October 21, 2019, 03:39:19 pm

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#### S_R_K

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##### Re: VCE Methods Question Thread!
« Reply #18180 on: October 10, 2019, 01:18:13 pm »
+2
pi t/8 = +/-ArcCos(1/2) + 2k pi
since cos is positive in quadrant 1 and 4?
Also shouldn’t it be -pi/3 not 5pi/3 since it’s cos not sin

Edit: mb am tripping about the -pi/3 comment but I still think adding 2kpi onto the second line is better. Correct me if I'm wrong though

You are correct. Inverse cos is one-to-one, so writing arccos(1/2) = x is effectively discarding all of the solutions outside 0 ≤ x ≤ pi.

I would not encourage Methods students who are unfamiliar with the domain/range of inverse circular functions to use them when solving equations with circular functions.

Much better is the usual approach of something like cos(x) = 1/2, so x = pi/3 + 2pik or x = 5pi/3 + 2pik, where k is an integer, and then make sure to include all and only solutions within the domain.

#### studyingg

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##### Re: VCE Methods Question Thread!
« Reply #18181 on: October 12, 2019, 01:31:23 pm »
0
Do we have to write our answer in the simplest form if the question does not explicitly specify to do so? Would an answer like 168/90 still be sufficient to get full marks, when it can be simplified further?

#### Sine

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##### Re: VCE Methods Question Thread!
« Reply #18182 on: October 12, 2019, 01:33:29 pm »
+1
Do we have to write our answer in the simplest form if the question does not explicitly specify to do so? Would an answer like 168/90 still be sufficient to get full marks, when it can be simplified further?
fully simplified and exact values are required unless stated otherwise e.g. Give the answer in a certain type of form or to a certain number of decimal places.

#### studyingg

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##### Re: VCE Methods Question Thread!
« Reply #18183 on: October 12, 2019, 01:56:35 pm »
+1
fully simplified and exact values are required unless stated otherwise e.g. Give the answer in a certain type of form or to a certain number of decimal places.

woah, I didn't even know this, thanks a lot!

#### Just another student

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##### Re: VCE Methods Question Thread!
« Reply #18184 on: October 12, 2019, 08:50:29 pm »
0
Can someone please help me with this question (image attached), i've spent so long trying to make sense of it. Thanks

#### DBA-144

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##### Re: VCE Methods Question Thread!
« Reply #18185 on: October 12, 2019, 09:05:53 pm »
+1
Can someone please help me with this question (image attached), i've spent so long trying to make sense of it. Thanks

Use that to find the anti derivative of the probability density function. multiply by x. evaluate the definite integral you have.

#### DrDusk

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##### Re: VCE Methods Question Thread!
« Reply #18186 on: October 12, 2019, 09:18:51 pm »
+2
Can someone please help me with this question (image attached), i've spent so long trying to make sense of it. Thanks

$\text{Firstly note the formula}\hspace{2mm}E(x) = \int_{-\infty}^{\infty}xf(x)dx \\ \therefore E(x) = \int_{0}^{2}\dfrac{\pi x}{4}\cos\bigg(\dfrac{\pi x}{4}\bigg)dx \\ \text{Now we know this as it's given hence we can just substitute}\hspace{2mm}\dfrac{\pi x}{4}\cos\bigg(\dfrac{\pi x}{4}\bigg) = \dfrac{d}{dx}\Bigg(x\sin \bigg(\dfrac{\pi x}{4}\bigg)\Bigg) - \sin \bigg(\dfrac{\pi x}{4}\bigg) \\ \therefore E(x) = \int_{0}^{2}\dfrac{d}{dx}\Bigg(x\sin \bigg(\dfrac{\pi x}{4}\bigg)\Bigg)dx -\int_{0}^{2}\sin\bigg(\dfrac{\pi x}{4}\bigg)dx \\ = x \sin \bigg(\dfrac{\pi x}{4}\bigg)\Bigg|_{0}^{2} + \dfrac{4}{\pi}\cos \bigg(\dfrac{\pi x}{4}\bigg) \Bigg |_{0}^{2} \\ = 2 - \dfrac{4}{\pi}$
« Last Edit: October 12, 2019, 09:25:07 pm by DrDusk »
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#### EllingtonFeint

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##### Re: VCE Methods Question Thread!
« Reply #18187 on: October 14, 2019, 02:01:48 pm »
0
Hey,
Soooo, I'm not sure how to complete the rest of this question.
I ~think~ it's something do with the unit circle maybe?? But I'm really not sure. How do I simplify it down? Could someone explain it to me, please?
Thanks!
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#### DrDusk

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##### Re: VCE Methods Question Thread!
« Reply #18188 on: October 14, 2019, 02:41:42 pm »
+2
Hey,
Soooo, I'm not sure how to complete the rest of this question.
I ~think~ it's something do with the unit circle maybe?? But I'm really not sure. How do I simplify it down? Could someone explain it to me, please?
Thanks!

