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March 30, 2024, 02:54:01 am

Author Topic: 3U Maths Question Thread  (Read 1230767 times)  Share 

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FallonXay

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Re: 3U Maths Question Thread
« Reply #810 on: October 22, 2016, 03:54:05 pm »
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We want to find where C(t)=0.3, right? Like, for what value of t will C(t)=0.3. If I SHIFT the graph down 0.3, then what used to be 0.3 is now going to be zero. Thus, it will be an x-intercept. But, just logically, if I want C(t) to equal 0.3, then in that case C(t)-0.3=0.3-0.3=0. Does that make sense? Definitely a difficult concept, but once you've seen it once you'll be able to answer any similar question :)

Ahh, rightyo. Got it - thanks!!!  :)
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RuiAce

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Re: 3U Maths Question Thread
« Reply #811 on: October 22, 2016, 04:05:09 pm »
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and how would you solve Q6/10?
For 6, you can go straight to your reference sheet. Instead of a theta, you have 2x, so just let 2x equal the general solution for arcsin(a) (again, this is on your formula sheet). Divide through by two, and you're done!

10 is an interesting one. I can't really think of a better method than guess and check to be honest. Nah, scratch that. We can see clearly that x=3 is a solution to our given equation. However, x=3 is NOT a solution to a) or b), and if it were we would be dividing by zero. So, c) and d) are left. Unfortunately, x=3 is a solution to both of them, so that's not very helpful. x=2 is also a solution to our given equation, but NOT for d), leaving us with c) as the answer :)

Jake
Question 10 was addressed in post #721

FallonXay

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Re: 3U Maths Question Thread
« Reply #812 on: October 22, 2016, 04:29:09 pm »
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I've got another 3 questions  ;D

Question f [attachment 1]:

Question f ii [attachment 2]:

Question 12 d i [attachment 3]:

P.s sorry for the spam of questions; and thank you for answering all of my questions - much appreciated!!!  :)
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RuiAce

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Re: 3U Maths Question Thread
« Reply #813 on: October 22, 2016, 04:46:28 pm »
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I've got another 3 questions  ;D

Question f [attachment 1]:

Question f ii [attachment 2]:

Question 12 d i [attachment 3]:

P.s sorry for the spam of questions; and thank you for answering all of my questions - much appreciated!!!  :)




RuiAce

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Re: 3U Maths Question Thread
« Reply #814 on: October 22, 2016, 04:50:41 pm »
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I've got another 3 questions  ;D

Question f [attachment 1]:

Question f ii [attachment 2]:

Question 12 d i [attachment 3]:

P.s sorry for the spam of questions; and thank you for answering all of my questions - much appreciated!!!  :)
Next one addressed in post #710

FallonXay

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Re: 3U Maths Question Thread
« Reply #815 on: October 22, 2016, 04:58:33 pm »
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Also, for this question. How come the answers don't adjust the domain so that x+pi/6 = 7pi/3?
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RuiAce

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Re: 3U Maths Question Thread
« Reply #816 on: October 22, 2016, 05:00:23 pm »
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Also, for this question. How come the answers don't adjust the domain so that x+pi/6 = 7pi/3?

FallonXay

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Re: 3U Maths Question Thread
« Reply #817 on: October 22, 2016, 05:03:14 pm »
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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #818 on: October 22, 2016, 05:13:02 pm »
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I've got another 3 questions  ;D

Question f [attachment 1]:

Question f ii [attachment 2]:

Question 12 d i [attachment 3]:

P.s sorry for the spam of questions; and thank you for answering all of my questions - much appreciated!!!  :)

And I'll tag in for the last one. Notice that when we have a right angle, it is almost definitely related to the product of the gradients being equal to -1 (since they are perpendicular). So:



Therefore we have the relationship:



Since P is in line with the points B and C, we know that we can simply say that P has coordinates \(\left(t,\frac{t^2}{k}\right)\). That's the parametric relationship specified in the question, we're done (it is also a parabola, if you wanted to prove that you could!) ;D

Key Point: If theres a right angle in a locus question, find the gradients, put their product equal to -1, and you'll almost always have what you need. :)

jamgoesbam

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Re: 3U Maths Question Thread
« Reply #819 on: October 22, 2016, 05:14:12 pm »
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Hi there!! I've seen multiple answers to this long binomial question and... nope still don't really get it :'D Could you pleaseee explain this question ('cause you guys have really good explanations)! Thanks heaps!! :)

RuiAce

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Re: 3U Maths Question Thread
« Reply #820 on: October 22, 2016, 05:15:54 pm »
+1
Hi there!! I've seen multiple answers to this long binomial question and... nope still don't really get it :'D Could you pleaseee explain this question ('cause you guys have really good explanations)! Thanks heaps!! :)
Addressed in post #253

WLalex

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Re: 3U Maths Question Thread
« Reply #821 on: October 22, 2016, 05:35:58 pm »
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Hey! wondering if someone can help me out with part (iv)

Tried to find the cartesian equation but finding a value for t and substituting that into y and as you can see (I've attached my working), it feels like I'm on a road to nowhere ahaha. So is this the right way to attack this question or am i missing something?

This is usually how I would do these but seems not to be working
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RuiAce

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Re: 3U Maths Question Thread
« Reply #822 on: October 22, 2016, 05:43:20 pm »
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Hey! wondering if someone can help me out with part (iv)

Tried to find the cartesian equation but finding a value for t and substituting that into y and as you can see (I've attached my working), it feels like I'm on a road to nowhere ahaha. So is this the right way to attack this question or am i missing something?

This is usually how I would do these but seems not to be working
Your photo's a bit small...

WLalex

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Re: 3U Maths Question Thread
« Reply #823 on: October 22, 2016, 05:45:26 pm »
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Your photo's a bit small...


yeah sorry, it was the only size that fit. Woah that is so much easier...how do we know what method to use?

Thanks also :)
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RuiAce

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Re: 3U Maths Question Thread
« Reply #824 on: October 22, 2016, 05:47:11 pm »
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yeah sorry, it was the only size that fit. Woah that is so much easier...how do we know what method to use?

Thanks also :)
t and 1/t, compared to t2 and 1/t2, is immediately suggestive of squaring.

But you need to be careful of what happens AFTER squaring.