VCAA EXAMS-Easiest to Hardest?

[faredcarsking123]

Q3 d)i) One stationary point implies that g'(x) only intersects the x-axis once. Hence, its discriminant is zero and you solve for the value of c.

ii) let x=a be the point where the stationary point occurs. g'(a)=0, and from this you can work out the value of a. Then, using g(a)=a you can also work out the value of k.

Q5 f) you want to solve dT/dx=0 for k, given that x=sqrt(7)/2
This gives you a value of k, however note from the equation in 5c that as k increases the time it takes to swim up the river also increases. Thus, for any value of k bigger than this, the fastest route will still be for Tasmania Jones to run directly to the desalination plant. Hence, k is greater than or equal to the value you found.

I think you're on the wrong exam bro

[psyxwar]

I think you're on the wrong exam bro

Awks thought it said 201, ill post up working for 2007 when i get home

[faredcarsking123]

Awks thought it said 201, ill post up working for 2007 when i get home

Thanks heaps!

[Edward Elric]

The 2013 exam 2 admittedly was pretty ridiculous. I just finished it and although my time and score on the exam wouldn't suggest I had much trouble with it, some questions are just so trippy. Like 4c; how on earth is that only two marks!? MC 16 tricked a lot of people as well; the percentage of people getting it right was close to the guess percentage of 20%

Just curious, what did you get on the 2013 exam 2 Izxnl?

[psyxwar]

The 2013 exam 2 admittedly was pretty ridiculous. I just finished it and although my time and score on the exam wouldn't suggest I had much trouble with it, some questions are just so trippy. Like 4c; how on earth is that only two marks!? MC 16 tricked a lot of people as well; the percentage of people getting it right was close to the guess percentage of 20%

Hmm I'm not sure, I thought it was a very accessible paper. Q16 could literally be done by evaluating the area of the region and then testing out every option; inverses and all that weren't necessary.

Q4c could definitely have done with more working space though. Its confusing how they give so many lines for binomial distribution questions but not for this one.

[lzxnl]

Hmm I'm not sure, I thought it was a very accessible paper. Q16 could literally be done by evaluating the area of the region and then testing out every option; inverses and all that weren't necessary.

Q4c could definitely have done with more working space though. Its confusing how they give so many lines for binomial distribution questions but not for this one.

It's not that it's hard to see how to do. It's more, are you going to be able to see that's how to do the question? For Q16, it's very easy to choose one of the wrong answers. People aren't going to think of the correct answer immediately. Once you see how to do it, it's easy.

Q4c had just enough working space if you crammed everything in.

[Yacoubb]

If we didn't know how to answer that "find the gradient" question on exam 2 of 2013, could we have just made up a gradient and used it to complete the remaining questions, getting consequential marks? Would we have to have picked a gradient WITHIN a particular range of values?

[vcestudent123]

I found Exam 1 2012 to be really easy.

[IndefatigableLover]

Just from memory

Last 2 of Q3
Last Q from Q5

Those I didn't even understand with solutions, but a lot more were tricky that I only got after I read solutions

I got 58/80 for that exam while in some others i got like 70,72,74 but on average i get like 67ish

Q3 di).
Pretty much just graph f(x) and just change it to fit h(x). You know you can flip it about the 'x-axis' (reflection in the 'x' axis)and then from there just compare the minimum/maximum points and subtract the distance between the two to get your translation effectively getting you a translation of '3' in the negative direction of 'x'.

Q3 dii).
This was a tricky question I felt since many people misinterpreted the question (like the maximum and minimum values of 'g'). You know your intercepts already from drawing out the graph and write it out in intercept form with some pro-numeral at the front (for the dilation). To find 'a', it's a bit hard to see but when you saw on the first question how they wrote g(x) in two different forms? You kind of have to see and use that dilation value as your final answer in the end..
So in the end you would have gotten some answer like: or add a reflection in the 'x-axis' for the same answer.

Q5) fi).

