Post by: Joseph41 on October 09, 2018, 03:25:52 pm
(https://i.imgur.com/be9RchO.png)
This thread is for all exam-related discussion. Was it easy? Was it hard? What did you get for each question? Feel free to post any and all of your thoughts below.
If you don't have an ATAR Notes account yet, no stress - you can register for free here!
Post by: crouchie on November 09, 2018, 01:19:21 pm
As usual, solutions will be posted here, and successively updated to be more thorough (such as CAS screenshots, etc).
Good luck everyone with the weekend's study.
Answers
SECTION A
1. E
2. B
3. D
4. A
5. D
6. C
7. C
8. E
9. D
10. A
11. C
12. A
13. B
14. C
15. E
16. B
17. E
18. D
19. E
20. B
Post by: AlphaZero on November 10, 2018, 11:05:51 am
Solutions by Daniel Traicos at TWM Publications
Fairly straight forward exam I thought. Some tricky multiple choice questions, but I think VCAA were quite generous with this exam.
Edit 1: Fixed solution to Q17. Fixed typo in Q1d.
Post by: Onyx on November 12, 2018, 11:51:23 am
And a two tail test for stat
Post by: HamConspiracy on November 12, 2018, 05:28:00 pm
Post by: AlphaZero on November 12, 2018, 05:39:32 pm
4. B
I get A for Q4. Can we just check?
Post by: jazzycab on November 12, 2018, 05:44:07 pm
MCQ
1. E
2. B
3. D
4. A
5. D
6. C
7. C
8. E
9. D
10. A
11. C
12. E
13. B
14. C
15. E
16. B
17. E
18. D
19. E
20. B
2. B
3. D
4. A
5. D
6. C
7. C
8. E
9. D
10. A
11. C
12. E
13. B
14. C
15. E
16. B
17. E
18. D
19. E
20. B
ERQ1
1a. \right):\left[-\sqrt{2},\sqrt{2}\right]\\\text{Ran}\left(f\left(x\right)\right):\left[-\pi,\pi\right])
1b: (https://i.imgur.com/LAw3W1U.png)
1c:=\frac{4}{\sqrt{2-x^2}})
1d:=-\frac{4}{\sqrt{2-x^2}})
1ei:\cup\left(0,\sqrt{2}\right))
1eii: \[g\left(x\right)=\lbrace\begin{array}{ll}
-4 & -\sqrt{2}<x<0 \\
4 & 0<x<\sqrt{2} \\
\end{array}\]
1eiii: (https://i.imgur.com/ghYT0Ll.png)
1b: (https://i.imgur.com/LAw3W1U.png)
1c:
1d:
1ei:
1eii: \[g\left(x\right)=\lbrace\begin{array}{ll}
-4 & -\sqrt{2}<x<0 \\
4 & 0<x<\sqrt{2} \\
\end{array}\]
1eiii: (https://i.imgur.com/ghYT0Ll.png)
ERQ2
2a: \\\text{Radius: }2)
2b:^2+y^2}&=\sqrt{2\left(x^2+\left(y-1\right)^2\right)}\\\left(x+1\right)^2+y^2&=2\left(x^2+\left(y-1\right)^2\right)\\x^2+2x+1+y^2&=2x^2+2y^2-4y+2\\0&=x^2+y^2-2x-4y+1\\0&=\left(x^2-2x+1\right)+\left(y^2-4y+4-4\right)\\0&=\left(x-1\right)^2+\left(y-2\right)^2-4\\4&=\left(x-1\right)^2+\left(y-2\right)^2\end{align*})
2c/d: (https://i.imgur.com/55VXDzt.png)
2e:
2b:
2c/d: (https://i.imgur.com/55VXDzt.png)
2e:
ERQ3
3a: \\&=\frac{\pi}{4}\left(\frac{4h^3}{3}+h\right)\end{align*})
3b:
3ci:\\\frac{dh}{dt}&=\frac{4}{\pi\left(4h^2+1\right)}\times\frac{4-5\sqrt{h}}{100}\\&=\frac{4-5\sqrt{h}}{25\pi\left(4h^2+1\right)}\end{align*})
3cii:
3d:
3e:
3f:
3b:
3ci:
3cii:
3d:
3e:
3f:
ERQ4
4a: 
4b:&=\frac{3\sqrt{5}+5}{2}\\y_{B}\left(\frac{1+\sqrt{5}}{2}\right)&=\frac{9+\sqrt{5}}{2}\end{align*})
4c:)
4d:)
4e:\text{ minutes}\\<br />4\text{ minutes total})
4b:
4c:
4d:
4e:
ERQ5
5a: (https://i.imgur.com/6bXy9qK.png)
5bi:}-20v=20a)
5bii:}-v&=a\\\frac{g}{2}-v&=a\\a&=\frac{g-2v}{2}\end{align*})
5c:})
5d:
5ei:
5eii:
5bi:
5bii:
5c:
5d:
5ei:
5eii:
ERQ6
6b:
6c:
6d: \(p<0.05\), therefore, \(H_{0}\) should be rejected
6e:
6f:
6g:
Post by: crouchie on November 12, 2018, 05:57:58 pm
I get A for Q4. Can we just check?You're right, my bad, a typo! Fixed.
