# Mark Scheme (Results) January 2007

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

2 January Core Mathematics C Mark Scheme Question Number. (a) f (x) = x + 6x f (x) = 6x + 6 Scheme B Marks, cao () Notes cao = correct answer only (a) Acceptable alternatives include x + 6x ; x + x; x + 6x + 0 Ignore LHS (e.g. use [whether correct or not] of d y dx and d y dx ) x + 6x + c or x + 6x + constant (i.e. the written word constant) is B0 Attempt to differentiate their f (x); x n x n. x n x n seen in at least one of the terms. Coefficient of x. ignored for the method mark. x x and x x 0 are acceptable. Acceptable alternatives include 6x + 6x 0 ; x + 6x c or 6x constant is A0 B cao Examples (a) f (x) = x + 6x B (a) f (x) = x + x B0 M0 A0 f (x) = x + A0 (a) f (x) = x + 6x B (a) x + x + 5 f (x) = 6x A0 = x + 6x B = 6x + 6 (a) y = x + x + 5 (a) f (x) = x + 6x + 5 B0 dy dx + x B0 f (x) = 6x + 6 d y dx A0 (a) f (x) = x + 6x B f (x) = 6x c A0 (a) f (x) = x + 6x + c B0 f (x) = 6x + 6

3 Question Number. (b) Notes Scheme x x ( x + x + 5) dx = + + 5x, Marks x + x + 5x = ( + + 5) () = 5 o.e. (7) o.e. = or equivalent (b) Attempt to integrate f(x); x n x n + Ignore incorrect notation (e.g. inclusion of integral sign) o.e. Acceptable alternatives include x x x x x x x + x + 5x ; + + 5x ; + + 5x + c ; + + 5x x x N.B. If the candidate has written the integral (either + + 5x or what they think is the integral) in part (a), it may not be rewritten in (b), but the marks may be awarded if the integral is used in (b). Substituting and into any function other than x + x + 5 and subtracting either way round. So using their f (x) or f (x) or their f (x) dx or their f (x) dx will gain the M mark (because none of these will give x + x + 5). Must substitute for all x s but could make a slip (for example) is acceptable for evidence of subtraction ( invisible brackets). 6 o.e. (e.g. 5, 5.75, ) Must be a single number (so 6 is A0). Answer only is M0A0M0A0 Examples x x (b) + x + 5x+ c (b) + x + 5x+ c c ( c) x =, + c = 5 x =, 6 + c M0 A0 (no subtraction) (b) f( x ) dx = ( + + 5) M0 A0, M0 = 5 9 = 6 A0 (Substituting and into x + x + 5, so nd M0)

4 (b) (6x + 6) dx = x + 6x M0 A0 (b) ( x + 6 x ) d x = x + x M0 A0 = + ( + 6) A0 = 8 + ( + ) A0 x (b) + x + 5x (one negative sign is sufficient for evidence of subtraction) = 6 = 5 (allow recovery, implying student was using invisible brackets ) (a) f(x) = x + x + 5 x f (x) = + x + 5x B0 M0 A0 (b) = 5 The candidate has written the integral in part (a). It is not rewritten in (b), but the marks may be awarded as the integral is used in (b).

5 Question Number. (a) Scheme 5 5!! 5 ( x) = + 5 ( x) + ( x) + ( x) +... = 0x + 0x 80x +... Marks B,,, (b) () ( + x)( x) 5 = ( + x)( 0x + ) = + x 0x + () 9x ( ) (6) Notes (a) 0x B 0x must be seen in this simplified form in (a). Correct structure: binomial coefficients (perhaps from Pascal s triangle), increasing powers of x. Allow slips. Accept other forms: 5 5 C,, also condone 5 but must be attempting to use 5. Condone use of invisible brackets and using x instead of x. Powers of x: at least powers of the type (x) a or x a seen for a. 0x (st ) 80x (nd ) Allow commas between terms. Terms may be listed rather than added Allow recovery from invisible brackets, so +. x +. x +. x = 0x + 0x 80x +... gains full marks. 5 5 x x x!! + 5 ( ) + ( ) + ( ) +... = + 0x+ 0x + 80 x +... gains B0A0 Misread: first terms, descending terms: if correct, would score B0,, st : one of 0x and 80x correct; nd : both 0x and 80x correct. (a) Long multiplication ( x) = x + x, ( x) = 6x + x 8x, ( x) = 8x + x x {+ 6x } ( x) 5 = 0x + 0x + 80x + 0x 0x must be seen in this simplified form in (a). Attempt repeated multiplication up to and including ( x) 5 B

