Post by: Joseph41 on October 22, 2019, 01:54:02 pm
(https://i.imgur.com/VFM9j89.png)
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Post by: AlphaZero on November 11, 2019, 04:04:27 pm
Sorry for the lateness...
Thoughts on the exam?
Post by: undefined on November 11, 2019, 05:33:12 pm
Post by: Tau on November 11, 2019, 05:35:33 pm
Anyone else think this has been by far the hardest specialist exam 2 by VCAA?
I don’t know, I thought it was a decently hard paper. Lots of show questions and some abstract maths. But of course there’s a tendency to overestimate the difficulty having just done it.
Post by: AnonymooseUser on November 11, 2019, 05:40:58 pm
Post by: undefined on November 11, 2019, 05:41:20 pm
I don’t know, I thought it was a decently hard paper. Lots of show questions and some abstract maths. But of course there’s a tendency to overestimate the difficulty having just done it.Fair enough. I thought it was relatively harder as there were a lot of variables which you had to work with.
Post by: Alexmaths on November 11, 2019, 05:49:25 pm
Post by: Purple_Mango on November 11, 2019, 05:56:38 pm
did anybody else just have parts of the quadratic formula for p and q?Yeah, my p and q ended up being part of the quadratic formula as well.
I wanted to die when attempting that 5 mark mechanics question. More especially so when the assessors didn't put on AC during this exam on the only hot day this week ;)
(also, I'm curious as to what the volume in terms of t could be, from question 1)
Post by: JeromeTT on November 11, 2019, 05:59:28 pm
Post by: Keekerz on November 11, 2019, 05:59:48 pm
Post by: randomnobody69420 on November 11, 2019, 06:01:41 pm
Post by: lolozo214 on November 11, 2019, 06:33:41 pm
Hey everyone, how did you do the first question of the vectors question? Isn't there 3 points which C could be? How do you arrive at one
A quadrilateral ABCD has C on the corner opposite A. Since AD and BC are parallel sides or a parallelogram, AD=BC
Post by: Edged on November 11, 2019, 06:43:41 pm
Yeah, my p and q ended up being part of the quadratic formula as well.
I wanted to die when attempting that 5 mark mechanics question. More especially so when the assessors didn't put on AC during this exam on the only hot day this week ;)
(also, I'm curious as to what the volume in terms of t could be, from question 1)
To find
and
Also considering the terminals,
I think that should be right, checking with the normal volume in terms of y.
Post by: S_R_K on November 11, 2019, 06:46:11 pm
Post by: Massimooo123 on November 11, 2019, 06:59:33 pm
Post by: Macrophagee on November 11, 2019, 07:12:17 pm
Yeah was not expecting that 5 marker....
Post by: Massimooo123 on November 11, 2019, 07:13:33 pm
Very tough 5 marker I thought.... Also was very confused finding the area and volume for the vectors question. Rest was okay? My answers for the complex seemed a bit too basic, did anybody else just have parts of the quadratic formula for p and q?
I forgot what the area of a parallelogram was so I just did the side lengths multiplied by each other (which, upon reflection, is very clearly wrong lol).
I think for the volume one you had to find the middle of the parallelogram (the "centroid" according to google), find the distance between that and the peak of the pyramid which gives the height, then apply the formula V=(1/3)A*h. I obviously had the area of the base wrong, but if this method of finding the volume is correct and I did everything else right, would I get full marks? I still don't really get how answer marks and consequential marks work tbh.
Post by: AnonymooseUser on November 11, 2019, 07:20:24 pm
Post by: jkay__ on November 11, 2019, 07:39:41 pm
So do people think the A+ cutoff will go down from 2018?
Here's hoping. Holy, I think it'll even be as low as like 65/80, or even lower
This exam was, immeasurably and irrefutably FUCKED. That's the only way I can describe this abomination. It was the HARDEST out of the past 20 goddamn years of Specialist exams, and that is in no way an exaggeration. It was comparable to Kilbaha / MAV papers (not sure if that's a fair comparison, haven't done much of them), but that's how incredibly gone it was.
If you feel like you did badly, no worries. A lot of kids that go Melbourne High said that they messed up, very terribly
Post by: S_R_K on November 11, 2019, 08:36:33 pm
I think for the volume one you had to find the middle of the parallelogram (the "centroid" according to google), find the distance between that and the peak of the pyramid which gives the height, then apply the formula V=(1/3)A*h. I obviously had the area of the base wrong, but if this method of finding the volume is correct and I did everything else right, would I get full marks? I still don't really get how answer marks and consequential marks work tbh.
