VCE Methods Question Thread!

[JellyBeanz]

I've been trying to do this question for like a week and I keep returning to it but I just can't seem to work it out! Could someone help me out please? ><

It's an integration by recognition problem. You have the derivative, so get the integral in the form of cos(PI*x/4) and then multiply by x since you know that Integral of x*f(x) over the given domain give E(X)

[TimGFranklin]

Can someone tell me if this is right, or even better where to find the 2016 sample exam solutions?

[JellyBeanz]

Can someone tell me if this is right, or even better where to find the 2016 sample exam solutions?

Yes that answer is correct, You can also find the sample solutions on itute.com

[MB_]

Can someone tell me if this is right, or even better where to find the 2016 sample exam solutions?

http://www.itute.com/wp-content/uploads/2016-VCAA-Mathematical-Methods-Sample-v2-April-Exam-1-Solutions.pdf

[blacksanta62]

I have a population of Xs. They can be red, or black. The population of 30 is shown below:

XXXXXXXXXX
XXXXXXXXXX
XXXXXXXXXX

Now from that population, I take a sample of 5 Xs.

XXXXX
Here, the proportion of red Xs is different to the proportion of red Xs in the population of 30, as I have only taken a sample of what the population is.

So the sample proportion is the proportion of a particular characteristic of a sample such as the sample of 5 from a population of 30 as above. The population proportion, on the other hand, is the proportion of the particular characteristic of the whole population, in this case of all 30.

Please note that samples can differ depending on where you take it. Like the above is only one sample of the 30 Xs. Another sample may have 2,3,4 or even all five of the red Xs.

I hope I make sense here.

So is the reason a larger sample more favourable because there's a better chance of that sample showing the population proportion?
Thank you

[eforlano]

Eek! Could someone show me worked solutions? For some reason I'm generally able to integrate by recognition but this one for some reason is really difficult!

[JellyBeanz]

Eek! Could someone show me worked solutions? For some reason I'm generally able to integrate by recognition but this one for some reason is really difficult!

I hope that's correct and you understand whats going on lol

[eforlano]

Thank you so much I finally get it!!

[blacksanta62]

With the attached question, whenever I solve them by matrix method (the determinant) I can never factorise and solve.
For the Q6, I end up getting:
(m-2) 3
2     (m-3)   
Cross multiply and get: (m^2 -5m +6) + 6 = 0
Then, I can't factorise, am I doing something wrong?

For Q5 I follow the same process and can't take out a factor of m when I get to the last step mentioned above. Could anyone who learnt the matrix method help me out please. Thank you

[MB_]

With the attached question, whenever I solve them by matrix method (the determinant) I can never factorise and solve.
For the Q6, I end up getting:
(m-2) 3
2     (m-3)   
Cross multiply and get: (m^2 -5m +6) + 6 = 0
Then, I can't factorise, am I doing something wrong?

For Q5 I follow the same process and can't take out a factor of m when I get to the last step mentioned above. Could anyone who learnt the matrix method help me out please. Thank you

The rule is ad-bc, you seem to be adding 6 on the end rather than subtracting it

[blacksanta62]

The rule is ad-bc, you seem to be adding 6 on the end rather than subtracting it

Legend! Thank you

[exit]

With the attached question, whenever I solve them by matrix method (the determinant) I can never factorise and solve.
For the Q6, I end up getting:
(m-2) 3
2     (m-3)   
Cross multiply and get: (m^2 -5m +6) + 6 = 0
Then, I can't factorise, am I doing something wrong?

For Q5 I follow the same process and can't take out a factor of m when I get to the last step mentioned above. Could anyone who learnt the matrix method help me out please. Thank you

ad-bc=0
Is the method allowed though? I learnt det in 1/2 but it's not in the course anymore....

[MandhreeE]

Hey guys. I have attached a graph from question 9 in the 2011 exam 1.

What I don't understand is why can you simply equate.

Despite g(x) being under the axis and above at another time. Also, why don't have to do it in parts like I was normally told to when this happens.
Thanks a lot, hope I didn't sound too confusing.

Hi i'm just doing this question but i can understand how the second simultaneous equation is found. I found the one through intergration but i can't find the second. Would someone please mind explaining it?

Thank you !!

[Pineapple66]

I don't think you're meant to use "cos of a negative = a positive" when solving trigonometric equations which is maybe why you are getting the questions wrong. Working is attached

thanks so much!!! makes sense now

Can someone please explain how to integrate 1/(2x-1)^3 ? thanks in advance ^^

[MB_]

Hi i'm just doing this question but i can understand how the second simultaneous equation is found. I found the one through intergration but i can't find the second. Would someone please mind explaining it?

Thank you !!

You need to find m in terms of a by finding the intersection between the two graphs and then use that as your upper value