Mark Scheme (Results) Summer 2009

Transcription

1 Mark (Results) Summer 009 GCE GCE Mathematics (6664/0)

2 June Core Mathematics C Mark Question Q x x x x dx 4 x x dx x x = 9 (9 + C scores A0) M AA M A (5) [5] st M for attempt to integrate st A for nd A for x x or a simplified version. x x or x kx or x kx. or a simplified version. Ignore + C, if seen, but two correct terms and an extra non-constant term scores MAA0. nd M for correct use of correct limits ('top' 'bottom'). Must be used in a 'changed function', not just the original. (The changed function may have been found by differentiation). Ignore 'poor notation' (e.g. missing integral signs) if the intention is clear. No working: The answer 9 with no working scores M0A0A0MA0 ( mark). 6664/0 GCE Mathematics June 009

3 Q... x or... x The or can be in unsimplified form. M kx kx k x k 6k = 8; 448 kx, 6k x [or 6( kx ) ] (If 6kx follows 6( kx ), isw and allow A) k = 4 (Ignore k = 0, if seen) The terms can be listed rather than added. Ignore any extra terms. M for either the x term or the x term. Requires correct binomial coefficient in any form with the correct power of x, but the other part of the coefficient (perhaps including powers of and/or k) may be wrong or missing. Allow binomial coefficients such as,,, C, C. However, 448 kx or similar is M0. B, A, A for the simplified versions seen above. Alternative: Note that a factor can be taken out first: Ignoring subsequent working (isw): Isw if necessary after correct working: e.g kx 6k x M B A A 4 4kx k x isw (Full marks are still available in part ). kx B; A, A (4) M, but the mark scheme still applies. M for equating their coefficient of x to 6 times that of x to get an equation in k, or equating their coefficient of x to 6 times that of x, to get an equation in k. Allow this M mark even if the equation is trivial, providing their coefficients from part have been used, e.g k 6k, but beware k = 4 following from this, which is A0. An equation in k alone is required for this M mark, so... e.g kx 6k x k 4 or similar is M0 A0 (equation in coefficients only is never seen), but... e.g kx 6k x 6 448k 6k k 4 will get M A (as coefficients rather than terms have now been considered). A () [6] The mistake kx would give a maximum of marks: MB0A0A0, MA 6664/0 GCE Mathematics June 009

4 Q f( k ) 8 f() 4 4 (6 )( k ) 8 So k = f( x) x k x k 8 x x 0 = (x 5)(x + ) M for substituting x = (not x ) and equating to 4 to form an equation in k. If the expression is expanded in this part, condone slips for this M mark. Treat the omission of the 8 here as a 'slip' and allow the M mark. Beware: Substituting x and equating to 0 (M0 A0) also gives k. Alternative; M for dividing by (x ), to get x (function of k), with remainder as a function of k, and equating the remainder to 4. [Should be x (4 k), remainder 4 k ]. No working: k with no working scores M0 A0. st M for multiplying out and substituting their (constant) value of k (in either order). The multiplying-out may occur earlier. Condone, for example, sign slips, but if the 4 (from part ) is included in the f(x) expression, this is M0. The nd M is still available. nd M for an attempt to factorise their three term quaatic (TQ). B () M A () M MA () [6] A The correct answer, as a product of factors, is required. 5 Allow x ( x ) Ignore following work (such as a solution to a quaatic equation). If the equation is solved but factors are never seen, the nd M is not scored. 6664/0 GCE Mathematics June 009 4

5 Q4 x = gives.6 (allow AWRT) Accept 5 x =.5 gives.580 (allow AWRT) Accept.58, (.44 ) ( ) = 6. (AWRT 6., even following minor slips) B B () B,[MAft] A (4) Overestimate B 'Since the trapezia lie above the curve', or an equivalent explanation, or sketch of (one or more) trapezia above the curve on a diagram (or on the given diagram, in which case there should be reference to this). (Note that there must be some reference to a trapezium or trapezia in the explanation or diagram). db () [8] B for or equivalent. For the M mark, the first bracket must contain the 'first and last' values, and the second bracket (which must be multiplied by ) must have no additional values. If the only mistake is to omit one of the values from the second bracket, this can be considered as a slip and the M mark can be allowed. Bracketing mistake: i.e. (.44 ) ( ) scores B M A0 A0 unless the final answer implies that the calculation has been done correctly (then full marks can be given). Alternative: Separate trapezia may be used, and this can be marked equivalently. ( ) (.554.)... ( st Aft for correct expression, ft their.6 and their.580 ) st B for 'overestimate', ignoring earlier mistakes and ignoring any reasons given. nd B is dependent upon the st B (overestimate). 6664/0 GCE Mathematics June 009 5