6 0x (st ) 80x (nd ) Misread: first terms, descending terms: if correct, would score B0,, st : one of 0x and 80x correct; nd : both 0x and 80x correct. 5

7 (b) Use their (a) and attempt to multiply out; terms (whether correct or incorrect) in x or higher can be ignored. If their (a) is correct an attempt to multiply out can be implied from the correct answer, so ( + x)( 0x) = 9x will gain. If their (a) is correct, the nd bracket must contain at least ( 0x) and an attempt to multiply out for the M mark. An attempt to multiply out is an attempt at out of the relevant terms (N.B. the terms in x may be combined but this will still count as terms). If their (a) is incorrect their nd bracket must contain all the terms in x 0 and x from their (a) AND an attempt to multiply all terms that produce terms in x 0 and x. N.B. ( + x)( x) 5 = ( + x)( x) [where x + is NOT the candidate s answer to (a)] = x i.e. candidate has ignored the power of 5: M0 N.B. The candidate may start again with the binomial expansion for ( x) 5 in (b). If correct (only needs 0x) may gain even if candidate did not gain B in part (a). N.B. Answer given in question. Example Answer in (a) is = x 0x 80 x... (b) ( + x)( + 0x) = + 0x + x = + x A0 6

8 Question Scheme Marks Number Centre,, i.e. (, 5), Notes r = ( ( )) + (6 ) or r = ( ( )) + (5 ) or r = ( ) + (6 5) o.e. (x ) + (y 5) = 5,, (6) Some use of correct formula in x or y coordinate. Can be implied. Use of ( ( ) xa xb, ( ya yb) ) (, ) or (, ) is M0 A0 but watch out for use of x ( ) A + xa xb etc which is okay. (, 5) (5, ) gains A0. Correct method to find r or r using given points or f.t. from their centre. Does not need to be simplified. (diameter) Attempting radius = is an incorrect method, so M0. N.B. Be careful of labelling: candidates may not use d for diameter and r for radius. Labelling should be ignored. Simplification may be incorrect mark awarded for correct method. Use of ( x x ) ( y y ) is M0. Write down (x ± a) + (y ± b) = any constant (a letter or a number). Numbers do not have to be substituted for a, b and if they are they can be wrong. LHS is (x ) + (y 5). Ignore RHS. RHS is 5. Ignore subsequent working. Condone use of decimals that leads to exact 5. Or correct equivalents, e.g. x + y x 0y + = 0. Alternative note the order of the marks needed for epen. As above. As above. x + y + (constant)x + (constant)y + constant = 0. Numbers do not have to be substituted for rd the constants and if they are they can be wrong. Attempt an appropriate substitution of the coordinates of their centre (i.e. working with nd coefficient of x and coefficient of y in equation of circle) and substitute (, ) or (, 6) into equation of circle. x 0y part of the equation x + y x 0y + = 0. + = 0 part of the equation x + y x 0y + = 0. Or correct equivalents, e.g. (x ) + (y 5) = 5. 7

