VCE Specialist 3/4 Question Thread!

[dandunu]

Hi guys,

I just finished the VCAA 2002 spesh exam 2, and I was wondering why the solutions said that only half the parabola should be sketched for question 5b(ii)? The assessment report doesn't really explain it.

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/02smexam2.pdf

Thanks in advance

[lzxnl]

Hi guys,

I just finished the VCAA 2002 spesh exam 2, and I was wondering why the solutions said that only half the parabola should be sketched for question 5b(ii)? The assessment report doesn't really explain it.

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/02smexam2.pdf

Thanks in advance

Read the question. It says 'positive Re(z) axis'.

Another complex numbers question (attached. VCAA Exam 2 2008). I'm able to do everything up to part f. I don't really know how to approach it. Are we able to use integration? I mean I guess we have the Cartesian equations of everything. Are there geometric ways to do this. What is the easiest way to do this?

No, you don't need integration. You're best off drawing out the circle and the shaded area. You'll find that you can relate that area to a combination of sectors/triangles. That's always the case for these questions (I haven't looked at the question much but I'd be surprised if you couldn't solve it that way).

[StupidProdigy]

How do I integrate this...?

[Escobar]

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths2.pdf
how do you do 4c?

[Floatzel98]

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths2.pdf
how do you do 4c?

I asked for help on this question a couple of days ago. It should be a couple of pages back, zsteve helped. Hopefully that helps.

[AndyCau]

can someone please help me with the question attatched

this was in the year 11 course, so no hate plz 

thanks in advance.

[cosine]

can someone please help me with the question attatched

this was in the year 11 course, so no hate plz 

thanks in advance.

Fourth roots of -16 just means to find

Now it does not specify which form to work/leave answer in, but honestly whenever using De Moivre's theorem, best to use polar form:









Now that is the first solution, and because it was raised to the power of four, then the total number of solutions will be four too. Because all the solutions are equally spaced out on the Argand diagram, we can divide 2 pi, which is the rotation of one whole argand diagram, by four, because thats how many solutions we have. (you can also do the method, but this is so much easier)

So because our base angle is , and we have four solutions that are evenly spaced out, add to the base angle and keep adding until you have a total of four solutions:



But, because we must restrict the domain of the last two solutions that are greater than pi (by simply minusing 2pi from the argument of each):

[lzxnl]

How do I integrate this...?

I feel sorry for students who actually attempted this integral by hand.

If you complete the square on the numerator, you get x^2 - 6x + 5 = (x-3)^2 - 4
Sub in y = x-3
Your integral is now 1/sqrt(y^2 - 4) dy
If you were as weird as I was in year 12, you'd know that this integral is arcosh(y/2), the inverse hyperbolic cosine (aka the inverse function of z = 1/2 * (e^y + e^-y))
Otherwise, you'd need to recall doing exercises that showed that the derivative of ln(x + sqrt(x^2+1)) = 1/sqrt(x^2 + 1) and replace the x^2 + 1 with a x^2 - 4

If you're really keen, you can make the substitution y = e^u + e^-u but I can't really explain where this comes from if you don't understand hyperbolic functions.

In either case, your answer would be ln((x-3) + sqrt((x-3)^2 - 4)) + c
No, this is not assessable without a calculator.

I'm sure this question is multi choice and they just want you to make a substitution or something.

[Floatzel98]

Does anyone have any idea why my calculator keeps getting this wrong (from VCAA 2009 Exam 2 Q5g) :



It only gives me the answer of -3.1043 and not 245.4. Even solving:



I only get the negative answer. Any idea why this is happening?

[alex1234]

Does anyone have any idea why my calculator keeps getting this wrong (from VCAA 2009 Exam 2 Q5g) :



It only gives me the answer of -3.1043 and not 245.4. Even solving:



I only get the negative answer. Any idea why this is happening?

working out with my CAS, i got two answers and i cannot find the reason why your CAS will not find two given answers, sorry maybe try updating your cas?

[Kel9901]

Does anyone have any idea why my calculator keeps getting this wrong (from VCAA 2009 Exam 2 Q5g) :



It only gives me the answer of -3.1043 and not 245.4. Even solving:



I only get the negative answer. Any idea why this is happening?

it's probably just how the calculator 'is'. when you see 'more solutions may exist' (as it does on mine), and the only answers given are not applicable (like the negative value here), you'll probably have to sketch the graph and find the value that way. not sure why that happens.

[Floatzel98]

Thanks guys. I found a couple ways to get the answer. I'll try updating my CAS though. Might help.

[zsteve]

Thanks guys. I found a couple ways to get the answer. I'll try updating my CAS though. Might help.

If this happens again (I think VCAA methods 2007 had something like this), place a domain restriction on the solution. That usually gets the correct result.

[lzxnl]

A good way to force it to find more solutions (I do this all the time on Wolfram Alpha even) is to take the solution they have, say x = 0, and then say
Solve equation for x > 0
and then 'Solve equation for x < 0'
Repeat until it says no more solutions. It should be able to find all solutions that there are if you keep making it look for new ones. Numerical methods have a tendency to get stuck at particular solutions.

[cosine]

The question asked to show all forces on the body, if I showed also the vertical and parallel forces made by the object, would full marks still be given assuming i labelled all other forces correctly?