June Further Pure Mathematics FP1 (new) Mark Scheme

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1 June Further Pure Mathematics FP (new) Mar Question Q (a) Mars B () (c) (d) z! # (" ) 5 (or awrt.4)! ) & ) & *! arctan' $ or arctan ' " $ ( % ( % arg z! "0.46 or 5.8 (awrt) (answer in degrees is A0 unless followed by correct conversion) " 8 # 9i # i + " i # i " 6 " 8i # 8i " 9!! " 5 # i i.e. a = -5 and b = or 5 " Alternative method to part (d) " 8 # 9i! ( " i)( a # bi), and so a# b!" 8 and b" a! 9 and attempt to solve as far as equation in one variable So a = -5 and b = (a) B needs both complex numbers as either points or vectors, in correct quadrants and with reasonably correct relative scale M Attempt at Pythagoras to find modulus of either complex number A condone correct answer even if negative sign not seen in (-) term A0 for, 5 (c) arctan is M0 unless followed by - # arctan or - " arctan Need to be clear that argz = or 5.8 for A (d) M Multiply numerator and denominator by conjugate of their denominator A for 5 and A for i (should be simplified) Alternative scheme for (d) Allow slips in woring for first M 5 a M A M A M A Aft M A Acao () () () [8] 87_974 GCE Mathematics January

2 Q (a) r( r # )( r # )! r # 4r # r, so use. # r 4. r #. r ) & ) &! n ( n # ) # 4' n( n # )(n # ) $ # ' n( n # ) $ 4 ( 6 % ( % M Mars A A! n n n n n n ( n # ) / n( n # ) # 8( # ) # 80 or! / # # 45 # 60 / n # 9n # 60! n( n # )( n # )( # ) ( ) 9 6 n n n n or =! # / # # 0! n ( n # ) n ( = ) ! ".. 0 M A M Acao (7) M! ( ) " ( )! 7640 " 560! 7070 (a) M expand and must start to use at least one standard formula A cao () [9] First A mars: One wrong term A A0, two wrong terms A0 A0. M: Tae out factor n(n + ) or n or (n + ) directly or from quartic A: See scheme (cubics must be simplified) M: Complete method including a quadratic factor and attempt to factorise it A Completely correct wor. Just gives =, no woring is 0 mars for the question. Alternative method Expands (n + )(n + )(n + ) and confirms that it equals / n n 45n 60 # # # together with statement = can earn last MA The previous MA can be implied if they are using a quartic. M is for substituting 40 and 0 into their answer to (a) and subtracting. (NB not 40 and ) Adding terms is M0A0 as the question said Hence 87_974 GCE Mathematics January

3 Q (a) x # 4! 0 x! i, x!,i Solving -term quadratic " 8, 64 " 00 x!! " 4 # i and - 4 -i Mars M, A M A Aft (5) i # (" i) # (" 4 # i) # (" 4 " i)! " 8 M Acso () [7] Alternative method : Expands f(x) as quartic and chooses, coefficient of x M -8 A cso (a) Just x = i is M A0 x!, is M0A0 M for solving quadratic follows usual conventions, then A for a correct root (simplified as here) and Aft for conjugate of first answer. Accept correct answers with no woring here. Do not give accuracy mars for factors unless followed by roots. M for adding four roots of which at least two are complex conjugates and getting a real answer. A for 8 following correct roots or the alternative method. If any incorrect woring in part (a) this A mar will be A0 87_974 GCE Mathematics January