Post by: DinWell on November 12, 2018, 06:05:49 pm
Why did we have to restrict the domain of g(x)? I thought that we consider it a different function on it's maximal domain. I interpreted it as, if g(x)=sign(x). Then g(x) = -4 for x<0 and 4 for x>0MCQ1. E
2. B
3. D
4. A
5. D
6. C
7. C
8. E
9. C
10. A
11. C
12. E
13. B
14. C
15. E
16. B
17. E
18. D
19. E
20. BERQ11a.
1b: GRAPH TO COME
1c:
1d:
1ei:
1eii: \[g\left(x\right)=\lbrace\begin{array}{ll}
-4 & -\sqrt{2}<x<0 \\
4 & 0<x<\sqrt{2} \\
\end{array}\]
1eiii: GRAPH TO COMEERQ22a:
2b:
2c/d: GRAPH TO COME
2e:ERQ33a:
3b:
3ci:
3cii:
3d:
3e:
3f:
Post by: mzhao on November 12, 2018, 06:10:51 pm
Why did we have to restrict the domain of g(x)? I thought that we consider it a different function on it's maximal domain. I interpreted it as, if g(x)=sign(x). Then g(x) = -4 for x<0 and 4 for x>0
Fair point. I think both answers are acceptable, as long as the hybrid function you defined matches the graph you sketched.
Post by: iroze007 on November 12, 2018, 06:12:36 pm
MCQ1. E
2. B
3. D
4. A
5. D
6. C
7. C
8. E
9. C
10. A
11. C
12. E
13. B
14. C
15. E
16. B
17. E
18. D
19. E
20. BERQ11a.
1b: GRAPH TO COME
1c:
1d:
1ei:
1eii: \[g\left(x\right)=\lbrace\begin{array}{ll}
-4 & -\sqrt{2}<x<0 \\
4 & 0<x<\sqrt{2} \\
\end{array}\]
1eiii: GRAPH TO COMEERQ22a:
2b:
2c/d: GRAPH TO COME
2e:ERQ33a:
3b:
3ci:
3cii:
3d:
3e:
3f:ERQ44a:
4b:
4c:
4d:
4e:
you sure about that segment q. I got 4pi/3 - root(3)
Post by: fluff on November 12, 2018, 06:21:13 pm
you sure about that segment q. I got 4pi/3 - root(3)
I got the same as you!
Post by: jazzycab on November 12, 2018, 06:23:36 pm
you sure about that segment q. I got 4pi/3 - root(3)
Not certain about any of them. Did them very quickly. Will check when I get a chance.
Post by: mzhao on November 12, 2018, 06:25:34 pm
MCQ5c:- NOTE: This is NOT the same form as specified in the question. I haven't had time to have a good look, but my feeling (after speaking with a number of other people) is that there was an error in this exam...
Nothing wrong with this question.
Spoiler
Post by: DinWell on November 12, 2018, 06:26:59 pm
Nothing wrong with this question.It's alright if you subbed in g=9.8 right? But still kept it in the correct form.Spoiler
Post by: AlphaZero on November 12, 2018, 06:29:16 pm
MCQ1. E
2. B
3. D
4. A
5. D
6. C
7. C
8. E
9. C
10. A
11. C
12. E
13. B
14. C
15. E
16. B
17. E
18. D
19. E
20. BERQ11a.
1b: GRAPH TO COME
1c:
1d:
1ei:
1eii: \[g\left(x\right)=\lbrace\begin{array}{ll}
-4 & -\sqrt{2}<x<0 \\
4 & 0<x<\sqrt{2} \\
\end{array}\]
1eiii: GRAPH TO COMEERQ22a:
2b:
2c/d: GRAPH TO COME
2e:ERQ33a:
3b:
3ci:
3cii:
3d:
3e:
3f:ERQ44a:
4b:
4c:
4d:
4e:ERQ55a: IMAGE TO COME
5bi:
5bii:
5c:- NOTE: This is NOT the same form as specified in the question. I haven't had time to have a good look, but my feeling (after speaking with a number of other people) is that there was an error in this exam...
