HSC 2012 MX1 Marathon #1 (archive) (1 Viewer)

AAEldar · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Timske said:
integral of e^x2 is 2xe^x2 right?

If you mean $\int e^{x^2}\;dx$ then you can't integrate that.

In regards to your attachment, their answer is correct. $\cos^2x=\frac{1}{2}(\cos 2x + 1)$ , hence the two and the half cancel.

Timske · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

AAEldar said:
If you mean $\int e^{x^2}\;dx$ then you can't integrate that.

what about (ex)^2?

Carrotsticks · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Timske said:
integral of e^x2 is 2xe^x2 right?

You just found the derivative, not the integral.

AAEldar said:
If you mean $\int e^{x^2}\;dx$ then you can't integrate that.

I think the solution has something to do with the Error Function. I remember doing a couple problems regarding it in Integral Calculus.

Timske said:
what about (ex)^2?

$\left (e^{x} \right )^{2} = e^{2x} \\\\ \int e^{2x}dx = \frac{1}{2}e^{2x} + C$

AAEldar · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Carrotsticks said:
I think the solution has something to do with the Error Function. I remember doing a couple problems regarding it in Integral Calculus.

$\left (e^{x} \right )^{2} = e^{2x} \\\\ \int e^{2x}dx = \frac{1}{2}e^{2x} + C$

Yea I meant in the HSC >.<

Also didn't even think of what he wrote being that integral hahah.

Timske · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

ohh i see... the curve was y=ex so y^2=(ex)^2

Carrotsticks · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Timske said:
ohh i see... the curve was y=ex so y^2=(ex)^2

As I said beforehand:

$\left (e^{x} \right )^{2} = e^{2x}$

It is easier to think of it this way.

Timske · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

its kind of confusing because i think its like (5x)^2 so it becomes 25x^2

Carrotsticks · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Timske said:
its kind of confusing because i think its like (5x)^2 so it becomes 25x^2

Not quite, refer to your surd laws. Perhaps that will clear a few things up.

Examine · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Need help with homework again guys.

cosx/1+sinx=secx-tanx

AND

sin(90-x)-cos^3x=cosx sin^2x

Aesytic · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

1. RHS = 1/cosx - sinx/cosx
= [1-sinx]/cosx
multiply top and bottom by 1+sinx,
=[1-sin^2x]/[cosx(1+sinx)]
=cos^2x/[cosx(1+sinx)]
=cosx/[1+sinx]
=LHS

2. LHS = cosx - cos^3x {sin(90-x) = cosx}
=cosx(1-cos^2x)
=cosx*sin^2x
=RHS

Nooblet94 · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Examine said:
Need help with homework again guys.

cosx/1+sinx=secx-tanx

AND

sin(90-x)-cos^3x=cosx sin^2x

First one:
<a href="http://www.codecogs.com/eqnedit.php?latex=LHS=\frac{\cos x}{1@plus;\sin x}\\~\\ =\frac{\cos x(1-\sin x)}{1-\sin^2x}\\~\\ =\frac{1-\sin x}{\cos x}\\~\\ =\frac{1}{\cos x}-\frac{\sin x}{\cos x}\\ ~\\ =\sec x-\tan x =RHS" target="_blank"><img src="http://latex.codecogs.com/gif.latex?LHS=\frac{\cos x}{1+\sin x}\\~\\ =\frac{\cos x(1-\sin x)}{1-\sin^2x}\\~\\ =\frac{1-\sin x}{\cos x}\\~\\ =\frac{1}{\cos x}-\frac{\sin x}{\cos x}\\ ~\\ =\sec x-\tan x =RHS" title="LHS=\frac{\cos x}{1+\sin x}\\~\\ =\frac{\cos x(1-\sin x)}{1-\sin^2x}\\~\\ =\frac{1-\sin x}{\cos x}\\~\\ =\frac{1}{\cos x}-\frac{\sin x}{\cos x}\\ ~\\ =\sec x-\tan x =RHS" /></a>

