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March 29, 2024, 01:13:41 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164232 times)  Share 

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TheEagle

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9630 on: March 27, 2020, 03:24:53 pm »
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Right right. Thank you!

How likely do you reckon it is for these types of questions to appear in an exam?

S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9631 on: March 27, 2020, 04:32:02 pm »
+2
Right right. Thank you!

How likely do you reckon it is for these types of questions to appear in an exam?

Hard to say. In the past there have been questions linking circular functions to complex numbers, and linking complex numbers to triangles and special quadrilaterals, so it's not unprecedented.

However, I would expect it to be broken up into a few more sub-questions than the one you posted, to give students more guidance on how to proceed and demonstrate their knowledge / skills. Otherwise too much of the cohort would get crushed.

TheEagle

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9632 on: March 27, 2020, 04:34:31 pm »
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Hard to say. In the past there have been questions linking circular functions to complex numbers, and linking complex numbers to triangles and special quadrilaterals, so it's not unprecedented.

However, I would expect it to be broken up into a few more sub-questions than the one you posted, to give students more guidance on how to proceed and demonstrate their knowledge / skills. Otherwise too much of the cohort would get crushed.


Makes sense! Thanks heaps

Mkiyar_1905

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9633 on: March 27, 2020, 06:24:39 pm »
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If anyone could help solve these it would be greatly appreciated.
Find all complex number solutions z of the equation 2z + 3z¯= 5.
Find all complex number solutions z of the equation 2z + 2iz¯= 0
sorry, I wasn't able to type the notation for z conjugate.
 

fun_jirachi

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9634 on: March 27, 2020, 06:47:08 pm »
+2
Hey there!

Recall that if \(z = x + iy\), then \(\bar{z} = x - iy\), and \(i \bar{z} = y + ix\).

Considering this, when we have \(2z + 3\bar{z} = 5\), we consider instead \(2(x + iy) + 3(x - iy) = 5 \implies 5x - iy = 5\). From here, we can deduce that \(\text{Re(}z) = 1 \text{ and } \text{Im(}z) = 0 \). Try a similar approach for the second question.

Hope this helps :)
« Last Edit: March 27, 2020, 10:03:14 pm by fun_jirachi »
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Mkiyar_1905

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9635 on: March 27, 2020, 08:33:32 pm »
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Thank you :)

TheEagle

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9636 on: March 29, 2020, 12:04:56 am »
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hey guys

how does this work? see pic attached

S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9637 on: March 29, 2020, 11:19:48 am »
+1
hey guys

how does this work? see pic attached

1. The image of the graph of y=sin(x), x in [0, 2pi] under a dilation from the y-axis by a factor of 1/2 is the graph of y=sin(2x), x in [0,pi]. Here we are considering two different functions / graphs, which have different domains.

2. The solutions to sin(2x) = k, x in [0,2pi] are the solutions to sin(2x) = k, 2x in [0,4pi]. Here we are only considering one function, but making use of the fact that 0 ≤ x ≤ 2pi -> 0 ≤ 2x ≤ 4pi.

TheEagle

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9638 on: April 05, 2020, 09:38:09 pm »
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Why do they choose to do modulus x instead of +/- x?

fun_jirachi

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9639 on: April 06, 2020, 12:11:20 am »
+3
Hey there!

\(\sqrt{x^2}\) doesn't necessarily equal \(\pm x\). In fact, \(\sqrt{x^2} = x, \ (\forall x \geq 0)\) and \( \sqrt{x^2} = -x \ (\forall x < 0)\), which is precisely the definition of the absolute value. In this case, we don't use plus-minus because it's important to note that regardless of the sign of x, \(\sqrt{x^2}\)  will be positive, and hence we need to consider the cases where \(x \geq 0\) and \(x < 0\) separately as opposed to saying 'okay, the derivative is positive and negative for different x'. Essentially, it helps us find the values of x where the expression is positive or negative :)
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HSC 2018: Mod Hist [88] | 2U Maths [98]
HSC 2019: Physics [92] | Chemistry [93] | English Adv [87] | 3U Maths [98] | 4U Maths [97]
ATAR: 99.05

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Guide Links:
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hellopanda

