Mark Scheme (Results) Summer 2007

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1 Mark (Results) Summer 007 GCE GCE Mathematics Core Mathematics C (6665) Edexcel Limited. Registered in England and Wales No Registered Office: One90 High Holborn, London WCV 7BH

2 June 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)

3 . (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.

4 . (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

5 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 , x 0.657, x 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) ( 7 f(0.655) 0.00 (0 At least one correct up to bracket, i.e 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 , x , 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) >0.655 and g(0.655) <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 ) and solving ; BA ( marks)

6 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 (Allow.7 ) A (4) (b) Greatest value ( ) 4 69, A () (c) sin( x ) ( ) sin(x + their α) their R ( x ) 0.8( 0 or 6. A (x ) π 0.80 Must be π their 0.8 or 80 their 6. or (x ) π Must be π + their 0.8 or 60 + their 6. x.7 or x (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 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 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 (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 (awrt) Both seen anywhere A Checking other values and discarding

7 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 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, ] 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 8. (a) 8 D 0, t 5, x 0e awrt A () (b) D 5 0 0e 8 +, t, x e x.549 ( ) A cso () Alt.(b) x 0e e x.549 ( ) A cso (c) T e T 8 e ln T T.06 or. or A () 5 Notes: (b) (main scheme) is for ( 0 0e ) 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 ) 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

Mark Scheme (Results) Summer 2010

Mark Scheme (Results) Summer 2010 Mark (Results) Summer 00 GCE Core Mathematics C3 (6665) Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WCV 7BH Edexcel is one of the leading