HSC 2016 MX1 Marathon (archive) (4 Viewers)

Drsoccerball · Mar 18, 2016

Re: HSC 2016 3U Marathon

leehuan said:
$$Use the inverse function theorem to find $\frac{d}{dx}\sec^{-1}{x} \\ $Verify your answer by redefining $y=\sec^{-1}{x}$

$$ By 'inverse function theorem' do you mean: $ sec^{-1}x=cos^{-1}(\frac{1}{x})?$

InteGrand · Mar 18, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
$$ By 'inverse function theorem' do you mean: $ sec^{-1}x=cos^{-1}(\frac{1}{x})?$

$\noindent He means the ``$\frac{\mathrm{d}x}{\mathrm{d}y} = \frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}}$ rule''.$

jathu123 · Mar 18, 2016

Re: HSC 2016 3U Marathon

I have 1 quick perms and combs question which I'm kinda stuck on, it may not be that hard but I don't want to create a new thread just for 1 question, so ill post it here.

Find the number of ways in which the letters of the word SQUARE can be arranged in a straight line so that exactly two vowels are next to each other.

The answer is 432 and since I got this from a past paper, the method they did was 12*3!*3!
Really appreciate if anyone could help me out on how they got that answer, thanks

ml125 · Mar 18, 2016

Re: HSC 2016 3U Marathon

jathu123 said:
I have 1 quick perms and combs question which I'm kinda stuck on, it may not be that hard but I don't want to create a new thread just for 1 question, so ill post it here.

Find the number of ways in which the letters of the word SQUARE can be arranged in a straight line so that exactly two vowels are next to each other.

The answer is 432 and since I got this from a past paper, the method they did was 12*3!*3!
Really appreciate if anyone could help me out on how they got that answer, thanks

3! x 3! is ordering the vowels and consonants as separate groups. You then multiply this by 12 as this is the possible number of arrangements of the two groups, as follows:

xx_x__
xx__x_
xx___x
_xx_x_
_xx__x
x_xx__
__xx_x
x__xx_
_x_xx_
x___xx
_x__xx
__x_xx

Pretty sure there might be a quicker way to do this but I just did it by listing lol

jathu123 · Mar 18, 2016

Re: HSC 2016 3U Marathon

ml125 said:
3! x 3! is ordering the vowels and consonants as separate groups. You then multiply this by 12 as this is the possible number of arrangements of the two groups, as follows:

xx_x__
xx__x_
xx___x
_xx_x_
_xx__x
x_xx__
__xx_x
x__xx_
_x_xx_
x___xx
_x__xx
__x_xx

Pretty sure there might be a quicker way to do this but I just did it by listing lol

You made it look so simple haha, thankyou!

Hooters · Mar 20, 2016

Re: HSC 2016 3U Marathon

Is my proof correct for this question? and can it be simplified?

leehuan · Mar 21, 2016

Re: HSC 2016 3U Marathon

Not sure if this can be done with HSC level maths but I'll just chuck it here.

$Prove that the exponential function increases faster than any polynomial ever will.$

Paradoxica · Mar 21, 2016

Re: HSC 2016 3U Marathon

leehuan said:
Not sure if this can be done with HSC level maths but I'll just chuck it here.

$Prove that the exponential function increases faster than any polynomial ever will.$

The higher order derivatives of any polynomial will eventually vanish.

This means that the values of the lesser derivatives are eventually overtaken by the exponential function.

But the lesser functions are arbitrary, and so we can jump inductively to the derivative of the polynomial in question.

Thus, the value of the derivative is eventually outstripped by the exponential function.

So the exponential function itself will eventually overtake the polynomial.

Can't be bothered constructing a formal argument, but if somebody wants to fill in the details, feel free to do so.

astroman · Mar 21, 2016

Re: HSC 2016 3U Marathon

help, forgot.differentiate

InteGrand · Mar 21, 2016

Re: HSC 2016 3U Marathon

astroman said:
help, forgot.

$\noindent Note $z=Ay^{-10}+B \mathrm{e}^{y}$. So $\frac{\mathrm{d}z}{\mathrm{d}y}=-10Ay^{-11}+B \mathrm{e}^{y}=-\frac{10A}{y^{11}}+B \mathrm{e}^{y}$.$

astroman · Mar 21, 2016

Re: HSC 2016 3U Marathon

cheers

parad0xica · Mar 28, 2016

Re: HSC 2016 3U Marathon

Let's get the ball rolling again with this question:

$P(2ap, ap^2)$ is a point on the parabola $x^2 = 4ay.$

$$Let $\theta$ be the acute angle between the tangent at $P$ and the line $SP$, which joins $P$ with the focus $S$. Show that $\cos \theta = \frac{|p|}{\sqrt{1+p^2}}.$

Modification: you cannot use $\tan ^2 x + 1 = \sec ^2 x$ or similar identities

Drsoccerball · Mar 29, 2016

Re: HSC 2016 3U Marathon

parad0xica said:
Let's get the ball rolling again with this question:

