Methods Exam Discussion!

[LOGCETERA]

How many marks do you guys think VCAA will detract if you wrote sin(theta) as sin^-1(theta) all throughout Q9?

[VCEStudent2034]

And if you let that equal zero, you get Pi/2. Plug into sin(theta), you get 1.

[Writing Overload]

Did anyone get q is between 0 and 1 not including 1 for the second part of that question?

[TnGn74]

Yeah, same. Then the derivative would be cos(theta).

But note that the domain of g is restricted, I believe to \( (0, \frac{ \pi }{3} ] \).

[VCEStudent2034]

Did anyone get q is between 0 and 1 not including 1 for the second part of that question?

I didn’t have time to write my answer...but that kinda sounds right.

[VCEStudent2034]

But note that the domain of g is restricted, I believe to \( (0, \frac{ \pi }{3} ] \).

Then what would the answer be?

[TnGn74]

I got maximum \( g( \frac{ \pi }{3} ) = \frac{ \sqrt{3} }{2} \) I think.

[VCEStudent2034]

I got maximum \( g( \frac{ \pi }{3} ) = \frac{ \sqrt{3} }{2} \) I think.

How did you know the angle was Pi/3?

[Deku]

How did you know the angle was Pi/3?

I think it’s cos of the domain. I was thinking about this in the car but the angle could never be Pi/2 because that would mean there are two 90 degrees angles in the triangle, I just didn’t know what I did wrong. Now we know it’s the domain.

[RaspberryTau]

I got maximum \( g( \frac{ \pi }{3} ) = \frac{ \sqrt{3} }{2} \) I think.

Same! You could also approach it using point P from the first part [(dy/dx = rise/run) solve for x], and use area of a triangle....

a) I got [-1,1]

b) I got [0,1) but may be wrong. Were you supposed to include 0? (ie. [0,1)

[daghgjhKSLJKad]

what did u guys get for the glazed doughnut question( in terms of g)?

[RaspberryTau]

Something like
(q-3)/6

I can't remember exactly - but it was /6.

[AlphaZero]

Been a while since I've been on ATAR Notes

Solutions

[RaspberryTau]

Legend!

[Deku]

What do you think the score range for 2021 grade distribution is?