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March 29, 2024, 02:05:14 am

Author Topic: VCE Methods Question Thread!  (Read 4802472 times)  Share 

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Justanotherhuman

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Re: VCE Methods Question Thread!
« Reply #17895 on: April 26, 2019, 11:39:37 pm »
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Hey guys!
I could use some help with the following questions:
1. In finding intersection points of inverse functions we can either equate equations or equate 1 equation to x.
For example:
3+ 1/x = 1/x-3 or
 3+ 1/x=x    <-- I'm unsure how you would solve this equation

2. Find inverse of function f: R / 0 --> R, f(x) = -3 + 1/2. Show points of intersections between graphs
I'm doing something wrong with the quadratic formula step of this question. When we get to equation via equating -3x +3 - 9x, could you step me through solving that?

3. In finding inverse via matrix, I'm learning the concept myself so it makes no sense:
In this example:
For matrix  [3   2] (that's one matrix lol), find X if AX is  [5   6]. What does IX mean? IX=X?
                  [1   6]                                                           [7   2]
Thanks

DBA-144

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Re: VCE Methods Question Thread!
« Reply #17896 on: April 27, 2019, 11:00:30 am »
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Hey guys!
I could use some help with the following questions:
1. In finding intersection points of inverse functions we can either equate equations or equate 1 equation to x.
For example:
3+ 1/x = 1/x-3 or
 3+ 1/x=x    <-- I'm unsure how you would solve this equation

2. Find inverse of function f: R / 0 --> R, f(x) = -3 + 1/2. Show points of intersections between graphs
I'm doing something wrong with the quadratic formula step of this question. When we get to equation via equating -3x +3 - 9x, could you step me through solving that?

3. In finding inverse via matrix, I'm learning the concept myself so it makes no sense:
In this example:
For matrix  [3   2] (that's one matrix lol), find X if AX is  [5   6]. What does IX mean? IX=X?
                  [1   6]                                                           [7   2]
Thanks


For 1, just get rid of all of the x denominators by multiplying both sides by whatever denominators are present. In your example, multiplying the entire equation by x and x and x-3 will give you what you need i.e quadratic that you can then easily solve.

for 2, can you write it again since there is no x in the equation, and I assume that you would not have trouble finding the inverse of a horizontal line :P

for 3, just multiply A and X together. I is the identity matrix, when you multiply a matrix by the I matrix, you get the original matrix. This is the matrix equivalent of multiplying my 1.

hope this helps.
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S_R_K

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Re: VCE Methods Question Thread!
« Reply #17897 on: April 27, 2019, 02:23:44 pm »
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Hey guys!
I could use some help with the following questions:
1. In finding intersection points of inverse functions we can either equate equations or equate 1 equation to x.

Not necessarily. You can only do this if certain assumptions are true. (In many cases you'll encounter in Methods, those assumptions will be true, but you should be aware of how they can fail, and make sure you know how to check if the assumptions are true).

See this thread for further discussion of this issue: https://atarnotes.com/forum/index.php?topic=185074.0

f0od

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Re: VCE Methods Question Thread!
« Reply #17898 on: April 27, 2019, 10:32:25 pm »
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This seems like a question for the Spesh Questions Thread. There are no functions in the course that would require you to use 'division of ordinates' to find horizontal asymptotes.

I think you need to include some brackets in your expression. Your inline typeset currently reads:  \(y=k+\dfrac{kx}{x}\),  which can obviously be simplified to just  \(y=2k\), which is a boring horizontal line with a hole at  \(x=0\).  Did you mean:  \(y=\dfrac{k+kx}{x}\) ? In either case, feel free to ask any questions  :)

Thanks for the reply :)

Yeah, I was kind of sure that division of ordinates isn't in the methods course, but it's the main focus in our firsts methods SAC pre-task (so far - possibly the actual sac as well), funnily enough :(

And yes, that was my big mistake – the darned brackets! I think I might end up asking my teacher, as the questions in this task are a bit odd haha but should I have any more questions, I will ask! Thank you! and might as well take this opportunity to thank everyone who has answered any of my (many) questions! I am so grateful <3
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VanessaS

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Re: VCE Methods Question Thread!
« Reply #17899 on: May 02, 2019, 04:59:53 pm »
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Hey guys, could you please help me with this question

ℎ(𝑥)=𝑚𝑥3+𝑥2+𝑛𝑥+𝑝
where 𝑚,𝑛 𝑎𝑛𝑑 𝑝∈𝑅\{0}
If ℎ(𝑥) has an inverse function express n in terms of m.

This was the answer but I am not sure how to get to it
1−3𝑚𝑛=0
𝑛=1/3𝑚
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AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17900 on: May 02, 2019, 07:11:44 pm »
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...

Hi there, the question and the provided answer to it is actually a little bit broken.

