Login

Welcome, Guest. Please login or register.

March 29, 2024, 04:51:55 am

Author Topic: VCE Methods Question Thread!  (Read 4802621 times)  Share 

0 Members and 6 Guests are viewing this topic.

DBA-144

  • MOTM: APR 19
  • Forum Obsessive
  • ***
  • Posts: 211
  • Respect: +35
Re: VCE Methods Question Thread!
« Reply #17910 on: May 04, 2019, 09:16:09 pm »
0
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?

Sorry! You need to express it in terms of y (question says so) and hence you need to write ky, as y=Ae^(kx), then ky gives you the derivative.
Not sure about your second one; what's the answers say? pretty sure if you have the derivative you have a 'rate of change'. Perhaps it requires the same method for q1?
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 :)

randomnobody69420

  • Trailblazer
  • *
  • Posts: 49
  • Respect: 0
Re: VCE Methods Question Thread!
« Reply #17911 on: May 06, 2019, 06:17:56 pm »
0
Not sure if this is relevant for this thread but in terms of maximising scores on SACs/exams etc. is it better to finish quickly and have more time to check answers at the end or to double/triple check your working out after answering a question and having less time to check again after you finish?
« Last Edit: May 06, 2019, 06:20:08 pm by randomnobody69420 »

colline

  • MOTM: NOV 19
  • Forum Obsessive
  • ***
  • Posts: 341
  • ♡ 2 Timothy 1:7 ♡
  • Respect: +512
Re: VCE Methods Question Thread!
« Reply #17912 on: May 06, 2019, 07:25:08 pm »
+4
Not sure if this is relevant for this thread but in terms of maximising scores on SACs/exams etc. is it better to finish quickly and have more time to check answers at the end or to double/triple check your working out after answering a question and having less time to check again after you finish?

Hey! :) I reckon it's best to try to get it right the first time. 'Checking' your work should just be making sure you're confident about your answers but if you finished the paper with a ton of blatant errors, going back to check will waste a lot of time. You're essentially doing the entire SAC again, you feel?

But I think it's best to DO leave some time at the end to check your working + answers. Maybe set yourself a time limit on how much time you should spend per question to make sure you get a balance between the two :)

VCE: Literature [50] Methods [50] Further [48] Chemistry [40] Biology [33]
2022: Bachelor of Science (Mathematical Economics) @ ANU

JR_StudyEd

  • MOTM: MAY 19
  • Forum Obsessive
  • ***
  • Posts: 379
  • Mental health is #1
  • Respect: +171
Re: VCE Methods Question Thread!
« Reply #17913 on: May 08, 2019, 02:51:14 pm »
0
Which exam is more difficult; Exam 1 or Exam 2? If one is indeed more difficult than the other, how so? If you think they are equally difficult, in what way?

EDIT: Should I make a new topic discussing this?
« Last Edit: May 08, 2019, 03:24:17 pm by JR_StudyEd »
Listens to K-Pop (Twice, Red Velvet, MAMAMOO) and Christmas music all year round.

fiona_atarnotes

  • Adventurer
  • *
  • Posts: 9
  • Respect: +2
Re: VCE Methods Question Thread!
« Reply #17914 on: May 09, 2019, 06:47:49 pm »
+2
Which exam is more difficult; Exam 1 or Exam 2? If one is indeed more difficult than the other, how so? If you think they are equally difficult, in what way?

EDIT: Should I make a new topic discussing this?
Hey!
So personally, exam 2 is the hardest due to a few things. 1. The short answer questions are much, much longer and worth a lot of marks (~10 marks each question). Each question takes a lot of effort because often you need to remember concepts raised in part a) for something like part f). 2. Exam 2 requires more extensive problem solving than exam 1 and this is reflected through the number of marks. 3. The multiple choice questions. Often these questions cannot be answered by having good skills in using your CAS but it also requires quite a bit of problem solving. 4. Lastly, the exam is 2.5 hours which is a very long time to stay focused and consciously aware of the tricks that VCAA could pull with the wording of question etc.

2017: Methods 50 + Premiere's | Physics 49 | Specialist 48 | Chemistry 46 | English 45
ATAR: 99.90
2018 - 2021: Research Science @ Monash

milanander

  • Trendsetter
  • **
  • Posts: 114
  • Nehemiah 8:10
  • Respect: +85
Re: VCE Methods Question Thread!
« Reply #17915 on: May 09, 2019, 07:50:31 pm »
0
Sorry if it's been asked before, but are we allowed to use concepts from outside the vce methods scope of study in our working out? For example: double differentiation?

