VCE Methods Question Thread!

[lzxnl]

Oh I'm retarded.

My function is completely wrong because when y = 2, x = 0.75 not 1.5. That changes things.

y = ax^2
2 = a*9/16
a=32/9
y = 32/9 x^2

Integrate from 0 to 0.75. Antiderivative is 32/27 x^3
So area is 32/27 * (0.75)^3
Half area = 32/54 * (0.75)^3
Setting this equal to 32/27 x^3
x=0.75*(1/2)^1/3
y= height = 32/9 * (0.75^2*0.5^2/3)
or 1.26

There we go.

[fleet street]

From Kilbaha 2006 Exam 2 (CAS),

If , find if has a domain and give the domain of .

The suggested solution is:

Interchange x and y





But
Domain

However, I am unsure as to how saying that y must be greater than zero justifies taking the positive square root. Also, I don't think that f(x) is one-to-one for the domain (0,0), so the inverse function would not exist.
Could someone clarify this please?
Thanks!

[BubbleWrapMan]

You're right, it's not invertible.

[fleet street]

Thanks! I just wanted to make sure.

[Zealous]

Just a quick question, has anyone bumped into any questions (trial exams, past exams etc.) requiring the change of base logarithm rule?

It's not actually in my textbook and I haven't put time to learning it (as simple as it is), so I had troubles on a recent Exam 1 I did.

[BubbleWrapMan]

Just a quick question, has anyone bumped into any questions (trial exams, past exams etc.) requiring the change of base logarithm rule?

It's not actually in my textbook and I haven't put time to learning it (as simple as it is), so I had troubles on a recent Exam 1 I did.

VCAA 2011 Exam 2, MC question 22

[pi]

VCAA 2011 Exam 2, MC question 22

Pretty sure I assigned each letter a number for that one when I did it LOL

/fail

[abcdqd]

"find the area of the region enclosed by the graph of 2x-3, the x-axis and the line x=2"
does this mean ONLY the region from the x-intercept of the graph to x=2?
would "find the area of the region enclosed by the graph of 2x-3, the coordinate axis and the line x=2" mean the region from x=0 to x=2?
thanks in advance

[Daenerys Targaryen]

"find the area of the region enclosed by the graph of 2x-3, the x-axis and the line x=2"

Essentially, yes.

Think of it as 'Find the area bounded by the three lines y=2x-3, y=0, x=2"

Draw them all, find the area enclosed by all three. And that's the area you are wanting to find.

[Homer]

enclosed
this means ONLY the region from the x-intercept of the graph to x=2

[abcdqd]

sorry i don't think i was clear enough
the question asked "find the area of the region enclosed by the graph of 2x-3, the x-axis and the line x=2"

the answers said to only find the green area, i thought that it would be the red area+the green area

[Homer]

the red region is bound by the y-axis. and the question only states the x-axis.

 

[abcdqd]

alright, thanks

[satya]

when a question asks for a domain that is strictly increasing or decreasing, would you include the turning points???

[lzxnl]

If you mean...asking for when sin x is strictly increasing on the interval [-2pi, 2pi], then you could safely say [-pi/2, pi/2]. The turning points are included because in every point in the domain from [-pi/2, pi/2], if a<b, then sin(a) < sin (b). The endpoints satisfy this too.