mts315114 - assessment report 2014 - tqa.tas.gov.au · mathematics specialised course code:...

Mathematics Specialised

Course Code: MTS315114

2014 Assessment Report

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To a large extent, the quality of candidate work in this examination was very pleasing. The examiners were also pleased to observe that, with the possible exception of Section C, candidates were generally much more thoughtful with the use of their calculators – with relatively little inappropriate use. SECTION A – Criterion 4 Question 1 Reasonably well done by those who recognised they were dealing with a series and not a sequence, although a surprising number failed to give an integral answer! Question 2 Not as well done as it should have been. It was sufficient to equate the expression to the fraction given and comment that since the solution for n was not a positive integer, then the given fraction was not a member of the sequence. Too many candidates equated and solved numerators and denominators separately. Question 3 Almost every candidate split the sum into three separate series, but quite a few tried to make one of these a geometric series by counting 42 twice! A surprising number also treated the separated series as arithmetic. The major problem, however, was careless use of algebra, particularly dealing with negatives or subtraction. Question 4 Reasonably well done. The vast majority used a substitution method, usually appropriately and adequately justified. Those who use the quadratic formula need to remember that their final expression should not contain ±. Candidates are required to make some kind of appropriate statement involving the definition of convergence. Question 5 Most set up the problem correctly using the standard method of differences, but often ended up with vr – vr+1 instead of vr – vr+3. Of those who completed this first step correctly, many did not apply the differences correctly and so failed to obtain six final terms.


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Question 6 The major difficulty candidates had with this problem was dealing with the ‘inner’ summation – not knowing how to handle a sum from k to n. Most simply used 1 to k or n. Those who used a difference generally took the sum from 1 to n and subtracted the sum from 1 to k, instead of k-‐1. Even when this first step was done correctly, very many candidates did not know how to handle the terms involving n in the second summation. There is a very simple way to solve this problem, but only two candidates found it! A few candidates who obtained the correct answer were surprised to find a familiar formula and perhaps wondered if they had made an error. A pleasingly small number of candidates simply used their calculators, gaining only one mark. SECTION B – Criterion 5 Question 7 Reasonably well done, although many candidates omitted reasons and many others gave reasons which were incomplete or vague. Question 8 Well done by most candidates, as they recognised the appropriate method. There was a variety of errors, including with the formula for the area of a circle, and not knowing that the area must be positive. Question 9 The best way to approach this question was not understood by many candidates. The key to success here is to leave the matrices as they are, and to recognise that matrix multiplication is not commutative, meaning that PQ is not the same as QP. Unfortunately, for those candidates seeing a difference of squares, (P-‐Q)(P+Q) is not the same as (P+Q)(P-‐Q). It was common for very able candidates to lose the majority of their marks on this question. Question 10 This question was generally answered very well. Candidates mostly found the correct transformation matrix and then knew what to do with it. It was not considered necessary for candidates to expand and simplify after substitution. It is worth noting that a majority of marks is still possible by using a correct method, even if an error is made early on.


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Question 11 Most candidates successfully obtained the correct matrix in reduced row-‐echelon form. Too many candidates gave incorrect or incomplete solutions and/or interpretation. These were both specifically asked for. It would be very wise at some stage to check your solution on your calculator. Whilst markers are always keen to give follow on marks after an error, the number of marks available is reduced when a candidate has changed the question into a much easier one! Question 12 There is an interesting number of ways to solve this question. Candidates generally did better when using a series of transformations. Candidates who first reflected in each axis had more success than those who first reflected in the line y=x. Many candidates who tried to obtain the required matrix by mapping from object to image points did not use appropriate pairs of points. As with question 10, candidates were not required to expand and simplify after substitution. As in previous years, a linear transformation was required to answer part (a). A matrix was not considered to be sufficient. Teachers might well use this question (asking candidates to come up with as many solutions as possible) as an analysis problem. SECTION C – Criterion 6 In this section, where explicit instructions were given (and where a standard solution was required) candidates, in general, presented their solutions well. However, where conceptual understanding was required to unpack a question, many candidates found difficulty in interpreting the requirements of the question and determining the methodology required to present an adequate solution. Question 13 A disappointing number of candidates appeared to be unfamiliar with the standard integral

cxdxx +=∫ tan sec2

, and rewrote dxx sec4

0

2∫π

π as ∫ +4

0

2 )tan1(π

π dxx for which the Casio Classpad

returned the result [ ]xxx −+ tanπ between the bounds 0 and 4π , and hence returned the appropriate

volume generated. As a ‘C’ standard question, it should not have been necessary to use a calculator to demonstrate achievement of the standard.

