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March 29, 2024, 11:56:43 am

Author Topic: Specialist 1/2 Question Thread!  (Read 119982 times)  Share 

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AlphaZero

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Re: Specialist 1/2 Question Thread!
« Reply #270 on: February 25, 2019, 11:03:10 pm »
+1
When all eight factors of 30 are multiplied together, the product is 30k. What is the value of k?


someone help me ......................... :-[
\begin{align*}30k&=(1\times 30)\times(2\times 15)\times (3\times 10)\times (5\times 6)\\
\implies 30k&=30^4\\
\implies \quad k&=30^3=27\,000\end{align*}
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DBA-144

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Re: Specialist 1/2 Question Thread!
« Reply #271 on: February 25, 2019, 11:05:19 pm »
0
\begin{align*}30k&=(1\times 30)\times(2\times 15)\times (3\times 10)\times (5\times 6)\\
\implies 30k&=30^4\\
\implies \quad k&=30^3=27\,000\end{align*}

Yeah I got the same answer, the user who posted the question said that it is 30^4,in spec 3/4 question thread. imo its an error on the question's/textbook's part- can you confirm?
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AlphaZero

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Re: Specialist 1/2 Question Thread!
« Reply #272 on: February 25, 2019, 11:09:09 pm »
0
Yeah I got the same answer, the user who posted the question said that it is 30^4,in spec 3/4 question thread. imo its an error on the question's/textbook's part- can you confirm?

It's an error. If the question had instead been  "When all eight factors of 30 are multiplied together, the product is \(k\). What is the value of \(k\)?",  then  \(k=30^4\)  would be the correct answer.
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Re: Specialist 1/2 Question Thread!
« Reply #273 on: February 26, 2019, 10:09:34 pm »
+1
I posted this in the 3/4 thread too, but it's useful to note that if there are only 8 factors of 30, and it's not a square number, this means you can pair the factors to multiply to give 30. Thus, without any multiplication or numerical arrangements at all, I know that the product must be 304.
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Ionic Doc

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Re: Specialist 1/2 Question Thread!
« Reply #274 on: March 13, 2019, 07:21:38 pm »
0
For the sequence 12, 6, 3, ..., S8 is equal to what?

Thnx in advance
im legit dying cause I got a methods SAT and a specialist SAT tomorrow ...and I had a 3/4 Psych SAC yesterday ... :-[
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AlphaZero

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Re: Specialist 1/2 Question Thread!
« Reply #275 on: March 14, 2019, 09:17:50 am »
+1
For the sequence 12, 6, 3, ..., S8 is equal to what?

Thnx in advance
im legit dying cause I got a methods SAT and a specialist SAT tomorrow ...and I had a 3/4 Psych SAC yesterday ... :-[

This is a geometric series with first term  \(a=12\)  and common ratio  \(r=1/2\).  Thus, the sum of the first \(8\) terms is given by \[S_8=\frac{12\big((1/2)^8-1\big)}{1/2-1}=\frac{765}{32}\]
« Last Edit: March 14, 2019, 09:25:25 am by AlphaZero »
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Evolio

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Re: Specialist 1/2 Question Thread!
« Reply #276 on: March 14, 2019, 12:08:53 pm »
0
Hi.
I was wondering how to find x when y=1/8

AlphaZero

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Re: Specialist 1/2 Question Thread!
« Reply #277 on: March 14, 2019, 06:47:54 pm »
+2
Hi.
I was wondering how to find x when y=1/8

We have  \[y=5\!\times\!2^{2/3}x,\]and so substituting  \(y=1/8\)  yields \[x=\frac{1}{40\!\times\!2^{2/3}}=\frac{2^{1/3}}{80}\]
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Evolio

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Re: Specialist 1/2 Question Thread!
« Reply #278 on: March 18, 2019, 10:21:45 am »
0
Okay, thanks!
 :)

Ionic Doc

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Re: Specialist 1/2 Question Thread!
« Reply #279 on: March 18, 2019, 01:58:44 pm »
0
hey is it me or is spec way more easier and interesting than methods
honestly methods is actually feeling more difficult than spec in year 11
I really  want to drop a math but I know I can't drop methods...so rip spec :-[
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coldairballoon

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Re: Specialist 1/2 Question Thread!
« Reply #280 on: March 30, 2019, 11:26:59 pm »
0
Hi guys, first time posting on a forum! I feel like an old man with new technology, so if I've formatted anything wrong...oops.
Anyway, I just wanted to know if anybody could help me with this question? I don't know what I'm doing wrong, but I can't seem to come up with an answer (I sort of get it, but I need to show working/formal proof):

Prove that the recurrence relation xn+1 = (xn)2 + 2 does not converge to a single value.

