(Finalised) Official BOS Mathematics Trial Examinations. (1 Viewer)

[Solution]

Carrot, just letting you know that I changed my mind and I am only doing the 3u paper. I don't wish to do the 4u paper since I think I will struggle A LOT.

 

[deswa1]

Fuark Carrot, I think I finally got that complex question... I'm going to get owned b ythis test (though the mechanics one I thought was way easier even though it was more marks). I would have missed something that made the complex easy I think- could you outline the method you used for this (can you PM me though so you don't spoli it for others). Thanks

 

[RealiseNothing]

Too straightforward, and didn't have enough space for it.

Warning: Possible solution to the complex numbers question, read at your own discretion.



Graph

Construct a triangle with vertices

Let the inclined angle be

Now:



Hence:



But since , the argument of makes a smaller angle, which can be seen as the triangle has the same "run" but a smaller "rise".

Hence:



Now substituting in what was equal to:



And since it is trivial that

Therefore:



As required.

 

[RealiseNothing]

Another way would be to use the fact that:



(for x>0).

And since the argument really comes from:



You can go on to prove the given inequality.

 

[deswa1]

Warning: Possible solution to the complex numbers question, read at your own discretion.



Graph

Construct a triangle with vertices

Let the inclined angle be

Now:



Hence:



But since , the argument of makes a smaller angle, which can be seen as the triangle has the same "run" but a smaller "rise".

Hence:



Now substituting in what was equal to:



And since it is trivial that

Therefore:



As required.

If the inclined angle is theta, how does sin(theta)=1/k? Isn't tan(theta) equal to 1/k because k is the length of the adjacent side not the hypotenuse?

 

[RealiseNothing]

If the inclined angle is theta, how does sin(theta)=1/k? Isn't tan(theta) equal to 1/k because k is the length of the adjacent side not the hypotenuse?

Shit yer I kinda of skipped a step in my working, but:



For all

So:



So it still works

(I'm hoping I'm not derping and this is right).

 

[Carrotsticks]

Is this question right? I ended up with unless i messed up somewhere..

Also with the complex number question, what if |z| > k? Wouldn't this mean the max arg would be pi? Not too sure about this one

Edit: Nvm inequality doesn't necessarily correspond to max, but just less than lol

You must have made a silly error somewhere.

Carrot, just letting you know that I changed my mind and I am only doing the 3u paper. I don't wish to do the 4u paper since I think I will struggle A LOT.

Sure thing, no worries.

Fuark Carrot, I think I finally got that complex question... I'm going to get owned b ythis test (though the mechanics one I thought was way easier even though it was more marks). I would have missed something that made the complex easy I think- could you outline the method you used for this (can you PM me though so you don't spoli it for others). Thanks

I'm not going to give you the complete answer, but here's a starting point that should help you finish the question in about 2 lines.

 

[deswa1]

I'm not going to give you the complete answer, but here's a starting point that should help you finish the question in about 2 lines.

Crap. Yes. Very nice. I didn't think of that- that's heaps neat (and WAY easier than what I did).

Lol I was trying to get that thing with the k as the hypotenuse but the only way I could think of was creating a circle with radius k (which is bigger than the unit circle) but that got messy too quickly so I stuffed that and just did ordinary inequalities stuff I guess I need to go over complex geometry again haha

 

[Sy123]

Any other scrapped 3U questions?

 

[Carrotsticks]

Any other scrapped 3U questions?

Here you go.

*Moves about in Simple Harmonic Motion.

 

[seanieg89]

@Carrot.

Didn't end up doing all of the exam on the bus, took a nap about 30 mins in and slept until sydney lol. But I think I found one or two issues. Will PM when I have looked more thoroughly.

 

[Carrotsticks]

Cheers =)

I think I know what you are talking about. I found some things earlier today.

 

[seanieg89]

From what I can remember, an early MC question was wrong, and the last MC question didn't have any choices written yet lol. I think some subtleties regarding whether inequalities were strict or not were missed in the last question but that was from a cursory glance, I will give you something more thorough in my PM. Might have time to do it tonight.

 

[Carrotsticks]

Sure thing. The ones you mentioned were fixed earlier today already, so should be all good. Let me know more later =)

 

[deswa1]

Just did that circle/simple harmonic one How many scrap questions do you have BTW lol? (not that I'm complaining- I love these questions)

 

[Carrotsticks]

Just did that circle/simple harmonic one How many scrap questions do you have BTW lol? (not that I'm complaining- I love these questions)

Care to show us your answer?

I have a whole bunch of scrapped questions.

 

[seanieg89]

Check 5 if you haven't already, in particular the meaning of the word "complex"...

edit. and after that I'm pretty happy with the mc.

 

[Carrotsticks]

Check 5 if you haven't already, in particular the meaning of the word "complex"...

edit. and after that I'm pretty happy with the mc.

*non-trivial

 

[deswa1]

Care to show us your answer?

I have a whole bunch of scrapped questions.

Umm it takes me ages to latex so I'll just run through my working.

1) is fairly straightforward- just use Pythag and trig in the triangle and its done.

2) First I differentiated the result in 1) to find d(theta)/dx. Now we have to find dx/dt (which is the velocity). But dx/dt=dx/d(theta) times d(theta)/dt. And it is given that d(theta)/dt=1 so that term pretty much cancels itself out so if we flip the derivative, we get dx/dt which is v. Then I squared both sides, halved them and differentiated with respect to x because a (acceleration)=d/dx (1/2v^2). And then this produced a result in the form of -n^2x which is simple harmonic. Hope you can follow this lol and I didn't make a mistake somewhere...

 

[seanieg89]

*non-trivial

Not sure if that is a commonly used phrase? Perhaps "non-real", or "with non-zero imaginary part" are clearer. Up to you .