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Well if the proofs for the irrationality was in the syllabus there's no point asking it because everyone would already know it. You aren't supposed to know the proofs

Re: HSC 2013 4U Marathon $i) A line segment of radius$ \ \ r \ \ $is rotated about the fixed origin by a small angle$ \delta \theta \ \ $sweeps out an area of$ \ \ \delta A \ \ \ $Show that$ \delta A = \frac{1}{2} r^2 \delta \theta $ii) Hence show that for some curve with the distance of...

Ah yes expression not equation haha

Re: MX2 Integration Marathon Interesting method, alternatively with the substitution, x=\sin^2 \theta , it ends up being the integral of (sin^2theta)

Re: MX2 Integration Marathon Yep

Re: MX2 Integration Marathon \int_0^1 \sqrt{\frac{1}{x} - 1} \ dx

shhh I don't do bio I'm trying to fit in

I'd like to see some hard statistics on employment rates from the respective Universities before we start calling superiority between universities, saying USYD is better because its old enough to be a failed Hogwarts isn't really a good indicator of quality with respect to the newer UWS (unless...

Q16 predictions? I'm hoping for inequalities and series for the whole 15 marks (and Q1 MC probability/permutations and that topic being non-existent everywhere else)

$Explain the relationship between replication of DNA and evolution$ \ \ \ \fbox{5}

I multiplied both sides of the equation by (z^2+1), this is because: (z^2+1)(z^4-z^2+1) = z^6 +1 Which becomes easier to deal with. NOTE: If you've done Series in 2U (if you haven't ignore), then notice that the sum 1 - z^2 + z^4 is a geometric series of ratio (-z^2) therefore we can use the...

$The quadratic formula approach isn't wrong$ \left(x - \frac{-b + \sqrt{b^2-4ac}}{2a} \right) \left(x - \frac{-b-\sqrt{b^2-4ac}}{2a} \right) = x^2 + \frac{b}{a} x + \frac{c}{a} \therefore \ \ a\left( x - \frac{-b + \sqrt{b^2-4ac}}{2a} \right) \left( x - \frac{-b-\sqrt{b^2-4ac}}{2a} \right) =...

I'll give you some hints: 7) You can do the sub u=x^2 then solve the quadratic, or: z^4 - z^2 + 1 = 0 Multiply both sides by (z^2+1) (z^2+1)(z^4-z^2+1) = 0 8)Similarly do a substitution or multiply both sides by (z^2-1) 9) We re-arrange: (z-1)^6 = - (z+1)^6 \therefore \ \...

This integral has no elementary indefinite integral, so I sincerely doubt it can be done without substitution. Moreover the bounds 2 and 1/2 are very special and designed for the specific sub u=1/x

Re: HSC 2013 2U Marathon Yea, you don't need to write full solutions, just the important steps will do. $Consider the function$ \ \ f(x) = \frac{A+x}{2\sqrt{Ax}} \ \ \ $where$ \ \ A>0 \ $is constant$ $i) Show that the minimum value of the function occurs at$ \ \ x=A \ \ \ \fbox{2} $ii)...

Re: HSC 2013 2U Marathon Fixed latex, that is the right idea anyway, if you didn't make any mistakes its right EDIT: Second integral is wrong EDIT2: First integral is wrong as well

Re: HSC 2013 2U Marathon $i) Differentiate$ \ \ x^{a+1} \ln x $ii) Hence find$ \ \ \int_1^e x^{a} \ln x \ dx

hahaha this thread

Re: HSC 2013 3U Marathon Thread Ahh yes that is true, thank you

Re: HSC 2013 3U Marathon Thread Well, essentially we take an arrangement of 3 couples that are already arranged so that no one sits with their couple. Then we add in another 2 people, (imagine this line to be a circle) So we have A _ B _ A _ C _ B _ C _ That is an arrangement of 3 couples...