GCE Edexcel GCE Core Mathematics C (666) January 006 Mark Scheme (Results) Edexcel GCE Core Mathematics C (666)
January 006 666 Core Mathematics C Mark Scheme. (a) +-5 + c = 0 or - + c = 0 c = A () (b) f(x) = ( x )( x x ) + (x ) B division x x + ) A () = ( )( (c) f ( ) 7 9 5 = + + c 8 Remainder = c +.5 =.5 ft their c Aft () (a) for evidence of substituting x = leading to linear equation in c 8 (b) B for identifying (x ) as a factor st for attempting to divide. Other factor must be at least ( x + one other term) nd for attempting to factorise a quadratic resulting from attempted division A for just ( x ) ( x+ ). (c) for attempting f( ± ). If not implied by.5 + c, we must see some A substitution of ±. follow through their c only, but it must be a.
+ px = + px + px ) B B () 9 9. (a) ( ) 9 ; ( N.B. (b) 9p = 6, so p = A 9 8 q = p or 6 p or 6p if that follows from their (a) So q = 576 Acao () 6 (a) nd B 9 for ( ) px or better. Condone, not +. (b) st for a linear equation for p. nd for either printed expression, follow through their p. 9 6 + px+ px leading to p =, q = scores BB0 AA0 i.e /6 ( ) ( ) ( ). (a) AB = + 5 [= 6] AB = 6 A () (b) + 5 p =, = 7 5, A () AB + = (c) ( x xp) ( y yp) ( x ) ( y ).5 +.5 = 6.5 LHS RHS oe A c.a.o () 7 (a) for an expression for AB or AB N.B. ( (b) for a full method for x p x x )... + + is M0 (c) st for using their x p and y p in LHS nd for using their AB in RHS N.B. x + y 7x 5y+ = 0 scores, of course, / for part (c). Condone use of calculator approximations that lead to correct answer given.
a. (a) 80 r = 0 80 0 80( ) r = = r r = r = * Acso () (b) (c) (d) u u ( ) ( ) 5 6 = 0 [ = 7.96875] 5 = 0 [ = 8.76565] either Difference = 9.9 (allow ± ) A () S 7 7 0( (0.75) ) = 0.75 = 5.977 (AWRT) 6 A () n 0( (0.75) ) > 00 0.75 n 00 (0.75) > 80 (or better) A n > log(0.75) log(0.75) (=.09 ) n = Acso () (a) st for use of S For Information nd substituting for a and moving (-r) to form linear equation in r. u = 0 u = 90 (b) n 5 for some correct use of ar.[0( ( ) 6 ) 0 is M0] u = 67.5 u = 50.65 (c) for a correct expression (need use of a and r) S = 0 (d) st for attempting S n > 00 [or = 00] (need use of a and some use of r) S = 77.5 nd n for valid attempt to solve r = p( r, p< ), must give linear eqn in n. S = 8.5 Any correct log form will do. S 5 = 66.09... Trial st for attempting at least values of S n, one n < and one n >. & nd for attempting S and S. Imp. st A for both values correct to s.f. or better. nd A for n =.
5. (a) ˆ 5 + 5 6 cos AOB = 5 5 or sin θ = 5 with use of cos θ = sin θ attempted = 7 5 * Acso () (b) AOB ˆ =.8700... radians.87 or better B () (c) Sector =, = 6.087 (AWRT) 6. A () 5 ( b) (d) Triangle = 5 sin( b) or 6 5 Segment = (their sector) their triangle d = (sector from c) = (AWRT). (ft their part(c)) Aft () 8 (a) for a full method leading to cos AOB ˆ [N.B. Use of calculator is M0] (usual rules about quoting formulae) (b) Use of (b) in degrees is M0 (d) st for full method for the area of triangle AOB nd for their sector their triangle. Dependent on st in part (d). Aft for their sector from part (c) [or. following a correct restart].
