Mark (Results) Summer 007 GCE GCE Mathematics Core Mathematics C (6665) Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WCV 7BH
June 007 6665 Core Mathematics C Mark. (a) 6 x ln x ln 6 or ln x ln [implied by 0.69 ] or ln 0 6 x (only this answer) A (cso) () (b) x x (e ) 4e + 0 (any term form) (e x )(e x ) 0 e x or e x Solving quadratic dep x ln, x 0 (or ln ) A (4) (6 marks) Notes: (a) Answer x with no working or no incorrect working seen: A ln 6 Beware x from ln x ln M0A0 ln (ln 6 ln ) ln x ln 6 ln x e allow, x (no wrong working) A (b) st for attempting to multiply through by e x : Allow y, X, even x, for e x x x x Be generous for e.g e + 4, e + 4e, y + y (from e -x x ), e x + 4e x e nd is for solving quadratic (may be by formula or completing the square) as far as getting two values for e x or y or X etc rd is for converting their answer(s) of the form e x k to x lnk (must be exact) A is for ln and ln or 0 (Both required and no further solutions)
. (a) x + x (x )(x + ) at any stage B (x + )(x ) (9 + x) f(x) f.t. on error in denominator factors (x )( x+ ), A (need not be single fraction) 4x + 4x 9 x Simplifying numerator to quadratic form [ ] (x )( x+ ) 4x + x Correct numerator A [( x )( x + )] (x )( x+ ) Factorising numerator, with a denominator o.e. (x )( x+ ) (x ) [ ] 4 x 6 ( ) A cso (7) x x Alt.(a) x + x (x )(x + ) at any stage B (x+ )(x + x ) (9 + x)( x+ ) f(x) A f.t. ( x+ )(x + x ) 4x + 0x 8x ( x+ )(x + x 4 ) ( x+ )(x + x 6) ( x+ )(x + x ) or (x )( x + 4x+ 4) ( x+ )(x + x+ ) Any one linear factor quadratic factor in numerator ( x+ )( x+ )(x ) ( x+ )(x + x ) o.e. (x ) 4x 6 x x ( ) o.e., A A (x ) 4 (4x 6) (b) Complete method for f ( x) ; e.g f( x) o.e A (x ) 8 or 8(x ) A () (x ) Not treating f (for f ) as misread (0 marks) Notes: (a) st in either version is for correct method st x+ ( x ) (9 + x) (x + )( x ) 9 + x x + ( x ) 9+ x A Allow or or ( fractions) (x )( x + ) (x )( x+ ) (x )( x+ ) nd in (main a) is for forming term quadratic in numerator rd is for factorising their quadratic (usual rules) ; factor of need not be extracted ( ) A is given answer so is cso Alt (a) rd is for factorising resulting quadratic Notice that B likely to be scored very late but on epen scored first (b) SC: For M allow ± given expression or one error in product rule Alt: Attempt at f(x) 4 (x ) and diff. ; k(x ) A; A as above Accept 8 ( 4x 4x + ). Differentiating original function mark as scheme.
. (a) dy x x x e + xe dx,a,a () d y (b) If 0, dx e x (x + x) 0 setting (a) 0 [e x 0] x(x + ) 0 ( x 0 ) or x A x 0, y 0 and x, y 4e ( 0.54 ) A () (c) d y d x x x x e xe xe e x + + + x x ( + 4x+ )e x, A () (d) d y d y x 0, > 0 () x, < 0 [ e ( 0.70 )] dx dx : Evaluate, or state sign of, candidate s (c) for at least one of candidate s x value(s) from (b) minimum maximum A (cso) () Alt.(d) For : Evaluate, or state sign of, d y at two appropriate values on either side of at dx least one of their answers from (b) or Evaluate y at two appropriate values on either side of at least one of their answers from (b) or Sketch curve (0 marks) Notes: (a) Generous M for attempt at f ( x) g ( x) + f ( x) g( x) st A for one correct, nd A for the other correct. x Note that x e on its own scores no marks (b) st A (x 0) may be omitted, but for nd A both sets of coordinates needed ; f.t only on candidate s x (c) requires complete method for candidate s (a), result may be unsimplified for A (d) A is cso; x 0, min, and x, max and no incorrect working seen., or (in alternative) sign of dy either side correct, or values of y appropriate to t.p. dx Need only consider the quadratic, as may assume e x > 0. If all marks gained in (a) and (c), and correct x values, give A for correct statements with no working
4. (a) x ( x) 0 o.e. (e.g. x ( x + ) ) x ( ) A (cso) () x Note( ), answer is given: need to see appropriate working and A is cso [Reverse process: Squaring and non-fractional equation, form f(x) A] Alt (i) (b) x 0.6455, x 0.657, x 4 0.656 st B is for one correct, nd B for other two correct If all three are to greater accuracy, award B0 B Alt (ii) (c) Choose values in interval (0.655, 0.655) or tighter and evaluate both f(0.655) 0.0005 ( 7 f(0.655) 0.00 (0 At least one correct up to bracket, i.e. -0.0005 or 0.00 Change of sign, x 0.65 is a root (correct) to d.p. Requires both correct up to bracket and conclusion as above Continued iterations at least as far as x 6 x 5 0.657, x 6 0.657, x 7 two correct to at least 4 s.f. A Conclusion : Two values correct to 4 d.p., so 0.65 is root to d.p. A If use g(0.655) 0.657..