ATAR Notes: Forum
Archived Discussion => Mathematics/Science/Technology => 2014 => Exam Discussion => Victoria => Mathematical Methods CAS => Topic started by: Zealous on November 05, 2014, 10:01:18 am
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So, how did you find it?
Any crazy curveball questions? How was it compared with the past VCAA exams?
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Well I'd be lucky to get 50% :-\
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There goes my chances of a decent score
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So harder than usual?? I'm sure you'll all be fine!!
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Don't stress so much! Remember, your SACs count more than that exam did.
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So harder than usual?? I'm sure you'll all be fine!!
I feel like this years was definitely harder than last years, the last two graph questions completely threw me off.
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Yeah feel like I'm going to cry now!
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It was ok, I'd say 70% of it was pretty straightforward, but q.10 required some pretty clever mathematical thinking (which I don't have) in addition to a knowledge of formulas etc.
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Damn that last page, would of got it if I did make a stupid error.
Overall though, beside the last page, I found it was all doable, it had many questions similar to previous years, just more difficult and annoying to work with
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Started off so well...nek minnut.
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It's definitely doable, dodgy anti-diff question was bound to make people make mistakes as well as the two probability questions (think Exam 1 of 2006), last question I felt was fine I felt (not as hard as 2011).. Felt it was on par with last years exam in difficulty I guess :P
Now onto Exam 2 guys so let's smash it :)
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I found it ok, until the last question, what did everyone get for it?
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Well that was easy :)
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It's definitely doable, dodgy anti-diff question was bound to make people make mistakes as well as the two probability questions (think Exam 1 of 2006), last question I felt was fine I felt (not as hard as 2011).. Felt it was on par with last years exam in difficulty I guess :P
Now onto Exam 2 guys so let's smash it :)
For the anti diff question with the area bound by the curve and y=4, did you do 12- anti diff of f(x) using the terminals -1 and 2?
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Here are my
preliminary solutions guys :)
I didn't check them, so I may have a few mistakes, but based on conversations with guys who did the exam they seem correct.
P.S. I have a blank copy of the exam too, so I'll upload that next.
Edit: I can now confirm my solutions are correct, except the graph is a bit revolting: I think it should go through (1,2) to get full marks.
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Here's the BLANK Exam 1 guys :)
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Here are my preliminary solutions guys :)
I didn't check them, so I may have a few mistakes, but based on conversations with guys who did the exam they seem correct.
P.S. I have a blank copy of the exam too, so I'll upload that next.
Thanks
Damn... I lost so many marks
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While you guys blow up, I've got a blank copy with me - now that I don't need to scan it, I'll do it on the train and get back to you. :P
It does look moderate in difficulty - I reckon anything above 60/80 could be an A. Particularly with that last question being worth almost 1/4 of the exam.
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I may be subject to exam bias but I found it to be the second most difficult exam to date only succeeded by 2010s cone of death.
I was very pressed for time and there were lots of opportunities to make silly mistakes. Lots of algebra and time consuming things like expanding integrals. The the area under curve question was really dodgy and just urgh. If they had specified it was between the turning point and the left end point it would have been a lot less ambiguous.
Hopefully a better exam 2 will follow.
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Awesome only lost last mark haha, but isn't 1b +-1/2......its the derivative equation of a hyperbola and no restriction on range?
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Urghh... could have done so much better if I read the questions properly. Area question in 5c I bound it with the y-axis and the final question I completely forgot about endpoints :(
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I forgot to minus 8 on each of the areas at the last question. O wow, had everything else the same as the solutions and that's where I stuff up....
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wasn't ridiculously bad but I felt it separated the boys (me) from the men (you guys) with the final two questions
FUCK just realised I forgot to simplify a few answers... is that bad?
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So quick question guys,
According to the answers everything was right except for my rule for f(x) from f'(x) and f(pi/2)=1/2
I accidentally wrote 1/2 x Cos(pi) = 1, not -1, so my c value came out to be -2 instead of -1.
Will this cost me 1 or 2 marks?
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(http://i1282.photobucket.com/albums/a531/Ovazealous/Picture2_zps61662ecf.png)
Looks familiar. :P
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So quick question guys,
According to the answers everything was right except for my rule for f(x) from f'(x) and f(pi/2)=1/2
I accidentally wrote 1/2 x Cos(pi) = 1, not -1, so my c value came out to be -2 instead of -1.
Will this cost me 1 or 2 marks?
probably only 1
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Ughhh I'm so pissed, got the last 3 parts of question 10 wrong.. that's 4 marks gone. And it was coz of running out of time. fml fml. Could've easily got 39 or 40
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Oh god I stuffed all my integration. All I needed was a 25
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That last question threw me way off. Goddammmmmmmnnnnnn
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m was 5ln(2) but would ln(32) be accepted or does it have to be simplified
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Ughhh I'm so pissed, got the last 3 parts of question 10 wrong.. that's 4 marks gone. And it was coz of running out of time. fml fml. Could've easily got 39 or 40
same here why didnt i think of similar triangles ffs
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What does everyone think the A/ A+ cut off will be this year?
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Could a 35 be an A+??
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m was 5ln(2) but would ln(32) be accepted or does it have to be simplified
That should be fine!
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Do you guys reckon it'll be marked up as much as last year? Or does it really depend on the difficulty of tomorrow's exam?
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Here are my preliminary solutions guys :)
I didn't check them, so I may have a few mistakes, but based on conversations with guys who did the exam they seem correct.
P.S. I have a blank copy of the exam too, so I'll upload that next.
32/40 for me. Hopefully that'll be a decent grade on the grade distribution :'(
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If you lose one mark on exam 1 do you reckon it's still possible to get a 50 raw?
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If you lose one mark on exam 1 do you reckon it's still possible to get a 50 raw?
I believe so.
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If you lose one mark on exam 1 do you reckon it's still possible to get a 50 raw?
definitely. i lost 1 mark in exam 1 and 1 mark in exam 2 and got 50. but it will depend on tomorrow's difficulty. if exam 2 is hard, then that is good for you. but if it easy, just losing one mark can be costly.
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32/40 for me. Hopefully that'll be a decent grade on the grade distribution :'(
I got the same....still 80%
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omfg I forgot to find the coordinates :-\ Just wrote the x-intercepts instead omg
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Question 10)b)i) could have also been done by finding the gradient in terms of u, and then subbing this into the equation y=mx+v using the point (2, 4) to find v.
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same here why didnt i think of similar triangles ffs
I let the gradient of (0,v) to (2,4) equal the gradient (2,4) to (u, 0). Ended up with the same thing.
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I let the gradient of (0,v) to (2,4) equal the gradient (2,4) to (u, 0). Ended up with the same thing.
Yup, that's what I did
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Question 10)b)i) could have also been done by finding the gradient in terms of u, and then subbing this into the equation y=mx+v using the point (2, 4) to find v.
Yep that's how i did it haha, i just hope +-1/2 (not just +1/2) was an acceptable answer for the second question, cos i don't know if it meant principal square root or just square root (no restrictions?) I drew the hyperbola and pointed to where each gradient would apply.........if i lost that gonna be so pissed haha
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Yep that's how i did it haha, i just hope +-1/2 (not just +1/2) was an acceptable answer for the second question, cos i don't know if it meant principal square root or just square root (no restrictions?) I drew the hyperbola and pointed to where each gradient would apply.........if i lost that gonna be so pissed haha
The calculator just gives 1/2
Edit: are you saying you don't know if it's +/-root(...)?
