HSC 2016 Maths Marathon (archive) (3 Viewers)

Drsoccerball · Apr 5, 2016

Re: HSC 2016 2U Marathon

Paradoxica said:
You haven't specified a variable.

He obviously means in terms of x...

Nailgun · Apr 5, 2016

Re: HSC 2016 2U Marathon

astroman said:
Differentiate
y = cos(a⁶+x⁶)

$\noindent$ Assuming $a$ is a constant$$

$\frac { d }{ dx } \cos { f\left( x \right) } =-f'\left( x \right) \sin { f\left( x \right) } \\ \\ y=\cos { ({ a }^{ 6 }+{ x }^{ 6 }) } \\ \\ \frac { dy }{ dx } =-6{ x }^{ 5 }\sin { ({ a }^{ 6 }+{ x }^{ 6 }) }$

InteGrand · Apr 5, 2016

Re: HSC 2016 2U Marathon

Here's another Q about averages.

$\noindent Let $n$ be an integer greater than $1$ and let $x_1,x_2,\ldots , x_n$ be $n$ real numbers. Let $w_1,w_2,\ldots , w_n$ be $n$ real numbers satisfying $w_i \geq 0$ for all $i$ and $\sum _{i=1} ^{n} w_i = 1$. Let $WA = \sum _{i=1}^{n} w_i x_i$ (this sort of sum is called a \textit{weighted average} of the $x_i$'s; the $w_i$'s are the weights). Show that $\min \left\{ x_1,x_2,\ldots x_n \right \} \leq WA \leq \max \left\{ x_1,x_2,\ldots x_n \right \}$.$

(min refers to the minimum of the set, and max refers to the maximum).

Is the above necessarily true if some of the w_i's are allowed to be negative (and still sum to 1)?

astroman · Apr 7, 2016

Re: HSC 2016 2U Marathon

differentiate:
$g(u)=\sqrt{10}u + \sqrt{3u}$

Nailgun · Apr 7, 2016

Re: HSC 2016 2U Marathon

astroman said:
differentiate:
$g(u)=\sqrt{10}u + \sqrt{3u}$

$g\left( u \right) =u\sqrt { 10 } +u^{ \frac { 1 }{ 2 } }\sqrt { 3 } \\ \\ g'\left( u \right) =\sqrt { 10 } +\frac { \sqrt { 3 } }{ 2\sqrt { u } } \\ \\ g'\left( u \right) =\sqrt { 10 } +\frac { \sqrt { 3u } }{ 2u }$

astroman · Apr 8, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
$g\left( u \right) =u\sqrt { 10 } +u^{ \frac { 1 }{ 2 } }\sqrt { 3 } \\ \\ g'\left( u \right) =\sqrt { 10 } +\frac { \sqrt { 3 } }{ 2\sqrt { u } } \\ \\ g'\left( u \right) =\sqrt { 10 } +\frac { \sqrt { 3u } }{ 2u }$

The answer has no u in the numerator and, sqrt u in the denominator.

leehuan · Apr 8, 2016

Re: HSC 2016 2U Marathon

astroman said:
The answer has no u in the numerator and, sqrt u in the denominator.

Which is the exact same thing according to the process of rationalising the denominator.

Nailgun · Apr 8, 2016

Re: HSC 2016 2U Marathon

astroman said:
The answer has no u in the numerator and, sqrt u in the denominator.

do you mean my second line lele

astroman · Apr 8, 2016

Re: HSC 2016 2U Marathon

dw all good

leehuan · Apr 10, 2016

Re: HSC 2016 2U Marathon

$$Find $A+B+C+D$ given that $\\x=\log_{D}{\left(\frac{A}{1+{B}^{C}}\right)}$ if and only if$\\ { 2 }^{ x }+{ 2 }^{ x+1 }+{ 2 }^{ x+2 }+\dots +{ 2 }^{ x+2015 }={ 4 }^{ x }+{ 4 }^{ x+1 }+{ 4 }^{ x+2 }+\dots +{ 4 }^{ x+2015 }$

Nailgun · Apr 10, 2016

Re: HSC 2016 2U Marathon

leehuan said:
$$Find $A+B+C+D$ given that $\\x=\log_{D}{\left(\frac{A}{1+{B}^{C}}\right)}$ if and only if$\\ { 2 }^{ x }+{ 2 }^{ x+1 }+{ 2 }^{ x+2 }+\dots +{ 2 }^{ x+2015 }={ 4 }^{ x }+{ 4 }^{ x+1 }+{ 4 }^{ x+2 }+\dots +{ 4 }^{ x+2015 }$

why didn't you post this earlier lelele
there was a question like this in half-yearly, and it was the only one i didnt understand
now i get it lelel

nobody else got it either though so lel

okie solutions coming

Paradoxica · Apr 10, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
why didn't you post this earlier lelele
there was a question like this in half-yearly, and it was the only one i didnt understand
now i get it lelel

nobody else got it either though so lel

okie solutions coming

You should have used "by inspection"

leehuan · Apr 10, 2016

Re: HSC 2016 2U Marathon

Paradoxica said:
You should have used "by inspection"

This is honestly more effective when you say it in real life than over the internet I swear

leehuan · Apr 10, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
why didn't you post this earlier lelele
there was a question like this in half-yearly, and it was the only one i didnt understand
now i get it lelel

nobody else got it either though so lel

okie solutions coming

Because I first saw the question this morning when I woke up

Nailgun · Apr 10, 2016

Re: HSC 2016 2U Marathon

just realised that i got answers through a mistake i made at the beginning so i have to redo
its not that much work lelel but i ceeebs

Ambility · Apr 10, 2016

Re: HSC 2016 2U Marathon

$\\2^x+2^{x+1}+2^{x+2}+\cdots +2^{x+2015}=4^x+4^{x+1}+4^{x+2}+\cdots +4^{x+2015}\\2^x(1+2+2^2+2^3+\cdots +2^{2015})=2^{2x}(1+4+4^2+4^3+\cdots 4^{2016})\\1+2+2^2+2^3+\cdots +2^{2015}=2^x(1+4+4^2+4^3+\cdots 4^{2016})\\\frac{1(2^{2016}-1)}{2-1}=2^x(\frac{4^{2016}-1}{4-1})\\2^x=\frac{2^{2016}-1}{\frac{4^{2016}-1}{3}}\\2^x=3\times\frac{2^{2016}-1}{2^{4032}-1}=3\times\frac{2^{2016}-1}{(2^{2016}-1)(2^{2016}+1)}\\2^x=\frac{3}{(2^{2016}+1)}\\$So, log$_2(\frac{3}{2^{2016}+1})=x\\A=3$, $B=2$, $C=2016$, $D=2\\\\\textbf{Next Question}\\$Find the derivative of $y=x^x$

Paradoxica · Apr 10, 2016

Re: HSC 2016 2U Marathon

Ambility said:
$$Find the derivative of $y=x^x$

$\noindent Power Rule: $ x \cdot x^{x-1} = x^x $ - FALSE$\\ $Exponential Rule: $ x^x \log{x} $ - FALSE$ \\ $Sum of these: $(1+\log{x})x^x $ - TRUE$

si2136 · Apr 10, 2016

Re: HSC 2016 2U Marathon

Ambility said:
$Find the derivative of $y=x^x$

leehuan · Apr 10, 2016

Re: HSC 2016 2U Marathon

Paradoxica said:
$Sum of these: $(1+\log{x})x^x $ - TRUE$

What. This isn't a coincidence right

Paradoxica · Apr 10, 2016

Re: HSC 2016 2U Marathon

leehuan said:
What. This isn't a coincidence right

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