Furthermore, some candidates attempted to integrate 2)cossin(xx by means of a substitution. It is probably

a good idea to be familiar with the standards document of the syllabus.


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Question 14

A number of candidates confused the 1cos− in the )sin(cos 1 x− with the reciprocal ratio xcos

1 , instead

of the inverse function 1cos−=y (x). Consequently, the function to be differentiated and simplified was unwieldy.

Of those who interpreted the function statement correctly, many gave the result to be x

x2sin1

cos

−− . It

was expected that this fraction would have been simplified using the trigonometric identities.

Since 2

0 π<< x , some candidates neatly used the fact that xx −=−

2)sin(cos 1 π . The differentiated

function 1−=dxdy followed succinctly.

Question 15 This question proved to be the most poorly answered in this section with only a small percentage of the candidature able to provide a viable proof. Once again the nomenclature for inverse trigonometric functions proved to be confusing. Many candidates rewrote )(cot 1 x− as -1)cot( x and proceeded from there to differentiate. Since the formula sheet (and calculators used in the coursework) use 111 tan and cos ,sin −−− (rather than the arcsin, arccos and arctan options), it is expected that candidates are familiar with this notation. In general, many candidates were not able to see the relationship between the given information and

proving dxd ( )(cot 1 x− ) =

211x+

− . In fact, the question implies that the given information be used.

Furthermore, since candidates were unfamiliar with the procedures for the derivation of

xyxyxy 111 tan and cos ,sin −−− === from known information, this exacerbated the problem and no solution strategy could be determined after candidates circuitously rewrote the given information of the question (many times). Question 16 While the majority of candidates answered this question well, there was a significant number of candidates who did not determine that the two curves intersected where 1=x . As a consequence, the challenging evaluation of literal expressions did not yield the required result. The arithmetic error of removing the parentheses of )24(2 −− xx to obtain 242 −− xx was a frequent disappointing one, while a significant number of candidates evaluated the area (inefficiently, but

correctly) by using ∫∫∫ −−−+21

21

21

0

1

0

2)-4())24(2(2 dxdxdx xxxx .


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There were, however, some candidates who added the definite integral ,dxx∫21

0

2)-4( while there were

others who did not consider the region below the x-‐axis as being between the curves. A few very adventurous candidates used the area between the curves and the y-‐axis. This proved to be more complex algebraically but some struggled through successfully. Question 17 This question generally was answered well by candidates and any errors were mainly of an algebraic nature only. However, a number of candidates made the error in perception regarding the points of contact of the tangents. At the points of intersection with the y-‐axis, the x-‐coordinate is zero implying the points of contact are (0, 3) and (0, -‐1), evaluated from the y-‐intercepts found in part a. All but one of the candidates who back-‐substituted the y-‐coordinates (which had been found in part a) into the equation of the curve took the non-‐zero solutions and hence found equations of tangents other than those that were required. A few candidates obtained a value of the gradient function at (0, 3) or (0, -‐1) that was undefined and hence stated that the tangent did not exist. [Although this was not the case here; a line through (0, 3) with a gradient that is undefined would have an equation of 0=x , (the y-‐axis).] One candidate found the equations of tangents at the points where the graph intersected the x-‐axis, obtaining exact surds for the intercepts. The examiner was most thankful to have a calculator to trace the given solution from that point on! Question 18 Many candidates ignored the instructions given within the question requiring exact values to be given, and the calculator not to be used in the first two parts. Although the preceding parts led candidates to the resultant sketch in part d, it was apparent that this leading information was ignored and a calculator was unnecessarily the main resource that was used to sketch the curve. Nevertheless, there was a good proportion of candidates whose solutions showed good promise. In part b, the question asks candidates to determine and classify the stationary points of )x(f . A good proportion of the candidates found the x-‐ coordinates of the stationary points, but not both coordinates of the points. Furthermore, a number of candidates were thrown by the instruction the ‘your calculator should not be used’. It was evident that in many instances candidates relied unduly on the use of calculators, and perhaps candidates would benefit from appropriate calculator-‐free times during lessons.


