PMT. Mark Scheme (Results) June GCE Further Pure Mathematics FP1 (6667/01) Original Paper

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1 Mark (Results) June 0 GCE Further Pure Mathematics FP (6667/0) Original Paper

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3 General Marking Guidance All candidates must receive the same treatment. Eaminers must mark the first candidate in eactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. Eaminers should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Eaminers should always award full marks if deserved, i.e. if the answer matches the mark scheme. Eaminers should also be prepared to award zero marks if the candidate s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and eemplification may be limited. When eaminers are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response.

4 (a) det M = a( a) (b) det M = 0 M a a+ = 0 ( a ) = 0 M a = A (a) M for ad bc (b) First M for their det M = 0 Second M for attempt to solve their term quadratic Method mark for solving term quadratic:. Factorisation ( + b + c) = ( + p)( + q), where pq = = c, leading to () () [5] ( a + b + c) = ( m + p)( n + q), where pq = c and mn =,leading to = a. Formula Attempt to use correct formula (with values for a, b and c).. Completing the square + b + c = 0 b Solving : ( ) leading to = ± ± q± c, q 0,

5 z = i is also a root B z i z i = z + z+ 5 ( ( ))( ( )) ( z )( z z 5) = M z = A (5) [5] Alternative f(-)=0 so z = is also a root z+ z + z+ 5 = 0 ( )( ) ( z ( i ))( z ( i )) = 0 z = i is also a root B First M for epanding their ( z α )( z β ) Second M for inspection or long division.

6 (a) z = + i π r = =, tanθ = so θ =, both r values z = + i r = + =, tanθ = so θ = 5π 6 (4) (b) zz = z z= () (c) Im M Aft Re () [8] (a) First M for use of Pythagoras, A for r = and π 5π Second M for use of tan or tan, A for θ = and 6 (b) M for their rr (c) M for either of their numbers plotted correctly on a single diagram. A for both their numbers plotted correctly on a single diagram

7 4(a) y = or y = dy y 0 d dy y dy = or = d d Gradient of normal is y or y = ( ) M M 8 y = + A (5) 4(b) At R, y = M 9 8 = + 8 9= 0 A ( + 9)( ) = 0 M = 9, y = A,A (5) [0] (a) (b) First M: Use of the product rule: The sum of two terms including dy/d, one of which is correct or dy = k d First A for correct derivative y or Second M for use of Perpendicular gradient rule mnm T = Third M for y = their mn ( ) or y= m+ c with their mn and (,) in an attempt tofind' c'. First M for substituting y = in their normal. First A for correct term quadratic Second M for attempt to solve their term quadratic

8 5 f()= + 7 = 6 = 8 B True for n = Assume true for n= k, k f( k) = + 7= 8p where p is a positive integer When n= k+ k + k f( k+ ) f( k) = M ( ) ( ) 9 k = dm = 8 k A k f( k+ ) = 8( + p) = 8q where q is a positive integer True for n= k+ dm True for n =, if true for n= kthen true for n= k+ Acso So n + 7 divisible by 8 for all n by Induction. (6) [6] B for f()= + 7 = 6 seen First M for substituting into f( k + ) f( k) or showing k f( k + ) = Second M for using f( k + ) f( k) or equivalent First A for f( k+ ) = f( k) + 8 k or equivalent. Third M for showing divisible by 8. Accept f( k) divisible by 8 and 8 k divisible by 8. Second A for conclusion with all 4 underlined elements that can be seen anywhere in the solution

9 6(a) y = 4 dy y = 4 d At P, d y = d p y p= p p ( ) A (5) 6(b)(i) At (-,) M p = ( p ) p p p = 0 A p = ± M p = +, q = A (4) 6(b)(ii) PR = + 6, QR = 6 Area of PQR = PR. QR = 8 (4) [] (a) (b)i (b)ii First M for or p or t. First A for accurate differentiation dy ky c d = or d y ; can be a function of dt d dt Second M applies y p their m( p ) ( their ) = or y = m + c using = p and y = p in an attempt to find c. Their m must be a function of p from calculus. First M substitute coordinates of the point R into their tangent Second M for solving term quadratic Second A for ± seen First M for attempt to find distance between their P and R or Q and R using formula or sketch and Pythagoras. Second M for using bh on their PQR Second A accept awrt.

10 7(a) n n n r= r= r= r ( r ) = r r M n ( n+ ) n( n+ )(n+ ) = 4 6 nn ( + = ) n n+ n+ ( ( ) ( )) A M A nn ( + )(n n ) = nn ( + )(n+ )( n ) = Acso (5) 7(b) r= r= 9 r ( r ) = r ( r ) r ( r ) M r= 0 r= r= ( ) ( ) = A = = A () [8] (a) (b) First M for epanding brackets First A for correct epressions for r and r Second M for factorising by nn+ ( ) Second A for (n n ) or equivalent factor First M for f(49 or 50) f(9 or 0) and attempt to use part (a).

11 8(a) (f () = ) 4( < 0) -4 B (f () = )( > 0) B Changes sign so root (in [,]) B () 8(b) a f( a ) b f( b ) a+ b a+ b f Interval is [.75,] BM A () 8(c) f( ) =.8.8 =.8.8 =.90 to sf. A (5) [] (b) B for awrt -.6 M for attempt to find f (.75) (c) A for f (.75) = awrt -. with.75 α or.75< α < or [.75, ] or (.75, ). First M for at least one of the two terms differentiated correctly. First A for correct derivative Second M for correct application of Newton-Raphson using their values. Second A for f(.8)= Third A for.90 cao

12 9(a) 0 = 5 () (b) det A = 7 0 so A is non-singular () (c) A = 7 () (d) k ( k ) ( k) = 7 k 7 ( k ) + ( k) M ( p lies on y = 4 ) 7 k = 4 7 k A,A () [9] (d) Alt k = then multiply out and attempt to solve y k simultaneous equations for or y in terms of k. M = k A k y = A

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