HSC 2016 MX1 Marathon (archive) (3 Viewers)

pikachu975 · Sep 29, 2016

Re: HSC 2016 3U Marathon

Oh yeah thanks didn't see that

Drongoski · Sep 29, 2016

Re: HSC 2016 3U Marathon

KingOfActing said:
$Find the value of $ \lim_{x \to 0 } \frac{1 - \cos{x}}{x^2} $ (using 3U methods @Paradoxica)$

$\frac {1-cos x}{x^2 }= \frac {2sin^2 \frac {x}{2}}{x^2} = \frac {2}{4} \times \left ( \frac {sin \frac {x}{2}}{\frac {x}{2}} \right )^2$

KingOfActing · Sep 29, 2016

Re: HSC 2016 3U Marathon

Drongoski said:
$\frac {1-cos x}{x^2 }= \frac {2sin^2 \frac {x}{2}}{x^2} = \frac {2}{4} \times \left ( \frac {sin \frac {x}{2}}{\frac {x}{2}} \right )^2$

Nice. What I had in mind was $\frac{1-\cos{x}}{x^2} = \frac{\sin^2{x}}{x^2}\times\frac{1}{1+\cos{x}} = \left(\frac{\sin{x}}{x}\right)^2\times\frac{1}{1+\cos{x}}$

leehuan · Sep 29, 2016

Re: HSC 2016 3U Marathon

Double angle is pretty common for that limit at a 3U level (although the question itself is uncommon)

That rationalising the numerator question brings back bad memories.

Paradoxica · Sep 29, 2016

Re: HSC 2016 3U Marathon

KingOfActing said:
$Find the value of $ \lim_{x \to 0 } \frac{1 - \cos{x}}{x^2} $ (using 3U methods @Paradoxica)$

Using the Symmetric Quotient Definition of the Second Derivative:

$f''(z) \equiv \lim_{x \to 0} \frac{f(z+x)-2f(z)+f(z-x)}{x^2}$

$f(z) = \cos{z}$

$-1 = \cos ''(0) = \lim_{x \to 0} \frac{\cos{(0+x)} - 2\cos{(0)} + \cos{(0-x)}}{x^2} = 2 \lim_{x \to 0} \frac{\cos{x}-1}{x^2}$

Rearrange both sides to obtain:

$\lim_{x \to 0} \frac{1-\cos{x}}{x^2} = \frac{1}{2}$

From the series expansion of cosx, we obtain:

$\lim_{x \to 0} \frac{1-(1-\frac{x^2}{2} + \mathcal{O}(x^4))}{x^2} = \lim_{x \to 0} \frac{1}{2}-\mathcal{O}(x^2) = \frac{1}{2}$

InteGrand · Sep 29, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
Using the Symmetric Quotient Definition of the Second Derivative:

$f''(z) \equiv \lim_{x \to 0} \frac{f(z+x)-2f(z)+f(z-x)}{x^2}$

$f(z) = \cos{z}$

$-1 = \cos ''(0) = \lim_{x \to 0} \frac{\cos{(0+x)} - 2\cos{(0)} + \cos{(0-x)}}{x^2} = 2 \lim_{x \to 0} \frac{\cos{x}-1}{x^2}$

Rearrange both sides to obtain:

$\lim_{x \to 0} \frac{1-\cos{x}}{x^2} = \frac{1}{2}$

From the series expansion of cosx, we obtain:

$\lim_{x \to 0} \frac{1-(1-\frac{x^2}{2} + \mathcal{O}(x^4))}{x^2} = \lim_{x \to 0} \frac{1}{2}-\mathcal{O}(x^2) = \frac{1}{2}$

Well so much for 3U methods.

trecex1 · Sep 29, 2016

Re: HSC 2016 3U Marathon

$\text{Prove that }\frac{1}{3}\binom{n}{0}-\frac{1}{4}\binom{n}{1}+\frac{1}{5}\binom{n}{2}-...-\frac{(-1)^{n+3}}{n+3}\binom{n}{n} \\ =\frac{2(n!)}{(n+3)!}$

InteGrand · Sep 29, 2016

Re: HSC 2016 3U Marathon

mini8658 said:
What do the brackets without the fraction line mean?

Those are binomial coefficients. You can read about them here: https://en.wikipedia.org/wiki/Binomial_coefficient .

KingOfActing · Sep 29, 2016

Re: HSC 2016 3U Marathon

mini8658 said:
Find the number of ways in which a pair of triangles can be drawn with 6 points as vertices, no three of the points being collinear.

