Methods Exam 1: Discussion!

[forchina]

I did the same wrong integration for 1/5e^-1/5 for question 8 and  b. Is that one mark or two marks off?

[Kaleidoscope]

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

[Sayf44]

Can i get an A+ with a 35?

[AngelWings]

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?

[keltingmeith]

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.

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.

[BLACKCATT]

Question 5c, does the area include the integral from 2 to 3 4-f(x) dx?

[Homer]

Dont think A+ would go below 36

[AngelWings]

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.

[keltingmeith]

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).

[Mieow]

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? 

[GeniDoi]

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?

[jonoz0r]

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

[jumcakes]

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.

[Velo]

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?

[keltingmeith]

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.