VCE Methods Question Thread!

[~T]

For the rounding issue, I would say round up because it is the minimum height.

You found that they had to be 176.46cm tall at least, so someone 176.45cm (after carefully patting down her hair and measuring very precisely) would not be accepted. So, following this logic someone 176cm would not be accepted either, so out of the option 177 should be the right answer.

I think VCAA has done quite a few questions like this before, but usually with people: for example you were required to realise that you had to round 15.1 to 16 people for whatever reason. But, there was one "tricky" question with Tasmania Jones asking for the maximum time he can take to get to the antidote (or whatever). You had to round down this time because you couldn't round up to the nearest whole number because by then Tasmania Jones would be dead even tho the answer was ___.52 or something.

Hope it kind of makes sense

Sorry, I never replied to this. Thank you for the reply, but your evidence is to the contrary - shadows made the same mistake a few posts ago. The 2008 exam 2 asked "Find the time, to the nearest minute, that Tasmania has to find an antidote" and the answers round up. The answer is 191.67, and they round to 192, even though he would be dead.

Would we presume that VCAA wouldn't make this kind of ambiguity again? If they did, what would you go with?

[jono88]

What happens when you antidiff a mod of cos(x)?

[Stevensmay]

What happens when you antidiff a mod of cos(x)?

Rewrite as a hybrid function and see if you can work it out.

Spoiler

Integrate as normal.

If we were to do this on a calculator, we will most likely get something like
, sgn could also be sign or signum.

Signum is a function that returns whether the number we gave it was positive or negative by outputting 1 or -1 respectively.
Sgn(3) will return 1.
Sgn(-2) will return -1.
Sign(0) will return +-1.

So by including sgn in our answer, we can do away with all the hybrid function stuff as it accounts for whether cos(x) will be positive or negative.

[aidana95]

Solve the equation 2log3(x+4)-log3(-x)=2 for allowable values of x


Anyone able to help with this one?

[BasicAcid]

Hint: 2 = log3(9)

Then basic log laws

[Stevensmay]

Solve the equation 2log3(x+4)-log3(-x)=2 for allowable values of x


Anyone able to help with this one?

Convert the 2 as BasicAcid said.





















Substitute back into our original equation to make sure these are still true.
x=-16 first.



So x=-16 is not a solution.

Test x=-1.




So x=-1 is the only solution.

If anyone could do an explanation as to why you cannot manipulate to make x=-16 work that would be much appreciated.
EDIT: If anyone could explain how false solutions come to exist that would be great.
Think I got it.

[KevinooBz]

Convert the 2 as BasicAcid said.





















Substitute back into our original equation to make sure these are still true.
x=-16 first.



So x=-16 is not a solution.

Test x=-1.




So x=-1 is the only solution.

If anyone could do an explanation as to why you cannot manipulate to make x=-16 work that would be much appreciated.

Can't have anything inside the log less than or equal to 0.
EDIT: okay nvm have no idea what that means haha.
EDIT2: actually i'm not sure if i'm right or wrong anymore cause i'm not sure on what a false solution is

[nerdgasm]

At a rough glance, it's because substituting x = -16 into the original equation yields 2log_3(-13) in the first term. x = -16 is a valid substitution in the second line, and if the original equation had been the second line of your working, it would be a valid solution.

I think the key idea is that the domain of 2log_3(x+4) is (-4, inf) while the domain of log_3(x+4)^2 is R\(-4). We need to be careful that we keep the original domain in mind when solving these sorts of questions. In this case, our algebraic manipulation to go from our first line to our second *introduced* new solutions, due to the expanded domain. However, these solutions must not be admitted.

That's just my (very rough) take on this situation.

[Phy124]

is defined for and is defined for therefore is defined over the domain and such; cannot be a solution to the equation.

Simply noting that is not a real number would also be sufficient grounds for excluding the solution.

[shadows]

loge(0) or log of any negative number is not defined (Therefore if you subbed x value into the log and it was 0 or negative, It must be rejected, (preferably state as log(-...) is not defined)

It is because if you were to take the e^x for example, the graph would never approach zero, therefore never be negative. as e^(1000000000) or e^(-1000000000) is still greater than zero.

[BasicAcid]

So to reiterate, when you add two graphs together, the intersection of their maximal domains become the new graph's domain?

[ahat]

What's the best way to factorise an equation such as:
16p² + 6p - 7 = 0?
The way I was taught, multiply the 16 by -7 and then find two numbers which will multiply to give -112 (i.e. 16 x -7) and add to give 6, i.e. -8 and 14. But, this is a lengthy solution and takes time especially with an equation like this. Is there a better method?

Thanks.

[shadows]

If you can't see the factors quickly, I wouldn't bother wasting time trying to figure it out.

Just use quadratic formula to find the solutions. Then turn those solutions into linear factors.

You probably won't see something like that in Exam 1. Usually in exam 1, their numbers are easy to work with

Well if its exam 2, then just use your cas

[Phy124]

So to reiterate, when you add two graphs together, the intersection of their maximal domains become the new graph's domain?

That is correct.

If you have two functions and , then if you add, subtract or multiply them (i.e. , or ) then the domain of the resultant function is the intersection of the domains of and . If you divide one by the other, say , then the result domain is the intersection of the domains of and , excluding where the denominator is zero (in this where case ).

[ahat]

If you can't see the factors quickly, I wouldn't bother wasting time trying to figure it out.

Just use quadratic formula to find the solutions. Then turn those solutions into linear factors.

You probably won't see something like that in Exam 1. Usually in exam 1, their numbers are easy to work with

Well if its exam 2, then just use your cas

I was asking because it was from 2010 exam 1 What you said though makes me think there must have been a quicker way (2010 E1 Q8  for reference) Thanks though, the quadratic formula would have been the best approach!

I'm wondering, does
2ln(x) = ln(x²)
I ask, because if you drew the graph of 2ln(x), then the domain would be x > 0, whereas the graph of ln(x²) would be R\{0}. According to the log laws though, they should be equivalent?

More specifically, if I was attempting to reject solutions based on this, and the equation had 2ln(x) then based on what I said above, what's the reason behind rejecting x = -1? Considering the fact if 2ln(x) = ln(x²) then all values, except 0, should be defined, right? (VCAA 2009 E1, Q9)

Mod Edit: Merged double post (Phy124)