VCE Specialist 3/4 Question Thread!

[psyxwar]

Anything from the study design is fair game for exam 1 and exam 2, unless otherwise stated in said design (but I don't think anything is exempted for specialist)

EDIT: LEL three in a row, whoops. .__. I think the point is clear now, at least.

we're just showing our love

[Vividdreamer]

How would I expand and simply

[keltingmeith]

How would I expand and simply (Image removed from quote.)

a) Notice we have a difference of two squares:




b) Go the long way:

[Vividdreamer]

a) Notice we have a difference of two squares:




b) Go the long way:

Thank you EulerFan101!, The second method feels more comfortable for me. I don't think this method of expansion was explained to me in Methods or Specialist.

[lzxnl]

Thank you EulerFan101!, The second method feels more comfortable for me. I don't think this method of expansion was explained to me in Methods or Specialist.

It's just using difference of two squares. If you ever see (a+b+ci)(a+b-ci), note that you're adding and subtracting ci in each bracket. Hence use DOTS.

[Robert123]

From vcaa 2006 exam 2, Q5aii
Given that z=cis(Pi/4), find the complex equation which passes through z and -z in terms of the conjugate of z.

I know the Cartesian equation is simple y=x but how do I transform it into a complex equation?

[psyxwar]

From vcaa 2006 exam 2, Q5aii
Given that z=cis(Pi/4), find the complex equation which passes through z and -z in terms of the conjugate of z.

I know the Cartesian equation is simple y=x but how do I transform it into a complex equation?

z bar = cis (-pi/4)

y=x is the perpendicular bisector of z bar and -z bar

Therefore |z-z bar| = |z+z bar|

[Professor_Oak]

Does anyone know if VCAA still does real world rounding for spesh? i.e. one calculates 6.8 goats, but you round down to 6 as one cannot have .8 of a goat.

[atom]

Hi, I'm having trouble with the attached question, would be awesome if someone could explain their worked solution. Thanks in advance

[Zealous]

After letting z=x+yi, expand out on your CAS calculator (this is multiple choice!):



Take the real component, then complete the square:



So the graph is a circle with center (2,0) and radius 2. However, because we are diving by 'z', we cannot include the point z=0+0i because diving by 0 is undefined, so the answer is E.

[keltingmeith]

Here's another method using the linearity of Re(z) operator:



So, we have a circle of radius 2 centred at (2, 0), as confirmed by Zealous. Once again, exclude (0, 0) using Zealous' logic.

[allstar]

a helicopter is rising with constant velocity 20m/s. When the helicopter is 60m above the ground a parcel is dropped. Find an equation in the form gt^2 + bt +c = 0 where g is the magnitude of the acceleration due to gravity and b and c are integers which when solved will give the exact time the parcel hits ground.

This is what I did but the ans says c = -120, i had +120?

[tiff_tiff]

how would i do this question:

a graph y = a/(x^2 +2ax +b)  where a,b are a subset of R has range (-infinity, 0) in union with [1/4, infinity) and one of the vertical asymptotes is x=-1
Find a and b

ive attached the answers i get everything up to the blue box, why did they make it equal 1/4, isn't that making an assumption that the maximum of the original will touch the minimum of the reciprical, but thats not always the case for example with y = x^2 + 2x + 3?

[lzxnl]

how would i do this question:

a graph y = a/(x^2 +2ax +b)  where a,b are a subset of R has range (-infinity, 0) in union with [1/4, infinity) and one of the vertical asymptotes is x=-1
Find a and b

ive attached the answers i get everything up to the blue box, why did they make it equal 1/4, isn't that making an assumption that the maximum of the original will touch the minimum of the reciprical, but thats not always the case for example with y = x^2 + 2x + 3?

So, you know the vertical asymptote => x = -1, bottom equals zero
In addition, you do know that the denominator has a max value of 4 because the minimum value for its reciprocal when positive is 1/4. Think about what the graph of this thing looks like. There are three parts to the graph and the middle section corresponds to your [1/4, infinity) range

[lucas.vang]

Hi this is from VCAA 2007 EXAM 1

I wasn't quite sure but is this enough for 4 marks?