VCE Methods Question Thread!

[keltingmeith]

I've been doing some past exams, and I've come across "linear approximation" questions a few times. Is it in the current Methods Study Design Still?

Thanks,
James

Oh man, haven't heard those words since I did VCE. Linear approximation is well and truly gone (note: it's still technically in Specialist as Euler's method, so maybe not "well and truly")

[Oynx]

Hey need some help with a question:

If cosx=−cos(π6) and π2 < x < π find the value of x.

Please explain, don't really get it. Thanks in advance

[1729]

Hey need some help with a question:

If cosx=−cos(π6) and π2 < x < π find the value of x.

Please explain, don't really get it. Thanks in advance

Hey there Oynx!

Did you mean or

I'm guessing the second one, for one thing, cos(u) = cos(-u); for another thing, cos(u) = -cos(u + pi)

So Assuming I didn't make any silly mistakes.

(and then it's pretty obvious x is 5pi/6) both properties used there should be fairly obvious from just, like, looking at the unit circle.

Hope this makes sense, if it doesn't don't hesitate to ask!

EDIT: Just in case if anybody is confused.
The reason behind it is because of the different quadrants, because if you did 180 - 60 = x for the second question you would get 120, which doesn't satisfy 180 < x < 270.

Start by thinking about what -cos(60) is on the unit circle; cos(60) is the horizontal distance at 60 degrees, so "negative" of that is reflecting 60 degrees across the middle which gives you cos(180 - 60) = cos(120), which is the first one but cos is horizontal distance, and there are two points on the other side of the unit circle with that. Because if you change the vertical distance only, the horizontal distance is the same and the point with the same horizontal distance but different vertical distance you can get by reflecting it over the other axis, the x axis. Reflecting over the x axis you can think of as negating the thing on the inside, so cos(-120), and then add 360 because negative numbers are inconvenient and you get 240.

[MoonChild1234]

is x^2 - (integral from 0 to 2 of x^3+4) considered an integral expression, or do both components need to be integrals?

[1729]

is x^2 - (integral from 0 to 2 of x^3+4) considered an integral expression, or do both components need to be integrals?

Is an integral expression.

However, x^2 by itself is not an integral expression because it has no integrals in it. An integral expression is an expression with an integral in it, so yes that entire thing is an integral expression.

[Corey King]

Hey guys,
I gave this as my answer to a revision question and it is incorrect. I cannot see what I did that is an error. Do you guys know what Im doing wrong?
Many thanks,
Corey

[Corey King]

Last attachment was too low res

[1729]

Last attachment was too low res

Hey Corey King. Welcome to the forums!

You messed your answer up with the third step. You are actually meant to expand the minus sign.

So it should be :

12y-15 -(2y+1) = 6y. In this case you can look it as 12y-15 -1(2y+1)=6y, try and expand the -1. So you should get
12y-15-2y+1=6y

You should be able to solve from there, and get the answer to be 3.5.

If you do not understand don't hesitate to ask!

[Corey King]

Fantastic, that helps
Thank you 1729!
Gratefully,
Corey

[Sherlock.Holmes]

Hey y'all,

Can someone please explain what I'm doing wrong, I don't really understand what the absolute function is and those two lines, and why they switch the signs in the answer

Thanks

[james.358]

Hey Sherlock.Holmes!

When you anti-differentiate 1/u, where u is a linear expression, you have to add absolute values. This ensures that the log argument (whatever goes into the log) is positive, so the result would be valid for all values of x.

You would still be correct if you didn't switch signs in the final answer as |3x - 4| = |4 - 3x|

Hope this helps!
James

[Bri MT]

Hey y'all,

Can someone please explain what I'm doing wrong, I don't really understand what the absolute function is and those two lines, and why they switch the signs in the answer

Thanks

Hey!

The absolute value signs can be used to indicate magnitude or "always make what's inside these positive".

For example:

| -2 | = | 2 | = 2

| 8 | = | -8 | = 8

|-x| = |x| = x [if x is positive] or -x [if x is negative]


Edit: beaten by James

[Sherlock.Holmes]

Hey!

The absolute value signs can be used to indicate magnitude or "always make what's inside these positive".

For example:

| -2 | = | 2 | = 2

| 8 | = | -8 | = 8

|-x| = |x| = x [if x is positive] or -x [if x is negative]


Edit: beaten by James

Hey Sherlock.Holmes!

When you anti-differentiate 1/u, where u is a linear expression, you have to add absolute values. This ensures that the log argument (whatever goes into the log) is positive, so the result would be valid for all values of x.

You would still be correct if you didn't switch signs in the final answer as |3x - 4| = |4 - 3x|

Hope this helps!
James

Makes sense. so whenever I put the absolute modulus (2 lines), the inside will always be positive?

[1729]

Hey y'all,

Can someone please explain what I'm doing wrong, I don't really understand what the absolute function is and those two lines, and why they switch the signs in the answer

Thanks

Your answer is wrong because you don't have absolute values
You have However you should have had either Or
With those two options being equivalent remember,

[Bri MT]

Makes sense. so whenever I put the absolute modulus (2 lines), the inside will always be positive?

It could also be zero as |0| = 0; it will always be non-negative. log(0) is undefined anyway