VCE Methods Question Thread!

[Phy124]

The function f has the rule f(x)= (x-2)²+1. Which of the following sets is a possible domain for f if the inverse function f ^-1 exists?
A [1,∞)
B (-∞,0]
C [0,5]
D (-2,∞)
E [0,∞)

Can someone explain to me why the answer is option B?

For the inverse function to exist the original function f must be one-to-one.

A parabola will not be a one-to-one function if it "exists" on both sides of its turning point.

As you would know, the turning point occurs at x = 2

A, C, D & E all have values on both sides of x = 2, thus will not have an inverse function.

The only suitable option for which the function f is one-to-one is B.

Additionally, for future reference, if the question asks to find the maximal domain of a parabola, that will still produce an inverse function, it will be either or where x = a is the turning point.

[soccerboi]

^ Thanks

Next question: How do we sketch the graph of y=root(1-x²) without using a calculator? Or do we just have to know that it forms a semicircle?

[pi]

That's something you should recognise.

[Hutchoo]

^ Thanks

Next question: How do we sketch the graph of y=root(1-x²) without using a calculator? Or do we just have to know that it forms a semicircle?

Yep^^,  you should memorise what all basic graphs look like.

[#1procrastinator]

If you translate y=x using the vector [0 2], then it would be (x,y) -> (x,y+2) so

x'=x
y'=y+2

But how/why does it become y'-2=y and then y'-2=x'?

Isn't y' already y+2 and so y'=x'?

[WhoTookMyUsername]

no. don't use it explicitly in any of methods sac/exam, unless of course your teacher gives you permission to do otherwise in a sac.

D:
My methods teacher said we could

[pi]

no. don't use it explicitly in any of methods sac/exam, unless of course your teacher gives you permission to do otherwise in a sac.

D:
My methods teacher said we could

Don't use it and you shouldn't need too anyway.

[Jenny_2108]

If you translate y=x using the vector [0 2], then it would be (x,y) -> (x,y+2) so

x'=x
y'=y+2

But how/why does it become y'-2=y and then y'-2=x'?

Isn't y' already y+2 and so y'=x'?

After translating: x'=x (1)
                          y'=y+2 => y'-2=y (2)

Given: y=x (3)
You substitute (1) and (2) into (3): y'-2=x'

[#1procrastinator]

^ Why do you write y'-2=y? Why can't we just write y'=x' if it's known that y'=y+2 and x'=x?

man these are some really noob questions lol

[Jenny_2108]

^ Why do you write y'-2=y? Why can't we just write y'=x' if it's known that y'=y+2 and x'=x?

man these are some really noob questions lol

y' is not equal to x', only y=x

[soccerboi]

When finding a general solution to a trig eqn, we write at the end "n E Z"
I forgot, what does "Z" stand for? integer??
Thanks

[Phy124]

When finding a general solution to a trig eqn, we write at the end "n E Z"
I forgot, what does "Z" stand for? integer??
Thanks

The set of integers.

Similar to how R is used to represent the set of real numbers and C for the set of complex numbers etc.

[soccerboi]

How do you find the general solution for 3sin(2(t-(pi/4)))=3/2?

[Phy124]

for and where

Therefore;



and



Divide both sides by 2 leaving;



and



Hence;



and

which is the same as

Meaning the general solution is;



edit: left out a two

[soccerboi]

Thanks you, you make it very clear