4 Q4 (a) f (.)!. ". " 6! " 0.9 f (.)!. ". " 6! Change of sign Root f 4 ( x)! x " x f 4 (.)! 0. f ( x0 ) " 0.9 x! x0 "!. " f 4( x0 ) 0.!.9 (c) * ".."! *,'0.9','0.877' need numerical values correct (to s.f.). 0." (or equivalent such as!.),'0.9','0.877' Mars M A () B B M Aft Acao (5) M *(0.877 # 0.9)! # A or (0.877 # 0.9)! , where *!. # so * 5. 8 (.796 ) (Allow awrt) A () [0] Alternative Uses equation of line joining (., -0.9) to (., 0.877) and substitutes y = # y# 0.9! ( x".) and y = 0, so * 5. 8 or awrt as before 0. (NB Gradient = 0.69) (a) M for attempt at f(.) and f(.) A need indication that there is a change of sign (could be 0.9<0, 0.88>0) and need conclusion. (These mars may be awarded in other parts of the question if not done in part (a)) B for seeing correct derivative (but may be implied by later correct wor) B for seeing 0. or this may be implied by later wor M Attempt Newton-Raphson with their values Aft may be implied by the following answer (but does not require an evaluation) Final A must.9 exactly as shown. So answer of.897 would get 4/5 If done twice ignore second attempt (c) M Attempt at ratio with their values of, f(.) and, f(.). N.B. If you see 0.9 "* or "* in the fraction then this is M0 A correct linear expression and definition of variable if not! (may be implied by final correct answer- does not need dp accuracy) A for awrt.8 If done twice ignore second attempt M A, A 87_974 GCE Mathematics January

5 Q5 (a) Mars ) a # a a # b& R! ' $ M A A ( a # ab a # b % () or their Puts their a # a = 5 or their a # b = 5 ( a a)( a b ) ( a ab)(a b) # # " # # = 5 ( or to 5), M, Alternative for Puts their a Solve to find either a or b a!, b! " # ab! 0 or their a + b = 0 Uses R + column vector = 5+ column vector, and equates rows to give two equations in a and b only Solves to find either a or b as above method (a) term correct: M A0 A0 or terms correct: M A A0 M M A, A M, M M A A (5) [8] M M as described in scheme (In the alternative scheme column vector can be general or specific for first M but must be specific for nd M) M requires solving equations to find a and/or b (though checing that correct answer satisfies the equations will earn this mar) This mar can be given independently of the first two method mars. So solving M!5Mfor example gives M0M0MA0A0 in part Also putting leading diagonal = 0 and other diagonal = 5 is M0M0MA0A0 (No possible solutions as a >0) A A for correct answers only Any Extra answers given, e.g. a = -5 and b = 5 or wrong answers deduct last A awarded So the two sets of answers would be A A0 Just the answer. a = -5 and b = 5 is A0 A0 Stopping at two values for a or for b no attempt at other is A0A0 Answer with no woring at all is 0 mars 87_974 GCE Mathematics January 009 5

6 Q6 (a) y! ( 8t)! 64t and (c) 6x! 6 + 4t! 64t Or identifies that a = 4 and uses general coordinates ( at, at ) (4, 0) B dy y! 4x! x Replaces x by " Uses Gradient of normal is 4t to give gradient [ (4 t ) " [ " t ] gradient of curve "!! ] t t y " 8t! " t( x " 4t ) y # tx! 8t # 4t (*) B B M, M Mars () () M Acso (5) (d) 8t# 4t At N, y = 0, so x! 8 # 4t or t Base SN! # t "! # t (8 4 ) 4 ( 4 4 ) B Bft Area of 6 PSN = (4 # 4t )(8t)! 6t( # t ) or 6t# 6t for t > 0 {Also Area of 6 PSN = (4 # 4 t )( " 8 t )! " 6 t ( # t ) for t < 0 } this is not required Alternatives: (c) d x d y! 8t and! 8 B dt dt dy dy! 7! M, then as in main scheme. dt dt t dy y (c) y! 6 B (or uses x! to give d x y 8 dy! 8 ) dy 8 8!!! M, then as in main scheme. y 8t t M A (4) [] (c) Second M need not be function of t Third M requires linear equation (not fraction) and should include the parameter t but could be given for equation of tangent (So tangent equation loses mars only and could gain BMM0MA0) (d) Second B does not require simplification and may be a constant rather than an expression in t. M needs correct area of triangle formula using ½ their SN + 8t Or may use two triangles in which case need (4t " 4) and (4t # 8 " 4 t ) for Bft Then Area of 6 PSN = (4t " 4)(8 t) # (4t # 8 " 4 t )(8 t)! 6 t( # t ) or 6t# 6t 87_974 GCE Mathematics January 009 5

