MATHEMATICS Unit Further Pure 2

Transcription

1 General Certificate of Education June 7 Advanced Level Eamination MATHEMATICS Unit Further Pure MFP Tuesday 6 June 7.3 pm to 3. 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 3 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. P9398/Jun7/MFP 6/6/6/ MFP

2 Answer all questions. (a) Given that fðrþ ¼ðr Þr, show that fðr þ Þ fðrþ ¼rð3r þ Þ (3 marks) (b) Use the method of differences to find the value of X99 r¼5 rð3r þ Þ ( marks) The cubic equation z 3 þ pz þ 6z þ q ¼ has roots a, b and g. (a) Write down the value of ab þ bg þ ga. ( mark) (b) Given that p and q are real and that a þ b þ g ¼ : (i) eplain why the cubic equation has two non-real roots and one real root; ( marks) (ii) find the value of p. ( marks) (c) One root of the cubic equation is þ 3i. Find: (i) the other two roots; (3 marks) (ii) the value of q. ( marks) 3 Use De Moivre s Theorem to find the smallest positive angle y for which ðcos y þ i sin yþ 5 ¼ i (5 marks) P9398/Jun7/MFP

3 s 3 (a) Differentiate tan with respect to. ( marks) (b) Show that ð tan d ¼ p p ln ffiffiffi (5 marks) 5 The sketch shows an Argand diagram. The points A and B represent the comple numbers z and z respectively. The angle AOB ¼ 9 and OA ¼ OB. Bðz Þ Im(z) Aðz Þ O Re(z) (a) Eplain why z ¼ iz. ( marks) (b) On a single copy of the diagram, draw: (i) the locus L of points satisfying jz z j¼jz z j ; ( marks) (ii) the locus L of points satisfying argðz z Þ¼arg z. (3 marks) (c) Find, in terms of z, the comple number representing the point of intersection of L and L. ( marks) 6 (a) Show that! ðk þ Þ k þ ¼ k þ k ðk þ Þ (3 marks) (b) Prove by induction that for all integers n 5 3 ::: n ¼ n þ n ( marks) Turn over for the net question Turn over P9398/Jun7/MFP

4 p 7 A curve has equation y ¼ ffiffi. (a) Show that the length of arc s of the curve between the points where ¼ and ¼ is given by s ¼ ð rffiffiffiffiffiffiffiffiffiffi þ d ( marks) (b) (i) Use the substitution ¼ sinh y to show that ð rffiffiffiffiffiffiffiffiffiffiffi ð þ d ¼ 8 cosh y dy (5 marks) (ii) Hence show that s ¼ sinh :5 þ p ffiffiffi 5 (6 marks) 8 (a) (i) Given that z 6 z 3 þ 8 ¼, show that z 3 ¼ i. ( marks) (ii) Hence solve the equation z 6 z 3 þ 8 ¼ giving your answers in the form re iy, where r> and p < yp. (6 marks) (b) Show that, for any real values of k and y, ðz ke iy Þðz ke iy Þ¼z kz cos y þ k ( marks) (c) Epress z 6 z 3 þ 8 as the product of three quadratic factors with real coefficients. (3 marks) END OF QUESTIONS Copyright Ó 7 AQA and its licensors. All rights reserved. P9398/Jun7/MFP

