Mark Scheme Summer 2009

Transcription

1 Mark Summer 009 GCE GCE Mathematics (6668/0)

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3 June Further Pure Mathematics FP (new) Mark Question Q (a) = rr ( + ) r ( r+ ) r ( r+ ) B aef () (b) n n r = r = = rr ( + ) r r+ = n n + n n + List the first two terms and the last two terms Includes the first two underlined terms and includes the final two = + ; underlined terms. n + n + + n + n + = n + n + = = ( n + )( n + ) ( n + ) ( n + ) ( n + )( n + ) n + 9n + 6 n n ( n + )( n + ) Attempt to combine to an at least term fraction to a single fraction and an attempt to take out the brackets from their numerator. = n + 5n ( n + )( n + ) = n(n + 5) ( n + )( n + ) Correct Result cso AG (5) [6] 87_97 GCE Mathematics January

4 Q (a) z = i, < θ y O α arg z (, ) r = ( ) + ( ) = + = 6 = 8 θ = tan ( ) = ( ( ) isin ( ) ) z = 8cos + ( ) So, z = 8 cos + isin A valid attempt to find the modulus and argument of i. Taking the cube root of the modulus and dividing the argument by. z = cos ( ( ) + isin( )) cos ( ( ) isin( )) + Also, ( ( 7 ) ( 7 z = 8cos + isin )) or ( ( 9 ) ( 9 z = 8cos + isin )) Adding or subtracting to the argument for z in order to find other roots. 7 7 ( isin ) = + z cos ( ( ) ( )) Any one of the final two roots and z = cos + isin Both of the final two roots. Special Case : Award SC: A0A0 for ALL three of ( + ) ( isin ) cos and cos( ) isin( ) ( ) +. cos isin, [6] Special Case : If r is incorrect (and not equal to 8) and candidate states the brackets ( )correctly then give the first accuracy mark ONLY where this is applicable. 87_97 GCE Mathematics January

5 Q dy sin ycos sin sin = d y cos sin sin y = sin sin An attempt to divide every term in the differential equation by sin. Can be implied. dy ycos = sin sin Integrating factor cos ± ( ) sin cos ± e or e d sin ln sin = e = e ln sin e or their P( ) ( ) ln cosec e d aef = sin sin or (sin ) or cosec aef dy ycos sin = sin d sin sin d y = sin sin sin d ( y ) their I.F. = sin their I.F d y = sin cos d y = cos or sin y cos ( ) sin = y cos sin = y sin K sin = + A credible attempt to integrate the RHS with/without + K ddd = sin + sin y K = sin + sin cao y K [8] 87_97 GCE Mathematics January

6 Q A = + 0 ( a cosθ ) dθ with Applies r ( dθ) 0 correct limits. Ignore dθ. B ( a + cos ) = a + 6acos + 9cos θ θ θ + cosθ = a + 6acosθ + 9 ± ± cosθ cos θ = Correct underlined epression. 9 9 = + 6 cosθ + + cosθ dθ A a a 0 Integrated epression with at least out of terms of the form ± Aθ ± Bsinθ ± Cθ ± Dsin θ. * 9 9 Ignore the θ 6 sinθ θ sinθ. Ignore limits. 0 a θ + 6asinθ + correct ft integration. ft Ignore the. Ignore limits. = a + a + + = ( a 0 9 0) ( 0) = a + 9 a + 9 Hence, 9 07 a + = Integrated epression equal to a + = d* a = 9 As a > 0, a = 7 a = 7 Some candidates may achieve a = 7 from incorrect working. Such candidates will not get full marks cso [8] 87_97 GCE Mathematics January

