Mark Scheme (Results) June GCE Further Pure FP1 (6667) Paper 1

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1 Mark Scheme (Results) June 0 GCE Further Pure FP (6667) Paper

2 Edexcel is one of the leading examining 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, Edexcel 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 or visit our website at If you have any subject specific questions about the content of this Mark Scheme that require the help of a subject specialist, you may find our Ask The Expert service helpful. Ask The Expert can be accessed online at the following link: June 0 Publications Code UA07965 All the material in this publication is copyright Edexcel Ltd 0

3 EDEXCEL GCE MATHEMATICS General Instructions for Marking. The total number of marks for the paper is 75.. The Edexcel Mathematics mark schemes use the following types of marks: M marks: method marks are awarded for knowing a method and attempting to apply it, unless otherwise indicated. A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned. B marks are unconditional accuracy marks (independent of M marks) Marks should not be subdivided.. Abbreviations These are some of the traditional marking abbreviations that will appear in the mark schemes and can be used if you are using the annotation facility on epen. bod benefit of doubt ft follow through the symbol will be used for correct ft cao correct answer only cso - correct solution only. There must be no errors in this part of the question to obtain this mark isw ignore subsequent working awrt answers which round to SC: special case oe or equivalent (and appropriate) dep dependent indep independent dp decimal places sf significant figures The answer is printed on the paper The second mark is dependent on gaining the first mark

4 June Further Pure Mathematics FP Mark Scheme Question x. f( x) = + x 7 (a) f() = f() = 8 Sign change (positive, negative) (and f( x) is continuous) therefore (a root) α is between x = and x =. Either any one of f() = - or f() = 8. Both values correct, sign change and conclusion M A () = f(.5) = awrt.7 (or truncated to.6) B (b) f (.5) { α.5} f (.5) = α.5 Attempt to find f (.5). f(.5) = awrt 0.7 with α.5 or < α <.5 or [,.5] or (,.5 ). or equivalent in words. In (b) there is no credit for linear interpolation and a correct answer with no working scores no marks. M A () 5 GCE Further Pure Mathematics FP (6667) June 0 FP 6667/0 June 0

5 . (a) z = ( ) + = 5 = or awrt.4 B (b) arg z = π tan ( ) tan ( ) or ( ) tan or cos ( ) 5 sin ( ) or sin ( ) or cos ( ) or M () (c) z = =.68 ( dp) awrt.68 A oe Can work in degrees for the method mark (arg z = o ) () - arg z = tan =-0.46 on itsown is M0 0z + 8 = 0 but π + tan ( - ) - ( - ) ( ) ( ) =.68scores MA π-tan =is M0 as is π-tan (.60) ± 00 4()(8) z = () 0 ± 00 = 0 ± = An attempt to use the quadratic formula (usual rules) M (d) 0 ± i = So, 5 i Attempt to simplify their in terms of i. E.g. i or i 4 If their b -4ac >0 then only the first M is available. p = 5, q = 5 ± i A oe z = ±. { } Correct answers with no working scores full marks. See appendix for alternative solution by completing the square y Note that the points are (, ), ( 5, ) and ( ) 5,. M () x The point (,) plotted correctly on the Argand diagram with/without label. The distinct points z and z plotted correctly and symmetrically about the x- axis on the Argand diagram with/without label. B B The points must be correctly placed relative to each other. If you are in doubt about awarding the marks then consult your team leader or use review. NB the second B mark in (d) depends on having obtained complex numbers in (c) () 8 GCE Further Pure Mathematics FP (6667) June 0 4 FP 6667/0 June 0

6 y=reflection; -B; Question. (a) (i) Ö A = Ö Ö Ö A = Ö Ö + Ö Ö = Ö Ö + A correct method to multiply out two matrices. Can be implied by two out of four correct elements. M 0 = 0 Correct answer A () (ii) Enlargement; scale factor, centre (0, 0). Enlargement; B; scale factor, centre (0, 0) B Allow from or about for centre and O or origin for (0, 0) () (b) 0 B = 0 Reflection; in the line y = x. Allow in the axis about the line y = -x etc. The question does not specify a single transformation so we would need to accept any combinations that are correct e.g. Anticlockwise rotation of 90 o about the origin followed by a reflection in the x-axis is acceptable. In cases like these, the combination has to be completely correct and scored as B (no part marks). If in doubt consult your Team Leader. xb () (c) k + C =, k is a constant. k 9 C is singular det C = 0. (Can be implied) det C = 0 B =0Special Case 9kkB(implied)M0A0 ( )9( k+ ) k = 0 Applies 9( k+ ) k M 9k+ 9 = k 9 = k ( ) +-k = k = A k = with no working can score full marks () GCE Further Pure Mathematics FP (6667) June FP 6667/0 June 0