>0.655 and g(0.655) 0.658..<0.655 A Conclusion : Both results correct, so 0.65 is root to d.p. A B; B () A A () (7 marks) 5. 4 (a) Finding g(4) k and f(k). or fg(x) ln x [ f() ln(x ) fg(4) ln(4 )] ln A () y (b) y ln(x ) e x or e x y, A (c) f (x) (e x + ) Allow y (e x + ) A Domain x R [Allow R, all reals, (-, ) ] independent B (4) y Shape, and x-axis should appear to be B x O x asymptote Equation x needed, may see in diagram (ignore others) Intercept (0, ) no other; accept y ⅔ (0.67) or on graph B ind. B ind () (d) x or exact equiv. B x, x or exact equiv. x, A () Note: (x + ) or ( x ) o.e. is M0A0 Alt: Squaring to quadratic (9 x 54x + 77 0) and solving ; BA ( marks)
6. (a) Complete method for R: e.g. Rcosα, Rsinα, R ( + ) R or.6 (or more accurate) A Complete method for tanα [Allow tan α ] α 0.588 (Allow.7 ) A (4) (b) Greatest value ( ) 4 69, A () (c) sin( x + 0.588) ( 0.775 ) sin(x + their α) their R ( x + 0.588) 0.8( 0 or 6. A (x + 0.588) π 0.80 Must be π their 0.8 or 80 their 6. or (x + 0.588) π + 0.80 Must be π + their 0.8 or 60 + their 6. x.7 or x 5.976 (awrt) Both (radians only) A (5) If 0.8 or 6. not seen, correct answers imply this A mark ( marks) Notes: (a) st on Epen for correct method for R, even if found second nd for correct method for tanα No working at all: A for, A for 0.588 or.7. N.B. Rcos α, Rsin α used, can still score A for R, but loses the A mark for α. cosα, sin α : apply the same marking. (b) for realising sin(x + α ) ±, so finding R 4. (c) Working in mixed degrees/rads : first two marks available Working consistently in degrees: Possible to score first 4 marks [Degree answers, just for reference, Only are 0. and 4.4 ] Third can be gained for candidate s 0.8 candidate s 0.588 + π or equiv. in degrees One of the answers correct in radians or degrees implies the corresponding M mark. Alt: (c) (i) Squaring to form quadratic in sin x or cos x [cos x 4cos x 8 0, sin x 6sin x 0] Correct values for cos x 0.95, 0.646; or sin x 0.767,.7 awrt A For any one value of cos x or sinx, correct method for two values of x x.7 or x 5.976 (awrt) Both seen anywhere A Checking other values (0.07, 4.0 or 0.869,.449) and discarding (ii) Squaring and forming equation of form a cosx + bsinx c 9sin x + 4cos x + sin x sin x + 5cos x Setting up to solve using R formula e.g. cos( x.76) (x.76) cos 0.56(0... ( α) A ( x.76) π α, π + α,... x.7 or x 5.976 (awrt) Both seen anywhere A Checking other values and discarding
sinθ cosθ sin θ + cos θ + 7. (a) cosθ sinθ cosθ sinθ Use of common denominator to obtain single fraction cosθ sinθ Use of appropriate trig identity (in this case sin θ + cos θ ) Use of sin θ sinθcosθ sin θ Alt.(a) (b) cosecθ ( ) A cso (4) sinθ cosθ tan θ + + tanθ + cosθ sinθ tanθ tanθ sec θ tanθ cosθ sinθ sin θ cosecθ ( ) (cso) A If show two expressions are equal, need conclusion such as QED, tick, true. y Shape (May be translated but need to see 4 sections ) B O 90 80 70 60 θ (c) cosecθ sin θ Allow sin θ T.P.s at y ±, asymptotic at correct x- values (dotted lines not required) [ for equation in sin θ ] (θ ) [ 4.80, 8.89 ; 40.80, 498.89 ] st for α, 80 α ; nd adding 60 to at least one of values θ 0.9, 69., 00.9, 49. ( d.p.) awrt B dep. (), A ; Note Alt.(c) st A for any two correct, nd A for other two Extra solutions in range lose final A only SC: Final 4 marks:θ 0.9, after M0M0 is B; record as M0M0AA0 tanθ + and form quadratic, tan θ tanθ + 0, A tanθ ( for attempt to multiply through by tanθ, A for correct equation above) ± 5 Solving quadratic [ ta nθ.68 or 0.89 ] θ 69., 49. θ 0.9, 00.9 ( d.p.), A, A ( is for one use of 80 + α, AA as for main scheme) A,A (6) ( marks)
8. (a) 8 D 0, t 5, x 0e 5 5.5 awrt A () (b) D 5 0 0e 8 +, t, x 5.56 8 e x.549 ( ) A cso () Alt.(b) x 0e 8 6 8 + 0e x.549 ( ) A cso (c) T 8 5.56...e T 8 e 0.954... 5.56... ln 0.954... 8 T T.06 or. or A () 5 Notes: (b) (main scheme) is for ( 0 0e 8 8 + ) e, or {0 + their(a)}e -(/8) N.B. The answer is given. There are many correct answers seen which deserve M0A0 or A0 (If adding two values, these should be 4.74 and 8.85) (7 marks) (c) st M is for ( 0 0e 8 8 + ) e 5 T T nd 8 M is for converting e k (k > 0) to T ln k 8. This is independent of st M. Trial and improvement: as scheme, correct process for their equation (two equal to s.f.) A as scheme