Going off of the question above, when finding the area under a graph, where part of it is below the x axis, we're usually taught to split them into different areas and solve them separately. If we just used |f(x)|, would that be allowed? Leads to the same answer obviously, and 10x quicker.

Pls help, SAC's on Wednesday :-\
— 2019 • 2020 —
UMEP 4.0, Systems 41, Methods 47, Specialist 46, Physics 46, Viscom 40, English 37
ATAR 99.20

— 2021 • 2023 —
Bachelor of Design (Mechanical Systems & Graphic Design)
University of Melbourne

f0od

  • Forum Regular
  • **
  • Posts: 61
  • Respect: 0
Re: VCE Methods Question Thread!
« Reply #17916 on: May 09, 2019, 10:38:57 pm »
0
Hi, I was wondering if someone would be able to help me out with this question (attached)
Thanks! :)
class of 2019

colline

  • MOTM: NOV 19
  • Forum Obsessive
  • ***
  • Posts: 341
  • ♡ 2 Timothy 1:7 ♡
  • Respect: +512
Re: VCE Methods Question Thread!
« Reply #17917 on: May 10, 2019, 12:11:43 am »
+1
Sorry if it's been asked before, but are we allowed to use concepts from outside the vce methods scope of study in our working out? For example: double differentiation?

Going off of the question above, when finding the area under a graph, where part of it is below the x axis, we're usually taught to split them into different areas and solve them separately. If we just used |f(x)|, would that be allowed? Leads to the same answer obviously, and 10x quicker.

Pls help, SAC's on Wednesday :-\

I think it's advisable to always remain within the Study Design. After all, you're being marked on not just your answer but also your working out. Can't be too certain, but I think that if you used a method from say, spesh, and you made a mistake in your working which led to the wrong answer even if it's clear you know what to do, you wouldn't get a method mark.

Best to play safe and do methods in methods is my stance.

Hi, I was wondering if someone would be able to help me out with this question (attached)
Thanks! :)

Remember that when a hybrid function is differentiable, it means that not only is it continuous (no breaks in between), the gradient of both parts must also be the same at (in this case) x=2. In this case we have two unknowns, a and k so we have to set out two equations. First, try subbing x=2 into both equations and make them equal to each other (this would be making sure that it is continuous). Next, diff both equations, sub x=2, and make them equal to each other again (here would be the 'differentiable' bit). Once you've done that, you've got yourself two simultaneous equations to work with, and you'll be able to solve for a and k.

Hope that clears it up! :)

VCE: Literature [50] Methods [50] Further [48] Chemistry [40] Biology [33]
2022: Bachelor of Science (Mathematical Economics) @ ANU

f0od

  • Forum Regular
  • **
  • Posts: 61
  • Respect: 0
Re: VCE Methods Question Thread!
« Reply #17918 on: May 10, 2019, 07:18:32 am »
+1
I think it's advisable to always remain within the Study Design. After all, you're being marked on not just your answer but also your working out. Can't be too certain, but I think that if you used a method from say, spesh, and you made a mistake in your working which led to the wrong answer even if it's clear you know what to do, you wouldn't get a method mark.

Best to play safe and do methods in methods is my stance.

Remember that when a hybrid function is differentiable, it means that not only is it continuous (no breaks in between), the gradient of both parts must also be the same at (in this case) x=2. In this case we have two unknowns, a and k so we have to set out two equations. First, try subbing x=2 into both equations and make them equal to each other (this would be making sure that it is continuous). Next, diff both equations, sub x=2, and make them equal to each other again (here would be the 'differentiable' bit). Once you've done that, you've got yourself two simultaneous equations to work with, and you'll be able to solve for a and k.

Hope that clears it up! :)

thanks so much! got it :D
class of 2019

peachxmh

  • Forum Regular
  • **
  • Posts: 84
  • ¯\_(ツ)_/¯
  • Respect: 0
Re: VCE Methods Question Thread!
« Reply #17919 on: May 10, 2019, 09:35:29 pm »
0
I need help with the following question:

Let f : R->R  where f(x) = ex + k, where k is a real constant. If f and f -1 have two points of intersection then:
A. k < -1
B. k < 0
C. k > 1
D. k ≤ 0
E. k ≤ 1

The answer is A. I'm thinking you need to equate f and f -1 and then use the discriminant (set it as > 0) to find k, however I'm having trouble rearranging the resulting equation into a quadratic form. I've tried equating f and f -1 to x respectively (since they should share a point of intersection on the line y=x), as well as the change of base rules for logarithms and exponential with no result.

