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March 28, 2024, 08:24:59 pm

Author Topic: 3U Maths Question Thread  (Read 1230184 times)  Share 

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spnmox

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Re: 3U Maths Question Thread
« Reply #4230 on: October 09, 2019, 02:39:18 pm »
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hey guys, for the attached question, the answer is A, but I'm not sure why. How do you get the value for f'(1)?

RuiAce

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Re: 3U Maths Question Thread
« Reply #4231 on: October 09, 2019, 03:25:28 pm »
+1
hey guys, for the attached question, the answer is A, but I'm not sure why. How do you get the value for f'(1)?
You can find a solution to this in the compilation.

spnmox

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Re: 3U Maths Question Thread
« Reply #4232 on: October 09, 2019, 03:35:10 pm »
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You can find a solution to this in the compilation.

thanks rui! i read the solutions. i'm still not sure about the f'(1) value. when we sketch the tangent line, it doesn't pass through (0,0), so how are you able to compare it to y=x? or does it pass through (0,0)?

RuiAce

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Re: 3U Maths Question Thread
« Reply #4233 on: October 09, 2019, 03:39:02 pm »
+2
thanks rui! i read the solutions. i'm still not sure about the f'(1) value. when we sketch the tangent line, it doesn't pass through (0,0), so how are you able to compare it to y=x? or does it pass through (0,0)?
Basically just sketch the line \(y=x\) alongside it.

Note that this is in fact doable, because the point \( (1,1) \) was plotted on the diagram. So \(y=x\) will just be the line through \((0,0)\) and \((1,1)\).

And then sketch the tangent. It should be steeper than that line.

Edit 2: Actually, if you sketch the tangent carefully, you'll find that despite how it misses the point \( (0,0)\), you wouldn't actually need that fact! The diagram not NOT being to scale plays a huge role here.
« Last Edit: October 09, 2019, 03:42:25 pm by RuiAce »

spnmox

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Re: 3U Maths Question Thread
« Reply #4234 on: October 09, 2019, 04:00:34 pm »
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Basically just sketch the line \(y=x\) alongside it.

Note that this is in fact doable, because the point \( (1,1) \) was plotted on the diagram. So \(y=x\) will just be the line through \((0,0)\) and \((1,1)\).

And then sketch the tangent. It should be steeper than that line.

Edit 2: Actually, if you sketch the tangent carefully, you'll find that despite how it misses the point \( (0,0)\), you wouldn't actually need that fact! The diagram not NOT being to scale plays a huge role here.

okay, so even if y=x passes through (0,0), and the tangent does not pass through (0,0), we can still compare them and say that the tangent is steeper?

diggity

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Re: 3U Maths Question Thread
« Reply #4235 on: October 09, 2019, 04:48:22 pm »
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Hey, I managed to get every question on the 2004 paper except for 7) a) iii). Was hoping someone could help me out with it, thank you.



Also, for question 5) b) iv), the solutions i found assumed that the intersection lies on y=x. When I did the question, I just let f(x) = f^-1(x) and factorised to get the required equation. They just let x = f(x). We can do this? I can see  the intuition but I don't know if it's correct mathematical reasoning.



Thank you!
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spnmox

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Re: 3U Maths Question Thread
« Reply #4236 on: October 09, 2019, 05:30:07 pm »
+1
Hey, I managed to get every question on the 2004 paper except for 7) a) iii). Was hoping someone could help me out with it, thank you.

(Image removed from quote.)

Also, for question 5) b) iv), the solutions i found assumed that the intersection lies on y=x. When I did the question, I just let f(x) = f^-1(x) and factorised to get the required equation. They just let x = f(x). We can do this? I can see  the intuition but I don't know if it's correct mathematical reasoning.

(Image removed from quote.)

Thank you!

hi, i'm not a moderator but a function and its inverse always intersect on y=x, it's just one of the properties of inverse functions:) and i have no idea for your first one.

fun_jirachi

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Re: 3U Maths Question Thread
« Reply #4237 on: October 09, 2019, 05:41:46 pm »
+1
Hey, I managed to get every question on the 2004 paper except for 7) a) iii). Was hoping someone could help me out with it, thank you.

(Image removed from quote.)

Also, for question 5) b) iv), the solutions i found assumed that the intersection lies on y=x. When I did the question, I just let f(x) = f^-1(x) and factorised to get the required equation. They just let x = f(x). We can do this? I can see  the intuition but I don't know if it's correct mathematical reasoning.

(Image removed from quote.)

Thank you!

Hey there!

