VCE Methods Question Thread!

[Qwerty1234qwerty]

Do we need to know how to solve cos(x)+cos(3x)=1/2 in exam 1?

No. You would have to use CAS

If a graph  has only a stationary point of inflection is it still considered to be one to one?

Yes
See https://mathbitsnotebook.com/Algebra2/Functions/FNOneOnto2.html

[minhalgill]

can we stop at the point where ive selected the text or do we have to continue till where they have, giving the answer as a simplified fraction?

[minhalgill]

What particularly are you struggling with? Recall that the average value is 1/x2-x1 multiplied the area under a function. Try applying it here.

i dont understand this forumla. where do we get it from? is it indepent of the shape of the graph? if it was a curve or any other graph would we calcualte it using this same formula?

[Bri MT]

i dont understand this forumla. where do we get it from? is it indepent of the shape of the graph? if it was a curve or any other graph would we calcualte it using this same formula?

The average value is the average y-value right? So if you got the graph and smoothed the area into a rectangle shape it would be the height of the rectangle.
The area of that rectangle by definition is the area under graph, or the integral of f(x)

To get the height of the rectangle, we divide the area by it's length.


Hope this helps

[S_R_K]

can we stop at the point where ive selected the text or do we have to continue till where they have, giving the answer as a simplified fraction?

I don't have a definitive answer, but I suspect that stopping at your point would be considered correct.

Firstly: you have written v in terms of u, and you have written the fraction in reduced form. There's no instruction to write your answer using a common denominator.

Secondly: if you continue reading the examiner's report for 10(b)(ii), one of the suggested answers involves using v = 8/(u-2) + 4, rather than the version with a common denominator. It would be odd for the examiners to suggest that you should use an "intermediate" step, rather than the correct final answer, from the previous question.

[sailinginwater]

Is the binomial theorem assessable in the methods study design? Has it ever appeared on past exams?

[Lear]

Is the binomial theorem assessable in the methods study design? Has it ever appeared on past exams?

It’s assessable and has appeared very often.

[sailinginwater]

It’s assessable and has appeared very often.

Do you know which exams it's appeared on?

[Lear]

Just have a look at last year's VCAA Exam 2 Question 3F

[sailinginwater]

2010 cone of death question methods exam 1
Are we likely to get consequential marks for part b and c if we got part a wrong?
It's not a "show that" question either

[Bri MT]

2010 cone of death question methods exam 1
Are we likely to get consequential marks for part b and c if we got part a wrong?
It's not a "show that" question either

I wouldn't count on consequential marks

[sailinginwater]

I wouldn't count on consequential marks

But part b and c are dependent on part a answer, and it isn't a "show that" either, so wouldn't it be unfair to not get consequentials?
It seems unfair that getting part a wrong automatically locks you out of part b and c

[darkz]

But part b and c are dependent on part a answer, and it isn't a "show that" either, so wouldn't it be unfair to not get consequentials?
It seems unfair that getting part a wrong automatically locks you out of part b and c

Well, yes, as students, we think its really unfair, but methods is a test of your mathematical abilities, so VCAA don't really care

[sailinginwater]

Methods exam 1 2010

2.b. find the antiderivative of 1(1-x) and evaluate it from 3 to 2

How would we find the antiderivative without using absolute values?

[darkz]

Methods exam 1 2010

2.b. find the antiderivative of 1(1-x) and evaluate it from 3 to 2

How would we find the antiderivative without using absolute values?

Just remember that absolute values were in previous study designs, and so the 2010 exam is no exception. With this being noted, a simple understanding of the absolute value concept is just all negative values -> positive. e.g. if you get |-3| -> 3