HSC 2015 Maths Marathon (archive) (4 Viewers)

leehuan · Sep 28, 2015

Re: HSC 2015 2U Marathon

flavurr said:
hey dr wouldn't it be 20/52 * 4/51 * 3/50 * 2/49 *1/48 as once you pick the first card, there is only a possibility of 4/51 to pick the 2nd of that same suit then 3/50 to pick the 3rd of that same suit ect

5 times 4 equals 20.

Drsoccerball · Sep 28, 2015

Re: HSC 2015 2U Marathon

leehuan said:
5 times 4 equals 20.

Firstly is my answer correct?

Drsoccerball · Sep 28, 2015

Re: HSC 2015 2U Marathon

flavurr said:
hey dr wouldn't it be 20/52 * 4/51 * 3/50 * 2/49 *1/48 as once you pick the first card, there is only a possibility of 4/51 to pick the 2nd of that same suit then 3/50 to pick the 3rd of that same suit ect

You may be right.

Flop21 · Sep 28, 2015

Re: HSC 2015 2U Marathon

Another question:

ln(x) - 3 / ln(x) = 2

Solve

psyc1011 · Sep 28, 2015

Re: HSC 2015 2U Marathon

By inspection

ln(x) = -1
ln(x) = 3

BlueGas · Sep 28, 2015

Re: HSC 2015 2U Marathon

psyc1011 said:
By inspection

ln(x) = -1
ln(x) = 3

So there's no need for working out?

InteGrand · Sep 28, 2015

Re: HSC 2015 2U Marathon

BlueGas said:
So there's no need for working out?

You should show working out, and you may not always be able to see the solutions by inspection anyway.

Flop21 · Sep 28, 2015

Re: HSC 2015 2U Marathon

psyc1011 said:
By inspection

ln(x) = -1
ln(x) = 3

It is a question worth 3 marks, I would assume they want to see how you've done it to get the full 3 marks. Plus that's not the correct answer.

So 2015ers have a go at doing it the proper way.

psyc1011 · Sep 28, 2015

Re: HSC 2015 2U Marathon

Flop21 said:
It is a question worth 3 marks, I would assume they want to see how you've done it to get the full 3 marks. Plus that's not the correct answer.

So 2015ers have a go at doing it the proper way.

Flop21 said:
Another question:

ln(x) - 3 / ln(x) = 2

Solve

solve?

InteGrand · Sep 28, 2015

Re: HSC 2015 2U Marathon

Flop21 said:
Plus that's not the correct answer.

$Those are correct values for $\ln x$, just need to exponentiate them to the base $e$ to get the final answers for $x$.$

Flop21 · Sep 28, 2015

Re: HSC 2015 2U Marathon

psyc1011 said:
solve?

What? Yes solve, that's what the question asks.

I think it's good practice to learn and do it the non 'inspection way' and to show full working out.

InteGrand said:
$Those are correct values for $\ln x$, just need to exponentiate them to the base $e$ to get the final answers for $x$.$

Yes but you wouldn't get full marks just putting ln x = 3 and -1 would you??? I assume 'solve' means find x.

InteGrand · Sep 28, 2015

Re: HSC 2015 2U Marathon

Flop21 said:
Yes but you wouldn't get full marks just putting ln x = 3 and -1 would you??? I assume 'solve' means find x.

That's right.

kawaiipotato · Sep 28, 2015

Re: HSC 2015 2U Marathon

ln(x) - 3 / ln(x) = 2

(ln(x))^2 - 2ln(x) - 3 = 0

(lnx - 3)(lnx + 1) = 0
lnx = 3 , lnx = -1
x = e^3 , e^-1

BlueGas · Sep 28, 2015

Re: HSC 2015 2U Marathon

kawaiipotato said:
ln(x) - 3 / ln(x) = 2

(ln(x))^2 - 2ln(x) - 3 = 0

(lnx - 3)(lnx + 1) = 0
lnx = 3 , lnx = -1
x = e^3 , e^-1

How did you get this line?

BlueGas · Sep 28, 2015

Re: HSC 2015 2U Marathon

BlueGas said:
How did you get this line?

I do understand multiplying ln(x) by 2 and then using the log law nlnx = ln(x)^n but how did you also get 2ln(x) - 3?

InteGrand · Sep 28, 2015

Re: HSC 2015 2U Marathon

BlueGas said:
I do understand multiplying ln(x) by 2 and then using the log law nlnx = ln(x)^n but how did you also get 2ln(x) - 3?

$No log laws are used. If it helps, just write $u=\ln x$ in the original equation, then multiply both sides by $u$ and rearrange to get a quadratic in $u$. Then solve the quadratic for $u$ and then get the solutions for $x$ by using $x=e^u$.$

BlueGas · Sep 28, 2015

Re: HSC 2015 2U Marathon

InteGrand said:
$No log laws are used. If it helps, just write $u=\ln x$ in the original equation, then multiply both sides by $u$ and rearrange to get a quadratic in $u$. Then solve the quadratic for $u$ and then get the solutions for $x$ by using $x=e^u$.$

Can you write the full working out if you don't mind?

InteGrand · Sep 28, 2015

Re: HSC 2015 2U Marathon

BlueGas said:
Can you write the full working out if you don't mind?

$\ln x - \frac{3}{\ln x}=2, x>0.$

$Multiplying through by $\ln x$, we have $\left( \ln x \right) ^2 - 3 = 2\ln x \Longleftrightarrow \left( \ln x \right) ^2 - 2\ln x - 3 = 0$.$

$Factorising, we have $\left(\ln x -3 \right)\left(\ln x +1 \right)=0\Rightarrow \ln x = 3$ or $\ln x = -1$. So $x=e^3$ or $e^{-1}$.$

flavurr · Sep 28, 2015

Re: HSC 2015 2U Marathon

Q. If the continuous function f(x) is an even function, prove that f'(x) is an odd function

BlueGas · Sep 28, 2015

Re: HSC 2015 2U Marathon

InteGrand said:
$\ln x - \frac{3}{\ln x}=2, x>0.$

$Multiplying through by $\ln x$, we have $\left( \ln x \right) ^2 - 3 = 2\ln x \Longleftrightarrow \left( \ln x \right) ^2 - 2\ln x - 3 = 0$.$

$Factorising, we have $\left(\ln x -3 \right)\left(\ln x +1 \right)=0\Rightarrow \ln x = 3$ or $\ln x = -1$. So $x=e^3$ or $e^{-1}$.$

LOL I thought it was (lnx - 3)/lnx damn it haha.

HSC 2015 Maths Marathon (archive) (4 Viewers)

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

#truth

Well-Known Member

Well-Known Member

Well-Known Member

#truth

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

New Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 4)