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

HeroicPandas · Sep 9, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$$A projectile is fired from the origin with initial velocity$ \ \ U$

$$to pass through the point$ \ (a,b)$

$Prove that there are 2 possible trajectories if$

$(U^2-gb)^2 > g^2(a^2+b^2) \ \ \ \ \fbox{5}$

Difficulty: Q14

$\rightarrow \\ \\ 1.$ Derive 6 equations of motion \\ \\ Trajectory: y = $\frac{-gx^2}{2U^2 cos^2\alpha} + xtan\alpha \\ $Since it passess through (a,b),\\ \\ \therefore b = $\frac{-ga^2}{2U^2 cos^2\alpha} + atan\alpha \\ \\ \therefore \frac{ga^2}{2U^2}tan^2\alpha - atan\alpha + \left ( \frac{ga^2}{2U^2} + b \right ) =0 \\ \\ $For 2 trajectories, there must be 2 distinct values of \alpha,$ $ hence \Delta >0$

$\bullet \\ \\ \therefore a^2 - 4\left ( \frac{ga^2}{2U^2}+b \right )\left ( \frac{ga^2}{2U^2} \right ) >0 \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\frac{1}{4} > \left ( \frac{g^2a^2}{4U^2}+ \frac{gb}{2U^2} \right ) \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~U^4 > g^2a^2 + 2gb \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~U^4 + g^2b^2 > g^2a^2 + 2gb + g^2b^2 \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~(U^2 - gb)^2 > g^2(a^2 + b^2)$

Sy123 · Sep 9, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
$\rightarrow \\ \\ 1.$ Derive 6 equations of motion \\ \\ Trajectory: y = $\frac{-gx^2}{2U^2 cos^2\alpha} + xtan\alpha \\ $Since it passess through (a,b),\\ \\ \therefore b = $\frac{-ga^2}{2U^2 cos^2\alpha} + atan\alpha \\ \\ \therefore \frac{ga^2}{2U^2}tan^2\alpha - atan\alpha + \left ( \frac{ga^2}{2U^2} + b \right ) =0 \\ \\ $For 2 trajectories, there must be 2 distinct values of \alpha,$ $ hence \Delta >0$

$\bullet \\ \\ \therefore a^2 - 4\left ( \frac{ga^2}{2U^2}+b \right )\left ( \frac{ga^2}{2U^2} \right ) >0 \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\frac{1}{4} > \left ( \frac{g^2a^2}{4U^2}+ \frac{gb}{2U^2} \right ) \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~U^4 > g^2a^2 + 2gb \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~U^4 + g^2b^2 > g^2a^2 + 2gb + g^2b^2 \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~(U^2 - gb)^2 > g^2(a^2 + b^2)$

Yep well done

$Show that$

$\sum_{k=1}^n (k^2+1)k! = n(n+1)!$

drunkenmonkee · Sep 9, 2013

Re: HSC 2013 3U Marathon Thread

.

HeroicPandas · Sep 9, 2013

Re: HSC 2013 3U Marathon Thread

can u prove it w/o induction?

drunkenmonkee · Sep 9, 2013

Re: HSC 2013 3U Marathon Thread

2+10+60+...+(n^2 +1)n! = n(n+1)!
@n=1, lhs=rhs=2
assume true for n=k, (k^2 +1)k! = k(k+1)!
prove true fro n= k+1, k(k+1)! + (k^2 +2k+2)(k+1)! (from prev step)
= (k+1)!(k^2 +3k +2) = (k+2)(k+1)(k+1)!
RHS= (k+1)(k+2)! = (k+2)(k+1)(k+1)!

drunkenmonkee · Sep 9, 2013

Re: HSC 2013 3U Marathon Thread

it was the first procedure that came to mind

Sy123 · Sep 9, 2013

Re: HSC 2013 3U Marathon Thread

Without induction, we take the sum from 1 to n of:

$k(k+1)! - (k-1)k! = (k^2+1)k!$

HeroicPandas · Sep 9, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Without induction, we take the sum from 1 to n of:

$k(k+1)! - (k-1)k! = (k^2+1)k!$

ah ok, thanks

omgiloverice · Sep 9, 2013

Re: HSC 2013 3U Marathon Thread

$$i)\ \ \ Given that $a=y+d $ use the binomial expansion of $(y+d)^{n},$ where $n$ is positive integer, to show that $a^{n}-y^{n-1}(y+nd)$ is divisible by $d^{2}$

$$ii)\ \ \ Hence show that if $a$ is the first term, $d$ the common difference, $L$ the $n$-th term of an arithmetic progression, then $a^{n}-L(a-d)^{n-1}$ is divisible by $d^{2}$

omgiloverice · Sep 16, 2013

Re: HSC 2013 3U Marathon Thread

Here is a fun question to burn off..

