HSC 2013-14 MX1 Marathon (archive) (2 Viewers)

HeroicPandas · Jun 9, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$Find the value of the sum, where$ \ \ n \ \ $is an integer$

$\tan^{-1} \frac{1}{n} + \dots + \tan^{-1} \frac{1}{3} + \tan^{-1}\frac{1}{2} + \tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3 + \dots + \tan^{-1} n$

$$ let tan^{-1}(x) + tan^{-1}(\frac{1}{x}) = y \\ \\ \therefore tan(tan^{-1}(x) + tan^{-1}(\frac{1}{x})) = tany \\ $\frac{x+ \frac{1}{x}}{1-1} = tany \\ \therefore y = tan^{-1}(\frac{\dots}{0}) \\ \therefore y = \frac{\pi}{2} \\ \\ \therefore tan^{-1} (x) + tan^{-1}(\frac{1}{x}) = \frac{\pi}{2}$

Group sum:

$tan^{-1}1 + tan^{-1}\frac{1}{1} + tan^{-1}2 + tan^{-1 }(\frac{1}{2}) + \dots + tan^{-1}n + tan^{-1}(\frac{1}{n})$

Use proven identity 'n' times

Therefore,

$\frac{n \pi}{2}$

Question:

$$Without evaluating the intergal,I = $\int_{\frac{1}{2}}^1 ln(x)dx,$ is it less or greater than zero?$

Sy123 · Jun 9, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
$$ let tan^{-1}(x) + tan^{-1}(\frac{1}{x}) = y \\ \\ \therefore tan(tan^{-1}(x) + tan^{-1}(\frac{1}{x})) = tany \\ $\frac{x+ \frac{1}{x}}{1-1} = tany \\ \therefore y = tan^{-1}(\frac{\dots}{0}) \\ \therefore y = \frac{\pi}{2} \\ \\ \therefore tan^{-1} (x) + tan^{-1}(\frac{1}{x}) = \frac{\pi}{2}$

Group sum:

$tan^{-1}1 + tan^{-1}\frac{1}{1} + tan^{-1}2 + tan^{-1 }(\frac{1}{2}) + \dots + tan^{-1}n + tan^{-1}(\frac{1}{n})$

Use proven identity 'n' times

Therefore,

$\frac{n \pi}{2}$

Question:

$$Without evaluating the intergal,I = $\int_{\frac{1}{2}}^1 ln(x)dx,$ is it less or greater than zero?$

Yep nice work.

AnimeX · Jun 9, 2013

Re: HSC 2013 3U Marathon Thread

Question:

$$Without evaluating the intergal,I = $\int_{\frac{1}{2}}^1 ln(x)dx,$ is it less or greater than zero?$ [/QUOTE]

Negative as the "area" (not sure how else to put it, but the region there..) between 0.5 and 1 would be negative.

Therefore less than zero.

NEw Question:

Show that the equation lnx + x = 0 has a root between x=0.5 and x=1.

HeroicPandas · Jun 9, 2013

Re: HSC 2013 3U Marathon Thread

AnimeX said:
Question:

$$Without evaluating the intergal,I = $\int_{\frac{1}{2}}^1 ln(x)dx,$ is it less or greater than zero?$

Negative as the "area" (not sure how else to put it, but the region there..) between 0.5 and 1 would be negative.

Therefore less than zero.

Question:

Show that the equation lnx + x = 0 has a root between x=0.5 and x=1.[/QUOTE]

Let f(x) = lnx + x

f(0.5) x f(1) <0

Therefore curve is continous and root lies between

Question:

$$Find $\int_0^{\frac{\pi}{2}} \sqrt{1-sin(x)}dx$

This is a 3u question

RealiseNothing · Jun 9, 2013

Re: HSC 2013 3U Marathon Thread

Amazing quotesmanship guys.

AnimeX · Jun 9, 2013

Re: HSC 2013 3U Marathon Thread

@heroicpandas,

was my answer correct for your question?

