HSC 2013-14 MX1 Marathon (archive) (1 Viewer)

Sy123 · Feb 17, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$Considering the identity$

$(2+x)^n = (1+(1+x))^n$

$Prove that$

$\binom{n}{k} \cdot \frac{\binom{n}{0}+\binom{n}{1}+\dots+\binom{n}{n}}{\binom{k}{0}+\binom{k}{1}+\dots+\binom{k}{k}} = \binom{n}{0} \binom{n}{k} + \binom{n}{1} \binom{n-1}{k} + \dots + \binom{n}{k} \binom{k}{k}$

$(2+x)^n = \binom{n}{0}+\binom{n}{1}(1+x) + \binom{n}{2} (1+x)^2 + \dots + \binom{n}{n}(1+x)^n$

Equate co-efficient of x^k

$\binom{n}{k} 2^{n-k} = \binom{n}{n} \binom{n}{k} + \binom{n}{n-1} \binom{n-1}{k} + \dots + \binom{n}{k} \binom{k}{k}$

$\binom{n}{k} \frac{2^n}{2^k} = \binom{n}{0} \binom{n}{k} + \dots + \binom{n}{k} \binom{k}{k}$

Express

$2^r = \binom{r}{0} + \dots + \binom{r}{r}$

Sub it in, and we are done

hayabusaboston · Feb 17, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$(2+x)^n = \binom{n}{0}+\binom{n}{1}(1+x) + \binom{n}{2} (1+x)^2 + \dots + \binom{n}{n}(1+x)^n$

Equate co-efficient of x^k

$\binom{n}{k} 2^{n-k} = \binom{n}{n} \binom{n}{k} + \binom{n}{n-1} \binom{n-1}{k} + \dots + \binom{n}{k} \binom{k}{k}$

$\binom{n}{k} \frac{2^n}{2^k} = \binom{n}{0} \binom{n}{k} + \dots + \binom{n}{k} \binom{k}{k}$

Express

$2^r = \binom{r}{0} + \dots + \binom{r}{r}$

Sub it in, and we are done

r u ignoring me sy? how come?

HeroicPandas · Feb 17, 2013

Re: HSC 2013 3U Marathon Thread

$$ Given that $\int $lnx.dx = $xlnx - x ($ignore the + C) \\ \\ $ $ Evaluate $\int $sinx .ln($\frac{1}{1+cos2x}).dx$

Sy123 · Feb 19, 2013

Re: HSC 2013 3U Marathon Thread

$$There exists a sequence of circular quadrants$ \ \ C_0, \ C_1, \ C_2, \ \dots$

$$And a sequence of lines$ \ \ l_1, \ l_2, \ l_3 \ \dots$

$C_0 \ \ $has a radius of 1$$

$Subsequent circular quadrants and lines all undergo the same operation as shown$

$\textbf{Find the area between} \ \ C_n \ \ $and$ \ \ l_n \ \ \textbf{in terms of n only}$

asianese · Feb 19, 2013

Re: HSC 2013 3U Marathon Thread

$$Find the value of the area under the curve $ y=e^x $ bound by the x-axis for $ -n\leq x \leq0 $ where $ n>0. \hfill{\fbox{1}} \\ \\$ Hence determine the area under the curve as $ n $ approaches infinity. \hfill{\fbox{1}} \\\\ \text{ Hence find the area bound by } \mathrm{y}=\ln x \text{ between } x=0 \text{ and } x=1 \text{ without evaluating a further integral. Explain your answer. } \hfill{\fbox{2}}$

HeroicPandas · Feb 19, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
$$Find the value of the area under the curve $ y=e^x $ bound by the x-axis for $ -n\leq x \leq0 $ where $ n>0. \hfill{\fbox{1}} \\ \\$ Hence determine the area under the curve as $ n $ approaches infinity. \hfill{\fbox{1}} \\\\ \text{ Hence find the area bound by } \mathrm{y}=\ln x \text{ between } x=0 \text{ and } x=1 \text{ without evaluating a further integral. Explain your answer. } \hfill{\fbox{2}}$

a) $A= 1 - \frac{1}{e^n}$ ?