Second one:
<a href="http://www.codecogs.com/eqnedit.php?latex=\\ LHS=\sin (90-x)-\cos^3x\\ =\cos x-\cos^3x\\ =\cos x(1-\cos^2x)\\ =\cos x\sin^2x\\ =RHS" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ LHS=\sin (90-x)-\cos^3x\\ =\cos x-\cos^3x\\ =\cos x(1-\cos^2x)\\ =\cos x\sin^2x\\ =RHS" title="\\ LHS=\sin (90-x)-\cos^3x\\ =\cos x-\cos^3x\\ =\cos x(1-\cos^2x)\\ =\cos x\sin^2x\\ =RHS" /></a>

Carrotsticks · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Examine said:
Need help with homework again guys.

cosx/1+sinx=secx-tanx

AND

sin(90-x)-cos^3x=cosx sin^2x

$\\ 1. $ Divide top and bottom by $ cos(x) \\\\ \begin{align*} LHS &= \frac{1}{\sec(x) + \tan(x)} \\\\ &= \frac{\sec(x) - \tan(x)}{\sec^{2}(x) - \tan^{2}(x)} \quad \\\\ &= \frac{\sec(x) - \tan(x)}{\tan^{2}(x) + 1 - \tan^{2}(x)} \\\\ &= \frac{\sec(x) - \tan(x)}{1} \\\\ &= \sec(x) - \tan(x) \\\\ &= RHS \end{align*} \\\\ 2. $Remember that $ \sin(90-x) = \cos(x): \\\\ \begin{align*} LHS &= \cos(x) - \cos^{3}(x) \\\\ &= \cos(x) \left ( 1 - \cos^{2}(x) \right ) \\\\ &= \cos(x) \times \sin^{2}(x) \\\\ &= RHS \end{align*}$

Examine · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

I'm not sure I follow the second step for the second question. How does -cos^3x turn into (1-cos^2)?

Carrotsticks · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Examine said:
I'm not sure I follow the second step for the second question. How does -cos^3x turn into (1-cos^2)?

I factorised cos(x) from the expression in the first line.

Examine · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Hmm, how do you guys figure it out so easily? Some of them are easy though some I just stare at and think, dafuq?

Carrotsticks · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Examine said:
Hmm, how do you guys figure it out so easily? Some of them are easy though some I just stare at and think, dafuq?

1. Learn the Unit Circle properly, so you need not 'memorise' any formulas.

2. Learn to 'recognise' expressions that can be simplified to become something else.

3. By doing nothing, you do something.

4. Do lots of questions obviously, and you will recognise patterns.

Peeik · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Carrotsticks said:
1. Learn the Unit Circle properly, so you need not 'memorise' any formulas.

2. Learn to 'recognise' expressions that can be simplified to become something else.

3. By doing nothing, you do something.

4. Do lots of questions obviously, and you will recognise patterns.

Pretty much this. Was completely stumped with these in year 11 but after doing numerous questions, it becomes 2nd nature by the end of year 11.

gurmies · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Carrotsticks said:
$\\ 1. $ Divide top and bottom by $ cos(x) \\\\ \begin{align*} LHS &= \frac{1}{\sec(x) + \tan(x)} \\\\ &= \frac{\sec(x) - \tan(x)}{\sec^{2}(x) - \tan^{2}(x)} \quad \\\\ &= \frac{\sec(x) - \tan(x)}{\tan^{2}(x) + 1 - \tan^{2}(x)} \\\\ &= \tan^{2}(x)+1 \\\\ &= \frac{\sec(x) - \tan(x)}{1} \\\\ &= \sec(x) - \tan(x) \\\\ &= RHS \end{align*} \\\\ 2. $Remember that $ \sin(90-x) = \cos(x): \\\\ \begin{align*} LHS &= \cos(x) - \cos^{3}(x) \\\\ &= \cos(x) \left ( 1 - \cos^{2}(x) \right ) \\\\ &= \cos(x) \times \sin^{2}(x) \\\\ &= RHS \end{align*}$

First proof, fourth line?

Carrotsticks · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

gurmies said:
First proof, fourth line?

Oh dear, it appears that I accidentally copy/pasted an extra line. Thank you for pointing this out.

Examine · Feb 9, 2012

Re: 2012 HSC MX1 Marathon

Peeik said:
Pretty much this. Was completely stumped with these in year 11 but after doing numerous questions, it becomes 2nd nature by the end of year 11.

I guess you're right, I was fully stumped at the start of this though I can do quite a few questions now.

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