VCE Specialist 3/4 Question Thread!

[lucas.vang]

which vcaa exam 2 was the hardest in your opinion?

[speedy]

Not all the time, they must share a point in common. Think of it like in methods with infinite solutions and no solutions for a  linear function and the line y=x. In both cases, they are parallel  however only the one with infinite solutions will go through the origin.
Does that clarify it for you?

Gotcha! Thanks

[eagles]

For 2009 Exam 2, can someone explain the region found for part f?

Because the region says "and |z| = 1", I'm interpreting the region to be the inner part and not the outer part as shown in the solutions.

Can someone point me in the right direction? Thanks.

[BLACKCATT]

For 2009 Exam 2, can someone explain the region found for part f?

Because the region says "and |z| = 1", I'm interpreting the region to be the inner part and not the outer part as shown in the solutions.

Can someone point me in the right direction? Thanks.

The region should be the outer part, the area enclosed by x^2+y^2=1 and x^2+y^2=2 is the outer portion
The area you are thinking about is only closed by x^2+y^2=1

[allstar]

hello
help on this question please?
thank you

I've already showed that b^2 +a^2c = 0

If P(2) = 0, show that b is an even number?

[eagles]

The region should be the outer part, the area enclosed by x^2+y^2=1 and x^2+y^2=2 is the outer portion
The area you are thinking about is only closed by x^2+y^2=1

Mmmm I still don't understand...

If x² + y² = 1 is a region, and "and" means intersection, why would we include the outside region as it is not included in the circle with radius 1?

[Mieow]

for b, how come they found the equations of the vertical forces and made it equal to zero? If it's on the point of moving down, shouldn't it be mgsin(30) - F - T = ma etc?

[eagles]

Since the object is on the point of moving down, it means it is on the verge of moving, and that it has not actually moved yet. So, it is stationary and we let a = 0, and hence the equation is equal to 0.

[speedy]

Mmmm I still don't understand...

If x² + y² = 1 is a region, and "and" means intersection, why would we include the outside region as it is not included in the circle with radius 1?

It is enclosed by |z|=2 and |z|=1 -> that there is a donut shape, excluding |z|<1 and |z|>2. Obviously the line then cuts the portion further.

Also, |z| = 1 does not imply the inside of the circle, it's just the circle, a single line.

[eagles]

Wowowowowow!!!!! That makes so much sense now thanks speedy + BLACKCATT

 

[eagles]

It is enclosed by |z|=2 and |z|=1 -> that there is a donut shape, excluding |z|<1 and |z|>2. Obviously the line then cuts the portion further.

Also, |z| = 1 does not imply the inside of the circle, it's just the circle, a single line.

Umm another question: if we were asked to shade it, would there be two possible solutions? For example, shade just like in the assessors' report but also shade the same section but below the line L and enclosed by the line y=0?

Thanks. 

[Robert123]

(Image removed from quote.)

for b, how come they found the equations of the vertical forces and made it equal to zero? If it's on the point of moving down, shouldn't it be mgsin(30) - F - T = ma etc?

If it's on the point on moving, then the acceleration is still going to be 0 and friction is at its maximum.

[speedy]

Umm another question: if we were asked to shade it, would there be two possible solutions? For example, shade just like in the assessors' report but also shade the same section but below the line L and enclosed by the line y=0?

Thanks.

I don't really know what you mean? Maybe draw it? Regardless, from the question in the exam, the only solution for that area is what is shaded in the exam report. You have four lines, which enclose an area.

[M_BONG]

How do I do this?


I subbed in an integer value ( n =4 ) and I got 4 solutions, so I chose D.

But answer says E.

Apparently, I have to take into account z = 0 is also a solution -> CAS doesn't spit this out. Someone explain? Cheers

[eagles]

I was thinking about the shaded red section: