HSC 2012 MX1 Marathon #2 (archive) (1 Viewer)

Sy123 · Jun 17, 2012

Re: HSC 2012 Marathon

barbernator said:
1 lol

Yes heh, Im lacking creativity right now, so I decided to make an easy one

Post a question!

barbernator · Jun 17, 2012

Re: HSC 2012 Marathon

find the inverse function of. and state its domain and range
<a href="http://www.codecogs.com/eqnedit.php?latex=y=e^{4x}@plus;5e^{2x}@plus;1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y=e^{4x}+5e^{2x}+1" title="y=e^{4x}+5e^{2x}+1" /></a>

bleakarcher · Jun 17, 2012

Re: HSC 2012 Marathon

barbernator said:
find the inverse function of. and state its domain and range
<a href="http://www.codecogs.com/eqnedit.php?latex=y=e^{4x}@plus;5e^{2x}@plus;6" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y=e^{4x}+5e^{2x}+6" title="y=e^{4x}+5e^{2x}+6" /></a>

Cbf getting a piece of paper and answering this right now but just to help out you have to restrict the function to a particular domain for there to exist an inverse function.

Sy123 · Jun 17, 2012

Re: HSC 2012 Marathon

barbernator said:
find the inverse function of. and state its domain and range
<a href="http://www.codecogs.com/eqnedit.php?latex=y=e^{4x}@plus;5e^{2x}@plus;6" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y=e^{4x}+5e^{2x}+6" title="y=e^{4x}+5e^{2x}+6" /></a>

$y=e^{4x}+5e^{2x}+1 \\ e^{4x}+5e^{2x}+1-y=0 \\ \\ e^{2x}=\frac{-5 \pm \sqrt{5^2-4(1-y)}}{2} \\ \\ e^{2x}=\frac{-5 \pm \sqrt{4y+21}}{2} \\ \\ x=\frac{1}{2}\ln{\frac{-5 \pm \sqrt{4y+21}}{2}} \\ \\ \ln{g(x)}>0 \\ \therefore x=\frac{1}{2}\ln{\frac{-5+\sqrt{4y+21}}{2}} \\ \\ f^{-1}(x)=\frac{1}{2}\ln{\frac{-5+\sqrt{4x+21}}{2}}$

I sketched the original function in Geogebra, and it turns out it is monotonic. So no need to restrict domain, and I have clarified it with ln(g(x))>0 since we cant have -5 - sqrt{4y+21} since that will always be negative

Sy123 · Jun 17, 2012

Re: HSC 2012 Marathon

$\text{Is the following statement true?} \\ \\ \int \frac{1}{\sqrt{1-x^2}} = \sin^{-1}{x}+C \\ \int \frac{-1}{\sqrt{1-x^2}}=\cos^{-1}{x}+C \\ \\ Since \ \ \ \int \frac{-1}{\sqrt{1-x^2}}=-\sin^{-1}{x}+C \\ \\ \therefore \cos^{-1}{x}=-sin^{-1}{x} \ \ \ \ \ \ \ \ ? \\ \\ \\ \text {If the statement is false, explain why}$

barbernator · Jun 17, 2012

Re: HSC 2012 Marathon

Sy123 said:
$\text{Is the following statement true?} \\ \\ \int \frac{1}{\sqrt{1-x^2}} = \sin^{-1}{x}+C \\ \int \frac{-1}{\sqrt{1-x^2}}=\cos^{-1}{x}+C \\ \\ Since \ \ \ \int \frac{-1}{\sqrt{1-x^2}}=-\sin^{-1}{x}+C \\ \\ \therefore \cos^{-1}{x}=-sin^{-1}{x} \ \ \ \ \ \ \ \ ? \\ \\ \\ \text {If the statement is false, explain why}$

different constants, as the restricted domains of the original functions vary by pi/2

simplify the following trig expression.

sin(x)+cos(2x)+sin(3x)+cos(4x) damnit that doesn't even come out good at all.

Here is a conceptual one

Find the volume created by rotating y=x^2 - 3 about the x axis. Find this volume using simpsons rule and 5 values. state why they are the same/

bleakarcher · Jun 17, 2012

Re: HSC 2012 Marathon

Sy123 said:
$y=e^{4x}+5e^{2x}+1 \\ e^{4x}+5e^{2x}+1-y=0 \\ \\ e^{2x}=\frac{-5 \pm \sqrt{5^2-4(1-y)}}{2} \\ \\ e^{2x}=\frac{-5 \pm \sqrt{4y+21}}{2} \\ \\ x=\frac{1}{2}\ln{\frac{-5 \pm \sqrt{4y+21}}{2}} \\ \\ \ln{g(x)}>0 \\ \therefore x=\frac{1}{2}\ln{\frac{-5+\sqrt{4y+21}}{2}} \\ \\ f^{-1}(x)=\frac{1}{2}\ln{\frac{-5+\sqrt{4x+21}}{2}}$

I sketched the original function in Geogebra, and it turns out it is monotonic. So no need to restrict domain, and I have clarified it with ln(g(x))>0 since we cant have -5 - sqrt{4y+21} since that will always be negative

With this question, how come the domain of the inverse function isnt identical to the range of the function?

