HSC 2015 MX1 Marathon (archive) (1 Viewer)

davidgoes4wce · Apr 30, 2015

Re: HSC 2015 3U Marathon

I'm actually a fan of the book despite there being errors in it. Any idea on what the recent edition is like(I think she released the 2015 paperback version)? Think she must have cut down on the errors.

photastic · Apr 30, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
I'm actually a fan of the book despite there being errors in it. Any idea on what the recent edition is like(I think she released the 2015 paperback version)? Think she must have cut down on the errors.

A fan of Maths In Focus??!!! enigma_1 help this poor soul.

enigma_1 · Apr 30, 2015

Re: HSC 2015 3U Marathon

wat

Are you like actually a fan?

:/

The questions are like band 2 material and compared to cambridge or fitzpatrick, nowhere near as much effort has been put into the book.

enigma_1 · Apr 30, 2015

Re: HSC 2015 3U Marathon

photastic said:
A fan of Maths In Focus??!!! enigma_1 help this poor soul.

Yeah I think he's been hypnotised by the font used in the textbook and the lame colour schemes.

photastic · Apr 30, 2015

Re: HSC 2015 3U Marathon

enigma_1 said:
Yeah I think he's been hypnotised by the font used in the textbook and the lame colour schemes.

Wth, the font is disgusting, especially the low definition diagrams.

enigma_1 · Apr 30, 2015

Re: HSC 2015 3U Marathon

photastic said:
Wth, the font is disgusting, especially the low definition diagrams.

Exactly. And those random pages where "perimeter" and "area" was defined...like is this fkn kindergarten??

matchalolz · May 6, 2015

Re: HSC 2015 3U Marathon

I'm not sure if this is 3u but it's part of a question in a 3u textbook so?
5t = 20ln(t+1)

how do i get an expression for t on its own so that i can just type it into a calculator to find an approximate value for t?
soz if this is stupid

Kaido · May 6, 2015

Re: HSC 2015 3U Marathon

so simplify first to get t=4ln(t+1)

use log laws to move the 4 then equate and solve using polynomials

matchalolz · May 6, 2015

Re: HSC 2015 3U Marathon

Kaido said:
so simplify first to get t=4ln(t+1)

use log laws to move the 4 then equate and solve using polynomials

i have no idea what you mean by this part

Librah · May 6, 2015

Re: HSC 2015 3U Marathon

Kaido said:
so simplify first to get t=4ln(t+1)

use log laws to move the 4 then equate and solve using polynomials

strong explanation.

Librah · May 6, 2015

Re: HSC 2015 3U Marathon

matchalolz said:
I'm not sure if this is 3u but it's part of a question in a 3u textbook so?
5t = 20ln(t+1)

how do i get an expression for t on its own so that i can just type it into a calculator to find an approximate value for t?
soz if this is stupid

Did the question actually ask to find an expression for t? If your trying to approximate, probably best to draw a graph or something.

FrankXie · May 6, 2015

Re: HSC 2015 3U Marathon

matchalolz said:
I'm not sure if this is 3u but it's part of a question in a 3u textbook so?
5t = 20ln(t+1)

how do i get an expression for t on its own so that i can just type it into a calculator to find an approximate value for t?
soz if this is stupid

there is an obvious solution which is t=0, by doing some calculus there is another one at around 10.35, but there is no formula for the solution thus you can not simply send it into calculator, but you can use Newton's method or other estimation approaches (even guess and check)

InteGrand · May 6, 2015

Re: HSC 2015 3U Marathon

matchalolz said:
I'm not sure if this is 3u but it's part of a question in a 3u textbook so?
5t = 20ln(t+1)

how do i get an expression for t on its own so that i can just type it into a calculator to find an approximate value for t?
soz if this is stupid

By inspection, t = 0 is a solution.

There is another root, but I don't know if we can find it by hand.

matchalolz · May 6, 2015

Re: HSC 2015 3U Marathon

thanks to everyone for trying anyway haha
for now i've just plugged it into wolfram alpha and used the answer for the further parts of the question

Librah · May 6, 2015

Re: HSC 2015 3U Marathon

matchalolz said:
thanks to everyone for trying anyway haha
for now i've just plugged it into wolfram alpha and used the answer for the further parts of the question

I believe it's impossible to solve for x for anything in the form lnx=x+c using algebraic methods.

Ekman · May 6, 2015

Re: HSC 2015 3U Marathon

matchalolz said:
I'm not sure if this is 3u but it's part of a question in a 3u textbook so?
5t = 20ln(t+1)

how do i get an expression for t on its own so that i can just type it into a calculator to find an approximate value for t?
soz if this is stupid

Well at least what I learnt in 2u about questions that asked you to solve x=e^x or x=lnx, you would always try guessing random values. Since this is from a 3u textbook, as Frank said, Newtons method of estimation may be required, however it still requires some form of random guessing for your starting value of course.

