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$\noindent Note that the angular velocity of the beacon in radians per second is $\omega = 4 \times \frac{2\pi}{60} = \frac{2}{15}\pi$. Now, clearly, when the light beacon is right at $P$, the rate (speed) at which it moves along the shoreline is just $v=r\omega = 3\times \frac{2}{15}\pi =...

Re: HSC 2016 4U Marathon - Advanced Level Here was my method (for obtaining the series). $\noindent Let $X$ be uniformly distributed on $[0,1]$, i.e. $X\sim \mathcal{U}(0,1)$. Let $A_n$ be the event that $\frac{1}{n+1}<X<\frac{1}{n}$ for $n=1,2,3,\ldots$. It is easy to see that the $A_n$'s...

Re: HSC 2016 4U Marathon - Advanced Level $$\sum _{n=1}^{\infty} \left[ 1-\frac{n}{2} \left(\sqrt{n^2 + 4}-n \right) \right] \cdot \frac{1}{n} =0.5355 \ldots$$

Re: HSC 2016 4U Marathon - Advanced Level Do you know whether there is a closed-form answer, or can we just give our answer as an infinite series?

$\noindent A quick way to do this without calculation is to just compare the vertical components of the initial velocity of each shot (the horizontal component doesn't affect the time of flight). Since they have the same speeds and the second one was shot up at a higher angle than the first, the...

Re: HSC 2016 4U Marathon $\noindent (Best to avoid using $P$ as a point on the ellipse, as $P$ is given in the question to be an exterior point $(x_0,y_0)$.) $\noindent Alternative method is to note that as $P(x_0,y_0)$ lies on the tangents to both $(x_1,y_1)$ and $(x_2,y_2)$ on the ellipse...

Re: HSC 2016 3U Marathon He meant "area" (it was a typo). He just named the curves y1 and y2 so that he could refer to the curves easily.

$\noindent Let $a = \beta + \gamma - \alpha$, $b = \gamma + \alpha - \beta$, and $c = \alpha + \beta -\gamma$. Then $a,b,c>0$ due to the inequality conditions on $\alpha, \beta, \gamma$. It also easy to check that these are all unequal. For instance, suppose $a=b$. Then $\beta +\gamma - \alpha...

Re: MX2 2016 Integration Marathon $$x+c$$

I think this was asked before. Check out this thread: http://community.boredofstudies.org/1003/maths/335733/another-hard-vectors-question.html

$\noindent 17. We are given $A=\begin{bmatrix}3 &2\\ 7&5\end{bmatrix}$. Hence$ $$\begin{align*}A^2 &= \begin{bmatrix}3 &2\\ 7&5\end{bmatrix}\begin{bmatrix}3 &2\\ 7&5\end{bmatrix} \\ &= \begin{bmatrix}3\cdot 3 + 2 \cdot 7 & 3 \cdot 2 + 2 \cdot 5 \\7\cdot 3 + 5\cdot 7 & 7\cdot 2 + 5 \cdot 5...

Re: Help i'm trapped in the title and can only speak once every 2 years Why did you write ''Re: Help i'm trapped in the title and can only speak once every 2 years'' in the reply title thing? Just for fun? Edit: oh, it's because Paradoxica had written ''Help i'm trapped in the title and...

Welcome! For the solutions, this is basically their reasoning (I'll just use the ball analogy again). a) We want the probability that the last five balls to be non-white. Since there are 20 balls to choose from to place in the last 5, we can choose which ones go in the last five in 20C5 ways...

This question is equivalent to thinking about it in terms of randomly arranging 15 white balls (quarters with no claims), 2 blue balls (quarters with one claim), and 3 black ball (quarters with 3 claims) in a straight line. The first Q. is asking us what's the probability that when we arrange...

$\noindent Note that we don't have the strict inequality hold at $x=0$, as we have equality instead. For $x>0$, the inequation is $x^2>x$, as $|x| = x$ for $x>0$. Since $x>0$, we can cancel it from either side of the inequality to get $x>1$ as the solution for this case.$ $\noindent The other...

$\noindent Because we said in the first paragraph that the inequation clearly holds for all $x<0$ (and $x$ can't be 0 since the RHS is undefined then), so we only needed to do the algebra for $x>0$.$

$\noindent In general, since we normally deal with arbitrary $x$, we need to multiply by $x^2$ instead. The whole point of this is so that we don't need to worry about flipping inequality signs due to potential multiplication by a negative number. But we didn't need to in the above working since...

$\noindent The inequation is $x^2 > |x|$. By sketching the graphs of $y=x^2$ and $y=|x|$ (note that they intersect at and only at $\left( \pm 1,1\right)$), we can easily see that the solution is $x<-1$ or $x>1$.$

$\noindent Note that $x\neq 0$. Also note that if $x$ is negative, the inequation holds always since the LHS is positive whilst the RHS is negative. Now, let's focus on $x>0$.$ $\noindent For $0<x<2$, we have $|x-2| = 2-x$, so the inequation becomes $2-x>\frac{1}{x}$. Since $x>0$, we can...

$\noindent Just find the area under the curve for each vehicle. This gives us the change in velocity for each vehicle. Adding this to the initial velocity gives us the final velocity. We don't need calculus to find the areas because they are just triangles (the curves are made up of straight...