Author Topic: BEC'S methods questions (Read 84424 times) Tweet Share

Mao · « **Reply #15 on:** December 31, 2007, 04:45:06 pm »

syntax on this forum:
http://freestudynotes.com/VCEforum/index.php/topic,807.0.html

a list of all the symbols:
http://omega.albany.edu:8008/Symbols.html
there are lists like this all over the place, this is the one i happen to have bookmarked...

humph · « **Reply #16 on:** December 31, 2007, 05:30:36 pm »

Quote from: bec on December 31, 2007, 03:50:39 pm

thanks! ahhh i love this site. if i could use the karma thing you'd all be +ing all over the place.

just wondering, how do you use the maths font where you can write fractions and pi etc?

read http://freestudynotes.com/VCEforum/index.php/topic,864.0.html for info about how to use LaTeX

Collin Li · « **Reply #17 on:** January 01, 2008, 02:13:36 am »

Quote from: bec on December 31, 2007, 03:50:39 pm

thanks! ahhh i love this site. if i could use the karma thing you'd all be +ing all over the place.

just wondering, how do you use the maths font where you can write fractions and pi etc?

You've got 3 more posts until you can use Karma (50 post minimum)

bec · « **Reply #18 on:** January 01, 2008, 01:56:16 pm »

Quote from: coblin on January 01, 2008, 02:13:36 am

You've got 3 more posts until you can use Karma (50 post minimum)

haha i know...2 more now I guess!

yet another question:
Find the values of m for which the equation f(x)=mx has exactly one solution, where $f(x) = \frac{2x}{x^{2}-1}$

The way I approached it was:
$\frac{2x}{x^{2}-1}=mx$
$2x= mx(x^{2}-1)$
$2 = m(x^{2}-1)$
$0 = m(x^{2}) - m - 2$
$\bigtriangleup = 0 - 4(m)(-m-2)$
$= 4m^{2} +8m$

One solution, $\therefore \bigtriangleup = 0$
$4m^{2} +8m$ = 0
$4m(m+2) = 0$
$m=0, m=-2$

The solution according to the book is $-2\le m \le 0$
I don't understand why it's a range of values and not just the m=0 and m=-2 that I got.

thank you so much for all this help everyone. i'd reciprocate but unless you need help in french there's not much i could help you with haha

Collin Li · « **Reply #19 on:** January 01, 2008, 03:12:04 pm »

The step that is not quite legal is: $2x = mx(x^2-1) \Rightarrow 2=m(x^2-1)$

You can't just divide both sides by $x$ because $x$ can equal zero (and you would divide by zero!). The correct progression would be:

$mx(x^2 - 1) - 2x = 0 \Rightarrow x(mx^2 - m - 2) = 0$

So now, you have the solutions:
$x=0\mbox{, and }mx^2 - m -2 =0$

You want to have 1 solution only! Regardless of whatever value of $m$ is chosen, there will be at least one solution: $x=0$ , so you want to ensure that the second set of solutions: $mx^2 - m -2 =0$ actually has no solutions.

Now, just solve for $\bigtriangleup < 0$ (zero solutions), and you should be fine.

bec · « **Reply #20 on:** January 01, 2008, 03:31:16 pm »

ahh I see...that's a tricky question actually! i never thought about dividing by x, it was just an automatic thing that I'd do if it wasn't a variable - so thanks for pointing it out, otherwise i'd probably make the same mistake again.

don't worry i won't post anymore questions (for now)....thanks for all your help!

enwiabe · « **Reply #21 on:** January 01, 2008, 05:02:47 pm »

Nooooo

I wanted to have a crack at one.

Oh well. Glad everyone else could help!

AppleXY · « **Reply #22 on:** January 01, 2008, 05:13:19 pm »

lol. Coblin is too fast

bec · « **Reply #23 on:** January 02, 2008, 08:52:40 pm »

Quote from: enwiabe on January 01, 2008, 05:02:47 pm

Nooooo I wanted to have a crack at one.

don't worry i have plenty more! share them round haha

ok so onto integrals and antidiff now...