7 4. (a) f( x) = x + x, x 0 x 5 5 = + f( x) x x x 5 = + { } f( x) x x 0 5 f( x) = x x At least two of the four terms differentiated correctly. Correct differentiation. (Allow any correct unsimplified form) M A () (b) 5 7 f (0.8) = (0.8) ( = 0.65) = (0.8) f (0.8) = = 60 A correct numerical expression for f(0.8) Attempt to insert x = 0.8 into their f (x). Does not require an evaluation. (If f(0.8) is incorrect for their derivative and there is no working score M0) B M α "0.65" = 0.8 " 5.065" Correct application of Newton-Raphson using their values. Does not require an evaluation. M = = (dp) A cao A correct answer only with no working scores no marks. N-R must be seen. (4) Ignore any further applications of N-R x 5 x is quite common and leads to f (0.8) = -.55 and a final answer of This would normally score MA0BMMA0 (4/6) Similarly for a derivative of x 0x where the corresponding values are f (0.8) = and answer A derivative of ( ) GCE Further Pure Mathematics FP (6667) June 0 6 FP 6667/0 June 0

8 5. (a) 4 a A =, where a and b are constants. b 4 A = a 4 Using the information in the question to form Therefore, = the matrix equation. Can be implied by both b 6 8 correct equations below. Do not allow this mark for other incorrect statements unless interpreted correctly later 4 4 a e.g. = 6 b 8 would be M0 unless followed by correct equations or 6 + 6a = 4b 8 So, 6 + 6a = and 4b = 8 Any one correct equation. Allow 6 + 6a = 4b 8 Any correct horizontal line M M giving a = and b =. Any one of a = or b =. A Both a = and b =. A (4) (b) det A = 8 ()() = 5 Finds determinant by applying 8 their ab. M det A = 5 A Special case: The equations b = and 4a = -8 give a = and b =. This comes from incorrect matrix multiplication. This will score nothing in (a) but allow all the marks in (b). Note that det A = 8 ab scores M0 here but the following marks are available. However, 0 beware det A = = area S = = 50 8 ab 5 5 This scores M0A0 MA0 Area S = (det A )(Area R) 0 Area S = 5 0 = 50 (units) their det A or 0 ( their det A ) M 50 or ft answer A If their det A < 0 then allow ft provided final answer > 0 (4) In (b) Candidates may take a more laborious route for the area scale factor and find the area of the unit square, for example, after the transformation represented by A. This needs to be a complete method to score any marks. Correctly establishing the area scale factor M. Correct answer 5 A. Then mark as original scheme. 8 GCE Further Pure Mathematics FP (6667) June 0 7 FP 6667/0 June 0

9 6. z + iz = + i ( x + iy ) + i( x iy ) z = x i y B Substituting z = x + i y and their z * into z + i z M x + iy + ix + y = + i Correct equation in x and y with i = -. Can be implied. A ( x y ) ( y x) + + i + = + i Re part : x + y = Im part : y + x = x + 9y = x + y = 8y = 6 y = An attempt to equate real and imaginary parts. Correct equations. Attempt to solve simultaneous equations to find one of x or y. At least one of the equations must contain both x and y terms. M A M x + y = x 6 = x = 5 Both x = 5 and y =. A { z = 5 i} (7) 7 GCE Further Pure Mathematics FP (6667) June 0 8 FP 6667/0 June 0