Indeed, if  \(n=\dfrac{1}{3m}\), then \(h^{-1}\) will exist, but the converse doesn't hold. That is, it's not necessarily true that \(n=\dfrac{1}{3m}\) if \(h^{-1}\) exists.

We can still learn from this question however.

Recall that:  the inverse of a function will exist (it may only be a partial function) if and only if it is one-to-one.

So, in the case of the cubic function  \(h(x)=mx^3+x^2+nx+p\),  where \(m,n,p\in\mathbb{R}\setminus\{0\}\),  you just need to ensure that there are either no stationary points, or that if there is one that it's a point of inflection.

We have \[h'(x)=3mx^2+2x+n=0\implies x=\frac{-1\pm\sqrt{1-3mn}}{3m},\ \ \text{for suitable }m\ \text{and }n,\] and so for \(h^{-1}\) to exist, we require \[1-3mn\leq 0\implies \boxed{n\geq\frac{1}{3m}\ \text{and}\ m>0\quad \mathbf{or}\quad n\leq\frac{1}{3m}\ \text{and}\ m<0.\ }\] The answer you gave is only a small subset of all the possible relationships between \(m\) and \(n\). For example, it's clear that  \(m=n=-2\)  does not satisfy  \(n=\dfrac{1}{3m}\),  yet if you sketch the graph of \(h\),  one would find that its inverse exists.

Hope this helps :)
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Donut

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Re: VCE Methods Question Thread!
« Reply #17901 on: May 03, 2019, 06:17:58 pm »
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If a function such as x^3 is restricted from [0,∞) is it still considered an odd function?

S_R_K

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Re: VCE Methods Question Thread!
« Reply #17902 on: May 04, 2019, 09:55:00 am »
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If a function such as x^3 is restricted from [0,∞) is it still considered an odd function?

No, because f(–x) is undefined for all x > 0, hence it is not true that f(x) = –f(–x).

rani_b

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Re: VCE Methods Question Thread!
« Reply #17903 on: May 04, 2019, 05:44:55 pm »
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Find the domain of the function h, where h(x) = cos(loga(x)), a>1 has an inverse function.

The options were the following, and C was correct:
A. ( a^(-pi/2), a^(pi/2) )

B. (0, pi)

C. [1, a^(pi/2) ]

D. [ a^(-pi/2), a^(pi/2) )

E. [ a^(-pi/2), a^(pi/2) ]

How do I approach this question?
Thanks :)
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AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17904 on: May 04, 2019, 06:58:03 pm »
+2
Find the domain of the function h, where h(x) = cos(loga(x)), a>1 has an inverse function.

The options were the following, and C was correct:
A. ( a^(-pi/2), a^(pi/2) )

B. (0, pi)

C. [1, a^(pi/2) ]

D. [ a^(-pi/2), a^(pi/2) )

E. [ a^(-pi/2), a^(pi/2) ]

How do I approach this question?
Thanks :)

Hint:  the inverse of a function exists if and only if it is one-to-one (noting that the inverse may only define a partial function)

One approach to this question is to graph \(h\) on your CAS for each of the domains given in the options and check if they are one-to-one or not. Upon doing so, you'll find that option C is the only possibility.

A more mathematically rigorous approach would to be to notice that the graph of  \(g(x)=\log_a(x)\)  is strictly increasing and so, for  \(h^{-1}\) to exist, we require the range of \(g\) to be such that  \(f(x)=\cos(x)\)  is monotonic.

Looking at option C, we see that  \(\log_a(1)=0\)  and  \(\log_a\left(a^{\pi/2}\right)=\dfrac{\pi}{2}\).  Clearly, for  \(0\leq x\leq a^{\pi/2}\),  the graph of  \(f(x)=\cos(x)\) is strictly decreasing, and so C is the correct answer.
« Last Edit: May 04, 2019, 08:14:25 pm by AlphaZero »
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DBA-144

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Re: VCE Methods Question Thread!
« Reply #17905 on: May 04, 2019, 07:15:04 pm »
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Hint:  the inverse of a function exists if and only if it is one-to-one (noting that the inverse may only define a partial function)

One approach to this question is to graph \(h\) on your CAS for each of the domains given in the options and check if they are one-to-one or not. Upon doing so, you'll find that option C is the only possibility.

A more mathematically rigorous method would to be to notice that the graph of  \(g(x)=\log_a(x)\)  is strictly increasing and so, for  \(h^{-1}\) to exist, we require the range of \(g\) to be such that  \(f(x)=\cos(x)\)  is monotonic.

Looking at option C, we see that  \(\log_a(1)=0\)  and  \(\log_a\left(a^{\pi/2}\right)=\dfrac{\pi}{2}\).  Clearly, for  \(0\leq x\leq a^{\pi/2}\),  the graph of  \(f(x)=\cos(x)\) is strictly decreasing, and so C is the correct answer.