Trebla · May 9, 2015

Re: HSC 2015 3U Marathon

$\noindent$Suppose that $f(x)\geq 0$ and is continuous for some $a\leq x \leq b$. Show algebraically that$\\\\\displaystyle\int_a^b f(x)\,dx \geq \displaystyle\int_c^d f(x)\,dx\\\\$where $c \geq a$ and $d\leq b$$

leehuan · May 9, 2015

Re: HSC 2015 3U Marathon

$RTP:\quad \int _{ a }^{ b }{ f(x)dx } \ge \int _{ c }^{ d }{ f(x)dx }$

$i.e.\quad \int _{ a }^{ c }{ f(x)dx } +\int _{ c }^{ d }{ f(x)dx } +\int _{ d }^{ b }{ f(x)dx } \ge \int _{ c }^{ d }{ f(x)dx }$

$i.e.\quad \int _{ a }^{ c }{ f(x)dx } +\int _{ d }^{ b }{ f(x)dx } \ge 0$

$Method\quad 1\quad (possibly\quad invalid\quad as\quad it\quad does\quad assume\quad the\quad graph)$

$As\quad c\quad and\quad d\quad both\quad lie\quad in\quad the\quad domain\quad a\le x\le b,$

$we\quad deduce\quad that\quad \int _{ a }^{ c }{ f(x)dx } \ge 0\quad and\quad \int _{ d }^{ b }{ f(x)dx } \ge 0$

$Adding\quad these\quad two\quad results\quad directly\quad shows\quad$

$\int _{ a }^{ c }{ f(x)dx } +\int _{ d }^{ b }{ f(x)dx } \ge 0\quad is\quad true.$

$Method\quad 2$

$Define\quad F(x)\quad to\quad be\quad a\quad primitive\quad of\quad f(x)$

$Then,\quad by\quad the\quad fundamental\quad theorem\quad of\quad calculus$

$\int _{ a }^{ c }{ f(x)dx } =F(c)-F(a)\quad and\quad \int _{ d }^{ b }{ f(x)dx } =F(b)-F(d)$

$Adding\quad these\quad results\quad turns\quad our\quad RTP\quad into:$

$\left[ F(c)-F(a) \right] +\left[ F(b)-F(d) \right] \ge 0$

$Given\quad f(x)\ge 0\quad for\quad for\quad a\le x\le b\quad we\quad know\quad that\quad F(x)\quad is\quad increasing\quad or\quad stationary\quad for\quad a\le x\le b$

$Thus,\quad for\quad a\le x\le c\quad and\quad d\le x\le b\quad F(x)\quad is\quad also\quad increasing\quad or\quad stationary.$

$Therefore,\quad since\quad c>a,\quad F(c)\ge F(a)\quad so\quad F(c)-F(a)\ge 0\quad and\quad similarly\quad F(b)-F(d)\ge 0$

$Adding\quad these\quad results\quad directly\quad shows\quad \left[ F(c)-F(a) \right] +\left[ F(b)-F(d) \right] \ge 0\quad is\quad true$

$Thus\quad the\quad aforementioned\quad statements\quad are\quad also\quad true\quad since\quad the\quad fundamental\quad theorem\quad is\quad reversible\\$

Next question is me bumping a previous.

leehuan said:
$A\quad particle\quad is\quad moving\quad in\quad a\quad straight\quad line\quad with\quad displacement\quad x\quad metres\quad at\quad time\quad t\quad seconds.\\ The\quad acceleration\quad of\quad the\quad particle\quad is\quad given:\\ \ddot { x } =\frac { x }{ 2 } ({ x }^{ 2 }-1)$

$\\ Initially,\quad the\quad particle\quad is\quad at\quad the\quad origin\quad moving\quad at\quad v=\frac { 1 }{ 2 } m{ s }^{ -1 }$

$Find\quad an\quad expression\quad for\quad the\quad displacement\quad of\quad the\quad particle\quad at\quad time\quad t,\quad given\\ \frac { 2 }{ 1-{ x }^{ 2 } } =\frac { 1 }{ 1+x } +\frac { 1 }{ 1-x }$

leehuan · May 16, 2015

Re: HSC 2015 3U Marathon

$$Describe the locus of the midpoint of PQ on the parabola ${ y }^{ 2 }=4ax$ given that PQ is a focal chord. You may use the variable point $T\left( a{ t }^{ 2 },2at \right)$ to define P and Q.$

dunjaaa · May 17, 2015

Re: HSC 2015 3U Marathon

Probability question

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