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2015 Assessment Report Page 1 of 5
2015 ASSESSMENT REPORT
OFFICE OF TASMANIAN
ASSESSMENT, STANDARDS
& CERTIFICATION
Mathematics Specialised Course Code: MTS415114 Overall the exam was successfully attempted by almost all students. Very few students left sections blank or answered a minimal number of questions. There was little evidence of students running out of time for the exam. Candidates need to be reminded of the importance of completing each section in a separate booklet in the exam. Many students wrote questions from multiple sections in a booklet which created great difficulty for the markers. Section A: Criterion 4 – Series and Sequences It appeared that this section was not favoured by many candidates and was left to be undertaken last. Consequently, many rash decisions were taken by many candidates. However, there were many others who had prepared, through practice, very well, and consequently presented high quality solutions. Candidates are reminded that markers assess all that is presented in determining a candidate’s knowledge and understanding, so it is unwise to scrub out, or liquid paper out, or erase solutions (or part thereof) for there may be worth in what can no longer be deciphered. A simple diagonal line through the work you wish to discard is a more time-efficient way of moving forward in your solution. Many skills that have been dealt with in previous years have not been maintained by many candidates. One such example is the solution of inequalities which aren’t linear. It is important to remember that a CAS calculator is not the panacea of all ‘ills’. A significant number of candidates confused the distinction between a series and a sequence: “A sequence is an ordered set of objects ………… which is able to be indexed”, while “A series is the sum of a finite or an infinite sequence of terms.” [Definition modified from Dictionary of Mathematics, E. J Borowski, and J. M. Borwein.] Question 1 This question generally caused no problems. However, candidates are reminded that, if an approximation is taken, the appropriate rounding technique needs to be applied. Question 2 Many candidates confused this question-type with a formal proof of convergence of a sequence. It is important to read the intent of the question, and, here, candidates were asked to find the conditions that need to be placed on x for the convergence of the series; that is, that a limiting sum exists. It was pleasing to come across those candidates who efficiently used graphical means to obtain
21or
21
−<> xx for necessary conditions of x for the series to have a limiting sum.
Two candidates received a ‘bonus’ for recognising that if 12 =x , a degenerate series exists, and so 1±≠x . Since
x = 0 was not contained in 2
1or 2
1−<> xx , few candidates mentioned this condition that needed to be
considered.
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Question 3 A number of candidates (successfully) used a proof by mathematical induction here; while, of the remaining candidature, a significant number of candidates failed to recognise the sum of differences (even though a difference equation was given). One candidate successfully used the fact that 1−−= nnn SSu in order to prove the statement. Question 4 The Oxford definition of ‘to rig’ which is relevant to the “rigging” method used by a number of candidates is “to hoax, play tricks on, befool; to manage or manipulate in some underhand or fraudulent manner”. It does not seem appropriate to use the term “rigging” within a proof. It is sufficient to give the mathematical justification of the provision for moving from one step to the next. This aside, most candidates had worked diligently in order to understand the form of this proof, and, as a consequence, generally this question was answered well. However, there were some who omitted the leading and post statements in the formal proof. Question 5 Once again it was obvious that most students had worked diligently on the understanding of the form of this proof and consequently presented logically sound solutions. Nevertheless, a significant number of candidates failed to mention the crucial statement in (n + 1) was true if the statement in n were true. Question 6 A small number of candidates displayed high level analysis and problem solving skills to present a succinct solution to this question while many others were confused that there were two series of n terms (to make 2n terms) embedded in the series statement. Some candidates decomposed the series further with little success. The majority of candidates substituted for 222 terms in their sum expression without taking into account, for example, the decomposition of the series in terms of n and substituting 111 for n. (222 for 2n.) Section B: Criterion 5 - Matrices and Linear Transformations Generally, students performed well on this section and showed a good understanding of the concepts being examined. Students are advised to ensure that they answer the question that is actually being asked. Unless otherwise stated, the use of CAS calculators in simplifying expressions was considered to be acceptable as long as the logical sequence of steps was recorded. Question 7
This question was well done. Most students solved by substituting ⎟⎟⎠
⎞⎜⎜⎝
⎛
+
−⎟⎟⎠
⎞⎜⎜⎝
⎛
−
−
42
22
dcba
dcba
, which turned
out to be an efficient method. Surprisingly, some students were unable to solve for cc =− ! The most common
error for students first finding 1)(2 −−= IBBA was not to use ⎟⎟⎠
⎞⎜⎜⎝
⎛=
1001
I .
Question 8 This question provided many students with their biggest challenge. More practice with this type of question is recommended. A minority of students gave a complete solution. Question 9 This question was generally well done. However, many students lost confidence when trying to find an
expression for 1−A and then did not give an expression for A1 . Students are advised to seek part marks,
especially when questions appear to be difficult.
