HSC 2012 MX1 Marathon #2 (archive) (5 Viewers)

[Timske]

Re: HSC 2012 Marathon

anyone?

 

[fortune_cookie]

Re: HSC 2012 Marathon

Difficulty: Easy

|x| is always positive, so multiply across

|x| >1/5

x>1/5 or -x>1/5

x>1/5 or x<-1/5

(-infty, -1/5) U (1/5, infty) in interval notation.

 

[Timske]

Re: HSC 2012 Marathon

nice your turn

 

[SpiralFlex]

Re: HSC 2012 Marathon

What difficult do you want?

 

[Timske]

Re: HSC 2012 Marathon

Medium

EDIT: EASY

 

[fortune_cookie]

Re: HSC 2012 Marathon

Medium

EDIT: EASY

Let f(x,y) be a function such that f(xy) = f(x)+f(y)

Prove by induction that f(a^n) = n x f(a) for all n>=1 (NOTE: It's n times f(a) )

It looks hard but it is actually very easy.

 

[Timske]

Re: HSC 2012 Marathon

Ill try once im home

 

[SpiralFlex]

Re: HSC 2012 Marathon

 

[Nooblet94]

Re: HSC 2012 Marathon

I didn't use a fraction substitution, nor the given result. I'll tex my working after I finish the second part. On second consideration, I guess I sort of did use a fraction...
Pt. 1
<a href="http://www.codecogs.com/eqnedit.php?latex=I= \int \frac{dx}{\sqrt{x}\sqrt{x@plus;1}}\\ $Let $u=2\sqrt{x} \Rightarrow du=\frac{dx}{\sqrt{x}}\\ x@plus;1=\frac{u^2}{4}@plus;1=\frac{1}{4}(u^2@plus;4)\\ ~\\ \therefore I=2\int \frac{du}{\sqrt{u^2@plus;4}}\\ =2\ln \left |u@plus;\sqrt{u^2@plus;4}\right |@plus;C\\ =2\ln \left |2\sqrt{x}@plus;\sqrt{4x@plus;4}\right |@plus;C\\ =2\ln \left |\sqrt{x}@plus;\sqrt{x@plus;1}\right | @plus;C~~$(Since $ \ln 2 $ is constant)$" target="_blank"><img src="http://latex.codecogs.com/gif.latex?I= \int \frac{dx}{\sqrt{x}\sqrt{x+1}}\\ $Let $u=2\sqrt{x} \Rightarrow du=\frac{dx}{\sqrt{x}}\\ x+1=\frac{u^2}{4}+1=\frac{1}{4}(u^2+4)\\ ~\\ \therefore I=2\int \frac{du}{\sqrt{u^2+4}}\\ =2\ln \left |u+\sqrt{u^2+4}\right |+C\\ =2\ln \left |2\sqrt{x}+\sqrt{4x+4}\right |+C\\ =2\ln \left |\sqrt{x}+\sqrt{x+1}\right | +C~~$(Since $ \ln 2 $ is constant)$" title="I= \int \frac{dx}{\sqrt{x}\sqrt{x+1}}\\ $Let $u=2\sqrt{x} \Rightarrow du=\frac{dx}{\sqrt{x}}\\ x+1=\frac{u^2}{4}+1=\frac{1}{4}(u^2+4)\\ ~\\ \therefore I=2\int \frac{du}{\sqrt{u^2+4}}\\ =2\ln \left |u+\sqrt{u^2+4}\right |+C\\ =2\ln \left |2\sqrt{x}+\sqrt{4x+4}\right |+C\\ =2\ln \left |\sqrt{x}+\sqrt{x+1}\right | +C~~$(Since $ \ln 2 $ is constant)$" /></a>

