Mark Scheme (Results) Summer 2007

Transcription

1 Mark Scheme (Results) Summer 007 GCE GCE Mathematics Core Mathematics C (666) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

2 June Core Mathematics C Mark Scheme Question Scheme Marks. 9 5 or or 5 or ( ) 5 5 = Acso () for an attempt to multiply out. There must be at least correct terms. Allow one sign slip only, no arithmetic errors. e.g ( 5) is A0 ( ) is A0 as indeed is 9± BUT ( = ) is M0A0 since there is more than a sign error is M0A0 since there is an arithmetic error. If all you see is that is but please check it has not come from incorrect working. Epansion of ( + 5)( + 5) is M0A0 Acso for only. Please check that no incorrect working is seen. Correct answer only scores both marks.

3 . (a) Attempt 8 or ( 8 ) = 6 A () (b) 5 5, B, B () (a) for: (on its own) or ( ) or ( ) 8 or 5 or ( 096) is M0 8 or 8 or or 8 or 096 A for 6 only (b) st B for 5 on its own or something. nd B for So e.g. 5 is B But 5 is B0 An epression showing cancelling is not sufficient (see first epression of QC the mark is scored for the second epression) Can use ISW (incorrect subsequent working) e.g 5 scores BB0 but it may lead to 5 which we ignore as ISW. Correct answers only score full marks in both parts.

4 y. (a) = 6 d + d or 6 + A () (b) 6 + or 6 + Aft () (c) C A: or A: both, simplified and + C A A () 7 (a) for some attempt to differentiate: Condone missing d y d n n A for both terms correct, as written or better. No + C here. Of course is acceptable. (b) for some attempt to differentiate again. Follow through their d y, at least one term correct d or correct follow through. Af.t. as written or better, follow through must have distinct terms and simplified e.g. =. (c) for some attempt to integrate: (+C alone is not sufficient) n n + dy d y. Condone misreading or for y. d d st A for either or (or better) is OK here too but not for nd A. 8 and 8 nd A for both or i.e. simplified terms and +C all on one line. 8 instead of is OK

5 . (a) Identify a = 5 and d = (May be implied) B ( ) u00 = a+ (00 ) d (= 5 + (00 ) ) = 0(p) or ( ).0 A () (b) ( S ) [ a (00 ) d] or ( a "their 0" ) = = [ 5 + (00 ) ] or ( 5 + "their 0" ) A = or 08 A () (a) B can be implied if the correct answer is obtained. If 0 is not obtained then the values of a and d must be clearly identified as a = 5 and d =. This mark can be awarded at any point. for attempt to use nth term formula with n = 00. Follow through their a and d. A Must have use of n = 00 and one of a or d correct or correct follow through. Must be 99 not 00. for 0 or.0 (i.e. condone missing sign here). Condone 0 here. N.B. a =, d = is B0 and a + 00d is M0 BUT + 00 is B and A if it leads to 0. Answer only of 0 (or.0) scores /. 6 (b) for use of correct sum formula with n = 00. Follow through their a and d and their 0. Must have some use of n = 00,and some of a, d or l correct or correct follow through. st A for any correct epression (i.e. must have a = 5 and d = ) but can f.t. their 0 still. nd A for 0800 or 08 (i.e. the sign is required before we accept 08 this time). 0800p is fine for A but 0800 is A0. ALT Listing (a) They might score B if a =5 and d = are clearly identified. Then award A together for 0. (b) 00 r= (r + ). Give for 00 (0) + k (with k >), A for k = 00 and A for 0800.

6 5. (a) 6 y Translation parallel to -ais Top branch intersects +ve y-ais Lower branch has no intersections A No obvious overlap 0, or marked on y- ais B () (b) =, y = 0 B, B () S.C. [Allow ft on first B for = when translated the wrong way but must be compatible with their sketch.] 5 (a) for a horizontal translation two branches with one branch cutting y ais only. If one of the branches cuts both aes (translation up and across) this is M0. A for a horizontal translation to left. Ignore any figures on aes for this mark. B for correct intersection on positive y-ais. More than intersection is B0. =0 and y =.5 in a table alone is insufficient unless intersection of their sketch is with +ve y-ais. A point marked on the graph overrides a point given elsewhere. (b) st B for =. NB is B0. Can accept = + if this is compatible with their sketch. Usually they will have A0 in part (a) (and usually B0 too) nd B for y = 0. S.C. If = - and y =0 and some other asymptotes are also given award BB0 The asymptote equations should be clearly stated in part (b). Simply marking =- or y =0 on the sketch is insufficient unless they are clearly marked asymptote = - etc.

