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March 29, 2024, 10:16:07 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164638 times)  Share 

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S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9795 on: February 18, 2021, 08:23:07 pm »
+3
Thanks to fun-jirachi for his excellent answer, but I thought I'd offer an alternative method which is less arithmetic heavy and doesn't rely on finding values of sin(x) and cos(x).

Note that
\(\left(\sin(x) + \cos(x)\right)^2 = \sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x) = 1 + \sin(2x)\)

Hence for part (i) \(\sin(2x) = \left(\frac{\sqrt{2}}{3}\right)^2 - 1 = -\frac{7}{9}\).

For part (ii), using a similar method to above, we have \(\left(\sin(x) - \cos(x)\right)^2 = 1 - \sin(2x) = \frac{16}{9}\)
Then we take the negative root to get \(sin(x) - \cos(x) = -\frac{4}{3}\). (We take the negative root since sin(x) < 0 and cos(x) > 0.)

For part (iii), we have \(\frac{\sin(x)+\cos(x)}{\sin(x)-\cos(x)} \times \frac{\sec(x)}{\sec(x)} = \frac{1 + \tan(x)}{1 - \tan(x)}\).

Then using the values for sin(x) + cos(x) and sin(x) - cos(x), we get \( \frac{1 + \tan(x)}{1 - \tan(x)} = -\frac{\sqrt{2}}{4}\), which gives a linear equation in tan(x).

Mahek755

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9796 on: March 02, 2021, 06:21:23 pm »
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Hi,

I have an analysis SAC coming up and I have no idea what this means. Could someone please explain it to me? Is it like a bunch of proof questions or exam-style 2 questions or something completely different?

Thank you in advance!!

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9797 on: March 08, 2021, 06:30:39 pm »
0
Hi,

I have an analysis SAC coming up and I have no idea what this means. Could someone please explain it to me? Is it like a bunch of proof questions or exam-style 2 questions or something completely different?

Thank you in advance!!

While VCAA set some general parameters, the format / design of a SAC is a school-based decision. You should ask your teacher about what your SAC will be like.

miyukiaura

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9798 on: March 21, 2021, 01:10:47 pm »
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Could anyone help with this circular function question I've attached? I was having trouble with finding the inverse of the sin function. Is there a way to do this on CAS?
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9799 on: March 21, 2021, 02:44:51 pm »
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Need help with bii + iii. Worked solutions use OP = λOA + (1 − λ)OB. Could someone please explain why ?? Thank you

fun_jirachi

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9800 on: March 21, 2021, 05:58:55 pm »
+4
Could anyone help with this circular function question I've attached? I was having trouble with finding the inverse of the sin function. Is there a way to do this on CAS?

You don't really have to use CAS - firstly notice that \(\sin (x-\pi) = -\sin (x)\) (followup: why is this the case?).

We then have that \(f(x) = -2\sin (x) + 1\). Since this is only stretched in the vertical direction and shifted upwards, we can keep the restriction of \(\{-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\}\) which implies that \(\{-1 \leq y \leq 3\}\). This can be used as verification of our inverse function when we have a look at the end points (endpoints on the original graph should be on the inverse functions graph, but reflected in y=x).

Now, our inverse function \(f^{-1}(x)\) will be a function of y in x that satisfies \(x = -2\sin (y) + 1\). Try rearranging this to get a function of y in x, and check your function is correct by verifying the endpoints (and perhaps their point of intersection, and/or any other point that piques your interest). If you need more help let us know :D

Need help with bii + iii. Worked solutions use OP = λOA + (1 − λ)OB. Could someone please explain why ?? Thank you


The worked solutions use that because that is the equation of a line segment between two points when lambda is strictly between 1 and 0. Notice if lambda is 0, P = B and if lambda is 1, P = A, and if lambda is half, then P is the midpoint. It's just convenient to think about it like that.