The range of the function is y>1 and the domain of the inverse function is x>=-21/4

Sy123 · Jun 17, 2012

Re: HSC 2012 Marathon

Hmm, the volume is not defined?

Because the parabola is fully above the x-axis, and there is no enclosed part to rotate about. Or is that the trick?

EDIT: I have no idea why the inverse function turned out like that, Ive rechecked my working out, and my algebra seems fine.

When you sketch both functions, they are near perfect reflections of each other except for the fact, that the inverse function is cut off at -21/4

But, it has something to do with the square root since,

when you test $\frac{-5+\sqrt{4x+21}}{2}>0$

You get x>1 (domain) while the range of the original function is indeed y>1

barbernator · Jun 17, 2012

Re: HSC 2012 Marathon

Sy123 said:
Hmm, the volume is not defined?

Because the parabola is fully above the x-axis, and there is no enclosed part to rotate about. Or is that the trick?

whoops, sorry it was supposed to be x^2 - 3 haha

bleakarcher · Jun 17, 2012

Re: HSC 2012 Marathon

Sy123 said:
Hmm, the volume is not defined?

Because the parabola is fully above the x-axis, and there is no enclosed part to rotate about. Or is that the trick?

EDIT: I have no idea why the inverse function turned out like that, Ive rechecked my working out, and my algebra seems fine.

When you sketch both functions, they are near perfect reflections of each other except for the fact, that the inverse function is cut off at -21/4

But, it has something to do with the square root since,

when you test $\frac{-5+\sqrt{4x+21}}{2}>0$

You get x>1 (domain) while the range of the original function is indeed y>1

oh yes of course. lol, I got confused when barbenator said the range was y>-21/4 and didnt bother working it out on my own..

barbernator · Jun 17, 2012

Re: HSC 2012 Marathon

bleakarcher said:
oh yes of course. lol, I got confused when barbenator said the range was y>-21/4 and didnt bother working it out on my own..

oh yeh whoops...

Timske · Jun 17, 2012

Re: HSC 2012 Marathon

barbernator said:
different constants, as the restricted domains of the original functions vary by pi/2

simplify the following trig expression.

sin(x)+cos(2x)+sin(3x)+cos(4x) damnit that doesn't even come out good at all.

Here is a conceptual one

Find the volume created by rotating y=x^2 - 3 about the x axis. Find this volume using simpsons rule and 5 values. state why they are the same
/

V = pi x^2 find intercepts y^2=(x^2-3)^2
Simpsons rule uses parabolic

Sy123 · Jun 17, 2012

Re: HSC 2012 Marathon

Also I dont understand your question, do you want me to find the volume, rotated by x-axis via normal method, AND simpsons rule, and explaining why they are the same volume?

I did that and I got $\frac{39\pi \sqrt{3}}{4}$

They are the same, because Simpsons Rule is approximating via sketching parabolas around the curve and the such, and since this is a perfect parabola, Simpson's Rule will give an exact answer?

Also I have a 3U test on Applications of Calculus to the Physical world and Inverse Functions, can someone post a question from those topics, if you could?
Thanks!

barbernator · Jun 17, 2012

Re: HSC 2012 Marathon

Sy123 said:
Also I dont understand your question, do you want me to find the volume, rotated by x-axis via normal method, AND simpsons rule, and explaining why they are the same volume?

I did that and I got $\frac{39\pi \sqrt{3}}{4}$

They are the same, because Simpsons Rule is approximating via sketching parabolas around the curve and the such, and since this is a perfect parabola, Simpson's Rule will give an exact answer?

yeh, that is correct. The question was aimed at just testing the understanding that simpsons rule uses parabolas to estimate that is all

barbernator · Jun 17, 2012

Re: HSC 2012 Marathon

Ok here is another question. Prove any product-sum or sum-product trig ratio from the startpoint of sin(x+y) and sin(x-y)

Sy123 · Jun 17, 2012

Re: HSC 2012 Marathon

barbernator said:
Ok here is another question. Prove any product-sum or sum-product trig ratio from the startpoint of sin(x+y) and sin(x-y)

Heh, what is the question asking?

I have not heard of product-sum or sum-product ratios, and when I google it, I get sinx cos y= 1/2(sin(x+y)+sin(x-y))
Is that what I have to prove?

Sanjeet · Jun 17, 2012

Re: HSC 2012 Marathon

barbernator said:
Ok here is another question. Prove any product-sum or sum-product trig ratio from the startpoint of sin(x+y) and sin(x-y)

Doesn't this require the use of diagrams?

barbernator · Jun 17, 2012

Re: HSC 2012 Marathon

nope, nothing but trig

nightweaver066 · Jun 17, 2012

Re: HSC 2012 Marathon

Sanjeet said:
Doesn't this require the use of diagrams?

You use the expansions and mess around to get the expressions.

SpiralFlex · Jun 17, 2012

Re: HSC 2012 Marathon

I'll join after English tomorrow, so regret playing TF2. I may find some awesome inverse questions.

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