2018 assessment report mathematics – specialised (mts415118) · 2018 assessment report...

2018 Assessment Report

Mathematics - Specialised (MTS415118) of 37

MATHEMATICS – SPECIALISED (MTS415118)

FEEDBACK FOR STUDENTS AND TEACHERS

GENERAL COMMENTS

Too many students used their calculators to solve problems, forgetting the description of the

criteria on the exam front cover… “The assessment for Mathematics Specialised Level 4 will be

based on the degree to which the learner can… solve problems and/or use techniques

involving”…the different skills learned in this course. Also, forgetting the instruction in the

‘Additional Instructions for Candidates’ and at the top of each section … “You must show the

method you used to solve a question. If you only show your answers you will get few, if any,

marks.”

SECTION A – SEQUENCES AND SERIES

QUESTION 1

No meaningful working required - students who tried to provide an explanation often made

errors dealing with the exponents oscillating between 1 and -1. Many students had an incorrect

understanding of odd and even.

QUESTION 2

Well done by most students. A lot of small algebraic errors in such a simple question.

QUESTION 3

(a) Most students did this perfectly

(b) Very few students saw the connection ln(1 + 2𝑥𝑥) = 2 ∫𝑓𝑓(𝑥𝑥) 𝑑𝑑𝑥𝑥

2018 ASSESSMENT REPORT



QUESTION 4

This question was reasonably well done as a formal proof. A few students established the limit

and did not prove L=1. However, working with the absolute value tripped quite a few students

|−(1 + 𝑛𝑛)| = 1 + 𝑛𝑛

QUESTION 5

(a) This was done very well with the vast majority of students gaining full marks.

(b) Most students recognised the connection to part (a) but many missed the factor of a 12

�1

4𝑟𝑟2 − 1

𝑛𝑛

𝑟𝑟=1

=12�

24𝑟𝑟2 − 1

𝑛𝑛

𝑟𝑟=1

= 12��

12𝑟𝑟 − 1

−1

2𝑟𝑟 + 1�

𝑛𝑛

𝑟𝑟=1

= ⋯

which meant they were not able to show the required result

(c) Done quite well. A good number of students recognised that they had to split the sum

into two different parts. Common errors were subtracting the sum to 100 terms (instead

of 99 terms) and not being able to show that the sum to infinity was 12

QUESTION 6

There was obvious confusion with the setting out of this question.

In fact, 𝑆𝑆1 = 12 − 02 + (−1)2 − (−2)2 + ⋯+ (−1)1−1 ≠ 1, and similarly 𝑆𝑆2 ≠ 3, etc

if 𝑛𝑛 = 1, 2, … are substituted directly into the given 𝑆𝑆𝑛𝑛 formula.

(a) This part was mostly well done by all students.

(b) This was poorly done. Many students unsuccessfully attempted to find 𝑆𝑆𝑛𝑛+1 , forgetting to give … + (−1)𝑛𝑛+1−1 = … + (−1)𝑛𝑛. Showing RHS = … = LHS was a much more successful strategy.

Also, almost every student who attempted this did not show that …

−(−1)𝑛𝑛−1 = ⋯+ (−1)𝑛𝑛



(c) This question was reasonably well done although many students could not do the

necessary algebraic working for the 𝑃𝑃𝑘𝑘+1 step. The importance of the structure of a proof

should be emphasised to students.

Common (and disappointing) errors were in the 𝑃𝑃𝑘𝑘+1 step, for example,

𝐿𝐿𝐿𝐿𝑆𝑆 = (𝑘𝑘 + 1)2 − 𝑘𝑘2 + (𝑘𝑘 − 1)2 − (𝑘𝑘 − 2)2 + ⋯+ (−1)𝑘𝑘 = (𝑘𝑘 + 1)2 − (−1)𝑘𝑘

SECTION B – MATRICES AND LINEAR TRANSFORMATIONS

GENERAL COMMENTS

Overall most students did well in this section, this may have been partly due to two very easy

questions. The question with the new 3D content had a lot of students getting no marks.

QUESTION 7

This was a very easy question done well by all.

QUESTION 8

This was a very easy question done well by all.

QUESTION 9

This was a straight forward question done well by most. The most common error in part (a)

was an algebraic mistake expanding a bracket and getting an incorrect minus sign.

Part b) drawing the graphs, many did not draw all the required lines.

QUESTION 10

Most knew what to do with this question but many made mistakes in the calculations involved in

the Gauss Jordan method. To get full marks in part a) it was required to reduce the matrix to

the simplest final stage �1 0 00 1 00 0 1

2𝑘𝑘 − 1𝑘𝑘1

�



Part c) had many making errors. Just substituting 𝑘𝑘 = 3 into the final matrix from part a) did not

give the correct solution. The correct method was to go back to the line assuming 𝑘𝑘 ≠ 3 then

substitute 𝑘𝑘 = 3 and proceed from there.

QUESTION 11

This question was either done very well or very poorly. Almost a third of students received no

marks for this question. For those that were familiar with this type of question it was straight

forward to get full marks. A few students used a vector approach to this question, though most

used matrices only. A common error was to find the equation of a line between two of the

points in symmetric form rather than find the equation of the plane.

QUESTION 12

(a) Done very well by most.

(b) The results were mixed, with a common error being to look at the given matrix and

assume incorrectly that there was a dilation of √3 both horizontally and vertically.

(c) This had many making a start but very few able to complete the algebra correctly to get to

the final solution.

SECTION C – DIFFERENTIAL CALCULUS + AREAS & VOLUMES

In general, it was obvious that this section would take a strong student a little longer than 36

minutes to complete. The nature of this section this year was one whereby candidates were

expected to be able show how results were arrived at. This meant that setting out played a

more prominent part than usual for this section. There were three questions that expected

candidates to show that LHS = RHS. There were also three instances where candidates were

expected to verify that a point lay on a curve. These questions are also effectively required an

LHS = RHS treatment.

Added to this was a Fundamental Theorem of Calculus question, which also requires a certain

treatment to establish the required result. Finally, there was a question which required



candidates to establish and classify stationary points and a point of inflection. In all of these there

are certain requirements in terms of setting out.

QUESTION 13

The major issue here was the nature of the LHS = RHS “proof” required.

QUESTION 14

This question was reasonably well done. Again, appropriate setting out was important here.

QUESTION 15

Both parts of this question were reasonably well done, however, a number of candidates just

substituted the co-ordinates into the equation and seemed to feel that this was all that was

required. The implicit differentiation in part a) was mostly very well done and candidates

seemed to be well prepared for this question.

QUESTION 16

(a) Mostly very well done in a fairly efficient manner.

(b) This was a Fundamental Theorem of Calculus question, with the word “hence” being most

important. Candidates were expected to treat it as such. There were a number of ways

that this could be shown. A number of students lost marks in this section for failing to

establish the link between the result in part a) and the required integral in part b). Many

candidates appeared to use their calculators. Some tried to integrate using parts or substitution.

These candidates lost a mark if they were successful, as the instruction was “hence”.

QUESTION 17

Most candidates made reasonable progress in this question. The requirement to locate and

establish the turning points was well done. The requirement to verify that there was a point of

inflection was where a number of candidates lost some marks. It is important to establish that

there was a change of concavity by showing and noting that there was a change of sign of the

second derivative and that the first derivative at the required point did not equal zero (i.e. there



was a non-stationary point of inflexion). Few candidates showed that although the function was

undefined when x = 0 the limit of f (x) did exist at x = 0.

QUESTION 18

(a) Many candidates lost the mark in part a) for failing to show the points of interest. In some

cases graphs were drawn without any scale on either axis.

(b) This was reasonably well done by a number of candidates. There was an expectation that

students needed to use an integration by parts and then integrate by substitution to score

full marks.

(c) Many candidates set up the wrong integral. Part credit was given for the candidates

showing appropriate working from this point.

SECTION D – INTEGRAL CALCULUS

Overall the performance on this section of the paper was OK with the normal mix of expected

behaviours. One common behaviour was to use the calculator to get over difficult parts of a

problem - usually to do the calculus which rather defeats the purpose of the question. So, if

there was one message to be conveyed; the questions in this section are meant to test whether

the students can integrate but many seemed to think it was OK if they could demonstrate that

their calculator can integrate.

