barbernator
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ok I will latex up some of the easier solutions and hopefully others can do some other ones. 
BOS MX2 Trials discussion thread. (1 Viewer)
- Thread starter Carrotsticks
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deswa1
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What I really took away from today was that my explanations and setting out for some questions is CRAP- I need to work on clarity and flow of argument a lot. And I've got my paper so I can scan my answers to questions people have (assuming I could do that one and you can read my writing)- scanning will take a bit of time though
iBibah
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Isn't it being marked?
Carrotsticks
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I marked his one early because he stayed back afterwards and wanted it marked.
Though for the most of the marking, he was watching me mark it =p
Though for the most of the marking, he was watching me mark it =p
deswa1
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Yeah this. I'm won't tell you my mark until everyone else gets it back but yeah, what he's saying is correct- I had to explain some of my reasoning and working to him. In the HSC though I won't have this luxury hence my post earlier about improving my clarity etc. So you could probs take about 3 marks off my score and that's what I'd get in the HSC. And I'm still a bit pissed because I didn't see 2 marks worth of questions lol.
someth1ng
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Still don't have a clue about any of the questions...LOL
Its just something to consider. I have a feeling that when the answers are released, most people will be kicking themselves because of what they missed.
barbernator
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Here is a start. 11 a)
<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{0}^{1}\frac{e^{2x}-1}{e^{2x}@plus;1}dx\\\\ \int_{0}^{1}(\frac{e^x}{e^x})\frac{e^{2x}-1}{e^{2x}@plus;1}dx\\\\\\ let~u=e^x,~du=e^xdx\\\\ x=0,~u=1\\ x=1,~u=e\\\\" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{0}^{1}\frac{e^{2x}-1}{e^{2x}+1}dx\\\\ \int_{0}^{1}(\frac{e^x}{e^x})\frac{e^{2x}-1}{e^{2x}+1}dx\\\\\\ let~u=e^x,~du=e^xdx\\\\ x=0,~u=1\\ x=1,~u=e\\\\" title="\int_{0}^{1}\frac{e^{2x}-1}{e^{2x}+1}dx\\\\ \int_{0}^{1}(\frac{e^x}{e^x})\frac{e^{2x}-1}{e^{2x}+1}dx\\\\\\ let~u=e^x,~du=e^xdx\\\\ x=0,~u=1\\ x=1,~u=e\\\\" /></a><a href="http://www.codecogs.com/eqnedit.php?latex=\int_{1}^{e}\frac{u^2-1}{u(u^2@plus;1)}\\\\ \int_{1}^{e}(\frac{1}{u}-\frac{-2}{u(u^2@plus;1)})\\\ \int_{1}^{e}(\frac{-1}{u}@plus;\frac{2u}{(u^2@plus;1)})\\\\ =\left [ ln(\frac{u^2@plus;1}{u}) \right ]_{1}^{e}\\\\ =-1@plus;ln(e^2@plus;1)-ln(2)" target="_blank">
