Mark Scheme (Results) Summer 2007 - Edexcel

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June 2007 6674 Further Pure Mathematics FP1 Mark Scheme Question Scheme Marks number 1. 1 1 and 3 are ‘critical values’, e.g. used in solution, or both seen as asymptotes B1 2 ( x + 1)( x − 3) = 2x − 3 ⇒ x( x − 4) = 0 x = 4, x = 0 M1: attempt to find at least one other critical value M1 A1, A1 1 1 0 < x < 1 , 3 < x < 4 M1: An inequality using 1 or 3 M1 A1, A1 (7) 2 2 First M mark can be implied by the two correct values, but otherwise a method must be seen. (The method may be graphical, but either (x = ) 4 or (x =) 0 needs to be clearly written or used in this case). Ignore ‘extra values’ which might arise through ‘squaring both sides’ methods. ≤ appearing: maximum one A mark penalty (final mark). 7
Question Scheme Marks number 2. Integrating factor ∫ −tan x dx ln(cos x) −ln(sec x) ⎛ ⎞ ( ) ⎟ ⎠ e ⎛ dy ⎜cos x − y sin x = ⎝ dx ∫ y cos x = 2sec x dx y cos x = 2 tan x ( + C) 2 2sec = e ⎞ x⎟ ⎠ (or equiv.) or e , = cos x ⎜or ⎝ 2 ⎛ ⎞ ( ) ⎟ ⎠ (or equiv.) 1 sec x M1, A1 d 2 ⎜Or : y cos x = 2sec x M1 A1(ft) ⎝ dx y = 3 at x = 0: C = 3 M1 2 tan x + 3 y = (Or equiv. in the form y = f(x)) A1 (7) cos x 1 st tan d ln(cos ) ln(sec ) M: Also scored for e ∫ x x − x x = e ( or e ) , then A0 for sec x . 2 nd M: Attempt to use their integrating factor (requires one side of the equation ‘correct’ for their integrating factor). 2 nd A: The follow-through is allowed only in the case where the integrating = 4 factor used is sec x or − sec x . ⎛ y sec x 2sec x dx ⎞ ⎜ ∫ ⎟ ⎝ ⎠ 3 rd M: Using y = 3 at x = 0 to find a value for C (dependent on an integration attempt, however poor, on the RHS). Alternative 1 st M: Multiply through the given equation by cos x . 1 st dy 2 A: Achieving cos x − y sin x = 2sec x . (Allowing the possibility of dx integrating by inspection). A1 7
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Page 5 and 6: 20 10 −2 −1 1 2 −10 −20 y x
Page 7 and 8: Question Scheme Marks number * 6. (

Mark Scheme (Results) Summer 2007 - Edexcel

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