With the first question, note that the equation of the height of the tide at the harbour and the height of the tide at the wharf is effectively the same function, but shifted.

ie. the equation of the tide's height at the harbour, where b is the midpoint of the minimum and the maximum, and z is the height of the tide is


Note that when t=0 (ie. 1am, since tides occur 1 hour earlier), the tide reaches a maximum ie. α=0, since z=b+3 and we know that the range is 6.
Hence, the full equation will just be


Now, since the water level will always be decreasing, we just need to know when the water level falls below 2 metres above low tide ie. 1 metre below the midpoint, and we just need to make sure the ship leaves before that time. ie. we have that

Solving for t,

ie. the ship must reach the harbour prior to 4:48, meaning it must leave by 4:28 am.

The intuition behind all this is just having some equation of motion, given a certain height and then solving for the time it reaches that height, if that makes sense.

Hope this makes sense :)

For the second question, spnmox is pretty much spot on; for any function, its inverse relation is effectively just the same thing, but flipped in y=x. So essentially wherever it intersects y=x, its relation will also do the same. Nowhere else does an intersection occur for a function and its inverse; hence we can just use x=f(x) as opposed to f(x)=f-1(x) which is generally harder to solve.
« Last Edit: October 09, 2019, 05:46:47 pm by fun_jirachi »
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HSC 2019: Physics [92] | Chemistry [93] | English Adv [87] | 3U Maths [98] | 4U Maths [97]
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Asking good questions

RuiAce

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Re: 3U Maths Question Thread
« Reply #4238 on: October 09, 2019, 06:29:03 pm »
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okay, so even if y=x passes through (0,0), and the tangent does not pass through (0,0), we can still compare them and say that the tangent is steeper?
Yes, because we don't have any issue with axis scale here.

(Otherwise this would be a concern.)

sady123

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Re: 3U Maths Question Thread
« Reply #4239 on: October 10, 2019, 03:57:01 pm »
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hi can i have help with this question? thanks!

RuiAce

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Re: 3U Maths Question Thread
« Reply #4240 on: October 13, 2019, 10:12:59 am »
+2
hi can i have help with this question? thanks!
The first statement can be, and in fact you should try to. But you'll find that the second statement cannot be.

Although I'm not sure what the context of the new Maths In Focus is, there are at least two ways of arguing that b) cannot be proven by induction. The first is that because we're not told anything about \(x\), in theory we have to just take for granted that it is a real number. But induction can only be applied on all positive integers, as far as MX1 is concerned.

For the other cases, we just assume \(x\) is a positive integer anyway. The second is that the base case cannot really be verified computationally, that is whether or not \(\sin 1 \cot 1 = \cos 1\). But unless we literally assume that \(\cot x = \frac{\cos x}{\sin x}\), which is basically what we're trying to prove in the first place, we cannot check that this statement holds. (Note that calculator verifications are not proofs.)

Note: There may be a third way in that the inductive step does not fall out nicely. Recall that we'd have to assume the statement holds for \(x=k\), and hence show that it holds for \(x=k+1\). You can use the \(\sin(a+b)\) and \(\tan(a+b)\) formulas to try verifying this, but you cannot assume \(\cot \theta = \frac{\cos\theta}{\sin\theta}\) or any equivalent results like \( \tan\theta = \frac{\sin\theta}{\cos\theta} \), which may potentially damage things. (Note that in the HSC, the definition we assume is that \(\cot\theta = \frac1{\tan\theta}\).

mani.s_

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Re: 3U Maths Question Thread
« Reply #4241 on: October 24, 2019, 09:34:38 am »
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Hi, can someone help me with question 17b???

Thanks :)

Youssefh_

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Re: 3U Maths Question Thread
« Reply #4242 on: October 28, 2019, 06:50:30 pm »
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I am very confused, I do not know how to calculate this, can please someone help me

DrDusk

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Re: 3U Maths Question Thread
« Reply #4243 on: October 28, 2019, 06:57:44 pm »
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I am very confused, I do not know how to calculate this, can please someone help me
Just type it into a calculator =) Or is it no calculators allowed?

spnmox

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Re: 3U Maths Question Thread
« Reply #4244 on: October 29, 2019, 05:59:40 pm »
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Hello,

A family of 6 adults and 3 children are randomly sitting around a circular table.
What is the number of possible seating arrangements if none of the children sit together?

I did 3!*6C2*2!*4! (putting 3 kids and 2 adults in a group, then arranging the 5 total 'things')
I got 4320, but the answer is 14400.

Why would my approach not work?