HSC 2015 MX2 Marathon (archive) (6 Viewers)

Ekman · Oct 18, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let$ \ a,b \ $be positive real numbers so that$ \ a + b = 1 \\ \\ $Prove that$ \ \frac{a}{1+a} + \frac{b}{1+b} \leq \frac{2}{3}$

Rearranging the inequality will reduce down: $ab \leq \frac{1}{4}$

In order to prove it is true:

$$ Assume: $ x+y \geq2\sqrt{xy} $ (And substituting x and y for a and b) $ \Rightarrow \frac{1}{2} \geq \sqrt{ab} \Rightarrow \frac{1}{4} \geq ab$

So knowing that $ab \leq \frac{1}{4}$ you can backtrack on the initial rearrangements and prove the final result. (Of course under exam conditions, all assumptions must be proven, such as the AM-GM inequality for 2 terms in the 2nd step)

InteGrand · Oct 18, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Rearranging the inequality will reduce down: $ab \leq \frac{1}{4}$

In order to prove it is true:

$$ Assume: $ x+y \geq2\sqrt{xy} $ (And substituting x and y for a and b) $ \Rightarrow \frac{1}{2} \geq \sqrt{ab} \Rightarrow \frac{1}{4} \geq ab$

So knowing that $ab \leq \frac{1}{4}$ you can backtrack on the initial rearrangements and prove the final result. (Of course under exam conditions, all assumptions must be proven, such as the AM-GM inequality for 2 terms in the 2nd step)

$One can prove that $ab\leq 4$ by noting that $b=1-a$, so $ab=a(1-a)$, and this is a concave down quadratic which is maximised at its vertex at $a=\frac{1}{2}$, where the value is $\frac{1}{4}$.$

Ekman · Oct 18, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$One can prove that $ab\leq 4$ by noting that $b=1-a$, so $ab=a(1-a)$, and this is a concave down quadratic which is maximised at its vertex at $a=\frac{1}{2}$, where the value is $\frac{1}{4}$.$

That is how I initially proved it, but I felt more comfortable using the AM-GM inequality

InteGrand · Oct 18, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Rearranging the inequality will reduce down: $ab \leq \frac{1}{4}$

In order to prove it is true:

$$ Assume: $ x+y \geq2\sqrt{xy} $ (And substituting x and y for a and b) $ \Rightarrow \frac{1}{2} \geq \sqrt{ab} \Rightarrow \frac{1}{4} \geq ab$

So knowing that $ab \leq \frac{1}{4}$ you can backtrack on the initial rearrangements and prove the final result. (Of course under exam conditions, all assumptions must be proven, such as the AM-GM inequality for 2 terms in the 2nd step)

$Also, provided all steps were biconditional in order to get from the required inequality to $ab\leq 4$, logically speaking, it is valid to simply prove $ab\leq 4$ and this proves the requirement, no need to retrace the steps. However, I'm not sure the HSC markers accept this, even though it is mathematically and logically valid.$

InteGrand · Oct 18, 2015

Re: HSC 2015 4U Marathon

Ekman said:
That is how I initially proved it, but I felt more comfortable using the AM-GM inequality

Saves time in the HSC exam having to prove the AM-GM inequality by using the parabola way haha. (Both ways are valid though of course.)

Sy123 · Oct 18, 2015

Re: HSC 2015 4U Marathon

$\\ $Prove for positive reals$ \ x_1,x_2, \dots , x_n \\ \\ \left(\frac{x_1 + x_2 + \dots + x_n}{n \sqrt[n]{x_1x_2\dots x_n}}\right)^n \geq \left( \frac{x_1 + x_2 + \dots + x_{n-1}}{(n-1) \sqrt[n-1]{x_1 x_2\dots x_{n-1}}} \right)^{n-1}$

kawaiipotato · Oct 18, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\frac{1}{k} \geq \ln \left(\frac{1+k}{k} \right) = \ln ( k+1) - \ln k$

$\sum_{k=1}^{n} \frac{1}{k} > \ln 2 - \ln 1 + \ln 3 - \ln 2 + \dots + \ln (n+1) - \ln n = \ln (1 + n)$

Just a minor thing, why does the equality get removed?

