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

[bleakarcher]

Re: HSC 2012 Marathon

A polynomials question.

 

[bleakarcher]

Re: HSC 2012 Marathon

^ a lot of you will probably recognise that one. If you do, please leave it for someone else lol.

 

[math man]

Re: HSC 2012 Marathon

hmm, if you write the polynomial as a product of n terms then take the derivative using the special nth product rule, then use a trig sub...you will get no where
so i will leave it for the hsc kids to figure out what they are properly meant to do

 

[bleakarcher]

Re: HSC 2012 Marathon

alright

anyone gonna give the question a go?

 

[math man]

Re: HSC 2012 Marathon

ok ill give it a go in words.

Express z^n -1 as a procduct of its n roots in factored form, divide both sides by z-1, which is a factor of z^n-1. now the hsc kids finish it for me... i i i...forgot what to do next.

 

[barbernator]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=z^n-1=0\\ \\ LHS=z^n-1=(z-1)(z-z_{1})(z-z_{2})...(z-z_{n-1})\\ \\ divide~both~sides~by~(z-1)\\ \\ z_{n-1}@plus;z_{n-2}@plus;z^{_{n-3}}@plus;...@plus;z@plus;1=(z-z_{1})(z-z_{2})...(z-z_{n-1})\\ \\ let~n=1\\ \\ n=(1-z_{1})(1-z_{2})(1-z_{3})...(1-z_{n-1})~as~required\\ \\ new~question\\ find~the~first~derivative~of~ln(\frac{1}{\sqrt{sin(x)}})" target="_blank"><img src="http://latex.codecogs.com/gif.latex?z^n-1=0\\ \\ LHS=z^n-1=(z-1)(z-z_{1})(z-z_{2})...(z-z_{n-1})\\ \\ divide~both~sides~by~(z-1)\\ \\ z_{n-1}+z_{n-2}+z^{_{n-3}}+...+z+1=(z-z_{1})(z-z_{2})...(z-z_{n-1})\\ \\ let~n=1\\ \\ n=(1-z_{1})(1-z_{2})(1-z_{3})...(1-z_{n-1})~as~required\\ \\ new~question\\ find~the~first~derivative~of~ln(\frac{1}{\sqrt{sin(x)}})" title="z^n-1=0\\ \\ LHS=z^n-1=(z-1)(z-z_{1})(z-z_{2})...(z-z_{n-1})\\ \\ divide~both~sides~by~(z-1)\\ \\ z_{n-1}+z_{n-2}+z^{_{n-3}}+...+z+1=(z-z_{1})(z-z_{2})...(z-z_{n-1})\\ \\ let~n=1\\ \\ n=(1-z_{1})(1-z_{2})(1-z_{3})...(1-z_{n-1})~as~required\\ \\ new~question\\ find~the~first~derivative~of~ln(\frac{1}{\sqrt{sin(x)}})" /></a>

edit: difficulty, medium

 

[deswa1]

Re: HSC 2012 Marathon



<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{Evaluate }\int_{0}^{1}2^xdx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{Evaluate }\int_{0}^{1}2^xdx" title="\textup{Evaluate }\int_{0}^{1}2^xdx" /></a>

Medium-hard

 

[zeebobDD]

Re: HSC 2012 Marathon

-cotx / 2 using the chain rule twice

 

[Carrotsticks]

Re: HSC 2012 Marathon

Medium-hard

You could have used log laws to simplify the expression before continuing =)

 

[Timske]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{d}{dx} ~ \ln(\tfrac{1}{\sqrt{sinx}}) , using ~ \ln f(x) = \frac{f'(x)}{f(x)} \\ \frac{d}{dx}~\tfrac{1}{\sqrt{sinx}} = \frac{d}{dx} (sinx)^{-1/2} \\\\ = -\frac{1}{2}(sinx)^{-3/2}*cosx = \frac{1}{2}*\frac{-cosx}{sin^{3/2}x} \\\\\therefore \frac{d}{dx} ~ \ln(\tfrac{1}{\sqrt{sinx}}) = \frac{1}{2}*\frac{-cosx}{sin^{3/2}x} ~ * \sqrt{sinx} \\\\ = \frac{1}{2} * \frac{-cosx}{sinx} \\\\ = \frac{-cotx}{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{d}{dx} ~ \ln(\tfrac{1}{\sqrt{sinx}}) , using ~ \ln f(x) = \frac{f'(x)}{f(x)} \\ \frac{d}{dx}~\tfrac{1}{\sqrt{sinx}} = \frac{d}{dx} (sinx)^{-1/2} \\\\ = -\frac{1}{2}(sinx)^{-3/2}*cosx = \frac{1}{2}*\frac{-cosx}{sin^{3/2}x} \\\\\therefore \frac{d}{dx} ~ \ln(\tfrac{1}{\sqrt{sinx}}) = \frac{1}{2}*\frac{-cosx}{sin^{3/2}x} ~ * \sqrt{sinx} \\\\ = \frac{1}{2} * \frac{-cosx}{sinx} \\\\ = \frac{-cotx}{2}" title="\frac{d}{dx} ~ \ln(\tfrac{1}{\sqrt{sinx}}) , using ~ \ln f(x) = \frac{f'(x)}{f(x)} \\ \frac{d}{dx}~\tfrac{1}{\sqrt{sinx}} = \frac{d}{dx} (sinx)^{-1/2} \\\\ = -\frac{1}{2}(sinx)^{-3/2}*cosx = \frac{1}{2}*\frac{-cosx}{sin^{3/2}x} \\\\\therefore \frac{d}{dx} ~ \ln(\tfrac{1}{\sqrt{sinx}}) = \frac{1}{2}*\frac{-cosx}{sin^{3/2}x} ~ * \sqrt{sinx} \\\\ = \frac{1}{2} * \frac{-cosx}{sinx} \\\\ = \frac{-cotx}{2}" /></a>