Hints for the other questions:
ii) \(\vec{u} \ \cdot\ \vec{v} = 0\) implies that the cosine of the angle between those vectors is zero. (Recall that \(\cos(\theta) = \frac{\vec{u} \ \cdot \ \vec{v}}{|\vec{u}||\vec{v}|}\))
iii) If P is where angle AOB bisects AB, then the cosines of AOP and POB should be equal. Use the same formula in the hint for part ii)

Hope this helps :)
« Last Edit: March 31, 2021, 08:58:59 am by fun_jirachi »
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miyukiaura

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9801 on: March 22, 2021, 11:35:48 am »
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Alright thank you fun_jirachi, that helps so much.

I was also wondering if there's a quick way to find the remainder when a complex polynomial is divided by a linear expression, without doing long division by hand. e.g.

What is the remainder when
is divided by

Thanks again :)
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fun_jirachi

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9802 on: March 22, 2021, 12:35:03 pm »
+2
Remainder theorem should still hold over complex numbers.

ie. for some \(f(z), z \in \mathbb{C}\), the remainder when dividing \(f(z)\) by \((z-a), a \in \mathbb{C}\) is just \(f(a)\). In this case, just find \(f(1+i)\) for the remainder.

Factorising is a whole different story, but remainders remain (haha) easy :D
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9803 on: March 30, 2021, 10:25:14 am »
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Hi guys just have a question regarding second derivatives,
I find it confusing to think about examples of how f''(x) can = 0 but the point at f(x) still be a turning point rather than a point of inflection. For the example of f(x) = x^4 f''(0) = 0 meaning that at (0,0) the acceleration of the graph x^4 would be zero. But shouldn't the acceleration then be positive? because if the graph went from going downwards into (0,0) and then going upwards from (0,0) that should indicate that overall, the graph should have a positive acceleration. I'm finding it difficult to have some intuition as to how f''(x) = 0 but it not being an inflection point. Am I missing something here? Thanks :)
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fun_jirachi

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9804 on: March 30, 2021, 11:12:42 am »
+4
Apologies if this doesn't answer your question, I may not have understood it properly.

The thing is, using the second derivative to calculate 'acceleration' is specific only to the point in question. You can't take the second derivative at x=0 then apply it/assume it applies for an interval including x=0. The acceleration changes with infinitesimal change in x. Also, such a function has no predefined direction - by saying 'shouldn't the acceleration then be positive? because if the graph went from going downwards into (0,0) and then going upwards from (0,0) that should indicate that overall, the graph should have a positive acceleration' you could just twist the words a bit and ask 'shouldn't the acceleration then be negative...the graph should have a negative acceleration'. It's a touchy topic - remember that derivatives work in terms of limits (think about differentiation from first principles - your value for h is just way too large for your perspective to make any sort of sense).

Also, strictly speaking, a point of inflection means that at a particular point, the function's derivative is zero while the derivative on some open interval including that point bar the point itself is either greater than or less than zero (cannot be neither (this implies a constant function) or both (not a function)).  For a turning point (as you describe it) to exist, values less than the point on the interval must have derivative of opposite sign to values greater than the point on the interval. This is a pretty wishywashy explanation, but imo enough to satiate what I think you're asking.
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9805 on: April 01, 2021, 10:19:45 pm »
+8
Hi guys just have a question regarding second derivatives,
I find it confusing to think about examples of how f''(x) can = 0 but the point at f(x) still be a turning point rather than a point of inflection. For the example of f(x) = x^4 f''(0) = 0 meaning that at (0,0) the acceleration of the graph x^4 would be zero. But shouldn't the acceleration then be positive? because if the graph went from going downwards into (0,0) and then going upwards from (0,0) that should indicate that overall, the graph should have a positive acceleration. I'm finding it difficult to have some intuition as to how f''(x) = 0 but it not being an inflection point. Am I missing something here? Thanks :)

I haven't been on this site for ages and got bored, so here I am. Anyway...

You need to understand what a derivative measures.



So, you can rewrite this thing as

where is some quantity such that $$\lim_{h\to 0} \frac{\eta}{h} = 0$$ For most functions, you can think of $$\eta$$ as being something times h2.