6. (a) t = 5 5 0 v =.80 9.7 5.7 B B B () (b) S 5;[0 + 5.7 + (. +.8 +.80 + 6.+ 9.7)] B [] = 5 [6.6] = 5.075 = AWRT 5 A () 6 (a) S.C. Penalise AWRT these values once at first offence, thus the following marks could be AWRT dp (Max /)
7. (a) (b) dy = 6x 0x A () dx 6x 0x =0 (x + ) ( x ) [=0] x = or (both x values) A 9 7 Points are (, - 0) and (, or or.70 or better) (both y values) A () 7 7 (c) d y dx = x 0 A () (d) d y x = ( = ) > 0 [(, - 0)] is a Min dx d y 7 x = ( = ) < 0 [ (, 7 )] is a Max dx A () 0 (a) n n for some correct attempt to differentiate x x (b) st for setting their d y dx nd for attempting to solve TQ but it must be based on their d y dx NO marks for answers only in part (b) (c) for attempting to differentiate their d y dx (d) for one correct use of their second derivative or a full method to determine the nature of one of their stationary points A both correct (= and = - ) are not required
8. (a) sin( θ + 0) = ( 5 on RHS) B 5 θ + 0 = 6.9 ( α = AWRT 7) B or =. (80 α ) θ = 6.9,. Acao () (b) tanθ =± or sin θ =± or cosθ =± B 5 5 ( tanθ = ) θ = 6. ( AWRT 6.) β = B or. ( 80 + β ) ( tanθ = ) θ = 6.6 ( 80 β ) or 96.6 (80 + their 6.6) (5) 9 (a) for 80 their first solution. Must be at the correct stage i.e. for θ + 0 (b) ALL M marks in (b) must be for θ =... st for 80 + their first solution nd for 80 their first solution rd for 80+ their 6.6 or 60 their first solution Answers Only can score full marks in both parts Not d.p.: loses A in part (a). In (b) all answers are AWRT. Ignore extra solutions outside range Radians Allow M marks for consistent work with radians only, but all A and B marks for angles must be in degrees. Mixing degrees and radians is M0.
9. (a) = x + x x 8x+ ( = 0) A x x = 0 ( )( ) x =, A () (b) Area of R = ( ) x + x d x (for ) B x + x dx= x + x ( ) ±, accept x ) [A] (Allow [ ] ( x + x) d x = +, + = 6 Area of R = = (Accept exact equivalent but not 0. ) Acao (6) 6 (a) st for forming a correct equation st A for a correct TQ (condone missing =0 but must have all terms on one side) nd for attempting to solve appropriate TQ (b) B for subtraction of. Either curve line or integral rectangle n n x x + st for some correct attempt at integration ( ) st A for x + x only i.e. can ignore x nd for some correct use of their as a limit in integral rd for some correct use of their as a limit in integral and subtraction either way round Special Case Line curve gets B0 but can have the other A marks provided final answer is +. 0
GENERAL PRINCIPLES FOR C & C MARKING Method mark for solving term quadratic:. Factorisation ( x + bx + c) = ( x + p)( x + q), where pq = c, leading to x = ( ax + bx + c) = ( mx + p)( nx + q), where pq = c and mn =. Formula Attempt to use correct formula (with values for a, b and c). a, leading to x =. Completing the square Solving x + bx + c = 0 : ( x ± p) ± q ± c, p 0, q 0, leading to x = Method marks for differentiation and integration:. Differentiation Power of at least one term decreased by. ( x n x n ). Integration Power of at least one term increased by. ( x n x n+ ) Use of a formula Where a method involves using a formula that has been learnt, the advice given in recent examiners reports is that the formula should be quoted first. Normal marking procedure is as follows: Method mark for quoting a correct formula and attempting to use it, even if there are mistakes in the substitution of values. Where the formula is not quoted, the method mark can be gained by implication from correct working with values, but will be lost if there is any mistake in the working. Exact answers Examiners reports have emphasised that where, for example, an exact answer is asked for, or working with surds is clearly required, marks will normally be lost if the candidate resorts to using rounded decimals. Answers without working The rubric says that these may gain no credit. Individual mark schemes will give details of what happens in particular cases. General policy is that if it could be done in your head, detailed working would not be required. Most candidates do show working, but there are occasional awkward cases and if the mark scheme does not cover this, please contact your team leader for advice. Misreads A misread must be consistent for the whole question to be interpreted as such. These are not common. In clear cases, please deduct the first A (or B) marks which would have been lost by following the scheme. (Note that marks is the maximum misread penalty, but that misreads which alter the nature or difficulty of the question cannot be treated so generously and it will usually be necessary here to follow the scheme as written). Sometimes following the scheme as written is more generous to the candidate than applying the misread rule, so in this case use the scheme as written. If in doubt please send to review or refer to Team Leader.