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

iStudent · Mar 1, 2014

Re: HSC 2013 3U Marathon Thread

nightweaver066 said:
$$Show that $ (x+1)(x^2+1)(x^4+1)(x^8+1)(x^{16}+1) = 1 + x + x^2 + x^3 + ... + x^{31}$

RHS = 1 + x + x^2 + ... + x^31
= (x^32-1)/(x-1)
= (x^16-1)(x^16+1)/(x-1)
= (x^8-1)(x^8+1)(x^16+1)/(x-1)
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= (x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)(x^16+1)/(x-1)
= (x+1)(x^2+1)(x^4+1)(x^8+1)(x^16+1)
= LHS

Axio · Mar 20, 2014

Re: HSC 2013 3U Marathon Thread

dunjaaa · Jul 12, 2014

Re: HSC 2013 3U Marathon Thread

Question:

Screen shot 2014-07-12 at 11.37.12 PM.png

, just remembered a good question that I did last year in SHM

Squar3root · Jul 13, 2014

Re: HSC 2013 3U Marathon Thread

Axio said:
1. Expand:

$\left(\frac{x^3+x^2-4x-4}{x^2-x-2}\right)^{9}$

fuck no

Squar3root · Jul 13, 2014

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$\\ $Let$ \\ \\ x = \frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)} \\ \\ $Find$ \ \ \ (x+1)^{48}$

put it into the calculator. 2ez

dunjaaa · Jul 13, 2014

Re: HSC 2013 3U Marathon Thread

Screen shot 2014-07-13 at 1.56.46 AM.png

pls

Sy123 · Jul 28, 2014

Re: HSC 2013 3U Marathon Thread

$\\ $Using the Binomial Theorem, find all real roots of the polynomial$ \\\\ x^3 + 4x^2 + 6x + 4$

cub3root · Jul 28, 2014

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$\\ $Using the Binomial Theorem, find all real roots of the polynomial$ \\\\ x^3 + 4x^2 + 6x + 4$

By inspection; $(x + 2)$ is a factor and using polynomial division we find that the quotient is $x^2 + 2x + 2$

hence: $x^3 + 4x^2 + 6x + 4 = (x + 2)(x^2 + 2x + 2)$

braintic · Jul 28, 2014

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$\\ $Using the Binomial Theorem, find all real roots of the polynomial$ \\\\ x^3 + 4x^2 + 6x + 4$

As you said 'using the Binomial theorem', I assume you want something different to the previous answer.

Not sure exactly what you're looking for, but perhaps:

x³ + 4x² + 6x + 4 = (x³ + 6x² + 12x + 8) - 2x² - 6x - 4
=(x+2)³ - 2(x² + 3x + 2)
=(x+2)³ - 2(x+2)(x+1)
=(x+2) [ (x+2)² - 2(x+1) ]
=(x+2) (x² + 2x + 2)

So x=-2

dunjaaa · Jul 29, 2014

Re: HSC 2013 3U Marathon Thread

I think what Sy meant was this:
x^3+4x^2+6x+4 = 0
Multiplying top and bottom by x,
[(x+1)^4-1]/x = 0
x+1 = -1 (Since x cannot = 0)
Therefore, x=-2 is the only real root of this polynomial.

Sy123 · Aug 3, 2014

Re: HSC 2013 3U Marathon Thread

$\\ $Find the domain and range of the following functions$ \\ $i)$ \ f(x) = \sin^{-1}(x) + \sin^{-1} \left( \frac{1}{x} \right) \\ \\ $ii)$ \ g(x) = \tan^{-1}(x) + \tan^{-1} \left( \frac{1}{x} \right)$

aDimitri · Aug 3, 2014

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$\\ $Find the domain and range of the following functions$ \\ $i)$ \ f(x) = \sin^{-1}(x) + \sin^{-1} \left( \frac{1}{x} \right) \\ \\ $ii)$ \ g(x) = \tan^{-1}(x) + \tan^{-1} \left( \frac{1}{x} \right)$

$\\ i)\text{ } x = \pm1 \text{, } y = \pm\frac{\pi}{2} \\ ii)\text{ } x \neq 0 \text{, } y = \pm\frac{\pi}{2}$

dunjaaa · Aug 3, 2014

Re: HSC 2013 3U Marathon Thread

Find: lim (n->infinity) {[e^(1/n) + e^(2/n) + e^(3/n) + ... + e^(n/n)]/n}

glittergal96 · Aug 4, 2014

Re: HSC 2013 3U Marathon Thread

dunjaaa said:
Find: lim (n->infinity) {[e^(1/n) + e^(2/n) + e^(3/n) + ... + e^(n/n)]/n}

This is just a Riemann sum.