 

[barbernator]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{0}^{1}2^xdx\\ let~u=2^x\\ du=ln(2)2^xdx\\ at~x=1,u=2\\ at~x=0,u=1\\ =\int_{1}^{2}\frac{1}{ln(2)}\\ =(x)ln(2)~dunno~how~to~do~numbered~box~brackets\\ =ln(2)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{0}^{1}2^xdx\\ let~u=2^x\\ du=ln(2)2^xdx\\ at~x=1,u=2\\ at~x=0,u=1\\ =\int_{1}^{2}\frac{1}{ln(2)}\\ =(x)ln(2)~dunno~how~to~do~numbered~box~brackets\\ =ln(2)" title="\int_{0}^{1}2^xdx\\ let~u=2^x\\ du=ln(2)2^xdx\\ at~x=1,u=2\\ at~x=0,u=1\\ =\int_{1}^{2}\frac{1}{ln(2)}\\ =(x)~/~ln(2)~dunno~how~to~do~numbered~box~brackets\\ =1/ln(2)" /></a>

whoops fail answer lol, it should be 1/ln(2)

new question, at a guess for not having done the question, medium to hard

<a href="http://www.codecogs.com/eqnedit.php?latex=2sin(\theta -\frac{\pi}{3})=cos(\theta @plus;\frac{\pi }{3})\\ express~tan(\theta )~in~the~form~a@plus;b\sqrt{3},~where~a~and~b~are~integers." target="_blank"><img src="http://latex.codecogs.com/gif.latex?2sin(\theta -\frac{\pi}{3})=cos(\theta +\frac{\pi }{3})\\ express~tan(\theta )~in~the~form~a+b\sqrt{3},~where~a~and~b~are~integers." title="2sin(\theta -\frac{\pi}{3})=cos(\theta +\frac{\pi }{3})\\ express~tan(\theta )~in~the~form~a+b\sqrt{3},~where~a~and~b~are~integers." /></a>

 

[Timske]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{0}^{1}2^xdx\\ let~u=2^x\\ du=ln(2)2^xdx\\ at~x=1,u=2\\ at~x=0,u=1\\ =\int_{1}^{2}\frac{1}{ln(2)}\\ =(x)ln(2)~dunno~how~to~do~numbered~box~brackets\\ =ln(2)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{0}^{1}2^xdx\\ let~u=2^x\\ du=ln(2)2^xdx\\ at~x=1,u=2\\ at~x=0,u=1\\ =\int_{1}^{2}\frac{1}{ln(2)}\\ =(x)ln(2)~dunno~how~to~do~numbered~box~brackets\\ =ln(2)" title="\int_{0}^{1}2^xdx\\ let~u=2^x\\ du=ln(2)2^xdx\\ at~x=1,u=2\\ at~x=0,u=1\\ =\int_{1}^{2}\frac{1}{ln(2)}\\ =(x)ln(2)~dunno~how~to~do~numbered~box~brackets\\ =ln(2)" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\left [(x)\ln(2) \right ]_{0}^{1}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\left [(x)\ln(2) \right ]_{0}^{1}" title="\left [(x)\ln(2) \right ]_{0}^{1}" /></a>

best i could get it

\left(\right) x_a^b

 

