VCE Methods Question Thread!

[clıppy]

(x+3)^2 = y^(-1/2*2) = y^-1
Reciprocal
(x+3)^-2 = y

I think I got lost at reciprocal. Do you just multiply by -1?

[lzxnl]

Consider f(x) + 1 nliu

You know exactly what I mean. That's like translating the x-axis down.

I think I got lost at reciprocal. Do you just multiply by -1?

Reciprocal means raise to power of -1, or have one divide it.

[Alwin]

I think I got lost at reciprocal. Do you just multiply by -1?

that's what nliu mean by reciprocal (in latex version)

[clıppy]

Ahh okay. I just couldn't see it before.
Cheers guys

[Alwin]

The only think I would add is:Consider f(x) + 1 nliu

You know exactly what I mean. That's like translating the x-axis down.

Umm, no? I'm not talking about changing terminals, just the translation.
eg int_0^1 f(x) dx = 1  -->  int_0^1 (f(x) + 1) dx = 2

Sorry, i wasn't very clear

[lzxnl]

No, I knew what you were trying to say.
Perhaps I should have made it clearer originally: translations themselves don't change the area, but if your integration region doesn't account for the translation, the area will change.

[Stick]

Original Question

I see in your signature that you do spesh Stick, so a hint for the algebraic way is substitution
for a, let u=3x, then you have to change the terminal

for b, i'll also show a non spesh way of doing it for those of you not doing spesh


and nliu's gone and answered the rest while I was LaTexing

The only think I would add is:Consider f(x) + 1 nliu

This makes sense now. Thanks.

[Jaswinder]

dunno how to. ans:a

[Alwin]

dunno how to. ans:a

You test each one =D

I'll do (B) for you, to quickly show you!
What we do is denote the sets. We try to prove P(X n Y) = P(X) * P(Y) which shows the events are independent
let P(X) be probability of getting something in the set {1, 3, 5 ,7, 9, 11}. So P(X) = 1/2
let P(Y) be probability of getting something in the set {2, 4, 6, 8, 10, 12}. So P(Y) = 1/2
This implies that P(X n Y) is the probability of getting something in the set {{1, 3, 5 ,7, 9, 11} n {2, 4, 6, 8, 10, 12}} = null set. So P(X n Y) = 0
Use the rule P(X n Y) = P(X) x P(Y) for independent events:
LHS = 0, but RHS = 1/2 * 1/2 = 1/4.  LHS ≠ RHS, this option is not an independent event.

You do the same for each of the other options

hope it helps!

[Jaswinder]

im still a lil confused. for example how would you test option a, how do you workout pr(a) and pr(b) and pr(a (intersection) b)?

[Alwin]

im still a lil confused. for example how would you test option a, how do you workout pr(a) and pr(b) and pr(a (intersection) b)?

Original Question

let Pr(A) be probability of getting something in the set {1, 3, 5 ,7, 9, 11}. So Pr(A) = 6/12 = 1/2
let Pr(B) be probability of getting something in the set {1, 4, 7, 10}. So Pr(B) = 4/12 = 1/3
Pr(A n B) is the probability of getting something in the set {{1, 3, 5 ,7, 9, 11} n {1, 4, 7, 10}} = {1, 7} So P(X n Y) = 2/12 = 1/6
Use the rule Pr(A n B) = Pr(A) * Pr(B) for independent events:
LHS = Pr(A n B) = 1/6
RHS = Pr(A) * Pr(B) = 1/2 *1/3 = 1/6
LHS = RHS ∴ YAYAYAYAYAYAYAYAYAY it's independent QED etc etc blah blah blah


Make more sense?

[clıppy]

Hopefully the picture is readable. Can someone help me with this question, I've got no clue where to begin.

[09Ti08]

Maximum value: f'(x)=0, so either x=-3 or x=2
The graph of f'(x) has a maximum, so it's an upside down parabola (coefficient a<0) => f(x) is a "negative" cubic function (coefficient a<0) with 2 turning points (generally), so going from left to right along the x axis, the first one is a local minimum (smaller value of x, which is -3), the second one is the local maximum (bigger value of x, which is 2).
I used a few inaccurate termonologies here, but hopefully you know what I mean.

[Professor Polonsky]

Since the original function is a cubic, we know the derivative function is a quadratic.

We know also know that its x-intercepts are at (-3, 0) and (2, 0) meaning that the stationary points of f(x) are at x=-3 and x=2.

The derivative function (the quadratic) has a maximum value, rather than a minimum value. This tells us that it's an 'upside-down parabola', if you like. So for .

If you just quickly sketch the derivative graph, you would see that at x=-3 you have a minimum turning point (derivative going from negative to positive) and at x=2 you have a maximum turning point (derivative going from maximum to positive).

[Smiley_]

A long trough whose cross-section is parabolic is 1 and /2

metres wide at the top and 2 metres
deep. Find the depth of water when it is half-full