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January 27, 2021, 05:55:47 am

### AuthorTopic: VCE Specialist 3/4 Question Thread!  (Read 1392544 times) Tweet Share

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#### keltingmeith

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9765 on: November 16, 2020, 08:10:44 pm »
+4
If the stem of a question writes an inverse circular function as say, arcsin(x), am I allowed to write it in my working as sin-1(x). I just prefer the second notation as its quicker to write.

Thanks

Look, it's probably fine, but if you can get into the habit of using the same notation as them, that would likely be better. For example, one time it wouldn't be fine:

$\text{If }y=x^2\text{ then evaluate }\frac{dy}{dx}$

And you respond with:

$f'(x)=2x$

This would not be fine, because you haven't defined what f(x) is. In the case of arcsin vs. $\sin^{-1}$, it's predefined so probably not an issue, but it's safer to just go with whatever they go with.
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#### Harrycc3000

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9766 on: December 05, 2020, 09:23:52 pm »
0
Hi guys,
Could someone explain this question to me please?

Let w = 2z. Describe the locus of w if z describes a circle with centre 1 + 2i and
soln is If w = 2z then w describes a circle with
centre (2, 4) and radius 6.

I don't get why u can just double everything and get the answer (esp the centre of the circle)
thanks!
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#### Opengangs

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9767 on: December 05, 2020, 10:17:23 pm »
+6
Hi guys,
Could someone explain this question to me please?

Let w = 2z. Describe the locus of w if z describes a circle with centre 1 + 2i and
soln is If w = 2z then w describes a circle with
centre (2, 4) and radius 6.

I don't get why u can just double everything and get the answer (esp the centre of the circle)
thanks!
Recall that if $z$ describes the locus of a circle with radius $r$, then we have the relation $\lvert z - z_0 \rvert = r,$ where $z_0$ represents the centre of the circle. So we have $\lvert z - (1 + 2i) \rvert = 3. \tag{1}$ We want to see what the transformation $w = 2z$ maps to.

Using $(1)$ and the fact that $w = 2z \iff z = \frac{1}{2}w$, it's not hard to see the result unfolding.
\begin{align*}
\lvert z - (1 + 2i) \rvert = 3 &\implies \left\lvert \frac{1}{2}w - (1 + 2i) \right\rvert = 3 \\
&\implies \left\lvert \frac{1}{2}\left(w - 2(1 + 2i)\right)\right\rvert = 3 \\
&\implies \frac{1}{2}\left\lvert w - 2(1 + 2i)\right\rvert = 3 \\
&\implies \lvert w - 2(1 + 2i) \rvert = 6.
\end{align*}

This represents a circle with centre $2(1 + 2i)$ and radius 6.
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#### TheEagle

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9768 on: December 06, 2020, 01:35:00 am »
+3
Perfectly said by Opengangs.

It might also be worthy to note that the centre of a circle is always described by the real and imaginary component of z.
i.e. if Z = x + yi,  then the centre of the circle will be (x,y)

#### cutiepie30

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9769 on: December 19, 2020, 09:02:59 pm »
+1
Hey Guys,

I just had a question regarding the Cambridge Specialist Maths Worked Example.

Can anyone please explain this to me, as attached in the image below.

Thanks

#### Danzorr

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9770 on: December 20, 2020, 11:17:04 am »
+6
Hey Guys,

I just had a question regarding the Cambridge Specialist Maths Worked Example.

Can anyone please explain this to me, as attached in the image below.

Thanks

I believe that this question is focusing on the symmetry property in the unit circle which means that there are two quadrants where sin is positive (1st and 2nd) and where sin is negative (3rd and 4th).

So firstly, from inspection of the graph, it can be seen that a would be in the first quadrant as it is before the turning point and it has a positive y coordinate. So from here, the question asks for all points from between [0, 2π], so you want to find all the other points around the unit circle for when a would be positive which would be in the 2nd quadrant. So to get to the 2nd quadrant, you would need a reference angle which is a. From there, you would need to do π - a (based on the symmetry property), thus providing the other solution for between [0, 2π].

Secondly, upon inspection of the graph, it can be seen that b is in the 2nd quadrant as it's positive and after the turning point. So for the other solution, you would want to get an answer in the first quadrant. since b is in the 2nd quadrant, you can just use b as the reference angle and do π - b to get the angle symmetric to b in the first quadrant.

Thirdly, for c, it can be inspected that c is negative but before the minimum turning point, therefore, it would be in the 3rd quadrant. So you would be looking for another answer in the 4th quadrant as sin is also negative in the 4th quadrant. To get an answer in the 4th quadrant, you would need to get a reference angle (the base angle). The answer gets this by doing c - π. Then based on the symmetry property, to get an answer in the 4th quadrant, you would do 2π - x, for this solution it would be 2π - (c - π) and that simplifies to 3π - c.

Lastly, for x = d, this is in the 4th quadrant as it's negative and past the minimum turning point. Just like for c, you would need to find a solution in the 3rd quadrant. So firstly, you would find a reference angle, which would be 2π - d. Then to get to the 3rd quadrant, you would do π + (2π - d), based on the symmetry property. Thus, this would simplify to 3π - d.

I believe that's how they got all the answers for the values, so overall, it's just based on the symmetry property.