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Question 10 Part (a) was generally very well done, although the structure of many solutions could have been more appropriate. Part (b) proved to be outside of the experience of many students. Many students left this part blank. Full marks were given for the two most obvious solutions. The algebra to find a complete set of solutions proved to be too tricky or time consuming for most students. A bonus mark was given to the few successful students. Question 11 This standard question was very well done. The most common issue was students failing to answer part (c). Students are advised to re-read a question once they have completed their solution. Question 12 This question was also very well done. Almost all students used the correct order for matrix multiplication in part (a). Errors in obtaining matrices A or C were quite common, but led to a small deduction as long as the rest of the solution was clear. Most students gained full marks for part (b). It was quite common for students to try to find the image, even though it had been given. A number of students used matrix B instead of matrix T. Again this led to a small deduction for an otherwise complete solution. Section C: Criterion 6 – Differential Calculus Question 13 Qualitatively 95% of students did not display an understanding of concavity. Only a few took the second derivative. Most took the first derivative and constructed the argument that because the first derivative was always positive that the function must be concave up, missing how a non-stationary inflection can create a graph with an always positive first derivative but not be always concave up. Yet later a few also highlighted that a non-stationary inflection existed, demonstrating only a procedural understanding. Question 14 Generally well done. Small numbers did not implicitly derive correctly. Of note is that while most students showed that the gradient of the function was zero at the prescribed point, and thus parallel to = 1 , many did not show that these curves were coincident and thus actually tangential. Question 15 Again generally well done. Many did not show appropriate working out for the derivatives involved. Also many students 'subbed in’ (1,0) rather than creating the argument that this was the only point of inflection. Question 16 This question was obviously a comfort question for most students, and appeared to be the gateway for a C level award. However, many students did not show adequate working in the calculation of what was a relatively straight forward derivative. Most had three lines for this.
𝑦 = 𝑥! + 1 𝑒! 𝑦! = 𝑥 + 1 !𝑒! 𝑜𝑟 𝑦! = 𝑥! + 2𝑥 + 1 𝑒!
𝑦!! = 𝑥 + 1 𝑥 + 3 𝑒! 𝑜𝑟 𝑦!! = 𝑥! + 4𝑥 + 3 𝑒! It is certainly the case that many students may have computed the right hand side example in their head, but they should show working in the exam. (if only to receive partial marks in case of algebraic error). More likely is that these were transferred from the calculator, leading to students attracting marks accordingly. Many did not discuss the nature of Points of inflection. 𝑦!! = 0 was solved and the existence of two Points of Inflection stated, but their local behaviour was not discussed. Occasionally a graph was missing.
2015 Assessment Report Page 4 of 5
Question 17 For roughly half of the students this was very well done and could be completed succinctly. Very few did not attempt the question and mistakes were varied. The most common was of course was accidental rotation about the x-axis. Other errors also displayed misconceptions in relating the calculations to the geometry of the problem. In particular some subtracted the line 𝑦 = −1 from their radius formula, even though in their working they rotated around the y-axis. This highlights a disconnect between the direction of integration and the variable of integration. A handful of students presented numerical answers based on calculator techniques which attracted only limited marks. Question 18 This question was attempted in earnest by few. Most who undertook the question understood they needed to relate the derivative to the equation for the line of a tangent. Many realised that the line had to pass through the origin and thus the line would take the form 𝑦 = 𝑚𝑥 but this is where most got stuck. A number of students made an attempt to find an intersection points but knew their answers to have missed the mark, and so in words or pseudo equations described the rest of the process. This attracted a fair portion of the marks for this question, but not so much as to disadvantage the high level ‘A’ students that had dropped one or two marks elsewhere but had successfully completed this challenging question. A number of students obviously used their calculator to find that they needed the tangent line 𝑦 = 12𝑥, which would abruptly appear in their work. This yielded a similar proportion of marks to the case above. It would have been good if students had used this problem solving technique to inform their algebra, but this was rarely the case, or if it were it was indeterminable. Overall, – and as was likely intended – successful completion of this question put students in good stead in order to receive an ‘A’. Of final note is that success in question 18 did not seem to relate to success in question 13. Many of those who did extremely well in question 18 erroneously used a first derivative argument to address question 13. This highlights that even the best students still commonly misunderstand concavity. This may be because question 16 has such a well-defined process that students may become competent in ‘using the second derivative’ without mentally digesting what is going on. Section D: Criterion 7 - Integral Calculus Overall quite good with some straightforward questions which were generally well answered and some challenging questions later in the section. Question19 Generally well done - a straightforward question involving integration by substitution. Question 20 Generally well done - this question was completed by a 𝑢 = 𝑥 − 1 substitution or by manipulating !
