HSC 2016 Maths Marathon (archive) (3 Viewers)

Paradoxica · Jan 11, 2016

Re: HSC 2016 2U Marathon

Well basically, the fractions are ratios, so it means the numerators and denominators are proportional. Equate using a constant of proportionality and then you can add them together.

InteGrand · Jan 11, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Okay I gave it a shot, but I'm pretty sure my logic is a bit faulty
I tried to solve it geometrically as well, I think. Is that what you meant by geometrically?
http://imgur.com/Mn29MRz

$$\noindent Your idea of letting $\frac{a}{b}=\frac{c}{d}=k$ was a good one! However, we shouldn't also let $\frac{a+c}{b+d}=k$, because then we are essentially letting $\frac{a+c}{b+d}=\frac{a}{b}=\frac{c}{d}$, which means we are assuming what we want to prove. Rather, we should start from $\frac{a}{b}=\frac{c}{d}=k$ and use this to \textit{show} that $\frac{a+c}{b+d}$ must also equal $k$. I.e. something like this$

$\noindent Let $\frac{a}{b}=\frac{c}{d}=k$, so $a=bk$ and $c=dk$. Then we have$

$\begin{align*}\frac{a+c}{b+d}&= \frac{bk+dk}{b+d} \\ &= \frac{k(b+d)}{b+d} \\ &= k \quad (\text{as }b+d\neq 0).\end{align*}$

$\noindent Hence $\frac{a+c}{b+d} =k= \frac{a}{b}=\frac{c}{d}$. $\blacksquare$$

Nailgun · Jan 11, 2016

Re: HSC 2016 2U Marathon

Roight, so basically you can't equate them from the beginning because that defeats the purpose of the question,
but you can sub in expressions for a and c, and then simplify to get it to equal k

InteGrand · Jan 11, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Roight, so basically you can't equate them from the beginning because that defeats the purpose of the question,
but you can sub in expressions for a and c, and then simplify to get it to equal k

Yeah

Green Yoda · Jan 11, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
$$\noindent Your idea of letting $\frac{a}{b}=\frac{c}{d}=k$ was a good one! However, we shouldn't also let $\frac{a+c}{b+d}=k$, because then we are essentially letting $\frac{a+c}{b+d}=\frac{a}{b}=\frac{c}{d}$, which means we are assuming what we want to prove. Rather, we should start from $\frac{a}{b}=\frac{c}{d}=k$ and use this to \textit{show} that $\frac{a+c}{b+d}$ must also equal $k$. I.e. something like this$

$\noindent Let $\frac{a}{b}=\frac{c}{d}=k$, so $a=bk$ and $c=dk$. Then we have$

$\begin{align*}\frac{a+c}{b+d}&= \frac{bk+dk}{b+d} \\ &= \frac{k(b+d)}{b+d} \\ &= k \quad (\text{as }b+d\neq 0).\end{align*}$

$\noindent Hence $\frac{a+c}{b+d} =k= \frac{a}{b}=\frac{c}{d}$. $\blacksquare$$

That's bloody amazing. Where can I find more questions like this? Also is this just normal algebra or some special topic?

InteGrand · Jan 11, 2016

Re: HSC 2016 2U Marathon

Rathin said:
That's bloody amazing. Where can I find more questions like this? Also is this just normal algebra or some special topic?

I don't know, I just thought it up. I suppose it's normal algebra. There's a lot of cool algebraic facts like this you can think up if you play around a bit, a lot of which you probably intuitively know are true actually. Here are some about averages:

$\noindent Suppose $x_1,x_2,\ldots, x_n$ is a list of $n$ real numbers. Suppose their average is $\mu$ (i.e. $\frac{x_1 + x_2 + \cdots + x_n}{n}=\mu$). Prove the following:$

$(a) Suppose we add a new real number $x_{n+1}$ to the list. Let the new average be $\mu_2$. Then if $x_{n+1}>\mu$, we have $\mu_2 > \mu$, if $x_{n+1}=\mu$, we have $\mu_2 = \mu$, and if $x_{n+1}<\mu$, we have $\mu_2 < \mu$.$

$(b) Suppose we consider the list $cx_1,cx_2,\ldots , cx_n$, where $c$ is a real number. The average of this list of numbers is $c\mu$.$

$(c) Suppose we consider the list $x_1+c,x_2+c,\ldots , x_n+c$, where $c$ is a real number. The average of this list of numbers is $\mu+c$.$

$\noindent A similar one: Show that in a game of cricket, if a team has a required run rate of $r$ (runs per over) at the start of the over, and hits more than $r$ runs in that over, their required run rate will go down, whilst if they hit fewer than $r$ runs that over, their required run rate will go up. Also show that if they hit exactly $r$ runs that over, their required run rate stays at $r$. You may use any of the results above.$

Also:

