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August 09, 2020, 10:13:18 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 1301685 times)  Share 

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Helish

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9645 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!

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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

Helish

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Re: Specialist 3/4 Question Thread!
« Reply #9646 on: April 26, 2020, 01:01:57 pm »
+1
That makes sense... actually I think I remember a similar problem when taking an integral which ended up with an absolute value and I was unsure whether to write ln(5 - x) or ln(x - 5) so I had to use the initial conditions to see which solution it was.
This is basically the same concept.
Thanks for your help :)
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

undefined

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9647 on: April 30, 2020, 04:29:11 am »
+1
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
depends which one you’re better at but id say methods as spesh generally scales about 40 if you’re above average and getting above 43 raw in it lowers the effect of scaling.
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coolguy246

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9648 on: May 30, 2020, 10:00:16 am »
+1
Do we need to know how to rotate solids of revolutions about lines other than the axis? In the study design it says x and y axis however on my school SACs, they are asking us to revolve it around other lines. (Like x=2)....
Cheers
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yfesshaye

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9649 on: June 01, 2020, 07:23:48 pm »
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Can some help me please, I am super confused
How would you convert Arg(z-a)-Arg(z)=pi/2 and Arg(z)-Arg(z-a)=pi/2, when a=i+0j, into a cartesian equation?

S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9650 on: June 01, 2020, 08:42:39 pm »
+1
Do we need to know how to rotate solids of revolutions about lines other than the axis? In the study design it says x and y axis however on my school SACs, they are asking us to revolve it around other lines. (Like x=2)....
Cheers

I don't recall seeing this asked on an exam, but it's not difficult. Just translate the graph so that the axis of rotation is the x or y axis, then find the volume of solid of revolution.

Can some help me please, I am super confused
How would you convert Arg(z-a)-Arg(z)=pi/2 and Arg(z)-Arg(z-a)=pi/2, when a=i+0j, into a cartesian equation?

Can you clarify - is "a=i+0j" meant to be a vector or a complex number?

In general Arg(z-u)-Arg(z)=pi/2, where u is a complex number, gives a semi-circle with endpoints at 0 and u. To see this, plot 0 and u on an Argand diagram, and then draw rays from 0 and u that intersect at a right-angle. The points of intersection lie on the semi-circle with endpoints at 0 and u. This uses the converse of Thales' theorem (a diameter subtends a right-angle at the circumference of a circle).

yfesshaye

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9651 on: June 01, 2020, 09:18:46 pm »
+1
I don't recall seeing this asked on an exam, but it's not difficult. Just translate the graph so that the axis of rotation is the x or y axis, then find the volume of solid of revolution.

Can you clarify - is "a=i+0j" meant to be a vector or a complex number?

In general Arg(z-u)-Arg(z)=pi/2, where u is a complex number, gives a semi-circle with endpoints at 0 and u. To see this, plot 0 and u on an Argand diagram, and then draw rays from 0 and u that intersect at a right-angle. The points of intersection lie on the semi-circle with endpoints at 0 and u. This uses the converse of Thales' theorem (a diameter subtends a right-angle at the circumference of a circle).

It is a vector, so it's coordinates on the Argand diagram would be (1, 0). Thanks btw :)

zhouzhennan

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9652 on: July 05, 2020, 10:23:56 pm »
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Can someone pls help me with these ques: Q1b, Q2 TYY!

1.  a) If z = x + (x + 1)i, find the real values of x for which lzl= 2. I've found the ans:(−1 ± √ 7 )/2
b) Find the value of x for which Arg (z) = π/3

2. Factorising this by completing the square:
 z^2 – (3 − 2i)z – (4 + 3i)

AlphaZero

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9653 on: July 05, 2020, 11:06:44 pm »
+1
Can someone pls help me with these ques: Q1b, Q2 TYY!

1.  a) If z = x + (x + 1)i, find the real values of x for which lzl= 2. I've found the ans:(−1 ± √ 7 )/2
b) Find the value of x for which Arg (z) = π/3

2. Factorising this by completing the square:
 z^2 – (3 − 2i)z – (4 + 3i)

Question 1b
I'm assuming that in this question, \(x\in\mathbb{R}\). For \(\text{Arg}(z)=\pi/3\), we would need \(x>0\), and so we can write \[\tan\left(\frac{\pi}{3}\right)=\frac{x+1}{x}.\] I'll let you solve this equation for \(x\). It might help if you draw a diagram.

