VCE Methods Question Thread!

[souka]

Interchange x and f(x) in the equation, and let's replace f(x) with f^-1(x). Then solve for f^-1(x). Hint: use the quadratic formula.

From there let f^-1(x)=f(x).

It seems that it is missing a condition, because I have seen a question which uses ad - bc =/= 0

[Phenomenol]

It seems that it is missing a condition, because I have seen a question which uses ad - bc =/= 0

Souka.

[souka]

Souka.

Doushio cannot in to sugaku

[alchemy]

Souka.

Doushio cannot in to sugaku

What?

[souka]

What?

Is that the full question? Idk just speculating oh maths

[Phenomenol]

f(x)=(ax+b)/(cx+d) where a,b,c,d are real numbers.
Find the values of a,b,c,d such that f^-1(x)=f(x)

Wait, did you amend the expression? I think it might be possible now.
Something along the lines of:
x = (af^-1(x)+b)/(cf^-1(x)+d)
x(cf^-1(x)+d) = (af^-1(x)+b)

You can do the rest. I believe in you

[alchemy]

help with these questions please

For the first question (4a), I would first find positive angles for which sin(x)=1/2 ---> note that I ignored the negative 1/2 here as it's easier to find the angles we are looking for in the given domain once you've found the positive equivalent of the angle first.
sin(pi/6)=1/2, using the table of values. sin is also positive in the second quadrant so (pi-(pi/6))=5pi/6 is your second angle you're looking for. Now multiply all the angles you've got by -1 to fit in the domain you're interested in.

For the second question, it's even easier than the first. Find the equivalent angle to sin(pi/6) that's in the second quadrant since they specify this already by saying it's within pi/2 and pi. So, just like we did in question 4a, the equivalent angle would be (pi-(pi/6))=5pi/6.

Is that the full question? Idk just speculating oh maths

Yes.

[alchemy]

Wait, did you amend the expression? I think it might be possible now.
Something along the lines of:
x = (af^-1(x)+b)/(cf^-1(x)+d)
x(cf^-1(x)+d) = (af^-1(x)+b)

You can do the rest. I believe in you

Hmm, I don't quite understand it yet...can someone please explain their working out?

[Phenomenol]

Hmm, I don't quite understand it yet...can someone please explain their working out?

Just rearrange it until f-1(x) is the subject.
x = (af^-1(x)+b)/(cf^-1(x)+d)
x(cf^-1(x)+d) = (af^-1(x)+b)
cxf^-1(x) + dx = af^-1(x)+b
f^-1(x)(cx-a)=b-dx
f^-1(x)=(b-dx)/(cx-a)

Now you need to compare between the original and inverse function.

[Einstein]

Sorry, but I've got a few questions here which i need clarifying.

1) if you look at the first screenshot, why does it say its worth trying P(5) and P(-5)? is it just trial and error, i know how to long divide with polynomials but just wondering if their was a way to finding the number first more quicker if that makes sense.

2) Im not getting how to use short division, if you look at screenshots 2 and 3, I've not got idea what to do.

Hope you can clarify things, Thanks

[b^3]

1) When the coefficient on highest power of is , then you try factors of the last value . So in this case you wouldn't try as isn't a factor of , but could try .

2) The first step is guessing a factor using 1). Then you know that you have to have that factor multiplied by a factor that has order of 1 less than what you started with (so if you have a cubic and find a linear factor you're left with  a quadratic). Then it's a matter of looking at the contributions to each term (so what gives , what gives e.t.c.) and then equating that back to your original cubic to find the coefficient.
e.g. The only two terms that will give the term is and , and we know that that must give , so you can find . The same goes for the other terms but you'll have more contributions, e.g. we can get a multiple of from (so quadratic and constant) and from (so linear and linear).

[Einstein]

Thanks b^3. Im still having trouble with  4 downwards to 10 in the third screenshot. I just done get where 2 and -12 came from.

[MNM101]

I tried substituting 1,-1,2,-2 but it just wouldn't work out

[IndefatigableLover]

(Image removed from quote.)
I tried substituting 1,-1,2,-2 but it just wouldn't work out

The numbers you could try are pretty much the factors of '12'. In this case you've missed numbers like ±3, ±4, ±6 and ±12 since these are also factors. One of those will work (and from there you can solve for 'x').

Spoiler

It's (x-3) or '3' is a factor

[IndefatigableLover]

Thanks b^3. Im still having trouble with  4 downwards to 10 in the third screenshot. I just done get where 2 and -12 came from.

Hey Photon I think it'd be easier if you just did polynomial long division to find out the answer to that question. I have no idea what they've done but if you know one factor of that polynomial then I'm sure you could just solve the quadratic once you've done long division.

However what they've basically done from Step 4 -> Step 10 is pretty much 'equated coefficients.' You'll see in Step 3 they wrote:

(with the original equation being: )
Basically when you expand the polynomial with the 'ax^2' in it, the power of that will move to 'ax^3' and that has to be equal to '2x^3' since they are the same power. Because of that, we can assume that a=2. This is also the same with the constant term where 'c' is the constant and when it's expanded (-2 times 'c') will give me -2c. This has to be equal to 24 so we just transpose for 'c' which happens to be -12.

They've just basically showed that for the next few steps as well (equating powers and the coefficients of each to find the factorised form of P(x)) but if you don't understand then just use long division and you should be alright imo!

EDIT: b^3 explains it really well down below!