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March 29, 2024, 12:56:41 am

Author Topic: 3U Maths Question Thread  (Read 1230262 times)  Share 

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Goodwil

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Re: 3U Maths Question Thread
« Reply #720 on: October 05, 2016, 04:19:01 pm »
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Hi, can someone please explain how to do this question?

RuiAce

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Re: 3U Maths Question Thread
« Reply #721 on: October 05, 2016, 04:30:02 pm »
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Hi, can someone please explain how to do this question?






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« Last Edit: October 05, 2016, 04:32:29 pm by RuiAce »

massive

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Re: 3U Maths Question Thread
« Reply #722 on: October 05, 2016, 04:53:59 pm »
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how do you do this guys?

massive

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Re: 3U Maths Question Thread
« Reply #723 on: October 05, 2016, 04:55:51 pm »
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sorry i have another question. How do you do part iv, it's one mark but wth :S

kevin217

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Re: 3U Maths Question Thread
« Reply #724 on: October 05, 2016, 05:29:54 pm »
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As an exercise, you may wish to expand (2x3-1/x)n on WolframAlpha. Use n=1,2,3,5,6,7,9,10,11 and you will find there is no constant term (or in other words, the constant term is 0). When n=4,8,12,... however, you'll find you get a constant

Edit: Actually Jake answered this one already. Refer to post 601.

On the reference sheet there is two general expansions for binomial theorem. However, if you used the other expansion you get a different answer? I get n = (4/3)k

MightyBeh

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Re: 3U Maths Question Thread
« Reply #725 on: October 05, 2016, 05:44:45 pm »
+1
how do you do this guys?
Not 100% confident on this one but if the answer is A I can share my working. ::)

sorry i have another question. How do you do part iv, it's one mark but wth :S

i) for an inverse to exist, f(x) must be a one to one function (there must be a unique y-value for every x-value). By restricting the domain to one side of the turning point at x =2 (after which y-values will begin to repeat) we turn f(x) into a one-to-one function, thus an inverse exists for values including and on one side of the point (2,0), in this case x >= 2.

ii) Domain and range of the inverse are the same as the original function, but swapped. This is because (x,y) → (y,x). The range of the inverse becomes [2,∞). The domain will be [0,∞) because that's the range of the original function.

iii) f(x)=f-1(x) implies that f(x)=x because any intersections between the function and its inverse will occur on the line y=x.



Edit: Make a mistake on this one, fixed it up. ;)
Edit Edit: Make a different mistake that I didn't fix up the first time.
iv) as the inverse function in this case is 2-sqrt(x) because k < 2.
« Last Edit: October 05, 2016, 06:38:19 pm by MightyBeh »
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RuiAce

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Re: 3U Maths Question Thread
« Reply #726 on: October 05, 2016, 06:18:15 pm »
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On the reference sheet there is two general expansions for binomial theorem. However, if you used the other expansion you get a different answer? I get n = (4/3)k



RuiAce

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Re: 3U Maths Question Thread
« Reply #727 on: October 05, 2016, 06:21:31 pm »
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how do you do this guys?


Inspiration: The tangent is there but we're only interested in the angles. Alternate segment theorem could work but it seems too dodgy here.




Edit: Actually, if you join BD the alternate segment theorem takes care of it more quickly.
« Last Edit: October 05, 2016, 06:32:23 pm by RuiAce »

RuiAce

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Re: 3U Maths Question Thread
« Reply #728 on: October 05, 2016, 06:24:46 pm »
+1
i) for an inverse to exist, f(x) must be a one to one function (there must be a unique y-value for every x-value). By restricting the domain to one side of the turning point at x =2 (after which y-values will begin to repeat) we turn f(x) into a one-to-one function, thus an inverse exists for values including and on one side of the point (2,0), in this case x >= 2.
Love the answer. Comments though.

1. They don't teach interval notation in the HSC. They only use x>=2 to denote [2, inf)
2. There's no definition of 'one-to-one' and 'onto' functions in the HSC either. If a function is one-to-one, it is just called 'invertible'.
No need for bijective functions in the HSC. This is because the HSC defines the range but absolutely ignores the concept of a 'codomain'
« Last Edit: October 05, 2016, 06:26:58 pm by RuiAce »

massive

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Re: 3U Maths Question Thread
« Reply #729 on: October 05, 2016, 06:31:42 pm »
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Not 100% confident on this one but if the answer is A I can share my working. ::)

Yhep the answers A, Howdu get it :O

----

And for part iv, the answer is actually 4-k :S

RuiAce

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Re: 3U Maths Question Thread
« Reply #730 on: October 05, 2016, 06:32:39 pm »
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Yhep the answers A, Howdu get it :O

----

And for part iv, the answer is actually 4-k :S
It is A. I just made a typo in my working.

massive

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Re: 3U Maths Question Thread
« Reply #731 on: October 05, 2016, 06:35:09 pm »
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It is A. I just made a typo in my working.

Oh yeaah thanks mate!  ;D

Also how do you do part iv for the second q attached (ans: 4-k)

RuiAce

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Re: 3U Maths Question Thread
« Reply #732 on: October 05, 2016, 06:39:55 pm »
+1
Oh yeaah thanks mate!  ;D

Also how do you do part iv for the second q attached (ans: 4-k)


MightyBeh

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Re: 3U Maths Question Thread
« Reply #733 on: October 05, 2016, 06:41:09 pm »
+1
Love the answer. Comments though.

1. They don't teach interval notation in the HSC. They only use x>=2 to denote [2, inf)
2. There's no definition of 'one-to-one' and 'onto' functions in the HSC either. If a function is one-to-one, it is just called 'invertible'.
No need for bijective functions in the HSC. This is because the HSC defines the range but absolutely ignores the concept of a 'codomain'

Thanks for that. :)

Oh yeaah thanks mate!  ;D

Also how do you do part iv for the second q attached (ans: 4-k)
I fixed it up, my bad. Because k < 2 the inverse is actually 2 - sqrt(x), not 2 + sqrt(x) like I had before.

 Edit: Rui beat me to it but yolo  8)
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massive

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Re: 3U Maths Question Thread
« Reply #734 on: October 05, 2016, 09:01:49 pm »
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I fixed it up, my bad. Because k < 2 the inverse is actually 2 - sqrt(x), not 2 + sqrt(x) like I had before.

 Edit: Rui beat me to it but yolo  8)

wait i still don't get why you take the negative root, even tho k<2