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A+ cutoff predictions? Could it go as low as 33-34?
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30/40. To get low 30s, is this an okay mark (provided I'm not absolutely massacred in exam 2 tomorrow :O )?
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I reckon the cutoff for A+ will be around 36
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For the conditional prob. question with m,
I accidently changed e^-x/5 to e^-x/2 after I wrote the correct integral statement
So, ended up with 2 - (2/sqrt(e))
How many marks would I lose for this?
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Yep that's how i did it haha, i just hope +-1/2 (not just +1/2) was an acceptable answer for the second question, cos i don't know if it meant principal square root or just square root (no restrictions?) I drew the hyperbola and pointed to where each gradient would apply.........if i lost that gonna be so pissed haha
We're given the function as f(x)= root(x^2+3), where the bottom half of the hyperbola is -root(x^2+3) so the function f(x) is only the upper branch :/ but depends on what VCAA has to say.
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Yep that's how i did it haha, i just hope +-1/2 (not just +1/2) was an acceptable answer for the second question, cos i don't know if it meant principal square root or just square root (no restrictions?) I drew the hyperbola and pointed to where each gradient would apply.........if i lost that gonna be so pissed haha
sqrt (.25) does equal +-1/2, however the gradient of that function at x=1 is increasing, meaning you want the positive one only.
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Yeah true, but i still don't know my spec teacher said if there were no range/domain restrictions you can't assume it means positive or negative (unless its like something that varies with time as it can be in two places at once), yeah so logically i'd say +- but everyone i spoke to wrote +1/2 only......haha can't afford to lose it since i already lost that last point (haha didn't realise f(5/2) was greater than f(6) haha)
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sqrt (.25) does equal +-1/2, however the gradient of that function at x=1 is increasing, meaning you want the positive one only.
Yeah look i think you guys are right, it's annoying though, cos if we had to write a square root function anytime they would expect us to explicitly mark the sign, but they didn't in their question so annoying
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Here are my preliminary solutions guys :)
I didn't check them, so I may have a few mistakes, but based on conversations with guys who did the exam they seem correct.
P.S. I have a blank copy of the exam too, so I'll upload that next.
your Q10. b. i).
Height isn't v, its v-4?
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Would you get any marks if you evaluated the integral between -1 and 3? Rather than from -1 to 2? As well as the area being subtracted from 16 of course.
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your Q10. b. i).
Height isn't v, its v-4?
Triangle OVU (big triangle) was used, not the small shaded triangle. I think there were multiple ways to do this question.
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For the question where we had to sketch the cubic, did we need to label the intercept at (0,0)? I figured since the question didn't say to then it doesn't matter, but I'm starting to get worried about it
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Triangle OVU (big triangle) was used, not the small shaded triangle. I think there were multiple ways to do this question.
Ahh I see, thank you
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You know i was just thinking, i spoke to a lot of people who were 'leaving exam two from last year till today" i felt so bad for em, because that last question on today's exam 1 was basically just last year's one on exam 2, unlucky!
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I did the same wrong integration for 1/5e^-1/5 for question 8 and b. Is that one mark or two marks off?
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For the question where we had to sketch the cubic, did we need to label the intercept at (0,0)? I figured since the question didn't say to then it doesn't matter, but I'm starting to get worried about it
It just said endpoints so I'm pretty sure you'll be okay :)
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Can i get an A+ with a 35?
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I think post-exam stress is going to be a real thing soon, that probably will be diagnosed by psychologists at the rate of all these VCAA exams.
Would you get any marks if you evaluated the integral between -1 and 3? Rather than from -1 to 2? As well as the area being subtracted from 16 of course.
With this question, would you have to subtract the section between x=2 and 3 from 16 or is that just the same thing since it's the top part?
Sorry guys, my brain's gone numb at the moment.
For the question where we had to sketch the cubic, did we need to label the intercept at (0,0)? I figured since the question didn't say to then it doesn't matter, but I'm starting to get worried about it
I did the same thing. Uh oh.
Damn, I think I got 29/40.. so many silly mistakes. I just differentiated the second question without solving it, I got -5loge(-2) instead of 5loge(2), I didn't notice the final graph had a domain, etc.. maybe ~32/40 if the examiners are lenient about silly working mistakes (mostly made during calculations of integrals)
Well, I guess I'm going to have to try and blitz exam 2 if I want that 35 >:(
Same thing here. I screwed so many marks for silly little things. I'm sort of hoping that, in order for VCAA to balance out the difficulty between today's Exam 1, that Exam 2 will be at a decent level. "Decent".
*mumbles to self* Why does "decent" and VCAA being the the same sentence sound like a joke?
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Here are my preliminary solutions guys :)
I didn't check them, so I may have a few mistakes, but based on conversations with guys who did the exam they seem correct.
P.S. I have a blank copy of the exam too, so I'll upload that next.
Edit: I can now confirm my solutions are correct, except the graph is a bit revolting: I think it should go through (1,2) to get full marks.
My answers match, so I reckon this exam is done and dusted. :P
A+ cut-off is at least 35 - it was a surprising exam, but an A+ student shouldn't have been too thrown. I also think an A would be about 30.
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Question 5c, does the area include the integral from 2 to 3 4-f(x) dx?
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Dont think A+ would go below 36
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What do you think the rest of the distribution will be like, guys?
Question 5c, does the area include the integral from 2 to 3 4-f(x) dx?
Same question as mine, reworded. Just putting it out there now.
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Question 5c, does the area include the integral from 2 to 3 4-f(x) dx?
Whoops, I missed that. .__. Yes, it should.
So the solutions have one mistake, unless chuck can explain himself.
EDIT: If anyone wants to agree with me, my answer is 8 - which was found by taking the area of the rectangle (16, so it's really a square) and taking away the area under the curve through its whole domain (-1 to 3).
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Whoops, I missed that. .__. Yes, it should.
So the solutions have one mistake, unless chuck can explain himself.
EDIT: If anyone wants to agree with me, my answer is 8 - which was found by taking the area of the rectangle (16, so it's really a square) and taking away the area under the curve through its whole domain (-1 to 3).
But wouldn't that technically mean it wasn't 'enclosed' between the graph of f(x) and the line y=4? Shouldn't that have included the line x-3 then? :-\
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Whoops, I missed that. .__. Yes, it should.
So the solutions have one mistake, unless chuck can explain himself.
EDIT: If anyone wants to agree with me, my answer is 8 - which was found by taking the area of the rectangle (16, so it's really a square) and taking away the area under the curve through its whole domain (-1 to 3).
Wait wait, can we please have someone confirm the area from 2 to 3 being included in the required area? Talking to classmates it seems that this is not the case, however I did include it. So is 2-3 included in the area?
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Wait wait, can we please have someone confirm the area from 2 to 3 being included in the required area? Talking to classmates it seems that this is not the case, however I did include it. So is 2-3 included in the area?
I just sketched the area on my CAS and it included from x=2 to x=3 in the area
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Wait wait, can we please have someone confirm the area from 2 to 3 being included in the required area? Talking to classmates it seems that this is not the case, however I did include it. So is 2-3 included in the area?
I'm under the impression it does not include that area.
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I checked the suggested answers and I think I got around 32/40. what grade do you think that would be? also if you found Q10 " v in terms of u" by equating gradients of triangles, did u get a different answer then the solution shows?