Pt. 2
<a href="http://www.codecogs.com/eqnedit.php?latex=\\ y=\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x@plus;1}}\\ y^2=\frac{1}{x}-\frac{2}{\sqrt{x(x@plus;1)}}@plus;\frac{1}{1@plus;x}\\ ~\\ V=\pi \int ^{\frac{9}{16}}_{\frac{1}{8}}y^2dx\\ =\pi \int^{\frac{9}{16}}_{\frac{1}{8}} \frac{1}{x}-\frac{2}{\sqrt{x(x@plus;1)}}@plus;\frac{1}{1@plus;x}\\ =\pi\left [ \right \ln(x(x@plus;1))-4\ln(\sqrt{x}@plus;\sqrt{x@plus;1})]^{\frac{9}{16}}_{\frac{1}{8}}\\ =\pi\left [ \ln\left ( \frac{\frac{225}{256}}{\frac{9}{64}} \right )-4\ln\left (\frac{\frac{1}{2\sqrt{2}}@plus;\frac{3}{2\sqrt{2}}}{\frac{3}{4}@plus;\frac{5}{4}} \right )\right ]\\ =\pi \left (\ln \frac{25}{4} -\ln4\right )\\ =\pi\ln\frac{25}{16}\\ \approx 1.402$ units$^3" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ y=\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\\ y^2=\frac{1}{x}-\frac{2}{\sqrt{x(x+1)}}+\frac{1}{1+x}\\ ~\\ V=\pi \int ^{\frac{9}{16}}_{\frac{1}{8}}y^2dx\\ =\pi \int^{\frac{9}{16}}_{\frac{1}{8}} \frac{1}{x}-\frac{2}{\sqrt{x(x+1)}}+\frac{1}{1+x}\\ =\pi\left [ \right \ln(x(x+1))-4\ln(\sqrt{x}+\sqrt{x+1})]^{\frac{9}{16}}_{\frac{1}{8}}\\ =\pi\left [ \ln\left ( \frac{\frac{225}{256}}{\frac{9}{64}} \right )-4\ln\left (\frac{\frac{1}{2\sqrt{2}}+\frac{3}{2\sqrt{2}}}{\frac{3}{4}+\frac{5}{4}} \right )\right ]\\ =\pi \left (\ln \frac{25}{4} -\ln4\right )\\ =\pi\ln\frac{25}{16}\\ \approx 1.402$ units$^3" title="\\ y=\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\\ y^2=\frac{1}{x}-\frac{2}{\sqrt{x(x+1)}}+\frac{1}{1+x}\\ ~\\ V=\pi \int ^{\frac{9}{16}}_{\frac{1}{8}}y^2dx\\ =\pi \int^{\frac{9}{16}}_{\frac{1}{8}} \frac{1}{x}-\frac{2}{\sqrt{x(x+1)}}+\frac{1}{1+x}\\ =\pi\left [ \right \ln(x(x+1))-4\ln(\sqrt{x}+\sqrt{x+1})]^{\frac{9}{16}}_{\frac{1}{8}}\\ =\pi\left [ \ln\left ( \frac{\frac{225}{256}}{\frac{9}{64}} \right )-4\ln\left (\frac{\frac{1}{2\sqrt{2}}+\frac{3}{2\sqrt{2}}}{\frac{3}{4}+\frac{5}{4}} \right )\right ]\\ =\pi \left (\ln \frac{25}{4} -\ln4\right )\\ =\pi\ln\frac{25}{16}\\ \approx 1.402$ units$^3" /></a>

No idea why the fraction in the fourth last line isn't displaying... Now, my question;

 

[Sanjeet]

Re: HSC 2012 Marathon

I didn't use a fraction substitution, nor the given result. I'll tex my working after I finish the second part. On second consideration, I guess I sort of did use a fraction...

<a href="http://www.codecogs.com/eqnedit.php?latex=I= \int \frac{dx}{\sqrt{x}\sqrt{x@plus;1}}\\ $Let $u=2\sqrt{x} \Rightarrow du=\frac{dx}{\sqrt{x}}\\ x@plus;1=\frac{u^2}{4}@plus;1=\frac{1}{4}(u^2@plus;4)\\ ~\\ \therefore I=2\int \frac{du}{\sqrt{u^2@plus;4}}\\ =2\ln \left |u@plus;\sqrt{u^2@plus;4}\right |@plus;C\\ =2\ln \left |2\sqrt{x}@plus;\sqrt{4x@plus;4}\right |@plus;C\\ =2\ln \left |\sqrt{x}@plus;\sqrt{x@plus;1}\right | @plus;C~~$(Since $ \ln 2 $ is constant)$" target="_blank"><img src="http://latex.codecogs.com/gif.latex?I= \int \frac{dx}{\sqrt{x}\sqrt{x+1}}\\ $Let $u=2\sqrt{x} \Rightarrow du=\frac{dx}{\sqrt{x}}\\ x+1=\frac{u^2}{4}+1=\frac{1}{4}(u^2+4)\\ ~\\ \therefore I=2\int \frac{du}{\sqrt{u^2+4}}\\ =2\ln \left |u+\sqrt{u^2+4}\right |+C\\ =2\ln \left |2\sqrt{x}+\sqrt{4x+4}\right |+C\\ =2\ln \left |\sqrt{x}+\sqrt{x+1}\right | +C~~$(Since $ \ln 2 $ is constant)$" title="I= \int \frac{dx}{\sqrt{x}\sqrt{x+1}}\\ $Let $u=2\sqrt{x} \Rightarrow du=\frac{dx}{\sqrt{x}}\\ x+1=\frac{u^2}{4}+1=\frac{1}{4}(u^2+4)\\ ~\\ \therefore I=2\int \frac{du}{\sqrt{u^2+4}}\\ =2\ln \left |u+\sqrt{u^2+4}\right |+C\\ =2\ln \left |2\sqrt{x}+\sqrt{4x+4}\right |+C\\ =2\ln \left |\sqrt{x}+\sqrt{x+1}\right | +C~~$(Since $ \ln 2 $ is constant)$" /></a>

An alternate solution with a different substitution:

 

[SpiralFlex]

Re: HSC 2012 Marathon

The question said fraction substitution but oh well...I should have added part 3, then the fraction substitution would make more sense.

