mathematics specialised - department of education · mathematics specialised course code: mts315109...
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Mathematics Specialised Course Code: MTS315109
2013 Assessment Report
Tasmanian Qualifications Authority Page 1 of 7
GENERAL COMMENTS On the whole, students performed reasonably well on this paper, but there remain too many instances of inappropriate use of CAS calculators. There were a couple of instances in this paper where a proof was required, and there were too many instances of proofs being very poorly set out. Students need to be reminded that they should not work on both sides of an equation simultaneously and hence arrive at a self-‐evident result such as 0 = 0, or something similar. SECTION A (SEQUENCES AND SERIES) Generally, students demonstrated a good level of understanding in this section. Many students made errors with algebraic manipulation. Question 1: Most students were not able to answer this 3 mark question completely. The numerator caused a lot of trouble as there was poor understanding of −1 !±! generating an oscillating sequence +1, -‐1, +1, -‐1 … also few students recognised the 2𝑛 !. Question 2: Many students failed to recognise the significance of 𝑒! > 0 ℎ𝑒𝑛𝑐𝑒 0 < 𝑒! < 2 only required x<𝑙𝑛2 Question 3: This question was reasonably well done although few students correctly justified any rigging, nor correctly saw the best method using 𝑛! + 1 > 𝑛! − 1 . Most students made a successful finish. Question 4: This question was full of errors: • Not reading the question correctly … terms 1, 2, 3 required. • 𝑆!""-‐𝑆!" • Computational errors by students attempting to making the numerical calculations without a
calculator
2013 Assessment Report
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Question 5: Whilst most students made reasonable progress with this inductive proof, many lost marks for imprecise proof structure and too many students could not complete the algebraic manipulations correctly. Question 6: This question was generally well done and most students made good progress. Students struggled to complete the GP summation because of errors with their manipulation of fractions. SECTION B (MATRICES AND LINEAR TRANSFORMATIONS) Overall, students performed very well on this section and mostly showed a good understanding of the concepts being examined. Question 7: Students who solved the equation by post-‐multiplying by an inverse matrix generally had no problems – although a few incorrectly tried to pre-‐multiply. Students who used a general matrix generally got lost in the algebra involved in solving the resulting simultaneous equations. Question 8: This should have been a very simple question, but frequently proved difficult. The simplest approach was to take two points on the y-‐axis (such as (0,0) and (0,1)), find their images under the given transformation, and then find the equation of the resulting line. It was disappointing to see how many students could not write down the equation of the y-‐axis, if that was the approach they attempted. Question 9: This should be a standard proof but was not always done well. Many students clearly did not recognise the notation for the determinant of a matrix, but instead thought that absolute values of the elements were required. Proofs were often poorly set out. Question 10: Most students knew that a simple substitution was required here to find the equation of the original curve, although algebraic errors were very frequent. Students were expected to simplify the equation of that curve.
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Question 11: This question was generally very well done, with the majority gaining full marks. There were the usual arithmetic and transcription errors. A few students were confused by having equations involving a, b and c instead of x, y and z, and had the former variables within their augmented matrices. Question 12: Students generally knew what was required in this question, but errors were fairly frequent. The most common error was an incorrect matrix representing the reflection transformation. The vast majority multiplied their three matrices in the correct order. The simplest approach from that point was to find the inverse of the resulting matrix and substitute from it into the given equation – students who tried to set up and solve simultaneous equations rarely succeeded. Some simplification of the final equation was required for full marks. SECTION C (DIFFERENTIAL AND INTEGRAL CALCULUS) In general, candidates demonstrated a sound conceptual understanding yet were frequently plagued by algebraic issues, particularly those involving indices and logarithms. Clearly structured and presented solutions were the exception, possibly contributing to candidates’ algebraic problems. Question 13: The majority of candidates scored full marks on this problem, showing appropriate working to differentiate the relation implicitly or using the chain rule. A significant fraction of candidates did not differentiate the constant term. Candidates who attempted to rearrange the equation and differentiate explicitly invariably discarded solutions when taking the square root. Question 14: This question was very well answered by the majority of candidates. The most common error resulted from incorrectly evaluating 2! = 0 . The definite integral was inconsistently accompanied by units, however this was not required. Candidates are encouraged to refer to the information sheet for standard integrals such as 2! 𝑑𝑥. Question 15: This question proved to be quite difficult for a large number of candidates. Roughly half of the responses neglected to show that the curves share a common point, whilst some candidates focused purely on the common point and did not consider the derivatives.
2013 Assessment Report
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Many candidates clearly made use of their calculator to find the derivatives, evident in frequent
transcription errors resulting in ! !!/!
