Author Topic: Dekoyl's question thread (Read 23330 times) Tweet Share

/0 · « **Reply #45 on:** October 23, 2008, 09:44:00 pm »

Quote from: dekoyl on October 23, 2008, 09:35:10 pm

Hello once again. I have a mental block. Could someone help me unblock this block?
How does:

$p^{2} - 6e^{\frac{p}{3}} + 2p + 6$

simplify to:

$p^{2} - p$

Thank you.

I don't think it does, actually...

There is no way to get rid of the $e$ or the six, unless p = 0

dekoyl · « **Reply #46 on:** October 23, 2008, 09:47:32 pm »

Quote from: DivideBy0 on October 23, 2008, 09:44:00 pm

I don't think it does, actually...

There is no way to get rid of the $e$ or the six, unless p = 0

Hmm really?
This is the absolute last question of Kilbaha 2008 Exam 2.
If you can be bothered, you could pull it out.

Sorry I don't have the e-form to show you.

orsel · « **Reply #47 on:** October 23, 2008, 10:12:21 pm »

sorry, cbf latex but:

coordinates of intersection (p,p) are on curve
therefore f^-1(p)=p
therefore 2e^(p/3)-2=p
substitute in and simplify

dekoyl · « **Reply #48 on:** October 25, 2008, 11:31:39 am »

Quote from: orsel on October 23, 2008, 10:12:21 pm

sorry, cbf latex but:

coordinates of intersection (p,p) are on curve
therefore f^-1(p)=p
therefore 2e^(p/3)-2=p
substitute in and simplify

Ah thank you very much.. I've got it now

+1
---------------------------------------------------------------------------------
Sometimes I really have difficulty "seeing through" questions.

If the point $(log_e 3, \frac{82\sqrt{3}}{9})$ lies on the curve of $y = 3e^{x/a}+e^{-x/a}$ what is $a$ equal to?

By substituting the choices (multiple choice), $a = \frac{2}{3}$ but I can't seem to get this answer algebraically.

Thanks!

Mao · « **Reply #49 on:** October 25, 2008, 04:09:04 pm »

what's p?

dekoyl · « **Reply #50 on:** October 25, 2008, 05:42:43 pm »

Quote from: Mao on October 25, 2008, 04:09:04 pm

what's p?

~~P was an irrational number which wasn't relevant to that part of the question.~~

~~Anyone able to help me out on the last question?~~

shinny · « **Reply #51 on:** October 25, 2008, 05:48:39 pm »

...2 unknowns and 1 piece of info? How's this work. Are you sure that p isn't a typo <_<

dekoyl · « **Reply #52 on:** October 25, 2008, 06:10:46 pm »

Oh shit. Sorry Mao (thought you were referring to the question before that) and shinjitsuzx. Yes that is indeed a typo.

p = 9

I'll do it again since I have another question..

1.If the point $(log_e 3, \frac{82\sqrt{3}}{9})$ lies on the curve of $y = 3e^{x/a}+e^{-x/a}$ what is $a$ equal to?

2. $f(x) = \frac{1}{2} - \frac{1}{4}|x-1|$ if $-1 \leq x \leq 3$
The above is a probability density function. What is the variance of X?
I'm sure I'm doing everything right but I keep getting $\frac{-7}{9}$ . I've checked my working over and over.

Pandemonium · « **Reply #53 on:** October 25, 2008, 06:50:01 pm »

well. for question 1... i got a = 2/3 and a = -2/7 without any restrictions...
perhaps i should check. it takes quite a few steps to get there though

dekoyl · « **Reply #54 on:** October 25, 2008, 06:53:51 pm »

Quote from: Pandemonium on October 25, 2008, 06:50:01 pm

well. for question 1... i got a = 2/3 and a = -2/7 without any restrictions...
perhaps i should check. it takes quite a few steps to get there though

a = 2/3 is correct.
I couldn't get it by hand though (had to use calculator).

shinny · « **Reply #55 on:** October 25, 2008, 06:56:30 pm »

I got 2/3 and -2/5. Don't see why that second solution can't be considered... I'll post the solutions in a sec.

dekoyl · « **Reply #56 on:** October 25, 2008, 06:59:37 pm »

Quote from: shinjitsuzx on October 25, 2008, 06:56:30 pm

I got 2/3 and -2/5. Don't see why that second solution can't be considered... I'll post the solutions in a sec.

Thanks in advance.

The 2nd solution just isn't one of the choices. It probably isn't the best written exam as it's a complimentary one.

Pandemonium · « **Reply #57 on:** October 25, 2008, 07:00:19 pm »

how'd you get -2/5? hmm i'm questioning my working out now

after the quadratic formula, i got like 3^(1/a) = 3sqrt3 and 3^(1/a) = 1/27sqrt3.

shinny · « **Reply #58 on:** October 25, 2008, 07:07:37 pm »

Sub in y and x and simplify to get: (too lazy to type out all the simplification steps...its quite tedious; just use your log laws)
$\frac{82 \sqrt{3}}{9}=3 \times 3^{\frac{1}{a}}+3^{-\frac{1}{a}}$

$\mbox{Let A=3^{\frac{1}{a}}}$

$3A^{2}-\frac{82 \sqrt{3}}{9}A+1=0$
Use the quadratic formula to get: (yet again too lazy)
$A=\frac{\sqrt{3}}{27} or 3 \sqrt{3}$

$3^{\frac{1}{a}}=3^{\frac{3}{2}} or 3^{\frac{-5}{2}}$

$\therefore a=\frac{2}{3} or \frac{-2}{5}$

Pandemonium · « **Reply #59 on:** October 25, 2008, 07:14:02 pm »

oops lol yeah you're right. that part. i suck. thanks for that.

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