The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

The Proof is Trivial!

Scroll to see replies

Original post by Renzhi10122
Nicely done.


Speaking of winning strategies, any luck with the painting problem I posted earlier? :biggrin: It looks like it has not been solved yet.
Original post by Lord of the Flies
Speaking of winning strategies, any luck with the painting problem I posted earlier? :biggrin: It looks like it has not been solved yet.


Nope, I'll give it another try now.

For reference,
http://www.thestudentroom.co.uk/showthread.php?t=2313384&page=157&p=56356303#post56356303
Original post by atsruser
Problem 559

I put this up in a separate thread with little response so here it is again. Not very difficult - maybe something that may appear on a MAT paper.


Consider the point A(k,e−k)A(k,e^{-k}) for large k>0k >0.

Find the value of m<km < k for which the point B(m,e−m)B(m, e^{-m}) is approximately k2k\sqrt{2} distant from A.


Hmm, no takers for this, a mere bagatelle?

Hint:

Spoiler

Original post by atsruser
Problem 558 (*)

Solve:

xy23=916−x3\sqrt[3]{xy^2} = \frac{91}{6}-\frac{x}{3}

x2y3=376−y3\sqrt[3]{x^2y} = \frac{37}{6}-\frac{y}{3}


Hint:

Spoiler

Original post by atsruser
Hint:

Spoiler



If no-one's done this already...

Solution 558

Spoiler

(edited 8 years ago)
Original post by StrangeBanana
If no-one's done this already...

Solution 558

Spoiler



Right idea, but sadly (u+v)3=64(u+v)^3 = 64 does not imply that u+v=8 u+v = 8 :frown:
Original post by atsruser
Right idea, but sadly (u+v)3=64(u+v)^3 = 64 does not imply that u+v=8 u+v = 8 :frown:


what is life

i can't even
Original post by joostan
Solution 546:

In=∫0π211+tan⁡n(x) dx=∫0π2cos⁡n(x)sin⁡n(x)+cos⁡n(x) dxI_n=\displaystyle\int_0^{\frac{ \pi}{2}} \dfrac{1}{1+\tan^n(x)} \ dx = \displaystyle\int_0^{\frac{\pi}{2}} \dfrac{\cos^n(x)}{\sin^n(x)+\cos^n(x)} \ dx
Sending x↦π2−xx \mapsto \dfrac{\pi}{2}-x yields:
In=∫0π2sin⁡n(x)sin⁡n(x)+cos⁡n(x) dx[br]⇒2In=∫0π21 dx[br]⇒In=π4I_n=\displaystyle\int_0^{\frac{\pi}{2}} \dfrac{\sin^n(x)}{\sin^n(x)+\cos^n(x)} \ dx[br]\Rightarrow 2I_n=\displaystyle\int_0^{\frac{\pi}{2}} 1 \ dx[br]\Rightarrow I_n=\dfrac{\pi}{4}.

Oh snap, 2nd place :s-smilie:.


May I ask at what point during the A-level syllabus one will the methods used in this proof? I'm almost certain it's not covered in AS. Perhaps C4/FP2?
(edited 8 years ago)
Original post by ETbuymilkandeggs
May I ask at what point during the A-level syllabus one will the methods used in this proof? I'm almost certain it's not covered in AS. Perhaps C4/FP2?


When do you cover substitution as an integration method? It's either C3 or C4.
Original post by ETbuymilkandeggs
May I ask at what point during the A-level syllabus one will the methods used in this proof? I'm almost certain it's not covered in AS. Perhaps C4/FP2?


This is C4. With some ingenuity required.


Posted from TSR Mobile
In an infinite sequence of integers, a word is a block of consecutive terms. A triplet is a block consisting of three consecutive, identical words.

Problem 560

Can a binary sequence be triplet-free?
Original post by EmptyMathsBox
...


Nice. If you are interested a neat alternate solution (not my own) proceeds as follows: Divide through by 2x2^x and observe that x=2 x = 2 is a solution. Call the LHS f(x)f(x), then since 2−32<1 \dfrac{\sqrt{2-\sqrt{3}}}{2} < 1 and 2+32<1 \dfrac{\sqrt{2+\sqrt{3}}}{2} < 1 , ff is decreasing, so x=2 x = 2 is the only solution.
Original post by StrangeBanana
If no-one's done this already...

Solution 558

Spoiler



Nicely done
I have a very similar in one of my hard papers.
Original post by TeeEm
Nicely done
I have a very similar in one of my hard papers.


Mistake from u+v=8


Posted from TSR Mobile
Original post by physicsmaths
Mistake from u+v=8


Posted from TSR Mobile


The numbers turn out to be nice and easy sans mistake.
Original post by physicsmaths
Mistake from u+v=8


I think I'd better edit this already :oops:
Original post by StrangeBanana
I think I'd better edit this already :oops:


get on it banana boy
This is one of my all time favourite questions. A brilliant example of "it's not obvious if you can't prove it".

Problem 561 [from a IA example sheet - if you understand the question, you know enough to solve it]

Take kk points in the plane, not all collinear. Show that there is a line going through exactly two of them.
Original post by metaltron
Well done, the trick was in factorising the polynomial, though you didn't show how you did it in the solution. Also, how can you guarantee that all the roots of the cubic are real?


Hi metaltron, please could I have a hint on how to factorise the polynomial in this way? Is there a well known technique I need to use or just look a bit closer? :smile:
Original post by Lord of the Flies
This is one of my all time favourite questions. A brilliant example of "it's not obvious if you can't prove it".

Problem 561 [from a IA example sheet - if you understand the question, you know enough to solve it]

Take kk points in the plane, not all collinear. Show that there is a line going through exactly two of them.

I am sad that there doesn't seem to be a nice proof of this :frown:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you