!"= 𝑒
!!!! or 𝑒!!!! − 1. Another common error was the assertion
that the two curves were tangential due to the property 𝑚!×𝑚! = 1 or in some cases, −1. The majority of candidates did not use appropriate structure, notation or commentary to clearly communicate mathematical ideas and discriminate between the two curves. Further, appropriate reasoning to support the proof was frequently absent. Question 16: This straight-‐forward application of the trapezoidal rule gave an easy five marks to many candidates. The most common error was the incorrect calculation of the interval width. Some candidates made a sketch to assist them to identify the correct number of intervals. Arithmetic errors in simplifying the fraction or evaluating the sum were also relatively common. Question 17: Despite the explicit instruction to not use calculators and the implication that working was expected, too many candidates simply stated the derivatives. Candidates who used the product rule to evaluate 𝑓′′′(𝑥) were generally more successful than candidates using the quotient rule who frequently got lost in the algebra. Candidates’ command of index laws was disappointing as were notational issues such as 1− 𝑥! !/! = 1− 𝑥!! . Even those candidates who were clearly well versed in the procedure for obtaining a MacLaurin series often did not recognise the role of a MacLaurin series as an approximation of the function
𝑓 𝑥 = sin!! 𝑥 and simply evaluated 𝑥 + !!
! !
!.!to approximate the integral or neglected to respond to
this part of the problem entirely. Question 18: This problem posed significant problems for many candidates. Correctly identifying the limits of integration was a challenge for many candidates. Clearly some candidates were uncomfortable with logarithmic functions, as expressions such as log! 𝑎! were common and apparently only resolved with the aid of a calculator or not at all. The constant 𝑎 also caused issues and was commonly treated as a variable whilst integrating or evaluating definite integrals. A common approach to the problem was to simply launch into the integral 𝑉 = 𝑥!𝑑𝑦!
! without considering the composition of the solid and the necessity to subtract this volume from another solid. Candidates also struggled to calculate the annular volume, frequently representing such a volume with 𝜋 𝑅 − 𝑟 !ℎ or its integral equivalent. Those candidates who spent time setting up the problem with the aid of a clearly labelled diagram and appropriate notation 𝑉!,𝑉!,𝑉!… etc. were usually successful in obtaining the correct result.
2013 Assessment Report
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SECTION D (TECHNIQUES OF INTEGRATION) Generally, students demonstrated a good level of understanding in this section. Many students need to take greater care with signs and algebraic manipulation. Question 19:
Most students recognised the need to separate the variables to obtain ∫∫ =+
dxdyy
111
2 , leading to
success. Answers did not have to be written in the form )4
tan( π−= xy in order to gain full marks.
Question 20: The main issue here was to provide a reasonable format for the solution. Question 21: This question was well done. Most students made a successful start. After that students needed to take care, especially with signs. A constant of integration was expected. Question 22: This question was generally well done. However, many students made errors. Take care when solving
C915 = ! Include the symbol when appropriate. Students who used the format
)2()1()1( 2 −+
++
+ xc
xb
xa generally had more success than those who didn’t.
Question 23: Whilst most students made reasonable progress, many lost marks for errors or incomplete solutions.
)5ln(51 IdII
−−=−∫ as 5<I . Take care with the sign and give an explanation as to why the symbol
is not required. Question 24: Most students made some progress with this question, but not so many managed a complete solution.
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SECTION E (COMPLEX NUMBERS) Generally, candidates performed well on this section and demonstrated a good understanding of the operations involved in working with complex numbers. However, while a CAS calculator is a wonderful calculating assistant; its use should not be to the detriment of demonstrating a candidate’s understanding of complex numbers (the criterion being assessed in this section). Question 25: Most candidates were able to simplify the given fraction to i33 −− ; and, in so doing, showed an appropriate recording sequence which demonstrated the method by which the result was obtained. Question 26: It is important to label axes when providing a graphical representation of a region: otherwise, the tick marks could be taken to mark off any scale. Nevertheless, the question was relatively well done, with the required region clearly marked and labelled. A few candidates omitted showing the required boundary of the desired region and, in many instances, the argument of 4
π did not intercept the y-‐axis at (0, 2). Question 27: Many candidates were unable to write an appropriate symbolic expression in part (a) to represent the cube roots of i. Those candidates who used 3)( ibai += opened themselves to a more complicated
algebraic solution than those who used ............. )( 31
2233 ππ ciszcisziz =⇒=⇒= to find the three
cube roots. A disturbing number of candidates began with the statement .03
1=i
Question 28: a) Almost all candidates were able to give 2
π as the smallest positive value of θ . However, there was
a number of candidates who resorted to logarithms to the base e, and then calculated iiln to
determine the smallest positive value of θ .
(The marker was able to find some wonderfully (obscure) results (for example, 2
ln π=
ii ) when
checking through candidates’ responses to determine the correctness, or otherwise, of given statements.)
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b) Instead of writing iii ei )( 2π
= , many candidates got stumped by writing ,)( 22
ieii eiπ
π
= and were then unable to transform the statement obtained to write .0)( 1)1(
2 icis i −==−π
c) Many candidates were unable to use the result in (b) to write iii )( as ie )( 2
π− , and hence express the result in Cartesian form.