Find the derivative of $\log_{e}(x^{2}+2)$ and hence find $\int\frac{6x}{x^{2}+2} dx$

I figured out the derivative alright ( $\frac{2x}{x^{2}+2}$ , not that i needed to tell you because you can probably work it out in your head ahha)
So doesn't that mean that $\int\frac{6x}{x^{2}+2} dx=3\log_{e}(x^{2}+2) + c?$
But the answers say that it should be $3\log_{e}(x^{2}-2) + c$
why minus and not plus?

Also, on another note... if I get a question like this, $\int\frac{3}{(4x+1)^{3/2}} dx$ is there a better way of working it out than rewriting the expression as $3(4x+1)^{\frac{-3}{2}}$ ? Let me rephrase, is there a better way that is VCE-LEVEL that I can work it out w/out doing that?

thanks heaps

Collin Li · « **Reply #24 on:** January 02, 2008, 09:01:25 pm »

For your second question, there isn't a better way. I think you mistyped your integral, because you've made it look like it's $(4x + 1^{\frac{3}{2}})$ instead of what I'm sure you meant to be $(4x+1)^{\frac{3}{2}}$ .

For your first question, you are correct. I suspect it is a typo of the book.

bec · « **Reply #25 on:** January 02, 2008, 09:05:22 pm »

thank you! and yes, i did mean for there to be brackets in that fraction (i've changed it now)

and here's another question. yay!

Find an antiderivative of $sin\frac{x}{2}+\frac{1}{(2x+1)^{2}}$
I'd write out my clumsy working but it'll take me half an hour to do it in latex...and it's all wrong anyway...

Collin Li · « **Reply #26 on:** January 02, 2008, 09:06:52 pm »

The anti-derivative is: $-2\cos\frac{x}{2} -\frac{1}{2(2x+1)}$

You get the first term by commonsense: you know that the derivative of cosine is sine, so you start off with $\cos\frac{x}{2}$ , then you put a $-2$ in front to compensate for what happens when you derive cosine. (Remember, the derivative of the anti-derivative should be what you started with)

The second term you get by re-expressing the original term as $(2x+1)^{-2}$ then using your anti-differentiation laws. It will be $(2x+1)^{-1}\times \frac{1}{-1}\times \frac{1}{2}$ when anti-differentiated.

bec · « **Reply #27 on:** January 02, 2008, 09:18:16 pm »

thank you very much once again!!

and another....
₂
If $\int f(x)dx=-3$ find the value of:
⁰

₂
a) $\int (3f(x)-1) dx$
⁰

₂
b) $\int (4-f(x)) dx$
⁰

I really don't understand how to work these out - I've never been taught anything like it and there's no specific examples like these that I can find in my book. Any tips?

Collin Li · « **Reply #28 on:** January 02, 2008, 09:22:34 pm »

The strategy is to break up the integral term by term, and then take out the constant factor, so you get something you know: $\displaystyle\int^2_0 f(x)\,dx$ so you can substitute it in.

a)
$\displaystyle\int^2_0 (3f(x)-1)\,dx = 3\displaystyle\int^2_0 f(x)\,dx - \displaystyle\int^2_0 1\,dx = 3\cdot (-3) - [x]^2_0 = -9 - (2-0) = -11$

b)
$\displaystyle\int^2_0 (4-f(x))\,dx = \displaystyle\int^2_0 4\,dx-\displaystyle\int^2_0 f(x)\,dx = [4x]^2_0 - (-3) = (8-0) + 9 = 17$

bec · « **Reply #29 on:** January 02, 2008, 09:30:13 pm »

karma +10000 if i could
I get it! that is actually such a clear explanation, before you'd added the working for (b) i was trying to work it out myself (predictably, you beat me anyway)
the solns in my text say the answer is 11 but i got the same answer as you anyway and the book sucks so i'm guessing it's wrong (again)

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