QUESTION 19

Most students knew to take partial fractions and could do so accurately. Several candidates

omitted absolute value signs on the log functions and/or the constant. In the main this question

was not badly answered.



QUESTION 20

Most students recognised that integration by parts was required. The accuracy of the calculation

was mixed; several wrote that the integral of ln 𝑥𝑥 is 1𝑥𝑥. A few made the substitution 𝑢𝑢 = ln 𝑥𝑥 to

obtain an integral in terms of 𝑢𝑢 and 𝑒𝑒𝑢𝑢 but then did integration by parts properly.

QUESTION 21

This question was rather poorly answered. Many students did not sort out the first part

correctly with the result that the reduction did not occur. Some students seemed to get the

answer with little or no working; this suggested the calculator was used. The difficulty with part

(a) meant that part (b) proved to be a mystery. The impression given was that several

candidates realised that the integral in (b) should be related to the answer to (a) but as they had

got (a) wrong there was no obvious connection. They were then stumped. Quite a number of

candidates evaluated the integral by plugging into the calculator - that was easy to spot as the

calculator result comes up in terms of inverse sin rather than inverse cos.

QUESTION 22

Most students knew that variables needed to be separated. Integrating 𝐴𝐴− 32 led to all sorts of

problems. This question uncovered all sorts of algebraic difficulties when given an expression of

the type 𝐴𝐴− 12 = 1𝑡𝑡

+ 𝑐𝑐 , trying to solve for 𝐴𝐴 proved beyond most. The standard 'trick' was to

change the equation to 𝐴𝐴12 = 𝑡𝑡 + 1

𝑐𝑐 and then squaring the result. This was not well answered.

QUESTION 23

Most candidates did not get this question at all. There was little understanding as to how the

integral in terms of 𝑥𝑥 would transform to that in terms of 𝑢𝑢 with all sorts of fudges going on.

Many had no idea how to find 𝑎𝑎 and 𝑏𝑏. Interestingly though, many persevered with doing part

(b) and did realise they needed the sin of 𝜋𝜋3 and

𝜋𝜋4. Apparently the importance of these values

did not dawn on them for part (a). There was lots of fiddling with the logarithms that wasn't



exactly convincing. Some did solve (b) properly using partial fractions. However many candidates

just evaluated the integral by putting the integrand into the calculator.

QUESTION 24

This question was well done. Almost all students knew the correct substitution and obtained the

appropriate reduced equation. Some of the separation of variables was suspect with the

integrand being the wrong way up. Others integrated correctly but seemed to get the calculator

to do it. Some got confused with evaluating the constant and then isolating '𝑦𝑦'; many students

forgot the possibility of having the negative square root.

SECTION E – COMPLEX NUMBERS

There was evidence that some students ran out of time during this section. This would be

mostly due to the previous 2 sections each being a little too long.

QUESTION 25

(a) This question was a nice easy start and the vast majority of students scored full marks for

this part.

(b) This question was also very well done, however, some students did not state their answer

as 𝑛𝑛 = 4 (or similar), leaving the answer as (1 − 𝑖𝑖)4 = −4

QUESTION 26

The question asked students to “prove that …”. The number of students who did not use any

reasonably correct setting out for a proof was disappointing. Many students assumed the result

and manipulated both sides of the equation to reach their version of an answer.

QUESTION 27

(a) Those students who used the General Quadratic Formula to solve the quadratic equation

were very successful. Unfortunately, not all students chose this simple technique. Some

chose to use the sum of a GP formula and generally got to the correct answer but



obviously a more complicated method is more likely to cause careless errors to occur! A

common error was the difficulty in finding the correct 𝑟𝑟 value in the 𝑟𝑟𝑐𝑐𝑖𝑖𝑟𝑟𝑟𝑟 form.

(b) Quite well done. A few students left their answer as 𝑐𝑐𝑐𝑐𝑐𝑐 𝜋𝜋216

or similar. Some students

ignored the negative part of the index.

QUESTION 28

Too many students did not label the axes or attempt to label the points of intersection. A few

students showed their lack of understanding of angles in radian form by drawing 𝜋𝜋3 as a smaller

angle than 𝜋𝜋4 . Many did not draw their 𝐴𝐴𝑟𝑟𝐴𝐴 𝜋𝜋

4 line through the point (1, 1) or even close.

QUESTION 29

(a) This question was not done well. A large number of students substituted (𝑧𝑧 − 𝑘𝑘𝑖𝑖) and

(𝑧𝑧 + 𝑘𝑘𝑖𝑖) and got nowhere. Those who did approach this problem correctly often gave

their solution as 𝑘𝑘 = ±√2, ±7, 0 without ensuring that the solutions satisfied both

𝑅𝑅𝑒𝑒𝑎𝑎𝑅𝑅 𝑝𝑝𝑎𝑎𝑟𝑟𝑡𝑡 = 0 and 𝐼𝐼𝐼𝐼𝑎𝑎𝐴𝐴𝑖𝑖𝑛𝑛𝑎𝑎𝑟𝑟𝑦𝑦 𝑝𝑝𝑎𝑎𝑟𝑟𝑡𝑡 = 0.

(b) Some students solved this using their calculators.

Those who did give 𝑘𝑘 = ±√2, ±7, 0 as their solution for part (a) often gave 𝑧𝑧 =

±√2𝑖𝑖, ±7𝑖𝑖, 0 as their solution to 𝑃𝑃(𝑧𝑧) = 0

QUESTION 30

(a) This question was reasonably well done.

(b) A few students did this question perfectly, however, many did not. Many students went

round in circles (incorrectly) e.g.

cos4 𝑟𝑟 = �12�𝑒𝑒𝑐𝑐𝑖𝑖 + 𝑒𝑒−𝑐𝑐𝑖𝑖��

4= �1

2�4�𝑒𝑒𝑐𝑐𝑖𝑖 + 𝑒𝑒−𝑐𝑐𝑖𝑖�

4= �1

2�4

(cos𝑟𝑟)4 etc



MATHEMATICS – SPECIALISED (MTS415118)

SOLUTIONS

SECTION A – SEQUENCES AND SERIES

QUESTION 1

𝑢𝑢𝑛𝑛+1 = 1𝑢𝑢𝑛𝑛

,𝑢𝑢1 = 2 ⇒ 𝑢𝑢2 = 12

, 𝑢𝑢3 = 2 , 𝑢𝑢4 = 12

,𝑢𝑢5 = 2 , 𝑢𝑢6 = 12

, …

So 𝑢𝑢100 =12

QUESTION 2

𝑆𝑆𝑛𝑛 = 𝑛𝑛(𝑛𝑛 + 2)

𝑢𝑢𝑛𝑛 = 𝑆𝑆𝑛𝑛 − 𝑆𝑆𝑛𝑛−1

𝑢𝑢𝑛𝑛 = 𝑛𝑛(𝑛𝑛 + 2)− (𝑛𝑛 − 1)(𝑛𝑛 + 1)

𝑢𝑢𝑛𝑛 = 𝑛𝑛2 + 2𝑛𝑛 − 𝑛𝑛2 + 1

𝑢𝑢𝑛𝑛 = 2𝑛𝑛 + 1

QUESTION 3

(a) 𝑓𝑓(𝑥𝑥) = 1 1+2𝑥𝑥

𝑓𝑓(0) = 1

𝑓𝑓′(𝑥𝑥) = −2 (1+2𝑥𝑥)2

𝑓𝑓′(0) = −2

𝑓𝑓′′(𝑥𝑥) = 8 (1+2𝑥𝑥)3

𝑓𝑓′′(0) = 8

𝑓𝑓′′′(𝑥𝑥) = −48 (1+2𝑥𝑥)4

𝑓𝑓′′′(0) = −48



11 + 2𝑥𝑥 = 1 + (−2)𝑥𝑥 + 8

𝑥𝑥2

2! + (−48)𝑥𝑥3

3! + ⋯

= 1 − 2𝑥𝑥 + 4𝑥𝑥2 − 8𝑥𝑥3 + ⋯

(b)

So = 2 ∫(1 − 2𝑥𝑥 + 4𝑥𝑥2 − 8𝑥𝑥3 + ⋯ )𝑑𝑑𝑥𝑥

= 2 �𝑥𝑥 − 𝑥𝑥2 +4𝑥𝑥3

3 − 2𝑥𝑥4 + ⋯�

+ 𝑐𝑐 (assume 𝑐𝑐 = 0)

