man0005's specialist question thread :)

[iNerd]

Don't use long division on that..

Just recognise that there are 4 x^2 on the numerator, thus try and get 4 of the denominator separate to divide.

Wtf. That is insanely sexy/eloquent.

[man0005]

hm how would you sketch something like Arg z > -pi/2
nevermind ^^

what would the domain of this be?
|z – 6| – |z + 6| = 3

[Mao]

what would the domain of this be?
|z – 6| – |z + 6| = 3

This is a hyperbola. It is the left-hand side only, with a Re(z) intercept of -1.5, domain is (-infinity,-1.5]

[man0005]

Ah what about this one:
If Sec^-1 x = 2
then x is equal to
a) 1.047  b) 0.8776  c) 1.100 d) -2.403  e) 0.5

[luken93]

[man0005]

For 2Sin^-1 (1/x+1)
why do you have to exclude zero for the implied range?

[enpassant]

disagree

[man0005]

did you get -2.403 empassant? cause thats what the answers said
but i got what luken got as well :/

[enpassant]

yes

[man0005]

how?!

[luken93]

how?!

check mine now..

[man0005]

isnt there a way to do it without the use of calculator then?

[BubbleWrapMan]

For 2Sin^-1 (1/x+1)
why do you have to exclude zero for the implied range?

Is the inside part 1/x+1 or 1/(x+1)?
If it's the latter (which makes sense based on your question)
1/(x+1) is a hyperbola with range R\{0}, so since it isn't possible for the hyperbola to equal 0, and sin^-1(0) = 0, then 2sin^-1(1/(x+1)) can never equal 0.

[VCE247]

How would I go about doing a question like this:
Let Sec B = b, B is an element of [pi/2 , pi ]
Find in terms of B, two values in the range [-pi, pi ] which
satisfy sec x = -b and cosec x = b

[luken93]

So basically, you are trying to find two values where









I'm not sure if this is what you're asking though?

Subbing back into the other equations, we see that


Hence, in terms of B: