Search results

For 3 b, let it approach the origin along y=kx, for some constant k. for 4c, use part a, which says the limit as x-> pi/2 is 1, and that f(x) is essentially the same as the equation except for one point and that it is continuous. Sub in 3pi/2 then use intermediate value theorem. q5, by...

Well, it may look simple to get the values you want, but you must note that it involves differentiating B(x), which may get a bit difficult if it had a lot of factors in it, unless you expand the whole thing. Then you have to sub it in for each c_1, which may be quite time consuming also. Is...

How nifty. Well, by first principles, B'(a_1) = lim x -> a_1 [B(x) - B(a_1)]/(x-a_1), and R(x) -> R(a_1) as x -> a_1. So, as x -> a_1, R(x)(x - a_1)/[B(x) - B(a_1)] -> R(x)/B'(x).

Yep that looks good.

I'm quite certain you won't get asked to find volumes of unbounded regions in the HSC. Change x-axis to y-axis and it should be fine.

There isn't exactly an area that is 'bounded' by those three curves. It probably means the y-axis. Oh and Riviet, what stops you from taking the rotation of the thing on the left side of y=x^3?

can be any quadratic.

Called taking an educated/intuitive guess of what linear combination of those 4 'basis' polynomials yield 4. This was a lucky question anyway, it had a simple solution, it wouldn't work in general. And I also prefer to use as few lines as possible.

Partial fractions take too long. If there's a quicker way of splitting it up then go for it.

Or, by inspection, 4/[(x-1)^2(x+1)^2] = [1/(x-1)^2(x+1)^2]{(1+2x+x^2)+(1-2x+x^2)+(1+x-x^2-x^3)+(1-x-x^2+x^3)} = [1/(x-1)^2(x+1)^2]{(1+x)^2 + (1-x)^2 - (x+1)^2(x-1) + (x-1)(x-1)^2} = 1/(x-1)^2 + 1/(x+1)^2 - 1/(x-1) + 1/(x+1) now integrate each separately.

Squaring both sides of an inequality does not always preserve the inequality, eg. -2 < -1 4 < 1 is false.

For q1. There exist 3 distinct real roots if there are two stationary points on either side of the x-axis. ie, they have different signs. f'(x) = 3x^2 - 3a = 0 => x = +\-sqrt[a] therefore, f(sqrt[a])f(-sqrt[a]) < 0 , since they have different signs. (asqrt[a] - 3asqrt[a] +c)(-asqrt[a] +...

You have to do the cross product first because if you did the dot product first, you'll be crossing a scalar with a vector, which doesn't work.

Suppose b > a b - a > 0 b^2 + ab + a^2 >= b^2 - |ab| + a^2 = (|b| - |a|)^2 +|ab| > 0 (equality only occurs with a=b=0, so this is strict) therefore, (b-a)(b^2 + ab + a^2) >= 0 b^3 - a^3 >= 0 b^3 >= a^3.

"The result follows by induction" would suffice.

Oi... I just don't want to teach English. Oh, go ahead and edit it aye. And get off the net, you got an exam tomorrow.

I'll give a tip. Try parametrising S.

If you could find a simple closed form integral then you'd be quite rich, well, maybe not, but famous.

"Its a bell curve, with the mean being roughly 66.00 (this isnt permanent, and can go up and down)." No it's not, it's close to a uniform distribution, by the way the uai is defined.

Actually, there is the same number (or cardinality) for both sets, as you can form a one to one correspondence between the two sets. Google up stuff on aleph-null.