10 n r = S = (r ) 7. { } n (a) n 4r 4r = + r = = 4. nn ( + )(n + ) 4. nn ( + ) + n 6 Multiplying out brackets and an attempt to use at least one of the two standard M formulae correctly. First two terms correct. A + n B = nn ( )(n ) nn ( ) n { } = n n + n + n + + ( )( ) 6( ) { } = n n + n+ n + + ( ) 6( ) { } = n n + n+ n ( n ) = n 4 Attempt to factorise out n Correct expression with n factorised out with no errors seen. M A (b) = n n+ n Correct proof. No errors seen. A * ( )( ) n (r ) = Sn Sn r = n + Note that there are no marks for proof by induction. (6) Use of Sn Sn or Sn Sn+ with the =. n(6n+ )(6n ) n(n+ )(n ) result from (a) used at least once. Correct unsimplified expression. E.g. Allow (n) for 6n. Note that (b) says hence so they have to be using the result from (a) = n(6n ) n(4n ) M A = + Factorising out n ( or n ) dm n(08n 4n ) = n(04n ) = n(5n ) { a = 5, b = } n n A (5 ) (4) 0 GCE Further Pure Mathematics FP (6667) June 0 9 FP 6667/0 June 0

11 8. C: y 48x = with general point P( t t ), 4. (a) y = 4ax a = = Using 48 4 y = 4ax to find a. M So, directrix has the equation x + = 0 x + = 0 A oe Correct answer with no working allow full marks () (b) ( ) dy y = 48 x = 48 x = x dx dy or (implicitly) y = 48x y = 48 dx dy dy dt or (chain rule) = = 4 dx dt dx 4 t dy =± kx dx dy ky c dx = dy their dt dx their dt M dy 48 4 When x = t, = = = dx t t t dy or = = = dx y 48t t dy = A dx t t T: y 4t = ( x t ) Applies y 4t = their mt ( x t ) or ( their ) y = m x+ c using T x = t and y = 4t in an attempt to find c. Their m T must be a function of t. M T: ty 4t = x t T: x ty + t = 0 Correct solution. A cso * Special case: If the gradient is quoted as /t, this can score M0A0MA (4), 4, gives t = t = B (c) Compare ( ) P t t with ( ) + = 0 with x = and y = gives ( ). NB x ty t 4t 4t+ = 0= t t = t = into T gives x y + = 0 Substitutes their t into T. M See Appendix for an alternative approach to find the tangent At X, x = y + = 0 Substitutes their x from (a) into T. M So, 9 = y y = 8 So the coordinates of X are (, 8 ). (, 8) A The coordinates must be together at the end for the final A e.g. as above or x = -, y = -8 (4) 0 GCE Further Pure Mathematics FP (6667) June 0 0 FP 6667/0 June 0

12 9. (a) 0 0 n = ; LHS = = RHS = = ( ) 6 As LHS = RHS, the matrix result is true for n =. Check to see that the result is true for n =. B Assume that the matrix equation is true for n = k, ie. k k 0 0 = k 6 ( ) With n = k+ the matrix equation becomes k = k k k = or k k ( ) 6 6 ( ) k 0 k ( ) by 0 6 M k+ k = or k k k 9( ) ( ) 0 + Correct unsimplified matrix with no errors seen. A k + 0 = k 9( ) k + 0 = k () ( ) k + 0 = k + ( ) If the result is true for n = k, ()then it is now true for n = k+. () As the result has shown to be true for n =,() then the result is true for all n. (4) All 4 aspects need to be mentioned at some point for the last A. Manipulates so that k k+ on at least one term. Correct result with no errors seen with some working between this and the previous A Correct conclusion with all previous marks earned dm A A cso (6) GCE Further Pure Mathematics FP (6667) June 0 FP 6667/0 June 0

13 9. (b) f() = 7 + 5= 7+ 5=, Shows that f () =. B {which is divisible by }. { f( n) is divisible by when n =. } Assume that for n= k, k f( k) = + is divisible by for k. So, ( k + ) f( k + ) = Correct unsimplified expression for f( k + ). B giving, k + f( k + ) = k+ k ( ) ( ) f( k + ) f( k) = Applies f( k + ) f( k). No simplification is necessary and condone missing brackets. M = 7 7 k+ k ( ) k = 7 7 Attempting to isolate 48( 7 k = ) ( ) 7 k M 48 7 k Acso k ( ) k and ( ) f( k+ ) = f( ) k, which is divisible by as both f( ) 48 7 k are both divisible by.() If the result is true for n = k, () then it is now true for n = k+. () As the result has shown to be true for n =,(4) then the result is true for all n. (5). All 5 aspects need to be mentioned at some point for the last A. Correct conclusion with no incorrect work. Don t condone missing brackets. There are other ways of proving this by induction. See appendix for alternatives. If you are in any doubt consult your team leader and/or use the review system. A cso (6) GCE Further Pure Mathematics FP (6667) June 0 FP 6667/0 June 0