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But wouldn't that technically mean it wasn't 'enclosed' between the graph of f(x) and the line y=4? Shouldn't that have included the line x-3 then? :-\
... I don't know where you're getting this x-3 from. The question is asking for the area enclose by the graph and the line y=4 - there is some area between the curves along the interval (2,3), so I would say you should include it.
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I checked the suggested answers and I think I got around 32/40. what grade do you think that would be? also if you found Q10 " v in terms of u" by equating gradients of triangles, did u get a different answer then the solution shows?
4u/(u-2) and 4+8/(u-2) are both correct answers.
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Dont think A+ would go below 36
:-\
Nahh, that's what it was last year - and that exam was a lot easier, I reckon it'll be below 36 - but not far below since anyone who had done ex2 from 2013 could have done the last question - it was q9 that would've stumped people a bit i reckon probably a 35 will be an A+
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I checked the suggested answers and I think I got around 32/40. what grade do you think that would be? also if you found Q10 " v in terms of u" by equating gradients of triangles, did u get a different answer then the solution shows?
I equated the gradients, and it turned out the same as the solutions show. You should be fine, given you didn't make and arithmetic error(s)
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If the correct answer was
Antiderivative (4-(3x^2-x^3)dx.....=27/4
And I did (2-(3x^2-x^3)dx.....=3/4 (I accidentally put 2 instead of 4)
On a 3 mark question, how many will 1 lose?
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... I don't know where you're getting this x-3 from. The question is asking for the area enclose by the graph and the line y=4 - there is some area between the curves along the interval (2,3), so I would say you should include it.
If this is the case this question would have single digit percentage for full marks in the assessors report.
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I checked the suggested answers and I think I got around 32/40. what grade do you think that would be? also if you found Q10 " v in terms of u" by equating gradients of triangles, did u get a different answer then the solution shows?
Answer to end all "will I get marks?" question:
If you DO NOT SHOW SUFFICIENT WORKING, you WILL lose marks.
If you DO SHOW SUFFICIENT WORKING, BUT USE A DIFFERENT METHOD, you WILL NOT lose marks.
There will always be multiple ways to approach these questions, and for half of the questions chuck and myself have used different methods to get the answers. There is NOTHING wrong with this.
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If the correct answer was
Antiderivative (4-(3x^2-x^3)dx.....=27/4
On a 3 mark question, how many will 1 lose?
We don't even know if this is the correct answer
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For finding the area under the line y=4, I intergrated (4 - f(x)) from -1 to 2.
Is this correct?
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... I don't know where you're getting this x-3 from. The question is asking for the area enclose by the graph and the line y=4 - there is some area between the curves along the interval (2,3), so I would say you should include it.
I think that the integral on the domain [2,3] shouldn't be included, since it's not enclosed by the graph
For example, if the question instead asked you to find the area bound by the graph of f and the x-axis, the required integral would be from 0 to 3, if i'm not mistaken, and the -1 bit would not be included - the question here is no different in wording
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Whoops, I missed that. .__. Yes, it should.
So the solutions have one mistake, unless chuck can explain himself.
EDIT: If anyone wants to agree with me, my answer is 8 - which was found by taking the area of the rectangle (16, so it's really a square) and taking away the area under the curve through its whole domain (-1 to 3).
It said the area bounded by y=4 and the graph and this is only between -1 and 2
From 2 to 3 the area has to also be bounded by x=3
So I think chuck is right
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We don't even know if this is the correct answer
'if the correct answer was'
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Yeah, this question is worded like crap. "Enclosed" is up to interpretation, since the domain is restricted and allows for the area in the interval (2,3) to be included within the calculation.
It's frustrating when a methods exam becomes an English vocabulary test. A diagram should have been used instead to tell us what the required area was. I'm calling ambiguous.
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doesn't it only include 2 to -1 because otherwise it would have to say enclosed by the x-axis aswell? no?
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It said the area bounded by y=4 and the graph and this is only between -1 and 2
From 2 to 3 the area has to also be bounded by x=3
So I think chuck is right
Except the graph doesn't exist past 3 - so the area enclosed between the curve and the line y=4 is bounded by that line.
I'll make sure to discuss this with a methods teacher, this question is starting to sound more famous than the the "this graph has equation (x-b)(x-a)" when b was negative fiasco.
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I just cannot believe they put the concept of the last question of the 2013 Exam 2 on 2014 exam 1. Percentage of students that got those final four questions on exam 2 last year was 6% 3% 3% and 2% respectively so to put the same concept on an exam one was pretty unfair I thought. Would have been a whole different story if 10bi was a show that question so you could at least attempt the rest.
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I'll make sure to discuss this with a methods teacher, this question is starting to sound more famous than the the "this graph has equation (x-b)(x-a)" when b was negative fiasco.
'graph of death'
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I just cannot believe they put the concept of the last question of the 2013 Exam 2 on 2014 exam 1. Percentage of students that got those final four questions on exam 2 last year was 6% 3% 3% and 2% respectively so to put the same concept on an exam one was pretty unfair I thought. Would have been a whole different story if 10bi was a show that question so you could at least attempt the rest.
Finding an area/distance function and minimising/maximising it is actually a fairly straightforward concept, and I felt they guided students through it very well.
The fact that it appeared on exam 2 last year AND was done so poorly is honestly more reason for them to put it in - they knew people would prepare for that kind of a question, so they made sure those who had the answers in their reference book could've take guesses with worked solutions from last year.
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GeniDoi - your signature "On the morning of Wednesday the 5th of October 2014, several seconds after the 15 minutes reading time concludes at 9:15am, a new record will be set for the most amount of students applying the chain rule at the same time in the southern hemisphere."
was the only thing i was thinking about when doing the first two questions on the exam. Haha
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Don't see why you would include 2-->3 in the area, thats not enclosed. It's open, no where did it say "and x=3". I agree it was a poorly worded question, but it was understood if you looked at the graph.
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If the correct answer was
Antiderivative (4-(3x^2-x^3)dx.....=27/4
And I did (2-(3x^2-x^3)dx.....=3/4 (I accidentally put 2 instead of 4)
On a 3 mark question, how many will 1 lose?
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For those saying the area from 2 to 3 is included in 5c due to the function being restricted, the line y=4 is not restricted, therefore the enclosed area is between -1 and 2, meaning 27/4 is the correct answer
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Don't see why you would include 2-->3 in the area, thats not enclosed. It's open, no where did it say "and x=3". I agree it was a poorly worded question, but it was understood if you looked at the graph.
It's enclosed by the domain of the cubic. At least I think.
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Yeah, even if the graph didn't exist for x>3, you KNOW that AT x =3 they don't touch, hence there is no more area enclosed by the two graphs
It didn't mention the x or y axis, it only mentioned the two graphs, if you were to calculate area from 2->3, you'd be using the two graphs and the x axis
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I say that its not enclosed therefore not included. But who knows with VCAA.
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I say that its not enclosed therefore not included. But who knows with VCAA.
Yeah, it's really up to VCAA's interpretation of this.
If they had said a few things extra, e.g. y=4 AND x=3, it might clear things up.
VCAA, why you be so ambiguous? (Someone should make a meme of this and post it here.)
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I'm so sorry, I'm really dumb because I never use forums, but... I can't find the suggested solutions someone posted on the first page. How do I get the attachment?
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Sorry, got it - I just hadn't logged in but was viewing it as a guest!
