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!

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

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