$\frac{1}{n}\sum_{k=1}^n e^{k/n}\rightarrow \int_0^1 e^x \, dx = e-1.$

dunjaaa · Aug 4, 2014

Re: HSC 2013 3U Marathon Thread

There was another way that I preferred (3u wise -> differentiation via first principles), but yeah nice work haha

Carrotsticks · Aug 4, 2014

Re: HSC 2013 3U Marathon Thread

dunjaaa said:
There was another way that I preferred (3u wise via first principles), but yeah nice work haha

First principles? As in differentiation by first principles?

dunjaaa · Aug 4, 2014

Re: HSC 2013 3U Marathon Thread

yeah thats what I meant

Sy123 · Aug 11, 2014

Re: HSC 2013 3U Marathon Thread

$\\ $Prove$ \ \sum_{k=1}^n \frac{(-1)^{k-1}}{k+1}\binom{n}{k} = \frac{n}{n+1}$

hi im trash · Aug 11, 2014

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$\\ $Prove$ \ \sum_{k=1}^n \frac{(-1)^{k-1}}{k+1}\binom{n}{k} = \frac{n}{n+1}$

Consider
$\ \sum_{r=0}^n \binom{n}{r}(-1)^r(x)^r = \(1-x)^n$

Integrate both sides
$\(-1)\frac{(1-x)^{n+1}}{n+1}+c=\sum_{r=0}^n \binom{n}{r}(-1)^{r}(x)^{r+1}\frac{1}{r+1}$

$\At x = 0, c =\frac{1}{n+1}$

Therefore:
$\ (-1)\frac{(1-x)^{n+1}}{n+1}+\frac{1}{n+1}=\sum_{r=0}^n \binom{n}{r}(-1)^{r}(x)^{r+1}\frac{1}{r+1}$

divide both sides by (-1)
$\ \frac{(1-x)^{n+1}}{n+1}-\frac{1}{n+1}=\sum_{r=0}^n \binom{n}{r}(-1)^{r-1}(x)^{r+1}\frac{1}{r+1}$

Expand RHS
$\ \sum_{r=0}^n \binom{n}{r}(-1)^{r-1}(x)^{r+1}\frac{1}{r+1} \rightarrow \binom{n}{0}(-1)^{-1}x^{1}+\sum_{r=1}^n \binom{n}{r}(-1)^{r-1}(x)^{r+1}\frac{1}{r+1}$

At x = 1
$\ -\frac{1}{n+1} = \binom{n}{0}(-1)^{-1}+\sum_{r=1}^n \binom{n}{r}(-1)^{r-1}\frac{1}{r+1}$
But
$\ \binom{n}{0}(-1)^{-1} = -1$
Therefore
$\ 1-\frac{1}{n+1} = \sum_{r=1}^n \binom{n}{r}(-1)^{r-1}\frac{1}{r+1}$

$\ \frac{n+1-1}{n+1}= \sum_{r=1}^n \binom{n}{r}(-1)^{r-1}\frac{1}{r+1}$

$\ \frac{n}{n+1}= \sum_{r=1}^n \binom{n}{r}(-1)^{r-1}\frac{1}{r+1}$

Sy123 · Aug 11, 2014

Re: HSC 2013 3U Marathon Thread

hi im trash said:
Consider
$\ \sum_{r=0}^n \binom{n}{r}(-1)^r(x)^r = \(1-x)^n$

Integrate both sides
$\(-1)\frac{(1-x)^{n+1}}{n+1}+c=\sum_{r=0}^n \binom{n}{r}(-1)^{r}(x)^{r+1}\frac{1}{r+1}$

$\At x = 0, c =\frac{1}{n+1}$

Therefore:
$\ (-1)\frac{(1-x)^{n+1}}{n+1}+\frac{1}{n+1}=\sum_{r=0}^n \binom{n}{r}(-1)^{r}(x)^{r+1}\frac{1}{r+1}$

divide both sides by (-1)
$\ \frac{(1-x)^{n+1}}{n+1}-\frac{1}{n+1}=\sum_{r=0}^n \binom{n}{r}(-1)^{r-1}(x)^{r+1}\frac{1}{r+1}$

Expand RHS
$\ \sum_{r=0}^n \binom{n}{r}(-1)^{r-1}(x)^{r+1}\frac{1}{r+1} \rightarrow \binom{n}{0}(-1)^{-1}x^{1}+\sum_{r=1}^n \binom{n}{r}(-1)^{r-1}(x)^{r+1}\frac{1}{r+1}$

At x = 1
$\ -\frac{1}{n+1} = \binom{n}{0}(-1)^{-1}+\sum_{r=1}^n \binom{n}{r}(-1)^{r-1}\frac{1}{r+1}$
But
$\ \binom{n}{0}(-1)^{-1} = -1$
Therefore
$\ 1-\frac{1}{n+1} = \sum_{r=1}^n \binom{n}{r}(-1)^{r-1}\frac{1}{r+1}$

$\ \frac{n+1-1}{n+1}= \sum_{r=1}^n \binom{n}{r}(-1)^{r-1}\frac{1}{r+1}$

$\ \frac{n}{n+1}= \sum_{r=1}^n \binom{n}{r}(-1)^{r-1}\frac{1}{r+1}$

Yes well done

---

$\\ $A number is divisible by 3 if the sum of its digits are divisible by 3$ \\ \\ $Find how many 4 digit numbers are even and divisible by 3$$

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