VCE Specialist 3/4 Question Thread!

[lzxnl]

Right, how do you do this?
Find the solutions of the equation z^4-2z^2+4=0 in polar form.

If it's fine, mind if you check if I did something wrong? Kinda stuck. Let t=z^2.
Hence t^2-2t+4=0
Used quad formula
t = 1 +- sqrt 3 i. Where do I go from here if I approached it from this way?

Thanks.

Look at the equation. Geometric series initial term 4, 3 terms, common ratio -z^2/2. Geometric series formula of a(1-r^n)/(1-r) then yields z^4 - 2z^2 + 4 = 4 (1-(-z^2/2)^3)/(1+z^2/2) = (z^6+8)/(z^2+2) after simplifying. Or you could just note that this was a term in the difference of two cubes factorising. Try solving this equal to 0. Exclude when z^2 + 2 =0 and solve z^6 + 8 = 0 for the other values of z. Should be easier to solve. I posted this earlier somewhere...

[Thorium]

How do we work out implied domain of cos^-1(x^2)?

[lzxnl]

You can only plug in numbers from -1 to 1 for inverse cosine, so whatever x^2 spits out must be between -1 to 1. You'll find that x can only take the values [0,1] for that restriction to be met.

[Thorium]

You can only plug in numbers from -1 to 1 for inverse cosine, so whatever x^2 spits out must be between -1 to 1. You'll find that x can only take the values [0,1] for that restriction to be met.

Thanks for the help. However, the book says (-1.1), which has baffled me.

[Eugenet17]

Thanks for the help. However, the book says (-1.1), which has baffled me.

if im not mistaken, isn't it because if the values of x are [-1,1] , x^2 will be between [0,1] which satisfies the fact that -1<=x^2=<1?

idk, i was just doing this exercise as well, thats how i thought about it

[b^3]

Thanks for the help. However, the book says (-1.1), which has baffled me.

It should be [-1,1] (since we can include the endpoints). As |zxn| has said, you need the outputs of to only give out values between -1 and 1 (inclusive). That is . Now we can look at this visually to determine what values of this works for:
https://www.desmos.com/calculator/x8vgpa1zvy
So that is when the y values of that curve is between -1 and 1, which we see to be (you could solve for this algebraically then check your extreme points visually), as this is where our function takes the extreme values.

EDIT: Beaten by 5 seconds.

[Eugenet17]

Spoiler

Can i get some help with these? Don't understand these questions at all..

[RKTR]

Spoiler

Can i get some help with these? Don't understand these questions at all..

10. a) sec B=b, B is in second quadrant   

          let B=pi-A,A is the basic angle

          for sec x=-b ,x needs to be in 1st or 4th quadrant
                              A will be one of your answer which is pi-B    
                                another answer will be -A which is B-pi
     
      b)  using cos(A)=sin(pi/2 -A)
            one of the values of x is pi/2 -B
            pi/2 -B is between -pi/2 and 0 so it is the negative of your basic angle
            basic angle is B-pi/2
            -pi+B-pi/2 = -3pi/2 +B

11.a
tan(y) =c  y is in 3rd quadrant
tan(x)=-c  x is in 2nd or 4th quadrant
let y=pi + A ,A is the basic angle , A=y-pi

value of x in 2nd quadrant = pi -(y-pi) =2pi-y
value of x in 4th quadrant = 2pi- (y-pi) =3pi-y

b.
using tan(A)=cot(pi/2 -A)
basic angle for cot x=c is pi/2-(y-pi) = 3pi/2 -y
value for x in 3rd quadrant = pi+ 3pi/2 -y= 5pi/2 -y

[Cort]

You know, I've never been a pious man, but I now know why people become religious. I have no idea what it's asking and my small frail mind can't comprehend thinking outside the box. Would anyone kindly tell me what 'idea' and holistic thought I have to adopt when I'm doing these type of questions?

I've attached working out. I think my problem isn't the working out..it's more of approaching the questions. I'm absolutely clueless.

[alchemy]

Can someone explain what's incorrect in my working out, which I've attached. The question is very simple: "Express cosec (theta) in terms of tan (theta)"

The answer is: plus or minus sqrt(tan^2 (theta) + 1)/(tan(theta))

[rhinwarr]

The steps are pretty much right. Just do the same thing except the first line should be sin^2(theta)+cos^2(theta)=1. You forgot the ^2

[alchemy]

The steps are pretty much right. Just do the same thing except the first line should be sin^2(theta)+cos^2(theta)=1. You forgot the ^2

LMAO! Talk about silly mistakes....

[eagles]

Having trouble with question 8b), can I have some help? xD

[Eugenet17]

Range of sin^-1(x) must be a subset of domain of cos(x) for the composite function to exist.
Range of sin^-1(x) is [ -pi/2,pi/2] which is not a subset of [0,pi] (Domain of cos(x)) so we must restrict all x values of sin^-1(x) so that its range is a subset of [0,pi]
0<=sin^-1(x)<=pi/2 ( [0,pi/2] is a subset of [0,pi] )
Solve,  0<=x<=1. Hence the new domain is [0,1].

To find the range you can just substitute the endpoints of the domain. ( 0 and 1)

[eagles]

Thank you! Makes a lot more sense to manipulate the domain rather than just answer "composite function does not exist" haha