VCE Specialist 3/4 Question Thread!

[Alwin]

Can someone help me with Q2 f) Short Answer Exam 2 2009?

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2009specmaths2-w.pdf

Edit: Also, Q1 e, ii)Short answer Exam 2 2011 as well

Q2 (f)

Hint: It's the area between two circles, and by symmetry it's 1/12 of the 'donut' shape

Answer

Q1 (e) ii)

Hint: Overlap (blue) area = 2 x area of quarter circles - area of the red square

Answer

[Robert123]

If the second derivative of a stationary point is 0, does that make it a stationary point of inflection?

[Crystall97]

As far as I know, that is correct (:

[Thorium]

If the second derivative of a stationary point is 0, does that make it a stationary point of inflection?

Yes that is correct, when we have the first and second drivative zer at a point, then this is the stationery point of inflexion. However, when we only have the second derivative zero at a point, then this is only a point of inflexion (not stationery). This is where the gradient is maximum/minimum, depending on whether it is increasing or decreasing.

[lzxnl]

If the second derivative of a stationary point is 0, does that make it a stationary point of inflection?

No. It does not necessarily have to be a stat point of inflection. Think y=x^4 at the origin. Second derivative is zero but obviously no inflection.
A point of inflection occurs when the second derivative changes sign. That necessarily means the second derivative is zero (assuming it's continuous; one could argue x=y^3 has a point of inflection at the origin, even though d^2y/dx^2 is not defined there). However, it can be zero without changing sign. If the second derivative has a turning point and is zero at that turning point, you won't get a point of inflection. Just look at y = x^4

[Mieow]

How would I solve this integral?


I keep getting the wrong answer 

[alchemy]

How would I solve this integral?


I keep getting the wrong answer 

I am able to get  integral 4/sqrt(25-(1-2 x)^2) dx = (-2 sin^-1)(1/5 (1-2 x))+c. Is that the correct answer?

[Zealous]

How would I solve this integral?


I keep getting the wrong answer 

[alchemy]

How would I solve this integral?


I keep getting the wrong answer 

If you're interested, my working out is attached. It's slightly shorter than Zealous', but more or less the same thing really.

[kinslayer]

I'm bored so here's an "alternative" method. See if you can spot the error

[lzxnl]

I don't actually see an error, aside from perhaps failing to qualify why sqrt(sin^2 theta) = sin theta (which is true if you want to define an inverse cosine function)

[nhmn0301]

Quick question, in relation to the proving integral question above, in the exam, if the question asks me to "show...", which method would be preferable?
Thanks!

[alchemy]

Quick question, in relation to the proving integral question above, in the exam, if the question asks me to "show...", which method would be preferable?
Thanks!

Probably Zealous' method. I took a few shortcuts as you can see in my working out; those shortcuts also don't apply to every question of this type so there's a risk. From what I understand, kinslayer has taken a few shortcuts too; as lzxnl mentioned, he didn't explicitly mention something that might've been required. Personally, kinslayer's method seems cool and unique in a way! We need more of those solutions from now on  : D

[kinslayer]

I don't actually see an error, aside from perhaps failing to qualify why sqrt(sin^2 theta) = sin theta (which is true if you want to define an inverse cosine function)

Yeah it was a trick question  I wrote it down and initially thought I must have made a mistake, but obviously the derivatives of arccos and arcsin differ only by a factor of -1 so the answers are equivalent.

Quick question, in relation to the proving integral question above, in the exam, if the question asks me to "show...", which method would be preferable?
Thanks!

Each approach is essentially doing the same thing. If I were you I would focus on practicing the approach that Zealous and Alchemy took, since they are using a known result for the integral, which means there's less technical stuff required and fewer possibilities to make a mistake on an exam. However, it is good to know how those results are derived, so I think that doing a few examples using my method may be beneficial.

[lzxnl]

Probably Zealous' method. I took a few shortcuts as you can see in my working out; those shortcuts also don't apply to every question of this type so there's a risk. From what I understand, kinslayer has taken a few shortcuts too; as lzxnl mentioned, he didn't explicitly mention something that might've been required. Personally, kinslayer's method seems cool and unique in a way! We need more of those solutions from now on  : D

Here's why you don't get those sorts of answers:

Because it's not necessary at all for VCE.
They're cool but not necessary. You may as well use what's in the course, unless you really want to be confused, which I wouldn't advise.