The point is, the derivative f'(x) only tells you if $$f(x+h)$$ changes linearly with h, and if so, how much. With your example, you can easily see that $$f(x) = x^4$$ does NOT change linearly near the origin. The second derivative then tells you how much the function changes quadratically near the origin etc. Your function changes only quartically near the origin.

This might be a bit overkill, but it's a more rigorous explanation of your query. I hope that helps!
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9806 on: April 03, 2021, 01:58:13 pm »
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I haven't been on this site for ages and got bored, so here I am. Anyway...

You need to understand what a derivative measures.



So, you can rewrite this thing as

where is some quantity such that $$\lim_{h\to 0} \frac{\eta}{h} = 0$$ For most functions, you can think of $$\eta$$ as being something times h2.

The point is, the derivative f'(x) only tells you if $$f(x+h)$$ changes linearly with h, and if so, how much. With your example, you can easily see that $$f(x) = x^4$$ does NOT change linearly near the origin. The second derivative then tells you how much the function changes quadratically near the origin etc. Your function changes only quartically near the origin.

This might be a bit overkill, but it's a more rigorous explanation of your query. I hope that helps!
Hello thanks for the explanation! I'm assuming that by changing linearly you just mean the gradient and by changing 'quadratically' you mean in terms of concavity. I was just wondering what changing 'quartically' actually looks like? Unless you could only just represent that algebraically. Thanks!
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9807 on: April 03, 2021, 03:54:46 pm »
+5
Hello thanks for the explanation! I'm assuming that by changing linearly you just mean the gradient and by changing 'quadratically' you mean in terms of concavity. I was just wondering what changing 'quartically' actually looks like? Unless you could only just represent that algebraically. Thanks!

So, if you look at the expression f(x+h) = f(x) + hf'(x) + ..., the 'linear' part of the change in f(x+h) as you vary h is f'(x). If f'(x) vanishes, but f has a second derivative, you can show then that the appropriate approximation is f(x+h) = f(x) + h^2/2 f''(x) + ..., and so now f''(x) describes how much f(x+h) changes as you change h. Look up Taylor series and/or Taylor's theorem/generalised mean value theorem if you're interested.

You only need the lowest order non-vanishing derivative at a point x to describe, near x, how the function behaves. If the first derivative vanishes but the second doesn't, the approximate dependence of f(x+h) on h, for small h, is quadratic. If the lowest order non-vanishing derivative at x is the fourth derivative, then f(x+h) will depend on the fourth power of h.
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9808 on: April 06, 2021, 11:45:25 am »
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Hi guys I was taking a look at the 2020 spec exam (exam 2) and the questions were for the most part fairly easy, except I found the attached question to be confusing.

Part e(ii) is the one I’m confused about. So firstly using the equation of the second derivative, I find where it equals zero and switched sign. I get the first answer (0 points of inflection) correct, but somehow they get n=1 for the 1 point of inflection. That is my main point of confusion.

Is there a way to solve that purely analytically, or do you just have to plot several graphs to see how many inflection points there are?
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9809 on: April 06, 2021, 12:09:39 pm »
+1
In any case, we have the trivial solution x=0, and the two you've found in e(i).

Here are a few questions you might like to consider:
- What happens when one (or more!) of the two solutions you found coincides with the trivial solution? When does this occur? (Note that you have already done this for x = 0). Is it possible to limit these instances to a finite set, then generalise their behaviour around x = 0?
- Check the behaviour of the second derivative around x = 0 (both the left-hand and right-hand limits)
- As e-x gets large, depending on the parity of n, we approach a function similar to xn. Does the parity also have an impact on how many inflection points the function has?

Things that might tip you off to checking the above include the finite number of cases, the clear distinction between cases etc. It's also a lot easier to start at small values of n, like 2 or 3 and then generalise for the sets mentioned above that include them.

As for analytical vs. several graphs - I'd like to think this way of thinking is analytical enough (you only really need to draw 2-3 graphs, and only if you really do need them). Since n is just an integer, an exhaustive graphing method is not going to be good enough as you need to generalise for some subset of the integers.

Hope this makes sense - feel free to drop any other queries or followups :)
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