$\noindent Suppose there are $K$ classes in a cohort (where $K\in \mathbb{Z}^+$, i.e. $K$ is a positive integer), and the number of students in class $i$ is $n_i\in \mathbb{Z}^+$, for $i=1,2,\ldots ,K$. Let $N = \sum _{i=1}^{K} n_i$ (this is the number of students in the grade). Suppose that for a recent yearly exam, the average mark for class $i$ was $\mu_i$. Show that the average mark in the grade was $\mu = \frac{n_1}{N}\mu_1 + \frac{n_2}{N}\mu_2+\cdots +\frac{n_K}{N} \mu_K$ (this sort of sum is known as a \textsl{weighted average}.).$

Nailgun · Jan 12, 2016

Re: HSC 2016 2U Marathon

DatAtarLyfe said:
its k, join the club.
Next question, for 2016ers ONLY (wanna see how many can get it, found it relatively easy)
View attachment 32641

Sorry for pulling this out from a while ago
Is it necessary to prove that H and R are perpendicular? Or is it assumed because they're the height and radius..?

EDIT: For InteGrands q, for b) can you just state that multiplying by c/c gets you that result?

InteGrand · Jan 12, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Sorry for pulling this out from a while ago
Is it necessary to prove that H and R are perpendicular? Or is it assumed because they're the height and radius..?

$\noindent Don't need to prove it, it's assumed.$

InteGrand · Jan 12, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Sorry for pulling this out from a while ago
Is it necessary to prove that H and R are perpendicular? Or is it assumed because they're the height and radius..?

EDIT: For InteGrands q, for b) can you just state that multiplying by c/c gets you that result?

$\noindent We wouldn't multiple by $\frac{c}{c}$, but rather we'd multiply both sides of the equation by $c$ (I guess that's what you meant). Another way to structure the proof is to say that the average of the new list is:$

$\begin{align*}\mu _{\text{new}} &= \frac{cx_1 + cx_2 + \cdots+ cx_n}{n} \quad (\text{by definition of average}) \\ &= \frac{c\left(x_1 + x_2 + \cdots + x_n \right)}{n}\quad (\text{factorising}) \\ &= c\mu \quad \Big{(}\text{as }\mu = \frac{x_1 + x_2 + \cdots + x_n}{n}\Big{)}.\end{align*}$

Drsoccerball · Jan 12, 2016

Re: HSC 2016 2U Marathon

New Question:

$$ Find : $ \frac{d}{dx} (\pi^{\sin^n{x}}+e^{\sin^n{x}})^n \cdot \cos{(nx)}$

$If you're bothered find the double derivative.$

InteGrand · Jan 12, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
New Question:

$$ Find : $ \frac{d}{dx} (\pi^{\sin^n{x}}+e^{\sin^n{x}})^n \cdot \cos{(nx)}$

$If you're bothered find the double derivative.$

That has to be one of the most tedious 2U questions I've ever seen (especially the second derivative one).

Drsoccerball · Jan 12, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
That has to be one of the most tedious 2U questions I've ever seen (especially the second derivative one).

$Actually given $ 2\sin{(x)}\cos{(x)}=\sin{(2x)} $ find the double derivative.$

EDIT: My bad my writing was messy misread what I wrote

Nailgun · Jan 13, 2016

Re: HSC 2016 2U Marathon

Wait, what is the question lol

InteGrand · Jan 13, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Wait, what is the question lol

Here are the ones about averages, mostly unanswered:

InteGrand said:
I don't know, I just thought it up. I suppose it's normal algebra. There's a lot of cool algebraic facts like this you can think up if you play around a bit, a lot of which you probably intuitively know are true actually. Here are some about averages:

$\noindent Suppose $x_1,x_2,\ldots, x_n$ is a list of $n$ real numbers. Suppose their average is $\mu$ (i.e. $\frac{x_1 + x_2 + \cdots + x_n}{n}=\mu$). Prove the following:$

$(a) Suppose we add a new real number $x_{n+1}$ to the list. Let the new average be $\mu_2$. Then if $x_{n+1}>\mu$, we have $\mu_2 > \mu$, if $x_{n+1}=\mu$, we have $\mu_2 = \mu$, and if $x_{n+1}<\mu$, we have $\mu_2 < \mu$.$

$(b) Suppose we consider the list $cx_1,cx_2,\ldots , cx_n$, where $c$ is a real number. The average of this list of numbers is $c\mu$.$

$(c) Suppose we consider the list $x_1+c,x_2+c,\ldots , x_n+c$, where $c$ is a real number. The average of this list of numbers is $\mu+c$.$

$\noindent A similar one: Show that in a game of cricket, if a team has a required run rate of $r$ (runs per over) at the start of the over, and hits more than $r$ runs in that over, their required run rate will go down, whilst if they hit fewer than $r$ runs that over, their required run rate will go up. Also show that if they hit exactly $r$ runs that over, their required run rate stays at $r$. You may use any of the results above.$

Also:

$\noindent Suppose there are $K$ classes in a cohort (where $K\in \mathbb{Z}^+$, i.e. $K$ is a positive integer), and the number of students in class $i$ is $n_i\in \mathbb{Z}^+$, for $i=1,2,\ldots ,K$. Let $N = \sum _{i=1}^{K} n_i$ (this is the number of students in the grade). Suppose that for a recent yearly exam, the average mark for class $i$ was $\mu_i$. Show that the average mark in the grade was $\mu = \frac{n_1}{N}\mu_1 + \frac{n_2}{N}\mu_2+\cdots +\frac{n_K}{N} \mu_K$ (this sort of sum is known as a \textsl{weighted average}.).$

.