7 Q5 y = sec = (sec ) (a) d d y (sec ) (sec tan ) sec tan = = Either (sec ) (sec tan ) or sec tan B aef Apply product rule: u = sec v = tan du dv = sec tan = sec d y = sec tan + sec Two terms added with one of either Asec tan or Bsec in the correct form. Correct differentiation = sec (sec ) + sec (b) d y Hence, = 6sec sec dy y = ( ) =, ( ) d = () = Applies tan = sec leading to the correct result. Both dy y = and d = AG B () d y ( ) ( ) = 6 = 8= 6 Attempts to substitute = into both terms in the epression for d y. d y sec (sec tan ) 8sec (sec tan ) = Two terms differentiated with either sec tan or 8sec tan being correct = sec tan 8sec tan ( ) ( ) d y d y = () 8 () = 96 6 = 80 = 80 B 6 80 ( ) ( ) ( ) sec { ( ) ( ) ( ) } 0 sec Applies a Taylor epansion with at least out of terms ft correctly. Correct Taylor series epansion. (6) [0] 87_97 GCE Mathematics January

8 Q6 z w =, z = i z + i (a) wz ( + i) = z wz+ iw= z iw= z wz i w i w= z( w) z = ( w) Complete method of rearranging to make z the subject. i w z = ( w) aef z i w = = w Putting in terms of their z w = d iw = w w = w w = 9 w u + iv = 9 u + iv u + v = 9 ( u ) + v u + v = 9u 8u v 0= 8u 8u + 8v + 9 Applies w = u + iv, and uses Pythagoras correctly to get an equation in terms of u and v without any i's. Correct equation. dd 0 = u u + v Simplifies down to u + v ± αu ± βv ± δ = 0. ddd ( ) u v + + = ( ) u v + = (b) 9 {Circle} centre ( ) v, 0, radius One of centre or radius correct. 8 8 Both centre and radius correct. Circle indicated on the Argand diagram in the correct position in follow through quadrants. Ignore plotted coordinates. Bft (8) O u Region outside a circle indicated only. B () [0] 87_97 GCE Mathematics January

9 Q7 y = a, a > (a) y a Correct Shape. Ignore cusps. Correct coordinates. B B a O a (b) a a a =, > () { > a}, a = a a = a aef + a = 0 ()( a ) ± = Applies the quadratic formula or completes the square in order to find the roots. ± + 8a = Both correct simplified down solutions. { < a}, + a = a + a = a or a = a aef { = 0 ( ) = 0} = 0, = 0 = B (6) (c) a a a >, > + 8a < {or} + + 8a > is less than their least value is greater than their maimum value B ft B ft {or} 0< < For{ < a}, Lowest < < Highest 0< < () [] 87_97 GCE Mathematics January 009 6

10 Q8 d t = e, = 0, = at t = 0. (a) AE, m + 5m + 6 = 0 ( m + )( m + ) = 0 m =,. So, mt mt Ae + Be, where m m. t t CF = Ae + Be t t Ae + Be d = ke = ke = ke t t t ke + 5( ke ) + 6ke = e ke = e k = t t t t t t { So, PI = e t } Substitutes k e t into the differential equation given in the question. Finds k =. So, A B their + their * t t t = e + e + e CF PI == t t t Ae Be e Finds d by differentiating their and their CF PI d* t = 0, = 0 0 = A + B + t = 0, = = A B A + B = A B = A =, B = 0 Applies t = 0, = 0 to and t = 0, = to d to form simultaneous equations. dd* So, t = e + e t t t = e + e cao (8) 87_97 GCE Mathematics January 009 6

11 t = e + e t (b) = = t t e e 0 Differentiates their to give d and puts d equal to 0. t e = 0 t = ln A credible attempt to solve. d* t = ln or t = ln or awrt 0.55 ln ln ln ln So, = e + e = e + e = + Substitutes their t back into and an attempt to eliminate out the ln s. dd = + = = 9 uses eact values to give 9 AG d 9e t = + e t At t = d ln, = 9e + e ln ln d Finds d and substitutes their t into d* 9 = 9() + = + = + As d 9 = + = < 0 then is maimum. 9 + < 0 and maimum conclusion. (7) [5] 87_97 GCE Mathematics January 009 6

12 87_97 GCE Mathematics January 009 6