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Methods Exam Discussion!

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VCEStudent2034:

--- Quote from: TnGn74 on November 03, 2021, 11:14:48 am ---But note that the domain of g is restricted, I believe to \( (0, \frac{ \pi }{3} ] \).

--- End quote ---
Then what would the answer be?

TnGn74:
I got maximum \( g( \frac{ \pi }{3} ) = \frac{ \sqrt{3} }{2} \) I think.

VCEStudent2034:

--- Quote from: TnGn74 on November 03, 2021, 11:19:03 am ---I got maximum \( g( \frac{ \pi }{3} ) = \frac{ \sqrt{3} }{2} \) I think.

--- End quote ---
How did you know the angle was Pi/3?

Deku:

--- Quote from: VCEStudent2034 on November 03, 2021, 11:23:34 am ---How did you know the angle was Pi/3?

--- End quote ---
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:

--- Quote from: TnGn74 on November 03, 2021, 11:19:03 am ---I got maximum \( g( \frac{ \pi }{3} ) = \frac{ \sqrt{3} }{2} \) I think.

--- End quote ---
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)???

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