HSC 2016 MX1 Marathon (archive) (1 Viewer)

Flop21 · Feb 17, 2016

Re: HSC 2016 3U Marathon

"In how many different ways can 3 boys and 3 girls sit in a circle if:

a) the boys and girls alternate

"

Just need this one also explained, thanks.

Paradoxica · Feb 17, 2016

Re: HSC 2016 3U Marathon

Flop21 said:
"In how many different ways can 3 boys and 3 girls sit in a circle if:

a) the boys and girls alternate

"

Just need this one also explained, thanks.

Consider the boys and girls separately. Firstly, place the boys every second position, and consider their possible combinations. (Ignore the spaces, treat it as you would a normal circle of people) Secondly, consider the remaining places for the girls to be seated in, assuming the boys have all been seated.

Paradoxica · Feb 17, 2016

Re: HSC 2016 3U Marathon

$$\noindent$1. f(x) = \frac{x^2-6x+6}{2x-4} \,\,\, g(x) = \frac{ax^2+bx+c}{x-d} \\ $The following information is known about $g(x). \\ g(x) $ and $ f(x) $ have the same vertical asymptotes.$ \\ $The oblique asymptotes are perpendicular, and intersect on the $y$-axis. $ \\ f(x) = g(x) $ has a solution at $x = -2. \\ $Find $g(x)$ and hence find the other solution to $f(x) = g(x)\\$2. $x+\frac{1}{x}$ is an integer. Show by induction, or any other means, that $x^n + \frac{1}{x^n}$ is an integer $ \forall n \in \mathbb{N}\\ $3. a) Show $\tan{\theta}-\cot{\theta} \equiv 2\tan{2\theta}\\ $b) Prove the identity: $8\cot{8x}+4\tan{4x}+2\tan{2x}+\tan{x}-\cot{x} \equiv 0 \\$ c) Prove by induction: $ \forall n \in \mathbb{N}\\2^n \cot{2^n x} + 2^{n-1} \tan{2^{n-1} x} + 2^{n-2} \tan{2^{n-2} x} +\cdots 2\tan{2x}+\tan{x}-\cot{x} \equiv 0$

Please just attempt these, they are not that difficult, and they are all straightforward.

I feel like people aren't taught to be unafraid of new problems, instead are taught to rote everything, then freak out when they encounter something foreign.

InteGrand · Feb 17, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
$$\noindent$1. f(x) = \frac{x^2-6x+6}{2x-4} \,\,\, g(x) = \frac{ax^2+bx+c}{x-d} \\ $The following information is known about $g(x). \\ g(x) $ and $ f(x) $ have the same vertical asymptotes.$ \\ $The oblique asymptotes are perpendicular, and intersect on the $y$-axis. $ \\ f(x) = g(x) $ has a solution at $x = -2. \\ $Find $g(x)$ and hence find the other solution to $f(x) = g(x)\\$2. $x+\frac{1}{x}$ is an integer. Show by induction, or any other means, that $x^n + \frac{1}{x^n}$ is an integer $ \forall n \in \mathbb{N}\\ $3. a) Show $\tan{\theta}-\cot{\theta} \equiv 2\tan{2\theta}\\ $b) Prove the identity: $8\cot{8x}+4\tan{4x}+2\tan{2x}+\tan{x}-\cot{x} \equiv 0 \\$ c) Prove by induction: $ \forall n \in \mathbb{N}\\2^n \cot{2^n x} + 2^{n-1} \tan{2^{n-1} x} + 2^{n-2} \tan{2^{n-2} x} +\cdots 2\tan{2x}+\tan{x}-\cot{x} \equiv 0$

Please just attempt these, they are not that difficult, and they are all straightforward.

I feel like people aren't taught to be unafraid of new problems, instead are taught to rote everything, then freak out when they encounter something foreign.

Either that, or there's not too many 2016'ers who can be bothered to answer Q's on these marathons.

Flop21 · Feb 17, 2016

Re: HSC 2016 3U Marathon

Can someone use Newton's method on "sin(x) = x/5" starting at x = 2.5.

InteGrand · Feb 17, 2016

Re: HSC 2016 3U Marathon

Flop21 said:
Can someone use Newton's method on "sin(x) = x/5" starting at x = 2.5.

$\noindent Newton's method is used to find roots to an equation of the form $f(x) = 0$ (i.e. need to make the R.H.S. 0). So for your equation, it's equivalent to $\sin x -\frac{x}{5}=0$. So just use Newton's method with $f(x)=\sin x -\frac{x}{5}$, and initial guess the given starting value of 2.5.$

Flop21 · Feb 17, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
$\noindent Newton's method is used to find roots to an equation of the form $f(x) = 0$ (i.e. need to make the R.H.S. 0). So for your equation, it's equivalent to $\sin x -\frac{x}{5}=0$. So just use Newton's method with $f(x)=\sin x -\frac{x}{5}$, and initial guess the given starting value of 2.5.$

"form f(x)=0"

Oh right, that's an important bit to know. Thanks!

Flop21 · Feb 18, 2016

Re: HSC 2016 3U Marathon

Again with Newton's method:

"Consider the function y= -x^3 +x^2 +x +0.5"

"Starting with the initial value of Xo=0 use Newton's method to find the next 3 approximations.

So when I use the method and put x=0, I keep getting -0.5 which according to the answers is the second approximation. How do I get the initial one they're asking for? (Answer is 0).

InteGrand · Feb 18, 2016

Re: HSC 2016 3U Marathon

Flop21 said:
Again with Newton's method:

"Consider the function y= -x^3 +x^2 +x +0.5"

"Starting with the initial value of Xo=0 use Newton's method to find the next 3 approximations.

