VCE Specialist 3/4 Question Thread!

[lolalol]

Struggling to find the range in q8a, I understand that the domain is the domain of cos(x) but how do you find the range? Cheers in advance!

[brightsky]

okay so we have y = arcsin(cosx). the argument of arcsine can only be within the interval [-1, 1]. so we have cosx E [-1, 1], which is true for all x E R. but we are given that cosx is restricted to [0,pi], so the implied domain is [0,pi] as you stated. to find the range, we work backwards. x E [0,pi] so what is cosx between? sketch y = cosx over the interval [0,pi] and you'll find that cosx E [-1, 1]. so what is arcsin(cosx) between? we first sketch draw a special cartesian plane, with y as the vertical axis, and cosx as the horizontal axis. we sketch y = arcsin(cosx) over the domain cosx E [-1, 1]. your graph should look exactly the same as y = arcsin(x) except with cosx as the horizontal axis instead of x. as you can see from the graph, arcsin(cosx) E [-pi/2, pi/2]. so the implied range is [-pi/2,pi/2].

[Kuroyuki]

Struggling to find the range in q8a, I understand that the domain is the domain of cos(x) but how do you find the range? Cheers in advance!

so you have the domain of cos x = 0 to pi
0 < or = x < = pi
take cos of everything
cos0 < or = cosx < or = cos pi
which gives the principal range of cos x to be
-1 < or = cos x < or = 1
then take arcsin of everything
so
arcsin(-1) < or = arcsin(cosx) < or = arcsin(1)
which will give
-pi on 2 < or = arcsin(cosx) < or = pi on 2
which is the range as arcsin(cosx) is the original function.
EDIT - made it a bit clearer i think, and i think brightsky's method is more reliable.

[RKTR]

y=2-6/x  x=f(t) y=g(t) find f'(t) when y=1 given g'(t)=0.4

how to do this?o.o

[brightsky]

dy/dx = (dy/dt)*(dt/dx) = 6/x^2
when y = 1, g'(t) = dy/dt = 0.4 = 2/5
when y = 1, x = 6
plug into first equation:
(2/5)*(dt/dx) = 6/(6)^2
(2/5)*(dt/dx) = 1/6
dt/dx = 5/12
dx/dt = 12/5
since f'(t) = dx/dt, then f'(t) = 12/5.

[RKTR]

oh didnt know that y=g(t) then g'(t)=dy/dt

is it like y=f(x) so f'(x)=dy/dx ?

[brightsky]

indeed.

[alchemy]

Is there a quick way to do this: (3x^2-6x+2)/((x-1)^2)(x+2)

[kinslayer]

Is there a quick way to do this: (3x^2-6x+2)/((x-1)^2)(x+2)

What do you need to do with it?

Also the way you wrote the brackets in the denominator is a little weird. Do you mean this?



or

[alchemy]

What do you need to do with it?

Also the way you wrote the brackets in the denominator is a little weird. Do you mean this?



or

Oh sorry, I forgot to mention what the question was. It was asking to split that expression into partial fractions. I'm doing it how the book suggested, and it ends up being excruciatingly long

[kinslayer]

Okay yeah I thought that might be it. Partial Fractions are usually pretty long to work out, but that one looks pretty standard.

Have a read of this page and see if it makes things any easier/clearer:

http://tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx

Otherwise if you think what you are doing isn't right then feel free to post it and someone can look over it:)

[Sanguinne]

How do I work out the asymptote other than x = 0 of y= (-x^2+1)/-2x.

[lzxnl]

How do I work out the asymptote other than x = 0 of y= (-x^2+1)/-2x.

Your function is also x/2 - 1/2x if you divide through. As x becomes large positively or negatively, the -1/2x term becomes tiny until it becomes negligible and arbitrarily close to zero. Thus your asymptote is y = x/2

[Cort]

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.

[nhmn0301]

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.

Then you sub in each t value you have inside z^2 and solve again. From each value of t, you should able to get 2 values more. Hence, you end up with 4 values
Hope this helps!