AQA Further pure maths 1

Transcription

1 AQA Further pure maths Name:... Tutor group:...

2 Key dates Further pure eam : 7th May 0 am Term dates: Term : Thursday September 0 - Friday October 0 (7 teaching days) Term 4: Monday 0 February 0 - Friday 0 March 0 (0 teaching days) Term : Monday October 0 Friday 6 December 0 (5 teaching days) Term 5: Monday 6 April 0 Friday June 0 (4 teaching days) Term : Tuesday January 0- Friday 0 February 0 (9 teaching days) Term 6: Monday June 0 Thursday 9 July 0 (9 teaching days)

3 Scheme of Assessment Further Mathematics Advanced Subsidiary (AS) Advanced Level (AS + A) Candidates for AS and/or A Level Further Mathematics are epected to have already obtained (or to be obtaining concurrently) an AS and/or A Level award in Mathematics. The Advanced Subsidiary (AS) award comprises three units chosen from the full suite of units in this specification, ecept that the Core units cannot be included. One unit must be chosen from MFP, MFP, MFP and MFP4. All three units can be at AS standard; for eample, MFP, MMB and MSA could be chosen. All three units can be in Pure Mathematics; for eample, MFP, MFP and MFP4 could be chosen. The Advanced (A Level) award comprises si units chosen from the full suite of units in this specification, ecept that the Core units cannot be included. The si units must include at least two units from MFP, MFP, MFP and MFP4. All four of these units could be chosen. At least three of the si units counted towards A Level Further Mathematics must be at A standard. All the units count for /% of the total AS marks 6/% of the total A level marks Further Pure All questions are compulsory. A graphics calculator may be used Written Paper hour 0 minutes 75 marks Grading System The AS qualifications will be graded on a five-point scale: A, B, C, D and E. The full A level qualifications will be graded on a si-point scale: A, A, B, C, D and E. To be awarded an A in Further Mathematics, candidates will need to achieve grade A on the full A level qualification and 90% of the maimum uniform mark on the aggregate of the best three of the A units which contributed towards Further Mathematics. For all qualifications, candidates who fail to reach the minimum standard for grade E will be recorded as U (unclassified) and will not receive a qualification certificate. Further pure subject content Algebra and Graphs Comple Numbers Roots and Coefficients of a quadratic equation Series Calculus Numerical Methods Trigonometry Matrices and Transformations

4 Further pure specifications Candidates will be epected to be familiar with the knowledge, skills and understanding implicit in the modules Core and Core. Candidates will also be epected to know for section 6 that the roots of an equation f()=0 can be located by considering changes of sign of f() in an interval of in which f() is continuous. Candidates may use relevant formulae included in the formulae booklet without proof. ALGEBRA AND GRAPHS Graphs of rational functions of the form a + b a + b, or c + d c + d + e + a + b + c + d Graphs of parabolas, ellipses and hyperbolas with equations y y = 4 a, + =, a b y = and y = c a b COMPLEX NUMBERS Non-real roots of quadratic equations. Sum, difference and product of comple numbers in the form +iy. Comparing real and imaginary parts. Sketching the graphs. Finding the equations of the asymptotes which will always be parallel to the coordinate aes. Finding points of intersection with the coordinate aes or other straight lines. Solving associated inequalities. Using quadratic theory (not calculus) to find the possible values of the function and the coordinates of the maimum or minimum points on the graph. E.g. + for y =, y = k + = k 4 k, 4 which has real roots if 6k + 8k 8 0, i.. e if k or k ; Stationary points are(, ) and (, ) Sketching the graphs. Finding points of intersection with the coordinate aes or other straight lines. Candidates will be epected to interpret the geometrical implication of equal roots, distinct real roots or no real roots. Knowledge of the effects on these equations of single transformations of these graphs involving translations, stretches parallel to the - or y-aes, and reflections in the line y =. Including the use of the equations of the asymptotes of the hyperbolas given in the formulae booklet. Comple conjugates - awareness that non-real roots of quadratic equations with real coefficients occur in conjugate pairs. Including solving equations E.g. Solving z + z = + i where z is the conjugate of z

5 ROOTS AND COEFFICENTS OF A QUADRATIC EQUATION Manipulating epressions E.g. α + β = ( α + β) αβ( α + β) involving α + β and αβ Forming an equation with roots α, β or, or α +, β + etc. α β β α SERIES Use of formulae for the sum of the squares and the sum of the cubes of the natural numbers. CALCULUS Finding the gradient of the tangent to a curve at a point, by taking the limit as h tends to zero of the gradient of a chord joining two points whose -coordinates differ by h Evaluation of simple improper integrals. NUMERICAL METHODS Finding roots of equations by interval bisection, linear interpolation and the Newton-Raphson method. Solving differential equations of the dy form f( ) d = Reducing a relation to a linear law. TRIGONOMETRY General solutions of trigonometric equations including use of eact values for the sine, cosine and π π π tangent of,, 6 4 MATRICES and matrices; addition and subtraction, multiplication by a scalar. Multiplying a matri by a matri or by a matri. The identity matri I for a matri. Transformations of points in the -y plane represented by matrices. E.g. Find a polynomial epression for r ( r + ) or ( r r + ) The equation will be given as y = f( ), E.g. where f( )is a simple polynomial 4 such as or + E.g. 4 d, d 0 4 Graphical illustration of these methods. n r= r= Using a step-by-step method based on the linear approimations y y + hf ( ); = + h, with given values for, y and h n+ n n n+ n 0 0 E.g. ; ; n + = k y = a + b y = a ; y = ab y Use of logarithms to base 0 where appropriate. Given numerical values of (,y), drawing a linear graph and using it to estimate the values of the unknown constants. π π Sin =, Cos, E.g. + = Tan = 6 Sin = 0., Cos( ) = 0. Transformations will be restricted to rotations about the origin, reflections in a line through the origin, stretches parallel to the - and y- aes, and enlargements with centre the origin. Use of the standard transformation matrices given in the formulae booklet. Combinations of these transformations n

6 The formulae booklet

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12 Content Algebra and Graphs Comple Numbers Roots and Coefficients of a quadratic equation Series Calculus Numerical Methods Trigonometry Matrices and Transformations

13 ALGEBRA AND GRAPHS

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15 Conics

16 COMPLEX NUMBERS

17 Roots and Coefficients of a quadratic equation Series

18 Calculus

19 Numerical methods

20 Linear laws

21 Trigonometry

22 Matrices and transformations

23 Matri transformations

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25 General Certificate of Education January 006 Advanced Subsidiary Eamination MATHEMATICS Unit Further Pure MFP Monday January pm to.00 pm For this paper you must have: an 8-page answer book the blue AQA booklet of formulae and statistical tables an insert for use in Question 6 (enclosed) You may use a graphics calculator. Time allowed: hour 0 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Eamining Body for this paper is AQA. The Paper Reference is MFP. Answer all questions. All necessary working should be shown; otherwise marks for method may be lost. Fill in the boes at the top of the insert. Information The maimum mark for this paper is 75. The marks for questions are shown in brackets. Advice Unless stated otherwise, formulae may be quoted, without proof, from the booklet. P80785/Jan06/MFP 6/6/6/ MFP

26 Answer all questions. (a) Show that the equation ˆ 0 has a root between 0.5 and. ( marks) (b) Use linear interpolation once to find an estimate of this root. Give your answer to two decimal places. ( marks) (a) For each of the following improper integrals, find the value of the integral or eplain briefly why it does not have a value: (i) 9 0 p d ; ( marks) (ii) 9 0 p d. ( marks) (b) Eplain briefly why the integrals in part (a) are improper integrals. ( mark) Find the general solution, in degrees, for the equation sin 4 0 ˆsin 50 (5 marks) 4 A curve has equation y ˆ 6 (a) Write down the equations of the two asymptotes to the curve. ( marks) (b) Sketch the curve and the two asymptotes. (4 marks) (c) Solve the inequality 6 < (4 marks) P80785/Jan06/MFP

27 5 (a) (i) p p Calculate i 5 5 i. (ii) p Hence verify that 5 i is a root of the equation p i 5 z ˆ z where z is the conjugate of z. ( marks) ( marks) (b) The quadratic equation p q ˆ 0 in which the coefficients p and q are real, has a comple root p 5 i. (i) Write down the other root of the equation. ( mark) (ii) Find the sum and product of the two roots of the equation. ( marks) (iii) Hence state the values of p and q. ( marks) 6 [Figure and Figure, printed on the insert, are provided for use in this question.] The variables and y are known to be related by an equation of the form y ˆ k n where k and n are constants. Eperimental evidence has provided the following approimate values: y (a) Complete the table in Figure, showing values of X and Y, where X ˆ log 0 and Y ˆ log 0 y Give each value to two decimal places. ( marks) (b) Show that if y ˆ k n, then X and Y must satisfy an equation of the form Y ˆ ax b ( marks) (c) Draw on Figure a linear graph relating X and Y. ( marks) (d) Find an estimate for the value of n. ( marks) P80785/Jan06/MFP Turn over for the net question Turn over s

28 4 7 (a) The transformation T is defined by the matri A, where A ˆ 0 0 (i) Describe the transformation T geometrically. ( marks) (ii) Calculate the matri product A. ( marks) (iii) Eplain briefly why the transformation T followed by T is the identity transformation. ( mark) (b) The matri B is defined by B ˆ 0 (i) Calculate B A. ( marks) (ii) Calculate B A B A. ( marks) 8 A curve has equation y ˆ. (a) Sketch the curve. ( marks) (b) (i) The curve is translated by units in the positive y direction. Write down the equation of the curve after this translation. ( marks) (ii) The original curve is reflected in the line y ˆ. Write down the equation of the curve after this reflection. ( mark) (c) (i) Show that if the straight line y ˆ c, where c is a constant, intersects the curve y ˆ, then the -coordinates of the points of intersection satisfy the equation c c ˆ 0 ( marks) (ii) (iii) (iv) Hence find the value of c for which the straight line is a tangent to the curve. ( marks) Using this value of c, find the coordinates of the point where the line touches the curve. ( marks) In the case where c ˆ 4, determine whether the line intersects the curve or not. ( marks) END OF QUESTIONS Copyright Ó 006 AQA and its licensors. All rights reserved. P80785/Jan06/MFP

29 AQA GCE Mark Scheme, 006 January series MFP MFP Q Solution Marks Totals Comments (a) f(0.5) = 0.875, f() = B Change of sign, so root between E (b) Complete line interpolation method M, M for partially correct method Estimated root = A Allow 5 as answer (a)(i) d= (+ c) Total 5 MA M for k 9 0 d = 6 A ft wrong coeff of (ii) d= (+ c) MA M for as 0, so no value E Tending to infinity clearly implied (b) Denominator 0 as 0 E Total 7 One solution is = 0 B PI by general formula Use of sin 0º = sin 50º M OE Second solution is = 0 A OE Introduction of 90nº, or 60nº or 80nº M Or πn/ or πn or πn GS ( n ),(0 + 90n) A 5 OE; ft one numerical error or omission of nd soln Total 5 4(a) Asymptotes =, y = 6 BB (b) Curve (correct general shape) M SC Only one branch: Curve passing through origin A B for origin Both branches approaching = A B for approaching both asymptotes Both branches approaching y = 6 A 4 ( Ma /4) (c) Correct method M Critical values ± BB From graph or calculation Solution set < < A 4 ft one error in CVs; NMS 4/4 after a good graph Total 0 5(a)(i) Full epansion of product M Use of i = m ( + 5i)( 5 i) = 5 + i A 5 5 = 5 must be used Accept not fully simplified (ii) z = iy (= 5 + i) M k Hence result A Convincingly shown (AG) (b)(i) Other root is 5 + i B (ii) Sum of roots is 5 B Product is 6 MA (iii) p = 5, q = 6 B B ft wrong answers in (ii) Total

30 MFP AQA GCE Mark Scheme, 006 January series MFP Q Solution Marks Totals Comments 6(a) X values.,.8 Y values 0.70,.48 B,, for each error (b) lg y = lg k + lg n M lg n = n lg M So Y = nx + lg k A (c) Four points plotted B, B if one error here; ft wrong values in (a) Good straight line drawn B ft incorrect points (appro collinear) (d) Method for gradient M Estimate for n A Allow AWRT ; ft grad of candidate's graph Total 7(a)(i) Reflection... M... in y = A OE (ii) 0 MA MA0 for three correct entries A 0 (iii) A = I or geometrical reasoning E (b)(i) MA MA0 for three correct entries B = 0 B A 0 = 0 0 A ft errors, dependent on both M marks (ii) 0 (B + A)(B A) = B M... = A 0 ft one error; MA0 for three correct (ft) entries Total 8(a) Good attempt at sketch M Correct at origin A (b)(i) y replaced by y B Equation is (y ) = B ft y + for y (ii) Equation is = y B (c)(i) ( + c) = + c + c B... = M Hence result A convincingly shown (AG) (ii) Tangent if (c ) 4c = 0 M ie if 48c + 44 = 0 so c = A (iii) = 0 M =, y = 6 A (iv) c = 4 discriminant = 48 < 0 MA OE So line does not intersect curve A Total 5 TOTAL 75 4

31 Further pure - AQA - January 006 Question ) Call f ( ) = + B a ) f(0.5) = = 0.875< 0 f () = + = > 0 According to the sign change rule, 0.5 we can say that there is at least a root of f in the interval [0.5,] b) Let's call A(0.5,f(0.5)) and B(,f()) f() f(0.5) 5 The gradient of the line AB is = o 5 An equation of the line AB is : y = ( ) 4 This line croos the ais when y = 0 5 which gives = ( ) = 5 A = ( 0.7) - 5 Question : ) a) i) d = d = + c f ( ) = ' s domain is, 9 9 we can therefore work out d = ( 9 ) ( 0 ) 6 0 = = 0 ii) d = d = = when = 0, is not defined. This integral does not have a value. b)these integrals are improper because the function to integrate are not defined for = 0. Question : Sin(4 + 0) = Sin(50) so = n 4 = n o o = n or = (80 50) + 60n 4 = n o o = n n

32 Question 4: 6 6 a) y =, dividing by gives y = + + when, y 6 because 0 The line with equation y = 6 is asymptote to the curve When, y the line with equation = is a (vertical) asymptote to the curve b) With a table of values ( to plot a couple of " guide"point s) we will obtain: 6 6 c) < We solve = 6 = ( ) = =. On the graph, draw the line y =. 6 From this, we can conclude that the solutions of < areall the values with < < Question 5: ai ))+ i 5 5 i = 5 i+ 5i i 5= 5+ i ( )( ) ii) if z = 5 i then z = 5+ i and z = 5+ i so indeed ( + i 5) z = z bi ) ) If one comple root is 5 i, then the other root is the conjugate 5 + i ii) Sum : 5 + i + 5 = 5 Product:( 5 + i)( 5 i) = 6 iii) therefore q = 6 and p = 5 Question 6: X Y n n b) y = k then log ( y) = log ( k ) 0 0 Y = log k+ nlog ( ) Y = nx + log Y = ax + b c) An estimate of the gradient is n = = with a = n ad n b= log k 0 k

33 Question 7: it ) is the reflection in the line y = 0 ii) A = 0 iii)the reflection in y = followed by the same reflection return the same object you started with. b) ib ) = 0 B = and B A = 0 0 = ii)( B + A)( B A) = = 0 Question 8: a) bi y = ) )( ) ii) = y y = ) ) ( + ) = c i gives c y = + c + c + c = 0 + c + c = E ( ) 0 ( ) ii)the line is tangent to the curve when there is one unique solution to this equation meaning that the discriminant is equal to 0 Discriminant: ( ) = 0 ( ) 0 c c = c c+ c = c + 44 = 0 c = iii) When c =, the equation( E) becomes = so = and y = + c= 6 The coordinates of the point where the line and the curve touches: (,6) ) 4, ( ) iv When c = the equation E becomes + = The discriminant is ( 4) 4 6 = 48 < 0 The line and the curve do not intersect. Grade boundaries

34 General Certificate of Education June 006 Advanced Subsidiary Eamination MATHEMATICS Unit Further Pure MFP Monday June pm to.00 pm For this paper you must have: an 8-page answer book the blue AQA booklet of formulae and statistical tables You may use a graphics calculator. Time allowed: hour 0 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Eamining Body for this paper is AQA. The Paper Reference is MFP. Answer all questions. Show all necessary working; otherwise marks for method may be lost. Information The maimum mark for this paper is 75. The marks for questions are shown in brackets. Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. P859/Jun06/MFP 6/6/6/ MFP

35 Answer all questions. The quadratic equation 6 ˆ 0 has roots a and b. (a)write down the numerical values of a b and ab. (b)(i)epand a b. (ii)show that a b ˆ 4. ( marks) ( mark) ( marks) (c)find a quadratic equation with roots a and b, giving your answer in the form p q r ˆ 0, where p, q and r are integers. ( marks) A curve satisfies the differential equation dy d ˆ log 0 Starting at the point (, )on the curve, use a step-by-step method with a step length of 0. to estimate the value of y at ˆ :4. Give your answer to three decimal places. (6 marks) Show that Xn rˆ r r ˆkn n n where k is a rational number. (4 marks) 4 Find, in radians, the general solution of the equation p cos ˆ giving your answers in terms of p. (5 marks) P859/Jun06/MFP

36 5 The matri M is defined by (a)find the matri: M ˆ 6 4 p p p 7 5 p (i) M ; ( marks) (ii) M 4. ( mark) (b)describe fully the geometrical transformation represented by M. (c)find the matri M 006. ( marks) ( marks) 6 It is given that z ˆ iy, where and y are real numbers. (a)write down, in terms of and y, an epression for z i where z i denotes the comple conjugate of z i. ( marks) (b)solve the equation z i ˆ iz giving your answer in the form a bi. (5 marks) Turn over for the net question P859/Jun06/MFP Turn over s

37 4 7 (a)describe a geometrical transformation by which the hyperbola 4y ˆ can be obtained from the hyperbola y ˆ. ( marks) (b)the diagram shows the hyperbola H with equation y 4 ˆ 0 y O By completing the square, describe a geometrical transformation by which the hyperbola H can be obtained from the hyperbola y ˆ. (4 marks) P859/Jun06/MFP

38 5 8 (a)the function f is defined for all real values of by f ˆ (i)epress f h f in the form ph qh rh where p, q and r are integers. (ii)use your answer to part (a)(i)to find the value of f '. (4 marks) ( marks) (b)the diagram shows the graphs of y ˆ and y ˆ for > 0 y y ˆ P y ˆ O The graphs intersect at the point P. (i)show that the -coordinate of P satisfies the equation f ˆ0, where f is the function defined in part (a). ( mark) (ii)taking ˆ as a first approimation to the root of the equation f ˆ0, use the Newton±Raphson method to find a second approimation to the root. ( marks) (c)the region enclosed by the curve y ˆ, the line ˆ and the -ais is shaded on the diagram. By evaluating an improper integral, find the area of this region. ( marks) P859/Jun06/MFP Turn over s

39 6 9 A curve C has equation y ˆ (a)(i)write down the coordinates of the points where C intersects the -ais. (ii)write down the equations of all the asymptotes of C. ( marks) ( marks) (b)(i)show that, if the line y ˆ k intersects C, then k k (5 marks) (ii)given that there is only one stationary point on C, find the coordinates of this stationary point. No credit will be given for solutions based on differentiation. (c)sketch the curve C. ( marks) ( marks) END OFQUESTIONS P859/Jun06/MFP

40 AQA GCE Mark Scheme, 006 June series MFP MFP Q Solution Marks Total Comments (a) α + β =, αβ = BB SC / for answers 6 and (b)(i) (α + β) = α +α β +αβ +β B Accept unsimplified (ii) α + β = ( α + β ) αβ ( α + β ) M (c) Substitution of numerical values m α + β = 4 A convincingly shown AG α β = B 8 7 Equation of form p ± 4p + r = 0 Answer = 0 A ft wrong value for α β Total 9 st increment is 0. lg... M or 0. lg. or 0. lg A PI =. y.060 A PI; ft numerical error nd increment is 0. lg. m consistent with first one A PI =.4 y A 6 ft numerical error Total 6 Σ(r r) = Σr Σr M At least one linear factor found m Σ(r r) = 6 n(n + )(n + ) m OE M... = n(n + )(n ) A 4 Total 4 4 π cos = stated or used B Condone decimals and/or degrees until 6 final mark Appropriate use of ± B Introduction of nπ M Division by M Of α + knπ or ± α + knπ =± π + 8 A 5 Total 5 5(a)(i) 0 M M if entries correct M = A, MA if entries correct 0 (ii) 0 M 4 = 0 B ft error in M provided no surds in M (b) Rotation (about the origin) M... through 45 clockwise A (c) Awareness of M 8 = I M OE; NMS / 0 m complete valid method M 006 = A ft error in M as above 0 Total 9

41 MFP AQA GCE Mark Scheme, 006 June series MFP (cont) Q Solution Marks Total Comments 6(a) ( z+ i) = iy i B (b)... = i y + M i = used at some stage Equating R and I parts M involving at least 5 terms in all = y +, y = A ft one sign error in (a) z = + i ma 5 ditto; allow =, y = Total 7 7(a) Stretch parallel to y ais... B... scale-factor parallel to y ais B (b) ( ) y = MA Translation in direction... A... units in positive direction A 4 Total 6 8(a)(i) ( + h) = + h + h + h B f ( + h) = + 5h + 4h + h MA PI; ft wrong coefficients f ( + h) f() = 5h + 4h + h A 4 ft numerical errors (ii) Dividing by h M f'() = 5 A ft numerical errors (b)(i) ( + ) =, hence result B convincingly shown (AG) (ii) 4 MA = = 5 5 A ft c's value of f'() (c) Area = d M... = M Ignore limits here... = 0 = A Total 9(a)(i) Intersections at (, 0), (, 0) BB Allow =, = (ii) Asymptotes = 0, =, y = B (b)(i) y = k k k = MA M for clearing denominator... (k ) +( k+)+=0 A ft numerical error = 4(k )(k 4), hence result ma 5 convincingly shown (AG) (ii) y = 4 at SP B 6 + = 0, so = MA A0 if other point(s) given (c) Curve with three branches B approaching vertical asymptotes Middle branch correct B Coordinates of SP not needed Other two branches correct B asymptotes shown Total 6 TOTAL 75 4

42 Question : ) 6 + = 0 has roots α and β 6 a) α + β = = αβ = ( ) Further pure - AQA - June 006 bi )) α + β = α + α β + αβ + β ii ( ) ( ) ( α β) ) α + β = α + β α β αβ = α + β αβ + α + β = = 8 4= 4 ) If = α = β c u and v then ( αβ ) + = 4 = αβ = = u v and u v An equation with roots 0 and u v i = α Question : Euler formula : y = n y + + n hf ( n) = and h = 0. for for =., y + 0. log () = β =.4, y log (. ). 9 Question : = = r= r= r= 6 = nn ( + )[ n + ] = nn ( + )( n ) 6 Question 4: n n n ( r r) r r nn ( )(n ) nn ( ) s : 4+ = = 0 Cos = π π = + kπ or = + kπ 6 6 π π π π = + k or = + k k 8 8

43 Question 5: 0 M= ai )) M = M = 0 π o b) M represents the rotation centre (0,0) angle ( = 45 ) clockwise 4 c Because then M 0 8 ) M represents a rotation with angle 45 is a rotation with a M 8 = I Question 6: a) z = + iy )( ) ( ) M = M = M M = I M = M M M = ( z i) ( iy ( )) b z + i = iz + + = + + = i( y+ ) i( y + ) = i( + iy) + i( y + ) = i y + + i( y) = y + i therefore = y Real parts are equal y = Imaginary parts are equal We solve y = ( y) y = 4y y = y = and = y = = z = + i Question 7: a) 4y = ( y) = This hyperbola can be obtain by a stretch,scale factor, parallel to the y-ais y = of the hyperbola b) y 4+ = 0 4 y + = 0 y + = ( ) 4 0 y = ( ) ngle 45 8=60 This hyperbola can be obtain by a translation of vector of the hyperbola y = 0 o

44 Question 8: a) f( ) = + ( ) ( + ) ( h h h h h ) ( ) i) f( + h) f() = ( h) ( h) = = h h h f( + h) () 5h+ 4h + h ii h h h h so f '() = 5 ) = = h 0 bi ) )The -coordinate of P satisfy the equation = + multiply by : = + ) If is an approimation of the root then = is a better approimation. f '( ) =, f ( ) = f () = and f '() = 5 4 = = = = 0 ii so f( ) = 0 f( ) a c) If it eits, the area is d. d = a = + a Question 9: a) i)the curve C crosses the ais when y = 0 ( + )( ) We solve 0 = which gives ( + )( ) = 0 ( ) = and = The curve crosses the -ais at (,0) and (,0) ii) " Vertical asymptote" = 0 and = (roots of thedenominator) y = = so y = is an asymptote. b)the curve and the line y=k intersect if the equation ( y = ) = k has solutions. k gives k k = = ( k) + (k ) = 0 The discriminant : ( ) 4 ( ) ( a k k ) must be positive or = 0 4k 8k 4 k k k + + k+ 5k+ 4 0 ( k )( k 4) 0

45 ii) If the curve has only one stationary point, the line y = k will intersect the curve at only one point, which means ( k )( k 4) = 0 this gives two values for k : k = or k = 4 Case : k = then the equation ( k) + (k ) = 0becomes = 0 There is no solution Case : k = 4 k + k = becomes + = then the equation ( ) ( ) (+ )( ) so = and y = = 4 ( ) The stationary point has coordinates (,4) c) + = 0 = ( ) 0

46 General Certificate of Education January 007 Advanced Subsidiary Eamination MATHEMATICS Unit Further Pure MFP Friday 6 January pm to.00 pm For this paper you must have: an 8-page answer book the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed: hour 0 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Eamining Body for this paper is AQA. The Paper Reference is MFP. Answer all questions. Show all necessary working; otherwise marks for method may be lost. Information The maimum mark for this paper is 75. The marks for questions are shown in brackets. Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. P89695/Jan07/MFP 6/6/ MFP

47 Answer all questions. (a) Solve the following equations, giving each root in the form a þ bi: (i) þ 6 ¼ 0; ( marks) (ii) þ 7 ¼ 0. ( marks) (b) (i) Epand ð þ Þ. ( marks) (ii) Epress ð þ iþ in the form a þ bi. ( marks) (iii) Hence, or otherwise, verify that ¼ þ i satisfies the equation þ 4i ¼ 0 ( marks) The matrices A and B are given by pffiffi A ¼ 6 4 pffiffiffi pffiffi 7 5, B ¼ 6 4 pffiffi 7 5 (a) Calculate: (i) A þ B ; (ii) BA. ( marks) ( marks) (b) Describe fully the geometrical transformation represented by each of the following matrices: (i) A ; (ii) B ; (iii) BA. ( marks) ( marks) ( marks) P89695/Jan07/MFP

48 The quadratic equation þ 4 þ ¼ 0 has roots a and b. (a) Write down the values of a þ b and ab. ( marks) (b) Show that a þ b ¼. ( marks) (c) Find the value of a 4 þ b 4. ( marks) 4 The variables and y are related by an equation of the form where a and b are constants. y ¼ a b (a) Using logarithms to base 0, reduce the relation y ¼ a b to a linear law connecting log 0 and log 0 y. ( marks) (b) The diagram shows the linear graph that results from plotting log 0 y against log 0. log 0 y O log 0 Find the values of a and b. (4 marks) P89695/Jan07/MFP Turn over s

49 4 5 A curve has equation y ¼ (a) Write down the equations of the three asymptotes to the curve. ( marks) (b) Sketch the curve. (You are given that the curve has no stationary points.) (4 marks) (c) Solve the inequality > 0 ( marks) 6 (a) (i) Epand ðr Þ. ( mark) (ii) Hence show that X n r¼ ðr Þ ¼ nð4n Þ (5 marks) (b) Hence find the sum of the squares of the odd numbers between 00 and 00. (4 marks) 7 The function f is defined for all real numbers by fðþ ¼sin þ p 6 (a) Find the general solution of the equation f ðþ ¼0. ( marks) (b) The quadratic function g is defined for all real numbers by gðþ ¼ pffiffiffi þ 4 It can be shown that gðþ gives a good approimation to fðþ for small values of. (i) Show that gð0:05þ and f ð0:05þ are identical when rounded to four decimal places. ( marks) (ii) A chord joins the points on the curve y ¼ gðþ for which ¼ 0 and ¼ h. Find an epression in terms of h for the gradient of this chord. ( marks) (iii) Using your answer to part (b)(ii), find the value of g 0 ð0þ. ( mark) P89695/Jan07/MFP

50 5 8 A curve C has equation 5 y 9 ¼ (a) (b) (c) Find the y-coordinates of the points on C for which ¼ 0, giving each answer in the p form k ffiffi, where k is an integer. ( marks) Sketch the curve C, indicating the coordinates of any points where the curve intersects the coordinate aes. ( marks) Write down the equation of the tangent to C at the point where C intersects the positive -ais. ( mark) (d) (i) Show that, if the line y ¼ 4 intersects C, the -coordinates of the points of intersection must satisfy the equation 6 00 þ 65 ¼ 0 ( marks) (ii) Solve this equation and hence state the relationship between the line y ¼ 4 and the curve C. ( marks) END OF QUESTIONS P89695/Jan07/MFP

51 MFP - AQA GCE Mark Scheme 007 January series MFP Q Solution Marks Total Comments (a)(i) Roots are ± 4i MA M for one correct root or two correct factors (ii) Roots are ± 4i MA M for correct method (b)(i) ( + ) = MA MA0 if one small error (ii) ( + i) = + i i = + i MA M if i = used (iii) ( + i) + ( + i) 4i M with attempt to evaluate = ( + i ) + ( i) = 0 A convincingly shown (AG) Total 0 0 (a)(i) A+ B = MA MA0 if entries correct; 0 Condone for (ii) 0 BA = 0 B,, Deduct one for each error; SC B, for AB (b)(i) Rotation 0 anticlockwise (abt O) MA M for rotation (ii) " Reflection in y = ( tan5 ) MA M for reflection (iii) Reflection in -ais BF / for reflection in y-ais ft (MA) only for the SC Alt: Answer to (i) followed by answer to (ii) MAF () MA0 if in wrong order or if order not made clear Total (a) α + β =, αβ = BB (b) Use of epansion of ( α + β ) M ( ) α + β = = (c) 4 4 α + β given in terms of α + βαβ, and/or α + β ma convincingly shown (AG); ma0 if α + β = used MA α + β = A OE Total 8 MA0 if num error made 4

52 MFP - AQA GCE Mark Scheme 007 January series MFP (cont) Q Solution Marks Total Comments 4(a) lg y = lg a+ blg MA M for use of one log law (b) Use of above result M a = 0 A b = gradient m OE; PI by answer ± = A 4 Total 6 5(a) Asymptotes y = 0, =, = B (b) Three branches approaching two vertical asymptotes B Asymptotes not necessarily drawn Middle branch passing through O B with no stationary points Curve approaching y = 0 as ± B All correct B 4 with asymptotes shown and curve approaching all asymptotes correctly (c) Critical values =, 0 and B Solution set < < 0, > MA M if one part correct or consistent with c's graph Total 0 r = 4r 4r+ B 6(a)(i) ( ) = M 4 4 = n n+ ma = n B Result convincingly shown A 5 AG (ii) ( ) (b) Sum = f (00) f (50) MA M for 00 ± and 50 ± = A 4 SC f(00) f(5) = : /4 Total 0 5

53 MFP - AQA GCE Mark Scheme 007 January series MFP (cont) Q Solution Marks Total Comments 7(a) Particular solution, eg π or 5π B Degrees or decimals penalised in rd 6 6 mark only Introduction of nπ or nπ M GS = π + nπ AF OE(accept unsimplified); 6 ft incorrect first solution (b)(i) f (0.05) g (0.05) B B either value AWRT both values correct to 4DP (ii) g( h) g(0) = h MA MA0 if num error made h 4 (iii) As h 0 this gives g' (0) = AF AWRT 0.866; ft num error Total 8 y 8(a) = 0 4 = M 9 y = 7 A PI y =± A (b) One branch generally correct B Asymptotes not needed Both branches correct B With implied asymptotes Intersections at (± 5, 0) B (c) Required tangent is = 5 BF ft wrong value in (b) (d)(i) y correctly eliminated M Fractions correctly cleared m = 0 A convincingly shown (AG) (ii) = 5 4 B No need to mention repeated root, but B0 if other values given as well Equal roots tangency E Accept 'It's a tangent' Total TOTAL 75 6

54 Further pure - AQA - January 007 Question : ai ) ) + 6 = 0 = 6 ) 7 0 Discriminant: ( ) (8 ) + = bi ) ) ( ) ) ( ) = 4i or = 4i ii + = = = i + 8i The roots are = = + 4i or = 4i ii + i = + i i = + i iii + i + + i ) Let's work out ( ) ( ) 4 = + i + (+ i) 4i = + i+ + i 4i = 0 = + i is a solution to the equation + i = 4 0 Question : 0 aia )) = + B= = 0 0 ii) BA = 0 o o Cos0 Sin0 o bi ) ) A= Arepresents the rotation centre O, 0 anticlockwise o o Sin0 Cos0 o o Cos0 Sin0 o ii) B = o o B represents the reflection in the line y = (tan5 ) Sin0 Cos0 0 iii) BA = BA represents the reflection in the line y = 0 ( inthe ais) 0 i This opening question gave almost all the candidates the opportunity to score a high number of marks. Even when careless errors were made, for eample the omission of the plus-or-minus symbol in one or both sections of part (a), there was much correct work for the eaminers to reward. The epansion of the cube of a binomial epression in part (b)(i) seemed to be tackled more confidently than in the past. Almost all the candidates used i = in part (b)(ii), though some were unsure how to deal with i. Like Question, this question was very productive for the majority of candidates, who showed a good grasp of matrices and transformations. A strange error in part (a)(i) was a failure to simplify the epression, though on this occasion the error was condoned. The most common mistake in part (a)(ii) was to multiply the two matrices the wrong way round. Only one mark was lost by this as long as the candidate made no other errors and was able to interpret the resulting product matri as a transformation in part (b)(iii). Occasionally a candidate misinterpreted the θ occurring in the formula booklet for reflections, and gave the mirror line as y= tan 60 instead of y= tan5. Question : = 0 has roots α and β 4 a) α + β = = and αβ = b) α + β = ( α + β ) αβ = ( ) = 4 = c) α + β = ( α + β) αβ = = = 4 It was pleasing to note that almost all candidates were aware that the sum of the roots was and not +, and that they were able to tackle the sum of the squares of the roots correctly in part (b). Part (c) was not so well answered. Relatively few candidates saw the short method based on the use of ( α + β ). Of those who used the epansion of 4 ( α + β), many found the correct epansion but still had difficulty arranging the terms so that the appropriate substitutions could be made.

55 Question 4: b b a) y = a so log y = log ( a ) 0 0 log y = log a+ blog log y = blog + log a Y = bx + c b) When log = 0, log y = so log a = a = 0 When log0 =, log0 y = 0 so 0 = b + b = Question 5: a) y = = ( )( + ) " vertical asymptotes" = and = (roots of the denominator) y = 0 y = 0 is asymptote to the curve. b) Part (a) of this question was well answered, most candidates being familiar with the equation b y = a and the technique needed to convert it into linear form. Some candidates seemed less happy with part (b), but most managed to show enough knowledge to score well here. Errors often arose from confusion between the intercept on the vertical ais and the corresponding value of y, which required the taking of an antilogarithm. Most candidates scored well here, picking up marks in all three parts of the question. Those who failed to obtain full marks in part (a) were usually candidates who gave y = instead of y = 0 as the equation of the horizontal asymptote. The sketch was usually reasonable but some candidates showed a stationary point, usually at or near the origin, despite the helpful information given in the question. There were many correct attempts at solving the inequality in part (c), though some answers bore no relation to the candidate s graph. c) > 0 = 0 for = 0 By plotting the line y = 0 on the graph, the we conclude that > 0when < < 0or >

56 Question 6: ai ))(r ) 4 4 = r r+ n n n n n ) ( ) = = ii r r r r r r= r= r= r= r= = 4 nn ( + )(n+ ) 4 nn ( + ) + n 6 = n[ ( n+ )(n+ ) 6( n+ ) + ] = n n + n+ n + ( = n 4n ) bs ) = (r ) (r ) = 00 (4 00 ) 50 (4 50 ) r= r= = = The simple request in part (a)(i) seemed to have the desired effect of setting the candidates along the right road in part (a)(ii). As usual many candidates struggled with the algebra but made reasonable progress. They could not hope to reach the printed answer legitimately if they equated Σ to rather than to n. Very few, even among the strongest candidates, achieved anything worthwhile in part (b), most using n = 00 instead of n =00 at the top end. No credit was given for simply writing down the correct answer without any working, as the question required the candidates to use the formula previously established rather than summing the numbers directly on a calculator. Question 7: π a) Sin + = 0 6 π + = 0 + kπ 6 π = + kπ k 6 big ) ) (0.05) = (0.05) = rounded to 4 decimal places f (0.05) = rounded to 4 decimal places gh ( ) g(0) ii)the gradient is = + h h 0 h h 4 = h = 4 4 h h h gh ( ) g(0) iii) When h tends to 0, = h tends to h 0 4 so g '(0) = The trigonometric equation in part (a) was more straightforward than usual and was correctly and concisely answered by a good number of candidates. Some earned two marks by finding a correct particular solution and introducing a term nπ (or nπ ), but a common mistake was to use the formula nπ+(-) n α with α equated to the particular solution rather than to 0. Part (b)(i) was usually answered adequately. The best candidates gave more than four decimal places, showing that the two numbers were different, before rounding them both to four decimal places. The eaminers on this occasion tolerated a more casual approach. There was an encouraging response to the differentiation from first principles called for in parts (b)(ii) and (b)(iii). No credit could be given in part (b)(iii) to those who simply wrote down the answer, as by doing so they were not showing any knowledge of the required technique. Strictly speaking the candidates should have used the phrase as h tends to zero or as h 0, but the eaminers allowed the mark for when h equals zero or the equating of h to 0, even though zero is the one value of h for which the chord referred to in part (b)(ii) does not eist.

57 Question 8: y a) = when = y y = = y = y = ± 7 y = or y = b) When = 0, y = ± When y = 0, = ± 5 Part (a) of this question was generally well answered apart from the omission of the plus-or-minus symbol by a sizeable minority of candidates. The wording of the question should have made it very clear that there would be more than one point of intersection. In part (b) many candidates sketched an ellipse instead of a hyperbola. Those who realised that it should be a hyperbola often lost a mark by showing too much curvature in the parts where the curve should be approaching its asymptotes. Most candidates gave an appropriate answer (in the light of their graph) to part (c). No credit was given here to those who had drawn an incorrect curve, such as an ellipse, which just happened to have a vertical tangent at its intersection with the positive -ais. cc ) intersects the positive -ais at the point (5,0) so the tangent has equation = 5 d) i) If y = 4intersects the curve, the n the -coordinate of the point(s) of intersection ii ( 4) satisfies the following equation: = ( 5) 5 9 = ) Discriminant: ( 00) ( 4) 5 + = = 0 + 5= = = = and y = 4 4 This is a repeated root, so the line is TANGENT to the curve. It was good to see that, no doubt helped by the printed answer, the majority of candidates coped successfully with the clearing of fractions needed in part (d)(i). Some candidates seemed to ignore the instruction to solve the equation in part (d)(ii) and went straight on to the statement that the line must be a tangent to the curve. Others mentioned the equal roots of the equation but failed to mention any relationship between the line and the curve. Grade boundaries

58 General Certificate of Education June 007 Advanced Subsidiary Eamination MATHEMATICS Unit Further Pure MFP Wednesday 0 June pm to.00 pm For this paper you must have: an 8-page answer book the blue AQA booklet of formulae and statistical tables an insert for use in Questions 5 and 9 (enclosed). You may use a graphics calculator. Time allowed: hour 0 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Eamining Body for this paper is AQA. The Paper Reference is MFP. Answer all questions. Show all necessary working; otherwise marks for method may be lost. Fill in the boes at the top of the insert. Information The maimum mark for this paper is 75. The marks for questions are shown in brackets. Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. P995/Jun07/MFP 6/6/6/ MFP

59 Answer all questions. The matrices A and B are given by A ¼, B ¼ 8 4 The matri M ¼ A B. (a) 0 Show that M ¼ n, where n is a positive integer. ( marks) 0 (b) (c) The matri M represents a combination of an enlargement of scale factor p and a reflection in a line L. State the value of p and write down the equation of L. ( marks) Show that M ¼ qi where q is an integer and I is the identity matri. ( marks) (a) Show that the equation þ 7 ¼ 0 has a root between.6 and.8. ( marks) (b) Use interval bisection twice, starting with the interval in part (a), to give this root to one decimal place. (4 marks) It is given that z ¼ þ iy, where and y are real numbers. (a) Find, in terms of and y, the real and imaginary parts of z iz where z is the comple conjugate of z. ( marks) (b) Find the comple number z such that z iz ¼ 6 ( marks) P995/Jun07/MFP

60 s 4 The quadratic equation þ 4 ¼ 0 has roots a and b. (a) Write down the values of a þ b and ab. ( marks) (b) Show that a þ b ¼. ( marks) 4 (c) Find a quadratic equation with integer coefficients such that the roots of the equation are 4 a and 4 b ( marks) 5 [Figure and Figure, printed on the insert, are provided for use in this question.] The variables and y are known to be related by an equation of the form y ¼ ab where a and b are constants. The following approimate values of and y have been found. 4 y (a) Complete the table in Figure, showing values of and Y, where Y ¼ log 0 y. Give each value of Y to three decimal places. ( marks) (b) Show that, if y ¼ ab, then and Y must satisfy an equation of the form Y ¼ m þ c ( marks) (c) Draw on Figure a linear graph relating and Y. ( marks) (d) Hence find estimates for the values of a and b. (4 marks) P995/Jun07/MFP Turn over

61 4 6 Find the general solution of the equation sin p ¼ pffiffiffi giving your answer in terms of p. (6 marks) 7 A curve has equation y ¼ þ (a) Write down the equations of the two asymptotes to the curve. ( marks) (b) (c) Sketch the curve, indicating the coordinates of the points where the curve intersects the coordinate aes. (5 marks) Hence, or otherwise, solve the inequality 0 < þ < ( marks) 8 For each of the following improper integrals, find the value of the integral or eplain briefly why it does not have a value: (a) ð 0 ð þ Þ d ; (4 marks) (b) ð 0 þ d. (4 marks) P995/Jun07/MFP

62 5 9 [Figure, printed on the insert, is provided for use in this question.] The diagram shows the curve with equation and the straight line with equation þ y ¼ þ y ¼ y O (a) Write down the eact coordinates of the points where the curve with equation þ y ¼ intersects the coordinate aes. ( marks) (b) (c) The curve is translated k units in the positive direction, where k is a constant. Write down, in terms of k, the equation of the curve after this translation. ( marks) Show that, if the line þ y ¼ intersects the translated curve, the -coordinates of the points of intersection must satisfy the equation ðk þ 4Þ þðk þ 6Þ ¼0 (4 marks) (d) Hence find the two values of k for which the line þ y ¼ is a tangent to the p translated curve. Give your answer in the form p ffiffiffi q, where p and q are integers. (4 marks) (e) On Figure, show the translated curves corresponding to these two values of k. ( marks) END OF QUESTIONS P995/Jun07/MFP

63 Surname Other Names Centre Number Candidate Number Candidate Signature General Certificate of Education June 007 Advanced Subsidiary Eamination MATHEMATICS Unit Further Pure MFP Insert Insert for use in Questions 5 and 9. Fill in the boes at the top of this page. Fasten this insert securely to your answer book. Turn over for Figure P995/Jun07/MFP 6/6/6/ Turn over s

64 Figure (for use in Question 5) 4 Y Figure (for use in Question 5) Y ~ O 4 ~ Figure (for use in Question 9) y O Copyright Ó 007 AQA and its licensors. All rights reserved. P995/Jun07/MFP

65 MFP - AQA GCE Mark Scheme 007 June series MFP Q Solution Mark Total Comments (a) 0 M = 0 B, B if subtracted the wrong way round (b) p = BF ft after B in (a) L is y = B Allow p =, L is y = 9 0 (c) M = 0 9 BF Or by geometrical reasoning; ft as before... = 9I BF ft as before Total 6 (a) f(.6) =.04, f(.8) = 0.6 B,B Allow dp throughout Sign change, so root between E (b) f(.7) considered first M f(.7) = 0.87, so root >.7 A f(.75) = , so root.7 ma 4 m for f(.65) after error Total 7 (a) Use of z = iy M z iz = + iy i y m Condone sign error here R = y, I = + y A Condone inclusion of i in I Allow if correct in (b) (b) y = 6, + y = 0 M Elimination of or y m z = 6i AF Accept =, y = 6; ft + y for y Total 6 4(a) α + β =, αβ = BB (b) α + β + = α β αβ... = = 4 M A Convincingly shown (AG) (c) Sum of roots = BF PI by term ± ; ft error(s) in (a) Product of roots = 6 8 αβ = BF ft wrong value of αβ Equation is + 8 = 0 BF ft wrong sum/product; = 0 needed Total 7 4

66 MFP - AQA GCE Mark Scheme 007 June series MFP (cont) Q Solution Mark Total Comments 5(a) Values 0.788, 0.99,.96 in table B, B if one correct (or if wrong number of dp given) (b) lg ab = lg a + lg b M lg b = lg b M So Y = (lg b) + lg a A Allow NMS (c) BF BF Four points plotted; ft wrong values in (a) Good straight line drawn; ft incorrect points (d) a = antilog of y-intercept MA Accept. to.5 b = antilog of gradient MA 4 Accept.58 to.6 Total π 6 One value of is π OE (PI); degrees/decimals penalised in B 6th mark only π π Another value is π = BF OE (PI); ft wrong first value Introduction of nπ or nπ M General solution for m 5π 7π GS = + nπ or = + n π A, 6 OE; A if one part correct Total 6 7(a) Asymptotes =, y = B,B (b) B Curve approaching asymptotes B,B Passing through,0 and 0, B,B 5 Both branches generally correct B if two branches shown (c) Solution set is > B,F B for good attempt; ft wrong point of intersection Total 9 5

67 MFP - AQA GCE Mark Scheme 007 June series MFP (cont) Q Solution Marks Totals Comments 8(a) 4 + d= + 4 (+ c) MA M for adding to inde at least once 9... = + 0 = 4 4 ma 4 0 (b) Second term is 4 B Condone no mention of limiting process; m if 0 stated or implied Integral of this is MA M for correct inde as 0, so no value E 4 Total 8 9(a) Intersections (, 0) ±, (0, ± ) BB Allow B for ( ), 0, (0, ) ( k) (b) Equation is + y = MA M if only one small error, eg + k for k (c) Correct elimination of y M Correct epansion of squares M Correct removal of denominator M Answer convincingly established A 4 AG (d) Tgt 4( k + 4) ( k + 6) = 0 M... k 4k + = 0 ma OE... k = ± A 4 (e) B Curve to left of line Total 5 TOTAL 75 B Curve to right of line Curves must touch the line in appro correct positions SC / if both curves are incomplete but touch the line correctly 6

68 Further pure - AQA - June 007 Question : a) M = A B= = 0 = 0 0 b)the matri is the refleion in the line y 0 = The scale factor of the enlargement is p = c M = 0 = 0 ) 9 9 I q = 9 n = This question was generally well answered. The most common errors were in part (b), where the mirror line was given as y = rather than y =. It was, however, acceptable to give this reflection in conjunction with an enlargement with scale factor. In part (c) some candidates omitted the factor before carrying out the matri multiplication, while others multiplied incorrectly to obtain the 9s and 0s in all the wrong places. Question : a) + 7= 0 Let ' s call f ( ) = + 7 f (.6) =.04 < 0 f (.8) = 0.6 > 0 According to the "sign change" rule, we can say that there is at least one solution of the equation f( ) = 0 between.6 and.8 b) Let ' s work out f (.7) f (.7) = 0.87 < 0 the solution is between.7 and.8 Let ' s work out f (.75) f (.75) = > 0 The solution is between.70 and.75 This is.7 rounded to decimal place Question : z = + iy a) z iz = + iy ( i iy) = + iy i y Re( z iz) = y Im( z iz) = + y y = 6 y = 6 b) z iz = 6 means + y = 0 9+ y = 0 This gives 8 = 6 = and y = 6 The solution is z = 6i Almost all the candidates were able to make a good start to this question, though some were unable to draw a proper conclusion in part (a). In part (b) the response was again very good. The most common error was not understanding what eactly was being asked for at the end of the question. If it is known that the root lies between.7 and.75, then it must be.7 to one decimal place. But many candidates gave no value to one decimal place, or gave a value of the function (usually 0.) rather than a value of. Although most of the candidates who took this paper are good at algebra, there are still quite a number who made elementary sign errors. The fourth term of the epansion in part (a) frequently came out with a plus instead of a minus. The error of including an i in the imaginary part was condoned. In part (b) some candidates seemed not to realise that they needed to equate real and imaginary parts, despite the hint in part (a). Others used 6 for both the real and imaginary parts of the right-hand side of the equation. But a large number solved correctly to obtain full marks. Those who had made the sign error already referred to were given full marks in part (b) if their work was otherwise faultless.

69 Question 4: + 4= 0 has roots α and β 4 a) α + β = and αβ = = β + α ) b + = = = α β αβ c) Let's call u = and v = α β 4 4 we have u+ v= + = 4( + ) = 4 = α β α β uv= = = = 8 α β αβ 4 4 u = and v = is + = α β An equation with roots 8 0 Question 5: 4 Y = 0 b) y = ab so log0 y = log 0( ab ) log0 y = log0 a+ log0 b Y = m + c with m= log b and c = log c) d) When = 0, Y = c= log a= 0.5 a = log0 b = gradient = = b = Question 6: π π Sin = = Sin so π π 5π 5π = + kπ = + kπ = + kπ 6 or π π 7 7 = π + kπ = π + kπ = π + kπ 6 k a Most candidates answered this question very well. Only a small number of candidates gave the wrong sign for the sum of the roots at the beginning of the question. Rather more made the equivalent mistake at the end of the question, giving the term with the wrong sign. Another common mistake at the end of the question was to omit the equals zero, so that the equation asked for was not given as an equation at all. The great majority of candidates gave the correct values to three decimal places in part (a). Most of them coped efficiently with the logarithmic manipulation in part (b), but occasionally a candidate would treat the epression ab as if it were ( ab ), resulting in a loss of at least four marks, as it was now impossible to distinguish validly between a and b. The plotting of the points on the graph was usually well done, but some candidates misread the vertical scale here, and again in part (d) when they attempted to read off the intercept on the Y-ais. Some candidates failed to use this intercept, resorting instead to more indirect methods of finding a value for log a. Methods for finding the gradient of the linear graph were equally clumsy. Some candidates, even after obtaining a correct equation in part (b), did not realise that the taking of antilogs was needed in part (d). As in past MFP papers, the question on trigonometric equations was not as well answered as most of the other questions. The use of radians presented a difficulty to many candidates, who seemed to have met the idea but not to have become really familiar with it. Most candidates knew that for a general solution it was necessary to introduce a term nπ or something similar somewhere in the solution. Many, however, brought in this term at an inappropriate stage. Some had learnt by heart a formula for the general solution of the equation sin = sin a, but applied it incorrectly. Many candidates earned 4 marks out of 6 for working correctly from one particular solution to the corresponding general solution, but omitting the other particular solution or finding it incorrectly.

70 Question 7: y = + a)" vertical asymptote" y = + b) When = 0, y = y = 0 when = = y = is asymptote to the curve Many candidates scored well on parts (a) and (b) of this question, but relatively few candidates made a good attempt at part (c). In part (a) the asymptotes were usually found correctly, as were the coordinates of the required points of intersection in part (b). The graph in part (b) was usually recognisable as a hyperbola, or at least as one branch of a hyperbola, the other branch not being seen. In part (c) many candidates resorted to algebraic methods for solving inequalities rather than simply reading off the solution set from the graph. These algebraic methods, more often than not, were spurious. c) = 0 for = and y = is an asymptote. + So,0< < for > + Question 8: 4 a) + d = + + c 4 This function is defined between 0 and d = + = + 0 = b) d = + d = + c 0 is not defined for = 0. no value This question proved to be very largely a test of integration. Many candidates answered part (a) without any apparent awareness of a limiting process, but full marks were awarded if the answer was correct. The positive indices meant that the powers of would tend to zero as itself tended to zero. In part (b) a slightly more difficult integration led to one term having a negative inde. Three marks were awarded for the integration, but for the final mark it was necessary to give some indication as to which term tended to infinity. Many candidates did not gain this last mark, but it was sad to see how many did not gain any marks at all in part (b). This was usually for one of two reasons. Either they failed to simplify the integrand and carried out a totally invalid process of integration, or they saw that the denominator of the integrand would become zero and said that this made the integral improper and therefore incapable of having any value. Since it was stated in the question that both integrals were improper, this comment failed to attract any sympathy from the eaminers.

71 Question 9: Ellipse : + y = and straight line + y = a) When = 0, y = y = ± 0, when y = = = ± The curve intersects the aes a t (0,), (0,-), (-,0),(,0) ( k) b) + y = c) + y = so y =. The -coordinate of the point(s) of intersection satisfy: ( k) + = k + = ( ) ( ) ( ) k + k = ( k+ 4) + k + 6 = 0 d)the line is tangent to the curve when this equation has a repeated root, meaning the discriminant is 0 ( k ) k+ = 0 ( k + 6) = 0 k + k+ k = k + k k Using the quadratic formula, we have k = ± e) = This final question presented a varied challenge, mostly focused on the algebraic skills needed to cope with the quadratic equation printed in part (c) of the question deriving this equation from two other equations and then using the discriminant of the equation to find the cases where the ellipse would touch the line. Many candidates were well prepared for this type of question and scored heavily, though with occasional errors and omissions. Part (e) of the question was often not attempted at all, or the attempts were totally incorrect, involving transformations other than translations parallel to the -ais. In some cases only one of the two possible cases was illustrated, usually the one where the ellipse touched the straight line on the lower left side of the line, but an impressive minority of scripts ended with an accurate portrayal of both cases. Grade boundaries

72 General Certificate of Education January 008 Advanced Subsidiary Eamination MATHEMATICS Unit Further Pure MFP Friday 5 January pm to.00 pm For this paper you must have: an 8-page answer book the blue AQA booklet of formulae and statistical tables an insert for use in Question 7 (enclosed). You may use a graphics calculator. Time allowed: hour 0 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Eamining Body for this paper is AQA. The Paper Reference is MFP. Answer all questions. Show all necessary working; otherwise marks for method may be lost. Fill in the boes at the top of the insert. Information The maimum mark for this paper is 75. The marks for questions are shown in brackets. Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. P975/Jan08/MFP 6/6/6/ MFP

73 Answer all questions. It is given that z ¼ þ i and that z is the comple conjugate of z. Find the real numbers and y such that þ iy ¼ z þ 4iz (4 marks) A curve satisfies the differential equation dy d ¼ Starting at the point (, 4) on the curve, use a step-by-step method with a step length of 0.0 to estimate the value of y at ¼ :0. Give your answer to si significant figures. (5 marks) Find the general solution of the equation tan 4 p ¼ 8 giving your answer in terms of p. (5 marks) 4 (a) Find Xn r¼ ðr 6rÞ epressing your answer in the form knðn þ Þðn þ pþðn þ qþ where k is a fraction and p and q are integers. (5 marks) (b) It is given that 000 S ¼ X ðr 6rÞ r¼ Without calculating the value of S, show that S is a multiple of 008. ( marks) P975/Jan08/MFP

74 5 The diagram shows the hyperbola 4 y ¼ and its asymptotes. y A O B (a) Write down the equations of the two asymptotes. ( marks) (b) The points on the hyperbola for which ¼ 4 are denoted by A and B. Find, in surd form, the y-coordinates of A and B. ( marks) (c) The hyperbola and its asymptotes are translated by two units in the positive y direction. Write down: (i) the y-coordinates of the image points of A and B under this translation; ( mark) (ii) the equations of the hyperbola and the asymptotes after the translation. ( marks) Turn over for the net question P975/Jan08/MFP Turn over s

75 4 6 The matri M is defined by " pffiffiffi # M ¼ p ffiffi (a) (i) Show that M ¼ pi where p is an integer and I is the identity matri. ( marks) (ii) Show that the matri M can be written in the form cos 60 sin 60 q sin 60 cos 60 where q is a real number. Give the value of q in surd form. ( marks) (b) The matri M represents a combination of an enlargement and a reflection. Find: (i) the scale factor of the enlargement; ( mark) (ii) the equation of the mirror line of the reflection. ( mark) (c) Describe fully the geometrical transformation represented by M 4. ( marks) P975/Jan08/MFP

76 5 7 [Figure, printed on the insert, is provided for use in this question.] The diagram shows the curve y ¼ þ The points A and B on the curve have -coordinates and þ h respectively. y A B O (a) (i) Show that the y-coordinate of the point B is þ h h þ h ( marks) (ii) Find the gradient of the chord AB in the form p þ qh þ rh where p, q and r are integers. ( marks) (iii) Eplain how your answer to part (a)(ii) can be used to find the gradient of the tangent to the curve at A. State the value of this gradient. ( marks) (b) The equation þ ¼ 0 has one real root, a. P975/Jan08/MFP (i) (ii) Taking ¼ as a first approimation to a, use the Newton-Raphson method to find a second approimation,,toa. ( marks) On Figure, draw a straight line to illustrate the Newton-Raphson method as used in part (b)(i). Show the points ð,0þ and ða,0þ on your diagram. ( marks) Turn over s

77 6 8 (a) (i) It is given that a and b are the roots of the equation þ 4 ¼ 0 Without solving this equation, show that a and b are the roots of the equation þ 6 þ 64 ¼ 0 (6 marks) (ii) State, giving a reason, whether the roots of the equation þ 6 þ 64 ¼ 0 are real and equal, real and distinct, or non-real. ( marks) (b) Solve the equation (c) þ 4 ¼ 0 Use your answers to parts (a) and (b) to show that pffiffiffi ð þ i Þ pffiffiffi ¼ð i Þ ( marks) ( marks) 9 A curve C has equation y ¼ ð 4Þ (a) Write down the equations of the three asymptotes of C. ( marks) (b) The curve C has one stationary point. By considering an appropriate quadratic equation, find the coordinates of this stationary point. (No credit will be given for solutions based on differentiation.) (6 marks) (c) Sketch the curve C. ( marks) END OF QUESTIONS P975/Jan08/MFP

78 Surname Other Names Centre Number Candidate Number Candidate Signature General Certificate of Education January 008 Advanced Subsidiary Eamination MATHEMATICS Unit Further Pure MFP Insert Insert for use in Question 7. Fill in the boes at the top of this page. Fasten this insert securely to your answer book. Turn over for Figure P975/Jan08/MFP 6/6/6/ Turn over s

79 Figure (for use in Question 7) y A O Copyright Ó 008 AQA and its licensors. All rights reserved. P975/Jan08/MFP

80 MFP - AQA GCE Mark Scheme 008 January series MFP Q Solution Marks Totals Comments z + 4i z = ( + i) + 4i ( i) M Use of conjugate... = ( + i) + (8i + 4) M Use of i =... = 6 + 9i, so = 6 and y = MA 4 M for equating Real and imaginary parts Total 4 0.0( ) added to value of y M Variations possible here So y(.0) 4.0 A PI Second increment is 0.0(.0 ) m A So y(.0) A 5 Total 5 Use of tan π 4 = B Degrees or decimals penalised in last mark only Introduction of nπ M or kn at any stage Division of all terms by 4 m Addition of π/8 m OE GS = π + nπ 6 4 A 5 OE Total 5 4(a) Use of formula for r or r n is a factor of the epression m clearly shown So is (n + ) m ditto S n = 4 nn ( + )( n + n ) A... = 4 ( )( 4)( ) AF 5 ft wrong value for k (b) n = 000 substituted into epression m The factor 004, or , seen Conclusion convincingly shown Need 000 is even, hence conclusion 4 A not OE Total 7 5(a) Asymptotes are y =± MA OE; M for y = ± m (b) = 4 substituted into equation M y = so y =± A Allow NMS (c)(i) y-coords are ± BF ft wrong answer to (b) ( y ) = MA MA0 if y + used (ii) Hyperbola is 4 Asymptotes are y = ± BF ft wrong gradients in (a) Total 8 6(a)(i) 0 M = 0 MA M if zeroes appear in the right places = I AF ft provided of right form (ii) q cos 60 = q= q= MA OE SC q = NMS / Other entries verified E surd for sin 60º needed (b)(i) SF = q = BF ft wrong value for q (ii) Equation is y = tan 0º B (c) M 4 =44I BF PI; ft wrong value in (a)(i) M 4 gives enlargement SF 44 BF ft if c s M 4 = ki Total 0 4

81 MFP - AQA GCE Mark Scheme 008 January series MFP (cont) Q Solution Marks Totals Comments 7(a)(i) ( + h) = + h h + h B PI y B = ( + h h + h )+ h+ BF ft numerical error... = + h h + h B convincingly shown (AG) (ii) Subtraction of and division by h MM Gradient of chord = h + h A (iii) As h 0, gr(chord) gr(tgt) = EBF E0 if h = 0 used; ft wrong value of p (b)(i) = =.5 M AF ft wrong gradient (ii) Tangent at A drawn M α and shown correctly A dep't only on the last M Total 8(a)(i) α + β =, αβ = 4 BB α + β = () (4)() = 6 MA α β = (4) = 64, hence result MA 6 convincingly shown (AG) (ii) Discriminant 0, so roots equal BE or by factorisation (b) ± 4 6 = M or by completing square... = ± i A (c) α, β = ± i and α = β, hence result E Total 9(a) Asymptotes = 0, = 4, y = 0 B (b) y = k = k( 4) M (c)... 0= k 4k A Discriminant = (4k) + 8k m At SP y = A not just k =... 0= + m So = A 6 y O B Curve with three branches approaching vertical asymptotes correctly B Outer branches correct B Middle branch correct Total TOTAL 75 5

82 Further pure - AQA - January 008 Question : z = + i z = i + iy = z + 4 iz + iy = + i + 4 i( i) + iy = 6+ 9i so = 6 y = The great majority of candidates found this question a good starter and obtained full marks without too much trouble. Careless errors sometimes caused candidates to miss out on one of the method marks for the question. Question : Euler formula : y = n y + n hf ( n) with f ( ) = and h = =, y = 4 so = =.0, y = = = = + = for for.00, y correct to5 sig. fig. This question provided most candidates with a further five marks. The most common error was to carry out three iterations instead of only two, which would usually cause the loss of only one mark as long as the working was fully shown; though of course the candidate may well have lost valuable time carrying out the unwanted calculations. Question : π π Tan 4( ) = = Tan 8 4 π π so 4 = + kπ 4 π 4 = + kπ 4 π π = + k 6 4 k As usual in MFP, many candidates were not thoroughly prepared for the task of finding the general solution of a trigonometric equation. The most common approach was to find (usually correctly) one value of and then to add a term nπ to this value. Many candidates showed only a slight degree of familiarity with radians, and there were some cases of serious misunderstanding of the implied order of operations in the epression π tan 4 8 Question 4: a r r r r n n nn r= r= r= 4 = nn ( + ) ( + ) 4 ( ) = nn+ ( n + n ) 4 n r 6 r = nn ( + )( n+ 4)( n ) 4 n n n ) 6 = 6 = ( + ) 6 ( + ) r= [ nn ] 000 bs ) = r 6r = = r= 4 = = which is a multiple of 008. Part (a) As in the previous question, many candidates were not sufficiently familiar with the techniques needed to carry out the necessary manipulation efficiently. It was noticeable that many candidates still obtained full marks despite their failure to spot the quick method of taking out common factors at the earliest opportunity. A common mistake was to omit the numerical factor, or to replace it with some 4 other number such as 4. Part (b) Good attempts at this part of the question were few and far between. Many candidates made no attempt at all. Some came to a halt after replacing n by 000 in their answer to part (a). Those who found a factor 004 usually went on to try to eplain how this would lead to a multiple of 008, but more often than not their arguments lacked cogency.

83 Question 5: a) Asymptotes : y = and y = b y y ) when = 4, 4 = = so A(4, ) and B(4, ) ci )) Image of A: (4, + ) Image of B: (4, ) ii) Equation of the hyperbola after translation: 4 Equations of the asymptotes : y = ± + = ( y ) The oblique asymptotes of a hyperbola were not always known by the candidates, though many found the necessary general equations in the formula booklet and correctly applied them to this particular case. Part (b) was very well answered by almost all candidates, while part (c) usually provided some further marks, the most common error being to use y + instead of y in the equations of the translated hyperbola and asymptotes. Question 6: 0 a) i) M = and M = 0 = I p = o o o Cos60 Sin60 ii) Cos60 = so M = = o o Sin60 Cos60 q = bi )) Scale factor is q = o ii) Reflection in theline y = ( Tan0 ) y = 4 c) M = I so M = 44I This represent an enlargement scale factor 44. Part (a)(i) of this question was almost universally well answered, but part (a)(ii) led to some very slipshod and unclear reasoning. In part (b)(i) many candidates did not appreciate that the scale factor of the enlargement must be the same as the value of q obtained in the previous part. In part (b)(ii) a common mistake was to use tan 60 in finding the gradient of the mirror line instead of halving the angle to obtain tan 0. Answers to part (c) were mostly very good, even from candidates who had been struggling with the earlier parts of the question.

84 Question 7: a)) i y = + with = + h ( ) ( ) gives y = + h + h + (,) (, ) ii) Gradient of AB = B = + h h + h + h+ = + h h + h A and B + h + h h + h ( ) + h h + h ( + h) ( ) h h + h = = h+ h h iii)the gradient of the tangent is h+ h h 0 The gradient of the tangent is B the limit of the gradient of the chord when h tends to 0 b) i) If is an approimation of the root α, ii) then = =, f( ) = f( ) = f( ) is a better aproimation f '( ) '( ) '( ) f = and f = so = =. 5 = Most candidates performed well in this question. Part (a)(i) seemed to present a stiff challenge to their algebraic skill, though they often came through the challenge successfully after several lines of working. Part (a)(ii) again proved harder than the eaminers had intended, some candidates having to struggle to evaluate the y-coordinate of the point A. Again the outcome was usually successful, though a substantial minority of candidates used differentiation of the answer to the previous part. In part (a)(iii) it was pleasing to see that a very good proportion of the candidates correctly mentioned h tending to zero rather than being equal to zero, which was not allowed, though the correct value of the gradient could be obtained by this method and one mark was awarded for this. The responses to part (b) suggested that most candidates were familiar with the Newton-Raphson method and were able to apply it correctly, but that they lacked an understanding of the geometry underlying the method, so that they failed to draw a tangent at the point A on the insert as required, or failed to indicate correctly the relationship between this tangent and the -ais. Question 8: a)) i + 4= 0has roots α and β ii b) α + β = αβ = 4 α + β = ( α + β ) αβ ( α + β ) αβ = ( αβ ) = 4 = 64 = = 6 4 Therefore an equation with roots α and β is ) Discriminant = = 0 = = 0 The roots are real and equal discriminant = ( ) 4 4 = 4 6 = = (i ) = ± i = ± i c) We can call α = + i and β = i from the question a) ii), we know that α = β = 0 meaning (+ i ) = ( i ) This question proved to be an ecellent source of marks for the majority of candidates, apart from the discriminating test provided by part (c). Many candidates seemed to be familiar with the techniques relating to the cubes of the roots of a quadratic equation, though some struggled to find the correct epression for the sum of the cubes of the roots, while others lost marks by not showing enough evidence in view of the fact that the required equation was printed on the question paper. Parts (a)(ii) and (b) proved straightforward for most candidates, but only a few were able to give a clear eplanation in part (c), where many candidates claimed that the original roots α and β must be equal.

85 Question 9: y = ( 4) a)" Vertical asymptotes" = 0 and = 4 ( roots of the denominator) y = = 0 y = 0 is asymptote to the curve 4 4 b) y = = k can be re arranged to k 4k = 0 4 Discriminant:( 4 k) 4 k ( ) = 6k + 8k = 8 k(k+ ) the line y = k is tangent to the curve when the discriminant is 0 this gives = 0 (,, 0) or k impossible for all 4 And for k = + = 4+ 4= 0, the equation becomes 0 ( = The stationary point has coordinates (, ) c) ) = 0 k = Most candidates scored well in parts (a) and (c) of this question. Many wrote down the equations of the two vertical asymptotes without any apparent difficulty, but struggled to find the horizontal asymptote, although in most cases they were successful. The sketchgraph in part (c) was usually drawn correctly. Part (b) was a standard eercise for the more able candidates. No credit was given for asserting that because = 0 and = 4 were asymptotes, it followed that = must provide a stationary point. A carefully reasoned argument based on the symmetry of the function in the denominator would have earned credit, but most candidates quite reasonably preferred to adopt the standard approach. Grade boundaries

86 General Certificate of Education June 008 Advanced Subsidiary Eamination MATHEMATICS Unit Further Pure MFP Monday 6 June pm to.00 pm For this paper you must have: an 8-page answer book the blue AQA booklet of formulae and statistical tables an insert for use in Questions 4 and 8 (enclosed). You may use a graphics calculator. Time allowed: hour 0 minutes Instructions Use black ink or black ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Eamining Body for this paper is AQA. The Paper Reference is MFP. Answer all questions. Show all necessary working; otherwise marks for method may be lost. Fill in the boes at the top of the insert. Information The maimum mark for this paper is 75. The marks for questions are shown in brackets. Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. P55/Jun08/MFP 6/6/ MFP

87 Answer all questions. The equation þ þ 5 ¼ 0 has roots a and b. (a) Write down the values of a þ b and ab. ( marks) (b) Find the value of a þ b. ( marks) (c) Show that a b þ b a ¼ 9 5. ( marks) (d) Find a quadratic equation, with integer coefficients, which has roots a b and b a. ( marks) It is given that z ¼ þ iy, where and y are real numbers. (a) Find, in terms of and y, the real and imaginary parts of iz þ z where z is the comple conjugate of z. ( marks) (b) Find the comple number z such that iz þ z ¼ 7 þ 8i ( marks) For each of the following improper integrals, find the value of the integral or eplain briefly why it does not have a value: (a) (b) ð 9 ð 9 p ffiffi d ; p ffiffi d. ( marks) (4 marks) P55/Jun08/MFP

88 4 [Figure and Figure, printed on the insert, are provided for use in this question.] The variables and y are related by an equation of the form where a and b are constants. y ¼ a þ b þ (a) The variables X and Y are defined by X ¼ ð þ Þ, Y ¼ yð þ Þ. Show that Y ¼ ax þ b. ( marks) (b) The following approimate values of and y have been found: 4 y (i) Complete the table in Figure, showing values of X and Y. ( marks) (ii) Draw on Figure a linear graph relating X and Y. ( marks) (iii) Estimate the values of a and b. ( marks) 5 (a) Find, in radians, the general solution of the equation cos þ p ¼ p ffiffiffi giving your answer in terms of p. (5 marks) (b) Hence find the smallest positive value of which satisfies this equation. ( marks) 6 The matrices A and B are given by A ¼ 0, B ¼ (a) Calculate the matri AB. ( marks) (b) Show that A is of the form ki, where k is an integer and I is the identity matri. ( marks) (c) Show that ðabþ 6¼ A B. ( marks) P55/Jun08/MFP Turn over s

89 4 7 A curve C has equation y ¼ 7 þ þ (a) Define the translation which transforms the curve with equation y ¼ curve C. onto the ( marks) (b) (i) Write down the equations of the two asymptotes of C. ( marks) (ii) Find the coordinates of the points where the curve C intersects the coordinate aes. ( marks) (c) Sketch the curve C and its two asymptotes. ( marks) 8 [Figure, printed on the insert, is provided for use in this question.] The diagram shows two triangles, T and T. y ~ T T O ~ (a) Find the matri of the stretch which maps T to T. ( marks) (b) The triangle T is reflected in the line y ¼ to give a third triangle, T. On Figure, draw the triangle T. ( marks) (c) Find the matri of the transformation which maps T to T. ( marks) P55/Jun08/MFP

90 5 9 The diagram shows the parabola y ¼ 4 and the point A with coordinates (, 4). y Að, 4Þ O (a) (b) Find an equation of the straight line having gradient m and passing through the point A(, 4). ( marks) Show that, if this straight line intersects the parabola, then the y-coordinates of the points of intersection satisfy the equation my 4y þð6 mþ ¼0 ( marks) (c) By considering the discriminant of the equation in part (b), find the equations of the two tangents to the parabola which pass through A. (d) (No credit will be given for solutions based on differentiation.) Find the coordinates of the points at which these tangents touch the parabola. (5 marks) (4 marks) END OF QUESTIONS P55/Jun08/MFP

91 Surname Other Names Centre Number Candidate Number Candidate Signature General Certificate of Education June 008 Advanced Subsidiary Eamination MATHEMATICS Unit Further Pure MFP Insert Insert for use in Questions 4 and 8. Fill in the boes at the top of this page. Fasten this insert securely to your answer book. Turn over for Figure P55/Jun08/MFP 6/6/ Turn over s

92 Figure (for use in Question 4) 4 y X Y.0 Figure (for use in Question 4) Y 0 ~ 0 0 O ~ X 0 P55/Jun08/MFP

93 Figure (for use in Question 8) y ~ T T O ~ P55/Jun08/MFP

94 MFP - AQA GCE Mark Scheme 008 June series Q Solution Marks Total Comments (a) α + β =, αβ = 5 BB MFP (b) α + β = (α + β) αβ M with numbers substituted... = 0 = 9 AF ft sign error(s) in (a) (c) α β α + β + = β α αβ M... = 9 5 A AG: A0 if α + β = used (d) Product of new roots is B PI by constant term or 5 Eqn is = 0 BF ft wrong value for product Total 8 (a) Use of z = iy M Use of i = M iz + z = ( y) + i ( y) A Condone inclusion of i in I part (b) Equating R and I parts M y = 7, y = 8 m with attempt to solve z = i A Allow =, y = Total 6 (a) (b) d= ( + c) MA M for correct power in integral as, so no value E d= ( + c) MA M for correct power in integral 0 as E PI d (0 ) = = A 4 Allow A for correct answer even if not 9 fully eplained Total 7 4(a) Multiplication by + M applied to all terms Y = ax + b convincingly shown A AG (b)(i) X = 8, 5, 4 in table B Y = 5.7,, 0. in table B Allow correct to SF 4

95 MFP - AQA GCE Mark Scheme 008 June series MFP (cont) Q Solution Marks Total Comments 4(b)(ii) Four points plotted BF ft incorrect values in table Reasonable line drawn BF ft incorrect points (iii) Method for gradient M or algebraic method for a or b a = gradient 0.9 A Allow from 0.88 to 0.9 incl b = Y-intercept.5 BF Allow from to inclusive; ft incorrect points/line NMS B for a, B for b Total 9 5(a) cos π = stated or used B Degrees or decimals penalised in 4 5th mark only Appropriate use of ± B OE Introduction of nπ M OE Subtraction of π and multiplication by m All terms multiplied by = π ± π + 4nπ A 5 OE 5(b) n = gives min pos = 7π MA NMS / provided (a) correct 6 Total 7 6(a) 0 4 AB = 4 0 MA MA0 if entries correct (b) 4 0 A = 0 4 B... = 4I B (c) (AB) = 6I B B = 4I B PI so A B = 6I (hence result) B Condone absence of conclusion Total 7 5

96 MFP - AQA GCE Mark Scheme 008 June series MFP (cont) Q Solution Marks Total Comments 7(a) Curve translated 7 in y direction B... and in negative direction B or answer in vector form (b)(i) Asymptotes = and y = 7 BB (ii) Intersections at (0, 8)... B... and ( 8, 0) 7 MA Allow AWRT.4; NMS / (c) At least one branch B of correct shape Complete graph B translation of y = / All correct including asymptotes B in roughly correct positions 8(a) Matri is (b) 0 0 Total 0 MA M if zeros in correct positions; allow NMS Third triangle shown correctly MA MA0 if one point wrong 6

97 MFP - AQA GCE Mark Scheme 008 June series MFP (cont) Q Solution Marks Total Comments 0 Alt: calculating matri from the 8(c) Matri of reflection is 0 B coordinates: M A, Multiplication of above matrices M in correct order 0 Answer is 0 AF ft wrong answer to (a); NMS / Total 7 9(a) Equation is y 4 = m( ) MA OE; MA0 if one small error (b) Elimination of M 4y 6 = m(y ) A OE (no fractions) Hence result A convincingly shown (AG) (c) Discriminant equated to zero M (m )(m ) = 0 ma OE; m for attempt at solving Tangents y = +, y = + AA 5 OE (d) m y y so point of contact is (, ) A m= y 4y + = 0 M so point of contact is (9, 6) A 4 Total 4 TOTAL 75 = 4 + 4= 0 M OE; m = needed for this OE; m = needed for this 7

98 Further pure - AQA - June 008 Question : + + 5= 0 has roots α and β a) α + β = and αβ = 5 b) α + β = ( α + β ) αβ = ( ) 5= 0= 9 α β α + β 9 c) + = = β α αβ 5 α β 9 α β αβ d) + = and = = β α 5 β α αβ α β 9 so an equation with roots and is + + = 0 β α = 0 Question : z = + iy a)iz + z = i( + iy) + ( iy) = i y + iy = ( y) + i( y) Re(iz + z) = y Im(iz + z) = y y = 7 6 9y = 5y = 5 b)iz + z = 7+ 8i means y = 8 6 4y = 6 y = 7 y = and = The solution is i Question : a) d = d c = + This integral has no value. b) d = d = + c = + c 0 so d = 0 = 9 9 The great majority of candidates showed a confident grasp of the algebra needed to deal with the sum of the squares of the roots of a quadratic equation, and answered all parts of this question efficiently. The only widespread loss of credit came in the very last part, where many candidates failed to give integer coefficients or else found a quadratic epression but without the necessary = 0 to make it an equation. Here again most candidates answered confidently and accurately. In part (a) many failed to state clearly which was the real part and which the imaginary part, though they recovered ground by using the correct epressions in part (b). As already stated, the solution of the simultaneous equations was often attempted by a substitution method. Whichever method was used, numerical and sign errors were fairly common, but most candidates obtained the correct values for and y. Two faults which were condoned this time were, in part (a), the retention of the factor i in the imaginary part, and, in part (b) following correct values of and y, a failure to give the final value of z correctly There were many all-correct solutions to this question from the stronger candidates. Others were unsure of themselves when dealing with the behaviour of powers of as tended to infinity. Many others did not reach the stage of making that decision: either they failed to convert the integrands correctly into powers of, or they integrated their powers of incorrectly. A reasonable grasp of AS Pure Core Mathematics is essential for candidates taking this paper.