Here's Drsoccerball's one about finding a derivative:

Drsoccerball said:
New Question:

$$ Find : $ \frac{d}{dx} (\pi^{\sin^n{x}}+e^{\sin^n{x}})^n \cdot \cos{(nx)}$

$If you're bothered find the double derivative.$

.

Nailgun · Jan 13, 2016

Re: HSC 2016 2U Marathon

$Ok\quad so\quad I\quad tried\quad Drsoccerball's\quad question\quad (I\quad don't\quad know\quad where\quad to\quad start\quad for\quad the\quad averages\quad ones)\\ I\quad realised\quad while\quad doing\quad it\quad I\quad don't\quad know\quad how\quad to\quad differentiate\quad { a }^{ x }\quad or\quad in\quad the\quad question\quad { \pi }^{ \sin ^{ n }{ x } }\\ My\quad textbook\quad only\quad says\quad that\quad if\quad \frac { d }{ dx } { a }^{ x }=\quad { a }^{ x }\quad \times \quad m\quad where\quad m\quad =\quad gradient\quad at\quad y-intercept\quad of\quad y={ a }^{ x }\\ Also\quad I\quad tried\quad checking\quad my\quad answer\quad in\quad Wolfram\quad Alpha\quad and\quad I\quad don't\quad know\quad if\quad I\quad just\quad didn't\quad simplify\quad correctly\\ or\quad if\quad I'm\quad just\quad plain\quad wrong\quad (probably)\\ \\ Halp\quad pls$

http://imgur.com/YnI54Hb

whoo that post took a lot of effort
sorry for my retarded latex, ill try get that sorted out lel
also ignore the crossing out (except the scribbles and the last z=)(sorry its when i realised that i didnt know how to differentiate a^x)

Shuuya · Jan 13, 2016

Re: HSC 2016 2U Marathon

I think this is how you differentiate a^x

leehuan · Jan 13, 2016

Re: HSC 2016 2U Marathon

Shuuya said:
I think this is how you differentiate a^x

View attachment 32793

True.

Now prove it.

Nailgun · Jan 13, 2016

Re: HSC 2016 2U Marathon

Oh jokes now I remember, it was in the section after logarithmic functions
$a={ e }^{ \log _{ e }{ a } }\\ \therefore \quad { a }^{ x }={ e }^{ \log _{ e }{ { a }^{ x } } }\\ \therefore { \quad a }^{ x }={ e }^{ x\log _{ e }{ { a } } }\\ \therefore \quad \frac { dy }{ dx } { a }^{ x }=\frac { dy }{ dx } { e }^{ x\log _{ e }{ { a } } }\\ \frac { dy }{ dx } { a }^{ x }={ e }^{ x\log _{ e }{ { a } } }.\log _{ e }{ a } \\ ={ a }^{ x }.\log _{ e }{ a }$

But that still doesn't help me with $\frac { d }{ dx } { \pi }^{ \sin ^{ n }{ x } }$ because its not a coefficient of x?

InteGrand · Jan 13, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Oh jokes now I remember, it was in the section after logarithmic functions
$a={ e }^{ \log _{ e }{ a } }\\ \therefore \quad { a }^{ x }={ e }^{ \log _{ e }{ { a }^{ x } } }\\ \therefore { \quad a }^{ x }={ e }^{ x\log _{ e }{ { a } } }\\ \therefore \quad \frac { dy }{ dx } { a }^{ x }=\frac { dy }{ dx } { e }^{ x\log _{ e }{ { a } } }\\ \frac { dy }{ dx } { a }^{ x }={ e }^{ x\log _{ e }{ { a } } }.\log _{ e }{ a } \\ ={ a }^{ x }.\log _{ e }{ a }$

But that still doesn't help me with $\frac { d }{ dx } { \pi }^{ \sin ^{ n }{ x } }$ because its not a coefficient of x?

$\noindent To differentiate something like $\pi ^{\sin ^{n} {x}}$ with respect to $x$, we would use the chain rule. Try letting $u =\sin ^n x$ (and remember that this is just $\left(\sin x \right)^n$ when differentiating $u$).$

leehuan · Jan 13, 2016

Re: HSC 2016 2U Marathon

Note that for a question as tedious as this, you actually WOULD rather set out the full working of

dy/dx = dy/du du/dx

Rather than just be smart and use the function of a function shorthand. Because you get lost too easily.

HSC 2016 Maths Marathon (archive) (3 Viewers)

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