14 Appendix Question Aliter. (c) Way z 0z + 8 = 0 ( z 5) 5+ 8 = 0 ( ) z ± 5 ± 5+ 8= 0 M ( z ) 5 = z 5 = z 5 = i Attempt to express their in terms of i. M z = ±. { p 5, q } So, 5 i = = 5 ± i A oe () Question Aliter. (c) Way z Scheme 0z + 8 = 0 ( ( )) ( ) ( ) z p+ i q z p i q = z pz+ p + q Marks p =± 0 and p ± q = 8 Uses sum and product of roots M p=± 0 p= 5 Attempt to solve for p(or q) M p= 5 and q= A () GCE Further Pure Mathematics FP (6667) June 0 FP 6667/0 June 0

15 Aliter 8. (c) Way dy = x = = B dx Gives y = ( x ) Uses (, ) and their to find the equation of the tangent. M ( ) x= y = Substitutes their x from (a) into their tangent M y = 8 So the coordinates of X are (, 8 ). A (4) Question Aliter 9. (b) f() = 7 + 5= 7+ 5=, Shows that f () =. B Way {which is divisible by }. { f( n) is divisible by when n =. } Assume that for n= k, k f( k) = + is divisible by for k. So, + = + Correct expression for f( k + ). B ( k + ) f( k ) 7 5 k + giving, f( k + ) = k+ k 7 + 5= Attempt to isolate 7 k- M k = 49 ( 7 + 5) 40 M Attempt to isolate 7 k- + 5 M f( k + ) = 49 f( k) 40 Correct expression in terms of f(k) A As both f(k) and 40 are divisible by then so is f(k + ). If the result is true for n = k, then it is now true for n = k+. As the result has shown to be true for n =, then the result is true for all n. Correct conclusion A (6) GCE Further Pure Mathematics FP (6667) June 0 4 FP 6667/0 June 0

16 Aliter 9. (b) f() = 7 + 5= 7+ 5=, Shows that f () =. B Way {which is divisible by }. { f( n) is divisible by when n =. } Assume that for n= k, f(k) is divisible by k sof ( k) = = m So, + = + Correct expression for f( k + ). B ( k + ) f( k ) 7 5 k + giving, f( k + ) = k 7 k k + 5= = Attempt to isolate 7 k- M = 49 ( m 5) + 5 Substitute for m M f ( k + ) = 49 m 40 Correct expression in terms of m A As both 49 m and 40 are divisible by then so is f(k + ). If the result is true for n = k, then it is now true for n = k+. As the result has shown to be true for n =, then the result is true for all n. Correct conclusion A (6) GCE Further Pure Mathematics FP (6667) June 0 5 FP 6667/0 June 0

17 Aliter 9. (b) f() = 7 + 5= 7+ 5=, Shows that f () =. B Way 4 {which is divisible by }. { f( n) is divisible by when n =. } Assume that for n= k, k f( k) = + is divisible by for k. + + = Correct expression for f( k + ). B ( k+ ) k f( k ) 5f( k) 7 5 5(7 5) k+ k f( k + ) + 5f( k) = (7 + 5) giving, k k Add appropriate multiple of f(k) For 7 k this is likely to be 5 (9, 0,.) For 7 k- (, 5, 47,..) Attempt to isolate 7 k M k ( ) = k = Correct expression A M k k ( ) and ( 7 5) f( + ) = f( k). As both f(k) k + are divisible by then so is f(k + ). If the result is true for n = k, then it is now true for n = k+. As the result has shown to be true for n =, then the result is true for all n. Correct conclusion A (6) GCE Further Pure Mathematics FP (6667) June 0 6 FP 6667/0 June 0

18 Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG8 4FN Telephone Fax Order Code UA07965 June 0 For more information on Edexcel qualifications, please visit Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Essex CM0 JE

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