99 Question 4: b a) y = a + ( ( + )) + y( + ) = a( + ) + b Y = ax + b 4 y b) X Y iii) The gradient is: a a b = Y ax = =.5 b.5 Question 5: π π Cos( + ) = = Cos( ) 4 π π π π so + = + kπ or + = + kπ 4 4 π 7π = + kπ or = + kπ π 7π = + k4π or = + k4π 6 6 π π 7π 7π b) + 4π = and + 4π = π The smallest value of is 6 Question 6: 0 0 A = 0 and B a) AB = b) A = = I c)( AB) = 6I = 0 6 = 4 0 AB = 4I = 6I so ( AB) A B 0 4 In part (a) most candidates simply wrote down the given equation, multiplied through by ( + ), and converted the result into X and Y notation. This was all that was required for the award of the two marks, but some candidates thought that more was needed and presented some rather heavy algebra. Some candidates lost credit because of a confusion between the upper- and lower-case letters. Parts (b)(i) and (b)(ii) were usually answered correctly on the insert, after which most candidates knew how to find estimates for a and b, occasionally losing a mark through a loss of accuracy after a poor choice of coordinates to use in the calculation of the gradient. Another way of losing a mark was to write down the estimate for a without showing any working. As usual in MFP, many candidates made a poor effort at finding the general solution of a trigonometric equation. In this case the solution π was almost always found in part = 4 (a), but the second solution π (or = 4 alternative) was relatively rarely seen. Many candidates were aware that when dealing with a cosine they needed to put a plus or minus symbol somewhere but were not sure where eactly it should go. The general term nπ usually appeared, but often in the wrong place. In part (b) only the strongest candidates, and not even all of these, were able to obtain any credit here. This question was very well answered by the majority of candidates, who showed confidence and accuracy in manipulating these simple matrices.

100 Question 7: y = a) Translation vector 7 b) i)" Vertical asymptote" = y = 7+ 7 y = 7 is asymptote to the curve + ii) When = 0, y = 8 8 y = 0= 7+ = 7 + = = The curve crosses the aes at (0,8) and (,0) 7 c) Many candidates scored well on this question but, for some, marks were lost in a variety of places. In part (a) some candidates were careless in giving the two parts of the translation. In part (b) (i) the horizontal asymptote was often found to be the -ais, which usually caused difficulty in part (c). In part (b)(ii), as in Question 4 (b)(iii), a mark was sometimes lost by a failure to show some necessary working, in this case for the intersection of the curve with the -ais. In part (c) the sketches were sometimes wildly wrong, despite the information provided in part (a) that the curve must be a translated version of the wellknown hyperbola. In many cases y = the curve shown was basically correct but did not appear to approach the asymptotes in a satisfactory way. Despite a generous interpretation of this on the part of the eaminers, some candidates lost credit because of seriously faulty drawing. Question 8: a)this is a stretch in the -direction by a factor 0 It is represented by the matri 0 b) 0 c) The matri of the reflection is so The matri which maps T to Tis = This question was not generally well answered, ecept for part (b) where most candidates were able to draw the reflected triangle correctly on the insert. In part (a) many candidates seemed not to recognise the transformation as a simple one-way stretch, and even those who did were not always familiar with the corresponding matri. They could have worked out the matri by considering the effect of the stretch on the points (, 0) and (0, ), but in many cases candidates either gave up, made a wild guess, or embarked on a lengthy algebraic process involving four equations and four unknowns. In part (c) relatively few candidates saw the benefit of matri multiplication for the composition of two transformations, and even they often multiplied the matrices the wrong way round. Again it was common to see candidates trying to find the matri from four equations, but this method was doomed to failure if the candidate, as happened almost every time, paired off the vertices incorrectly.

101 Question 9: y = 4 A(, 4) a) y 4 = m ( ) y = m m + 4 b) If the line intersects the curve then the coordinates of the point of intersection satisfy both equations simultaneously y y = 4 = = y = my m + 6 y so y m m my 4y + 6 m = 0 c)the line is a tangent when this equation has a repeated root meaning that the discriminant is 0 Discriminant = ( 4) 4 m (6 m) = 6 64m+ 48m = 0 m 4 m+ = 0 (m )( m ) = 0 m= or m= The equation of the two tangents are y = + and y = + d) If m =, the equation becomes y 4y+ 4 = 0 m = y = ( ) 0 y y+ = y y + 6 = 0 If, the equation becomes 4 0 y = ( 6) 0 The tangents touch the curve at (, ) and (9,6) y = and = y = 6 and = 9 Most candidates know that there is likely to be a question involving quadratic theory and are well equipped to answer it. In this case it was possible to answer parts (c) and (d) without having been successful in the earlier parts of the question, and this was often seen. At the same time there were many candidates who did well in parts (a) and (b) but made an error in the discriminant in part (c) leading to a significant loss of marks thereafter. In part (a) many candidates found a particular value for m rather than finding an equation that would be valid for all values of m. In part (b), as mentioned above, the elimination was not always carried out by the most efficient method, but many candidates still managed to establish the required equation. In part (c) some candidates lost all credit by failing to indicate that the discriminant must be zero for the line to be a tangent to the parabola; while others found the two gradients correctly but omitted the actual equations asked for in the question. Those who did find the correct gradients nearly always went on to gain all or most of the marks available in part (d), from a quadratic in y (the more direct way) or from a quadratic in Grade boundaries

102 General Certificate of Education January 009 Advanced Subsidiary Eamination MATHEMATICS Unit Further Pure MFP Thursday 5 January am to 0.0 am For this paper you must have: an 8-page answer book the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed: hour 0 minutes Instructions Use black ink or black ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Eamining Body for this paper is AQA. The Paper Reference is MFP. Answer all questions. Show all necessary working; otherwise marks for method may be lost. Information The maimum mark for this paper is 75. The marks for questions are shown in brackets. Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. P0885/Jan09/MFP 6/6/6/6/ MFP

103 Answer all questions. A curve passes through the point ð0, Þ and satisfies the differential equation dy pffiffiffiffiffiffiffiffiffiffiffiffiffi d ¼ þ Starting at the point ð0, Þ, use a step-by-step method with a step length of 0. to estimate the value of y at ¼ 0:4. Give your answer to five decimal places. (5 marks) The comple number þ i is a root of the quadratic equation þ b þ c ¼ 0 where b and c are real numbers. (a) Write down the other root of this equation. ( mark) (b) Find the values of b and c. (4 marks) Find the general solution of the equation tan p p ¼ ffiffi (5 marks) 4 It is given that (a) S n ¼ X n ðr r þ Þ r¼ Use the formulae for X n r and X n r to show that S n ¼ n. r¼ r¼ (5 marks) (b) Hence show that Xn r¼nþ ðr r þ Þ ¼kn for some integer k. ( marks) P0885/Jan09/MFP

104 5 The matrices A and B are defined by A ¼ k k k k, B ¼ k k k k where k is a constant. (a) Find, in terms of k : (i) A þ B ; (ii) A. ( mark) ( marks) (b) Show that ða þ BÞ ¼ A þ B. (4 marks) (c) It is now given that k ¼. (i) Describe the geometrical transformation represented by the matri A. ( marks) (ii) The matri A represents a combination of an enlargement and a reflection. Find the scale factor of the enlargement and the equation of the mirror line of the reflection. ( marks) 6 A curve has equation y ¼ ð Þð Þ ð Þ (a) (i) Write down the equations of the three asymptotes of this curve. ( marks) (ii) (iii) State the coordinates of the points at which the curve intersects the -ais. ( mark) Sketch the curve. (You are given that the curve has no stationary points.) (4 marks) (b) Hence, or otherwise, solve the inequality ð Þð Þ < 0 ( marks) ð Þ P0885/Jan09/MFP Turn over s

105 4 7 The points Pða, cþ and Qðb, dþ lie on the curve with equation y ¼ fðþ. The straight line PQ intersects the -ais at the point Rðr, 0Þ. The curve y ¼ fðþ intersects the -ais at the point Sðb, 0Þ. y c P y ¼ fðþ O R S a r b b d Q (a) Show that r ¼ a þ c b a c d (4 marks) (b) Given that a ¼, b ¼ and fðþ ¼0 4 (i) find the value of r ; ( marks) (ii) show that b r 0:8. ( marks) P0885/Jan09/MFP

106 5 8 For each of the following improper integrals, find the value of the integral or eplain why it does not have a value: (a) (b) (c) ð ð ð 4 d ; 5 4 d ; ð 4 5 4Þ d. ( marks) ( marks) ( mark) 9 A hyperbola H has equation y ¼ (a) (b) Find the equations of the two asymptotes of H, giving each answer in the form y ¼ m. ( marks) Draw a sketch of the two asymptotes of H, using roughly equal scales on the two coordinate aes. Using the same aes, sketch the hyperbola H. ( marks) (c) (i) Show that, if the line y ¼ þ c intersects H, the -coordinates of the points of intersection must satisfy the equation c ðc þ Þ ¼0 (4 marks) (ii) Hence show that the line y ¼ þ c intersects H in two distinct points, whatever the value of c. ( marks) (iii) Find, in terms of c, the y-coordinates of these two points. ( marks) END OF QUESTIONS P0885/Jan09/MFP

107 MFP - AQA GCE Mark Scheme 009 January series MFP Q Solution Marks Total Comments First increment is 0., so y. BB PI; variations possible here Second increment is M , so y A,F 5 A if accuracy lost; ft num error Total 5 (a) Other root is i B (b) Sum of roots = 4 BF ft error in (a) So b = 4 BF ft wrong value for sum Product is B So c = BF 4 ft wrong value for product Alternative: Substituting + i into equation M Equating R and I parts m + b = 0, so b = 4 A 5 + b + c = 0, so c = AF (4) ft wrong value for b Total 5 tan π = B Decimals/degrees penalised at 5 th mark Introduction of nπ M (or nπ) at any stage Going from π to m Including dividing all terms by = π + nπ Allow +, or ±; A with dec/deg; A,F 5 8 ft wrong first solution Total 5 4(a) Sn = Σr Σ r+σ M Correct epressions substituted m At least for first two terms Correct epansions A Σ = n B Answer convincingly obtained A 5 AG (b) S n S n attempted M Condone S n S n+ here Answer 7n A Total 7 4

108 MFP - AQA GCE Mark Scheme 009 January series MFP (cont) Q Solution Marks Total Comments 0 k 5(a)(i) A + B = k 0 B (ii) A = k 0 0 k B, B if three entries correct (b) (A + B) 4k 0 = = B, B if three entries correct 0 4k B = A, hence result BB 4 (c)(i) A is an enlargement (centre O) M with SF A Condone k (ii) Scale factor is now B Condone k Mirror line is y = tan MA Total 6(a)(i) Asymptotes = 0, =, y = B (ii) Intersections at (, 0) and (, 0) B (iii) At least one branch approaching asymptotes B Each branch B 4 (b) 0 < <, < < B,B Allow B if one repeated error occurs, eg for < Alternative: Complete correct algebraic method MA () Total 0 7(a) Use of similar triangles or algebra M Some progress needed Correct relationship established ma eg r a b = a c c d Hence result convincingly shown A 4 AG (b)(i) c = f(a) = 4, d = f(b) = B,B r = 8 5 (.5) BF Allow AWRT.5; ft small error (ii) β = 0.74(4) MA Allow AWRT.7 So β r (AG) A Allow only dp if earlier values to dp Total 0 5

109 MFP - AQA GCE Mark Scheme 009 January series MFP (cont) Q Solution Marks Total Comments 8(a) 4 4 d= 4 (+ c) MA M if inde correct This tends to as, so no value AF ft wrong coefficient (b) d= 4 (+ c) MA M if inde correct 5 4 d = 0 ( 4) = 4 AF ft wrong coefficient (c) Subtracting 4 leaves, so no value BF ft if c has no value in (a) but has a finite answer in (b) Total 7 9(a) Asymptotes are y =± MA MA0 if correct but not in required form (b) Asymptotes correct on sketch BF With gradients steeper than ; ft from y = ± m with m > Two branches in roughly correct positions B Approaching asymptotes correctly B Asymptotes y = ± m needed here (c)(i) Elimination of y M Clearing denominator correctly M c (c + ) = 0 ma 4 Convincingly found (AG) (ii) Discriminant = 8c + 8 B Accept unsimplified > 0 for all c, hence result E OE (iii) Solving gives y = + c = c c = ± ( + ) MA c ( c ) ± + A Accept Total 4 TOTAL 75 y = c+ c± 8c + 8 6

110 Further pure - AQA - January 009 Question : Let's call f( ) = + we use the Euler formula: y = y + hf ( ) with h = 0. n+ n n = 0, y =, f ( ) = so y = + 0. =. y f = 0., =., ( ) = + 0. =.098 so y y = =.4096 Question : + b + c = 0 has root + i a)the other root is the conjugate i bb ) = (( + i) + ( i)) = b = 4 c= ( + i)( i) = = c = Question : π π Tan = = Tan π π = + kπ π = + kπ 6 π π = k 8 k Question 4: n n n n n S = r r+ = r r+ r= r= r= r= = nn ( + )(n+ ) nn ( + ) + n 6 = n[ ( n+ )(n+ ) ( n+ ) + ] = n( n + n+ n+ n + ) Sn = n( n ) = n n n n b) (r r+ ) = (r r+ ) (r r + ) r= n+ r= r= = ( n) n = 7n The great majority of candidates were able to make a good start to the paper by giving a correct numerical solution to the differential equation. As on past papers, some candidates used the value of the derivative at the upper end of the interval rather than the lower end. This was perfectly acceptable for full marks, though these candidates gave themselves slightly more calculation to carry out as they were not using the very simple value of the derivative at = 0. Some candidates carried out three iterations instead of two; these candidates lost a mark as well as giving themselves etra work. Almost every candidate gave the right answer to part (a), and many went on to use the relationships between the roots and coefficients of a quadratic equation to answer part (b) correctly, though quite a number did not insert a minus sign for the coefficient of. Some candidates followed other approaches, such as substituting the known root into the equation, epanding and equating real and imaginary parts, which if done carefully usually led to the right answers. The tangent function is the most friendly trigonometrical function for this type of equation, and it was noticeable that the great majority of candidates were aware that the period of the function was π, not π. Unfortunately a substantial number of candidates, after introducing the general term nπ, failed to divide it by and thus lost three of the five marks available. Part (a) of this question was very well answered. The algebra needed was not quite as heavy as is sometimes seen in questions on this topic, and the fact that the answer was given seemed to steer candidates smoothly to a correct solution. Only a small proportion of candidates wrote for Σ. In part (b), only a minority of candidates showed any awareness of the need for a subtraction, but those who did show this awareness usually found the right answer with very little difficulty.

111 Question 5: k k k k A = and B k k = k k 0 k ai )) A+ B= k 0 ii) A b)( A B) and A k 0 = 0 k = = + B k k k k 0 k 0 0 4k k 0 k 0 4k 0 = + = 0 k 0 k 0 4k so indeed ( + ) = + A B A B 0 ci ) ) k=, A = 0 represents an enlargement centre O, scale factor ii) A = The point A(,0)is transformed into A'(,) OA= and OA' = the scale factor of the enlargement is The mirror line is the bissector of the angle (AOA'), o the equation of this line is y = Tan(.5 ) The first nine marks in this question seemed to be found very easy for most candidates to earn. The presence of so many k s in the matrices did not prevent them from carrying out all the calculations correctly and confidently, though sometimes the flow of the reasoning in part (b) was not made totally clear. Part (c)(ii), on the other hand, proved too hard for many candidates, who seemed to resort to guesswork for their answers. Question 6: ( )( ) y = ( ) a) i)" vertical asymptotes" = 0 and = y = = y = is asymptote to the curve ii)the curve intersects the -ais when y = 0 which means ( )( ) = 0 = or = The curve crosses the -ais at (,0) and (,0) iii) c) y< 0 when the curve is below the -ais this happens whn e 0 < < or < < The first two parts of this question were generally well answered, and there were many good sketches in part (a)(iii), though the middle branch of the curve was often badly drawn, many candidates appearing to ignore the helpful information provided about there being no stationary points. Part (b) could be answered very easily by looking at the parts of the graph which were below the -ais, and reading off the two intervals in which this was the case. Instead, many candidates used a purely algebraic approach, and in most cases the inequality was rapidly reduced to the incorrect form ( )( ) < 0, though some candidates obtained the correct version, ( )( )( ) < 0, and after making a table of signs came up with the correct solution.

112 Question 7: a) Let's work out the equation of the line PQ d c the gradient is b a d c Eq: y c= ( a) b a This line intersects the -ais when y = 0 d c b a so c= ( r a) c = r a b a d c b a r = a+ c c d b) a =, b = and f ( ) = 0 Question 8: 4 i) c= f( a) = 0 = 40 6 = 4 4 d = f( b) = 0 = 60 8 = 8 so r = + 4 = ii Solve f = = 4 ) ( ) so β r 4 4 a) d = 4 + c = 4 The integral has no value. 4 = (0 ) 0 = 0 or = b) d = 4 + c when, 0 a 5 a so d = 4 = 4 a ( 4) 4 a d = c) d = c = c The integral has no value. There was a poor response to this question, particularly in part (a) where many candidates seemed to be unfamiliar with the background to the process of linear interpolation. Those who were completely or partially successful followed one of two approaches, a geometrical approach based on similar triangles or an algebraic approach based on the equation of the line PQ. In both approaches there were frequent sign errors which prevented the candidate from legitimately obtaining the required formula, but some candidates produced clear, concise and correct proofs. Part (b)(i) was more often productive for candidates, some of whom used an alternative correct formula for linear interpolation. In part (b)(ii), many candidates did not realise that they had to solve the equation f() = 0 to find the value of β, and those who did solve the equation sometimes lost accuracy in establishing the required approimate value. Most candidates were able to integrate the given powers of, and many were able to draw the correct conclusions, though sign errors were common in part (b). In part (c), most of those who had been successful so far were able to complete the task successfully, but it was noticeable that some candidates did not take a hint from the award of only one mark for this part of the question: instead of drawing a quick conclusion from the results already obtained they went through the whole process of integrating, substituting and taking limits, possibly losing valuable time.