[deswa1]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{0}^{1}2^xdx\\ let~u=2^x\\ du=ln(2)2^xdx\\ at~x=1,u=2\\ at~x=0,u=1\\ =\int_{1}^{2}\frac{1}{ln(2)}\\ =(x)ln(2)~dunno~how~to~do~numbered~box~brackets\\ =ln(2)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{0}^{1}2^xdx\\ let~u=2^x\\ du=ln(2)2^xdx\\ at~x=1,u=2\\ at~x=0,u=1\\ =\int_{1}^{2}\frac{1}{ln(2)}\\ =(x)ln(2)~dunno~how~to~do~numbered~box~brackets\\ =ln(2)" title="\int_{0}^{1}2^xdx\\ let~u=2^x\\ du=ln(2)2^xdx\\ at~x=1,u=2\\ at~x=0,u=1\\ =\int_{1}^{2}\frac{1}{ln(2)}\\ =(x)~/~ln(2)~dunno~how~to~do~numbered~box~brackets\\ =1/ln(2)" /></a>

whoops fail answer lol, it should be 1/ln(2)

new question, at a guess for not having done the question, medium to hard

<a href="http://www.codecogs.com/eqnedit.php?latex=2sin(\theta -\frac{\pi}{3})=cos(\theta @plus;\frac{\pi }{3})\\ express~tan(\theta )~in~the~form~a@plus;b\sqrt{3},~where~a~and~b~are~integers." target="_blank"><img src="http://latex.codecogs.com/gif.latex?2sin(\theta -\frac{\pi}{3})=cos(\theta +\frac{\pi }{3})\\ express~tan(\theta )~in~the~form~a+b\sqrt{3},~where~a~and~b~are~integers." title="2sin(\theta -\frac{\pi}{3})=cos(\theta +\frac{\pi }{3})\\ express~tan(\theta )~in~the~form~a+b\sqrt{3},~where~a~and~b~are~integers." /></a>

This isn't the answer- maybe try again or someone else can have a go.

 

[Timske]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{0}^{1}2^xdx\\ let~u=2^x\\ du=ln(2)2^xdx\\ at~x=1,u=2\\ at~x=0,u=1\\ =\int_{1}^{2}\frac{1}{ln(2)}\\ =(x)ln(2)~dunno~how~to~do~numbered~box~brackets\\ =ln(2)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{0}^{1}2^xdx\\ let~u=2^x\\ du=ln(2)2^xdx\\ at~x=1,u=2\\ at~x=0,u=1\\ =\int_{1}^{2}\frac{1}{ln(2)}\\ =(x)ln(2)~dunno~how~to~do~numbered~box~brackets\\ =ln(2)" title="\int_{0}^{1}2^xdx\\ let~u=2^x\\ du=ln(2)2^xdx\\ at~x=1,u=2\\ at~x=0,u=1\\ =\int_{1}^{2}\frac{1}{ln(2)}\\ =(x)~/~ln(2)~dunno~how~to~do~numbered~box~brackets\\ =1/ln(2)" /></a>

whoops fail answer lol, it should be 1/ln(2)

new question, at a guess for not having done the question, medium to hard

<a href="http://www.codecogs.com/eqnedit.php?latex=2sin(\theta -\frac{\pi}{3})=cos(\theta @plus;\frac{\pi }{3})\\ express~tan(\theta )~in~the~form~a@plus;b\sqrt{3},~where~a~and~b~are~integers." target="_blank"><img src="http://latex.codecogs.com/gif.latex?2sin(\theta -\frac{\pi}{3})=cos(\theta +\frac{\pi }{3})\\ express~tan(\theta )~in~the~form~a+b\sqrt{3},~where~a~and~b~are~integers." title="2sin(\theta -\frac{\pi}{3})=cos(\theta +\frac{\pi }{3})\\ express~tan(\theta )~in~the~form~a+b\sqrt{3},~where~a~and~b~are~integers." /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=tan(\theta ) = -3-\sqrt{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?tan(\theta ) = -3-\sqrt{3}" title="tan(\theta ) = -3-\sqrt{3}" /></a>

right??

 

[barbernator]

Re: HSC 2012 Marathon

This isn't the answer- maybe try again or someone else can have a go.

i am sure that is the answer

 

[barbernator]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=tan(\theta ) = -3-\sqrt{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?tan(\theta ) = -3-\sqrt{3}" title="tan(\theta ) = -3-\sqrt{3}" /></a>

right??

nope, go again post your working and i will check where you have gone wrong.

 

[deswa1]

Re: HSC 2012 Marathon

nope, go again post your working and i will check where you have gone wrong.

Is the answer 8-5root3? It was a nice question

 

[nightweaver066]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{Evaluate }\int_{0}^{1}2^xdx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{Evaluate }\int_{0}^{1}2^xdx" title="\textup{Evaluate }\int_{0}^{1}2^xdx" /></a>

Medium-hard

= [(2^x)/ln2] limits 0 to 1
= 2/ln(2) - 1/ln(2)
= 1/ln(2)

 

[Timske]

Re: HSC 2012 Marathon

i am sure that is the answer

nah its not

 

[barbernator]

Re: HSC 2012 Marathon

Is the answer 8-5root3? It was a nice question

nope, you guys have the root3 part correct hahah