Question 2
The process of competing the square over \(\mathbb{C}\) is same as that over \(\mathbb{R}\) since the rules of algebra are the same (provided we remember that \(i^2=-1\) and \(1/i=-i\)): \begin{align*}z^2-(3-2i)z-(4+3i)&=\left[z-\left(\frac32-i\right)\right]^2-\left(\frac32-i\right)^2-(4+3i)\\ &=\left[z-\left(\frac32-i\right)\right]^2-\frac{21}{4}.\end{align*} I'll let you verify that when you expand the last two terms, you indeed get the RHS.
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1729

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9654 on: July 05, 2020, 11:38:50 pm »
+2
Can someone pls help me with these ques: Q1b, Q2 TYY!

1.  a) If z = x + (x + 1)i, find the real values of x for which lzl= 2. I've found the ans:(−1 ± √ 7 )/2
b) Find the value of x for which Arg (z) = π/3

2. Factorising this by completing the square:
 z^2 – (3 − 2i)z – (4 + 3i)
Assuming you know how to convert complex numbers from cartesian form to modulus argument form. How do you find the argument of a complex number?

You can get a general expression for x + (x+1) i.
Using the triangle above can you see a way to find the bottom right angle?
The process of competing the square over \(\mathbb{C}\) is same as that over \(\mathbb{R}\) since the rules of algebra are the same (provided we remember that \(i^2=-1\) and \(1/i=-i\)): \begin{align*}z^2-(3-2i)z-(4+3i)&=\left[z-\left(\frac32-i\right)\right]^2-\left(\frac32-i\right)^2-(4+3i)\\ &=\left[z-\left(\frac32-i\right)\right]^2-\frac{21}{4}.\end{align*} I'll let you verify that when you expand the last two terms, you indeed get the RHS.
Do you know what to do to factor this, zhouzhennan?
Try this
cough cough difference of squares
« Last Edit: July 05, 2020, 11:44:45 pm by 1729 »
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zhouzhennan

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9655 on: July 06, 2020, 02:45:22 pm »
+1
Assuming you know how to convert complex numbers from cartesian form to modulus argument form. How do you find the argument of a complex number?
(Image removed from quote.)
You can get a general expression for x + (x+1) i.
Using the triangle above can you see a way to find the bottom right angle?Do you know what to do to factor this, zhouzhennan?
Try this
cough cough difference of squares
Question 1b
I'm assuming that in this question, \(x\in\mathbb{R}\). For \(\text{Arg}(z)=\pi/3\), we would need \(x>0\), and so we can write \[\tan\left(\frac{\pi}{3}\right)=\frac{x+1}{x}.\] I'll let you solve this equation for \(x\). It might help if you draw a diagram.

Question 2
The process of competing the square over \(\mathbb{C}\) is same as that over \(\mathbb{R}\) since the rules of algebra are the same (provided we remember that \(i^2=-1\) and \(1/i=-i\)): \begin{align*}z^2-(3-2i)z-(4+3i)&=\left[z-\left(\frac32-i\right)\right]^2-\left(\frac32-i\right)^2-(4+3i)\\ &=\left[z-\left(\frac32-i\right)\right]^2-\frac{21}{4}.\end{align*} I'll let you verify that when you expand the last two terms, you indeed get the RHS.

Thank u both so much!!! one more quick question:
Solve the equation (x + iy)^2 = i − 1 to find a real value for x^2 for which x, y are real.

Appreciate your help!

keltingmeith

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9656 on: July 06, 2020, 02:57:52 pm »
+1
Thank u both so much!!! one more quick question:
Solve the equation (x + iy)^2 = i − 1 to find a real value for x^2 for which x, y are real.

Appreciate your help!

This one should be fairly straightforward if you remember that you can equate real and imaginary parts to make two different equations. What are you struggling with?
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zhouzhennan

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9657 on: July 06, 2020, 03:14:43 pm »
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This one should be fairly straightforward if you remember that you can equate real and imaginary parts to make two different equations. What are you struggling with?

i got x^2=-1 +1/4x^2 after equating but i can't seem to solve for x by hand (this is a tech free)

keltingmeith

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9658 on: July 06, 2020, 03:24:48 pm »
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i got x^2=-1 +1/4x^2 after equating but i can't seem to solve for x by hand (this is a tech free)

Okay, so if multiply everything by x^2, you get:

x^4=-x^2+1/4

Which kind of looks like a quadratic, huh? What do you think will happen if you substitute u=x^2? Do you think you can take it from here?

Use x + iy = r e^{i theta}, if you can't solve it from there then you should review complex numbers

This is just straight up not going to help them, but also is not covered in the specialist curriculum
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zhouzhennan

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9659 on: July 06, 2020, 03:36:18 pm »
+1
Okay, so if multiply everything by x^2, you get:

x^4=-x^2+1/4

Which kind of looks like a quadratic, huh? What do you think will happen if you substitute u=x^2? Do you think you can take it from here?

omg thank u so much i finally got it!! Rly appreciate your help :-[ :o