Author Topic: 3U Maths Question Thread (Read 1240261 times) Tweet Share

bsdfjnlkasn · « **Reply #1740 on:** March 31, 2017, 01:40:41 pm »

Hey there,

Could someone please explain what happened in the last two lines of the provided working?
I'm just not sure how they changed -pi/6 to 11pi/6 I understand it's got to do with using pi/6 in the 4th quadrant but I don't know why they did this extra step

Thank you!

jamonwindeyer · « **Reply #1741 on:** March 31, 2017, 01:58:00 pm »

Quote from: bsdfjn;lkasn on March 31, 2017, 01:40:41 pm

Hey there,

Could someone please explain what happened in the last two lines of the provided working?
I'm just not sure how they changed -pi/6 to 11pi/6 I understand it's got to do with using pi/6 in the 4th quadrant but I don't know why they did this extra step

Thank you!

Hey! So presumably we are solving for 0 to 2pi, that negative answer is NOT in that range! So we need to write it differently, and how do we change the way an angle is written without changing its value? We add 2pi (0 and 2pi being the same thing is the easiest example of that) - So adding 2pi to -Pi/6 is what gives us that answer, by bringing the angle into the required range

Sukakadonkadonk · « **Reply #1742 on:** March 31, 2017, 02:10:39 pm »

Hi guys,

Could anyone help me with this problem?

jamonwindeyer · « **Reply #1743 on:** March 31, 2017, 02:13:49 pm »

Quote from: Sukakadonkadonk on March 31, 2017, 02:10:39 pm

Hi guys,

Could anyone help me with this problem?

Hey bud! Not at my computer right now, but hint in the meantime: it's the auxiliary angle method! Or the t method, alternatively!

Sukakadonkadonk · « **Reply #1744 on:** March 31, 2017, 02:17:45 pm »

Quote from: jamonwindeyer on March 31, 2017, 02:13:49 pm

Hey bud! Not at my computer right now, but hint in the meantime: it's the auxiliary angle method! Or the t method, alternatively!

Alright, thank you.

How would you use auxiliary angle for this?

RuiAce · « **Reply #1745 on:** March 31, 2017, 02:39:02 pm »

Quote from: Sukakadonkadonk on March 31, 2017, 02:17:45 pm

Alright, thank you.

How would you use auxiliary angle for this?

Jamon overlooked something

jamonwindeyer · « **Reply #1746 on:** March 31, 2017, 02:42:15 pm »

Quote from: Sukakadonkadonk on March 31, 2017, 02:17:45 pm

Alright, thank you.

How would you use auxiliary angle for this?

Alright I'm back!

$1+\cos{2x}=\sin{2x}\implies\sin{2x}-\cos{2x}=1$

Make the LHS of that equivalent to $A\sin(2x-\alpha)=A\sin{2x}\cos{\alpha}-A\cos{2x}\sin{\alpha}$. As is normal for this method, we get two equations:

$A\cos{\alpha}=1\\A\sin{\alpha}=1$

Solve them simultaneously to find $A=\sqrt{2},\alpha=\frac{\pi}{4}$. So we rewrite the equation as:

$\sqrt{2}\sin{\left(2x-\frac{\pi}{4}\right)}=1\\\sin{\left(2x-\frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}\\$

And then you solve that as a normal trig equation! Remember, solving for $0\le x\le2\pi$, you need to solve for double that for $2x$, that is: $0\le2x\le4\pi$

I've assumed a few bits of knowledge up there, not sure how familiar you are with the method and some of its peculiarities, anything about it unclear? I'm happy to clarify if so

Edit: Auxillary is probably the more standard approach for the question type, but Rui's given you an alternative that works on the double angle results, both valid, if you use this one do check your answers match

RuiAce · « **Reply #1747 on:** March 31, 2017, 03:23:35 pm »

Quote from: jamonwindeyer on March 31, 2017, 02:42:15 pm

Edit: Auxillary is probably the more standard approach for the question type, but Rui's given you an alternative that works on the double angle results, both valid, if you use this one do check your answers match