= 2𝑥𝑥 − 2𝑥𝑥2 +8𝑥𝑥3

3 − 4𝑥𝑥4 + ⋯

QUESTION 4

Prove: lim𝑛𝑛→∞

� 𝑛𝑛2

1+𝑛𝑛+𝑛𝑛2� = 1 i.e. for any 𝜖𝜖 >

0,∃ 𝑁𝑁(𝜖𝜖) 𝑟𝑟. 𝑡𝑡. � 𝑛𝑛2

1+𝑛𝑛+𝑛𝑛2− 1� < 𝜖𝜖 ∀ 𝑛𝑛 > 𝑁𝑁

Proof: � 𝑛𝑛2

1+𝑛𝑛+𝑛𝑛2− 1� < 𝜖𝜖

⟸ �𝑛𝑛2 − (1 + 𝑛𝑛 + 𝑛𝑛2)

1 + 𝑛𝑛 + 𝑛𝑛2 � < 𝜖𝜖

⟸ �−(1 + 𝑛𝑛)

1 + 𝑛𝑛 + 𝑛𝑛2�< 𝜖𝜖

⟸ 𝑛𝑛 + 1

1 + 𝑛𝑛 + 𝑛𝑛2 < 𝜖𝜖

⟸ 𝑛𝑛 + 1𝑛𝑛 + 𝑛𝑛2

< 𝜖𝜖 𝑎𝑎𝑟𝑟 𝑛𝑛 + 1

1 + 𝑛𝑛 + 𝑛𝑛2<

𝑛𝑛 + 1𝑛𝑛 + 𝑛𝑛2

⟸ 1𝑛𝑛

< 𝜖𝜖

�1

1 + 2𝑥𝑥 𝑑𝑑𝑥𝑥 =

12

ln(1 + 2𝑥𝑥) + 𝑐𝑐

ln(1 + 2𝑥𝑥) = 2�1

1 + 2𝑥𝑥 𝑑𝑑𝑥𝑥



⟸ 𝑛𝑛 >1𝜖𝜖

let 𝑁𝑁 =1𝜖𝜖

so ∀ 𝑛𝑛 > 𝑁𝑁, �𝑛𝑛2

1 + 𝑛𝑛 + 𝑛𝑛2− 1� < 𝜖𝜖 and lim

𝑛𝑛→∞�

𝑛𝑛2

1 + 𝑛𝑛 + 𝑛𝑛2� = 1

OR Prove: lim𝑛𝑛→∞

� 𝑛𝑛2

1+𝑛𝑛+𝑛𝑛2� = 1

i.e. for any 𝜖𝜖 > 0,∃ 𝑁𝑁(𝜖𝜖) 𝑟𝑟. 𝑡𝑡. � 𝑛𝑛2

1+𝑛𝑛+𝑛𝑛2− 1� < 𝜖𝜖 ∀ 𝑛𝑛 > 𝑁𝑁

Proof: � 𝑛𝑛2

1+𝑛𝑛+𝑛𝑛2− 1� < 𝜖𝜖

⟸ �𝑛𝑛2 − (1 + 𝑛𝑛 + 𝑛𝑛2)

1 + 𝑛𝑛 + 𝑛𝑛2� < 𝜖𝜖

⟸ �−(1 + 𝑛𝑛)

1 + 𝑛𝑛 + 𝑛𝑛2� < 𝜖𝜖

⟸ 𝑛𝑛 + 1

1 + 𝑛𝑛 + 𝑛𝑛2< 𝜖𝜖

⟸ 𝑛𝑛 + 1 < 𝜖𝜖 + 𝜖𝜖𝑛𝑛 + 𝜖𝜖𝑛𝑛2

⟸ 𝜖𝜖𝑛𝑛2 + (𝜖𝜖 − 1)𝑛𝑛 + 𝜖𝜖 − 1 < 0

⟸ 𝑛𝑛 >−(𝜖𝜖 − 1) + �(𝜖𝜖 − 1)2 − 4𝜖𝜖(𝜖𝜖 − 1)

2𝜖𝜖

=(1 − 𝜖𝜖) + √1 + 2𝜖𝜖 − 3𝜖𝜖2

2𝜖𝜖

let 𝑁𝑁 =(1 − 𝜖𝜖) + √1 + 2𝜖𝜖 − 3𝜖𝜖2

2𝜖𝜖

so ∀ 𝑛𝑛 > 𝑁𝑁, �𝑛𝑛2

1 + 𝑛𝑛 + 𝑛𝑛2− 1� < 𝜖𝜖

and lim𝑛𝑛→∞

�𝑛𝑛2

1 + 𝑛𝑛 + 𝑛𝑛2� = 1



QUESTION 5

(a) 1

2𝑟𝑟−1− 1

2𝑟𝑟+1= 2𝑟𝑟+1−(2𝑟𝑟−1)

(2𝑟𝑟−1)(2𝑟𝑟+1)= 2

4𝑟𝑟2−1

(b)

(c)

�1

4𝑟𝑟2 − 1

𝑛𝑛

𝑟𝑟=1

=12�

24𝑟𝑟2 − 1

𝑛𝑛

𝑟𝑟=1

=12��

12𝑟𝑟 − 1

−1

2𝑟𝑟 + 1�

𝑛𝑛

𝑟𝑟=1

=12�1 −

13

+13−

15

+15−

17

+ ⋯+1

2𝑛𝑛 − 1

−1

2𝑛𝑛 + 1�

=12�1 −

12𝑛𝑛 + 1

�

1�

2𝑛𝑛�

�

14𝑟𝑟2 − 1

∞

𝑟𝑟=100

= �1

4𝑟𝑟2 − 1

∞

𝑟𝑟=1

−�1

4𝑟𝑟2 − 1

99

𝑟𝑟=1

= lim𝑛𝑛→∞

�𝑛𝑛

2𝑛𝑛 + 1� −

992 × 99 + 1

=12−

99199

=1

398



QUESTION 6

(a) 𝑆𝑆4 = 42 − 32 + 22 − 12 = 10

(b) 𝑆𝑆𝑛𝑛+1 = (𝑛𝑛 + 1)2 − 𝑛𝑛2 + (𝑛𝑛 − 1)2 − (𝑛𝑛 − 2)2 + (𝑛𝑛 − 3)2 + ⋯+ (−1)𝑛𝑛

(𝑛𝑛 + 1)2 − 𝑆𝑆𝑛𝑛= (𝑛𝑛 + 1)2

− (𝑛𝑛2 − (𝑛𝑛 − 1)2 + (𝑛𝑛 − 2)2 − (𝑛𝑛 − 3)2 + ⋯+ (−1)𝑛𝑛−1)

= (𝑛𝑛 + 1)2 − 𝑛𝑛2 + (𝑛𝑛 − 1)2 − (𝑛𝑛 − 2)2 + (𝑛𝑛 − 3)2 + ⋯− (−1)𝑛𝑛−1

= (𝑛𝑛 + 1)2 − 𝑛𝑛2 + (𝑛𝑛 − 1)2 − (𝑛𝑛 − 2)2 + (𝑛𝑛 − 3)2 + ⋯+ (−1)𝑛𝑛

= 𝑆𝑆𝑛𝑛+1

(c) Prove: 𝑛𝑛2 − (𝑛𝑛 − 1)2 + (𝑛𝑛 − 2)2 − (𝑛𝑛 − 3)2 + ⋯+ (−1)𝑛𝑛−1 =𝑛𝑛(𝑛𝑛+1)

2 , 𝑛𝑛𝜖𝜖ℤ+

Proof: Let 𝑃𝑃𝑛𝑛 be the statement above

Consider 𝑃𝑃1 𝐿𝐿𝐿𝐿𝑆𝑆 = 12 = 1

𝑅𝑅𝐿𝐿𝑆𝑆 =1(1 + 1)

2= 1 = 𝐿𝐿𝐿𝐿𝑆𝑆

So 𝑃𝑃1 is true

Assume 𝑃𝑃𝑘𝑘 is true

𝑖𝑖. 𝑒𝑒. 𝑘𝑘2 − (𝑘𝑘 − 1)2 + (𝑘𝑘 − 2)2 − (𝑘𝑘 − 3)2 + ⋯+ (−1)𝑘𝑘−1

=𝑘𝑘(𝑘𝑘 + 1)