6 Q r 96 or r or r 4 r (*) 5 a 4 or a 96 a =..., 9 M Acso () M, A () (d) (d) S S , = 8, = 8.00 (AWRT 80) M for forming an equation for r based on 96 and 4 (e.g. 96r 4 scores M). The equation must involve multiplication/division rather than addition/subtraction. A Do not penalise solutions with working in decimals, providing these are correctly rounded or truncated to at least dp and the final answer / is seen. Alternative: (verification) M 8 Using r and multiplying 4 by this (or multiplying by r three times). A Obtaining 96 (cso). (A conclusion is not required). M 4 96 (no real evidence of calculation) is not quite enough and scores M A0. MAft, () M, A () for the use of a correct formula or for 'working back' by dividing by (or by their r) twice from 4 (or 5 times from 96). Exceptionally, allow M also for using ar 6 4 or ar 96 instead of ar 5 4 or ar 96, or for dividing by r three times from 4 (or 6 times from 96) but no other exceptions are allowed. M for use of sum to 5 terms formula with values of a and r. If the wrong power is used, e.g. 4, the M mark is scored only if the correct sum formula is stated. st Aft for a correct expression or correct ft their a with r. nd A for awrt 80, even following 'minor inaccuracies'. Condone missing brackets round the for the marks in part. Alternative: M for adding 5 terms and st Aft for adding the 5 terms that ft from their a and r. M for use of correct sum to infinity formula with their a. For this mark, if a value of r different from the given value is being used, M can still be allowed providing r. [9] 6664/0 GCE Mathematics June 009 6

7 Q6 x 9 y 4 Centre is (, ) x y "9" "4" r "9" "4" = 5 (or 5 ) PQ ( ) ( 5 ) or 8 6 = 0 = radius, diam. (N.B. For A, need a comment or conclusion) [ALT: midpt. of PQ ( ), ( 5) : M, = (, ) = centre: A] [ALT: eqn. of PQ x 4y 0: M, verify (, ) lies on this: A] [ALT: find two grads, e.g. PQ and P to centre: M, equal diameter: A] [ALT: show that point S (, 5) or (, ) lies on circle: M because PSQ = 90, semicircle diameter: A] R must lie on the circle (angle in a semicircle theorem) often implied by a diagram with R on the circle or by subsequent working) x 0 y 4y 0 (y )(y + 6) = 0 y... (M is dependent on previous M) y = 6 or (Ignore y = 6 if seen, and 'coordinates' are not required)) st M for attempt to complete square. Allow ( x ) k, or ( y ) k, k 0. st A x-coordinate, nd A y-coordinate nd M for a full method leading to r =, with their 9 and their 4, rd A 5 or 5 The st M can be implied by (, ) but a full method must be seen for the nd M. Where the 'diameter' in part has clearly been used to answer part, no marks in, but in this case the M (not the A) for part can be given for work seen in. Alternative st M for comparing with x y gx fy c 0 to write down centre ( g, f ) directly. Condone sign errors for this M mark. nd M for using r g f c. Condone sign errors for this M mark. st M for setting x = 0 and getting a TQ in y by using eqn. of circle. nd M (dep.) for attempt to solve a TQ leading to at least one solution for y. Alternative : (Requires the B mark as in the main scheme) st M for using (, 4, 5) triangle with vertices (, ),(0, ),(0, y ) to get a linear or quaatic equation in y (e.g. ( y ) 5). nd M (dep.) as in main scheme, but may be scored by simply solving a linear equation. Alternative : (Not requiring realisation that R is on the circle) B for attempt at m m PR QR, (NOT m ) or for attempt at Pythag. in triangle PQR. PQ st M for setting x = 0, i.e. (0, y), and proceeding to get a TQ in y. Then main scheme. Alternative by 'verification': B for attempt at m m PR QR, (NOT m ) or for attempt at Pythag. in triangle PQR. PQ st M for trying (0, ). nd M (dep.) for performing all required calculations. A for fully correct working and conclusion. M A, A M A (5) M A () B M dm A (4) [] 6664/0 GCE Mathematics June 009

