HSC 2013 Maths Marathon (archive) (3 Viewers)

omgiloverice · May 11, 2013

2 unit people deserve a marathon thread too...
I'll start off with a typical hsc troll question (but abit harder)

$A point P(x,y) moves such that it is equidistant from y = 1 and A(p , q)$

Find the equation at which P moves

Carrotsticks · May 12, 2013

Re: HSC 2013 2U Marathon

$\\ $By using the Trapezoidal Rule, with four function values, on the curve $ y=\frac{1}{x} $ in the domain $ 1 \leq x \leq 2 $, approximate $ \ln 2 $ correct to three decimal places.$$

Registered User · May 12, 2013

Re: HSC 2013 2U Marathon

$\int \sin(3x)dx$

I haven't done 2U so I don't know how difficult the questions are supposed to be. This is an easy one.

Registered User · May 12, 2013

Re: HSC 2013 2U Marathon

$$Solve$

$\cos(x)=\cos(3x)$

omgiloverice · May 12, 2013

Re: HSC 2013 2U Marathon

Registered User said:
$\int \sin(3x)dx$

I haven't done 2U so I don't know how difficult the questions are supposed to be. This is an easy one.

2U people ain't that stupid...
$-\frac{1}{3} cos(3x) +c$

Registered User said:
$$Solve$

$\cos(x)=\cos(3x)$

0?

Reloman · May 12, 2013

Re: HSC 2013 2U Marathon

Consider the two functions: y=mx-4 and y=x^3
Find the value of 'm' such that it is a tangent to the function y=x^3

Makematics · May 12, 2013

Re: HSC 2013 2U Marathon

Registered User said:
$$Solve$

$\cos(x)=\cos(3x)$

General solutions aren't in the 2 unit course?

Makematics · May 12, 2013

Re: HSC 2013 2U Marathon

Carrotsticks said:
$\\ $By using the Trapezoidal Rule, with four function values, on the curve $ y=\frac{1}{x} $ in the domain $ 1 \leq x \leq 2 $, approximate $ \ln 2 $ correct to three decimal places.$$

$$Let I$ ~=\int_{1}^{2}\frac{1}{x}~dx\\\\~~~~~~~~~~~~~=[\ln(x)]_1^2\\\\~~~~~~~~~~~~~=\ln(2)-\ln(1)\\\\~~~~~~~~~~~~~=\ln(2)\\\\$I$ ~\approx\frac{h}{2}~[y_{0}+y_{3}+2(y_{1}+y_{2})]\\\\x_{0}=1,~y_{0}=1\\\\x_{1}=\frac{4}{3},~y_{1}= \frac{3}{4} \\\\x_{2}=\frac{5}{3},~y_{2}=\frac{3}{5}\\\\x_{3}=2,~y_{3}=\frac{1}{2}\\\\h=\frac{2-1}{3}\\\\~~~~~~~=\frac{1}{3}\\\\I\approx \frac{1}{6}[1+\frac{1}{2}+2(\frac{3}{4}+\frac{3}{5})]\\\\~~~\approx \frac{7}{10}\\\\~~~\approx 0.700\\\\\therefore \ln(2)\approx0.700$

Yayyy first time using LaTex

asianese · May 12, 2013

Re: HSC 2013 2U Marathon

Look up the Align function in latex...it will save you all those tildes.

HeroicPandas · May 12, 2013

Re: HSC 2013 2U Marathon

$(a)$ If $y = x^x,$ find $\frac{dy}{dx} \\ \\ (b)$If z = x^{x^{x}}, $ $ hence using part a or otherwise find $\frac{dz}{dx}$

nightweaver066 · May 12, 2013

Re: HSC 2013 2U Marathon

HeroicPandas said:
$(a)$ If $y = x^x,$ find $\frac{dy}{dx} \\ \\ (b)$If z = x^{x^{x}}, $ $ hence using part a or otherwise find $\frac{dz}{dx}$

This question would trump 3U students lol.

Possibly the average 4U student too.

HeroicPandas · May 12, 2013

Re: HSC 2013 2U Marathon

nightweaver066 said:
This question would trump 3U students lol.

Possibly the average 4U student too.

haha

asianese · May 12, 2013

Re: HSC 2013 2U Marathon

I agree with nightweaver...it'd be too hard without guidance in 2u.

Here is a solution:

$y=x^x \\ $Rewriting this as an exponential, we get $ y=e^{\ln(x^x)}=e^{x\ln x} $ by log laws.$\\ \frac{dy}{dx} = \left(\frac{d}{dx}(x \ln x)\right) e^{x\ln x}\\= \left(\frac{x}{x}+\ln x\right)e^{x\ln x} \\=(1+\ln x)x^x.\\\\ z=x^{x^x}=x^y=e^{y\ln x}\\ \frac{dz}{dx}=\left(\frac{d}{dx}(y\ln x)\right)e^{y\ln x}\\ = \left(\frac{d}{dx}y\ln x+\frac{y}{x}\right)e^{y\ln x} \\ = \left[ x^x(1+\ln x)\ln x+\frac{x^x}{x}\right]x^{x^x} \\ = x^{x^x} \left[ \frac{x^x}{x}+x^x\ln x(1+\ln x)\right]$

Clearly this question is too involved for a 2u student, and would challenge 3u and 4u students as well.

