# Thread: HSC 2017-2018 Maths Marathon

1. ## Re: HSC 2017 Maths (Advanced) Marathon

Originally Posted by Mathew587
Oh you can do that? My teachers a bit weird and would want us to actually write the whole solution out. How could you prove Tn = Sn - Sn-1?? ^^D
\noindent \begin{align*} S_n &= T_1 + T_2 +T_3 + ... + T_{n-1} + T_n \\ S_{n-1} &= T_1 + T_2 + T_3 + ... + T_{n-1} \end{align*} \\ As can be seen, S_n - S_{n-1} = T_n

2. ## Re: HSC 2017 Maths (Advanced) Marathon

Just a reminder this is a marathon thread, not for posting textbook errata. I have removed those posts and put into a separate thread
http://community.boredofstudies.org/...ta-2017-a.html

3. ## Re: HSC 2017 Maths (Advanced) Marathon

$Given x= \frac{6}{5} and y=1+\frac{1}{x}$

$Express y as a simple fraction$

4. ## Re: HSC 2017 Maths (Advanced) Marathon

Originally Posted by davidgoes4wce
$Given x= \frac{6}{5} and y=1+\frac{1}{x}$

$Express y as a simple fraction$
$x=\frac{6}{5} \Rightarrow y = 1+\frac{1}{x} = 1 + \frac{1}{\frac{6}{5}} = 1 + \frac{5}{6} = \frac{11}{6}$

6. ## Re: HSC 2017 Maths Marathon

OK I got nothing else to do on a Friday night I am going to have an honest go at this question.

11. ## Re: HSC 2017 Maths Marathon

Originally Posted by davidgoes4wce
$\noindent \textbf{(a)} Now 0.52^{\circ} \equiv \frac{13\pi}{4500} radians. If we construct an altitude from earth viewpoint, this angle is halved, i.e. \frac{13\pi}{9000} radians. \\\\ Now \tan{\left(\frac{13\pi}{9000}\right)} = \frac{r_{moon}}{385000} \Rightarrow r_{moon} = 385000\tan{\left(\frac{13\pi}{9000}\right)}. And so we have r_{moon} = 1700 km (correct to 2 s.f.) \\\\ Similarly, \tan{\left(\frac{13\pi}{9000}\right)} = \frac{r_{sun}}{150000000} \Rightarrow r_{sun} = 150000000\tan{\left(\frac{13\pi}{9000}\right)}. And so r_{sun} = 680000 km (correct to 2 s.f.)$

$\noindent \textbf{(b)} 3.8 cm per year \Rightarrow (3.8 \times 10^9) cm per 10^9 years. That is, the moon moves 38000 kilometres away from earth viewpoint, a total of 423000 km. Noting that r_{moon} is constant, half of the new angle subtended by the moon \frac{\theta}{2} is given by \tan{\left(\frac{\theta}{2}\right)} = \frac{r_{moon}}{423000}. And so \frac{\theta}{2} \approx 0.237^{\circ} \Rightarrow \theta = 0.47^{\circ} (to 2 s.f.)$

$\noindent \textbf{(c)} 15 cm per year \Rightarrow (15 \times 10^9) cm per 10^9 years. That is, the earth moves 150000 kilometres away from the sun, a total of 150150000 km. Noting that r_{sun} is constant, half of the new angle subtended by the sun \frac{\psi}{2} is given by \tan{\left(\frac{\psi}{2}\right)} = \frac{r_{sun}}{150150000}. And so \frac{\psi}{2} \approx 0.259^{\circ} \Rightarrow \psi = 0.52^{\circ} (to 2 s.f.)$

$\noindent \textbf{(d)} The radius of the area obscured by the moon r is given by \tan{\left(\frac{0.47^{\circ}}{2}\right)} = \frac{r}{150150000}. That is, r = 150150000\tan{\left(\frac{47\pi}{36000}\right)}. Now call this obscured area A_1 and call the area of the disc represented by the sun A_2. Well A_1 = \pi r^2 = \pi \left[150150000 \tan{\left(\frac{47 \pi}{36000}\right)}\right]^2 and A_2 = \pi (r_{sun})^2 = \pi \left[150000000 \tan{\left( \frac{13 \pi}{9000}\right)}\right]^2 \\\\ Now \frac{A_1}{A_2} = \frac{\pi \left[150150000 \tan{\left(\frac{47 \pi}{36000}\right)}\right]^2}{\pi \left[150000000 \tan{\left( \frac{13 \pi}{9000}\right)}\right]^2} \approx 0.818. Thus, 82\% of the disc represented by the sun is obscured by the moon (to 2 s.f.)$

12. ## Re: HSC 2017 Maths Marathon

what happened to this thread o
and how do you guys know if you thoroughly know a topic in math?

