Maths Methods (Exam 1): Discussion, Questions & Potential Solutions

[zoefiggis]

yo if anyone has a copy of the exam could they please post how many marks each question was worth, trying to add up my marks
thanks!

[Lear]

It’s quite literally posted right above you.

[Dais_Deorum]

In the latter stages of the exam, I was running out of time and skipped a bit of important working to get to the answering. It was the right answer, but idk how strict the examiners are. What is the 'line' for when there is not enough working out?

[bea420]

yo if anyone has a copy of the exam could they please post how many marks each question was worth, trying to add up my marks
thanks!

hahah i see you zoe )

[georgehanna]

Hi Everyone
Any hope to find a copy of the MM Exam 1?

[Sine]

Hi Everyone
Any hope to find a copy of the MM Exam 1?

Sorry, from this year copies of VCAA exams will not be posted on ATAR Notes as these exams are subject to copyright.

Please have a read of the following post by K888 which reiterates this.
https://atarnotes.com/forum/index.php?topic=182246.0

[Mattjbr2]

My teacher confirmed
9ai: pi(4n+1)
9aii: -pi(4n+1).
Do you guys think that will be accepted? If not, will i at least get 1 mark of out 3 for the working?

[hums_student]

My teacher confirmed
9ai: pi(4n+1)
9aii: -pi(4n+1).
Do you guys think that will be accepted? If not, will i at least get 1 mark of out 3 for the working?

Hold up - every solutions thingo I've seen has +/-pi(2n+1) (also the answers I got)
Where did you get 4n from

[DinWell]

My teacher confirmed
9ai: pi(4n+1)
9aii: -pi(4n+1).
Do you guys think that will be accepted? If not, will i at least get 1 mark of out 3 for the working?

Yeah, I don't know why you're teacher confirmed those solutions, unless I'm missing something.

[Mattjbr2]

Yeah, I don't know why you're teacher confirmed those solutions, unless I'm missing something.

[WrongWong]

My teacher confirmed
9ai: pi(4n+1)
9aii: -pi(4n+1).
Do you guys think that will be accepted? If not, will i at least get 1 mark of out 3 for the working?

It would be 4n if n was any integer, but because the question explicitly states that n is even (part i) and n is odd (part ii) the correct answer is the one involving 2n.

I've done a mock of solutions here: https://drive.google.com/file/d/1hGcrpUJ8ZDRwrNhTRcALI2slH4exFqTL/view?usp=sharing

[Mattjbr2]

It would be 4n if n was any integer, but because the question explicitly states that n is even (part i) and n is odd (part ii) the correct answer is the one involving 2n.

I've done a mock of solutions here: https://drive.google.com/file/d/1hGcrpUJ8ZDRwrNhTRcALI2slH4exFqTL/view?usp=sharing

Thank you so much for typing that up. Absolute legend.

[DinWell]

It would be 4n if n was any integer, but because the question explicitly states that n is even (part i) and n is odd (part ii) the correct answer is the one involving 2n.

I've done a mock of solutions here: https://drive.google.com/file/d/1hGcrpUJ8ZDRwrNhTRcALI2slH4exFqTL/view?usp=sharing

Could you please explain how you'd get 4n if n was any integer? I was under the impression that you can't simplify it if they didn't explicitly say that it was odd or even.

[DinWell]

My teacher confirmed
9ai: pi(4n+1)
9aii: -pi(4n+1).
Do you guys think that will be accepted? If not, will i at least get 1 mark of out 3 for the working?

I'm curious as to how you arrived at 4n? Do you remember what you did?

[WrongWong]

Could you please explain how you'd get 4n if n was any integer? I was under the impression that you can't simplify it if they didn't explicitly say that it was odd or even.

If n was any integer and the question was still asking for the bounds to be even then odd (part i). Then we would have to integrate xsin(x) as:



As if k is an even integer, then k = 2n when n is an integer, if k is odd, k = 2n+1 where n is an integer.



We do the same process but with 2n+1 is as the lower bound and 2n as the upper bound for part ii resulting in -(4n+1) pi.

Mattjbr2's teach got it wrong