To get the height of the pyramid, I found the magnitude of the projection of AP onto the vector 6i + 2j + 5k (the vector from part d which is perpendicular to the parallelogram), this gave a pyramid volume of 24. (This is the point of finding a unit vector perpendicular to AB and AD).
Using one of the other corners of the base would have worked as well.
Post by: peds01 on November 11, 2019, 08:51:12 pm
Post by: Massimooo123 on November 11, 2019, 08:58:02 pm
To get the height of the pyramid, I found the magnitude of the projection of AP onto the vector 6i + 2j + 5k (the where from part d which is perpendicular to the parallelogram), this gave a pyramid volume of 24. (This is the point of finding a unit vector perpendicular to AB and AD).
Using one of the other corners of the base would have worked as well.
That makes more sense than what I did, I don't know why I just forgot about that random-ass question and moved on like it was nothing. I think my height had a square root in it so things aren't looking good to say the least lol. I don't understand why my method would give a wrong height, but I guess it could just be a silly mistake. The peak of the pyramid would be directly above the intersection of the two lines AC and BD right (not directly above as in having a greater z coordinate in the xyz plane, but directly above as in perpendicular to the base of the pyramid)?
Ahh, this sucks. I keep thinking I'm doing well on my exams and then awakening to a harsh reality and I've still been surprised every time it's happened :-[
Post by: Purple_Mango on November 11, 2019, 09:02:06 pm
If that is it, then I'm pretty happy ! I just had the expanded version of it, so it should be fine, I guess.
I think that should be right, checking with the normal volume in terms of y.
And based off of other comments, looks like I messed up the volume of the pyramid. I got the height by getting the scalar product of the unit vector of 6i+2j+5k with whatever P was, instead of AP (or other vertex --> P). No wonder I got a weird volume. :o
Although I found this exam difficult, personally, I consider MAV exams to be harder. 2019 MAV exam 2 especially bamboozled me
Post by: tx inspire on November 11, 2019, 09:13:18 pm
Post by: S_R_K on November 11, 2019, 09:15:19 pm
I don't understand why my method would give a wrong height, but I guess it could just be a silly mistake. The peak of the pyramid would be directly above the intersection of the two lines AC and BD right (not directly above as in having a greater z coordinate in the xyz plane, but directly above as in perpendicular to the base of the pyramid)?
The peak of a pyramid is not necessarily vertically above the centroid of the base.
Post by: AlphaZero on November 11, 2019, 09:31:35 pm
Sorry for taking so long!
Post by: mzhao on November 11, 2019, 09:54:40 pm
Solution are up!
Sorry for taking so long!
Great work as always, AlphaZero!
I believe there should be one correction to Q2d:
"All complex solutions... have non-zero... imaginary parts"
So
will yield an imaginary number, which would make q complex, where q should instead be a real number.
is my answer. No need for absolute value within the sqrt, as from the first sentence of the question, we assume solutions always have a complex component, and hence, the discriminant is always negative.
EDIT:
clarified that I am considering the possibility of a modulus within the sqrt
and thanks Tau for pointing out that a in the denominator requires modulus
Post by: Tau on November 11, 2019, 10:00:32 pm
Wouldn’t the modulus apply to the a in the denominator as well though?
Post by: mzhao on November 11, 2019, 10:01:58 pm
Wouldn’t the modulus apply to the a in the denominator as well though?
Sorry, I was referring to a potential absolute value within the sqrt.
But yes, that is a good pick up, will fix!
Post by: S_R_K on November 11, 2019, 10:07:37 pm
Solution are up!
Sorry for taking so long!
Some preliminary comments:
Multiple choice, Question 4: i^1! = i, and both i^2! and i^3! = -1, then i^n! = 1, for all n ≥ 4, since n! is a multiple of 4 for all n ≥ 4. This gives i – 2 + 97*1.
Extended response 3a ii, I'm confused about your inequalities. I got, when 0 ≤ b < a, we have r > s (so that a – b > 0 and r/s > 1).
Extended response 6a. I think your calculation gives the chance of getting at least one sample with mean mass less than 370 or greater than 375. I did 1 – (1 – Pr(370 ≤ X ≤ 375))^2.
Otherwise, thanks again.
Post by: Massimooo123 on November 12, 2019, 12:18:13 am
Extended response 6a. I think your calculation gives the chance of getting at least one sample with mean mass less than 370 or greater than 375. I did 1 – (1 – Pr(370 ≤ X ≤ 375))^2.
Otherwise, thanks again.
I did (Pr(370 ≤ X ≤ 375))^2 + 2*(Pr(370 ≤ X ≤ 375))*(1-Pr(370 ≤ X ≤ 375)) and got the same answer that your expression gives.
