2 – 4 × 8/3) – (0 – 4/3)2u1/2 (u ? 1) ?∫2u1/2 du A1[2u1/2 (u ? 1)?4u3/2/3] A1A1caoor but must be exact Examiner's CommentsMost candidates used integration by substitution, though a significant minority used integration by parts. In general, the former were more successful, with the main difficulty being in expanding (u – 1)u-1/2 as u1/2 – u-1/2. Some proceeded from here using integration by parts, with mixed success. When parts were used, the most common error was in deriving v = 2(1 + x)1/2 from v' = (1 + x)-1/2.substituting correct limits M1 8/3 A1caoTotal5 ................
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