5d:
5ei:
5eii:ERQ6
6b:
6c:
6d: \(p<0.05\), therefore, \(H_{0}\) should be rejected
6e:
6f:
6g:
Well done jazzycab!!
Post by: mzhao on November 12, 2018, 06:30:10 pm
It's alright if you subbed in g=9.8 right? But still kept it in the correct form.
VCAA usually doesn't discriminate between 'g' and '9.8' :)
Post by: DinWell on November 12, 2018, 06:39:31 pm
4e. wasn't it to a certain number of decimal places? 6e. Why 146.52? I got 146.51. I get that it has to be greater than 146.5107385.... but will they say 146.51 is wrong?MCQ1. E
2. B
3. D
4. A
5. D
6. C
7. C
8. E
9. C
10. A
11. C
12. E
13. B
14. C
15. E
16. B
17. E
18. D
19. E
20. BERQ11a.
1b: GRAPH TO COME
1c:
1d:
1ei:
1eii: \[g\left(x\right)=\lbrace\begin{array}{ll}
-4 & -\sqrt{2}<x<0 \\
4 & 0<x<\sqrt{2} \\
\end{array}\]
1eiii: GRAPH TO COMEERQ22a:
2b:
2c/d: GRAPH TO COME
2e:ERQ33a:
3b:
3ci:
3cii:
3d:
3e:
3f:ERQ44a:
4b:
4c:
4d:
4e:ERQ55a: IMAGE TO COME
5bi:
5bii:
5c:
5d:
5ei:
5eii:ERQ6
6b:
6c:
6d: \(p<0.05\), therefore, \(H_{0}\) should be rejected
6e:
6f:
6g:
Post by: iroze007 on November 12, 2018, 06:40:35 pm
MCQ1. E
2. B
3. D
4. A
5. D
6. C
7. C
8. E
9. C
10. A
11. C
12. E
13. B
14. C
15. E
16. B
17. E
18. D
19. E
20. BERQ11a.
1b: GRAPH TO COME
1c:
1d:
1ei:
1eii: \[g\left(x\right)=\lbrace\begin{array}{ll}
-4 & -\sqrt{2}<x<0 \\
4 & 0<x<\sqrt{2} \\
\end{array}\]
1eiii: GRAPH TO COMEERQ22a:
2b:
2c/d: GRAPH TO COME
2e:ERQ33a:
3b:
3ci:
3cii:
3d:
3e:
3f:ERQ44a:
4b:
4c:
4d:
4e:ERQ55a: IMAGE TO COME
5bi:
5bii:
5c:
5d:
5ei:
5eii:ERQ6
6b:
6c:
6d: \(p<0.05\), therefore, \(H_{0}\) should be rejected
6e:
6f:
6g:
6e) 146.51? instead of 146.52
Post by: mzhao on November 12, 2018, 06:43:07 pm
4e. wasn't it to a certain number of decimal places?
4.1 if I recall correctly!
Post by: fluff on November 12, 2018, 06:45:38 pm
Post by: DinWell on November 12, 2018, 06:46:14 pm
Crouchie and jazzycab, check your answers to Q12.I think I chose A for the Q. Whats correct?
Here are counter examples:
Select \(\vec{a}=-\vec{b}\).
They are parallel, but \(0=|\vec{a}+\vec{b}|\neq |\vec{a}|+|\vec{b}|=2|\vec{a}|\).
Select \(\vec{a}=(1, 0)\) and \(\vec{b}=(0, 1)\).
They are perpendicular but, \(\sqrt{2}=|\vec{a}+\vec{b}|\neq |\vec{a}|+|\vec{b}|=2\).
Post by: AlphaZero on November 12, 2018, 06:48:32 pm
Post by: DinWell on November 12, 2018, 06:52:46 pm
Sorry all! Excuse my stupidness. \(p\Rightarrow q\) does not necessarily mean \(q\Rightarrow p\). Whoops...Which was the correct ans? They have different ones :(
Post by: mzhao on November 12, 2018, 06:55:17 pm
Which was the correct ans? They have different ones :(
It appears as if there is no correct answer then, unless they imply the two vectors go in the same direction when they say "parallel", as opposed to antiparallel.
