VCE Methods Question Thread!

[IndefatigableLover]

2012 VCAA Mathematical Methods (CAS) Written examination 1 - Worked solutions

Woah these archived exam discussion/solutions from previous years are actually top-notch! Thanks for sharing that Phy124

[sparksfly]

Thanks

Are we required to know how to sketch graphs to the power of 2/3 and other weird looking ones on Exam 1 without a calculator? It's on a trial exam I'm doing- not sure if I need to go back and memorise all different forms of "weird" graphs?

[TrebleClef]

Can someone explain to me what reflecting about the y axis means? Thanks!

[IndefatigableLover]

Thanks

Are we required to know how to sketch graphs to the power of 2/3 and other weird looking ones on Exam 1 without a calculator? It's on a trial exam I'm doing- not sure if I need to go back and memorise all different forms of "weird" graphs?

The current study design only state:

drawing graphs by hand for y=x^n where n ∈ N and n = –1, –2, 1/2 and transformations of these to the form y = a (x + b)^n + c where a, b and c ∈ R;

- Page 66
and

sketch by hand graphs of polynomial functions up to degree 4

- Page 134

http://www.vcaa.vic.edu.au/documents/vce/mathematics/mathsstd.pdf

Have a deeper look yourself in case I missed anything but it doesn't look like they do unless someone can clarify about it

[IndefatigableLover]

Can someone explain to me what reflecting about the y axis means? Thanks!

I'm pretty sure reflecting about the y axis is the same as 'from' the y axis (basically in the 'x' axis)...

So for example, say you had a point (a,b) and you reflected it about the 'y' axis then the new point would be (-a,b).

[Phy124]

Woah these archived exam discussion/solutions from previous years are actually top-notch! Thanks for sharing that Phy124

No worries!

I agree, they're a great resource as you can see how students were feeling after the exam (and how they find the exam to be) and of course the suggested (and quite detailed) solutions that high scoring students race to make immediately after the exam

For those unaware, by navigating the boards by year and subject you'll find some great discussion regarding the exams that were made just after the exam as well as suggested solutions for chem/methods/spesh among others.

[sparksfly]

The current study design only state: - Page 66
and  - Page 134

http://www.vcaa.vic.edu.au/documents/vce/mathematics/mathsstd.pdf

Have a deeper look yourself in case I missed anything but it doesn't look like they do unless someone can clarify about it

Thank you so much!!


Another question- how do you work this out:
Finding the point of intersection between a graph and its inverse- f(x)= 1-e^-x and f-1(x) = -loge(1-x)?

The answer is (0,0) when you sketch it but how to you work it out by hand if you equate the two equations or sub y=x?

Thank you in advance!

[lzxnl]

Thank you so much!!


Another question- how do you work this out:
Finding the point of intersection between a graph and its inverse- f(x)= 1-e^-x and f-1(x) = -loge(1-x)?

The answer is (0,0) when you sketch it but how to you work it out by hand if you equate the two equations or sub y=x?

Thank you in advance!

You can't solve this one by hand. It's really just guess-work to find it.

[TrebleClef]

I'm pretty sure reflecting about the y axis is the same as 'from' the y axis (basically in the 'x' axis)...

So for example, say you had a point (a,b) and you reflected it about the 'y' axis then the new point would be (-a,b).

Thank you!

[Jason12]

I got this but the solutions said [15/4ln4 - (9/2)]. What am I doing wrong?

[Reus]

Could someone explain how thanks!!
Also @Jason12 what exam is this?

[achre]

Could someone explain how thanks!!
Also @Jason12 what exam is this?

So, let x=pi, and chuck it underneath 2:

[Reus]

So, let x=pi, and chuck it underneath 2:

Thanks so much!!

[Phy124]

I got this but the solutions said [15/4ln4 - (9/2)]. What am I doing wrong?

(Image removed from quote.)

From your last line you can simplify like so:



Additionally so I assume you've made a mistake somewhere further up by converting a negative to a positive or vice versa by accident. I think you should have in the earlier step.

Could someone explain how thanks!!
Also @Jason12 what exam is this?

[sparksfly]

When integrating a curve in the 3rd quadrant, under the x axis, how do I set up my definite integral? If its from x= -2 to -1, which value goes on top? Is -1 on top and -2 on the bottom correct? And do I need a negative sign outside the whole definite integral because its below the x axis?


Thanks