GCE Mathematics Advanced GCE Unit 7: Core Mathematics Mark Scheme for June 0 Oxford Cambridge and RSA Examinations
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7 Mark Scheme June 0 (i) integral of form ke x M any non-zero constant k different from 6; using substitution u x to obtain ke u earns M (but answer to be in terms of x ) correct e x 6 A or equiv such as (ii) integral of form ln( ) M any non-zero constant k ; allow if brackets absent; ln (after sub n) earns M 0 correct 5l n(x ) A or equiv such as ln( ) brackets rather than modulus signs but brackets or modulus signs must be present (so that 5lnx earns A0) Include + c at least once B 5 anywhere in the whole of question ; this mark available even if no marks awarded for integration 5 Apply one of the transformations correctly to their equation B correct ln x ln B or equiv Show at least one logarithm property M correctly applied to their equation of resulting curve (even if errors have been made earlier) y ln( x ) A or equiv of required form; ln x earns A; correct answer only earns /; condone absence of y (a) State sin cos sin B or unsimplified equiv such as 7(sin cos ) sin Attempt to find value of cos M by valid process; may be implied A exact answer required; ignore subsequent work to find angle (b) Attempt use of identity for cos M of form cos ; initial use of cos sin needs attempt to express e x sin in terms of cos to earn M 6cos 9cos 0 A or unsimplified equiv or equiv involving sec Attempt solution of -term quadratic eqn M for cos or (after adjustment) for sec Use sec at some stage cos M or equiv A 5 or equiv; and (finally) no other answer 8
7 Mark Scheme June 0 (i) Draw sketch of y ( x ) *B touching positive x-axis and extending at least as far as the y-axis; no need for or 6 to be marked; ignore wrong intercepts Draw straight line with positive gradient *B at least in first quadrant and reaching positive y-axis; assess the two graphs independently of each other Indicate two roots B AG; dep *B *B and two correct graphs which meet on the y-axis; indicated in words or by marks on sketch [SC: Draw sketch of y ( x ) x 6 and indicate the two roots : B (i.e. max mark)] (ii) State 0 or x 0 B not merely for coordinates (0, 6) (iii) correct first iterate B to at least dp; any starting value (> 6) Show correct iteration process M producing at least iterates in all; may be implied by plausible converging values at least correct iterates A allowing recovery after error; iterates given to only d.p. acceptable; values may be rounded or truncated.8 A answer required to exactly dp; A0 here if number of iterates is not enough to justify.8; attempt consisting of answer only earns 0/ [ 0. 7.7769.789 ;.005.559.7790.789 ;.059767.6.78.7850 ;.087 798.7 060.780.7850 ;.7.7 769.789.785 ; 5.0 695.8 5.7867.785] 8 5 Attempt use of product rule *M to produce ln( ) kx kx x x form xln(x ) A x x A or equiv Attempt second use of product rule *M Attempt use of quotient (or product) rule *M allow numerator the wrong way round 8x 8 x(x ) 6x ln(x ) x (x ) A or equiv Substitute into attempt at second deriv M dep *M *M *M 96 ln5 5 A 8 or exact equiv consisting of two terms 8
7 Mark Scheme June 0 6 Method : (Differentiation; assume value 0 ; eqn of tangent; through origin) Differentiate to obtain k(x 5) M any constant k ( x 5) A or equiv Attempt to find equation of tangent at P and attempt to show tangent passing through origin M assuming value 0 ; or equiv y x and confirm that 5 tangent passes through O A AG; necessary detail needed Method : (Differentiation; equate y change change x to deriv; solve for x) Differentiate to obtain k(x 5) M any constant k Equate y change change ( x 5) A or equiv x to deriv and attempt solution M x 5 x obtain 0 only (x 5) and solve to Method : (Differentiation; find x from y f( x) x and y x 5 ) Differentiate to obtain k(x 5) M any constant k A ( x 5) A or equiv State ( 5) x, y x 5, eliminate y and