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SECTION D – Criterion 7 Question 19 Reasonably well done. A large number candidates recognised that they were dealing with a substitution problem (u=x2-‐1) and not integration by parts. Integration by parts was attempted unsuccessfully by some candidates. It was pleasing that very few candidates used their calculator to just integrate the expression (and received no marks in that instance). Question 20 Generally well done. Finding the derivative and substituting in to show that the equation was satisfied was the most common, and simplest, solution. An alternative was to separate the variables and solve the resultant differential equation. Question 21 The method for solving separable equations was well understood by many candidates. There were some incorrect integrals used by a small number of candidates. Question 22 Homogeneous equations were generally well understood and completed fully by many candidates. There were still many answers that did not take into account the negative value when simplifying a y2

expression. Question 23 This question was attempted in two ways. The most common way was to express the repeated fractions with denominators of Ax+B and Cx+D as they were both quadratic denominators and this was overlooked by some candidates. Comparing coefficients was a much more suitable technique to find the constants rather than substituting numbers which was a laborious process. The alternative method to rearrange the numerator was much more efficient and setting up the fractions (4x2+4)/(1+x2)2 + 4x/(1+x2)2 made the question less laborious. The common factor could be simplified in the first fraction and with the use of integration by substitution this method was generally well done by those candidates who saw this method. Question 24 For candidates who made it to this question, the concept of integration by parts was generally well understood. There were a number of arithmetic and algebraic errors in the solutions of some candidates but the process was clearly presented by most candidates who had a serious attempt at this question.


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SECTION E – Criterion 8 Most candidates had a competent understanding of Complex Numbers and many scored highly. Of the candidates who demonstrated this competent understanding, marks were lost mainly in the first few questions. Question 25 Many candidates made an error with the Principal Argument – choosing –𝜋 6 instead of

5𝜋6.

Candidates needed to realise that tan 𝑧 = − 13

∴ 𝑧 = −𝜋 6 𝑜𝑟 5𝜋

6 ; position z and hence acknowledge that z was in the 3rd quadrant.

Question 26 Most candidates failed to realise that Arg 𝑤 = 𝜋

3 is a real number 𝜋 3 ≈ 1.047… . ie a point on the x axis. Most candidates instead drew a line representing {w: Arg 𝑤 = 𝜋

3 } Question 27 This question was well done with a variety of options for the diagonal line using regions defined by Arg 𝑧 ≥ −𝜋 4 ; Im 𝑧 ≥ −Re(𝑧) ; etc Question 28 (a) A straight forward expansion was all that was required ie 1 line of working. (b) Recognising this expression as a quadratic and solving it using the quadratic formula quickly

revealed where to use part (a) and hence only 3 or 4 lines of working were required to reveal the 2 solutions for z. (z=3+i or -‐2-‐i) Unfortunately many candidates tried to equate real and imaginary parts and could not complete without ‘fudge’ intervention from their calculator.

Question 29 Most candidates were able to gain full marks on this question. (a) No problem with this part (b) Most candidates used long division to identify the second quadratic factor (𝑧! + 4) and

immediately saw this as the difference of squares (𝑧! − (2𝑖)!).


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Question 30 This question was done successfully by most candidates who factorised initially 𝑃 𝑧 = 𝑧(𝑧! + 16) =0 ∴ 𝑧 = 0 or 𝑧! = −16 . Candidates who solved for z using the conventional method completed this successfully and mostly gained full marks:-‐ ie z= 2(!!!!"