In this video here five men has 5 different colours of clothing to choose from, what is the probability that all the men come out wearing a different colour.

asianese · Sep 22, 2013

Re: HSC 2013 3U Marathon Thread

Is the answer 1/35?

seanieg89 · Sep 22, 2013

Re: HSC 2013 3U Marathon Thread

5!/(5^5)=24/625.

omgiloverice · Sep 22, 2013

Re: HSC 2013 3U Marathon Thread

I'm going with seanieg89 answer.

$$Using first principles prove that $ \frac{d}{dx}(x^{n})=nx^{n-1}$

asianese · Sep 22, 2013

Oh Yeh oops. Was thinking 4c4/7c4 for some reason.

HeroicPandas · Sep 22, 2013

Re: HSC 2013 3U Marathon Thread

omgiloverice said:
I'm going with seanieg89 answer.

$$Using first principles prove that $ \frac{d}{dx}(x^{n})=nx^{n-1}$

the easiest way to do this: (the time i wrote this, my internet got cut but luckily i saved the URL for this image)
$\rightarrow \\ \frac{dy}{dx} = \lim_{h\rightarrow 0} \frac{f(x+h)-f(h)}{h} \\ \\ = \lim_{h\rightarrow 0} \frac{(x+h)^n - h^n}{h} \\ \\ = \lim_{h\rightarrow 0} \frac{\left ( \binom{n}{0}x^n + \binom{n}{1}x^{n-1} + \dots + \binom{n}{n}h^n \right )-h^n}{h} \\ \\ = \lim_{h\rightarrow 0} \frac{\left ( \binom{n}{0}x^n + \binom{n}{1}x^{n-1}h + \binom{n}{2}x^{n-2}h \dots + \binom{n}{n-1}xh^n \right )}{h} ($as $\binom{n}{n}=1) \\ \\ = \lim_{h\rightarrow 0} \binom{n}{1}x^{n-1} + \frac{\binom{n}{0}x^n + \binom{n}{2}x^{n-2}+ \dots + \binom{n}{n-1}xh^n}{h} \\ \\ = nx^{n-1} + 0 ($as $\binom{n}{1}=n)$

seanieg89 · Sep 22, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
the easiest way to do this: (the time i wrote this, my internet got cut but luckily i saved the URL for this image)
$\rightarrow \\ \frac{dy}{dx} = \lim_{h\rightarrow 0} \frac{f(x+h)-f(h)}{h} \\ \\ = \lim_{h\rightarrow 0} \frac{(x+h)^n - h^n}{h} \\ \\ = \lim_{h\rightarrow 0} \frac{\left ( \binom{n}{0}x^n + \binom{n}{1}x^{n-1} + \dots + \binom{n}{n}h^n \right )-h^n}{h} \\ \\ = \lim_{h\rightarrow 0} \frac{\left ( \binom{n}{0}x^n + \binom{n}{1}x^{n-1}h + \binom{n}{2}x^{n-2}h \dots + \binom{n}{n-1}xh^n \right )}{h} ($as $\binom{n}{n}=1) \\ \\ = \lim_{h\rightarrow 0} \binom{n}{1}x^{n-1} + \frac{\binom{n}{0}x^n + \binom{n}{2}x^{n-2}+ \dots + \binom{n}{n-1}xh^n}{h} \\ \\ = x^{n-1} + 0 ($as $\binom{n}{1}=1)$

Hence prove that

$\frac{d}{dx}\left(x^r\right)=rx^{r-1}$

for all rational $r$ .

Carrotsticks · Sep 22, 2013

Re: HSC 2013 3U Marathon Thread

seanieg89 said:
Hence prove that

$\frac{d}{dx}\left(x^r\right)=rx^{r-1}$

for all rational $r$ .

I was THIS |-| close to putting this question in the BOS Trial.

Sy123 · Sep 22, 2013

Re: HSC 2013 3U Marathon Thread

seanieg89 said:
Hence prove that

$\frac{d}{dx}\left(x^r\right)=rx^{r-1}$

for all rational $r$ .

$$Let$ \ \ r=\frac{p}{q} \ \ $for integers$ \ \ p, q \ \ $with no common divisor except$ \ \ \pm 1$

$y = x^{p/q}$

$y^q= x^p$

$Since we have proven$ \ \ \frac{d}{dz} z^n = nz^{n-1} \ \ $and we assume chain rule$

$\frac{d}{dx} y^q = \frac{d}{dx} x^p$

$q y^{q-1} y' = px^{p-1}$

$y' = \frac{p}{q} \times \frac{x^{p-1}}{(x^{\frac{p}{q}} )^{q-1}} = \frac{p}{q} x^{p-1 - p/q ( q-1)} = \frac{p}{q} x^{p/q - 1} = rx^{r-1}$

--------

$Show that$ \ \frac{d}{dx} \sin x = \cos x \ \ $using first principles$

seanieg89 · Sep 22, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$Show that$ \ \frac{d}{dx} \sin x = \cos x \ \ $using first principles$

Good question!

Carrotsticks · Sep 22, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$$Let$ \ \ r=\frac{p}{q} \ \ $for integers$ \ \ p, q \ \ $with no common divisor except$ \ \ \pm 1$

To shorten it, you can say that p and q are 'relatively prime' or 'coprime'.

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