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I don't think it was ambiguous. If you draw y=4 and f over the right domain. The only enclosed area is between -1 and 2.
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So quick question. I left 2 parts in q9 not in their simplest form. I left part bi as 57/96 rather than 19/32 and bii as 45/57 rather than 15/19. Will this cost me marks? Besides that I found the paper fairly straightforward.
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Yeah, even if the graph didn't exist for x>3, you KNOW that AT x =3 they don't touch, hence there is no more area enclosed by the two graphs
It didn't mention the x or y axis, it only mentioned the two graphs, if you were to calculate area from 2->3, you'd be using the two graphs and the x axis
The x or y axis has nothing to do with it. It's just a coincidence that the right endpoint lies on the x axis.
Imagine if the function was restricted to not include the maximum TP that lies at (2,4). Eg, the domain is (-1,1). In this case, I'm sure everyone would still calculate the area between -1 to 1, despite the are not being "bounded". This is the exact same scenario, except we are calculating the area in the local max/min and the (2,3) area.
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So quick question. I left 2 parts in q9 not in their simplest form. I left part bi as 57/96 rather than 19/32 and bii as 45/57 rather than 15/19. Will this cost me marks? Besides that I found the paper fairly straightforward.
I did the same. Thought I got it wrong but you have just alerted me to the fact that it was just unsimplified!
Cheers.
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I did the same. Thought I got it wrong but you have just alerted me to the fact that it was just unsimplified!
Cheers.
They are generally really picky about it so it may cost you marks. Really depends on the
assessors marking your paper (ie luck)
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So quick question. I left 2 parts in q9 not in their simplest form. I left part bi as 57/96 rather than 19/32 and bii as 45/57 rather than 15/19. Will this cost me marks? Besides that I found the paper fairly straightforward.
I'm sorry, I think this will cost you marks due to the simple fact that not many students will have done it - they generally only award equivalent answers when NOT doing so would fuck up their grade distribution. In this case, I don't even know how you got the answer /96 in the first place - I think most people would have had it in its simplest form straight off.
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They are generally really picky about it so it may cost you marks. Really depends on the
assessors marking your paper (ie luck)
Yes depends on how the rest of the state go depending on how picky they are. I assume I would lose at least 1 out of 4 minimum for not simplifying. Could quite possibly be 2 out of 4 though. Really just depends.
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The x or y axis has nothing to do with it. It's just a coincidence that the right endpoint lies on the x axis.
Imagine if the function was restricted to not include the maximum TP that lies at (2,4). Eg, the domain is (-1,1). In this case, I'm sure everyone would still calculate the area between -1 to 1, despite the are not being "bounded". This is the exact same scenario, except we are calculating the area in the local max/min and the (2,3) area.
There is nothing stopping it, they would have said "and x=3". 27/4 is the only enclosed area.
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(http://i61.tinypic.com/29y6hi.png)
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Yo guys, so I asked for another booklet and the supervisor told me that I could do my workings (for the last question) on the following page which was, of course, completely blank
Is this alright or am I stuffed?
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So it looks like my score is 30-32/40 (can't completely remember my answers). Do you guys think that this will fall into the A range?
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Yo guys, so I asked for another booklet and the supervisor told me that I could do my workings (for the last question) on the following page which was, of course, completely blank
Is this alright or am I stuffed?
The markers get the whole booklet in it's raw form. It'll be fine :)
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It was a question I just realised at the end I had stuffed up so in the rush I somehow didn't figure that 3/24 was 1/8. Oh well besides that I only lossed one other mark on the last part of the last question because I used the wrong endpoint.
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So it looks like my score is 30-32/40 (can't completely remember my answers). Do you guys think that this will fall into the A range?
If I'm not wrong, according to last year 30-31 would've been a B+ , 32 would be A
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There is nothing stopping it, they would have said "and x=3". 27/4 is the only enclosed area.
Would I recieve consequential marks for using incorrect terminals, but having correct integration and area (even though it's incorrect?)
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The x or y axis has nothing to do with it. It's just a coincidence that the right endpoint lies on the x axis.
Imagine if the function was restricted to not include the maximum TP that lies at (2,4). Eg, the domain is (-1,1). In this case, I'm sure everyone would still calculate the area between -1 to 1, despite the are not being "bounded". This is the exact same scenario, except we are calculating the area in the local max/min and the (2,3) area.
They wouldn't ask you to do that because it is not valid, nothing is enclosed then
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I wrote my integrals from -1 to 3 and got the answer 8. That said the answer seems to be 27/4 so I am wondering how many marks I would get for my working. I'm thinking either one or two because I identified 4-f(x) and then anti diffed correctly. Despite my terminals and answer being wrong, I am pretty optimistic that I will still get 2 marks.
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If I'm not wrong, according to last year 30-31 would've been a B+ , 32 would be A
[/quote]
Okay thanks. Hopefully it's an A :)
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Would I recieve consequential marks for correct integration and area (even though it's incorrect?)
Well, two marks must be for the setup and answer, which you obviously wouldn't get. So I think the 3rd mark must be for the integration, which you would get. So probably 1/3.
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This seems straightforward, but where are the two marks on the very first question allocated? I wrote just the answer - was any more working required? Thanks! :)
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Could someone confirm this for me please..!!! :(
For question 5c... I did something completely different from the suggested solutions but somehow I got the same answer?
I assumed its symmetrical so I said the area enclosed is the same as the area under the curve from 0 to 3...... but yeh, somehow it gave me 27/4 as well.. which is the same as what everyone else got?
Will I get any mark at all?....I think what I did was wrong but somehow got the right answer? hahahahaha
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This seems straightforward, but where are the two marks on the very first question allocated? I wrote just the answer - was any more working required? Thanks! :)
Yes - you also should've shown some some indication of using the product rule.
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Awesome only lost last mark haha, but isn't 1b +-1/2......its the derivative equation of a hyperbola and no restriction on range?
That's what I got. Can anyone clarify whether +-1/2 is accepted?
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Can we lose marks for not simplifying?...
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Could someone confirm this for me please..!!! :(
For question 5c... I did something completely different from the suggested solutions but somehow I got the same answer?
I assumed its symmetrical so I said the area enclosed is the same as the area under the curve from 0 to 3...... but yeh, somehow it gave me 27/4 as well.. which is the same as what everyone else got?
Will I get any mark at all?....I think what I did was wrong but somehow got the right answer? hahahahaha
No you're 100% right! It is symmetrical! Nice solution, VCAA will froth over that - very elegant. (As long as you specified that it was symmetrical on the exam)
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Can we lose marks for not simplifying?...
Yes for numerical simplification, no for algebraic simplification.
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It said the area bounded by y=4 and the graph and this is only between -1 and 2
From 2 to 3 the area has to also be bounded by x=3
So I think chuck is right
I agree.
Did anyone else do 4x4 - the integral between -1 and 3?
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That's what I got. Can anyone clarify whether +-1/2 is accepted?
I'd say no, because the question defines you to take the positive root in the initial function. Looking at the function, that slope has a positive gradient, so your answer should be positive.
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This seems straightforward, but where are the two marks on the very first question allocated? I wrote just the answer - was any more working required? Thanks! :)
No it'll be fine. They're highly valued because they're easy: VCAA needs to try to get a 20/40 mean somehow hahaha
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Yes for numerical simplification, no for algebraic simplification.
What............................