!), k= -‐2, -‐1, 0, 1 AND remembered the extra

factor of z from the original factorising.

2014 Mathematics Specialised Solutions

1. ∑ involves a geometric series of n terms with and common

ratio 4, and thus requires

.

Thus the lowest integral value of n required is 15.

2. Consider

Since there is no positive integral solution,

is not a member of the sequence.

3. Sum

∑ ∑

{ }

{ }

4. It is required that for any such that |

| .

This is true if |

|

i.e. if

i.e. if

[

]

i.e. if

i.e. if

i.e. if

Thus

is a suitable N and so the sequence converges to 1.

5. Let

Sum to n terms

Sum to infinity

6. This can be done several ways. Perhaps the simplest is what follows.

Sum

and so on

Another approach is:

∑(∑

) ∑(∑

∑

)

∑ (

)

which leads to the same solution.

7. (a) X = AB does not exist because the number of columns in A does not match the

number of rows in B.

(b) Y = BA does exist because the number of columns in B matches the number of

rows in A.

(c) does not exist because does not exist since A is not a square

matrix.

8. Matrix T (

) and its determinant = . Hence the area of the image is

.

9.

10. Let the transformation be T.

Hence matrix T (

√

√

) or ( √

√

)

Thus the original curve is √

(

√

)

√

( √ )

or √ √

11. (

)

(

)

(

)

(

) ⇒

, which is the equation of a straight line

12. (a) There are several ways in which a suitable transformation T can be found – the

following is one:

Reflect Ellipse A in the line ⇒ (

), and then dilate by

vertically

and 2 horizontally ⇒ (

).

Thus (

) (

) (

) (

) | |

(

) (

)

(

)

(

)

13. Volume ∫ [ ]

14.

√

15. Let

[Differentiating with respect to x]

16. ⇒

⇒ ⇒

Hence .

Area ∫ [

]

(

) (

)

17. (a) The y-intercepts of the curve occur when

. Hence ⇒

Thus the y-intercepts are .

(b) Differentiating the equation of the curve with respect to x:

Thus at

Hence the tangents are

.

(c) To find the intersection point:

⇒

⇒

Hence the tangents intersect at the point .

18. (a) ⇒

⇒

(b) ⇒

If ⇒ minimum

If ⇒ maximum

(c) As

(d) Zeros: ⇒

⇒

y-intercept:

Endpoint:

19. ∫

√ ∫

[

]

when

√ √ √ √

20.

⇒

, as required.

21.

⇒ ∫

∫

but when

⇒

(

)

22.

[

⇒ ⇒

]

∫ ∫

| |

| |

but when ⇒

⇒

| |

23. Let

⇒ ⇒

∫

∫ (

)

[

]

24. ∫

∫

(

∫

)

∫

(

)∫

∫

25. √ √ ( √

) | | √

26.

27. | |

28. (a)

(b) ⇒ √

√

29. (a)

⇒ is a root of the polynomial equation

(b) Since the equation has real coefficients, is also a root, and so the

polynomial is divisible by .

Hence

Clearly , and by expanding and equating like terms it is easily

seen that .

30.

Consider ⇒

for

(

) (

) (

) (

)

√ √ √ √ √ √ √ √

(

) (

)

( √ )( √ )

TASMANIAN QUALIFICATIONS AUTHORITY

ASSESSMENT PANEL REPORT

MTS315114 Mathematics Specialised

15% (27) 25% (45) 42% (75) 18% (33) 180

20% (39) 24% (48) 38% (74) 18% (36) 197

10 % 19 % 39 % 32 %

20 % 25 % 36 % 19 %

11 % 19 % 39 % 30 %

64% (116) 36% (64) 3% (6) 97% (174)

72% (142) 28% (55) 2% (3) 98% (194)

70% 30% 1% 99%

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EA HA CA SA Total

Previous 5 years (all examined subjects)!

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mts315114 - assessment report 2014 - tqa.tas.gov.au · mathematics specialised course code:...

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