VCE Methods Question Thread!

[sailinginwater]

Some of the marks I lost in sacs were due to overthinking questions, thus getting them wrong, and having variations of the same question in a way which throws me off. But since I am doing more practise exams than I had for my school's sac revision, I would have covered nearly all the possible questions that can be asked on exams, thus increasing my chances of getting full marks on exam 1?

Im finding the vcaa exams slightly harder than my school sacs (I think my school's sacs were considered quite easy actually)

[Sine]

Some of the marks I lost in sacs were due to overthinking questions, thus getting them wrong, and having variations of the same question in a way which throws me off. But since I am doing more practise exams than I had for my school's sac revision, I would have covered nearly all the possible questions that can be asked on exams, thus increasing my chances of getting full marks on exam 1?

Im finding the vcaa exams slightly harder than my school sacs (I think my school's sacs were considered quite easy actually)

If you do your vcaa exams under full exam conditions you will likely get close to that score in the actual exam so if you are scoring highly on exam 1 it's possible to full mark. I think on all my VCAA practice exams (which I did under strict exam conditions) I was able to get 39 or 40/40 on all of them and ended up getting 40/40 on the actual exam. 

However, it's really easy to drop a single mark on exam 1 so even the best students who have prepared relentlessly over the whole year end up getting 38-39/40 on the actual exam so it's great to get yourself in a position to score as high as possible but really anything can happen in the actual exam.

[sailinginwater]

If you do your vcaa exams under full exam conditions you will likely get close to that score in the actual exam so if you are scoring highly on exam 1 it's possible to full mark. I think on all my VCAA practice exams (which I did under strict exam conditions) I was able to get 39 or 40/40 on all of them and ended up getting 40/40 on the actual exam. 

However, it's really easy to drop a single mark on exam 1 so even the best students who have prepared relentlessly over the whole year end up getting 38-39/40 on the actual exam so it's great to get yourself in a position to score as high as possible but really anything can happen in the actual exam.

Is it generally true that the more practise exams you do, provided that you learn from each exam, the higher chance you have of getting full marks since you would have covered nearly all the possible questions and tricks that can show up in the exam?

[Sine]

Is it generally true that the more practise exams you do, provided that you learn from each exam, the higher chance you have of getting full marks since you would have covered nearly all the possible questions and tricks that can show up in the exam?

hmm kinda, I generally believe your study score for maths subjects will correlate to the amount of practice exams that you do. But, of course exceptions will obviously be possible.

[sailinginwater]

Are graphs like (x-1)^(2/3) + 2 still in the study design?
2011 mav methods exam 1 question 2

[DBA-144]

Are graphs like (x-1)^(2/3) + 2 still in the study design?
2011 mav methods exam 1 question 2

Yeah I am quite sure they can still come up. You might find that it helps to know the basic shape of these graphs. If you dont, just sub in values to find the basic shape of the graph and then use transformations to help you out with the rest.

Eg. With your question, but just with the basic, simple graph.

Sub in x =0 y=0
x=1, y=1
x=-1 y=1, as u have to square the negtive then you take the square root of that.

From this, you should be able to understand that the graph resembles the square root graph or the cube root graph for when x is greater than zero, znd when it is less than 0, it is just a reflection in the y axis for the basic square root graph. Then, from this basic shape, just apply the transformations, the turning point will move up 2 units, and 1 to the right. There will be no x intercepts, but the y intercept will be at (0,3)

Sketch this on you cas to make sure. Hope this helps, feel free to clarify any misconceptions!

[sailinginwater]

Yeah I am quite sure they can still come up. You might find that it helps to know the basic shape of these graphs. If you dont, just sub in values to find the basic shape of the graph and then use transformations to help you out with the rest.

Eg. With your question, but just with the basic, simple graph.

Sub in x =0 y=0
x=1, y=1
x=-1 y=1, as u have to square the negtive then you take the square root of that.

From this, you should be able to understand that the graph resembles the square root graph or the cube root graph for when x is greater than zero, znd when it is less than 0, it is just a reflection in the y axis for the basic square root graph. Then, from this basic shape, just apply the transformations, the turning point will move up 2 units, and 1 to the right. There will be no x intercepts, but the y intercept will be at (0,3)

Sketch this on you cas to make sure. Hope this helps, feel free to clarify any misconceptions!

Exam 1 is non cas

[DBA-144]

I meant sketch it on your cas now, so you can see the shape.

[sailinginwater]

Can someone please explain how to do question 4 of the 2013 mav trial exam 1?
Find the equation of the normal time the curve y = log₂(x+2) at the y intercept
I tried doing this question as if it were a log_e graph but it didn't work

[DBA-144]

Can someone please explain how to do question 4 of the 2013 mav trial exam 1?

 

What s the question? Just modify your current post by adding the question into it and I can just post the solution here.

[TTanner01]

Can anyone help me on logarithms?
1. log₅(x) divided by log₅(x^0.5)

2. 2(logx)^2 - logx = 6

You're trying to find 'x'.
Thank you <3

[MB_]

Can someone please explain how to do question 4 of the 2013 mav trial exam 1?
Find the equation of the normal time the curve y = log₂(x+2) at the y intercept
I tried doing this question as if it were a log_e graph but it didn't work

Can you show your working so we can see where you're stuck

Can anyone help me on logarithms?
1. log₅(x) divided by log₅(x^0.5)

2. 2(logx)^2 - logx = 6

You're trying to find 'x'.
Thank you <3

1. Recall that log_a(b^c)=clog_ab
2. Substitute logx for u and solve for u like you usually would in a quadratic and then sub your answers for u back into u = logx and solve for x

[darkz]

Can anyone help me on logarithms?
1. log₅(x) divided by log₅(x^0.5)

2. 2(logx)^2 - logx = 6

You're trying to find 'x'.
Thank you <3

Please refer to the image for the solutions

Edit: Oops, second part is wrong, cause I managed to read it as -6ln(x) and its meant to be log10(x) rip 

[fun_jirachi]

Q1 is just 2. Recognise from your log laws that

So from that, you get that
For Q2, solve it as a quadratic in (logx). When you see a term that's squared, another that's the term by itself, and a constant, usually its a quadratic. If it makes it easier, let logx equal some constant a, then solve for a, substituting logx=a once you've found your solutions for a.

And from there I'm pretty sure you should be able to find the values of x.

[TTanner01]

Please refer to the image for the solutions

Thanks for the first solution!
For the second one, where did you get 'e' from (btw what even does that mean)
I was told by my tutor that if you don't write a base number in front of a log, then it's assumed that it's 10, in this case, that's what I did.