June 009 6667 Further Pure Mathematics FP (new) Mar Q (a) Mars B () (c) (d) z + ( ) 5 (or awrt.4) α 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 π + arctan or π arctan Need to be clear that argz - 0.46 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 009 47
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( + ) + 8} or { + + 45 + 6} { n + 9n + 6} n( n + )( n + )( + ) ( ) 9 6 n n n n or + { + + } n ( n + ) n ( ) 40 40 0 M A M Acao (7) M (40 4 4 ) (0 7) 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 6} + + + 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 009 48
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 009 49
Q4 (a) f (.).. 6 ( 0.9) f (.).. 6 ( 0.877) Change of sign Root f ( x) x x f (.) 0. f ( x0 ) 0.9 x x0. f ( x0 ) 0..9 (c) α.. α ± '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 α (0.877 + 0.9). 0.9 +. 0.877 A or (0.877 + 0.9) 0. 0.9, where α. + so α. 8 (.796 ) (Allow awrt) A () [0] Alternative Uses equation of line joining (., -0.9) to (., 0.877) and substitutes y 0 M 0.9 + 0.877 y+ 0.9 ( x.) and y 0, so α. 8 or awrt as before 0. A, A (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 α or 0.877 α in the fraction then this is M0 A correct linear expression and definition of variable if not α (may be implied by final correct answer- does not need dp accuracy) A for awrt.8 If done twice ignore second attempt 87_974 GCE Mathematics January 009 50
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
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 PSN (4 + 4t )(8t) 6t( + t ) or 6t+ 6t for t > 0 {Also Area of 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 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 PSN (4t 4)(8 t) + (4t + 8 4 t )(8 t) 6 t( + t ) or 6t+ 6t 87_974 GCE Mathematics January 009 5
Q7 (a) Use 4 a ( ) 0 a, Determinant: ( 4) ( ) 0 ( ) 4 B 0 (c) 4 6 4( 6) + ( + ), 0 + 0 ( 6) + ( + ) + Lies on y x + Alternatives: (c) x x ( x+ ),, 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
Q8 (a) f () 5 + 8 + 6, (which is divisible by 4). ( True 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 True for n + if true for n. True for n, true for all n. B Mars M A M A Aft Acso (7) n + For n, n n n ( True for n.) B (a) Alternative for nd M: Part Alternative + + + + ( + ) + ( + ) ( + ) ( + ) True for n + if true for n. True for n, true for all n M A A M A A cso 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) (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 (7) [4] 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 + + 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 009 54