If triangles can overlap and use the same vertices : ${6 \choose 3}^2 = 400$

If triangles must be adjacent and not overlap: ${6 \choose 4}{4 \choose 2} = 90$

KingOfActing · Sep 29, 2016

Re: HSC 2016 3U Marathon

trecex1 said:
$\text{Prove that }\frac{1}{3}\binom{n}{0}-\frac{1}{4}\binom{n}{1}+\frac{1}{5}\binom{n}{2}-...-\frac{(-1)^{n+3}}{n+3}\binom{n}{n} \\ =\frac{2(n!)}{(n+3)!}$

$$Consider the function $ f(x) = \sum_{k = 0}^{n} {n \choose k}\frac{x^{k+3}}{k+3}\\$It is clear that $ f(0) = 0$ and that $f(-1) = -S$ where $S$ is the sum we seek to find. Differentiating both sides, we get\\$f'(x) = \sum_{k = 0}^{n} {n \choose k}x^{k+2} = x^2\sum_{k = 0}^{n} {n \choose k}x^k = x^2(1+x)^n\\$Integrating both sides from $-1$ to $0$:$\\\begin{align*}f(0) - f(-1) &= \int_{-1}^{0}x^2(1+x)^n\,dx\\S &=\int_{0}^{1} (1-u)^2u^n\,du\qquad\qquad (u = x+1)\\&=\int_0^1u^n - 2u^{n+1} + u^{n+2}\,du\\&=\frac{1}{n+1} - \frac{2}{n+2} + \frac{1}{n+3}\\&=\frac{2}{(n+1)(n+2)(n+3)}\\&=\frac{2\times n!}{(n+3)!}\end{align*}$

Drsoccerball · Sep 29, 2016

Re: HSC 2016 3U Marathon

New Question:

$$ Prove by induction : $ \sqrt{1} + \sqrt{2} + ... + \sqrt{n} \geq \frac{2n}{3} \cdot \sqrt{n}$

* Also think of an alternate way of proving this *

KingOfActing · Sep 29, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
New Question:

$$ Prove by induction : $ \sqrt{1} + \sqrt{2} + ... + \sqrt{n} \geq \frac{2n}{3} \cdot \sqrt{n}$

* Also think of an alternate way of proving this *

$$Alt. Method: Consider the function $ f(x) = \sqrt{x} $. Draw rectangles 1 width apart from $x = 0$ to $x = n$. Since the function is monotonically increasing, the sum of these rectangles is greater than the area under the curve.$\sum_{k= 0}^n 1\times x \geq \int_0^n \sqrt{x}\,dx\\\sqrt{1} + \sqrt{2} + \dots + \sqrt{n} \geq \frac{2n\sqrt{n}}{3}$

mini865 said:
Find the exact values of tan(sin^-1(-2/3))

$$Let $\sin{\theta} = \frac{-2}{3}\\$We want to find $ \tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}} =\frac{\sin{\theta}}{\sqrt{1-\sin^2{\theta}}} = \frac{\frac{-2}{3}}{\frac{\sqrt{5}}{3}} = \frac{-2}{\sqrt{5}}$

We know that this value is right because arcsin(-2/3) lies in the fourth quadrant, yadda yadda

Drsoccerball · Sep 29, 2016

Re: HSC 2016 3U Marathon

The induction is left uncompleted but heres a follow up :
$$ Prove that: $ \frac{2n}{3} \cdot \sqrt{n} \leq \sqrt{1} + \sqrt{2} + ... + \sqrt{n} \leq \frac{(4n+3)}{6} \cdot \sqrt{n}$

KingOfActing · Sep 29, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
The induction is left uncompleted but heres a follow up :
$$ Prove that: $ \frac{2n}{3} \cdot \sqrt{n} \leq \sqrt{1} + \sqrt{2} + ... + \sqrt{n} \leq \frac{(4n+3)}{6} \cdot \sqrt{n}$