Question 29: In part (b), the wording of the question stated specifically that the candidate was to write
12580428 234 +−+− zzzz as a product of linear factors without using a calculator. Hence it was a requirement of this question that the factors of 2562 +− zz be found (and a recording sequence shown) by a means other than that of a CAS calculation: that is, the quadratic formula or a difference of two squares needed to be used, for example, to assist factorisation. Using the reasons “by inspection”, or “by observation” to perform multi-‐step operations (using a calculator) was not, in the view of the examiner, sufficient methodology to obtain the linear factors from the initial roots 1 + 2i and 1 – 2i. Question 30: While most candidates recognised that 54321 zzzzz +−+−+− could be expressed as a sum of a geometric series, 6)(1( z−− ) was disappointingly written as 6)1( z+ by many candidates. Also, while
solving 011 6
=+−zz , many omitted considering that z cannot take the value -‐1. This led to incorrectly
obtaining six zeros to the quintic equation. Another common error involved the transposition of 01 6 =− z to 16 −=z which then gave the incorrect argument of the complex number. Finally, since the arguments involved were ‘exact values’, it was expected that the factors were given free of the trigonometric ratios of the arguments.
TASMANIAN QUALIFICATIONS AUTHORITY
ASSESSMENT PANEL REPORT
MTS315109 Mathematics Specialised
20% (39) 24% (48) 38% (74) 18% (36) 197
20% (37) 23% (44) 34% (64) 23% (43) 188
11 % 19 % 39 % 31 %
18 % 24 % 36 % 22 %
11 % 19 % 39 % 30 %
72% (142) 28% (55) 2% (3) 98% (194)
73% (138) 27% (50) 3% (6) 97% (181)
68% 32% 1% 99%
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Mathematics Specialised – 2013 Examination Solutions
Question 1
nth
term
.
Question 2
Convergence to 0 requires and hence .
Thus since , convergence requires .
Question 3
It is required that for any |
| .
This is true if
i.e. if
[for
]
i.e. if
i.e. if
i.e. if
i.e. if
Hence
is a suitable N and so {
} converges to 0.
Question 4
(a) Denote the sum to n terms by
Hence the first term,
Now
And
(b) ∑ ∑
∑
Question 5
To prove: ∑
Proof: Denote the proposition by ;
thus
LHS
Assume
Hence remains to be proven, that
.
LHS
⇒
Thus since and ⇒ then is true for all positive integers n by the
Principle of Mathematical Induction.
Question 6
The series
to 2n terms can be divided into
series 1:
to n terms
plus series 2:
to n terms.
Denote the rth
term of series 1 by
Sum to n terms
Now series 2 is geometric, with first term
and common ratio
.
Sum to n terms
( (
) )
( (
) )
( (
)
)
(a) In the original series, sum to 2n terms
( (
)
).
(b) As
(
)
.
Question 7
If X(
) (
) (
) (
)
(
)
(
)
(
) .
Question 8
The y-axis contains the points .
Given , then
- .
Question 9
Let (
) (
) | | | |
(
)
| |
| || |
Question 10
Let
. Hence if is a point on the original curve, then
must satisfy the equation of the image. Hence the original curve is
given by (
)
(
) (
) (
)
Question 11
(a) The required equations are:
(b) (
)
→ (
) → (
) →
(
)
→ (
) → (
)
Hence .
Question 12
Rotation = (
) (
√
√
)
Reflection = ( (
) (
)
(
) (
)) (
)
Dilation = (
)
Transformation, (
) (
) (
√
√
)
(
) (
√
√
)
(
√
√ )
(√
√
) (
√
√
) ( √
√
)
If lies on the image, then ( √
√
) must satisfy the equation of the
original curve. Hence the equation of the image is given by:
( √
)
( √
)
√
( √ ) ( √ )
( √ )
√ √ √
√ √
Question 13
If , then
.
or
.
Question 14
Area ∫ (
)
[
]
(
) (
)
Question 15
If
then when
⇒
and so when
⇒
and so when
⇒
Hence both curves meet at the point with gradient 1
the curves are tangential to each other.
Question 16
Electric power generated
-
Question 17
(a)
√
( )
( )
( )
[ ]
(b)
∫
[
]
Question 18
(a) (b) ⇒
∫
{∫ ∫ ∫
}
{
[
]
}
{
}
(
)
Question 19
⇒ ∫
∫ and so
Given that when
⇒
⇒
(
)
Question 20
⇒
⇒
Question 21
∫ ∫
∫
Question 22
Let
If ⇒ ⇒
If ⇒ ⇒
Equating coeff. of ⇒ ⇒
∫
∫(
(
))
[
|
|]
Question 23
The equation
, with , becomes
, which can be rearranged to
.
∫
∫
[Given , no absolute value is necessary]
But when ⇒
Question 24
where
∫
∫
| | [Using substitution ]
(
)
Question 25
Question 26
Question 27
Let (
)
(
)
(
)
√
√
Question 28
(a)
(b) (
)
(c) ( ) (
)
Question 29
(a)
is a root of the equation
(b) By the conjugate root theorem, is also a root of the equation
⇒
By long division, or otherwise, it can be shown that
and the zeros of are √
Question 30
(a)
Consider
(
)
(
) (
)
( (
)) ( (
)) (
) (
)
(
√
) (
√
) (
√
) (
√
)
(b) (
) (
)