Drsoccerball · Oct 18, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
Just a minor thing, why does the equality get removed?

I think it's the fact that we subbed $x = \frac{1}{k}$ ?

godofindolence · Oct 18, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$Set the speed of rotation to $\omega = 45 \text{ rpm} = 45 \times 2\pi \times \frac{1}{60}\text{ rad}/\text{s}=1.5\pi \text{ rad}/\text{s}$ (SI units)$

Sorry, this is from this question awhile ago:

A 3 metre string AB has a mass of 5kg attached at point B. The string is rotated in a horizontal circle about A and breaks as soon as it exceeds a speed of rotation of 45 revolutions per minute. Find the maximum possible tension in the string.

I don't think I learnt mechanics very well, so I have no idea what what is going on with the 2pi?
I did get 1.5 pi rad/s but I used v=rw and calculated v with distance over time (2*pi*3*45)/60, so is your method of finding w just a simplification of this?

InteGrand · Oct 18, 2015

Re: HSC 2015 4U Marathon

godofindolence said:
Sorry, this is from this question awhile ago:

A 3 metre string AB has a mass of 5kg attached at point B. The string is rotated in a horizontal circle about A and breaks as soon as it exceeds a speed of rotation of 45 revolutions per minute. Find the maximum possible tension in the string.

I don't think I learnt mechanics very well, so I have no idea what what is going on with the 2pi?
I did get 1.5 pi rad/s but I used v=rw and calculated v with distance over time (2*pi*3*45)/60, so is your method of finding w just a simplification of this?

$My method is simply converting rpm's (which is the units we are given the speed of rotation in) to radians per second:$

$$45$ rpm $= 45\text{ }\frac{\text{rev}}{\text{min}}$$

$$=45\times \frac{2\pi\text{ rad}}{60\text{ s}}$ (since one revolution is $2\pi$ radians and one minute is $60$ seconds)$

$=45\times 2\pi \times \frac{1}{60}\text{ rad}/\text{s}$

$$=1.5 \pi\text{ rad}/\text{s}$.$

godofindolence · Oct 18, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$My method is simply converting rpm's (which is the units we are given the speed of rotation in) to radians per second:$

$$45$ rpm $= 45\text{ }\frac{\text{rev}}{\text{min}}$$

$$=45\times \frac{2\pi\text{ rad}}{60\text{ s}}$ (since one revolution is $2\pi$ radians and one minute is $60$ seconds)$

$=45\times 2\pi \times \frac{1}{60}\text{ rad}/\text{s}$

$$=1.5 \pi\text{ rad}/\text{s}$.$

Oh thank you, that makes sense now!!

glittergal96 · Oct 18, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Prove for positive reals$ \ x_1,x_2, \dots , x_n \\ \\ \left(\frac{x_1 + x_2 + \dots + x_n}{n \sqrt[n]{x_1x_2\dots x_n}}\right)^n \geq \left( \frac{x_1 + x_2 + \dots + x_{n-1}}{(n-1) \sqrt[n-1]{x_1 x_2\dots x_{n-1}}} \right)^{n-1}$

$AM-GM in $n$ variables tells us:\\ \\ $\textrm{AM}(x_1,\ldots,x_{n-1})^{(n-1)/n}\cdot x_n^{1/n}\leq \frac{(n-1)\cdot \textrm{AM}(x_1,\ldots,x_{n-1}) +x_n}{n}=\textrm{AM}(x_1,\ldots,x_n).$\\ \\ Dividing both sides by $\textrm{GM}(x_1,\ldots,x_n)$ and raising to the power of $n$ them tells us \\ \\ $\left(\frac{\textrm{AM}(x_1,\ldots,x_n)}{\textrm{GM}(x_1,\ldots,x_n)}\right)^n\geq \frac{\textrm{AM}(x_1,\ldots,x_{n-1})^{n-1}\cdot x_n}{GM(x_1,\ldots,x_n)^n}=\left(\frac{\textrm{AM}(x_1,\ldots,x_{n-1})}{\textrm{GM}(x_1,\ldots,x_{n-1})}\right)^{n-1}$\\ \\ as required.$

omegadot · Oct 18, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let$ \ a,b \ $be positive real numbers so that$ \ a + b = 1 \\ \\ $Prove that$ \ \frac{a}{1+a} + \frac{b}{1+b} \leq \frac{2}{3}$

