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March 29, 2024, 03:36:01 am

Author Topic: VCE Methods Question Thread!  (Read 4802551 times)  Share 

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jasmine24

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Re: VCE Methods Question Thread!
« Reply #19110 on: March 13, 2021, 04:52:43 pm »
+2
Hi :D

The answer is wrong, so it makes total sense that this doesn't make sense! What's in the spoiler is correct :D

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Hope this helps!
Thank u so much!

parieeelol

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Re: VCE Methods Question Thread!
« Reply #19111 on: March 19, 2021, 07:37:39 pm »
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Hey guys :) Just a quick question about the Methods exams:
I was working through this question, and got the answer (log4(5) - 1) / 3
The answer that the solutions have provided is different, however, even though they both calculate to the same decimal value.
Would examiners take away marks for not writing the answer in the form that the textbook has?
Thanks!
« Last Edit: March 19, 2021, 07:39:21 pm by parieeelol »

SS1314

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Re: VCE Methods Question Thread!
« Reply #19112 on: March 19, 2021, 08:08:03 pm »
0
Hey guys :) Just a quick question about the Methods exams:
I was working through this question, and got the answer (log4(5) - 1) / 3
The answer that the solutions have provided is different, however, even though they both calculate to the same decimal value.
Would examiners take away marks for not writing the answer in the form that the textbook has?
Thanks!

No, examiners cannot take marks off for having answer in different form IF a certain form is not specified in question, e.g. if an exam question asks to give answer in terms of log2 than you need to change your answer in order to match that form.

For that question, the log4 has changed to log2 through the change of base formula.
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Corey King

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Re: VCE Methods Question Thread!
« Reply #19113 on: April 05, 2021, 01:47:28 pm »
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Hey guys,
The textbook has the answer as +4/9 and -4/9
I get 4/9 twice as my answer.
What am I misunderstanding here?
Many thanks,
Corey
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« Last Edit: April 05, 2021, 01:49:07 pm by Corey King »

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19114 on: April 05, 2021, 02:42:26 pm »
+1
The only solution to that equation in the link is 4/9; there might be something else we’re missing in the rest of the question. I suspect either you’ve done something wrong prior or the solutions are in fact incorrect. If there is any additional info we should know about the rest of the question, please post it up!

Hope this helps :)
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Corey King

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Re: VCE Methods Question Thread!
« Reply #19115 on: April 05, 2021, 07:31:25 pm »
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The only solution to that equation in the link is 4/9; there might be something else we’re missing in the rest of the question. I suspect either you’ve done something wrong prior or the solutions are in fact incorrect. If there is any additional info we should know about the rest of the question, please post it up!

Hope this helps :)

Sure, thankyou :)
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fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19116 on: April 05, 2021, 08:06:16 pm »
+1
That makes a lot more sense (turns out both my suspicions were true :/).

Your first step here is what's causing the issue - clearly, \((9x-4)^2 \neq 81x^4-72x^2+16\). I'm not quite sure how you got that as the expression for \(h(x)\) as the square of a quadratic expression as this is the square of a linear expression. If you explain to us how you got here, we may be able to rectify some of your confusion :)

In the meantime, try using the correct equivalent expression for \(h(x)\), which is \((9x^2-4)^2\). Note that your answers should be \(\pm \frac{2}{3}\).

Hope this helps :)
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a weaponized ikea chair

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Re: VCE Methods Question Thread!
« Reply #19117 on: April 08, 2021, 06:33:29 pm »
0
Hello. Question B.

Any help is appreciated with this question. I do not understand any of the solutions given. Looking at the graph of the quadratic in the solution and comparing them to the given domains does not make much sense to me. I want to know how they got each solution.

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19118 on: April 08, 2021, 06:47:36 pm »
+2
The discriminant of a parabola \(ax^2 + bx + c\) is denoted by \(\Delta = b^2 - 4ac\). If there are two distinct roots, the discriminant will be greater than zero, if there is a single real root (it will be a double root) the discriminant is zero, otherwise the discriminant is less than zero and there are two complex roots.

In this case, we want to have the discriminant greater than zero. Substituting the relevant coefficients and simplifying the expression will yield the second line in the solution.

Does this make sense?
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Re: VCE Methods Question Thread!
« Reply #19119 on: April 08, 2021, 06:55:57 pm »
0
The discriminant of a parabola \(ax^2 + bx + c\) is denoted by \(\Delta = b^2 - 4ac\). If there are two distinct roots, the discriminant will be greater than zero, if there is a single real root (it will be a double root) the discriminant is zero, otherwise the discriminant is less than zero and there are two complex roots.

In this case, we want to have the discriminant greater than zero. Substituting the relevant coefficients and simplifying the expression will yield the second line in the solution.

Does this make sense?

Why can c<8? Why is the second solution greater than 8?

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19120 on: April 08, 2021, 07:01:50 pm »
+5
It asks you to consider where the parabola \(c^2 - 56c + 400\) is greater than zero. This will be at \(c < 28-8\sqrt{6}, c > 28 + 8\sqrt{6}\) (it's easier for you to understand and visualise if you draw this parabola out and check the inequality). Now, imagine that \(c = 8\), our function then becomes \(-2x+8\) which is a linear function which has exactly one intercept (and is not what we want in this case). Unfortunately, this solution lies in the subset of the reals \(c < 28-8\sqrt{6}\), so we omit \(c = 8\) only from that subset. Note that \(c < 8\) was already part of this subset of the real numbers, but we don't want to exclude it because it is only \(c = 8\) which is invalid.

If you like, you can think about it as \((c < 28-8\sqrt{6} \cup c > 28 + 8\sqrt{6}) \backslash \{c = 8\}\).
« Last Edit: April 08, 2021, 07:03:28 pm by fun_jirachi »
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ABB0005

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Re: VCE Methods Question Thread!
« Reply #19121 on: April 11, 2021, 08:16:36 pm »
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Anyone know how people who make online bound references print and bind them for the exam?

ArtyDreams

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Re: VCE Methods Question Thread!
« Reply #19122 on: April 11, 2021, 09:36:28 pm »
+4
Anyone know how people who make online bound references print and bind them for the exam?

I've seen people doing it on various platforms such as Microsoft Word, OneNote, Noteability, etc!
You can get them binded at places such as Officeworks.

arnavg2207

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Re: VCE Methods Question Thread!
« Reply #19123 on: April 13, 2021, 02:40:53 pm »
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Hey guys,
Just wondering what the best way to find the range of a composite function is. Do you need to graph it?
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Re: VCE Methods Question Thread!
« Reply #19124 on: April 13, 2021, 08:10:14 pm »
+4
This is a tough question to answer (simply because of the scope, not because it's vague or in any other way a bad question!). So I have some questions to pose in turn:

- What would you usually do to find the range of a composite function? Note that strictly there is no real best way (as long the answer is correct and it works quickly and accurately enough for you).
- For your method, what optimisations would you like to make? Are there any examples that demonstrate why your method may not work as well as you'd like?
- Are there any concepts about composite functions you're struggling to understand?

To answer your last question, you don't always need to graph it. If on inspection the graph is easy enough to visualise or if the range of the inner function maintains set equality with the domain of the outer function, then you shouldn't need to.
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