VCE Methods Question Thread!

[DisaFear]

Hopefully this is right

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Yea, sorry about the last bit, died a little inside. But I'm sure a capable person like you could do that

(I really need to learn latex)

[dinosaur93]

lol, to be honest, it helped alot thx...

but remember the 'do the rest' bit, thaqts the bit where I dont really know how to do it by hand so yeah, thats unsure of how to go about with that hence is a crucial part for a non-tech SAC...

[DisaFear]

CAS confirmed. One of the signs in the lower workings is also wrong, forgot to fix it. Be wary

[link125]

Hi I was wondering if someone could help explain Question 3 of the Extended Response for Exam 2

http://www.vcaa.vic.edu.au/vcaa/vce/studies/mathematics/cas/pastexams/2011/2011mmcas2-w.pdf

Particularly: 3a ii and 3b

I asked my teacher today to explain it but he didn't make much sense...

Hopefully I have more luck here

[Lasercookie]

Hi I was wondering if someone could help explain Question 3 of the Extended Response for Exam 2

http://www.vcaa.vic.edu.au/vcaa/vce/studies/mathematics/cas/pastexams/2011/2011mmcas2-w.pdf

Particularly: 3a ii and 3b

I asked my teacher today to explain it but he didn't make much sense...

Hopefully I have more luck here

Q3a
ii. So you've found the derivative:

The questions is telling you for any value of x. In this case, it's all real numbers (since the domain of the original function is defined to be that) - in other words, you can put in any number and you'll get a number greater or equal to 5.

It's only a 1 mark question, so as an answer probably want a sentence just pointing out:
- , therefore

(e.g. , any other number you input will be 'something positive' + 5')

For 3b. you're given a general cubic.
i. For cubics, you can have 0, 1, 2 stationary points

I'm not sure of the best way to explain this - I'm also not sure if my explanation is entirely accurate, giving it my best shot:

I've set up this here to provide a visual example: https://www.desmos.com/calculator/iyhn6nstby
It's just the cubic, it's derivative, the quadratic discriminant and a bunch of sliders to change the a, b, c, k values

I assume you already know that cubics can have 1 or 2 stationary points. I also assume you know that stationary points occur when dy/dx = 0.

So for cubics with zero stationary points:

So we know that dy/dx is a quadratic
And that there's also this thing called the discriminant:

http://www.mathwords.com/d/discriminant_quadratic.htm

When this discriminant is less than 0, there are no real solutions for the derivative, and it is for these situations where the cubics will be the ones that have zero stationary points.

For an example of what it looks like,  try setting a = 2, b = 1, c = 1, k =1

So yeah, m = 0, 1, 2

ii. Well having down part i now, we have three possible cubics right?
Inverse functions only exist for 1:1 functions, so just take a look at our three possible answers and we can reject anything that isn't 1:1.

In this case, reject the one with 2 stationary points, leaving the answer as:
m = 0, 1


Edit: fixed error with calculating derivative 0_o :/

[pi]

I remember that question, good times... Learnt on that day during the exam that a cubic can have no stationary points LOL



btw, why are you doing the VCAA exam now? :O

[Phy124]

I remember that question, good times... Learnt on that day during the exam that a cubic can have no stationary points LOL



btw, why are you doing the VCAA exam now? :O

Still can't believe I wrote m = 1, 2 for i and m = 1 for ii, despite having a parabola with no intercepts in front of my eyes. The shit I do in Exams... *facepalm*

By the way, link125, you know it's April, yeah? 

[link125]

Thanks laser, really good explanations

LOL I know it's April, tell my Methods teacher that. On the holidays he wanted us to do as much as Exam 1 and 2 as we could. Now he's telling us we have a SAC this Friday with possible similar questions like that and others that are on the exams :S

Now I also know about desmos. Looks like a handy resource, so thanks for that as well

[dinosaur93]

1. Given that for could be:

a. ln

b. ln

c. ln

d.

e.



2. Find the derivative of by hand show all working..


3.Given that and a, b R, find the exact values of a and b.


4. Given , show that the derivative of y with respect to x is

A.

B.

C.

D.

E.

[kensan]

What method can I use to work out questions like these? I don't even know where to start lol
Find the exact derivative of      and   

[pi]

Tip (you'll never get this in VCAA btw), but take the derivative of to be and then apply the chain rule appropriately

[rife168]

What method can I use to work out questions like these? I don't even know where to start lol
Find the exact derivative of      and   

Just to clear up how VegemitePi got that result:






     you could leave it at this, but from the original question we know that





edit: I should've used instead of to make it look the same as what VegemitePi wrote, but it's exactly the same principle

[kensan]

awesome, thanks guys, appreciate the help
glad it's not in the exams haha

[brightsky]

awesome, thanks guys, appreciate the help
glad it's not in the exams haha

another way to look at it:

remember that a^x = e^ln(a^x) = e^(lna * x)
we know that e^(stuff) = d/dx(stuff) e^(stuff), so d/dx (e^(ln a * x)) = ln a e^(lna * x) which can be simplified to ln a (a^x).

[Deceitful Wings]

The lambert W function is defined to be the inverse of xe^x . In other words, if f(x) = x e^x then f^-1(x) = W(x). Note that the Lambert function is not a 1 to 1 function.

Sketch a graph of y=W(x) on a set of axes below, showing all important features. in particular, label all important points with their exact coordinates and label any asymptote with its equation.

I have no idea how to do this question... I tried doing it on the CAS and I attempted to find the inverse algebraically, but none of them seemed to work. Please help!