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That's what I got. Can anyone clarify whether +-1/2 is accepted?
The original graph was just a square root graph so te derivative would only be half a hyperbola wouldnt it?
Worry not! U couldve done something way stupider. like me and change 5in2 to in25.... :(
Also stuffed up the last one makrer and made e^0 equal 0 instead of 1. Gonna chop my self.
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What............................
They like you to simplify fractions, surds and logs but not to factorise expressions or rationalise surds. So like 1/sqrt(2) is acceptable, 2/4 is not, sqrt(12) is not, x^2+x^3 is acceptable, -ln(1/2) is not, etc.
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No you're 100% right! It is symmetrical! Nice solution, VCAA will froth over that - very elegant. (As long as you specified that it was symmetrical on the exam)
REALLY? friends keep telling me it's not symmetrical so I've been stressing over it. I'm glad it's correct, and yeah I did state that it was symmetrical on the paper.
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I had another solution for the infamous 5c. I took the integral of the function-4 and flipped the terminals to get a positive area.
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What............................
You lose marks for not turning 8/32 into 1/4
You don't lose marks if you leave 3x(x+2) as your answer, as opposed to having left it as 3x^2 + 6x
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What............................
My teacher marked exam one last yr and saud that they only penalised one question for lack of simplification. Tbh probably the probability qs.
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For the sake of hypotheticals, is it still theoretically possible to get a 45+ raw with an A for the exam 1? (I think I got somewhere in the range of 36-38 which is an A in some years)
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They like you to simplify fractions, surds and logs but not to factorise expressions or rationalise surds. So like 1/sqrt(2) is acceptable, 2/4 is not, sqrt(12) is not, x^2+x^3 is acceptable, -ln(1/2) is not, etc.
So if I did ln(32) instead of 5ln(2) would that be okay?
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REALLY? friends keep telling me it's not symmetrical so I've been stressing over it. I'm glad it's correct, and yeah I did state that it was symmetrical on the paper.
No don't worry about them! It's totally symmetrical about (1,2) - you just can't see it on my hideous graph :'( It's actually a really beautiful way of doing it - VCAA may have even intended it to be done that way. I highly applaud you, and kick myself for not seeing it :P
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If the correct answer was
Antiderivative (4-(3x^2-x^3)dx.....=27/4
And I did (2-(3x^2-x^3)dx.....=3/4 (I accidentally put 2 instead of 4)
On a 3 mark question, how many will i lose?
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If the correct answer was
Antiderivative (4-(3x^2-x^3)dx.....=27/4
And I did (2-(3x^2-x^3)dx.....=3/4 (I accidentally put 2 instead of 4)
On a 3 mark question, how many will i lose?
2.
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2.
1 for wrong answer and 1 for wrong equation yeah?
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So if I did ln(32) instead of 5ln(2) would that be okay?
Generally, no. However, they make exceptions depending on how many other students didn't simplify. Basically, the examiners all have a meeting where they mark a few papers together to get a feel for how students went and the different methods they used. If a sufficient number of people wrote ln(32) they will accept it for the sake of the grade distribution. However, I feel that it is unlikely in this case because it's not especially complex and given that the 2 was there originally - you needed to raise 2^5 to get the 32 in the first place (i.e. you overcomplicated, rather than neglecting to simplify). In the exam last year on Question 10 they accepted all variations of the solution, but that was a long expression with hideous fractions and logs - so it's a different scenario.
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Nah fuq da h8ers. It's totally symmetrical about (1,2) - you just can't see it on my hideous graph :'( It's actually a really beautiful way of doing it - VCAA may have even intended it to be done that way. I highly applaud you, and kick myself for not seeing it :P
Haha, when I saw all these complex workings that everyones been posting I was so scared. But yeah, lucky it's right... I'm still sad for getting a few other questions wrong due to silly mistakes though, there goes my chance for a 40. Hopefully, I can do better on exam 2 tmr.
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1 for wrong answer and 1 for wrong equation yeah?
Yes :)
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Did anyone else get the correct answer for 10bi using Pythagorus rather than similar triangles?
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They like you to simplify fractions, surds and logs but not to factorise expressions or rationalise surds. So like 1/sqrt(2) is acceptable, 2/4 is not, sqrt(12) is not, x^2+x^3 is acceptable, -ln(1/2) is not, etc.
Does this apply to exam 2 as well?
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Hey all,
regarding Question 1.b), I forgot to sub in 1 all together. It was a 3 mark question. How many marks will I lose if I differentiated correctly?
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Hey all,
regarding Question 1.b), I forgot to sub in 1 all together. It was a 3 mark question. How many marks will I lose if I differentiated correctly?
You would have lost 2 marks, one for correct substitution into the derivative and one for the answer.
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Did anyone else get the correct answer for 10bi using Pythagorus rather than similar triangles?
Hypothetically you could have used Pythagorus' theorum… can you post what you did?
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Hypothetically you could have used Pythagorus' theorum… can you post what you did?
I have no idea, I tried to work it out again but couldn't. I definitely got the same answer though.
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I have no idea, I tried to work it out again but couldn't. I definitely got the same answer though.
Presumably if you got the same answer, you're fine.
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Hi guys,
Would I score a method mark for 9.b.ii) for writing the formula to find the probability but not subbing in any values?
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Hi guys,
Would I score a method mark for 9.b.ii) for writing the formula to find the probability but not subbing in any values?
Yes.
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Presumably if you got the same answer, you're fine.
Not necessarily: remember that comment on the 2013 Specialist Maths Exam 1 report? "There were several cases where incorrect working fortuitously led to a correct answer. Students should be reminded that in such cases, the final answer mark will not be awarded if it is not supported by relevant and correct working."
Just curious: did you follow the sort of reasoning I show in the attached photo? (just the 1st step)
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I got 27/40
and 80+ % in all my SACS
is it still possible to get 35 raw study score?
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Not necessarily: remember that supremely bitchy comment on the 2013 Specialist Maths Exam 1 report? "There were several cases where incorrect working fortuitously led to a correct answer. Students should be reminded that in such cases, the final answer mark will not be awarded if it is not supported by relevant and correct working."
Well there goes Q10bii... made a logical guess at 4 and therefore the area. Thought I'd still get an answer mark
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I can't believe it. Using the solutions that Chuck posted, I got about 13/40!
What do you think that will net me?
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Well there goes Q10bii... made a logical guess at 4 and therefore the area. Thought I'd still get an answer mark
Have a look at the photo I posted, is it similar to what you did? I don't think what you did was wrong, per se, I just can't definitively say it was right until I understand what you did :P
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Think I got 33/40, made silly mistakes but overall this exam was alright, required a bit more thinking than past exam 1s
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I got 27/40
and 80+ % in all my SACS
is it still possible to get 35 raw study score?
It depends on how the rest of ur cohort did aswell. Say u go to a very competitive school abd everyone else aces the exam all ur sacs will get scaled aswell. And you also have exam two to go!! Theres still a possibility as long as u dont give up!!
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I can't believe it. Using the solutions that Chuck posted, I got about 13/40!
What do you think that will net me?
Are you sure you're counting all the method marks properly? If you did get 13, that's been pretty consistently in the middle of the D+ band, although last year it was at the top of the D range. Forget about it, relax, and focus on tomorrow :)
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Have a look at the photo I posted, is it similar to what you did? I don't think what you did was wrong, per se, I just can't definitively say it was right until I understand what you did :P
I wasn't the one who used Pythag Chuck :) Was just a random comment :)
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got 33-35 for this exam
about what mark do i need tomorrow for a 40+?
rough estimate
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hey guys,
for part 2 of the last question, If I found the wrong value of v in terms of u, on the second question would I get a mark for finding the area by subbing in the wrong value of v?