HeroicPandas · Jun 9, 2013

Re: HSC 2013 3U Marathon Thread

AnimeX said:
@heroicpandas,

was my answer correct for your question?

oh yeh sorry, yes it was correct!

AnimeX · Jun 9, 2013

Re: HSC 2013 3U Marathon Thread

Show that the equation lnx + x = 0 has a root between x=0.5 and x=1.

HeroicPandas said:
Let f(x) = lnx + x

f(0.5) x f(1) <0

Therefore curve is continous and root lies between

yup this is correct, but I think it would be better if you subbed in f(0.5) and f(1)

LOL sorry, idk why this quoting stuff isn't working...

Edit: works now..

Sy123 · Jun 9, 2013

Re: HSC 2013 3U Marathon Thread

AnimeX said:
@heroicpandas,

was my answer correct for your question?

I think it would be more correct to say that the region bounded by the ln(x) curve from x=1 to x=0.5 lies below the x-axis since,

$\ln(x) < 0 \ \ $for$ \ \ 0<x<1$

Therefore the region lies below x-axis, and hence the integral must be negative.

=====

$By using the Double Angle formula, evaluate$

$\int_0^{\frac{\pi}{2}} \cos^{4}x + \sin^4 x \ dx$

HeroicPandas · Jun 9, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
I think it would be more correct to say that the region bounded by the ln(x) curve from x=1 to x=0.5 lies below the x-axis since,

$\ln(x) < 0 \ \ $for$ \ \ 0<x<1$

Therefore the region lies below x-axis, and hence the integral must be negative.

=====

$By using the Double Angle formula, evaluate$

$\int_0^{\frac{\pi}{2}} \cos^{4}x + \sin^4 x \ dx$

$\int_0^{\frac{\pi}{2}} \cos^{4}x + \sin^4 x \ dx \\ =\int ((c^2 + s^2)^2 - 2s^2c^2)dx \\ = \int (1 - 2(\frac{1}{2}(1-cos2x)(\frac{1}{2}(1+cos2x))))dx \\ = \int (1- (\frac{1}{2})(1 - cos^22x))dx \\ = \int (1 - \frac{1}{2}(\frac{1}{2}(1-cos4x)))dx \\ = \int (\frac{3}{4} + \frac{1}{4}cos4x)dx \\ = \left [ \frac{3x}{4}+ \frac{sin4x}{14}\right ]_0^{\frac{\pi}{2}} \\ = \frac{3 \pi}{8}$

AnimeX · Jun 9, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
I think it would be more correct to say that the region bounded by the ln(x) curve from x=1 to x=0.5 lies below the x-axis since,

$\ln(x) < 0 \ \ $for$ \ \ 0<x<1$

Therefore the region lies below x-axis, and hence the integral must be negative.

=====

$By using the Double Angle formula, evaluate$

$\int_0^{\frac{\pi}{2}} \cos^{4}x + \sin^4 x \ dx$

Oh yep, I agree, "area" wasn't the correct word.

Now your question:

$\int_0^{\frac{\pi}{2}} \cos^{4}x + \sin^4 x \ dx = \int_0^{\frac{\pi}{2}}\frac{1}{4} (1+cos2x)^2 + \frac{1}{4} (1-cos2x)^2 = \frac{1}{4}\int_0^{\frac{\pi}{2}} 3+cos4x = \frac{3pi}{8}$

how do you make a new line in Latex?

HeroicPandas · Jun 9, 2013

Re: HSC 2013 3U Marathon Thread

AnimeX said:
Oh yep, I agree, "area" wasn't the correct word.

Now your question:

$\int_0^{\frac{\pi}{2}} \cos^{4}x + \sin^4 x \ dx = \int_0^{\frac{\pi}{2}}\frac{1}{4} (1+cos2x)^2 + \frac{1}{4} (1-cos2x)^2 = \frac{1}{4}\int_0^{\frac{\pi}{2}} 3+cos4x = \frac{3pi}{8}$

how do you make a new line in Latex?