2 , 𝑘𝑘𝜖𝜖ℤ+

Consider 𝑃𝑃𝑘𝑘+1

𝑖𝑖. 𝑒𝑒. (𝑘𝑘 + 1)2 − 𝑘𝑘2 + (𝑘𝑘 − 1)2 − (𝑘𝑘 − 2)2 + ⋯+ (−1)𝑘𝑘

=(𝑘𝑘 + 1)(𝑘𝑘 + 2)

2 ,𝑘𝑘𝜖𝜖ℤ+



𝐿𝐿𝐿𝐿𝑆𝑆 = (𝑘𝑘 + 1)2

−k(k+1)

2 from part (b) and assuming 𝑃𝑃𝑘𝑘 is true

=2(𝑘𝑘 + 1)2 − 𝑘𝑘(𝑘𝑘 + 1)

2

=𝑘𝑘2 + 3𝑘𝑘 + 2

2

=(𝑘𝑘 + 1)(𝑘𝑘 + 2)

2= 𝑅𝑅𝐿𝐿𝑆𝑆

So 𝑃𝑃𝑘𝑘 true ⟹ 𝑃𝑃𝑘𝑘+1 true

Since 𝑃𝑃1 is true and 𝑃𝑃𝑘𝑘 true ⟹ 𝑃𝑃𝑘𝑘+1 true then by the Principle of Mathematical Induction 𝑃𝑃𝑛𝑛 is true ∀𝑛𝑛𝜖𝜖ℤ+.



SECTION B – MATRICES AND LINEAR TRANSFORMATIONS

QUESTION 7

(a) 𝐴𝐴2 = �2 01 8� �

2 01 8� = � 4 0

10 64�

(b) det𝐴𝐴 = 2 × 8 − 0 × 1 = 16

𝐴𝐴−1 =1

16� 8 0−1 2� = �

12

0

−1

1618

�

QUESTION 8

(a) |𝐴𝐴| = 𝑎𝑎𝑑𝑑 − 𝑏𝑏𝑐𝑐

|𝐴𝐴𝑇𝑇| = 𝑎𝑎𝑑𝑑 − 𝑐𝑐𝑏𝑏 = |𝐴𝐴|

(b) 𝑅𝑅 = �cos 𝑟𝑟 − sin 𝑟𝑟sin 𝑟𝑟 cos 𝑟𝑟 � and 𝑅𝑅𝑇𝑇 = � cos 𝑟𝑟 sin 𝑟𝑟

−sin 𝑟𝑟 cos 𝑟𝑟�

𝑅𝑅𝑅𝑅𝑇𝑇 = �cos 𝑟𝑟 − sin 𝑟𝑟sin 𝑟𝑟 cos 𝑟𝑟 � � cos 𝑟𝑟 sin 𝑟𝑟

−sin 𝑟𝑟 cos 𝑟𝑟�

= � cos2 𝑟𝑟 + sin2 𝑟𝑟 cos 𝑟𝑟 sin 𝑟𝑟 − cos 𝑟𝑟 sin 𝑟𝑟cos 𝑟𝑟 sin 𝑟𝑟 − cos 𝑟𝑟 sin 𝑟𝑟 sin2 𝑟𝑟 + cos2 𝑟𝑟

�

= �1 00 1�

= 𝐼𝐼



QUESTION 9

(a) 𝑦𝑦 = 1√3𝑥𝑥 so tan𝑟𝑟 = 1

√3 ∴ 𝑟𝑟 = 𝜋𝜋

6 ⇒

�cos 2 𝜋𝜋

6sin 2 𝜋𝜋

6

sin 2 𝜋𝜋6

− cos 2 𝜋𝜋6

� = �12

√32

√32

− 12

�

𝑥𝑥′ = 12𝑥𝑥 + √3

2𝑦𝑦 and 𝑦𝑦′ = √3

2𝑥𝑥 − 1

2𝑦𝑦

image line is 𝑥𝑥′√3 − 𝑦𝑦′ = 2

So �12𝑥𝑥 + √3

2𝑦𝑦�√3 − �√3

2𝑥𝑥 − 1

2𝑦𝑦� = 2

√32𝑥𝑥 + 3

2𝑦𝑦 − √3

2𝑥𝑥 + 1

2𝑦𝑦 = 2

2𝑦𝑦 = 2 ⇒ 𝑦𝑦 = 1

(b)

�√3 , 1�

�2√3

3 , 0� 𝑥𝑥

𝑦𝑦 = 1

𝑦𝑦 =1√3

𝑥𝑥

𝑦𝑦 = √3𝑥𝑥 − 2



QUESTION 10

(a) �1 −2 2−1 3 −12 −5 𝑘𝑘

1 𝑘𝑘−2

� = �1 −2 20 1 10 −1 𝑘𝑘 − 4

1

𝑘𝑘 + 1−4

� 𝑅𝑅1 + 𝑅𝑅2𝑅𝑅3 − 2𝑅𝑅1

= �1 −2 20 1 10 0 𝑘𝑘 − 3

1

𝑘𝑘 + 1𝑘𝑘 − 3

� 𝑅𝑅3 + 𝑅𝑅2

= �1 −2 20 1 10 0 1

1

𝑘𝑘 + 11

� 𝑅𝑅3 ÷ (𝑘𝑘 − 3), 𝑘𝑘 ≠ 3

= �1 −2 20 1 00 0 1

1 𝑘𝑘1� 𝑅𝑅2 − 𝑅𝑅3

= �1 −2 00 1 00 0 1

−1 𝑘𝑘1� 𝑅𝑅1 − 2𝑅𝑅3

= �1 0 00 1 00 0 1

2𝑘𝑘 − 1

𝑘𝑘1

� 𝑅𝑅1 + 2𝑅𝑅2

(b) for 𝑘𝑘 ≠ 3, 𝑥𝑥 = 2𝑘𝑘 − 1, 𝑦𝑦 = 𝑘𝑘, 𝑧𝑧 = 1

(c) for 𝑘𝑘 = 3, matrix becomes �1 −2 20 1 10 0 0

1 40�

let 𝑧𝑧 = 𝑡𝑡, 𝑦𝑦 + 𝑧𝑧 = 4 ⇒ 𝑦𝑦 = 4 − 𝑡𝑡

𝑥𝑥 − 2𝑦𝑦 + 2𝑧𝑧 = 1 ⇒ 𝑥𝑥 = 9 − 4𝑡𝑡

i.e. the 3 planes meet in a straight line given by 𝑥𝑥 = 9 − 4𝑡𝑡,𝑦𝑦 = 4 − 𝑡𝑡, 𝑧𝑧 = 𝑡𝑡



QUESTION 11

(a) Let the plane be 𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑦𝑦 + 𝑐𝑐𝑧𝑧 = 1

(1, 4, 2) ⇒ 𝑎𝑎 + 4𝑏𝑏 + 2𝑐𝑐 = 1

(3, 2, 1) ⇒ 3𝑎𝑎 + 2𝑏𝑏 + 𝑐𝑐 = 1

(5, 3, 5) ⇒ 5𝑎𝑎 + 3𝑏𝑏 + 5𝑐𝑐 = 1

�1 4 23 2 15 3 5

111� = �

1 0 00 1 00 0 1

15�

2 7�

−6 35�

�

𝑎𝑎 = 15, 𝑏𝑏 = 27 , 𝑐𝑐 = −

635

So 15𝑥𝑥 + 2

7𝑦𝑦 − 6

35𝑧𝑧 = 1

⇒ 7𝑥𝑥 + 10𝑦𝑦 − 6𝑧𝑧 = 35

(b) 𝑥𝑥 = 1 + 2𝑡𝑡, 𝑦𝑦 = 1 + 𝑡𝑡, 𝑧𝑧 = −3 + 4𝑡𝑡

⇒ 7𝑥𝑥 + 10𝑦𝑦 − 6𝑧𝑧 = 7(1 + 2𝑡𝑡) + 10(1 + 𝑡𝑡) − 6(−3 + 4𝑡𝑡)

= 7 + 14𝑡𝑡 + 10 + 10𝑡𝑡 + 18 − 24𝑡𝑡

= 35 as required

So the plane 7𝑥𝑥 + 10𝑦𝑦 − 6𝑧𝑧 = 35 contains the line given by

𝑥𝑥 = 1 + 2𝑡𝑡, 𝑦𝑦 = 1 + 𝑡𝑡, 𝑧𝑧 = −3 + 4𝑡𝑡



QUESTION 12

(a) �√3 1−1 √3

� �00� = �0

0� = 𝑂𝑂′

�√3 1−1 √3

� �10� = �√3

−1� = 𝐴𝐴′

�√3 1−1 √3

� �11� = �√3 + 1

√3 − 1� = 𝐵𝐵′

�√3 1−1 √3

� �01� = � 1

√3� = 𝐶𝐶′

(b) M is a dilation by a factor of 2 in both the 𝑥𝑥 and the 𝑦𝑦 directions, followed by a rotation clockwise through 𝜋𝜋

6

radians (by inspection!)

Dilation �2 00 2�

Rotation �√32

12

−12

√32

�

OR using an algebraic approach …..

dilation by a factor of 𝑎𝑎 in the 𝑥𝑥 direction and by a factor of 𝑏𝑏 in the 𝑦𝑦

direction ⇒ �𝑎𝑎 00 𝑏𝑏�

Rotation matrix ⇒ �cos 𝑟𝑟 − sin 𝑟𝑟sin 𝑟𝑟 cos 𝑟𝑟 �

�𝑎𝑎 00 𝑏𝑏� �

cos 𝑟𝑟 − sin 𝑟𝑟sin 𝑟𝑟 cos 𝑟𝑟 � = �𝑎𝑎 cos 𝑟𝑟 −𝑎𝑎 sin 𝑟𝑟

𝑏𝑏 sin 𝑟𝑟 𝑏𝑏 cos 𝑟𝑟 � = �√3 1−1 √3

�

O

A’

A

B B’

C

C’



∴ 𝑎𝑎 cos𝑟𝑟 = √3 and − 𝑎𝑎 sin 𝑟𝑟 = 1 ⇒ tan𝑟𝑟 = −1√3

so 𝑟𝑟 = −𝜋𝜋6

𝑎𝑎 cos �−𝜋𝜋6� = √3 ⇒ 𝑎𝑎 =

√3√3

2�= 2

𝑏𝑏 sin �−𝜋𝜋6� = −1 ⇒ 𝑏𝑏 =

−1−1

2�= 2

So M is a dilation by a factor of 2 in both the 𝑥𝑥 and the 𝑦𝑦 directions, followed by a rotation clockwise through 𝜋𝜋

6 radians.

(c) 𝑲𝑲 = � 𝑘𝑘 1−1 𝑘𝑘�

𝑲𝑲 has a similar effect to M = dilation by a factor of √𝑘𝑘2 + 1 in both the 𝑥𝑥 and the 𝑦𝑦

directions and a rotation (clockwise through

arctan �1𝑘𝑘� )

Original circle 𝑥𝑥2 + 𝑦𝑦2 = 1 circle with centre (0, 0) and 𝑟𝑟 = 1

⟶ image 𝑥𝑥2 + 𝑦𝑦2 = 𝑘𝑘2 + 1 circle with centre (0, 0) and 𝑟𝑟 = √𝑘𝑘2 + 1



SECTION C – DIFFERENTIAL CALCULUS AND AREA/VOLUME

QUESTION 13

𝑦𝑦 = arctan𝑥𝑥

𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

=1

1 + 𝑥𝑥2 and

𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2

= −2𝑥𝑥

(1 + 𝑥𝑥2)2

(1 + 𝑥𝑥2)𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2

+ 2𝑥𝑥𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

= (1 + 𝑥𝑥2) ⋅−2𝑥𝑥

(1 + 𝑥𝑥2)2 + 2𝑥𝑥 ⋅1

1 + 𝑥𝑥2

=−2𝑥𝑥

1 + 𝑥𝑥2+

2𝑥𝑥1 + 𝑥𝑥2

= 0 as required

QUESTION 14

𝑥𝑥2𝑦𝑦2 = 1

2𝑥𝑥𝑦𝑦2 + 𝑥𝑥2 ⋅ 2𝑦𝑦𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

= 0


=−2𝑥𝑥𝑦𝑦2

2𝑥𝑥2𝑦𝑦= −

𝑦𝑦𝑥𝑥

𝑑𝑑2𝑦𝑦𝑑𝑑𝑥𝑥2

=−1𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥 ⋅ 𝑥𝑥 − (−𝑦𝑦) ⋅ 1

𝑥𝑥2

=𝑦𝑦 − 𝑑𝑑𝑦𝑦

𝑑𝑑𝑥𝑥 ⋅ 𝑥𝑥𝑥𝑥2

=𝑦𝑦 − �−𝑦𝑦𝑥𝑥� ⋅ 𝑥𝑥

𝑥𝑥2



=2𝑦𝑦𝑥𝑥2

as required

QUESTION15

(a) cos 𝑦𝑦 + 𝑦𝑦 sin 𝑥𝑥 = 𝑥𝑥2

If 𝑥𝑥 = −1 and 𝑦𝑦 = 0 then cos 𝑦𝑦 + 𝑦𝑦 sin 𝑥𝑥 = cos 0 + 0 sin(−1) = 1

and 𝑥𝑥2 = (−1)2 = 1

If 𝑥𝑥 = 0 and 𝑦𝑦 =𝜋𝜋2

then cos 𝑦𝑦 + 𝑦𝑦 sin 𝑥𝑥 = cos𝜋𝜋2

+𝜋𝜋2

sin 0 = 0

and 𝑥𝑥2 = (0)2 = 0

So (−1, 0) and �0, 𝜋𝜋2� both lie on the curve.

(b) cos 𝑦𝑦 + 𝑦𝑦 sin 𝑥𝑥 = 𝑥𝑥2

− sin 𝑦𝑦𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

+𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

sin 𝑥𝑥 + 𝑦𝑦 cos 𝑥𝑥 = 2𝑥𝑥

�0,𝜋𝜋2� ⇒ − sin

𝜋𝜋2⋅𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

+𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

sin 0 +𝜋𝜋2

cos 0 = 0


=𝜋𝜋2

So tangent has gradient of 𝜋𝜋2

(−1, 0) and �0,𝜋𝜋2� ⇒ 𝐼𝐼 =

𝜋𝜋2 − 0

0 − (−1)=𝜋𝜋2

Same gradient with a common point so the tangent to the curve at �0, 𝜋𝜋2�

also passes through (−1, 0).



QUESTION 16

(a) 𝑑𝑑𝑑𝑑𝑥𝑥�2√1−𝑥𝑥� = 2√1−𝑥𝑥 ⋅ ln 2 ⋅ − 1

2(1 − 𝑥𝑥)−

12

= −2√1−𝑥𝑥 ln 2

2√1 − 𝑥𝑥

(b) 𝑦𝑦 = 2√1−𝑥𝑥

√1−𝑥𝑥 ,−3 ≤ 𝑥𝑥 ≤ 0

𝑎𝑎𝑟𝑟𝑒𝑒𝑎𝑎 = �2√1−𝑥𝑥

√1 − 𝑥𝑥𝑑𝑑𝑥𝑥

0

−3

= −2ln 2

�−2√1−𝑥𝑥 ln 2

2√1 − 𝑥𝑥𝑑𝑑𝑥𝑥

0

−3

=−2ln 2

�2√1−𝑥𝑥�−30

=−2ln 2

(2 − 22)

=4

ln 2 𝑢𝑢𝑛𝑛𝑖𝑖𝑡𝑡𝑟𝑟2

QUESTION 17

(a) 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 (ln 𝑥𝑥)2 , 𝑥𝑥 > 0

𝑓𝑓′(𝑥𝑥) = (ln 𝑥𝑥)2 + 𝑥𝑥 ⋅ 2 ln 𝑥𝑥 ⋅ 1𝑥𝑥 = (ln 𝑥𝑥)2 + 2 ln 𝑥𝑥 = ln 𝑥𝑥 (ln 𝑥𝑥 + 2)

