VCE Methods Question Thread!

[eliseeeeee]

Just double checking that we won't be examined on general solutions? I can't find it on the study design but we covered it in class and I keep getting practice exam questions so I thought I'd check. Thanks

[Pineapple66]

use a scalar, so times everything by e^t
then it is essentially a hidden quadratic so sub u=e^t
and solve it from there

btw i am horrid at methods so someone correct me if i am wrong

Thanks so much!! you're a gem.
With Q10d (trapezium question in exam 1 2015) how would you differentiate 4-2cos(theta)/sin(theta) ?? I tried using the Quotient rule and i got f(x) = 4 - 2cos(theta), f'(x) = 2 sin (theta), g(x) = sin (theta) and g'(x) = cos (theta) .... when I put it into the quotient rule formula i get a weird answer...

[flawsnllama]

Thanks so much!! you're a gem.
With Q10d (trapezium question in exam 1 2015) how would you differentiate 4-2cos(theta)/sin(theta) ?? I tried using the Quotient rule and i got f(x) = 4 - 2cos(theta), f'(x) = 2 sin (theta), g(x) = sin (theta) and g'(x) = cos (theta) .... when I put it into the quotient rule formula i get a weird answer...

Your differentiated answer should be 2sin squared(theta) - (4-2cos(theta) * cos(theta) / sin squared (theta)
https://www.youtube.com/watch?v=OlDkigg3Q5g
Hopefully this helps! 

[Syndicate]

Thanks so much!! you're a gem.
With Q10d (trapezium question in exam 1 2015) how would you differentiate 4-2cos(theta)/sin(theta) ?? I tried using the Quotient rule and i got f(x) = 4 - 2cos(theta), f'(x) = 2 sin (theta), g(x) = sin (theta) and g'(x) = cos (theta) .... when I put it into the quotient rule formula i get a weird answer...

using the quotient rule, you would get:


This can be simplified down.

[Pineapple66]

Your differentiated answer should be 2sin squared(theta) - (4-2cos(theta) * cos(theta) / sin squared (theta)
https://www.youtube.com/watch?v=OlDkigg3Q5g
Hopefully this helps! 

omg i didn't even know they had videos on youtube going through exam questions, thank you !! 

using the quotient rule, you would get:


This can be simplified down.

thank you!!! Just one thing though, so I get 2sin squared(theta) - (4-2cos(theta) * cos(theta) / sin squared (theta) like flawsnllama said but then I think I'm stuffing up the simplification or something. In both of your answers + the vid, why does -4 become -4cos(x) ? 

This is what I end up doing :/

--> 2sin squared(theta) - (4-2cos(theta) * cos(theta) / sin squared (theta)
--> 2 sin squared (theta) - (4 - 2 cos squared (theta) / sin squared (Theta)
--> 2 sin squared (theta) + 2 cos squared (theta) - 4 / sin squared (theta)
fin.

Why am I wrong?

[Syndicate]

omg i didn't even know they had videos on youtube going through exam questions, thank you !! 

thank you!!! Just one thing though, so I get 2sin squared(theta) - (4-2cos(theta) * cos(theta) / sin squared (theta) like flawsnllama said but then I think I'm stuffing up the simplification or something. In both of your answers + the vid, why does -4 become -4cos(x) ? 

This is what I end up doing :/

> 2sin squared(theta) - (4-2cos(theta) * cos(theta) / sin squared (theta)
--> 2 sin squared (theta) - (4 - 2 cos squared (theta) / sin squared (Theta)
--> 2 sin squared (theta) + 2 cos squared (theta) - 4 / sin squared (theta)
fin.

Why am I wrong?

You are forgetting to multiply the 4 with the cosx.



For example:
4(x+2) = 4x+8 not 4x+2 or x+8

You multiply everything in the brackets with whatever is outside.

[Pineapple66]

You are forgetting to multiply the 4 with the cosx.



For example:
4(x+2) = 4x+8 not 4x+2 or x+8

You multiply everything in the brackets with whatever is outside.

yesss i got it! thank you so much again

[samuelbeattie76]

Find the probability that a number selected at random from the first 500 positive integers (1 to 500) will be divisible by 3,7 and 9.

The title of the chapter is the addition rule so we need to use it to solve.

[blacksanta62]

How do I do questions like the one attached?
Thank you

[Escobar]

How do I do questions like the one attached?
Thank you

attached solution (i didnt write notation out properly, lazy)

[Syndicate]

Find the probability that a number selected at random from the first 500 positive integers (1 to 500) will be divisible by 3,7 and 9.

The title of the chapter is the addition rule so we need to use it to solve.

In this question we are basically trying to calculate the probability of \( A \cap B \cap C \). In this case, we will assume that: event A is the probability of obtaining a number divisible by 3, event B is the probability of obtaining a number divisible by 7, and event C is the probability of obtaining a number divisible by 9.

Pr(A) = 166/500
Pr(B) = 71/500
Pr(C) = 55/500
Pr( \( A \cup B \cup C \) ) = 285/500







Edit: minor layout fix

[exit]

Am i allowed to use items in the old course (what i learnt in 1/2) to solve problems? WIll i still be awarded full marks? (methods) thx

bump

[HasibA]

bump

  yeah thats fine, in fact, all 1/2 is assumed knowledge, and they might throw in a 1/2 q to seem how rmbr what etc.

[nadiaaa]

How do we solve this question? Its from MAV 2016

[YellowTongue]

How do we solve this question? Its from MAV 2016

Is the answer B? I think it should be, as the fact that we are generating 90% confidence intervals means that 90% of them would be expected to contain p.