So when I use the method and put x=0, I keep getting -0.5 which according to the answers is the second approximation. How do I get the initial one they're asking for? (Answer is 0).

$\noindent What do you mean about how to find the initial one? The initial one is given to us. Just keep iterating that process (do it three times) so that you obtain the next three approximations, $x_1, x_2$ and $x_3$. So once you find the first approximation $x_1$, sub. this into the Newton's Method formula to get $x_2$. Then sub. \emph{this} in to obtain $x_3$. The formula is $x_{n} = x_{n-1} - \frac{f\left(x_{n-1}\right)}{f^\prime \left(x_{n-1}\right)}$, $n=1,2,3,\ldots$.$

Flop21 · Feb 18, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
$\noindent What do you mean about how to find the initial one? The initial one is given to us. Just keep iterating that process (do it three times) so that you obtain the next three approximations, $x_1, x_2$ and $x_3$. So once you find the first approximation $x_1$, sub. this into the Newton's Method formula to get $x_2$. Then sub. \emph{this} in to obtain $x_3$.$

Ohh. I did not realise Xo=0 was them telling us that's the first value. I thought they meant sub that in. But you do that to get the next one anyway.

Thanks I understand now.

leehuan · Feb 18, 2016

Re: HSC 2016 3U Marathon

Flop21 said:
Ohh. I did not realise Xo=0 was them telling us that's the first value. I thought they meant sub that in. But you do that to get the next one anyway.

Thanks I understand now.

It's convention for 0 in the subscript to be the initial. Kinda like series actually, except T1 is the first term there cause you don't have a zero-th term

davidgoes4wce · Feb 18, 2016

Re: HSC 2016 3U Marathon

Flop21 said:
Again with Newton's method:

"Consider the function y= -x^3 +x^2 +x +0.5"

"Starting with the initial value of Xo=0 use Newton's method to find the next 3 approximations.

So when I use the method and put x=0, I keep getting -0.5 which according to the answers is the second approximation. How do I get the initial one they're asking for? (Answer is 0).

$$ The iterative method is based on Newton Raphson Method which states that $ x_{n+1}=x_{n}-\frac{f(x_{n})}{f' (x_{n})}$

$Given \ x_{1}=0 , f(x)=-x^3+x^2+x+0.50, f'(x)=-3x^2+2x+1 , $we substitute all terms into $ x_{2}=x_{1}-\frac{f(x)}{f' (x)}$

$x_{2}=0 - [\frac{-x^3+x^2+x+0.50}{-3x^2+2x+1}]$

$x_{2}=0 - [\frac{-(0.00)^3+(0)^2+(0)+0.50}{-3(0)^2+2(0)+1}] =-0.50$

$$ Do the same thing for $ x_{3}, x_{3}=x_{2}-\frac{f(x)}{f' (x)} , $ except we know use $ x_{2}=-0.50$

$x_{3}=0 - [\frac{-(-0.50)^3+(-0.50)^2+(-0.50)+0.50}{-3(-0.50)^2+2(-0.50)+1}] =0. $We have found $ x_{3}=0$

Flop21 · Feb 18, 2016

Re: HSC 2016 3U Marathon

Differentiate sin^-1 x/3

I thought I knew how to do these, but am not getting the answer for this one.

InteGrand · Feb 18, 2016

Re: HSC 2016 3U Marathon

Flop21 said:
Differentiate sin^-1 x/3

I thought I knew how to do these, but am not getting the answer for this one.

$\frac{1}{\sqrt{9-x^2}}$

davidgoes4wce · Feb 18, 2016

Re: HSC 2016 3U Marathon

Flop21 said:
Differentiate sin^-1 x/3

I thought I knew how to do these, but am not getting the answer for this one.

$$ The rule for the inverse sine derivative $ \frac{d}{dx} (sin^{-1} \frac{x}{a})= \frac{1}{\sqrt{a^2-x^2}}$

$\frac{d}{dx} (sin^{-1} \frac{x}{3})=\frac{1}{\sqrt{3^2-x^2}}$

$= \frac{1}{\sqrt{9-x^2}}$

Flop21 · Feb 18, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
$\frac{1}{\sqrt{9-x^2}}$

Yes but can someone show working out please, I know the answer, just don't know how to reach the answer.

InteGrand · Feb 18, 2016

Re: HSC 2016 3U Marathon

Flop21 said:
Yes but can someone show working out please, I know the answer, just don't know how to reach the answer.

It's a standard differentiation formula as davidgoes4wce showed in his post.

Flop21 · Feb 18, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
$$ The rule for the inverse sine derivative $ \frac{d}{dx} (sin^{-1} \frac{x}{a})= \frac{1}{\sqrt{a^2-x^2}}$

$\frac{d}{dx} (sin^{-1} \frac{x}{3})=\frac{1}{\sqrt{3^2-x^2}}$

$= \frac{1}{\sqrt{9-x^2}}$

Oh thanks. I don't remember being taught that rule. Only that sin^-1 x = 1/root(1-x^2).

Makes sense knowing that though now.

InteGrand · Feb 18, 2016

Re: HSC 2016 3U Marathon

Flop21 said:
Oh thanks. I don't remember being taught that rule. Only that sin^-1 x = 1/root(1-x^2).

Makes sense knowing that though now.

$\noindent It could still be done via the chain rule if one only knew the rule for $\frac{\mathrm{d}}{\mathrm{d}x}\left(\sin^{-1} x \right)$.$

davidgoes4wce · Feb 18, 2016

Re: HSC 2016 3U Marathon

Flop21 said:
Oh thanks. I don't remember being taught that rule. Only that sin^-1 x = 1/root(1-x^2).

Makes sense knowing that though now.

Need to memorise that, that's one of the formulas not listed on the 2016 Formulae Sheet.

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