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    HSC 2014 MX2 Marathon (archive)

    Re: HSC 2014 4U Marathon do i know u? :p
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    HSC 2014 MX2 Marathon (archive)

    Re: HSC 2014 4U Marathon (ii) arg[(1+i)(1+2i)(1+3i)]=arg(-10) Using arg(xy)=arg(x)+arg(y), u get the required result noting that the argument of any negative real number is pi and also since the arguments of each individual complex numbers are acute
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    HSC 2014 MX2 Marathon (archive)

    Re: HSC 2014 4U Marathon Asymptotes are both y=x (forgot to label in ss) by poly division
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    HSC 2012-14 MX2 Integration Marathon (archive)

    Re: MX2 Integration Marathon change into half angles -> int{[x+2sin(x/2)cos(x/2)]/2cos^2(x/2) dx} =int{1/2 [xsec^2(x/2)]+tan(x/2) dx} =[xtan(x/2)] between x=0 and x=pi/4 by reverse product rule =pi/4tan(pi/8)
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    HSC 2012-14 MX2 Integration Marathon (archive)

    Re: MX2 Integration Marathon Not sure if correct
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    HSC 2012-14 MX2 Integration Marathon (archive)

    Re: MX2 Integration Marathon
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    HSC 2014 MX2 Marathon (archive)

    Re: HSC 2014 4U Marathon Domain: x=±1 Range: y=±pi I drew each separate graph and they shared the common points of (1,pi/2),(-1,-pi/2) and hence by the summing the graphs, y-ordinate doubles. Everywhere else is undefined, you get 2 dots at (x=1,y=pi) and (x=-1,y=-pi)
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    nhantacu

    nhantacu
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    HSC 2012-14 MX2 Integration Marathon (archive)

    Re: MX2 Integration Marathon chucking a carrot I see I got I(n)={[k(n-m+1)]/[m+k(n+1)]}I(n-m)
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    HSC 2012-14 MX2 Integration Marathon (archive)

    Re: MX2 Integration Marathon Just sat my half yearly for MX2 today, this was my recurrence question. I reckon it was a good one :D
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    HSC 2012-14 MX2 Integration Marathon (archive)

    Re: MX2 Integration Marathon
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    HSC 2014 MX2 Marathon (archive)

    Re: HSC 2014 4U Marathon so careless tonight sigh.. (ii) First case: B>=3A, B<=-A (both positive) Second Case: B>=-A, B<=3A (both negative) (iii)First Case: A>=-1, B>=-1, B>=3A, B<=-A Second Case: A<=1, B<=1, B>=-A, B<=3A
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    Why did the GFC cause Australia's CAD to improve?

    The GFC caused a downturn in the international business cycle, thus world growth contracted but more importantly consumer confidence declined. As a result, foreign investment also declines, reducing the Australia's capital inflows which in turn slows the CAD hence why it improved. Consumer...
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    HSC 2014 MX2 Marathon (archive)

    Re: HSC 2014 4U Marathon B>=3A?
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    HSC 2014 MX2 Marathon (archive)

    Re: HSC 2014 4U Marathon (i)By inspection x=-1 is a factor of P(x), by polynomial division we get the required result (ii) For 3 real roots, we set the discriminant >=0 and we get B<=-A (iii) The region is bounded by lines B>=-1, A>=-1, B<=-A
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    HSC 2012-14 MX2 Integration Marathon (archive)

    Re: MX2 Integration Marathon trying to think of another way of doing this without involving substitution
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    HSC 2014 MX2 Marathon (archive)

    Re: HSC 2014 4U Marathon
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    HSC 2014 MX2 Marathon (archive)

    Re: HSC 2014 4U Marathon woops forgot about that haha
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    HSC 2012-14 MX2 Integration Marathon (archive)

    Re: MX2 Integration Marathon Oh, well the first thing that sprung in my head was IBP twice lol
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    HSC 2014 MX2 Marathon (archive)

    Re: HSC 2014 4U Marathon Next Question:
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