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March 28, 2024, 10:33:22 pm

Author Topic: 3U Maths Question Thread  (Read 1230224 times)  Share 

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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #510 on: August 08, 2016, 11:45:53 am »
+1
1. y=-3sin^-1 x

For domain I got -1≤x≤1

For range I was not too sure???

Quick correction for above as well, the range of this function:



We take the normal range for inverse sine and multiply everything by three ;D

You also won't need to find the point of inflexion for that last one to do the question, the domain and range should be enough to do it properly ;D (though Rui has shown you how to do it if you want to) ;D

RuiAce

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Re: 3U Maths Question Thread
« Reply #511 on: August 08, 2016, 12:02:41 pm »
+1
hey, how do you do this guys?




I have a lecture now so I will resume this later. I have set most of part (i) up in the meantime.
_____________


I'll hand it over to you from here. Tell me if everything puzzles out eventually.
« Last Edit: August 08, 2016, 01:21:15 pm by RuiAce »

conic curve

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Re: 3U Maths Question Thread
« Reply #512 on: August 08, 2016, 01:14:17 pm »
0
How would you do this

for i and ii would you have to use your circle geometry theorems

RuiAce

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Re: 3U Maths Question Thread
« Reply #513 on: August 08, 2016, 01:20:21 pm »
+1
How would you do this

for i and ii would you have to use your circle geometry theorems
Theorems to use:

i) Alternate segment theorem
ii) Vertically opposite
iii) Alternate segment theorem (in the OTHER circle)
Base angles + matching sides on isosceles triangle
iv) <ACT = <ART (angle standing on minor arc AT)
<ACT = <TBC (base angles on isos ∆)
Therefore <ART = <TBC
Therefore <ART = ABR (vertically opposite)
 so you have a new isosceles triangle.

massive

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Re: 3U Maths Question Thread
« Reply #514 on: August 08, 2016, 03:45:19 pm »
0

_____________




What is this magic :O??
Can/should you do this for all parametrics qs?
wait i just realised, howdu get p=2 :S, the discriminant equals 0, then what??

Also howdu find the coordinates for q without using p=2?
« Last Edit: August 08, 2016, 03:52:33 pm by massive »

RuiAce

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Re: 3U Maths Question Thread
« Reply #515 on: August 08, 2016, 07:15:23 pm »
+1
Yeah I had a play around with that and it did nothing but show 0=0 lol. Ok so:





massive

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Re: 3U Maths Question Thread
« Reply #516 on: August 08, 2016, 08:54:05 pm »
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hey guys howdu do part c, ive been trying for ages idk why i cant get it  >:(

RuiAce

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Re: 3U Maths Question Thread
« Reply #517 on: August 08, 2016, 09:06:38 pm »
+1
hey guys howdu do part c, ive been trying for ages idk why i cant get it  >:(


Tell me if I did something wrong.

massive

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Re: 3U Maths Question Thread
« Reply #518 on: August 09, 2016, 12:40:53 pm »
0
hey guys how do you do part b?

Also, in general, how do you prove that a coefficient(s) is/are the greatest in an expansion. do you just have to proves the the term in front divided by that term have a value of less than one, thus it is the largest?
« Last Edit: August 09, 2016, 01:01:37 pm by massive »

massive

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Re: 3U Maths Question Thread
« Reply #519 on: August 09, 2016, 01:55:16 pm »
0
Also how do you do this?? (part ii)

RuiAce

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Re: 3U Maths Question Thread
« Reply #520 on: August 09, 2016, 02:05:02 pm »
+2
Also how do you do this?? (part ii)
This is what I think:

You can either include the digit in the group, or you cannot.

There are 10 digits. You can say yes to that digit, or no to that digit. (And note there can't be repetition.)

So that's a total of 210 outcomes.

(But, if you have to have at LEAST one digit, you can't have no no no no no no no no no no)

So in total 210 - 1
______________

If you were to use part (i), the trick is that you want to find 10C1 + 10C2 + 10C3 + ... + 10C10

I.e. 210 MINUS 10C0

Cheers Jamon :)
« Last Edit: August 09, 2016, 02:28:13 pm by RuiAce »

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #521 on: August 09, 2016, 02:24:00 pm »
+2
This is what I think:

You can either include the digit in the group, or you cannot.

There are 9 digits. You can say yes to that digit, or no to that digit. (And note there can't be repetition.)

So that's a total of 29 outcomes.

(But, if you have to have at LEAST one digit, you can't have no no no no no no no no no)

So in total 29 - 1
______________

If you were to use part (i), the trick is that you want to find 9C1 + 9C2 + 9C3 + ... + 9C1

I.e. 29 MINUS 9C0

Perfect, except there are 10 digits including 0, so that would be:



hey guys how do you do part b?

Also, in general, how do you prove that a coefficient(s) is/are the greatest in an expansion. do you just have to proves the the term in front divided by that term have a value of less than one, thus it is the largest?

In general, that method is correct! The usual idea is to take the general form of the k-th coefficient, and the general form of the (k+1)th coefficient, and take the ratio (Tk)/T_(k+1)). As soon as this is greater than one you have the k-th coefficient as greater (remember k can only be a positive integer, no decimals) ;D

Don't quite have time to punch out a full solution for (ii), however, I think it would be the exact same thing, except instead of considering the general form of coefficients, you'd consider the general form of TERMS! So:



Do the same thing and find a ratio. There should be a value of k which makes it EQUAL to 1, and that will give your equal terms ;D

massive

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Re: 3U Maths Question Thread
« Reply #522 on: August 09, 2016, 02:53:27 pm »
+1
Thanks Jamon and Rui!!!!  ;D

massive

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Re: 3U Maths Question Thread
« Reply #523 on: August 09, 2016, 03:36:22 pm »
0
guys for this question how do you do it? i just cant figure out the last lines.

massive

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Re: 3U Maths Question Thread
« Reply #524 on: August 09, 2016, 03:56:37 pm »
0
guys for this i got 124, but the answer is 9420??!?!