HSC 2013 Maths Marathon (archive) (1 Viewer)

[girlworld_club]

Re: HSC 2013 2U Marathon

thanks. haha silly me

 

[mahmoudali]

Re: HSC 2013 2U Marathon

yes i did that i get to

-12 sin (2x + pi/4) = 0

And a take the basic angle as 0 degrees? So i end up with 0, 180, 360, etc then minus pi/4 and divide by 2 for each.

feel like i'm doing something wrong can't seem to get answer.

 

[Menomaths]

Re: HSC 2013 2U Marathon

Give a reasoning with your answer so I know you know how to do it

 

[Shazer2]

Re: HSC 2013 2U Marathon

Give a reasoning with your answer so I know you know how to do it

Answer is D. Since concavity at a is upwards, so f''(a) > 0, and the gradient at a is negative, so f'(a) < 0.

 

[Menomaths]

Re: HSC 2013 2U Marathon

Answer is D. Since concavity at a is upwards, so f''(a) > 0, and the gradient at a is negative, so f'(a) < 0.

Yep

 

[Menomaths]

Re: HSC 2013 2U Marathon

My method:
2cos^2(x)+1=3-2sin^2(x)

2cos^2(x) = 2-2sin^2(x)

cos^2(x) = 1-sin^2(x)

cos^2(x) = cos^2(x)

Since 1-sin^2(x) = cos^2(x)

Is this a viable method to solve this question or do I have to muck around with one side w/o touching the other?

 

[Shazer2]

Re: HSC 2013 2U Marathon

 

[girlworld_club]

Re: HSC 2013 2U Marathon

Differentiate:

Kt / ( bt + (K - bt)e^-Kt )

where T is a constant.

 

[Shazer2]

Re: HSC 2013 2U Marathon

I made this so you can check your LaTeX output: JSLaTeX

 

[Sy123]

Re: HSC 2013 2U Marathon

I made this so you can check your LaTeX output: JSLaTeX

Nice!

 

[leesh95]

Re: HSC 2013 2U Marathon

Can someone explain how do you find the distance traveled in calculus. So when they give you the formula for velocity and displacement and ask you to find the total distance, is it just the displacement? I get really confused with this.

 

[leesh95]

Re: HSC 2013 2U Marathon

My method:
2cos^2(x)+1=3-2sin^2(x)

2cos^2(x) = 2-2sin^2(x)

cos^2(x) = 1-sin^2(x)

cos^2(x) = cos^2(x)

Since 1-sin^2(x) = cos^2(x)

Is this a viable method to solve this question or do I have to muck around with one side w/o touching the other?

I would like to know this too? Do you have to only work on one side or can you solve both sides at once?

 

[Carrotsticks]

Re: HSC 2013 2U Marathon

I would like to know this too? Do you have to only work on one side or can you solve both sides at once?

You can do one of the following.

LHS = ... = ... = ... = RHS

--------------

RHS = ... = ... = ... = LHS-

--------------

LHS = ... = X

RHS = ... = X

Therefore LHS = RHS.

 

[RealiseNothing]

Re: HSC 2013 2U Marathon

You can do one of the following.

LHS = ... = ... = ... = RHS

--------------

RHS = ... = ... = ... = LHS-

--------------

LHS = ... = X

RHS = ... = X

Therefore LHS = RHS.

How about adding a variable (say X) to the RHS then you can play around with both sides at the same time, as long as you prove that X=0.

 

[Shazer2]

Re: HSC 2013 2U Marathon

Nice!

Thanks! Hope it's useful

 

[Menomaths]

Re: HSC 2013 2U Marathon

So my method's incorrect?

 

[Carrotsticks]

Re: HSC 2013 2U Marathon

How about adding a variable (say X) to the RHS then you can play around with both sides at the same time, as long as you prove that X=0.

That is valid, but I've never really seen a problem that needed that.

 

[Carrotsticks]

Re: HSC 2013 2U Marathon

So my method's incorrect?

For purposes of the HSC yes.

 

[leesh95]

Re: HSC 2013 2U Marathon

Is this an arithmetic series and we just apply the formula don't we?

 

[Carrotsticks]

Re: HSC 2013 2U Marathon

Is this an arithmetic series and we just apply the formula don't we?

Not quite, the angles are in AP but the actual terms may not necessarily be in AP.

Hint: Consider the identity sin(x) = cos(90-x)