Would greatly appreciate if you could explain in detail how to get the answer! Thank youuuu :)
2019: VCE
2020: Med @ Monash

DBA-144

  • MOTM: APR 19
  • Forum Obsessive
  • ***
  • Posts: 211
  • Respect: +35
Re: VCE Methods Question Thread!
« Reply #17920 on: May 10, 2019, 10:42:42 pm »
+1
I need help with the following question:

Let f : R->R  where f(x) = ex + k, where k is a real constant. If f and f -1 have two points of intersection then:
A. k < -1
B. k < 0
C. k > 1
D. k ≤ 0
E. k ≤ 1

The answer is A. I'm thinking you need to equate f and f -1 and then use the discriminant (set it as > 0) to find k, however I'm having trouble rearranging the resulting equation into a quadratic form. I've tried equating f and f -1 to x respectively (since they should share a point of intersection on the line y=x), as well as the change of base rules for logarithms and exponential with no result.

Would greatly appreciate if you could explain in detail how to get the answer! Thank youuuu :)

This is not worth the trouble of finding a mathematical solution. Sketch it on your cas for the k values given and you should be able to get the answer. Intuitively, it cannot include 0, as e^x and in(x) do not have 2 solutions, and the same thing applies for any value of k that is greater than 0- there are not 2 solutions. If you sketch this, you can discern this. Hence, we have either B or A. Now, sketching again we get that there are 2 sols when k<-1. Hence A.
Like I said, there is almost certainly a mathematical solution, but this is not recommended given how long this would take. Another way would just be to sketch y=x and y=e^x + k. From here, it is really easy to see that if I move the exponential up, it won't intersect with the linear graph, but moving it down will cause it to intersect. Then just eliminate B like I did above.
Quadratic unlikely to work as you can't really convert between e^x and x and get a nice quadratic/other expression.

I hope this helps. I know that I did not provide a full detailed solution, but I hope this helps nonetheless.
Edit: y intercept of e^x is at 1,  of y=x is 0 hence must move down by one unit for one sol and by more than one for 2. hence k<-1.
« Last Edit: May 11, 2019, 12:02:06 pm by DBA-144 »
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 :)

AlphaZero

  • MOTM: DEC 18
  • Forum Obsessive
  • ***
  • Posts: 352
  • \[\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}\]
  • Respect: +160
Re: VCE Methods Question Thread!
« Reply #17921 on: May 11, 2019, 03:52:18 pm »
+1
I'm thinking you need to equate f and f -1 and then use the discriminant (set it as > 0) to find k, however I'm having trouble rearranging the resulting equation into a quadratic form. I've tried equating f and f -1 to x respectively (since they should share a point of intersection on the line y=x), as well as the change of base rules for logarithms and exponential with no result.

You actually can't write a "quadratic equation" resulting from any of those equations since it's not possible to isolate \(x\).

This is not worth the trouble of finding a mathematical solution.

I actually believe the following solution is quicker and more economical given that it can be applied to most increasing functions.

The equation  \(f(x)=f^{-1}(x)\), where  \(f(x)=e^x+k\),  will have one solution for \(x\) if the graphs of \(f\) and \(f^{-1}\) are tangential to each other, and so we have \[\begin{cases} f(x)=x\\ f'(x)=1\end{cases} \implies k=-1\ \text{ and }\ x=0.\] Since the equation  \(f(x)=f^{-1}(x)\)  has no solution for  \(k=0\),  we have two solutions only for  \(k<-1\).

Edit: y intercept of e^x is at 1,  of y=x is 0 hence must move down by one unit for one sol and by more than one for 2. hence k<-1.

On its own, this actually isn't a sufficient reason for the answer and requires more discussion.

For example, if we instead had  \(f(x)=e^{2x}+k\),  then the correct answer would be  \(k<\dfrac{-2}{e}\)  even though the \(y\)-axis intercept of  \(y=e^{2x}\) is at  \((0,\,1)\).

