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Would you lose marks in the HSC if you didn't prove the base case (if using transfinite induction)?

Re: HSC 2016 MX2 Combinatorics Marathon What was the method you used? The same as the one you posted on the thread where you asked this Q before (in the unweighted case)?

Re: HSC 2016 MX2 Combinatorics Marathon Yes this one looks like it'd probably be much nicer done without having to sum a series. And I haven't thought about the pmf problem since that time lol. Maybe I'll try it again later.

Re: HSC 2016 MX2 Combinatorics Marathon The one glittergal96 asked was with probability p for the coin iirc. The one you're referring to is from a few years ago (OK, I think it was from 2 years ago, after checking through some threads) I think, the one by glittergal96 was more recent, I think...

Re: HSC 2016 MX2 Combinatorics Marathon A somewhat related Q (which I'm not sure if there is a nice answer to): Let K be a predetermined positive integer. Let the coin have probability p in (0, 1) of landing heads on a given toss. Let X be the no. of tosses required to first obtain K heads in...

Re: HSC 2016 MX2 Combinatorics Marathon I think it does; I think glittergal96 (lol) asked a similar Q before, about expected no. of tosses required to get k heads in a row for a weighted coin. This Q. is basically same as that because here the profit is just equal to the no. of tosses required.

$\noindent Write $A=\begin{bmatrix}a &b \\ c& d \end{bmatrix}$, $a,b,c,d\in \mathbb{R}$. We are told that $AX = XA$ for any choice of real two-by-two matrix $X$. So for all $w,x,y,z\in \mathbb{R}$, we have $\begin{bmatrix}a &b \\ c& d \end{bmatrix}\begin{bmatrix}w &x \\ y& z...

This essentially assumes that A and B are invertible (which is indeed true as can be proved by consideration of determinants, or definition of inverses etc., and if we are allowed to use knowledge of matrix inverses, we can do the proof like that or other ways using inverses).

Note AB = A I B = A (A2 AB AB B2) B (using the given equalities involving I, noting that AB AB = I, as (AB)2 = AB AB) = A A A A (B A) B B B B = I I BA I I (as A2 = I = B2) = BA. (Of course we have used associativity of matrix multiplication a lot here.)

In = I can be proved by induction using that fact, yes.

$\noindent Correct! This is because $\frac{1}{2}\left(x-\sqrt{x^2-4}\right)$ will always be less than or equal to 1 in the domain of $f^{-1}$. (The other one is always greater than or equal to 1 in this domain.)$

Yeah pretty much.

$\noindent Recall $y$ needs to be less than or equal to $0$. Assuming you did everything correctly so far (didn't check over it), we need to take the solution that'll be less than 1, since to make a natural log be negative, the argument has to be less than 1. So, which of the solutions be the...

$\noindent For the inverse function $y=f^{-1}(x)$, we have $y\leq 0$, and $x=e^y + \frac{1}{e^y}$. Now solve this for $y$ to get the solution, and take the solution with $y\leq 0$. To solve this, note that it becomes a quadratic in $e^y$ when we multiply through by $e^y$ and rearrange etc.$

Let A be m x n. Since AB is defined, B is n x p for some p. Since BA is defined, n x p is compatible with m x n, which implies p = m. Now, AB is m x p (since A is m x n and B is n x p), i.e. m x m. Similarly, BA is n x n. (m,n,p positive integers)

Expand using the distributive law and the given fact that AB = BA. So: (A – B)(A + B) = A(A + B) – B(A+B) (by distributivity of matrix multiplication) = A2 + AB – BA – B2 (by distributivity of matrix multiplication and since X*X = X2 for any square matrix X, by definition) = A2 + O –...

To solve tan(x) = 1 (x between -360 deg and 360 deg), we get the solution of x = 45 deg. in the first quadrant. So the solutions will be: x = 45 deg, 225 deg, -135 deg, -315 deg. (Basically add multiples 180 deg. from the 45 deg. solution (multiples include negative multiples), because the...

$\noindent Let $x = 2\theta$. Since $\theta$ is between $-180^\circ$ and $180^\circ$, $2\theta$ (i.e. $x$) is between $-360^\circ$ and $360^\circ$. So first solve $\tan x=1$ with solutions between $-360^\circ$ and $360^\circ$ (you should obtain exactly four solutions). Then find the solutions in...

$\noindent If we're solving $\tan 2\theta = 1$ for $\theta$ between $-180^\circ$ and $180^\circ$, note that this is four complete cycles of the function $\tan 2\theta$, so there will be four solutions.$

Re: HSC 2016 3U Marathon