113 Question 9: y y = = a) Asymptotes : y = ± b) ci ) ) y= + c intersects the hyperbola so the -coordinate of the point of intersection satisfy: ( + c) = ( ) c c = c ( c + ) = 0 ii) Let's work out the discriminant: ( c) 4 ( c + ) 4c + 4c + 8= 8c + 8 > 0 The discriminant is positive for all values of c, meaning that the line y = + c crosses the curve at two points for all the values of c. c± 8c + 8 c± c + iii) Solving the equation : = = = c± c + y = + c so y = c± c + As in the June 008 paper, there were many candidates who appeared to struggle with the earlier parts of the final question but then came into their own when they reached the epected test of quadratic theory. Although the asymptotes of a hyperbola had appeared before on MFP papers, many candidates seemed unfamiliar with them, and even with the hyperbola itself. A common error in part (a) was to fail to take a square root. In part (b), even those who drew the asymptotes and the two branches in roughly the right positions rarely showed the curve actually approaching the supposed asymptotes. Part (c)(i) was etremely well answered, the printed answer guiding most candidates unerringly towards a correct piece of algebra, though some made a sign error and then made another one in order to reach the answer. The minus signs caused many candidates to go wrong in finding the discriminant for part (c)(ii), and the fact that the discriminant needed to be strictly positive was not stated in many cases. The simplest approach to part (c)(iii) was to solve the given equation for in terms of c and then to add another c to obtain the values of y, but most candidates did not seem to realise this. Many candidates obtained a valid quadratic equation for y but still failed to write down the quadratic formula as applied to their equation, and thus failed to earn any marks in this part. Those who attempted to substitute values into the formula, either for or for y, often omitted the factor c from the coefficient of and/or ignored the minus sign in front of this coefficient. Grade boundaries

114 General Certificate of Education June 009 Advanced Subsidiary Eamination MATHEMATICS Unit Further Pure MFP Monday June am to 0.0 am For this paper you must have: an 8-page answer book the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed: hour 0 minutes Instructions Use black ink or black ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Eamining Body for this paper is AQA. The Paper Reference is MFP. Answer all questions. Show all necessary working; otherwise marks for method may be lost. Information The maimum mark for this paper is 75. The marks for questions are shown in brackets. Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. P575/Jun09/MFP 6/6/ MFP

115 Answer all questions. The equation þ 8 ¼ 0 has roots a and b. (a) Write down the values of a þ b and ab. ( marks) (b) Find the value of a þ b. ( marks) (c) Find a quadratic equation which has roots 4a and 4b. Give your answer in the form þ p þ q ¼ 0, where p and q are integers. ( marks) A curve has equation y ¼ 6 þ 5 The points A and B on the curve have -coordinates and þ h respectively. (a) Find, in terms of h, the gradient of the line AB, giving your answer in its simplest form. (5 marks) (b) Eplain how the result of part (a) can be used to find the gradient of the curve at A. State the value of this gradient. ( marks) The comple number z is defined by where is real. z ¼ þ i (a) Find, in terms of, the real and imaginary parts of: (i) z ; ( marks) (ii) z þ z. ( marks) (b) Show that there is eactly one value of for which z þ z is real. ( marks) P575/Jun09/MFP

116 4 The variables and y are known to be related by an equation of the form where a and b are constants. y ¼ ab (a) Given that Y ¼ log 0 y, show that and Y must satisfy an equation of the form Y ¼ m þ c ( marks) (b) The diagram shows the linear graph which has equation Y ¼ m þ c. Y ~ Use this graph to calculate: 0 O ~ (i) (ii) an approimate value of y when ¼ :, giving your answer to one decimal place; an approimate value of when y ¼ 80, giving your answer to one decimal place. (You are not required to find the values of m and c.) (4 marks) 5 (a) Find the general solution of the equation cosð pþ ¼ giving your answer in terms of p. (6 marks) (b) From your general solution, find all the solutions of the equation which lie between 0p and p. ( marks) P575/Jun09/MFP Turn over s

117 4 6 An ellipse E has equation þ y 4 ¼ (a) (b) (c) Sketch the ellipse E, showing the coordinates of the points of intersection of the ellipse with the coordinate aes. ( marks) The ellipse E is stretched with scale factor parallel to the y-ais. Find and simplify the equation of the curve after the stretch. The original ellipse, E, is translated by the vector ellipse is ( marks) a. The equation of the translated b 4 þ y 8 þ 6y ¼ 5 Find the values of a and b. (5 marks) 7 (a) Using surd forms where appropriate, find the matri which represents: (i) a rotation about the origin through 0 anticlockwise; ( marks) (ii) a reflection in the line y ¼ p ffiffi. ( marks) (b) The matri A, where " A ¼ pffiffi # pffiffi represents a combination of an enlargement and a reflection. Find the scale factor of the enlargement and the equation of the mirror line of the reflection. ( marks) (c) The transformation represented by A is followed by the transformation represented by B, where " pffiffiffi # B ¼ pffiffiffi Find the matri of the combined transformation and give a full geometrical description of this combined transformation. (5 marks) P575/Jun09/MFP

118 5 8 A curve has equation y ¼ ð Þð 5Þ (a) Write down the equations of the three asymptotes to the curve. ( marks) (b) Show that the curve has no point of intersection with the line y ¼. ( marks) (c) (i) Show that, if the curve intersects the line y ¼ k, then the -coordinates of the points of intersection must satisfy the equation ðk Þ 6k þ 5k ¼ 0 ( marks) (ii) Show that, if this equation has equal roots, then kð4k þ 5Þ ¼0 ( marks) (d) Hence find the coordinates of the two stationary points on the curve. (5 marks) END OF QUESTIONS P575/Jun09/MFP

119 MFP - AQA GCE Mark Scheme 009 June series MFP Q Solution Marks Totals Comments (a) α + β =, αβ = 4 BB (b) α + β = ( ) ( 4) = 8 MAF M for substituting in correct formula; 4 ft wrong answer(s) in (a) (c) Sum of roots = 4(8 4 ) = BF ft wrong answer in (b) Product = 6(αβ) = 56 BF ft wrong answer in (a) Equation is + 56 = 0 BF ft wrong sum and/or product; allow p =, q = 56 ; condone omission of = 0 Total 7 (a) When =, y = B PI Use of ( + h) = 4 + 4h + h M Correct method for gradient M Gradient = h h h h A, 5 A if only one small error made (b) As h tends to 0, E, E for h = 0... the gradient tends to BF dependent on at least E ft small error in (a) Total 8 (a)(i) z = ( 4) + i(4 ) MA M for use of i = R and I parts clearly indicated AF Condone inclusion of i in I part ft one numerical error z + z = i 4 4 MAF M for correct use of conjugate ft numerical error in (i) (ii) ( ) ( ) (b) z + z real if imaginary part zero M... ie if = AF ft provided imaginary part linear Total 7 4(a) lg(ab ) = lg a + lg(b ) M Use of one log law... = lg a + lg b M Use of another log law A Correct relationship established [ SCAfter M0M0, B for correct form ] (b)(i) When =., Y., so y.6 MA Allow.7; allow NMS (ii) When y = 80, Y.90, so. MA 4 M for Y.9, allow NMS Total 7 4

120 MFP - AQA GCE Mark Scheme 009 June series MFP (cont) Q Solution Marks Totals Comments 5(a) cos π = Decimals/degrees penalised at 6th B mark only Appropriate use of ± B OE Introduction of nπ M (or nπ) at any stage Going from π to m including dividing all terms by = π ± π + n π A,F 6 OE; A with decimals and/or degrees; ft 9 wrong first solution (b) At least one value in given range M compatible with c s GS Correct values 9 π, 94 π, 98 π A, A if one omitted or wrong values included; A0 if only one correct value given Total 9 6(a) Ellipse with centre of origin B Allow unequal scales on aes ( ±,0) and (0 ± ) shown on diagram B, Condone AWRT.7 for ; B for incomplete attempt (b) y replaced by Equation is now y MA MA0 for y instead of y y + = A 6 (c) Attempt at completing the square ( y ) 4( ) M AA [Alt: replace by a and y by y b (M) M if one replacement correct 4 8a + y 6by...] (ma) Condone errors in constant terms a = and b = AA 5 Total 5

121 MFP - AQA GCE Mark Scheme 009 June series MFP (cont) Q Solution Marks Totals Comments 7(a)(i) cos0 sin 0 Matri is MA M for sin 0 cos0 (PI) (ii) Matri is cos60 sin 60 MA M for sin 60 cos60 (PI) (b) SF, line y = BB OE (c) Attempt at BA or AB M 0 4 BA = 4 0 ma m if zeros in correct positions Enlargement SF 4 BF ft use of AB (answer still 4) 0 k or after BA = k 0... and reflection in line y = BF 5 0 k ft only from BA = k 0 Total 8(a) Asymptotes =, = 5, y = B (b) y = ( )( 5) = M OE = 0 m OE Disc t = 6 40 < 0, so no pt of int n A convincingly shown (AG) (c)(i) y = k = k( 6 5) M OE... ( k ) 6k+ 5k = 0 A convincingly shown (AG) (ii) Discriminant = 6k 0k(k ) M OE... = 0 when k(4k + 5) = 0 A convincingly shown (AG) (d) k = 0 gives = 0, y = 0 B 5 k = gives = MA OE ( 5) = 0, so = 5 A y = 5 4 B 5 Total 5 TOTAL 75 6

122 Further pure - AQA - June 009 Question : = has roots α and β 8 a) α + β = and αβ = = 4 ) ( ) 4 8 b α + β = α + β αβ = = + 4 = 8 c ) 4α + 4β = 4( α + β ) = = 4α 4β = 6( αβ ) = 6 6 = 56 An equation with roots 4α and 4β is = 0 Question : y = 6+ 5 for =, y = = for = + h, y = ( + h) 6( + h) + 5 = 4 + 4h + h 6h + 5 h h A(, ) and B( + h, h h ) ( h h ) ( ) h h a) The gradient of the line AB : = = h + h h b) When h tends to 0, the chord tends to the tangent and its gradient is Question : z = + i ) ) ( ) 4 4 ( 4) 4 aiz = + i = + i = + i Re( z ) 4 Im( z ) 4 ) ( 4) (4 4) Re( ) 4 z Im( ) 4 4 ) ii z + z = + i + i = + + i b z z = = + z = + + z = + z is real when = = Most candidates found this a very good introduction to the paper and gained full marks, or very nearly so. Errors in parts (a) and (b) were rare, and were nearly always sign errors rather than those caused by the use of an incorrect procedure. In part (c) the most common mistake was to have a 4 instead of a 6 in the product of the roots of the required equation. Some candidates failed to see the connection with part (b) when finding the sum of the roots, but more often than not they still found the right value for this sum. Almost all candidates knew how to find the gradient of the line in part (a) of this question, but a distressingly large minority of them made a sign error in subtracting the y-values of A and B, which led to the introduction of an unwanted term -6/h in the answer. In part (b) this should have caused difficulties, but almost invariably the candidates took this term as tending to zero as h tended to zero. Those candidates who wrote h = 0 instead of letting h tend to zero lost a mark here, but it is pleasing to report that this error was comparatively rare. In part (a) (i) of this question most candidates showed some knowledge of comple numbers but failed to display clearly the real and imaginary parts asked for in the question. Part (a) (ii) appeared to cause candidates no trouble at all; but by way of contrast, very few candidates saw what was required in part (b): many equated the real part to zero, while others equated the real and imaginary parts to each other.

123 Question 4: y = ab a) log y = log ( ab ) 0 0 Y = log a+ log b 0 0 Y = log b+ log a Y = m + c 0 0 b) i) when =., Y =. so ii) y = 80, Y = log so. m = log b and c= log a y = Question 5: π Cos( π ) = = Cos π π π = + kπ or π = + kπ 4π π = + kπ or = + kπ 4π π π π = + k or = + k π 99π b)0π < < π < < 9 9 4π + k6π 94 and = k = 5 = π 9 9 π + k4π 94π or = k = = 9 98π k = 4 = Question 6: y + = 4 a) When = 0, y = ± When y = 0, = ± y b)the equation of the ellipse after the transformation is : + = 4 y + = 6 ( a) ( y b) c)the equation of the ellipse after the translation is : + = 4 4( a) + ( y b) = 8a = 8 gives a = 6b = 6 gives b = a + a + y by + b = Part (a) was very familiar to the majority of candidates perhaps too familiar in many cases as the result lg (ab ) = lg a + lg b was quoted rather than properly shown by the use of the laws of logarithms. Part (b) presented a less familiar type of challenge. It was not necessary or indeed helpful to try to find values for the constants m and c. All that was needed was to read values directly from the graph and to convert as appropriate between y and Y. Part (a) was a standard type of question on this paper, but as on past papers it was common to see candidates introducing the general term nπ after they had divided both sides of the equation by. Another very frequent source of confusion came from an inappropriate use of the ± symbol. Candidates who wrote out the whole equation twice, once with a plus sign and once with a minus sign, at an early stage, usually obtained a correct general solution, whereas those who used the plus-or-minus sign often ran into errors when adding π to each side of the equation. Part (b) was very poorly answered. Many candidates used n = 0 or n = in their general solution and did not seem to notice that the resulting values of were well below the minimum value of 0π required by the question. Some candidates wrote down the correct answers but failed to indicate how they had found them from their general solutions as specified in the wording of the question. The sketches of the ellipse in part (a) of this question were mostly satisfactory, though some candidates failed to take square roots when working out the required coordinates. Part (b) was one of those parts of questions where the absence of any eplanation made it hard for the eaminers to see what the candidates were trying to do. Some gave the answer in simplified form without any preliminary working. On this occasion the eaminers condoned this poor eamination technique where it looked as if the candidate had done the right thing. But there were many errors here: some modified the term rather than the y term, some multiplied the y by instead of dividing it by, and many failed to include this in the squaring process when simplifying their equations. Part (c) was found to be rather hard, but many candidates realised that completing the square was needed and at least made an effort, often failing to see the connection between their correct equations and the required numbers a and b. The alternative approach of applying the translation from the outset, using the letters a and b, was about as popular as the completing the square method, but the algebra became too heavy for many candidates and the necessary comparing of terms did not take place.

124 Question 7: o a) i) rotation about the origin 0 anticlockwise : cos0 sin 0 sin 0 cos0 = ii) reflection in the line y = = ( Tan0) : cos 60 sin 60 sin 60 cos 60 = ) b A= = Arepresents a combination of an enlargement scale factor and the reflection in the line y = 0 4 c) Let's work out BA: BA= 4 0 This represents an enlargement scale factor 4 and a reflection in the line y = In this question part (a) was very well answered in general, candidates being able to choose the correct form of matri for each type of transformation and to insert the correct forms for the necessary sines and cosines. In part (b), however, the majority of candidates failed to see the close connection between the given matri and the one that they had found in part (a) (ii). The answers given often appeared to be wild guesses. In part (c) many candidates saw the need to multiply the two matrices but often in the wrong order. Those who obtained the correct matri often confused it with a scalar matri when giving the geometrical interpretation.

125 Question 8: y = ( )( 5) a)" vertical asymptotes" = and = 5 y = = y = b) Let ' s solve = = = = 40 = 4< 0 Discriminant:(-6) There is no solution, theline y = does not cross the curve c) i) If y = kcrosses the curve, then is solution to = k 6+ 5 = k 6k + 5k ) ii)the equation has equal roots when the discriminant is 0 Discriminan ( k k + k = k t = ( 6k) 4 ( ) 5 = k k k + 5k = 0 k(4k+ 5) = 0 k k = k + k = Part (a) was usually well answered, with most candidates writing down the equations of the two vertical asymptotes but then struggling to establish the equation of the horizontal asymptote; although the correct answer y = appeared more frequently than the most common wrong answer, y = 0. Part (b) was slightly unfamiliar, but most candidates tackled it confidently. A sign error often led to the disappearance of the term, while in other cases the quadratic and its discriminant were correctly found but the candidates omitted to mention the essential feature of the discriminant, ie the fact that it was negative, indicating the absence of any points of intersection. Candidates were mostly on familiar ground with part (c), but marks were sometimes lost in part (c) (i) by a failure to show enough steps to justify the printed answer, and in part (c) (ii) by a sign error which prevented the candidate from reaching the required equation legitimately. Most candidates were able to make a good attempt at part (d) in the time remaining for them. The origin was usually given correctly as one stationary point, and many found the -coordinate of the other stationary point after a more or less lengthy calculation. Many failed to see that the y-coordinate was simply the value of k used in this calculation, but they were able to calculate it correctly from the equation of the curve. 5 d) This equation gives k = 0or k = 4 for k = we have = = and y = 0, for k = we have + = =, ( 5) = 0 = and y = The stationary points are (0,0) and (, ) 4 Grade boundaries

126 General Certificate of Education Advanced Subsidiary Eamination January 00 Mathematics MFP Unit Further Pure Wednesday January 00.0 pm to.00 pm For this paper you must have: an 8-page answer book the blue AQA booklet of formulae and statistical tables an insert for use in Question 6 (enclosed). You may use a graphics calculator. Time allowed hour 0 minutes Instructions Use black ink or black ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Eamining Body for this paper is AQA. The Paper Reference is MFP. Answer all questions. Show all necessary working; otherwise marks for method may be lost. Fill in the boes at the top of the insert. Information The marks for questions are shown in brackets. The maimum mark for this paper is 75. Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. P60/Jan0/MFP 6/6/6/ MFP

127 Answer all questions. The quadratic equation 6 þ ¼ 0 has roots a and b. (a) Write down the values of a þ b and ab. ( marks) (b) Show that a þ b ¼ 6. ( marks) (c) Find a quadratic equation, with integer coefficients, which has roots a b and b a. (4 marks) The comple number z is defined by z ¼ þ i (a) Find the value of z, giving your answer in its simplest form. ( marks) (b) Hence show that z 8 ¼ 6. ( marks) (c) Show that ðzþ ¼ z. ( marks) Find the general solution of the equation sin 4 þ p ¼ 4 (4 marks) P60/Jan0/MFP

128 4 It is given that A ¼ 4 and that I is the identity matri. (a) Show that ða IÞ ¼ ki for some integer k. ( marks) (b) Given further that find the integer p such that B ¼ p ða BÞ ¼ðA IÞ (4 marks) 5 (a) Eplain why ð 6 0 d is an improper integral. ( mark) (b) For each of the following improper integrals, find the value of the integral or eplain briefly why it does not have a value: (i) (ii) ð 6 0 ð 6 0 d ; 5 4 d. ( marks) ( marks) Turn over for the net question P60/Jan0/MFP Turn over s

129 4 6 [Figure, printed on the insert, is provided for use in this question.] The diagram shows a rectangle R. y ~ 5 R O 5 0 ~ (a) The rectangle R is mapped onto a second rectangle, R, by a transformation with 0 matri. 0 (i) Calculate the coordinates of the vertices of the rectangle R. ( marks) (ii) On Figure, draw the rectangle R. ( mark) (b) The rectangle R is rotated through 90 clockwise about the origin to give a third rectangle, R. (i) On Figure, draw the rectangle R. ( marks) (ii) Write down the matri of the rotation which maps R onto R. ( mark) (c) Find the matri of the transformation which maps R onto R. ( marks) P60/Jan0/MFP

130 5 7 A curve C has equation y ¼ ð Þ. (a) (i) Write down the equations of the asymptotes of the curve C. ( marks) (ii) Sketch the curve C. ( marks) (b) The line y ¼ intersects the curve C at a point which has -coordinate a. (i) Show that a lies within the interval <<4. ( marks) (ii) Starting from the interval <<4, use interval bisection twice to obtain an interval of width 0.5 within which a must lie. ( marks) 8 (a) Show that Xn r¼ r þ X n r¼ r can be epressed in the form knðn þ Þðan þ bn þ cþ where k is a rational number and a, b and c are integers. (4 marks) (b) Show that there is eactly one positive integer n for which Xn r¼ r þ X n r r¼ r¼ r ¼ 8 X n (5 marks) Turn over for the net question P60/Jan0/MFP Turn over s

131 6 9 The diagram shows the hyperbola y a b ¼ and its asymptotes. y P Að, 0Þ The constants a and b are positive integers. The point A on the hyperbola has coordinates ð, 0Þ. The equations of the asymptotes are y ¼ and y ¼. (a) Show that a ¼ and b ¼ 4. (4 marks) (b) The point P has coordinates ð, 0Þ. A straight line passes through P and has gradient m. Show that, if this line intersects the hyperbola, the -coordinates of the points of intersection satisfy the equation ðm 4Þ m þðm þ 6Þ ¼0 (4 marks) (c) Show that this equation has equal roots if m ¼ 6. ( marks) (d) There are two tangents to the hyperbola which pass through P. Find the coordinates of the points at which these tangents touch the hyperbola. (No credit will be given for solutions based on differentiation.) (5 marks) END OF QUESTIONS P60/Jan0/MFP

132 MFP - AQA GCE Mark Scheme 00 January series Q Solution Mark Total Comments (a) α + β =, αβ = BB (b) α + β = ( α + β ) αβ ( α + β ) M or other appropriate formula... = 8 ( )() = 6 ma m for substn of numerical values; A for result shown (AG) α + β (c) Sum of roots = αβ 6... = = 8 AF ft wrong value for αβ Product = αβ = BF ditto Equation is 54 + = 0 AF 4 Integer coeffs and = 0 needed; ft wrong sum and/or product Total 9 (a) z = + i + i = i MA M for use of i = (b) z 8 = (i) 4 M or equivalent complete method... = 6i 4 = 6 A convincingly shown (AG) (c) (z) = ( i) M for use of z = i... = i = z A convincingly shown (AG) Total 6 sin π = stated or used B Deg/dec penalised in 4th mark Introduction of nπ M (or nπ) at any stage Going from 4 + π to 4 m incl division of all terms by 4 = π + n π 6 A 4 or equivalent unsimplified form Total 4 4(a) 0 I = 0 B stated or used at any stage M Attempt at (A I) M with at most one numerical error ( A - I ) = = 0 0 I A (b) 0 A- B = p 0 B p 0 ( A- B ) = 0 p MA M A0 if entries correct... = (A I) for p = 9 AF 4 ft wrong value of k Total 7 4