U are too pro...

$$Find with FULL working out: \\ \\ a) I = $\int_{-1}^1 \sin(-x)dx \\ \\ $b) Hence analysing the above integral, find J = $\int_{-\infty}^{\infty}$ x\left ( sin(x^{x^{x^{x \dots}}})\right )^{2012}dx$

^Sy's great idea!

Sy123 · Jun 9, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
$\int_0^{\frac{\pi}{2}} \cos^{4}x + \sin^4 x \ dx \\ =\int ((c^2 + s^2)^2 - 2s^2c^2)dx \\ = \int (1 - 2(\frac{1}{2}(1-cos2x)(\frac{1}{2}(1+cos2x))))dx \\ = \int (1- (\frac{1}{2})(1 - cos^22x))dx \\ = \int (1 - \frac{1}{2}(\frac{1}{2}(1-cos4x)))dx \\ = \int (\frac{3}{4} + \frac{1}{4}cos4x)dx \\ = \left [ \frac{3x}{4}+ \frac{sin4x}{14}\right ]_0^{\frac{\pi}{2}} \\ = \frac{3 \pi}{8}$

AnimeX said:
Oh yep, I agree, "area" wasn't the correct word.

Now your question:

$\int_0^{\frac{\pi}{2}} \cos^{4}x + \sin^4 x \ dx = \int_0^{\frac{\pi}{2}}\frac{1}{4} (1+cos2x)^2 + \frac{1}{4} (1-cos2x)^2 = \frac{1}{4}\int_0^{\frac{\pi}{2}} 3+cos4x = \frac{3pi}{8}$

how do you make a new line in Latex?

Nice work.

You can make a new lines by typing \\

i.e. [tex ]$Line 1$ \\ $Line 2$ [/ tex]

======

$Solve$

$\sin^{-1}(3x+1) = \cos^{-1}(x)$

HeroicPandas · Jun 9, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Nice work.

You can make a new lines by typing \\

i.e. [tex ]$Line 1$ \\ $Line 2$ [/ tex]

======

$Solve$

$\sin^{-1}(3x+1) = \cos^{-1}(x)$

Add both sides with arcsin(x) and i'll let someone else do this

Makematics · Jun 9, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
$$Find $\int_0^{\frac{\pi}{2}} \sqrt{1-sin(x)}dx$

$\\\sqrt{1-\sin(x)}=\sqrt{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}}\\\\~~~~~~~~~~~~~~=\sqrt{(\cos\frac{x}{2}-\sin\frac{x}{2})^2}\\\\~~~~~~~~~~~~~~=\cos\frac{x}{2}-\sin\frac{x}{2}\\\\\therefore \int_{0}^{\frac{\pi}{2}}\sqrt{1-\sin(x)}~dx=\int_{0}^{\frac{\pi}{2}}(\cos\frac{x}{2}-\sin\frac{x}{2})~dx\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~=[2\sin\frac{x}{2}+2\cos\frac{x}{2}]_{0}^{\frac{\pi}{2}}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~=2\sqrt{2}-2$

I didn't mess around with any absolute values because the area is clearly positive from a sketch of the function being integrated.