b) A =1

c) A =1, y=lnx is the inverse function of y=e^x, using shading and inspection $\int_{0}^{1} lnx.dx \equiv \int_{-n}^{0} e^x.dx$ AS n-> INFINTY

asianese · Feb 19, 2013

Re: HSC 2013 3U Marathon Thread

Yep

Sy123 · Feb 22, 2013

Re: HSC 2013 3U Marathon Thread

Great question I came across:

$$The three sides in a triangle are in ratio$ \ \ 4 \ : 5 \ : \ 6$

$Prove that one angle is twice the other$

asianese · Feb 22, 2013

Re: HSC 2013 3U Marathon Thread

Sy, that is a very VERY nice question. I will post solution tomorrow if no one has it.

nightweaver066 · Feb 23, 2013

Re: HSC 2013 3U Marathon Thread

I wouldn't say it's great... unless i'm doing something wrong.

asianese · Feb 23, 2013

Re: HSC 2013 3U Marathon Thread

$We shall prove that $ 2\alpha=\gamma.\\ \\ $Using the sine rule: $\\ \frac{\sin\alpha}{4}=\frac{\sin\beta}{5}= \frac{\sin \gamma}{6}.\\ $Using the cosine rule: $ 4^2=5^2+6^2-2\times5\times6\cos\alpha\\16=61-60\cos\alpha\\ \cos\alpha=\frac{3}{4} \hfill{(1)}\\\\ $Now, using$ \frac{\sin\alpha}{4}=\frac{\sin \gamma}{6}, $ multiply by $ (1).\\ \frac{\cos \alpha \sin\alpha}{4}=\frac{\cos \alpha \sin \alpha}{6}\\ \frac{2\cos\alpha\sin\alpha}{8}=\frac{3}{4} \times\frac{\sin \gamma}{6}\\ \begin{align*} \Rightarrow \sin2\alpha&=\frac{8\times3}{4\times6}\sin\gamma\\ &=\sin\gamma \end{align*} \\ \Rightarrow 2\alpha = \gamma $ since both angles are within a triangle.$

Whoever made this question up must have tried so many numbers...or they're just geniuses. Or this is related to some higher mathematical idea.

Sy123 · Feb 24, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
$We shall prove that $ 2\alpha=\gamma.\\ \\ $Using the sine rule: $\\ \frac{\sin\alpha}{4}=\frac{\sin\beta}{5}= \frac{\sin \gamma}{6}.\\ $Using the cosine rule: $ 4^2=5^2+6^2-2\times5\times6\cos\alpha\\16=61-60\cos\alpha\\ \cos\alpha=\frac{3}{4} \hfill{(1)}\\\\ $Now, using$ \frac{\sin\alpha}{4}=\frac{\sin \gamma}{6}, $ multiply by $ (1).\\ \frac{\cos \alpha \sin\alpha}{4}=\frac{\cos \alpha \sin \alpha}{6}\\ \frac{2\cos\alpha\sin\alpha}{8}=\frac{3}{4} \times\frac{\sin \gamma}{6}\\ \begin{align*} \Rightarrow \sin2\alpha&=\frac{8\times3}{4\times6}\sin\gamma\\ &=\sin\gamma \end{align*} \\ \Rightarrow 2\alpha = \gamma $ since both angles are within a triangle.$

Whoever made this question up must have tried so many numbers...or they're just geniuses. Or this is related to some higher mathematical idea.

What I did was find cosalpha cosbeta and cosgamma (in terms of numbers)
Then show that 2 of them satisfy cos2theta =2cos^2 theta -1

======

As for how they made that question, if we choose to find the ratio of sides in a triangle where 1 angle is twice the other
We just use the relation:

$\frac{\sin \alpha}{a} = \frac{\sin 2\alpha}{b} = \frac{\sin (\pi-3\alpha)}{c}$

Then after some manipulation

$b^2=a(a+c)$

We just pick a 'reference number' such as a=1

$b^2=1+c$

Then we find a pair of b and c such that they match the equation, so b=6/4 c=5/4

Therefore the sides are in ratio 4 : 5 : 6 since similar triangles make it so that the actual numbers doesn't matter as long as we keep the same angles.
Also I'm sure they had to do some constructions by hand and notice that similarity

Sy123 · Feb 24, 2013

Re: HSC 2013 3U Marathon Thread

$An$ \ \ n-$sided regular polygon exists$

$All interior angles of the polygon are below 180^{\circ}$

$Prove by process of mathematical induction that the sum of all interior angles of this polygon is$

$S=180(n-2)^{\circ}$

asianese · Feb 24, 2013

Re: HSC 2013 3U Marathon Thread

$$a) A circle has the equation $ x^2+y^2=1. $ A tangent is drawn from the point $P(\cos\theta,\sin\theta) $ where $\theta $ is the angle between the positive $x$-axis and the radius that meets the circle at $P.\\\\ $i) By considering $ y $ as a positive function of $x$, find the equation of the tangent at $P. \hfill{\fbox{3}}\\\\ $ii) Suppose another tangent is drawn from $ Q(\cos\theta,-\sin\theta). $ Write down the equation of this tangent and hence find the point of intersection, $T$, of the two tangents. $ \hfill{\fbox{2}}\\\\ $iii) Find the angle $ \theta $ such that $ T $ has co-ordinates $(2,0). \hfill{\fbox{1}}\\\\$

$$b) Now suppose that $ T $ moves in simple harmonic motion along the $x$-axis with its acceleration given by $ \ddot{x}=-4x.\\\\ $i) Find its velocity in terms of its distance from the origin.$ \hfill{\fbox{3}}\\\\ $ii) By considering the velocity equation and the construction of $T$, find the restriction of the position of $T. \hfill{\fbox{2}} \\\\ $iii) By considering the graph of $f(\alpha)=\frac{1}{\cos\alpha}$ or otherwise, deduce the possible values of $\theta.\hfill{\fbox{2}}\\\\$

$$c) Suppose now that $T$ is free to move along $x\geq1. $ What happens to the angle $ \theta $ as $T$ approaches infinity? Comment on the physical restriction of $\theta$. \hfill{\fbox{2}}$

nightweaver066 · Feb 24, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Great question I came across:

$$The three sides in a triangle are in ratio$ \ \ 4 \ : 5 \ : \ 6$

$Prove that one angle is twice the other$

Since we know the sides are in that ratio, we know that all triangles following that are similar so they all have the same corresponding angles.

Lets use the cos rule to determine the smallest angle (angle opposite smallest side) and to determine the largest angle (angle opposite largest side).

You find the largest angle is twice the smallest angle.

Don't think you guys would like this answer

asianese · Feb 24, 2013

Re: HSC 2013 3U Marathon Thread

nightweaver066 said:
Since we know the sides are in that ratio, we know that all triangles following that are similar so they all have the same corresponding angles.

Lets use the cos rule to determine the smallest angle (angle opposite smallest side) and to determine the largest angle (angle opposite largest side).

You find the largest angle is twice the smallest angle.

Don't think you guys would like this answer

That's the boring short approach

asianese · Feb 24, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$An$ \ \ n-$sided regular polygon exists$

$All interior angles of the polygon are below 180^{\circ}$

$Prove by process of mathematical induction that the sum of all interior angles of this polygon is$

$S=180(n-2)^{\circ}$

Just nit-picking - should there be a restriction on n?

nightweaver066 · Feb 24, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
That's the boring short approach

It works.

asianese · Feb 24, 2013

Re: HSC 2013 3U Marathon Thread

Most definitely haha.

Sy123 · Feb 24, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
Just nit-picking - should there be a restriction on n?

Naturally any n-sided polygon has 3 or more sides otherwise it wouldn't be a polygon

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