Post by: DinWell on November 12, 2018, 06:57:07 pm
It appears as if there is no correct answer then, unless they imply the two vectors go in the same direction when they say "parallel", as opposed to antiparallel.Oh no! What a waste of time lmao. I spent like 2 and a half minutes on that question :'(
Post by: popcorndudette on November 12, 2018, 06:58:47 pm
6e) 146.51? instead of 146.52
When you sub in 146.51 you get 0.049... so you still reject. 146.52 gives p-value of 0.05.... so answer is 146.52
Post by: joon on November 12, 2018, 07:02:27 pm
Post by: DinWell on November 12, 2018, 07:03:41 pm
Does anyone have a copy of the question booklet?We're not allowed to upload here due to copyright issues.
Post by: AlphaZero on November 12, 2018, 07:05:53 pm
It appears as if there is no correct answer then, unless they imply the two vectors go in the same direction when they say "parallel", as opposed to antiparallel.
No, you made the same mistake as I did! The answer is A. The question asks us: If p, then q? and gives us 5 options for q. We do NOT require "If q, then p" to be also true.
Post by: mzhao on November 12, 2018, 07:08:23 pm
No, you made the same mistake as I did! The answer is A. The question asks us: If p, then q? and gives us 5 options for q. We do NOT require "If q, then p" to be also true.
Oh right my apologies. I actually wrote A, so that is assuaging :)
Post by: HamConspiracy on November 12, 2018, 07:11:54 pm
Post by: Dais_Deorum on November 12, 2018, 07:33:18 pm
Post by: iroze007 on November 12, 2018, 07:35:17 pm
When you sub in 146.51 you get 0.049... so you still reject. 146.52 gives p-value of 0.05.... so answer is 146.52
surely they will accept 146.51
Post by: HamConspiracy on November 12, 2018, 07:52:21 pm
Post by: hom0sexual on November 12, 2018, 07:53:32 pm
surely they will accept 146.51
It asked you to find the value of the sample mean and then to round it, so surely it would be 146.511... --> 146.51.
Post by: jazzycab on November 12, 2018, 08:02:25 pm
Nothing wrong with this question.Spoiler
Of course... I must've missed it in my rush to get them done...
Post by: popcorndudette on November 12, 2018, 08:05:41 pm
It asked you to find the value of the sample mean and then to round it, so surely it would be 146.511... --> 146.51.
I'm pretty sure it asked for the minimum value of sample mean so that you don't reject the null hypothesis
Post by: hom0sexual on November 12, 2018, 08:15:16 pm
I'm pretty sure it asked for the minimum value of sample mean so that you don't reject the null hypothesis
Yeah, the minimum value that's right. The point is they asked you to round it not to find the minimum sample mean with 2 d.p. that would lead to H0 not being rejected.
Post by: Onyx on November 12, 2018, 08:21:47 pm
Of course... I must've missed it in my rush to get them done...
Left it in the first equation that u provided, will I get a mark? Should do right?
Post by: DinWell on November 12, 2018, 08:23:32 pm
Left it in the first equation that u provided, will I get a mark? Should do right?You wouldn't get full marks if that's what your asking. It asked specifically for a certain form.
Post by: HamConspiracy on November 12, 2018, 08:38:18 pm
Post by: TheAspiringDoc on November 12, 2018, 08:38:33 pm
Post by: DinWell on November 12, 2018, 09:06:11 pm
2e. isn't it 4pi/3 instead of 2pi/3?MCQ1. E
2. B
3. D
4. A
5. D
6. C
7. C
8. E
9. C
10. A
11. C
12. E
13. B
14. C
15. E
16. B
17. E
18. D
19. E
20. BERQ11a.
1b: GRAPH TO COME
1c:
1d:
1ei:
1eii: \[g\left(x\right)=\lbrace\begin{array}{ll}
-4 & -\sqrt{2}<x<0 \\
4 & 0<x<\sqrt{2} \\
\end{array}\]
1eiii: GRAPH TO COMEERQ22a:
2b:
2c/d: GRAPH TO COME
2e:ERQ33a:
3b:
3ci:
3cii:
3d:
3e:
3f:ERQ44a:
4b:
4c:
4d:
4e:ERQ55a: IMAGE TO COME
5bi:
5bii:
5c:
5d:
5ei:
5eii:ERQ6
6b:
6c:
6d: \(p<0.05\), therefore, \(H_{0}\) should be rejected
6e:
6f:
6g:
Post by: Mattjbr2 on November 12, 2018, 09:07:10 pm
2e. isn't it 4pi/3 instead of 2pi/3?