Makematics · Jun 9, 2013

Re: HSC 2013 3U Marathon Thread

Alternatively,

$\\$Let$~ \sin x=\sin^2\theta$, since we want to eliminate the square root$\\\\x=\sin^{-1}(sin^2\theta) \\\\x=0,~\theta=0\\\\x=\frac{\pi}{2},~\theta=\frac{\pi}{2}\\\\dx=\frac{2\sin\theta\cos\theta}{\sqrt{1-\sin^4\theta}}~d\theta\\\\\\~~~~=\frac{2\sin\theta\cos\theta}{\sqrt{1-\sin^2\theta}\sqrt{1+\sin^2\theta}}~d\theta\\\\\\~~~~=\frac{2\sin\theta}{\sqrt{1+\sin^2\theta}}~d \theta\\\\\sqrt{1-\sin(x)}=\sqrt{1-\sin^2\theta}\\\\~~~~~~~~~~~~~~~~~=\cos\theta\\\\ \therefore\int_{0}^{\frac{\pi}{2}}\sqrt{1-\sin(x)}~dx=\int_{0}^{\frac{\pi}{2}}\frac{2\sin \theta \cos\theta}{\sqrt{1+\sin^2\theta}}~d\theta\\\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2[\sqrt{1+\sin^2\theta}~]_{0}^{\frac{\pi}{2}}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2\sqrt{2}-2$

HeroicPandas · Jun 10, 2013

Re: HSC 2013 3U Marathon Thread

Makematics said:
$\\\sqrt{1-\sin(x)}=\sqrt{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}}\\\\~~~~~~~~~~~~~~=\sqrt{(\cos\frac{x}{2}-\sin\frac{x}{2})^2}\\\\~~~~~~~~~~~~~~=\cos\frac{x}{2}-\sin\frac{x}{2}\\\\\therefore \int_{0}^{\frac{\pi}{2}}\sqrt{1-\sin(x)}~dx=\int_{0}^{\frac{\pi}{2}}(\cos\frac{x}{2}-\sin\frac{x}{2})~dx\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~=[2\sin\frac{x}{2}+2\cos\frac{x}{2}]_{0}^{\frac{\pi}{2}}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~=2\sqrt{2}-2$

I didn't mess around with any absolute values because the area is clearly positive from a sketch of the function being integrated.

Makematics said:
Alternatively,

$\\$Let$~ \sin x=\sin^2\theta$, since we want to eliminate the square root$\\\\x=\sin^{-1}(sin^2\theta) \\\\x=0,~\theta=0\\\\x=\frac{\pi}{2},~\theta=\frac{\pi}{2}\\\\dx=\frac{2\sin\theta\cos\theta}{\sqrt{1-\sin^4\theta}}~d\theta\\\\\\~~~~=\frac{2\sin\theta\cos\theta}{\sqrt{1-\sin^2\theta}\sqrt{1+\sin^2\theta}}~d\theta\\\\\\~~~~=\frac{2\sin\theta}{\sqrt{1+\sin^2\theta}}~d \theta\\\\\sqrt{1-\sin(x)}=\sqrt{1-\sin^2\theta}\\\\~~~~~~~~~~~~~~~~~=\cos\theta\\\\ \therefore\int_{0}^{\frac{\pi}{2}}\sqrt{1-\sin(x)}~dx=\int_{0}^{\frac{\pi}{2}}\frac{2\sin \theta \cos\theta}{\sqrt{1+\sin^2\theta}}~d\theta\\\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2[\sqrt{1+\sin^2\theta}~]_{0}^{\frac{\pi}{2}}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2\sqrt{2}-2$

very nice!!! u can also convert sine into cosine then apply double angles (or is it half angles?)

Makematics · Jun 10, 2013

Re: HSC 2013 3U Marathon Thread

true! which method were you thinking of when you posted it?

HeroicPandas · Jun 10, 2013

Re: HSC 2013 3U Marathon Thread

Makematics said:
true! which method were you thinking of when you posted it?

First Dooblaye angles (or half angles), then second method i thought of was ur first method and lastly, let u = sinx

Makematics · Jun 10, 2013

Re: HSC 2013 3U Marathon Thread

that's the beauty of integration

so many methods, and they all work!

Make a Difference – Donate to Bored of Studies!

HSC 2013-14 MX1 Marathon (archive) (2 Viewers)

Heroic!

This too shall pass

Member

Heroic!

what is that?It is Cowpea

Member

Heroic!

Member

This too shall pass

Heroic!

Member

Heroic!

This too shall pass

Heroic!

Well-Known Member

Well-Known Member

Heroic!

Well-Known Member

Heroic!

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)