Mark Scheme (Results) June GCE Further Pure FP3 (6669) Paper 1

Transcription

1 Mark (Results) June 0 GCE Further Pure FP (6669) Paper

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3 EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75.. The Edexcel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epen. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark

4 . June 0 Further Pure Mathematics FP 6669 Mark dy 6x dx = and so surface area = π x ( + (6 x ) dx B 4 = 4 π ( 6 x ) π D = 806 (to sf) 6 Use limits and 0 to give [ ] Notes: B Both bits CAO but condone lack of π M Integrating y their dy 4 + dx, getting k(+ 6 x ), condone lack of π dx If they use a substitution it must be a complete method. A CAO DM Correct use of and 0 as limits A CAO. dy x (a) (i) = + arcsin x d x ( x ) (ii) At given value derivative π + π = + = 6 6 (b) x dy 6e = 4x dx + 9e 6 M = x x e + 9e M = 5 x x 4 x x ( e + e ) + ( e e ) dy A cso = dx 5coshx + 4sinhx (a) M Notes: Differentiating getting an arcsinx term and a A CAO B CAO any correct form ± x term B 5 () () (5) 8 GCE Further Pure Mathematics FP (6669) June 0

5 (b) M x ae Of correct form 4x ± be A CAO M Getting from expression in 4 x to x and x e e e only M Using sinhx and coshx in terms of ( e x + e x ) and ( e x e x ) A CSO answer given. (a) x 0x+ 4 = ( x 5) + 9 so = = B x 0x+ 4 ( x 5) + 9 u + 9 (mark can be earned in either part (a) or (b)) u I = x 5 du = arctan u + 9 I = du = arctan ( x 5) + 9 π π D Uses limits and 0 to give Uses limits 8 and 5 to give (5) (b) Alt (b) Alt (b) Alt I = x 5 x 5 ln + + or x 5+ ( x 5) + 9 I = ln or I = ln (( x 5) + ( x 5) + 9 ) Uses limits 5 and 8 to give ln( + ). D 9 Uses u = x-5 to get I = u du = ar sinh ln = { u+ u + 9} u 9 + Uses limits and 0 and ln expression to give ln( + ). D Use substitution x 5= tan θ, dx sec θ dθ and so I = secθdθ = ln(secθ + tan θ) π D Uses limits 0 and to get ln( + ). 4 (a) B CAO allow recovery in (b) M Integrating getting k arctan term A CAO DM Correctly using limits. A CAO GCE Further Pure Mathematics FP (6669) June 0 Notes:

6 (b) M Integrating to get a ln or hyperbolic term A CAO DM Correctly using limits. A CAO 4. (a) n x n x n(ln x) In = (ln x) dx x e e n n (ln ) x nx x = (ln x) e n In = In dx DM Acso (b) I I 0 e e x e x dx = = = or e e e I = = e e e 4e = I, I = I and I = I so I = Notes: (a)m Using integration by parts, integrating x, differentiating (ln x ) n A CAO DM Correctly using limits and e A CSO answer given (b)m Evaluating I0 or Iby an attempt to integrate something A CAO M Finding I (also probably I and I ) If n s left in M0 A I CAO 8 GCE Further Pure Mathematics FP (6669) June 0

7 5. (a) Graph of y = sinhx B Shape of graph x e B Asymptote: y = B (b) Use definition ( x x ) x 9 4x 6 x 0 x e = or 9 x = ln() Value 0 on y axis and value 0.7 or ln on x axis ( ) e e = e e e = to form quadratic Notes: (a) B y = sinhx first and third quadrant. x B Shape of y = e correct intersects on positive axes. B Equation of asymptote, y =, given. Penlise extra asymptotes here 4B Intercepts correct both x (b) M Getting a three terms quadratic in e A Correct three term quadratic x DM Solving for e x A CAO for e condone omission of negative value. B CAO one answer only B D B (5) 9 GCE Further Pure Mathematics FP (6669) June 0 4

8 6. (a) n = ( j-k) (i + j + k) = 6 i j 6 k o.a.e. (e.g. i j k ) (b) Line l has direction i j k B Angle between line l and normal is given by (cos β or sin α) = = ft α = 90 β = 6 degrees to nearest degree. A awrt (c) Alt Plane P has equation r.(i j k ) = Perpendicular distance is ( 7) = 8 9 i j k 7 (c) Alt Parallel plane through A has equation r. = i j k M Plane P has equation r. = So O lies between the two and perpendicular distance is + 7 = 8 A (c) Alt Distance A to (,,) = Perpendicular distance is sin α = 8 9 = 8 (c) Alt 4 Finding Cartesian equation of plane P: x y z = 0 nα + nβ + nγ + d () () () 8 d = = = n + n + n = MA MA MA () 0 Notes: (a) M Cross product of the correct vectors A CAO o.e. (b) B CAO M Angle between i j k and i j k, formula of correct form Aft 8/9ft A CAO awrt (c) M Eqn of plane using i j k or dist of A from O or finding length of AP A Correct equation (must have = ) or A to (,,) = M Using correct method to find perpendicular distance A CAO GCE Further Pure Mathematics FP (6669) June 0 5