Would you mind giving more of an explanation for the formal reasoning behind the answer? I can't quite understand it lol
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f0od

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Re: VCE Methods Question Thread!
« Reply #17906 on: May 04, 2019, 08:26:12 pm »
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Hey, I need some help with these questions
any help would be greatly appreciated!

1. Find dy/dx in terms of y
y = Aekx

2. The mass, m kg, of radioactive lead remaining in a sample t hours after observations began is given by m = 2e−0.2t. Express the rate of decay as a function of m.
« Last Edit: May 04, 2019, 08:31:42 pm by f0od »
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AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17907 on: May 04, 2019, 08:47:27 pm »
+2
Would you mind giving more of an explanation for the formal reasoning behind the answer? I can't quite understand it lol

It's probably important that we look at the original wording of the question (VCAA 2014 Exam 2 - Question 13). Also note that in the original question, you are told that the domain is chosen so that \(h\) is one-to-one, instead of being told that the inverse of \(h\) exists.


Notice that it actually says "Which one of the following could be the domain?" The word "could" is actually pretty significant in that there are actually infinitely many domains that could be picked to make \(h\) a one-to-one function. So, the approach I showed, although should be understood, isn't actually that important. This question is more about just checking each option, which can be done with a graph.

However, if you didn't want to draw any graphs, what I wrote above leads you to the correct answer, so I'll go through it in more detail.

For clarity, I'm defining  \(g(x)=\log_a(x)\), where \(a>1\),  and  \(f(x)=\cos(x)\),  so that  \(h(x)=f(g(x))\).

The fact that \(g\) is strictly increasing means essentially: the bigger the input, the bigger the output.  This would guarantee therefore, that given \(a<x<b\), we would have \(g(a)<g(x)<g(b)\).

So, if  \(1\leq x\leq a^{\pi/2}\),  we know then that  \(0\leq g(x)\leq \pi/2\).  Since \(g(x)\) is being inputted into \(f\), the only question that remains is:  Is \(f(x)\) a one-to-one function (or monotone) on the interval \([0,\,\pi/2]\)? By either, imagining the graph of  \(f(x)=\cos(x)\), or using the unit circle, the answer is yes.

Hence, C is the correct answer.
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DBA-144

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Re: VCE Methods Question Thread!
« Reply #17908 on: May 04, 2019, 09:01:19 pm »
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Hey, I need some help with these questions
any help would be greatly appreciated!

1. Find dy/dx in terms of y
y = Aekx

2. The mass, m kg, of radioactive lead remaining in a sample t hours after observations began is given by m = 2e−0.2t. Express the rate of decay as a function of m.

For 1, it is just that you need to derive e^kx, giving ke^(kx) and hence you answer is dy/dx= Ake^(kx).
In your second question, you need to derive a function of the same form. (use cas if you are unsure, the working is the exact same as what is used above).
However, as the question is asking for rate of decay. therefore, you must write the 'negative' of whatever your derivative is.
i.e that your derivative is dm/dt=-0.4e^(-0.2t) but as the rate of decay is required, you need to write dm/dt= 0.4e(-0.2t) [this is the answer]
The reasoning behind this is: 
derivative obtained is positive, at a value --> the material is decaying
derivative at avalue is negative --> the material is growing/increasing.
PLease note that the material never grows, since the derivative that we have obtained is always >0.

If this does not make sense, please feel free to ask more questions!

PM me for Methods (raw 46) and Chemistry (raw 48) resources (notes, practice SACs, etc.)

I also offer tutoring for these subjects, units 1-4 :)

f0od

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Re: VCE Methods Question Thread!
« Reply #17909 on: May 04, 2019, 09:09:56 pm »
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For 1, it is just that you need to derive e^kx, giving ke^(kx) and hence you answer is dy/dx= Ake^(kx).
In your second question, you need to derive a function of the same form. (use cas if you are unsure, the working is the exact same as what is used above).
However, as the question is asking for rate of decay. therefore, you must write the 'negative' of whatever your derivative is.
i.e that your derivative is dm/dt=-0.4e^(-0.2t) but as the rate of decay is required, you need to write dm/dt= 0.4e(-0.2t) [this is the answer]
The reasoning behind this is: 
derivative obtained is positive, at a value --> the material is decaying
derivative at avalue is negative --> the material is growing/increasing.
PLease note that the material never grows, since the derivative that we have obtained is always >0.

If this does not make sense, please feel free to ask more questions!
Thanks for replying! :)

For q1, I got kAe^(kx) as well, but the answer is ky, and I don't know if this has to do with deriving it by y?

and for q2, I got to -2/5(e^(-1/5t) but how do I write that as a rate?
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