HeroicPandas · May 12, 2013

Re: HSC 2013 2U Marathon

asianese said:
I agree with nightweaver...it'd be too hard without guidance in 2u.

Here is a solution:

$y=x^x \\ $Rewriting this as an exponential, we get $ y=e^{\ln(x^x)}=e^{x\ln x} $ by log laws.$\\ \frac{dy}{dx} = \left(\frac{d}{dx}(x \ln x)\right) e^{x\ln x}\\= \left(\frac{x}{x}+\ln x\right)e^{x\lnx} \\=(1+\ln x)x^x.\\\\ z=x^{x^x}=x^y=e^{y\lnx}\\ \frac{dz}{dx}=\frac{d}{dx}(y\ln x)e^{y\ln x}\\ = \left(\frac{d}{dx}y\ln x+\frac{y}{x}\right)e^{y\ln x} \\ = \left[ x^x(1+\ln x)\ln x+\frac{x^x}{x}\right]x^x^x \\ = x^x^x \left[ \frac{x^x}{x}+x^x\ln x(1+\ln x)\right]$

Clearly this question is too involved for a 2u student, and would challenge 3u and 4u students as well.

Nice!

EDIT: misread ur post xD

Makematics · May 12, 2013

Re: HSC 2013 2U Marathon

asianese said:
Look up the Align function in latex...it will save you all those tildes.

Yeah that took a while LOL! thank you!

asianese · May 12, 2013

Re: HSC 2013 2U Marathon

HeroicPandas said:
Nice!

Why would it be hard without guidance? this question is similar to differentiating $y = 2^x$

You will see in all HSCs that 2u students are guided through questions with 'hints' to get to the final answer. You'd probably need to set the question up:

ai) Show that y=x^x=e^xlnx by log laws (1)
ii) Find dy/dx (2 or 3, depends on marking scheme. things to note here are: 1 mark for realising you have a chain rule in the exponential, and then differentiating xlnx with a product rule successfully. so this is 3 marks, really)

b) let z=x^x^x. By considering z=x^y, and using a), find dz/dx (4? ok so you'll give 1 mark for using part a at least. then another mark for doing the chain rule and realising the derivative of y is dy/dx (you'd be surprised), product rule again, and back substitution)

HeroicPandas · May 12, 2013

Re: HSC 2013 2U Marathon

asianese said:
You will see in all HSCs that 2u students are guided through questions with 'hints' to get to the final answer. You'd probably need to set the question up:

ai) Show that y=x^x=e^xlnx by log laws (1)
ii) Find dy/dx (2 or 3, depends on marking scheme. things to note here are: 1 mark for realising you have a chain rule in the exponential, and then differentiating xlnx with a product rule successfully. so this is 3 marks, really)

b) let z=x^x^x. By considering z=x^y, and using a), find dz/dx (4? ok so you'll give 1 mark for using part a at least. then another mark for doing the chain rule and realising the derivative of y is dy/dx (you'd be surprised), product rule again, and back substitution)

yeh yeh i understand, i misread ur post haha

Finding the derivative of $x^{x^{x}}$ would be hard for 2U students

Sy123 · May 12, 2013

Re: HSC 2013 2U Marathon

$Consider a$ \ \ n \ \ $sided regular polygon$

$$There exists a point in this regular polygon, that is equidistant from all other points in the polygon, call this point$ \ \ O$

$$The distance from$ \ \ O \ \ $to any corner of the polygon is given by$ \ \ r$

$$Let$ \ \ K = \frac{1}{2} r^2 \sin \frac{2\pi}{n}$

$i) Prove that the area of the regular polygon$ \ \ A \ \ $is given by$

$A = nK \ \ \ \ \fbox{2}$

$$ii) It is known that, for some small$ \ \ \theta \ \ \sin \theta \approx \theta$

$$From this it follows that$ \ \ \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$

$$By making the substitution$ \ \ n = \frac{1}{m}$

$Prove that$

$\lim_{n \to \infty} A = \pi r^2 \ \ \ \ \fbox{3}$

$$iii) Interpret this result geometrically$ \ \ \ \fbox{1}$

omgiloverice · May 12, 2013

Re: HSC 2013 2U Marathon

HeroicPandas said:
$(a)$ If $y = x^x,$ find $\frac{dy}{dx} \\ \\ (b)$If z = x^{x^{x}}, $ $ hence using part a or otherwise find $\frac{dz}{dx}$

Am I right?

EDIT: damn got solved while I was solving :L

Kostie100 · May 12, 2013

Re: HSC 2013 2U Marathon

asianese said:
$y=x^x \\ $Rewriting this as an exponential, we get $ y=e^{\ln(x^x)}[/QUOTE] Can you explain his to me. I have probably just forgotten something really simple lol$

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