Mark Summer 009 GCE GCE Mathematics (6668/0)
Edecel is one of the leading eamining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. Through a network of UK and overseas offices, Edecel s centres receive the support they need to help them deliver their education and training programmes to learners. For further information, please call our GCE line on 08 576 005, our GCSE team on 08 576 007, or visit our website at www.edecel.com. If you have any subject specific questions about the content of this Mark that require the help of a subject specialist, you may find our Ask The Epert email service helpful. Ask The Epert can be accessed online at the following link: http://www.edecel.com/aboutus/contact-us/ Summer 009 Publications Code UA05 All the material in this publication is copyright Edecel Ltd 009
June 009 6668 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 009 55
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 7 7 + 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 009 56
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 009 57
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 07. 9 07 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 009 58
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 + + + +... 6 { ( ) ( ) ( ) } 0 sec + + 8 + +... Applies a Taylor epansion with at least out of terms ft correctly. Correct Taylor series epansion. (6) [0] 87_97 GCE Mathematics January 009 59
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 + 9 + 9v 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 + 9 9 8 Simplifies down to u + v ± αu ± βv ± δ = 0. ddd ( ) u v + + = 9 8 9 8 6 8 0 ( ) u v + = 9 9 8 6 (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 009 60
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
Q8 d t + 5 + 6 = 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
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
87_97 GCE Mathematics January 009 6