HSC 2015 Maths Marathon (archive) (1 Viewer)

Drsoccerball · Mar 12, 2015

Re: HSC 2015 2U Marathon

FrankXie said:
sorry but all tries are wrong. you do need to draw a diagram

Frank is the solution on 3u marathon correct?

Ekman · Mar 12, 2015

Re: HSC 2015 2U Marathon

Im not in the position to solve it, due to other exams, but ill give you the integration formula:

$\int^2_1 \frac{1}{x} - (2-x) dx + \int^4_2 \frac{1}{x} dx + 2$

FrankXie · Mar 12, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
Frank is the solution on 3u marathon correct?

correctly represented by the integration.
but this one different

FrankXie · Mar 12, 2015

Re: HSC 2015 2U Marathon

Ekman said:
Im not in the position to solve it, due to other exams, but ill give you the integration formula:

$\int^2_1 \frac{1}{x} - (2-x) dx + \int^4_2 \frac{1}{x} dx + 2$

why the plus 2 at the end?

Kaido · Mar 12, 2015

Re: HSC 2015 2U Marathon

FrankXie said:
new question

$$ Find the area enclosed by the curve $ y=\frac{1}{x}, $ the line $ y=2-x, $ the $ x-$axis, and the line $ x=4.$

this is troll to the next level

Ekman · Mar 12, 2015

Re: HSC 2015 2U Marathon

FrankXie said:
why the plus 2 at the end?

Whoops I just re-read the question, I didn't see that it was bounded between the x-axis as well. In that case there is no need for the +2

integral95 · Mar 12, 2015

Re: HSC 2015 2U Marathon

FrankXie said:
new question

$$ Find the area enclosed by the curve $ y=\frac{1}{x}, $ the line $ y=2-x, $ the $ x-$axis, and the line $ x=4.$

The line is a tangent to the curve.....

FrankXie · Mar 12, 2015

Re: HSC 2015 2U Marathon

Kaido said:
this is troll to the next level

Ekman said:
Whoops I just re-read the question, I didn't see that it was bounded between the x-axis as well. In that case there is no need for the +2

integral95 said:
The line is a tangent to the curve.....

lol if this question appeared in 2U exam, the diagram would have been given. Ekman, you got it right.

photastic · Mar 13, 2015

Re: HSC 2015 2U Marathon

My favourite logarithm question.

Differentiate (ln(ln(ln(lnx))))

Drsoccerball · Mar 14, 2015

Re: HSC 2015 2U Marathon

photastic said:
My favourite logarithm question.

Differentiate (ln(ln(ln(lnx))))

It cant simply be -6/x^4?

InteGrand · Mar 14, 2015

Re: HSC 2015 2U Marathon

photastic said:
My favourite logarithm question.

Differentiate (ln(ln(ln(lnx))))

Let $f_n (x):=\ln (\ln (\ln (...(\ln x)...)))$ , where the ln appears n times, $n\in \mathbb{Z}^+$ , be denoted by $\ln _n x$ . (Note that $\ln _n x = \ln (\ln _{n-1}x)$ .)

By the chain rule, $f_n ^\prime (x)=\frac{(\ln _{n-1}x)^\prime}{\ln _{n-1}x}=\frac{f_{n-1}^\prime (x)}{\ln _{n-1}x}$ .

Carrying through with recursive formula for $f_n ^\prime$ , we can find that $f_n ^\prime (x)=\frac{\frac{1}{x}}{\prod_{i=1}^{n-1}\limits \ln _i x}$ , using $(\ln x)^\prime = \frac{1}{x}$ at the end of the process.

Plugging in n = 4 gives us our desired derivative:

$f_4 ^\prime (x)=\frac{\frac{1}{x}}{\prod_{i=1}^{3}\limits \ln _i x}$ .

photastic · Mar 15, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
It cant simply be -6/x^4?

Here's an alternative.

Drsoccerball · Mar 28, 2015

Re: HSC 2015 2U Marathon

$$ The acceleration of a particle is given by $ a=\frac{2}{x} +x + \frac{1}{x+1}$

$If the particle is at rest at the origin when t =3, Find the displacement when t=14$

InteGrand · Mar 28, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
$$ The acceleration of a particle is given by $ a=\frac{2}{x} +x + \frac{1}{x+1}$

Do you mean t's instead of x's? Otherwise we get infinite acceleration when the particle is at the origin.

Drsoccerball · Mar 28, 2015

Re: HSC 2015 2U Marathon

InteGrand said:
Do you mean t's instead of x's? Otherwise we get infinite acceleration when the particle is at the origin.

damn it didnt see that when making it D: let me change the question too
$$ The acceleration of a particle is given by $ a=\frac{2}{t} +t + \frac{1}{t+1}$

$If the particle is at rest at x=5 when t =3, Find the displacement when t=14$

InteGrand · Mar 28, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
damn it didnt see that when making it D: let me change the question too
$$ The acceleration of a particle is given by $ a=\frac{2}{t} +t + \frac{1}{t+1}$

$If the particle is at rest at x=5 when t =3, Find the displacement when t=14$

You have to end up integrating logs to get x as a function of t, which is beyond 2U level.

This integral would probably be provided then if this question came up in 2U: $\int \ln u \text{ d}u = u\ln u - u +c$ .

Drsoccerball · Mar 28, 2015

Re: HSC 2015 2U Marathon

InteGrand said:
You have to end up integrating logs to get x as a function of t, which is beyond 2U level.

This integral would probably be provided then if this question came up in 2U: $\int \ln u \text{ d}u = u\ln u - u +c$ .

I was thinking that they use the diagram to find the area so the integral isnt really needed

Ekman · Mar 28, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
I was thinking that they use the diagram to find the area so the integral isnt really needed

Sketching a diagram for $v= 2\ln{t} +\frac{t^2}{2} + \ln{(t+1)}$ is still beyond the scope of 2u...

Drsoccerball · Mar 28, 2015

Re: HSC 2015 2U Marathon

Ekman said:
Sketching a diagram for $v= 2\ln{t} +\frac{t^2}{2} + \ln{(t+1)}$ is still beyond the scope of 2u...

Draw them seperately

Ekman · Mar 28, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
Draw them seperately

Also just to add, the question is a bit faulty as the displacement, velocity and acceleration is undefined at t=0. So when this particle starts moving, its initially location, velocity and acceleration is undefined... I understand that the question is asking about the integration between specific intervals, but in a mathematical and physical sense, the question seems a bit faulty.

Also drawing them out separately still wont help...

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