9 Question Number Scheme Marks. x log 5 = log7 or x = log 5 7 log7 x = log5 =.76 () Notes N.B. It is never possible to award an A mark after giving M0. If M0 is given then the marks will be M0 A0 A0. Acceptable alternatives include st x log 5 = log7 ; x log0 5 = log0 7 ; x loge5 = loge7 ; x ln 5 = ln7 ; x = log 5 7 Can be implied by a correct exact expression as shown on the first mark An exact expression for x that can be evaluated on a calculator. Acceptable alternatives include st log7 log0 7 loge 7 ln7 logq 7 x = ; x = ; x = ; x = ; x = where q is a number log5 log0 5 loge 5 ln 5 logq 5 This may not be seen (as, for example, log 5 7 can be worked out directly on many calculators) so this A mark can be implied by the correct final answer or the right answer corrected to or truncated to a greater accuracy than significant figures or.8 a number Alternative: x = where this fraction, when worked out as a decimal rounds to.76. a number (N.B. remember that this A mark cannot be awarded without the M mark). If the line for the M mark is missing but this line is seen (with or without the x =) and is correct the method can be assumed and st given..76 cao nd N.B. 5 7 =.76 and x 5 = 7, x =.76 are both M0 A0 A0 Answer only.76: full marks ( ) Answer only to a greater accuracy but which rounds to.76: A0 (e.g..760,.760,.760,.7607 etc) Answer only.8: A0 Trial and improvement: award marks as for answer only. 8

10 Examples 7. x = log5 M0 A = 7 =.76 A0 Answer only but clear that x =.76 Working seen, so scheme applied. 5.8 = 7 A M0 A0 A0 Answer only but clear that x =.8. log57 = x. log57 = x x =.760 A0 x =.76. x log 5 = log7. x ln 5 = ln7 x =.0... x = x =.76 x =.76. x log 5 = log7. log7 5 = x M0 x = log 5 x =.697 log7 A0 x =.8 A0 x = A = 8., 5.75 = 6.7. x = 5.76 M0 A0 A = 7 A0. x log 5 = log7. log7 x = log5 x =.8 A0 x =.8 A0 N.B.. x 5 = 7 M0 A M0 A0 x =.76 A0 =.76 A0 9

11 Question Number Scheme Marks 5. (a) f( ) = ( ) + ( ) + ( ) 6 { = } = 0, x + is a factor () (b) x + x + x 6 = (x + )(x + x ), = (x + )(x + )(x ), () (c),, B () (7) Notes Line in mark scheme in { } does not need to be seen. 5(a) Attempting f(±): No x s; allow invisible brackets for M mark Long division: M0 A0. = 0 and minimal conclusion (e.g. factor, hence result, QED,, ). If result is stated first [i.e. If x + is a factor, f( ) = 0] conclusion is not needed. Invisible brackets used as brackets can get, so f( ) = { = } = 0, x + is a factor, but f( ) = = = 0, x + is a factor A0 Acceptable alternatives include: x = is a factor, f( ) is a factor. 5(b) st requires division by (x + ) to get x + ax + b where a 0 and b 0 or equivalent with division by (x + ) or (x ). (x + )(x + x ) or (x + )(x +x ) or (x )(x + 5x + 6) [If long division has been done in (a), minimum seen in (b) to get first is to make some reference to their quotient x + ax + b.] Attempt to factorise their quadratic (usual rules). Combining all factors is not required. Answer only: Correct Answer only with one sign slip: (x + )(x + )(x + ) scores st st nd M0 nd A0 (x + )(x )(x ) scores st M0 st A0 nd nd Answer to (b) can be seen in (c). 5(b) Alternative comparing coefficients (x + )(x + ax + b) = x + ( + a)x + (a + b)x + b Attempt to compare coefficients of two terms to find values of a and b a =, b = Or (x + )(ax + bx + c) = ax + (a + b)x + (b + c)x + c Attempt to compare coefficients of three terms to find values of a, b and c. a =, b =, c = Then apply scheme as above 0

12 5(b) Alternative using factor theorem Show f( ) = 0; allow invisible brackets x + is a factor Show f() = 0 x is a factor 5(c),, or (, 0), (, 0), (, 0) only. Do not ignore subsequent working. Ignore any working in previous parts of the question. Can be seen in (b) B

14 Finding answers by trying different values (e.g. trying multiples of π) in cos x + = 5sinx : as for answer only.

15 Question Number Scheme Marks 7. y = x(x 6x + 5) = x 6x + 5x, x 6x 5x ( x 6x + 5x) dx = +, ft x x 5x = + 0 = x 5x x + = ( 6 + 0) =, (both) total area = + 7 = o.e. cso (9)