𝑓𝑓′′(𝑥𝑥) = 2 ln 𝑥𝑥 ⋅1𝑥𝑥

+2𝑥𝑥

=2𝑥𝑥

(ln 𝑥𝑥 + 1)

Stat pts when 𝑓𝑓′(𝑥𝑥) = 0

ln 𝑥𝑥 (ln 𝑥𝑥 + 2) = 0

ln 𝑥𝑥 = 0 and ln 𝑥𝑥 + 2 = 0

𝑥𝑥 = 1 and 𝑥𝑥 = 𝑒𝑒−2



at 𝑥𝑥 = 1, 𝑓𝑓(1) = 1 (ln 1)2 = 0, 𝑓𝑓′(1) = 0, 𝑓𝑓′′(1) =

21

(ln 1 + 1) = 2 > 0 so concave up

∴ (1, 0) is a local min pt

at 𝑥𝑥 = 𝑒𝑒−2, 𝑓𝑓(𝑒𝑒−2) = 𝑒𝑒−2 (ln 𝑒𝑒−2)2 = 4𝑒𝑒−2, 𝑓𝑓′(𝑒𝑒−2) = 0

𝑓𝑓′′(𝑒𝑒−2) = 2𝑒𝑒−2

(ln 𝑒𝑒−2 + 1) = −2𝑒𝑒−2

< 0 so concave down

∴ (𝑒𝑒−2, 4𝑒𝑒−2) is a local max pt

Pts of inflection when 𝑓𝑓′′(𝑥𝑥) = 0

2𝑥𝑥 (ln𝑥𝑥 + 1) = 0

2𝑥𝑥 = 0 and ln 𝑥𝑥 + 1 = 0

no soln and 𝑥𝑥 = 𝑒𝑒−1

at 𝑥𝑥 = 𝑒𝑒−1, 𝑓𝑓(𝑒𝑒−1) = 𝑒𝑒−1 (ln 𝑒𝑒−1)2 = 𝑒𝑒−1

𝑓𝑓′(𝑒𝑒−1) = ln 𝑒𝑒−1 (ln 𝑒𝑒−1 + 2) = −1 < 0

𝑓𝑓′′(𝑒𝑒−1) =2𝑒𝑒−1 (ln 𝑒𝑒−1 + 1) = 0

𝑥𝑥 0.2 𝑒𝑒−1 0.5

𝑓𝑓′′(𝑥𝑥) < 0 = 0 > 0

as 𝑥𝑥 ↑ 𝑒𝑒−1, 𝑓𝑓′′(𝑥𝑥) changes from concave down to concave up

∴ (𝑒𝑒−1, 𝑒𝑒−1) is a pt of inflection (decreasing curve)



(b) 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 (ln 𝑥𝑥)2 , 𝑥𝑥 > 0

QUESTION 18

(a) 𝑦𝑦 = arcsin 𝑥𝑥 , 0 ≤ 𝑥𝑥 ≤ 1

(b)

𝑎𝑎𝑟𝑟𝑒𝑒𝑎𝑎 = � arcsin 𝑥𝑥 𝑑𝑑𝑥𝑥1

0

let 𝑓𝑓 = arcsin 𝑥𝑥 ⇒ 𝑓𝑓′

=1

√1 − 𝑥𝑥2 and 𝐴𝐴′ = 1 ⇒ 𝐴𝐴 = 𝑥𝑥

= � 𝑥𝑥 arcsin 𝑥𝑥 �01− �

𝑥𝑥√1 − 𝑥𝑥2

𝑑𝑑𝑥𝑥1

0

�1,𝜋𝜋2�

(𝑒𝑒−1, 𝑒𝑒−1)

(𝑒𝑒−2, 4𝑒𝑒−2)

(1, 0)

(0, 0)



=𝜋𝜋2

+ 12�𝑢𝑢−

12𝑑𝑑𝑥𝑥

0

1

𝑅𝑅𝑒𝑒𝑡𝑡 𝑢𝑢 = 1 − 𝑥𝑥2 ⇒ 𝑑𝑑𝑢𝑢

= −2𝑥𝑥 𝑑𝑑𝑥𝑥 𝑥𝑥 = 0 ⇒ 𝑢𝑢 = 1

−12𝑑𝑑𝑢𝑢 = 𝑥𝑥 𝑑𝑑𝑥𝑥 𝑥𝑥 = 1

⇒ 𝑢𝑢 = 0

=𝜋𝜋2

+ 1 2�2𝑢𝑢

12�1

0

=𝜋𝜋2

+ 12

(0 − 2)

=𝜋𝜋2

− 1 𝑢𝑢𝑛𝑛𝑖𝑖𝑡𝑡𝑟𝑟2

(c)

𝑣𝑣𝑣𝑣𝑅𝑅𝑢𝑢𝐼𝐼𝑒𝑒 = 𝜋𝜋� 12𝑑𝑑𝑦𝑦

𝜋𝜋2

0

− 𝜋𝜋�(sin 𝑦𝑦)2𝑑𝑑𝑦𝑦

𝜋𝜋2

0

𝑦𝑦 = arcsin 𝑥𝑥 ⇒ 𝑥𝑥

= sin 𝑦𝑦

= 𝜋𝜋 [𝑦𝑦]0𝜋𝜋2 − 𝜋𝜋� �

12−

12

cos 2𝑦𝑦�2

𝑑𝑑𝑦𝑦 cos 2𝑦𝑦 = 1 − 2 sin2 𝑦𝑦

𝜋𝜋2

0

= 𝜋𝜋 �𝜋𝜋2− 0� − 𝜋𝜋 �

𝑦𝑦2−

14

sin 2𝑦𝑦�0

𝜋𝜋2

=𝜋𝜋2

2 − 𝜋𝜋 ��

𝜋𝜋4−

14

sin𝜋𝜋� − �0 −14

sin 0��

=𝜋𝜋2

2 −𝜋𝜋2

4

=𝜋𝜋2

4 𝑢𝑢𝑛𝑛𝑖𝑖𝑡𝑡𝑟𝑟3



SECTION D – INTEGRAL CALCULUS

QUESTION 19

𝑥𝑥+3

𝑥𝑥 (𝑥𝑥−1) = 𝐴𝐴𝑥𝑥

+ 𝐵𝐵𝑥𝑥−1

𝑥𝑥 + 3 = 𝐴𝐴(𝑥𝑥 − 1) + 𝐵𝐵𝑥𝑥

let 𝑥𝑥 = 0 3 = −𝐴𝐴 ⇒ 𝐴𝐴 = −3

let 𝑥𝑥 = 1 4 = 𝐵𝐵

�𝑥𝑥 + 3

𝑥𝑥 (𝑥𝑥 − 1) 𝑑𝑑𝑥𝑥 = �4

𝑥𝑥 − 1−

3𝑥𝑥

𝑑𝑑𝑥𝑥

= 4 ln|𝑥𝑥 − 1| − 3 ln|𝑥𝑥| + 𝑐𝑐

= ln �(𝑥𝑥−1)4

𝑥𝑥3� + 𝑐𝑐

QUESTION 20

∫ 𝑥𝑥3 ln 𝑥𝑥 𝑑𝑑𝑥𝑥 = �𝑥𝑥4

4ln 𝑥𝑥�

1

𝑒𝑒− 1

4 ∫ 𝑥𝑥3 𝑑𝑑𝑥𝑥𝑒𝑒1

𝑒𝑒1

=𝑒𝑒4

4ln 𝑒𝑒 − 0 −

14

�𝑥𝑥4

4�1

𝑒𝑒

=𝑒𝑒4

4−

14

�𝑒𝑒4

4−

14�

=𝑒𝑒4

4−𝑒𝑒4

16+

14

= 3𝑒𝑒4 + 1

4

let 𝑓𝑓(𝑥𝑥) = ln𝑥𝑥 𝑓𝑓′(𝑥𝑥) = 1𝑥𝑥

𝐴𝐴′(𝑥𝑥) = 𝑥𝑥3 𝐴𝐴(𝑥𝑥) = 𝑥𝑥4

4



QUESTION 21

(a) 𝑦𝑦 = 𝑥𝑥 √1 − 𝑥𝑥2 − arccos 𝑥𝑥


= �1 − 𝑥𝑥2 + 𝑥𝑥 ⋅12

(1 − 𝑥𝑥2)−12 ⋅ −2𝑥𝑥 −

1−√1 − 𝑥𝑥2

= �1 − 𝑥𝑥2 −𝑥𝑥2

√1 − 𝑥𝑥2+

1√1 − 𝑥𝑥2

= �1 − 𝑥𝑥2 +1 − 𝑥𝑥2

√1 − 𝑥𝑥2

= 2 �1 − 𝑥𝑥2

(b)

