Mark Scheme (Results) Summer 2007

Transcription

1 Mark Scheme (Results) Summer 007 GCE GCE Mathematics Core Mathematics C (4) Edecel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

2 June 007 Mark Scheme. = d (Or equivalent, such as, or 8 = 8 = + 4, st M: k k 0. ) M A [or 4, or ( ), or ( + ) ] M A nd M: Substituting limits 8 and into a changed function (i.e. not or ), and subtracting, either way round. nd A: This final mark is still scored if + 4 is reached via a decimal. (4) 4 N.B. Integration constant +C may appear, e.g. + C 8 = ( 8 + C) ( + C) = + 4 (Still full marks) But a final answer such as C is A0. N.B. It will sometimes be necessary to ignore subsequent working (isw) after a correct form is seen, e.g. = d (M A), followed by incorrect simplification d = = (still M A). The second M mark is still available for substituting 8 and into and subtracting. June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics

3 . (a) f() = = (M: Attempt f() or f( )) M A () (If continues to say remainder =, isw) Answer must be seen in part (a), not part (b). (b) ( + )( + ) M A ( + )( )( ) (If continues to solve an equation, isw) M A (4) (a) Answer only (if correct) scores both marks. ( as answer only is M0 A0). Alternative (long division): Divide by ( ) to get ( + a + b), a 0, b 0. [M] ( + 4), and seen. [A] (If continues to say remainder =, isw) (b) First M requires division by ( + ) to get ( + a + b), a 0, b 0. Second M for attempt to factorise their quadratic, even if wrongly obtained, perhaps with a remainder from their division. Usual rule: ( k + a + b) = ( p + c)( q + d), where pq = k and cd = b. Just solving their quadratic by the formula is M0. Combining all factors is not required. Alternative (first marks): ( + )( + a + b) = + ( + a) + (a + b) + b = 0, then compare coefficients to find values of a and b. [M] a =, b = [A] Alternative: Factor theorem: Finding that f ( ) = 0 factor is, ( ) [M, A] Finding that f = 0 factor is, ( ) [M, A] If just one of these is found, score the first marks M A M0 A0. Losing a factor of : ( + ) ( ) scores M A M A0. Answer only, one sign wrong: e.g. ( + )( )( + ) scores M A M A0. June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics

4 . (a) + k [Allow unsimplified versions, e.g. 5 + ( ) k, C0 + Ck ] B 5 ( ) k + ( k) [See below for acceptable versions] M A () N.B. THIS NEED NOT BE SIMPLIFIED FOR THE A (isw is applied) (b) k = 5k k = (or equiv. fraction, or 0.4) (Ignore k = 0, if seen) M Acso () (c) c = = 5 5 (or equiv. fraction, or.8) Acso () (Ignore, so is fine) 5 (a) The terms can be listed rather than added. M: Requires correct structure: binomial coefficients (perhaps from Pascal s triangle), increasing powers of. Allow a slip or slips such as: k + k, ( ) 5 + k + ( k) k + k, + + But: 5 + k k or similar is M0. Both and terms must be seen. and or equivalent such as C and C are acceptable, and even and are acceptable for the method mark. A: Any correct (possibly unsimplified) version of these terms. and or equivalent such as and are acceptable. C C Descending powers of : Can score the M mark if the required first 4 terms are not seen. Multiplying out ( + k)( + k)( + k)( + k)( + k)( + k) : M: A full attempt to multiply out (power ) B and A as on the main scheme. (b) M: Equating the coefficients of and (even if trivial, e.g. k = 5k). Allow this mark also for the misread : equating the coefficients of and. An equation in k alone is required for this M mark, although condone = 5k k = 5k k =. 5 k ( ) 4 June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics

5 ( ) 4. (a) 4 = cosθ M cosθ = A 5 45 = = (*) Acso () 0 4 (b) sin A + = (or equiv. Pythag. method) M 4 7 sin A = sin A = 7 or equivalent eact form, e.g. 4 (a) M: Is also scored for 5 = 4 + ( 4 cosθ ) or = ( 5 4cosθ ) or cosθ = or cosθ = st A: Rearranged correctly and numerically correct (possibly unsimplified), in the form cos θ =... or 0 cosθ = 45 (or equiv. in the form p cos θ = q ). Alternative (verification): 4 = [M] 4 Evaluate correctly, at least to = [A] Conclusion (perhaps as simple as a tick). [Acso] (Just achieving = is insufficient without at least a tick). (b) M: Using a correct method to find an equation in sin A or sin A which would give an eact value. Correct answer without working (or with unclear working or decimals): Still scores both marks. 7, A () 5 5 June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics

6 5. (a).44 (allow also eact answer ),.7 Allow awrt B, B () (b) (0.5).. { ( ) } M Aft = 4.04 (Must be s.f.) A (4) (c) Area of triangle = ( ) B (Could also be found by integration, or even by the trapezium rule on y = ) Area required = Area of triangle Answer to (b) (Subtract either way round) 4.04 =.9 Allow awrt Aft () (ft from (b), dependent on the B, and on answer to (b) less than ) (a) If answers are given to only d.p. (.4 and.4), this is B0 B0, but full marks can be given in part (b) if 4.04 is achieved. (b) Bracketing mistake: i.e. (0.5)(0 + ) + ( ) scores B M A0 A0 unless the final answer implies that the calculation has been done correctly (then full marks can be given). Alternative (finding and adding separate areas): (Triangle/trapezium formulae, and height of triangle/trapezium) [B] Fully correct method for total area, with values from table. [M, Aft] 4.04 [A] (c) B: Can be given for with no working, but should not be given for obtained from wrong working. Aft: This is a dependent follow-through: the B for must have been scored, and the answer to (b) must be less than. B M 9 June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics

7 . log 0.8 (a) = or log8 0.8, log8 = Allow awrt M, A () (b) log = log B log log7 = log 7 M Remove logs to form equation in, using the base correctly: = 7 M = (Ignore = 0, if seen) Acso (4) (a) Allow also the implicit answer Answer only: 0.07 or awrt: Full marks. (M A). Answer only: 0. or awrt (insufficient accuracy): M A0 Trial and improvement: Award marks as for answer only. (b) Alternative: log = log B log 7 + = log 7 + log = log M Remove logs to form equation in : = M = (Ignore = 0, if seen) A Alternative: log 7 = log 7 + log B log (log 7 + log ) = log = + log 7 M + log 7) = ( ) = or log = log log 7 M = A Attempts using change of base will usually require the same steps as in the main scheme or alternatives, so can be marked equivalently. A common mistake: log log log7 = log7 B M0 = 7 = M( Recovery ), but A0 7 June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics

8 7. (a) Gradient of AM: Gradient of l: y = ( ) or ( ) = or = M: use of m m =, or equiv. M y = [ = + 9] B y (Any equiv. form) M A (4) (b) = : y = + 9 = y = (or show that for y =, = ) (*) B () (A conclusion is not required). (c) ( = ) ( ) + ( ( )) r M: Attempt N.B. Simplification is not required to score M A r or r M A ( ± ) + ( y ± ) = k, k 0 ( Value for k not needed, could be r or r) M ( ) + ( y + ) = (or equiv.) A (4) Allow ( ) or other eact equivalents for. (But ( ) + ( y ) = scores M A0) (Correct answer with no working scores full marks) (a) nd M: eqn. of a straight line through (, ) with any gradient ecept 0 or. Alternative: Using (, ) in y = m + c to find a value of c scores M, but an equation (general or specific) must be seen. Having coords the wrong way round, e.g. y = ( ), loses the nd M mark unless a correct general formula is seen, e.g. y y = m ). ( 9 If the point P (, ) is used to find the gradient of MP, maimum marks are (a) B0 M0 M A (b) B0. (c) st M: Condone one slip, numerical or sign, inside a bracket. Must be attempting to use points P(, ) and A(, ), or perhaps P and B. (Correct coordinates for B are (5, 4)). st M alternative is to use a complete Pythag. method on triangle MAP, n.b. MP = MA =. Special case: If candidate persists in using their value for the y-coordinate of P instead of the given, allow the M marks in part (c) if earned. 8 June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics

