June 2011 Further Pure Mathematics FP Mark Scheme

Transcription

1 . June 0 Further Pure Mthemtics FP 6669 Mrk dy 6x dx = nd so surfce re = π x ( + (6 x ) dx B 4 = 4 π ( 6 x ) π D = 806 (to sf) 6 Use limits nd 0 to give [ ] B Both bits CAO but condone lck of π M Integrting y their dy 4 + dx, getting k(+ 6 x ), condone lck of π dx If they use substitution it must be complete method. A CAO DM Correct use of nd 0 s limits A CAO. dy x () (i) = + rcsin x d x ( x ) (ii) At given vlue derivtive 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 () M Differentiting getting n rcsinx term nd A CAO B CAO ny correct form ± x term B 5 () () 8 GCE Further Pure Mthemtics FP (6669) June 0

2 (b) M x e Of correct form 4x ± be A CAO M Getting from expression in 4 x to x nd x e e e only M Using sinhx nd coshx in terms of ( e x + e x ) nd ( e x e x ) A CSO nswer given. () x 0x+ 4 = ( x 5) + 9 so = = B x 0x+ 4 ( x 5) + 9 u + 9 (mrk cn be erned in either prt () or (b)) u I = x 5 du = rctn u + 9 I = du = rctn ( x 5) + 9 π π D Uses limits nd 0 to give Uses limits 8 nd 5 to give (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 nd 8 to give ln( + ). D 9 Uses u = x-5 to get I = u du = r sinh ln = { u+ u + 9} u 9 + Uses limits nd 0 nd ln expression to give ln( + ). D Use substitution x 5= tn θ, dx sec θ dθ nd so I = secθdθ = ln(secθ + tn θ) π D Uses limits 0 nd to get ln( + ). 4 () B CAO llow recovery in (b) M Integrting getting k rctn term A CAO DM Correctly using limits. A CAO GCE Further Pure Mthemtics FP (6669) June 0

3 (b) M Integrting to get ln or hyperbolic term A CAO DM Correctly using limits. A CAO 4. () 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) e e x e x dx I0 = = = or e e e I = = e e e 4e I = I0, I = I nd I = I so I = ()M Using integrtion by prts, integrting x, differentiting (ln x ) n A CAO DM Correctly using limits nd e A CSO nswer given (b)m Evluting I0 or Iby n ttempt to integrte something A CAO M Finding I (lso probbly I nd I ) If n s left in M0 A I CAO 8 GCE Further Pure Mthemtics FP (6669) June 0

4 5. () Grph of y = sinhx B Shpe of grph x e B Asymptote: y = B (b) Use definition ( x x ) x 9 4x 6 x 0 x e = or 9 x = ln() Vlue 0 on y xis nd vlue 0.7 or ln on x xis ( ) e e = e e e = to form qudrtic () B y = sinhx first nd third qudrnt. x B Shpe of y = e correct intersects on positive xes. B Eqution of symptote, y =, given. Penlise extr symptotes here 4B Intercepts correct both x (b) M Getting three terms qudrtic in e A Correct three term qudrtic x DM Solving for e x A CAO for e condone omission of negtive vlue. B CAO one nswer only B D B 9 GCE Further Pure Mthemtics FP (6669) June 0 4

5 6. () n = ( j-k) (i + j + k) = 6 i j 6 k o..e. (e.g. i j k ) (b) Line l hs direction i j k B Angle between line l nd norml is given by (cos β or sin α) = = ft α = 90 β = 6 degrees to nerest degree. A wrt (c) Alt Plne P hs eqution r.(i j k ) = Perpendiculr distnce is ( 7) = 8 9 i j k 7 (c) Alt Prllel plne through A hs eqution r. = i j k M Plne P hs eqution r. = So O lies between the two nd perpendiculr distnce is + 7 = 8 A (c) Alt Distnce A to (,,) = Perpendiculr distnce is sin α = 8 9 = 8 (c) Alt 4 Finding Crtesin eqution of plne P: x y z = 0 nα + nβ + nγ + d () () () 8 d = = = n + n + n = MA MA MA () 0 () M Cross product of the correct vectors A CAO o.e. (b) B CAO M Angle between i j k nd i j k, formul of correct form Aft 8/9ft A CAO wrt (c) M Eqn of plne using i j k or dist of A from O or finding length of AP A Correct eqution (must hve = ) or A to (,,) = M Using correct method to find perpendiculr distnce A CAO GCE Further Pure Mthemtics FP (6669) June 0 5