You were lucky that it so happens to be that  \(e^0=\left.\dfrac{d}{dx}\big[e^x\big]\right|_{x=0}=1\).
« Last Edit: May 11, 2019, 10:08:55 pm by AlphaZero »
2015\(-\)2017:  VCE
2018\(-\)2021:  Bachelor of Biomedicine and Mathematical Sciences Diploma, University of Melbourne


I\'m Not A Robot

  • Trailblazer
  • *
  • Posts: 26
  • Respect: 0
Re: VCE Methods Question Thread!
« Reply #17922 on: May 12, 2019, 12:55:41 pm »
0
Does anyone know how to answer this Q worth 9MKS:
A friend bets you $100 on a game involving two six-sided dice, one red and one green.
You choose the number of times the pair of dice will be rolled. You win if the number of
times a red 6 is rolled is at most 2 and the number of times a green 6 is rolled is at least 2.
a) How many times should the dice be rolled to maximise your chance of winning?
b) With that number of rolls, what are your expected winnings?

Thanks in advance

AlphaZero

  • MOTM: DEC 18
  • Forum Obsessive
  • ***
  • Posts: 352
  • \[\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}\]
  • Respect: +160
Re: VCE Methods Question Thread!
« Reply #17923 on: May 12, 2019, 05:21:57 pm »
+2
Does anyone know how to answer this Q worth 9MKS:
A friend bets you $100 on a game involving two six-sided dice, one red and one green.
You choose the number of times the pair of dice will be rolled. You win if the number of
times a red 6 is rolled is at most 2 and the number of times a green 6 is rolled is at least 2.
a) How many times should the dice be rolled to maximise your chance of winning?
b) With that number of rolls, what are your expected winnings?

Thanks in advance

This is quite an interesting question.

Part a
First, it's important to realise that the results of each dice are independent of each other, and that the probability of rolling a 6 on any given roll on either dice is constant.

Let \(n\) be the number of times the pair of dice are rolled, where \(n\geq 2\).
Let \(R\) be the number of observed \(6\)'s on the red dice from \(n\) rolls.
Let \(G\) be the number of observed \(6\)'s on the green dice from \(n\) rolls.

Note: \(R\) and \(G\) have the same distribution, so you could just define a single variable \(X\), but I chose to separate them for clarity.

That is, \[R\sim \text{Bi}(n,\ 1/6)\quad\text{and}\quad G\sim\text{Bi}(n,\ 1/6).\] We are told that you win the game so long as \(R\leq 2\) and \(G\geq 2\).  Hence, \[\text{Pr}(\text{win})=\text{Pr}(R\leq 2)\times \text{Pr}(G\geq 2)\] From here, you could using your CAS define a function say \[b(n)=\texttt{binomCdf}(n,\,1/6,\,0,\,2)\cdot \texttt{binomCdf}(n,\,1/6,\,2,\,n)\] and then use trial and error to find the value of \(n\) that gives you the highest probability of winning. Or, you could continue as follows: \begin{align*}\text{Pr}(\text{win})&=\text{Pr}(R\leq 2)\times\Big[1-\text{Pr}(G\leq 1)\Big]\\
&=\left[\sum_{k=0}^2\binom{n}{k}\left(\frac16\right)^k\left(\frac56\right)^{n-k}\right]\left[1-\sum_{k=0}^1\binom{n}{k}\left(\frac16\right)^k\left(\frac56\right)^{n-k}\right]\\
&\qquad \vdots\\
&=\frac{1}{250}\left(\frac{5}{36}\right)^n(n^2+9n+50)\big[5\!\times\!6^n-5^n(n+5)\big] \end{align*}
Using a graph of \(\text{Pr}(\text{win})\) against \(n\), it is quite easy to see that  \(\boxed{n=12\,}\)  rolls will maximise the chance of winning.


Part b
When  \(n=12\), \[\text{Pr}(\text{win})\approx 0.419101\quad\text{(6DP)}\] and so your expected earnings from the game is \[E=\$100 \times 0.419101=\$41.91\quad\text{(nearest cent)}.\] Assuming you also bet \(\$100\) to play the game, you expected winnings is \[W=\$41.91-\$100=\boxed{-\$58.09\,}\quad \text{(nearest cent)}\]
2015\(-\)2017:  VCE
2018\(-\)2021:  Bachelor of Biomedicine and Mathematical Sciences Diploma, University of Melbourne


-_-zzz

  • Trailblazer
  • *
  • Posts: 26
  • Respect: 0
Re: VCE Methods Question Thread!
« Reply #17924 on: May 17, 2019, 11:44:24 pm »
0
Hey guys,

Just wondering what your opinions are on ExamPro for methods, I'm finding their questions very difficult and wanted to know if they're comparable to VCAA questions in any way in terms of difficulty.

Cheers :)
« Last Edit: May 17, 2019, 11:46:00 pm by -_-zzz »