9 n n n 8. (a) r (or equiv.) (Allow ar if r is seen in (b)) B () (b) r n > (Using answer to (a), which must include r and n, and ) (Allow equals sign or the wrong inequality sign) (Condone slips such as omitting a zero) r n > 4 ( n )logr > log4 M (Introducing logs and dealing correctly with the power) (Allow equals sign or the wrong inequality sign) log 4 n > + log r (*) Acso () log 4 log 4 (c) r =.09: n > + or n > log.09 log.09 ( n > ) (Allow equality) M Year 8 or 0 (If one of these is correct, ignore the other) A () n 0 a( r ) 50000(.09 ) (d) Sn = = r.09 M A (Must be this answer nearest 0000) A () 9 (b) Incorrect inequality sign at any stage loses the A mark. Condone missing brackets if otherwise correct, e.g n log r > log 4. A common mistake: r > M ( n ) log r > log M0 ( Recovery from here is not possible). (c) Correct answer with no working scores full marks. Year 7 (or 0) with no working scores M A0. Treat other methods (e.g. year by year calculation) as if there is no working. (d) M: Use of the correct formula with a = 50000, 5000 or , and n = 9, 0, or 5. M can also be scored by a year by year method, with terms added. (Allow the M mark if there is evidence of adding 9, 0, or 5 terms). st A is scored if 0 correct terms have been added (allow nearest 00 ). (50000, 54500, 59405, 475, 70579, 79, 8855, 940, 998, 08595) No working shown: Special case: scores mark, scored as, 0, 0. (Other answers with no working score no marks). M 9 June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics

10 9. (a) y Sine wave (anywhere) with at least turning points. M 0.5 Starting on positive y-ais, going up to a ma., then min. below -ais, no further π/ π π/ π turning points in range, finishing above -ais at = π or 0. There must be 0.5 some indication of scale on the y-ais (e.g., or 0.5) A () Ignore parts of graph outside 0 to π. n.b. Give credit if necessary for what is seen on an initial sketch (before any transformation has been performed). 5π π (b) 0,,,0,,0 (Ignore any etra solutions) (Not 50, 0 ) B, B, B () π π π and π are insufficient, but if both are seen allow B B0. (c) awrt 0.7 radians ( ), or awrt 40.5 (40.54 ) (α) B π ( π α ) (.4 ) or ( 80 α ) if α is in degrees. NOT π α M π Subtract from α (or from ( π α ) ) or subtract 0 if α is in degrees M 0.8 (or 0.0π),.9 (or 0.π) Allow awrt A, A (5) (The st A mark is dependent on just the nd M mark) (b) The zeros are not required, i.e. allow 0.5, etc. (and also allow coordinates the wrong way round). These marks are also awarded if the eact intercept values are seen in part (a), but if values in (b) and (a) are contradictory, (b) takes precedence. (c) B: If the required value of α is not seen, this mark can be given by implication if a final answer rounding to 0.8 or 0.9 (or a final answer rounding to.9 or.90) is achieved. (Also see premature appro. note*). π π Special case: sin + = 0.5 sin + sin = 0.5 sin = 0. 5 = arcsin 0.5 = and = π =. 99 (B0 M M0 A0 A0) (This special case mark is also available for degrees ) Etra solutions outside 0 to π : Ignore. Etra solutions between 0 and π : Loses the final A mark. *Premature approimation in part (c): e.g. α = 4, 80 4 = 9, 4 0 = and 9 0 = 09 Changing to radians: 0.9 and.90 This would score B (required value of α not seen, but there is a final answer 0.9 (or.90)), M M A0 A0. 0 June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics 0

11 0. (a) 4 + y = 00 M A V = y = V = 00 (*) M Acso (4) dv (b) = 00 4 d B Equate their dv d to 0 and solve for or : = 50 or = 50 (7.07 ) M A Evaluate V: 4 V = 00( 50) (50 50) = 94 cm Allow awrt M A (5) d V (c) d = 8 Negative, Maimum M, Aft () (a) st M: Attempting an epression in terms of and y for the total surface area (the epression should be dimensionally correct). st A: Correct epression (not necessarily simplified), equated to 00. nd M: Substituting their y into y to form an epression in terms of only. (Or substituting y from y into their area equation). (b) st A: Ignore = 50, if seen. The nd M mark (for substituting their value into the given epression for V) is dependent on the st M Final A: Allow also eact value or or equiv. single term. (c) Allow marks if the work for (c) is seen in (b) (or vice-versa). M: Find second derivative and consider its sign. dv A: Second derivative following through correctly from their, and correct d reason/conclusion (it must be a maimum, not a minimum). An actual value of does not have to be used this mark can still be awarded if no value has been found or if a wrong value is used. Alternative: dv M: Find value of on each side of = 50 and consider sign. d dv A: Indicate sign change of positive to negative for, and conclude ma. d Alternative: M: Find value of V on each side of = 50 and compare with 94. A: Indicate that both values are less than 94, and conclude ma. June 007 Advanced Subsidiary/Advanced Level in GCE Mathematics