��1 − 𝑥𝑥2 𝑑𝑑𝑥𝑥 =12�2 �1 − 𝑥𝑥2 𝑑𝑑𝑥𝑥1

0

1

0

=12�𝑥𝑥 �1 − 𝑥𝑥2 − arccos 𝑥𝑥�

0

1

=12�0 − �0 −

𝜋𝜋2��

= 𝜋𝜋4



QUESTION 22

(a) 𝑑𝑑𝐴𝐴𝑑𝑑𝑡𝑡

= 𝐴𝐴32

𝑡𝑡2

� 𝐴𝐴−32𝑑𝑑𝐴𝐴 = �𝑡𝑡−2𝑑𝑑𝑡𝑡

−2𝐴𝐴−12 = −

1𝑡𝑡

+ 𝑐𝑐

2√𝐴𝐴

=1𝑡𝑡

+ 𝑐𝑐

𝑡𝑡 = 1,𝐴𝐴 = 1 ⇒ 2 = 1 + 𝑐𝑐 ∴ 𝑐𝑐 = 1

So 2√𝐴𝐴

=1𝑡𝑡

+ 1

2√𝐴𝐴

=1 + 𝑡𝑡𝑡𝑡

√𝐴𝐴 = 2𝑡𝑡𝑡𝑡 + 1

𝐴𝐴 = �2𝑡𝑡𝑡𝑡 + 1

�2

(b) 𝑑𝑑𝐴𝐴𝑑𝑑𝑡𝑡

> 0, so 𝐴𝐴 is increasing for all 𝑡𝑡

𝑎𝑎𝑟𝑟 𝑡𝑡 → ∞,2𝑡𝑡𝑡𝑡 + 1

→ 2 so 𝐴𝐴 → 22 = 4𝑘𝑘𝐼𝐼2



QUESTION 23

(a) Method 1 Method 2

(b)

�𝑑𝑑𝑢𝑢

tan𝑢𝑢

𝜋𝜋3

𝜋𝜋4

= �cos𝑢𝑢sin 𝑢𝑢

𝑑𝑑𝑢𝑢

𝜋𝜋3

𝜋𝜋4

let 𝑡𝑡 = sin 𝑢𝑢 ⇒ 𝑑𝑑𝑡𝑡 = cos𝑢𝑢 𝑑𝑑𝑢𝑢

𝑢𝑢 =𝜋𝜋3

⇒ 𝑡𝑡 =√32

𝑎𝑎𝑛𝑛𝑑𝑑 𝑢𝑢 =𝜋𝜋4

⇒ 𝑡𝑡 =√22

= �1𝑡𝑡

𝑑𝑑𝑡𝑡

√32

√22

let 𝑥𝑥 = tan𝑢𝑢 ⇒ 𝑑𝑑𝑥𝑥 = sec2 𝑢𝑢 𝑑𝑑𝑢𝑢

𝑥𝑥 = 1 ⇒ 𝑢𝑢 =𝜋𝜋4

𝑥𝑥 = √3 ⇒ 𝑢𝑢 =𝜋𝜋3

�1

𝑥𝑥 (1 + 𝑥𝑥2)

√3

1

𝑑𝑑𝑥𝑥 = �sec2 𝑢𝑢 𝑑𝑑𝑢𝑢

tan𝑢𝑢 (1 + tan2 𝑢𝑢 )

𝜋𝜋3

𝜋𝜋4

= �𝑑𝑑𝑢𝑢

tan𝑢𝑢

𝜋𝜋3

𝜋𝜋4

with 𝑎𝑎 =𝜋𝜋4

and 𝑏𝑏 =𝜋𝜋3

let 𝑥𝑥 = tan𝑢𝑢 ⇒ 𝑢𝑢 =arctan 𝑥𝑥 so 𝑑𝑑𝑢𝑢 = 𝑑𝑑𝑥𝑥

1+𝑥𝑥2

𝑥𝑥 = 1 ⇒ 𝑢𝑢 =𝜋𝜋4

𝑥𝑥 = √3 ⇒ 𝑢𝑢 =𝜋𝜋3

�1

𝑥𝑥 (1 + 𝑥𝑥2)

√3

1

𝑑𝑑𝑥𝑥 = �𝑑𝑑𝑢𝑢

tan𝑢𝑢

𝜋𝜋3

𝜋𝜋4

with 𝑎𝑎 =𝜋𝜋4

and 𝑏𝑏 =𝜋𝜋3



= [ln 𝑡𝑡]√22

√32

= ln√32− ln

√22

= ln√3√2

=12

ln32

QUESTION 24

𝑥𝑥𝑦𝑦 𝑑𝑑𝑑𝑑𝑑𝑑𝑥𝑥

= 𝑥𝑥2 + 2𝑦𝑦2


= 𝑥𝑥2 + 2𝑦𝑦2

𝑥𝑥𝑦𝑦 𝑅𝑅𝑒𝑒𝑡𝑡 𝑉𝑉 =

𝑦𝑦𝑥𝑥

⇒ 𝑦𝑦 = 𝑉𝑉𝑥𝑥 𝑎𝑎𝑛𝑛𝑑𝑑 𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥

= 𝑉𝑉 + 𝑥𝑥𝑑𝑑𝑉𝑉𝑑𝑑𝑥𝑥

𝑉𝑉 + 𝑥𝑥𝑑𝑑𝑉𝑉𝑑𝑑𝑥𝑥

=𝑥𝑥2 + 2𝑉𝑉2𝑥𝑥2

𝑉𝑉𝑥𝑥2

=1 + 2𝑉𝑉2

𝑉𝑉

𝑥𝑥𝑑𝑑𝑉𝑉𝑑𝑑𝑥𝑥

=1 + 𝑉𝑉2

𝑉𝑉

�𝑉𝑉

1 + 𝑉𝑉2 𝑑𝑑𝑉𝑉 = �

1𝑥𝑥

𝑑𝑑𝑥𝑥

12

ln(1 + 𝑉𝑉2) = ln𝑥𝑥 + 𝑐𝑐

12

ln�1 +𝑦𝑦2

𝑥𝑥2�= ln𝑥𝑥 + 𝑐𝑐

(1, 0) ⇒ 12

ln�1 +02

12�= ln 1 + 𝑐𝑐

∴ 𝑐𝑐 = 0

�𝑉𝑉

1 + 𝑉𝑉2 𝑑𝑑𝑉𝑉 =

12�

1𝑢𝑢

𝑑𝑑𝑢𝑢

where 𝑢𝑢 = 1 + 𝑉𝑉2 and 𝑑𝑑𝑢𝑢 = 2𝑉𝑉𝑑𝑑𝑉𝑉



12

ln�1 +𝑦𝑦2

𝑥𝑥2� = ln 𝑥𝑥

ln�1 +𝑦𝑦2

𝑥𝑥2� = 2 ln 𝑥𝑥 = ln 𝑥𝑥2

1 +𝑦𝑦2

𝑥𝑥2= 𝑥𝑥2

𝑦𝑦2 = 𝑥𝑥2 (𝑥𝑥2 − 1)

𝑦𝑦 = ±�𝑥𝑥2(𝑥𝑥2 − 1) = ± 𝑥𝑥 �(𝑥𝑥2 − 1)



