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March 19, 2024, 10:51:28 pm

Author Topic: 3U Maths Question Thread  (Read 1227559 times)  Share 

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RuiAce

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Re: 3U Maths Question Thread
« Reply #3540 on: July 16, 2018, 10:21:08 pm »
+4
Hi!
I was just wondering if someone could show me the working out for this question. I can't get to the answer, but I don't know where I'm going wrong in my working out.
(the answer is apparently C?)
Thanks :)
\begin{align*}v&= 2e^{-\frac{x}{2}}\\ v^2 &= 4e^{-x}\\ \frac12 v^2 &= 2e^{-x}\\ \frac{d}{dx} \left( \frac12 v^2 \right) &= \frac{d}{dx} 2e^{-x}\\ \ddot{x} &= -2e^{-x} \end{align*}
Now sub \(x=-2\).

martinarena_

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Re: 3U Maths Question Thread
« Reply #3541 on: July 16, 2018, 10:50:28 pm »
+1
\begin{align*}v&= 2e^{-\frac{x}{2}}\\ v^2 &= 4e^{-x}\\ \frac12 v^2 &= 2e^{-x}\\ \frac{d}{dx} \left( \frac12 v^2 \right) &= \frac{d}{dx} 2e^{-x}\\ \ddot{x} &= -2e^{-x} \end{align*}
Now sub \(x=-2\).

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arii

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Re: 3U Maths Question Thread
« Reply #3542 on: July 21, 2018, 06:34:23 pm »
0
Solve sin^-1 (2x-1) = cos^-1(x)
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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #3543 on: July 21, 2018, 07:12:16 pm »
+1
Solve sin^-1 (2x-1) = cos^-1(x)

Hey! I think easiest way is to draw a diagram of a triangle with an unknown angle. You know the sine of that angle is 2x-1, and the cosine of that angle is x, so I'd make a triangle with three sides:

1: 2x-1
2: x
3: 1, which is also the hypotenuse

This means the unknown angle will be the angle referenced by the two sides of the equation above. Then it is just a matter of using Pythagoras:



However, \(x\neq0\), because if you substitute it in:



It clearly doesn't work, there is the error there caused by the differing ranges of the two inverse trig functions. So, the answer is \(x=\frac{4}{5}\) ;D

supR

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Re: 3U Maths Question Thread
« Reply #3544 on: July 21, 2018, 07:29:14 pm »
0
Hey, does anyone have a guide or any instructions on how to find the restrictions of loci. It seems like a common 1-2 mark question and I'm unsure how to solve it.

Thanks!
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RuiAce

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Re: 3U Maths Question Thread
« Reply #3545 on: July 21, 2018, 07:44:52 pm »
+3
Hey, does anyone have a guide or any instructions on how to find the restrictions of loci. It seems like a common 1-2 mark question and I'm unsure how to solve it.

Thanks!
The only real "rule of thumb" there is to a restriction on loci is to think about when things start breaking down (e.g. square of a number is negative). Ultimately it still depends on the question itself, which you should post if you have trouble with it.
« Last Edit: July 21, 2018, 07:47:21 pm by RuiAce »

_laurenteague

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Re: 3U Maths Question Thread
« Reply #3546 on: July 22, 2018, 05:23:20 pm »
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Hey Jake, I was wondering how you differentiate this: y = cos(inverse sinx)

Opengangs

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Re: 3U Maths Question Thread
« Reply #3547 on: July 22, 2018, 05:48:59 pm »
+1
Hey Jake, I was wondering how you differentiate this: y = cos(inverse sinx)
Hey, _laurenteague.

This is a composite function! We can see this because we have an "outer" function and an "inner" function. And when we have a composite function or a function of the form \( f(g(x)) \), we use the chain rule to differentiate!

So, the chain rule basically tells us that:
\[ y = f(g(x)) \Rightarrow \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]

In this case, our "\(f(x)\)" is the "outer" function \(\cos(x)\) and our \(g(x)\) is the "inner" function \(\sin^{-1}(x)\).

Thus, our derivative becomes:
\[ \begin{align*}\frac{dy}{dx} &= -\sin(\sin^{-1}(x)) \cdot \frac{1}{\sqrt{1 - x^2}} \\ &= -\frac{x}{\sqrt{1-x^2}}.\end{align*}\]

infectmarshroom

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Re: 3U Maths Question Thread
« Reply #3548 on: July 23, 2018, 11:53:23 pm »
0
Hey, I can’t really seem to grasp what the last part of this question is asking. I seem to conclude an answer of 2, however this does not fit the range of the inverse of g^-1(x).

Q12b) iii)
https://thsconline.github.io/s/?view=5338&n=Fort%20St%202017%20w.%20sol

RuiAce

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Re: 3U Maths Question Thread
« Reply #3549 on: July 24, 2018, 12:01:31 am »
+2
Hey, I can’t really seem to grasp what the last part of this question is asking. I seem to conclude an answer of 2, however this does not fit the range of the inverse of g^-1(x).

Q12b) iii)
https://thsconline.github.io/s/?view=5338&n=Fort%20St%202017%20w.%20sol
Did you mean 12d) iii)?