16 Notes Attempt to multiply out, must be a cubic. Award A mark for their final version of expansion (but final version does not need to have like terms collected). Attempt to integrate; x n x n +. Generous mark for some use of integration, so e.g. x x x x( x )( x 5) dx= x 5x would gain method mark. p q Ft on their final version of expansion provided it is in the form ax + bx +... Integrand must have at least two terms and all terms must be integrated correctly. If they integrate twice (e.g. and ) and get different answers, take the better of the 0 two. Substitutes and subtracts (either way round) for one integral. Integral must be a changed function. Either and 0, and or and 0. : 0 for bottom limit can be implied (provided that it is 0). For [ ] 0 Substitutes and subtracts (either way round) for two integrals. Integral must be a changed function. Must have and 0 and and (or and ). The two integrals do not need to be the same, but they must have come from attempts to integrate the same function. and o.e. (if using f( x )) where f(x) = f( x) ) or x 5x x +. and o.e. (if using f( x) or The answer must be consistent with the integral they are using (so f( x) = loses this A and the final A). may not be seen explicitly. Can be implied by a subsequent line of working. 5th their value for [ ] 0 + their value for [ ] Dependent on at least one of the values coming from integration (other may come from e.g. trapezium rules). This can be awarded even if both values already positive. 7 cso o.e. N.B. c.s.o. f( x) or ft 5

17 Question Scheme Marks Number 8. C = 00v (a) 7, 00v + = 0 7 v = 900 d v = 70 cso (5) (b) = 800v v = 70, > 0 { minimum} ft () or v = 70, = { = = } { minimum} 5 (c) v = 70, C = C = 0 () (9) Notes 8(a) Attempt to differentiate v n v n. Must be seen and marked in part (a) not part (b). Must be differentiating a function of the form av + bv. o.e. ( 00v + + c is A0) 7 dc Their = 0. Can be implied by their = P + Q P = ± Q. Dependent on both of the previous Ms. d Attempt to rearrange their into the form vn = number or v n number = 0, n 0. v = 70 cso but allow v = ±70. v = 70 km per h also acceptable. cso Answer only is 0 out of 5. Method of completing the square: send to review. 6

18 8(a) Trial and improvement f(v) = 00 v + v 7 Attempts to evaluate f(v) for values a, b, c where (i) a < 70, b = 70 and c > 70 or (ii) a, b < 70 and c > 70 or (iii) a < 70 and b, c > 70. All correct and states v = 70 (exact) Then nd M0, rd M0, nd A0. 8(a) Graph Correct shape (ignore anything drawn for v < 0). v = 70 (exact) Then nd M0, rd M0, nd A0. 8(b) Attempt to differentiate their ; v n v n (including v 0 0). must be correct. Ft only from their value of v and provided their value of v is +ve. Must be some (minimal) indication that their value of v is being used. Statement: When v = their value of v, > 0 is sufficient provided 800v > 0 for their value of v. If substitution of their v seen: correct substitution of their v into 800v, but, provided evaluation is +ve, ignore incorrect evaluation. N.B. Parts in mark scheme in { } do not need to be seen. ft 8(c) Substitute their value of v that they think will give C min (independent of the method of obtaining this value of v and independent of which part of the question it comes from). 0 or 0 Must have part (a) completely correct (i.e. all 5 marks) to gain this. Answer only gains provided part (a) is completely correct.. 7

19 Examples 8(b) 8(b) = 800v v = 70, > 0 8(b) = 800v > 0 A0 (no indication that a value of v is being used) 8(b) Answer from (a): v = 0 = 800v v = 0, ft 8(b) 8(b) = 800v v = 70, = = 8.6 (correct substitution of 70 seen, evaluation wrong but positive) = 800v v = 70, = A0 (correct substitution of 70 not seen) 8

20 Question Number Scheme (6 ) cos PQR = (a) =, 6 6 π PQR = () (b) Area = 6 π m = π m cso ( ) () (c) Area of = 6 6 sin π m Marks = 9 m cso () (d) Area of segment = π 9 m =. m () (e) π Perimeter = m =.6 m ft () () Notes 9(a) N.B. a = b + c bccosa is in the formulae book. Use of cosine rule for cos PQR. Allow A, θ or other symbol for angle. (i) (6 ) = cos PQR : Apply usual rules for formulae: (a) formula not stated, must be correct, (b) correct formula stated, allow one sign slip when substituting. 6 6 (6 ) or (ii) cos PQR = ± ± ± ± 6 6 Also allow invisible brackets [so allow Correct expression π (6 ) ] in (i) or (ii) o.e. (e.g. 6 or 7 ) 9

21 9(a) Alternative a sinθ = where θ is any symbol and a < 6. 6 sinθ = where θ is any symbol. 6 π 9(b) Use of r θ with r = 6 and θ = their (a). For M mark θ does not have to be exact. M0 if using degrees. π c.s.o. ( (a) correct exact or decimal value) N.B. Answer given in question Special case: Can come from an inexact value in (a) PQR =.09 Area = 6.09 = 7.6 (or 7.7) = values used on calculator) gets. PQR =.09 Area = 6.09 = 7.6 (or 7.7) =.97π = 9(c) Use of r sin θ with r = 6 and their (a). π (no errors seen, assume full π gets A0. θ = cos (their PQR) in degrees or radians Method can be implied by correct decimal provided decimal is correct (corrected or truncated to at least decimal places) c.s.o. Must be exact, but correct approx. followed by 9 is okay (e.g. = = 9 ) 9(c) Alternative (using bh ) Attempt to find h using trig. or Pythagoras and use this h in bh form to find the area of triangle PQR 9 c.s.o. Must be exact, but correct approx. followed by 9 is okay (e.g. = = 9 ) cso cso 9(d) Use of area of sector area of or use of r ( θ sin θ ) Any value to decimal place or more which rounds to.. 9(e) [6 their (a)]. Correct for their (a) to decimal place or more ft 0

22 Question Number 0. (a) (b) Scheme {S n = } a + ar + + ar n B {rs n = } ar + ar + + ar n ( r)s n = a( r n ) d n a( r ) cso S n = ( ) r () a = 00, r =, n = 0, 0 00( ), S 0 = Marks (c) = 0,600 () 5 a =, r = 6 B 5 a 6 S =, S = r 5 = o.e. () B () (d) < r < (or r < ) () Notes 0(a) S n not required. The following must be seen: at least one + sign, a, ar n and one other intermediate term. No extra terms (usually ar n ). Multiply by r; rs n not required. At least of their terms on RHS correctly multiplied by r. Subtract both sides: LHS must be ±( r)s n, RHS must be in the form ±a( r pn + q ). Only award this mark if the line for S n = or the line for rs n = contains a term of the form ar cn + d Method mark, so may contain a slip but not awarded if last term of their S n = last term of their rs n. B d Completion c.s.o. N.B. Answer given in question cso 0(a) S n not required. The following must be seen: at least one + sign, a, ar n and one other intermediate term. No extra terms (usually ar n ). On RHS, multiply by r r Or Multiply LHS and RHS by ( r) B

23 Multiply by ( r) convincingly (RHS) and take out factor of a. d Method mark, so may contain a slip. Completion c.s.o. N.B. Answer given in question cso 0(b) Substitute r = with a = 00 or 00 and n = 9 or 0 into formula for S n. 0 00( ) or equivalent. 0,600 0(b) Alternative method: adding 0 terms (i) Answer only: full marks. ( ) (ii) {+ 0,00} = 0,600 or 00( {+,0)} = 0,600 for two correct terms (as above o.e.) and an indication that the sum is needed (e.g. + sign or the word sum). 0,00 o.e. as final term. Can be implied by a correct final answer. 0,600. a 0(c) N.B. S = r is in the formulae book. B r = seen or implied anywhere. 5 a Substitute a = and their r into 6 r. Usual rules about quoting formula. 5 o.e. a 0(d) N.B. S = r for r < is in the formulae book. < r < or r < In words or symbols. Take symbols if words and symbols are contradictory. Must be < not. B

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