HSC Mathematics Marathon (5 Viewers)

jyu · Jan 30, 2011

If you are giving up the gradient question, try the next one that you asked for.

Sorry, it was Drongoski asking for another question.

Drongoski · Jan 30, 2011

jyu said:
If you are giving up the gradient question, try the next one that you asked for.

Sorry, it was Drongoski asking for another question.

No I wasn't asking for another question for myself. I thought once a question was done another question need to be posted. I myself have no question to post.

Drongoski · Jan 30, 2011

jyu said:

x can only = 1 so that y = 0 and so log₁₀(x+y) = 0.

Bored Of Fail 2 · Jan 30, 2011

Drongoski said:
x can only = 1 so that y = 0 and so log₁₀(x+y) = 0.

thats a good one

jyu · Jan 30, 2011

$f\left ( x \right )$
is an odd function and
$f\left ( x+2 \right )=-f\left ( x \right )$ .
When
$x\epsilon \left ( 0,1 \right )$ , $f\left ( x \right )=2^{x}$ .
Evaluate
$f\left ( log_{\frac{1}{2}}18 \right )$ .

cutemouse · Jan 31, 2011

jyu said:
$f\left ( x \right )$
is an odd function and
$f\left ( x+2 \right )=-f\left ( x \right )$ .
When
$x\epsilon \left ( 0,1 \right )$ , $f\left ( x \right )=2^{x}$ .
Evaluate
$f\left ( log_{\frac{1}{2}}18 \right )$ .

Lol, if you're going to be all smart using notation not used in the HSC then at least get it all right. f(x) isn't a function; it's a rule.

Bored Of Fail 2 · Jan 31, 2011

y = log (1/2) 18
(1/2)^y = 18
ln ( 1/2)^y = ln(18)
y= ln(18) / ln(1/2) = -4.1699

now we know that f (-x) = -f(x) ( because its odd ) = f(x+2) = ( given to us )

therefore we add or subtract multiples of 2 to the value of y until it gets inside the domain [0,1] , then we can use the rule f(x) = 2^(x) , but it will never get in that domain :|

something similar to that, cnt be bothered right now

UnrealAnchovies · Jan 31, 2011

Can someone please answer my conics question on like page 2 or 3, so I can get some questions going too.

jyu · Jan 31, 2011

the answers

hscishard said:
Oh god. Noones going to type all of the proof.
a) is x(x1)/a^2 - y(y1)/b^2 = 1
b) PS/PS' = (x1 -a/e) / (x1+a/e)
TS/TS' = (ae-a^2/x1) / (ae + a^2/x1)
=(x1 -a/e) / (x1+a/e).

jyu · Jan 31, 2011

jm01 said:
Lol, if you're going to be all smart using notation not used in the HSC then at least get it all right. f(x) isn't a function; it's a rule.

Usually we just say function f(x), we don't say the rule f(x) of the function

hscishard · Jan 31, 2011

jyu said:
the answers

He prob wants someone to post the solution step by step. I'm definitely not going to do that

funnytomato · Jan 31, 2011

Could some one teach me how to use LaTex? cos this link http://community.boredofstudies.org/showthread.php?t=234259 isn't working.

ohexploitable · Jan 31, 2011

funnytomato said:
Could some one teach me how to use LaTex? cos this link http://community.boredofstudies.org/showthread.php?t=234259 isn't working.

Read this: http://en.wikibooks.org/wiki/LaTeX

UnrealAnchovies · Jan 31, 2011

Hscishard have you done integration? I know ohexploitable hasn't. :-\

hscishard · Jan 31, 2011

4unit integration? Na, i've only read parts of it.

AAEldar · Jan 31, 2011

Find:
$\int{e^xsinx\;dx}$

jyu · Jan 31, 2011

AAEldar said:
Find:
$\int{e^xsinx\;dx}$

$\int e^{x}sinxdx=\frac{e^{x}}{2}\left ( sinx-cosx \right )+c$

jyu · Jan 31, 2011

Evaluate
$\int_{-\pi }^{\pi }x^{3}cos\left ( 3x \right )dx$

Bored Of Fail 2 · Jan 31, 2011

zero -its an odd function

a product of an odd function (x^3) and even function (cos(3x) ) gives an odd function

Drongoski · Jan 31, 2011

jyu said:
Evaluate
$\int_{-\pi }^{\pi }x^{3}cos\left ( 3x \right )dx$

= 0

Oh I didn't see Bored's answer. Again it shows he knows his maths.

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