6 7. () ( ) ( ) (b) Det M = k( 0 ) = k + 4 = k k - - T M = so cofctors = -4 k- k+ - - k - M ( - A mrk for ech 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. Eqution of l cn be written s B M () x λ + y = 4λ z λ nd so (r - ) b = 0 where = i - j nd b = i +4 j + k or equivlent or r=+λb where = i - j nd b = i +4 j + k or equivlent () M Finding determinnt t lest one component correct. A CAO (b) M Finding mtrix of cofctors or its trnspose M Finding inverse mtrix, /(det) cofctors + trnspose A At lest seven terms correct (so t most incorrect) condone missing det A At lest eight terms correct (so t most incorrect) condone missing det A All nine terms correct, condone missing det (c) B Eqution of l M Using inverse trnsformtion mtrix correctly M Finding generl point in terms of λ. A CAO for generl point in terms of one prmeter B ft for vector eqution of their l Bft GCE Further Pure Mthemtics FP (6669) June 0 6

7 8. () dy dy dθ bcoshθ x yy xb bcoshθ Uses = = or = 0 y = = d dx x dθ sinh θ b y sinhθ bcoshθ So y bsinh θ = ( x cosh θ ) sinhθ b(cosh θ sinh θ) = xb coshθ y sinhθ nd s (cosh θ sinh θ ) = xbcoshθ ysinhθ = b M Acso (b) P is the point (,0) coshθ (c) l hs eqution x = nd meets l t Q b(coshθ ) (, ) sinhθ () () (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 ) = b coshθ sinh θ Simplify frction by using cosh θ sinh θ = to give ft M M 4M x(4 y + b ) = b Acso (6) (d) Alt First line of solution s before MAft 4Y + b = b coth θ + cosech θ cothθ cosechθ + M ( ) ( coth coth cosech ) = b θ θ θ M ( ) X(4 Y + b ) = b coth θ (cothθ cosech θ ( + sech θ)) 4M Simplify expnsion by using coth θ cosech θ = to give x(4 y + b ) = b Acso (6) 4 GCE Further Pure Mthemtics FP (6669) June 0 7

8 8. ()M Finding grdient in terms of θ A CAO M Finding eqution of tngent A CSO (nswer given) look for ± (cosh θ sinh θ) (b)m Putting y = 0into their tngent Aft P found, ft for their tngent o.e. (c) M Putting x = into their tngent. A CAO Q found o.e. (d) For Alt nd M Finding expressions, in terms of sinh θ nd coshθ but must be dding A Ft on their P nd Q, M Finding 4y + b M Simplified, fctorised, mximum of terms per brcket 4M Finding x(4 y + b ), completely fctorised, mximum of terms per brcket A CSO (d) For Alts, 4 nd 5 M Finding expressions, in terms of sinh θ nd coshθ but must be dding A Ft on their P nd Q M Getting coshθ in terms of x M y or y in terms of coshθ or sinhθ in terms of x nd y 4M Getting eqution in terms of x nd y only. No squre roots. A CSO GCE Further Pure Mthemtics FP (6669) June 0 8

9 8(d) Alt (coshθ + ) b(coshθ ) As min scheme ft X =, Y = coshθ sinhθ coshθ in terms of x M cosh θ = x b(coshθ -) b( x) sinhθ in terms of x nd y M sinh θ = = y ( x ) y b( x) Using cosh θ - sinh θ = 4M = x ( x ) y Simplifies to give required eqution y 4 x( x) = b ( x), x(4 y + b ) = b Acso (6) Alt 4 (coshθ + ) b(coshθ ) X =, Y = coshθ sinhθ cosh θ = x b (cosh θ - ) b (cosh θ - ) y = = 4(cosh θ - ) 4(cosh θ + ) x b x y = o.e x 4 x Simplifies to give required eqution As min scheme coshθ in terms of x y in terms of coshθ only Forms eqution in x nd y only ft M M 4M A cso (6) Alt 5 (coshθ + ) b(coshθ ) X =, Y = coshθ sinhθ cosh θ = x b(cosh θ - ) b(cosh θ - ) y = = sinhθ cosh θ - As min scheme coshθ in terms of x y in terms of coshθ only ft M M Eliminte nd forms eqution in x nd y 4M Simplifies to give required eqution Acso GCE Further Pure Mthemtics FP (6669) June 0 9