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9640 on: April 18, 2020, 02:20:04 pm »
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Could someone help with these complex number questions please!
1) Find all solutions of (z-3-i)^3 = 27
2) Describe and sketch the set of complex numbers satisfying the conditions (/z/-1)(/z/-2)<=0 and Im(z)>0

S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9641 on: April 18, 2020, 02:32:37 pm »
+2
Could someone help with these complex number questions please!
1) Find all solutions of (z-3-i)^3 = 27
2) Describe and sketch the set of complex numbers satisfying the conditions (/z/-1)(/z/-2)<=0 and Im(z)>0

1) Solve \(w^3=27\) by writing both sides in mod-arg form and using de Moivre's theorem, then solve \(w=z-3-i\) for each of the solutions found in the first part.

2) \(\lvert z \rvert \) is a real number, so solving  \( (\lvert z \rvert - 1)( \lvert z \rvert - 2) \leq 0 \Rightarrow 1 \leq \lvert z \rvert \leq 2 \). Then sketch the region in the plane that is the intersection of \( 1 \leq \lvert z \rvert \leq 2 \) and \(\text{Im}(z) >0 \) (It should be a half-annulus in the upper half-plane)
« Last Edit: April 18, 2020, 02:45:50 pm by S_R_K »

lonelywhale._

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9642 on: April 18, 2020, 09:16:48 pm »
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Hello,
could someone please explain how to do the following question

4. A circle of radius 5 has its centre at the point C with position vector 2i+6j relative
to the origin O. A general point P on the circle has position r relative to O. The angle
between i and CP, measured anticlockwise from i to CP, is denoted by θ.
a). Give the vector function for P

Thank you

S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9643 on: April 19, 2020, 09:38:42 am »
+1
Hello,
could someone please explain how to do the following question

4. A circle of radius 5 has its centre at the point C with position vector 2i+6j relative
to the origin O. A general point P on the circle has position r relative to O. The angle
between i and CP, measured anticlockwise from i to CP, is denoted by θ.
a). Give the vector function for P

Thank you

A vector function for the circle with radius 1 centred at (0, 0) is \(\cos \theta \vec{i} + \sin \theta \vec{j}\). To see this, notice that letting \(x = \cos \theta\) and \(y = \sin \theta\) gives \(x^2 + y^2 = 1\).

What could be the vector function for the circle with radius r centred at (0, 0)? (Idea: define x and y such that \(x^2 + y^2 = r^2\)

Then consider, what could be the vector function for the circle with radius r centred at (a, b)? Use the same idea as before, but consider what you want the cartesian equation to be.

Helish

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9644 on: April 26, 2020, 01:01:01 pm »
+1
VCE SPECIALIST MATHS Q&A THREAD

To go straight to posts from 2020, click here.

What is this thread for?
If you have general questions about the VCE Specialist Maths course or how to improve in certain areas, this is the place to ask!


Who can/will answer questions?
Everyone is welcome to contribute; even if you're unsure of yourself, providing different perspectives is incredibly valuable.

Please don't be dissuaded by the fact that you haven't finished Year 12, or didn't score as highly as others, or your advice contradicts something else you've seen on this thread, or whatever; none of this disqualifies you from helping others. And if you're worried you do have some sort of misconception, put it out there and someone else can clarify and modify your understanding! 

There'll be a whole bunch of other high-scoring students with their own wealths of wisdom to share with you, including TuteSmart tutors! So you may even get multiple answers from different people offering their insights - very cool.


To ask a question or make a post, you will first need an ATAR Notes account. You probably already have one, but if you don't, it takes about four seconds to sign up - and completely free!

OTHER SPESH RESOURCES

Original post.
similar to the methods one Methods [3/4] Summer Holidays Question Thread! post away your questions from your summer holidays self-studying, everyone can discuss and benefit! I'll try answer as much questions as possible too ^^


ok guys I don't really know how to ask a question on this, so I just pressed the quote button. I am stuck in a dilemma and hope someone can really help me. I am currently in year 11 doing methods and spesh and was considering going to a tutor in y 12. I know this guy who gives guarantee of raw 43+ in spesh and 45+ in methods. However he is very expensive and I cannot afford to go to both. Which one should I join. Currently I am ranked 3 in the entire methods cohort but for spesh IDK what im ranked but I  know I got 58% and the average was 62% plus there are only 6 people doing spesh in the entire year. Which tution class should I join