$$The result clearly holds for $n = 1\qquad \left(1 \geq \frac23\right)\\$Suppose the result holds for $n = k$, where $k$ is some integer. Then, in the case of $n = k+1$, we have:$\\\sqrt{1}+\sqrt{2} + \dots + \sqrt{k} + \sqrt{k+1} \geq \frac23k\sqrt{k} + \sqrt{k+1}\\$We would like to prove that:$\\\frac23k\sqrt{k} + \sqrt{k+1} \geq \frac23(k+1)\sqrt{k+1} \\\iff\frac23k\sqrt{k} + \sqrt{k+1} - \frac23(k+1)\sqrt{k+1} \geq 0\label{a}{\qquad(*)}\\$Let $f(k) = \frac23k\sqrt{k} + \sqrt{k+1} - \frac23(k+1)\sqrt{k+1}\\f(0) = \frac13\\f'(k) = \sqrt{k} + \frac{1}{2\sqrt{k+1}} -\sqrt{k+1} = \frac{2\sqrt{k^2+k}+ 1-2k}{2\sqrt{k+1}} \geq \frac{2k+ 1 - 2k}{2\sqrt{k+1}} > 0\\$Hence (for positive $k$)$\\ f(k) = \frac23k\sqrt{k} + \sqrt{k+1} - \frac23(k+1)\sqrt{k+1} \geq f(0) = \frac13 \geq 0\\$And so $(*)$ holds true for all $k \geq 1$.$\\\\$Since the inequality holds for $n=1$, it must hold for $n=2$, which proves the case of $n=3$ and so on, and as such the result holds true for all $n \geq 1$.$

#WhenYouCan'tSeeTheIntendedMethod

penapplepineapplepen · Sep 29, 2016

Re: HSC 2016 3U Marathon

oi i need help with dis

can we just use:

Acceleration = v * dv/dx?

or do we have to use:

2) dx/dt = blah

therefore dt/dx = blah1
therefore t = blah3
and then differentiate two times to find the acceleratoin?

trecex1 · Sep 29, 2016

Re: HSC 2016 3U Marathon

penapplepineapplepen said:
oi i need help with dis

View attachment 33483

can we just use:

Acceleration = v * dv/dx?

or do we have to use:

2) dx/dt = blah

therefore dt/dx = blah1
therefore t = blah3
and then differentiate two times to find the acceleratoin?

a=d(v^2/2)/dx you can quote that

leehuan · Sep 29, 2016

Re: HSC 2016 3U Marathon

penapplepineapplepen said:
oi i need help with dis

View attachment 33483

can we just use:

Acceleration = v * dv/dx?

or do we have to use:

2) dx/dt = blah

therefore dt/dx = blah1
therefore t = blah3
and then differentiate two times to find the acceleratoin?

trecex1 said:
a=d(v^2/2)/dx you can quote that

You can quote a = v.dv/dx

a = d/dx(v^2/2) is probably preferred in 3U because it's actually a part of the 3U course. But v.dv/dx is a very commonly used result in 4U so that's something worth noting.

(To my knowledge, though, I don't believe it's 'banned' in 3U - use of v dv/dx.)

Paradoxica · Sep 29, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
New Question:

$$ Prove by induction : $ \sqrt{1} + \sqrt{2} + ... + \sqrt{n} \geq \frac{2n}{3} \cdot \sqrt{n}$

* Also think of an alternate way of proving this *

Preliminary Result:

$1. \, 3k^2 > 1 \quad \forall k \in \mathbb{N}$

Proof: Obvious

Now we begin.

$$\noindent$ 3k^2 - 1 > 0 \\\\ 64k^6 -48k^4 + 12k^2 -1 > 64k^6 -48k^4 + 9k^2 \\\\ (4k^2-1)^3 > (8k^2-3)^2k^2 \\ (4k^2-1)\sqrt{4k^2-1)} > 8k^3-3k \\\\ 3k > 8k^3 - 8k^2 \sqrt{k^2-\frac{1}{4}} + 2\sqrt{k^2-\frac{1}{4}} \\\\ 9k > 4k^2 \left( 2k - 2\sqrt{k^2-\frac{1}{4}} \right) + 4k + \left( 2k + 2\sqrt{k^2-\frac{1}{4}} \right) \\\\ (3\sqrt{k})^2 > (2k)^2 \left( \sqrt{k+\frac{1}{2}} - \sqrt{k-\frac{1}{2}} \right)^2 + 2\times 2k \times \left( \sqrt{k+\frac{1}{2}} - \sqrt{k-\frac{1}{2}} \right)\left( \sqrt{k+\frac{1}{2}} + \sqrt{k-\frac{1}{2}} \right) + \left( \sqrt{k+\frac{1}{2}} + \sqrt{k-\frac{1}{2}} \right)^2 \\\\ 3\sqrt{k} > 2k \left( \sqrt{k+\frac{1}{2}} - \sqrt{k-\frac{1}{2}} \right) + \left( \sqrt{k+\frac{1}{2}} + \sqrt{k-\frac{1}{2}} \right) \\\\ \frac{3}{2} \sqrt{k} > \left ( k+\frac{1}{2} \right ) \sqrt{k+\frac{1}{2}} - \left ( k-\frac{1}{2} \right ) \sqrt{k-\frac{1}{2}}$