13. ## Re: HSC 2017-2018 Maths Marathon

Im going to Year 10 in 2018 and am learning differential calculus, to answer OP's question I sketched the graph, any advanced criticism is appreciated as I am learning

14. ## Re: HSC 2017-2018 Maths Marathon

Originally Posted by HeroWise
Im going to Year 10 in 2018 and am learning differential calculus, to answer OP's question I sketched the graph, any advanced criticism is appreciated as I am learning
When you do working in 2u make sure to write out obvious statements, they may seem stupid but they are necessary to get the marks, particularly for curve sketching. For example above- include in your working a phrase like "testing nature," and a concluding statement "therefore horizontal point of inflexion at P(X,Y) etc.

15. ## Re: HSC 2017-2018 Maths Marathon

Originally Posted by HeroWise
Im going to Year 10 in 2018 and am learning differential calculus, to answer OP's question I sketched the graph, any advanced criticism is appreciated as I am learning
It doesn't matter if your graph's scale isn't perfect, you don't need to measure out centimetres on your ruler. As long as you sketch a reasonable graph, showing any necessary features (including any that were asked for) e.g. intercepts, maxima/minima, pts. of inflexion you'll get the marks. The intent is not to test you on your measuring skills.

Just as a quick example, I would sketch it something like this. Just note that this graph shows the y-intercept more clearly. It's possible to find any x intercepts but this isn't expected in the 2U course, although it should be in the 3U course. I was taught to label my points of inflexion with a line going through the point but I don't think it matters as long as you make it clear that the point IS one of inflexion. In an exam I would draw the graph larger, though.

A point of inflexion means a change in concavity, i.e. from positive to negative or vice versa. so to prove that a point is one of inflexion, you need to show that the sign of the second derivative changes around that point.

$y'' = 18x - 12 \\ x = \frac{2}{3}, y'' = 0 \\ x = 1, y'' = 6 \\ x = 0, y ''= -12$

Because $y''$ changes signs around $x=\frac{2}{3}$, a point of inflexion exists at $x = \frac{2}{3}$.

Just to elaborate on why you need to test the second derivative around the point - first note that any quadratic, such as $y = x^2$, has no points of inflexion because $y''$ is always a constant (in this case, $2$) and so the equation $y'' = 0$ has no solution.

Consider the curve $y = x^4$. The curvature of the graph is identical to $y = x^2$ (you can graph it digitally to check). Here, $y'' = 12x^2$, so the equation $y'' = 0$ has a solution at $x = 0$. But from the previous paragraph, clearly this curve can't have a point of inflexion. This is because $y'' \geq 0$ for all real $x$, so the second derivative never changes signs and because of this, no points of inflexion exist on this curve.

16. ## Re: HSC 2017-2018 Maths Marathon

Originally Posted by fan96
It doesn't matter if your graph's scale isn't perfect, you don't need to measure out centimetres on your ruler. As long as you sketch a reasonable graph, showing any necessary features (including any that were asked for) e.g. intercepts, maxima/minima, pts. of inflexion you'll get the marks. The intent is not to test you on your measuring skills.

Just as a quick example, I would sketch it something like this. Just note that this graph shows the y-intercept more clearly. It's possible to find any x intercepts but this isn't expected in the 2U course, although it should be in the 3U course. I was taught to label my points of inflexion with a line going through the point but I don't think it matters as long as you make it clear that the point IS one of inflexion. In an exam I would draw the graph larger, though.

A point of inflexion means a change in concavity, i.e. from positive to negative or vice versa. so to prove that a point is one of inflexion, you need to show that the sign of the second derivative changes around that point.

$y'' = 18x - 12 \\ x = \frac{2}{3}, y'' = 0 \\ x = 1, y'' = 6 \\ x = 0, y ''= -12$

Because $y''$ changes signs around $x=\frac{2}{3}$, a point of inflexion exists at $x = \frac{2}{3}$.

Just to elaborate on why you need to test the second derivative around the point - first note that any quadratic, such as $y = x^2$, has no points of inflexion because $y''$ is always a constant (in this case, $2$) and so the equation $y'' = 0$ has no solution.

Consider the curve $y = x^4$. The curvature of the graph is identical to $y = x^2$ (you can graph it digitally to check). Here, $y'' = 12x^2$, so the equation $y'' = 0$ has a solution at $x = 0$. But from the previous paragraph, clearly this curve can't have a point of inflexion. This is because $y'' \geq 0$ for all real $x$, so the second derivative never changes signs and because of this, no points of inflexion exist on this curve.
Just a tiny tiny comment but make sure your horizontal point of inflection looks really horizontal so that if you drew a tangent to it the tangent would be horizontal.

17. ## Re: HSC 2017-2018 Maths Marathon

Originally Posted by pikachu975
Just a tiny tiny comment but make sure your horizontal point of inflection looks really horizontal so that if you drew a tangent to it the tangent would be horizontal.
I asked my teacher he says about 3mm run in HPOI is good
Well in reality no one is gonna actually measure for 3mm run, around that should bbe fine

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