5 AQA Further pure Jun 7 Answers Question : a) f( r) ( r ) r f( r+ ) f( r) r( r+ ) ( r ) r r ( r+ ) rr ( ) r r + r+ r + r r(3r+ ) (5) b) r(3r+ ) f( r+ ) f( r) f (5) f 99 r 5 r 5 r 5 + f (5) f (5) + f (53) f (5) f (99) f (98) + f () f (99) All the terms cancel ecept f() f(5) r(3r + ) 8675 Eam report Almost all candidates were successful with part (a) However, in part (b) a number of candidates used 99 r and r to evaluate r(3r+ ) contrary r to the requirement of the question and so, even with a correct answer, scored no marks. The most successful candidates for this part of the question were those who carefully wrote out a number of rows including the first and last row, to illustrate the cancellations. Some candidates went awry when writing down the first or last terms of the series. Question : a) αβ + βγ + αγ 6 biα + β + γ < ) ) This can only happens if one of the root is not a real number so if αis a comple number, th en β α because p and q are real numbers and γ is real (because otherwise γ would be a root too, making roots instead of the epected 3) ii ) α + β + γ ( α + β + γ ) ( αβ + βγ + αγ ) + + ( α β γ) 6 + ( α β + γ) α + β + γ So p ( α + β + γ) p c) α + i β α 3 3 α + β + γ + 3i 3i+ γ γ ii) q αβγ ( + 3 i)( 3i)() ( + 9) i Eam report Whilst part (a) was usually correctly done, part (b)(i) was poorly answered. Some candidates were able to comment on the condition that as the sum of the squares of the roots was less than zero there would have to be comple roots, but few stated the conditions that the coefficients of the cubic equation were all real. The value of p in part (b)(ii) was very often correct but in part (c)(i) a very common error as to use α in order to find the third root. This method led to α from which almost all candidates using this method wrote α without even considering the possibility that α could equal.. Part (c)(ii) was usually worked correctly although αβγ + q appeared from time to time. Question 3: 5 ( θ θ) ( θ) ( θ) Cos + isin Cos 5 + isin 5 i Cos(5 θ) and Sin(5 θ) 3 3 so 5θ θ 3 Eam report There were many incomplete solutions to this question. Whilst most candidates used the de Moivre's Theorem correctly, many candidates either equated real parts only to arrive at an incorrect answer, or equated imaginary parts. In this latter case, the solution θ appeared frequently in spite of the 3 request in the question that θ should be positive, or the correct answer appeared but from an incomplete solution. Some candidates solved cosθ and sinθ - but gave two different values of θ as their answer, one from each equation.

6 Question : dy a) y Tan Tan + d + dy Tan + d + b) Tan d ( Tan ) d Tan d + Tan ln( + ) l n ln Eam report This is the first time that a question has been set on inverse trigonometrical functions since this topic was included in the MFP specification. It was clear that many candidates did not know what tan. was. They were able to complete part (a) with the help of the formulae booklet although even then there was confusion between the derivatives of tan - and tanh - as the derivative of tan - was given as. However it was part (b) that revealed the true lack of understanding of inverse trigonometrical functions. Part (b) was either abandoned altogether or when attempted tan - was frequently written as tan. Question 5: o a) Angle AOB 9 and OA OB In comple terms this means that z and Arg( z) Arg( z) z i This gives e i z b) i) A( z ), B( z ) and M ( z) L is the perpendicular bisector of AB. z iz z z z z is equivalent to AM BM ii)arg( z z ) arg( ) z L is the half line from B, parallel to OA. c) Let's call I the point of intersection and I(z ) OBIAis a square : Because OAbeing perpendicular to OB we knowthat IBis also perpendicular toob and by symmetry about the line L, IAis perpendicular toao. OBIAis a quadrilateral with right angles and OAOB so OBIAis a square In comple term, z z + z z + iz ( + iz ) I z I Eam report Eplanations in part (a) were very unclear and generally far from convincing. Candidates generally referred to what had happened to the coordinates of the points represented by z and z, but few made allusion to the significance of i in the iz. The neatest solutions came from candidates who considered multiplication of a comple number by i as a rotation anticlockwise of / Inaccurate copying of the diagram in part (b) caused loss of marks. For instance, although candidates knew that the locus L was the perpendicular bisector of AB, poor diagrams meant that their line did not pass through the origin. Again, for the locus L, although the majority of candidates drew a half line through B, their line was not always parallel to OA. Part (c) proved to be beyond most candidates probably because few realised that the point of intersection of L and L was, in fact, the fourth verte of the square whose three other vertices were A, O and B Question 6: k+ k+ k+ ( ) ( ) ( ) k+ k k+ k k + k k+ kk ( + ) ( k + ) ( k+ ) k kk ( + ) ( k + ) Eam report Part (a) was usually answered correctly although there were many very long-winded algebraic methods employed including the multiplication out of just about every bracket followed immediately by their re-factorisation.