My intuition (and by consequence my recommendation) is based off the fact that the constant term vanishes when applying double angle results. Provided the constant term is set to vanish it is superior; else auxiliary angle is the best way to go

legorgo18 · « **Reply #1748 on:** March 31, 2017, 07:21:50 pm »

Hey! I dont understand this perms and coms q... at all (esp part a)

From an unlimited supply of 14 different books, it is desired to make up parcels containing r books, all different.

a) For what value of r will there be as many packets as possible, all with different contents?
b) How many such packets will there be?
c) How many of these would contain one particular kind of book?

Thanks for any help! I just hate this topic so much zz

RuiAce · « **Reply #1749 on:** March 31, 2017, 08:21:11 pm »

Quote from: legorgo18 on March 31, 2017, 07:21:50 pm

Hey! I dont understand this perms and coms q... at all (esp part a)

From an unlimited supply of 14 different books, it is desired to make up parcels containing r books, all different.

a) For what value of r will there be as many packets as possible, all with different contents?
b) How many such packets will there be?
c) How many of these would contain one particular kind of book?

Thanks for any help! I just hate this topic so much zz

Assuming packets and parcels are the same thing.
$\text{Let }n\text{ be the number of parcels we have}\\ \text{For the sake of demonstration, let's call the books }A,B,C,\dots, N$
____________________________
$\text{For the sake of demonstration, let's analyse the cases }n=1, 2, 3\\ \text{When }n=1\text{ we may have the following packets}:\\ \{A\}, \{B\}, \dots, \{ N\}\\ \text{You can check that there will be }14=\binom{14}1\text{ possible packets}$
$\text{When }n=2\text{ we may have the following packets}:\\ \{A,B\}, \{A,C\}, \dots, \{A,N\}, \{B,C\}, \{ B,D\}, \dots \{B, N\} \dots, \{M,N\}\\ \text{You can check that there will be }(14+13+...+1)=\binom{14}{2}\text{ possible packets}$
$\text{When }n=3\text{ things get bizarre, and we may have the following packets:}\\ \{A,B,C\}, \dots, \{A,B,N\} , \{A,C,D\}, \dots, \{A,C,N\}\dots \{A,M,N\}, \{B,C,D\}, \dots, \{L,M,N\}\\ \text{It becomes harder to see that there will be }\binom{14}{3}\text{ possible packets}$
____________________________
$\text{Claim: If }r\text{ is the number of books in a packet}\\ \text{Then }\binom{14}{r}\text{ is the number of packets we will have}$
$\text{Proof: The number of packets is equal to the number of ways}\\ \text{in which we can choose }r\text{ letters}\text{ out of }A,B,\dots, N\\ \text{in any order we like}\\ \text{Basically, we are just putting }r\text{ books into every packet}$
$\text{This explanation should make it clear that the answer is }\binom{14}{r}$
$\text{It now falls to finding the value of }r\text{ that maximises }\binom{14}{r}\\ \text{This question is now a binomial theorem question}\\ \text{One may consider the greatest coefficients approach of }\frac{T_{k+1}}{T_k} \ge 1\text{ or }\frac{T_{k+1}}{T_k} < 1$
$\text{However, this example is easier, as }\binom{14}r\\ \text{is the coefficient of }x^r\text{ on the standard expansion of }(1+x)^{14}$
$\text{Because we are considering the expansion of just }(1+x)^{14}\\ \text{The greatest coefficient is going to be the middle term}\\ \text{So }r=7$
$\text{b) easily follows from a). }\binom{14}7=3432$
I will let you consider c) first by yourself. It may help to once again consider examples. Or maybe the subcases of r=1, r=2, r=3, ...

To make life easier, you may, without loss of generality, consider only book A. Because books B, C, D, etc. will follow the same rule.

Sukakadonkadonk · « **Reply #1750 on:** March 31, 2017, 10:31:28 pm »

Quote from: jamonwindeyer on March 31, 2017, 02:42:15 pm

Alright I'm back!

