 October 19, 2019, 05:23:09 am AuthorTopic: VCE Methods Question Thread!  (Read 2481518 times) Tweet Share

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JR_StudyEd Re: VCE Methods Question Thread!
« Reply #17955 on: June 07, 2019, 04:39:03 pm »
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How do solutions of trig equations relate to sin, cos and tan graphs?
VCE Class of 2019
Subjects: English, Psychology (2018), Maths Methods, Chemistry, Biology, Health and Human Development

Good luck for exams everyone! <3

AlphaZero Re: VCE Methods Question Thread!
« Reply #17956 on: June 08, 2019, 01:46:30 pm »
+1
How do solutions of trig equations relate to sin, cos and tan graphs?

Graphically, solving the equation  $f(x)=k$  will give the $x$-coordinates of the point(s) of intersection between the graph of  $y=f(x)$  and the graph of  $y=k$.

This is true for any function $f$, not just circular functions of course.

For example, consider solving  $\sin(x)=\dfrac{1}{2}$  for  $x\in[-\pi,\,3\pi]$. 2015$-$2017:  VCE
2018$-$2021:  Bachelor of Biomedicine (Human Structure & Function) and Applied Mathematics, University of Melbourne

Donut

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« Reply #17957 on: June 08, 2019, 06:10:51 pm »
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Is chapter 7 in the Cambridge 3/4 textbook needed? My school completely skipped it.

DBA-144

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« Reply #17958 on: June 08, 2019, 06:56:33 pm »
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Is chapter 7 in the Cambridge 3/4 textbook needed? My school completely skipped it.

I would do it or at least look over it. Some parts eg. the non standard functions like x^2/3 are worth knowing as they may come up on mcq (i think, at least) the odd and even functions may be worth revising as well. Otherwise, it's mainly the conclusion to the functions and graphs section of the course. Overall, you should at least look over it. Who knows what will help on the sac/exam AlphaZero Re: VCE Methods Question Thread!
« Reply #17959 on: June 08, 2019, 06:59:11 pm »
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Is chapter 7 in the Cambridge 3/4 textbook needed? My school completely skipped it.

Exercises 7B through 7E are fairly straight forward extensions of what you've already looked at in chapter 1. I think your school may have assumed that you already know the content in it.

Of course, I would still recommend reading through the those exercises on your own just to make sure you understand what is going on.

However, I especially recommend that you go through 7A since it covers some concepts that are not so easy and would require some investigation 2015$-$2017:  VCE
2018$-$2021:  Bachelor of Biomedicine (Human Structure & Function) and Applied Mathematics, University of Melbourne

AlphaZero Re: VCE Methods Question Thread!
« Reply #17960 on: June 09, 2019, 10:23:52 pm »
+1
If a question asks for an answer correct to two decimal places, do you give your answer using an equal sign (=) or an approximately equal sign (≈)?

Both are acceptable.

Although, strictly speaking, using "$\approx$" is more correct since using "$=$" is for elements that are identical, but I guess people generally accept the fact that "$=$" can be used with the understanding that a finite degree of accuracy is involved. For example, the number $\sqrt{2}$ is equal to $1.4142$, correct to $4$ decimal places.
2015$-$2017:  VCE
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DBA-144

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« Reply #17961 on: June 10, 2019, 02:27:16 pm »
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Suppose that we have a function sin (x) with domain [0.2pi]. If the function undergoes a dilation of factor 2 from the y axis, what is the functions new domain? would it change? Why/why not?

AlphaZero Re: VCE Methods Question Thread!
« Reply #17962 on: June 10, 2019, 03:49:45 pm »
+3
Suppose that we have a function sin (x) with domain [0.2pi]. If the function undergoes a dilation of factor 2 from the y axis, what is the functions new domain? would it change? Why/why not?

A dilation by factor $2$ from the $y$-axis is defined 'precisely' by $(x',\,y')=(2x,\,y).$ Essentially, every point gets mapped to a new one under the transformation, so, if  $x\in [0,\,2\pi]$,  then  $x'=2x\in [0,\, 4\pi]$,  and so the answer to your question is yes - the domain changes.

If we let  $f:[0,\,2\pi]\to\mathbb{R},\ f(x)=\sin(x)$,  then shown below are the graphs of  $y=f(x)$  and  $y=f\left(\dfrac{x}{2}\right)$. Additional note:  the domain is modified in the same way the range is modified under a dilation by factor $2$ from the $y$-axis.
« Last Edit: June 10, 2019, 03:52:29 pm by AlphaZero »
2015$-$2017:  VCE
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Jackson.Sprigg

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« Reply #17963 on: June 11, 2019, 12:34:38 pm »
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For this question it does to the power of x. I remember from unit 1 spec that this is a geometric sequence, but that's the only reason I know how they got the answer. Is there a more intuitive way to think about this relationship?

Thank You!

AlphaZero Re: VCE Methods Question Thread!
« Reply #17964 on: June 11, 2019, 02:28:36 pm »
+1
For this question it does to the power of x. I remember from unit 1 spec that this is a geometric sequence, but that's the only reason I know how they got the answer. Is there a more intuitive way to think about this relationship?

Thank You!

I can see how one can draw connections to geometric sequences - well done.

Indeed, this is a bit of a weird question. I don't really have a great way of explaining this intuitively, so perhaps someone else could post their ideas, but I'll give it a shot.

First, let's ignore the fact that they butchered the question wording. $0.92^{1/10}<1$,  so to decrease the area by this factor as $x$ increases actually increases the area lol.

Let's think about it this way. For every meter further from $B$, you need to multiply by that factor of  $0.92^{1/10}$.  For example: $\text{At }x=1 \text{ m},\ \ A=0.02\times (0.92^{1/10})^1\ \text{mm}^2\\ \text{At }x=2 \text{ m},\ \ A=0.02\times(0.92^{1/10})^2\ \text{mm}^2\\ \text{At }x=1/2 \text{ m},\ \ A=0.02\times(0.92^{1/10})^{1/2}\ \text{mm}^2,$ and so drawing out this idea, at $x$ metres, you must multiply by the factor $0.92^{1/10}$  '$x$ times': $A(x)=0.02\times (0.92^{1/10})^x=0.02\times 0.92^{x/10}$
2015$-$2017:  VCE
2018$-$2021:  Bachelor of Biomedicine (Human Structure & Function) and Applied Mathematics, University of Melbourne

JR_StudyEd Re: VCE Methods Question Thread!
« Reply #17965 on: June 12, 2019, 06:15:07 pm »
0
Graphically, solving the equation  $f(x)=k$  will give the $x$-coordinates of the point(s) of intersection between the graph of  $y=f(x)$  and the graph of  $y=k$.

This is true for any function $f$, not just circular functions of course.

For example, consider solving  $\sin(x)=\dfrac{1}{2}$  for  $x\in[-\pi,\,3\pi]$.

(Image removed from quote.)
So the solutions of sin(x) = 1/2 are just the points of intersection between the base sin function and the horizontal line y=1/2, within the specified domain?

Can you provide another example of this with another type of function?
VCE Class of 2019
Subjects: English, Psychology (2018), Maths Methods, Chemistry, Biology, Health and Human Development

Good luck for exams everyone! <3

AlphaZero Re: VCE Methods Question Thread!
« Reply #17966 on: June 13, 2019, 10:23:37 am »
+3
So the solutions of sin(x) = 1/2 are just the points of intersection between the base sin function and the horizontal line y=1/2, within the specified domain?

Can you provide another example of this with another type of function?

It's probably best to think about this generally. Don't get too hung up on the functions involved.

The $x$-coordinate(s) of the point(s) of intersection between the graphs of  $y=f(x)$  and  $y=g(x)$  is/are obtained by solving  $f(x)=g(x)$.

It just happens to be that one of the functions is a constant function.

Another Example:  $y=\sin(x)+\cos(2x)$  and  $y=0$
(Note: Methods students do not need to know how to solve this by hand. Specialist students do) 2015$-$2017:  VCE
2018$-$2021:  Bachelor of Biomedicine (Human Structure & Function) and Applied Mathematics, University of Melbourne

Monkeymafia

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« Reply #17967 on: June 23, 2019, 05:56:20 pm »
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What is the probability of getting exactly 3 kings when drawing 5 cards from a deck of 52 cards?

K     K      K      Not K     Not K

4/52 x 3/51 x 2/50 x 48/49 x 47/48

I got 47/270725. Is this correct?

AlphaZero Re: VCE Methods Question Thread!
« Reply #17968 on: June 23, 2019, 06:51:49 pm »
+2
What is the probability of getting exactly 3 kings when drawing 5 cards from a deck of 52 cards?

K     K      K      Not K     Not K

4/52 x 3/51 x 2/50 x 48/49 x 47/48

I got 47/270725. Is this correct?

Assuming sampling without replacement, no, this answer is not correct since there are actually several different ways you could sample to obtain 3 kings from drawing 5 cards. The one you provided is actually only one of these possible sequences. For example, another possibility is  $(K,K,N,K,N)$.

Essentially, you need to find all the possible combinations where 3 kings are obtained, find their individual probabilities, and then add them all up. To help, you could draw a tree diagram (although it would get messy). Here are some combinations to get you started: $\text{Pr}(K,K,K,N,N)=\dots\\ \text{Pr}(K,K,N,K,N)=\dots\\ \text{Pr}(K,K,N,N,K)=\dots\\ \vdots\ \ \ \text{etc.}$
Another way you could answer the problem, which isn't strictly in the study design is to use the hypergeometric distribution, which gives $\text{Pr}(\text{event})=\frac{\displaystyle\binom{4}{3}\binom{48}{2}}{\displaystyle\binom{52}{5}}=\frac{4\times 1128}{2\,598\,960}=\frac{94}{54145}$
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Monkeymafia

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« Reply #17969 on: June 23, 2019, 07:19:13 pm »
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Assuming sampling without replacement, no, this answer is not correct since there are actually several different ways you could sample to obtain 3 kings from drawing 5 cards. The one you provided is actually only one of these possible sequences. For example, another possibility is  $(K,K,N,K,N)$.

Essentially, you need to find all the possible combinations where 3 kings are obtained, find their individual probabilities, and then add them all up. To help, you could draw a tree diagram (although it would get messy). Here are some combinations to get you started: $\text{Pr}(K,K,K,N,N)=\dots\\ \text{Pr}(K,K,N,K,N)=\dots\\ \text{Pr}(K,K,N,N,K)=\dots\\ \vdots\ \ \ \text{etc.}$
Another way you could answer the problem, which isn't strictly in the study design is to use the hypergeometric distribution, which gives $\text{Pr}(\text{event})=\frac{\displaystyle\binom{4}{3}\binom{48}{2}}{\displaystyle\binom{52}{5}}=\frac{4\times 1128}{2\,598\,960}=\frac{94}{54145}$

So if my answer of 47/270725 represents one way the kings could be selected, then why would it be wrong to just multiply my answer by 5! (120)?