7 6. (a) ( ) = 8 (b) + 8 = 0 ± ( 8) = or ( ) = - + (any correct epression) (*) Acso () + ± 8= 0 8 = 6 = or = = B ( ± ) y = M: Attempt at least one y value = +, y = 6 + =, y = 6 A (5) (a) for correct attempt to form an equation in only. Condone sign errors/slips but attempt at A 7 (b) this line must be seen. E.g. ± = 8 is OK for. Acso for correctly simplifying to printed form. No incorrect working seen. The = 0 is required. These two marks can be scored in part (b). For multiple attempts pick best. st for use of correct formula. If formula is not quoted then a fully correct substitution is required. Condone missing = or just + or instead of + for. For completing the square must have as printed or better. If they have 8 then can be given for ( ) =0 ± 8= 0. st A for - + any correct epression. (The + is required but = is not) B for simplifying the surd e.g. 8 =. Must reduce to b so 6 or are OK. nd for attempting to find at least one y value. Substitution into one of the given equations and an attempt to solve for y. nd A for correct y answers. Pairings need not be eplicit but they must say which is and which y. Mis-labelling and y loses final A only.

8 7. (a) Attempt to use discriminant b ac k ( k + ) > 0 k k > 0 (*) Acso () (b) k k = 0 ± ± ±, with = or = or ± ( k a)( k b) ab ( k ) ( k ) k = and 6 (both) A k <, k > 6 or (, );( 6, ) M: choosing outside Aft () 6 (a) for use of b ac, one of b or c must be correct. Or full attempt using completing the square that leads to a TQ in k k k e.g. + = ( k+ ) Acso Correct argument to printed result. Need to state (or imply) that b ac >0 and no incorrect working seen. Must have >0. If > 0 just appears with k k+ > 0that is OK. If >0 appears on last line only with no eplanation give A0. b ac followed by k k > 0 only is insufficient so M0A0 ( ) e.g. k k+ (missing brackets) can get A0 but Using b a c > 0 is M0. k + ( k+ ) is M0A0 (wrong formula) (b) st for attempting to find critical regions. Factors, formula or completing the square. st A for k = 6 and only nd for choosing the outside regions nd Af.t. as printed or f.t. their (non identical) critical values 6 < k < is A0 but ignore if it follows a correct version < k < 6 is M0A0 whatever their diagram looks like Condone use of instead of k for critical values and final answers in (b). Treat this question as two mark parts. If part (a) is seen in (b) or vice versa marks can be awarded.

9 8. (a) ( = )k + 5 [must be seen in part (a) or labelled a = ] B () a (b) ( a = )(k + 5) + 5 = 9k + 0 (*) Acso () (c)(i) a = (9k + 0) + 5 ( = 7k 65) = r + a r = k + (k + 5) + (9k + 0) + (7k + 65) (ii) = 0k + 90 A = 0(k + 9) (or eplain why divisible by 0) Aft () 7 (b) for attempting to find a, follow through their a k. Acso for simplifying to printed result with no incorrect working seen. (c) st for attempting to find a. Can allow a slip here e.g. (9k + 0) [i.e. forgot +5] nd for attempting sum of relevant terms, follow through their (a) and (b). Must have terms starting with k. Use of arithmetic series formulae at this point is M0A0A0 st A for simplifying to 0k + 90 or better nd Aft for taking out a factor of 0 or dividing by 0 or an eplanation in words true k. Follow through their sum of terms provided that both Ms are scored and their sum is divisible by 0. A comment is not required. e.g. 0 k + 90 = k +9 is OK for this final A. 0 S.C. 5 a r r= = 0k + 90 = 0(k + 9) can have M0A0Aft.