SECTION E – COMPLEX NUMBERS

QUESTION 25

𝑧𝑧 = 1 − 𝑖𝑖

(a) 𝑧𝑧2 = (1 − 𝑖𝑖)2 = 1 − 2𝑖𝑖 + 𝑖𝑖2 = −2𝑖𝑖

which is purely imaginary

(b) (𝑧𝑧2)2 = (−2𝑖𝑖)2 = 4𝑖𝑖2 = −4 which is real

So 𝑛𝑛 = 4 is a positive integer such that 𝑧𝑧𝑛𝑛 is real

QUESTION 26

𝑧𝑧 = 𝑎𝑎 + 𝑏𝑏𝑖𝑖 for 𝑎𝑎, 𝑏𝑏 𝜖𝜖 ℝ

1𝑧𝑧

= 1

𝑎𝑎 + 𝑏𝑏𝑖𝑖 ×

𝑎𝑎 − 𝑏𝑏𝑖𝑖𝑎𝑎 − 𝑏𝑏𝑖𝑖

= 𝑎𝑎 − 𝑏𝑏𝑖𝑖 𝑎𝑎2 + 𝑏𝑏2

𝑧𝑧̅ |𝑧𝑧|2 =

𝑎𝑎 − 𝑏𝑏𝑖𝑖

�√𝑎𝑎2 + 𝑏𝑏2�2 =

𝑎𝑎 − 𝑏𝑏𝑖𝑖 𝑎𝑎2 + 𝑏𝑏2

= 1𝑧𝑧

QUESTION 27

(a) 𝑧𝑧2 − 6𝑧𝑧 + 36 = 0

𝑧𝑧 =6 ± √62 − 4 × 1 × 36

2 × 1

𝑧𝑧 =6 ± √−108

2=

6 ± 6√3 𝑖𝑖2

= 3 ± 3√3 𝑖𝑖

𝑧𝑧 = 3 + 3√3 𝑖𝑖 = 6𝑐𝑐𝑖𝑖𝑟𝑟 �𝜋𝜋3�

𝑧𝑧 = 3 − 3√3 𝑖𝑖 = 6𝑐𝑐𝑖𝑖𝑟𝑟 �−𝜋𝜋3�



(b) 𝑧𝑧−3 = �6𝑐𝑐𝑖𝑖𝑟𝑟 �± 𝜋𝜋3��

−3

= 6−3𝑐𝑐𝑖𝑖𝑟𝑟 �∓3𝜋𝜋3�

=1

216𝑐𝑐𝑖𝑖𝑟𝑟(∓𝜋𝜋)

= −1

216

QUESTION 28

�𝑍𝑍: 𝜋𝜋4≤ 𝐴𝐴𝑟𝑟𝐴𝐴 𝑍𝑍 ≤ 𝜋𝜋

3� ∩ {𝑍𝑍: |𝑍𝑍 − 1| ≤ 1}

QUESTION 29

(a) 𝑃𝑃(𝑧𝑧) = 𝑧𝑧4 − 2𝑧𝑧3 + 51𝑧𝑧2 − 98𝑧𝑧 + 98

𝑃𝑃(𝑘𝑘𝑖𝑖) = (𝑘𝑘𝑖𝑖)4 − 2(𝑘𝑘𝑖𝑖)3 + 51(𝑘𝑘𝑖𝑖)2 − 98𝑘𝑘𝑖𝑖 + 98 = 0

⇒ 𝑘𝑘4 + 2𝑘𝑘3𝑖𝑖 − 51𝑘𝑘2 − 98𝑘𝑘𝑖𝑖 + 98 = 0

(𝑘𝑘4 − 51𝑘𝑘2 + 98) + 𝑖𝑖(2𝑘𝑘3 − 98𝑘𝑘) = 0

So, real: 𝑘𝑘4 − 51𝑘𝑘2 + 98 = 0 imaginary: 2𝑘𝑘3 − 98𝑘𝑘 = 0

(𝑘𝑘2 − 49)(𝑘𝑘2 − 2) = 0 2𝑘𝑘(𝑘𝑘2 − 49) = 0

𝑹𝑹𝑹𝑹 𝒁𝒁

𝑰𝑰𝑰𝑰 𝒁𝒁 �

12 ,√32�

(1,1)

(1,0) (2,0)

−1

1

𝑂𝑂 𝜋𝜋3

𝜋𝜋4



𝑘𝑘 = ±7, ±√2 𝑘𝑘 = 0, ±7

∴ 𝑘𝑘 = ±7 (satisfies both conditions)

(b) from part (a) we know that 𝑧𝑧 − 7𝑖𝑖 and 𝑧𝑧 + 7𝑖𝑖 are both factors

(𝑧𝑧 − 7𝑖𝑖)(𝑧𝑧 + 7𝑖𝑖) = 𝑧𝑧2 + 49

𝑃𝑃(𝑧𝑧) = 𝑧𝑧4 − 2𝑧𝑧3 + 51𝑧𝑧2 − 98𝑧𝑧 + 98

= (𝑧𝑧2 + 49)(𝑧𝑧2 + 𝑎𝑎𝑧𝑧 + 𝑏𝑏)

= 𝑧𝑧4 + 𝑎𝑎𝑧𝑧3 + (𝑏𝑏 + 49)𝑧𝑧2 + 49𝑎𝑎𝑧𝑧 + 49𝑏𝑏

Comparing coefficients gives

𝑎𝑎 = −2 and 𝑏𝑏 + 49 = 51 ⇒ 𝑏𝑏 = 2

So 𝑃𝑃(𝑧𝑧) = (𝑧𝑧2 + 49)(𝑧𝑧2 − 2𝑧𝑧 + 2)

= (𝑧𝑧 − 7𝑖𝑖)(𝑧𝑧 + 7𝑖𝑖)(𝑧𝑧 − 1 − 𝑖𝑖)(𝑧𝑧 − 1 + 𝑖𝑖)

𝑃𝑃(𝑧𝑧) = 0 ⇒ 𝑧𝑧 = ±7𝑖𝑖, 1 ± 𝑖𝑖

QUESTION 30

(a) 12�𝑒𝑒𝑐𝑐𝑖𝑖 + 𝑒𝑒−𝑐𝑐𝑖𝑖� = 1

2(cos𝑟𝑟 + 𝑖𝑖 sin 𝑟𝑟 + cos(−𝑟𝑟) + 𝑖𝑖 sin(−𝑟𝑟))

= 12

(cos𝑟𝑟 + 𝑖𝑖 sin 𝑟𝑟 + cos 𝑟𝑟 − 𝑖𝑖 sin 𝑟𝑟)

= 12

(2 cos𝑟𝑟)

= cos 𝑟𝑟

(b) cos4 𝑟𝑟 = �12�𝑒𝑒𝑐𝑐𝑖𝑖 + 𝑒𝑒−𝑐𝑐𝑖𝑖��

4

=1

16��𝑒𝑒𝑐𝑐𝑖𝑖�

4+ 4�𝑒𝑒𝑐𝑐𝑖𝑖�

3 𝑒𝑒−𝑐𝑐𝑖𝑖 + 6�𝑒𝑒𝑐𝑐𝑖𝑖�

2�𝑒𝑒−𝑐𝑐𝑖𝑖�2

+ 4𝑒𝑒𝑐𝑐𝑖𝑖�𝑒𝑒−𝑐𝑐𝑖𝑖�3

+ �𝑒𝑒−𝑐𝑐𝑖𝑖�4�

=1

16�𝑒𝑒𝑐𝑐4𝑖𝑖 + 4𝑒𝑒𝑐𝑐3𝑖𝑖 𝑒𝑒−𝑐𝑐𝑖𝑖 + 6𝑒𝑒𝑐𝑐2𝑖𝑖𝑒𝑒−𝑐𝑐2𝑖𝑖 + 4𝑒𝑒𝑐𝑐𝑖𝑖𝑒𝑒−𝑐𝑐3𝑖𝑖 + 𝑒𝑒−𝑐𝑐4𝑖𝑖�

=1

16�𝑒𝑒𝑐𝑐4𝑖𝑖 + 4𝑒𝑒𝑐𝑐2𝑖𝑖 + 6 + 4𝑒𝑒−𝑐𝑐2𝑖𝑖 + 𝑒𝑒−𝑐𝑐4𝑖𝑖�



=1

16��𝑒𝑒𝑐𝑐4𝑖𝑖 + 𝑒𝑒−𝑐𝑐4𝑖𝑖� + 4�𝑒𝑒𝑐𝑐2𝑖𝑖 + 𝑒𝑒−𝑐𝑐2𝑖𝑖� + 6�

=1

16(2 cos 4𝑟𝑟 + 4(2 cos 2𝑟𝑟) + 6)

=18

(cos 4𝑟𝑟 + 4 cos 2𝑟𝑟 + 3)

as required

2018 assessment report mathematics – specialised (mts415118) · 2018 assessment report...

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