VCE Methods Question Thread!

[Blondie21]

In 4b, you would've worked out that f(x) would have stationary points when  . This is saying that if we change the 'a' and 'b' values inside of f(x), we can get up to two different stationary points.

If    and    are different values, then we will get two different x-values and hence two different stationary points.

But here's the thing, the question wants you to find 'a' in terms of 'b' so that there is only 1 stationary point. If we can get both and to equal the exact same thing, there would only be one x-value for a stationary point.

That's why we equate them - hopefully that makes sense!

So according to your working out, if you equate any two x values which are stationary points, you will find another stationary point??

[Phy124]

You are trying to find the values of and that will result in the two x-values being the same (because we only want one).

Say for example and then we have and

Therefore we only have one solution as both values are the same and hence we only have one stationary point in this case.

This will be the same for any values of and that satisfy (what you get when you simplify the expression )

e.g. and satisfies and gives only the solution

[M_BONG]

Ok this is a general question

I am intending to do some Methods revision for the remainder of holidays (since I have not touched Methods for like 3 weeks).

We have finished integration in class and I have two options for myself:

1. Start probability
OR
2. Do past exam papers for Methods; everything except Probability.

Should I start Probability ahead of class? Term 3 is busy so it would be good to finish the Methods course ahead but I am worried that
a) Probability may be too confusing, difficult or time-consuming to complete
or b) I won't retain much by starting ahead of class.

Is Methods probability difficult to learn? Do I need a teacher to guide me through?

Should I start probability or revise Unit 3? Thanks everyone!

[keltingmeith]

Completely depends on how you feel - do you feel confident with the course so far? Do you think you can handle learning a bunch of new material?

3/4 probability is very different to 1/2 probability, there are a lot of new concepts to learn. They're not difficult,  but they're new, so if you feel like you're not confident with the course so far, I'd suggest revising that. If you are, get a head start - having an idea of what's to come can only help.

[IndefatigableLover]

You have plenty of time for Methods revision as time goes by so you don't have to stress about starting practice exams just yet. Maybe pick out some questions from Checkpoints as you go through the course would be a better idea of revision but personally I would start on Probability and get a head start however if you found anything that you had trouble with in the previous topics for Methods then I suggest going over that first and then continuing on with Probability

[Blondie21]

You are trying to find the values of and that will result in the two x-values being the same (because we only want one).

Say for example and then we have and

Therefore we only have one solution as both values are the same and hence we only have one stationary point in this case.

This will be the same for any values of and that satisfy (what you get when you simplify the expression )

e.g. and satisfies and gives only the solution

Ohhhhh that makes sense. Thank you!!

[soNasty]

can someone help me simplify

please lol.. im doing the 2013 exam 1 and im struggling to get an answer with this beast. This is the last question btw.

[brightsky]

-300(ln(2))^2/(40ln(2)) + 20ln(2) = 20ln(2) - 15/2 ln(2) = 25/2 ln(2)

[nhmn0301]

Anyone here using ClassPad 330? Do you guys know how to get rid of the "E" in for example  2.31232423E+28?
Thanks !

[Bronzebottom64]

Anyone here using ClassPad 330? Do you guys know how to get rid of the "E" in for example  2.31232423E+28?
Thanks !

Capital E in calculator notation is the same as saying 'by ten to the.' In your example the answer is 2.31232423 x 10²⁸

[BLACKCATT]

How do you change the E showing up to an exact value in the graphs section of Ti-nspire cas?

[keltingmeith]

It's a bit of an arduous process, but thankfully b^3 tells you how to do it in here.

[skeletalclown]

The mass, X of a packet of potato chips is such that X~N(150,40). A packet with mass less than 143g is rejected as being underweight, and packets greater than 162g are rejected as being overweight.
In a batch of 500 packets, what is the probability of less than 66 packets being rejected?

This is the fourth part to a question, I'm fine with all the other parts but just cant seem to figure this out?? Any help would be very appreciated.

p.s: (part 1 was "what proportion rejected for being underweight", a simple normCDF question, part 2 was the same for overweight packets, bart 3 was finding the probability of rejection (Pr(underweight)+Pr(overweight))), and part 5, which i also was able to do, was finding which weights between which 90% of packets lie.

[Conic]

There is a set number of bags of chips (set number of "trials"), and the the weight of each bag is independent of the others. A bag is either rejected or accepted (each trial is a success or failure). This means that the number of bags that are rejected is a Binomial variable, with n=500 and p=Pr(Bag is rejected). So we can say B~Bi(500,Pr(Bag is rejected)), where B is the number of rejected bags. We want to find the probability that less than 66 of these are rejected. This means we have 0≤B≤65. You can find this using the binomcdf function on your CAS.

[skeletalclown]

Thank you, this is one of the first things i tried and gives an answer similar to the one stated at the back of my text book, but it is several significant figures out. I entered binomCDF(500,0.1631,0,65), and this gave 0.0236, and yet my texbook states 0.0299, have i entered something wrong, is the textbook, wrong, are we both wrong?? And thank you very much for the help and prompt reply!