VCE Specialist 3/4 Question Thread!

[itsangelan]

Sorry dumb question how do you know the third linear factor is cz+d?

[Shadowxo]

Sorry dumb question how do you know the third linear factor is cz+d?

since we already have a quadratic factor, we need a linear factor as it's cubic. This means it has z^1 and no higher, and so it's either z - a - bi or just z- a, as there's no imaginary component to P(z), all real. So it's z-a, or cz+d as we multiply both by c (which we can because it's a zero)
Struggled explaining this, but does this help a bit?

[itsangelan]

Yes it does thankyou!

[deStudent]

http://m.imgur.com/a/yUGDc

For f) on the worked answer's 3rd to final line, why does it seem like they're adding +2pi to the basic angles to find the required additional solutions?

Shouldn't it just be +1pi since they're essentially saying "theta = 2x - pi/3” by adjusting the domain. So the period is still Pi?

Thanks

[de]

http://m.imgur.com/a/yUGDc

For f) on the worked answer's 3rd to final line, why does it seem like they're adding +2pi to the basic angles to find the required additional solutions?

Shouldn't it just be +1pi since they're essentially saying "theta = 2x - pi/3” by adjusting the domain. So the period is still Pi?

Thanks

You're right about the period of tan being pi.
However within each period there is only ONE solution to tan=-1. So -pi/4 is the "basic angle" so to speak, not "angles" for tan- there is only one (which makes sense when you think about the fact that the graph is always increasing).
Now look and you'll see that all the solutions in that third line are just adding the period pi to that basic angle.
Basically there's only one original solution, so they're adding pi to that, NOT 2pi to both AND

[deStudent]

Thanks de ^^^

Need some more help (last of circular functions, thank jesus. probs needa do more on this topic tho fml)

Q11a) I don't understand why x = +-pi/2 + gamma is incorrect? It fits the domain and its in the correct quadrant isn't it?

Q10a) how is sec(beta) equal to positive 'b' when it's in quadrant 2? Isn't sin and cosec only positive here?

http://m.imgur.com/a/UgFH4

[Shadowxo]

Thanks de ^^^

Need some more help (last of circular functions, thank jesus. probs needa do more on this topic tho fml)

Q11a) I don't understand why x = +-pi/2 + gamma is incorrect? It fits the domain and its in the correct quadrant isn't it?

Q10a) how is sec(beta) equal to positive 'b' when it's in quadrant 2? Isn't sin and cosec only positive here?

http://m.imgur.com/a/UgFH4

Hi,
10a, b doesn't have to be a positive number, in this case b is negative
I'd do these kinds of questions by finding the angle in the first quadrant and going from there
11a, adding and subtracting pi/2  won't keep the same value. eg sin(5º) doesn't equal sin (95º). Sinx = cos (pi/2-x)
Way I'd do it would be to say tan(gamma) = tan(pi + alpha), alpha = gamma - pi
if tanx = -c, tan (pi-alpha) = -c and tan (2pi-alpha) = -c
x = pi - alpha = 2pi -gamma or
x = 2pi - alpha = 3pi - gamma

[geminii]

Hi everyone!

I'm not sure how to tell if a set of matrices is linearly dependent or linearly independent. The textbook doesn't help and the online explanations talk about scalar multiples, and I have no idea what those are.

Here is the question I'm working on:

Determine whether the following set of vectors is linearly dependent.

a = [4]
      [1]
      [3]

b = [2]
      [-1]
      [3]

c = [-4]
      [2]
      [6]

It'd be great if someone could help with this! TIA

[Shadowxo]

If they're linearly dependent, then they can be represented as a triangle, so Aa +Bb = c (a b and c being the vectors)
Represent the vectors in i,j,k format (eg the first one is 4i +j +3k)
Then do Aa + Bb = c and equate the i j and k parts to see if A and B exist such that they can make the expression be true.
Hope this helps

[Gogo14]

What happens to the asymptote of f(x)=x-25+1/(x-5)
When you take the f(abs(x)) ? Is it still an asymptote, even tho it intersects the graph?

[Shadowxo]

What happens to the asymptote of f(x)=x-25+1/(x-5)
When you take the f(abs(x)) ? Is it still an asymptote, even tho it intersects the graph?

In some cases, the graph intersects the asymptote, so as far as I know, yes it's still an asymptote even though it intersects the graph.

[Sine]

yeah some graphs will have an asymptote that cuts a part of the graph, that is okay. Even when graphing you can draw the asymptote to be cutting the graph and you won't lose any marks.

Just make sure the actual part of the asymptote is accurate, this is very easy to think you have correct but harsh assessors/teachers take off marks for any conceptual error.

[lzxnl]

An asymptote to a function f(x) is a curve such that the distance between f(x) and said curve becomes arbitrarily small as at least one of x, y becomes infinitely positive or negative.

To illustrate my point, consider the hyperbola y = 1/x and the x axis. By arbitrarily small, I mean that if you give me ANY positive number q, I can find an x value p such that for all x > p, the distance between the hyperbola and the x axis is smaller than q. How do I do this?

Let q > 0 be given.
Then, I want to find p such that for all x > p, |1/x - 0| < q.
Suppose I'm only looking at x > 0 as the other side is done the same way.
1/x < q -> x > 1/q
So if I take p to be 1/q, then for all x values greater than p, the graph 1/x is less than q units away from the x axis. This means that no matter how close you want 1/x to get to the x axis, 1/x can and will ALWAYS be closer than than that distance to the x axis.

In your example, it doesn't matter that the graph happens to intersect the asymptote, as long as the graph approaches the asymptote as x -> infinity

[Gogo14]

An asymptote to a function f(x) is a curve such that the distance between f(x) and said curve becomes arbitrarily small as at least one of x, y becomes infinitely positive or negative.

To illustrate my point, consider the hyperbola y = 1/x and the x axis. By arbitrarily small, I mean that if you give me ANY positive number q, I can find an x value p such that for all x > p, the distance between the hyperbola and the x axis is smaller than q. How do I do this?

Let q > 0 be given.
Then, I want to find p such that for all x > p, |1/x - 0| < q.
Suppose I'm only looking at x > 0 as the other side is done the same way.
1/x < q -> x > 1/q
So if I take p to be 1/q, then for all x values greater than p, the graph 1/x is less than q units away from the x axis. This means that no matter how close you want 1/x to get to the x axis, 1/x can and will ALWAYS be closer than than that distance to the x axis.

In your example, it doesn't matter that the graph happens to intersect the asymptote, as long as the graph approaches the asymptote as x -> infinity

Sorry I got a bit lost, but now understand that he graph can cross the asymptote.
second question
I know that the definition of an asymptote is x reaching +/- infinity, but for this graph, when it does the oblique asymptote changes as the graph is reflected in y axis. So the oblique asy as x approaches -infinity does not equal the oblique asy as x approaches +infinity.

[deStudent]

For these 2 questions http://m.imgur.com/a/GnFVp

Q7b) I'm abit lost, because looking at it graphically does not make any sense. How does finding the angles the lines make with the x-axis, then taking the difference allow us to determine the angle made between the graphs?

Q5c) My working is what the worked solutions had. Only determined asymptotes and TP. How does this let us be confident in sketching the graph with only this information? I know it'll always be positive and ascend, but what makes us so sure that the graph doesn't do anything dodgy between the -ve asymptote and y-axis and +ve asymptote and the y-axis?

Also, something unrelated. If we're asked to differentiate arctan(2/5x) in the exam,  do we have to also write the restriction of x or is the answer fine?