7 Q7 (a) Use 4 a" (" + " )! 0 a,! Determinant: ( + 4) " (" + " )! 0 ( 6) " ) 4 & B! ' $ 0 ( % (c) ) 4 & ) " 6 & ) 4( " 6) # ( # ) & ' $ ' $,! ' $ 0 ( % ( # % 0 ( ( " 6) # ( # ) % ) & ' $ ( # % Lies on y! x # Alternatives: ) " &) x & ) x " ( x # ) & (c) ' $' $,! ' $, ( " 4 %( x # % ( " x # 4( x # ) % ) x " 6 &! ' $, which was of the form ( - 6, +) ( x # % ) " &) x & ) x " y & ) " 6 & Or ' $' $,!' $!' $, and solves simultaneous equations ( " 4 %( y% ( " x # 4y% ( # % Both equations correct and eliminate one letter to get x = or y = + or 0x" 0y! " 0 or equivalent. Completely correct wor ( to x = and y = + ), and conclusion lies on y! x # Mars M, A M () M Acso () M, Aft A () [8] M, A, A M A A (a) Allow sign slips for first M Allow sign slip for determinant for first M (This mar may be awarded for /0 appearing in inverse matrix.) ) 4 & Second M is for correctly treating the by matrix, ie for ' $ ( % Watch out for determinant ( + 4) ( - + -) = 0 M0 then final answer is A0 (c) M for multiplying matrix by appropriate column vector A correct wor (ft wrong determinant) A for conclusion 87_974 GCE Mathematics January 009 5

8 Q8 (a) f ()! 5 # 8 #! 6, (which is divisible by 4). (8True for n = ). Using the formula to write down f( + ), f ( # )! 5 # 8( # ) # f ( # ) " f ( )! 5 # # 8( # ) # " 5 " 8 " #! 5(5 ) # 8 # 8 # " 5 " 8 "! 4(5 ) # 8 f ( # )! 4(5 # ) # f ( ), which is divisible by 4 8True for n = + if true for n =. True for n =, 8true for all n. B Mars M A M A Aft Acso (7) ) n # For n =, ' ( n " n & ) $! ' " n% ( " & ) $! ' " % ( " & $ " % (8True for n =.) B (a) Alternative for nd M: Part Alternative # ) " & ) # " &) " & ) # " " & ' $!' $' $!' $ ( " % ( " %( " % ( # " " % ) ( # ) # " ( # ) &! ' $ ( ( # ) " ( # ) % 8True for n = + if true for n =. True for n =, 8true for all n f ( # )! 5(5 ) # 8 # 8 # M! 4(5 ) # 8 # (5 # 8 # ) A or! 5(5 # 8# ) " " 4! 4(5 # ) # f ( ), or! 5f ( ) " 4(8# ) which is divisible by 4 A (or similar methods) M A A M A A cso (a) B Correct values of 6 or 4 for n = or for n = 0 (Accept is a multiple of ) M Using the formula to write down f( + ) A Correct expression of f(+) (or for f(n +) M Start method to connect f(+) with f() as shown A correct woring toward multiples of 4, A ft result including f( +) as subject, Acso conclusion B correct statement for n = or n = 0 First M: Set up product of two appropriate matrices product can be either way round A A0 for one or two slips in simplified result A A all correct simplified A0 A0 more than two slips M: States in terms of ( + ) A Correct statement A for induction conclusion # (7) [4] ) " & ) # " " & May write ' $! ' $. Then may or may not complete the proof. ( " % ( # " " % This can be awarded the second M (substituting + )and following A (simplification) in part. The first three mars are awarded as before. Concluding that they have reached the same matrix and therefore a result will then be part of final A cso but also need other statements as in the first method. 87_974 GCE Mathematics January

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