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March 28, 2024, 11:32:29 pm

Author Topic: 3U Maths Question Thread  (Read 1230233 times)  Share 

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RuiAce

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Re: 3U Maths Question Thread
« Reply #840 on: October 23, 2016, 11:46:24 am »
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Help with part (iii) please..I got an expression in terms of n, n+1, n+2 and (1/2) but can't figure out how to manipulate it into what the answers wants
Already addressed a while back in post #199

WLalex

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Re: 3U Maths Question Thread
« Reply #841 on: October 23, 2016, 12:36:38 pm »
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Thanks for that Rui!!

Another one that makes no sense... I got an answers but it was wrong and I don't quite understand what they did (part ii)


Thanks again
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FallonXay

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Re: 3U Maths Question Thread
« Reply #842 on: October 23, 2016, 12:54:22 pm »
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Heyyo again, How do you do both parts of this question? Thanks!  :)
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FallonXay

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Re: 3U Maths Question Thread
« Reply #843 on: October 23, 2016, 12:59:33 pm »
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and part iii of this q  ~ thanks :)
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RuiAce

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Re: 3U Maths Question Thread
« Reply #844 on: October 23, 2016, 01:00:49 pm »
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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #845 on: October 23, 2016, 01:06:42 pm »
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Thanks for that Rui!!

Another one that makes no sense... I got an answers but it was wrong and I don't quite understand what they did (part ii)

Thanks again

Sure! So let's start by doing what we always do to find tangents to curves, differentiate. I'll call the two graphs \(y_1\) and \(y_2\):



So we'll come back to that later; we also need the point of intersection, and at that point of intersection:



But remember the gradients from above will need to be equal too (common tangent), so we also have \(re^{rx}=\frac{1}{x}\).

What we have here are two equations linking \(r\) and \(x\); we're going to solve these simultaneously! The goal will be to use the result we were just given somewhere too. With that in mind, let's rearrange one of those equations:



Aha, theres the result from Part (i); we can then say that \(rx\approx0.56\)!

Therefore back in \(*\), we have: \(\log_e{x}=e^{0.56}\), and that's an equation you can solve for \(x\), and then substitute back to find \(r\); does that make sense? :)

WLalex

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Re: 3U Maths Question Thread
« Reply #846 on: October 23, 2016, 01:14:31 pm »
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Sure! So let's start by doing what we always do to find tangents to curves, differentiate. I'll call the two graphs \(y_1\) and \(y_2\):



So we'll come back to that later; we also need the point of intersection, and at that point of intersection:



But remember the gradients from above will need to be equal too (common tangent), so we also have \(re^{rx}=\frac{1}{x}\).

What we have here are two equations linking \(r\) and \(x\); we're going to solve these simultaneously! The goal will be to use the result we were just given somewhere too. With that in mind, let's rearrange one of those equations:



Aha, theres the result from Part (i); we can then say that \(rx\approx0.56\)!

Therefore back in \(*\), we have: \(\log_e{x}=e^{0.56}\), and that's an equation you can solve for \(x\), and then substitute back to find \(r\); does that make sense? :)

Yes that makes sense thank you! Except one part, how can we assume that rx = 0.56 (I know we found it as an approximate about but howdy you know that it equals rx as there was not mention of that above??)
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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #847 on: October 23, 2016, 01:20:20 pm »
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Heyyo again, How do you do both parts of this question? Thanks!  :)

Hey! First bit:



Clearly this last part is less than zero; what is inside the bracket is always negative as long as \(k\) is a positive integer (which it is) ;D

For your next question, I'll refer you to the HSC solutions and give you a bit of an explanation, since I'd type the same thing:



So we start with the proof for the lowest case and an assumption, as usual. Basically what they do then is consider one additional term on the LHS, and then they tie in the induction assumption where indicated! Remember, if we've assumed that the previous LHS is less than zero, it's kind of like just adding the \(\frac{1}{(k+1)^2}\) term onto both sides; the inequality is maintained. Then some algebra using Part (i) is applied ;D

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #848 on: October 23, 2016, 01:21:23 pm »
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Yes that makes sense thank you! Except one part, how can we assume that rx = 0.56 (I know we found it as an approximate about but howdy you know that it equals rx as there was not mention of that above??)

Part (i) said:



Now we have:



Essentially, put \(t=rx\) ;D

WLalex

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Re: 3U Maths Question Thread
« Reply #849 on: October 23, 2016, 01:24:12 pm »
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ooooooo completely missed that! makes much more sense now thank you greatly
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FallonXay

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Re: 3U Maths Question Thread
« Reply #850 on: October 23, 2016, 01:29:48 pm »
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Hey! First bit:



Clearly this last part is less than zero; what is inside the bracket is always negative as long as \(k\) is a positive integer (which it is) ;D

For your next question, I'll refer you to the HSC solutions and give you a bit of an explanation, since I'd type the same thing:

(Image removed from quote.)

So we start with the proof for the lowest case and an assumption, as usual. Basically what they do then is consider one additional term on the LHS, and then they tie in the induction assumption where indicated! Remember, if we've assumed that the previous LHS is less than zero, it's kind of like just adding the \(\frac{1}{(k+1)^2}\) term onto both sides; the inequality is maintained. Then some algebra using Part (i) is applied ;D

Sorry, would you be able to elaborate on the second last line? - I don't really understand the application of part i in the induction.
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RuiAce

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Re: 3U Maths Question Thread
« Reply #851 on: October 23, 2016, 01:32:32 pm »
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Sorry, would you be able to elaborate on the second last line? - I don't really understand the application of part i in the induction.


« Last Edit: October 23, 2016, 01:38:56 pm by RuiAce »

FallonXay

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Re: 3U Maths Question Thread
« Reply #852 on: October 23, 2016, 01:37:33 pm »
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FallonXay

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Re: 3U Maths Question Thread
« Reply #853 on: October 23, 2016, 01:59:08 pm »
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Hiya, how do you solve these 2 questions? (I'm assuming there's a more efficient way of solving Q9 than expanding it all out haha  :P)
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RuiAce

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Re: 3U Maths Question Thread
« Reply #854 on: October 23, 2016, 02:03:23 pm »
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Hiya, how do you solve these 2 questions? (I'm assuming there's a more efficient way of solving Q9 than expanding it all out haha  :P)
Question 5 is best done by guess and check.

The product of roots is clearly -42
Hence A and D are out.

The sum of 2 roots at a time is -41
This may be used to deduce that B is the correct answer.

Note that the presence of the a just means to ignore the sum of 1 root a time.