!!!= 1 −
!!!!
. Question 21 Generally well done - the homogeneous question was well tackled although solutions should be written as 𝑦 = ⋯ if possible rather than left in other forms. Question 22 The question presented many difficulties for candidates and was not attempted by many. Some candidates did use the graphics calculator to find the integral. Those that did this and then found the value by substituting values did get some credit.
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Question 23 Generally well done although there were many errors in (b) with incorrect expansions of the brackets. Question 24 This question could be done as a type 2 DE or as a type 3 DE (separable). Splitting into partial fractions was the most common way to solve this question after separating the variables. Log errors were much too common. 22 out of 150 scripts scored higher than 25/30 15 out of 150 scored less than 10/30 Section E: Criterion 8 – Complex Numbers Question 25: There were a remarkable number of successful ways to solve this problem. The question was mostly well done apart from some students showing insufficient working or others trying to invert w, i-1 and i on the first line. Question 26 Many students simplified 𝑧 + 𝑖𝑧 then went no further, achieving 1 mark. A number of students tried to compute the modulus rather than the Argument. Many students gave the single answer !
! without mentioning
that 𝐴𝑟𝑔 𝑧 + 𝑖𝑧 could be undefined or equal to − !!!
. They were not penalised for this. Question 27 For correctly using De Moivre’s Theorem students achieved 1 mark. For expressing that the sin component of 𝑧! would have to be 0 two more marks were achieved. Most students failed to identify all the possible solutions. Some also gave the answer 𝜃 = !"
!, for 𝑛𝜖ℤ which was incorrect as 𝐴𝑟𝑔(𝑧) was used explicitly in the
question. Question 28 Almost all students successfully showed that 𝑖 was a solution of the equation, though some students missed this part of the question. The rest of the question was done well too, though common mistakes were forgetting to factor out z and not combining (𝑧 − 𝑖) and (𝑧 + 𝑖) into a real factor. Very few students checked to see if 𝑧! − 4𝑧 + 6 could be further factorised into linear factors, but students were not penalised here. Question 29 Most students did not represent clearly on their diagram that !
! makes an angle of 60º with the x-axis. Very few
realised that the intersection of the boundary was at 1, 3 and represented this on the diagram. Many students struggled with simplifying 0 ≤ 𝑧 + 𝑧 ≤ 2. Overall this question was more than half correct with most students. Question 30(a) This question was attempted by using the sum of a GP formula or factorising by hand. Those using the GP formula often arrived incorrectly at 𝑧! = −1 which cost them 2 marks by the end of the question. Those factorising by hand often fared better, but some wrote the answers from the calculator without showing full working. Question 30(b) This question was successfully completed by very few students. Even less used part (a) in their answer and they were not penalised for this. Most correct answers here were achieved by redoing the question over again.
−2 −1 1 2
−1
1
2
3
x
y
⎟⎟⎠
⎞⎜⎜⎝
⎛−
22,2
⎟⎟⎠
⎞⎜⎜⎝
⎛
22,2
Maths Specialised (MTS415114) 2015 Exam Solutions Section A – Sequences and Series Q1: 𝑆! = 5 = !
!!!= !
!!!
∴ 3 = 5 − 5𝑟 ∴ 𝑟 = !!.
∴ 𝑆!" =! !! !
!
!"
!!
= 5 1 − !!
!"= 4.999895.
Q2: Seriesconvergesifandonlyif 𝑎! < 1.
∴ −1 < !!!− 1 < 1
∴ 0 < !!!< 2.
Frompicture𝑥 < − !!or𝑥 > !
!.
Q3: 𝑈! =!!!!!!!(!!!)
= !!
(!!!)− (!!!)!
!(usingresult)
= 𝑉! − 𝑉!!!,where𝑉! =!!
!!!.
∴ 𝑆! = 𝑈! + 𝑈! +⋯𝑈!!! + 𝑈!
= 𝑉! − 𝑉! + 𝑉! − 𝑉! +⋯+ 𝑉!!! − 𝑉!!! + 𝑉! − 𝑉!!!
= 𝑉! − 𝑉!
= !!
!!!− !
!= !!
!!!.
Q4: !!!!!!
!!convergesto1ifgivenany𝜀 > 0∃ 𝑁(𝜀)s.t. !!!!!
!
!!− 1 < 𝜀
⇐ !!!!!!
!!− 1 < 𝜀
⇐ !!!!!!!!!
!!< 𝜀
⇐ !!!!!
< 𝜀
⇐ !!!!!
