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See the answers here: http://www.thenakedscientists.com/HTML/questions/question/1000139/

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $Just do what I did, except actually show they are roots by substituting in those values and saying they give $0$. E.g. to show that $3$ was a root, you could just do this (calling the polynomial $P$):$ $$P(3) = 3^3 -6\times...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $Observe that $-1,3$ and $4$ are roots, so $y = (x+1)(x-3)(x-4)$.$

Re: MX2 2016 Integration Marathon There shouldn't be brackets of any sorts at all actually. Also, the integral diverges.

Re: MX2 2016 Integration Marathon $\noindent Well, I suppose the ``humorous'' answer would be $\frac{1}{d}\cdot \ln \left|x\right|+C$.

Re: MX2 2016 Integration Marathon $By the way, an easier way to integrate the $\frac{1}{1+\sin ^2 x}$ is to divide top and bottom by $\cos^2 x$ to get $\frac{\sec^2 x}{\sec^2 x + \tan^2 x} = \frac{\sec ^2 x}{1+2\tan^2 x}$, which results in the same integral as before upon substituting $\tau...

Re: MX2 2016 Integration Marathon $At the end for the last integral $\Big{(}$the $\frac{1}{1+\sin^2 x}$ one$\Big{)}$, I think you accidentally substituted wrongly. It should be like this.$ $Let $\tau = \tan x, \mathrm{d}x =\frac{1}{1+\tau^2} \text{ d}\tau, \sin^2 x =\frac{\tau^2}{1+\tau^2}$...

Re: MX2 2016 Integration Marathon $This is nice, but at the very end, your $t$ is $\tan x$, so instead of $\tan^{-1} x$'s in the final answer, some of those should be $\tan x$'s $\ddot{\smile}$.$

Re: MX2 2016 Integration Marathon $\noindent Let $g(x) = \frac{3x-4}{3x+4}$, then to find $g^{-1}$, we have$ $\noindent $\begin{align*}x&=\frac{3y-4}{3y+4}\quad (\text{where }y=g^{-1}(x))\\ \Rightarrow x\cdot 3y + 4x &= 3y-4 \\ \Rightarrow 3y \left(1-x \right)&= 4 \left(x+1\right) \\...

The jump between HSC 2U Maths and Maths Extension 1 isn't that big I think; at any rate, for most people it's much smaller than the jump from Maths Extension 1 to Maths Extension 2.

Re: HSC 2016 MX2 Combinatorics Marathon $\noindent \textbf{NEW QUESTION}$ $\noindent Alice and Bob play a game where they each toss a coin repeatedly. Alice tosses a coin that has a probability of coming up heads of $p_1 \in (0,1)$. Bob tosses a coin that comes up heads with probability $p_2...

$Oh yeah, didn't see the part where it said the coefficient of 2 for $x^4$.$

$\underline{FIRST QUESTION}$ $The constant term is just $-2k^2$. This is because the only way to obtain a constant term from expanding the brackets is to multiply a constant term from the first bracket with a constant term in the second bracket (if we instead took an $x$ or higher power term...

Oh yeah, (15) could be Hasse diagram or tree.

$This is the rest of my interpretation (explanations of the maths at the bottom). Starting off from Picture number (7) (not sure what (8) means really; not sure about (11) either).$ $At the present their lives were independent (7) of each other, but they knew they could not remain like this...

HSC Conics is mainly just messy algebra.

Re: HSC 2016 4U Marathon Oh sorry, didn't see you answered it. Yes, you answered it correctly. :)

Most HSC students probably don't know what the floor function is. So it'd be unlikely to be asked in the HSC probably.

Re: HSC 2016 4U Marathon $\textbf{UNANSWERED QUESTION}$ $\noindent Sketch (or for this Marathon, describe) the locus given by $\arg \left(z^3\right) = \frac{\pi}{3}$.$

Both.