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I think I got about 35/36 (idek) out of 40and like I'm stressing so much now since so many people seem to have gotten 100% you know, my SAC marks are around 100% scaled, so if I did well on Exam 2, is there any chance at all for like 46/47? :((
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Here are my preliminary solutions guys :)
I didn't check them, so I may have a few mistakes, but based on conversations with guys who did the exam they seem correct.
P.S. I have a blank copy of the exam too, so I'll upload that next.
Edit: I can now confirm my solutions are correct, except the graph is a bit revolting: I think it should go through (1,2) to get full marks.
Hey Chuck, can I ask how you knew to use OVU and PQU as similar triangles, not RVQ and PQU? Thanks. :)
EDIT: I'm talking about q10bi (as you've probably realised)
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Generally, no. However, they make exceptions depending on how many other students didn't simplify. Basically, the examiners all have a meeting where they mark a few papers together to get a feel for how students went and the different methods they used. If a sufficient number of people wrote ln(32) they will accept it for the sake of the grade distribution. However, I feel that it is unlikely in this case because it's not especially complex and given that the 2 was there originally - you needed to raise 2^5 to get the 32 in the first place (i.e. you overcomplicated, rather than neglecting to simplify). In the exam last year on Question 10 they accepted all variations of the solution, but that was a long expression with hideous fractions and logs - so it's a different scenario.
I've never seen in the assessors reports for either methods or spesh where they penalise students for not applying log laws to change the final answer to look more elegant. In32 = 5In2
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Hey Chuck, can I ask how you knew to use OVU and PQU as similar triangles, not RVQ and PQU? Thanks. :)
EDIT: I'm talking about q10bi (as you've probably realised)
I could have used RVQ and PQU, or even RVQ and OVU - it doesn't matter, they're all similar. The reason I chose the pair I did was because OVU had v by itself and PQU didn't have a v, which made it only 1 step to simplify :)
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Random question can someone please tell me if, on 2011's exam 2, the co-ordinate for the endpoint of the parabola (-1, 0) is an error? I thought it would be (0, -1)...maybe I'm losing my mind.
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I've never seen in the assessors reports for either methods or spesh where they penalise students for not applying log laws to change the final answer to look more elegant. In32 = 5In2
Well, to be fair, it hasn't rally come up all that often. The last time I can remember there was a (numerical) log in an answer was 2013, and I explained that case. This is just based off what the chief assessor said when they came to my school, but I could very well be wrong.
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I could have used RVQ and PQU, or even RVQ and OVU - it doesn't matter, they're all similar. The reason I chose the pair I did was because OVU had v by itself and PQU didn't have a v, which made it only 1 step to simplify :)
Ah gotcha, thank you! :)
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Here are my preliminary solutions guys :)
I didn't check them, so I may have a few mistakes, but based on conversations with guys who did the exam they seem correct.
P.S. I have a blank copy of the exam too, so I'll upload that next.
Edit: I can now confirm my solutions are correct, except the graph is a bit revolting: I think it should go through (1,2) to get full marks.
Hey Chuck, could you please post your solutions and your scanned exam as a separate thread so I can sticky it? Thanks :)
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I have B+ SACs in a medium-strong cohort, and got 31-33 for exam 1. Roughly what would I need to get for exam 2 to get around a 35 SS?
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Hey Chuck, could you please post your solutions and your scanned exam as a separate thread so I can sticky it? Thanks :)
Done. Thanks ninwa :)
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Done. Thanks ninwa :)
Thanks! Appreciate your efforts - and gave you a shout out on the Facebook page :)
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Hi everyone sorry if this sounds like a really obnoxious question, but what would be the highest number of marks you would need to get on exams 1 and 2 to get a ss of 48+?
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Hi everyone sorry if this sounds like a really obnoxious question, but what would be the highest number of marks you would need to get on exams 1 and 2 to get a ss of 48+?
For last year it was somewhere in the range of 114/120
It varies every year, to be truly safe, somewhere like 118/120 would guarantee you 48+
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For last year it was somewhere in the range of 114/120
It varies every year, to be truly safe, somewhere like 118/120 would guarantee you 48+
What about for a 40+?
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i got 19/40 for exam 1 :( what would i need for get out of 80 on exam 2 to get at least a 30ss?
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I think I got about 35/36 (idek) out of 40and like I'm stressing so much now since so many people seem to have gotten 100% you know, my SAC marks are around 100% scaled, so if I did well on Exam 2, is there any chance at all for like 46/47? :((
well some guy last year got
100/100 sacs
36/40
72.5/80
45 SS
so yeah its possible to get 46/47
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If you write the wrong equation (similar to what it should be but 1 number accidentally different
and work correctly to get an answer that is correct for your working, do you get the answer mark or not?
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t'was a day for comical errors, indeed. Somehow in my stress I calculated the endpoint in 5.b as (-1, 2) instead of
(-1,4) and that means 5c is screwed as well =////////
And for q6 I rooted both sides instead of squaring both sides in the last step, ffs. I hope I get some mercy from the examiners.
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So i checked old threads about simplifying numerical values.. apparently it depends on each year to whether the exam was hard or not. So do you think i would penalised for not simplifying 10/12 --> 5/6?? :(
Do we also need to simplify in exam 2
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t'was a day for comical errors, indeed. Somehow in my stress I calculated the endpoint in 5.b as (-1, 2) instead of
(-1,4) and that means 5c is screwed as well =////////
And for q6 I rooted both sides instead of squaring both sides in the last step, ffs. I hope I get some mercy from the examiners.
I did the same thing for Q6, just had a brainfart =/.
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I made a stupid arithmetic error in the probably question giving me 2/3 instead of 19/34 so I know I lost that mark. If I used 2/3 in the conditional probability question would I still get at least 1 mark?
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Just compared my answers to the answers in the other thread. I am laughing so hard right now. I got 4/40. Thank god I stopped caring about this subject midway through the year. Pumped for my 5 SS now :D
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Just compared my answers to the answers in the other thread. I am laughing so hard right now. I got 4/40. Thank god I stopped caring about this subject midway through the year. Pumped for my 5 SS now :D
same here bud ;D ;D ;D ;D ;D ;D ;D ;D ;D ;D ;D ;D ;D ;D ;D ;D ;D ;D ;D
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I'm so confused, like i thought all 'sac' marks that you get are external (come from an exam), so like if you were ranked second in sacs, youd get the second highest exam mark as your sac mark - isnt that how it works, like even if you aced your sacs, if the highest exam mark wasn't 100, doesn't that mean your sac score is smaller than 100?
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t'was a day for comical errors, indeed. Somehow in my stress I calculated the endpoint in 5.b as (-1, 2) instead of
(-1,4) and that means 5c is screwed as well =////////
And for q6 I rooted both sides instead of squaring both sides in the last step, ffs. I hope I get some mercy from the examiners.
I did the same thing for Q6, just had a brainfart =/.
