4u Mathematics Marathon V 1.0 (1 Viewer)

[who_loves_maths]

Same idea as the 3u Marathon, thanks to Slide Rule who's initiated it.

here's the first question to start off:

Question 1:

(a) Show that d[cos(x)]/dx = - sin(x)

(b) Prove that d[sec(x)]/dx = sec(x).tan(x)

(c) Hence, prove that:
(i) For all odd positive integral 'n', where limits of integration are from x = pi/3 to x = 0:

___________________________________ n ________________ /_n_\
DIntegral[tan(x).[cos(x) + sec(x)]^n dx] = _SUM (2^j - 1)(2^j + 1).|(n+j)|
__________________________________ j = 1 ____ j.2^j______ \ _2 /

where 'j' is, like n, odd {ie. increases by 2}

(ii) For all even positive integral 'n', limits of integration same as (i):

________________n ________________ /_n_\
I = [nC(n/2)].ln2 + SUM (2^j - 1)(2^j + 1).|(n+j)|
_______________j = 2_____j.2^j______ \ _2 /

where 'j' is, like n, even.

(iii) Furthermore, prove the definite integrals in (i) and (ii) by using the method of Mathematical Induction.


Note: for the above question, 1) "DIntegral" denotes "definite integral".

 

[香港!]

^
hahaha, u wan to "break the stagnancy of this thread" and then u put an even harder question in, making it more 'stagnant'

 

[Riviet]

^
hahaha, u wan to "break the stagnancy of this thread" and then u put an even harder question in, making it more 'stagnant'

That is so true, rofl!

 

[Stan..]

I'll bite.
1st one.
lim(h->O) Cos(x+h) - Cosx/h
lim cosxcosh - sinxsinh - cosx/h
lim cosx(cosh - 1)/h - sinxsinh/h
d(cosx)/dx = -sinx

2nd one.
d/dx(secx) = -sinx * -1/cos^2x
= Sinx/cos^2x
= Tanxsecx

3rd and 4th, Haven't done reduction techniques. So, I cannot attempt.

 

[Dumsum]

Shit I'm glad that kind of question wasn't in our exam.

 

[who_loves_maths]

Originally Posted by Stan...
3rd and 4th, Haven't done reduction techniques. So, I cannot attempt.

yes you can, and should, definitely attempt them Stan. since after all, they haven't a thing to do with reduction-formulae srsli.

Originally Posted by Dumsum
Shit I'm glad that kind of question wasn't in our exam.

Dumsum, don't worry, these types of questions will hardly ever be in a 4u hsc exam - because the question is only 3u stuff (belive it or not).

it's probably just that i haven't given away too much in the lead-up questions to the last two parts that make them "seem" hard - like 4u level - when they are actually not

anyways, give them a go


P.S. it doesn't even belong under the "harder 3u" chapter of the 4u course - it's actually "normal" 3u stuff.

 

[icycloud]

Sorry for my ignorance. What is the last part of the sum:

/_n_\
|(n+j)|
\ _2 /

meant to mean?

 

[lucifel]

see this is where LaTeX would've really come in handy, sums on this type set is just weird.

 

[Stan..]

Alright, I'll give it a shot.

 

[who_loves_maths]

Originally Posted by icycloud
Sorry for my ignorance. What is the last part of the sum:
/_n_\
|(n+j)|
\ _2 /
meant to mean?

hi icycloud,

no you are not ignorant - i'm just a bad typist
it is the combination notation. ie. it means [nC((n+j)/2)] ; that is, the number of subsets containing (n+j)/2 elements of an n-set.

 

[icycloud]

hi icycloud,

no you are not ignorant - i'm just a bad typist
it is the combination notation. ie. it means [nC((n+j)/2)] ; that is, the number of subsets containing (n+j)/2 elements of an n-set.