$\text{Well}\hspace{2mm} \sin(\pi) = 0\hspace{2mm}\text{and}\hspace{2mm}\cos(\pi) = -1 \\ \therefore f'(\pi) = -e^\pi$
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#### ketchup21

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##### Re: VCE Methods Question Thread!
« Reply #18189 on: October 16, 2019, 02:54:46 pm »
0
I have a problem.

So we've been doing past exams and all lately and almost every question (except probability), i just get stuck and think i don't know how to do this. But when the teacher explains it, it makes sense and is something i could easily do.
My question is, how do i understand what it is that the question is asking me to do? like half the time, i miss that i have to factor first. Basically, i don't know how the question is expecting me to approach it as there can be many ways, but they don't give the correct answer.

#### redpanda83

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##### Re: VCE Methods Question Thread!
« Reply #18190 on: October 16, 2019, 09:53:51 pm »
0
I have a problem.

So we've been doing past exams and all lately and almost every question (except probability), i just get stuck and think i don't know how to do this. But when the teacher explains it, it makes sense and is something i could easily do.
My question is, how do i understand what it is that the question is asking me to do? like half the time, i miss that i have to factor first. Basically, i don't know how the question is expecting me to approach it as there can be many ways, but they don't give the correct answer.

I always underline key words, and most of the time question will literally tell you "find the" "Find the area". just look out for those. Especially in exam 1, 9/10 are always straight forward.
ways to approach a certain question - so what i have seen is that a lot of ppl find it helpful when they have a certain plan written down, or general steps.
e.g. How to solve trig equations - 1. Get the trig function (sin, cos, tan) by itself
2. Find the basic angle. (use exact triangles)
3. If necessary modify the given domain.
4. Determine which quadrant exist within given domain.
5. Decide which quadrant is available. (clue is in the answer+ve or -ve; sin(2x) =+ve or negative number)
6. Solve the equation for the variable: find the angle in relation to the correct quadrant.
Having a plan like this in your head can keep you on track. like spend 10 sec thinking of how you will approach the question, if you can write down information from question and it will start to make sense. Most importantly Dont get scared!! it doesnt bite
hope it helps

#### jnlfs2010

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##### Re: VCE Methods Question Thread!
« Reply #18191 on: October 17, 2019, 12:52:14 pm »
0
I have a problem.

So we've been doing past exams and all lately and almost every question (except probability), i just get stuck and think i don't know how to do this. But when the teacher explains it, it makes sense and is something i could easily do.
My question is, how do i understand what it is that the question is asking me to do? like half the time, i miss that i have to factor first. Basically, i don't know how the question is expecting me to approach it as there can be many ways, but they don't give the correct answer.

I feel you just need to get more practice on exam-style questions, and if you do enough and keep a record of questions that you get wrong, it should help in the future. Recording all your mistakes and where you get wrong into like a "Logbook" allows you to keep track of what you may make errors. Over time and the gathering of experience will allow you to better see through what the question is asking you.
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#### sarah15

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##### Re: VCE Methods Question Thread!
« Reply #18192 on: October 17, 2019, 07:33:23 pm »
0
Hi! I am having trouble with q4d from VCAA 2016 Methods exam 2. The examination report doesn't really help.
I was thinking to find the derivative of g(x), let k equal a value > 1 and show that it is an increasing function.
But it turns out to be a decreasing function...
I would appreciate someone's help!

#### redpanda83

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##### Re: VCE Methods Question Thread!
« Reply #18193 on: October 18, 2019, 08:44:31 pm »
0
Hi! I am having trouble with q4d from VCAA 2016 Methods exam 2. The examination report doesn't really help.
I was thinking to find the derivative of g(x), let k equal a value > 1 and show that it is an increasing function.
But it turns out to be a decreasing function...
I would appreciate someone's help!
Its actually strictly increasing. writing kx+1/x+k in rational form may help with the explanation.
g(x) = k-((k^2)-1)/x+k
as x1<x2,  ((k^2)-1)/x1+k >((k^2)-1)/x2+k
thus k-((k^2)-1)/x1+k < k-((k^2)-1)/x2+k
therefore g(x2)>g(x1) hence strictly increasing
you can also try using gradient method - strictly increasing function has +ve gradient over that domain which is much simpler to do