So then it follows that:

$\frac{3}{2} \sum_{k=1}^{n} \sqrt{k} > \sum_{k=1}^n \left ( k+\frac{1}{2} \right ) \sqrt{k+\frac{1}{2}} - \sum_{k=1}^n \left ( k-\frac{1}{2} \right ) \sqrt{k-\frac{1}{2}}$

The right hand side yields to Telescoping and is equal to:

$n \sqrt{n+\frac{1}{2}}+\frac{1}{2} \left(\sqrt{n+\frac{1}{2}}-\frac{1}{\sqrt{2}} \right)$

which is obviously greater than

$n\sqrt{n}$

So we can therefore conclude:

$\sum_{k=0}^n \sqrt{k} > \frac{2}{3}n\sqrt{n}$

As desired.

trecex1 · Sep 29, 2016

Re: HSC 2016 3U Marathon

$\\ \text{Prove that }x\leq e^{x-1}\\ \text{Hence or otherwise:} \\\\ i)\text{ For }a_1,a_2...a_n > 0 \text{ then } \frac{a_1+a_2+...+a_n}{n}\geq \sqrt[n]{a_1a_2..a_n}\\ ii)\text{For }a,b,c,d >0 \text{ then }(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq 9 \\ iii) \text{For }a,b,c,> 0 \text{ then } \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq \frac{3}{2} \\ \\ \text{Bonus: more methods to prove }i)$

Drsoccerball · Sep 29, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
Preliminary Result:

$1. \, 3k^2 > 1 \quad \forall k \in \mathbb{N}$

Proof: Obvious

Now we begin.

$$\noindent$ 3k^2 - 1 > 0 \\\\ 64k^6 -48k^4 + 12k^2 -1 > 64k^6 -48k^4 + 9k^2 \\\\ (4k^2-1)^3 > (8k^2-3)^2k^2 \\ (4k^2-1)\sqrt{4k^2-1)} > 8k^3-3k \\\\ 3k > 8k^3 - 8k^2 \sqrt{k^2-\frac{1}{4}} + 2\sqrt{k^2-\frac{1}{4}} \\\\ 9k > 4k^2 \left( 2k - 2\sqrt{k^2-\frac{1}{4}} \right) + 4k + \left( 2k + 2\sqrt{k^2-\frac{1}{4}} \right) \\\\ (3\sqrt{k})^2 > (2k)^2 \left( \sqrt{k+\frac{1}{2}} - \sqrt{k-\frac{1}{2}} \right)^2 + 2\times 2k \times \left( \sqrt{k+\frac{1}{2}} - \sqrt{k-\frac{1}{2}} \right)\left( \sqrt{k+\frac{1}{2}} + \sqrt{k-\frac{1}{2}} \right) + \left( \sqrt{k+\frac{1}{2}} + \sqrt{k-\frac{1}{2}} \right)^2 \\\\ 3\sqrt{k} > 2k \left( \sqrt{k+\frac{1}{2}} - \sqrt{k-\frac{1}{2}} \right) + \left( \sqrt{k+\frac{1}{2}} + \sqrt{k-\frac{1}{2}} \right) \\\\ \frac{3}{2} \sqrt{k} > \left ( k+\frac{1}{2} \right ) \sqrt{k+\frac{1}{2}} - \left ( k-\frac{1}{2} \right ) \sqrt{k-\frac{1}{2}}$

So then it follows that:

$\frac{3}{2} \sum_{k=1}^{n} \sqrt{k} > \sum_{k=1}^n \left ( k+\frac{1}{2} \right ) \sqrt{k+\frac{1}{2}} - \sum_{k=1}^n \left ( k-\frac{1}{2} \right ) \sqrt{k-\frac{1}{2}}$

The right hand side yields to Telescoping and is equal to:

$n \sqrt{n+\frac{1}{2}}+\frac{1}{2} \left(\sqrt{n+\frac{1}{2}}-\frac{1}{\sqrt{2}} \right)$

which is obviously greater than

$n\sqrt{n}$

So we can therefore conclude:

$\sum_{k=0}^n \sqrt{k} > \frac{2}{3}n\sqrt{n}$

As desired.

Damn just do rectangles lel... Someone do the second limit.

HSC 2016 MX1 Marathon (archive) (3 Viewers)

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