10 (a) f ( ) = ( + C) A = 5: C = 65 C = 0 A () (b) ( 5 ) or ( + )( ) or ( + )( ) = ( + )( ) (*) Acso () (a) (c) 0 y st for attempting to integrate, n n + Shape Through origin st A for all terms correct, need not be simplified. Ignore + C here. B B,0 and (,0) B () nd for some use of = 5 and f(5)=65 to form an equation in C based on their integration. There must be some visible attempt to use = 5 and f(5)=65. No +C is M0. nd A for C = 0. This mark cannot be scored unless a suitable equation is seen. 9 (b) for attempting to take out a correct factor or to verify. Allow usual errors on signs. They must get to the equivalent of one of the given partially factorised epressions or, if verifying, ( + 8 ) i.e. with no errors in signs. Acso for proceeding to printed answer with no incorrect working seen. Comment not required. This mark is dependent upon a fully correct solution to part (a) so AM0A0A0 for (a) & (b). Will be common or AA0A0. To score in (b) they must score in (a). (c) st B for positive shaped curve (with a ma and a min) positioned anywhere. nd B for any curve that passes through the origin (B0 if it only touches at the origin) rd B for the two points clearly given as coords or values marked in appropriate places on ais. Ignore any etra crossing points (they should have lost first B). Condone (.5, 0) if clearly marked on ve -ais. Condone (0, ) etc if marked on +ve ais. Curve can stop (i.e. not pass through) at (-.5, 0) and (, 0). A point on the graph overrides coordinates given elsewhere.

11 0. (a) = : y = 5 + =, = : y = 6 + = (can be given st B for ( ) + ( ( ) ) = 70 in (b) or (c)) nd B for - PQ = (*) Acso () (b) y = 6 + dy d = dy = : = = M: Evaluate at one of the points d A dy = : = = Parallel A: Both correct + conclusion A (5) d (c) Finding gradient of normal m = y = ( ) Aft y = 0 o.e. Acso () (a) for attempting PQ or PQ using their P and their Q. Usual rules about quoting formulae. We must see attempt at ( y y ) P Q + for. PQ =... etc could be A0. Acso for proceeding to the correct answer with no incorrect working seen. (b) st for multiplying by, the or 6 must be correct. nd for some correct differentiation, at least one term must be correct as printed. st A for a fully correct derivative. These marks can be awarded anywhere when first seen. rd for attempting to substitute = or = in their derivative. Substituting in y is M0. nd A for - from both substitutions and a brief comment. The must come from their derivative. (c) st for use of the perpendicular gradient rule. Follow through their. nd for full method to find the equation of the normal or tangent at P. If formula is quoted allow slips in substitution, otherwise a correct substitution is required. st Aft for a correct epression. Follow through their and their changed gradient. nd Acso for a correct equation with = 0 and integer coefficients. This mark is dependent upon the coming from their derivative in (b) hence cso. Tangent can get M0A0A0, changed gradient can get M0AA0orAA0. Condone confusion over terminology of tangent and normal, mark gradient and equation. MR Allow for or (+6) but not omitting or treating it as.

12 . (a) y = ( + ) Gradient = (b) + = + =, 9 A (), A 0 y = + = = A () 9 (c) Where y =, l : A = l : B = M: Attempt one of these A Area = ( B A )( y P ) = = = o.e. A () 8 8 (a) for an attempt to write + y 8 =0 in the form y = m + c y y or a full method that leads to m =, e.g find points, and attempt gradient using e.g. finding y = alone can score (even if they go on to say m = ) A dy for m = (can ignore the +c) or = d (b) for forming a suitable equation in one variable and attempting to solve leading to =..or y= st A for any eact correct value for nd A for any eact correct value for y (These marks can be scored anywhere, they may treat (a) and (b) as a single part) (c) st for attempting the coordinate of A or B. One correct value seen scores. st A for A = and B = nd for a full method for the area of the triangle follow through their,, y. e.g. determinant approach nd A for 9 or an eact equivalent =... (...) A B P All accuracy marks require answers as single fractions or mied s not necessarily in lowest terms. 9