Me three for Q6, don't know what was going through my head :l
Other than that, the last question was awful, only until there was like 2 minutes before the end that I realised I had to use similar triangles but by then it was too late to finish the question :c
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I'm so confused, like i thought all 'sac' marks that you get are external (come from an exam), so like if you were ranked second in sacs, youd get the second highest exam mark as your sac mark - isnt that how it works, like even if you aced your sacs, if the highest exam mark wasn't 100, doesn't that mean your sac score is smaller than 100?
Yes, that's how it works as far as I know. Getting a 100/100 sac score just guarantees that you topped your class and thus got your own exam marks as sac scores unless someone in your class beat you (in which case you get their exam marks as sac scores). So for the example posted above the guy didn't actually get 100% for sac scores unless someone else in his class aced both exams.
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I'm so confused, like i thought all 'sac' marks that you get are external (come from an exam), so like if you were ranked second in sacs, youd get the second highest exam mark as your sac mark - isnt that how it works, like even if you aced your sacs, if the highest exam mark wasn't 100, doesn't that mean your sac score is smaller than 100?
vcaa gets your internally scaled sac marks. Highest being 100 for your cohort (ie. top rank for your school) and then theyll base the external scaling of your sacs on how well your cohort as a whole does on the exam. Assuming that if you all did really well you just had hard sacs and so youll get scaled up alot. If you all do poorly your sac will get scaled down assuming you had easy sacs. So it really is team work haha. But ofcourse there are gonna be outliers so dont worry too much as long as you do well on younr exam. They try and make it as fair as possible. (All heard from me teachaaaar dont know if its actually tru)
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i spent so much time on this subject but i lost everything to careless mistakes :'( :'(
is there a chance i can get 45+ss if i got 32-34/40??? :-\
(in a strong cohort so likely 95-100% scaled sac scores)
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Here are my preliminary solutions guys :)
I didn't check them, so I may have a few mistakes, but based on conversations with guys who did the exam they seem correct.
P.S. I have a blank copy of the exam too, so I'll upload that next.
Edit: I can now confirm my solutions are correct, except the graph is a bit revolting: I think it should go through (1,2) to get full marks.
You are my hero! My hero whi has pointed out that I just scraped past 50%... but a hero nonetheless!
Get pumped for tomorrow's exam everyone!
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Think i got 38/40 and was a silly mistake from 40/40
Big big shame considering theres no way ill get A+ in exam 2, i needed to make the most of exam 1 :(
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can anyone explain question 6 from paper 1? Even though its done already, I still don't get the worked solution.
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Here are my preliminary solutions guys :)
I didn't check them, so I may have a few mistakes, but based on conversations with guys who did the exam they seem correct.
P.S. I have a blank copy of the exam too, so I'll upload that next.
Edit: I can now confirm my solutions are correct, except the graph is a bit revolting: I think it should go through (1,2) to get full marks.
Thank you so much for this! :D Feel like I did pretty well... good luck everyone for tomorrow!!
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can anyone explain question 6 from paper 1? Even though its done already, I still don't get the worked solution.
Logex-logex^1/2=3
Logex-1/2logex=3
1/2logex=3
Logex=6
e^6=x
Does it make more sense like that? :D
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Logex-1/2logex=3
1/2logex=3
Does it make more sense like that? :D
sorry. I'm still unsure what happened between these two steps
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sorry. I'm still unsure what happened between these two steps
Thats okay! Think of logex as one term like a and then if u take it out as a common its 1-1/2 which is 1/2
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sorry. I'm still unsure what happened between these two steps
Let ln(x)=a,
ln(x)-(1/2)ln(x)=3
a-(1/2)a=3
(1/2)a=3
(1/2)ln(x)=3
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thanks I get it now once you let a = loge^(x)
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i deadset have lost 14+ marks on this exam, all my bloody effort throughout the year has gone down the drain. stuff this shit, dont even want to think about tomorrows exam at all
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Well I barely passed 50%, if it is still possible, how much would I be looking to get for exam 2 if I'm trying to get a study score of 33+? --
Thanks
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The first 7 or 8 questions seemed fairly normal to me. Questions 9 and 10 required a bit more deeper level thinking but judging by the suggested solutions I have full marked exam 1. Now all I need is a solid performance in exam 2 to back it up :)
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I got 80+ % on all my SACS, in a pretty strong cohort...
got 26-27/40 for exam 1
say I get 55/80 for exam 2
what study score am I looking at?
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Will I lose points because I didnt line up the equals (=) symbols
i.e..
blah blah blah = something else
something else = another thing and a bit
= x+ y
I was told to always line up the = signs but I was rushing and as far as I can remember, I didn't line them up on at least two of the questions (exam brain kicked in where people tend to make silly mistakes)
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t'was a day for comical errors, indeed. Somehow in my stress I calculated the endpoint in 5.b as (-1, 2) instead of
(-1,4) and that means 5c is screwed as well =////////
And for q6 I rooted both sides instead of squaring both sides in the last step, ffs. I hope I get some mercy from the examiners.
rooting both side is and easy mistake, don't worry I did it to and I have no idea why, I guess the exam nerves got to me anyway :)
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Will I lose points because I didnt line up the equals (=) symbols
i.e..
blah blah blah = something else
something else = another thing and a bit
= x+ y
I was told to always line up the = signs but I was rushing and as far as I can remember, I didn't line them up on at least two of the questions (exam brain kicked in where people tend to make silly mistakes)
That's... An incredibly stupid reason to lose marks. I think you'll be fine.
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That's... An incredibly stupid reason to lose marks. I think you'll be fine.
I know it is a stupid reason but at my previous school, we would lose marks on SAC's and tests if we didn't.
Fingers crossed the exam markers aren't as picky with such small things..
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How many students actually get 100% (40/40 and 80/80) on their methods exams?
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How many students actually get 100% (40/40 and 80/80) on their methods exams?
Somewhere between none and all of them.
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Somewhere between none and all of them.
Looooool
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I got 80+ % on all my SACS, in a pretty strong cohort...
got 26-27/40 for exam 1
say I get 55/80 for exam 2
what study score am I looking at?
Modify message
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I got 80+ % on all my SACS, in a pretty strong cohort...
got 26-27/40 for exam 1
say I get 55/80 for exam 2
what study score am I looking at?
Modify message
None of us know, hence the lack of response to this. Most likely from 33-36 ish depending on how others find the exams. But then again I'm not completely sure.
People may flame me for my opinion on the exam, but tbh, I really didn't think it was an exam with any tricks to it. I felt lots of the questions could have analogs found in textbooks and it was really a matter of whether you read the question properly and whether you made a stupid mistake.
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None of us know, hence the lack of response to this. Most likely from 33-36 ish depending on how others find the exams. But then again I'm not completely sure.
People may flame me for my opinion on the exam, but tbh, I really didn't think it was an exam with any tricks to it. I felt lots of the questions could have analogs found in textbooks and it was really a matter of whether you read the question properly and whether you made a stupid mistake.
I agree. Fairly standard exam. Nothing novel. 2 stars - wouldn't do again.
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^And I still struggled LOL
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i know you are all gonna hate me for saying this, but ive got my fingers crossed for an extremely hard exam 2
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I got 80+ % on all my SACS, in a pretty strong cohort...
got 26-27/40 for exam 1
say I get 55/80 for exam 2
what study score am I looking at?
Modify message
its pretty impossible to determine how well your cohort is in comparison. for all we know you could be getting really easy sacs in comparison to other schools which is why it seems like a strong cohort. ya feeelll. nonetheless if you do have a strong cohort im sure you can still get mid 30s atleast worry not you will land on your feet! (if ur a cat)
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i know you are all gonna hate me for saying this, but ive got my fingers crossed for an extremely hard exam 2
me too :-* so i wont feel as bad about losing 79 marks on stupid mistakes if id have lost them anyway.