#### cutiepie30

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9771 on: December 20, 2020, 04:19:40 pm »
0
Thanks so much, @Danzorr, it makes so much sense now

#### a weaponized ikea chair

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9772 on: January 02, 2021, 05:40:25 pm »
0
Hello,

Please see attached. The question is on left (Question 4C) and the solution on the right. I have a few questions:

1. Wouldn't the general solution be x = 76 - 5t since the original equation is 3x -5y =38?

2. Wouldn't the general solution be y = 38 - 3t since the general rule for a diophantine equation is y =y_0 -(a/d)t. In this case a = 3, and d = 1, so I do not understand how they got + 3t.

3. The entire simplifying part makes no sense. If you replaced t with t - 15, are not you changing the value of the equation without balancing it? I don't get this part at all. Why not t - 10, t - 200, or t - 0.37919?  I don't understand.

Thanks.

#### keltingmeith

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9773 on: January 04, 2021, 01:33:30 am »
+2
Hello,

Please see attached. The question is on left (Question 4C) and the solution on the right. I have a few questions:

1. Wouldn't the general solution be x = 76 - 5t since the original equation is 3x -5y =38?

2. Wouldn't the general solution be y = 38 - 3t since the general rule for a diophantine equation is y =y_0 -(a/d)t. In this case a = 3, and d = 1, so I do not understand how they got + 3t.

3. The entire simplifying part makes no sense. If you replaced t with t - 15, are not you changing the value of the equation without balancing it? I don't get this part at all. Why not t - 10, t - 200, or t - 0.37919?  I don't understand.

Thanks.

Preface: I'm not that great with my integer algebra, so sorry if this explanation feels a bit funny. Lmk if you need clarifications

For 1 and 2 - you're right, but it looks to me like there's some symmetry. For example:

If t=1:
x=76+5
y=38+3

If t=-1:
x=76+(-5)=76-5
y=38+(-3)=38-3

You can verify that these both satisfy the original equation of you'd like. Now, you're proposing the REAL solutions should be these:

x=76-5t
y=38-3t

Well, for t=1:
x=76-5
y=38-3

And t=-1:
x=76-(-5)=76+5
y=38-(-3)=38+3

Notice something? The values x and y take are exactly the same - just for reversed values of t! There's a symmetry that means everything still works, as long as you make the change for both values of t. This actually leads to your question number 3:

You could make any of those substitutions that you described. t is what we call a dummy variable, and all it does is control which answers correspond to t=0,1,2,3, etc. For example, look at what happens if I plug in t=16 with the new equation they provided:

x=1+5(16)=81
y=-7+3(16)=41

Which match the same values we got for t=1 in the original equation. Remember - these equations define a line of solutions for x and y. If you change the value of t by the same amount in both equations, it won't change the actual line that x and y make, it'll just change where the first value of t puts you on that line. I can write up a desmos graph later if this idea confuses you, still.

You might find it easier to think of this as a substitution - instead of changing t with t-15, think of it as making the substitution t=u-15. Then your dummy variable becomes u, and when u=16, t=16-15=1, and so u=16 will give the same (x,y) values as t=1 (which is what we saw)
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#### a weaponized ikea chair

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9774 on: January 04, 2021, 07:25:22 pm »
0

Thank you for your response. I understand it now

#### miyukiaura

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9775 on: January 08, 2021, 01:22:09 pm »
0
Does anyone know how to factorise this:
$z^3 - (2-i)z^2 + z - 2 + i$

it's one of the examples from the study design. I tried expanding (z+a)(z+b)(z_c) to get a general expression for the trinomial and then equated coefficients to get 3 simultaneous equations, but that didn't work
2020 - Biology [49]

#### SS1314

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9776 on: January 08, 2021, 01:58:33 pm »
+1
Does anyone know how to factorise this:
$z^3 - (2-i)z^2 + z - 2 + i$

it's one of the examples from the study design. I tried expanding (z+a)(z+b)(z_c) to get a general expression for the trinomial and then equated coefficients to get 3 simultaneous equations, but that didn't work

In order to solve this, I just let b=2-i in order to turn the equation into a simpler form. Then it becomes clear that z-b is a factor. Hope this helps
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#### miyukiaura

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9777 on: January 08, 2021, 03:44:13 pm »
0
Thank you! Idk how I didn't think of that hahah
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#### keltingmeith

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9778 on: January 08, 2021, 03:55:02 pm »
0
Thank you! Idk how I didn't think of that hahah

TBF, your approach should have worked, it's just difficult since you haven't effectively covered solving simultaneous equations of 3 unknowns in VCE. The other approach requires you to recognise that z=2-i is a solution to that equation, which I wouldn't expect people to get instantly (I certainly didn't notice it!)
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#### lach_chau04

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9779 on: January 22, 2021, 05:12:24 pm »
+1
Hi, I have some questions, any help would be appreciated

It's from the Exercise 4.3 of Maths Quest 12 Specialist, if that helps, but the questions are here below.

11 From a hot air balloon rising vertically upward with a speed of 8 m/s, a sandbag is
dropped which hits the ground in 4 seconds. Determine the height of the balloon
when the sandbag was dropped.

12 A missile is projected vertically upward with a speed of 73.5 m/s, and 3 seconds
later a second missile is projected vertically upward from the same point with the
same speed. Find when and where the two missiles collide.

13 A flare, A, is fired vertically upwards with a velocity of 35 m/s from a boat. Four
seconds later, another flare, B, is fired vertically upwards from the same point with
a velocity of 75 m/s. Find when and where the flares collide.

Thanks guys!!!
Nani?
2019 - VCE Mathematical Methods Units 1 & 2