8 Q (i) (ii) (i) (ii) tan 45, 5 sin 5.6, 56.4 (AWRT: 4, 56) 4sin x sin x cos x 4sin xcos x sin x sin x(4cos x ) 0 Other possibilities (after squaring): sin x (6sin x ) 0, (6cos x 9)(cos x ) x = 0, 80 seen x = 4.4, 8.6 (AWRT: 4, 9) st B for 45 seen (, where 90) nd B for 5 seen, or ft (80 + ) if is negative, or ( 80) if is positive. If tan k is obtained from wrong working, nd Bft is still available. rd B for awrt 4 (, where 90 ) 4 th B for awrt 56, or ft (80 ) if is positive, or (80 + ) if is negative. If sin k is obtained from wrong working, 4 th Bft is still available. st sin x M for use of tan x cos x. Condone sin x. cosx nd M for correct work leading to factors (may be implied). st B for 0, nd B for 80. rd B for awrt 4 (, where 80) 0 B, Bft B, Bft (4) M M B, B B, Bft (6) 4 th B for awrt 9, or ft (60 ). If cos k is obtained from wrong working, 4 th Bft is still available. N.B. Losing sin x 0 usually gives a maximum of marks MM0B0B0BB Alternative: (squaring both sides) st M for squaring both sides and using a 'quaatic' identity. e.g. 6sin 9(sec ) nd M for reaching a factorised form. e.g. (6cos 9)(cos ) 0 Then marks are equivalent to the main scheme. Extra solutions, if not rejected, are penalised as in the main scheme. For both parts of the question: Extra solutions outside required range: Ignore Extra solutions inside required range: For each pair of B marks, the nd B mark is lost if there are two correct values and one or more extra solution(s), e.g. tan 45, 45, 5 is B B0 Answers in radians: Loses a maximum of B marks in the whole question (to be deducted at the first and second occurrence). [0] 6664/0 GCE Mathematics June 009 8

9 Q8 log y y y 8 or 0.5 M A () 5 or 4 6 or 5 9 M [or log 5log or log 6 4log or log 5 9log ] [or log0 log or log 0 log0 6 log 6 or log 0 log 5 log log ] log + log6 9 (logx)... or (logx)(logx)... (May not be seen explicitly, so M may be implied by later work, and the base may be 0 rather than ) log x x = 8 log x x 8 M for getting out of logs correctly. If done by change of base, log 0 y is insufficient for the M, but y scores M. A for the exact answer, e.g. log 0 y 0.90 y scores M (implied) A0. Correct answer with no working scores both marks. Allow both marks for implicit statements such as log 0.5. st M for expressing or 6 or 5 as a power of, or for a change of base enabling evaluation of log, log 6 or log 5 by calculator. (Can be implied by 5, 4 or 9 respectively). st A for 9 (exact). nd M for getting log x = constant. The constant can be a log or a sum of logs. If written as log x instead of log x, allow the M mark only if subsequent work implies correct interpretation. nd A for 8 (exact). Change of base methods leading to a non-exact answer score A0. rd Aft for an answer of. An ft answer may be non-exact. their 8 Possible mistakes: 9 9 log log x x x... scores MA(implied by 9)M0A0A0 log 5 log x log x x 5... scores M0A0(9 never seen)ma0a0 x log 48 log x log x x 5.45, x 0.94 scores M0A0MA0Aft No working (or trial and improvement ): x = 8 scores M0 A0 M A A0 A M A Aft (5) [] 6664/0 GCE Mathematics June 009 9

10 Q9 (d) (Arc length =) r r r. Can be awarded by implication from later work, e.g. rh or (rh + rh) in the S formula. (Requires use of = ). r (Sector area =) r r. Can be awarded by implication from later work, e.g. the correct volume formula. (Requires use of = ). Surface area = sectors + rectangles + curved face (= r rh ) (See notes below for what is allowed here) Volume = 00 = rh Sub for h: S r = r (*) r r ds 800 r or r 800r or r 800r r d S 0 r..., r = 900, or AWRT 9. (NOT 9. or 9. ) d S d S and consider sign, > 0 so point is a minimum r 800 S min (Using their value of r, however found, in the given S formula) = 9.65 (AWRT: 80) (Dependent on full marks in part ) B B M B Acso (5) MA M, A (4) M, Aft () M A () [] M for attempting a formula (with terms added) for surface area. May be incomplete or wrong and may have extra term(s), but must have an r (or r ) term and an rh (or rh ) term. In parts, and (d), ignore labelling of parts st n n M for attempt at differentiation (one term is sufficient) r kr nd M for setting their derivative (a 'changed function') = 0 and solving as far as r... (depending upon their 'changed function', this could be r... or r..., etc., but the algebra must deal with a negative power of r and should be sound apart from n possible sign errors, so that r... is consistent with their derivative). n n M for attempting second derivative (one term is sufficient) r kr, and considering its sign. Substitution of a value of r is not required. (Equating it to zero is M0). Aft for a correct second derivative (or correct ft from their first derivative) and a valid reason (e.g. > 0), and conclusion. The actual value of the second derivative, if found, can be ignored. To score this mark as ft, their second derivative must indicate a minimum. Alternative: ds M: Find value of on each side of their value of r and consider sign. d S Aft: Indicate sign change of negative to positive for, and conclude minimum. Alternative: M: Find value of S on each side of their value of r and compare with their Aft: Indicate that both values are more than 9.65, and conclude minimum. 6664/0 GCE Mathematics June 009 0