VCE Methods Question Thread!

[Shadowxo]

Hi

For this http://imgur.com/a/Bl2Ew

I don't understand how 9dii and iii can have a different answer? ii is dependent and iii is independent but what they're both saying is "given the sum of the 2 numbers is at least "x", what is the probability of A occurring?".

If we know C or D occurs then wouldn't this affect the probability of A occurring? this makes me believe that they are both dependent.

note: I know that for iii Pr(A|D) = Pr(A) which means it's independent but it doesn't seem very intuitive otherwise.

Thanks

The thing about this question is that it is unintuitive.
If D happens, A happens 1/6 of the time. If you look at the sample space you'll see that out of the 6 possible outcomes that would lead to a number of at least 10, only one of them has a 4 rolled on the first die. So they're independent, as the fact that the number was at least 10 doesn't affect the probability of the first die being a 4.
If C happens, however, A doesn't happen 1/6 of the time. Looking at all the possible outcomes, A does not occur in 1/6th of them
I know it's confusing, probability often is. In this case, it just so happens that the probability of the number 4 being on the first dice is 1/6 when the sum is at least 10, and not 1/6 when the number is at least 8.

Hope this helps a bit and good luck

[TheCommando]

\[4^2^x^-^1=36\]

\[\log_{e}36/\log_{e}4=2x-1\]

\[\frac{1}{2}(\log_{e}36/\log_{e}4 +1) =x\]

\[(\log_{e}6/\log_{e}4 +0.5) =x\]

How did u get rid of that 1/2 and go from step two to step 3?

[MisterNeo]

How did u get rid of that 1/2 and go from step two to step 3?

You can simplify it further.

[TheCommando]

You can simplify it further.

Thank you
That 1/2 times the 2ln6/ln4 cancels out the 1/2 right
Cause in my cas it comes out with ln6/2ln2

[Shadowxo]

n*log_ab=log_abⁿ
They just used that to move the 1/2 to a power inside the log.
1/2*ln36=ln(36^1/2) = ln6
ln = log_e
Also, I just wanted to mention there are multiple solutions to this question. The way I solved it gave x= ln3/ln4 + 1 giving different answers for a and b, but these answers are equivalent. Hope this helps

Thank you
That 1/2 times the 2ln6/ln4 cancels out the 1/2 right
Cause in my cas it comes out with ln6/2ln2

2*ln2 = ln2²=ln4 so it's the same

[TheCommando]

just one more but how come that +1 turns into 1/2
And why exactly does is the form in the loge36/loge4 =2x-1
Like is there a rule where u cant take loge2x-1 etc

[MisterNeo]

Thank you
That 1/2 times the 2ln6/ln4 cancels out the 1/2 right
Cause in my cas it comes out with ln6/2ln2

Yeah, I forgot to expand log 4 into log 2². I fixed my working out.

just one more but how come that +1 turns into 1/2
And why exactly does is the form in the loge36/loge4 =2x-1
Like is there a rule where u cant take loge2x-1 etc

If you're solving for x, then you move everything to the left hand side. So you add 1, and divide by 2 to isolate x by itself.

[TheCommando]

http://imgur.com/a/pmwUR
Hey, how do i do 1e and 1f
As well how do i do q5 and q6 of tthe second image
Thanks in advance

[MisterNeo]

http://imgur.com/a/pmwUR
Hey, how do i do 1e and 1f
As well how do i do q5 and q6 of tthe second image
Thanks in advance

Q1(e): You need to simplify the fraction before differentiating.




Q1(f): e⁴ is a constant and becomes 0 when derived.



Q5: a is a constant. And, assume that it has plus-minus because I don't know how to do that symbol.





You do the same for part b. The term "a" is a constant, so treat it like the number 7 or something.

Q6: You just need to sub y into that 2y numerator, and try to get that as a derivative.


Prove that y = that.
Hope this helps

[TheCommando]

So for q6 how do i prove y= that dy/dx form? And for 1e how did u simplyfy it, i dont get it

[TheCommando]

Thank you! That helped

[MisterNeo]

So for q6 how do i prove y= that dy/dx form? And for 1e how did u simplyfy it, i dont get it

Simplifying the fraction is just dividing each the whole thing by the denominator.




Tbh, that's all I know in Q6.
Best that you ask Rui to help with it. 

[TheCommando]

Simplifying the fraction is just dividing each the whole thing by the denominator.




Tbh, that's all I know in Q6.
Best that you ask Rui to help with it. 

Ah ok i get it, thanks man!

[VanillaRice]

So for q6 how do i prove y= that dy/dx form?

With 'show that' questions, I like to pretend that anything that is written after "show that" isn't actually there - we need to get to that point.
 Therefore, we need to begin by finding dy/dx:

Spoiler

Assuming you have already differentiated using the chain rule, we would get something that looks like (unsimplified):

Now expanding:


Now find a common denominator to get everything under the one fraction:


Let's put this aside for now. So, we now need to get this part (the numerator):

and get it into a form which looks like (the original function):

I'm going to do something a bit cheaty (but not really  ) and expand the above:


This looks very familiar!
So now:



This contains our original function! So now:


as required

[2352300]

hi sorry to ask, i was wondering how you would anti-differentiate

-x^3 + 4sin^2x + 4cos^2x

the answer is -1/4x^2 + 4x

i understand the first part of the answer but i dont understand how the circular functions would anti-differentiate into 4x

thank you