$Here is another approach, though it is a little longer compared to other suggested solutions.$\\ \begin{align*}\frac{a}{1+ a} + \frac{b}{1 + b} &= \frac{1 + 2ab}{2 + ab}, \quad \mbox{since} \,\, a + b = 1\\&= \frac{2a^2 - 2a - 1}{a^2 - a - 2}, \quad \mbox{since} \,\, b = 1 - a\\&= 2 \cdot \frac{2a^2 - 2a - 4 + 3}{2a^2 - 2a - 4}\\&= 2 + \frac{3}{a^2 - a - 2}\end{align*}\\$Now$ f(a) = a^2 - a - 2$ is a quadratic function with a local minimum at $a = 1/2. \,\,$Hence$\\ \begin{align*}a^2 - a -2 &\geqslant \left (\frac{1}{2} \right )^2 - \left (\frac{1}{2} \right ) - 2 = - \frac{9}{4}\\\Rightarrow \frac{1}{a^2 - a - 2} &\leqslant -\frac{4}{9}\end{align*}\\$Giving$\\\frac{a}{1+a} + \frac{b}{1 + b} \leqslant 2 - \frac{4}{3} = \frac{2}{3},\,\,$as required to show.$

omegadot · Oct 18, 2015

Re: HSC 2015 4U Marathon

$$If$\,\ z \in \mathbb{C}\,\, $such that$\,\, \frac{1}{2} \leqslant |z| \leqslant 4, \,\,$find the sum of the greatest and least value of\,\,$ \left |z + \frac{1}{z} \right |.$

Drsoccerball · Oct 18, 2015

Re: HSC 2015 4U Marathon

omegadot said:
$$If$\,\ z \in \mathbb{C}\,\, $such that$\,\, \frac{1}{2} \leqslant |z| \leqslant 4, \,\,$find the sum of the greatest and least value of\,\,$ \left |z + \frac{1}{z} \right |.$

This has been asked before as well

omegadot · Oct 18, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
This has been asked before as well

$\noindent$Mmmm....What about if$\,\, z \in \mathbb{C}, z \neq 0,\,\,$such that$\\ \mbox{arg} \left (-\frac{1}{z} \right ) = - \frac{2\pi}{3} \quad \mbox{and} \quad \left |1 - \frac{2}{z} \right | = 1,\,\,$find$\,\,z.$

Silly Sausage · Oct 18, 2015

Re: HSC 2015 4U Marathon

University mechanics.
Given $v(t)=\frac{g}{k}(1-e^{-kt})$
When x=0, t=0, find x(t).
Using series, prove that as k tends towards 0, $v=gt$ and $x=\frac{gt^2}{2}$ .

kawaiipotato · Oct 18, 2015

Re: HSC 2015 4U Marathon

omegadot said:
$$Mmmm....What about if$\,\, z \in \mathbb{C}, z \neq 0,\,\,$such that$\\ \mbox{arg} \left (-\frac{1}{z} \right ) = - \frac{2\pi}{3} \quad \mbox{and} \quad \left |1 - \frac{2}{z} \right | = 1,\,\,$find$\,\,z.$

I got z = 1 - sqrt3 i?

Paradoxica · Oct 18, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
I got z = 1 - sqrt3 i?

That looks correct.

Glyde · Oct 18, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
I got z = 1 - sqrt3 i?

Did you guys look at these questions at the start of your hsc year and wwant to crawl in a hole?

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