HSC 2013 Maths Marathon (archive) (3 Viewers)

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

HSC 2013 Maths Marathon (archive) (3 Viewers)

Exaı̸̸̸̸̸̸̸̸lted Member

Member

Active Member

Member

Exaı̸̸̸̸̸̸̸̸lted Member

Member

This too shall pass

Member

Retired

This too shall pass

what is that?It is Cowpea

Member

Member

Well-Known Member

Retired

what is that?It is Cowpea

This too shall pass

Member

New Member

New Member

Users Who Are Viewing This Thread (Users: 0, Guests: 3)