HSC 2013 Maths Marathon (archive) (1 Viewer)

Menomaths · Oct 21, 2013

Re: HSC 2013 2U Marathon

ocatal said:
Back to the marathon guys?

$$Solve the equation$ \ 2\ln(x\sqrt{3}) = \ln(6 - 7x).$

is it x= 2/3 and x= -3?

leesh95 · Oct 21, 2013

Re: HSC 2013 2U Marathon

ocatal said:
Back to the marathon guys?

$$Solve the equation$ \ 2\ln(x\sqrt{3}) = \ln(6 - 7x).$

x=2/3

ocatal · Oct 21, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
is it x= 2/3 and x= -3?

Only 2/3

hamster-lol · Oct 21, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
is it x= 2/3 and x= -3?

Don't take x=-3 otherwise $\ln(f(x)) < 0$

Menomaths · Oct 21, 2013

Re: HSC 2013 2U Marathon

hamster-lol said:
Don't take x=-3 otherwise $\ln(f(x)) < 0$

oh yeh

Badsiii · Oct 21, 2013

Re: HSC 2013 2U Marathon

ocatal said:
Back to the marathon guys?

$$Solve the equation$ \ 2\ln(x\sqrt{3}) = \ln(6 - 7x).$

Ln(xroot3)^2=ln(6-7x)
3x^2=6-7x
3x^2+7x-6=0
X=2/3 or x=/=-3
Therefore x=2/3

Sy123 · Oct 21, 2013

Re: HSC 2013 2U Marathon

Youi_ said:
In the interest of maximizing skills please explain the solution to your question I differentiated g(x) and lost it after that

$g'(x) = f'(x) (\ln f(x) + 1)$

$\therefore \ \ g'(a) = f'(a) (\ln f(a) + 1) \ \ \fbox{1}$

$From the original equation$

$g(a) = f(a) \ln f(a)$

$g(a) + f(a) = f(a) (\ln f(a) +1)$

$$therefore, due to$ \ \ \fbox{1} \ \ \frac{g'(a)}{f'(a)} = \ln f(a) + 1 \ \ $if$ \ \ f'(a) \neq 0$

$\therefore \ \ g(a) + f(a) = f(a) \times \frac{g'(a)}{f'(a)}$

$g'(a) = 0$

$\therefore \ \ f(a) + g(a) = 0 \ \ $if$ \ \ f'(a) \neq 0$

I don't think the result holds if f'(a) = 0 though, an example is

$y= (x^2+1) \ln (x^2+1)$

hamster-lol · Oct 21, 2013

Re: HSC 2013 2U Marathon

Simplify:

$(x-a)(x-b)(x-c)...(x-y)(x-z)$

Carrotsticks · Oct 21, 2013

Re: HSC 2013 2U Marathon

lol that problem

Sy123 · Oct 21, 2013

Re: HSC 2013 2U Marathon

$$Prove that the volume of a sphere with radius$ \ \ r \ \ $is$ \ \ \frac{4\pi r^3}{3}$

RealiseNothing · Oct 21, 2013

Re: HSC 2013 2U Marathon

Carrotsticks said:
lol that problem

+1

Badsiii · Oct 21, 2013

Re: HSC 2013 2U Marathon

hamster-lol said:
Simplify:

$(x-a)(x-b)(x-c)...(x-y)(x-z)$

0 lol (x-x)

Youi_ · Oct 21, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$g'(x) = f'(x) (\ln f(x) + 1)$

$\therefore \ \ g'(a) = f'(a) (\ln f(a) + 1) \ \ \fbox{1}$

$From the original equation$

$g(a) = f(a) \ln f(a)$

$g(a) + f(a) = f(a) (\ln f(a) +1)$

$$therefore, due to$ \ \ \fbox{1} \ \ \frac{g'(a)}{f'(a)} = \ln f(a) + 1 \ \ $if$ \ \ f'(a) \neq 0$

$\therefore \ \ g(a) + f(a) = f(a) \times \frac{g'(a)}{f'(a)}$

$g'(a) = 0$

$\therefore \ \ f(a) + g(a) = 0 \ \ $if$ \ \ f'(a) \neq 0$

I don't think the result holds if f'(a) = 0 though, an example is

$y= (x^2+1) \ln (x^2+1)$

Thanks

SpiralFlex · Oct 21, 2013

Re: HSC 2013 2U Marathon

hamster-lol said:
Simplify:

$(x-a)(x-b)(x-c)...(x-y)(x-z)$

This brings back memories of high school. That was one of the questions that was posted during my early years on the forum

Carrotsticks · Oct 21, 2013

Re: HSC 2013 2U Marathon

RealiseNothing said:
+1

Remember the time you actually tried to expand it?

RealiseNothing · Oct 21, 2013

Re: HSC 2013 2U Marathon

Carrotsticks said:
Remember the time you actually tried to expand it?

I try to forget these things.

Sy123 · Oct 21, 2013

Re: HSC 2013 2U Marathon

$Prove the following$

$$i)$ \ \ \sin^2 x + \cos^2 x = 1$

$$ii)$ \ \ \log_a b = \frac{\log_c b}{\log_c a}$

$iii) A triangle constructed in a circle with one side$

$as a diameter is a right angled triangle$

Badsiii · Oct 21, 2013

Re: HSC 2013 2U Marathon

$State the range and domain for$

$$i)$ \ \ \ \sqrt{4-x^2}$

mahoo_mk · Oct 21, 2013

Re: HSC 2013 2U Marathon

Badsiii said:
$State the range and domain for$

$$i)$ \ \ \ \sqrt{4-x^2}$

fify

mahoo_mk · Oct 21, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$$Prove that the volume of a sphere with radius$ \ \ r \ \ $is$ \ \ \frac{4\pi r^3}{3}$

volume of semi circle rotated about the either y or x axis from r to -r
$V= \pi \int_{-r}^{r} -x^2 +r^2 dx ....... $after doing it $ V= \frac{4}{3} \pi r^3$

now give what i want

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