Post by: MubMurshed on November 12, 2019, 06:59:24 am
Because we have sqrt(2d-4), and we have already taken out i from the expression. Therefore we can’t make it negative otherwise we create an another i and the restriction that it has to be lower than sqrt(6)/2 no longer applies since it is not an imaginary value. Therefore d should be > 2.
Would I be correct in thinking that?
Post by: S_R_K on November 12, 2019, 08:46:59 am
With q2c about -1 < d < 5, doesn’t d have to be be greater than 2? I did 2 < d < 5.
Because we have sqrt(2d-4), and we have already taken out i from the expression. Therefore we can’t make it negative otherwise we create an another i and the restriction that it has to be lower than sqrt(6)/2 no longer applies since it is not an imaginary value. Therefore d should be > 2.
Would I be correct in thinking that?
If 2 < d ≤ 5, then the solutions have non-zero imaginary component, and lie somewhere on the vertical line segment joining –1 – i*sqrt(6)/2 and –1 + i*sqrt(6)/2.
If d = 2, the solution is –1.
If –1 ≤ d < 2, the solutions have zero imaginary component, and lie on the real axis between -1 – sqrt(6)/2 and -1 + sqrt(6)/2.
So –1 ≤ d ≤ 5 gives all values of d that satisfy |z + 1| ≤ sqrt(6)/2.
Post by: AlphaZero on November 12, 2019, 01:34:57 pm
"All complex solutions... have non-zero... imaginary parts"
So
will yield an imaginary number, which would make q complex, where q should instead be a real number.
While \(\dfrac{\sqrt{b^2-4ac}}{2a}\) might be non-real, \(\left|\dfrac{\sqrt{b^2-4ac}}{2a}\right|\) is most certainly non-negative and real. Your answer is equivalent to mine, and the modulus brackets take care of it.
Multiple choice, Question 4: i^1! = i, and both i^2! and i^3! = -1, then i^n! = 1, for all n ≥ 4, since n! is a multiple of 4 for all n ≥ 4. This gives i – 2 + 97*1.
I knew this doing the question, but I couldn't be bothered carefully counting when it's an MCQ.
Extended response 3a ii, I'm confused about your inequalities. I got, when 0 ≤ b < a, we have r > s (so that a – b > 0 and r/s > 1).
\(k\) is also positive when \(a-b<0\) and \(0<r/s<1\), making the log negative as well. There are two possibilities, and I think both should be stated.
Extended response 6a. I think your calculation gives the chance of getting at least one sample with mean mass less than 370 or greater than 375. I did 1 – (1 – Pr(370 ≤ X ≤ 375))^2.
Yep, I screwed this up. My bad. In my defence, I had my Human Structure and Function exam 1 just before I did the solutions (so I was pretty tired) and I had my Differential Equations exam this morning, so I wasn't exactly taking my time with the solutions :)
Post by: MubMurshed on November 12, 2019, 02:03:35 pm
If 2 < d ≤ 5, then the solutions have non-zero imaginary component, and lie somewhere on the vertical line segment joining –1 – i*sqrt(6)/2 and –1 + i*sqrt(6)/2.
If d = 2, the solution is –1.
If –1 ≤ d < 2, the solutions have zero imaginary component, and lie on the real axis between -1 – sqrt(6)/2 and -1 + sqrt(6)/2.
So –1 ≤ d ≤ 5 gives all values of d that satisfy |z + 1| ≤ sqrt(6)/2.
Ahh I see, completely forgot about that it could be just real too haha. Thanks!
Post by: S_R_K on November 12, 2019, 02:30:44 pm
\(k\) is also positive when \(a-b<0\) and \(0<r/s<1\), making the log negative as well. There are two possibilities, and I think both should be stated.
Agreed, but if a – b > 0, then we require r/s > 1, which gives a > b and r > s; in your solutions you've got a > b and s > r. Similarly, if b > a we need s > r, but you've got r > s.
Also: If a > b, can we have (and if not, why not) r < s < 0? This would also give r/s > 1. I don't see any reason to rule out negative values of P (eg. perhaps P is a monetary balance).
Post by: AlphaZero on November 12, 2019, 04:15:25 pm
Agreed, but if a – b > 0, then we require r/s > 1, which gives a > b and r > s; in your solutions you've got a > b and s > r. Similarly, if b > a we need s > r, but you've got r > s.
Also: If a > b, can we have (and if not, why not) r < s < 0? This would also give r/s > 1. I don't see any reason to rule out negative values of P (eg. perhaps P is a monetary balance).
*sigh yuck too many errors (2 now rip) in the solutions. I'll fix Q3a.ii now.
In the previous part, we required \(P>0\) to arrive at the result, so I think it's safe to assume \(P>0\) then.