You know when you see the word 'maximum' (or minimum) that calculus will be used. In this case they gave you the answer for which you differentiate (that is the equation of 'Q'). Essentially you differentiate that (making sure you use ) and let it equal zero. Solving for 'p' will yield you 3 answers however you reject two of them as 'p' must be in between zero and one. Now sub it back into the equation of 'Q' and you've got your final answer

Q5 fii).
Here you pretty much have to sub it back into the f(t) but in the domain of between 20 and 30 since the maximum value appears in this domain rather than the other one. As a result, you know:




This is the value in which the maximum occurs. Now you just have to sub it into the integral (with 30 being on top of the terminal and b on the bottom since it's less than 30), and you'll find that your answer gives you b=20.9

[faredcarsking123]

Q3 di).
Pretty much just graph f(x) and just change it to fit h(x). You know you can flip it about the 'x-axis' (reflection in the 'x' axis)and then from there just compare the minimum/maximum points and subtract the distance between the two to get your translation effectively getting you a translation of '3' in the negative direction of 'x'.

Q3 dii).
This was a tricky question I felt since many people misinterpreted the question (like the maximum and minimum values of 'g'). You know your intercepts already from drawing out the graph and write it out in intercept form with some pro-numeral at the front (for the dilation). To find 'a', it's a bit hard to see but when you saw on the first question how they wrote g(x) in two different forms? You kind of have to see and use that dilation value as your final answer in the end..
So in the end you would have gotten some answer like: or add a reflection in the 'x-axis' for the same answer.

Q5) fi).

You know when you see the word 'maximum' (or minimum) that calculus will be used. In this case they gave you the answer for which you differentiate (that is the equation of 'Q'). Essentially you differentiate that (making sure you use ) and let it equal zero. Solving for 'p' will yield you 3 answers however you reject two of them as 'p' must be in between zero and one. Now sub it back into the equation of 'Q' and you've got your final answer

Q5 fii).
Here you pretty much have to sub it back into the f(t) but in the domain of between 20 and 30 since the maximum value appears in this domain rather than the other one. As a result, you know:




This is the value in which the maximum occurs. Now you just have to sub it into the integral (with 30 being on top of the terminal and b on the bottom since it's less than 30), and you'll find that your answer gives you b=20.9

For 3di) is one of their solutions incorrect?
They said 1 unit left and reflection in Y, but shouldn't 1 unit left already complete the transformation without a reflection?

Also for 5ii) How do we know that it is in the Domain 20 to 30 and why do we get 1-(P)?

Thanks a lot for your help

[psyxwar]

If we didn't know how to answer that "find the gradient" question on exam 2 of 2013, could we have just made up a gradient and used it to complete the remaining questions, getting consequential marks? Would we have to have picked a gradient WITHIN a particular range of values?

I don't think they awarded consequential marks, but you should do it anyway because it's better than nothing and they may be feeling lenient.

[Yacoubb]

I don't think they awarded consequential marks, but you should do it anyway because it's better than nothing and they may be feeling lenient.

:O WHAT?!
How evil could they possibly be? That's so unfair. No, are you sure? When have they ever NOT awarded consequentials, unless of course you don't show working out?

[lzxnl]

:O WHAT?!
How evil could they possibly be? That's so unfair. No, are you sure? When have they ever NOT awarded consequentials, unless of course you don't show working out?

As Zezima has mentioned, the question was made to be a troll. Although honestly, I still found it pretty doable 2013 and 2012 exam 2s took me pretty much the same length of time
And ironically less time than other exam 2s I've done

[Robert123]

I think the way that question was created was meant to separate students. The exam panel creates separate marking schemes for each exams - so they don't always have to give you method marks, all the marks can go to the answer only -  and so if you didn't get the gradient right, the remaining questions remained inaccessible according to my teacher, who marked last year's exam.

It stops sneaky people making the gradient be something easy like "1" or "a" and do the rest of the question hoping to get conseqs when if you did it using the actual correct answer, it would be a lot harder.

How would it be much harder considering you have access to a CAS to do all the annoying algebra. And the sneaky people are still going to lose a mark from the first part so I can't see the justification of making other questions marginally easier by losing marks on more difficult questions.
Anyway, they should award method marks should be awarded considered the subject is called methods   

[spectroscopy]

It stops sneaky people making the gradient be something easy like "1" or "a" and do the rest of the question hoping to get conseqs when if you did it using the actual correct answer, it would be a lot harder.

damn they figured out my strategy LOL