3U Maths Question Thread

[RuiAce]

Sorry I have another question.
for part iv, how do I prove it is the shortest beam? I managed to get this value when dy/da=0, but i don't know how to prove minimisation with this question


can't thank you all enough for the help!

Mod edit: Posts merged. You're free to ask a few questions in succession, but please resort to the 'modify' function at the top right corner of a post to refrain from double posting.

With that second one, both the second derivative and table method lack any appeal whatsoever. How many marks was it worth? (Unless it's worth at least 3 marks, the test should not be necessary for this question, whatever it's source was.)

[junzhang]

ahh thank you!! that question was only worth 2 marks and the solution itself didn't prove that it was a minimum. I was just checking because i usually lose marks for forgetting to justify max/min.

[RuiAce]

ahh thank you!! that question was only worth 2 marks and the solution itself didn't prove that it was a minimum. I was just checking because i usually lose marks for forgetting to justify max/min.

Usually you should. That particular question was just taking it too far

[J.B]

Hi,
I was just wondering if someone could explain Question 6 to me:
You need to find the general solution of cos2x=√2 / 2
I solved this by finding x= π/8, and (7π)/8
so I assumed the answer would be
B,

But the answer kept 2x= π/4, (7π/4)
& created a general solution out of this and then divided everything by 2, giving A as the answer.

So I'm just a bit confused why one way is right and the other is not?
(I have attached the question, its from the James Ruse 2016 Trial.)

I am also unsure what the answers have done for Question 3, and 12 d) ii).

If you need me to post the answers, just let me know.
Thanks

[RuiAce]

Hi,
I was just wondering if someone could explain Question 6 to me:
You need to find the general solution of cos2x=√2 / 2
I solved this by finding x= π/8, and (7π)/8
so I assumed the answer would be
B,

But the answer kept 2x= π/4, (7π/4)
& created a general solution out of this and then divided everything by 2, giving A as the answer.

So I'm just a bit confused why one way is right and the other is not?
(I have attached the question, its from the James Ruse 2016 Trial.)

I am also unsure what the answers have done for Question 3, and 12 d) ii).

If you need me to post the answers, just let me know.
Thanks

__________________________________

Typing a circle on this forum takes way too much effort so I'm gonna represent the circle as a line

[Opengangs]

Q12 d) ii)
Use Binomial probability:
To calculate the four odd numbers, we get: (1/2)^4, since in each stage, there are only 3 odd numbers.
To calculate the 2 sixes, we get (1/6)^2

Using Binomial probability, we get 6C2 * (1/6)^2 * (1/2)^4
= 5/192

[J.B]

Thank you,

I'm still just a bit confused with the binomial probability as with:
nCr p^r q^n-r
where p= probability of success, and q= probability of failure.
I thought that p+q had to =1 ?

[RuiAce]

Thank you,

I'm still just a bit confused with the binomial probability as with:
nCr p^r q^n-r
where p= probability of success, and q= probability of failure.
I thought that p+q had to =1 ?

It's not actually a binomial probability - he misused the terminology there. It's similar to binomial probability, but it isn't exactly that. Intuitively, this is because he didn't associate a success and failure outcome; if a success is getting a 6, a fail is not getting a 6; not specifically an odd number.

I just didn't comment because I thought his final answer was still correct anyhow (unless it wasn't).

[J.B]

Oh ok thank you,
and yes it's correct.

[Opengangs]

It's not actually a binomial probability - he misused the terminology there. It's similar to binomial probability, but it isn't exactly that. Intuitively, this is because he didn't associate a success and failure outcome; if a success is getting a 6, a fail is not getting a 6; not specifically an odd number.

I just didn't comment because I thought his final answer was still correct anyhow (unless it wasn't).

Yeah, my mistake.

Method is still correct, but I'll explain it a bit clearer.
Since we want two sixes, chances of that happening is (1/6)^2.
Since we also want four odds, chances of that is (1/2)^4.

Now, the question doesn't ask for a particular order in which they may occur. So, we use the combination form: 6C2 or 6C4 -> they yield the same outcome, due to binomial symmetry.
This just means there are 6C2 ways of selecting two sixes, and 4 odd numbers.

[RuiAce]

Yeah, my mistake.

Method is still correct, but I'll explain it a bit clearer.
Since we want two sixes, chances of that happening is (1/6)^2.
Since we also want four odds, chances of that is (1/2)^4.

Now, the question doesn't ask for a particular order in which they may occur. So, we use the combination form: 6C2 or 6C4 -> they yield the same outcome, due to binomial symmetry.
This just means there are 6C2 ways of selecting two sixes, and 4 odd numbers.

Yeah. That looks more accurate.

[junzhang]

Hi,
could you help me with this? thanks

[Opengangs]

Hi,
could you help me with this? thanks

Trapezoidal rule: h/2(f(a) + f(b) + 2f((a+b)/2), where h is the interval between two function values.

[J.B]

Hi,
I was just wondering if this is correct?
Thanks

[Opengangs]

Hi,
I was just wondering if this is correct?
Thanks

Yeah, that looks correct.
If you're unsure, plug some values for t