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Edexcel A2 C4 Mathematics June 2016 - Official Thread

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Original post by XxKingSniprxX


Have you seen 'Good will hunting'?


YES YES YES! It's such a good movie!! :biggrin:
Original post by XxKingSniprxX
Omg, I remember watching 'A Brilliant Young Mind'... is that the one where he leaves the Maths exam hall running after the Chinese girl he has a crush on in the end. :biggrin: :lol:

I've got to see 'A Beautiful Mind' sometime soon....

Have you seen 'Good will hunting'?


Those three movies are incredible, and all about mathematicians which is even better :u:

Spoiler

Anyhow this isn't entirely related to C4 so we should move on, feel free to PM if you want to chat about it a bit more :biggrin:

To make this post a little more C4 related, here's a few problems:

1. Differentiate y=xxy = x^{x} writing your answer in terms of xx

2. What lengths does the rectangle of greatest perimeter which can be inscribed within a given circle of radius rr have? Prove it
(edited 8 years ago)
Original post by Euclidean

To make this post a little more C4 related, here's a few problems:

1. Differentiate xxx^{x} writing your answer in terms of xx


Spoiler



2. What lengths does the rectangle of greatest perimeter which can be inscribed within a given circle of radius rr have? Prove it


Spoiler

(edited 8 years ago)
Original post by Zacken
...


Spoiler spoiler spoiler :tongue:
(edited 8 years ago)
Original post by Euclidean


2. What lengths does the rectangle of greatest perimeter which can be inscribed within a given circle of radius rr have? Prove it


As an extension: what happens if we decide to maximise the sum of the square or cube of the lengths?
Original post by Euclidean
Spoiler spoiler spoiler :tongue:


My apologies, spoilered! You might want to edit your quoting. :tongue:
Original post by Euclidean


1. Differentiate y=xxy = x^{x} writing your answer in terms of xx


Another extension: Differentiate y=x(xx)y = x^{\left(x^x\right)} and y=(xx)xy = \left(x^{x}\right)^x.
3. Using θπ4θ\theta \mapsto \frac{\pi}{4} - \theta, evaluate: 0π/4ln(1+tanθ)dθ\displaystyle \int_0^{\pi/4} \ln (1 + \tan \theta) \, \mathrm{d}\theta.

Extension: Evaluate 01ln(1+x)1+x2dx\displaystyle \int_0^{1} \frac{\ln (1 + x)}{1+x^2} \, \mathrm{d}x.
Reply 368
Original post by Zacken
3. Using θπ4θ\theta \mapsto \frac{\pi}{4} - \theta, evaluate: 0π/4ln(1+tanθ)dθ\displaystyle \int_0^{\pi/4} \ln (1 + \tan \theta) \, \mathrm{d}\theta.

Extension: Evaluate 01ln(1+x)1+x2dx\displaystyle \int_0^{1} \frac{\ln (1 + x)}{1+x^2} \, \mathrm{d}x.


Not sure what that arrow means (I'm a noob) but I assumed I could use the addition formulae stuff and got to the integral of ln(2) - ln(1+tanx) which is actually more terms than the original question....I have no idea what to do lol
(edited 8 years ago)
Original post by Lilly1234567890
Are we meant to know how to integrate this for c4?


Posted from TSR Mobile



Yea I think so :smile:

Original post by Lilly1234567890
How does 2sinx become (sinx+cosx)+ (sinx+cosx)


Posted from TSR Mobile


2sinx=(sinx+cosx)+(sinx-cosx)
Original post by Fudge2
Not sure what that arrow means (I'm a noob) but I assumed I could use the addition formulae stuff and got to the integral of ln(2) - ln(1+tanx) which is actually more terms than the original question....I have no idea what to do lol


If I say xf(x)x \mapsto f(x), I'm really just making a substitution u=f(x)u = f(x) then u=xu=x. :-)

Well if you call your integral II - then you have I=ln2II = \ln 2 - I. :wink:

(although that's not quite correct, use that idea).
(edited 8 years ago)
Original post by Zacken
As an extension: what happens if we decide to maximise the sum of the square or cube of the lengths?


Problem didn't take so long, the solution in LaTeX took a year:

Spoiler

Original post by Zacken
3. Using θπ4θ\theta \mapsto \frac{\pi}{4} - \theta, evaluate: 0π/4ln(1+tanθ)dθ\displaystyle \int_0^{\pi/4} \ln (1 + \tan \theta) \, \mathrm{d}\theta.

Extension: Evaluate 01ln(1+x)1+x2dx\displaystyle \int_0^{1} \frac{\ln (1 + x)}{1+x^2} \, \mathrm{d}x.

Spoiler

Original post by BBeyond

Spoiler



Yep.
Original post by Euclidean
Problem didn't take so long, the solution in LaTeX took a year


Good work! :biggrin:
Original post by Euclidean
Problem didn't take so long, the solution in LaTeX took a year:

Spoiler



That is very impressive :redface: What are you going to be doing at uni?
Original post by BBeyond
That is very impressive :redface: What are you going to be doing at uni?


Maths of course! :biggrin:
4. Evaluate 0π/2lnsinθdθ\displaystyle \int_0^{\pi/2} \ln \sin \theta \, \mathrm{d}\theta using a similar substitution to my other integral.

5. Generalise this substitution and prove that it holds for all smoothly behaved ff when evaluating abf(x)dx\displaystyle \int_a^b f(x) \, \mathrm{d}x
(edited 8 years ago)
Original post by Euclidean
Maths of course! :biggrin:


Fair probs should have guessed :tongue: Camb?
Original post by BBeyond
Fair probs should have guessed :tongue: Camb?


Oxford :wink:

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