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I know exam marks can be "pulled up" if your GAT score was much better than your exam score but can your SAC results pull down your exam results?
I felt like this exam was much easier than any of the sacs i have completed this year (still receiving fairly average/good sac results) and going by the suggested answers, I have potentially only lost about 3 marks on the first methods exam. I know i still have to factor in tomorrows exam but what would happen if my exam results were 15% (say 95%) better than my sac results (all around 80%)
Would they lower my exam mark because of my not so good sac marks?
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That should be fine!
how about -5ln(1/2) its the same thing just didn't apply .5 to power of -1 which is 2
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how about -5ln(1/2) its the same thing just didn't apply .5 to power of -1 which is 2
yeah that's all right, past vcaa assessment reports have had it in both forms
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I know exam marks can be "pulled up" if your GAT score was much better than your exam score but can your SAC results pull down your exam results?
I felt like this exam was much easier than any of the sacs i have completed this year (still receiving fairly average/good sac results) and going by the suggested answers, I have potentially only lost about 3 marks on the first methods exam. I know i still have to factor in tomorrows exam but what would happen if my exam results were 15% (say 95%) better than my sac results (all around 80%)
Would they lower my exam mark because of my not so good sac marks?
nup exams are raw, dont change, stay da same, mhmmm 8) 8) 8) 8)
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Here are my preliminary solutions guys :)
I didn't check them, so I may have a few mistakes, but based on conversations with guys who did the exam they seem correct.
P.S. I have a blank copy of the exam too, so I'll upload that next.
Edit: I can now confirm my solutions are correct, except the graph is a bit revolting: I think it should go through (1,2) to get full marks.
Thanks for the solutions!
But I've always wondered, where do people get the exams from?
Especially to have the solutions up like not much after the exam finished.
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I know exam marks can be "pulled up" if your GAT score was much better than your exam score but can your SAC results pull down your exam results?
I felt like this exam was much easier than any of the sacs i have completed this year (still receiving fairly average/good sac results) and going by the suggested answers, I have potentially only lost about 3 marks on the first methods exam. I know i still have to factor in tomorrows exam but what would happen if my exam results were 15% (say 95%) better than my sac results (all around 80%)
Would they lower my exam mark because of my not so good sac marks?
GATs do...absolutely NOTHING...I swear...
People keep saying that GAT scores affect your English score. A friend of mine got 36 on the writing part of the GAT and got 49 for English. I got a 36 for the writing part of the GAT too and got a 50 in English Language. Correlation? Zilch. Unless maybe if you missed the exam for some reason.
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GATs do...absolutely NOTHING...I swear...
People keep saying that GAT scores affect your English score. A friend of mine got 36 on the writing part of the GAT and got 49 for English. I got a 36 for the writing part of the GAT too and got a 50 in English Language. Correlation? Zilch. Unless maybe if you missed the exam for some reason.
Okay, good because I am freaking out that if I have done as well as I think I have compared to my not so flash efforts during my sacs, that they are going to pull me up on the 15% improvement and somehow change my exam score to better relate to my sac scores.
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Okay, good because I am freaking out that if I have done as well as I think I have compared to my not so flash efforts during my sacs, that they are going to pull me up on the 15% improvement and somehow change my exam score to better relate to my sac scores.
if your GAT and exam marks don't correlate all they do is mark it a third time. likewise if your two first marks don't correlate they mark it a third time. What you get on the exam is what you got without any influence from sacs or GAT.
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Thanks for the solutions!
But I've always wondered, where do people get the exams from?
Especially to have the solutions up like not much after the exam finished.
I went to school with the intention of only wanting to see the exam in the morning before I did what I had to do there. Teachers get any exam 30mins after writing time, so I got it then while fetching it for one of my teachers and photocopied it. It was an easy paper (imo, sorry for the guys who found it hard) so I thought "what the heck" and decided to do it. The I did it in about 20mins and had the answers ready for kids coming out of the exams and like counsouled my friends before I uploaded it here :P
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I went to school with the intention of only wanting to see the exam in the morning before I did what I had to do there. Teachers get any exam 30mins after writing time, so I got it then while fetching it for one of my teachers and photocopied it. It was an easy paper (imo, sorry for the guys who found it hard) so I thought "what the heck" and decided to do it. The I did it in about 20mins and had the answers ready for kids coming out of the exams and like counsouled my friends before I uploaded it here :P
I'm not surprised. It was of comparable difficulty to the infamous 2011 Specialist Maths exam 1 in which the A+ cutoff was 38.5/40. I mean come on; you had like six full questions or something worth 2 marks each.
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I'm not surprised. It was of comparable difficulty to the infamous 2011 Specialist Maths exam 1 in which the A+ cutoff was 38.5/40. I mean come on; you had like six full questions or something worth 2 marks each.
Exactly. I think this was the easiest methods exam yet, unfortunately for everyone. Today will probably be a killer.
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It was of comparable difficulty to the infamous 2011 Specialist Maths exam 1 in which the A+ cutoff was 38.5/40.
Spesh Exam this year better not be like 2011. Hoping for something more like 2012, I make a crazy amount of mistakes when I do things by hand.
Anyway, I'm very interested as to what the exam 2 will be like today - it is apparently 5 questions according to exam cover so questions will be a bit shorter overall.
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Can your bound reference book just be a booklet of practice exams that you have done and marked yourself?
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Can your bound reference book just be a booklet of practice exams that you have done and marked yourself?
if it's all bounded correctly then I don't see why not
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Given the far greater number of students who do methods than spec (and the overall difference in maths ability) if an A+ was 38.5 for a comparable spec exam, wouldn't it basically be guaranteed to be lower than that in this year's exam 1?
And i don't know about the a+ cut off being higher than 36, since yesterday's exam wasn't hard - but it was definitely not the easiest exam 1 vcaa have done in the past
i dare say it was harder than all of vcaa's past ex 1 except for 2010 and a lot harder than last years, when an a+ was a 36
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Found the exam fairly straight forward. But unfortunately I messed up my dA/du, so I got a quadratic that never equalled zero :/ I rubbed that out and then just assumed the minimum was when u=6.... Hopefully I still get 1/2 for finding A(u).
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Don't stress so much! Remember, your SACs count more than that exam did.
I highly doubt it. Even so, as long as you did well at the first half and attempted the full second half to the best of your ability, the bell curve will fix it up. This exam certainly looks harder than last years.
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Don't stress so much! Remember, your SACs count more than that exam did.
I highly doubt it. Even so, as long as you did well at the first half and attempted the full second half to the best of your ability, the bell curve will fix it up. This exam certainly looks harder than last years.
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Don't stress so much! Remember, your SACs count more than that exam did.
I highly doubt it. Even so, as long as you did well at the first half and attempted the full second half to the best of your ability, the bell curve will fix it up. This exam certainly looks harder than last years.
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I highly doubt it. Even so, as long as you did well at the first half and attempted the full second half to the best of your ability, the bell curve will fix it up. This exam certainly looks harder than last years.
Your SACs are worth 34% of your mark - that exam was worth 22%.
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Your SACs are worth 34% of your mark - that exam was worth 22%.
This was necro'd from a year ago.