attempt solution M condone this attempt at eqn of tangent 0 only A Method : (No differentiation; general line through origin to meet curve at one point only) Eliminate y from equations y kx and y x 5 and attempt formation of quadratic eqn M k x x 5 0 A or equiv Equate discriminant to zero to find k M 0 k or equiv and confirm x A 5 Method 5: (No differentiation; use coords of P to find eqn of OP; confirm meets curve once) 0 5 Use coordinates (, 5) to obtain y x 0 or equiv as equation of OP B Eliminate y from this eqn and eqn of curve and attempt quadratic eqn M should be 9x 60x 00 0or equiv Attempt solution or attempt discriminant M Confirm 0 only or discriminant = 0 A
7 Mark Scheme June 0 Either: Integrate to obtain correct 9 k(x 5) *M any constant k ( x 5) A Apply limits 5 and 0 M dep *M; the right way round Make sound attempt at triangle area and calculate (triangle area) minus (their area under curve) M or equiv 0 0 5 6 9 5 and hence 5 9 5 A 9 or exact equiv involving single term Or: Arrange to x = and integrate to obtain ky ky form *M 5 y y A 9 Apply limits 0 and 5 M dep *M; the right way round Make sound attempt at triangle area and calculate (their area from integration) minus (triangle area) M 0 5 5 5 and hence 5 5 A (9) or exact equiv involving single term 9 9 9 7 (i) Either: Attempt solution of at least one linear eq n of form ax b M A and (finally) no other answer Or: Attempt solution of -term quadratic eq n obtained by squaring attempt at g( x ) on LHS and squaring or on RHS M A () and (finally) no other answer (ii) Either: (x 5) 5 for h B Attempt to find inverse function M of function of form ax b ( x 0) 9 A or equiv in terms of x Or: State or imply g is ( x 5) g g Attempt composition of with M 5 9 A () or more simplified equiv in terms of x (iii) State x 0 B give B for answer x 0 8 B
7 Mark Scheme June 0 0.0t k 8 (i) Differentiate to obtain form e M any constant k different from 00 0.0t 0.0t 5.6e or 5.6e A or (unsimplified) equiv.9 or.9 or.87 or.87 A but not greater accuracy; allow if final statement seems contradictory; answer only earns 0/ differentiation is needed (ii) Either: State or imply M 75e kt B or equiv Attempt to find formula for M M 0.07t ( ln 8 ) t M 75e A 0 or equiv such as 75e 5 Equate masses and attempt rearrangement M as far as equation with e appearing once 0.06t 6 e A 5 or equiv of required form which might involve 5. or greater accuracy on RHS; final two marks might be earned in part iii 0.t Or: State or imply M 75 r B for positive value r 0. 75.6 t B Attempt to find M in terms of e M Equate masses and attempt rearrangement M 0.06t 6 e A 5 or equiv of required form which might involve 5. or greater accuracy on RHS; final two marks might be earned in part iii (iii) Attempt solution involving logarithm of any equation of form e mt c M whether the conclusion of part ii or not 7. A or greater accuracy 7. ; correct answer only earns both marks 0 5
7 Mark Scheme June 0 9 (i) Use at least one identity correctly B angle-sum or angle-difference identity Attempt use of relevant identities in single rational expression M not earned if identities used in expression where step equiv to A B C A B C D E F D E F or similar has been carried out; condone (for MA) if signs of identities apparently switched (so that, for example, denominator appears as cos cos sin sin cos cos cos sin sin ) sin cos sin cos cos cos A or equiv but with the other two terms from each of num r and den r absent Attempt factorisation of num r and den r M sin and hence tan cos A 5 AG; necessary detail needed (ii) State or imply form k tan50 M obtained without any wrong method seen State or imply sin50 tan50 A or equiv such as 9cos50 A or exact equiv (such as ); correct 9 answer only earns / (iii) State or imply tan 6 k B State k 6 B Attempt second value of M using 6 tan k (multiple of 80) 6 tan k 0 A and no other value 6
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