September 28, 2020, 10:06:58 pm

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#### mashal hameed

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##### Re: 3U Maths Question Thread
« Reply #4275 on: March 24, 2020, 12:48:28 pm »
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#### HELPER2020

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##### Re: 3U Maths Question Thread
« Reply #4276 on: March 25, 2020, 03:41:26 pm »
+1

#### shekhar.patel

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##### Re: 3U Maths Question Thread
« Reply #4277 on: June 10, 2020, 11:01:38 pm »
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Hi,
How do we do this question.
Thank you.

#### wsdm

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##### Re: 3U Maths Question Thread
« Reply #4278 on: June 11, 2020, 12:05:53 am »
+2
Hi,
How do we do this question.
Thank you.
To start off, we're going to utilise the double angle formulae, i.e. $\cos(\alpha + \beta) = \cos\,\alpha\,\cos\,\beta-\sin\,\alpha\,\sin\,\beta\\
\cos(\alpha - \beta) = \cos\,\alpha\,\cos\,\beta+\sin\,\alpha\,\sin\,\beta$

Next, we add those two together to give $\cos(\alpha + \beta)+\cos(\alpha-\beta)=2\cos\,\alpha\,cos\,\beta$
For now, we'll let $P$ and $Q$ equal to $P=\alpha+\beta\\
Q=\alpha-\beta$

Now, adding $P$ and $Q$ together to get $P+Q=2\alpha\\
\alpha=\frac{P+Q}{2}$

We can also multiply the equation for $Q$ to get $P-Q=2\beta\\
\beta=\frac{P-Q}{2}$

Substituting the values we have found for $\alpha$ and $\beta$ $\cos(P)+\cos(Q)=2\cos\,\frac{P+Q}{2}\,cos\,\frac{P-Q}{2}, \text{as required.}$
For the second part of question, try and attempt this own your own by substituting values as you would for $P$ and $Q$ ($P=x$ and $Q=5x$) and then solve for $x$.

Don't hesitate to ask for further elaboration.
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#### twelftholmes

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##### Re: 3U Maths Question Thread
« Reply #4279 on: August 04, 2020, 08:56:20 am »
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hey! this is a question from the 3unit sample paper

Question 8 from multiple choice:
A team of 11 students is to be chosen from a group of 18 students. Among the 18 students are 3 students who are left-handed. What is the number of possible teams containing at least 1 student who is left-handed?

Thanks in advance for any help!

#### jaccs

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##### Re: 3U Maths Question Thread
« Reply #4280 on: August 04, 2020, 09:31:32 am »
+1
consider the unrestricted choices:  18C11 =31 824
if we remove the 3 left-handed from the pool to choose from, that leaves 15 right-handed to select 11
then subtract the case of no left handed: 15C11 = 1365
at least one left handed is 31 824 - 1365 = 30 459

the alternative is to find all the cases: 1, 2, 3, left handed students and add them
3C1 x 15C10 + 3C2 x 15C9 + 3C3 x 15C8 = 30 459
« Last Edit: August 04, 2020, 09:38:58 am by jaccs »

#### jaccs

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##### Re: 3U Maths Question Thread
« Reply #4281 on: August 06, 2020, 04:15:13 pm »
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Question 1 from the mock trial in the Atar notes extension 1 topic tests
Is there something missing from the information in this question?

Gerald is conducting an experiment at a golf course. He needs to guess the lowest amount of golf balls on the golf course at any given time. He is given the following information:
– There are 3 holes that a golfer can choose from
– It is guaranteed that one hole contains at least 4 balls.
What is the minimum number of balls Gerald should guess?

the solution says 3 x 3 + 1 = 10
do we assume a certain number of balls in any hole - at least 3 and then one has to have at least 4?
or am i just completely misinterpreting it
thanks

#### Opengangs

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##### Re: 3U Maths Question Thread
« Reply #4282 on: August 13, 2020, 09:21:24 am »
+1
Question 1 from the mock trial in the Atar notes extension 1 topic tests
Is there something missing from the information in this question?

