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March 29, 2024, 04:16:30 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164542 times)  Share 

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amyzzwq

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9825 on: September 05, 2021, 08:24:08 pm »
0
For the TI, if you input expressions using decimal notation, it gives answers using decimal notation. So for exact answers, convert those decimals to fractions of integers.

It worked! Thanks

amyzzwq

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9826 on: September 05, 2021, 08:25:28 pm »
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Hi, can someone explain the answer in part c please? I don't really understand why it's done that way

S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9827 on: September 06, 2021, 12:03:17 pm »
+5
The key idea is that the geometric image of \( \lvert z - a \rvert = \lvert z - b \rvert\) is the perpendicular bisector of \(a\) and \(b\). Using this fact there are a few ways to find \(b\).

In this case, since L passes through the origin at an angle of \(\frac{\pi}{3}\) from the positive real axis, we know that L bisects the angle between \(a\) and \(b\). Since \(a\) is on the negative real axis, and the angle between the negative real axis and L is \(\frac{\pi}{3}\), we can find \(b\) by multipling \(a\) by \(\frac{2\pi}{3}\). This gives \(b = \text{cis}\left(-\pi\right) \text{cis}\left(\frac{2\pi}{3}\right) = \text{cis}\left(-\frac{\pi}{3}\right)\)

amyzzwq

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9828 on: September 06, 2021, 06:13:36 pm »
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The key idea is that the geometric image of \( \lvert z - a \rvert = \lvert z - b \rvert\) is the perpendicular bisector of \(a\) and \(b\). Using this fact there are a few ways to find \(b\).

In this case, since L passes through the origin at an angle of \(\frac{\pi}{3}\) from the positive real axis, we know that L bisects the angle between \(a\) and \(b\). Since \(a\) is on the negative real axis, and the angle between the negative real axis and L is \(\frac{\pi}{3}\), we can find \(b\) by multipling \(a\) by \(\frac{2\pi}{3}\). This gives \(b = \text{cis}\left(-\pi\right) \text{cis}\left(\frac{2\pi}{3}\right) = \text{cis}\left(-\frac{\pi}{3}\right)\)

Can I jsut confirm a is -1 in this case right?

S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9829 on: September 06, 2021, 07:05:15 pm »
+2
Yes, sorry I should have made that clear. I was taking \(a=-1=\text{cis}\left(-\pi\right)\)

amyzzwq

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9830 on: September 06, 2021, 08:40:38 pm »
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Yes, sorry I should have made that clear. I was taking \(a=-1=\text{cis}\left(-\pi\right)\)

All good! Thanks for the explanation

a weaponized ikea chair

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9831 on: September 14, 2021, 05:58:53 pm »
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Is it possible to have an equation that cannot be expressed as 'y='?

For instance the equation can be rearranged to 'y=', but do equations exist where this is impossible to do?

S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9832 on: September 14, 2021, 07:35:05 pm »
+4
Is it possible to have an equation that cannot be expressed as 'y='?

For instance the equation can be rearranged to 'y=', but do equations exist where this is impossible to do?

Yes, of course. It is pretty clear for equations like \(y\sin(x)=x\sin(y)\).

But let's ignore transcendental functions (exponentials, logarithms, circular functions, hyperbolic functions, etc.) and just stick to polynomial equations, and let's also suppose that we want to express \(y\) as an algebraic function of \(x\) - ie. we can only use the standard arithmetic operations and take rational exponents. In this case, an important result is the Abel-Ruffini theorem, which states that for any polynomial equation of degree five or greater, there is no general formula for the solutions of the equation in terms of the coefficients of the polynomial. Applied to this context, this means that a polynomial equation like \(ay^5 + by + c = 0\) can't always be rearranged - for arbitrary choices of \(a, b, c\) - to get \(y\) in terms of its coefficients. (This is also a special case of a very beautiful topic in advanced algebra - Galois theory - where the solutions to polynomial equations are studied by means of symmetries of the roots).

miyukiaura

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9833 on: October 04, 2021, 12:37:19 pm »
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Does anyone know if we are required to check concavity on either side of a point when determining the point of inflection? i.e. after solving f''(x) = 0 for x, do we need to check x values below and above to see if it really is a point of inflection? My teacher says to do this but the vcaa reports never mention this...
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1729

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9834 on: October 04, 2021, 12:53:58 pm »
+2
Does anyone know if we are required to check concavity on either side of a point when determining the point of inflection? i.e. after solving f''(x) = 0 for x, do we need to check x values below and above to see if it really is a point of inflection? My teacher says to do this but the vcaa reports never mention this...
Hi, yes you would be required to check concavity on either side of the point. This is because \(f''\left(x\right)=0\) does not necessarily indicate a point of inflection, it only indicates that it is a point that is neither concave up or down. When determining nature of points it can represent a point of local minima, local maxima or point of inflection.