Oh alright. Here's my crack at the first Q:

I=∫tan(x) [cos(x) + sec(x)]^n dx

Let u = sec(x)
Thus, du = sec(x)tan(x) dx

Limits become 1 --> 2

I=∫(u+1/u)^n / u du

Expand using Binomial Theorem:

I=∫[nC0 u^(n-1) + nC1 u^(n-3) + ... + nC(n-1) u^(-n-3) + nCn u^(-n-1)] du

= [nC0 u^n/n + nC1 u^(n-2)/(n-2) + ... + nC(n-1) u^(-n+2)/(-n+2) + nCn u^(-n)/-n] (u = 1 --> 2)

Since n is odd, nCk = nC(n-k)

Thus,

I = nC0/n [u^n - u^(-n)] + nC1/(n-2) [u^(n-2) - u^(-n+2)] + ... nC([n-1]/2) [u - u^-1]

Sub limits u = 1 ---> 2, we get: (the part u = 1 yields I = 0, thus is discounted)

I = nC0/n [2^n - 2^(-n)] + nC1/(n-2) [2^(n-2) - 2^(-n+2)] + ... nC([n-1]/2) [2 - 1/2]

= nC0/n [(2^2n - 1)/(2^n)] + nC1/(n-2) [{2^2(n-2) - 1}/2^(n-2)] + ... nC([n-1]/2) [(2^2 -1)/2]

= nΣ(k=1) nC[(n+k)/2]/k * (2^2k-1)/2^k

= nΣ(k=1) nC[(n+k)/2] * (2^k-1)(2^k+1)/{k*2^k}

, as required.

For the second part, for even integers one would use a similar method, except that the middle term is singled out to be

nC(n/2) * u^-1,
integrating gives nC(n/2) ln(u)
and subbing in u = 2, we get nC(n/2) * ln(2)

which would yield the required answer. (Don't want to type the whole solution up =D)

 

[who_loves_maths]

^ i skim-read that entire post but i know you've solved that question since the methods you used were all correct. so well done icycloud
pretty good for a yr 11 only just beginning yr 12.

oh and don't forget, as the person who solved the last question in this thread, you now have the "privilege" of setting the next question, Question 2

hey btw, what school do you go to icycloud? and what was your maths rank in your grade for yr 11?

 

[Stan..]

Well Done Icy, I now see where I went wrong.

 

[icycloud]

^ i skim-read that entire post but i know you've solved that question since the methods you used were all correct. so well done icycloud
pretty good for a yr 11 only just beginning yr 12.

oh and don't forget, as the person who solved the last question in this thread, you now have the "privilege" of setting the next question, Question 2

hey btw, what school do you go to icycloud? and what was your maths rank in your grade for yr 11?

Hehe I'm from Sydney Boys . Was ranked 3rd in Maths, missed out on equal 1st by 0.5 marks (0.5 marks which I could have contested [long story], but couldn't be bothered since it was year 11 only).

Anyway, Question 2
Find:

∫(Ln[x + Sqrt[x^2+1]])^2 dx

Hint:

Spoiler

First find ∫Ln[x + Sqrt[x^2+1]] dx

 

[pLuvia]

What chapter you done in 4u maths icycloud?

 

[Estel]

Q2, without using the spoiler (I'll leave that method to someone in Yr 11)
∫(Ln[x + Sqrt[x^2+1]])^2 dx

Let u = Ln[Sqrt[x^2+1] + x]
Then obviously -u = Ln[Sqrt[x^2+1] - x]
du = dx/Sqrt[x^2+1] = 2dx/(e^ + e^[-u])
And then the substitution is made possible.

Q3. (Quite difficult, but accessible to everyone since it is circle geometry, a Yr 10 topic)

ABC is acute-angled. The circle diameter AC meets AB again at F, and the circle diameter AB meets AC again at E. BE meets the circle diameter AC at P, and CF meets the circle diameter AB at Q.
Prove AP = AQ (hint: if it was a HSC q, it would probably be worth at least 6 marks)

 

[sikeveo]

What chapter you done in 4u maths icycloud?

he's probably finished the book. In my grade, the top 5 or so had finished 4u a while back.

 

[pLuvia]

Oh, alrights maybe, damn

 

[Riviet]

icycloud is a genius :uhhuh:

 

[who_loves_maths]

Q3. (Quite difficult, but accessible to everyone since it is circle geometry, a Yr 10 topic)

ABC is acute-angled. The circle diameter AC meets AB again at F, and the circle diameter AB meets AC again at E. BE meets the circle diameter AC at P, and CF meets the circle diameter AB at Q.
Prove AP = AQ (hint: if it was a HSC q, it would probably be worth at least 6 marks)

yes, a good question Estel
but that wasn't much of a hint you gave. i thought hints were supposed to help ppl solve the problem?

{i won't use spoiler tags since solution is quite long, it would be inconvenient.}

Question 3:

new question, Question 4:
hmm ...
can't think of one right now off top of my head... will need time to conjure one up since i have no resources now (returned all textbooks) ...
i will post the new question up before end of day.