Methods Exam 1: Discussion!

[BLACKCATT]

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'

[keltingmeith]

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.

[theshunpo]

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

[DanielJ]

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.

[faredcarsking123]

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?

[EasyMoney]

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

[GeniDoi]

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.

[Aman476]

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

[ChazRose]

I say that its not enclosed therefore not included. But who knows with VCAA.

[AngelWings]

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.)

[Mrs_Honeythunder]

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?

[Mrs_Honeythunder]

Sorry, got it - I just hadn't logged in but was viewing it as a guest!

[de]

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.

[SeanC]

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.

[GeniDoi]

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.