HSC 2013-14 MX1 Marathon (archive) (2 Viewers)

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)
.
.
.
.
= (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$$

HSC 2013-14 MX1 Marathon (archive) (2 Viewers)

Well-Known Member

=o

Active Member

realest nigga

realest nigga

Active Member

This too shall pass

Well-Known Member

Well-Known Member

Active Member

This too shall pass

i'm the cook

Active Member

Active Member

Active Member

Retired

Active Member

This too shall pass

Member

This too shall pass

Users Who Are Viewing This Thread (Users: 0, Guests: 2)