(Will look at it in the meantime. Edit: A quick glance suggests that it just wants you to find an expression for the function \( g^{-1}(f(x)) \). I haven't properly read their provided solutions yet.)
« Last Edit: July 24, 2018, 12:06:12 am by RuiAce »

infectmarshroom

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Re: 3U Maths Question Thread
« Reply #3550 on: July 24, 2018, 12:09:14 am »
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Yes part d) sorry, it has a monetary value, and talks about it being symmetrical about the point 1. Don’t really understand what it’s getting at.

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #3551 on: July 24, 2018, 01:00:26 am »
+3
Yes part d) sorry, it has a monetary value, and talks about it being symmetrical about the point 1. Don’t really understand what it’s getting at.

Hey!

We know \(f(\phi)\) will be some value between 2 and 3, since \(x=\phi\) puts it somewhere along the original semicircle between B and C. Now, the value that \(g^{-1}\) will spit out in response to that value is the \(x\) value corresponding to that \(y\) value for the function \(g\), which is the quarter circle from A to B. If you can imagine a horizontal line drawn for \(y=f(\phi\), we want the x-coordinate of the second POI with the semicircle, besides that for \(x=\phi\).

So, the question becomes how to generalise it. If \(\phi=1.5\) then \(g^{-1}f(\phi)=0.5\), because of that symmetry about \(x=1\) that \(f\) possesses. You could pick a few other test cases, but you end up figuring out that the answer is \(2-\phi\).

This is a hard thing to explain without actually pointing to a diagram as I say it, but hopefully this helps some! If not I'll try again tomorrow ;D


RuiAce

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Re: 3U Maths Question Thread
« Reply #3552 on: July 24, 2018, 10:11:36 am »
+3
Hmm, this is what I make of the question. (I mean, what Jamon said seems to make sense but I wanted to give a go myself anyway.)

1. \( f(x)\) is not invertible. It fails the horizontal line test since it involves the entire semi-circle.

2. \(g(x)\), however, is invertible. This falls out from the domain restriction.

Note that \(g(x)\) is the quadrant of the circle with domain \( 0\leq x \leq 1\) and range \(2 \leq x \leq 3\). The graph of the inverse can be, of course, determined by a simple reflection about \(y=x\), however at the very least we know that it has domain \(2 \leq x \leq 3\) and range \( 0\leq x \leq 1\).

Picture

3. I will temporarily overkill the question by also considering what happens when \( 0 \leq \phi \leq 1\). For such a fixed value of \(\phi\), we note that \(f(x)\) and \(g(x)\) represent the same function. Therefore, since \(g^{-1}(x)\) is the inverse function of \(g(x)\) over the relevant domains, it is also the inverse of \( f(x)\) here. Thus \(g^{-1}(f(x)) = x\) for all \(0\leq x \leq 1\). Therefore, in THIS case, \(g^{-1}(f(\phi)) = \phi\).

4. Now we consider what the question is asking for, i.e. when \( 1\leq \phi \leq 2\). Here, it is not immediately true that \( g^{-1} (f(\phi)) = \phi \) anymore. Reason being, \(\phi\) does not lie in the correct interval \(0 \leq x \leq 1\).

5. As it turns out, we can rewrite a (linear) expression in terms of \(\phi\), that falls in the correct interval:
\[ 1\leq \phi \leq 2\\ -1\leq \phi-2\leq 0\\ 0\leq 2-\phi \leq 1 \]
So we see that \( \boxed{0\leq 2-\phi \leq 1} \). Therefore, we can deduce that \( \boxed{g^{-1}(f(2-\phi)) = 2-\phi }\).

6. Finally, we deal with the issue of symmetry. The above expression is still kinda useless, because we want \(\phi\) under the function \(f\), not \(2-\phi\). It turns out that it's a stepping stone.
\[ \text{The function }f(x) = 2+\sqrt{1-(x-1)^2}\text{, defined over the natural domain }0\leq x\leq 2\\ \text{is symmetric about }x=1\\ \text{thanks to the properties of the semi-circle.} \]
Picture
I mean, they give you a picture in the question, but here's one from desmos:

\[ \text{We know from our study of even functions in Year 11 that}\\ \text{if a curve is symmetric about the line }x=0\\ \text{then }f(x) = f(-x). \]
\[ \text{As it turns out, there is a theorem that states}\\ \text{if a curve is symmetric about the line}\boxed{x=a}\\ \text{then }\boxed{f(x)=f(2a-x)} \]
The proof of this theorem is typically done pictorially. It involves measuring the (perpendicular) distances of a curve that is symmetric about the line \(x=a\). Here's a rough idea:
Picture



« Last Edit: July 24, 2018, 11:46:40 am by RuiAce »

infectmarshroom

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Re: 3U Maths Question Thread
« Reply #3553 on: July 24, 2018, 05:57:46 pm »
+1
Thanks a lot Rui. Parts 1-5 of your explanation really got it over the line for my brain to comprehend   ;D ;D ;D

martinarena_

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Re: 3U Maths Question Thread
« Reply #3554 on: July 26, 2018, 08:41:33 pm »
0
Hi, how do I solve q24 and 25 for permutations? Thank you in advance :))
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