I got 4pi/3
Post by: Onyx on November 12, 2018, 09:10:26 pm
i got 4pi/3 - root3 for the area of the complex part.
for 6e, I got 146.51, I did invNorm(0.05,0,1) to get -1.64485362591 or something like that, then (root2(x-150))/3 = -1.64485362591, to get 146.51 to 2d.p.
Mod Edit: Post merge, use the Modify button to edit your first post to avoid posting twice/three times in a row! :)
Post by: jazzycab on November 12, 2018, 09:36:56 pm
Why did we have to restrict the domain of g(x)? I thought that we consider it a different function on it's maximal domain. I interpreted it as, if g(x)=sign(x). Then g(x) = -4 for x<0 and 4 for x>0I think the wording of this question is very ambiguous - \(f'\left(x\right)\) is over its maximal domain, which could mean \(g\left(x\right)\) could have any domain of which \(\left(-\sqrt{2},0\right)\cup\left(0,\sqrt{2}\right)\) is a subset. Given that the graph question didn't ask for any specific points to be labelled, I'm now more inclined to think that the intention was for \(g\left(x\right)\) to have its maximal domain, \(\mathbb{R}\setminus\lbrace 0\rbrace\).
you sure about that segment q. I got 4pi/3 - root(3)You're correct, I've made some error somewhere that gave me an angle of \(\frac{\pi}{3}\) instead of \(\frac{2\pi}{3}\)
6e) 146.51? instead of 146.52As has been stated, I rounded up as 146.51 is the largest value for which \(H_{0}\) is rejected, as opposed to 146.52, which is the smallest value for which \(H_{0}\) is not rejected.
Did 4e ask for the total time (4.1) or the time interval?4e asked for 'what period of time' which I think is ambiguous - I only included the total time as an afterthought. I realise now that this is going to be incorrect to the nearest minute as I rounded the lower end-point up to 91.8 (as the two yachts are not within 0.2 km of one another at \(t=91.7\)
Is question 9 C or D? Two answers have been provided so far.I've made an algebraic error on this one - it's D
Post by: AlphaZero on November 13, 2018, 12:11:11 am
As has been stated, I rounded up as 146.51 is the largest value for which \(H_{0}\) is rejected, as opposed to 146.52, which is the smallest value for which \(H_{0}\) is not rejected.
I believe you are not meant to directional round your answer here. What you've done here makes sense if the variable we are concerned with is discrete.
We see this occur a lot in Methods with sample sizes. For eg: "Find the smallest value of \(n\) for which \(\text{Pr}(X>1)>0.99\)").
However, here the variable we are concerned here with is continuous. The question states to find the smallest \(c\) such that \(\text{Pr}(\overline{X}<c)>0.05\), correct to two decimal places. The value of \(c\) is 146.51073854119...., which, correct to two decimal places, is 146.51, not 146.52.
I understand the logic you're trying to apply (that \(\text{Pr}(\overline{X}<146.510000)<0.05\), so we should answer with 146.52), but 146.510000 is not the answer we are supplying. We are saying that the answer is 146.51, correct to two decimal places.
Post by: AlphaZero on November 13, 2018, 12:32:57 am
Get them here: https://atarnotes.com/forum/index.php?topic=181370.msg1084174#msg1084174
Post by: Mattjbr2 on November 13, 2018, 06:39:20 am
Also, my solutions are up (finally).
Get them here: https://atarnotes.com/forum/index.php?topic=181370.msg1084174#msg1084174
You have 17 as D. You sure it's not E?
Post by: AlphaZero on November 13, 2018, 08:48:24 am
You have 17 as D. You sure it's not E?
Arrrrghghgh. Why am I so bad at this? Yes, the answer is E, not D...
Fixing it now
Post by: DinWell on November 13, 2018, 12:06:55 pm
Also, my solutions are up (finally).Brilliant work as always, dantraicos.
Get them here: https://atarnotes.com/forum/index.php?topic=181370.msg1084174#msg1084174
Post by: JamesMaths on November 20, 2018, 01:33:32 pm
(Part 1)
PDF on my web site:
https://unimelb.academia.edu/JamesCui
Provide Tutorials in Maths Methods and Specialist Maths in Balwyn.
Regards
James.
Post by: JamesMaths on November 20, 2018, 01:35:07 pm
(Part 2)
PDF on my web site:
https://unimelb.academia.edu/JamesCui
Provide Tutorials in Maths Methods and Specialist Maths in Balwyn.
Regards
James.
Post by: JamesMaths on November 20, 2018, 01:36:18 pm
(Part 3)
PDF on my web site:
https://unimelb.academia.edu/JamesCui
Provide Tutorials in Maths Methods and Specialist Maths in Balwyn.
Regards
James.