HSC 2015 MX1 Marathon (archive) (5 Viewers)

InteGrand · Oct 10, 2015

Re: HSC 2015 3U Marathon

Paradoxica said:
Now that everyone knows the answer to this question...

$Hence, find the least possible value of the radius.$

$What do you mean? The radius is $r=\frac{x^2 + y^2}{2x}$. If we take $y=x$ and tend $x$ to 0, the radius tends to 0.$

Paradoxica · Oct 10, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
$What do you mean? The radius is $r=\frac{x^2 + y^2}{2x}$. If we take $y=x$ and tend $x$ to 0, the radius tends to 0.$

I meant the least possible algebraic value.... I shall clarify.

InteGrand · Oct 10, 2015

Re: HSC 2015 3U Marathon

Paradoxica said:
I meant the least possible algebraic value.... I shall clarify.

$For any given $x$ and $y$ though, there is one and only one value $r$ may take, and it was found by several users on the previous page, i.e. $r=\frac{x^2 + y^2}{2x}$. Is the answer you are after that $r\geq y$?$

Paradoxica · Oct 10, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
$For any given $x$ and $y$ though, there is one and only one value $r$ may take, and it was found by several users on the previous page, i.e. $r=\frac{x^2 + y^2}{2x}$. Is the answer you are after that $r\geq y$?$

Yes.... but don't tell people how you got that... Should be easy though (changed the question again to lead to the correct answer).

InteGrand · Oct 10, 2015

Re: HSC 2015 3U Marathon

Paradoxica said:
Yes.... but don't tell people how you got that... Should be easy though.

Could change the wording to 'find a lower bound on the radius in terms of x and y'.

Crisium · Oct 10, 2015

Re: HSC 2015 3U Marathon

Paradoxica said:
Now that everyone knows the answer to this question...

$Hence, find any restrictions on the value of r$

Don't they only teach restrictions at ruse and other top tier selective schools for parametrics?

kawaiipotato · Oct 10, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
$For any given $x$ and $y$ though, there is one and only one value $r$ may take, and it was found by several users on the previous page, i.e. $r=\frac{x^2 + y^2}{2x}$. Is the answer you are after that $r\geq y$?$

was this how it was done?
y-x >= 0
(y-x)^2 >= 0
y^2 + x^2 >= 2xy
(x^2 + y^2)/2x >= y
r >= y

InteGrand · Oct 10, 2015

Re: HSC 2015 3U Marathon

kawaiipotato said:
was this how it was done?
y-x >= 0
(y-x)^2 >= 0
y^2 + x^2 >= 2xy
(x^2 + y^2)/2x >= y
r >= y

First line is unnecessary, since the second line is true always, as $y$ and $x$ are real numbers. But otherwise, yes. (Basically AM-GM Inequality)

InteGrand · Oct 10, 2015

Re: HSC 2015 3U Marathon

Crisium said:
Don't they only teach restrictions at ruse and other top tier selective schools for parametrics?

It comes up in the HSC and is assessable, e.g. look at Question 3(b) (iv) on Page 5 of the 2011 HSC Mathematics Extension 1 paper: http://www.boardofstudies.nsw.edu.a...ms/pdf_doc/2011-hsc-exam-mathematics-ext1.pdf

rand_althor · Oct 10, 2015

Re: HSC 2015 3U Marathon

$$Evaluate $\int^2_0 \left (\sqrt{\frac{4-x}{x}}-\sqrt{\frac{x}{4-x}} \right )~\mathrm{d}x$ using the substitution $u=4x-x^2$ or $x=4\sin^2{\theta}.$

Crisium · Oct 10, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$Evaluate $\int^2_0 \left (\sqrt{\frac{4-x}{x}}-\sqrt{\frac{x}{4-x}} \right )~\mathrm{d}x$ using the substitution $u=4x-x^2$ or $x=4\sin^2{\theta}.$

Using the x = 4sin^2(O) substitution you should get

[ cos^2(O) - sin^2(O) ] / sin(O)cos(O)

It's in the form of f'(x) / f(x)

So it should result in log(sin(O)cos(O)) and sub in the limits of integration - But these limits have to change to account for the substitution

rand_althor · Oct 10, 2015

Re: HSC 2015 3U Marathon

Crisium said:
Using the x = 4sin^2(O) substitution you should get

[ cos^2(O) - sin^2(O) ] / sin(O)cos(O)

It's in the form of f'(x) / f(x)

So it should result in log(sin(O)cos(O)) and sub in the limits of integration - But these limits have to change to account for the substitution

You've made a mistake in the bold part. Perhaps you have incorrectly substituted?

Crisium · Oct 10, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
You've made a mistake in the bold part. Perhaps you subbed in u and du incorrectly?

crap

I forgot to take into account the "differentiate with respect to part"

mega dept

photastic · Oct 10, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$Evaluate $\int^2_0 \left (\sqrt{\frac{4-x}{x}}-\sqrt{\frac{x}{4-x}} \right )~\mathrm{d}x$ using the substitution $u=4x-x^2$ or $x=4\sin^2{\theta}.$

By inspection the answer is 4.

kawaiipotato · Oct 11, 2015

Re: HSC 2015 3U Marathon

photastic said:
By inspection the answer is 4.

How did you inspect it?

Drsoccerball · Oct 11, 2015

Re: HSC 2015 3U Marathon

kawaiipotato said:
How did you inspect it?

Wolfram is the best way to inspect.

Paradoxica · Oct 11, 2015

Re: HSC 2015 3U Marathon

Find the least possible value of 4x^2 + 1/x when x is positive.

Zlatman · Oct 11, 2015

Re: HSC 2015 3U Marathon

Paradoxica said:
Find the least possible value of 4x^2 + 1/x when x is positive.

$$Differentiating: $ y = 4x^2 + \frac{1}{x}$

$\frac{dy}{dx} = 8x - \frac{1}{x^2}$

$When \frac{dy}{dx} = 0 \\ 8x^3 = 1 \\ x = \frac{1}{2}$

$$The second derivative is positive when $ x = \frac{1}{2}$

$$Therefore, the minimum value occurs when $ x = \frac{1}{2} $, i.e. $ y = 3$

$The least possible value of $ 4x^2 + \frac{1}{x} $ is $ 3 $ (when x is positive).$

EDIT: a bit of extra info may be needed.

As x approaches 0, y approaches 1/x, which tends towards positive infinity as x approaches 0.
As x approaches positive infinity, y approaches 4x^2, which tends towards positive infinity as x approaches positive infinity.

Since there are no vertical asymptotes for positive x, there must be a local minimum point between the two places the curve approaches positive infinity.

integral95 · Oct 11, 2015

Re: HSC 2015 3U Marathon

Using the symmetrical properties, find

$\int_{-a}^{a} cos^{-1}x \ \ dx \ \ \ \ \ \ -1\leq\ a \leq 1$

Paradoxica · Oct 11, 2015

Re: HSC 2015 3U Marathon

integral95 said:
Using the symmetrical properties, find

$\int_{-a}^{a} \cos^{-1}x \ \ dx \ \ \ \ \ \ -1\leq\ a \leq 1$

$= \int_{-a}^{a} \cos^{-1}(a+(-a)-x) dx = \int_{-a}^{a} \cos^{-1}(-x) dx = \int_{-a}^{a} \pi - \cos^{-1}(x) dx$

$2I = \int_{-a}^{a} \pi dx = 2a\pi$

$\therefore I = a\pi$

Ok, now where is my actual pie?

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