133 MFP - AQA GCE Mark Scheme 00 January series MFP Q Solution Mark Total Comments 5(a) as 0 E Condone has no value at = 0 (b)(i) d= (+c) 6 0 d= MA AF M for correct power of ft wrong coefficient of (ii) d= 4 (+c) MA M for correct power of 4 as 0, so no value EF ft wrong coefficient of 4 Total 7 6(a)(i) Coords (, ), (9, ), (9, 4), (, 4) MA M for multn of by or y by (PI) (ii) R shown correctly on insert B (b)(i) R shown correctly on insert B,F B for rectangle with vertices correct; ft if c s R is a rectangle in st quad 0 (ii) Matri of rotation is 0 B (c) Multiplication of matrices M (either way) or other complete method 0 Required matri is 0 A Total 8 7(a)(i) Asymptotes =, y = 0 BB (ii) One correct branch B Both branches correct B no etra branches; = shown (b)(i) f() =, f(4) = B where f() = ( )( ) ; OE Sign change, so α between and 4 E (ii) f(.5) considered first M OE but must consider =.5 f(.5) > 0 so < α <.5 A Some numerical value(s) needed f(.5) < 0 so.5 < α <.5 A Condone absence of values here Total 9 5

134 MFP - AQA GCE Mark Scheme 00 January series MFP Q Solution Mark Total Comments 8(a) Σr + Σr = M at least one term correct n ( n+ ) + n( n+ ) 4 Factor n clearly shown m or n + clearly shown to be a factor... = ( )( ) 4 nn+ n n+ AA 4 OE; A for, A for quadratic 4 (b) Valid equation formed M Factors n, n + removed m n 9n 0 = 0 A OE Valid factorisation or solution m of the correct quadratic n = 0 is the only pos int solution A 5 SC / for n = 0 after correct quad Total 9 9(a) 4 E, E for verif n or incomplete proof =, y = 0 0= so a = a b Asymps ± = ± so b = a = 4 E, 4 ditto a (b) Line is y 0 = m( ) B OE Elimination of y M 4 m ( + ) = 6 A OE (no fractions) So (m 4) m + (m + 6) = 0 A 4 convincingly shown (AG) (c) Discriminant equated to zero M 4m 4 4m 4 64m + 6m + 56 = 0 A OE m + 6 = 0, hence result A convincingly shown (AG) (d) 6 m = = 0 M = 0, so = 4 ma Method for y-coordinates m using m = ± 4 or from equation of hyperbola; dep t on previous m y = ± 4 A 5 Total 6 TOTAL 75 6

135 Further pure - AQA - January 00 Question : 6 + = 0 has roots α and β 6 a) α + β = = and αβ = b α + β = α + β αβ α + β = = = α + β = 6 ) ( ) ( ) () 8 6 α β α + β 6 α β c) + = = = 8 and = αβ = β α αβ β α α β An equation with the roots and is 8 + = 0 β α Question : z = + i a) z = ( + i) = + i+ i ( ) ( ) b) z = z = i = 6i = i = 6 c) ( z) = ( i) = i = i = z Question : π π Sin 4 + = = Sin 4 π π 4+ = + kπ 4 π 4= + kπ 4 π π = + k 6 Question 4: a) ( A I) = = = I p 0 b)( A B) = I p 0 = = 0 p so p = 9 = 0 Most candidates seemed to be very familiar with the techniques needed in this question. The formula for the sum of the cubes of the roots was either quoted confidently and correctly or worked out from first principles. Errors occurred mainly in part (c): algebraic errors in finding the sum or even the product of the roots of the required equation; errors in choosing which numerical values to substitute for αβ or α + β ; and, very frequently, a failure to present the final answer in an acceptable form, with integer coefficients and the = 0 at the end. Full marks were usually awarded in this question. Answers to part (b) were sometimes very laborious but eventually correct, but by contrast some answers were so brief as to be not totally convincing, earning one mark out of two. A few candidates fell short of full credit in part (c) by working on (z) but not mentioning z. This trigonometric equation was slightly more straightforward than usual, in that there was only one solution of the equation between 0 and π. For many candidates, this did not appear to make things simpler at all: they applied a general formula for sin θ = sin α and did not always realise that their two solutions were equivalent. They were not penalised as long as the second solution was correct, but this was not always the case. What was etremely pleasing to see from an eaminer s point of view was that the majority of candidates carried out the necessary operations in the right order, so that all the terms, including the nπ term, were divided by 4. As usual on this paper, the work on matrices was very good indeed, with most candidates working out all the steps efficiently. Some tried to epand the epressions (A I) and (A B) but almost invariably assumed commutativity of multiplication. A rather silly way to lose a mark was to work correctly to the equation p = and then to solve this equation incorrectly, which happened quite frequently.

136 Question 5: 6 0 a) d is an improper because is not defined when = b) i) d = + c d = = 6 0 = ii) d = 4 + c when 0, The integral has no value Question 6: ai )) 0 = 4 4 The coordinates of the vertices of R are(, ),(, 4),(9, 4),(9, ) ii) bi )) o ii)90 clockwise is equivalent o cos 70 sin 70 0 to 70 anticlockwise : = sin 70 cos c ) = Most candidates showed confidence with the integration needed in this question but were much less confident with the concept of an improper integral. The eplanations in part (a) were often very wide of the mark, and indeed quite absurd, while in other cases the statements made were too vague to be worthy of the mark, using the word it without making it clear whether this referred to the integrand or to the integrated function. Another mark was often lost at the end of the question, where candidates thought that 0 -/4 was equal to zero. As the candidates turned the page to tackle this question, there was a noticeable dip in the level of their performance. The majority of candidates showed a surprising lack of ability to work out the coordinates of image points under a transformation given by a matri. A common misunderstanding was to carry out a two-way stretch with centre (, ) instead of with centre (0, 0). Luckily the candidates still had the chance to carry out the required rotation in part (b)(i) using their rectangle from part (a). Part (b)(ii) was often answered poorly, some candidates being confused by the clockwise rotation, when the formula booklet assumes an anticlockwise rotation, and many candidates failing to give numerical values for cos 70 and sin 70. Most candidates realised that a matri multiplication was needed in part (c), but many used the wrong matrices or multiplied the matrices in the wrong Question 7 y = ( ) a) i)" Vertical asymptote" = When, y 0 y = 0 is asymptote to the curve. ii) b) The line y = intersects the curve. satisfy both equations Let's call f( ) = ( )( ) = = ( ) ( )( ) 0 f () = < 0 and f (4) = > 0 According the the change sign rule, we know that there is a least one solution between and 4. ii) f (.5) = 0.5 > 0 < α <.5 f (.5) = < 0.5 < α <.5 Most candidates started well by writing down = as the equation of one asymptote to the given curve, and then struggled to find the horizontal asymptote, though most were ultimately successful. The graph was often drawn correctly but almost equally often it appeared with one of its branches below the -ais. In part (b), most candidates went to some trouble establishing a function which would have the value 0, or, at the point of intersection. The most popular technique for this was to clear denominators to obtain a cubic in factorised form, often converted unhelpfully into epanded form. Other candidates often used subtraction to obtain a suitable function. Once this was done, the way was clear for a candidate to earn 5 marks, but in some cases only of the 5 were gained as interval bisection was not used as required by the question.

137 Question 8 n n a) r + r = n( n+ ) + nn ( + ) r= r= 4 = nn ( + ) [ nn ( + ) + ] 4 ( ) = n n+ ( n + n+ ) 4 4 b r = nn+ n+ = nn+ n+ 6 n ) 8 8 ( )( ) ( )( ) r= n n n therefore, r + r = 8 r when r= r= r= 4 n + n+ = n+ 4 ( n + n + ) = 6(n+ ) ( ) ( ) ( ) 6 n + n+ = n n n = (n+ )( n 0) = 0 n = 0 is the only positive integer solution Candidates who were accustomed to look for common factors an approach almost always needed in questions on this topic were able to obtain high marks in both parts, though some of these candidates surprisingly failed to solve the quadratic in part (b). Candidates who preferred to epand and simplify everything and then hope to spot some factors were often successful in part (a) but could not realistically hope for more than one mark in part (b).

138 Question 9 y = a b 4 a A belongs to the curve so a a b b The asymptotes are y =± so = a a b) P(, 0) gradient m ) (,0) = = 4 The equation of the line is y 0 = m ( ) This line intersects the parabola so the -coordinate of the points of intersection satisfies: y = m m ( m m) = 4 6 m + m m = 4 6 m + m m + = ( 4) ( b = 4 (4 ) ( 6) 0 m m+ m + 6) = 0 c)the equation has equal roots when the discriminant is 0 m m + = 0 ( m ) 4 ( 4) ( 6) m d m gives m m 48m + 56 = ) = 6 = =± =± a = ( E) The equation E becomes + + = = = 0 ( ) The coordinates of the points ( 4) 0 = 4 4 y = m( ) so y =± = ± 4 m m m = = = are (4, 4 ) and (4, 4 ) Part (a) was found very hard by most candidates. Many failed to use both pieces of information supplied just before part (a), so that they could establish a = or b = a but could not hope to complete the two requests. Whether they were attempting one half or both halves of the question, they often wrote down the results they were supposed to be proving, possibly earning some credit for verifying these results, though the reasoning was sometimes very hard to follow. Part (b) was much more familiar to well-prepared candidates, but marks were often lost either by a failure to form a correct equation for the straight line or by sign errors after the elimination of y. The solutions to parts (c) and (d) were often presented in the reverse order, but full credit was given for all correct working shown. In part (c), many candidates made a good attempt to deal with the discriminant of the quadratic equation printed in part (b), but were careless about indicating that this discriminant should be equal to zero for equal roots. Once again, sign errors often caused a loss of marks. In part (d), the unique value of was often found correctly by the stronger candidates, but relatively few of these went on to find the values of y, and those who did sometimes did so via a rather roundabout approach. Grade boundaries

139 General Certificate of Education Advanced Subsidiary Eamination June 00 Mathematics MFP Unit Further Pure Thursday 7 May am to 0.0 am For this paper you must have: the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed hour 0 minutes Instructions Use black ink or black ball-point pen. Pencil should only be used for drawing. Fill in the boes at the top of this page. Answer all questions. Write the question part reference (eg (a), (b)(i) etc) in the left-hand margin. You must answer the questions in the spaces provided. Do not write outside the bo around each page. Show all necessary working; otherwise marks for method may be lost. Do all rough work in this book. Cross through any work that you do not want to be marked. Condensed Information The marks for questions are shown in brackets. The maimum mark for this paper is 75. Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. P79/Jun0/MFP 6/6/ MFP

140 A curve passes through the point ð, Þ and satisfies the differential equation dy d ¼ þ Starting at the point ð, Þ, use a step-by-step method with a step length of 0. to estimate the y-coordinate of the point on the curve for which ¼ :. Give your answer to three decimal places. (No credit will be given for methods involving integration.) (6 marks) It is given that z ¼ þ iy, where and y are real numbers. (a) (b) Find, in terms of and y, the real and imaginary parts of ð iþz z Hence find the comple number z such that ð iþz z ¼ 0ð þ iþ (4 marks) ( marks) Find the general solution, in degrees, of the equation cosð5 0 Þ ¼cos 40 (5 marks) 4 The variables and y are related by an equation of the form y ¼ a þ b where a and b are constants. The following approimate values of and y have been found y (a) Complete the table on page, showing values of X, where X ¼. ( mark) (b) On the diagram on page, draw a linear graph relating X and y. ( marks) (c) Use your graph to find estimates, to two significant figures, for: (i) the value of when y ¼ 5 ; ( marks) (ii) the values of a and b. ( marks) P79/Jun0/MFP

141 4 6 8 X y y 0 ~ 0 0 O X ~ 5 A curve has equation y ¼. The point A on the curve has coordinates ð, 6Þ. The point B on the curve has -coordinate þ h. (a) Show that the gradient of the line AB is 6h þ h. (4 marks) (b) Eplain how the result of part (a) can be used to show that A is a stationary point on the curve. ( marks) Turn over s P79/Jun0/MFP

142 4 6 The matrices A and B are defined by p ffiffi A ¼ 6 4 p ffiffi pffiffi p ffiffiffi 7 5, B ¼ 6 4 p ffiffi p ffiffiffi p ffiffi p 7 5 ffiffi Describe fully the geometrical transformation represented by each of the following matrices: (a) A; (b) B; (c) A ; (d) B ; ( marks) ( marks) ( marks) ( marks) (e) AB. ( marks) 7 (a) (i) Write down the equations of the two asymptotes of the curve y ¼. ( marks) (ii) Sketch the curve y ¼, showing the coordinates of any points of intersection with the coordinate aes. ( marks) (iii) On the same aes, again showing the coordinates of any points of intersection with the coordinate aes, sketch the line y ¼ 5. ( mark) (b) (i) Solve the equation ¼ 5 ( marks) (ii) Find the solution of the inequality < 5 ( marks) P79/Jun0/MFP

143 5 8 The quadratic equation 4 þ 0 ¼ 0 has roots a and b. (a) Write down the values of a þ b and ab. ( marks) (b) Show that a þ b ¼. ( marks) 5 (c) Find a quadratic equation, with integer coefficients, which has roots a þ b and b þ a. (6 marks) 9 A parabola P has equation y ¼. (a) (i) Sketch the parabola P. ( marks) (ii) On your sketch, draw the two tangents to P which pass through the point ð, 0Þ. ( marks) (b) (i) Show that, if the line y ¼ mð þ Þ intersects P, then the -coordinates of the points of intersection must satisfy the equation m þð4m Þ þð4m þ Þ ¼0 ( marks) (ii) Show that, if this equation has equal roots, then 6m ¼ ( marks) (iii) Hence find the coordinates of the points at which the tangents to P from the point ð, 0Þ touch the parabola P. ( marks) END OF QUESTIONS Copyright Ó 00 AQA and its licensors. All rights reserved. P79/Jun0/MFP

144 MFP - AQA GCE Mark Scheme 00 June series MFP Q Solution Marks Total Comments First increment is 0. (= 0.) M variations possible here So net value of y is. A PI Second inc t is 0.( +. ) = 0. ma PI Third inc t is 0.( +. ) = 0.78 A PI So y AF 6 ft one numerical error Total 6 (a) Use of z = iy M Use of i = M ( i)z z = y + i(y ) A, 4 A if one numerical error made (b) y = 0, y = 0 M equate and attempt to solve so z = 5 + 0i A allow = 5, y = 0 Total 6 Introduction of 60n M (or 80n ) at any stage; condone nπ (or nπ) 5 0 = ±40 (+60n ) B OE, eg RHS 40 or 0 Going from 5 0 to m including division of all terms by 5 GS is = 4 ± 8 + 7n A, 5 OE; A if radians present in answer Total 5 4(a) 4, 6, 6, 64 entered in table B (b) Four points plotted accurately BF ft wrong values in (a) Linear graph drawn B (c)(i) Finding X for y = 5 and taking sq root M 5. A AWRT 5. or 5.; NMS / (ii) Calculation of gradient M a = gradient 0.7 A AWRT 0.6 to 0.8; NMS / b = y-intercept 4.5 BF can be found by calculation; ft c s y-intercept Total 8 4

145 MFP - AQA GCE Mark Scheme 00 June series MFP (cont) Q Solution Marks Total Comments 5(a) At B, y = ( + h) ( + h) M with attempt to epand and simplify Grad AB = = (8 + h + 6h + h ) (4 + h) B correct epansion of ( + h) (= 6 + 6h + h ) = ( 6+ 6 h + h ) ( 6) 6h + h h ( + h) m = 6h+ h A 4 convincingly shown (AG) (b) As h 0 this gradient 0 so gradient of curve at A is 0 E, E for h = 0 Total 6 6(a) Rotation 45 (anticlockwise)(about O) MA M for rotation (b) Reflection in y = tan.5 MA M for reflection (c) Rotation 90 (anticlockwise)(about O) MAF M for rotation or correct matri; ft wrong angle in (a) (d) Identity transformation B,F ft wrong mirror line in (b); B for B = I 0 (e) AB = 0 MA allow M if two entries correct Reflection in y = A Total 7(a)(i) Asymptotes = and y = 0 B,B may appear on graph (ii) Complete graph with correct shape B Coordinates 0, shown B (iii) Correct line, (0, 5) and (.5, 0) shown B (b)(i) + 4 = 0 B = or =.5 MA M for valid method for quadratic (ii) < <, >.5 B,F B for partially correct solution; ft incorrect roots of quadratic (one above, one below ) Total 0 5

146 MFP - AQA GCE Mark Scheme 00 June series MFP (cont) Q Solution Marks Total Comments 8(a) α + β = 4, αβ = 0 B,B (b) α + β + = α β αβ M 4 = = 0 5 A convincingly shown (AG) (c) Sum of roots = (α + β) + (ans to (b)) M = AF ft wrong value for α + β 4 M for attempt to epand product Product = αβ MA αβ (at least two terms correct) = 4 AF ft wrong value for αβ 5 Equation is = 0 AF 6 integer coeffs and = 0 needed here; ft one numerical error Total 0 9(a)(i) Parabola drawn M with -ais as line of symmetry passing through (, 0) A (ii) Two tangents passing through (, 0) BB to c s parabola (b)(i) Elimination of y Correct epansion of ( + ) M B Result A convincingly shown (AG) (ii) Correct discriminant B 6m 4 8m + = 6m 4 + 8m M OE Result A convincingly shown (AG) (iii) 9 + = 0 M OE = 6, y = ± A,A Total TOTAL 75 6

147 Further pure - AQA - June 00 Question : Euler formula : y = n y + n hf ( n) with h = 0. and f ( ) = + + =, y =, f ( ) = so y = + 0. =. =., y =., f ( ) =. so y = =.4 =., y =.4, f ( ) =.78 so y = = Question : z = + iy a i z z = i + iy iy ) ( ) ( )( ) ( ) = + iy i + y + iy = ( y) + i(y ) Re= y Im = y y = 0 y = 0 b)( iz ) z = 0( + i) = 0 + 0i when y = 0 = 5 z = 5 + 0i Question : o 0 Cos(5 0 ) = cos = 40 + k60 or 5 0 = 40 + k60 5 = 60 + k60 or 5 = 0 + k60 = + k7 or = 4 + k7 o o o o Question 4: y = a + b a) X y b) c) i) When y = 5, X = ii) Gradient a = = b= y ax = b 4.5 Most candidates scored high marks on this opening question, though there were some who seemed to have no idea what to do. A mark was often lost by carrying out an unwanted fourth iteration. A small number of candidates used the upper boundaries of the intervals rather than the lower boundaries. This was accepted for full marks, but credit was lost if the candidate switched from one method to the other in the course of working through the solution. The great majority of candidates showed that they had the necessary knowledge of comple numbers to cope with this very straightforward question. In a distressingly high number of instances the work was marred by elementary errors in the algebra, most commonly by a sign error causing z to appear as iy. Many candidates also failed to indicate clearly in part (a) which were the real and imaginary parts, though many recovered the mark by using the real and imaginary parts correctly in part (b) of the question. Most candidates introduced a term 60n into their work at some stage, sometimes at a very late stage indeed, but credit was given for having some awareness of general solutions. A number of candidates gave the equivalent in radians, even though the question specified that degrees were to be used in this case. Marks were often lost by the omission or misuse of the plus-or-minus symbol. In some cases this was introduced too late, after the candidate had reached the stage of writing 5 = 60. In other cases the symbol appeared correctly but then ± became ± 60. High marks were almost invariably gained in this question. In particular the first three marks were earned by almost all the candidates. Part (c)(i) was often answered without any sign of awareness of a distinction between and X, a distinction which is of the utmost importance in this type of question. In part (c)(ii) many candidates used calculations based on pairs of coordinates found in the table, but this was accepted as these coordinates could equally have been found from the graph. The value of b often emerged inaccurately from these calculations, though the candidate could so easily have used the y-intercept.

