VCE Methods Question Thread!

[Unsplash]

Thanks so much guys! A lot of work for a multi choice question lol.

[andytime]

Hello!
I do not understand how 7pi/6 and 11pi/6 was obtained for this 'find the general solution' question. I thought you were supposed to find sin^-1(-1/2) and then subtract pi/3 but that is incorrect. Could someone please help me break down the steps to get the answer?

Thank you in advance!

[S200]

Hello!
I do not understand how 7pi/6 and 11pi/6 was obtained for this 'find the general solution' question. I thought you were supposed to find sin^-1(-1/2) and then subtract pi/3 but that is incorrect. Could someone please help me break down the steps to get the answer?

Thank you in advance!

So you do go inverse, as you said. But you are solving the inverse of a negative half, so your answer needs to be in the third or fourth quadrant.
So, you add the answer

to pi, (for 3rd quadrant, and you minus it from 2pi, (for 4th quadrant). So you end up with

Which you can then solve for x.

[secretweapon]

y = 3sin(2x) - 4 for [0 to pi/2]
find the domain over which the gradient is positive
i found it to be (0, pi/4) but the answers said [0, pi/4)
Why does the zero have the rectangle brackets?

[RuiAce]

y = 3sin(2x) - 4 for [0 to pi/2]
find the domain over which the gradient is positive
i found it to be (0, pi/4) but the answers said [0, pi/4)
Why does the zero have the rectangle brackets?

Because the gradient at 0 is equal to 6, which is also positive. So 0 should be included in the final answer.

What was your working out? Solving f’(x)>0 should have left 0 in the final answer

Edited in later on

\begin{align*}f^prime(x) &> 0\\ 6\cos 2x &> 0\\ cos 2x &> 0\end{align*}

[secretweapon]

how hard is it to get between 30-35 raw for methods? I'm starting to struggle a bit with the textbook questions, any tips for improvement?

[secretweapon]

how hard is it to get between 30-35 raw for methods? I'm starting to struggle a bit with the textbook questions, any tips for improvement?

bump

[S200]

how hard is it to get between 30-35 raw for methods? I'm starting to struggle a bit with the textbook questions, any tips for improvement?

Hey there.
My top tip right now would be to try some practise exams, as that has been an emense help to me, when I was in that boat hardly a week ago...
So yeah, head over to the VCAA website rn and grab the 2016 or 2017 exam and try a few practise questions.
If you aren't getting ahead, you could maybe try asking your teacher for a bit of extra tutoring?
Hopefully this helps. Incidentally, if any area in particular is bothering you, you could post a question on here, and myself or one of th eothers will try to answer it in a way that explains the concepts to you.
But bear in mind that no-one is on here 24/7, so you could maybe try a little patience, and resist bumping the thread for at least 24h

Cheers.

EDIT
I realised I didn't actually address your first question...
I would say that a 30-35 should be somewhat easy, averaging around 70% on your SAC's and similar on exam...
Obviously the exam weighting is heavier, so by doing the above and familiariseing yourself with the VCAA style of exam, you should do fairly well...
Exam 1 is 22% of your raw SS, and Exam 2 is 44%.

So overall, the exam is 66%, so if you focus a heap on acing that, your lack of proficiency I couldn't word that nicer, sorry.. on the textbook questions (which I totally feel btw...) can be overcome, and your overall SS should  reflect your expertise, with a 40 easily within reach.

[secretweapon]

Can someone please explain how to do this integration by recognition problem?
if y = x sin(x), find dy/dx
hence, find the value of ∫(2*xcos(x),x,-π,π/2)
don't know how to type the bold thing in latex, but it's asking to find the value of 2x cos(x) dx with the top terminal at pi/2 and the bottom terminal at -pi

i've found dy/dx for y = x sin(x) to be x cos(x) + sin(x) but i'm struggling from here on. Could someone please explain how to do this process step by step?
Thanks

[Lear]

Try separating the derivative of xsin(x) and then dividing the integral with xcos(x) by 1/2 and moving the integral of sin(x) to the right side.

[S200]

How do you differentiate this?

_________________

And can this be simplified any further?

________________
And this attachment...
I literally don't know where to start. 

Sorry, this is just getting longer and longer...

[Lear]

- -

Hey,
For your first question I just resort to using a formula provided by my teacher. There's an explanation for it but unless you really want to know why the formula works it is unnecessary. For a function u^x the derivative is simply ln(u) * u^x.

Second question - I do not think so but am not sure

Third question - So to find the maximum area you firstly need to find the area in terms of x. Area is base * height. We can make the base = x as it is a certain distance from the origin. We can also make the height be f(x) as we know the top right corner must lie on the function provided and thus its y value must be the height. Using this information the area of the square in terms of x is just 4x-x^3. Using this you can use the derivative to find the value of x

[S200]

Hey,
For your first question I just resort to using a formula provided by my teacher. There's an explanation for it but unless you really want to know why the formula works it is unnecessary. For a function u^x the derivative is simply ln(u) * u^x.

Thanks heaps Bro!

Second question - I do not think so but am not sure

Third question - So to find the maximum area you firstly need to find the area in terms of x. Area is base * height. We can make the base = x as it is a certain distance from the origin. We can also make the height be f(x) as we know the top right corner must lie on the function provided and thus its y value must be the height. Using this information the area of the square in terms of x is just 4x-x^3. Using this you can use the derivative to find the value of x

Genius!!!

Thank you SO much.

[secretweapon]

Try separating the derivative of xsin(x) and then dividing the integral with xcos(x) by 1/2 and moving the integral of sin(x) to the right side.

Hi, could you please explain with a picture if possible? Also, do you know how to find the integral of 3/(5x-2), x > 2/5
since it isn't possible to have powers of 0
Thanks

[Lear]

Hi, could you please explain with a picture if possible? Also, do you know how to find the integral of 3/(5x-2), x > 2/5
since it isn't possible to have powers of 0
Thanks

Have a solution attached. If I made a mistake somewhere please let me know.
Also for the integral of 3/(5x-2), I am not sure if this is required in the methods course. I may be wrong, however.