$1+\cos{2x}=\sin{2x}\implies\sin{2x}-\cos{2x}=1$

Make the LHS of that equivalent to $A\sin(2x-\alpha)=A\sin{2x}\cos{\alpha}-A\cos{2x}\sin{\alpha}$. As is normal for this method, we get two equations:

$A\cos{\alpha}=1\\A\sin{\alpha}=1$

Solve them simultaneously to find $A=\sqrt{2},\alpha=\frac{\pi}{4}$. So we rewrite the equation as:

$\sqrt{2}\sin{\left(2x-\frac{\pi}{4}\right)}=1\\\sin{\left(2x-\frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}\\$

And then you solve that as a normal trig equation! Remember, solving for $0\le x\le2\pi$, you need to solve for double that for $2x$, that is: $0\le2x\le4\pi$

I've assumed a few bits of knowledge up there, not sure how familiar you are with the method and some of its peculiarities, anything about it unclear? I'm happy to clarify if so

Edit: Auxillary is probably the more standard approach for the question type, but Rui's given you an alternative that works on the double angle results, both valid, if you use this one do check your answers match

OHH I see.
Yes the solutions did do it Rui's way but I didn't really understand it at first.
Lol, didn't think a question could have so many different solutions. Even t method is available!

RuiAce · « **Reply #1751 on:** March 31, 2017, 10:35:52 pm »

Quote from: Sukakadonkadonk on March 31, 2017, 10:31:28 pm

OHH I see.
Yes the solutions did do it Rui's way but I didn't really understand it at first.
Lol, didn't think a question could have so many different solutions. Even t method is available!

The formulas I applied are all on the reference sheet should you forget them in the exam.

Rest is just factorising and a typical process.

Sukakadonkadonk · « **Reply #1752 on:** March 31, 2017, 10:42:31 pm »

Quote from: RuiAce on March 31, 2017, 10:35:52 pm

The formulas I applied are all on the reference sheet should you forget them in the exam.

Rest is just factorising and a typical process.

Yes, you explained it well the first time. I meant the solutions attached to the past paper didn't really suffice for me.
Thanks.

Sukakadonkadonk · « **Reply #1753 on:** April 01, 2017, 08:24:20 pm »

Hi again,

I'm confused with how the 2θ/sin(2θ) plays in these solution. Does it equal one or something?
Question was to prove θ^2 times cosec^2(2θ) = 1/4
Could someone please help?

jakesilove · « **Reply #1754 on:** April 01, 2017, 08:28:11 pm »

Quote from: Sukakadonkadonk on April 01, 2017, 08:24:20 pm

Hi again,

I'm confused with how the 2θ/sin(2θ) plays in these solution. Does it equal one or something?
Question was to prove θ^2 times cosec^2(2θ) = 1/4
Could someone please help?

Yep, you're absolutely right! There is a relationship that, for very small theta,

$\lim_{x \rightarrow 0} \frac{\theta}{\sin \theta} =1$

So, if we just let theta equal two alpha, the relationship is also true for

$\lim_{x \rightarrow 0} \frac{2 \alpha}{\sin 2 \alpha} =1$

Which is the same as the working out that you posted. Does that make sense?

Search the forums now!

We have moved!

Author Topic: 3U Maths Question Thread (Read 1240261 times) Tweet Share

bsdfjnlkasn

Re: 3U Maths Question Thread

jamonwindeyer

Re: 3U Maths Question Thread

Sukakadonkadonk

Re: 3U Maths Question Thread

jamonwindeyer

3U Maths Question Thread

Sukakadonkadonk

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

jamonwindeyer

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

legorgo18

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

Sukakadonkadonk

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

Sukakadonkadonk

Re: 3U Maths Question Thread

Sukakadonkadonk

Re: 3U Maths Question Thread

jakesilove

Re: 3U Maths Question Thread

Recent Posts

User:
Password: