VCE Methods Question Thread!

[keltingmeith]

For the question attached, I got the same answer but had it in fraction, VCAA 2008 exam 1, would full marks be given If i showed all working out and included my answer in fraction? (They don't say whether to use decimal or fraction)

Many thanks guys

Yes, that's fine. Some people play with the rule, "use whatever they did in the question", but it really doesn't matter.

[HopefulLawStudent]

Yes, that's fine. Some people play with the rule, "use whatever they did in the question", but it really doesn't matter.

VCAA 2012 Assessors' Report: ‘In all questions where a numerical answer is required, an exact value must be given unless otherwise specified’.

I always interpreted as if they don't ask for a rounded answer, you MUST give an exact answer? My teacher's been pounding it into my head since the beginning of the year; so much so that I can now quote word for word that one sentence in the 2012 assessors report.

[HopefulLawStudent]

How do I do 6b? English isn't my first language so I'm having a bit of trouble understanding the question...

[cosine]

How on earth can this question be done? VCAA 2008 Exam: 2

[Adiamond]

Since g(x) is f(x-a) you can just sub in (x-a) into the original f(x) which turns into tan((x-a)/2), which you can then plug into your calculator and solve it such that when x is equal to 1, the equation equals 1.
Meaning the calculator syntax would be solve(tan((1-a)/2)=1,a

You then get a whacked general equation which substitutes 'n1' for any integer here and the answer will be true but because the domain for a is (-1,1) non inclusive it can be gathered that the integer here will be 0 (because if it was anything else the answer would be outside the domain), meaning that the answer is or, just simply .

[cosine]

Since g(x) is f(x-a) you can just sub in (x-a) into the original f(x) which turns into tan((x-a)/2), which you can then plug into your calculator and solve it such that when x is equal to 1, the equation equals 1.
Meaning the calculator syntax would be solve(tan((1-a)/2)=1,a

You then get a whacked general equation which substitutes 'n1' for any integer here and the answer will be true but because the domain for a is (-1,1) non inclusive it can be gathered that the integer here will be 0 (because if it was anything else the answer would be outside the domain), meaning that the answer is or, just simply .

I don't understand the second part of your explanation, I don't get the domain part..

[Adiamond]

Yeah I didn't explain fully why it would be outside the domain, the general equation that the calculator gives you should have '4n₁/2' in it, meaning that any other integer value that you sub in for n₁ will make the equation go outside of the domain as shown by subbing in -1 and 1, getting (4(-1)-pi+2)/2 and (4(1)-pi+2)/2 respectively, both being outside of the (-1,1) domain that is specified for a.

[cosine]

For the attached question, the question asks to find a rule for the graph above, in which I answered:

f(x)  = Hybrid:









The answer is a bit different than mine, but I was wondering if mine is technically incorrect? (the whole thing is a hybrid, unsure how to use the latex for it, apologies).

[keltingmeith]

For the attached question, the question asks to find a rule for the graph above, in which I answered:

f(x)  = Hybrid:









The answer is a bit different than mine, but I was wondering if mine is technically incorrect? (the whole thing is a hybrid, unsure how to use the latex for it, apologies).

Nope - your endpoints are wrong and so is your second equation.

[cosine]

Nope - your endpoints are wrong and so is your second equation.

Well yeah the end points are meant to be round brackets, typed them in wrong. But besides these two errors, would this still be correct?

Cheers

[knightrider]

How would you solve this equation?

[StupidProdigy]

How would you solve this equation?

Pretty sure you can't. If you come across something like this in an exam there'll most likely be a convenient value you can sub in that makes the equation work (usually a simple value like -1, 0, 1). In this case just by inspection x=0 is the only solution!

[cosine]

How would you solve this equation?

You can apply the null factor law here, just like (x-1)(x+2) = 0 you can split into two:

x-1 = 0
x+2 = 0

Same with this, so you have:

x = 0


Solve these separately:

x=0 is a solution


We are saying that if we raise e to the power of any number, it will equal zero. But no matter what number we raise e by, or any number raised to any other number, the answer will always be positive and non-zero, and hence this has no solutions

only

[knightrider]

Pretty sure you can't. If you come across something like this in an exam there'll most likely be a convenient value you can sub in that makes the equation work (usually a simple value like -1, 0, 1). In this case just by inspection x=0 is the only solution!

You can apply the null factor law here, just like (x-1)(x+2) = 0 you can split into two:

x-1 = 0
x+2 = 0

Same with this, so you have:

x = 0


Solve these separately:

x=0 is a solution


We are saying that if we raise e to the power of any number, it will equal zero. But no matter what number we raise e by, or any number raised to any other number, the answer will always be positive and non-zero, and hence this has no solutions

only

Thanks so much StupidProdigy  and cosine  really helped.

[knightrider]

How do you know what x and h values to use for questions like these?