7 Question 6:continues n + Pr oposition Pn : For n,... 3 n n is to be proven by induction. Eam report 3 3 Base case: n and n The proposition is true for n n + + Let's suppose that the propostion is true for nk, k + meaning... 3 k k Let's show that the proposition is true for nk+, k + Let's show that... ( k ) + k + 3 ( ) ( ) k 3 k ( k ) + + k+ k+ from part a) k ( k+ ) ( k+ ) Conclusion :If the proposition is true for nk, then it is true for nk+ because the proposition is true for n, according to te induction principle I can conclude that the proposition is true for all n : n + for all n,... 3 n n There was however much muddled thinking in part (b). Whilst most candidates had some outline of the method of induction many candidates attempted this part with no reference whatever to the series product in question, whilst others tried to add the (k + ) th term to the sum of k products. Candidates who did consider the series usually used rather than Π but this was not penalised. Question 7: y dy a) s + d from the formulae booklet d dy dy d d s + d + d Eam report This question was generally answered well and many candidates were able to score out of the available 5 marks. Part (a) was well answered apart from a few candidates who dy wrote followed by d

8 Question 7:continues d b)) i Sinh θ Coshθ Sinhθ dθ d 8Coshθ Sinhθ dθ when Sinh so Sinh and when Sinh, θ θ.5, θ + Sinh.5 Sinh θ + 8 o s d Coshθ Sinhθ dθ Sinh θ Sinh.5 Cosh θ s 8Coshθ Sinhθ dθ Sinh θ Sinh.5 [ θ θ ] Sinh.5 s 8Cosh θ dθ b) ii) Coshθ Cosh θ so Cosh θ + Coshθ Sinh.5 Sinh.5 so s 8( + Cosh θ) dθ + Coshθdθ Sinh.5 + h s + Sinh Sinh.5 Sinh(Sin.5) Cosθ 8Coshθ Sinhθ dθ Sinhθ Sn i hθ SinhθCosθ Sinhθ + Sinh θ 5 with θ Sinh.5 we have Sinh(Sinh.5) + 5 s Sinh + Sinh Sinh S (.5) inh.5 Sinh.5 5 Eam report In part (b) there were two main sources of error. The first was to interchange d with dθ without any consideration of d ; and the second dθ was to write Sinh θ + Sinhθ + as. Sinh θ Sinhθ There were also a few candidates who were unable to differentiate Sinh θ. In part (b)(i), most candidates were able to integrate 8cosh θ correctly but few were able to arrive at the printed result in part (b)(ii). Two factors contributed to this. Candidates either failed to change the limits for to the corresponding limits for θ or else wrote the answer with no evident method. This was unacceptable as the answer for the arc lengths was given. Question 8: 6 3 ai )) z z Let z be t, the equation becomes t t + 8 discriminant:(-)- 8-6(i) 3 ± i 3 So t z z ± i ii) Let's write z ( re ) r e 3 iθ 3 3 i3θ ± i ± i e These comple are equal when ± i 3 r and 3θ ± + k r and θ ± + k k,, 3 This gives 6 soutions : 3 7 ± i ± i ± i e or e or e b)( z ke )( z ke ) z zk( e + e ) + k e c z 6 3 ) z 8 iθ iθ iθ iθ + z zkcosθ + k i i i i i i ( z e )( z e )( z e )( z e )( z e )( z e ) 7 3 z Cos z Cos z Cos Eam report Candidates were usually able to establish the result in part (a) although the methods used were sometimes somewhat inelegant. Part (a)(ii) was reasonably well done although some carelessness was in evidence in this part. For instance, some candidates although showing that the argument of z 3 was ± continued their solution with only + and so arrived at a total of three roots. Others having reached 3 z 8 then thought that z 8 also. A few candidates used a method which, although possible, was not really suitable. They replaced the z in z + z + 8 with ±i and so arrived at z 6 ± 8i. This latter equation gave the twelve roots of z -.6 and the method was incomplete unless 6 of the roots were rejected. Part (b) was generally well done, but part (c) was really only completed by candidates who had correctly answered part (a)(ii).

