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September 20, 2019, 02:48:57 am

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#### S_R_K

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##### Re: VCE Methods Question Thread!
« Reply #17895 on: April 25, 2019, 09:54:17 am »
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Any tips for making Methods more fun (or less boring/monotonous/painful) to study? Just in general, not referring to any topic.

Methods can be an extremely dull subject if it is approached as a list of basic skills / knowledge to memorise or rote learn (not suggesting this is what you are doing, of course, it sounds like you want to avoid this). The strengths of the subject lie in the many opportunities for connecting the various concepts (which is often tested in tougher exam questions). So if possible, try to minimise the amount of routine textbook questions you are doing, and prioritise exam qs and qs from good quality commercial papers. The course will feel more rewarding if you are spending your time doing questions that require more application and understanding.

#### f0od

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##### Re: VCE Methods Question Thread!
« Reply #17896 on: April 26, 2019, 09:52:30 pm »
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Is it possible to find the y asymptote through the division of ordinates, or is it only possible to find the x asymptote using this method?

Also: quick question: has anyone done anything related to the 'investigation of graphs of the form y = k + kx/x , k > 0? If so, would I be able to ask you a few questions?

Thanks!
« Last Edit: April 26, 2019, 10:14:55 pm by f0od »
class of 2019

#### AlphaZero

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##### Re: VCE Methods Question Thread!
« Reply #17897 on: April 26, 2019, 11:14:58 pm »
0
Is it possible to find the y asymptote through the division of ordinates, or is it only possible to find the x asymptote using this method?

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.

Also: quick question: has anyone done anything related to the 'investigation of graphs of the form y = k + kx/x , k > 0? If so, would I be able to ask you a few questions?
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
« Last Edit: April 26, 2019, 11:19:22 pm by AlphaZero »
2015$-$2017:  VCE
2018$-$2021:  Bachelor of Biomedicine and Concurrent Diploma in Mathematical Sciences, University of Melbourne

#### Justanotherhuman

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##### Re: VCE Methods Question Thread!
« Reply #17898 on: April 26, 2019, 11:39:37 pm »
0
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 #17899 on: April 27, 2019, 11:00:30 am »
0
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

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.

#### S_R_K

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##### Re: VCE Methods Question Thread!
« Reply #17900 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 #17901 on: April 27, 2019, 10:32:25 pm »
0
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

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
class of 2019

#### VanessaS

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##### Re: VCE Methods Question Thread!
« Reply #17902 on: May 02, 2019, 04:59:53 pm »
0

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

#### AlphaZero

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##### Re: VCE Methods Question Thread!
« Reply #17903 on: May 02, 2019, 07:11:44 pm »
0
...

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
2015$-$2017:  VCE
2018$-$2021:  Bachelor of Biomedicine and Concurrent Diploma in Mathematical Sciences, University of Melbourne

#### Donut

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##### Re: VCE Methods Question Thread!
« Reply #17904 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 #17905 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 #17906 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
2018: Psychology [50]

#### AlphaZero

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##### Re: VCE Methods Question Thread!
« Reply #17907 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 »
2015$-$2017:  VCE
2018$-$2021:  Bachelor of Biomedicine and Concurrent Diploma in Mathematical Sciences, University of Melbourne

#### DBA-144

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##### Re: VCE Methods Question Thread!
« Reply #17908 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

#### f0od

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##### Re: VCE Methods Question Thread!
« Reply #17909 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 »
class of 2019