Gerald is conducting an experiment at a golf course. He needs to guess the lowest amount of golf balls on the golf course at any given time. He is given the following information:
– There are 3 holes that a golfer can choose from
– It is guaranteed that one hole contains at least 4 balls.
What is the minimum number of balls Gerald should guess?

the solution says 3 x 3 + 1 = 10
do we assume a certain number of balls in any hole - at least 3 and then one has to have at least 4?
or am i just completely misinterpreting it
thanks

Hi there!
I admit that I may not have framed the question as well as I should have but the intention was to apply the pigeonhole principle but in reverse! You have 3 holes and you want to guarantee that one hole contains 4 balls. We observe that that if we stack 3 balls in each hole, then the next ball MUST guarantee that one of these holes contains 4 balls. So the minimum number of balls that guarantees 4 balls must be (3 x 3 + 1 = 10).
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#### jaccs

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##### Re: 3U Maths Question Thread
« Reply #4283 on: August 13, 2020, 03:43:14 pm »
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Thanks for the clarification, that makes sense, if the golfer is sinking balls in the three holes.
Thanks again.

#### twelftholmes

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##### Re: 3U Maths Question Thread
« Reply #4284 on: September 26, 2020, 07:50:30 pm »
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hey! with solving this differential equation I'm confused about a step in the worked solution.

Why did they move the constant C from the power to the front and turn it into a k? I know this is a log law but I would initially think to take the whole power (x^2/2 +C) to the front and turn it into k, not just C. Is this just something you are able to do (like a rule) or is there a line of working out that they grouped into one (which frustratingly this textbook usually does haha).

Thanks in advance for any assistance!

#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4285 on: September 26, 2020, 07:59:01 pm »
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Hey there!

'I would initially think to take the whole power (x^2/2 +C) to the front and turn it into k, not just C' - note you can't actually do this (if I'm interpreting this correctly?)

The crux of what they're doing is using the fact that you can split $e^{f(x)+g(x)}$ into $e^{f(x)} \times e^{g(x)}$. Note that since C is a constant, we also have that $e^C$ is also a constant - they just replace it with k to simplify things (ie. a combination of the two aforementioned ideas). It's a lot cleaner to look at and typically multiplication is easier to deal with in computation than addition (exceptions exist, of course). You can't simplify the polynomial (because no such constant exists) and it's often redundant to assign a variable that refers to it when you often need the polynomial for other things (differentiation, computing its value at a point, etc.)

Hope this helps
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#### twelftholmes

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##### Re: 3U Maths Question Thread
« Reply #4286 on: September 26, 2020, 08:06:36 pm »
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of course!! i just relaised that x^2/2 isn't a constant so you can't group that with the c.

#### twelftholmes

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##### Re: 3U Maths Question Thread
« Reply #4287 on: September 27, 2020, 02:45:36 pm »
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hey! back at it again with another differential equation question,

It's only a minor thing but in case it's important I'll ask just in case.
In the last step when they took the square root, why is it only negative instead of both plus/minus?

I know this is super simple but thanks for any clarification!

#### fun_jirachi

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##### Re: 3U Maths Question Thread
« Reply #4288 on: September 27, 2020, 02:59:28 pm »
+1
Hey

They took the negative root because of the initial condition $x = \frac{\pi}{2}, y = -1$. It's a little problematic to express $f(y)$ as a function of x (in this case $f(y) = y^2$), so it's actually really important to simplify and make the distinction between the positive and negative root. Pay attention especially to the initial condition as it is (clearly, from this example) not only used to find constants of integration
« Last Edit: September 27, 2020, 05:27:07 pm by fun_jirachi »
Failing everything, but I'm still Flareon up.
Spoiler
HSC 2018: Modern History [88] | 2U Maths [98]
HSC 2019: Physics [92] | Chemistry [93] | English Advanced [87] | Maths Extension 1 [98] | Maths Extension 2 [97]
ATAR: 99.05

UCAT: 3310 - Verbal Reasoning [740] | Decision Making [890] | Quantitative Reasoning [880] | Abstract Reasoning [800]