$P(2ap, ap^2)$ is a point on the parabola $x^2 = 4ay.$

$$Let $\theta$ be the acute angle between the tangent at $P$ and the line $SP$, which joins $P$ with the focus $S$. Show that $\cos \theta = \frac{|p|}{\sqrt{1+p^2}}.$

Hint: $\tan^2{x} +1 = sec^2{x} $ May be needed. $$

parad0xica · Mar 29, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
Hint: $\tan^2{x} +1 = sec^2{x} $ May be needed. $$

I'm going to modify this question: you cannot use this identity (or similar ones) to help you

There is a method without using that trig identity and I want that person to think of it instead of using memory

Paradoxica · Mar 29, 2016

Re: HSC 2016 3U Marathon

parad0xica said:
Let's get the ball rolling again with this question:

$P(2ap, ap^2)$ is a point on the parabola $x^2 = 4ay.$

$$Let $\theta$ be the acute angle between the tangent at $P$ and the line $SP$, which joins $P$ with the focus $S$. Show that $\cos \theta = \frac{|p|}{\sqrt{1+p^2}}.$

Use the following diagram as a reference image:

By the reflective property of the parabola, the angle between the tangent and the focal chord is equal to the angle between the abscissa line and the tangent.

From the reference diagram, solve for the points B and D.

$$\noindent$B(ap,0)\\D(2ap,0)$

The lengths BE, DE are respectively:

$$\noindent$ |BE| = a|p|\sqrt{p^2+1}\\|DE|=ap^2$

$$\noindent$ \cos{\theta} =\left |\frac{DE}{BE}\right |= \frac{|p|}{\sqrt{p^2+1}}$

Paradoxica · Mar 29, 2016

Re: HSC 2016 3U Marathon

$$\noindent 1. $x+\frac{1}{x}$ is an integer. Show by induction, or any other means, that $x^n + \frac{1}{x^n}$\\ \indent is an integer $ \forall n \in \mathbb{N}\\ $2. a) Show $\tan{\theta}-\cot{\theta} \equiv 2\tan{2\theta}\\ $\indent b) Prove the identity: $8\cot{8x}+4\tan{4x}+2\tan{2x}+\tan{x}-\cot{x} \equiv 0 \\$\indent c) Prove by induction: $ \forall n \in \mathbb{N}\\ \indent 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$

InteGrand · Apr 1, 2016

Re: HSC 2016 3U Marathon

$\noindent Let $n$ be a positive integer. Using mathematical induction or otherwise, show that $\frac{\mathrm{d} ^n}{\mathrm{d} x^{n}} \left(x^n\right) = n!$ (i.e. the $n$-th derivative is $n!$).$

leehuan · Apr 1, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
$\noindent Let $n$ be a positive integer. Using mathematical induction or otherwise, show that $\frac{\mathrm{d} ^n}{\mathrm{d} x^{n}} \left(x^n\right) = n!$ (i.e. the $n$-th derivative is $n!$).$

I remember this one.

Use a derivative rule

leehuan · Apr 5, 2016

Re: HSC 2016 3U Marathon

Thought about this question just now. Not sure if it can be done.

$$Let ${L}_{n}\left(x\right)=\underbrace{\ln \left({\ln\left({\ln{\left( \dots \ln{x}\right)}}\right)}\right)}_{n\text{ applications of log}}\\ $Use mathematical induction to show that $\\ \frac{d}{dx} {L}_{n}\left(x\right)=\frac{1}{{ L }_{ 0 }\left( x \right) { L }_{ 1 }\left( x \right) { L }_{ 2 }\left( x \right) { L }_{ 3 }\left( x \right) \dots { L }_{ n-1 }\left( x \right)} \\ $ for integers $n\ge 1$

KingOfActing · Apr 5, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
$\noindent Let $n$ be a positive integer. Using mathematical induction or otherwise, show that $\frac{\mathrm{d} ^n}{\mathrm{d} x^{n}} \left(x^n\right) = n!$ (i.e. the $n$-th derivative is $n!$).$

$Base case $ n = 1 $: $\\\frac{d}{dx}x = 1 = 1!\\$Assume the result for $n = k \geq 1\\\frac{d^{k+1}}{dx^{k+1}} x^{k+1} = \frac{d^k}{dx^k}\frac{d}{dx}x^{k+1} = \frac{d^k}{dx^k}(k+1)x^k = (k+1)\frac{d^k}{dx^k}x^k = (k+1)k! = (k+1)!\\$Q.E.D.$

HSC 2016 MX1 Marathon (archive) (4 Viewers)

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Active Member

New Member

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Active Member

-insert title here-

-insert title here-

Well-Known Member

Well-Known Member

Well-Known Member

lukewarm mess

Users Who Are Viewing This Thread (Users: 0, Guests: 4)