Version.0 General Certificate of Education (A-level) June 0 Mathematics MPC3 (Specification 6360) Pure Core 3 Final Mark Scheme
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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 0) 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.
MPC3 June 0 (a) (b) 6 or,0 6 dy = d B condone 0.67 AWRT M k where k =, 6 or 6 A k = (c) y ln 6 =.798 ln =.4849 3 ln8 =.8904 4 ln 4 = 3.78 5 ln 30 = 3.40 6 ln 36 = 3.5835 7 ln 4 = 3.7377 M A 5+ y-values correct, either eact or correct to 3SF (rounded or truncated) or better all 7 y-values correct (and only these 7 values), either eact or correct to 3SF (rounded or truncated) or better A = (.798 + 3.7377) 3 + 4.4849 + 3.78+ 3.5835 + (.8904 + 3.40) M correct use of Simpson s rule on their 7 y-values, condone missing square brackets = 8.4 A 4 CAO this value only Total 7
(a)(i) y = e dy = e + e d M A A ISW 3 ke + le where k and l are s or s k = Independent of each other l = ( = e ( + ) ) (ii) = dy d = 3e M tangent: y e 3e ( ) correct substitution of = into their d y d but must have earned M in part (i) = OE A CSO (no ISW), must have scored first 4 marks common correct answer: y= 3e e (b) sin3 y = + cos3 dy ( + cos3 )6cos3 sin 3 ( 3sin 3 ) = d ( + cos 3 ) M ± p( + cos 3 ) cos3± qsin 3 (sin 3 ) ( + cos3 ) where p and q are rational numbers condone poor use/omission of brackets PI by further working = 6cos3 + 6cos 3 + 6sin 3 ( + cos3 ) A this line must be seen in this form (ie in terms of cos 3 and sin 3), but allow sin 3 replaced by cos 3 condone denominator correctly epanded 6cos3 + 6 = ( + cos3 ) 6 = A 4 CSO + cos3 m Total 9 correct use of ksin 3 kcos 3 k ksin 3= k cos 3 + = or
note: if degrees used then no marks in (a) and (c) f = cos e or reverse 3(a) f( 0.4) = 0.3 f ( 0.5 ) 0. = sight of ±0.3 (AWRT) AND 0. M (AWRT) < < A CSO, note f () must be defined, condone 0.4 α 0.5 change of sign 0.4 α 0.5 (M) alternative method e 0.4 =.5, cos ( 0.4 ) =.8 e 0.5 =.65, cos ( 0.5 ) =.57 at 0.4 e < cos ( ) at 0.5 e > cos ( ) cos = e (b) = cos(e ) (A) = ( cos(e ) + ) = + cos(e ) B 0.4 < α < 0.5 AG must see middle line, and no errors seen, but condone cos e (c) = 0.4 = 0.539 B CAO 3 = 0.48 B CAO Total 5
sin ± 0.5 = ± 4.5 PI by sight of 94.5 etc M condone ± 4.4 θ = 94.5, 345.5 (AWRT) A no etras in interval, ignore answers outside interval 4(a)(i) (ii) cot (+ 30) = 7 cosec(+ 30) condone replacing + 30 by Y (cosec ( 30) ) 7 cosec( 30) + = + M correct use of cosec Y = + cot Y cosec ( 30) 7cosec( 30) 4 0 + + + = A must be in this form (cosec(+ 30) ± )(cosec(+ 30) ± 4) = 0 m attempt at factorisation cosec( + 30) = or 4 A must be this line using f ( + 30) + 30 = 94.5, 345.5 = 8., 57.8 (AWRT) B one correct answer, allow 8.3, ignore etra solutions B 6 CAO both answers correct and no etras in interval, ignore answers outside interval (b) stretch (I) scale factor (II) parallel to -ais (III) M I and either II or III A I + II + III translate E 5 B 4 condone 5 to left or 5 in 0 (direction) alternative method translate (E) 30 0 (B) stretch (M) as above scale factor parallel to -ais (A) as above Total
5(a) f not E OE (b) y = + = y + y + = M M swap and y a correct net line g = OE A 3 [ ] y = either order (c) g 0.5 B sight of 0.5 OE (d) = + + ( + ) = ( + ) + = 4 + 4+ or M or = + or + = = 0 A 3 CSO B sight of + or ( + ) one correct step, must be one of these four lines Total 8 6(a) 3ln = 4 4 ln = 3 4 = e 3 B ISW. Condone 3 e 4 (b) 0 3ln + = 9 ln 3(ln ) + 0 = 9ln M correctly multiplying by ln. 3(ln ) 9ln 0 0 + = A (3ln ± 4)(ln ± 5) = 0 m 4 ln =, 5 A 3 4 3 5 = e, e A 5 condone 3 4 use of formula, or completing the square must be correct e Total 6
7(a)(i) 3 M A modulus graph, approimate V shape, touching negative -ais and crossing y- ais, 3 marked, graph symmetrical, straight lines (ii) (b)(i) 3+ 3 = ( 3+ 3 = ) modulus graph in 3 sections, touching M -ais and crossing positive y-ais correct curvature A their >, their < independent correct curve A 3 and = ±, y = marked 0 = 3 4 A M either A or B seen, all terms on one side = 4, A,A 3 + 3 = + 3+ = 0 B =, A,A 5 =,, 4 SC NMS or partial method correct value /5 correct values /5 independent of 3 correct values 5/5 method mark more than 3 distinct values ma /5 (ii) > 4, < M,A > their largest, < their smallest; CAO Total
cos ( + tan ) u = + tan 8 du = sec d du u OE d M = m u = = u = ( + tan ) + ( c) du condone = asec where a is a d constant k (d u u ), where k is a constant d u u correct integral of their epression but must have scored M m A correct, or AF A 5 CSO, no ISW Total 5
9 (a) ln d u = ln du ( d) l = dv = ( d ) v = = ln ( d) = ln ( + c) 4 A 3 A M correct direction and sight of, (b) y = (ln ) dy = ln d M k ln A k = where k =,or (c) y = ln e V = π ln d B dv u = = ( ln ) ( d) du = ln v = d M = ( ln ) ln ( d) m all correct, incl brackets, π, limits and d (but d may be seen BEFORE this line) du k = ln where d correct direction with k =,or and sight of correct substitution of their terms into the parts formula = ( ln ) ln ( d) A integral needs to be simplified to ln = ( ln ) ( ln ) OE 4 e V = (π) ( ln ) ( ln ) 4 e correct substitution of and e into their = (π) e 0 + m epressions of the form 4 4 p(ln ) + qln + rwhere p, q and r are non-zero rational numbers, and an intention to subtract Do not condone F() F(e) π = e 4 OE A 6 e π etc 4 4 Total TOTAL 75