VCE Methods Question Thread!

[atar.notes.user]

Nah you need to multiply the function by x to find E(x)

omg i feel so stupid 😅😅😅 i forgot about that bit

[uhoh]

omg i feel so stupid 😅😅😅 i forgot about that bit

Ahah all good! I didn't realise either

Random question: if integral 1 to 4 f(x) dx=3, what does integral of 2 to 8 f(x/2)dx equal? 6?

For Q18
----> 1. I used discriminant <0 to get K>1 for no real solutions

Then since -1< (or =) sin(x) < (or =)1, where sinx=1+(1-k)^1/2 or sinx=1-(1-k)^1/2, I got the following two points

----> 2. sinx=1+(1-k)^1/2 is never true since -1< (or =) sin(x) < (or =)1 and sinx=1+(1-k)^1/2> (or=) to 1
So for no real solution, I solved 1+(1-k)^1/2> (or=) to get K< (or=) to 1

----> 3. sinx=1-(1-k)^1/2 only works if -1<(or =)1-(1-k)^1/2< (or =)1 since -1< (or =) sin(x) < (or =)1
So for solutions, I got -3< (or =)k< (or =)1. That means for no real solutions, k is R/(-3,1)

The ans is d (-infinity,3) and (1,infinity) though, which doesn't take into account what I wrote following the 2nd arrow. Shouldn't we take the INTERSECTION of what was found in the 3 points?

Q20) From the matrices, I got the equations of x' and y'. I rearranged them for x in terms of x' and y in terms of y'. I subbed this is into the original eqn, then let it equal the transformed equation (y=2sin(x+pi/2)-3).

I don't get the answer, which is B though

[atar.notes.user]

are we allowed to write in pencil in the exam? bc i remember something about it being scanned??

[Sine]

are we allowed to write in pencil in the exam? bc i remember something about it being scanned??

i wrote both exam 1 and exam 2 in pencil (mechanical pencil) and recieved a study score last year

[atar.notes.user]

Aside from multiple choice, are you allowed to write the exam in pencil?

i wrote both exam 1 and exam 2 in pencil (mechanical pencil) and recieved a study score last year

alright thank you😅😅

[uhoh]

Does smooth mean the y-value has to be the same AND the gradient is the same, or just that the y-value is the same?

And in the next question, it asks us to draw the graph of the function? the examiner's report says that the graph must be smooth i.e. at t=8 and t=16. Why though? The function is continuous, not smooth

[atar.notes.user]

Ahah all good! I didn't realise either

Random question: if integral 1 to 4 f(x) dx=3, what does integral of 2 to 8 f(x/2)dx equal? 6?

For Q18
----> 1. I used discriminant <0 to get K>1 for no real solutions

Then since -1< (or =) sin(x) < (or =)1, where sinx=1+(1-k)^1/2 or sinx=1-(1-k)^1/2, I got the following two points

----> 2. sinx=1+(1-k)^1/2 is never true since -1< (or =) sin(x) < (or =)1 and sinx=1+(1-k)^1/2> (or=) to 1
So for no real solution, I solved 1+(1-k)^1/2> (or=) to get K< (or=) to 1

----> 3. sinx=1-(1-k)^1/2 only works if -1<(or =)1-(1-k)^1/2< (or =)1 since -1< (or =) sin(x) < (or =)1
So for solutions, I got -3< (or =)k< (or =)1. That means for no real solutions, k is R/(-3,1)

The ans is d (-infinity,3) and (1,infinity) though, which doesn't take into account what I wrote following the 2nd arrow. Shouldn't we take the INTERSECTION of what was found in the 3 points?

Q20) From the matrices, I got the equations of x' and y'. I rearranged them for x in terms of x' and y in terms of y'. I subbed this is into the original eqn, then let it equal the transformed equation (y=2sin(x+pi/2)-3).

I don't get the answer, which is B though

18:
so what i just did was graphed it in my cas including the k. used the slider to see when it would touch the x-axis, and from there its pretty simple, u can just see the answer

20:
the curve with a, b, c is ORIGINAL
the curve without a, b, c is TRANSFORMED

so when we get x' and y', if we sub these transformed x' and y' back (dont transpose to make x and y the subject) in the TRANSFORMED curve, we'll get the initial curve.
thats bascially it, B will be ur answer

Does smooth mean the y-value has to be the same AND the gradient is the same, or just that the y-value is the same?

And in the next question, it asks us to draw the graph of the function? the examiner's report says that the graph must be smooth i.e. at t=8 and t=16. Why though? The function is continuous, not smooth

when u draw the graph in ur calculator, c(8 )= 8000 and c(16)=8000
so obviously, the concentration isnt abrupty gonna go from 0 to 8000; so from logic m must be 8000

[ZNormal]

Hi, I was just wondering if we still need to know when to put absolute lines when integrating log functions?

If so can someone please teach me because I don’t know when to put it on.

Thank you so much in advance !

[atar.notes.user]

Hi, I was just wondering if we still need to know when to put absolute lines when integrating log functions?

If so can someone please teach me because I don’t know when to put it on.

Thank you so much in advance !

so pretty much, u cant log a negative number yeah?
thats why whenever u integrate a reciprocal function, u have to put absolutue lines in ur log
bc although ur original function could take any x value, we have to make sure the numbers that are will be log(ed?) MUST be negative

[VanillaRice]

Hi, I was just wondering if we still need to know when to put absolute lines when integrating log functions?

If so can someone please teach me because I don’t know when to put it on.

Thank you so much in advance !

As of the new study design, I don't believe that you are required to use the absolute value in the log term when integrating functions of the form a/(bx+d).
However, consider why we must add an absolute value term. An antiderivative of 1/x is ln(|x|). Sketch both of these in your calculator to see what they look like. Now sketch ln(x) instead of ln|x|. What's the difference? The left half has disappeared.
Consider the domain of 1/x, which is R\{0}. Therefore, the domain of the antiderivative must also be R\{0}. That is the function of the absolute value - to copy and paste the right (positive x values) side of the function onto the left (negative x-values) side. How can we get around this without using the absolute value? VCAA questions should now specify a domain for the function in which you are integrating. (If not, you can also use hybrid functions to define your antiderivative.)

Example: if we are finding an antiderivative of f(x) = 1/x, where x > 0, we get ln(x). Both functions have a domain x > 0, so the absolute value problem does not need to be considered.
If we are finding an antiderivative of f(x) = 1/x where x < 0, we get ln(-x). Both functions have a domain of x < 0, so we once again have no need to consider the absolute value.

Where a domain is specified, you must consider the sign (positive/negative) of what is in the brackets of the logarithm.

Hope this helps

[captkirk]

Should I memorise how to do first principles for exam 1? Is it worth knowing for exam 1

[Sine]

Should I memorise how to do first principles for exam 1? Is it worth knowing for exam 1

it's not too much to memorise so i'd think it's worth it though the chance of it coming up is <1%

learn it via the proof and you don't really need to memorise but understand (you can prove the formula in the exam if you understand it)

[QueenSmarty]

Are related rates equations still part of the study design?

[Sine]

Are related rates equations still part of the study design?

it is not explictly on the study design. However it is just an application of the chain rule (which is on the study design) so imo could still come up albeit a low probability of doing so.

[QueenSmarty]

Thanks Sine!

With 2012 exam 2, would an answer of 405/64 for question 4ci get the mark or would it need to be completely simplified to 6 21/64