Methods Exam 1: Discussion!

[jumcakes]

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.

[Aman476]

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+

[theshunpo]

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)

[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?

[GeniDoi]

... 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.

[keltingmeith]

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.

[BLACKCATT]

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

[Sayf44]

For finding the area under the line y=4, I intergrated (4 - f(x)) from -1 to 2.

Is this correct?

[jumcakes]

... 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

[S61778]

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

[faredcarsking123]

We don't even know if this is the correct answer

'if the correct answer was'

[GeniDoi]

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.

[rosieee]

doesn't it only include 2 to -1 because otherwise it would have to say enclosed by the x-axis aswell? no?

[keltingmeith]

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.

[Jnf17]

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.