If anyone could take a look at it it'd be so helpful for me! Thanks.
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S_R_K

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Re: Specialist 1/2 Question Thread!
« Reply #281 on: March 31, 2019, 09:38:11 am »
0
hey is it me or is spec way more easier and interesting than methods
honestly methods is actually feeling more difficult than spec in year 11
I really  want to drop a math but I know I can't drop methods...so rip spec :-[

Spec is definitely more interesting, and it's pretty common to find Units 1 & 2 Spec easier than Units 1 & 2 Methods. A lot of that is due to the crowded curriculum in Methods, where you have to learn the content at a very fast pace, but in Spec you have more time to absorb the concepts and consolidate what you've learned. It's worth persevering with both, they complement each other very well in year 12, with quite a lot of overlapping content.

Hi guys, first time posting on a forum! I feel like an old man with new technology, so if I've formatted anything wrong...oops.
Anyway, I just wanted to know if anybody could help me with this question? I don't know what I'm doing wrong, but I can't seem to come up with an answer (I sort of get it, but I need to show working/formal proof):

Prove that the recurrence relation xn+1 = (xn)2 + 2 does not converge to a single value.

If anyone could take a look at it it'd be so helpful for me! Thanks.

Firstly, the recurrence relation is only well defined if we have a value for x0 (the initial value). But in this case, the proof of divergence can be done for any value of x0.

There are a couple of approaches. One is to simply prove that the sequence does not converge. We can do this by supposing that the sequence converges to L and deriving a contradiction. Given that xn+1 = (xn)^2 + 2, if we assume that the limit is L, then we'll have L = L^2 + 2 as n approaches infinity. This equation has no solution, hence the limit does not exist.

A stronger argument is to prove not only that the sequence does not converge, but that it is unbounded (ie. the terms approach infinity). This can be shown by noting that (xn)^2 > xn when xn > 1, but for this sequence xn > 1 for all n ≥ 1. Hence x(n+1) > xn + 2. So for any number L, we can find always find a term in the sequence that is larger than L.
« Last Edit: March 31, 2019, 09:40:45 am by S_R_K »

RuiAce

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Re: Specialist 1/2 Question Thread!
« Reply #282 on: March 31, 2019, 09:40:57 am »
+4
A stronger argument is to prove not only that the sequence does not converge, but that it is unbounded (ie. the terms approach infinity). This can be shown by noting that (xn)^2 > xn when n ≥ 1. Hence x(n+1) > xn + 2. So for any number L, we can find always find a term in the sequence that is larger than L.
Tiny subtlety. \((x_n)^2 > x_n\) only if we can assume that \( |x_n| > 1 \). Just need to add in that if \(|x_n| \leq 1\), then \( x_{n+1} = (x_n)^2 + 2 > 0+ 2 > 1 \geq x_n\).

Otherwise makes sense. Is this stuff actually in the spesh 1/2 course?

S_R_K

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Re: Specialist 1/2 Question Thread!
« Reply #283 on: March 31, 2019, 09:47:00 am »
+1
Tiny subtlety. \((x_n)^2 > x_n\) only if we can assume that \( |x_n| > 1 \). Just need to add in that if \(|x_n| \leq 1\), then \( x_{n+1} = (x_n)^2 + 2 > 0+ 2 > 1 \geq x_n\).

Otherwise makes sense. Is this stuff actually in the spesh 1/2 course?

Thanks, I think I fixed up that in the edit – for any x0 we know that x1 = (x0)^2 + 2 > 1.

This is not really part of the course. There is a little bit on convergence / divergence of geometric series, but students aren't required to do any reasonably rigorous proofs. And in general, most of the sequences / series part of the course is on finite sequences / series.

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Re: Specialist 1/2 Question Thread!
« Reply #284 on: April 23, 2019, 06:16:49 pm »
0
I'm confused with a few questions regarding sequences and series

1) I need to find the next 3 terms of,
    2,-1,4,3,6,-6

2) I'm so confused about the conversion from explicit to recursive relations, for example,
   how do you change u(n)=n^2-4n=7 into a recursive relation

3) find the value of a for which (3,a+2,a^2) defines an arithmetic sequence