< 𝜀,since1 − 𝑛 < 0fornlargeenoughand𝑛! > 0
⇐ !!!< 𝜀,since!!!
!!< !
!!
⇐ !!< 𝜀
⇐ 𝑛 > !!
∴ choose𝑁 = !!
∴ !!!!!!
!!convergesto1.
Q5: Provethat2!" + 2! +⋯+ 2!!!! = 2!! − 2!!!!forall Nn∈ .
Let𝑃!be2!" + 2! +⋯+ 2!!!! = 2!! − 2!!!!forall Nn∈ .
Prove𝑃!istrue:
LHS=2!" RHS=2!! − 2!!!!
= 2!! − 2!"
= 2!"=LHS
∴𝑃!istrue.
Assume𝑃! istrue.
ie.2!" + 2! +⋯+ 2!!!! = Equation(*)
RTP𝑃!!!istrue.
ie.2!" + 2! +⋯+ 2!!!(!!!) = 2!! − 2!!!(!!!)
LHS=
= 2!! − 2!!!! + 2!!!!!! Using(*)
= 2!! − 2×2!!!!!! + 2!!!!!!
= 2!! − 2!!!!!!
= 2!! − 2!!!(!!!)
=RHS
∴𝑃! istrue⇒ 𝑃!!!istrue
∴bytheprincipleofmathematicalinduction,𝑃!istrueforall Nn∈ .
∴2!" + 2! +⋯+ 2!!!! = 2!! − 2!!!!forall Nn∈ .
Q6: 1! − 1 + 2! + 2 + 3! − 3 + 4! + 4 + 5! − 5 +⋯to2nterms
= 1! + 2! + 3! +⋯ 𝑡𝑜 𝑛 𝑡𝑒𝑟𝑚𝑠 + −1 + 2 − 3 + 4 − 5 +⋯ 𝑡𝑜 𝑛 𝑡𝑒𝑟𝑚𝑠
=∑=
n
rr
1
2 + (2 + 4 + 6 +⋯ 𝑡𝑜 !!!!
𝑡𝑒𝑟𝑚𝑠) + −1 − 3 − 5 +⋯ 𝑡𝑜 !!!!
𝑡𝑒𝑟𝑚𝑠
=∑=
n
rr
1
2 + !!!!!
4 + 2× !!!!− 1 +
!!!!!
2 + 2× !!!!− 1
= !(!!!)(!!!!)!
+ !!!!
4 + 𝑛 − 1 − 2 − !!!!
2 + 𝑛 + 1 − 2
= !(!!!)(!!!!)!
+ (!!!)(!!!)!
− (!!!)(!!!)!
= !!!!"
2(2𝑛 + 1) + 3(𝑛 − 1) − 3(𝑛 + 1)
= !!!!"
4𝑛! + 2𝑛 − 6
= !!!!
2𝑛! + 𝑛 − 3
= (!!!)(!!!)(!!!!)!
Tofind𝑆!!!substitute2𝑛 = 222so𝑛 = 111:
𝑆!!! =!!!!! !×!!!!! !!!!!
!
∴ 𝑆!!! = 462000.
Section B – Matrices and Linear Transformations Q7: 𝐴𝐵 = 𝐴 + 2𝐵
∴ 𝐴𝐵 − 𝐴 = 2𝐵
∴ 𝐴(𝐵 − 𝐼) = 2𝐵
∴ 𝐴 = 2𝐵(𝐵 − 𝐼)!!
∴ 𝐴 = 1 00 4 .
Q8: 𝑥!𝑦! =
− !!
!!
!!
!!
𝑥𝑦
∴ 𝑇 =− !!
!!
!!
!!
=−𝑐𝑜𝑠 !
!𝑠𝑖𝑛 !
!
𝑠𝑖𝑛 !!
𝑐𝑜𝑠 !!
=𝑐𝑜𝑠 !!
!𝑠𝑖𝑛 !!
!
𝑠𝑖𝑛 !!!
−𝑐𝑜𝑠 !!!
Thisrepresentsareflectionontheline𝑦 = 𝑡𝑎𝑛 !!𝑥or𝑦 = 3𝑥.
Q9: Let𝐴 = 𝑎 𝑏𝑐 𝑑
LHS=𝑑𝑒𝑡 𝑎 𝑏𝑐 𝑑
!! RHS= !
!"# ! !! !
= 𝑑𝑒𝑡 !!"!!"
𝑑 −𝑏−𝑐 𝑎 = !
!"!!"
= 𝑑𝑒𝑡!
!"!!"− !!"!!"
− !!"!!"
!!"!!"
= 𝐿𝐻𝑆
= !!"!!"
× !!"!!"
− − !!"!!"
×− !!"!!"
∴ 𝐴!! = !!.