<a href="http://www.codecogs.com/eqnedit.php?latex==\int_{1}^{e}\frac{u^2-1}{u(u^2@plus;1)}dx\\\\ =\int_{1}^{e}(\frac{1}{u}-\frac{-2}{u(u^2@plus;1)})dx\\\ =\int_{1}^{e}(\frac{-1}{u}@plus;\frac{2u}{(u^2@plus;1)})dx\\\\ =\left [ ln(\frac{u^2@plus;1}{u}) \right ]_{1}^{e}\\\\ =-1@plus;ln(e^2@plus;1)-ln(2)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=\int_{1}^{e}\frac{u^2-1}{u(u^2+1)}dx\\\\ =\int_{1}^{e}(\frac{1}{u}-\frac{-2}{u(u^2+1)})dx\\\ =\int_{1}^{e}(\frac{-1}{u}+\frac{2u}{(u^2+1)})dx\\\\ =\left [ ln(\frac{u^2+1}{u}) \right ]_{1}^{e}\\\\ =-1+ln(e^2+1)-ln(2)" title="=\int_{1}^{e}\frac{u^2-1}{u(u^2+1)}dx\\\\ =\int_{1}^{e}(\frac{1}{u}-\frac{-2}{u(u^2+1)})dx\\\ =\int_{1}^{e}(\frac{-1}{u}+\frac{2u}{(u^2+1)})dx\\\\ =\left [ ln(\frac{u^2+1}{u}) \right ]_{1}^{e}\\\\ =-1+ln(e^2+1)-ln(2)" /></a>
11 b) i)
<a href="http://www.codecogs.com/eqnedit.php?latex=1\equiv A(x)(3cosx@plus;2sinx)@plus;B(x)(3cosx-2sinx)\\\\ 1\equiv cosx(3A(x)@plus;3B(x))@plus;sinx(2A(x)-2B(x))\\\\ \therefore 3A(x)@plus;3B(x)=cosx~(1)\\ \therefore 2A(x)-2B(x)=sinx~(2)\\\\ from~(1),~B(x)=\frac{cosx}{3}-A(x)~sub~into~(2)\\\\ 2A(x)-\frac{2cosx}{3}@plus;2A(x)=sinx\\\\ \therefore A(x)=\frac{cosx}{6}@plus;\frac{sinx}{4}\\\\ \therefore B(x)=\frac{cosx}{6}-\frac{sinx}{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?1\equiv A(x)(3cosx+2sinx)+B(x)(3cosx-2sinx)\\\\ 1\equiv cosx(3A(x)+3B(x))+sinx(2A(x)-2B(x))\\\\ \therefore 3A(x)+3B(x)=cosx~(1)\\ \therefore 2A(x)-2B(x)=sinx~(2)\\\\ from~(1),~B(x)=\frac{cosx}{3}-A(x)~sub~into~(2)\\\\ 2A(x)-\frac{2cosx}{3}+2A(x)=sinx\\\\ \therefore A(x)=\frac{cosx}{6}+\frac{sinx}{4}\\\\ \therefore B(x)=\frac{cosx}{6}-\frac{sinx}{4}" title="1\equiv A(x)(3cosx+2sinx)+B(x)(3cosx-2sinx)\\\\ 1\equiv cosx(3A(x)+3B(x))+sinx(2A(x)-2B(x))\\\\ \therefore 3A(x)+3B(x)=cosx~(1)\\ \therefore 2A(x)-2B(x)=sinx~(2)\\\\ from~(1),~B(x)=\frac{cosx}{3}-A(x)~sub~into~(2)\\\\ 2A(x)-\frac{2cosx}{3}+2A(x)=sinx\\\\ \therefore A(x)=\frac{cosx}{6}+\frac{sinx}{4}\\\\ \therefore B(x)=\frac{cosx}{6}-\frac{sinx}{4}" /></a>
ii)
<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{0}^{\frac{\pi}{4}}\left ( \frac{\frac{cosx}{6}@plus;\frac{sinx}{4}}{3cosx-2sinx}@plus;\frac{\frac{cosx}{6}-\frac{sinx}{4}}{3cosx@plus;2sinx} \right )dx~\\\\\\ \frac{1}{12}\int_{0}^{\frac{\pi}{4}}\left ( \frac{2cosx@plus;3sinx}{3cosx-2sinx}@plus;\frac{2cosx-3sinx}{3cosx@plus;2sinx}\right )dx\\\\\\ \frac{1}{12}\left [ ln(\frac{3cosx@plus;2sinx}{3cosx-2sinx}) \right ]_{0}^{\frac{\pi}{4}}\\\\ =\frac{ln(5)}{12}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{0}^{\frac{\pi}{4}}\left ( \frac{\frac{cosx}{6}+\frac{sinx}{4}}{3cosx-2sinx}+\frac{\frac{cosx}{6}-\frac{sinx}{4}}{3cosx+2sinx} \right )dx~\\\\\\ \frac{1}{12}\int_{0}^{\frac{\pi}{4}}\left ( \frac{2cosx+3sinx}{3cosx-2sinx}+\frac{2cosx-3sinx}{3cosx+2sinx}\right )dx\\\\\\ \frac{1}{12}\left [ ln(\frac{3cosx+2sinx}{3cosx-2sinx}) \right ]_{0}^{\frac{\pi}{4}}\\\\ =\frac{ln(5)}{12}" title="\int_{0}^{\frac{\pi}{4}}\left ( \frac{\frac{cosx}{6}+\frac{sinx}{4}}{3cosx-2sinx}+\frac{\frac{cosx}{6}-\frac{sinx}{4}}{3cosx+2sinx} \right )dx~\\\\\\ \frac{1}{12}\int_{0}^{\frac{\pi}{4}}\left ( \frac{2cosx+3sinx}{3cosx-2sinx}+\frac{2cosx-3sinx}{3cosx+2sinx}\right )dx\\\\\\ \frac{1}{12}\left [ ln(\frac{3cosx+2sinx}{3cosx-2sinx}) \right ]_{0}^{\frac{\pi}{4}}\\\\ =\frac{ln(5)}{12}" /></a>
<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{0}^{1}\frac{e^{2x}-1}{e^{2x}@plus;1}dx\\\\ \int_{0}^{1}(\frac{e^x}{e^x})\frac{e^{2x}-1}{e^{2x}@plus;1}dx\\\\\\ let~u=e^x,~du=e^xdx\\\\ x=0,~u=1\\ x=1,~u=e\\\\" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{0}^{1}\frac{e^{2x}-1}{e^{2x}+1}dx\\\\ \int_{0}^{1}(\frac{e^x}{e^x})\frac{e^{2x}-1}{e^{2x}+1}dx\\\\\\ let~u=e^x,~du=e^xdx\\\\ x=0,~u=1\\ x=1,~u=e\\\\" title="\int_{0}^{1}\frac{e^{2x}-1}{e^{2x}+1}dx\\\\ \int_{0}^{1}(\frac{e^x}{e^x})\frac{e^{2x}-1}{e^{2x}+1}dx\\\\\\ let~u=e^x,~du=e^xdx\\\\ x=0,~u=1\\ x=1,~u=e\\\\" /></a><a href="http://www.codecogs.com/eqnedit.php?latex=\int_{1}^{e}\frac{u^2-1}{u(u^2@plus;1)}\\\\ \int_{1}^{e}(\frac{1}{u}-\frac{-2}{u(u^2@plus;1)})\\\ \int_{1}^{e}(\frac{-1}{u}@plus;\frac{2u}{(u^2@plus;1)})\\\\ =\left [ ln(\frac{u^2@plus;1}{u}) \right ]_{1}^{e}\\\\ =-1@plus;ln(e^2@plus;1)-ln(2)" target="_blank">
<a href="http://www.codecogs.com/eqnedit.php?latex==\int_{1}^{e}\frac{u^2-1}{u(u^2@plus;1)}dx\\\\ =\int_{1}^{e}(\frac{1}{u}-\frac{-2}{u(u^2@plus;1)})dx\\\ =\int_{1}^{e}(\frac{-1}{u}@plus;\frac{2u}{(u^2@plus;1)})dx\\\\ =\left [ ln(\frac{u^2@plus;1}{u}) \right ]_{1}^{e}\\\\ =-1@plus;ln(e^2@plus;1)-ln(2)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?=\int_{1}^{e}\frac{u^2-1}{u(u^2+1)}dx\\\\ =\int_{1}^{e}(\frac{1}{u}-\frac{-2}{u(u^2+1)})dx\\\ =\int_{1}^{e}(\frac{-1}{u}+\frac{2u}{(u^2+1)})dx\\\\ =\left [ ln(\frac{u^2+1}{u}) \right ]_{1}^{e}\\\\ =-1+ln(e^2+1)-ln(2)" title="=\int_{1}^{e}\frac{u^2-1}{u(u^2+1)}dx\\\\ =\int_{1}^{e}(\frac{1}{u}-\frac{-2}{u(u^2+1)})dx\\\ =\int_{1}^{e}(\frac{-1}{u}+\frac{2u}{(u^2+1)})dx\\\\ =\left [ ln(\frac{u^2+1}{u}) \right ]_{1}^{e}\\\\ =-1+ln(e^2+1)-ln(2)" /></a>
11 b) i)
<a href="http://www.codecogs.com/eqnedit.php?latex=1\equiv A(x)(3cosx@plus;2sinx)@plus;B(x)(3cosx-2sinx)\\\\ 1\equiv cosx(3A(x)@plus;3B(x))@plus;sinx(2A(x)-2B(x))\\\\ \therefore 3A(x)@plus;3B(x)=cosx~(1)\\ \therefore 2A(x)-2B(x)=sinx~(2)\\\\ from~(1),~B(x)=\frac{cosx}{3}-A(x)~sub~into~(2)\\\\ 2A(x)-\frac{2cosx}{3}@plus;2A(x)=sinx\\\\ \therefore A(x)=\frac{cosx}{6}@plus;\frac{sinx}{4}\\\\ \therefore B(x)=\frac{cosx}{6}-\frac{sinx}{4}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?1\equiv A(x)(3cosx+2sinx)+B(x)(3cosx-2sinx)\\\\ 1\equiv cosx(3A(x)+3B(x))+sinx(2A(x)-2B(x))\\\\ \therefore 3A(x)+3B(x)=cosx~(1)\\ \therefore 2A(x)-2B(x)=sinx~(2)\\\\ from~(1),~B(x)=\frac{cosx}{3}-A(x)~sub~into~(2)\\\\ 2A(x)-\frac{2cosx}{3}+2A(x)=sinx\\\\ \therefore A(x)=\frac{cosx}{6}+\frac{sinx}{4}\\\\ \therefore B(x)=\frac{cosx}{6}-\frac{sinx}{4}" title="1\equiv A(x)(3cosx+2sinx)+B(x)(3cosx-2sinx)\\\\ 1\equiv cosx(3A(x)+3B(x))+sinx(2A(x)-2B(x))\\\\ \therefore 3A(x)+3B(x)=cosx~(1)\\ \therefore 2A(x)-2B(x)=sinx~(2)\\\\ from~(1),~B(x)=\frac{cosx}{3}-A(x)~sub~into~(2)\\\\ 2A(x)-\frac{2cosx}{3}+2A(x)=sinx\\\\ \therefore A(x)=\frac{cosx}{6}+\frac{sinx}{4}\\\\ \therefore B(x)=\frac{cosx}{6}-\frac{sinx}{4}" /></a>