9 7. (a) ( ) ( ) (b) Det M = k( 0 ) = k + 4 = k k - - T M = so cofactors = -4 k- k+ - - k - M ( - A mark for each term wrong) - - M = -4 k- k k + M A - k - x 4 4 y = + λ. z 7 x x x 4+ 4λ Use y = y M with k =. i.e. y = 4 λ z z + z 7+ λ (c) Let ( x, y, z) be on l. Equation of l can be written as B M () (5) x λ + y = 4λ z λ and so (r - a) b = 0 where a = i - j and b = i +4 j + k or equivalent or r=a+λb where a = i - j and b = i +4 j + k or equivalent Notes: (a) M Finding determinant at least one component correct. A CAO (b) M Finding matrix of cofactors or its transpose M Finding inverse matrix, /(det) cofactors + transpose A At least seven terms correct (so at most incorrect) condone missing det A At least eight terms correct (so at most incorrect) condone missing det A All nine terms correct, condone missing det (c) B Equation of l M Using inverse transformation matrix correctly M Finding general point in terms of λ. A CAO for general point in terms of one parameter B ft for vector equation of their l Bft (5) GCE Further Pure Mathematics FP (6669) June 0 6

10 8. (a) dy dy dθ bcoshθ x yy xb bcoshθ Uses = = or = 0 y = = d dx x dθ a sinh θ a b ya asinhθ bcoshθ So y bsinh θ = ( x acosh θ ) asinhθ ab(cosh θ sinh θ) = xb coshθ ya sinhθ and as (cosh θ sinh θ ) = xbcoshθ yasinhθ = ab M Acso a (b) P is the point (,0) coshθ (c) l has equation x = a and meets l at Q b(coshθ ) ( a, ) sinhθ () () a(coshθ + ) b(coshθ ) (d) Alt The mid point of PQ is given by X =, Y = coshθ sinhθ cosh θ + coshθ + sinh θ 4Y + b = b sinh θ cosh θ coshθ = b sinh θ (coshθ + )(coshθ )coshθ X(4 Y + b ) = ab coshθ sinh θ Simplify fraction by using cosh θ sinh θ = to give ft M M 4M x(4 y + b ) = ab Acso (6) (d) Alt First line of solution as before MAft 4Y + b = b coth θ + cosech θ cothθ cosechθ + M ( ) ( coth coth cosech ) = b θ θ θ M ( ) X(4 Y + b ) = ab coth θ (cothθ cosech θ ( + sech θ)) 4M Simplify expansion by using coth θ cosech θ = to give x(4 y + b ) = ab Acso (6) 4 GCE Further Pure Mathematics FP (6669) June 0 7

11 8. (a)m Finding gradient in terms of θ A CAO M Finding equation of tangent A CSO (answer given) look for ± (cosh θ sinh θ) (b)m Putting y = 0into their tangent Aft P found, ft for their tangent o.e. (c) M Putting x = a into their tangent. A CAO Q found o.e. (d) For Alt and M Finding expressions, in terms of sinh θ and coshθ but must be adding A Ft on their P and Q, M Finding 4y + b M Simplified, factorised, maximum of terms per bracket 4M Finding x(4 y + b ), completely factorised, maximum of terms per bracket A CSO (d) For Alts, 4 and 5 M Finding expressions, in terms of sinh θ and coshθ but must be adding A Ft on their P and Q M Getting coshθ in terms of x M y or y in terms of coshθ or sinhθ in terms of x and y 4M Getting equation in terms of x and y only. No square roots. A CSO GCE Further Pure Mathematics FP (6669) June 0 8

12 8(d) Alt a(coshθ + ) b(coshθ ) As main scheme ft X =, Y = coshθ sinhθ a coshθ in terms of x M cosh θ = x a b(coshθ -) b( a x) sinhθ in terms of x and y M sinh θ = = y ( x a) y a b( a x) Using cosh θ - sinh θ = 4M = x a ( x a) y Simplifies to give required equation y 4 x( a x) = b ( a x), x(4 y + b ) = ab Acso (6) Alt 4 a(coshθ + ) b(coshθ ) X =, Y = coshθ sinhθ a cosh θ = x a b (cosh θ - ) b (cosh θ - ) y = = 4(cosh θ - ) 4(cosh θ + ) a x b x a y = o.e x 4 x a Simplifies to give required equation As main scheme coshθ in terms of x y in terms of coshθ only Forms equation in x and y only ft M M 4M A cso (6) Alt 5 a(coshθ + ) b(coshθ ) X =, Y = coshθ sinhθ a cosh θ = x a b(cosh θ - ) b(cosh θ - ) y = = sinhθ cosh θ - As main scheme coshθ in terms of x y in terms of coshθ only ft M M Eliminate and forms equation in x and y 4M Simplifies to give required equation Acso GCE Further Pure Mathematics FP (6669) June 0 9

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