GROWL. I am off, keep this thread going. Be back at 6. *Assuming I don't get distracted*

 

[Sanjeet]

Re: HSC 2012 Marathon

But the *hint* was a fraction.. it didn't say you had to use one

 

[Timske]

Re: HSC 2012 Marathon

Let f(x,y) be a function such that f(xy) = f(x)+f(y)

Prove by induction that f(a^n) = n x f(a) for all n>=1 (NOTE: It's n times f(a) )

It looks hard but it is actually very easy.

I have a solution gotta latex it

<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{80} \textup{Given} ~f\left ( xy \right ) = f(x) @plus; f(y)\\~\textup{Prove~ by ~induction ~that}~ f(a^n) = nf(a) ~\textup{for ~all} ~n\geq 1\\\\ Step~1rove ~true~for~ n=1\\LHS=a^n=a^1=a\\RHS = n*a=1*a=a\\\therefore~ True ~for~ n=1\\\\Step~2: Assume true for n=k\\i.e~f(a^k)=kf(a)\\\\ Step 3:~Prove~ true~ for~ n=k@plus;1\\i.e~f(a^{k@plus;1}) = (k@plus;1)f(a)\\LHS=f(a^{k@plus;1})=f(a^k*a)\\=f(a^k)@plus;f(a)\\=kf(a)@plus;f(a)\\=(k@plus;1)f(a)=RHS" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\dpi{80} \textup{Given} ~f\left ( xy \right ) = f(x) + f(y)\\~\textup{Prove~ by ~induction ~that}~ f(a^n) = nf(a) ~\textup{for ~all} ~n\geq 1\\\\ Step~1rove ~true~for~ n=1\\LHS=a^n=a^1=a\\RHS = n*a=1*a=a\\\therefore~ True ~for~ n=1\\\\Step~2: Assume true for n=k\\i.e~f(a^k)=kf(a)\\\\ Step 3:~Prove~ true~ for~ n=k+1\\i.e~f(a^{k+1}) = (k+1)f(a)\\LHS=f(a^{k+1})=f(a^k*a)\\=f(a^k)+f(a)\\=kf(a)+f(a)\\=(k+1)f(a)=RHS" title="\dpi{80} \textup{Given} ~f\left ( xy \right ) = f(x) + f(y)\\~\textup{Prove~ by ~induction ~that}~ f(a^n) = nf(a) ~\textup{for ~all} ~n\geq 1\\\\ Step~1rove ~true~for~ n=1\\LHS=a^n=a^1=a\\RHS = n*a=1*a=a\\\therefore~ True ~for~ n=1\\\\Step~2: Assume true for n=k\\i.e~f(a^k)=kf(a)\\\\ Step 3:~Prove~ true~ for~ n=k+1\\i.e~f(a^{k+1}) = (k+1)f(a)\\LHS=f(a^{k+1})=f(a^k*a)\\=f(a^k)+f(a)\\=kf(a)+f(a)\\=(k+1)f(a)=RHS" /></a>

 

[Nooblet94]

Re: HSC 2012 Marathon

The question said fraction substitution but oh well...I should have added part 3, then the fraction substitution would make more sense.

GROWL. I am off, keep this thread going. Be back at 6. *Assuming I don't get distracted*

I'm curious, what was the fraction you had in mind?

 

[Timske]

Re: HSC 2012 Marathon

new Q spiral i want a hard one

 

[barbernator]

Re: HSC 2012 Marathon

off the top of my head, the fraction could be x=1/u^2

 

[Sanjeet]

Re: HSC 2012 Marathon

Wouldn't that be u = 1/sqrt(x) ?

 

[barbernator]

Re: HSC 2012 Marathon

Wouldn't that be u = 1/sqrt(x) ?

same thing, its an equation.

 

[Timske]

Re: HSC 2012 Marathon

barbs you got any hard questions

 

[barbernator]

Re: HSC 2012 Marathon

barbs you got any hard questions

i have a booklet of past CSSA trials, im sure i can find something

have u done SHM yet?