VCE Methods Question Thread!

[IndefatigableLover]

didnt quite understand that, sorry. Could you show me how 'D' is done possibly? thanks

Alright so with 'D' you can see that the basic form of the equation will be:

(we know that it is 'x-2' inside due to the translation of the point of inflection resting on (2,0)).

From here we just need to check the dilation and we can see that the graph intersects the 'y' axis (that is 'x') at (0,-8). Sub that in and you'll find that 'a=1'.

I'm sure you can work out the derivative of: but yeah the derivative is:





Sketch the graph of that and you've just drawn the velocity-time graph for 'D'

[kinslayer]

Good advice.

My questions are attached and circled. My calculator gave weird answers that got me nowhere.
For question 1c., my calculator gave: .
Please help by answering the questions and briefly tell me how to do it on the calculator correctly, if I was doing it wrong.
(Solving gave the same answer.)
Thanks in advance.

Doesn't look like you have done anything wrong, you are just not all the way there.

The median, m, is a number such that

In this case we have that



Now set P(X<m) = 1/2 and you get the equation that your calculator gave you.

There is no solution in terms of elementary functions, so you need to get your calculator to solve that equation for m and you get an approximate numerical answer. I get . There are two solutions for m, make sure to pick the one that is greater than 1.

d) is the same thing, just set P(x<m) = 0.95 instead.

[BLACKCATT]

When the mod is around the function, this implies that the y-values are taken as absolute and are always positive. When the mod is only around the x, the x-value  is taken as always positive, so negative values of x would in fact spit out a y-value that correspond to the positive x-value, which suggests that this graph is symmetrical about the y-axis. If you're still confused try graphing the function f(x)=ln(x) and then put the absolute sign in the two places.

graphing In(|x|) gives the reflection of In(x) in the y axis. Why does the positive side of In(x) reflect over the y-axis ending up with negative x values?

[keltingmeith]

graphing In(|x|) gives the reflection of In(x) in the y axis. Why does the positive side of In(x) reflect over the y-axis ending up with negative x values?

Basically, every number that comes out of the mod signs is going to be positive. So, even though you can't do ln(-1), if you do ln(|-1|), it becomes ln(1). This will happen with every single negative number, and effectively you get a mirror-image over the y-axis, because all the negative x-values are going to give the exact same answer as their positive counter-part.

[LiquidPaperz]

does doing methods 3/4 before you actually do the course give much advantage?

[LiquidPaperz]

same goes with specialist

[keltingmeith]

Not really, granted a lot of methods 3/4 is contained in 1/2, so there's not really much point in "reading ahead" for 3/4. The one place there's little overlap is in probability, so your might find it advantageous to read ahead for that a bit, and try to get a handle on the kinds of applications you can get with random variables. This would be particularly helpful since most people seem to find random variables difficult, and often hate probability (although it's my personal favourite area of study).

For specialist... I dunno, make sure you know what complex numbers are and the difference between a vector and scalar quantity? There's actually very little content in it, you'll be fine if you do no work beforehand.

[LiquidPaperz]

alright, thanks for that Euler! Im doing 1/2 methods atm, and there are a few people in my cohort who have done the 3/4 methods course (and even started exams) and they are beginning with 3/4 specialist course, seeming i wanted to be rank 1 i was wondering what i need to do; to do so?

[kinslayer]

alright, thanks for that Euler! Im doing 1/2 methods atm, and there are a few people in my cohort who have done the 3/4 methods course (and even started exams) and they are beginning with 3/4 specialist course, seeming i wanted to be rank 1 i was wondering what i need to do; to do so?

If someone's ranked higher than you for SACs, that only impacts you negatively (relatively speaking) if you score higher than they do on the external assessment. All those people spending the extra time to read ahead are helping you by making your cohort stronger. Don't worry about being rank 1, just worry about doing your best.

If you have a lot of free time now to do 3/4 exercises then buy the book and go nuts, it can't hurt. Otherwise, focus on learning what's presented in class now. It can be tempting to read ahead, but if it is at the expense of something else then it may not be worth it.

[keltingmeith]

alright, thanks for that Euler! Im doing 1/2 methods atm, and there are a few people in my cohort who have done the 3/4 methods course (and even started exams) and they are beginning with 3/4 specialist course, seeming i wanted to be rank 1 i was wondering what i need to do; to do so?

Some may disagree, but I believe that you can prepare too much. Sometimes it's better to learn things at the normal pace - and I reckon with specialist, maybe being a little ahead (doing pre-readings) could be beneficial, but being up to mechanics when everyone is only learning about parametric equations could hurt you. Particularly if you learn something wrong.

[LiquidPaperz]

I see, thanks for that!

btw with this Q, answers A how?

edit: E

[kinslayer]

I see, thanks for that!

btw with this Q, answers A how?

If C is supply charge in dollars and E is energy in megajoules then C = 30 + 0.02 E

Answer can't be A. The supply charge for 5 megajoules is $30.10. The answer is E.

C(0) = 30 and C(5000) = 130.

[MNM101]

How do u find the differentiation of x^4+4x^2/2x

[Orb]

How do u find the differentiation of x^4+4x^2/2x

I'm assuming it's x^4 + (4x^2/2x) not the entire equation over 2x.

Either way, simplify the equation first, obtaining x^4 + 2x from the first equation.

Then we just differentiate it, so using the formula dy/dx x^n becomes nx^(n-1),

We obtain 4x^3 + 2

[kinslayer]

How do u find the differentiation of x^4+4x^2/2x