9 Mathematics - AQA GCE Mark Scheme 7June series Key to mark scheme and abbreviations used in marking M m or dm A B E mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for eplanation or ft or F follow through from previous incorrect result MC mis-copy CAO correct answer only MR mis-read CSO correct solution only RA required accuracy AWFW anything which falls within FW further work AWRT anything which rounds to ISW ignore subsequent work ACF any correct form FIW from incorrect work AG answer given BOD given benefit of doubt SC special case WR work replaced by candidate OE or equivalent FB formulae book A, or (or ) accuracy marks NOS not on scheme EE deduct marks for each error G graph NMS no method shown c candidate PI possibly implied sf significant figure(s) SCA substantially correct approach dp decimal place(s) No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. However, there are situations in some units where part marks would be appropriate, particularly when similar techniques are involved. Your Principal Eaminer will alert you to these and details will be provided on the mark scheme. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded. June 7 3

10 Mathematics - AQA GCE Mark Scheme 7June series MFP Q Solution Marks Total Comments (a) f ( r + ) f ( r) r( r + ) ( r ) r M r( r + r+ r + r) A any epanded form r( 3r+ ) A 3 AG (b) r 5 f( 5 ) f( 5 ) OE r 5 f ( 5 ) f ( 5) PI 99 9 MA clearly shown. Accept r 99 f ( ) f ( 99) 99 r( 3r+ ) f ( ) f ( 5) m clear cancellation r AF cao Total 7 (a) αβ 6 B (b)(i) Sum of squares < not all real E Coefficients real conjugate pair E (ii) ( ) α α + αβ MA A for numerical values inserted ( α ) AF p AF cao (c)(i) 3i is a root B Use of appropriate relationship eg α M M if checked Third root AF 3 incorrect p (ii) q ( 3i )( + 3i) M allow even if sign error AF ft incorrect 3 rd root Total 5i cosθ + i sinθ cos5θ + i sin5θ M or e θ 3 ( ) 5 cos5θ α 3i used unless the root is sin5 θ ma or i e 3 5θ or 7! AF m for both R&I parts written down θ or 8! AF 5 ft provided the value of 5θ is a correct value SC cos5θ + isin5θ i (M) sin5θ (B) or for cos5θ θ (B) (3) Total 5

11 Mathematics - AQA GCE Mark Scheme 7June series MFP (cont) Q Solution Marks Total Comments (a) tan + + BB (b) d tan d tan either use of part (a) or integration by M + parts. Allow if sign error d ln ( ) + MAF ft on d I tan ln M ln A 5 AG Total 7 5(a) Eplanation E,, i E for i e or iz y+ i (b)(i) Perpendicular bisector of AB B through O B (ii) half-line B If L is taken to be the line AB give B from B B parallel to OA B 3 (c) ( + i) z MA ft if L taken as line AB Total 9 k + ( k ) k + ( k + ) k ( k + ) k M 6(a) k + k k + k ( k + ) k + ( k + ) A A 3 AG (b) Assume true for n k, then... 3 ( k + ) M k + k + A ( ) 3 True for n shown B Pn Pn+ and Ptrue E only if the other 3 marks earned Total 7 5

12 Mathematics - AQA GCE Mark Scheme 7June series MFP (cont) Q Solution Marks Total Comments 7(a) dy d B accept etc dy + + d + MAF A ft sign error in d y d AG (b)(i) sinh θ, d 8sinhθcoshθdθ MA M for any attempt at d d I sinh θ + 8sinh cosh d sinh θ θ θ θ M θ coshθ 8sinhθcoshθdθ sinhθ m ie use of cosh θ sinh θ 8cosh θ dθ A 5 AG (ii) Use of cosh θ + coshθ M allow if sign error ( + θ) I cosh dθ A oe θ + sinhθ AF oe Use of sinh θ sinhθcoshθ m sinh + + AF sinh + 5 A 6 AG Total 5 6

13 Mathematics - AQA GCE Mark Scheme 7June series MFP (cont) Q Solution Marks Total Comments 8(a)(i) 3 ± 6 3 z M ± i A AG (ii) i i + i e, i e z i ki M AA i ki e or e 3 M M for either z ± i ± 3i ± 7i e, e, e A,, F 6 M for either result or for one of r, θ ± r A, θ ± A allow A for any 3 correct ft errors in ± (b) Multiplication of brackets M i θ iθ Use of e + e cosθ A AG (c) i i z e z e Product is z cos z + MAF PI ( z cos z+ ) 7 ( z cos z+ ) 3 ( z cos z+ ) AF 3 ( or z + z + ) Total 3 TOTAL 75 7