= !"!!"!"!!" !
= !!"!!"
Q10(a):𝑀 = 2 00 2
𝑀! − 5𝑀 + 6𝐼 = 2 00 2
!− 5 2 0
0 2 + 6 1 00 1
= 4 00 4 − 10 0
0 10 + 6 00 6
= 0 00 0 .
Q10(b):𝑀! − 5𝑀 + 6𝐼 = 𝑂
∴ 𝑀 − 2𝐼 𝑀 − 3𝐼 = 𝑂
⇐ 𝑀 = 2𝐼 𝑜𝑟 𝑀 = 3𝐼
So𝑀 = 3𝐼isanothersolution.
Itcanbeshownthat 2 00 3 , 3 0
0 2 and𝑎 𝑏
!!!!!!!!!
5 − 𝑎 , 𝑎, 𝑏𝜖ℝarealsosolutions.
Q11(a):3 4 21 1 44 5 7
1710
Q11(b):~𝑅! ↔ 𝑅! 1 1 4
3 4 24 5 7
7110
~𝑅! − 3𝑅!𝑅! − 4𝑅!
1 1 40 1 −100 1 −9
7
−20−18
~𝑅! − 𝑅!
𝑅! − 𝑅!
1 0 140 1 −100 0 1
27−202
~𝑅! − 14𝑅!𝑅! + 10𝑅!
1 0 00 1 00 0 1
−102
∴ 𝑥 = 2, 𝑦 = 0, 𝑧 = −1.
Q11(c):Thethreeplanesintersectatthesinglepoint(2,0,-1)in3Dspace.
Section C – Differential Calculus Q13: 𝑦 = 𝑥 + 2!.
Weneed!!!
!!!> 0 ∀ 𝑥 ∈ ℝ.
!"!"= 1 + 2!×𝑙𝑛2. !!!
!!!= 𝑙𝑛2×2!×𝑙𝑛2 = 𝑙𝑛2 !2!.
Since 𝑙𝑛2 ! > 0and2! > 0∀ 𝑥 ∈ ℝthen𝑦 = 𝑥 + 2!isalwaysconcaveup.
Q14: Firstshowthattheline𝑦 = 1touchesthecurveat𝑥 = 1:
Sub.𝑦 = 1: 𝑥 + 1 ! = 4𝑥 ∴ 𝑥! + 2𝑥 + 1 − 4𝑥 = 0 ∴ 𝑥! − 2𝑥 + 1 = 0
∴ (𝑥 − 1)! = 0 ∴ 𝑥 = 1.
Nextshowthat!"!"= 0at𝑥 = 1forthecurve(since𝑦 = 1hasagradientof0):
Byimplicitdifferentiation:
2× 𝑥 + 𝑦 × 1 + !"!"
= 4 ∴ 1 + !"!"= !
!!! ∴ !"
!"= !
!!!− 1.
Sub.𝑥 = 1and𝑦 = 1:!"!"= !
!!!− 1 = 1 − 1 = 0.
∴ 𝑦 = 1isatangenttothecurve 𝑥 + 𝑦 ! = 4𝑥where𝑥 = 1.
Q15: 𝑦 = arcsin (1 − 𝑥).
Astationarypointofinflectionoccurswhen!"!"= 0,!
!!!!!
= 0and!!!
!!!doesnotchangesign
eitherside.
!"!"= !
!! !!! !×−1 = − !!!!!!!!!!
= − !!!!!!!
. ∴ !"!"≠ 0forany 𝑥 ∈ ℝ.
Anon-stationarypointofinflectionoccurswhen!"!"≠ 0,!
!!!!!
= 0and!!!
!!!changessigneither
side.
!!!!!!
= − − !!× −𝑥! + 2𝑥 !!!× −2𝑥 + 2 = − !!!
!!!!!!!.
∴ !!!
!!!= 0when𝑥 = 1.When𝑥 = !
!,!
!!!!!
= ! !!> 0.When𝑥 = !
!,!
!!!!!
= − ! !!< 0.
Since∴ !!!
!!!= 0onlywhen𝑥 = 1,itistheonlypointofinflection.
Q16: 𝑦 = 𝑥! + 1 𝑒! ∴ !"!"= 2𝑥𝑒! + 𝑥! + 1 𝑒! = 𝑥 + 1 !𝑒!
!!!!!!
= 2 𝑥 + 1 𝑒! + 𝑥 + 1 !𝑒! = 𝑥! + 4𝑥 + 3 𝑒! = 𝑥 + 1 𝑥 + 3 𝑒!.
Stat.pointsoccurwhen!"!"= 0.Solve 𝑥 + 1 !𝑒! = 0.
∴ 𝑥 = −1or𝑒! = 0 (𝑖𝑚𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒!).
When𝑥 = −1,𝑦 = !!and!
!!!!!
= 0.
When𝑥 = −2,!"!"= !
!!> 0andwhen𝑥 = 0,!"
!"= 𝑒 > 0,so −1, !
!isastationarypointof
inflectiononarisingcurve.
NonStat.pointsoccurwhen!!!
!!!= 0.Solve 𝑥 + 1 𝑥 + 3 𝑒! = 0.
∴ 𝑥 = −1or𝑥 = −3or𝑒! = 0 (𝑖𝑚𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒!).
𝑥 = −1isalreadyconsidered.When𝑥 = −3,𝑦 = !"!!and!"
!"= !
!!> 0.
When𝑥 = −4,!!!
!!!= !
!!> 0andwhen𝑥 = −2,!
!!!!!
= − !!< 0.∴ !
!!!!!
changessignaround
𝑥 = −3.
So −3, !"!!
isanon-stationarypointofinflectiononarisingcurve.
lim!→!! 𝑥! + 1 𝑒! = 0andlim!→! 𝑥! + 1 𝑒! =∞.
Q17: Volumeofwholecontainer=!!𝜋𝑟! = !
!𝜋×2! = !"!
! 𝑚!.
Volumeofwater= ∫−
−
1
2
2dyxπ 𝑥! + 𝑦! = 4
= ( )∫−
−
−1
2
24 dyyπ ∴ 𝑥! = 4 − 𝑦!
= 4𝑦 − !!
! !!
!!
=4 −1 − !! !
!− 4 −2 − !! !
!
=−4 + !!+ 8 − !
!
=!!
𝑚!.
∴ 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = !!
!"!= !
!".
Q18: Letupperintersectionpointoccurat𝑥 = 𝑘.𝑦 = 𝑥! + 16,so!"!"= 3𝑥!
Slopeoftangentis3𝑘!.Itpassesthrough 𝑘, 𝑘! + 16 sotheequationis:
𝑦 − (𝑘! + 16) = 3𝑘!(𝑥 − 𝑘)
∴ 𝑦 = 3𝑘!𝑥 − 3𝑘! + 𝑘! + 16
−4 −3 −2 −1 1
−1
1
2
3
x
y
⎟⎠
⎞⎜⎝
⎛−e2,1
⎟⎠
⎞⎜⎝
⎛− 3
10,3e
∴ 𝑦 = 3𝑘!𝑥 − 2𝑘! + 16.
Sincethetangentpassesthrough(0,0),sub.𝑥 = 0and𝑦 = 0.
∴ 0 = −2𝑘! + 16so𝑘 = ±2.Fromthediagram𝑘 = 2.
Sothetangentlinebecomes𝑦 = 12𝑥.
Tofindtheotherpointofintersectionsolve𝑦 = 𝑥! + 16and𝑦 = 12𝑥.
Thisgives𝑥 = −4.
∴ 𝐼 = ( ) ( )( )dxxx∫−
−+2
4
3 1216
= !!
!− 6𝑥! + 16𝑥
!!
!
= !!
!− 6×2! + 16×2 − !! !
!− 6 −4 ! + 16(−4)
= 4 − 24 + 32 − 64 + 96 + 64
= 108 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠.
Section D – Integral Calculus Q19: 𝐼 = 𝑠𝑖𝑛𝑥𝑐𝑜𝑠!𝑥𝑑𝑥 Let𝑢 = 𝑐𝑜𝑠𝑥
= − 𝑢!𝑑𝑢 ∴ 𝑑𝑢 = −𝑠𝑖𝑛𝑥𝑑𝑥
= − !!
!+ 𝐶
∴ ∫π
o
xdxx 4cossin = − !"#!!! !
!= − − !
!− − !
!= !
!.
Q20: 𝐼 = !!!!
𝑑𝑥 Let𝑢 = 1 + 𝑥 ∴ 𝑑𝑢 = 𝑑𝑥 ∴ 𝑥 = 𝑢 − 1
= !!!!𝑑𝑢
= 1 − !!𝑑𝑢
= 𝑢 − 𝑙𝑛 𝑢 + 𝐶
∴ ∫ +
1
0 1dxxx
= 1 + 𝑥 − 𝑙𝑛 1 + 𝑥 !! = 2 − 𝑙𝑛2 − 1 − 𝑙𝑛1 = 1 − 𝑙𝑛2
Q21: 𝑥 !"!"= 𝑥 + 𝑦 Let𝑣 = !
!
∴ !"!"= !!!
!= 1 + !
! ∴ 𝑦 = 𝑣𝑥
∴ 𝑣 + 𝑥 !"!"= 1 + 𝑣 ∴ !"
!"= 𝑣 + 𝑥 !"
!"
∴ 𝑥 !"!"= 1
∴ 1𝑑𝑣 = !!𝑑𝑥
∴ 𝑣 = 𝑙𝑛 𝑥 + 𝐶
∴ !!= 𝑙𝑛 𝑥 + 𝐶
∴ 𝑦 = 𝑥(𝑙𝑛 𝑥 + 𝐶)
Substituting𝑦 = 0when𝑥 = 2:
∴ 0 = 2(𝑙𝑛2 + 𝐶)
∴ 𝐶 = −𝑙𝑛2
∴ 𝑦 = 𝑥(ln 𝑥 − 𝑙𝑛2)
Q22: ∫ +
2
3
cos11
π
π
dxx
𝑐𝑜𝑠2𝑥 = 2𝑐𝑜𝑠!𝑥 − 1
∫=2
3
2
2cos2
1π
π
dxx
∴ 1 + 𝑐𝑜𝑠2𝑥 = 2𝑐𝑜𝑠!𝑥
∫=2
3
2
2cos
121
π
π
dxx
∴ 1 + 𝑐𝑜𝑠𝑥 = 2𝑐𝑜𝑠! !!
∫=2
3
2
2sec
21
π
π
dxx
2
3
212
tan
21
π
π⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡⎟⎠
⎞⎜⎝
⎛
=
x
= 𝑡𝑎𝑛 !!− 𝑡𝑎𝑛 !
!
= 1 − !!
Q23(a): 𝑓 𝑥 = 𝑙𝑛𝑥 ! 𝑔! 𝑥 = 1
𝑓′ 𝑥 = !(!"#)!!!
! 𝑔 𝑥 = 𝑥
= 𝑥 𝑙𝑛𝑥 ! − 𝑥𝑛(𝑙𝑛𝑥)!!!
𝑥𝑑𝑥
= 𝑥 𝑙𝑛𝑥 ! − 𝑛 (𝑙𝑛𝑥)!!!𝑑𝑥
Q23(b):
𝐼 = 𝑥 𝑙𝑛𝑥 ! − 3 𝑙𝑛𝑥 !𝑑𝑥 Using23(a)when𝑛 = 3
= 𝑥 𝑙𝑛𝑥 ! − 3 𝑥 𝑙𝑛𝑥 ! − 2𝑙𝑛𝑥𝑑𝑥 + 𝐶 Using23(a)when𝑛 = 2
= 𝑥 𝑙𝑛𝑥 ! − 3 𝑥 𝑙𝑛𝑥 ! − 2(𝑥𝑙𝑛𝑥 − 𝑥) + 𝐶 FormulaSheet
= 𝑥 𝑙𝑛𝑥 ! − 3𝑥 𝑙𝑛𝑥 ! + 6𝑥𝑙𝑛𝑥 − 6𝑥 + 𝐶
Q24: !!(!"""!!)
𝑑𝑥 = 𝑘𝑑𝑡 !!(!"""!!)
≡ !!+ !
!!!!!!
∴ !!"""
!!+ !
!"""!!𝑑𝑥 = 𝑘𝑑𝑡(*) ∴ 1 = 1000𝐴 + 𝐵𝑥 − 𝐴𝑥
∴ !!"""
𝑙𝑛 𝑥 − 𝑙𝑛 1000 − 𝑥 = 𝑘𝑡 + 𝐶 sub.𝑥 = 0:𝐴 = !!"""
Sub.𝑡 = 0,𝑥 = 250: sub.𝑥 = 1000:𝐵 = !!"""
(*)
!!"""
𝑙𝑛250 − 𝑙𝑛750 = 𝐶
∴ 𝐶 = − !"!!"""
Sub.𝑡 = 1,𝑥 = 500
∴ !!"""
𝑙𝑛500 − 𝑙𝑛500 = 𝑘 − !"!!"""
∴ 𝑘 = !"!!"""
∴ !!"""
𝑙𝑛 𝑥 − 𝑙𝑛 1000 − 𝑥 = !"!!"""
𝑡 − !"!!"""
∴ 𝑙𝑛 !!"""!!
= 𝑙𝑛3. 𝑡 − 𝑙𝑛3
∴ !!"""!!
= 𝑒!"!.!!!"!
∴ !!"""!!
= 3!× !!
When𝑡 = 3, !!"""!!
= 3!× !!
∴ 𝑥 = 900.So900peoplehavethediseaseafter3months.
Section E – Complex Numbers Q25: !
!= !
!!!+ !
!= !!!!!
!!! != !
!!!!= !
!!!.
∴ 𝑤 = 1 + 𝑖.
Q26: 𝑧 + 𝑖𝑧 = 𝑥 + 𝑖𝑦 + 𝑖 𝑥 − 𝑖𝑦 = 𝑥 + 𝑖𝑦 + 𝑖𝑥 − 𝑖!𝑦 = 𝑥 + 𝑦 + 𝑖(𝑥 + 𝑦)
∴ 𝐴𝑟𝑔 𝑧 + 𝑖𝑧 = 𝑎𝑟𝑐𝑡𝑎𝑛 !!!!!!
= arctan (1)provided𝑥 + 𝑦 ≠ 0.
Thisis!!if𝑥 + 𝑦 > 0,!!!
!if𝑥 + 𝑦 < 0andundefinedif𝑥 + 𝑦 = 0.
Q27: 𝑧! = (𝑟𝑐𝑖𝑠𝜃)! = 𝑟!𝑐𝑖𝑠 2𝜃 = 𝑟! cos 2𝜃 + 𝑖𝑠𝑖𝑛(2𝜃) .
Hence𝑟!isrealifsin 2𝜃 = 0.
∴ 2𝜃 = ⋯ ,−𝜋, 0,𝜋, 2𝜋,…
∴ 𝜃 = ⋯ ,− !!, 0, !
!,𝜋,…
Hence𝐴𝑟𝑔 𝑧 = − !!, 0, !
!,𝜋.
Q28: 𝑧 = 𝑖isasolutionto𝐹 𝑧 = 0ifandonlyif𝐹 𝑖 = 0.
𝐹 𝑖 = 𝑖! − 4𝑖! + 7𝑖! − 4𝑖! + 6𝑖 = 𝑖 − 4 − 7𝑖 + 4 + 6𝑖 = 0.
∴ iisasolutionbythefactortheorem.
Since𝑧 = 𝑖isasolution,then𝑧 = −𝑖isasolutionbytheconjugateroottheorem.
∴ 𝐹 𝑧 = 𝑧 − 𝑖 𝑧 + 𝑖 … = 𝑧! + 1 … = 𝑧 𝑧! + 1 (𝑎𝑧! + 𝑏𝑧 + 𝑐)
Byinspection𝑎 = 1,𝑐 = 6.Also−4 = 𝑏(comparing𝑧!terms).
∴ 𝐹 𝑧 = 𝑧(𝑧! + 1)(𝑧! − 4𝑧 + 6).
Notethatforthequadratic∆= −4 ! − 4×1×6 = −8 < 0sothiscannotbefactorisedintolinearfactors.
Q30(a):Let𝑃 𝑧 = 𝑧! − 𝑧! + 𝑧! − 1.
Thisis𝑆!ofaGPwhere𝑎 = −1,𝑟 = −𝑧!.
∴ 𝑃 𝑧 =!! !! !!! !
!! !!!= 0.
∴ !!!!
!!!!= 0.
∴ 𝑧! = 1where𝑧! ≠ −1
∴ 𝑧! = 𝑐𝑖𝑠(0 + 2𝑛𝜋) !,where𝑧 ≠ ±𝑖,𝑛 ∈ ℤ.
∴ 𝑧 = 𝑐𝑖𝑠 !"!
,𝑧 ≠ ±𝑖
∴ 𝑧 = 𝑐𝑖𝑠 − !!!
, 𝑐𝑖𝑠 − !!, 𝑐𝑖𝑠 0 , 𝑐𝑖𝑠 !
!, 𝑐𝑖𝑠 !!
!, cis (𝜋).
∴ 𝑧 = 𝑐𝑖𝑠 − !!!
, 𝑐𝑖𝑠 − !!, 1 , 𝑐𝑖𝑠 !
!, 𝑐𝑖𝑠 !!
!,−1.
Q30(b):𝑧! − 4𝑧! + 16𝑧! − 64 = 64 !!
!"− !!
!"+ !!
!− 1 = 0
∴ 𝑠𝑜𝑙𝑣𝑒 𝑤! − 𝑤! + 𝑤! − 1 = 0,where𝑤 = !!.
∴ 𝑤 = 𝑐𝑖𝑠 − !!!
, 𝑐𝑖𝑠 − !!, 1 , 𝑐𝑖𝑠 !
!, 𝑐𝑖𝑠 !!
!,−1(from30(a)
∴ 𝑧 = 2𝑐𝑖𝑠 − !!!
, 2𝑐𝑖𝑠 − !!, 2 , 2𝑐𝑖𝑠 !
!, 2𝑐𝑖𝑠 !!
!,−2
∴ 𝑧 = − 2 − 2𝑖, 2 − 2𝑖, 2, 2 + 2𝑖,− 2 + 2𝑖,−2.