4U Revising Game (1 Viewer)

[ronnknee]

Back when you were a youngster? Hahahha I'm sure you haven't aged THAT much, afterall you finished HSC only last year =p

 

[Undermyskin]

Can I just post the answers? Typing the whole solution is tedious.

1. ln6
2. use t= tan@/2
3. 4/11
4. pqr= 100*p +10*q+r = 99*p+9*q+p+q+r = 99*p+9*q+3A = 3(33*p+3*q+A) is obviously divisible by 3.

Please check my calculations. hix

 

[Poad]

i got 4/7 for 3.

Aaand I got 2/11 for 3. >_>

 

[Mark576]

i got 4/7 for 3.

:wave: I got 4/7 as well.

 

[Trebla]

By considering the graph of the hyperbola y = 1/x, show that
lim (1 + 1/n)ⁿ = e
^n->∞

 

[Undermyskin]

Crap. Ouch. It's 4/7.

 

[Rhanoct]

Find a reduction formula for [integral sign]e^x cos^n x

 

[ronnknee]

It's best to show working out so that those who struggle can see the processes involved

 

[Rhanoct]

Well done sir.

 

[Grammar Police]

Here's some maths for you;

And MY favourite;


Let x = 0.999∞

1x = 0.999∞
10x = 9.999∞
10x-1x = 9x = 9.999∞-0.999∞
9x = 9
x = 1


A POX ON YOUR FOOLISH 'MATHS'

 

[waxwing]

complex numbers!

prove that the roots of

z^3 + 3pz^2 + 3qz + r = 0

form an equilateral triangle if and only if p^2 = q

I have a curious way of answering this:

Spoiler

We can map the cube roots of unity (the solutions to z³=1) to any other three points which also form an equilateral triangle on the Argand diagram by the following sequence of transformations:
1. A shrink/enlarge
2. a rotation - multiply by complex number a, with |a|=1
3. a translation

So, we start with z³=1
Then stretch: z³=r (r is a real)
Then rotate: (az)³=r
Then translate: (az-k)³=r
Expanding (and let a=1, I'm just skipping an obvious step there to avoid all the typing), the form is:
z³-3kz²+3k²z+k³-r=0
from which the result follows.

 

[waxwing]

full marks mr. waxwing

Actually -1 mark for writing +k³ instead of -k³ at the end there

 

[ronnknee]

I should really try and keep my own thread alive but right now there are so many questions that I can't keep track of haha

Anyway,


The region bounded by the curves y = x², y = (x - 2)² and the x-axis is rotated about the line x = 2. Use the method of cylindrical shells to find the volume of the solid

 

[DivJx]

Oi um.

I appreciated the first one. Even though time AND money ---> time + money...not time x money

sorry to disappoint but time AND money is actually time x money not time + money, as AND in probability means x and OR is what means +

 

[pLuvia]

sorry to disappoint but time AND money is actually time x money not time + money, as AND in probability means x and OR is what means +

It wasn't dealing with probability lol

 

[Undermyskin]

this is not completely correct. there are points in your equation which do not satisfy the locus.

Why is it? I think his solution is typical and similar to what we've been taught...

So you mean that it's not absolutely correct to put the Re = 0? Or you mean we have to include x =/= 2 and y =/= 0?

 

[3unitz]

Why is it? I think his solution is typical and similar to what we've been taught...

So you mean that it's not absolutely correct to put the Re = 0? Or you mean we have to include x =/= 2 and y =/= 0?

the obvious is z =/= 2 because of the 0 denominator, the other is z =/= i as w is purely imaginary.

 

[le91]

The locus there is correct. It's easier to do arg(w)=1/2pi + k pi and then expand and find the locus geometrically though.

 

[cheney31]

i already gave my answer to your intergal.. its just 1 by inspection !

nice inspection LOL


the better way (or more "idiotic" way than inspection) is to make the top as

1-e^-x + e^-x

then u can solve it by INSPECTION...

 

[conics2008]

I should really try and keep my own thread alive but right now there are so many questions that I can't keep track of haha

Anyway,


The region bounded by the curves y = x², y = (x - 2)² and the x-axis is rotated about the line x = 2. Use the method of cylindrical shells to find the volume of the solid

hey about time i found this thread..

ill post the solution

find p where it is the intersection of y=x^2 and y=(x-2)^2

p(1,1)

the shaded reigion is sys therefore just calculate the vol from 1 to 0

equation of volume =

pi { (x-2)^2 -(2-x-dx)^2)(x^2)

after expanding and simplyfing..

you get

2pi { 2-x }{x^2} dx

there fore total volume of solid is

lim dx->0 4pi sigam from 0 to 1 S 2x^2-x^3 dx

= 5pi/3

can some confirm this please, because im having second thoughts..

here is a question to keep this game rolling =]

Q) The sequence (X(n)) is given by

X(1) = 1 and X(n+1) = 4+X(n) / 1+ X(n)

prove by induction that for n>= 1 that X(n) = 2{1+a^n/1-a^n} where a=-1/3 ( 4 marks )