ii)
<a href="http://www.codecogs.com/eqnedit.php?latex=\int_{0}^{\frac{\pi}{4}}\left ( \frac{\frac{cosx}{6}@plus;\frac{sinx}{4}}{3cosx-2sinx}@plus;\frac{\frac{cosx}{6}-\frac{sinx}{4}}{3cosx@plus;2sinx} \right )dx~\\\\\\ \frac{1}{12}\int_{0}^{\frac{\pi}{4}}\left ( \frac{2cosx@plus;3sinx}{3cosx-2sinx}@plus;\frac{2cosx-3sinx}{3cosx@plus;2sinx}\right )dx\\\\\\ \frac{1}{12}\left [ ln(\frac{3cosx@plus;2sinx}{3cosx-2sinx}) \right ]_{0}^{\frac{\pi}{4}}\\\\ =\frac{ln(5)}{12}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int_{0}^{\frac{\pi}{4}}\left ( \frac{\frac{cosx}{6}+\frac{sinx}{4}}{3cosx-2sinx}+\frac{\frac{cosx}{6}-\frac{sinx}{4}}{3cosx+2sinx} \right )dx~\\\\\\ \frac{1}{12}\int_{0}^{\frac{\pi}{4}}\left ( \frac{2cosx+3sinx}{3cosx-2sinx}+\frac{2cosx-3sinx}{3cosx+2sinx}\right )dx\\\\\\ \frac{1}{12}\left [ ln(\frac{3cosx+2sinx}{3cosx-2sinx}) \right ]_{0}^{\frac{\pi}{4}}\\\\ =\frac{ln(5)}{12}" title="\int_{0}^{\frac{\pi}{4}}\left ( \frac{\frac{cosx}{6}+\frac{sinx}{4}}{3cosx-2sinx}+\frac{\frac{cosx}{6}-\frac{sinx}{4}}{3cosx+2sinx} \right )dx~\\\\\\ \frac{1}{12}\int_{0}^{\frac{\pi}{4}}\left ( \frac{2cosx+3sinx}{3cosx-2sinx}+\frac{2cosx-3sinx}{3cosx+2sinx}\right )dx\\\\\\ \frac{1}{12}\left [ ln(\frac{3cosx+2sinx}{3cosx-2sinx}) \right ]_{0}^{\frac{\pi}{4}}\\\\ =\frac{ln(5)}{12}" /></a>
Last edited:
Genius
Can't believe I didn't see that
Spent ages trying to manipulate it and figure out the substitution lol
Carrotsticks
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Haha, that's what I meant by 'you just need to do 1 thing to the original expression'.Genius
Can't believe I didn't see that
Spent ages trying to manipulate it and figure out the substitution lol
deswa1
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Yeah its one that you either see it or you don't. You're only year 11 though so you've got time to learn those things.Genius
Can't believe I didn't see that
Spent ages trying to manipulate it and figure out the substitution lol
deswa1
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I don't think very many people at all could do the whole thing in 3 hours.
someth1ng
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That bit would be suffice.
Carrotsticks
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Did anybody work out Q16?
iBibah
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Why is ln using absolute values?
I have never seen absolute values used with it except on here ?
someth1ng
Retired Nov '14
ln is supposed to be absolute value (in 4U).