A point of inflection is a point at which there is a transfer in concavity, (it goes from concave up to concave down or vice versa). 2020 Exam 2 Question 3 E ii is a question where you need to carefully consider this.

If you draw the graph of \(f''\left(x\right)\) and observe that it intersects \(y=0\) without actually coming from the negative side to positive side (ie. touches \(x\) axis as turning pint) then it will not be a point of inflection. So what we essentially are doing is evalulating when \(f''\left(x\right)\) is zero or undefined and double checking to see that it does actually change sign.

I would assume VCAA would want to see how you know that \(f''\left(x\right)=0\) indicates point of inflection so to be safe state with reasoning why it is indeed a point of inflection in any assessment that asks about it.

jasperray

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9835 on: October 30, 2021, 10:04:55 pm »
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Hey guys, does anyone have a copy of the worked solutions to the 2019 NHT VCAA Exam 2? Particularly for question 6. iTute doesn't have any worked solutions for this.

Thanks :)

JamesMaths

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9836 on: November 05, 2021, 05:14:33 pm »
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I do have VCAA 2019 NHT Exam 2 Question 6 hand writing solutions.

By the way, there seems no discussion about this morning's SM Exam 1 yet.

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9837 on: November 06, 2021, 10:37:16 pm »
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Hello,

I am stuck on this question:

Does the expression have a finite or infinite positive pair of integer solutions that satisfy the relation? Prove either way.

The only example I can think of is

mabajas76

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9838 on: November 07, 2021, 01:01:11 am »
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Hello,

I am stuck on this question:

Does the expression have a finite or infinite positive pair of integer solutions that satisfy the relation? Prove either way.

The only example I can think of is
I am assuming x doesn't equal y.
If x=1 and y=0, that is a solution. As for proof, maybe contradicition?
So assume that x^y=x+y for an infinite number of paired values belonging to the set N. With x and y being different, well one thing you could say is since the numbers x=3 and y=4 don't work, there has to be a limit to the amount of responses and thus saying infinite pairs of soloutions work is wrong. Alternitvely, an odd + an odd= an even, but given how exponents work, an odd times an odd gives an odd, therefore x and y can't BOTH belong to odd positive integers, an even and an even add to an even, and an even times an even also makes an even so if both x and y are divisible by 2 it works. Adding an odd and an even make an odd number, and if x is the odd number and y is the even it works, but if y is odd and x is even it does not, so it only works when x is even.
Therefore, only when both x and y are divisible by 2 can this equation be true, or if x is odd and y is even. As a result, I would use that to say that there is a finite amount of positive integers. But I am probably wrong so sorry.
« Last Edit: November 07, 2021, 01:04:25 am by mabajas76 »
"Don't give up, and don't put too much effort into things that don't matter"-Albert Einstein, probably.

fun_jirachi

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9839 on: November 07, 2021, 01:35:49 am »
+2
Hello,

I am stuck on this question:

Does the expression have a finite or infinite positive pair of integer solutions that satisfy the relation? Prove either way.

The only example I can think of is

I think for a question like this, it's better to get you thinking, so a question in return:
- Rewrite the question as 'Find all integer solutions to the function \(f(x) = x^k - x - k, k \in \mathbb{Z}\).'. Use brute force for small values of \(k\) and see what you can deduce.

With x and y being different, well one thing you could say is since the numbers x=3 and y=4 don't work, there has to be a limit to the amount of responses and thus saying infinite pairs of soloutions work is wrong.

This isn't true. Taking a single solution out of an infinite set doesn't make the set finite. Consider for example, \(\mathbb{R}\) and \(\mathbb{R} \ \backslash \ \{0\}\). Decent attempt nonetheless - nothing wrong with being unsure :D

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