VCE Specialist 3/4 Question Thread!

[soccerboi]

The function f(x)=A+Bsin^-1(x), where A and B are integers, passes through the points with coordinates (1/2 , (18-pi)/6 ) and ( -1, (6+pi)/2 .
Find the values of A and B.

When i subbed -1 in to the eqn, the sin^-1(-1), does that become 3pi/2 or -pi/2 ? Cause they give different values for A and B, and the solutions used pi/2 but why?
Thanks

[soccerboi]

Can someone show me how to solve z³-(2-i)z²+z-2+i=0 for z E C ,without using calculator. I forgot how to start, do we sub in random numbers to find a factor?

[Hancock]

Can someone show me how to solve z³-(2-i)z²+z-2+i=0 for z E C ,without using calculator. I forgot how to start, do we sub in random numbers to find a factor?

You need to use factorization by recognition, or something like that. It's called factorisation by grouping(?). First:

z^3 - (2-i)z^2 + z - 2 + i = 0
z^2(z-2+i)+1(z-2+i)=0
(z^2+1)(z-2+i) = 0
(z-i)(z+i)(z-2+i)=0

z = i, -i, 2-i

[Hancock]

The function f(x)=A+Bsin^-1(x), where A and B are integers, passes through the points with coordinates (1/2 , (18-pi)/6 ) and ( -1, (6+pi)/2 .
Find the values of A and B.

When i subbed -1 in to the eqn, the sin^-1(-1), does that become 3pi/2 or -pi/2 ? Cause they give different values for A and B, and the solutions used pi/2 but why?
Thanks

You use -pi/2 because the arcsin (inverse sine function) is restricted to the domain of -pi/2 to pi/2

[Special At Specialist]

All inverse trig functions (arcsin, arccos, arctan, arcsec, arccsc and arccot) are within the domain [-pi/2, pi] and as close to 0 as possible.
So with arcsin(-1), it is -pi/2, since that is the only value within that domain to make sin(-pi/2) = -1.

[soccerboi]

Oh ok, thanks a lot guys

[soccerboi]

How do i antidiff tan²xsec²x?

[Lasercookie]

How do i antidiff tan²xsec²x?

edit: fixed mistake, forgot to power of 3 :/

[soccerboi]

Thank you

[soccerboi]

How do i antidiff secx?
Why cant i do this: root(sec²x), so antidiff is root(tanx)?

Also, why is secx=1/cox not equal to cos^-1x?
Thanks 

[b^3]

Remember that is the inverse of that is if then . is the reciprocal of that is 1 divided by . Draw the two graphs out and you will see (they aren't the same).
As for the first one, this is quite a hard question (doubt you would ever get it on a VCAA exam) but the hint is
then see what subsitution you can do.

EDIT: pit's way for int sec(x) dx is easier, follow that instead

Moderator action: removed real name, sorry for the inconvenience

[pi]

As explained by b^3, ignore the "-1" as a power/index, it's not, it shows inverse and inverse only (you will NEVER see negative powers being written there for trig functions). It's just notation you'll have to get used to. Such is maths I suppose


I've got a simpler way for integral of sec(x), write it as 1/cos(x) and then multiply both sides by cos(x)/cos(x). Then let u=sin(x) after substituting 1-sin^2(x)=cos^2(x). And the rest is elementary.

[soccerboi]

How do you find the volume of a shape rotated about the y axis?

e.g The region S is in the first quadrant of the x-y plane is bounded by the axes, the line x=3 and the curve . Find the volume of the solid when S is rotated around about the y axis.

Thanks

[Hancock]

Sorry, me being a noob 

[b^3]

Also be careful of which region you are rotating, with what Hancook has you will get the area rotated that is left of the curve (and with y=root(10)) in the graph below, but we want the volume of the solid of the area on the right that is between the curve, the axis and x=3 to be rotated.

So we can split it into two volumes, one from the top part that os rotated on the right side of the graph and one for the ccylinder below it.

So you would have to do
So that would be
Then you need to add the cylinder thats still left for y=0 to y=1.
So that would be
Total voulme would then be
Which gives cubic units

EDIT: Should be fixed now.

EDIT2: You could of also have rotated the line x= aroud the y-axis. then taken away the voulme that Hancook found.