VCE Methods Question Thread!

[Special At Specialist]

I'm very surprised that there isn't some sort of program on the calculator that can solve this like any other equation (there probably is and I just don't know about it).
Nevertheless, wouldn't a quicker way be to just do this?
(55.8 - 51.2) / invNorm(0.8, 0, 1) = 5.47

[#1procrastinator]

If Pr(A)=3/5, Pr(B)=1/4 and Pr(A U B), how do you find Pr(A'^B) (^ = intersection) without using a Karnaugh map. What's an intuitive way of finding it from first principles?

[dinosaur93]

sinx(sin(x)-cos(x))

How do we find the period of this?

The answer is just /pi but I have no clue as how to derive to that answer

[dfgjgddjidfg]

Why do they use the inverse function to find the area. and why the area between x=3 and x=0. Can someone explain the general rule about when we should use invere functions to find the area. thanks

[Homer]

when the region bound by the line with eqations is y=something rather than x=something

[soccerboi]

VCAA 2008 EX2 Q3a) It says "find the time to the nearest minute" i put 3 hours 12 minutes, but report says 192 minutes. Would i still get the mark?

[hongs-]

Can anyone help me with the VCAA 2009 exam  2 question 2cde?
I read the solutions, but I still don't understand why v is w and whatnot

[BubbleWrapMan]

Can anyone help me with the VCAA 2009 exam  2 question 2cde?
I read the solutions, but I still don't understand why v is w and whatnot

w is the speed when the train leaves the tunnel at P, and at this point d = 0 (since d is the distance from P). So, when d = 0, v = w. This gives an equation relating k and w, so you can solve for k.

If you replace k in the formula with what you found in c, the equation becomes a relation between v, w and d. But since they give you v and d, you can plug those in and solve for w.

Now that you have w, you can find k, and then you can sub k back into the original equation, so now it relates v and d only. The train stops at v = 0, so you can solve for d.

2009 Exam 2 Q1e)

Why is the domain of f(g(x)) not R⁺U{0}, the domain of f?

The domain of a composite is the domain of the 'inside' function after it has been restricted, but since ran(g) was already a subset of dom(f), there was no restriction on ran(g), and therefore no restriction on dom(g) which is R

[D.H]

I think I broke my own thread since I can't even load the page but I can load everything else..
ANYWAY I'll just ask here:


VCAA Answer:
Hypergeometric distribution
1 – (Pr (W = 0) + Pr (W = 1)) = 0.672 where W is the number of rods in the sample with a size fault

What on earth is a hypergeometric distribution?!
(This is VCAA 2003 btw)

[al3x]

Hey guys, I'm guessing this is a pretty easy question, but I don't know how to do it.

It says:

For f(x) = x^3 + 2x, the average rate of change with respect to x over the interval [1,5] is:

Answer being 33


Also:

The average value of the function f(x) = e^(2x)cos(3x) for 0<=x<=PI is closest to:

with the answer being -26.3

Explanation on this form of rate of change would be great.
Cheeers

[Shenz0r]

I think I broke my own thread since I can't even load the page but I can load everything else..
ANYWAY I'll just ask here:
(Image removed from quote.)

VCAA Answer:
Hypergeometric distribution
1 – (Pr (W = 0) + Pr (W = 1)) = 0.672 where W is the number of rods in the sample with a size fault

What on earth is a hypergeometric distribution?!
(This is VCAA 2003 btw)

Old study design, don't worry about it.

[Holmes]

So the range of g has to be (0, infinity)
Now find the x values for the extreme y values.
So 3-x goes to infinity for x approching -infinity.
3-x=0
then x=3
So this domain restirction will give the required range.
i.e. dom g*=(-infinity,3)
Then use that to obtain the fog* function.
fog*(x)=(3-x)^(-1/2) for x is a member of (-infinity,3)

EDIT:small edit
EDIT2: that was a bad explanantion sorry. If you want me to clear it up, just ask.

Nah, that was a good explanation!

[Hutchoo]

Hey guys, I'm guessing this is a pretty easy question, but I don't know how to do it.

It says:

For f(x) = x^3 + 2x, the average rate of change with respect to x over the interval [1,5] is:

Answer being 33


Also:

The average value of the function f(x) = e^(2x)cos(3x) for 0<=x<=PI is closest to:

with the answer being -26.3

Explanation on this form of rate of change would be great.
Cheeers

Average rate of change is just talking about the gradient across two points. (This is poorly worded, but look up a definition from your textbook).

Essentially:
y = x^3 + 2x.

Since they tell us that they want the value to be from 1 to 5 (since x is an element of [1,5]), what you do is this:

1) Sub in the first point (1) into y.
2) Get your y value. (So, in this case, when x = 1, y = 3, when x = 5, y = 135).
3) y2-y1/(x2-x1) = m (gradient)

Sub in your values and there's your answer.

[D.H]

I'm really stuck on  c)iii and I don't understand that first line of the solutions
I put the VCAA answers next to the corresponding question to save you guys some time

Oops I probably should include that f(x)= x^3e^-2x

[monkeywantsabanana]

This may be strange... but can you look through your theory book BEFORE reading time? (i.e. when you go sit down in the exam room and 15 minutes reading hasn't started)

And Calculators - are we allowed to use that as well during this time?