148 Question 5: y = A(, 6) B, B ( ) ( ) = + h y = + h + h = + h + h+ h h B (, 6 6) + hh+ h h 6 6 = h + h yb ya h + h a)the gradient of the line AB = = + h h + 6h = = 6h+ h h b) when h 0, the gradient of the line AB tends to the gradient of the tangent at A 6h+ h 0 h 0 B A ( 6 6) ( 6) The tangent has gradient 0, A is a stationary point. Question 6: π π Cos Sin 4 4 a) A= π π Sin Cos 4 4 π o represents the rotation centre O, rad ( or 45 ) anticlockwise 4 π π Cos Sin ) 4 4 b B= π π Sin Cos 4 4 π represent the reflection in the line y = ( Tan ) 8 c) A = A A π o represents the rotation centre O, ( or 90 ) anticlockwise. d) B = I identity transformation π π Cos Sin 0 e) AB = 0 = π π Sin Cos π y = ( Tan ) represents the reflection in the line 4 y = As has happened in past papers on MFP, the epansion of the cube of a binomial epression involved some lengthy pieces of algebra for many candidates, though the correct answer was often legitimately obtained. Most candidates were then able to put all the necessary terms into the formula for the gradient of a straight line and obtain the required answer correctly. There was a good response to part (b), where many candidates stated correctly that h must tend to zero. Only rarely did they say, inappropriately, that it must be equal to zero. High marks were often earned in this question, generally from the multiplication of the matrices rather than from the geometrical eplanations, which tended to be shaky. In parts (c) and (d) the vast majority of candidates calculated a matri product rather than base their answers purely on the transformations already found in parts (a) and (b). The transformation in part (c) was often given as a reflection rather than a rotation, and in part (d) many candidates stated that the matri was the identity matri but did not make any geometrical statement as to what this matri represented. In part (e) the correct matri was often obtained but the candidates failed to give the correct geometrical interpretation, or in some cases resorted to a full description of the transformation as a combination of the reflection and rotation found in parts (b) and (a). When this was done correctly, full credit was given.

149 Question 7: a)) i y = = is a " vertical " asymptote when, y 0, ii) when = 0, y = The curve crosses the y-ais at (0, ) for all, y 0 The curve does not cross the -ais iii) y = 0 is asymptote to the curve b) i) = 5 gives = ( 5)( ) + 5 = + 4 = 0 ( 7)( ) = 0 =.5 or = ii) Plot the line y = 5. < 5 :the part of the graph "below the line" is obtained for < < and >.5 Question 8: = 0 has roots α and β a) α + β = 4 and αβ = 0 α + β 4 b) + = = = α β αβ 0 5 c) α + + β + = α + β + ( + ) = 4 + = β α α β 5 and 4 ( α + ) ( β + ) = αβ β α αβ = = = = An equation with these roots is : + = = The first eight marks out of the ten available in this question were gained without much difficulty by the majority of candidates, apart from some careless errors such as omitting to indicate the coordinates asked for on the sketch. By contrast the inequality in part (b)(ii) was badly answered. Few candidates seemed to think of reading off the answers from the graph, the majority preferring an algebraic approach, which if done properly would have been worth much more than the two marks on offer. The algebraic method usually failed at the first step with an illegitimate multiplication of both sides of the inequality by. Some candidates multiplied by ( ) but could not cope with the resulting cubic epression. This was another well-answered question. The first two parts presented no difficulty to any reasonably competent candidate. In part (c) some candidates, faced with the task of finding the sum of the roots of the required equation, repeated the work done in part (b) rather than quoting the result obtained there. The epansion of the product of the new roots caused some unepected difficulties, some candidates failing to deal properly with two terms which should have given them constant values. The final mark was often lost by a failure to observe the technical requirements spelt out in the question.

150 Question 9: P: y = ai )) ii) b) i) y = m( + )intersects P, then the - coordinates of the point of intersection y = m ( + ) satisfies both equations: y = this gives m ( + ) = m + 4m 4m m + + = 0 + ( 4m ) ( m ) = 0 ( Eq) ii) this equations has equal roots if the discriminant is 0 ( m ) m ( m ) = m 8m + 6m 8m = 0 6m iii) y = m( + ) is the equation of the line going through P with gradient m This line is a tangent when 6m = m = ± 4 9 If m =, the equation ( Eq) becomes : + = = 0 = = ( 6) 0 = 6 The sketch of the parabola P was generally well attempted. When asked to sketch two tangents to this parabola, many candidates revealed a poor understanding of the idea of a tangent to a curve. Part (b) was found familiar by all good candidates, and parts (b)(i) and (b)(ii) yielded full marks provided that a little care was taken to avoid sign errors. Part (b)(iii) was more demanding but many candidates found their way to earning at least some credit, either by substituting the value of m into the quadratic found in part (b)(i) or by some more roundabout method. then y =± ( + ) = ± 4 The tangents to the parabola from P touch it at (6, ) and (6, ) Grade boundaries

151 General Certificate of Education Advanced Subsidiary Eamination January 0 Mathematics MFP Unit Further Pure Friday 4 January 0.0 pm to.00 pm For this paper you must have: the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed hour 0 minutes Instructions Use black ink or black ball-point pen. Pencil should only be used for drawing. Fill in the boes at the top of this page. Answer all questions. Write the question part reference (eg (a), (b)(i) etc) in the left-hand margin. You must answer the questions in the spaces provided. Do not write outside the bo around each page. Show all necessary working; otherwise marks for method may be lost. Do all rough work in this book. Cross through any work that you do not want to be marked. Condensed Information The marks for questions are shown in brackets. The maimum mark for this paper is 75. Advice Unless stated otherwise, you may quote formulae, without proof, from the booklet. P865/Jan/MFP 6/6/ MFP

152 The quadratic equation 6 þ 8 ¼ 0 has roots a and b. (a) Write down the values of a þ b and ab. ( marks) (b) Find a quadratic equation, with integer coefficients, which has roots a and b. (4 marks) (c) Hence find the values of a and b. ( mark) (a) Find, in terms of p and q, the value of the integral ð q p d. ( marks) (b) Show that only one of the following improper integrals has a finite value, and find that value: (i) (ii) ð 0 ð d ; d. ( marks) (a) Write down the matri corresponding to each of the following transformations: (i) a rotation about the origin through 90 clockwise; ( mark) (ii) a rotation about the origin through 80. ( mark) (b) The matrices A and B are defined by " # " 4 A ¼, B ¼ # 4 (i) Calculate the matri AB. ( marks) (ii) Show that ða þ BÞ ¼ ki, where I is the identity matri, for some integer k. ( marks) (c) Describe the single geometrical transformation, or combination of two geometrical transformations, represented by each of the following matrices: (i) A þ B ; (ii) ða þ BÞ ; (iii) ða þ BÞ 4. ( marks) ( marks) ( marks) P865/Jan/MFP

153 4 Find the general solution of the equation sin 4 p ¼ giving your answer in terms of p. (6 marks) 5 (a) It is given that z ¼ i. (i) Calculate the value of z, giving your answer in the form a þ bi. ( marks) (ii) Hence verify that z is a root of the equation z þ z þ 4 ¼ 0 ( marks) (b) Show that z ¼ þ i also satisfies the equation in part (a)(ii). ( marks) (c) Show that the equation in part (a)(ii) has two equal real roots. ( marks) Turn over s P865/Jan/MFP

154 4 6 The diagram shows a circle C and a line L, which is the tangent to C at the point ð, Þ. The equations of C and L are þ y ¼ and þ y ¼ respectively. y ~ L C O ~ The circle C is now transformed by a stretch with scale factor parallel to the -ais. The image of C under this stretch is an ellipse E. (a) (b) On the diagram below, sketch the ellipse E, indicating the coordinates of the points where it intersects the coordinate aes. (4 marks) Find equations of: (i) the ellipse E ; ( marks) (ii) the tangent to E at the point ð, Þ. ( marks) y ~ O ~ P865/Jan/MFP

155 5 7 A graph has equation y ¼ 4 þ 9 (a) (b) Eplain why the graph has no vertical asymptote and give the equation of the horizontal asymptote. ( marks) Show that, if the line y ¼ k intersects the graph, the -coordinates of the points of intersection of the line with the graph must satisfy the equation k þð9k þ 4Þ ¼0 ( marks) (c) Show that this equation has real roots if 4 k 4. (5 marks) 8 (d) Hence find the coordinates of the two stationary points on the curve. (No credit will be given for methods involving differentiation.) (6 marks) 8 (a) The equation þ þ ¼ 0 has one real root. Taking ¼ 50 as a first approimation to this root, use the Newton-Raphson method to find a second approimation,, to the root. ( marks) (b) (i) Given that S n ¼ P n rðr þ Þ, use the formulae for P n r and P n r¼ r¼ r¼ r to show that S n ¼ nðn þ Þ (5 marks) (ii) The lowest integer n for which S n > is denoted by N. Show that N þ N þ N > 0 ( mark) (c) Find the value of N, justifying your answer. ( marks) Copyright ª 0 AQA and its licensors. All rights reserved. P865/Jan/MFP

156 Mark Scheme General Certificate of Education (A-level) Mathematics Further Pure January 0 MFP Q Solution Marks Total Comments (a) α + β = 6, αβ = 8 BB (b) Sum of new roots = 6 (8) = 0 MAF ft wrong value(s) in (a) Product = 8 = 4 BF ditto Equation + 4 = 0 AF 4 = 0 needed here; ft wrong value(s) for sum/product (c) α and β are ±8i B Total 7 (a) d = (+ c) MA M for correct inde q p d = p q (b)(i) As p 0, p, so no value AF OE; ft wrong coefficient of (ii) As q, q 0, so value is ¼ MAF ft wrong coefficient of or reversal of limits Total 6 (a)(i) 0 0 B (ii) 0 0 B (b)(i) 0 4 AB = 4 0 MA MA0 if entries correct (ii) 0 5 A + B = ( A + B) = 0 5 B B... = 5I BF ft if c s (A + B) is of the form ki (c)(i) Rot n 90 clockwise, enlargem t SF 5 B, OE (ii) Rotation 80, enlargement SF 5 B, F Accept enlargement SF 5 ; ft wrong value of k (iii) Enlargement SF 65 B, F B for pure enlargement; ft ditto Total 4 sin π = B OE; dec/deg penalised at 6th mark sin ( ) 6 ( 5π ) 6 = B BF OE; ft wrong first value Use of nπ M (or nπ) at any stage π Going from 4 to m including division of all terms by 4 π π GS = + 8 nπ or = 4 + nπ AA 6 OE Total 6 4

157 Mark Scheme General Certificate of Education (A-level) Mathematics Further Pure January 0 MFP(cont) Q Solution Marks Total Comments 5(a)(i) z = + i = i MA M for use of i = 4 i 4 4 i 4 (ii) LHS = + + i + = 0 MA AG; M for z correct (b) LHS = i i + = 0 MA 4 AG; M for z correct (c) z real z = z M Clearly stated Discr t zero or correct factorisation A AG Total 8 6(a) Sketch of ellipse M centred at origin Correct relationship to circle A Coords ±, 0, 0, ± B, 4 Accept 8 for ; B for any of = ±, y = ± allow B if all correct ecept for use of decimals (at least one DP) (b)(i) Replacing by E is ( ) ( ) () + y = M A or by (ii) Tangent is + y = MA M for complete valid method Total 8 7(a) Denom never zero, so no vert asymp E Horizontal asymptote is y = 0 B (b) 4 = k( + 9) M Hence result clearly shown A AG (c) Real roots if b 4ac 0 E PI (at any stage) Discriminant = 4k(9k + 4) M... = (6k + 6k ) m m for epansion... = (8k )(k + ) m m for correct factorisation Result (AG) clearly justified A 5 eg by sketch or sign diagram (d) = = 0 MA or equivalent using k = k 8... ( + ) = 0 = A k = 9 8 = ( 9) = 0 = 9 A SPs are,, 9, A 6 correctly paired ( ) ( ) 8 A Total 5 OE 5

158 Mark Scheme General Certificate of Education (A-level) Mathematics Further Pure January 0 MFP(cont) Q Solution Marks Total Comments 8(a) 50 + (50 ) B For numerator (PI by value 0050) = 50 (50 ) + 4(50) + B For denominator (PI by value 770) 46. B Allow AWRT 46. 8(b)(i) Σr(r + ) = Σr + Σr M... = ( ) m 6 n ( n + )(n + ) + n( n + ) correct formulae substituted... = n( n + )(n + ) mm m for each factor (n and n + )... = n(n + ) convincingly shown A 5 AG (ii) Correct epansion of n(n + ) B and conclusion drawn (AG) (c) Attempt at value of S 46 Attempt at value of S 45 S 45 < < S 46, so N = 46 M m A Alternative method Root of equation in (a) is 45.8 So lowest integer value is 46 (B) Total TOTAL 75 Allow AWRT 45.7 or

159 Further pure - AQA - January 0 Question : a + = has roots α and β ) a) α + β = 6 and αβ = 8 b α + β = α + β αβ = = ) ( ) α β = ( αβ ) = 8 = 4 An equation with roots α and β is = = o α = 8i and β = 8i c) 4 (8 i) s + 4 = 0 Question : a q ) d q q d q p = p = = + = p p b)) i When p 0,, so d has no value 0 p p q ii) When q, 0 so d q = 4 Question : o o Cos70 Sin70 a) i)90 clockwise = 70 anticlockwise: Sin70 Cos70 0 = 0 o Cos80 Sin80 0 ii) Rotation80 : Sin80 Cos80 = b)) i AB = 4 = ii)( A + B) = = 5I 0 5 = cia )) + B= = represents the rotation 90 by an enlargement scale factor 5 0 ii)( A + B) = 5I = 5 0 o represents the rotation 80 followed by an enlargement scale factor 5 4 )( ) iii A + B = I I = I o anticlockwise followed represents an enlargement scale factor 65 Most candidates started the paper in confident fashion, earning the first si marks with apparent ease, though some lost the sith mark through failing to write = 0 to complete their equation in part (b). Relatively few candidates realised that they were being tested on their knowledge of comple numbers in part (c), and even those who did sometimes failed to obtain the one mark on offer by misidentifying the two correct roots of their equation. This question again provided a good number of marks for the vast majority of candidates, who showed an adequate knowledge of integration, but in an unepectedly large number of cases sign errors were made in part (a), causing the letters p and q to be interchanged. In part (b), most candidates were able to identify correctly which integral had the finite value. Like the preceding questions, this one produced a good opportunity for most candidates to score high marks. The first two marks were sometimes lost because the candidate failed to provide numerical values for the sines and cosines in their matrices. Also the first matri was often that of a 90 anticlockwise rotation, rather than a clockwise one as required. Most candidates earned two marks in part (b)(i), relatively few finding BA instead of AB. Full marks were very common in part (b)(ii). The geometrical interpretations asked for in part (c) were mostly correct, though some floundered somewhat in the first part. In part (c)(ii), the answer enlargement with scale-factor 5 was acceptable, and very common.

160 Question 4: π π Sin 4 = = Sin 6 π π π π so 4 = + kπ or 4 = π + + kπ 6 6 π π 4= + kπ or 4 = + kπ 6 π π π π = + k or = + k k 8 4 Question 5: z = i i i) z = i = i+ i = ( i = ) 4 4 ii) z + z + = i + + i + = + = z is a solution to the equation z + z + = 0 4 b) z = + i = z (we are going to use the property: ( u+ v) = u + v ) the equation become 0 ( z ) + ( z ) + = ( z ) + ( z ) + = z + z = 0 = 0 z is also a solution to the equation z + z + = 0 4 c) If z is real then z = zand Question 6: + y = + y = z + z+ = 4 ( z + ) = 0 z = is a real repeated root a)the circle crosses the aes at (0, ),(0,- ),(,0), (-,0) therefore The ellipse crosses the aes at (0, ),(0,- ),(,0), (-,0) ) )The equation of the ellipse is : bi + y = + y = 4 ii)the point of contact and the tangent are affected by the same tranformation The equation of the new tangent at (,) is + y = As mentioned above, there was a pleasing improvement in the way candidates approached this trigonometrical equation, in contrast to what has been seen over the years. Marks were lost, however, by a failure in many cases to find a correct second particular solution before the general term was added in. The fact that the sine of the angle in the equation was negative clearly made this task a little harder than usual. The first si marks in this question were obtained with apparent ease in the great majority of scripts. Part (c) usually produced no further marks as the candidates omitted the star from z without eplaining why this was legitimate. Many candidates seemed to treat the given equation as a quadratic, despite the fact that two roots had already been found and two more were now being asked for. In part (a) of this question, most candidates managed to draw an acceptable attempt at an ellipse touching the given circle in the appropriate places. Occasionally the stretch would be applied parallel to the y-ais rather than the -ais. The required coordinates were indicated with various levels of accuracy, sometimes appearing as integers, sometimes with minus signs omitted. In part (b), the most successful candidates were often the ones who wrote the least all that was needed was to replace by / in each of the two given equations. Many candidates answered part (b)(i) concisely and correctly but then went into a variety of long methods to find the equation of the tangent in part (b)(ii). Some used implicit differentiation and were usually successful. Others used chain-rule differentiation and usually made errors. Yet others embarked on a very complicated piece of work based on quadratic theory and using a general gradient m. This must have consumed a large amount of time and was l bl f l

161 Question 7: 4 y = + 9 a) For all, + 9> 0so there is no vertical asymptote. 4 4 y = = y = 0 is asymptote tothecurve b) y = kintersects the curve so the -coordinate of the point of intersection 4 satisfies( y = ) k = + 9 k+ = ( 9) 4 k k Eq c)the equation has real roots when the discriminant is 0 (-) 4 (9 4) = 0 ( ) k k+ k 6k 0 k + k (8k )(k+ ) 0 Draw a sketch to support your answer k 8 d) When k = or, the discriminant is 0 8 and the equation has two equal roots which means that the line y = k is tangent the curve, corresponding to a maimum or minimum value of y(a stationary point) if k,( Eq ) = becomes = = 0 + = = ( ) 0 Onestationary point is (, ) 9 if k =,( Eq) becomes + = = 0 = 0 = ( 9) 9 Another stationary point is(9, ) 8 Part (a) of this question was not always answered as well as epected. Many candidates gave a good eplanation for the absence of a vertical asymptote but omitted to attempt the equation of the horizontal asymptote. When both parts were answered, the attempts were usually successful, but an equation y = instead of y = 0 was quite common. Part (b) was absolutely straightforward and afforded two easy marks to almost all the candidates. Part (c) involved inequalities, and while most candidates were familiar with the need to work on the discriminant at this stage, many lost a mark through not clearly and correctly stating the condition for real roots; another mark was lost when the candidate, having legitimately obtained the two critical values, failed to justify the inequalities in the final answer. Again, a sign error in the manipulation of the discriminant often spoiled the attempt to find the two critical values. It was good to see many candidates gaining full credit in part (d) even when they had struggled unsuccessfully in part (c). The majority of candidates had clearly practised their techniques in this type of question. Sometimes the finding of the y-values required an unwarranted amount of effort, since they were known from the outset, and some candidates lost the final mark because of a failure to give the correct y-coordinates.

162 Question 8: a f = + + = has one real root ) ( ) If = 50is an approimation of the root f( ) then = is a better appro. f '( ) f( ) = = 0050 f '( ) = + 4+ f '( ) = = so = 50 = 46. to D. P 770 n n n n bs ) = r(r+ ) = r + r= r + r n r= r= r= r= Sn = nn ( + )(n+ ) + nn ( + ) 6 = nn ( + ) [ n+ + ] = nn ( + )(n+ ) = nn ( + ) ii) N( N + ) > ( ) N N N + N + > + N + N > 0 c) According to part a),an approimation of the root is 46. Let try N=45 then N + N + N = 46 then N = N + N + N = N = 46 is the lowest interger so that S > N 80 4 Most candidates applied the Newton Raphson method correctly in part (a) of this question and obtained a correct value. They then proceeded in many cases to prove the result given in part (b)(i), often by epanding fully before attempting to find the factorised form. Luckily in this case it was not very hard to obtain the factors from the epanded form, particularly as the answer was given. Some candidates, however, lost credit through not showing steps at the end, which they may have thought unnecessary as the answer was obvious, but this overlooked the importance of good eamination technique when the answer is printed on the question paper. For a similar reason, some candidates lost the one mark available in part (b)(ii), no doubt thinking that the result was obvious and not writing enough steps. Part (c) required an awareness that N must be an integer, and also that the answer to part (a) was not necessarily an accurate guide to the root of the equation given there. Many candidates did a lot of work to find that root more accurately, or indeed quoted the value found on a calculator (45.75), but failed to draw a correct conclusion about the value of N. A more successful approach, in general, was to evaluate the lefthand side of the inequality for N = 46 and then for N = 45, thus showing that the former value was the lowest integer for which the inequality was satisfied. Grade boundaries

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